COMPENDIUM HMMT

The Harvard-MIT Mathematics Tournament

Gerard Romo Garrido

Toomates Coolección

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Se agradecerá cualquier observación, comentario o colaboración a toomates@gmail.com

La biblioteca Toomates Coolección consta de los siguientes documentos:

Problem-solving Libros de texto (en catalán)

Geometría Axiomática pdf 1 2 → 23 Geometria analítica pdf 1 2

Problemas de Geometría pdf 1 2 → 9

Introducción a la Geometría pdf doc

Álgebra pdf 1 2 Nombres (Preàlgebra) pdf doc

Teoría de números pdf 1 2 3 4 5 6 7 Àlgebra pdf 1 2 → 24

Combinatoria pdf doc Combinatòria i Probabilitat pdf doc

Probabilidad pdf doc Estadística pdf doc

Trigonometría pdf doc Trigonometria pdf doc

Desigualdades pdf doc Funcions pdf doc

Números complejos pdf doc Nombres Complexos pdf doc

Guía del estudiante de Olimpiadas pdf Àlgebra Lineal 2n batx.

pdf doc

Geometria Lineal 2n batx.

pdf 1 2

Càlcul Infinitesimal 2n batx.

pdf 1 2

Programació Lineal 2n batx.

pdf doc

Recopilaciones de pruebas PAU:

Catalunya TEC , Catalunya CCSS , Galicia , País Vasco , Portugal A , Portugal B

Recopilaciones de problemas olímpicos y preolímpicos (España):

OME , OMEFL , OMEC , OMEM , Canguro , Cangur

Recopilaciones de problemas olímpicos y preolímpicos (Internacional):

IMO , OMI , USAMO , AIME , AMC 8 , AMC 10, AMC 12 , SMT , Kangourou, Kangaroo

Mathcounts , Archimede , HMMT .

Versión de este documento: 01/12/2021

Todos estos documentos se actualizan constantemente. ¡No utilices una versión anticuada! Descarga totalmente gratis la última

versión de los documentos en los enlaces superiores.

www.toomates.net

Índice.

Cada apartado de cada año viene acompañado de sus respectivas soluciones.

1998 5

1999 39

2000 71

2001 116

2002 157

2003 212

2004 272

2005 336

2006 390

2007 476

2008 532

2009 618

2010 697

2011 799

2012 975

2013 1067

2014 1166

2015 1279

2016 1372

2017 1453

2018 1537

2019 1626

2020 1718

2021 1829

Los apartados en los que se estructura esta prueba han ido cambiando a lo largo de los

años:

1998 Advanced Topics, Algebra, Calculus, Geometry, Team

1999 Advanced Topics, Algebra, Calculus, Geometry, Oral, Team 2000 Advanced Topics, Algebra, Calculus, General, Geometry, Guts, Oral, Power, Team

2001, 2002Advanced Topics, Algebra, Calculus, General part 1, General part 2, Geometry, Guts, Team

2003, 2007 Algebra, Calculus, Combinatorics, General part 1, General part 2, Geometry, Guts, Team

Desde el 2008 en adelante se celebran dos competiciones: Febrero y Noviembre.

2008 – 2010 Febrero: Algebra, Calculus, Combinatorics, General part 1, General part 2 ,Geometry, Guts, Team

Noviembre: General part 1, General part 2, Guts, Team

2011 hasta la actualidad: Noviembre: General, Theme, Team, Guts

Febrero: Algebra and Calculus, Algebra and Combinatorics , Algebra and Geometry, Calculus and

Combinatorics, Calculus and Geometry, Combinatorics and Geometry, Guts, Team A, Team B 2012 Febrero: Team A, Team B, Algebra, Combinatorics, Geometry, Guts.,

2017 Febrero: Team, Algebra and Number Theory, Combinatorics, Geometry, Guts

Actualmente:

Noviembre Febrero

Individual General Algebra, Geometry,

Combinatorics (50 minutos cada

una)

Team Respuesta corta (60 min.) Prueba (60 minutos)

Guts 36 problemas en grupos de

3 (80 min)

36 problemas en grupos de 4 (80

min.)

Dificultad Entre AMC y AIME Entre AIME y Olimpiadas

Concursantes Estudiantes bachillerato de

todo el mundo

Estudiantes bachillerato de todo el

mundo

Desde el 2013 se celebra, además, una competición denominada “Invitational” (Para

los 50 mejores clasificados en la fase de Febrero, una prueba de 4 horas con problemas

de nivel olímpico).

Este documento se basa en los materiales que se encuentran en la página web

https://www.hmmt.org/www/archive/problems

Agrupados en un único archivo “pdf” mediante la aplicación www.ilovepdf.com

1998 HMMT Advanced Topics Event

1. Evaluate sin( )sin( )1998 237 1998 1653°+ ° °− ° .

2. How many values of x, -19 < x < 98 satisfy cos sin2 22 1x x+ = ?

3. Find the sum of the infinite series 1 2 3 411998

11998

2 11998

3+ + + +( ) ( ) ( ) Κ

4. Find the range of f AA A A A A A

A A A A( )

sin ( cos cos sin sin cos )

tan (sec sin tan )=

+ + +

−

3 32 4 2 2 2

if

An

≠π

2.

5. How many positive integers less than 1998 are relatively prime to 1547? (Two

integers are relatively prime if they have no common factors besides 1.)

6. In the diagram below, how many distinct paths are there from January 1 to December

31, moving from one adjacent dot to the next either to the right, down, or diagonally

down to the right?

Jan. 1 -> * * * * * * * * * *

* * * * * * *

* * * * * * * * * *

* * * * *

* * * * * * * * * * <-Dec. 31

7. The Houson Association of Mathematics Educators decides to hold a grand forum on

mathematics education and invites a number of politicians from around the United

States to participate. Around lunch time the politicians decide to play a game. In this

game, players can score 19 points for pegging the coordinator of the gathering with a

spit ball, 9 points for downing an entire cup of the forum’s interpretation of coffee, or

8 points for quoting more than three consecutive words from the speech Senator

Bobbo delivered before lunch. What is the product of the two greatest scores that a

player cannot score in this game?

8. Given any two positive real numbers x and y, then x y◊ is a positive real number

defined in terms of x and y by some fixed rule. Suppose the operation x y◊ satisfies

the equations

( ) ( )x y y x y y⋅ ◊ = ◊ and ( )x x x◊ ◊ = ◊1 1 for all x y, > 0 . Given that 1 1 1◊ = , find

19 98◊ .

9. Bob’s Rice ID number has six digits, each a number from 1 to 9, and any digit can be

used any number of times. The ID number satifies the following property: the first

two digits is a number divisible by 2, the first three digits is a number divisible by 3,

etc. so that the ID number itself is divisible by 6. One ID number that satifies this

condition is 123252. How many different possibilities are there for Bob’s ID number?

10. In the fourth annual Swirled Series, the Oakland Alphas are playing the San Francisco

Gammas. The first game is played in San Francisco and succeeding games alternate in

location. San Francisco has a 50% chance of winning their home games, while

Oakland has a probability of 60% of winning at home. Normally, the series will

stretch on forever until one team gets a three game lead, in which case they are

declared the winners. However, after each game in San Francisco there is a 50%

chance of an earthquake, which will cause the series to end with the team that has

won more games declared the winner. What is the probability that the Gammas will

win?

Advanced Topics

1. Evaluatesin(1998◦ + 237◦) sin(1998◦ − 1653◦).

Answer: −14. We have sin(1998◦+237◦) sin(1998◦−1653◦) = sin(2235◦) sin(345◦) = sin(75◦) sin(−15◦) =

− sin(75◦) sin(15◦) = − sin(15◦) cos(15◦) = − sin(30◦)2 = − 1

4 .

2. How many values of x, −19 < x < 98, satisfy

cos2 x+ 2 sin2 x = 1?

Answer: 38. For any x, sin2 x+cos2 x = 1. Subtracting this from the given equation gives sin2 x = 0,or sinx = 0. Thus x must be a multiple of π, so −19 < kπ < 98 for some integer k, or approximately−6.1 < k < 31.2. There are 32 values of k that satisfy this, so there are 38 values of x that satisfycos2 x+ 2 sin2 x = 1.

3. Find the sum of the infinite series

1 + 2(1

1998) + 3(

11998

)2 + 4(1

1998)3 + . . . .

Answer:(

19981997

)2or 3992004

3988009 . We can rewrite the sum as(1 +

11998

+(

11998

)2

+ . . .

)+

(1

1998+(

11998

)2

+(

11998

)3

+ . . .

)+

((1

1998

)2

+(

11998

)3

+ . . .

)+. . . .

Evaluating each of the infinite sums gives

11− 1

1998

+1

1998

1− 11998

+( 11998 )2

1− 11998

+. . . =19981997

·

(1 +

11998

+(

11998

)2

+ . . .

)=

19981997

·(

1 +1

1998+ (

11998

)2 + . . .

),

which is equal to(

19981997

)2, or 39920043988009 , as desired.

4. Find the range of

f(A) =(sinA)

(3 cos2A+ cos4A+ 3 sin2A+ (sin2A)(cos2A)

)(tanA) (secA− (sinA)(tanA))

if A 6= nπ2 .

Answer: (3, 4). We factor the numerator and write the denominator in term of fractions to get

(sinA)(3 + cos2A)(sin2A+ cos2A)(sinAcosA

) (1

cosA −sin2 AcosA

) =(sinA)(3 + cos2A)(sin2A+ cos2A)

(sinA)(1−sin2 A)cos2 A

.

Because sin2A+ cos2A = 1, 1− sin2A = cos2A, so the expression is simply equal to 3 + cos2A. Therange of cos2A is (0, 1) (0 and 1 are not included because A 6= nπ

2 , so the range of 3 + cos2A is (3, 4).

5. How many positive integers less than 1998 are relatively prime to 1547? (Two integers are relativelyprime if they have no common factors besides 1.)

Answer: 1487. The factorization of 1547 is 7 · 13 · 17, so we wish to find the number of positiveintegers less than 1998 that are not divisible by 7, 13, or 17. By the Principle of Inclusion-Exclusion, wefirst subtract the numbers that are divisible by one of 7, 13, and 17, add back those that are divisibleby two of 7, 13, and 17, then subtract those divisible by three of them. That is,

1997−⌊

19977

⌋−⌊

199713

⌋−⌊

199717

⌋+⌊

19977 · 13

⌋+⌊

19977 · 17

⌋+⌊

199713 · 17

⌋−⌊

19977 · 13 · 17

⌋,

or 1487.

1

6. In the diagram below, how many distinct paths are there from January 1 to December 31, mov-ing from one adjacent dot to the next either to the right, down, or diagonally down to the right?

Answer: 372. For each dot in the diagram, we can count the number of paths from January 1to it by adding the number of ways to get to the dots to the left of it, above it, and above and tothe left of it, starting from the topmost leftmost dot. This yields the following numbers of paths:

So the number of paths from January 1 to December 31 is 372.

7. The Houson Association of Mathematics Educators decides to hold a grand forum on mathematicseducation and invites a number of politicians from the United States to participate. Around lunchtime the politicians decide to play a game. In this game, players can score 19 points for peggingthe coordinator of the gathering with a spit ball, 9 points for downing an entire cup of the forum’sinterpretation of coffee, or 8 points for quoting more than three consecutive words from the speechSenator Bobbo delivered before lunch. What is the product of the two greatest scores that a playercannot score in this game?

Answer: 1209. Attainable scores are positive integers that can be written in the form 8a+ 9b+ 19c,where a, b, and c are nonnegative integers. Consider attainable number of points modulo 8.

Scores that are 0 (mod 8) can be obtained with 8a for positive a.

Scores that are 1 (mod 8) greater than or equal to 9 can be obtained with 9 + 8a for nonnegative a.

Scores that are 2 (mod 8) greater than or equal to 18 can be obtained with 9 · 2 + 8a.

Scores that are 3 (mod 8) greater than or equal to 19 can be obtained with 19 + 8a.

Scores that are 4 (mod 8) greater than or equal to 19 + 9 = 28 can be obtained with 19 + 9 + 8a.

Scores that are 5 (mod 8) greater than or equal to 19 + 9 · 2 = 37 can be obtained with 19 + 9 · 2 + 8a.

Scores that are 6 (mod 8) greater than or equal to 19 · 2 = 38 can be obtained with 19 · 2 + 8a.

Scores that are 7 (mod 8) greater than or equal to 19 · 2 + 9 = 47 can be obtained with 19 · 2 + 9 + 8a.

So the largest two unachievable values are 39 and 31. Multiplying them gives 1209.

2

8. Given any two positive real numbers x and y, then x � y is a positive real number defined in terms ofx and y by some fixed rule. Suppose the operation x � y satisfies the equations (x · y) � y = x(y � y)and (x � 1) � x = x � 1 for all x, y > 0. Given that 1 � 1 = 1, find 19 � 98.

Answer: 19. Note first that x�1 = (x·1)�1 = x·(1�1) = x·1 = x. Also, x�x = (x�1)�x = x�1 = x.Now, we have (x · y) � y = x · (y � y) = x · y. So 19 � 98 = (19

98 · 98) � 98 = 1998 · (98 � 98) = 19

98 · 98 = 19.

9. Bob’s Rice ID number has six digits, each a number from 1 to 9, and any digit can be used any numberof times. The ID number satisfies the following property: the first two digits is a number divisible by2, the first three digits is a number divisible by 3, etc. so that the ID number itself is divisible by 6.One ID number that satisfies this condition is 123252. How many different possibilities are there forBob’s ID number?

Answer: 324. We will count the number of possibilities for each digit in Bob’s ID number, thenmultiply them to find the total number of possibilities for Bob’s ID number. There are 3 possibilitiesfor the first digit given any last 5 digits, because the entire number must be divisible by 3, so thesum of the digits must be divisible by 3. Because the first two digits are a number divisible by 2, thesecond digit must be 2, 4, 6, or 8, which is 4 possibilities. Because the first five digits are a numberdivisible by 5, the fifth digit must be a 5. Now, if the fourth digit is a 2, then the last digit has twochoices, 2, 8, and the third digit has 5 choices, 1, 3, 5, 7, 9. If the fourth digit is a 4, then the lastdigit must be a 6, and the third digit has 4 choices, 2, 4, 6, 8. If the fourth digit is a 6, then the lastdigit must be a 4, and the third digit has 5 choices, 1, 3, 5, 7, 9. If the fourth digit is an 8, then thelast digit has two choices, 2, 8, and the third digit has 4 choices, 2, 4, 6, 8. So there are a total of3 · 4 (2 · 5 + 4 + 5 + 2 · 4) = 3 · 4 · 27 = 324 possibilities for Bob’s ID number.

10. In the fourth annual Swirled Series, the Oakland Alphas are playing the San Francisco Gammas. Thefirst game is played in San Francisco and succeeding games alternate in location. San Francisco hasa 50% chance of winning their home games, while Oakland has a probability of 60% of winning athome. Normally, the serios will stretch on forever until one team gets a three-game lead, in which casethey are declared the winners. However, after each game in San Francisco there is a 50% chance of anearthquake, which will cause the series to end with the team that has won more games declared thewinner. What is the probability that the Gammas will win?

Answer: 3473 . Let F (x) be the probability that the Gammas will win the series if they are ahead by

x games and are about to play in San Francisco, and let A(x) be the probability that the Gammas willwin the series if they are ahead by x games and are about to play in Oakland. Then we have

F (2) =34

+A(1)

4

A(1) =6F (0)

10+

4F (2)10

F (0) =14

+A(1)

4+A(−1)

4

A(−1) =6F (−2)

10+

4F (0)10

F (−2) =A(−1)

4

Plugging A(1) = 6F (0)10 + 4F (2)

10 into F (2) = 34 + A(1)

4 , we get

F (2) =34

+14

(6F (0)

10+

4F (2)10

)9F (2)

10=

34

+6F (0)

40⇔ F (2) =

56

+F (0)

6

3

Plugging A(−1) = 6F (−2)10 + 4F (0)

10 into F (−2) = A(−1)4 , we get

34A(−1)40

=4F (0)

10⇔ F (−2) =

2F (0)17

Now,

F (0) =14

+14

(6F (0)

10+

4F (2)10

)+

14

(6F (−2)

10+

4F (0)10

)This simplifies to F (0) = 1

4 + F (0)4 + F (2)

10 + 6F (−2)40 . Then, plugging our formulas in, we get

F (0) =14

+F (0)

4+

110

(56

+F (0)

6

)+

3F (0)170

.

73F (0)102

=13⇔ F (0) =

3473.

Since F (0) is the situation before the Series has started, the probability that the Gammas will win is 3473 .

4

1998 HMMT Algebra Event

1. The cost of 3 hamburgers, 5 milk shakes, and 1 order of fries at a certain fast food restaurant is $23.50. At the same restaurant, the cost of 5 hamburgers, 9 milk shakes, and 1 order of fries is $39.50. What is the cost of 2 hamburgers, 2 milk shakes and 2 orders of fries at this restaurant? 2. Bobbo starts swimming at 2 feet/s across a 100 foot wide river with a current of 5

feet/s. Bobbo doesn’t know that there is a waterfall 175 feet from where he entered the river. He realizes his predicament midway across the river. What is the minimum speed that Bobbo must increase to make it to the other side of the river safely?

3. Find the sum of every even positive integer less than 233 not divisible by 10.

4. Given that r and s are relatively prime positive integers such that r

s=

+

+

2 2 10

5 3 5

( )

( ( ),

find r and s. 5. A man named Juan has three rectangular solids, each having volume 128. Two of the

faces of one solid have areas 4 and 32. Two faces of another solid have areas 64 and 16. Finally, two faces of the last solid have areas 8 and 32. What is the minimum possible exposed surface area of the tallest tower Juan can construct by stacking his solids one on top of the other, face to face? (Assume that the base of the tower is not exposed.)

6. How many pairs of positive integers ( , )a b with a b≤ satisfy 1 1 1

6a b+ = ?

7. Given that three roots of f x x ax bx c( ) = + + +4 2 are 2, -3, and 5, what is the value

of a b c+ + ?

8. Find the set of solutions for x in the inequality x

x

x

x

+

+

>+

+

1

2

3 4

2 9 when x x≠ − ≠ −2 9

2, .

9. Suppose f x( ) is a rational function such that 32

1 2f

f x

xx

x( )

( )+ = for x ≠ 0 .

Find f ( )−2 .

10. G. H. Hardy once went to visit Srinivasa Ramanujan in the hospital, and he started the

conversation with: “I came here in taxi-cab number 1729. That number seems dull to

me, which I hope isn’t a bad omen.” “Nonsense,” said Ramanujan. “The number isn’t

dull at all. It’s quite interesting. It’s the smallest number that can be expressed as the

sum of two cubes in two different ways.” Ramujan had immediately seen that

1729 12 1 10 93 3 3 3= + = + . What is the smallest positive integer representable as the

sum of the cubes of three positive integers in two different ways?

HMMT 1998: Algebra Solutions

1. Problem: The cost of 3 hamburgers, 5 milk shakes, and 1 order of fries at a certain fast food restaurantis $23.50. At the same restaurant, the cost of 5 hamburgers, 9 milk shakes, and 1 order of fries is $39.50.What is the cost of 2 hamburgers, 2 milk shakes ,and 2 orders of fries at this restaurant?

Solution: Let H = hamburger, M = milk shake, and F = order of fries. Then 3H + 5M + F = $23.50.Multiplying the equation by 2 yields 6H + 10M + 2F = $47. Also, it is given that 5H + 9M + F = $39.50.Then subtracting the following equations

6H +10M +2F = $47.005H +9M +F = $39.50

yields H + M + F = $7.50. Multiplying the equation by 2 yields 2H + 2M + 2F = $15 .

2. Problem: Bobbo starts swimming at 2 feet/s across a 100 foot wide river with a current of 5 feet/s.Bobbo doesn’t know that there is a waterfall 175 feet from where he entered the river. He realizes hispredicament midway across the river. What is the minimum speed that Bobbo must increase to make it tothe other side of the river safely?

Solution: When Bobbo is midway across the river, he has travelled 50 feet. Going at a speed of 2feet/s, this means that Bobbo has already been in the river for 50 feet

20 feet/s = 25 s. Then he has traveled5 feet/s · 25s = 125 feet down the river. Then he has 175 feet-125 feet = 50 feet left to travel downstreambefore he hits the waterfall.

Bobbo travels at a rate of 5 feet/s downstream. Thus there are 50 feet5 feet/s = 10s before he hits the waterfall.

He still has to travel 50 feet horizontally across the river. Thus he must travel at a speed of 50 feet10s = 5

feet/s. This is a 3 feet/s difference from Bobbo’s original speed of 2 feet/s.

3. Problem: Find the sum of every even positive integer less than 233 not divisible by 10.

Solution: We find the sum of all positive even integers less than 233 and then subtract all the positiveintegers less than 233 that are divisible by 10.

2+4+. . .+232 = 2(1+2+. . .+116) = 116·117 = 13572. The sum of all positive integers less than 233 thatare divisible by 10 is 10+20+. . .+230 = 10(1+2+. . .+23) = 2760. Then our answer is 13572-2760 = 10812 .

4. Problem: Given that r and s are relatively prime positive integers such that rs = 2(

√2+√

10)

5(√

3+√

5), find r

and s.

Solution: Squaring both sides of the given equation yields r2

s2 = 4(12+4√

5)

25(3+√

5)= 16(3+

√5)

25(3+√

5)= 16

25 . Because r

and s are positive and relatively prime, then by inspection, r = 4 and s = 5 .

5. Problem: A man named Juan has three rectangular solids, each having volume 128. Two of the facesof one solid have areas 4 and 32. Two faces of another solid have areas 64 and 16. Finally, two faces of thelast solid have areas 8 and 32. What is the minimum possible exposed surface area of the tallest tower Juancan construct by stacking his solids one on top of the other, face to face? (Assume that the base of the toweris not exposed).

Solution: Suppose that x, y, z are the sides of the following solids. Then Volume = xyz = 128. For thefirst solid, without loss of generality (with respect to assigning lengths to x, y, z), xy = 4 and yz = 32. Thenxy2z = 128. Then y = 1. Solving the remaining equations yields x = 4 and z = 32. Then the first solid hasdimensions 4× 1× 32.

For the second solid, without loss of generality, xy = 64 and yz = 16. Then xy2z = 1024. Then y = 8.Solving the remaining equations yields x = 8 and z = 2. Then the second solid has dimensions 8× 8× 2.

For the third solid, without loss of generality, xy = 8 and yz = 32. Then y = 2. Solving the remainingequations yields x = 4 and z = 16. Then the third solid has dimensions 4× 2× 16.

To obtain the tallest structure, Juan must stack the boxes such that the longest side of each solid isoriented vertically. Then for the first solid, the base must be 1 × 4, so that the side of length 32 can

1

contribute to the height of the structure. Similarly, for the second solid, the base must be 8× 2, so that theside of length 8 can contribute to the height. Finally, for the third solid, the base must be 4× 2. Thus thestructure is stacked, from bottom to top: second solid, third solid, first solid. This order is necessary, sothat the base of each solid will fit entirely on the top of the solid directly beneath it.

All the side faces of the solids contribute to the surface area of the final solid. The side faces of the bottomsolid have area 8 · (8 + 2 + 8 + 2) = 160. The side faces of the middle solid have area 16 · (4 + 2 + 4 + 2) = 192.The sides faces of the top solid have area 32 · (4 + 1 + 4 + 1) = 320.

Furthermore, the top faces of each of the solids are exposed. The top face of the bottom solid is partiallyobscured by the middle solid. Thus the total exposed area of the top face of the bottom solid is 8 ·2−4 ·2 = 8.The top face of the middle solid is partially obscured by the top solid. Thus the total exposed area of thetop face of the middle solid is 4 · 2− 4 · 1 = 4. The top face of the top solid is fully exposed. Thus the totalexposed area of the top face of the top solid is 4 · 1 = 4.

Then the total surface area of the entire structure is 160 + 192 + 320 + 8 + 4 + 4 = 688 .

6. Problem: How many pairs of positive integers (a, b) with ≤ b satisfy 1a + 1

b = 16?

Solution: 1a + 1

b = 16 ⇒

a+bab = 1

6 ⇒ ab = 6a+6b⇒ ab−6a−6b = 0. Factoring yields (a−b)(b−6)−36 = 0.Then (a−6)(b−6) = 36. Because a and b are positive integers, only the factor pairs of 36 are possible valuesof a− 6 and b− 6. The possible pairs are:

a− 6 = 1, b− 6 = 36a− 6 = 2, b− 6 = 18a− 6 = 3, b− 6 = 12a− 6 = 4, b− 6 = 9a− 6 = 6, b− 6 = 6

Because a ≤ b, the symmetric cases, such as a− 6 = 12, b− 6 = 3 are not applicable. Then there are 5possible pairs.

7. Problem: Given that three roots of f(x) = x4+ax2+bx+c are 2, -3, and 5, what is the value of a+b+c?

Solution: By definition, the coefficient of x3 is negative the sum of the roots. In f(x), the coefficient of x3

is 0. Thus the sum of the roots of f(x) is 0. Then the fourth root is -4. Then f(x) = (x−2)(x+3)(x−5)(x+4).Notice that f(1) is 1 + a + b + c. Thus our answer is f(1)− 1 = (1− 2)(1 + 3)(1− 5)(1 + 4)− 1 = 79 .

8. Problem: Find the set of solutions for x in the inequality x+1x+2 > 3x+4

2x+9 when x 6= −2, x 6= 92 .

Solution: There are 3 possible cases of x: 1) − 92 < x, 2) 9

2 ≤ x ≤ −2, 3) −2 < x. For the cases (1) and(3), x + 2 and 2x + 9 are both positive or negative, so the following operation can be carried out withoutchanging the inequality sign:

x + 1x + 2

>3x + 42x + 9

⇒ 2x2 + 11x + 9 > 3x2 + 10x + 8

⇒ 0 > x2 − x− 1

The inequality holds for all 1−√

52 < x < 1+

√5

2 . The initial conditions were − 92 < x or −2 < x. The

intersection of these three conditions occurs when 1−√

52 < x < 1+

√5

2 .Case (2) is 9

2 ≤ x ≤ −2. For all x satisfying these conditions, x + 2 < 0 and 2x + 9 > 0. Then thefollowing operations will change the direction of the inequality:

2

x + 1x + 2

>3x + 42x + 9

⇒ 2x2 + 11x + 9 < 3x2 + 10x + 8

⇒ 0 < x2 − x− 1

The inequality holds for all x < 1−√

52 and 1+

√5

2 < x. The initial condition was −92 ≤ x ≤ −2. Hence

the intersection of these conditions yields all x such that −92 ≤ x ≤ −2. Then all possible cases of x are

−92 ≤ x ≤ −2 ∪ 1−

√5

2 < x < 1+√

52 .

9. Problem: Suppose f(x) is a rational function such that 3f( 1x ) + 2f(x)

x = x2 for x 6= 0. Find f(−2).

Solution: Let x = −12 . Then

3f(−2) +2f(−1

2 )−12

=14

⇒ 3f(−2)− 4f(−12

) =14

(1)

Let x = −2. Then

3f(−12

) +2f(−2)−2

= 4

⇒ 3f(−12

)− f(−2) = 4 (2)

Solving this system of equations {(1), (2)} for f(−2) yields f(−2) = 6720 .

10. Problem: G.H. Hardy once went to visit Srinivasa Ramanujan in the hospital, and he started theconversation with: “I came here in taxi-cab number 1729. That number seems dull to me, which I hopeisn’t a bad omen.” “Nonsense,” said Ramanujan. “The number isn’t dull at all. It’s quite interesting. It’sthe smallest number that can be expressed as the sum of two cubes in two different ways.” Ramanujan hadimmediately seen that 1729 = 123 + 13 = 103 + 93. What is the smallest positive integer representable asthe sum of the cubes of three positive integers in two different ways?

Solution: Let this smallest positive integer be represented as a3 + b3 + c3 = d3 + e3 + f3. By inspection,a solution is not possible with the first 4 cubes. We prove that it is impossible to write the same numberas two different sums of the first 5 cubes. Because we necessarily need to use the 5th cube (otherwise, thisproof would be for the first 4 cubes), we have 53 + b3 + c3 = d3 + e3 +f3. Without loss of generality, supposed = 5. By inspection, there is no solution to b3 + c3 = e3 + f3, such that b, c, e, f ≤ 5 and b, c and e, f areunique.

Then none of d, e, f are 5. Then at least two must be 4, otherwise the RHS would be too small. Withoutloss of generality, suppose d = e = 4. Then b3 + c3 = 3 + f3. By inspection, there are no possible solutionsif b, c, f ≤ 4.

Thus if a = 5, there are no solutions.Suppose that there is a solution within the first 6 cubes. Then a = 6. By the same analysis as above,

d = e = 5, otherwise the RHS would be too small. Then b3 + c3 = 34 + f3. By inspection, we see that apossible solution is b = 3, c = 2, f = 1. Then the desired integer is 63 + 33 + 23 = 251 .

3

1998 Harvard/MIT Math Tournament

CALCULUS Answer Sheet

Name:

School: Grade:

1 6

2 7

3 8

4 9

5 10

TOTAL:

CALCULUS

Question One. [3 points]

Farmer Tim is lost in the densely-forested Cartesian plane. Starting

from the origin he walks a sinusoidal path in search of home; that

is, after t minutes he is at position (t,sint).

Five minutes after he sets out, Alex enters the forest at the origin

and sets out in search of Tim. He walks in such a way that after he

has been in the forest for m minutes, his position is (m,cos t).

What is the greatest distance between Alex and Farmer Tim while

they are walking in these paths?

Question Two. [3 points]

A cube with sides 1m in length is filled with water, and has a tiny

hole through which the water drains into a cylinder of radius 1m.

If the water level in the cube is falling at a rate of 1 cm

s , at what

rate is the water level in the cylinder rising?

Question Three. [4 points]

Find the area of the region bounded by the graphs of y = x2, y = x ,

and x = 2 .

Question Four. [4 points]

Let f(x) = 1+

x

2+x2

4+x 3

8+…, for −1≤ x ≤ 1. Find e

f(x)dx

0

1

∫.

Question Five. [5 points]

Evaluate limx→1

x

x

sin(1−x).

Question Six. [5 points]

Edward, the author of this test, had to escape from prison to work

in the grading room today. He stopped to rest at a place 1,875 feet

from the prison and was spotted by a guard with a crossbow.

The guard fired an arrow with an initial velocity of 100 ft

s . At the

same time, Edward started running away with an acceleration of

1 ft

s 2. Assuming that air resistance causes the arrow to decelerate at

1 ft

s 2 and that it does hit Edward, how fast was the arrow moving at

the moment of impact (in ft

s )?

Question Seven. [5 points]

A parabola is inscribed in equilateral triangle ABC of side length 1 in

the sense that AC and BC are tangent to the parabola at A and B,

respectively:

Find the area between AB and the parabola.

Question Eight. [6 points]

Find the slopes of all lines passing through the origin and tangent to

the curve y2

= x3

+ 39x − 35.

Question Nine. [7 points]

Evaluate

1

n ⋅ 2n−1n=1

∞

∑

Question Ten. [8 points]

Let S be the locus of all points (x,y) in the first quadrant such that

x

t+

y

1− t= 1 for some t with 0<t<1. Find the area of S.

HMMT 1998: Calculus Solutions

1. Problem: Farmer Tim is lost in the densely-forested Cartesian plane. Starting from the origin he walksa sinusoidal path in search of home; that is, after t minutes he is at position (t, sin t).

Five minutes after he sets out, Alex enters the forest at the origin and sets out in search of Tim. Hewalks in such a way that after he has been in the forest for m minutes, his position is (m, cos t).

What is the greatest distance between Alex and Farmer Tim while they are walking in these paths?

Solution: At arbitrary time t, Farmer Tim is at position (t, sin t) and Alex is at position (t − 5, cos t).Hence at time t, the distance, d, between Tim and Alex is d =

√(sin t− cos t)2 + 25. To find the maximum

value of d, we solve for t such that dddt = 0.

dddt = (sin t−cos t)(cos t+sin t)√

(sin t−cos t)2+25. Then dd

dt = 0 ⇒ sin2 t − cos2 t = 0 ⇒ sin2 t = cos2 t. Equality happens if t is

any constant multiple of π4 .

Notice that to maximize d, we need to maximize (sin t − cos t)2. This is achieved when cos t = − sin t.Because we determined earlier that t is a constant multiple of π

4 , then with this new condition, we see thatt must be a constant multiple of 3π

4 .

Then (sin t− cos t)2 = 2⇒ d =√

29 .

2. Problem: A cube with sides 1m in length is filled with water, and has a tiny hole through which thewater drains into a cylinder of radius 1m. If the water level in the cube is falling at a rate of 1 cm/s, at whatrate is the water level in the cylinder rising?

Solution: The magnitude of the change in volume per unit time of the two solids is the same. The changein volume per unit time of the cube is 1 cm·m2/s. The change in volume per unit time of the cylinder isπ · dhdt ·m

2, where dhdt is the rate at which the water level in the cylinder is rising.

Solving the equation π · dhdt ·m2 = 1 cm ·m2/s yields 1

π cm/s .

3. Problem: Find the area of the region bounded by the graphs of y = x2, y = x, and x = 2.

Solution: There are two regions to consider. First, there is the region bounded by y = x2 and y = x, inthe interval [0, 1]. In this interval, the values of y = x are greater than the values of y = x2, thus the area is

calculated by∫ 1

0

(x− x2

)dx.

Second, there is the region bounded by y = x2 and y = x and x = 2, in the interval [1, 2]. In this interval,

the values of y = x2 are greater than the values of y = x, thus the area is calculated by∫ 2

1

(x2 − x

)dx.

Then the total area of the region bounded by the three graphs is∫ 1

0

(x− x2

)dx+

∫ 2

1

(x2 − x

)dx = 1 .

4. Problem: Let f(x) = 1 + x2 + x2

4 + x3

8 + . . ., for −1 ≤ x ≤ 1. Find

√e

∫ 1

0

f(x)dx.

Solution: Observe that f(x) is merely an infinite geometric series. Thus f(x) = 11− x

2= 2

2−x . Then∫ 1

0

22− x

= 2 ln 2. Then√e2 ln 2 =

√22 = 2 .

5. Problem: Evaluate limx→1

xx

sin(1−x) .

Solution: Rewrite the expression to evaluate as eln xx

sin(1−x) . Then we must evaluate limx→1

eln xx

sin(1−x).

limx→1

lnxx

sin(1−x) = limx→1

(x

sin(1− x)lnx)

. Because direct calculation of the limit results in indeterminate

form ( 10 · 0), we can use L’Hopital’s rule to evaluate the limit. By L’Hopital’s rule, lim

x→1

(x

sin(1− x)lnx)

=

limx→1

lnx+ 1− cos(1− x)

. This limit is simply -1.

1

Hence limx→1

eln xx

sin(1−x)= e−1 = 1

e .

6. Problem: Edward, the author of this test, had to escape from prison to work in the grading roomtoday. He stopped to rest at a place 1,875 feet from the prison and was spotted by a guard with a crossbow.The guard fired an arrow with an initial velocity of 100 ft/s. At the same time, Edward started runningaway with an acceleration of 1 ft/s2. Assuming that air resistance causes the arrow to decelerate at 1 ft/s2

and that it does hit Edward, how fast was the arrow moving at the moment of impact (in ft/s)?

Solution: We use the formula for distance, d = 12at

2 + vt + d0. Then after t seconds, Edward is atlocation 1875 + 1

2 (1)(t2) from the prison. After t seconds, the arrow is at location 12 (−1)(t2) + 100t from the

prison. When the arrow hits Edward, both objects are at the same distance away from the tower. Hence1875 + 1

2 (1)(t2) = 12 (−1)(t2) + 100t. Solving for t yields t2 − 100t + 1875 = 0 ⇒ t = 25 or t = 75. Then it

must be t = 25, because after the arrow hits Edward, he will stop running.After 25 seconds, the arrow is moving at a velocity of 100-25(1) = 75 ft/s .

7. Problem: A parabola is inscribed in equilateral triangle ABC of side length 1 in the sense that ACand BC are tangent to the parabola at A and B, respectively. Find the area between AB and the parabola.

Solution: Suppose A = (0, 0), B = (1, 0), and C = ( 12 ,√

32 ). Then the parabola in question goes through

(0, 0) and (1, 0) and has tangents with slopes of√

3 and −√

3, respectively, at these points. Suppose theparabola has equation y = ax2 + bx+ c. Then dy

dx = 2ax+ b.At point (0, 0), dydx = b. Also the slope at (0, 0), as we determined earlier, is

√3. Hence b =

√3. Similarly,

at point (1, 0), dydx = 2a+ b. The slope at (1, 0), as we determined earlier, is −

√3. Then a = −

√3.

Since the parabola goes through (0, 0), c = 0. Hence the equation of the parabola is y = −√

3x2 +√

3x.The desired area is simply the area under the parabolic curve in the interval [0, 1].

Hence∫ 1

0

(−√

3x2 +√

3x)dx =

√3

6 .

8. Problem: Find the slopes of all lines passing through the origin and tangent to the curvey2 = x3 + 39x− 35.

Solution: Any line passing throug the origin has equation y = mx, where m is the slope of the line. If aline is tangent to the given curve, then at the point of tangency, (x, y), dy

dx = m.First, we calculate dy

dx of the curve: 2ydy = 3x2dx+ 39dx⇒ dydx = 3x2+39

2y . Substituting mx for y, we getthe following system of equations:

m2x2 = x3 + 39x− 35

m =3x2 + 39

2mx

Solving for x yields the equation x3 − 39x+ 70 = 0 ⇒ (x− 2)(x+ 7)(x− 5) = 0 ⇒ x = 2 or x = −7 orx = 5. These solutions indicate the x-coordinate of the points at which the desired lines are tangent to thecurve. Solving for the slopes of these lines, we get m = ±

√512 for x = 2, no real solutions for x = −7, and

m = ±√

2855 for x = 5. Thus m = ±

√512 ,±

√2855 .

9. Problem: Evaluate∞∑n=1

1n · 2n−1

.

Solution: Note that if we take the integral of f(x) in problem 4, we get the function F (x) = x + x2

2·2 +x3

3·22 + . . .. Evaluating this integral in the interval [0, 1], we get 1 + 12·2 + 1

3·22 + . . ., which is the desired sum.

Hence∫ 1

0

22− x

dx = 2 ln 2.

2

10. Problem: Let S be the locus of all points (x, y) in the first quadrant such that xt + y

1−t = 1 for somet with 0 < t < 1. Find the area of S.

Solution: Solving for t in the given equation, we get t2 + (y − x − 1)t + x = 0. Using the quadratic

equation, we get t = (x+1−y)±√

(y−x−1)2−4x

2 . For all valid combinations of (x, y), t is positive and less than 1(this is easy to see by inspection). All valid combinations of (x, y) are those that make (y−x−1)2−4x ≥ 0.

Solving for y in the equation (y−x−1)2−4x = 0 yields y2−(2x+2)y+(x−1)2 ≥ 0⇒ y = (x+1)±2√x.

In the original equation, it is given that xt + y

1−t = 1, and 0 < t < 1. This implies that x, y < 1. Then theonly possible y < 1 that satisfies (y − x− 1)2 − 4x = 0 is y = x+ 1− 2

√x.

Then to satisfy the inequality (y − x− 1)2 − 4x ≥ 0, we must have y ≤ x+ 1− 2√x. Recall that this is

when 0 < y < 1. Hence we integrate in the interval [0, 1]: ∈10 x+ 1− 2

√x = 1

6 .

3

1998 Harvard/MIT Math Tournament

GEOMETRY Answer Sheet

Name:

School: Grade:

1 7

2 8

3 9

4 10a

5 10b

6 10c

TOTAL:

GEOMETRY

Question One. [3 points]

Quadrilateral ALEX, pictured below (but not necessarily to scale!)

can be inscribed in a circle; with m∠LAX = 20° and m∠AXE = 100°:

A

L

E

X D

N

Calculate m∠EDX .

Question Two. [3 points]

Anne and Lisa enter a park that has

two concentric circular paths joined

by two radial paths, one of which is at

the point where they enter. Anne goes

in to the inner circle along the first

radial path, around by the shorter way

to the second radial path and out

along it to the exit. Walking at a the

same rate, Lisa goes around the outer

circle to the exit, taking the shorter

of the two directions around the park. enter

exit

They arrive at the exit at the same time. The radial paths meet at

the center of the park; what is the angle between them?

Question Three. [4 points]

MD is a chord of length 2 in a circle of radius 1, and L is chosen on

the circle so that the area of triangle MLD is the maximized.

Find m∠MLD.

Question Four. [4 points]

A cube with side length 100cm is filled with water and has a hole

through which the water drains into a cylinder of radius 100cm. If

the water level in the cube is falling at a rate of 1 cm

s, how fast is

the water level in the cylinder rising?

Question Five. [5 points]

Square SEAN has side length 2 and a

quarter-circle of radius 1 around E is cut

out.

Find the radius of the largest circle that

can be inscribed in the remaining figure.

S

EA

N

Question Six. [5 points]

A circle is inscribed in an equilateral

triangle of side length 1. Tangents to

the circle are drawn that cut off

equilateral triangles at each corner.

Circles are inscribed in each of these

equilateral triangles. If this process is

repeated infinitely many times, what is

the sum of the areas of all the circles?

Question Seven. [5 points]

Pyramid EARLY has rectangular base EARL and apex Y, and all of its

edges are of integer length. The four edges from the apex have

lengths 1, 4, 7, 8 (in no particular order), and EY is perpendicular to

YR.

Find the area of rectangle EARL.

Question Eight. [6 points]

It is not possible to construct a segment of length π using a

straightedge, compass, and a given segment of length 1. The

following construction, given in 1685 by Adam Kochansky, yields a

segment whose length agrees with π to five decimal places:

Construct a circle of radius 1 and call its center O. Construct a

diameter AB of this circle and a line l tangent to the circle at A.

Next, draw a circle with radius 1 centered at A, and call one of the

intersections with the original circle C. Now from C draw an arc of

radius 1 intersecting the circle around A at D, where D lies outside

of the circle centered at O. Draw OD and let E be its point of

intersection with l . Construct H on AE such that A is between

H and E, and HE=3.

The distance between B and H is then close to π; calculate its exact

value.

Question Nine. [7 points]

Let T be the intersection of the common internal tangents of circles

C1, C2 with centers O1, O2 respectively. Let P be one of the points

of tangency on C1 and let line l bisect angle O1TP . Label the

intersection of l with C1 that is farthest from T, R, and label the

intersection of l with C2 that is closest to T, S. If C1 has radius 4,

C2 has radius 6, and O1O2 = 20 , calculate (TR)(TS).

O2O1

C1

C2

T

R

S

l

P

Question Ten [8 points].

Lukas is playing pool on a table shaped like an equilateral triangle.

The pockets are at the corners of the triangle and are labeled C, H,

and T. Each side of the table is 16 feet long. Lukas shoots a ball

from corner C of the table in such a way that on the second bounce,

the ball hits 2 feet away from him along side CH.

a. [5 points]

How many times will the ball bounce before hitting a pocket?

b. [2 points]

Which pocket will the ball hit?

c. [1 points]

How far will the ball travel before hitting the pocket?

Geometry SolutionsHarvard-MIT Math Tournament

Written by Anders Kaseorg

1. m6 EDX = 180◦ −m6 LAX −m6 ELA = 180◦ −m 6 LAX − (180◦ −m 6 AXE) = 80◦ .

2. If the angle in radians is θ, then Anne travels Rθ and Lisa travels (R − r) + rθ + (R − r).Setting these equal yields R(θ − 2) = r(θ − 2), so θ = 2 radians .

3. m6 MLD = 12

_

AB = 45◦ .

4. The cylinder has a cross-sectional area π times greater than the cube, so the water raises 1π

times as quickly in the cylinder as it lowers in the cube; that is, at 1π

cms .

5. If the new circle has radius r, then the distance from its center to E can be computed either

as 1 + r or (2− r)√

2. Setting these equal yields r = 2√

2−1√2+1

= 5− 3√

2 .

6. The central circle has area π(

12√

3

)2 = π12 , and each of the three small triangles are copies of

the entire figure dilated by 13 . Therefore, the total area is given by K = π

12 + 3 ·(

13

)2K ⇐⇒

23K = π

12 ⇐⇒ K = π8 .

7. Let E = (a, b, 0), A = (−c, b, 0), R = (−c,−d, 0), L = (a,−d, 0), Y = (0, 0, h), and observethat EY 2 + RY 2 = a2 + b2 + c2 + d2 + 2h2 = AY 2 + LY 2, which can only be satisfied byEY = 1, AY = 4, RY = 8, LY = 7 (or the symmetric configurations). Since EA is anintegral side of a triangle whose other sides are 1 and 4, we must have EA = 4; similarly,EL = 7. Therefore, the area of rectangle EARL is 28 . (Such a pyramid may be constructedby taking a = 1

8 , b = 114 , c = 31

8 , d = 9714 , h =

√307156 .)

8. Since OD ⊥ AC and 4AOC is equilateral, we have 6 AOD = 30◦. So AE = 1√3, and

BH =√

AB2 + AH2 =√

22 +(3− 1√

3

)2 =√

403 − 2

√3 ≈ 3.141533339.

9. The dilation of ratio − 23 about T sends C2 to C1, O2 to O1, and S to the other intersection

of s with C1, which we shall call U . We can now compute TR · TS = 32TR · TU = 3

2TP 2 =32 (O1T

2 −O1P2) = 3

2

((O1O21+3/2

)2

−O1P2

)= 3

2

((205/2

)2

− 42

)= 72 .

10. Suppose that when the ball hits a side of the table, instead of reflecting the ball’s path, wereflect the entire table over this side so that the path remains straight. If we repeatedlyreflect the table over its sides in all possible ways, we get a triangular grid that tiles theplane. Whenever the path crosses n lines in this grid parallel to CT , it will cross 7

8n linesparallel to CH and 15

8 n lines parallel to HT . After crosssing 8 + 7 + 15 = 30 grid lines it willhave crossed three lines simultaneously again, which means that the ball will have landed ina pocket after bouncing 27 times. By picturing the grid it is easy to see that the pocketin question is H . The distance the ball travels during the 1

8 of its trip described in theproblem is the third side of a triangle with an 120◦ angle between two sides 16 and 14, whichis√

162 + 142 − 2 · 16 · 14 cos 120◦ = 26, so length of the entire trip is 208 .

1

Power Question - Coloring Graphs

Perhaps you have heard of the Four Color Theorem (if not, don’t panic!), which essentially says that any

map (e.g. a map of the United States) can be colored with four or fewer colors without giving neighboring

regions (e.g. Massachusetts and New Hampshire) the same color. It was, until 22 years ago, one of the

most famous unsolved problems in mathematics. The only known proof, however, is so long that it

requires a computer to carry it out. This power question will lead you through the basic definitions and

theorems of graph theory necessary to prove the Five Color Theorem, a weaker and much easier (though

still quite challenging) statement.

Part I - Graphs

Definitions: A graph is a collection of points and lines (or curves), called

vertices and edges, respectively, where each edge connects exactly two

distinct vertices and any two vertices are connected by at most one edge. We

say that two vertices are adjacent if they are connected by an edge.

Notation: For this problem we will denote vertices by capital letters and the

edge connecting vertices X and Y by XY. Vertices will be drawn as dots so that

they will be distinguishable from edge crossings. We will denote graphs by

capital script letters, usually G or H.

Example: Here is a graph with vertices A, B, C, D, and edges AB, AC, BC.

B D

A C

a. Which of the following are graphs? List the vertices and edges of each graph

and explain why the others are not graphs. [1 point each]

i. A B iv. A B C

D C D E

ii. v. B C

A B C A

iii. B vi. A B

A C E

C

D

Definition: Graph G is said to be isomorphic (from Greek roots meaning “same

structure”) to H, written G ≅ H, if the vertices of G can be given the names of

the vertices of H in such a way that G and H have the same edges. This

relabling is called an isomorphism and is reversible, thus implying H ≅ G as

well.

Example: A B B L G ≅ H since we can relable the

vertices of G as follows:

A↔B, B↔L, C↔U, D↔E

D C E U (we write the double arrows to

emphasize the reversibility of

G H the isomorphism).

b. [1 point each]

i. List the edges of H.

ii. Show that H ≅ G in the example by writing an isomorphism from H to G.

From now on the diagrams in the problem will not have the vertices explicitly

labled unless necessary. Thus to show two unlabled graphs are isomorphic we

can arbitrarily lable the vertices of one and then show that the other can be

labled with the same names to yield the same edges.

c. For each of the following pairs of graphs, state whether or not they are

isomorphic (you do not need to justify your answer). [parts i-iii are 1 point

each, parts iv-vi are 3 points each]

i. iv.

and and

ii. v.

and and

iii. vi.

and and

A couple more definitions are best given in this section, though there won’t be

any questions specifically about them until later.

Definitions: A graph H is a subgraph of a graph G if every vertex of H is a

vertex of G and every edge of H is an edge of G.

The degree of a vertex is the number of edges connected to it.

Example: B B

H is a subgraph of G. Vertex B has

degree 2 in G and degree 1 in H.

A C A C

G H

Part II - Planar Graphs

There are not many nontrivial properties possessed by all graphs, so we will restrict our attention to

graphs with certain nice properties that we can exploit to prove cool theorems. So now we need one

more round of definitions, then the real fun begins!

Definitions: A graph is planar if it is isomorphic to a graph that has been drawn

in a plane without edge crossings. Otherwise a graph is nonplanar. It shouldn’t

be too hard to see that every subgraph of a planar graph is planar.

A walk in a graph is a sequence A1 A2 ... An of not necessarily distinct vertices

in which Ak is adjacent to Ak+1 for k = 1, 2, ..., n-1.

A graph is connected if every pair of vertices is joined by a walk, or

equivalently if there is a walk that passes through every vertex at least once.

Otherwise a graph is said to be disconnected.

When a planar graph is actually drawn in a plane without edge crossings, it

cuts the plane into regions called faces of the graph.

Example: A R L Is connected since the walk

EDWARDEARLY passes through every

Y vertex at least once. It has 3 faces

(don’t forget the exterior region

when counting).

W D E

a. How many vertices, edges, and faces do each of the following graphs have?

[3 points each]

i. ii.

b. For each of the following graphs, state whether or not it is planar. You do

not need to justify your answer. [3 points each]

i. iii.

ii. iv.

Euler’s formula states if a connected planar graph has v vertices, e edges, and

f faces, then v-e+f=2. Proving this would go beyond the scope of this problem,

so you may take this formula as given.

c. [15 points] Prove that if G is planar and connected with v≥3, then 3

2f ≤ e ≤ 3v − 6.

d. Prove that the following graphs are nonplanar.

i. [5 points] ii. [10 points]

e. [12 points] Prove that every planar graph has at least one vertex of degree 5

or less.

Part III - Coloring

Now we can begin talking about coloring planar graphs.

Definition: We say a graph has been colored if a color has been assigned to

each vertex in such a way that adjacent vertices have different colors.

The chromatic number χ (the Greek letter chi, pronounced like sky without the

s) of a graph is the smallest number of colors with which it can be colored.

Example: Here is a graph with chromatic number 3, colored in black, green,

and red.

b g

r

b b

a. Calculate χ for each of the following graphs. [3 points each]

i. iii.

ii. iv.

b. [14 points] Given a graph G, let χ be the chromatic number of the graph

obtained by taking the vertices of G and drawing edges only between those

that are not adjacent in G. Prove that χ + χ ≥ 2 v .

c. Assume G is a planar graph with χ>5.

i. [5 points] Prove that at least one of the following must be isomorphic to a

subgraph of G (where the vertex V is only adjacent to the vertices shown).

V

V V

V

V V

ii. [20 points] Prove that if one of the above is a subgraph of G (where the

vertex V is only adjacent to the vertices shown) then we can remove one

vertex (and all edges touching it) from G to obtain a graph with fewer vertices

which also has χ>5 (you may use the Jordan Curve Theorem, which states that

a closed loop that does not intersect itself divides the plane into an inside and

outside, and if a continuous curve joins a point on the inside to one on the

outside then it must cross the loop).

iii. [5 points] Prove the Five Color Theorem: Every planar graph has χ≤5.

1998 Power Question Solutions

I. Graphs, total of 20 points

a. completely correct gets 1 point, total of 6 points

i. yes. vertices A,B,C,D, edges AB,AC,AD,BD

ii. no. A and B are connected twice

iii. yes. vertices A,B,C, edges ABC

iv. yes. vertices A,B,C,D,E, edges AB,AD,AE,BE,CD,CE

v. no. the edge from B does not connect to another vertex

vi. no. E is connected to itself

b. completely correct gets 1 point, total of 2 points

i. BL,BE,BU,EL,EU

ii. A↔B, B↔L, C↔U, D↔E

c. 1 point each for i-iii, 3 for iv-vi, total of 12 points

i. no

ii. no

iii. no

iv. yes

v. no

vi. yes

note that the second graph in v has more vertices of degree 3 than the first

II. Planar Graphs, total of 60 points

a. 1 point each for v, e, f, total of 6 points

i. v=9, e=16, f=9

ii. v=7, e=14, f=9

note that v-e+f=2 by Euler’s formula

b. 3 points each, total of 12 points note that this is a planar drawing of ii:

i. yes

ii. yes

iii. no

iv. yes

for iii, note that if we remove the two vertices of degree two and make the vertices they

were adjacent to adjacent to each other then we get a graph isomorphic to the one in d.ii.

c. total of 15 points

partial credit: up to 5 for effort, 5 for insight, otherwise as described below

[1 point] If G has 3 vertices and two edges then the assertion is easy since f=1.

[6 points] Otherwise each face is bounded by 3 or more edges, hence counting the number

of edges bounding each faces and summing, then using the fact that we at most double

counted [-2 for asserting we exactly double counted] each face we get 3f ≤ 2e, thus

proving the left inequality.

[8 points] Now f ≤ (2/3)e and we want to eliminate f, so we can add v-e to both sides to

get v-e+f ≤ v-e+(2/3)e, thus by Euler’s formula 2 ≤ v-e/3, and rearranging yields the right

inequality.

d. 5 points for i, 10 for ii, total of 15 points

partial credit: if Jordan curve thm is used then 3 points for i, 5 for ii.

on part ii: 3 for effort, 3 for insight, otherwise as described

i. Assume it is planar. v=5, e=10, 3v-6=9<e, contradicting c, thus nonplanar

ii. [1 point] Note that v=6, e=9, so 3v-6=12>e is not a contradiction

[2 points] There are no triangles in this graph, thus every face is bordered by at least 4

edges. This will allow us to prove a better inequality as in c with the assumption the

graph is planar, and that will provide the contradiction.

[2 points] Thus as in the proof of c we get 4f ≤ 2e.

[4 points] Hence f ≤ e/2, and again we want to eliminate f so adding v-e to both sides and

applying Euler’s formula yields 2 ≤ v-e/2. (or e ≤ 2v-4)

[1 point] 6-9/2<2, contradicting the inequality just proved, thus the graph is not planar.

e. total of 12 points

partial credit: 4 for effort, 4 for insight, otherwise as described

[2 points] Without loss of generality we may assume the graph is connected.

[2 points] If the graph has fewer than 3 vertices then the assertion is obvious.

[5 points] Otherwise we can apply the inequality in c. First assume every vertex has

degree at least 6, then adding the degrees of each vertex double counts the edges so 6v ≤

2e.

[3 points] Now by c. e ≤ 3v-6, and combining these inequalities we get 6v ≤ 6v-12, a

contradiction, thus the graph must have at least one vertex of degree 5 or less.

III. Coloring, total of 56 points

a. 3 points each, total of 12 points

i. 4

ii. 5

iii. 4

iv. 3

b. total of 14 points

partial credit: 5 for effort, 5 for insight, otherwise as described

[8 points] Number the colors used in G 1, 2, ..., χ. Let v1 be the number of vertices

colored with color 1. Then since no pair of them are adjacent, the are all adjacent in the

new graph and thus χ ≥ v1. Repeating this procedure for v2, ..., vχ and adding we get χχ

≥ v.

[6 points] By the AM-GM inequality χ+χ ≥ 2 χχ ≥ 2 v .

c. 5 points for i and iii, 20 for ii, total of 30 points

partial credit: as described below

i. By II.e. G must have a vertex of degree 5 or less

ii. [2 points] We want to remove the vertex V without decreasing the number of

colors needed. In the cases where V has degree < 5 this is obviously possible.

[10 points] In the case where V has degree 5, color the adjacent vertices with colors 1, 2,

3, 4, 5 in clockwise order. If the vertices colored 1 and 3 are not connected by a walk with

all vertices colored 1 or 3 then change the vertex colored 3 to 1, change any 1 vertices

adjacent to it to 3, then change any 3 vertices adjacent to those to 1, and so on. This

recoloring does not affect any of the other vertices adjacent to V and thus allows us to

make V color 3, i.e. if a sixth color is needed for G then it is still needed after we remove

V.

[8 points] If the vertices colored 1 and 3 are connected by a walk of vertices colored 1 or

3 then look at the vertices colored 2 and 4. If they are connected by a walk then by the

Jordan Curve Theorem [-4 for not using this] this walk must cross the closed curve

formed by the 1-3 walk and the edges from V to the vertices colored 1 and 3. The graph is

planar, so we can assume it is drawn without edge crossings, thus the walk from the

vertex colored 2 to the one colored 4 must cross this curve at a vertex, hence they are not

connected by a walk with all vertices colored 2 or 4, and by the previous argument we can

change the vertex colored 2 to 4 and make V color 2, so it can be removed without

removing the need for a sixth color.

iii. [4 points] We have shown that for any graph with χ>5 we can find one with fewer

vertices, and since the proof did not depend on the number of vertices we can keep doing

this until we get a graph with 5 vertices, which can clearly be colored with only 5 colors,

thus G could not require more than 5 colors.

[1 points] To make this rigorous some mention should be made of the well-ordering

principal or infinite descent.

Historical notes and inspiration for further study of the subject:

The Four Color theorem was first conjectured in 1852. Many “proofs” were given,

including one in 1879 by A. B. Kempe that was believed to be correct until P. J. Heawood

found a flaw in 1890. The proof of the Four Color theorem, given in 1976 by Haken and

Appel, uses a similar technique to what we just used, but instead of using 6 possible

subgraphs it uses 1482. The way these theorems relate to coloring maps is that we can

consider the dual graph, formed by putting a vertex in every face and connecting the

vertices whose faces share an edge. Coloring the dual graph is the same as coloring the

faces of the original graph in such a way that no two faces sharing an edge have the same

color, which is what one does when coloring a map.

While it may seem that we used many definitions in this power question, there are many

more that had to be omitted. Some of the other interesting definitions used in graph

theory are expansion, which is what you get if you add vertices along an edge of a graph,

and supergraph, which is what you get if you add new vertices anywhere and connect

them arbitrarily to the other vertices. Kuratowski’s Theorem states that every nonplanar

graph is a supergraph of an expansion of one of the two graphs in II.d. A corollary of this

is that a graph is nonplanar iff it is a supergraph of an expansion of one of those two

graphs, thus we can classify planar and nonplanar graphs in terms of just two particular

graphs.

A planar graph can alo be described as one that can be drawn on a sphere with no edge

crossings. A torus is essentially the surface of a solid sphere with a hole drilled all the

way through it. The genus g of a graph is the minimum number of holes that must be

drilled in a sphere in order to be able to draw it without edge crossings. In other words a

graph of genus g can be drawn on a g holed torus without edge crossings, but not on a g-1

holed torus. The Heawood Coloring Theorem states that if G has genus g > 0 then G can

be colored with 7 + 1+ 48g

2

colors. Note that plugging in g=0 yields 4, but no proof

of this theorem is known that does not depend on the condition g > 0 (of course we can

combine the proof for g>0 with the four color theorem to conclude that the result holds

for all g, but it would be nice to have just one proof that works equally well for all g).

Another thing we can study about graphs is what kind of walks exist in them. An

eulerian walk, named after the great Swiss mathematician Leonhard Euler (1707-1783,

pronounced Oiler), is one that uses every edge exactly once. A hamiltonian walk, named

after the Irish mathematician Sir William Rowan Hamilton (1805-1865), is one that uses

every vertex exactly once. Try to classify which graphs have such walks, with or without

the condition that the starting and ending points must be the same (in the case of a

hamiltonian walk the end vertex is thus counted twice and it is called a closed

hamiltonian walk to distinguish from the open walk).

This power question was barely an introduction to graph theory. It is a very broad field of

mathematics, closely related to topology and knot theory. I hope you enjoyed this test,

learned something from it, and that you will continue your studies of mathematics for

many years.

Edward Early

2/14/98

Advanced Topics

Harvard-MIT Math TournamentFebruary 27, 1999

1. One of the receipts for a math tournament showed that 72 identical trophies were purchased for$-99.9-, where the first and last digits were illegible. How much did each trophy cost?

2. Stacy has d dollars. She enters a mall with 10 shops and a lottery stall. First she goes to thelottery and her money is doubled, then she goes into the first shop and spends 1024 dollars. Afterthat she alternates playing the lottery and getting her money doubled (Stacy always wins) thengoing into a new shop and spending $1024. When she comes out of the last shop she has no moneyleft. What is the minimum possible value of d?

3. An unfair coin has the property that when flipped four times, it has the same nonzero probabilityof turning up 2 heads and 2 tails (in any order) as 3 heads and 1 tail (in any order). What is theprobability of getting a head in any one flip?

4. You are given 16 pieces of paper numbered 16, 15, ..., 2, 1 in that order. You want to put themin the order 1, 2, ..., 15, 16 switching only two adjacent pieces of paper at a time. What is theminimum number of switches necessary?

5. For any finite set S, let f(S) be the sum of the elements of S (if S is empty then f(S) = 0).Find the sum over all subsets E of S of f(E)

f(S) for S = {1, 2, ..., 1999}.

6. Matt has somewhere between 1000 and 2000 pieces of paper he’s trying to divide into piles ofthe same size (but not all in one pile or piles of one sheet each). He tries 2, 3, 4, 5, 6, 7, and 8 pilesbut ends up with one sheet left over each time. How many piles does he need?

7. Find an ordered pair (a, b) of real numbers for which x2 + ax+ b has a non-real root whose cubeis 343.

8. Let C be a circle with two diameters intersecting at an angle of 30 degrees. A circle S is tangentto both diameters and to C, and has radius 1. Find the largest possible radius of C.

9. As part of his effort to take over the world, Edward starts producing his own currency. As partof an effort to stop Edward, Alex works in the mint and produces 1 counterfeit coin for every 99real ones. Alex isn’t very good at this, so none of the counterfeit coins are the right weight. Sincethe mint is not perfect, each coin is weighed before leaving. If the coin is not the right weight, thenit is sent to a lab for testing. The scale is accurate 95% of the time, 5% of all the coins minted aresent to the lab, and the lab’s test is accurate 90% of the time. If the lab says a coin is counterfeit,what is the probability that it really is?

10. Find the minimum possible value of the largest of xy, 1 − x − y + xy, and x + y − 2xy if0 ≤ x ≤ y ≤ 1.

Advanced Topics Solutions

Harvard-MIT Math Tournament

February 27, 1999

Problem AT1 [3 points]

One of the receipts for a math tournament showed that 72 identical trophies were purchased for$-99.9-, where the first and last digits were illegible. How much did each trophy cost?

Solution: The price must be divisible by 8 and 9. Thus the last 3 digits must be divisible by8, so the price ends with 992, and the first digit must be 7 to make the total divisible by 9.$799.92/72 = $11.11.

Problem AT2 [3 points]

Stacy has d dollars. She enters a mall with 10 shops and a lottery stall. First she goes to the lotteryand her money is doubled, then she goes into the first shop and spends 1024 dollars. After thatshe alternates playing the lottery and getting her money doubled (Stacy always wins) then goinginto a new shop and spending $1024. When she comes out of the last shop she has no money left.What is the minimum possible value of d?

Solution: Work backwards. Before going into the last shop she had $1024, before the lottery shehad $512, then $1536, $768, . . . . We can easily prove by induction that if she ran out of moneyafter n shops, 0 ≤ n ≤ 10, she must have started with 1024− 210−n dollars. Therefore d is 1023.

Problem AT3 [4 points]

An unfair coin has the property that when flipped four times, it has the same probability of turningup 2 heads and 2 tails (in any order) as 3 heads and 1 tail (in any order). What is the probabilityof getting a head in any one flip?

Solution: Let p be the probability of getting a head in one flip. There are 6 ways to get 2 heads and2 tails, each with probability p2(1−p)2, and 4 ways to get 3 heads and 1 tail, each with probabilityp3(1− p). We are given that 6p2(1− p)2 = 4p3(1− p). Clearly p is not 0 or 1, so we can divide byp2(1− p) to get 6(1− p) = 4p. Therefore p is 3

5.

Problem AT4 [4 points]

You are given 16 pieces of paper numbered 16, 15, ..., 2, 1 in that order. You want to put themin the order 1, 2, ..., 15, 16 switching only two adjacent pieces of paper at a time. What is theminimum number of switches necessary?

Solution: Piece 16 has to move to the back 15 times, piece 15 has to move to the back 14 times,. . . , piece 2 has to move to the back 1 time, piece 1 has to move to the back 0 times. Since onlyone piece can move back in each switch, we must have at least 15 + 14 + . . .+ 1 = 120 switches.

1

Problem AT5 [5 points]

For any finite set S, let f(S) be the sum of the elements of S (if S is empty then f(S) = 0). Find

the sum over all subsets E of S of f(E)f(S) for S = {1, 2, ..., 1999}.

Solution: An n element set has 2n subsets, so each element of S appears in 21998 subsets E, so oursum is 21998 · 1+2+...+1999

1+2+...+1999 = 21998.

Problem AT6 [5 points]

Matt has somewhere between 1000 and 2000 pieces of paper he’s trying to divide into piles of thesame size (but not all in one pile or piles of one sheet each). He tries 2, 3, 4, 5, 6, 7, and 8 pilesbut ends up with one sheet left over each time. How many piles does he need?

Solution: The number of sheets will leave a remainder of 1 when divided by the least commonmultiple of 2, 3, 4, 5, 6, 7, and 8, which is 8 · 3 · 5 · 7 = 840. Since the number of sheets is between1000 and 2000, the only possibility is 1681. The number of piles must be a divisor of 1681 = 412,hence it must be 41.

Problem AT7 [5 points]

Find an ordered pair (a, b) of real numbers for which x2 + ax + b has a non-real root whose cubeis 343.

Solution: The cube roots of 343 are the roots of x3− 343, which is (x− 7)(x2+7x+49). Thereforethe ordered pair we want is (7,49).

Problem AT8 [6 points]

Let C be a circle with two diameters intersecting at an angle of 30 degrees. A circle S is tangentto both diameters and to C, and has radius 1. Find the largest possible radius of C.

Solution: For C to be as large as possible we want S to be as small as possible. It is not hard to seethat this happens in the situation shown below. Then the radius of C is 1+csc 15 = 1+

√2+

√6.

The computation of sin 15 can be done via the half angle formula.

C

S

15

1

2

Problem AT9 [7 points]

As part of his effort to take over the world, Edward starts producing his own currency. As part ofan effort to stop Edward, Alex works in the mint and produces 1 counterfeit coin for every 99 realones. Alex isn’t very good at this, so none of the counterfeit coins are the right weight. Since themint is not perfect, each coin is weighed before leaving. If the coin is not the right weight, then itis sent to a lab for testing. The scale is accurate 95% of the time, 5% of all the coins minted aresent to the lab, and the lab’s test is accurate 90% of the time. If the lab says a coin is counterfeit,what is the probability that it really is?

Solution: 5% of the coins are sent to the lab, and only .95% of the coins are sent to the lab andcounterfeit, so there is a 19% chance that a coin sent to the lab is counterfeit and an 81% chancethat it is real. The lab could correctly detect a counterfeit coin or falsely accuse a real one ofbeing counterfeit, so the probability that a coin the lab says is counterfeit really is counterfeit is

19/100·9/1019/100·9/10+81/100·1/10 = 19

28.

Problem AT10 [8 points]

Find the minimum possible value of the largest of xy, 1−x−y+xy, and x+y−2xy if 0 ≤ x ≤ y ≤ 1.

Solution: I claim the answer is 4/9. Let s = x + y, p = xy, so x and y ares±√

s2−4p2 . Since x

and y are real, s2 − 4p ≥ 0. If one of the three quantities is less than or equal to 1/9, then atleast one of the others is at least 4/9 by the pigeonhole principle since they add up to 1. Assumethat s − 2p < 4/9, then s2 − 4p < (4/9 + 2p)2 − 4p, and since the left side is non-negative we get0 ≤ p2 − 5

9p+481 = (p − 1

9)(p− 49). This implies that either p ≤ 1

9 or p ≥ 49 , and either way we’re

done. This minimum is achieved if x and y are both 1/3, so the answer is 4

9, as claimed.

3

Algebra

Harvard-MIT Math TournamentFebruary 27, 1999

1. If a@b = a3−b3a−b , for how many real values of a does a@1 = 0?

2. For what single digit n does 91 divide the 9-digit number 12345n789?

3. Alex is stuck on a platform floating over an abyss at 1 ft/s. An evil physicist has arranged forthe platform to fall in (taking Alex with it) after traveling 100ft. One minute after the platformwas launched, Edward arrives with a second platform capable of floating all the way across theabyss. He calculates for 5 seconds, then launches the second platform in such a way as to maximizethe time that one end of Alex’s platform is between the two ends of the new platform, thus givingAlex as much time as possible to switch. If both platforms are 5 ft long and move with constantvelocity once launched, what is the speed of the second platform (in ft/s)?

4. Find all possible values of da where a2 − 6ad+ 8d2 = 0, a 6= 0.

5. You are trapped in a room with only one exit, a long hallway with a series of doors and landmines. To get out you must open all the doors and disarm all the mines. In the room is a panelwith 3 buttons, which conveniently contains an instruction manual. The red button arms a mine,the yellow button disarms two mines and closes a door, and the green button opens two doors.Initially 3 doors are closed and 3 mines are armed. The manual warns that attempting to disarmtwo mines or open two doors when only one is armed/closed will reset the system to its initial state.What is the minimum number of buttons you must push to get out?

6. Carl and Bob can demolish a building in 6 days, Anne and Bob can do it in 3, Anne and Carlin 5. How many days does it take all of them working together if Carl gets injured at the end ofthe first day and can’t come back? Express your answer as a fraction in lowest terms.

7. Matt has somewhere between 1000 and 2000 pieces of paper he’s trying to divide into piles ofthe same size (but not all in one pile or piles of one sheet each). He tries 2, 3, 4, 5, 6, 7, and 8 pilesbut ends up with one sheet left over each time. How many piles does he need?

8. If f(x) is a monic quartic polynomial such that f(−1) = −1, f(2) = −4, f(−3) = −9, andf(4) = −16, find f(1).

9. How many ways are there to cover a 3× 8 rectangle with 12 identical dominoes?

10. Pyramid EARLY is placed in (x, y, z) coordinates so that E = (10, 10, 0), A = (10,−10, 0),R = (−10,−10, 0), L = (−10, 10, 0), and Y = (0, 0, 10). Tunnels are drilled through the pyramidin such a way that one can move from (x, y, z) to any of the 9 points (x, y, z − 1), (x± 1, y, z − 1),(x, y± 1, z − 1),(x± 1, y± 1, z − 1). Sean starts at Y and moves randomly down to the base of thepyramid, choosing each of the possible paths with probability 1

9 each time. What is the probabilitythat he ends up at the point (8, 9, 0)?

Algebra Solutions

Harvard-MIT Math Tournament

February 27, 1999

Problem A1 [3 points]

If a@b = a3−b3

a−b , for how many real values of a does a@1 = 0?

Solution: If a3−1a−1 = 0, then a3 − 1 = 0, or (a − 1)(a2 + a + 1) = 0. Thus a = 1, which is an

extraneous solution since that makes the denominator of the original expression 0, or a is a root

of a2 + a + 1. But this quadratic has no real roots, in particular its roots are −1±√−3

2 . Thereforethere are no such real values of a, so the answer is 0.

Problem A2 [3 points]

For what single digit n does 91 divide the 9-digit number 12345n789?

Solution 1: 123450789 leaves a remainder of 7 when divided by 91, and 1000 leaves a remainder of90, or -1, so adding 7 multiples of 1000 will give us a multiple of 91.

Solution 2: For those who don’t like long division, there is a quicker way. First notice that 91 = 7·13,and 7·11·13 = 1001. Observe that 12345n789 = 123·1001000+45n·1001−123·1001+123−45n+789It follows that 91 will divide 12345n789 iff 91 divides 123− 45n+ 789 = 462− n. The number 462is divisible by 7 and leaves a remainder of 7 when divided by 13.

Problem A3 [4 points]

Alex is stuck on a platform floating over an abyss at 1 ft/s. An evil physicist has arranged for theplatform to fall in (taking Alex with it) after traveling 100ft. One minute after the platform waslaunched, Edward arrives with a second platform capable of floating all the way across the abyss.He calculates for 5 seconds, then launches the second platform in such a way as to maximize thetime that one end of Alex’s platform is between the two ends of the new platform, thus giving Alexas much time as possible to switch. If both platforms are 5 ft long and move with constant velocityonce launched, what is the speed of the second platform (in ft/s)?

Solution: The slower the second platform is moving, the longer it will stay next to the first platform.However, it needs to be moving fast enough to reach the first platform before it’s too late. Let vbe the velocity of the second platform. It starts 65 feet behind the first platform, so it reaches theback of the first platform at 60

v−1 seconds, and passes the front at 70v−1 seconds, so the time to switch

is 10v−1 . Hence we want v to be as small as possible while still allowing the switch before the first

platform falls. Therefore the time to switch will be maximized if the back of the second platformlines up with the front of the first platform at the instant that the first platform has travelled 100ft,which occurs after 100 seconds. Since the second platform is launched 65 seconds later and has totravel 105 feet, its speed is 105/35 = 3ft/s.

1

Problem A4 [4 points]

Find all possible values of da where a2 − 6ad+ 8d2 = 0, a 6= 0.

Solution: Dividing a2−6ad+8d2 = 0 by a2, we get 1−6da +8(da)

2 = 0. The roots of this quadraticare 1

2, 14.

Problem A5 [5 points]

You are trapped in a room with only one exit, a long hallway with a series of doors and land mines.To get out you must open all the doors and disarm all the mines. In the room is a panel with3 buttons, which conveniently contains an instruction manual. The red button arms a mine, theyellow button disarms two mines and closes a door, and the green button opens two doors. Initially3 doors are closed and 3 mines are armed. The manual warns that attempting to disarm two minesor open two doors when only one is armed/closed will reset the system to its initial state. What isthe minimum number of buttons you must push to get out?

Solution: Clearly we do not want to reset the system at any time. After pressing the red buttonr times, the yellow button y times, and the green button g times, there will be 3 + r − 2y armedmines and 3 + y − 2g closed doors, so we want the values of r, y, and g that make both of thesequantities 0 while minimizing r + y + g. From the number of doors we see that y must be odd,from the number of mines we see y = (3 + r)/2 ≥ 3/2, so y ≥ 3. Then g = (3 + y)/2 ≥ 3, andr = 2y−3 ≥ 3, so r+ y+ g ≥ 9. Call the red, yellow, and green buttons 1, 2, and 3 respectively fornotational convenience, then a sequence of buttons that would get you out is 123123123. Anotherpossibility is 111222333, and of course there are others. Therefore the answer is 9.

Problem A6 [5 points]

Carl and Bob can demolish a building in 6 days, Anne and Bob can do it in 3, Anne and Carl in5. How many days does it take all of them working together if Carl gets injured at the end of thefirst day and can’t come back? Express your answer as a fraction in lowest terms.

Solution: Let a be the portion of the work that Anne does in one day, similarly b for Bob and c forCarl. Then what we are given is the system of equations b+ c = 1/6, a+ b = 1/3, and a+ c = 1/5.Thus in the first day they complete a+ b+ c = 1

2(1/6+ 1/3+ 1/5) = 7/20, leaving 13/20 for Anne

and Bob to complete. This takes 13/201/3 = 39/20 days, for a total of 59

20.

Problem A7 [5 points]

Matt has somewhere between 1000 and 2000 pieces of paper he’s trying to divide into piles of thesame size (but not all in one pile or piles of one sheet each). He tries 2, 3, 4, 5, 6, 7, and 8 pilesbut ends up with one sheet left over each time. How many piles does he need?

Solution: The number of sheets will leave a remainder of 1 when divided by the least commonmultiple of 2, 3, 4, 5, 6, 7, and 8, which is 8 · 3 · 5 · 7 = 840. Since the number of sheets is between1000 and 2000, the only possibility is 1681. The number of piles must be a divisor of 1681 = 412,hence it must be 41.

2

Problem A8 [6 points]

If f(x) is a monic quartic polynomial such that f(−1) = −1, f(2) = −4, f(−3) = −9, andf(4) = −16, find f(1).

Solution: The given data tells us that the roots of f(x) + x2 are -1, 2, -3, and 4. Combining withthe fact that f is monic and quartic we get f(x) + x2 = (x + 1)(x − 2)(x + 3)(x − 4). Hencef(1) = (2)(−1)(4)(−3)− 1 = 23.

Problem A9 [7 points]

How many ways are there to cover a 3× 8 rectangle with 12 identical dominoes?

Solution: Trivially there is 1 way to tile a 3× 0 rectangle, and it is not hard to see there are 3 waysto tile a 3 × 2. Let Tn be the number of tilings of a 3 × n rectangle, where n is even. From thediagram below we see the recursion Tn = 3Tn−2 + 2(Tn−4 + Tn−6 + . . .+ T2 + T0). Given that, wecan just calculate T4 = 11, T6 = 41, and T8 is 153.

. . . . . . etc...

Problem A10 [8 points]

Pyramid EARLY is placed in (x, y, z) coordinates so that E = (10, 10, 0), A = (10,−10, 0), R =(−10,−10, 0), L = (−10, 10, 0), and Y = (0, 0, 10). Tunnels are drilled through the pyramid insuch a way that one can move from (x, y, z) to any of the 9 points (x, y, z − 1), (x ± 1, y, z − 1),(x, y± 1, z− 1),(x± 1, y± 1, z− 1). Sean starts at Y and moves randomly down to the base of thepyramid, choosing each of the possible paths with probability 1

9 each time. What is the probabilitythat he ends up at the point (8, 9, 0)?

Solution 1: Start by figuring out the probabilities of ending up at each point on the way down thepyramid. Obviously we start at the top vertex with probability 1, and each point on the next leveldown with probability 1/9. Since each probability after n steps will be some integer over 9n, wewill look only at those numerators. The third level down has probabilities as shown below. Thinkof this as what you would see if you looked at the pyramid from above, and peeled off the top twolayers.

1 2 3 2 12 4 6 4 23 6 9 6 32 4 6 4 21 2 3 2 1

3

What we can observe here is not only the symmetry along vertical, horizontal, and diagonal axes,but also that each number is the product of the numbers at the ends of its row and column (e.g.6 = 2 ·3). This comes from the notion of independence of events, i.e. that if we east and then south,we end up in the same place as if we had moved south and then east. Since we are only looking forthe probability of ending up at (8, 9, 0), we need only know that this is true for the top two rowsof the square of probabilities, which depend only on the top two rows of the previous layer. Thiswill follow from the calculation of the top row of each square, which we can do via an algorithmsimilar to Pascal’s triangle. In the diagram below, each element is the sum of the 3 above it.

1

1 1 1

1 2 3 2 1

1 3 6 7 6 3 1

1 4 10 16 19 16 10 4 1

1 5 15 30 45 51 45 30 15 5 1

Now observe that the first 3 numbers in row n, where the top is row 0, are 1, n, n(n+1)2 . This fact

is easily proved by induction on n, so the details are left to the reader. Now we can calculate thetop two rows of each square via another induction argument, or by independence, to establish thatthe second row is always n times the first row. Therefore the probability of ending up at the point(8,9,0) is 550

910 .

Solution 2: At each move, the x and y coordinates can each increase by 1, decrease by 1, or staythe same. The y coordinate must increase 9 times and stay the same 1 times, the x coordinate caneither increase 8 times and stay the same 1 time or decrease 1 time and increase 9 times. Now weconsider every possible case. First consider the cases where the x coordinate decreases once. If thex coordinate decreases while the y coordinate increases, then we have 8 moves that are the sameand 2 that are different, which can be done in 10!

8! = 90 ways. If the x coordinate decreases whilethe y coordinate stays the same, then we have 9 moves that are the same and 1 other, which can bedone in 10!

9! = 10 ways. Now consider the cases where the x coordinate stays the same twice. If they coordinate stays the same while the x coordinate increases, then we have 7 moves that are thesame, 2 that are the same, and 1 other, which can be done in 10!

7!2! = 360 ways. If the y coordinatestays the same while the x coordinate stays the same, then we have 8 moves that are the same and 2that are different, which can be done in 10!

8! = 90 ways. Therefore there are 360+90+90+10 = 550paths to (8,9,0), out of 910 possible paths to the bottom, so the probability of ending up at thepoint (8,9,0) is 550

910 .

4

Calculus

Harvard-MIT Math TournamentFebruary 27, 1999

1. Find all twice differentiable functions f(x) such that f ′′(x) = 0, f(0) = 19, and f(1) = 99.

2. A rectangle has sides of length sinx and cosx for some x. What is the largest possible area ofsuch a rectangle?

3. Find ∫ 4π√

2

−4π√

2(

sinx1 + x4

+ 1)dx.

4. f is a continuous real-valued function such that f(x+y) = f(x)f(y) for all real x, y. If f(2) = 5,find f(5).

5. Let f(x) = x+ 12x+ 1

2x+ 1

2x+...

for x > 0. Find f(99)f ′(99).

6. Evaluate ddx(sinx− 4

3 sin3 x) when x = 15.

7. If a right triangle is drawn in a semicircle of radius 1/2 with one leg (not the hypotenuse) alongthe diameter, what is the triangle’s maximum possible area?

8. A circle is randomly chosen in a circle of radius 1 in the sense that a point is randomly chosenfor its center, then a radius is chosen at random so that the new circle is contained in the originalcircle. What is the probability that the new circle contains the center of the original circle?

9. What fraction of the Earth’s volume lies above the 45 degrees north parallel? You may assumethe Earth is a perfect sphere. The volume in question is the smaller piece that we would get if thesphere were sliced into two pieces by a plane.

10. Let An be the area outside a regular n-gon of side length 1 but inside its circumscribed circle,let Bn be the area inside the n-gon but outside its inscribed circle. Find the limit as n tends toinfinity of An

Bn.

Calculus Solutions

Harvard-MIT Math Tournament

February 27, 1999

Problem C1 [3 points]

Find all twice differentiable functions f(x) such that f ′′(x) = 0, f(0) = 19, and f(1) = 99.

Solution: Since f ′′(x) = 0 we must have f(x) = ax + b for some real numbers a, b. Thus f(0) =b = 19 and f(1) = a+ 19 = 99, so a = 80. Therefore f(x) = 80x+ 19.

Problem C2 [3 points]

A rectangle has sides of length sinx and cosx for some x. What is the largest possible area of sucha rectangle?

Solution: We wish to maximize sinx · cosx = 12 sin 2x. But sin 2x ≤ 1, with equality holding for

x = π/4, so the maximum is 1

2.

Problem C3 [4 points]

Find∫ 4π

√2

−4π√2(sinx

1 + x4+ 1)dx.

Solution: The function sinx1+x4 is odd, so its integral over this interval is 0. Thus we get the same

answer if we just integrate dx, namely, 8π√2.

Problem C4 [4 points]

f is a continuous real-valued function such that f(x+ y) = f(x)f(y) for all real x, y. If f(2) = 5,find f(5).

Solution 1: Since f(nx) = f(x)n for all integers n, f(5) = f(1)5 and f(2) = f(1)2, so f(5) =f(2)5/2 = 25

√5.

Solution 2: More generally, since f(nx) = f(x)n for all integers n, f(1) = c = f(1/n)n for someconstant c and all integers n. Thus f(k/n) = f(1/n)k = f(1)k/n = ck/n for all rational numbersk/n. By continuity, it follows that f(x) = cx for all real numbers x. Since f(2) = 5, c =

√5, so

f(5) = 25√5.

1

Problem C5 [5 points]

Let f(x) = x+ 12x+ 1

2x+ 1

2x+...

for x > 0. Find f(99)f ′(99).

Solution: Assume that the continued fraction converges (it does) so that f(x) is well defined. Noticethat f(x) − x = 1

x+f(x) , so f(x)2 − x2 = 1, or f(x) =√1 + x2 (we need the positive square root

since x > 0). Thus f ′(x) = x√1+x2

, so f(x)f ′(x) = x. In particular, f(99)f ′(99) = 99.

Problem C6 [5 points]

Evaluate ddx(sinx− 4

3 sin3 x) when x = 15.

Solution: Of course this problem can be done by brute force, differentiating and then using the halfangle formula to find sin and cos of 15, but there is a quicker way. eix = cosx+ i sinx, so sin(3x)is the imaginary part of (cosx + i sinx)3, which is 3 cos2 x sinx − sin3 x = 3 sinx − 4 sin3 x, so the

expression we are differentiating is just 13 sin(3x). Hence the derivative is cos(3x), and cos 45 =

√2

2.

Problem C7 [5 points]

If a right triangle is drawn in a semicircle of radius 1/2 with one leg (not the hypotenuse) alongthe diameter, what is the triangle’s maximum possible area?

Solution: It is easy to see that we will want one vertex of the triangle to be where the diametermeets the semicircle, so the diameter is divided into segments of length x and 1−x, where x is thelength of the leg on the diameter. The other leg of the triangle will be the geometric mean of these

two numbers,√

x(1− x). Therefore the area of the triangle isx√

x(1−x)

2 , so it will be maximized

when ddx(x

3 − x4) = 3x2 − 4x3 = 0, or x = 3/4. Therefore the maximum area is 3√3

32.

Problem C8 [6 points]

A circle is randomly chosen in a circle of radius 1 in the sense that a point is randomly chosenfor its center, then a radius is chosen at random so that the new circle is contained in the originalcircle. What is the probability that the new circle contains the center of the original circle?

Solution: If the center of the new circle is more than 1/2 away from the center of the originalcircle then the new circle cannot possibly contain the center of the original one. Let x be thedistance between the centers (by symmetry this is all we need to consider), then for 0 ≤ x ≤ 1/2the probability of the new circle containing the center of the original one is 1− x

1−x . Hence we need

to compute∫ 1/20 (1− x

1−x)dx = 12 −

∫ 1/20

x1−xdx. To evaluate the integral, we can integrate by parts

to get

−x ln(1− x)|1/20 −∫ 1/2

0− ln(1− x)dx = −1

2ln(

1

2)− [(1− x) ln(1− x)− (1− x)]

1/20 = ln 2− 1

2.

2

Alternatively, we can use polynomial division to find that x1−x = −1 + 1

1−x , so∫ 1/20

x1−xdx =

∫ 1/20 (−1 + 1

1−x)dx = ln 2− 12 . Therefore the probability is 1

2 − (ln 2− 12) = 1− ln2.

Problem C9 [7 points]

What fraction of the Earth’s volume lies above the 45 degrees north parallel? You may assumethe Earth is a perfect sphere. The volume in question is the smaller piece that we would get if thesphere were sliced into two pieces by a plane.

Solution 1: Without loss of generality, look at a sphere of radius 1 centered at the origin. If you likecartesian coordinates, then you can slice the sphere into discs with the same z coordinate, which

have radius√1− z2, so the region we are considering has volume

∫ 1√2/2 π(1− z2)dz = π(23 − 5

√2

12 ),

and dividing by 4π/3 we get 8−5√2

16.

Solution 2: For those who prefer spherical coordinates, we can find the volume of the spherical capplus a cone whose vertex is the center of the sphere. This region is where r ranges from 0 to 1, θranges from 0 to 2π, and φ ranges from 0 to π/4. Remembering we need to subtract off the volume ofthe cone, which has height 1√

2and a circular base of radius 1√

2, then divide by 4

3π to get the fraction

of the volume of the sphere, we find that we need to evaluate

∫

1

0

∫

2π

0

∫ π/4

0r2 sinφdφdθdr− 1

3π( 1

√

2)2 1

√

2

4π/3 . The

integral is just 2π 13(− cosπ/4 + cos 0) = 4π−2π

√2

6 . Putting this back in the answer and simplifying

yields 8−5√2

16.

Solution 3: Cavalieri’s Principle states that if two solids have the same cross-sectional areas at everyheight, then they have the same volume. This principle is very familiar in the plane, where we knowthat the area of a triangle depends only on the base and height, not the precise position of the apex.To apply it to a sphere, consider a cylinder with radius 1 and height 1. Now cut out a cone whosebase is the upper circle of the cylinder and whose apex is the center of the lower circle. Then at aheight z the area is π(12 − z2), exactly the same as for the upper hemisphere! The portion lyingabove the the 45 degrees north parallel is that which ranges from height 1√

2to 1. The volume of

the cylinder in this range is π ·12(1− 1√2). The volume of the cone in this range is the volume of the

entire cone minus the portion from height 0 to 1√2, i.e., 1

3π(12 ·1− ( 1√

2)2 1√

2). Therefore the fraction

of the Earth’s volume that lies above the 45 degrees north parallel isπ(1− 1

√

2)− 1

3π(1− 1

2√

2)

4π/3 = 8−5√2

16.

Solution 4: Another way to approach this problem is to integrate the function√

1− x2 − y2 overthe region x2+ y2 ≤ 1√

2, subtract off a cylinder of radius and height 1√

2, then divide by the volume

of the sphere. One could also use the nontrivial fact that the surface area of a portion of a sphere ofradius r between two parallel planes separated by a distance z is 2πr2z, so in particular the surfacearea of this cap is 2π(1− 1√

2). Now, the ratio of the surface area of the cap to the surface area of

the sphere is the same as the ratio of the volume of the cap plus the cone considered in Solution 2to the volume of the whole sphere, so this allows us to avoid integration entirely.

3

Problem C10 [8 points]

Let An be the area outside a regular n-gon of side length 1 but inside its circumscribed circle, letBn be the area inside the n-gon but outside its inscribed circle. Find the limit as n tends to infinityof An

Bn.

Solution: The radius of the inscribed circle is 12 cot

πn , the radius of the circumscribed circle is

12 csc

πn , and the area of the n-gon is n

4 cotπn . The diagram below should help you verify that these

are correct.

π / n

r

R

1/2

Then An = π(12 cscπn)

2 − n4 cot

πn , and Bn = n

4 cotπn − π(12 cot

πn)

2, so AnBn

=π(csc π

n)2−n cot π

nn cot π

n−π(cot π

n)2. Let s

denote sin πn and c denote cos π

n . Multiply numerator and denominator by s2 to get AnBn

= π−ncsncs−πc2

.

Now use Taylor series to replace s by πn − (π

n)3

6 + . . . and c by 1 − (πn)2

2 + . . .. By l’Hôpital’s ruleit will suffice to take just enough terms so that the highest power of n in the numerator anddenominator is determined, and that turns out to be n−2 in each case. In particular, we get the

limit AnBn

=π−nπ

n+n 2

3(πn)3+...

nπn−n 2

3(πn)3−π+π(π

n)2+...

=2

3

π3

n2+...

1

3

π3

n2+...

→ 2.

4

Geometry

Harvard-MIT Math Tournament

February 27, 1999

1. Two 10× 24 rectangles are inscribed in a circle as shown. Find the shaded area.

2. A semicircle is inscribed in a semicircle of radius 2 as shown. Find the radius of the smaller

semicircle.

3. In a cube with side length 6, what is the volume of the tetrahedron formed by any vertex and

the three vertices connected to that vertex by edges of the cube?

4. A cross-section of a river is a trapezoid with bases 10 and 16 and slanted sides of length 5. At

this section the water is flowing at π mph. A little ways downstream is a dam where the water

flows through 4 identical circular holes at 16 mph. What is the radius of the holes?

5. In triangle BEN shown below with its altitudes intersecting at X, NA = 7, EA = 3, AX = 4,

and NS = 8. Find the area of BEN .

B

N

H

S A

E

X

6. A sphere of radius 1 is covered in ink and rolling around between concentric spheres of radii 3

and 5. If this process traces a region of area 1 on the larger sphere, what is the area of the region

traced on the smaller sphere?

7. A dart is thrown at a square dartboard of side length 2 so that it hits completely randomly.

What is the probability that it hits closer to the center than any corner, but within a distance 1 of

a corner?

8. Squares ABKL,BCMN,CAOP are drawn externally on the sides of a triangle ABC. The line

segments KL,MN,OP , when extended, form a triangle A′B′C ′. Find the area of A′B′C ′ if ABC

is an equilateral triangle of side length 2.

1

9. A regular tetrahedron has two vertices on the body diagonal of a cube with side length 12. The

other two vertices lie on one of the face diagonals not intersecting that body diagonal. Find the

side length of the tetrahedron.

10. In the figure below, AB = 15, BD = 18, AF = 15, DF = 12, BE = 24, and CF = 17. Find

BG : FG.

B

D

E

F

G

A

C

2

Geometry Solutions

Harvard-MIT Math Tournament

February 27, 1999

Problem G1 [3]

Two 10× 24 rectangles are inscribed in a circle as shown. Find the shaded area.

Solution: The rectangles are 10 × 24, so their diagonals, which are diameters of the circle, havelength 26. Therefore the area of the circle is π132, and the overlap is a 10 × 10 square, so theshaded area is π132 − 2 · 10 · 24 + 102 = 169π − 380.

Problem G2 [3]

A semicircle is inscribed in a semicircle of radius 2 as shown. Find the radius of the smallersemicircle.

Solution: Draw a line from the center of the smaller semicircle to the center of the larger one, anda line from the center of the larger semicircle to one of the other points of intersection of the twosemicircles. We now have a right triangle whose legs are both the radius of the smaller semicircleand whose hypotenuse is 2, therefore the radius of the smaller semicircle is

√2.

Problem G3 [4]

In a cube with side length 6, what is the volume of the tetrahedron formed by any vertex and thethree vertices connected to that vertex by edges of the cube?

Solution: We have a tetrahedron whose base is half a face of the cube and whose height is the sidelength of the cube, so its volume is 1

3· (1

2· 62) · 6 = 36.

1

Problem G4 [4]

A cross-section of a river is a trapezoid with bases 10 and 16 and slanted sides of length 5. At thissection the water is flowing at π mph. A little ways downstream is a dam where the water flowsthrough 4 identical circular holes at 16 mph. What is the radius of the holes?

Solution: The volume of water going through any cross-section of the river in an hour (assumingthe cross-sections are parallel) is the area times the velocity. The trapezoid has height 4, hencearea 52, so the volume of water going through at any hour is 52π. Let r be the radius of the holes,

then the total area is 4πr2, so the volume of water is 64πr2. Therefore 64πr2 = 52π, so r =√13

4.

Problem G5 [5]

In triangle BEN shown below with its altitudes intersecting at X, NA = 7, EA = 3, AX = 4, andNS = 8. Find the area of BEN .

B

N

H

S A

E

X

Solution: The idea is to try to find a base and height for the triangle so that we can find the area.By the Pythagorean theorem, EX = 5, NX =

√65, and SX = 1. Triangles AXE and BXS are

similar since they have the same angles. The ratio of their side lengths is 4:1, so BS = 3/4 andBX = 5/4. Now using either NE or NB as a base, we get that the area of BEN is 1

2· (8 + 3

4) · 6

or 1

2· (4 + 5

4) · 10, both of which simplify to 105

4.

Problem G6 [5]

A sphere of radius 1 is covered in ink and rolling around between concentric spheres of radii 3 and5. If this process traces a region of area 1 on the larger sphere, what is the area of the region tracedon the smaller sphere?

Solution: The figure drawn on the smaller sphere is just a scaled down version of what was drawnon the larger sphere, so the ratio of the areas is the ratio of the surface area of the spheres. Thisis the same as the ratio of the squares of the radii, which is 9

25.

Problem G7 [5]

A dart is thrown at a square dartboard of side length 2 so that it hits completely randomly. Whatis the probability that it hits closer to the center than any corner, but within a distance 1 of acorner?

2

Solution: By symmetry it will suffice to consider one quarter of the dartboard, which is a squareof side length 1. Therefore the probability is the area of the desired region in this square. Thedesired region is the part of the circle of radius 1 centered at a corner that is closer to the oppositecorner. The points closer to the opposite corner are those that are on the other side of the diagonalthrough the other two corners, so the desired region is a quarter of a circle of radius 1 minus a righttriangle with legs of length 1. Therefore the area (and hence the probability) is π−2

4.

Problem G8 [6]

Squares ABKL,BCMN,CAOP are drawn externally on the sides of a triangle ABC. The linesegments KL,MN,OP , when extended, form a triangle A′B′C ′. Find the area of A′B′C ′ if ABCis an equilateral triangle of side length 2.

Solution: Triangle ABC has area√3, and each of the three squares has area 4. The three remaining

regions are congruent, so just consider the one that includes vertex B. Triangle KBN has two sidesof length 2 and an angle of 120◦ between them, to bisecting that angle we get two halves of anequilateral triangle of side length 2, so the area is again

√3. The remaining region is an equilateral

triangle of side length 2√3, so its area is (2

√3)2

√3/4 = 3

√3. Therefore the area of A′B′C ′ is√

3 + 3 · 4 + 3 ·√3 + 3 · 3

√3 = 12+ 13

√3.

Note that this problem is still solvable, but much harder, if the first triangle is not equilateral.

Problem G9 [7]

A regular tetrahedron has two vertices on the body diagonal of a cube with side length 12. Theother two vertices lie on one of the face diagonals not intersecting that body diagonal. Find theside length of the tetrahedron.

Solution: Let ABCD be a tetrahedron of side s. We want to find the distance between two ofits opposite sides. Let E be the midpoint of AD, F the midpoint of BC. Then AE = s/2,AF = s

√3/2, and angle AEF = 90◦. So the distance between the two opposite sides is EF =√

AF 2 −AE2 =√

3s2/4− s2/4 = s/√2.

Now we find the distance between a body diagonal and a face diagonal of a cube of side a. Let Obe the center of the cube and P be the midpoint of the face diagonal. Then the plane containingP and the body diagonal is perpendicular to the face diagonal. So the distance between the body

and face diagonals is the distance between P and the body diagonal, which is a

2

√

2

3(the altitude

from P of right triangle OPQ, where Q is the appropriate vertex of the cube). So now s√2= a

2

√

2

3,

thus s = a/√3 = 12/

√3 = 4

√3.

Problem G10 [8]

In the figure below, AB = 15, BD = 18, AF = 15, DF = 12, BE = 24, and CF = 17. FindBG : FG.

Solution: Our goal is to find the lengths BG and FG. There are several ways to go about doingthis, but we will show only one here. We will make several uses of Stewart’s theorem, which can

3

B

D

E

F

G

A

C

be proved using the law of cosines twice. By Stewart’s theorem on triangle ABD and line BF ,152 · 12 + 182 · 15 = BF 2 · 27 + 15 · 12 · 27, so BF = 10 and EF = 14. By Stewart’s theoremon triangle ABE and line AF , AE2 · 10 + 152 · 14 = 152 · 24 + 14 · 10 · 24, so AE =

√561. By

Stewart’s theorem on triangle AED and line EF , ED2 · 15 + 561 · 12 = 142 · 27 + 12 · 15 · 27,so ED = 2

√57. By Stewart’s theorem on triangle CFE and line FD, 142 · CD + 172 · 2

√57 =

122 · (CD + 2√57) + 2

√57 · CD · (CD + 2

√57), so CD =

√57 and CE = 3

√57. Note that

DG = 18−BG and apply Menelaus’ theorem to triangle BED and the line through C, G, and Fto get 3 · 18−BG

BG· 1014

= 1, so BG = 135

11. Similarly CG = 17−FG, so applying Menelaus’ theorem to

triangle CFE and the line through B, G, and D we get 24

10· FG

17−FG· 1

2= 1, so FG = 85

11. Therefore

BG : FG = 27 : 17.

4

Oral Event

Harvard-MIT Math TournamentFebruary 27, 1999

1. [25] Start with an angle of 60◦ and bisect it, then bisect the lower 30◦ angle, then the upper15◦ angle, and so on, always alternating between the upper and lower of the previous two anglesconstructed. This process approaches a limiting line that divides the original 60◦ angle into twoangles. Find the measure (degrees) of the smaller angle.

2. [25] Alex, Pei-Hsin, and Edward got together before the contest to send a mailing to all theinvited schools. Pei-Hsin usually just stuffs the envelopes, but if Alex leaves the room she has tolick them as well and has a 25% chance of dying from an allergic reaction before he gets back.Licking the glue makes Edward a bit psychotic, so if Alex leaves the room there is a 20% chancethat Edward will kill Pei-Hsin before she can start licking envelopes. Alex leaves the room andcomes back to find Pei-Hsin dead. What is the probability that Edward was responsible?

3. [30] If x, y, and z are distinct positive integers such that x2 + y2 = z3, what is the smallestpossible value of x+ y + z.

4. [35] Evaluate∑∞n=0

cosnθ2n , where cos θ = 1

5 .

5. [45] Let, r be the inradius of triangle ABC. Take a point D on side BC, and let r1 and r2 bethe inradii of triangles ABD and ACD. Prove that r, r1, and r2 can always be the side lengths ofa triangle.

6. [45] You want to sort the numbers 5 4 3 2 1 using block moves. In other words, you can takeany set of numbers that appear consecutively and put them back in at any spot as a block. Forexample, 6 5 3 4 2 1 → 4 2 6 5 3 1 is a valid block move for 6 numbers. What is the minimumnumber of block moves necessary to get 1 2 3 4 5?

7. [55] Evaluate∑∞n=1

n5

n! .

8. [55] What is the smallest square-free composite number that can divide a number of the form4242 . . . 42± 1?

9. [60] You are somewhere on a ladder with 5 rungs. You have a fair coin and an envelope thatcontains either a double-headed coin or a double-tailed coin, each with probability 1/2. Everyminute you flip a coin. If it lands heads you go up a rung, if it lands tails you go down a rung. Ifyou move up from the top rung you win, if you move down from the bottom rung you lose. Youcan open the envelope at any time, but if you do then you must immediately flip that coin once,after which you can use it or the fair coin whenever you want. What is the best strategy (i.e. onwhat rung(s) should you open the envelope)?

10. [75] A, B, C, D, and E are relatively prime integers (i.e., have no single common factor) suchthat the polynomials 5Ax4 + 4Bx3 + 3Cx2 + 2Dx+E and 10Ax3 + 6Bx2 + 3Cx+D together have7 distinct integer roots. What are all possible values of A? Your team has been given a sealedenvelope that contains a hint for this problem. If you open the envelope, the value of this problemdecreases by 20 points. To get full credit, give the sealed envelope to the judge before presentingyour solution.

Oral Solutions

Harvard-MIT Math TournamentFebruary 27, 1999

Problem O1 [25 points]

Start with an angle of 60◦ and bisect it, then bisect the lower 30◦ angle, then the upper 15◦ angle,and so on, always alternating between the upper and lower of the previous two angles constructed.This process approaches a limiting line that divides the original 60◦ angle into two angles. Findthe measure (degrees) of the smaller angle.

Solution: The fraction of the original angle is 12 −

14 + 1

8 −+ · · ·. This is just a geometric series withfirst term 1/2 and ratio -1/2, so the sum is 1/3. Therefore the smaller angle is 20◦.

Score 20 points for the correct answer, 5 points for a correct justification. Just setting up the sumis worth 6 points, getting 1/3 is worth 9 more.

Problem O2 [25 points]

Alex, Pei-Hsin, and Edward got together before the contest to send a mailing to all the invitedschools. Pei-Hsin usually just stuffs the envelopes, but if Alex leaves the room she has to lick themas well and has a 25% chance of dying from an allergic reaction before he gets back. Licking theglue makes Edward a bit psychotic, so if Alex leaves the room there is a 20% chance that Edwardwill kill Pei-Hsin before she can start licking envelopes. Alex leaves the room and comes back tofind Pei-Hsin dead. What is the probability that Edward was responsible?

Solution: There are two possibilities: either Edward killed Pei-Hsin or the envelopes did. Theenvelope could only be responsible if Edward was not, so the chances of that would be 4/5·1/4 = 1/5.This is the same as the probability that Edward killed her, so the events are equally likely and theanswer is 50%, or 1/2.

Score 20 points for the correct answer, 5 points for a correct justification. Not many places to givepartial credit.

Problem O3 [30 points]

If x, y, and z are distinct positive integers such that x2 + y2 = z3, what is the smallest possiblevalue of x+ y + z.

Solution 1: Without loss of generality let x > y. We must have z3 expressible as the sum of twosquares, and this first happens when z = 5. Then x and y can be 10 and 5 or 11 and 2. If z > 5then z ≥ 10 for z3 to be a sum of two distinct squares, so x2 > 500, x > 22, so x + y + z > 32.Thus the smallest possible value of x+ y + z is 11 + 2 + 5 = 18.

Solution 2: If z > 5, then z ≥ 6, so z3 ≥ 216. Now x2 + y2 ≥ 216, so x ≥ 11 and y ≥ 1, thusx + y + z ≥ 18. Since x = 11, y = 1, z = 6 does not work, we must have x + y + z > 18, and thesolution given is the best possible.

1

Score 20 points for the correct answer, 5 points for justifying that we can’t do better with z ≤ 5, 5points for justifying that we can’t do better with z > 5.

Problem O4 [35 points]

Evaluate∑∞n=0

cosnθ2n , where cos θ = 1

5 .

Solution: cosnθ is the real part of einθ, so the sum is the real part of∑∞n=0

einθ

2n . This is a geometricseries with initial term 1 and ratio eiθ

2 , so its sum is 11−eiθ/2 . We are given cos θ = 1

5 , so sin θ = ±2√

65 .

Thus the sum is 1010−1∓2i

√6

= 90±20i√

6105 , and the real part is 6

7 .

Score 20 points for the correct answer, 15 points for a correct justification.

Problem O5 [45 points]

Let r be the radius of the inscribed circle of triangle ABC. Take a point D on side BC, and let r1

and r2 be the inradii of triangles ABD and ACD. Prove that r, r1, and r2 can always be the sidelengths of a triangle.

Solution: We must show that r, r1, and r2 satisfy the triangle inequality, i.e. that the sum of anytwo of them exceeds the third. Clearly r is the largest of the three, so we need only verify thatr1 + r2 > r. Let K and s be the area and semiperimeter of triangle ABC. Similarly define K1,K2, s1, and s2. Observe that s is larger than s1 or s2 and that K1 + K2 = K. While these factsare almost trivial to verify, they must be stated. Then r = K/s, r1 = K1/s1, and r2 = K2/s2, sor1 + r2 = K1/s1 +K2/s2 > K1/s+K2/s = K/s = r.

The correct use of areas and semiperimeters is worth 25 points, each of the critical facts is worth 10points. I don’t know of any other way to do this problem, so attempts at alternate proofs shouldget at most 15 points for effort unless they really on the right track to another solution.

Problem O6 [45 points]

You want to sort the numbers 5 4 3 2 1 using block moves. In other words, you can take any setof numbers that appear consecutively and put them back in at any spot as a block. For example,6 5 3 4 2 1 → 4 2 6 5 3 1 is a valid block move for 6 numbers. What is the minimum number ofblock moves necessary to get 1 2 3 4 5?

Solution 1: Here is a sequence of 3 moves that works: 54321→32541→34125→12345. But how dowe know we can’t do it in 2 moves? From any position there are 20 possible permutations via blockmoves, 16 from moving a block of size 1 and 4 from moving a block of size 2. One could simplywrite the 20 permutations of 54321 and the 20 permutations of 12345 and try to see that they havenothing in common, which would suffice since the inverse of a block move is also a block move. Amore clever method is to notice that if we could sort 54321 in 2 moves then we could sort 4321in 2 moves as well by simply deleting the 5 from each step. But 4321 has only 10 permutationsfrom block moves, namely 3421, 3241, 3214, 4231, 4213, 2431, 4312, 1432, 4132, and 2143. The 10

2

permutations of 1234 are 2134, 2314, 2341, 1324, 1342, 3124, 1243, 4123, 1423, and 3412. Thesetwo sets of permutations have nothing in common, thus it takes at least 3 moves to sort 4321, andhence at least 3 moves to sort 54321.

Solution 2: There is a more elegant way to show we need at least 3 moves. Given a permutationof {1, 2, 3, 4, 5} (or any ordered set), define a descent to be an adjacent pair of numbers in thepermutation such that the left number is greater than the right one. For example, 12345, 34215,and 54321 have 0, 2, and 4 descents, respectively. Any permutation obtained from 12345 byone block move has (at most) one descent, at the left edge of the moved block. Similarly, anypermutation obtained from 54321 by one block move has (at least) three descents, so that we can’tget from 54321 to 12345 by two block moves.

Score 20 points for the correct answer, 10 points for a numerical example proving that is attainable,and 15 points for proving it can’t be done in 2 or fewer moves.

Problem O7 [55 points]

Evaluate∑∞n=1

n5

n! .

Solution: We start by noticing that∑∞n=1

nn! =

∑∞n=1

1(n−1)! =

∑∞n=0

1n! = e. Next we see that∑∞

n=1n2

n! =∑∞n=1

n(n−1)! =

∑∞n=0

1+nn! =

∑∞n=0

1n! +

∑∞n=0

nn! = e+e = 2e. Let f(k) =

∑∞n=1

nk

n! , then∑∞n=1

nk

n! =∑∞n=1

nk−1

(n−1)! =∑∞n=0

(1+n)k−1

n! , so by the binomial theorem f(k) =∑k−1j=0

(k−1j

)· f(j).

Armed with this formula, we can easily compute f(3) = f(0) + 2f(1) + f(2) = e + 2e + 2e = 5e,f(4) = 1 · e+ 3 · e+ 3 · 2e+ 1 · 5e = 15e, and f(5) = 1 · e+ 4 · e+ 6 · 2e+ 4 · 5e+ 1 · 15e = 52e.

Score 30 points for the correct answer, 25 points for a correct justification. If on the right trackwith a good justification, but an arithmetic error is made along the way, score 5 points for eachf(j), j = 0, 1, 2, 3, 4, correctly computed.

Problem O8 [55 points]

What is the smallest square-free composite number that can divide a number of the form4242 . . . 42± 1?

Solution: It is easy to see that such a number can never be divisible by 2, 3, 5, or 7. They canbe divisible by 11, the smallest example being 4242424241 = 11 · 547 · 705073. What makes thisproblem hard is finding the next prime that can divide such a number. Let Tn =

∑ni=0 42 · 102i.

Then the numbers Tn modulo a prime p will always be periodic, since Tn = 100Tn1 + 42, so we justneed to compute one period and see if it contains ±1. Thus we find that modulo 13 we get 3, 4, 0,3, . . . , modulo 17 we get 8, 9, 7, 11, 3, 2, 4, 0, 8, . . . , modulo 19 we get 4, 5, 10, 16, 8, 6, 15, 3, 0, 4,. . . , and modulo 23 we get 19, 10, 7, 6, 21, 3, 20, 18, 2, 12, 0, 19, . . . , so none of these primes canever divide Tn ± 1. But 424241 = 29 · 14629, so 29 can also divide numbers of this form. Thereforethe smallest composite number that can divide Tn ± 1 for some n is 319, and the smallest such nis 83.

Score 30 points for the correct answer, 15 points for showing it is the smallest, 10 points for showingit does work. Just seeing that 11 is the smallest prime divisor is worth 5 points, finding for exactlywhich n is worth 5 more.

3

Problem O9 [60 points]

You are somewhere on a ladder with 5 rungs. You have a fair coin and an envelope that containseither a double-headed coin or a double-tailed coin, each with probability 1/2. Every minute youflip a coin. If it lands heads you go up a rung, if it lands tails you go down a rung. If you moveup from the top rung you win, if you move down from the bottom rung you lose. You can openthe envelope at any time, but if you do then you must immediately flip that coin once, after whichyou can use it or the fair coin whenever you want. What is the best strategy (i.e. on what rung(s)should you open the envelope)?

Solution: First consider the probability of winning if you never open the envelope. Let q(n) be theprobability of winning from the nth rung with just the fair coin, then q(n) = q(n−1)+q(n+1)

2 , so itis not hard to calculate that q(n) = n/6. If we open the envelope, then there’s a 1/2 chance thatit is heads and we win, and a 1/2 chance that it is tails and we end up one rung down with justthe fair coin (obviously we keep using the double sided coin iff it is double headed). Let us startby analyzing rung 1. If we don’t open the envelope, then we have a 1/2 chance of losing and a 1/2chance of ending up on rung 2 with the envelope. If we do open the envelope, then we have a 1/2chance of losing and a 1/2 chance of winning, which is a better outcome, so we should open theenvelope on rung 1. Next we look at rung 5. If we don’t open the envelope, then we have a 1/2chance of winning and a 1/2 chance of moving down to rung 4 with the envelope. If we do openthe envelope, then we we have a 1/2 chance of winning and a 1/2 chance of moving down to rung4 without the envelope. Let p(n) be the probability of winning from rung n if we are there withthe envelope still unopened. Then clearly p(n) ≥ q(n) for all n if we’re using optimal strategy, sowe should not open the envelope on rung 5. Next we look at rung 4. If we open the envelope,then our chance of winning is 1/2 + q(3)/2 = 3/4. If we don’t, then our chance of winning isp(5)/2 + p(3)/2. We do know that p(5) = 1/2 + p(4)/2, but this is not enough to tell us what todo on rung 4. Looking at rung 3, we can open the envelope for a probability 1/2 + q(2)/2 = 2/3 ofwinning, and we can not open the envelope for a probability p(4)/2 + p(2)/2 of winning. On rung2, we can open the envelope for a probability 1/2 + q(1)/2 = 7/12 of winning, and we can not openthe envelope for a probability p(3)/2 + p(1)/2 = p(3)/2 + 1/4 of winning.Now we can use all this information together for the complete answer. We know p(2) ≥ 7/12,therefore p(3) ≥ p(4)/2 + 7/24, and we know p(4) ≥ p(5)/2 + p(3)/2 ≥ 1/2+p(4)/2

2 + p(3)/2 ≥1/2+p(4)/2

2 + p(4)/2+7/242 . Isolating p(4) in this inequality, we get p(4) ≥ 1/2+7/24 > 3/4, therefore we

should not open the envelope on rung 4. Now from p(4) = p(5)/2+p(3)/2 and p(5) = 1/2+p(4)/2we have p(4) = 1/2+p(4)/2

2 +p(3)/2, so p(3) = 3p(4)/2−1/2 ≥ 11/16 > 2/3, so we should not openthe envelope on rung 3. Now p(2) ≥ p(3)/2 + 1/4 ≥ 19/32 > 7/12, so we should not open theenvelope on rung 2. Therefore the best strategy is to open the envelope iff we are on the bottomrung.

For each rung, score 4 points for the correct answer and 8 more for a correct justification.

Problem O10 [75 points]

A,B,C,D, and E are relatively prime integers (i.e., have no single common factor) such that thepolynomials 5Ax4 +4Bx3 +3Cx2 +2Dx+E and 10Ax3 +6Bx2 +3Cx+D together have 7 distinctinteger roots. What are all possible values of A? Your team has been given a sealed envelope thatcontains a hint for this problem. If you open the envelope, the value of this problem decreases by20 points. To get full credit, give the sealed envelope to the judge before presenting your solution.

4

Hint: Consider A = 1, B = D = 0, C = 750, and E = 19845.

Solution: Call the negatives of the roots of the first polynomial a, b, c, d, and the negatives of theroots of the second polynomial e, f, g (using the negatives avoids negative signs for the rest of theproof, thus preventing the possibility of dropping a sign). Then 5Ax4 + 4Bx3 + 3Cx2 + 2Dx+E =5A(x + a)(x + b)(x + c)(x + d) and 10Ax3 + 6Bx2 + 3Cx + D = 10A(x + e)(x + f)(x + g). ThusB = 5

4A(a+ b+ c+ d) = 53A(e+ f + g), C = 5

3A(ab+ ac+ ad+ bc+ bd+ cd) = 103 A(ef + eg+ fg),

D = 52A(abc+ abd+ acd+ bcd) = 10A(efg), and E = 5A(abcd). From these equations we see that

since all the variables are integers, it must be the case that A|B, A|3C, A|D, and A|E, thereforeA,B,C,D, and E can only be relatively prime if A is ±1 or ±3. Now we need numerical examplesto show that both of these are possible. Without loss of generality let g = 0, so D = 0. Then e

and f are −3B±√

9B2−30AC10A , so 9B2 − 30AC must be a perfect square. Let B = 0 in the hope that

solutions will still exist to this simplified problem. First let us try to find an example with A = −1,so we need 30C to be a perfect square. This first happens for C = 30, and in that case e and fare ±3. We need a + b + c + d = 0, so let’s try to look for a = −b, c = −d. This doesn’t workfor C = 30 since 3C/5 = 18 is not the sum of two distinct squares. For that we will need to tryC = 52 ·30 = 750, for which we get e, f = ±15, a, b = ±3, and c, d = ±21. Thus for A = ±1 we canuse B = D = 0, C = ∓750, and E = ±19845. Similarly we can find A = ±3, B = D = 0, C = ∓250,and E = ±735, for which (a, b, c, d, e, f, g) = (1,−1, 7,−7, 5,−5, 0).

Note that these give us quintic polynomials with integer coefficients possessing 4 relative extremaand 3 points of inflection at lattice points, such as 3x5 − 250x3 + 735x.

Score 20 points for the correct answer, 15 points for a correct justification, 20 points for a numericalexample for ±1, 20 points for a numerical example for ±3. If the sealed envelope isn’t presentedat the beginning of the solution, no credit is given for the ±1 example. If the ± is forgotten, give10 points for the answer, 10 for the justification, and 10 for each numerical example. There areinfinitely many possible examples, so anything other than the two given above must be checked foraccuracy.

5

Team

Harvard-MIT Math TournamentFebruary 27, 1999

1. A combination lock has a 3 number combination, with each number an integer between 0 and 39inclusive. Call the numbers n1, n2, and n3. If you know that n1 and n3 leave the same remainderwhen divided by 4, and n2 and n1 + 2 leave the same remainder when divided by 4, how manypossible combinations are there?

2. A ladder is leaning against a house with its lower end 15 feet from the house. When the lowerend is pulled 9 feet farther from the house, the upper end slides 13 feet down. How long is theladder (in feet)?

3. How many non-empty subsets of {1, 2, 3, 4, 5, 6, 7, 8} have exactly k elements and do not containthe element k for some k = 1, 2, ..., 8.

4. Consider the equation FORTY + TEN + TEN = SIXTY , where each of the ten lettersrepresents a distinct digit from 0 to 9. Find all possible values of SIXTY .

5. If a and b are randomly selected real numbers between 0 and 1, find the probability that thenearest integer to a−b

a+b is odd.

6. Reduce the number 3√

2 +√

5 + 3√

2−√

5.

7. Let 11−x−x2−x3 =

∑∞i=0 anx

n, for what positive integers n does an−1 = n2?

8. Find all the roots of (x2 + 3x+ 2)(x2 − 7x+ 12)(x2 − 2x− 1) + 24 = 0.

9. Evaluate∑17n=2

n2+n+1n4+2n3−n2−2n

.

10. If 5 points are placed in the plane at lattice points (i.e. points (x, y) where x and y are bothintegers) such that no three are collinear, then there are 10 triangles whose vertices are amongthese points. What is the minimum possible number of these triangles that have area greater than1/2?

11. Circles C1, C2, C3 have radius 1 and centers O,P,Q respectively. C1 and C2 intersect at A, C2

and C3 intersect at B, C3 and C1 intersect at C, in such a way that 6 APB = 60◦, 6 BQC = 36◦,and 6 COA = 72◦. Find angle ABC (degrees).

12. A fair coin is flipped every second and the results are recorded with 1 meaning heads and 0meaning tails. What is the probability that the sequence 10101 occurs before the first occuranceof the sequence 010101?

Team Solutions

Harvard-MIT Math Tournament

February 27, 1999

Problem T1 [15]

A combination lock has a 3 number combination, with each number an integer between 0 and 39inclusive. Call the numbers n1, n2, and n3. If you know that n1 and n3 leave the same remainderwhen divided by 4, and n2 and n1 + 2 leave the same remainder when divided by 4, how manypossible combinations are there?

Solution: There are 40 choices for the last number, and for each of these we have 10 choices foreach of the first two numbers, thus giving us a total of 4000 possible combinations. It is interestingto note that these restrictions are actually true for Master locks.

Problem T2 [15]

A ladder is leaning against a house with its lower end 15 feet from the house. When the lower endis pulled 9 feet farther from the house, the upper end slides 13 feet down. How long is the ladder(in feet)?

Solution: Of course the house makes a right angle with the ground, so we can use the Pythagoreantheorem. Let x be the length of the ladder and y be the original height at which it touched thehouse. Then we are given x2 = 152 + y2 = 242 + (y − 13)2. Isolating y in the second equation weget y = 20, thus x is 25.

Problem T3 [20]

How many non-empty subsets of {1, 2, 3, 4, 5, 6, 7, 8} have exactly k elements and do not containthe element k for some k = 1, 2, ..., 8.

Solution: Probably the easiest way to do this problem is to count how many non-empty subsets of{1, 2, . . . , n} have k elements and do contain the element k for some k. The element k must havek − 1 other elements with it to be in a subset of k elements, so there are

(

n−1

k−1

)

such subsets. Now∑

n

k=1

(

n−1

k−1

)

= (1 + 1)n−1 = 2n−1, so that is how many non-empty sets contain some k and havek elements. The set {1, 2, . . . , n} has 2n subsets (each element either is or is not in a particularsubset), one of which is the empty set, so the number of non-empty subsets of {1, 2, 3, 4, 5, 6, 7, 8}have exactly k elements and do not contain the element k for some k is 2n − 2n−1 − 1 = 2n−1 − 1.In the case n = 8, this yields 127.

Problem T4 [20]

Consider the equation FORTY +TEN+TEN = SIXTY , where each of the ten letters representsa distinct digit from 0 to 9. Find all possible values of SIXTY .

1

Solution: Since Y +N +N ends in Y , N must be 0 or 5. But if N = 5 then T +E+E+1 ends inT, which is impossible, so N = 0 and E = 5. Since F 6= S we must have O = 9, R+T +T +1 > 10,and S = F + 1. Now I 6= 0, so it must be that I = 1 and R + T + T + 1 > 20. Thus R and T are6 and 7, 6 and 8, or 7 and 8 in some order. But X can’t be 0 or 1 since those are taken, and Xcannot be 3 since F and S have to be consecutive, so it must be that R + T + T + 1 is 21 or 23.This is satisfied only for R = 7, T = 8, so F = 2, S = 3, and Y = 6. This SIXTY = 31486.

Problem T5 [30]

If a and b are randomly selected real numbers between 0 and 1, find the probability that the nearestinteger to a−b

a+bis odd.

Solution: The only reasonable way I know of to do this problem is geometrically (yes, you can useintegrals to find the areas of the triangles involved, but I don’t consider that reasonable). First letus find the points (a, b) in the plane for which the nearest integer to a−b

a+bis 0, i.e. −1

2≤ a−b

a+b≤ 1

2.

Taking the inequalities one at a time, −1

2≤ a−b

a+bimplies that a+ b ≥ 2(b− a), or b ≤ 3a, so these

points must lie below the line y = 3x. Similarly, a−b

a+b≤ 1

2implies that (a, b) must lie above the

line y = 1

3x. Now we can look for the points (a, b) for which the nearest integer to a−b

a+bis 1, i.e.

1

2≤ a−b

a+b≤ 3

2, and we find that all points in the first quadrant that lie above the line y = 3x satisfy

this inequality. Similarly, the closest integer to a−b

a+bis -1 for all points in the first quadrant below

the line y = 1

3x. For a and b between 0 and 1, the locus of points (a, b) for which the nearest integer

to a−b

a+bis odd is two right triangles with legs of length 1 and 1

3, so together they have area 1

3. The

locus of all points (a, b) with a and b between 0 and 1 is a square of side length 1, and thus hasarea 1. Therefore the probability that the nearest integer to a−b

a+bis odd is 1

3.

Problem T6 [30]

Reduce the number3

√

2 +√5 +

3

√

2−√5.

Solution: Observe that (3

√

2 +√5+

3

√

2−√5)3 = (2+

√5)−3(

3

√

2 +√5)−3(

3

√

2−√5)+(2−

√5) =

4 − 3(3

√

2 +√5 +

3

√

2−√5). Hence

3

√

2 +√5 +

3

√

2−√5 is a root of the cubic x3 + 3x − 4 =

(x− 1)(x2 + x+ 4). The roots of x2 + x+ 4 are imaginary, so3

√

2 +√5 +

3

√

2−√5 = 1.

Problem T7 [30]

Let 1

1−x−x2−x3 =∑∞

i=0 anxn, for what positive integers n does an−1 = n2?

Solution: Multiplying both sides by 1−x−x2−x3 the right hand side becomes a0+(a1−a0)x+(a2−a1−a0)x

2+ . . ., and setting coefficients of xn equal to each other we find that a0 = 1, a1 = 1, a2 = 2,and an = an−1+an−2+an−3 for n ≥ 3. Thus the sequence of an’s starts 1, 1, 2, 4, 7, 13, 24, 44, 81,149, . . . . So we now see that a0 = 12 and a8 = 92. What makes it impossible for this to happen againis that the sequence is growing exponentially. It will suffice to show that an > 1.5n for n > 2, sincen2/(n− 1)2 < 1.5 for n ≥ 6, thus when an−1 exceeds n2 at n = 10 there can be no more solutionsto an−1 = n2. Observe that an > 1.5an−1 for n = 3, 4, 5. By way of induction, assume it for n− 2,

2

n−1, and n, then an+1 = an+an−1+an−2 > 1.5n+1.5n−1+1.5n−2 = 1.5n−2(1+1.5+1.52) > 1.5n+1.Thus, by induction, an > 1.5n for n > 2, so the only solutions are 1, 9.

Problem T8 [35]

Find all the roots of (x2 + 3x+ 2)(x2 − 7x+ 12)(x2 − 2x− 1) + 24 = 0.

Solution: We re-factor as (x + 1)(x − 3)(x + 2)(x − 4)(x2 − 2x − 1) + 24, or (x2 − 2x − 3)(x2 −2x − 8)(x2 − 2x− 1) + 24, and this becomes (y − 4)(y − 9)(y − 2) + 24 where y = (x − 1)2. Now,(y − 4)(y − 9)(y − 2) + 24 = (y − 8)(y − 6)(y − 1), so y is 1, 6, or 8. Thus the roots of the originalpolynomial are 0,2,1±

√6,1± 2

√2.

Problem T9 [40]

Evaluate∑

17n=2

n2+n+1

n4+2n3−n2−2n.

Solution: Observe that the denominator n4+2n3−n2− 2n = n(n− 1)(n+1)(n+2). Thus we can

rewrite the fraction as n2−n+1

n4+2n3−n2−2n= a

n−1+ b

n+ c

n+1+ d

n+2for some real numbers a, b, c, and d. This

method is called partial fractions. Condensing the right hand side as a fraction over n4+2n3−n2−2nwe get n2−n+1 = a(n3+3n2+2n)+ b(n3+2n2−n−2)+ c(n3+n2−2n)+d(n3−n). Comparingcoefficients of each power of n we get a+b+c+d = 0, 3a+2b+c = 2, 2a−b−2c−d = 2, and −2b = 2.This is a system of 4 equations in 4 variables, and its solution is a = 1/2, b = −1/2, c = 1/2, and d =−1/2. Thus the summation becomes 1

2(1− 1

2+ 1

3− 1

4+ 1

2− 1

3+ 1

4− 1

5+ 1

3− 1

4+ 1

5− 1

6+· · ·+ 1

16− 1

17+ 1

18− 1

19).

Notice that almost everything cancels to leave us with 1

2(1 + 1

3− 1

17− 1

19) = 592

969.

Problem T10 [45]

If 5 points are placed in the plane at lattice points (i.e. points (x, y) where x and y are bothintegers) such that no three are collinear, then there are 10 triangles whose vertices are amongthese points. What is the minimum possible number of these triangles that have area greater than1/2?

Solution: By the pigeonhole principle, the 5 points cannot all be distinct modulo 2, so two of themmust have a midpoint that is also a lattice point. This midpoint is not one of the 5 since no 3are collinear. Pick’s theorem states that the area of a polygon whose vertices are lattice points isB/2 + I − 1 where B is the number of lattice points on the boundary and I is the number in theinterior. Thus those two points form the base of 3 triangles whose area will be greater than 1/2 byPick’s theorem since there are 4 lattice points on the boundary. Now it also turns out that at leastone of the triangles must contain a lattice point, thus giving us a fourth triangle with area greaterthan 1/2. This is actually pretty easy to show with the aid of a picture or some visualization.Suppose we have 4 points and we’re trying to find a 5th one so that no triangle will contain aninterior lattice point. The 4 lattice points must form a quadrilateral of area 1, so in fact it is aparallelogram (think deeply about it). Draw the four sides, extending them throughout the plain.Each vertex is now the tip of an infinite triangular region of the plane, and if the 5th lattice pointis chosen in that region then the triangle formed by the 5th point and the two vertices of theparallelogram adjacent to the one we are considering will form a triangle containing the vertex we

3

are considering. But the part of the plane that isn’t in one of these 4 regions contains no latticepoints or else we could draw a parallelogram congruent to the first one with lattice point verticesand containing that lattice point, but that would violate Pick’s theorem since the parallelogram hasarea 1. Therefore we must have a fourth triangle with area greater than 1/2 (one must justify thatthis really is in addition to the 3 triangles we already knew we’d get). An example that achievesthis minimum is the points (0,0), (1,0), (1,1), (2,1), and (2,-1). Therefore the minumum possiblenumber of these triangles that have area greater than 1/2 is 4.A less trivial example that achieves the minimum is (0,0), (1,1), (2,1), (3,2), and (7,5).

Problem T11 [55]

Circles C1, C2, C3 have radius 1 and centers O,P,Q respectively. C1 and C2 intersect at A, C2 andC3 intersect at B, C3 and C1 intersect at C, in such a way that 6 APB = 60◦, 6 BQC = 36◦, and6 COA = 72◦. Find angle ABC (degrees).

Solution: Using a little trig, we have BC = 2 sin 18, AC = 2 sin 36, and AB = 2 sin 30 (see leftdiagram). Call these a, b, and c, respectively. By the law of cosines, b2 = a2 + c2 − 2ac cosABC,

therefore cosABC = sin2 18+sin2 30−sin2 36

2 sin 18 sin 30. In the right diagram below we let x = 2 sin 18 and see

that x+ x2 = 1, hence sin 18 = −1+√5

4. Using whatever trig identities you prefer you can find that

sin2 36 = 5−√5

4, and of course sin 30 = 1

2. Now simplification yields sin2 18 + sin2 30− sin2 36 = 0,

so 6 ABC = 90◦.Note that this means that if a regular pentagon, hexagon, and decagon are inscribed in a circle,then we can take one side from each and form a right triangle.

1

36

72

3636

72

1

x2x

x

x

Problem T12 [65]

A fair coin is flipped every second and the results are recorded with 1 meaning heads and 0meaning tails. What is the probability that the sequence 10101 occurs before the first occuranceof the sequence 010101?

Solution: Call it a win if we reach 10101, a loss if we reach 010101. Let x be the probability ofwinning if the first flip is a 1, let y be the probability of winning if the first flip is a 0. Then theprobability of winning is (x+ y)/2 since the first flip is 1 or 0, each with probability 1/2. If we everget two 1’s in a row, that is the same as starting with a 1 as far as the probability is concerned.Similarly, if we get two 0’s in a row, then we might as well have started with a single 0.

start

1 0

11 10... 01 00...

101 100... 011...010

1010 1011... 0101 0100...

10101 10100... 01011...

010101 010100...

01010

4

From the tree of all possible sequences, shown above, in which the probability of moving along anyparticular line is 1/2, we see that x = x(1/2+1/8)+y(1/4+1/16)+1/16, and y = x(1/4+1/16)+y(1/2 + 1/8 + 1/32). Solving these two equations in two unknowns we get x = 11/16 and y = 5/8.Therefore the probablity that the sequence 10101 occurs before the first occurance of the sequence010101 is 21/32.

5

Advan ed Topi sRi e Mathemati s Tournament 20001. How many di�erent ways are there to paint the sides of a tetrahedron with exa tly4 olors? Ea h side gets its own olor, and two olorings are the same if one an berotated to get the other.2. Simplify (�1+ip32 )6 + (�1�ip32 )6 to the form a + bi.3. Evaluate P1n=1 1n2+2n .4. Five positive integers from 1 to 15 are hosen without repla ement. What is the prob-ability that their sum is divisible by 3?5. Find all 3-digit numbers whi h are the sums of the ubes of their digits.6. 6 people ea h have a hat. If they shu�e their hats and redistribute them, what is theprobability that exa tly one person gets their own hat ba k?7. Assume that a; b; ; d are positive integers, and a = bd = 34 , pa2 + 2 � pb2 + d2 = 15.Find a + bd� ad� b .8. How many non-isomorphi graphs with 9 verti es, with ea h vertex onne ted to exa tly6 other verti es, are there? (Two graphs are isomorphi if one an relabel the verti esof one graph to make all edges be exa tly the same.)9. The Cin innati Reals are playing the Houston Alphas in the last game of the SwirledSeries. The Alphas are leading by 1 run in the bottom of the 9th (last) inning, andthe Reals are at bat. Ea h batter has a 13 han e of hitting a single and a 23 han e ofmaking an out. If the Reals hit 5 or more singles before they make 3 outs, they willwin. If the Reals hit exa tly 4 singles before making 3 outs, they will tie the game andsend it into extra innings, and they will have a 35 han e of eventually winning the game(sin e they have the added momentum of oming from behind). If the Reals hit fewerthan 4 singles, they will LOSE! What is the probability that the Alphas hold o� theReals and win, sending the pa ked Alphadome into a frenzy? Express the answer as afra tion.10. I all two people A and B and think of a natural number n. Then I give the numbern to A and the number n+ 1 to B. I tell them that they have both been given naturalnumbers, and further that they are onse utive natural numbers. However, I don't tellA what B's number is and vi e versa. I start by asking A if he knows B's number. Hesays \no". Then I ask B if he knows A's number, and he says \no" too. I go ba k toA and ask, and so on. A and B an both hear ea h other's responses. Do I ever get a\yes" in response? If so, who responds �rst with \yes" and how many times does hesay \no" before this? Assume that both A and B are very intelligent and logi al. Youmay need to onsider multiple ases.

Advan ed Topi s SolutionsRi e Mathemati s Tournament 20001. Assume we have 4 olors - 1, 2, 3, and 4. Fix the bottom as olor 1. On the remainingsides you an have olors 2, 3, 4 (in that order), or 2, 4, 3, whi h are not rotationallyidenti al. So, there are 2 ways to olor it.2. Sin e os 2�3 = �12 and sin 2�3 = p32 , we an write the �rst term as ( os 2�3 + i sin 2�3 )6.Sin e os 4�3 = �12 and sin 4�3 = �p32 , we an write the se ond term as ( os 4�3 + i sin 4�3 )6.Now, we apply DeMoivre's Theorem to simplify the �rst expression to ( os 6 � 2�3 + sin 6 � 2�3 ) =( os 4� + sin 4�) = 1+0 = 1. Similarly, we simplify the se ond expression to ( os 6 � 4�3 + sin 6 � 4�3 ) =( os 8� + sin 8�) = 1 + 0 = 1. Thus, the total sum is 1 + 1 = 2.3. We know that 1n2+2n = 1n(n+2) = 1n� 1n+22 . So, if we sum this from 1 to1, all terms ex eptfor 112 + 122 will an el out (a "teles oping" series). Therefore, the sum will be 34 .4. The possibilities for the numbers are:� all �ve are divisible by 3� three are divisible by 3, one is � 1 (mod 3) and one is � 2 (mod 3)� two are divisible by 3, and the other three are either� 1 (mod 3) or � 2 (mod 3)� one is divisible by 3, two are � 1 (mod 3) and two are � 2 (mod 3)� four are � 1 (mod 3) and one is � 2 (mod 3)� four are � 2 (mod 3) and one is � 1 (mod 3)This gives us 1001 possible ombinations out of �155 � or 3003. So, the probability is10013003 =13 .5. 153,370,371,4076. There are 6 people that ould get their hat ba k, so we must multiply 6 by the numberof ways that the other 5 people an arrange their hats su h that no one gets his/her hatba k. So, the number of ways this will happen is (6 � derangement of 5), or 6�44 = 264.Sin e there are 6! = 720 possible arrangements of hats, the probability of exa tly oneperson getting their hat ba k is 264720 =1130 .7. We an view these onditions as a geometry diagram as seen below. So, we know thatef = 34 (sin e e = a� b = 34 � 34d = 34f and we know that pe2 + f 2 = 15 (sin e this ispa2 + 2�pb2 + d2). Also, note that a +bd�ad�b = (a�b)( �d) = ef . So, solvingfor e and f , we �nd that e2+ f 2 = 225, so 16e2+16f 2 = 3600, so (4e)2+(4f)2 = 3600,so (3f)2+(4f)2 = 3600, so f 2(32 + 42) = 3600, so 25f 2 = 3600, so f 2 = 144 and f = 12.Thus, e = 34 � 12 = 9. Therefore, ef = 9 � 12 = 108.

ba

d

c

e

f

8. It suÆ es to onsider the omplements of the graphs, so we are looking for graphs with9 verti es, where ea h vertex is onne ted to 2 others. There are 4 di�erent graphs -see below.

9. The probability of the Reals hitting 0 singles is (23)3. The probability of the Reals hittingexa tly 1 single is �32� �(23)3 � 13 , sin e there are 3 spots to put the two outs (the last spotmust be an out, sin e the inning has to end on an out). The probability of the Realshitting exa tly 2 singles is �42� �(23)3 � (13)3. The probability of the Reals hitting exa tly3 singles is �52� �(23)3 � (13)3. If any of these happen, the Alphas win right away. Addingthese gives us a 656729 han e of this happening. If exa tly 4 singles o ur (with probability�62� �(23)3 � (13)4), then there is a 25 han e that the Alphas win. The probability of thishappening is 25 � 40729 . Thus, the total probability of the Alphas winning is the sum ofthese two probabilities, or 656729 + 16729 =224243 .10. A will say yes when B says no to n � 1 or n, as A will then know B's number is onegreater than A's number. Thus, A responds �rst, after n�12 "no" responses if n isodd, after n2 "no" responses if n is even.

AlgebraRi e Mathemati s Tournament 20001. How many integers x satisfy jxj + 5 < 7 and jx� 3j > 2?2. Evaluate 20003 � 1999 � 20002 � 19992 � 2000 + 19993.3. Five students take a test on whi h any integer s ore from 0 to 100 in lusive is possible.What is the largest possible di�eren e between the median and the mean of the s ores?4. What is the fewest number of multipli ations required to rea h x2000 from x, using onlypreviously generated powers of x? For example, x ! x2 ! x4 ! x8 ! x16 ! x32 !x64 ! x128 ! x256 ! x512 ! x1024 ! x1536 ! x1792 ! x1920 ! x1984 ! x2000 uses 15multipli ations.5. A ja ket was originally pri ed $100. The pri e was redu ed by 10% three times andin reased by 10% four times in some order. To the nearest ent, what was the �nalpri e?6. Barbara, Edward, Abhinav, and Alex took turns writing this test. Working alone, they ould �nish it in 10, 9, 11, and 12 days, respe tively. If only one person works on thetest per day, and nobody works on it unless everyone else has spent at least as manydays working on it, how many days (an integer) did it take to write this test?7. A number n is alled multipli atively perfe t if the produ t of all the positive divisorsof n is n2. Determine the number of positive multipli atively perfe t numbers less than100.8. A man has three daughters. The produ t of their ages is 168, and he remembers thatthe sum of their ages is the number of trees in his yard. He ounts the trees but annotdetermine any of their ages. What are all possible ages of his oldest daughter?9. a = bd = 34 , pa2 + 2 �pb2 + d2 = 15. Find a + bd� ad� b .10. Find the smallest positive integer a su h that x4 + a2 is not prime for any integer x.

Algebra SolutionsRi e Mathemati s Tournament 20001. The only integers that satisfy jxj + 5 < 7 are the ones that satisfy jxj < 2 - namely,�1; 0; 1. The integers that satisfy jx� 3j > 2 are 6; 7; 8; : : : and 0;�1;�2; : : :. So, theintegers that satisfy both are 0;�1, and there are 2 of them.2. 20003 � 1999 � 20002 � 19992 � 2000 + 19993 an be fa tored into (2000� 1999)20002 +19992(�2000 + 1999), whi h redu es to 20002�19992. This fa tors into (2000 + 1999)(2000� 1999),whi h is equal to 3999.3. Let the s ores be a,b, ,d,e, where 0 � a � b � � d � e � 100. So, the mean is15(a+ b + + d+ e); and the median is . So, we want to maximize 15(a+ b+ + d+ e)� . To do this, we must maximize d and e and minimize or maximize . One way todo this is to let a = b = = 0 and d = e = 100, so the di�eren e between the meanand the median is 15(0 + 0 + 0 + 100 + 100) � 0) = 2005 = 40. If we maximize , then = d = e = 100, and then the mean is 15(0 + 0 + 100 + 100 + 100) = 60, and themedian is 60, with a di�eren e of 40 as well.4. The pri e starts at $100. Clearly, the order of pri e hanges does not matter. It isredu ed by 10% three times ( $100 ! $90 ! $81 ! $72:90), and the new pri e is$72.90. It is in reased by 10% four times ($72:90 ! $80:19 ! $88:209 ! $97:0299 !$106:73289) , and the new pri e is $106.73289. Rounded to the nearest ent, this is$106:73.5. Every day Edward works, he gets 19 of the test done. Similarly, every day Barbaraworks, she gets 110 of the test done, every day Abhinav works, he gets 111 of the testdone, and every day Alex works, he gets 112 of the test done. So, after 4 days (aftereveryone has worked on the test one day, they have ompleted 19+ 110+ 111+ 112 = 38:535%of the test. After 8 days, they have ompleted twi e that, or 77:0707% of the test. AfterEdward, Barbara, and Abhinav ea h work one more day, the test will be omplete inthe minimum amount of time, so the test will take 11 days to omplete. If the leasteÆ ient workers work after the 8th day, the test still takes 11 days to omplete.6. A shortest path is x ! x2 ! x4 ! x8 ! x12 ! x24 ! x25 ! x50 ! x100 ! x200 !x400 ! x800 ! x1600 ! x2000, using 13 multipli ations.7. All multipli atively perfe t numbers have exa tly 4 distin t positive divisors, or 1. So,we must look for numbers that are either� 1� a produ t of two distin t primes� a ube of a primeNumbers satisfying one of these onditions less than 100 are: 1, 6, 8, 10, 14, 15, 21, 22,26, 27, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94,95. There are 33 of these.

8. 168 = 23 �3 �7. There are only 2 ombinations of these whose sums allow indistinguisha-bility of the ages. If there are 27 trees, 2, 4, 21 and 1, 12, 14 years are possible. Ifthere are 21 trees, 2, 7, 12 and 3, 4, 14 are possible. So, the possible ages of the oldestdaughter are 12, 14, 21.9. We an view these onditions as a geometry diagram as seen below. So, we know thatef = 34 (sin e e = a� b = 34 � 34d = 34f and we know that pe2 + f 2 = 15 (sin e this ispa2 + 2�pb2 + d2). Also, note that a +bd�ad�b = (a�b)( �d) = ef . So, solvingfor e and f , we �nd that e2+ f 2 = 225, so 16e2+16f 2 = 3600, so (4e)2+(4f)2 = 3600,so (3f)2+(4f)2 = 3600, so f 2(32 + 42) = 3600, so 25f 2 = 3600, so f 2 = 144 and f = 12.Thus, e = 3412 = 9. Therefore, ef = 9 � 12 = 108.b

a

d

c

e

f

10. a = 1 learly does not work, sin e if x = 1, then x4 + a2 = 2, whi h is prime. a = 2 learly does not work, sin e if x = 1, then x4 + a2 = 5, whi h is also prime. Here is atable for a's and values of x that show they do not work:a x a4 + x23 10 100094 1 175 2 416 1 377 20 160049So, let us onsider a = 8 - i.e. the sum x4 + 64. This is the same as (x2 + 8)2 � 16x2 =(x2 + 4x+ 8)(x2 � 4x+ 8) by the di�eren e of squares. This is learly not prime forany integer x. So, the answer is a = 8.

Cal ulusRi e Mathemati s Tournament 20001. Find the slope of the tangent at the point of in e tion of y = x3 � 3x2 + 6x+ 2000.2. Karen is attempting to limb a rope that is not se urely fastened. If she pulls herselfup x feet at on e, then the rope slips x3 feet down. How many feet at a time must shepull herself up to limb as eÆ iently as possible?3. A re tangle of length 14� and height 4 is bise ted by the x-axis and is in the �rst andfourth quadrants. The graph of y = sin(x) + C divides the area of the square in half.What is C?4. For what value of x (0 < x < �2 ) does tan x+ ot x a hieve its minimum?5. For �1 < x < 1, let f(x) = P1i=1 xii . Find a losed form expression (a losed formexpression is one not involving summation) for f .6. A hallway of width 6 feet meets a hallway of width 6p5 feet at right angles. Find thelength of the longest pipe that an be arried horizontally around this orner.7. An envelope of a set of lines is a urve tangent to all of them. What is the envelope ofthe family of lines y = 2x0 + x(1� 1x20 ), with x0 ranging over the positive real numbers?Hint: slope at a point P on a urve is dydx jP .8. Find R �20 ln sin �d�.9. Let f(x) = vuutx +s0 +rx+q0 +px + : : :. If f(a) = 4, then �nd f 0(a).10. A mirror is onstru ted in the shape of y equals �px for 0 � x � 1, and �1 for1 < x < 9. A ray of light enters at (10,1) with slope 1. How many times does it boun ebefore leaving?

Cal ulus SolutionsRi e Mathemati s Tournament 20001. y = x3�3x2+6x+2000, so y0 = 3x2�6x+6 and y00 = 6x�6, so the point of in e tionis the solution to 6x�6 = 0, or x = 1. At x = 1, the slope is f 0jx=1 = 3(1)2�6(1)+6 =3.2. The hange in Karen's position is x � x3. The optimal length to limb is at a riti alpoint. The only realisti riti al point is at the solution to 1� 3x2 = 0 or x = p33 .3. R 14�0 sinx+C = 0, from the statement of the problem. So, [� os x + Cx℄j�40 = 0. Thus, os �4 + �4C+1 = 0. So, �p22 + �4C+1 = 0, and solving for C, we �nd that C = 2p2�4� .4. Let y = tan x. So, we want to �nd the minimum of y+ 1y , where 0 � y � 1. Taking thederivative and minimizing, we �nd that the minimum o urs at y = 1, so the minimumof the given fun tion o urs at ar tan 1 = �4 .5. f(x) = P1i=1 xii . So, f 0(x) = P1i=0 xi = 11�x . Thus, f(x) = � ln(1� x).6. Assume the pipe barely �ts around the orner (i.e. it is in onta t with the orner).The lower orner is at (0; 0) and the upper orner is at (6; 6p5). Call x0 the point onthe lower wall it hits at the tightest spot. Given an x0, the longest a pipe ould be withone end at x0 and leaning against the (6; 6p5) orner is rx20 + (6p5 + 36p5x0�6)2. We wantthe minimum of all of these "longest pipes", be ause the pipe needs to �t at all anglesaround the orner. Taking the derivative (without the square root for simpli ity) andsetting it equal to 0, we need to solve x30 � 6x20 + 36x0 � 1296 = 0. We an qui kly �ndthat x0 = 12 is the only good solution, so the maximum length is 12p6.7. Using the hint, take dydx . Set this equal to 0 and solve for x relative to x0. Plug this infor x0 in the given family of lines to obtain the envelope y = x+ 1x ; x > 0.8. Let I denote the given integral. Under the transformation � ! �2 � �, I transforms toR �20 ln( os(�))d�. So,2I = R �20 ln(sin � os �)d�= R �0 (ln(sin 2�)� ln 2)d(2�)=2= � �2 ln 2 + 12 R �0 sin(�)d�= �pi2 � ln 2 + R �20 sin(�)d�= �pi2 � ln 2 + I giving I = ��2 ln 2.9. Note �rst that ([f(x)℄2 � x)2 = f(x), so if f(a) = 4, then (16� a)2 = 4, so a = 14.Now, f 0(x) = 1+ f 0(x)2([f(x)℄2�x)2f(x) , so f 0(14) = 431 .10. Solution: Throughout this solution we will use the fa t that when light boun es o� amirror, the angle of in iden e is equal to the angle of re e tion. First the beam hitsthe point (8,-1), then (6,1), (4,-1), (2,1), and then is travelling along the line y = x� 1.Thus the beam hits the parabola at the point (1+ 1�p52 ; 1�p52 ). To estimate p5, noti ethat 222 = 484 and 232 = 529, so p5 = p50010 = 2:2 : : :. Thus 1�p52 = �:6 : : :, so the light

hits the parabola at approximately (.4,-.6). The slope of the tangent to the parabolaat this point is �12 (:4)�1=2, whi h is about -.8, so we need to �nd the slope of the beamafter it re e ts o� of this tangent. For purposes of �nding this slope, hange oordinatesso that the point of interse tion is the origin. The beam is oming in along y = x, andy = 1:2x is perpendi ular to the tangent. The diagram below should larify the setup.reflected

beam

y=x

perpendicular totangent

tangentWe will �nd the new path of the light by �nding the re e tion about the line y = 1:2xof a point on its in oming path. We know the point (1,1.2) is on the line y = 1:2x, so aperpendi ular through this point is y � 1:2 = �:8(x� 1), whi h interse ts y = x at thepoint (1.1,1.1). Thus the new path goes through the point (.9,1.3), so it has slope 1.4(all values rounded to one de imal pla e). Going ba k to our original oordinate system,the light is now travelling along the line y + :6 = 1:4(x� :4), so it next hits the mirrorat (1.5,1). After that the x oordinate in reases by 2=1:4 = 1:4 between boun es, so ithits (2.9,-1), (4.3,1), (5.7,-1), (7.1,1), (8.5,-1), and �nally (9.9,1). A loser examinationof the approximations made (e.g. by re�ning them to two de imal pla es) reveals thatthe last boun e is a tually further to the left (at (9.21,1), to be more pre ise), so indeedthe light does boun e 12 times.

GeneralRi e Mathemati s Tournament 20001. If a = 2b + , b = 2 + d, 2 = d+ a� 1, d = a� , what is b?2. The temperatures f ÆF and ÆC are equal when f = 95 + 32. What temperature is thesame in both ÆF and ÆC?3. A twelve foot tree asts a �ve foot shadow. How long is Henry's shadow (at the sametime of day) if he is �ve and a half feet tall?4. Ti kets for the football game are $10 for students and $15 for non-students. If 3000fans attend and pay $36250, how many students went?5. Find the interior angle between two sides of a regular o tagon (degrees).6. Three ards, only one of whi h is an a e, are pla ed fa e down on a table. You sele tone, but do not look at it. The dealer turns over one of the other ards, whi h is notthe a e (if neither are, he pi ks one of them randomly to turn over). You get a han eto hange your hoi e and pi k either of the remaining two fa e-down ards. If yousele ted the ards so as to maximize the han e of �nding the a e on the se ond try,what is the probability that you sele ted it on the(a) �rst try?(b) se ond try?7. Find [p19992000℄ where [x℄ is the greatest integer less than or equal to x.8. Bobo the lown was juggling his spheri al ows again when he realized that when hedrops a ow is related to how many ows he started o� juggling. If he juggles 1, hedrops it after 64 se onds. When juggling 2, he drops one after 55 se onds, and the other55 se onds later. In fa t, he was able to reate the following table: ows started juggling 1 2 3 4 5 6 7 8 9 10 11se onds he drops after 64 55 47 40 33 27 22 18 14 13 12 ows started juggling 12 13 14 15 16 17 18 19 20 21 22se onds he drops after 11 10 9 8 7 6 5 4 3 2 1He an only juggle up to 22 ows. To juggle the ows the longest, what number of owsshould he start o� juggling? How long (in minutes) an he juggle for?9. Edward's formula for the sto k market predi ts orre tly that the pri e of HMMT isdire tly proportional to a se ret quantity x and inversely proportional to y, the numberof hours he slept the night before. If the pri e of HMMT is $12 when x = 8 and y = 4,how many dollars does it ost when x = 4 and y = 8?10. Bob has a 12 foot by 20 foot garden. He wants to put fen ing around it to keep outthe neighbor's dog. Normal fen eposts ost $2 ea h while strong ones ost $3 ea h. Ifhe needs one fen epost for every 2 feet and has $70 to spend on the fen eposts, what isthe largest number of strong fen eposts he an buy?11. If a�b = a+ba�b , �nd n su h that 3�n = 3.

12. In 2020, the United States admits North Mathemati a as the 51st state. It onsistsof 5 islands joined by bridges as shown. Is it possible to ross all the bridges withoutdoubling over? If so, what is the di�eren e (positive) between the number of the startisland and the number of the end island?1 32

5413. How many permutations of 123456 have exa tly one number in the orre t pla e?14. The author of this question was born on April 24, 1977. What day of the week wasthat?15. Whi h is greater: (35)(53) or (53)(35)?16. Joe bikes x miles East at 20 mph to his friend's house. He then turns South and bikesx miles at 20 mph to the store. Then, Joe turns East again and goes to his grandma'shouse at 14 mph. On this last leg, he has to arry our he bought for her at the store.Her house is 2 more miles from the store than Joe's friend's house is from the store.Joe spends a total of 1 hour on the bike to get to his grandma's house. If Joe then ridesstraight home in his grandma's heli opter at 78 mph, how many minutes does it takeJoe to get home from his grandma's house?17. In how many distin t ways an the letters of STANTON be arranged?18. You use a lo k with four dials, ea h of whi h is set to a number between 0 and 9(in lusive). You an never remember your ode, so normally you just leave the lo kwith ea h dial one higher than the orre t value. Unfortunately, last night someone hanged all the values to 5. All you remember about your ode is that none of thedigits are prime, 0, or 1, and that the average value of the digits is 5. How many ombinations will you have to try?19. Eleven pirates �nd a treasure hest. When they split up the oins in it, they �nd thatthere are 5 oins left. They throw one pirate overboard and split the oins again, onlyto �nd that there are 3 oins left over. So, they throw another pirate over and try again.This time, the oins split evenly. What is the least number of oins there ould havebeen?20. Given: AC has length 5, semi ir le AB has radius 1, semi ir le BC has diameter 3.What per ent of the the big ir le is shaded?A B C

21. Find the area of the six-pointed star if all edges are of length s, all a ute angles are 60 Æand all obtuse angles are 240 Æ.22. An equilateral triangle with sides of length 4 has an isos eles triangle with the samebase and half the height ut out of it. Find the remaining area.23. What are the last two digits of 7777?24. Peter is randomly �lling boxes with andy. If he has 10 pie es of andy and 5 boxes ina row labeled A, B, C, D, and E, how many ways an he distribute the andy so thatno two adja ent boxes are empty?25. How many points does one have to pla e on a unit square to guarantee that two of themare stri tly less than 1/2 unit apart?26. Janet is trying to �nd Tim in a Cartesian forest. Janet is 5p2 miles from (0,0), p41miles from (1,0), and p61 miles from (0,1). Tim is p65 miles from (0,0), 2p13 milesfrom (1,0), and p58 miles from (0,1). How many miles apart are Janet and Tim?

General SolutionsRi e Mathemati s Tournament 20001. Sin e d = a � , substitute into the equation b = 2 + d to get b = 2 + a� = a + .Also, substitute into 2 = d+a� 1 to get 2 = a� +a� 1, or 3 = 2a� 1. Now, sin eb = a+ , we an substitute into a = 2b+ to get a = 2a+2 + , or a = �3 . Sin e weknow 3 = 2a� 1 from above, we substitute in to get 3 = �6 � 1, or = �19 . Thus,we �nd that a = 13 , d = 49 , and b = 29 .2. Let x be the temperature we are looking for, so x = 95 + 32. So, �45x = 32, sox = �54 � 32 =�40.3. Let x be the length of Henry's shadow in feet. Using similar triangles, we �nd that5:512 = x5 , so x = 5:512 � 5 = 1124 � 5 =5524 .4. Let x be the number of students and y be the number of non-students. We thenhave the equations x + y = 3000 and 10x + 15y = 36250. Substituting, we �nd that10(3000� y)+ 15y = 36250, or 30000� 10y+15y = 36250, so 30000+ 5y = 36250. So,5y = 6250, and y = 1250, so x =1750.5. The total number of degrees in an o tagon is (8� 2) �180 = 1080. Sin e the degrees areevenly distributed among the angles, the measure of one interior angle is 10808 =135 Æ.6. Pi k any ard �rst, then pi k the other fa e-down ard.(a) 13(b) 237. p19992000 � 4471:241, and [4471:241℄ =4471.8. The time that Bobo an juggle is the number of ows times se onds. So, we get thefollowing table: ows started juggling 1 2 3 4 5 6 7 8 9 10 11total time 64 110 141 160 165 162 154 144 126 130 132 ows started juggling 12 13 14 15 16 17 18 19 20 21 22total time 132 130 126 120 112 102 90 76 60 42 22Thus, we see that the maximum o urs with 5 ows, and the total time is 165 se onds= 234 minutes.9. Let p be the pri e of HMMT. So p = k � xy , where k is a onstant to be determined. Weknow that when x = 8 and y = 4 that p = 12, so, solving for k, we �nd that k = 6. So,when x = 4 and y = 8, we �nd that p = 6 � 48 =3.10. On the 12 foot sides, he needs 7 posts, and on the 20 foot sides, he needs 9 posts, so heneeds 7 + 9 + 7 + 9 = 32 total posts. Let x be the number of normal fen eposts and ybe the number of strong fen eposts, so x+ y = 32. To spend $70, we have the equation2x+3y = 70. Substituting, we �nd that 2(32� y)+ 3y = 70, so 64� 2y+3y = 70, and64 + y = 70, so y =6.11. Substituting, we �nd that 3+n3�n = 3, so 3 + n = 9� 3n, thus 4n = 6. So, n =32 .

12. Yes - one possible path is 4! 1! 2! 4! 5! 2! 3! 5, so the di�eren e betweenthe start island and end island is 1.13. The number of rearrangements keeping 1 number in its spot and rearranging the other5 su h that none are in the right spot is 44. There are 6 numbers to �x, this gives usan answer of 44 � 6 =264.14. Feburary 26, 2000 is a Saturday. April 24, 2000 is 23 � 365+1+1+1+1+1+1 = 8401days away from April 24, 1977 (in luding the leap years). So, Feburary 26, 2000 is8401�3�31�24 = 8343 days after April 24, 1977. Now, 83437 = 1191 with a remainderof 6. So, 6 days before Saturday is Sunday.15. Noti e that 36 = 729 while 54 = 625, and sin e ln 5 > ln 3, it follows that 54 ln 3 < 36 ln 5,so 53�5 ln 3 < 35�3 ln5, and so by laws of logarithms, 53 ln 35 < 35 ln 53. Again applyinglaws of logarithms, it follows that ln (35)(53) < ln (53)(35). So, sin e lnx is an in reasingfun tion, it follows that (35)(53) < (53)(35), and so it follows that (53)(35) is greater.One an also use the laws of exponents to redu e the values to 3(54) and 5(36). These ond is learly larger.16. Expressing the total time Joe has biked in hours leads to the equation x20+ x20+ x+214 = 1.So, x = 5. Thus, we an onstru t the diagram below, and �nd the total time that ittakes to get ba k: 13 � 178 = 16 hours, or 10 minutes.5 mi. Friend’s house

shore 7 mi. Gma’s house

13 mi.5 mi.

12 mi.17. There are 7! ways to arrange those letters. However, for every distin t arrangement,there are 2! � 2! = 4 total arrangements of the 2 T's and 2 N's. Therefore, the totalnumber of distin t ways to arrange the letters is 7!2!2! = 1260.18. The only digits possible are 4, 6, 8, and 9. The only groups of numbers allowed keepingthe average at 5 are 8444 are 6464. There are 4 ways to arrange 8444 and 6 ways toarrange 6464 so there are only 10 ombinations to try.19. Let y be the number of oins in the hest. From the problem, we know that y � 5(mod 11), y � 3 (mod 10), and y � 0 (mod 9). Combining these gives us thaty � 423 (mod 990), so the answer is 423.20. The area of the big ir le is (52)2� = 254 �. The area of the ir le with diameter AB is�, and the area of the ir le with diameter BC is 94�. Thus, the per entage of the big ir le that is shaded is 254 ���� 94�254 � = 25�4�925 = 1225 = 48%.21. It is lear that we an split the �gure into 12 equilateral triangles, all of whi h have sidelength s. So, sin e the area of one of these triangles is s2p34 , the total area is 3s2p3.

22. Aiso = Aeq2 so �A = Aeq �Aiso = 12Aeq. Also, Aeq = 42�p34 = 4p3, so �A = 2p3. Thus,the area is 2p3.23. Noti e that 72k+1 � 3 (mod 4) for k 2 Z (k is an integer). Also, 71 � 7 (mod 100),72 � 49 (mod 100), 73 � 43 (mod 100), 74 � 1 (mod 100), with the y le repeatingafterwards. So, learly 7777 � 3 (mod 4), and sin e the y le has period 4 (mod 100),we an on lude that 7777 � 43 (mod 100).24. If there are no empty boxes, there are 126 ways of distributing the identi al andy.If there is one empty box, there are 84 ways of distributing the andy and 5 ways of hoosing the empty box, so there are 84 � 5 = 420 ways. If there are two empty boxes,there are 36 ways of distributing the andy and 6 ombinations of assigning the emptyboxes so that they are not adja ent, so there are 36 � 6 = 216 ways. If there are threeempty boxes, there are 9 ways of distributing the andy, and only 1 way of arrangingthe empty boxes. Having four or �ve empty boxes is impossible, sin e some two wouldhave to be adja ent. So, the total number of ways is 126 + 420 + 216 + 9 = 771.25. We an pla e 9 points as shown, all at least 12 unit apart, but the next point must beless than 12 , so 10 points must be pla ed. There is no arrangement of 10 points withdistan e at least 12 . The proof of this is a simple appli ation of the Pigeonhole Prin iple.

26. Janet is at (5;�5) and Tim is at (7; 4). They are p85 miles apart.

GeometryRi e Mathemati s Tournament 20001. How many re tangles are there on an 8x8 he kerboard?2. In a triangle the sum of squares of the sides is 96. What is the maximum possible valueof the sum of the medians?3. Find PB, given that PA = 15; PC = 20; PD = 7, and ABCD is a square.

D CBA�� aaaaaCCCC #####P4. Find the total area of the non-triangle regions in the �gure below (the shaded area).

b/3

a/4

a/3

a

b

5. Side AB = 3. 4ABF is an equilateral triangle. Side DE = AB = AF = GE.6 FED = 60 Æ. FG = 1. Cal ulate the area of ABCDE.E

F

G

D

A

BC

6. What is the area of the largest ir le ontained in an equilateral triangle of area 8p3?7. Let ABC be a triangle ins ribed in the ellipse x24 + y29 = 1. If its entroid is the origin(0,0), �nd its area.8. A sphere is ins ribed inside a pyramid with a square as a base whose height is p152 timesthe length of one edge of the base. A ube is ins ribed inside the sphere. What is theratio of the volume of the pyramid to the volume of the ube?9. How many hexagons are in the �gure below with verti es on the given verti es? (Notethat a hexagon need not be onvex, and edges may ross!)10. Let C1 and C2 be two on entri re e tive hollow metal spheres of radius R and Rp3respe tively. From a point P on the surfa e of C2, a ray of light is emitted inward at 30Æfrom the radial dire tion. The ray eventually returns to P. How many total re e tionso� of C1 and C2 does it take?

Geometry SolutionsRi e Mathemati s Tournament 20001. Consider the board labeled as below, with labels for olumns and rows. To hoose anyre tangle on the board, it is suÆ ient to hoose some number (1-8) of adja ent olumns,and some number (1-8) of adja ent rows, sin e the re tangle an be reated by formingthe interse tion of the olumns and rows. For instan e, the interse tion of olumns 2,3and rows 3,4,5 is the re tangle shaded below. So, there are 8 ways to hoose 1 adja ent olumn, 7 ways to hoose 2 adja ent olumns, : : :, 1 way to hoose 8 adja ent olumns,so there are 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 total ways to hoose the olumns, and36 ways to hoose the rows. Thus, the total number of ways to hoose a re tangle (i.e.the total number of re tangles) is 362 = 1296.1 2 3 4 5 6 7 8

1

2

3

4

5

7

6

8

2. The maximum o urs in an equilateral triangle, in whi h ase the sides a = b � aregiven by a2 + b2 + 2 = 3a2 = 96, so a = 4p2. Thus, the medians are 3 � (ap32 ) =6p6.3. Let s be the side length of ABCD. Sin e ABCD is a square, we an write 2s2 =(PA)2+(PC)2 = (PB)2+(PC)2. So, we an substitute in for PA,PC, and PD to getthat PB =24.4. Noti e that in general, when there is a re tangle of side length x and y, the area ofthe non-triangle regions ( reated by drawing a line onne ting the midpoint of twoopposite lines and a line onne ting two opposite orners - see diagram for examples)is simply 34 of the original area of the box, sin e the area of the ex luded triangles are12 � y2 � x2 + 12 � y2 � x2 = 2 � xy2 = xy4 , so the desired area is simply ab� ab4 = 34ab, as desired.So, if we onsider the four re tangles making up the large re tangle that are dividedthis way (the one with dimensions a4x b3 , et .), we an say that the total shaded area is34 of the total area of the re tangle - that is, 34ab.5. The area of 4ABF = 12bh = 12 � 3h = 32(3 sin 60 Æ) = 92 sin 60 Æ. The area of 4FCDE =area of 4ABF - area of 4FGC = 92 sin 60 Æ � 12 sin 60 Æ = 4 sin 60 Æ. Therefore, area ofABCDE = area of4ABF+ area of FCDE = 92 sin 60 Æ+4 sin 60 Æ = 172 sin 60 Æ = 17p34 .6. The area of an equilateral triangle with side length s is p34 s2. Therefore, the side lengthis 4p2 and height h = 2p6. Now, if r is the radius of the ins ribed ir le, then r = h3 ,sin e we have an equilateral triangle. Thus the area is �r2 =8�3 .7. Let's "resize" the oordinates to be x0 = x; y0 = 2y3 . This keeps the origin at (0,0), butturns our ellipse into a ir le of radius 2. Thus now, the triangle is equilateral, and we

E

F

G

D

A

BC

h1

2

h2

b

an see it now has area 3p3. On e we expand ba k, we an see we are just multiplyingthe area by 32 and so the answer is 9p32 .8. By standard formula, we have that the radius of the ins ibed ir le, r, is r = ab sinA2a+b(iso eles triangle that is formed by utting the pyramid verti ally in half ( uts the baseinto 2 equal re tangles)). h2+( b2)2 = a2 gives a = qh2 + b24 . Also sinA = ha . Therefore,r = bh2qh2+ b24 +b . Note that the diameter of the ube is the diameter of the sphere. Let lbe the length of the side of the ube, so the diameter of the ube is lp3 = 2r, so l = 2rp3 .So, the volume of the pyramid is 13b2h and the ube volume if l3. So, the ratio is 25p36 .9. A hexagon an be formed by removing any vertex, removing all verti es onne ted tothat vertex, and then removing any edges onne ted to any of the removed verti es,and these are the only hexagons in the diagram. Thus, sin e there are 10 verti es, thereare 10 hexagons in the �gure below.

10. Consider the attened version of the situation. Then let O be the enter fo the spheres,A be the �rst re e tion point, B be the point of C1 su h that OB is perpendi ular toOP .Then sin e OB = 1,OP = p3, 6 PBO = 60 Æ and P;B;A are ollinear, 6 BAO = 60 Æimplies that 6 POA = 30 Æ. Therefore, ea h re e tion takes the ray 112 around the ir le,so there are 11 re e tions.

Guts

HMMT 2000

School:

Problem Gu1 [4]

The sum of 3 real numbers is known to be zero. If the sum of their cubes is πe, what is theirproduct equal to?

Problem Gu2 [5]

If X = 1 + x + x2 + x3 + ... and Y = 1 + y + y2 + y3 + ..., what is 1 + xy + x2y2 + x3y3 + ...in terms of X and Y only?

School:

Problem Gu3 [±7]

Using 3 colors, red, blue and yellow, how many different ways can you color a cube (modulorigid rotations)?

Problem Gu4 [5]

Let ABC be a triangle and H be its orthocentre. If it is given that B is (0, 0), C is (1, 2)and H is (5, 0), find A.

School:

Problem Gu5 [3]

Find all natural numbers n such that n equals the cube of the sum of its digits.

Problem Gu6 [±10]

If integers m, n, k satisfy m2 + n2 + 1 = kmn, what values can k have?

1

School:

Problem Gu7 [7]

Suppose you are give a fair coin and a sheet of paper with the polynomial xm written on it.Now for each toss of the coin, if heads show up, you much erase the polynomial xr (where ris going to change with time – initially it is m) written on the paper and replace it by xr−1.If tails show up, replace it by xr+1. What is the expected value of the polynomial I get afterm such tosses? (Note: this is a different concept from the most probable value)

Problem Gu8 [±4]

Johny’s father tells him : “I am twice as old as you will be seven years from the time I wasthrice as old as you were”. What is Johny’s age?

School:

Problem Gu9 [6]

A cubic polynomial f satisfies f(0) = 0, f(1) = 1, f(2) = 2, f(3) = 4. What is f(5)?

Problem Gu10 [7]

What is the total surface area of an ice cream cone, radius R, height H, with a sphericalscoop of ice cream of radius r on top? (Given R < r)

School:

Problem Gu11 [6]

Let M be the maximum possible value of x1x2 + x2x3 + ... + x5x1 where x1, x2, ..., x5 isa permutation of (1, 2, 3, 4, 5) and let N be the number of permutations for which thismaximum is attained. Evaluate M + N .

Problem Gu12 [9]

Calculate the number of ways of choosing 4 numbers from the set {1, 2, ..., 11} such that atleast 2 of the numbers are consecutive.

2

School:

Problem Gu13 [±4]

Determine the remainder when (x4 − 1)(x2 − 1) is divided by 1 + x + x2.

Problem Gu14 [7]

ABCD is a cyclic quadrilateral inscribed in a circle of radius 5, with AB = 6, BC = 7, CD =8. Find AD.

School:

Problem Gu15 [8]

Find the number of ways of filling a 8× 8 grid with 0’s and X’s so that the number of 0’s ineach row and each column is odd.

Problem Gu16 [5]

Solve for real x, y:

x + y = 2x5 + y5 = 82

School:

Problem Gu17 [5]

Find the highest power of 3 dividing(

666333

)

Problem Gu18 [±5]

What is the value of∞∑

n=1(tan−1

√n− tan−1

√n + 1)?

3

School:

Problem Gu19 [3]

Define a ∗ b = a−b1−ab

. What is (1 ∗ (2 ∗ (3 ∗ ...(n ∗ (n + 1))...)))?

Problem Gu20 [6]

What is the minimum possible perimeter of a triangle two of whose sides are along the x-andy-axes and such that the third contains the point (1,2)?

School:

Problem Gu21 [8]

How many ways can you color a necklace of 7 beads with 4 colors so that no two adjacentbeads have the same color?

Problem Gu22 [6]

Find the smallest n such that 22000 divides n!

School:

Problem Gu23 [5]

How many 7-digit numbers with distinct digits can be made that are divisible by 3?

Problem Gu24 [±3]

At least how many moves must a knight make to get from one corner of a chessboard to theopposite corner?

4

School:

Problem Gu25 [4]

Find the next number in the sequence 131, 111311, 311321, 1321131211,

Problem Gu26 [5]

What are the last 3 digits of 1! + 2! + ... + 100!

School:

Problem Gu27 [±6]

What is the smallest number that can be written as a sum of 2 squares in 3 ways?

Problem Gu28 [8]

What is the smallest possible volume to surface ratio of a solid cone with height = 1 unit?

School:

Problem Gu29 [±9]

What is the value of

√√√√1 + 2

√1 + 3

√1 + 4

√1 + 5

√1 + ...

Problem Gu30 [7]

ABCD is a unit square. If 6 PAC = 6 PCD, find the length BP .

5

School:

Problem Gu31 [10]

Given collinear points A, B, C such that AB = BC. How can you construct a point D onAB such that AD = 2DB, using only a straightedge? (You are not allowed to measuredistances)

Problem Gu32 [7]

How many (nondegenerate) tetrahedrons can be formed from the vertices of an n-dimensionalhypercube?

School:

Problem Gu33 [±5]

Characterise all numbers that cannot be written as a sum of 1 or more consecutive oddnumbers.

Problem Gu34 [±6]

What is the largest n such that n! + 1 is a square?

School:

Problem Gu35 [4]

If 1 + 2x + 3x2 + ... = 9, find x.

Problem Gu36 [6]

If, in a triangle of sides a, b, c, the incircle has radius b+c−a2

, what is the magnitude of angleA?

6

School:

Problem Gu37 [9]

A cone with semivertical angle 30◦ is half filled with water. What is the angle it must betilted by so that water starts spilling?

Problem Gu38 [4]

What is the largest number you can write with three 3’s and three 8’s, using only symbols+,-,/,× and exponentiation?

School:

Problem Gu39 [±8]

If r = 1/3, what is the value, rounded to 100 decimal digits, of

7∑n=0

2n

1 + x2n

Problem Gu40 [±10]

Let φ(n) denote the number of positive integers less than equal to n and relatively prime ton. find all natural numbers n and primes p such that φ(n) = φ(np).

School:

Problem Gu41 [7]

A observes a building of height h at an angle of inclination α from a point on the ground.After walking a distance a toward it, the angle is now 2α, and walking a further distance bcauses to increase to 3α. Find h in terms of a and b.

Problem Gu42 [4]

A n × n magic square contains numbers from 1 to n2 such that the sum of every row andevery column is the same. What is this sum?

7

School:

Problem Gu43 [6]

Box A contains 3 black and 4 blue marbles. Box B has 7 black and 1 blue, whereas Box Chas 2 black, 3 blue and 1 green marble. I close my eyes and pick two marbles from 2 differentboxes. If it turns out that I get 1 black and 1 blue marble, what is the probability that theblack marble is from box A and the blue one is from C?

Problem Gu44 [6]

A function f : Z → Z satisfies

f(x + 4)− f(x) = 8x + 20

f(x2 − 1) = (f(x)− x)2 + x2 − 2

(1)

Find f(0) and f(1).

School:

Problem Gu45 [7]

Find all positive integers x for which there exists a positive integer y such that(

xy

)= 1999000

Problem Gu46 [6]

For what integer values of n is 1 + n + n2/2 + ... + nn/n! an integer?

School:

Problem Gu47

Find an n < 100 such that n · 2n − 1 is prime. Score will be n − 5 for correct n, 5 − n forincorrect n (0 points for answer < 5).

8

Guts Solutions

HMMT 2000

1. πe/3

2. XYX+Y −1

3. 57

4. (1,2)

5. 1, 512, 4913, 5832

6. 3

7. (x2+12

)m.

8. 14

9. 15

10. πR√

R2 + H2 + 2πr2(1 +√

1− R2

r2 )

11. 58

12. 274

13. 3

14.√

51.

15. 249.

16. (-1,3) and (3,-1)

17. 1 = 30

18. −π/4

19. 1

20. 3 + 2√

2 +√

3 +√

6

21. 546

22. 2008

23. 224640

24. 6

25. 1113122113111221

26. 313

27. 325

28. 1/6

29. 3

30. 1

31. D is harmonic conjugate of C, standard construction.

32.

33. Of the form 4n + 2.

34.

35. 2/3

36. 90 degrees

37.

38. 333888

39. 254.50000... (more than 100 zeros).

40. n any odd number, p = 2

41. h = 14b

√(a + b)(3b− a)

42. n(n2 + 1)/2

43. 120/1147

44. f(0) = -1, f(1) = 1

45. 1999000, 2000

46. 1,2

47.

Oral Event

HMMT 2000

1. [25] Find all integer solutions to m2 = n6 + 1.

2. [30] How many positive solutions are there to x10 + 7x9 + 14x8 + 1729x7 − 1379x6 = 0 ?How many positive integer solutions ?

3. [35] Suppose the positive integers a, b, c satisfy an + bn = cn, where n is a positive integergreater than 1. Prove that a, b, c > n. (Note: Fermat’s Last Theorem may not be used)

4. [40] On an n× n chessboard, numbers are written on each square so that the number ina square is the average of the numbers on the adjacent squares. Show that all the numbersare the same.

5. [45] Show that it is impossible to find a triangle in the plane with all integer coordinatessuch that the lengths of the sides are all odd.

6. [45] Prove that every multiple of 3 can be written as a sum of four cubes (positive ornegative).

7. [45] A regular tetrahedron of volume 1 is filled with water of total volume 7/16. Is itpossible that the center of the tetrahedron lies on the surface of the water? How about in acube of volume 1?

8. [55] f is a polynomial of degree n with integer coefficients and f(x) = x2 + 1 for x =1, 2, ..., n. What are the possible values for f(0) ?

9. [60] Let −→v1 ,−→v2 ,−→v3 ,−→v4 and −→v5 be vectors in three dimensions. Show that for some i, j in1, 2, 3, 4, 5, −→vi · −→vj ≥ 0.

10. [75] 23 frat brothers are sitting in a circle. One, call him Alex, starts with a gallon ofwater. On the first turn, Alex gives each person in the circle some rational fraction of hiswater. On each subsequent turn, every person with water uses the same scheme as Alexdid to distribute his water, but in relation to themselves. For instance, suppose Alex gave1/2 and 1/6 of his water to his left and right neighbors respectively on the first turn andkept 1/3 for himself. On each subsequent turn everyone gives 1/2 and 1/6 of the waterthey started the turn with to their left and right neighbours (resp.) and keep the final thirdfor themselves. After 23 turns, Alex again has a gallon of water. What possibilities arethere for the scheme he used in the first turn? (Note: you may find it useful to know that1 + x + x2 + ... + x22 has no polynomial factors with rational coefficients.)

Oral Round Solutions

HMMT 2000

1. The equation involves only even powers, so we may assume m, n are positive (negativesolutions may be generated by switching signs). n6 ≤ n6 + 1 ≤ n6 + 2n3 + 1, so n6 + 1lies between two squares, with strict inequality unless n = 0. So the only solutions arem = ±1, n = 0.

2. Dividing by x6, we get the equation x4 + 7x3 + 14x2 + 1729x − 1379 = 0. This is amonotonically increasing function of x on [0,∞). From f(0) < 0 and f(x) → ∞ asx → ∞ we see that there’s only one positive solution. In fact, f(1) > 0, so there’s nopositive integer solution.

3. Assume w.lo.g that c > b ≥ a. Then cn − bn = (c − b)(cn−1 + ... + bn−1) > 1(nbn−1) ≥nan−1. So an > nan−1, hence a > n.

4. Let m be the smallest number, written on some square S. Then clearly all of theadjacent squares to S must have m on them (they can’t have anything smaller, since mis the smallest, and can’t have anything larger since m is the average of those adjacentnumbers). Since this applies to every square on which m is written, all of the squaresmust have m on them. Otherwise there will be some “boundary” square with m on it,which will not have m written on one of its neighbours, a contradiction.

5. Suppose there is such a triangle with vertices at (m1, n1), (m2, n2), (m3, n3). Then we

are given√

(m1 −m2)2 + (n1 − n2)2 is an odd integer, so that (m1 −m2)2 + (n1 − n2)

2

is 1 (mod 4). Hence exactly one of m1 −m2 and n1 − n2 must be odd and the othermust be even. Suppose w.lo.g m1 −m2 is odd and n1 − n2 is even. Similarly, exactlyone of m1−m3 and n1−n3 must be odd and the other must be even. If m1−m3 is oddand n1 − n3 even, then m2 −m3 and n2 − n3 are both even, which is a contradictionsice we want (m2−m3)

2 + (n2−n3)2 to be 1 (mod 4). Similarly if m1−m3 is even and

n1 − n3 is odd then m2 −m3 and n2 − n3 are both odd, and (m2 −m3)2 + (n2 − n3)

2 is2 (mod 4), a contradiction.

Alternate solution: By Pick’s theorem (A = I + 12B − 1), the area is an integer or a

half-integer. But by Hero’s formula, the area is 14

√product of 4 odd integers. this is a

contradiction.

6. Any multiple of 3 can be written as 6k or 6k + 15. Now, 6k = (k + 1)3 + (k − 1)3 +(−k)3 + (−k)3. 6k + 15 = (k + 2)3 + (−k + 2)3 + (2k − 1)3 + (−2k)3.

1

7. If the tetrahedron is oriented so that two opposite edges of the tetrahedron are parallelto the ground, it is clear from symmetry that the plane parallel to the ground andpassing through the center of the tetrahedron splits it into equal spaces of volume1/2. Let’s call this position 1. If, instead, it is oriented with the point down and thetop face parallel to the ground, the plane through the center splits it into regions ofvolume 27/64 and 37/64. To see this, note that the center of a regular tetrahedron is3/4 of the way down the altitude from each vertex. Hence the tetrahedral section hasvolume (3/4)3 = 27/64. Let’s call this position position 2. If the tetrahedron is turnedcontinuously from position 1 to position 2, because 27/64 < 7/16 < 1/2, it must passthrough an orientation s.t. the fraction of volume below the center is 7/16.

The cube has 180◦ rotational symmetry, so any plane through the center splits it intosections of equal volume. Hence it is impossible to get a volume of 7/16.

8. Let f be one such polynomial, for example, the obvious one: f(x) = x2 + 1. Then allsuch polynomials must be of the form f(x)+g(x) where g(x) = 0 at x = 1, 2, ..., n. g(x)must of the form C(x−1)(x−2)...(x−n) and since f +g has to have integer coefficients,C must be an integer. At 0 this gives us (f +g)(0) = f(0)+g(0) = 1+C(−1)nn!. SinceC can be any positive or negative integer, the answer is that f(0) can be any integercongruent to 1(mod n!), i.e. of the form 1 + Cn!, C an integer.

9. This is a special case of the problem with n + 2 vectors in n dimensions. First it isclear that we can take all the vectors to be of length 1. Then we induct on n. The firstcase is n=1. Here the statement is that given a, b, c real numbers, the at least one ofab, bc, and ac is nonnegative. Without loss of generality, we can assume that a and bare of the same sign, but then ab ≥ 0. Now assume the statement is false for the n-dimensional case. Choose some vector, say −−→vn+2, and project the other vectors onto the

space perpendicular to −−→vn+2 to get−→v′

i = −→vi − (−→vi · −−→vn+2)−−→vn+2. This is essentially takingout the space parallel to −−→vn+2 and reducing the problem by one dimension. The only

thing left to check is that if −→vi ·−→vj < 0 for all i,j then−→v′

i ·−→v′

j < 0 for all i,j. This is just a

calculation:−→v′

i ·−→v′

j = −→vi ·−→vj −2(−→vj ·−−→vn+2)(−→vi ·−−→vn+2)+(−−→vn+2 ·−−→vn+2)(−→vj ·−−→vn+2)(−→vi ·−−→vn+2) andsince (−−→vn+2 ·−−→vn+2) = 1, this is just −→vi ·−→vj −(−→vj ·−−→vn+2)(−→vi ·−−→vn+2). (−→vj ·−−→vn+2) and (−→vi ·−−→vn+2)

are both negative by assumption so their product is positive and−→v′

i ·−→v′

j < −→vi · −→vj < 0.

10. Let ω be a 23rd root of unity, a solution to the equation x23 = 1. We can encode thestate of the game as 23 rational numbers a0, ..., a22 representing the amount of watereach person has, with Alex having the frction a0 of the water and continuing to hisright. Then we can look at the polynomial f(x) = a22x

22 + ... + a0 which still encodesthe state of the game. In this schema the original state is the constant 1, and we haveat all times the condition f(1) = 1 (conservation of water). However, we want a cyclicrepresentation in order to encode the play of the game, so we let the state of the gamebe f(ω) = a22ω

22 + ... + a0. In this representation, a step in the game can be encodedas multiplication by another polynomial in ω, g(ω) = b22ω

22 + ... + b0 where the bi

are the fraction of water that Alex initially gives to the i-th person on his right. Theexample in the problem is represented by 1

6ω + 1

3+ 1

2ω22. So we would like to know

which g are solutions to g(ω)23 = 1. We can take 23rd roots of both sides and findthat g(ω) = b22ω

22 + ... + b0 = ωk for some integer k. To simplify this we need a smalllemma.

2

Lemma: If a22ω22 + ... + a0 is some rational linear combination of the powers of ω and

is equal to 0, then a22 = a21 = ... = a0.

Proof: Assume this is false. So there exists some polynomial A(x) = a22x22 + ... + a0

with ω as a root that is not a constant multiple of B(x) = x22 +x21 + ...+1. Now applyEuclid’s algorithm to find the G.C.D. of A(x) and B(x) and call it D(x). But D(x)is a rational polynomial that divides B(x), which has no non-trivial rational divisors.Hence D(x) must be either a constant or of degree 22. Since A(ω) = B(ω) = 0 allthe terms in the algorithm have a root at ω and so does D(x). So since 0 is not anacceptable value for a G.C.D., D(x) must have degree 22. Since the degree of eachsuccessive term in Euclid’s algorithm decreases by at least one, this means that D(x)must be A(x) and hence that B(x) is a constant multiple of A(x). This contradicts theoriginal assumption and the Lemma is proved.

So the Lemma tells us that b22 = b21 = ... = bk − 1 = ... = b0. Since we already knowthat g(1) = 1 and hence b22 + b21 + ... + b0 = 1, and that all the bi are nonnegative,we get that bi = 0 for i 6= k and bk = 1. Since this is a solution for all k, the possibleschemes for Alex to use are just those involving giving all his water to any one otherperson.

3

Power TestRi e Mathemati s Tournament 20001. A derangement of a string of distin t elements is a rearrangement of the string su hthat no element appears in its original position. For example, BCA is a derangement ofABC. Dn represents the number of derangements of any string omposed of n distin telements. D2 = 1 and D3 = 2.(a) What are D4 and D5?(b) How many derangments are there of the string ABCDEFG?( ) Find a re ursive relationship for Dn in terms of the previous two terms (Dn�1 andDn�2).(d) Find a re ursive relationship for Dn in terms of only the previous term, Dn�1.2. Find the number of 3-letter "words" that use letters from the 10-letter set fA;B;C; :::; Jgin whi h all letters are di�erent and the letters appear in alphabeti al order.3. Assume that a hand of thirteen ards is dealt from a randomized de k of 52 ards. LetA be the probability that the hand ontains two a es. Let B be the probability thatit ontains two a es if you already know it ontains at least one a e. Let C be theprobability that it ontains at least two a es if you already know it ontains an a e ofhearts. Write down the inequality relationship between A,B and C, i.e. one possibilityis A = B > C.4. Find the number of rearrangments of 12345 (in luding 12345) su h that none of thefollowing is true: 1 is in position 5, 2 is in position 1, 3 is in position 2, 4 is in position4, and 5 is in position 3.5. Assume that nobody has a birthday on February 29th. How large must a group be sothat there is a greater than 50% han e that at least 2 members have the same birthday?6. Find the number of ombinations of length k that use elements from a set of n distin telements, allowing repetition.7. Find the number of ombinations of length k that use elements from a given set of ndistin t elements, allowing repetition and with no missing elements. (Obviously, k mustbe greater than n)8. An ele tion takes pla e between two andidates. Candidate A wins by a vote of 1032to 971. If the votes are ounted one at a time and in a random order, determine theprobability that the winner was never behind at any point in the ounting.9. Evaluate the sum �1000 � +12 �1001 � +13 �1002 � + : : :+ 1101 �100100�.10. Find the number of distributions of a given set of m identi al balls into a given set ofn distin t boxes.11. Find the number of distributions of a given set of m distin t balls into a given set ofn distin t boxes if ea h box must ontain a spe i� number of balls (mi is the numberof balls to be put into box i). Please state the answer with only fa torials (not in ombinatorial notation).

12. Find the oeÆ ient of X2Y 3 in ea h of the following.(a) (X + Y + 1)7(b) (X2 + Y � 1)713. Find the number of "words" of length m from a set of n letters, if ea h letter musto ur at least on e in ea h word.14. Find the number of ways to distribute seven distin t balls into three distin t boxes ifea h box must ontain a di�erent number of balls, allowing an empty box.15. How many ways an a lass of 10 students be divided into two groups of 3 and 1 groupof 4?16. Find the number of subsets A of the set of digits f0; 1; 2; 3; : : : ; 9g su h that A ontainsno two onse utive digits. Hint: Find a better statement of the problem; �nd a re ursiveformula, and then attempt to solve the problem for the number of digits given.17. If we are trying to �nd the number of words of length m from a given set of n letters,with ea h letter o uring at least on e in ea h word, let us all the answer T (m;n).This is equivalent to �nding the number of distribution of a set of m distin t balls intoa set of n distin t boxes, if no boxes an be empty. T (m;n) is the sum of all possiblepartitions of the balls (i.e. we sum all possible ways of putting the balls into boxes (4in box 1, 2 in box 2, 1 in box 3 for example)). More pre isely, if we all mi to be thenumber of balls in box i, then T (m;n) = Pm1+m2+m3=mm1;m2;m3>=1 m!m1!m2!m3!:::mn! . For example,T (3; 2) = 3!1!2! + 3!2!1! = 3 + 3 = 6. Find a re ursive pattern for T (m;n) in terms ofprevious terms (previous meaning a smaller m, a smaller n, or both). Hint: set up asort of "Pas al's Triangle" for T (m;n). Prove your answer using words.18. You have an in�nite number of 1 ent, 2 ent, and 5 ent stamps. You are trying topost a letter that requires n ents of postage stamps, where n > 8. Let a(n) be thenumber of sequen es of stamps that give exa tly the required postage of n ents (i.e.order matters). Find a(n) in terms of previous terms of the sequen e of a's, using asfew previous terms as possible.19. Suppose we have n lines in a plane in general position, whi h means that none areparallel to ea h other and that no three of these lines interse t at a single point. Findthe number of regions that these lines divide the plane into...(a) in a re ursive form.(b) in a nonre ursive formula.20. Find the 2000th positive integer that is not the di�eren e between any two integersquares.

Power Test SolutionsRi e Mathemati s Tournament 20001. (a) D4 = 9; D5 = 44(b) This is D7 =1854( ) Dn = (n� 1)(Dn�1 +Dn�2)(d) Dn = n �Dn�1 + (�1)n2. 10 � 9 � 8 give the number of all di�erent 3 letter words. One sixth of these are inalphabeti al order. 10�9�86 = 120.3. A < B < C. More spe i� ally, A = :26; B = :37; C = :56.4. This is just derangements of 23541. D5 =44.5. We want 1� 365 � 364 � 363 : : : (365� n + 1)=365n > 50%. Just guestimating gives theanswer of 23 people.6. Sin e we want to hoose k times from n distin t elements, this is equivalent to hoos-ing where to put n� 1 "dividers" that separate the hoi es. For example, if we wantedto hoose 3 s oops of i e ream from the avors ho olate, vanilla, strawberry, and o�ee,we an represent the hoi e of 2 vanilla and 1 o�ee by (divider), hoi e, hoi e,(divider),(divider), hoi e.Noti e that the hoi es of positions for the dividers ompletely determines whi h ele-ments we hoose. Therefore, we have n+ k� 1 spa es to �ll with n� 1 dividers, so thenumber of ways of doing this is �n+k�1n�1 � or �n+k�1k �7. We an use the same method as the above argument, ex ept that we know every elemento urs at least on e. So, this is the same as hoosing k � n times from the n distin tobje ts, so the answer is �k�n�n+1n�1 �, or �k�1n�1� or �k�1k�n�.8. FIX????????? 1� 9711033 = 6210339. Let S be the sum desired. Then 101 � S = 1011 �1000 � + 1012 �1001 � + 1013 �1002 � + : : :+ 101101�100100�. Now, onsider a general term in this expression - i.e. 101i+1 �100i �. This is equal to101i+1 � 100!(100�i)!i! = 101�100!(101�(i+1))!i!�(i+1) = 101!(101�(i+1))!i+1! = �101i+1�. So, we an simplify the termsto get 101 � S = �1011 � + �1012 � + �1013 � + : : :+ �101100� + �101101�. Thus, 101 � S = 2101 � 1,so S = 2101�1101 .10. �n+m�1m � or �n+m�1n�1 �. This an be seen from the fa t that ea h distribution an bedes ribed by a ombination of letters, where ea h letter represents a di�erent box. Thenumber of times ea h letter o urs in the ombination is determined by the number ofballs in the orresponding box.11. m![m1!m2!m3!:::mn!℄ . This is also equal to �mm1� �m�m1m2 � �m�m1�m2m3 � : : :12. This is very similiar to 11. A) 7!2!3!2! =210 B)�7!=(3!3!) =-14013. This is the sum of all the distribution numbers in whi h the numbers m1 : : :mn runthrough all possible sequen es of n positive integers adding up tom: Pm1+:::mn=mmi>=1 m!m1!m2!:::mn!

14. The boxes an be in any order, so we have a fa tor of 6. Seven an be partitioned intothree distin t parts as 0,1,6; 0,2,5; 0,3,4; or 1,2,4. So, the answer is 6 h 7!0!1!6! + 7!0!2!5! + 7!0!3!4! + 7!1!2!4!i =1008.15. This is partitioning 10 into 3 partitions, 2 of 1 type and one of the other. Thus, theanswer is 10!3!3!4!2! =2100.16. This is equivalent to �nding the number of sequen es of length 10 omposed of 0's and1's. (0 in a spot orresponds to that spot's number (0-9) not being in the subset.)However, we an't have two onse utive 1's. If we try to generalize, let 0 = A, 1 = Band we are doing an n-letter "word" instead of ten. Set wn = number of n-letter words(satisfying the onditions); set an = number of words ounted by wn that begin with A;set bn = number of words ounted by wn that begin with B. wn = an + bn. an = bn�1.bn = wn�1. Combining these we get the re ursive relationship wn = wn�1+wn�2. Thenwe an build up to �nd that w10 = 144.17. T (m;n) = n(T (m � 1; n � 1) + T (m � 1; n)) for 1 < n < m. To prove this,look at T (5; 3) and think of it as the number of 5-letter words from fA;B;Cg with nomissing letters. There are 3 hoi es for the �rst letter. After this, the remaining fourletters must be �lled in, and the �rst letter ( all it X) does not have to be used again.There are two ases:� If X does not o ur again, then the word an be ompleted in T (4; 2) ways.� If X does o ur again, then the number of ways to omplete the word is T (4; 3).As we have n hoi es for the letter X (�rst letter), we get that T (5; 3) = 3 � (T (4; 2) +T (4; 3)), or the above, in general.18. an = an�1+an�2+an�5. Every way to make n ents ends in either a 1 ent, 2 ents,or 5 ents (sin e order matters, there is a distin t last stamp). There are an�1 ways tomake n ents ending in a 1 ent stamp, sin e this is the number of ways to make n� 1 ents. Similarly, there are an�2 ways to make n ents ending in a 2 ents stamp, andan�5 ways to make n ents ending in a 5 ents stamp. Sin e every way to make n entsends in one of these stamps, and there is no overlap, the total number of ways to maken ents is the sum of these, or an�1 + an�2 + an�5.19. (a) rn = rn�1 + n(b) �n+12 � + 120. First, we will show that a positive integer x is not a di�eren e of squares if and only ifx � 2 (mod 4).First, suppose x = a2�b2 for some integers a; b, so x = (a� b)(a+ b). Now, if a�b = 1,then a+ b = 2n+ 1 for some integer n, so x = (a� b)(a+ b) = 2n+ 1, and x is odd.Conversely, if x is odd, then x = 2n + 1 for some integer n, so x = 1(2n + 1) =(n+ 1� n)(n + 1 + n) = (n + 1)2 � n2.Now, if a � b = 2, then a + b = 2n + 2 for some integer n, so x = (a� b)(a+ b) =2(2n+ 2) = 4n+ 4Conversely, if x � 0 (mod 4), then x = 4n+4 for some integer n, so x = (n+ 2)2�n2.So, for all other ases, a�b � 2. If a�b is even, then we know a�b = 2n for some integern, and a + b = 2n + 2m for some integer m, so x = (a� b)(a + b) = 2n � (2n + 2m) =4nm+ 4n2, so x � 0 (mod 4), and we have already overed this ase.

If a� b is odd, then we know a� b = 2n+1 for some integer n, and a+ b = 2n+2m+1,where m is some integer, so x = (a� b)(a+ b) = (2n+1) � (2n+2m+1) = 4(n2+nm+n) + 2n+ 2m+ 1 � 1 (mod 4), whi h is a ase that we have already overed.Therefore, x is not a di�eren e of two squares if and only if x � 2 (mod 4). So, the2000th number is 4 � 1999 + 2 =7998.

Team TestRi e Mathemati s Tournament 20001. You are given a number, and round it to the nearest thousandth, round this result tonearest hundredth, and round this result to the nearest tenth. If the �nal result is .7,what is the smallest number you ould have been given? As is ustomary, 5's are alwaysrounded up. Give the answer as a de imal.2. The pri e of a gold ring in a ertain universe is proportional to the square of its purityand the ube of its diameter. The purity is inversely proportional to the square of thedepth of the gold mine and dire tly proportional to the square of the pri e, while thediameter is determined so that it is proportional to the ube root of the pri e and alsodire tly proportional to the depth of the mine. How does the pri e vary solely in termsof the depth of the gold mine?3. Find the sum of all integers from 1 to 1000 in lusive whi h ontain at least one 7 intheir digits, i.e. �nd 7 + 17 + :::+ 979 + 987 + 997.4. All arrangements of letters VNNWHTAAIE are listed in lexi ographi (di tionary) or-der. If AAEHINNTVW is the �rst entry, what entry number is VANNAWHITE?5. Given os (� + �) + sin (�� �) = 0, tan� = 12000 , �nd tan�.6. If � is a root of x3 � x � 1 = 0, ompute the value of �10 + 2�8 � �7 � 3�6 � 3�5 +4�4 + 2�3 � 4�4 � 6�� 17.7. 8712 is an integral multiple of its reversal, 2178, as 8712=4*2178. Find another 4-digitnumber whi h is a non-trivial integral multiple of its reversal.8. A woman has $1.58 in pennies, ni kels, dimes, quarters, half-dollars and silver dollars.If she has a di�erent number of oins of ea h denomination, how many oins does shehave?9. Find all positive primes of the form 4x4 + 1, for x an integer.10. How many times per day do at least two of the three hands on a lo k oin ide?11. Find all polynomials f(x) with integer oeÆ ients su h that the oeÆ ients of bothf(x) and [f(x)℄3 lie in the set f0; 1;�1g12. At a dan e, Abhinav starts from point (a; 0) and moves along the negative x dire tionwith speed va, while Pei-Hsin starts from (0; b) and glides in the negative y-dire tionwith speed vb. What is the distan e of losest approa h between the two?13. Let P1; P2; : : : ; Pn be a onvex n-gon. If all lines PiPj are joined, what is the maximumpossible number of interse tions in terms of n obtained from stri tly inside the polygon?14. De�ne a sequen e <xn> of real numbers by spe ifying an initial x0 and by the re urren exn+1 = 1+xn1�xn . Find xn as a fun tion of x0 and n, in losed form. There may be multiple ases.15. limn!1 nr 2q1� os 2�n =?

Team Test SolutionsRi e Mathemati s Tournament 20001. .6445 rounds to .645 to .65 to .7. Otherwise .6444... rounds to .644. So the smallestnumber is .6445.2. Let =pri e, p=purity, d =diameter, h=depth of gold mine, ki= onstant. We are given = k1p2d3, p = k2 2h2 , and d = k3 3p h. So, = k1k22 4h4k33 h3 = k4 5 1h . Thus, k4 4 = h,and = k5h 14 . Thus, p varies as h14 .3. The sum of the numbers from 700 to 799 is 799�8002 � 699�7002 = 74950. The sum ofthe numbers from 70 to 79 is 79�802 � 69�702 = 745. So, all numbers that end from70 to 79 (ex luding those starting with 7, sin e we ounted those from 700 to 799) is745 �9+10(100+ 200 + : : :+ 600 + 800 + 900) = 44705. The sum of all numbers endingin 7 is 9(7+17+27+37+47+57+67+87+97)+9(100+200+: : :+600+800+900) = 38187.So, the total sum of numbers ontaining a 7 is 74950 + 44705 + 38186 = 157842.4. 738,826. This an be arrived at by stepping down, starting with �nding how many ombinations are there that begin with a letter other than V or W, and so forth. Theanswer is 8�9!2�2 + 4�7!2 + 4 � 6! + 4 � 4! + 3! + 2! + 2! = 738826.5. 0 = os (�+ �) + sin (�� �) = os� � os � � sin� � sin� + sin� � os � � sin� os� =( os� + sin�)�( os � � sin�). So os�+sin� = 0 or os ��sin � = 0. Then tan� = �1or tan� = 1. Sin e tan � is given as 12000 , tan� =�1.6. Sin e �3 � � � 1 = 0, then �10 = �8 + �7. So, we an redu e our expression to3�8 � 3�6 � 3�5 + 4�4 + 2�3 � 4�2 � 6� � 17. Also, 3�8 � 3�6 � 3�5 = 0, so ourexpression redu es to 4�4 + 2�3 � 4�2 � 6� � 17. Also, 4�4 � 4�2 � 4�0, so ourexpression redu es to 2�3� 2�� 17. Now, 2�3� 2�� 2 = 0, so our expression redu esto -15, whi h is our answer.7. Another 4-digit number that satis�es this property is 9801, sin e 9801=9*1089.8. If she has a silver dollar, then she would have too many other oins, as 0 half dollars,2 quarters, 3 dimes, et . would be greater than the total. So she has no silver dollars,and at least one of every other denomination. Continuing, it turns out the only feasiblesolution is 0 silver dollars, 1 half dollar, 2 quarters, 3 dimes, 4 ni kels, 8 pennies, for atotal of 18 oins.9. It suÆ es to onsider x � 1, sin e 4(�x)4+1 = 4(x)4+1, and 4(0)+1 = 1 is not prime.So, 4x4+1 = (4x4 + 4x2 + 1)�4x2 = (2x2 + 1)2�(2x)2 = (2x2 + 1� 2x)(2x2 + 1 + 2x).For integers x, both 2x2 � 2x+ 1 and 2x2 + 2x+ 1 are integers, so this fa tors 4x4 + 1unless 2x2 � 2x+ 1 = �1 or 2x2 + 2x+ 1 = �1. Sin e x > 0, then 2x2 + 2x+ 1 > 1, sowe must have 2x2�2x+1 = �1. 2x2�2x+1 = �1 is absurd (4x4+1; 2x2+2x+1 > 0,so 2x2 � 2x + 1 = 4x4+12x2+2x+1 > 0), so we solve 2x2 � 2x + 1 = 1, or 2x2 � 2x = 0, sox(x� 1) = 0, and x = 0 or x = 1. We have already reje ted x = 0, so the only ase leftis x = 1, or 4(1)4 + 1 = 5.10. The se ond hand rosses the minute hand 59 times an hour. The se ond hand rosses thehour hand 60 times an hour, ex ept for 2 of the hours, due to the movement of the hourhand. The minute hand and the hour hand ross 22 times total, be ause the hour hand

ompletes 2 rotations in a day, and the minute hand ompletes 24. The se ond, hour,and minute hand all oin ide only at noon and midnight, but we've ounted ea h ofthese 12:00's 3 times instead of on e. Therefore, the answer is 59�60+60�24�2+22�2�2,giving us 2872 rossings.11. f(x) is either 0 or something of the form �xm, where m � 0.12. A's position is (a� Vat; 0) and P 's position is (0; b� Vbt). So, at time t, the distan ebetween them is q(a� Vat)2 + (b� Vbt)2. Noti e that this distan e is the same as thedistan e between the point (a; b) and the line (Vat; Vbt), whi h is the same as the lineVbx� Vay = 0. The distan e from a line Ax + By + C = 0 and (x0; y0) is jAx0+By0+CjpA2+B2 ,so the answer is jbVa�aVbjpV 2a+V 2b13. Given any 4 verti es, there is exa tly one interse tion of all the diagonals onne tingthem. So, the answer is �n4�.14. x0 if n � 0 (mod 4),1+x01�x0 if n � 1 (mod 4), � 1x0 if n � 2 (mod 4), x0�1x0+1 if n � 3 (mod4).15. Consider a regular n-gon with radius r. Let x be the side length of the n-gon. So,sin e the entral angle is 2�n (see diagram below), use the Law of Cosines to �nd thatx2 = r2 + r2 � 2r � r os 2�n , so x2 = 2r2(1� os 2�n ). Thus, x = rp2q1� os 2�n .So, the total perimeter of the n-gon is nx = nrp2q1� os 2�n . Now, if we takelimn!1 of the perimeter, the result will be 2�n, sin e the n-gon approa hes a irle,so limn!1 nrp2q1� os 2�n = 2�r, and so limn!1 nrq1� os 2�n = �rp2.rr

x

2πn

Advanced Topics TestHarvard-MIT Math Tournament

March 3, 2001

1. Find x− y, given that x4 = y4 + 24, x2 + y2 = 6, and x+ y = 3.

2. Find logn(

12

)logn−1

(13

)· · · log2

(1n

)in terms of n.

3. Calculate the sum of the coefficients of P (x) if (20x27 + 2x2 + 1)P (x) = 2001x2001.

4. Boris was given a Connect Four game set for his birthday, but his color-blindness makesit hard to play the game. Still, he enjoys the shapes he can make by dropping checkers intothe set. If the number of shapes possible modulo (horizontal) flips about the vertical axis ofsymmetry is expressed as 9(1 + 2 + · · ·+ n), find n. (Note: the board is a vertical grid withseven columns and eight rows. A checker is placed into the grid by dropping it from the topof a column, and it falls until it hits either the bottom of the grid or another checker alreadyin that column. Also, 9(1 + 2 + · · · + n) is the number of shapes possible, with two shapesthat are horizontal flips of each other counted as one. In other words, the shape that consistssolely of 3 checkers in the rightmost row and the shape that consists solely of 3 checkers inthe leftmost row are to be considered the same shape.)

5. Find the 6-digit number beginning and ending in the digit 2 that is the product ofthree consecutive even integers.

6. There are two red, two black, two white, and a positive but unknown number of bluesocks in a drawer. It is empirically determined that if two socks are taken from the drawerwithout replacement, the probability they are of the same color is 1

5. How many blue socks

are there in the drawer?

7. Order these four numbers from least to greatest: 556, 1051, 1735, 3128.

8. Find the number of positive integer solutions to nx + ny = nz with nz < 2001.

9. Find the real solutions of (2x+ 1)(3x+ 1)(5x+ 1)(30x+ 1) = 10.

10. Alex picks his favorite point (x, y) in the first quadrant on the unit circle x2 +y2 = 1,such that a ray from the origin through (x, y) is θ radians counterclockwise from the positivex-axis. He then computes cos−1

(4x+3y

5

)and is surprised to get θ. What is tan(θ)?

Advanced Topics Test SolutionsHarvard-MIT Math Tournament

March 3, 2001

1. Find x− y, given that x4 = y4 + 24, x2 + y2 = 6, and x+ y = 3.

Solution: 246·3 = x4−y4

(x2+y2)(x+y)= (x2+y2)(x+y)(x−y)

(x2+y2)(x+y)= x− y = 4

3.

2. Find logn(

12

)logn−1

(13

)· · · log2

(1n

)in terms of n.

Solution: Using log 1x

= − log x and logb a = log alog b

, we get that the product equals(− log 2)(− log 3)···(− logn)

logn··· log 3 log 2= (−1)n−1 .

3. Calculate the sum of the coefficients of P (x) if (20x27 + 2x2 + 1)P (x) = 2001x2001.Solution: The sum of coefficients of f(x) is the value of f(1) for any polynomial f .

Plugging in 1 to the above equation, P (1) = 200123

= 87 .

4. Boris was given a Connect Four game set for his birthday, but his color-blindness makesit hard to play the game. Still, he enjoys the shapes he can make by dropping checkers intothe set. If the number of shapes possible modulo (horizontal) flips about the vertical axis ofsymmetry is expressed as 9(1 + 2 + · · ·+ n), find n. (Note: the board is a vertical grid withseven columns and eight rows. A checker is placed into the grid by dropping it from the topof a column, and it falls until it hits either the bottom of the grid or another checker alreadyin that column. Also, 9(1 + 2 + · · · + n) is the number of shapes possible, with two shapesthat are horizontal flips of each other counted as one. In other words, the shape that consistssolely of 3 checkers in the rightmost row and the shape that consists solely of 3 checkers inthe leftmost row are to be considered the same shape.)

Solution: There are 97 total shapes possible, since each of the 7 columns can containanywhere from 0 to 8 checkers. The number of shapes symmetric with respect to a horizontalflip is the number of shapes of the leftmost four columns, since the configuration of thesefour columns uniquely determines the configuration of the remaining columns if it is knownthe shape is symmetric: 94. Now we know there are 97 − 94 non-symmetric shapes, so thereare 97−94

2non-symmetric shapes modulo flips. Thus the total number of shapes modulo flips

is 94 + 97−94

2= 94

(1 + 93−1

2

)= 94

(93+1

2

)= (38(36−1)

2= 936(36+1)

2= 9(1 + 2 + · · · + 36), so

n = 36 = 729 .

5. Find the 6-digit number beginning and ending in the digit 2 that is the product ofthree consecutive even integers.

Solution: Because the last digit of the product is 2, none of the three consecutive evenintegers end in 0. Thus they must end in 2, 4, 6 or 4, 6, 8, so they must end in 4, 6, 8 since 2·4·6

does not end in 2. Call the middle integer n. Then the product is (n−2)n(n+ 2) = n3−4n,so n > 3

√200000 =

3√

200 · 103 ≈ 60, but clearly n < 3√

300000 =3√

300 · 103 < 70. Thusn = 66, and the product is 663 − 4 · 66 = 287232 .

6. There are two red, two black, two white, and a positive but unknown number of bluesocks in a drawer. It is empirically determined that if two socks are taken from the drawerwithout replacement, the probability they are of the same color is 1

5. How many blue socks

are there in the drawer?Solution: Let the number of blue socks be x > 0. Then the probability of drawing a

red sock from the drawer is 26+x

and the probability of drawing a second red sock from the

drawer is 16+x−1

= 15+x

, so the probability of drawing two red socks from the drawer without

replacement is 2(6+x)(5+x)

. This is the same as the probability of drawing two black socksfrom the drawer and the same as the probability of drawing two white socks from the drawer.The probability of drawing two blue socks from the drawer, similarly, is x(x−1)

(6+x)(5+x). Thus the

probability of drawing two socks of the same color is the sum of the probabilities of drawingtwo red, two black, two white, and two blue socks from the drawer: 3 2

(6+x)(5+x)+ x(x−1)

(6+x)(5+x)=

x2−x+6(6+x)(5+x)

= 15. Cross-multiplying and distributing gives 5x2 − 5x + 30 = x2 + 11x + 30, so

4x2 − 16x = 0, and x = 0 or 4. But since x > 0, there are 4 blue socks.

7. Order these four numbers from least to greatest: 556, 1051, 1735, 3128.Solution: 1051 > 951 = 3102 = 2734 > 1735 > 1635 = 3228 > 3128 > 2528 = 556, so the

ordering is 556, 3128, 1735, 1051 .

8. Find the number of positive integer solutions to nx + ny = nz with nz < 2001.Solution: If n = 1, the relation can not hold, so assume otherwise. If x > y, the left

hand side factors as ny(nx−y + 1) so nx−y + 1 is a power of n. But it leaves a remainder of 1when divided by n and is greater than 1, a contradiction. We reach a similar contradictionif y > x. So y = x and 2nx = nz, so 2 is a power of n and n = 2. So all solutions are of theform 2x + 2x = 2x+1, which holds for all x. 2x+1 < 2001 implies x < 11, so there are 10solutions.

9. Find the real solutions of (2x+ 1)(3x+ 1)(5x+ 1)(30x+ 1) = 10.Solution: (2x + 1)(3x + 1)(5x + 1)(30x + 1) = [(2x + 1)(30x + 1)][(3x + 1)(5x + 1)] =

(60x2 + 32x+ 1)(15x2 + 8x+ 1) = (4y+ 1)(y+ 1) = 10, where y = 15x2 + 8x. The quadratic

equation in y yields y = 1 and y = −94. For y = 1, we have 15x2 +8x−1 = 0, so x = −4±

√31

15.

For y = −94, we have 15x2 + 8x+ 9

4, which yields only complex solutions for x. Thus the real

solutions are −4±√

3115

.

10. Alex picks his favorite point (x, y) in the first quadrant on the unit circle x2 +y2 = 1,such that a ray from the origin through (x, y) is θ radians counterclockwise from the positivex-axis. He then computes cos−1

(4x+3y

5

)and is surprised to get θ. What is tan(θ)?

Solution: x = cos(θ), y = sin(θ). By the trig identity you never thought you’d need,4x+3y

5= cos(θ− φ), where φ has sine 3/5 and cosine 4/5. Now θ− φ = θ is impossible, since

φ 6= 0, so we must have θ− φ = −θ, hence θ = φ/2. Now use the trusty half-angle identities

to get tan(θ) = 13

.

Algebra TestHarvard-MIT Math Tournament

March 3, 2001

1. Find x− y, given that x4 = y4 + 24, x2 + y2 = 6, and x+ y = 3.

2. Find (x+ 1)(x2 + 1)(x4 + 1)(x8 + 1) · · · , where |x| < 1.

3. How many times does 24 divide into 100! (factorial)?

4. Given that 7, 999, 999, 999 has at most two prime factors, find its largest prime factor.

5. Find the 6-digit number beginning and ending in the digit 2 that is the product ofthree consecutive even integers.

6. What is the last digit of 11 + 22 + 33 + · · ·+ 100100?

7. A polynomial P has four roots, 14, 1

2, 2, 4. The product of the roots is 1, and P (1) = 1.

Find P (0).

8. How many integers between 1 and 2000 inclusive share no common factors with 2001?

9. Find the number of positive integer solutions to nx + ny = nz with nz < 2001.

10. Find the real solutions of (2x+ 1)(3x+ 1)(5x+ 1)(30x+ 1) = 10.

Algebra Test SolutionsHarvard-MIT Math Tournament

March 3, 2001

1. Find x− y, given that x4 = y4 + 24, x2 + y2 = 6, and x+ y = 3.

Solution: 246·3 = x4−y4

(x2+y2)(x+y)= (x2+y2)(x+y)(x−y)

(x2+y2)(x+y)= x− y = 4

3.

2. Find (x+ 1)(x2 + 1)(x4 + 1)(x8 + 1) · · · , where |x| < 1.Solution: Let S = (x + 1)(x2 + 1)(x4 + 1)(x8 + 1) · · · = 1 + x + x2 + x3 + · · · . Since

xS = x+ x2 + x3 + x4 + · · · , we have (1− x)S = 1, so S = 11−x .

3. How many times does 24 divide into 100! (factorial)?Solution: We first determine the number of times 2 and 3 divide into 100! = 1·2·3 · · · 100.

Let 〈N〉n be the number of times n divides into N (i.e. we want to find 〈100!〉24). Since 2only divides into even integers, 〈100!〉2 = 〈2 · 4 · 6 · · · 100〉. Factoring out 2 once from eachof these multiples, we get that 〈100!〉2 = 〈250 · 1 · 2 · 3 · · · 50〉2. Repeating this process, wefind that 〈100!〉2 = 〈2050+25+12+6+3+1 · 1〉2 = 97. Similarly, 〈100!〉3 = 〈333+11+3+1〉3 = 48.Now 24 = 23 · 3, so for each factor of 24 in 100! there needs to be three multiples of 2 andone multiple of 3 in 100!. Thus 〈100!〉24 = ([〈100!〉2 /3] + 〈100!〉3) = 32 , where [N ] is thegreatest integer less than or equal to N .

4. Given that 7, 999, 999, 999 has at most two prime factors, find its largest prime factor.Solution: 7, 999, 999, 999 = 8 · 109 − 1 = 20003 − 1 = (2000 − 1)(20002 + 2000 + 1), so

(20002 + 2000 + 1) = 4,002,001 is its largest prime factor.

5. Find the 6-digit number beginning and ending in the digit 2 that is the product ofthree consecutive even integers.

Solution: Because the last digit of the product is 2, none of the three consecutive evenintegers end in 0. Thus they must end in 2, 4, 6 or 4, 6, 8, so they must end in 4, 6, 8 since 2·4·6does not end in 2. Call the middle integer n. Then the product is (n−2)n(n+ 2) = n3−4n,so n > 3

√200000 =

3√

200 · 103 ≈ 60, but clearly n < 3√

300000 =3√

300 · 103 < 70. Thusn = 66, and the product is 663 − 4 · 66 = 287232 .

6. What is the last digit of 11 + 22 + 33 + · · ·+ 100100?Solution: Let L(d, n) be the last digit of a number ending in d to the nth power. For

n ≥ 1, we know that L(0, n) = 0, L(1, n) = 1, L(5, n) = 5, L(6, n) = 6. All numbersending in odd digits in this series are raised to odd powers; for odd n, L(3, n) = 3 or 7,L(7, n) = 3 or 7, L(9, n) = 9. All numbers ending in even digits are raised to even powers;for even n, L(2, n) = 4 or 6, L(4, n) = L(6, n) = 6, L(8, n) = 6 or 4. Further, for each

last digit that has two possible values, the possible values will be present equally as often.Now define S(d) such that S(0)=0 and for 1 ≤ d ≤ 9, S(d) = L(d, d) + L(d, d + 10) +L(d, d+ 20) +L(d, d+ 30) + · · ·+L(d, d+ 90), so that the sum we want to calculate becomesS(0) + S(1) + S(2) + · · ·+ S(9). But by the above calculations all S(d) are divisible by 10,so their sum is divisible by 10, which means its last digit is 0 .

7. A polynomial P has four roots, 14, 1

2, 2, 4. The product of the roots is 1, and P (1) = 1.

Find P (0).Solution: A polynomial Q with n roots, x1, . . . , xn, and Q(x0) = 1 is given by Q(x) =

(x−x1)(x−x2)···(x−xn)(x0−x1)(x0−x2)···(x0−x4)

, so P (0) = 134· 12·(−1)·(−3)

= 89

.

8. How many integers between 1 and 2000 inclusive share no common factors with 2001?Solution: Two integers are said to be relatively prime if they share no common factors,

that is if there is no integer greater than 1 that divides evenly into both of them. Note that1 is relatively prime to all integers. Let ϕ(n) be the number of integers less than n thatare relatively prime to n. Since ϕ(mn) = ϕ(m)ϕ(n) for m and n relatively prime , we haveϕ(2001) = ϕ(3 · 23 · 29) = (3− 1)(23− 1)(29− 1) = 1232 .

9. Find the number of positive integer solutions to nx + ny = nz with nz < 2001.Solution: If n = 1, the relation cannot hold, so assume otherwise. If x > y, the left

hand side factors as ny(nx−y + 1) so nx−y + 1 is a power of n. But it leaves a remainder of 1when divided by n and is greater than 1, a contradiction. We reach a similar contradictionif y > x. So y = x and 2nx = nz, so 2 is a power of n and n = 2. So all solutions are of theform 2x + 2x = 2x+1, which holds for all x. 2x+1 < 2001 implies x < 11, so there are 10solutions.

10. Find the real solutions of (2x+ 1)(3x+ 1)(5x+ 1)(30x+ 1) = 10.Solution: (2x + 1)(3x + 1)(5x + 1)(30x + 1) = [(2x + 1)(30x + 1)][(3x + 1)(5x + 1)] =

(60x2 + 32x+ 1)(15x2 + 8x+ 1) = (4y+ 1)(y+ 1) = 10, where y = 15x2 + 8x. The quadratic

equation in y yields y = 1 and y = −94. For y = 1, we have 15x2 +8x−1 = 0, so x = −4±

√31

15.

For y = −94, we have 15x2 + 8x+ 9

4, which yields only complex solutions for x. Thus the real

solutions are −4±√

3115

.

Calculus TestHarvard-MIT Math Tournament

March 3, 2001

1. A sequence of ants walk from (0, 0) to (1, 0) in the plane. The nth ant walks alongn semicircles of radius 1

nwith diameters lying along the line from (0, 0) to (1, 0). Let Ln be

the length of the path walked by the nth ant. Compute limn→∞

Ln.

2. The polynomial 3x5−250x3 +735x is interesting because it has the maximum possiblenumber of relative extrema and points of inflection at integer lattice points for a quinticpolynomial. What is the sum of the x-coordinates of these points?

3. A balloon that blows up in the shape of a perfect cube is being blown up at a rate suchthat at time t fortnights, it has surface area 6t square furlongs. At how many cubic furlongsper fortnight is the air being pumped in when the surface area is 144 square furlongs?

4. What is the size of the largest rectangle that can be drawn inside of a 3-4-5 righttriangle with one of the rectangle’s sides along one of the legs of the triangle?

5. Same as question 4, but now we want one of the rectangle’s sides to be along thehypotenuse.

6. The graph of x2 − (y − 1)2 = 1 has one tangent line with positive slope that passesthrough (x, y) = (0, 0). If the point of tangency is (a, b), find sin−1(a

b) in radians.

7. Find the coefficient of x12 in the Maclaurin series (i.e. Taylor series around x = 0) for1

1−3x+2x2 .

8. Evaluate∞∑n=0

cot−1(n2 + n+ 1).

9. On the planet Lemniscate, the people use the elliptic table of elements, a far moreadvanced version of our periodic table. They’re not very good at calculus, though, so they’veasked for your help. They know that Kr is somewhat radioactive and deteriorates into Pl, avery unstable element that deteriorates to form the stable element As. They started with ablock of Kr of size 10 and nothing else. (Their units don’t translate into English, sorry.) and

nothing else. At time t, they let x(t) be the amount of Kr, y(t) the amount of Pl, and z(t)the amount of As. They know that x′(t) = −x, and that, in the absence of Kr, y′(t) = −2y.Your job is to find at what time t the quantity of Pl will be largest. You should assume thatthe entire amount of Kr that deteriorates has turned into Pl.

10. Evaluate the definite integral+1∫−1

2u332+u998+4u1664 sinu691

1+u666 du.

Calculus Test SolutionsHarvard-MIT Math Tournament

March 3, 2001

1. A sequence of ants walk from (0, 0) to (1, 0) in the plane. The nth ant walks alongn semicircles of radius 1

nwith diameters lying along the line from (0, 0) to (1, 0). Let Ln be

the length of the path walked by the nth ant. Compute limn→∞

Ln.

Solution: A semicircle of radius 1n

has length 12π(

2n

)= π

n, so n such semicircles have

total length π .

2. The polynomial 3x5−250x3 +735x is interesting because it has the maximum possiblenumber of relative extrema and points of inflection at integer lattice points for a quinticpolynomial. What is the sum of the x-coordinates of these points?

Solution: The first derivative is 15x4−750x2 +735, whose roots (which give the relativeextrema) sum to 750/15 = 50. The second derivative is 60x3 − 1500x, whose roots (whichgive the points of inflection) sum to 1500/60 = 25, for a grand total of 75 .

3. A balloon that blows up in the shape of a perfect cube is being blown up at a rate suchthat at time t fortnights, it has surface area 6t square furlongs. At how many cubic furlongsper fortnight is the air being pumped in when the surface area is 144 square furlongs?

Solution: The surface area at time t is 6t, so the volume is t3/2. Hence the air is being

pumped in at a rate of 32

√t. When the surface area is 144, t = 24, so the answer is 3

√6 .

4. What is the size of the largest rectangle that can be drawn inside of a 3-4-5 righttriangle with one of the rectangle’s sides along one of the legs of the triangle?

Solution: Clearly one vertex of the rectangle will be at the right angle. Position thetriangle with the leg of length 4 along the x-axis and the leg of length 3 along the y-axis.Then the hypotenuse is along the line y = 3− (3/4)x.Suppose the rectangle has a side of length y along the leg of length 3. Then the area isy(4/3)(3 − y) = 4y − (4/3)y2. The derivative of this is 0 when 4 − (8/3)y = 0, or y = 3/2,giving an area of 3.Or, if you prefer, suppose the rectangle has a side of length x along the leg of length 4. Thenthe area is x(3 − (3/4)x) = 3x − (3/4)x2. The derivative of this is 0 when 3 − (3/2)x = 0,or x = 2, again giving an area of 3 .

5. Same as question 4, but now we want one of the rectangle’s sides to be along thehypotenuse.

Solution: Put the hypotenuse along the x-axis, with the short leg starting at the originso that the right angle is at the point (9/5, 12/5). For notational convenience, let’s just scaleeverything by a factor of 5 and then remember to divide the final area by 25, so now the toppoint is at (9, 12).Let (a, 0) be the point where the edge of the rectangle along the hypotenuse starts. Thenthe height is h = (4/3)a since the leg of length 3 is along the line y = (4/3)x. The leg oflength 4 is along the line x = 25 − (4/3)y, so the horizontal edge of the rectangle ends atb = 25 − (4/3)h = 25 − (16/9)a. The area of the rectangle is (b − a)h = (25 − (16/9)a −a)(4/3)a = 100

3a− 100

27a2. The derivative of this is 0 when 100

3= 200

27a, or a = 9/2. Thus the

maximum area is (100/3)(9/2)−(100/27)(81/4)25

= 3 .

6. The graph of x2 − (y − 1)2 = 1 has one tangent line with positive slope that passesthrough (x, y) = (0, 0). If the point of tangency is (a, b), find sin−1(a

b) in radians.

Solution: Differentiating both sides of the equation, we find that 2x − 2(y − 1)dydx

= 0,

and so dydx

= xy−1

= ab−1

. The line passing through (0, 0) and (a, b) has slope ba, so b

a= a

b−1.

Solving simultaneously with a2 − (b − 1)2 = 1, we get b2 − b − [(b− 1)2 + 1] = 0, and so

b = 2, a =√

(2). Finally, sin−1(ab) = π

4.

7. Find the coefficient of x12 in the Maclaurin series (i.e. Taylor series around x = 0) for1

1−3x+2x2 .Solution: If you know formal power series, then this is not such a hard question, but

since this is a calculus test... Use partial fractions to get 11−3x+2x2 = 1/2

1−2x− 1

1−x . Now eachof these can be expanded as a geometric series (or take derivatives and get the same result)to get 1

2(1 + 2x+ 4x2 + 8x3 + · · · )− (1 +x+x2 +x3 + · · · ), so the coefficient of xn is 2n−1−1.

When n = 12, that’s 2047 .

8. Evaluate∞∑n=0

cot−1(n2 + n+ 1).

Solution:∞∑n=0

cot(n2 + n + 1) =∞∑n=0

arctan( 1n2+n+1

) =∞∑n=0

arctan(n + 1) − arctan(n)

by the sum/difference formula for tangent. This sum, taken out to n = N , telescopes to

− arctan(0) + arctan(N + 1). So as N goes to infinity, the sum goes to π/2 .

9. On the planet Lemniscate, the people use the elliptic table of elements, a far moreadvanced version of our periodic table. They’re not very good at calculus, though, so they’veasked for your help. They know that Kr is somewhat radioactive and deteriorates into Pl, avery unstable element that deteriorates to form the stable element As. They started with ablock of Kr of size 10 and nothing else. (Their units don’t translate into English, sorry.) andnothing else. At time t, they let x(t) be the amount of Kr, y(t) the amount of Pl, and z(t)

the amount of As. They know that x′(t) = −x, and that, in the absence of Kr, y′(t) = −2y.Your job is to find at what time t the quantity of Pl will be largest. You should assume thatthe entire amount of Kr that deteriorates has turned into Pl.

Solution: This problem is long-winded since it’s giving an autonomous linear system ofdifferential equations without using any such language (and it includes a number of subtlereferences). The system we have is x′ = −x, y′ = x− 2y. It’s not hard to see that x = 10e−t

satisfies the first equation and the initial condition. Plugging this into the second equationand using the integrating factor e2t (or using eigenvalues and eigenvectors to solve the systemdirectly, though I don’t want to begin to explain what that means) lets us solve for y. Moreprecisely, we want to solve y′ + 2y = 10e−t. Multiply by e2t and simplify the left hand sideto get (ye2t)′ = 10et. Integrating both sides with respect to t then yields ye2t = 10et +C, ory = 10e−t+Ce−2t. Since y(0) = 0, we find C = −10. Now to maximize y, we solve y′(t) = 0,

or −10e−t + 20e−2t = 0, or t = ln 2 .

10. Evaluate the definite integral+1∫−1

2u332+u998+4u1664 sinu691

1+u666 du.

Solution: The term 4u1664 sinu691

1+u666 is odd in u, so its integral is 0. Now make the sub-

stitution u = v1/333 ⇒ du = 1333v−332/333dv to find that

+1∫−1

2u332+u998

1+u666 du = 1333

+1∫−1

2+v2

1+v2 dv =

1333

+1∫−1

(1 + 1

1+v2

)dv = 2

333

1∫0

(1 + 1

1+v2

)dv = 2

333

(1 +

1∫0

11+v2 dv

)= 2

333(1 + tan−1 1) = 2

333

(1 + π

4

).

General Test Solutions (First Half)Harvard-MIT Math Tournament

March 3, 2001

1. What is the last digit of 17103 + 5?Solution: Let 〈N〉 be the last digit of N . 〈172〉 = 9, 〈173〉 = 3, 〈174〉 = 1, and〈175〉 = 7.

Since this pattern keeps on repeating itself,⟨174N

⟩= 1 for any integer N . Thus 〈172·25〉 =

〈17100〉 = 1, so 〈17103〉 = 3, and 〈17103 + 5〉 = 〈3 + 5〉 = 8 .

2. Find x+ y, given that x2 − y2 = 10 and x− y = 2.Solution: x2 − y2 = (x− y)(x+ y) = 2(x+ y) = 10, so x+ y = 5 .

3. There are some red and blue marbles in a box. We are told that there are twelvemore red marbles than blue marbles, and we experimentally determine that when we picka marble randomly we get a blue marble one quarter of the time. How many marbles arethere in the box?

Solution: Call the number of blue marbles x, so the number of red marbles is x + 12and the total number of marbles is 2x + 12. The probability of picking a red marble isx

2x+12= 1

4⇒ x = 6, so 2x+ 12 = 24 .

4. Find a+ b+ c+ d+ e if

3a+ 2b+ 4d = 10,6a+ 5b+ 4c+ 3d+ 2e = 8,

a+ b+ 2c+ 5e = 3,2c+ 3d+ 3e = 4, anda+ 2b+ 3c+ d = 7.

Solution: Adding the first, third, and fifth equations, we get 5a + 5b + 5c + 5d + 5e =10 + 3 + 7⇒ a+ b+ c+ d+ e = 4 .

5. What is the sum of the coefficients of the expansion (x+ 2y − 1)6?Solution: The sum of the coefficients of a polynomial is that polynomial evaluated at

1, which for the question at hand is (1 + 2 · 1− 1)6 = 26 = 64 .

6. A right triangle has a hypotenuse of length 2, and one of its legs has length 1. Thealtitude to its hypotenuse is drawn. What is the area of the rectangle whose diagonal is thisaltitude?

Solution: Call the triangle ABC, with AC = 2 and BC = 1. By the Pythagoreantheorem, AB =

√3. Call the point at which the altitude intersects the hypotenuse D. Let

E 6= B be the vertex of the rectangle on AB and F 6= B be the vertex of the rectangle onBC. Triangle BDC is similar to triangle ABC, so BD =

√3

2. Triangle DBF is similar to

triangle ABC, so DF =√

34

and BF = 34. The area of the rectangle is thus

√3

434

= 3√

316

.

7. Find (x+ 1)(x2 + 1)(x4 + 1)(x8 + 1) · · · , where |x| < 1.Solution: Let S = (x + 1)(x2 + 1)(x4 + 1)(x8 + 1) · · · = 1 + x + x2 + x3 + · · · . Since

xS = x+ x2 + x3 + x4 + · · · , we have (1− x)S = 1, so S = 11−x .

8. How many times does 24 divide into 100!?Solution: We first determine the number of times 2 and 3 divide into 100! = 1·2·3 · · · 100.

Let 〈N〉n be the number of times n divides into N (i.e. we want to find 〈100!〉24). Since 2only divides into even integers, 〈100!〉2 = 〈2 · 4 · 6 · · · 100〉. Factoring out 2 once from eachof these multiples, we get that 〈100!〉2 = 〈250 · 1 · 2 · 3 · · · 50〉2. Repeating this process, wefind that 〈100!〉2 = 〈2050+25+12+6+3+1 · 1〉2 = 97. Similarly, 〈100!〉3 = 〈333+11+3+1〉3 = 48.Now 24 = 23 · 3, so for each factor of 24 in 100! there needs to be three multiples of 2 andone multiple of 3 in 100!. Thus 〈100!〉24 = ([〈100!〉2 /3] + 〈100!〉3) = 32 , where [N ] is thegreatest integer less than or equal to N .

9. Boris was given a Connect Four game set for his birthday, but his color-blindness makesit hard to play the game. Still, he enjoys the shapes he can make by dropping checkers intothe set. If the number of shapes possible modulo (horizontal) flips about the vertical axis ofsymmetry is expressed as 9(1 + 2 + · · ·+ n), find n. (Note: the board is a vertical grid withseven columns and eight rows. A checker is placed into the grid by dropping it from the topof a column, and it falls until it hits either the bottom of the grid or another checker alreadyin that column. Also, 9(1 + 2 + · · · + n) is the number of shapes possible, with two shapesthat are horizontal flips of each other counted as one. In other words, the shape that consistssolely of 3 checkers in the rightmost row and the shape that consists solely of 3 checkers inthe leftmost row are to be considered the same shape.)

Solution: There are 97 total shapes possible, since each of the 7 columns can containanywhere from 0 to 8 checkers. The number of shapes symmetric with respect to a horizontalflip is the number of shapes of the leftmost four columns, since the configuration of thesefour columns uniquely determines the configuration of the remaining columns if it is knownthe shape is symmetric: 94. Now we know there are 97 − 94 non-symmetric shapes, so thereare 97−94

2non-symmetric shapes modulo flips. Thus the total number of shapes modulo flips

is 94 + 97−94

2= 94

(1 + 93−1

2

)= 94

(93+1

2

)= (38(36−1)

2= 936(36+1)

2= 9(1 + 2 + · · · + 36), so

n = 36 = 729 .

10. Find the 6-digit number beginning and ending in the digit 2 that is the product ofthree consecutive even integers.

Solution: Because the last digit of the product is 2, none of the three consecutive evenintegers end in 0. Thus they must end in 2, 4, 6 or 4, 6, 8, so they must end in 4, 6, 8 since 2·4·6does not end in 2. Call the middle integer n. Then the product is (n−2)n(n+ 2) = n3−4n,so n > 3

√200000 =

3√

200 · 103 ≈ 60, but clearly n < 3√

300000 =3√

300 · 103 < 70. Thusn = 66, and the product is 663 − 4 · 66 = 287232 .

General Test Solutions (Second Half)Harvard-MIT Math Tournament

March 3, 2001

1. A circle of radius 3 crosses the center of a square of side length 2. Find the differencebetween the areas of the nonoverlapping portions of the figures.

Solution: Call the area of the square s, the area of the circle c, and the area of theoverlapping portion x. The area of the circle not overlapped by the square is c− x and theare of the square not overlapped by the circle is s − x, so the difference between these two

is (c− x)− (s− x) = c− s = 9π2 − 4 .

2. Call three sides of an opaque cube adjacent if someone can see them all at once. Drawa plane through the centers of each triple of adjacent sides of a cube with edge length 1.Find the volume of the closed figure bounded by the resulting planes.

Solution: The volume of the figure is half the volume of the cube (which can be seenby cutting the cube into 8 equal cubes and realizing that the planes cut each of these cubes

in half), namely 12

.

3. Find x if xxx...

= 2.

Solution:√

2√

2√

2...

= 212

212

2...

= 211...

= 2, so x =√

2 .

4. Some students are taking a math contest, in which each student takes one of fourtests. One third of the students take one test, one fourth take another test, one fifth takethe next test, and 26 students take the last test. How many students are taking the contestin total?

Solution: Call the total number of students n. We know n = n3

+ n4

+ n5

+26, so n = 120 .

5. What is the area of a square inscribed in a semicircle of radius 1, with one of its sidesflush with the diameter of the semicircle?

Solution: Call the center of the semicircle O, a point of contact of the square and thecircular part of the semicircle A, the closer vertex of the square on the diameter B, and theside length of the square x. We know OA = 1, AB = x, OB = x

2, and ∠ABO is right. By

the Pythagorean theorem, x2 = 45

.

6. You take a wrong turn on the way to MIT and end up in Transylvania, where 99% ofthe inhabitants are vampires and the rest are regular humans. For obvious reasons, you want

to be able to figure out who’s who. On average, nine-tenths of the vampires are correctlyidentified as vampires and nine-tenths of humans are correct identified as humans. What isthe probability that someone identified as a human is actually a human?

Solution: Consider a sample of 1000 inhabitants. On average, 990 are vampires and 10are people. 99 vampires are identified as human and 9 humans are identified as human. So

out of the 108 who pass, only 112

are human.

7. A real numbers x is randomly chosen in the interval[−151

2, 151

2

]. Find the probability

that the closest integer to x is odd.Solution: By using a graphical method, we can see that, for real x on

[−n− 1

2, n+ 1

2

],

n an even integer, the probability that the closest integer to x is odd is n2n+1

. The desired

probability is(

1531

).

8. A point on a circle inscribed in a square is 1 and 2 units from the two closest sides ofthe square. Find the area of the square.

Solution: Call the point in question A, the center of the circle O, and its radius r.Consider a right triangle BOA with hypotenuse OA: OA has length r, and BO and BAhave lengths r − 1 and r − 2. By the Pythagorean theorem, (r − 1)2 + (r − 2)2 = r2 ⇒r2 − 6r + 5 = 0⇒ r = 5 since r > 4. The area of the square is (2r)2 = 100 .

9. Two circles are concentric. The area of the ring between them is A. In terms of A,find the length of the longest chord contained entirely within the ring.

Solution: Let the radii of the circles be r and R > r, so A = π(R2 − r2). By the

Pythagorean theorem, the length of the chord is 2√R2 − r2 = 2

√Aπ

.

10. Find the volume of the tetrahedron with vertices (5, 8, 10), (10, 10, 17), (4, 45, 46), (2, 5, 4).Solution: Each vertex (x, y, z) obeys x+ y = z + 3, so all the vertices are coplanar and

the volume of the tetrahedron is 0 .

General Test, First HalfHarvard-MIT Math Tournament

March 3, 2001

1. What is the last digit of 17103 + 5?

2. Find x+ y, given that x2 − y2 = 10 and x− y = 2.

3. There are some red and blue marbles in a box. We are told that there are twelvemore red marbles than blue marbles, and we experimentally determine that when we picka marble randomly we get a blue marble one quarter of the time. How many marbles arethere in the box?

4. Find a+ b+ c+ d+ e if

3a+ 2b+ 4d = 10,6a+ 5b+ 4c+ 3d+ 2e = 8,

a+ b+ 2c+ 5e = 3,2c+ 3d+ 3e = 4, anda+ 2b+ 3c+ d = 7.

5. What is the sum of the coefficients of the expansion (x+ 2y − 1)6?

6. A right triangle has a hypotenuse of length 2, and one of its legs has length 1. Thealtitude to its hypotenuse is drawn. What is the area of the rectangle whose diagonal is thisaltitude?

7. Find (x+ 1)(x2 + 1)(x4 + 1)(x8 + 1) · · · , where |x| < 1.

8. How many times does 24 divide into 100!?

9. Boris was given a Connect Four game set for his birthday, but his color-blindness makesit hard to play the game. Still, he enjoys the shapes he can make by dropping checkers intothe set. If the number of shapes possible modulo (horizontal) flips about the vertical axis ofsymmetry is expressed as 9(1 + 2 + · · ·+ n), find n. (Note: the board is a vertical grid withseven columns and eight rows. A checker is placed into the grid by dropping it from the top

of a column, and it falls until it hits either the bottom of the grid or another checker alreadyin that column. Also, 9(1 + 2 + · · · + n) is the number of shapes possible, with two shapesthat are horizontal flips of each other counted as one. In other words, the shape that consistssolely of 3 checkers in the rightmost row and the shape that consists solely of 3 checkers inthe leftmost row are to be considered the same shape.)

10. Find the 6-digit number beginning and ending in the digit 2 that is the product ofthree consecutive even integers.

General Test Solutions (Second Half)Harvard-MIT Math Tournament

March 3, 2001

1. A circle of radius 3 crosses the center of a square of side length 2. Find the differencebetween the areas of the nonoverlapping portions of the figures.

2. Call three sides of an opaque cube adjacent if someone can see them all at once. Drawa plane through the centers of each triple of adjacent sides of a cube with edge length 1.Find the volume of the closed figure bounded by the resulting planes.

3. Find x if xxx...

= 2.4. Some students are taking a math contest, in which each student takes one of four

tests. One third of the students take one test, one fourth take another test, one fifth takethe next test, and 26 students take the last test. How many students are taking the contestin total?

5. What is the area of a square inscribed in a semicircle of radius 1, with one of its sidesflush with the diameter of the semicircle?

6. You take a wrong turn on the way to MIT and end up in Transylvania, where 99% ofthe inhabitants are vampires and the rest are regular humans. For obvious reasons, you wantto be able to figure out who’s who. On average, nine-tenths of the vampires are correctlyidentified as vampires and nine-tenths of humans are correct identified as humans. What isthe probability that someone identified as a human is actually a human?

7. A real numbers x is randomly chosen in the interval[−151

2, 151

2

]. Find the probability

that the closest integer to x is odd.

8. A point on a circle inscribed in a square is 1 and 2 units from the two closest sides ofthe square. Find the area of the square.

9. Two circles are concentric. The area of the ring between them is A. In terms of A,find the length of the longest chord contained entirely within the ring.

10. Find the volume of the tetrahedron with vertices (5, 8, 10), (10, 10, 17), (4, 45, 46), (2, 5, 4).

Geometry TestHarvard-MIT Math Tournament

March 3, 2001

1. A circle of radius 3 crosses the center of a square of side length 2. Find the positivedifference between the areas of the nonoverlapping portions of the figures.

2. Call three sides of an opaque cube adjacent if someone can see them all at once. Drawa plane through the centers of each triple of adjacent sides of a cube with edge length 1.Find the volume of the closed figure bounded by the resulting planes.

3. Square ABCD is drawn. Isosceles Triangle CDE is drawn with E a right angle.Square DEFG is drawn. Isosceles triangle FGH is drawn with H a right angle. Thisprocess is repeated infinitely so that no two figures overlap each other. If square ABCD hasarea 1, compute the area of the entire figure.

4. A circle has two parallel chords of length x that are x units apart. If the part of thecircle included between the chords has area 2 + π, find x.

5. Find the volume of the tetrahedron with vertices (5, 8, 10), (10, 10, 17), (4, 45, 46), (2, 5, 4).

6. A point on a circle inscribed in a square is 1 and 2 units from the two closest sides ofthe square. Find the area of the square.

7. Equilateral triangle ABC with side length 1 is drawn. A square is drawn such that itsvertex at A is opposite to its vertex at the midpoint of BC. Find the area enclosed within

the intersection of the insides of the triangle and square. Hint: sin 75 =√

2(√

3+1)4

.

8. Point D is drawn on side BC of equilateral triangle ABC, and AD is extended pastD to E such that angles EAC and EBC are equal. If BE = 5 and CE = 12, determine thelength of AE.

9. Parallelogram AECF is inscribed in square ABCD. It is reflected across diagonalAC to form another parallelogram AE ′CF ′. The region common to both parallelograms hasarea m and perimeter n. Compute the value of m

n2 if AF : AD = 1 : 4.

10. A is the center of a semicircle, with radius AD lying on the base. B lies on the basebetween A and D, and E is on the circular portion of the semicircle such that EBA is aright angle. Extend EA through A to C, and put F on line CD such that EBF is a line.Now EA = 1, AC =

√2, BF = 2−

√2

4, CF = 2

√5+√

104

, and DF = 2√

5−√

104

. Find DE.

Geometry Test SolutionsHarvard-MIT Math Tournament

March 3, 2001

1. A circle of radius 3 crosses the center of a square of side length 2. Find the positivedifference between the areas of the nonoverlapping portions of the figures.

Solution: Call the area of the square s, the area of the circle c, and the area of theoverlapping portion x. The area of the circle not overlapped by the square is c− x and theare of the square not overlapped by the circle is s − x, so the difference between these two

is (c− x)− (s− x) = c− s = 9π2 − 4 .

2. Call three sides of an opaque cube adjacent if someone can see them all at once. Drawa plane through the centers of each triple of adjacent sides of a cube with edge length 1.Find the volume of the closed figure bounded by the resulting planes.

Solution: The volume of the figure is half the volume of the cube (which can be seenby cutting the cube into 8 equal cubes and realizing that the planes cut each of these cubes

in half), namely 12

.

3. Square ABCD is drawn. Isosceles Triangle CDE is drawn with E a right angle.Square DEFG is drawn. Isosceles triangle FGH is drawn with H a right angle. Thisprocess is repeated infinitely so that no two figures overlap each other. If square ABCD hasarea 1, compute the area of the entire figure.

Solution: Let the area of the nth square drawn be Sn and the area of the nth trianglebe Tn. Since the hypotenuse of the nth triangle is of length

√Sn, its legs are of length

l =√

Sn2

, so Sn+1 = l2 = Sn2

and Tn = l2

2= Sn

4. Using the recursion relations, Sn = 1

2n−1 and

Tn = 12n+1 , so Sn + Tn = 1

2n−1 + 12n+1 =

(12

+ 2)

12n

= 52

12n

. Thus the total area of the figure is∞∑n=1

Sn + Tn = 52

∞∑n=1

12n

= 52

.

4. A circle has two parallel chords of length x that are x units apart. If the part of thecircle included between the chords has area 2 + π, find x.

Solution: Let C be the area of the circle, S be the area of the square two of whoseedges are the chords, and A be the area of the part of the circle included between the chords.The radius of the circle is

√2

2x, so C = π

2x2, and S = x2. Then the area A is the area

of the square plus one half of the difference between the areas of the circle and square:

A = C−S2

+ S = C+S2

=1+π

2

2x2, so x =

√2A

1+π2

= 2 .

5. Find the volume of the tetrahedron with vertices (5, 8, 10), (10, 10, 17), (4, 45, 46), (2, 5, 4).

Solution: Each vertex (x, y, z) obeys x+ y = z + 3, so all the vertices are coplanar andthe volume of the tetrahedron is 0 .

6. A point on a circle inscribed in a square is 1 and 2 units from the two closest sides ofthe square. Find the area of the square.

Solution: Call the point in question A, the center of the circle O, and its radius r.Consider a right triangle BOA with hypotenuse OA: OA has length r, and BO and BAhave lengths r − 1 and r − 2. By the Pythagorean theorem, (r − 1)2 + (r − 2)2 = r2 ⇒r2 − 6r + 5 = 0⇒ r = 5 since r > 4. The area of the square is (2r)2 = 100 .

7. Equilateral triangle ABC with side length 1 is drawn. A square is drawn such that itsvertex at A is opposite to its vertex at the midpoint of BC. Find the area enclosed within

the intersection of the insides of the triangle and square. Hint: sin 75 =√

2(√

3+1)4

.Solution: Let D be the midpoint of BC, F 6= A be the point of intersection of the

square and triangle lying on AC, b be the length of FC, x be the side length of the triangle,and y be the length of AD. By the law of sines on triangle CDF , we have 2 sin 75

x= sin 45

b, so

b = x sin 452 sin 75

=√

2x4 sin 75

. The area of the desired figure can easily be seen to be 12(x− b)y since it

can be seen as two triangles of width y and height x−b2

. This reduces to 12

(1−

√2

4 sin 75

)xy.

Then by the Pythagorean theorem on triangle ABD, x2 =(x2

)2+ y2, so y =

√3

2x, and the

area becomes√

34

(1−

√2

4 sin 75

)x2 =

√3

4

(1−

√2

4 sin 75

)= 3

4(√

3+1).

8. Point D is drawn on side BC of equilateral triangle ABC, and AD is extended pastD to E such that angles EAC and EBC are equal. If BE = 5 and CE = 12, determine thelength of AE.

Solution:By construction, ABEC is a cyclic quadrilateral. Ptolemy’s theorem says thatfor cyclic quadrilaterals, the sum of the products of the lengths of the opposite sides equals theproduct of the lengths of the diagonals. This yields (BC)(AE) = (BA)(CE) + (BE)(AC).Since ABC is equilateral, BC = AC = AB, so dividing out by this common value we getAE = CE +BE = 17 .

9. Parallelogram AECF is inscribed in square ABCD. It is reflected across diagonalAC to form another parallelogram AE ′CF ′. The region common to both parallelograms hasarea m and perimeter n. Compute the value of m

n2 if AF : AD = 1 : 4.Solution: By symmetry, the region is a rhombus, AXCY, centered at the center of the

square, O. Consider isoceles right triangle ACD. By the technique of mass points, we findthat DO : Y O = 7 : 1. Therefore, the rhombus is composed of four triangles, whose sidesare in the ratio 1 : 7 : 5

√2. The perimeter of the rhombus is 20

√2N , and the area is 14N2.

The required ratio is thus 7400

.

10. A is the center of a semicircle, with radius AD lying on the base. B lies on the basebetween A and D, and E is on the circular portion of the semicircle such that EBA is aright angle. Extend EA through A to C, and put F on line CD such that EBF is a line.Now EA = 1, AC =

√2, BF = 2−

√2

4, CF = 2

√5+√

104

, and DF = 2√

5−√

104

. Find DE.Solution: Let θ = ∠AED and x = DE. By the law of cosines on triangle ADE, we have

1 = 1+x2−2x cos θ ⇒ 2x cos θ = x2. Then by the law of cosines on triangle CDE (note that

CD =√

5), we have 5 =(1 +√

2)2

+x2−2(1 +√

2)x cos θ =

(1 +√

2)2

+x2−(1 +√

2)x2.

Solving the quadratic equation gives x =√

2−√

2 .

Guts Round QuestionsHarvard-MIT Math Tournament

March 3, 2001

1. [5] January 3, 1911 was an odd date as its abbreviated representation, 1/3/1911, canbe written using only odd digits (note all four digits are written for the year). To the nearestmonth, how many months will have elapsed between the most recent odd date and the nextodd date (today is 3/3/2001, an even date).

2. [4] Ken is the best sugar cube retailer in the nation. Trevor, who loves sugar, iscoming over to make an order. Ken knows Trevor cannot afford more than 127 sugar cubes,but might ask for any number of cubes less than or equal to that. Ken prepares seven cupsof cubes, with which he can satisfy any order Trevor might make. How many cubes are inthe cup with the most sugar?

3. [7] Find the number of triangulations of a general convex 7-gon into 5 triangles by 4diagonals that do not intersect in their interiors.

4. [7] Find∞∏n=2

(1− 1

n2

).

5. [± 6] Let ABC be a triangle with incenter I and circumcenter O. Let the circumradiusbe R. What is the least upper bound of all possible values of IO?

6. [8] Six students taking a test sit in a row of seats with aisles only on the two sides ofthe row. If they finish the test at random times, what is the probability that some studentwill have to pass by another student to get to an aisle?

7. [5] Suppose a, b, c, d, and e are objects that we can multiply together, but themultiplication doesn’t necessarily satisfy the associative law, i.e. (xy)z does not necessarilyequal x(yz). How many different ways are there to interpret the product abcde?

8. [10] Compute 1 · 2 + 2 · 3 + · · ·+ (n− 1)n.

9. [5] Suppose x satisfies x3 + x2 + x+ 1 = 0. What are all possible values of x4 + 2x3 +2x2 + 2x+ 1?

10. [8] Two concentric circles have radii r and R > r. Three new circles are drawn sothat they are each tangent to the big two circles and tangent to the other two new circles.Find R

r.

11. [8] 12 points are placed around the circumference of a circle. How many ways arethere to draw 6 non-intersecting chords joining these points in pairs?

12. [± 6] How many distinct sets of 8 positive odd integers sum to 20?

13. [5] Find the number of real zeros of x3 − x2 − x+ 2.

14. [8] Find the exact value of 1 + 11+ 2

1+ 1

1+ 21+···

.

15. [6] A beaver walks from (0, 0) to (4, 4) in the plane, walking one unit in the positivex direction or one unit in the positive y direction at each step. Moreover, he never goes toa point (x, y) with y > x. How many different paths can he walk?

16. [6] After walking so much that his feet get really tired, the beaver staggers so that,at each step, his coordinates change by either (+1,+1) or (+1,−1). Now he walks from(0, 0) to (8, 0) without ever going below the x-axis. How many such paths are there?

17. [4] Frank and Joe are playing ping pong. For each game, there is a 30% chance thatFrank wins and a 70% chance Joe wins. During a match, they play games until someonewins a total of 21 games. What is the expected value of number of games played per match?

18. [5] Find the largest prime factor of −x10 − x8 − x6 − x4 − x2 − 1, where x = 2i,i =√−1.

19. [9] Calculate2001∑n=1

n3.

20. [± 4] Karen has seven envelopes and seven letters of congratulations to variousHMMT coaches. If she places the letters in the envelopes at random with each possibleconfiguration having an equal probability, what is the probability that exactly six of theletters are in the correct envelopes?

21. [10] Evaluate∞∑i=1

(i+1)(i+2)(i+3)(−2)i

.

22. [6] A man is standing on a platform and sees his train move such that after tseconds it is 2t2 + d0 feet from his original position, where d0 is some number. Call thesmallest (constant) speed at which the man have to run so that he catches the train v. Interms of n, find the nth smallest value of d0 that makes v a perfect square.

23. [5] Alice, Bob, and Charlie each pick a 2-digit number at random. What is theprobability that all of their numbers’ tens’ digits are different from each others’ tens’ digitsand all of their numbers’ ones digits are different from each others’ ones’ digits?

24. [6] Square ABCD has side length 1. A dilation is performed about point A, creatingsquare AB′C ′D′. If BC ′ = 29, determine the area of triangle BDC ′.

25. [± 10] What is the remainder when 100! is divided by 101?

26. [6] A circle with center at O has radius 1. Points P and Q outside the circle areplaced such that PQ passes through O. Tangent lines to the circle through P hit the circleat P1 and P2, and tangent lines to the circle through Q hit the circle at Q1 and Q2. If∠P1PP2 = 45◦ and angleQ1QQ2 = 30◦, find the minimum possible length of arc P2Q2.

27. [5] Mona has 12 match sticks of length 1, and she has to use them to make regularpolygons, with each match being a side or a fraction of a side of a polygon, and no twomatches overlapping or crossing each other. What is the smallest total area of the polygonsMona can make?

28. [4] How many different combinations of 4 marbles can be made from 5 indistinguish-able red marbles, 4 indistinguishable blue marbles, and 2 indistinguishable black marbles?

29. [10] Count the number of sequences a1, a2, a3, a4, a5 of integers such that ai ≤ 1 forall i and all partial sums (a1, a1 + a2, etc.) are non-negative.

30. [± 4] How many roots does arctanx = x2 − 1.6 have, where the arctan function isdefined in the range −pi

2< arctanx < pi

2?

31. [5] If two fair dice are tossed, what is the probability that their sum is divisible by5?

32. [10] Count the number of permutations a1a2 . . . a7 of 1234567 with longest decreasingsubsequence of length at most two (i.e. there does not exist i < j < k such that ai > aj > ak).

33. [± 5] A line of soldiers 1 mile long is jogging. The drill sergeant, in a car, movingat twice their speed, repeatedly drives from the back of the line to the front of the line andback again. When each soldier has marched 15 miles, how much mileage has been added tothe car, to the nearest mile?

34. [8] Find all the values of m for which the zeros of 2x2 −mx− 8 differ by m− 1.

35. [7] Find the largest integer that divides m5 − 5m3 + 4m for all m ≥ 5.. 5! = 120 .

36. [4] Count the number of sequences 1 ≤ a1 ≤ a2 ≤ · · · ≤ a5 of integers with ai ≤ ifor all i.

37. [5] Alex and Bob have 30 matches. Alex picks up somewhere between one and sixmatches (inclusive), then Bob picks up somewhere between one and six matches, and so on.The player who picks up the last match wins. How many matches should Alex pick up atthe beginning to guarantee that he will be able to win?

38. [9] The cafeteria in a certain laboratory is open from noon until 2 in the afternoonevery Monday for lunch. Two professors eat 15 minute lunches sometime between noonand 2. What is the probability that they are in the cafeteria simultaneously on any givenMonday?

39. [9] What is the remainder when 22001 is divided by 27 − 1?

40. [5] A product of five primes is of the form ABC,ABC, where A, B, and C representdigits. If one of the primes is 491, find the product ABC,ABC.

41. [4] If(a+ 1

a

)2= 3, find

(a+ 1

a

)3in terms of a.

42. [10] Solve x =√x− 1

x+√

1− 1x

for x.

43. [4] When a single number is added to each member of the seqence 20, 50, 100, thesequence becomes expressable as x, ax, a2x. Find a.

44. [7] Through a point in the interior of a triangle ABC, three lines are drawn, oneparallel to each side. These lines divide the sides of the triangle into three regions each.Let a, b, and c be the lengths of the sides opposite ∠A, ∠B, and ∠C, respectively, and leta′, b′, and c′ be the lengths of the middle regions of the sides opposite ∠A, ∠B, and ∠C,respectively. Find the numerical value of a′/a+ b′/b+ c′/c.

45. [4] A stacking of circles in the plane consists of a base, or some number of unitcircles centered on the x-axis in a row without overlap or gaps, and circles above the x-axisthat must be tangent to two circles below them (so that if the ends of the base were securedand gravity were applied from below, then nothing would move). How many stackings ofcircles in the plane have 4 circles in the base?

46. [± 5] Draw a rectangle. Connect the midpoints of the opposite sides to get 4congruent rectangles. Connect the midpoints of the lower right rectangle for a total of 7rectangles. Repeat this process infinitely. Let n be the minimum number of colors we canassign to the rectangles so that no two rectangles sharing an edge have the same color and mbe the minimum number of colors we can assign to the rectangles so that no two rectanglessharing a corner have the same color. Find the ordered pair (n,m).

47. [7] For the sequence of numbers n1, n2, n3, . . . , the relation ni = 2ni−1 + a holds forall i > 1. If n2 = 5 and n8 = 257, what is n5?

48. [8] What is the smallest positive integer x for which x2 + x+ 41 is not a prime?

49. [5] If 19

of 60 is 5, what is 120

of 80?

50. [9] The Fibonacci sequence F1, F2, F3, . . . is defined by F1 = F2 = 1 and Fn+2 =Fn+1 + Fn. Find the least positive integer t such that for all n > 0, Fn = Fn+t.

51. [5] Some people like to write with larger pencils than others. Ed, for instance, likesto write with the longest pencils he can find. However, the halls of MIT are of limited heightL and width L. What is the longest pencil Ed can bring through the halls so that he cannegotiate a square turn?

52. [6] Find all ordered pairs (m,n) of integers such that 231m2 = 130n2.

53. [7] Find the sum of the infinite series 132−12

(112 − 1

32

)+ 1

52−32

(132 − 1

52

)+ 1

72−52

(152 − 1

72

)+

· · · .

54. [10] The set of points (x1, x2, x3, x4) in R4 such that x1 ≥ x2 ≥ x3 ≥ x4 is a cone(or hypercone, if you insist). Into how many regions is this cone sliced by the hyperplanesxi − xj = 1 for 1 ≤ i < j ≤ n?

55. [7] How many multiples of 7 between 106 and 109 are perfect squares?

56. [6] A triangle has sides of length 888, 925, and x > 0. Find the value of x thatminimizes the area of the circle circumscribed about the triangle.

57. [5] Let x = 20011002 − 2001−1002 and y = 20011002 + 2001−1002. Find x2 − y2.

58. [9] Let (x, y) be a point in the cartesian plane, x, y > 0. Find a formula in terms ofx and y for the minimal area of a right triangle with hypotenuse passing through (x, y) andlegs contained in the x and y axes.

59. [10] Trevor and Edward play a game in which they take turns adding or removingbeans from a pile. On each turn, a player must either add or remove the largest perfectsquare number of beans that is in the heap. The player who empties the pile wins. Forexample, if Trevor goes first with a pile of 5 beans, he can either add 4 to make the total9, or remove 4 to make the total 1, and either way Edward wins by removing all the beans.There is no limit to how large the pile can grow; it just starts with some finite number ofbeans in it, say fewer than 1000.

Before the game begins, Edward dispatches a spy to find out how many beans will be inthe opening pile, call this n, then “graciously” offers to let Trevor go first. Knowing that thefirst player is more likely to win, but not knowing n, Trevor logically but unwisely accepts,and Edward goes on to win the game. Find a number n less than 1000 that would promptthis scenario, assuming both players are perfect logicians. A correct answer is worth thenearest integer to log2(n− 4) points.

60. [∞] Find an n such that n!− (n − 1)! + (n − 2)!− (n − 3)! + · · · ± 1! is prime. Be

prepared to justify your answer for{

n, n≤25

[n+22510 ], n>25

points, where [N ] is the greatest integer

less than N .

1. [5] January 3, 1911 was an odd date as its abbreviated representation, 1/3/1911, canbe written using only odd digits (note all four digits are written for the year). To the nearestmonth, how many months will have elapsed between the most recent odd date and the nextodd date (today is 3/3/2001, an even date).

Solution: The most recent odd date was 11/19/1999 (November has 30 days, but theassumption that it has 31 days does not change the answer), and the next odd date willbe 1/1/3111. From 11/19/1999 to 1/1/2000 is about 1 month. From 2000 to 3111 is 1111years, or 12 · 1111 = 13332 months, so the total number of months is 13333 .

2. [4] Ken is the best sugar cube retailer in the nation. Trevor, who loves sugar, iscoming over to make an order. Ken knows Trevor cannot afford more than 127 sugar cubes,but might ask for any number of cubes less than or equal to that. Ken prepares seven cupsof cubes, with which he can satisfy any order Trevor might make. How many cubes are inthe cup with the most sugar?

Solution: The only way to fill seven cups to satisfy the above condition is to use a binaryscheme, so the cups must contain 1, 2, 4, 8, 16, 32, and 64 cubes of sugar.

3. [7] Find the number of triangulations of a general convex 7-gon into 5 triangles by 4diagonals that do not intersect in their interiors.

Solution: Define the Catalan numbers by C(n) = 1n+1

(2nn

). The current solution is the

C(number of triangles) = C(5) = 42 .

4. [7] Find∞∏n=2

(1− 1

n2

).

Solution:∞∏n=2

(1− 1

n2

)=

∞∏n=2

n2−1n2 =

∞∏n=2

(n−1)(n+1)n·n = 1·3

2·22·43·3

3·54·4

4·65·5

5·76·6 · · · = 1·2·3·3·4·4·5·5···

2·2·3·3·4·4·5·5··· =

12

.

5. [± 6] Let ABC be a triangle with incenter I and circumcenter O. Let the circumradiusbe R. What is the least upper bound of all possible values of IO?

Solution: I always lies inside the convex hull of ABC, which in turn always lies in thecircumcircle of ABC, so IO < R. On the other hand, if we first draw the circle Ω of radiusR about O and then pick A, B, and C very close together on it, we can force the convex hullof ABC to lie outside the circle of radius R− ε about O for any ε. Thus the answer is R .

6. [8] Six students taking a test sit in a row of seats with aisles only on the two sides ofthe row. If they finish the test at random times, what is the probability that some studentwill have to pass by another student to get to an aisle?

1

Solution: The probability p that no student will have to pass by another student to getto an aisle is the probability that the first student to leave is one of the students on the end,the next student to leave is on one of the ends of the remaining students, etc.: p = 2

6· 2

5· 2

4· 2

3,

so the desired probability is 1− p = 4345

.

7. [5] Suppose a, b, c, d, and e are objects that we can multiply together, but themultiplication doesn’t necessarily satisfy the associative law, i.e. (xy)z does not necessarilyequal x(yz). How many different ways are there to interpret the product abcde?

Solution: C(number of letters− 1) = C(4) = 14 .

8. [10] Compute 1 · 2 + 2 · 3 + · · ·+ (n− 1)n.

Solution: Let S = 1 · 2 + 2 · 3 + · · · + (n − 1)n. We known∑i=1

i = n(n+1)2

andn∑i=1

i2 =

n(n+1)(2n+1)6

. So S = 1(1 + 1) + 2(2 + 1) + · · · + (n − 1)n = (12 + 22 + · · ·+ (n− 1)2) +

(1 + 2 + · · ·+ (n− 1)) = (n−1)(n)(2n−1)6

+ (n−1)(n)2

= (n−1)n(n+1)3

.

We can also arrive at the solution by realizing thatn∑i=1

i2 =n∑i=1

i2 +n∑i=2

i2 +n∑i=3

i2 + · · · +n∑i=n

i2 = nn∑i=1

i2−(

1∑i=1

i2 +2∑i=1

i2 +3∑i=1

i2 + · · ·n−1∑n=1

i2)

= nn(n+1)2−(

1·22

+ 2·32

+ · · ·+ (n−1)n2

)=

nn(n+1)2− 1

2S = n(n+1)(2n+1)

6, so S = (n−1)n(n+1)

3.

9. [5] Suppose x satisfies x3 + x2 + x+ 1 = 0. What are all possible values of x4 + 2x3 +2x2 + 2x+ 1?

Solution: x4 + 2x3 + 2x2 + 2x + 1 = (x + 1)(x3 + x2 + x + 1) = 0 is the only possiblesolution.

10. [8] Two concentric circles have radii r and R > r. Three new circles are drawn sothat they are each tangent to the big two circles and tangent to the other two new circles.Find R

r.

Solution: The centers of the three new circles form a triangle. The diameter of the newcircles is R− r, so the side length of the triangle is R− r. Call the center of the concentriccircle O, two vertices of the triangle A and B, and AB’s midpoint D. OA is the average Rand r, namely R+r

2. Using the law of sines on triangle DAO, we get sin(30)

AD= sin(90)

AO⇒ R = 3r,

so Rr

= 3 .

11. [8] 12 points are placed around the circumference of a circle. How many ways arethere to draw 6 non-intersecting chords joining these points in pairs?

Solution: C(number of chords) = C(6) = 132 .

2

12. [± 6] How many distinct sets of 8 positive odd integers sum to 20?Solution: This is the same as the number of ways 8 nonnegative even integers sum to

12 (we subtract 1 from each integer in the above sum). All 11 possibilities are (leaving out0s): 12, 10 + 2, 8 + 4, 8 + 2 + 2, 6 + 6, 6 + 4 + 2, 6 + 2 + 2 + 2 + 2, 4 + 4 + 4, 4 + 4 + 2 + 2,4 + 2 + 2 + 2 + 2, 2 + 2 + 2 + 2 + 2 + 2.

13. [5] Find the number of real zeros of x3 − x2 − x+ 2.Solution: Let f(x) = x3 − x2 − x+ 2, so f ′(x) = 3x2 − 2x− 1. The slope is zero when

3x2 − 2x − 1 = 0, where x = −13

and x = 1. Now f(

13

)> 0 and f(1) > 0, so there are no

zeros between x = −13

and x = 1. Since limx→+∞

f(x) > 0, there are no zeros for x > 1. Since

limx→−∞

f(x) < 0, there is one zero for x < −13, for a total of 1 zero.

14. [8] Find the exact value of 1 + 11+ 2

1+ 1

1+ 21+···

.

Solution: Let x be what we are trying to find. x − 1 = 11+ 2

1+ 1

1+ 21+···

⇒ 1x−1− 1 =

21+ 1

1+ 21+···

⇒ 21

x−1−1

= x⇒ x2 − 2 = 0, so x =√

2 since x > 0.

15. [6] A beaver walks from (0, 0) to (4, 4) in the plane, walking one unit in the positivex direction or one unit in the positive y direction at each step. Moreover, he never goes toa point (x, y) with y > x. How many different paths can he walk?

Solution: C(4) = 14 .

16. [6] After walking so much that his feet get really tired, the beaver staggers so that,at each step, his coordinates change by either (+1,+1) or (+1,−1). Now he walks from(0, 0) to (8, 0) without ever going below the x-axis. How many such paths are there?

Solution: C(4) = 14 .

17. [4] Frank and Joe are playing ping pong. For each game, there is a 30% chance thatFrank wins and a 70% chance Joe wins. During a match, they play games until someonewins a total of 21 games. What is the expected value of number of games played per match?

Solution: The expected value of the ratio of Frank’s to Joe’s score is 3:7, so Frank isexpected to win 9 games for each of Frank’s 21. Thus the expected number of games in amatch is 30 .

18. [5] Find the largest prime factor of −x10 − x8 − x6 − x4 − x2 − 1, where x = 2i,i =√−1.

3

Solution: 13 .

19. [9] Calculate2001∑n=1

n3.

Solution:2001∑n=1

n3 =

(2001∑n=1

n

)2

=(

2001·20022

)2= 4012013006001 .

20. [± 4] Karen has seven envelopes and seven letters of congratulations to variousHMMT coaches. If she places the letters in the envelopes at random with each possibleconfiguration having an equal probability, what is the probability that exactly six of theletters are in the correct envelopes?

Solution: 0 , since if six letters are in their correct envelopes the seventh is as well.

21. [10] Evaluate∞∑i=1

(i+1)(i+2)(i+3)(−2)i

.

Solution: This is the power series of 6(1+x)4 expanded about x = 0 and evaluated at

x = −12, so the solution is 96 .

22. [6] A man is standing on a platform and sees his train move such that after tseconds it is 2t2 + d0 feet from his original position, where d0 is some number. Call thesmallest (constant) speed at which the man have to run so that he catches the train v. Interms of n, find the nth smallest value of d0 that makes v a perfect square.

Solution: The train’s distance from the man’s original position is t2 +d0, and the man’sdistance from his original position if he runs at speed v is vt at time t. We need to findwhere t2 + d0 = vt has a solution. Note that this is a quadratic equation with discriminantD =

√v2 − 4d0, so it has solutions for real D, i.e. where v ≥

√4d0, so 4d0 must be a perfect

square. This happens when 4d0 is an even power of 2: the smallest value is 20, the second

smallest is 22, the third smallest is 24, and in general the nth smallest is 22(n−1) , or 4n−1 .

23. [5] Alice, Bob, and Charlie each pick a 2-digit number at random. What is theprobability that all of their numbers’ tens’ digits are different from each others’ tens’ digitsand all of their numbers’ ones digits are different from each others’ ones’ digits?

Solution: 910

810

89

79

= 112225

.

24. [6] Square ABCD has side length 1. A dilation is performed about point A, creatingsquare AB′C ′D′. If BC ′ = 29, determine the area of triangle BDC ′.

Solution: 292 − 2 · 12(29)

(292

)− 1

2= 420 .

4

25. [± 10] What is the remainder when 100! is divided by 101?Solution: Wilson’s theorem says that for p a prime, (p− 1)! ≡ −1(p), so the remainder

is 100 .

26. [6] A circle with center at O has radius 1. Points P and Q outside the circle areplaced such that PQ passes through O. Tangent lines to the circle through P hit the circleat P1 and P2, and tangent lines to the circle through Q hit the circle at Q1 and Q2. If∠P1PP2 = 45◦ and angleQ1QQ2 = 30◦, find the minimum possible length of arc P2Q2.

Solution: (45− 30)◦ = π12

.

27. [5] Mona has 12 match sticks of length 1, and she has to use them to make regularpolygons, with each match being a side or a fraction of a side of a polygon, and no twomatches overlapping or crossing each other. What is the smallest total area of the polygonsMona can make?

Solution: 4√

34

=√

3 .

28. [4] How many different combinations of 4 marbles can be made from 5 indistinguish-able red marbles, 4 indistinguishable blue marbles, and 2 indistinguishable black marbles?

Solution: 5 + 4 + 3 = 12 .

29. [10] Count the number of sequences a1, a2, a3, a4, a5 of integers such that ai ≤ 1 forall i and all partial sums (a1, a1 + a2, etc.) are non-negative.

Solution: C(length + 1) = C(6) = 132 .

30. [± 4] How many roots does arctanx = x2 − 1.6 have, where the arctan function isdefined in the range −pi

2< arctanx < pi

2?

Solution: 2 .

31. [5] If two fair dice are tossed, what is the probability that their sum is divisible by5?

Solution: 14

.

32. [10] Count the number of permutations a1a2 . . . a7 of 1234567 with longest decreasingsubsequence of length at most two (i.e. there does not exist i < j < k such that ai > aj > ak).

Solution: C(7) = 429 .

33. [± 5] A line of soldiers 1 mile long is jogging. The drill sergeant, in a car, movingat twice their speed, repeatedly drives from the back of the line to the front of the line and

5

back again. When each soldier has marched 15 miles, how much mileage has been added tothe car, to the nearest mile?

Solution: 30 .

34. [8] Find all the values of m for which the zeros of 2x2 −mx− 8 differ by m− 1.

Solution: 6,−103

.

35. [7] Find the largest integer that divides m5 − 5m3 + 4m for all m ≥ 5.. 5! = 120 .

36. [4] Count the number of sequences 1 ≤ a1 ≤ a2 ≤ · · · ≤ a5 of integers with ai ≤ ifor all i.

Solution: C(number of terms) = C(5) = 42 .

37. [5] Alex and Bob have 30 matches. Alex picks up somewhere between one and sixmatches (inclusive), then Bob picks up somewhere between one and six matches, and so on.The player who picks up the last match wins. How many matches should Alex pick up atthe beginning to guarantee that he will be able to win?

Solution: 2 .

38. [9] The cafeteria in a certain laboratory is open from noon until 2 in the afternoonevery Monday for lunch. Two professors eat 15 minute lunches sometime between noonand 2. What is the probability that they are in the cafeteria simultaneously on any givenMonday?

Solution: 1564

.

39. [9] What is the remainder when 22001 is divided by 27 − 1?Solution: 22001(mod7) = 26 = 64 .

40. [5] A product of five primes is of the form ABC,ABC, where A, B, and C representdigits. If one of the primes is 491, find the product ABC,ABC.

Solution: 491 · 1001 · 2 = 982,982 .

41. [4] If(a+ 1

a

)2= 3, find

(a+ 1

a

)3in terms of a.

Solution: 0 .

6

42. [10] Solve x =√x− 1

x+√

1− 1x

for x.

Solution: 1+√

52

.

43. [4] When a single number is added to each member of the seqence 20, 50, 100, thesequence becomes expressable as x, ax, a2x. Find a.

Solution: 53

.

44. [7] Through a point in the interior of a triangle ABC, three lines are drawn, oneparallel to each side. These lines divide the sides of the triangle into three regions each.Let a, b, and c be the lengths of the sides opposite ∠A, ∠B, and ∠C, respectively, and leta′, b′, and c′ be the lengths of the middle regions of the sides opposite ∠A, ∠B, and ∠C,respectively. Find the numerical value of a′/a+ b′/b+ c′/c.

Solution: 1 .

45. [4] A stacking of circles in the plane consists of a base, or some number of unitcircles centered on the x-axis in a row without overlap or gaps, and circles above the x-axisthat must be tangent to two circles below them (so that if the ends of the base were securedand gravity were applied from below, then nothing would move). How many stackings ofcircles in the plane have 4 circles in the base?

Solution: C(4) = 14 .

46. [± 5] Draw a rectangle. Connect the midpoints of the opposite sides to get 4congruent rectangles. Connect the midpoints of the lower right rectangle for a total of 7rectangles. Repeat this process infinitely. Let n be the minimum number of colors we canassign to the rectangles so that no two rectangles sharing an edge have the same color and mbe the minimum number of colors we can assign to the rectangles so that no two rectanglessharing a corner have the same color. Find the ordered pair (n,m).

Solution: (3, 4) .

47. [7] For the sequence of numbers n1, n2, n3, . . . , the relation ni = 2ni−1 + a holds forall i > 1. If n2 = 5 and n8 = 257, what is n5?

Solution: 33 .

48. [8] What is the smallest positive integer x for which x2 + x+ 41 is not a prime?Solution: 40 .

49. [5] If 19

of 60 is 5, what is 120

of 80?

7

Solution: In base 15, 6 .

50. [9] The Fibonacci sequence F1, F2, F3, . . . is defined by F1 = F2 = 1 and Fn+2 =Fn+1 + Fn. Find the least positive integer t such that for all n > 0, Fn = Fn+t.

Solution: 60 .

51. [5] Some people like to write with larger pencils than others. Ed, for instance, likesto write with the longest pencils he can find. However, the halls of MIT are of limited heightL and width L. What is the longest pencil Ed can bring through the halls so that he cannegotiate a square turn?

Solution: 3L .

52. [6] Find all ordered pairs (m,n) of integers such that 231m2 = 130n2.

Solution: The unique solution is (0, 0) .

53. [7] Find the sum of the infinite series 132−12

(112 − 1

32

)+ 1

52−32

(132 − 1

52

)+ 1

72−52

(152 − 1

72

)+

· · · .Solution: 1− 1

3+ 1

3− 1

5+ · · · = 1 .

54. [10] The set of points (x1, x2, x3, x4) in R4 such that x1 ≥ x2 ≥ x3 ≥ x4 is a cone(or hypercone, if you insist). Into how many regions is this cone sliced by the hyperplanesxi − xj = 1 for 1 ≤ i < j ≤ n?

Solution: C(4) = 14 .

55. [7] How many multiples of 7 between 106 and 109 are perfect squares?

Solution:

[√109

72

]−[√

106

72

]= 4517− 142 = 4375 .

56. [6] A triangle has sides of length 888, 925, and x > 0. Find the value of x thatminimizes the area of the circle circumscribed about the triangle.

Solution: 259 .

57. [5] Let x = 20011002 − 2001−1002 and y = 20011002 + 2001−1002. Find x2 − y2.Solution: -4 .

58. [9] Let (x, y) be a point in the cartesian plane, x, y > 0. Find a formula in terms ofx and y for the minimal area of a right triangle with hypotenuse passing through (x, y) andlegs contained in the x and y axes.

8

Solution: 2xy .

59. [10] Trevor and Edward play a game in which they take turns adding or removingbeans from a pile. On each turn, a player must either add or remove the largest perfectsquare number of beans that is in the heap. The player who empties the pile wins. Forexample, if Trevor goes first with a pile of 5 beans, he can either add 4 to make the total9, or remove 4 to make the total 1, and either way Edward wins by removing all the beans.There is no limit to how large the pile can grow; it just starts with some finite number ofbeans in it, say fewer than 1000.

Before the game begins, Edward dispatches a spy to find out how many beans will be inthe opening pile, call this n, then “graciously” offers to let Trevor go first. Knowing that thefirst player is more likely to win, but not knowing n, Trevor logically but unwisely accepts,and Edward goes on to win the game. Find a number n less than 1000 that would promptthis scenario, assuming both players are perfect logicians. A correct answer is worth thenearest integer to log2(n− 4) points.

Solution: The correct answers are 0 (worth imaginary points), 5 (worth 0 points), 20(4 points), 29, 45 (5 points), 80 (6 points), 101, 116, 135, 145, 165, 173 (7 points), 236, 257(8 points), 397, 404, 445, 477, 540, 565, 580, 629, 666 (9 points), 836, 845, 885, 909, 944,949, 954, 975 (10 points). This game is called Epstein’s Put-or-Take-a-Square game. It isunknown whether or not these numbers (or the first player’s win positions) have positivedensity.

60. [∞] Find an n such that n!− (n − 1)! + (n − 2)!− (n − 3)! + · · · ± 1! is prime. Be

prepared to justify your answer for{

n, n≤25

[n+22510 ], n>25

points, where [N ] is the greatest integer

less than N .Solution: 3, 4, 5, 6, 7, 8, 10, 15, 19, 41 (26 points), 59, 61 (28 points), 105 (33 points),

160 (38 points) are the only ones less than or equal to 335. If anyone produces an answerlarger than 335, then we ask for justification to call their bluff. It is not known whether ornot there are infinitely many such n.

9

Team Round SolutionsHarvard-MIT Math Tournament

March 3, 2001

1. How many digits are in the base two representation of 10! (factorial)?

2. On a certain unidirectional highway, trucks move steadily at 60 miles per hour spaced1/4 of a mile apart. Cars move steadily at 75 miles per hour spaced 3 seconds apart. A lonesports car weaving through traffic at a steady forward speed passes two cars between eachtruck it passes. How quickly is it moving in miles per hour?

3. What is the 18th digit after the decimal point of 100009899

?

4. P is a polynomial. When P is divided by x − 1, the remainder is −4. When P isdivided by x − 2, the remainder is −1. When P is divided by x − 3, the remainder is 4.Determine the remainder when P is divided by x3 − 6x2 + 11x− 6.

5. Find all x between −π2

and π2

such that 1− sin4 x− cos2 x = 116

.

6. What is the radius of the smallest sphere in which 4 spheres of radius 1 will fit?

7. The Fibonacci numbers are defined by F1 = F2 = 1 and Fn+2 = Fn+1 + Fn for n ≥ 1.The Lucas numbers are defined by L1 = 1, L2 = 2, and Ln+2 = Ln+1 + Ln for n ≥ 1.

Calculate

15∏n=1

F2nFn

13∏n=1

Ln

.

8. Express sin 10+sin 20+sin 30+sin 40+sin 50+sin 60+sin 70+sin 80cos 5 cos 10 cos 20

without using trigonometric func-tions.

9. Compute∞∑i=1

aiai

for a > 1.

10. Define a monic irreducible polynomial with integral coefficients to be a polynomialwith leading coefficient 1 that cannot be factored, and the prime factorization of a polynomialwith leading coefficient 1 as the factorization into monic irreducible polynomials. How many

not necessarily distinct monic irreducible polynomials are there in the prime factorization of(x8 + x4 + 1)(x8 + x+ 1) (for instance, (x+ 1)2 has two prime factors)?

11. Define a? = (a − 1)/(a + 1) for a 6= −1. Determine all real values N for which(N?)? = tan 15.

12. All subscripts in this problem are to be considered modulo 6, that means for examplethat ω7 is the same as ω1. Let ω1, . . . ω6 be circles of radius r, whose centers lie on a regularhexagon of side length 1. Let Pi be the intersection of ωi and ωi+1 that lies further from thecenter of the hexagon, for i = 1, . . . 6. Let Qi, i = 1 . . . 6, lie on ωi such that Qi, Pi, Qi+1

are colinear. Find the number of possible values of r.

Team Round SolutionsHarvard-MIT Math Tournament

March 3, 2001

1. How many digits are in the base two representation of 10! (factorial)?Solution: We write 10! = 28 · 34 · 52 · 7. The number of digits (base 2) of 10! is equal to

[log2 10!] = 8 + log2(34 · 52 · 7). Since 213 < 32 · 52 · 7 < 214, the number of digits is 8 + 13 =21 .

2. On a certain unidirectional highway, trucks move steadily at 60 miles per hour spaced1/4 of a mile apart. Cars move steadily at 75 miles per hour spaced 3 seconds apart. A lonesports car weaving through traffic at a steady forward speed passes two cars between eachtruck it passes. How quickly is it moving in miles per hour?

Solution: The cars are 1/8 of a mile apart. Consider the reference frame in which thetrucks move at 0 velocity (and the cars move at 15). Call the speed of the sports car in thisreference frame v. The amount of time for the sports car to move from one truck to thenext is 1/4 miles

v, and the amount of time for two regular cars to pass the truck is 1/8 miles

15mph.

Equating these, we get v = 30, and v + 60 = 90 mph.

3. What is the 18th digit after the decimal point of 100009899

?Solution: 10000

9899satisfies 100(x− 1) = 1.01x, so each pair of adjacent digits is generated

by adding the previous two pairs of digits. So the decimal is 1.01020305081321345590 . . . ,and the 18th digit is 5 .

4. P is a polynomial. When P is divided by x − 1, the remainder is −4. When P isdivided by x − 2, the remainder is −1. When P is divided by x − 3, the remainder is 4.Determine the remainder when P is divided by x3 − 6x2 + 11x− 6.

Solution: The remainder polynomial is simply the order two polynomial that goes

through the points (1,−4), (2,−1), and (3, 4): x2 − 5 .

5. Find all x between −π2

and π2

such that 1− sin4 x− cos2 x = 116

.Solution: 1 − sin4 x − cos2 x = 1

16⇒ (16 − 16 cos2 x) − sin4 x − 1 = 0 ⇒ 16 sin4 x −

16 sin2 x + 1 = 0. Use the quadratic formula in sinx to obtain sin2 x = 12±√

34

. Since

cos 2x = 1− 2 sin2 x = ±√

32

, we get x = ± π12,±5π

12.

6. What is the radius of the smallest sphere in which 4 spheres of radius 1 will fit?

Solution: The centers of the smaller spheres lie on a tetrahedron. Let the points of thetetrahedron be (1, 1, 1), (−1,−1, 1), (−1, 1,−1), and (1,−1,−1). These points have distance√

(3) from the center, and√

(2) from each other, so the radius of the smallest sphere in

which 4 spheres of radius√

(2) will fit is√

(2) +√

(3). Scale this to the correct answer by

dividing by√

(2): 2+√

62

.

7. The Fibonacci numbers are defined by F1 = F2 = 1 and Fn+2 = Fn+1 + Fn for n ≥ 1.The Lucas numbers are defined by L1 = 1, L2 = 2, and Ln+2 = Ln+1 + Ln for n ≥ 1.

Calculate

15∏n=1

F2nFn

13∏n=1

Ln

.

Solution: It is easy to show that Ln = F2n

Fn, so the product above is L14L15 = 843 ·

1364 = 1149852 .

8. Express sin 10+sin 20+sin 30+sin 40+sin 50+sin 60+sin 70+sin 80cos 5 cos 10 cos 20

without using trigonometric func-tions.

Solution: We will use the identities cos a + cos b = 2 cos a+b2

cos a−b2

and sin a + sin b =2 sin a+b

2cos a+b

2. The numerator is (sin 10+sin 80)+(sin 20+sin 70)+(sin 30+sin 60)+(sin 40+

sin 60) = 2 sin 45 (cos 35 + cos 25 + cos 15 + cos 35) = 2 sin 45 ((cos 35 + cos 5) + (cos 25 + cos 15)) =

4 sin 45 cos 20 (cos 15 + cos 5) = 8 sin 45 cos 20 cos 10 cos 5, so the fraction equals 8 sin 45 = 4√

2 .

9. Compute∞∑i=1

aiai

for a > 1.

Solution: The sum S = a+ax+ax2 +ax3 + · · · for x < 1 can be determined by realizing

that xS = ax+ ax2 + ax3 + · · · and (1− x)S = a, so S = a1−x . Using this, we have

∞∑i=1

aiai

=

a∞∑i=1

iai

= a[

1a

+ 2a2 + 3

a3 + · · ·]

= a[(

1a

+ 1a2 + 1

a3 + · · ·)

+(

1a2 + 1

a3 + 1a4 + · · ·

)+ · · ·

]=

a[

11−a + 1

a1

1−a + 1a2

11−a + · · ·

]= a

1−a

[1 + 1

a+ 1

a2 + · · ·]

=(

a1−a

)2.

10. Define a monic irreducible polynomial with integral coefficients to be a polynomialwith leading coefficient 1 that cannot be factored, and the prime factorization of a polynomialwith leading coefficient 1 as the factorization into monic irreducible polynomials. How manynot necessarily distinct monic irreducible polynomials are there in the prime factorization of(x8 + x4 + 1)(x8 + x+ 1) (for instance, (x+ 1)2 has two prime factors)?

Solution: x8+x4+1 = (x8+2x4+1)−x4 = (x4+1)2−(x2)2 = (x4−x2+1)(x4+x2+1) =(x4−x2+1)(x2+x+1)(x2−x+1), and x8+x+1 = (x2+x+1)(x6−x5+x3−x2+1). If an integerpolynomial f(x) = anx

n + · · ·+ a0(modp), where p does not divide an, has no zeros, then f

has no rational roots. Taking p = 2, we find x6 − x5 + x3 − x2 + 1 is irreducible. The primefactorization of our polynomial is thus (x4−x2+1)(x2−x+1)(x2+x+1)2(x6−x5+x3−x2+1),so the answer is 5 .

11. Define a? = (a − 1)/(a + 1) for a 6= −1. Determine all real values N for which(N?)? = tan 15.

Solution:Let x = N?. Then (x − 1) cos 15 = (x + 1) sin 15. Squaring and rearranging

terms, and using the fact that cos2 15− sin2 15 = cos 30 =√

32

, we have 3x2 − 4√

3x+ 3 = 0.

Solving, we find that x =√

3 or√

33

. However, we may reject the second root because it

yields a negative value for (N?)?. Therefore x =√

3 and N = 1+x1−x = 1+

√3

1−√

3= −2−

√3 .

12. All subscripts in this problem are to be considered modulo 6, that means for examplethat ω7 is the same as ω1. Let ω1, . . . ω6 be circles of radius r, whose centers lie on a regularhexagon of side length 1. Let Pi be the intersection of ωi and ωi+1 that lies further from thecenter of the hexagon, for i = 1, . . . 6. Let Qi, i = 1 . . . 6, lie on ωi such that Qi, Pi, Qi+1

are colinear. Find the number of possible values of r.Solution: Consider two consecutive circles ωi and ωi+1. Let Qi, Q

′i be two points on ωi

and Qi+1, Q′i+1 on ωi+1 such that Qi, Pi and Qi+1 are colinear and also Q′i, Pi and Q′i+1.Then QiQ

′i = 2∠QiPiQ

′i = 2∠Qi+1PiQ

′i+1 = ∠Qi+1Q

′i+1. Refer to the center of ωi as Oi.

The previous result shows that the lines OiQi and Oi+1Qi+1 meet at the same angle as thelines OiQ

′i and Oi+1Q

′i+1, call this angle ψi. ψi is a function solely of the circles ωi and ωi+1

and the distance between them (we have just showed that any two points Qi and Q′i on ωigive the same value of ψi, so ψi can’t depend on this.) Now, the geometry of ωi and ωi+1

is the same for every i, so ψi is simply a constant ψ which depends only on r. We know6ψ = 0 mod 2π because Q7 = Q1.

We now compute ψ. It suffices to do the computaiton for some specific choice of Qi.Take Qi to be the intersection of OiOi+1 and ωi which is further from Oi+1. We are tocompute the angle between OiQi and Oi+1Qi+1 which is the same as ∠OiOi+1Qi+1. Note thetriangle 4OiPiOi+1 is isosceles, call the base angle ξ. We have ∠OiOi+1Qi+1 = ∠OiOi+1Pi+∠PiOi+1Qi+1 = ξ + (π − 2∠Oi+1PiQi+1) = ξ + (π − 2(π − ∠QiOi+1Pi − ∠PiQiOi+1)) =ξ − π + 2(ξ + (1/2)∠PiOiOi+1) = ξ − π + 2(ξ + (1/2)ξ) = 4ξ − π.

So we get 6(4ξ−π) = 0 mod 2π. Noting that ξ must be acute, ξ = π/12, π/6, π/4, π/3 or 5π/12.r is uniquely determined as (1/2) sec ξ so there are 5 possible values of r.

Harvard-MIT Math TournamentMarch 17, 2002

Individual Subject Test: Advanced Topics

1. Eight knights are randomly placed on a chessboard (not necessarily on distinct squares).A knight on a given square attacks all the squares that can be reached by moving either (1)two squares up or down followed by one squares left or right, or (2) two squares left or rightfollowed by one square up or down. Find the probability that every square, occupied or not,is attacked by some knight.

2. A certain cafeteria serves ham and cheese sandwiches, ham and tomato sandwiches, andtomato and cheese sandwiches. It is common for one meal to include multiple types ofsandwiches. On a certain day, it was found that 80 customers had meals which containedboth ham and cheese; 90 had meals containing both ham and tomatoes; 100 had mealscontaining both tomatoes and cheese. 20 customers’ meals included all three ingredients.How many customers were there?

3. How many four-digit numbers are there in which at least one digit occurs more thanonce?

4. Two fair coins are simultaneously flipped. This is done repeatedly until at least one ofthe coins comes up heads, at which point the process stops. What is the probability thatthe other coin also came up heads on this last flip?

5. Determine the number of subsets S of {1, 2, 3, . . . , 10} with the following property: thereexist integers a < b < c with a ∈ S, b /∈ S, c ∈ S.

6. In how many ways can the numbers 1, 2, . . . , 2002 be placed at the vertices of a regular2002-gon so that no two adjacent numbers differ by more than 2? (Rotations and reflectionsare considered distinct.)

7. A manufacturer of airplane parts makes a certain engine that has a probability p of failingon any given flight. Their are two planes that can be made with this sort of engine, one thathas 3 engines and one that has 5. A plane crashes if more than half its engines fail. Forwhat values of p do the two plane models have the same probability of crashing?

8. Given a 9 × 9 chess board, we consider all the rectangles whose edges lie along gridlines (the board consists of 81 unit squares, and the grid lines lie on the borders of the unitsquares). For each such rectangle, we put a mark in every one of the unit squares insideit. When this process is completed, how many unit squares will contain an even number ofmarks?

9. Given that a, b, c are positive real numbers and loga b + logb c + logc a = 0, find the valueof (loga b)3 + (logb c)3 + (logc a)3.

10. One fair die is rolled; let a denote the number that comes up. We then roll a dice; letthe sum of the resulting a numbers be b. Finally, we roll b dice, and let c be the sum of theresulting b numbers. Find the expected (average) value of c.

1

Harvard-MIT Math TournamentMarch 17, 2002

Individual Subject Test: Advanced Topics

1. Eight knights are randomly placed on a chessboard (not necessarily on distinctsquares). A knight on a given square attacks all the squares that can be reached by movingeither (1) two squares up or down followed by one squares left or right, or (2) two squaresleft or right followed by one square up or down. Find the probability that every square,occupied or not, is attacked by some knight.

Solution: 0 . Since every knight attacks at most eight squares, the event can only occurif every knight attacks exactly eight squares. However, each corner square must be attacked,and some experimentation readily finds that it is impossible to place a knight so as to attacka corner and seven other squares as well.

2. A certain cafeteria serves ham and cheese sandwiches, ham and tomato sandwiches,and tomato and cheese sandwiches. It is common for one meal to include multiple types ofsandwiches. On a certain day, it was found that 80 customers had meals which containedboth ham and cheese; 90 had meals containing both ham and tomatoes; 100 had mealscontaining both tomatoes and cheese. 20 customers’ meals included all three ingredients.How many customers were there?

Solution: 230 . Everyone who ate just one sandwich is included in exactly one of thefirst three counts, while everyone who ate more than one sandwich is included in all fourcounts. Thus, to count each customer exactly once, we must add the first three figures andsubtract the fourth twice: 80 + 90 + 100− 2 · 20 = 230.

3. How many four-digit numbers are there in which at least one digit occurs more thanonce?

Solution: 4464 . There are 9000 four-digit numbers altogether. If we consider howmany four-digit numbers have all their digits distinct, there are 9 choices for the first digit(since we exclude leading zeroes), and then 9 remaining choices for the second digit, then8 for the third, and 7 for the fourth, for a total of 9 · 9 · 8 · 7 = 4536. Thus the remaining9000− 4536 = 4464 numbers have a repeated digit.

4. Two fair coins are simultaneously flipped. This is done repeatedly until at least oneof the coins comes up heads, at which point the process stops. What is the probability thatthe other coin also came up heads on this last flip?

Solution: 1/3 . Let the desired probability be p. There is a 1/4 chance that both coins

will come up heads on the first toss. Otherwise, both can come up heads simultaneouslyonly if both are tails on the first toss, and then the process restarts as if from the beginning;thus this situation occurs with probability p/4. Thus p = 1/4 + p/4; solving, p = 1/3.Alternate Solution: The desired event is equivalent to both coins coming up tails for nsuccessive turns (for some n ≥ 0), then both coins coming up heads. For any fixed value of

1

n, the probability of this occurring is 1/4n+1. Since all these events are disjoint, the totalprobability is 1/4 + 1/42 + 1/43 + · · · = 1/3.

5. Determine the number of subsets S of {1, 2, 3, . . . , 10} with the following property:there exist integers a < b < c with a ∈ S, b /∈ S, c ∈ S.

Solution: 968 There are 210 = 1024 subsets of {1, 2, . . . , 10} altogether. Any subsetwithout the specified property must be either the empty set or a block of consecutive integers.To specify a block of consecutive integers, we either have just one element (10 choices) or apair of distinct endpoints (

(102

)= 45 choices). So the number of sets with our property is

1024− (1 + 10 + 45) = 968.

6. In how many ways can the numbers 1, 2, . . . , 2002 be placed at the vertices of a regular2002-gon so that no two adjacent numbers differ by more than 2? (Rotations and reflectionsare considered distinct.)

Solution: 4004 . There are 2002 possible positions for the 1. The two numbers adjacentto the 1 must be 2 and 3; there are two possible ways of placing these. The positions ofthese numbers uniquely determine the rest: for example, if 3 lies clockwise from 1, then thenumber lying counterclockwise from 2 must be 4; the number lying clockwise from 3 mustbe 5; the number lying counterclockwise from 4 must now be 6; and so forth. Eventually,2002 is placed adjacent to 2000 and 2001, so we do get a valid configuration. Thus there are2002 · 2 possible arrangements.

7. A manufacturer of airplane parts makes a certain engine that has a probability p offailing on any given flight. Their are two planes that can be made with this sort of engine,one that has 3 engines and one that has 5. A plane crashes if more than half its engines fail.For what values of p do the two plane models have the same probability of crashing?

Solution: They have the same probability of failing if(52

)p3(1−p)2 +

(51

)p4(1−p)+p5 =(

31

)p2(1−p)+p3, which is true iff p2(6p3−15p2 +12p−3) = 0. This is clearly true for p = 0.

We know it is true for p = 1, since both probabilities would be 1 in this case, so we knowp − 1 is a factor of 6p3 − 15p2 + 12p − 3. Thus, factoring gives that the engines have thesame probability of failing if p2(p − 1)(6p2 − 9p + 3) = 0. By the quadratic formula (or by

factoring), the quadratic has roots p = 12, 1, so the answer is 0, 1

2, 1 .

8. Given a 9 × 9 chess board, we consider all the rectangles whose edges lie along gridlines (the board consists of 81 unit squares, and the grid lines lie on the borders of the unitsquares). For each such rectangle, we put a mark in every one of the unit squares insideit. When this process is completed, how many unit squares will contain an even number ofmarks?

Solution: 56 . Consider the rectangles which contain the square in the ith row and jthcolumn. There are i possible positions for the upper edge of such a rectangle, 10− i for thelower edge, j for the left edge, and 10− j for the right edge; thus we have i(10− i)j(10− j)rectangles altogether, which is odd iff i, j are both odd, i.e. iff i, j ∈ {1, 3, 5, 7, 9}. There arethus 25 unit squares which lie in an odd number of rectangles, so the answer is 81−25 = 56.

2

9. Given that a, b, c are positive real numbers and loga b + logb c + logc a = 0, find thevalue of (loga b)3 + (logb c)3 + (logc a)3.

Solution: 3 . Let x = loga b and y = logb c; then logc a = −(x + y). Thus we want tocompute the value of x3 + y3− (x + y)3 = −3x2y− 3xy2 = −3xy(x + y). On the other hand,−xy(x + y) = (loga b)(logb c)(logc a) = 1, so the answer is 3.

10. One fair die is rolled; let a denote the number that comes up. We then roll a dice;let the sum of the resulting a numbers be b. Finally, we roll b dice, and let c be the sum ofthe resulting b numbers. Find the expected (average) value of c.

Solution: 343/8 The expected result of an individual die roll is (1+2+3+4+5+6)/6 =

7/2. For any particular value of b, if b dice are rolled independently, then the expected sumis (7/2)b. Likewise, when we roll a dice, the expected value of their sum b is (7/2)a, so theexpected value of c is (7/2)2a. Similar reasoning again shows us that the expected value ofa is 7/2 and so the expected value of c overall is (7/2)3 = 343/8.

3

Harvard-MIT Math TournamentMarch 17, 2002

Individual Subject Test: Algebra

1. Nine nonnegative numbers have average 10. What is the greatest possible value for theirmedian?

2. p and q are primes such that the numbers p + q and p + 7q are both squares. Find thevalue of p.

3. Real numbers a, b, c satisfy the equations a + b + c = 26, 1/a + 1/b + 1/c = 28. Find thevalue of

a

b+

b

c+

c

a+

a

c+

c

b+

b

a.

4. If a positive integer multiple of 864 is picked randomly, with each multiple having thesame probability of being picked, what is the probability that it is divisible by 1944?

5. Find the greatest common divisor of the numbers 2002 + 2, 20022 + 2, 20023 + 2, . . ..

6. Find the sum of the even positive divisors of 1000.

7. The real numbers x, y, z, w satisfy

2x + y + z + w = 1

x + 3y + z + w = 2

x + y + 4z + w = 3

x + y + z + 5w = 25.

Find the value of w.

8. Determine the value of the sum

3

12 · 22+

5

22 · 32+

7

32 · 42+ · · ·+ 29

142 · 152.

9. For any positive integer n, let f(n) denote the number of 1’s in the base-2 representationof n. For how many values of n with 1 ≤ n ≤ 2002 do we have f(n) = f(n + 1)?

10. Determine the value of

2002 +1

2(2001 +

1

2(2000 + · · ·+ 1

2(3 +

1

2· 2)) · · · ).

1

Harvard-MIT Math TournamentMarch 17, 2002

Individual Subject Test: Algebra

1. Nine nonnegative numbers have average 10. What is the greatest possible value fortheir median?

Solution: 18 If the median is m, then the five highest numbers are all at least m, sothe sum of all the numbers is at least 5m. Thus 90 ≥ 5m ⇒ m ≤ 18. Conversely, we canachieve m = 18 by taking four 0’s and five 18’s.

2. p and q are primes such that the numbers p + q and p + 7q are both squares. Findthe value of p.

Solution: 2 . Writing x2 = p + q, y2 = p + 7q, we have 6q = y2 − x2 = (y − x)(y + x).Since 6q is even, one of the factors y − x, y + x is even, and then the other is as well; thus6q is divisible by 4 ⇒ q is even ⇒ q = 2 and 6q = 12. We may assume x, y are both takento be positive; then we must have y − x = 2, y + x = 6⇒ x = 2, so p + 2 = 22 = 4⇒ p = 2also.

3. Real numbers a, b, c satisfy the equations a + b + c = 26, 1/a + 1/b + 1/c = 28. Findthe value of

a

b+

b

c+

c

a+

a

c+

c

b+

b

a.

Solution: 725 Multiplying the two given equations gives

a

a+

a

b+

a

c+

b

a+

b

b+

b

c+

c

a+

c

b+

c

c= 26 · 28 = 728,

and subtracting 3 from both sides gives the answer, 725.

4. If a positive integer multiple of 864 is picked randomly, with each multiple having thesame probability of being picked, what is the probability that it is divisible by 1944?

Solution: The probability that a multiple of 864 = 2533 is divisible by 2744 = 2335 isthe same as the probability that a multiple of 22 is divisible by 32, which since 4 and 9 are

relatively prime is 19

.

5. Find the greatest common divisor of the numbers 2002 + 2, 20022 + 2, 20023 + 2, . . ..Solution: 6 . Notice that 2002+2 divides 20022−22, so any common divisor of 2002+2

and 20022 + 2 must divide (20022 + 2)− (20022− 22) = 6. On the other hand, every numberin the sequence is even, and the nth number is always congruent to 1n + 2 ≡ 0 modulo 3.Thus, 6 divides every number in the sequence.

6. Find the sum of the even positive divisors of 1000.

1

Solution: 2184 . Notice that 2k is a divisor of 1000 iff k is a divisor of 500, so we needonly find the sum of the divisors of 500 and multiply by 2. This can be done by enumeratingthe divisors individually, or simply by using the formula: σ(22 · 53) = (1 + 2 + 22)(1 + 5 +52 + 53) = 1092, and then doubling gives 2184. Alternate Solution: The sum of all thedivisors of 1000 is (1 + 2 + 22 + 23)(1 + 5 + 52 + 53) = 2340. The odd divisors of 1000 aresimply the divisors of 125, whose sum is 1 + 5 + 52 + 53 = 156; subtracting this from 2340,we are left with the sum of the even divisors of 1000, which is 2184.

7. The real numbers x, y, z, w satisfy

2x + y + z + w = 1

x + 3y + z + w = 2

x + y + 4z + w = 3

x + y + z + 5w = 25.

Find the value of w.Solution: 11/2 . Multiplying the four equations by 12, 6, 4, 3 respectively, we get

24x + 12y + 12z + 12w = 12

6x + 18y + 6z + 6w = 12

4x + 4y + 16z + 4w = 12

3x + 3y + 3z + 15w = 75.

Adding these yields 37x + 37y + 37z + 37w = 111, or x + y + z + w = 3. Subtract this fromthe fourth given equation to obtain 4w = 22, or w = 11/2.

8. Determine the value of the sum

3

12 · 22+

5

22 · 32+

7

32 · 42+ · · ·+ 29

142 · 152.

Solution: 224/225 The sum telescopes as( 1

12− 1

22

)+( 1

22− 1

32

)+ · · ·+

( 1

142− 1

152

)=

1

12− 1

152=

224

225.

9. For any positive integer n, let f(n) denote the number of 1’s in the base-2 represen-tation of n. For how many values of n with 1 ≤ n ≤ 2002 do we have f(n) = f(n + 1)?

Solution: 501 . If n is even, then n + 1 is obtained from n in binary by changing thefinal 0 to a 1; thus f(n + 1) = f(n) + 1. If n is odd, then n + 1 is obtained by changingthe last 0 to a 1, the ensuing string of 1s to 0s, and then changing the next rightmost 0 toa 1. This produces no net change in the number of 1’s iff n ends in 01 in base 2. Thus,

2

f(n + 1) = f(n) if and only if n is congruent to 1 mod 4, and there are 501 such numbers inthe specified range.

10. Determine the value of

2002 +1

2(2001 +

1

2(2000 + · · ·+ 1

2(3 +

1

2· 2)) · · · ).

Solution: 4002 . We can show by induction that n+12([n−1]+1

2(· · ·+1

2·2) · · · ) = 2(n−1).

For n = 3 we have 3 + 12· 2 = 4, giving the base case, and if the result holds for n, then

(n + 1) + 122(n− 1) = 2n = 2(n + 1)− 2. Thus the claim holds, and now plug in n = 2002.

Alternate Solution: Expand the given expression as 2002+2001/2+2000/22+· · ·+2/22000.Letting S denote this sum, we have S/2 = 2002/2 + 2001/22 + · · · + 2/22001, so S − S/2 =2002 − (1/2 + 1/4 + · · · + 1/22000) − 2/22001 = 2002 − (1 − 1/22000) − 1/22000 = 2001, soS = 4002.

3

Harvard-MIT Math TournamentMarch 17, 2002

Individual Subject Test: Calculus

1. Two circles have centers that are d units apart, and each has diameter√

d. For any d,let A(d) be the area of the smallest circle that contains both of these circles. Find lim

d→∞A(d)d2 .

2. Find limh→0

x2−(x+h)2

h.

3. We are given the values of the differentiable real functions f, g, h, as well as the derivativesof their pairwise products, at x = 0:

f(0) = 1; g(0) = 2; h(0) = 3; (gh)′(0) = 4; (hf)′(0) = 5; (fg)′(0) = 6.

Find the value of (fgh)′(0).

4. Find the area of the region in the first quadrant x > 0, y > 0 bounded above the graphof y = arcsin(x) and below the graph of the y = arccos(x).

5. What is the minimum vertical distance between the graphs of 2 + sin(x) and cos(x)?

6. Determine the positive value of a such that the parabola y = x2 + 1 bisects the area ofthe rectangle with vertices (0, 0), (a, 0), (0, a2 + 1), and (a, a2 + 1).

7. Denote by 〈x〉 the fractional part of the real number x (for instance, 〈3.2〉 = 0.2). Apositive integer N is selected randomly from the set {1, 2, 3, . . . ,M}, with each integer havingthe same probability of being picked, and

⟨87303

N⟩

is calculated. This procedure is repeatedM times and the average value A(M) is obtained. What is lim

M→∞A(M)?

8. Evaluate(√

2−1)/2∫0

dx(2x+1)

√x2+x

.

9. Suppose f is a differentiable real function such that f(x) + f ′(x) ≤ 1 for all x, andf(0) = 0. What is the largest possible value of f(1)? (Hint: consider the function exf(x).)

10. A continuous real function f satisfies the identity f(2x) = 3f(x) for all x. If∫ 1

0f(x) dx =

1, what is∫ 2

1f(x) dx?

1

Harvard-MIT Math TournamentMarch 17, 2002

Individual Subject Test: Calculus

1. Two circles have centers that are d units apart, and each has diameter√

d. For any d,let A(d) be the area of the smallest circle that contains both of these circles. Find lim

d→∞A(d)d2 .

Solution: This equals limd→∞

π(

d+√

d2

)2

d2 = π4

.

2. Find limh→0

x2−(x+h)2

h.

Solution: This equals limh→0

x2−x2−h2−2hxh

= limh→0−h−2x = −2x . Alternate Solution: This

is the definition of the derivative of −x2 with respect to x, which is −2x.

3. We are given the values of the differentiable real functions f, g, h, as well as thederivatives of their pairwise products, at x = 0:

f(0) = 1; g(0) = 2; h(0) = 3; (gh)′(0) = 4; (hf)′(0) = 5; (fg)′(0) = 6.

Find the value of (fgh)′(0).Solution: 16 By the product rule, (fgh)′ = f ′gh + fg′h + fgh′ = ((fg)′h + (gh)′f +

(hf)′g)/2. Evaluated at 0, this gives 16.

4. Find the area of the region in the first quadrant x > 0, y > 0 bounded above thegraph of y = arcsin(x) and below the graph of the y = arccos(x).

Solution: We can integrate over y rather than x. In particular, the solution isπ/4∫0

sin y dy+

π/2∫π/4

cos y dy =(1− 1√

2

)2 = 2−

√2 .

5. What is the minimum vertical distance between the graphs of 2 + sin(x) and cos(x)?Solution: The derivative of 2 + sin(x) − cos(x) is cosx + sinx, which in the interval

0 ≤ x < 2π is zero at x = 3π4

, 7π4

. At 7π4

, when sin(x) is negative and cos(x) is positive, the

distance reaches its minimal value of 2−√

2 .

6. Determine the positive value of a such that the parabola y = x2 + 1 bisects the areaof the rectangle with vertices (0, 0), (a, 0), (0, a2 + 1), and (a, a2 + 1).

Solution:√

3 The area of the rectangle is a3 + a. The portion under the parabola has

area∫ a

0x2 + 1dx = a3/3 + a. Thus we wish to solve the equation a3 + a = 2(a3/3 + a);

dividing by a and rearranging gives a2/3 = 1, so a =√

3.

1

7. Denote by 〈x〉 the fractional part of the real number x (for instance, 〈3.2〉 = 0.2).A positive integer N is selected randomly from the set {1, 2, 3, . . . ,M}, with each integerhaving the same probability of being picked, and

⟨87303

N⟩

is calculated. This procedure isrepeated M times and the average value A(M) is obtained. What is lim

M→∞A(M)?

Solution: This method of picking N is equivalent to uniformly randomly selecting apositive integer. Call this the average value of

⟨87303

N⟩

for N a positive integer. In lowestterms, 87

303= 29

101, so the answer is the same as the average value of 0

101, 1

101, . . . , 100

101, which is

1+2+···+100101·101

= 100·101/2101·101

= 50101

.

8. Evaluate(√

2−1)/2∫0

dx(2x+1)

√x2+x

.

Solution: Let u =√

x2 + x. Then du = 2x+12√

x2+xdx. So the integral becomes 2

∫du

(4x2+4x+1),

or 2∫

du4u2+1

. This is tan−1(2u), yielding a final answer of tan−1(2√

x2 + x) + C for the in-

definite integral. The definite integral becomes tan−1(1)− tan−1(0) = π4

.

9. Suppose f is a differentiable real function such that f(x) + f ′(x) ≤ 1 for all x, andf(0) = 0. What is the largest possible value of f(1)? (Hint: consider the function exf(x).)

Solution: 1− 1/e Let g(x) = exf(x); then g′(x) = ex(f(x) + f ′(x)) ≤ ex. Integrating

from 0 to 1, we have g(1)− g(0) =∫ 1

0g′(x)dx ≤

∫ 1

0exdx = e− 1. But g(1)− g(0) = e · f(1),

so we get f(1) ≤ (e − 1)/e. This maximum is attained if we actually have g′(x) = ex

everywhere; this entails the requirement f(x) + f ′(x) = 1, which is met by f(x) = 1− e−x.

10. A continuous real function f satisfies the identity f(2x) = 3f(x) for all x. If∫ 1

0f(x) dx = 1, what is

∫ 2

1f(x) dx?

Solution: 5 Let S =∫ 2

1f(x) dx. By setting u = 2x, we see that

∫ 1

1/2f(x) dx =∫ 1

1/2f(2x)/3 dx =

∫ 2

1f(u)/6 du = S/6. Similarly,

∫ 1/2

1/4f(x) dx = S/36, and in general∫ 1/2n−1

1/2n f(x) dx = S/6n. Adding finitely many of these, we have∫ 1

1/2n f(x) dx = S/6+S/36+

· · ·+ S/6n = S · (1− 1/6n)/5. Taking the limit as n→∞, we have∫ 1

0f(x) dx = S/5. Thus

S = 5, the answer.

2

Harvard-MIT Math TournamentMarch 17, 2002

Individual General Test: Part 1

1. What is the maximum number of lattice points (i.e. points with integer coordinates) inthe plane that can be contained strictly inside a circle of radius 1?

2. Eight knights are randomly placed on a chessboard (not necessarily on distinct squares).A knight on a given square attacks all the squares that can be reached by moving either (1)two squares up or down followed by one squares left or right, or (2) two squares left or rightfollowed by one square up or down. Find the probability that every square, occupied or not,is attacked by some knight.

3. How many triples (A, B, C) of positive integers (positive integers are the numbers1, 2, 3, 4, . . .) are there such that A + B + C = 10, where order does not matter (for in-stance the triples (2, 3, 5) and (3, 2, 5) are considered to be the same triple) and where twoof the integers in a triple could be the same (for instance (3, 3, 4) is a valid triple).

4. We call a set of professors and committees on which they serve a university if(1) given two distinct professors there is one and only one committee on which they both

serve,(2) given any committee, C, and any professor, P , not on that committee, there is exactly

one committee on which P serves and no professors on committee C serve, and(3) there are at least two professors on each committee; there are at least two committees.

What is the smallest number of committees a university can have?

5. A square and a regular hexagon are drawn with the same side length. If the area of thesquare is

√3, what is the area of the hexagon?

6. A man, standing on a lawn, is wearing a circular sombrero of radius 3 feet. Unfortunately,the hat blocks the sunlight so effectively that the grass directly under it dies instantly. Ifthe man walks in a circle of radius 5 feet, what area of dead grass will result?

7. A circle is inscribed in a square dartboard. If a dart is thrown at the dartboard and hitsthe dartboard in a random location, with all locations having the same probability of beinghit, what is the probability that it lands within the circle?

8. Count the number of triangles with positive area whose vertices are points whose (x, y)-coordinates lie in the set {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}.

9. Real numbers a, b, c satisfy the equations a + b + c = 26, 1/a + 1/b + 1/c = 28. Find thevalue of

a

b+

b

c+

c

a+

a

c+

c

b+

b

a.

10. A certain cafeteria serves ham and cheese sandwiches, ham and tomato sandwiches,and tomato and cheese sandwiches. It is common for one meal to include multiple types ofsandwiches. On a certain day, it was found that 80 customers had meals which containedboth ham and cheese; 90 had meals containing both ham and tomatoes; 100 had mealscontaining both tomatoes and cheese. 20 customers’ meals included all three ingredients.How many customers were there?

1

Harvard-MIT Math TournamentMarch 17, 2002

Individual General Test: Part 1

1. What is the maximum number of lattice points (i.e. points with integer coordinates)in the plane that can be contained strictly inside a circle of radius 1?

Solution: 4 . The circle centered at (1/2, 1/2) shows that 4 is achievable. On the otherhand, no two points within the circle can be at a mutual distance of 2 or greater. If thereare more than four lattice points, classify all such points by the parity of their coordinates:(even, even), (even, odd), (odd, even), or (odd, odd). Then some two points lie in the sameclass. Since they are distinct, this means either their first or second coordinates must differby at least 2, so their distance is at least 2, a contradiction.

2. Eight knights are randomly placed on a chessboard (not necessarily on distinctsquares). A knight on a given square attacks all the squares that can be reached by movingeither (1) two squares up or down followed by one squares left or right, or (2) two squaresleft or right followed by one square up or down. Find the probability that every square,occupied or not, is attacked by some knight.

Solution: 0 . Since every knight attacks at most eight squares, the event can only occurif every knight attacks exactly eight squares. However, each corner square must be attacked,and some experimentation readily finds that it is impossible to place a knight so as to attacka corner and seven other squares as well.

3. How many triples (A, B, C) of positive integers (positive integers are the numbers1, 2, 3, 4, . . .) are there such that A + B + C = 10, where order does not matter (for instancethe triples (2, 3, 5) and (3, 2, 5) are considered to be the same triple) and where two of theintegers in a triple could be the same (for instance (3, 3, 4) is a valid triple).

Solution: The triples can merely be enumerated: (1, 1, 8), (1, 2, 7), (1, 3, 6), (1, 4, 5),(2, 2, 6), (2, 3, 5), (2, 4, 4), and (3, 3, 4). There are 8 elements.

4. We call a set of professors and committees on which they serve a university if(1) given two distinct professors there is one and only one committee on which they both

serve,(2) given any committee, C, and any professor, P , not on that committee, there is exactly

one committee on which P serves and no professors on committee C serve, and(3) there are at least two professors on each committee; there are at least two committees.What is the smallest number of committees a university can have?Solution: Let C be any committee. Then there exists a professor P not on C (or else

there would be no other committees). By axiom 2, P serves on a committee D having nocommon members with C. Each of these committees has at least two members, and foreach Q ∈ C, R ∈ D, there exists (by axiom 1) a committee containing Q and R, which(again by axiom 1) has no other common members with C or D. Thus we have at least

1

2 + 2 · 2 = 6 committees. This minimum is attainable - just take four professors and let anytwo professors form a committee.

5. A square and a regular hexagon are drawn with the same side length. If the area ofthe square is

√3, what is the area of the hexagon?

Solution: The hexagon is composed of six equilateral triangles each of side length 4√

3

(with base b = 4√

3 and height√

32

b), so the total area is 92

.

6. A man, standing on a lawn, is wearing a circular sombrero of radius 3 feet. Unfortu-nately, the hat blocks the sunlight so effectively that the grass directly under it dies instantly.If the man walks in a circle of radius 5 feet, what area of dead grass will result?

Solution: 60π ft2 Let O be the center of the man’s circular trajectory. The sombrerokills all the grass that is within 3 feet of any point that is 5 feet away from O — i.e. allthe grass at points P with 2 ≤ OP ≤ 8. The area of this annulus is then π(82 − 22) = 60πsquare feet.

7. A circle is inscribed in a square dartboard. If a dart is thrown at the dartboard andhits the dartboard in a random location, with all locations having the same probability ofbeing hit, what is the probability that it lands within the circle?

Solution: The answer is the area of the circle over the area of the square, which is π4

.

8. Count the number of triangles with positive area whose vertices are points whose(x, y)-coordinates lie in the set {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}.

Solution: There are(93

)= 84 triples of points. 8 of them form degenerate triangles (the

ones that lie on a line), so there are 84− 8 = 76 nondegenerate triangles.

9. Real numbers a, b, c satisfy the equations a + b + c = 26, 1/a + 1/b + 1/c = 28. Findthe value of

a

b+

b

c+

c

a+

a

c+

c

b+

b

a.

Solution: 725 Multiplying the two given equations gives

a

a+

a

b+

a

c+

b

a+

b

b+

b

c+

c

a+

c

b+

c

c= 26 · 28 = 728,

and subtracting 3 from both sides gives the answer, 725.

10. A certain cafeteria serves ham and cheese sandwiches, ham and tomato sandwiches,and tomato and cheese sandwiches. It is common for one meal to include multiple types ofsandwiches. On a certain day, it was found that 80 customers had meals which containedboth ham and cheese; 90 had meals containing both ham and tomatoes; 100 had mealscontaining both tomatoes and cheese. 20 customers’ meals included all three ingredients.How many customers were there?

2

Solution: 230 . Everyone who ate just one sandwich is included in exactly one of thefirst three counts, while everyone who ate more than one sandwich is included in all fourcounts. Thus, to count each customer exactly once, we must add the first three figures andsubtract the fourth twice: 80 + 90 + 100− 2 · 20 = 230.

3

Harvard-MIT Math TournamentMarch 17, 2002

Individual General Test: Part 2

1. The squares of a chessboard are numbered from left to right and top to bottom (so thatthe first row reads 1, 2, . . . , 8, the second reads 9, 10, . . . , 16, and so forth). The number 1 ison a black square. How many black squares contain odd numbers?

2. You are in a completely dark room with a drawer containing 10 red, 20 blue, 30 green,and 40 khaki socks. What is the smallest number of socks you must randomly pull out inorder to be sure of having at least one of each color?

3. Solve for x in 3 =

√x +

√x +√

x + · · ·.

4. Dan is holding one end of a 26 inch long piece of light string that has a heavy bead on itwith each hand (so that the string lies along two straight lines). If he starts with his handstogether at the start and leaves his hands at the same height, how far does he need to pullhis hands apart so that the bead moves upward by 8 inches?

5. A square and a regular hexagon are drawn with the same side length. If the area of thesquare is

√3, what is the area of the hexagon?

6. Nine nonnegative numbers have average 10. What is the greatest possible value for theirmedian?

7. p and q are primes such that the numbers p + q and p + 7q are both squares. Find thevalue of p.

8. Two fair coins are simultaneously flipped. This is done repeatedly until at least one ofthe coins comes up heads, at which point the process stops. What is the probability thatthe other coin also came up heads on this last flip?

9. A and B are two points on a circle with center O, and C lies outside the circle, on rayAB. Given that AB = 24, BC = 28, OA = 15, find OC.

10. How many four-digit numbers are there in which at least one digit occurs more thanonce?

1

Harvard-MIT Math TournamentMarch 17, 2002

Individual General Test: Part 2

1. The squares of a chessboard are numbered from left to right and top to bottom (sothat the first row reads 1, 2, . . . , 8, the second reads 9, 10, . . . , 16, and so forth). The number1 is on a black square. How many black squares contain odd numbers?

Solution: 16 . The black squares in the nth row contain odd numbers when n is oddand even numbers when n is even; thus there are four rows where the black squares containodd numbers, and each such row contributes four black squares.

2. You are in a completely dark room with a drawer containing 10 red, 20 blue, 30 green,and 40 khaki socks. What is the smallest number of socks you must randomly pull out inorder to be sure of having at least one of each color?

Solution: 91 . The maximum number of socks that can be pulled out without repre-senting every color is 20 blue + 30 green + 40 khaki = 90, so 91 is the minimum needed toensure that this doesn’t happen.

3. Solve for x in 3 =

√x +

√x +√

x + · · ·.Solution: Squaring both sides and subtracting x from both sides, we get 9− x = 3, or

x = 6 .

4. Dan is holding one end of a 26 inch long piece of light string that has a heavy beadon it with each hand (so that the string lies along two straight lines). If he starts with hishands together at the start and leaves his hands at the same height, how far does he needto pull his hands apart so that the bead moves upward by 8 inches?

Solution: After he pulls the bead is 5 inches below his hands, and it is 13 inches fromeach hand. Using the Pythagorean theorem, his hands must be 2 · 12 = 24 inches apart.

5. A square and a regular hexagon are drawn with the same side length. If the area ofthe square is

√3, what is the area of the hexagon?

Solution: The hexagon is composed of six equilateral triangles each of side length 4√

3

(with base b = 4√

3 and height√

32

b), so the total area is 92

.

6. Nine nonnegative numbers have average 10. What is the greatest possible value fortheir median?

Solution: 18 If the median is m, then the five highest numbers are all at least m, sothe sum of all the numbers is at least 5m. Thus 90 ≥ 5m ⇒ m ≤ 18. Conversely, we canachieve m = 18 by taking four 0’s and five 18’s.

7. p and q are primes such that the numbers p + q and p + 7q are both squares. Findthe value of p.

1

Solution: 2 . Writing x2 = p + q, y2 = p + 7q, we have 6q = y2 − x2 = (y − x)(y + x).Since 6q is even, one of the factors y − x, y + x is even, and then the other is as well; thus6q is divisible by 4 ⇒ q is even ⇒ q = 2 and 6q = 12. We may assume x, y are both takento be positive; then we must have y − x = 2, y + x = 6⇒ x = 2, so p + 2 = 22 = 4⇒ p = 2also.

8. Two fair coins are simultaneously flipped. This is done repeatedly until at least oneof the coins comes up heads, at which point the process stops. What is the probability thatthe other coin also came up heads on this last flip?

Solution: 1/3 . Let the desired probability be p. There is a 1/4 chance that both coins

will come up heads on the first toss. Otherwise, both can come up heads simultaneouslyonly if both are tails on the first toss, and then the process restarts as if from the beginning;thus this situation occurs with probability p/4. Thus p = 1/4 + p/4; solving, p = 1/3.Alternate Solution: The desired event is equivalent to both coins coming up tails for nsuccessive turns (for some n ≥ 0), then both coins coming up heads. For any fixed value ofn, the probability of this occurring is 1/4n+1. Since all these events are disjoint, the totalprobability is 1/4 + 1/42 + 1/43 + · · · = 1/3.

9. A and B are two points on a circle with center O, and C lies outside the circle, onray AB. Given that AB = 24, BC = 28, OA = 15, find OC.

Solution: 41 . Let M be the midpoint of AB; then 4OMB is a right triangle withOB = 15, MB = 12, so OM = 9. Now 4OMC is a right triangle with OM = 9, MC = 40,so OC = 41.

10. How many four-digit numbers are there in which at least one digit occurs more thanonce?

Solution: 4464 . There are 9000 four-digit numbers altogether. If we consider howmany four-digit numbers have all their digits distinct, there are 9 choices for the first digit(since we exclude leading zeroes), and then 9 remaining choices for the second digit, then8 for the third, and 7 for the fourth, for a total of 9 · 9 · 8 · 7 = 4536. Thus the remaining9000− 4536 = 4464 numbers have a repeated digit.

2

Harvard-MIT Math TournamentMarch 17, 2002

Individual Subject Test: Geometry

1. A man, standing on a lawn, is wearing a circular sombrero of radius 3 feet. Unfortunately,the hat blocks the sunlight so effectively that the grass directly under it dies instantly. Ifthe man walks in a circle of radius 5 feet, what area of dead grass will result?

2. Dan is holding one end of a 26 inch long piece of light string that has a heavy bead onit with each hand (so that the string lies along straight lines). If he starts with his handstogether at the start and leaves his hands at the same height, how far does he need to pullhis hands apart so that the bead moves upward by 8 inches?

3. A square and a regular hexagon are drawn with the same side length. If the area of thesquare is

√3, what is the area of the hexagon?

4. We call a set of professors and committees on which they serve a university if(1) given two distinct professors there is one and only one committee on which they both

serve,(2) given any committee, C, and any professor, P , not on that committee, there is exactly

one committee on which P serves and no professors on committee C serve, and(3) there are at least two professors on each committee; there are at least two committees.

What is the smallest number of committees a university can have?

5. Consider a square of side length 1. Draw four lines that each connect a midpoint of aside with a corner not on that side, such that each midpoint and each corner is touched byonly one line. Find the area of the region completely bounded by these lines.

6. If we pick (uniformly) a random square of area 1 with sides parallel to the x− and y−axesthat lies entirely within the 5-by-5 square bounded by the lines x = 0, x = 5, y = 0, y = 5(the corners of the square need not have integer coordinates), what is the probability thatthe point (x, y) = (4.5, 0.5) lies within the square of area 1?

7. Equilateral triangle ABC of side length 2 is drawn. Three squares external to thetriangle, ABDE, BCFG, and CAHI, are drawn. What is the area of the smallest trianglethat contains these squares?

8. Equilateral triangle ABC of side length 2 is drawn. Three squares containing the triangle,ABDE, BCFG, and CAHI, are drawn. What is the area of the smallest triangle thatcontains these squares?

9. A and B are two points on a circle with center O, and C lies outside the circle, on rayAB. Given that AB = 24, BC = 28, OA = 15, find OC.

10. Let 4ABC be equilateral, and let D, E, F be points on sides BC, CA, AB respectively,with FA = 9, AE = EC = 6, CD = 4. Determine the measure (in degrees) of ∠DEF .

1

Harvard-MIT Math TournamentMarch 17, 2002

Individual Subject Test: Geometry

1. A man, standing on a lawn, is wearing a circular sombrero of radius 3 feet. Unfortu-nately, the hat blocks the sunlight so effectively that the grass directly under it dies instantly.If the man walks in a circle of radius 5 feet, what area of dead grass will result?

Solution: 60π ft2 Let O be the center of the man’s circular trajectory. The sombrerokills all the grass that is within 3 feet of any point that is 5 feet away from O — i.e. allthe grass at points P with 2 ≤ OP ≤ 8. The area of this annulus is then π(82 − 22) = 60πsquare feet.

2. Dan is holding one end of a 26 inch long piece of light string that has a heavy beadon it with each hand (so that the string lies along straight lines). If he starts with his handstogether at the start and leaves his hands at the same height, how far does he need to pullhis hands apart so that the bead moves upward by 8 inches?

Solution: After he pulls the bead is 5 inches below his hands, and it is 13 inches fromeach hand. Using the Pythagorean theorem, his hands must be 2 · 12 = 24 inches apart.

3. A square and a regular hexagon are drawn with the same side length. If the area ofthe square is

√3, what is the area of the hexagon?

Solution: The hexagon is composed of six equilateral triangles each of side length 4√

3

(with base b = 4√

3 and height√

32

b), so the total area is 92

.

4. We call a set of professors and committees on which they serve a university if(1) given two distinct professors there is one and only one committee on which they both

serve,(2) given any committee, C, and any professor, P , not on that committee, there is exactly

one committee on which P serves and no professors on committee C serve, and(3) there are at least two professors on each committee; there are at least two committees.What is the smallest number of committees a university can have?Solution: Let C be any committee. Then there exists a professor P not on C (or else

there would be no other committees). By axiom 2, P serves on a committee D having nocommon members with C. Each of these committees has at least two members, and foreach Q ∈ C, R ∈ D, there exists (by axiom 1) a committee containing Q and R, which(again by axiom 1) has no other common members with C or D. Thus we have at least2 + 2 · 2 = 6 committees. This minimum is attainable - just take four professors and let anytwo professors form a committee.

5. Consider a square of side length 1. Draw four lines that each connect a midpoint of aside with a corner not on that side, such that each midpoint and each corner is touched byonly one line. Find the area of the region completely bounded by these lines.

1

Solution: In unit square ABCD, denote by E, F, G, H the respective midpoints of sidesAB, BC, CD,DA. Let I be the intersection of AF and DE, let J be the intersection of BGand AF , let K be the intersection of CH and BG, and let L be the intersection of DE andCH. We want to find the area of square IJKL. The area of ABF is 1

4, which is equal to

12AF · BJ =

√5

4BJ , so BJ = 1√

5. Using similar triangles, GK = JF = 1

2BJ . Thus the

length of a side of IJKL is JK =√

52− 1√

5− 1

21√5

= 1√5, and the area of IJKL is 1

5.

6. If we pick (uniformly) a random square of area 1 with sides parallel to the x− andy−axes that lies entirely within the 5-by-5 square bounded by the lines x = 0, x = 5, y =0, y = 5 (the corners of the square need not have integer coordinates), what is the probabilitythat the point (x, y) = (4.5, 0.5) lies within the square of area 1?

Solution: The upper-left corner of the unit square is picked uniformly from the square0 ≤ x ≤ 4; 1 ≤ y ≤ 5, and for it to contain the desired point it must lie in the square

3.5 ≤ x ≤ 4; 1 ≤ y ≤ 1.5. The answer is the ratio of the squares’ areas, 14/16 = 1

64.

7. Equilateral triangle ABC of side length 2 is drawn. Three squares external to thetriangle, ABDE, BCFG, and CAHI, are drawn. What is the area of the smallest trianglethat contains these squares?

Solution: The equilateral triangle with sides lying on lines DG, EH, and FI has minimalarea. (The only other reasonable candidate is the triangle with sides along DE,FG, HI, buta quick sketch shows that it is larger.) Let J , K, and L be the vertices of this triangle closestto D, H, and F , respectively. Clearly, KI = FL = 2. Triangle FCI is a 30◦ − 30◦ − 120◦

triangle, so we can calculate the length of FI as 2√

3, making the side length of 4JKL

4 + 2√

3, and its area 12 + 7√

3 .

8. Equilateral triangle ABC of side length 2 is drawn. Three squares containing thetriangle, ABDE, BCFG, and CAHI, are drawn. What is the area of the smallest trianglethat contains these squares?

Solution: The equilateral triangle with sides lying on lines DE, FG, and HI has minimalarea. Let J , K, and L be the vertices of this triangle closest to D, H, and F , respectively.Clearly, DE = 2. Denote by M the intersection of CI and BD. Using the 30◦ − 30◦ − 120◦

triangle BCM , we get BM = 2/√

3, and thus MD = 2− 2/√

3. By symmetry, MJ bisectsangle DMI, from which we see that4JMD is a 30◦−60◦−90◦. We then get JD = 2

√3−2,

making the side length of JKL 4√

3− 2, and its area 13√

3− 12 .

9. A and B are two points on a circle with center O, and C lies outside the circle, onray AB. Given that AB = 24, BC = 28, OA = 15, find OC.

Solution: 41 . Let M be the midpoint of AB; then 4OMB is a right triangle withOB = 15, MB = 12, so OM = 9. Now 4OMC is a right triangle with OM = 9, MC = 40,so OC = 41.

2

10. Let 4ABC be equilateral, and let D, E, F be points on sides BC, CA, AB respec-tively, with FA = 9, AE = EC = 6, CD = 4. Determine the measure (in degrees) of∠DEF .

Solution: 60 . Let H, I be the respective midpoints of sides BC, AB, and also extendCB and EF to intersect at J . By equal angles, 4EIF ∼ 4JBF . However, BF = 12− 9 =3 = 9 − 6 = IF , so in fact 4EIF ∼= 4JBF , and then JB = 6. Now let HI intersect EFat K, and notice that 4EIK ∼ 4JHK ⇒ IK/HK = EI/JH = 6/12 = 1/2 ⇒ HK = 4,since IK + HK = HI = 6. Now consider the 60◦ rotation about E carrying triangle CHEto triangle HIE; we see that it also takes D to K, and thus ∠DEF = ∠DEK = 60◦.

3

1. [4] An (l, a)-design of a set is a collection of subsets of that set such that each subsetcontains exactly l elements and that no two of the subsets share more than a elements. Howmany (2,1)-designs are there of a set containing 8 elements?

2. [5] A lattice point in the plane is a point of the form (n, m), where n and m are integers.Consider a set S of lattice points. We construct the transform of S, denoted by S ′, by thefollowing rule: the pair (n,m) is in S ′ if and only if any of (n, m− 1), (n, m + 1), (n− 1, m),(n + 1, m), and (n, m) is in S. How many elements are in the set obtained by successivelytransforming {(0, 0)} 14 times?

3. [5] How many elements are in the set obtained by transforming {(0, 0), (2, 0)} 14 times?

4. [4] How many ways are there of using diagonals to divide a regular 6-sided polygon intotriangles such that at least one side of each triangle is a side of the original polygon and thateach vertex of each triangle is a vertex of the original polygon?

5. [6] Two 4× 4 squares are randomly placed on an 8× 8 chessboard so that their sides liealong the grid lines of the board. What is the probability that the two squares overlap?

6. [±6] Find all values of x that satisfy x = 1− x + x2− x3 + x4− x5 + · · · (be careful; thisis tricky).

7. [5] A rubber band is 4 inches long. An ant begins at the left end. Every minute, theant walks one inch along rightwards along the rubber band, but then the band is stretched(uniformly) by one inch. For what value of n will the ant reach the right end during the nthminute?

8. [7] Draw a square of side length 1. Connect its sides’ midpoints to form a second square.Connect the midpoints of the sides of the second square to form a third square. Connect themidpoints of the sides of the third square to form a fourth square. And so forth. What isthe sum of the areas of all the squares in this infinite series?

9. [4] Find all values of x with 0 ≤ x < 2π that satisfy sin x + cos x =√

2.

1

10. [6] The mathematician John is having trouble remembering his girlfriend Alicia’s 7-digit phone number. He remembers that the first four digits consist of one 1, one 2, and two3s. He also remembers that the fifth digit is either a 4 or 5. While he has no memory of thesixth digit, he remembers that the seventh digit is 9 minus the sixth digit. If this is all theinformation he has, how many phone numbers does he have to try if he is to make sure hedials the correct number?

11. [7] How many real solutions are there to the equation

||||x| − 2| − 2| − 2| = ||||x| − 3| − 3| − 3| ?

12. [±7] This question forms a three question multiple choice test. After each question,there are 4 choices, each preceded by a letter. Please write down your answer as the orderedtriple (letter of the answer of Question #1, letter of the answer of Question #2, letter of theanswer of Question #3). If you find that all such ordered triples are logically impossible,then write “no answer” as your answer. If you find more than one possible sets of answers,then provide all ordered triples as your answer.

When we refer to “the correct answer to Question X” it is the actual answer, not theletter, to which we refer. When we refer to “the letter of the correct answer to question X”it is the letter contained in parentheses that precedes the answer to which we refer.

You are given the following condition: No two correct answers to questions on the testmay have the same letter.Question 1. If a fourth question were added to this test, and if the letter of its correct answerwere (C), then:

(A) This test would have no logically possible set of answers.(B) This test would have one logically possible set of answers.(C) This test would have more than one logically possible set of answers.(D) This test would have more than one logically possible set of answers.

Question 2. If the answer to Question 2 were “Letter (D)” and if Question 1 were not onthis multiple-choice test (still keeping Questions 2 and 3 on the test), then the letter of theanswer to Question 3 would be:

(A) Letter (B)(B) Letter (C)(C) Letter (D)(D) Letter (A)

Question 3. Let P1 = 1. Let P2 = 3. For all i > 2, define Pi = Pi−1Pi−2 − Pi−2. Which is afactor of P2002?

(A) 3(B) 4(C) 7(D) 9

2

13. [7] A domino is a 1-by-2 or 2-by-1 rectangle. A domino tiling of a region of the planeis a way of covering it (and only it) completely by nonoverlapping dominoes. For instance,there is one domino tiling of a 2-by-1 rectangle and there are 2 tilings of a 2-by-2 rectangle(one consisting of two horizontal dominoes and one consisting of two vertical dominoes).How many domino tilings are there of a 2-by-10 rectangle?

14. [7] An omino is a 1-by-1 square or a 1-by-2 horizontal rectangle. An omino tiling of aregion of the plane is a way of covering it (and only it) by ominoes. How many omino tilingsare there of a 2-by-10 horizontal rectangle?

15. [7] How many sequences of 0s and 1s are there of length 10 such that there are no three0s or 1s consecutively anywhere in the sequence?

16. [5] Divide an m-by-n rectangle into mn nonoverlapping 1-by-1 squares. A polyominoof this rectangle is a subset of these unit squares such that for any two unit squares S, T inthe polyomino, either

(1) S and T share an edge or(2) there exists a positive integer n such that the polyomino contains unit squares

S1, S2, S3, . . . , Sn such that S and S1 share an edge, Sn and T share an edge, and for allpositive integers k < n, Sk and Sk+1 share an edge.

We say a polyomino of a given rectangle spans the rectangle if for each of the four edgesof the rectangle the polyomino contains a square whose edge lies on it.

What is the minimum number of unit squares a polyomino can have if it spans a 128-by-343 rectangle?

17. [±8] Find the number of pentominoes (5-square polyominoes) that span a 3-by-3 rect-angle, where polyominoes that are flips or rotations of each other are considered the samepolyomino.

18. [±5] Call the pentominoes found in the last problem square pentominoes. Just likedominos and ominos can be used to tile regions of the plane, so can square pentominoes. Inparticular, a square pentomino tiling of a region of the plane is a way of covering it (and onlyit) completely by nonoverlapping square pentominoes. How many square pentomino tilingsare there of a 12-by-12 rectangle?

3

19. [8] For how many integers a (1 ≤ a ≤ 200) is the number aa a square?

20. [7] The Antarctican language has an alphabet of just 16 letters. Interestingly, everyword in the language has exactly 3 letters, and it is known that no word’s first letter equalsany word’s last letter (for instance, if the alphabet were {a, b} then aab and aaa could notboth be words in the language because a is the first letter of a word and the last letterof a word; in fact, just aaa alone couldn’t be in the language). Given this, determine themaximum possible number of words in the language.

21. [7] The Dyslexian alphabet consists of consonants and vowels. It so happens that afinite sequence of letters is a word in Dyslexian precisely if it alternates between consonantsand vowels (it may begin with either). There are 4800 five-letter words in Dyslexian. Howmany letters are in the alphabet?

22. [5] A path of length n is a sequence of points (x1, y1), (x2, y2), . . . , (xn, yn) with integercoordinates such that for all i between 1 and n− 1 inclusive, either

(1) xi+1 = xi + 1 and yi+1 = yi (in which case we say the ith step is rightward) or(2) xi+1 = xi and yi+1 = yi + 1 (in which case we say that the ith step is upward).This path is said to start at (x1, y1) and end at (xn, yn). Let P (a, b), for a and b nonneg-

ative integers, be the number of paths that start at (0, 0) and end at (a, b).

Find10∑i=0

P (i, 10− i).

23. [5] Find P (7, 3).

24. [7] A restricted path of length n is a path of length n such that for all i between 1 andn− 2 inclusive, if the ith step is upward, the i + 1st step must be rightward.

Find the number of restricted paths that start at (0, 0) and end at (7, 3).

4

25. [±4] A math professor stands up in front of a room containing 100 very smart mathstudents and says, “Each of you has to write down an integer between 0 and 100, inclusive, toguess ‘two-thirds of the average of all the responses.’ Each student who guesses the highestinteger that is not higher than two-thirds of the average of all responses will receive a prize.”If among all the students it is common knowledge that everyone will write down the bestresponse, and there is no communication between students, what single integer should eachof the 100 students write down?

26. [±4] Another professor enters the same room and says, “Each of you has to write downan integer between 0 and 200. I will then compute X, the number that is 3 greater than halfthe average of all the numbers that you will have written down. Each student who writesdown the number closest to X (either above or below X) will receive a prize.” One student,who misunderstood the question, announces to the class that he will write the number 107.If among the other 99 students it is common knowledge that all 99 of them will write downthe best response, and there is no further communication between students, what singleinteger should each of the 99 students write down?

27. [7] Consider the two hands of an analog clock, each of which moves with constantangular velocity. Certain positions of these hands are possible (e.g. the hour hand halfwaybetween the 5 and 6 and the minute hand exactly at the 6), while others are impossible (e.g.the hour hand exactly at the 5 and the minute hand exactly at the 6). How many differentpositions are there that would remain possible if the hour and minute hands were switched?

28. [6] Count how many 8-digit numbers there are that contain exactly four nines as digits.

29. [8] A sequence s0, s1, s2, s3, . . . is defined by s0 = s1 = 1 and, for every positive integern, s2n = sn, s4n+1 = s2n+1, s4n−1 = s2n−1 + s2

2n−1/sn−1. What is the value of s1000?

30. [9] A conical flask contains some water. When the flask is oriented so that its base ishorizontal and lies at the bottom (so that the vertex is at the top), the water is 1 inch deep.When the flask is turned upside-down, so that the vertex is at the bottom, the water is 2inches deep. What is the height of the cone?

5

31. [6] Express, as concisely as possible, the value of the product

(03 − 350)(13 − 349)(23 − 348)(33 − 347) · · · (3493 − 1)(3503 − 0).

32. [9] Two circles have radii 13 and 30, and their centers are 41 units apart. The linethrough the centers of the two circles intersects the smaller circle at two points; let A be theone outside the larger circle. Suppose B is a point on the smaller circle and C a point onthe larger circle such that B is the midpoint of AC. Compute the distance AC.

33. [8] The expression bxc denotes the greatest integer less than or equal to x. Find thevalue of ⌊

2002!

2001! + 2000! + 1999! + · · ·+ 1!

⌋.

34. [7] Points P and Q are 3 units apart. A circle centered at P with a radius of√

3 unitsintersects a circle centered at Q with a radius of 3 units at points A and B. Find the areaof quadrilateral APBQ.

35. [±7] Suppose a, b, c, d are real numbers such that

|a− b|+ |c− d| = 99; |a− c|+ |b− d| = 1.

Determine all possible values of |a− d|+ |b− c|.

36. [±6] Find the set consisting of all real values of x such that the three numbers 2x, 2x2, 2x3

form a non-constant arithmetic progression (in that order).

6

37. [8] Call a positive integer “mild” if its base-3 representation never contains the digit 2.How many values of n (1 ≤ n ≤ 1000) have the property that n and n2 are both mild?

38. [6] Massachusetts Avenue is ten blocks long. One boy and one girl live on each block.They want to form friendships such that each boy is friends with exactly one girl and vice-versa. Nobody wants a friend living more than one block away (but they may be on thesame block). How many pairings are possible?

39. [7] In the x-y plane, draw a circle of radius 2 centered at (0, 0). Color the circle redabove the line y = 1, color the circle blue below the line y = −1, and color the rest of thecircle white. Now consider an arbitrary straight line at distance 1 from the circle. We coloreach point P of the line with the color of the closest point to P on the circle. If we pick suchan arbitrary line, randomly oriented, what is the probability that it contains red, white, andblue points?

40. [9] Find the volume of the three-dimensional solid given by the inequality√

x2 + y2 +|z| ≤ 1.

41. [9] For any integer n, define bnc as the greatest integer less than or equal to n. For anypositive integer n, let

f(n) = bnc+⌊n

2

⌋+⌊n

3

⌋+ · · ·+

⌊n

n

⌋.

For how many values of n, 1 ≤ n ≤ 100, is f(n) odd?

42. [±10] Find all the integers n > 1 with the following property: the numbers 1, 2, . . . , ncan be arranged in a line so that, of any two adjacent numbers, one is divisible by the other.

7

43. [9] Given that a, b, c are positive integers satisfying

a + b + c = gcd(a, b) + gcd(b, c) + gcd(c, a) + 120,

determine the maximum possible value of a.

44. [5] The unknown real numbers x, y, z satisfy the equations

x + y

1 + z=

1− z + z2

x2 − xy + y2;

x− y

3− z=

9 + 3z + z2

x2 + xy + y2.

Find x.

45. [9] Find the number of sequences a1, a2, . . . , a10 of positive integers with the propertythat an+2 = an+1 + an for n = 1, 2, . . . , 8, and a10 = 2002.

46. [±6] Points A, B, C in the plane satisfy AB = 2002, AC = 9999. The circles withdiameters AB and AC intersect at A and D. If AD = 37, what is the shortest distance frompoint A to line BC?

47. [9] The real function f has the property that, whenever a, b, n are positive integers suchthat a + b = 2n, the equation f(a) + f(b) = n2 holds. What is f(2002)?

48. [9] A permutation of a finite set is a one-to-one function from the set to itself; forinstance, one permutation of {1, 2, 3, 4} is the function π defined such that π(1) = 1, π(2) = 3,π(3) = 4, and π(4) = 2. How many permutations π of the set {1, 2, . . . , 10} have the propertythat π(i) 6= i for each i = 1, 2, . . . , 10, but π(π(i)) = i for each i?

49. [7] Two integers are relatively prime if they don’t share any common factors, i.e. iftheir greatest common divisor is 1. Define ϕ(n) as the number of positive integers that areless than n and relatively prime to n. Define ϕd(n) as the number of positive integers thatare less than dn and relatively prime to n.

What is the least n such that ϕx(n) = 64000, where x = ϕy(n), where y = ϕ(n)?

50. [6] Give the set of all positive integers n such that ϕ(n) = 20022 − 1.

51. [10] Define ϕk(n) as the number of positive integers that are less than or equal to n/kand relatively prime to n. Find φ2001(20022 − 1). (Hint: φ(2003) = 2002.)

8

52. [±8] Let ABCD be a quadrilateral, and let E, F, G, H be the respective midpointsof AB, BC, CD,DA. If EG = 12 and FH = 15, what is the maximum possible area ofABCD?

53. [10] ABC is a triangle with points E, F on sides AC, AB, respectively. Suppose thatBE,CF intersect at X. It is given that AF/FB = (AE/EC)2 and that X is the midpointof BE. Find the ratio CX/XF .

54. [10] How many pairs of integers (a, b), with 1 ≤ a ≤ b ≤ 60, have the property that bis divisible by a and b + 1 is divisible by a + 1?

55. [10] A sequence of positive integers is given by a1 = 1 and an = gcd(an−1, n) + 1 forn > 1. Calculate a2002.

56. [±6] x, y are positive real numbers such that x+ y2 = xy. What is the smallest possiblevalue of x?

57. [9] How many ways, without taking order into consideration, can 2002 be expressedas the sum of 3 positive integers (for instance, 1000 + 1000 + 2 and 1000 + 2 + 1000 areconsidered to be the same way)?

58. [8] A sequence is defined by a0 = 1 and an = 2an−1 for n ≥ 1. What is the last digit (inbase 10) of a15?

59. [7] Determine the value of

1 · 2− 2 · 3 + 3 · 4− 4 · 5 + · · ·+ 2001 · 2002.

60. [10] A 5 × 5 square grid has the number −3 written in the upper-left square and thenumber 3 written in the lower-right square. In how many ways can the remaining squaresbe filled in with integers so that any two adjacent numbers differ by 1, where two squaresare adjacent if they share a common edge (but not if they share only a corner)?

9

61. Bob Barker went back to school for a PhD in math, and decided to raise the intellectuallevel of The Price is Right by having contestants guess how many objects exist of a certaintype, without going over. The number of points you will get is the percentage of the correctanswer, divided by 10, with no points for going over (i.e. a maximum of 10 points).

Let’s see the first object for our contestants...a table of shape (5, 4, 3, 2, 1) is an arrange-ment of the integers 1 through 15 with five numbers in the top row, four in the next, threein the next, two in the next, and one in the last, such that each row and each column isincreasing (from left to right, and top to bottom, respectively). For instance:

1 2 3 4 56 7 8 910 11 1213 1415

is one table. How many tables are there?

62. Our next object up for bid is an arithmetic progression of primes. For example, theprimes 3, 5, and 7 form an arithmetic progression of length 3. What is the largest possi-ble length of an arithmetic progression formed of positive primes less than 1,000,000? Beprepared to justify your answer.

63. Our third and final item comes to us from Germany, I mean Geometry. It is knownthat a regular n-gon can be constructed with straightedge and compass if n is a primethat is 1 plus a power of 2. It is also possible to construct a 2n-gon whenever an n-gon isconstructible, or a p1p2 · · · pm-gon where the pi’s are distinct primes of the above form. Whatis really interesting is that these conditions, together with the fact that we can construct asquare, is that they give us all constructible regular n-gons. What is the largest n less than4,300,000,000 such that a regular n-gon is constructible?

Help control the pet population. Have your pets spayed or neutered. Bye-bye.

10

Harvard-MIT Math TournamentMarch 17, 2002Guts Round

1. An (l, a)-design of a set is a collection of subsets of that set such that each subsetcontains exactly l elements and that no two of the subsets share more than a elements. Howmany (2,1)-designs are there of a set containing 8 elements?

Solution: There are(82

)= 28 2-element subsets. Any two distinct such subsets have at

most 1 common element; hence, for each subset, we can decide independently whether or

not it belongs to the design, and we thus obtain 228 designs.

2. A lattice point in the plane is a point of the form (n,m), where n and m are integers.Consider a set S of lattice points. We construct the transform of S, denoted by S ′, by thefollowing rule: the pair (n,m) is in S ′ if and only if any of (n, m− 1), (n, m + 1), (n− 1, m),(n + 1, m), and (n, m) is in S. How many elements are in the set obtained by successivelytransforming {(0, 0)} 14 times?

Solution: Transforming it k ≥ 1 times yields the “diamond” of points (n, m) such that|n|+ |m| ≤ k. The diamond contains (k+1)2 +k2 lattice points (this can be seen by rotatingthe plane 45 degrees and noticing the lattice points in the transforms form two squares, oneof which is contained in the other), so the answer is 421 .

3. How many elements are in the set obtained by transforming {(0, 0), (2, 0)} 14 times?Solution: Transforming it k ≥ 1 times yields the diamond {(n,m) : |n−1|+|m| ≤ k+1}

with the points (1, k), (1, k + 1), (1,−k), (1,−k − 1) removed (this can be seen inductively).So we get (k + 1)2 + k2 − 4 lattice points, making the answer 477 .

4. How many ways are there of using diagonals to divide a regular 6-sided polygon intotriangles such that at least one side of each triangle is a side of the original polygon and thateach vertex of each triangle is a vertex of the original polygon?

Solution: The number of ways of triangulating a convex (n+2)-sided polygon is(2nn

)1

n+1,

which is 14 in this case. However, there are two triangulations of a hexagon which produceone triangle sharing no sides with the original polygon, so the answer is 14− 2 = 12 .

5. Two 4× 4 squares are randomly placed on an 8× 8 chessboard so that their sides liealong the grid lines of the board. What is the probability that the two squares overlap?

Solution: 529/625 . Each square has 5 horizontal · 5 vertical = 25 possible positions,

so there are 625 possible placements of the squares. If they do not overlap, then eitherone square lies in the top four rows and the other square lies in the bottom four rows, orone square lies in the left four columns and the other lies in the right four columns. Thefirst possibility can happen in 2 · 5 · 5 = 50 ways (two choices of which square goes ontop, and five horizontal positions for each square); likewise, so can the second. However,

1

this double-counts the 4 cases in which the two squares are in opposite corners, so we have50 + 50 − 4 = 96 possible non-overlapping arrangements ⇒ 252 − 96 = 529 overlappingarrangements.

6. Find all values of x that satisfy x = 1− x + x2 − x3 + x4 − x5 + · · · (be careful; thisis tricky).

Solution: Multiplying both sides by 1 + x gives (1 + x)x = 1, or x = −1±√

52

. However,

the series only converges for |x| < 1, so only the answer x = −1+√

52

makes sense.

7. A rubber band is 4 inches long. An ant begins at the left end. Every minute, theant walks one inch along rightwards along the rubber band, but then the band is stretched(uniformly) by one inch. For what value of n will the ant reach the right end during the nthminute?

Solution: 7 The ant traverses 1/4 of the band’s length in the first minute, 1/5 of thelength in the second minute (the stretching does not affect its position as a fraction of theband’s length), 1/6 of the length in the third minute, and so on. Since

1/4 + 1/5 + · · ·+ 1/9 < 0.25 + 0.20 + 0.167 + 0.143 + 0.125 + 0.112 = 0.997 < 1,

the ant does not cover the entire band in six minutes. However,

1/4 + · · ·+ 1/10 > 0.25 + 0.20 + 0.16 + 0.14 + 0.12 + 0.11 + 0.10 = 1.08 > 1,

so seven minutes suffice.

8. Draw a square of side length 1. Connect its sides’ midpoints to form a second square.Connect the midpoints of the sides of the second square to form a third square. Connect themidpoints of the sides of the third square to form a fourth square. And so forth. What isthe sum of the areas of all the squares in this infinite series?

Solution: The area of the first square is 1, the area of the second is 12, the area of the

third is 14, etc., so the answer is 1 + 1

2+ 1

4+ 1

8+ · · · = 2 .

9. Find all values of x with 0 ≤ x < 2π that satisfy sin x + cos x =√

2.Solution: Squaring both sides gives sin2 x + cos2 x + 2 sin x cos x = 1 + sin 2x = 2, so

x = π4, 5π

4.

10. The mathematician John is having trouble remembering his girlfriend Alicia’s 7-digitphone number. He remembers that the first four digits consist of one 1, one 2, and two 3s.He also remembers that the fifth digit is either a 4 or 5. While he has no memory of thesixth digit, he remembers that the seventh digit is 9 minus the sixth digit. If this is all theinformation he has, how many phone numbers does he have to try if he is to make sure hedials the correct number?

2

Solution: There are 4!2!

= 12 possibilities for the first four digits. There are two pos-sibilities for the fifth digit. There are 10 possibilities for the sixth digit, and this uniquelydetermines the seventh digit. So he has to dial 12 · 2 · 10 = 240 numbers.

11. How many real solutions are there to the equation

||||x| − 2| − 2| − 2| = ||||x| − 3| − 3| − 3| ?

Solution: 6 . The graphs of the two sides of the equation can be graphed on the sameplot to reveal six intersection points.

12. This question forms a three question multiple choice test. After each question, thereare 4 choices, each preceded by a letter. Please write down your answer as the ordered triple(letter of the answer of Question #1, letter of the answer of Question #2, letter of the answerof Question #3). If you find that all such ordered triples are logically impossible, then write“no answer” as your answer. If you find more than one possible set of answers, then provideall ordered triples as your answer.

When we refer to “the correct answer to Question X” it is the actual answer, not theletter, to which we refer. When we refer to “the letter of the correct answer to question X”it is the letter contained in parentheses that precedes the answer to which we refer.

You are given the following condition: No two correct answers to questions on the testmay have the same letter.Question 1. If a fourth question were added to this test, and if the letter of its correct answerwere (C), then:

(A) This test would have no logically possible set of answers.(B) This test would have one logically possible set of answers.(C) This test would have more than one logically possible set of answers.(D) This test would have more than one logically possible set of answers.

Question 2. If the answer to Question 2 were “Letter (D)” and if Question 1 were not onthis multiple-choice test (still keeping Questions 2 and 3 on the test), then the letter of theanswer to Question 3 would be:

(A) Letter (B)(B) Letter (C)(C) Letter (D)(D) Letter (A)

Question 3. Let P1 = 1. Let P2 = 3. For all i > 2, define Pi = Pi−1Pi−2 − Pi−2. Which is afactor of P2002?

(A) 3(B) 4(C) 7(D) 9

Solution: (A, C, D) . Question 2: In order for the answer to be consistent with the

condition, “If the answer to Question 2 were Letter (D),” the answer to this question actuallymust be “Letter (D).” The letter of this answer is (C).

3

Question 1: If a fourth question had an answer with letter (C), then at least two answerswould have letter (C) (the answers to Questions 2 and 4). This is impossible. So, (A) mustbe the letter of the answer to Question 1.

Question 3: If we inspect the sequence Pi modulo 3, 4, 7, and 9 (the sequences quicklybecome periodic), we find that 3, 7, and 9 are each factors of P2002. We know that letters(A) and (C) cannot be repeated, so the letter of this answer must be (D).

13. A domino is a 1-by-2 or 2-by-1 rectangle. A domino tiling of a region of the planeis a way of covering it (and only it) completely by nonoverlapping dominoes. For instance,there is one domino tiling of a 2-by-1 rectangle and there are 2 tilings of a 2-by-2 rectangle(one consisting of two horizontal dominoes and one consisting of two vertical dominoes).How many domino tilings are there of a 2-by-10 rectangle?

Solution: The number of tilings of a 2-by-n, rectangle is the nth Fibonacci number Fn,where F0 = F1 = 1 and Fn = Fn−1 +Fn−1 for n ≥ 2. (This is not hard to show by induction.)The answer is 89 .

14. An omino is a 1-by-1 square or a 1-by-2 horizontal rectangle. An omino tiling of aregion of the plane is a way of covering it (and only it) by ominoes. How many omino tilingsare there of a 2-by-10 horizontal rectangle?

Solution: There are exactly as many omino tilings of a 1-by-n rectangle as there aredomino tilings of a 2-by-n rectangle. Since the rows don’t interact at all, the number ofomino tilings of an m-by-n rectangle is the number of omino tilings of a 1-by-n rectangleraised to the mth power, Fm

n . The answer is 892 = 7921 .

15. How many sequences of 0s and 1s are there of length 10 such that there are no three0s or 1s consecutively anywhere in the sequence?

Solution: We can have blocks of either 1 or 2 0s and 1s, and these blocks must bealternating between 0s and 1s. The number of ways of arranging blocks to form a sequenceof length n is the same as the number of omino tilings of a 1-by-n rectangle, and we maystart each sequence with a 0 or a 1, making 2Fn or, in this case, 178 sequences.

16. Divide an m-by-n rectangle into mn nonoverlapping 1-by-1 squares. A polyomino ofthis rectangle is a subset of these unit squares such that for any two unit squares S, T in thepolyomino, either

(1) S and T share an edge or(2) there exists a positive integer n such that the polyomino contains unit squares

S1, S2, S3, . . . , Sn such that S and S1 share an edge, Sn and T share an edge, and for allpositive integers k < n, Sk and Sk+1 share an edge.

We say a polyomino of a given rectangle spans the rectangle if for each of the four edgesof the rectangle the polyomino contains a square whose edge lies on it.

What is the minimum number of unit squares a polyomino can have if it spans a 128-by-343 rectangle?

4

Solution: To span an a×b rectangle, we need at least a+b−1 squares. Indeed, considera square of the polyomino bordering the left edge of the rectangle and one bordering theright edge. There exists a path connecting these squares; suppose it runs through c differentrows. Then the path requires at least b− 1 horizontal and c− 1 vertical steps, so it uses atleast b + c− 1 different squares. However, since the polyomino also hits the top and bottomedges of the rectangle, it must run into the remaining a − c rows as well, so altogether weneed at least a+ b−1 squares. On the other hand, this many squares suffice — just considerall the squares bordering the lower or right edges of the rectangle. So, in our case, the answeris 128 + 343− 1 = 470 .

17. Find the number of pentominoes (5-square polyominoes) that span a 3-by-3 rect-angle, where polyominoes that are flips or rotations of each other are considered the samepolyomino.

Solution: By enumeration, the answer is 6 .

18. Call the pentominoes found in the last problem square pentominoes. Just likedominos and ominos can be used to tile regions of the plane, so can square pentominoes. Inparticular, a square pentomino tiling of a region of the plane is a way of covering it (and onlyit) completely by nonoverlapping square pentominoes. How many square pentomino tilingsare there of a 12-by-12 rectangle?

Solution: Since 5 does not divide 144, there are 0 .

19. For how many integers a (1 ≤ a ≤ 200) is the number aa a square?Solution: 107 If a is even, we have aa = (aa/2)2. If a is odd, aa = (a(a−1)/2)2 · a, which

is a square precisely when a is. Thus we have 100 even values of a and 7 odd square values(12, 32, . . . , 132) for a total of 107.

20. The Antarctican language has an alphabet of just 16 letters. Interestingly, everyword in the language has exactly 3 letters, and it is known that no word’s first letter equalsany word’s last letter (for instance, if the alphabet were {a, b} then aab and aaa could notboth be words in the language because a is the first letter of a word and the last letterof a word; in fact, just aaa alone couldn’t be in the language). Given this, determine themaximum possible number of words in the language.

Solution: 1024 Every letter can be the first letter of a word, or the last letter of aword, or possibly neither, but not both. If there are a different first letters and b differentlast letters, then we can form a · 16 · b different words (and the desired conditions will bemet). Given the constraints 0 ≤ a, b; a+ b ≤ 16, this product is maximized when a = b = 8,giving the answer.

21. The Dyslexian alphabet consists of consonants and vowels. It so happens that afinite sequence of letters is a word in Dyslexian precisely if it alternates between consonantsand vowels (it may begin with either). There are 4800 five-letter words in Dyslexian. Howmany letters are in the alphabet?

5

Solution: 12 Suppose there are c consonants, v vowels. Then there are c · v · c · v · c +v · c · v · c · v = (cv)2(c + v) five-letter words. Thus, c + v = 4800/(cv)2 = 3 · (40/cv)2, so cv isa divisor of 40. If cv ≤ 10, we have c + v ≥ 48, impossible for c, v integers; if cv = 40, thenc + v = 3 which is again impossible. So cv = 20, giving c + v = 12, the answer. As a check,this does have integer solutions: (c, v) = (2, 10) or (10, 2).

22. A path of length n is a sequence of points (x1, y1), (x2, y2), . . . , (xn, yn) with integercoordinates such that for all i between 1 and n− 1 inclusive, either

(1) xi+1 = xi + 1 and yi+1 = yi (in which case we say the ith step is rightward) or(2) xi+1 = xi and yi+1 = yi + 1 (in which case we say that the ith step is upward).This path is said to start at (x1, y1) and end at (xn, yn). Let P (a, b), for a and b nonneg-

ative integers, be the number of paths that start at (0, 0) and end at (a, b).

Find10∑i=0

P (i, 10− i).

Solution: This is just the number of paths of length 10. The ith step can be eitherupward or rightward, so there are 210 = 1024 such paths.

23. Find P (7, 3).Solution: The number of paths that start at (0, 0) and end at (n,m) is

(n+m

n

), since we

must choose n of our n + m steps to be rightward steps. In this case, the answer is 120 .

24. A restricted path of length n is a path of length n such that for all i between 1 andn− 2 inclusive, if the ith step is upward, the i + 1st step must be rightward.

Find the number of restricted paths that start at (0, 0) and end at (7, 3).Solution: This is equal to the number of lattice paths from (0, 0) to (7, 3) that use only

rightward and diagonal (upward+rightward) steps plus the number of lattice paths from(0, 0) to (7, 2) that use only rightward and diagonal steps, which is equal to the number ofpaths (as defined above) from (0, 0) to (4, 3) plus the number of paths from (0, 0) to (5, 2),or(4+33

)+(5+22

)= 56 .

25. A math professor stands up in front of a room containing 100 very smart mathstudents and says, “Each of you has to write down an integer between 0 and 100, inclusive,to guess ‘two-thirds of the average of all the responses.’ Each student who guesses the highestinteger that is not higher than two-thirds of the average of all responses will receive a prize.”If among all the students it is common knowledge that everyone will write down the bestresponse, and there is no communication between students, what single integer should eachof the 100 students write down?

Solution: Since the average cannot be greater than 100, no student will write downa number greater than 2

3· 100. But then the average cannot be greater than 2

3· 100, and,

realizing this, each student will write down a number no greater than (23)2 · 100. Continuing

in this manner, we eventually see that no student will write down an integer greater than0 , so this is the answer.

6

26. Another professor enters the same room and says, “Each of you has to write down aninteger between 0 and 200. I will then compute X, the number that is 3 greater than half theaverage of all the numbers that you will have written down. Each student who writes downthe number closest to X (either above or below X) will receive a prize.” One student, whomisunderstood the question, announces to the class that he will write the number 107. Ifamong the other 99 students it is common knowledge that all 99 of them will write down thebest response, and there is no further communication between students, what single integershould each of the 99 students write down?

Solution: Use the same logic to get 7 . Note 6 and 8 do not work.

27. Consider the two hands of an analog clock, each of which moves with constantangular velocity. Certain positions of these hands are possible (e.g. the hour hand halfwaybetween the 5 and 6 and the minute hand exactly at the 6), while others are impossible (e.g.the hour hand exactly at the 5 and the minute hand exactly at the 6). How many differentpositions are there that would remain possible if the hour and minute hands were switched?

Solution: 143 We can look at the twelve-hour cycle beginning at midnight and endingjust before noon, since during this time, the clock goes through each possible position exactlyonce. The minute hand has twelve times the angular velocity of the hour hand, so if thehour hand has made t revolutions from its initial position (0 ≤ t < 1), the minute handhas made 12t revolutions. If the hour hand were to have made 12t revolutions, the minutehand would have made 144t. So we get a valid configuration by reversing the hands preciselywhen 144t revolutions land the hour hand in the same place as t revolutions — i.e. when143t = 144t− t is an integer, which clearly occurs for exactly 143 values of t correspondingto distinct positions on the clock (144− 1 = 143).

28. Count how many 8-digit numbers there are that contain exactly four nines as digits.Solution: There are

(84

)· 94 sequences of 8 numbers with exactly four nines. A sequence

of digits of length 8 is not an 8-digit number, however, if and only if the first digit iszero. There are

(74

)93 8-digit sequences that are not 8-digit numbers. The answer is thus(

84

)· 94 −

(74

)93 = 433755 .

29. A sequence s0, s1, s2, s3, . . . is defined by s0 = s1 = 1 and, for every positive integern, s2n = sn, s4n+1 = s2n+1, s4n−1 = s2n−1 + s2

2n−1/sn−1. What is the value of s1000?

Solution: 720 Some experimentation with small values may suggest that sn = k!, wherek is the number of ones in the binary representation of n, and this formula is in fact provableby a straightforward induction. Since 100010 = 11111010002, with six ones, s1000 = 6! = 720.

30. A conical flask contains some water. When the flask is oriented so that its base ishorizontal and lies at the bottom (so that the vertex is at the top), the water is 1 inch deep.When the flask is turned upside-down, so that the vertex is at the bottom, the water is 2inches deep. What is the height of the cone?

7

Solution: 12

+√

936

. Let h be the height, and let V be such that V h3 equals the volume

of the flask. When the base is at the bottom, the portion of the flask not occupied by waterforms a cone similar to the entire flask, with a height of h− 1; thus its volume is V (h− 1)3.When the base is at the top, the water occupies a cone with a height of 2, so its volume isV · 23. Since the water’s volume does not change,

V h3 − V (h− 1)3 = 8V

⇒ 3h2 − 3h + 1 = h3 − (h− 1)3 = 8

⇒ 3h2 − 3h− 7 = 0.

Solving via the quadratic formula and taking the positive root gives h = 12

+√

936

.

31. Express, as concisely as possible, the value of the product

(03 − 350)(13 − 349)(23 − 348)(33 − 347) · · · (3493 − 1)(3503 − 0).

Solution: 0 . One of the factors is 73 − 343 = 0, so the whole product is zero.

32. Two circles have radii 13 and 30, and their centers are 41 units apart. The linethrough the centers of the two circles intersects the smaller circle at two points; let A be theone outside the larger circle. Suppose B is a point on the smaller circle and C a point onthe larger circle such that B is the midpoint of AC. Compute the distance AC.

Solution: 12√

13 Call the large circle’s center O1. Scale the small circle by a factor of2 about A; we obtain a new circle whose center O2 is at a distance of 41− 13 = 28 from O1,and whose radius is 26. Also, the dilation sends B to C, which thus lies on circles O1 and O2.So points O1, O2, C form a 26-28-30 triangle. Let H be the foot of the altitude from C toO1O2; we have CH = 24 and HO2 = 10. Thus, HA = 36, and AC =

√242 + 362 = 12

√13.

33. The expression bxc denotes the greatest integer less than or equal to x. Find thevalue of ⌊

2002!

2001! + 2000! + 1999! + · · ·+ 1!

⌋.

Solution: 2000 We break up 2002! = 2002(2001)! as

2000(2001!) + 2 · 2001(2000!) = 2000(2001!) + 2000(2000!) + 2002 · 2000(1999!)

> 2000(2001! + 2000! + 1999! + · · ·+ 1!).

On the other hand,

2001(2001! + 2000! + · · ·+ 1!) > 2001(2001! + 2000!) = 2001(2001!) + 2001! = 2002!.

Thus we have 2000 < 2002!/(2001! + · · ·+ 1!) < 2001, so the answer is 2000.

8

34. Points P and Q are 3 units apart. A circle centered at P with a radius of√

3 unitsintersects a circle centered at Q with a radius of 3 units at points A and B. Find the areaof quadrilateral APBQ.

Solution: The area is twice the area of triangle APQ, which is isosceles with side lengths

3, 3,√

3. By Pythagoras, the altitude to the base has length√

32 − (√

3/2)2 =√

33/2, so

the triangle has area√

994

. Double this to get 3√

112

.

35. Suppose a, b, c, d are real numbers such that

|a− b|+ |c− d| = 99; |a− c|+ |b− d| = 1.

Determine all possible values of |a− d|+ |b− c|.Solution: 99 If w ≥ x ≥ y ≥ z are four arbitrary real numbers, then |w− z|+ |x−y| =

|w − y| + |x − z| = w + x − y − z ≥ w − x + y − z = |w − x| + |y − z|. Thus, in our case,two of the three numbers |a− b|+ |c− d|, |a− c|+ |b− d|, |a− d|+ |b− c| are equal, and thethird one is less than or equal to these two. Since we have a 99 and a 1, the third numbermust be 99.

36. Find the set consisting of all real values of x such that the three numbers 2x, 2x2, 2x3

form a non-constant arithmetic progression (in that order).

Solution: The empty set, Ø . Trivially, x = 0, 1 yield constant arithmetic progressions;we show that there are no other possibilities. If these numbers do form a progression, then,by the AM-GM (arithmetic mean-geometric mean) inequality,

2 · 2x2

= 2x + 2x3 ≥ 2√

2x · 2x3

⇒ 2x2 ≥ 2(x+x3)/2 ⇒ x2 ≥ (x + x3)/2

⇒ x(x− 1)2 = x3 − 2x2 + x ≤ 0.

Assuming x 6= 0, 1, we can divide by (x−1)2 > 0 and obtain x < 0. However, then 2x, 2x3are

less than 1, while 2x2is more than 1, so the given sequence cannot possibly be an arithmetic

progression.

37. Call a positive integer “mild” if its base-3 representation never contains the digit 2.How many values of n (1 ≤ n ≤ 1000) have the property that n and n2 are both mild?

Solution: 7 Such a number, which must consist entirely of 0’s and 1’s in base 3, cannever have more than one 1. Indeed, if n = 3a + 3b + higher powers where b > a, thenn2 = 32a + 2 · 3a+b + higher powers which will not be mild. On the other hand, if n does justhave one 1 in base 3, then clearly n and n2 are mild. So the values of n ≤ 1000 that workare 30, 31, . . . , 36; there are 7 of them.

9

38. Massachusetts Avenue is ten blocks long. One boy and one girl live on each block.They want to form friendships such that each boy is friends with exactly one girl and vice-versa. Nobody wants a friend living more than one block away (but they may be on thesame block). How many pairings are possible?

Solution: 89 Let an be the number of pairings if there are n blocks; we have a1 =1, a2 = 2, and we claim the Fibonacci recurrence is satisfied. Indeed, if there are n blocks,either the boy on block 1 is friends with the girl on block 1, leaving an−1 possible pairingsfor the people on the remaining n − 1 blocks, or he is friends with the girl on block 2, inwhich case the girl on block 1 must be friends with the boy on block 2, and then there arean−2 possibilities for the friendships among the remaining n−2 blocks. So an = an−1 +an−2,and we compute: a3 = 3, a4 = 5, . . . , a10 = 89.

39. In the x-y plane, draw a circle of radius 2 centered at (0, 0). Color the circle redabove the line y = 1, color the circle blue below the line y = −1, and color the rest of thecircle white. Now consider an arbitrary straight line at distance 1 from the circle. We coloreach point P of the line with the color of the closest point to P on the circle. If we pick suchan arbitrary line, randomly oriented, what is the probability that it contains red, white, andblue points?

Solution: Let O = (0, 0), P = (1, 0), and H the foot of the perpendicular from O tothe line. If ∠POH (as measured counterclockwise) lies between π/3 and 2π/3, the line willfail to contain blue points; if it lies between 4π/3 and 5π/3, the line will fail to contain red

points. Otherwise, it has points of every color. Thus, the answer is 1− 2π3

/2π = 23

.

40. Find the volume of the three-dimensional solid given by the inequality√

x2 + y2 +|z| ≤ 1.

Solution: 2π/3 . The solid consists of two cones, one whose base is the circle x2+y2 = 1

in the xy-plane and whose vertex is (0, 0, 1), and the other with the same base but vertex(0, 0,−1). Each cone has a base area of π and a height of 1, for a volume of π/3, so theanswer is 2π/3.

41. For any integer n, define bnc as the greatest integer less than or equal to n. For anypositive integer n, let

f(n) = bnc+⌊n

2

⌋+⌊n

3

⌋+ · · ·+

⌊n

n

⌋.

For how many values of n, 1 ≤ n ≤ 100, is f(n) odd?Solution: 55 Notice that, for fixed a, bn/ac counts the number of integers b ∈

{1, 2, . . . , n} which are divisible by a; hence, f(n) counts the number of pairs (a, b), a, b ∈{1, 2, . . . , n} with b divisible by a. For any fixed b, the number of such pairs is d(b) (thenumber of divisors of b), so the total number of pairs f(n) equals d(1)+d(2)+ · · ·+d(n). Butd(b) is odd precisely when b is a square, so f(n) is odd precisely when there are an odd num-ber of squares in {1, 2, . . . , n}. This happens for 1 ≤ n < 4; 9 ≤ n < 16; . . . ; 81 ≤ n < 100.Adding these up gives 55 values of n.

10

42. Find all the integers n > 1 with the following property: the numbers 1, 2, . . . , n canbe arranged in a line so that, of any two adjacent numbers, one is divisible by the other.

Solution: 2, 3, 4, 6 The values n = 2, 3, 4, 6 work, as shown by respective examples1, 2; 2, 1, 3; 2, 4, 1, 3; 3, 6, 2, 4, 1, 5. We shall show that there are no other possibilities. Ifn = 2k + 1 is odd, then none of the numbers k + 1, k + 2, . . . , 2k + 1 can divide any other,so no two of these numbers are adjacent. This is only possible if they occupy the 1st, 3rd,. . . , (2k + 1)th positions in the line, which means every number ≤ k is adjacent to two ofthese and hence divides two of them. But k only divides one of these numbers when k ≥ 2.Thus no odd n ≥ 5 works. If n = 2k is even, the numbers k + 1, k + 2, . . . , 2k again mustbe mutually nonadjacent, but now this means we can have up to two numbers ≤ k each ofwhich is adjacent to only one number > k, and if there are two such numbers, they mustbe adjacent. If k ≥ 4, then each of k − 1, k divides only one of the numbers k + 1, . . . , 2k,so k − 1, k must be adjacent, but this is impossible. Thus no even k ≥ 8 works, and we aredone.

43. Given that a, b, c are positive integers satisfying

a + b + c = gcd(a, b) + gcd(b, c) + gcd(c, a) + 120,

determine the maximum possible value of a.Solution: 240 . Notice that (a, b, c) = (240, 120, 120) achieves a value of 240. To see

that this is maximal, first suppose that a > b. Notice that a + b + c = gcd(a, b) + gcd(b, c) +gcd(c, a) + 120 ≤ gcd(a, b) + b + c + 120, or a ≤ gcd(a, b) + 120. However, gcd(a, b) is aproper divisor of a, so a ≥ 2 · gcd(a, b). Thus, a− 120 ≤ gcd(a, b) ≤ a/2, yielding a ≤ 240.Now, if instead a ≤ b, then either b > c and the same logic shows that b ≤ 240⇒ a ≤ 240,or b ≤ c, c > a (since a, b, c cannot all be equal) and then c ≤ 240⇒ a ≤ b ≤ c ≤ 240.

44. The unknown real numbers x, y, z satisfy the equations

x + y

1 + z=

1− z + z2

x2 − xy + y2;

x− y

3− z=

9 + 3z + z2

x2 + xy + y2.

Find x.Solution: 3

√14 Cross-multiplying in both equations, we get, respectively, x3 + y3 =

1 + z3, x3 − y3 = 27− z3. Now adding gives 2x3 = 28, or x = 3√

14.

45. Find the number of sequences a1, a2, . . . , a10 of positive integers with the propertythat an+2 = an+1 + an for n = 1, 2, . . . , 8, and a10 = 2002.

Solution: 3 Let a1 = a, a2 = b; we successively compute a3 = a + b; a4 = a +2b; . . . ; a10 = 21a + 34b. The equation 2002 = 21a + 34b has three positive integersolutions, namely (84, 7), (50, 28), (16, 49), and each of these gives a unique sequence.

46. Points A, B, C in the plane satisfy AB = 2002, AC = 9999. The circles withdiameters AB and AC intersect at A and D. If AD = 37, what is the shortest distance frompoint A to line BC?

11

Solution: ∠ADB = ∠ADC = π/2 since D lies on the circles with AB and AC asdiameters, so D is the foot of the perpendicular from A to line BC, and the answer is thegiven 37 .

47. The real function f has the property that, whenever a, b, n are positive integers suchthat a + b = 2n, the equation f(a) + f(b) = n2 holds. What is f(2002)?

Solution: We know f(a) = n2−f(2n−a) for any a, n with 2n > a; repeated applicationgives

f(2002) = 112 − f(46) = 112 − (62 − f(18)) = 112 − (62 − (52 − f(14)))

= 112 − (62 − (52 − (42 − f(2)))).

But f(2) = 22 − f(2), giving f(2) = 2, so the above simplifies to 112 − (62 − (52 − (42 −2))) = 96 .

48. A permutation of a finite set is a one-to-one function from the set to itself; forinstance, one permutation of {1, 2, 3, 4} is the function π defined such that π(1) = 1, π(2) = 3,π(3) = 4, and π(4) = 2. How many permutations π of the set {1, 2, . . . , 10} have the propertythat π(i) 6= i for each i = 1, 2, . . . , 10, but π(π(i)) = i for each i?

Solution: For each such π, the elements of {1, 2, . . . , 10} can be arranged into pairs {i, j}such that π(i) = j; π(j) = i. Choosing a permutation π is thus tantamount to choosing apartition of {1, 2, . . . , 10} into five disjoint pairs. There are 9 ways to pair off the number1, then 7 ways to pair off the smallest number not yet paired, and so forth, so we have9 · 7 · 5 · 3 · 1 = 945 partitions into pairs.

49. Two integers are relatively prime if they don’t share any common factors, i.e. if theirgreatest common divisor is 1. Define ϕ(n) as the number of positive integers that are lessthan n and relatively prime to n. Define ϕd(n) as the number of positive integers that areless than dn and relatively prime to n.

What is the least n such that ϕx(n) = 64000, where x = ϕy(n), where y = ϕ(n)?Solution: For fixed n, the pattern of integers relatively prime to n repeats every n

integers, so ϕd(n) = dϕ(n). Therefore the expression in the problem equals ϕ(n)3. The cuberoot of 64000 is 40. ϕ(p) = p−1 for any prime p. Since 40 is one less than a prime, the leastn such that ϕ(n) = 40 is 41 .

50. Give the set of all positive integers n such that ϕ(n) = 20022 − 1.

Solution: The empty set, Ø . If m is relatively prime to n and m < n, then n − mmust likewise be relatively prime to n, and these are distinct for n > 2 since n/2, n arenot relatively prime. Therefore, for all n > 2, ϕ(n) must be even. 20022 − 1 is odd, andϕ(2) = 1 6= 20022 − 1, so no numbers n fulfill the equation.

51. Define ϕk(n) as the number of positive integers that are less than or equal to n/kand relatively prime to n. Find φ2001(20022 − 1). (Hint: φ(2003) = 2002.)

12

Solution: ϕ2001(20022 − 1) = ϕ2001(2001 · 2003) = the number of m that are relativelyprime to both 2001 and 2003, where m ≤ 2003. Since φ(n) = n − 1 implies that n isprime, we must only check for those m relatively prime to 2001, except for 2002, which isrelatively prime to 20022 − 1. So ϕ2001(20022 − 1) = ϕ(2001) + 1 = ϕ(3 · 23 · 29) + 1 =(3− 1)(23− 1)(29− 1) + 1 = 1233 .

52. Let ABCD be a quadrilateral, and let E, F, G, H be the respective midpoints ofAB, BC, CD,DA. If EG = 12 and FH = 15, what is the maximum possible area ofABCD?

Solution: The area of EFGH is EG ·FH sin θ/2, where θ is the angle between EG andFH. This is at most 90. However, we claim the area of ABCD is twice that of EFGH. Tosee this, notice that EF = AC/2 = GH, FG = BD/2 = HE, so EFGH is a parallelogram.The half of this parallelogram lying inside triangle DAB has area (BD/2)(h/2), where h isthe height from A to BD, and triangle DAB itself has area BD · h/2 = 2 · (BD/2)(h/2). Asimilar computation holds in triangle BCD, proving the claim. Thus, the area of ABCD isat most 180 . And this maximum is attainable — just take a rectangle with AB = CD =15, BC = DA = 12.

53. ABC is a triangle with points E, F on sides AC, AB, respectively. Suppose thatBE,CF intersect at X. It is given that AF/FB = (AE/EC)2 and that X is the midpointof BE. Find the ratio CX/XF .

Solution: Let x = AE/EC. By Menelaus’s theorem applied to triangle ABE and lineCXF ,

1 =AF

FB· BX

XE· EC

CA=

x2

x + 1.

Thus, x2 = x + 1, and x must be positive, so x = (1 +√

5)/2. Now apply Menelaus totriangle ACF and line BXE, obtaining

1 =AE

EC· CX

XF· FB

BA=

CX

XF· x

x2 + 1,

so CX/XF = (x2 + 1)/x = (2x2 − x)/x = 2x− 1 =√

5 .

54. How many pairs of integers (a, b), with 1 ≤ a ≤ b ≤ 60, have the property that b isdivisible by a and b + 1 is divisible by a + 1?

Solution: The divisibility condition is equivalent to b − a being divisible by both aand a + 1, or, equivalently (since these are relatively prime), by a(a + 1). Any b satisfyingthe condition is automatically ≥ a, so it suffices to count the number of values b − a ∈{1 − a, 2 − a, . . . , 60 − a} that are divisible by a(a + 1) and sum over all a. The numberof such values will be precisely 60/[a(a + 1)] whenever this quantity is an integer, whichfortunately happens for every a ≤ 5; we count:a = 1 gives 30 values of b;

13

a = 2 gives 10 values of b;a = 3 gives 5 values of b;a = 4 gives 3 values of b;a = 5 gives 2 values of b;a = 6 gives 2 values (b = 6 or 48);any a ≥ 7 gives only one value, namely b = a, since b > a implies b ≥ a + a(a + 1) > 60.

Adding these up, we get a total of 106 pairs.

55. A sequence of positive integers is given by a1 = 1 and an = gcd(an−1, n) + 1 forn > 1. Calculate a2002.

Solution: 3 . It is readily seen by induction that an ≤ n for all n. On the other hand,a1999 is one greater than a divisor of 1999. Since 1999 is prime, we have a1999 = 2 or 2000;the latter is not possible since 2000 > 1999, so we have a1999 = 2. Now we straightforwardlycompute a2000 = 3, a2001 = 4, and a2002 = 3.

56. x, y are positive real numbers such that x + y2 = xy. What is the smallest possiblevalue of x?

Solution: 4 Notice that x = y2/(y−1) = 2+(y−1)+1/(y−1) ≥ 2+2 = 4. Conversely,x = 4 is achievable, by taking y = 2.

57. How many ways, without taking order into consideration, can 2002 be expressedas the sum of 3 positive integers (for instance, 1000 + 1000 + 2 and 1000 + 2 + 1000 areconsidered to be the same way)?

Solution: Call the three numbers that sum to 2002 A, B, and C. In order to preventredundancy, we will consider only cases where A ≤ B ≤ C. Then A can range from 1 to667, inclusive. For odd A, there are 1000 − 3(A−1)

2possible values for B. For each choice

of A and B, there can only be one possible C, since the three numbers must add up to afixed value. We can add up this arithmetic progression to find that there are 167167 possiblecombinations of A, B, C, for odd A. For each even A, there are 1002 − 3A

2possible values

for B. Therefore, there are 166833 possible combinations for even A. In total, this makes334000 possibilities.

58. A sequence is defined by a0 = 1 and an = 2an−1 for n ≥ 1. What is the last digit (inbase 10) of a15?

Solution: 6 . Certainly a13 ≥ 2, so a14 is divisible by 22 = 4. Writing a14 = 4k, we havea15 = 24k = 16k. But every power of 16 ends in 6, so this is the answer.

59. Determine the value of

1 · 2− 2 · 3 + 3 · 4− 4 · 5 + · · ·+ 2001 · 2002.

Solution: 2004002 . Rewrite the expression as

2 + 3 · (4− 2) + 5 · (6− 4) + · · ·+ 2001 · (2002− 2000)

14

= 2 + 6 + 10 + · · ·+ 4002.

This is an arithmetic progression with (4002− 2)/4 + 1 = 1001 terms and average 2002, soits sum is 1001 · 2002 = 2004002.

60. A 5 × 5 square grid has the number −3 written in the upper-left square and thenumber 3 written in the lower-right square. In how many ways can the remaining squaresbe filled in with integers so that any two adjacent numbers differ by 1, where two squaresare adjacent if they share a common edge (but not if they share only a corner)?

Solution: 250 If the square in row i, column j contains the number k, let its “index”be i + j − k. The constraint on adjacent squares now says that if a square has index r,the squares to its right and below it each have index r or r + 2. The upper-left square hasindex 5, and the lower-right square has index 7, so every square must have index 5 or 7. Theboundary separating the two types of squares is a path consisting of upward and rightwardsteps; it can be extended along the grid’s border so as to obtain a path between the lower-leftand upper-right corners. Conversely, any such path uniquely determines each square’s indexand hence the entire array of numbers — except that the two paths lying entirely along theborder of the grid fail to separate the upper-left from the lower-right square and thus do notcreate valid arrays (since these two squares should have different indices). Each path consistsof 5 upward and 5 rightward steps, so there are

(105

)= 252 paths, but two are impossible, so

the answer is 250.

61. Bob Barker went back to school for a PhD in math, and decided to raise theintellectual level of The Price is Right by having contestants guess how many objects existof a certain type, without going over. The number of points you will get is the percentageof the correct answer, divided by 10, with no points for going over (i.e. a maximum of 10points).

Let’s see the first object for our contestants...a table of shape (5, 4, 3, 2, 1) is an arrange-ment of the integers 1 through 15 with five numbers in the top row, four in the next, threein the next, two in the next, and one in the last, such that each row and each column isincreasing (from left to right, and top to bottom, respectively). For instance:

1 2 3 4 56 7 8 910 11 1213 1415

is one table. How many tables are there?Solution: 15!/(34 · 53 · 72 · 9) = 292864 . These are Standard Young Tableaux.

62. Our next object up for bid is an arithmetic progression of primes. For example,the primes 3, 5, and 7 form an arithmetic progression of length 3. What is the largestpossible length of an arithmetic progression formed of positive primes less than 1,000,000?Be prepared to justify your answer.

15

Solution: 12 . We can get 12 with 110437 and difference 13860.

63. Our third and final item comes to us from Germany, I mean Geometry. It is knownthat a regular n-gon can be constructed with straightedge and compass if n is a primethat is 1 plus a power of 2. It is also possible to construct a 2n-gon whenever an n-gon isconstructible, or a p1p2 · · · pm-gon where the pi’s are distinct primes of the above form. Whatis really interesting is that these conditions, together with the fact that we can construct asquare, is that they give us all constructible regular n-gons. What is the largest n less than4,300,000,000 such that a regular n-gon is constructible?

Solution: The known primes of this form (Fermat primes) are 3, 5, 17, 257, and 65537,and the result is due to Gauss (German). If there are other such primes (unknown), thenthey are much bigger than 1010. So for each product of these primes, we can divide 4.3 · 109

by that number and take log2 to find the largest power of 2 to multiply by, then compare theresulting numbers. There are 32 cases to check, or just observe that 232 = 4, 294, 967, 296 isso close that there’s likely a shortcut. Note that 232 + 1 is divisible by 641, and hence notprime. 3 · 5 · 17 · 257 · 65537 = 232 − 1 is smaller; replacing any of the factors by the closestpower of 2 only decreases the product, and there’s not enough room to squeeze in an extra

factor of 2 without replacing all of them, and that gives us 232 , so indeed that it is theanswer.

Help control the pet population. Have your pets spayed or neutered. Bye-bye.

16

Team EventHMMT 2002

Palindromes. A palindrome is a positive integer n not divisible by 10 such that if you write thedecimal digits of n in reverse order, the number you get is n itself. For instance, the numbers 4and 25752 are palindromes.

1. [15] Determine the number of palindromes that are less than 1000.

2. [30] Determine the number of four-digit integers n such that n and 2n are both palindromes.

3. [40] Suppose that a positive integer n has the property that n, 2n, 3n, . . ., 9n are all palindromes.Prove that the decimal digits of n are all zeros or ones.

Floor functions. The notation bxc stands for the largest integer less than or equal to x.

4. [15] Let n be an integer. Prove that⌊n

2

⌋+⌊

n + 12

⌋= n.

5. [20] Prove for integers n that ⌊n

2

⌋⌊n + 1

2

⌋=⌊

n2

4

⌋.

In problems 6–7 you may use without proof the known summations

L∑n=1

n = n(n + 1)/2 andL∑

n=1

n3 = n2(n + 1)2/4 for positive integers L.

6. [20] For positive integers L, let SL =∑L

n=1 bn/2c. Determine all L for which SL is a squarenumber.

7. [45] Let TL =∑L

n=1

⌊n3/9

⌋for positive integers L. Determine all L for which TL is a square

number.

Luck of the dice. Problems 8–12 concern a two-player game played on a board consisting of fourteenspaces in a row. The leftmost space is labeled START, and the rightmost space is labeled END.Each of the twelve other squares, which we number 1 through 12 from left to right, may be blankor may be labeled with an arrow pointing to the right. The term blank square will refer to one ofthese twelve squares that is not labeled with an arrow. The set of blank squares on the board willbe called a board configuration; the board below uses the configuration {1, 2, 3, 4, 7, 8, 10, 11, 12}.

START ⇒ ⇒ ⇒ END1 2 3 4 5 6 7 8 9 10 11 12

For i ∈ {1, 2}, player i has a die that produces each integer from 1 to si with probability 1/si. Heres1 and s2 are positive integers fixed before the game begins. The game rules are as follows:

1. The players take turns alternately, and player 1 takes the first turn.

2. On each of his turns, player i rolls his die and moves his piece to the right by the numberof squares that he rolled. If his move ends on a square marked with an arrow, he moves hispiece forward another si squares. If that move ends on an arrow, he moves another si squares,repeating until his piece comes to rest on a square without an arrow.

3. If a player’s move would take him past the END square, instead he lands on the END square.

4. Whichever player reaches the END square first wins.

As an example, suppose that s1 = 3 and the first player is on square 4 in the sample board shownabove. If the first player rolls a 2, he moves to square 6, then to square 9, finally coming to rest onsquare 12. If the second player does not reach the END square on her next turn, the first playerwill necessarily win on his next turn, as he must roll at least a 1.

8. [35] In this problem only, assume that s1 = 4 and that exactly one board square, say squarenumber n, is marked with an arrow. Determine all choices of n that maximize the average distancein squares the first player will travel in his first two turns.

9. [30] In this problem suppose that s1 = s2. Prove that for each board configuration, the firstplayer wins with probability strictly greater than 1

2 .

10. [30] Exhibit a configuration of the board and a choice of s1 and s2 so that s1 > s2, yet thesecond player wins with probability strictly greater than 1

2 .

11. [55] In this problem assume s1 = 3 and s2 = 2. Determine, with proof, the nonnegative integerk with the following property:

1. For every board configuration with strictly fewer than k blank squares, the first player winswith probability strictly greater than 1

2 ; but

2. there exists a board configuration with exactly k blank squares for which the second playerwins with probability strictly greater than 1

2 .

12. [65] Now suppose that before the game begins, the players choose the initial game state asfollows:

1. The first player chooses s1 subject to the constraint that 2 ≤ s1 ≤ 5; then

2. the second player chooses s2 subject to the constraint that 2 ≤ s2 ≤ 5 and then specifies theboard configuration.

Prove that the second player can always make her decisions so that she will win the game withprobability strictly greater than 1

2 .

Team Event SolutionsHMMT 2002

Palindromes. A palindrome is a positive integer n not divisible by 10 such that if you write thedecimal digits of n in reverse order, the number you get is n itself. For instance, the numbers 4and 25752 are palindromes.

1. [15] Determine the number of palindromes that are less than 1000.

Solution. Every one-digit number (there are nine) is a palindrome. The two-digit palindromeshave the form a a for a nonzero digit a, so there are nine of them. A three-digit palindrome is a b awith a a nonzero digit and b any digit, so there are 9 × 10 = 90 of these. Thus the number ofpalindromes less than 1000 is 9 + 9 + 90 = 108.

2. [30] Determine the number of four-digit integers n such that n and 2n are both palindromes.

Solution. Let n = a b b a. If a, b ≤ 4 then there are no carries in the multiplication n × 2, and2n = (2a) (2b) (2b) (2a) is a palindrome. We shall show conversely that if n and 2n are palindromes,then necessarily a, b ≤ 4. Hence the answer to the problem is 4 × 5 = 20 (because a cannot bezero).

If a ≥ 5 then 2n is a five-digit number whose most significant digit is 1, but because 2n is even,its least significant digit is even, contradicting the assumption that 2n is a palindrome. Thereforea ≤ 4. Consequently 2n is a four-digit number, and its tens and hundreds digits must be equal.Because a ≤ 4, there is no carry out of the ones place in the multiplication n× 2, and therefore thetens digit of 2n is the ones digit of 2b. In particular, the tens digit of 2n is even. But if b ≥ 5, thecarry out of the tens place makes the hundreds digit of 2n odd, which is impossible. Hence b ≤ 4as well.

3. [40] Suppose that a positive integer n has the property that n, 2n, 3n, . . ., 9n are all palindromes.Prove that the decimal digits of n are all zeros or ones.

Solution. First consider the ones digit a of n; we claim that a = 1. Certainly a cannot be even,for then 5n would be divisible by 10. If a is 5, 7, or 9, then 2n has an even ones digit, while itsmost significant digit is 1. If a is 3, then 4n has an even ones digit but most significant digit 1.Thus a = 1 is the only possibility. Moreover 9n has the same number of digits as n, for otherwise9n would have most significant digit 1 but least significant digit 9, which is forbidden.

Now suppose n has at least one digit that is neither a zero nor a one. Let b be the leftmost (i.e.,most significant) such digit, so that the left end of the decimal representation of n looks like

a1 . . . ar b . . .

for some r ≥ 1 and digits ai ∈ {0, 1}. When n is multiplied by 9, there will be a carry out of thecolumn containing b. In particular, the rth digit from the left in 9n will not be 9ar. But the rightend of the decimal representation of n is

. . . ar . . . a1 ;

because each ai is 0 or 1, there are no carries out of the first r−1 columns, so the rth digit from theright in 9n will be 9ar. Thus 9n is not a palindrome, a contradiction. This completes the proof.

1

Floor functions. The notation bxc stands for the largest integer less than or equal to x.

4. [15] Let n be an integer. Prove that⌊n

2

⌋+⌊

n + 12

⌋= n.

Solution. Suppose n = 2m is even. Then bn/2c = bmc = m and b(n + 1)/2c = bm + 1/2c = m,whose sum is m+m = 2m = n. Otherwise n = 2m+1 is odd. In this case bn/2c = bm + 1/2c = mand b(n + 1)/2c = bm + 1c = m + 1, whose sum is m + (m + 1) = 2m + 1 = n, as desired.

5. [20] Prove for integers n that ⌊n

2

⌋⌊n + 1

2

⌋=⌊

n2

4

⌋.

Solution. Suppose n = 2m is even; then bn/2c = bmc = m and b(n + 1)/2c = bm + 1/2c = m,whose product is m2 =

⌊m2⌋

=⌊(2m)2/4

⌋. Otherwise n = 2m + 1 is odd, so that bn/2c =

bm + 1/2c = m and b(n + 1)/2c = bm + 1c = m + 1, whose product is m2 + m. On the other side,we find that ⌊

n2

4

⌋=⌊

4m2 + 4m + 14

⌋=⌊m2 + m +

14

⌋= m2 + m,

as desired.

In problems 6–7 you may use without proof the known summations

L∑n=1

n = n(n + 1)/2 andL∑

n=1

n3 = n2(n + 1)2/4 for positive integers L.

6. [20] For positive integers L, let SL =∑L

n=1 bn/2c. Determine all L for which SL is a squarenumber.

Solution. We distinguish two cases depending on the parity of L. Suppose first that L = 2k − 1is odd, where k ≥ 1. Then

SL =∑

1≤n≤2k−1

⌊n

2

⌋= 2

∑0≤m<k

m = 2 · k(k − 1)2

= k(k − 1).

If k = 1, this is the square number 0. If k > 1 then (k − 1)2 < k(k − 1) < k2, so k(k − 1) is notsquare. Now suppose L = 2k is even, where k ≥ 1. Then SL = SL−1 + k = k2 is always square.Hence SL is square exactly when L = 1 or L is even.

7. [45] Let TL =∑L

n=1

⌊n3/9

⌋for positive integers L. Determine all L for which TL is a square

number.

Solution. Since TL is square if and only if 9TL is square, we may consider 9TL instead of TL.

It is well known that n3 is congruent to 0, 1, or 8 modulo 9 according as n is congruent to 0, 1,or 2 modulo 3. (Proof: (3m + k)3 = 27m3 + 3(9m2)k + 3(3m)k2 + k3 ≡ k3 (mod 9).) Therefore

2

n3− 9⌊n3/9

⌋is 0, 1, or 8 according as n is congruent to 0, 1, or 2 modulo 3. We find therefore that

9TL =∑

1≤n≤L

9⌊

n3

9

⌋=

∑1≤n≤L

n3 −#{1 ≤ n ≤ L : n ≡ 1 (mod 3)} − 8#{1 ≤ n ≤ L : n ≡ 2 (mod 3)}

=(

12L(L + 1)

)2

−⌊

L + 23

⌋− 8

⌊L + 1

3

⌋.

Clearly 9TL <(L(L + 1)/2

)2for L ≥ 1. We shall prove that 9TL >

(L(L + 1)/2 − 1

)2for L ≥ 4,

whence 9TL is not square for L ≥ 4. Because(L(L + 1)/2− 1

)2=(L(L + 1)/2

)2 − L(L + 1) + 1,

we need only show that ⌊L + 2

3

⌋+ 8

⌊L + 1

3

⌋≤ L2 + L− 2.

But the left-hand side of this is bounded above by 3L+10/3, and the inequality 3L+10/3 ≤ L2+L−2means exactly L2 − 2L− 16/3 ≥ 0 or (L− 1)2 ≥ 19/3, which is true for L ≥ 4, as desired.

Hence TL is not square for L ≥ 4. By direct computation we find T1 = T2 = 0 and T3 = 3, so TL issquare only for L ∈ {1,2}.

Luck of the dice. Problems 8–12 concern a two-player game played on a board consisting of fourteenspaces in a row. The leftmost space is labeled START, and the rightmost space is labeled END.Each of the twelve other squares, which we number 1 through 12 from left to right, may be blankor may be labeled with an arrow pointing to the right. The term blank square will refer to one ofthese twelve squares that is not labeled with an arrow. The set of blank squares on the board willbe called a board configuration; the board below uses the configuration {1, 2, 3, 4, 7, 8, 10, 11, 12}.

START ⇒ ⇒ ⇒ END1 2 3 4 5 6 7 8 9 10 11 12

For i ∈ {1, 2}, player i has a die that produces each integer from 1 to si with probability 1/si. Heres1 and s2 are positive integers fixed before the game begins. The game rules are as follows:

1. The players take turns alternately, and player 1 takes the first turn.

2. On each of his turns, player i rolls his die and moves his piece to the right by the numberof squares that he rolled. If his move ends on a square marked with an arrow, he moves hispiece forward another si squares. If that move ends on an arrow, he moves another si squares,repeating until his piece comes to rest on a square without an arrow.

3. If a player’s move would take him past the END square, instead he lands on the END square.

4. Whichever player reaches the END square first wins.

As an example, suppose that s1 = 3 and the first player is on square 4 in the sample board shownabove. If the first player rolls a 2, he moves to square 6, then to square 9, finally coming to rest on

3

square 12. If the second player does not reach the END square on her next turn, the first playerwill necessarily win on his next turn, as he must roll at least a 1.

8. [35] In this problem only, assume that s1 = 4 and that exactly one board square, say squarenumber n, is marked with an arrow. Determine all choices of n that maximize the average distancein squares the first player will travel in his first two turns.

Solution. Because expectation is linear, the average distance the first player travels in his first twoturns is the average sum of two rolls of his die (which does not depend on the board configuration)plus four times the probability that he lands on the arrow on one of his first two turns. Thus we justneed to maximize the probability that player 1 lands on the arrow in his first two turns. If n ≥ 5,player 1 cannot land on the arrow in his first turn, so he encounters the arrow with probability atmost 1/4. If instead n ≤ 4, player 1 has a 1/4 chance of landing on the arrow on his first turn.If he misses, then he has a 1/4 chance of hitting the arrow on his second turn provided that heis not beyond square n already. The chance that player 1’s first roll left him on square n − 1 orfarther left is (n− 1)/4. Hence his probability of benefiting from the arrow in his first two turns is1/4 + (1/4)(n− 1)/4, which is maximized for n = 4, where it is greater than the value of 1/4 thatwe get from n ≥ 5. Hence the answer is n = 4.

9. [30] In this problem suppose that s1 = s2. Prove that for each board configuration, the firstplayer wins with probability strictly greater than 1

2 .

Solution. Let σ1 and σ2 denote the sequence of the next twelve die rolls that players 1 and 2respectively will make. The outcome of the game is completely determined by the σi. Now player1 wins in all cases in which σ1 = σ2, for then each of player 2’s moves bring her piece to a squarealready occupied by player 1’s piece. It is sufficient, therefore, to show that player 1 wins at leasthalf the cases in which σ1 6= σ2. But all these cases can be partitioned into disjoint pairs

{(σ1, σ2), (σ2, σ1)},

and player 1 wins in at least one case in each pair. For if player 2 wins in the case (σ1, σ2), sayon her nth turn, the first n elements of σ1 do not take player 1 beyond space 12, while the first nelements of σ2 must take player 2 beyond space 12. Clearly, then, player 1 wins (σ2, σ1) on his nth

turn.

10. [30] Exhibit a configuration of the board and a choice of s1 and s2 so that s1 > s2, yet thesecond player wins with probability strictly greater than 1

2 .

Solution. Let s1 = 3 and s2 = 2 and place an arrow on all the even-numbered squares. In thisconfiguration, player 1 can move at most six squares in a turn: up to three from his roll and anadditional three if his roll landed him on an arrow. Hence player 1 cannot win on his first orsecond turn. Player 2, however, wins immediately if she ever lands on an arrow. Thus player 2 hasprobability 1/2 of winning on her first turn, and failing that, she has probability 1/2 of winning onher second turn. Hence player 2 wins with probability at least 1/2 + (1/2)(1/2) = 3/4.

11. [55] In this problem assume s1 = 3 and s2 = 2. Determine, with proof, the nonnegative integerk with the following property:

1. For every board configuration with strictly fewer than k blank squares, the first player winswith probability strictly greater than 1

2 ; but

4

2. there exists a board configuration with exactly k blank squares for which the second playerwins with probability strictly greater than 1

2 .

Solution. The answer is k = 3. Consider the configuration whose blank squares are 2, 6, and 10.Because these numbers represent all congruence classes modulo 3, player 1 cannot win on his firstturn: he will come to rest on one of the blank squares. But player 2 will win on her first turn if sherolls a 1, for 2, 6, and 10 are all even. Thus player 2 wins on her first turn with probability 1/2.Failing this, player 1 may fail to win on his second turn, for instance, if he rolled a 2 previouslyand now rolls a 1, ending up on square 6. Then player 2 will again have probability 1/2 of winningon her next turn. Thus player 2 wins the game with probability exceeding 1/2.

We must now prove that all configurations with fewer than three blanks favor player 1. If thenumbers of the blank squares represent at most one residue class modulo 3, then clearly player1 wins on his first turn with probability at least 2/3. This disposes of the cases of no blanks,just one blank, and two blanks that are congruent modulo 3. In the remaining case, there aretwo blank squares whose indices are incongruent modulo 3. Then player 1 wins on his first turnwith probability only 1/3. If he does not win immediately, player 2 wins on her first turn withprobability at most 1/2, for there is a blank in at least one congruence class modulo 2. If player2 does not win on her first turn, then player 1 wins on his second turn with probability at least2/3, for there is only one blank square in front of him now. Thus player 1 wins the game withprobability at least 1/3 + (2/3)(1/2)(2/3) = 5/9 > 1/2, as desired.

12. [65] Now suppose that before the game begins, the players choose the initial game state asfollows:

1. The first player chooses s1 subject to the constraint that 2 ≤ s1 ≤ 5; then

2. the second player chooses s2 subject to the constraint that 2 ≤ s2 ≤ 5 and then specifies theboard configuration.

Prove that the second player can always make her decisions so that she will win the game withprobability strictly greater than 1

2 .

Solution. If s1 ∈ {3, 5}, take s2 = 2 and put arrows on the even-numbered squares. Player 1cannot win on his first turn because he can move at most 2s1 spaces in a turn. Player 2 wins onher first turn with probability 1/2. Failing that, player 1 might fail to win on his second turn,and player 2 will again have probability 1/2 of winning on her second turn, so her probability ofwinning the game is certainly greater than 1/2.

If s1 = 4, take s2 = 3 and leave blank only squares 1, 4, 7, and 10. These occupy all congruenceclasses modulo 4, so player 1 cannot win on his first turn. But the blank squares lie in the samecongruence class modulo 3, so player 2 then wins on her first turn with probability 2/3.

Finally, if s1 = 2, take s2 = 5 and leave all squares blank. Then player 2 moves 3 squares in a turnon average, hence covers 15 squares on average in her first five turns. Moreover, the distributionof player 2’s distance traveled in five turns is symmetric about 15. Thus player 2 has probabilitygreater than 1/2 of reaching END by the end of her fifth turn. Player 1, on the other hand, cannotwin in five turns because he can move at most 10 squares in those turns.

5

Harvard-MIT Mathematics TournamentMarch 15, 2003

Individual Round: Algebra Subject Test

1. Find the smallest value of x such that a ≥ 14√a− x for all nonnegative a.

2. Compute tan2(20◦)−sin2(20◦)tan2(20◦) sin2(20◦)

.

3. Find the smallest n such that n! ends in 290 zeroes.

4. Simplify: 2√

1.5 +√

2− (1.5 +√

2).

5. Several positive integers are given, not necessarily all different. Their sum is 2003.Suppose that n1 of the given numbers are equal to 1, n2 of them are equal to 2, . . .,n2003 of them are equal to 2003. Find the largest possible value of

n2 + 2n3 + 3n4 + · · ·+ 2002n2003.

6. Let a1 = 1, and let an = bn3/an−1c for n > 1. Determine the value of a999.

7. Let a, b, c be the three roots of p(x) = x3 + x2 − 333x− 1001. Find a3 + b3 + c3.

8. Find the value of 132+1

+ 142+2

+ 152+3

+ · · ·.

9. For how many integers n, for 1 ≤ n ≤ 1000, is the number 12

(2nn

)even?

10. Suppose P (x) is a polynomial such that P (1) = 1 and

P (2x)

P (x+ 1)= 8− 56

x+ 7

for all real x for which both sides are defined. Find P (−1).

1

Harvard-MIT Mathematics TournamentMarch 15, 2003

Individual Round: Algebra Subject Test — Solutions

1. Find the smallest value of x such that a ≥ 14√a− x for all nonnegative a.

Solution: 49

We want to find the smallest value of x such that x ≥ 14sqrta − a for all a. This isjust the maximum possible value of 14

√a − a = 49 − (

√a − 7)2, which is clearly 49,

achieved when a = 49.

2. Compute tan2(20◦)−sin2(20◦)tan2(20◦) sin2(20◦)

.

Solution: 1

If we multiply top and bottom by cos2(20◦), the numerator becomes sin2(20◦) · (1 −cos2 20◦) = sin4(20◦), while the denominator becomes sin4(20◦) also. So they are equal,and the ratio is 1.

3. Find the smallest n such that n! ends in 290 zeroes.

Solution: 1170

Each 0 represents a factor of 10 = 2 · 5. Thus, we wish to find the smallest factorialthat contains at least 290 2’s and 290 5’s in its prime factorization. Let this numberbe n!, so the factorization of n! contains 2 to the power p and 5 to the power q, where

p =⌊n

2

⌋+

⌊n

22

⌋+

⌊n

23

⌋+ · · · and q =

⌊n

5

⌋+

⌊n

52

⌋+

⌊n

53

⌋+ · · ·

(this takes into account one factor for each single multiple of 2 or 5 that is ≤ n,an additional factor for each multiple of 22 or 52, and so on). Naturally, p ≥ qbecause 2 is smaller than 5. Thus, we want to bring q as low to 290 as possible.If q = bn

5c + b n

52 c + b n53 c + · · ·, we form a rough geometric sequence (by taking away

the floor function) whose sum is represented by 290 ≈ n/51−1/5

. Hence we estimaten = 1160, and this gives us q = 288. Adding 10 to the value of n gives the necessarytwo additional factors of 5, and so the answer is 1170.

4. Simplify: 2√

1.5 +√

2− (1.5 +√

2).

Solution: 1/2

The given expression equals√

6 + 4√

2− (1.5+√

2) =√

6 + 2√

8− (1.5+√

2). But on

inspection, we see that (√

2+√

4)2 = 6+2√

8, so the answer is (√

2+√

4)−(1.5+√

2) =2− 3/2 = 1/2.

5. Several positive integers are given, not necessarily all different. Their sum is 2003.Suppose that n1 of the given numbers are equal to 1, n2 of them are equal to 2, . . .,n2003 of them are equal to 2003. Find the largest possible value of

n2 + 2n3 + 3n4 + · · ·+ 2002n2003.

1

Solution: 2002

The sum of all the numbers is n1 +2n2 + · · ·+2003n2003, while the number of numbersis n1 + n2 + · · ·+ n2003. Hence, the desired quantity equals

(n1 + 2n2 + · · ·+ 2003n2003)− (n1 + n2 + · · ·+ n2003)

= (sum of the numbers)− (number of numbers)

= 2003− (number of numbers),

which is maximized when the number of numbers is minimized. Hence, we should havejust one number, equal to 2003, and then the specified sum is 2003− 1 = 2002.

Comment: On the day of the contest, a protest was lodged (successfully) on thegrounds that the use of the words “several” and “their” in the problem statementimplies there must be at least 2 numbers. Then the answer is 2001, and this maximumis achieved by any two numbers whose sum is 2003.me way.)

6. Let a1 = 1, and let an = bn3/an−1c for n > 1. Determine the value of a999.

Solution: 999

We claim that for any odd n, an = n. The proof is by induction. To get the basecases n = 1, 3, we compute a1 = 1, a2 = b23/1c = 8, a3 = b33/8c = 3. And ifthe claim holds for odd n ≥ 3, then an+1 = b(n + 1)3/nc = n2 + 3n + 3, so an+2 =b(n+2)3/(n2+3n+3)c = b(n3+6n2+12n+8)/(n2+3n+2)c = bn+2+ n2+3n+2

n2+3n+3c = n+2.

So the claim holds, and in particular, a999 = 999.

7. Let a, b, c be the three roots of p(x) = x3 + x2 − 333x− 1001. Find a3 + b3 + c3.

Solution: 2003

We know that x3 + x2 − 333x − 1001 = (x − a)(x − b)(x − c) = x3 − (a + b + c)x2 +(ab+ bc+ ca)x−abc. Also, (a+ b+ c)3−3(a+ b+ c)(ab+ bc+ ca)+3abc = a3 + b3 + c3.Thus, a3 + b3 + c3 = (−1)3 − 3(−1)(−333) + 3 · 1001 = 2003.

8. Find the value of 132+1

+ 142+2

+ 152+3

+ · · ·.Solution: 13/36

Each term takes the form

1

n2 + (n− 2)=

1

(n+ 2) · (n− 1).

Using the method of partial fractions, we can write (for some constants A,B)

1

(n+ 2) · (n− 1)=

A

(n+ 2)+

B

(n− 1)

⇒ 1 = A · (n− 1) +B · (n+ 2)

Setting n = 1 we get B = 13, and similarly with n = −2 we get A = −1

3. Hence the

sum becomes

1

3·[(

1

2− 1

5

)+

(1

3− 1

6

)+

(1

4− 1

7

)+

(1

5− 1

8

)+ · · ·

].

Thus, it telescopes, and the only terms that do not cancel produce a sum of 13· (1

2+

13

+ 14) = 13

36.

2

9. For how many integers n, for 1 ≤ n ≤ 1000, is the number 12

(2nn

)even?

Solution: 990

In fact, the expression(

2nn

)is always even, and it is not a multiple of four if and only

if n is a power of 2, and there are 10 powers of 2 between 1 and 1000.

Let f(N) denote the number of factors of 2 in N . Thus,

f(n!) =⌊n

2

⌋+

⌊n

4

⌋+

⌊n

8

⌋+ · · · =

∞∑

k=1

⌊n

2k

⌋.

Also, it is clear that f(ab) = f(a) + f(b) and f(ab) = f(a)− f(b) for integers a, b. Now

for any positive integer n, let m be the integer such that 2m ≤ n < 2m+1. Then

f

((2n

n

))= f

((2n)!

n!n!

)=

∞∑

k=1

⌊2n

2k

⌋− 2

( ∞∑

k=1

⌊n

2k

⌋)

=∞∑

k=1

⌊n

2k−1

⌋− 2

( ∞∑

k=1

⌊n

2k

⌋)

= bnc −( ∞∑

k=1

⌊n

2k

⌋)

= n−(

m∑

k=1

⌊n

2k

⌋)

≥ n−(

m∑

k=1

n

2k

)

= n− n(

2m − 1

2m

)=

n

2m≥ 1.

Both equalities hold when n = 2m, and otherwise, f((

2nn

)) > 1.

10. Suppose P (x) is a polynomial such that P (1) = 1 and

P (2x)

P (x+ 1)= 8− 56

x+ 7

for all real x for which both sides are defined. Find P (−1).

Solution: −5/21

Cross-multiplying gives (x + 7)P (2x) = 8xP (x + 1). If P has degree n and leadingcoefficient c, then the leading coefficients of the two sides are 2nc and 8c, so n = 3.Now x = 0 is a root of the right-hand side, so it’s a root of the left-hand side, sothat P (x) = xQ(x) for some polynomial Q⇒ 2x(x+ 7)Q(2x) = 8x(x+ 1)Q(x+ 1) or(x+7)Q(2x) = 4(x+1)Q(x+1). Similarly, we see that x = −1 is a root of the left-handside, giving Q(x) = (x + 2)R(x) for some polynomial R ⇒ 2(x + 1)(x + 7)R(2x) =4(x + 1)(x + 3)R(x + 1), or (x + 7)R(2x) = 2(x + 3)R(x + 1). Now x = −3 is a rootof the left-hand side, so R(x) = (x+ 6)S(x) for some polynomial S.

At this point, P (x) = x(x+2)(x+6)S(x), but P has degree 3, so S must be a constant.Since P (1) = 1, we get S = 1/21, and then P (−1) = (−1)(1)(5)/21 = −5/21.

3

Harvard-MIT Mathematics TournamentMarch 15, 2003

Individual Round: Calculus Subject Test

1. A point is chosen randomly with uniform distribution in the interior of a circle of radius1. What is its expected distance from the center of the circle?

2. A particle moves along the x-axis in such a way that its velocity at position x is givenby the formula v(x) = 2 + sin x. What is its acceleration at x = π

6?

3. What is the area of the region bounded by the curves y = x2003 and y = x1/2003 andlying above the x-axis?

4. The sequence of real numbers x1, x2, x3, . . . satisfies limn→∞(x2n + x2n+1) = 315 andlimn→∞(x2n + x2n−1) = 2003. Evaluate limn→∞(x2n/x2n+1).

5. Find the minimum distance from the point (0, 5/2) to the graph of y = x4/8.

6. For n an integer, evaluate

limn→∞

(1√

n2 − 02+

1√n2 − 12

+ · · ·+ 1√n2 − (n− 1)2

).

7. For what value of a > 1 is ∫ a2

a

1

xlog

x− 1

32dx

minimum?

8. A right circular cone with a height of 12 inches and a base radius of 3 inches is filledwith water and held with its vertex pointing downward. Water flows out through ahole at the vertex at a rate in cubic inches per second numerically equal to the heightof the water in the cone. (For example, when the height of the water in the cone is 4inches, water flows out at a rate of 4 cubic inches per second.) Determine how manyseconds it will take for all of the water to flow out of the cone.

9. Two differentiable real functions f(x) and g(x) satisfy

f ′(x)g′(x)

= ef(x)−g(x)

for all x, and f(0) = g(2003) = 1. Find the largest constant c such that f(2003) > cfor all such functions f, g.

10. Evaluate ∫ ∞

−∞1− x2

1 + x4dx.

1

Harvard-MIT Mathematics TournamentMarch 15, 2003

Individual Round: Calculus Subject Test — Solutions

1. A point is chosen randomly with uniform distribution in the interior of a circle of radius1. What is its expected distance from the center of the circle?

Solution: 2/3

The probability of the point falling between a distance r and r + dr from the centeris the ratio of the area of the corresponding annulus to the area of the whole circle:π[(r+dr)2−r2]

π→ 2πr dr

π= 2r dr for small values of dr. Then the expected distance is∫ 1

0 r cot 2r dr = 23.

2. A particle moves along the x-axis in such a way that its velocity at position x is givenby the formula v(x) = 2 + sin x. What is its acceleration at x = π

6?

Solution: 5√

3/4

Acceleration is given by a = dvdt

= dvdx· dx

dt= dv

dx· v = cos x · (2 + sin x) = 5

√3/4.

3. What is the area of the region bounded by the curves y = x2003 and y = x1/2003 andlying above the x-axis?

Solution: 1001/1002

The two curves intersect at (0, 0) and (1, 1), so the desired area is∫ 1

0

(x1/2003 − x2003

)dx =

[x2004/2003

2004/2003− x2004

2004

]1

0=

1001

1002.

4. The sequence of real numbers x1, x2, x3, . . . satisfies limn→∞(x2n + x2n+1) = 315 andlimn→∞(x2n + x2n−1) = 2003. Evaluate limn→∞(x2n/x2n+1).

Solution: −1

We have limn→∞(x2n+1−x2n−1) = limn→∞[(x2n+x2n+1)−(x2n+x2n−1)] = 315−2003 =−1688; it follows that x2n+1 → −∞ as n→∞. Then

limn→∞

x2n

x2n+1

= limn→∞

x2n + x2n+1

x2n+1

− 1 = −1,

since x2n + x2n+1 → 315 while x2n+1 → −∞.

5. Find the minimum distance from the point (0, 5/2) to the graph of y = x4/8.

Solution:√

17/2

We want to minimize x2+(x4/8−5/2)2 = x8/64−5x4/8+x2+25/4, which is equivalentto minimizing z4/4 − 10z2 + 16z, where we have set z = x2. The derivative of thisexpression is z3 − 20z + 16, which is seen on inspection to have 4 as a root, leading tothe factorization (z − 4)(z + 2− 2

√2)(z + 2− 2

√2). Since z = x2 ranges over [0,∞),

the possible minima are at z = 0, z = −2 + 2√

2, and z = 4. However, the derivativeis positive on (0,−2 + 2

√2), so this leaves only 0 and 4 to be tried. We find that the

minimum is in fact achieved at z = 4, so the closest point on the graph is given by

x = ±2, with distance√

22 + (24/8− 5/2)2 =√

17/2.

1

6. For n an integer, evaluate

limn→∞

(1√

n2 − 02+

1√n2 − 12

+ · · ·+ 1√n2 − (n− 1)2

).

Solution: π/2

Note that 1√n2−i2

= 1n· 1√

1−( in

)2, so that the sum we wish to evaluate is just a Riemann

sum. Then,

limn→∞

(1

n

n−1∑

i=0

1√1− ( i

n)2

)=

∫ 1

0

1√1− x2

dx = [sin−1 x]10 =π

2.

7. For what value of a > 1 is ∫ a2

a

1

xlog

x− 1

32dx

minimum?

Solution: 3

Let f(a) =∫ a2

a1x

log x−132

dx. Then we want dfda

= 0; by the Fundamental Theorem ofCalculus and the chain rule, this implies that

2a

(1

a2log

a2 − 1

32

)− 1

alog

a− 1

32=

d

da

(∫ a2

c

1

xlog

x− 1

32dx−

∫ a

c

1

xlog

x− 1

32dx

)= 0,

where c is any constant with 1 < c < a. Then 2 log a2−132

= log a−132

, so that (a2−132

)2 =a−132

. After canceling factors of (a−1)/32 (since a > 1), this simplifies to (a2−1)(a+1) =32 ⇒ a3 + a2 − a − 33 = 0, which in turn factors as (a − 3)(a2 + 4a + 11) = 0. Thequadratic factor has no real solutions, so this leaves only a = 3. However, we have thata > 1, and we can check that f(1) = 0, lima→∞ f(a) > 0, and f(3) < 0, so the globalminimum does occur at a = 3.

8. A right circular cone with a height of 12 inches and a base radius of 3 inches is filledwith water and held with its vertex pointing downward. Water flows out through ahole at the vertex at a rate in cubic inches per second numerically equal to the heightof the water in the cone. (For example, when the height of the water in the cone is 4inches, water flows out at a rate of 4 cubic inches per second.) Determine how manyseconds it will take for all of the water to flow out of the cone.

Solution: 9π/2

When the water in the cone is h inches high, it forms a cone similar to the original, sothat its base has radius h/4 and its volume is hence πh3/48. The given condition thenstates that

d

dt

(πh3

48

)= −h⇒ πh2

16· dhdt

= −h⇒ 2h · dhdt

= −32

π.

Integrating with respect to t, we get that h2 = −32t/π + C; setting t = 0, h = 12, weget C = 144. The cone empties when h = 0, so 0 = −32t/π + 144⇒ t = 9π/2.

2

9. Two differentiable real functions f(x) and g(x) satisfy

f ′(x)g′(x)

= ef(x)−g(x)

for all x, and f(0) = g(2003) = 1. Find the largest constant c such that f(2003) > cfor all such functions f, g.

Solution: 1− ln 2

Rearranging the given equation gives f ′(x)e−f(x) = g′(x)e−g(x) for all x, so ddx

(e−f(x) −e−g(x)) = −f ′(x)e−f(x) +g′(x)e−g(x) = 0. Thus, e−f(x)−e−g(x) is a constant, and it mustbe less than e−f(0) = e−1. Thus, e−f(2003) < e−g(2003)+e−1 = 2e−1 = eln 2−1 ⇒ f(2003) >1 − ln 2. On the other hand, we can find positive-valued functions e−f(x), e−g(x) thattake on the required values at 0 and 2003 and have constant difference arbitrarily closeto e−1. For example, for arbitrarily large t, we can set e−f(x) = e−(t(2003−x)+1) + e−1 −e−(2003t+1) and e−g(x) = e−(t(2003−x)+1), and we can check that the resulting functionsf, g satisfy the required conditions. Thus, we can make f(2003) arbitrarily close to1− ln 2, so this is the answer.

10. Evaluate ∫ ∞

−∞1− x2

1 + x4dx.

Solution: 0

Let S =∫ ∞0 1/(x4 + 1) dx; note that the integral converges absolutely. Substituting

x = 1/u, so that dx = −1/u2 du, we have

S =∫ ∞

0

1

1 + x4dx =

∫ 0

∞1

1 + u−4

du

−u2=

∫ 0

∞−u2

u4 + 1du

=∫ ∞

0

u2

1 + u4du =

∫ ∞

0

x2

1 + x4dx

(the manipulations are justified by absolute convergence), from which we see that∫ ∞0 (1−x2)/(1+x4) dx = 0. Since the integrand is an even function, it follows that the

integral from −∞ to ∞ is zero as well.

3

Harvard-MIT Mathematics TournamentMarch 15, 2003

Individual Round: Combinatorics Subject Test

1. You have 2003 switches, numbered from 1 to 2003, arranged in a circle. Initially, eachswitch is either ON or OFF, and all configurations of switches are equally likely. Youperform the following operation: for each switch S, if the two switches next to S wereinitially in the same position, then you set S to ON; otherwise, you set S to OFF.What is the probability that all switches will now be ON?

2. You are given a 10× 2 grid of unit squares. Two different squares are adjacent if theyshare a side. How many ways can one mark exactly nine of the squares so that no twomarked squares are adjacent?

3. Daniel and Scott are playing a game where a player wins as soon as he has two pointsmore than his opponent. Both players start at par, and points are earned one at atime. If Daniel has a 60% chance of winning each point, what is the probability thathe will win the game?

4. In a certain country, there are 100 senators, each of whom has 4 aides. These senatorsand aides serve on various committees. A committee may consist either of 5 senators,of 4 senators and 4 aides, or of 2 senators and 12 aides. Every senator serves on 5committees, and every aide serves on 3 committees. How many committees are therealtogether?

5. We wish to color the integers 1, 2, 3, . . . , 10 in red, green, and blue, so that no twonumbers a and b, with a − b odd, have the same color. (We do not require that allthree colors be used.) In how many ways can this be done?

6. In a classroom, 34 students are seated in 5 rows of 7 chairs. The place at the center ofthe room is unoccupied. A teacher decides to reassign the seats such that each studentwill occupy a chair adjacent to his/her present one (i.e. move one desk forward, back,left or right). In how many ways can this reassignment be made?

7. You have infinitely many boxes, and you randomly put 3 balls into them. The boxesare labeled 1, 2, . . .. Each ball has probability 1/2n of being put into box n. The ballsare placed independently of each other. What is the probability that some box willcontain at least 2 balls?

8. For any subset S ⊆ {1, 2, . . . , 15}, a number n is called an “anchor” for S if n andn+ |S| are both members of S, where |S| denotes the number of members of S. Findthe average number of anchors over all possible subsets S ⊆ {1, 2, . . . , 15}.

9. At a certain college, there are 10 clubs and some number of students. For any twodifferent students, there is some club such that exactly one of the two belongs to thatclub. For any three different students, there is some club such that either exactly oneor all three belong to that club. What is the largest possible number of students?

1

10. A calculator has a display, which shows a nonnegative integer N , and a button, whichreplaces N by a random integer chosen uniformly from the set {0, 1, . . . , N − 1}, pro-vided that N > 0. Initially, the display holds the number N = 2003. If the button ispressed repeatedly until N = 0, what is the probability that the numbers 1, 10, 100,and 1000 will each show up on the display at some point?

2

Harvard-MIT Mathematics TournamentMarch 15, 2003

Individual Round: Combinatorics Subject Test — Solutions

1. You have 2003 switches, numbered from 1 to 2003, arranged in a circle. Initially, eachswitch is either ON or OFF, and all configurations of switches are equally likely. Youperform the following operation: for each switch S, if the two switches next to S wereinitially in the same position, then you set S to ON; otherwise, you set S to OFF.What is the probability that all switches will now be ON?

Solution: 1/22002

There are 22003 equally likely starting configurations. All switches end up ON if andonly if switches 1, 3, 5, 7, . . . , 2003, 2, 4, . . . , 2002 — i.e. all 2003 of them — were initiallyin the same position. This initial position can be ON or OFF, so this situation occurswith probability 2/22003 = 1/22002.

2. You are given a 10× 2 grid of unit squares. Two different squares are adjacent if theyshare a side. How many ways can one mark exactly nine of the squares so that no twomarked squares are adjacent?

Solution: 36

Since each row has only two squares, it is impossible for two marked squares to bein the same row. Therefore, exactly nine of the ten rows contain marked squares.Consider two cases:

Case 1: The first or last row is empty. These two cases are symmetrical, so assumewithout loss of generality that the first row is empty. There are two possibilities for thesecond row: either the first square is marked, or the second square is marked. Sincethe third row must contain a marked square, and it cannot be in the same columnas the marked square in the second row, the third row is determined by the second.Similarly, all the remaining rows are determined. This leaves two possibilities if thefirst row is empty. Thus, there are four possibilities if the first or last row is empty.

Case 2: The empty row is not the first or last. Then, there are two blocks of (one ofmore) consecutive rows of marked squares. As above, the configuration of the rows ineach of the two blocks is determined by the position of the marked square in the firstof its rows. That makes 2× 2 = 4 possible configurations. There are eight possibilitiesfor the empty row, making a total of 32 possibilities in this case.

Together, there are 36 possible configurations of marked squares.

3. Daniel and Scott are playing a game where a player wins as soon as he has two pointsmore than his opponent. Both players start at par, and points are earned one at atime. If Daniel has a 60% chance of winning each point, what is the probability thathe will win the game?

Solution: 9/13

Consider the situation after two points. Daniel has a 9/25 chance of winning, Scott,4/25, and there is a 12/25 chance that the players will be tied. In the latter case, werevert to the original situation. In particular, after every two points, either the game

1

returns to the original situation, or one player wins. If it is given that the game lasts2k rounds, then the players must be at par after 2(k−1) rounds, and then Daniel winswith probability (9/25)/(9/25 + 4/25) = 9/13. Since this holds for any k, we concludethat Daniel wins the game with probability 9/13.

4. In a certain country, there are 100 senators, each of whom has 4 aides. These senatorsand aides serve on various committees. A committee may consist either of 5 senators,of 4 senators and 4 aides, or of 2 senators and 12 aides. Every senator serves on 5committees, and every aide serves on 3 committees. How many committees are therealtogether?

Solution: 160

If each senator gets a point for every committee on which she serves, and every aidegets 1/4 point for every committee on which he serves, then the 100 senators get 500points altogether, and the 400 aides get 300 points altogether, for a total of 800 points.On the other hand, each committee contributes 5 points, so there must be 800/5 = 160committees.

5. We wish to color the integers 1, 2, 3, . . . , 10 in red, green, and blue, so that no twonumbers a and b, with a − b odd, have the same color. (We do not require that allthree colors be used.) In how many ways can this be done?

Solution: 186

The condition is equivalent to never having an odd number and an even number inthe same color. We can choose one of the three colors for the odd numbers anddistribute the other two colors freely among the 5 even numbers; this can be done in3 · 25 = 96 ways. We can also choose one color for the even numbers and distributethe other two colors among the 5 odd numbers, again in 96 ways. This gives a totalof 192 possibilities. However, we have double-counted the 3 · 2 = 6 cases where allodd numbers are the same color and all even numbers are the same color, so there areactually 192− 6 = 186 possible colorings.

6. In a classroom, 34 students are seated in 5 rows of 7 chairs. The place at the center ofthe room is unoccupied. A teacher decides to reassign the seats such that each studentwill occupy a chair adjacent to his/her present one (i.e. move one desk forward, back,left or right). In how many ways can this reassignment be made?

Solution: 0

Color the chairs red and black in checkerboard fashion, with the center chair black.Then all 18 red chairs are initially occupied. Also notice that adjacent chairs have dif-ferent colors. It follows that we need 18 black chairs to accommodate the reassignment,but there are only 17 of them. Thus, the answer is 0.

7. You have infinitely many boxes, and you randomly put 3 balls into them. The boxesare labeled 1, 2, . . .. Each ball has probability 1/2n of being put into box n. The ballsare placed independently of each other. What is the probability that some box willcontain at least 2 balls?

Solution: 5/7

2

Notice that the answer is the sum of the probabilities that boxes 1, 2, . . ., respectively,contain at least 2 balls, since those events are mutually exclusive. For box n, theprobability of having at least 2 balls is

3[(1/2n)2(1− 1/2n)] + (1/2n)3 = 3/22n − 2/23n = 3/4n − 2/8n.

Summing to infinity using the geometric series formula, we get the answer (3/4)/(1−1/4)− (2/8)/(1− 1/8), which is equal to 5/7.

8. For any subset S ⊆ {1, 2, . . . , 15}, a number n is called an “anchor” for S if n andn+ |S| are both members of S, where |S| denotes the number of members of S. Findthe average number of anchors over all possible subsets S ⊆ {1, 2, . . . , 15}.Solution: 13/8

We first find the sum of the numbers of anchors of all subsets S; this is equivalent tofinding, for each n, the number of sets for which n is an anchor, and then summingover all n. Suppose that n is an anchor for S, and S has k elements. Then n, n+ k ∈S ⇒ k ≥ 2, and also n + k ≤ 15, or k ≤ 15 − n. The remaining k − 2 elements ofS (other than n and n + k) may be freely chosen from the remaining 13 members of

{1, 2, . . . , 15}, so we get(

13k−2

)possible sets S. Summing over all allowed values of k,

we then have(

130

)+

(131

)+

(132

)+ · · · +

(13

13−n

)sets with n as an anchor. If we sum

over all n = 1, 2, . . . , 13 (since there are no possible values of k when n > 13), we geta total of

13

(13

0

)+ 12

(13

1

)+ 11

(13

2

)+ · · ·+

(13

12

).

If we call this quantity A, then, by symmetry, 2A equals

13(

130

)+ 12

(131

)+ 11

(132

)+ · · · +

(1312

)

+(

131

)+ 2

(132

)+ · · · + 12

(1312

)+ 13

(1313

)

= 13[(

130

)+

(131

)+

(132

)+ · · · +

(1312

)+

(1313

)]= 13 · 213.

So A = 13 · 212 is the total number of anchors over all possible sets S. Finally, to findthe average number of anchors, we divide by the number of sets, which is 215; thus,the answer is 13 · 212/215 = 13/8.

9. At a certain college, there are 10 clubs and some number of students. For any twodifferent students, there is some club such that exactly one of the two belongs to thatclub. For any three different students, there is some club such that either exactly oneor all three belong to that club. What is the largest possible number of students?

Solution: 513

Let C be the set of clubs; each student then corresponds to a subset of C (the clubsto which that student belongs). The two-student condition implies that these subsetsmust be all distinct. Now (assuming there is more than one student) some studentbelongs to a nonempty set S of clubs. For every subset T ⊆ C, let f(T ) be the subsetof C consisting of those clubs that are in exactly one of S and T (so that f(T ) =(S ∪ T )− (S ∩ T )). It is straightforward to check that f(f(T )) = T and f(T ) 6= T , sothat the collection of all 210 subsets of C is partitioned into pairs {T, f(T )}. Moreover,

3

as long as S is distinct from T and f(T ), every club is in either none or exactly two ofthe sets S, T , and f(T ), so we cannot have a student corresponding to T and anothercorresponding to f(T ). This puts an upper bound of 513 possible students (one for S,one for ∅ = f(S), and one for each of the 511 other pairs). On the other hand, if wetake some club c, we can have one student belonging to no clubs and 512 other studentsall belonging to c and to the 512 possible subsets of the other 9 clubs, respectively. Itis readily checked that this arrangement meets the conditions — for the three-studentcondition, either all three students are in c, or one is the student who belongs to noclubs and we reduce to the two-student condition — so 513 is achievable.

10. A calculator has a display, which shows a nonnegative integer N , and a button, whichreplaces N by a random integer chosen uniformly from the set {0, 1, . . . , N − 1}, pro-vided that N > 0. Initially, the display holds the number N = 2003. If the button ispressed repeatedly until N = 0, what is the probability that the numbers 1, 10, 100,and 1000 will each show up on the display at some point?

Solution: 1/2224222

First, we claim that if the display starts at some N , the probability that any givennumber M < N will appear at some point is 1/(M+1). We can show this by inductionon N . If N = M +1 (the base case), M can only be reached if it appears after the firststep, and this occurs with probability 1/N = 1/(M + 1). If N > M + 1 and the claimholds for N − 1, then there are two possibilities starting from N . If the first step leadsto N − 1 (this occurs with probability 1/N), the probability of seeing M subsequentlyis 1/(M +1) by the induction hypothesis. If the first step leads to something less thanN−1 (probability (N−1)/N), then it leads to any of the integers {0, 1, . . . , N−2} withequal probability. But this is exactly what the first step would have been if we hadstarted from N−1; hence, the probability of seeing M is again 1/(M+1) by induction.Thus, the overall probability of seeing M is 1

N· 1

M+1+ N−1

N· 1

M+1= 1/(M +1), proving

the induction step and the claim.

Now let P (N,M) (M < N) be the probability of eventually seeing the number M ifwe start at N ; note that this is the same as the conditional probability of seeing Mgiven that we see N . Hence, the desired probability is

P (2003, 1000) · P (1000, 100) · P (100, 10) · P (10, 1) =1

1001· 1

101· 1

11· 12

=1

2224222.

4

Harvard-MIT Mathematics TournamentMarch 15, 2003

Individual Round: General Test, Part 1

1. 10 people are playing musical chairs with n chairs in a circle. They can be seatedin 7! ways (assuming only one person fits on each chair, of course), where differentarrangements of the same people on chairs, even rotations, are considered different.Find n.

2. OPEN is a square, and T is a point on side NO, such that triangle TOP has area 62and triangle TEN has area 10. What is the length of a side of the square?

3. There are 16 members on the Height-Measurement Matching Team. Each member wasasked, “How many other people on the team — not counting yourself — are exactlythe same height as you?” The answers included six 1’s, six 2’s, and three 3’s. Whatwas the sixteenth answer? (Assume that everyone answered truthfully.)

4. How many 2-digit positive integers have an even number of positive divisors?

5. A room is built in the shape of the region between two semicircles with the same centerand parallel diameters. The farthest distance between two points with a clear line ofsight is 12m. What is the area (in m2) of the room?

6. In how many ways can 3 bottles of ketchup and 7 bottles of mustard be arranged in arow so that no bottle of ketchup is immediately between two bottles of mustard? (Thebottles of ketchup are mutually indistinguishable, as are the bottles of mustard.)

7. Find the real value of x such that x3 + 3x2 + 3x+ 7 = 0.

8. A broken calculator has the + and × keys switched. For how many ordered pairs (a, b)of integers will it correctly calculate a+ b using the labelled + key?

9. Consider a 2003-gon inscribed in a circle and a triangulation of it with diagonalsintersecting only at vertices. What is the smallest possible number of obtuse trianglesin the triangulation?

10. Bessie the cow is trying to navigate her way through a field. She can travel onlyfrom lattice point to adjacent lattice point, can turn only at lattice points, and cantravel only to the east or north. (A lattice point is a point whose coordinates are bothintegers.) (0, 0) is the southwest corner of the field. (5, 5) is the northeast corner ofthe field. Due to large rocks, Bessie is unable to walk on the points (1, 1), (2, 3), or(3, 2). How many ways are there for Bessie to travel from (0, 0) to (5, 5) under theseconstraints?

1

Harvard-MIT Mathematics TournamentMarch 15, 2003

Individual Round: General Test, Part 1 — Solutions

1. 10 people are playing musical chairs with n chairs in a circle. They can be seatedin 7! ways (assuming only one person fits on each chair, of course), where differentarrangements of the same people on chairs, even rotations, are considered different.Find n.

Solution: 4

The number of ways 10 people can be seated on n chairs is n! multiplied by the numberof ways one can choose n people out of 10. Hence we must solve 7! = n! ·10!/(n! · (10−n)!). This is equivalent to (10− n)! = 10!/7! = 8 · 9 · 10 = 720 = 6!. We therefore haven = 4.

2. OPEN is a square, and T is a point on side NO, such that triangle TOP has area 62and triangle TEN has area 10. What is the length of a side of the square?

Solution: 12

62 = PO ·OT/2 and 10 = EN ·TN/2 = PO ·TN/2, so adding gives 72 = PO · (OT +TN)/2 = PO ·ON/2 = PO2/2⇒ PO = 12.

3. There are 16 members on the Height-Measurement Matching Team. Each member wasasked, “How many other people on the team — not counting yourself — are exactlythe same height as you?” The answers included six 1’s, six 2’s, and three 3’s. Whatwas the sixteenth answer? (Assume that everyone answered truthfully.)

Solution: 3

For anyone to have answered 3, there must have been exactly 4 people with the sameheight, and then each of them would have given the answer 3. Thus, we need at leastfour 3’s, so 3 is the remaining answer. (More generally, a similar argument shows thatthe number of members answering n must be divisible by n+ 1.)

4. How many 2-digit positive integers have an even number of positive divisors?

Solution: 84

An integer has an odd number of divisors precisely if it is a square. So we take the 902-digit numbers (10, 11, . . . , 99) and remove the 6 squares (42, 52, . . . , 92), for a total of84.

5. A room is built in the shape of the region between two semicircles with the same centerand parallel diameters. The farthest distance between two points with a clear line ofsight is 12m. What is the area (in m2) of the room?

Solution: 18π

The maximal distance is as shown in the figure (next page). Call the radii R and r,R > r. Then R2−r2 = 62 by the Pythagorean theorem, so the area is (π/2)·(R2−r2) =18π.

1

6. In how many ways can 3 bottles of ketchup and 7 bottles of mustard be arranged in arow so that no bottle of ketchup is immediately between two bottles of mustard? (Thebottles of ketchup are mutually indistinguishable, as are the bottles of mustard.)

Solution: 22

Consider the blocks of consecutive bottles of ketchup in such an arrangement. A blockof just one bottle must occur at the beginning or the end of the row, or else it wouldbe between two bottles of mustard. However, a block of two or three bottles can occuranywhere. We cannot have three blocks of one bottle each, since there are only twopossible locations for such blocks. Thus, we either have a block of one bottle and ablock of two, or one block of all three bottles. In the first case, if the single bottleoccurs at the beginning of the row, then anywhere from 1 to 7 bottles of mustard mayintervene before the block of 2 ketchup bottles, giving 7 possible arrangements. Welikewise have 7 arrangements if the single bottle occurs at the end of the row. Finally,if there is just one block of three bottles, anywhere from 0 to 7 mustard bottles mayprecede it, giving 8 possible arrangements. So, altogether, we have 7 + 7 + 8 = 22configurations.

7. Find the real value of x such that x3 + 3x2 + 3x+ 7 = 0.

Solution: −1− 3√

6

Rewrite the equation as (x+ 1)3 + 6 = 0 to get (x+ 1)3 = −6⇒ x+ 1 = 3√−6⇒ x =

−1− 3√

6.

8. A broken calculator has the + and × keys switched. For how many ordered pairs (a, b)of integers will it correctly calculate a+ b using the labelled + key?

Solution: 2

We need a+ b = ab, or a = bb−1

= 1− 1b−1

, so 1/(b− 1) is an integer. Thus b must be0 or 2, and a is 0 or 2, respectively. So there are 2.

9. Consider a 2003-gon inscribed in a circle and a triangulation of it with diagonalsintersecting only at vertices. What is the smallest possible number of obtuse trianglesin the triangulation?

Solution: 1999

By induction, it follows easily that any triangulation of an n-gon inscribed in a circlehas n− 2 triangles. A triangle is obtuse unless it contains the center of the circle in itsinterior (in which case it is acute) or on one of its edges (in which case it is right). Itis then clear that there are at most 2 non-obtuse triangles, and 2 is achieved when thecenter of the circle is on one of the diagonals of the triangulation. So the minimumnumber of obtuse triangles is 2001− 2 = 1999.

2

10. Bessie the cow is trying to navigate her way through a field. She can travel onlyfrom lattice point to adjacent lattice point, can turn only at lattice points, and cantravel only to the east or north. (A lattice point is a point whose coordinates are bothintegers.) (0, 0) is the southwest corner of the field. (5, 5) is the northeast corner ofthe field. Due to large rocks, Bessie is unable to walk on the points (1, 1), (2, 3), or(3, 2). How many ways are there for Bessie to travel from (0, 0) to (5, 5) under theseconstraints?

Solution: 32

In the figure, each point is labeled with the number of ways to reach that point. Thenumbers are successively computed as follows: The point (0, 0) can trivially be reachedin 1 way. When Bessie reaches any subsequent point (x, y) (other than a rock), shecan arrive either via a northward or an eastward step, so the number of ways she canreach that point equals the number of ways of reaching (x − 1, y) plus the number ofways of reaching (x, y− 1). By iterating this calculation, we eventually find that (5, 5)can be reached in 32 ways.

(0,0)

1

1

1

1

1

1

1

2

1

3

4

1

2

1

3

7

2

0

1

3

10

3

3

3

1

6

16

4

7

10

1

16

32

(5,5)

3

Harvard-MIT Mathematics TournamentMarch 15, 2003

Individual Round: General Test, Part 2

1. A compact disc has the shape of a circle of diameter 5 inches with a 1-inch-diametercircular hole in the center. Assuming the capacity of the CD is proportional to itsarea, how many inches would need to be added to the outer diameter to double thecapacity?

2. You have a list of real numbers, whose sum is 40. If you replace every number x onthe list by 1− x, the sum of the new numbers will be 20. If instead you had replacedevery number x by 1 + x, what would the sum then be?

3. How many positive rational numbers less than π have denominator at most 7 whenwritten in lowest terms? (Integers have denominator 1.)

4. In triangle ABC with area 51, points D and E trisect AB and points F and G trisectBC. Find the largest possible area of quadrilateral DEFG.

5. You are given a 10× 2 grid of unit squares. Two different squares are adjacent if theyshare a side. How many ways can one mark exactly nine of the squares so that no twomarked squares are adjacent?

6. The numbers 112, 121, 123, 153, 243, 313, and 322 are among the rows, columns, anddiagonals of a 3 × 3 square grid of digits (rows and diagonals read left-to-right, andcolumns read top-to-bottom). What 3-digit number completes the list?

7. Daniel and Scott are playing a game where a player wins as soon as he has two pointsmore than his opponent. Both players start at par, and points are earned one at atime. If Daniel has a 60% chance of winning each point, what is the probability thathe will win the game?

8. If x ≥ 0, y ≥ 0 are integers, randomly chosen with the constraint x + y ≤ 10, what isthe probability that x+ y is even?

9. In a classroom, 34 students are seated in 5 rows of 7 chairs. The place at the center ofthe room is unoccupied. A teacher decides to reassign the seats such that each studentwill occupy a chair adjacent to his/her present one (i.e. move one desk forward, back,left or right). In how many ways can this reassignment be made?

10. Several positive integers are given, not necessarily all different. Their sum is 2003.Suppose that n1 of the given numbers are equal to 1, n2 of them are equal to 2, . . .,n2003 of them are equal to 2003. Find the largest possible value of

n2 + 2n3 + 3n4 + · · ·+ 2002n2003.

1

Harvard-MIT Mathematics TournamentMarch 15, 2003

Individual Round: General Test, Part 2 — Solutions

1. A compact disc has the shape of a circle of diameter 5 inches with a 1-inch-diametercircular hole in the center. Assuming the capacity of the CD is proportional to itsarea, how many inches would need to be added to the outer diameter to double thecapacity?

Solution: 2

Doubling the capacity is equivalent to doubling the area, which is initially π[(5/2)2 −(1/2)2] = 6π. Thus we want to achieve an area of 12π, so if the new diameter is d, wewant π[(d/2)2− (1/2)2] = 12π ⇒ d = 7. Thus we need to add 2 inches to the diameter.

2. You have a list of real numbers, whose sum is 40. If you replace every number x onthe list by 1− x, the sum of the new numbers will be 20. If instead you had replacedevery number x by 1 + x, what would the sum then be?

Solution: 100

Let n be the number of numbers on the list. If each initial number is replaced byits negative, the sum will then be −40, and adding 1 to every number on this listincreases the sum by n, so n − 40 = 20 ⇒ n = 60. Then, if we had simply added1 to each of the initial numbers (without negating first), the sum would increase to40 + n = 40 + 60 = 100.

3. How many positive rational numbers less than π have denominator at most 7 whenwritten in lowest terms? (Integers have denominator 1.)

Solution: 54

We can simply list them. The table shows that there are 3+3+6+6+12+6+18 = 54.

Denominator Values

1 11, 2

1, 3

1

2 12, 3

2, 5

2

3 13, 2

3, 4

3, 5

3, 7

3, 8

3

4 14, 3

4, 5

4, 7

4, 9

4, 11

4

5 15, 2

5, 3

5, 4

5, 6

5, 7

5, 8

5, 9

5, 11

5, 12

5, 13

5, 14

5

6 16, 5

6, 7

6, 11

6, 13

6, 17

6

7 17, 2

7, . . . , 6

7, 8

7, 9

7, . . . , 13

7, 15

7, 16

7, . . . , 20

7

4. In triangle ABC with area 51, points D and E trisect AB and points F and G trisectBC. Find the largest possible area of quadrilateral DEFG.

Solution: 17

Assume E is between D and B, and F is between G and B (the alternative is to switchtwo points, say D and E, which clearly gives a non-convex quadrilateral with smaller

1

area). If two triangles have their bases on the same line and the same opposite vertex,then it follows from the 1

2bh formula that their areas are in the same ratio as their

bases. In particular (brackets denote areas),

[DBG]

[ABC]=

[DBG]

[ABG]· [ABG]

[ABC]=DB

AB· BGBC

=2

3· 23

=4

9,

and similarly [EBF ]/[ABC] = 1/9. Subtracting gives [DEFG]/[ABC] = 1/3, so theanswer is [ABC]/3 = 17.

5. You are given a 10× 2 grid of unit squares. Two different squares are adjacent if theyshare a side. How many ways can one mark exactly nine of the squares so that no twomarked squares are adjacent?

Solution: 36

Since each row has only two squares, it is impossible for two marked squares to bein the same row. Therefore, exactly nine of the ten rows contain marked squares.Consider two cases:

Case 1: The first or last row is empty. These two cases are symmetrical, so assumewithout loss of generality that the first row is empty. There are two possibilities for thesecond row: either the first square is marked, or the second square is marked. Sincethe third row must contain a marked square, and it cannot be in the same columnas the marked square in the second row, the third row is determined by the second.Similarly, all the remaining rows are determined. This leaves two possibilities if thefirst row is empty. Thus, there are four possibilities if the first or last row is empty.

Case 2: The empty row is not the first or last. Then, there are two blocks of (one ofmore) consecutive rows of marked squares. As above, the configuration of the rows ineach of the two blocks is determined by the position of the marked square in the firstof its rows. That makes 2× 2 = 4 possible configurations. There are eight possibilitiesfor the empty row, making a total of 32 possibilities in this case.

Together, there are 36 possible configurations of marked squares.

6. The numbers 112, 121, 123, 153, 243, 313, and 322 are among the rows, columns, anddiagonals of a 3 × 3 square grid of digits (rows and diagonals read left-to-right, andcolumns read top-to-bottom). What 3-digit number completes the list?

Solution: 524

1 1 25 2 43 1 3

The center digit is the middle digit of 4 numbers (hence at least 3 members of theabove list), so it must be 2. The top-left digit begins at least 2 members of the abovelist, so it must be 1 or 3. If it is 3, then after placing 313 we see that we need threemore numbers starting with 3, impossible; hence, it is 1. So 243 and 313 must (in someorder) be the last row and the last column, and now it is easy to complete the grid asshown; the answer is 524.

2

7. Daniel and Scott are playing a game where a player wins as soon as he has two pointsmore than his opponent. Both players start at par, and points are earned one at atime. If Daniel has a 60% chance of winning each point, what is the probability thathe will win the game?

Solution: 9/13

Consider the situation after two points. Daniel has a 9/25 chance of winning, Scott,4/25, and there is a 12/25 chance that the players will be tied. In the latter case, werevert to the original situation. In particular, after every two points, either the gamereturns to the original situation, or one player wins. If it is given that the game lasts2k rounds, then the players must be at par after 2(k−1) rounds, and then Daniel winswith probability (9/25)/(9/25 + 4/25) = 9/13. Since this holds for any k, we concludethat Daniel wins the game with probability 9/13.

8. If x ≥ 0, y ≥ 0 are integers, randomly chosen with the constraint x + y ≤ 10, what isthe probability that x+ y is even?

Solution: 6/11

For each p ≤ 10, if x + y = p, x can range from 0 to p, yielding p + 1 ordered pairs(x, y). Thus there are a total of 1 + 2 + 3 + · · ·+ 11 allowable ordered pairs (x, y), but1 + 3 + 5 + · · ·+ 11 of these pairs have an even sum. So the desired probability is

1 + 3 + 5 + · · ·+ 11

1 + 2 + 3 + · · ·+ 11=

62

11 · 6 =6

11.

9. In a classroom, 34 students are seated in 5 rows of 7 chairs. The place at the center ofthe room is unoccupied. A teacher decides to reassign the seats such that each studentwill occupy a chair adjacent to his/her present one (i.e. move one desk forward, back,left or right). In how many ways can this reassignment be made?

Solution: 0

Color the chairs red and black in checkerboard fashion, with the center chair black.Then all 18 red chairs are initially occupied. Also notice that adjacent chairs have dif-ferent colors. It follows that we need 18 black chairs to accommodate the reassignment,but there are only 17 of them. Thus, the answer is 0.

10. Several positive integers are given, not necessarily all different. Their sum is 2003.Suppose that n1 of the given numbers are equal to 1, n2 of them are equal to 2, . . .,n2003 of them are equal to 2003. Find the largest possible value of

n2 + 2n3 + 3n4 + · · ·+ 2002n2003.

Solution: 2002

The sum of all the numbers is n1 +2n2 + · · ·+2003n2003, while the number of numbersis n1 + n2 + · · ·+ n2003. Hence, the desired quantity equals

(n1 + 2n2 + · · ·+ 2003n2003)− (n1 + n2 + · · ·+ n2003)

= (sum of the numbers)− (number of numbers)

3

= 2003− (number of numbers),

which is maximized when the number of numbers is minimized. Hence, we should havejust one number, equal to 2003, and then the specified sum is 2003− 1 = 2002.

Comment: On the day of the contest, a protest was lodged (successfully) on thegrounds that the use of the words “several” and “their” in the problem statementimplies there must be at least 2 numbers. Then the answer is 2001, and this maximumis achieved by any two numbers whose sum is 2003.

4

Harvard-MIT Mathematics TournamentMarch 15, 2003

Individual Round: Geometry Subject Test

1. AD and BC are both perpendicular to AB, and CD is perpendicular to AC. If AB = 4and BC = 3, find CD.

C

A

D

B

2. As shown, U and C are points on the sides of triangle MNH such that MU = s,UN = 6, NC = 20, CH = s, HM = 25. If triangle UNC and quadrilateral MUCHhave equal areas, what is s?

M

N

C

H

U

6

s

25

20

s

3. A room is built in the shape of the region between two semicircles with the same centerand parallel diameters. The farthest distance between two points with a clear line ofsight is 12m. What is the area (in m2) of the room?

4. Farmer John is inside of an ellipse with reflective sides, given by the equation x2/a2 +y2/b2 = 1, with a > b > 0. He is standing at the point (3, 0), and he shines a laserpointer in the y-direciton. The light reflects off the ellipse and proceeds directly towardFarmer Brown, traveling a distance of 10 before reaching him. Farmer John then spinsaround in a circle; wherever he points the laser, the light reflects off the wall and hitsFarmer Brown. What is the ordered pair (a, b)?

1

5. Consider a 2003-gon inscribed in a circle and a triangulation of it with diagonalsintersecting only at vertices. What is the smallest possible number of obtuse trianglesin the triangulation?

6. Take a clay sphere of radius 13, and drill a circular hole of radius 5 through its center.Take the remaining “bead” and mold it into a new sphere. What is this sphere’sradius?

7. Let RSTUV be a regular pentagon. Construct an equilateral triangle PRS with pointP inside the pentagon. Find the measure (in degrees) of angle PTV .

8. Let ABC be an equilateral triangle of side length 2. Let ω be its circumcircle, and letωA, ωB, ωC be circles congruent to ω centered at each of its vertices. Let R be the setof all points in the plane contained in exactly two of these four circles. What is thearea of R?

9. In triangle ABC, 6 ABC = 50◦ and 6 ACB = 70◦. Let D be the midpoint of sideBC. A circle is tangent to BC at B and is also tangent to segment AD; this circleinstersects AB again at P . Another circle is tangent to BC at C and is also tangentto segment AD; this circle intersects AC again at Q. Find 6 APQ (in degrees).

10. Convex quadrilateral MATH is given with HM/MT = 3/4, and 6 ATM = 6 MAT =6 AHM = 60◦. N is the midpoint of MA, and O is a point on TH such that linesMT,AH,NO are concurrent. Find the ratio HO/OT .

2

Harvard-MIT Mathematics TournamentMarch 15, 2003

Individual Round: Geometry Subject Test — Solutions

1. AD and BC are both perpendicular to AB, and CD is perpendicular to AC. If AB = 4and BC = 3, find CD.

C

A

D

B

Solution: 20/3

By Pythagoras in 4ABC, AC = 5. But 6 CAD = 90◦ − 6 BAC = 6 ACB, so righttriangles CAD,BCA are similar, and CD/AC = BA/CB = 4/3⇒ CD = 20/3.

2. As shown, U and C are points on the sides of triangle MNH such that MU = s,UN = 6, NC = 20, CH = s, HM = 25. If triangle UNC and quadrilateral MUCHhave equal areas, what is s?

M

N

C

H

U

6

s

25

20

s

Solution: 4

Using brackets to denote areas, we have [MCH] = [UNC] + [MUCH] = 2[UNC]. Onthe other hand, triangles with equal altitudes have their areas in the same ratio astheir bases, so

2 =[MNH]

[UNC]=

[MNH]

[MNC]· [MNC]

[UNC]=NH

NC· MN

UN=s+ 20

20· s+ 6

6.

Clearing the denominator gives (s+ 20)(s+ 6) = 240, and solving the quadratic givess = 4 or −30. Since s > 0, we must have s = 4.

1

3. A room is built in the shape of the region between two semicircles with the same centerand parallel diameters. The farthest distance between two points with a clear line ofsight is 12m. What is the area (in m2) of the room?

Solution: 18π

The maximal distance is as shown in the figure. Call the radii R and r, R > r. ThenR2 − r2 = 62 by the Pythagorean theorem, so the area is (π/2) · (R2 − r2) = 18π.

4. Farmer John is inside of an ellipse with reflective sides, given by the equation x2/a2 +y2/b2 = 1, with a > b > 0. He is standing at the point (3, 0), and he shines a laserpointer in the y-direciton. The light reflects off the ellipse and proceeds directly towardFarmer Brown, traveling a distance of 10 before reaching him. Farmer John then spinsaround in a circle; wherever he points the laser, the light reflects off the wall and hitsFarmer Brown. What is the ordered pair (a, b)?

Solution: (5, 4)

The points where the farmers are standing must be the foci of the ellipse, so they are(3, 0) and (−3, 0). If the total distance traveled is 10, then a must be half of that, or5, since the distance traveled by a ray reflecting off the wall from when it leaves onefocus to when it reaches the other focus is 2a, the length of the major axis. If a focusis at x = 3, then we have 3 =

√a2 − b2 =

√25− b2, yielding b = 4.

5. Consider a 2003-gon inscribed in a circle and a triangulation of it with diagonalsintersecting only at vertices. What is the smallest possible number of obtuse trianglesin the triangulation?

Solution: 1999

By induction, it follows easily that any triangulation of an n-gon inscribed in a circlehas n− 2 triangles. A triangle is obtuse unless it contains the center of the circle in itsinterior (in which case it is acute) or on one of its edges (in which case it is right). Itis then clear that there are at most 2 non-obtuse triangles, and 2 is achieved when thecenter of the circle is on one of the diagonals of the triangulation. So the minimumnumber of obtuse triangles is 2001− 2 = 1999.

6. Take a clay sphere of radius 13, and drill a circular hole of radius 5 through its center.Take the remaining “bead” and mold it into a new sphere. What is this sphere’sradius?

Solution: 12

Let r be the radius of the sphere. We take cross sections of the bead perpendicularto the line of the drill and compare them to cross sections of the sphere at the same

2

distance from its center. At a height h, the cross section of the sphere is a circle withradius

√r2 − h2 and thus area π(r2 − h2). At the same height, the cross section of

the bead is an annulus with outer radius√

132 − h2 and inner radius 5, for an area ofπ(132 − h2)− π(52) = π(122 − h2) (since 132 − 52 = 122). Thus, if r = 12, the sphereand the bead will have the same cross-sectional area π(122 − h2) for |h| ≤ 12 and 0for |h| > 12. Since all the cross sections have the same area, the two clay figures thenhave the same volume. And certainly there is only one value of r for which the twovolumes are equal, so r = 12 is the answer.

7. Let RSTUV be a regular pentagon. Construct an equilateral triangle PRS with pointP inside the pentagon. Find the measure (in degrees) of angle PTV .

Solution: 6

We have 6 PRV = 6 SRV − 6 SRP = 108◦−60◦ = 48◦. Since PR = RS = RV , trianglePRV is isosceles, so that 6 V PR = 6 RV P = (180◦ − 6 PRV )/2 = 66◦. Likewise, wehave 6 TPS = 66◦, so that

6 TPV = 360◦ − (6 V PR + 6 RPS + 6 SPT ) = 360◦ − (66◦ + 60◦ + 66◦) = 168◦.

Finally, by symmetry, triangle PTV is isosceles (PT = TV ), so 6 PTV = 6 TV P =(180◦ − 6 TPV )/2 = 6◦. (See the figure.)

R S

U

TP

V

8. Let ABC be an equilateral triangle of side length 2. Let ω be its circumcircle, and letωA, ωB, ωC be circles congruent to ω centered at each of its vertices. Let R be the setof all points in the plane contained in exactly two of these four circles. What is thearea of R?

Solution: 2√

3

ωA, ωB, ωC intersect at the circumcenter; thus, every point within the circumcircle, andno point outside of it, is in two or more circles. The area inside exactly two circlesis shaded in the figure. The two intersection points of ωA and ωB, together with A,form the vertices of an equilateral triangle. As shown, this equilateral triangle cuts offa “lip” of ω (bounded by a 60◦ arc of ω and the corresponding chord) and another,congruent lip of ωB that is not part of the region of interest. By rotating the first lipto the position of the second, we can reassemble the equilateral triangle. Doing this foreach of the 6 such triangles, we see that the desired area equals the area of a regularhexagon inscribed in ω. The side length of this hexagon is (2/3) · (√3/2) · 2 = 2

√3/3,

so its area is 6 · (√3/4) · (2√3/3)2 = 2√

3, and this is the answer.

3

ω

A

B C

B

A

Cω ω

ω

9. In triangle ABC, 6 ABC = 50◦ and 6 ACB = 70◦. Let D be the midpoint of sideBC. A circle is tangent to BC at B and is also tangent to segment AD; this circleinstersects AB again at P . Another circle is tangent to BC at C and is also tangentto segment AD; this circle intersects AC again at Q. Find 6 APQ (in degrees).

Solution: 70

Suppose the circles are tangent to AD at E,F , respectively; then, by equal tangents,DE = DB = DC = DF ⇒ E = F (as shown). So, by the Power of a Point Theorem,AP · AB = AE2 = AF 2 = AQ · AC ⇒ AP/AQ = AC/AB ⇒ 4APQ ∼ 4ACB,giving 6 APQ = 6 ACB = 70◦.

A

B

F

CD

PQ

E

10. Convex quadrilateral MATH is given with HM/MT = 3/4, and 6 ATM = 6 MAT =6 AHM = 60◦. N is the midpoint of MA, and O is a point on TH such that linesMT,AH,NO are concurrent. Find the ratio HO/OT .

Solution: 9/16

Triangle MAT is equilateral, so HM/AT = HM/MT = 3/4. Also, 6 AHM = 6 ATM ,so the quadrilateral is cyclic. Now, let P be the intersection of MT,AH,NO. ExtendMH and NO to intersect at point Q. Then by Menelaus’s theorem, applied to triangle

4

AHM and line QNP , we have

HQ

QM· MN

NA· APPH

= 1,

while applying the same theorem to triangle THM and line QPO gives

HQ

QM· MP

PT· TOOH

= 1.

Combining givesHO/OT = (MP/PT )·(AN/NM)·(HP/PA) = (MP/PA)·(HP/PT )(because AN/NM = 1). But sinceMATH is cyclic,4APT ∼ 4MPH, soMP/PA =HP/PT = HM/AT = 3/4, and the answer is (3/4)2 = 9/16. (See figure.)

A M

H

T

O

P

Q

N

5

Harvard-MIT Mathematics TournamentMarch 15, 2003

Guts Round

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HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

1. [5] Simplify2003

√2√

11− 3√

5 · 4006√

89 + 12√

55.

2. [5] The graph of x4 = x2y2 is a union of n different lines. What is the value of n?

3. [5] If a and b are positive integers that can each be written as a sum of two squares,then ab is also a sum of two squares. Find the smallest positive integer c such thatc = ab, where a = x3 + y3 and b = x3 + y3 each have solutions in integers (x, y), butc = x3 + y3 does not.

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HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

4. [6] Let z = 1− 2i. Find 1z

+ 2z2 + 3

z3 + · · ·.5. [6] Compute the surface area of a cube inscribed in a sphere of surface area π.

6. [6] Define the Fibonacci numbers by F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2 for n ≥ 2. Forhow many n, 0 ≤ n ≤ 100, is Fn a multiple of 13?

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HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

7. [6] a and b are integers such that a+√b =

√15 +

√216. Compute a/b.

8. [6] How many solutions in nonnegative integers (a, b, c) are there to the equation

2a + 2b = c! ?

9. [6] For x a real number, let f(x) = 0 if x < 1 and f(x) = 2x− 2 if x ≥ 1. How manysolutions are there to the equation

f(f(f(f(x)))) = x?

1

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HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

10. [7] Suppose that A,B,C,D are four points in the plane, and let Q,R, S, T, U, V be therespective midpoints of AB,AC,AD,BC,BD,CD. If QR = 2001, SU = 2002, TV =2003, find the distance between the midpoints of QU and RV .

11. [7] Find the smallest positive integer n such that 12 +22 +32 +42 + · · ·+n2 is divisibleby 100.

12. [7] As shown in the figure, a circle of radius 1 has two equal circles whose diameterscover a chosen diameter of the larger circle. In each of these smaller circles we similarlydraw three equal circles, then four in each of those, and so on. Compute the area ofthe region enclosed by a positive even number of circles.

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HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

13. [7] If xy = 5 and x2 + y2 = 21, compute x4 + y4.

14. [7] A positive integer will be called “sparkly” if its smallest (positive) divisor, otherthan 1, equals the total number of divisors (including 1). How many of the numbers2, 3, . . . , 2003 are sparkly?

15. [7] The product of the digits of a 5-digit number is 180. How many such numbersexist?

2

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HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

16. [8] What fraction of the area of a regular hexagon of side length 1 is within distance12

of at least one of the vertices?

17. [8] There are 10 cities in a state, and some pairs of cities are connected by roads.There are 40 roads altogether. A city is called a “hub” if it is directly connected toevery other city. What is the largest possible number of hubs?

18. [8] Find the sum of the reciprocals of all the (positive) divisors of 144.

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HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

19. [8] Let r, s, t be the solutions to the equation x3 + ax2 + bx+ c = 0. What is the valueof (rs)2 + (st)2 + (rt)2 in terms of a, b, and c?

20. [8] What is the smallest number of regular hexagons of side length 1 needed to com-pletely cover a disc of radius 1?

21. [8] r and s are integers such that

3r ≥ 2s− 3 and 4s ≥ r + 12.

What is the smallest possible value of r/s?

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HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

22. [9] There are 100 houses in a row on a street. A painter comes and paints every housered. Then, another painter comes and paints every third house (starting with housenumber 3) blue. Another painter comes and paints every fifth house red (even if itis already red), then another painter paints every seventh house blue, and so forth,alternating between red and blue, until 50 painters have been by. After this is finished,how many houses will be red?

23. [9] How many lattice points are enclosed by the triangle with vertices (0, 99), (5, 100),and (2003, 500)? Don’t count boundary points.

24. [9] Compute the radius of the inscribed circle of a triangle with sides 15, 16, and 17.

3

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HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

25. [9] Let ABC be an isosceles triangle with apex A. Let I be the incenter. If AI = 3and the distance from I to BC is 2, then what is the length of BC?

26. [9] Find all integers x such that x2 + 6x+ 28 is a perfect square.

27. [9] The rational numbers x and y, when written in lowest terms, have denominators60 and 70, respectively. What is the smallest possible denominator of x+ y?

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HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

28. [10] A point in three-space has distances 2, 6, 7, 8, 9 from five of the vertices of a regularoctahedron. What is its distance from the sixth vertex?

29. [10] A palindrome is a positive integer that reads the same backwards as forwards,such as 82328. What is the smallest 5-digit palindrome that is a multiple of 99?

30. [10] The sequence a1, a2, a3, . . . of real numbers satisfies the recurrence

an+1 =a2

n − an−1 + 2an

an−1 + 1.

Given that a1 = 1 and a9 = 7, find a5.

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HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

31. [10] A cylinder of base radius 1 is cut into two equal parts along a plane passingthrough the center of the cylinder and tangent to the two base circles. Suppose thateach piece’s surface area is m times its volume. Find the greatest lower bound for allpossible values of m as the height of the cylinder varies.

32. [10] If x, y, and z are real numbers such that 2x2 + y2 + z2 = 2x− 4y + 2xz − 5, findthe maximum possible value of x− y + z.

33. [10] We are given triangle ABC, with AB = 9, AC = 10, and BC = 12, and a pointD on BC. B and C are reflected in AD to B′ and C ′, respectively. Suppose that linesBC ′ and B′C never meet (i.e., are parallel and distinct). Find BD.

4

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HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

34. [12] OKRA is a trapezoid with OK parallel to RA. If OK = 12 and RA is a positiveinteger, how many integer values can be taken on by the length of the segment in thetrapezoid, parallel to OK, through the intersection of the diagonals?

35. [12] A certain lottery has tickets labeled with the numbers 1, 2, 3, . . . , 1000. The lotteryis run as follows: First, a ticket is drawn at random. If the number on the ticket is odd,the drawing ends; if it is even, another ticket is randomly drawn (without replacement).If this new ticket has an odd number, the drawing ends; if it is even, another ticketis randomly drawn (again without replacement), and so forth, until an odd numberis drawn. Then, every person whose ticket number was drawn (at any point in theprocess) wins a prize.

You have ticket number 1000. What is the probability that you get a prize?

36. [12] A teacher must divide 221 apples evenly among 403 students. What is the minimalnumber of pieces into which she must cut the apples? (A whole uncut apple counts asone piece.)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

37. [15] A quagga is an extinct chess piece whose move is like a knight’s, but much longer:it can move 6 squares in any direction (up, down, left, or right) and then 5 squares ina perpendicular direction. Find the number of ways to place 51 quaggas on an 8 × 8chessboard in such a way that no quagga attacks another. (Since quaggas are naturallybelligerent creatures, a quagga is considered to attack quaggas on any squares it canmove to, as well as any other quaggas on the same square.)

38. [15] Given are real numbers x, y. For any pair of real numbers a0, a1, define a sequenceby an+2 = xan+1 +yan for n ≥ 0. Suppose that there exists a fixed nonnegative integerm such that, for every choice of a0 and a1, the numbers am, am+1, am+3, in this order,form an arithmetic progression. Find all possible values of y.

39. [15] In the figure, if AE = 3, CE = 1, BD = CD = 2, and AB = 5, find AG.

FG

A

BD

C

E

5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

40. [18] All the sequences consisting of five letters from the set {T, U,R,N, I, P} (withrepetitions allowed) are arranged in alphabetical order in a dictionary. Two sequencesare called “anagrams” of each other if one can be obtained by rearranging the letters ofthe other. How many pairs of anagrams are there that have exactly 100 other sequencesbetween them in the dictionary?

41. [18] A hotel consists of a 2× 8 square grid of rooms, each occupied by one guest. Allthe guests are uncomfortable, so each guest would like to move to one of the adjoiningrooms (horizontally or vertically). Of course, they should do this simultaneously, insuch a way that each room will again have one guest. In how many different ways canthey collectively move?

42. [18] A tightrope walker stands in the center of a rope of length 32 meters. Everyminute she walks forward one meter with probability 3/4 and backward one meterwith probability 1/4. What is the probability that she reaches the end in front of herbefore the end behind her?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

43. Write down an integer N between 0 and 10, inclusive. You will receive N points —unless some other team writes down the same N , in which case you receive nothing.

44. A partition of a number n is a sequence of positive integers, arranged in nonincreasingorder, whose sum is n. For example, n = 4 has 5 partitions: 1 + 1 + 1 + 1 = 2 +1 + 1 = 2 + 2 = 3 + 1 = 4. Given two different partitions of the same number,n = a1 + a2 + · · · + ak = b1 + b2 + · · · + bl, where k ≤ l, the first partition is said todominate the second if all of the following inequalities hold:

a1 ≥ b1;

a1 + a2 ≥ b1 + b2;

a1 + a2 + a3 ≥ b1 + b2 + b3;...

a1 + a2 + · · ·+ ak ≥ b1 + b2 + · · ·+ bk.

Find as many partitions of the number n = 20 as possible such that none of thepartitions dominates any other. Your score will be the number of partitions you find.If you make a mistake and one of your partitions does dominate another, your score isthe largest m such that the first m partitions you list constitute a valid answer.

45. Find a set S of positive integers such that no two distinct subsets of S have the samesum. Your score will be b20(2n/r−2)c, where n is the number of elements in the set S,and r is the largest element of S (assuming, of course, that this number is nonnegative).

Hej d̊a!

6

Harvard-MIT Mathematics TournamentMarch 15, 2003

Guts Round — Solutions

1. Simplify2003

√2√

11− 3√

5 · 4006√

89 + 12√

55.

Solution: −1

Note that (2√

11 + 3√

5)2 = 89 + 12√

55. So, we have

2003√

2√

11− 3√

5 · 4006√

89 + 12√

55 =2003

√2√

11− 3√

5 · 2003√

2√

11 + 3√

5

=2003

√(2√

11)2 − (3√

5)2 = 2003√−1 = −1.

2. The graph of x4 = x2y2 is a union of n different lines. What is the value of n?

Solution: 3

The equation x4 − x2y2 = 0 factors as x2(x + y)(x− y) = 0, so its graph is the unionof the three lines x = 0, x+ y = 0, and x− y = 0.

3. If a and b are positive integers that can each be written as a sum of two squares, thenab is also a sum of two squares. Find the smallest positive integer c such that c = ab,where a = x3 +y3 and b = x3 +y3 each have solutions in integers (x, y), but c = x3 +y3

does not.

Solution: 4

We can’t have c = 1 = 13 + 03 or c = 2 = 13 + 13, and if c = 3, then a or b = ±3 whichis not a sum of two cubes (otherwise, flipping signs of x and y if necessary, we wouldget either a sum of two nonnegative cubes to equal 3, which clearly does not happen,or a difference of two nonnegative cubes to equal 3, but the smallest difference betweentwo successive cubes ≥ 1 is 23 − 13 = 7). However, c = 4 does meet the conditions,with a = b = 2 = 13 + 13 (an argument similar to the above shows that there are nox, y with 4 = x3 + y3), so 4 is the answer.

4. Let z = 1− 2i. Find 1z

+ 2z2 + 3

z3 + · · ·.Solution: (2i− 1)/4

Let x = 1z

+ 2z2 + 3

z3 + · · ·, so z · x =(1 + 2

z+ 3

z2 + 4z3 + · · ·

). Then z · x − x =

1 + 1z

+ 1z2 + 1

z3 + · · · = 11−1/z

= zz−1

. Solving for x in terms of z, we obtain x = z(z−1)2

.

Plugging in the original value of z produces x = (2i− 1)/4.

5. Compute the surface area of a cube inscribed in a sphere of surface area π.

Solution: 2

The sphere’s radius r satisfies 4πr2 = π ⇒ r = 1/2, so the cube has body diagonal 1,hence side length 1/

√3. So, its surface area is 6(1/

√3)2 = 2.

6. Define the Fibonacci numbers by F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2 for n ≥ 2. Forhow many n, 0 ≤ n ≤ 100, is Fn a multiple of 13?

1

Solution: 15

The sequence of remainders modulo 13 begins 0, 1, 1, 2, 3, 5, 8, 0, and then we haveFn+7 ≡ 8Fn modulo 13 by a straightforward induction. In particular, Fn is a multipleof 13 if and only if 7 | n, so there are 15 such n.

7. a and b are integers such that a+√b =

√15 +

√216. Compute a/b.

Solution: 1/2

Squaring both sides gives a2+b+2a√b = 15+

√216; separating rational from irrational

parts, we get a2 + b = 15, 4a2b = 216, so a2 and b equal 6 and 9. a is an integer, soa2 = 9, b = 6⇒ a/b = 3/6 = 1/2. (We cannot have a = −3, since a+

√b is positive.)

8. How many solutions in nonnegative integers (a, b, c) are there to the equation

2a + 2b = c! ?

Solution: 5

We can check that 2a + 2b is never divisible by 7, so we must have c < 7. The binaryrepresentation of 2a +2b has at most two 1’s. Writing 0!, 1!, 2!, . . . , 6! in binary, we cancheck that the only possibilities are c = 2, 3, 4, giving solutions (0, 0, 2), (1, 2, 3), (2, 1, 3),(3, 4, 4), (4, 3, 4).

9. For x a real number, let f(x) = 0 if x < 1 and f(x) = 2x − 2 if x ≥ 1. How manysolutions are there to the equation

f(f(f(f(x)))) = x?

Solution: 2

Certainly 0, 2 are fixed points of f and therefore solutions. On the other hand, therecan be no solutions for x < 0, since f is nonnegative-valued; for 0 < x < 2, we have0 ≤ f(x) < x < 2 (and f(0) = 0), so iteration only produces values below x, and forx > 2, f(x) > x, and iteration produces higher values. So there are no other solutions.

10. Suppose that A,B,C,D are four points in the plane, and let Q,R, S, T, U, V be therespective midpoints of AB,AC,AD,BC,BD,CD. If QR = 2001, SU = 2002, TV =2003, find the distance between the midpoints of QU and RV .

Solution: 2001

This problem has far more information than necessary: QR and UV are both parallelto BC, and QU and RV are both parallel to AD. Hence, QUV R is a parallelogram,and the desired distance is simply the same as the side length QR, namely 2001. (Seefigure, next page.)

11. Find the smallest positive integer n such that 12 + 22 + 32 + 42 + · · · + n2 is divisibleby 100.

Solution: 24

The sum of the first n squares equals n(n+1)(2n+1)/6, so we require n(n+1)(2n+1)to be divisible by 600 = 24 · 25. The three factors are pairwise relatively prime, so

2

Q

V

U

R

A

B

D

C

one of them must be divisible by 25. The smallest n for which this happens is n = 12(2n + 1 = 25), but then we do not have enough factors of 2. The next smallest isn = 24 (n+ 1 = 25), and this works, so 24 is the answer.

12. As shown in the figure, a circle of radius 1 has two equal circles whose diameters covera chosen diameter of the larger circle. In each of these smaller circles we similarly drawthree equal circles, then four in each of those, and so on. Compute the area of theregion enclosed by a positive even number of circles.

Solution: π/e

At the nth step, we have n! circles of radius 1/n! each, for a total area of n! ·π/(n!)2 =π/n!. The desired area is obtained by adding the areas of the circles at step 2, thensubtracting those at step 3, then adding those at step 4, then subtracting those at step5, and so forth. Thus, the answer is

π

2!− π

3!+π

4!− · · · = π

(1

0!− 1

1!+

1

2!− 1

3!+ · · ·

)= πe−1.

13. If xy = 5 and x2 + y2 = 21, compute x4 + y4.

Solution: 391

We have 441 = (x2 + y2)2 = x4 + y4 + 2(xy)2 = x4 + y4 + 50, yielding x4 + y4 = 391.

3

14. A positive integer will be called “sparkly” if its smallest (positive) divisor, other than1, equals the total number of divisors (including 1). How many of the numbers2, 3, . . . , 2003 are sparkly?

Solution: 3

Suppose n is sparkly; then its smallest divisor other than 1 is some prime p. Hence,n has p divisors. However, if the full prime factorization of n is pe1

1 pe22 · · · per

r , thenumber of divisors is (e1 + 1)(e2 + 1) · · · (er + 1). For this to equal p, only one factorcan be greater than 1, so n has only one prime divisor — namely p — and we gete1 = p− 1⇒ n = pp−1. Conversely, any number of the form pp−1 is sparkly. There arejust three such numbers in the desired range (21, 32, 54), so the answer is 3.

15. The product of the digits of a 5-digit number is 180. How many such numbers exist?

Solution: 360

Let the digits be a, b, c, d, e. Then abcde = 180 = 22 · 32 · 5. We observe that thereare 6 ways to factor 180 into digits a, b, c, d, e (ignoring differences in ordering): 180 =1 · 1 · 4 · 5 · 9 = 1 · 1 · 5 · 6 · 6 = 1 · 2 · 2 · 5 · 9 = 1 · 2 · 3 · 5 · 6 = 1 · 3 · 3 · 4 · 5 = 2 · 2 · 3 · 3 · 5.There are (respectively) 60, 30, 60, 120, 60, and 30 permutations of these breakdowns,for a total of 360 numbers.

16. What fraction of the area of a regular hexagon of side length 1 is within distance 12

ofat least one of the vertices?

Solution: π√

3/9

The hexagon has area 6(√

3/4)(1)2 = 3√

3/2. The region we want consists of six 120◦

arcs of circles of radius 1/2, whichcan be reassembled into two circles of radius 1/2. Soits area is π/2, and the ratio of areas is π

√3/9.

17. There are 10 cities in a state, and some pairs of cities are connected by roads. Thereare 40 roads altogether. A city is called a “hub” if it is directly connected to everyother city. What is the largest possible number of hubs?

Solution: 6

If there are h hubs, then(

h2

)roads connect the hubs to each other, and each hub is

connected to the other 10 − h cities; we thus get(

h2

)+ h(10 − h) distinct roads. So,

40 ≥(

h2

)+h(10−h) = −h2/2+19h/2, or 80 ≥ h(19−h). The largest h ≤ 10 satisfying

this condition is h = 6, and conversely, if we connect each of 6 cities to every othercity and place the remaining 40− [

(62

)+ 6(10− 6)] = 1 road wherever we wish, we can

achieve 6 hubs. So 6 is the answer.

18. Find the sum of the reciprocals of all the (positive) divisors of 144.

Solution: 403/144

As d ranges over the divisors of 144, so does 144/d, so the sum of 1/d is 1/144 times thesum of the divisors of 144. Using the formula for the sum of the divisors of a number(or just counting them out by hand), we get that this sum is 403, so the answer is403/144.

4

19. Let r, s, t be the solutions to the equation x3 + ax2 + bx+ c = 0. What is the value of(rs)2 + (st)2 + (rt)2 in terms of a, b, and c?

Solution: b2 − 2ac

We have (x− r)(x− s)(x− t) = x3 + ax2 + bx+ c, so

a = −(r + s+ t), b = rs+ st+ rt, c = −rst.

So we have

(rs)2 + (st)2 + (rt)2 = (rs+ st+ rt)2 − 2rst(r + s+ t) = b2 − 2ac.

20. What is the smallest number of regular hexagons of side length 1 needed to completelycover a disc of radius 1?

Solution: 3

First, we show that two hexagons do not suffice. Specifically, we claim that a hexagoncovers less than half of the disc’s boundary. First, a hexagon of side length 1 may beinscribed in a circle, and this covers just 6 points. Translating the hexagon verticallyupward (regardless of its orientation) will cause it to no longer touch any point onthe lower half of the circle, so that it now covers less than half of the boundary. Byrotational symmetry, the same argument applies to translation in any other direction,proving the claim. Then, two hexagons cannot possibly cover the disc.

The disc can be covered by three hexagons as follows. Let P be the center of the circle.Put three non-overlapping hexagons together at point P . This will cover the circle,since each hexagon will cover a 120◦ sector of the circle.

21. r and s are integers such that

3r ≥ 2s− 3 and 4s ≥ r + 12.

What is the smallest possible value of r/s?

Solution: 1/2

We simply plot the two inequalities in the sr-plane and find the lattice point satisfyingboth inequalities such that the slope from it to the origin is as low as possible. We findthat this point is (2, 4) (or (3, 6)), as circled in the figure, so the answer is 2/4 = 1/2.

5

4s = r + 12

s

r

3r = 2s - 3

22. There are 100 houses in a row on a street. A painter comes and paints every house red.Then, another painter comes and paints every third house (starting with house number3) blue. Another painter comes and paints every fifth house red (even if it is alreadyred), then another painter paints every seventh house blue, and so forth, alternatingbetween red and blue, until 50 painters have been by. After this is finished, how manyhouses will be red?

Solution: 52

House n ends up red if and only if the largest odd divisor of n is of the form 4k+1. Wehave 25 values of n = 4k + 1; 13 values of n = 2(4k + 1) (given by k = 0, 1, 2, . . . , 12);7 values of n = 4(4k + 1) (k = 0, 1, . . . , 6); 3 values of n = 8(4k + 1) (k = 0, 1, 2); 2 ofthe form n = 16(4k+ 1) (for k = 0, 1); 1 of the form n = 32(4k+ 1); and 1 of the formn = 64(4k + 1). Thus we have a total of 25 + 13 + 7 + 3 + 2 + 1 + 1 = 52 red houses.

23. How many lattice points are enclosed by the triangle with vertices (0, 99), (5, 100), and(2003, 500)? Don’t count boundary points.

Solution: 0

Using the determinant formula, we get that the area of the triangle is∣∣∣∣∣

5 12003 401

∣∣∣∣∣/

2 = 1.

There are 4 lattice points on the boundary of the triangle (the three vertices and(1004, 300)), so it follows from Pick’s Theorem that there are 0 in the interior.

24. Compute the radius of the inscribed circle of a triangle with sides 15, 16, and 17.

Solution:√

21

Hero’s formula gives that the area is√

24 · 9 · 8 · 7 = 24√

21. Then, using the resultthat the area of a triangle equals the inradius times half the perimeter, we see that theradius is

√21.

6

25. Let ABC be an isosceles triangle with apex A. Let I be the incenter. If AI = 3 andthe distance from I to BC is 2, then what is the length of BC?

Solution: 4√

5

Let X and Y be the points where the incircle touches AB and BC, respectively.Then AXI and AY B are similar right triangles. Since I is the incenter, we haveIX = IY = 2. Using the Pythagorean theorem on triangle AXI, we find AX =

√5.

By similarity, AY/AX = BY/IX. Plugging in the numbers given, 5/√

5 = BY/2, soBY = 2

√5. Y is the midpoint of BC, so BC = 4

√5.

26. Find all integers m such that m2 + 6m+ 28 is a perfect square.

Solution: 6,−12

We must have m2+6m+28 = n2, where n is an integer. Rewrite this as (m+3)2+19 =n2 ⇒ n2 − (m + 3)2 = 19 ⇒ (n −m − 3)(n + m + 3) = 19. Let a = n −m − 3 andb = n+m+ 3, so we want ab = 19. This leaves only 4 cases:

• a = 1, b = 19. Solve the system n−m− 3 = 1 and n+m+ 3 = 19 to get n = 10and m = 6, giving one possible solution.

• a = 19, b = 1. Solve the system, as above, to get n = 10 and m = −12.

• a = −1, b = −19. We get n = −10 and m = −12.

• a = −19, b = −1. We get n = −10 and m = 6.

Thus the only m are 6 and −12.

27. The rational numbers x and y, when written in lowest terms, have denominators 60and 70, respectively. What is the smallest possible denominator of x+ y?

Solution: 84

Write x + y = a/60 + b/70 = (7a + 6b)/420. Since a is relatively prime to 60 and b isrelatively prime to 70, it follows that none of the primes 2, 3, 7 can divide 7a+6b, so wewon’t be able to cancel any of these factors in the denominator. Thus, after reducingto lowest terms, the denominator will still be at least 22 · 3 · 7 = 84 (the product ofthe powers of 2, 3, and 7 dividing 420). On the other hand, 84 is achievable, by taking(e.g.) 1/60 + 3/70 = 25/420 = 5/84. So 84 is the answer.

28. A point in three-space has distances 2, 6, 7, 8, 9 from five of the vertices of a regularoctahedron. What is its distance from the sixth vertex?

Solution:√

21

By a simple variant of the British Flag Theorem, if ABCD is a square and P any pointin space, AP 2 +CP 2 = BP 2 +DP 2. Four of the five given vertices must form a squareABCD, and by experimentation we find their distances to the given point P must beAP = 2, BP = 6, CP = 9, DP = 7. Then A,C, and the other two vertices E,F alsoform a square AECF , so 85 = AP 2 + CP 2 = EP 2 + FP 2 = 82 + FP 2 ⇒ FP =

√21.

29. A palindrome is a positive integer that reads the same backwards as forwards, such as82328. What is the smallest 5-digit palindrome that is a multiple of 99?

Solution: 54945

7

Write the number asXY ZY X. This is the same as 10000X+1000Y +100Z+10Y +X =99(101X+10Y +Z)+20Y +2X+Z. We thus want 20Y +2X+Z to be a multiple of 99,with X as small as possible. This expression cannot be larger than 20·9+2·9+9 = 207,and it is greater than 0 (since X 6= 0), so for this to be a multiple of 99, it must equal99 or 198. Consider these two cases.

To get 198, we must have Y = 9, which then leaves 2X+Z = 18. The smallest possibleX is 5, and then Z becomes 8 and we have the number 59895.

To get 99, we must have Y = 4. Then, 2X + Z = 19, and, as above, we find theminimal X is 5 and then Z = 9. This gives us the number 54945. This is smaller thanthe other number, so it is the smallest number satisfying the conditions of the problem.

30. The sequence a1, a2, a3, . . . of real numbers satisfies the recurrence

an+1 =a2

n − an−1 + 2an

an−1 + 1.

Given that a1 = 1 and a9 = 7, find a5.

Solution: 3

Let bn = an+1. Then the recurrence becomes bn+1−1 = (b2n−bn−1)/bn−1 = b2n/bn−1−1,so bn+1 = b2n/bn−1. It follows that the sequence (bn) is a geometric progression, fromwhich b25 = b1b9 = 2 · 8 = 16 ⇒ b5 = ±4. However, since all bn are real, they eitheralternate in sign or all have the same sign (depending on the sign of the progression’scommon ratio); either way, b5 has the same sign as b1, so b5 = 4⇒ a5 = 3.

31. A cylinder of base radius 1 is cut into two equal parts along a plane passing throughthe center of the cylinder and tangent to the two base circles. Suppose that each piece’ssurface area is m times its volume. Find the greatest lower bound for all possible valuesof m as the height of the cylinder varies.

Solution: 3

Let h be the height of the cylinder. Then the volume of each piece is half the volumeof the cylinder, so it is 1

2πh. The base of the piece has area π, and the ellipse formed

by the cut has area π · 1 ·√

1 + h2

4because its area is the product of the semiaxes times

π. The rest of the area of the piece is half the lateral area of the cylinder, so it is πh.Thus, the value of m is

π + π√

1 + h2/4 + πh

πh/2=

2 + 2h+√

4 + h2

h

=2

h+ 2 +

√4

h2+ 1,

a decreasing function of h whose limit as h → ∞ is 3. Therefore the greatest lowerbound of m is 3.

32. If x, y, and z are real numbers such that 2x2 + y2 + z2 = 2x− 4y + 2xz − 5, find themaximum possible value of x− y + z.

Solution: 4

The equation rearranges as (x− 1)2 + (y + 2)2 + (x− z)2 = 0, so we must have x = 1,y = −2, z = 1, giving us 4.

8

33. We are given triangle ABC, with AB = 9, AC = 10, and BC = 12, and a point D onBC. B and C are reflected in AD to B′ and C ′, respectively. Suppose that lines BC ′

and B′C never meet (i.e., are parallel and distinct). Find BD.

Solution: 6

The lengths of AB and AC are irrelevant. Because the figure is symmetric about AD,lines BC ′ and B′C meet if and only if they meet at a point on line AD. So, if theynever meet, they must be parallel to AD. Because AD and BC ′ are parallel, trianglesABD and ADC ′ have the same area. Then ABD and ADC also have the same area.Hence, BD and CD must have the same length, so BD = 1

2BC = 6.

34. OKRA is a trapezoid with OK parallel to RA. If OK = 12 and RA is a positiveinteger, how many integer values can be taken on by the length of the segment in thetrapezoid, parallel to OK, through the intersection of the diagonals?

Solution: 10

Let RA = x. If the diagonals intersect at X, and the segment is PQ with P on KR,then 4PKX ∼ 4RKA and 4OKX ∼ 4RAX (by equal angles), giving RA/PX =AK/XK = 1 + AX/XK = 1 + AR/OK = (x + 12)/12, so PX = 12x/(12 + x).Similarly XQ = 12x/(12 + x) also, so PQ = 24x/(12 + x) = 24− 288

12+x. This has to be

an integer. 288 = 2532, so it has (5 + 1)(3 + 1) = 18 divisors. 12 + x must be one ofthese. We also exclude the 8 divisors that don’t exceed 12, so our final answer is 10.

35. A certain lottery has tickets labeled with the numbers 1, 2, 3, . . . , 1000. The lottery isrun as follows: First, a ticket is drawn at random. If the number on the ticket is odd,the drawing ends; if it is even, another ticket is randomly drawn (without replacement).If this new ticket has an odd number, the drawing ends; if it is even, another ticketis randomly drawn (again without replacement), and so forth, until an odd numberis drawn. Then, every person whose ticket number was drawn (at any point in theprocess) wins a prize.

You have ticket number 1000. What is the probability that you get a prize?

Solution: 1/501

Notice that the outcome is the same as if the lottery instead draws all the tickets, inrandom order, and awards a prize to the holder of the odd ticket drawn earliest and eacheven ticket drawn before it. Thus, the probability of your winning is the probabilitythat, in a random ordering of the tickets, your ticket precedes all the odd tickets. This,in turn, equals the probability that your ticket is the first in the setconsisting of yourown and all odd-numbered tickets, irrespective of the other even-numbered tickets.Since all 501! orderings of these tickets are equally likely, the desired probability is1/501.

36. A teacher must divide 221 apples evenly among 403 students. What is the minimalnumber of pieces into which she must cut the apples? (A whole uncut apple counts asone piece.)

Solution: 611

Consider a bipartite graph, with 221 vertices representing the apples and 403 verticesrepresenting the students; each student is connected to each apple that she gets a

9

piece of. The number of pieces then equals the number of edges in the graph. Eachstudent gets a total of 221/403 = 17/31 apple, but each component of the graphrepresents a complete distribution of an integer number of apples to an integer numberof students and therefore uses at least 17 apple vertices and 31 student vertices. Thenwe have at most 221/17 = 403/31 = 13 components in the graph, so there are atleast 221 + 403 − 13 = 611 edges. On the other hand, if we simply distribute in thestraightforward manner — proceeding through the students, cutting up a new applewhenever necessary but never returning to a previous apple or student — we can createa graph without cycles, and each component does involve 17 apples and 31 students.Thus, we get 13 trees, and 611 edges is attainable.

37. A quagga is an extinct chess piece whose move is like a knight’s, but much longer: itcan move 6 squares in any direction (up, down, left, or right) and then 5 squares ina perpendicular direction. Find the number of ways to place 51 quaggas on an 8 × 8chessboard in such a way that no quagga attacks another. (Since quaggas are naturallybelligerent creatures, a quagga is considered to attack quaggas on any squares it canmove to, as well as any other quaggas on the same square.)

Solution: 68

Represent the 64 squares of the board as vertices of a graph, and connect two verticesby an edge if a quagga can move from one to the other. The resulting graph consists of4 paths of length 5 and 4 paths of length 3 (given by the four rotations of the two pathsshown, next page), and 32 isolated vertices. Each path of length 5 can accommodateat most 3 nonattacking quaggas in a unique way (the first, middle, and last vertices),and each path of length 3 can accommodate at most 2 nonattacking quaggas in aunique way; thus, the maximum total number of nonattacking quaggas we can have is4 · 3 + 4 · 2 + 32 = 52. For 51 quaggas to fit, then, just one component of the graphmust contain one less quagga than its maximum.

If this component is a path of length 5, there are(

52

)− 4 = 6 ways to place the two

quaggas on nonadjacent vertices, and then all the other locations are forced; the 4 suchpaths then give us 4 · 6 = 24 possibilities this way. If it is a path of length 3, there are3 ways to place one quagga, and the rest of the board is forced, so we have 4 · 3 = 12possibilities here. Finally, if it is one of the 32 isolated vertices, we simply leave thissquare empty, and the rest of the board is forced, so we have 32 possibilities here. Sothe total is 24 + 12 + 32 = 68 different arrangements.

38. Given are real numbers x, y. For any pair of real numbers a0, a1, define a sequence byan+2 = xan+1 + yan for n ≥ 0. Suppose that there exists a fixed nonnegative integerm such that, for every choice of a0 and a1, the numbers am, am+1, am+3, in this order,form an arithmetic progression. Find all possible values of y.

Solution: 0, 1, (1±√5)/2

Note that x = 1 (or x = 0), y = 0 gives a constant sequence, so it will always havethe desired property. Thus, y = 0 is one possibility. For the rest of the proof, assumey 6= 0.

We will prove that am and am+1 may take on any pair of values, for an appropriatechoice of a0 and a1. Use induction on m. The case m = 0 is trivial. Suppose thatam and am+1 can take on any value. Let p and q be any real numbers. By setting

10

am = q−xpy

(remembering that y 6= 0) and am+1 = p, we get am+1 = p and am+2 = q.Therefore, am+1 and am+2 can have any values if am and am+1 can. That completesthe induction.

Now we determine the nonzero y such that am, am+1, am+3 form an arithmetic sequence;that is, such that am+3−am+1 = am+1−am. But because am+3 = (x2 +y)am+1 +xyam

by the recursion formula, we can eliminate am+3 from the equation, obtaining theequivalent condition (x2 + y − 2)am+1 + (xy + 1)am = 0. Because the pair am, am+1

can take on any values, this condition means exactly that x2 + y − 2 = xy + 1 = 0.Then x = −1/y, and 1/y2 + y − 2 = 0, or y3 − 2y2 + 1 = 0. One root of this cubic isy = 1, and the remaining quadratic factor y2 − y − 1 has the roots (1±√5)/2. Sinceeach such y gives an x for which the condition holds, we conclude that the answer tothe problem is y = 0, 1, or (1±√5)/2.

39. In the figure, if AE = 3, CE = 1, BD = CD = 2, and AB = 5, find AG.

FG

A

BD

C

E

Solution: 3√

66/7

By Stewart’s Theorem, AD2 · BC + CD · BD · BC = AB2 · CD + AC2 · BD, soAD2 = (52 · 2 + 42 · 2 − 2 · 2 · 4)/4 = (50 + 32 − 16)/4 = 33/2. By Menelaus’sTheorem applied to line BGE and triangle ACD, DG/GA · AE/EC · CB/BD = 1,so DG/GA = 1/6⇒ AD/AG = 7/6. Thus AG = 6 · AD/7 = 3

√66/7.

40. All the sequences consisting of five letters from the set {T, U,R,N, I, P} (with repeti-tions allowed) are arranged in alphabetical order in a dictionary. Two sequences arecalled “anagrams” of each other if one can be obtained by rearranging the letters of the

11

other. How many pairs of anagrams are there that have exactly 100 other sequencesbetween them in the dictionary?

Solution: 0

Convert each letter to a digit in base 6: I 7→ 0, N 7→ 1, P 7→ 2, R 7→ 3, T 7→ 4, U 7→ 5.Then the dictionary simply consists of all base-6 integers from 000006 to 555556 innumerical order. If one number can be obtained from another by a rearrangement ofdigits, then the numbers are congruent modulo 5 (this holds because a number abcde6

= 64 · a + 63 · b + 62 · c + 6 · d + e is congruent modulo 5 to a + b + c + d + e), but ifthere are 100 other numbers between them, then their difference is 101, which is notdivisible by 5. So there are no such pairs.

41. A hotel consists of a 2 × 8 square grid of rooms, each occupied by one guest. All theguests are uncomfortable, so each guest would like to move to one of the adjoiningrooms (horizontally or vertically). Of course, they should do this simultaneously, insuch a way that each room will again have one guest. In how many different ways canthey collectively move?

Solution: 1156

Imagine that the rooms are colored black and white, checkerboard-style. Each guest ina black room moves to an adjacent white room (and vice versa). If, for each such guest,we place a domino over the original room and the new room, we obtain a covering ofthe 2 × n grid by n dominoes, since each black square is used once and each whitesquare is used once. Applying a similar procedure to each guest who begins in a whiteroom and moves to a black room, we obtain a second domino tiling. Conversely, it isreadily verified that any pair of such tilings uniquely determines a movement pattern.Also, it is easy to prove by induction that the number of domino tilings of a 2 × ngrid is the (n + 1)th Fibonacci number (this holds for the base cases n = 1, 2, andfor a 2× n rectangle, the two rightmost squares either belong to one vertical domino,leaving a 2× (n− 1) rectangle to be covered arbitrarily, or to two horizontal dominoeswhich also occupy the adjoining squares, leaving a 2× (n− 2) rectangle to be coveredfreely; hence, the numbers of tilings satisfy the Fibonacci recurrence). So the numberof domino tilings of a 2 × 8 grid is 34, and the number of pairs of such tilings is342 = 1156, the answer.

42. A tightrope walker stands in the center of a rope of length 32 meters. Every minute shewalks forward one meter with probability 3/4 and backward one meter with probability1/4. What is the probability that she reaches the end in front of her before the endbehind her?

Solution: 316/(316 + 1)

After one minute, she is three times as likely to be one meter forward as one meterback. After two minutes she is either in the same place, two meters forward, or twometers back. The chance of being two meters forward is clearly (3/4)2 and thus 32 = 9times greater than the chance of being two back ((1/4)2). We can group the minutesinto two-minute periods and ignore the periods of no net movement, so we can considerher to be moving 2 meters forward or backward each period, where forward movementis 32 times as likely as backward movement. Repeating the argument inductively, weeventually find that she is 316 times more likely to move 16 meters forward than 16

12

meters backward, and thus the probability is 316/(316 + 1) that she will meet the frontend of the rope first.

43. Write down an integer N between 0 and 10, inclusive. You will receive N points —unless some other team writes down the same N , in which case you receive nothing.

Comments: Somewhat surprisingly, this game does not appear to be well-documentedin the literature. A better-known relative is Undercut, described by Douglas Hofstadterin his book Metamagical Themas: Questing for the Essence of Mind and Pattern. Inthis game, two players each choose an integer in a specified interval and receive thatmany points — unless the two numbers differ by 1, in which case the lower playerreceives a number of points equal to the sum of the two numbers. See, for example,http://mathforum.org/mam/96/undercut/ .

44. A partition of a number n is a sequence of positive integers, arranged in descendingorder, whose sum is n. For example, n = 4 has 5 partitions: 1 + 1 + 1 + 1 = 2 +1 + 1 = 2 + 2 = 3 + 1 = 4. Given two different partitions of the same number,n = a1 + a2 + · · · + ak = b1 + b2 + · · · + bl, where k ≤ l, the first partition is said todominate the second if all of the following inequalities hold:

a1 ≥ b1;

a1 + a2 ≥ b1 + b2;

a1 + a2 + a3 ≥ b1 + b2 + b3;...

a1 + a2 + · · ·+ ak ≥ b1 + b2 + · · ·+ bk.

Find as many partitions of the number n = 20 as possible such that none of thepartitions dominates any other. Your score will be the number of partitions you find.If you make a mistake and one of your partitions does dominate another, your score isthe largest m such that the first m partitions you list constitute a valid answer.

Comments: Computer searches have found that the maximum possible number ofpartitions is 20, achieved e.g. by the following:

9 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 6 + 3 + 3 + 2 + 2 + 2 + 28 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 5 + 5 + 4 + 1 + 1 + 1 + 1 + 1 + 17 + 4 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 5 + 5 + 3 + 2 + 2 + 1 + 1 + 1

7 + 3 + 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 5 + 5 + 2 + 2 + 2 + 2 + 27 + 3 + 2 + 2 + 2 + 1 + 1 + 1 + 1 5 + 4 + 4 + 3 + 1 + 1 + 1 + 1

7 + 2 + 2 + 2 + 2 + 2 + 2 + 1 5 + 4 + 4 + 2 + 2 + 2 + 16 + 5 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 5 + 4 + 3 + 3 + 3 + 1 + 1

6 + 4 + 3 + 2 + 1 + 1 + 1 + 1 + 1 5 + 3 + 3 + 3 + 3 + 36 + 4 + 2 + 2 + 2 + 2 + 1 + 1 4 + 4 + 4 + 4 + 2 + 1 + 16 + 3 + 3 + 3 + 2 + 1 + 1 + 1 4 + 4 + 4 + 3 + 3 + 2

In general, as shown in

http://www-math.mit.edu/~efedula/chains.ps

“Chain Lengths in the Dominance Lattice,” by Edward Early, 2002

13

the maximum number of mutually nondominating partitions of n is roughly on the

order of eπ√

2n/3 (in fact, the ratio of the total number of partitions to n to the maximumnumber of nondominating partitions has polynomial order of magnitude), although aproof by explicit construction is not known.

45. Find a set S of positive integers such that no two distinct subsets of S have the samesum. Your score will be b20(2n/r−2)c, where n is the number of elements in the set S,and r is the largest element of S (assuming, of course, that this number is nonnegative).

Comments: The obvious such sets are those of the form {1, 2, 4, . . . , 2n−1}, but theseachieve a score of 0. A bit of experimentation finds that the set {3, 5, 6, 7} also works;this answer achieves 5 points. The set {6, 9, 11, 12, 13} achieves 9 points. Doing signif-icantly better than this by hand computation becomes difficult. The reference

http://www.combinatorics.org/Volume 5/PDF/v5i1r3.pdf

“A Construction for Sets of Integers with Distinct Subset Sums,”by Tom Bohman, 1997

(from whence the above examples) provides a construction that achieves a score of 50.Erdős conjectured in effect that a maximum possible score exists, but this conjectureremains open. The strongest known result in this direction is that, if S has n elements,the largest element is at least a constant times 2n/

√n (proven by Erdős and Moser in

1955, with an improvement on the constant by Elkies in 1986).

Hej d̊a!

14

Harvard-MIT Mathematics TournamentMarch 15, 2003

Team Round

Completions and Configurations

Given a set A and a nonnegative integer k, the k-completion of A is the collection of allk-element subsets of A, and a k-configuration of A is any subset of the k-completion of A(including the empty set and the entire k-completion). For instance, the 2-completion ofA = {1, 2, 3} is {{1, 2}, {1, 3}, {2, 3}}, and the 2-configurations of A are

{} {{1, 2}}{{1, 3}} {{2, 3}}

{{1, 2}, {1, 3}} {{1, 2}, {2, 3}}{{1, 3}, {2, 3}} {{1, 2}, {1, 3}, {2, 3}}

The order of an element a of A with respect to a given k-configuration of A is the numberof subsets in the k-configuration that contain a. A k-configuration of a set A is consistent ifthe order of every element of A is the same, and the order of a consistent k-configuration isthis common value.

1. (a) [10] How many k-configurations are there of a set that has n elements?

(b) [10] How many k-configurations that have m elements are there of a set that hasn elements?

2. [15] Suppose A is a set with n elements, and k is a divisor of n. Find the number ofconsistent k-configurations of A of order 1.

3. (a) [15] Let An = {a1, a2, a3, . . . , an, b}, for n ≥ 3, and let Cn be the 2-configurationconsisting of {ai, ai+1} for all 1 ≤ i ≤ n − 1, {a1, an}, and {ai, b} for 1 ≤ i ≤ n.Let Se(n) be the number of subsets of Cn that are consistent of order e. FindSe(101) for e = 1, 2, and 3.

(b) [20] Let A = {V,W,X, Y, Z, v, w, x, y, z}. Find the number of subsets of the2-configuration

{ {V,W}, {W,X}, {X, Y }, {Y, Z}, {Z, V }, {v, x}, {v, y}, {w, y}, {w, z}, {x, z},{V, v}, {W,w}, {X, x}, {Y, y}, {Z, z} }

that are consistent of order 1.

(c) [30] Let A = {a1, b1, a2, b2, . . . , a10, b10}, and consider the 2-configuration C con-sisting of {ai, bi} for all 1 ≤ i ≤ 10, {ai, ai+1} for all 1 ≤ i ≤ 9, and {bi, bi+1} forall 1 ≤ i ≤ 9. Find the number of subsets of C that are consistent of order 1.

Define a k-configuration of A to be m-separable if we can label each element of A with aninteger from 1 to m (inclusive) so that there is no element E of the k-configuration all ofwhose elements are assigned the same integer. If C is any subset of A, then C is m-separableif we can assign an integer from 1 to m to each element of C so that there is no element Eof the k-configuration such that E ⊆ C and all elements of E are assigned the same integer.

1

4. (a) [15] Suppose A has n elements, where n ≥ 2, and C is a 2-configuration of A thatis not m-separable for any m < n. What is (in terms of n) the smallest numberof elements that C can have?

(b) [15] Show that every 3-configuration of an n-element set A is m-separable forevery integer m ≥ n/2.

(c) [25] Fix k ≥ 2, and suppose A has k2 elements. Show that any k-configuration

of A with fewer than(

k2−1k−1

)elements is k-separable.

5. [30] Let Bk(n) be the largest number of elements in a 2-separable k-configuration ofa set with 2n elements (2 ≤ k ≤ n). Find a closed-form expression (i.e. an expressionnot involving any sums or products with an variable number of terms) for Bk(n).

6. [40] Prove that any 2-configuration containing e elements is m-separable for some

m ≤ 12

+√

2e+ 14.

A cell of a 2-configuration of a set A is a nonempty subset C of A such that

i. for any two distinct elements a, b of C, there exists a sequence c0, c1, . . . , cn of elementsof A with c0 = a, cn = b, and such that {c0, c1}, {c1, c2}, . . . , {cn−1, cn} are all elementsof the 2-configuration, and

ii. if a is an element of C and b is an element of A but not of C, there does NOTexist a sequence c0, c1, . . . , cn of elements of A with c0 = a, cn = b, and such that{c0, c1}, {c1, c2}, . . ., {cn−1, cn} are all elements of the 2-configuration.

Also, we define a 2-configuration of A to be barren if there is no subset {a0, a1, . . . , an} of A,with n ≥ 2, such that {a0, a1}, {a1, a2}, . . . , {an−1, an} and {an, a0} are all elements of the2-configuration.

7. [20] Show that, given any 2-configuration of a set A, every element of A belongs toexactly one cell.

8. (a) [15] Given a set A with n ≥ 1 elements, find the number of consistent 2-configurations of A of order 1 with exactly 1 cell.

(b) [25] Given a set A with 10 elements, find the number of consistent 2-configurationsof A of order 2 with exactly 1 cell.

(c) [25] Given a set A with 10 elements, find the number of consistent 2-configurationsof order 2 with exactly 2 cells.

9. (a) [15] Show that if every cell of a 2-configuration of a finite set A is m-separable,then the whole 2-configuration is m-separable.

(b) [30] Show that any barren 2-configuration of a finite set A is 2-separable.

10. [45] Show that every consistent 2-configuration of order 4 on a finite set A has a subsetthat is a consistent 2-configuration of order 2.

2

Harvard-MIT Mathematics TournamentMarch 15, 2003

Team Round — Solutions

Completions and Configurations

Given a set A and a nonnegative integer k, the k-completion of A is the collection of allk-element subsets of A, and a k-configuration of A is any subset of the k-completion of A(including the empty set and the entire k-completion). For instance, the 2-completion ofA = {1, 2, 3} is {{1, 2}, {1, 3}, {2, 3}}, and the 2-configurations of A are

{} {{1, 2}}{{1, 3}} {{2, 3}}

{{1, 2}, {1, 3}} {{1, 2}, {2, 3}}{{1, 3}, {2, 3}} {{1, 2}, {1, 3}, {2, 3}}

The order of an element a of A with respect to a given k-configuration of A is the numberof subsets in the k-configuration that contain a. A k-configuration of a set A is consistent ifthe order of every element of A is the same, and the order of a consistent k-configuration isthis common value.

1. (a) How many k-configurations are there of a set that has n elements?

Solution: An n-element set has(

nk

)subsets of size k, and we can construct a k-

configuration by independently choosing, for each subset, whether or not to include it,

so there are 2(nk) k-configurations.

(b) How many k-configurations that have m elements are there of a set that has nelements?

Solution: Again, an n-element set has(

nk

)subsets of size k, so there are

((n

k)m

)k-

configurations with m elements.

2. Suppose A is a set with n elements, and k is a divisor of n. Find the number ofconsistent k-configurations of A of order 1.

Solution: Given such a k-configuration, we can write out all the elements of one ofthe k-element subsets, then all the elements of another subset, and so forth, eventuallyobtaining an ordering of all n elements of A. Conversely, given any ordering of theelements of A, we can construct a consistent k-configuration of order 1 from it bygrouping together the first k elements, then the next k elements, and so forth. In fact,each consistent k-configuration of order 1 corresponds to (n/k)!(k!)n/k different suchorderings, since the elements of A within each of the n/k k-element subsets can beordered in k! ways, and the various subsets can also be ordered with respect to eachother in (n/k)! different ways. Thus, since there are n! orderings of the elements of A,we get n!

(n/k)!(k!)n/k different consistent k-configurations of order 1.

3. (a) Let An = {a1, a2, a3, . . . , an, b}, for n ≥ 3, and let Cn be the 2-configurationconsisting of {ai, ai+1} for all 1 ≤ i ≤ n − 1, {a1, an}, and {ai, b} for 1 ≤ i ≤ n.Let Se(n) be the number of subsets of Cn that are consistent of order e. FindSe(101) for e = 1, 2, and 3.

1

Solution: For convenience, we assume the ai are indexed modulo 101, so that ai+1 = a1

when ai = a101.In any consistent subset of C101 of order 1, b must be paired with exactly one ai, say

a1. Then, a2 cannot be paired with a1, so it must be paired with a3, and likewise we findwe use the pairs {a4, a5}, {a6, a7}, . . . , {a100, a101} — and this does give us a consistentsubset of order 1. Similarly, pairing b with any other ai would give us a unique extensionto a consistent configuration of order 1. Thus, we have one such 2-configuration for eachi, giving S1(101) = 101 altogether.

In a consistent subset of order 2, b must be paired with two other elements. Supposeone of them is ai. Then ai is also paired with either ai−1 or ai+1, say ai+1. But thenai−1 needs to be paired up with two other elements, and ai is not available, so it mustbe paired with ai−2 and b. Now b has its two pairs determined, so nothing else can bepaired with b. Thus, for j 6= i − 1, i, we have that aj must be paired with aj−1 andaj+1. So our subset must be of the form

{{b, ai}, {ai, ai+1}, {ai+1, ai+2}, . . . , {a101, a1}, . . . , {ai−2, ai−1}, {ai−1, b}}

for some i. On the other hand, for any i = 1, . . . , 101, this gives a subset meeting ourrequirements. So, we have 101 possibilities, and S2(101) = 101.

Finally, in a consistent subset of order 3, each ai must be paired with ai−1, ai+1,and b. But then b occurs in 101 pairs, not just 3, so we have a contradiction. Thus, nosuch subset exists, so S3(101) = 0.

(b) Let A = {V,W,X, Y, Z, v, w, x, y, z}. Find the number of subsets of the 2-configuration

{ {V,W}, {W,X}, {X, Y }, {Y, Z}, {Z, V }, {v, x}, {v, y}, {w, y}, {w, z}, {x, z},

{V, v}, {W,w}, {X, x}, {Y, y}, {Z, z} }that are consistent of order 1.

Solution: No more than two of the pairs {v, x}, {v, y}, {w, y}, {w, z}, {x, z} may beincluded in a 2-configuration of order 1, since otherwise at least one of v, w, x, y, z wouldoccur more than once. If exactly one is included, say {v, x}, then w, y, z must be pairedwith W,Y, Z, respectively, and then V and X cannot be paired. So either none orexactly two of the five pairs above must be used. If none, then v, w, x, y, z must bepaired with V,W,X, Y, Z, respectively, and we have 1 2-configuration arising in thismanner. If exactly two are used, we can check that there are 5 ways to do this withoutduplicating an element:

{v, x}, {w, y} {v, x}, {w, z} {v, y}, {w, z} {v, y}, {x, z} {w, y}, {x, z}

In each case, it is straightforward to check that there is a unique way of pairing up theremaining elements of A. So we get 5 2-configurations in this way, and the total is 6.

(c) Let A = {a1, b1, a2, b2, . . . , a10, b10}, and consider the 2-configuration C consistingof {ai, bi} for all 1 ≤ i ≤ 10, {ai, ai+1} for all 1 ≤ i ≤ 9, and {bi, bi+1} for all1 ≤ i ≤ 9. Find the number of subsets of C that are consistent of order 1.

Solution: LetAn = {a1, b1, a2, b2, . . . , an, bn} for n ≥ 1, and consider the 2-configurationCn consisting of {ai, bi} for all 1 ≤ i ≤ n, {ai, ai+1} for all 1 ≤ i ≤ n− 1, and {bi, bi+1}

2

for all 1 ≤ i ≤ n−1. Let Nn be the number of subsets of Cn that are consistent of order1 (call these “matchings” of Cn). Consider any matching of Cn+2. Either an+2 is pairedwith bn+2, in which case the remaining elements of our matching form a matching ofCn+1; or an+2 is paired with an+1, in which case bn+2 must be paired with bn+1, andthe remaining elements form a matching of Cn. It follows that Nn+2 = Nn+1 +Nn. Bydirect calculation, N1 = 1 and N2 = 2, and now computing successive values of Nn

using the recurrence yields N10 = 89.

Define a k-configuration of A to be m-separable if we can label each element of A with aninteger from 1 to m (inclusive) so that there is no element E of the k-configuration all ofwhose elements are assigned the same integer. If C is any subset of A, then C is m-separableif we can assign an integer from 1 to m to each element of C so that there is no element Eof the k-configuration such that E ⊆ C and all elements of E are assigned the same integer.

4. (a) Suppose A has n elements, where n ≥ 2, and C is a 2-configuration of A that isnot m-separable for any m < n. What is (in terms of n) the smallest number ofelements that C can have?

Solution: We claim that every pair of elements of A must belong to C, so that theanswer is

(n2

). Indeed, if a, b ∈ A and {a, b} is not in the 2-configuration, then we can

assign the other elements of A the numbers 1, 2, . . . , n − 2 and assign a and b boththe number n − 1, so that C is (n − 1)-separable. On the other hand, if every pair ofelements of A is in the configuration, then A cannot be m-separable for m < n, sincethis would require assigning the same number to at least two elements, and then wewould have a pair whose elements have the same number.

(b) Show that every 3-configuration of an n-element set A is m-separable for everyinteger m ≥ n/2.

Solution: We can successively label the elements of A with the numbers 1, 1, 2, 2,3, 3, . . . , dn/2e. Then surely no 3-element subset can have all its elements labeled withthe same number, since no label is assigned to more than two elements. Thus, whenm ≥ n/2⇒ m ≥ dn/2e, this labeling shows that any 3-configuration is m-separable.

(c) Fix k ≥ 2, and suppose A has k2 elements. Show that any k-configuration of A

with fewer than(

k2−1k−1

)elements is k-separable.

Solution: The argument is similar to that used in problem 2. Suppose the config-uration is not k-separable. Consider all possible orderings of the k2 elements of A.For each ordering, assign the first k elements the number 1, the next k elements thenumber 2, and so forth. By assumption, for each such assignment, there exists someelement of the k-configuration whose elements are all assigned the same number. Nowconsider any given element E of the k-configuration. For each i, we count the orderingsin which all k elements of E receive the number i: there are k! possible orderings forthe elements of E, and there are (k2−k)! possible orderings for the remaining elementsof A. Altogether, this gives k · k! · (k2 − k)! orderings in which the elements of E allreceive the same label. So if, in each of the (k2)! orderings of the elements of A, thereis some E all of whose members receive the same label, then there must be at least

(k2)!

k · k! · (k2 − k)! =(k2 − 1)!

(k − 1)!(k2 − k)! =

(k2 − 1

k − 1

)

3

elements E of the k-configuration. Hence, if there are fewer elements, the k-configurationis k-separable, as desired.

5. Let Bk(n) be the largest possible number of elements in a 2-separable k-configuration ofa set with 2n elements (2 ≤ k ≤ n). Find a closed-form expression (i.e. an expressionnot involving any sums or products with a variable number of terms) for Bk(n).

Solution: First, a lemma: For any a with 0 ≤ a ≤ 2n,(

ak

)+

(2n−a

k

)≥ 2

(nk

). (By

convention, we set(

ak

)= 0 when a < k.) Proof: We may assume a ≥ n, since otherwise

we can replace a with 2n−a. Now we prove the result by induction on a. For the basecase, if a = n, then the lemma states that 2

(nk

)≥ 2

(nk

), which is trivial. If the lemma

holds for some a > 0, then by the familiar identity,[(a+ 1

k

)+

(2n− a− 1

k

)]−

[(a

k

)+

(2n− ak

)]

=

[(a+ 1

k

)−

(a

k

)]−

[(2n− ak

)+

(2n− a− 1

k

)]

=

(a

k − 1

)−

(2n− a− 1

k − 1

)> 0

(since a > 2n−a− 1), so(

a+1k

)+

(2n−a−1

k

)>

(ak

)+

(2n−a

k

)≥ 2

(nk

), giving the induction

step. The lemma follows.

Now suppose that the elements of A are labeled such that a elements of the setA receive the number 1 and 2n − a elements receive the number 2. Then the k-configuration can include all k-element subsets of A except those contained among thea elements numbered 1 or the 2n − a elements numbered 2. Thus, we have at most(

2nk

)−

(ak

)−

(2n−a

k

)elements in the k-configuration, and by the lemma, this is at most

(2n

k

)− 2

(n

k

).

On the other hand, we can achieve(

2nk

)− 2

(nk

)via the recipe above — take all the

k-element subsets of A, except those contained entirely within the first n elementsor entirely within the last n elements. Then, labeling the first n elements with thenumber 1 and the last n elements with the number 2 shows that the configuration is2-separable. So, Bk(n) =

(2nk

)− 2

(nk

).

6. Prove that any 2-configuration containing e elements is m-separable for some m ≤12

+√

2e+ 14.

Solution: Suppose m is the minimum integer for which the given configuration C onset A is m-separable, and fix a corresponding labeling of the elements of A. Let Ai bethe set of all elements with the label i. Then, for any i, j with 1 ≤ i < j ≤ m, theremust exist ai ∈ Ai, aj ∈ Aj with {ai, aj} ∈ C, since otherwise the elements of Aj couldhave been reassigned the label i, decreasing the number of distinct labels necessary andthus contradicting the minimality of m. We thus get at least

(m2

)different elements of

C. Therefore, e ≥(

m2

)= m(m− 1)/2 = [(m− 1

2)2 − 1

4]/2, and solving for m gives the

desired result.

4

A cell of a 2-configuration of a set A is a nonempty subset C of A such that

i. for any two distinct elements a, b of C, there exists a sequence c0, c1, . . . , cn of elementsof A with c0 = a, cn = b, and such that {c0, c1}, {c1, c2}, . . . , {cn−1, cn} are all elementsof the 2-configuration, and

ii. if a is an element of C and b is an element of A but not of C, there does NOTexist a sequence c0, c1, . . . , cn of elements of A with c0 = a, cn = b, and such that{c0, c1}, {c1, c2}, . . ., {cn−1, cn} are all elements of the 2-configuration.

Also, we define a 2-configuration of A to be barren if there is no subset {a0, a1, . . . , an} of A,with n ≥ 2, such that {a0, a1}, {a1, a2}, . . . , {an−1, an} and {an, a0} are all elements of the2-configuration.

7. Show that, given any 2-configuration of a set A, every element of A belongs to exactlyone cell.

Solution: First, given a, let Ca be the set of all b ∈ A for which there exists a sequencea = c0, c1, . . . , cn = b as in the definition of a cell. Certainly a ∈ Ca (take n = 0); weclaim that Ca is a cell. If b, b′ ∈ Ca, then there exist sequences a = c0, c1, . . . , cn = band a = c′0, c

′1, . . . , c

′m = b′, so the sequence b = cn, cn−1, . . . , c1, c0, c

′1, . . . , c

′m = b′ shows

that the first condition is met. For the second, suppose that there does exist a sequenceb = c0, c1, . . . , cn = b′ with b ∈ Ca, b

′ /∈ Ca. Then, concatenating with our sequencefrom a to b, we get a sequence from a to b′, contradicting the assumption b′ /∈ Ca.Thus, the second condition holds, and Ca is a cell. So a lies in at least one cell.

But now, note that if C is a cell containing a, then all b for which such a sequencefrom a to b exists must lie in C (or the second condition is violated), and if no suchsequence exists, then b cannot lie in C (or the first condition is violated). Thus, theelements of C are uniquely determined, so there is exactly one cell containing a, andthe proof is complete.

8. (a) Given a set A with n ≥ 1 elements, find the number of consistent 2-configurationsof A of order 1 with exactly 1 cell.

Solution: There must be some pair {a, b} in the 2-configuration, since each elementa ∈ A must belong to one pair. Since neither a nor b can now belong to any other pair,this must be the entire cell. Thus, there is 1 such 2-configuration when n = 2, andthere are none when n 6= 2.

(b) Given a set A with 10 elements, find the number of consistent 2-configurations ofA of order 2 with exactly 1 cell.

Solution: Consider such a configuration; let {a1, a2} be an element of it. Then a2 be-longs to exactly one other pair; call it {a2, a3}. Likewise, a3 belongs to exactly one otherpair {a3, a4}, and so forth; since we have finitely many elements, we must eventuallyreach some pair {am, ak} that revisits a previously used element (m > k). But this isonly possible if k = 1, since each other ak with k < m is already used in two pairs. Now,{a1, . . . , am} constitutes a complete cell, because none of these elements can be usedin any more pairs, so m = 10. Thus, every consistent 2-configuration of order 2 withexactly 1 cell gives rise to a permutation a1, . . . , a10 of the elements of A, and conversely,each such permutation gives us a 2-configuration {{a1, a2}, . . . , {a9, a10}, {a10, a1}}. Infact, each configuration corresponds to exactly 20 permutations, depending which of the

5

10 elements of the 2-configuration we choose as {a1, a2} and which of these two elementsof A we in turn choose to designate as a1 (as opposed to a2). Therefore, the number ofsuch 2-configurations is 10!/20 = 9!/2 = 181440. (More generally, the same argumentshows that, when A has n ≥ 3 elements, there are (n− 1)!/2 such 2-configurations.)

(c) Given a set A with 10 elements, find the number of consistent 2-configurations oforder 2 with exactly 2 cells.

Solution: Notice that if we look only at the pairs contained within any fixed cell, eachelement of that cell still lies in 2 such pairs, since all the pairs it belongs to are containedwithin that cell. Thus we have an induced consistent 2-configuration of order 2 of eachcell.

Now, each cell must have at least 3 elements for the configuration to be 2-consistent.So we can have either two 5-element cells, a 4-element cell and a 6-element cell, or a3-element cell and a 7-element cell. If there are two 5-element cells, we can choosethe members of the first cell in

(105

)ways, and then (by the reasoning in the previous

problem) we have 4!/2 ways to build a consistent 2-configuration of order 2 of eachcell. However, choosing 5 elements for the first cell is equivalent to choosing the other5 elements for the first cell, since the two cells are indistinguishable; thus, we haveovercounted by a factor of 2. So we have

(105

)· (4!/2)2/2 = 252 · 144/2 = 18144 ways to

form our configuration if we require it to have two cells of 5 elements each.If we have one 4-element cell and one 6-element cell, then there are

(104

)ways

to determine which 4 elements go in the smaller cell, and then 3!/2 ways and 5!/2ways, respectively, to construct the 2-configurations of the two cells, for a total of(

104

)· (3!/2) · (5!/2) = 210 · 3 · 60 = 37800 configurations (no overcounting here), and

by similar reasoning, we have(

103

)· (2!/2) · (6!/2) = 120 · 1 · 360 = 43200 configurations

with one 3-element cell and one 7-element cell. Thus, altogether, we have a total of18144 + 37800 + 43200 = 99144 consistent 2-configurations of order 2 with exactly 2cells.

9. (a) Show that if every cell of a 2-configuration of a finite set A is m-separable, thenthe whole 2-configuration is m-separable.

Solution: Let C be a 2-configuration of A with cells A1, . . . , An, so that there isno element of C with one element in Ai and another in Aj for i 6= j. Suppose thateach cell is m-separable, so that for each i, 1 ≤ i ≤ n, there is a labeling functionfi : Ai → {1, . . . ,m} such that no two elements in the same pair of C are assigned thesame number. Then, by combining, we get a function f on all of A whose restrictionto Ai is fi for each i. By the definition of fi, within each Ai there is no element of Cboth of whose elements are mapped to the same integer; and as above, we know thatthere are no elements of C not contained inside any Ai. Thus, C is m-separable, by theexistence of f .

(b) Show that any barren 2-configuration of a finite set A is 2-separable.

Solution: It is sufficient to show each cell is 2-separable, by part (a). A barren 2-configuration by definition cannot have any cycles (i.e. subsets {a0, . . . , an}, n ≥ 2,where each {ai, ai+1} and {an, a0} all belong to the 2-configuration). For any twodistinct elements a, b of A in the same cell of a 2-configuration C, define the distancebetween them to be the smallest n such that there exists a sequence a = a0, a1, . . . , an =

6

b with {a0, a1}, {a1, a2}, . . . , {an−1, an} all belonging to the 2-configuration. Notice thatthe terms of this sequence are all distinct: if ai = aj for i < j, then we have the shortersequence a0, a1, . . . , ai, aj+1, . . . , an, contradicting minimality.

Now let C be a barren 2-configuration of A. Pick any element a of A; label it andall elements at even distance from it with the integer 1, and label all elements at odddistance from it with the integer 2. We claim no two different elements with the samelabel appear in the same element of C. Otherwise, let b and c be such elements, andlet a = a0, a1, . . . , an = b and a = a′0, a

′1, . . . , a

′m = c be the corresponding minimal

sequences. Consider the largest p such that ap ∈ {a′0, . . . , a′m}; write ap = a′q. We claimthe set {an, an−1, . . . , ap, a

′q+1, . . . , a

′m} is then a cycle. It is straightforward to check

that all its elements are distinct; the only issue is whether it has at least 3 elements. Ifnot, we would have ap = an or a′m. Assume that ap = an ⇒ p = n ⇒ q = m − 1 (byminimality of our sequence (a′i)), but this means that m = n+ 1, so the distances of band c from a have opposite parities, contradicting the assumption that they have thesame label. The case a′q = a′m is similar. Thus, our set really does have at least threeelements, and it is a cycle. But since A is barren, it contains no cycles, and we have acontradiction.

Thus, after all, no two elements with the same label appear in the same pair of C,so the cell containing a is 2-separable, and we are done.

10. Show that every consistent 2-configuration of order 4 on a finite set A has a subsetthat is a consistent 2-configuration of order 2.

Solution: First, assume the 2-configuration has just one cell. We claim there exists asequence a0, a1, . . . , an of elements of A (not necessarily all distinct) such that the list

{a0, a1}, {a1, a2}, . . . , {an−1, an}, {an, a0}

contains each element of the 2-configuration exactly once. To see this, consider thelongest sequence such that {a0, a1}, . . . , {an−1, an}, {an, a0} are all distinct elements ofthe 2-configuration. (We may take n = 0 if necessary. Note that the finiteness conditionensures such a maximal sequence exists.) Each element of A occurs an even numberof times among these pairs (since each occurrence in the sequence contributes to twopairs). If every element occurs 4 times or 0 times, then the elements occurring in thesequence form a cell, since they cannot occur in any other pairs in the 2-configuration.Hence, they are all of A, and our sequence uses all the pairs in the 2-configuration,so the claim follows. Otherwise, there is some element ai occurring exactly twice.Choose b1 so that {ai, b1} is one of the two pairs in the 2-configuration not used by oursequence. Then choose b2 so that {b1, b2} be another pair not used thus far. Continuein this manner, choosing new elements bk with {bk, bk+1} a pair not already used, untilwe reach a point where finding another unused pair is impossible. Now, our pairs sofar are

{a0, a1}, . . . , {an−1, an}, {an, a0},{ai, b1}, {b1, b2}, . . . , {bk−1, bk}.

Every element is used in an even number of these pairs, except possibly ai, which isused in three pairs, and bk, which is used in an odd number of pairs (so one or three) —unless ai = bk, in which case this element occurs four times. But since it is impossibleto continue the sequence, bk must indeed have been used four times, so bk = ai.

7

But now we can construct the following sequence of distinct elements of the 2-configuration:

{a0, a1}, . . . , {ai−1, ai}, {ai, b1}, {b1, b2}, . . . , {bk−1, ai},

{ai, ai+1}, . . . , {an−1, an}, {an, a0}.This contradicts the maximality of our original sequence. This contradiction meansthat our original sequence must have used all the pairs in the 2-configuration, after all.

So we can express the 2-configuration via such a sequence of pairs, where each pair’ssecond element equals the first element of the next pair. If A has n elements, then(since each element appears in four pairs) we have 2n pairs. So we can choose the 1st,3rd, 5th, . . . , (2n− 1)th pairs, and then each element of A belongs to just two of thesepairs, because each occurrence of the element as an ai contributes to two consecutivepairs from our original sequence (or the first and last such pairs). Thus, we have ourconsistent 2-configuration of order 2, as desired.

Finally, if A consists of more than one cell, then the pairs within any given cell forma consistent 2-configuration of order 4 on that cell. So we simply apply the aboveprocedure to obtain a consistent 2-configuration of order 2 on each cell, and thencombining these gives a consistent 2-configuration of order 2 on A, as desired.

Comments: A note for those who might have found these problems rather foreign— the objects described here are actually of considerable importance; they constitute theelements of graph theory, one of the major research areas of modern mathematics. Whatwe have called a “2-configuration” is generally called a graph, and what we have called a“k-configuration” (k > 2) is generally called a hypergraph. The graph in problem 3b is thePetersen graph, a ubiquitous counterexample in graph theory. A consistent 2-configurationof order n is an n-regular graph; a cell is a component; a barren 2-configuration is a forest(and a forest with one component is a tree); and an m-separable configuration is m-colorable(and the minimum m for which a graph is m-colorable is its chromatic number).

8

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Individual Round: Algebra Subject Test

1. How many ordered pairs of integers (a,b) satisfy all of the following inequalities?

a2 + b2 < 16

a2 + b2 < 8a

a2 + b2 < 8b

2. Find the largest number n such that (2004!)! is divisible by ((n!)!)!.

3. Compute: ⌊20053

2003 · 2004− 20033

2004 · 2005

⌋.

4. Evaluate the sum

1

2b√1c+ 1+

1

2b√2c+ 1+

1

2b√3c+ 1+ · · ·+ 1

2b√100c+ 1.

5. There exists a positive real number x such that cos(tan−1(x)) = x. Find the value ofx2.

6. Find all real solutions to x4 + (2− x)4 = 34.

7. If x, y, k are positive reals such that

3 = k2

(x2

y2+y2

x2

)+ k

(x

y+y

x

),

find the maximum possible value of k.

8. Let x be a real number such that x3 + 4x = 8. Determine the value of x7 + 64x2.

9. A sequence of positive integers is defined by a0 = 1 and an+1 = a2n + 1 for each n ≥ 0.

Find gcd(a999, a2004).

10. There exists a polynomial P of degree 5 with the following property: if z is a complexnumber such that z5+2004z = 1, then P (z2) = 0. Calculate the quotient P (1)/P (−1).

1

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Individual Round: Algebra Subject Test — Solutions

1. How many ordered pairs of integers (a,b) satisfy all of the following inequalities?

a2 + b2 < 16

a2 + b2 < 8a

a2 + b2 < 8b

Solution: 6

This is easiest to see by simply graphing the inequalities. They correspond to the(strict) interiors of circles of radius 4 and centers at (0, 0), (4, 0), (0, 4), respectively. Sowe can see that there are 6 lattice points in their intersection (circled in the figure).

2. Find the largest number n such that (2004!)! is divisible by ((n!)!)!.

Solution: 6

For positive integers a, b, we have

a! | b! ⇔ a! ≤ b! ⇔ a ≤ b.

Thus,

((n!)!)! | (2004!)! ⇔ (n!)! ≤ 2004! ⇔ n! ≤ 2004 ⇔ n ≤ 6.

3. Compute: ⌊20053

2003 · 2004− 20033

2004 · 2005

⌋.

Solution: 8

Let x = 2004. Then the expression inside the floor brackets is

(x+ 1)3

(x− 1)x− (x− 1)3

x(x+ 1)=

(x+ 1)4 − (x− 1)4

(x− 1)x(x+ 1)=

8x3 + 8x

x3 − x = 8 +16x

x3 − x.

Since x is certainly large enough that 0 < 16x/(x3 − x) < 1, the answer is 8.

1

4. Evaluate the sum

1

2b√1c+ 1+

1

2b√2c+ 1+

1

2b√3c+ 1+ · · ·+ 1

2b√100c+ 1.

Solution: 190/21

The first three terms all equal 1/3, then the next five all equal 1/5; more generally,

for each a = 1, 2, . . . , 9, the terms 1/(2⌊√

a2⌋

+ 1) to 1/(2⌊√

a2 + 2a⌋

+ 1) all equal

1/(2a + 1), and there are 2a + 1 such terms. Thus our terms can be arranged into

9 groups, each with sum 1, and only the last term 1/(2⌊√

100⌋

+ 1) remains, so the

answer is 9 + 1/21 = 190/21.

5. There exists a positive real number x such that cos(tan−1(x)) = x. Find the value ofx2.

Solution: (−1 +√

5)/2

Draw a right triangle with legs 1, x; then the angle θ opposite x is tan−1 x, and we cancompute cos(θ) = 1/

√x2 + 1. Thus, we only need to solve x = 1/

√x2 + 1. This is

equivalent to x√x2 + 1 = 1. Square both sides to get x4 + x2 = 1⇒ x4 + x2 − 1 = 0.

Use the quadratic formula to get the solution x2 = (−1+√

5)/2 (unique since x2 mustbe positive).

6. Find all real solutions to x4 + (2− x)4 = 34.

Solution: 1±√2

Let y = 2− x, so x+ y = 2 and x4 + y4 = 34. We know

(x+ y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 = x4 + y4 + 2xy(2x2 + 2y2 + 3xy).

Moreover, x2+y2 = (x+y)2−2xy, so the preceding equation becomes 24 = 34+2xy(2 ·22−xy), or (xy)2−8xy−9 = 0. Hence xy = 9 or −1. Solving xy = 9, x+ y = 2 producescomplex solutions, and solving xy = −1, x+ y = 2 produces (x, y) = (1 +

√2, 1−√2)

or (1−√2, 1 +√

2). Thus, x = 1±√2.

7. If x, y, k are positive reals such that

3 = k2

(x2

y2+y2

x2

)+ k

(x

y+y

x

),

find the maximum possible value of k.

Solution: (−1 +√

7)/2

We have 3 = k2(x2/y2 + y2/x2) + k(x/y + y/x) ≥ 2k2 + 2k, hence 7 ≥ 4k2 + 4k + 1 =(2k+1)2, hence k ≤ (

√7−1)/2. Obviously k can assume this value, if we let x = y = 1.

8. Let x be a real number such that x3 + 4x = 8. Determine the value of x7 + 64x2.

Solution: 128

2

For any integer n ≥ 0, the given implies xn+3 = −4xn+1 + 8xn, so we can rewrite anysuch power of x in terms of lower powers. Carrying out this process iteratively gives

x7 = −4x5 + 8x4

= 8x4 + 16x3 − 32x2

= 16x3 − 64x2 + 64x

= −64x2 + 128.

Thus, our answer is 128.

9. A sequence of positive integers is defined by a0 = 1 and an+1 = a2n + 1 for each n ≥ 0.

Find gcd(a999, a2004).

Solution: 677

If d is the relevant greatest common divisor, then a1000 = a2999 + 1 ≡ 1 = a0 (mod d),

which implies (by induction) that the sequence is periodic modulo d, with period 1000.In particular, a4 ≡ a2004 ≡ 0. So d must divide a4. Conversely, we can see thata5 = a2

4 + 1 ≡ 1 = a0 modulo a4, so (again by induction) the sequence is periodicmodulo a4 with period 5, and hence a999, a2004 are indeed both divisible by a4. So theanswer is a4, which we can compute directly; it is 677.

10. There exists a polynomial P of degree 5 with the following property: if z is a complexnumber such that z5+2004z = 1, then P (z2) = 0. Calculate the quotient P (1)/P (−1).

Solution: −2010012/2010013

Let z1, . . . , z5 be the roots of Q(z) = z5+2004z−1. We can check these are distinct (byusing the fact that there’s one in a small neighborhood of each root of z5 +2004z, or bynoting that Q(z) is relatively prime to its derivative). And certainly none of the rootsof Q is the negative of another, since z5 + 2004z = 1 implies (−z)5 + 2004(−z) = −1,so their squares are distinct as well. Then, z2

1 , . . . , z25 are the roots of P , so if we write

C for the leading coefficient of P , we have

P (1)

P (−1)=

C(1− z21) · · · (1− z2

5)

C(−1− z21) · · · (−1− z2

5)

=[(1− z1) · · · (1− z5)] · [(1 + z1) · · · (1 + z5)]

[(i− z1) · · · (i− z5)] · [(i+ z1) · · · (i+ z5)]

=[(1− z1) · · · (1− z5)] · [(−1− z1) · · · (−1− z5)]

[(i− z1) · · · (i− z5)] · [(−i− z1) · · · (−i− z5)]

=(15 + 2004 · 1− 1)(−15 + 2004 · (−1)− 1)

(i5 + 2004 · i− 1)(−i5 + 2004 · (−i)− 1)

=(2004)(−2006)

(−1 + 2005i)(−1− 2005i)

= −20052 − 1

20052 + 1= −4020024/4020026 = −2010012/2010013.

Alternative Solution: In fact, we can construct the polynomial P explicitly (up tomultiplication by a constant). We write P (z2) as a polynomial in z; it must use only

3

even powers of z and be divisible by z5+2004z−1, so we are inspired to try a differenceof squares,

P (z2) = (z5 + 2004z − 1)(z5 + 2004z + 1) = (z5 + 2004z)2 − 12 = z2(z4 + 2004)2 − 1,

givingP (z) = z(z2 + 2004)2 − 1.

Now plugging in z = 1 and z = −1 rapidly gives (20052 − 1)/(−20052 − 1) as before.

4

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Individual Round: Calculus Subject Test

1. Let f(x) = sin(sin x). Evaluate limh→0f(x+h)−f(h)

xat x = π.

2. Suppose the function f(x)− f(2x) has derivative 5 at x = 1 and derivative 7 at x = 2.Find the derivative of f(x)− f(4x) at x = 1.

3. Find limx→∞( 3√x3 + x2 − 3

√x3 − x2).

4. Let f(x) = cos(cos(cos(cos(cos(cos(cos(cos x))))))), and suppose that the number asatisfies the equation a = cos a. Express f ′(a) as a polynomial in a.

5. A mouse is sitting in a toy car on a negligibly small turntable. The car cannot turn onits own, but the mouse can control when the car is launched and when the car stops(the car has brakes). When the mouse chooses to launch, the car will immediatelyleave the turntable on a straight trajectory at 1 meter per second.

Suddenly someone turns on the turntable; it spins at 30 rpm. Consider the set S ofpoints the mouse can reach in his car within 1 second after the turntable is set inmotion. (For example, the arrows in the figure below represent two possible paths themouse can take.) What is the area of S, in square meters?

30 rpm

.5 m

1 m

6. For x > 0, let f(x) = xx. Find all values of x for which f(x) = f ′(x).

7. Find the area of the region in the xy-plane satisfying x6 − x2 + y2 ≤ 0.

8. If x and y are real numbers with (x+ y)4 = x− y, what is the maximum possible valueof y?

9. Find the positive constant c0 such that the series

∞∑

n=0

n!

(cn)n

converges for c > c0 and diverges for 0 < c < c0.

10. Let P (x) = x3 − 32x2 + x + 1

4. Let P [1](x) = P (x), and for n ≥ 1, let P [n+1](x) =

P [n](P (x)). Evaluate∫ 10 P

[2004](x) dx.

1

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Individual Round: Calculus Subject Test — Solutions

1. Let f(x) = sin(sin x). Evaluate limh→0f(x+h)−f(h)

xat x = π.

Solution: 0

The expression f(x+h)−f(h)x

is continuous at h = 0, so the limit is just f(x)−f(0)x

. Letting

x = π yields sin(sin π)−sin(sin 0)π

= 0.

2. Suppose the function f(x)− f(2x) has derivative 5 at x = 1 and derivative 7 at x = 2.Find the derivative of f(x)− f(4x) at x = 1.

Solution: 19

Let g(x) = f(x)− f(2x). Then we want the derivative of

f(x)− f(4x) = (f(x)− f(2x)) + (f(2x)− f(4x)) = g(x) + g(2x)

at x = 1. This is g′(x) + 2g′(2x) at x = 1, or 5 + 2 · 7 = 19.

3. Find limx→∞( 3√x3 + x2 − 3

√x3 − x2).

Solution: 2/3

Observe that

limx→∞

[(x+ 1/3)− 3

√x3 + x2

]= lim

x→∞x/3 + 1/27

( 3√x3 + x2)2 + ( 3

√x3 + x2)(x+ 1/3) + (x+ 1/3)2

,

by factoring the numerator as a difference of cubes. The numerator is linear in x, whilethe denominator is at least 3x2, so the limit as x → ∞ is 0. By similar arguments,limx→∞[(x− 1/3)− 3

√x3 − x2] = 0. So, the desired limit equals

2/3 + limx→∞[(x− 1/3)− 3

√x3 − x2]− lim

x→∞[(x+ 1/3)− 3√x3 + x2] = 2/3.

4. Let f(x) = cos(cos(cos(cos(cos(cos(cos(cos x))))))), and suppose that the number asatisfies the equation a = cos a. Express f ′(a) as a polynomial in a.

Solution: a8 − 4a6 + 6a4 − 4a2 + 1

This is an exercise using the chain rule. Define f0(x) = x and fn(x) = cos fn−1(x) forn ≥ 0. We will show by induction that fn(a) = a and f ′

n(a) = (− sin a)n for all n. Thecase n = 0 is clear. Then fn(a) = cos fn−1(a) = cos a = a, and

f ′n(a) = f ′

n−1(a) · (− sin fn−1(a)) = (− sin a)n−1 · (− sin a) = (− sin a)n

by induction. Now, f(x) = f8(x), so f ′(a) = (− sin a)8 = sin8 a. But sin2 a = 1 −cos2 a = 1− a2, so f ′(a) = (1− a2)4 = a8 − 4a6 + 6a4 − 4a2 + 1.

1

5. A mouse is sitting in a toy car on a negligibly small turntable. The car cannot turn onits own, but the mouse can control when the car is launched and when the car stops(the car has brakes). When the mouse chooses to launch, the car will immediatelyleave the turntable on a straight trajectory at 1 meter per second.

Suddenly someone turns on the turntable; it spins at 30 rpm. Consider the set S ofpoints the mouse can reach in his car within 1 second after the turntable is set inmotion. (For example, the arrows in the figure below represent two possible paths themouse can take.) What is the area of S, in square meters?

30 rpm

.5 m

1 m

Solution: π/6

The mouse can wait while the table rotates through some angle θ and then spend theremainder of the time moving along that ray at 1 m/s. He can reach any point betweenthe starting point and the furthest reachable point along the ray, (1−θ/π) meters out.So the area is given by the polar integral

∫ π

0

(1− θ/π)2

2dθ =

1

2· 1

π2

∫ π

0φ2 dφ = π/6

(where we have used the change of variables φ = π − θ).6. For x > 0, let f(x) = xx. Find all values of x for which f(x) = f ′(x).

Solution: 1

Let g(x) = log f(x) = x log x. Then f ′(x)/f(x) = g′(x) = 1 + log x. Thereforef(x) = f ′(x) when 1 + log x = 1, that is, when x = 1.

7. Find the area of the region in the xy-plane satisfying x6 − x2 + y2 ≤ 0.

Solution: π/2

Rewrite the condition as |y| ≤ √x2 − x6. The right side is zero when x is −1, 0, or 1,and it bounds an area symmetric about the x- and y-axes. Therefore, we can calculatethe area by the integral

2∫ 1

−1

√x2 − x6 dx = 4

∫ 1

0x√

1− x4 dx = 2∫ 1

0

√1− u2 du = π/2.

8. If x and y are real numbers with (x+ y)4 = x− y, what is the maximum possible valueof y?

Solution: 3 3√

2/16

2

By drawing the graph of the curve (as shown), which is just a 135◦ clockwise rotationand scaling of y = x4, we see that the maximum is achieved at the unique point wheredy/dx = 0. Implicit differentiation gives 4(dy/dx+ 1)(x+ y)3 = 1− dy/dx, so settingdy/dx = 0 gives 4(x+y)3 = 1. So x+y = 1/ 3

√4 = 3√

2/2, and x−y = (x+y)4 = 3√

2/8.Subtracting and dividing by 2 gives y = ( 3

√2/2− 3

√2/8)/2 = 3 3

√2/16.

0.5 1 1.5

-0.5

-1

-1.5

9. Find the positive constant c0 such that the series

∞∑

n=0

n!

(cn)n

converges for c > c0 and diverges for 0 < c < c0.

Solution: 1/e

The ratio test tells us that the series converges if

limn→∞

(n+ 1)!/(c(n+ 1))n+1

n!/(cn)n=

1

c· lim

n→∞

(n

n+ 1

)n

is less than one and diverges if it is greater than one. But

limn→∞

(n

n+ 1

)n

= limn→∞

(1 +

1

n

)−n

=1

e.

Then the limit above is just 1/ce, so the series converges for c > 1/e and diverges for0 < c < 1/e.

10. Let P (x) = x3 − 32x2 + x + 1

4. Let P [1](x) = P (x), and for n ≥ 1, let P [n+1](x) =

P [n](P (x)). Evaluate∫ 10 P

[2004](x) dx.

Solution: 1/2

By Note that P (1 − x) = 1 − P (x). It follows easily by induction that P [k](1 − x) =1− P [k](x) for all positive integers k. Hence

∫ 1

0P [2004](x) dx =

∫ 1

01− P [2004](1− x) dx

= 1−∫ 1

0P [2004](1− x) dx

= 1−∫ 1

0P [2004](u) du (u = 1− x).

Hence∫ 10 P

[2004](x) dx = 1/2.

3

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Individual Round: Combinatorics Subject Test

1. There are 1000 rooms in a row along a long corridor. Initially the first room contains1000 people and the remaining rooms are empty. Each minute, the following happens:for each room containing more than one person, someone in that room decides it istoo crowded and moves to the next room. All these movements are simultaneous (sonobody moves more than once within a minute). After one hour, how many differentrooms will have people in them?

2. How many ways can you mark 8 squares of an 8× 8 chessboard so that no two markedsquares are in the same row or column, and none of the four corner squares is marked?(Rotations and reflections are considered different.)

3. A class of 10 students took a math test. Each problem was solved by exactly 7 of thestudents. If the first nine students each solved 4 problems, how many problems didthe tenth student solve?

4. Andrea flips a fair coin repeatedly, continuing until she either flips two heads in a row(the sequence HH) or flips tails followed by heads (the sequence TH). What is theprobability that she will stop after flipping HH?

5. A best-of-9 series is to be played between two teams; that is, the first team to win 5games is the winner. The Mathletes have a chance of 2/3 of winning any given game.What is the probability that exactly 7 games will need to be played to determine awinner?

6. A committee of 5 is to be chosen from a group of 9 people. How many ways can it bechosen, if Bill and Karl must serve together or not at all, and Alice and Jane refuse toserve with each other?

7. We have a polyhedron such that an ant can walk from one vertex to another, travelingonly along edges, and traversing every edge exactly once. What is the smallest possibletotal number of vertices, edges, and faces of this polyhedron?

8. Urn A contains 4 white balls and 2 red balls. Urn B contains 3 red balls and 3 blackballs. An urn is randomly selected, and then a ball inside of that urn is removed. Wethen repeat the process of selecting an urn and drawing out a ball, without returningthe first ball. What is the probability that the first ball drawn was red, given that thesecond ball drawn was black?

9. A classroom consists of a 5 × 5 array of desks, to be filled by anywhere from 0 to 25students, inclusive. No student will sit at a desk unless either all other desks in its rowor all others in its column are filled (or both). Considering only the set of desks that areoccupied (and not which student sits at each desk), how many possible arrangementsare there?

1

10. In a game similar to three card monte, the dealer places three cards on the table: thequeen of spades and two red cards. The cards are placed in a row, and the queenstarts in the center; the card configuration is thus RQR. The dealer proceeds to move.With each move, the dealer randomly switches the center card with one of the twoedge cards (so the configuration after the first move is either RRQ or QRR). What isthe probability that, after 2004 moves, the center card is the queen?

2

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Individual Round: Combinatorics Subject Test — Solutions

1. There are 1000 rooms in a row along a long corridor. Initially the first room contains1000 people and the remaining rooms are empty. Each minute, the following happens:for each room containing more than one person, someone in that room decides it istoo crowded and moves to the next room. All these movements are simultaneous (sonobody moves more than once within a minute). After one hour, how many differentrooms will have people in them?

Solution: 31

We can prove by induction on n that the following pattern holds for 0 ≤ n ≤ 499:after 2n minutes, the first room contains 1000− 2n people and the next n rooms eachcontain 2 people, and after 2n + 1 minutes, the first room contains 1000 − (2n + 1)people, the next n rooms each contain 2 people, and the next room after that contains1 person. So, after 60 minutes, we have one room with 940 people and 30 rooms with2 people each.

2. How many ways can you mark 8 squares of an 8× 8 chessboard so that no two markedsquares are in the same row or column, and none of the four corner squares is marked?(Rotations and reflections are considered different.)

Solution: 21600

In the top row, you can mark any of the 6 squares that is not a corner. In thebottom row, you can then mark any of the 5 squares that is not a corner and notin the same column as the square just marked. Then, in the second row, you have6 choices for a square not in the same column as either of the two squares alreadymarked; then there are 5 choices remaining for the third row, and so on down to 1for the seventh row, in which you make the last mark. Thus, altogether, there are6 · 5 · (6 · 5 · · · 1) = 30 · 6! = 30 · 720 = 21600 possible sets of squares.

3. A class of 10 students took a math test. Each problem was solved by exactly 7 of thestudents. If the first nine students each solved 4 problems, how many problems didthe tenth student solve?

Solution: 6

Suppose the last student solved n problems, and the total number of problems onthe test was p. Then the total number of correct solutions written was 7p (seven perproblem), and also equal to 36+n (the sum of the students’ scores), so p = (36+n)/7.The smallest n ≥ 0 for which this is an integer is n = 6. But we also must have n ≤ p,so 7n ≤ 36 + n, and solving gives n ≤ 6. Thus n = 6 is the answer.

4. Andrea flips a fair coin repeatedly, continuing until she either flips two heads in a row(the sequence HH) or flips tails followed by heads (the sequence TH). What is theprobability that she will stop after flipping HH?

Solution: 1/4

1

The only way that Andrea can ever flip HH is if she never flips T , in which case shemust flip two heads immediately at the beginning. This happens with probability 1

4.

5. A best-of-9 series is to be played between two teams; that is, the first team to win 5games is the winner. The Mathletes have a chance of 2/3 of winning any given game.What is the probability that exactly 7 games will need to be played to determine awinner?

Solution: 20/81

If the Mathletes are to win, they must win exactly 5 out of the 7 games. One ofthe 5 games they win must be the 7th game, because otherwise they would win thetournament before 7 games are completed. Thus, in the first 6 games, the Mathletesmust win 4 games and lose 2. The probability of this happening and the Mathleteswinning the last game is

[(6

2

)·(

2

3

)4

·(

1

3

)2]·(

2

3

).

Likewise, the probability of the other team winning on the 7th game is

[(6

2

)·(

1

3

)4

·(

2

3

)2]·(

1

3

).

Summing these values, we obtain 160/729 + 20/729 = 20/81.

6. A committee of 5 is to be chosen from a group of 9 people. How many ways can it bechosen, if Bill and Karl must serve together or not at all, and Alice and Jane refuse toserve with each other?

Solution: 41

If Bill and Karl are on the committee, there are(73

)= 35 ways for the other group

members to be chosen. However, if Alice and Jane are on the committee with Bill andKarl, there are

(51

)= 5 ways for the last member to be chosen, yielding 5 unacceptable

committees. If Bill and Karl are not on the committee, there are(75

)= 21 ways for

the 5 members to be chosen, but again if Alice and Jane were to be on the committee,there would be

(53

)= 10 ways to choose the other three members, yielding 10 more

unacceptable committees. So, we obtain (35− 5)+ (21− 10) = 41 ways the committeecan be chosen.

7. We have a polyhedron such that an ant can walk from one vertex to another, travelingonly along edges, and traversing every edge exactly once. What is the smallest possibletotal number of vertices, edges, and faces of this polyhedron?

Solution: 20

This is obtainable by construction. Consider two tetrahedrons glued along a face; thisgives us 5 vertices, 9 edges, and 6 faces, for a total of 20, and one readily checks thatthe required Eulerian path exists. Now, to see that we cannot do better, first noticethat the number v of vertices is at least 5, since otherwise we must have a tetrahedron,which does not have an Eulerian path. Each vertex is incident to at least 3 edges, and

2

in fact, since there is an Eulerian path, all except possibly two vertices are incident toan even number of edges. So the number of edges is at least (3+3+4+4+4)/2 (sinceeach edge meets two vertices) = 9. Finally, if f = 4 then each face must be a triangle,because there are only 3 other faces for it to share edges with, and we are again in thecase of a tetrahedron, which is impossible; therefore f ≥ 5. So f+v+e ≥ 5+5+9 = 19.But since f+v−e = 2−2g (where g is the number of holes in the polyhedron), f+v+emust be even. This strengthens our bound to 20 as needed.

8. Urn A contains 4 white balls and 2 red balls. Urn B contains 3 red balls and 3 blackballs. An urn is randomly selected, and then a ball inside of that urn is removed. Wethen repeat the process of selecting an urn and drawing out a ball, without returningthe first ball. What is the probability that the first ball drawn was red, given that thesecond ball drawn was black?

Solution: 7/15

This is a case of conditional probability; the answer is the probability that the firstball is red and the second ball is black, divided by the probability that the second ballis black.

First, we compute the numerator. If the first ball is drawn from Urn A, we have aprobability of 2/6 of getting a red ball, then a probability of 1/2 of drawing the secondball from Urn B, and a further probability of 3/6 of drawing a black ball. If the firstball is drawn from Urn B, we have probability 3/6 of getting a red ball, then 1/2 ofdrawing the second ball from Urn B, and 3/5 of getting a black ball. So our numeratoris

1

2

(2

6· 12· 36

+3

6· 12· 35

)=

7

60.

We similarly compute the denominator: if the first ball is drawn from Urn A, we havea probability of 1/2 of drawing the second ball from Urn B, and 3/6 of drawing ablack ball. If the first ball is drawn from Urn B, then we have probability 3/6 thatit is red, in which case the second ball will be black with probability (1/2) · (3/5),and probability 3/6 that the first ball is black, in which case the second is black withprobability (1/2) · (2/5). So overall, our denominator is

1

2

(1

2· 36

+3

6

[1

2· 35

+1

2· 25

])=

1

4.

Thus, the desired conditional probability is (7/60) / (1/4) = 7/15.

9. A classroom consists of a 5 × 5 array of desks, to be filled by anywhere from 0 to 25students, inclusive. No student will sit at a desk unless either all other desks in its rowor all others in its column are filled (or both). Considering only the set of desks that areoccupied (and not which student sits at each desk), how many possible arrangementsare there?

Solution: 962

The set of empty desks must be of the form (non-full rows) × (non-full columns): eachempty desk is in a non-full column and a non-full row, and the given condition impliesthat each desk in such a position is empty. So if there are fewer than 25 students, then

3

both of these sets are nonempty; we have 25−1 = 31 possible sets of non-full rows, and31 sets of non-full columns, for 961 possible arrangements. Alternatively, there may be25 students, and then only 1 arrangement is possible. Thus there are 962 possibilitiesaltogether.

10. In a game similar to three card monte, the dealer places three cards on the table: thequeen of spades and two red cards. The cards are placed in a row, and the queenstarts in the center; the card configuration is thus RQR. The dealer proceeds to move.With each move, the dealer randomly switches the center card with one of the twoedge cards (so the configuration after the first move is either RRQ or QRR). What isthe probability that, after 2004 moves, the center card is the queen?

Solution: 1/3 + 1/(3 · 22003)

If the probability that the queen is the center card after move n is pn, then the prob-ability that the queen is an edge card is 1 − pn, and the probability that the queenis the center card after move n + 1 is pn+1 = (1 − pn)/2. This recursion allows usto calculate the first few values of pn. We might then notice in 1, 0, 1

2, 1

4, 3

8, 5

16, 11

32, · · · ,

that the value of each fraction is close to 1/3, and getting closer for larger n. In factsubtracting 1/3 from each fraction yields 2

3,−1

3, 1

6,− 1

12, 1

24,− 1

48, · · · . This suggests the

formula pn = 13

+ 23(−1

2)n, and one can then prove that this formula is in fact correct

by induction. Thus, p(2004) = 13

+ 23(−1

2)2004 = 1

3+ 1

3·22003 .

The recurrence can also be solved without guessing — by generating functions, forexample, or by using the fundamental theorem of linear recurrences, which ensuresthat the solution is of the form pn = a+ b(−1/2)n for some constants a, b.

4

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Individual Round: General Test, Part 1

1. There are 1000 rooms in a row along a long corridor. Initially the first room contains1000 people and the remaining rooms are empty. Each minute, the following happens:for each room containing more than one person, someone in that room decides it istoo crowded and moves to the next room. All these movements are simultaneous (sonobody moves more than once within a minute). After one hour, how many differentrooms will have people in them?

2. What is the largest whole number that is equal to the product of its digits?

3. Suppose f is a function that assigns to each real number x a value f(x), and supposethe equation

f(x1 + x2 + x3 + x4 + x5) = f(x1) + f(x2) + f(x3) + f(x4) + f(x5)− 8

holds for all real numbers x1, x2, x3, x4, x5. What is f(0)?

4. How many ways can you mark 8 squares of an 8× 8 chessboard so that no two markedsquares are in the same row or column, and none of the four corner squares is marked?(Rotations and reflections are considered different.)

5. A rectangle has perimeter 10 and diagonal√

15. What is its area?

6. Find the ordered quadruple of digits (A,B,C,D), with A > B > C > D, such that

ABCD

− DCBA

= BDAC.

7. Let ACE be a triangle with a point B on segment AC and a point D on segment CEsuch that BD is parallel to AE. A point Y is chosen on segment AE, and segmentCY is drawn. Let X be the intersection of CY and BD. If CX = 5, XY = 3, what isthe ratio of the area of trapezoid ABDE to the area of triangle BCD?

A

Y

X

E

D

C

B

1

8. You have a 10× 10 grid of squares. You write a number in each square as follows: youwrite 1, 2, 3, . . . , 10 from left to right across the top row, then 11, 12, . . . , 20 across thesecond row, and so on, ending with a 100 in the bottom right square. You then writea second number in each square, writing 1, 2, . . . , 10 in the first column (from top tobottom), then 11, 12, . . . , 20 in the second column, and so forth.

When this process is finished, how many squares will have the property that their twonumbers sum to 101?

9. Urn A contains 4 white balls and 2 red balls. Urn B contains 3 red balls and 3 blackballs. An urn is randomly selected, and then a ball inside of that urn is removed. Wethen repeat the process of selecting an urn and drawing out a ball, without returningthe first ball. What is the probability that the first ball drawn was red, given that thesecond ball drawn was black?

10. A floor is tiled with equilateral triangles of side length 1, as shown. If you drop aneedle of length 2 somewhere on the floor, what is the largest number of triangles itcould end up intersecting? (Only count the triangles whose interiors are met by theneedle — touching along edges or at corners doesn’t qualify.)

2

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Individual Round: General Test, Part 1 — Solutions

1. There are 1000 rooms in a row along a long corridor. Initially the first room contains1000 people and the remaining rooms are empty. Each minute, the following happens:for each room containing more than one person, someone in that room decides it istoo crowded and moves to the next room. All these movements are simultaneous (sonobody moves more than once within a minute). After one hour, how many differentrooms will have people in them?

Solution: 31

We can prove by induction on n that the following pattern holds for 0 ≤ n ≤ 499:after 2n minutes, the first room contains 1000− 2n people and the next n rooms eachcontain 2 people, and after 2n + 1 minutes, the first room contains 1000 − (2n + 1)people, the next n rooms each contain 2 people, and the next room after that contains1 person. So, after 60 minutes, we have one room with 940 people and 30 rooms with2 people each.

2. What is the largest whole number that is equal to the product of its digits?

Solution: 9

Suppose the number n has k + 1 digits, the first of which is d. Then the number is atleast d · 10k. On the other hand, each of the digits after the first is at most 9, so theproduct of the digits is at most d · 9k. Thus, if n equals the product of its digits, then

d · 10k ≤ n ≤ d · 9k

which forces k = 0, i.e., the number has only one digit. So n = 9 is clearly the largestpossible value.

3. Suppose f is a function that assigns to each real number x a value f(x), and supposethe equation

f(x1 + x2 + x3 + x4 + x5) = f(x1) + f(x2) + f(x3) + f(x4) + f(x5)− 8

holds for all real numbers x1, x2, x3, x4, x5. What is f(0)?

Solution: 2

Plug in x1 = x2 = x3 = x4 = x5 = 0. Then the equation reads f(0) = 5f(0) − 8, so4f(0) = 8, so f(0) = 2.

4. How many ways can you mark 8 squares of an 8× 8 chessboard so that no two markedsquares are in the same row or column, and none of the four corner squares is marked?(Rotations and reflections are considered different.)

Solution: 21600

In the top row, you can mark any of the 6 squares that is not a corner. In thebottom row, you can then mark any of the 5 squares that is not a corner and notin the same column as the square just marked. Then, in the second row, you have

1

6 choices for a square not in the same column as either of the two squares alreadymarked; then there are 5 choices remaining for the third row, and so on down to 1for the seventh row, in which you make the last mark. Thus, altogether, there are6 · 5 · (6 · 5 · · · 1) = 30 · 6! = 30 · 720 = 21600 possible sets of squares.

5. A rectangle has perimeter 10 and diagonal√

15. What is its area?

Solution: 5

If the sides are x and y, we have 2x+ 2y = 10, so x+ y = 5, and√x2 + y2 =

√15, so

x2 + y2 = 15. Squaring the first equation gives x2 + 2xy + y2 = 25, and subtractingthe second equation gives 2xy = 10, so the area is xy = 5.

6. Find the ordered quadruple of digits (A,B,C,D), with A > B > C > D, such that

ABCD

− DCBA

= BDAC.

Solution: (7, 6, 4, 1)

Since D < A, when A is subtracted from D in the ones’ column, there will be a borrowfrom C in the tens’ column. Thus, D+ 10−A = C. Next, consider the subtraction inthe tens’ column, (C−1)−B. Since C < B, there will be a borrow from the hundreds’column, so (C − 1 + 10) − B = A. In the hundreds’ column, B − 1 ≥ C, so we donot need to borrow from A in the thousands’ column. Thus, (B − 1) − C = D andA−D = B. We thus have a system of four equations in four variables A,B,C,D, andsolving by standard methods (e.g. substitution) produces (A,B,C,D) = (7, 6, 4, 1).

7. Let ACE be a triangle with a point B on segment AC and a point D on segment CEsuch that BD is parallel to AE. A point Y is chosen on segment AE, and segmentCY is drawn. Let X be the intersection of CY and BD. If CX = 5, XY = 3, what isthe ratio of the area of trapezoid ABDE to the area of triangle BCD?

Solution: 39/25

A

L

K

Y

X

E

D

C

B

Draw the altitude from C to AE, intersecting line BD at K and line AE at L. ThenCK is the altitude of triangle BCD, so triangles CKX and CLY are similar. SinceCY/CX = 8/5, CL/CK = 8/5. Also triangles CKB and CLA are similar, so thatCA/CB = 8/5, and triangles BCD and ACE are similar, so that AE/BD = 8/5.The area of ACE is (1/2)(AE)(CL), and the area of BCD is (1/2)(BD)(CK), so the

2

ratio of the area of ACE to the area of BCD is 64/25. Therefore, the ratio of the areaof ABDE to the area of BCD is 39/25.

8. You have a 10× 10 grid of squares. You write a number in each square as follows: youwrite 1, 2, 3, . . . , 10 from left to right across the top row, then 11, 12, . . . , 20 across thesecond row, and so on, ending with a 100 in the bottom right square. You then writea second number in each square, writing 1, 2, . . . , 10 in the first column (from top tobottom), then 11, 12, . . . , 20 in the second column, and so forth.

When this process is finished, how many squares will have the property that their twonumbers sum to 101?

Solution: 10

The number in the ith row, jth column will receive the numbers 10(i − 1) + j and10(j − 1) + i, so the question is how many pairs (i, j) (1 ≤ i, j ≤ 10) will have

101 = [10(i− 1) + j] + [10(j − 1) + i] ⇔ 121 = 11i+ 11j = 11(i+ j).

Now it is clear that this is achieved by the ten pairs (1, 10), (2, 9), (3, 8), . . . , (10, 1) andno others.

9. Urn A contains 4 white balls and 2 red balls. Urn B contains 3 red balls and 3 blackballs. An urn is randomly selected, and then a ball inside of that urn is removed. Wethen repeat the process of selecting an urn and drawing out a ball, without returningthe first ball. What is the probability that the first ball drawn was red, given that thesecond ball drawn was black?

Solution: 7/15

This is a case of conditional probability; the answer is the probability that the firstball is red and the second ball is black, divided by the probability that the second ballis black.

First, we compute the numerator. If the first ball is drawn from Urn A, we have aprobability of 2/6 of getting a red ball, then a probability of 1/2 of drawing the secondball from Urn B, and a further probability of 3/6 of drawing a black ball. If the firstball is drawn from Urn B, we have probability 3/6 of getting a red ball, then 1/2 ofdrawing the second ball from Urn B, and 3/5 of getting a black ball. So our numeratoris

1

2

(2

6· 12· 36

+3

6· 12· 35

)=

7

60.

We similarly compute the denominator: if the first ball is drawn from Urn A, we havea probability of 1/2 of drawing the second ball from Urn B, and 3/6 of drawing ablack ball. If the first ball is drawn from Urn B, then we have probability 3/6 thatit is red, in which case the second ball will be black with probability (1/2) · (3/5),and probability 3/6 that the first ball is black, in which case the second is black withprobability (1/2) · (2/5). So overall, our denominator is

1

2

(1

2· 36

+3

6

[1

2· 35

+1

2· 25

])=

1

4.

Thus, the desired conditional probability is (7/60) / (1/4) = 7/15.

3

10. A floor is tiled with equilateral triangles of side length 1, as shown. If you drop aneedle of length 2 somewhere on the floor, what is the largest number of triangles itcould end up intersecting? (Only count the triangles whose interiors are met by theneedle — touching along edges or at corners doesn’t qualify.)

Solution: 8

Let L be the union of all the lines of the tiling. Imagine walking from one end of theneedle to the other. We enter a new triangle precisely when we cross one of the linesof the tiling. Therefore, the problem is equivalent to maximizing the number of timesthe needle crosses L. Now, the lines of the tiling each run in one of three directions.It is clear that the needle cannot cross more than three lines in any given direction,since the lines are a distance

√3/2 apart and the needle would therefore have to be of

length greater than 3√

3/2 > 2. Moreover, it cannot cross three lines in each of twodifferent directions. To see this, notice that its endpoints would have to lie in eitherthe two light-shaded regions or the two dark-shaded regions shown, but the closest twopoints of such opposite regions are at a distance of 2 (twice the length of a side of atriangle), so the needle cannot penetrate both regions.

Therefore, the needle can cross at most three lines in one direction and two lines ineach of the other two directions, making for a maximum of 3 + 2 + 2 = 7 crossings and7 + 1 = 8 triangles intersected. The example shows that 8 is achievable, as long as theneedle has length greater than

√3 < 2.

4

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Individual Round: General Test, Part 2

1. Find the largest number n such that (2004!)! is divisible by ((n!)!)!.

2. Andrea flips a fair coin repeatedly, continuing until she either flips two heads in a row(the sequence HH) or flips tails followed by heads (the sequence TH). What is theprobability that she will stop after flipping HH?

3. How many ordered pairs of integers (a, b) satisfy all of the following inequalities?

a2 + b2 < 16

a2 + b2 < 8a

a2 + b2 < 8b

4. A horse stands at the corner of a chessboard, a white square. With each jump, thehorse can move either two squares horizontally and one vertically or two vertically andone horizontally (like a knight moves). The horse earns two carrots every time it landson a black square, but it must pay a carrot in rent to rabbit who owns the chessboardfor every move it makes. When the horse reaches the square on which it began, it canleave. What is the maximum number of carrots the horse can earn without touchingany square more than twice?

5. Eight strangers are preparing to play bridge. How many ways can they be groupedinto two bridge games — that is, into unordered pairs of unordered pairs of people?

6. a and b are positive integers. When written in binary, a has 2004 1’s, and b has 2005 1’s(not necessarily consecutive). What is the smallest number of 1’s a+ b could possiblyhave?

7. Farmer John is grazing his cows at the origin. There is a river that runs east to west50 feet north of the origin. The barn is 100 feet to the south and 80 feet to the eastof the origin. Farmer John leads his cows to the river to take a swim, then the cowsleave the river from the same place they entered and Farmer John leads them to thebarn. He does this using the shortest path possible, and the total distance he travelsis d feet. Find the value of d.

1

8. A freight train leaves the town of Jenkinsville at 1:00 PM traveling due east at constantspeed. Jim, a hobo, sneaks onto the train and falls asleep. At the same time, Julieleaves Jenkinsville on her bicycle, traveling along a straight road in a northeasterlydirection (but not due northeast) at 10 miles per hour. At 1:12 PM, Jim rolls overin his sleep and falls from the train onto the side of the tracks. He wakes up andimmediately begins walking at 3.5 miles per hour directly towards the road on whichJulie is riding. Jim reaches the road at 2:12 PM, just as Julie is riding by. What is thespeed of the train in miles per hour?

9. Given is a regular tetrahedron of volume 1. We obtain a second regular tetrahedron byreflecting the given one through its center. What is the volume of their intersection?

10. A lattice point is a point whose coordinates are both integers. Suppose Johann walksin a line from the point (0, 2004) to a random lattice point in the interior (not onthe boundary) of the square with vertices (0, 0), (0, 99), (99, 99), (99, 0). What is theprobability that his path, including the endpoints, contains an even number of latticepoints?

2

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Individual Round: General Test, Part 2 — Solutions

1. Find the largest number n such that (2004!)! is divisible by ((n!)!)!.

Solution: 6

For positive integers a, b, we have

a! | b! ⇔ a! ≤ b! ⇔ a ≤ b.

Thus,

((n!)!)! | (2004!)! ⇔ (n!)! ≤ 2004! ⇔ n! ≤ 2004 ⇔ n ≤ 6.

2. Andrea flips a fair coin repeatedly, continuing until she either flips two heads in a row(the sequence HH) or flips tails followed by heads (the sequence TH). What is theprobability that she will stop after flipping HH?

Solution: 1/4

The only way that Andrea can ever flip HH is if she never flips T , in which case shemust flip two heads immediately at the beginning. This happens with probability 1

4.

3. How many ordered pairs of integers (a, b) satisfy all of the following inequalities?

a2 + b2 < 16

a2 + b2 < 8a

a2 + b2 < 8b

Solution: 6

This is easiest to see by simply graphing the inequalities. They correspond to the(strict) interiors of circles of radius 4 and centers at (0, 0), (4, 0), (0, 4), respectively. Sowe can see that there are 6 lattice points in their intersection (circled in the figure).

1

4. A horse stands at the corner of a chessboard, a white square. With each jump, thehorse can move either two squares horizontally and one vertically or two vertically andone horizontally (like a knight moves). The horse earns two carrots every time it landson a black square, but it must pay a carrot in rent to rabbit who owns the chessboardfor every move it makes. When the horse reaches the square on which it began, it canleave. What is the maximum number of carrots the horse can earn without touchingany square more than twice?

Solution: 0

The horse must alternate white and black squares, and it ends on the same squarewhere it started. Thus it lands on the same number of black squares (b) as whitesquares (w). Thus, its net earnings will be 2b− (b+w) = b−w = 0 carrots, regardlessof its path.

5. Eight strangers are preparing to play bridge. How many ways can they be groupedinto two bridge games — that is, into unordered pairs of unordered pairs of people?

Solution: 315

Putting 8 people into 4 pairs and putting those 4 pairs into 2 pairs of pairs are inde-pendent. If the people are numbered from 1 to 8, there are 7 ways to choose the personto pair with person 1. Then there are 5 ways to choose the person to pair with theperson who has the lowest remaining number, 3 ways to choose the next, and 1 way tochoose the last (because there are only 2 people remaining). Thus, there are 7 · 5 · 3 · 1ways to assign 8 people to pairs and similarly there are 3 · 1 ways to assign 4 pairs to2 pairs of pairs, so there are 7 · 5 · 3 · 3 = 315 ways.

6. a and b are positive integers. When written in binary, a has 2004 1’s, and b has 2005 1’s(not necessarily consecutive). What is the smallest number of 1’s a+ b could possiblyhave?

Solution: 1

Consider the following addition:

111 · · · 100 · · · 01+ 11 · · · 11= 1000 · · · · · · · · · 00

2

By making the blocks of 1’s and 0’s appropriately long, we can ensure that the addendsrespectively contain 2004 and 2005 1’s. (To be precise, we get a = 24008− 22005 +1 andb = 22005 − 1.) Then the sum has only one 1. And clearly it is not possible to get anyless than one 1.

7. Farmer John is grazing his cows at the origin. There is a river that runs east to west50 feet north of the origin. The barn is 100 feet to the south and 80 feet to the eastof the origin. Farmer John leads his cows to the river to take a swim, then the cowsleave the river from the same place they entered and Farmer John leads them to thebarn. He does this using the shortest path possible, and the total distance he travelsis d feet. Find the value of d.

Solution: 40√

29

Suppose we move the barn to its reflection across the river’s edge. Then paths fromthe origin to the river and then to the old barn location correspond to paths fromthe origin to the river and then to the new barn location, by reflecting the secondpart of the path across the river, and corresponding paths have the same length. Nowthe shortest path from the origin to the river and then to the new barn location is astraight line. The new barn location is 200 feet north and 80 feet east of the origin, sothe value of d is

√2002 + 802 = 40

√29.

(0,0)

(20,50)

(80,−100)

8. A freight train leaves the town of Jenkinsville at 1:00 PM traveling due east at constantspeed. Jim, a hobo, sneaks onto the train and falls asleep. At the same time, Julieleaves Jenkinsville on her bicycle, traveling along a straight road in a northeasterlydirection (but not due northeast) at 10 miles per hour. At 1:12 PM, Jim rolls overin his sleep and falls from the train onto the side of the tracks. He wakes up andimmediately begins walking at 3.5 miles per hour directly towards the road on whichJulie is riding. Jim reaches the road at 2:12 PM, just as Julie is riding by. What is thespeed of the train in miles per hour?

Solution: 62.5

Julie’s distance is (10 mph) · (6/5 hrs) = 12 miles. Jim’s walking distance, afterfalling off the train, is (3.5 mph) · (1 hr) = 3.5 miles at a right angle to the road.Therefore, Jim rode the train for

√122 + 3.52 = 1

2

√242 + 72 = 25/2 miles, and its

speed is (25/2 mi)/(1/5 hr) = 62.5 mph.

3

9. Given is a regular tetrahedron of volume 1. We obtain a second regular tetrahedron byreflecting the given one through its center. What is the volume of their intersection?

Solution: 1/2

Imagine placing the tetrahedron ABCD flat on a table with vertex A at the top.By vectors or otherwise, we see that the center is 3/4 of the way from A to thebottom face, so the reflection of this face lies in a horizontal plane halfway between Aand BCD. In particular, it cuts off the smaller tetrahedron obtained by scaling theoriginal tetrahedron by a factor of 1/2 about A. Similarly, the reflections of the otherthree faces cut off tetrahedra obtained by scaling ABCD by 1/2 about B, C, and D.On the other hand, the octahedral piece remaining remaining after we remove thesefour smaller tetrahedra is in the intersection of ABCD with its reflection, since thereflection sends this piece to itself. So the answer we seek is just the volume of thispiece, which is

(volume of ABCD)− 4 · (volume of ABCD scaled by a factor of 1/2)

= 1− 4(1/2)3 = 1/2.

10. A lattice point is a point whose coordinates are both integers. Suppose Johann walksin a line from the point (0, 2004) to a random lattice point in the interior (not onthe boundary) of the square with vertices (0, 0), (0, 99), (99, 99), (99, 0). What is theprobability that his path, including the endpoints, contains an even number of latticepoints?

Solution: 3/4

If Johann picks the point (a, b), the path will contain gcd(a, 2004−b)+1 points. Therewill be an odd number of points in the path if gcd(a, 2004− b) is even, which is true ifand only if a and b are both even. Since there are 492 points with a, b both even and982 total points, the probability that the path contains an even number of points is

982 − 492

982=

492(22 − 12)

492(22)=

3

4.

4

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Individual Round: Geometry Subject Test

1. In trapezoid ABCD, AD is parallel to BC. 6 A = 6 D = 45◦, while 6 B = 6 C = 135◦.If AB = 6 and the area of ABCD is 30, find BC.

? C

A45 45

135135B

D

6

Area = 30

2. A parallelogram has 3 of its vertices at (1, 2), (3,8), and (4, 1). Compute the sum ofthe possible x-coordinates for the 4th vertex.

3. A swimming pool is in the shape of a circle with diameter 60 ft. The depth varieslinearly along the east-west direction from 3 ft at the shallow end in the east to 15 ftat the diving end in the west (this is so that divers look impressive against the sunset)but does not vary at all along the north-south direction. What is the volume of thepool, in ft3?

4. P is inside rectangle ABCD. PA = 2, PB = 3, and PC = 10. Find PD.

5. Find the area of the region of the xy-plane defined by the inequality |x|+|y|+|x+y| ≤ 1.

6. In trapezoid ABCD shown, AD is parallel to BC, and AB = 6, BC = 7, CD =8, AD = 17. If sides AB and CD are extended to meet at E, find the resulting angleat E (in degrees).

A

E

CB

17

8

7

6

D

7. Yet another trapezoid ABCD has AD parallel to BC. AC and BD intersect at P .If [ADP ]/[BCP ] = 1/2, find [ADP ]/[ABCD]. (Here the notation [P1 · · ·Pn] denotesthe area of the polygon P1 · · ·Pn.)

8. A triangle has side lengths 18, 24, and 30. Find the area of the triangle whose verticesare the incenter, circumcenter, and centroid of the original triangle.

9. Given is a regular tetrahedron of volume 1. We obtain a second regular tetrahedron byreflecting the given one through its center. What is the volume of their intersection?

10. Right triangle XY Z has right angle at Y and XY = 228, Y Z = 2004. Angle Y istrisected, and the angle trisectors intersect XZ at P and Q so that X,P,Q, Z lie onXZ in that order. Find the value of (PY + Y Z)(QY +XY ).

1

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Individual Round: Geometry Subject Test — Solutions

1. In trapezoid ABCD, AD is parallel to BC. 6 A = 6 D = 45◦, while 6 B = 6 C = 135◦.If AB = 6 and the area of ABCD is 30, find BC.

Solution: 2√

2

Draw altitudes from B and C to AD and label the points of intersection X and Y ,respectively. Then ABX and CDY are 45◦−45◦−90◦ triangles with BX = CY = 3

√2.

So, the area ofABX and the area of CDY are each 9, meaning that the area of rectangleBCYX is 12. Since BX = 3

√2, BC = 12/(3

√2) = 2

√2.

Area BCYX = 30 − 18 = 12✓

3 2✓

2 2✓

A

B C

DX Y

69 9

3 2

2. A parallelogram has 3 of its vertices at (1, 2), (3,8), and (4, 1). Compute the sum ofthe possible x-coordinates for the 4th vertex.

Solution: 8

There are 3 possible locations for the 4th vertex. Let (a, b) be its coordinates. If it isopposite to vertex (1, 2), then since the midpoints of the diagonals of a parallelogramcoincide, we get (a+1

2, b+2

2) = (3+4

2, 8+1

2). Thus (a, b) = (6, 7). By similar reasoning for

the other possible choices of opposite vertex, the other possible positions for the fourthvertex are (0, 9) and (2,−5), and all of these choices do give parallelograms. So theanswer is 6 + 0 + 2 = 8.

3. A swimming pool is in the shape of a circle with diameter 60 ft. The depth varieslinearly along the east-west direction from 3 ft at the shallow end in the east to 15 ftat the diving end in the west (this is so that divers look impressive against the sunset)but does not vary at all along the north-south direction. What is the volume of thepool, in ft3?

Solution: 8100π

Take another copy of the pool, turn it upside-down, and put the two together to forma cylinder. It has height 18 ft and radius 30 ft, so the volume is π(30 ft)2 · 18 ft =16200π ft3; since our pool is only half of that, the answer is 8100π ft3.

4. P is inside rectangle ABCD. PA = 2, PB = 3, and PC = 10. Find PD.

Solution:√

95

Draw perpendiculars from P to E on AB, F on BC, G on CD, and H on DA, andlet AH = BF = w, HD = FC = x, AE = DG = y, and EB = GC = z. ThenPA2 = w2 + y2, PB2 = w2 + z2, PC2 = x2 + z2, and PD2 = x2 + y2. Adding andsubtracting, we see that PD2 = PA2 − PB2 + PC2 = 95, so PD =

√95.

1

y z

x

w

zy

w

x

H

G

F

E

P

D C

BA

5. Find the area of the region of the xy-plane defined by the inequality |x|+|y|+|x+y| ≤ 1.

Solution: 3/4

To graph this region we divide the xy-plane into six sectors depending on which ofx, y, x+ y are ≥ 0, or ≤ 0. The inequality simplifies in each case:

Sector Inequality Simplified inequalityx ≥ 0, y ≥ 0, x+ y ≥ 0 x+ y + (x+ y) ≤ 1 x+ y ≤ 1/2x ≥ 0, y ≤ 0, x+ y ≥ 0 x− y + (x+ y) ≤ 1 x ≤ 1/2x ≥ 0, y ≤ 0, x+ y ≤ 0 x− y − (x+ y) ≤ 1 y ≥ −1/2x ≤ 0, y ≥ 0, x+ y ≥ 0 −x+ y + (x+ y) ≤ 1 y ≤ 1/2x ≤ 0, y ≥ 0, x+ y ≤ 0 −x+ y − (x+ y) ≤ 1 x ≥ −1/2x ≤ 0, y ≤ 0, x+ y ≤ 0 −x− y − (x+ y) ≤ 1 x+ y ≥ −1/2

We then draw the region; we get a hexagon as shown. The hexagon intersects eachregion in an isosceles right triangle of area 1/8, so the total area is 6 · 1/8 = 3/4.

6. In trapezoid ABCD shown, AD is parallel to BC, and AB = 6, BC = 7, CD =8, AD = 17. If sides AB and CD are extended to meet at E, find the resulting angleat E (in degrees).

Solution: 90

Choose point F on AD so that BCDF is a parallelogram. Then BF = CD = 8, andAF = AD−DF = AD−BC = 10, so 4ABF is a 6-8-10 right triangle. The requiredangle is equal to 6 ABF = 90◦.

2

A D

6

7

8

B C

7

8

10 F

7. Yet another trapezoid ABCD has AD parallel to BC. AC and BD intersect at P .If [ADP ]/[BCP ] = 1/2, find [ADP ]/[ABCD]. (Here the notation [P1 · · ·Pn] denotesthe area of the polygon P1 · · ·Pn.)

Solution: 3− 2√

2

A homothety (scaling) about P takes triangle ADP into BCP , since AD,BC areparallel and A,P,C; B,P,D are collinear. The ratio of homothety is thus

√2. It

follows that, if we rescale to put [ADP ] = 1, then [ABP ] = [CDP ] =√

2, just by theratios of lengths of bases. So [ABCD] = 3 + 2

√2, so [ADP ]/[ABCD] = 1/(3 + 2

√2).

Simplifying this, we get 3− 2√

2.

8. A triangle has side lengths 18, 24, and 30. Find the area of the triangle whose verticesare the incenter, circumcenter, and centroid of the original triangle.

Solution: 3

There are many solutions to this problem, which is straightforward. The given triangleis a right 3-4-5 triangle, so the circumcenter is the midpoint of the hypotenuse. Co-ordinatizing for convenience, put the vertex at (0, 0) and the other vertices at (0, 18)and (24, 0). Then the circumcenter is (12, 9). The centroid is at one-third the sum ofthe three vertices, which is (8, 6). Finally, since the area equals the inradius times halfthe perimeter, we can see that the inradius is (18 · 24/2) / ([18 + 24 + 30]/2) = 6. Sothe incenter of the triangle is (6, 6). So the small triangle has a base of length 2 and aheight of 3, hence its area is 3.

9. Given is a regular tetrahedron of volume 1. We obtain a second regular tetrahedron byreflecting the given one through its center. What is the volume of their intersection?

Solution: 1/2

Imagine placing the tetrahedron ABCD flat on a table with vertex A at the top.By vectors or otherwise, we see that the center is 3/4 of the way from A to thebottom face, so the reflection of this face lies in a horizontal plane halfway between Aand BCD. In particular, it cuts off the smaller tetrahedron obtained by scaling theoriginal tetrahedron by a factor of 1/2 about A. Similarly, the reflections of the otherthree faces cut off tetrahedra obtained by scaling ABCD by 1/2 about B, C, and D.On the other hand, the octahedral piece remaining remaining after we remove thesefour smaller tetrahedra is in the intersection of ABCD with its reflection, since thereflection sends this piece to itself. So the answer we seek is just the volume of thispiece, which is

(volume of ABCD)− 4 · (volume of ABCD scaled by a factor of 1/2)

= 1− 4(1/2)3 = 1/2.

3

10. Right triangle XY Z has right angle at Y and XY = 228, Y Z = 2004. Angle Y istrisected, and the angle trisectors intersect XZ at P and Q so that X,P,Q, Z lie onXZ in that order. Find the value of (PY + Y Z)(QY +XY ).

Solution: 1370736

The triangle’s area is (228 · 2004)/2 = 228456. All the angles at Y are 30 degrees,so by the sine area formula, the areas of the three small triangles in the diagram areQY · Y Z/4, PY · QY/4, and XY · PY/4, which sum to the area of the triangle. Soexpanding (PY + Y Z)(QY +XY ), we see that it equals

4 · 228456 +XY · Y Z = 6 · 228456 = 1370736.

P

Q

ZY

X

4

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Guts Round

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

1. [5] Find the value of

(6

1

)21 +

(6

2

)22 +

(6

3

)23 +

(6

4

)24 +

(6

5

)25 +

(6

6

)26.

2. [5] If the three points(1, a, b)

(a, 2, b)

(a, b, 3)

are collinear (in 3-space), what is the value of a+ b?

3. [5] If the system of equations

|x+ y| = 99

|x− y| = c

has exactly two real solutions (x, y), find the value of c.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

4. [6] A tree grows in a rather peculiar manner. Lateral cross-sections of the trunk,leaves, branches, twigs, and so forth are circles. The trunk is 1 meter in diameter toa height of 1 meter, at which point it splits into two sections, each with diameter .5meter. These sections are each one meter long, at which point they each split into twosections, each with diameter .25 meter. This continues indefinitely: every section oftree is 1 meter long and splits into two smaller sections, each with half the diameter ofthe previous.

What is the total volume of the tree?

5. [6] Augustin has six 1× 2× π bricks. He stacks them, one on top of another, to forma tower six bricks high. Each brick can be in any orientation so long as it rests flat ontop of the next brick below it (or on the floor). How many distinct heights of towerscan he make?

6. [6] Find the smallest integer n such that√n+ 99−√n < 1.

1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

7. [6] Find the shortest distance from the line 3x+4y = 25 to the circle x2 +y2 = 6x−8y.

8. [6] I have chosen five of the numbers {1, 2, 3, 4, 5, 6, 7}. If I told you what their productwas, that would not be enough information for you to figure out whether their sumwas even or odd. What is their product?

9. [6] A positive integer n is picante if n! ends in the same number of zeroes whetherwritten in base 7 or in base 8. How many of the numbers 1, 2, . . . , 2004 are picante?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

10. [7] Let f(x) = x2 + x4 + x6 + x8 + · · · , for all real x such that the sum converges. Forhow many real numbers x does f(x) = x?

11. [7] Find all numbers n with the following property: there is exactly one set of 8 differentpositive integers whose sum is n.

12. [7] A convex quadrilateral is drawn in the coordinate plane such that each of itsvertices (x, y) satisfies the equations x2 + y2 = 73 and xy = 24. What is the area ofthis quadrilateral?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

13. [7] Find all positive integer solutions (m,n) to the following equation:

m2 = 1! + 2! + · · ·+ n!.

14. [7] If a1 = 1, a2 = 0, and an+1 = an + an+2

2for all n ≥ 1, compute a2004.

15. [7] A regular decagon A0A1A2 · · ·A9 is given in the plane. Compute ∠A0A3A7 indegrees.

2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

16. [8] An n-string is a string of digits formed by writing the numbers 1, 2, . . . , n in someorder (in base ten). For example, one possible 10-string is

35728910461

What is the smallest n > 1 such that there exists a palindromic n-string?

17. [8] Kate has four red socks and four blue socks. If she randomly divides these eightsocks into four pairs, what is the probability that none of the pairs will be mismatched?That is, what is the probability that each pair will consist either of two red socks orof two blue socks?

18. [8] On a spherical planet with diameter 10, 000 km, powerful explosives are placed atthe north and south poles. The explosives are designed to vaporize all matter within5, 000 km of ground zero and leave anything beyond 5, 000 km untouched. After theexplosives are set off, what is the new surface area of the planet, in square kilometers?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

19. [8] The Fibonacci numbers are defined by F1 = F2 = 1, and Fn = Fn−1 + Fn−2 forn ≥ 3. If the number

F2003

F2002

− F2004

F2003

is written as a fraction in lowest terms, what is the numerator?

20. [8] Two positive rational numbers x and y, when written in lowest terms, have theproperty that the sum of their numerators is 9 and the sum of their denominators is10. What is the largest possible value of x+ y?

21. [8] Find all ordered pairs of integers (x, y) such that 3x4y = 2x+y + 22(x+y)−1.

3

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

22. [9] I have written a strictly increasing sequence of six positive integers, such that eachnumber (besides the first) is a multiple of the one before it, and the sum of all sixnumbers is 79. What is the largest number in my sequence?

23. [9] Find the largest integer n such that 3512 − 1 is divisible by 2n.

24. [9] We say a point is contained in a square if it is in its interior or on its boundary.Three unit squares are given in the plane such that there is a point contained in allthree. Furthermore, three points A, B, C, are given, each contained in at least one ofthe squares. Find the maximum area of triangle ABC.

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

25. [9] Suppose x3 − ax2 + bx − 48 is a polynomial with three positive roots p, q, and rsuch that p < q < r. What is the minimum possible value of 1/p+ 2/q + 3/r?

26. [9] How many of the integers 1, 2, . . . , 2004 can be represented as (mn + 1)/(m + n)for positive integers m and n?

27. [9] A regular hexagon has one side along the diameter of a semicircle, and the twoopposite vertices on the semicircle. Find the area of the hexagon if the diameter of thesemicircle is 1.

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

28. [10] Find the value of

(2003

1

)+

(2003

4

)+

(2003

7

)+ · · ·+

(2003

2002

).

29. [10] A regular dodecahedron is projected orthogonally onto a plane, and its image isan n-sided polygon. What is the smallest possible value of n?

30. [10] We have an n-gon, and each of its vertices is labeled with a number from theset {1, . . . , 10}. We know that for any pair of distinct numbers from this set there isat least one side of the polygon whose endpoints have these two numbers. Find thesmallest possible value of n.

4

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

31. [10] P is a point inside triangle ABC, and lines AP,BP,CP intersect the oppositesides BC,CA,AB in points D,E, F , respectively. It is given that ∠APB = 90◦, andthat AC = BC and AB = BD. We also know that BF = 1, and that BC = 999.Find AF .

32. [10] Define the sequence b0, b1, . . . , b59 by

bi =

{1 if i is a multiple of 3

0 otherwise.

Let {ai} be a sequence of elements of {0, 1} such that

bn ≡ an−1 + an + an+1 (mod 2)

for 0 ≤ n ≤ 59 (a0 = a60 and a−1 = a59). Find all possible values of 4a0 + 2a1 + a2.

33. [10] A plane P slices through a cube of volume 1 with a cross-section in the shape ofa regular hexagon. This cube also has an inscribed sphere, whose intersection with Pis a circle. What is the area of the region inside the regular hexagon but outside thecircle?

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

34. [12] Find the number of 20-tuples of integers x1, . . . , x10, y1, . . . , y10 with the followingproperties:

• 1 ≤ xi ≤ 10 and 1 ≤ yi ≤ 10 for each i;

• xi ≤ xi+1 for i = 1, . . . , 9;

• if xi = xi+1, then yi ≤ yi+1.

35. [12] There are eleven positive integers n such that there exists a convex polygon with nsides whose angles, in degrees, are unequal integers that are in arithmetic progression.Find the sum of these values of n.

36. [12] For a string of P ’s and Q’s, the value is defined to be the product of the positionsof the P ’s. For example, the string PPQPQQ has value 1 · 2 · 4 = 8.

Also, a string is called antipalindromic if writing it backwards, then turning all theP ’s into Q’s and vice versa, produces the original string. For example, PPQPQQ isantipalindromic.

There are 21002 antipalindromic strings of length 2004. Find the sum of the reciprocalsof their values.

5

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

37. [15] Simplify∏2004

k=1 sin(2πk/4009).

38. [15] Let S = {p1p2 · · · pn | p1, p2, . . . , pn are distinct primes and p1, . . . , pn < 30}. As-sume 1 is in S. Let a1 be an element of S. We define, for all positive integers n:

an+1 = an/(n+ 1) if an is divisible by n+ 1;

an+1 = (n+ 2)an if an is not divisible by n+ 1.

How many distinct possible values of a1 are there such that aj = a1 for infinitely manyj’s?

39. [15] You want to arrange the numbers 1, 2, 3, . . . , 25 in a sequence with the followingproperty: if n is divisible by m, then the nth number is divisible by the mth number.How many such sequences are there?

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

40. [18] You would like to provide airline service to the 10 cities in the nation of Schizophre-nia, by instituting a certain number of two-way routes between cities. Unfortunately,the government is about to divide Schizophrenia into two warring countries of five citieseach, and you don’t know which cities will be in each new country. All airplane servicebetween the two new countries will be discontinued. However, you want to make surethat you set up your routes so that, for any two cities in the same new country, it willbe possible to get from one city to the other (without leaving the country).

What is the minimum number of routes you must set up to be assured of doing this,no matter how the government divides up the country?

41. [18] A tetrahedron has all its faces triangles with sides 13, 14, 15. What is its volume?

42. [18] S is a set of complex numbers such that if u, v ∈ S, then uv ∈ S and u2 + v2 ∈ S.Suppose that the number N of elements of S with absolute value at most 1 is finite.What is the largest possible value of N?

6

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

43. Write down an integer from 0 to 20 inclusive. This problem will be scored as follows: ifN is the second-largest number from among the responses submitted, then each teamthat submits N gets N points, and everyone else gets zero. (If every team picks thesame number then nobody gets any points.)

44. Shown on your answer sheet is a 20 × 20 grid. Place as many queens as you can sothat each of them attacks at most one other queen. (A queen is a chess piece that canmove any number of squares horizontally, vertically, or diagonally.) It’s not very hardto get 20 queens, so you get no points for that, but you get 5 points for each furtherqueen beyond 20. You can mark the grid by placing a dot in each square that containsa queen.

45. A binary string of length n is a sequence of n digits, each of which is 0 or 1. Thedistance between two binary strings of the same length is the number of positionsin which they disagree; for example, the distance between the strings 01101011 and00101110 is 3 since they differ in the second, sixth, and eighth positions.

Find as many binary strings of length 8 as you can, such that the distance betweenany two of them is at least 3. You get one point per string.

7

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Guts Round — Solutions

1. Find the value of(6

1

)21 +

(6

2

)22 +

(6

3

)23 +

(6

4

)24 +

(6

5

)25 +

(6

6

)26.

Solution: 728

This sum is the binomial expansion of (1+2)6, except that it is missing the first term,(60

)20 = 1. So we get 36 − 1 = 728.

2. If the three points(1, a, b)

(a, 2, b)

(a, b, 3)

are collinear (in 3-space), what is the value of a+ b?

Solution: 4

The first two points are distinct (otherwise we would have a = 1 and a = 2 simulta-neously), and they both lie on the plane z = b, so the whole line is in this plane andb = 3. Reasoning similarly with the last two points gives a = 1, so a+ b = 4.

3. If the system of equations

|x+ y| = 99

|x− y| = c

has exactly two real solutions (x, y), find the value of c.

Solution: 0

If c < 0, there are no solutions. If c > 0 then we have four possible systems of linearequations given by x+ y = ±99, x− y = ±c, giving four solutions (x, y). So we musthave c = 0, and then we do get two solutions (x = y, so they must both equal ±99/2).

4. A tree grows in a rather peculiar manner. Lateral cross-sections of the trunk, leaves,branches, twigs, and so forth are circles. The trunk is 1 meter in diameter to a height of1 meter, at which point it splits into two sections, each with diameter .5 meter. Thesesections are each one meter long, at which point they each split into two sections, eachwith diameter .25 meter. This continues indefinitely: every section of tree is 1 meterlong and splits into two smaller sections, each with half the diameter of the previous.

What is the total volume of the tree?

Solution: π/2

If we count the trunk as level 0, the two sections emerging from it as level 1, and soforth, then the nth level consists of 2n sections each with diameter 1/2n, for a volumeof 2n(π/4 · 2−2n) = (π/4) · 2−n. So the total volume is given by a simple infinite sum,

.25π · (1 + 1/2 + 1/4 + ...) = .25π · 2 = π/2.

1

5. Augustin has six 1 × 2 × π bricks. He stacks them, one on top of another, to form atower six bricks high. Each brick can be in any orientation so long as it rests flat ontop of the next brick below it (or on the floor). How many distinct heights of towerscan he make?

Solution: 28

If there are k bricks which are placed so that they contribute either 1 or 2 height,then the height of these k bricks can be any integer from k to 2k. Furthermore,towers with different values of k cannot have the same height. Thus, for each k thereare k + 1 possible tower heights, and since k is any integer from 0 to 6, there are1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 possible heights.

6. Find the smallest integer n such that√n+ 99−√n < 1.

Solution: 2402

This is equivalent to

√n+ 99 <

√n+ 1

n+ 99 < n+ 1 + 2√n

49 <√n

So the smallest integer n with this property is 492 + 1 = 2402.

7. Find the shortest distance from the line 3x+ 4y = 25 to the circle x2 + y2 = 6x− 8y.

Solution: 7/5

The circle is (x− 3)2 + (y + 4)2 = 52. The center (3,−4) is a distance of

|3 · 3 + 4 · −4− 25|√32 + 42

=32

5

from the line, so we subtract 5 for the radius of the circle and get 7/5.

8. I have chosen five of the numbers {1, 2, 3, 4, 5, 6, 7}. If I told you what their productwas, that would not be enough information for you to figure out whether their sumwas even or odd. What is their product?

Solution: 420

Giving you the product of the five numbers is equivalent to telling you the productof the two numbers I didn’t choose. The only possible products that are achieved bymore than one pair of numbers are 12 ({3, 4} and {2, 6}) and 6 ({1, 6} and {2, 3}). Butin the second case, you at least know that the two unchosen numbers have odd sum(and so the five chosen numbers have odd sum also). Therefore, the first case musthold, and the product of the five chosen numbers is

1 · 2 · 5 · 6 · 7 = 1 · 3 · 4 · 5 · 7 = 420.

9. A positive integer n is picante if n! ends in the same number of zeroes whether writtenin base 7 or in base 8. How many of the numbers 1, 2, . . . , 2004 are picante?

Solution: 4

2

The number of zeroes in base 7 is the total number of factors of 7 in 1 ·2 · · ·n, which is

bn/7c+⌊n/72

⌋+

⌊n/73

⌋+ · · · .

The number of zeroes in base 8 is bac, where

a = (bn/2c+⌊n/22

⌋+

⌊n/23

⌋+ · · · )/3

is one-third the number of factors of 2 in the product n!. Now⌊n/2k

⌋/3 ≥ ⌊

n/7k⌋

for allk, since (n/2k)/3 ≥ n/7k. But n can only be picante if the two sums differ by at most2/3, so in particular this requires (bn/22c)/3 ≤ bn/72c+ 2/3⇔ bn/4c ≤ 3 bn/49c+ 2.This cannot happen for n ≥ 12; checking the remaining few cases by hand, we findn = 1, 2, 3, 7 are picante, for a total of 4 values.

10. Let f(x) = x2 +x4 +x6 +x8 + · · · , for all real x such that the sum converges. For howmany real numbers x does f(x) = x?

Solution: 2

Clearly x = 0 works. Otherwise, we want x = x2/(1− x2), or x2 + x− 1 = 0. Discardthe negative root (since the sum doesn’t converge there), but (−1 +

√5)/2 works, for

a total of 2 values.

11. Find all numbers n with the following property: there is exactly one set of 8 differentpositive integers whose sum is n.

Solution: 36, 37

The sum of 8 different positive integers is at least 1 + 2 + 3 + · · ·+ 8 = 36, so we musthave n ≥ 36. Now n = 36 satisfies the desired property, since in this case we musthave equality — the eight numbers must be 1, . . . , 8. And if n = 37 the eight numbersmust be 1, 2, . . . , 7, 9: if the highest number is 8 then the sum is 36 < n, while if thehighest number is more than 9 the sum is > 1 + 2 + · · · + 7 + 9 = 37 = n. So thehighest number must be 9, and then the remaining numbers must be 1, 2, . . . , 7. Thusn = 37 also has the desired property.

However, no other values of n work: if n > 37 then {1, 2, 3, . . . , 7, n − 28} and{1, 2, . . . , 6, 8, n − 29} are both sets of 8 distinct positive integers whose sum is n.So n = 36, 37 are the only solutions.

12. A convex quadrilateral is drawn in the coordinate plane such that each of its vertices(x, y) satisfies the equations x2 + y2 = 73 and xy = 24. What is the area of thisquadrilateral?

Solution: 110

The vertices all satisfy (x + y)2 = x2 + y2 + 2xy = 73 + 2 · 24 = 121, so x+ y = ±11.Similarly, (x−y)2 = x2+y2−2xy = 73−2·24 = 25, so x−y = ±5. Thus, there are foursolutions: (x, y) = (8, 3), (3, 8), (−3,−8), (−8,−3). All four of these solutions satisfythe original equations. The quadrilateral is therefore a rectangle with side lengths of5√

2 and 11√

2, so its area is 110.

13. Find all positive integer solutions (m,n) to the following equation:

m2 = 1! + 2! + · · ·+ n!.

3

Solution: (1, 1), (3, 3)

A square must end in the digit 0, 1, 4, 5, 6, or 9. If n ≥ 4, then 1! + 2! + · · · + n! endsin the digit 3, so cannot be a square. A simple check for the remaining cases revealsthat the only solutions are (1, 1) and (3, 3).

14. If a1 = 1, a2 = 0, and an+1 = an + an+2

2for all n ≥ 1, compute a2004.

Solution: −21002

By writing out the first few terms, we find that an+4 = −4an. Indeed,

an+4 = 2(an+3 − an+2) = 2(an+2 − 2an+1) = 2(−2an) = −4an.

Then, by induction, we get a4k = (−4)k for all positive integers k, and setting k = 501gives the answer.

15. A regular decagon A0A1A2 · · ·A9 is given in the plane. Compute ∠A0A3A7 in degrees.

Solution: 54◦

Put the decagon in a circle. Each side subtends an arc of 360◦/10 = 36◦. The in-scribed angle ∠A0A3A7 contains 3 segments, namely A7A8, A8A9, A9A0, so the angleis 108◦/2 = 54◦.

16. An n-string is a string of digits formed by writing the numbers 1, 2, . . . , n in some order(in base ten). For example, one possible 10-string is

35728910461

What is the smallest n > 1 such that there exists a palindromic n-string?

Solution: 19

The following is such a string for n = 19:

9|18|7|16|5|14|3|12|1|10|11|2|13|4|15|6|17|8|19

where the vertical bars indicate breaks between the numbers. On the other hand, tosee that n = 19 is the minimum, notice that only one digit can occur an odd numberof times in a palindromic n-string (namely the center digit). If n ≤ 9, then (say)the digits 1, 2 each appear once in any n-string, so we cannot have a palindrome. If10 ≤ n ≤ 18, then 0, 9 each appear once, and we again cannot have a palindrome. So19 is the smallest possible n.

17. Kate has four red socks and four blue socks. If she randomly divides these eight socksinto four pairs, what is the probability that none of the pairs will be mismatched?That is, what is the probability that each pair will consist either of two red socks orof two blue socks?

Solution: 3/35

The number of ways Kate can divide the four red socks into two pairs is(42

)/2 = 3. The

number of ways she can divide the four blue socks into two pairs is also 3. Therefore,the number of ways she can form two pairs of red socks and two pairs of blue socks is

4

3 · 3 = 9. The total number of ways she can divide the eight socks into four pairs is[8!/(2! ·2! ·2! ·2!)]/4! = 105, so the probability that the socks come out paired correctlyis 9/105 = 3/35.

To see why 105 is the correct denominator, we can look at each 2! term as representingthe double counting of pair (ab) and pair (ba), while the 4! term represents the numberof different orders in which we can select the same four pairs. Alternatively, we knowthat there are three ways to select two pairs from four socks. To select three pairs fromsix socks, there are five different choices for the first sock’s partner and then three waysto pair up the remaining four socks, for a total of 5 · 3 = 15 pairings. To select fourpairs from eight socks, there are seven different choices for the first sock’s partner andthen fifteen ways to pair up the remaining six socks, for a total of 7 ·15 = 105 pairings.

18. On a spherical planet with diameter 10, 000 km, powerful explosives are placed atthe north and south poles. The explosives are designed to vaporize all matter within5, 000 km of ground zero and leave anything beyond 5, 000 km untouched. After theexplosives are set off, what is the new surface area of the planet, in square kilometers?

Solution: 100, 000, 000π

The explosives have the same radius as the planet, so the surface area of the “cap”removed is the same as the new surface area revealed in the resulting “dimple.” Thusthe area is preserved by the explosion and remains π · (10, 000)2.

19. The Fibonacci numbers are defined by F1 = F2 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 3.If the number

F2003

F2002

− F2004

F2003

is written as a fraction in lowest terms, what is the numerator?

Solution: 1

Before reducing, the numerator is F 22003 − F2002F2004. We claim F 2

n − Fn−1Fn+1 =(−1)n+1, which will immediately imply that the answer is 1 (no reducing required).This claim is straightforward to prove by induction on n: it holds for n = 2, and if itholds for some n, then

F 2n+1−FnFn+2 = Fn+1(Fn−1+Fn)−Fn(Fn+Fn+1) = Fn+1Fn−1−F 2

n = −(−1)n+1 = (−1)n+2.

20. Two positive rational numbers x and y, when written in lowest terms, have the propertythat the sum of their numerators is 9 and the sum of their denominators is 10. Whatis the largest possible value of x+ y?

Solution: 73/9

For fixed denominators a < b (with sum 10), we maximize the sum of the fractions bygiving the smaller denominator as large a numerator as possible: 8/a + 1/b. Then, ifa ≥ 2, this quantity is at most 8/2 + 1/1 = 5, which is clearly smaller than the sumwe get by setting a = 1, namely 8/1 + 1/9 = 73/9. So this is the answer.

21. Find all ordered pairs of integers (x, y) such that 3x4y = 2x+y + 22(x+y)−1.

Solution: (0, 1), (1, 1), (2, 2)

5

The right side is 2x+y(1 + 2x+y−1). If the second factor is odd, it needs to be a powerof 3, so the only options are x + y = 2 and x + y = 4. This leads to two solutions,namely (1,1) and (2,2). The second factor can also be even, if x + y − 1 = 0. Thenx+ y = 1 and 3x4y = 2 + 2, giving (0, 1) as the only other solution.

22. I have written a strictly increasing sequence of six positive integers, such that eachnumber (besides the first) is a multiple of the one before it, and the sum of all sixnumbers is 79. What is the largest number in my sequence?

Solution: 48

If the fourth number is ≥ 12, then the last three numbers must sum to at least 12 +2 · 12 + 22 · 12 = 84 > 79. This is impossible, so the fourth number must be less than12. Then the only way we can have the required divisibilities among the first fournumbers is if they are 1, 2, 4, 8. So the last two numbers now sum to 79− 15 = 64. Ifwe call these numbers 8a, 8ab (a, b > 1) then we get a(1+ b) = a+ab = 8, which forcesa = 2, b = 3. So the last two numbers are 16, 48.

23. Find the largest integer n such that 3512 − 1 is divisible by 2n.

Solution: 11

Write3512 − 1 = (3256 + 1)(3256 − 1) = (3256 + 1)(3128 + 1)(3128 − 1)

= · · · = (3256 + 1)(3128 + 1) · · · (3 + 1)(3− 1).

Now each factor 32k+ 1, k ≥ 1, is divisible by just one factor of 2, since 32k

+ 1 =(32)2k−1

+1 ≡ 12k−1+1 = 2 (mod 4). Thus we get 8 factors of 2 here, and the remaining

terms (3 + 1)(3− 1) = 8 give us 3 more factors of 2, for a total of 11.

24. We say a point is contained in a square if it is in its interior or on its boundary. Threeunit squares are given in the plane such that there is a point contained in all three.Furthermore, three points A, B, C, are given, each contained in at least one of thesquares. Find the maximum area of triangle ABC.

Solution: 3√

3/2

Let X be a point contained in all three squares. The distance from X to any pointin any of the three squares is at most

√2, the length of the diagonal of the squares.

Therefore, triangle ABC is contained in a circle of radius√

2, so its circumradius is atmost

√2. The triangle with greatest area that satisfies this property is the equilateral

triangle in a circle of radius√

2. (This can be proved, for example, by considering thatthe maximum altitude to any given side is obtained by putting the opposite vertexat the midpoint of its arc, and it follows that all the vertices are equidistant.) Theequilateral triangle is also attainable, since making X the circumcenter and positioningthe squares such that AX, BX, and CX are diagonals (of the three squares) and ABCis equilateral, leads to such a triangle. This triangle has area 3

√3/2, which may be

calculated, for example, using the sine formula for area applied to ABX, ACX, andBCX, to get 3/2(

√2)2 sin 120◦. (See diagram, next page.)

25. Suppose x3 − ax2 + bx− 48 is a polynomial with three positive roots p, q, and r suchthat p < q < r. What is the minimum possible value of 1/p+ 2/q + 3/r?

6

X

C

BA

Solution: 3/2

We know pqr = 48 since the product of the roots of a cubic is the constant term. Now,

1

p+

2

q+

3

r≥ 3 3

√6

pqr=

3

2

by AM-GM, with equality when 1/p = 2/q = 3/r. This occurs when p = 2, q = 4,r = 6, so 3/2 is in fact the minimum possible value.

26. How many of the integers 1, 2, . . . , 2004 can be represented as (mn + 1)/(m + n) forpositive integers m and n?

Solution: 2004

For any positive integer a, we can let m = a2 + a − 1, n = a + 1 to see that everypositive integer has this property, so the answer is 2004.

27. A regular hexagon has one side along the diameter of a semicircle, and the two oppositevertices on the semicircle. Find the area of the hexagon if the diameter of the semicircleis 1.

Solution: 3√

3/26

The midpoint of the side of the hexagon on the diameter is the center of the circle.Draw the segment from this center to a vertex of the hexagon on the circle. Thissegment, whose length is 1/2, is the hypotenuse of a right triangle whose legs havelengths a/2 and a

√3, where a is a side of the hexagon. So 1/4 = a2(1/4 + 3), so

a2 = 1/13. The hexagon consists of 6 equilateral triangles of side length a, so the areaof the hexagon is 3a2

√3/2 = 3

√3/26.

a

7

28. Find the value of(

2003

1

)+

(2003

4

)+

(2003

7

)+ · · ·+

(2003

2002

).

Solution: (22003 − 2)/3

Let ω = −1/2+i√

3/2 be a complex cube root of unity. Then, by the binomial theorem,we have

ω2(ω + 1)2003 =

(2003

0

)ω2 +

(2003

1

)ω3 +

(2003

2

)ω4 + · · ·+

(2003

2003

)ω2005

22003 =

(2003

0

)+

(2003

1

)+

(2003

2

)+ · · ·+

(2003

2003

)

ω−2(ω−1 + 1)2003 =

(2003

0

)ω−2 +

(2003

1

)ω−3 +

(2003

2

)ω−4 + · · ·+

(2003

2003

)ω−2005

If we add these together, then the terms(2003

n

)for n ≡ 1 (mod 3) appear with coefficient

3, while the remaining terms appear with coefficient 1 + ω + ω2 = 0. Thus the desiredsum is just (ω2(ω+1)2003 +22003 +ω−2(ω−1 +1)2003)/3. Simplifying using ω+1 = −ω2

and ω−1 + 1 = −ω gives (−1 + 22003 + −1)/3 = (22003 − 2)/3.

29. A regular dodecahedron is projected orthogonally onto a plane, and its image is ann-sided polygon. What is the smallest possible value of n?

Solution: 6

We can achieve 6 by projecting onto a plane perpendicular to an edge of the dodeca-heron. Indeed, if we imagine viewing the dodecahedron in such a direction, then 4 ofthe faces are projected to line segments (namely, the two faces adjacent to the edgeand the two opposite faces), and of the remaining 8 faces, 4 appear on the front of thedodecahedron and the other 4 are on the back. Thus, the dodecahedron appears asshown.

To see that we cannot do better, note that, by central symmetry, the number of edgesof the projection must be even. So we just need to show that the answer cannot be 4.But if the projection had 4 sides, one of the vertices would give a projection formingan acute angle, which is not possible. So 6 is the answer.

30. We have an n-gon, and each of its vertices is labeled with a number from the set{1, . . . , 10}. We know that for any pair of distinct numbers from this set there is

8

at least one side of the polygon whose endpoints have these two numbers. Find thesmallest possible value of n.

Solution: 50

Each number be paired with each of the 9 other numbers, but each vertex can be usedin at most 2 different pairs, so each number must occur on at least d9/2e = 5 differentvertices. Thus, we need at least 10 · 5 = 50 vertices, so n ≥ 50.

To see that n = 50 is feasible, let the numbers 1, . . . , 10 be the vertices of a completegraph. Then each vertex has degree 9, and there are

(102

)= 45 edges. If we attach extra

copies of the edges 1-2, 3-4, 5-6, 7-8, and 9-10, then every vertex will have degree 10.In particular, the graph has an Eulerian tour, so we can follow this tour, successivelynumbering vertices of the 50-gon according to the vertices of the graph we visit. Then,for each edge of the graph, there will be a corresponding edge of the polygon with thesame two vertex labels on its endpoints. It follows that every pair of distinct numbersoccurs at the endpoints of some edge of the polygon, and so n = 50 is the answer.

31. P is a point inside triangle ABC, and lines AP,BP,CP intersect the opposite sidesBC,CA,AB in points D,E, F , respectively. It is given that ∠APB = 90◦, and thatAC = BC and AB = BD. We also know that BF = 1, and that BC = 999. FindAF .

Solution: 499/500

A P

F

E

D

C

B

Let AC = BC = s, AB = BD = t. Since BP is the altitude in isosceles triangleABD, it bisects angle B. So, the Angle Bisector Theorem in triangle ABC givenAE/EC = AB/BC = t/s. Meanwhile, CD/DB = (s − t)/t. Now Ceva’s theoremgives us

AF

FB=

(AE

EC

)·(CD

DB

)=s− ts

⇒ AB

FB= 1 +

s− ts

=2s− ts

⇒ FB =st

2s− t .Now we know s = 999, but we need to find t given that st/(2s − t) = FB = 1. Sost = 2s− t⇒ t = 2s/(s+ 1), and then

AF = FB · AFFB

= 1 · s− ts

=(s2 − s)/(s+ 1)

s=s− 1

s+ 1=

499

500.

9

32. Define the sequence b0, b1, . . . , b59 by

bi =

{1 if i is a multiple of 3

0 otherwise.

Let {ai} be a sequence of elements of {0, 1} such that

bn ≡ an−1 + an + an+1 (mod 2)

for 0 ≤ n ≤ 59 (a0 = a60 and a−1 = a59). Find all possible values of 4a0 + 2a1 + a2.

Solution: 0, 3, 5, 6

Try the four possible combinations of values for a0 and a1. Since we can write an ≡bn−1−an−2−an−1, these two numbers completely determine the solution {ai} beginningwith them (if there is one). For a0 = a1 = 0, we can check that the sequence beginning0, 0, 0, 0, 1, 1 and repeating every 6 indices is a possible solution for {ai}, so one possiblevalue for 4a0 + 2a1 + a2 is 0. The other three combinations for a0 and a1 similarly leadto valid sequences (produced by repeating the sextuples 0, 1, 1, 1, 0, 1; 1, 0, 1, 1, 1, 0;1, 1, 0, 1, 0, 1, respectively); we thus obtain the values 3, 5, and 6.

33. A plane P slices through a cube of volume 1 with a cross-section in the shape of aregular hexagon. This cube also has an inscribed sphere, whose intersection with Pis a circle. What is the area of the region inside the regular hexagon but outside thecircle?

Solution: (3√

3− π)/4

One can show that the hexagon must have as its vertices the midpoints of six edges ofthe cube, as illustrated; for example, this readily follows from the fact that oppositesides of the hexagons and the medians between them are parallel. We then concludethat the side of the hexagon is

√2/2 (since it cuts off an isosceles triangle of leg 1/2

from each face), so the area is (3/2)(√

2/2)2(√

3) = 3√

3/4. Also, the plane passesthrough the center of the sphere by symmetry, so it cuts out a cross section of radius1/2, whose area (which is contained entirely inside the hexagon) is then π/4. Thesought area is thus (3

√3− π)/4.

34. Find the number of 20-tuples of integers x1, . . . , x10, y1, . . . , y10 with the following prop-erties:

• 1 ≤ xi ≤ 10 and 1 ≤ yi ≤ 10 for each i;

10

• xi ≤ xi+1 for i = 1, . . . , 9;

• if xi = xi+1, then yi ≤ yi+1.

Solution:(10910

)

By setting zi = 10xi + yi, we see that the problem is equivalent to choosing a nonde-creasing sequence of numbers z1, z2, . . . , z10 from the values 11, 12, . . . , 110. Making afurther substitution by setting wi = zi−11+ i, we see that the problem is equivalent tochoosing a strictly increasing sequence of numbers w1, . . . , w10 from among the values1, 2, . . . , 109. There are

(10910

)ways to do this.

35. There are eleven positive integers n such that there exists a convex polygon with nsides whose angles, in degrees, are unequal integers that are in arithmetic progression.Find the sum of these values of n.

Solution: 106

The sum of the angles of an n-gon is (n − 2)180, so the average angle measure is(n − 2)180/n. The common difference in this arithmetic progression is at least 1, sothe difference between the largest and smallest angles is at least n− 1. So the largestangle is at least (n− 1)/2+ (n− 2)180/n. Since the polygon is convex, this quantity isno larger than 179: (n− 1)/2− 360/n ≤ −1, so that 360/n− n/2 ≥ 1/2. Multiplyingby 2n gives 720−n2 ≥ n. So n(n+1) ≤ 720, which forces n ≤ 26. Of course, since thecommon difference is an integer, and the angle measures are integers, (n−2)180/nmustbe an integer or a half integer, so (n− 2)360/n = 360− 720/n is an integer, and then720/n must be an integer. This leaves only n = 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24 aspossibilities. When n is even, (n−2)180/n is not an angle of the polygon, but the meanof the two middle angles. So the common difference is at least 2 when (n − 2)180/nis an integer. For n = 20, the middle angle is 162, so the largest angle is at least162 + 38/2 = 181, since 38 is no larger than the difference between the smallest andlargest angles. For n = 24, the middle angle is 165, again leading to a contradiction.So no solution exists for n = 20, 24. All of the others possess solutions:

n angles3 59, 60, 614 87, 89, 91, 935 106, 107, 108, 109, 1106 115, 117, 119, 121, 123, 1258 128, 130, 132, 134, 136, 138, 140, 1429 136, . . . , 14410 135, 137, 139, . . . , 15312 139, 141, 143, . . . , 16115 149, 150, . . . , 16316 150, 151, . . . , 16518 143, 145, . . . , 177

(These solutions are quite easy to construct.) The desired value is then 3+4+5+6+8 + 9 + 10 + 12 + 15 + 16 + 18 = 106.

11

36. For a string of P ’s and Q’s, the value is defined to be the product of the positions ofthe P ’s. For example, the string PPQPQQ has value 1 · 2 · 4 = 8.

Also, a string is called antipalindromic if writing it backwards, then turning all theP ’s into Q’s and vice versa, produces the original string. For example, PPQPQQ isantipalindromic.

There are 21002 antipalindromic strings of length 2004. Find the sum of the reciprocalsof their values.

Solution: 20051002/2004!

Consider the product(

1

1+

1

2004

)(1

2+

1

2003

)(1

3+

1

2002

)· · ·

(1

1002+

1

1003

).

This product expands to 21002 terms, and each term gives the reciprocal of the valueof a corresponding antipalindromic string of P ’s and Q’s as follows: if we choosethe term 1/n for the nth factor, then our string has a P in position n and Q inposition 2005 − n; if we choose the term 1/(2005 − n), then we get a Q in positionn and P in position 2005 − n. Conversely, each antipalindromic string has its valuerepresented by exactly one of our 21002 terms. So the value of the product is thenumber we are looking for. But when we simplify this product, the nth factor becomes1/n + 1/(2005 − n) = 2005/n(2005 − n). Multiplying these together, we get 1002factors of 2005 in the numerator and each integer from 1 to 2004 exactly once in thedenominator, for a total of 20051002/2004!.

37. Simplify∏2004

k=1 sin(2πk/4009).

Solution:√

400922004

Let ζ = e2πi/4009 so that sin(2πk/4009) = ζk−ζ−k

2iand x4009− 1 =

∏4008k=0 (x− ζk). Hence

1+x+ · · ·+x4008 =∏4008

k=1 (x−ζk). Comparing constant coefficients gives∏4008

k=1 ζk = 1,

setting x = 1 gives∏4008

k=1 (1− ζk) = 4009, and setting x = −1 gives∏4008

k=1 (1 + ζk) = 1.Now, note that sin(2π(4009− k)/4009) = − sin(2πk/4009), so

(2004∏

k=1

sin(2πk/4009)

)2

= (−1)2004

4008∏

k=1

sin(2πk/4009)

=4008∏

k=1

ζk − ζ−k

2i

=1

(2i)4008

4008∏

k=1

ζ2k − 1

ζk

=1

24008

4008∏

k=1

(ζ2k − 1)

=1

24008

4008∏

k=1

(ζk − 1)(ζk + 1)

=4009 · 124008

.

12

However, sin(x) is nonnegative on the interval [0, π], so our product is positive. Hence

it is√

400922004 .

38. Let S = {p1p2 · · · pn | p1, p2, . . . , pn are distinct primes and p1, . . . , pn < 30}. Assume1 is in S. Let a1 be an element of S. We define, for all positive integers n:

an+1 = an/(n+ 1) if an is divisible by n+ 1;

an+1 = (n+ 2)an if an is not divisible by n+ 1.

How many distinct possible values of a1 are there such that aj = a1 for infinitely manyj’s?

Solution: 512

If a1 is odd, then we can see by induction that aj = (j+1)a1 when j is even and aj = a1

when j is odd (using the fact that no even j can divide a1). So we have infinitely manyj’s for which aj = a1.

If a1 > 2 is even, then a2 is odd, since a2 = a1/2, and a1 may have only one factor of2. Now, in general, let p = min({p1, . . . , pn} \ {2}). Suppose 1 < j < p. By induction,we have aj = (j + 1)a1/2 when j is odd, and aj = a1/2 when j is even. So ai 6= a1 forall 1 < j < p. It follows that ap = a1/2p. Then, again using induction, we get for allnonnegative integers k that ap+k = ap if k is even, and ap+k = (p+ k+1)ap if k is odd.Clearly, ap 6= a1 and p + k + 1 6= 2p when k is odd (the left side is odd, and the rightside even). It follows that aj = a1 for no j > 1. Finally, when a1 = 2, we can checkinductively that aj = j + 1 for j odd and aj = 1 for j even.

So our answer is just the number of odd elements in S. There are 9 odd prime numberssmaller than 30, so the answer is 29 = 512.

39. You want to arrange the numbers 1, 2, 3, . . . , 25 in a sequence with the following prop-erty: if n is divisible by m, then the nth number is divisible by the mth number. Howmany such sequences are there?

Solution: 24

Let the rearranged numbers be a1, . . . , a25. The number of pairs (n,m) with n | mmust equal the number of pairs with an | am, but since each pair of the former typeis also of the latter type, the converse must be true as well. Thus, n | m if and onlyif an | am. Now for each n = 1, 2, . . . , 6, the number of values divisible by n uniquelydetermines n, so n = an. Similarly, 7, 8 must either be kept fixed by the rearrangementor interchanged, because they are the only values that divide exactly 2 other numbersin the sequence; since 7 is prime and 8 is not, we conclude they are kept fixed. Then wecan easily check by induction that n = an for all larger composite numbers n ≤ 25 (byusing m = am for all proper factors m of n) and n = 11 (because it is the only primethat divides exactly 1 other number). So we have only the primes n = 13, 17, 19, 23left to rearrange, and it is easily seen that these can be permuted arbitrarily, leaving4! possible orderings altogether.

40. You would like to provide airline service to the 10 cities in the nation of Schizophrenia,by instituting a certain number of two-way routes between cities. Unfortunately, thegovernment is about to divide Schizophrenia into two warring countries of five cities

13

each, and you don’t know which cities will be in each new country. All airplane servicebetween the two new countries will be discontinued. However, you want to make surethat you set up your routes so that, for any two cities in the same new country, it willbe possible to get from one city to the other (without leaving the country).

What is the minimum number of routes you must set up to be assured of doing this,no matter how the government divides up the country?

Solution: 30

Each city C must be directly connected to at least 6 other cities, since otherwisethe government could put C in one country and all its connecting cities in the othercountry, and there would be no way out of C. This means that we have 6 routes foreach of 10 cities, counted twice (since each route has two endpoints) ⇒ 6 · 10/2 = 30routes. On the other hand, this is enough: picture the cities arranged around a circle,and each city connected to its 3 closest neighbors in either direction. Then if C and Dare in the same country but mutually inaccessible, this means that on each arc of thecircle between C and D, there must be (at least) three consecutive cities in the othercountry. Then this second country would have 6 cities, which is impossible. So ourarrangement achieves the goal with 30 routes.

41. A tetrahedron has all its faces triangles with sides 13, 14, 15. What is its volume?

Solution: 42√

55

Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Let AD,BE be altitudes.Then BD = 5, CD = 9. (If you don’t already know this, it can be deduced from thePythagorean Theorem: CD2−BD2 = (CD2+AD2)−(BD2+AD2) = AC2−AB2 = 56,while CD +BD = BC = 14, giving CD −BD = 56/14 = 4, and now solve the linearsystem.) Also, AD =

√AB2 −BD2 = 12. Similar reasoning gives AE = 33/5,

EC = 42/5.

A

H

I

G

F

E

DC B

Now let F be the point on BC such that CF = BD = 5, and let G be on AC suchthat CG = AE = 33/5. Imagine placing face ABC flat on the table, and letting Xbe a point in space with CX = 13, BX = 14. By mentally rotating triangle BCX

14

about line BC, we can see that X lies on the plane perpendicular to BC through F .In particular, this holds if X is the fourth vertex of our tetrahedron ABCX. Similarly,X lies on the plane perpendicular to AC through G. Let the mutual intersection ofthese two planes and plane ABC be H. Then XH is the altitude of the tetrahedron.

To find XH, extend FH to meet AC at I. Then 4CFI ∼ 4CDA, a 3-4-5 triangle,so FI = CF · 4/3 = 20/3, and CI = CF · 5/3 = 25/3. Then IG = CI −CG = 26/15,and HI = IG · 5/4 = 13/6. This leads to HF = FI − HI = 9/2, and finallyXH =

√XF 2 −HF 2 =

√AD2 −HF 2 = 3

√55/2.

Now XABC is a tetrahedron whose base 4ABC has area AD ·BC/2 = 12 ·14/2 = 84,and whose height XH is 3

√55/2, so its volume is (84)(3

√55/2)/3 = 42

√55.

42. S is a set of complex numbers such that if u, v ∈ S, then uv ∈ S and u2 + v2 ∈ S.Suppose that the number N of elements of S with absolute value at most 1 is finite.What is the largest possible value of N?

Solution: 13

First, if S contained some u 6= 0 with absolute value < 1, then (by the first condition)every power of u would be in S, and S would contain infinitely many different numbersof absolute value < 1. This is a contradiction. Now suppose S contains some numberu of absolute value 1 and argument θ. If θ is not an integer multiple of π/6, then uhas some power v whose argument lies strictly between θ + π/3 and θ + π/2. Thenu2 + v2 = u2(1 + (v/u)2) has absolute value between 0 and 1, since (v/u)2 lies onthe unit circle with angle strictly between 2π/3 and π. But u2 + v2 ∈ S, so this is acontradiction.

This shows that the only possible elements of S with absolute value ≤ 1 are 0 and thepoints on the unit circle whose arguments are multiples of π/6, giving N ≤ 1+12 = 13.To show that N = 13 is attainable, we need to show that there exists a possible set Scontaining all these points. Let T be the set of all numbers of the form a+ bω, wherea, b are integers are ω is a complex cube root of 1. Since ω2 = −1 − ω, T is closedunder multiplication and addition. Then, if we let S be the set of numbers u suchthat u2 ∈ T , S has the required properties, and it contains the 13 complex numbersspecified, so we’re in business.

43. Write down an integer from 0 to 20 inclusive. This problem will be scored as follows: ifN is the second-largest number from among the responses submitted, then each teamthat submits N gets N points, and everyone else gets zero. (If every team picks thesame number then nobody gets any points.)

Solution: The only Nash equilibria of this game (where each team plays its bestpossible move given the other teams’ choices) are fairly degenerate: every team butone plays 1, and the remaining team is more likely to choose 2 than any higher number.Of course, we cannot assume perfectly rational play in reality — nor are the utilityfunctions the same, since the goal is to score higher than other teams, not to maximizeone’s own expected number of points. It will be interesting to see what the submissionsare.

44. Shown on your answer sheet is a 20 × 20 grid. Place as many queens as you can sothat each of them attacks at most one other queen. (A queen is a chess piece that can

15

move any number of squares horizontally, vertically, or diagonally.) It’s not very hardto get 20 queens, so you get no points for that, but you get 5 points for each furtherqueen beyond 20. You can mark the grid by placing a dot in each square that containsa queen.

Solution: An elementary argument shows there cannot be more than 26 queens: wecannot have more than 2 in a row or column (or else the middle queen would attackthe other two), so if we had 27 queens, there would be at least 7 columns with morethan one queen and thus at most 13 queens that are alone in their respective columns.Similarly, there would be at most 13 queens that are alone in their respective rows.This leaves 27 − 13 − 13 = 1 queen who is not alone in her row or column, and shetherefore attacks two other queens, contradiction.

Of course, this is not a very strong argument since it makes no use of the diagonals.The best possible number of queens is not known to us; the following constructiongives 23:

45. A binary string of length n is a sequence of n digits, each of which is 0 or 1. Thedistance between two binary strings of the same length is the number of positionsin which they disagree; for example, the distance between the strings 01101011 and00101110 is 3 since they differ in the second, sixth, and eighth positions.

Find as many binary strings of length 8 as you can, such that the distance betweenany two of them is at least 3. You get one point per string.

Solution: The maximum possible number of such strings is 20. An example of a set

16

attaining this bound is

00000000 00110101

11001010 10011110

11100001 01101011

11010100 01100110

10111001 10010011

01111100 11001101

00111010 10101100

01010111 11110010

00001111 01011001

10100111 11111111

This example is taken from page 57 of F. J. MacWilliams and N. J. A. Sloane, TheTheory of Error Correcting Codes (New York: Elsevier Publishing, 1977). The proofthat 20 is the best possible is elementary but too long to reproduce here; see pages537–541 of MacWilliams and Sloane for details.

In general, a set of M strings of length n such that any two have a distance of at leastd is called an (n,M, d)-code. These objects are of basic importance in coding theory,which studies how to transmit information through a channel with a known error rate.For example, since the code given above has minimum distance 3, I can transmit toyou a message consisting of strings in this code, and even if there is a possible errorrate of one digit in each string, you will still be able to determine the intended messageuniquely.

17

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Team Round

A Build-It-Yourself Table [150 points]

An infinite table of nonnegative integers is constructed as follows: in the top row, somenumber is 1 and all other numbers are 0’s; in each subsequent row, every number is the sumof some two of the three closest numbers in the preceding row. An example of such a tableis shown below.

· · · 0 0 0 0 1 0 0 0 0 · · ·· · · 0 0 0 0 1 1 0 0 0 · · ·· · · 0 0 0 1 1 2 1 0 0 · · ·· · · 0 0 1 1 3 3 2 0 0 · · ·· · · 0 1 2 4 4 6 3 2 0 · · ·. .

. ......

......

......

......

.... . .

The top row (with the one 1) is called row 0; the next row is row 1; the next row is row2, and so forth.

Note that the following problems require you to prove the statements for every table thatcan be constructed by the process described above, not just for the example shown.

1. [10] Show that any number in row n (for n > 0) is at most 2n−1.

2. [20] What is the earliest row in which the number 2004 may appear?

3. [35] Let

S(n, r) =

(n− 1

r − 1

)+

(n− 1

r

)+

(n− 1

r + 1

)+ · · ·+

(n− 1

n− 1

)

for all n, r > 0, and in particular S(n, r) = 0 if r > n > 0. Prove that the number inrow n of the table, r columns to the left of the 1 in the top row, is at most S(n, r).(Hint: First prove that S(n− 1, r − 1) + S(n− 1, r) = S(n, r).)

4. [25] Show that the sum of all the numbers in row n is at most (n+ 2)2n−1.

A pair of successive numbers in the same row is called a switch pair if one number in thepair is even and the other is odd.

5. [15] Prove that the number of switch pairs in row n is at most twice the number ofodd numbers in row n.

6. [20] Prove that the number of odd numbers in row n is at most twice the number ofswitch pairs in row n− 1.

7. [25] Prove that the number of switch pairs in row n is at most twice the number ofswitch pairs in row n− 1.

1

Written In The Stars [125 points]

Suppose S is a finite set with a binary operation ? — that is, for any elements a, b of S,there is defined an element a ? b of S. It is given that (a ? b) ? (a ? b) = b ? a for all a, b ∈ S.

8. [20] Prove that a ? b = b ? a for all a, b ∈ S.

Let T be the set of elements of the form a ? a for a ∈ S.

9. [15] If b is any element of T , prove that b ? b = b.

Now suppose further that (a ? b) ? c = a ? (b ? c) for all a, b, c ∈ S. (Thus we can write anexpression like a ? b ? c ? d without ambiguity.)

10. [25] Let a be an element of T . Let the image of a be the set of all elements of T thatcan be represented as a ? b for some b ∈ T . Prove that if c is in the image of a, thena ? c = c.

11. [40] Prove that there exists an element a ∈ T such that the equation a ? b = a holdsfor all b ∈ T .

12. [25] Prove that there exists an element a ∈ S such that the equation a ? b = a holdsfor all b ∈ S.

Sigma City [125 points]

13. [25] Let n be a positive odd integer. Prove that

blog2 nc+ blog2(n/3)c+ blog2(n/5)c+ blog2(n/7)c+ · · ·+ blog2(n/n)c = (n− 1)/2.

Let σ(n) denote the sum of the (positive) divisors of n, including 1 and n itself.

14. [30] Prove thatσ(1) + σ(2) + σ(3) + · · ·+ σ(n) ≤ n2

for every positive integer n.

15. [30] Prove thatσ(1)

1+σ(2)

2+σ(3)

3+ · · ·+ σ(n)

n≤ 2n

for every positive integer n.

16. [40] Now suppose again that n is odd. Prove that

σ(1)blog2 nc+ σ(3)blog2(n/3)c+ σ(5)blog2(n/5)c+ · · ·+ σ(n)blog2(n/n)c < n2/8.

2

Harvard-MIT Mathematics TournamentFebruary 28, 2004

Team Round — Solutions

A Build-It-Yourself Table

An infinite table of nonnegative integers is constructed as follows: in the top row, somenumber is 1 and all other numbers are 0’s; in each subsequent row, every number is the sumof some two of the three closest numbers in the preceding row. An example of such a tableis shown below.

· · · 0 0 0 0 1 0 0 0 0 · · ·· · · 0 0 0 0 1 1 0 0 0 · · ·· · · 0 0 0 1 1 2 1 0 0 · · ·· · · 0 0 1 1 3 3 2 0 0 · · ·· · · 0 1 2 4 4 6 3 2 0 · · ·. .

. ......

......

......

......

.... . .

The top row (with the one 1) is called row 0; the next row is row 1; the next row is row2, and so forth.

Note that the following problems require you to prove the statements for every table thatcan be constructed by the process described above, not just for the example shown.

1. Show that any number in row n (for n > 0) is at most 2n−1.

Solution: We use induction on n. It is clear that any number in row 1 is at most1 = 20. Now, if every number in row n is at most 2n−1, then every number in row n+1is the sum of two numbers in row n and so is at most 2n−1 + 2n−1 = 2n. This gives theinduction step, and the result follows.

2. What is the earliest row in which the number 2004 may appear?

Solution:

· · · 0 0 1 0 0 · · ·· · · 0 0 1 1 0 · · ·· · · 0 1 2 2 0 · · ·· · · 0 3 4 4 0 · · ·· · · 0 7 8 8 0 · · ·· · · 0 15 16 16 0 · · ·· · · 0 31 31 32 0 · · ·· · · 0 62 63 63 0 · · ·· · · 0 125 125 126 0 · · ·· · · 0 250 251 251 0 · · ·· · · 0 501 501 502 0 · · ·· · · 0 1002 1002 1003 0 · · ·· · · 0 2004 2004 2005 0 · · ·

By the previous problem, it cannot appear before row 12. By starting off the table asshown above, we see that row 12 is possible, so this is the answer.

1

3. Let

S(n, r) =

(n− 1

r − 1

)+

(n− 1

r

)+

(n− 1

r + 1

)+ · · ·+

(n− 1

n− 1

)

for all n, r > 0, and in particular S(n, r) = 0 if r > n > 0. Prove that the number inrow n of the table, r columns to the left of the 1 in the top row, is at most S(n, r).(Hint: First prove that S(n− 1, r − 1) + S(n− 1, r) = S(n, r).)

Solution: First, we prove the statement in the hint: adding the ith term of the sumfor S(n− 1, r − 1) to the ith term for S(n− 1, r), for each i, we get that S(n− 1, r −1) + S(n− 1, r) equals

((n− 2

r − 2

)+

(n− 2

r − 1

))+

((n− 2

r − 1

)+

(n− 2

r

))+ · · ·

· · ·+((

n− 2

n− 3

)+

(n− 2

n− 2

))+

(n− 2

n− 2

)

=

(n− 1

r − 1

)+

(n− 1

r

)+ · · ·+

(n− 1

n− 1

)= S(n, r).

Now we can prove the main statement by induction on n. The base case n = 1 is clear.If the statement holds for n− 1, then first suppose r > 1. Then the number in row n,r columns to the left, is the sum of two of the three numbers above it, which, by theinduction hypothesis, are at most S(n−1, r−1), S(n−1, r), S(n−1, r+1) respectively.Since the first two of these are greater than the last (because the summation formulagives S(n− 1, r − 1) = S(n− 1, r) +

(n−1r−2

)and S(n− 1, r) = S(n− 1, r + 1) +

(n−1r−1

)),

we have an upper bound of S(n − 1, r − 1) + S(n − 1, r) = S(n, r) by the above. Sothe result follows by induction. Finally, in the case r = 1, the quantity in question isjust 2n−1, and the result holds by Problem 1.

4. Show that the sum of all the numbers in row n is at most (n+ 2)2n−1.

Solution: The previous problem gives an upper bound on the number located rcolumns to the left of the initial 1; adding over all r = 1, 2, . . . , n gives

n−1∑s=0

(s+ 1)

(n− 1

s

)

since the term(

n−1s

)occurs for the s+1 values r = 1, 2, . . . , s+1. But this sum equals

(n + 1)2n−2. For example, add the sum to itself and reverse the terms of the secondsum to get

n−1∑s=0

(s+ 1)

(n− 1

s

)+

n−1∑s=0

([n− 1− s] + 1)

(n− 1

n− 1− s)

=n−1∑s=0

(n+ 1)

(n− 1

s

)= (n+ 1)

n−1∑s=0

(n− 1

s

)= (n+ 1)2n−1,

and our original sum is half of this.

So the sum of the terms in row n to the left of the central column is at most (n+1)2n−2.Similarly, the sum of the terms to the right of the central column is at most (n+1)2n−2.Adding these together, plus the upper bound of 2n−1 for the central number (Problem1), gives our upper bound of (n+ 2)2n−1 for the sum of all the numbers in the row.

2

A pair of successive numbers in the same row is called a switch pair if one number in thepair is even and the other is odd.

5. Prove that the number of switch pairs in row n is at most twice the number of oddnumbers in row n.

Solution: Each switch pair contains an odd number, and each odd number can belongto at most two switch pairs (since it has only two neighbors).

6. Prove that the number of odd numbers in row n is at most twice the number of switchpairs in row n− 1.

Solution: Each odd number in row n is the sum of two of the three numbers above itin row n− 1; these three numbers cannot all have the same parity (or else any sum oftwo of them would be even), so somewhere among them is a switch pair. Since eachswitch pair in row n − 1 can contribute to at most two odd numbers in row n in thismanner (namely, the two numbers immediately below the members of the pair), theresult follows.

7. Prove that the number of switch pairs in row n is at most twice the number of switchpairs in row n− 1.

Solution: If we go sufficiently far to the left or right in row n, we get to zeroes.Therefore, row n consists of a finite number of “odd blocks” of the form

EOOO · · ·OE

(where E represents an even number and O an odd number), which are separated byeven numbers, except that one even number may simultaneously be an endpoint of twoodd blocks. Each odd block contributes two switch pairs to row n, so it is enough toshow that each odd block has a switch pair somewhere above it in row n− 1. But theodd block consists of at least one O between two E’s, making for at least three terms.If there were no switch pairs above the block, then in particular, the first three termsimmediately above it would be all odd or all even, and then the second term in ourblock would have to be even, contradicting the assumption that it was O. This provesthe result.

Written In The Stars

Suppose S is a finite set with a binary operation ? — that is, for any elements a, b of S,there is defined an element a ? b of S. It is given that (a ? b) ? (a ? b) = b ? a for all a, b ∈ S.

8. Prove that a ? b = b ? a for all a, b ∈ S.

Solution: We have

a ? b = (b ? a) ? (b ? a)

= ([a ? b] ? [a ? b]) ? ([a ? b] ? [a ? b])

= [a ? b] ? [a ? b]

= b ? a.

3

Let T be the set of elements of the form a ? a for a ∈ S.

9. If b is any element of T , prove that b ? b = b.

Solution: If b ∈ T , then b = a ? a for some a, so b ? b = (a ? a) ? (a ? a) = a ? a (bythe given property) = b.

Now suppose further that (a ? b) ? c = a ? (b ? c) for all a, b, c ∈ S. (Thus we can write anexpression like a ? b ? c ? d without ambiguity.)

10. Let a be an element of T . Let the image of a be the set of all elements of T that can berepresented as a ? b for some b ∈ T . Prove that if c is in the image of a, then a ? c = c.

Solution: Write c = a ? b, and then a ? c = a ? a ? b = a ? b (since a ? a = a) = c.

11. Prove that there exists an element a ∈ T such that the equation a ? b = a holds for allb ∈ T .

Solution: Choose a whose image contains as few elements as possible — we knowwe can do this, since T , being a subset of S, is finite. We claim that this a works.Indeed, suppose c is in the image of a. Then, for any d in the image of c, a ? (c ? d) =(a ? c) ? d = c ? d = d, so d is also in the image of a. So the image of c is contained inthe image of a. But a was chosen to have image as small as possible, so the two imagesmust coincide. In particular, a ? a = a is in the image of c. So

a = c ? a = a ? c = c.

This argument shows that a is the only element of the image of a, which gives whatwe wanted.

Alternative Solution: This can also be solved without using Problem 10: Theproduct of any two elements of T is also in T , since commutativity and associativitygive (b ? b) ? (c ? c) = (b ? c) ? (b ? c) for b, c ∈ S. Then let a1, a2, . . . , an be all theelements of T , and put a = a1 ?a2 ? · · ·?an; this value does not depend on the orderingof the elements. If b ∈ T , then a = c ? b, where c is the ?-product of all elements of Tdifferent from b, and consequently a ? b = (c ? b) ? b = c ? (b ? b) = c ? b = a.

12. Prove that there exists an element a ∈ S such that the equation a ? b = a holds for allb ∈ S.

Solution: The same a as in the previous problem will do the trick. Indeed, for anyb ∈ S, we have

a ? b = (b ? a) ? (b ? a) = (a ? a) ? (b ? b)

(we have used commutativity and associativity). But a ? a = a, and b ? b ∈ T , so thisexpression equals a ? (b ? b) = a, as required.

Sigma City

13. Let n be a positive odd integer. Prove that

blog2 nc+ blog2(n/3)c+ blog2(n/5)c+ blog2(n/7)c+ · · ·+ blog2(n/n)c = (n− 1)/2.

4

Solution: Note that blog2 kc is the cardinality of the set {2, 4, 8, . . . , 2blog2 kc}, i.e.,the number of powers of 2 that are even and are at most k. Then blog2(n/k)c is thenumber of even powers of 2 that are at most n/k, or equivalently (multiplying each suchnumber by k) the number of positive even numbers ≤ n whose greatest odd divisor isk. Summing over all odd k, we get the number of even numbers ≤ n, which is just(n− 1)/2.

Let σ(n) denote the sum of the (positive) divisors of n, including 1 and n itself.

14. Prove thatσ(1) + σ(2) + σ(3) + · · ·+ σ(n) ≤ n2

for every positive integer n.

Solution: The ith term on the left is the sum of all d dividing i. If we write this sumout explicitly, then each term d = 1, 2, . . . , n appears bn/dc times — once for eachmultiple of d that is ≤ n. Thus, the sum equals

bn/1c+ 2 bn/2c+ 3 bn/3c+ · · ·+ n bn/nc ≤ n/1 + 2n/2 + 3n/3 + · · ·+ n/n

= n+ n+ · · ·+ n

= n2.

15. Prove thatσ(1)

1+σ(2)

2+σ(3)

3+ · · ·+ σ(n)

n≤ 2n

for every positive integer n.

Solution: This is similar to the previous solution. If d is a divisor of i, then so is i/d,and (i/d)/i = 1/d. Summing over all d, we see that σ(i)/i is the sum of the reciprocalsof the divisors of i, for each positive integer i. So, summing over all i from 1 to n, weget the value 1/d appearing bn/dc times, once for each multiple of d that is at most n.In particular, the sum is

1

1

⌊n1

⌋+

1

2

⌊n2

⌋+

1

3

⌊n3

⌋+ · · ·+ 1

n

⌊nn

⌋<

n

12+n

22+ · · ·+ n

n2.

So now all we need is 1/12 + 1/22 + · · · + 1/n2 < 2. This can be obtained from theclassic formula 1/12 + 1/22 + · · · = π2/6, or from the more elementary estimate

1/22 + 1/32 + · · ·+ 1/n2 < 1/(1 · 2) + 1/(2 · 3) + · · ·+ 1/((n− 1) · n)

= (1/1− 1/2) + (1/2− 1/3) + · · ·+ (1/(n− 1)− 1/n

)

= 1− 1/n

< 1.

16. Now suppose again that n is odd. Prove that

σ(1) blog2 nc+ σ(3) blog2(n/3)c+ σ(5) blog2(n/5)c+ · · ·+ σ(n) blog2(n/n)c < n2/8.

Solution: The term σ(i) blog2(n/i)c is the sum of the divisors of i times the numberof even numbers ≤ n whose greatest odd divisor is i. Thus, summing over all odd i,

5

we get the sum of d over all pairs (d, j), where j < n is even and d is an odd divisorof j. Each odd number d then appears bn/2dc times, since this is the number of evennumbers < n that have d as a divisor. So the sum equals

bn/2c+ 3 bn/6c+ 5 bn/10c+ · · ·+ n bn/2nc≤ (n− 1)/2 + 3(n− 1)/6 + · · ·+m(n− 1)/2m,

where m is the greatest odd integer less than n/2. (We can ignore the terms d bn/2dcfor d > m because these floors are zero.) This expression equals

(n− 1)/2 + (n− 1)/2 + · · ·+ (n− 1)/2 = (n− 1)(m+ 1)/4 ≤ (n− 1)(n+ 1)/8,

which is less than n2/8, as required.

6

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Individual Round: Algebra Subject Test

1. How many real numbers x are solutions to the following equation?

|x− 1| = |x− 2|+ |x− 3|

2. How many real numbers x are solutions to the following equation?

2003x + 2004x = 2005x

3. Let x, y, and z be distinct real numbers that sum to 0. Find the maximum possiblevalue of

xy + yz + zx

x2 + y2 + z2.

4. If a, b, c > 0, what is the smallest possible value of⌊

a+bc

⌋+

⌊b+ca

⌋+

⌊c+a

b

⌋? (Note that

bxc denotes the greatest integer less than or equal to x.)

5. Ten positive integers are arranged around a circle. Each number is one more than thegreatest common divisor of its two neighbors. What is the sum of the ten numbers?

6. Find the sum of the x-coordinates of the distinct points of intersection of the planecurves given by x2 = x+ y + 4 and y2 = y − 15x+ 36.

7. Let x be a positive real number. Find the maximum possible value of

x2 + 2−√x4 + 4

x.

8. Compute∞∑

n=0

n

n4 + n2 + 1.

9. The number 27,000,001 has exactly four prime factors. Find their sum.

10. Find the sum of the absolute values of the roots of x4 − 4x3 − 4x2 + 16x− 8 = 0.

1

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Individual Round: Algebra Subject Test — Solutions

1. How many real numbers x are solutions to the following equation?

|x− 1| = |x− 2|+ |x− 3|

Solution: 2

If x < 1, the equation becomes (1 − x) = (2− x) + (3− x) which simplifies to x = 4,contradicting the assumption x < 1. If 1 ≤ x ≤ 2, we get (x− 1) = (2− x) + (3− x),which gives x = 2. If 2 ≤ x ≤ 3, we get (x− 1) = (x− 2) + (3− x), which again givesx = 2. If x ≥ 3, we get (x− 1) = (x− 2) + (x− 3), or x = 4. So 2 and 4 are the onlysolutions, and the answer is 2.

2. How many real numbers x are solutions to the following equation?

2003x + 2004x = 2005x

Solution: 1

Rewrite the equation as (2003/2005)x + (2004/2005)x = 1. The left side is strictlydecreasing in x, so there cannot be more than one solution. On the other hand, theleft side equals 2 > 1 when x = 0 and goes to 0 when x is very large, so it must equal1 somewhere in between. Therefore there is one solution.

3. Let x, y, and z be distinct real numbers that sum to 0. Find the maximum possiblevalue of

xy + yz + zx

x2 + y2 + z2.

Solution: −1/2

Note that 0 = (x+ y+ z)2 = x2 + y2 + z2 + 2xy+ 2yz+ 2zx. Rearranging, we get thatxy+ yz + zx = −1

2(x2 + y2 + z2), so that in fact the quantity is always equal to −1/2.

4. If a, b, c > 0, what is the smallest possible value of⌊

a+bc

⌋+

⌊b+ca

⌋+

⌊c+a

b

⌋? (Note that

bxc denotes the greatest integer less than or equal to x.)

Solution: 4

Since bxc > x− 1 for all x, we have that

⌊a+ b

c

⌋+

⌊b+ c

a

⌋+

⌊c+ a

b

⌋>

a+ b

c+b+ c

a+c+ a

b− 3

=

(a

b+b

a

)+

(b

c+c

b

)+

( ca

+a

c

)− 3.

But by the AM-GM inequality, each of the first three terms in the last line is at least2. Therefore, the lefthand side is greater than 2 + 2 + 2− 3 = 3. Since it is an integer,the smallest value it can be is 4. This is in fact attainable by letting (a, b, c) = (6, 8, 9).

1

5. Ten positive integers are arranged around a circle. Each number is one more than thegreatest common divisor of its two neighbors. What is the sum of the ten numbers?

Solution: 28

First note that all the integers must be at least 2, because the greatest common divisorof any two positive integers is at least 1. Let n be the largest integer in the circle.The greatest common divisor of its two neighbors is n− 1. Therefore, each of the twoneighbors is at least n − 1 but at most n, so since n − 1 - n for n − 1 ≥ 2, they mustboth be equal to n− 1. Let m be one of the numbers on the other side of n− 1 fromn. Then gcd(n,m) = n − 2. Since n − 2 ≥ 0, n − 2 | n only for n = 3 or 4. If n = 3,each number must be 2 or 3, and it is easy to check that there is no solution. If n = 4,then it is again not hard to find that there is a unique solution up to rotation, namely4322343223. The only possible sum is therefore 28.

6. Find the sum of the x-coordinates of the distinct points of intersection of the planecurves given by x2 = x+ y + 4 and y2 = y − 15x+ 36.

Solution: 0

Substituting y = x2 − x− 4 into the second equation yields

0 = (x2 − x− 4)2 − (x2 − x− 4) + 15x− 36

= x4 − 2x3 − 7x2 + 8x+ 16− x2 + x+ 4 + 15x− 36

= x4 − 2x3 − 8x2 + 24x− 16

= (x− 2)(x3 − 8x+ 8) = (x− 2)2(x2 + 2x− 4).

This quartic has three distinct real roots at x = 2,−1 ± √5. Each of these yields adistinct point of intersection, so the answer is their sum, 0.

-4 -2 2 4

-10

-5

5

10

7. Let x be a positive real number. Find the maximum possible value of

x2 + 2−√x4 + 4

x.

2

Solution: 2√

2− 2

Rationalizing the numerator, we get

x2 + 2−√x4 + 4

x· x

2 + 2 +√x4 + 4

x2 + 2 +√x4 + 4

=(x2 + 2)2 − (x4 + 4)

x(x2 + 2 +√x4 + 4)

=4x2

x(x2 + 2 +√x4 + 4)

=4

1x(x2 + 2 +

√x4 + 4)

=4

x+ 2x

+√x2 + 4

x2

.

Since we wish to maximize this quantity, we wish to minimize the denominator. ByAM-GM, x + 2

x≥ 2√

2 and x2 + 4x2 ≥ 4, so that the denominator is at least 2

√2 + 2.

Therefore,x2 + 2−√x4 + 4

x≤ 4

2√

2 + 2= 2√

2− 2,

with equality when x =√

2.

8. Compute∞∑

n=0

n

n4 + n2 + 1.

Solution: 1/2

Note that

n4 + n2 + 1 = (n4 + 2n2 + 1)− n2 = (n2 + 1)2 − n2 = (n2 + n+ 1)(n2 − n+ 1).

Decomposing into partial fractions, we find that

n

n4 + n2 + 1=

1

2

(1

n2 − n+ 1− 1

n2 + n+ 1

).

Now, note that if f(n) = 1n2−n+1

, then f(n + 1) = 1(n+1)2−(n+1)+1

= 1n2+n+1

. It followsthat

∞∑n=0

n

n4 + n2 + 1=

1

2

((f(0)− f(1)) + (f(1)− f(2)) + (f(2)− f(3)) + · · ·

).

Since f(n) tends towards 0 as n gets large, this sum telescopes to f(0)/2 = 1/2.

9. The number 27,000,001 has exactly four prime factors. Find their sum.

Solution: 652

3

First, we factor

27x6 + 1 = (3x2)3 + 1

= (3x2 + 1)(9x4 − 3x2 + 1)

= (3x2 + 1)((9x4 + 6x2 + 1)− 9x2)

= (3x2 + 1)((3x2 + 1)2 − (3x)2)

= (3x2 + 1)(3x2 + 3x+ 1)(3x2 − 3x+ 1).

Letting x = 10, we get that 27000001 = 301 · 331 · 271. A quick check shows that301 = 7 · 43, so that 27000001 = 7 · 43 · 271 · 331. Each factor here is prime, and theirsum is 652.

10. Find the sum of the absolute values of the roots of x4 − 4x3 − 4x2 + 16x− 8 = 0.

Solution: 2 + 2√

2 + 2√

3

x4 − 4x3 − 4x2 + 16x− 8 = (x4 − 4x3 + 4x2)− (8x2 − 16x+ 8)

= x2(x− 2)2 − 8(x− 1)2

= (x2 − 2x)2 − (2√

2x− 2√

2)2

= (x2 − (2 + 2√

2)x+ 2√

2)(x2 − (2− 2√

2)x− 2√

2).

But noting that (1 +√

2)2 = 3 + 2√

2 and completing the square,

x2 − (2 + 2√

2)x+ 2√

2 = x2 − (2 + 2√

2)x+ 3 + 2√

2− 3

= (x− (1 +√

2))2 − (√

3)2

= (x− 1−√

2 +√

3)(x− 1−√

2−√

3).

Likewise,

x2 − (2− 2√

2)x− 2√

2 = (x− 1 +√

2 +√

3)(x− 1 +√

2−√

3),

so the roots of the quartic are 1 ± √2 ± √3. Only one of these is negative, namely1−√2−√3, so the sum of the absolute values of the roots is

(1 +√

2 +√

3) + (1 +√

2−√

3) + (1−√

2 +√

3)− (1−√

2−√

3) = 2 + 2√

2 + 2√

3.

4

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Individual Round: Calculus Subject Test

1. Let f(x) = x3 + ax+ b, with a 6= b, and suppose the tangent lines to the graph of f atx = a and x = b are parallel. Find f(1).

2. A plane curve is parameterized by x(t) =∫ ∞t

cos uudu and y(t) =

∫ ∞t

sin uudu for 1 ≤ t ≤

2. What is the length of the curve?

3. Let f : R→ R be a continuous function with∫ 10 f(x)f ′(x)dx = 0 and

∫ 10 f(x)2f ′(x)dx =

18. What is∫ 10 f(x)4f ′(x)dx?

4. Let f : R→ R be a smooth function such that f ′(x)2 = f(x)f ′′(x) for all x. Supposef(0) = 1 and f (4)(0) = 9. Find all possible values of f ′(0).

5. Calculatelim

x→0+

(xxx − xx

).

6. The graph of r = 2 + cos 2θ and its reflection over the line y = x bound five regions inthe plane. Find the area of the region containing the origin.

7. Two ants, one starting at (−1, 1), the other at (1, 1), walk to the right along theparabola y = x2 such that their midpoint moves along the line y = 1 with constantspeed 1. When the left ant first hits the line y = 1

2, what is its speed?

8. If f is a continuous real function such that f(x − 1) + f(x + 1) ≥ x + f(x) for all x,what is the minimum possible value of

∫ 20051 f(x)dx?

9. Compute∞∑

k=0

4

(4k)!.

10. Let f : R→ R be a smooth function such that f ′(x) = f(1−x) for all x and f(0) = 1.Find f(1).

1

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Individual Round: Calculus Subject Test — Solutions

1. Let f(x) = x3 + ax+ b, with a 6= b, and suppose the tangent lines to the graph of f atx = a and x = b are parallel. Find f(1).

Solution: 1

Since f ′(x) = 3x2 + a, we must have 3a2 + a = 3b2 + a. Then a2 = b2, and since a 6= b,a = −b. Thus f(1) = 1 + a+ b = 1.

2. A plane curve is parameterized by x(t) =∫ ∞t

cos uudu and y(t) =

∫ ∞t

sin uudu for 1 ≤ t ≤

2. What is the length of the curve?

Solution: ln 2

By the Second Fundamental Theorem of Calculus, dxdt

= − cos tt

and dydt

= − sin tt

. There-fore, the length of the curve is

∫ 2

1

√√√√(dx

dt

)2

+

(dy

dt

)2

dt =∫ 2

1

√cos2 t

t2+

sin2 t

t2dt =

∫ 2

1

1

tdt = [ln t]21 = ln 2.

3. Let f : R→ R be a continuous function with∫ 10 f(x)f ′(x)dx = 0 and

∫ 10 f(x)2f ′(x)dx =

18. What is∫ 10 f(x)4f ′(x)dx?

Solution: 486/5

0 =∫ 1

0f(x)f ′(x)dx =

∫ f(1)

f(0)u du =

1

2(f(1)2 − f(0)2), and

18 =∫ 1

0f(x)2f ′(x)dx =

∫ f(1)

f(0)u2du =

1

3(f(1)3 − f(0)3).

The first equation implies f(0) = ±f(1). The second equation shows that f(0) 6= f(1),and in fact 54 = f(1)3 − f(0)3 = 2f(1)3, so f(1) = 3 and f(0) = −3. Then

∫ 1

0f(x)4f ′(x)dx =

∫ f(1)

f(0)u4du =

1

5(f(1)5 − f(0)5) =

1

5(243 + 243) =

486

5.

4. Let f : R→ R be a smooth function such that f ′(x)2 = f(x)f ′′(x) for all x. Supposef(0) = 1 and f (4)(0) = 9. Find all possible values of f ′(0).

Solution: ±√3

Let f ′(0) = a. Then the equation gives f ′′(0) = a2. Differentiating the given equationgives

2f ′(x)f ′′(x) = f(x)f ′′′(x) + f ′(x)f ′′(x),

or f ′(x)f ′′(x) = f(x)f ′′′(x). Differentiating once more gives

f ′(x)f ′′′(x) + f ′′(x)2 = f(x)f (4)(x) + f ′(x)f ′′′(x)

1

or f ′′(x)2 = f(x)f (4)(x), giving 9 = f (4)(0) = a4. Thus a = ±√3. These are indeed

both attainable by f(x) = e±x√

3.

Alternative Solution: Rewrite the given equation as f ′′(x)f ′(x)

= f ′(x)f(x)

. Integrating both

sides gives ln f ′(x) = ln f(x) + C1, and exponentiating gives f ′(x) = Cf(x). This hassolution f(x) = AeCx for constants A and C. Since f(0) = 1, A = 1, and differentiatingwe find that C4 = f (4)(0) = 9, yielding f ′(0) = C = ±√3.

5. Calculatelim

x→0+

(xxx − xx

).

Solution: −1

We first calculate limx→0+ xx: it is just exp(limx→0+ x ln x). But

limx→0+

x ln x = limx→0+

lnx

1/x= lim

x→0+

1/x

−1/x2= lim

x→0+−x = 0

by L’Hôpital’s Rule. Therefore limx→0+ xx = 1. Then limx→0+ xxx= 01 = 0, so

limx→0+

(xxx − xx

)= −1.

6. The graph of r = 2 + cos 2θ and its reflection over the line y = x bound five regions inthe plane. Find the area of the region containing the origin.

Solution: 9π/2− 8

The original graph is closer to the origin than its reflection for θ ∈ (π4, 3π

4) ∪ (5π

4, 7π

4),

and the region is symmetric about the origin. Therefore the area we wish to find isthe polar integral

4∫ 3π

4

π4

1

2(2 + cos 2θ)2dθ = 2

∫ 3π4

π4

(4 + 4 cos 2θ + cos2 2θ)dθ

= 2∫ 3π

4

π4

(4 + 4 cos 2θ +

1

2(1 + cos 4θ)

)dθ

=[9θ + 4 sin 2θ +

1

4sin 4θ

] 3π4

π4

=(

27π

4− 4

)−

(9π

4+ 4

)=

9π

2− 8.

2

7. Two ants, one starting at (−1, 1), the other at (1, 1), walk to the right along theparabola y = x2 such that their midpoint moves along the line y = 1 with constantspeed 1. When the left ant first hits the line y = 1

2, what is its speed?

Solution: 3√

3− 3

When the left ant first hits the line y = 12, the right ant hits the line y = 3

2. The

left ant is then at (−√

22, 1

2), and the right ant is at (

√6

2, 3

2). Let the left ant have

velocity with components vx and vy, the right ant velocity with components wx andwy. Since dy

dx= 2x, vy

vx= −√2 and wy

wx=√

6. Since the midpoint of the ants moves

at speed 1 along the line y = 1, 12(vx + wx) = 1 and 1

2(vy + wy) = 0. Therefore,√

2vx = −vy = wy =√

6wx =√

6(2− vx). Solving for vx gives 2√

6√6+

√2

= 3−√3. Thenthe speed of the left ant is

√v2

x + v2y =

√v2

x + (−√

2vx)2 =√

3|vx| = 3√

3− 3.

8. If f is a continuous real function such that f(x − 1) + f(x + 1) ≥ x + f(x) for all x,what is the minimum possible value of

∫ 20051 f(x)dx?

Solution: 2010012

Let g(x) = f(x)− x. Then

g(x− 1) + x− 1 + g(x+ 1) + x+ 1 ≥ x+ g(x) + x,

or g(x− 1) + g(x+ 1) ≥ g(x). But now,

g(x+ 3) ≥ g(x+ 2)− g(x+ 1) ≥ −g(x).

Therefore∫ a+6

ag(x)dx =

∫ a+3

ag(x)dx+

∫ a+6

a+3g(x)dx

=∫ a+3

a(g(x) + g(x+ 3))dx ≥ 0.

It follows that ∫ 2005

1g(x) =

333∑

n=0

∫ 6n+7

6n+1g(x)dx ≥ 0,

so that

∫ 2005

1f(x)dx =

∫ 2005

1(g(x) + x)dx ≥

∫ 2005

1x dx =

[x2

2

]2005

1

=20052 − 1

2= 2010012.

Equality holds for f(x) = x.

9. Compute∞∑

k=0

4

(4k)!.

Solution: e+ 1/e+ 2 cos 1

3

This is the power series

4 +4x4

4!+

4x8

8!+ · · ·

evaluated at x = 1. But this power series can be written as the sum

(1 +

x

1!+x2

2!+x3

3!+x4

4!+x5

5!+x6

6!+x7

7!+ · · ·

)

+

(1− x

1!+x2

2!− x3

3!+x4

4!− x5

5!+x6

6!− x7

7!+ · · ·

)

+ 2

(1− x2

2!+x4

4!− x6

6!+ · · ·

)

= ex + e−x + 2 cos x.

It follows that the quantity is e+ 1/e+ 2 cos 1.

10. Let f : R→ R be a smooth function such that f ′(x) = f(1−x) for all x and f(0) = 1.Find f(1).

Solution: sec 1 + tan 1

Differentiating the given equation gives f ′′(x) = −f(x). This has solutions of the formA cos(x) +B sin(x). Since f(0) = 1, A = 1. Then f ′(x) = B cos(x)− sin(x) and

f(1− x) = cos(1− x) +B sin(1− x)= cos 1 cos x+ sin 1 sin x+B sin 1 cos x−B cos 1 sin x

= (cos 1 + B sin 1) cos x+ (sin 1−B cos 1) sin x.

Therefore, B = cos 1+B sin 1 and −1 = sin 1−B cos 1, both of which yield as solutions

B =cos 1

1− sin 1=

1 + sin 1

cos 1= sec 1 + tan 1.

4

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Individual Round: Combinatorics Subject Test

1. A true-false test has ten questions. If you answer five questions “true” and five “false,”your score is guaranteed to be at least four. How many answer keys are there for whichthis is true?

2. How many nonempty subsets of {1, 2, 3, . . . , 12} have the property that the sum of thelargest element and the smallest element is 13?

3. The Red Sox play the Yankees in a best-of-seven series that ends as soon as one teamwins four games. Suppose that the probability that the Red Sox win Game n is n−1

6.

What is the probability that the Red Sox will win the series?

4. In how many ways can 4 purple balls and 4 green balls be placed into a 4× 4 grid suchthat every row and column contains one purple ball and one green ball? Only one ballmay be placed in each box, and rotations and reflections of a single configuration areconsidered different.

5. Doug and Ryan are competing in the 2005 Wiffle Ball Home Run Derby. In each round,each player takes a series of swings. Each swing results in either a home run or an out,and an out ends the series. When Doug swings, the probability that he will hit a homerun is 1/3. When Ryan swings, the probability that he will hit a home run is 1/2. Inone round, what is the probability that Doug will hit more home runs than Ryan hits?

6. Three fair six-sided dice, each numbered 1 through 6, are rolled. What is the probabilitythat the three numbers that come up can form the sides of a triangle?

7. What is the maximum number of bishops that can be placed on an 8 × 8 chessboardsuch that at most three bishops lie on any diagonal?

8. Every second, Andrea writes down a random digit uniformly chosen from the set{1, 2, 3, 4}. She stops when the last two numbers she has written sum to a primenumber. What is the probability that the last number she writes down is 1?

9. Eight coins are arranged in a circle heads up. A move consists of flipping over twoadjacent coins. How many different sequences of six moves leave the coins alternatingheads up and tails up?

10. You start out with a big pile of 32004 cards, with the numbers 1, 2, 3, . . . , 32004 writtenon them. You arrange the cards into groups of three any way you like; from each group,you keep the card with the largest number and discard the other two. You now againarrange these 32003 remaining cards into groups of three any way you like, and in eachgroup, keep the card with the smallest number and discard the other two. You nowhave 32002 cards, and you again arrange these into groups of three and keep the largestnumber in each group. You proceed in this manner, alternating between keeping thelargest number and keeping the smallest number in each group, until you have just onecard left.

How many different values are possible for the number on this final card?

1

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Individual Round: Combinatorics Subject Test — Solutions

1. A true-false test has ten questions. If you answer five questions “true” and five “false,”your score is guaranteed to be at least four. How many answer keys are there for whichthis is true?

Solution: 22

Suppose that either nine or ten of the questions have the same answer. Then no matterwhich five questions we pick to have this answer, we will be right at least four times.Conversely, suppose that there are at least two questions with each answer; we willshow that we can get a score less than four. By symmetry, assume there are at leastfive questions whose answer is true. Then if we label five of these false, not only willwe get these five wrong, but we will also have answered all the false questions withtrue, for a total of at least seven incorrect. There are 2 ways for all the questions tohave the same answer, and 2 · 10 = 20 ways for one question to have a different answerfrom the others, for a total of 22 ways.

2. How many nonempty subsets of {1, 2, 3, . . . , 12} have the property that the sum of thelargest element and the smallest element is 13?

Solution: 1365

If a is the smallest element of such a set, then 13 − a is the largest element, andfor the remaining elements we may choose any (or none) of the 12 − 2a elementsa + 1, a + 2, . . . , (13− a)− 1. Thus there are 212−2a such sets whose smallest elementis a. Also, 13− a ≥ a clearly implies a < 7. Summing over all a = 1, 2, . . . , 6, we get atotal of

210 + 28 + 26 + · · ·+ 20 = 45 + 44 + · · ·+ 40 = (46 − 1)/(4− 1) = 4095/3 = 1365

possible sets.

3. The Red Sox play the Yankees in a best-of-seven series that ends as soon as one teamwins four games. Suppose that the probability that the Red Sox win Game n is n−1

6.

What is the probability that the Red Sox will win the series?

Solution: 1/2

Note that if we imagine that the series always continues to seven games even after oneteam has won four, this will never change the winner of the series. Notice also thatthe probability that the Red Sox will win Game n is precisely the probability that theYankees will win Game 8−n. Therefore, the probability that the Yankees win at leastfour games is the same as the probability that the Red Sox win at least four games,namely 1/2.

4. In how many ways can 4 purple balls and 4 green balls be placed into a 4× 4 grid suchthat every row and column contains one purple ball and one green ball? Only one ballmay be placed in each box, and rotations and reflections of a single configuration areconsidered different.

1

Solution: 216

There are 4! = 24 ways to place the four purple balls into the grid. Choose any purpleball, and place two green balls, one in its row and the other in its column. Thereare four boxes that do not yet lie in the same row or column as a green ball, and atleast one of these contains a purple ball (otherwise the two rows containing green ballswould contain the original purple ball as well as the two in the columns not containinggreen balls). It is then easy to see that there is a unique way to place the remaininggreen balls. Therefore, there are a total of 24 · 9 = 216 ways.

5. Doug and Ryan are competing in the 2005 Wiffle Ball Home Run Derby. In each round,each player takes a series of swings. Each swing results in either a home run or an out,and an out ends the series. When Doug swings, the probability that he will hit a homerun is 1/3. When Ryan swings, the probability that he will hit a home run is 1/2. Inone round, what is the probability that Doug will hit more home runs than Ryan hits?

Solution: 1/5

Denote this probability by p. Doug hits more home runs if he hits a home run on hisfirst try when Ryan does not, or if they both hit home runs on their first try and Doughits more home runs thereafter. The probability of the first case occurring is 1

3· 1

2= 1

6,

and the probability of the second case occurring is 13· 1

2· p = p

6. Therefore p = 1

6+ p

6,

which we solve to find p = 15.

6. Three fair six-sided dice, each numbered 1 through 6, are rolled. What is the probabilitythat the three numbers that come up can form the sides of a triangle?

Solution: 37/72

Denote this probability by p, and let the three numbers that come up be x, y, and z. Wewill calculate 1−p instead: 1−p is the probability that x ≥ y+z, y ≥ z+x, or z ≥ x+y.Since these three events are mutually exclusive, 1 − p is just 3 times the probabilitythat x ≥ y+ z. This happens with probability (0 + 1 + 3 + 6 + 10 + 15)/216 = 35/216,so the answer is 1− 3 · (35/216) = 1− 35/72 = 37/72.

7. What is the maximum number of bishops that can be placed on an 8 × 8 chessboardsuch that at most three bishops lie on any diagonal?

Solution: 38

If the chessboard is colored black and white as usual, then any diagonal is a solid color,so we may consider bishops on black and white squares separately. In one direction,the lengths of the black diagonals are 2, 4, 6, 8, 6, 4, and 2. Each of these can haveat most three bishops, except the first and last which can have at most two, giving atotal of at most 2+3+3+3+3+3+2 = 19 bishops on black squares. Likewise therecan be at most 19 bishops on white squares for a total of at most 38 bishops. This isindeed attainable as in the diagram below.

2

XX X XX X X X

X

X

XXX

X

X

X

X X X X XX

X

X

X

X

X

X X

XXXX

X

X

X

X X

8. Every second, Andrea writes down a random digit uniformly chosen from the set{1, 2, 3, 4}. She stops when the last two numbers she has written sum to a primenumber. What is the probability that the last number she writes down is 1?

Solution: 15/44

Let pn be the probability that the last number she writes down is 1 when the firstnumber she writes down is n. Suppose she starts by writing 2 or 4. Then she cancontinue writing either 2 or 4, but the first time she writes 1 or 3, she stops. Thereforep2 = p4 = 1

2. Suppose she starts by writing 1. Then she stops if she writes 1,

2, or 4, but continues if she writes 3. Therefore p1 = 14(1 + p3). If she starts by

writing 3, then she stops if she writes 2 or 4 and otherwise continues. Thereforep3 = 1

4(p1 + p3) = 1

16(1 + 5p3). Solving gives p3 = 1

11and p1 = 3

11. The probability we

want to find is therefore 14(p1 + p2 + p3 + p4) = 15

44.

9. Eight coins are arranged in a circle heads up. A move consists of flipping over twoadjacent coins. How many different sequences of six moves leave the coins alternatingheads up and tails up?

Solution: 7680

Imagine we flip over two adjacent coins by pushing a button halfway between them.Then the outcome depends only on the parities of the number of times that each buttonis pushed. To flip any coin, we must push the two buttons adjacent to that coin a totalof an odd number of times. To flip every other coin, the parities must then progressaround the circle as even, even, odd, odd, even, even, odd, odd. There are 4 ways toassign these parities. If we assume each button is pressed either once or not at all,this accounts for only four presses, so some button is also pressed twice more. Supposethis button was already pushed once. There are 4 of these, and the number of possiblesequences of presses is then 6!/3! = 120. Suppose it has not already been pressed.There are 4 of these as well, and the number of possible sequences is 6!/2! = 360. Thetotal number of sequences is then 4(4 · 120 + 4 · 360) = 7680.

10. You start out with a big pile of 32004 cards, with the numbers 1, 2, 3, . . . , 32004 writtenon them. You arrange the cards into groups of three any way you like; from each group,you keep the card with the largest number and discard the other two. You now againarrange these 32003 remaining cards into groups of three any way you like, and in eachgroup, keep the card with the smallest number and discard the other two. You nowhave 32002 cards, and you again arrange these into groups of three and keep the largestnumber in each group. You proceed in this manner, alternating between keeping thelargest number and keeping the smallest number in each group, until you have just onecard left.

3

How many different values are possible for the number on this final card?

Solution: 32004 − 2 · 31002 + 2

We claim that if you have cards numbered 1, 2, . . . , 32n and perform 2n successivegrouping operations, then c is a possible value for your last remaining card if and onlyif

3n ≤ c ≤ 32n − 3n + 1.

This gives 32n− 2 · 3n + 2 possible values of c, for a final answer of 32004− 2 · 31002 + 2.

Indeed, notice that the last remaining card c must have been the largest of some set ofthree at the (2n− 1)th step; each of these was in turn the largest of some set of three(and so c was the largest of some set of 9 cards) remaining at the (2n− 3)th step; eachof these was in turn the largest of some set of three (and so c was the largest of someset of 27) remaining at the (2n − 5)th step; continuing in this manner, we get that cwas the largest of some 3n cards at the first step, so c ≥ 3n. A similar analysis of allof the steps in which we save the smallest card gives that c is the smallest of some setof 3n initial cards, so c ≤ 32n − 3n + 1.

To see that any c in this interval is indeed possible, we will carry out the groupingsinductively so that, after 2i steps, the following condition is satisfied: if the numbersremaining are a1 < a2 < · · · < a32(n−i) , then c is one of these, and there are at least3n−i − 1 numbers smaller than c and at least 3n−i − 1 numbers larger than c. This iscertainly true when i = 0, so it suffices to show that if it holds for some i < n, wecan perform the grouping so that the condition will still hold for i + 1. Well, we firstgroup the smallest numbers as {a1, a2, a3}, {a4, a5, a6}, . . . , {a3n−i−5, a3n−i−4, a3n−i−3}.We then group the remaining numbers in such a way that c and the largest 3n−i − 1numbers are each the largest in its respective group; it is easy to see that we can dothis. After retaining the largest number in each group, we will then have at least3n−i−1 − 1 numbers smaller than c remaining and at least 3n−i − 1 numbers largerthan c remaining. And for the next grouping, we similarly group the largest 3n−i − 3numbers into 3n−i−1−1 groups, and arrange the remaining numbers so that the smallest3n−i−1 − 1 numbers and c are all the smallest in their groups. After this round ofdiscarding, then c will be retained, and we will still have at least 3n−i−1 − 1 numberslarger than c and 3n−i−1 numbers smaller than c. This proves the induction step, andnow the solution is complete.

4

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Individual Round: General Test, Part 1

1. How many real numbers x are solutions to the following equation?

|x− 1| = |x− 2|+ |x− 3|

2. A true-false test has ten questions. If you answer five questions “true” and five “false,”your score is guaranteed to be at least four. How many answer keys are there for whichthis is true?

3. Let ABCD be a regular tetrahedron with side length 2. The plane parallel to edgesAB and CD and lying halfway between them cuts ABCD into two pieces. Find thesurface area of one of these pieces.

4. Find all real solutions to x3 + (x+ 1)3 + (x+ 2)3 = (x+ 3)3.

5. In how many ways can 4 purple balls and 4 green balls be placed into a 4× 4 grid suchthat every row and column contains one purple ball and one green ball? Only one ballmay be placed in each box, and rotations and reflections of a single configuration areconsidered different.

6. In an election, there are two candidates, A and B, who each have 5 supporters. Eachsupporter, independent of other supporters, has a 1

2probability of voting for his or her

candidate and a 12

probability of being lazy and not voting. What is the probability ofa tie (which includes the case in which no one votes)?

7. If a, b, c > 0, what is the smallest possible value of⌊

a+bc

⌋+

⌊b+ca

⌋+

⌊c+a

b

⌋? (Note that

bxc denotes the greatest integer less than or equal to x.)

8. Ten positive integers are arranged around a circle. Each number is one more than thegreatest common divisor of its two neighbors. What is the sum of the ten numbers?

9. A triangular piece of paper of area 1 is folded along a line parallel to one of the sidesand pressed flat. What is the minimum possible area of the resulting figure?

10. What is the smallest integer x larger than 1 such that x2 ends in the same three digitsas x does?

1

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Individual Round: General Test, Part 1 — Solutions

1. How many real numbers x are solutions to the following equation?

|x− 1| = |x− 2|+ |x− 3|

Solution: 2

If x < 1, the equation becomes (1 − x) = (2− x) + (3− x) which simplifies to x = 4,contradicting the assumption x < 1. If 1 ≤ x ≤ 2, we get (x− 1) = (2− x) + (3− x),which gives x = 2. If 2 ≤ x ≤ 3, we get (x− 1) = (x− 2) + (3− x), which again givesx = 2. If x ≥ 3, we get (x− 1) = (x− 2) + (x− 3), or x = 4. So 2 and 4 are the onlysolutions, and the answer is 2.

2. A true-false test has ten questions. If you answer five questions “true” and five “false,”your score is guaranteed to be at least four. How many answer keys are there for whichthis is true?

Solution: 22

Suppose that either nine or ten of the questions have the same answer. Then no matterwhich five questions we pick to have this answer, we will be right at least four times.Conversely, suppose that there are at least two questions with each answer; we willshow that we can get a score less than four. By symmetry, assume there are at leastfive questions whose answer is true. Then if we label five of these false, not only willwe get these five wrong, but we will also have answered all the false questions withtrue, for a total of at least seven incorrect. There are 2 ways for all the questions tohave the same answer, and 2 · 10 = 20 ways for one question to have a different answerfrom the others, for a total of 22 ways.

3. Let ABCD be a regular tetrahedron with side length 2. The plane parallel to edgesAB and CD and lying halfway between them cuts ABCD into two pieces. Find thesurface area of one of these pieces.

Solution: 1 + 2√

3

The plane intersects each face of the tetrahedron in a midline of the face; by symmetryit follows that the intersection of the plane with the tetrahedron is a square of sidelength 1. The surface area of each piece is half the total surface area of the tetrahedronplus the area of the square, that is, 1

2· 4 · 22

√3

4+ 1 = 1 + 2

√3.

4. Find all real solutions to x3 + (x+ 1)3 + (x+ 2)3 = (x+ 3)3.

Solution: 3

The equation simplifies to 3x3 + 9x2 + 15x+ 9 = x3 + 9x2 + 27x+ 27, or equivalently,2x3− 12x− 18 = 2(x− 3)(x2 +3x+3) = 0. The discriminant of x2 +3x+3 is −3 < 0,so the only real solution is x = 3.

1

5. In how many ways can 4 purple balls and 4 green balls be placed into a 4× 4 grid suchthat every row and column contains one purple ball and one green ball? Only one ballmay be placed in each box, and rotations and reflections of a single configuration areconsidered different.

Solution: 216

There are 4! = 24 ways to place the four purple balls into the grid. Choose any purpleball, and place two green balls, one in its row and the other in its column. Thereare four boxes that do not yet lie in the same row or column as a green ball, and atleast one of these contains a purple ball (otherwise the two rows containing green ballswould contain the original purple ball as well as the two in the columns not containinggreen balls). It is then easy to see that there is a unique way to place the remaininggreen balls. Therefore, there are a total of 24 · 9 = 216 ways.

6. In an election, there are two candidates, A and B, who each have 5 supporters. Eachsupporter, independent of other supporters, has a 1

2probability of voting for his or her

candidate and a 12

probability of being lazy and not voting. What is the probability ofa tie (which includes the case in which no one votes)?

Solution: 63/256

The probability that exactly k supporters of A vote and exactly k supporters of B vote

is(

5k

)2 · 1210 . Summing over k from 0 to 5 gives

(1

210

)(1 + 25 + 100 + 100 + 25 + 1) =

252

1024=

63

256.

7. If a, b, c > 0, what is the smallest possible value of⌊

a+bc

⌋+

⌊b+ca

⌋+

⌊c+a

b

⌋? (Note that

bxc denotes the greatest integer less than or equal to x.)

Solution: 4

Since bxc > x− 1 for all x, we have that

⌊a+ b

c

⌋+

⌊b+ c

a

⌋+

⌊c+ a

b

⌋>

a+ b

c+b+ c

a+c+ a

b− 3

=

(a

b+b

a

)+

(b

c+c

b

)+

( ca

+a

c

)− 3.

But by the AM-GM inequality, each of the first three terms in the last line is at least2. Therefore, the lefthand side is greater than 2 + 2 + 2 − 3 = 3. Since it is aninteger, the smallest value it can be is therefore 4. This is in fact attainable by letting(a, b, c) = (6, 8, 9).

8. Ten positive integers are arranged around a circle. Each number is one more than thegreatest common divisor of its two neighbors. What is the sum of the ten numbers?

Solution: 28

First note that all the integers must be at least 2, because the greatest common divisorof any two positive integers is at least 1. Let n be the largest integer in the circle.The greatest common divisor of its two neighbors is n− 1. Therefore, each of the two

2

neighbors is at least n − 1 but at most n, so since n − 1 - n for n − 1 ≥ 2, they mustboth be equal to n− 1. Let m be one of the numbers on the other side of n− 1 fromn. Then gcd(n,m) = n − 2. Since n − 2 ≥ 0, n − 2 | n only for n = 3 or 4. If n = 3,each number must be 2 or 3, and it is easy to check that there is no solution. If n = 4,then it is again not hard to find that there is a unique solution up to rotation, namely4322343223. The only possible sum is therefore 28.

9. A triangular piece of paper of area 1 is folded along a line parallel to one of the sidesand pressed flat. What is the minimum possible area of the resulting figure?

Solution: 2/3

Let the triangle be denoted ABC, and suppose we fold parallel to BC. Let the distancefrom A to BC be h, and suppose we fold along a line at a distance of ch from A. Wewill assume that neither angle B nor C is obtuse, for the area of overlap will only besmaller if either is obtuse. If c ≤ 1

2, then A does not fold past the edge BC, so the

overlap is a triangle similar to the original with height ch; the area of the figure isthen 1 − c2 ≥ 3

4. Suppose c > 1

2, so that A does fold past BC. Then the overlap is a

trapezoid formed by taking a triangle of height ch similar to the original and removinga triangle of height (2c− 1)h similar to the original. The area of the resulting figure isthus 1 − c2 + (2c − 1)2 = 3c2 − 4c + 2. This is minimized when c = 2

3, when the area

is 23< 3

4; the minimum possible area is therefore 2

3.

10. What is the smallest integer x larger than 1 such that x2 ends in the same three digitsas x does?

Solution: 376

The condition is that 1000 | x2 − x = x(x − 1). Since 1000 = 23 · 53, and 2 cannotdivide both x and x − 1, 23 = 8 must divide one of them. Similarly, 53 = 125 mustdivide either x or x − 1. We try successive values of x that are congruent to 0 or 1modulo 125 and see which ones have the property that x or x− 1 is divisible by 8. Itis easy to check that 125, 126, 250, 251, and 375 do not work, but the next value, 376,does, so this is the answer.

3

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Individual Round: General Test, Part 2

1. The volume of a cube (in cubic inches) plus three times the total length of its edges(in inches) is equal to twice its surface area (in square inches). How many inches longis its long diagonal?

2. Find three real numbers a < b < c satisfying:

a+ b+ c = 21/4

1/a+ 1/b+ 1/c = 21/4

abc = 1.

3. Working together, Jack and Jill can paint a house in 3 days; Jill and Joe can paint thesame house in 4 days; or Joe and Jack can paint the house in 6 days. If Jill, Joe, andJack all work together, how many days will it take them?

4. In how many ways can 8 people be arranged in a line if Alice and Bob must be nextto each other, and Carol must be somewhere behind Dan?

5. You and I play the following game on an 8 × 8 square grid of boxes: Initially, everybox is empty. On your turn, you choose an empty box and draw an X in it; if anyof the four adjacent boxes are empty, you mark them with an X as well. (Two boxesare adjacent if they share an edge.) We alternate turns, with you moving first, andwhoever draws the last X wins. How many choices do you have for a first move thatwill enable you to guarantee a win no matter how I play?

6. A cube with side length 2 is inscribed in a sphere. A second cube, with faces parallelto the first, is inscribed between the sphere and one face of the first cube. What is thelength of a side of the smaller cube?

7. Three distinct lines are drawn in the plane. Suppose there exist exactly n circles inthe plane tangent to all three lines. Find all possible values of n.

8. What is the maximum number of bishops that can be placed on an 8 × 8 chessboardsuch that at most three bishops lie on any diagonal?

9. In how many ways can the cells of a 4 × 4 table be filled in with the digits 1, 2, . . . , 9so that each of the 4-digit numbers formed by the columns is divisible by each of the4-digit numbers formed by the rows?

10. Let bxc denote the greatest integer less than or equal to x. How many positive integersless than 2005 can be expressed in the form bx bxcc for some positive real x?

1

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Individual Round: General Test, Part 2 — Solutions

1. The volume of a cube (in cubic inches) plus three times the total length of its edges(in inches) is equal to twice its surface area (in square inches). How many inches longis its long diagonal?

Solution: 6√

3

If the side length of the cube is s inches, then the condition implies s3 +3 ·12s = 2 ·6s2,or s(s2− 12s+ 36) = s(s− 6)2 = 0. Therefore s = 6, and the long diagonal has lengths√

3 = 6√

3.

2. Find three real numbers a < b < c satisfying:

a+ b+ c = 21/4

1/a+ 1/b+ 1/c = 21/4

abc = 1.

Solution: 1/4, 1, 4

By inspection, one notices that if b is a number such that b+1/b = 17/4, then a = 1, c =1/b will work. Again by inspection (or by solving the quadratic b2 − 17b/4 + 1 = 0),one finds b = 1/4 or 4, so the numbers are 1/4, 1, and 4.

Alternative Solution: Note that ab + bc + ca = abc(1/a + 1/b + 1/c) = 21/4, soa, b, and c are the roots of the polynomial x3 − 21x2/4 + 21x/4− 1, which factors as14(x− 1)(x− 4)(4x− 1), giving the same answer.

3. Working together, Jack and Jill can paint a house in 3 days; Jill and Joe can paint thesame house in 4 days; or Joe and Jack can paint the house in 6 days. If Jill, Joe, andJack all work together, how many days will it take them?

Solution: 8/3

Suppose that Jack paints x houses per day, Jill paints y houses per day, and Joe paintsz houses per day. Together, Jack and Jill paint 1/3 of a house in a day — that is,

x+ y = 1/3.

Similarly,y + z = 1/4,

andz + x = 1/6.

Adding all three equations and dividing by 2 gives

x+ y + z = 3/8.

So, working together, the three folks can paint 3/8 houses in a day, or 8/3 days perhouse.

1

4. In how many ways can 8 people be arranged in a line if Alice and Bob must be nextto each other, and Carol must be somewhere behind Dan?

Solution: 5040

Let us place Alice and Bob as a single person; there are then 7! = 5040 differentarrangements. Alice can be in front of Bob or vice versa, multiplying the number ofpossibilities by 2, but Carol is behind Dan in exactly half of those, so that the answeris just 5040.

5. You and I play the following game on an 8 × 8 square grid of boxes: Initially, everybox is empty. On your turn, you choose an empty box and draw an X in it; if anyof the four adjacent boxes are empty, you mark them with an X as well. (Two boxesare adjacent if they share an edge.) We alternate turns, with you moving first, andwhoever draws the last X wins. How many choices do you have for a first move thatwill enable you to guarantee a win no matter how I play?

Solution: 0

I can follow a symmetry strategy: whenever you play in the box S, I play in the imageof S under the 180◦ rotation about the center of the board. This ensures that the boardwill always be centrally symmetric at the beginning of your turn. Thus, if you play inan empty box S, its symmetric image S ′ is also empty at the beginning of your turn,and it remains so after your turn, since the even size of the board ensures that S canbe neither equal to nor adjacent to S ′. In particular, I always have a move available.Since the first person without an available move loses, you are guaranteed to lose. Sothe answer is that you have 0 choices for a first move that will guarantee your win.

6. A cube with side length 2 is inscribed in a sphere. A second cube, with faces parallelto the first, is inscribed between the sphere and one face of the first cube. What is thelength of a side of the smaller cube?

Solution: 2/3

First note that the long diagonal of the cube has length 2√

3, so the radius of thesphere is

√3. Let x be the side length of the smaller cube. Then the distance from the

center of the sphere to the far face of the smaller cube is 1+x, while the distance fromthe center of the far face to a vertex lying on the sphere is x

√2

2. Therefore, the square

of the radius is 3 = (1 + x)2 + x2

2, or 3x2 + 4x− 4 = (3x− 2)(x+ 2) = 0, so x = 2

3.

7. Three distinct lines are drawn in the plane. Suppose there exist exactly n circles inthe plane tangent to all three lines. Find all possible values of n.

Solution: 0, 2, 4

If the three lines form a triangle, then there are 4 circles, namely the incircle and thethree excircles. If the three lines concur or are all parallel, then there are 0 circles. Iftwo lines are parallel and the third is not, then there are 2 circles lying between thetwo parallel lines, one on each side of the transverse line. These are the only possibleconfigurations, so the answers are 0, 2, and 4.

2

8. What is the maximum number of bishops that can be placed on an 8 × 8 chessboardsuch that at most three bishops lie on any diagonal?

Solution: 38

If the chessboard is colored black and white as usual, then any diagonal is a solid color,so we may consider bishops on black and white squares separately. In one direction,the lengths of the black diagonals are 2, 4, 6, 8, 6, 4, and 2. Each of these can haveat most three bishops, except the first and last which can have at most two, giving atotal of at most 2+3+3+3+3+3+2 = 19 bishops on black squares. Likewise therecan be at most 19 bishops on white squares for a total of at most 38 bishops. This isindeed attainable as in the diagram below.

XX X XX X X X

X

X

XXX

X

X

X

X X X X XX

X

X

X

X

X

X X

XXXX

X

X

X

X X

9. In how many ways can the cells of a 4 × 4 table be filled in with the digits 1, 2, . . . , 9so that each of the 4-digit numbers formed by the columns is divisible by each of the4-digit numbers formed by the rows?

Solution: 9

If a and b are 4-digit numbers with the same first digit, and a divides b, then sinceb < a+ 1000 ≤ 2a, b must equal a. In particular, since the number formed by the firstrow of the table divides the number in the first column (and both have the same firstdigit), these numbers must be equal; call their common value n. Then, for k = 2, 3, or4, we find that the number in the kth column and the number in the kth row have thesame first digit (namely the kth digit of n), so by the same reasoning, they are equal.Also, the smallest number b formed by any column is divisible by the largest numbera formed by any row, but by the symmetry just proven, a is also the largest numberformed by any column, so a ≥ b. Since b is divisible by a, we must have equality. Thenall columns contain the same number — and hence all rows also contain the same

3

number — which is only possible if all 16 cells contain the same digit. Conversely, foreach d = 1, . . . , 9, filling in all 16 cells with the digit d clearly gives a table meetingthe required condition, so we have exactly 9 such tables, one for each digit.

10. Let bxc denote the greatest integer less than or equal to x. How many positive integersless than 2005 can be expressed in the form bx bxcc for some positive real x?

Solution: 990

Let {x} = x− bxc be the fractional part of x. Note that

bx bxcc = b(bxc+ {x}) bxcc = bxc2 + b{x} bxcc .

Because {x} may take on any value in the half-open interval [0, 1), the quantityb{x} bxcc can take on any integer value between 0 and bxc − 1, inclusive.

If bxc = n, then bx bxcc can be any of the numbers n2, n2 + 1, . . . , n2 + n− 1. In otherwords, there are precisely n possible values that bx bxcc can take, and moreover, all ofthem are less than (n+ 1)2. Because 442 + 43 = 1979 < 2005 and 452 = 2025 > 2005,n can range between 1 and 44, inclusive. Therefore, the answer is

44∑n=1

n =44 · 45

2= 990.

4

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Individual Round: Geometry Subject Test

1. The volume of a cube (in cubic inches) plus three times the total length of its edges(in inches) is equal to twice its surface area (in square inches). How many inches longis its long diagonal?

2. Let ABCD be a regular tetrahedron with side length 2. The plane parallel to edgesAB and CD and lying halfway between them cuts ABCD into two pieces. Find thesurface area of one of these pieces.

3. Let ABCD be a rectangle with area 1, and let E lie on side CD. What is the area ofthe triangle formed by the centroids of triangles ABE, BCE, and ADE?

4. Let XY Z be a triangle with 6 X = 60◦ and 6 Y = 45◦. A circle with center P passesthrough points A and B on side XY , C and D on side Y Z, and E and F on side ZX.Suppose AB = CD = EF . Find 6 XPY in degrees.

5. A cube with side length 2 is inscribed in a sphere. A second cube, with faces parallelto the first, is inscribed between the sphere and one face of the first cube. What is thelength of a side of the smaller cube?

6. A triangular piece of paper of area 1 is folded along a line parallel to one of the sidesand pressed flat. What is the minimum possible area of the resulting figure?

7. Let ABCD be a tetrahedron such that edges AB, AC, and AD are mutually perpen-dicular. Let the areas of triangles ABC, ACD, and ADB be denoted by x, y, and z,respectively. In terms of x, y, and z, find the area of triangle BCD.

8. Let T be a triangle with side lengths 26, 51, and 73. Let S be the set of points insideT which do not lie within a distance of 5 of any side of T . Find the area of S.

9. Let AC be a diameter of a circle ω of radius 1, and let D be the point on AC suchthat CD = 1/5. Let B be the point on ω such that DB is perpendicular to AC, andlet E be the midpoint of DB. The line tangent to ω at B intersects line CE at thepoint X. Compute AX.

10. Let AB be the diameter of a semicircle Γ. Two circles, ω1 and ω2, externally tangentto each other and internally tangent to Γ, are tangent to the line AB at P and Q,respectively, and to semicircular arc AB at C and D, respectively, with AP < AQ.Suppose F lies on Γ such that 6 FQB = 6 CQA and that 6 ABF = 80◦. Find 6 PDQin degrees.

1

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Individual Round: Geometry Subject Test — Solutions

1. The volume of a cube (in cubic inches) plus three times the total length of its edges(in inches) is equal to twice its surface area (in square inches). How many inches longis its long diagonal?

Solution: 6√

3

If the side length of the cube is s inches, then the condition implies s3 +3 ·12s = 2 ·6s2,or s(s2− 12s+ 36) = s(s− 6)2 = 0. Therefore s = 6, and the long diagonal has lengths√

3 = 6√

3.

2. Let ABCD be a regular tetrahedron with side length 2. The plane parallel to edgesAB and CD and lying halfway between them cuts ABCD into two pieces. Find thesurface area of one of these pieces.

Solution: 1 + 2√

3

The plane intersects each face of the tetrahedron in a midline of the face; by symmetryit follows that the intersection of the plane with the tetrahedron is a square of sidelength 1. The surface area of each piece is half the total surface area of the tetrahedronplus the area of the square, that is, 1

2· 4 · 22

√3

4+ 1 = 1 + 2

√3.

3. Let ABCD be a rectangle with area 1, and let E lie on side CD. What is the area ofthe triangle formed by the centroids of triangles ABE, BCE, and ADE?

Solution: 1/9

Let the centroids of ABE, BCE, and ADE be denoted by X, Y , and Z, respectively.Let d(P,QR) denote the distance from P to line QR. Since the centroid lies two-thirdsof the distance from each vertex to the midpoint of the opposite edge, d(X,AB) =d(Y,CD) = d(Z,CD) = 1

3BC, so Y Z is parallel to CD and d(X,Y Z) = BC −

23BC = 1

3BC. Likewise, d(Z,AD) = 1

3DE and d(Y,BC) = 1

3CE, so that since Y Z

is perpendicular to AD and BC, we have that Y Z = CD − 13(DE + CE) = 2

3CD.

Therefore, the area of XY Z is 12(1

3BC)(2

3CD) = 1

9BC · CD = 1

9.

A B

CD E

4. Let XY Z be a triangle with 6 X = 60◦ and 6 Y = 45◦. A circle with center P passesthrough points A and B on side XY , C and D on side Y Z, and E and F on side ZX.Suppose AB = CD = EF . Find 6 XPY in degrees.

1

Solution: 255/2

Since PAB, PCD, and PEF are all isosceles triangles with equal legs and equalbases, they are congruent. It follows that the heights of each are the same, so that P isequidistant from the sides of XY Z. Therefore, P is the incenter and therefore lies onthe angle bisectors of XY Z. Thus 6 Y XP = 1

26 Y XZ = 30◦ and 6 PY X = 1

26 ZY X =

452

◦. Therefore 6 XPY = 180◦ − 30◦ − 45

2

◦= 255

2

◦.

5. A cube with side length 2 is inscribed in a sphere. A second cube, with faces parallelto the first, is inscribed between the sphere and one face of the first cube. What is thelength of a side of the smaller cube?

Solution: 2/3

First note that the long diagonal of the cube has length 2√

3, so the radius of thesphere is

√3. Let x be the side length of the smaller cube. Then the distance from the

center of the sphere to the far face of the smaller cube is 1+x, while the distance fromthe center of the far face to a vertex lying on the sphere is x

√2

2. Therefore, the square

of the radius is 3 = (1 + x)2 + x2

2, or 3x2 + 4x− 4 = (3x− 2)(x+ 2) = 0, so x = 2

3.

6. A triangular piece of paper of area 1 is folded along a line parallel to one of the sidesand pressed flat. What is the minimum possible area of the resulting figure?

Solution: 2/3

Let the triangle be denoted ABC, and suppose we fold parallel to BC. Let the distancefrom A to BC be h, and suppose we fold along a line at a distance of ch from A. Wewill assume that neither angle B nor C is obtuse, for the area of overlap will only besmaller if either is obtuse. If c ≤ 1

2, then A does not fold past the edge BC, so the

overlap is a triangle similar to the original with height ch; the area of the figure isthen 1 − c2 ≥ 3

4. Suppose c > 1

2, so that A does fold past BC. Then the overlap is a

trapezoid formed by taking a triangle of height ch similar to the original and removinga triangle of height (2c− 1)h similar to the original. The area of the resulting figure isthus 1 − c2 + (2c − 1)2 = 3c2 − 4c + 2. This is minimized when c = 2

3, when the area

is 23< 3

4; the minimum possible area is therefore 2

3.

7. Let ABCD be a tetrahedron such that edges AB, AC, and AD are mutually perpen-dicular. Let the areas of triangles ABC, ACD, and ADB be denoted by x, y, and z,respectively. In terms of x, y, and z, find the area of triangle BCD.

Solution:√x2 + y2 + z2

Place A, B, C, and D at (0, 0, 0), (b, 0, 0), (0, c, 0), and (0, 0, d) in Cartesian coordinatespace, with b, c, and d positive. Then the plane through B, C, and D is given by theequation x

b+ y

c+ z

d= 1. The distance from the origin to this plane is then

1√1b2

+ 1c2

+ 1d2

=bcd√

b2c2 + c2d2 + d2b2=

bcd

2√x2 + y2 + z2

.

Then if the area of BCD is K, the volume of the tetrahedron is

bcd

6=

bcdK

6√x2 + y2 + z2

,

2

implying K =√x2 + y2 + z2.

Alternative Solution: The area of BCD is also half the length of the cross prod-uct of the vectors

−−→BC = (0,−c, d) and

−−→BD = (−b, 0, d). This cross product is

(−cd,−db,−bc) = −2(y, z, x), which has length 2√x2 + y2 + z2. Thus the area of

BCD is√x2 + y2 + z2.

8. Let T be a triangle with side lengths 26, 51, and 73. Let S be the set of points insideT which do not lie within a distance of 5 of any side of T . Find the area of S.

Solution: 135/28

Note that the sides of S are parallel to the sides of T , so S is a triangle similar to T .The semiperimeter of T is s = 1

2(26 + 51 + 73) = 75. By Heron’s formula, the area

of T is√

75 · 49 · 24 · 2 = 420. If r is the inradius of T , then the area of T is rs, sor = 420/75 = 28/5. It follows that the inradius of S is r − 5 = 3/5, and the ratio ofsimilitude between S and T is 3/28. Therefore, the area of S is 420 · (3/28)2 = 135/28.

9. Let AC be a diameter of a circle ω of radius 1, and let D be the point on AC suchthat CD = 1/5. Let B be the point on ω such that DB is perpendicular to AC, andlet E be the midpoint of DB. The line tangent to ω at B intersects line CE at thepoint X. Compute AX.

Solution: 3

We first show that AX is perpendicular to AC. Let the tangent to ω at A intersectCB at Z and CE at X ′. Since ZA is parallel to BD and BE = ED, ZX ′ = X ′A.Therefore, X ′ is the midpoint of the hypotenuse of the right triangle ABZ, so it is alsoits circumcenter. Thus X ′A = X ′B, and since X ′A is tangent to ω and B lies on ω,we must have that X ′B is tangent to ω, so X = X ′.

Let O be the center of ω. Then OD = 45, so BD = 3

5and DE = 3

10. Then AX =

DE · ACDC

= 310· 2

1/5= 3.

A

B

C

E

DO

X’

Z

3

10. Let AB be the diameter of a semicircle Γ. Two circles, ω1 and ω2, externally tangentto each other and internally tangent to Γ, are tangent to the line AB at P and Q,respectively, and to semicircular arc AB at C and D, respectively, with AP < AQ.Suppose F lies on Γ such that 6 FQB = 6 CQA and that 6 ABF = 80◦. Find 6 PDQin degrees.

Solution: 35

Extend the semicircle centered at O to an entire circle ω, and let the reflection of Fover AB be F ′. Then CQF ′ is a straight line. Also, the homothety centered at Ctaking ω1 into ω takes P to a point X on ω and AB to the parallel line tangent to ωat X. Therefore, X is the midpoint of semicircle AXB, and C, P , and X lie on a line.Similarly, D, Q, and X lie on a line. So,

45◦ = 6 XCB = 6 PCB = 6 PCQ+ 6 QCB = 6 PCQ+ 10◦,

since 6 QCB = 6 F ′CB = 6 F ′AB = 6 FAB = 90◦ − 6 ABF = 10◦. Thus 6 PCQ = 35◦.We will show that 6 PCQ = 6 PDQ to get that 6 PDQ = 35◦.

Note that 6 XPQ subtends the sum of arcs AC and BX, which is equal to arc XC.Therefore 6 XPQ = 6 CDX, so CDQP is cyclic and 6 PCQ = 6 PDQ. The conclusionfollows.

C

D

A BP

Q

F’

F

X

O

4

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Guts Round

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

1. [5] Find the largest positive integer n such that 1 + 2 + 3 + · · · + n2 is divisible by1 + 2 + 3 + · · ·+ n.

2. [5] Let x, y, and z be positive real numbers such that (x · y) + z = (x + z) · (y + z).What is the maximum possible value of xyz?

3. [5] Find the sum21

41 − 1+

22

42 − 1+

24

44 − 1+

28

48 − 1+ · · · .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

4. [6] What is the probability that in a randomly chosen arrangement of the numbers andletters in “HMMT2005,” one can read either “HMMT” or “2005” from left to right?(For example, in “5HM0M20T,” one can read “HMMT.”)

5. [6] For how many integers n between 1 and 2005, inclusive, is 2 · 6 · 10 · · · (4n − 2)divisible by n!?

6. [6] Let m ◦ n = (m + n)/(mn + 4). Compute ((· · · ((2005 ◦ 2004) ◦ 2003) ◦ · · · ◦ 1) ◦ 0).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

7. [6] Five people of different heights are standing in line from shortest to tallest. As ithappens, the tops of their heads are all collinear; also, for any two successive people,the horizontal distance between them equals the height of the shorter person. If theshortest person is 3 feet tall and the tallest person is 7 feet tall, how tall is the middleperson, in feet?

8. [6] Let ABCD be a convex quadrilateral inscribed in a circle with shortest side AB.The ratio [BCD]/[ABD] is an integer (where [XY Z] denotes the area of triangleXY Z.) If the lengths of AB, BC, CD, and DA are distinct integers no greater than10, find the largest possible value of AB.

9. [6] Farmer Bill’s 1000 animals — ducks, cows, and rabbits — are standing in a circle.In order to feel safe, every duck must either be standing next to at least one cow orbetween two rabbits. If there are 600 ducks, what is the least number of cows therecan be for this to be possible?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

10. [7] You are given a set of cards labeled from 1 to 100. You wish to make piles ofthree cards such that in any pile, the number on one of the cards is the product ofthe numbers on the other two cards. However, no card can be in more than one pile.What is the maximum number of piles you can form at once?

11. [7] The Dingoberry Farm is a 10 mile by 10 mile square, broken up into 1 mile by 1 milepatches. Each patch is farmed either by Farmer Keith or by Farmer Ann. WheneverAnn farms a patch, she also farms all the patches due west of it and all the patchesdue south of it. Ann puts up a scarecrow on each of her patches that is adjacent toexactly two of Keith’s patches (and nowhere else). If Ann farms a total of 30 patches,what is the largest number of scarecrows she could put up?

12. [7] Two vertices of a cube are given in space. The locus of points that could be a thirdvertex of the cube is the union of n circles. Find n.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

13. [7] Triangle ABC has AB = 1, BC =√

7, and CA =√

3. Let ℓ1 be the line throughA perpendicular to AB, ℓ2 the line through B perpendicular to AC, and P the pointof intersection of ℓ1 and ℓ2. Find PC.

14. [7] Three noncollinear points and a line ℓ are given in the plane. Suppose no two of thepoints lie on a line parallel to ℓ (or ℓ itself). There are exactly n lines perpendicularto ℓ with the following property: the three circles with centers at the given points andtangent to the line all concur at some point. Find all possible values of n.

15. [7] Let S be the set of lattice points inside the circle x2 + y2 = 11. Let M be thegreatest area of any triangle with vertices in S. How many triangles with vertices inS have area M?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

16. [8] A regular octahedron has a side length of 1. What is the distance between twoopposite faces?

17. [8] Compute

2

√

23

√

24

√

25√

2 · · ·.

18. [8] If a, b, and c are random real numbers from 0 to 1, independently and uniformlychosen, what is the average (expected) value of the smallest of a, b, and c?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

19. [8] Regular tetrahedron ABCD is projected onto a plane sending A, B, C, and Dto A′, B′, C ′, and D′ respectively. Suppose A′B′C ′D′ is a convex quadrilateral withA′B′ = B′C’ and C ′D′ = D′A′, and suppose that the area of A′B′C ′D′ = 4. Giventhese conditions, the set of possible lengths of AB consists of all real numbers in theinterval [a, b). Compute b.

20. [8] If n is a positive integer, let s(n) denote the sum of the digits of n. We say thatn is zesty if there exist positive integers x and y greater than 1 such that xy = n ands(x)s(y) = s(n). How many zesty two-digit numbers are there?

21. [8] In triangle ABC with altitude AD, ∠BAC = 45◦, DB = 3, and CD = 2. Find thearea of triangle ABC.

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

22. [9] Find

{ln(1 + e)} + {ln(1 + e2)} + {ln(1 + e4)} + {ln(1 + e8)} + · · · ,

where {x} = x − ⌊x⌋ denotes the fractional part of x.

23. [9] The sides of a regular hexagon are trisected, resulting in 18 points, including ver-tices. These points, starting with a vertex, are numbered clockwise as A1, A2, . . . , A18.The line segment AkAk+4 is drawn for k = 1, 4, 7, 10, 13, 16, where indices are takenmodulo 18. These segments define a region containing the center of the hexagon. Findthe ratio of the area of this region to the area of the large hexagon.

24. [9] In the base 10 arithmetic problem HMMT + GUTS = ROUND, each distinctletter represents a different digit, and leading zeroes are not allowed. What is themaximum possible value of ROUND?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

25. [9] An ant starts at one vertex of a tetrahedron. Each minute it walks along a randomedge to an adjacent vertex. What is the probability that after one hour the ant windsup at the same vertex it started at?

26. [9] In triangle ABC, AC = 3AB. Let AD bisect angle A with D lying on BC, andlet E be the foot of the perpendicular from C to AD. Find [ABD]/[CDE]. (Here,[XY Z] denotes the area of triangle XY Z).

27. [9] In a chess-playing club, some of the players take lessons from other players. It ispossible (but not necessary) for two players both to take lessons from each other. Itso happens that for any three distinct members of the club, A, B, and C, exactly oneof the following three statements is true: A takes lessons from B; B takes lessons fromC; C takes lessons from A. What is the largest number of players there can be?

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

28. [10] There are three pairs of real numbers (x1, y1), (x2, y2), and (x3, y3) that satisfy

both x3 − 3xy2 = 2005 and y3 − 3x2y = 2004. Compute(

1 − x1

y1

)(

1 − x2

y2

) (

1 − x3

y3

)

.

29. [10] Let n > 0 be an integer. Each face of a regular tetrahedron is painted in oneof n colors (the faces are not necessarily painted different colors.) Suppose there aren3 possible colorings, where rotations, but not reflections, of the same coloring areconsidered the same. Find all possible values of n.

30. [10] A cuboctahedron is a polyhedron whose faces are squares and equilateral trianglessuch that two squares and two triangles alternate around each vertex, as shown.

What is the volume of a cuboctahedron of side length 1?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

31. [10] The L shape made by adjoining three congruent squares can be subdivided intofour smaller L shapes.

Each of these can in turn be subdivided, and so forth. If we perform 2005 successivesubdivisions, how many of the 42005 L’s left at the end will be in the same orientationas the original one?

32. [10] Let a1 = 3, and for n ≥ 1, let an+1 = (n + 1)an − n. Find the smallest m ≥ 2005such that am+1 − 1 | a2

m − 1.

33. [10] Triangle ABC has incircle ω which touches AB at C1, BC at A1, and CA at B1.Let A2 be the reflection of A1 over the midpoint of BC, and define B2 and C2 similarly.Let A3 be the intersection of AA2 with ω that is closer to A, and define B3 and C3

similarly. If AB = 9, BC = 10, and CA = 13, find [A3B3C3]/[ABC]. (Here [XY Z]denotes the area of triangle XY Z.)

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

34. [12] A regular octahedron ABCDEF is given such that AD, BE, and CF are per-pendicular. Let G, H , and I lie on edges AB, BC, and CA respectively such thatAGGB

= BHHC

= CIIA

= ρ. For some choice of ρ > 1, GH , HI, and IG are three edges of aregular icosahedron, eight of whose faces are inscribed in the faces of ABCDEF . Findρ.

35. [12] Let p = 224036583 − 1, the largest prime currently known. For how many positiveintegers c do each of the quadratics ±x2 ± px ± c have rational roots?

36. [12] One hundred people are in line to see a movie. Each person wants to sit inthe front row, which contains one hundred seats, and each has a favorite seat, chosenrandomly. They enter the row one at a time from the far right. As they walk, if theyreach their favorite seat, they sit, but to avoid stepping over people, if they encountera person already seated, they sit to that person’s right. If the seat furthest to the rightis already taken, they sit in a different row. What is the most likely number of peoplethat will get to sit in the first row?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

37. [15] Let a1, a2, . . . , a2005 be real numbers such that

a1 · 1 + a2 · 2 + a3 · 3 + · · · + a2005 · 2005 = 0

a1 · 12 + a2 · 22 + a3 · 32 + · · ·+ a2005 · 20052 = 0

a1 · 13 + a2 · 23 + a3 · 33 + · · ·+ a2005 · 20053 = 0...

a1 · 12004 + a2 · 22004 + a3 · 32004 + · · · + a2005 · 20052004 = 0

and

a1 · 12005 + a2 · 22005 + a3 · 32005 + · · ·+ a2005 · 20052005 = 1.

What is the value of a1?

38. [15] In how many ways can the set of ordered pairs of integers be colored red and bluesuch that for all a and b, the points (a, b), (−1− b, a + 1), and (1− b, a− 1) are all thesame color?

39. [15] How many regions of the plane are bounded by the graph of

x6 − x5 + 3x4y2 + 10x3y2 + 3x2y4 − 5xy4 + y6 = 0?

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

40. [18] In a town of n people, a governing council is elected as follows: each person castsone vote for some person in the town, and anyone that receives at least five votes iselected to council. Let c(n) denote the expected number of people elected to council ifeveryone votes randomly. Find limn→∞ c(n)/n.

41. [18] There are 42 stepping stones in a pond, arranged along a circle. You are standingon one of the stones. You would like to jump among the stones so that you movecounterclockwise by either 1 stone or 7 stones at each jump. Moreover, you would liketo do this in such a way that you visit each stone (except for the starting spot) exactlyonce before returning to your initial stone for the first time. In how many ways canyou do this?

42. [18] In how many ways can 6 purple balls and 6 green balls be placed into a 4× 4 gridsuch that every row and column contains two balls of one color and one ball of theother color? Only one ball may be placed in each box, and rotations and reflections ofa single configuration are considered different.

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6

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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

43. Write down an integer N between 0 and 20 inclusive. If at least N teams write downN , your score is N ; otherwise it is 0.

44. Write down a set S of positive integers, all greater than 1, whose product is P , suchthat for each x ∈ S, x is a proper divisor of (P/x)+1. Your score is 2n, where n = |S|.

45. A binary word is a finite sequence of 0’s and 1’s. A square subword is a subsequenceconsisting of two identical chunks next to each other. For example, the word 100101011contains the square subwords 00, 0101 (twice), 1010, and 11.

Find a long binary word containing a small number of square subwords. Specifically,write down a binary word of any length n ≤ 50. Your score will be max{0, n − s},where s is the number of occurrences of square subwords. (That is, each differentsquare subword will be counted according to the number of times it appears.)

7

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Guts Round — Solutions

1. Find the largest positive integer n such that 1 + 2 + 3 + · · ·+ n2 is divisible by 1 + 2 +3 + · · ·+ n.

Solution: 1

The statement is

n(n+ 1)

2| n

2(n2 + 1)

2⇔ n+ 1 | n(n2 + 1) = n3 + n.

But n+ 1 also divides (n+ 1)(n2−n+ 2) = n3 +n+ 2, so n+ 1 must divide 2. Hence,n cannot be greater than 1. And n = 1 clearly works, so that is the answer.

2. Let x, y, and z be positive real numbers such that (x · y) + z = (x+ z) · (y+ z). Whatis the maximum possible value of xyz?

Solution: 1/27

The condition is equivalent to z2 + (x+ y− 1)z = 0. Since z is positive, z = 1− x− y,so x+ y + z = 1. By the AM-GM inequality,

xyz ≤(x+ y + z

3

)3

=1

27,

with equality when x = y = z = 13.

3. Find the sum21

41 − 1+

22

42 − 1+

24

44 − 1+

28

48 − 1+ · · · .

Solution: 1

Notice that

22k

42k − 1=

22k+ 1

42k − 1− 1

42k − 1=

1

22k − 1− 1

42k − 1=

1

42k−1 − 1− 1

42k − 1.

Therefore, the sum telescopes as

(1

42−1 − 1− 1

420 − 1

)+

(1

420 − 1− 1

421 − 1

)+

(1

421 − 1− 1

422 − 1

)+ · · ·

and evaluates to 1/(42−1 − 1) = 1.

4. What is the probability that in a randomly chosen arrangement of the numbers andletters in “HMMT2005,” one can read either “HMMT” or “2005” from left to right?(For example, in “5HM0M20T,” one can read “HMMT.”)

Solution: 23/144

1

To read “HMMT,” there are(84

)ways to place the letters, and 4!

2ways to place the

numbers. Similarly, there are(84

)4!2

arrangements where one can read ”2005.” The

number of arrangements in which one can read both is just(84

). The total number of

arrangements is 8!4, thus the answer is

(84

)4!2

+(84

)4!2− (

84

)8!4

=

(8

4

)4

8!· 23 =

23

144.

5. For how many integers n between 1 and 2005, inclusive, is 2 · 6 · 10 · · · (4n− 2) divisibleby n!?

Solution: 2005

Note that

2 · 6 · 10 · · · (4n− 2) = 2n · 1 · 3 · 5 · · · (2n− 1)

= 2n · 1 · 2 · 3 · · · 2n2 · 4 · 6 · · · 2n

=1 · 2 · 3 · · · 2n1 · 2 · 3 · · ·n ,

that is, it is just (2n)!/n!. Therefore, since (2n)!/(n!)2 =(2nn

)is always an integer, the

answer is 2005.

6. Let m ◦ n = (m+ n)/(mn+ 4). Compute ((· · · ((2005 ◦ 2004) ◦ 2003) ◦ · · · ◦ 1) ◦ 0).

Solution: 1/12

Note that m ◦ 2 = (m + 2)/(2m + 4) = 12, so the quantity we wish to find is just

(12◦ 1) ◦ 0 = 1

3◦ 0 = 1/12.

7. Five people of different heights are standing in line from shortest to tallest. As ithappens, the tops of their heads are all collinear; also, for any two successive people,the horizontal distance between them equals the height of the shorter person. If theshortest person is 3 feet tall and the tallest person is 7 feet tall, how tall is the middleperson, in feet?

Solution:√

21

If A, B, and C are the tops of the heads of three successive people and D, E, and F aretheir respective feet, let P be the foot of the perpendicular from A to BE and let Q bethe foot of the perpendicular from B to CF . Then, by equal angles, 4ABP ∼ 4BCQ,so

CF

BE=CF

BQ=CQ

BQ+ 1 =

BP

AP+ 1 =

BE

AP=BE

AD.

Therefore the heights of successive people are in geometric progression. Hence, theheights of all five people are in geometric progression, so the middle height is

√3 · 7 =√

21 feet.

2

D

A

E

B

F

C

P

Q

8. Let ABCD be a convex quadrilateral inscribed in a circle with shortest side AB. Theratio [BCD]/[ABD] is an integer (where [XY Z] denotes the area of triangle XY Z.)If the lengths of AB, BC, CD, and DA are distinct integers no greater than 10, findthe largest possible value of AB.

Solution: 5

Note that[BCD]

[ABD]=

12BC · CD · sinC

12DA · AB · sinA =

BC · CDDA · AB

since ∠A and ∠C are supplementary. If AB ≥ 6, it is easy to check that no assignmentof lengths to the four sides yields an integer ratio, but if AB = 5, we can let BC = 10,CD = 9, and DA = 6 for a ratio of 3. The maximum value for AB is therefore 5.

9. Farmer Bill’s 1000 animals — ducks, cows, and rabbits — are standing in a circle.In order to feel safe, every duck must either be standing next to at least one cow orbetween two rabbits. If there are 600 ducks, what is the least number of cows therecan be for this to be possible?

Solution: 201

Suppose Bill has r rabbits and c cows. At most r−1 ducks can be between two rabbits:each rabbit can serve up to two such ducks, so at most 2r/2 = r ducks will each beserved by two rabbits, but we cannot have equality, since this would require alternatingbetween rabbits and ducks all the way around the circle, contradicting the fact thatmore than half the animals are ducks. Also, at most 2c ducks can each be adjacent toa cow. So we need 600 ≤ r − 1 + 2c = (400− c)− 1 + 2c, giving c ≥ 201. Conversely,an arrangement with 201 cows is possible:

RDRDR · · ·DR︸ ︷︷ ︸199 R, 198 D

DCD DCD DCD · · · DCD︸ ︷︷ ︸201 C, 402 D

.

So 201 is the answer.

10. You are given a set of cards labeled from 1 to 100. You wish to make piles of three cardssuch that in any pile, the number on one of the cards is the product of the numberson the other two cards. However, no card can be in more than one pile. What is themaximum number of piles you can form at once?

Solution: 8

Certainly, the two factors in any pile cannot both be at least 10, since then the productwould be at least 10× 11 > 100. Also, the number 1 can not appear in any pile, sincethen the other two cards in the pile would have to be the same. So each pile must use

3

one of the numbers 2, 3, . . . , 9 as one of the factors, meaning we have at most 8 piles.Conversely, it is easy to construct a set of 8 such piles, for example:

{9, 11, 99} {8, 12, 96} {7, 13, 91} {6, 14, 84}

{5, 15, 75} {4, 16, 64} {3, 17, 51} {2, 18, 36}

11. The Dingoberry Farm is a 10 mile by 10 mile square, broken up into 1 mile by 1 milepatches. Each patch is farmed either by Farmer Keith or by Farmer Ann. WheneverAnn farms a patch, she also farms all the patches due west of it and all the patchesdue south of it. Ann puts up a scarecrow on each of her patches that is adjacent toexactly two of Keith’s patches (and nowhere else). If Ann farms a total of 30 patches,what is the largest number of scarecrows she could put up?

Solution: 7

Whenever Ann farms a patch P , she also farms all the patches due west of P anddue south of P . So, the only way she can put a scarecrow on P is if Keith farms thepatch immediately north of P and the patch immediately east of P , in which case Anncannot farm any of the patches due north of P or due east of P . That is, Ann canonly put a scarecrow on P if it is the easternmost patch she farms in its east-west row,and the northernmost in its north-south column. In particular, all of her scarecrowpatches are in different rows and columns. Suppose that she puts up n scarecrows.The farthest south of these must be in the 10th row or above, so she farms at least 1patch in that column; the second-farthest south must be in the 9th row above, so shefarms at least 2 patches in that column; the third-farthest south must be in the 8throw or above, so she farms at least 3 patches in that column, and so forth, for a totalof at least

1 + 2 + · · ·+ n = n(n+ 1)/2

patches. If Ann farms a total of 30 < 8 · 9/2 patches, then we have n < 8. On theother hand, n = 7 scarecrows are possible, as shown:

S

S

S

S

S

S

S

12. Two vertices of a cube are given in space. The locus of points that could be a thirdvertex of the cube is the union of n circles. Find n.

Solution: 10

Let the distance between the two given vertices be 1. If the two given vertices areadjacent, then the other vertices lie on four circles, two of radius 1 and two of radius√

2. If the two vertices are separated by a diagonal of a face of the cube, then thelocus of possible vertices adjacent to both of them is a circle of radius 1

2, the locus of

4

possible vertices adjacent to exactly one of them is two circles of radius√

22

, and the

locus of possible vertices adjacent to neither of them is a circle of radius√

32

. If the twogiven vertices are separated by a long diagonal, then each of the other vertices lie onone of two circles of radius

√2

3, for a total of 10 circles.

13. Triangle ABC has AB = 1, BC =√

7, and CA =√

3. Let `1 be the line through Aperpendicular to AB, `2 the line through B perpendicular to AC, and P the point ofintersection of `1 and `2. Find PC.

Solution: 3

By the Law of Cosines, ∠BAC = cos−1 3+1−72√

3= cos−1(−

√3

2) = 150◦. If we let Q be the

intersection of `2 and AC, we notice that ∠QBA = 90◦ − ∠QAB = 90◦ − 30◦ = 60◦.It follows that triangle ABP is a 30-60-90 triangle and thus PB = 2 and PA =

√3.

Finally, we have ∠PAC = 360◦ − (90◦ + 150◦) = 120◦, and

PC = (PA2 + AC2 − 2PA · AC cos 120◦)1/2 = (3 + 3 + 3)1/2 = 3.

A B

P

C

Q

14. Three noncollinear points and a line ` are given in the plane. Suppose no two of thepoints lie on a line parallel to ` (or ` itself). There are exactly n lines perpendicularto ` with the following property: the three circles with centers at the given points andtangent to the line all concur at some point. Find all possible values of n.

Solution: 1

The condition for the line is that each of the three points lies at an equal distance fromthe line as from some fixed point; in other words, the line is the directrix of a parabolacontaining the three points. Three noncollinear points in the coordinate plane deter-mine a quadratic polynomial in x unless two of the points have the same x-coordinate.Therefore, given the direction of the directrix, three noncollinear points determine aparabola, unless two of the points lie on a line perpendicular to the directrix. Thiscase is ruled out by the given condition, so the answer is 1.

15. Let S be the set of lattice points inside the circle x2 + y2 = 11. Let M be the greatestarea of any triangle with vertices in S. How many triangles with vertices in S havearea M?

Solution: 16

The boundary of the convex hull of S consists of points with (x, y) or (y, x) = (0,±3),(±1,±3), and (±2,±2). For any triangle T with vertices in S, we can increase itsarea by moving a vertex not on the boundary to some point on the boundary. Thus,

5

if T has area M , its vertices are all on the boundary of S. The next step is to see(either by inspection or by noting that T has area no larger than that of an equilateraltriangle inscribed in a circle of radius

√10, which has area less than 13) that M = 12.

There are 16 triangles with area 12, all congruent to one of the following three: vertices(2, 2), (1,−3), and (−3, 1); vertices (3,−1), (−3,−1), and (1, 3); or vertices (3,−1),(−3,−1), and (0, 3).

16. A regular octahedron has a side length of 1. What is the distance between two oppositefaces?

Solution:√

6/3

Imagine orienting the octahedron so that the two opposite faces are horizontal. Projectonto a horizontal plane; these two faces are congruent equilateral triangles which (whenprojected) have the same center and opposite orientations. Hence, the vertices of theoctahedron project to the vertices of a regular hexagon ABCDEF .

A

B

CD

E

F

O M

Let O be the center of the hexagon and M the midpoint of AC. Now ABM is a30-60-90 triangle, so AB = AM/(

√3/2) = (1/2)/(

√3/2) =

√3/3. If we let d denote

the desired vertical distance between the opposite faces (which project to ACE andBDF ), then by the Pythagorean Theorem, AB2 + d2 = 12, so d =

√1− AB2 =

√6/3.

17. Compute

2

√

23

√2

4

√2

5√

2 · · ·.

Solution: 2e−1

Taking the base 2 logarithm of the expression gives

1 +1

2

(1 +

1

3

(1 +

1

4(1 + · · ·)

))= 1 +

1

2!+

1

3!+

1

4!+ · · · = e− 1.

Therefore the expression is just 2e−1.

18. If a, b, and c are random real numbers from 0 to 1, independently and uniformly chosen,what is the average (expected) value of the smallest of a, b, and c?

Solution: 1/4

Let d be a fourth random variable, also chosen uniformly from [0, 1]. For fixed a, b,and c, the probability that d < min{a, b, c} is evidently equal to min{a, b, c}. Hence,

6

if we average over all choices of a, b, c, the average value of min{a, b, c} is equal tothe probability that, when a, b, c, and d are independently randomly chosen, d <min{a, b, c}, i.e., that d is the smallest of the four variables. On the other hand, bysymmetry, the probability that d is the smallest of the four is simply equal to 1/4, sothat is our answer.

19. Regular tetrahedron ABCD is projected onto a plane sending A, B, C, and D toA′, B′, C ′, and D′ respectively. Suppose A′B′C ′D′ is a convex quadrilateral withA′B′ = A′D′ and C ′B′ = C ′D′, and suppose that the area of A′B′C ′D′ = 4. Giventhese conditions, the set of possible lengths of AB consists of all real numbers in theinterval [a, b). Compute b.

Solution: 2 4√

6

The value of b occurs when the quadrilateral A′B′C ′D′ degenerates to an isoscelestriangle. This occurs when the altitude from A to BCD is parallel to the plane. Lets = AB. Then the altitude from A intersects the center E of face BCD. Since

EB = s√3, it follows that A′C ′ = AE =

√s2 − s2

3= s

√6

3. Then since BD is parallel to

the plane, B′D′ = s. Then the area of A′B′C ′D′ is 4 = 12· s2

√6

3, implying s2 = 4

√6, or

s = 2 4√

6.

20. If n is a positive integer, let s(n) denote the sum of the digits of n. We say that nis zesty if there exist positive integers x and y greater than 1 such that xy = n ands(x)s(y) = s(n). How many zesty two-digit numbers are there?

Solution: 34

Let n be a zesty two-digit number, and let x and y be as in the problem statement.Clearly if both x and y are one-digit numbers, then s(x)s(y) = n 6= s(n). Thus eitherx is a two-digit number or y is. Assume without loss of generality that it is x. Ifx = 10a + b, 1 ≤ a ≤ 9 and 0 ≤ b ≤ 9, then n = 10ay + by. If both ay and by areless than 10, then s(n) = ay + by, but if either is at least 10, then s(n) < ay + by. Itfollows that the two digits of n share a common factor greater than 1, namely y. It isnow easy to count the zesty two-digit numbers by first digit starting with 2; there area total of 5 + 4 + 5 + 2 + 7 + 2 + 5 + 4 = 34.

21. In triangle ABC with altitude AD, ∠BAC = 45◦, DB = 3, and CD = 2. Find thearea of triangle ABC.

Solution: 15

Suppose first that D lies between B and C. Let ABC be inscribed in circle ω, andextend AD to intersect ω again at E. Note that A subtends a quarter of the circle, soin particular, the chord through C perpendicular to BC and parallel to AD has lengthBC = 5. Therefore, AD = 5 +DE. By power of a point, 6 = BD ·DC = AD ·DE =AD2 − 5AD, implying AD = 6, so the area of ABC is 1

2BC · AD = 15.

If D does not lie between B and C, then BC = 1, so A lies on a circle of radius√

2/2through B and C. But then it is easy to check that the perpendicular to BC throughD cannot intersect the circle, a contradiction.

7

B

C

DE

A

22. Find{ln(1 + e)}+ {ln(1 + e2)}+ {ln(1 + e4)}+ {ln(1 + e8)}+ · · · ,

where {x} = x− bxc denotes the fractional part of x.

Solution: 1− ln(e− 1)

Since ln(1 + e2k) is just larger than 2k, its fractional part is ln(1 + e2k

) − ln e2k

=ln(1 + e−2k

). But now notice that

n∏

k=0

(1 + x2k

) = 1 + x+ x2 + · · ·+ x2n+1−1.

(This is easily proven by induction or by noting that every nonnegative integer lessthan 2n+1 has a unique (n+1)-bit binary expansion.) If |x| < 1, this product convergesto 1

1−xas n goes to infinity. Therefore,

∞∑

k=0

ln(1 + e−2k

) = ln∞∏

k=0

(1 + (e−1)2k

) = ln1

1− e−1= ln

e

e− 1= 1− ln(e− 1).

23. The sides of a regular hexagon are trisected, resulting in 18 points, including vertices.These points, starting with a vertex, are numbered clockwise as A1, A2, . . . , A18. Theline segment AkAk+4 is drawn for k = 1, 4, 7, 10, 13, 16, where indices are taken modulo18. These segments define a region containing the center of the hexagon. Find theratio of the area of this region to the area of the large hexagon.

Solution: 9/13

Let us assume all sides are of side length 3. Consider the triangle A1A4A5. LetP be the point of intersection of A1A5 with A4A8. This is a vertex of the innerhexagon. Then ∠A4A1A5 = ∠A5A4P , by symmetry. It follows that A1A4A5 ∼ A4PA5.Also, ∠A1A4A5 = 120◦, so by the Law of Cosines A1A5 =

√13. It follows that

PA5 = (A4A5) · (A4A5)/(A1A5) = 1/√

13. Let Q be the intersection of A1A5 andA16A2. By similar reasoning, A1Q = 3/

√13, so PQ = A1A5 − A1Q − PA5 = 9/

√13.

By symmetry, the inner region is a regular hexagon with side length 9/√

13. Hencethe ratio of the area of the smaller to larger hexagon is (3/

√13)2 = 9/13.

8

24. In the base 10 arithmetic problem HMMT +GUTS = ROUND, each distinct letterrepresents a different digit, and leading zeroes are not allowed. What is the maximumpossible value of ROUND?

Solution: 16352

Clearly R = 1, and from the hundreds column, M = 0 or 9. Since H +G = 9 + O or10 + O, it is easy to see that O can be at most 7, in which case H and G must be 8and 9, so M = 0. But because of the tens column, we must have S + T ≥ 10, and infact since D cannot be 0 or 1, S + T ≥ 12, which is impossible given the remainingchoices. Therefore, O is at most 6.

Suppose O = 6 and M = 9. Then we must have H and G be 7 and 8. With theremaining digits 0, 2, 3, 4, and 5, we must have in the ones column that T and S are2 and 3, which leaves no possibility for N . If instead M = 0, then H and G are 7 and9. Since again S + T ≥ 12 and N = T + 1, the only possibility is S = 8, T = 4, andN = 5, giving ROUND = 16352 = 7004 + 9348 = 9004 + 7348.

25. An ant starts at one vertex of a tetrahedron. Each minute it walks along a randomedge to an adjacent vertex. What is the probability that after one hour the ant windsup at the same vertex it started at?

Solution: (359 + 1)/(4 · 359)

Let pn be the probability that the ant is at the original vertex after n minutes; thenp0 = 1. The chance that the ant is at each of the other three vertices after n minutes is13(1−pn). Since the ant can only walk to the original vertex from one of the three others,

and at each there is a 13

probability of doing so, we have that pn+1 = 13(1 − pn). Let

qn = pn− 14. Substituting this into the recurrence, we find that qn+1 = 1

4+ 1

3(−qn− 3

4) =

−13qn. Since q0 = 3

4, qn = 3

4· (−1

3

)n. In particular, this implies that

p60 =1

4+ q60 =

1

4+

3

4· 1

360=

359 + 1

4 · 359.

26. In triangle ABC, AC = 3AB. Let AD bisect angle A with D lying on BC, and let Ebe the foot of the perpendicular from C to AD. Find [ABD]/[CDE]. (Here, [XY Z]denotes the area of triangle XY Z).

Solution: 1/3

By the Angle Bisector Theorem, DC/DB = AC/AB = 3. We will show that AD =DE. Let CE intersect AB at F . Then since AE bisects angle A, AF = AC = 3AB,and EF = EC. Let G be the midpoint of BF . Then BG = GF , so GE ‖ BC. Butthen since B is the midpoint of AG, D must be the midpoint of AE, as desired. Then[ABD]/[CDE] = (AD ·BD)/(ED · CD) = 1/3.

9

A B G F

E

C

D

27. In a chess-playing club, some of the players take lessons from other players. It ispossible (but not necessary) for two players both to take lessons from each other. Itso happens that for any three distinct members of the club, A, B, and C, exactly oneof the following three statements is true: A takes lessons from B; B takes lessons fromC; C takes lessons from A. What is the largest number of players there can be?

Solution: 4

If P , Q, R, S, and T are any five distinct players, then consider all pairs A,B ∈{P,Q,R, S, T} such that A takes lessons from B. Each pair contributes to exactlythree triples (A,B,C) (one for each of the choices of C distinct from A and B); threetriples (C,A,B); and three triples (B,C,A). On the other hand, there are 5×4×3 = 60ordered triples of distinct players among these five, and each includes exactly one ofour lesson-taking pairs. That means that there are 60/9 such pairs. But this numberisn’t an integer, so there cannot be five distinct people in the club.

On the other hand, there can be four people, P , Q, R, and S: let P and Q both takelessons from each other, and let R and S both take lessons from each other; it is easyto check that this meets the conditions. Thus the maximum number of players is 4.

28. There are three pairs of real numbers (x1, y1), (x2, y2), and (x3, y3) that satisfy both

x3 − 3xy2 = 2005 and y3 − 3x2y = 2004. Compute(1− x1

y1

)(1− x2

y2

)(1− x3

y3

).

Solution: 1/1002

By the given, 2004(x3 − 3xy2) − 2005(y3 − 3x2y) = 0. Dividing both sides by y3 andsetting t = x

yyields 2004(t3 − 3t) − 2005(1 − 3t2) = 0. A quick check shows that this

cubic has three real roots. Since the three roots are precisely x1

y1, x2

y2, and x3

y3, we must

have 2004(t3 − 3t)− 2005(1− 3t2) = 2004(t− x1

y1

)(t− x2

y2

)(t− x3

y3

). Therefore,

(1− x1

y1

)(1− x2

y2

)(1− x3

y3

)=

2004(13 − 3(1))− 2005(1− 3(1)2)

2004=

1

1002.

29. Let n > 0 be an integer. Each face of a regular tetrahedron is painted in one of n colors(the faces are not necessarily painted different colors.) Suppose there are n3 possiblecolorings, where rotations, but not reflections, of the same coloring are considered thesame. Find all possible values of n.

Solution: 1, 11

10

We count the possible number of colorings. If four colors are used, there are twodifferent colorings that are mirror images of each other, for a total of 2

(n4

)colorings. If

three colors are used, we choose one color to use twice (which determines the coloring),for a total of 3

(n3

)colorings. If two colors are used, we can either choose one of those

colors and color three faces with it, or we can color two faces each color, for a total of3(

n2

)colorings. Finally, we can also use only one color, for

(n1

)colorings. This gives a

total of

2

(n

4

)+ 3

(n

3

)+ 3

(n

2

)+

(n

1

)=

1

12n2(n2 + 11)

colorings. Setting this equal to n3, we get the equation n2(n2 + 11) = 12n3, or equiva-lently n2(n− 1)(n− 11) = 0, giving the answers 1 and 11.

30. A cuboctahedron is a polyhedron whose faces are squares and equilateral triangles suchthat two squares and two triangles alternate around each vertex, as shown.

What is the volume of a cuboctahedron of side length 1?

Solution: 5√

2/3

We can construct a cube such that the vertices of the cuboctahedron are the midpointsof the edges of the cube.

Let s be the side length of this cube. Now, the cuboctahedron is obtained from thecube by cutting a tetrahedron from each corner. Each such tetrahedron has a base inthe form of an isosceles right triangle of area (s/2)2/2 and height s/2 for a volume of(s/2)3/6. The total volume of the cuboctahedron is therefore

s3 − 8 · (s/2)3/6 = 5s3/6.

Now, the side of the cuboctahedron is the hypotenuse of an isosceles right triangleof leg s/2; thus 1 = (s/2)

√2, giving s =

√2, so the volume of the cuboctahedron is

5√

2/3.

11

31. The L shape made by adjoining three congruent squares can be subdivided into foursmaller L shapes.

Each of these can in turn be subdivided, and so forth. If we perform 2005 successivesubdivisions, how many of the 42005 L’s left at the end will be in the same orientationas the original one?

Solution: 42004 + 22004

After n successive subdivisions, let an be the number of small L’s in the same orientationas the original one; let bn be the number of small L’s that have this orientation rotatedcounterclockwise 90◦; let cn be the number of small L’s that are rotated 180◦; andlet dn be the number of small L’s that are rotated 270◦. When an L is subdivided,it produces two smaller L’s of the same orientation, one of each of the neighboringorientations, and none of the opposite orientation. Therefore,

(an+1, bn+1, cn+1, dn+1) = (dn + 2an + bn, an + 2bn + cn, bn + 2cn + dn, cn + 2dn + an).

It is now straightforward to show by induction that

(an, bn, cn, dn) = (4n−1 + 2n−1, 4n−1, 4n−1 − 2n−1, 4n−1)

for each n ≥ 1. In particular, our desired answer is a2005 = 42004 + 22004.

32. Let a1 = 3, and for n ≥ 1, let an+1 = (n+ 1)an − n. Find the smallest m ≥ 2005 suchthat am+1 − 1 | a2

m − 1.

Solution: 2010

We will show that an = 2 · n! + 1 by induction. Indeed, the claim is obvious for n = 1,and (n+ 1)(2 · n! + 1)− n = 2 · (n+ 1)! + 1. Then we wish to find m ≥ 2005 such that2(m + 1)! | 4(m!)2 + 4m!, or dividing by 2 ·m!, we want m + 1 | 2(m! + 1). Supposem+1 is composite. Then it has a proper divisor d > 2, and since d | m!, we must haved | 2, which is impossible. Therefore, m+1 must be prime, and if this is the case, thenm+ 1 | m! + 1 by Wilson’s Theorem. Therefore, since the smallest prime greater than2005 is 2011, the smallest possible value of m is 2010.

33. Triangle ABC has incircle ω which touches AB at C1, BC at A1, and CA at B1. LetA2 be the reflection of A1 over the midpoint of BC, and define B2 and C2 similarly.Let A3 be the intersection of AA2 with ω that is closer to A, and define B3 and C3

similarly. If AB = 9, BC = 10, and CA = 13, find [A3B3C3]/[ABC]. (Here [XY Z]denotes the area of triangle XY Z.)

Solution: 14/65

Notice that A2 is the point of tangency of the excircle opposite A to BC. Therefore, byconsidering the homothety centered at A taking the excircle to the incircle, we noticethat A3 is the intersection of ω and the tangent line parallel to BC. It follows that

12

A1B1C1 is congruent to A3B3C3 by reflecting through the center of ω. We thereforeneed only find [A1B1C1]/[ABC]. Since

[A1BC1]

[ABC]=A1B ·BC1

AB ·BC =((9 + 10− 13)/2)2

9 · 10=

1

10,

and likewise [A1B1C]/[ABC] = 49/130 and [AB1C1]/[ABC] = 4/13, we get that

[A3B3C3]

[ABC]= 1− 1

10− 49

130− 4

13=

14

65.

34. A regular octahedron ABCDEF is given such that AD, BE, and CF are perpen-dicular. Let G, H, and I lie on edges AB, BC, and CA respectively such thatAGGB

= BHHC

= CIIA

= ρ. For some choice of ρ > 1, GH, HI, and IG are three edgesof a regular icosahedron, eight of whose faces are inscribed in the faces of ABCDEF .Find ρ.

Solution: (1 +√

5)/2

Let J lie on edge CE such that EJJC

= ρ. Then we must have that HIJ is another face ofthe icosahedron, so in particular, HI = HJ . But since BC and CE are perpendicular,HJ = HC

√2. By the Law of Cosines, HI2 = HC2 + CI2 − 2HC · CI cos 60◦ =

HC2(1 + ρ2 − ρ). Therefore, 2 = 1 + ρ2 − ρ, or ρ2 − ρ− 1 = 0, giving ρ = 1+√

52

.

A

C

B

D

E

I

HJ

G

35. Let p = 224036583−1, the largest prime currently known. For how many positive integersc do the quadratics ±x2 ± px± c all have rational roots?

Solution: 0

This is equivalent to both discriminants p2±4c being squares. In other words, p2 mustbe the average of two squares a2 and b2. Note that a and b must have the same parity,and that (a+b

2)2 + (a−b

2)2 = a2+b2

2= p2. Therefore, p must be the hypotenuse in a

Pythagorean triple. Such triples are parametrized by k(m2 − n2, 2mn,m2 + n2). Butp ≡ 3 (mod 4) and is therefore not the sum of two squares. This implies that p is notthe hypotenuse of any Pythagorean triple, so the answer is 0.

13

36. One hundred people are in line to see a movie. Each person wants to sit in the frontrow, which contains one hundred seats, and each has a favorite seat, chosen randomlyand independently. They enter the row one at a time from the far right. As they walk,if they reach their favorite seat, they sit, but to avoid stepping over people, if theyencounter a person already seated, they sit to that person’s right. If the seat furthestto the right is already taken, they sit in a different row. What is the most likely numberof people that will get to sit in the first row?

Solution: 10

Let S(i) be the favorite seat of the ith person, counting from the right. Let P (n) bethe probability that at least n people get to sit. At least n people sit if and only ifS(1) ≥ n, S(2) ≥ n− 1, . . ., S(n) ≥ 1. This has probability:

P (n) =100− (n− 1)

100· 100− (n− 2)

100· · · 100

100=

100!

(100− n)! · 100n.

The probability, Q(n), that exactly n people sit is

P (n)− P (n+ 1) =100!

(100− n)! · 100n− 100!

(99− n)! · 100n+1=

100! · n(100− n)! · 100n+1

.

Now,

Q(n)

Q(n− 1)=

100! · n(100− n)! · 100n+1

· (101− n)! · 100n

100! · (n− 1)=n(101− n)

100(n− 1)=

101n− n2

100n− 100,

which is greater than 1 exactly when n2 − n− 100 < 0, that is, for n ≤ 10. Therefore,the maximum value of Q(n) occurs for n = 10.

37. Let a1, a2, . . . , a2005 be real numbers such that

a1 · 1 + a2 · 2 + a3 · 3 + · · · + a2005 · 2005 = 0a1 · 12 + a2 · 22 + a3 · 32 + · · · + a2005 · 20052 = 0a1 · 13 + a2 · 23 + a3 · 33 + · · · + a2005 · 20053 = 0

......

......

...a1 · 12004 + a2 · 22004 + a3 · 32004 + · · · + a2005 · 20052004 = 0

and

a1 · 12005 + a2 · 22005 + a3 · 32005 + · · · + a2005 · 20052005 = 1.

What is the value of a1?

Solution: 1/2004!

The polynomial p(x) = x(x − 2)(x − 3) · · · (x − 2005)/2004! has zero constant term,has the numbers 2, 3, . . . , 2005 as roots, and satisfies p(1) = 1. Multiplying the nthequation by the coefficient of xn in the polynomial p(x) and summing over all n gives

a1p(1) + a2p(2) + a3p(3) + · · ·+ a2005p(2005) = 1/2004!

(since the leading coefficient is 1/2004!). The left side just reduces to a1, so 1/2004! isthe answer.

14

38. In how many ways can the set of ordered pairs of integers be colored red and blue suchthat for all a and b, the points (a, b), (−1− b, a+1), and (1− b, a− 1) are all the samecolor?

Solution: 16

Let ϕ1 and ϕ2 be 90◦ counterclockwise rotations about (−1, 0) and (1, 0), respectively.Then ϕ1(a, b) = (−1 − b, a + 1), and ϕ2(a, b) = (1 − b, a − 1). Therefore, the possiblecolorings are precisely those preserved under these rotations. Since ϕ1(1, 0) = (−1, 2),the colorings must also be preserved under 90◦ rotations about (−1, 2). Similarly, onecan show that they must be preserved under rotations about any point (x, y), where xis odd and y is even. Decompose the lattice points as follows:

L1 = {(x, y) | x+ y ≡ 0 (mod 2)}L2 = {(x, y) | x ≡ y − 1 ≡ 0 (mod 2)}L3 = {(x, y) | x+ y − 1 ≡ y − x+ 1 ≡ 0 (mod 4)}L4 = {(x, y) | x+ y + 1 ≡ y − x− 1 ≡ 0 (mod 4)}

Within any of these sublattices, any point can be brought to any other through appro-priate rotations, but no point can be brought to any point in a different sublattice. Itfollows that every sublattice must be colored in one color, but that different sublatticescan be colored differently. Since each of these sublattices can be colored in one of twocolors, there are 24 = 16 possible colorings.

1

1

1 1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

2

2

2 2

2

2

2

2

2

2

2 21 1 1

3

33

3

34

4

4

4

1

1

1

3

4

4

1

39. How many regions of the plane are bounded by the graph of

x6 − x5 + 3x4y2 + 10x3y2 + 3x2y4 − 5xy4 + y6 = 0?

Solution: 5

The left-hand side decomposes as

(x6 + 3x4y2 + 3x2y4 + y6)− (x5 − 10x3y2 + 5xy4) = (x2 + y2)3 − (x5 − 10x3y2 + 5xy4).

Now, note that

(x+ iy)5 = x5 + 5ix4y − 10x3y2 − 10ix2y3 + 5xy4 + iy5,

so that our function is just (x2 + y2)3 − <((x+ iy)5). Switching to polar coordinates,this is r6−<(r5(cos θ+ i sin θ)5) = r6− r5 cos 5θ by de Moivre’s rule. The graph of ourfunction is then the graph of r6 − r5 cos 5θ = 0, or, more suitably, of r = cos 5θ. Thisis a five-petal rose, so the answer is 5.

15

40. In a town of n people, a governing council is elected as follows: each person casts onevote for some person in the town, and anyone that receives at least five votes is electedto council. Let c(n) denote the average number of people elected to council if everyonevotes randomly. Find limn→∞ c(n)/n.

Solution: 1− 65/24e

Let ck(n) denote the expected number of people that will receive exactly k votes. Wewill show that limn→∞ ck(n)/n = 1/(e · k!). The probability that any given personreceives exactly k votes, which is the same as the average proportion of people thatreceive exactly k votes, is

(n

k

)·(

1

n

)k

·(n− 1

n

)n−k

=

(n− 1

n

)n

· n(n− 1) · · · (n− k + 1)

k! · (n− 1)k.

Taking the limit as n→∞ and noting that limn→∞(1− 1

n

)n= 1

egives that the limit

is 1/(e · k!), as desired. Therefore, the limit of the average proportion of the town thatreceives at least five votes is

1− 1

e

(1

0!+

1

1!+

1

2!+

1

3!+

1

4!

)= 1− 65

24e.

41. There are 42 stepping stones in a pond, arranged along a circle. You are standingon one of the stones. You would like to jump among the stones so that you movecounterclockwise by either 1 stone or 7 stones at each jump. Moreover, you would liketo do this in such a way that you visit each stone (except for the starting spot) exactlyonce before returning to your initial stone for the first time. In how many ways canyou do this?

Solution: 63

Number the stones 0, 1, . . . , 41, treating the numbers as values modulo 42, and let rn

be the length of your jump from stone n. If you jump from stone n to n+ 7, then youcannot jump from stone n+ 6 to n+ 7 and so must jump from n+ 6 to n+ 13. Thatis, if rn = 7, then rn+6 = 7 also. It follows that the 7 values rn, rn+6, rn+12, . . . , rn+36

are all equal: if one of them is 7, then by the preceding argument applied repeatedly,all of them must be 7, and otherwise all of them are 1. Now, for n = 0, 1, 2, . . . , 42,let sn be the stone you are on after n jumps. Then sn+1 = sn + rsn , and we havesn+1 = sn + rsn ≡ sn + 1 (mod 6). By induction, sn+i ≡ sn + i (mod 6); in particular

16

sn+6 ≡ sn, so rsn+6 = rsn . That is, the sequence of jump lengths is periodic with period6 and so is uniquely determined by the first 6 jumps. So this gives us at most 26 = 64possible sequences of jumps rs0 , rs1 , . . . , rs41 .

Now, the condition that you visit each stone exactly once before returning to theoriginal stone just means that s0, s1, . . . , s41 are distinct and s42 = s0. If all jumps arelength 7, then s6 = s0, so this cannot happen. On the other hand, if the jumps are notall of length 7, then we claim s0, . . . , s41 are indeed all distinct. Indeed, suppose si = sj

for some 0 ≤ i < j < 42. Since sj ≡ si + (j − i) (mod 6), we have j ≡ i (mod 6), soj − i = 6k for some k. Moreover, since the sequence of jump lengths has period 6, wehave

si+6 − si = si+12 − si+6 = · · · = si+6k − si+6(k−1).

Calling this common value l, we have kl ≡ 0 mod 42. But l is divisible by 6, andj − i < 42⇒ k < 7 means that k is not divisible by 7, so l must be. So l, the sum ofsix successive jump lengths, is divisible by 42. Hence the jumps must all be of length7, as claimed.

This shows that, for the 64−1 = 63 sequences of jumps that have period 6 and are notall of length 7, you do indeed reach every stone once before returning to the startingpoint.

42. In how many ways can 6 purple balls and 6 green balls be placed into a 4 × 4 grid ofboxes such that every row and column contains two balls of one color and one ball ofthe other color? Only one ball may be placed in each box, and rotations and reflectionsof a single configuration are considered different.

Solution: 5184

In each row or column, exactly one box is left empty. There are 4! = 24 ways to choosethe empty spots. Once that has been done, there are 6 ways to choose which two rowshave 2 purple balls each. Now, assume without loss of generality that boxes (1, 1),(2, 2), (3, 3), and (4, 4) are the empty ones, and that rows 1 and 2 have two purpleballs each. Let A, B, C, and D denote the 2 × 2 squares in the top left, top right,bottom left, and bottom right corners, respectively (so A is formed by the first tworows and first two columns, etc.). Let a, b, c, and d denote the number of purple ballsin A, B, C, and D, respectively. Then 0 ≤ a, d ≤ 2, a+ b = 4, and b+ d ≤ 4, so a ≥ d.

Now suppose we are given the numbers a and d, satisfying 0 ≤ d ≤ a ≤ 2. Fortunately,the numbers of ways to color the balls in A, B, C, and D are independent of eachother. For example, given a = 1 and d = 0, there are 2 ways to color A and 1 way tocolor D and, no matter how the coloring of A is done, there are always 2 ways to colorB and 3 ways to color C. The numbers of ways to choose the colors of all the balls isas follows:

a\d 0 1 20 1 · (1 · 2) · 1 = 2 0 01 2 · (2 · 3) · 1 = 12 2 · (1 · 1) · 2 = 4 02 1 · (2 · 2) · 1 = 4 1 · (3 · 2) · 2 = 12 1 · (2 · 1) · 1 = 2

In each square above, the four factors are the number of ways of arranging the balls in A,B, C, and D, respectively. Summing this over all pairs (a, d) satisfying 0 ≤ d ≤ a ≤ 2gives a total of 36. The answer is therefore 24 · 6 · 36 = 5184.

17

43. Write down an integer N between 0 and 20 inclusive. If more than N teams writedown N , your score is N ; otherwise it is 0.

Remark: Well, maybe you can get 1 point...

44. Write down a set S of positive integers, all greater than 1, such that for each x ∈ S, xis a proper divisor of (P/x) + 1, where P is the product of all the elements of S. Yourscore is 2n, where n = |S|.Remark: This question was posed by Stefan Znám in 1972. Solutions exist for |S| ≥ 5,but an explicit solution has been found only for |S| ≤ 13. The simplest solution is S ={2, 3, 11, 23, 31}. A solution for |S| = 7 is given by S = {2, 3, 11, 17, 101, 149, 3109}.For more information, see http://mathworld.wolfram.com/ZnamsProblem.html.

45. A binary word is a finite sequence of 0’s and 1’s. A square subword is a subsequenceconsisting of two identical chunks next to each other. For example, the word 100101011contains the square subwords 00, 0101 (twice), 1010, and 11.

Find a long binary word containing a small number of square subwords. Specifically,write down a binary word of any length n ≤ 50. Your score will be max{0, n − s},where s is the number of occurrences of square subwords. (That is, each differentsquare subword will be counted according to the number of times it appears.)

Remark: See http://www.combinatorics.org/Volume 10/PDF/v10i1r12.pdf foranalysis of this problem. The maximum possible score is on the order of 25. In general,the minimum number of square subwords of a word of length n tends to roughly 0.551nas n→∞.

18

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Team Round A

Disconnected Domino Rally [175]

On an infinite checkerboard, the union of any two distinct unit squares is called a (dis-connected) domino. A domino is said to be of type (a, b), with a ≤ b integers not both zero,if the centers of the two squares are separated by a distance of a in one orthogonal directionand b in the other. (For instance, an ordinary connected domino is of type (0, 1), and adomino of type (1, 2) contains two squares separated by a knight’s move.)

Each of the three pairs of squares above forms a domino of type (1, 2).

Two dominoes are said to be congruent if they are of the same type. A rectangle is said tobe (a, b)-tileable if it can be partitioned into dominoes of type (a, b).

1. [15] Prove that for any two types of dominoes, there exists a rectangle that can betiled by dominoes of either type.

2. [25] Suppose 0 < a ≤ b and 4 - mn. Prove that the number of ways in which an m×nrectangle can be partitioned into dominoes of type (a, b) is even.

3. [10] Show that no rectangle of the form 1× k or 2× n, where 4 - n, is (1, 2)-tileable.

4. [35] Show that all other rectangles of even area are (1, 2)-tileable.

5. [25] Show that for b even, there exists some M such that for every n > M , a 2b × nrectangle is (1, b)-tileable.

6. [40] Show that for b even, there exists some M such that for every m,n > M with mneven, an m× n rectangle is (1, b)-tileable.

7. [25] Prove that neither of the previous two problems holds if b is odd.

An Interlude — Discovering One’s Roots [100]

A kth root of unity is any complex number ω such that ωk = 1. You may use the followingfacts: if ω 6= 1, then

1 + ω + ω2 + · · ·+ ωk−1 = 0,

and if 1, ω, . . . , ωk−1 are distinct, then

(xk − 1) = (x− 1)(x− ω)(x− ω2) · · · (x− ωk−1).

1

8. [25] Suppose x is a fifth root of unity. Find, in radical form, all possible values of

2x+1

1 + x+

x

1 + x2+

x2

1 + x3+

x3

1 + x4.

9. [25] Let A1A2 . . . Ak be a regular k-gon inscribed in a circle of radius 1, and let Pbe a point lying on or inside the circumcircle. Find the maximum possible value of(PA1)(PA2) · · · (PAk).

10. [25] Let P be a regular k-gon inscribed in a circle of radius 1. Find the sum of thesquares of the lengths of all the sides and diagonals of P .

11. [25] Let P (x) = anxn + an−1x

n−1 + · · · + a0 be a polynomial with real coefficients,an 6= 0. Suppose every root of P is a root of unity, but P (1) 6= 0. Show that thecoefficients of P are symmetric; that is, show that an = a0, an−1 = a1, . . .

Early Re-tile-ment [125]

Let S = {s0, . . . , sn} be a finite set of integers, and define S + k = {s0 + k, . . . , sn + k}. Wesay that two sets S and T are equivalent, written S ∼ T , if T = S + k for some k. Given a(possibly infinite) set of integers A, we say that S tiles A if A can be partitioned into subsetsequivalent to S. Such a partition is called a tiling of A by S.

12. [20] Suppose the elements of A are either bounded below or bounded above. Showthat if S tiles A, then it does so uniquely, i.e., there is a unique tiling of A by S.

13. [35] Let B be a set of integers either bounded below or bounded above. Then showthat if S tiles all other integers Z\B, then S tiles all integers Z.

14. [35] Suppose S tiles the natural numbers N. Show that S tiles the set {1, 2, . . . , k} forsome positive integer k.

15. [35] Suppose S tiles N. Show that S is symmetric; that is, if −S = {−sn, . . . ,−s0},show that S ∼ −S.

2

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Team Round A — Solutions

Disconnected Domino Rally [175]

On an infinite checkerboard, the union of any two distinct unit squares is called a (dis-connected) domino. A domino is said to be of type (a, b), with a ≤ b integers not both zero,if the centers of the two squares are separated by a distance of a in one orthogonal directionand b in the other. (For instance, an ordinary connected domino is of type (0, 1), and adomino of type (1, 2) contains two squares separated by a knight’s move.)

Each of the three pairs of squares above forms a domino of type (1, 2).

Two dominoes are said to be congruent if they are of the same type. A rectangle is said tobe (a, b)-tileable if it can be partitioned into dominoes of type (a, b).

1. [15] Prove that for any two types of dominoes, there exists a rectangle that can betiled by dominoes of either type.

Solution: Note that a type (a, b) domino tiles a max{1, 2a}×2b rectangle (see diagramfor a > 0). Then both type (a, b) and type (a′, b′) dominoes tile a (max{1, 2a} ·max{1, 2a′})× (2b · 2b′) rectangle.

b b

a

a

2. [25] Suppose 0 < a ≤ b and 4 - mn. Prove that the number of ways in which an m×nrectangle can be partitioned into dominoes of type (a, b) is even.

Solution: If the rectangle is tileable, it can be partitioned into an odd number ofdominoes. Consider the reflection of the partitioned rectangle over one axis. Thisgives another partition of the rectangle. In fact, it cannot be the same partition, forsuppose it were. Then we can pair each domino with its reflected image, but sincethere are an odd number of dominoes, one must reflect into itself. Since a > 0, this isnot possible. Therefore, we can pair off partitions and their reflections, and it followsthat the total number of partitions is even.

1

3. [10] Show that no rectangle of the form 1× k or 2× n, where 4 - n, is (1, 2)-tileable.

Solution: The claim is obvious for 1×k rectangles. For the others, color the first twocolumns black, the next two white, the next two black, etc. Each (1, 2) domino willcontain one square of each color, so in order to be tileable, the rectangle must containthe same number of black and white squares. This is the case only when 4 | n.

4. [35] Show that all other rectangles of even area are (1, 2)-tileable.

Solution: First, we demonstrate that there exist (1, 2)-tilings of 2 × 4, 3 × 4, 3 × 6,and 5× 6 rectangles.

Now, notice that by combining these rectangles, we can form any rectangle of evenarea other than those described in the previous problem: using the first rectangle, wecan form any 2× n rectangle with 4 | n. By combining the first two, we can form anym× 4 rectangle with m ≥ 2, and by combining the last three, we can form any m× 6rectangle with m ≥ 3. From these, we can form any m× n rectangle with m ≥ 3 andn even and greater than 2, completing the proof.

5. [25] Show that for b even, there exists some M such that for every n > M , a 2b × nrectangle is (1, b)-tileable.

Solution: Recall from above that we can tile a 2 × 2b rectangle. Four columns of a(b + 1) × 2b rectangle can be tiled as shown below, and repeating this b

2times tiles

the entire rectangle. Since any integer at least b can be written as a positive linearcombination of 2 and b+ 1, we can tile any 2b× n rectangle for n ≥ b.

6. [40] Show that for b even, there exists some M such that for every m,n > M with mneven, an m× n rectangle is (1, b)-tileable.

Solution: By the diagram below, it is possible to tile a (2b+ 2)× (4b+ 1) rectangle.Since we can already tile a (2b+ 2)× 2b rectangle by above, and 2b is relatively primeto 4b + 1, this will allow us to tile any (2b + 2) × n rectangle for n sufficiently large.Combining this with the previous problem, this will allow us to tile any m×n rectanglefor m and n sufficiently large and m even, completing the proof.

To tile the (2b+ 2)× (4b+ 1) rectangle, we first tile the following piece:

2

2b-1 2b

b+1

2

This is then combined with two 2×2b rectangles, a 2b× b rectangle, and a 2b× (2b+1)rectangle as follows:

2 x 2b

2b x 2b+1

2 x 2b

2b x b

7. [25] Prove that neither of the previous two problems holds if b is odd.

Solution: Color the grid black and white in checkerboard fashion. Then if b is odd,the two squares that make up a (1, b) domino always have the same color. Therefore,for an m × n rectangle to be (1, b)-tileable, it must have an even number of squaresof each color. Then for any M , we can choose m and n larger than M such that n isodd and 4 - m. A 2b× n rectangle and an m× n rectangle then contain bn and mn/2squares of each color, respectively. Since both bn and mn/2 are odd, neither of theserectangles is (1, b)-tileable.

An Interlude — Discovering One’s Roots [100]

A kth root of unity is any complex number ω such that ωk = 1. You may use the followingfacts: if ω 6= 1, then

1 + ω + ω2 + · · ·+ ωk−1 = 0,

and if 1, ω, . . . , ωk−1 are distinct, then

(xk − 1) = (x− 1)(x− ω)(x− ω2) · · · (x− ωk−1).

8. [25] Suppose x is a fifth root of unity. Find, in radical form, all possible values of

2x+1

1 + x+

x

1 + x2+

x2

1 + x3+

x3

1 + x4.

3

Solution: Note that

x

1 + x2+

x2

1 + x3=

x6

1 + x2+

x4

x2 + x5=x4 + x6

1 + x2= x4 =

1

x, and

1

1 + x+

x3

1 + x4=

x5

1 + x+

x4

x+ x5=x4 + x5

1 + x= x4 =

1

x.

Therefore, the sum is just 2x + 2x. If x = 1, this is 4. Otherwise, let y = x + 1

x. Then

x satisfies

0 = 1 + x+ x2 + x3 + x4 =

(x2 +

1

x2+ 2

)+

(x+

1

x

)− 1 = y2 + y − 1,

so solving this quadratic yields y = −1±√5

2, or 2y = −1±√5. Since each value of y can

correspond to only 2 possible values of x, and there are 4 possible values of x besides1, both of these values for y are possible, which yields the answers, 4 and −1±√5.

9. [25] Let A1A2 . . . Ak be a regular k-gon inscribed in a circle of radius 1, and let Pbe a point lying on or inside the circumcircle. Find the maximum possible value of(PA1)(PA2) · · · (PAk).

Solution: Place the vertices at the kth roots of unity, 1, ω, . . . , ωk−1, and place P atsome complex number p. Then

((PA1)(PA2) · · · (PAk))2 =

k−1∏i=0

|p− ωi|2

= |pk − 1|2,since xk − 1 = (x − 1)(x − ω) · · · (x − ωk−1). This is maximized when pk is as far aspossible from 1, which occurs when pk = −1. Therefore, the maximum possible valueof (PA1)(PA2) · · · (PAk) is 2.

10. [25] Let P be a regular k-gon inscribed in a circle of radius 1. Find the sum of thesquares of the lengths of all the sides and diagonals of P .

Solution: Place the vertices of P at the kth roots of unity, 1, ω, ω2, . . . , ωk−1 . Wewill first calculate the sum of the squares of the lengths of the sides and diagonals thatcontain the vertex 1. This is

k−1∑i=0

|1− ωi|2 =k−1∑i=0

(1− ωi)(1− ω̄i)

=k−1∑i=0

(2− ωi − ω̄i)

= 2k − 2k−1∑i=0

ωi

= 2k,

using the fact that 1 + ω + · · · + ωk−1 = 0. Now, by symmetry, this is the sum of thesquares of the lengths of the sides and diagonals emanating from any vertex. Sincethere are k vertices and each segment has two endpoints, the total sum is 2k ·k/2 = k2.

4

11. [25] Let P (x) = anxn + an−1x

n−1 + · · · + a0 be a polynomial with real coefficients,an 6= 0. Suppose every root of P is a root of unity, but P (1) 6= 0. Show that thecoefficients of P are symmetric; that is, show that an = a0, an−1 = a1, . . .

Solution: Since the coefficients of P are real, the complex conjugates of the roots ofP are also roots of P . Now, if x is a root of unity, then x−1 = x̄. But the roots of

xnP (x−1) = a0xn + a1x

n−1 + · · ·+ an

are then just the complex conjugates of the roots of P , so they are the roots of P .Therefore, P (x) and xnP (x−1) differ by a constant multiple c. Since an = ca0 anda0 = can, c is either 1 or −1. But if it were −1, then

P (1) = an + an−1 + · · ·+ a0 =1

2((an + a0) + (an−1 + a1) + · · ·+ (a0 + an)) = 0,

a contradiction. Therefore c = 1, giving the result.

Early Re-tile-ment [125]

Let S = {s0, . . . , sn} be a finite set of integers, and define S + k = {s0 + k, . . . , sn + k}. Wesay that two sets S and T are equivalent, written S ∼ T , if T = S + k for some k. Given a(possibly infinite) set of integers A, we say that S tiles A if A can be partitioned into subsetsequivalent to S. Such a partition is called a tiling of A by S.

12. [20] Suppose the elements of A are either bounded below or bounded above. Showthat if S tiles A, then it does so uniquely, i.e., there is a unique tiling of A by S.

Solution: Assume A is bounded below; the other case is analogous. In choosingthe tiling of A, note that there is a unique choice for the set S0 that contains theminimum element of A. But then there is a unique choice for the set S1 that containsthe minimum element of A\S0. Continuing in this manner, there is a unique choicefor the set containing the minimum element not yet covered, so we see that the tilingis uniquely determined.

13. [35] Let B be a set of integers either bounded below or bounded above. Then showthat if S tiles all other integers Z\B, then S tiles all integers Z.

Solution: Assume B is bounded above; the other case is analogous. Let a be the dif-ference between the largest and smallest element of S. Denote the sets in the partitionof Z\B by Sk, k ∈ Z, such that the minimum element of Sk, which we will denote ck,is strictly increasing as k increases. Since B is bounded above, there exists some k0

such that ck0 is larger than all the elements of B. Let

Tl =∞⋃

k=l

Sk.

Suppose l ≥ k0. Note that any element in Sk, k < l, is at most cl − 1 + a. Therefore,Tl contains all integers that are at least cl + a. Since the minimum element of Tl iscl, Tl is completely determined by which of the integers cl + 1, cl + 2, . . . , cl + a− 1 itcontains. This implies that there are at most 2a−1 possible nonequivalent sets Tl whenl ≥ k0 (here we extend the notion of equivalence to infinite sets in the natural way.)By the Pigeonhole Principle, there must then be some l2 > l1 ≥ k0 such that Tl1 ∼ Tl2 .But then it is easy to see that the set Sl1 ∪ Sl1+1 ∪ · · · ∪ Sl2−1 tiles Z, so S tiles Z.

5

14. [35] Suppose S tiles the natural numbers N. Show that S tiles the set {1, 2, . . . , k} forsome positive integer k.

Solution: Using the notation from above, we can find l1 < l2 such that Tl1 ∼ Tl2 . Bythe same argument as in problem 12, as long as Tl1 6= N, there is a unique choice forSl1−1 that contains the largest integer not in Tl1 . Since the same can be said for Tl2 ,we must have that Tl1−1 ∼ Tl2−1. Continuing in this manner, we find that there mustexist some l for which N ∼ Tl; then S tiles N\Tl = {1, 2, · · · , cl − 1}.

15. [35] Suppose S tiles N. Show that S is symmetric; that is, if −S = {−sn, . . . ,−s0},show that S ∼ −S.

Solution: Assume without loss of generality that the minimum element of S is 0. Bythe previous problem, S tiles the set {1, 2, . . . , k} for some positive integer k. Then letP (x) be the polynomial

∑ni=0 x

si . To say that the set {1, 2, . . . , k}, or equivalently theset {0, 1, . . . , k − 1}, is tiled by S is to say that there is some polynomial Q(x) withcoefficients 0 or 1 such that P (x)Q(x) = 1+x+ · · ·+xk−1 = (xk−1)/(x−1). It followsthat all the roots of P (x) are roots of unity, but P (1) 6= 0. By question 11 above, thisimplies that P (x) is symmetric. Therefore, s0 + sn = s1 + sn−1 = · · · = sn + s0, so Sis symmetric.

6

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Team Round B

Disconnected Domino Rally [150]

On an infinite checkerboard, the union of any two distinct unit squares is called a (dis-connected) domino. A domino is said to be of type (a, b), with a ≤ b integers not both zero,if the centers of the two squares are separated by a distance of a in one orthogonal directionand b in the other. (For instance, an ordinary connected domino is of type (0, 1), and adomino of type (1, 2) contains two squares separated by a knight’s move.)

Each of the three pairs of squares above forms a domino of type (1, 2).

Two dominoes are said to be congruent if they are of the same type. A rectangle is said tobe (a, b)-tileable if it can be partitioned into dominoes of type (a, b).

1. [15] Let 0 < m ≤ n be integers. How many different (i.e., noncongruent) dominoescan be formed by choosing two squares of an m× n array?

2. [10] What are the dimensions of the rectangle of smallest area that is (a, b)-tileable?

3. [20] Prove that every (a, b)-tileable rectangle contains a rectangle of these dimensions.

4. [30] Prove that an m× n rectangle is (b, b)-tileable if and only if 2b | m and 2b | n.

5. [35] Prove that an m× n rectangle is (0, b)-tileable if and only if 2b | m or 2b | n.

6. [40] Let k be an integer such that k | a and k | b. Prove that if an m× n rectangle is(a, b)-tileable, then 2k | m or 2k | n.

An Interlude — Discovering One’s Roots [100]

A kth root of unity is any complex number ω such that ωk = 1.

7. [15] Find a real, irreducible quartic polynomial with leading coefficient 1 whose rootsare all twelfth roots of unity.

8. [25] Let x and y be two kth roots of unity. Prove that (x+ y)k is real.

9. [30] Let x and y be two distinct roots of unity. Prove that x+ y is also a root of unityif and only if y

xis a cube root of unity.

10. [30] Let x, y, and z be three roots of unity. Prove that x+ y+ z is also a root of unityif and only if x+ y = 0, y + z = 0, or z + x = 0.

1

Early Re-tile-ment [150]

Let S = {s0, . . . , sn} be a finite set of integers, and define S + k = {s0 + k, . . . , sn + k}.We say that S and T are equivalent, written S ∼ T , if T = S + k for some k. Given a(possibly infinite) set of integers A, we say that S tiles A if A can be partitioned into subsetsequivalent to S. Such a partition is called a tiling of A by S.

11. [20] Find all sets S with minimum element 1 that tile A = {1, . . . , 12}.12. [35] Let A be a finite set with more than one element. Prove that the number of

nonequivalent sets S which tile A is always even.

13. [25] Exhibit a set S which tiles the integers Z but not the natural numbers N.

14. [30] Suppose that S tiles the set of all integer cubes. Prove that S has only oneelement.

15. [40] Suppose that S tiles the set of odd prime numbers. Prove that S has only oneelement.

2

Harvard-MIT Mathematics TournamentFebruary 19, 2005

Team Round B — Solutions

Disconnected Domino Rally [150]

On an infinite checkerboard, the union of any two distinct unit squares is called a (dis-connected) domino. A domino is said to be of type (a, b), with a ≤ b integers not both zero,if the centers of the two squares are separated by a distance of a in one orthogonal directionand b in the other. (For instance, an ordinary connected domino is of type (0, 1), and adomino of type (1, 2) contains two squares separated by a knight’s move.)

Each of the three pairs of squares above forms a domino of type (1, 2).

Two dominoes are said to be congruent if they are of the same type. A rectangle is said tobe (a, b)-tileable if it can be partitioned into dominoes of type (a, b).

1. [15] Let 0 < m ≤ n be integers. How many different (i.e., noncongruent) dominoescan be formed by choosing two squares of an m× n array?

Solution: We must have 0 ≤ a < m, 0 ≤ b < n, a ≤ b, and a and b not both 0. Thenumber of pairs (a, b) with b < a < m is m(m− 1)/2, so the answer is

mn− m(m− 1)

2− 1 = mn− m2 −m+ 2

2.

2. [10] What are the dimensions of the rectangle of smallest area that is (a, b)-tileable?

Solution: If a = 0, then a 1×2b rectangle is tileable in an obvious way. If a > 0, thena 2a× 2b rectangle is tileable by dividing the rectangle into quarters and pairing eachsquare with a square from the diagonally opposite quarter. The answer is thereforemax{1, 2a} × 2b. Minimality of area follows from the next question.

b b

a

a

1

3. [20] Prove that every (a, b)-tileable rectangle contains a rectangle of these dimensions.

Solution: An m×n rectangle, m ≤ n, does not contain a max{1, 2a}×2b rectangle ifand only if m < 2a or n < 2b. But if either of these is the case, then the square closestto the center of the rectangle cannot be paired with any other square of the rectangleto form a domino of type (a, b), so the rectangle cannot be (a, b)-tileable.

4. [30] Prove that an m× n rectangle is (b, b)-tileable if and only if 2b | m and 2b | n.

Solution: Color the first b rows of an m×n rectangle black, the next b white, the nextb black, etc. Any (b, b) domino covers one square of each color, so for the rectangle tobe (b, b)-tileable, there must be the same number of black squares as white squares.This is possible only when 2b | m. Similarly, we must have 2b | n. It is easy to exhibita tiling of all such rectangles, proving the claim. (It is also possible to prove this usingthe lemma described below.)

5. [35] Prove that an m× n rectangle is (0, b)-tileable if and only if 2b | m or 2b | n.

Solution: It is easy to exhibit a tiling of such a rectangle. The other direction followsfrom below. (It is also possible to prove this using a coloring argument as above:starting in one corner, divide the board into b× b blocks and color them checkerboardfashion. The details are left to the reader.)

6. [40] Let k be an integer such that k | a and k | b. Prove that if an m× n rectangle is(a, b)-tileable, then 2k | m or 2k | n.

Solution: We prove the following lemma.

Lemma. Let k be a positive integer such that k|a and k|b. Then an m×n rectangle is(a, b)-tileable if and only if an m′ × n′ rectangle is (a

k, b

k)-tileable for

⌊mk

⌋ ≤ m′ ≤ ⌈mk

⌉and

⌊nk

⌋ ≤ n′ ≤ ⌈nk

⌉. (Here, bxc denotes the greatest integer less than or equal to x,

while dxe denotes the least integer greater than or equal to x.)

Proof. Number the rows and columns in order. For each pair 0 ≤ i, j < k, considerthe set of squares in a row congruent to i modulo k and in a column congruent to jmodulo k. If one square of a type (a, b) domino lies in this set, then so does the other.We can therefore partition the rectangle into these sets and then tile these sets instead.Each such set is a rectangular array of dimensions m′× n′, with

⌊mk

⌋ ≤ m′ ≤ ⌈mk

⌉and⌊

nk

⌋ ≤ n′ ≤ ⌈nk

⌉, and a type (a, b) domino on the original rectangle is a type (a

k, b

k)

domino on this new array. Since all possible pairs (m′, n′) occur, the result follows.

Suppose 2k - m and 2k - n. Then at least one of⌊

mk

⌋and

⌈mk

⌉is odd, so we can choose

m′ odd. Likewise we can choose n′ odd. But then an m′ × n′ rectangle has odd areaand so cannot be tileable, implying that the m× n rectangle is not tileable.

An Interlude — Discovering One’s Roots [100]

A kth root of unity is any complex number ω such that ωk = 1.

2

7. [15] Find a real, irreducible quartic polynomial with leading coefficient 1 whose rootsare all twelfth roots of unity.

Solution: All twelfth roots of unity are roots of

x12 − 1 = (x6 − 1)(x6 + 1)

= (x3 − 1)(x3 + 1)(x6 + 1)

= (x− 1)(x2 + x+ 1)(x+ 1)(x2 − x+ 1)(x2 + 1)(x4 − x2 + 1),

so the answer is x4 − x2 + 1.

8. [25] Let x and y be two kth roots of unity. Prove that (x+ y)k is real.

Solution: Note that

(x+ y)k =k∑

i=0

(k

i

)xiyk−i

=1

2

k∑i=0

(k

i

)(xiyk−i + xk−iyi)

by pairing the ith and (k − i)th terms. But xk−iyi = (xiyk−i)−1 since x and y are kthroots of unity. Moreover, since x and y have absolute value 1, so does xiyk−i, so xk−iyi

is in fact its complex conjugate. It follows that their sum is real, thus so is (x+ y)k.

This can also be shown geometrically. The argument of x (the angle between the vectorx and the positive x-axis) is an integer multiple of 2π

k, as is the argument of y. Since

x + y bisects the angle between x and y, its argument is an integer multiple of πk.

Multiplying this angle by k gives a multiple of π, so (x+ y)k is real.

9. [30] Let x and y be two distinct roots of unity. Prove that x+ y is also a root of unityif and only if y

xis a cube root of unity.

Solution: This is easiest to see geometrically. The vectors corresponding to x, y, and−x − y sum to 0, so they form a triangle. In order for them all to be roots of unity,they must all have length one, so the triangle must be equilateral. Therefore the anglebetween x and y is ±2π

3, that is, y

xis a cube root of unity.

10. [30] Let x, y, and z be three roots of unity. Prove that x+ y+ z is also a root of unityif and only if x+ y = 0, y + z = 0, or z + x = 0.

Solution: Again, we consider the geometric picture. Arrange the vectors x, y, z, and−x − y − z so as to form a quadrilateral. If they are all roots of unity, they form aquadrilateral all of whose side lengths are 1. If the quadrilateral is degenerate, then twoof the vectors sum to 0, which implies the result. But even if it is not degenerate, thequadrilateral must be a rhombus, and since opposite sides of a rhombus are parallel,this again implies that two of the four roots of unity sum to 0.

Early Re-tile-ment [150]

Let S = {s0, . . . , sn} be a finite set of integers, and define S + k = {s0 + k, . . . , sn + k}.We say that S and T are equivalent, written S ∼ T , if T = S + k for some k. Given a(possibly infinite) set of integers A, we say that S tiles A if A can be partitioned into subsetsequivalent to S. Such a partition is called a tiling of A by S.

3

11. [20] Find all sets S with minimum element 1 that tile A = {1, . . . , 12}.Solution: This can be done by brute force. Alternatively, note that if P (x) and Q(x)are polynomials with coefficients either 0 or 1 with P (x)Q(x) = x + x2 + · · · + x12,then the set consisting of the exponents of nonzero terms in P tiles A. Either way,we find that S is one of the following: {1}, {1, 2}, {1, 3}, {1, 4}, {1, 7}, {1, 2, 3},{1, 3, 5}, {1, 5, 9}, {1, 2, 3, 4}, {1, 2, 7, 8}, {1, 4, 7, 10}, {1, 2, 3, 4, 5, 6}, {1, 2, 3, 7, 8, 9},{1, 2, 5, 6, 9, 10}, {1, 3, 5, 7, 9, 11}, or A itself.

12. [35] Let A be a finite set with more than one element. Prove that the number ofnonequivalent sets S which tile A is always even.

Solution: Suppose A can be partitioned into sets S0, . . . , Sm, each equivalent to S.(This partition is unique, simply by choosing S0 to contain the smallest element of A,S1 the smallest element of A not in S0, etc.) Then if Sj = S + tj, each element of Acan be written uniquely as si + tj for some i and j. But then the set T containing alltj also tiles A by translation by the si. We cannot have S and T equivalent, for if so,since A has more than one element, both S and T would as well. This would implythat s0 + t1 = s1 + t0, an overlap in the tiling of A. We can thus pair together S andT , each of which tile A, so that the total number of sets tiling A must be even.

13. [25] Exhibit a set S which tiles the integers Z but not the natural numbers N.

Solution: One example is {1, 3, 4, 6}. Since its elements are all distinct modulo 4, ittiles Z by translation by multiples of 4. On the other hand, it is easy to see that itcannot tile N: 1 is contained in {1, 3, 4, 6}, but then there is no possible set for 2 tobe contained in that does not overlap. Another example is {1, 2, 6}.

14. [30] Suppose that S tiles the set of all integer cubes. Prove that S has only oneelement.

Solution: Let the difference between the smallest and largest element of S be a. Thenthe set equivalent to S that contains b3 can only contain integers between b3 − a andb3 + a, inclusive. But for sufficiently large b, b3 is the only cube in this range, so S canonly have one element.

15. [40] Suppose that S tiles the set of odd prime numbers. Prove that S has only oneelement.

Solution: Consider the set S0 equivalent to S that contains 3. If it contains 5 butnot 7, then the set S1 equivalent to S containing 7 must contain 9, which is not prime.Likewise, S0 cannot contain 7 but not 5, because then the set S1 containing 5 mustcontain 9. Suppose S0 contains 3, 5, and 7. Then any other set S1 of the tiling containselements p, p+ 2, and p+ 4. But not all of these can be prime, because one of them isdivisible by 3. Finally, suppose S0 contains 3 and has second-smallest element p > 7.Then the set S1 containing 5 does not contain 7 but does contain p+ 2, and the set S2

containing 7 contains p+ 4. But as before, not all of p, p+ 2, and p+ 4 can be prime.Therefore S has no second-smallest element, so it has only one element.

4

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Individual Round: Algebra Test

1. Larry can swim from Harvard to MIT (with the current of the Charles River) in 40 minutes,or back (against the current) in 45 minutes. How long does it take him to row from Harvardto MIT, if he rows the return trip in 15 minutes? (Assume that the speed of the current andLarry’s swimming and rowing speeds relative to the current are all constant.) Express youranswer in the format mm:ss.

2. Find all real solutions (x, y) of the system x2 + y = 12 = y2 + x.

3. The train schedule in Hummut is hopelessly unreliable. Train A will enter Intersection Xfrom the west at a random time between 9:00 am and 2:30 pm; each moment in that intervalis equally likely. Train B will enter the same intersection from the north at a random timebetween 9:30 am and 12:30 pm, independent of Train A; again, each moment in the interval isequally likely. If each train takes 45 minutes to clear the intersection, what is the probabilityof a collision today?

4. Let a1, a2, . . . be a sequence defined by a1 = a2 = 1 and an+2 = an+1 + an for n ≥ 1. Find

∞∑n=1

an

4n+1.

5. Tim has a working analog 12-hour clock with two hands that run continuously (instead of,say, jumping on the minute). He also has a clock that runs really slow—at half the correctrate, to be exact. At noon one day, both clocks happen to show the exact time. At any giveninstant, the hands on each clock form an angle between 0◦ and 180◦ inclusive. At how manytimes during that day are the angles on the two clocks equal?

6. Let a, b, c be the roots of x3−9x2+11x−1 = 0, and let s =√

a+√

b+√

c. Find s4−18s2−8s.

7. Letf(x) = x4 − 6x3 + 26x2 − 46x + 65.

Let the roots of f(x) be ak + ibk for k = 1, 2, 3, 4. Given that the ak, bk are all integers, find|b1|+ |b2|+ |b3|+ |b4|.

8. Solve for all complex numbers z such that z4 + 4z2 + 6 = z.

9. Compute the value of the infinite series

∞∑n=2

n4 + 3n2 + 10n + 102n · (n4 + 4)

10. Determine the maximum value attained by

x4 − x2

x6 + 2x3 − 1

over real numbers x > 1.

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Algebra Test: Solutions

1. Larry can swim from Harvard to MIT (with the current of the Charles River) in 40minutes, or back (against the current) in 45 minutes. How long does it take him torow from Harvard to MIT, if he rows the return trip in 15 minutes? (Assume that thespeed of the current and Larry’s swimming and rowing speeds relative to the currentare all constant.) Express your answer in the format mm:ss.

Answer: 14:24Solution: Let the distance between Harvard and MIT be 1, and let c, s, r denotethe speeds of the current and Larry’s swimming and rowing, respectively. Then we aregiven

s + c =1

40=

9

360, s− c =

1

45=

8

360, r − c =

1

15=

24

360,

so

r + c = (s + c)− (s− c) + (r − c) =9− 8 + 24

360=

25

360,

and it takes Larry 360/25 = 14.4 minutes, or 14:24, to row from Harvard to MIT.

2. Find all real solutions (x, y) of the system x2 + y = 12 = y2 + x.

Answer: (3, 3), (−4,−4),

(1 + 3

√5

2,1− 3

√5

2

),

(1− 3

√5

2,1 + 3

√5

2

)Solution: We have x2+y = y2+x which can be written as (x−y)(x+y−1) = 0. Thecase x = y yields x2 +x−12 = 0, hence (x, y) = (3, 3) or (−4,−4). The case y = 1−x

yields x2 + 1 − x − 12 = x2 − x − 11 = 0 which has solutions x = 1±√

1+442

= 1±3√

52

.The other two solutions follow.

3. The train schedule in Hummut is hopelessly unreliable. Train A will enter IntersectionX from the west at a random time between 9:00 am and 2:30 pm; each moment in thatinterval is equally likely. Train B will enter the same intersection from the north ata random time between 9:30 am and 12:30 pm, independent of Train A; again, eachmoment in the interval is equally likely. If each train takes 45 minutes to clear theintersection, what is the probability of a collision today?

Answer: 1348

Solution: Suppose we fix the time at which Train B arrives at Intersection X; thencall the interval during which Train A could arrive (given its schedule) and collide withTrain B the “disaster window.”

We consider two cases:

(i) Train B enters Intersection X between 9:30 and 9:45. If Train B arrives at 9:30,the disaster window is from 9:00 to 10:15, an interval of 11

4hours. If Train B

arrives at 9:45, the disaster window is 112

hours long. Thus, the disaster windowhas an average length of (11

4+ 11

2)÷ 2 = 11

8. From 9:00 to 2:30 is 51

2hours. The

probability of a collision is thus 118÷ 51

2= 1

4.

1

(ii) Train B enters Intersection X between 9:45 and 12:30. Here the disaster windowis always 11

2hours long, so the probability of a collision is 11

2÷ 51

2= 3

11.

From 9:30 to 12:30 is 3 hours. Now case (i) occurs with probability 14÷ 3 = 1

12, and

case (ii) occurs with probability 1112

. The overall probability of a collision is therefore112· 1

4+ 11

12· 3

11= 1

48+ 1

4= 13

48.

4. Let a1, a2, . . . be a sequence defined by a1 = a2 = 1 and an+2 = an+1 + an for n ≥ 1.Find

∞∑n=1

an

4n+1.

Answer: 111

Solution: Let X denote the desired sum. Note that

X =1

42+

1

43+

2

44+

3

45+

5

46+ . . .

4X =1

41+

1

42+

2

43+

3

44+

5

45+

8

46+ . . .

16X =1

40+

1

41+

2

42+

3

43+

5

44+

8

45+

13

46+ . . .

so that X + 4X = 16X − 1, and X = 1/11.

5. Tim has a working analog 12-hour clock with two hands that run continuously (insteadof, say, jumping on the minute). He also has a clock that runs really slow—at halfthe correct rate, to be exact. At noon one day, both clocks happen to show the exacttime. At any given instant, the hands on each clock form an angle between 0◦ and180◦ inclusive. At how many times during that day are the angles on the two clocksequal?

Answer: 33Solution: A tricky thing about this problem may be that the angles on the twoclocks might be reversed and would still count as being the same (for example, bothangles could be 90◦, but the hour hand may be ahead of the minute hand on one clockand behind on the other).

Let x, −12 ≤ x < 12, denote the number of hours since noon. If we take 0◦ to meanupwards to the “XII” and count angles clockwise, then the hour and minute hands ofthe correct clock are at 30x◦ and 360x◦, and those of the slow clock are at 15x◦ and180x◦. The two angles are thus 330x◦ and 165x◦, of course after removing multiples of360◦ and possibly flipping sign; we are looking for solutions to

330x◦ ≡ 165x◦ (mod 360◦) or 330x◦ ≡ −165x◦ (mod 360◦).

In other words,360 | 165x or 360 | 495x.

Or, better yet,165

360x =

11

24x and/or

495

360x =

11

8x

2

must be an integer. Now x is any real number in the range [−12, 12), so 11x/8 rangesin [−16.5, 16.5), an interval that contains 33 integers. For any value of x such that11x/24 is an integer, of course 11x/8 = 3× (11x/24) is also an integer, so the answeris just 33.

6. Let a, b, c be the roots of x3 − 9x2 + 11x − 1 = 0, and let s =√

a +√

b +√

c. Finds4 − 18s2 − 8s.

Answer: −37Solution: First of all, as the left side of the first given equation takes values −1, 2,−7, and 32 when x = 0, 1, 2, and 3, respectively, we know that a, b, and c are distinctpositive reals. Let t =

√ab +

√bc +

√ca, and note that

s2 = a + b + c + 2t = 9 + 2t,

t2 = ab + bc + ca + 2√

abcs = 11 + 2s,

s4 = (9 + 2t)2 = 81 + 36t + 4t2 = 81 + 36t + 44 + 8s = 125 + 36t + 8s,

18s2 = 162 + 36t,

so that s4 − 18s2 − 8s = −37.

7. Letf(x) = x4 − 6x3 + 26x2 − 46x + 65.

Let the roots of f(x) be ak + ibk for k = 1, 2, 3, 4. Given that the ak, bk are all integers,find |b1|+ |b2|+ |b3|+ |b4|.

Answer: 10Solution: The roots of f(x) must come in complex-conjugate pairs. We can thensay that a1 = a2 and b1 = −b2; a3 = a4 and b3 = −b4. The constant term of f(x) isthe product of these, so 5 · 13 = (a1

2 + b12)(a3

2 + b32). Since ak and bk are integers

for all k, and it is simple to check that 1 and i are not roots of f(x), we must havea1

2 + b12 = 5 and a3

2 + b32 = 13. The only possible ways to write these sums with

positive integers is 12 + 22 = 5 and 22 + 32 = 13, so the values of a1 and b1 up to signare 1 and 2; and a3 and b3 up to sign are 2 and 3. From the x3 coefficient of f(x), weget that a1 + a2 + a3 + a4 = 6, so a1 + a3 = 3. From the limits we already have, thistells us that a1 = 1 and a3 = 2. Therefore b1, b2 = ±2 and b3, b4 = ±3, so the requiredsum is 2 + 2 + 3 + 3 = 10.

8. Solve for all complex numbers z such that z4 + 4z2 + 6 = z.

Answer:1± i

√7

2,−1± i

√11

2Solution: Rewrite the given equation as (z2 + 2)

2+ 2 = z. Observe that a solution

to z2 +2 = z is a solution of the quartic by substitution of the left hand side into itself.This gives z = 1±i

√7

2. But now, we know that z2−z +2 divides into (z2 + 2)

2−z +2 =z4 +4z2−z+6. Factoring it out, we obtain (z2 − z + 2) (z2 + z + 3) = z4 +4z2−z+6.

Finally, the second term leads to the solutions z = −1±i√

112

.

9. Compute the value of the infinite series∞∑

n=2

n4 + 3n2 + 10n + 10

2n · (n4 + 4)

3

Answer: 1110

Solution: We employ the difference of squares identity, uncovering the factorizationof the denominator: n4 + 4 = (n2 + 2)2 − (2n)2 = (n2 − 2n + 2)(n2 + 2n + 2). Now,

n4 + 3n2 + 10n + 10

n4 + 4= 1 +

3n2 + 10n + 6

n4 + 4

= 1 +4

n2 − 2n + 2− 1

n2 + 2n + 2

=⇒∞∑

n=2

n4 + 3n2 + 10n + 10

2n · (n4 + 4)=

∞∑n=2

1

2n+

4

2n · (n2 − 2n + 2)− 1

2n · (n2 + 2n + 2)

=1

2+

∞∑n=2

1

2n−2 · ((n− 1)2 + 1)− 1

2n · ((n + 1)2 + 1)

The last series telescopes to 12

+ 110

, which leads to an answer of 12

+ 12

+ 110

= 1110

.

10. Determine the maximum value attained by

x4 − x2

x6 + 2x3 − 1

over real numbers x > 1.

Answer: 16

Solution: We have the following algebra:

x4 − x2

x6 + 2x3 − 1=

x− 1x

x3 + 2− 1x3

=x− 1

x(x− 1

x

)3+ 2 + 3

(x− 1

x

)≤

x− 1x

3(x− 1

x

)+ 3

(x− 1

x

) =1

6

where(x− 1

x

)3+ 1 + 1 ≥ 3

(x− 1

x

)in the denominator was deduced by the AM-GM

inequality. As a quick check, equality holds where x− 1x

= 1 or when x = 1+√

52

.

4

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Individual Round: Calculus Test

1. A nonzero polynomial f(x) with real coefficients has the property that f(x) = f ′(x)f ′′(x).What is the leading coefficient of f(x)?

2. Compute limx→0

ex cos x − 1− x

sin(x2).

3. At time 0, an ant is at (1, 0) and a spider is at (−1, 0). The ant starts walkingcounterclockwise along the unit circle, and the spider starts creeping to the right alongthe x-axis. It so happens that the ant’s horizontal speed is always half the spider’s.What will the shortest distance ever between the ant and the spider be?

4. Compute∞∑

k=1

k4

k!.

5. Compute

∫ 1

0

dx√x + 3√

x.

6. A triangle with vertices at (1003, 0), (1004, 3), and (1005, 1) in the xy-plane is revolvedall the way around the y-axis. Find the volume of the solid thus obtained.

7. Find all positive real numbers c such that the graph of f : R → R given by f(x) =x3− cx has the property that the circle of curvature at any local extremum is centeredat a point on the x-axis.

8. Compute

∫ π/3

0

x tan2(x)dx.

9. Compute the sum of all real numbers x such that

2x6 − 3x5 + 3x4 + x3 − 3x2 + 3x− 1 = 0.

10. Suppose f and g are differentiable functions such that

xg(f(x))f ′(g(x))g′(x) = f(g(x))g′(f(x))f ′(x)

for all real x. Moreover, f is nonnegative and g is positive. Furthermore,∫ a

0

f(g(x))dx = 1− e−2a

2

for all reals a. Given that g(f(0)) = 1, compute the value of g(f(4)).

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Calculus Test: Solutions

1. A nonzero polynomial f(x) with real coefficients has the property that f(x) = f ′(x)f ′′(x).What is the leading coefficient of f(x)?

Answer: 118

Solution: Suppose that the leading term of f(x) is cxn, where c 6= 0. Then theleading terms of f ′(x) and of f ′′(x) are cnxn−1 and cn(n − 1)xn−2, respectively, socxn = cnxn−1 · cn(n− 1)xn−2, which implies that n = (n− 1) + (n− 2), or n = 3, andc = cn · cn(n− 1) = 18c2, or c = 1

18.

2. Compute limx→0

ex cos x − 1− x

sin(x2).

Answer: 12

Solution: Let’s compute all the relevant Maclaurin series expansions, up to thequadratic terms:

x cos x = x + . . . , ex cos x = 1 + x +1

2x2 + . . . , sin(x2) = x2 + . . . ,

so

limx→0

ex cos x − 1− x

sin(x2)= lim

x→0

12x2 + . . .

x2 + . . .=

1

2.

3. At time 0, an ant is at (1, 0) and a spider is at (−1, 0). The ant starts walkingcounterclockwise along the unit circle, and the spider starts creeping to the right alongthe x-axis. It so happens that the ant’s horizontal speed is always half the spider’s.What will the shortest distance ever between the ant and the spider be?

Answer:√

144

Solution: Picture an instant in time where the ant and spider have x-coordinates aand s, respectively. If 1 ≤ s ≤ 3, then a ≤ 0, and the distance between the bugs is atleast 1. If s > 3, then, needless to say the distance between the bugs is at least 2. If−1 ≤ s ≤ 1, then s = 1− 2a, and the distance between the bugs is

√(a− (1− 2a))2 + (1− a2) =

√8a2 − 6a + 2 =

√(8a− 3)2 + 7

8,

which attains the minimum value of√

7/8 when a = 3/8.

4. Compute∞∑

k=1

k4

k!.

Answer: 15eSolution: Define, for non-negative integers n,

Sn :=∞∑

k=0

kn

k!,

1

where 00 = 1 when it occurs. Then S0 = e, and, for n ≥ 1,

Sn =∞∑

k=0

kn

k!=

∞∑k=1

kn

k!=

∞∑k=0

(k + 1)n

(k + 1)!=

∞∑k=0

(k + 1)n−1

k!=

n−1∑i=0

(n− 1

i

)Si,

so we can compute inductively that S1 = e, S2 = 2e, S3 = 5e, and S4 = 15e.

5. Compute ∫ 1

0

dx√x + 3√

x

Answer: 5− 6 ln 2Solution: Writing x = u6 so that dx = 6u5du, we have∫ 1

0

dx√x + 3√

x=

∫ 1

0

6u5du

u3 + u2

= 6

∫ 1

0

u3du

u + 1

= 6

∫ 1

0

(u2 − u + 1− 1

u + 1

)du

= 6

(u3

3− u2

2+ u− ln |u + 1|

∣∣∣∣10

)= 5− 6 ln(2)

6. A triangle with vertices at (1003, 0), (1004, 3), and (1005, 1) in the xy-plane is revolvedall the way around the y-axis. Find the volume of the solid thus obtained.

Answer: 5020πSolution: Let T ⊂ R2 denote the triangle, including its interior. Then T ’s area is5/2, and its centroid is (1004, 4/3), so∫

(x,y)∈T

x dx dy =5

2· 1004 = 2510.

We are interested in the volume ∫(x,y)∈T

2πx dx dy,

but this is just 2π · 2510 = 5020π.

7. Find all positive real numbers c such that the graph of f : R → R given by f(x) =x3− cx has the property that the circle of curvature at any local extremum is centeredat a point on the x-axis.

Answer:√

32

Solution: The equation 0 = f ′(x) = 3x2 − c has two real roots: ±√

c/3. Let

a :=√

c/3. As f ′′(−a) = −6√

c/3 < 0, f has a unique local maximum at x = −a.

2

Because f has half-turn symmetry about the origin, it suffices to consider this localextremum. The radius of curvature at any local extremum is

r(x) =1

|f ′′(x)|=

1

6|x|,

so the condition in the problem is equivalent to

r(−a) = f(−a)

1

6a= −a(a2 − c)

1 = 6a2(c− a2) = 2c(2c/3)

c =√

3/2.

8. Compute ∫ π/3

0

x tan2(x)dx

Answer: π√

33− π2

18− ln 2

Solution: We have∫ π/3

0

x tan2(x)dx =

∫ π/3

0

x

(−1 +

1

cos2(x)

)dx

= −x2

2

∣∣∣∣π/3

0

+

∫ π/3

0

xdx

cos2(x)

= −x2

2

∣∣∣∣π/3

0

+

(x tan(x)

∣∣∣∣π/3

0

−∫ π/3

0

tan(x)dx

)(u = x; dv =

dx

cos2(x))

= −x2

2+ x tan(x) + ln |cos(x)|

∣∣∣∣π/3

0

= −π2

18+

π√

3

3− ln(2)

9. Compute the sum of all real numbers x such that

2x6 − 3x5 + 3x4 + x3 − 3x2 + 3x− 1 = 0

Answer: −12

Solution: The carefully worded problem statement suggests that repeated rootsmight be involved (not to be double counted), as well as complex roots (not to becounted). Let P (x) = 2x6 − 3x5 + 3x4 + x3 − 3x2 + 3x− 1. Now, a is a double root ofthe polynomial P (x) if and only if P (a) = P ′(a) = 0. Hence, we consider the system

P (a) = 2a6 − 3a5 + 3a3 + a3 − 3a2 + 3a− 1 = 0

P ′(a) = 12a5 − 15a4 + 12a3 + 3a2 − 6a + 3 = 0

=⇒ 3a4 + 8a3 − 15a2 + 18a− 7 = 0

37a3 − 57a2 + 57a− 20 = 0

a2 − a + 1 = 0

3

We have used polynomial long division to deduce that any double root must be a rootof a2−a+1! With this information, we can see that P (x) = (x2−x+1)2(2x2 +x−1).The real roots are easily computed via the quadratic formula, leading to an answer of−1

2. In fact the repeated roots were complex.

10. Suppose f and g are differentiable functions such that

xg(f(x))f ′(g(x))g′(x) = f(g(x))g′(f(x))f ′(x)

for all real x. Moreover, f is nonnegative and g is positive. Furthermore,∫ a

0

f(g(x))dx = 1− e−2a

2

for all reals a. Given that g(f(0)) = 1, compute the value of g(f(4)).

Answer: e−16 or 1e16

Solution: Differentiating the given integral with respect to a gives f(g(a)) = e−2a.Now

xd [ln (f(g(x)))]

dx= x

f ′(g(x))g′(x)

f(g(x))=

g′(f(x))f ′(x)

g(f(x))=

d [ln (g(f(x)))]

dx

where the second equals sign follows from the given. Since ln (f(g(x))) = −2x, we have−x2+C = ln (g(f(x))), so g(f(x)) = Ke−x2

. It follows that K = 1 and g(f(4)) = e−16.

4

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Individual Round: Combinatorics Test

1. Vernonia High School has 85 seniors, each of whom plays on at least one of the school’s three varsitysports teams: football, baseball, and lacrosse. It so happens that 74 are on the football team; 26 areon the baseball team; 17 are on both the football and lacrosse teams; 18 are on both the baseball andfootball teams; and 13 are on both the baseball and lacrosse teams. Compute the number of seniorsplaying all three sports, given that twice this number are members of the lacrosse team.

2. Compute2∑

n60=0

n60∑n59=0

· · ·n3∑

n2=0

n2∑n1=0

n1∑n0=0

1.

3. A moth starts at vertex A of a certain cube and is trying to get to vertex B, which is opposite A, infive or fewer “steps,” where a step consists in traveling along an edge from one vertex to another. Themoth will stop as soon as it reaches B. How many ways can the moth achieve its objective?

4. A dot is marked at each vertex of a triangle ABC. Then, 2, 3, and 7 more dots are marked on thesides AB, BC, and CA, respectively. How many triangles have their vertices at these dots?

5. Fifteen freshmen are sitting in a circle around a table, but the course assistant (who remains standing)has made only six copies of today’s handout. No freshman should get more than one handout, and anyfreshman who does not get one should be able to read a neighbor’s. If the freshmen are distinguishablebut the handouts are not, how many ways are there to distribute the six handouts subject to the aboveconditions?

6. For how many ordered triplets (a, b, c) of positive integers less than 10 is the product a× b× c divisibleby 20?

7. Let n be a positive integer, and let Pushover be a game played by two players, standing squarelyfacing each other, pushing each other, where the first person to lose balance loses. At the HMPT,2n+1 competitors, numbered 1 through 2n+1 clockwise, stand in a circle. They are equals in Pushover:whenever two of them face off, each has a 50% probability of victory. The tournament unfolds in n+1rounds. In each round, the referee randomly chooses one of the surviving players, and the players pairoff going clockwise, starting from the chosen one. Each pair faces off in Pushover, and the losers leavethe circle. What is the probability that players 1 and 2n face each other in the last round? Expressyour answer in terms of n.

8. In how many ways can we enter numbers from the set {1, 2, 3, 4} into a 4 × 4 array so that all of thefollowing conditions hold?

(a) Each row contains all four numbers.

(b) Each column contains all four numbers.

(c) Each “quadrant” contains all four numbers. (The quadrants are the four corner 2× 2 squares.)

9. Eight celebrities meet at a party. It so happens that each celebrity shakes hands with exactly twoothers. A fan makes a list of all unordered pairs of celebrities who shook hands with each other. Iforder does not matter, how many different lists are possible?

10. Somewhere in the universe, n students are taking a 10-question math competition. Their collectiveperformance is called laughable if, for some pair of questions, there exist 57 students such that eitherall of them answered both questions correctly or none of them answered both questions correctly.Compute the smallest n such that the performance is necessarily laughable.

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Combinatorics Test: Solutions

Combinatorics Test

1. Vernonia High School has 85 seniors, each of whom plays on at least one of the school’sthree varsity sports teams: football, baseball, and lacrosse. It so happens that 74 areon the football team; 26 are on the baseball team; 17 are on both the football andlacrosse teams; 18 are on both the baseball and football teams; and 13 are on both thebaseball and lacrosse teams. Compute the number of seniors playing all three sports,given that twice this number are members of the lacrosse team.

Answer: 11Solution: Suppose that n seniors play all three sports and that 2n are on thelacrosse team. Then, by the principle of inclusion-exclusion, 85 = (74 + 26 + 2n) −(17 + 18 + 13) + (n) = 100 + 2n− 48 + n = 52 + 3n. It is easily seen that n = 11.

2. Compute2∑

n60=0

n60∑n59=0

· · ·n3∑

n2=0

n2∑n1=0

n1∑n0=0

1.

Answer: 1953Solution: The given sum counts the number of non-decreasing 61-tuples of integers(n0, . . . , n60) from the set {0, 1, 2}. Such 61-tuples are in one-to-one correspondencewith strictly increasing 61-tuples of integers (m0, . . . ,m60) from the set {0, 1, 2, . . . , 62}:simply let mk = nk + k. But the number of such (m0, . . . ,m60) is almost by definition(6361

)=(632

)= 1953.

3. A moth starts at vertex A of a certain cube and is trying to get to vertex B, which isopposite A, in five or fewer “steps,” where a step consists in traveling along an edgefrom one vertex to another. The moth will stop as soon as it reaches B. How manyways can the moth achieve its objective?

Answer: 48Solution: Let X, Y, Z be the three directions in which the moth can intially go. Wecan symbolize the trajectory of the moth by a sequence of stuff from Xs, Y s, and Zsin the obvious way: whenever the moth takes a step in a direction parallel or oppositeto X, we write down X, and so on.

The moth can reach B in either exactly 3 or exactly 5 steps. A path of length 3 mustbe symbolized by XY Z in some order. There are 3! = 6 such orders. A trajectoryof length 5 must by symbolized by XY ZXX, XY ZY Y , or XY ZZZ, in some order,There are 3 · 5!

3!1!1!= 3 · 20 = 60 possibilities here. However, we must remember to

subtract out those trajectories that already arrive at B by the 3rd step: there are3 · 6 = 18 of those. The answer is thus 60− 18 + 6 = 48.

1

4. A dot is marked at each vertex of a triangle ABC. Then, 2, 3, and 7 more dots aremarked on the sides AB, BC, and CA, respectively. How many triangles have theirvertices at these dots?

Answer: 357Solution: Altogether there are 3 + 2 + 3 + 7 = 15 dots, and thus

(153

)= 455

combinations of 3 dots. Of these combinations,(2+23

)+(2+33

)+(2+73

)= 4+10+84 = 98

do not give triangles because they are collinear (the rest do give triangles). Thus455− 98 = 357 different triangles can be formed.

5. Fifteen freshmen are sitting in a circle around a table, but the course assistant (whoremains standing) has made only six copies of today’s handout. No freshman shouldget more than one handout, and any freshman who does not get one should be able toread a neighbor’s. If the freshmen are distinguishable but the handouts are not, howmany ways are there to distribute the six handouts subject to the above conditions?

Answer: 125Solution: Suppose that you are one of the freshmen; then there’s a 6/15 chance thatyou’ll get one of the handouts. We may ask, given that you do get a handout, howmany ways are there to distribute the rest? We need only multiply the answer to thatquestion by 15/6 to answer the original question.

Going clockwise around the table from you, one might write down the sizes of the gapsbetween people with handouts. There are six such gaps, each of size 0–2, and the sumof their sizes must be 15 − 6 = 11. So the gap sizes are either 1, 1, 1, 2, 2, 2 in someorder, or 0, 1, 2, 2, 2, 2 in some order. In the former case, 6!

3!3!= 20 orders are possible;

in the latter, 6!1!1!4!

= 30 are. Altogether, then, there are 20 + 30 = 50 possibilities.

Multiplying this by 15/6, or 5/2, gives 125.

6. For how many ordered triplets (a, b, c) of positive integers less than 10 is the producta× b× c divisible by 20?

Answer: 102Solution: One number must be 5. The other two must have a product divisible by4. Either both are even, or one is divisible by 4 and the other is odd. In the formercase, there are 48 = 3 × 4 × 4 possibilities: 3 positions for the 5, and any of 4 evennumbers to fill the other two. In the latter case, there are 54 = 3×2×9 possibilites: 3positions and 2 choices for the multiple of 4, and 9 ways to fill the other two positionsusing at least one 5.

7. Let n be a positive integer, and let Pushover be a game played by two players, standingsquarely facing each other, pushing each other, where the first person to lose balanceloses. At the HMPT, 2n+1 competitors, numbered 1 through 2n+1 clockwise, stand ina circle. They are equals in Pushover: whenever two of them face off, each has a 50%probability of victory. The tournament unfolds in n + 1 rounds. In each round, thereferee randomly chooses one of the surviving players, and the players pair off goingclockwise, starting from the chosen one. Each pair faces off in Pushover, and the losersleave the circle. What is the probability that players 1 and 2n face each other in thelast round? Express your answer in terms of n.

2

Answer:2n − 1

8n

Solution: At any point during this competition, we shall say that the situation isliving if both players 1 and 2n are still in the running. A living situation is far ifthose two players are diametrically opposite each other, and near otherwise, in whichcase (as one can check inductively) they must be just one person shy of that maximalseparation. At the start of the tournament, the situation is living and near. In eachof rounds 1 to n, a far situation can never become near, and a near situation can staynear or become far with equal likelihood.

In each of rounds 1 to n− 1, a living situation has a 1/4 probability of staying living.Therefore, at the end of round k, where 1 ≤ k ≤ n − 1, the situation is near withprobability 1/8k, and far with probability 1/4k − 1/8k. In round n, a far situation hasa 1/4 probability of staying living, whereas a near situation has only a 1/8 probabilityof staying living. But if the situation is living at the beginning of the last round, itcan only be far, so we can say with complete generality that, at the end of round k,where 1 ≤ k ≤ n, the situation is living and far with probability 1/4k − 1/8k. We areinterested in finding the probability that the situation is living at the end of round n(and hence far); that probability is thus 1

4n − 18n = 2n−1

8n .

8. In how many ways can we enter numbers from the set {1, 2, 3, 4} into a 4× 4 array sothat all of the following conditions hold?

(a) Each row contains all four numbers.

(b) Each column contains all four numbers.

(c) Each “quadrant” contains all four numbers. (The quadrants are the four corner2× 2 squares.)

Answer: 288Solution: Call a filled 4× 4 array satisfying the given conditions cool. There are 4!possibilities for the first row; WLOG, let it be 1 2 3 4. Since each quadrant has tocontain all four numbers, we have exactly four possibilities for the second row, namely:

(i) 3 4 1 2

(ii) 3 4 2 1

(iii) 4 3 1 2

(iv) 4 3 2 1

I claim that the number of cool arrays with (i) is equal to those with (iv), and thatthe number of cool arrays with (ii) is equal to those with (iii). Let’s first consider (i)and (iv). Now, (i) is

1 2 3 4

3 4 1 2

while (iv) is

1 2 3 4

4 3 2 1

3

In (iv), switch 3 and 4 (relabeling doesn’t affect the coolness of the array); then, itbecomes

1 2 4 3

3 4 2 1

Now, interchange the last two columns, which also does not affect the coolness. Thisgives us (i). Hence, the cool arrays with (i) and the cool arrays with (iv) have a 1:1correspondence. Using the exact same argument, we can show that the number of coolarrays with (ii) equals those with (iii).

So we only need consider cases (i) and (ii). It is easy to verify that there are four coolarrays with (i), determined precisely by, say, the first two entries of the third row; andtwo with (ii), determined precisely by, say the first entry of the third row. Hence, theanswer is 4!× (4 + 2)× 2 = 288.

9. Eight celebrities meet at a party. It so happens that each celebrity shakes hands withexactly two others. A fan makes a list of all unordered pairs of celebrities who shookhands with each other. If order does not matter, how many different lists are possible?

Answer: 3507Solution: Let the celebrities get into one or more circles so that each circle hasat least three celebrities, and each celebrity shook hands precisely with his or herneighbors in the circle.

Let’s consider the possible circle sizes:

• There’s one big circle with all 8 celebrities. Depending on the ordering of thepeople in the circle, the fan’s list can still vary. Literally speaking, there are 7!different circles 8 people can make: fix one of the people, and then there are 7choices for the person to the right, 6 for the person after that, and so on. Butthis would be double-counting because, as far as the fan’s list goes, it makes nodifference if we “reverse” the order of all the people. Thus, there are 7!/2 = 2520different possible lists here.

• 5+3. In this case there are(85

)ways to split into the two circles, 4!

2essentially

different ways of ordering the 5-circle, and 2!2

ways for the 3-circle, giving a totalcount of 56 · 12 · 1 = 672.

• 4+4. In this case there are(84

)/2 = 35 ways to split into the two circles (we divide

by 2 because here, unlike in the 5 + 3 case, it does not matter which circle iswhich), and 3!

2= 3 ways of ordering each, giving a total count of 35 · 3 · 3 = 315.

Adding them up, we get 2520 + 672 + 315 = 3507.

10. Somewhere in the universe, n students are taking a 10-question math competition.Their collective performance is called laughable if, for some pair of questions, thereexist 57 students such that either all of them answered both questions correctly ornone of them answered both questions correctly. Compute the smallest n such thatthe performance is necessarily laughable.

Answer: 253Solution: Let ci,j denote the number of students correctly answering questions i andj (1 ≤ i < j ≤ 10), and let wi,j denote the number of students getting both questions

4

wrong. An individual student answers k questions correctly and 10 − k questionsincorrectly. This student answers

(k2

)pairs of questions correctly and

(10−k

2

)pairs of

questions incorrectly. Now observe that(k

2

)+

(10− k

2

)= k2 − 10k + 45 = (k − 5)2 + 20 ≥ 20

Therefore, ∑1≤i<j≤10

ci,j + wi,j ≥ 20n

Now if the performance is not laughable, then ci,j ≤ 56 and wi,j ≤ 56 for all 1 ≤i < j ≤ 10. Observe that there are 2

(102

)= 90 of these variables. Hence, in a boring

performance,

20n ≤∑

1≤i<j≤10

ci,j + wi,j ≤ 90 · 56 = 5040

or n ≤ 252. In particular this implies that if n ≥ 253, the performance is laughable.This is the best bound because

(105

)= 252, and if each of 252 students correctly

answers a different 5 element subset of the 10 questions, then ci,j = wi,j = 56 for all1 ≤ i < j ≤ 10.

5

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Individual Round: General Test, Part 1

1. How many positive integers x are there such that 3x has 3 digits and 4x has four digits?

2. What is the probability that two cards randomly selected (without replacement) froma standard 52-card deck are neither of the same value nor the same suit?

3. A square and an equaliteral triangle together have the property that the area of eachis the perimeter of the other. Find the square’s area.

4. Find √31 +

√31 +

√31 + . . .√

1 +√

1 +√

1 + . . .

.

5. In the plane, what is the length of the shortest path from (−2, 0) to (2, 0) that avoidsthe interior of the unit circle (i.e., circle of radius 1) centered at the origin?

6. Six celebrities meet at a party. It so happens that each celebrity shakes hands withexactly two others. A fan makes a list of all unordered pairs of celebrities who shookhands with each other. If order does not matter, how many different lists are possible?

7. The train schedule in Hummut is hopelessly unreliable. Train A will enter IntersectionX from the west at a random time between 9:00 am and 2:30 pm; each moment in thatinterval is equally likely. Train B will enter the same intersection from the north ata random time between 9:30 am and 12:30 pm, independent of Train A; again, eachmoment in the interval is equally likely. If each train takes 45 minutes to clear theintersection, what is the probability of a collision today?

8. A dot is marked at each vertex of a triangle ABC. Then, 2, 3, and 7 more dots aremarked on the sides AB, BC, and CA, respectively. How many triangles have theirvertices at these dots?

9. Take a unit sphere S, i.e., a sphere with radius 1. Circumscribe a cube C about S,and inscribe a cube D in S, so that every edge of cube C is parallel to some edge ofcube D. What is the shortest possible distance from a point on a face of C to a pointon a face of D?

10. A positive integer n is called “flippant” if n does not end in 0 (when written in decimalnotation) and, moreover, n and the number obtained by reversing the digits of n areboth divisible by 7. How many flippant integers are there between 10 and 1000?

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

General Test, Part 1: Solutions

1. How many positive integers x are there such that 3x has 3 digits and 4x has four digits?

Answer: 84Solution: Note that x must be between 250 and 333, inclusive. There are 84 integersin that interval.

2. What is the probability that two cards randomly selected (without replacement) froma standard 52-card deck are neither of the same value nor the same suit?

Answer: 1217

Solution: After choosing a first card, the second needs to be in one of the otherthree suits and of a different value. Hence, the answer is 3·12

52−1= 12

17.

3. A square and an equaliteral triangle together have the property that the area of eachis the perimeter of the other. Find the square’s area.

Answer: 12 3√

4Solution: Let s, t be the side lengths of the square and triangle, respectively. Then4s =

√3t2

4, and so

3t = s2 =

(√3t2

16

)2

=3t4

28,

so that t3 = 28 and s2 = 3t = 33√

28 = 12 3√

4.

4. Find √31 +

√31 +

√31 + . . .√

1 +√

1 +√

1 + . . .

.

Answer: 6−√

5Solution: Let the numerator be x and the denominator y. Then x2 = 31 + x, so, asx > 0, we have

x =1 +√

1 + 4 · 31

2=

1 + 5√

5

2.

Similarly we compute that

y =1 +√

1 + 4 · 12

=1 +√

5

2,

so thatx

y=

1 + 5√

5

1 +√

5=

1 + 5√

5

1 +√

5· 1−

√5

1−√

5=−24 + 4

√5

−4= 6−

√5.

1

5. In the plane, what is the length of the shortest path from (−2, 0) to (2, 0) that avoidsthe interior of the unit circle (i.e., circle of radius 1) centered at the origin?

Answer: 2√

3 +π

3Solution: The path goes in a line segment tangent to the circle, then an arc of thecircle, then another line segment tangent to the circle. Since one of these tangent linesand a radius of the circle give two legs of a right triangle with hypotenuse the line from(0,0) to (-2,0) or (2,0), the length of each tangent line is

√22 − 12 =

√3. Also, because

these are 30◦-60◦-90◦ right triangles, the angle of the arc is 60◦ and has length π/3.

6. Six celebrities meet at a party. It so happens that each celebrity shakes hands withexactly two others. A fan makes a list of all unordered pairs of celebrities who shookhands with each other. If order does not matter, how many different lists are possible?

Answer: 70Solution: Let the celebrities get into one or more circles so that each circle hasat least three celebrities, and each celebrity shook hands precisely with his or herneighbors in the circle.

Either there is one big circle of all 6 celebrities or else there are two small circles of 3celebrities each.

If there is one big circle of 6, then depending on the ordering of the people in the circle,the fan’s list can still vary. Literally speaking, there are 5! different circles 6 people canmake: fix one of the people, and then there are 5 choices for the person to the right, 4for the person after that, and so on. But this would be double-counting because, as faras the fan’s list goes, it makes no difference if we “reverse” the order of all the people.There are thus 5!/2 = 60 different possible lists here.

If there are two small circles of 3, then there are(63

)different ways the members of

the “first” small circle may be selected. But this, too, is double-counting, because itmakes no difference which circle is termed the “first” and which the “second.” Thereare therefore

(63

)/2 = 10 essentially different ways to split up the people into the two

circles. In a circle of just three, each person shakes hands with both the others, sonaturally the order of people in the circle doesn’t matter. There are thus 10 differentpossible lists here.

(Note that, translated into the language of graph theory, the problem is asking for thenumber of graphs on six labeled vertices such that each vertex has degree two.)

7. The train schedule in Hummut is hopelessly unreliable. Train A will enter IntersectionX from the west at a random time between 9:00 am and 2:30 pm; each moment in thatinterval is equally likely. Train B will enter the same intersection from the north ata random time between 9:30 am and 12:30 pm, independent of Train A; again, eachmoment in the interval is equally likely. If each train takes 45 minutes to clear theintersection, what is the probability of a collision today?

Answer: 1348

Solution: Suppose we fix the time at which Train B arrives at Intersection X; thencall the interval during which Train A could arrive (given its schedule) and collide withTrain B the “disaster window.”

We consider two cases:

2

(i) Train B enters Intersection X between 9:30 and 9:45. If Train B arrives at 9:30,the disaster window is from 9:00 to 10:15, an interval of 11

4hours. If Train B

arrives at 9:45, the disaster window is 112

hours long. Thus, the disaster windowhas an average length of (11

4+ 11

2)÷ 2 = 11

8. From 9:00 to 2:30 is 51

2hours. The

probability of a collision is thus 118÷ 51

2= 1

4.

(ii) Train B enters Intersection X between 9:45 and 12:30. Here the disaster windowis always 11

2hours long, so the probability of a collision is 11

2÷ 51

2= 3

11.

From 9:30 to 12:30 is 3 hours. Now case (i) occurs with probability 14÷ 3 = 1

12, and

case (ii) occurs with probability 1112

. The overall probability of a collision is therefore112· 1

4+ 11

12· 3

11= 1

48+ 1

4= 13

48.

8. A dot is marked at each vertex of a triangle ABC. Then, 2, 3, and 7 more dots aremarked on the sides AB, BC, and CA, respectively. How many triangles have theirvertices at these dots?

Answer: 357Solution: Altogether there are 3 + 2 + 3 + 7 = 15 dots, and thus

(153

)= 455

combinations of 3 dots. Of these combinations,(2+23

)+(2+33

)+(2+73

)= 4+10+84 = 98

do not give triangles because they are collinear (the rest do give triangles). Thus455− 98 = 357 different triangles can be formed.

9. Take a unit sphere S, i.e., a sphere with radius 1. Circumscribe a cube C about S,and inscribe a cube D in S, so that every edge of cube C is parallel to some edge ofcube D. What is the shortest possible distance from a point on a face of C to a pointon a face of D?

Answer:3−√

3

3= 1−

√3

3, or equivalent

Solution: Using the Pythagorean theorem, we know that the length of a diagonal ofa cube of edge length s is s

√3. Since D is inscribed in a sphere that has diameter 2,

this means that its side length is 2/√

3.

The distance from a face of D to a face of C will be the distance between them alongany line perpendicular to both of them; take such a line passing through the center ofS. The distance from the center to any face of D along this line will be half the sidelength of D, or 1/

√3. The distance from the center to the edge of C is the radius of

S, which is 1. Therefore the desired distance is 1− 1/√

3 = 1−√

33

.

10. A positive integer n is called “flippant” if n does not end in 0 (when written in decimalnotation) and, moreover, n and the number obtained by reversing the digits of n areboth divisible by 7. How many flippant integers are there between 10 and 1000?

Answer: 17Solution: We use the notation “|” to mean “divides.”

There is only one flippant 2-digit number, namely 77. Indeed, if 10a + b is flippant(where a, b are integers 1–9), then 7 | 10a + b and 7 | 10b + a. Thus,

7 | 3(10a + b)− (10b + a) = 29a− 7b = a + 7(4a− b),

so that 7 | a, and similarly 7 | b, so we’d better have a = b = 7.

3

There are 16 flippant 3-digit numbers. First consider the 12 palindromic ones (oneswhere the hundreds and units digits are the same): 161, 252, 343, 434, 525, 595, 616,686, 707, 777, 868, and 959. Now consider the general case: suppose 100a + 10b + c isflippant, where a, b, c are integers 1–9. Then 7 | 100a + 10b + c and 7 | 100c + 10b + a,so 7 | (100a + 10b + c)− (100c + 10b + a) = 99(a− c), and so 7 | a− c. In order for thisnot to result in a palindromic integer, we must have a − c = ±7 and, moreover, both100a + 10b + a and 100c + 10b + c must be palindromic flippant integers. Consultingour list above, we find 4 more flippant integers: 168, 259, 861, and 952.

4

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Individual Round: General Test, Part 2

1. Larry can swim from Harvard to MIT (with the current of the Charles River) in 40 minutes, or back(against the current) in 45 minutes. How long does it take him to row from Harvard to MIT, if herows the return trip in 15 minutes? (Assume that the speed of the current and Larry’s swimming androwing speeds relative to the current are all constant.) Express your answer in the format mm:ss.

2. Find22

22 − 1· 32

32 − 1· 42

42 − 1· · · · · 20062

20062 − 1.

3. Let C be the unit circle. Four distinct, smaller congruent circles C1, C2, C3, C4 are internally tangentto C such that Ci is externally tangent to Ci−1 and Ci+1 for i = 1, . . . , 4 where C5 denotes C1 and C0

represents C4. Compute the radius of C1.

4. Vernonia High School has 85 seniors, each of whom plays on at least one of the school’s three varsitysports teams: football, baseball, and lacrosse. It so happens that 74 are on the football team; 26 areon the baseball team; 17 are on both the football and lacrosse teams; 18 are on both the baseball andfootball teams; and 13 are on both the baseball and lacrosse teams. Compute the number of seniorsplaying all three sports, given that twice this number are members of the lacrosse team.

5. If a, b are nonzero real numbers such that a2 + b2 = 8ab, find the value of∣∣∣∣a + b

a− b

∣∣∣∣.6. Octagon ABCDEFGH is equiangular. Given that AB = 1, BC = 2, CD = 3, DE = 4, and

EF = FG = 2, compute the perimeter of the octagon.

7. What is the smallest positive integer n such that n2 and (n + 1)2 both contain the digit 7 but (n + 2)2

does not?

8. Six people, all of different weights, are trying to build a human pyramid: that is, they get into theformation

A

B C

D E F

We say that someone not in the bottom row is “supported by” each of the two closest people beneathher or him. How many different pyramids are possible, if nobody can be supported by anybody oflower weight?

9. Tim has a working analog 12-hour clock with two hands that run continuously (instead of, say, jumpingon the minute). He also has a clock that runs really slow—at half the correct rate, to be exact. Atnoon one day, both clocks happen to show the exact time. At any given instant, the hands on eachclock form an angle between 0◦ and 180◦ inclusive. At how many times during that day are the angleson the two clocks equal?

10. Fifteen freshmen are sitting in a circle around a table, but the course assistant (who remains standing)has made only six copies of today’s handout. No freshman should get more than one handout, and anyfreshman who does not get one should be able to read a neighbor’s. If the freshmen are distinguishablebut the handouts are not, how many ways are there to distribute the six handouts subject to the aboveconditions?

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

General Test, Part 2: Solutions

1. Larry can swim from Harvard to MIT (with the current of the Charles River) in 40minutes, or back (against the current) in 45 minutes. How long does it take him torow from Harvard to MIT, if he rows the return trip in 15 minutes? (Assume that thespeed of the current and Larry’s swimming and rowing speeds relative to the currentare all constant.) Express your answer in the format mm:ss.

Answer: 14:24Solution: Let the distance between Harvard and MIT be 1, and let c, s, r denotethe speeds of the current and Larry’s swimming and rowing, respectively. Then we aregiven

s + c =1

40=

9

360, s− c =

1

45=

8

360, r − c =

1

15=

24

360,

so

r + c = (s + c)− (s− c) + (r − c) =9− 8 + 24

360=

25

360,

and it takes Larry 360/25 = 14.4 minutes, or 14:24, to row from Harvard to MIT.

2. Find22

22 − 1· 32

32 − 1· 42

42 − 1· · · · · 20062

20062 − 1.

Answer:4012

2007Solution:

2006∏k=2

k2

k2 − 1=

2006∏k=2

k2

(k − 1)(k + 1)=

2006∏k=2

k

k − 1

2006∏k=2

k

k + 1=

2006

1· 2

2007=

4012

2007.

3. Let C be the unit circle. Four distinct, smaller congruent circles C1, C2, C3, C4 areinternally tangent to C such that Ci is externally tangent to Ci−1 and Ci+1 for i =1, . . . , 4 where C5 denotes C1 and C0 represents C4. Compute the radius of C1.

Answer:√

2− 1Solution: Let O and O′ be the centers of C and C1 respectively, and let C1 betangent to C, C2, C4 at P , Q, and R respectively. Observe that QORO′ is a square andthat P, O′, and O are collinear. Thus, if r is the desired radius, 1 = r+OO′ = r+r

√2,

so that r = 1√2+1

=√

2− 1.

4. Vernonia High School has 85 seniors, each of whom plays on at least one of the school’sthree varsity sports teams: football, baseball, and lacrosse. It so happens that 74 areon the football team; 26 are on the baseball team; 17 are on both the football andlacrosse teams; 18 are on both the baseball and football teams; and 13 are on both thebaseball and lacrosse teams. Compute the number of seniors playing all three sports,given that twice this number are members of the lacrosse team.

1

Answer: 11Solution: Suppose that n seniors play all three sports and that 2n are on thelacrosse team. Then, by the principle of inclusion-exclusion, 85 = (74 + 26 + 2n) −(17 + 18 + 13) + (n) = 100 + 2n− 48 + n = 52 + 3n. It is easily seen that n = 11.

5. If a, b are nonzero real numbers such that a2 + b2 = 8ab, find the value of

∣∣∣∣a + b

a− b

∣∣∣∣.Answer:

√15

3Solution: Note that∣∣∣∣a + b

a− b

∣∣∣∣ =

√(a + b)2

(a− b)2=

√a2 + b2 + 2ab

a2 + b2 − 2ab=

√10ab

6ab=

√15

3.

6. Octagon ABCDEFGH is equiangular. Given that AB = 1, BC = 2, CD = 3,DE = 4, and EF = FG = 2, compute the perimeter of the octagon.

Answer: 20 +√

2Solution: Extend sides AB, CD,EF,GH to form a rectangle: let X be the intersec-tion of lines GH and AB; Y that of AB and CD; Z that of CD and EF ; and W thatof EF and GH.

As BC = 2, we have BY = Y C =√

2. As DE = 4, we have DZ = ZE = 2√

2. AsFG = 2, we have FW = WG =

√2.

We can compute the dimensions of the rectangle: WX = Y Z = Y C + CD + DZ =3 + 3

√2, and XY = ZW = ZE + EF + FW = 2 + 3

√2. Thus, HX = XA = XY −

AB−BY = 1+2√

2, and so AH =√

2HX = 4+√

2, and GH = WX−WG−HX = 2.The perimeter of the octagon can now be computed by adding up all its sides.

7. What is the smallest positive integer n such that n2 and (n+1)2 both contain the digit7 but (n + 2)2 does not?

Answer: 27Solution: The last digit of a square is never 7. No two-digit squares begin with 7.There are no 3-digit squares beginning with the digits 17, 27, 37, or 47. In fact, thesmallest square containing the digit 7 is 576 = 242. Checking the next few numbers,we see that 252 = 625, 262 = 676, 272 = 729, 282 = 784, and 292 = 841, so the answeris 27.

8. Six people, all of different weights, are trying to build a human pyramid: that is, theyget into the formation

A

B C

D E F

We say that someone not in the bottom row is “supported by” each of the two closestpeople beneath her or him. How many different pyramids are possible, if nobody canbe supported by anybody of lower weight?

2

Answer: 16Solution: Without loss of generality, let the weights of the people be 1, 2, 3, 4, 5,and 6. Clearly we must have A = 1. Then, equally clearly, either B or C must be 2.

Suppose B = 2: Then either C or D must be 3. If C = 3, we have 3! = 6 possibilitiesto fill the bottom row. If D = 3, then C = 4 and we have 2! = 2 possibilities to fill Eand F . Altogether there are 6 + 2 = 8 possibilities in this case.

Suppose C = 2: then, similarly, there are 8 possibilities here.

Altogether there are 8 + 8 = 16 possibilities.

9. Tim has a working analog 12-hour clock with two hands that run continuously (insteadof, say, jumping on the minute). He also has a clock that runs really slow—at halfthe correct rate, to be exact. At noon one day, both clocks happen to show the exacttime. At any given instant, the hands on each clock form an angle between 0◦ and180◦ inclusive. At how many times during that day are the angles on the two clocksequal?

Answer: 33Solution: A tricky thing about this problem may be that the angles on the twoclocks might be reversed and would still count as being the same (for example, bothangles could be 90◦, but the hour hand may be ahead of the minute hand on one clockand behind on the other).

Let x, −12 ≤ x < 12, denote the number of hours since noon. If we take 0◦ to meanupwards to the “XII” and count angles clockwise, then the hour and minute hands ofthe correct clock are at 30x◦ and 360x◦, and those of the slow clock are at 15x◦ and180x◦. The two angles are thus 330x◦ and 165x◦, of course after removing multiples of360◦ and possibly flipping sign; we are looking for solutions to

330x◦ ≡ 165x◦ (mod 360◦) or 330x◦ ≡ −165x◦ (mod 360◦).

In other words,360 | 165x or 360 | 495x.

Or, better yet,165

360x =

11

24x and/or

495

360x =

11

8x

must be an integer. Now x is any real number in the range [−12, 12), so 11x/8 rangesin [−16.5, 16.5), an interval that contains 33 integers. For any value of x such that11x/24 is an integer, of course 11x/8 = 3× (11x/24) is also an integer, so the answeris just 33.

10. Fifteen freshmen are sitting in a circle around a table, but the course assistant (whoremains standing) has made only six copies of today’s handout. No freshman shouldget more than one handout, and any freshman who does not get one should be able toread a neighbor’s. If the freshmen are distinguishable but the handouts are not, howmany ways are there to distribute the six handouts subject to the above conditions?

Answer: 125Solution: Suppose that you are one of the freshmen; then there’s a 6/15 chance thatyou’ll get one of the handouts. We may ask, given that you do get a handout, how

3

many ways are there to distribute the rest? We need only multiply the answer to thatquestion by 15/6 to answer the original question.

Going clockwise around the table from you, one might write down the sizes of the gapsbetween people with handouts. There are six such gaps, each of size 0–2, and the sumof their sizes must be 15 − 6 = 11. So the gap sizes are either 1, 1, 1, 2, 2, 2 in someorder, or 0, 1, 2, 2, 2, 2 in some order. In the former case, 6!

3!3!= 20 orders are possible;

in the latter, 6!1!1!4!

= 30 are. Altogether, then, there are 20 + 30 = 50 possibilities.

Multiplying this by 15/6, or 5/2, gives 125.

4

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Individual Round: Geometry Test

1. Octagon ABCDEFGH is equiangular. Given that AB = 1, BC = 2, CD = 3,DE = 4, and EF = FG = 2, compute the perimeter of the octagon.

2. Suppose ABC is a scalene right triangle, and P is the point on hypotenuse AC suchthat ∠ABP = 45◦. Given that AP = 1 and CP = 2, compute the area of ABC.

3. Let A, B, C, and D be points on a circle such that AB = 11 and CD = 19. PointP is on segment AB with AP = 6, and Q is on segment CD with CQ = 7. The linethrough P and Q intersects the circle at X and Y . If PQ = 27, find XY .

4. Let ABC be a triangle such that AB = 2, CA = 3, and BC = 4. A semicircle with itsdiameter on BC is tangent to AB and AC. Compute the area of the semicircle.

5. Triangle ABC has side lengths AB = 2√

5, BC = 1, and CA = 5. Point D is on sideAC such that CD = 1, and F is a point such that BF = 2 and CF = 3. Let E be theintersection of lines AB and DF . Find the area of CDEB.

6. A circle of radius t is tangent to the hypotenuse, the incircle, and one leg of an isoscelesright triangle with inradius r = 1 + sin π

8. Find rt.

7. Suppose ABCD is an isosceles trapezoid in which AB ‖ CD. Two mutually externallytangent circles ω1 and ω2 are inscribed in ABCD such that ω1 is tangent to AB, BC,and CD while ω2 is tangent to AB, DA, and CD. Given that AB = 1, CD = 6,compute the radius of either circle.

8. Triangle ABC has a right angle at B. Point D lies on side BC such that 3∠BAD =∠BAC. Given AC = 2 and CD = 1, compute BD.

9. Four spheres, each of radius r, lie inside a regular tetrahedron with side length 1 suchthat each sphere is tangent to three faces of the tetrahedron and to the other threespheres. Find r.

10. Triangle ABC has side lengths AB = 65, BC = 33, and AC = 56. Find the radius ofthe circle tangent to sides AC and BC and to the circumcircle of triangle ABC.

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Geometry Test: Solutions

1. Octagon ABCDEFGH is equiangular. Given that AB = 1, BC = 2, CD = 3,DE = 4, and EF = FG = 2, compute the perimeter of the octagon.

Answer: 20 +√

2Solution: Extend sides AB, CD,EF,GH to form a rectangle: let X be the intersec-tion of lines GH and AB; Y that of AB and CD; Z that of CD and EF ; and W thatof EF and GH.

As BC = 2, we have BY = Y C =√

2. As DE = 4, we have DZ = ZE = 2√

2. AsFG = 2, we have FW = WG =

√2.

We can compute the dimensions of the rectangle: WX = Y Z = Y C + CD + DZ =3 + 3

√2, and XY = ZW = ZE + EF + FW = 2 + 3

√2. Thus, HX = XA = XY −

AB−BY = 1+2√

2, and so AH =√

2HX = 4+√

2, and GH = WX−WG−HX = 2.The perimeter of the octagon can now be computed by adding up all its sides.

2. Suppose ABC is a scalene right triangle, and P is the point on hypotenuse AC suchthat ∠ABP = 45◦. Given that AP = 1 and CP = 2, compute the area of ABC.

Answer: 95

Solution: Notice that BP bisects the right angle at B. Thus, we write AB = 2x,BC = x. By the Pythagorean theorem, 5x2 = 9, from which the area 1

2(x)(2x) = x2 =

95.

3. Let A, B, C, and D be points on a circle such that AB = 11 and CD = 19. PointP is on segment AB with AP = 6, and Q is on segment CD with CQ = 7. The linethrough P and Q intersects the circle at X and Y . If PQ = 27, find XY .

Answer: 31Solution: Suppose X,P,Q, Y lie in that order. Let PX = x and QY = y. By powerof a point from P , x · (27 + y) = 30, and by power of a point from Q, y · (27 + x) =84. Subtracting the first from the second, 27 · (y − x) = 54, so y = x + 2. Now,x · (29 + x) = 30, and we find x = 1,−30. Since −30 makes no sense, we take x = 1and obtain XY = 1 + 27 + 3 = 31.

4. Let ABC be a triangle such that AB = 2, CA = 3, and BC = 4. A semicircle with itsdiameter on BC is tangent to AB and AC. Compute the area of the semicircle.

Answer: 27π40

Solution: Let O, D, and E be the midpoint of the diameter and the points oftangency with AB and AC respectively. Then [ABC] = [AOB] + [AOC] = 1

2(AB +

AC)r, where r is the radius of the semicircle. Now by Heron’s formula, [ABC] =√92· 1

2· 3

2· 5

2= 3

√15

4. We solve for r = 3

√15

10and compute 1

2πr2 = 27π

40.

5. Triangle ABC has side lengths AB = 2√

5, BC = 1, and CA = 5. Point D is on sideAC such that CD = 1, and F is a point such that BF = 2 and CF = 3. Let E be theintersection of lines AB and DF . Find the area of CDEB.

1

Answer: 2235

Solution: Draw segment AF . Then notice AF = 4, and we have a right triangle.

Now draw line CE, let it intersect AF at G. By Ceva, FG =4

3and AG =

8

3. Using

mass points we find thatAE

EB= 6 so

[AEF ]

[BEF ]= 6, and since [ABF ] = 4, [BEF ] =

4

7.

It’s easy to see that [CDF ] =1

5[ACF ] =

6

5, so

[BCDE] = [CDF ]− [BEF ] =6

5− 4

7=

22

35.

6. A circle of radius t is tangent to the hypotenuse, the incircle, and one leg of an isoscelesright triangle with inradius r = 1 + sin π

8. Find rt.

Answer:2 +√

2

4Solution: The distance between the point of tangency of the two circles and thenearest vertex of the triangle is seen to be both r(csc π

8− 1) and t(csc π

8+ 1), so

rt =r2(csc π

8− 1)

csc π8

+ 1=

(1 + sin π8)2(1− sin π

8)

1 + sin π8

= 1− sin2 π

8

=1

2+

1− 2 sin2 π8

2=

1

2+

cos π4

2=

1

2+

√2

4=

2 +√

2

4.

7. Suppose ABCD is an isosceles trapezoid in which AB ‖ CD. Two mutually externallytangent circles ω1 and ω2 are inscribed in ABCD such that ω1 is tangent to AB, BC,and CD while ω2 is tangent to AB, DA, and CD. Given that AB = 1, CD = 6,compute the radius of either circle.

Answer: 37

Solution: Let the radius of both circles be r, and let ω1 be centered at O1. Let ω1 betangent to AB, BC, and CD at P , Q, and R respectively. Then, by symmetry, PB =12− r and RC = 3− r. By equal tangents from B and C, BQ = 1

2− r and QC = 3− r.

Now, ∠BO1C is right because m∠O1BC+m∠BCO1 = 12(m∠PBC+m∠BCR) = 90◦.

Since O1Q ⊥ BC, r2 = O1Q2 = BQ ·QC = (1

2− r)(3− r) = r2 − 7

2r + 3

2. Solving, we

find r = 37.

8. Triangle ABC has a right angle at B. Point D lies on side BC such that 3∠BAD =∠BAC. Given AC = 2 and CD = 1, compute BD.

Answer: 38

Solution: Let BD = x. We reflect D over AB to D′. Then DD′ = 2x, but ADbisects CAD′, so 4x = AD′ = AD. Also, AD =

√x2 + AB2 =

√x2 + AC2 −BC2 =√

x2 + 4− (x + 1)2 =√

3− 2x. We have the quadratic 16x2 = 3 − 2x which givesx = 3/8.

9. Four spheres, each of radius r, lie inside a regular tetrahedron with side length 1 suchthat each sphere is tangent to three faces of the tetrahedron and to the other threespheres. Find r.

2

Answer:

√6− 1

10Solution: Let O be the center of the sphere that is tangent to the faces ABC, ABD,and BCD. Let P , Q be the feet of the perpendiculars from O to ABC and ABDrespectively. Let R be the foot of the perpendicular from P to AB. Then, OPRQis a quadrilateral such that ∠P , ∠Q are right angles and OP = OQ = r. Also, ∠Ris the dihedral angle between faces ABC and ABD, so cos ∠R = 1/3. We can thencompute QR =

√2r, so BR =

√6r. Hence, 1 = AB = 2(

√6r) + 2r = 2r(

√6 + 1), so

r = (√

6− 1)/10.

10. Triangle ABC has side lengths AB = 65, BC = 33, and AC = 56. Find the radius ofthe circle tangent to sides AC and BC and to the circumcircle of triangle ABC.

Answer: 24Solution: Let Γ be the circumcircle of triangle ABC, and let E be the center ofthe circle tangent to Γ and the sides AC and BC. Notice that ∠C = 90◦ because332 + 562 = 652. Let D be the second intersection of line CE with Γ, so that D isthe midpoint of the arc AB away from C. Because ∠BCD = 45◦, one can easilycalculate CD = 89

√2/2. The power of E with respect to Γ is both r(65 − r) and

r√

2 · (89√

2/2− r√

2) = r(89− 2r), so r = 89− 65 = 24.

3

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Guts Round

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

1. [5] A bear walks one mile south, one mile east, and one mile north, only to find itself whereit started. Another bear, more energetic than the first, walks two miles south, two miles east,and two miles north, only to find itself where it started. However, the bears are not whiteand did not start at the north pole. At most how many miles apart, to the nearest .001 mile,are the two bears’ starting points?

2. [5] Compute the positive integer less than 1000 which has exactly 29 positive proper divisors.(Here we refer to positive integer divisors other than the number itself.)

3. [5] At a nursey, 2006 babies sit in a circle. Suddenly each baby pokes the baby immediatelyto either its left or its right, with equal probability. What is the expected number of unpokedbabies?

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

4. [6] Ann and Anne are in bumper cars starting 50 meters apart. Each one approaches theother at a constant ground speed of 10 km/hr. A fly starts at Ann, flies to Anne, then backto Ann, and so on, back and forth until it gets crushed when the two bumper cars collide.When going from Ann to Anne, the fly flies at 20 km/hr; when going in the opposite directionthe fly flies at 30 km/hr (thanks to a breeze). How many meters does the fly fly?

5. [6] Find the number of solutions in positive integers (k; a1, a2, . . . , ak; b1, b2, . . . , bk) to theequation a1(b1) + a2(b1 + b2) + · · ·+ ak(b1 + b2 + · · ·+ bk) = 7.

6. [6] Suppose ABC is a triangle such that AB = 13, BC = 15, and CA = 14. Say D is themidpoint of BC, E is the midpoint of AD, F is the midpoint of BE, and G is the midpointof DF . Compute the area of triangle EFG.

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

7. [6] Find all real numbers x such that x2 +⌊x

2

⌋+⌊x

3

⌋= 10.

8. [6] How many ways are there to label the faces of a regular octahedron with the integers1–8, using each exactly once, so that any two faces that share an edge have numbers thatare relatively prime? Physically realizable rotations are considered indistinguishable, butphysically unrealizable reflections are considered different.

9. [6] Four unit circles are centered at the vertices of a unit square, one circle at each vertex.What is the area of the region common to all four circles?

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

10. [7] Let f(x) = x2 − 2x. How many distinct real numbers c satisfy f(f(f(f(c)))) = 3?

11. [7] Find all positive integers n > 1 for which n2+7n+136n−1 is the square of a positive integer.

12. [7] For each positive integer n let Sn denote the set {1, 2, 3, . . . , n}. Compute the numberof triples of subsets A,B, C of S2006 (not necessarily nonempty or proper) such that A is asubset of B and S2006 −A is a subset of C.

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

The problems in this batch all depend on each other. If you solve them correctly, you willproduce a triple of mutually consistent answers. There is only one such triple. Your scorewill be determined by how many of your answers match that triple.

13. [7] Let Z be as in problem 15. Let X be the greatest integer such that |XZ| ≤ 5. Find X.

14. [7] Let X be as in problem 13. Let Y be the number of ways to order X crimson flowers, Xscarlet flowers, and X vermillion flowers in a row so that no two flowers of the same hue areadjacent. (Flowers of the same hue are mutually indistinguishable.) Find Y .

15. [7] Let Y be as in problem 14. Find the maximum Z such that three circles of radius√

Z cansimultaneously fit inside an equilateral triangle of area Y without overlapping each other.

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

16. [8] A sequence a1, a2, a3, . . . of positive reals satisfies an+1 =√

1+an2 . Determine all a1 such

that ai =√

6+√

24 for some positive integer i.

17. [8] Begining at a vertex, an ant is crawls between the vertices of a regular octahedron. Afterreaching a vertex, it randomly picks a neighboring vertex (sharing an edge) and walks to thatvertex along the adjoining edge (with all possibilities equally likely.) What is the probabilitythat after walking along 2006 edges, the ant returns to the vertex where it began?

18. [8] Cyclic quadrilateral ABCD has side lengths AB = 1, BC = 2, CD = 3 and DA = 4.Points P and Q are the midpoints of BC and DA. Compute PQ2.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

19. [8] Let ABC be a triangle with AB = 2, CA = 3, BC = 4. Let D be the point diametricallyopposite A on the circumcircle of ABC, and let E lie on line AD such that D is the midpointof AE. Line l passes through E perpendicular to AE, and F and G are the intersections ofthe extensions of AB and AC with l. Compute FG.

20. [8] Compute the number of real solutions (x, y, z, w) to the system of equations:

x = z + w + zwx z = x + y + xyz

y = w + x + wxy w = y + z + yzw

21. [8] Find the smallest positive integer k such that z10 + z9 + z6 + z5 + z4 + z +1 divides zk−1.

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

22. [9] Let f(x) be a degree 2006 polynomial with complex roots c1, c2, . . . , c2006, such that the set{|c1|, |c2|, . . . , |c2006|} consists of exactly 1006 distinct values. What is the minimum numberof real roots of f(x)?

23. [9] Let a0, a1, a2, . . . be a sequence of real numbers defined by a0 = 21, a1 = 35, and an+2 =4an+1 − 4an + n2 for n ≥ 2. Compute the remainder obtained when a2006 is divided by 100.

24. [9] Two 18-24-30 triangles in the plane share the same circumcircle as well as the same incircle.What’s the area of the region common to both the triangles?

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

25. [9] Points A, C, and B lie on a line in that order such that AC = 4 and BC = 2. Circlesω1, ω2, and ω3 have BC,AC, and AB as diameters. Circle Γ is externally tangent to ω1 andω2 at D and E respectively, and is internally tangent to ω3. Compute the circumradius oftriangle CDE.

26. [9] Let a ≥ b ≥ c be real numbers such that

a2bc + ab2c + abc2 + 8 = a + b + c

a2b + a2c + b2c + b2a + c2a + c2b + 3abc = −4a2b2c + ab2c2 + a2bc2 = 2 + ab + bc + ca

If a + b + c > 0, then compute the integer nearest to a5.

27. [9] Let N denote the number of subsets of {1, 2, 3, . . . , 100} that contain more prime numbersthan multiples of 4. Compute the largest integer k such that 2k divides N .

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

28. [10] A pebble is shaped as the intersection of a cube of side length 1 with the solid spheretangent to all of the cube’s edges. What is the surface area of this pebble?

29. [10] Find the area in the first quadrant bounded by the hyperbola x2 − y2 = 1, the x-axis,and the line 3x = 4y.

30. [10] ABC is an acute triangle with incircle ω. ω is tangent to sides BC, CA, and AB atD, E, and F respectively. P is a point on the altitude from A such that Γ, the circle withdiameter AP , is tangent to ω. Γ intersects AC and AB at X and Y respectively. GivenXY = 8, AE = 15, and that the radius of Γ is 5, compute BD ·DC.

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

The problems in this batch all depend on each other. If you solve them correctly, you willproduce a triple of mutually consistent answers. There is only one such triple. Your scorewill be determined by how many of your answers match that triple.

31. [10] Let A be as in problem 33. Let W be the sum of all positive integers that divide A.Find W .

32. [10] In the alphametic WE × EY E = SCENE, each different letter stands for a differentdigit, and no word begins with a 0. The W in this problem has the same value as the W inproblem 31. Find S.

33. [10] Let W , S be as in problem 32. Let A be the least positive integer such that an acutetriangle with side lengths S, A, and W exists. Find A.

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

34. [12] In bridge, a standard 52-card deck is dealt in the usual way to 4 players. By convention,each hand is assigned a number of “points” based on the formula

4× (# A’s) + 3× (# K’s) + 2× (# Q’s) + 1× (# J’s).

Given that a particular hand has exactly 4 cards that are A, K, Q, or J, find the probabilitythat its point value is 13 or higher.

35. [12] A sequence is defined by A0 = 0, A1 = 1, A2 = 2, and, for integers n ≥ 3,

An =An−1 + An−2 + An−3

3+

1n4 − n2

Compute limN→∞ AN .

36. [12] Four points are independently chosen uniformly at random from the interior of a regulardodecahedron. What is the probability that they form a tetrahedron whose interior containsthe dodecahedron’s center?

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

37. [15] Compute∞∑

n=1

2n + 52n · (n3 + 7n2 + 14n + 8)

38. [15] Suppose ABC is a triangle with incircle ω, and ω is tangent to BC and CA at D andE respectively. The bisectors of ∠A and ∠B intersect line DE at F and G respectively, suchthat BF = 1 and FG = GA = 6. Compute the radius of ω.

39. [15] A fat coin is one which, when tossed, has a 2/5 probability of being heads, 2/5 of beingtails, and 1/5 of landing on its edge. Mr. Fat starts at 0 on the real line. Every minute, hetosses a fat coin. If it’s heads, he moves left, decreasing his coordinate by 1; if it’s tails, hemoves right, increasing his coordinate by 1. If the coin lands on its edge, he moves back to0. If Mr. Fat does this ad infinitum, what fraction of his time will he spend at 0?

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

40. [18] Compute∞∑

k=1

3k + 12k3 + k2

· (−1)k+1.

41. [18] Let Γ denote the circumcircle of triangle ABC. Point D is on AB such that CD bisects∠ACB. Points P and Q are on Γ such that PQ passes through D and is perpendicular toCD. Compute PQ, given that BC = 20, CA = 80, AB = 65.

42. [18] Suppose hypothetically that a certain, very corrupt political entity in a universe holdsan election with two candidates, say A and B. A total of 5,825,043 votes are cast, but, in asudden rainstorm, all the ballots get soaked. Undaunted, the election officials decide to guesswhat the ballots say. Each ballot has a 51% chance of being deemed a vote for A, and a 49%chance of being deemed a vote for B. The probability that B will win is 10−X . What is Xrounded to the nearest 10?

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IXth HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

43. Write down at least one, and up to ten, different 3-digit prime numbers. If you somehow failto do this, we will ignore your submission for this problem. Otherwise, you’re entered into agame with other teams. In this game, you start with 10 points, and each number you writedown is like a bet: if no one else writes that number, you gain 1 point, but if anyone elsewrites that number, you lose 1 point. Thus, your score on this problem can be anything from0 to 20.

44. On the Euclidean plane are given 14 points:

A = (0, 428) B = (9, 85) C = (42, 865) D = (192, 875)E = (193, 219) F = (204, 108) G = (292, 219) H = (316, 378)I = (375, 688) J = (597, 498) K = (679, 766) L = (739, 641)

M = (772, 307) N = (793, 0)

A fly starts at A, visits all the other points, and comes back to A in such a way as to minimizethe total distance covered. What path did the fly take? Give the names of the points it visitsin order. Your score will be

20 + bthe optimal distancec − byour distancec

or 0, whichever is greater.

45. On your answer sheet, clearly mark at least seven points, as long as

(i) No three are collinear.

(ii) No seven form a convex heptagon.

Please do not cross out any points; erase if you can do so neatly. If the graders deem thatyour paper is too messy, or if they determine that you violated one of those conditions, yoursubmission for this problem will be disqualified. Otherwise, your score will be the number ofpoints you marked minus 6, even if you actually violated one of the conditions but were ableto fool the graders.

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Guts Round Solutions

1. A bear walks one mile south, one mile east, and one mile north, only to find itselfwhere it started. Another bear, more energetic than the first, walks two miles south,two miles east, and two miles north, only to find itself where it started. However, thebears are not white and did not start at the north pole. At most how many milesapart, to the nearest .001 mile, are the two bears’ starting points?

Answer: 3.477Solution: Say the first bear walks a mile south, an integer n > 0 times aroundthe south pole, and then a mile north. The middle leg of the first bear’s journey isa circle of circumference 1/n around the south pole, and therefore about 1

2nπmiles

north of the south pole. (This is not exact even if we assume the Earth is perfectlyspherical, but it is correct to about a micron.) Adding this to the mile that the bearwalked south/north, we find that it started about 1 + 1

2nπmiles from the south pole.

Similarly, the second bear started about 2 + 22mπ

miles from the south pole for someinteger m > 0, so they must have started at most

3 +1

2nπ+

2

2mπ≤ 3 +

3

2π≈ 3.477

miles apart.

2. Compute the positive integer less than 1000 which has exactly 29 positive properdivisors. (Here we refer to positive integer divisors other than the number itself.)

Answer: 720Solution: Recall that the number N = pe1

1 pe22 · · · p

ekk (where the pi are distinct

primes) has exactly (e1 +1)(e2 +1) · · · (ek +1) positive integer divisors including itself.We seek N < 1000 such that this expression is 30. Since 30 = 2 · 3 · 5, we takee1 = 1, e2 = 2, e3 = 4. Then we see that N = 513242 = 720 is satisfactory.

3. At a nursey, 2006 babies sit in a circle. Suddenly each baby pokes the baby immediatelyto either its left or its right, with equal probability. What is the expected number ofunpoked babies?

Answer: 10032

Solution: The probability that any given baby goes unpoked is 1/4. So the answeris 2006/4 = 1003/2.

4. Ann and Anne are in bumper cars starting 50 meters apart. Each one approaches theother at a constant ground speed of 10 km/hr. A fly starts at Ann, flies to Anne, thenback to Ann, and so on, back and forth until it gets crushed when the two bumpercars collide. When going from Ann to Anne, the fly flies at 20 km/hr; when going inthe opposite direction the fly flies at 30 km/hr (thanks to a breeze). How many metersdoes the fly fly?

1

Answer: 55Solution: Suppose that at a given instant the fly is at Ann and the two cars are12d apart. Then, while each of the cars travels 4d, the fly travels 8d and meets Anne.Then the fly turns around, and while each of the cars travels d, the fly travels 3d andmeets Ann again. So, in this process described, each car travels a total of 5d while thefly travels 11d. So the fly will travel 11

5times the distance traveled by each bumper

car: 115· 50

2= 55 meters.

5. Find the number of solutions in positive integers (k; a1, a2, . . . , ak; b1, b2, . . . , bk) to theequation

a1(b1) + a2(b1 + b2) + · · ·+ ak(b1 + b2 + · · ·+ bk) = 7.

Answer: 15Solution: Let k, a1, . . . , ak, b1, . . . , bk be a solution. Then b1, b1+b2, . . . , b1+· · ·+bk isjust some increasing sequence of positive integers. Considering the ai as multiplicities,the ai’s and bi’s uniquely determine a partition of 7. Likewise, we can determine ai’sand bi’s from any partition of 7, so the number of solutions is p(7) = 15.

6. Suppose ABC is a triangle such that AB = 13, BC = 15, and CA = 14. Say D isthe midpoint of BC, E is the midpoint of AD, F is the midpoint of BE, and G is themidpoint of DF . Compute the area of triangle EFG.

Answer: 214

Solution: By Heron’s formula, [ABC] =√

21(21− 15)(21− 14)(21− 13) = 84.

Now, unwinding the midpoint conditions yields [EFG] = [DEF ]2

= [BDE]4

= [ABD]8

=[ABC]

16= 84

16= 21

4.

7. Find all real numbers x such that

x2 +⌊x

2

⌋+⌊x

3

⌋= 10.

Answer: −√

14Solution: Evidently x2 must be an integer. Well, there aren’t that many things tocheck, are there? Among positive x,

√8 is too small and

√9 is too big; among negative

x, −√

15 is too small and −√

13 is too big.

8. How many ways are there to label the faces of a regular octahedron with the integers 1–8, using each exactly once, so that any two faces that share an edge have numbers thatare relatively prime? Physically realizable rotations are considered indistinguishable,but physically unrealizable reflections are considered different.

Answer: 12Solution: Well, instead of labeling the faces of a regular octahedron, we may labelthe vertices of a cube. Then, as no two even numbers may be adjacent, the evennumbers better form a regular tetrahedron, which can be done in 2 ways (because

2

rotations are indistiguishable but reflections are different). Then 3 must be opposite6, and the remaining numbers — 1, 5, 7 — may be filled in at will, in 3! = 6 ways.The answer is thus 2× 6 = 12.

9. Four unit circles are centered at the vertices of a unit square, one circle at each vertex.What is the area of the region common to all four circles?

Answer: π3

+ 1−√

3Solution: The desired region consists of a small square and four “circle segments,”i.e. regions of a circle bounded by a chord and an arc. The side of this small squareis just the chord of a unit circle that cuts off an angle of 30◦, and the circle segmentsare bounded by that chord and the circle. Using the law of cosines (in an isoscelestriangle with unit leg length and vertex angle 30◦), we find that the square of thelength of the chord is equal to 2 −

√3. We can also compute the area of each circle

segment, namelyπ

12− 1

2(1)(1) sin 30◦ =

π

12− 1

4. Hence, the desired region has area

2−√

3 + 4

(π

12− 1

4

)=

π

3+ 1−

√3.

10. Let f(x) = x2 − 2x. How many distinct real numbers c satisfy f(f(f(f(c)))) = 3?

Answer: 9Solution: We see the size of the set f−1(f−1(f−1(f−1(3)))). Note that f(x) =(x − 1)2 − 1 = 3 has two solutions: x = 3 and x = −1, and that the fixed pointsf(x) = x are x = 3 and x = 0. Therefore, the number of real solutions is equal to thenumber of distinct real numbers c such that c = 3, c = −1, f(c) = −1 or f(f(c)) = −1,or f(f(f(c))) = −1. The equation f(x) = −1 has exactly one root x = 1. Thus, thelast three equations are equivalent to c = 1, f(c) = 1, and f(f(c)) = 1. f(c) = 1 hastwo solutions, c = 1±

√2, and for each of these two values c there are two preimages.

It follows that the answer is 1 + 1 + 1 + 2 + 4 = 9.

11. Find all positive integers n > 1 for which n2+7n+136n−1

is the square of a positive integer.

Answer: 5, 37Solution: Write n2+7n+136

n−1= n + 8n+136

n−1= n + 8 + 144

n−1= 9 + (n− 1) + 144

(n−1). We seek

to find p and q such that pq = 144 and p + q + 9 = k2. The possibilities are seen to be1 + 144 + 9 = 154, 2 + 72 + 9 = 83, 3 + 48 + 9 = 60, 4 + 36 + 9 = 49, 6 + 24 + 9 = 39,8 + 18 + 9 = 35, 9 + 16 + 9 = 34, and 12 + 12 + 9 = 33. Of these, {p, q} = {4, 36} isthe only solution to both equations. Hence n− 1 = 4, 36 and n = 5, 37.

12. For each positive integer n let Sn denote the set {1, 2, 3, . . . , n}. Compute the numberof triples of subsets A, B, C of S2006 (not necessarily nonempty or proper) such that Ais a subset of B and S2006 − A is a subset of C.

Answer: 24012

Solution: Let Ao, Bo, Co be sets satisfying the said conditions. Note that 1 ∈ Ao

implies that 1 ∈ Bo and 1 /∈ S2006 − Ao so that 1 may or may not be in Co. Also,

3

1 /∈ Ao implies that 1 ∈ S2006 − Ao ⊂ Co while 1 may or may not be in Bo. Thusthere are four possibilities for the distribution of 1, and since the same argument holdsindependently for 2, 3, . . . , 2006, the answer is 42006 or 24012.

13. Let Z be as in problem 15. Let X be the greatest integer such that |XZ| ≤ 5. FindX.

Answer: 2Solution: Problems 13–15 go together. See below.

14. Let X be as in problem 13. Let Y be the number of ways to order X crimson flowers,X scarlet flowers, and X vermillion flowers in a row so that no two flowers of the samehue are adjacent. (Flowers of the same hue are mutually indistinguishable.) Find Y .

Answer: 30Solution: Problems 13–15 go together. See below.

15. Let Y be as in problem 14. Find the maximum Z such that three circles of radius√

Zcan simultaneously fit inside an equilateral triangle of area Y without overlapping eachother.

Answer: 10√

3− 15Solution: We first find that, in problem 15, each of the circles of radius

√Z is the

incircle of a 30◦-60◦-90◦ triangle formed by cutting the equilateral one in half. The

equilateral triangle itself has sidelength 2√

Y4√3

, so the said inradius is

√Z =

1 +√

3− 2

2· 12· 2√

Y4√

3,

so that

Z =(−1 +

√3)2

4√

3Y =

4− 2√

3

4√

3Y =

2√

3− 3

6Y.

Now we guess that X = 2 and see that, miraculously, everything works: in the problem14, say a crimson flower is placed first. Then there are 2 possibilities for C C , 4for C C , 2 for C C , and 2 for C C, giving a total of 10. Of course, thefirst flower can be of any of the three hues, so Y = 3 · 10 = 30. We compute Z andcheck X in a straightforward manner.

If X > 2, then Y > 30, and Z > 10√

3−15, with the result that X ≤ 2, a contradiction.Assuming X < 2 results in a similar contradiction.

16. A sequence a1, a2, a3, . . . of positive reals satisfies an+1 =√

1+an

2. Determine all a1 such

that ai =√

6+√

24

for some positive integer i.

Answer:

√2 +√

6

2,

√3

2,1

2Solution: Clearly a1 < 1, or else 1 ≤ a1 ≤ a2 ≤ a3 ≤ . . . . We can therefore

write a1 = cos θ for some 0 < θ < 90◦. Note that cosθ

2=

√1 + cos θ

2, and cos 15◦ =

4

√6 +√

2

4. Hence, the possibilities for a1 are cos 15◦, cos 30◦, and cos 60◦, which are

√2 +√

6

2,

√3

2, and

1

2.

17. Begining at a vertex, an ant is crawls between the vertices of a regular octahedron.After reaching a vertex, it randomly picks a neighboring vertex (sharing an edge) andwalks to that vertex along the adjoining edge (with all possibilities equally likely.)What is the probability that after walking along 2006 edges, the ant returns to thevertex where it began?

Answer: 22005+13·22006

Solution: For each nonnegative integer n, let an, bn, and cn denote the respectiveprobabilities that the ant is where it began, at a neighbor of where it began, or isopposite where it began after moving along n edges. We seek a2006. We have a0 = 1and b0 = c0 = 0. We also have the recursive system

an =bn−1

4

bn = an−1 +bn−1

2+ cn−1

cn =bn−1

4

for integers n ≥ 1. Substituting into the second equation we have bn = bn−1

2+ bn−2

2

for n ≥ 2. Solving the characteristic equation x2 − x2− 1

2= 0 for x = 1, −1

2, we write

bn = a · 1n + b(−1/2)n. Using b0 = 0, b1 = 1, we compute

bn =2

3· (1− (−1/2)n)

From which we find a2006 = b20054

= 16

(1 + 1

22005

)= 22005+1

3·22006 .

18. Cyclic quadrilateral ABCD has side lengths AB = 1, BC = 2, CD = 3 and DA = 4.Points P and Q are the midpoints of BC and DA. Compute PQ2.

Answer: 11635

Solution: Construct AC, AQ,BQ,BD, and let R denote the intersection of ACand BD. Because ABCD is cyclic, we have that 4ABR ∼ 4DCR and 4ADR ∼4BCR. Thus, we may write AR = 4x, BR = 2x, CR = 6x, DR = 12x. Now, Ptolemyapplied to ABCD yields 140x2 = 1 · 3 + 2 · 4 = 11. Now BQ is a median in triangleABD. Hence, BQ2 = 2BA2+2BD2−AD2

4. Likewise, CQ2 = 2CA2+2CD2−DA2

4. But PQ is a

median in triangle BQC, so PQ2 = 2BQ2+2CQ2−BC2

4= AB2+BD2+CD2+CA2−BC2−AD2

4=

(196+100)x2+12+32−22−42

4= 148x2−5

2=

148· 11140

−5

2= 116

35.

Another solution is possible. Extend AD and BC past A and B to their intersection S.Use similar triangles SAB and SCD, and similar triangles SAC and SBD to computeSA and SB, then apply the Law of Cosines twice, first to compute the cosine of ∠Aand then to compute PQ2.

5

19. Let ABC be a triangle with AB = 2, CA = 3, BC = 4. Let D be the point diametri-cally opposite A on the circumcircle of ABC, and let E lie on line AD such that D isthe midpoint of AE. Line l passes through E perpendicular to AE, and F and G arethe intersections of the extensions of AB and AC with l. Compute FG.

Answer: 102445

Solution: Using Heron’s formula we arrive at [ABC] = 3√

154

. Now invoking the rela-

tion [ABC] = abc4R

where R is the circumradius of ABC, we compute R2 =(

2·3[ABC]2

)=

6415

. Now observe that ∠ABD is right, so that BDEF is a cyclic quadrilateral. HenceAB · AF = AD · AE = 2R · 4R = 512

15. Similarly, AC · AG = 512

15. It follows that

BCGF is a cyclic quadrilateral, so that triangles ABC and AGF are similar. ThenFG = BC · AF

AC= 4 · 512

2·15·3 = 102445

20. Compute the number of real solutions (x, y, z, w) to the system of equations:

x = z + w + zwx z = x + y + xyz

y = w + x + wxy w = y + z + yzw

Answer: 5Solution: The first equation rewrites as x = w+z

1−wz, which is a fairly strong reason

to consider trigonometric substitution. Let x = tan(a), y = tan(b), z = tan(c), andw = tan(d), where −90◦ < a, b, c, d < 90◦. Under modulo 180◦, we find a ≡ c + d; b ≡d + a; c ≡ a + b; d ≡ b + c. Adding all of these together yields a + b + c + d ≡ 0.Then a ≡ c + d ≡ −a− b so b ≡ −2a. Similarly, c ≡ −2b; d ≡ −2c; d ≡ −2a. Hence,c ≡ −2b ≡ 4a, d ≡ −2c ≡ −8a, and a ≡ −2d ≡ 16a, so the only possible solutionsare (a, b, c, d) ≡ (t,−2t, 4t,−8t) where 15t ≡ 0. Checking, these, we see that actually5t ≡ 0, which yields 5 solutions. Our division by 1 − yz is valid since 1 − yz = 0iff yz = 1, but x = y + z + xyz so y = −z, which implies that yz ≤ 0 < 1, whichis impossible. (The solutions we have computed are in fact (0,0,0,0) and the cyclicpermutations of (tan(36◦), tan(−72◦), tan(−36◦), tan(72◦)).)

21. Find the smallest positive integer k such that z10 + z9 + z6 + z5 + z4 + z + 1 divideszk − 1.

Answer: 84Solution: Let Q(z) denote the polynomial divisor. We need that the roots of Q arek-th roots of unity. With this in mind, we might observe that solutions to z7 = 1 andz 6= 1 are roots of Q, which leads to its factorization. Alternatively, we note that

(z − 1)Q(z) = z11 − z9 + z7 − z4 + z2 − 1 = (z4 − z2 + 1)(z7 − 1)

Solving for the roots of the first factor, z2 = 1+i√

32

= ±cisπ/3 (we use the notationcis(x) = cos(x) + i sin(x)) so that z = ±cis(±π/6). These are primitive 12-th rootsof unity. The other roots of Q(z) are the primitive 7-th roots of unity (we introducedz = 1 by multiplication.) It follows that the answer is lcm[12, 7] = 84.

6

22. Let f(x) be a degree 2006 polynomial with complex roots c1, c2, . . . , c2006, such thatthe set

{|c1|, |c2|, . . . , |c2006|}

consists of exactly 1006 distinct values. What is the minimum number of real roots off(x)?

Answer: 6Solution: The complex roots of the polynomial must come in pairs, ci and ci, both ofwhich have the same absolute value. If n is the number of distinct absolute values |ci|corresponding to those of non-real roots, then there are at least 2n non-real roots off(x). Thus f(x) can have at most 2006−2n real roots. However, it must have at least1006− n real roots, as |ci| takes on 1006− n more values. By definition of n, these allcorrespond to real roots. Therefore 1006−n ≤ # real roots ≤ 2006− 2n, so n ≤ 1000,and # real roots ≥ 1006− n ≥ 6. It is easy to see that equality is attainable.

23. Let a0, a1, a2, . . . be a sequence of real numbers defined by a0 = 21, a1 = 35, andan+2 = 4an+1 − 4an + n2 for n ≥ 2. Compute the remainder obtained when a2006 isdivided by 100.

Answer: 0Solution: No pattern is evident in the first few terms, so we look for a formula foran. If we write an = An2 + Bn + C + bn and put bn+2 = 4bn+1 − 4bn. Rewriting theoriginal recurrence, we find

An2 + (4A + B)n + (4A + 2B + C) + bn+2

= 4(An2 + (2A + B)n + (A + B + C) + bn+1

)− 4

(An2 + Bn + C + bn

)+ n2

= n2 + 8An + (4A + 4B) + 4bn+1 − 4bn

Solving, A = 1, B = 4, C = 8. With this information, we can solve for b0 = 1 andb1 = 6. Since the characteristic equation of the recurrence of the bi is x2 − 4x + 4 =(x − 2)2 = 0, we have bn = (Dn + E) · 2n for some constants D and E. Using theknown values b0 and b1, we compute D = 2 and E = 1, and finally

an = n2 + 4n + 8 + (2n + 1) · 2n

Now, taking modulo 100, we have a2006 ≡ 62 +4 ·6+8+13 ·22006 (mod 100). Evidently22006 ≡ 0 (mod 4), but by Euler’s theorem 2φ(25) ≡ 220 ≡ 1 (mod 25), and so 22006 ≡26 ≡ 14 (mod 25). Now the Chinese remainder theorem yields 22006 ≡ 64 (mod 100),and we compute a2006 ≡ 36 + 24 + 8 + 13 · 64 ≡ 0 (mod 100).

24. Two 18-24-30 triangles in the plane share the same circumcircle as well as the sameincircle. What’s the area of the region common to both the triangles?

Answer: 132Solution: Notice, first of all, that 18-24-30 is 6 times 3-4-5, so the triangles areright. Thus, the midpoint of the hypotenuse of each is the center of their commoncircumcircle, and the inradius is 1

2(18+24−30) = 6. Let one of the triangles be ABC,

7

where ∠A < ∠B < ∠C = 90◦. Now the line ` joining the midpoints of sides AB andAC is tangent to the incircle, because it is the right distance (12) from line BC. So,the hypotenuse of the other triangle lies along `. We may formulate this thus: Thehypotenuse of each triangle is parallel to the shorter leg, and therefore perpendicularto the longer leg, of the other. Now it is not hard to see, as a result of these parallel-and perpendicularisms, that the other triangle “cuts off” at each vertex of 4ABC asmaller, similar right triangle. If we compute the dimensions of these smaller triangles,we find that they are as follows: 9-12-15 at A, 6-8-10 at B, and 3-4-5 at C. The totalarea chopped off of 4ABC is thus

9 · 12

2+

6 · 82

+3 · 42

= 54 + 24 + 6 = 84.

The area of 4ABC is 18 · 24/2 = 216. The area of the region common to both theoriginal triangles is thus 216− 84 = 132.

25. Points A, C, and B lie on a line in that order such that AC = 4 and BC = 2. Circlesω1, ω2, and ω3 have BC,AC, and AB as diameters. Circle Γ is externally tangentto ω1 and ω2 at D and E respectively, and is internally tangent to ω3. Compute thecircumradius of triangle CDE.

Answer: 23

Solution: Let the center of ωi be Oi for i = 1, 2, 3 and let O denote the center ofΓ. Then O,D, and O1 are collinear, as are O, E, and O2. Denote by F the point oftangency between Γ and ω3; then F, O, and O3 are collinear. Writing r for the radiusof Γ we have OO1 = r + 2, OO2 = r + 1, OO3 = 3 − r. Now since O1O3 = 1 andO3O2 = 2, we apply Stewart’s theorem:

OO21 ·O2O3 + OO2

2 ·O1O3 = OO23 ·O1O2 + O1O3 ·O3O2 ·O1O2

2(r + 2)2 + (r + 1)2 = 3(3− r)2 + 1 · 2 · 3

We find r = 67. Now the key observation is that the circumcircle of triangle CDE is

the incircle of triangle OO1O2. We easily compute the sides of OO1O2 to be 137, 20

7, and

3. By Heron’s formula, the area of OO1O2 is 187, but the semiperimeter is 27

7, so the

desired radius is 23.

26. Let a ≥ b ≥ c be real numbers such that

a2bc + ab2c + abc2 + 8 = a + b + c

a2b + a2c + b2c + b2a + c2a + c2b + 3abc = −4

a2b2c + ab2c2 + a2bc2 = 2 + ab + bc + ca

If a + b + c > 0, then compute the integer nearest to a5.

Answer: 1279Solution: We factor the first and third givens, obtaining the system

a2bc + ab2c + abc2 − a− b− c = (abc− 1)(a + b + c) = −8

a2b + a2c + b2c + b2a + c2a + c2b + 3abc = (ab + bc + ca)(a + b + c) = −4

a2b2c + ab2c2 + a2bc2 − ab− bc− ca = (abc− 1)(ab + bc + ca) = 2

8

Writing X = a+b+c, Y = ab+bc+ca, Z = abc−1, we have XZ = −8, XY = −4, Y Z =2. Multiplying the three yields (XY Z)2 = 64 from which XY Z = ±8. Since we aregiven X > 0, multiplying the last equation by X we have 2X = XY Z = ±8. EvidentlyXY Z = 8 from which X = 4, Y = −1, Z = −2. We conclude that a, b, c are the rootsof the polynomial P (t) = t3 − 4t2 − t + 1. Thus, P (a) = a3 − 4a2 − a + 1 = 0, andalso P (b) = P (c) = 0. Now since P (1/2) = −3

8, P (0) = 1 and P (−2/3) = −11

27, we

conclude that −2/3 < c < 0 < b < 1/2 < a. It follows that |b5 + c5| < 12. Thus, we

compute a5 + b5 + c5.

Defining Sn = an + bn + cn, we have Sn+3 = 4Sn+2 + Sn+1 − Sn for n ≥ 0. EvidentlyS0 = 3, S1 = 4, S2 = (a + b + c)2− 2(ab + bc + ca) = 18. Then S3 = 4 · 18 + 4− 3 = 73,S4 = 4 · 73 + 18− 4 = 306, and S5 = 4 · 306 + 73− 18 = 1279. Since |b5 + c5| < 1

2, we

conclude that |S5 − a5| < 12, or that 1279 is the integer nearest to a5.

27. Let N denote the number of subsets of {1, 2, 3, . . . , 100} that contain more primenumbers than multiples of 4. Compute the largest integer k such that 2k divides N .

Answer: 52Solution: Let S denote a subset with the said property. Note that there are 25multiples of 4 and 25 primes in the set {1, 2, 3, . . . , 100}, with no overlap between thetwo. Let T denote the subset of 50 numbers that are neither prime nor a multiple of 4,and let U denote the 50 other numbers. Elements of T can be arbitrarily included inor excluded by S. Consider S ∩ U = S1 and U − S = S2 (the set difference is definedto be all elements of U that are not in S.) S1 and S2 are two disjoint sets such thatU = S1 ∩ S2. If S1 contains more multiples of 4 than primes, then S2 contains moreprimes than multiples of 4, and conversely. Furthermore, S1 contains an equal numberof primes and multiples of 4 if and only if S2 contains equal numbers as well. Let Vdenote an arbitrary subset of T . It follows from examining pairs of sets V ∪ S1 andV ∪ S2 that

N = 250 · 12

(250 −

25∑k=0

(25

k

)2)

= 249 ·(

250 −(

50

25

))Since 50! is divisible by 2 exactly 25 + 12 + 6 + 3 + 1 = 47 times while 25! is divisibleby 2 exactly 12 + 6 + 3 + 1 = 22 times, it follows that

(5025

)is divisible by 2 exactly 3

times, so the answer is 49 + 3 = 52.

28. A pebble is shaped as the intersection of a cube of side length 1 with the solid spheretangent to all of the cube’s edges. What is the surface area of this pebble?

Answer:6√

2− 5

2π

Solution: Imagine drawing the sphere and the cube. Take a cross section, with aplane parallel to two of the cube’s faces, passing through the sphere’s center. In thiscross section, the sphere looks like a circle, and the cube looks like a square (of sidelength 1) inscribed in that circle. We can now calculate that the sphere has diameter

9

d :=√

2 and surface area S := πd2 = 2π, and that the sphere protrudes a distance ofx :=

√2−12

out from any given face of the cube.

It is known that the surface area chopped off from a sphere by any plane is proportionalto the perpendicular distance thus chopped off. Thus, each face of the cube chops ofa fraction x

dof the sphere’s surface. The surface area of the pebble contributed by the

sphere is thus S · (1 − 6 · xd), whereas the cube contributes 6 circles of radius 1

2, with

total area 6 · π(

12

)2= 3

2π. The pebble’s surface area is therefore

S ·(1− 6 · x

d

)+

3

2π = 2π ·

(1− 6 ·

√2− 1

2√

2

)+

3

2π =

6√

2− 5

2π.

29. Find the area in the first quadrant bounded by the hyperbola x2 − y2 = 1, the x-axis,and the line 3x = 4y.

Answer: ln 74

Solution: Convert to polar coordinates: the hyperbola becomes

1 = r2(cos2 θ − sin2 θ) = r2 cos(2θ),

so, letting α := arctan(3/4), the area is

S :=

∫ α

0

r2

2dθ =

1

2

∫ α

0

sec(2θ)dθ =1

4ln |sec(2θ) + tan(2θ)|

∣∣∣∣α0

.

Now

tan(2α) =2 tan α

1− tan2 α=

3/2

7/16=

24

7,

sec(2α) =√

1 + tan2(2α) =25

7,

so

S =1

4

(ln

∣∣∣∣25

7+

24

7

∣∣∣∣− ln |1 + 0|)

=ln 7

4.

30. ABC is an acute triangle with incircle ω. ω is tangent to sides BC, CA, and AB atD, E, and F respectively. P is a point on the altitude from A such that Γ, the circlewith diameter AP , is tangent to ω. Γ intersects AC and AB at X and Y respectively.Given XY = 8, AE = 15, and that the radius of Γ is 5, compute BD ·DC.

Answer: 6754

Solution: By the Law of Sines we have sin ∠A = XYAP

= 45. Let I, T , and Q denote the

center of ω, the point of tangency between ω and Γ, and the center of Γ respectively.Since we are told ABC is acute, we can compute tan ∠A

2= 1

2. Since ∠EAI = A

2

and AE is tangent to ω, we find r = AE2

= 152. Let H be the foot of the altitude

from A to BC. Define hT to be the homothety about T which sends Γ to ω. Wehave hT (AQ) = DI, and conclude that A, T , and D are collinear. Now since AP isa diameter of Γ, ∠PAT is right, implying that DTHP is cyclic. Invoking Power of

10

a Point twice, we have 225 = AE2 = AT · AD = AP · AH. Because we are givenradius of Γ we can find AP = 10 and AH = 45

2= ha. If we write a, b, c, s in the usual

manner with respect to triangle ABC, we seek BD ·DC = (s − b)(s − c). But recallthat Heron’s formula gives us√

s(s− a)(s− b)(s− c) = K

where K is the area of triangle ABC. Writing K = rs, we have (s− b)(s− c) = r2ss−a

.

Knowing r = 152, we need only compute the ratio s

a. By writing K = 1

2aha = rs, we

find sa

= ha

2r= 3

2. Now we compute our answer, r2s

s−a=(

152

)2 · sa

sa−1

= 6754

.

31. Let A be as in problem 33. Let W be the sum of all positive integers that divide A.Find W .

Answer: 8Solution: Problems 31–33 go together. See below.

32. In the alphametic WE × EY E = SCENE, each different letter stands for a differentdigit, and no word begins with a 0. The W in this problem has the same value as theW in problem 31. Find S.

Answer: 5Solution: Problems 31–33 go together. See below.

33. Let W , S be as in problem 32. Let A be the least positive integer such that an acutetriangle with side lengths S, A, and W exists. Find A.

Answer: 7Solution: There are two solutions to the alphametic in problem 32: 36×686 = 24696and 86 × 636 = 54696. So (W, S) may be (3, 2) or (8, 5). If (W, S) = (3, 2), then byproblem (3) A = 3, but then by problem 31 W = 4, a contradiction. So, (W, S) mustbe (8, 5). By problem 33, A = 7, and this indeed checks in problem 31.

34. In bridge, a standard 52-card deck is dealt in the usual way to 4 players. By convention,each hand is assigned a number of “points” based on the formula

4× (# A’s) + 3× (# K’s) + 2× (# Q’s) + 1× (# J’s).

Given that a particular hand has exactly 4 cards that are A, K, Q, or J, find theprobability that its point value is 13 or higher.

Answer: 1971820

Solution: Obviously, we can ignore the cards lower than J. Simply enumerate theways to get at least 13 points: AAAA (1), AAAK (16), AAAQ (16), AAAJ (16),AAKK (36), AAKQ (96), AKKK (16). The numbers in parentheses represent thenumber of ways to choose the suits, given the choices for the values. We see that thereare a total of 1 + 16 + 16 + 16 + 36 + 96 + 16 = 197 ways to get at least 13. There area total of

(164

)= 1820 possible ways to choose 4 cards from the 16 total A’s, K’s, Q’s,

and J’s. Hence the answer is 197/1820.

11

35. A sequence is defined by A0 = 0, A1 = 1, A2 = 2, and, for integers n ≥ 3,

An =An−1 + An−2 + An−3

3+

1

n4 − n2

Compute limN→∞ AN .

Answer: 136− π2

12.

Solution: If we sum the given equation for n = 3, 4, 5, . . . , N , we obtain

N∑n=3

An =N∑

n=3

An−1 + An−2 + An−3

3+

1

n4 − n2

This reduces dramatically to

AN +2AN−1

3+

AN−2

3= A2 +

2A1

3+

A0

3+

N∑n=3

1

n4 − n2(*)

Let limN→∞ AN = L. Under this limit, the left hand side of (*) is simply 2L. Wecompute the sum on the right with the help of partial fractions

limN→∞

N∑n=3

1

n4 − n2=

∞∑n=3

1

n2 − 1− 1

n2

=

(∞∑

n=3

1

2

(1

n− 1− 1

n + 1

))+

1

12+

1

22−

∞∑n=1

1

n2

=1

2

(1

2+

1

3

)+

5

4− π2

6

=5

3− π2

6

With this we easily find L = 12·(2 + 2

3· 1 + 1

3· 0 + 5

3− π2

6

)= 13

6− π2

12, and we are done.

36. Four points are independently chosen uniformly at random from the interior of a regulardodecahedron. What is the probability that they form a tetrahedron whose interiorcontains the dodecahedron’s center?

Answer: 18

Solution: Situate the origin O at the dodecahedron’s center, and call the four randompoints Pi, where 1 ≤ i ≤ 4.

To any tetrahedron P1P2P3P4 we can associate a quadruple (ε(ijk)), where (ijk) rangesover all conjugates of the cycle (123) in the alternating group A4: εijk is the sign ofthe directed volume [OPiPjPk]. Assume that, for a given tetrahedron P1P2P3P4, allmembers of its quadruple are nonzero (this happens with probability 1). For 1 ≤ i ≤4, if we replace Pi with its reflection through the origin, the three members of thetetrahedron’s quadruple that involve Pi all flip sign, because each [OPiPjPk] is a linear

12

function of the vector−−→OPi. Thus, if we consider the 16 sister tetrahedra obtained

by choosing independently whether to flip each Pi through the origin, the quadruplesrange through all 16 possibilities (namely, all the quadruples consisting of ±1s). Twoof these 16 tetrahedra, namely those with quadruples (1, 1, 1, 1) and (−1,−1,−1,−1),will contain the origin.

So the answer is 2/16 = 1/8.

37. Compute∞∑

n=1

2n + 5

2n · (n3 + 7n2 + 14n + 8)

Answer: 13724− 8 ln 2

Solution: First, we manipulate using partial fractions and telescoping:

∞∑n=1

2n + 5

2n · (n3 + 7n2 + 14n + 8)=

1

2·

∞∑n=1

1

2n

(2

n + 1− 1

n + 2− 1

n + 4

)=

1

4− 1

2

∞∑n=1

1

2n · (n + 4)

Now, consider the function f(r, k) :=∑∞

n=1rn

nk . We have

∂f(r, k)

∂r=

∂

∂r

∞∑n=1

rn

nk=

∞∑n=1

∂

∂r

[rn

nk

]=

∞∑n=1

rn−1

nk−1=

1

rf(r, k − 1)

df(r, 1)

dr=

1

r

∞∑n=1

rn

n0=

1

r· r

1− r=

1

1− r

f(r, 1) =

∫dr

1− r= − ln(1− r) + f(0, 1)

By inspection, f(0, 1) = 0, so f(

12, 1)

=∑∞

n=11

n·2n = ln(2). It is easy to compute thedesired sum in terms of f

(12, 1), and we find

∑∞n=1

12n(n+4)

= 16 ln(2)− 13112

. Hence, our

final answer is 13724− 8 ln(2).

38. Suppose ABC is a triangle with incircle ω, and ω is tangent to BC and CA at D andE respectively. The bisectors of ∠A and ∠B intersect line DE at F and G respectively,such that BF = 1 and FG = GA = 6. Compute the radius of ω.

Answer: 2√

55

Solution: Let α, β, γ denote the measures of 12∠A, 1

2∠B, 1

2∠C, respectively. We have

m∠CEF = 90◦−γ, m∠FEA = 90◦+γ, m∠AFG = m∠AFE = 180◦−α−(90◦+γ) =β = m∠ABG, so ABFG is cyclic. Now AG = GF implies that BG bisects ∠ABF .Since BG by definition bisects ∠ABC, we see that F must lie on BC. Hence, F = D.If I denotes the incenter of triangle ABC, then ID is perpendicular to BC, but sinceA, I, F are collinear, we have that AD ⊥ BC. Hence, ABC is isoceles with AB = AC.Furthermore, BC = 2BF = 2. Moreover, since ABFG is cyclic, ∠BGA is a right angle.Construct F ′ on minor arc GF such that BF ′ = 6 and F ′G = 1, and let AB = x. By

13

the Pythagorean theorem, AF ′ = BG =√

x2 − 36, so that Ptolemy applied to ABF ′Gyields x2 − 36 = x + 36. We have (x − 9)(x + 8) = 0. Since x is a length we findx = 9. Now we have AB = AC = 9. Pythagoras applied to triangle ABD now yieldsAD =

√92 − 12 = 4

√5, which enables us to compute [ABC] = 1

2· 2 · 4

√5 = 4

√5.

Since the area of a triangle is also equal to its semiperimeter times its inradius, wehave 4

√5 = 10r or r = 2

√5

5.

REMARK. In fact, ABFG is always a cyclic quadrilateral for which AB plays a di-ameter. That is, we could have proven this fact without using FG = GA.

39. A fat coin is one which, when tossed, has a 2/5 probability of being heads, 2/5 ofbeing tails, and 1/5 of landing on its edge. Mr. Fat starts at 0 on the real line. Everyminute, he tosses a fat coin. If it’s heads, he moves left, decreasing his coordinate by1; if it’s tails, he moves right, increasing his coordinate by 1. If the coin lands on itsedge, he moves back to 0. If Mr. Fat does this ad infinitum, what fraction of his timewill he spend at 0?

Answer: 13

Solution: For n ∈ Z, let an be the fraction of the time Mr. Fat spends at n. Bysymmetry, an = a−n for all n.

For n > 0, we have an = 25an−1 + 2

5an+1, or an+1 = 5

2an − an−1. This Fibonacci-like

recurrence can be solved explicitly to obtain

an = α · 2|n| + β · 2−|n|

for all n ∈ Z. Now we also have ∑n∈Z

an = 1,

so we better have α = 0, so that a0 = β and a±1 = β2. Now we also have a0 =

25a−1 + 2

5a1 + 1

5, so β = 1

3. This matches perfectly with

∑n∈Z an = 1.

40. Compute∞∑

k=1

3k + 1

2k3 + k2· (−1)k+1.

Answer: π2

12+ π

2− 2 + ln 2

Solution: Via partial fraction decomposition we write the sum as∞∑

k=1

(1

k− 2

1 + 2k+

1

k2

)(−1)k+1

Now recall that∞∑

k=1

1

k2=

π2

6= S1

∞∑k=1

(−1)k+1

k= ln(2) = S2

∞∑k=1

(−1)k+1

2k − 1=

π

4= S3

14

Manipulating (1), we deduce

∞∑k=1

(−1)k+1

k2=

(∞∑

k=1

1

k2

)− 2 ·

∞∑k=1

1

(2k)2

=π2

6− 2/4 · π

2

6=

π2

12= S4

It is then easily seen that the answer is equal to S2 + 2 · S3 − 2 + S4.

41. Let Γ denote the circumcircle of triangle ABC. Point D is on AB such that CD bisects∠ACB. Points P and Q are on Γ such that PQ passes through D and is perpendicularto CD. Compute PQ, given that BC = 20, CA = 80, AB = 65.

Answer: 4√

745Solution: Suppose that P lies between A and B and Q lies between A and C, and letline PQ intersect lines AC and BC at E and F respectively. As usual, we write a, b, cfor the lengths of BC, CA, AB. By the angle bisector theorem, AD/DB = AC/CB sothat AD = bc

a+band BD = ac

a+b. Now by Stewart’s theorem, c · CD2 +

(ac

a+b

) (bc

a+b

)c =

a2bca+b

+ ab2ca+b

from which CD2 = ab((a+b)2−c2)(a+b)2

. Now observe that triangles CDE and

CDF are congruent, so ED = DF . By Menelaus’ theorem, CAAE

EDDF

FBBC

= 1 so thatCABC

= AEFB

. Since CF = CE while b > a, it follows that AE = b(b−a)a+b

so that EC = 2aba+b

.

Finally, DE =√

CE2 − CD2 =

√ab(c2−(a−b)2)

a+b. Plugging in a = 20, b = 80, c = 65,

we see that AE = 48, EC = 32, DE = 10 as well as AD = 52, BD = 13. Now letPD = x, QE = y. By power of a point about D and E, we have x(y + 10) = 676 andy(x + 10) = 1536. Subtracting one from the other, we see that y = x + 86. Therefore,x2 +96x− 676 = 0, from which x = −48+2

√745. Finally, PQ = x+ y +10 = 4

√745.

42. Suppose hypothetically that a certain, very corrupt political entity in a universe holdsan election with two candidates, say A and B. A total of 5,825,043 votes are cast,but, in a sudden rainstorm, all the ballots get soaked. Undaunted, the election officialsdecide to guess what the ballots say. Each ballot has a 51% chance of being deemed avote for A, and a 49% chance of being deemed a vote for B. The probability that Bwill win is 10−X . What is X rounded to the nearest 10?

Answer: 510Solution: Let N = 2912521, so that the number of ballots cast is 2N + 1. Let Pbe the probability that B wins, and let α = 51% and β = 49% and γ = β/α < 1. Wehave

10−X = P =N∑

i=0

(2N + 1

N − i

)αN−iβN+1+i = αNβN+1

N∑i=0

(2N + 1

N − i

)γi

(think of 2i + 1 as representing B’s margin of victory). Now

22N+1

2N + 1<

(2N + 1

N

)<

N∑i=0

(2N + 1

N − i

)γi < 22N+1,

15

So

−X = log P = N log α+(N+1) log β+(2N+1) log 2−ε = N log(2α)+(N+1) log(2β)−ε,

where 0 < ε < log(2N + 1) < 7. With a calculator, we find that

−X ≈ 25048.2− 25554.2− ε = −506.0− ε,

so X ≈ 510.

43. Write down at least one, and up to ten, different 3-digit prime numbers. If you somehowfail to do this, we will ignore your submission for this problem. Otherwise, you’reentered into a game with other teams. In this game, you start with 10 points, andeach number you write down is like a bet: if no one else writes that number, you gain1 point, but if anyone else writes that number, you lose 1 point. Thus, your score onthis problem can be anything from 0 to 20.

Solution: There are 143 three-digit primes. None of the following necessarily appliesto the actual contest, but it might be useful to think about. Suppose that you’retrying to maximize your expected score on this problem. Then you should write downa number if you think the probability that someone else is writing it is less than 1/2(of course, limit yourself to 10 numbers). You should avoid writting down any numberif you think the probability that someone else is writing it is more than 1/2 (of course,write down at least 1 number). Suppose that you expect a total of M different numbersare going to be written down, but have no idea what numbers they might be. If youthink M ≥ 72, you should write down 10 numbers at random; if M ≤ 71, you shouldwrite just 1 number.

44. On the Euclidean plane are given 14 points:

A = (0, 428) B = (9, 85) C = (42, 865) D = (192, 875)

E = (193, 219) F = (204, 108) G = (292, 219) H = (316, 378)

I = (375, 688) J = (597, 498) K = (679, 766) L = (739, 641)

M = (772, 307) N = (793, 0)

A fly starts at A, visits all the other points, and comes back to A in such a way as tominimize the total distance covered. What path did the fly take? Give the names ofthe points it visits in order. Your score will be

20 + bthe optimal distancec − byour distancec

or 0, whichever is greater.

Answer: The optimal path is ACDIKLJMNHGEFB(A), or the reverse, of course.In this way the total distance covered by the fly is just over 3591.22.Solution: This problem is an instance of the Traveling Salesman Problem, whichis NP-hard. There is an obvious algorithm in O(n!) time (where n is the number ofpoints), but faster algorithms exist. Nonetheless, the best strategy for solving thisproblem is probably to draw the points and exercise your geometric intuition.

16

45. On your answer sheet, clearly mark at least seven points, as long as

(i) No three are collinear.

(ii) No seven form a convex heptagon.

Please do not cross out any points; erase if you can do so neatly. If the gradersdeem that your paper is too messy, or if they determine that you violated one of thoseconditions, your submission for this problem will be disqualified. Otherwise, your scorewill be the number of points you marked minus 6, even if you actually violated one ofthe conditions but were able to fool the graders.

Solution: This is the heptagon case of what is known as the “Happy Ending” or“Erdős-Szekeres” problem, which in general asks, For any integer n ≥ 3, what is thesmallest N(n), such that any N(n) points in the plane in general position determinea convex n-gon? It is known that such an N(n) always exists and is finite (in fact aspecific upper bound has been found). The best known lower bound is N(n) ≥ 2n−2+1;Erdős and Szekeres conjectured that this bound is tight. The n ≤ 5 cases have beenknown for some time. According to the Wikipedia, the n = 6 case is solved butunpublished, and for n ≥ 7, the problem remains open.

For a discussion, see

W. Morris and V. Soltan. The Erdős-Szekeres Problem on Points in ConvexPostion—A Survey, Bulletin of the American Math Monthly. 37 (2000),437–458.

This article is available at

http://www.ams.org/bull/2000-37-04/S0273-0979-00-00877-6/home.html.

If N(7) = 33, the highest sure score on this problem would be 32 − 6 = 26. It is notknown whether there exist arbitrarily large sets of points that will fool the graders.

The unexamined life is not worth living.

17

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Team Round A

Mobotics [120]

Spring is finally here in Cambridge, and it’s time to mow our lawn. For the purpose of these problems,our lawn consists of little clumps of grass arranged at the points of a certain grid (to be specified later).Our machinery consists of a fleet of identical mowbots (or “mobots” for short). A mobot is a lawn-mowingmachine. To mow our lawn, we begin by choosing a formation: we place as many mobots as we want atvarious clumps of grass and orient each mobot’s head in a certain direction. At the blow of a whistle, eachmobot starts moving in the direction we’ve chosen, mowing every clump of grass in its path (including theclump it starts on) until it goes off the lawn.

Because the spring is so young, our lawn is rather delicate. Consequently, we want to make sure thatevery clump of grass is mowed once and only once. We will not consider formations that do not meet thiscriterion.

One more thing: two formations are considered “different” if there exists a clump of grass for which either(1) for exactly one of the formations does a mobot starts on that clump, or (2) there are mobots starting onthis clump for both the formations, but they’re oriented in different directions.

1. [15] For this problem, our lawn consists of a row of n clumps of grass. This row runs in an east-westdirection. In our formation, each mobot may be oriented toward the north, south, east, or west. Oneexample of an allowable formation if n = 6 is symbolized below:

· · ← ↑ ↑↓

(The mobot on the third clump will move westward, mowing the first three clumps. Each of the lastthree clumps is mowed by a different mobot.) Here’s another allowable formation for n = 6, considereddifferent from the first:

· · ← ↑ ↑ →Compute the number of different allowable formations for any given n.

2. [25] For this problem, our lawn is an m× n rectangular grid of clumps, that is, with m rows runningeast-west and n columns running north-south. To be even more explicit, we might say our clumps areat the lattice points

{(x, y) ∈ Z2 | 0 ≤ x < n and 0 ≤ y < m}.However, mobots are now allowed to be oriented to go either north or east only. So one allowableformation for m = 2, n = 3 might be as follows:

· → ·↑ → ·

Prove that the number of allowable formations for given m and n is(m + n)!

m!n!.

3. [40] In this problem, we stipulate that m ≥ n, and the lawn is shaped differently. The clumps are nowat the lattice points in a trapezoid:

{(x, y) ∈ Z2 | 0 ≤ x < n and 0 ≤ y < m + 1− n + x},

As in problem 2, mobots can be set to move either north or east. For given m and n, determine withproof the number of allowable formations.

4. [15] In this problem and the next, the lawn consists of points in a triangular grid of size n, so that forn = 3 the lawn looks like

·· ·· · ·

1

Mobots are allowed to be oriented to the east, 30◦ west of north, or 30◦ west of south. Under theseconditions, for any given n, what is the minimum number of mobots needed to now the lawn?

5. [25] With the same lawn and the same allowable mobot orientations as in the previous problem, letus call a formation “happy” if it is invariant under 120◦ rotations. (A rotation applies both to thepositions of the mobots and to their orientations.) An example of a happy formation for n = 2 mightbe

↖↙ →

Find the number of happy formations for a given n.

Polygons [110]

6. [15] Let n be an integer at least 5. At most how many diagonals of a regular n-gon can be simultaneouslydrawn so that no two are parallel? Prove your answer.

7. [25] Given a convex n-gon, n ≥ 4, at most how many diagonals can be drawn such that each drawndiagonal intersects every other drawn diagonal either in the interior of the n-gon or at a vertex? Proveyour answer.

8. [15] Given a regular n-gon with sides of length 1, what is the smallest radius r such that there isa non-empty intersection of n circles of radius r centered at the vertices of the n-gon? Give r as aformula in terms of n. Be sure to prove your answer.

9. [40] Let n ≥ 3 be a positive integer. Prove that given any n angles 0 < θ1, θ2, . . . , θn < 180◦, such thattheir sum is 180(n− 2) degrees, there exists a convex n-gon having exactly those angles, in that order.

10. [15] Suppose we have an n-gon such that each interior angle, measured in degrees, is a positive integer.Suppose further that all angles are less than 180◦, and that all angles are different sizes. What is themaximum possible value of n? Prove your answer.

What do the following problems have in common? [170]

11. [15] The lottery cards of a certain lottery contain all nine-digit numbers that can be formed with thedigits 1, 2 and 3. There is exactly one number on each lottery card. There are only red, yellow andblue lottery cards. Two lottery numbers that differ from each other in all nine digits always appear oncards of different color. Someone draws a red card and a yellow card. The red card has the number122 222 222 and the yellow card has the number 222 222 222. The first prize goes to the lottery cardwith the number 123 123 123. What color(s) can it possibly have? Prove your answer.

12. [25] A 3× 3× 3 cube is built from 27 unit cubes. Suddenly five of those cubes mysteriously teleportaway. What is the minimum possible surface area of the remaining solid? Prove your answer.

13. [40] Having lost a game of checkers and my temper, I dash all the pieces to the ground but one. Thislast checker, which is perfectly circular in shape, remains completely on the board, and happens tocover equal areas of red and black squares. Prove that the center of this piece must lie on a boundarybetween two squares (or at a junction of four).

14. [40] A number n is called bummed out if there is exactly one ordered pair of positive integers (x, y)such that

bx2/yc+ by2/xc = n.

Find all bummed out numbers.

15. [50] Find, with proof, all positive integer palindromes whose square is also a palindrome.

2

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Team Round A: Solutions

Mobotics [120]

Spring is finally here in Cambridge, and it’s time to mow our lawn. For the purpose of theseproblems, our lawn consists of little clumps of grass arranged at the points of a certain grid(to be specified later). Our machinery consists of a fleet of identical mowbots (or “mobots”for short). A mobot is a lawn-mowing machine. To mow our lawn, we begin by choosing aformation: we place as many mobots as we want at various clumps of grass and orient eachmobot’s head in a certain direction. At the blow of a whistle, each mobot starts moving inthe direction we’ve chosen, mowing every clump of grass in its path (including the clump itstarts on) until it goes off the lawn.

Because the spring is so young, our lawn is rather delicate. Consequently, we want tomake sure that every clump of grass is mowed once and only once. We will not considerformations that do not meet this criterion.

One more thing: two formations are considered “different” if there exists a clump of grassfor which either (1) for exactly one of the formations does a mobot starts on that clump, or(2) there are mobots starting on this clump for both the formations, but they’re oriented indifferent directions.

1. [15] For this problem, our lawn consists of a row of n clumps of grass. This rowruns in an east-west direction. In our formation, each mobot may be oriented towardthe north, south, east, or west. One example of an allowable formation if n = 6 issymbolized below:

· · ← ↑ ↑↓

(The mobot on the third clump will move westward, mowing the first three clumps.Each of the last three clumps is mowed by a different mobot.) Here’s another allowableformation for n = 6, considered different from the first:

· · ← ↑ ↑ →

Compute the number of different allowable formations for any given n.

Solution: Let a be the number of clumps mowed by a mobot oriented to go west,and let b be the number of clumps mowed by a mobot oriented to go east. (So in thetwo examples given, (a, b) would be (3, 0) and (3, 1), respectively.) We may ask howmany allowable formations with a given ordered pair (a, b) there are, and then sumour answers up over all possible ordered pairs (a, b).

Given any particular (a, b), first of all, a and b had better be non-negative integerssumming to at most n. As long as that’s true, there will be n− a− b clumps of grasswhich are each mowed by a single mobot oriented to go either north or south: there

1

are thus 2n−a−b possibilities. So our answer is

n∑a=0

n−a∑b=0

2n−a−b =n∑

a=0

(2n−a + 2n−a−1 + · · ·+ 20) =n∑

a=0

(2n−a+1 − 1)

= −(n + 1) +n∑

a=0

2n−a+1 = −(n + 1) + (2n+1 + 2n + 2n−1 + · · ·+ 21)

= −(n + 1) + (2n+2 − 2) = 2n+2 − n− 3.

2. [25] For this problem, our lawn is an m×n rectangular grid of clumps, that is, with mrows running east-west and n columns running north-south. To be even more explicit,we might say our clumps are at the lattice points

{(x, y) ∈ Z2 | 0 ≤ x < n and 0 ≤ y < m}.

However, mobots are now allowed to be oriented to go either north or east only. Soone allowable formation for m = 2, n = 3 might be as follows:

· → ·↑ → ·

Prove that the number of allowable formations for given m and n is(m + n)!

m! n!.

Solution: There is a one-to-one correspondence between allowable formations andpaths from (0, 0) to (n,m) made up of n moves 1 unit to the right and m moves 1unit up. The correspondence works as follows: There must be a mobot at (0, 0), sostart the path there. If that mobot is oriented to move up, then move to the right;if that mobot is oriented to move to the right, then move up. In doing so, you willmeet another mobot, upon which you can repeat the above process, until you leavethe lawn. Once you leave the lawn, there will be only one way to proceed to (n, m) —either keep going right, or keep going up. (For the example in the problem, our pathwould be (0, 0)–(1, 0)–(1, 1)–(1, 2)–(2, 2)–(3, 2).)

Conversely, given any path from (0, 0) to (n,m), we can derive back a mobot formation:place a mobot at every lattice point of the lawn that the path touches, and don’t orientthat mobot in the same direction as the path takes when leaving that point.

It is easy to check that this is indeed a one-to-one correspondence as claimed. Everypath from (0, 0) to (n,m) consists of n moves to the right and m moves up done in an

arbitrary order, and there are precisely(

m+nm

)= (m+n)!

m! n!orders.

3. [40] In this problem, we stipulate that m ≥ n, and the lawn is shaped differently. Theclumps are now at the lattice points in a trapezoid:

{(x, y) ∈ Z2 | 0 ≤ x < n and 0 ≤ y < m + 1− n + x},

As in problem 2, mobots can be set to move either north or east. For given m and n,determine with proof the number of allowable formations.

2

Solution: For exactly the same reasons as in problem 2, we have a one-to-onecorrespondence between the allowable formations and paths (going 1 unit up or rightat a time) from (0, 0) to (n, m) avoiding points (x, y) with y > m + 1− n + x.

The number of these paths equals the total number of up/right paths from (0, 0) to(n,m) minus the number of up/right paths from (0, 0) to (n, m) that do pass throughat least one point (x, y) with y > m + 1− n + x. The first of these numbers is

(m+n

m

),

as before. It remains to calculate the second of these numbers.

To that end, we first note that the reflection of (n, m) across the line y = m+2−n+xis (n − 2, m + 2). Now there is a one-to-one correspondence between up/right pathsfrom (0, 0) to (m,n) that pass through something with y ≥ m+2−n+x with up/rightpaths from (0, 0) to (n− 2, m + 2): indeed, taking a path of one kind, we may isolatethe first point on it that lies on the line y = m + 2− n + x, and reflect the rest of thepath through that line, to obtain a path of the other kind. There are therefore

(m+nm+2

)paths of either kind.

The answer is therefore (m + n

m

)−(

m + n

m + 2

)or, if you happen to prefer,

(m + n)! [(m + 2)(m + 1)− n(n− 1)]

(m + 2)! n!.

4. [15] In this problem and the next, the lawn consists of points in a triangular grid ofsize n, so that for n = 3 the lawn looks like

·· ·· · ·

Mobots are allowed to be oriented to the east, 30◦ west of north, or 30◦ west of south.Under these conditions, for any given n, what is the minimum number of mobotsneeded to now the lawn?

Answer: nSolution: It is evident that n mobots are enough; just place one at the left edge ofeach row, and set them to move to the right. Suppose for the sake of contradictionthat, for some n, it is possible to mow the lawn with fewer than n mobots. Considerthe minimum n for which this is the case. Clearly n > 1. Now consider the mobotthat mows the southwest corner of the lawn. This mobot is confined to either thebottom row or the left edge of the lawn; let’s assume it’s the former (the latter case isvery similar). This mobot starts in the bottom row. Let’s throw it away. Now if anyremaining mobots start in the bottom row, they must not be set to move east, andwe may advance them by one step in the direction they’re set to go in, so that theyland either off the lawn or in the second-to-last row. In doing so we have produced astarting formation of fewer than n − 1 mobots in the first n − 1 rows that will mowthose rows. This contradicts the assumed minimality of n.

3

5. [25] With the same lawn and the same allowable mobot orientations as in the previousproblem, let us call a formation “happy” if it is invariant under 120◦ rotations. (Arotation applies both to the positions of the mobots and to their orientations.) Anexample of a happy formation for n = 2 might be

↖↙ →

Find the number of happy formations for a given n.

Solution: If n ≡ 1 (mod 3), then there is a clump of grass at the center of the lawn;otherwise there are 3 blades of grass equally closest to the center. In the former case,whatever mobot mows this center blade of grass cannot possibly have a counterpartunder a 120◦ rotation: if this mobot starts in the center, it must be oriented a certainway, and whatever way that is, cannot remain invariant under such a rotation; butif the mobot does not start in the center, then it will collide in the center with itscounterparts under the 120◦ rotations.

In the case that n 6≡ 1 (mod 3), by considering how the three central blades of grasscan be mowed, we easily see that there are two possibilities, both of which involvestarting one mobot at each of these three positions. Both possibilities distinctly havethe effect of cutting up the rest of the lawn (i.e., the part of the lawn not mowed byany of these three mobots) into three congruent pieces. The point is that these piecesare equivalent to trapezoids as in problem 3, both in their shape, and in the alloweddirections of mobot motion. Luckily, we know how to count the ways to configure eachsuch trapezoid. Call the trapezoid defined in problem 3 an “(m, n)-trapezoid.”

If n ≡ 0 (mod 3), then these pieces are either all (2n3

, n−33

)-trapezoids or all (2n−33

, n3)-

trapezoids, depending on how the central mobots are configured. In this case, theanswer is ((

n− 12n3

)−(

n− 12n+6

3

))3

+

((n− 12n−3

3

)−(

n− 12n+3

3

))3

.

If, on the other hand, n ≡ 2 (mod 3), then these pieces are either all (2n−13

, n−23

)-trapezoids or all (2n−4

3, n+1

3)-trapezoids, again depending on how the central mobots

are configured. In this case, the answer is((n− 12n−1

3

)−(

n− 12n+1

3

))3

+

((n− 12n−4

3

)−(

n− 12n−2

3

))3

.

Polygons [110]

6. [15] Let n be an integer at least 5. At most how many diagonals of a regular n-goncan be simultaneously drawn so that no two are parallel? Prove your answer.

Answer: nSolution: Let O be the center of the n-gon. Let us consider two cases, based on theparity of n:

4

• n is odd. In this case, for each diagonal d, there is exactly one vertex D of then-gon, such that d is perpendicular to line OD; and of course, for each vertexD, there is at least one diagonal d perpendicular to OD, because n ≥ 5. Theproblem of picking a bunch of d’s so that no two are parallel is thus transmutedinto one of picking a bunch of d’s so that none of the corresponding D’s are thesame. Well, go figure.

• n is even. What can I say? For each diagonal d, the perpendicular dropped fromO to d either passes through two opposite vertices of the n-gon, or else bisectstwo opposite sides. Conversely, for each line joining opposite vertices or bisectingopposite sides, there is at least one diagonal perpendicular to it, because n ≥ 6.By reasoning similar to the odd case, we find the answer to be n.

7. [25] Given a convex n-gon, n ≥ 4, at most how many diagonals can be drawn suchthat each drawn diagonal intersects every other drawn diagonal either in the interiorof the n-gon or at a vertex? Prove your answer.

Answer: If n = 4, then 2; otherwise, n.Solution: First of all, assume without loss of generality that the n-gon is regular(this has no effect as far as diagonal intersection is concerned). Also, treat n = 4 as aspecial case; obviously the answer is 2 here.

If n is odd, simply draw n diagonals, connecting each vertex to the ones (n − 1)/2vertices away (in either direction).

If n is even, first draw the n/2 diagonals connecting pairs of vertices n/2 verticesapart. Then, there are n diagonals connecting pairs of vertices n/2− 1 vertices apart;they come in n/2 pairs of parallel diagonals; from each such pair, randomly pick onediagonal and draw it.

To see that these constructions work, note that two diagonals, each connecting pairsof vertices at least n/2− 1 vertices apart, can fail to intersect or share a vertex only ifthey are parallel.

The previous problem shows that these constructions are optimal.

8. [15] Given a regular n-gon with sides of length 1, what is the smallest radius r suchthat there is a non-empty intersection of n circles of radius r centered at the verticesof the n-gon? Give r as a formula in terms of n. Be sure to prove your answer.

Answer: r =1

2csc

180◦

nSolution: It is easy to see that, with this r, all the circles pass through the centerof the n-gon. The following proves that this r is necessary even if the word “circle” isreplaced by the word “disk.”

For n even, it is easy to see using symmetry that containing the center point is neces-sary and sufficient. For n odd, there is more work to do. Again, containing the centerpoint is sufficient. To see its necessity, consider three circles: a circle at a vertex A,and the two circles on the segment BC opposite A. Circles B and C intersect in aregion R symmetric about the perpendicular bisector of BC, with the closest pointof R to A being on this line. Hence, circle A must intersect R at some point on theperpendicular bisector of BC; and thus we see the entire perpendicular bisector of BC

5

inside of the n-gon is contained in the circles. Now, this bisector contains the centerof the n-gon, so some circle must contain the center. But by symmetry, if one circlecontains the center, all do. Thus, in any case, it is necessary and sufficient for r tobe large enough so the center is contained in a circle. Basic trigonometry gives theanswer, which equals the distance between a vertex and the center.

9. [40] Let n ≥ 3 be a positive integer. Prove that given any n angles 0 < θ1, θ2, . . . , θn <180◦, such that their sum is 180(n − 2) degrees, there exists a convex n-gon havingexactly those angles, in that order.

Solution: We induct on n. The statement holds trivially for n = 3, as all trianglesare convex. Now, suppose that the statement is true for n − 1, where n ≥ 4. Letθ1, θ2, . . . , θn be n angles less than 180◦ whose sum equals 180(n − 2) degrees. Thestatement is clearly true if n = 4 and θ1 = θ3 = 180◦ − θ2 = 180◦ − θ4 since we caneasily form a parallelogram, so assume otherwise.

I claim that there exist two adjacent angles whose sum is greater than 180◦. Assumeotherwise. Then, we have θi+θi+1 ≤ 180 for i = 1, 2, . . . , n, where θn+1 = θ1. Summingthese inequalities over all i yields 2 · 180(n− 2) ≤ 180n, which is equivalent to n ≤ 4.Of course, we can have n = 4 if and only if we have equality in each of the aboveinequalities, forcing us to have a parallelogram contrary to our assumption.

Hence, we have two adjacent angles with sum greater than 180◦. Without loss ofgenerality, let these angles be θn−1 and θn, relabeling if necessary. By the inductivehypothesis, we may construct an (n−1)-gon with angles θ1, θ2, . . . , θn−2, θn−1+θn−180◦,as these angles are each less than 180◦ and their sum equals 180(n−3) degrees. Considerthe vertex with angle θn−1 + θn − 180◦. Note that we can “clip off” a triangle withangles θn−1 + θn − 180◦, 180◦ − θn−1, and 180◦ − θn at this vertex, yielding an n-gonwith the desired angles, completing the inductive step.

10. [15] Suppose we have an n-gon such that each interior angle, measured in degrees, is apositive integer. Suppose further that all angles are less than 180◦, and that all anglesare different sizes. What is the maximum possible value of n? Prove your answer.

Answer: 26Solution: Let’s work with the exterior angles (each is 180 minus the interior angle).Then the conditions on the exterior angles are identical to the conditions on the interiorangles: each is a positive integer between 1 and 179 inclusive. The sum of the exteriorangles is exactly 360. However, the sum of 1 through 27 is 27 · 28/2 = 378, which istoo large. We can get 26 using angles of 1 through 25 (sum 325) and an angle of 35.The previous problem shows that this is actually possible.

What do the following problems have in common? [170]

11. [15] The lottery cards of a certain lottery contain all nine-digit numbers that can beformed with the digits 1, 2 and 3. There is exactly one number on each lottery card.There are only red, yellow and blue lottery cards. Two lottery numbers that differfrom each other in all nine digits always appear on cards of different color. Someone

6

draws a red card and a yellow card. The red card has the number 122 222 222 and theyellow card has the number 222 222 222. The first prize goes to the lottery card withthe number 123 123 123. What color(s) can it possibly have? Prove your answer.

Answer: The card with the number 123 123 123 is red.Solution: First, it can in fact be red, if, say, cards are colored based on the first digitonly (1 = red, 2 = yellow, 3 = blue). We now endeavor to show it must be red.

Consider the cards 333 133 133 and 331 331 331: they each differ in all their digitsfrom 122 222 222 and from 222 222 222, so they must both be blue. Now 211 311 311differs in all its digits from both 122 222 222 and 333 133 133, so it must be yellow.Finally, 123 123 123 differs in all its digits from both 331 331 331 and 211 311 311, soit must be red.

12. [25] A 3 × 3 × 3 cube is built from 27 unit cubes. Suddenly five of those cubes mys-teriously teleport away. What is the minimum possible surface area of the remainingsolid? Prove your answer.

Answer: 50Solution: Orient the cube so that its edges are parallel to the x-, y-, and z-axes. Aset of three unit cubes whose centers differ only in their x-coordinate will be termedan “x-row”; there are thus nine x-rows. Define “y-row” and “z-row” similarly.

To achieve 50, simply take away one x-row and one y-row (their union consists ofprecisely five unit cubes).

To show that 50 is the minimum: Note that there cannot be two x-rows that areboth completely removed, as that would imply removing six unit cubes. (Similarstatements apply for y- and z-rows, of course.) It is also impossible for there to be onex-row, one y-row, and one z-row that are all removed, as that would imply removingseven unit cubes. Every x-, y-, or z-row that is not completely removed contributesat least 2 square units to the surface area. Thus, the total surface area is at least9 · 2 + 8 · 2 + 8 · 2 = 50.

13. [40] Having lost a game of checkers and my temper, I dash all the pieces to the groundbut one. This last checker, which is perfectly circular in shape, remains completely onthe board, and happens to cover equal areas of red and black squares. Prove that thecenter of this piece must lie on a boundary between two squares (or at a junction offour).

Solution: Suppose, for the sake of contradiction, that the problem is false. Evidently,at least one boundary between adjacent squares must lie within our checker, or else thechecker would exist entirely within one square, meaning it would cover only one color.Note also that a checker’s diameter is smaller than the side of any square of the board,so there are at most two such boundaries within our checker (one in each direction).Let ` be this, or one of these, boundaries. Draw a diameter d of the checker parallelto `. Presumably, the strip of the checker between ` and d is part red, part black.These red and black areas are unequal, however, because the center of the checker doesnot lie on any boundary between squares. But, if we were to swap colors within thisstrip, then the checker would have equal red and black areas, because then it would be

7

colored in a way such that flipping it across d swaps the colors. This shows that, theway it is currently colored, the checker does not have equal red and black areas. Thisgives us the desired contradiction.

14. [40] A number n is called bummed out if there is exactly one ordered pair of positiveintegers (x, y) such that

bx2/yc+ by2/xc = n.

Find all bummed out numbers.

Answer: 2, 6, 8, 10Solution: Suppose n is bummed out. If (a, b) is one solution for (x, y) to the givenequation bx2/yc+by2/xc = n, then (b, a) is another, so the unique solution (a, b) betterhave the property that a = b and n = 2a ≥ 2. In particular, n is an even positiveinteger.

Now, if n = 2a ≥ 12, then setting x = a− 1 ≥ 5, y = a + 1 ≥ 7, we have⌊x2

y

⌋+

⌊y2

x

⌋=

⌊a− 3 +

4

a + 1

⌋+

⌊a + 3 +

4

a− 1

⌋= 2a = n,

so n cannot be bummed out.

Moreover, b12/2c+ b22/1c = 4, so 4 is not bummed out. The only possibilities left are2, 6, 8, and 10.

To check these, note that

n =

⌊x2

y

⌋+

⌊y2

x

⌋> −2 +

x2

2y+

x2

2y+

y2

x≥ −2 +

3x3√

4

so

x <3√

4

3(n + 2) < .53(n + 2),

and similarly for y. So we only have to check x, y ≤ b.53(10 + 2)c = 6:

x \ y 1 2 3 4 5 61 2 4 9 16 25 362 4 4 5 9 12 183 9 5 6 7 9 134 16 9 7 8 9 115 25 12 9 9 10 116 36 18 13 11 11 12

15. [50] Find, with proof, all positive integer palindromes whose square is also a palin-drome.

Answer: A palindrome satisfies the requirement if and only if the sum of the squaresof its digits is less than 10. We may categorize these numbers this way:

• 3

• Any palindromic combination of 1s and 0s with at most nine 1s.

8

• Any palindrome consisting of a single 2 in the middle and 1s and 0s elsewhere,with at most four 1s.

• 2000 . . . 0002

• 2000 . . . 0001000 . . . 0002

Solution: Let n :=d∑

i=0

ai ·10i be a palindrome, where the ai are digits with ai = ad−i

and ad 6= 0. Then, if we let

bk :=∑

i+j=k

aiaj

for all 0 ≤ k ≤ 2d, then

n2 =2d∑

k=0

bk · 10k

(this is not necessarily the decimal expansion of n2, however). We have to show that∑di=0 a2

i < 10 if and only if n2 is a palindrome.

Suppose∑d

i=0 a2i < 10. Then, by the AM-GM inequality, we have

bk =∑

i+j=k

aiaj ≤∑

i+j=k

a2i + a2

j

2≤

d∑i=0

a2i

2+

d∑j=0

a2j

2<

10

2+

10

2= 10.

Thus, loosely speaking, no carrying is ever done in computing n × n by long multi-plication, so the digit in the 10k place in n2 is precisely bk, and it’s easy to see thatbk = b2d−k and that b2d = a2

d 6= 0. So n2 is indeed a palindrome, as desired.

Now suppose∑d

i=0 a2i ≥ 10. Here note that

bd =∑

i+j=d

aiaj =d∑

i=0

aiad−i =d∑

i=0

a2i ≥ 10.

Thus, it cannot be true that, for all k, bk represents the 10k digit of n2, because nodigit can be greater than or equal to 10. Let ` be the greatest such that b` does notrepresent the 10` digit of n2. We are trying to prove that n2 cannot be a palindrome.Consider three cases:

• ad = a0 ≥ 4. In this case we must have ` ≥ 2d, because b2d = a2d > 10.

If a0 = 4, then n2 ends in the digit 6, but lies in the interval [16 · 102d, 25 · 102d),and so starts with either a 1 or a 2; thus, n2 cannot be a palindrome. Similarly,if a0 = 5, then n2 ends in 5 but starts with 2 or 3; if a0 = 6, then n2 ends in 6but starts with 3 or 4; if a0 = 7, then n2 ends in 9 but starts with 4, 5, or 6; ifa0 = 8, then n2 ends in 4 but starts with 6, 7 or 8; if a0 = 9, then n2 ends in 1but starts with 8 or 9.

• ` ≥ 2d and ad = a0 ≤ 3.

Here we do something similar, but with a slight twist. The units digit of n2 isa2

0. Because ` ≥ 2d, n2 must be in the interval [(a20 + 1) · 102d, (a0 + 1)2 · 102d),

9

which is certainly a subset of the interval [(a20 + 1) · 102d, a2

0 · 102d+1). No integerin even this larger interval manages to start with the digit a2

0, so n2 cannot bepalindromic.

• ` < 2d.

Here we can rest assured that n2 does have (2d + 1) digits — that is, the firstdigit is in the 102d place. In order for n2 to be a palindrome, the digits in the 10k

and 102d−k places must always be the same.

Now b`, b`+1, . . . , b2d had all better be less than 10, or else ` would be greater thanwhat it is. Thus, the numbers just listed do appear as the lowest digits of n2 inleft-to-right order, although they don’t appear as the highest (2d + 1 − `) digitsof n2 in right-to-left order. Thus, n2 cannot be a palindrome.

10

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Team Round B

Mobotics [135]

Spring is finally here in Cambridge, and it’s time to mow our lawn. For the purpose of these problems, ourlawn consists of little clumps of grass arranged in an m× n rectangular grid, that is, with m rows runningeast-west and n columns running north-south. To be even more explicit, we might say our clumps are at thelattice points

{(x, y) ∈ Z2 | 0 ≤ x < n and 0 ≤ y < m}.Our machinery consists of a fleet of identical mowbots (or “mobots” for short). A mobot is a lawn-mowingmachine. To mow our lawn, we begin by choosing a formation: we place as many mobots as we want atvarious clumps of grass and orient each mobot’s head in a certain direction, either north or east (not southor west). At the blow of a whistle, each mobot starts moving in the direction we’ve chosen, mowing everyclump of grass in its path (including the clump it starts on) until it goes off the lawn.

Because the spring is so young, our lawn is rather delicate. Consequently, we want to make sure thatevery clump of grass is mowed once and only once. We will not consider formations that do not meet thiscriterion.

One more thing: two formations are considered “different” if there exists a clump of grass for which either(1) for exactly one of the formations does a mobot starts on that clump, or (2) there are mobots starting onthis clump for both the formations, but they’re oriented in different directions.

As an example, one allowable formation for m = 2, n = 3 might be as follows:

· → ·↑ → ·

1. [25] Prove that the maximum number of mobots you need to mow your lawn is m + n− 1.

2. [40] Prove that the minimum number of mobots you need to mow your lawn is min{m,n}.

3. [15] Prove that, given any formation, each mobot may be colored in one of three colors — say, white,black, and blue — such that no two adjacent clumps of grass are mowed by different mobots of thesame color. Two clumps of grass are adjacent if the distance between them is 1. In your proof, youmay use the Four-Color Theorem if you’re familiar with it.

4. [15] For n = m = 4, find a formation with 6 mobots for which there are exactly 12 ways to color themobots in three colors as in problem 3. (No proof is necessary.)

5. [40] For n, m ≥ 3, prove that a formation has exactly six possible colorings satisfying the conditionsin problem 3 if and only if there is a mobot that starts at (1, 1).

Polygons [130]

6. [15] Suppose we have a regular hexagon and draw all its sides and diagonals. Into how many regionsdo the segments divide the hexagon? (No proof is necessary.)

7. [25] Suppose we have an octagon with all angles of 135◦, and consecutive sides of alternating length1 and

√2. We draw all its sides and diagonals. Into how many regions do the segments divide the

octagon? (No proof is necessary.)

8. [25] A regular 12-sided polygon is inscribed in a circle of radius 1. How many chords of the circle thatjoin two of the vertices of the 12-gon have lengths whose squares are rational? (No proof is necessary.)

9. [25] Show a way to construct an equiangular hexagon with side lengths 1, 2, 3, 4, 5, and 6 (notnecessarily in that order).

10. [40] Given a convex n-gon, n ≥ 4, at most how many diagonals can be drawn such that each drawndiagonal intersects every other drawn diagonal strictly in the interior of the n-gon? Prove that youranswer is correct.

1

What do the following problems have in common? [135]

11. [15] Find the largest positive integer n such that 1! + 2! + 3! + · · ·+ n! is a perfect square. Prove thatyour answer is correct.

12. [15] Find all ordered triples (x, y, z) of positive reals such that x + y + z = 27 and x2 + y2 + z2 − xy−yz − zx = 0. Prove that your answer is correct.

13. [25] Four circles with radii 1, 2, 3, and r are externally tangent to one another. Compute r. (No proofis necessary.)

14. [40] Find the prime factorization of

20062 · 2262− 6692 · 3599 + 15932 · 1337.

(No proof is necessary.)

15. [40] Let a, b, c, d be real numbers so that c, d are not both 0. Define the function

m(x) =ax + b

cx + d

on all real numbers x except possibly −d/c, in the event that c 6= 0. Suppose that the equationx = m(m(x)) has at least one solution that is not a solution of x = m(x). Find all possible values ofa + d. Prove that your answer is correct.

2

IXth Annual Harvard-MIT Mathematics TournamentSaturday 25 February 2006

Team Round B: Solutions

Mobotics [135]

Spring is finally here in Cambridge, and it’s time to mow our lawn. For the purpose of theseproblems, our lawn consists of little clumps of grass arranged in an m× n rectangular grid,that is, with m rows running east-west and n columns running north-south. To be even moreexplicit, we might say our clumps are at the lattice points

{(x, y) ∈ Z2 | 0 ≤ x < n and 0 ≤ y < m}.

Our machinery consists of a fleet of identical mowbots (or “mobots” for short). A mobotis a lawn-mowing machine. To mow our lawn, we begin by choosing a formation: we placeas many mobots as we want at various clumps of grass and orient each mobot’s head in acertain direction, either north or east (not south or west). At the blow of a whistle, eachmobot starts moving in the direction we’ve chosen, mowing every clump of grass in its path(including the clump it starts on) until it goes off the lawn.

Because the spring is so young, our lawn is rather delicate. Consequently, we want tomake sure that every clump of grass is mowed once and only once. We will not considerformations that do not meet this criterion.

One more thing: two formations are considered “different” if there exists a clump of grassfor which either (1) for exactly one of the formations does a mobot starts on that clump, or(2) there are mobots starting on this clump for both the formations, but they’re oriented indifferent directions.

As an example, one allowable formation for m = 2, n = 3 might be as follows:

· → ·↑ → ·

1. [25] Prove that the maximum number of mobots you need to mow your lawn is m+n−1.

Solution: This is attainable if we place one mobot at each clump in the first row,oriented north, and one mobot in the first column of each row except the first, orientedeast.

To show that at most m+n− 1 mobots can be used, note that each mobot must mowat least one of the m + n− 1 clumps in union of the first row and last column.

2. [40] Prove that the minimum number of mobots you need to mow your lawn ismin{m, n}.

Solution: This is attainable if we place one mobot at each clump in the first column,oriented toward the east, or one mobot at each clump in the last row, oriented towardthe north (whichever of the two is more efficient).

To show that at least min{m, n} mobots must be used, note that each of the clumps(0, 0), (1, 1), (2, 2), . . . , (min{m, n} − 1, min{m, n} − 1) must be mowed by a differentmobot.

1

3. [15] Prove that, given any formation, each mobot may be colored in one of three colors— say, white, black, and blue — such that no two adjacent clumps of grass are mowedby different mobots of the same color. Two clumps of grass are adjacent if the distancebetween them is 1. In your proof, you may use the Four-Color Theorem if you’refamiliar with it.

Solution: We can divide the coordinate plane into regions: Let’s say a point belongsto Region 0 if the closest lattice point to it is not on the lawn, and each mobot Mowns a region that is the set of points for which the closest lattice point is on the lawnand mowed by M . Applying the Four-Color Theorem to these regions, we note that allthe conditions demanded in the problem are satisfied. In particular at most 3 colorsare used on the mobots because every mobot region borders Region 0 and hence is notcolored the same color as Region 0.

4. [15] For n = m = 4, find a formation with 6 mobots for which there are exactly 12ways to color the mobots in three colors as in problem 3. (No proof is necessary.)

Solution: Place north-oriented mobots at (0, 0), (1, 0), (2, 2), and (3, 2), and east-oriented mobots at (2, 0) and (2, 1). Other formations are possible.

5. [40] For n,m ≥ 3, prove that a formation has exactly six possible colorings satisfyingthe conditions in problem 3 if and only if there is a mobot that starts at (1, 1).

Solution: Let’s trace a path through all the mobots. There must be a mobot at(0, 0), so start the path there. If that mobot is oriented to move up, then move tothe right; if that mobot is oriented to move to the right, then move up. In doing so,you will meet another mobot, upon which you can repeat the above process, until youleave the lawn.

In tracing this path, we’ve effectively ordered the mobots, starting with (0, 0). At anypoint in this ordering, we have a number of choices for how to color the current mobotwithout violating the condition with any previously colored mobot. Specifically, thereare three ways to color the mobot at (0, 0). If a mobot has x- or y-coordinate 0, thereare 2 ways to color it: it need only not be the same color as the previous mobot onour trail. If a mobot has both x- and y-coordinates positive, then there is only 1 wayto color it: it cannot be the same color as whatever mobots mows the clump directlysouth or directly west of its starting point.

Thus, if there is a mobot at (1, 1), our path must begin with either (0, 0)–(0, 1)–(1, 1)or (0, 0)–(1, 0)–(1, 1), and in either case there are only 6 ways to do the coloring.Otherwise, if there is no mobot at (1, 1), our path must begin with either (0, 0)–(0, 1)–(0, 2) or (0, 0)–(1, 0)–(2, 0), and in either case there are already 3× 2× 2 = 12 ways todo the coloring thus far.

Polygons [130]

6. [15] Suppose we have a regular hexagon and draw all its sides and diagonals. Into howmany regions do the segments divide the hexagon? (No proof is necessary.)

2

Answer: 24Solution: An accurate diagram and a careful count yields the answer.

7. [25] Suppose we have an octagon with all angles of 135◦, and consecutive sides ofalternating length 1 and

√2. We draw all its sides and diagonals. Into how many

regions do the segments divide the octagon? (No proof is necessary.)

Answer: 84Solution: The easiest way to see the answer is to view the octagon as five unitsquares in a cross arrangement, with four half-squares wedged at the corners. Thecenter square is divided into 8 regions. The other 4 squares are each divided into15 regions. The 4 half-squares are each divided into 4 regions. The answer is thus8 + 4× 15 + 4× 4 = 84.

8. [25] A regular 12-sided polygon is inscribed in a circle of radius 1. How many chordsof the circle that join two of the vertices of the 12-gon have lengths whose squares arerational? (No proof is necessary.)

Answer: 42Solution: The chords joining vertices subtend minor arcs of 30◦, 60◦, 90◦, 120◦, 150◦,or 180◦. There are 12 chords of each of the first five kinds and 6 diameters. For achord with central angle θ, we can draw radii from the two endpoints of the chord tothe center of the circle. By the law of cosines, the square of the length of the chord is1 + 1− 2 cos θ, which is rational when θ is 60◦, 90◦, 120◦, or 180◦. The answer is thus12 + 12 + 12 + 6 = 42.

9. [25] Show a way to construct an equiangular hexagon with side lengths 1, 2, 3, 4, 5,and 6 (not necessarily in that order).

Solution: The trick is to view an equiangular hexagon as an equilateral trianglewith its corners cut off. Consider an equilateral triangle with side length 9, and cutoff equilateral triangles of side length 1, 2, and 3 from its corners. This yields anequiangular hexagon with sides of length 1, 6, 2, 4, 3, 5 in that order.

10. [40] Given a convex n-gon, n ≥ 4, at most how many diagonals can be drawn suchthat each drawn diagonal intersects every other drawn diagonal strictly in the interiorof the n-gon? Prove that your answer is correct.

Answer: bn/2cSolution: If n is even, simply draw all n/2 diagonals connecting a vertex to the onen/2 vertices away. If n is odd, pretend one of the vertices does not exist, and do theabove for the (n− 1)-gon remaining.

To show this is optimal, consider any given drawn diagonal: it divides the remainingn− 2 vertices into two camps, one of which therefore has size at most bn/2c − 1, andone cannot draw two diagonals sharing a vertex.

What do the following problems have in common? [135]

11. [15] Find the largest positive integer n such that 1! + 2! + 3! + · · · + n! is a perfectsquare. Prove that your answer is correct.

3

Answer: 3Solution: Clearly 1! + 2! + 3! = 9 works. For n ≥ 4, we have

1! + 2! + 3! + · · ·+ n! ≡ 1! + 2! + 3! + 4! ≡ 3 (mod 5),

but there are no squares congruent to 3 modulo 5.

12. [15] Find all ordered triples (x, y, z) of positive reals such that x + y + z = 27 andx2 + y2 + z2 − xy − yz − zx = 0. Prove that your answer is correct.

Answer: (9, 9, 9)

Solution: We have x2 + y2 + z2−xy− yz− zx =(x− y)2 + (y − z)2 + (z − x)2

2= 0,

which implies x = y = z =27

3= 9.

13. [25] Four circles with radii 1, 2, 3, and r are externally tangent to one another. Computer. (No proof is necessary.)

Answer: 6/23Solution: Let A, B, C, P be the centers of the circles with radii 1, 2, 3, and r,respectively. Then, ABC is a 3-4-5 right triangle. Using the law of cosines in 4PAByields

cos ∠PAB =32 + (1 + r)2 − (2 + r)2

2 · 3 · (1 + r)=

3− r

3(1 + r)

Similarly,

cos ∠PAC =42 + (1 + r)2 − (3 + r)2

2 · 4 · (1 + r)=

2− r

2(1 + r)

We can now use the equation (cos ∠PAB)2 + (cos ∠PAC)2 = 1, which yields 0 =23r2 + 132r − 36 = (23r − 6)(r + 6), or r = 6/23.

14. [40] Find the prime factorization of

20062 · 2262− 6692 · 3599 + 15932 · 1337.

(No proof is necessary.)

Answer: 2 · 3 · 7 · 13 · 29 · 59 · 61 · 191Solution: Upon observing that 2262 = 669 + 1593, 3599 = 1593 + 2006, and 1337 =2006− 669, we are inspired to write a = 2006, b = 669, c = −1593. The expression inquestion then rewrites as a2(b − c) + b2(c − a) + c2(a − b). But, by experimenting inthe general case (e.g. setting a = b), we find that this polynomial is zero when twoof a, b, c are equal. Immediately we see that it factors as (b − a)(c − b)(a − c), so theoriginal expression is a way of writing (−1337) · (−2262) · (3599). Now, 1337 = 7 · 191,2262 = 2 · 3 · 13 · 29, and 3599 = 602 − 12 = 59 · 61.

15. [40] Let a, b, c, d be real numbers so that c, d are not both 0. Define the function

m(x) =ax + b

cx + d

4

on all real numbers x except possibly −d/c, in the event that c 6= 0. Suppose that theequation x = m(m(x)) has at least one solution that is not a solution of x = m(x).Find all possible values of a + d. Prove that your answer is correct.

Answer: 0Solution: That 0 is a possible value of a + d can be seen by taking m(x) = −x, i.e.,a = −d = 1, b = c = 0. We will now show that 0 is the only possible value of a + d.

The equation x = m(m(x)) implies x =(a2 + bc)x + (a + d)b

(a + d)cx + (bc + d2), which in turn implies

(a + d)[cx2 + (−a + d)x− b] = 0.

Suppose for the sake of contradiction that a + d 6= 0. Then the above equation wouldfurther imply

cx2 + (−a + d)x− b = 0, x(cx + d) = ax + b,

which would imply x = m(x) for any x except possibly −d/c. But of course −d/c isnot a root of x = m(m(x)) anyway, so in this case, all solutions of x = m(m(x)) arealso solutions of x = m(x), a contradiction. So our assumption was wrong, and in facta + d = 0, as claimed.

5

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Individual Round: Algebra Test

1. [3] Compute ⌊2007! + 2004!2006! + 2005!

⌋.

(Note that bxc denotes the greatest integer less than or equal to x.)

2. [3] Two reals x and y are such that x− y = 4 and x3 − y3 = 28. Compute xy.

3. [4] Three real numbers x, y, and z are such that (x+4)/2 = (y+9)/(z−3) = (x+5)/(z−5). Determinethe value of x/y.

4. [4] Compute23 − 123 + 1

· 33 − 1

33 + 1· 4

3 − 143 + 1

· 53 − 1

53 + 1· 6

3 − 163 + 1

.

5. [5] A convex quadrilateral is determined by the points of intersection of the curves x4 + y4 = 100 andxy = 4; determine its area.

6. [5] Consider the polynomial P (x) = x3 + x2 − x + 2. Determine all real numbers r for which thereexists a complex number z not in the reals such that P (z) = r.

7. [5] An infinite sequence of positive real numbers is defined by a0 = 1 and an+2 = 6an − an+1 forn = 0, 1, 2, · · · Find the possible value(s) of a2007.

8. [6] Let A := Q \ {0, 1} denote the set of all rationals other than 0 and 1. A function f : A → R hasthe property that for all x ∈ A,

f (x) + f

(1− 1

x

)= log |x|.

Compute the value of f(2007).

9. [7] The complex numbers α1, α2, α3, and α4 are the four distinct roots of the equation x4 +2x3 +2 = 0.Determine the unordered set

{α1α2 + α3α4, α1α3 + α2α4, α1α4 + α2α3}.

10. [8] The polynomial f(x) = x2007+17x2006+1 has distinct zeroes r1, . . . , r2007. A polynomial P of degree2007 has the property that P

(rj + 1

rj

)= 0 for j = 1, . . . , 2007. Determine the value of P (1)/P (−1).

1

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Individual Round: Algebra Test

1. [3] Compute ⌊2007! + 2004!2006! + 2005!

⌋.

(Note that bxc denotes the greatest integer less than or equal to x.)

Answer: 2006. We have⌊2007! + 2004!2006! + 2005!

⌋=

⌊(2007 · 2006 + 1

2005

)· 2005!

(2006 + 1) · 2005!

⌋=⌊

2007 · 2006 + 12005

2007

⌋=⌊2006 +

12005 · 2007

⌋.

2. [3] Two reals x and y are such that x− y = 4 and x3 − y3 = 28. Compute xy.

Answer: −3. We have 28 = x3−y3 = (x−y)(x2 +xy+y2) = (x−y)((x−y)2 +3xy) = 4 · (16+3xy),from which xy = −3.

3. [4] Three real numbers x, y, and z are such that (x+4)/2 = (y+9)/(z−3) = (x+5)/(z−5). Determinethe value of x/y.

Answer: 1/2. Because the first and third fractions are equal, adding their numerators and denomi-nators produces another fraction equal to the others: ((x+4)+(x+5))/(2+(z−5)) = (2x+9)/(z−3).Then y + 9 = 2x + 9, etc.

4. [4] Compute23 − 123 + 1

· 33 − 1

33 + 1· 4

3 − 143 + 1

· 53 − 1

53 + 1· 6

3 − 163 + 1

.

Answer: 4363 . Use the factorizations n3 − 1 = (n− 1)(n2 + n + 1) and n3 + 1 = (n + 1)(n2 − n + 1)

to write1 · 73 · 3

· 2 · 134 · 7

· 3 · 215 · 13

· 4 · 316 · 21

· 5 · 437 · 31

=1 · 2 · 433 · 6 · 7

=4363

.

5. [5] A convex quadrilateral is determined by the points of intersection of the curves x4 + y4 = 100 andxy = 4; determine its area.

Answer: 4√

17. By symmetry, the quadrilateral is a rectangle having x = y and x = −y as axes ofsymmetry. Let (a, b) with a > b > 0 be one of the vertices. Then the desired area is(√

2(a− b))·(√

2(a + b))

= 2(a2 − b2) = 2√

a4 − 2a2b2 + b4 = 2√

100− 2 · 42 = 4√

17.

6. [5] Consider the polynomial P (x) = x3 + x2 − x + 2. Determine all real numbers r for which thereexists a complex number z not in the reals such that P (z) = r.

Answer: r > 3, r < 4927 . Because such roots to polynomial equations come in conjugate pairs, we

seek the values r such that P (x) = r has just one real root x. Considering the shape of a cubic, we areinterested in the boundary values r such that P (x)− r has a repeated zero. Thus, we write

P (x)− r = x3 + x2 − x + (2− r) = (x− p)2(x− q) = x3 − (2p + q)x2 + p(p + 2q)x− p2q.

Then q = −2p−1 and 1 = p(p+2q) = p(−3p−2) so that p = 1/3 or p = −1. It follows that the graphof P (x) is horizontal at x = 1/3 (a maximum) and x = −1 (a minimum), so the desired values r arer > P (−1) = 3 and r < P (1/3) = 1/27 + 1/9− 1/3 + 2 = 49/27.

1

7. [5] An infinite sequence of positive real numbers is defined by a0 = 1 and an+2 = 6an − an+1 forn = 0, 1, 2, · · · Find the possible value(s) of a2007.

Answer: 22007. The characteristic equation of the linear homogeneous equation is m2 + m − 6 =(m + 3)(m − 2) = 0 with solutions m = −3 and m = 2. Hence the general solution is given byan = A(2)n +B(−3)n where A and B are constants to be determined. Then we have an > 0 for n ≥ 0,so necessarily B = 0, and a0 = 1⇒ A = 1. Therefore, the unique solution to the recurrence is an = 2n

for all n.

8. [6] Let A := Q \ {0, 1} denote the set of all rationals other than 0 and 1. A function f : A → R hasthe property that for all x ∈ A,

f (x) + f

(1− 1

x

)= log |x|.

Compute the value of f(2007).

Answer: log (2007/2006) . Let g : A→ A be defined by g(x) := 1− 1/x; the key property is that

g(g(g(x))) = 1− 11− 1

1− 1x

= x.

The given equation rewrites as f(x) + f(g(x)) = log |x|. Substituting x = g(y) and x = g(g(z)) givesthe further equations f(g(y)) + f (g (g(y))) = log |g(x)| and f (g (g(z))) + f(z) = log |g(g(x))|. Settingy and z to x and solving the system of three equations for f(x) gives

f(x) =12· (log |x| − log |g(x)|+ log |g(g(x))|) .

For x = 2007, we have g(x) = 20062007 and g(g(x)) = −1

2006 , so that

f(2007) =log |2007| − log

∣∣ 20062007

∣∣+ log∣∣ −12006

∣∣2

= log (2007/2006) .

9. [7] The complex numbers α1, α2, α3, and α4 are the four distinct roots of the equation x4 +2x3 +2 = 0.Determine the unordered set

{α1α2 + α3α4, α1α3 + α2α4, α1α4 + α2α3}.

Answer: {1±√

5,−2}. Employing the elementary symmetric polynomials (s1 = α1+α2+α3+α4 =−2, s2 = α1α2+α1α3+α1α4+α2α3+α2α4+α3α4 = 0, s3 = α1α2α3+α2α3α4+α3α4α1+α4α1α2 = 0,and s4 = α1α2α3α4 = 2) we consider the polynomial

P (x) = (x− (α1α2 + α3α4))(x− (α1α3 + α2α4))(x− (α1α4 + α2α3))

Because P is symmetric with respect to α1, α2, α3, α4, we can express the coefficients of its expandedform in terms of the elementary symmetric polynomials. We compute

P (x) = x3 − s2x2 + (s3s1 − 4s4)x + (−s2

3 − s4s21 + s4s2)

= x3 − 8x− 8= (x + 2)(x2 − 2x− 4)

The roots of P (x) are −2 and 1±√

5, so the answer is {1±√

5,−2}.Remarks. It is easy to find the coefficients of x2 and x by expansion, and the constant term can becomputed without the complete expansion and decomposition of (α1α2 + α3α4)(α1α3 + α2α4)(α1α4 +α2α3) by noting that the only nonzero 6th degree expressions in s1, s2, s3, and s4 are s6

1 and s4s21. The

general polynomial P constructed here is called the cubic resolvent and arises in Galois theory.

2

10. [8] The polynomial f(x) = x2007+17x2006+1 has distinct zeroes r1, . . . , r2007. A polynomial P of degree2007 has the property that P

(rj + 1

rj

)= 0 for j = 1, . . . , 2007. Determine the value of P (1)/P (−1).

Answer: 289259 . For some constant k, we have

P (z) = k2007∏j=1

(z −

(rj +

1rj

)).

Now writing ω3 = 1 with ω 6= 1, we have ω2 + ω = −1. Then

P (1)/P (−1) =k

Q2007j=1

“1−

“rj+

1rj

””k

Q2007j=1

“−1−

“rj+

1rj

”” =∏2007

j=1

r2j−rj+1

r2j +rj+1

=∏2007

j=1(−ω−rj)(−ω2−rj)

(ω−rj)(ω2−rj)

= f(−ω)f(−ω2)f(ω)f(ω2) = (−ω2007+17ω2006+1)(−(ω2)2007+17(ω2)2006+1)

(ω2007+17ω2006+1)((ω2)2007+17(ω2)2006+1) = (17ω2)(17ω)(2+17ω2)(2+17ω)

= 2894+34(ω+ω2)+289 = 289

259 .

3

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Individual Round: Calculus Test

1. [3] Compute:

limx→0

x2

1− cos(x)

2. [3] Determine the real number a having the property that f(a) = a is a relative minimum of f(x) =x4 − x3 − x2 + ax + 1.

3. [4] Let a be a positive real number. Find the value of a such that the definite integral∫ a2

a

dx

x +√

x

achieves its smallest possible value.

4. [4] Find the real number α such that the curve f(x) = ex is tangent to the curve g(x) = αx2.

5. [5] The function f : R→ R satisfies f(x2)f ′′(x) = f ′(x)f ′(x2) for all real x. Given that f(1) = 1 andf ′′′(1) = 8, determine f ′(1) + f ′′(1).

6. [5] The elliptic curve y2 = x3+1 is tangent to a circle centered at (4, 0) at the point (x0, y0). Determinethe sum of all possible values of x0.

7. [5] Compute∞∑

n=1

1n · (n + 1) · (n + 1)!

.

8. [6] Suppose that ω is a primitive 2007th root of unity. Find(22007 − 1

) 2006∑j=1

12− ωj

.

For this problem only, you may express your answer in the form m · nk + p, where m,n, k, and p arepositive integers. Note that a number z is a primitive nth root of unity if zn = 1 and n is the smallestnumber amongst k = 1, 2, . . . , n such that zk = 1.

9. [7] g is a twice differentiable function over the positive reals such that

g(x) + 2x3g′(x) + x4g′′(x) = 0 for all positive reals x. (1)lim

x→∞xg(x) = 1 (2)

Find the real number α > 1 such that g(α) = 1/2.

10. [8] Compute ∫ ∞

0

e−x sin(x)x

dx

1

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Individual Round: Calculus Test

1. [3] Compute:

limx→0

x2

1− cos(x)

Answer: 2. Since sin2(x) = 1− cos2(x), we multiply the numerator and denominator by 1 + cos(x)and use the fact that x/ sin(x)→ 1, obtaining

limx→0

x2

1− cos(x)= lim

x→0

x2(1 + cos(x))1− cos2(x)

= limx→0

(x

sin(x)

)2

· 2 = 2

Remarks. Another solution, using L’Hôpital’s rule, is possible: limx→0x2

1−cos(x) = limx→02x

sin(x) = 2.

2. [3] Determine the real number a having the property that f(a) = a is a relative minimum of f(x) =x4 − x3 − x2 + ax + 1.

Answer: 1. Being a relative minimum, we have 0 = f ′(a) = 4a3 − 3a2 − 2a + a = a(4a + 1)(a− 1).Then a = 0, 1,−1/4 are the only possibilities. However, it is easily seen that a = 1 is the only valuesatisfying f(a) = a.

3. [4] Let a be a positive real number. Find the value of a such that the definite integral∫ a2

a

dx

x +√

x

achieves its smallest possible value.

Answer: 3− 2√

2. Let F (a) denote the given definite integral. Then

F ′(a) =dda

∫ a2

a

dx

x +√

x= 2a · 1

a2 +√

a2− 1

a +√

a.

Setting F ′(a) = 0, we find that 2a + 2√

a = a + 1 or (√

a + 1)2 = 2. We find√

a = ±√

2 − 1, andbecause

√a > 0, a = (

√2− 1)2 = 3− 2

√2.

4. [4] Find the real number α such that the curve f(x) = ex is tangent to the curve g(x) = αx2.

Answer: e2/4. Suppose tangency occurs at x = x0. Then ex0 = αx20 and f ′(x0) = 2αx0. On the

other hand, f ′(x) = f(x), so αx20 = 2αx0. Clearly, α = 0 and x0 = 0 are impossible, so it must be that

x0 = 2. Then α = ex0/(x20) = e2/4.

5. [5] The function f : R→ R satisfies f(x2)f ′′(x) = f ′(x)f ′(x2) for all real x. Given that f(1) = 1 andf ′′′(1) = 8, determine f ′(1) + f ′′(1).

Answer: 6. Let f ′(1) = a and f ′′(1) = b. Then setting x = 1 in the given equation, b = a2.Differentiating the given yields

2xf ′(x2)f ′′(x) + f(x2)f ′′′(x) = f ′′(x)f ′(x2) + 2xf ′(x)f ′′(x2).

Plugging x = 1 into this equation gives 2ab + 8 = ab + 2ab, or ab = 8. Then because a and b are real,we obtain the solution (a, b) = (2, 4).

Remarks. A priori, the function needn’t exist, but one possibility is f(x) = e2x−2.

1

6. [5] The elliptic curve y2 = x3+1 is tangent to a circle centered at (4, 0) at the point (x0, y0). Determinethe sum of all possible values of x0.

Answer: 13 . Note that y2 ≥ 0, so x3 ≥ −1 and x ≥ −1. Let the circle be defined by (x−4)2 +y2 = c

for some c ≥ 0. Now differentiate the equations with respect to x, obtaining 2y dydx = 3x2 from the

given and 2y dydx = −2x + 8 from the circle. For tangency, the two expressions dy

dx must be equal if theyare well-defined, and this is almost always the case. Thus, −2x0 + 8 = 3x2

0 so x0 = −2 or x0 = 4/3,but only the latter corresponds to a point on y2 = x3 + 1. Otherwise, y0 = 0, and this gives the trivialsolution x0 = −1.

7. [5] Compute∞∑

n=1

1n · (n + 1) · (n + 1)!

.

Answer: 3− e. We write

∞∑n=1

1n · (n + 1) · (n + 1)!

=∞∑

n=1

(1n− 1

n + 1

)1

(n + 1)!=

∞∑n=1

1n · (n + 1)!

−∞∑

n=1

1(n + 1) · (n + 1)!

12

+∞∑

n=2

1n · (n + 1)!

−∞∑

n=1

1(n + 1) · (n + 1)!

=12

+∞∑

n=1

1(n + 1) · (n + 2)!

− 1(n + 1) · (n + 1)!

12

+∞∑

n=1

1− (n + 2)(n + 1) · (n + 2)!

=12−(

13!

+14!

+ · · ·)

= 3−(

10!

+11!

+12!

+ · · ·)

= 3− e.

Alternatively, but with considerably less motivation, we can induce telescoping by adding and sub-tracting e− 2 = 1/2! + 1/3! + · · · , obtaining

2− e +∞∑

n=1

n(n + 1) + 1n · (n + 1) · (n + 1)!

= 2− e +∞∑

n=1

(n + 1)2 − n

n · (n + 1) · (n + 1)!

2− e +∞∑

n=1

1n · n!

− 1(n + 1) · (n + 1)!

= 3− e.

8. [6] Suppose that ω is a primitive 2007th root of unity. Find(22007 − 1

) 2006∑j=1

12− ωj

.

For this problem only, you may express your answer in the form m · nk + p, where m,n, k, and p arepositive integers. Note that a number z is a primitive nth root of unity if zn = 1 and n is the smallestnumber amongst k = 1, 2, . . . , n such that zk = 1.

Answer: 2005 · 22006 + 1. Note that

1z − ω

+ · · ·+ 1z − ω2006

=

∑2006j=1

∏i 6=j(z − ωi)

(z − ω) · · · (z − ω2006)

=ddz

[z2006 + z2005 + · · ·+ 1

]z2006 + z2005 + · · ·+ 1

=2006z2005 + 2005z2004 + · · ·+ 1

z2006 + z2005 + · · ·+ 1· z − 1z − 1

=2006z2006 − z2005 − z2004 − · · · − 1

z2007 − 1· z − 1z − 1

=2006z2007 − 2007z2006 + 1

(z2007 − 1)(z − 1).

Plugging in z = 2 gives 2005·22006+122007−1 ; whence the answer.

9. [7] g is a twice differentiable function over the positive reals such that

g(x) + 2x3g′(x) + x4g′′(x) = 0 for all positive reals x. (1)lim

x→∞xg(x) = 1 (2)

2

Find the real number α > 1 such that g(α) = 1/2.

Answer: 6π . In the first equation, we can convert the expression 2x3g′(x)+x4g′′(x) into the derivative

of a product, and in fact a second derivative, by writing y = 1/x. Specifically,

0 = g(x) + 2x3g′(x) + x4g′′(x) = g

(1y

)+ 2y−3g′

(1y

)+ y−4g′′

(1y

)= g

(1y

)+

ddy

[−y−2g′

(1y

)]= g

(1y

)+

d2

dy2

[g

(1y

)]Thus g

(1y

)= c1 cos(y) + c2 sin(y) or g(x) = c1 cos(1/x) + c2 sin(1/x). Now the second condition gives

1 = limx→∞

c1x + c2 ·sin(1/x)

1/x= c2 + lim

x→∞c1x

It must be that c1 = 0, c2 = 1. Now since 0 < 1/α < 1, the value of α such that g(α) = sin(1/α) = 1/2is given by 1/α = π/6 and so α = 6/π.

10. [8] Compute ∫ ∞

0

e−x sin(x)x

dx

Answer: π4 . We can compute the integral by introducing a parameter and exchanging the order of

integration:∫ ∞

0

e−x

(sin(x)

x

)dx =

∫ ∞

0

e−x

(∫ 1

0

cos(ax)da

)dx =

∫ 1

0

(∫ ∞

0

e−x cos(ax)dx

)da

=∫ 1

0

Re[∫ ∞

0

e(−1+ai)xdx

]da =

∫ 1

0

Re

[e(−1+ai)x

−1 + ai

∣∣∣∣∣∞

x=0

]da

=∫ 1

0

Re[

11− ai

]da =

∫ 1

0

Re[

1 + ai

1 + a2

]da

=∫ 1

0

11 + a2

da = tan−1(a)

∣∣∣∣∣1

a=0

=π

4

3

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Individual Round: Combinatorics Test

1. [3] A committee of 5 is to be chosen from a group of 9 people. How many ways can it be chosen, ifBiff and Jacob must serve together or not at all, and Alice and Jane refuse to serve with each other?

2. [3] How many 5-digit numbers abcde exist such that digits b and d are each the sum of the digits totheir immediate left and right? (That is, b = a + c and d = c + e.)

3. [4] Jack, Jill, and John play a game in which each randomly picks and then replaces a card from astandard 52 card deck, until a spades card is drawn. What is the probability that Jill draws the spade?(Jack, Jill, and John draw in that order, and the game repeats if no spade is drawn.)

4. [4] On the Cartesian grid, Johnny wants to travel from (0, 0) to (5, 1), and he wants to pass throughall twelve points in the set S = {(i, j) | 0 ≤ i ≤ 1, 0 ≤ j ≤ 5, i, j ∈ Z}. Each step, Johnny may go fromone point in S to another point in S by a line segment connecting the two points. How many ways arethere for Johnny to start at (0, 0) and end at (5, 1) so that he never crosses his own path?

5. [5] Determine the number of ways to select a positive number of squares on an 8× 8 chessboard suchthat no two lie in the same row or the same column and no chosen square lies to the left of and belowanother chosen square.

6. [5] Kevin has four red marbles and eight blue marbles. He arranges these twelve marbles randomly, ina ring. Determine the probability that no two red marbles are adjacent.

7. [5] Forty two cards are labeled with the natural numbers 1 through 42 and randomly shuffled into astack. One by one, cards are taken off of the top of the stack until a card labeled with a prime numberis removed. How many cards are removed on average?

8. [6] A set of six edges of a regular octahedron is called Hamiltonian cycle if the edges in some orderconstitute a single continuous loop that visits each vertex exactly once. How many ways are there topartition the twelve edges into two Hamiltonian cycles?

9. [7] Let S denote the set of all triples (i, j, k) of positive integers where i + j + k = 17. Compute∑(i,j,k)∈S

ijk.

10. [8] A subset S of the nonnegative integers is called supported if it contains 0, and k + 8, k + 9 ∈ S forall k ∈ S. How many supported sets are there?

1

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Individual Round: Combinatorics Test

1. [3] A committee of 5 is to be chosen from a group of 9 people. How many ways can it be chosen, ifBiff and Jacob must serve together or not at all, and Alice and Jane refuse to serve with each other?

Answer: 41. If Biff and Jacob are on the committee, there are(73

)= 35 ways for the other members

to be chosen. Amongst these 35 possibilities, we reject the(51

)= 5 choices where both Alice and Jane

are also serving. If Biff and Jacob are not serving, then there are(75

)= 21 ways to choose the remaining

5 members. Again, we reject the(53

)= 10 instances where Alice and Jane are chosen, so the total is

(35− 5) + (21− 10) = 41.

2. [3] How many 5-digit numbers abcde exist such that digits b and d are each the sum of the digits totheir immediate left and right? (That is, b = a + c and d = c + e.)

Answer: 330. Note that a > 0, so that b > c, and e ≥ 0 so that d ≥ c. Conversely, for each choiceof (b, c, d) with b > c and d ≥ c, there exists a unique pair (a, e) such that abcde is a number havingthe desired property. Thus, we compute

9∑c=0

(9− c)(10− c) =9∑

c=0

c2 − 19c + 90 = 330.

3. [4] Jack, Jill, and John play a game in which each randomly picks and then replaces a card from astandard 52 card deck, until a spades card is drawn. What is the probability that Jill draws the spade?(Jack, Jill, and John draw in that order, and the game repeats if no spade is drawn.)

Answer: 1237 . The desired probability is the relative probability that Jill draws the spade. In the first

round, Jack, Jill, and John draw a spade with probability 1/4, 3/4 · 1/4, and (3/4)2 · 1/4 respectively.Thus, the probability that Jill draws the spade is

3/4 · 1/41/4 + 3/4 · 1/4 + (3/4)2 · 1/4

=1237

.

4. [4] On the Cartesian grid, Johnny wants to travel from (0, 0) to (5, 1), and he wants to pass throughall twelve points in the set S = {(i, j) | 0 ≤ i ≤ 1, 0 ≤ j ≤ 5, i, j ∈ Z}. Each step, Johnny may go fromone point in S to another point in S by a line segment connecting the two points. How many ways arethere for Johnny to start at (0, 0) and end at (5, 1) so that he never crosses his own path?

Answer: 252. Observe that Johnny needs to pass through the points (0, 0), (1, 0), (2, 0), . . . , (5, 0)in that order, and he needs to pass through (0, 1), (1, 1), (2, 1), . . . , (5, 1) in that order, or else he willintersect his own path. Then, the problem is equivalent to interlacing those two sequence together, sothat the first term is (0, 0) and the final term is (5, 1). To do this, we need to select 5 positions out of10 to have points with x-coordinate 0. Hence the answer is

(105

)= 252.

5. [5] Determine the number of ways to select a positive number of squares on an 8× 8 chessboard suchthat no two lie in the same row or the same column and no chosen square lies to the left of and belowanother chosen square.

1

Answer: 12869 =(168

)− 1. If k is the number of squares chosen, then there are

(8k

)ways to choose k

columns, and(8k

)ways to choose k rows, and this would uniquely determine the set of squares selected.

Thus the answer is

8∑k=1

(8k

)(8k

)= −1 +

8∑k=0

(8k

)(8k

)= −1 +

(168

)= 12869

6. [5] Kevin has four red marbles and eight blue marbles. He arranges these twelve marbles randomly, ina ring. Determine the probability that no two red marbles are adjacent.

Answer: 733 . Select any blue marble and consider the remaining eleven marbles, arranged in a line.

The proportion of arrangement for which no two red marbles are adjacent will be the same as for theoriginal twelve marbles, arranged in a ring. The total number of ways of arranging 4 red marbles outof 11 is

(114

)= 330. To count the number of arrangements such that no two red marbles are adjacent,

there must be one red marble between each two would-be adjacent red marbles. Having fixed thepositions of three blue marbles we have four blue marbles to play with. So that we can arrange theremaining four marbles is

(84

)= 70 ways. This yields a probability of 70/330 = 7/33 as our final

answer.

7. [5] Forty two cards are labeled with the natural numbers 1 through 42 and randomly shuffled into astack. One by one, cards are taken off of the top of the stack until a card labeled with a prime numberis removed. How many cards are removed on average?

Answer: 4314 . Note that there are 13 prime numbers amongst the cards. We may view these as

separating the remaining 29 cards into 14 groups of nonprimes - those appearing before the first prime,between the first and second, etc. Each of these groups is equally likely to appear first, so 29/14nonprimes are removed on average. We are done since exactly one prime is always drawn.

8. [6] A set of six edges of a regular octahedron is called Hamiltonian cycle if the edges in some orderconstitute a single continuous loop that visits each vertex exactly once. How many ways are there topartition the twelve edges into two Hamiltonian cycles?

Answer: 6. Call the octahedron ABCDEF , where A,B, and C are opposite D,E, and F, respec-tively. Note that each Hamiltonian cycle can be described in terms of the order it visits vertices inexactly 12 different ways. Conversely, listing the six vertices in some order determines a Hamiltoniancycle precisely when no pair of opposite vertices are listed consecutively or first-and-last. Suppose webegin with AB. If D is listed third, then the final three letters are CEF or FEC. Otherwise, C or Fis listed next, and each gives three possibilities for the final three. For example ABC is be followed byDEF, DFE, or EDF. Thus, there are 6 ·4 · (2+3+3) = 192 listings. These correspond to 192/12 = 16Hamiltonian cycles. Finally, the complement of all but four Hamiltonian cycles is a Hamiltonian cycle.For, each vertex has degree four, so is an endpoint of two edges in the complement of a Hamiltoniancycle, so is also a Hamiltonian cycle unless it describes two opposite faces. It follows that there are sixpairs of disjoint Hamiltonian cycles.

9. [7] Let S denote the set of all triples (i, j, k) of positive integers where i + j + k = 17. Compute∑(i,j,k)∈S

ijk.

2

Answer: 11628 =(195

). We view choosing five objects from a row of 19 objects in an unusual way.

First, remove two of the chosen objects, the second and fourth, which are not adjacent nor at eitherend, forming three nonempty groups of consecutive objects. We then have i, j, and k choices for thefirst, third, and fifth objects. Because this is a reversible process taking a triple (i, j, k) to ijk choices,the answer is

(195

)= 11628.

A simple generating functions argument is also possible. Let sn =∑

i+j+k=n ijk. Then

∑n≥0

snxn =

∑n≥0

nxn

3

=(

x

(1− x)2

)3

=x3

(1− x)6,

and so sn =((

6n− 3

))=(

n + 25

), yielding s17 =

(195

).

10. [8] A subset S of the nonnegative integers is called supported if it contains 0, and k + 8, k + 9 ∈ S forall k ∈ S. How many supported sets are there?

Answer: 1430. Note that every supported set S contains 0, 8, 9, 16, 17, 18, 24-27, 32-36, 40-45,48-54, and all n ≥ 55. Now define S := Z+ \ S, which is a subset of {1 − 7, 10 − 15, 19 − 23, 28 −31, 37, 38, 39, 46, 47, 55} satisfying the opposite property that k ∈ S =⇒ k − 8, k − 9 ∈ S.

5546 47

37 38 3928 29 30 31

19 20 21 22 2310 11 12 13 14 15

1 2 3 4 5 6 7∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗

(0, 0) ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ (16, 0)

Consider the above arrangement after removing the numbers not in S. The condition that S besupported ensures that sets S are in bijective correspondence with paths from (0,0) to (16,0) consistingof discrete steps of 〈1, 1〉 and 〈1,−1〉 and lying above the x-axis: from the modified version of the abovediagram, a unique path passes through the top items left in each column. The number of such paths isthe 8th Catalan number, so the answer is C8 = 1

8+1

(8·28

)= 12870

9 = 1430. (Incidentally, 16 choose 8 wascomputed in an earlier problem.) Without the explicit formula for Catalan numbers, the answer canbe computed recursively by filling in the number of ways a path can reach (16,0) from each positionin the figure. One works right to left, obtaining the following:

18 1

35 7 1110 27 6 1

275 75 20 5 1572 165 48 14 4 1

1001 297 90 28 9 3 11430 429 132 42 14 5 2 1

1430 429 132 42 14 5 2 1 1

One can exploit symmetry and, having determined the middle column, sum the squares: 12 + 72 +202 + 282 + 142 = 1430.

3

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Individual Round: General Test, Part 1

1. [2] Michael has 16 white socks, 3 blue socks, and 6 red socks in a drawer. Ever the lazy college student,he has overslept and is late for his favorite team’s season-opener. Because he is now in such a rush toget from Harvard to Foxborough, he randomly takes socks from the drawer (one at a time) until hehas a pair of the same color. What is the largest number of socks he could possibly withdraw in thisfashion?

2. [2] Rectangle ABCD has side lengths AB = 12 and BC = 5. Let P and Q denote the midpoints ofsegments AB and DP , respectively. Determine the area of triangle CDQ.

3. [3] A, B, C, and D are points on a circle, and segments AC and BD intersect at P , such that AP = 8,PC = 1, and BD = 6. Find BP , given that BP < DP .

4. [3] Let a and b be integer solutions to 17a + 6b = 13. What is the smallest possible positive value fora− b?

5. [4] Find the smallest positive integer that is twice a perfect square and three times a perfect cube.

6. [4] The positive integer n is such that the numbers 2n and 5n start with the same digit when writtenin decimal notation; determine this common leading digit.

7. [4] Jack, Jill, and John play a game in which each randomly picks and then replaces a card from astandard 52 card deck, until a spades card is drawn. What is the probability that Jill draws the spade?(Jack, Jill, and John draw in that order, and the game repeats if no spade is drawn.)

8. [5] Determine the largest positive integer n such that there exist positive integers x, y, z so that

n2 = x2 + y2 + z2 + 2xy + 2yz + 2zx + 3x + 3y + 3z − 6

9. [6] I have four distinct rings that I want to wear on my right hand hand (five distinct fingers.) One ofthese rings is a Canadian ring that must be worn on a finger by itself, the rest I can arrange howeverI want. If I have two or more rings on the same finger, then I consider different orders of rings alongthe same finger to be different arrangements. How many different ways can I wear the rings on myfingers?

10. [7] α1, α2, α3, and α4 are the complex roots of the equation x4 +2x3 +2 = 0. Determine the unorderedset

{α1α2 + α3α4, α1α3 + α2α4, α1α4 + α2α3}.

1

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Individual Round: General Test, Part 1

1. [2] Michael has 16 white socks, 3 blue socks, and 6 red socks in a drawer. Ever the lazy college student,he has overslept and is late for his favorite team’s season-opener. Because he is now in such a rush toget from Harvard to Foxborough, he randomly takes socks from the drawer (one at a time) until hehas a pair of the same color. What is the largest number of socks he could possibly withdraw in thisfashion?

Answer: 4. It is possible for him to begin with three socks of different colors, but an instance of thePigeon Hole Principle is that among any four objects of three types some two are the same type.

2. [2] Rectangle ABCD has side lengths AB = 12 and BC = 5. Let P and Q denote the midpoints ofsegments AB and DP , respectively. Determine the area of triangle CDQ.

Answer: 15. Note that [CDP ] = 12 · 5 · 12 = 30, while the area of triangle CDQ is half of the area

of triangle CDP .

3. [3] A, B, C, and D are points on a circle, and segments AC and BD intersect at P , such that AP = 8,PC = 1, and BD = 6. Find BP , given that BP < DP .

Answer: 2. Same as Geometry #2.

4. [3] Let a and b be integer solutions to 17a + 6b = 13. What is the smallest possible positive value fora− b?

Answer: 17. First group as 17(a−b)+23b = 13. Taking this equation modulo 23, we get −6(a−b) ≡−10 (mod 23). Since -4 is an inverse of -6 modulo 23, then we multiply to get (a− b) ≡ 17 (mod 23).Therefore, the smallest possible positive value for (a−b) is 17. This can be satisfied by a = 5, b = −12.

5. [4] Find the smallest positive integer that is twice a perfect square and three times a perfect cube.

Answer: 648. Let n be such a number. If n is divisible by 2 and 3 exactly e2 and e3 times, thene2 is odd and a multiple of three, and e3 is even and one more than a multiple of three. The smallestpossible exponents are n2 = 3 and n3 = 4. The answer is then 23 · 34 = 648.

6. [4] The positive integer n is such that the numbers 2n and 5n start with the same digit when writtenin decimal notation; determine this common leading digit.

Answer: 3. Note 1 = 12 < 22 < 32 < 10 < 42 < · · · < 92 < 102 = 100. Divide 2n and 5n by 10repeatedly until each is reduced to a decimal number less than 10 but at least 1; call the resultingnumbers x and y. Since (5n)(2n) = 10n, either xy = 1 or xy = 10. Because 2n and 5n begin with thesame digit, x and y are bounded by the same pair of adjacent integers. It follows that either x = y = 1or 3 ≤ x, y < 4. Because n is positive, neither 2n nor 5n is a perfect power of 10, so the former isimpossible.

7. [4] Jack, Jill, and John play a game in which each randomly picks and then replaces a card from astandard 52 card deck, until a spades card is drawn. What is the probability that Jill draws the spade?(Jack, Jill, and John draw in that order, and the game repeats if no spade is drawn.)

Answer: 1237 . Same as Combo #3.

8. [5] Determine the largest positive integer n such that there exist positive integers x, y, z so that

n2 = x2 + y2 + z2 + 2xy + 2yz + 2zx + 3x + 3y + 3z − 6

Answer: 8. The given equation rewrites as n2 = (x + y + z + 1)2 + (x + y + z + 1) − 8. Writingr = x + y + z + 1, we have n2 = r2 + r − 8. Clearly, one possibility is n = r = 8, which is realized byx = y = 1, z = 6. On the other hand, for r > 8, we have r2 < r2 + r − 8 < (r + 1)2.

1

9. [6] I have four distinct rings that I want to wear on my right hand hand (five distinct fingers.) One ofthese rings is a Canadian ring that must be worn on a finger by itself, the rest I can arrange howeverI want. If I have two or more rings on the same finger, then I consider different orders of rings alongthe same finger to be different arrangements. How many different ways can I wear the rings on myfingers?

Answer: 600. First we pick the finger for the Canadian ring. This gives a multiplicative factor of 5.For distributing the remaining 3 rings among 4 fingers, they can either be all on the same finger (4 · 3!ways), all on different fingers (

(43

)· 3! ways), or two on one finger and one on another (4 ·

(32

)· 2! · 3

ways.) Therefore, I have 5 · (24 + 24 + 72) = 600 choices.

10. [7] α1, α2, α3, and α4 are the complex roots of the equation x4 +2x3 +2 = 0. Determine the unorderedset

{α1α2 + α3α4, α1α3 + α2α4, α1α4 + α2α3}.

Answer: {1±√

5,−2}. Same as Algebra #9.

2

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Individual Round: General Test, Part 2

1. [2] A cube of edge length s > 0 has the property that its surface area is equal to the sum of its volumeand five times its edge length. Compute all possible values of s.

2. [2] A parallelogram has 3 of its vertices at (1,2), (3,8), and (4,1). Compute the sum of all possible xcoordinates of the 4th vertex.

3. [3] Compute ⌊2007! + 2004!2006! + 2005!

⌋.

(Note that bxc denotes the greatest integer less than or equal to x.)

4. [3] Three brothers Abel, Banach, and Gauss each have portable music players that can share musicwith each other. Initially, Abel has 9 songs, Banach has 6 songs, and Gauss has 3 songs, and none ofthese songs are the same. One day, Abel flips a coin to randomly choose one of his brothers and headds all of that brother’s songs to his collection. The next day, Banach flips a coin to randomly chooseone of his brothers and he adds all of that brother’s collection of songs to his collection. Finally, eachbrother randomly plays a song from his collection with each song in his collection being equally likelyto be chosen. What is the probability that they all play the same song?

5. [4] A best of 9 series is to be played between two teams. That is, the first team to win 5 games is thewinner. One of the teams, the Mathletes, has a 2/3 chance of winning any given game. What is theprobability that the winner is determined in the 7th game?

6. [4] Circle ω has radius 5 and is centered at O. Point A lies outside ω such that OA = 13. The twotangents to ω passing through A are drawn, and points B and C are chosen on them (one on eachtangent), such that line BC is tangent to ω and ω lies outside triangle ABC. Compute AB +AC giventhat BC = 7.

7. [4] My friend and I are playing a game with the following rules: If one of us says an integer n, theopponent then says an integer of their choice between 2n and 3n, inclusive. Whoever first says 2007 orgreater loses the game, and their opponent wins. I must begin the game by saying a positive integerless than 10. With how many of them can I guarantee a win?

8. [5] Compute the number of sequences of numbers a1, a2, . . . , a10 such that

I. ai = 0 or 1 for all i

II. ai · ai+1 = 0 for i = 1, 2, . . . , 9III. ai · ai+2 = 0 for i = 1, 2, . . . , 8.

9. [6] Let A := Q \ {0, 1} denote the set of all rationals other than 0 and 1. A function f : A → R hasthe property that for all x ∈ A,

f (x) + f

(1− 1

x

)= log |x|.

Compute the value of f(2007).

10. [7] ABCD is a convex quadrilateral such that AB = 2, BC = 3, CD = 7, and AD = 6. It also has anincircle. Given that ∠ABC is right, determine the radius of this incircle.

1

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Individual Round: General Test, Part 2

1. [2] A cube of edge length s > 0 has the property that its surface area is equal to the sum of its volumeand five times its edge length. Compute all possible values of s.

Answer: 1,5. Same as Geometry #1.

2. [2] A parallelogram has 3 of its vertices at (1,2), (3,8), and (4,1). Compute the sum of all possible xcoordinates of the 4th vertex.

Answer: 8. There are three possibilities: the 4th vertex must be opposite one of the three givenvertices. These three possibilities have as a medial triangle the three given vertices, so the sum of theirx coordinates is the same as the sum of the x coordinates of the given triangle.

3. [3] Compute ⌊2007! + 2004!2006! + 2005!

⌋.

(Note that bxc denotes the greatest integer less than or equal to x.)

Answer: 2006. Same as Algebra #1.

4. [3] Three brothers Abel, Banach, and Gauss each have portable music players that can share musicwith each other. Initially, Abel has 9 songs, Banach has 6 songs, and Gauss has 3 songs, and none ofthese songs are the same. One day, Abel flips a coin to randomly choose one of his brothers and headds all of that brother’s songs to his collection. The next day, Banach flips a coin to randomly chooseone of his brothers and he adds all of that brother’s collection of songs to his collection. Finally, eachbrother randomly plays a song from his collection with each song in his collection being equally likelyto be chosen. What is the probability that they all play the same song?

Answer: 1288 . If Abel copies Banach’s songs, this can never happen. Therefore, we consider only the

cases where Abel copies Gauss’s songs. Since all brothers have Gauss’s set of songs, the probability thatthey play the same song is equivalent to the probability that they independently match whichever songGauss chooses. Case 1: Abel copies Gauss and Banach copies Gauss (1/4 chance) – The probability ofsongs matching is then 1/12 ·1/9. Case 2: Abel copies Gauss and Banach copies Abel (1/4 probability)– The probability of songs matching is then 1/12 · 1/18. We add the two probabilities together to get1/4 · 1/12 · (1/9 + 1/18) = 1/288.

5. [4] A best of 9 series is to be played between two teams. That is, the first team to win 5 games is thewinner. One of the teams, the Mathletes, has a 2/3 chance of winning any given game. What is theprobability that the winner is determined in the 7th game?

Answer: 2081 . If the Mathletes are the winners, they must win the 7th game and have won exactly

four of the previous 6 games. The probability of this occurring is((2/3)4 · (1/3)2 ·

(62

))· (2/3).

Analogously, the other team wins with probability((1/3)4 · (2/3)2 ·

(62

))· (1/3). Summing, the prob-

ability is (62

)· 22 ·

(23 · 12 + 15

)37

=5 · 434

=2081

.

6. [4] Circle ω has radius 5 and is centered at O. Point A lies outside ω such that OA = 13. The twotangents to ω passing through A are drawn, and points B and C are chosen on them (one on each

1

tangent), such that line BC is tangent to ω and ω lies outside triangle ABC. Compute AB +AC giventhat BC = 7.

Answer: 17. Same as Geometry #4.

7. [4] My friend and I are playing a game with the following rules: If one of us says an integer n, theopponent then says an integer of their choice between 2n and 3n, inclusive. Whoever first says 2007 orgreater loses the game, and their opponent wins. I must begin the game by saying a positive integerless than 10. With how many of them can I guarantee a win?

Answer: 6. We assume optimal play and begin working backward. I win if I say any number between1004 and 2006. Thus, by saying such a number, my friend can force a win for himself if I ever saya number between 335 and 1003. Then I win if I say any number between 168 and 334, because myfriend must then say one of the losing numbers just considered. Similarly, I lose by saying 56 through167, win by saying 28 through 55, lose with 10 through 17, win with 5 through 9, lose with 2 through4, and win with 1.

8. [5] Compute the number of sequences of numbers a1, a2, . . . , a10 such that

I. ai = 0 or 1 for all i

II. ai · ai+1 = 0 for i = 1, 2, . . . , 9III. ai · ai+2 = 0 for i = 1, 2, . . . , 8.

Answer: 60. Call such a sequence “good,” and let An be the number of good sequences of lengthn. Let a1, a2, . . . , an be a good sequence. If a1 = 0, then a1, a2, . . . , an is a good sequence if and onlyif a2, . . . , an is a good sequence, so there are An−1 of them. If a1 = 1, then we must have a2 = a3 = 0,and in this case, a1, a2, . . . , an is a good sequence if and only if a4, a5, . . . , an is a good sequence,so there are An−3 of them. We thus obtain the recursive relation An = An−1 + An−3. Note thatA1 = 2, A2 = 3, A3 = 4. Plugging these into the recursion eventually yields A10 = 60.

9. [6] Let A := Q \ {0, 1} denote the set of all rationals other than 0 and 1. A function f : A → R hasthe property that for all x ∈ A,

f (x) + f

(1− 1

x

)= log |x|.

Compute the value of f(2007).

Answer: log (2007/2006) . Same as Algebra #8.

10. [7] ABCD is a convex quadrilateral such that AB = 2, BC = 3, CD = 7, and AD = 6. It also has anincircle. Given that ∠ABC is right, determine the radius of this incircle.

Answer: 1+√

133 . Same as Geometry #10.

2

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Individual Round: Geometry Test

1. [3] A cube of edge length s > 0 has the property that its surface area is equal to the sum of its volumeand five times its edge length. Compute all possible values of s.

2. [3] A, B, C, and D are points on a circle, and segments AC and BD intersect at P , such that AP = 8,PC = 1, and BD = 6. Find BP , given that BP < DP.

3. [4] Circles ω1, ω2, and ω3 are centered at M,N, and O, respectively. The points of tangency betweenω2 and ω3, ω3 and ω1, and ω1 and ω2 are tangent at A, B, and C, respectively. Line MO intersectsω3 and ω1 again at P and Q respectively, and line AP intersects ω2 again at R. Given that ABC isan equilateral triangle of side length 1, compute the area of PQR.

4. [4] Circle ω has radius 5 and is centered at O. Point A lies outside ω such that OA = 13. The twotangents to ω passing through A are drawn, and points B and C are chosen on them (one on eachtangent), such that line BC is tangent to ω and ω lies outside triangle ABC. Compute AB +AC giventhat BC = 7.

5. [5] Five marbles of various sizes are placed in a conical funnel. Each marble is in contact with theadjacent marble(s). Also, each marble is in contact all around the funnel wall. The smallest marblehas a radius of 8, and the largest marble has a radius of 18. What is the radius of the middle marble?

6. [5] Triangle ABC has ∠A = 90◦, side BC = 25, AB > AC, and area 150. Circle ω is inscribed inABC, with M its point of tangency on AC. Line BM meets ω a second time at point L. Find thelength of segment BL.

7. [5] Convex quadrilateral ABCD has sides AB = BC = 7, CD = 5, and AD = 3. Given additionallythat m∠ABC = 60◦, find BD.

8. [6] ABCD is a convex quadrilateral such that AB < AD. The diagonal AC bisects ∠BAD, andm∠ABD = 130◦. Let E be a point on the interior of AD, and m∠BAD = 40◦. Given that BC =CD = DE, determine m∠ACE in degrees.

9. [7] 4ABC is right angled at A. D is a point on AB such that CD = 1. AE is the altitude from A toBC. If BD = BE = 1, what is the length of AD?

10. [8] ABCD is a convex quadrilateral such that AB = 2, BC = 3, CD = 7, and AD = 6. It also has anincircle. Given that ∠ABC is right, determine the radius of this incircle.

1

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Individual Round: Geometry Test

1. [3] A cube of edge length s > 0 has the property that its surface area is equal to the sum of its volumeand five times its edge length. Compute all possible values of s.

Answer: 1,5. The volume of the cube is s3 and its surface area is 6s2, so we have 6s2 = s3 + 5s, or0 = s3 − 6s2 + 5s = s(s− 1)(s− 5).

2. [3] A, B, C, and D are points on a circle, and segments AC and BD intersect at P , such that AP = 8,PC = 1, and BD = 6. Find BP , given that BP < DP.

Answer: 2. Writing BP = x and PD = 6 − x, we have that BP < 3. Power of a point at P givesAP · PC = BP · PD or 8 = x(6 − x). This can be solved for x = 2 and x = 4, and we discard thelatter.

3. [4] Circles ω1, ω2, and ω3 are centered at M,N, and O, respectively. The points of tangency betweenω2 and ω3, ω3 and ω1, and ω1 and ω2 are tangent at A, B, and C, respectively. Line MO intersectsω3 and ω1 again at P and Q respectively, and line AP intersects ω2 again at R. Given that ABC isan equilateral triangle of side length 1, compute the area of PQR.

Answer: 2√

3. Note that ONM is an equilateral triangle of side length 2, so m∠BPA = m∠BOA/2 =π/6. Now BPA is a 30-60-90 triangle with short side length 1, so AP =

√3. Now A and B are the

midpoints of segments PR and PQ, so [PQR] = PRPA ·

PQPB [PBA] = 2 · 2[PBA] = 2

√3.

4. [4] Circle ω has radius 5 and is centered at O. Point A lies outside ω such that OA = 13. The twotangents to ω passing through A are drawn, and points B and C are chosen on them (one on eachtangent), such that line BC is tangent to ω and ω lies outside triangle ABC. Compute AB +AC giventhat BC = 7.

1

Answer: 17. Let T1, T2, and T3 denote the points of tangency of AB,AC, and BC with ω, respec-tively. Then 7 = BC = BT3 + T3C = BT1 + CT2. By Pythagoras, AT1 = AT2 =

√132 − 52 = 12.

Now note that 24 = AT1 + AT2 = AB + BT1 + AC + CT2 = AB + AC + 7.

5. [5] Five marbles of various sizes are placed in a conical funnel. Each marble is in contact with theadjacent marble(s). Also, each marble is in contact all around the funnel wall. The smallest marblehas a radius of 8, and the largest marble has a radius of 18. What is the radius of the middle marble?

Answer: 12. One can either go through all of the algebra, find the slope of the funnel wall and gofrom there to figure out the radius of the middle marble. Or one can notice that the answer will justbe the geometric mean of 18 and 8 which is 12.

6. [5] Triangle ABC has ∠A = 90◦, side BC = 25, AB > AC, and area 150. Circle ω is inscribed inABC, with M its point of tangency on AC. Line BM meets ω a second time at point L. Find thelength of segment BL.

Answer: 45√

17/17. Let D be the foot of the altitude from A to side BC. The length of AD is2 · 150/25 = 12. Triangles ADC and BDA are similar, so CD · DB = AD2 = 144 ⇒ BD = 16 andCD = 9 ⇒ AB = 20 and AC = 15. Using equal tangents or the formula inradius as area divided bysemiperimeter, we can find the radius of ω to be 5. Now, let N be the tangency point of ω on AB.By power of a point, we have BL · BM = BN2. Since the center of ω together with M,A, and Ndetermines a square, BN = 15 and BM = 5

√17, and we have BL = 45

√17/17.

7. [5] Convex quadrilateral ABCD has sides AB = BC = 7, CD = 5, and AD = 3. Given additionallythat m∠ABC = 60◦, find BD.

Answer: 8. Triangle ABC is equilateral, so AC = 7 as well. Now the law of cosines shows thatm∠CDA = 120◦; i.e., ABCD is cyclic. Ptolemy’s theorem now gives AC ·BD = AB ·CD + AD ·BC,or simply BD = CD + AD = 8.

8. [6] ABCD is a convex quadrilateral such that AB < AD. The diagonal AC bisects ∠BAD, andm∠ABD = 130◦. Let E be a point on the interior of AD, and m∠BAD = 40◦. Given that BC =CD = DE, determine m∠ACE in degrees.

2

Answer: 55◦. First, we check that ABCD is cyclic. Reflect B over AC to B′ on AD, and note thatB′C = CD. Therefore, m∠ADC = m∠B′DC = m∠CB′D = 180◦ −m∠AB′C = 180◦ −m∠CBA.Now m∠CBD = m∠CAD = 20◦ and m∠ADC = 180◦ −m∠CBA = 30◦. Triangle CDE is isosceles,so m∠CED = 75◦ and m∠AEC = 105◦. It follows that m∠ECA = 180◦−m∠AEC−m∠CAE = 55◦.

9. [7] 4ABC is right angled at A. D is a point on AB such that CD = 1. AE is the altitude from A toBC. If BD = BE = 1, what is the length of AD?

Answer: 3√

2− 1. Let AD = x, angle ABC = t. We also have ∠ BCA = 90−t and ∠ DCA = 90−2tso that ∠ ADC = 2t. Considering triangles ABE and ADC, we obtain, respectively,

cos(t) = 1/(1 + x) and cos(2t) = x. By the double angle formula we get, (1 + x)3 = 2.

Alternatively, construct M, the midpoint of segment BC, and note that triangles ABC, EBA, andMBD are similar. Thus, AB2 = BC ·BE = BC. In particular,

AB =BC

AB=

AB

BE=

BD

BM=

2BD

BC=

2AB2

,

from which AB = 3√

2 and AD = 3√

2− 1.

10. [8] ABCD is a convex quadrilateral such that AB = 2, BC = 3, CD = 7, and AD = 6. It also has anincircle. Given that ∠ABC is right, determine the radius of this incircle.

Answer: 1+√

133 . Note that AC2 = AB2 +BC2 = 13 = CD2−DA2. It follows that ∠DAC is right,

and so[ABCD] = [ABC] + [DAC] = 2 · 3/2 + 6 ·

√13/2 = 3 + 3

√13

On the other hand, if I denotes the incenter and r denotes the inradius,

[ABCD] = [AIB] + [BIC] + [CID] + [DIA] = AB · r/2 + BC · r/2 + CD · r/2 + DA · r/2 = 9r

Therefore, r = (3 + 3√

13)/9 = 1+√

133 .

3

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Guts Round

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

Note that there are just 36 problems in the Guts round this year.

1. [5] Define the sequence of positive integers an recursively by a1 = 7 and an = 7an−1 for all n ≥ 2.Determine the last two digits of a2007.

2. [5] A candy company makes 5 colors of jellybeans, which come in equal proportions. If I grab a randomsample of 5 jellybeans, what is the probability that I get exactly 2 distinct colors?

3. [5] The equation x2 + 2x = i has two complex solutions. Determine the product of their real parts.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

4. [6] A sequence consists of the digits 122333444455555 . . . such that the each positive integer n isrepeated n times, in increasing order. Find the sum of the 4501st and 4052nd digits of this sequence.

5. [6] Compute the largest positive integer such that 2007!2007n is an integer.

6. [6] There are three video game systems: the Paystation, the WHAT, and the ZBoz2π, and none ofthese systems will play games for the other systems. Uncle Riemann has three nephews: Bernoulli,Galois, and Dirac. Bernoulli owns a Paystation and a WHAT, Galois owns a WHAT and a ZBoz2π, andDirac owns a ZBoz2π and a Paystation. A store sells 4 different games for the Paystation, 6 differentgames for the WHAT, and 10 different games for the ZBoz2π. Uncle Riemann does not understand thedifference between the systems, so he walks into the store and buys 3 random games (not necessarilydistinct) and randomly hands them to his nephews. What is the probability that each nephew receivesa game he can play?

1

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

7. [7] A student at Harvard named KevinWas counting his stones by 11

He messed up n timesAnd instead counted 9s

And wound up at 2007.

How many values of n could make this limerick true?

8. [7] A circle inscribed in a square,Has two chords as shown in a pair.

It has radius 2,And P bisects TU.

The chords’ intersection is where?

Answer the question by giving the distance of the point of intersection from the center of the circle.

9. [7] I ponder some numbers in bed,All products of three primes I’ve said, n = 372 · 3 . . .

Apply φ they’re still fun: φ(n) =now Elev’n cubed plus one. 113 + 1?

What numbers could be in my head?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

10. [8] Let A12 denote the answer to problem 12. There exists a unique triple of digits (B,C,D) such that10 > A12 > B > C > D > 0 and

A12BCD −DCBA12 = BDA12C,

where A12BCD denotes the four digit base 10 integer. Compute B + C + D.

11. [8] Let A10 denote the answer to problem 10. Two circles lie in the plane; denote the lengths of theinternal and external tangents between these two circles by x and y, respectively. Given that theproduct of the radii of these two circles is 15/2, and that the distance between their centers is A10,determine y2 − x2.

12. [8] Let A11 denote the answer to problem 11. Determine the smallest prime p such that the arithmeticsequence p, p + A11, p + 2A11, . . . begins with the largest possible number of primes.

There is just one triple of possible (A10, A11, A12) of answers to these three problems. Your teamwill receive credit only for answers matching these. (So, for example, submitting a wrong answer forproblem 11 will not alter the correctness of your answer to problem 12.)

2

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

13. [9] Determine the largest integer n such that 72048 − 1 is divisible by 2n.

14. [9] We are given some similar triangles. Their areas are 12, 32, 52 . . . , and 492. If the smallest trianglehas a perimeter of 4, what is the sum of all the triangles’ perimeters?

15. [9] Points A,B, and C lie in that order on line `, such that AB = 3 and BC = 2. Point H is such thatCH is perpendicular to `. Determine the length CH such that ∠AHB is as large as possible.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

16. [10] Let ABC be a triangle with AB = 7, BC = 9, and CA = 4. Let D be the point such thatAB ‖ CD and CA ‖ BD. Let R be a point within triangle BCD. Lines ` and m going through R areparallel to CA and AB respectively. Line ` meets AB and BC at P and P ′ respectively, and m meetsCA and BC at Q and Q′ respectively. If S denotes the largest possible sum of the areas of trianglesBPP ′, RP ′Q′, and CQQ′, determine the value of S2.

17. [10] During the regular season, Washington Redskins achieve a record of 10 wins and 6 losses. Computethe probability that their wins came in three streaks of consecutive wins, assuming that all possiblearrangements of wins and losses are equally likely. (For example, the record LLWWWWWLWWL-WWWLL contains three winning streaks, while WWWWWWWLLLLLLWWW has just two.)

18. [10] Convex quadrilateral ABCD has right angles ∠A and ∠C and is such that AB = BC andAD = CD. The diagonals AC and BD intersect at point M . Points P and Q lie on the circumcircleof triangle AMB and segment CD, respectively, such that points P,M , and Q are collinear. Supposethat m∠ABC = 160◦ and m∠QMC = 40◦. Find MP ·MQ, given that MC = 6.

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

19. [10] Define x ? y =√

x2+3xy+y2−2x−2y+4

xy+4 . Compute

((· · · ((2007 ? 2006) ? 2005) ? · · · ) ? 1).

20. [10] For a a positive real number, let x1, x2, x3 be the roots of the equation x3 − ax2 + ax − a = 0.Determine the smallest possible value of x3

1 + x32 + x3

3 − 3x1x2x3.

21. [10] Bob the bomb-defuser has stumbled upon an active bomb. He opens it up, and finds the red andgreen wires conveniently located for him to cut. Being a seasoned member of the bomb-squad, Bobquickly determines that it is the green wire that he should cut, and puts his wirecutters on the greenwire. But just before he starts to cut, the bomb starts to count down, ticking every second. Eachtime the bomb ticks, starting at time t = 15 seconds, Bob panics and has a certain chance to move hiswirecutters to the other wire. However, he is a rational man even when panicking, and has a 1

2t2 chanceof switching wires at time t, regardless of which wire he is about to cut. When the bomb ticks at t = 1,Bob cuts whatever wire his wirecutters are on, without switching wires. What is the probability thatBob cuts the green wire?

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

22. [12] The sequence {an}n≥1 is defined by an+2 = 7an+1 − an for positive integers n with initial valuesa1 = 1 and a2 = 8. Another sequence, {bn}, is defined by the rule bn+2 = 3bn+1 − bn for positiveintegers n together with the values b1 = 1 and b2 = 2. Find gcd(a5000, b501).

23. [12] In triangle ABC, ∠ABC is obtuse. Point D lies on side AC such that ∠ABD is right, and point Elies on side AC between A and D such that BD bisects ∠EBC. Find CE, given that AC = 35, BC = 7,and BE = 5.

24. [12] Let x, y, n be positive integers with n > 1. How many ordered triples (x, y, n) of solutions arethere to the equation xn − yn = 2100 ?

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

25. [12] Two real numbers x and y are such that 8y4 + 4x2y2 + 4xy2 + 2x3 + 2y2 + 2x = x2 + 1. Find allpossible values of x + 2y2.

26. [12] ABCD is a cyclic quadrilateral in which AB = 4, BC = 3, CD = 2, and AD = 5. Diagonals ACand BD intersect at X. A circle ω passes through A and is tangent to BD at X. ω intersects AB andAD at Y and Z respectively. Compute Y Z/BD.

27. [12] Find the number of 7-tuples (n1, . . . , n7) of integers such that

7∑i=1

n6i = 96957.

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

28. [15] Compute the circumradius of cyclic hexagon ABCDEF , which has side lengths AB = BC =2, CD = DE = 9, and EF = FA = 12.

29. [15] A sequence {an}n≥1 of positive reals is defined by the rule an+1a5n−1 = a4

na2n−2 for integers n > 2

together with the initial values a1 = 8 and a2 = 64 and a3 = 1024. Compute√a1 +

√a2 +

√a3 + · · ·

30. [15] ABCD is a cyclic quadrilateral in which AB = 3, BC = 5, CD = 6, and AD = 10. M , I, and Tare the feet of the perpendiculars from D to lines AB,AC, and BC respectively. Determine the valueof MI/IT .

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

31. [18] A sequence {an}n≥0 of real numbers satisfies the recursion an+1 = a3n − 3a2

n + 3 for all positiveintegers n. For how many values of a0 does a2007 = a0?

32. [18] Triangle ABC has AB = 4, BC = 6, and AC = 5. Let O denote the circumcenter of ABC. Thecircle Γ is tangent to and surrounds the circumcircles of triangles AOB, BOC, and AOC. Determinethe diameter of Γ.

33. [18] Compute ∫ 2

1

9x + 4x5 + 3x2 + x

dx.

(No, your TI-89 doesn’t know how to do this one. Yes, the end is near.)

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

34. [?] The Game. Eric and Greg are watching their new favorite TV show, The Price is Right. Bob Barkerrecently raised the intellectual level of his program, and he begins the latest installment with biddingon following question: How many Carmichael numbers are there less than 100,000?

Each team is to list one nonnegative integer not greater than 100,000. Let X denote the answer toBob’s question. The teams listing N , a maximal bid (of those submitted) not greater than X, willreceive N points, and all other teams will neither receive nor lose points. (A Carmichael number isan odd composite integer n such that n divides an−1 − 1 for all integers a relatively prime to n with1 < a < n.)

35. [≤ 25] The Algorithm. There are thirteen broken computers situated at the following set S of thirteenpoints in the plane:

A = (1, 10) B = (976, 9) C = (666, 87)D = (377, 422) E = (535, 488) F = (775, 488)G = (941, 500) H = (225, 583) I = (388, 696)J = (3, 713) K = (504, 872) L = (560, 934)

M = (22, 997)

At time t = 0, a repairman begins moving from one computer to the next, traveling continuously instraight lines at unit speed. Assuming the repairman begins and A and fixes computers instantly, whatpath does he take to minimize the total downtime of the computers? List the points he visits in order.Your score will be bN

40c, where

N = 1000 + bthe optimal downtimec − byour downtimec ,

or 0, whichever is greater. By total downtime we mean the sum∑P∈S

tP ,

where tP is the time at which the repairman reaches P .

36. [25] The Marathon. Let ω denote the incircle of triangle ABC. The segments BC, CA, and AB aretangent to ω at D, E, and F , respectively. Point P lies on EF such that segment PD is perpendicularto BC. The line AP intersects BC at Q. The circles ω1 and ω2 pass through B and C, respectively,and are tangent to AQ at Q; the former meets AB again at X, and the latter meets AC again at Y .The line XY intersects BC at Z. Given that AB = 15, BC = 14, and CA = 13, find bXZ · Y Zc.

5

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Guts Round

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

Note that there are just 36 problems in the Guts round this year.

1. [5] Define the sequence of positive integers an recursively by a1 = 7 and an = 7an−1 for all n ≥ 2.Determine the last two digits of a2007.

Answer: 43. Note that the last two digits of 74 are 01. Also, a2006 = 7a2005 = (−1)a2005 = −1 = 3(mod 4) since a2005 is odd. Therefore, a2007 = 7a2006 = 73 = 43 (mod 100).

2. [5] A candy company makes 5 colors of jellybeans, which come in equal proportions. If I grab a randomsample of 5 jellybeans, what is the probability that I get exactly 2 distinct colors?

Answer: 12125 . There are

(52

)= 10 possible pairs of colors. Each pair of colors contributes 25−2 = 30

sequences of beans that use both colors. Thus, the answer is 10 · 30/55 = 12/125.

3. [5] The equation x2 + 2x = i has two complex solutions. Determine the product of their real parts.

Answer: 1−√

22 . Complete the square by adding 1 to each side. Then (x + 1)2 = 1 + i = e

iπ4√

2, so

x + 1 = ±eiπ8

4√

2. The desired product is then(−1 + cos

(π

8

)4√

2)(−1− cos

(π

8

)4√

2)

= 1− cos2(π

8

)√2 = 1−

(1 + cos

(π4

))2

√2 =

1−√

22

.

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

4. [6] A sequence consists of the digits 122333444455555 . . . such that the each positive integer n isrepeated n times, in increasing order. Find the sum of the 4501st and 4052nd digits of this sequence.

Answer: 13. Note that n contributes n · d(n) digits, where d(n) is the number of digits of n. Thenbecause 1 + · · · + 99 = 4950, we know that the digits of interest appear amongst copies of two digitnumbers. Now for 10 ≤ n ≤ 99, the number of digits in the subsequence up to the last copy of n is

1 + 2 + 3 + · · ·+ 9 + 2 · (10 + · · ·+ n) = 2 · (1 + · · ·+ n)− 45 = n2 + n− 45.

Since 672 + 67− 45 = 4511, the two digits are 6 and 7 in some order, so have sum 13.

5. [6] Compute the largest positive integer such that 2007!2007n is an integer.

Answer: 9. Note that 2007 = 32 · 223. Using the fact that the number of times a prime p divides n!is given by ⌊

n

p

⌋+⌊

n

p2

⌋+⌊

n

p3

⌋+ · · · ,

it follows that the answer is 9.

6. [6] There are three video game systems: the Paystation, the WHAT, and the ZBoz2π, and none ofthese systems will play games for the other systems. Uncle Riemann has three nephews: Bernoulli,Galois, and Dirac. Bernoulli owns a Paystation and a WHAT, Galois owns a WHAT and a ZBoz2π, andDirac owns a ZBoz2π and a Paystation. A store sells 4 different games for the Paystation, 6 differentgames for the WHAT, and 10 different games for the ZBoz2π. Uncle Riemann does not understand the

1

difference between the systems, so he walks into the store and buys 3 random games (not necessarilydistinct) and randomly hands them to his nephews. What is the probability that each nephew receivesa game he can play?

Answer: 725 . Since the games are not necessarily distinct, probabilities are independent. Multiplying

the odds that each nephew receives a game he can play, we get 10/20 · 14/20 · 16/20 = 7/25.

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

7. [7] A student at Harvard named KevinWas counting his stones by 11

He messed up n timesAnd instead counted 9s

And wound up at 2007.

How many values of n could make this limerick true?

Answer: 21. The mathematical content is that 9n + 11k = 2007, for some nonnegative integers nand k. As 2007 = 9 · 223, k must be divisible by 9. Using modulo 11, we see that n is 3 more than amultiple of 11. Thus, the possibilities are n = 223, 212, 201, . . . , 3, which are 21 in number.

8. [7] A circle inscribed in a square,Has two chords as shown in a pair.

It has radius 2,And P bisects TU.

The chords’ intersection is where?

Answer the question by giving the distance of the point of intersection from the center of the circle.

Answer: 2√

2− 2. Let OB intersect the circle at X and Y , and the chord PM at Q, such that O

lies between X and Q. Then MNXQ is a parallelogram. For, OB ‖ NM by homothety at C andPM ‖ NX because MNXP is an isoceles trapezoid. It follows that QX = MN. Considering that thecenter of the circle together with points M,C, and N determines a square of side length 2, it followsthat MN = 2

√2, so the answer is 2

√2− 2.

9. [7] I ponder some numbers in bed,All products of three primes I’ve said, n = 372 · 3 . . .

Apply φ they’re still fun: φ(n) =now Elev’n cubed plus one. 113 + 1?

What numbers could be in my head?

Answer: 2007,2738,3122. The numbers expressible as a product of three primes are each ofthe form p3, p2q, or pqr, where p, q, and r are distinct primes. Now, φ(p3) = p2(p − 1), φ(p2q) =p(p−1)(q−1), and φ(pqr) = (p−1)(q−1)(r−1). We require 113+1 = 12·111 = 223237. The first case iseasy to rule out, since necessarily p = 2 or p = 3, which both fail. The second case requires p = 2, p = 3,or p = 37. These give q = 667, 223, and 2, respectively. As 667 = 23 · 29, we reject 22 · 667, but32233 = 2007 and 3722 = 2738. In the third case, exactly one of the primes is 2, since all other primesare odd. So say p = 2. There are three possibilities for (q, r): (2·1+1, 2·32 ·37+1), (2·3+1, 2·3·37+1),and (2 · 32 + 1, 2 · 37 + 1). Those are (3, 667), (7, 223), and (19, 75), respectively, of which only (7, 223)is a pair of primes. So the third and final possibility is 2 · 7 · 223 = 3122.

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

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10. [8] Let A12 denote the answer to problem 12. There exists a unique triple of digits (B,C,D) such that10 > A12 > B > C > D > 0 and

A12BCD −DCBA12 = BDA12C,

where A12BCD denotes the four digit base 10 integer. Compute B + C + D.

Answer: 11. Since D < A12, when A is subtracted from D we must carry over from C. Thus,D + 10 − A12 = C. Next, since C − 1 < C < B, we must carry over from the tens digit, so that(C − 1 + 10)− B = A12. Now B > C so B − 1 ≥ C, and (B − 1)− C = D. Similarly, A12 −D = B.Solving this system of four equations produces (A12, B,C, D) = (7, 6, 4, 1).

11. [8] Let A10 denote the answer to problem 10. Two circles lie in the plane; denote the lengths of theinternal and external tangents between these two circles by x and y, respectively. Given that theproduct of the radii of these two circles is 15/2, and that the distance between their centers is A10,determine y2 − x2.

Answer: 30. Suppose the circles have radii r1 and r2. Then using the tangents to build righttriangles, we have x2 + (r1 + r2)2 = A2

10 = y2 + (r1 − r2)2. Thus, y2 − x2 = (r1 + r2)2 − (r1 − r2)2 =4r1r2 = 30.

12. [8] Let A11 denote the answer to problem 11. Determine the smallest prime p such that the arithmeticsequence p, p + A11, p + 2A11, . . . begins with the largest possible number of primes.

Answer: 7. First, note that the maximal number of initial primes is bounded above by the smallestprime not dividing A11, with equality possible only if p is this prime. For, if q is the smallest primenot dividing A11, then the first q terms of the arithmetic sequence determine a complete residue classmodulo q, and the multiple of q is nonprime unless it equals q. If q < A11, then q must appear first inthe sequence, and thus divide the (q + 1)st term. If q > A11, then A11 = 2 and q = 3 by Bertrand’spostulate, so q must appear first by inspection.

Now since A11 = 30, the bound is 7. In fact, 7, 37, 67, 97, 127, and 157 are prime, but 187 is not.Then on the one hand, our bound of seven initial primes is not realizable. On the other hand, thisimplies an upper bound of six, and this bound is achieved by p = 7. Smaller primes p yield only oneinitial prime, so 7 is the answer.

Remarks. A number of famous theorems are concerned with the distribution of prime numbers. Fortwo relatively prime positive integers a and b, the arithmetic progression a, a + b, a + 2b, . . . containsinfinitely many primes, a result known as Dirichlet’s theorem. It was shown recently (c. 2004) thatthere exist arbitrarily long arithmetic progressions consisting of primes only. Bertrand’s postulate statesthat for any positive integer n, there exists a prime p such that n < p ≤ 2n. This is an unfortunatemisnomer, as the statement is known to be true. As with many theorems concerning the distributionsof primes, these results are easily stated in elementary terms, concealing elaborate proofs.

There is just one triple of possible (A10, A11, A12) of answers to these three problems. Your teamwill receive credit only for answers matching these. (So, for example, submitting a wrong answer forproblem 11 will not alter the correctness of your answer to problem 12.)

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

13. [9] Determine the largest integer n such that 72048 − 1 is divisible by 2n.

Answer: 14. We have

72048 − 1 = (7− 1)(7 + 1)(72 + 1

) (74 + 1

)· · ·(71024 + 1

).

In the expansion, the eleven terms other than 7 + 1 are divisible by 2 exactly once, as can be checkedeasily with modulo 4.

14. [9] We are given some similar triangles. Their areas are 12, 32, 52 . . . , and 492. If the smallest trianglehas a perimeter of 4, what is the sum of all the triangles’ perimeters?

Answer: 2500. Because the triangles are all similar, they all have the same ratio of perimetersquared to area, or, equivalently, the same ratio of perimeter to the square root of area. Because thelatter ratio is 4 for the smallest triangle, it is 4 for all the triangles, and thus their perimeters are4 · 1, 4 · 3, 4 · 5, . . . , 4 · 49, and the sum of these numbers is [4(1 + 3 + 5 + · · ·+ 49) = 4(252) = 2500.

15. [9] Points A,B, and C lie in that order on line `, such that AB = 3 and BC = 2. Point H is such thatCH is perpendicular to `. Determine the length CH such that ∠AHB is as large as possible.

Answer:√

10. Let ω denote the circumcircle of triangle ABH. Since AB is fixed, the smaller theradius of ω, the bigger the angle AHB. If ω crosses the line CH in more than one point, thenthere exists a smaller circle that goes through A and B that crosses CH at a point H ′. But angleAH ′B is greater than AHB, contradicting our assumption that H is the optimal spot. Thus thecircle ω crosses the line CH at exactly one spot: ie, ω is tangent to CH at H. By Power of a Point,CH2 = CA CB = 5 · 2 = 10, so CH =

√10.

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

16. [10] Let ABC be a triangle with AB = 7, BC = 9, and CA = 4. Let D be the point such thatAB ‖ CD and CA ‖ BD. Let R be a point within triangle BCD. Lines ` and m going through R areparallel to CA and AB respectively. Line ` meets AB and BC at P and P ′ respectively, and m meetsCA and BC at Q and Q′ respectively. If S denotes the largest possible sum of the areas of trianglesBPP ′, RP ′Q′, and CQQ′, determine the value of S2.

4

Answer: 180. Let R′ denote the intersection of the lines through Q′ and P ′ parallel to ` and mrespectively. Then [RP ′Q′] = [R′P ′Q′]. Triangles BPP ′, R′P ′Q′, and CQQ′ lie in ABC withoutoverlap, so that on the one hand, S ≤ ABC. On the other, this bound is realizable by taking R to be avertex of triangle BCD. We compute the square of the area of ABC to be 10·(10−9)·(10−7)·(10−4) =180.

17. [10] During the regular season, Washington Redskins achieve a record of 10 wins and 6 losses. Computethe probability that their wins came in three streaks of consecutive wins, assuming that all possiblearrangements of wins and losses are equally likely. (For example, the record LLWWWWWLWWL-WWWLL contains three winning streaks, while WWWWWWWLLLLLLWWW has just two.)

Answer: 45286 . Suppse the winning streaks consist of w1, w2, and w3 wins, in chronological order,

where the first winning streak is preceded by l0 consecutive losses and the i winning streak is imme-diately succeeded by li losses. Then w1, w2, w3, l1, l2 > 0 are positive and l0, l3 ≥ 0 are nonnegative.The equations

w1 + w2 + w3 = 10 and (l0 + 1) + l1 + l2 + (l3 + 1) = 8,

are independent, and have(92

)and

(73

)solutions, respectively. It follows that the answer is(

92

)(73

)(166

) =3152002

.

18. [10] Convex quadrilateral ABCD has right angles ∠A and ∠C and is such that AB = BC andAD = CD. The diagonals AC and BD intersect at point M . Points P and Q lie on the circumcircleof triangle AMB and segment CD, respectively, such that points P,M , and Q are collinear. Supposethat m∠ABC = 160◦ and m∠QMC = 40◦. Find MP ·MQ, given that MC = 6.

Answer: 36. Note that m∠QPB = m∠MPB = m∠MAB = m∠CAB = ∠BCA = ∠CDB. Thus,MP ·MQ = MB ·MD. On the other hand, segment CM is an altitude of right triangle BCD, soMB ·MD = MC2 = 36.

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

19. [10] Define x ? y =√

x2+3xy+y2−2x−2y+4

xy+4 . Compute

((· · · ((2007 ? 2006) ? 2005) ? · · · ) ? 1).

5

Answer:√

159 . Note that x ? 2 =

√x2+6x+4−2x−4+4

2x+4 =√

(x+2)2

2(x+2) = 12 for x > −2. Because x ? y > 0 if

x, y > 0, we need only compute 12 ? 1 =

√14+ 3

2+1−3+412+4

=√

159 .

20. [10] For a a positive real number, let x1, x2, x3 be the roots of the equation x3 − ax2 + ax − a = 0.Determine the smallest possible value of x3

1 + x32 + x3

3 − 3x1x2x3.

Answer: −4. Note that x1 + x2 + x3 = x1x2 + x2x3 + x3x1 = a. Then

x31 + x3

2 + x33 − 3x1x2x3 = (x1 + x2 + x3)(x2

1 + x22 + x2

3 − (x1x2 + x2x3 + x3x1))

= (x1 + x2 + x3)((x1 + x2 + x3)

2 − 3 (x1x2 + x2x3 + x3x1))

= a · (a2 − 3a) = a3 − 3a2.

The expression is negative only where 0 < a < 3, so we need only consider these values of a. Finally,AM-GM gives 3

√(6− 2a)(a)(a) ≤ (6−2a)+a+a

3 = 2, with equality where a = 2, and this rewrites as(a− 3)a2 ≥ −4.

21. [10] Bob the bomb-defuser has stumbled upon an active bomb. He opens it up, and finds the red andgreen wires conveniently located for him to cut. Being a seasoned member of the bomb-squad, Bobquickly determines that it is the green wire that he should cut, and puts his wirecutters on the greenwire. But just before he starts to cut, the bomb starts to count down, ticking every second. Eachtime the bomb ticks, starting at time t = 15 seconds, Bob panics and has a certain chance to move hiswirecutters to the other wire. However, he is a rational man even when panicking, and has a 1

2t2 chanceof switching wires at time t, regardless of which wire he is about to cut. When the bomb ticks at t = 1,Bob cuts whatever wire his wirecutters are on, without switching wires. What is the probability thatBob cuts the green wire?

Answer: 2330 . Suppose Bob makes n independent decisions, with probabilities of switching p1, p2, . . . , pn.

Then in the expansion of the product

P (x) = (p1 + (1− p1)x)(p2 + (1− p2)x) · · · (pn + (1− pn)x),

the sum of the coefficients of even powers of x gives the probability that Bob makes his original decision.This is just (P (1) + P (−1))/2, so the probability is just

1 +(1− 1

15 15

) (1− 1

14 14

)· · ·(1− 1

2 2

)2

=1 + 14 16

15 1513 1514 14 · · ·

1 32 2

2=

1 + 815

2=

2330

.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

22. [12] The sequence {an}n≥1 is defined by an+2 = 7an+1 − an for positive integers n with initial valuesa1 = 1 and a2 = 8. Another sequence, {bn}, is defined by the rule bn+2 = 3bn+1 − bn for positiveintegers n together with the values b1 = 1 and b2 = 2. Find gcd(a5000, b501).

Answer: 89. We show by induction that an = F4n−2 and bn = F2n−1, where Fk is the kth Fibonaccinumber. The base cases are clear. As for the inductive steps, note that

Fk+2 = Fk+1 + Fk = 2Fk + Fk−1 = 3Fk − Fk−2

andFk+4 = 3Fk+2 − Fk = 8Fk + 3Fk−2 = 7Fk − Fk−4.

We wish to compute the greatest common denominator of F19998 and F1001. The Fibonacci numberssatisfy the property that gcd(Fm, Fn) = Fgcd(m,n), which can be proven by noting that they are periodicmodulo any positive integer. So since gcd(19998, 1001) = 11, the answer is F11 = 89.

6

23. [12] In triangle ABC, ∠ABC is obtuse. Point D lies on side AC such that ∠ABD is right, and point Elies on side AC between A and D such that BD bisects ∠EBC. Find CE, given that AC = 35, BC = 7,and BE = 5.

Answer: 10. Reflect A and E over BD to A′ and E′ respectively. Note that the angle conditionsshow that A′ and E′ lie on AB and BC respectively. B is the midpoint of segment AA′ and CE′ =BC −BE′ = 2. Menelaus’ theorem now gives

CD

DA· AA′

A′B· BE′

E′C= 1,

from which DA = 5CD or CD = AC/6. By the angle bisector theorem, DE = 5CD/7, so thatCE = 12CD/7 = 10.

24. [12] Let x, y, n be positive integers with n > 1. How many ordered triples (x, y, n) of solutions arethere to the equation xn − yn = 2100 ?Answer: 49. Break all possible values of n into the four cases: n = 2, n = 4, n > 4 and n odd. ByFermat’s theorem, no solutions exist for the n = 4 case because we may write y4 +

(225)4 = x4.

We show that for n odd, no solutions exist to the more general equation xn − yn = 2k where k is apositive integer. Assume otherwise for contradiction’s sake, and suppose on the grounds of well orderingthat k is the least exponent for which a solution exists. Clearly x and y must both be even or bothodd. If both are odd, we have (x− y)(xn−1 + .... + yn−1). The right factor of this expression containsan odd number of odd terms whose sum is an odd number greater than 1, impossible. Similarly if xand y are even, write x = 2u and y = 2v. The equation becomes un − vn = 2k−n. If k − n is greaterthan 0, then our choice k could not have been minimal. Otherwise, k− n = 0, so that two consecutivepositive integers are perfect nth powers, which is also absurd.For the case that n is even and greater than 4, consider the same generalization and hypotheses.Writing n = 2m, we find (xm − ym)(xm + ym) = 2k. Then xm − ym = 2a < 2k. By our previous work,we see that m cannot be an odd integer greater than 1. But then m must also be even, contrary tothe minimality of k.Finally, for n = 2 we get x2 − y2 = 2100. Factoring the left hand side gives x− y = 2a and x + y = 2b,where implicit is a < b. Solving, we get x = 2b−1 +2a−1 and y = 2b−1−2a−1, for a total of 49 solutions.Namely, those corresponding to (a, b) = (1, 99), (2, 98), · · · , (49, 51).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

25. [12] Two real numbers x and y are such that 8y4 + 4x2y2 + 4xy2 + 2x3 + 2y2 + 2x = x2 + 1. Find allpossible values of x + 2y2.

Answer: 12 . Writing a = x + 2y2, the given quickly becomes 4y2a + 2x2a + a + x = x2 + 1. We can

rewrite 4y2a for further reduction to a(2a− 2x) + 2x2a + a + x = x2 + 1, or

2a2 + (2x2 − 2x + 1)a + (−x2 + x− 1) = 0. (∗)

The quadratic formula produces the discriminant

(2x2 − 2x + 1)2 + 8(x2 − x + 1) = (2x2 − 2x + 3)2,

an identity that can be treated with the difference of squares, so that a = −2x2+2x−1±(2x2−2x+3)4 =

12 ,−x2 +x− 1. Now a was constructed from x and y, so is not free. Indeed, the second expression fliesin the face of the trivial inequality: a = −x2 + x − 1 < −x2 + x ≤ x + 2y2 = a. On the other hand,a = 1/2 is a bona fide solution to (*), which is identical to the original equation.

7

26. [12] ABCD is a cyclic quadrilateral in which AB = 4, BC = 3, CD = 2, and AD = 5. Diagonals ACand BD intersect at X. A circle ω passes through A and is tangent to BD at X. ω intersects AB andAD at Y and Z respectively. Compute Y Z/BD.

Answer: 115143 . Denote the lengths AB,BC, CD, and DA by a, b, c, and d respectively. Because

ABCD is cyclic, 4ABX ∼ 4DCX and 4ADX ∼ 4BCX. It follows that AXDX = BX

CX = ac and

AXBX = DX

CX = db . Therefore we may write AX = adk, BX = abk, CX = bck, and DX = cdk for some k.

Now, ∠XDC = ∠BAX = ∠Y XB and ∠DCX = ∠XBY , so 4BXY ∼ 4CDX. Thus, XY =DX · BX

CD = cdk · abkc = abdk2. Analogously, XY = acdk2. Note that XY/XZ = CB/CD. Since

∠Y XZ = π − ∠ZAY = ∠BCD, we have that 4XY Z ∼ 4CBD. Thus, Y Z/BD = XY/CB = adk2.

Finally, Ptolemy’s theorem applied to ABCD gives

(ad + bc)k · (ab + cd)k = ac + bd

It follows that the answer isad(ac + bd)

(ab + cd)(ad + bc)=

20 · 2322 · 26

=115143

27. [12] Find the number of 7-tuples (n1, . . . , n7) of integers such that

7∑i=1

n6i = 96957.

Answer: 2688. Consider the equation in modulo 9. All perfect 6th powers are either 0 or 1. Since 9divides 96957, it must be that each ni is a multiple of 3. Writing 3ai = ni and dividing both sides by36, we have a6

1 + · · ·+a67 = 133. Since sixth powers are nonnegative, |ai| ≤ 2. Again considering modulo

9, we see that ai 6= 0. Thus, a6i ∈ {1, 64}. The only possibility is 133 = 64 + 64 + 1 + 1 + 1 + 1 + 1, so

|a1|, . . . , |a7| consists of 2 2’s and 5 1’s. It follows that the answer is(72

)· 27 = 2688.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

28. [15] Compute the circumradius of cyclic hexagon ABCDEF , which has side lengths AB = BC =2, CD = DE = 9, and EF = FA = 12.

8

Answer: 8. Construct point E′ on the circumcircle of ABCDEF such that DE′ = EF = 12 andE′F = DE = 9; then BE′ is a diameter. Let BE′ = d. Then CE′ =

√BE′2 −BC2 =

√d2 − 4 and

BD =√

BE′2 −DE′2 =√

d2 − 144. Applying Ptolemy’s theorem to BCDE′ now yields

9 · d + 2 · 12 =√

(d2 − 4)(d2 − 144)

Squaring and rearranging, we find 0 = d4 − 229d2 − 432d = d(d − 16)(d2 + 16d + 27). Since d is apositive real number, d = 16, and the circumradius is 8.

29. [15] A sequence {an}n≥1 of positive reals is defined by the rule an+1a5n−1 = a4

na2n−2 for integers n > 2

together with the initial values a1 = 8 and a2 = 64 and a3 = 1024. Compute√a1 +

√a2 +

√a3 + · · ·

Answer: 3√

2. Taking the base-2 log of the sequence {an} converts the multiplicative rule to a morefamiliar additive rule: log2(an+1) − 4 log2(an) + 5 log2(an−1) − 2 log2(an−2) = 0. The characteristicequation is 0 = x3 − 4x2 + 5x− 2 = (x− 1)2(x− 2), so log2(an) is of the form a · n + b + c · 2n and wefind an = 22n+2n−1

. Now,√a1 +

√a2 +

√a3 + · · · =

√2 ·

√4 +

√16 +

√64 + · · ·.

We can estimate the new nested radical expression as 3, which expands thus

3 =√

4 + 5 =√

4 +√

16 + 9 =

√4 +

√16 +

√64 + 17 = · · ·

As a rigorous confirmation, we have 2k + 1 =√

4k + (2k+1 + 1), as desired. It follows that the answeris 3√

2.

30. [15] ABCD is a cyclic quadrilateral in which AB = 3, BC = 5, CD = 6, and AD = 10. M , I, and Tare the feet of the perpendiculars from D to lines AB,AC, and BC respectively. Determine the valueof MI/IT .

9

Answer: 259 . Quadrilaterals AMID and DICT are cyclic, having right angles ∠AMD,∠AID, and

∠CID,∠CTD respectively. We see that M , I, and T are collinear. For, m∠MID = π −m∠DAM =π −m∠DAB = m∠BCD = π −m∠DCT = π −m∠DIT . Therefore, Menelaus’ theorem applied totriangle MTB and line ICA gives

MI

IT· TC

CB· BA

AM= 1

On the other hand, triangle ADM is similar to triangle CDT since ∠AMD ∼= ∠CTD and ∠DAM ∼=∠DCT and thus AM/CT = AD/CD. It follows that

MI

IT=

BC ·AM

AB · CT=

BC ·AD

AB · CD=

5 · 103 · 6

=259

Remarks. The line MIT , constructed in this problem by taking perpendiculars from a point on thecircumcircle of ABC, is known as the Simson line. It is often helpful for us to use directed angles whileangle chasing to avoid supplementary configuration issues, such as those arising while establishing thecollinearity of M, I, and T .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

31. [18] A sequence {an}n≥0 of real numbers satisfies the recursion an+1 = a3n − 3a2

n + 3 for all positiveintegers n. For how many values of a0 does a2007 = a0?

Answer: 32007. If x appears in the sequence, the next term x3 − 3x2 + 3 is the same if and only if0 = x3 − 3x2 − x + 3 = (x− 3)(x− 1)(x + 1). Moreover, that next term is strictly larger if x > 3 andstrictly smaller if x < −1. It follows that no values of a0 with |a0 − 1| > 2 yield a0 = a2007.

Now suppose a0 = a2007 and write a0 = 1 + eαi + e−αi; the values a0 we seek will be in bijectivecorrespondence with solutions α where 0 ≤ α ≤ π. Then

a1 = (a0 − 1)3 − 3a0 + 4 = e3αi + 3eαi + 3e−αi + e−3αi − 3eαi − 3e−αi − 3 + 4 = e3αi + e−3αi + 1,

and an easy inductive argument gives a2007 = e32007αi + e−32007αi + 1. It follows that a0 = a2007 isequivalent to cos(α) = cos

(32007α

). Now,

cos(32007α

)− cos(α) = 2 sin

((32007 + 1

2

)α

)sin((

32007 − 12

)α

),

so since sin(kx) = 0 for a positive integer k if and only if x is a multiple of πk , the solutions α are

{0, 2π32007−1 , 4π

32007−1 , . . . , π}∪{0, 2π32007+1 , . . . , π}. Because our values k are consecutive, these sets overlap

only at 0 and π, so there are 32007 distinct α.

10

32. [18] Triangle ABC has AB = 4, BC = 6, and AC = 5. Let O denote the circumcenter of ABC. Thecircle Γ is tangent to and surrounds the circumcircles of triangles AOB, BOC, and AOC. Determinethe diameter of Γ.

Answer: 256√

717 . Denote by ω, Γ1,Γ2, and Γ3 the circumcenters of triangles ABC, BOC,COA, and

AOB, respectively. An inversion about ω interchanges Γ1 and line BC, Γ2 and line CA, and Γ3 andline AB. This inversion also preserves tangency between generalized circles, so the image of Γ is a circletangent to AB, BC, and CA. It is the incircle of ABC because it is closer to O than these lines andABC is acute.

Now we run a few standard calculations. Where s, r, and R denote the semiperimeter, inradius, andcircumradius of ABC, respectively, we have the following:

[ABC] =√

s(s− a)(s− b)(s− c) =15√

74

;

r = [ABC]/s =√

7/2;

R =abc

4[ABC]=

8√7

OI2 = R(R− 2r) =87.

Let OI intersect the incircle of ABC at P and Q, with I between P and O. Then OP = r + OI andOQ = r − OI, and PQ is a diameter. Under the inversion, P and Q map to P ′ and Q′ respectively.Because P, I, O, and Q are collinear in that order, P ′ and Q′ are diametrically opposed on Γ. It followsthat the diameter of Γ is

P ′Q′ = OP ′ + OQ′ =R2

OP+

R2

OQ= R2

(1

r + OI+

1r −OI

)=

2rR2

r2 −OI2.

We plug in the values found above to arrive at 256√

717 .

33. [18] Compute ∫ 2

1

9x + 4x5 + 3x2 + x

dx.

(No, your TI-89 doesn’t know how to do this one. Yes, the end is near.)

Answer: ln 8023 . We break the given integral into two pieces:

∫ 2

1

9x + 4x5 + 3x2 + x

dx = 5∫ 2

1

x4 + 3x + 1x5 + 3x2 + x

dx−∫ 2

1

5x4 + 6x + 1x5 + 3x2 + x

dx.

11

These two new integrals are easily computed; for, the first integrand reduces to 1/x and the second isof the form f ′(x)/f(x). We obtain[

5 ln |x| − ln |x5 + 3x2 + x|]21

= ln 32− ln 46 + ln 5 = ln8023

Motivation. Writing f(x) = 9x + 4 and g(x) = x5 + 3x2 + x = x(x4 + 3x + 1), we wish to findthe antiderivative of f(x)/g(x). It makes sense to consider other rational functions with denominatorg(x) that have an exact antiderivative. Clearly, if the numerator were f1(x) = x4 + 3x + 1 or aconstant multiple, then we can integrate the function. Another trivial case is if the numerator weref2(x) = g′(x) = 5x4 + 6x + 1 or a constant multiple. Guessing that f(x) is a linear combination off1(x) and f2(x), we easily find that f(x) = 9x + 4 = 5f1(x)− f2(x).

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10th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

34. [?] The Game. Eric and Greg are watching their new favorite TV show, The Price is Right. Bob Barkerrecently raised the intellectual level of his program, and he begins the latest installment with biddingon following question: How many Carmichael numbers are there less than 100,000?

Each team is to list one nonnegative integer not greater than 100,000. Let X denote the answer to Bob’squestion. The teams listing N , a maximal bid (of those submitted) not greater than X, will receiveN points, and all other teams will neither receive nor lose points. (A Carmichael number is an oddcomposite integer n such that n divides an−1−1 for all integers a relatively prime to n with 1 < a < n.)Answer: 16? There are 16 such numbers: 561, 1105, 1729, 2465, 2821, 6601, 8911, 10585, 15841,29341, 41041, 46657, 52633, 62745, 63973, and 75361. The next, 101101, is too large to be counted.Their distribution is considerably more subtle than that of the primes, and it was only recently proventhat there are infinitely many Carmichael numbers. For sufficiently large n, C(n) > n2/7, although thisbound has been subsequently improved. (For details, see Alford, W. R.; Granville, A.; and Pomerance,C. “There are Infinitely Many Carmichael Numbers.” Annals of Mathematics. 139 (1994), 703-722.)The expectation is that teams are unable to determine that X = 16; otherwise, the obvious dominantplay is listing 16. The Problem Czar, anticipating that teams will attempt to deduce X by consideringthe point values of the other problems in the triplet, has chosen a value X that is considerably lower.Teams may of course speculate whether this action has been taken, and to what extent, etc... On theactual TV show, many contestants win by guessing prices of 1, or other numbers dramatically lowerthan the actual price. This strategy is enhanced because of the show’s ordered bidding, and will bemore difficult here. It will be interesting to see the submissions.

35. [≤ 25] The Algorithm. There are thirteen broken computers situated at the following set S of thirteenpoints in the plane:

A = (1, 10) B = (976, 9) C = (666, 87)D = (377, 422) E = (535, 488) F = (775, 488)G = (941, 500) H = (225, 583) I = (388, 696)J = (3, 713) K = (504, 872) L = (560, 934)

M = (22, 997)

At time t = 0, a repairman begins moving from one computer to the next, traveling continuously instraight lines at unit speed. Assuming the repairman begins and A and fixes computers instantly, whatpath does he take to minimize the total downtime of the computers? List the points he visits in order.Your score will be bN

40c, where

N = 1000 + bthe optimal downtimec − byour downtimec ,

or 0, whichever is greater. By total downtime we mean the sum∑P∈S

tP ,

12

where tP is the time at which the repairman reaches P .

Answer: ADHIKLEFGBCJM. This is an instance of the minimum-latency problem, which is atleast NP-hard. There is an easy O(n!) algorithm, but this is unavailable to teams on computationalgrounds (100MHz calculators used to seem fast...) The best strategy may be drawing an accuratepicture and exercising geometric intuition. The distribution of the points somewhat resembles a short,four-pronged fork with its outermost prongs bent apart; it is plausible to assume that the optimal orderrespects this shape. The optimal downtime is 24113.147907, realized by ADHIKLEFGBCJM, thougha number of others also receive positive marks.

36. [25] The Marathon. Let ω denote the incircle of triangle ABC. The segments BC, CA, and AB aretangent to ω at D, E, and F , respectively. Point P lies on EF such that segment PD is perpendicularto BC. The line AP intersects BC at Q. The circles ω1 and ω2 pass through B and C, respectively,and are tangent to AQ at Q; the former meets AB again at X, and the latter meets AC again at Y .The line XY intersects BC at Z. Given that AB = 15, BC = 14, and CA = 13, find bXZ · Y Zc.

Answer: 101. Construct D′ diametrically opposed to D, so that ∠DFD′ and ∠DED′ are right, andnote that P lies on DD′. By standard angle chasing, m∠FDD′ = β (half angle B) and m∠D′DE = γ.Thus, m∠DD′F = 90◦ − β and m∠ED′D = 90◦ − γ. Then by the law of sines, DE : ED′ : D′F :FD = cos(γ) : sin(γ) : sin(β) : sin(γ). Now using 4DEP ∼ 4FD′P and 4DFP ∼ 4ED′P , we have

EP

PF=

ED · ED′

FD · FD′ =sin(γ) cos(γ)sin(β) sin(β)

=c

b.

Let the dilation centered at A sending E to C map P and F to P ′ and F ′, respectively. Note thatAF ′ = AC as AE and AF are equal tangents, and CP ′ : P ′F ′ = EP : PF = c : b by similarity. Thenby Menelaus’ theorem,

1 =BQ

QC

CP ′

P ′F ′F ′A

AB=

BQ

QC

c

b

b

c,

so that BQ = QC and AQ is actually a median. So, AQ2 = 14

(2b2 + 2c2 − a2

)= 148. Now by Power

of a Point, AB · AX = AQ2 = AC · AY, so AX = 148/15 and AY = 148/13. Moreover, BXCY iscyclic as 4ABC ∼ 4AY X. Thus, XZ ·Y Z = BZ ·CZ, and it suffices to compute BZ/CZ. Menelausonce more gives

1 =BZ

ZC

CY

Y A

AX

XB,

13

whence, BZ/CZ = (AY/AX)(BX/CY ) = (15/13)((77 · 13)/(21 · 15)) = 11/3. We write CZ = 3d andBZ = 11d. Because AX < AB and AY < AC, Z does not lie on segment BC. Given the configurationinformation, BC = 8d = 14, so d = 7/4, and finally bBZ · CZc = b33d2c = b 161716 c = 101.

14

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Team Round: A Division

Σ, τ, and You: Fun at Fraternities? [270]

A number theoretic function is a function whose domain is the set of positive integers. A multiplicativenumber theoretic function is a number theoretic function f such that f(mn) = f(m)f(n) for all pairs ofrelatively prime positive integers m and n. Examples of multiplicative number theoretic functions includeσ, τ, φ, and µ, defined as follows. For each positive integer n,

• The sum-of-divisors function, σ(n), is the sum of all positive integer divisors of n. If p1, . . . , pi aredistinct primes and e1, . . . , ei are positive integers,

σ (pe11 · · · p

eii ) =

i∏k=1

(1 + pk + · · ·+ pek

k ) =i∏

k=1

pek+1k − 1pk − 1

.

• The divisor function, τ(n), is the number of positive integer divisors of n. It can be computed by theformula

τ (pe11 · · · p

eii ) = (e1 + 1) · · · (ei + 1),

where p1, . . . , pi and e1, . . . , ei are as above.

• Euler’s totient function, φ(n), is the number of positive integers k ≤ n such that k and n are relativelyprime. For p1, . . . , pi and e1, . . . , ei as above, the phi function satisfies

φ (pe11 · · · p

eii ) =

i∏k=1

pek−1k (pk − 1).

• The Möbius function, µ(n), is equal to either 1, -1, or 0. An integer is called square-free if it is notdivisible by the square of any prime. If n is a square-free positive integer having an even number ofdistinct prime divisors, µ(n) = 1. If n is a square-free positive integer having an odd number of distinctprime divisors, µ(n) = −1. Otherwise, µ(n) = 0.

The Möbius function has a number of peculiar properties. For example, if f and g are number theoreticfunctions such that

g(n) =∑d|n

f(d),

for all positive integers n, thenf(n) =

∑d|n

g(d)µ(n

d

).

This is known as Möbius inversion. In proving the following problems, you may use any of the precedingassertions without proving them. You may also cite the results of previous problems, even if you were unableto prove them.

1. [15] Evaluate the functions φ(n), σ(n), and τ(n) for n = 12, n = 2007, and n = 22007.

2. [20] Solve for the positive integer(s) n such that φ(n2) = 1000φ(n).

3. [25] Prove that for every integer n greater than 1,

σ(n)φ(n) ≤ n2 − 1.

When does equality hold?

1

4. [25] Let F and G be two multiplicative functions, and define for positive integers n,

H(n) =∑d|n

F (d)G(n

d

).

The number theoretic function H is called the convolution of F and G. Prove that H is multiplicative.

5. [30] Prove the identity ∑d|n

τ(d)3 =

∑d|n

τ(d)

2

.

6. [25] Show that for positive integers n, ∑d|n

φ(d) = n.

7. [25] Show that for positive integers n, ∑d|n

µ(d)d

=φ(n)

n.

8. [30] Determine with proof, a simple closed form expression for∑d|n

φ(d)τ(n

d

).

9. [35] Find all positive integers n such that

n∑k=1

φ(k) =3n2 + 5

8.

10. [40] Find all pairs (n, k) of positive integers such that

σ(n)φ(n) =n2

k.

Grab Bag - Miscellaneous Problems [130]

11. [30] Find all functions f : Q→ Q such that

f(x)f(y) = f(x) + f(y)− f(xy)1 + f(x + y) = f(xy) + f(x)f(y)

for all rational numbers x, y.

12. [30] Let ABCD be a cyclic quadrilateral, and let P be the intersection of its two diagonals. PointsR,S, T, and U are feet of the perpendiculars from P to sides AB,BC, CD, and AD, respectively. Showthat quadrilateral RSTU is bicentric if and only if AC ⊥ BD. (Note that a quadrilateral is calledinscriptible if it has an incircle; a quadrilateral is called bicentric if it is both cyclic and inscriptible.)

13. [30] Find all nonconstant polynomials P (x), with real coefficients and having only real zeros, such thatP (x + 1)P (x2 − x + 1) = P (x3 + 1) for all real numbers x.

14. [40] Find an explicit, closed form formula for

n∑k=1

k · (−1)k ·(nk

)n + k + 1

.

2

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Team Round: A Division

Σ, τ, and You: Fun at Fraternities? [270]

A number theoretic function is a function whose domain is the set of positive integers. A multiplicativenumber theoretic function is a number theoretic function f such that f(mn) = f(m)f(n) for all pairs ofrelatively prime positive integers m and n. Examples of multiplicative number theoretic functions includeσ, τ, φ, and µ, defined as follows. For each positive integer n,

• The sum-of-divisors function, σ(n), is the sum of all positive integer divisors of n. If p1, . . . , pi aredistinct primes and e1, . . . , ei are positive integers,

σ (pe11 · · · p

eii ) =

i∏k=1

(1 + pk + · · ·+ pek

k ) =i∏

k=1

pek+1k − 1pk − 1

.

• The divisor function, τ(n), is the number of positive integer divisors of n. It can be computed by theformula

τ (pe11 · · · p

eii ) = (e1 + 1) · · · (ei + 1),

where p1, . . . , pi and e1, . . . , ei are as above.

• Euler’s totient function, φ(n), is the number of positive integers k ≤ n such that k and n are relativelyprime. For p1, . . . , pi and e1, . . . , ei as above, the phi function satisfies

φ (pe11 · · · p

eii ) =

i∏k=1

pek−1k (pk − 1).

• The Möbius function, µ(n), is equal to either 1, -1, or 0. An integer is called square-free if it is notdivisible by the square of any prime. If n is a square-free positive integer having an even number ofdistinct prime divisors, µ(n) = 1. If n is a square-free positive integer having an odd number of distinctprime divisors, µ(n) = −1. Otherwise, µ(n) = 0.

The Möbius function has a number of peculiar properties. For example, if f and g are number theoreticfunctions such that

g(n) =∑d|n

f(d),

for all positive integers n, thenf(n) =

∑d|n

g(d)µ(n

d

).

This is known as Möbius inversion. In proving the following problems, you may use any of the precedingassertions without proving them. You may also cite the results of previous problems, even if you were unableto prove them.

1. [15] Evaluate the functions φ(n), σ(n), and τ(n) for n = 12, n = 2007, and n = 22007.

Solution. For n = 12 = 22 · 31,

φ(12) = 2(2− 1)(3− 1) = 4, σ(12) = (1 + 2 + 4)(1 + 3) = 28, τ(12) = (2 + 1)(1 + 1) = 6;

for n = 2007 = 32 · 223,

φ(2007) = 3(3−1)(223−1) = 1332, σ(2007) = (1+3+9)(1+223) = 2912, τ(2007) = (2+1)(1+1) = 6;

and for n = 22007,

φ(22007

)= 22006, σ

(22007

)=(1 + 2 + · · ·+ 22007

)= 22008 − 1, τ

(22007

)= 2007 + 1 = 2008. �

1

2. [20] Solve for the positive integer(s) n such that φ(n2) = 1000φ(n).

Answer: 1000.

Solution. The unique solution is n = 1000. For, φ(pn) = pφ(n) for every prime p dividing n, so thatφ(n2) = nφ(n) for all positive integers n. �

3. [25] Prove that for every integer n greater than 1,

σ(n)φ(n) ≤ n2 − 1.

When does equality hold?Solution. Note that

σ(mn)φ(mn) = σ(m)φ(m)σ(n)φ(n) ≤ (m2 − 1)(n2 − 1) = (mn)2 − (m2 + n2 − 1) < (mn)2 − 1

for any pair of relatively prime positive integers (m,n) other than (1, 1). Now, for p a prime and k apositive integer, σ

(pk)

= 1 + p + · · ·+ pk = pk+1−1p−1 and φ

(pk)

= pk − 1p · p

k = (p− 1)pk−1. Thus,

σ(pk)φ(pk)

=pk+1 − 1

p− 1· (p− 1)pk−1 = (pk+1 − 1)pk−1 = p2k − pk−1 ≤ p2k − 1,

with equality where k = 1. It follows that equality holds in the given inequality if and only if n isprime. �

4. [25] Let F and G be two multiplicative functions, and define for positive integers n,

H(n) =∑d|n

F (d)G(n

d

).

The number theoretic function H is called the convolution of F and G. Prove that H is multiplicative.Solution. Let m and n be relatively prime positive integers. We have

H(m)H(n) =

∑d|m

F (d)G(m

d

)∑d′|n

F (d′)G( n

d′

)=

∑d|m, d′|n

F (d)F (d′)G(m

d′

)G(n

d

)=

∑d|m, d′|n

F (dd′)G(mn

dd′

)=∑d|mn

F (d)G(mn

d

)= H(mn). �

5. [30] Prove the identity ∑d|n

τ(d)3 =

∑d|n

τ(d)

2

.

Solution. Note that τ3 is multiplicative; in light of the convolution property just shown, it followsthat both sides of the posed equality are multiplicative. Thus, it would suffice to prove the claim forn a power of a prime. So, write n = pk where p is a prime and k is a nonnegative integer. Then∑

d|n

τ(d)3 =k∑

i=0

τ(pi)3

=k∑

i=0

(i + 1)3

= 13 + · · ·+ (k + 1)3 =(k + 1)2(k + 2)2

4=(

(k + 1)(k + 2)2

)2

=

(k∑

i=0

τ(pi))2

=

∑d|n

τ(d)

2

,

as required. �

2

6. [25] Show that for positive integers n, ∑d|n

φ(d) = n.

Solution. Both sides are multiplicative functions of n, the right side trivially and the left because forrelatively prime positive integers n and n′,∑

d|n

φ(d)

∑d′|n′

φ(d′)

=∑

d|n, d′|n′

φ(d)φ(d′),

and φ(d)φ(d′) = φ(dd′). The identity is then easy to check; since φ(pk) = pk−1(p−1) for positive integersk and φ(1) = 1, we have φ(1) + φ(p) + · · ·+ φ(pk) = 1 + (p− 1) + (p2 − p) + · · ·+

(pk − pk−1

)= pk, as

desired. �

7. [25] Show that for positive integers n, ∑d|n

µ(d)d

=φ(n)

n.

Solution. On the grounds of the previous problem, Möbius inversion with f(k) = φ(k) and g(k) = kgives:

φ(n) = f(n) =∑d|n

g(d)µ(n

d

)=∑d′|n

g( n

d′

)µ(d′) =

∑d′|n

n

d′µ(d′). �

Alternatively, one uses the convolution of the functions f(k) = n and g(k) = µ(d)d . The strategy is the

same as the previous convolution proof. For n = pk with k a positive integer, we have φ(n) = pk−pk−1,while the series reduces to pk · µ(1) + pk · µ(p)/p = pk − pk−1.

8. [30] Determine with proof, a simple closed form expression for∑d|n

φ(d)τ(n

d

).

Solution. We claim the series reduces to σ(n). The series counts the ordered triples (d, x, y) withd|n;x|d; 0 < y ≤ n/d; and (y, n/d) = 1. To see this, write∑

d|n

φ(d)τ(n

d

)=∑d′|n

φ(n

d′)τ(d′),

so that for a given d′|n we may choose x and y as described above. On the other hand, we can countthese triples by groups sharing a given x. Fixing x as a divisor of n fixes an integer n

x . Then d variessuch that n

d is a divisor of nx . For each divisor n

d of nx there are precisely φ

(nd

)choices y, so that by

the lemma from the previous problem, there are nx triples (d, x, y) for a given x. It follows that there

are precisely σ(n) such triples (d, x, y). �

Again, an alternative is to use the multiplicativity of the convolution, although it is now a little moredifficult. Write n = pk so that

∑d|n

φ(d)τ(n

d

)=

k∑m=0

φ (pm) τ(pk−m

)= k + 1 +

k∑m=1

pm−1(p− 1)(k −m + 1)

= k + 1 +

(k∑

m=1

pm(k −m + 1)

)−

(k∑

m=1

pm−1(k −m + 1)

)= k + 1 + pk − k +

k−1∑m′=1

pm′

= 1 + p + · · ·+ pk = σ(pk).

3

9. [35] Find all positive integers n such that

n∑k=1

φ(k) =3n2 + 5

8.

Answer: 1,3,5.

Solution. We contend that the proper relation is

n∑k=1

φ(k) ≤ 3n2 + 58

. (∗)

Let Φ(k) denote the left hand side of (∗). It is trivial to see that for n ≤ 7 the posed inequality holds,has equality where n = 1, 3, 5, and holds strictly for n = 7. Note that φ(2k) ≤ k and φ(2k + 1) ≤ 2k,the former because 2, 4, . . . , 2k share a common divisor. It follows that φ(2k)+φ(2k+1) ≤ 3k. Supposefor the sake of induction that Φ(2k − 1) < 3(2k−1)2+5

8 . Then

Φ(2k + 1) = Φ(2k − 1) + φ(2k) + φ(2k + 1) <3(2k − 1)2 + 5

8+ 3k =

3(2k + 1)2 + 58

.

To complete the proof, it is enough to note that for a positive integer k,

3(2k − 1)2 + 58

+ k <3(2k)2 + 5

8. �

10. [40] Find all pairs (n, k) of positive integers such that

σ(n)φ(n) =n2

k.

Answer: (1,1).

Solution. It is clear that for a given integer n, there is at most one integer k for which the equationholds. For n = 1 this is k = 1. But, for n > 1, problem 1 asserts that σ(n)φ(n) ≤ n2 − 1 < n2, so thatk ≥ 2. We now claim that 2 > n2

σ(n)φ(n) . Write n = pe11 · · · p

ek

k , where the pi are distinct primes andei ≥ 1 for all i, and let q1 < q2 < · · · be the primes in ascending order. Then

n2

σ(n)φ(n)=

k∏i=1

p2eii

pei+1i −1

pi−1 · (pi − 1)pei−1i

=k∏

i=1

p2eii

p2eii − pei−1

i

=k∏

i=1

11− p−1−ei

i

≤k∏

i=1

11− p−2

i

<∞∏

i=1

11− q−2

i

=∞∏

i=1

∞∑j=0

1q2ji

=∞∑

n=1

1n2

< 1 +∞∑

n=2

1n2 − 1

= 1 +12

((11− 1

3

)+(

12− 1

4

)+(

13− 1

5

)+ · · ·

)=

74

< 2.

It follows that there can be no solutions to k = n2

σ(n)φ(n) other than n = k = 1. �

Grab Bag - Miscellaneous Problems [130]

4

11. [30] Find all functions f : Q→ Q such that

f(x)f(y) = f(x) + f(y)− f(xy)1 + f(x + y) = f(xy) + f(x)f(y)

for all rational numbers x, y.

Answer: f(x) = 1∀x, and f(x) = 1− x∀x.

Solution. Considering the first equation, either side of the second equation is equal to f(x) + f(y).Now write g(x) = 1− f(x), so that

g(xy) = 1− f(xy) = 1− f(x)− f(y) + f(x)f(y) = (1− f(x))(1− f(y)) = g(x)g(y)g(x + y) = 1− f(x + y) = 1− f(x) + 1− f(y) = g(x) + g(y),

By induction, g(nx) = ng(x) for all integers n, so that g(p/q) = (p/q)g(1) for integers p and q with qnonzero; i.e., g(x) = xg(1). As g is multiplicative, g(1) = g(1)2, so the only possibilities are g(1) = 1and g(1) = 0. These give g(x) = x and g(x) = 0, or f(x) = 1 − x and f(x) = 1, respectively. Oneeasily checks that these functions are satisfactory. �

12. [30] Let ABCD be a cyclic quadrilateral, and let P be the intersection of its two diagonals. PointsR,S, T, and U are feet of the perpendiculars from P to sides AB,BC, CD, and AD, respectively. Showthat quadrilateral RSTU is bicentric if and only if AC ⊥ BD. (Note that a quadrilateral is calledinscriptible if it has an incircle; a quadrilateral is called bicentric if it is both cyclic and inscriptible.)

Solution. First we show that RSTU is always inscriptible. Note that in addition to ABCD, we havecyclic quadrilaterals ARPU and BSPR. Thus,

∠PRU = ∠PAU = ∠CAD = ∠CBD = ∠SBP = ∠SRP,

and it follows that P lies on the bisector of ∠SRU. Analogously, P lies on the bisectors of ∠TSR and∠UTS, so is equidistant from lines UR,RS, ST, and TU, and RSTU is inscriptible having incenter P.Now we show that RSTU is cyclic if and only if the diagonals of ABCD are orthogonal. We have

∠APB = π − ∠BAP − ∠PBA = π − ∠RAP − ∠PBR = π − ∠RUP − ∠PSR

= π − 12

(∠RUT + ∠TSR) .

It follows that ∠APB = π2 if and only if ∠RUT + ∠TSR = π, as desired. �

13. [30] Find all nonconstant polynomials P (x), with real coefficients and having only real zeros, such thatP (x + 1)P (x2 − x + 1) = P (x3 + 1) for all real numbers x.

5

Answer: {P(x) = xk |k ∈ Z+}.

Solution. Note that if P (α) = 0, then by setting x = α− 1 in the given equation, we find 0 = P (x3 +1) = P (α3−3α2 +3α). Because P is nonconstant, it has at least one zero. Because P has finite degree,there exist minimal and maximal roots of P . Writing α3 − 3α2 + 3α ≥ α ⇐⇒ α(α − 1)(α − 2) ≥ 0,we see that the largest zero of P cannot exceed 2. Likewise, the smallest zero cannot be negative, soall of the zeroes of P lie in [0, 2]. Moreover, if α /∈ {0, 1, 2} is a zero of P , then α′ = α3 − 3α2 + 3α isanother zero of P that lies strictly between α and 1. Because P has only finitely many zeroes, all ofits zeroes must lie in {0, 1, 2}. Now write P (x) = kxp(x− 1)q(x− 2)r for nonnegative integers p, q andr having a positive sum. The given equation becomes

k2(x + 1)pxq(x− 1)r(x2 − x + 1)p(x2 − x)q(x2 − x− 1)r = P (x + 1)P (x2 − x + 1)= P (x3 + 1) = k(x3 + 1)px3q(x3 − 1)r. (∗)

For the leading coefficients to agree, we require k = k2. Because the leading coefficient is nonzero, Pmust be monic. In (∗), r must be zero lest P (x3 + 1) = 0 have complex roots. Then q must be zero aswell. For, if q is positive, then P (x2−x + 1) = 0 has 1 as a root while P (x3 + 1) does not. Finally, theremaining possibilities are P (x) = xp for p an arbitrary positive integer. It is easily seen that thesepolynomials are satisfactory. �

14. [40] Find an explicit, closed form formula for

n∑k=1

k · (−1)k ·(nk

)n + k + 1

.

Answer: −1

(2n+1n ) or − n!(n+1)!

(2n+1)! or obvious equivalent.

Solution. Consider the interpolation of the polynomial P (x) = x · n! at x = 0, 1, . . . , n. We obtainthe identity

P (x) = x · n! =n∑

k=0

k · n!∏j 6=k

x− j

k − j

=n∑

k=0

k · n! · x(x− 1) · · · (x− k + 1)(x− k − 1) · · · (x− n)k!(n− k)!(−1)n−k

=n∑

k=1

k · (−1)n−k ·(

n

k

)· x(x− 1) · · · (x− k + 1)(x− k − 1) · · · (x− n).

This identity is valid for all complex numbers x, but, to extract a factor 1n+k+1 from the valid product

of each summand, we set x = −n− 1, so that

−(n+1)! =n∑

k=1

k(−1)n−k

(n

k

)(−n−1) · · · (−n−k)(−n−k−2) · · · (−2n−1) =

n∑k=1

k(−1)k(nk

)(2n + 1)!

n!(n + k + 1).

Finally,n∑

k=1

k · (−1)k ·(nk

)n + k + 1

=−n!(n + 1)!(2n + 1)!

=−1(

2n+1n

) .�

6

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Team Round: B Division

Compute (x− a)(x− b) · · · (x− z) - Short Answer [200]

For this section, your team should give only the anwers to the problems.

1. [20] Find the sum of the positive integer divisors of 22007.

2. [20] The four sides of quadrilateral ABCD are equal in length. Determine the perimeter of ABCDgiven that it has area 120 and AC = 10.

3. [20] Five people are crowding into a booth against a wall at a noisy restaurant. If at most three canfit on one side, how many seating arrangements accomodate them all?

4. [20] Thomas and Michael are just two people in a large pool of well qualified candidates for appointmentto a problem writing committee for a prestigious college math contest. It is 40 times more likely thatboth will serve if the size of the committee is increased from its traditional 3 members to a whoppingn members. Determine n. (Each person in the pool is equally likely to be chosen.)

5. [20] The curves y = x2(x − 3)2 and y = (x2 − 1)(x − 2) intersect at a number of points in the realplane. Determine the sum of the x-coordinates of these points of intersection.

6. [20] Andrew has a fair six sided die labeled with 1 through 6 as usual. He tosses it repeatedly, and onevery third roll writes down the number facing up as long as it is not the 6. He stops as soon as thelast two numbers he has written down are squares or one is a prime and the other is a square. Whatis the probability that he stops after writing squares consecutively?

7. [20] Three positive reals x, y, and z are such that

x2 + 2(y − 1)(z − 1) = 85y2 + 2(z − 1)(x− 1) = 84z2 + 2(x− 1)(y − 1) = 89.

Compute x + y + z.

8. [20] Find the positive real number(s) x such that 12

(3x2 − 1

)=(x2 − 50x− 10

) (x2 + 25x + 5

).

9. [20] Cyclic quadrilateral ABCD has side lengths AB = 1, BC = 2, CD = 3, and AD = 4. DetermineAC/BD.

10. [20] A positive real number x is such that

3√

1− x3 + 3√

1 + x3 = 1.

Find x2.

Adult Acorns - Gee, I’m a Tree! [200]

In this section of the team round, your team will derive some basic results concerning tangential quadri-laterals. Tangential quadrilaterals have an incircle, or a circle lying within them that is tangent to all foursides. If a quadrilateral has an incircle, then the center of this circle is the incenter of the quadrilateral. Asyou shall see, tangential quadrilaterals are related to cyclic quadrilaterals. For reference, a review of cyclicquadrilaterals is given at the end of this section.

Your answers for this section of the team test should be proofs. Note that you may use anystandard facts about cyclic quadrilaterals, such as those listed at the end of this test, without proving them.Additionally, you may cite the results of previous problems, even if you were unable to prove them.

For these problems, ABCD is a tangential quadrilateral having incenter I. For the first three problems,the point P is constructed such that triangle PAB is similar to triangle IDC and lies outside ABCD.

1

1. [30] Show that PAIB is cyclic by proving that ∠IAP is supplementary to ∠PBI.

2. [40] Show that triangle PAI is similar to triangle BIC. Then conclude that

PA =PI

BC·BI.

3. [25] Deduce from the above thatBC

AD· AI

BI· DI

CI= 1.

4. [25] Show that AB + CD = AD + BC. Use the above to conclude that for some positive number α,

AB = α ·(

AI

CI+

BI

DI

)BC = α ·

(BI

DI+

CI

AI

)CD = α ·

(CI

AI+

DI

BI

)DA = α ·

(DI

BI+

AI

CI

).

5. [40] Show that

AB ·BC = BI2 +AI ·BI · CI

DI.

6. [40] Let the incircle of ABCD be tangent to sides AB,BC, CD, and AD at points P,Q,R, and S,respectively. Show that ABCD is cyclic if and only if PR ⊥ QS.

A brief review of cyclic Quadrilaterals.

The following discussion of cyclic quadrilaterals is included for reference. Any of the results given heremay be cited without proof in your writeups.

A cyclic quadrilateral is a quadrilateral whose four vertices lie on a circle called the circumcircle (thecircle is unique if it exists.) If a quadrilateral has a circumcircle, then the center of this circumcircleis called the circumcenter of the quadrilateral. For a convex quadrilateral ABCD, the following areequivalent:

• Quadrilateral ABCD is cyclic;

• ∠ABD = ∠ACD (or ∠BCA = ∠BDA, etc.);

• Angles ∠ABC and ∠CDA are supplementary, that is, m∠ABC + m∠CDA = 180◦ (or angles∠BCD and ∠BAD are supplementary);

Cyclic quadrilaterals have a number of interesting properties. A cyclic quadrilateral ABCD satisfies

AC ·BD = AB · CD + AD ·BC,

a result known as Ptolemy’s theorem. Another result, typically called Power of a Point, asserts thatgiven a circle ω, a point P anywhere in the plane of ω, and a line ` through P intersecting ω at pointsA and B, the value of AP · BP is independent of `; i.e., if a second line `′ through P intersects ω atA′ and B′, then AP ·BP = A′P ·B′P. This second theorem is proved via similar triangles. Say P liesoutside of ω, that ` and `′ are as before and that A and A′ lie on segments BP and B′P respectively.Then triangle AA′P is similar to triangle B′BP because the triangles share an angle at P and we have

m∠AA′P = 180◦ −m∠B′A′A = m∠ABB′ = m∠PBB′.

The case where A = B is valid and describes the tangents to ω. A similar proof works for P inside ω.

2

10th Annual Harvard-MIT Mathematics TournamentSaturday 24 February 2007

Team Round: B Division

Compute (x− a)(x− b) · · · (x− z) - Short Answer [200]

For this section, your team should give only the anwers to the problems.

1. [20] Find the sum of the positive integer divisors of 22007.

Answer: 22008 − 1. The divisors are the powers of two not exceeding 22007. So the sum is

1 + 2 + 22 + · · ·+ 22007 = −1 + 2 + 2 + 22 + · · ·+ 22007 = −1 + 22 + 22 + · · ·+ 22007 = · · · = −1 + 22008.

2. [20] The four sides of quadrilateral ABCD are equal in length. Determine the perimeter of ABCDgiven that it has area 120 and AC = 10.

Answer: 52. Let M be the midpoint of AC. Then triangles AMB,BMC,CMD, and DMA are allright triangles having legs 5 and h for some h. The area of ABCD is 120, but also 4 · ( 1

2 · 5 · h) = 10h,

so h = 12. Then AB = BC = CD = DA =√

122 + 52 = 13, and the perimeter of ABCD is 52.

3. [20] Five people are crowding into a booth against a wall at a noisy restaurant. If at most three canfit on one side, how many seating arrangements accomodate them all?

Answer: 240. Three people will sit on one side and two sit on the other, giving a factor of two.Then there are 5! ways to permute the people.

4. [20] Thomas and Michael are just two people in a large pool of well qualified candidates for appointmentto a problem writing committee for a prestigious college math contest. It is 40 times more likely thatboth will serve if the size of the committee is increased from its traditional 3 members to a whoppingn members. Determine n. (Each person in the pool is equally likely to be chosen.)

Answer: 16. Suppose there are k candidates. Then the probability that both serve on a 3 memberedcommittee is (k − 2)/

(k3

), and the odds that both serve on an n membered committee are

(k−2n−2

)/(

kn

).

The ratio of the latter to the former is(k3

)(k−2n−2

)(k − 2)

(kn

) =k!(k − 2)!1!(k − 3)!n!(k − n)!

k!(k − 2)!(n− 2)!(k − n)!3!(k − 3)!=

n · (n− 1)3!

.

Solving n · (n− 1) = 240 produces n = 16,−15, and we discard the latter.

5. [20] The curves y = x2(x − 3)2 and y = (x2 − 1)(x − 2) intersect at a number of points in the realplane. Determine the sum of the x-coordinates of these points of intersection.

Answer: 7. Because the first curve touches the x-axis at x = 0 and x = 3 while the second curvecrosses the x-axis at x = ±1 and x = 2, there are four points of intersection. In particular, the pointsof intersection have x-coordinates determined by the difference of the two curves:

0 = x2(x− 3)2 − (x2 − 1)(x− 2) = (x4 − 6x3 + · · · )− (x3 + · · · ) = x4 − 7x3 + · · · .

We need only the first two coefficients to determine x1 + x2 + x3 + x4 = −(−7

1

)= 7.

1

6. [20] Andrew has a fair six sided die labeled with 1 through 6 as usual. He tosses it repeatedly, and onevery third roll writes down the number facing up as long as it is not the 6. He stops as soon as thelast two numbers he has written down are squares or one is a prime and the other is a square. Whatis the probability that he stops after writing squares consecutively?

Answer: 4/25. We can safely ignore all of the rolls he doesn’t record. The probability that he stopsafter writing two squares consecutively is the same as the probability that he never rolls a prime. For,as soon as the first prime is written, either it must have been preceded by a square or it will be followedby a nonnegative number of additional primes and then a square. So we want the probability that twonumbers chosen uniformly with replacement from {1, 2, 3, 4, 5} are both squares, which is (2/5)2.

7. [20] Three positive reals x, y, and z are such that

x2 + 2(y − 1)(z − 1) = 85y2 + 2(z − 1)(x− 1) = 84z2 + 2(x− 1)(y − 1) = 89.

Compute x + y + z.

Answer: 18. Add the three equations to obtain

x2 + y2 + z2 + 2xy + 2yz + 2zx− 4x− 4y − 4z + 6 = 258,

which rewrites as (x + y + z − 2)2 = 256. Evidently, x + y + z = 2± 16. Since x, y, and z are positive,x + y + z > 0 so x + y + z = 2 + 16 = 18.

8. [20] Find the positive real number(s) x such that 12

(3x2 − 1

)=(x2 − 50x− 10

) (x2 + 25x + 5

).

Answer: 25 + 2√

159. Write a = x2−50x−10 and b = x2+25x+5; the given becomes a+2b−12 = ab,

so 0 = 2ab−a−2b+1 = (a−1)(2b−1). Then a−1 = x2−50x−11 = 0 or 2b−1 = 2x2 +50x+9 = 0.The former has a positive root, x = 25 + 2

√159, while the latter cannot, for obvious reasons.

9. [20] Cyclic quadrilateral ABCD has side lengths AB = 1, BC = 2, CD = 3, and AD = 4. DetermineAC/BD.

Answer: 5/7. Let the diagonals intersect at P. Note that triangles ABP and DCP are similar,so that 3AP = DP and 3BP = CP. Additionally, triangles BCP and ADP are similar, so that2BP = AP. It follows that

AC

BD=

AP + PC

BP + PD=

2BP + 3BP

BP + 6BP=

57.

2

10. [20] A positive real number x is such that

3√

1− x3 + 3√

1 + x3 = 1.

Find x2.

Answer:3√283 . Cubing the given equation yields

1 = (1− x3) + 3 3√

(1− x3)(1 + x3)(

3√

1− x3 + 3√

1 + x3)

+ (1 + x3) = 2 + 3 3√

1− x6.

Then −13 = 3

√1− x6, so −1

27 = 1− x6 and x6 = 2827 and x2 =

3√283 .

Adult Acorns - Gee, I’m a Tree! [200]

In this section of the team round, your team will derive some basic results concerning tangential quadri-laterals. Tangential quadrilaterals have an incircle, or a circle lying within them that is tangent to all foursides. If a quadrilateral has an incircle, then the center of this circle is the incenter of the quadrilateral. Asyou shall see, tangential quadrilaterals are related to cyclic quadrilaterals. For reference, a review of cyclicquadrilaterals is given at the end of this section.

Your answers for this section of the team test should be proofs. Note that you may use anystandard facts about cyclic quadrilaterals, such as those listed at the end of this test, without proving them.Additionally, you may cite the results of previous problems, even if you were unable to prove them.

For these problems, ABCD is a tangential quadrilateral having incenter I. For the first three problems,the point P is constructed such that triangle PAB is similar to triangle IDC and lies outside ABCD.

1. [30] Show that PAIB is cyclic by proving that ∠IAP is supplementary to ∠PBI.

Solution. Note that I lies on the angle bisectors of the angles of quadrilateral ABCD. So writing∠DAB = 2α, ∠ABC = 2β, ∠BCD = 2γ, and ∠CDA = 2δ, we have

∠IAP + ∠PBI = ∠IAB + ∠BAP + ∠PBA + ∠ABI

= ∠IAB + ∠CDI + ∠ICD + ∠ABI

= α + β + γ + δ.

We are done because the angles in quadrilateral ABCD add up to 360◦. �

2. [40] Show that triangle PAI is similar to triangle BIC. Then conclude that

PA =PI

BC·BI.

Solution. We have ∠IBC = ∠ABI because I lies on the angle bisector, and ∠ABI = ∠API becausePAIB is cyclic. Additionally,

∠BCI = ∠ICD = ∠PBA = ∠PIA,

by the angle bisector CI, that triangles PAB and IDC are similar, and the fact that PAIB iscyclic, respectively. It follows that triangles PAI and BIC are similar. In particular, it follows thatIP/PA = BC/BI, as required. �

3

3. [25] Deduce from the above thatBC

AD· AI

BI· DI

CI= 1.

Solution. Exchanging the roles of A and D with B and C, respectively, converts the formula fromproblem 2 into another formula:

PB =PI

AD·AD.

Then one the one hand, dividing the two gives PA/PB = (AD · BI)/(BC · AI). On the other hand,PA/PB = DI/CI because triangles PAB and IDC are similar. Clearing the denominators in theequation

DI

CI=

AD ·BI

BC ·AI

yields the desired form. �

4. [25] Show that AB + CD = AD + BC. Use the above to conclude that for some positive number α,

AB = α ·(

AI

CI+

BI

DI

)BC = α ·

(BI

DI+

CI

AI

)CD = α ·

(CI

AI+

DI

BI

)DA = α ·

(DI

BI+

AI

CI

).

Solution. Draw in the points of tangency P,Q,R, and S, of the incircle with sides AB,BC, CD, andAD, as shown. Then we have equal tangents AP = AS, BP = BQ, CQ = CR, and DR = DS. Then

AB + CD = AP + BP + CR + DR = AS + (BQ + CQ) + DS = BC + AD.

Using the result of problem 3, we set BC = x · BI · CI and AD = x · AI · DI for some x, andAB = y ·AI ·BI and CD = y · CI ·DI for some y. Now because AB + CD = BC + AD, we obtain

y(AI ·BI + CI ·DI) = x(BI · CI + AI ·DI).

So it follows that the ratio AB : BC : CD : DA is uniquely determined. One easily checks that theposed ratio satisfies the three required relations. �

5. [40] Show that

AB ·BC = BI2 +AI ·BI · CI

DI.

Solution. Returning to the original set up, Ptolemy’s theorem applied to quadrilateral PAIB givesAB · PI = PA · BI + PB · AI. Substituting equation PA = PI

BC · BI from problem 2 and its cousinPB = PI

AD ·AI allows us to write

AB · PI =PI

BC·BI2 +

PI

AD·AI2,

4

orAB ·BC = BI2 +

BC

AD·AI2.

Substituting the formula BC/AD = BI·CIAI·DI from problem 3 finishes the problem. �

6. [40] Let the incircle of ABCD be tangent to sides AB,BC, CD, and AD at points P,Q,R, and S,respectively. Show that ABCD is cyclic if and only if PR ⊥ QS.

Solution. Let the diagonals of PQRS intersect at T. Because AP and AS are tangent to ω at Pand S, we may write α = ∠ASP = ∠SPA = ∠SQP and β = ∠CQR = ∠QRC = ∠QPR. Then∠PTQ = π − α − β. On the other hand, ∠PAS = π − 2α and ∠RCQ = π − 2β, so that ABCD iscyclic if and only if

π = ∠BAD + ∠DCB = 2π − 2α− 2β,

or simplyπ/2 = π − α− β = ∠PTQ,

as desired. �

A brief review of cyclic Quadrilaterals.

The following discussion of cyclic quadrilaterals is included for reference. Any of the results given heremay be cited without proof in your writeups.

A cyclic quadrilateral is a quadrilateral whose four vertices lie on a circle called the circumcircle (thecircle is unique if it exists.) If a quadrilateral has a circumcircle, then the center of this circumcircleis called the circumcenter of the quadrilateral. For a convex quadrilateral ABCD, the following areequivalent:

• Quadrilateral ABCD is cyclic;

• ∠ABD = ∠ACD (or ∠BCA = ∠BDA, etc.);

• Angles ∠ABC and ∠CDA are supplementary, that is, m∠ABC + m∠CDA = 180◦ (or angles∠BCD and ∠BAD are supplementary);

Cyclic quadrilaterals have a number of interesting properties. A cyclic quadrilateral ABCD satisfies

AC ·BD = AB · CD + AD ·BC,

a result known as Ptolemy’s theorem. Another result, typically called Power of a Point, asserts thatgiven a circle ω, a point P anywhere in the plane of ω, and a line ` through P intersecting ω at pointsA and B, the value of AP · BP is independent of `; i.e., if a second line `′ through P intersects ω atA′ and B′, then AP ·BP = A′P ·B′P. This second theorem is proved via similar triangles. Say P liesoutside of ω, that ` and `′ are as before and that A and A′ lie on segments BP and B′P respectively.Then triangle AA′P is similar to triangle B′BP because the triangles share an angle at P and we have

m∠AA′P = 180◦ −m∠B′A′A = m∠ABB′ = m∠PBB′.

The case where A = B is valid and describes the tangents to ω. A similar proof works for P inside ω.

5

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Individual Round: Algebra Test

1. [3] Positive real numbers x, y satisfy the equations x2 + y2 = 1 and x4 + y4 = 1718 . Find xy.

2. [3] Let f(n) be the number of times you have to hit the √ key on a calculator to get a number lessthan 2 starting from n. For instance, f(2) = 1, f(5) = 2. For how many 1 < m < 2008 is f(m) odd?

3. [4] Determine all real numbers a such that the inequality |x2 + 2ax + 3a| ≤ 2 has exactly one solutionin x.

4. [4] The function f satisfies

f(x) + f(2x + y) + 5xy = f(3x− y) + 2x2 + 1

for all real numbers x, y. Determine the value of f(10).

5. [5] Let f(x) = x3 + x + 1. Suppose g is a cubic polynomial such that g(0) = −1, and the roots of g arethe squares of the roots of f . Find g(9).

6. [5] A root of unity is a complex number that is a solution to zn = 1 for some positive integer n.Determine the number of roots of unity that are also roots of z2 +az + b = 0 for some integers a and b.

7. [5] Compute∞∑

n=1

n−1∑k=1

k

2n+k.

8. [6] Compute arctan (tan 65◦ − 2 tan 40◦). (Express your answer in degrees.)

9. [7] Let S be the set of points (a, b) with 0 ≤ a, b ≤ 1 such that the equation

x4 + ax3 − bx2 + ax + 1 = 0

has at least one real root. Determine the area of the graph of S.

10. [8] Evaluate the infinite sum∞∑

n=0

(2n

n

)15n

.

1

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Individual Round: Algebra Test

1. [3] Positive real numbers x, y satisfy the equations x2 + y2 = 1 and x4 + y4 = 1718 . Find xy.

Answer: 16 We have 2x2y2 = (x2 + y2)2 − (x4 + y4) = 1

18 , so xy = 16 .

2. [3] Let f(n) be the number of times you have to hit the √ key on a calculator to get a number lessthan 2 starting from n. For instance, f(2) = 1, f(5) = 2. For how many 1 < m < 2008 is f(m) odd?

Answer: 242 This is [21, 22) ∪ [24, 28) ∪ [216, 232) . . . , and 28 < 2008 < 216 so we have exactly thefirst two intervals.

3. [4] Determine all real numbers a such that the inequality |x2 + 2ax+ 3a| ≤ 2 has exactly one solutionin x.

Answer: 1, 2 Let f(x) = x2 + 2ax+ 3a. Note that f(−3/2) = 9/4, so the graph of f is a parabolathat goes through (−3/2, 9/4). Then, the condition that |x2 + 2ax+ 3a| ≤ 2 has exactly one solutionmeans that the parabola has exactly one point in the strip −1 ≤ y ≤ 1, which is possible if and onlyif the parabola is tangent to y = 1. That is, x2 + 2ax + 3a = 2 has exactly one solution. Then, thediscriminant ∆ = 4a2 − 4(3a− 2) = 4a2 − 12a+ 8 must be zero. Solving the equation yields a = 1, 2.

4. [4] The function f satisfies

f(x) + f(2x+ y) + 5xy = f(3x− y) + 2x2 + 1

for all real numbers x, y. Determine the value of f(10).

Answer: −49 Setting x = 10 and y = 5 gives f(10) + f(25) + 250 = f(25) + 200 + 1, from whichwe get f(10) = −49.

Remark: By setting y = x2 , we see that the function is f(x) = − 1

2x2 + 1, and it can be checked that

this function indeed satisfies the given equation.

5. [5] Let f(x) = x3 +x+ 1. Suppose g is a cubic polynomial such that g(0) = −1, and the roots of g arethe squares of the roots of f . Find g(9).

Answer: 899 Let a, b, c be the zeros of f . Then f(x) = (x − a)(x − b)(x − c). Then, the roots ofg are a2, b2, c2, so g(x) = k(x− a2)(x− b2)(x− c2) for some constant k. Since abc = −f(0) = −1, wehave k = ka2b2c2 = −g(0) = 1. Thus,

g(x2) = (x2 − a2)(x2 − b2)(x2 − c2) = (x− a)(x− b)(x− c)(x+ a)(x+ b)(x+ c) = −f(x)f(−x).

Setting x = 3 gives g(9) = −f(3)f(−3) = −(31)(−29) = 899.

6. [5] A root of unity is a complex number that is a solution to zn = 1 for some positive integer n.Determine the number of roots of unity that are also roots of z2 +az+ b = 0 for some integers a and b.

Answer: 8 The only real roots of unity are 1 and −1. If ζ is a complex root of unity that is also a rootof the equation z2+az+b, then its conjugate ζ̄ must also be a root. In this case, |a| = |ζ+ζ̄| ≤ |ζ|+|ζ̄| =2 and b = ζζ̄ = 1. So we only need to check the quadratics z2+2z+1, z2+z+1, z2+1, z2−z+1, z2−2z+1.We find 8 roots of unity: ±1, ±i, 1

2 (±1±√

3i).

7. [5] Compute∞∑

n=1

n−1∑k=1

k

2n+k.

1

Answer: 49 We change the order of summation:

∞∑n=1

n−1∑k=1

k

2n+k=∞∑

k=1

k

2k

∞∑n=k+1

12n

=∞∑

k=1

k

4k=

49.

(The last two steps involve the summation of an infinite geometric series, and what is sometimes calledan infinite arithmetico-geometric series. These summations are quite standard, and thus we omit thedetails here.)

8. [6] Compute arctan (tan 65◦ − 2 tan 40◦). (Express your answer in degrees as an angle between 0◦ and180◦.)

Answer: 25◦ First Solution: We have

tan 65◦ − 2 tan 40◦ = cot 25◦ − 2 cot 50◦ = cot 25◦ − cot2 25◦ − 1cot 25◦

=1

cot 25◦= tan 25◦.

Therefore, the answer is 25◦.

Second Solution: We have

tan 65◦−2 tan 40◦ =1 + tan 20◦

1− tan 20◦− 4 tan 20◦

1− tan2 20◦=

(1− tan 20◦)2

(1− tan 20◦)(1 + tan 20◦)= tan(45◦−20◦) = tan 25◦.

Again, the answer is 25◦.

9. [7] Let S be the set of points (a, b) with 0 ≤ a, b ≤ 1 such that the equation

x4 + ax3 − bx2 + ax+ 1 = 0

has at least one real root. Determine the area of the graph of S.

Answer: 14 After dividing the equation by x2, we can rearrange it as

(x+

1x

)2

+ a

(x+

1x

)− b− 2 = 0

Let y = x+ 1x . We can check that the range of x+ 1

x as x varies over the nonzero reals is (−∞,−2]∪[2,∞).Thus, the following equation needs to have a real root:

y2 + ay − b− 2 = 0.

Its discriminant, a2 + 4(b+ 2), is always positive since a, b ≥ 0. Then, the maximum absolute value ofthe two roots is

a+√a2 + 4(b+ 2)

2.

We need this value to be at least 2. This is equivalent to√a2 + 4(b+ 2) ≥ 4− a.

We can square both sides and simplify to obtain

2a ≥ 2− b

This equation defines the region inside [0, 1]× [0, 1] that is occupied by S, from which we deduce thatthe desired area is 1/4.

2

10. [8] Evaluate the infinite sum∞∑

n=0

(2nn

)15n.

Answer:√

5 First Solution: Note that

(2nn

)=

(2n)!n! · n!

=(2n)(2n− 2)(2n− 4) · · · (2)

n!· (2n− 1)(2n− 3)(2n− 5) · · · (1)

n!

= 2n · (−2)n

n!

(−1

2

)(−1

2− 1)(−1

2− 2)· · ·(−1

2− n+ 1

)= (−4)n

(− 1

2

n

).

Then, by the binomial theorem, for any real x with |x| < 14 , we have

(1− 4x)−1/2 =∞∑

n=0

(− 1

2

n

)(−4x)n =

∞∑n=0

(2nn

)xn.

Therefore,

∞∑n=0

(2nn

)(15

)n

=1√

1− 45

=√

5.

Second Solution: Consider the generating function

f(x) =∞∑

n=0

(2nn

)xn.

It has formal integral given by

g(x) = I(f(x)) =∞∑

n=0

1n+ 1

(2nn

)xn+1 =

∞∑n=0

Cnxn+1 = x

∞∑n=0

Cnxn,

where Cn = 1n+1

(2nn

)is the nth Catalan number. Let h(x) =

∑∞n=0 Cnx

n; it suffices to compute thisgenerating function. Note that

1 + xh(x)2 = 1 + x∑

i,j≥0

CiCjxi+j = 1 + x

∑k≥0

(k∑

i=0

CiCk−i

)xk = 1 +

∑k≥0

Ck+1xk+1 = h(x),

where we’ve used the recurrence relation for the Catalan numbers. We now solve for h(x) with thequadratic equation to obtain

h(x) =1/x±

√1/x2 − 4/x2

=1±√

1− 4x2x

.

Note that we must choose the − sign in the ±, since the + would lead to a leading term of 1x for h (by

expanding√

1− 4x into a power series). Therefore, we see that

f(x) = D(g(x)) = D(xh(x)) = D

(1−√

1− 4x2

)=

1√1− 4x

and our answer is hence f(1/5) =√

5.

3

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Individual Round: Calculus Test

1. [3] Let f(x) = 1 + x + x2 + · · ·+ x100. Find f ′(1).

2. [3] Let ` be the line through (0, 0) and tangent to the curve y = x3 + x + 16. Find the slope of `.

3. [4] Find all y > 1 satisfying∫ y

1

x ln x dx =14

.

4. [4] Let a, b be constants such that limx→1

(ln(2− x))2

x2 + ax + b= 1. Determine the pair (a, b).

5. [4] Let f(x) = sin6(

x4

)+ cos6

(x4

)for all real numbers x. Determine f (2008)(0) (i.e., f differentiated

2008 times and then evaluated at x = 0).

6. [5] Determine the value of limn→∞

n∑k=0

(n

k

)−1

.

7. [5] Find p so that limx→∞ xp(

3√

x + 1 + 3√

x− 1− 2 3√

x)

is some non-zero real number.

8. [7] Let T =∫ ln 2

0

2e3x + e2x − 1e3x + e2x − ex + 1

dx. Evaluate eT .

9. [7] Evaluate the limit limn→∞

n−12 (1+ 1

n ) (11 · 22 · · · · · nn) 1

n2 .

10. [8] Evaluate the integral∫ 1

0

ln x ln(1− x) dx.

1

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Individual Round: Calculus Test

1. [3] Let f(x) = 1 + x + x2 + · · ·+ x100. Find f ′(1).

Answer: 5050 Note that f ′(x) = 1+2x+3x2 + · · ·+100x99, so f ′(1) = 1+2+ · · ·+100 = 100·1012 =

5050.

2. [3] Let ` be the line through (0, 0) and tangent to the curve y = x3 + x + 16. Find the slope of `.

Answer: 13 Let the point of tangency be (t, t3 + t + 16), then the slope of ` is (t3 + t + 16)/t. Onthe other hand, since dy/dx = 3x2 + 1, the slope of ` is 3t2 + 1. Therefore,

t3 + t + 16t

= 3t2 + 1.

Simplifying, we get t3 = 8, so t = 2. It follows that the slope is 3(2)2 + 1 = 13.

3. [4] Find all y > 1 satisfying∫ y

1

x ln x dx =14

.

Answer:√

e Applying integration by parts with u = ln x and v = 12x2, we get∫ y

1

x ln x dx =12x2 ln x

∣∣∣y1− 1

2

∫ y

1

x dx =12y2 ln y − 1

4y2 +

14.

So y2 ln y = 12y2. Since y > 1, we obtain ln y = 1

2 , and thus y =√

e.

4. [4] Let a, b be constants such that limx→1

(ln(2− x))2

x2 + ax + b= 1. Determine the pair (a, b).

Answer: (−2, 1) When x = 1, the numerator is 0, so the denominator must be zero as well, so1 + a + b = 0. Using l’Hôpital’s rule, we must have

1 = limx→1

(ln(2− x))2

x2 + ax + b= limx→1

2 ln(2− x)(x− 2)(2x + a)

,

and by the same argument we find that 2 + a = 0. Thus, a = −2 and b = 1. This is indeed a solution,as can be seen by finishing the computation.

5. [4] Let f(x) = sin6(x4

)+ cos6

(x4

)for all real numbers x. Determine f (2008)(0) (i.e., f differentiated

2008 times and then evaluated at x = 0).

Answer: 38 We have

sin6 x + cos6 x = (sin2 x + cos2 x)3 − 3 sin2 x cos2 x(sin2 x + cos2 x)

= 1− 3 sin2 x cos2 x = 1− 34

sin2 2x = 1− 34

(1− cos 4x

2

)=

58

+38

cos 4x.

It follows that f(x) = 58 + 3

8 cos x. Thus f (2008)(x) = 38 cos x. Evaluating at x = 0 gives 3

8 .

6. [5] Determine the value of limn→∞

n∑k=0

(n

k

)−1

.

1

Answer: 2 Let Sn denote the sum in the limit. For n ≥ 1, we have Sn ≥(n0

)−1 +(nn

)−1 = 2. Onthe other hand, for n ≥ 3, we have

Sn =(

n

0

)−1

+(

n

1

)−1

+(

n

n− 1

)−1

+(

n

n

)−1

+n−2∑k=2

(n

k

)−1

≤ 2 +2n

+ (n− 3)(

n

2

)−1

which goes to 2 as n→∞. Therefore, Sn → 2.

7. [5] Find p so that limx→∞ xp(

3√

x + 1 + 3√

x− 1− 2 3√

x)

is some non-zero real number.

Answer: 53 Make the substitution t = 1

x . Then the limit equals to

limt→0

t−p

(3

√1t

+ 1 + 3

√1t− 1− 2 3

√1t

)= limt→0

t−p−13(

3√

1 + t + 3√

1− t− 2).

We need the degree of the first nonzero term in the MacLaurin expansion of 3√

1 + t + 3√

1− t− 2. Wehave

3√

1 + t = 1 +13t− 1

9t2 + o(t2), 3

√1− t = 1− 1

3t− 1

9t2 + o(t2).

It follows that 3√

1 + t + 3√

1− t − 2 = − 29 t2 + o(t2). By consider the degree of the leading term, it

follows that −p− 13 = −2. So p = 5

3 .

8. [7] Let T =∫ ln 2

0

2e3x + e2x − 1e3x + e2x − ex + 1

dx. Evaluate eT .

Answer: 114 Divide the top and bottom by ex to obtain that

T =∫ ln 2

0

2e2x + ex − e−x

e2x + ex − 1 + e−xdx

Notice that 2e2x + ex − e−x is the derivative of e2x + ex − 1 + e−x, and so

T =[

ln |e2x + ex − 1 + e−x|]ln 2

0= ln

(4 + 2− 1 +

12

)− ln 2 = ln

(114

)Therefore, eT = 11

4 .

9. [7] Evaluate the limit limn→∞

n−12 (1+ 1

n ) (11 · 22 · · · · · nn) 1

n2 .

Answer: e−1/4 Taking the logarithm of the expression inside the limit, we find that it is

−12

(1 +

1n

)ln n +

1n2

n∑k=1

k ln k =1n

n∑k=1

k

nln(

k

n

).

We can recognize this as the as Riemann sum expansion for the integral∫ 1

0x ln x dx, and thus the

limit of the above sum as n → ∞ equals to the value of this integral. Evaluating this integral usingintegration by parts, we find that∫ 1

0

x ln x dx =12x2 ln x

∣∣∣10−∫ 1

0

x

2dx = −1

4.

Therefore, the original limit is e−1/4.

2

10. [8] Evaluate the integral∫ 1

0

ln x ln(1− x) dx.

Answer: 2− π2

6 We have the MacLaurin expansion ln(1− x) = −x− x2

2 −x3

3 − · · · . So

∫ 1

0

ln x ln(1− x) dx = −∫ 1

0

ln x

∞∑n=1

xn

ndx = −

∞∑n=1

1n

∫ 1

0

xn ln x dx.

Using integration by parts, we get

∫ 1

0

xn ln x dx =xn+1 ln x

n + 1

∣∣∣∣∣1

0

−∫ 1

0

xn

n + 1dx = − 1

(n + 1)2.

(We used the fact that limx→0 xn ln x = 0 for n > 0, which can be proven using l’Hôpital’s rule.)Therefore, the original integral equals to

∞∑n=1

1n(n + 1)2

=∞∑n=1

(1n− 1

n + 1− 1

(n + 1)2

).

Telescoping the sum and using the well-known identity∑∞n=0

1n2 = π2

6 , we see that the above sum isequal to 2− π2

6 .

3

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Individual Round: Combinatorics Test

1. [3] A 3 × 3 × 3 cube composed of 27 unit cubes rests on a horizontal plane. Determine the numberof ways of selecting two distinct unit cubes from a 3 × 3 × 1 block (the order is irrelevant) with theproperty that the line joining the centers of the two cubes makes a 45◦ angle with the horizontal plane.

2. [3] Let S = {1, 2, . . . , 2008}. For any nonempty subset A ⊂ S, define m(A) to be the median of A(when A has an even number of elements, m(A) is the average of the middle two elements). Determinethe average of m(A), when A is taken over all nonempty subsets of S.

3. [4] Farmer John has 5 cows, 4 pigs, and 7 horses. How many ways can he pair up the animals so thatevery pair consists of animals of different species? (Assume that all animals are distinguishable fromeach other.)

4. [4] Kermit the frog enjoys hopping around the infinite square grid in his backyard. It takes him 1 Jouleof energy to hop one step north or one step south, and 1 Joule of energy to hop one step east or onestep west. He wakes up one morning on the grid with 100 Joules of energy, and hops till he falls asleepwith 0 energy. How many different places could he have gone to sleep?

5. [5] Let S be the smallest subset of the integers with the property that 0 ∈ S and for any x ∈ S, wehave 3x ∈ S and 3x + 1 ∈ S. Determine the number of non-negative integers in S less than 2008.

6. [5] A Sudoku matrix is defined as a 9×9 array with entries from {1, 2, . . . , 9}and with the constraint that each row, each column, and each of the nine3× 3 boxes that tile the array contains each digit from 1 to 9 exactly once.A Sudoku matrix is chosen at random (so that every Sudoku matrix hasequal probability of being chosen). We know two of squares in this matrix,as shown. What is the probability that the square marked by ? containsthe digit 3?

12

?

7. [6] Let P1, P2, . . . , P8 be 8 distinct points on a circle. Determine the number of possible configurationsmade by drawing a set of line segments connecting pairs of these 8 points, such that: (1) each Pi isthe endpoint of at most one segment and (2) two no segments intersect. (The configuration with noedges drawn is allowed. An example of a valid configuration is shown below.)

P1

P2

P3

P4

P5

P6

P7

P8

8. [6] Determine the number of ways to select a sequence of 8 sets A1, A2, . . . , A8, such that each is asubset (possibly empty) of {1, 2}, and Am contains An if m divides n.

9. [7] On an infinite chessboard (whose squares are labeled by (x, y), where x and y range over all integers),a king is placed at (0, 0). On each turn, it has probability of 0.1 of moving to each of the four edge-neighboring squares, and a probability of 0.05 of moving to each of the four diagonally-neighboringsquares, and a probability of 0.4 of not moving. After 2008 turns, determine the probability that theking is on a square with both coordinates even. An exact answer is required.

10. [7] Determine the number of 8-tuples of nonnegative integers (a1, a2, a3, a4, b1, b2, b3, b4) satisfying 0 ≤ak ≤ k, for each k = 1, 2, 3, 4, and a1 + a2 + a3 + a4 + 2b1 + 3b2 + 4b3 + 5b4 = 19.

1

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Individual Round: Combinatorics Test

1. [3] A 3 × 3 × 3 cube composed of 27 unit cubes rests on a horizontal plane. Determine the numberof ways of selecting two distinct unit cubes from a 3 × 3 × 1 block (the order is irrelevant) with theproperty that the line joining the centers of the two cubes makes a 45◦ angle with the horizontal plane.

Answer: 60 There are 6 such slices, and each slice gives 10 valid pairs (with no overcounting).Therefore, there are 60 such pairs.

2. [3] Let S = {1, 2, . . . , 2008}. For any nonempty subset A ⊂ S, define m(A) to be the median of A(when A has an even number of elements, m(A) is the average of the middle two elements). Determinethe average of m(A), when A is taken over all nonempty subsets of S.

Answer: 20092 For any subset A, we can define the “reflected subset” A′ = {i | 2009− i ∈ A}. Then

m(A) = 2009 −m(A′). Note that as A is taken over all nonempty subsets of S, A′ goes through allthe nonempty subsets of S as well. Thus, the average of m(A) is equal to the average of m(A)+m(A′)

2 ,which is the constant 2009

2 .

Remark: : This argument is very analogous to the famous argument that Gauss used to sum the series1 + 2 + · · ·+ 100.

3. [4] Farmer John has 5 cows, 4 pigs, and 7 horses. How many ways can he pair up the animals so thatevery pair consists of animals of different species? Assume that all animals are distinguishable fromeach other. (Please write your answer as an integer, without any incomplete computations.)

Answer: 100800 Since there are 9 cow and pigs combined and 7 horses, there must be a pair with 1cow and 1 pig, and all the other pairs must contain a horse. There are 4×5 ways of selecting the cow-pigpair, and 7! ways to select the partners for the horses. It follows that the answer is 4×5×7! = 100800.

4. [4] Kermit the frog enjoys hopping around the infinite square grid in his backyard. It takes him 1 Jouleof energy to hop one step north or one step south, and 1 Joule of energy to hop one step east or onestep west. He wakes up one morning on the grid with 100 Joules of energy, and hops till he falls asleepwith 0 energy. How many different places could he have gone to sleep?

Answer: 10201

It is easy to see that the coordinates of the frog’s final position must have the same parity. Supposethat the frog went to sleep at (x, y). Then, we have that −100 ≤ y ≤ 100 and |x| ≤ 100 − |y|, so xcan take on the values −100 + |y|,−98 + |y|, . . . , 100− |y|. There are 101− |y| such values, so the totalnumber of such locations is

100∑y=−100

101− |y| = 201 · 101− 2 · 100(100 + 1)2

= 1012 = 10201.

5. [5] Let S be the smallest subset of the integers with the property that 0 ∈ S and for any x ∈ S, wehave 3x ∈ S and 3x + 1 ∈ S. Determine the number of non-negative integers in S less than 2008.

Answer: 128 Write the elements of S in their ternary expansion (i.e. base 3). Then the secondcondition translates into, if d1d2 · · · dk ∈ S, then d1d2 · · · dk0 and d1d2 · · · dk1 are also in S. It followsthat S is the set of nonnegative integers whose tertiary representation contains only the digits 0 and1. Since 2 · 36 < 2008 < 37, there are 27 = 128 such elements less than 2008. Therefore, there are 128such non-negative elements.

1

6. [5] A Sudoku matrix is defined as a 9×9 array with entries from {1, 2, . . . , 9}and with the constraint that each row, each column, and each of the nine3× 3 boxes that tile the array contains each digit from 1 to 9 exactly once.A Sudoku matrix is chosen at random (so that every Sudoku matrix hasequal probability of being chosen). We know two of squares in this matrix,as shown. What is the probability that the square marked by ? containsthe digit 3?

12

?

Answer: 221 The third row must contain the digit 1, and it cannot appear in the leftmost three

squares. Therefore, the digit 1 must fall into one of the six squares shown below that are marked with ?.By symmetry, each starred square has an equal probability of containing the digit 1 (To see this moreprecisely, note that swapping columns 4 and 5 gives another Sudoku matrix, so the probability thatthe 4th column ? has the 1 is the same as the probability that the 5th column ? has the 1. Similarly,switching the 4-5-6th columns with the 7-8-9th columns yields another Sudoku matrix, which impliesin particular that the probability that the 4th column ? has the 1 is the same as the probability thatthe 7th column ? has the 1. The rest of the argument follows analogously.) Therefore, the probabilitythat the ? square contains 1 is 1/6.

12

? ? ? ? ? ?

Similarly the probability that the digit 2 appears at ? is also 1/6. By symmetry, the square ? has equalprobability of containing the digits 3, 4, 5, 6, 7, 8, 9. It follows that this probability is

(1− 1

6 −16

)/7 =

221 .

7. [6] Let P1, P2, . . . , P8 be 8 distinct points on a circle. Determine the number of possible configurationsmade by drawing a set of line segments connecting pairs of these 8 points, such that: (1) each Pi isthe endpoint of at most one segment and (2) two no segments intersect. (The configuration with noedges drawn is allowed. An example of a valid configuration is shown below.)

P1

P2

P3

P4

P5

P6

P7

P8

Answer: 323 Let f(n) denote the number of valid configurations when there are n points on thecircle. Let P be one of the points. If P is not the end point of an edge, then there are f(n− 1) waysto connect the remaining n− 1 points. If P belongs to an edge that separates the circle so that thereare k points on one side and n− k − 2 points on the other side, then there are f(k)f(n− k − 2) waysof finishing the configuration. Thus, f(n) satisfies the recurrence relation

f(n) = f(n− 1) + f(0)f(n− 2) + f(1)f(n− 3) + f(2)f(n− 4) + · · ·+ f(n− 2)f(0), n ≥ 2.

The initial conditions are f(0) = f(1) = 1. Using the recursion, we find that f(2) = 2, f(3) = 4, f(4) =9, f(5) = 21, f(6) = 51, f(7) = 127, f(8) = 323.

Remark: These numbers are known as the Motzkin numbers. This is sequence A001006 in the On-LineEncyclopedia of Integer Sequences (http://www.research.att.com/~njas/sequences/A001006). In

2

Richard Stanley’s Enumerative Combinatorics Volume 2, one can find 13 different interpretations ofMotzkin numbers in exercise 6.38.

8. [6] Determine the number of ways to select a sequence of 8 sets A1, A2, . . . , A8, such that each is asubset (possibly empty) of {1, 2}, and Am contains An if m divides n.

Answer: 2025 Consider an arbitrary x ∈ {1, 2}, and let us consider the number of ways for x to bein some of the sets so that the constraints are satisfied. We divide into a few cases:

• Case: x /∈ A1. Then x cannot be in any of the sets. So there is one possibility.

• Case: x ∈ A1 but x /∈ A2. Then the only other sets that x could be in are A3, A5, A7, and xcould be in some collection of them. There are 8 possibilities in this case.

• Case: x ∈ A2. Then x ∈ A1 automatically. There are 4 independent choices to be make here:(1) whether x ∈ A5; (2) whether x ∈ A7; (3) whether x ∈ A3, and if yes, whether x ∈ A6; (4)whether x ∈ A4, and if yes, whether x ∈ A8. There are 2× 2× 3× 3 = 36 choices here.

Therefore, there are 1 + 8 + 36 = 45 ways to place x into some of the sets. Since the choices for x = 1and x = 2 are made independently, we see that the total number of possibilities is 452 = 2025.

Remark: The solution could be guided by the following diagram. Set A is above B and connected to Bif and only if A ⊂ B. Such diagrams are known as Hasse diagrams, which are used to depict partiallyordered sets.

A8

A4 A6

||||

||||

A5

PPPPPPPPPPPPPPP A7

BBBB

BBBB

A2 A3

||||

||||

A1

9. [7] On an infinite chessboard (whose squares are labeled by (x, y), where x and y range over all integers),a king is placed at (0, 0). On each turn, it has probability of 0.1 of moving to each of the four edge-neighboring squares, and a probability of 0.05 of moving to each of the four diagonally-neighboringsquares, and a probability of 0.4 of not moving. After 2008 turns, determine the probability that theking is on a square with both coordinates even. An exact answer is required.

Answer:14

+3

4 · 52008Since only the parity of the coordinates are relevant, it is equivalent to

consider a situation where the king moves (1, 0) with probability 0.2, moves (0, 1) with probability 0.2,moves (1, 1) with probability 0.2, and stays put with probability 0.4. This can be analyzed using thegenerating function

f(x, y) = (0.4 + 2× 0.1x + 2× 0.1y + 4× 0.05xy)2008 =(2 + x + y + xy)2008

52008.

We wish to find the sum of the coefficients of the terms xayb, where both a and b are even. Thisis simply equal to 1

4 (f(1, 1) + f(1,−1) + f(−1, 1) + f(−1,−1)). We have f(1, 1) = 1 and f(1,−1) =f(−1, 1) = f(−1,−1) = 1/52008. Therefore, the answer is

14

(f(1, 1) + f(1,−1) + f(−1, 1) + f(−1,−1)) =14

(1 +

352008

)=

14

+3

4 · 52008.

3

10. [7] Determine the number of 8-tuples of nonnegative integers (a1, a2, a3, a4, b1, b2, b3, b4) satisfying 0 ≤ak ≤ k, for each k = 1, 2, 3, 4, and a1 + a2 + a3 + a4 + 2b1 + 3b2 + 4b3 + 5b4 = 19.

Answer: 1540 For each k = 1, 2, 3, 4, note that set of pairs (ak, bk) with 0 ≤ ak ≤ k mapsbijectively to the set of nonnegative integers through the map (ak, bk) 7→ ak +(k +1)bk, as ak is simplythe remainder of ak + (k + 1)bk upon division by k + 1. By letting xk = ak + (k + 1)bk, we see thatthe problem is equivalent to finding the number of quadruples of nonnegative integers (x1, x2, x3, x4)such that x1 + x2 + x3 + x4 = 19. This is the same as finding the number of quadruples of positiveintegers (x1 + 1, x2 + 1, x3 + 1, x4 + 1) such that x1 + x2 + x3 + x4 = 23. By a standard “dots andbars” argument, we see that the answer is

(223

)= 1540.

A generating functions solution is also available. It’s not hard to see that the answer is the coefficientof x19 in

(1 + x)(1 + x + x2

) (1 + x + x2 + x3

) (1 + x + x2 + x3 + x4

)(1 + x2 + x4 + · · ·

) (1 + x3 + x6 + · · ·

) (1 + x4 + x8 + · · ·

) (1 + x5 + x10 + · · ·

)=(

1− x2

1− x

)(1− x3

1− x

)(1− x4

1− x

)(1− x5

1− x

)(1

1− x2

)(1

1− x3

)(1

1− x4

)(1

1− x5

)=

1(1− x)4

= (1− x)−4.

Using binomial theorem, we find that the coefficient of x19 in (1− x)−4 is (−1)19(−4

19

)=(2219

)= 1540.

4

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Individual Round: General Test, Part 1

1. [2] Let ABCD be a unit square (that is, the labels A, B,C, D appear in that order around the square).Let X be a point outside of the square such that the distance from X to AC is equal to the distancefrom X to BD, and also that AX =

√2

2 . Determine the value of CX2.

2. [3] Find the smallest positive integer n such that 107n has the same last two digits as n.

3. [3] There are 5 dogs, 4 cats, and 7 bowls of milk at an animal gathering. Dogs and cats are distin-guishable, but all bowls of milk are the same. In how many ways can every dog and cat be paired witheither a member of the other species or a bowl of milk such that all the bowls of milk are taken?

4. [3] Positive real numbers x, y satisfy the equations x2 + y2 = 1 and x4 + y4 = 1718 . Find xy.

5. [4] The function f satisfies

f(x) + f(2x + y) + 5xy = f(3x− y) + 2x2 + 1

for all real numbers x, y. Determine the value of f(10).

6. [4] In a triangle ABC, take point D on BC such that DB = 14, DA = 13, DC = 4, and the circumcircleof ADB is congruent to the circumcircle of ADC. What is the area of triangle ABC?

7. [5] The equation x3 − 9x2 + 8x + 2 = 0 has three real roots p, q, r. Find 1p2 + 1

q2 + 1r2 .

8. [5] Let S be the smallest subset of the integers with the property that 0 ∈ S and for any x ∈ S, wehave 3x ∈ S and 3x + 1 ∈ S. Determine the number of positive integers in S less than 2008.

9. [5] A Sudoku matrix is defined as a 9×9 array with entries from {1, 2, . . . , 9}and with the constraint that each row, each column, and each of the nine3× 3 boxes that tile the array contains each digit from 1 to 9 exactly once.A Sudoku matrix is chosen at random (so that every Sudoku matrix hasequal probability of being chosen). We know two of squares in this matrix,as shown. What is the probability that the square marked by ? containsthe digit 3?

12

?

10. [6] Let ABC be an equilateral triangle with side length 2, and let Γ be a circle with radius 12 centered at

the center of the equilateral triangle. Determine the length of the shortest path that starts somewhereon Γ, visits all three sides of ABC, and ends somewhere on Γ (not necessarily at the starting point).Express your answer in the form of

√p − q, where p and q are rational numbers written as reduced

fractions.

1

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Individual Round: General Test, Part 1

1. [2] Let ABCD be a unit square (that is, the labels A, B,C, D appear in that order around the square).Let X be a point outside of the square such that the distance from X to AC is equal to the distancefrom X to BD, and also that AX =

√2

2 . Determine the value of CX2.

Answer: 52

A

B C

D

M NX

Since X is equidistant from AC and BD, it must lie on either the perpendicular bisector of AB or theperpendicular bisector of AD. It turns that the two cases yield the same answer, so we will just assumethe first case. Let M be the midpoint of AB and N the midpoint of CD. Then, XM is perpendicularto AB, so XM = 1

2 and thus XN = 32 , NC = 1

2 . By the Pythagorean Theorem we find XC =√

102

and the answer follows.

2. [3] Find the smallest positive integer n such that 107n has the same last two digits as n.

Answer: 50 The two numbers have the same last two digits if and only if 100 divides their difference106n, which happens if and only if 50 divides n.

3. [3] There are 5 dogs, 4 cats, and 7 bowls of milk at an animal gathering. Dogs and cats are distin-guishable, but all bowls of milk are the same. In how many ways can every dog and cat be paired witheither a member of the other species or a bowl of milk such that all the bowls of milk are taken?

Answer: 20 Since there are 9 dogs and cats combined and 7 bowls of milk, there can only be onedog-cat pair, and all the other pairs must contain a bowl of milk. There are 4× 5 ways of selecting thedog-cat pair, and only one way of picking the other pairs, since the bowls of milk are indistinguishable,so the answer is 4× 5 = 20.

4. [3] Positive real numbers x, y satisfy the equations x2 + y2 = 1 and x4 + y4 = 1718 . Find xy.

Answer: 16 Same as Algebra Test problem 1.

5. [4] The function f satisfies

f(x) + f(2x + y) + 5xy = f(3x− y) + 2x2 + 1

for all real numbers x, y. Determine the value of f(10).

Answer: −49 Same as Algebra Test problem 4.

6. [4] In a triangle ABC, take point D on BC such that DB = 14, DA = 13, DC = 4, and the circumcircleof ADB is congruent to the circumcircle of ADC. What is the area of triangle ABC?

Answer: 108 Same as Geometry Test problem 4.

1

7. [5] The equation x3 − 9x2 + 8x + 2 = 0 has three real roots p, q, r. Find 1p2 + 1

q2 + 1r2 .

Answer: 25 From Vieta’s relations, we have p + q + r = 9, pq + qr + pr = 8 and pqr = −2. So

1p2

+1q2

+1r2

=(pq + qr + rp)2 − 2(p + q + r)(pqr)

(pqr)2=

82 − 2 · 9 · (−2)(−2)2

= 25 .

8. [5] Let S be the smallest subset of the integers with the property that 0 ∈ S and for any x ∈ S, wehave 3x ∈ S and 3x + 1 ∈ S. Determine the number of positive integers in S less than 2008.

Answer: 127 Same as Combinatorics Problem 5.

9. [5] A Sudoku matrix is defined as a 9×9 array with entries from {1, 2, . . . , 9}and with the constraint that each row, each column, and each of the nine3× 3 boxes that tile the array contains each digit from 1 to 9 exactly once.A Sudoku matrix is chosen at random (so that every Sudoku matrix hasequal probability of being chosen). We know two of squares in this matrix,as shown. What is the probability that the square marked by ? containsthe digit 3?

12

?

Answer: 221 Same as Combinatorics Test problem 6.

10. [6] Let ABC be an equilateral triangle with side length 2, and let Γ be a circle with radius 12 centered at

the center of the equilateral triangle. Determine the length of the shortest path that starts somewhereon Γ, visits all three sides of ABC, and ends somewhere on Γ (not necessarily at the starting point).Express your answer in the form of

√p − q, where p and q are rational numbers written as reduced

fractions.

Answer:

√283− 1 Same as Geometry Test problem 8.

2

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Individual Round: General Test, Part 2

1. [2] Four students from Harvard, one of them named Jack, and five students from MIT, one of themnamed Jill, are going to see a Boston Celtics game. However, they found out that only 5 tickets remain,so 4 of them must go back. Suppose that at least one student from each school must go see the game,and at least one of Jack and Jill must go see the game, how many ways are there of choosing which 5people can see the game?

2. [2] Let ABC be an equilateral triangle. Let Ω be a circle inscribed in ABC and let ω be a circletangent externally to Ω as well as to sides AB and AC. Determine the ratio of the radius of Ω to theradius of ω.

3. [3] A 3× 3× 3 cube composed of 27 unit cubes rests on a horizontal plane. Determine the number ofways of selecting two distinct unit cubes (order is irrelevant) from a 3× 3× 1 block with the propertythat the line joining the centers of the two cubes makes a 45◦ angle with the horizontal plane.

4. [3] Suppose that a, b, c, d are real numbers satisfying a ≥ b ≥ c ≥ d ≥ 0, a2 + d2 = 1, b2 + c2 = 1, andac+ bd = 1/3. Find the value of ab− cd.

5. [4] Kermit the frog enjoys hopping around the infinite square grid in his backyard. It takes him 1 Jouleof energy to hop one step north or one step south, and 1 Joule of energy to hop one step east or onestep west. He wakes up one morning on the grid with 100 Joules of energy, and hops till he falls asleepwith 0 energy. How many different places could he have gone to sleep?

6. [4] Determine all real numbers a such that the inequality |x2 + 2ax+ 3a| ≤ 2 has exactly one solutionin x.

7. [5] A root of unity is a complex number that is a solution to zn = 1 for some positive integer n.Determine the number of roots of unity that are also roots of z2 +az+ b = 0 for some integers a and b.

8. [5] A piece of paper is folded in half. A second fold is made such that the angle marked below hasmeasure φ (0◦ < φ < 90◦), and a cut is made as shown below.

Á

When the piece of paper is unfolded, the resulting hole is a polygon. Let O be one of its vertices.Suppose that all the other vertices of the hole lie on a circle centered at O, and also that ∠XOY = 144◦,where X and Y are the the vertices of the hole adjacent to O. Find the value(s) of φ (in degrees).

9. [6] Let ABC be a triangle, and I its incenter. Let the incircle of ABC touch side BC at D, andlet lines BI and CI meet the circle with diameter AI at points P and Q, respectively. Given BI =6, CI = 5, DI = 3, determine the value of (DP/DQ)2.

10. [6] Determine the number of 8-tuples of nonnegative integers (a1, a2, a3, a4, b1, b2, b3, b4) satisfying 0 ≤ak ≤ k, for each k = 1, 2, 3, 4, and a1 + a2 + a3 + a4 + 2b1 + 3b2 + 4b3 + 5b4 = 19.

1

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Individual Round: General Test, Part 2

1. [2] Four students from Harvard, one of them named Jack, and five students from MIT, one of themnamed Jill, are going to see a Boston Celtics game. However, they found out that only 5 tickets remain,so 4 of them must go back. Suppose that at least one student from each school must go see the game,and at least one of Jack and Jill must go see the game, how many ways are there of choosing which 5people can see the game?

Answer: 104 Let us count the number of way of distributing the tickets so that one of the conditionsis violated. There is 1 way to give all the tickets to MIT students, and

(75

)ways to give all the tickets to

the 7 students other than Jack and Jill. Therefore, the total number of valid ways is(95

)−1−

(75

)= 104.

2. [2] Let ABC be an equilateral triangle. Let Ω be a circle inscribed in ABC and let ω be a circletangent externally to Ω as well as to sides AB and AC. Determine the ratio of the radius of Ω to theradius of ω.

Answer: 3 Same as Geometry Test problem 2.

3. [3] A 3× 3× 3 cube composed of 27 unit cubes rests on a horizontal plane. Determine the number ofways of selecting two distinct unit cubes (order is irrelevant) from a 3× 3× 1 block with the propertythat the line joining the centers of the two cubes makes a 45◦ angle with the horizontal plane.

Answer: 60 Same as Combinatorics Test problem 1.

4. [3] Suppose that a, b, c, d are real numbers satisfying a ≥ b ≥ c ≥ d ≥ 0, a2 + d2 = 1, b2 + c2 = 1, andac+ bd = 1/3. Find the value of ab− cd.

Answer: 2√

23 We have

(ab− cd)2 = (a2 + d2)(b2 + c2)− (ac+ bd)2 = (1)(1)−(

13

)2

=89.

Since a ≥ b ≥ c ≥ d ≥ 0, ab− cd ≥ 0, so ab− cd = 2√

23 .

Comment: Another way to solve this problem is to use the trigonometric substitutions a = sin θ,b = sinφ, c = cosφ, d = cos θ.

5. [4] Kermit the frog enjoys hopping around the infinite square grid in his backyard. It takes him 1 Jouleof energy to hop one step north or one step south, and 1 Joule of energy to hop one step east or onestep west. He wakes up one morning on the grid with 100 Joules of energy, and hops till he falls asleepwith 0 energy. How many different places could he have gone to sleep?

Answer: 10201 Same as Combinatorics Test problem 4.

6. [4] Determine all real numbers a such that the inequality |x2 + 2ax+ 3a| ≤ 2 has exactly one solutionin x.

Answer: 1, 2 Same as Algebra Test problem 3.

7. [5] A root of unity is a complex number that is a solution to zn = 1 for some positive integer n.Determine the number of roots of unity that are also roots of z2 +az+ b = 0 for some integers a and b.

Answer: 8 Same as Algebra Test problem 6.

8. [5] A piece of paper is folded in half. A second fold is made such that the angle marked below hasmeasure φ (0◦ < φ < 90◦), and a cut is made as shown below.

1

Á

When the piece of paper is unfolded, the resulting hole is a polygon. Let O be one of its vertices.Suppose that all the other vertices of the hole lie on a circle centered at O, and also that ∠XOY = 144◦,where X and Y are the the vertices of the hole adjacent to O. Find the value(s) of φ (in degrees).

Answer: 81◦ Same as Geometry Test problem 5.

9. [6] Let ABC be a triangle, and I its incenter. Let the incircle of ABC touch side BC at D, andlet lines BI and CI meet the circle with diameter AI at points P and Q, respectively. Given BI =6, CI = 5, DI = 3, determine the value of (DP/DQ)2.

Answer: 7564 Same as Geometry Test problem 9.

10. [6] Determine the number of 8-tuples of nonnegative integers (a1, a2, a3, a4, b1, b2, b3, b4) satisfying 0 ≤ak ≤ k, for each k = 1, 2, 3, 4, and a1 + a2 + a3 + a4 + 2b1 + 3b2 + 4b3 + 5b4 = 19.

Answer: 1540 Same as Combinatorics Test problem 10.

2

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Individual Round: Geometry Test

1. [3] How many different values can ∠ABC take, where A,B,C are distinct vertices of a cube?

2. [3] Let ABC be an equilateral triangle. Let Ω be its incircle (circle inscribed in the triangle) and let ωbe a circle tangent externally to Ω as well as to sides AB and AC. Determine the ratio of the radiusof Ω to the radius of ω.

3. [4] Let ABC be a triangle with ∠BAC = 90◦. A circle is tangent to the sides AB and AC at X andY respectively, such that the points on the circle diametrically opposite X and Y both lie on the sideBC. Given that AB = 6, find the area of the portion of the circle that lies outside the triangle.

A B

C

X

Y

4. [4] In a triangle ABC, take point D on BC such that DB = 14, DA = 13, DC = 4, and the circumcircleof ADB is congruent to the circumcircle of ADC. What is the area of triangle ABC?

5. [5] A piece of paper is folded in half. A second fold is made at an angle φ (0◦ < φ < 90◦) to the first,and a cut is made as shown below.

Á

When the piece of paper is unfolded, the resulting hole is a polygon. Let O be one of its vertices.Suppose that all the other vertices of the hole lie on a circle centered at O, and also that ∠XOY = 144◦,where X and Y are the the vertices of the hole adjacent to O. Find the value(s) of φ (in degrees).

6. [5] Let ABC be a triangle with ∠A = 45◦. Let P be a point on side BC with PB = 3 and PC = 5.Let O be the circumcenter of ABC. Determine the length OP .

7. [6] Let C1 and C2 be externally tangent circles with radius 2 and 3, respectively. Let C3 be a circleinternally tangent to both C1 and C2 at points A and B, respectively. The tangents to C3 at A and Bmeet at T , and TA = 4. Determine the radius of C3.

8. [6] Let ABC be an equilateral triangle with side length 2, and let Γ be a circle with radius 12 centered at

the center of the equilateral triangle. Determine the length of the shortest path that starts somewhereon Γ, visits all three sides of ABC, and ends somewhere on Γ (not necessarily at the starting point).Express your answer in the form of

√p − q, where p and q are rational numbers written as reduced

fractions.

9. [7] Let ABC be a triangle, and I its incenter. Let the incircle of ABC touch side BC at D, andlet lines BI and CI meet the circle with diameter AI at points P and Q, respectively. Given BI =6, CI = 5, DI = 3, determine the value of (DP/DQ)2.

10. [7] Let ABC be a triangle with BC = 2007, CA = 2008, AB = 2009. Let ω be an excircle of ABCthat touches the line segment BC at D, and touches extensions of lines AC and AB at E and F ,respectively (so that C lies on segment AE and B lies on segment AF ). Let O be the center of ω. Let` be the line through O perpendicular to AD. Let ` meet line EF at G. Compute the length DG.

1

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Individual Round: Geometry Test

1. [3] How many different values can ∠ABC take, where A,B,C are distinct vertices of a cube?

Answer: 5 . In a unit cube, there are 3 types of triangles, with side lengths (1, 1,√

2), (1,√

2,√

3)and (

√2,√

2,√

2). Together they generate 5 different angle values.

2. [3] Let ABC be an equilateral triangle. Let Ω be its incircle (circle inscribed in the triangle) and let ωbe a circle tangent externally to Ω as well as to sides AB and AC. Determine the ratio of the radiusof Ω to the radius of ω.

Answer: 3 Label the diagram as shown below, where Ω and ω also denote the center of thecorresponding circles. Note that AM is a median and Ω is the centroid of the equilateral triangle.So AM = 3MΩ. Since MΩ = NΩ, it follows that AM/AN = 3, and triangle ABC is the image oftriangle AB′C ′ after a scaling by a factor of 3, and so the two incircles must also be related by a scalefactor of 3.

A

B C

B′ C ′

Ω

ω

M

N

3. [4] Let ABC be a triangle with ∠BAC = 90◦. A circle is tangent to the sides AB and AC at X andY respectively, such that the points on the circle diametrically opposite X and Y both lie on the sideBC. Given that AB = 6, find the area of the portion of the circle that lies outside the triangle.

A B

C

X

Y

Answer: π − 2 Let O be the center of the circle, and r its radius, and let X ′ and Y ′ be the pointsdiametrically opposite X and Y , respectively. We have OX ′ = OY ′ = r, and ∠X ′OY ′ = 90◦. Sincetriangles X ′OY ′ and BAC are similar, we see that AB = AC. Let X ′′ be the projection of Y ′ ontoAB. Since X ′′BY ′ is similar to ABC, and X ′′Y ′ = r, we have X ′′B = r. It follows that AB = 3r, sor = 2.

A B

C

X

Y

X ′

Y ′

X ′′

O

1

Then, the desired area is the area of the quarter circle minus that of the triangle X ′OY ′. And theanswer is 1

4πr2 − 1

2r2 = π − 2.

4. [4] In a triangle ABC, take point D on BC such that DB = 14, DA = 13, DC = 4, and the circumcircleof ADB is congruent to the circumcircle of ADC. What is the area of triangle ABC?

Answer: 108

A

B CDM

The fact that the two circumcircles are congruent means that the chord AD must subtend the sameangle in both circles. That is, ∠ABC = ∠ACB, so ABC is isosceles. Drop the perpendicular M fromA to BC; we know MC = 9 and so MD = 5 and by Pythagoras on AMD, AM = 12. Therefore, thearea of ABC is 1

2 (AM)(BC) = 12 (12)(18) = 108.

5. [5] A piece of paper is folded in half. A second fold is made such that the angle marked below hasmeasure φ (0◦ < φ < 90◦), and a cut is made as shown below.

Á

When the piece of paper is unfolded, the resulting hole is a polygon. Let O be one of its vertices.Suppose that all the other vertices of the hole lie on a circle centered at O, and also that ∠XOY = 144◦,where X and Y are the the vertices of the hole adjacent to O. Find the value(s) of φ (in degrees).

Answer: 81◦ Try actually folding a piece of paper. We see that the cut out area is a kite, as shownbelow. The fold was made on AC, and then BE and DE. Since DC was folded onto DA, we have∠ADE = ∠CDE.

A

B

C

D

E

Either A or C is the center of the circle. If it’s A, then ∠BAD = 144◦, so ∠CAD = 72◦. UsingCA = DA, we see that ∠ACD = ∠ADC = 54◦. So ∠EDA = 27◦, and thus φ = 72◦ + 27◦ = 99◦,which is inadmissible, as φ < 90◦.

So C is the center of the circle. Then, ∠CAD = ∠CDA = 54◦, ∠ADE = 27◦, and φ = 54◦+27◦ = 81◦.

6. [5] Let ABC be a triangle with ∠A = 45◦. Let P be a point on side BC with PB = 3 and PC = 5.Let O be the circumcenter of ABC. Determine the length OP .

2

Answer:√

17 Using extended Sine law, we find the circumradius of ABC to be R = BC2 sin A = 4

√2.

By considering the power of point P , we find that R2 −OP 2 = PB · PC = 15. So OP =√R2 − 15 =√

16 · 2− 15 =√

17.

7. [6] Let C1 and C2 be externally tangent circles with radius 2 and 3, respectively. Let C3 be a circleinternally tangent to both C1 and C2 at points A and B, respectively. The tangents to C3 at A and Bmeet at T , and TA = 4. Determine the radius of C3.

Answer: 8 Let D be the point of tangency between C1 and C2. We see that T is the radical centerof the three circles, and so it must lie on the radical axis of C1 and C2, which happens to be theircommon tangent TD. So TD = 4.

T

A

B

D

C1

C2

C3

We havetan

∠ATD2

=2TD

=12, and tan

∠BTD2

=3TD

=34.

Thus, the radius of C3 equals to

TA tan∠ATB

2= 4 tan

(∠ATD + ∠BTD

2

)= 4 ·

tan ∠ATD2 + tan ∠BTD

2

1− tan ∠ATD2 tan ∠BTD

2

= 4 ·12 + 3

4

1− 12 ·

34

= 8.

8. [6] Let ABC be an equilateral triangle with side length 2, and let Γ be a circle with radius 12 centered at

the center of the equilateral triangle. Determine the length of the shortest path that starts somewhereon Γ, visits all three sides of ABC, and ends somewhere on Γ (not necessarily at the starting point).Express your answer in the form of

√p − q, where p and q are rational numbers written as reduced

fractions.

Answer:√

283 − 1 Suppose that the path visits sides AB,BC,CA in this order. Construct points

A′, B′, C ′ so that C ′ is the reflection of C across AB, A′ is the reflection of A across BC ′, and B′ isthe reflection of B across A′C ′. Finally, let Γ′ be the circle with radius 1

2 centered at the center ofA′B′C ′. Note that Γ′ is the image of Γ after the three reflections: AB,BC ′, C ′A′.

3

A

B CA′

B′ C ′

When the path hits AB, let us reflect the rest of the path across AB and follow this reflected path.When we hit BC ′, let us reflect the rest of the path across BC ′, and follow the new path. And whenwe hit A′C ′, reflect the rest of the path across A′C ′ and follow the new path. We must eventually endup at Γ′.

It is easy to see that the shortest path connecting some point on Γ to some point on Γ′ lies on theline connecting the centers of the two circles. We can easily find the distance between the two centers

to be

√32 +

(1√3

)2

=√

283 . Therefore, the length of the shortest path connecting Γ to Γ′ has length√

283 −1. By reflecting this path three times back into ABC, we get a path that satisfies our conditions.

9. [7] Let ABC be a triangle, and I its incenter. Let the incircle of ABC touch side BC at D, andlet lines BI and CI meet the circle with diameter AI at points P and Q, respectively. Given BI =6, CI = 5, DI = 3, determine the value of (DP/DQ)2.

Answer: 7564

A

B CD

EF

PQ

I

Let the incircle touch sides AC and AB at E and F respectively. Note that E and F both lie onthe circle with diameter AI since ∠AEI = ∠AFI = 90◦. The key observation is that D,E, P arecollinear. To prove this, suppose that P lies outside the triangle (the other case is analogous), then∠PEA = ∠PIA = ∠IBA+∠IAB = 1

2 (∠B+∠A) = 90◦− 12∠C = ∠DEC, which implies that D,E, P

are collinear. Similarly D,F,Q are collinear. Then, by Power of a Point, DE · DP = DF · DQ. SoDP/DQ = DF/DE.

Now we compute DF/DE. Note that DF = 2DB sin ∠DBI = 2√

62 − 32(

36

)= 3√

3, and DE =2DC sin ∠DCI = 2

√52 − 32

(35

)= 24

5 . Therefore, DF/DE = 5√

38 .

10. [7] Let ABC be a triangle with BC = 2007, CA = 2008, AB = 2009. Let ω be an excircle of ABCthat touches the line segment BC at D, and touches extensions of lines AC and AB at E and F ,respectively (so that C lies on segment AE and B lies on segment AF ). Let O be the center of ω. Let` be the line through O perpendicular to AD. Let ` meet line EF at G. Compute the length DG.

Answer: 2014024 Let line AD meet ω again at H. Since AF and AE are tangents to ω and ADHis a secant, we see that DEHF is a harmonic quadrilateral. This implies that the pole of AD withrespect to ω lies on EF . Since ` ⊥ AD, the pole of AD lies on `. It follows that the pole of AD is G.

4

A

B CD

E

F

G

H

O

Thus, G must lie on the tangent to ω at D, so C,D,B,G are collinear. Furthermore, since the pencilof lines (AE,AF ;AD,AG) is harmonic, by intersecting it with the line BC, we see that (C,B;D,G)is harmonic as well. This means that

BD

DC· CGGB

= −1.

(where the lengths are directed.) The semiperimeter of ABC is s = 12 (2007 + 2008 + 2009) = 3012. So

BD = s− 2009 = 1003 and CD = s− 2008 = 1004. Let x = DG, then the above equations gives

10031004

· x+ 1004x− 1003

= 1.

Solving gives x = 2014024.

Remark: If you are interested to learn about projective geometry, check out the last chapter of GeometryRevisited by Coxeter and Greitzer or Geometric Transformations III by Yaglom.

5

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Guts Round

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

1. [5] Determine all pairs (a, b) of real numbers such that 10, a, b, ab is an arithmetic progression.

2. [5] Given right triangle ABC, with AB = 4, BC = 3, and CA = 5. Circle ω passes through A and istangent to BC at C. What is the radius of ω?

3. [5] How many ways can you color the squares of a 2× 2008 grid in 3 colors such that no two squaresof the same color share an edge?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

4. [6] Find the real solution(s) to the equation (x+ y)2 = (x+ 1)(y − 1).

5. [6] A Vandal and a Moderator are editing a Wikipedia article. The article originally is error-free.Each day, the Vandal introduces one new error into the Wikipedia article. At the end of the day, themoderator checks the article and has a 2/3 chance of catching each individual error still in the article.After 3 days, what is the probability that the article is error-free?

6. [6] Determine the number of non-degenerate rectangles whose edges lie completely on the grid lines ofthe following figure.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

7. [6] Given that x+ sin y = 2008 and x+ 2008 cos y = 2007, where 0 ≤ y ≤ π/2, find the value of x+ y.

8. [6] Trodgor the dragon is burning down a village consisting of 90 cottages. At time t = 0 an angrypeasant arises from each cottage, and every 8 minutes (480 seconds) thereafter another angry peasantspontaneously generates from each non-burned cottage. It takes Trodgor 5 seconds to either burn apeasant or to burn a cottage, but Trodgor cannot begin burning cottages until all the peasants aroundhim have been burned. How many seconds does it take Trodgor to burn down the entire village?

9. [6] Consider a circular cone with vertex V , and let ABC be a triangle inscribed in the base of the cone,such that AB is a diameter and AC = BC. Let L be a point on BV such that the volume of the coneis 4 times the volume of the tetrahedron ABCL. Find the value of BL/LV .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

10. [7] Find the number of subsets S of {1, 2, . . . 63} the sum of whose elements is 2008.

11. [7] Let f(r) =∑2008

j=21jr = 1

2r + 13r + · · ·+ 1

2008r . Find∑∞

k=2 f(k).

12. [7] Suppose we have an (infinite) cone C with apex A and a plane π. The intersection of π and C isan ellipse E with major axis BC, such that B is closer to A than C, and BC = 4, AC = 5, AB = 3.Suppose we inscribe a sphere in each part of C cut up by E with both spheres tangent to E . What isthe ratio of the radii of the spheres (smaller to larger)?

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

13. [8] Let P (x) be a polynomial with degree 2008 and leading coefficient 1 such that

P (0) = 2007, P (1) = 2006, P (2) = 2005, . . . , P (2007) = 0.

Determine the value of P (2008). You may use factorials in your answer.

14. [8] Evaluate the infinite sum∑∞

n=1n

n4+4 .

15. [8] In a game show, Bob is faced with 7 doors, 2 of which hide prizes. After he chooses a door, thehost opens three other doors, of which one is hiding a prize. Bob chooses to switch to another door.What is the probability that his new door is hiding a prize?

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

16. [9] Point A lies at (0, 4) and point B lies at (3, 8). Find the x-coordinate of the point X on the x-axismaximizing ∠AXB.

17. [9] Solve the equation √√√√x+

√4x+

√16x+

√. . .+

√42008x+ 3−√x = 1.

Express your answer as a reduced fraction with the numerator and denominator written in their primefactorization.

18. [9] Let ABC be a right triangle with ∠A = 90◦. Let D be the midpoint of AB and let E be a pointon segment AC such that AD = AE. Let BE meet CD at F . If ∠BFC = 135◦, determine BC/AB.

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2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

19. [10] Let ABCD be a regular tetrahedron, and let O be the centroid of triangle BCD. Consider thepoint P on AO such that P minimizes PA+ 2(PB + PC + PD). Find sin ∠PBO.

20. [10] For how many ordered triples (a, b, c) of positive integers are the equations abc+ 9 = ab+ bc+ caand a+ b+ c = 10 satisfied?

21. [10] Let ABC be a triangle with AB = 5, BC = 4 and AC = 3. Let P and Q be squares inside ABCwith disjoint interiors such that they both have one side lying on AB. Also, the two squares each havean edge lying on a common line perpendicular to AB, and P has one vertex on AC and Q has onevertex on BC. Determine the minimum value of the sum of the areas of the two squares.

A B

C

PQ

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

22. [10] For a positive integer n, let θ(n) denote the number of integers 0 ≤ x < 2010 such that x2 − n is

divisible by 2010. Determine the remainder when2009∑n=0

n · θ(n) is divided by 2010.

23. [10] Two mathematicians, Kelly and Jason, play a cooperative game. The computer selects some secretpositive integer n < 60 (both Kelly and Jason know that n < 60, but that they don’t know what thevalue of n is). The computer tells Kelly the unit digit of n, and it tells Jason the number of divisorsof n. Then, Kelly and Jason have the following dialogue:

Kelly: I don’t know what n is, and I’m sure that you don’t know either. However, I know that n isdivisible by at least two different primes.

Jason: Oh, then I know what the value of n is.

Kelly: Now I also know what n is.

Assuming that both Kelly and Jason speak truthfully and to the best of their knowledge, what are allthe possible values of n?

24. [10] Suppose that ABC is an isosceles triangle with AB = AC. Let P be the point on side AC so thatAP = 2CP . Given that BP = 1, determine the maximum possible area of ABC.

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3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

25. [12] Alice and the Cheshire Cat play a game. At each step, Alice either (1) gives the cat a penny,which causes the cat to change the number of (magic) beans that Alice has from n to 5n or (2) givesthe cat a nickel, which causes the cat to give Alice another bean. Alice wins (and the cat disappears)as soon as the number of beans Alice has is greater than 2008 and has last two digits 42. What is theminimum number of cents Alice can spend to win the game, assuming she starts with 0 beans?

26. [12] Let P be a parabola, and let V1 and F1 be its vertex and focus, respectively. Let A and B bepoints on P so that ∠AV1B = 90◦. Let Q be the locus of the midpoint of AB. It turns out that Qis also a parabola, and let V2 and F2 denote its vertex and focus, respectively. Determine the ratioF1F2/V1V2.

27. [12] Cyclic pentagon ABCDE has a right angle ∠ABC = 90◦ and side lengths AB = 15 and BC = 20.Supposing that AB = DE = EA, find CD.

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

28. [15] Let P be a polyhedron where every face is a regular polygon, and every edge has length 1. Eachvertex of P is incident to two regular hexagons and one square. Choose a vertex V of the polyhedron.Find the volume of the set of all points contained in P that are closer to V than to any other vertex.

29. [15] Let (x, y) be a pair of real numbers satisfying

56x+ 33y =−y

x2 + y2, and 33x− 56y =

x

x2 + y2.

Determine the value of |x|+ |y|.

30. [15] Triangle ABC obeys AB = 2AC and ∠BAC = 120◦. Points P and Q lie on segment BC suchthat

AB2 +BC · CP = BC2

3AC2 + 2BC · CQ = BC2

Find ∠PAQ in degrees.

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

31. [18] Let C be the hyperbola y2 − x2 = 1. Given a point P0 on the x-axis, we construct a sequenceof points (Pn) on the x-axis in the following manner: let `n be the line with slope 1 passing passingthrough Pn, then Pn+1 is the orthogonal projection of the point of intersection of `n and C onto thex-axis. (If Pn = 0, then the sequence simply terminates.)

Let N be the number of starting positions P0 on the x-axis such that P0 = P2008. Determine theremainder of N when divided by 2008.

32. [18] Cyclic pentagon ABCDE has side lengths AB = BC = 5, CD = DE = 12, and AE = 14.Determine the radius of its circumcircle.

33. [18] Let a, b, c be nonzero real numbers such that a+ b+ c = 0 and a3 + b3 + c3 = a5 + b5 + c5. Findthe value of a2 + b2 + c2.

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4

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

34. Who Wants to Be a Millionaire. In 2000, the Clay Mathematics Institute named seven MillenniumPrize Problems, with each carrying a prize of $1 Million for its solution. Write down the name of ONEof the seven Clay Millennium Problems. If your submission is incorrect or misspelled, then yoursubmission is disqualified. If another team wrote down the same Millennium Problem as you, then youget 0 points, otherwise you get 20 points.

35. NUMB3RS. The RSA Factoring Challenge, which ended in 2007, challenged computational mathe-maticians to factor extremely large numbers that were the product of two prime numbers. The largestnumber successfully factored in this challenge was RSA-640, which has 193 decimal digits and carrieda prize of $20, 000. The next challenge number carried prize of $30, 000, and contains N decimal digits.Your task is to submit a guess for N . Only the team(s) that have the closest guess(es) receives points.If k teams all have the closest guesses, then each of them receives

⌈20k

⌉points.

36. The History Channel. Below is a list of famous mathematicians. Your task is to list a subset ofthem in the chronological order of their birth dates. Your submission should be a sequence of letters.If your sequence is not in the correct order, then you get 0 points. Otherwise your score will bemin{max{5(N − 4), 0}, 25}, where N is the number of letters in your sequence.

(A) Niels Abel (B) Arthur Cayley (C) Augustus De Morgan (D) Gustav Dirichlet (E) Leonhard Euler(F) Joseph Fourier (G) Évariste Galois (H) Carl Friedrich Gauss (I) Marie-Sophie Germain (J) JosephLouis Lagrange (K) Pierre-Simon Laplace (L) Henri Poincaré (N) Bernhard Riemann

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5

11th Annual Harvard-MIT Mathematics TournamentSaturday 23 February 2008

Guts Round

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

1. [5] Determine all pairs (a, b) of real numbers such that 10, a, b, ab is an arithmetic progression.

Answer: (4,−2), ( 52 ,−5) Since 10, a, b is an arithmetic progression, we have a = 1

2 (10 + b). Also,

we have a+ ab = 2b, and so a(1 + b) = 2b. Substituting the expression for a gives (10 + b)(1 + b) = 4b.Solving this quadratic equation gives the solutions b = −2 and b = −5. The corresponding values fora can be found by a = 1

2 (10 + b).

2. [5] Given right triangle ABC, with AB = 4, BC = 3, and CA = 5. Circle ω passes through A and istangent to BC at C. What is the radius of ω?

Answer: 258 Let O be the center of ω, and let M be the midpoint of AC. Since OA = OC,

OM ⊥ AC. Also, ∠OCM = ∠BAC, and so triangles ABC and CMO are similar. Then, CO/CM =AC/AB, from which we obtain that the radius of ω is CO = 25

8 .

3. [5] How many ways can you color the squares of a 2× 2008 grid in 3 colors such that no two squaresof the same color share an edge?

Answer: 2 · 32008 Denote the colors A,B,C. The left-most column can be colored in 6 ways. Foreach subsequent column, if the kth column is colored with AB, then the (k+ 1)th column can only becolored with one of BA,BC,CA. That is, if we have colored the first k columns, then there are 3 waysto color the (k + 1)th column. It follows that the number of ways of coloring the board is 6× 32007.

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

4. [6] Find the real solution(s) to the equation (x+ y)2 = (x+ 1)(y − 1).

Answer: (−1, 1) Set p = x + 1 and q = y − 1, then we get (p + q)2 = pq, which simplifies to

p2 + pq + q2 = 0. Then we have (p+ q2 )2 + 3q2

4 , and so p = q = 0. Thus (x, y) = (−1, 1).

5. [6] A Vandal and a Moderator are editing a Wikipedia article. The article originally is error-free.Each day, the Vandal introduces one new error into the Wikipedia article. At the end of the day, themoderator checks the article and has a 2/3 chance of catching each individual error still in the article.After 3 days, what is the probability that the article is error-free?

Answer: 416729 Consider the error that was introduced on day 1. The probability that the Moderator

misses this error on all three checks is 1/33, so the probability that this error gets removed is 1− 133 .

Similarly, the probability that the moderator misses the other two errors are 1− 132 and 1− 1

3 . So theprobability that the article is error-free is(

1− 133

)(1− 1

32

)(1− 1

3

)=

416729

.

6. [6] Determine the number of non-degenerate rectangles whose edges lie completely on the grid lines ofthe following figure.

1

Answer: 297 First, let us count the total number of rectangles in the grid without the hole in themiddle. There are

(72

)= 21 ways to choose the two vertical boundaries of the rectangle, and there are

21 ways to choose the two horizontal boundaries of the rectangles. This makes 212 = 441 rectangles.However, we must exclude those rectangles whose boundary passes through the center point. We cancount these rectangles as follows: the number of rectangles with the center of the grid lying in theinterior of its south edge is 3 × 3 × 3 = 27 (there are three choices for each of the three other edges);the number of rectangles whose south-west vertex coincides with the center is 3×3 = 9. Summing overall 4 orientations, we see that the total number of rectangles to exclude is 4(27 + 9) = 144. Therefore,the answer is 441− 144 = 297.

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

7. [6] Given that x+ sin y = 2008 and x+ 2008 cos y = 2007, where 0 ≤ y ≤ π/2, find the value of x+ y.

Answer: 2007 + π2 Subtracting the two equations gives sin y−2008 cos y = 1. But since 0 ≤ y ≤ π/2,

the maximum of sin y is 1 and the minimum of cos y is 0, so we must have sin y = 1, so y = π/2 andx+ y = 2007 + π

2 .

8. [6] Trodgor the dragon is burning down a village consisting of 90 cottages. At time t = 0 an angrypeasant arises from each cottage, and every 8 minutes (480 seconds) thereafter another angry peasantspontaneously generates from each non-burned cottage. It takes Trodgor 5 seconds to either burn apeasant or to burn a cottage, but Trodgor cannot begin burning cottages until all the peasants aroundhim have been burned. How many seconds does it take Trodgor to burn down the entire village?

Answer: 1920 We look at the number of cottages after each wave of peasants. Let An be thenumber of cottages remaining after 8n minutes. During each 8 minute interval, Trodgor burns a totalof 480/5 = 96 peasants and cottages. Trodgor first burns An peasants and spends the remainingtime burning 96 − An cottages. Therefore, as long as we do not reach negative cottages, we have therecurrence relation An+1 = An − (96−An), which is equivalent to An+1 = 2An − 96. Computing thefirst few terms of the series, we get that A1 = 84, A2 = 72, A3 = 48, and A4 = 0. Therefore, it takesTrodgor 32 minutes, which is 1920 seconds.

9. [6] Consider a circular cone with vertex V , and let ABC be a triangle inscribed in the base of the cone,such that AB is a diameter and AC = BC. Let L be a point on BV such that the volume of the coneis 4 times the volume of the tetrahedron ABCL. Find the value of BL/LV .

Answer: π4−π Let R be the radius of the base, H the height of the cone, h the height of the pyramid

and let BL/LV = x/y. Let [·] denote volume. Then [cone] = 13πR

2H and [ABCL] = 13πR

2h andh = x

x+yH. We are given that [cone] = 4[ABCL], so x/y = π4−π .

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

10. [7] Find the number of subsets S of {1, 2, . . . 63} the sum of whose elements is 2008.

Answer: 6 Note that 1 + 2 + · · ·+ 63 = 2016. So the problem is equivalent to finding the numberof subsets of {1, 2, · · · 63} whose sum of elements is 8. We can count this by hand: {8}, {1, 7}, {2, 6},{3, 5}, {1, 2, 5}, {1, 3, 4}.

2

11. [7] Let f(r) =∑2008j=2

1jr = 1

2r + 13r + · · ·+ 1

2008r . Find∑∞k=2 f(k).

Answer: 20072008 We change the order of summation:

∞∑k=2

2008∑j=2

1jk

=2008∑j=2

∞∑k=2

1jk

=2008∑j=2

1j2(1− 1

j )=

2008∑j=2

1j(j − 1)

=2008∑j=2

(1

j − 1− 1j

)= 1− 1

2008=

20072008

.

12. [7] Suppose we have an (infinite) cone C with apex A and a plane π. The intersection of π and C isan ellipse E with major axis BC, such that B is closer to A than C, and BC = 4, AC = 5, AB = 3.Suppose we inscribe a sphere in each part of C cut up by E with both spheres tangent to E . What isthe ratio of the radii of the spheres (smaller to larger)?

Answer: 13 It can be seen that the points of tangency of the spheres with E must lie on its major

axis due to symmetry. Hence, we consider the two-dimensional cross-section with plane ABC. Thenthe two spheres become the incentre and the excentre of the triangle ABC, and we are looking for theratio of the inradius to the exradius. Let s, r, ra denote the semiperimeter, inradius, and exradius(opposite to A) of the triangle ABC. We know that the area of ABC can be expressed as both rs andra(s− |BC|), and so r

ra= s−|BC|

s . For the given triangle, s = 6 and a = 4, so the required ratio is 13 .

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

13. [8] Let P (x) be a polynomial with degree 2008 and leading coefficient 1 such that

P (0) = 2007, P (1) = 2006, P (2) = 2005, . . . , P (2007) = 0.

Determine the value of P (2008). You may use factorials in your answer.

Answer: 2008!− 1 Consider the polynomial Q(x) = P (x) + x − 2007. The given conditionstell us that Q(x) = 0 for x = 0, 1, 2, . . . , 2007, so these are the roots of Q(x). On the other hand,we know that Q(x) is also a polynomial with degree 2008 and leading coefficient 1. It follows thatQ(x) = x(x− 1)(x− 2)(x− 3) · · · (x− 2007). Thus

P (x) = x(x− 1)(x− 2)(x− 3) · · · (x− 2007)− x+ 2007.

Setting x = 2008 gives the answer.

14. [8] Evaluate the infinite sum∑∞n=1

nn4+4 .

Answer: 38 We have

∞∑n=1

n

n4 + 4=∞∑n=1

n

(n2 + 2n+ 2)(n2 − 2n+ 2)

=14

∞∑n=1

(1

n2 − 2n+ 2− 1n2 + 2n+ 2

)

=14

∞∑n=1

(1

(n− 1)2 + 1− 1

(n+ 1)2 + 1

).

Observe that the sum telescopes. From this we find that the answer is 14

(1

02+1 + 112+1

)= 3

8 .

15. [8] In a game show, Bob is faced with 7 doors, 2 of which hide prizes. After he chooses a door, thehost opens three other doors, of which one is hiding a prize. Bob chooses to switch to another door.What is the probability that his new door is hiding a prize?

3

Answer: 521 If Bob initially chooses a door with a prize, then he will not find a prize by switching.

With probability 5/7 his original door does not hide the prize. After the host opens the three doors,the remaining three doors have equal probability of hiding the prize. Therefore, the probability thatBob finds the prize is 5

7 × 13 = 5

21 .

Remark: This problem can be easily recognized as a variation of the classic Monty Hall problem.

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16. [9] Point A lies at (0, 4) and point B lies at (3, 8). Find the x-coordinate of the point X on the x-axismaximizing ∠AXB.

Answer: 5√

2− 3 Let X be a point on the x-axis and let θ = ∠AXB. We can easily see thatthe circle with diameter AB does not meet the x-axis, so θ ≤ π. Thus, maximizing θ is equivalent tomaximizing sin θ. By the Law of Sines, this in turn is equivalent to minimizing the circumradius oftriangle ABX. This will occur when the circumcircle of ABX is the smaller of the two circles throughA and B tangent to the x-axis. So let X now be this point of tangency. Extend line AB to meet thex-axis at C = (−3, 0); by Power of a Point CX2 = CA ·CB = 50 so CX = 5

√2. Clearly X has larger

x-coordinate than C, so the x-coordinate of X is 5√

2− 3.

17. [9] Solve the equation √√√√x+

√4x+

√16x+

√. . .+

√42008x+ 3−√x = 1.

Express your answer as a reduced fraction with the numerator and denominator written in their primefactorization.

Answer: 124016 Rewrite the equation to get√√√√

x+

√4x+

√16x+

√. . .+

√42008x+ 3 =

√x+ 1.

Squaring both sides yields √4x+

√. . .+

√42008x+ 3 = 2

√x+ 1.

Squaring again yields √16x+

√. . .+

√42008x+ 3 = 4

√x+ 1.

One can see that by continuing this process one gets√42008x+ 3 = 22008

√x+ 1,

so that 2 · 22008√x = 2. Hence x = 4−2008. It is also easy to check that this is indeed a solution to the

original equation.

18. [9] Let ABC be a right triangle with ∠A = 90◦. Let D be the midpoint of AB and let E be a pointon segment AC such that AD = AE. Let BE meet CD at F . If ∠BFC = 135◦, determine BC/AB.

Answer:√

132 Let α = ∠ADC and β = ∠ABE. By exterior angle theorem, α = ∠BFD + β =

45◦ + β. Also, note that tanβ = AE/AB = AD/AB = 1/2. Thus,

1 = tan 45◦ = tan(α− β) =tanα− tanβ

1 + tanα tanβ=

tanα− 12

1 + 12 tanα

.

4

Solving for tanα gives tanα = 3. Therefore, AC = 3AD = 32AB. Using Pythagorean Theorem, we

find that BC =√

132 AB. So the answer is

√132 .

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

19. [10] Let ABCD be a regular tetrahedron, and let O be the centroid of triangle BCD. Consider thepoint P on AO such that P minimizes PA+ 2(PB + PC + PD). Find sin ∠PBO.

Answer: 16 We translate the problem into one about 2-D geometry. Consider the right triangle

ABO, and P is some point on AO. Then, the choice of P minimizes PA + 6PB. Construct the line` through A but outside the triangle ABO so that sin ∠(AO, `) = 1

6 . For whichever P chosen, letQ be the projection of P onto `, then PQ = 1

6AP . Then, since PA + 6PB = 6(PQ + PB), it isequivalent to minimize PQ+ PB. Observe that this sum is minimized when B,P,Q are collinear andthe line through them is perpendicular to ` (so that PQ + PB is simply the distance from B to `).Then, ∠AQB = 90◦, and since ∠AOB = 90◦ as well, we see that A,Q,P,B are concyclic. Therefore,∠PBO = ∠OPA = ∠(AO, `), and the sine of this angle is therefore 1

6 .

20. [10] For how many ordered triples (a, b, c) of positive integers are the equations abc+ 9 = ab+ bc+ caand a+ b+ c = 10 satisfied?

Answer: 21 Subtracting the first equation from the second, we obtain 1−a−b−c+ab+bc+ca−abc =(1− a)(1− b)(1− c) = 0. Since a, b, and c are positive integers, at least one must equal 1. Note thata = b = c = 1 is not a valid triple, so it suffices to consider the cases where exactly two or one of a, b, care equal to 1. If a = b = 1, we obtain c = 8 and similarly for the other two cases, so this gives 3ordered triples. If a = 1, then we need b+ c = 9, which has 6 solutions for b, c 6= 1; a similar argumentfor b and c gives a total of 18 such solutions. It is easy to check that all the solutions we found areactually solutions to the original equations. Adding, we find 18 + 3 = 21 total triples.

21. [10] Let ABC be a triangle with AB = 5, BC = 4 and AC = 3. Let P and Q be squares inside ABCwith disjoint interiors such that they both have one side lying on AB. Also, the two squares each havean edge lying on a common line perpendicular to AB, and P has one vertex on AC and Q has onevertex on BC. Determine the minimum value of the sum of the areas of the two squares.

A B

C

PQ

Answer: 14449 Let the side lengths of P and Q be a and b, respectively. Label two of the vertices of

P as D and E so that D lies on AB and E lies on AC, and so that DE is perpendicular to AB. Thetriangle ADE is similar to ACB. So AD = 3

4a. Using similar arguments, we find that

3a4

+ a+ b+4b3

= AB = 5

soa

4+b

3=

57.

Using Cauchy-Schwarz inequality, we get

(a2 + b2

)( 142

+132

)≥(a

4+b

3

)2

=2549.

5

It follows thata2 + b2 ≥ 144

49.

Equality occurs at a = 3635 and b = 48

35 .

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22. [10] For a positive integer n, let θ(n) denote the number of integers 0 ≤ x < 2010 such that x2 − n is

divisible by 2010. Determine the remainder when2009∑n=0

n · θ(n) is divided by 2010.

Answer: 335 Let us consider the sum∑2009n=0 n · θ(n) (mod 2010) in a another way. Consider the

sum 02 + 12 + 22 + · · · + 20072 (mod 2010). For each 0 ≤ n < 2010, in the latter sum, the term n

appears θ(n) times, so the sum is congruent to∑2009n=0 n · θ(n). In other words,

2009∑n=0

n · θ(n) =2009∑n=0

n2 =(2009)(2009 + 1)(2 · 2009 + 1)

6≡ (−1) · 2010

6· (−1) = 335 (mod 2010).

23. [10] Two mathematicians, Kelly and Jason, play a cooperative game. The computer selects some secretpositive integer n < 60 (both Kelly and Jason know that n < 60, but that they don’t know what thevalue of n is). The computer tells Kelly the unit digit of n, and it tells Jason the number of divisorsof n. Then, Kelly and Jason have the following dialogue:

Kelly: I don’t know what n is, and I’m sure that you don’t know either. However, I know that n isdivisible by at least two different primes.

Jason: Oh, then I know what the value of n is.

Kelly: Now I also know what n is.

Assuming that both Kelly and Jason speak truthfully and to the best of their knowledge, what are allthe possible values of n?

Answer: 10 The only way in which Kelly can know that n is divisible by at least two differentprimes is if she is given 0 as the unit digit of n, since if she received anything else, then there is somenumber with that unit digit and not divisible by two primes (i.e., 1, 2, 3, 4, 5, 16, 7, 8, 9). Then, afterKelly says the first line, Jason too knows that n is divisible by 10.

The number of divisors of 10, 20, 30, 40, 50 are 4, 6, 8, 8, 6, respectively. So unless Jason received 4, hecannot otherwise be certain of what n is. It follows that Jason received 4, and thus n = 10.

24. [10] Suppose that ABC is an isosceles triangle with AB = AC. Let P be the point on side AC so thatAP = 2CP . Given that BP = 1, determine the maximum possible area of ABC.

Answer: 910 Let Q be the point on AB so that AQ = 2BQ, and let X be the intersection of BP

and CQ. The key observation that, as we will show, BX and CX are fixed lengths, and the ratio ofareas [ABC]/[BCX] is constant. So, to maximize [ABC], it is equivalent to maximize [BCX].

Using Menelaus’ theorem on ABP , we have

BX · PC ·AQXP · CA ·QB = 1.

Since PC/CA = 1/3 and AQ/QB = 2, we get BX/XP = 3/2. It follows that BX = 3/5. Bysymmetry, CX = 3/5.

Also, we have

[ABC] = 3[BPC] = 3 · 53

[BXC] = 5[BXC].

6

Note that [BXC] is maximized when ∠BXC = 90◦ (one can check that this configuration is indeedpossible). Thus, the maximum value of [BXC] is 1

2BX · CX = 12

(35

)2 = 950 . It follows that the

maximum value of [ABC] is 910 .

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11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

25. [12] Alice and the Cheshire Cat play a game. At each step, Alice either (1) gives the cat a penny,which causes the cat to change the number of (magic) beans that Alice has from n to 5n or (2) givesthe cat a nickel, which causes the cat to give Alice another bean. Alice wins (and the cat disappears)as soon as the number of beans Alice has is greater than 2008 and has last two digits 42. What is theminimum number of cents Alice can spend to win the game, assuming she starts with 0 beans?

Answer: 35 Consider the number of beans Alice has in base 5. Note that 2008 = 310135, 42 = 1325,and 100 = 4005. Now, suppose Alice has dk · · · d2d1 beans when she wins; the conditions for winningmean that these digits must satisfy d2d1 = 32, dk · · · d3 ≥ 310, and dk · · · d3 = 4i + 1 for some i. Togain these dk · · · d2d1 beans, Alice must spend at least 5(d1 + d2 + · · · + dk) + k − 1 cents (5 centsto get each bean in the “units digit” and k − 1 cents to promote all the beans). We now must havek ≥ 5 because dk · · · d2d1 > 2008. If k = 5, then dk ≥ 3 since dk · · · d3 ≥ 3100; otherwise, we havedk ≥ 1. Therefore, if k = 5, we have 5(d1 + d2 + · · · + dk) + k − 1 ≥ 44 > 36; if k > 5, we have5(d1 + d2 + · · ·+ dk) + k− 1 ≥ 30 + k− 1 ≥ 35. But we can attain 36 cents by taking dk · · · d3 = 1000,so this is indeed the minimum.

26. [12] Let P be a parabola, and let V1 and F1 be its vertex and focus, respectively. Let A and B bepoints on P so that ∠AV1B = 90◦. Let Q be the locus of the midpoint of AB. It turns out that Qis also a parabola, and let V2 and F2 denote its vertex and focus, respectively. Determine the ratioF1F2/V1V2.

Answer: 78 Since all parabolas are similar, we may assume that P is the curve y = x2. Then, if

A = (a, a2) and B = (b, b2), the condition that ∠AV1B = 90◦ gives ab+ a2b2 = 0, or ab = −1. Then,the midpoint of AB is

A+B

2=(a+ b

2,a2 + b2

2

)=(a+ b

2,

(a+ b)2 − 2ab2

)=(a+ b

2,

(a+ b)2

2+ 1).

(Note that a + b can range over all real numbers under the constraint ab = −1.) It follows that thelocus of the midpoint of AB is the curve y = 2x2 + 1.

Recall that the focus of y = ax2 is (0, 14a ). We find that V1 = (0, 0), V2 = (0, 1), F1 = (0, 1

4 ),F2 = (0, 1 + 1

8 ). Therefore, F1F2/V1V2 = 78 .

27. [12] Cyclic pentagon ABCDE has a right angle ∠ABC = 90◦ and side lengths AB = 15 and BC = 20.Supposing that AB = DE = EA, find CD.

Answer: 7 By Pythagoras, AC = 25. Since AC is a diameter, angles ∠ADC and ∠AEC are alsoright, so that CE = 20 and AD2 + CD2 = AC2 as well. Beginning with Ptolemy’s theorem,

(AE · CD +AC ·DE)2 = AD2 · EC2 =(AC2 − CD2

)EC2

=⇒ CD2(AE2 + EC2

)+ 2 · CD ·AE2 ·AC +AC2

(DE2 − EC2

)= 0

=⇒ CD2 + 2CD(AE2

AC

)+DE2 − EC2 = 0.

It follows that CD2 + 18CD − 175 = 0, from which CD = 7.

Remark: A simple trigonometric solution is possible. One writes α = ∠ACE = ∠ECD =⇒ ∠DAC =90◦ − 2α and applies double angle formula.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

28. [15] Let P be a polyhedron where every face is a regular polygon, and every edge has length 1. Eachvertex of P is incident to two regular hexagons and one square. Choose a vertex V of the polyhedron.Find the volume of the set of all points contained in P that are closer to V than to any other vertex.

Answer:√

23 Observe that P is a truncated octahedron, formed by cutting off the corners from

a regular octahedron with edge length 3. So, to compute the value of P , we can find the volume ofthe octahedron, and then subtract off the volume of truncated corners. Given a square pyramid whereeach triangular face an equilateral triangle, and whose side length is s, the height of the pyramid is√

22 s, and thus the volume is 1

3 · s2 ·√

22 s =

√2

6 s3. The side length of the octahedron is 3, and noting

that the octahedron is made up of two square pyramids, its volume must be is 2 ·√

2(3)3

6 = 9√

2.The six “corners” that we remove are all square pyramids, each with volume

√2

6 , and so the resultingpolyhedron P has volume 9

√2− 6 ·

√2

6 = 8√

2.

Finally, to find the volume of all points closer to one particular vertex than any other vertex, notethat due to symmetry, every point in P (except for a set with zero volume), is closest to one of the 24vertices. Due to symmetry, it doesn’t matter which V is picked, so we can just divide the volume of Pby 24 to obtain the answer

√2

3 .

29. [15] Let (x, y) be a pair of real numbers satisfying

56x+ 33y =−y

x2 + y2, and 33x− 56y =

x

x2 + y2.

Determine the value of |x|+ |y|.Answer: 11

65 Observe that

1x+ yi

=x− yix2 + y2

= 33x− 56y + (56x+ 33y)i = (33 + 56i)(x+ yi).

So

(x+ yi)2 =1

33 + 56i=

1(7 + 4i)2

=(

7− 4i65

)2

.

It follows that (x, y) = ± ( 765 ,− 4

65

).

30. [15] Triangle ABC obeys AB = 2AC and ∠BAC = 120◦. Points P and Q lie on segment BC suchthat

AB2 +BC · CP = BC2

3AC2 + 2BC · CQ = BC2

Find ∠PAQ in degrees.

Answer: 40◦ We have AB2 = BC(BC − CP ) = BC · BP, so triangle ABC is similar to trianglePBA. Also, AB2 = BC(BC−2CQ)+AC2 = (BC−CQ)2−CQ2+AC2, which rewrites as AB2+CQ2 =BQ2 +AC2. We deduce that Q is the foot of the altitude from A. Thus, ∠PAQ = 90◦−∠QPA = 90◦−∠ABP −∠BAP. Using the similar triangles, ∠PAQ = 90◦ −∠ABC −∠BCA = ∠BAC − 90◦ = 40◦.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

31. [18] Let C be the hyperbola y2 − x2 = 1. Given a point P0 on the x-axis, we construct a sequenceof points (Pn) on the x-axis in the following manner: let `n be the line with slope 1 passing passing

8

through Pn, then Pn+1 is the orthogonal projection of the point of intersection of `n and C onto thex-axis. (If Pn = 0, then the sequence simply terminates.)

Let N be the number of starting positions P0 on the x-axis such that P0 = P2008. Determine theremainder of N when divided by 2008.

Answer: 254 Let Pn = (xn, 0). Then the `n meet C at (xn+1, xn+1 − xn). Since this point lies onthe hyperbola, we have (xn+1 − xn)2 − x2

n+1 = 1. Rearranging this equation gives

xn+1 =x2n − 12xn

.

Choose a θ0 ∈ (0, π) with cot θ0 = x0, and define θn = 2nθ0. Using the double-angle formula, we have

cot θn+1 = cot(2θn) =cot2 θn − 1

2 cot θn.

It follows by induction that xn = cot θn. Then, P0 = P2008 corresponds to cot θ0 = cot(22008θ0

)(assuming that P0 is never at the origin, or equivalently, 2nθ is never an integer multiple of π). So, weneed to find the number of θ0 ∈ (0, π) with the property that 22008θ0− θ0 = kπ for some integer k. Wehave θ0 = kπ

22008−1 , so k can be any integer between 1 and 22008 − 2 inclusive (and note that since thedenominator is odd, the sequence never terminates). It follows that the number of starting positionsis N = 22008 − 2.

Finally, we need to compute the remainder when N is divided by 2008. We have 2008 = 23 × 251.Using Fermat’s Little Theorem with 251, we get 22008 ≡ (2250

)4 · 256 ≡ 14 · 5 = 5 (mod 251). So wehave N ≡ 3 (mod 251) and N ≡ −2 (mod 8). Using Chinese Remainder Theorem, we get N ≡ 254(mod 2008).

32. [18] Cyclic pentagon ABCDE has side lengths AB = BC = 5, CD = DE = 12, and AE = 14.Determine the radius of its circumcircle.

Answer: 225√

1188 Let C ′ be the point on minor arc BCD such that BC ′ = 12 and C ′D = 5, and write

AC ′ = BD = C ′E = x, AD = y, and BD = z. Ptolemy applied to quadrilaterals ABC ′D,BC ′DE,and ABDE gives

x2 = 12y + 52

x2 = 5z + 122

yz = 14x+ 5 · 12

Then(x2 − 52)(x2 − 122) = 5 · 12yz = 5 · 12 · 14x+ 52 · 122,

from which x3 − 169x− 5 · 12 · 14 = 0. Noting that x > 13, the rational root theorem leads quickly tothe root x = 15. Then triangle BCD has area

√16 · 1 · 4 · 11 = 8

√11 and circumradius R = 5·12·15

4·8√

11=

225√

1188 .

33. [18] Let a, b, c be nonzero real numbers such that a+ b+ c = 0 and a3 + b3 + c3 = a5 + b5 + c5. Findthe value of a2 + b2 + c2.

Answer: 65 Let σ1 = a+ b+ c, σ2 = ab+ bc+ ca and σ3 = abc be the three elementary symmetric

polynomials. Since a3 + b3 + c3 is a symmetric polynomial, it can be written as a polynomial in σ1, σ2

and σ3. Now, observe that σ1 = 0, and so we only need to worry about the terms not containing σ1. Byconsidering the degrees of the terms, we see that the only possibility is σ3. That is, a3 + b3 + c3 = kσ3

for some constant k. By setting a = b = 1, c = −2, we see that k = 3.

By similar reasoning, we find that a5 + b5 + c5 = hσ2σ3 for some constant h. By setting a = b = 1 andc = −2, we get h = −5.

9

So, we now know that a+ b+ c = 0 implies

a3 + b3 + c3 = 3abc and a5 + b5 + c5 = −5abc(ab+ bc+ ca)

Then a3 + b3 + c3 = a5 + b5 + c5 implies that 3abc = −5abc(ab+ bc+ ca). Given that a, b, c are nonzero,we get ab+ bc+ ca = − 3

5 . Then, a2 + b2 + c2 = (a+ b+ c)2 − 2(ab+ bc+ ca) = 65 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

34. Who Wants to Be a Millionaire. In 2000, the Clay Mathematics Institute named seven MillenniumPrize Problems, with each carrying a prize of $1 Million for its solution. Write down the name of ONEof the seven Clay Millennium Problems. If your submission is incorrect or misspelled, then yoursubmission is disqualified. If another team wrote down the same Millennium Problem as you, then youget 0 points, otherwise you get 20 points.

Solution: The seven Millennium Prize Problems are:

(a) Birch and Swinnerton-Dyer Conjecture

(b) Hodge Conjecture

(c) Navier-Stokes Equations

(d) P vs NP

(e) Poincaré Conjecture

(f) Riemann Hypothesis

(g) Yang-Mills Theory

More information can be found on its official website http://www.claymath.org/millennium/.

As far as this as an HMMT problem goes, it’s probably a good idea to submit something that youthink is least likely for another team to think of (or to spell correctly). Though, this may easily turninto a contest of who can still remember the names of the user ranks from the Art of Problem Solvingforum.

35. NUMB3RS. The RSA Factoring Challenge, which ended in 2007, challenged computational mathe-maticians to factor extremely large numbers that were the product of two prime numbers. The largestnumber successfully factored in this challenge was RSA-640, which has 193 decimal digits and carrieda prize of $20, 000. The next challenge number carried prize of $30, 000, and contains N decimal digits.Your task is to submit a guess for N . Only the team(s) that have the closest guess(es) receives points.If k teams all have the closest guesses, then each of them receives

⌈20k

⌉points.

Answer: 212 For more information, see the Wikipedia entry at http://en.wikipedia.org/wiki/RSA_Factoring_Challenge.

RSA-640 was factored in November 2005, and the effort took approximately 30 2.2GHz-Opteron-CPUyears over five months of calendar time.

36. The History Channel. Below is a list of famous mathematicians. Your task is to list a subset ofthem in the chronological order of their birth dates. Your submission should be a sequence of letters.If your sequence is not in the correct order, then you get 0 points. Otherwise your score will bemin{max{5(N − 4), 0}, 25}, where N is the number of letters in your sequence.

(A) Niels Abel (B) Arthur Cayley (C) Augustus De Morgan (D) Gustav Dirichlet (E) Leonhard Euler(F) Joseph Fourier (G) Évariste Galois (H) Carl Friedrich Gauss (I) Marie-Sophie Germain (J) JosephLouis Lagrange (K) Pierre-Simon Laplace (L) Henri Poincaré (N) Bernhard Riemann

Answer: any subsequence of EJKFIHADCGBNL The corresponding birth dates are listed below:

(A) Niels Abel (1802–1829)

10

(B) Arthur Cayley (1821–1895)

(C) Augustus De Morgan (1806–1871)

(D) Gustav Dirichlet (1805–1859)

(E) Leonhard Euler (1707–1783)

(F) Joseph Fourier (1768–1830)

(G) Évariste Galois (1811–1832)

(H) Carl Friedrich Gauss (1777–1855)

(I) Marie-Sophie Germain (1776–1831)

(J) Joseph Louis Lagrange (1736–1813)

(K) Pierre-Simon Laplace (1749–1827)

(L) Henri Poincaré (1854–1912)

(N) Bernhard Riemann (1826–1866)

11

11th Annual Harvard-MIT Mathematics Tournament

Saturday 23 February 2008

Team Round: A Division

Lattice Walks [90]

1. [20] Determine the number of ways of walking from (0, 0) to (5, 5) using only up and right unitsteps such that the path does not pass through any of the following points: (1, 1), (1, 4), (4, 1), (4, 4).

2. [20] Let n > 2 be a positive integer. Prove that there are 12(n− 2)(n + 1) ways to walk from

(0, 0) to (n, 2) using only up and right unit steps such that the walk never visits the line y = xafter it leaves the origin.

3. [20] Let n > 4 be a positive integer. Determine the number of ways to walk from (0, 0) to(n, 2) using only up and right unit steps such that the path does not meet the lines y = x ory = x− n + 2 except at the start and at the end.

4. [30] Let n > 6 be a positive integer. Determine the number of ways to walk from (0, 0) to(n, 3) using only up and right unit steps such that the path does not meet the lines y = x ory = x− n + 3 except at the start and at the end.

Lattice and Centroids [130]

A d-dimensional lattice point is a point of the form (x1, x2, . . . , xd) where x1, x2, . . . , xd are all inte-gers. For a set of d-dimensional points, their centroid is the point found by taking the coordinate-wise average of the given set of points.

Let f(n, d) denote the minimal number f such that any set of f lattice points in the d-dimensionalEuclidean space contains a subset of size n whose centroid is also a lattice point.

5. [10] Let S be a set of 5 points in the 2-dimensional lattice. Show that we can always choosea pair of points in S whose midpoint is also a lattice point.

6. [10] Construct a set of 2d d-dimensional lattice points so that for any two chosen points A, B,the line segment AB does not pass through any other lattice point.

7. [35] Show that for positive integers n and d,

(n− 1)2d + 1 ≤ f(n, d) ≤ (n− 1)nd + 1.

1

8. [40] Show that for positive integers n1, n2 and d,

f(n1n2, d) ≤ f(n1, d) + n1 (f(n2, d)− 1) .

9. [35] Determine, with proof, a simple closed-form expression for f(2a, d).

Incircles [180]

In the following problems, ABC is a triangle with incenter I. Let D,E, F denote the points wherethe incircle of ABC touches sides BC, CA, AB, respectively.

A

B CD

E

F

I

At the end of this section you can find some terminology and theorems that may be helpful to you.

10. On the circumcircle of ABC, let A′ be the midpoint of arc BC (not containing A).

(a) [10] Show that A, I, A′ are collinear.

(b) [20] Show that A′ is the circumcenter of BIC.

11. [30] Let lines BI and EF meet at K. Show that I, K, E, C, D are concyclic.

12. [40] Let K be as in the previous problem. Let M be the midpoint of BC and N the midpointof AC. Show that K lies on line MN .

13. [40] Let M be the midpoint of BC, and T diametrically opposite to D on the incircle ofABC. Show that DT, AM, EF are concurrent.

14. [40] Let P be a point inside the incircle of ABC. Let lines DP,EP, FP meet the incircleagain at D′, E′, F ′. Show that AD′, BE′, CF ′ are concurrent.

Glossary and some possibly useful facts

• A set of points is collinear if they lie on a common line. A set of lines is concurrent if theypass through a common point. A set of points are concyclic if they lie on a common circle.

• Given ABC a triangle, the three angle bisectors are concurrent at the incenter of the triangle.The incenter is the center of the incircle, which is the unique circle inscribed in ABC, tangentto all three sides.

2

• Ceva’s theorem states that given ABC a triangle, and points X, Y, Z on sides BC, CA, AB,respectively, the lines AX, BY, CZ are concurrent if and only if

BX

XB· CY

Y A· AZ

ZB= 1.

• “Trig” Ceva states that given ABC a triangle, and points X, Y, Z inside the triangle, thelines AX, BY, CZ are concurrent if and only if

sin ∠BAX

sin ∠XAC· sin ∠CBY

sin ∠Y BA· sin ∠ACZ

sin ∠ZCB= 1.

3

11th Annual Harvard-MIT Mathematics Tournament

Saturday 23 February 2008

Team Round: A Division

Lattice Walks [90]

1. [20] Determine the number of ways of walking from (0, 0) to (5, 5) using only up and right unitsteps such that the path does not pass through any of the following points: (1, 1), (1, 4), (4, 1), (4, 4).

Answer: 34

Solution: In the following figure, each lattice point (with the bottom-left-most point (0, 0))is labeled with the number of ways of reaching there from (0, 0). With the exception of theforbidden points, the labels satisfy the recursion formula f(x, y) = f(x − 1, y) + f(x, y − 1).We see from the diagram that there are 34 ways to reach (5, 5).

1

1

1

1

1

1 34

17

17

5

1

111111

2

1

1

1

4

2

4

5

2

8

4

12

17

12

4

17

0

0

0

0

2. [20] Let n > 2 be a positive integer. Prove that there are 12(n− 2)(n + 1) ways to walk from

(0, 0) to (n, 2) using only up and right unit steps such that the walk never visits the line y = xafter it leaves the origin.

Solution: The first two steps can only go to the right. Then we need to compute thenumber of ways of walking from (2, 0) to (n, 2) which does not pass through the point (2, 2).There are

(n2

)ways to walk from (2, 0) to (n, 2), and exactly one of those paths passes through

the point (2, 2). So the number of valid paths is(n2

)− 1 = 1

2n(n− 1)− 1 = 12(n− 2)(n + 1).

Remark: We used the well-known fact that there are(a+ba

)ways to walk from (0, 0) to (a, b)

using only up and right unit steps. This is true because there are a + b steps, and we need tochoose a of them to be right steps, and the rest up steps.

3. [20] Let n > 4 be a positive integer. Determine the number of ways to walk from (0, 0) to(n, 2) using only up and right unit steps such that the path does not meet the lines y = x ory = x− n + 2 except at the start and at the end.

1

Answer: 12(n2 − 5n + 2)

Solution: It is easy to see the the first two steps and the last two steps must all be rightsteps. So we need to compute the number of walks from (2, 0) to (n−2, 0) that does not passthrough (2, 2) and (n− 2, 0). There are

(n−2

2

)paths from (2, 0) to (n− 2, 0), and exactly two

of them are invalid. So the answer is(n−2

2

)− 2 = 1

2(n− 2)(n− 3)− 2 = 12(n2 − 5n + 2).

4. [30] Let n > 6 be a positive integer. Determine the number of ways to walk from (0, 0) to(n, 3) using only up and right unit steps such that the path does not meet the lines y = x ory = x− n + 3 except at the start and at the end.

Answer: 16(n− 6)(n− 1)(n + 1) Consider the first point of the path that lies on x = 3.

There are two possibilities for this point: (3, 0) and (3, 1), and there is exactly one valid wayof getting to each point from the origin. Similarly, consider the last point of the path thatlies on x = n − 3. There are two possibilities: (n − 3, 2) and (n − 3, 3), and there is exactlyone valid way of getting to the destination from each of the two points. Now we count thenumber of valid paths from each of (3, 0) and (3, 1), to each of (n− 3, 2) and (n− 3, 3), andthe answer will be the sum.

• From (3, 1) to (n− 3, 2): there are no forbidden points along the way, so there are n− 5ways.

• From (3, 0) to (n− 3, 2): the path must not pass through (n− 3, 0), and there is exactlyone invalid path. So there are

(n−4

2

)− 1 ways.

• From (3, 1) to (n− 3, 3): the path must not pass through (3, 3), and there is exactly oneinvalid path. So there are

(n−4

2

)− 1 ways.

• From (3, 0) to (n− 3, 3): the path must not pass through (n− 3, 0) and (3, 3), and thereare exactly two invalid paths. So there are

(n−3

3

)− 2 ways.

Summing, we obtain the answer:

n−5+(

n− 42

)−1+

(n− 4

2

)−1+

(n− 3

3

)−2 =

n3 − 6n2 − n + 66

=(n− 6)(n− 1)(n + 1)

6.

Lattice and Centroids [130]

A d-dimensional lattice point is a point of the form (x1, x2, . . . , xd) where x1, x2, . . . , xd are all inte-gers. For a set of d-dimensional points, their centroid is the point found by taking the coordinate-wise average of the given set of points.

Let f(n, d) denote the minimal number f such that any set of f lattice points in the d-dimensionalEuclidean space contains a subset of size n whose centroid is also a lattice point.

5. [10] Let S be a set of 5 points in the 2-dimensional lattice. Show that we can always choosea pair of points in S whose midpoint is also a lattice point.

Solution: Consider the parities of the coordinates. There are four possibilities: (odd, odd),(odd, even), (even, odd), (even, even). By the pigeonhole principle, two of the points musthave the same parity in both coordinates (i.e., they are congruent in mod 2). Then, themidpoint of these two points must be a lattice point.

2

6. [10] Construct a set of 2d d-dimensional lattice points so that for any two chosen points A, B,the line segment AB does not pass through any other lattice point.

Solution: The simplest example is the set of 2d points of the form (a1, a2, . . . , ad), whereak ∈ {0, 1} for each k. This is the set of vertices of a d-dimensional cube.

7. [35] Show that for positive integers n and d,

(n− 1)2d + 1 ≤ f(n, d) ≤ (n− 1)nd + 1.

Solution: Note that taking the set of points to be a multiset does not affect f(n, d) asadding multiples of n to any of the coordinate values does not change the result. The lowerbound is obtained by considering the multiset consisting of n − 1 copies of each of the 2d

0, 1-vectors of length d, as it contains no submultiset of size n whose centroid is also a latticepoint. By the pigeonhole principle, any multiset of (n− 1)nd + 1 lattice points must containn points whose coordinates are congruent modulo n. The centroid of these n points is also alattice point, thus proving the upper bound.

8. [40] Show that for positive integers n1, n2 and d,

f(n1n2, d) ≤ f(n1, d) + n1 (f(n2, d)− 1) .

Solution: Given a multiset of f(n1, d) + n1 (f(n2, d)− 1) lattice points, we may selectl = f(n2, d) pairwise disjoint submultisets S1, S2, . . . , Sl, each consisting of n1 points, whosecentroid is a lattice point. Let ϕ map each multiset Si to its centroid gi. By the definition off(n2, d), there exists a submultiset T ⊂ {g1, g2, . . . , gl} satisfying |T | = n2 whose centroid isa lattice point. Then

⋃i∈T ϕ−1(gi) is a multiset of n1n2 lattice points whose centroid is also

a lattice point.

9. [35] Determine, with proof, a simple closed-form expression for f(2a, d).

Answer: (2a − 1)2d + 1

Solution: From Problem ??, f(2a, d) ≥ (2a − 1)2d + 1. We prove by induction on a thatf(2a, d) ≤ (2a − 1)2d + 1. When a = 1, Problem ?? shows that f(2, d) ≤ 2d + 1. Fix a > 1and suppose that the assertion holds for smaller values of a. Using Problem ??,

f(2a, d) ≤ f(2, d) + 2(f(2a−1, d)− 1

)≤ 2d + 1 + 2 · (2a−1 − 1)2d

= (2a − 1)2d + 1.

Thus f(2a, d) = (2a − 1)2d + 1.

Incircles [180]

In the following problems, ABC is a triangle with incenter I. Let D,E, F denote the points wherethe incircle of ABC touches sides BC, CA, AB, respectively.

3

A

B CD

E

F

I

At the end of this section you can find some terminology and theorems that may be helpful to you.

10. On the circumcircle of ABC, let A′ be the midpoint of arc BC (not containing A).

(a) [10] Show that A, I, A′ are collinear.(b) [20] Show that A′ is the circumcenter of BIC.

Solution:

A

B C

I

A′

(a) Since A′ bisectors the arc BC, the two arcs A′B and A′C are equal, and so ∠BAA′ =∠CAA′. Thus, A′ lies on the angle bisector of BAC. Since I also lies on the anglebisector of BAC, we see that A, I, A′ are collinear.

(b) We have

∠CIA′ = ∠A′AC + ∠ICA = ∠A′AB + ∠ICB = ∠A′CB + ∠ICB = ∠ICA′.

Therefore, A′I = A′C. By similar arguments, A′I = A′B. So, A′ is equidistant fromB, I, C, and thus is its circumcenter.

11. [30] Let lines BI and EF meet at K. Show that I, K, E, C, D are concyclic.

Solution: First, note that there are two possible configurations, as K could lie insidesegment EF , or on its extension. The following proof works for both cases. We have

A

B CD

E

F

IK

4

∠KIC = ∠IBC + ∠ICB =12∠ABC +

12∠ACB = 90◦ − 1

2∠BAC = ∠AEF.

It follows that I, K, E, C are concyclic. The point D also lies on this circle because ∠IDC =∠IEC = 90◦. Thus, all five points are concyclic.

12. [40] Let K be as in the previous problem. Let M be the midpoint of BC and N the midpointof AC. Show that K lies on line MN .

Solution: Since I, K, E, C are concyclic, we have ∠IKC = ∠IEC = 90◦. Let C ′ be thereflection of C across BI, then C ′ must lie on AB. Then, K is the midpoint of CC ′. Considera dilation centered at C with factor 1

2 . Since C ′ lies on AB, it follows that K lies on MN .

A

B C

C ′

IK

M

N

13. [40] Let M be the midpoint of BC, and T diametrically opposite to D on the incircle ofABC. Show that DT, AM, EF are concurrent.

Solution: If AB = AC, then the result is clear as AM and DT coincide. So, assume thatAB 6= AC.

A

B CD

E

F

T

I

X YZ

M

Let lines DT and EF meet at Z. Construct a line through Z parallel to BC, and let itmeet AB and AC at X and Y , respectively. We have ∠XZI = 90◦, and ∠XFI = 90◦.Therefore, F,Z, I, X are concyclic, and thus ∠IXZ = ∠IFZ. By similar arguments, we alsohave ∠IY Z = ∠IEZ. Thus, triangles IFE and IXY are similar. Since IE = IF , we mustalso have IX = IY . Since IZ is an altitude of the isosceles triangle IXY , Z is the midpointof XY .

5

Since XY and BC are parallel, there is a dilation centered at A that sends XY to BC. Soit must send the midpoint Z to the midpoint M . Therefore, A, Z,M are collinear. It followsthat DT, AM, EF are concurrent.

14. [40] Let P be a point inside the incircle of ABC. Let lines DP,EP, FP meet the incircleagain at D′, E′, F ′. Show that AD′, BE′, CF ′ are concurrent.

Solution: Using the trigonometric version of Ceva’s theorem, it suffices to prove that

sin ∠BAD′

sin ∠D′AC· sin ∠CBE′

sin ∠E′BA· sin ∠ACF ′

sin ∠F ′CB= 1. (†)

A

B CD

E

F D′

E ′

F ′

P

Using sine law, we have

sin ∠BAD′ =FD′

AD′· sin ∠AFD′ =

FD′

AD′· sin ∠FDD′

Let r be the inradius of ABC. Using the extended sine law, we have FD′ = 2r sin ∠FDD′.Therefore,

sin ∠BAD′ =2r

AD′· sin2 ∠FDD′.

Do this for all the factors in (†), and we get

sin ∠BAD′

sin ∠D′AC· sin ∠CBE′

sin ∠E′BA· sin ∠ACF ′

sin ∠F ′CB=(

sin ∠FDD′

sin ∠D′DE· sin ∠DEE′

sin ∠E′EF· sin ∠EFF ′

sin ∠F ′FD

)2

Since DD′, EE′, FF ′ are concurrent, the above expression equals to 1 by using trig Ceva ontriangle DEF . The result follows.

Remark: This result is known as Steinbart Theorem. Beware that its converse is not com-pletely true. For more information and discussion, see Darij Grinberg’s paper “Variations ofthe Steinbart Theorem” at http://de.geocities.com/darij_grinberg/.

Glossary and some possibly useful facts

• A set of points is collinear if they lie on a common line. A set of lines is concurrent if theypass through a common point. A set of points are concyclic if they lie on a common circle.

6

• Given ABC a triangle, the three angle bisectors are concurrent at the incenter of the triangle.The incenter is the center of the incircle, which is the unique circle inscribed in ABC, tangentto all three sides.

• Ceva’s theorem states that given ABC a triangle, and points X, Y, Z on sides BC, CA, AB,respectively, the lines AX, BY, CZ are concurrent if and only if

BX

XB· CY

Y A· AZ

ZB= 1.

• “Trig” Ceva states that given ABC a triangle, and points X, Y, Z inside the triangle, thelines AX, BY, CZ are concurrent if and only if

sin ∠BAX

sin ∠XAC· sin ∠CBY

sin ∠Y BA· sin ∠ACZ

sin ∠ZCB= 1.

7

11th Annual Harvard-MIT Mathematics Tournament

Saturday 23 February 2008

Team Round: B Division

Tropical Mathematics [95]

For real numbers x and y, let us consider the two operations ⊕ and � defined by

x⊕ y = min(x, y) and x� y = x + y.

We also include∞ in our set, and it satisfies x⊕∞ = x and x�∞ =∞ for all x. When unspecified,� precedes ⊕ in the order of operations.

1. [10] (Distributive law) Prove that (x⊕ y)� z = x� z ⊕ y � z for all x, y, z ∈ R ∪ {∞}.

2. [10] (Freshman’s Dream) Let zn denote z � z � z � · · · � z with z appearing n times. Provethat (x⊕ y)n = xn ⊕ yn for all x, y ∈ R ∪ {∞} and positive integer n.

3. [35] By a tropical polynomial we mean a function of the form

p(x) = an � xn ⊕ an−1 � xn−1 ⊕ · · · ⊕ a1 � x⊕ a0,

where exponentiation is as defined in the previous problem.

Let p be a tropical polynomial. Prove that

p

(x + y

2

)≥ p(x) + p(y)

2

for all x, y ∈ R ∪ {∞}. (This means that all tropical polynomials are concave.)

4. [40] (Fundamental Theorem of Algebra) Let p be a tropical polynomial:

p(x) = an � xn ⊕ an−1 � xn−1 ⊕ · · · ⊕ a1 � x⊕ a0, an 6=∞

Prove that we can find r1, r2, . . . , rn ∈ R ∪ {∞} so that

p(x) = an � (x⊕ r1)� (x⊕ r2)� · · · � (x⊕ rn)

for all x.

Juggling [125]

A juggling sequence of length n is a sequence j(·) of n nonnegative integers, usually written as astring

j(0)j(1) . . . j(n− 1)

such that the mapping f : Z→ Z defined by

f(t) = t + j(t)

1

is a permutation of the integers. Here t denotes the remainder of t when divided by n. In this case,we say that f is the corresponding juggling pattern.

For a juggling pattern f (or its corresponding juggling sequence), we say that it has b balls if the per-mutation induces b infinite orbits on the set of integers. Equivalently, b is the maximum number suchthat we can find a set of b integers {t1, t2, . . . , tb} so that the sets {ti, f(ti), f(f(ti)), f(f(f(ti))), . . . }are all infinite and mutually disjoint (i.e. non-overlapping) for i = 1, 2, . . . , b. (This definition willbecome clear in a second.)

Now is probably a good time to pause and think about what all this has to do with juggling.Imagine that we are juggling a number of balls, and at time t, we toss a ball from our hand up toa height j(t). This ball stays up in the air for j(t) units of time, so that it comes back to our handat time f(t) = t + j(t). Then, the juggling pattern presents a simplified model of how balls arejuggled (for instance, we ignore information such as which hand we use to toss the ball). A throwheight of 0 (i.e., j(t) = 0 and f(t) = t) represents that no thrown takes place at time t, which couldcorrespond to an empty hand. Then, b is simply the minimum number of balls needed to carry outthe juggling.

The following graphical representation may be helpful to you. On a horizontal line, an curve isdrawn from t to f(t). For instance, the following diagram depicts the juggling sequence 441 (or thejuggling sequences 414 and 144). Then b is simply the number of contiguous “paths” drawn, whichis 3 in this case.

14 4 1 4 4 1 4 4 4 4

Figure 1: Juggling diagram of 441.

5. [10] Prove that 572 is not a juggling sequence.

6. [40] Suppose that j(0)j(1) · · · j(n − 1) is a valid juggling sequence. For i = 0, 1, . . . , n − 1,Let ai denote the remainder of j(i) + i when divided by n. Prove that (a0, a1, . . . , an−1) is apermutation of (0, 1, . . . , n− 1).

7. [30] Determine the number of juggling sequences of length n with exactly 1 ball.

8. [40] Prove that the number of balls b in a juggling sequence j(0)j(1) · · · j(n− 1) is simply theaverage

b =j(0) + j(1) + · · ·+ j(n− 1)

n.

9. [5] Show that the converse of the previous statement is false by providing a non-jugglingsequence j(0)j(1)j(2) of length 3 where the average 1

3(j(0) + j(1) + j(2)) is an integer. Showthat your example works.

2

Incircles [180]

In the following problems, ABC is a triangle with incenter I. Let D,E, F denote the points wherethe incircle of ABC touches sides BC, CA, AB, respectively.

A

B CD

E

F

I

At the end of this section you can find some terminology and theorems that may be helpful to you.

10. [15] Let a, b, c denote the side lengths of BC, CA, AB. Find the lengths of AE,BF, CD interms of a, b, c.

11. [15] Show that lines AD,BE, CF pass through a common point.

12. [35] Show that the incenter of triangle AEF lies on the incircle of ABC.

13. [35] Let A1, B1, C1 be the incenters of triangle AEF, BDF,CDE, respectively. Show thatA1D,B1E,C1F all pass through the orthocenter of A1B1C1.

14. [40] Let X be the point on side BC such that BX = CD. Show that the excircle ABCopposite of vertex A touches segment BC at X.

15. [40] Let X be as in the previous problem. Let T be the point diametrically opposite to D onon the incircle of ABC . Show that A, T, X are collinear.

Glossary and some possibly useful facts

• A set of points is collinear if they lie on a common line. A set of lines is concurrent if theypass through a common point.

• Given ABC a triangle, the three angle bisectors are concurrent at the incenter of the triangle.The incenter is the center of the incircle, which is the unique circle inscribed in ABC, tangentto all three sides.

• The excircles of a triangle ABC are the three circles on the exterior the triangle but tangentto all three lines AB, BC, CA.

3

• The orthocenter of a triangle is the point of concurrency of the three altitudes.

• Ceva’s theorem states that given ABC a triangle, and points X, Y, Z on sides BC, CA, AB,respectively, the lines AX, BY, CZ are concurrent if and only if

BX

XB· CY

Y A· AZ

ZB= 1.

4

11th Annual Harvard-MIT Mathematics Tournament

Saturday 23 February 2008

Team Round: B Division

Tropical Mathematics [95]

For real numbers x and y, let us consider the two operations ⊕ and � defined by

x⊕ y = min(x, y) and x� y = x + y.

We also include∞ in our set, and it satisfies x⊕∞ = x and x�∞ =∞ for all x. When unspecified,� precedes ⊕ in the order of operations.

1. [10] (Distributive law) Prove that (x⊕ y)� z = x� z ⊕ y � z for all x, y, z ∈ R ∪ {∞}.

Solution: This is equivalent to proving that

min(x, y) + z = min(x + z, y + z).

Consider two cases. If x ≤ y, then LHS = x+z and RHS = x+z. If x > y, then LHS = y+zand RHS = y + z. It follows that LHS = RHS.

2. [10] (Freshman’s Dream) Let zn denote z � z � z � · · · � z with z appearing n times. Provethat (x⊕ y)n = xn ⊕ yn for all x, y ∈ R ∪ {∞} and positive integer n.

Solution: Without loss of generality, suppose that x ≤ y, then LHS = min(x, y)n = xn =nx, and RHS = min(xn, yn) = min(nx, ny) = nx.

3. [35] By a tropical polynomial we mean a function of the form

p(x) = an � xn ⊕ an−1 � xn−1 ⊕ · · · ⊕ a1 � x⊕ a0,

where exponentiation is as defined in the previous problem.

Let p be a tropical polynomial. Prove that

p

(x + y

2

)≥ p(x) + p(y)

2

for all x, y ∈ R ∪ {∞}. (This means that all tropical polynomials are concave.)

Solution: First, note that for any x1, . . . , xn, y1, . . . , yn, we have

min{x1 + y1, x2 + y2, . . . , xn + yn} ≥ min{x1, x2, . . . , xn}+ min{y1, y2, . . . , yn}.

Indeed, suppose that xm + ym = mini{xi + yi}, then xm ≥ mini xi and ym ≥ mini yi, and somini{xi + yi} = xm + ym ≥ mini xi + mini yi.

Now, let us write a tropical polynomial in a more familiar notation. We have

p(x) = min0≤k≤n

{ak + kx}.

1

So

p

(x + y

2

)= min

0≤k≤n

{ak + k

(x + y

2

)}=

12

min0≤k≤n

{(ak + kx) + (ak + ky)}

≥ 12

(min

0≤k≤n{ak + kx}+ min

0≤k≤n{ak + ky}

)=

12

(p(x) + p(y)) .

4. [40] (Fundamental Theorem of Algebra) Let p be a tropical polynomial:

p(x) = an � xn ⊕ an−1 � xn−1 ⊕ · · · ⊕ a1 � x⊕ a0, an 6=∞

Prove that we can find r1, r2, . . . , rn ∈ R ∪ {∞} so that

p(x) = an � (x⊕ r1)� (x⊕ r2)� · · · � (x⊕ rn)

for all x.

Solution: Again, we havep(x) = min

0≤k≤n{ak + kx}.

So the graph of y = p(x) can be drawn as follows: first, draw all the lines y = ak + kx,k = 0, 1, . . . , n, then trace out the lowest broken line, which then is the graph of y = p(x).

So p(x) is piecewise linear and continuous, and has slopes from the set {0, 1, 2, . . . , n}. Weknow from the previous problem that p(x) is concave, and so its slope must be decreasing(this can also be observed simply from the drawing of the graph of y = p(x)). Then, let rk

denote the x-coordinate of the leftmost kink such that the slope of the graph is less than kto the right of this kink. Then, rn ≤ rn−1 ≤ · · · ≤ r1, and for rk−1 ≤ x ≤ rk, the graph ofp is linear with slope k. Note that is if possible that rk−1 = rk, if no segment of p has slopek. Also, since an 6=∞, the leftmost piece of p(x) must have slope n, and thus rn exists, andthus all ri exist.

Now, compare p(x) with

q(x) = an � (x⊕ r1)� (x⊕ r2)� · · · � (x⊕ rn)= an + min(x, r1) + min(x, r2) + · · ·+ min(x, rn).

For rk−1 ≤ x ≤ rk, the slope of q(x) is k, and for x ≤ rn the slope of q is n and for x ≥ r1

the slope of q is 0. So q is piecewise linear, and of course it is continuous. It follows that thegraph of q coincides with that of p up to a translation. By taking any x < rn, we see thatq(x) = an + nx = p(x), we see that the graphs of p and q coincide, and thus they must bethe same function.

2

Juggling [125]

A juggling sequence of length n is a sequence j(·) of n nonnegative integers, usually written as astring

j(0)j(1) . . . j(n− 1)

such that the mapping f : Z→ Z defined by

f(t) = t + j(t)

is a permutation of the integers. Here t denotes the remainder of t when divided by n. In this case,we say that f is the corresponding juggling pattern.

For a juggling pattern f (or its corresponding juggling sequence), we say that it has b balls if the per-mutation induces b infinite orbits on the set of integers. Equivalently, b is the maximum number suchthat we can find a set of b integers {t1, t2, . . . , tb} so that the sets {ti, f(ti), f(f(ti)), f(f(f(ti))), . . . }are all infinite and mutually disjoint (i.e. non-overlapping) for i = 1, 2, . . . , b. (This definition willbecome clear in a second.)

Now is probably a good time to pause and think about what all this has to do with juggling.Imagine that we are juggling a number of balls, and at time t, we toss a ball from our hand up toa height j(t). This ball stays up in the air for j(t) units of time, so that it comes back to our handat time f(t) = t + j(t). Then, the juggling pattern presents a simplified model of how balls arejuggled (for instance, we ignore information such as which hand we use to toss the ball). A throwheight of 0 (i.e., j(t) = 0 and f(t) = t) represents that no thrown takes place at time t, which couldcorrespond to an empty hand. Then, b is simply the minimum number of balls needed to carry outthe juggling.

The following graphical representation may be helpful to you. On a horizontal line, an curve isdrawn from t to f(t). For instance, the following diagram depicts the juggling sequence 441 (or thejuggling sequences 414 and 144). Then b is simply the number of contiguous “paths” drawn, whichis 3 in this case.

14 4 1 4 4 1 4 4 4 4

Figure 1: Juggling diagram of 441.

5. [10] Prove that 572 is not a juggling sequence.

Solution: We are given j(0) = 5, j(1) = 7 and j(2) = 2. So f(3) = 3 + j(0) = 8 andf(1) = 1 + j(1) = 8. Thus f(3) = f(1) and so f is not a permutation of Z, and hence 572 isnot a juggling pattern. (In other words, there is a “collision” at times t ≡ 2 (mod 3).)

3

6. [40] Suppose that j(0)j(1) · · · j(n − 1) is a valid juggling sequence. For i = 0, 1, . . . , n − 1,Let ai denote the remainder of j(i) + i when divided by n. Prove that (a0, a1, . . . , an−1) is apermutation of (0, 1, . . . , n− 1).

Solution: Suppose that ai = j(i) + i− bin, where bi is an integer. Note that f(i− bin) =i − bin + j(i) = ai. Since {i − bin | i = 0, 1, . . . , n − 1} contains n distinct integers (as theirresidue mod n are all distinct), and f is a permutation, we see that after applying the mapf , the resulting set {a0, a1, . . . , an−1} is a set of n distinct integers. Since 0 ≤ ai < n fromdefinition, we see that (a0, a1, . . . , an−1) is a permutation of (0, 1, . . . , n− 1).

7. [30] Determine the number of juggling sequences of length n with exactly 1 ball.

Answer: 2n − 1. Solution: With 1 ball, we simply need to decide at times should the

ball land in our hand. That is, we need to choose a non-empty subset of {0, 1, 2, . . . , n − 1}where the ball lands. It follows that the answer is 2n − 1.

8. [40] Prove that the number of balls b in a juggling sequence j(0)j(1) · · · j(n− 1) is simply theaverage

b =j(0) + j(1) + · · ·+ j(n− 1)

n.

Solution: Consider the corresponding juggling diagram. Say the length of an curve from tto f(t) is f(t) − t. Let us draw only the curves whose left endpoint lies inside [0, Mn − 1].For every single ball, the sum of the lengths of the arrows drawn corresponding to thatball is between Mn − J and Mn + J , where J = max{j(0), j(1), . . . , j(n − 1)}. It followsthat the sum of the lengths of the arrows drawn is between b(Mn − J) and b(Mn + J).Since the arrow drawn at t has length j(t), the sum of the lengths of the arrows drawn isM(j(0) + j(1) + · · ·+ j(n− 1)). It follows that

b(Mn− J) ≤M(j(0) + j(1) + · · ·+ j(n− 1)) ≤ b(Mn + J).

Dividing by Mn, we get

b

(1− J

nM

)≤ j(0) + j(1) + · · ·+ j(n− 1)

n≤ b

(1 +

J

nM

).

Since we can take M to be arbitrarily large, we must have

b =j(0) + j(1) + · · ·+ j(n− 1)

n,

as desired.

9. [5] Show that the converse of the previous statement is false by providing a non-jugglingsequence j(0)j(1)j(2) of length 3 where the average 1

3(j(0) + j(1) + j(2)) is an integer. Showthat your example works.

Solution: One such example is 210. This is not a juggling sequence since f(0) = f(1) = 2.

4

Incircles [180]

In the following problems, ABC is a triangle with incenter I. Let D,E, F denote the points wherethe incircle of ABC touches sides BC, CA, AB, respectively.

A

B CD

E

F

I

At the end of this section you can find some terminology and theorems that may be helpful to you.

10. [15] Let a, b, c denote the side lengths of BC, CA, AB. Find the lengths of AE,BF, CD interms of a, b, c.

Solution: Let x = AE = AF , y = BD = BF , z = CD = CE. Since BC = BD + CD, wehave a = x+y. Similarly with the other sides, we arrive at the following system of equations:

a = y + z, b = x + z, c = x + y.

Solving this system gives us

AE = x =b + c− a

2,

BF = y =a + c− b

2,

CD = z =a + b− c

2.

11. [15] Show that lines AD,BE, CF pass through a common point.

Solution: Using Ceva’s theorem on triangle ABC, we see that it suffices to show that

BD

DC· CE

EA· AF

FB= 1.

Since AF = AE, BD = BF , and CD = CE (due to equal tangents), we see that the LHS isindeed 1.

Remark: The point of concurrency is known as the Gergonne point.

12. [35] Show that the incenter of triangle AEF lies on the incircle of ABC.

Solution: Let segment AI meet the incircle at A1. Let us show that A1 is the incenter ofAEF .

5

A

B CD

E

F

A1

I

Since AE = AF and AA′ is the angle bisector of ∠EAF , we find that A1E = A1F . Usingtangent-chord, we see that ∠AFA1 = ∠A1EF = ∠A1FE. Therefore, A1 lies on the anglebisector of ∠AFE. Since A1 also lies on the angle bisector of ∠EAF , A1 must be the incenterof AEF , as desired.

13. [35] Let A1, B1, C1 be the incenters of triangle AEF, BDF,CDE, respectively. Show thatA1D,B1E,C1F all pass through the orthocenter of A1B1C1.

Solution: Using the result from the previous problem, we see that A1, B1, C1 are respectivelythe midpoints of the arc FE, FD,DF of the incircle. We have

D

E

F

A1

B1C1

I

∠DA1C1 + ∠B1C1A1 =12∠DIC1 +

12∠B1IF +

12∠FIA1

=14

(∠EID + ∠DIF + ∠FIE)

=14· 360◦

= 90◦.

It follows that A1D is perpendicular to B1C1, and thus A1D passes through the orthocenterof A1B1C1. Similarly, A1D,B1E,C1F all pass through the orthocenter of A1B1C1.

14. [40] Let X be the point on side BC such that BX = CD. Show that the excircle ABCopposite of vertex A touches segment BC at X.

Solution: Let the excircle touch lines BC, AC and AB at X ′, Y and Z, respectively. Usingthe equal tangent property repeatedly, we have

BX ′ −X ′C = BZ − CY = (EY − CY )− (FZ −BZ) = CE −BF = CD −BD.

It follows that BX ′ = CD, and thus X ′ = X. So the excircle touches BC at X.

6

A

BC

D

E

F

X

Y

Z

T

15. [40] Let X be as in the previous problem. Let T be the point diametrically opposite to D onon the incircle of ABC . Show that A, T, X are collinear.

Solution: Consider a dilation centered at A that carries the incircle to the excircle. Thisdilation must send the diameter DT to some the diameter of excircle that is perpendicularto BC. The only such diameter is the one goes through X. It follows that T gets carried toX. Therefore, A, T, X are collinear.

Glossary and some possibly useful facts

• A set of points is collinear if they lie on a common line. A set of lines is concurrent if theypass through a common point.

• Given ABC a triangle, the three angle bisectors are concurrent at the incenter of the triangle.The incenter is the center of the incircle, which is the unique circle inscribed in ABC, tangentto all three sides.

• The excircles of a triangle ABC are the three circles on the exterior the triangle but tangentto all three lines AB, BC, CA.

• The orthocenter of a triangle is the point of concurrency of the three altitudes.

7

• Ceva’s theorem states that given ABC a triangle, and points X, Y, Z on sides BC, CA, AB,respectively, the lines AX, BY, CZ are concurrent if and only if

BX

XB· CY

Y A· AZ

ZB= 1.

8

1st

Annual Harvard-MIT November Tournament

Saturday 8 November 2008

Individual Round

1. [2] Find the minimum of x2− 2x over all real numbers x.

2. [3] What is the units digit of 72009?

3. [3] How many diagonals does a regular undecagon (11-sided polygon) have?

4. [4] How many numbers between 1 and 1, 000, 000 are perfect squares but not perfect cubes?

5. [5] Joe has a triangle with area√

3. What’s the smallest perimeter it could have?

6. [5] We say “s grows to r” if there exists some integer n > 0 such that sn = r. Call a real number r

“sparse” if there are only finitely many real numbers s that grow to r. Find all real numbers that aresparse.

7. [6] Find all ordered pairs (x, y) such that

(x − 2y)2 + (y − 1)2 = 0.

8. [7] How many integers between 2 and 100 inclusive cannot be written as m · n, where m and n haveno common factors and neither m nor n is equal to 1? Note that there are 25 primes less than 100.

9. [7] Find the product of all real x for which

23x+1− 17 · 22x + 2x+3 = 0.

10. [8] Find the largest positive integer n such that n3 + 4n2− 15n − 18 is the cube of an integer.

1st

Annual Harvard-MIT November Tournament

Saturday 8 November 2008

Individual Round

1. [2] Find the minimum of x2 − 2x over all real numbers x.

Answer: -1 Write x2 − 2x = x2 − 2x + 1− 1 = (x− 1)2 − 1. Since (x− 1)2 ≥ 0, it is clear that theminimum is −1.

Alternate method: The graph of y = x2 − 2x is a parabola that opens up. Therefore, the minimum

occurs at its vertex, which is at −b

2a= −(−2)

2 = 1. But 12 − 2 · 1 = −1, so the minimum is −1.

2. [3] What is the units digit of 72009?

Answer: 7 Note that the units digits of 71, 72, 73, 74, 75, 76, . . . follows the pattern 7, 9, 3, 1, 7, 9, 3, 1, . . ..The 2009th term in this sequence should be 7.

Alternate method: Note that the units digit of 74 is equal to 1, so the units digit of (74)502 is also1. But (74)502 = 72008, so the units digit of 72008 is 1, and therefore the units digit of 72009 is 7.

3. [3] How many diagonals does a regular undecagon (11-sided polygon) have?

Answer: 44 There are 8 diagonals coming from the first vertex, 8 more from the next, 7 from thenext, 6 from the next, 5 from the next, etc., and 1 from the last, for 8+8+7+6+5+4+3+2+1 = 44total.

Third method: Each vertex has 8 diagonals touching it. There are 11 vertices. Since each diagonaltouches two vertices, this counts every diagonal twice, so there are 8·11

2 = 44 diagonals.

4. [4] How many numbers between 1 and 1, 000, 000 are perfect squares but not perfect cubes?

Answer: 990 1000000 = 10002 = 106. A number is both a perfect square and a perfect cube if andonly if it is exactly a perfect sixth power. So, the answer is the number of perfect squares, minus thenumber of perfect sixth powers, which is 1000 − 10 = 990.

5. [5] Joe has a triangle with area√

3. What’s the smallest perimeter it could have?

Answer: 6 The minimum occurs for an equilateral triangle. The area of an equilateral triangle

with side-length s is√

34 s2, so if the area is

√3 then s =

√√3 4√

3= 2. Multiplying by 3 to get the

perimeter yields the answer.

6. [5] We say “s grows to r” if there exists some integer n > 0 such that sn = r. Call a real number r

“sparse” if there are only finitely many real numbers s that grow to r. Find all real numbers that aresparse.

Answer: -1,0,1 For any number x, other than these 3, x, 3√

x, 5√

x, 7√

x, . . . provide infinitely many

possible values of s, so these are the only possible sparse numbers. On the other hand, −1 is the onlypossible value of s for r = −1, 0 is the only value for r = 0, and −1 and 1 are the only values for r = 1.Therefore, −1, 0, and 1 are all sparse.

7. [6] Find all ordered pairs (x, y) such that

(x − 2y)2 + (y − 1)2 = 0.

Answer: (2,1) The square of a real number is always at least 0, so to have equality we must have

(x − 2y)2 = 0 and (y − 1)2 = 0. Then y = 1 and x = 2y = 2.

1

8. [7] How many integers between 2 and 100 inclusive cannot be written as m · n, where m and n haveno common factors and neither m nor n is equal to 1? Note that there are 25 primes less than 100.

Answer: 35 A number cannot be written in the given form if and only if it is a power of a prime.We can see this by considering the prime factorization. Suppose that k = pe1

1 pe2

2 · · · pen

n, with p1, . . . , pn

primes. Then we can write m = pe1

1 and n = pe2

2 · · · pen

n. So, we want to find the powers of primes that

are less than or equal to 100. There are 25 primes, as given in the problem statement. The squaresof primes are 22, 32, 52, 72. The cubes of primes are 23, 33. The fourth powers of primes are 24, 34.The fifth powers of primes are 25, The sixth powers of primes are 26. There are no seventh or higherpowers of primes between 2 and 100. This adds 10 non-primes to the list, so that in total there are10 + 25 = 35 such integers.

9. [7] Find the product of all real x for which

23x+1 − 17 · 22x + 2x+3 = 0.

Answer: -3 We can re-write the equation as 2x(2·(2x)2−17·(2x)+8) = 0, or 2·(2x)2−17·(2x)+8 = 0.Make the substitution y = 2x. Then we have 2y2 − 17y + 8 = 0, which has solutions (by the quadratic

formula) y = 17±√

289−644 = 17±15

4 = 8, 12 , so 2x = 8, 1

2 and x = 3,−1. The product of these numbers is−3.

10. [8] Find the largest positive integer n such that n3 + 4n2 − 15n − 18 is the cube of an integer.

Answer: 19 Note that the next cube after n3 is (n + 1)3 = n3 + 3n2 + 3n + 1. After that, it is(n+2)3 = n3+6n2+12n+8. n3+6n3+12n+8 is definitely bigger than n3+4n2−15n−18, so the largestcube that n3 +4n2−15n−18 could be is (n+1)3. On the other hand, for n ≥ 4, n3 +4n2−15n−18 islarger than (n−2)3 = n3−6n2+12n−8 (as 4n2−15n−18 > −6n2+12n−8 ⇐⇒ 10n2−27n−10 > 0,which is true for n ≥ 4).

So, we will check for all solutions to n3 + 4n2 − 15n − 18 = (n− 1)3, n3, (n + 1)3. The first case yields

n3 + 4n2 − 15n − 18 = n3 − 3n2 + 3n − 1 ⇐⇒ 7n2 − 18n − 17 = 0

which has no integer solutions. The second case yields

n3 + 4n2 − 15n − 18 = n3 ⇐⇒ 4n2 − 15n − 18 = 0

which also has no integer solutions. The final case yields

n3 + 4n2 − 15n − 18 = n3 + 3n2 + 3n + 1 ⇐⇒ n2 − 18n − 19 = 0

which has integer solutions n = −1, 19. So, the largest possible n is 19.

Remark: The easiest way to see that the first two polynomials have no integer solutions is using theRational Root Theorem, which states that the rational solutions of a polynomial axn + . . . + b are allof the form ± b

′

a′, where b′ divides b and a′ divides a.

2

1st Annual Harvard-MIT November TournamentSaturday 8 November 2008

Theme Round

Triangular Areas [25]

Triangles have many properties. Calculate some things about their area.

1. [3] A triangle has sides of length 9, 40, and 41. What is its area?

2. [4] Let ABC be a triangle, and let M be the midpoint of side AB. If AB is 17 units long and CM is8 units long, find the maximum possible value of the area of ABC.

3. [5] Let DEF be a triangle and H the foot of the altitude from D to EF . If DE = 60, DF = 35, andDH = 21, what is the difference between the minimum and the maximum possible values for the areaof DEF?

4. [6] Right triangle XY Z, with hypotenuse Y Z, has an incircle of radius 38 and one leg of length 3. Find

the area of the triangle.

5. [7] A triangle has altitudes of length 15, 21, and 35. Find its area.

Chessboards [25]

Joe B. is playing with some chess pieces on a 6× 6 chessboard. Help him find out some things.

1. [3] Joe B. first places the black king in one corner of the board. In how many of the 35 remainingsquares can he place a white bishop so that it does not check the black king?

2. [4] Joe B. then places a white king in the opposite corner of the board. How many total ways can heplace one black bishop and one white bishop so that neither checks the king of the opposite color?

3. [5] Joe B. now clears the board. How many ways can he place 3 white rooks and 3 black rooks on theboard so that no two rooks of opposite color can attack each other?

4. [6] Joe B. is frustrated with chess. He breaks the board, leaving a 4 × 4 board, and throws 3 blackknights and 3 white kings at the board. Miraculously, they all land in distinct squares! What is theexpected number of checks in the resulting position? (Note that a knight can administer multiplechecks and a king can be checked by multiple knights.)

5. [7] Suppose that at some point Joe B. has placed 2 black knights on the original board, but gets boredof chess. He now decides to cover the 34 remaining squares with 17 dominos so that no two overlapand the dominos cover the entire rest of the board. For how many initial arrangements of the twopieces is this possible?

Note: Chess is a game played with pieces of two colors, black and white, that players can move between squares on a rectangular grid. Some of the piecesmove in the following ways:

• Bishop: This piece can move any number of squares diagonally if there are no other pieces along its path.

• Rook: This piece can move any number of squares either vertically or horizontally if there are no other pieces along its path.

• Knight: This piece can move either two squares along a row and one square along a column or two squares along a column and one square along arow.

• King: This piece can move to any open adjacent square (including diagonally).

If a piece can move to a square occupied by a king of the opposite color, we say that it is checking the king.

If a piece moves to a square occupied by another piece, this is called attacking.

1

1st Annual Harvard-MIT November TournamentSaturday 8 November 2008

Theme Round

Triangular Areas [25]

Triangles have many properties. Calculate some things about their area.

1. [3] A triangle has sides of length 9, 40, and 41. What is its area?

Answer: 180 Observe that 92 +402 = 412, so this triangle is right and therefore has area 12 ·9 ·40 =

180.

2. [4] Let ABC be a triangle, and let M be the midpoint of side AB. If AB is 17 units long and CM is8 units long, find the maximum possible value of the area of ABC.

Answer: 68 Let h be the length of the altitude from C to AB. Observe that K = 12 · h · AB ≤

12 · CM ·AB = 68 and that equality is achieved when CM ⊥ AB.

3. [5] Let DEF be a triangle and H the foot of the altitude from D to EF . If DE = 60, DF = 35, andDH = 21, what is the difference between the minimum and the maximum possible values for the areaof DEF?

Answer: 588 Observe that the two possible configurations come from DEF obtuse and DEF acute.In either case, we have that HF =

√352 − 212 = 28 and EH =

√602 − 212 = 9

√39. This means that

HF −EH, so in the two cases the values of EF are FH + EH and EH −FH. The difference in areais hence 1

2 · 2 · FH ·DH = 28 · 21 = 588.

4. [6] Right triangle XY Z, with hypotenuse Y Z, has an incircle of radius 38 and one leg of length 3. Find

the area of the triangle.

Answer: 2116 Let the other leg have length x. Then the tangents from Y and Z to the incircle have

length x− 38 and 3− 3

8 . So the hypotenuse has length x+ 94 , the semiperimeter of the triangle is x+ 21

8 ,and the area of the triangle is 3

8 (x + 218 ). But the area can also be calculated as 3x

2 . Setting theseexpressions equal, we find x = 7

8 and the area is 2116 .

5. [7] A triangle has altitudes of length 15, 21, and 35. Find its area.

Answer: 245√

3 If A is the area of the triangle, the sides are 2A15 , 2A

21 , and 2A35 . So the triangle is

similar to a 115 , 1

21 , 135 triangle, which is similar to a 3, 5, 7 triangle. Let the sides be 3k, 5k, and 7k.

Then the angle between the sides of length 3k and 7k is 120◦, so the area is 15√

34 k2. But the area can

also be calculated as (3k)(35)2 = 105k

2 . Setting these values equal, k = 14√

33 and the area is 245

√3.

Chessboards [25]

Joe B. is playing with some chess pieces on a 6× 6 chessboard. Help him find out some things.

1. [3] Joe B. first places the black king in one corner of the board. In how many of the 35 remainingsquares can he place a white bishop so that it does not check the black king?

Answer: 30 Any square not on the diagonal containing the corner is a possible location, and thereare 36− 6 = 30 such squares.

1

2. [4] Joe B. then places a white king in the opposite corner of the board. How many total ways can heplace one black bishop and one white bishop so that neither checks the king of the opposite color?

Answer: 876 Observe that either both bishops are on the diagonal containing both kings or neitherare. If both are on the diagonal, each of the

(42

)= 6 choices of pairs of squares yields one possible

configuration, so there are 6 possibilities in this case. Off the diagonal, any pair of locations works,giving 30 · 29 = 870 possibilities in this case. Summing, we obtain 870 + 6 = 876 total possibilities.

3. [5] Joe B. now clears the board. How many ways can he place 3 white rooks and 3 black rooks on theboard so that no two rooks of opposite color can attack each other?

Answer: 608 Consider first placing the white rooks. They will occupy either 3 columns and 1row, 3 columns and 2 rows, 3 columns and 3 rows, 2 rows and 2 columns, 2 columns and 3 rows, or1 column and 3 rows. First note that placing the black rooks is impossible in the second, third, andfifth cases. The first case can happen in 4 · 4 ways, and each leads to a unique way to place the blackrooks. In the fourth case, we can choose the row with 2 rooks in 4 ways, place the rooks in

(42

)ways,

choose the column of the other rook in 2 ways, and place it in 3 ways, for a total of 4 ·(42

)· 2 · 3 = 144

ways to place the white rooks in this configuration; the black rooks can then be placed in any of 4possible locations, and there are 4 ways to do this, leading to 576 possibilities in this case. Finally, thesixth case is analogous to the first, giving 16 possibilities. Summing, we find 16 + 576 + 16 = 608 totalplacements.

4. [6] Joe B. is frustrated with chess. He breaks the board, leaving a 4 × 4 board, and throws 3 blackknights and 3 white kings at the board. Miraculously, they all land in distinct squares! What is theexpected number of checks in the resulting position? (Note that a knight can administer multiplechecks and a king can be checked by multiple knights.)

Answer: 95 We first compute the expected number of checks between a single knight-king pair. If

the king is located at any of the 4 corners, the knight has 2 possible checks. If the king is located inone of the 8 squares on the side of the board but not in the corner, the knight has 3 possible checks.If the king is located in any of the 4 central squares, the knight has 4 possible checks. Summing up,4 · 2 + 8 · 3 + 4 · 4 = 48 of the 16 · 15 knight-king positions yield a single check, so each pair yields

4816·15 = 1

5 expected checks. Now, note that each of the 9 knight-king pairs is in each of 16 · 15 possiblepositions with equal probability, so by linearity of expectation the answer is 9 · 1

5 = 95 .

5. [7] Suppose that at some point Joe B. has placed 2 black knights on the original board, but gets boredof chess. He now decides to cover the 34 remaining squares with 17 dominos so that no two overlapand the dominos cover the entire rest of the board. For how many initial arrangements of the twopieces is this possible?

Answer: 324 Color the squares of the board red and blue in a checkerboard pattern, and observethat any domino will cover exactly one red square and one blue square. Therefore, if the two knightscover squares of the same color, this is impossible. We now claim that it is always possible if theycover squares of opposite colors, which will give an answer of 182 = 324. Consider the rectangle R withthe knights at its corners. Because the knights cover differently colored squares, R must have one sidelength odd and one side length even. Therefore, the 4 lines bounding R cut the original board into Rand up to 8 other rectangles, which can be put together into rectangles with at least one side even.These rectangles can be tiled, and it is easy to see that R can be tiled, proving the claim.

Note: Chess is a game played with pieces of two colors, black and white, that players can move between squares on a rectangular grid. Some of the piecesmove in the following ways:

• Bishop: This piece can move any number of squares diagonally if there are no other pieces along its path.

• Rook: This piece can move any number of squares either vertically or horizontally if there are no other pieces along its path.

• Knight: This piece can move either two squares along a row and one square along a column or two squares along a column and one square along arow.

• King: This piece can move to any open adjacent square (including diagonally).

If a piece can move to a square occupied by a king of the opposite color, we say that it is checking the king.

If a piece moves to a square occupied by another piece, this is called attacking.

2

1st

Annual Harvard-MIT November Tournament

Saturday 8 November 2008

Team Round

Unit Fractions [100]

A unit fraction is a fraction of the form 1

n, where n is a positive integer. In this problem, you

will find out how rational numbers can be expressed as sums of these unit fractions. Even if youdo not solve a problem, you may apply its result to later problems.

We say we decompose a rational number q into unit fractions if we write q as a sum of 2 ormore distinct unit fractions. In particular, if we write q as a sum of k distinct unit fractions, wesay we have decomposed q into k fractions. As an example, we can decompose 2

3into 3 fractions:

2

3= 1

3+ 1

4+ 1

12.

1. (a) Decompose 1 into unit fractions.

(b) Decompose 1

4into unit fractions.

(c) Decompose 2

5into unit fractions.

2. Explain how any unit fraction 1

ncan be decomposed into other unit fractions.

3. (a) Write 1 as a sum of 4 distinct unit fractions.

(b) Write 1 as a sum of 5 distinct unit fractions.

(c) Show that, for any integer k > 3, 1 can be decomposed into k unit fractions.

4. Say that ab

is a positive rational number in simplest form, with a 6= 1. Further, say that n isan integer such that:

1

n>

a

b>

1

n + 1

Show that when ab− 1

n+1is written in simplest form, its numerator is smaller than a.

5. An aside: the sum of all the unit fractions

It is possible to show that, given any real M, there exists a positive integer k large enoughthat:

k∑

n=1

1

n=

1

1+

1

2+

1

3. . . > M

Note that this statement means that the infinite harmonic series,∑

∞

n=11

n, grows without

bound, or diverges. For the specific example M = 5, find a value of k, not necessarily the

smallest, such that the inequality holds. Justify your answer.

6. Now, using information from problems 4 and 5, prove that the following method to decomposeany positive rational number will always terminate:

Step 1. Start with the fraction ab. Let t1 be the largest unit fraction 1

nwhich is less than or

equal to ab.

Step 2. If we have already chosen t1 through tk, and if t1 + t2 + . . . + tk is still less than ab, then

let tk+1 be the largest unit fraction less than both tk and ab.

1

Step 3. If t1 + . . . + tk+1 equals ab, the decomposition is found. Otherwise, repeat step 2.

Why does this method never result in an infinite sequence of ti?

Juicy Numbers [100]

A juicy number is an integer j > 1 for which there is a sequence a1 < a2 < . . . < ak of positiveintegers such that ak = j and such that the sum of the reciprocals of all the ai is 1. For example,6 is a juicy number because 1

2+ 1

3+ 1

6= 1, but 2 is not juicy.

In this part, you will investigate some of the properties of juicy numbers. Remember that ifyou do not solve a question, you can still use its result on later questions.

1. Explain why 4 is not a juicy number.

2. It turns out that 6 is the smallest juicy integer. Find the next two smallest juicy numbers,and show a decomposition of 1 into unit fractions for each of these numbers. You do not needto prove that no smaller numbers are juicy.

3. Let p be a prime. Given a sequence of positive integers b1 through bn, exactly one of whichis divisible by p, show that when

1

b1

+1

b2

+ . . . +1

bn

is written as a fraction in lowest terms, then its denominator is divisible by p. Use this factto explain why no prime p is ever juicy.

4. Show that if j is a juicy integer, then 2j is juicy as well.

5. Prove that the product of two juicy numbers (not necessarily distinct) is always a juicynumber. Hint: if j1 and j2 are the two numbers, how can you change the decompositions of1 ending in 1

j1or 1

j2to make them end in 1

j1j2?

2

1st

Annual Harvard-MIT November Tournament

Saturday 8 November 2008

Team Round

Unit Fractions [100]

A unit fraction is a fraction of the form 1n, where n is a positive integer. In this problem, you

will find out how rational numbers can be expressed as sums of these unit fractions. Even if youdo not solve a problem, you may apply its result to later problems.

We say we decompose a rational number q into unit fractions if we write q as a sum of 2 ormore distinct unit fractions. In particular, if we write q as a sum of k distinct unit fractions, wesay we have decomposed q into k fractions. As an example, we can decompose 2

3 into 3 fractions:23 = 1

3 + 14 + 1

12 .

1. (a) Decompose 1 into unit fractions.

Answer: 12 + 1

3 + 16

(b) Decompose 14 into unit fractions.

Answer: 18 + 1

12 + 124

(c) Decompose 25 into unit fractions.

Answer: 15 + 1

10 + 115 + 1

30

2. Explain how any unit fraction 1n

can be decomposed into other unit fractions.

Answer: 12n

+ 13n

+ 16n

3. (a) Write 1 as a sum of 4 distinct unit fractions.

Answer: 12 + 1

3 + 17 + 1

42

(b) Write 1 as a sum of 5 distinct unit fractions.

Answer: 12 + 1

3 + 17 + 1

43 + 143·42

(c) Show that, for any integer k > 3, 1 can be decomposed into k unit fractions.

Solution: If we can do it for k fractions, simply replace the last one (say 1n) with

1n+1 + 1

n(n+1) . Then we can do it for k + 1 fractions. So, since we can do it for k = 3, wecan do it for any k > 3.

4. Say that ab

is a positive rational number in simplest form, with a 6= 1. Further, say that n isan integer such that:

1

n>

a

b>

1

n + 1

Show that when ab− 1

n+1 is written in simplest form, its numerator is smaller than a.

Solution: ab− 1

n+1 = a(n+1)−b

b(n+1) . Therefore, when we write it in simplest form, its numerator

will be at most a(n + 1) − b. We claim that a(n + 1) − b < a. Indeed, this is the same asan − b < 0 ⇐⇒ an < b ⇐⇒ b

a> n, which is given.

1

5. An aside: the sum of all the unit fractions

It is possible to show that, given any real M, there exists a positive integer k large enoughthat:

k∑

n=1

1

n=

1

1+

1

2+

1

3. . . > M

Note that this statement means that the infinite harmonic series,∑

∞

n=11n, grows without

bound, or diverges. For the specific example M = 5, find a value of k, not necessarily the

smallest, such that the inequality holds. Justify your answer.

Solution: Note that 1n+1 + 1

n+2 + . . . + 12n

> 12n

+ . . . + 12n

= 12 . Therefore, if we apply this

to n = 1, 2, 4, 8, 16, 32, 64, 128, we get

(

1

2

)

+

(

1

3+

1

4

)

+

(

1

5+

1

6+

1

7+

1

8

)

+ . . . +

(

1

129+ . . . +

1

256

)

>1

2+ . . . +

1

2= 4

so, adding in 11 , we get

256∑

n=1

1

n> 5

so k = 256 will suffice.

6. Now, using information from problems 4 and 5, prove that the following method to decomposeany positive rational number will always terminate:

Step 1. Start with the fraction ab. Let t1 be the largest unit fraction 1

nwhich is less than or

equal to ab.

Step 2. If we have already chosen t1 through tk, and if t1 + t2 + . . . + tk is still less than ab, then

let tk+1 be the largest unit fraction less than both tk and ab.

Step 3. If t1 + . . . + tk+1 equals ab, the decomposition is found. Otherwise, repeat step 2.

Why does this method never result in an infinite sequence of ti?

Solution: Let ak

bk

= ab− t1 − . . .− tk, where ak

bk

is a fraction in simplest terms. Initially, this

algorithm will have t1 = 1, t2 = 12 , t3 = 1

3 , etc. until ak

bk

< 1k+1 . This will eventually happen

by problem 5, since there exists a k such that 11 + . . . + 1

k+1 > ak

bk

. At that point, there is

some n with 1n

< tk such that 1n

> ak

bk

> 1n+1 . In this case, tk+1 = 1

n+1 .

Suppose that there exists nk such that 1nk

> ak

bk> 1

nk+1 for some k. Then we have tk+1 = 1nk+1

andak+1

bk+1< 1

nk(nk+1) . This shows that once we have found nk such that 1nk

> ak

bk> 1

nk+1 and1

nk≤ tk, we no longer have to worry about tk+1 being less than tk, since tk+1 = 1

nk+1 < 1nk

<

tk, and also nk+1 ≥ nk(nk + 1) while 1nk(nk+1) ≤ 1

nk+1 = tk+1.

On the other hand, once we have found such an nk, the sequence {ak} must be decreasingby problem 4. Since the ak are all integers, we eventually have to get to 0 (as there is noinfinite decreasing sequence of positive integers). Therefore, after some finite number of stepsthe algorithm terminates with ak+1 = 0, so 0 = ak

bk= a

b− t1 − . . . − tk, so a

b= t1 + . . . + tk,

which is what we wanted.

2

Juicy Numbers [100]

A juicy number is an integer j > 1 for which there is a sequence a1 < a2 < . . . < ak of positiveintegers such that ak = j and such that the sum of the reciprocals of all the ai is 1. For example,6 is a juicy number because 1

2 + 13 + 1

6 = 1, but 2 is not juicy.

In this part, you will investigate some of the properties of juicy numbers. Remember that ifyou do not solve a question, you can still use its result on later questions.

1. Explain why 4 is not a juicy number.

Solution: If 4 were juicy, then we would have 1 = . . .+ 14 . The . . . can only possible contain

12 and 1

3 , but it is clear that 12 + 1

4 , 13 + 1

4 , and 12 + 1

3 + 14 are all not equal to 1.

2. It turns out that 6 is the smallest juicy integer. Find the next two smallest juicy numbers,and show a decomposition of 1 into unit fractions for each of these numbers. You do not needto prove that no smaller numbers are juicy.

Answer: 12 and 15 1 = 12 + 1

4 + 16 + 1

12 , 1 = 12 + 1

3 + 110 + 1

15 .

3. Let p be a prime. Given a sequence of positive integers b1 through bn, exactly one of whichis divisible by p, show that when

1

b1+

1

b2+ . . . +

1

bn

is written as a fraction in lowest terms, then its denominator is divisible by p. Use this factto explain why no prime p is ever juicy.

Solution: We can assume that bn is the term divisible by p (i.e. bn = kp) since the orderof addition doesn’t matter. We can then write

1

b1+

1

b2+ . . . +

1

bn−1=

a

b

where b is not divisible by p (since none of the bi are). But then ab

+ 1kp

= kpa+bkpb

. Since b

is not divisible by p, kpa + b is not divisible by p, so we cannot remove the factor of p fromthe denominator. In particular, p cannot be juicy as 1 can be written as 1

1 , which has adenominator not divisible by p, whereas being juicy means we have a sum 1

b1+ . . . + 1

bn= 1,

where b1 < b2 < . . . < bn = p, and so in particular none of the bi with i < n are divisible byp.

4. Show that if j is a juicy integer, then 2j is juicy as well.

Solution: Just replace 1b1

+ . . . + 1bn

with 12 + 1

2b1+ 1

2b2+ . . . + 1

2bn. Since n > 1, 2b1 > 2.

5. Prove that the product of two juicy numbers (not necessarily distinct) is always a juicynumber. Hint: if j1 and j2 are the two numbers, how can you change the decompositions of1 ending in 1

j1or 1

j2to make them end in 1

j1j2?

Solution: Let 1 = 1b1

+ . . . + 1bn

= 1c1

+ . . . + 1cm

, where bn = j1 and cm = j2. Then

3

1 =1

b1+ . . . +

1

bn−1+

(

1

bnc1+

1

bnc2+ . . . +

1

bncm

)

and so j1j2 is juicy.

4

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

1. [5] Find the sum of all solutions for x:

xy = 1x + y = 3

2. [5] Evaluate the sum1− 2 + 3− 4 + · · ·+ 2007− 2008.

3. [5] What is the largest x such that x2 divides 24 · 35 · 46 · 57?

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

4. [6] What is the smallest prime divisor of 5710710

+ 1?

5. [6] What is the sum of all integers x such that |x + 2| ≤ 10?

6. [6] Sarah is deciding whether to visit Russia or Washington, DC for the holidays. She makes herdecision by rolling a regular 6-sided die. If she gets a 1 or 2, she goes to DC. If she rolls a 3, 4, or 5,she goes to Russia. If she rolls a 6, she rolls again. What is the probability that she goes to DC?

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

7. [7] Compute 20092 − 20082.

8. [7] Alice rolls two octahedral dice with the numbers 2, 3, 4, 5, 6, 7, 8, 9. What’s the probability the twodice sum to 11?

9. [7] Let a0 = 67 , and

an+1 =

{2an if an < 1

2

2an − 1 if an ≥ 12 .

Find a2008.

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

10. [8] Find the sum of all positive integers n such that n divides n2 + n + 2.

11. [8] Al has a rectangle of integer side lengths a and b, and area 1000. What is the smallest perimeterit could have?

12. [8] Solve the following system of equations for w.

2w + x + y + z = 1w + 2x + y + z = 2w + x + 2y + z = 2w + x + y + 2z = 1.

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

13. [9] Find the number of distinct primes dividing 1 · 2 · 3 · · · 9 · 10.

14. [9] You have a 2 × 3 grid filled with integers between 1 and 9. The numbers in each row and columnare distinct, the first row sums to 23, and the columns sum to 14, 16, and 17 respectively.

14 16 1723 a b c

x y z

What is x + 2y + 3z?

15. [9] A cat is going up a stairwell with ten stairs. However, instead of walking up the stairs one at atime, the cat jumps, going either two or three stairs up at each step (though if necessary, it will justwalk the last step). How many different ways can the cat go from the bottom to the top?

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

16. [10] If p and q are positive integers and 20082009 < p

q < 20092010 , what is the minimum value of p?

17. [10] Determine the last two digits of 1717, written in base 10.

18. [10] Find the coefficient of x6 in the expansion of

(x + 1)6 ·6∑

i=0

xi

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

19. [11] Let P be a polynomial with P (1) = P (2) = · · · = P (2007) = 0 and P (0) = 2009!. P (x) hasleading coefficient 1 and degree 2008. Find the largest root of P (x).

20. [11] You have a die with faces labelled 1 through 6. On each face, you draw an arrow to an adjacentface, such that if you start on a face and follow the arrows, after 6 steps you will have passed throughevery face once and will be back on your starting face. How many ways are there to draw the arrowsso that this is true?

21. [11] Call a number overweight if it has at least three positive integer divisors (including 1 and thenumber), and call a number obese if it has at least four positive integer divisors (including 1 and thenumber). How many positive integers between 1 and 200 are overweight, but not obese?

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

22. [12] Sandra the Maverick has 5 pairs of shoes in a drawer, each pair a different color. Every day for5 days, Sandra takes two shoes out and throws them out the window. If they are the same color, shetreats herself to a practice problem from a past HMMT. What is the expected value (average number)of practice problems she gets to do?

23. [12] If x and y are real numbers such that (x−4)2

4 + y2

9 = 1, find the largest possible value of x2

4 + y2

9 .

24. [12] Let f(x) = 11−x . Let fk+1(x) = f(fk(x)), with f1(x) = f(x). What is f2008(2008)?

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

25. [13] Evaluate the sum

cos(

2π

18

)+ cos

(4π

18

)+ · · ·+ cos

(34π

18

).

26. [13] John M. is sitting at (0, 0), looking across the aisle at his friends sitting at (i, j) for each 1 ≤ i ≤ 10and 0 ≤ j ≤ 5. Unfortunately, John can only see a friend if the line connecting them doesn’t passthrough any other friend. How many friends can John see?

27. [13] ABCDE is a regular pentagon inscribed in a circle of radius 1. What is the area of the set ofpoints inside the circle that are farther from A than they are from any other vertex?

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

28. [14] Johnny the grad student is typing all the integers from 1 to ∞, in order. The 2 on his computeris broken however, so he just skips any number with a 2. What’s the 2008th number he types?

29. [14] Let p(x) be the polynomial of degree 4 with roots 1, 2, 3, 4 and leading coefficient 1. Let q(x) bethe polynomial of degree 4 with roots 1, 1

2 , 13 , and 1

4 and leading coefficient 1. Find limx→1p(x)q(x) .

30. [14] Alice has an equilateral triangle ABC of area 1. Put D on BC, E on CA, and F on AB, withBD = DC, CE = 2EA, and 2AF = FB. Note that AD,BE, and CF pass through a single point M .What is the area of triangle EMC?

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

31. [15] Find the sum of all primes p for which there exists a prime q such that p2 + pq + q2 is a square.

32. [15] Pirate ships Somy and Lia are having a tough time. At the end of the year, they are both onepillage short of the minimum required for maintaining membership in the Pirate Guild, so they decideto pillage each other to bring their counts up. Somy by tradition only pillages 28 · 3k coins for integersk, and Lia by tradition only pillages 82 · 3j coins for integers j. Note that each pillage can have adifferent k or j. Soma and Lia work out a system where Somy pillages Lia n times, Lia pillages Somyn times, and after both sets of pillages Somy and Lia are financially even.

What is the smallest n can be?

33. [15] The polynomial ax2 − bx + c has two distinct roots p and q, with a, b, and c positive integers andwith 0 < p, q < 1. Find the minimum possible value of a.

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

34. [20] How many hits did math tournament get on Google the morning of November 8, 2008? If yousubmit integer N , and the correct answer is A, you will receive b20·min{N

A , AN }c points for this problem.

35. [25] Find max{Perimeter(T )} for T a triangle contained in a regular heptagon (7-sided figure) of unitedge length. Write your answer N to 2 places after the decimal. If the correct answer rounded to 2decimal places is A, you will receive 0 points if N > A and bmax{0, 25−50 ·(A−N)}c points otherwise.

36. [25] How many numbers less than 1,000,000 are the product of exactly 2 distinct primes? You willreceive b25− 50 · |NA − 1|c points, if you submit N and the correct answer is A.

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1st Annual Harvard-MIT November TournamentSaturday 8 November 2008

Guts Round

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND

1. [5] Find the sum of all solutions for x:

xy = 1x + y = 3

Answer: 3 Substitute 3− x in for y into the first equation:

x(3− x) = 1⇔ x2 − 3x + 1 = 0

This equation has two distinct roots, each of which corresponds to a possible solution x. The sum ofthe roots of the quadratic equation ax2 + bx + c = 0 is −b

a , which in this case is 3.

2. [5] Evaluate the sum1− 2 + 3− 4 + · · ·+ 2007− 2008.

Answer: −1004 Every odd integer term can be paired with the next even integer, and this pairsums to −1. There are 1004 such pairs, so the total sum is −1004.

3. [5] What is the largest x such that x2 divides 24 · 35 · 46 · 57?

Answer: 12 We factor the product as 24 · 32 · 5 · 7 · 19 · 23. If x2 divides this product, x can have atmost 2 factors of 2, 1 factor of 3, and no factors of any other prime. So 22 · 3 = 12 is the largest valueof x.

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND

4. [6] What is the smallest prime divisor of 5710710

+ 1?

Answer: 2 Notice that 5 to any power is odd, so this number is even. Then 2 is a prime divisor.It also happens to be the smallest prime.

5. [6] What is the sum of all integers x such that |x + 2| ≤ 10?

Answer: −42 The inequality |x + 2| ≤ 10 holds if and only if x + 2 ≤ 10 and x + 2 ≥ −10. Sox must be in the range −12 ≤ x ≤ 8. If we add up the integers in this range, each positive integercancels with its additive inverse, so the sum is equal to −12− 11− 10− 9 = −42.

6. [6] Sarah is deciding whether to visit Russia or Washington, DC for the holidays. She makes herdecision by rolling a regular 6-sided die. If she gets a 1 or 2, she goes to DC. If she rolls a 3, 4, or 5,she goes to Russia. If she rolls a 6, she rolls again. What is the probability that she goes to DC?

Answer: 25 On each roll, the probability that Sarah decides to go to Russia is 3/2 times the

probability she decides to go to DC. So, the total probability that she goes to Russia is 3/2 timesthe total probability that she goes to DC. Since these probabilities sum to 1 (they are the only twoeventual outcomes) Sarah goes to DC with probability 2

5 and Russia with probability 35 .

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1

1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND

7. [7] Compute 20092 − 20082.

Answer: 4017 Factoring this product with difference of squares, we find it equals:

(2009 + 2008)(2009− 2008) = (4017)(1) = 4017

.

8. [7] Alice rolls two octahedral dice with the numbers 2, 3, 4, 5, 6, 7, 8, 9. What’s the probability the twodice sum to 11?

Answer: 18 No matter what comes up on the first die, there is exactly one number that could

appear on the second die to make the sum 11, because 2 can be paired with 9, 3 with 8, and so on.So, there is a 1

8 chance of getting the correct number on the second die.

9. [7] Let a0 = 67 , and

an+1 =

{2an if an < 1

2

2an − 1 if an ≥ 12 .

Find a2008.

Answer: 57 We calculate the first few ai:

a1 =57, a2 =

37, a3 =

67

= a0

So this sequence repeats every three terms, so a2007 = a0 = 67 . Then a2008 = 5

7 .

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND

10. [8] Find the sum of all positive integers n such that n divides n2 + n + 2.

Answer: 3 Since n always divides n2 +n, the only n that work are divisors of 2, because if n dividesa and n divides b, then n divides a + b. So the solutions are 1 and 2 which sum to 3.

11. [8] Al has a rectangle of integer side lengths a and b, and area 1000. What is the smallest perimeterit could have?

Answer: 130 To minimize the sum of the side lengths, we need to keep the height and width asclose as possible, because the square has the smallest perimeter of all rectangles with a fixed area. So,40 and 25 multiply to 1000 and are as close as possible - the 40x25 rectangle has perimeter 130.

12. [8] Solve the following system of equations for w.

2w + x + y + z = 1w + 2x + y + z = 2w + x + 2y + z = 2w + x + y + 2z = 1.

Answer: −15 Add all the equations together to find that 5x+5y+5z+5w = 6, or x+y+z+w = 6

5 .

We can now subtract this equation from the first equation to see that w = −15 .

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND

2

13. [9] Find the number of distinct primes dividing 1 · 2 · 3 · · · 9 · 10.

Answer: 4 A prime divides this product if and only if it divides one of the multiplicands, so primedivisors of this product must be less than 10. There are 4 primes less than 10, namely, 2, 3, 5, and 7.

14. [9] You have a 2 × 3 grid filled with integers between 1 and 9. The numbers in each row and columnare distinct, the first row sums to 23, and the columns sum to 14, 16, and 17 respectively.

14 16 1723 a b c

x y z

What is x + 2y + 3z?

Answer: 49 The sum of all 6 numbers is 14 + 16 + 17 = 47, so x + y + z = 47− 23 = 24. If threedistinct digits sum to 24, they must be 7, 8, and 9, because any other triple of digits would have asmaller sum. So, we try placing these digits in for x, y, and z, and the only arrangement that does notforce equal digits in any row or column is x = 8, y = 7, z = 9. In this case, x + 2y + 3z = 49.

15. [9] A cat is going up a stairwell with ten stairs. However, instead of walking up the stairs one at atime, the cat jumps, going either two or three stairs up at each step (though if necessary, it will justwalk the last step). How many different ways can the cat go from the bottom to the top?

Answer: 12 The number of ways for the cat to get to the ith step is the number of ways for thecat to get to step i − 2 plus the number of ways to get to step i − 3, because for each way to get tostep i, we can undo the last move the cat made to go back to one of these two steps. The cat can getto step 1 in 0 ways, to step 2 in 1 way, and to step 3 in 1 way. Now we repeatedly use our formula forcalculating the number of ways to get to the ith step to see that the cat gets to:

Step 1 2 3 4 5 6 7 8 9 10Number of ways 0 1 1 1 2 2 3 4 5 7

So our answer is 5 + 7 = 12.

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND

16. [10] If p and q are positive integers and 20082009 < p

q < 20092010 , what is the minimum value of p?

Answer: 4017 By multiplying out the fraction inequalities, we find that 2008q + 1 ≤ 2009p and2010p+ ≤ 2009q. Adding 2009 times the first inequality to 2008 times the second, we find that2008 · 2009q + 4017 ≤ 2008 · 2009q + p, or p ≥ 4017. This minumum is attained when p

q = 40174019 .

17. [10] Determine the last two digits of 1717, written in base 10.

Answer: 77 We are asked to find the remainder when 1717 is divided by 100. Write the power as(7 + 10)17 and expand with the binomial theorem:

(7 + 10)17 = 717 + 17 · 716 · 10 + . . .

We can ignore terms with more than one factor of 10 because these terms are divisible by 100, soadding them does not change the last two digits. Now, 74 = 2401, which has remainder 1 mod 100, so717 has last two digits 07 and 716 · 10 has last two digits 70. We add these together.

18. [10] Find the coefficient of x6 in the expansion of

(x + 1)6 ·6∑

i=0

xi

3

Answer: 64 Each term of (x + 1)6 can be multiplied by a unique power xi, 0 ≤ i ≤ 6 to get a sixthdegree term. So the answer is the sum of the coefficients of the terms of (x + 1)6, which is the same assubstituting x = 1 into this power to get 26 = 64.

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19. [11] Let P be a polynomial with P (1) = P (2) = · · · = P (2007) = 0 and P (0) = 2009!. P (x) hasleading coefficient 1 and degree 2008. Find the largest root of P (x).

Answer: 4034072 P (0) is the constant term of P (x), which is the product of all the roots of thepolynomial, because its degree is even. So the product of all 2008 roots is 2009! and the product ofthe first 2007 is 2007!, which means the last root is 2009!

2007! = 2009 · 2008 = 4034072.

20. [11] You have a die with faces labelled 1 through 6. On each face, you draw an arrow to an adjacentface, such that if you start on a face and follow the arrows, after 6 steps you will have passed throughevery face once and will be back on your starting face. How many ways are there to draw the arrowsso that this is true?

Answer: 32 There are 4 choices for where to go from face 1. Consider the 4 faces adjacent to 1.We can visit either 1, 2, or 3 of them before visiting the face opposite 1. If we only visit one of theseadjacent faces, we have 4 choices for which one, then we visit face 6, opposite face 1, then we visit theremaining 3 faces in one of two orders - for a total of 8 ways. If we visit 2 adjacent faces first, there is8 choices for these two faces, then 2 choices for the path back from face 6 to face 1. Lastly, there are 8ways to visit three of the adjacent faces before visiting the opposite face. These choices give 32 total.

21. [11] Call a number overweight if it has at least three positive integer divisors (including 1 and thenumber), and call a number obese if it has at least four positive integer divisors (including 1 and thenumber). How many positive integers between 1 and 200 are overweight, but not obese?

Answer: 6 A positive integer is overweight, but not obese, if it has exactly 3 factors - this can onlyhappen if that integer is the square of a prime. (If two primes, p and q, divide the number, then p, q,pq, and 1 all divide it, making it at least obese). So, the integers less than 200 which are squares of aprime are the squares of 2, 3, 5, 7, 11, and 13.

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1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND

22. [12] Sandra the Maverick has 5 pairs of shoes in a drawer, each pair a different color. Every day for5 days, Sandra takes two shoes out and throws them out the window. If they are the same color, shetreats herself to a practice problem from a past HMMT. What is the expected value (average number)of practice problems she gets to do?

Answer: 59 On any given day, there is a 1

9 chance that the second shoe that Sandra chooses makes

a pair with the first shoe she chose. Thus the average number of problems she does in a day is 19 , so,

by the linearity of expectation, she does 59 problems total, on average.

23. [12] If x and y are real numbers such that (x−4)2

4 + y2

9 = 1, find the largest possible value of x2

4 + y2

9 .

Answer: 9 The first equation is an ellipse with major axis parallel to the y-axis. If the secondexpression is set equal to a certain value c, then it is also the equation of an ellipse with major axisparallel to the y-axis; further, it is similar to the first ellipse. So the largest value of c occurs whenboth ellipse are tangent on the x-axis, at x = 6, y = 0, which gives 9 as the largest value of c.

24. [12] Let f(x) = 11−x . Let fk+1(x) = f(fk(x)), with f1(x) = f(x). What is f2008(2008)?

4

Answer: −12007 Notice that, if x 6= 0, 1, then f2(x) = 1

1− 11−x

= x−1x , which means that f3(x) =

11− x−1

x

= x. So fn is periodic with period n = 3, which means that f2007(x) = x so f2008(2008) =

f(2008) = −12007 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND

25. [13] Evaluate the sum

cos(

2π

18

)+ cos

(4π

18

)+ · · ·+ cos

(34π

18

).

Answer: −1 If k < 18, then we can pair cos(

kπ18

)with cos

((18−k)π

18

), and these two terms sum to

0. If k > 18, then the pair cos(

kπ18

)and cos

((36−k)π

18

)also sums to 0. So, the only term in this series

that is left over is cos(

18π18

)= −1.

26. [13] John M. is sitting at (0, 0), looking across the aisle at his friends sitting at (i, j) for each 1 ≤ i ≤ 10and 0 ≤ j ≤ 5. Unfortunately, John can only see a friend if the line connecting them doesn’t passthrough any other friend. How many friends can John see?

Answer: 36 The simplest method is to draw a picture and count which friends he can see. Johncan see the friend on point (i, j) if and only if i and j are relatively prime.

27. [13] ABCDE is a regular pentagon inscribed in a circle of radius 1. What is the area of the set ofpoints inside the circle that are farther from A than they are from any other vertex?

Answer: π5 Draw the perpendicular bisectors of all the sides and diagonals of the pentagon with

one endpoint at A. These lines all intersect in the center of the circle, because they are the set ofpoints equidistant from two points on the circle. Now, a given point is farther from A than from pointX if it is on the X side of the perpendicular bisector of segment AX. So, we want to find the area ofthe set of all points which are separated from A by all of these perpendicular bisectors, which turnsout to be a single 72o sector of the circle, which has area π

5 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND

28. [14] Johnny the grad student is typing all the integers from 1 to ∞, in order. The 2 on his computeris broken however, so he just skips any number with a 2. What’s the 2008th number he types?

Answer: 3781 Write 2008 in base 9 as 2671, and interpret the result as a base 10 number such thatthe base 9 digits 2, 3, . . . 8 correspond to the base 10 digits 3, 4, . . . 9. This gives an answer of 3781.

29. [14] Let p(x) be the polynomial of degree 4 with roots 1, 2, 3, 4 and leading coefficient 1. Let q(x) bethe polynomial of degree 4 with roots 1, 1

2 , 13 , and 1

4 and leading coefficient 1. Find limx→1p(x)q(x) .

Answer: −24 Consider the polynomial f(x) = x4q( (1 )(x) – it has the same roots, 1, 2, 3, and 4,

as p(x). But this polynomial also has the same coefficients as q(x), just in reverse order. Its leadingcoefficient is q(0) = 1 · 1

2 ·13 ·

14 = 1

24 . So f(x) is p(x) scaled by 124 , which means that p(x)/f(x) goes

to 24 as x goes to 1, and f(x)/q(x) goes to −1.

30. [14] Alice has an equilateral triangle ABC of area 1. Put D on BC, E on CA, and F on AB, withBD = DC, CE = 2EA, and 2AF = FB. Note that AD,BE, and CF pass through a single point M .What is the area of triangle EMC?

Answer: 16 Triangles ACF and BCF share a height, so the ratio of their areas is AF/BF = 1/2.

By the same method, the ratio of the areas of AMF and BMF is 1/2. So, the ratio of the areas of

5

ACM and BCM is also 1/2. Similarly, the ratio of the areas of ABM and BCM is 1/2. But the sumof the areas of ACM , BCM , and ABM is 1, so the area of ACM is 1

4 . Then the area of EMC is 2/3the area of ACM , because they share heights, so their areas are in the same ratio as their bases. Thearea of EMC is then 2·1

3·4 = 16 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND

31. [15] Find the sum of all primes p for which there exists a prime q such that p2 + pq + q2 is a square.

Answer: 8 3 and 5 both work, because 32 + 3 · 5 + 52 = 49. Now, say p2 + pq + q2 = k2, for apositive integer k. Then (p + q)2 − k2 = pq, or:

(p + q + k)(p + q − k) = pq

Since p+ q +k is a divisor of pq, and it is greater than p and q, p+ q +k = pq. Then p+ q−k = 1. So:

2p + 2q = pq + 1⇔ pq − 2p− 2q + 4 = 3⇔ (p− 2)(q − 2) = 3

This shows that one of p and q is 3 and the other is 5.

32. [15] Pirate ships Somy and Lia are having a tough time. At the end of the year, they are both onepillage short of the minimum required for maintaining membership in the Pirate Guild, so they decideto pillage each other to bring their counts up. Somy by tradition only pillages 28 · 3k coins for integersk, and Lia by tradition only pillages 82 · 3j coins for integers j. Note that each pillage can have adifferent k or j. Soma and Lia work out a system where Somy pillages Lia n times, Lia pillages Somyn times, and after both sets of pillages Somy and Lia are financially even.

What is the smallest n can be?

Answer: 2 Clearly, n = 1 cannot be acheived, because 28 · 3k is never a multiple of 82. However,two pillages is enough: Somy pillages 28 and 28 · 81 from Lia, and Lia pillages 81 and 81 · 27 fromSomy. As is easily checked, both pillage 28 · 82.

33. [15] The polynomial ax2 − bx + c has two distinct roots p and q, with a, b, and c positive integers andwith 0 < p, q < 1. Find the minimum possible value of a.

Answer: 5 Let x and y be the roots. Then:

b

a= x + y < 2⇒ b < 2a

c

a= xy < 1⇒ c < a⇒ a > 1

b2 > 4ac > 4c2 ⇒ b > 2c

Evaluated at 1, the polynomial must be greater than 0, so a + c > b. Then:

2c < b < a + c

2c + 1 ≤ b ≤ a + c− 1

a ≥ c + 2 ≥ 3

If a = 3, then c = 1 and b = 3, by the above bounds, but this polynomial has complex roots. Similarly,if a = 4, then c = 1 and b is forced to be either 3 or 4, again giving either 0 or 1 distinct real roots. Soa ≥ 5. But the polynomial 5x2 − 5x + 1 satisfies the condition, so 5 is the answer.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

1st HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND

34. [20] How many hits did “math tournament” get on Google the morning of November 8, 2008? Ifyou submit integer N , and the correct answer is A, you will receive b20 ·min{N

A , AN }c points for this

problem.

35. [25] Find max{Perimeter(T )} for T a triangle contained in a regular septagon (7-sided figure) of unitedge length. Write your answer N to 2 places after the decimal. If the correct answer rounded to 2decimal places is A, you will receive 0 points if N < A and bmax{0, 25−50 ·(N−A)}c points otherwise.

Answer: 5.85086 Let the septagon be A0A1 . . . A6.

If x is a point that can move along the x-axis, the distance from x to a fixed point p is a convexfunction in the x-coordinate. Therefore, the sum of the distances from x to two other points is convextoo, so if x is constrained to lie on a closed line segment, its maximum value is attained at an endpoint.Therefore, the triangle of maximal perimeter has vertices at the vertices of the pentagon. The trianglewith the largest such perimeter has almost evenly spaced vertices, so triangle A0A2A4 has the maximalperimeter. It has area 5.85 . . ..

36. [25] How many numbers less than 1,000,000 are the product of exactly 2 distinct primes? You willreceive b25− 50 · |NA − 1|c points, if you submit N and the correct answer is A.

Answer: 209867 While it is difficult to compute this answer without writing a program or using acalculator, it can be approximated using the fact that the number of primes less than a positive integern is about n

log n .

7

12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Individual Round: Algebra Test

1. [3] If a and b are positive integers such that a2 − b4 = 2009, find a+ b.

2. [3] Let S be the sum of all the real coefficients of the expansion of (1 + ix)2009. What is log2(S)?

3. [4] If tanx+ tan y = 4 and cotx+ cot y = 5, compute tan(x+ y).

4. [4] Suppose a, b and c are integers such that the greatest common divisor of x2 +ax+ b and x2 + bx+ cis x + 1 (in the set of polynomials in x with integer coefficients), and the least common multiple ofx2 + ax+ b and x2 + bx+ c is x3 − 4x2 + x+ 6. Find a+ b+ c.

5. [4] Let a, b, and c be the 3 roots of x3 − x+ 1 = 0. Find 1a+1 + 1

b+1 + 1c+1 .

6. [5] Let x and y be positive real numbers and θ an angle such that θ 6= π2n for any integer n. Suppose

sin θx

=cos θy

andcos4 θx4

+sin4 θ

y4=

97 sin 2θx3y + y3x

.

Compute xy + y

x .

7. [5] Simplify the product100∏m=1

100∏n=1

xn+m + xn+m+2 + x2n+1 + x2m+1

x2n + 2xn+m + x2m.

Express your answer in terms of x.

8. [7] If a, b, x and y are real numbers such that ax + by = 3, ax2 + by2 = 7, ax3 + by3 = 16, andax4 + by4 = 42, find ax5 + by5.

9. [7] Let f(x) = x4 + 14x3 + 52x2 + 56x+ 16. Let z1, z2, z3, z4 be the four roots of f . Find the smallestpossible value of |zazb + zczd| where {a, b, c, d} = {1, 2, 3, 4}.

10. [8] Let f(x) = 2x3− 2x. For what positive values of a do there exist distinct b, c, d such that (a, f(a)),(b, f(b)), (c, f(c)), (d, f(d)) is a rectangle?

12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Individual Round: Algebra Test

1. [3] If a and b are positive integers such that a2 − b4 = 2009, find a+ b.

Answer: 47

Solution: We can factor the equation as (a − b2)(a + b2) = 41 · 49, from which it is evident thata = 45 and b = 2 is a possible solution. By examining the factors of 2009, one can see that there areno other solutions.

2. [3] Let S be the sum of all the real coefficients of the expansion of (1 + ix)2009. What is log2(S)?

Answer: 1004

Solution: The sum of all the coefficients is (1 + i)2009, and the sum of the real coefficients is the realpart of this, which is 1

2

((1 + i)2009 + (1− i)2009

)= 21004. Thus log2(S) = 1004.

3. [4] If tanx+ tan y = 4 and cotx+ cot y = 5, compute tan(x+ y).

Answer: 20

Solution: We have cotx + cot y = tan x+tan ytan x tan y , so tanx tan y = 4

5 . Thus, by the tan sum formula,tan(x+ y) = tan x+tan y

1−tan x tan y = 20.

4. [4] Suppose a, b and c are integers such that the greatest common divisor of x2 +ax+ b and x2 + bx+ cis x + 1 (in the ring of polynomials in x with integer coefficients), and the least common multiple ofx2 + ax+ b and x2 + bx+ c is x3 − 4x2 + x+ 6. Find a+ b+ c.

Answer: −6

Solution: Since x+1 divides x2+ax+b and the constant term is b, we have x2+ax+b = (x+1)(x+b),and similarly x2 +bx+c = (x+1)(x+c). Therefore, a = b+1 = c+2. Furthermore, the least commonmultiple of the two polynomials is (x+ 1)(x+ b)(x+ b−1) = x3−4x2 +x+ 6, so b = −2. Thus a = −1and c = −3, and a+ b+ c = −6.

5. [4] Let a, b, and c be the 3 roots of x3 − x+ 1 = 0. Find 1a+1 + 1

b+1 + 1c+1 .

Answer: −2

Solution: We can substitute x = y−1 to obtain a polynomial having roots a+1, b+1, c+1, namely,(y − 1)3 − (y − 1) + 1 = y3 − 3y2 + 2y + 1. The sum of the reciprocals of the roots of this polynomialis, by Viete’s formulas, 2

−1 = −2.

6. [5] Let x and y be positive real numbers and θ an angle such that θ 6= π2n for any integer n. Suppose

sin θx

=cos θy

andcos4 θx4

+sin4 θ

y4=

97 sin 2θx3y + y3x

.

1

Compute xy + y

x .

Answer: 4

Solution: From the first relation, there exists a real number k such that x = k sin θ and y = k cos θ.Then we have

cos4 θsin4 θ

+sin4 θ

cos4 θ=

194 sin θ cos θsin θ cos θ(cos2 θ + sin2 θ)

= 194.

Notice that if t = xy + y

x then (t2 − 2)2 − 2 = cos4 θsin4 θ

+ sin4 θcos4 θ = 194 and so t = 4.

7. [5] Simplify the product100∏m=1

100∏n=1

xn+m + xn+m+2 + x2n+1 + x2m+1

x2n + 2xn+m + x2m.

Express your answer in terms of x.

Answer: x9900(

1+x100

2

)2

(OR 14x

9900 + 12x

10000 + 14x

10100)

Solution: We notice that the numerator and denominator of each term factors, so the product isequal to

100∏m=1

100∏n=1

(xm + xn+1)(xm+1 + xn)(xm + xn)2

.

Each term of the numerator cancels with a term of the denominator except for those of the form(xm +x101) and (x101 +xn) for m,n = 1, . . . , 100, and the terms in the denominator which remain areof the form (x1 + xn) and (x1 + xm) for m,n = 1, . . . , 100. Thus the product simplifies to(

100∏m=1

xm + x101

x1 + xm

)2

Reversing the order of the factors of the numerator, we find this is equal to

(100∏m=1

x101−m + x101

x1 + xm

)2

=

(100∏m=1

x100−mx1 + xm+1

x1 + xm

)2

=

(x1 + x101x1 + x1

100∏m=1

x100−m

)2

= (x99·100

2 )2(

1 + x100

2

)2

as desired.

8. [7] If a, b, x, and y are real numbers such that ax + by = 3, ax2 + by2 = 7, ax3 + by3 = 16, andax4 + by4 = 42, find ax5 + by5.

Answer: 20.

Solution: We have ax3 + by3 = 16, so (ax3 + by3)(x+ y) = 16(x+ y) and thus

ax4 + by4 + xy(ax2 + by2) = 16(x+ y)

2

It follows that42 + 7xy = 16(x+ y) (1)

From ax2 + by2 = 7, we have (ax2 + by2)(x+ y) = 7(x+ y) so ax3 + by3 + xy(ax2 + by2) = 7(x+ y).This simplifies to

16 + 3xy = 7(x+ y) (2)

We can now solve for x+ y and xy from (1) and (2) to find x+ y = −14 and xy = −38. Thus we have(ax4 + by4)(x+ y) = 42(x+ y), and so ax5 + by5 + xy(ax3 + by3) = 42(x+ y). Finally, it follows thatax5 + by5 = 42(x+ y)− 16xy = 20 as desired.

9. [7] Let f(x) = x4 + 14x3 + 52x2 + 56x+ 16. Let z1, z2, z3, z4 be the four roots of f . Find the smallestpossible value of |zazb + zczd| where {a, b, c, d} = {1, 2, 3, 4}.

Answer: 8

Solution: Note that 116f(2x) = x4 + 7x3 + 13x2 + 7x+ 1. Because the coefficients of this polynomial

are symmetric, if r is a root of f(x) then 4r is as well. Further, f(−1) = −1 and f(−2) = 16 so f(x)

has two distinct roots on (−2, 0) and two more roots on (−∞,−2). Now, if σ is a permutation of{1, 2, 3, 4}:|zσ(1)zσ(2) + zσ(3)zσ(4)| ≤ 1

2 (zσ(1)zσ(2) + zσ(3)zσ(4) + zσ(4)zσ(3) + zσ(2)zσ(1))

Let the roots be ordered z1 ≤ z2 ≤ z3 ≤ z4, then by rearrangement the last expression is at least:12 (z1z4 + z2z3 + z3z2 + z4z1)

Since the roots come in pairs z1z4 = z2z3 = 4, our expression is minimized when σ(1) = 1, σ(2) =4, σ(3) = 3, σ(4) = 2 and its minimum value is 8.

10. [8] Let f(x) = 2x3− 2x. For what positive values of a do there exist distinct b, c, d such that (a, f(a)),(b, f(b)), (c, f(c)), (d, f(d)) is a rectangle?

Answer: [√

33 , 1]

Solution: Say we have four points (a, f(a)), (b, f(b)), (c, f(c)), (d, f(d)) on the curve which forma rectangle. If we interpolate a cubic through these points, that cubic will be symmetric around thecenter of the rectangle. But the unique cubic through the four points is f(x), and f(x) has only onepoint of symmetry, the point (0, 0).

So every rectangle with all four points on f(x) is of the form (a, f(a)), (b, f(b)), (−a, f(−a)), (−b, f(−b)),and without loss of generality we let a, b > 0. Then for any choice of a and b these points form a paral-lelogram, which is a rectangle if and only if the distance from (a, f(a)) to (0, 0) is equal to the distancefrom (b, f(b)) to (0, 0). Let g(x) = x2 + (f(x))2 = 4x6 − 8x4 + 5x2, and consider g(x) restricted tox ≥ 0. We are looking for all the values of a such that g(x) = g(a) has solutions other than a.

Note that g(x) = h(x2) where h(x) = 4x3 − 8x2 + 5x. This polynomial h(x) has a relative maximumof 1 at x = 1

2 and a relative minimum of 25/27 at x = 56 . Thus the polynomial h(x) − h(1/2) has

the double root 1/2 and factors as (4x2 − 4x + 1)(x − 1), the largest possible value of a2 for whichh(x2) = h(a2) is a2 = 1, or a = 1. The smallest such value is that which evaluates to 25/27 other than5/6, which is similarly found to be a2 = 1/3, or a =

√3

3 . Thus, for a in the range√

33 ≤ a ≤ 1 the

equation g(x) = g(a) has nontrivial solutions and hence an inscribed rectangle exists.

3

12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Individual Round: Calculus Test

1. [3] Let f be a differentiable real-valued function defined on the positive real numbers. The tangentlines to the graph of f always meet the y-axis 1 unit lower than where they meet the function. Iff(1) = 0, what is f(2)?

2. [3] The differentiable function F : R→ R satisfies F (0) = −1 and

d

dxF (x) = sin(sin(sin(sin(x)))) · cos(sin(sin(x))) · cos(sin(x)) · cos(x).

Find F (x) as a function of x.

3. [4] Compute eA where A is defined as ∫ 4/3

3/4

2x2 + x+ 1x3 + x2 + x+ 1

dx.

4. [4] Let P be a fourth degree polynomial, with derivative P ′, such that P (1) = P (3) = P (5) = P ′(7) = 0.Find the real number x 6= 1, 3, 5 such that P (x) = 0.

5. [4] Compute

limh→0

sin(π3 + 4h

)− 4 sin

(π3 + 3h

)+ 6 sin

(π3 + 2h

)− 4 sin

(π3 + h

)+ sin

(π3

)h4

.

6. [5] Let p0(x), p1(x), p2(x), . . . be polynomials such that p0(x) = x and for all positive integers n,ddxpn(x) = pn−1(x). Define the function p(x) : [0,∞)→ R by p(x) = pn(x) for all x ∈ [n, n+ 1). Giventhat p(x) is continuous on [0,∞), compute

∞∑n=0

pn(2009).

7. [5] A line in the plane is called strange if it passes through (a, 0) and (0, 10 − a) for some a in theinterval [0, 10]. A point in the plane is called charming if it lies in the first quadrant and also lies belowsome strange line. What is the area of the set of all charming points?

8. [7] Compute ∫ √3

1

x2x2+1 + ln(x2x2x2+1

)dx.

9. [7] Let R be the region in the plane bounded by the graphs of y = x and y = x2. Compute the volumeof the region formed by revolving R around the line y = x.

10. [8] Let a and b be real numbers satisfying a > b > 0. Evaluate∫ 2π

0

1a+ b cos(θ)

dθ.

Express your answer in terms of a and b.

12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Individual Round: Calculus Test Solutions

1. [3] Let f be a differentiable real-valued function defined on the positive real numbers. The tangentlines to the graph of f always meet the y-axis 1 unit lower than where they meet the function. Iff(1) = 0, what is f(2)?

Answer: ln 2

Solution: The tangent line to f at x meets the y-axis at f(x) − 1 for any x, so the slope of thetangent line is f ′(x) = 1

x , and so f(x) = ln(x) + C for some a. Since f(1) = 0, we have C = 0, and sof(x) = ln(x). Thus f(2) = ln(2).

2. [3] The differentiable function F : R→ R satisfies F (0) = −1 and

d

dxF (x) = sin(sin(sin(sin(x)))) · cos(sin(sin(x))) · cos(sin(x)) · cos(x).

Find F (x) as a function of x.

Answer: − cos(sin(sin(sin(x))))

Solution: Substituting u = sin(sin(sin(x))), we find

F (x) =∫

sin(u)du = − cos(u) + C.

for some C. Since F (0) = 1 we find C = 0.

3. [4] Compute eA where A is defined as ∫ 4/3

3/4

2x2 + x+ 1x3 + x2 + x+ 1

dx.

Answer: 169

Solution: We can use partial fractions to decompose the integrand to 1x+1 + x

x2+1 , and then integratethe addends separately by substituting u = x + 1 for the former and u = x2 + 1 for latter, to obtainln(x+ 1) + 1

2 ln(x2 + 1)∣∣4/33/4

= ln((x+ 1)√x2 + 1)

∣∣4/33/4

= ln 169 . Thus eA = 16/9.

Alternate solution: Substituting u = 1/x, we find

A =∫ 3/4

4/3

2u+ u2 + u3

1 + u+ u2 + u3(− 1u2

)du =∫ 4/3

3/4

2/u+ 1 + u

1 + u+ u2 + u3du

Adding this to the original integral, we find

2A =∫ 4/3

3/4

2/u+ 2 + 2u+ 2u2

1 + u+ u2 + u3du =

∫ 4/3

3/4

2udu

Thus A = ln 169 and eA = 16

9 .

1

4. [4] Let P be a fourth degree polynomial, with derivative P ′, such that P (1) = P (3) = P (5) = P ′(7) = 0.Find the real number x 6= 1, 3, 5 such that P (x) = 0.

Answer: 8911

Solution: Observe that 7 is not a root of P . If r1, r2, r3, r4 are the roots of P , then P ′(7)P (7) =∑

i1

7−ri= 0. Thus r4 = 7−

(∑i 6=4

17−ri

)−1

= 7 +(

16 + 1

4 + 12

)−1 = 7 + 12/11 = 89/11.

5. [4] Compute

limh→0

sin(π3 + 4h

)− 4 sin

(π3 + 3h

)+ 6 sin

(π3 + 2h

)− 4 sin

(π3 + h

)+ sin

(π3

)h4

Answer:√

32

Solution: The derivative of a function is defined as f ′(x) = limh→0f(x+h)−f(x)

h . Iterating thisformula four times yields

f (4)(x) = limh→0

f(x+ 4h)− 4f(x+ 3h) + 6f(x+ 2h)− 4f(x+ h) + f(x)h4

.

Substituting f = sin and x = π/3, the expression is equal to sin(4)(π/3) = sin(π/3) =√

32 .

6. [5] Let p0(x), p1(x), p2(x), . . . be polynomials such that p0(x) = x and for all positive integers n,ddxpn(x) = pn−1(x). Define the function p(x) : [0,∞) → R x by p(x) = pn(x) for all x ∈ [n, n + 1].Given that p(x) is continuous on [0,∞), compute

∞∑n=0

pn(2009).

Answer: e2010 − e2009 − 1

Solution: By writing out the first few polynomials, one can guess and then show by induction thatpn(x) = 1

(n+1)! (x+ 1)n+1 − 1n!x

n. Thus the sum evaluates to e2010 − e2009 − 1 by the series expansionof ex.

7. [5] A line in the plane is called strange if it passes through (a, 0) and (0, 10 − a) for some a in theinterval [0, 10]. A point in the plane is called charming if it lies in the first quadrant and also lies belowsome strange line. What is the area of the set of all charming points?

Answer: 50/3

Solution: The strange lines form an envelope (set of tangent lines) of a curve f(x), and we first findthe equation for f on [0, 10]. Assuming the derivative f ′ is continuous, the point of tangency of theline ` through (a, 0) and (0, b) to f is the limit of the intersection points of this line with the lines `εpassing through (a + ε, 0) and (0, b − ε) as ε → 0. If these limits exist, then the derivative is indeedcontinuous and we can calculate the function from the points of tangency.

The intersection point of ` and `ε can be calculated to have x-coordinate a(a−ε)a+b , so the tangent point

of ` has x-coordinate limε→0a(a−ε)a+b = a2

a+b = a2

10 . Similarly, the y-coordinate is b2

10 = (10−a)210 . Thus,

2

solving for the y coordinate in terms of the x coordinate for a ∈ [0, 10], we find f(x) = 10−2√

10√x+x,

and so the area of the set of charming points is∫ 10

0

(10− 2

√10√x+ x

)dx = 50/3.

8. [7] Compute ∫ √3

1

x2x2+1 + ln(x2x2x2+1

)dx.

Answer: 13

Solution: Using the fact that x = eln(x), we evaluate the integral as follows:∫x2x2+1 + ln

(x2x2x2+1

)dx =

∫x2x2+1 + x2x2+1 ln(x2)dx

=∫eln(x)(2x2+1)(1 + ln(x2))dx

=∫xex

2 ln(x2)(1 + ln(x2))dx

Noticing that the derivative of x2 ln(x2) is 2x(1 + ln(x2)), it follows that the integral evaluates to

12ex

2 ln(x2) =12x2x2

.

Evaluating this from 1 to√

3 we obtain the answer.

9. [7] let R be the region in the plane bounded by the graphs of y = x and y = x2. Compute the volumeof the region formed by revolving R around the line y = x.

Answer:√

2π60

Solution: We integrate from 0 to 1 using the method of washers. Fix d between 0 and 1. Let theline x = d intersect the graph of y = x2 at Q, and let the line x = d intersect the graph of y = x atP . Then P = (d, d), and Q = (d, d2). Now drop a perpendicular from Q to the line y = x, and let Rbe the foot of this perpendicular. Because PQR is a 45− 45− 90 triangle, QR = (d− d2)/

√2. So the

differential washer has a radius of (d − d2)/√

2 and a height of√

2dx. So we integrate (from 0 to 1)the expression [(x− x2)/

√2]2√

2dx, and the answer follows.

10. [8] Let a and b be real numbers satisfying a > b > 0. Evaluate∫ 2π

0

1a+ b cos(θ)

dθ.

Express your answer in terms of a and b.

Answer: 2π√a2−b2

3

Solution: Using the geometric series formula, we can expand the integral as follows:∫ 2π

0

1a+ b cos(θ)

dθ =1a

∫ 2π

0

1 +b

acos(θ) +

(b

a

)2

cos2(θ)dθ

=1a

∞∑n=0

∫ 2π

0

(b

a

)n(eiθ + e−iθ

2

)ndθ

=2πa

∞∑n=0

(b2

a2

)n (2nn

)22n

dθ

To evaluate this sum, recall that Cn = 1n+1

(2nn

)is the nth Catalan number. The generating function

for the Catalan numbers is∞∑n=0

Cnxn =

1−√

1− 4x2x

,

and taking the derivative of x times this generating function yields∑(

2nn

)xn = 1√

1−4x. Thus the

integral evaluates to 2π√a2−b2 , as desired.

4

12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Individual Round: Combinatorics Test

1. [3] How many ways can the integers from −7 to 7 inclusive be arranged in a sequence such that theabsolute value of the numbers in the sequence does not decrease?

2. [3] Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. Whatis the expected number of cards that will be strictly between the two jokers?

3. [4] How many rearrangements of the letters of “HMMTHMMT” do not contain the substring “HMMT”?(For instance, one such arrangement is HMMHMTMT.)

4. [4] How many functions f : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} satisfy f(f(x)) = f(x) for all x ∈ {1, 2, 3, 4, 5}?

5. [4] Let s(n) denote the number of 1’s in the binary representation of n. Compute

1255

∑0≤n<16

2n(−1)s(n).

6. [5] How many sequences of 5 positive integers (a, b, c, d, e) satisfy abcde ≤ a + b + c + d + e ≤ 10?

7. [5] Paul fills in a 7× 7 grid with the numbers 1 through 49 in a random arrangement. He then eraseshis work and does the same thing again, to obtain two different random arrangements of the numbersin the grid. What is the expected number of pairs of numbers that occur in either the same row aseach other or the same column as each other in both of the two arrangements?

8. [7] There are 5 students on a team for a math competition. The math competition has 5 subject tests.Each student on the team must choose 2 distinct tests, and each test must be taken by exactly twopeople. In how many ways can this be done?

9. [7] The squares of a 3 × 3 grid are filled with positive integers such that 1 is the label of the upper-leftmost square, 2009 is the label of the lower-rightmost square, and the label of each square dividesthe one directly to the right of it and the one directly below it. How many such labelings are possible?

10. [8] Given a rearrangement of the numbers from 1 to n, each pair of consecutive elements a and b ofthe sequence can be either increasing (if a < b) or decreasing (if b < a). How many rearrangementsof the numbers from 1 to n have exactly two increasing pairs of consecutive elements? Express youranswer in terms of n.

12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Individual Round: Combinatorics Test

1. [3] How many ways can the integers from −7 to 7 be arranged in a sequence such that the absolutevalue of the numbers in the sequence is nondecreasing?

Answer: 128

Solution: Each of the pairs a, −a must occur in increasing order of a for a = 1, . . . , 7, but a caneither occur before or after −a, for a total of 27 = 128 possible sequences.

2. [3] Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. Whatis the expected number of cards that will be between the two jokers?

Answer: 52/3

Solution: Each card has an equal likelihood of being either on top of the jokers, in between them,or below the jokers. Thus, on average, 1/3 of them will land between the two jokers.

3. [4] In how many ways can you rearrange the letters of “HMMTHMMT” such that the consecutivesubstring “HMMT” does not appear?

Answer: 361

Solution: There are 8!/(4!2!2!) = 420 ways to order the letters. If the permuted letters contain“HMMT”, there are 5 · 4!/2! = 60 ways to order the other letters, so we subtract these. However, wehave subtracted “HMMTHMMT” twice, so we add it back once to obtain 361 possibilities.

4. [4] How many functions f : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} satisfy f(f(x)) = f(x) for all x ∈ {1, 2, 3, 4, 5}?Answer: 196

Solution: A fixed point of a function f is an element a such that f(a) = a. The condition is equivalentto the property that f maps every number to a fixed point. Counting by the number of fixed pointsof f , the total number of such functions is

5∑k=1

(5k

)k5−k = 1 · (50) + 5 · (14 + 41) + 10 · (23 + 32)

= 1 + 25 + 10 · 17= 196.

5. [4] Let s(n) denote the number of 1’s in the binary representation of n. Compute

1255

∑0≤n<16

2n(−1)s(n).

Answer: 45

Solution: Notice that if n < 8, (−1)s(n) = (−1)·(−1)s(n+8) so the sum becomes 1255 (1−28)

∑0≤n<8 2n(−1)s(n) =

45.

1

6. [5] How many sequences of 5 positive integers (a, b, c, d, e) satisfy abcde ≤ a + b + c + d + e ≤ 10?

Answer: 116

Solution: We count based on how many 1’s the sequence contains. If a = b = c = d = e = 1 thenthis gives us 1 possibility. If a = b = c = d = 1 and e 6= 1, e can be 2, 3, 4, 5, 6. Each such sequence(1, 1, 1, 1, e) can be arranged in 5 different ways, for a total of 5 · 5 = 25 ways in this case.

If three of the numbers are 1, the last two can be (2, 2), (3, 3), (2, 3), (2, 4), or (2, 5). Counting ordering,this gives a total of 2 · 10 + 3 · 20 = 80 possibilities.

If two of the numbers are 1, the other three must be equal to 2 for the product to be under 10, andthis yields 10 more possibilities.

Thus there are 1 + 25 + 80 + 10 = 116 such sequences.

7. [7] Paul fills in a 7× 7 grid with the numbers 1 through 49 in a random arrangement. He then eraseshis work and does the same thing again (to obtain two different random arrangements of the numbersin the grid). What is the expected number of pairs of numbers that occur in either the same row aseach other or the same column as each other in both of the two arrangements?

Answer: 147/2

Solution: Each of the(492

)pairs of numbers has a probability of

14·(72)

(492 ) = 1/4 of being in the same

row or column in one of the arrangements, so the expected number that are in the same row or columnin both arrangements is (

492

)· (1/4)2 =

1472

.

8. [7] There are 5 students on a team for a math competition. The math competition has 5 subject tests.Each student on the team must choose 2 distinct tests, and each test must be taken by exactly twopeople. In how many ways can this be done?

Answer: 2040

Solution: We can model the situation as a bipartite graph on 10 vertices, with 5 nodes representingthe students and the other 5 representing the tests. We now simply want to count the number ofbipartite graphs on these two sets such that there are two edges incident on each vertex.

Notice that in such a graph, we can start at any vertex and follow one of the edges eminating fromit, then follow the other edge eminating from the second vertex, etc, and in this manner we musteventually end up back at the starting vertex, so the graph is partitioned into even cycles. Since eachvertex has degree two, we cannot have a 2-cycle, so we must have either a 10-cycle or a 4-cycle and a6-cycle.

In the former case, starting with Person A, there are 5 ways to choose one of his tests. This test can betaken by one of 4 other people, who can take one of 4 other tests, which can be taken by one of 3 otherpeople, etc, so the number of 10-cycles we obtain in this way is 5! · 4!. However, it does not matterwhich of the first person’s tests we choose first in a given 10-cycle, so we overcounted by a factor of 2.Thus there are 5! · 4!/2 = 1440 possibilities in this case.

In the latter case, there are(53

)2= 100 ways to choose which three people and which three tests are

in the 6-cycle. After choosing this, a similar argument to that above shows there are 2! · 1!/2 possible4-cycles and 3! · 2!/2 possible 6-cycles, for a total of 100 · 1 · 6 = 600 possibilities in this case.

Thus there are a total of 2040 ways they can take the tests.

2

9. [5] The squares of a 3 × 3 grid are filled with positive integers such that 1 is the label of the upper-leftmost square, 2009 is the label of the lower-rightmost square, and the label of each square dividesthe one directly to the right of it and the one directly below it. How many such labelings are possible?

Answer: 2448

Solution: We factor 2009 as 72 · 41 and place the 41’s and the 7’s in the squares separately. Thenumber of ways to fill the grid with 1’s and 41’s so that the divisibility property is satisfied is equal tothe number of nondecreasing sequences a1, a2, a3 where each ai ∈ {0, 1, 2, 3} and the sequence is not0, 0, 0 and not 1, 1, 1 (here ai corresponds to the number of 41’s in the ith column.) Thus there are(3+4−1

3

)− 2 = 18 ways to choose which squares are divisible by 41.

To count the arrangements of divisibility by 7 and 49, we consider three cases.

If 49 divides the middle square, then each of the squares to the right and below it are divisible 49.The two squares in the top row (besides the upper left) can be (1, 1), (1, 7), (1, 49), (7, 7), (7, 49), or(49, 49) (in terms of the highest power of 7 dividing the square). The same is true, independently, forthe two blank squares on the left column, for a total of 62 = 36 possibilities in this case.

If 1 is the highest power of 7 dividing the middle square, there are also 36 possibilities by a similarargument.

If 7 is the highest power of 7 dividing the middle square, there are 8 possibilities for the upper rightthree squares. Thus there are 64 possibilities in this case.

Thus there are a total of 136 options for the divisibility of each number by 7 and 72, and 18 optionsfor the divisibility of the numbers by 41. Since each number divides 2009, this uniquely determinesthe numbers, and so there are a total of 18 · 136 = 2448 possibilities.

10. [8] Given a rearrangement of the numbers from 1 to n, each pair of consecutive elements a and b ofthe sequence can be either increasing (if a < b) or decreasing (if b < a). How many rearrangements ofthe numbers from 1 to n have exactly two increasing pairs of consecutive elements?

Answer: 3n − (n + 1) · 2n + n(n + 1)/2 or equivalent

Solution: Notice that each such permutation consists of 3 disjoint subsets of {1, . . . , n} whose union is{1, . . . , n}, each arranged in decreasing order. For instance, if n = 6, in the permutation 415326 (whichhas the two increasing pairs 15 and 26), the three sets are {4, 1}, {5, 3, 2}, and 6. There are 3n ways tochoose which of the first, second, or third set each element is in. However, we have overcounted: somechoices of these subsets result in permutations with 1 or 0 increasing pairs, such as {6, 5, 4}, {3, 2}, {1}.Thus, we must subtract the number of ordered partitions of {1, 2, . . . , n} into 3 subsets for which theminimum value of the first is not less than the maximum of the second, or the minimum value of thesecond is not less than the maximum of the third.

We first prove that the number of permutations having exactly one increasing consecutive pair ofelements is 2n − (n + 1). To do so, note that there are 2n ways to choose which elements occur beforethe increasing pair, and upon choosing this set we must arrange them in decreasing order, followed bythe remaining elements arranged in decreasing order. The resulting permutation will have either oneincreasing pair or none. There are exactly n + 1 subsets for which the resulting permutation has none,namely, {}, {n}, {n, n− 1}, {n, n− 1, n− 2}, etc. Thus the total number of permutations having oneincreasing pair is 2n − (n + 1) as desired.

We now count the partitions of {1, 2, . . . , n} whose associated permutation has exactly one increasingpair. For each of the 2n − (n + 1) permutations p having exacly one increasing pair, there are n + 1partitions of {1, 2, . . . , n} into 3 subsets whose associated permutation is p. This is because there aren + 1 ways to choose the ”breaking point” to split one of the subsets into two. Thus there are a totalof (n + 1)(2n − (n + 1)) partitions whose associaated permutation has exactly one increasing pair.

Finally, we must count the number of partitions whose associated permutation is n, n − 1, . . . , 3, 2, 1,i.e. has no increasing pair. There are n+2

2 ways of placing two barriers between these elements to splitthe numbers into three subsets, and so there are n+2

2 such partitions of {1, 2, . . . , n} into three subsets.

3

Thus, subtracting off the partitions we did not want to count, the answer is 3n − (n + 1)(2n − (n +1))−

(n+2

2

)= 3n − (n + 1) · 2n + n(n + 1)/2.

4

12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Individual Round: General Test, Part 1

1. [2] If a and b are positive integers such that a2 − b4 = 2009, find a + b.

2. [2] Suppose N is a 6-digit number having base-10 representation a b c d e f . If N is 6/7 of the numberhaving base-10 representation d e f a b c , find N .

3. [3] A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below,so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of theresulting trapezoid.

5

8

4. [3] If tan x + tan y = 4 and cot x + cot y = 5, compute tan(x + y).

5. [4] Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. Whatis the expected number of cards that will be strictly between the two jokers?

6. [4] The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacentto the vertex of the chosen corner. What is the height of the cube when the freshly-cut face is placedon a table?

7. [5] Let s(n) denote the number of 1’s in the binary representation of n. Compute

1255

∑0≤n<16

2n(−1)s(n).

8. [5] Let a, b, and c be the 3 roots of x3 − x + 1 = 0. Find 1a+1 + 1

b+1 + 1c+1 .

9. [6] How many functions f : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} satisfy f(f(x)) = f(x) for all x ∈ {1, 2, 3, 4, 5}?

10. [6] A kite is a quadrilateral whose diagonals are perpendicular. Let kite ABCD be such that ∠B =∠D = 90◦. Let M and N be the points of tangency of the incircle of ABCD to AB and BC respectively.Let ω be the circle centered at C and tangent to AB and AD. Construct another kite AB′C ′D′ that issimilar to ABCD and whose incircle is ω. Let N ′ be the point of tangency of B′C ′ to ω. If MN ′ ‖ AC,then what is the ratio of AB : BC?

12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Individual Round: General Test, Part 1

1. [2] If a and b are positive integers such that a2 − b4 = 2009, find a + b.

Answer: 47

Solution: We can factor the equation as (a − b2)(a + b2) = 41 · 49, from which it is evident thata = 45 and b = 2 is a possible solution. By examining the factors of 2009, one can see that there areno other solutions.

2. [2] Suppose N is a 6-digit number having base-10 representation a b c d e f . If N is 6/7 of the numberhaving base-10 representation d e f a b c , find N .

Answer: 461538

Solution: We have 7(abcdef)10 = 6(defabc)10, so 699400a + 69940b + 6994c = 599300d + 59930e +5993f . We can factor this equation as 6994(100a + 10b + c) = 5993(100d + 10e + f), which yields538(abc)10 = 461(def)10. Since gcd(538, 461) = 1, we must have (abc)10 = 461 and (def)10 = 538.

3. [3] A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below,so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of theresulting trapezoid.

5

8

Answer: 55/2

Solution: Drawing the perpendiculars from the point of intersection of the corners to the bases ofthe trapezoid, we see that we have similar 3−4−5 right triangles, and we can calculate that the lengthof the smaller base is 3. Thus the area of the trapezoid is 8+3

2 · 5 = 55/2.

4. [3] If tan x + tan y = 4 and cot x + cot y = 5, compute tan(x + y).

Answer: 20

Solution: We have cot x+cot y = tan x+tan ytan x tan y , so tan x tan y = 4

5 . Thus, by the tan addition formula,tan(x + y) = tan x+tan y

1−tan x tan y = 20.

5. [4] Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. Whatis the expected number of cards that will be strictly between the two jokers?

Answer: 52/3

1

Solution: Each card has an equal likelihood of being either on top of the jokers, in between them,or below the jokers. Thus, on average, 1/3 of them will land between the two jokers.

6. [4] The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacentto the vertex of the chosen corner. What is the height of the cube when the freshly-cut face is placedon a table?

Answer: 2√

3/3

Solution: The major diagonal has a length of√

3. The volume of the pyramid is 1/6, and so itsheight h satisfies 1

3 ·h ·√

34 (√

2)2 = 1/6 since the freshly cut face is an equilateral triangle of side length√2. Thus h =

√3/3, and the answer follows.

7. [5] Let s(n) denote the number of 1’s in the binary representation of n. Compute

1255

∑0≤n<16

2n(−1)s(n).

Answer: 45

Solution: Notice that if n < 8, (−1)s(n) = (−1)·(−1)s(n+8) so the sum becomes 1255 (1−28)

∑0≤n<8 2n(−1)s(n) =

45.

8. [5] Let a, b, and c be the 3 roots of x3 − x + 1 = 0. Find 1a+1 + 1

b+1 + 1c+1 .

Answer: −2

Solution: We can substitute x = y−1 to obtain a polynomial having roots a+1, b+1, c+1, namely,(y − 1)3 − (y − 1) + 1 = y3 − 3y2 + 2y + 1. The sum of the reciprocals of the roots of this polynomialis, by Viete’s formulas, 2

−1 = −2.

9. [6] How many functions f : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} satisfy f(f(x)) = f(x) for all x ∈ {1, 2, 3, 4, 5}?

Answer: 196

Solution: A fixed point of a function f is an element a such that f(a) = a. The condition is equivalentto the property that f maps every number to a fixed point. Counting by the number of fixed pointsof f , the total number of such functions is

5∑k=1

(5k

)k5−k = 1 · (05 + 50) + 5 · (14 + 41) + 10 · (23 + 32)

= 1 + 25 + 10 · 17= 196.

10. [6] A kite is a quadrilateral whose diagonals are perpendicular. Let kite ABCD be such that ∠B =∠D = 90◦. Let M and N be the points of tangency of the incircle of ABCD to AB and BC respectively.Let ω be the circle centered at C and tangent to AB and AD. Construct another kite AB′C ′D′ that issimilar to ABCD and whose incircle is ω. Let N ′ be the point of tangency of B′C ′ to ω. If MN ′ ‖ AC,then what is the ratio of AB : BC?

2

Answer: 1+√

52

Solution: Let’s focus on the right triangle ABC and the semicircle inscribed in it since the situationis symmetric about AC. First we find the radius a of circle O. Let AB = x and BC = y. Drawing theradii OM and ON , we see that AM = x− a and 4AMO ∼ 4ABC. In other words,

AM

MO=

AB

BCx− a

a=

x

y

a =xy

x + y.

Now we notice that the situation is homothetic about A. In particular,

4AMO ∼ 4ONC ∼ 4CN ′C ′.

Also, CB and CN ′ are both radii of circle C. Thus, when MN ′ ‖ AC ′, we have

AM = CN ′ = CB

x− a = y

a =xy

x + y= x− y

x2 − xy − y2 = 0

x =y

2±√

y2

4+ y2

AB

BC=

x

y=

1 +√

52

.

3

12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Individual Round: General Test, Part 2

1. [2] How many ways can the integers from −7 to 7 be arranged in a sequence such that the absolutevalues of the numbers in the sequence are nonincreasing?

2. [2] How many ways can you tile the white squares of following 2× 24 grid with dominoes? (A dominocovers two adjacent squares, and a tiling is a non-overlapping arrangement of dominoes that coversevery white square and does not intersect any black square.)

3. [3] Let S be the sum of all the real coefficients of the expansion of (1 + ix)2009. What is log2(S)?

4. [3] A torus (donut) having inner radius 2 and outer radius 4 sits on a flat table. What is the radiusof the largest spherical ball that can be placed on top of the center torus so that the ball still touchesthe horizontal plane? (If the x – y plane is the table, the torus is formed by revolving the circle in thex – z plane centered at (3, 0, 1) with radius 1 about the z axis. The spherical ball has its center on thez-axis and rests on either the table or the donut.)

5. [4] Suppose a, b and c are integers such that the greatest common divisor of x2 +ax+ b and x2 + bx+ cis x + 1 (in the set of polynomials in x with integer coefficients), and the least common multiple ofx2 + ax + b and x2 + bx + c is x3 − 4x2 + x + 6. Find a + b + c.

6. [4] In how many ways can you rearrange the letters of “HMMTHMMT” such that the substring“HMMT” does not appear? (For instance, one such rearrangement is HMMHMTMT.)

7. [5] Let Fn be the Fibonacci sequence, that is, F0 = 0, F1 = 1, and Fn+2 = Fn+1 + Fn. Compute∑∞n=0 Fn/10n.

8. [5] The incircle ω of equilateral triangle ABC has radius 1. Three smaller circles are inscribed tangentto ω and the sides of ABC, as shown. Three smaller circles are then inscribed tangent to the previouscircles and to each of two sides of ABC. This process is repeated an infinite number of times. Whatis the total length of the circumferences of all the circles?

9. [6] How many sequences of 5 positive integers (a, b, c, d, e) satisfy abcde ≤ a + b + c + d + e ≤ 10?

10. [6] Let T be a right triangle with sides having lengths 3, 4, and 5. A point P is called “awesome” if Pis the center of a parallelogram whose vertices all lie on the boundary of T . What is the area of theset of awesome points?

12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Individual Round: General Test, Part 2

1. [2] How many ways can the integers from −7 to 7 be arranged in a sequence such that the absolutevalues of the numbers in the sequence are nonincreasing?

Answer: 128

Solution: Each of the pairs a, −a must occur in increasing order of a for a = 1, . . . , 7, but a caneither occur before or after −a, for a total of 27 = 128 possible sequences.

2. [2] How many ways can you tile the white squares of following 2× 24 grid with dominoes? (A dominocovers two adjacent squares, and a tiling is a non-overlapping arrangement of dominoes that coversevery white square and does not intersect any black square.)

Answer: 27

Solution: Divide the rectangle into three 3 × 8 sub-rectangles. It is easy to count that there are 3ways of tiling each of these sub-rectangles independently, for a total of 33 = 27 possibilities.

3. [3] Let S be the sum of all the real coefficients of the expansion of (1 + ix)2009. What is log2(S)?

Answer: 1004

Solution: The sum of all the coefficients is (1 + i)2009, and the sum of the real coefficients is the realpart of this, which is 1

2

((1 + i)2009 + (1− i)2009

)= 21004. Thus log2(S) = 1004.

4. [3] A torus (donut) having inner radius 2 and outer radius 4 sits on a flat table. What is the radiusof the largest spherical ball that can be placed on top of the center torus so that the ball still touchesthe horizontal plane? (If the x− y plane is the table, the torus is formed by revolving the circle in thex− z plane centered at (3, 0, 1) with radius 1 about the z axis. The spherical ball has its center on thez-axis and rests on either the table or the donut.)

Answer: 9/4

Solution: Let r be the radius of the sphere. One can see that it satisfies (r + 1)2 = (r − 1)2 + 32 bythe Pythagorean Theorem, so r = 9/4.

5. [4] Suppose a, b and c are integers such that the greatest common divisor of x2 +ax+ b and x2 + bx+ cis x + 1 (in the set of polynomials in x with integer coefficients), and the least common multiple ofx2 + ax+ b and x2 + bx+ c is x3 − 4x2 + x+ 6. Find a+ b+ c.

Answer: −6

Solution: Since x+1 divides x2+ax+b and the constant term is b, we have x2+ax+b = (x+1)(x+b),and similarly x2 +bx+c = (x+1)(x+c). Therefore, a = b+1 = c+2. Furthermore, the least common

1

multiple of the two polynomials is (x+ 1)(x+ b)(x+ b−1) = x3−4x2 +x+ 6, so b = −2. Thus a = −1and c = −3, and a+ b+ c = −6.

6. [4] In how many ways can you rearrange the letters of “HMMTHMMT” such that the consecutivesubstring “HMMT” does not appear?

Answer: 361

Solution: There are 8!/(4!2!2!) = 420 ways to order the letters. If the permuted letters contain“HMMT”, there are 5 · 4!/2! = 60 ways to order the other letters, so we subtract these. However, wehave subtracted “HMMTHMMT” twice, so we add it back once to obtain 361 possibilities.

7. [5] Let Fn be the Fibonacci sequence, that is, F0 = 0, F1 = 1, and Fn+2 = Fn+1 + Fn. Compute∑∞n=0 Fn/10n.

Answer: 10/89

Solution: Write F (x) =∑∞

n=0 Fnxn. Then the Fibonacci recursion tells us that F (x) − xF (x) −

x2F (x) = x, so F (x) = x/(1− x− x2). Plugging in x = 1/10 gives the answer.

8. [5] The incircle ω of equilateral triangle ABC has radius 1. Three smaller circles are inscribed tangentto ω and the sides of ABC, as shown. Three smaller circles are then inscribed tangent to the previouscircles and to each of two sides of ABC. This process is repeated an infinite number of times. Whatis the total length of the circumferences of all the circles?

Answer: 5π

Solution: One can find using the Pythagorean Theorem that, in each iteration, the new circles haveradius 1/3 of that of the previously drawn circles. Thus the total circumference is 2π+3·2π( 1

1−1/3−1) =5π.

9. [6] How many sequences of 5 positive integers (a, b, c, d, e) satisfy abcde ≤ a+ b+ c+ d+ e ≤ 10?

Answer: 116

Solution: We count based on how many 1’s the sequence contains. If a = b = c = d = e = 1 thenthis gives us 1 possibility. If a = b = c = d = 1 and e 6= 1, e can be 2, 3, 4, 5, 6. Each such sequence(1, 1, 1, 1, e) can be arranged in 5 different ways, for a total of 5 · 5 = 25 ways in this case.

If three of the numbers are 1, the last two can be (2, 2), (3, 3), (2, 3), (2, 4), or (2, 5). Counting ordering,this gives a total of 2 · 10 + 3 · 20 = 80 possibilities.

If two of the numbers are 1, the other three must be equal to 2 for the product to be under 10, andthis yields 10 more possibilities.

Thus there are 1 + 25 + 80 + 10 = 116 such sequences.

2

10. [6] Let T be a right triangle with sides having lengths 3, 4, and 5. A point P is called awesome if P isthe center of a parallelogram whose vertices all lie on the boundary of T . What is the area of the setof awesome points?

Answer: 3/2

Solution: The set of awesome points is the medial triangle, which has area 6/4 = 3/2.

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12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Individual Round: Geometry Test

1. [3] A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below,so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of theresulting trapezoid.

5

82. [3] The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacent

to the vertex of the chosen corner. What is the height of the cube when the freshly-cut face is placedon a table?

3. [4] Let T be a right triangle with sides having lengths 3, 4, and 5. A point P is called awesome if P isthe center of a parallelogram whose vertices all lie on the boundary of T . What is the area of the setof awesome points?

4. [4] A kite is a quadrilateral whose diagonals are perpendicular. Let kite ABCD be such that ∠B =∠D = 90◦. Let M and N be the points of tangency of the incircle of ABCD to AB and BC respectively.Let ω be the circle centered at C and tangent to AB and AD. Construct another kite AB′C ′D′ that issimilar to ABCD and whose incircle is ω. Let N ′ be the point of tangency of B′C ′ to ω. If MN ′ ‖ AC,then what is the ratio of AB : BC?

5. [4] Circle B has radius 6√

7. Circle A, centered at point C, has radius√

7 and is contained in B. LetL be the locus of centers C such that there exists a point D on the boundary of B with the followingproperty: if the tangents from D to circle A intersect circle B again at X and Y , then XY is alsotangent to A. Find the area contained by the boundary of L.

6. [5] Let ABC be a triangle in the coordinate plane with vertices on lattice points and with AB = 1.Suppose the perimeter of ABC is less than 17. Find the largest possible value of 1/r, where r is theinradius of ABC.

7. [5] In triangle ABC, D is the midpoint of BC, E is the foot of the perpendicular from A to BC, and Fis the foot of the perpendicular from D to AC. Given that BE = 5, EC = 9, and the area of triangleABC is 84, compute |EF |.

8. [7] Triangle ABC has side lengths AB = 231, BC = 160, and AC = 281. Point D is constructed onthe opposite side of line AC as point B such that AD = 178 and CD = 153. Compute the distancefrom B to the midpoint of segment AD.

9. [7] Let ABC be a triangle with AB = 16 and AC = 5. Suppose the bisectors of angles ∠ABC and∠BCA meet at point P in the triangle’s interior. Given that AP = 4, compute BC.

10. [8] Points A and B lie on circle ω. Point P lies on the extension of segment AB past B. Line ` passesthrough P and is tangent to ω. The tangents to ω at points A and B intersect ` at points D and Crespectively. Given that AB = 7, BC = 2, and AD = 3, compute BP.

12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Individual Round: Geometry Test

1. [3] A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below,so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of theresulting trapezoid.

5

8

Answer: 55/2

Solution: Drawing the perpendiculars from the point of intersection of the corners to the bases ofthe trapezoid, we see that we have similar 3−4−5 right triangles, and we can calculate that the lengthof the smaller base is 3. Thus the area of the trapezoid is 8+3

2 · 5 = 55/2.

2. [3] The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacentto the vertex of the chosen corner. What is the height of the cube when the freshly-cut face is placedon a table?

Answer: 2√

3/3

Solution: The major diagonal has a length of√

3. The volume of the pyramid is 1/6, and so itsheight h satisfies 1

3 ·h ·√

34 (√

2)2 = 1/6 since the freshly cut face is an equilateral triangle of side length√2. Thus h =

√3/3, and the answer follows.

3. [4] Let T be a right triangle with sides having lengths 3, 4, and 5. A point P is called awesome if P isthe center of a parallelogram whose vertices all lie on the boundary of T . What is the area of the setof awesome points?

Answer: 3/2

Solution: The set of awesome points is the medial triangle, which has area 6/4 = 3/2.

4. [4] A kite is a quadrilateral whose diagonals are perpendicular. Let kite ABCD be such that ∠B =∠D = 90◦. Let M and N be the points of tangency of the incircle of ABCD to AB and BC respectively.Let ω be the circle centered at C and tangent to AB and AD. Construct another kite AB′C ′D′ that issimilar to ABCD and whose incircle is ω. Let N ′ be the point of tangency of B′C ′ to ω. If MN ′ ‖ AC,then what is the ratio of AB : BC?

Answer: 1+√

52

1

Solution: Let’s focus on the right triangle ABC and the semicircle inscribed in it since the situationis symmetric about AC. First we find the radius a of circle O. Let AB = x and BC = y. Drawing theradii OM and ON , we see that AM = x− a and 4AMO ∼ 4ABC. In other words,

AM

MO=AB

BCx− aa

=x

y

a =xy

x+ y.

Now we notice that the situation is homothetic about A. In particular,

4AMO ∼ 4ONC ∼ 4CN ′C ′.

Also, CB and CN ′ are both radii of circle C. Thus, when MN ′ ‖ AC ′, we have

AM = CN ′ = CB

x− a = y

a =xy

x+ y= x− y

x2 − xy − y2 = 0

x =y

2±√y2

4+ y2

AB

BC=x

y=

1 +√

52

.

5. [4] Circle B has radius 6√

7. Circle A, centered at point C, has radius√

7 and is contained in B. LetL be the locus of centers C such that there exists a point D on the boundary of B with the followingproperty: if the tangents from D to circle A intersect circle B again at X and Y , then XY is alsotangent to A. Find the area contained by the boundary of L.

Answer: 168π

Solution: The conditions imply that there exists a triangle such that B is the circumcircle and Ais the incircle for the position of A. The distance between the circumcenter and incenter is given by√

(R− 2r)R, where R, r are the circumradius and inradius, respectively. Thus the locus of C is acircle concentric to B with radius 2

√42. The conclusion follows.

6. [4] Let ABC be a triangle in the coordinate plane with vertices on lattice points and with AB = 1.Suppose the perimeter of ABC is less than 17. Find the largest possible value of 1/r, where r is theinradius of ABC.

Answer: 1 + 5√

2 +√

65

Solution: Let a denote the area of the triangle, r the inradius, and p the perimeter. Then a = rp/2,so r = 2a/p > 2a/17. Notice that a = h/2 where h is the height of the triangle from C to AB, and h isan integer since the vertices are lattice points. Thus we first guess that the inradius is minimized whenh = 1 and the area is 1/2. In this case, we can now assume WLOG that A = (0, 0), B = (1, 0), andC = (n+ 1, 1) for some nonnegative integer n. The perimeter of ABC is

√n2 + 2n+ 2 +

√n2 + 1 + 1.

Since n = 8 yields a perimeter greater than 17, the required triangle has n = 7 and inradius r = 1/p =1

1+5√

2+√

65which yields the answer of 1/r = 1 + 5

√2 +√

65. We can now verify that this is indeed

2

minimal over all h by noting that its perimeter is greater than 17/2, which is the upper bound in thecase h ≥ 2.

7. [5] In 4ABC, D is the midpoint of BC, E is the foot of the perpendicular from A to BC, and F isthe foot of the perpendicular from D to AC. Given that BE = 5, EC = 9, and the area of triangleABC is 84, compute |EF |.

Answer: 6√

375 , 21

205

√7585

Solution: There are two possibilities for the triangle ABC based on whether E is between B andC or not. We first consider the former case.

We find from the area and the Pythagorean theorem that AE = 12, AB = 13, and AC = 15. We canthen use Stewart’s theorem to obtain AD = 2

√37.

Since the area of 4ADC is half that of ABC, we have 12AC · DF = 42, so DF = 14/5. Also,

DC = 14/2 = 7 so ED = 9− 7 = 2.

Notice that AEDF is a cyclic quadrilateral. By Ptolemy’s theorem, we have EF ·2√

37 = (28/5) ·12 +2 · (54/5). Thus EF = 6

√37

5 as desired.

The latter case is similar.

8. [7] Triangle ABC has side lengths AB = 231, BC = 160, and AC = 281. Point D is constructed onthe opposite side of line AC as point B such that AD = 178 and CD = 153. Compute the distancefrom B to the midpoint of segment AD.

Answer: 208

Solution: Note that ∠ABC is right since

BC2 = 1602 = 50 · 512 = (AC −AB) · (AC +AB) = AC2 −AB2.

Construct point B′ such that ABCB′ is a rectangle, and construct D′ on segment B′C such thatAD = AD′. Then

B′D′2 = AD′2 −AB′2 = AD2 −BC2 = (AD −BC)(AD +BC) = 18 · 338 = 782.

It follows that CD′ = B′C − B′D′ = 153 = CD; thus, points D and D′ coincide, and AB ‖ CD. LetM denote the midpoint of segment AD, and denote the orthogonal projections M to lines AB and BCby P and Q respectively. Then Q is the midpoint of BC and AP = 39, so that PB = AB−AP = 192and

BM = PQ =√

802 + 1922 = 16√

52 + 122 = 208.

9. [7] Let ABC be a triangle with AB = 16 and AC = 5. Suppose the bisectors of angles ∠ABC and∠BCA meet at point P in the triangle’s interior. Given that AP = 4, compute BC.

Answer: 14

Solution: As the incenter of triangle ABC, point P has many properties. Extend AP past P to itsintersection with the circumcircle of triangle ABC, and call this intersection M. Now observe that

∠PBM = ∠PBC + ∠CBM = ∠PBC + ∠CAM = β + α = 90− γ,

where α, β, and γ are the half-angles of triangle ABC. Since

∠BMP = ∠BMA = ∠BCA = 2γ,

3

it follows that BM = MP = CM. Let Q denote the intersection of AM and BC, and observe that4AQB ∼ 4CQM and 4AQC ∼ 4BQM ; some easy algebra gives

AM/BC = (AB ·AC +BM · CM)/(AC · CM +AB ·BM).

Writing (a, b, c, d, x) = (BC,AC,AB,MP,AP ), this is (x + d)/a = (bc + d2)/((b + c)d). Ptolemy’stheorem applied to ABCD gives a(d+ x) = d(b+ c). Multiplying the two gives (d+ x)2 = bc+ d2. Weeasily solve for d = (bc− x2)/(2x) = 8 and a = d(b+ c)/(d+ x) = 14.

10. [8] Points A and B lie on circle ω. Point P lies on the extension of segment AB past B. Line ` passesthrough P and is tangent to ω. The tangents to ω at points A and B intersect ` at points D and Crespectively. Given that AB = 7, BC = 2, and AD = 3, compute BP.

Answer: 9

Solution: Say that ` be tangent to ω at point T. Observing equal tangents, write

CD = CT +DT = BC +AD = 5.

Let the tangents to ω at A and B intersect each other at Q. Working from Menelaus applied to triangleCDQ and line AB gives

−1 =DA

AQ· QBBC· CPPD

=DA

BC· CP

PC + CD

=32· CP

PC + 5,

from which PC = 10. By power of a point, PT 2 = AP · BP, or 122 = BP · (BP + 7), from whichBP = 9.

4

12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Guts Round

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12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

1. [5] Compute1 · 22 + 2 · 32 + 3 · 42 + · · ·+ 19 · 202.

2. [5] Given that sinA+ sinB = 1 and cosA+ cosB = 3/2, what is the value of cos(A−B)?

3. [5] Find all pairs of integer solutions (n,m) to

23n

= 32m

− 1.

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12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

4. [6] Simplify: i0 + i1 + · · ·+ i2009.

5. [6] In how many distinct ways can you color each of the vertices of a tetrahedron either red, blue, orgreen such that no face has all three vertices the same color? (Two colorings are considered the sameif one coloring can be rotated in three dimensions to obtain the other.)

6. [6] Let ABC be a right triangle with hypotenuse AC. Let B′ be the reflection of point B across AC,and let C ′ be the reflection of C across AB′. Find the ratio of [BCB′] to [BC ′B′].

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12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

7. [6] How many perfect squares divide 23 · 35 · 57 · 79?

8. [6] Which is greater, log2008(2009) or log2009(2010)?

9. [6] An icosidodecahedron is a convex polyhedron with 20 triangular faces and 12 pentagonal faces.How many vertices does it have?

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1

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12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

10. [7] Let a, b, and c be real numbers. Consider the system of simultaneous equations in x and y:

ax+ by = c− 1(a+ 5)x+ (b+ 3)y = c+ 1

Determine the value(s) of c, in terms of a, such that the system has a solution for any a and b.

11. [7] There are 2008 distinct points on a circle. If you connect two of these points to form a line and thenconnect another two points (distinct from the first two) to form another line, what is the probabilitythat the two lines intersect inside the circle?

12. [7] Bob is writing a sequence of letters of the alphabet, each of which can be either uppercase orlowercase, according to the following two rules:

• If he had just written an uppercase letter, he can either write the same letter in lowercase afterit, or the next letter of the alphabet in uppercase.

• If he had just written a lowercase letter, he can either write the same letter in uppercase afterit, or the preceding letter of the alphabet in lowercase.

For instance, one such sequence is aAaABCDdcbBC. How many sequences of 32 letters can he writethat start at (lowercase) a and end at (lowercase) z? (The alphabet contains 26 letters from a to z,without wrapping around, so that z does not precede a.)

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12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

13. [8] How many ordered quadruples (a, b, c, d) of four distinct numbers chosen from the set {1, 2, 3, . . . , 9}satisfy b < a, b < c, and d < c?

14. [8] Compute2009∑k=1

k

(⌊2009k

⌋−⌊

2008k

⌋).

Here bxc denotes the largest integer that is less than or equal to x.

15. [8] Stan has a stack of 100 blocks and starts with a score of 0, and plays a game in which he iteratesthe following two-step procedure:

(a) Stan picks a stack of blocks and splits it into 2 smaller stacks each with a positive number ofblocks, say a and b. (The order in which the new piles are placed does not matter.)

(b) Stan adds the product of the two piles’ sizes, ab, to his score.

The game ends when there are only 1-block stacks left. What is the expected value of Stan’s score atthe end of the game?

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2

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12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

16. [9] A spider is making a web between n > 1 distinct leaves which are equally spaced around a circle.He chooses a leaf to start at, and to make the base layer he travels to each leaf one at a time, makingstraight lines of silk from one leaf to another, such that no two of the lines of silk cross each other andhe visits every leaf exactly once. In how many ways can the spider make the base layer of the web?Express your answer in terms of n.

17. [9] How many positive integers n ≤ 2009 have the property that blog2(n)c is odd?

18. [9] Find the positive integer n such that n3 + 2n2 + 9n+ 8 is the cube of an integer.

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12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

19. [10] Shelly writes down a vector v = (a, b, c, d), where 0 < a < b < c < d are integers. Let σ(v) denotethe set of 24 vectors whose coordinates are a, b, c, and d in some order. For instance, σ(v) contains(b, c, d, a). Shelly notes that there are 3 vectors in σ(v) whose sum is of the form (s, s, s, s) for some s.What is the smallest possible value of d?

20. [10] A positive integer is called jubilant if the number of 1’s in its binary representation is even. Forexample, 6 = 1102 is a jubilant number. What is the 2009th smallest jubilant number?

21. [10] Simplify2 cos2(ln(2009)i) + i sin(ln(4036081)i).

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12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

22. [10] A circle having radius r1 centered at point N is externally tangent to a circle of radius r2 centeredat M . Let l and j be the two common external tangent lines to the two circles. A circle centered atP with radius r2 is externally tangent to circle N at the point at which l coincides with circle N , andline k is externally tangent to P and N such that points M , N , and P all lie on the same side of k.For what ratio r1/r2 are j and k parallel?

23. [10] The roots of z6 + z4 + z2 + 1 = 0 are the vertices of a convex polygon in the complex plane. Findthe sum of the squares of the side lengths of the polygon.

24. [10] Compute, in terms of n,n∑

k=0

(n− kk

)2k.

Note that whenever s < t,(st

)= 0.

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3

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12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

25. [12] Four points, A, B, C, andD, are chosen randomly on the circumference of a circle with independentuniform probability. What is the expected number of sides of triangle ABC for which the projectionof D onto the line containing the side lies between the two vertices?

26. [12] Define the sequence {xi}i≥0 by x0 = 2009 and xn = − 2009n

∑n−1k=0 xk for all n ≥ 1. Compute the

value of∑2009

k=0 2nxn.

27. [12] Circle Ω has radius 5. Points A and B lie on Ω such that chord AB has length 6. A unit circle ωis tangent to chord AB at point T. Given that ω is also internally tangent to Ω, find AT ·BT.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

28. [15] The vertices of a regular hexagon are labeled cos(θ), cos(2θ), . . . , cos(6θ). For every pair of ver-tices, Benjamin draws a blue line through the vertices if one of these functions can be expressed as apolynomial function of the other (that holds for all real θ), and otherwise Roberta draws a red linethrough the vertices. In the resulting graph, how many triangles whose vertices lie on the hexagonhave at least one red and at least one blue edge?

29. [15] The average of a set of distinct primes is 27. What is the largest prime that can be in this set?

30. [15] Let f be a polynomial with integer coefficients such that the greatest common divisor of all itscoefficients is 1. For any n ∈ N, f(n) is a multiple of 85. Find the smallest possible degree of f .

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31. [18] How many ways are there to win tic-tac-toe in Rn? (That is, how many lines pass through threeof the lattice points (a1, . . . , an) in Rn with each coordinate ai in {1, 2, 3}?) Express your answer interms of n.

32. [18] Circle Ω has radius 13. Circle ω has radius 14 and its center P lies on the boundary of circleΩ. Points A and B lie on Ω such that chord AB has length 24 and is tangent to ω at point T. FindAT ·BT.

33. [18] Let m be a positive integer. Let d(n) denote the number of divisors of n, and define the function

F (x) =105m∑n=1

d(n)nx

.

Define the numbers a(n) to be the positive integers for which

F (x)2 =1052m∑n=1

a(n)nx

for all real x. Express a(105m) in terms of m.

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34. [≤ 25] How many integer lattice points (points of the form (m,n) for integers m and n) lie inside oron the boundary of the disk of radius 2009 centered at the origin?

If your answer is higher than the correct answer, you will receive 0 points. If your answer is d less thanthe correct answer, your score on this problem will be the larger of 0 and 25− bd/10c.

35. [≤ 25] Von Neumann’s Poker: The first step in Von Neumann’s game is selecting a random numberon [0, 1]. To generate this number, Chebby uses the factorial base: the number 0.A1A2A3A4 . . . standsfor∑∞

n=0An

(n+1)! , where each An is an integer between 0 and n, inclusive.

Chebby has an infinite number of cards labeled 0,1,2, . . .. He begins by putting cards 0 and 1 into ahat and drawing randomly to determine A1. The card assigned A1 does not get reused. Chebby thenadds in card 2 and draws for A2, and continues in this manner to determine the random number. Ateach step, he only draws one card from two in the hat.

Unfortunately, this method does not result in a uniform distribution. What is the expected value ofChebby’s final number?

Your score on this problem will be the larger of 0 and b25(1 − d)c, where d is the positive differencebetween your answer and the correct answer.

36. [≤ 25] Euler’s Bridge: The following figure is the graph of the city of Konigsburg in 1736 - verticesrepresent sections of the cities, edges are bridges. An Eulerian path through the graph is a path whichmoves from vertex to vertex, crossing each edge exactly once. How many ways could World War IIbombers have knocked out some of the bridges of Konigsburg such that the Allied victory parade couldtrace an Eulerian path through the graph? (The order in which the bridges are destroyed matters.)

Your score on this problem will be the larger of 0 and 25 − bd/10c, where d is the positive differencebetween your answer and the correct answer.

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12th Annual Harvard-MIT Mathematics TournamentSaturday 21 February 2009

Guts Round

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12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

1. [5] Compute1 · 22 + 2 · 32 + 3 · 42 + · · ·+ 19 · 202.

Answer: 41230 y Solution: We can write this as (13 + 23 + · · · + 203) − (12 + 22 + · · · + 202),

which is equal to 44100− 2870 = 41230.

2. [5] Given that sinA+ sinB = 1 and cosA+ cosB = 3/2, what is the value of cos(A−B)?

Answer: 5/8

Solution: Squaring both equations and add them together, one obtains 1+9/4 = 2+2(cos(A) cos(B)+sin(A) sin(B)) = 2 + 2 cos(A−B). Thus cosA−B = 5/8.

3. [5] Find all pairs of integer solutions (n,m) to

23n

= 32m

− 1.

Answer: (0, 0) and (1, 1)

Solution: We find all solutions of 2x = 3y − 1 for positive integers x and y. If x = 1, we obtain thesolution x = 1, y = 1, which corresponds to (n,m) = (0, 0) in the original problem. If x > 1, considerthe equation modulo 4. The left hand side is 0, and the right hand side is (−1)y−1, so y is even. Thuswe can write y = 2z for some positive integer z, and so 2x = (3z − 1)(3z + 1). Thus each of 3z − 1 and3z + 1 is a power of 2, but they differ by 2, so they must equal 2 and 4 respectively. Therefore, theonly other solution is x = 3 and y = 2, which corresponds to (n,m) = (1, 1) in the original problem.

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4. [6] Simplify: i0 + i1 + · · ·+ i2009.

Answer: 1 + i

Solution: By the geometric series formula, the sum is equal to i2010−1i−1 = −2

i−1 = 1 + i.

5. [6] In how many distinct ways can you color each of the vertices of a tetrahedron either red, blue, orgreen such that no face has all three vertices the same color? (Two colorings are considered the sameif one coloring can be rotated in three dimensions to obtain the other.)

Answer: 6

Solution: If only two colors are used, there is only one possible arrangement up to rotation, so thisgives 3 possibilities. If all three colors are used, then one is used twice. There are 3 ways to choosethe color that is used twice. Say this color is red. Then the red vertices are on a common edge, andthe green and blue vertices are on another edge. We see that either choice of arrangement of the greenand blue vertices is the same up to rotation. Thus there are 6 possibilities total.

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6. [6] Let ABC be a right triangle with hypotenuse AC. Let B′ be the reflection of point B across AC,and let C ′ be the reflection of C across AB′. Find the ratio of [BCB′] to [BC ′B′].

Answer: 1 Solution: Since C, B′, and C ′ are collinear, it is evident that [BCB′] = 12 [BCC ′]. It

immediately follows that [BCB′] = [BC ′B′]. Thus, the ratio is 1 .

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7. [6] How many perfect squares divide 23 · 35 · 57 · 79?

Answer: 120

Solution: The number of such perfect squares is 2 · 3 · 4 · 5, since the exponent of each prime can beany nonnegative even number less than the given exponent.

8. [6] Which is greater, log2008(2009) or log2009(2010)?

Answer: log2008 2009 .

Solution: Let f(x) = logx(x + 1). Then f ′(x) = x ln x−(x+1) ln(x+1)x(x+1) ln2 x

< 0 for any x > 1, so f isdecreasing. Thus log2008(2009) is greater.

9. [6] An icosidodecahedron is a convex polyhedron with 20 triangular faces and 12 pentagonal faces.How many vertices does it have?

Answer: 30

Solution: Since every edge is shared by exactly two faces, there are (20 · 3 + 12 · 5)/2 = 60 edges.Using Euler’s formula v − e+ f = 2, we see that there are 30 vertices.

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10. [7] Let a, b, and c be real numbers. Consider the system of simultaneous equations in variables x andy:

ax+ by = c− 1(a+ 5)x+ (b+ 3)y = c+ 1

Determine the value(s) of c in terms of a such that the system always has a solution for any a and b.

Answer: 2a/5 + 1. (or 2a+55 )

Solution: We have to only consider when the determinant of(

a ba+5 b+3

)is zero. That is, when

b = 3a/5. Plugging in b = 3a/5, we find that (a+ 5)(c− 1) = a(c+ 1) or that c = 2a/5 + 1.

11. [7] There are 2008 distinct points on a circle. If you connect two of these points to form a line and thenconnect another two points (distinct from the first two) to form another line, what is the probabilitythat the two lines intersect inside the circle?

Answer: 1/3

Solution: Given four of these points, there are 3 ways in which to connect two of them and thenconnect the other two, and of these possibilities exactly one will intersect inside the circle. Thus 1/3of all the ways to connect two lines and then connect two others have an intersection point inside thecircle.

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12. [7] Bob is writing a sequence of letters of the alphabet, each of which can be either uppercase orlowercase, according to the following two rules:

• If he had just written an uppercase letter, he can either write the same letter in lowercase afterit, or the next letter of the alphabet in uppercase.

• If he had just written a lowercase letter, he can either write the same letter in uppercase afterit, or the preceding letter of the alphabet in lowercase.

For instance, one such sequence is aAaABCDdcbBC. How many sequences of 32 letters can he writethat start at (lowercase) a and end at (lowercase) z? (The alphabet contains 26 letters from a to z.)

Answer: 376

Solution: The smallest possible sequence from a to z is aABCD . . . Zz, which has 28 letters. Toinsert 4 more letters, we can either switch two (not necessarily distinct) letters to lowercase and backagain (as in aABCcCDEFfFGH . . . Zz), or we can insert a lowercase letter after its correspondinguppercase letter, insert the previous letter of the alphabet, switch back to uppercase, and continue thesequence (as in aABCcbBCDE . . . Zz). There are

(272

)= 13 · 27 sequences of the former type and 25

of the latter, for a total of 376 such sequences.

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13. [8] How many ordered quadruples (a, b, c, d) of four distinct numbers chosen from the set {1, 2, 3, . . . , 9}satisfy b < a, b < c, and d < c?

Answer: 630

Solution: Given any 4 elements p < q < r < s of {1, 2, . . . , 9}, there are 5 ways of rearrangingthem to satisfy the inequality: prqs, psqr, qspr, qrps, and rspq. This gives a total of

(94

)· 5 = 630

quadruples.

14. [8] Compute2009∑k=1

k

(⌊2009k

⌋−⌊

2008k

⌋).

Answer: 2394

Solution: The summand is equal to k if k divides 2009 and 0 otherwise. Thus the sum is equal tothe sum of the divisors of 2009, or 2394.

15. [8] Stan has a stack of 100 blocks and starts with a score of 0, and plays a game in which he iteratesthe following two-step procedure:

(a) Stan picks a stack of blocks and splits it into 2 smaller stacks each with a positive number ofblocks, say a and b. (The order in which the new piles are placed does not matter.)

(b) Stan adds the product of the two piles’ sizes, ab, to his score.

The game ends when there are only 1-block stacks left. What is the expected value of Stan’s score atthe end of the game?

Answer: 4950

Solution: Let E(n) be the expected value of the score for an n-block game. It suffices to show thatthe score is invariant regardless of how the game is played. We proceed by induction. We have E(1) = 0and E(2) = 1. We require that E(n) = E(n−k)+E(k)+(n−k)k for all k. Setting k = 1, we hypothesizethat E(n) = n(n− 1)/2. This satisfies the recursion and base cases so E(100) = 100 · 99/2 = 4950.

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12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

16. [9] A spider is making a web between n > 1 distinct leaves which are equally spaced around a circle.He chooses a leaf to start at, and to make the base layer he travels to each leaf one at a time, makinga straight line of silk between each consecutive pair of leaves, such that no two of the lines of silk crosseach other and he visits every leaf exactly once. In how many ways can the spider make the base layerof the web? Express your answer in terms of n.

Answer: n2n−2

Solution: There are n ways to choose a starting vertex, and at each vertex he has only two choicesfor where to go next: the nearest untouched leaf in the clockwise direction, and the nearest untouchedleaf in the counterclockwise direction. For, if the spider visited a leaf which is not nearest in somedirection, there are two untouched leaves which are separated by this line of silk, and so the silk wouldeventually cross itself. Thus, for the first n−2 choices there are 2 possibilities, and the (n−1)st choiceis then determined.

Note: This formula can also be derived recursively.

17. [9] How many positive integers n ≤ 2009 have the property that blog2(n)c is odd?

Answer: 682

Solution: We wish to find n such that there is some natural number k for which 2k − 1 ≤ log2 n <2k. Since n ≤ 2009 we must have k ≤ 5. This is equivalent to finding the number of positiveintegers n ≤ 2009 satisfying 22k−1 ≤ n < 22k for some k ≤ 5, so the number of such integers is2 + 23 + 25 + 27 + 29 = 682.

18. [9] If n is a positive integer such that n3 + 2n2 + 9n+ 8 is the cube of an integer, find n.

Answer: 7

Solution: Since n3 < n3 + 2n2 + 9n + 8 < (n + 2)3, we must have n3 + 2n2 + 9n + 8 = (n + 1)3.Thus n2 = 6n+ 7, so n = 7.

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19. [10] Shelly writes down a vector v = (a, b, c, d), where 0 < a < b < c < d are integers. Let σ(v) denotethe set of 24 vectors whose coordinates are a, b, c, and d in some order. For instance, σ(v) contains(b, c, d, a). Shelly notes that there are 3 vectors in σ(v) whose sum is of the form (s, s, s, s) for some s.What is the smallest possible value of d?

Answer: 6

Solution: If k = a + b + c + d, first you notice 4 | 3k, and k ≥ 10. So we try k = 12, which workswith a, b, c, d = 1, 2, 3, 6 and not 1, 2, 4, 5.

20. [10] A positive integer is called jubilant if the number of 1’s in its binary representation is even. Forexample, 6 = 1102 is a jubilant number. What is the 2009th smallest jubilant number?

Answer: 4018

Solution: Notice that for each pair of consecutive positive integers 2k and 2k + 1, their binaryrepresentation differs by exactly one 1 (in the units digit), so exactly one of 2 and 3 is jubilant, exactlyone of 4 and 5 is jubilant, etc. It follows that there are exactly 2009 jubilant numbers less than or equalto 4019. We now simply need to check whether 4018 or 4019 is jubilant. Since the binary representationof 4018 is 111110110010, 4018 is the 2009th jubilant number.

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21. [10] Simplify2 cos2(ln(2009)i) + i sin(ln(4036081)i).

Answer: 40360824036081 Solution: We have

2 cos2(ln(2009)i) + i sin(ln(4036081)i) = 1 + cos(2 ln(2009)i) + i sin(ln(4036081)i)= 1 + cos(ln(4036081)i) + i sin(ln(4036081)i)

= 1 + ei2 ln(4036081)

= 1 +1

4036081

=40360824036081

as desired.

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22. [10] A circle having radius r1 centered at point N is tangent to a circle of radius r2 centered at M .Let l and j be the two common external tangent lines to the two circles. A circle centered at P withradius r2 is externally tangent to circle N at the point at which l coincides with circle N , and line k isexternally tangent to P and N such that points M , N , and P all lie on the same side of k. For whatratio r1/r2 are j and k parallel?

Answer: 3

Solution: Suppose the lines are parallel. Draw the other tangent line to N and P - since M andP have the same radius, it is tangent to all three circles. Let j and k meet circle N at A and B,respectively. Then by symmetry we see that ∠ANM = ∠MNP = ∠PNB = 60◦ since A, N , and Bare collinear (perpendicular to j and k). Let D be the foot of the perpendicular from M to AN . In4MDN , we have MN = 2DN , so r1 + r2 = 2(r1 − r2), and so r1/r2 = 3.

23. [10] The roots of z6 + z4 + z2 + 1 = 0 are the vertices of a convex polygon in the complex plane. Findthe sum of the squares of the side lengths of the polygon.

Answer: 12− 4√

2

Solution: Factoring the polynomial as (z4 + 1)(z2 + 1) = 0, we find that the 6 roots are e±iπ/4,e±iπ/2, e±i3π/4. The calculation then follows from the Law of Cosines or the distance formula.

24. [10] Compute, in terms of n,n∑k=0

(n− kk

)2k.

Note that whenever s < t,(st

)= 0.

Answer: 2·2n+(−1)n

3

Solution: Let Tn =∑nk=0

(n−kk

)2k. From Pascal’s recursion for binomial coefficients, we can find

Tn = 2Tn−2 + Tn−1, with T0 = 1 and T1 = 1. The characteristic polynomial of this recursion isx2 − x − 2 = 0, which has roots 2 and −1. Thus Tn = a · 2n + b · (−1)n for some a and b. From theinitial conditions we have a + b = 1 and 2a − b = 1. It follows that a = 2/3 and b = 1/3, from whichthe conclusion follows.

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25. [12] Four points, A, B, C, andD, are chosen randomly on the circumference of a circle with independentuniform probability. What is the expected number of sides of triangle ABC for which the projectionof D onto the line containing the side lies between the two vertices?

Answer: 3/2

Solution: By linearity of expectations, the answer is exactly 3 times the probability that the orthog-onal projection of D onto AB lies interior to the segment. This happens exactly when either ∠DABor ∠DBA is obtuse, which is equivalent to saying that A and B lie on the same side of the diameterthrough D. This happens with probability 1/2. Therefore, desired answer is 3/2.

26. [12] Define the sequence {xi}i≥0 by x0 = 2009 and xn = − 2009n

∑n−1k=0 xk for all n ≥ 1. Compute the

value of∑2009n=0 2nxn.

Answer: 2009

Solution: We have

− nxn2009

= xn−1 + xn−2 + ...+ x0 = xn−1 +(n− 1)xn−1

2009

, which yields the recursion xn = n−2010n xn−1. Unwinding this recursion, we find xn = (−1)n · 2009 ·(

2008n

). Thus

2009∑k=0

2nxn =2009∑k=0

(−2)n · 2009 ·(

2008n

)

= 20092008∑k=0

(−2)n(

2008n

)= 2009(−2 + 1)2008

as desired.

27. [12] Circle Ω has radius 5. Points A and B lie on Ω such that chord AB has length 6. A unit circle ωis tangent to chord AB at point T. Given that ω is also internally tangent to Ω, find AT ·BT.Answer: 2

Solution: Let M be the midpoint of chord AB and let O be the center of Ω. Since AM = BM = 3,Pythagoras on triangle AMO gives OM = 4. Now let ω be centered at P and say that ω and Ω aretangent at Q. Because the diameter of ω exceeds 1, points P and Q lie on the same side of AB. Bytangency, O,P , and Q are collinear, so that OP = OQ−PQ = 4. Let H be the orthogonal projectionof P onto OM ; then OH = OM−HM = OM−PT = 3. Pythagoras on OHP gives HP 2 = 7. Finally,

AT ·BT = AM2 −MT 2 = AM2 −HP 2 = 9− 7 = 2.

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28. [15] The vertices of a regular hexagon are labeled cos(θ), cos(2θ), . . . , cos(6θ). For every pair of vertices,Bob draws a blue line through the vertices if one of these functions can be expressed as a polynomialfunction of the other (that holds for all real θ), and otherwise Roberta draws a red line through the

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vertices. In the resulting graph, how many triangles whose vertices lie on the hexagon have at leastone red and at least one blue edge?

Answer: 14 Solution: The existence of the Chebyshev polynomials, which express cos(nθ) as

a polynomial in cos(θ), imply that Bob draws a blue line between cos(θ) and each other vertex, andalso between cos(2θ) and cos(4θ), between cos(2θ) and cos(6θ), and between cos(3θ) and cos(6θ) (bysubstituting θ′ = 2θ or 3θ as necessary). We now show that Roberta draws a red line through eachother pair of vertices.

Let m and n be positive integers. Notice that cos(nθ) is a periodic function with period 2πn , and

cos(mθ) is periodic with period 2πm . Thus, any polynomial in cos(mθ) is also periodic of period 2π

m .This may not be the minimum period of the polynomial, however, so the minimum period is 2π

mk forsome k. Therefore, if cos(nθ) can be expressed as a polynomial in cos(mθ) then 2π

n = 2πmk for some k,

so m | n. This shows that there is a blue line between two vertices cos(aθ) and cos(bθ) if and only ifone of a or b divides the other.

Drawing the graph, one can easily count that there are 3 triangles with all blue edges, 3 triangles withall red edges, and

(63

)= 20 triangles total. Thus there are 20− 3− 3 = 14 triangles having at least one

red and at least one blue edge.

29. [15] The average of a set of distinct primes is 27. What is the largest prime that can be in this set?

Answer: 139 Solution: Denote the set of these primes by A and the number of elements in A by

n. There are 9 primes smaller than 27, namely 2,3,5,7,11,13,17,19 and 23. Since 27 is odd and all primesexcept 2 are odd, 2 /∈ A. Thus the largest prime p is at most 27·9−3−5−7−11−13−17−19−23 = 145,so p ≤ 141. When the primes are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 139, their average is 27. Thereforep = 139.

30. [15] Let f be a polynomial with integer coefficients such that the greatest common divisor of all itscoefficients is 1. For any n ∈ N, f(n) is a multiple of 85. Find the smallest possible degree of f .

Answer: 17 Solution: Notice that, if p is a prime and g is a polynomial with integer coefficients

such that g(n) ≡ 0 (mod p) for some n, then g(n + mp) is divisible by p as well for any integermultiple mp of p. Therefore, it suffices to find the smallest possible degree of a polynomial f for whichf(0), f(1), f(2), . . . , f(16) are divisible by 17 and by 5.

There is a polynomial of degree 17 with integer coefficients having f(0) = f(1) = · · · = f(16) = 0,namely f(x) = (x)(x− 1)(x− 2) · · · (x− 16). Thus the minimal degree is no larger than 17.

Now, let f be such a polynomial and consider f modulo 17. The polynomial has 17 roots, so it mustbe at least degree 17 when taken modulo 17. Thus f has degree at least 17 as well.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

31. [18] How many ways are there to win tic-tac-toe in Rn? (That is, how many lines pass through threeof the lattice points (a1, . . . , an) in Rn with each coordinate ai in {1, 2, 3}?) Express your answer interms of n.

Answer: (5n − 3n)/2 Solution: A line consists of three points. Each coordinate can do one

of three things passing from the first point to the last point: increase by 1 each time, stay the same,or decrease by 1 each time. There are three ways to stay the same (three coordinates), one way toincrease by 1, and one way to decrease by 1, so there are 5n possible types of behavior. Determiningthis behavior uniquely determines the end point and start point except that we have traced every lineexactly twice (forwards and backwards) and incorrectly counted the 3n “lines” where each coordinatestays the same, so we subtract 3n and divide by 2.

7

32. [18] Circle Ω has radius 13. Circle ω has radius 14 and its center P lies on the boundary of circleΩ. Points A and B lie on Ω such that chord AB has length 24 and is tangent to ω at point T. FindAT ·BT.Answer: 56 Solution: LetM be the midpoint of chordAB; thenAM = BM = 12 and Pythagoras

on triangle AMO gives MO = 5. Note that ∠AOM = ∠AOB/2 = ∠APB = ∠APT + ∠TPB ortan (∠AOM) = tan (∠APT + ∠TPB). Applying the tangent addition formula,

AM

MO=

ATTP + BT

TP

1− ATTP ·

BTTP

=AB · TP

TP 2 −AT ·BT,

from which AT ·BT = TP 2 −AB · TP ·MO/AM = 142 − 24 · 14 · 5/12 = 56.

33. [18] Let m be a positive integer. Let d(n) denote the number of divisors of n, and define the function

F (x) =105m∑n=1

d(n)nx

.

Define the numbers a(n) to be the positive integers for which

F (x)2 =1052m∑n=1

a(n)nx

for all real x. Express a(105m) in terms of m.

Answer:(m3+6m2+11m+6

6

)3

OR(m+3

3

)3(The expanded polynomial 1

216 (216+1188m+2826m2+3815m3+3222m4+1767m5+630m6+141m7+18m8 +m9) is also an acceptable answer.)

Solution: The denominator of a term in the expansion of F (x)2 is equal to nx if and only if it isa product of two terms of F of the form d(n/k)

(n/k)x and d(k)kx for some divisor k of n. Thus a(105m) =∑

k|105m d(k)d( 105m

k ). We can write k = 3a5b7c with a, b, c ≤ m for any divisor k of 105m, and in thiscase d(k) = (a+ 1)(b+ 1)(c+ 1). Thus the sum becomes∑

0≤a,b,c≤m

(a+ 1)(b+ 1)(c+ 1)(m− a+ 1)(m− b+ 1)(m− c+ 1).

For a fixed b and c, we can factor out (b+ 1)(c+ 1)(m− b+ 1)(m− c+ 1) from the terms having thisb and c and find that the sum is equal to

a(105m) =∑

0≤b,c≤m

(b+ 1)(c+ 1)(m− b+ 1)(m− c+ 1)

(m+1∑a=1

a(m− a+ 2)

)

=∑

0≤b,c≤m

(b+ 1)(c+ 1)(m− b+ 1)(m− c+ 1)(

(m+ 2)(m+ 1)(m+ 2)

2− (m+ 1)(m+ 2)(2m+ 3)

6

)

=∑

0≤b,c≤m

(b+ 1)(c+ 1)(m− b+ 1)(m− c+ 1)(

(3m+ 6− 2m− 3)(m+ 1)(m+ 2)6

)

=∑

0≤b,c≤m

(b+ 1)(c+ 1)(m− b+ 1)(m− c+ 1)(m+ 3

3

)

8

Fixing c and factoring out terms again, we find by a similar argument that a(105m) =(m+3

3

)3.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12th HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

34. [≤ 25] Descartes’s Blackjack: How many integer lattice points (points of the form (m,n) for integersm and n) lie inside or on the boundary of the disk of radius 2009 centered at the origin?

If your answer is higher than the correct answer, you will receive 0 points. If your answer is d less thanthe correct answer, your score on this problem will be the larger of 0 and 25− bd/10c.

Answer: 12679605

35. [≤ 25] Von Neumann’s Poker: The first step in Von Neumann’s game is selecting a random numberon [0, 1]. To generate this number, Chebby uses the factorial base: the number 0.A1A2A3A4 . . . standsfor∑∞n=0

An

(n+1)! , where each An is an integer between 0 and n, inclusive.

Chebby has an infinite number of cards labeled 0,1,2, . . .. He begins by putting cards 0 and 1 into ahat and drawing randomly to determine A1. The card assigned A1 does not get reused. Chebby thenadds in card 2 and draws for A2, and continues in this manner to determine the random number. Ateach step, he only draws one card from two in the hat.

Unfortunately, this method does not result in a uniform distribution. What is the expected value ofChebby’s final number?

Your score on this problem will be the larger of 0 and b25(1 − d)c, where d is the positive differencebetween your answer and the correct answer.

Answer: .57196

36. [≤ 25] Euler’s Bridge: The following figure is the graph of the city of Konigsburg in 1736 - verticesrepresent sections of the cities, edges are bridges. An Eulerian path through the graph is a path whichmoves from vertex to vertex, crossing each edge exactly once. How many ways could World War IIbombers have knocked out some of the bridges of Konigsburg such that the Allied victory parade couldtrace an Eulerian path through the graph? (The order in which the bridges are destroyed matters.)

Your score on this problem will be the larger of 0 and 25 − bd/10c, where d is the positive differencebetween your answer and the correct answer.

Answer: 13023

9

12th Annual Harvard-MIT Math Tournament

Saturday 21 February 2009

Team Round - Division A

A graph consists of a set of vertices and a set of edges connecting pairs of vertices. If verticesu and v are connected by an edge, we say they are adjacent. The degree of a vertex is the numberof vertices adjacent to it.

A coloring of a graph is an assignment of a color to each vertex of the graph. A coloring is goodif no two adjacent vertices are the same color. The chromatic number of a graph is the minimumnumber of colors needed in any good coloring of that graph.

For example, the chromatic number of the following graph is 4. The good coloring shown belowuses four colors, and there is no good coloring using only three colors because four of the verticesare mutually adjacent.

A D

B C

A

1. [8] Let n ≥ 3 be a positive integer. A triangulation of a convex n-gon is a set of n− 3 of itsdiagonals which do not intersect in the interior of the polygon. Along with the n sides, thesediagonals separate the polygon into n− 2 disjoint triangles. Any triangulation can be viewedas a graph: the vertices of the graph are the corners of the polygon, and the n sides and n−3diagonals are the edges.

For a fixed n-gon, different triangulations correspond to different graphs. Prove that all ofthese graphs have the same chromatic number.

2. (a) [4] Let P be a graph with one vertex vn for each positive integer n. If a < b, then an edgeconnects vertices va and vb if and only if b

a is a prime number. What is the chromaticnumber of P? Prove your answer.

(b) [6] Let T be a graph with one vertex vn for every integer n. An edge connects va and vb

if |a− b| is a power of two. What is the chromatic number of T? Prove your answer.

3. A graph is finite if it has a finite number of vertices.

(a) [6] Let G be a finite graph in which every vertex has degree k. Prove that the chromaticnumber of G is at most k + 1.

(b) [10] In terms of n, what is the minimum number of edges a finite graph with chromaticnumber n could have? Prove your answer.

4. A k-clique of a graph is a set of k vertices such that all pairs of vertices in the clique areadjacent.

(a) [6] Find a graph with chromatic number 3 that does not contain any 3-cliques.

Page 1 of 2

(b) [10] Prove that, for all n > 3, there exists a graph with chromatic number n that doesnot contain any n-cliques.

5. The size of a finite graph is the number of vertices in the graph.

(a) [15] Show that, for any n > 2, and any positive integer N , there are finite graphs withsize at least N and with chromatic number n such that removing any vertex (and all itsincident edges) from the graph decreases its chromatic number.

(b) [15] Show that, for any positive integers n and r, there exists a positive integer N suchthat for any finite graph having size at least N and chromatic number equal to n, itis possible to remove r vertices (and all their incident edges) in such a way that theremaining vertices form a graph with chromatic number at least n− 1.

6. For any set of graphs G1, G2, . . . , Gn all having the same set of vertices V , define their overlap,denoted G1 ∪ G2 ∪ · · · ∪ Gn, to be the graph having vertex set V for which two vertices areadjacent in the overlap if and only if they are adjacent in at least one of the graphs Gi. Inother words, the edge set of the overlap of the graphs Gi is the union of their edge sets.

(a) [10] Let G and H be graphs having the same vertex set and let a be the chromaticnumber of G and b the chromatic number of H. Find, in terms of a and b, the largestpossible chromatic number of G ∪H. Prove your answer.

(b) [10] Suppose G is a graph with chromatic number n. Suppose there exist k graphsG1, G2, . . . , Gk having the same vertex set as G such that G1 ∪ G2 ∪ · · · ∪ Gk = G andeach Gi has chromatic number at most 2. Show that k ≥ dlog2(n)e.

7. [20] Let n be a positive integer. Let Vn be the set of all sequences of 0’s and 1’s of length n.Define Gn to be the graph having vertex set Vn, such that two sequences are adjacent in Gn

if and only if they differ in either 1 or 2 places. For instance, if n = 3, the sequences (1, 0, 0),(1, 1, 0), and (1, 1, 1) are mutually adjacent, but (1, 0, 0) is not adjacent to (0, 1, 1).

Show that, if n + 1 is not a power of 2, then the chromatic number of Gn is at least n + 2.

8. [30] Two colorings are distinct if there is no way to relabel the colors to transform one intothe other. Equivalently, they are distinct if and only if there is some pair of vertices whichare the same color in one coloring but different colors in the other. For what pairs (n, k) ofpositive integers does there exist a finite graph with chromatic number n which has exactlyk distinct good colorings using n colors?

Page 2 of 2

12th Annual Harvard-MIT Math Tournament

Saturday 21 February 2009

Solutions: Team Round - Division A

1. [8] Let n ≥ 3 be a positive integer. A triangulation of a convex n-gon is a set of n− 3 of itsdiagonals which do not intersect in the interior of the polygon. Along with the n sides, thesediagonals separate the polygon into n− 2 disjoint triangles. Any triangulation can be viewedas a graph: the vertices of the graph are the corners of the polygon, and the n sides and n−3diagonals are the edges.

For a fixed n-gon, different triangulations correspond to different graphs. Prove that all ofthese graphs have the same chromatic number.

Solution: We will show that all triangulations have chromatic number 3, by induction onn. As a base case, if n = 3, a triangle has chromatic number 3. Now, given a triangulation ofan n-gon for n > 3, every edge is either a side or a diagonal of the polygon. There are n sidesand only n − 3 diagonals in the edge-set, so the Pigeonhole Principle guarentees a trianglewith two side edges. These two sides must be adjacent, so we can remove this triangle to leavea triangulation of an (n−1)-gon, which has chromatic number 3 by the inductive hypothesis.Adding the last triangle adds only one new vertex with two neighbors, so we can color thisvertex with one of the three colors not used on its neighbors.

2. (a) [6] Let P be a graph with one vertex vn for each positive integer n. If a < b, then an edgeconnects vertices va and vb if and only if b

a is a prime number. What is the chromaticnumber of P? Prove your answer.Answer: 2

Solution: At least two colors are needed in a good coloring of P . We show that two issufficient. Write the positive integer n as pe1

1 pe22 . . . pek

k , for distinct primes p1, p2, . . . pk,and let f(n) = e1 + e2 + . . . + ek. Notice that if va and vb are connected, then f(a) andf(b) have opposite parity. So, if we color vn red if f(n) is odd and blue otherwise, thetwo-coloring is good.

(b) [6] Let T be a graph with one vertex vn for every integer n. An edge connects va and vb

if |a− b| is a power of two. What is the chromatic number of T? Prove your answer.Answer: 3

Solution: Since v0, v1, and v2 are all connected to each other, three colors is necessary.Now, color vn red if n ≡ 0 (mod 3), blue if n ≡ 1 (mod 3), and green otherwise. Sinceva and vb are the same color only if 3|(a − b), no two connected vertices are the samecolor.

3. A graph is finite if it has a finite number of vertices.

(a) [6] Let G be a finite graph in which every vertex has degree k. Prove that the chromaticnumber of G is at most k + 1.

Solution: We find a good coloring with k + 1 colors. Order the vertices and colorthem one by one. Since each vertex has at most k neighbors, one of the k + 1 colors hasnot been used on a neighbor, so there is always a good color for that vertex. In fact, wehave shows that any graph in which every vertex has degree at most k can be coloredwith k + 1 colors.

1

(b) [10] In terms of n, what is the minimum number of edges a finite graph with chromaticnumber n could have? Prove your answer.

Answer: n(n−1)2

Solution: We prove this claim by induction - it holds for n = 1. Now assume theclaim holds for n, and consider a graph of chromatic number n + 1. This graph musthave at least one vertex of degree n, or else, by part a), it could be colored with only ncolors.Now, if we remove this vertex, the remaining graph must have chromatic number n orn + 1 - if the chromatic number is n− 1 or less, we can add the vertex back and give it anew color, creating a good coloring with only n colors. By the inductive hypothesis, thenew graph has at least n(n−1)

2 edges, so the original graph had at least n(n−1)2 +n = n(n+1)

2edges.The complete graph on n + 1 vertices has exactly n(n+1)

2 edges, so the lower bound istight and the inductive step is complete.

4. A k-clique of a graph is a set of k vertices such that all pairs of vertices in the clique areadjacent.

(a) [4] Find a graph with chromatic number 3 that does not contain any 3-cliques.

Solution: Consider a graph with 5 vertices arranged in a circle, with each vertexconnected to its two neighbors. If only two colors are used, it is impossible to alternatecolors to avoid using the same color on two adjacent vertices, so the chromatic numberis 3.

(b) [10] Prove that, for all n > 3, there exists a graph with chromatic number n that doesnot contain any n-cliques.

Solution: We prove the claim by induction on n. The case n = 3 was adressed in (a).Let n ≥ 3 and suppose G is a graph with chromatic number n containing no n-cliques.We produce a graph G′ with chromatic number n + 1 containing no (n + 1)-cliques asfollows. Add a vertex v to G, and add an edge from v to each vertex of G.To see this graph has chromatic number n + 1, observe that any coloring of the verticesof G′ restricts to a valid coloring of the vertices of G. So at least n distinct colors mustbe used among the vertices of G. In addition, another color must be used for v. Bycoloring v a new color, we have constructed a coloring of G′ having n + 1 colors.Lastly, any (n+1)-clique in G′ must have at least n vertices in G which form an n-clique,which is impossible. Therefore, G′ has no (n + 1)-cliques.

5. The size of a finite graph is the number of vertices in the graph.

(a) [15] Show that, for any n > 2, and any positive integer N , there are finite graphs withsize at least N and with chromatic number n such that removing any vertex (and all itsincident edges) from the graph decreases its chromatic number.

Solution: Let k > 1 be an odd number, and let G be a graph with k vertices arrangedin a circle, with each vertex connected to its two neighbors. If n = 3, these graphs canbe arbitrarily large, and are the graphs we need. If n > 3, let H be a complete graphon n − 3 vertices, and let J be the graph created by adding an edge from every vertexin G to every vertex in H. Then n − 3 colors are needed to color H and another 3 are

2

needed to color G, so n colors is both necessary and sufficient for a good coloring of J .Now, say a vertex is removed from J . There are two cases:If the vertex was removed from G, then the remaining vertices in G can be colored with2 colors, because the cycle has been broken. A set of n− 3 different colors can be usedto color H, so only n − 1 colors are needed to color the reduced graph. On the otherhand, if the vertex was removed from H, then n − 4 colors are used to color H and 3used to color G. So removing any vertex decreases the chromatic number of J .

(b) [15] Show that, for any positive integers n and r, there exists a positive integer N suchthat for any finite graph having size at least N and chromatic number equal to n, itis possible to remove r vertices (and all their incident edges) in such a way that theremaining vertices form a graph with chromatic number at least n− 1.

Solution: We claim that N = nr is large enough. Take a graph with at least nrvertices and chromatic number n, and take a good n-coloring of the graph. Then byPigeonhole, at least r of the vertices are the same color, which means that no pair ofthese r vertices is adjacent.Remove this r vertices. If the resulting graph can be colored with only n−2 colors, thenwe can add the r vertices back in and color them with a new (n− 1)st color, creating agood coloring of the graph with only n−1 colors. Since the original graph has chromaticnumber n, it must be impossible to color the smaller graph with n−2 colors, so we haveremoved r vertices without decreasing the chromatic number by 2 or more.

6. For any set of graphs G1, G2, . . . , Gn all having the same set of vertices V , define their overlap,denoted G1 ∪ G2 ∪ · · · ∪ Gn, to be the graph having vertex set V for which two vertices areadjacent in the overlap if and only if they are adjacent in at least one of the graphs Gi.

(a) [10] Let G and H be graphs having the same vertex set and let a be the chromaticnumber of G and b the chromatic number of H. Find, in terms of a and b, the largestpossible chromatic number of G ∪H. Prove your answer.Answer: ab

Solution: First, we show that we can always color G∪H using ab colors. Given a goodcoloring of G in a colors c1, . . . , ca and a good coloring of H using b colors d1, . . . , db,define ab new colors to be the ordered pairs (ci, dj). Label a vertex of G ∪H with thecolor (ci, dj) if it is colored ci in G and di in H. This gives a good coloring of G ∪H.Now, it only remains to find graphs G and H such that G∪H has chromatic number ab.Consider the complete graph Kab having ab vertices v1, . . . , vab (every pair of vertices isadjacent). Let G be the graph with vertices v1, . . . , vab such that vi is connected to vj

if and only if i− j is a multiple of a. Also, let H be the graph on v1, . . . , vab such thattwo vertices are adjacent if and only if they are not adjacent in G. Then G ∪H = Kab.Note that G has chromatic number a since it is the disjoint union of b complete graphson a vertices. Also, H has chromatic number b since we can color each set of verticesvi with a color corresponding to i modulo b to obtain a good coloring. The chromaticnumber of G ∪H is clearly ab, and so we have found such a pair of graphs.

(b) [10] Suppose G is a graph with chromatic number n. Suppose there exist k graphsG1, G2, . . . , Gk having the same vertex set as G such that G1 ∪ G2 ∪ · · · ∪ Gk = G andeach Gi has chromatic number at most 2. Show that k ≥ dlog2(n)e, and show that onecan always find such a decomposition of G into dlog2(n)e graphs.

3

Solution: [NOTE: This problem differs from the problem statement in the test asadministered at the 2009 HMMT. The reader is encouraged to try it before reading thesolution.]The bound on k follows from iterating part (a).Let G be a graph with chromatic number n. Consider a coloring of G using n colorslabeled 1, 2, . . . , n. For i from 1 to dlog2(n)e, define Gi to be the graph on the verticesof G for which two vertices are connected by an edge if and only if the ith digit fromthe right in the binary expansions of their colors do not match. Clearly each of thegraphs Gi have chromatic number at most 2, by coloring each node with the ith digitof the binary expansion of their color in G. Moreover, each edge occurs in some Gi,since if two vertices match in every digit they are not connected by an edge. ThereforeG1 ∪G2 ∪ · · · ∪Gdlog2(n)e = G, and so we have found such a decomposition of G.

7. [20] Let n be a positive integer. Let Vn be the set of all sequences of 0’s and 1’s of length n.Define Gn to be the graph having vertex set Vn, such that two sequences are adjacent in Gn

if and only if they differ in either 1 or 2 places. For instance, if n = 3, the sequences (1, 0, 0),(1, 1, 0), and (1, 1, 1) are mutually adjacent, but (1, 0, 0) is not adjacent to (0, 1, 1).

Show that, if n + 1 is not a power of 2, then the chromatic number of Gn is at least n + 2.

Solution: We will assume that there is a coloring with n+1 colors and derive a contradiction.For each string s, let Ts be the set consisting of all strings that differ from s in at most 1place. Thus Ts has size n + 1 and all vertices in Ts are adjacent. In particular, if there is an(n + 1)-coloring, then each color is used exactly once in Ts. Let c be one of the colors that weused. We will determine how many vertices are colored with c. We will do this by countingin two ways.

Let k be the number of vertices colored with color c. Each such vertex is part of Ts for exactlyn + 1 values of s. On the other hand, each Ts contains exactly one vertex with color c. Itfollows that k(n + 1) = 2n. In particular, since k is an integer, n + 1 divides 2n. This isa contradiction since n + 1 is now a power of 2 by assumption, so actually there can be non + 1-coloring, as claimed.

8. [30] Two colorings are distinct if there is no way to relabel the colors to transform one intothe other. Equivalently, they are distinct if and only if there is some pair of vertices whichare the same color in one coloring but different colors in the other. For what pairs (n, k) ofpositive integers does there exist a finite graph with chromatic number n which has exactlyk distinct good colorings using n colors?

Answer: (1, 1), (2, 2k) for integers k ≥ 0, and (n, k) for n > 2, k > 0

Solution: If n = 1, there is only one coloring. If n = 2, then each connected component ofthe graph can be colored in two ways, because the color of any vertex in the graph determinesthe colors of all vertices connected to it. If the color scheme in one component is fixed, andthere are k components, then there are 2k−1 ways to finish the coloring.

Now say n > 2. We construct a graph with k different colorings. We begin with a completegraph G on n vertices, which can be colored in exactly one way. Let v1, v2, and v3 be threevertices in the complete graph. If k > 1, add to the graph a row of vertices w1, w2, . . . wk−1,such that wi is connected to wi+1 for 1 ≤ k − 2. Now, if i ≡ 0 (3), connect wi to all the

4

vertices in G except v1 and v2. If i ≡ 1 (mod 3), connect wi to all the vertices in G exceptv2 and v3, and if i ≡ 2 (mod 3), connect wi to all the vertices in G except v1 and v3.

We need to show that this graph can be colored with n colors in exactly k different ways.Say that v1 is colored red, v2 blue, and v3 green. Then each of the wi can be colored one ofexactly two colors. Further, there is exactly one possible color that wi and wi+1 could bothbe. Call the color wi and wi+1 could both be wi’s leading color, and call wi’s other color itslagging color. Notice that wi’s lagging color is wi+1’s leading color. So, if any wi is coloredwith its lagging color, then all wj with j > i are also colored with their lagging colors.

So one possibility is that all the wi are colored with their leading colors. Otherwise, some ofthem are colored with their lagging colors - these colorings are completely defined by whichone of the k− 1 wi is the first vertex colored with its lagging color. So there are 1 + k− 1 ork colorings of this graph, as needed.

5

12th Annual Harvard-MIT Math Tournament

Saturday 21 February 2009

Team Round - Division B

A graph consists of a set of vertices and a set of edges connecting pairs of vertices. If vertices uand v are connected by an edge, we say they are adjacent. The degree of a vertex u is the numberof vertices v which are adjacent to u. A finite graph has a finite number of vertices.

A coloring of a graph is an assignment of a color to each vertex of the graph. A coloring is goodif no two adjacent vertices are the same color. The chromatic number of a graph is the minimumnumber of colors needed in any good coloring of that graph.

For example, the chromatic number of the following graph is 4. The good coloring shown belowuses four colors, and there is no good coloring using only three colors because four of the verticesare mutually adjacent.

A D

B C

A

1. [6] What are the chromatic numbers of each of the three graphs shown below? Draw a coloringhaving the minimum number of colors for each. Label the vertices with symbols to indicatethe color of each vertex. (For example, you may mark a vertex “R” if you wish to indicatethat it is colored red.)

2. [6] In a connected graph, it is possible to reach any vertex from any other vertex by followingthe edges. A tree is a connected graph with n vertices and n − 1 edges for some positiveinteger n. Suppose n ≥ 2. What is the chromatic number of a tree having n vertices? Proveyour answer.

3. [8] Let n ≥ 3 be a positive integer. A triangulation of a convex n-gon is a set of n− 3 of itsdiagonals which do not intersect in the interior of the polygon. Along with the n sides, thesediagonals separate the polygon into n− 2 disjoint triangles. Any triangulation can be viewedas a graph: the vertices of the graph are the corners of the polygon, and the n sides and n−3diagonals are the edges.

For a fixed n-gon, different triangulations correspond to different graphs. Prove that all ofthese graphs have the same chromatic number.

4. [10] Let G be a finite graph in which every vertex has degree less than or equal to k. Provethat the chromatic number of G is less than or equal to k + 1.

Page 1 of 2

5. [10] A k-clique of a graph is a set of k vertices such that all pairs of vertices in the clique areadjacent. The clique number of a graph is the size of the largest clique in the graph. Doesthere exist a graph which has a clique number smaller than its chromatic number?

6. (a) [5] If a single vertex is removed from a finite graph, show that the graph’s chromaticnumber cannot decrease by more than 1.

(b) [15] Show that, for any n > 2, there are infinitely many graphs with chromatic numbern such that removing any vertex from the graph decreases its chromatic number.

Page 2 of 2

12th Annual Harvard-MIT Math Tournament

Saturday 21 February 2009

Solutions: Team Round - Division B

1. [6] What are the chromatic numbers of each of the three graphs shown below? Draw a coloringhaving the minimum number of colors for each. Label the vertices with symbols to indicatethe color of each vertex. (For example, you may mark a vertex “R” if you wish to indicatethat it is colored red.)

Solution: The chromatic numbers of the three graphs are 3, 5, and 2, respectively.

2. [6] In a connected graph, it is possible to reach any vertex from any other vertex by followingthe edges. A tree is a connected graph with n vertices and n − 1 edges for some positiveinteger n. Suppose n ≥ 2. What is the chromatic number of a tree having n vertices? Proveyour answer.

Solution: The chromatic number of any tree is 2. We show this by induction on the sizeof the tree. A tree with 2 nodes can clearly be 2-colored. Now, suppose a tree of size n − 1can be colored in 2 colors. Given a tree of size n, choose any leaf (node with only one edgecoming out of it), say l, of the tree, and color it red. The node it is adjacent to, say x, mustbe colored a different color, say blue. Using the inductive hypothesis, we can 2-color the treeformed by removing l, and we can choose the color of x to be blue and the other color to bered. Thus the entire tree on n vertices can be two-colored, completing the induction.

3. [8] Let n ≥ 3 be a positive integer. A triangulation of a convex n-gon is a set of n− 3 of itsdiagonals which do not intersect in the interior of the polygon. Along with the n sides, thesediagonals separate the polygon into n− 2 disjoint triangles. Any triangulation can be viewedas a graph: the vertices of the graph are the corners of the polygon, and the n sides and n−3diagonals are the edges.

For a fixed n-gon, different triangulations correspond to different graphs. Prove that all ofthese graphs have the same chromatic number.

Solution: We will show that all triangulations have chromatic number 3, by induction onn. As a base case, if n = 3, a triangle has chromatic number 3. Now, given a triangulation ofan n-gon for n > 3, every edge is either a side or a diagonal of the polygon. There are n sidesand only n − 3 diagonals in the edge-set, so the Pigeonhole Principle guarentees a trianglewith two side edges. These two sides must be adjacent, so we can remove this triangle to leavea triangulation of an n− 1-gon, which has chromatic number 3 by the inductive hypothesis.Adding the last triangle adds only one new vertex with two neighbors, so we can color thisvertex with one of the three colors not used on its neighbors.

4. [10] Let G be a finite graph in which every vertex has degree less than or equal to k. Provethat the chromatic number of G is less than or equal to k + 1.

1

Solution: Using a greedy algorithm we find a good coloring with k + 1 colors. Order thevertices and color them one by one - since each vertex has at most k neighbors, one of thek + 1 colors has not been used on a neighbor, so there is always a good color for that vertex.In fact, we have shows that any graph in which every vertex has degree at most k can becolored with k + 1 colors.

5. [10] A k-clique of a graph is a set of k vertices such that all pairs of vertices in the clique areadjacent. The clique number of a graph is the size of the largest clique in the graph. Doesthere exist a graph which has a clique number smaller than its chromatic number?

Solution: Consider a graph with 5 vertices arranged in a circle, with each vertex connectedto its two neighbors. If only two colors are used, it is impossible to alternate colors to avoidusing the same color on two adjacent vertices, so the chromatic number is 3. Its clique numberis 2, so we have found such a graph.

6. (a) [5] If a single vertex (and all its incident edges) is removed from a finite graph, showthat the graph’s chromatic number cannot decrease by more than 1.

Solution: Suppose the chromatic number of the graph was C, and removing a singlevertex resulted in a graph with chromatic number at most C−2. Then we can color theremaining graph with at most C − 2 colors. Replacing the vertex and its edges, we canthen choose any color not already used to form a coloring of the original graph using atmost C − 1 colors, contradicting the fact that C is the chromatic number of the graph.

(b) [15] Show that, for any n > 2, there are infinitely many graphs with chromatic numbern such that removing any vertex (and all its incident edges) from the graph decreasesits chromatic number.

Solution: Let k > 0 be an odd number, and let G be a graph with k vertices arrangedin a circle, with each vertex connected to its two neighbors. If n = 3, these graphs canbe arbitrarily large, and are the graphs we need. If n > 3, let H be a complete graphon n − 3 vertices, and let J be the graph created by adding an edge from every vertexin G to every vertex in H. Then n − 3 colors are needed to color H and another 3 areneeded to color G, so n colors is both necessary and sufficient for a good coloring of J .Now, say a vertex is removed from J . There are two cases:If the vertex was removed from G, then the remaining vertices in G can be colored with2 colors, because the cycle has been broken. A set of n− 3 different colors can be usedto color H, so only n − 1 colors are needed to color the reduced graph. On the otherhand, if the vertex was removed from H, then n − 4 colors are used to color H and 3used to color G. So removing any vertex decreases the chromatic number of J .

2

2nd Annual Harvard-MIT November TournamentSaturday 7 November 2009

General Test

1. [2] Evaluate the sum:

112 − 12 + 122 − 22 + 132 − 32 + . . . + 202 − 102.

2. [3] Given that a+ b+ c = 5 and that 1 ≤ a, b, c ≤ 2, what is the minimum possible value of 1a+b + 1

b+c?

3. [3] What is the period of the function f(x) = cos(cos(x))?

4. [4] How many subsets A of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} have the property that no two elements of A sumto 11?

5. [5] A polyhedron has faces that are all either triangles or squares. No two square faces share an edge,and no two triangular faces share an edge. What is the ratio of the number of triangular faces to thenumber of square faces?

6. [5] Find the maximum value of x + y, given that x2 + y2 − 3y − 1 = 0.

7. [6] There are 15 stones placed in a line. In how many ways can you mark 5 of these stones so thatthere are an odd number of stones between any two of the stones you marked?

8. [7] Let 4ABC be an equilateral triangle with height 13, and let O be its center. Point X is chosenat random from all points inside 4ABC. Given that the circle of radius 1 centered at X lies entirelyinside 4ABC, what is the probability that this circle contains O?

9. [7] A set of points is convex if the points are the vertices of a convex polygon (that is, a non-self-intersecting polygon with all angles less than or equal to 180◦). Let S be the set of points (x, y) suchthat x and y are integers and 1 ≤ x, y ≤ 26. Find the number of ways to choose a convex subset of Sthat contains exactly 98 points.

10. [8] Compute

∞∏n=0

(1−

(12

)3n

+(

14

)3n).

2nd Annual Harvard-MIT November TournamentSaturday 7 November 2009

General Test

1. [2] Evaluate the sum:

112 − 12 + 122 − 22 + 132 − 32 + . . .+ 202 − 102.

Answer: 2100 This sum can be written as∑10a=1 (a+ 10)2 − a2 =

∑10a=1 10(2a+ 10) = 10 ∗ 10 ∗

11 + 10 ∗ 10 ∗ 10 = 2100.

2. [3] Given that a+ b+ c = 5 and that 1 ≤ a, b, c ≤ 2, what is the minimum possible value of 1a+b + 1

b+c?

Answer: 47 If a > 1 and b < 2, we can decrease the sum by decreasing a and increasing b. You can

follow a similar procedure if c > 1 and b < 2. Therefore, the sum is minimized when b = 2. We canthen cross-multiply the two fractions and see that we are trying to minimize a+c+4

(a+2)(c+2) = 7(a+2)(c+2) .

The product of two numbers with a fixed sum is maximized when those two numbers are equal, so7

(a+2)(c+2) is minimized for a = c = 32 , which gives us an answer of 4

7 .

3. [3] What is the period of the function f(x) = cos(cos(x))?

Answer: π Since f(x) never equals cos(1) for x ∈ (0, π) but f(0) = cos(1), the period is at least π.However, cos(x+ π) = − cos(x), so cos(cos(x+ π)) = cos(cos(x)).

4. [4] How many subsets A of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} have the property that no two elements of A sumto 11?

Answer: 243 For each element listed, there is exactly one other element such that the two elementssum to 11. Thus, we can list all the 10 numbers above as 5 pairs of numbers, such that each pair sumsto 11. The problem then can be solved as follows: in any given subset with no two elements summingto 11, at most one element from each pair can be present. Thus, there are 3 ways in which each paircan contribute to a given subset (no element, the first element in the pair, or the second element inthe pair). Since there are 5 pairs, the total number of ways to construct a subset with no two elementssumming to 11 is 35 = 243.

5. [5] A polyhedron has faces that are all either triangles or squares. No two square faces share an edge,and no two triangular faces share an edge. What is the ratio of the number of triangular faces to thenumber of square faces?

Answer: 43 Let s be the number of square faces and t be the number of triangular faces. Every

edge is adjacent to exactly one square face and one triangular face. Therefore, the number of edges isequal to 4s, and it is also equal to 3t. Thus 4s = 3t and t

s = 43

6. [5] Find the maximum value of x+ y, given that x2 + y2 − 3y − 1 = 0.

Answer:√

26+32 We can rewrite x2 +y2−3y−1 = 0 as x2 +(y− 3

2 )2 = 134 . We then see that the set

of solutions to x2− y2− 3y− 1 = 0 is the circle of radius√

132 and center (0, 3

2 ). This can be written asx =

√132 cos(θ) and y =

√132 sin(θ)+ 3

2 . Thus, x+y = 32 +√

132 (cos(θ)+sin(θ)) = 3

2 +√

132

√2 sin(θ+45◦),

which is maximized for θ = 45◦ and gives√

26+32 . (We could also solve this geometrically by noting

that if x+ y attains a maximum value of s then the line x+ y = s is tangent to the circle.)

7. [6] There are 15 stones placed in a line. In how many ways can you mark 5 of these stones so thatthere are an odd number of stones between any two of the stones you marked?

Answer: 77 Number the stones 1 through 15 in order. We note that the condition is equivalentto stipulating that the stones have either all odd numbers or all even numbers. There are

(85

)ways to

choose 5 odd-numbered stones, and(75

)ways to choose all even-numbered stones, so the total number

of ways to pick the stones is(85

)+(75

)= 77. (

(nk

)is the number of ways to choose k out of n items. It

equals n!k!(n−k)! ).

8. [7] Let 4ABC be an equilateral triangle with height 13, and let O be its center. Point X is chosenat random from all points inside 4ABC. Given that the circle of radius 1 centered at X lies entirelyinside 4ABC, what is the probability that this circle contains O?

Answer:√

3π100 The set of points X such that the circle of radius 1 centered at X lies entirely inside

4ABC is itself a triangle, A′B′C ′, such that AB is parallel to A′B′, BC is parallel to B′C ′, and CA isparallel to C ′A′, and furthermore AB and A′B′, BC and B′C ′, and CA and C ′A′ are all 1 unit apart.We can use this to calculate that A′B′C ′ is an equilateral triangle with height 10, and hence has area100√

3. On the other hand, the set of points X such that the circle of radius 1 centered at X contains O

is a circle of radius 1, centered at O, and hence has area π. The probability that the circle centered atX contains O given that it also lies in ABC is then the ratio of the two areas, that is, π

100√3

=√

3π100 .

9. [7] A set of points is convex if the points are the vertices of a convex polygon (that is, a non-self-intersecting polygon with all angles less than or equal to 180◦). Let S be the set of points (x, y) suchthat x and y are integers and 1 ≤ x, y ≤ 26. Find the number of ways to choose a convex subset of Sthat contains exactly 98 points.

Answer: 4958 For this problem, let n = 26. A convex set may be divided into four subsets: a set ofpoints with maximal y coordinate, a set of points with minimal y coordinate, the points to the left ofone of these subsets, and the points to the right of one of these subsets (the left, top, right, and bottomof the corresponding convex polygon). Each of these four parts contains at most n points. (All pointsin the top or bottom have distinct x coordinates while all points in the left or right have distinct ycoordinates.) Moreover, there are four corners each of which is contained in two of these regions. Thisimplies that at most 4n − 4 distinct points are in any convex set. To find a set of size 4n − 6 we canremove 2 additional points. Either exactly one of the top, bottom, left, or right contains exactly n− 2points or some two of them each contain exactly n− 1 points.

Any of the(10098

)= 4950 sets of 98 points with either x or y coordinate either 1 or 26 have this property.

Suppose instead that some of the points have x coordinate and y coordinate both different from 1 andfrom 26. In this case we can check that it is impossible for one side to have n−2 points. If two oppositesides (top/bottom or left/right) have n−1 points, then we obtain all the points on the boundary of ann− 1 by n rectangle (of which there are four). If two adjacent sides (any of the other pairs) have n− 1points, then we obtain the points on the boundary of an n by n square with the points (1, 1), (1, 2),(2, 1) missing and the point (2, 2) added (or one of its rotations). There are an additional 4 such sets,for a total of 4958.

10. [8] Compute

∞∏n=0

(1−

(12

)3n

+(

14

)3n).

Answer: 23 We can rewrite each term as

1+( 12 )3n+1

1+( 12 )3n . In the infinite product, each term of the form

1+(

12

)3n

with n > 0 appears once in the numerator and once in the denominator. The only remainingterm is 1 +

(12

)1 in the first denominator.

2nd Annual Harvard-MIT November TournamentSaturday 7 November 2009

Theme Round

Shortest Paths

1. [3] Paul starts with the number 19. In one step, he can add 1 to his number, divide his number by 2,or divide his number by 3. What is the minimum number of steps Paul needs to get to 1?

2. [4] You start with a number. Every second, you can add or subtract any number of the form n! toyour current number to get a new number. In how many ways can you get from 0 to 100 in 4 seconds?(n! is defined as n× (n− 1)× (n− 2)× · · · × 2× 1, so 1! = 1, 2! = 2, 3! = 6, 4! = 24, etc.)

3. [5] Let C be the circle of radius 12 centered at (0, 0). What is the length of the shortest path in theplane between (8

√3, 0) and (0, 12

√2) that does not pass through the interior of C?

4. [6] You are given a 5× 6 checkerboard with squares alternately shaded black and white. The bottom-left square is white. Each square has side length 1 unit. You can normally travel on this board at aspeed of 2 units per second, but while you travel through the interior (not the boundary) of a blacksquare, you are slowed down to 1 unit per second. What is the shortest time it takes to travel fromthe bottom-left corner to the top-right corner of the board?

5. [7] The following grid represents a mountain range; the number in each cell represents the height of themountain located there. Moving from a mountain of height a to a mountain of height b takes (b− a)2

time. Suppose that you start on the mountain of height 1 and that you can move up, down, left, orright to get from one mountain to the next. What is the minimum amount of time you need to get tothe mountain of height 49?

1 3 6 10 15 21 282 5 9 14 20 27 344 8 13 19 26 33 397 12 18 25 32 38 4311 17 24 31 37 42 4616 23 30 36 41 45 4822 29 35 40 44 47 49

Five Guys

6. [3] There are five guys named Alan, Bob, Casey, Dan, and Eric. Each one either always tells the truthor always lies. You overhear the following discussion between them:

Alan: "All of us are truth-tellers."Bob: "No, only Alan and I are truth-tellers."Casey: "You are both liars."Dan: "If Casey is a truth-teller, then Eric is too."Eric: "An odd number of us are liars."

Who are the liars?

7. [4] Five guys are eating hamburgers. Each one puts a top half and a bottom half of a hamburger bunon the grill. When the buns are toasted, each guy randomly takes two pieces of bread off of the grill.What is the probability that each guy gets a top half and a bottom half?

Page 1 of 2

8. [5] A single burger is not enough to satisfy a guy’s hunger. The five guys go to Five Guys’ Restaurant,which has 20 different meals on the menu. Each meal costs a different integer dollar amount between$1 and $20. The five guys have $20 to split between them, and they want to use all the money to orderfive different meals. How many sets of five meals can the guys choose?

9. [6] Five guys each have a positive integer (the integers are not necessarily distinct). The greatestcommon divisor of any two guys’ numbers is always more than 1, but the greatest common divisor ofall the numbers is 1. What is the minimum possible value of the product of the numbers?

10. [7] Five guys join five girls for a night of bridge. Bridge games are always played by a team of twoguys against a team of two girls. The guys and girls want to make sure that every guy and girl playagainst each other an equal number of times. Given that at least one game is played, what is the leastnumber of games necessary to accomplish this?

Page 2 of 2

2nd Annual Harvard-MIT November TournamentSaturday 7 November 2009

Theme Round

Shortest Paths

1. [3] Paul starts with the number 19. In one step, he can add 1 to his number, divide his number by 2,or divide his number by 3. What is the minimum number of steps Paul needs to get to 1?

Answer: 6 One possible path is 19→ 20→ 10→ 5→ 6→ 2→ 1.

2. [4] You start with a number. Every second, you can add or subtract any number of the form n! toyour current number to get a new number. In how many ways can you get from 0 to 100 in 4 seconds?(n! is defined as n× (n− 1)× (n− 2)× · · · × 2× 1, so 1! = 1, 2! = 2, 3! = 6, 4! = 24, etc.)

Answer: 36 To get to 100, you have to use one number which is at least 5! = 120, because24× 4 = 96, which is less than 100. If you use 6! = 720 or anything larger, you need to get back from720 to 100 (or further) in three seconds. Since 3 · 5! < 620, there is no way to do this in 3 seconds.This means you have to use 5! at least once. The remaining numbers must get you from 120 to 100.If you use three numbers all at most 3!, you can move by at most 3 · 3! = 18 < 120− 100. This meansyou have to use 4!. From 120− 24 = 96, there are two ways to get to 100: adding 6 then subtracting2, or adding 2 twice. So, to get to 100 from 0 in four seconds, you must either add 120, subtract 24,add 6, and subtract 2, or add 120, subtract 24, and add 2 twice. You can do these steps in any order,so the first sequence yields 24 paths and the second sequence yields 12.

3. [5] Let C be the circle of radius 12 centered at (0, 0). What is the length of the shortest path in theplane between (8

√3, 0) and (0, 12

√2) that does not pass through the interior of C?

Answer: 12 + 4√

3 + π The shortest path consists of a tangent to the circle, a circular arc, andthen another tangent. The first tangent, from (8

√3, 0) to the circle, has length 4

√3, because it is a leg

of a 30-60-90 right triangle. The 15◦ arc has length 15360 (24π), or π, and the final tangent, to (0, 12

√2),

has length 12.

4. [6] You are given a 5× 6 checkerboard with squares alternately shaded black and white. The bottom-left square is white. Each square has side length 1 unit. You can normally travel on this board at aspeed of 2 units per second, but while you travel through the interior (not the boundary) of a blacksquare, you are slowed down to 1 unit per second. What is the shortest time it takes to travel fromthe bottom-left corner to the top-right corner of the board?

Answer: 1+5√

22 It is always faster to take a path around a black square than through it, since the

length of the hypotenuse of any right triangle is greater than half the sum of the length of its legs.Therefore, an optimal path always stays on white squares or on boundaries, and the shortest such pathhas length 1 + 5

√2.

5. [7] The following grid represents a mountain range; the number in each cell represents the height of themountain located there. Moving from a mountain of height a to a mountain of height b takes (b− a)2

time. Suppose that you start on the mountain of height 1 and that you can move up, down, left, orright to get from one mountain to the next. What is the minimum amount of time you need to get tothe mountain of height 49?

1 3 6 10 15 21 282 5 9 14 20 27 344 8 13 19 26 33 397 12 18 25 32 38 4311 17 24 31 37 42 4616 23 30 36 41 45 4822 29 35 40 44 47 49

Answer: 212 Consider the diagonals of the board running up and to the right - so the first diagonalis the square 1, the second diagonal is the squares 2 and 3, and so on. The ith ascent is the largest steptaken from a square in the ith diagonal to a square in the i + 1st. Since you must climb from square1 to square 49, the sum of the ascents is at least 48. Since there are 12 ascents, the average ascent isat least 4.

The 1st and 12th ascents are at most 2, and the 2nd and 11th ascents are at most 3. The 6thand 7th ascents are at least 6, and the 5th and 8th ascents are at least 5. Because f(x) = x2 is convex,the sum of squares of the ascents is minimized when they are as close together as possible. One possibleshortest path is then 1 → 3 → 6 → 10 → 14 → 19 → 25 → 31 → 36 → 40 → 44 → 47 → 49, whichhas ascents of size 2, 3, 4, 4, 5, 6, 6, 5, 4, 4, 3, and 2. Thus, our answer is 212, the sums of the squares ofthese ascents. There are other solutions to this problem. One alternative problem involves computingthe shortest path to each square of the graph, recursively, starting from squares 2 and 3.

Five Guys

6. [3] There are five guys named Alan, Bob, Casey, Dan, and Eric. Each one either always tells the truthor always lies. You overhear the following discussion between them:

Alan: "All of us are truth-tellers."Bob: "No, only Alan and I are truth-tellers."Casey: "You are both liars."Dan: "If Casey is a truth-teller, then Eric is too."Eric: "An odd number of us are liars."

Who are the liars?

Answer: Alan, Bob, Dan, and Eric Alan and Bob each claim that both of them are telling thetruth, but they disagree on the others. Therefore, they must both be liars, and Casey must be atruth-teller. If Dan is a truth-teller, then so is Eric, but then there would only be two truth-tellers,contradicting Eric’s claim. Therefore, Dan is a liar, and so is Eric.

7. [4] Five guys are eating hamburgers. Each one puts a top half and a bottom half of a hamburger bunon the grill. When the buns are toasted, each guy randomly takes two pieces of bread off of the grill.What is the probability that each guy gets a top half and a bottom half?

Answer: 863 Say a guy is content if he gets a top half and a bottom half. Suppose, without loss of

generality, that the first guy’s first piece of bread is a top. Then there is a 59 chance that his second

piece of bread is a bottom. By the same reasoning, given that the first guy is content, there is a 47

chance that the second guy is content. Given that the first two guys are content, there is a 35 chance

that the third guy is content, and so on. Our final answer is 59 ·

47 ·

35 ·

23 ·

11 = 8

63 .

8. [5] A single burger is not enough to satisfy a guy’s hunger. The five guys go to Five Guys’ Restaurant,which has 20 different meals on the menu. Each meal costs a different integer dollar amount between$1 and $20. The five guys have $20 to split between them, and they want to use all the money to orderfive different meals. How many sets of five meals can the guys choose?

Answer: 7 Suppose the meals, sorted in descending order, cost 5 +x1, 4 +x2, . . . , 1 +x5. To satisfythe conditions in the problem, the xi must be a non-increasing sequence of non-negative integers whichsums to 5. Therefore, there is exactly one order for each partition of 5: order the elements of thepartition from largest to smallest and use these parts as the xi. For example, the partition 3 + 2corresponds to the order 5 + 3, 4 + 2, 3, 2, 1. There are thus 7 orders, corresponding to the 7 partitionsof 5 below.

1 + 1 + 1 + 1 + 1, 1 + 1 + 1 + 2, 1 + 2 + 2, 1 + 1 + 3, 2 + 3, 1 + 4, 5

These partitions yield the following seven orders:

(2, 3, 4, 5, 6), (1, 3, 4, 5, 7), (1, 2, 4, 6, 7), (1, 2, 3, 5, 7),

(1, 2, 3, 6, 8), (1, 2, 3, 5, 9), (1, 2, 3, 4, 10)

9. [6] Five guys each have a positive integer (the integers are not necessarily distinct). The greatestcommon divisor of any two guys’ numbers is always more than 1, but the greatest common divisor ofall the numbers is 1. What is the minimum possible value of the product of the numbers?

Answer: 32400 Let ω(n) be the number of distinct prime divisors of a number. Each of the guys’numbers must have ω(n) ≥ 2, since no prime divides all the numbers. Therefore, if the answer hasprime factorization pe1

1 pe22 . . . pek

k , then e1 + e2 + . . .+ ek ≥ 10. If p2 divided any of the guys’ numbers,we could divide their number by p to reduce the product. Therefore we may assume ei ≤ 4 for eachi, so the smallest possible product is 243452. This bound is achievable: give the guys the numbers10, 6, 6, 6, and 15.

10. [7] Five guys join five girls for a night of bridge. Bridge games are always played by a team of twoguys against a team of two girls. The guys and girls want to make sure that every guy and girl playagainst each other an equal number of times. Given that at least one game is played, what is the leastnumber of games necessary to accomplish this?

Answer: 25 Suppose that each guy plays each girl t times. Since each guy plays against two girlsin one game, the total number of games each guy plays is 5t

2 . Then the total number of games is 25t4 ,

which is a multiple of 25 and therefore at least 25. To check that 25 games is enough, we arrange theguys and girls in two circles. A good pair of guys is a pair of guys who are adjacent in the circle; agood pair of girls is defined similarly. There are 5 good pairs of guys and girls — making each goodpair of guys play each good pair of girls works.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

1. [5] If f(x) = x/(x + 1), what is f(f(f(f(2009))))?

2. [5] A knight begins on the lower-left square of a standard chessboard. How many squares couldthe knight end up at after exactly 2009 legal knight’s moves? (A knight’s move is 2 squares eitherhorizontally or vertically, followed by 1 square in a direction perpendicular to the first.)

3. [5] Consider a square, inside which is inscribed a circle, inside which is inscribed a square, inside whichis inscribed a circle, and so on, with the outermost square having side length 1. Find the differencebetween the sum of the areas of the squares and the sum of the areas of the circles.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

4. [6] A cube has side length 1. Find the product of the lengths of the diagonals of this cube (a diagonalis a line between two vertices that is not an edge).

5. [6] Tanks has a pile of 5 blue cards and 5 red cards. Every morning, he takes a card and throws itdown a well. What is the probability that the first card he throws down and the last card he throwsdown are the same color?

6. [6] Find the last two digits of 10321032. Express your answer as a two-digit number.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

7. [7] A computer program is a function that takes in 4 bits, where each bit is either a 0 or a 1, andoutputs TRUE or FALSE. How many computer programs are there?

8. [7] The angles of a convex n-sided polygon form an arithmetic progression whose common difference(in degrees) is a non-zero integer. Find the largest possible value of n for which this is possible. (Apolygon is convex if its interior angles are all less than 180◦.)

9. [7] Daniel wrote all the positive integers from 1 to n inclusive on a piece of paper. After carefulobservation, he realized that the sum of all the digits that he wrote was exactly 10,000. Find n.

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

10. [8] Admiral Ackbar needs to send a 5-character message through hyperspace to the Rebels. Eachcharacter is a lowercase letter, and the same letter may appear more than once in a message. Whenthe message is beamed through hyperspace, the characters come out in a random order. Ackbar chooseshis message so that the Rebels have at least a 1

2 chance of getting the same message he sent. Howmany distinct messages could he send?

11. [8] Lily and Sarah are playing a game. They each choose a real number at random between -1 and 1.They then add the squares of their numbers together. If the result is greater than or equal to 1, Lilywins, and if the result is less than 1, Sarah wins. What is the probability that Sarah wins?

12. [8] Let ω be a circle of radius 1 centered at O. Let B be a point on ω, and let l be the line tangent toω at B. Let A be on l such that ∠AOB = 60◦. Let C be the foot of the perpendicular from B to OA.Find the length of line segment OC.

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

13. [8] 8 students are practicing for a math contest, and they divide into pairs to take a practice test. Inhow many ways can they be split up?

14. [8] Let f(x) = x4 + ax3 + bx2 + cx + d be a polynomial whose roots are all negative integers. Ifa + b + c + d = 2009, find d.

15. [8] The curves x2 + y2 = 36 and y = x2 − 7 intersect at four points. Find the sum of the squares ofthe x-coordinates of these points.

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

16. [9] Pick a random digit in the decimal expansion of 199999 . What is the probability that it is 0?

17. [9] A circle passes through the points (2, 0) and (4, 0) and is tangent to the line y = x. Find the sumof all possible values for the y-coordinate of the center of the circle.

18. [9] Let f be a function that takes in a triple of integers and outputs a real number. Suppose that fsatisfies the equations

f(a, b, c) =f(a + 1, b, c) + f(a− 1, b, c)

2

f(a, b, c) =f(a, b + 1, c) + f(a, b− 1, c)

2

f(a, b, c) =f(a, b, c + 1) + f(a, b, c− 1)

2

for all integers a, b, c. What is the minimum number of triples at which we need to evaluate f in orderto know its value everywhere?

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

19. [11] You are trapped in ancient Japan, and a giant enemy crab is approaching! You must defeat it bycutting off its two claws and six legs and attacking its weak point for massive damage. You cannotcut off any of its claws until you cut off at least three of its legs, and you cannot attack its weak pointuntil you have cut off all of its claws and legs. In how many ways can you defeat the giant enemy crab?(Note that the legs are distinguishable, as are the claws.)

20. [11] Consider an equilateral triangle and a square both inscribed in a unit circle such that one side ofthe square is parallel to one side of the triangle. Compute the area of the convex heptagon formed bythe vertices of both the triangle and the square.

21. [11] Let f(x) = x2 + 2x + 1. Let g(x) = f(f(· · · f(x))), where there are 2009 fs in the expression forg(x). Then g(x) can be written as

g(x) = x22009+ a22009−1x

22009−1 + · · ·+ a1x + a0,

where the ai are constants. Compute a22009−1.

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

22. [12] Five cards labeled A, B, C, D, and E are placed consecutively in a row. How many ways can theybe re-arranged so that no card is moved more than one position away from where it started? (Notmoving the cards at all counts as a valid re-arrangement.)

23. [12] Let a0, a1, . . . be a sequence such that a0 = 3, a1 = 2, and an+2 = an+1 + an for all n ≥ 0. Find

8∑n=0

an

an+1an+2.

24. [12] Penta chooses 5 of the vertices of a unit cube. What is the maximum possible volume of the figurewhose vertices are the 5 chosen points?

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

25. [14] Find all solutions to x4 + 2x3 + 2x2 + 2x + 1 = 0 (including non-real solutions).

26. [14] In how many ways can the positive integers from 1 to 100 be arranged in a circle such that thesum of every two integers placed opposite each other is the same? (Arrangements that are rotationsof each other count as the same.) Express your answer in the form a! · bc.

27. [14] ABCD is a regular tetrahedron of volume 1. Maria glues regular tetrahedra A′BCD, AB′CD,ABC ′D, and ABCD′ to the faces of ABCD. What is the volume of the tetrahedron A′B′C ′D′?

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

28. [17] Six men and their wives are sitting at a round table with 12 seats. These men and women arevery jealous — no man will allow his wife to sit next to any man except for himself, and no womanwill allow her husband to sit next to any woman except for herself. In how many distinct ways canthese 12 people be seated such that these conditions are satisfied? (Rotations of a valid seating areconsidered distinct.)

29. [17] For how many integer values of b does there exist a polynomial function with integer coefficientssuch that f(2) = 2010 and f(b) = 8?

30. [17] Regular hexagon ABCDEF has side length 2. A laser beam is fired inside the hexagon from pointA and hits BC at point G. The laser then reflects off BC and hits the midpoint of DE. Find BG.

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

31. [20] There are two buildings facing each other, each 5 stories high. How many ways can Kevin stringziplines between the buildings so that:

(a) each zipline starts and ends in the middle of a floor.

(b) ziplines can go up, stay flat, or go down, but can’t touch each other (this includes touching attheir endpoints).

Note that you can’t string a zipline between two floors of the same building.

32. [20] A circle ω1 of radius 15 intersects a circle ω2 of radius 13 at points P and Q. Point A is on line PQsuch that P is between A and Q. R and S are the points of tangency from A to ω1 and ω2, respectively,such that the line AS does not intersect ω1 and the line AR does not intersect ω2. If PQ = 24 and∠RAS has a measure of 90◦, compute the length of AR.

33. [20] Compute

∞∑n=2009

1(n

2009

) .Note that

(nk

)is defined as n!

k!(n−k)! .

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

34. [25] How many hits does “3.1415” get on Google? Quotes are for clarity only, and not part of thesearch phrase. Also note that Google does not search substrings, so a webpage with 3.14159 on it willnot match 3.1415. If A is your answer, and S is the correct answer, then you will get max(25−| ln(A)−ln(S)|, 0) points, rounded to the nearest integer.

35. [25] Call an integer n > 1 radical if 2n − 1 is prime. What is the 20th smallest radical number? If A

is your answer, and S is the correct answer, you will get max(

25(

1− |A−S|S

), 0)

points, rounded tothe nearest integer.

36. [25] Write down a pair of integers (a, b), where −100000 < a < b < 100000. You will get max(25, k)points, where k is the number of other teams’ pairs that you interleave. (Two pairs (a, b) and (c, d) ofintegers interleave each other if a < c < b < d or c < a < d < b.)

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2nd Annual Harvard-MIT November TournamentSaturday 7 November 2009

Guts Round

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

1. [5] If f(x) = x/(x+ 1), what is f(f(f(f(2009))))?

Answer: 20098037 f(f(x)) = (x/(x+1))

(x/(x+1))+1 = x/2x+ 1, f(f(f(f(x)))) = x/4x+ 1 = 20098037

2. [5] A knight begins on the lower-left square of a standard chessboard. How many squares couldthe knight end up at after exactly 2009 legal knight’s moves? (A knight’s move is 2 squares eitherhorizontally or vertically, followed by 1 square in a direction perpendicular to the first.)

Answer: 32 The knight goes from a black square to a white square on every move, or vice versa,so after 2009 moves he must be on a square whose color is opposite of what he started on. So he canonly land on half the squares after 2009 moves. Note that he can access any of the 32 squares (thereare no other parity issues) because any single jump can also be accomplished in 3 jumps, so with 2009jumps, he can land on any of the squares of the right color.

3. [5] Consider a square, inside which is inscribed a circle, inside which is inscribed a square, inside whichis inscribed a circle, and so on, with the outermost square having side length 1. Find the differencebetween the sum of the areas of the squares and the sum of the areas of the circles.

Answer: 2− π2

The ratio of the area of each square and the circle immediately inside it is 4π . The total sum of the

areas of the squares is 1 + 12 + 1

4 + . . . = 2. Difference in area is then 2− 2 · 4π .

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

4. [6] A cube has side length 1. Find the product of the lengths of the diagonals of this cube (a diagonalis a line between two vertices that is not an edge).

Answer: 576 There are 12 diagonals that go along a face and 4 that go through the center of the

cube, so the answer is√

212 ·√

34

= 576.

5. [6] Tanks has a pile of 5 blue cards and 5 red cards. Every morning, he takes a card and throws itdown a well. What is the probability that the first card he throws down and the last card he throwsdown are the same color?

Answer: 49 Once he has thrown the first card down the well, there are 9 remaining cards, and only

4 have the same color as the card that was thrown down. Therefore, the probability that the last cardhe throws down has the same color is 4

9 .

6. [6] Find the last two digits of 10321032. Express your answer as a two-digit number.

Answer: 76 The last two digits of 10321032 is the same as the last two digits of 321032. The lasttwo digits of 32n repeat with a period of four as 32, 24, 68, 76, 32, 24, 68, 76, . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

7. [7] A computer program is a function that takes in 4 bits, where each bit is either a 0 or a 1, andoutputs TRUE or FALSE. How many computer programs are there?Answer: 65536 The function has 24 inputs and 2 outputs for each possible input, so the answer is224

= 216 = 65536.

8. [7] The angles of a convex n-sided polygon form an arithmetic progression whose common difference(in degrees) is a non-zero integer. Find the largest possible value of n for which this is possible. (Apolygon is convex if its interior angles are all less than 180◦.)

Answer: 27 The exterior angles form an arithmetic sequence too (since they are each 180◦ minusthe corresponding interior angle). The sum of this sequence must be 360◦. Let the smallest exteriorangle be x and the common difference be d. The sum of the exterior angles is then x+ (x+ a) + (x+2a)+ . . .+(x+(n−1)a) = n(n−1)

2 ·a+nx. Setting this to 360, and using nx > 0, we get n(n−1) < 720,so n ≤ 27.

9. [7] Daniel wrote all the positive integers from 1 to n inclusive on a piece of paper. After carefulobservation, he realized that the sum of all the digits that he wrote was exactly 10,000. Find n.

Answer: 799 Let S(n) denote the sum of the digits of n, and let f(x) =x∑n=0

S(n). (We may add

n = 0 because S(0) = 0.) Observe that:

f(99) =9∑a=0

(9∑b=0

(a+ b)

)= 10

9∑b=0

b+ 109∑a=0

a = 900

If a is an integer between 1 and 9 inclusive, then:

100a+99∑n=100a

S(n) =100a+99∑n=100a

(a+ S(n− 100a)) = 100a+ f(99) = 100a+ 900

Summing, we get:

f(100a+ 99) =a∑

n=0

(100a+ 900) = 900(a+ 1) + 50a(a+ 1)

This formula can be used to find benchmarks. However, it turns out that this formula alone will suffice,as things turn out rather nicely:

900(a+ 1) + 50a(a+ 1) = 1000050a2 + 950a+ 900 = 10000

50a2 + 950a− 9100 = 050(a+ 26)(a− 7) = 0

a = 7

Therefore f(799) = 10000, and our answer is 799.

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

10. [8] Admiral Ackbar needs to send a 5-character message through hyperspace to the Rebels. Eachcharacter is a lowercase letter, and the same letter may appear more than once in a message. Whenthe message is beamed through hyperspace, the characters come out in a random order. Ackbar chooseshis message so that the Rebels have at least a 1

2 chance of getting the same message he sent. Howmany distinct messages could he send?

Answer: 26 If there is more than one distinct letter sent in the message, then there will be at mosta 1/5 chance of transmitting the right message. So the message must consist of one letter repeated fivetimes, so there are 26 possible messages.

11. [8] Lily and Sarah are playing a game. They each choose a real number at random between -1 and 1.They then add the squares of their numbers together. If the result is greater than or equal to 1, Lilywins, and if the result is less than 1, Sarah wins. What is the probability that Sarah wins?

Answer: π4 If we let x denote Lily’s choice of number and y denote Sarah’s, then all possible

outcomes are represented by the square with vertices (−1,−1), (−1, 1), (1,−1), and (1, 1). Sarah winsif x2 + y2 ≤ 1, which is the area inside the unit circle. Since this has an area of π and the entire squarehas an area of 4, the probability that Sarah wins is π

4 .

12. [8] Let ω be a circle of radius 1 centered at O. Let B be a point on ω, and let l be the line tangent toω at B. Let A be on l such that ∠AOB = 60◦. Let C be the foot of the perpendicular from B to OA.Find the length of line segment OC.

Answer: 12 We have OC/OB = cos(60◦). Since OB = 1, OC = 1

2 .

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

13. [8] 8 students are practicing for a math contest, and they divide into pairs to take a practice test. Inhow many ways can they be split up?

Answer: 105 We create the pairs one at a time. The first person has 7 possible partners. Set thispair aside. Of the remaining six people, pick a person. He or she has 5 possible partners. Set this pairaside. Of the remaining four people, pick a person. He or she has 3 possible partners. Set this pairaside. Then the last two must be partners. So there are 7 ·5 ·3 = 105 possible groupings. Alternatively,we can consider the 8! permutations of the students in a line, where the first two are a pair, the nexttwo are a pair, etc. Given a grouping, there are 4! ways to arrange the four pairs in order, and in eachpair, 2 ways to order the students. So our answer is 8!

4!24 = 7 · 5 · 3 = 105.

14. [8] Let f(x) = x4 + ax3 + bx2 + cx + d be a polynomial whose roots are all negative integers. Ifa+ b+ c+ d = 2009, find d.

Answer: 528 Call the roots −x1,−x2,−x3, and −x4. Then f(x) must factor as (x+x1)(x+x2)(x+x3)(x + x4). If we evaluate f at 1, we get (1 + x1)(1 + x2)(1 + x3)(1 + x4) = a + b + c + d + 1 =2009+1 = 2010. 2010 = 2 ·3 ·5 ·67. d is the product of the four roots, so d = (−1) · (−2) · (−4) · (−66).

15. [8] The curves x2 + y2 = 36 and y = x2 − 7 intersect at four points. Find the sum of the squares ofthe x-coordinates of these points.

Answer: 26 If we use the system of equations to solve for y, we get y2+y−29 = 0 (since x2 = y+7).The sum of the roots of this equation is −1. Combine this with x2 = y + 7 to see that the sum of thesquare of the possible values of x is 2 · (−1 + 7 · 2) = 26.

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

16. [9] Pick a random digit in the decimal expansion of 199999 . What is the probability that it is 0?

Answer: 45 The decimal expansion of 1

99999 is 0.00001.

17. [9] A circle passes through the points (2, 0) and (4, 0) and is tangent to the line y = x. Find the sumof all possible values for the y-coordinate of the center of the circle.

Answer: −6 First, we see that the x coordinate must be 3. Let the y coordinate be y. Now, we seethat the radius is r =

√1 + y2. The line from the center of the circle to the point of tangency with

the line x = y is perpendicular to the line x = y. Hence, the distance from the center of the circle tothe line x = y is d =

√2(3− y). Letting r = d we see that the two possible values for y are 1 and −7,

which sum to −6.

18. [9] Let f be a function that takes in a triple of integers and outputs a real number. Suppose that fsatisfies the equations

f(a, b, c) =f(a+ 1, b, c) + f(a− 1, b, c)

2

f(a, b, c) =f(a, b+ 1, c) + f(a, b− 1, c)

2

f(a, b, c) =f(a, b, c+ 1) + f(a, b, c− 1)

2

for all integers a, b, c. What is the minimum number of triples at which we need to evaluate f in orderto know its value everywhere?

Answer: 8 Note that if we have the value of f at the 8 points:(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 1, 1), we can calculate the value for anytriple of points because we have that f(a + 1, b, c) − (a, b, c) constant for any a, if b and c are fixed(and similarly for the other coordinates). To see why we cannot do this with less points, notice thatwe need to determine what the value of these 8 points anyways, and there is no ”more efficient” wayto determine them all in fewer evaluations.

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2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

19. [11] You are trapped in ancient Japan, and a giant enemy crab is approaching! You must defeat it bycutting off its two claws and six legs and attacking its weak point for massive damage. You cannotcut off any of its claws until you cut off at least three of its legs, and you cannot attack its weak pointuntil you have cut off all of its claws and legs. In how many ways can you defeat the giant enemy crab?(Note that the legs are distinguishable, as are the claws.)

Answer: 14400 The answer is given by 6!2!(52

), because we can cut off the claws and legs in any

order and there are(52

)ways to decide when to cut off the two claws (since we can do it at any time

among the last 5 cuts).

20. [11] Consider an equilateral triangle and a square both inscribed in a unit circle such that one side ofthe square is parallel to one side of the triangle. Compute the area of the convex heptagon formed bythe vertices of both the triangle and the square.

Answer: 3+√

32

Consider the diagram above. We see that the shape is a square plus 3 triangles. The top and bottomtriangles have base

√2 and height 1

2 (√

3 −√

2), and the triangle on the side has the same base andheight 1−

√2

2 Adding their areas, we get the answer.

21. [11] Let f(x) = x2 + 2x+ 1. Let g(x) = f(f(· · · f(x))), where there are 2009 fs in the expression forg(x). Then g(x) can be written as

g(x) = x22009+ a22009−1x

22009−1 + · · ·+ a1x+ a0,

where the ai are constants. Compute a22009−1.

Answer: 22009 f(x) = (x+1)2, so f(xn+cxn−1+. . .) = (xn+cxn−1+. . .+1)2 = x2n+2cx2n−1+. . ..Applying the preceding formula repeatedly shows us that the coefficient of the term of second highestdegree in the polynomial doubles each time, so after 2009 applications of f it is 22009.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

22. [12] Five cards labeled A, B, C, D, and E are placed consecutively in a row. How many ways can theybe re-arranged so that no card is moved more than one position away from where it started? (Notmoving the cards at all counts as a valid re-arrangement.)

Answer: 8 The only things we can do is leave cards where they are or switch them with adjacentcards. There is 1 way to leave them all where they are, 4 ways to switch just one adjacent pair, and 3ways to switch two different adjacent pairs, for 8 possibilities total.

23. [12] Let a0, a1, . . . be a sequence such that a0 = 3, a1 = 2, and an+2 = an+1 + an for all n ≥ 0. Find

8∑n=0

anan+1an+2

.

Answer: 105212 We can re-write an

an+1an+2as an+2−an+1

an+1an+2= 1

an+1− 1

an+2. We can thus re-write the sum

as

(1a1− 1a2

)+(

1a2− 1a3

)+(

1a4− 1a3

)+ . . .+

(1a9− 1a10

)=

1a1− 1a10

=12− 1

212=

105212

.

24. [12] Penta chooses 5 of the vertices of a unit cube. What is the maximum possible volume of the figurewhose vertices are the 5 chosen points?

Answer: 12 Label the vertices of the cube A, B, C, D, E, F , G, H, such that ABCD is the top

face of the cube, E is directly below A, F is directly below B, G is directly below C, and H is directly

below D. We can obtain a volume of 12 by taking the vertices A, B, C, F , and H. To compute the

volume of ACBFH, we will instead compute the volume of the parts of the cube that are not part ofABCFH. This is just the three tetrahedrons CFGH, AEFH, and ACDH, which each have volume 1

6(by using the 1

3bh formula for the area of a pyramid). Therefore, the volume not contained in ACBFHis 3 · 1

6 = 12 , so the volume contained in ACBFH is 1− 1

2 = 12 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

25. [14] Find all solutions to x4 + 2x3 + 2x2 + 2x+ 1 = 0 (including non-real solutions).

Answer: −1, i,−i We can factor the polynomial as (x+ 1)2(x2 + 1).

26. [14] In how many ways can the positive integers from 1 to 100 be arranged in a circle such that thesum of every two integers placed opposite each other is the same? (Arrangements that are rotationsof each other count as the same.) Express your answer in the form a! · bc.

Answer: 49! · 249 Split the integers up into pairs of the form (x, 101−x). In the top half of the circle,exactly one element from each pair occurs, and there are thus 50! ways to arrange them. and also 250

ways to decide whether the larger or smaller number in each pair occurs in the top half of the circle.We then need to divide by 100 since rotations are not considered distinct, so we get 50!250

100 = 49! · 249.

27. [14] ABCD is a regular tetrahedron of volume 1. Maria glues regular tetrahedra A′BCD, AB′CD,ABC ′D, and ABCD′ to the faces of ABCD. What is the volume of the tetrahedron A′B′C ′D′?

Answer: 12527 Consider the tetrahedron with vertices at W = (1, 0, 0), X = (0, 1, 0), Y = (0, 0, 1),

and Z = (1, 1, 1). This tetrahedron is similar to ABCD. It has center O = ( 12 ,

12 ,

12 ). We can

construct a tetrahedron W ′X ′Y ′Z ′ in the same way that A′B′C ′D′ was constructed by letting W ′

be the reflection of W across XY Z and so forth. Then we see that Z ′ = (− 13 ,−

13 ,−

13 ), so OZ ′ has

length 56

√3, whereas OZ has length 1

2

√3. We thus see that W ′X ′Y ′Z ′ has a side length

5612

= 53 that

of WXY Z, so by similarity the same is true of A′B′C ′D′ and ABCD. In particular, the volume ofA′B′C ′D′ is

(53

)3 that of ABCD, so it is 12527 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

28. [17] Six men and their wives are sitting at a round table with 12 seats. These men and women arevery jealous — no man will allow his wife to sit next to any man except for himself, and no womanwill allow her husband to sit next to any woman except for herself. In how many distinct ways canthese 12 people be seated such that these conditions are satisfied? (Rotations of a valid seating areconsidered distinct.)

Answer: 288000 Think of this problem in terms of “blocks” of men and women, that is, groups ofmen and women sitting together. Each block must contain at least two people; otherwise you have aman sitting next to two women (or vice-versa).

We will define the notation [a1, b1, a2, b2, . . .] to mean a seating arrangement consisting of, going inorder, a1 men, b1 women, a2 men, b2 women, and so on.

Split the problem into three cases, each based on the number of blocks of men and women:

Case 1: One block of each, [6, 6].

There are 12 ways to choose the seats where the men sit, and 6! ways to arrange those men. The twowomen on either side of the men are uniquely determined by the men they sit next to. There are 4!ways to arrange the other four women. This gives 6! · 288 ways.

Case 2: Two blocks of each.

The arrangement is [a, b, c, d], where a + c = b + d = 6. There are five distinct block schematics:[2, 2, 4, 4], [2, 3, 4, 3], [2, 4, 4, 2], [3, 2, 3, 4], and [3, 3, 3, 3]. (The others are rotations of one of the above.)

For each of these, there are 6! ways to arrange the men. In addition, four women are uniquely deter-mined because they sit next to a man. There are 2 ways to arrange the other two women. Each of thefirst four possible block schematics gives 12 distinct rotations, while the fifth one gives 6. This gives6!(2)(12 + 12 + 12 + 12 + 6) = 6! · 108 ways.

Case 3: Three blocks of each, [2, 2, 2, 2, 2, 2].

There are 4 ways to choose where the men sit and 6! ways to arrange those men. Each placement ofmen will uniquely determine the placement of each women. This gives 6! · 4 ways.

Then we have a grand total of 6! · (288 + 108 + 4) = 6! · 400 = 288000 seating arrangements.

29. [17] For how many integer values of b does there exist a polynomial function with integer coefficientssuch that f(2) = 2010 and f(b) = 8?

Answer: 32 We can take f(x) = − 2002d (x − b) + 2010 for all divisors d of −2002. To see that we

can’t get any others, note that b − 2 must divide f(b) − f(2), so b − 2 divides −2002 (this is becauseb− 2 divides bn − 2n and hence any sum of numbers of the form bn − 2n).

30. [17] Regular hexagon ABCDEF has side length 2. A laser beam is fired inside the hexagon from pointA and hits BC at point G. The laser then reflects off BC and hits the midpoint of DE. Find BG.

Answer: 25 Look at the diagram below, in which points J , K, M , T , and X have been defined. M

is the midpoint of DE, BCJK is a rhombus with J lying on the extension of CD, T is the intersectionof lines CD and GM when extended, and X is on JT such that XM ‖ JK.

It can be shown that m∠MDX = m∠MXD = 60◦, so 4DMX is equilateral, which yields XM = 1.The diagram indicates that JX = 5. One can show by angle-angle similarity that 4TXM ∼ 4TJK,which yields TX = 5.

One can also show by angle-angle similarity that 4TJK ∼ 4TCG, which yields the proportionTJJK = TC

CG . We know everything except CG, which we can solve for. This yields CG = 85 , so BG = 2

5 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

31. [20] There are two buildings facing each other, each 5 stories high. How many ways can Kevin stringziplines between the buildings so that:

(a) each zipline starts and ends in the middle of a floor.

(b) ziplines can go up, stay flat, or go down, but can’t touch each other (this includes touching attheir endpoints).

Note that you can’t string a zipline between two floors of the same building.

Answer: 252 Associate with each configuration of ziplines a path in the plane as follows: Supposethere are k ziplines. Let a0, . . . , ak be the distances between consecutive ziplines on the left building

(a0 is the floor on which the first zipline starts, and ak is the distance from the last zipline to the topof the building). Define b0, . . . , bk analogously for the right building. The path in the plane consists ofstarting at (0, 0) and going a distance a0 to the right, b0 up, a1 to the right, b1 up, etc. We thus gofrom (0, 0) to (5, 5) while traveling only up and to the right between integer coordinates. We can checkthat there is exactly one configuration of ziplines for each such path, so the answer is the number ofpaths from (0, 0) to (5, 5) where you only travel up and to the right. This is equal to

(105

)= 252, since

there are 10 total steps to make, and we must choose which 5 of them go to the right.

32. [20] A circle ω1 of radius 15 intersects a circle ω2 of radius 13 at points P and Q. Point A is on line PQsuch that P is between A and Q. R and S are the points of tangency from A to ω1 and ω2, respectively,such that the line AS does not intersect ω1 and the line AR does not intersect ω2. If PQ = 24 and∠RAS has a measure of 90◦, compute the length of AR.

Answer: 14 +√

97 Let O1 be the center of ω1 and O2 be the center of ω2. Then O1O2 and PQare perpendicular. Let their point of intersection be X. Using the Pythagorean theorem, the fact thatPQ = 24, and our knowledge of the radii of the circles, we can compute that O1X = 9 and O2X = 5,so O1O2 = 14. Let SO1 and RO2 meet at Y . Then SARY is a square, say of side length s. ThenO1Y = s− 15 and O2Y = s− 13. So, O1O2Y is a right triangle with sides 14, s− 15, and s− 13. Bythe Pythagorean theorem, (s− 13)2 + (s− 15)2 = 142. We can write this as 2s2 − 4 · 14s+ 198 = 0, ors2 − 28s + 99 = 0. The quadratic formula then gives s = 28±

√388

2 = 14 ±√

97. Since 14 −√

97 < 15and Y O1 > 15, we can discard the root of 14−

√97, and the answer is therefore 14 +

√97.

33. [20] Compute

∞∑n=2009

1(n

2009

) .Note that

(nk

)is defined as n!

k!(n−k)! .

Answer: 20092008 Observe that

k + 1k

(1(n−1k

) − 1(nk

)) =k + 1k

(nk

)−(n−1k

)(nk

)(n−1k

)=

k + 1k

(n−1k−1

)(nk

)(n−1k

)=

k + 1k

(n− 1)!k!k!(n− k − 1)!(n− k)!n!(n− 1)!(k − 1)!(n− k)!

=k + 1k

k · k!(n− k − 1)!n!

=(k + 1)!(n− k − 1)!

n!

=1(nk+1

)Now apply this with k = 2008 and sum across all n from 2009 to ∞. We get

∞∑n=2009

1(n

2009

) =20092008

∞∑n=2009

1(n−12008

) − 1(n

2008

) .All terms from the sum on the right-hand-side cancel, except for the initial 1

(20082008)

, which is equal to 1,

so we get∑∞n=2009

1

( n2009)

= 20092008 .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

34. [25] How many hits does “3.1415” get on Google? Quotes are for clarity only, and not part of thesearch phrase. Also note that Google does not search substrings, so a webpage with 3.14159 on it willnot match 3.1415. If A is your answer, and S is the correct answer, then you will get max(25−| ln(A)−ln(S)|, 0) points, rounded to the nearest integer.

Answer: 422000

35. [25] Call an integer n > 1 radical if 2n − 1 is prime. What is the 20th smallest radical number? If Ais your answer, and S is the correct answer, you will get max

(25(

1− |A−S|S

), 0)

points, rounded tothe nearest integer.

Answer: 4423

36. [25] Write down a pair of integers (a, b), where −100000 < a < b < 100000. You will get max(25, k)points, where k is the number of other teams’ pairs that you interleave. (Two pairs (a, b) and (c, d) ofintegers interleave each other if a < c < b < d or c < a < d < b.)

2nd Annual Harvard-MIT November TournamentSaturday 7 November 2009

Team Round

Down the Infinite Corridor

Consider an isosceles triangle T with base 10 and height 12. Define a sequence ω1, ω2, . . . of circles suchthat ω1 is the incircle of T and ωi+1 is tangent to ωi and both legs of the isosceles triangle for i > 1.

1. [3] Find the radius of ω1.

2. [3] Find the ratio of the radius of ωi+1 to the radius of ωi.

3. [3] Find the total area contained in all the circles.

Bouncy Balls

In the following problems, you will consider the trajectories of balls moving and bouncing off of theboundaries of various containers. The balls are small enough that you can treat them as points. Let ussuppose that a ball starts at a point X, strikes a boundary (indicated by the line segment AB) at Y , andthen continues, moving along the ray Y Z. Balls always bounce in such a way that ∠XY A = ∠BY Z. Thisis indicated in the above diagram.

X

Y

Z

A

B

ball bounces off of AB at the point Y

Balls bounce off of boundaries in the same way light reflects off of mirrors - if the ball hits the boundaryat point P , the trajectory after P is the reflection of the trajectory before P through the perpendicular tothe boundary at P .

A ball inside a rectangular container of width 7 and height 12 is launched from the lower-left vertex ofthe container. It first strikes the right side of the container after traveling a distance of

√53 (and strikes no

other sides between its launch and its impact with the right side).

4. [2] Find the height at which the ball first contacts the right side.

5. [3] How many times does the ball bounce before it returns to a vertex? (The final contact with a vertexdoes not count as a bounce.)

Now a ball is launched from a vertex of an equilateral triangle with side length 5. It strikes the opposite sideafter traveling a distance of

√19.

6. [4] Find the distance from the ball’s point of first contact with a wall to the nearest vertex.

7. [4] How many times does the ball bounce before it returns to a vertex? (The final contact with a vertexdoes not count as a bounce.)

Page 1 of 2

In this final problem, a ball is again launched from the vertex of an equilateral triangle with side length 5.

8. [6] In how many ways can the ball be launched so that it will return again to a vertex for the first timeafter 2009 bounces?

Super Mario 64!

Mario is once again on a quest to save Princess Peach. Mario enters Peach’s castle and finds himself in aroom with 4 doors. This room is the first in a sequence of 2 indistinugishable rooms. In each room, 1 doorleads to the next room in the sequence (or, for the second room, into Bowser’s level), while the other 3 doorslead to the first room.

9. [3] Suppose that in every room, Mario randomly picks a door to walk through. What is the expectednumber of doors (not including Mario’s initial entrance to the first room) through which Mario willpass before he reaches Bowser’s level?

10. [4] Suppose that instead there are 6 rooms with 4 doors. In each room, 1 door leads to the next roomin the sequence (or, for the last room, Bowser’s level), while the other 3 doors lead to the first room.Now what is the expected number of doors through which Mario will pass before he reaches Bowser’slevel?

11. [5] In general, if there are d doors in every room (but still only 1 correct door) and r rooms, the last ofwhich leads into Bowser’s level, what is the expected number of doors through which Mario will passbefore he reaches Bowser’s level?

Page 2 of 2

2nd Annual Harvard-MIT November TournamentSaturday 7 November 2009

Team Round

Down the Infinite Corridor

Consider an isosceles triangle T with base 10 and height 12. Define a sequence ω1, ω2, . . . of circles suchthat ω1 is the incircle of T and ωi+1 is tangent to ωi and both legs of the isosceles triangle for i > 1.

1. [3] Find the radius of ω1.

Answer: 103 Using the Pythagorean theorem, we see that the legs of T each have length 13. Let r

be the radius of ω1. We can divide T into three triangles, each with two vertices at vertices of T andone vertex at the center of ω1. These triangles all have height r and have bases 13, 13, and 10. Thustheir total area is 13r

2 + 13r2 + 10r

2 = 18r. However, T has height 12 and base 10, so its area is 60. Thus18r = 60, so r = 10

3 .

2. [3] Find the ratio of the radius of ωi+1 to the radius of ωi.

3. [3] Find the total area contained in all the circles.

Answer: 180π13 Using the notation from the previous solution, the area contained in the ith circle is

equal to πr2i . Since the radii form a geometric sequence, the areas do as well. Specifically, the areasform a sequence with initial term π · 100

9 and common ratio 1681 , so their sum is then π · 100

96581

= 180π13 .

Bouncy Balls

In the following problems, you will consider the trajectories of balls moving and bouncing off of theboundaries of various containers. The balls are small enough that you can treat them as points. Let ussuppose that a ball starts at a point X, strikes a boundary (indicated by the line segment AB) at Y , andthen continues, moving along the ray Y Z. Balls always bounce in such a way that ∠XY A = ∠BY Z. Thisis indicated in the above diagram.

X

Y

Z

A

B

ball bounces off of AB at the point Y

Balls bounce off of boundaries in the same way light reflects off of mirrors - if the ball hits the boundaryat point P , the trajectory after P is the reflection of the trajectory before P through the perpendicular tothe boundary at P .

A ball inside a rectangular container of width 7 and height 12 is launched from the lower-left vertex ofthe container. It first strikes the right side of the container after traveling a distance of

√53 (and strikes no

other sides between its launch and its impact with the right side).

4. [2] Find the height at which the ball first contacts the right side.

Answer: 2 Let h be this height. Then, using the Pythagorean theorem, we see that h2 + 72 = 53,so h = 2.

5. [3] How many times does the ball bounce before it returns to a vertex? (The final contact with a vertexdoes not count as a bounce.)

Answer: 5 Every segment the ball traverses between bounces takes it 7 units horizontally and 2units up. Thus, after 5 bounces it has traveled up 10 units, and the final segment traversed takes itdirectly to the upper right vertex of the rectangle.

Now a ball is launched from a vertex of an equilateral triangle with side length 5. It strikes the opposite sideafter traveling a distance of

√19.

6. [4] Find the distance from the ball’s point of first contact with a wall to the nearest vertex.

Answer: 2

A

B

C

M

Y

√19

5√

32

52

5

Consider the diagram above, where M is the midpoint of BC. Then AM is perpendicular to BC sinceABC is equilateral, so by the Pythagorean theorem AM = 5

√3

2 . Then, using the Pythagorean theoremagain, we see that MY = 1

2 , so that BY = 2.

7. [4] How many times does the ball bounce before it returns to a vertex? (The final contact with a vertexdoes not count as a bounce.)

Answer: 7 The key idea is that, instead of reflecting the line AY off of BC, we will reflect ABCabout BC and extend AY beyond 4ABC. We keep doing this until the extension of AY hits a vertexof one of our reflected triangles. This is illustrated in the diagram below:

We can calculate that the line AY has slope3√

3272

= 3√

37 , so that (as indicated in the diagram), AY

first intersects a vertex at the point(

352 ,

15√

32

). To get there, it has to travel through 2 horizontal

lines, 1 upward sloping line, and 4 downward sloping lines, so it bounces 2 + 1 + 4 = 7 times total.

In this final problem, a ball is again launched from the vertex of an equilateral triangle with side length 5.

8. [6] In how many ways can the ball be launched so that it will return again to a vertex for the first timeafter 2009 bounces?

Answer: 502 We will use the same idea as in the previous problem. We first note that every vertex

of a triangle can be written uniquely in the form a(5, 0) + b(

52 ,

5√

32

), where a and b are non-negative

integers. Furthermore, if a ball ends at a(5, 0) + b(

52 ,

5√

32

), then it bounches off of a wall 2(a+ b)− 3

times. Therefore, the possible directions that you can launch the ball in correspond to solutions to2(a+b)−3 = 2009, or a+b = 1006. However, if a and b have a common factor, say, k, then the ball willpass through the vertex corresponding to a

k and bk before it passes through the vertex corresponding

to a and b. Therefore, we must discount all such pairs a, b. This corresponds to when a is even ora = 503, so after removing these we are left with 502 remaining possible values of a, hence 502 possibledirections in which to launch the ball.

Super Mario 64!

Mario is once again on a quest to save Princess Peach. Mario enters Peach’s castle and finds himself in aroom with 4 doors. This room is the first in a sequence of 2 indistinugishable rooms. In each room, 1 doorleads to the next room in the sequence (or, for the second room, into Bowser’s level), while the other 3 doorslead to the first room.

9. [3] Suppose that in every room, Mario randomly picks a door to walk through. What is the expectednumber of doors (not including Mario’s initial entrance to the first room) through which Mario willpass before he reaches Bowser’s level?

Answer: 20 Let Ei be the expected number of doors through which Mario will pass in the future if heis currently in room i for i = 1, 2, 3 (we will set E3 = 0). We claim that Ei = 1+ 3

4E1 + 14Ei+1. Indeed,

the 1 at the beginning comes from the fact that we need to pass through a door to leave the room, the34E1 comes from the fact that there is a 3

4 chance of ending up in room 1, and the 14Ei+1 corresponds

to the fact that there is a 14 chance of ending up in Ei+1. Using this, we get E1 = 1 + 3

4E1 + 14E2, or

E1 = 4 + E2. We also get E2 = 1 + 34E1. Solving this system of equations yields E1 = 20.

10. [4] Suppose that instead there are 6 rooms with 4 doors. In each room, 1 door leads to the next roomin the sequence (or, for the last room, Bowser’s level), while the other 3 doors lead to the first room.Now what is the expected number of doors through which Mario will pass before he reaches Bowser’slevel?

Answer: 5460 This problem works in the same general way as the last problem, but it can bemore succintly solved using the general formula, which is provided below in the solution to the nextproblem.

11. [5] In general, if there are d doors in every room (but still only 1 correct door) and r rooms, the last ofwhich leads into Bowser’s level, what is the expected number of doors through which Mario will passbefore he reaches Bowser’s level?

Answer:d(dr − 1)d− 1

Let Ei be the expected number of doors through which Mario will pass in

the future if he is currently in room i for i = 1, 2, . . . , r + 1 (we will set Er+1 = 0). We claimthat Ei = 1 + d−1

d E1 + 1dEi+1. This is because, as before, there is a d−1

d chance of ending up inroom 1, and a 1

d chance of ending up in room i + 1. Note that we can re-write this equation asdEi = d+ (d− 1)E1 + Ei+1.

We can solve this system of equation as follows: let E1 = Ei + ci. Then we can re-write each equationas dE1 − d · ci = d + (d − 1)E1 + E1 − ci+1, so ci+1 = d · ci + d, and c1 = 0. We can then see thatci = dd

i−1−1d−1 (either by finding the pattern or by using more advanced techniques, such as those given

at http://en.wikipedia.org/wiki/Recurrence_relation). Solving for E1 using E1 = Er + cr andEr = 1 + d−1

d E1, we get E1 = d(dr−1)d−1 .

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Algebra Subject Test

1. [3] Suppose that x and y are positive reals such that

x− y2 = 3, x2 + y4 = 13.

Find x.

2. [3] The rank of a rational number q is the unique k for which q = 1a1

+ · · ·+ 1ak

, where each ai is thesmallest positive integer such that q ≥ 1

a1+ · · ·+ 1

ai. Let q be the largest rational number less than 1

4

with rank 3, and suppose the expression for q is 1a1

+ 1a2

+ 1a3

. Find the ordered triple (a1, a2, a3).

3. [4] Let S0 = 0 and let Sk equal a1 + 2a2 + . . . + kak for k ≥ 1. Define ai to be 1 if Si−1 < i and -1 ifSi−1 ≥ i. What is the largest k ≤ 2010 such that Sk = 0?

4. [4] Suppose that there exist nonzero complex numbers a, b, c, and d such that k is a root of both theequations ax3 + bx2 + cx + d = 0 and bx3 + cx2 + dx + a = 0. Find all possible values of k (includingcomplex values).

5. [5] Suppose that x and y are complex numbers such that x + y = 1 and that x20 + y20 = 20. Find thesum of all possible values of x2 + y2.

6. [5] Suppose that a polynomial of the form p(x) = x2010 ± x2009 ± · · · ± x± 1 has no real roots. Whatis the maximum possible number of coefficients of −1 in p?

7. [5] Let a, b, c, x, y, and z be complex numbers such that

a =b + c

x− 2, b =

c + a

y − 2, c =

a + b

z − 2.

If xy + yz + zx = 67 and x + y + z = 2010, find the value of xyz.

8. [6] How many polynomials of degree exactly 5 with real coefficients send the set {1, 2, 3, 4, 5, 6} to apermutation of itself?

9. [7] Let f(x) = cx(x − 1), where c is a positive real number. We use fn(x) to denote the polynomialobtained by composing f with itself n times. For every positive integer n, all the roots of fn(x) arereal. What is the smallest possible value of c?

10. [8] Let p(x) and q(x) be two cubic polynomials such that p(0) = −24, q(0) = 30, and

p(q(x)) = q(p(x))

for all real numbers x. Find the ordered pair (p(3), q(6)).

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Algebra Subject Test

1. [3] Suppose that x and y are positive reals such that

x− y2 = 3, x2 + y4 = 13.

Find x.

Answer: 3+√

172 Squaring both sides of x− y2 = 3 gives x2 + y4− 2xy2 = 9. Subtract this equation

from twice the second given to get x2 + 2xy2 + y4 = 17 =⇒ x+ y2 = ±17. Combining this equationwith the first given, we see that x = 3±

√17

2 . Since x is a positive real, x must be 3+√

172 .

2. [3] The rank of a rational number q is the unique k for which q = 1a1

+ · · ·+ 1ak

, where each ai is thesmallest positive integer such that q ≥ 1

a1+ · · ·+ 1

ai. Let q be the largest rational number less than 1

4

with rank 3, and suppose the expression for q is 1a1

+ 1a2

+ 1a3

. Find the ordered triple (a1, a2, a3).

Answer: (5, 21, 421) Suppose that A and B were rational numbers of rank 3 less than 14 , and

let a1, a2, a3, b1, b2, b3 be positive integers so that A = 1a1

+ 1a2

+ 1a3

and B = 1b1

+ 1b2

+ 1b3

are theexpressions for A and B as stated in the problem. If b1 < a1 then A < 1

a1−1 ≤1b1< B. In other words,

of all the rationals less than 14 with rank 3, those that have a1 = 5 are greater than those that have

a1 = 6, 7, 8, . . . Therefore we can “build” q greedily, adding the largest unit fraction that keeps q lessthan 1

4 :15 is the largest unit fraction less than 1

4 , hence a1 = 5;121 is the largest unit fraction less than 1

4 −15 , hence a2 = 21;

1421 is the largest unit fraction less than 1

4 −15 −

121 , hence a3 = 421.

3. [4] Let S0 = 0 and let Sk equal a1 + 2a2 + . . .+ kak for k ≥ 1. Define ai to be 1 if Si−1 < i and -1 ifSi−1 ≥ i. What is the largest k ≤ 2010 such that Sk = 0?

Answer: 1092 Suppose that SN = 0 for some N ≥ 0. Then aN+1 = 1 because N + 1 ≥ SN . Thefollowing table lists the values of ak and Sk for a few k ≥ N :

k ak SkN 0N + 1 1 N + 1N + 2 1 2N + 3N + 3 −1 NN + 4 1 2N + 4N + 5 −1 N − 1N + 6 1 2N + 5N + 7 −1 N − 2

We see inductively that, for every i ≥ 1,

SN+2i = 2N + 2 + i

andSN+1+2i = N + 1− i

thus S3N+3 = 0 is the next k for which Sk = 0. The values of k for which Sk = 0 satisfy the recurrencerelation pn+1 = 3pn+3, and we compute that the first terms of the sequence are 0, 3, 12, 39, 120, 363, 1092;hence 1092 is our answer.

Algebra Subject Test

4. [4] Suppose that there exist nonzero complex numbers a, b, c, and d such that k is a root of both theequations ax3 + bx2 + cx+ d = 0 and bx3 + cx2 + dx+ a = 0. Find all possible values of k (includingcomplex values).

Answer: 1,−1,i,−i Let k be a root of both polynomials. Multiplying the first polynomial by k

and subtracting the second, we have ak4 − a = 0, which means that k is either 1, −1, i, or −i. Ifa = b = c = d = 1, then −1, i, and −i are roots of both polynomials. If a = b = c = 1 and d = −3,then 1 is a root of both polynomials. So k can be 1, −1, i, and −i.

5. [5] Suppose that x and y are complex numbers such that x+ y = 1 and that x20 + y20 = 20. Find thesum of all possible values of x2 + y2.

Answer: −90 We have x2 + y2 + 2xy = 1. Define a = 2xy and b = x2 + y2 for convenience. Then

a+ b = 1 and b− a = x2 + y2 − 2xy = (x− y)2 = 2b− 1 so that x, y =√

2b−1±12 . Then

x20 + y20 =(√

2b− 1 + 12

)20

+(√

2b− 1− 12

)20

=1

220[(√

2b− 1 + 1)20 + (√

2b− 1− 1)20]

=2

220

[(√

2b− 1)20 +(

202

)(√

2b− 1)18 +(

204

)(√

2b− 1)16 + . . .

]=

2220

[(2b− 1)10 +

(202

)(2b− 1)9 +

(204

)(2b− 1)8 + . . .

]= 20

We want to find the sum of distinct roots of the above polynomial in b; we first prove that the originalpolynomial is square-free. The conditions x+y = 1 and x20+y20 = 20 imply that x20+(1−x)20−20 = 0;let p(x) = x20 + (1− x)20 − 20. p is square-free if and only if GCD(p, p′) = c for some constant c:

GCD(p, p′) = GCD(x20 + (1− x)20 − 20, 20(x19 − (1− x)19))= GCD(x20 − x(1− x)19 + (1− x)19 − 20, 20(x19 − (1− x)19))= GCD((1− x)19 − 20, x19 − (1− x)19)= GCD((1− x)19 − 20, x19 − 20)

The roots of x19 − 20 are 19√

20k exp( 2πik19 ) for some k = 0, 1, . . . , 18; the roots of (1 − x)19 − 20 are

1 − 19√

20k exp( 2πik19 ) for some k = 0, 1, . . . , 18. If x19 − 20 and (1 − x)19 − 20 share a common root,

then there exist integers m,n such that 19√

20m exp( 2πim19 ) = 1− 19

√20n exp( 2πin

19 ); since the imaginaryparts of both sides must be the same, we have m = n and 19

√20m exp( 2πim

19 ) = 12 =⇒ 20m = 1

219 , acontradiction. Thus we have proved that the polynomial in x has no double roots. Since for each bthere exists a unique pair (x, y) (up to permutations) that satisfies x2 + y2 = b and (x+ y)2 = 1, thepolynomial in b has no double roots.

Let the coefficient of bn in the above equation be [bn]. By Vieta’s Formulas, the sum of all possiblevalues of b = x2 + y2 is equal to − [b9]

[b10] . [b10] = 2220

(210)

and [b9] = 2220

(−(101

)29 +

(202

)29); thus

− [b9][b10] = − (10

1 )29−(202 )29

210 = −90.

6. [5] Suppose that a polynomial of the form p(x) = x2010 ± x2009 ± · · · ± x± 1 has no real roots. Whatis the maximum possible number of coefficients of −1 in p?

Answer: 1005 Let p(x) be a polynomial with the maximum number of minus signs.p(x) cannot have more than 1005 minus signs, otherwise p(1) < 0 and p(2) ≥ 22010−22009−. . .−2−1 =1, which implies, by the Intermediate Value Theorem, that p must have a root greater than 1.

Let p(x) =x2011 + 1x+ 1

= x2010−x2009 +x2008− . . .−x+ 1. −1 is the only real root of x2011 + 1 = 0 but

p(−1) = 2011; therefore p has no real roots. Since p has 1005 minus signs, it is the desired polynomial.

Algebra Subject Test

7. [5] Let a, b, c, x, y, and z be complex numbers such that

a =b+ c

x− 2, b =

c+ a

y − 2, c =

a+ b

z − 2.

If xy + yz + zx = 67 and x+ y + z = 2010, find the value of xyz.

Answer: −5892 Manipulate the equations to get a common denominator: a = b+cx−2 =⇒ x − 2 =

b+ca =⇒ x− 1 = a+b+c

a =⇒ 1x−1 = a

a+b+c ; similarly, 1y−1 = b

a+b+c and 1z−1 = c

a+b+c . Thus

1x− 1

+1

y − 1+

1z − 1

= 1

(y − 1)(z − 1) + (x− 1)(z − 1) + (x− 1)(y − 1) = (x− 1)(y − 1)(z − 1)xy + yz + zx− 2(x+ y + z) + 3 = xyz − (xy + yz + zx) + (x+ y + z)− 1

xyz − 2(xy + yz + zx) + 3(x+ y + z)− 4 = 0xyz − 2(67) + 3(2010)− 4 = 0

xyz = −5892

8. [6] How many polynomials of degree exactly 5 with real coefficients send the set {1, 2, 3, 4, 5, 6} to apermutation of itself?

Answer: 714 For every permutation σ of {1, 2, 3, 4, 5, 6}, Lagrange Interpolation1 gives a polynomialof degree at most 5 with p(x) = σ(x) for every x = 1, 2, 3, 4, 5, 6. Additionally, this polynomial isunique: assume that there exist two polynomials p, q of degree ≤ 5 such that they map {1, 2, 3, 4, 5, 6}to the same permutation. Then p − q is a nonzero polynomial of degree ≤ 5 with 6 distinct roots, acontradiction. Thus an upper bound for the answer is 6! = 720 polynomials.

However, not every polynomial obtained by Lagrange interpolation is of degree 5 (for example, p(x) =x). We can count the number of invalid polynomials using finite differences.2 A polynomial hasdegree less than 5 if and only if the sequence of 5th finite differences is 0. The 5th finite difference ofp(1), p(2), p(3), p(4), p(5), p(6) is p(1) − 5p(2) + 10p(3) − 10p(4) + 5p(5) − p(6); thus we want to solvep(1)−5p(2)+10p(3)−10p(4)+5p(5)−p(6) = 0 with {p(1), p(2), p(3), p(4), p(5), p(6)} = {1, 2, 3, 4, 5, 6}.Taking the above equation modulo 5, we get p(1) = p(6) (mod 5) =⇒ {p(1), p(6)} = {1, 6}. Note that1−5p(2)+10p(3)−10p(4)+5p(5)−6 = 0 if and only if 6−5p(5)+10p(4)−10p(3)+5p(2)−1 = 0, so we mayassume that p(1) = 1 and double our result later. Then we have {p(2), p(3), p(4), p(5)} = {2, 3, 4, 5}and

−p(2) + 2p(3)− 2p(4) + p(5) = 1.

The above equation taken modulo 2 implies that p(2), p(5) are of opposite parity, so p(3), p(4) are ofopposite parity. We do casework on {p(2), p(5)}:

(a) p(2) = 2, p(5) = 3; 2p(3)− 2p(4) = 0 is a contradiction

(b) p(2) = 2, p(5) = 5; 2p(3)− 2p(4) = −2 =⇒ p(3)− p(4) = −1 =⇒ p(3) = 3, p(4) = 4

(c) p(2) = 3, p(5) = 2; 2p(3)− 2p(4) = −2 =⇒ p(3)− p(4) = −1 =⇒ p(3) = 4, p(4) = 5

(d) p(2) = 3, p(5) = 4; 2p(3)− 2p(4) = 0 is a contradiction

(e) p(2) = 4, p(5) = 3; 2p(3)− 2p(4) = 2 =⇒ p(3)− p(4) = 1 but {p(3), p(4)} = {2, 5}, contradiction

(f) p(2) = 4, p(5) = 5; 2p(3)− 2p(4) = 0 is a contradiction

(g) p(2) = 5, p(5) = 2; 2p(3)− 2p(4) = 4 =⇒ p(3)− p(4) = 2, contradiction

(h) p(2) = 5, p(5) = 4; 2p(3)− 2p(4) = 2 =⇒ p(3)− p(4) = 1 =⇒ p(3) = 3, p(4) = 2

Hence there are a total of 720− 2(3) = 714 polynomials.

1See http://en.wikipedia.org/wiki/Lagrange_interpolation.2See http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=1263378.

Algebra Subject Test

9. [7] Let f(x) = cx(x − 1), where c is a positive real number. We use fn(x) to denote the polynomialobtained by composing f with itself n times. For every positive integer n, all the roots of fn(x) arereal. What is the smallest possible value of c?

Answer: 2 We first prove that all roots of fn(x) are greater than or equal to − c4 and less than or

equal to 1 + c4 . Suppose that r is a root of fn(x). If r = − c

4 , f−1(r) = { 12} and − c

4 <12 < 1 + c

4 sincec is positive. Suppose r 6= − c

4 ; by the quadratic formula, there exist two complex numbers r1, r2 suchthat r1 + r2 = 1 and f(r1) = f(r2) = r. Thus all the roots of fn(x) (except 1

2 ) come in pairs that sumto 1. No root r of fn(x) can be less than − c

4 , otherwise fn+1(x) has an imaginary root, f−1(r). Also,no root r of fn(x) can be greater than 1 + c

4 , otherwise its “conjugate” root will be less than − c4 .

Define g(x) = 12

(1 +

√1 + 4x

c

), the larger inverse of f(x). Note that gn(x) is the largest element of

f−n(x) (which is a set). gn(0) should be less than or equal to 1 + c4 for all n. Let x0 be the nonzero

real number such that g(x0) = x0; then cx0(x0 − 1) = x0 =⇒ xo = 1 + 1c . x0 < g(x) < x if x > x0

and x < g(x) < x0 if x < x0; it can be proved that gn converges to x0. Hence we have the requirementthat x0 = 1 + 1

c ≤ 1 + c4 =⇒ c ≥ 2.

We verify that c = 2 is possible. All the roots of f−n(x) will be real if g(0) ≤ 1 + c4 = 3

2 . We knowthat 0 < 3

2 =⇒ g(0) < 32 , so g2(0) < 3

2 and gn(0) < gn+1(0) < 32 for all n. Therefore all the roots of

fn(x) are real.

10. [8] Let p(x) and q(x) be two cubic polynomials such that p(0) = −24, q(0) = 30, and

p(q(x)) = q(p(x))

for all real numbers x. Find the ordered pair (p(3), q(6)).

Answer: (3,−24) Note that the polynomials f(x) = ax3 and g(x) = −ax3 commute under com-position. Let h(x) = x + b be a linear polynomial, and note that its inverse h−1(x) = x − b is also alinear polynomial. The composite polynomials h−1fh and h−1gh commute, since function compositionis associative, and these polynomials are also cubic.

We solve for the a and b such that (h−1fh)(0) = −24 and (h−1gh)(0) = 30. We must have:

ab3 − b = −24, −ab3 − b = 30⇒ a = 1, b = −3

These values of a and b yield the polynomials p(x) = (x − 3)3 + 3 and q(x) = −(x − 3)3 + 3. Thepolynomials take on the values p(3) = 3 and q(6) = −24.

Remark: The pair of polynomials found in the solution is not unique. There is, in fact, an entirefamily of commuting cubic polynomials with p(0) = −24 and q(0) = 30. They are of the form

p(x) = tx(x− 3)(x− 6)− 24, q(x) = −tx(x− 3)(x− 6) + 30

where t is any real number. However, the values of p(3) and q(6) are the same for all polynomials inthis family. In fact, if we give the initial conditions p(0) = k1 and q(0) = k2, then we get a generalsolution of

p(x) = t

(x3 − 3

2(k1 + k2)x2 +

12

(k1 + k2)2 x)

+k2 − k1

k2 + k1x+ k1

q (x) = −t(x3 − 3

2(k1 + k2)x2 +

12

(k1 + k2)2 x)− k2 − k1

k2 + k1x+ k2.

Algebra Subject Test

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Calculus Subject Test

1. [3] Suppose that p(x) is a polynomial and that p(x)− p′(x) = x2 + 2x+ 1. Compute p(5).

2. [3] Let f be a function such that f(0) = 1, f ′(0) = 2, and

f ′′(t) = 4f ′(t)− 3f(t) + 1

for all t. Compute the 4th derivative of f , evaluated at 0.

3. [4] Let p be a monic cubic polynomial such that p(0) = 1 and such that all the zeros of p′(x) are alsozeros of p(x). Find p. Note: monic means that the leading coefficient is 1.

4. [4] Compute limn→∞

∑nk=1 | cos(k)|

n.

5. [4] Let the functions f(α, x) and g(α) be defined as

f(α, x) =

(x2

)αx− 1

g(α) =d4f

dx4

∣∣∣∣x=2

Then g(α) is a polynomial in α. Find the leading coefficient of g(α).

6. [5] Let f(x) = x3 − x2. For a given value of c, the graph of f(x), together with the graph of the linec + x, split the plane up into regions. Suppose that c is such that exactly two of these regions havefinite area. Find the value of c that minimizes the sum of the areas of these two regions.

7. [6] Let a1, a2, and a3 be nonzero complex numbers with non-negative real and imaginary parts. Findthe minimum possible value of

|a1 + a2 + a3|3√|a1a2a3|

.

8. [6] Let f(n) =∞∑k=2

1kn · k!

. Calculate∞∑n=2

f(n).

9. [7] Let x(t) be a solution to the differential equation

(x+ x′)2 + x · x′′ = cos t

with x(0) = x′(0) =√

25 . Compute x

(π4

).

10. [8] Let f(n) =n∑k=1

1k

. Then there exists constants γ, c, and d such that

f(n) = ln(n) + γ +c

n+

d

n2+O(

1n3

),

where the O( 1n3 ) means terms of order 1

n3 or lower. Compute the ordered pair (c, d).

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Calculus Subject Test

1. [3] Suppose that p(x) is a polynomial and that p(x)− p′(x) = x2 + 2x+ 1. Compute p(5).

Answer: 50 Observe that p(x) must be quadratic. Let p(x) = ax2 + bx+ c. Comparing coefficientsgives a = 1, b−2a = 2, and c− b = 1. So b = 4, c = 5, p(x) = x2 +4x+5 and p(5) = 25+20+5 = 50.

2. [3] Let f be a function such that f(0) = 1, f ′(0) = 2, and

f ′′(t) = 4f ′(t)− 3f(t) + 1

for all t. Compute the 4th derivative of f , evaluated at 0.

Answer: 54 Putting t = 0 gives f ′′(0) = 6. By differentiating both sides, we get f (3)(t) = 4f ′′(t)−3f ′(t) and f (3)(0) = 4·6−3·2 = 18. Similarly, f (4)(t) = 4f (3)(t)−3f ′′(t) and f (4)(0) = 4·18−3·6 = 54.

3. [4] Let p be a monic cubic polynomial such that p(0) = 1 and such that all the zeros of p′(x) are alsozeros of p(x). Find p. Note: monic means that the leading coefficient is 1.

Answer: (x+ 1)3 A root of a polynomial p will be a double root if and only if it is also a root ofp′. Let a and b be the roots of p′. Since a and b are also roots of p, they are double roots of p. But pcan have only three roots, so a = b and a becomes a double root of p′. This makes p′(x) = 3c(x− a)2

for some constant 3c, and thus p(x) = c(x − a)3 + d. Because a is a root of p and p is monic, d = 0and c = 1. From p(0) = 1 we get p(x) = (x+ 1)3.

4. [4] Compute limn→∞

∑nk=1 | cos(k)|

n.

Answer: 2π The main idea lies on the fact that positive integers are uniformly distributed modulo

π. (In the other words, if each integer n is written as qπ + r where q is an integer and 0 ≤ r < π, thevalue of r will distribute uniformly in the interval [0, π].) Using this fact, the summation is equivalentto the average value (using the Riemann summation) of the function | cos(k)| over the interval [0, π].Therefore, the answer is 1

π

∫ π0| cos(k)| = 2

π .

5. [4] Let the functions f(α, x) and g(α) be defined as

f(α, x) =

(x2

)αx− 1

g(α) =d4f

dx4

∣∣∣∣x=2

Then g(α) is a polynomial in α. Find the leading coefficient of g(α).

Answer: 116 Write the first equation as (x− 1)f =

(x2

)α. For now, treat α as a constant. From thisequation, repeatedly applying derivative with respect to x gives

(x− 1)f ′ + f =(α

2

)(x2

)α−1

(x− 1)f ′′ + 2f ′ =(α

2

)(α− 12

)(x2

)α−2

(x− 1)f (3) + 3f ′′ =(α

2

)(α− 12

)(α− 2

2

)(x2

)α−3

(x− 1)f (4) + 4f (3) =(α

2

)(α− 12

)(α− 2

2

)(α− 3

2

)(x2

)α−4

Substituting x = 2 to all equations gives g(α) = f (4)(α, 2) =(α2

) (α−1

2

) (α−2

2

) (α−3

2

)− 4f (3)(α, 2).

Because f (3)(α, 2) is a cubic polynomial in α, the leading coefficient of g(α) is 116 .

Calculus Subject Test

6. [5] Let f(x) = x3 − x2. For a given value of c, the graph of f(x), together with the graph of the linec + x, split the plane up into regions. Suppose that c is such that exactly two of these regions havefinite area. Find the value of c that minimizes the sum of the areas of these two regions.

Answer: − 1127 Observe that f(x) can be written as (x − 1

3 )3 − 13 (x − 1

3 ) − 227 , which has 180◦

symmetry around the point ( 13 ,−

227 ). Suppose the graph of f cuts the line y = c+x into two segments

of lengths a and b. When we move the line toward point P with a small distance ∆x (measured alongthe line perpendicular to y = x+ c), the sum of the enclosed areas will increase by |a− b|(∆x). As longas the line x + c does not passes through P , we can find a new line x + c∗ that increases the sum ofthe enclosed areas. Therefore, the sum of the areas reaches its maximum when the line passes throughP . For that line, we can find that c = y − x = − 2

27 −13 = − 11

27 .

7. [6] Let a1, a2, and a3 be nonzero complex numbers with non-negative real and imaginary parts. Findthe minimum possible value of

|a1 + a2 + a3|3√|a1a2a3|

.

Answer:√

3 3√

2 Write a1 in its polar form reiθ where 0 ≤ θ ≤ π2 . Suppose a2, a3 and r are fixed so

that the denominator is constant. Write a2 + a3 as seiφ. Since a2 and a3 have non-negative real andimaginary parts, the angle φ lies between 0 and π

2 . Consider the function

f(θ) = |a1 + a2 + a3|2 = |reiθ + seiφ|2 = r2 + 2rs cos(θ − φ) + s2.

Its second derivative is f ′′(θ) = −2rs(cos(θ − φ))). Since −π2 ≤ (θ − φ) ≤ π2 , we know that f ′′(θ) < 0

and f is concave. Therefore, to minimize f , the angle θ must be either 0 or π2 . Similarly, each of a1, a2

and a3 must be either purely real or purely imaginary to minimize f and the original fraction.

By the AM-GM inequality, if a1, a2 and a3 are all real or all imaginary, then the minimum value ofthe fraction is 3. Now suppose only two of the ai’s, say, a1 and a2 are real. Since the fraction ishomogenous, we may fix a1 + a2 - let the sum be 2. The term a1a2 in the denominator acheives itsmaximum only when a1 and a2 are equal, i.e. when a1 = a2 = 1. Then, if a3 = ki for some realnumber k, then the expression equals √

k2 + 43√k

.

Squaring and taking the derivative, we find that the minimum value of the fraction is√

3 3√

2, attainedwhen k =

√2. With similar reasoning, the case where only one of the ai’s is real yields the same

minimum value.

8. [6] Let f(n) =∞∑k=2

1kn · k!

. Calculate∞∑n=2

f(n).

Answer: 3− e

∞∑n=2

f(n) =∞∑k=2

∞∑n=2

1kn · k!

=∞∑k=2

1k!

∞∑n=2

1kn

=∞∑k=2

1k!· 1k(k − 1)

=∞∑k=2

1(k − 1)!

· 1k2(k − 1)

Calculus Subject Test

=∞∑k=2

1(k − 1)!

(1

k − 1− 1k2− 1k

)

=∞∑k=2

(1

(k − 1)(k − 1)!− 1k · k!

− 1k!

)

=∞∑k=2

(1

(k − 1)(k − 1)!− 1k · k!

)−∞∑k=2

1k!

=1

1 · 1!−(e− 1

0!− 1

1!

)= 3− e

9. [7] Let x(t) be a solution to the differential equation

(x+ x′)2 + x · x′′ = cos t

with x(0) = x′(0) =√

25 . Compute x

(π4

).

Answer:4√450

5 Rewrite the equation as x2 + 2xx′ + (xx′)′ = cos t. Let y = x2, so y′ = 2xx′ and

the equation becomes y+ y′ + 12y′′ = cos t. The term cos t suggests that the particular solution should

be in the form A sin t+B cos t. By substitution and coefficient comparison, we get A = 45 and B = 2

5 .Since the function y(t) = 4

5 sin t + 25 cos t already satisfies the initial conditions y(0) = x(0)2 = 2

5 andy′(0) = 2x(0)x′(0) = 4

5 , the function y also solves the initial value problem. Note that since x ispositive at t = 0 and y = x2 never reaches zero before t reaches π

4 , the value of x(π4

)must be positive.

Therefore, x(π4

)= +

√y(π4

)=√

65 ·√

22 =

4√4505 .

10. [8] Let f(n) =n∑k=1

1k

. Then there exists constants γ, c, and d such that

f(n) = ln(n) + γ +c

n+

d

n2+O(

1n3

),

where the O( 1n3 ) means terms of order 1

n3 or lower. Compute the ordered pair (c, d).

Answer: ( 12 ,−

112 ) From the given formula, we pull out the term k

n3 from O( 1n4 ), making f(n) =

log(n) + γ + cn + d

n2 + kn3 +O( 1

n4 ). Therefore,

f(n+1)−f(n) = log(n+ 1n

)−c(

1n− 1n+ 1

)−d(

1n2− 1

(n+ 1)2

)−k(

1n3− 1

(n+ 1)3

)+O

(1n4

).

For the left hand side, f(n+ 1)− f(n) = 1n+1 . By substituting x = 1

n , the formula above becomes

x

x+ 1= log (1 + x)− cx2 · 1

x+ 1− dx3 · x+ 2

(x+ 1)2− kx4 · x

2 + 3x+ 3(x+ 1)3

+O(x4).

Because x is on the order of 1n , 1

(x+1)3 is on the order of a constant. Therefore, all the terms in the

expansion of kx4 · x2+3x+3(x+1)3 are of order x4 or higher, so we can collapse it into O(x4). Using the Taylor

expansions, we get

x(1− x+ x2

)+O

(x4)

=(x− 1

2x2 +

13x3

)− cx2(1− x)− dx3(2) +O

(x4).

Coefficient comparison gives c = 12 and d = − 1

12 .

Calculus Subject Test

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Combinatorics Subject Test

1. [2] Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. How many (potentially empty) subsets T of S are there suchthat, for all x, if x is in T and 2x is in S then 2x is also in T?

2. [3] How many positive integers less than or equal to 240 can be expressed as a sum of distinct factorials?Consider 0! and 1! to be distinct.

3. [4] How many ways are there to choose 2010 functions f1, . . . , f2010 from {0, 1} to {0, 1} such thatf2010 ◦f2009 ◦ · · · ◦f1 is constant? Note: a function g is constant if g(a) = g(b) for all a, b in the domainof g.

4. [4] Manya has a stack of 85 = 1 + 4 + 16 + 64 blocks comprised of 4 layers (the kth layer from the tophas 4k−1 blocks; see the diagram below). Each block rests on 4 smaller blocks, each with dimensionshalf those of the larger block. Laura removes blocks one at a time from this stack, removing only blocksthat currently have no blocks on top of them. Find the number of ways Laura can remove precisely 5blocks from Manya’s stack (the order in which they are removed matters).

5. [5] John needs to pay 2010 dollars for his dinner. He has an unlimited supply of 2, 5, and 10 dollarnotes. In how many ways can he pay?

6. [5] An ant starts out at (0, 0). Each second, if it is currently at the square (x, y), it can move to(x − 1, y − 1), (x − 1, y + 1), (x + 1, y − 1), or (x + 1, y + 1). In how many ways can it end up at(2010, 2010) after 4020 seconds?

7. [6] For each integer x with 1 ≤ x ≤ 10, a point is randomly placed at either (x, 1) or (x,−1) with equalprobability. What is the expected area of the convex hull of these points? Note: the convex hull of afinite set is the smallest convex polygon containing it.

8. [6] How many functions f from {−1005, . . . , 1005} to {−2010, . . . , 2010} are there such that the fol-lowing two conditions are satisfied?

• If a < b then f(a) < f(b).

• There is no n in {−1005, . . . , 1005} such that |f(n)| = |n|.

9. [7] Rosencrantz and Guildenstern are playing a game where they repeatedly flip coins. Rosencrantzwins if 1 heads followed by 2009 tails appears. Guildenstern wins if 2010 heads come in a row. Theywill flip coins until someone wins. What is the probability that Rosencrantz wins?

10. [8] In a 16 × 16 table of integers, each row and column contains at most 4 distinct integers. What isthe maximum number of distinct integers that there can be in the whole table?

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Combinatorics Subject Test

1. [2] Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. How many (potentially empty) subsets T of S are there suchthat, for all x, if x is in T and 2x is in S then 2x is also in T?

Answer: 180 We partition the elements of S into the following subsets: {1, 2, 4, 8}, {3, 6}, {5, 10},{7}, {9}. Consider the first subset, {1, 2, 4, 8}. Say 2 is an element of T . Because 2 · 2 = 4 is in S, 4must also be in T . Furthermore, since 4 · 2 = 8 is in S, 8 must also be in T . So if T contains 2, it mustalso contain 4 and 8. Similarly, if T contains 1, it must also contain 2, 4, and 8. So T can contain thefollowing subsets of the subset {1, 2, 4, 8}: the empty set, {8}, {4, 8}, {2, 4, 8}, or {1, 2, 4, 8}. This gives5 possibilities for the first subset. In general, we see that if T contains an element q of one of thesesubsets, it must also contain the elements in that subset that are larger than q, because we created thesubsets for this to be true. So there are 3 possibilities for {3, 6}, 3 for {5, 10}, 2 for {7}, and 2 for {9}.This gives a total of 5 · 3 · 3 · 2 · 2 = 180 possible subsets T .

2. [3] How many positive integers less than or equal to 240 can be expressed as a sum of distinct factorials?Consider 0! and 1! to be distinct.

Answer: 39 Note that 1 = 0!, 2 = 0! + 1!, 3 = 0! + 2!, and 4 = 0! + 1! + 2!. These are the onlynumbers less than 6 that can be written as the sum of factorials. The only other factorials less than240 are 3! = 6, 4! = 24, and 5! = 120. So a positive integer less than or equal to 240 can only contain3!, 4!, 5!, and/or one of 1, 2, 3, or 4 in its sum. If it contains any factorial larger than 5!, it will belarger than 240. So a sum less than or equal to 240 will will either include 3! or not (2 ways), 4! or not(2 ways), 5! or not (2 ways), and add an additional 0, 1, 2, 3 or 4 (5 ways). This gives 2 · 2 · 2 · 5 = 40integers less than 240. However, we want only positive integers, so we must not count 0. So there are39 such positive integers.

3. [4] How many ways are there to choose 2010 functions f1, . . . , f2010 from {0, 1} to {0, 1} such thatf2010 ◦f2009 ◦ · · · ◦f1 is constant? Note: a function g is constant if g(a) = g(b) for all a, b in the domainof g.

Answer: 42010 〉 22010 If all 2010 functions are bijective1, then the composition f2010 ◦f2009 ◦ · · · ◦f1will be bijective also, and therefore not constant. If, however, one of f1, . . . , f2010 is not bijective, sayfk, then fk(0) = fk(1) = q, so f2010 ◦ f2009 ◦ · · · ◦ fk+1 ◦ fk ◦ · · · f1(0) = f2010 ◦ f2009 ◦ · · · ◦ fk+1(q) =f2010 ◦ f2009 ◦ · · · ◦ fk+1 ◦ fk ◦ · · · f1(1). So the composition will be constant unless all fi are bijective.Since there are 4 possible functions2 from {0, 1} to {0, 1} and 2 of them are bijective, we subtract thecases where all the functions are bijective from the total to get 42010 〉 22010.

4. [4] Manya has a stack of 85 = 1 + 4 + 16 + 64 blocks comprised of 4 layers (the kth layer from the tophas 4k∈1 blocks; see the diagram below). Each block rests on 4 smaller blocks, each with dimensionshalf those of the larger block. Laura removes blocks one at a time from this stack, removing only blocksthat currently have no blocks on top of them. Find the number of ways Laura can remove precisely 5blocks from Manya’s stack (the order in which they are removed matters).

1See http://en.wikipedia.org/wiki/Bijection.2The four functions are a, b, c, and d, defined as follows: a(0) = 0, a(1) = 0; b(0) = 0, b(1) = 1; c(0) = 1, c(1) = 0; d(0) = 1,

d(1) = 1.

Combinatorics Subject Test

Answer: 3384 Each time Laura removes a block, 4 additional blocks are exposed, increasing thetotal number of exposed blocks by 3. She removes 5 blocks, for a total of 1 ·4 ·7 ·10 ·13 ways. However,the stack originally only has 4 layers, so we must subtract the cases where removing a block on thebottom layer does not expose any new blocks. There are 1 · 4 · 4 · 4 · 4 = 256 of these (the last factor of4 is from the 4 blocks that we counted as being exposed, but were not actually). So our final answeris 1 · 4 · 7 · 10 · 13〉 1 · 4 · 4 · 4 · 4 = 3384.

5. [5] John needs to pay 2010 dollars for his dinner. He has an unlimited supply of 2, 5, and 10 dollarnotes. In how many ways can he pay?

Answer: 20503

Let the number of 2, 5, and 10 dollar notes John can use be x, y, and z respectively. We wish tofind the number of nonnegative integer solutions to 2x + 5y + 10z = 2010. Consider this equationmod 2. Because 2x, 10z, and 2010 are even, 5y must also be even, so y must be even. Now considerthe equation mod 5. Because 5y, 10z, and 2010 are divisible by 5, 2x must also be divisible by 5,so x must be divisible by 5. So both 2x and 5y are divisible by 10. So the equation is equivalent to10x′ + 10y′ + 10z = 2010, or x′ + y′ + z = 201, with x′, y′, and z nonnegative integers. There is awell-known bijection between solutions of this equation and picking 2 of 203 balls in a row on the table(explained in further detail below), so there are

2032

)= 20503 ways.

The bijection between solutions of x′+y′+z = 201 and arrangements of 203 balls in a row is as follows.Given a solution of the equation, we put x′ white balls in a row, then a black ball, then y′ white balls,then a black ball, then z white balls. This is like having 203 balls in a row on a table and picking twoof them to be black. To go from an arrangement of balls to a solution of the equation, we just read offx′, y′, and z from the number of white balls in a row. There are

2032

)ways to choose 2 of 203 balls to

be black, so there are2032

)solutions to x′ + y′ + z = 201.

6. [5] An ant starts out at (0, 0). Each second, if it is currently at the square (x, y), it can move to(x 〉 1, y 〉 1), (x 〉 1, y + 1), (x + 1, y 〉 1), or (x + 1, y + 1). In how many ways can it end up at(2010, 2010) after 4020 seconds?

Answer:40201005

)2Note that each of the coordinates either increases or decreases the x and y-

coordinates by 1. In order to reach 2010 after 4020 steps, each of the coordinates must be increased3015 times and decreased 1005 times. A permutation of 3015 plusses and 1005 minuses for each of xand y uniquely corresponds to a path the ant could take to (2010, 2010), because we can take orderedpairs from the two lists and match them up to a valid step the ant can take. So the number of waysthe ant can end up at (2010, 2010) after 4020 seconds is equal to the number of ways to arrange plussesand minuses for both x and y, or (

40201005

))2.

7. [6] For each integer x with 1 ≤ x ≤ 10, a point is randomly placed at either (x, 1) or (x,〉1) with equalprobability. What is the expected area of the convex hull of these points? Note: the convex hull of afinite set is the smallest convex polygon containing it.

Answer: 1793128 Let n = 10. Given a random variable X, let E(X) denote its expected value. If all

points are collinear, then the convex hull has area zero. This happens with probability 22n (either all

points are at y = 1 or all points are at y = 〉1). Otherwise, the points form a trapezoid with height2 (the trapezoid is possibly degenerate, but this won’t matter for our calculation). Let x1,l be thex-coordinate of the left-most point at y = 1 and x1,r be the x-coordinate of the right-most point aty = 1. Define x∈1,l and x∈1,r similarly for y = 〉1. Then the area of the trapezoid is

2 · (x1,r 〉 x1,l) + (x∈1,r 〉 x∈1,l)

2= x1,r + x∈1,r 〉 x1,l 〉 x∈1,l.

The expected area of the convex hull (assuming the points are not all collinear) is then, by linearity ofexpectation,

E(x1,r + x∈1,r 〉 x1,l 〉 x∈1,l) = E(x1,r) + E(x∈1,r)〉 E(x1,l)〉 E(x∈1,l).

Combinatorics Subject Test

We need only compute the expected values given in the above equation. Note that x1,r is equal to

k with probability 2k−1

2n∈2 , except that it is equal to n with probability 2n−1∈12n∈2 (the denominator is

2n〉 2 instead of 2n because we need to exclude the case where all points are collinear). Therefore, theexpected value of x1,r is equal to

1

2n 〉 2

((n∑

k=1

k · 2k∈1)〉 n · 1

)

=1

2n 〉 2

1 + 2 + · · ·+ 2n∈1

)+2 + 4 + · · ·+ 2n∈1

)+ · · ·+ 2n∈1 〉 n

)=

1

2n 〉 2

(2n 〉 1) + (2n 〉 2) + · · ·+

2n 〉 2n∈1

)〉 n

)=

1

2n 〉 2(n · 2n 〉 (2n 〉 1)〉 n)

= (n〉 1)2n 〉 1

2n 〉 2

Similarly, the expected value of x∈1,r is also (n 〉 1) 2n∈12n∈2 . By symmetry, the expected value of both

x1,l and x∈1,l is n+1〉 (n〉1) 2n∈12n∈2 . This says that if the points are not all collinear then the expected

area is 2 ·(

(n〉 1) 2n∈12n−1∈1 〉 (n + 1)

). So, the expected area is

2

2n· 0 +

(1〉 2

2n

)· 2 ·

((n〉 1)

2n 〉 1

2n∈1 〉 1〉 (n + 1)

)

= 2 · 2n∈1 〉 1

2n∈1·(

(n〉 1)2n 〉 1

2n∈1 〉 1〉 (n + 1)

)= 2 · (n〉 1)(2n 〉 1)〉 (n + 1)(2n∈1 〉 1)

2n∈1

= 2 · ((2n〉 2)〉 (n + 1))2n∈1 + 2

2n∈1

= 2n〉 6 +1

2n∈3

Plugging in n = 10, we get 14 + 1128 = 1793

128 .

8. [6] How many functions f from {〉1005, . . . , 1005} to {〉2010, . . . , 2010} are there such that the fol-lowing two conditions are satisfied?

• If a < b then f(a) < f(b).

• There is no n in {〉1005, . . . , 1005} such that |f(n)| = |n|.

Answer: · · · Note: the intended answer was40192011

), but the original answer was incorrect. The

correct answer is:

11733467826666773000724417738143880005531795870067107864012250438426995524609421666308605302966355504513409792805200762540756742811158611534813828022157596601875355477425764387233393584166695775000921640409535245687759455481741935349426766583008743635349407582844600705064877936286986176650915007126065996533696012706527852653952524215262304533916630291476263072382369363170971857101590310272130771639046414860423440232291348986940615141526024728199828817542362875717775477730951963033440695688189065502901813036762704306742550223341513844812312983802280527897951362595751647771568390543466492616362963283875803634852904329986459861362633348204891967272842242778625137520975558407856496002297523759366027

Combinatorics Subject Test

1506637984075036473724713869804364399766664507880042495122618597629613572449327653716600671574771752928091064660762269356178948295992047879612800838053160730032437457679147756158814950350323343872212037598984941717082402228562569617570267467242529665983280657359336668742613422094179386207330487537984173936781232801614775355365060827617078032786368164886083912495458822261016691599286765781539448097306313975219520659873979836562387314290328539769699667459275254643229234106717245366005816917271187760792

This obviously cannot be computed by hand, but there is a polynomial-time dynamic programmingalgorithm that will compute it.

9. [7] Rosencrantz and Guildenstern are playing a game where they repeatedly flip coins. Rosencrantzwins if 1 heads followed by 2009 tails appears. Guildenstern wins if 2010 heads come in a row. Theywill flip coins until someone wins. What is the probability that Rosencrantz wins?

Answer: 22009∈13·22008∈1 We can assume the first throw is heads (because neither player can win starting

from a string of only tails). Let x be the probability that Rosencrantz wins. Let y be the probabilitythat Rosencrantz wins after HT.

Whenever there is a string of less than 2009 tails followed by a heads, the heads basically means thetwo are starting from the beginning, where Rosencrantz has probability x of winning.

We also know that x = y(1〉 122009 ). This is because from the initial heads there is a (1〉 1

22009 ) chanceRosencrantz doesn’t lose, and in this case the last two flips are HT, in which case Rosencrantz hasprobability y of winning.

If the first two throws are HT, there is a 122008 chance Rosencrantz wins; otherwise, there is eventually

a heads, and so we are back in the case of starting from a heads, which corresponds to x. Therefore,y = 1

22008 + x(1〉 122008 ). Putting this together with the previous equation, we get:

x =

122008 + x(1〉 1

22008 ))

1〉 122009

)=⇒ x =

(1+22008x∈x

22008

)(22009∈122009

)=⇒ 24017x = x

24017 〉 22009 〉 22008 + 1

)+ 22009 〉 1

=⇒ x = 22009∈122009+22008∈1 ,

so the answer is 22009∈122009+22008∈1 = 22009∈1

3·22008∈1 .

10. [8] In a 16 × 16 table of integers, each row and column contains at most 4 distinct integers. What isthe maximum number of distinct integers that there can be in the whole table?

Answer: 49 First, we show that 50 is too big. Assume for sake of contradiction that a labelingwith at least 50 distinct integers exists. By the Pigeonhole Principle, there must be at least one row,say the first row, with at least 4 distinct integers in it; in this case, that is exactly 4, since that is themaximum number of distinct integers in one row. Then, in the remaining 15 rows there must be atleast 46 distinct integers (these 46 will also be distinct from the 4 in the first row). Using Pigeonholeagain, there will be another row, say the second row, with 4 distinct integers in it. Call the set ofintegers in the first and second rows S. Because the 4 distinct integers in the second row are distinctfrom the 4 in the first row, there are 8 distinct values in the first two rows, so |S| = 8. Now considerthe subcolumns containing the cells in rows 3 to 16. In each subcolumn, there are at most 2 values notin S, because there are already two distinct values in that column from the cells in the first two rows.So, the maximum number of distinct values in the table is 16 · 2 + 8 = 40, a contradiction. So a validlabeling must have fewer than 50 distinct integers. Below, we show by example that 49 is attainable.

Combinatorics Subject Test

1 17 33 - - - - - - - - - - - - -- 2 18 34 - - - - - - - - - - - -- - 3 19 35 - - - - - - - - - - -- - - 4 20 36 - - - - - - - - - -- - - - 5 21 37 - - - - - - - - -- - - - - 6 22 38 - - - - - - - -- - - - - - 7 23 39 - - - - - - -- - - - - - - 8 24 40 - - - - - -- - - - - - - - 9 25 41 - - - - -- - - - - - - - - 10 26 42 - - - -- - - - - - - - - - 11 27 43 - - -- - - - - - - - - - - 12 28 44 - -- - - - - - - - - - - - 13 29 45 -- - - - - - - - - - - - - 14 30 46

47 - - - - - - - - - - - - - 15 3132 48 - - - - - - - - - - - - - 16

Cells that do not contain a number are colored with color 49.

Combinatorics Subject Test

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

General Test, Part 1

1. [3] Suppose that x and y are positive reals such that

x− y2 = 3, x2 + y4 = 13.

Find x.

2. [3] Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. How many (potentially empty) subsets T of S are there suchthat, for all x, if x is in T and 2x is in S then 2x is also in T?

3. [4] A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convexpentagon that has an area of 7

10 of the area of the original rectangle. Find the ratio of the longer sideof the rectangle to the shorter side of the rectangle.

4. [4] Let S0 = 0 and let Sk equal a1 + 2a2 + . . . + kak for k ≥ 1. Define ai to be 1 if Si−1 < i and -1 ifSi−1 ≥ i. What is the largest k ≤ 2010 such that Sk = 0?

5. [4] Manya has a stack of 85 = 1 + 4 + 16 + 64 blocks comprised of 4 layers (the kth layer from the tophas 4k−1 blocks; see the diagram below). Each block rests on 4 smaller blocks, each with dimensionshalf those of the larger block. Laura removes blocks one at a time from this stack, removing only blocksthat currently have no blocks on top of them. Find the number of ways Laura can remove precisely 5blocks from Manya’s stack (the order in which they are removed matters).

6. [5] John needs to pay 2010 dollars for his dinner. He has an unlimited supply of 2, 5, and 10 dollarnotes. In how many ways can he pay?

7. [6] Suppose that a polynomial of the form p(x) = x2010 ± x2009 ± · · · ± x± 1 has no real roots. Whatis the maximum possible number of coefficients of −1 in p?

8. [6] A sphere is the set of points at a fixed positive distance r from its center. Let S be a set of 2010-dimensional spheres. Suppose that the number of points lying on every element of S is a finite numbern. Find the maximum possible value of n.

9. [7] Three unit circles ω1, ω2, and ω3 in the plane have the property that each circle passes through thecenters of the other two. A square S surrounds the three circles in such a way that each of its foursides is tangent to at least one of ω1, ω2 and ω3. Find the side length of the square S.

10. [8] Let a, b, c, x, y, and z be complex numbers such that

a =b + c

x− 2, b =

c + a

y − 2, c =

a + b

z − 2.

If xy + yz + zx = 67 and x + y + z = 2010, find the value of xyz.

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

General Test, Part 1

1. [3] Suppose that x and y are positive reals such that

x− y2 = 3, x2 + y4 = 13.

Find x.

Answer: 3+√

172 Squaring both sides of x− y2 = 3 gives x2 + y4− 2xy2 = 9. Subtract this equation

from twice the second given to get x2 + 2xy2 + y4 = 17 =⇒ x + y2 = ±17. Combining this equationwith the first given, we see that x = 3±

√17

2 . Since x is a positive real, x must be 3+√

172 .

2. [3] Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. How many (potentially empty) subsets T of S are there suchthat, for all x, if x is in T and 2x is in S then 2x is also in T?

Answer: 180 We partition the elements of S into the following subsets: {1, 2, 4, 8}, {3, 6}, {5, 10},{7}, {9}. Consider the first subset, {1, 2, 4, 8}. Say 2 is an element of T . Because 2 · 2 = 4 is in S, 4must also be in T . Furthermore, since 4 · 2 = 8 is in S, 8 must also be in T . So if T contains 2, it mustalso contain 4 and 8. Similarly, if T contains 1, it must also contain 2, 4, and 8. So T can contain thefollowing subsets of the subset {1, 2, 4, 8}: the empty set, {8}, {4, 8}, {2, 4, 8}, or {1, 2, 4, 8}. This gives5 possibilities for the first subset. In general, we see that if T contains an element q of one of thesesubsets, it must also contain the elements in that subset that are larger than q, because we created thesubsets for this to be true. So there are 3 possibilities for {3, 6}, 3 for {5, 10}, 2 for {7}, and 2 for {9}.This gives a total of 5 · 3 · 3 · 2 · 2 = 180 possible subsets T .

3. [4] A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convexpentagon that has an area of 7

10 of the area of the original rectangle. Find the ratio of the longer sideof the rectangle to the shorter side of the rectangle.

Answer:√

5

A

D

C

B

B′

E

Given a polygon P1P2 · · ·Pk, let [P1P2 · · ·Pk] denote its area. Let ABCD be the rectangle. Supposewe fold B across AC, and let E be the intersection of AD and B′C. Then we end up with the pentagonACDEB′, depicted above. Let’s suppose, without loss of generality, that ABCD has area 1. Then4AEC must have area 3

10 , since

General Test, Part 1

[ABCD] = [ABC] + [ACD]= [AB′C] + [ACD]= [AB′E] + 2[AEC] + [EDC]= [ACDEB′] + [AEC]

=710

[ABCD] + [AEC],

That is, [AEC] = 310 [ABCD] = 3

10 .

Since 4ECD is congruent to 4EAB′, both triangles have area 15 . Note that 4AB′C, 4ABC,

and 4CDA are all congruent, and all have area 12 . Since 4AEC and 4EDC share altitude DC,

DEEA = [DEC]

[AEC] = 23 . Because 4CAE is isosceles, CE = EA. Let AE = 3x. The CE = 3x, DE = 2x,

and CD = x√

9− 4 = x√

5. Then ADDC = AE+ED

DC = 3+2√5

=√

5.

4. [4] Let S0 = 0 and let Sk equal a1 + 2a2 + . . . + kak for k ≥ 1. Define ai to be 1 if Si−1 < i and -1 ifSi−1 ≥ i. What is the largest k ≤ 2010 such that Sk = 0?

Answer: 1092 Suppose that SN = 0 for some N ≥ 0. Then aN+1 = 1 because N + 1 ≥ SN . Thefollowing table lists the values of ak and Sk for a few k ≥ N :

k ak Sk

N 0N + 1 1 N + 1N + 2 1 2N + 3N + 3 −1 NN + 4 1 2N + 4N + 5 −1 N − 1N + 6 1 2N + 5N + 7 −1 N − 2

We see inductively that, for every i ≥ 1,

SN+2i = 2N + 2 + i

andSN+1+2i = N + 1− i

thus S3N+3 = 0 is the next k for which Sk = 0. The values of k for which Sk = 0 satisfy the recurrencerelation pn+1 = 3pn+3, and we compute that the first terms of the sequence are 0, 3, 12, 39, 120, 363, 1092;hence 1092 is our answer.

5. [4] Manya has a stack of 85 = 1 + 4 + 16 + 64 blocks comprised of 4 layers (the kth layer from the tophas 4k−1 blocks; see the diagram below). Each block rests on 4 smaller blocks, each with dimensionshalf those of the larger block. Laura removes blocks one at a time from this stack, removing only blocksthat currently have no blocks on top of them. Find the number of ways Laura can remove precisely 5blocks from Manya’s stack (the order in which they are removed matters).

General Test, Part 1

Answer: 3384 Each time Laura removes a block, 4 additional blocks are exposed, increasing thetotal number of exposed blocks by 3. She removes 5 blocks, for a total of 1 ·4 ·7 ·10 ·13 ways. However,the stack originally only has 4 layers, so we must subtract the cases where removing a block on thebottom layer does not expose any new blocks. There are 1 · 4 · 4 · 4 · 4 = 256 of these (the last factor of4 is from the 4 blocks that we counted as being exposed, but were not actually). So our final answeris 1 · 4 · 7 · 10 · 13− 1 · 4 · 4 · 4 · 4 = 3384.

6. [5] John needs to pay 2010 dollars for his dinner. He has an unlimited supply of 2, 5, and 10 dollarnotes. In how many ways can he pay?

Answer: 20503

Let the number of 2, 5, and 10 dollar notes John can use be x, y, and z respectively. We wish tofind the number of nonnegative integer solutions to 2x + 5y + 10z = 2010. Consider this equationmod 2. Because 2x, 10z, and 2010 are even, 5y must also be even, so y must be even. Now considerthe equation mod 5. Because 5y, 10z, and 2010 are divisible by 5, 2x must also be divisible by 5,so x must be divisible by 5. So both 2x and 5y are divisible by 10. So the equation is equivalent to10x′ + 10y′ + 10z = 2010, or x′ + y′ + z = 201, with x′, y′, and z nonnegative integers. There is awell-known bijection between solutions of this equation and picking 2 of 203 balls in a row on the table(explained in further detail below), so there are

(2032

)= 20503 ways.

The bijection between solutions of x′+y′+z = 201 and arrangements of 203 balls in a row is as follows.Given a solution of the equation, we put x′ white balls in a row, then a black ball, then y′ white balls,then a black ball, then z white balls. This is like having 203 balls in a row on a table and picking twoof them to be black. To go from an arrangement of balls to a solution of the equation, we just read offx′, y′, and z from the number of white balls in a row. There are

(2032

)ways to choose 2 of 203 balls to

be black, so there are(2032

)solutions to x′ + y′ + z = 201.

7. [6] Suppose that a polynomial of the form p(x) = x2010 ± x2009 ± · · · ± x± 1 has no real roots. Whatis the maximum possible number of coefficients of −1 in p?

Answer: 1005 Let p(x) be a polynomial with the maximum number of minus signs.p(x) cannot have more than 1005 minus signs, otherwise p(1) < 0 and p(2) ≥ 22010−22009−. . .−2−1 =1, which implies, by the Intermediate Value Theorem, that p must have a root greater than 1.

Let p(x) =x2011 + 1

x + 1= x2010−x2009 + x2008− . . .−x + 1. −1 is the only real root of x2011 + 1 = 0 but

p(−1) = 2011; therefore p has no real roots. Since p has 1005 minus signs, it is the desired polynomial.

8. [6] A sphere is the set of points at a fixed positive distance r from its center. Let S be a set of 2010-dimensional spheres. Suppose that the number of points lying on every element of S is a finite numbern. Find the maximum possible value of n.

Answer: 2 The answer is 2 for any number of dimensions. We prove this by induction on thedimension.

Note that 1-dimensional spheres are pairs of points, and 2-dimensional spheres are circles.

Base case, d = 2: The intersection of two circles is either a circle (if the original circles are identical, andin the same place), a pair of points, a single point (if the circles are tangent), or the empty set. Thus,in dimension 2, the largest finite number of intersection points is 2, because the number of pairwiseintersection points is 0, 1, or 2 for distinct circles.

We now prove that the intersection of two k-dimensional spheres is either the empty set, a (k − 1)-dimensional sphere, a k-dimensional sphere (which only occurs if the original spheres are identical andcoincident). Consider two spheres in k-dimensional space, and impose a coordinate system such thatthe centers of the two spheres lie on one coordinate axis. Then the equations for the two spheresbecome identical in all but one coordinate:

(x1 − a1)2 + x22 + . . . + x2

k = r21

(x1 − a2)2 + x22 + . . . + x2

k = r22

General Test, Part 1

If a1 = a2, the spheres are concentric, and so they are either nonintersecting or coincident, intersectingin a k-dimensional sphere. If a1 6= a2, then subtracting the equations and solving for x1 yieldsx1 = r2

1−a21−r2

2+a22

2(a2−a2). Plugging this in to either equation above yields a single equation equation that

describes a (k − 1)-dimensional sphere.

Assume we are in dimension d, and suppose for induction that for all k less than d, any two distinctk-dimensional spheres intersecting in a finite number of points intersect in at most two points. Supposewe have a collection of d-dimensional spheres s1, s2, . . . , sm. Without loss of generality, suppose the si

are distinct. Let ti be the intersection of si and si+1 for 1 ≤ i < m. If any ti are the empty set, thenthe intersection of the ti is empty. None of the ti is a d-dimensional sphere because the si are distinct.Thus each of t1, t2, . . . , tm−1 is a (d− 1)-dimensional sphere, and the intersection of all of them is thesame as the intersection of the d-dimensional spheres. We can then apply the inductive hypothesisto find that t1, . . . , tm−1 intersect in at most two points. Thus, by induction, a set of spheres in anydimension which intersect at only finitely many points intersect at at most two points.

We now exhibit a set of 22009 2010-dimensional spheres, and prove that their intersection containsexactly two points. Take the spheres with radii

√2013 and centers (0,±1,±1, . . . ,±1), where the sign

of each coordinate is independent from the sign of every other coordinate. Because of our choice ofradius, all these spheres pass through the points (±2, 0, 0, . . . 0). Then the intersection is the set ofpoints (x1, x2, . . . , x2010) which satisfy the equations x2

1 + (x2 ± 1)2 + · · · + (x2010 ± 1)2 = 2013. Theonly solutions to these equations are the points (±2, 0, 0, . . . , 0) (since (xi + 1)2 must be the same as(xi − 1)2 for all i > 1, because we may hold all but one of the ± choices constant, and change theremaining one).

9. [7] Three unit circles ω1, ω2, and ω3 in the plane have the property that each circle passes through thecenters of the other two. A square S surrounds the three circles in such a way that each of its foursides is tangent to at least one of ω1, ω2 and ω3. Find the side length of the square S.

Answer:√

6+√

2+84

Pω1

Q

ω2

R

ω3

M

E

A

F

C

D

B

N

By the Pigeonhole Principle, two of the sides must be tangent to the same circle, say ω1. Since Ssurrounds the circles, these two sides must be adjacent, so we can let A denote the common vertexof the two sides tangent to ω1. Let B, C, and D be the other vertices of S in clockwise order, andlet P , Q, and R be the centers of ω1, ω2, and ω3 respectively, and suppose WLOG that they are alsoin clockwise order. Then AC passes through the center of ω1, and by symmetry (since AB = AD) it

General Test, Part 1

must also pass through the other intersection point of ω2 and ω3. That is, AC is the radical axis of ω2

and ω3.

Now, let M and N be the feet of the perpendiculars from P and R, respectively, to side AD. Let Eand F be the feet of the perpendiculars from P to AB and from R to DC, respectively. Then PEAMand NRFD are rectangles, and PE and RF are radii of ω1 and ω2 respectively. Thus AM = EP = 1and ND = RF = 1. Finally, we have

MN = PR · cos(180◦ − ∠EPR)= cos(180◦ − EPQ−RPQ)= − cos ((270◦ − 60◦)/2 + 60◦)= − cos(165◦)= cos(15◦)

=√

6 +√

24

.

Thus AD = AM + MN + ND = 1 +√

6+√

24 + 1 =

√6+√

2+84 as claimed.

10. [8] Let a, b, c, x, y, and z be complex numbers such that

a =b + c

x− 2, b =

c + a

y − 2, c =

a + b

z − 2.

If xy + yz + zx = 67 and x + y + z = 2010, find the value of xyz.

Answer: −5892 Manipulate the equations to get a common denominator: a = b+cx−2 =⇒ x − 2 =

b+ca =⇒ x− 1 = a+b+c

a =⇒ 1x−1 = a

a+b+c ; similarly, 1y−1 = b

a+b+c and 1z−1 = c

a+b+c . Thus

1x− 1

+1

y − 1+

1z − 1

= 1

(y − 1)(z − 1) + (x− 1)(z − 1) + (x− 1)(y − 1) = (x− 1)(y − 1)(z − 1)xy + yz + zx− 2(x + y + z) + 3 = xyz − (xy + yz + zx) + (x + y + z)− 1

xyz − 2(xy + yz + zx) + 3(x + y + z)− 4 = 0xyz − 2(67) + 3(2010)− 4 = 0

xyz = −5892

General Test, Part 1

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

General Test, Part 2

1. [3] Below is pictured a regular seven-pointed star. Find the measure of angle a in radians.

a

2. [3] The rank of a rational number q is the unique k for which q = 1a1

+ · · ·+ 1ak

, where each ai is thesmallest positive integer such that q ≥ 1

a1+ · · ·+ 1

ai. Let q be the largest rational number less than 1

4

with rank 3, and suppose the expression for q is 1a1

+ 1a2

+ 1a3

. Find the ordered triple (a1, a2, a3).

3. [4] How many positive integers less than or equal to 240 can be expressed as a sum of distinct factorials?Consider 0! and 1! to be distinct.

4. [4] For 0 ≤ y ≤ 2, let Dy be the half-disk of diameter 2 with one vertex at (0, y), the other vertex onthe positive x-axis, and the curved boundary further from the origin than the straight boundary. Findthe area of the union of Dy for all 0 ≤ y ≤ 2.

5. [5] Suppose that there exist nonzero complex numbers a, b, c, and d such that k is a root of both theequations ax3 + bx2 + cx + d = 0 and bx3 + cx2 + dx + a = 0. Find all possible values of k (includingcomplex values).

6. [5] Let ABCD be an isosceles trapezoid such that AB = 10, BC = 15, CD = 28, and DA = 15. Thereis a point E such that 4AED and 4AEB have the same area and such that EC is minimal. FindEC.

7. [5] Suppose that x and y are complex numbers such that x + y = 1 and that x20 + y20 = 20. Find thesum of all possible values of x2 + y2.

8. [6] An ant starts out at (0, 0). Each second, if it is currently at the square (x, y), it can move to(x − 1, y − 1), (x − 1, y + 1), (x + 1, y − 1), or (x + 1, y + 1). In how many ways can it end up at(2010, 2010) after 4020 seconds?

9. [7] You are standing in an infinitely long hallway with sides given by the lines x = 0 and x = 6. Youstart at (3, 0) and want to get to (3, 6). Furthermore, at each instant you want your distance to (3, 6)to either decrease or stay the same. What is the area of the set of points that you could pass throughon your journey from (3, 0) to (3, 6)?

10. [8] In a 16 × 16 table of integers, each row and column contains at most 4 distinct integers. What isthe maximum number of distinct integers that there can be in the whole table?

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

General Test, Part 2

1. [3] Below is pictured a regular seven-pointed star. Find the measure of angle a in radians.

a

Answer: 3π7 The measure of the interior angle of a point of the star is π

7 because it is an inscribed

angle on the circumcircle which intercepts a seventh of the circle.1

a

π7

ϕπ7

ϕπ7

Consider the triangle shown above in bold. Because the sum of the angles in any triangle is π,

2ϕ+ 3(π

7

)= π = 2ϕ+ a

Canceling the 2ϕ on the right-hand side and on the left-hand side, we obtain

a =3π7.

2. [3] The rank of a rational number q is the unique k for which q = 1a1

+ · · ·+ 1ak

, where each ai is thesmallest positive integer such that q ≥ 1

a1+ · · ·+ 1

ai. Let q be the largest rational number less than 1

4

with rank 3, and suppose the expression for q is 1a1

+ 1a2

+ 1a3

. Find the ordered triple (a1, a2, a3).

Answer: (5, 21, 421) Suppose that A and B were rational numbers of rank 3 less than 14 , and

let a1, a2, a3, b1, b2, b3 be positive integers so that A = 1a1

+ 1a2

+ 1a3

and B = 1b1

+ 1b2

+ 1b3

are theexpressions for A and B as stated in the problem. If b1 < a1 then A < 1

a1−1 ≤1b1< B. In other words,

of all the rationals less than 14 with rank 3, those that have a1 = 5 are greater than those that have

a1 = 6, 7, 8, . . . Therefore we can “build” q greedily, adding the largest unit fraction that keeps q lessthan 1

4 :15 is the largest unit fraction less than 1

4 , hence a1 = 5;121 is the largest unit fraction less than 1

4 −15 , hence a2 = 21;

1421 is the largest unit fraction less than 1

4 −15 −

121 , hence a3 = 421.

1http://en.wikipedia.org/wiki/Inscribed_angle_theorem

General Test, Part 2

3. [4] How many positive integers less than or equal to 240 can be expressed as a sum of distinct factorials?Consider 0! and 1! to be distinct.

Answer: 39 Note that 1 = 0!, 2 = 0! + 1!, 3 = 0! + 2!, and 4 = 0! + 1! + 2!. These are the onlynumbers less than 6 that can be written as the sum of factorials. The only other factorials less than240 are 3! = 6, 4! = 24, and 5! = 120. So a positive integer less than or equal to 240 can only contain3!, 4!, 5!, and/or one of 1, 2, 3, or 4 in its sum. If it contains any factorial larger than 5!, it will belarger than 240. So a sum less than or equal to 240 will will either include 3! or not (2 ways), 4! or not(2 ways), 5! or not (2 ways), and add an additional 0, 1, 2, 3 or 4 (5 ways). This gives 2 · 2 · 2 · 5 = 40integers less than 240. However, we want only positive integers, so we must not count 0. So there are39 such positive integers.

4. [4] For 0 ≤ y ≤ 2, let Dy be the half-disk of diameter 2 with one vertex at (0, y), the other vertex onthe positive x-axis, and the curved boundary further from the origin than the straight boundary. Findthe area of the union of Dy for all 0 ≤ y ≤ 2.

Answer: π

0.5 1 1.5 2

0.5

1

1.5

2

0

From the picture above, we see that the union of the half-disks will be a quarter-circle with radius2, and therefore area π. To prove that this is the case, we first prove that the boundary of everyhalf-disk intersects the quarter-circle with radius 2, and then that the half-disk is internally tangentto the quarter-circle at that point. This is sufficient because it is clear from the diagram that we neednot worry about covering the interior of the quarter-circle.

Let O be the origin. For a given half-disk Dy, label the vertex on the y-axis A and the vertex onthe x-axis B. Let M be the midpoint of line segment AB. Draw segment OM , and extend it until itintersects the curved boundary of Dy. Label the intersection point C. This construction is shown inthe diagram below.

2

2

O B

A

Dy

C

M

We first prove that C lies on the quarter-circle, centered at the origin, with radius 2. Since M is themidpoint of AB, and A is on the y-axis, M is horizontally halfway between B and the y-axis. Since

General Test, Part 2

O and B are on the x-axis (which is perpendicular to the y-axis), segments OM and MB have thesame length. Since M is the midpoint of AB, and AB = 2, OM = 1. Since Dy is a half-disk withradius 1, all points on its curved boundary are 1 away from its center, M . Then C is 2 away from theorigin, and the quarter-circle consists of all points which are 2 away from the origin. Thus, C is anintersection of the half-disk Dy with the positive quarter-circle of radius 2.

It remains to show that the half-disk Dy is internally tangent to the quarter-circle. Since OC is aradius of the quarter-circle, it is perpendicular to the tangent of the quarter-circle at C. Since MC isa radius of the half-disk, it is perpendicular to the tangent of the half-disk at C. Then the tangentslines of the half-disk and the quarter-circle coincide, and the half-disk is tangent to the quarter-circle.It is obvious from the diagram that the half-disk lies at least partially inside of the quarter-circle, thehalf-disk Dy is internally tangent to the quarter-circle.

Then the union of the half-disks is be a quarter-circle with radius 2, and has area π.

5. [5] Suppose that there exist nonzero complex numbers a, b, c, and d such that k is a root of both theequations ax3 + bx2 + cx+ d = 0 and bx3 + cx2 + dx+ a = 0. Find all possible values of k (includingcomplex values).

Answer: 1,−1,i,−i Let k be a root of both polynomials. Multiplying the first polynomial by k

and subtracting the second, we have ak4 − a = 0, which means that k is either 1, −1, i, or −i. Ifa = b = c = d = 1, then −1, i, and −i are roots of both polynomials. If a = b = c = 1 and d = −3,then 1 is a root of both polynomials. So k can be 1, −1, i, and −i.

6. [5] Let ABCD be an isosceles trapezoid such that AB = 10, BC = 15, CD = 28, and DA = 15. Thereis a point E such that 4AED and 4AEB have the same area and such that EC is minimal. FindEC.

Answer: 216√145

A B

D C

E

X

M

Y

The locus of points E such that [AED] = [AEB] forms a line, since area is a linear function of thecoordinates of E; setting the areas equal gives a linear equation in the coordinates E2. Note that Aand M , the midpoint of DB, are on this line; A because both areas are 0, and M because the trianglesshare an altitude, and bases DM and MB are equal in length. Then AM is the set of points satisfyingthe area condition. The point E, then, is such that 4AEC is a right angle (to make the distanceminimal) and E lies on AM .

Let X be the point of intersection of AM and CD. Then 4AMB ∼ 4XMD, and since MD = BM ,they are in fact congruent. Thus DX = AB = 10, and XC = 18. Similarly, BX = 15, so ABXD is aparallelogram. Let Y be the foot of the perpendicular from A to DC, so that DY = DC−AB

2 = 9. Then

2The notation [A1A2 . . . An] means the area of polygon A1A2 . . . An.

General Test, Part 2

AY =√AD2 −DY 2 =

√225− 81 = 12. Then Y X = DX − DY = 1 and AX =

√AY 2 + Y X2 =√

144 + 1 =√

145. Since both 4AXY and 4CXE have a right angle, and ∠EXC and ∠Y XA arecongruent because they are vertical angles, 4AXY ∼ 4CXE. Then CE

AY = CXAX , so CE = 12 · 18√

145=

216√145

.

7. [5] Suppose that x and y are complex numbers such that x+ y = 1 and that x20 + y20 = 20. Find thesum of all possible values of x2 + y2.

Answer: −90 We have x2 + y2 + 2xy = 1. Define a = 2xy and b = x2 + y2 for convenience. Then

a+ b = 1 and b− a = x2 + y2 − 2xy = (x− y)2 = 2b− 1 so that x, y =√

2b−1±12 . Then

x20 + y20 =(√

2b− 1 + 12

)20

+(√

2b− 1− 12

)20

=1

220[(√

2b− 1 + 1)20 + (√

2b− 1− 1)20]

=2

220

[(√

2b− 1)20 +(

202

)(√

2b− 1)18 +(

204

)(√

2b− 1)16 + . . .

]=

2220

[(2b− 1)10 +

(202

)(2b− 1)9 +

(204

)(2b− 1)8 + . . .

]= 20

We want to find the sum of distinct roots of the above polynomial in b; we first prove that the originalpolynomial is square-free. The conditions x+y = 1 and x20+y20 = 20 imply that x20+(1−x)20−20 = 0;let p(x) = x20 + (1− x)20 − 20. p is square-free if and only if GCD(p, p′) = c for some constant c:

GCD(p, p′) = GCD(x20 + (1− x)20 − 20, 20(x19 − (1− x)19))= GCD(x20 − x(1− x)19 + (1− x)19 − 20, 20(x19 − (1− x)19))= GCD((1− x)19 − 20, x19 − (1− x)19)= GCD((1− x)19 − 20, x19 − 20)

The roots of x19 − 20 are 19√

20k exp( 2πik19 ) for some k = 0, 1, . . . , 18; the roots of (1 − x)19 − 20 are

1 − 19√

20k exp( 2πik19 ) for some k = 0, 1, . . . , 18. If x19 − 20 and (1 − x)19 − 20 share a common root,

then there exist integers m,n such that 19√

20m exp( 2πim19 ) = 1− 19

√20n exp( 2πin

19 ); since the imaginaryparts of both sides must be the same, we have m = n and 19

√20m exp( 2πim

19 ) = 12 =⇒ 20m = 1

219 , acontradiction. Thus we have proved that the polynomial in x has no double roots. Since for each bthere exists a unique pair (x, y) (up to permutations) that satisfies x2 + y2 = b and (x+ y)2 = 1, thepolynomial in b has no double roots.

Let the coefficient of bn in the above equation be [bn]. By Vieta’s Formulas, the sum of all possiblevalues of b = x2 + y2 is equal to − [b9]

[b10] . [b10] = 2220

(210)

and [b9] = 2220

(−(101

)29 +

(202

)29); thus

− [b9][b10] = − (10

1 )29−(202 )29

210 = −90.

8. [6] An ant starts out at (0, 0). Each second, if it is currently at the square (x, y), it can move to(x − 1, y − 1), (x − 1, y + 1), (x + 1, y − 1), or (x + 1, y + 1). In how many ways can it end up at(2010, 2010) after 4020 seconds?

Answer:(40201005

)2Note that each of the coordinates either increases or decreases the x and y-

coordinates by 1. In order to reach 2010 after 4020 steps, each of the coordinates must be increased3015 times and decreased 1005 times. A permutation of 3015 plusses and 1005 minuses for each of xand y uniquely corresponds to a path the ant could take to (2010, 2010), because we can take orderedpairs from the two lists and match them up to a valid step the ant can take. So the number of waysthe ant can end up at (2010, 2010) after 4020 seconds is equal to the number of ways to arrange plussesand minuses for both x and y, or (

(40201005

))2.

General Test, Part 2

9. [7] You are standing in an infinitely long hallway with sides given by the lines x = 0 and x = 6. Youstart at (3, 0) and want to get to (3, 6). Furthermore, at each instant you want your distance to (3, 6)to either decrease or stay the same. What is the area of the set of points that you could pass throughon your journey from (3, 0) to (3, 6)?

Answer: 9√

3 + 21π2

3 60

3√

3 B

C

A

3

A

6

If you draw concentric circles around the destination point, the condition is equivalent to the restrictionthat you must always go inwards towards the destination. In the diagram above, the regions throughwhich you might pass are shaded.

We find the areas of regions A, B, and C separately, and add them up (doubling the area of region A,because there are two of them).

The hypotenuse of triangle A is of length 6, and the base is of length 3, so it is a π6 -π3 -π2 triangle

(30-60-90 triangle) with area 9√

32 . Then the total area of the regions labeled A is 9

√3.

Since the angle of triangle A nearest the center of the circle (the destination point) is π3 , sector B has

central angle π3 . Then the area of sector B is 1

2r2θ = 1

2 · 36 · π3 = 6π.

Region C is a half-disc of radius 3, so its area is 9π2 .

Thus, the total area is 9√

3 + 21π2 .

10. [8] In a 16 × 16 table of integers, each row and column contains at most 4 distinct integers. What isthe maximum number of distinct integers that there can be in the whole table?

Answer: 49 First, we show that 50 is too big. Assume for sake of contradiction that a labelingwith at least 50 distinct integers exists. By the Pigeonhole Principle, there must be at least one row,say the first row, with at least 4 distinct integers in it; in this case, that is exactly 4, since that is themaximum number of distinct integers in one row. Then, in the remaining 15 rows there must be at

General Test, Part 2

least 46 distinct integers (these 46 will also be distinct from the 4 in the first row). Using Pigeonholeagain, there will be another row, say the second row, with 4 distinct integers in it. Call the set ofintegers in the first and second rows S. Because the 4 distinct integers in the second row are distinctfrom the 4 in the first row, there are 8 distinct values in the first two rows, so |S| = 8. Now considerthe subcolumns containing the cells in rows 3 to 16. In each subcolumn, there are at most 2 values notin S, because there are already two distinct values in that column from the cells in the first two rows.So, the maximum number of distinct values in the table is 16 · 2 + 8 = 40, a contradiction. So a validlabeling must have fewer than 50 distinct integers. Below, we show by example that 49 is attainable.

1 17 33 - - - - - - - - - - - - -- 2 18 34 - - - - - - - - - - - -- - 3 19 35 - - - - - - - - - - -- - - 4 20 36 - - - - - - - - - -- - - - 5 21 37 - - - - - - - - -- - - - - 6 22 38 - - - - - - - -- - - - - - 7 23 39 - - - - - - -- - - - - - - 8 24 40 - - - - - -- - - - - - - - 9 25 41 - - - - -- - - - - - - - - 10 26 42 - - - -- - - - - - - - - - 11 27 43 - - -- - - - - - - - - - - 12 28 44 - -- - - - - - - - - - - - 13 29 45 -- - - - - - - - - - - - - 14 30 46

47 - - - - - - - - - - - - - 15 3132 48 - - - - - - - - - - - - - 16

Cells that do not contain a number are colored with color 49.

General Test, Part 2

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Geometry Subject Test

1. [3] Below is pictured a regular seven-pointed star. Find the measure of angle a in radians.

a

2. [3] A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convexpentagon that has an area of 7

10 of the area of the original rectangle. Find the ratio of the longer sideof the rectangle to the shorter side of the rectangle.

3. [4] For 0 ≤ y ≤ 2, let Dy be the half-disk of diameter 2 with one vertex at (0, y), the other vertex onthe positive x-axis, and the curved boundary further from the origin than the straight boundary. Findthe area of the union of Dy for all 0 ≤ y ≤ 2.

4. [4] Let ABCD be an isosceles trapezoid such that AB = 10, BC = 15, CD = 28, and DA = 15. Thereis a point E such that 4AED and 4AEB have the same area and such that EC is minimal. FindEC.

5. [4] A sphere is the set of points at a fixed positive distance r from its center. Let S be a set of 2010-dimensional spheres. Suppose that the number of points lying on every element of S is a finite numbern. Find the maximum possible value of n.

6. [5] Three unit circles ω1, ω2, and ω3 in the plane have the property that each circle passes through thecenters of the other two. A square S surrounds the three circles in such a way that each of its foursides is tangent to at least one of ω1, ω2 and ω3. Find the side length of the square S.

7. [6] You are standing in an infinitely long hallway with sides given by the lines x = 0 and x = 6. Youstart at (3, 0) and want to get to (3, 6). Furthermore, at each instant you want your distance to (3, 6)to either decrease or stay the same. What is the area of the set of points that you could pass throughon your journey from (3, 0) to (3, 6)?

8. [6] Let O be the point (0, 0). Let A, B,C be three points in the plane such that AO = 15, BO = 15,and CO = 7, and such that the area of triangle ABC is maximal. What is the length of the shortestside of ABC?

9. [7] Let ABCD be a quadrilateral with an inscribed circle centered at I. Let CI intersect AB at E. If∠IDE = 35◦, ∠ABC = 70◦, and ∠BCD = 60◦, then what are all possible measures of ∠CDA?

10. [8] Circles ω1 and ω2 intersect at points A and B. Segment PQ is tangent to ω1 at P and to ω2 at Q,and A is closer to PQ than B. Point X is on ω1 such that PX ‖ QB, and point Y is on ω2 such thatQY ‖ PB. Given that ∠APQ = 30◦ and ∠PQA = 15◦, find the ratio AX/AY .

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Geometry Subject Test

1. [3] Below is pictured a regular seven-pointed star. Find the measure of angle a in radians.

a

Answer: 3π7 The measure of the interior angle of a point of the star is π

7 because it is an inscribed

angle on the circumcircle which intercepts a seventh of the circle.1

a

π7

ϕπ7

ϕπ7

Consider the triangle shown above in bold. Because the sum of the angles in any triangle is π,

2ϕ+ 3(π

7

)= π = 2ϕ+ a

Canceling the 2ϕ on the right-hand side and on the left-hand side, we obtain

a =3π7.

2. [3] A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convexpentagon that has an area of 7

10 of the area of the original rectangle. Find the ratio of the longer sideof the rectangle to the shorter side of the rectangle.

Answer:√

5

1http://en.wikipedia.org/wiki/Inscribed_angle_theorem

Geometry Subject Test

A

D

C

B

B′

E

Given a polygon P1P2 · · ·Pk, let [P1P2 · · ·Pk] denote its area. Let ABCD be the rectangle. Supposewe fold B across AC, and let E be the intersection of AD and B′C. Then we end up with the pentagonACDEB′, depicted above. Let’s suppose, without loss of generality, that ABCD has area 1. Then4AEC must have area 3

10 , since

[ABCD] = [ABC] + [ACD]= [AB′C] + [ACD]= [AB′E] + 2[AEC] + [EDC]= [ACDEB′] + [AEC]

=710

[ABCD] + [AEC],

That is, [AEC] = 310 [ABCD] = 3

10 .

Since 4ECD is congruent to 4EAB′, both triangles have area 15 . Note that 4AB′C, 4ABC,

and 4CDA are all congruent, and all have area 12 . Since 4AEC and 4EDC share altitude DC,

DEEA = [DEC]

[AEC] = 23 . Because 4CAE is isosceles, CE = EA. Let AE = 3x. The CE = 3x, DE = 2x,

and CD = x√

9− 4 = x√

5. Then ADDC = AE+ED

DC = 3+2√5

=√

5.

3. [4] For 0 ≤ y ≤ 2, let Dy be the half-disk of diameter 2 with one vertex at (0, y), the other vertex onthe positive x-axis, and the curved boundary further from the origin than the straight boundary. Findthe area of the union of Dy for all 0 ≤ y ≤ 2.

Answer: π

0.5 1 1.5 2

0.5

1

1.5

2

0

From the picture above, we see that the union of the half-disks will be a quarter-circle with radius2, and therefore area π. To prove that this is the case, we first prove that the boundary of everyhalf-disk intersects the quarter-circle with radius 2, and then that the half-disk is internally tangentto the quarter-circle at that point. This is sufficient because it is clear from the diagram that we neednot worry about covering the interior of the quarter-circle.

Geometry Subject Test

Let O be the origin. For a given half-disk Dy, label the vertex on the y-axis A and the vertex onthe x-axis B. Let M be the midpoint of line segment AB. Draw segment OM , and extend it until itintersects the curved boundary of Dy. Label the intersection point C. This construction is shown inthe diagram below.

2

2

O B

A

Dy

C

M

We first prove that C lies on the quarter-circle, centered at the origin, with radius 2. Since M is themidpoint of AB, and A is on the y-axis, M is horizontally halfway between B and the y-axis. SinceO and B are on the x-axis (which is perpendicular to the y-axis), segments OM and MB have thesame length. Since M is the midpoint of AB, and AB = 2, OM = 1. Since Dy is a half-disk withradius 1, all points on its curved boundary are 1 away from its center, M . Then C is 2 away from theorigin, and the quarter-circle consists of all points which are 2 away from the origin. Thus, C is anintersection of the half-disk Dy with the positive quarter-circle of radius 2.

It remains to show that the half-disk Dy is internally tangent to the quarter-circle. Since OC is aradius of the quarter-circle, it is perpendicular to the tangent of the quarter-circle at C. Since MC isa radius of the half-disk, it is perpendicular to the tangent of the half-disk at C. Then the tangentslines of the half-disk and the quarter-circle coincide, and the half-disk is tangent to the quarter-circle.It is obvious from the diagram that the half-disk lies at least partially inside of the quarter-circle, thehalf-disk Dy is internally tangent to the quarter-circle.

Then the union of the half-disks is be a quarter-circle with radius 2, and has area π.

4. [4] Let ABCD be an isosceles trapezoid such that AB = 10, BC = 15, CD = 28, and DA = 15. Thereis a point E such that 4AED and 4AEB have the same area and such that EC is minimal. FindEC.

Answer: 216√145

A B

D C

E

X

M

Y

Geometry Subject Test

The locus of points E such that [AED] = [AEB] forms a line, since area is a linear function of thecoordinates of E; setting the areas equal gives a linear equation in the coordinates E2. Note that Aand M , the midpoint of DB, are on this line; A because both areas are 0, and M because the trianglesshare an altitude, and bases DM and MB are equal in length. Then AM is the set of points satisfyingthe area condition. The point E, then, is such that 4AEC is a right angle (to make the distanceminimal) and E lies on AM .

Let X be the point of intersection of AM and CD. Then 4AMB ∼ 4XMD, and since MD = BM ,they are in fact congruent. Thus DX = AB = 10, and XC = 18. Similarly, BX = 15, so ABXD is aparallelogram. Let Y be the foot of the perpendicular from A to DC, so that DY = DC−AB

2 = 9. ThenAY =

√AD2 −DY 2 =

√225− 81 = 12. Then Y X = DX − DY = 1 and AX =

√AY 2 + Y X2 =√

144 + 1 =√

145. Since both 4AXY and 4CXE have a right angle, and ∠EXC and ∠Y XA arecongruent because they are vertical angles, 4AXY ∼ 4CXE. Then CE

AY = CXAX , so CE = 12 · 18√

145=

216√145

.

5. [4] A sphere is the set of points at a fixed positive distance r from its center. Let S be a set of 2010-dimensional spheres. Suppose that the number of points lying on every element of S is a finite numbern. Find the maximum possible value of n.

Answer: 2 The answer is 2 for any number of dimensions. We prove this by induction on thedimension.

Note that 1-dimensional spheres are pairs of points, and 2-dimensional spheres are circles.

Base case, d = 2: The intersection of two circles is either a circle (if the original circles are identical, andin the same place), a pair of points, a single point (if the circles are tangent), or the empty set. Thus,in dimension 2, the largest finite number of intersection points is 2, because the number of pairwiseintersection points is 0, 1, or 2 for distinct circles.

We now prove that the intersection of two k-dimensional spheres is either the empty set, a (k − 1)-dimensional sphere, a k-dimensional sphere (which only occurs if the original spheres are identical andcoincident). Consider two spheres in k-dimensional space, and impose a coordinate system such thatthe centers of the two spheres lie on one coordinate axis. Then the equations for the two spheresbecome identical in all but one coordinate:

(x1 − a1)2 + x22 + . . .+ x2

k = r21

(x1 − a2)2 + x22 + . . .+ x2

k = r22

If a1 = a2, the spheres are concentric, and so they are either nonintersecting or coincident, intersectingin a k-dimensional sphere. If a1 6= a2, then subtracting the equations and solving for x1 yieldsx1 = r21−a

21−r

22+a2

22(a2−a2)

. Plugging this in to either equation above yields a single equation equation thatdescribes a (k − 1)-dimensional sphere.

Assume we are in dimension d, and suppose for induction that for all k less than d, any two distinctk-dimensional spheres intersecting in a finite number of points intersect in at most two points. Supposewe have a collection of d-dimensional spheres s1, s2, . . . , sm. Without loss of generality, suppose the siare distinct. Let ti be the intersection of si and si+1 for 1 ≤ i < m. If any ti are the empty set, thenthe intersection of the ti is empty. None of the ti is a d-dimensional sphere because the si are distinct.Thus each of t1, t2, . . . , tm−1 is a (d− 1)-dimensional sphere, and the intersection of all of them is thesame as the intersection of the d-dimensional spheres. We can then apply the inductive hypothesisto find that t1, . . . , tm−1 intersect in at most two points. Thus, by induction, a set of spheres in anydimension which intersect at only finitely many points intersect at at most two points.

We now exhibit a set of 22009 2010-dimensional spheres, and prove that their intersection containsexactly two points. Take the spheres with radii

√2013 and centers (0,±1,±1, . . . ,±1), where the sign

of each coordinate is independent from the sign of every other coordinate. Because of our choice ofradius, all these spheres pass through the points (±2, 0, 0, . . . 0). Then the intersection is the set of

2The notation [A1A2 . . . An] means the area of polygon A1A2 . . . An.

Geometry Subject Test

points (x1, x2, . . . , x2010) which satisfy the equations x21 + (x2 ± 1)2 + · · · + (x2010 ± 1)2 = 2013. The

only solutions to these equations are the points (±2, 0, 0, . . . , 0) (since (xi + 1)2 must be the same as(xi − 1)2 for all i > 1, because we may hold all but one of the ± choices constant, and change theremaining one).

6. [5] Three unit circles ω1, ω2, and ω3 in the plane have the property that each circle passes through thecenters of the other two. A square S surrounds the three circles in such a way that each of its foursides is tangent to at least one of ω1, ω2 and ω3. Find the side length of the square S.

Answer:√

6+√

2+84

Pω1

Q

ω2

R

ω3

M

E

A

F

C

D

B

N

By the Pigeonhole Principle, two of the sides must be tangent to the same circle, say ω1. Since Ssurrounds the circles, these two sides must be adjacent, so we can let A denote the common vertexof the two sides tangent to ω1. Let B, C, and D be the other vertices of S in clockwise order, andlet P , Q, and R be the centers of ω1, ω2, and ω3 respectively, and suppose WLOG that they are alsoin clockwise order. Then AC passes through the center of ω1, and by symmetry (since AB = AD) itmust also pass through the other intersection point of ω2 and ω3. That is, AC is the radical axis of ω2

and ω3.

Now, let M and N be the feet of the perpendiculars from P and R, respectively, to side AD. Let Eand F be the feet of the perpendiculars from P to AB and from R to DC, respectively. Then PEAMand NRFD are rectangles, and PE and RF are radii of ω1 and ω2 respectively. Thus AM = EP = 1and ND = RF = 1. Finally, we have

MN = PR · cos(180◦ − ∠EPR)= cos(180◦ − EPQ−RPQ)= − cos ((270◦ − 60◦)/2 + 60◦)= − cos(165◦)= cos(15◦)

=√

6 +√

24

.

Thus AD = AM +MN +ND = 1 +√

6+√

24 + 1 =

√6+√

2+84 as claimed.

Geometry Subject Test

7. [6] You are standing in an infinitely long hallway with sides given by the lines x = 0 and x = 6. Youstart at (3, 0) and want to get to (3, 6). Furthermore, at each instant you want your distance to (3, 6)to either decrease or stay the same. What is the area of the set of points that you could pass throughon your journey from (3, 0) to (3, 6)?

Answer: 9√

3 + 21π2

3 60

3√

3 B

C

A

3

A

6

If you draw concentric circles around the destination point, the condition is equivalent to the restrictionthat you must always go inwards towards the destination. In the diagram above, the regions throughwhich you might pass are shaded.

We find the areas of regions A, B, and C separately, and add them up (doubling the area of region A,because there are two of them).

The hypotenuse of triangle A is of length 6, and the base is of length 3, so it is a π6 -π3 -π2 triangle

(30-60-90 triangle) with area 9√

32 . Then the total area of the regions labeled A is 9

√3.

Since the angle of triangle A nearest the center of the circle (the destination point) is π3 , sector B has

central angle π3 . Then the area of sector B is 1

2r2θ = 1

2 · 36 · π3 = 6π.

Region C is a half-disc of radius 3, so its area is 9π2 .

Thus, the total area is 9√

3 + 21π2 .

8. [6] Let O be the point (0, 0). Let A,B,C be three points in the plane such that AO = 15, BO = 15,and CO = 7, and such that the area of triangle ABC is maximal. What is the length of the shortestside of ABC?

Answer: 20 We claim that O should be the orthocenter of the triangle ABC. If O is not on analtitude of 4ABC, suppose (without loss of generality) that AO is not perpendicular to BC. We can

Geometry Subject Test

rotate A around O, leaving B and C fixed, to make AO perpendicular to BC, which strictly increasesthe area. Therefore, if [ABC] is maximal then 4ABC is an isosceles triangle with orthocenter O andbase AB.

O

C

A BF

D E

Let F be the foot of the perpendicular from C to AB. Since ∠FOA and ∠COE are vertical, ∠FAO =∠OCE. Then 4FAO is similar to 4FCB, so we have AF

OF = CFBF = OF+7

AF , so AF 2 = OF 2 + 7 · OF .Since AF 2 = 225 − OF 2, 2 · OF 2 + 7 · OF − 225 = 0, so OF = 9. Then AF = 12, so AB = 24 andBC = 20. Thus, the length of the shortest side of 4ABC is 20.

9. [7] Let ABCD be a quadrilateral with an inscribed circle centered at I. Let CI intersect AB at E. If∠IDE = 35◦, ∠ABC = 70◦, and ∠BCD = 60◦, then what are all possible measures of ∠CDA?

Answer: 70◦ and 160◦

Arbitrarily defining B and C determines I and E up to reflections across BC. D lies on both the circledetermined by ∠EDI = 35◦ and the line through C tangent to the circle (and on the opposite side ofB); since the intersection of a line and a circle has at most two points, there are only two cases forABCD. The diagram below on the left shows the construction made in this solution, containing bothcases. The diagram below on the right shows only the degenerate case.

B C

E

B′

I

D

A

70◦ 60◦B C

A

D

I

70◦ 60◦

Geometry Subject Test

Reflect B across EC to B′; then CB = CB′. Since BA and BC are tangent to the circle centeredat I, IB is the angle bisector of ∠ABC. Then ∠IBE = ∠IB′E = 35◦. If B′ = D, then ∠ADC =∠EB′C = 70◦. Otherwise, since ∠IB′E = 35◦ = ∠IDE (given), EB′DI is a cyclic quadrilateral.Then ∠IED = ∠IB′D = 35◦ and ∠BCI = ∠ECD = 30◦, so 4CED ∼ 4CBI.

Since ∠CID is exterior to 4DIE, ∠CID = ∠IDE + ∠DEI = 70◦. Then 4CDI ∼ 4CEB. BecauseEB′DI is cyclic, ∠IDC = ∠IEB′ = ∠IEB = 180◦−70◦−30◦ = 80◦. Then ∠ADC = 2∠IDC = 160◦.

Thus, the two possible measures are 70◦ and 160◦.

10. [8] Circles ω1 and ω2 intersect at points A and B. Segment PQ is tangent to ω1 at P and to ω2 at Q,and A is closer to PQ than B. Point X is on ω1 such that PX ‖ QB, and point Y is on ω2 such thatQY ‖ PB. Given that ∠APQ = 30◦ and ∠PQA = 15◦, find the ratio AX/AY .

Answer: 2−√

3

A

P

ω1

Q

ω2

B

X

Y

CM

Let C be the fourth vertex of parallelogram APCQ. The midpoint M of PQ is the intersection of thediagonals of this parallelogram. Because M has equal power3 with respect to the two circles ω1 andω2, it lies on

←→AB, the circles’ radical axis4. Therefore, C lies on

←→AB as well.

Using a series of parallel lines and inscribed arcs, we have:

∠APC = ∠APQ+ ∠CPQ = ∠APQ+ ∠PQA = ∠ABP + ∠QBA = ∠PBQ = ∠XPB,

where the last equality follows from the fact that PX ‖ QB.

3http://en.wikipedia.org/wiki/Power_of_a_point4http://en.wikipedia.org/wiki/Radical_axis

Geometry Subject Test

We also know that ∠BXP = 180◦ − ∠PAB = ∠CAP , so triangles BXP and CAP are similar. Bythe spiral similarity theorem5, triangles BPC and XPA are similar, too.

By analogous reasoning, triangles BQC and Y QA are similar. Then we have:

AX

AY=AX/BC

AY/BC=AP/CP

AQ/CQ=AP 2

AQ2

where the last inequality holds because APCQ is a parallelogram. Using the Law of Sines, the lastexpression equals sin2 15◦

sin2 30◦ = 2−√

3.

5This theorem states that if 4PAB and 4PXY are similar and oriented the same way, then 4PAX and 4PBY are similartoo. It is true because the first similarity implies that AP/BP = XP/Y P and ∠APB = ∠XPY , which proves the secondsimilarity.

Geometry Subject Test

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Guts Round

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13th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

1. [4] If A = 109 − 987654321 and B = 123456789+110 , what is the value of

√AB?

2. [4] Suppose that x, y, and z are non-negative real numbers such that x + y + z = 1. What is themaximum possible value of x + y2 + z3?

3. [4] In a group of people, there are 13 who like apples, 9 who like blueberries, 15 who like cantaloupe,and 6 who like dates. (A person can like more than 1 kind of fruit.) Each person who likes blueberriesalso likes exactly one of apples and cantaloupe. Each person who likes cantaloupe also likes exactlyone of blueberries and dates. Find the minimum possible number of people in the group.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

4. [5] To set up for a Fourth of July party, David is making a string of red, white, and blue balloons. Heplaces them according to the following rules:

• No red balloon is adjacent to another red balloon.

• White balloons appear in groups of exactly two, and groups of white balloons are separated byat least two non-white balloons.

• Blue balloons appear in groups of exactly three, and groups of blue balloons are separated by atleast three non-blue balloons.

If David uses over 600 balloons, determine the smallest number of red balloons that he can use.

5. [5] You have a length of string and 7 beads in the 7 colors of the rainbow. You place the beads onthe string as follows – you randomly pick a bead that you haven’t used yet, then randomly add it toeither the left end or the right end of the string. What is the probability that, at the end, the colors ofthe beads are the colors of the rainbow in order? (The string cannot be flipped, so the red bead mustappear on the left side and the violet bead on the right side.)

6. [5] How many different numbers are obtainable from five 5s by first concatenating some of the 5s, thenmultiplying them together? For example, we could do 5 ·55 ·55, 555 ·55, or 55555, but not 5 ·5 or 2525.

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13th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

7. [6] What are the last 8 digits of

11× 101× 1001× 10001× 100001× 1000001× 111?

8. [6] Each square in the following hexomino has side length 1. Find the minimum area of any rectanglethat contains the entire hexomino.

9. [6] Indecisive Andy starts out at the midpoint of the 1-unit-long segment HT . He flips 2010 coins. Oneach flip, if the coin is heads, he moves halfway towards endpoint H, and if the coin is tails, he moveshalfway towards endpoint T . After his 2010 moves, what is the expected distance between Andy andthe midpoint of HT?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

10. [7] Let ABC be a triangle with AB = 8, BC = 15, and AC = 17. Point X is chosen at random online segment AB. Point Y is chosen at random on line segment BC. Point Z is chosen at random online segment CA. What is the expected area of triangle XY Z?

11. [7] From the point (x, y), a legal move is a move to (x3 + u, y

3 + v), where u and v are real numberssuch that u2 + v2 ≤ 1. What is the area of the set of points that can be reached from (0, 0) in a finitenumber of legal moves?

12. [7] How many different collections of 9 letters are there? A letter can appear multiple times in acollection. Two collections are equal if each letter appears the same number of times in both collections.

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13th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

13. [8] A triangle in the xy-plane is such that when projected onto the x-axis, y-axis, and the line y = x,the results are line segments whose endpoints are (1, 0) and (5, 0), (0, 8) and (0, 13), and (5, 5) and(7.5, 7.5), respectively. What is the triangle’s area?

14. [8] In how many ways can you fill a 3 × 3 table with the numbers 1 through 9 (each used once) suchthat all pairs of adjacent numbers (sharing one side) are relatively prime?

15. [8] Pick a random integer between 0 and 4095, inclusive. Write it in base 2 (without any leading zeroes).What is the expected number of consecutive digits that are not the same (that is, the expected numberof occurrences of either 01 or 10 in the base 2 representation)?

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13th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

16. [9] Jessica has three marbles colored red, green, and blue. She randomly selects a non-empty subset ofthem (such that each subset is equally likely) and puts them in a bag. You then draw three marblesfrom the bag with replacement. The colors you see are red, blue, red. What is the probability that theonly marbles in the bag are red and blue?

17. [9] An ant starts at the origin, facing in the positive x-direction. Each second, it moves 1 unit forward,then turns counterclockwise by sin−1( 3

5 ) degrees. What is the least upper bound on the distancebetween the ant and the origin? (The least upper bound is the smallest real number r that is at leastas big as every distance that the ant ever is from the origin.)

18. [9] Find two lines of symmetry of the graph of the function y = x + 1x . Express your answer as two

equations of the form y = ax + b.

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13th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

19. [10] A 5-dimensional ant starts at one vertex of a 5-dimensional hypercube of side length 1. A moveis when the ant travels from one vertex to another vertex at a distance of

√2 away. How many ways

can the ant make 5 moves and end up on the same vertex it started at?

20. [10] Find the volume of the set of points (x, y, z) satisfying

x, y, z ≥ 0x + y ≤ 1y + z ≤ 1z + x ≤ 1

21. [10] Let4ABC be a scalene triangle. Let ha be the locus of points P such that |PB−PC| = |AB−AC|.Let hb be the locus of points P such that |PC − PA| = |BC − BA|. Let hc be the locus of points Psuch that |PA− PB| = |CA− CB|. In how many points do all of ha, hb, and hc concur?

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22. [12] You are the general of an army. You and the opposing general both have an equal number of troopsto distribute among three battlefields. Whoever has more troops on a battlefield always wins (you winties). An order is an ordered triple of non-negative real numbers (x, y, z) such that x + y + z = 1, andcorresponds to sending a fraction x of the troops to the first field, y to the second, and z to the third.Suppose that you give the order

(14 , 1

4 , 12

)and that the other general issues an order chosen uniformly

at random from all possible orders. What is the probability that you win two out of the three battles?

23. [12] In the country of Francisca, there are 2010 cities, some of which are connected by roads. Betweenany two cities, there is a unique path which runs along the roads and which does not pass through anycity twice. What is the maximum possible number of cities in Francisca which have at least 3 roadsrunning out of them?

24. [12] Define a sequence of polynomials as follows: let a1 = 3x2−x, let a2 = 3x2−7x+ 3, and for n ≥ 1,let an+2 = 5

2an+1 − an. As n tends to infinity, what is the limit of the sum of the roots of an?

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25. [15] How many functions f : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} have the property that f({1, 2, 3}) andf(f({1, 2, 3})) are disjoint?

26. [15] Express the following in closed form, as a function of x:

sin2(x)+sin2(2x) cos2(x)+sin2(4x) cos2(2x) cos2(x)+· · ·+sin2(22010x) cos2(22009x) · · · cos2(2x) cos2(x).

27. [15] Suppose that there are real numbers a, b, c ≥ 1 and that there are positive reals x, y, z such that

ax + by + cz = 4xax + yby + zcz = 6

x2ax + y2by + z2cz = 9.

What is the maximum possible value of c?

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28. [18] Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament willhave 4 rounds, and in each round the 9 players are split into 3 groups of 3. During the tournament,each player plays every other player exactly once. How many different ways can Danielle divide the 9people into three groups in each round to satisfy these requirements?

29. [18] Compute the remainder when30303∑k=1

kk

is divided by 101.

30. [18] A monomial term xi1xi2 . . . xikin the variables x1, x2, . . . x8 is square-free if i1, i2, . . . ik are distinct.

(A constant term such as 1 is considered square-free.) What is the sum of the coefficients of the square-free terms in the following product? ∏

1≤i<j≤8

(1 + xixj)

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31. [21] In the Democratic Republic of Irun, 5 people are voting in an election among 5 candidates. Ifeach person votes for a single candidate at random, what is the expected number of candidates thatwill be voted for?

32. [21] There are 101 people participating in a Secret Santa gift exchange. As usual each person israndomly assigned another person for whom (s)he has to get a gift, such that each person gives andreceives exactly one gift and no one gives a gift to themself. What is the probability that the firstperson neither gives gifts to or receives gifts from the second or third person? Express your answer asa decimal rounded to five decimal places.

33. [21] Let a1 = 3, and for n > 1, let an be the largest real number such that

4(a2

n−1 + a2n

)= 10an−1an − 9.

What is the largest positive integer less than a8?

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Note: Problems 34 and 36 are estimation problems. You will receive a number of points (between 0and 25) based on how close your answer is to the correct answer. Throughout, we will let C denote thecorrect answer and A denote your answer. If a described scoring method would ever assign a negativenumber of points, you will receive zero points instead.

34. [25] 3000 people each go into one of three rooms randomly. What is the most likely value for themaximum number of people in any of the rooms? Your score for this problem will be 0 if you writedown a number less than or equal to 1000. Otherwise, it will be 25− 27 |A−C|

min(A,C)−1000 .

35. [25] Call an positive integer almost-square if it can be written as a · b, where a and b are integers anda ≤ b ≤ 4

3a. How many almost-square positive integers are less than or equal to 1000000? Your scorewill be equal to 25− 65 |A−C|

min(A,C) .

36. [25] Consider an infinite grid of unit squares. An n-omino is a subset of n squares that is connected.Below are depicted examples of 8-ominoes. Two n-ominoes are considered equivalent if one can beobtained from the other by translations and rotations. What is the number of distinct 15-ominoes?Your score will be equal to 25− 13| ln(A)− ln(C)|.

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13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Guts Round

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1. [4] If A = 109 − 987654321 and B = 123456789+110 , what is the value of

√AB?

Answer: 12345679 Both A and B equal 12345679, so√AB = 12345679 as well.

2. [4] Suppose that x, y, and z are non-negative real numbers such that x + y + z = 1. What is themaximum possible value of x+ y2 + z3?

Answer: 1 Since 0 ≤ y, z ≤ 1, we have y2 ≤ y and z3 ≤ z. Therefore x+ y2 + z3 ≤ x+ y + z = 1.We can get x+ y2 + z3 = 1 by setting (x, y, z) = (1, 0, 0).

3. [4] In a group of people, there are 13 who like apples, 9 who like blueberries, 15 who like cantaloupe,and 6 who like dates. (A person can like more than 1 kind of fruit.) Each person who likes blueberriesalso likes exactly one of apples and cantaloupe. Each person who likes cantaloupe also likes exactlyone of blueberries and dates. Find the minimum possible number of people in the group.

Answer: 22 Everyone who likes cantaloupe likes exactly one of blueberries and dates. However,there are 15 people who like cantaloupe, 9 who like blueberries, and 6 who like dates. Thus, everyonewho likes blueberries or dates must also like cantaloupes (because if any of them didn’t, we would endup with less than 15 people who like cantaloupe).

Since everyone who likes blueberries likes cantaloupes, none of them can like apples. However, the 6people who like both cantaloupe and dates can also like apples. So, we could have a group where 7people like apples alone, 9 like blueberries and cantaloupe, and 6 like apples, cantaloupe, and dates.This gives 22 people in the group, which is optimal.

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4. [5] To set up for a Fourth of July party, David is making a string of red, white, and blue balloons. Heplaces them according to the following rules:

• No red balloon is adjacent to another red balloon.

• White balloons appear in groups of exactly two, and groups of white balloons are separated byat least two non-white balloons.

• Blue balloons appear in groups of exactly three, and groups of blue balloons are separated by atleast three non-blue balloons.

If David uses over 600 balloons, determine the smallest number of red balloons that he can use.

Answer: 99 It is possible to achieve 99 red balloons with the arrangement

WWBBBWW RBBBWWRBBBWW . . .RBBBWW︸ ︷︷ ︸99 RBBBWW’s

,

which contains 99 · 6 + 7 = 601 balloons.

Now assume that one can construct a chain with 98 or fewer red balloons. Then there can be 99blocks of non-red balloons, which in total must contain more than 502 balloons. The only validcombinations of white and blue balloons are WWBBB, BBBWW, and WWBBBWW (Any otherscontain the subsequence BBBWWBBB, which is invalid). The sequence . . .WWR must be followedby BBBWW; otherwise two groups of white balloons would be too close. Similarly, the sequence

Guts Round

RWW . . . must be preceded by WWBBB. It follows that WWBBBWW can be used at most once in avalid sequence, meaning that there can be at most 98 · 5 + 7 = 497 non-red balloons. Contradiction.Therefore the minimum is 99 red balloons. (Better if the party’s outdoors; then we’d have 99 redballoons floating in the summer sky. :-p)

5. [5] You have a length of string and 7 beads in the 7 colors of the rainbow. You place the beads onthe string as follows – you randomly pick a bead that you haven’t used yet, then randomly add it toeither the left end or the right end of the string. What is the probability that, at the end, the colors ofthe beads are the colors of the rainbow in order? (The string cannot be flipped, so the red bead mustappear on the left side and the violet bead on the right side.)

Answer: 15040 The threading method does not depend on the colors of the beads, so at the end all

configurations are equally likely. Since there are 7! = 5040 configurations in total, the probability ofany particular configuration is 1

5040 .

6. [5] How many different numbers are obtainable from five 5s by first concatenating some of the 5s, thenmultiplying them together? For example, we could do 5 ·55 ·55, 555 ·55, or 55555, but not 5 ·5 or 2525.

Answer: 7 If we do 55555, then we’re done.

Note that 5, 55, 555, and 5555 all have completely distinguishable prime factorizations. This meansthat if we are given a product of them, we can obtain the individual terms. The number of 5555’s isthe exponent of 101, the number of 555’s is the exponent of 37, the number of 55’s is the exponentof 11 minus the exponent of 101, and the number of 5’s is just whatever we need to get the properexponent of 5. Then the answer is the number of ways we can split the five 5’s into groups of at leastone. This is the number of unordered partitions of 5, which is 7.

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7. [6] What are the last 8 digits of

11× 101× 1001× 10001× 100001× 1000001× 111?

Answer: 19754321 Multiply terms in a clever order.

11 · 101 · 10001 = 11, 111, 111,111 · 1001 · 1000001 = 111, 111, 111, 111.

The last eight digits of 11, 111, 111 · 111, 111, 111, 111 are 87654321. We then just need to compute thelast 8 digits of 87654321 · 100001 = 87654321 + . . . 32100000, which are 19754321.

8. [6] Each square in the following hexomino has side length 1. Find the minimum area of any rectanglethat contains the entire hexomino.

Answer: 212 If a rectangle contains the entire hexomino, it must also contain its convex hull, which is

an origin-symmetric hexagon. It is fairly clear that the smallest rectangle that contains such a hexagonmust share one set of parallel sides with the hexagon. There are three such rectangles, and checkingthem all, we find that the one shown below is the smallest. It has area 21

2 .

Guts Round

9. [6] Indecisive Andy starts out at the midpoint of the 1-unit-long segment HT . He flips 2010 coins. Oneach flip, if the coin is heads, he moves halfway towards endpoint H, and if the coin is tails, he moveshalfway towards endpoint T . After his 2010 moves, what is the expected distance between Andy andthe midpoint of HT?

Answer: 14 Let Andy’s position be x units from the H end after 2009 flips. If Any moves towards

the H end, he ends up at x2 , a distance of 1−x

2 from the midpoint. If Andy moves towards the T end,he ends up at 1+x

2 , a distance of x2 from the midpoint. His expected distance from the midpoint is then

1−x2 + x

2

2=

14.

Since this does not depend on x, 14 is the answer.

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10. [7] Let ABC be a triangle with AB = 8, BC = 15, and AC = 17. Point X is chosen at random online segment AB. Point Y is chosen at random on line segment BC. Point Z is chosen at random online segment CA. What is the expected area of triangle XY Z?

Answer: 15 Let E(X) denote the expected value of X, and let [S] denote the area of S. Then

E([4XY Z]) = E([4ABC]− [4XY B]− [4ZY C]− [4XBZ])= [4ABC]− E([4XY B])− E([4ZY C])− [4XBZ]),

where the last step follows from linearity of expectation1. But [4XY B] = 12 · BX · BY · sin(B).

The 12 sin(B) term is constant, and BX and BY are both independent with expected values AB

2 andBC2 , respectively. Thus E([4XY B]) = 1

8AB · BC · sin(B) = 14 [4ABC]. Similarly, E([4ZY C]) =

E([4ZBX]) = 14 [4ABC].

Then we have E([4XY Z]) = (1− 14 − 1

4 − 14 )[4ABC] = 1

4 [4ABC] = 15.

Note: We can also solve this problem (and the more general case of polygons) by noting that thearea of XY Z is linear in the coordinates of X, Y , and Z, so the expected area of XY Z is the sameas the area of X ′Y ′Z ′, where X ′ is the expected location of X, Y ′ is the expected location of Y , andZ ′ is the expected location of Z. In our case, this corresponds to the midpoints of the three sides AB,BC, and CA.

1See http://www-math.mit.edu/~spielman/AdvComplexity/handout.ps for an introduction to linearity of expectation andother important tools in probability

Guts Round

11. [7] From the point (x, y), a legal move is a move to (x3 + u, y3 + v), where u and v are real numberssuch that u2 + v2 ≤ 1. What is the area of the set of points that can be reached from (0, 0) in a finitenumber of legal moves?

Answer: 9π4 We claim that the set of points is the disc with radius 3

2 centered at the origin, which

clearly has area 9π4 .

First, we show that the set is contained in this disc. This is because if we are currently at a distanceof r from the origin, then we can’t end up at a distance of greater than r

3 + 1 from the origin after asingle move. Since r

3 + 1 < 32 if r < 3

2 , we will always end up in the disc of radius 32 if we start in it.

Since the origin is inside this disc, any finite number of moves will leave us inside this disc.

Next, we show that all points in this disc can be reached in a finite number of moves. Indeed, afterone move we can get all points within a distance of 1. After two moves, we can get all points withina distance of 4

3 . After three moves, we can get all points within a distance of 139 . In general, after n

moves we can get all points within a distance of 32 − 1

2·3k−1 . This means that for any distance d < 32 ,

we will eventually get all points within a distance of d, so all points in the disc of radius 32 can be

reached after some number of moves.

12. [7] How many different collections of 9 letters are there? A letter can appear multiple times in acollection. Two collections are equal if each letter appears the same number of times in both collections.

Answer:(349

)We put these collections in bijections with binary strings of length 34 containing 9

zeroes and 25 ones. Take any such string - the 9 zeroes will correspond to the 9 letters in the collection.If there are n ones before a zero, then that zero corresponds to the (n + 1)st letter of the alphabet.This scheme is an injective map from the binary strings to the collections, and it has an inverse, so thenumber of collections is

(349

).

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13. [8] A triangle in the xy-plane is such that when projected onto the x-axis, y-axis, and the line y = x,the results are line segments whose endpoints are (1, 0) and (5, 0), (0, 8) and (0, 13), and (5, 5) and(7.5, 7.5), respectively. What is the triangle’s area?

Answer: 172 Sketch the lines x = 1, x = 5, y = 8, y = 13,y = 10 − x, and y = 15 − x. The

triangle has to be contained in the hexagonal region contained in all these lines. If all the projectionsare correct, every other vertex of the hexagon must be a vertex of the triangle, which gives us twopossibilities for the triangle. One of these triangles has vertices at (2, 8), (1, 13), and (5, 10), and hasan area of 17

2 . It is easy to check that the other triangle has the same area, so the answer is unique.

14. [8] In how many ways can you fill a 3 × 3 table with the numbers 1 through 9 (each used once) suchthat all pairs of adjacent numbers (sharing one side) are relatively prime?

Answer: 2016 The numbers can be separated into four sets. Numbers in the set A = {1, 5, 7} canbe placed next to anything. The next two sets are B = {2, 4, 8} and C = {3, 9}. The number 6, whichforms the final set D, can only be placed next to elements of A. The elements of each group can beinterchanged without violating the condition, so without loss of generality, we can pretend we havethree 1’s, three 2’s, two 3’s, and one 6, as long as we multiply our answer by 3!3!2! at the end. Theavailable arrangements are, grouped by the position of the 6, are:

When 6 is in contact with three numbers:1 2 36 1 21 2 3

When 6 is in contact with two numbers:

Guts Round

6 1 21 2 32 3 1

6 1 21 1 32 3 2

The next two can be flipped diagonally to create different arrangements:

6 1 21 2 31 3 2

6 1 21 2 33 1 2

Those seven arrangements can be rotated 90, 180, and 270 degrees about the center to generate a totalof 28 arrangements. 28 · 3!3!2! = 2016.

15. [8] Pick a random integer between 0 and 4095, inclusive. Write it in base 2 (without any leading zeroes).What is the expected number of consecutive digits that are not the same (that is, the expected numberof occurrences of either 01 or 10 in the base 2 representation)?

Answer: 204814096 Note that every number in the range can be written as a 12-digit binary string. For

i = 1, 2, . . . 11, let Ri be a random variable which is 1 if the ith and (i+ 1)st digits differ in a randomlychosen number in the range. By linearity of expectation, E(

∑iRi) =

∑E(Ri).

Since we choose every binary string of length 12 with equal probability, the sum of the expectationsis 11

2 . However, this is not the expected number of 01s and 10s - we need to subtract the occasionswhere the leading digit is zero. There is a 1

2 chance that the number starts with a 0, in which case wemust ignore the first digit change - unless the number was 0, in which case there are no digit changes.Therefore, our answer is 11

2 − 12 + 1

4096 = 204814096 .

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16. [9] Jessica has three marbles colored red, green, and blue. She randomly selects a non-empty subset ofthem (such that each subset is equally likely) and puts them in a bag. You then draw three marblesfrom the bag with replacement. The colors you see are red, blue, red. What is the probability that theonly marbles in the bag are red and blue?

Answer: 2735 There are two possible sets of marbles in the bag, {red,blue} and {red,blue,green}.

Initially, both these sets are equally likely to be in the bag. However, the probability of red, blue, redbeing drawn from a set S of marbles is proportional to |S|−3, as long as red and blue are both in S. ByBayes’s Rule, we must weight the probability of these two sets by |S|−3. The answer is (1/2)3

(1/2)3+(1/3)3 .

17. [9] An ant starts at the origin, facing in the positive x-direction. Each second, it moves 1 unit forward,then turns counterclockwise by sin−1( 3

5 ) degrees. What is the least upper bound on the distancebetween the ant and the origin? (The least upper bound is the smallest real number r that is at leastas big as every distance that the ant ever is from the origin.)

Answer:√

10 We claim that the points the ant visits lie on a circle of radius√

102 . We show this

by saying that the ant stays a constant distance√

102 from the point ( 1

2 ,32 ).

Suppose the ant moves on a plane P . Consider a transformation of the plane P ′ such that after the firstmove, the ant is at the origin of P ′ and facing in the direction of the x′ axis (on P ′). The transformationto get from P to P ′ can be gotten by rotating P about the origin counterclockwise through an anglesin−1( 3

5 ) and then translating it 1 unit to the right. Observe that the point ( 12 ,

32 ) is fixed under this

transformation, which can be shown through the expression ( 12 + 3

2 i)(45 + 3

5 i) + 1 = 12 + 3

2 i. It followsthat at every point the ant stops, it will always be the same distance from ( 1

2 ,32 ). Since it starts at

(0, 0), this fixed distance is√

102 .

Guts Round

(12, 3

2)

x

x′

Since sin−1( 35 ) is not a rational multiple of π, the points the ant stops at form a dense subset of the

circle in question. As a result, the least upper bound on the distance between the ant and the originis the diameter of the circle, which is

√10.

18. [9] Find two lines of symmetry of the graph of the function y = x + 1x . Express your answer as two

equations of the form y = ax+ b.

Answer: y = (1 +√

2)x and y = (1−√2)x The graph of the function y = x+ 1x is a hyperbola. We

can see this more clearly by writing it out in the standard form x2−xy+1 = 0 or(y2

)2−(x− 12y)2 = 1.

The hyperbola has asymptotes given by x = 0 and y = x, so the lines of symmetry will be the (interiorand exterior) angle bisectors of these two lines. This means that they will be y = tan(67.5◦)x and

y = − cot(67.5◦)x, which, using the tangent half-angle formula tan(x2

)=√

1+cos(x)1−cos(x) , gives the two

lines y = (1 +√

2)x and y = (1−√2)x.

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19. [10] A 5-dimensional ant starts at one vertex of a 5-dimensional hypercube of side length 1. A moveis when the ant travels from one vertex to another vertex at a distance of

√2 away. How many ways

can the ant make 5 moves and end up on the same vertex it started at?

Answer: 6240 We let the cube lie in R5 with each corner with coordinates 1 or 0. Assume the antstarts at (0, 0, 0, 0, 0). Every move the ant adds or subtracts 1 to two of the places. Note that thismeans the ant can only land on a vertex with the sum of its coordinates an even number. Every movethe ant has

(52

)= 10 choices.

From any vertex there are 10 two-move sequences that put the ant at the same vertex it started at.

There are 6 two-move sequences to move from one vertex to a different, chosen vertex. If your chosenvertex differs from your current vertex by 2 of the 5 coordinates, your first move corrects for one ofthese two. There are 2 ways to choose which coordinate to correct for on the first move, and thereare 3 ways to choose the second coordinate you change, yielding 6 sequences. If your chosen vertexdiffers from your current vertex by 4 of the 5 coordinates, each move corrects for two of these four.This yields

(42

)= 6 sequences.

Finally, there are 60 three-move sequences that put the ant at the same vertex it started at. There are10 ways to choose the first move, and there are 6 ways to make two moves to return to your originalposition.

The motion of the ant can be split into two cases.

Case 1: After the 3rd move the ant is on the vertex it started at. There are (60)(10) = 600 differentpossible paths.

Case 2: After the third move the ant is on a vertex different from the one it started on. There are(103 − 60)(6) = (940)(6) = 5640 different possible paths.

So there are 6240 total possible paths.

Guts Round

20. [10] Find the volume of the set of points (x, y, z) satisfying

x, y, z ≥ 0x+ y ≤ 1y + z ≤ 1z + x ≤ 1

Answer: 14 Without loss of generality, assume that x ≥ y - half the volume of the solid is on this

side of the plane x = y. For each value of c from 0 to 12 , the region of the intersection of this half of

the solid with the plane y = c is a trapezoid. The trapezoid has height 1− 2c and average base 12 , so

it has an area of 12 − c.

The total volume of this region is 12 times the average area of the trapezoids, which is 1

2 · 14 = 18 . Double

that to get the total volume, which is 14 .

21. [10] Let4ABC be a scalene triangle. Let ha be the locus of points P such that |PB−PC| = |AB−AC|.Let hb be the locus of points P such that |PC − PA| = |BC − BA|. Let hc be the locus of points Psuch that |PA− PB| = |CA− CB|. In how many points do all of ha, hb, and hc concur?

Answer: 2 The idea is similar to the proof that the angle bisectors concur or that the perpendicularbisectors concur. Assume WLOG that BC > AB > CA. Note that ha and hb are both hyperbolas.Therefore, ha and hb intersect in four points (each branch of ha intersects exactly once with each branchof hb). Note that the branches of ha correspond to the cases when PB > PC and when PB < PC.Similarly, the branches of hb correspond to the cases when PC > PA and PC < PA.

If either PA < PB < PC or PC < PB < PA (which each happens for exactly one point of intersectionof ha and hb), then |PC − PA| = |PC − PB|+ |PB − PA| = |AB −AC|+ |BC −BA| = |BC −AC|,and so P also lies on hc. So, exactly two of the four points of intersection of ha and hb lie on hc,meaning that ha, hb, and hc concur in four points.

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13th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

22. [12] You are the general of an army. You and the opposing general both have an equal number of troopsto distribute among three battlefields. Whoever has more troops on a battlefield always wins (you winties). An order is an ordered triple of non-negative real numbers (x, y, z) such that x+ y+ z = 1, andcorresponds to sending a fraction x of the troops to the first field, y to the second, and z to the third.Suppose that you give the order

(14 ,

14 ,

12

)and that the other general issues an order chosen uniformly

at random from all possible orders. What is the probability that you win two out of the three battles?

Answer: 58 Let x be the portion of soldiers the opposing general sends to the first battlefield, and

y the portion he sends to the second. Then 1− x− y is the portion he sends to the third. Then x ≥ 0,y ≥ 0, and x + y ≤ 1. Furthermore, you win if one of the three conditions is satisfied: x ≤ 1

4 andy ≤ 1

4 , x ≤ 14 and 1− x− y ≤ 1

2 , or y ≤ 14 and 1− x− y ≤ 1

2 . This is illustrated in the picture below.

Guts Round

This triangle is a linear projection of the region of feasible orders, so it preserves area and probabilityratios. The probability that you win, then is given by the portion of the triangle that satisfies one ofthe three above constraints — in other words, the area of the shaded region divided by the area of theentire triangle. We can easily calculate this to be

51612

= 58 .

23. [12] In the country of Francisca, there are 2010 cities, some of which are connected by roads. Betweenany two cities, there is a unique path which runs along the roads and which does not pass through anycity twice. What is the maximum possible number of cities in Francisca which have at least 3 roadsrunning out of them?

Answer: 1004 The restrictions on how roads connect cities directly imply that the graph of thecities of Francisca with the roads as edges is a tree. Therefore the sum of the degrees of all the verticesis 2009 · 2 = 4018. Suppose that b vertices have degree ≥ 3. The other 2010 − b vertices must have adegree of at least 1, so 3b + (2010 − b) ≤ 4018 or 2b ≤ 2008. So b is at most 1004. We can achieveb = 1004 with the following graph:

· · ·

24. [12] Define a sequence of polynomials as follows: let a1 = 3x2−x, let a2 = 3x2−7x+ 3, and for n ≥ 1,let an+2 = 5

2an+1 − an. As n tends to infinity, what is the limit of the sum of the roots of an?

Answer: 133 By using standard methods for solving linear recurrences2, we see that this recurrence

has a characteristic polynomial of x2 − 52x + 1 = (x − 1

2 )(x − 2), hence an(x) = c(x) · 2n + d(x) · 2−nfor some polynomials c and d. Plugging in n = 1 and n = 2 gives

2c(x) +12d(x) = 3x2 − x

and

4c(x) +14d(x) = 3x2 − 7x+ 3.

Subtracting the first equation from two times the second equation gives 6c(x) = 3x2 − 13x + 6, soc(x) = 3x2−13x+6

6 . As n grows large, the c(x)2n term dominates compared to the d(x)2−n term, sothe roots of an(x) converge to the roots of c(x). Thus the roots of an(x) converge to the roots of3x2 − 13x+ 6, which by Vieta’s formula3 have a sum of 13

3 .

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13th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

25. [15] How many functions f : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} have the property that f({1, 2, 3}) andf(f({1, 2, 3})) are disjoint?

Answer: 94 Let f({1, 2, 3}) be A. Then A∩f(A) = ∅, so A must be a subset of {4, 5}. If B = {4, 5},there are 23 − 2 ways to assign each element in {1, 2, 3} to a value in {4, 5}, and 9 ways to assign eachelement of {4, 5} to a value in {1, 2, 3}, for a total of 54 choices of f . If A = {4}, there is 1 possiblevalue for each element of {1, 2, 3}, 4 ways to assign {4} with a value from {1, 2, 3, 5}, and 5 ways toassign a value to {5}. Similarly, if A = {5}, there are 4 · 5 = 20 choices for f . In total, there are54 + 20 · 2 = 94 possible functions.

2See the sections “Linear homogeneous recurrence relations with constant coefficients” and “Theorem” at http://en.

wikipedia.org/wiki/Recurrence_relation.3See http://mathworld.wolfram.com/VietasFormulas.html.

Guts Round

26. [15] Express the following in closed form, as a function of x:

sin2(x)+sin2(2x) cos2(x)+sin2(4x) cos2(2x) cos2(x)+· · ·+sin2(22010x) cos2(22009x) · · · cos2(2x) cos2(x).

Answer: 1− sin2(22011x)42011 sin2(x)

Note that

sin2(x) + sin2(2x) cos2(x) + · · ·+ sin2(22010x) cos2(22009x) · · · cos2(2x) cos2(x)= (1− cos2(x)) + (1− cos2(2x)) cos2(x) + · · ·+ (1− cos2(22010x)) cos2(22009x) · · · cos2(x),

which telescopes to 1−cos2(x) cos2(2x) cos2(4x) · · · cos2(22010x). To evaluate cos2(x) cos2(2x) · · · cos2(22010x),multiply and divide by sin2(x). We then get

1− sin2(x) cos2(x) cos2(2x) · · · cos2(22010x)sin2(x)

.

Using the double-angle formula for sin, we get that sin2(y) cos2(y) = sin2(2y)4 . Applying this 2011 times

makes the above expression

1− sin2(22011x)42011 sin2(x)

,

which is in closed form.

27. [15] Suppose that there are real numbers a, b, c ≥ 1 and that there are positive reals x, y, z such that

ax + by + cz = 4xax + yby + zcz = 6

x2ax + y2by + z2cz = 9.

What is the maximum possible value of c?

Answer: 3√

4 The Cauchy-Schwarz inequality states that given 2 sequences of n real numbersx1, x2, . . . , xn and y1, y2, . . . , yn, then (x2

1+x22+. . .+x2

n)(y21 +y2

2 +. . .+y2n) ≥ (x1y1+x2y2+. . .+xnyn)2

with equality holding if and only if x1y1

= x2y2

= . . . = xn

yn. Applying this to {ax/2, by/2, cz/2} and

{xax/2, yby/2, zcz/2} yields (ax+by+cz)(x2ax+y2by+z2cz) ≥ (xax+yby+zbz)2 with equality holdingif and only if x = y = z.

However, equality does hold (both sides evaluate to 36), so x = y = z. The second equation thenbecomes x(ax + bx + cx) = 6, which implies x = 3

2 . Then we have a3/2 + b3/2 + c3/2 = 4. To maximizec, we minimize a and b by setting a = b = 1. Then c3/2 = 2 or c = 3

√4.

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13th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

28. [18] Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament willhave 4 rounds, and in each round the 9 players are split into 3 groups of 3. During the tournament,each player plays every other player exactly once. How many different ways can Danielle divide the 9people into three groups in each round to satisfy these requirements?

Answer: 20160 We first split the 9 people up arbitrarily into groups of 3. There are (93)(6

3)(33)

3! = 280ways of doing this. Without loss of generality, label the people 1 through 9 so that the first roundgroups are {1, 2, 3}, {4, 5, 6}, and {7, 8, 9}. We will use this numbering to count the number of waysDBR can divide the 9 players in rounds 2, 3, and 4.

Guts Round

In round 2, because players 1, 2, and 3 are together in the first round, they must be in separate groups,and likewise for {4, 5, 6} and {7, 8, 9}. Disregarding ordering of the three groups in a single round,we will first place 1, 2, and 3 into their groups, then count the number of ways to place {4, 5, 6} and{7, 8, 9} in the groups with them. We do this by placing one member from each of {4, 5, 6} and {7, 8, 9}into each group. There are (3!)2 ways to do this. Now, because of symmetry, we can use the round 2grouping {1, 4, 7}, {2, 5, 8}, {3, 6, 9} to list out the remaining possibilities for round 3 and 4 groupings.

Casework shows that there are 2 ways to group the players in the remaining two rounds. We multiply280 · (3!)2 · 2 to get 20160.

29. [18] Compute the remainder when30303∑k=1

kk

is divided by 101.

Answer: 29 The main idea is the following lemma:

Lemma. For any non-negative integer n and prime p,∑n+p2−pk=n+1 kk ≡ 1 (mod p).

Proof. Note that ab depends only on the value of a (mod p) and the value of b (mod p − 1). Since pand p− 1 are relatively prime, the Chinese Remainder Theorem4 implies that any p2 − p consecutiveintegers will take on each possible pair of a residue mod p and a residue mod p− 1. In other words, ifwe let (a, b) = (k (mod p), k (mod p− 1)), then as k ranges through p2 − p consecutive integers, (a, b)will range through all p2− p ordered pairs of residues mod p and residues mod p− 1. This implies that

n+p2−p∑k=n+1

kk ≡p−1∑b=1

p∑a=1

ab.

It is well-known that∑pa=1 a

b ={ −1 p− 1 | b

0 p− 1 - b . We will sketch a proof here. When p − 1 | b, the

result follows from Fermat’s Little Theorem5. When p − 1 - b, it suffices to consider the case whenb | p− 1, since the bth powers mod p are the same as the gcd(b, p− 1)th powers mod p, and there arean equal number of every non-zero bth power. But in this case, the bth powers are just the solutionsto x

p−1b − 1, which add up to zero by Vieta’s formulas.

Now, using the formula for∑ab, we get that

p−1∑b=1

p∑a=1

ab ≡ −1 (mod p),

which completes the lemma.

We now apply the lemma with p = 101 and n = 3, 10103, and 20103 to get that∑30303k=1 kk ≡(∑3

k=1 kk)− 3. But

∑3k=1 k

k = 11 + 22 + 33 = 1 + 4 + 27 = 32, so the answer is 32− 3 = 29.

30. [18] A monomial term xi1xi2 . . . xik in the variables x1, x2, . . . x8 is square-free if i1, i2, . . . ik are distinct.(A constant term such as 1 is considered square-free.) What is the sum of the coefficients of the square-free terms in the following product? ∏

1≤i<j≤8

(1 + xixj)

4See http://en.wikipedia.org/wiki/Chinese_remainder_theorem.5See http://en.wikipedia.org/wiki/Fermat’s_little_theorem.

Guts Round

Answer: 764 Let an be the sum of the coefficients of the square-terms in the product∏

1≤i<j≤n(1+xixj). Square-free terms in this product come in two types: either they include xn, or they do not.The sum of the coefficients of the terms that include xn is (n − 1)an−2, since we can choose anyof the n − 1 other variables to be paired with xn, and then choose any square-free term from theproduct taken over the other n − 2 variables. The sum of the coefficients of the terms that do notinclude xn are an−1, because they all come from the product over the other n− 1 variables. Therefore,an = an−1 + (n− 1)an−2.

We use this recursion to find a8. As base cases, a0 and a1 are both 1. Then a2 = 2, a3 = 4, a4 = 10,a5 = 26, a6 = 76, a7 = 232, and finally, a8 = 764.

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13th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

31. [21] In the Democratic Republic of Irun, 5 people are voting in an election among 5 candidates. Ifeach person votes for a single candidate at random, what is the expected number of candidates thatwill be voted for?

Answer: 2101625 The probability that a chosen candidate will receive no votes at all is ( 4

5 )5, or theprobability that every person will vote for someone other than that one candidate. Then the probabilitythat a chosen candidate will receive at least one vote is 1− ( 4

5 )5 = 21013125 . By linearity of expectations,

the final answer is this probability times the number of candidates, or 5 · 21013125 = 2101

625 .

32. [21] There are 101 people participating in a Secret Santa gift exchange. As usual each person israndomly assigned another person for whom (s)he has to get a gift, such that each person gives andreceives exactly one gift and no one gives a gift to themself. What is the probability that the firstperson neither gives gifts to or receives gifts from the second or third person? Express your answer asa decimal rounded to five decimal places.

Answer: 0.96039 Let Dk denote the number of derangements of {1, 2, . . . , k}. (A derangement is apermutation in which no element appears in its original position.)

Call the first three people A, B, and C. Let X → Y denote that X gives a gift to Y and let X 6→ Ydenote that X gives a gift to anyone other than Y . We are fine unless we have A → B, B → A,A→ C, or C → A. We will compute the number of ways for various things to occur, then combine itinto what we want.

There are Dn−1 ways to have A → B 6→ A. This is because if A → B, we can treat A and B asa single vertex, and since B 6→ A, we have a derangement. Similarly, there are Dn−1 ways to haveB → A 6→ B. Thirdly, there are Dn−2 ways to have A → B → A, since Dn−2 says how many waysthe other n − 2 people can exchange their gifts. So there are 2Dn−1 + Dn−2 ways to have a conflictbetween A and B.

Similarly, there are 2Dn−1 +Dn−2 ways to have a conflict between A and C.

Using similar arguments, we can show that there are Dn−2 ways for B → A → C 6→ B to occur andDn−3 ways for B → A→ C → B to occur. We get the same results when B and C are reversed. Thisgives 2Dn−2 + 2Dn−3 ways for a conflict to occur within all three people.

By the Principle of Inclusion-Exclusion, the total number of ways to have a conflict is

(# conflicts between A and B)+(# conflicts between A and C)−(# conflicts between A, B, and C),

which evaluates to 4Dn−1 − 2Dn−3.

Approximating Dn as n!e (the actual formula is this quantity rounded to the nearest integer, so this is

a great approximation), we find that the probability of no conflicts is

1− 4Dn−1 − 2Dn−3

Dn≈ 1−4

((n− 1)!/en!/e

)−2(

(n− 3)!/en!/e

)=n(n− 1)(n− 2)− 4(n− 1)(n− 2)− 2

n(n− 1)(n− 2).

Guts Round

Substitute m = n− 1 (this makes m = 100, so the expression is easier to evaluate) to get a probabilityof

m3 −m− 4m2 + 4m− 2m3 −m =

m3 − 4m2 + 3m− 2m3 −m =

1,000,000− 40,000 + 300− 2100 · 9999

=960298

100 · 9999

= 0.960208 · (1.000100010001 . . .) = 0.960208 + 0.0000960208 + . . . = 0.9603940 . . .

To five decimal places, the desired probability is 0.96039.

33. [21] Let a1 = 3, and for n > 1, let an be the largest real number such that

4(a2n−1 + a2

n

)= 10an−1an − 9.

What is the largest positive integer less than a8?

Answer: 335 Let tn be the larger real such that an = tn + 1tn

. Then t1 = 3+√

52 . We claim that

tn = 2tn−1. Writing the recurrence as a quadratic polynomial in an, we have:

4a2n − 10an−1an + 4a2

n−1 + 9 = 0.

Using the quadratic formula, we see that an = 54an−1 + 3

4

√a2n−1 − 4. (We ignore the negative square

root, since an is the largest real number satisfying the polynomial.) Substituting tn−1 + 1tn−1

for an−1,

we see that√a2n−1 − 4 =

√t2n−1 − 2 + 1

t2n−1, so we have:

an =54

(tn−1 +

1tn−1

)+

34

√(tn−1 − 1

tn−1

)2

= 2tn−1 +1

2tn−1

so tn = 2tn−1, as claimed. Then a8 = 128(3+√

5)2 + 2

128(3+√

5). The second term is vanishingly small,

so ba8c = b64(3 +√

5)c. We approximate√

5 to two decimal places as 2.24, making this expressionb335.36c = 335. Since our value of

√5 is correct to within 0.005, the decimal is correct to within 0.32,

which means the final answer is exact.

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13th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

34. [25] 3000 people each go into one of three rooms randomly. What is the most likely value for themaximum number of people in any of the rooms? Your score for this problem will be 0 if you writedown a number less than or equal to 1000. Otherwise, it will be 25− 27 |A−C|

min(A,C)−1000 .

Answer: 1019 To get a rough approximation, we can use the fact that a sum of identical randomvariables converges to a Gaussian distribution,6 in this case with a mean of 1000 and a variance of3000 · 29 = 667. Since

√667 ≈ 26, 1026 is a good guess, as Gaussians tend to differ from their mean by

approximately their variance.

The actual answer was computed with the following python program:

facts = [0]*3001facts[0]=1for a in range(1,3001):facts[a]=a*facts[a-1]

def binom(n,k):return facts[n]/(facts[k]*facts[n-k])

6See http://en.wikipedia.org/wiki/Central_limit_theorem.

Guts Round

maxes = [0]*3001

M = 1075

for a in range(0,3001):for b in range(0,3001-a):c = 3000-a-bm = max(a,max(b,c))if m < M:

maxes[m] += facts[3000]/(facts[a]*facts[b]*facts[c])print [a,b]

best = 1000for a in range(1000,1050):print maxes[a],aif maxes[best] <= maxes[a]:best = a

print maxes[best]print best

We can use arguments involving the Chernoff bound7 to show that the answer is necessarily less than1075. Alternately, if we wanted to be really careful, we could just set M = 3001, but then we’d haveto wait a while for the script to finish.

35. [25] Call an positive integer almost-square if it can be written as a · b, where a and b are integers anda ≤ b ≤ 4

3a. How many almost-square positive integers are less than or equal to 1000000? Your scorewill be equal to 25− 65 |A−C|

min(A,C) .

Answer: 130348 To get a good estimate for the number of almost-square integers, note that anynumber of the form a · b, with b ≤ 4

3a, will be by definition almost-square. Let’s assume that it’srelatively unlikely that a number is almost-square in more than one way. Then the number of almost-square numbers less than n will be approximately

√n∑

a=1

43a∑b=a

1 =13

√n∑

a=1

a =16√n(√n+ 1

),

which is about n6 . So, n

6 will be a fairly good estimate for the number of almost-square numbers lessthan n, making 160000 a reasonable guess.

We can do better, though. For example, we summed a3 all the way up to

√n, but we are really

overcounting here because when a is close to√n, a · b will be less than n only when b ≤ n

a , as opposedto b ≤ 4a

3 . So we should really be taking the sum

7See http://en.wikipedia.org/wiki/Chernoff_bound.

Guts Round

√3n4∑

a=1

4a3∑

b=a

1 +

√n∑

a=√

3n4

na∑b=a

1

=

√3n4∑

a=1

a

3+

√n∑

a=√

3n4

(na− a)

≈ 16

3n4

+ n

(log(√n)− log(

√3n4

)

)−(n

2− 3n

8

)=

n

8+ n

log(4)− log(3)2

− n

8

= nlog(4)− log(3)

2.

In the process of taking the sum, we saw that we had something between n8 and n

6 , so we could alsoguess something between 166000 and 125000, which would give us about 145000, an even better answer.If we actually calculate log(4)−log(3)

2 , we see that it’s about 0.14384, so 143840 would be the best guessif we were to use this strategy. In reality, we would want to round down a bit in both cases, sincewe are overcounting (because numbers could be square-free in multiple ways), so we should probablyanswer something like 140000.

A final refinement to our calculation (and perhaps easier than the previous one), is to assume that theproducts a ·b that we consider are randomly distributed between 1 and n, and to compute the expectednumber of distinct numbers we end up with. This is the same type of problem as number 31 on thiscontest, and we compute that if we randomly distribute k numbers between 1 and n then we expectto end up with n

(1− (1− 1

n

)k) distinct numbers. When k = n log(4)−log(3)2 , we get that this equals

n

1−((

1− 1n

)n) log(4)−log(3)2

= n(

1−√elog(3)−log(4)

)

= n

(1−

√34

)

= n

(1−√

32

)≈ 0.134n,

Giving us an answer of 134000, which is very close to the correct answer.

The actual answer was found by computer, using the following C++ program:

#include <stdio.h>

using namespace std;

bool isAlmostSquare(int n){for(int k=1;k*k<=n;k++)if(n%k==0 && 3*(n/k) <= 4*k) return true;

return false;}

Guts Round

int main(){int c = 0;for(int n=1;n<=1000000;n++)if(isAlmostSquare(n)) c++;

printf("%d\n",c);return 0;

}

36. [25] Consider an infinite grid of unit squares. An n-omino is a subset of n squares that is connected.Below are depicted examples of 8-ominoes. Two n-ominoes are considered equivalent if one can beobtained from the other by translations and rotations. What is the number of distinct 15-ominoes?Your score will be equal to 25− 13| ln(A)− ln(C)|.

Answer: 3426576 We claim that there are approximately 3n−1

4 n-ominoes. First, we define an orderon the squares in an n-omino, as follows: we order the squares from left to right, and within a column,we order the squares from top to bottom.

We construct an n-omino by starting with a single square and attaching squares, one by one, as closeto order we just defined as possible. After the first square is placed, the next square can be attached toits bottom or right side. If the second square is right of the first, then the third square can be attachedto the top, right, or bottom edges of the second. The third square cannot be attached to the firstsquare, because the first column must be completed before the second is begun. If the second squareis below the first, the third square can be attached to the bottom or right sides of the second square,or the right side of the first way. In either case, there are 3 choices for the third square.

Suppose that m squares have been added, and that k squares are in the right-most column right now.If k = 1, then the next square can be attached to the top, right, or bottom side of this square. If k > 1,then the next square can be added to the bottom or right side of the last square in this column. Thenext square can also be added to the right of the top square in this column, so there are still 3 choicesfor the next square. Therefore, there are 3n−1 ways to build up the entire n-omino.

Almost all n-omines have no rotational symmetry, so they can be built in 4 different ways by thismethod. This correction gives us our estimate of 3n−1

4 for the number of n-ominoes. This reasoningis not exactly correct, because some n-ominoes are highly non-convex, and do not admit an in-ordertraversal like the one described above. If some columns in the n-ominoes have gaps, they are notenumerated by this construction, so 3n−1

4 is an underestimate. We estimate that there are 119574215-ominoes. The actual value of 3426576, which can be found in the Sloane Encyclopedia of IntegerSequences, differes from the estimate by less than a factor of 3.

Guts Round

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Team Round A

1. You are trying to sink a submarine. Every second, you launch a missile at a point of your choosing onthe x-axis. If the submarine is at that point at that time, you sink it. A firing sequence is a sequenceof real numbers that specify where you will fire at each second. For example, the firing sequence2, 3, 5, 6, . . . means that you will fire at 2 after one second, 3 after two seconds, 5 after three seconds,6 after four seconds, and so on.

(a) [5] Suppose that the submarine starts at the origin and travels along the positive x-axis with an(unknown) positive integer velocity. Show that there is a firing sequence that is guaranteed to hitthe submarine eventually.

(b) [10] Suppose now that the submarine starts at an unknown integer point on the non-negativex-axis and again travels with an unknown positive integer velocity. Show that there is still afiring sequence that is guaranteed to hit the submarine eventually.

2. [15] Consider the following two-player game. Player 1 starts with a number, N . He then subtracts aproper divisor of N from N and gives the result to player 2 (a proper divisor of N is a positive divisorof N that is not equal to 1 or N). Player 2 does the same thing with the number she gets from player 1,and gives the result back to player 1. The two players continue until a player is given a prime numberor 1, at which point that player loses. For which values of N does player 1 have a winning strategy?

3. [15] Call a positive integer in base 10 k-good if we can split it into two integers y and z, such that yis all digits on the left and z is all digits on the right, and such that y = k · z. For example, 2010 is2-good because we can split it into 20 and 10 and 20 = 2 · 10. 20010 is also 2-good, because we cansplit it into 20 and 010. In addition, it is 20-good, because we can split it into 200 and 10.

Show that there exists a 48-good perfect square.

4. [20] Letex + ey = A

xex + yey = B

x2ex + y2ey = C

x3ex + y3ey = D

x4ex + y4ey = E.

Prove that if A, B, C, and D are all rational, then so is E.

5. [20] Show that, for every positive integer n, there exists a monic polynomial of degree n with integercoefficients such that the coefficients are decreasing and the roots of the polynomial are all integers.

6. [20] Let S be a convex set in the plane with a finite area a. Prove that either a = 0 or S is bounded.Note: a set is bounded if it is contained in a circle of finite radius. Note: a set is convex if, whenevertwo points A and B are in the set, the line segment between them is also in the set.

7. [25] Point P lies inside a convex pentagon AFQDC such that FPDQ is a parallelogram. Given that∠FAQ = ∠PAC = 10◦, and ∠PFA = ∠PDC = 15◦. What is ∠AQC?

8. [30] A knight moves on a two-dimensional grid. From any square, it can move 2 units in one axis-parallel direction, then move 1 unit in an orthogonal direction, the way a regular knight moves in agame of chess. The knight starts at the origin. As it moves, it keeps track of a number t, which isinitially 0. When the knight lands at the point (a, b), the number is changed from x to ax + b.

Show that, for any integers a and b, it is possible for the knight to land at the points (1, a) and (−1, a)with t equal to b.

Page 1 of 2

9. [30] Let p(x) = anxn +an−1xn−1 + . . .+a0 be a polynomial with complex coefficients such that ai 6= 0

for all i. Prove that |r| ≤ 2 maxn−1i=0 |

ai−1ai| for all roots r of all such polynomials p. Here we let |z|

denote the absolute value of the complex number z.

10. Call an 2n-digit base-10 number special if we can split its digits into two sets of size n such that thesum of the numbers in the two sets is the same. Let pn be the probability that a randomly-chosen2n-digit number is special. (We allow leading zeros in 2n-digit numbers).

(a) [20] The sequence pn converges to a constant c. Find c.

(b) [45] Let qn = pn − c. There exists a unique positive constant r such that qn

rn converges to aconstant d. Find r and d.

Page 2 of 2

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Team Round A

1. You are trying to sink a submarine. Every second, you launch a missile at a point of your choosing onthe x-axis. If the submarine is at that point at that time, you sink it. A firing sequence is a sequenceof real numbers that specify where you will fire at each second. For example, the firing sequence2, 3, 5, 6, . . . means that you will fire at 2 after one second, 3 after two seconds, 5 after three seconds,6 after four seconds, and so on.

(a) [5] Suppose that the submarine starts at the origin and travels along the positive x-axis with an(unknown) positive integer velocity. Show that there is a firing sequence that is guaranteed to hitthe submarine eventually.Solution: The firing sequence 1, 4, 9, . . . , n2, . . . works. If the velocity of the submarine is v,then after v seconds it will be at x = v2, the same location where the mine explodes at time v.

(b) [10] Suppose now that the submarine starts at an unknown integer point on the non-negativex-axis and again travels with an unknown positive integer velocity. Show that there is still afiring sequence that is guaranteed to hit the submarine eventually.Solution: Represent the submarine’s motion by an ordered pair (a, b), where a is the startingpoint of the submarine and b is its velocity. We want to find a way to map each positive integerto a possible ordered pair so that every ordered pair is covered. This way, if we fire at bnn + anat time n, where (an, bn) is the point that n maps to, then we will eventually hit the submarine.(Keep in mind that bnn + an would be the location of the submarine at time n.) There are manysuch ways to map the positive integers to possible points; here is one way:

1→ (1, 1), 2→ (2, 1), 3→ (1, 2), 4→ (3, 1), 5→ (2, 2), 6→ (1, 3), 7→ (4, 1), 8→ (3, 2),

9→ (2, 3), 10→ (1, 4), 11→ (5, 1), 12→ (4, 2), 13→ (3, 3), 14→ (2, 4), 15→ (1, 5), . . .

(The path of points trace out diagonal lines that sweep every lattice point in the coordinate plane.)Since we cover every point, we will eventually hit the submarine.Remark: The mapping shown above is known as a bijection between the positive integers andordered pairs of integers (a, b) where b > 0.

2. [15] Consider the following two-player game. Player 1 starts with a number, N . He then subtracts aproper divisor of N from N and gives the result to player 2 (a proper divisor of N is a positive divisorof N that is not equal to 1 or N). Player 2 does the same thing with the number she gets from player 1,and gives the result back to player 1. The two players continue until a player is given a prime numberor 1, at which point that player loses. For which values of N does player 1 have a winning strategy?

Answer: All even numbers except for odd powers of 2. First we show that if you are stuck with anodd number, then you are guaranteed to lose. Suppose you have an odd number ab, where a and b areodd numbers, and you choose to subtract a. You pass your opponent the number a(b−1). This cannotbe a power of 2 (otherwise a is a power of 2 and hence a = 1, which is not allowed), so your opponentcan find an odd proper divisor of a(b − 1), and you will have a smaller odd number. Eventually youwill get to an odd prime and lose.

Now consider even numbers that aren’t powers of 2. As with before, you can find an odd proper divisorof N and pass your opponent an odd number, so you are guaranteed to win.

Finally consider powers of 2. If you have the number N = 2k, it would be unwise to choose a properdivisor other than 2k−1; otherwise you would give your opponent an even number that isn’t a powerof 2. Therefore if k is odd, you will end up with 2 and lose. If k is even, though, your opponent willend up with 2 and you will win.

Therefore player 1 has a winning strategy for all even numbers except for odd powers of 2.

Team Round A

3. [15] Call a positive integer in base 10 k-good if we can split it into two integers y and z, such that yis all digits on the left and z is all digits on the right, and such that y = k · z. For example, 2010 is2-good because we can split it into 20 and 10 and 20 = 2 · 10. 20010 is also 2-good, because we cansplit it into 20 and 010. In addition, it is 20-good, because we can split it into 200 and 10.

Show that there exists a 48-good perfect square.

Solution: We wish to find integers a, z such that 48z · 10a + z = z(48 · 10a + 1) a perfect square,where z < 10a. This would prove that there exists a 48-good perfect square because we are pullingoff the last a digits of the number and get two integers 48z and z. To make z small by keeping theproduct a perfect square, we’d like 48 · 10a + 1 to be divisible by some reasonably large square. Takea = 42 = ϕ(49). By Euler’s theorem, 1042 ≡ 1 (mod 49), so 48 · 10a + 1 is a multiple of 49. Then we

can take z = 48·10a+149 . (Clearly z < 10a, so we’re fine.) Then we have z(48 · 10a + 1) =

(48·1042+1

7

)2

.

4. [20] Letex + ey = A

xex + yey = B

x2ex + y2ey = C

x3ex + y3ey = D

x4ex + y4ey = E.

Prove that if A, B, C, and D are all rational, then so is E.

Solution: We can express x + y in two ways:

x + y =AD −BC

AC −B2

x + y =AE − C2

AD −BC

(We have to be careful if AC −B2 or AD −BC is zero. We’ll deal with that case later.) It is easy tocheck that these equations hold by substituting the expressions for A, B, C, D, and E. Setting thesetwo expressions for x + y equal to each other, we get

AD −BC

AC −B2=

AE − C2

AD −BC,

which we can easily solve for E as a rational function of A, B, C, and D. Therefore if A, B, C, andD are all rational, then E will be rational as well.

Now, we have to check what happens if AC −B2 = 0 or AD−BC = 0. If AC −B2 = 0, then writingdown the expressions for A, B, and C gives us that (x − y)2ex+y = 0, meaning that x = y. If x = y,and x 6= 0, A and D are also non-zero, and B

A = ED = x. Since B

A is rational and D is rational, thisimplies that E is rational. If x = y = 0, then E = 0 and so is certainly rational.

We finally must check what happens if AD − BC = 0. Since AD − BC = (x + y)(AC − B2), eitherAC − B2 = 0 (a case we have already dealt with), or x + y = 0. But if x + y = 0 then AE − C2 = 0,which implies that E = C2

A (we know that A 6= 0 because ex and ey are both positive). Since A and Care rational, this implies that E is also rational.

So, we have shown E to be rational in all cases, as desired.

5. [20] Show that, for every positive integer n, there exists a monic polynomial of degree n with integercoefficients such that the coefficients are decreasing and the roots of the polynomial are all integers.

Solution: We claim we can find values a and b such that p(x) = (x− a)(x + b)n is a polynomial ofdegree n + 1 that satisfies these constraints. We show that its coefficients are decreasing by finding ageneral formula for the coefficient of xk.

Team Round A

The coefficient of xk is bk(nk

)− abk−1

(nk−1

), which can be seen by expanding out (x + b)n and then

multiplying by (x− a). Then we must prove that

bk+1

(n

k + 1

)− abk

(n

k

)< bk

(n

k

)− abk−1

(n

k − 1

),

or

abk−1

(b

(n

k

)−(

n

k − 1

))> bk

(b

(n

k + 1

)−(

n

k

)).

Choose b > max(

(nk)

( nk−1)

)in order to make sure the right-hand term in each product on each side of

the inequality sign is positive (we’ll be dividing by it, so this makes things much easier), and choose

a > max(b(b( n

k+1)−(nk))

b(nk)−( n

k−1)

)to make sure the inequality always holds. Since there are only finite values

that k can take on given a fixed n (namely, integers between 0 and n inclusive), we can always findvalues of a and b that satisfy these constraints.

6. [20] Let S be a convex set in the plane with a finite area a. Prove that either a = 0 or S is bounded.Note: a set is bounded if it is contained in a circle of finite radius. Note: a set is convex if, whenevertwo points A and B are in the set, the line segment between them is also in the set.

Solution: If all points in S lie on a straight line, then a = 0.

Otherwise we may pick three points A, B, and C that are not collinear. Let ω be the incircle of4ABC,with I its center and r its radius. Since S is convex, S must contain ω.

A

BC

I

ω

X

Z

Y

M

Suppose S also contains a point X at a distance d from I, with d > R. We will show that d ≤√

r2 + a2

r2 ,which implies that the S is bounded since all points are contained within the circle centered at I of

radius√

r2 + a2

r2 .

Let Y and Z be on ω such that XY and XZ are tangents to ω. Because S is convex, it must containkite IY XZ, whose area we can compute in terms of d and r.

Let M be the midpoint of Y Z. Since 4IY X ∼ 4IMY , we know that IMIY = IY

IX , that is, IM =(IY )2

IX = r2

d . Then MY =√

r2 − r4

d2 = r√

1− ( rd )2 = 12Y Z.

The area of IY XZ is 12 (Y Z)(IX) = rd

√1− ( rd )2 = r

√d2 − r2. This must be less than or equal to a,

the area of S. This yields a2 ≥ r2d2 − r4 or d2 ≤ r2 + a2

r2 . It follows that d ≤√

r2 + a2

r2 , as desired.

7. [25] Point P lies inside a convex pentagon AFQDC such that FPDQ is a parallelogram. Given that∠FAQ = ∠PAC = 10◦, and ∠PFA = ∠PDC = 15◦. What is ∠AQC?

Answer: π12 Let C ′ be the point such that there is a spiral similarity between 4AFP and 4AQC ′.

In other words, one triangle can be formed from the other by dilating and rotating about one of the

Team Round A

triangle’s vertices (in this case, A). We will show that C ′ is C, so our answer will be ∠AQC =∠AQC ′ = ∠AFP = 15◦. By the spiral similarity theorem, 4AFQ ∼ 4APC ′ (this is intuitive bylooking at a diagram), so ∠PAC = ∠FAQ = 10◦, so to show that C ′ is C, it is sufficient to show that∠PDC ′ = 15◦.

Q D

F P

A

C ′X

15◦

10◦

10◦

Let X be the fourth point of the parallelogram FPC ′X (see the above diagram). The angle betweenlines FP and FA is 15◦. Since XC ′ ‖ FP , the angle between FA and XC ′ is 15◦ as well. In addition,the angle between QA and QC ′ is ∠AQC ′ = 15◦, so ∠XC ′Q = ∠FAQ. Further, because FPC ′X is aparallelogram, QC′

XC′ = QC′

FP . By similar triangles 4AFP and 4AQC ′, QC′

FP = QAFA . By SAS similarity,

there is a spiral similarity between 4XC ′Q and 4FAQ, so ∠FQX = ∠AQC ′ = 15◦.

Note that the segments FP , XC ′, and QD are all parallel and equal in length. Therefore, 4FQX ∼=4PDC ′ are congruent, and ∠PDC ′ = 15◦ as well. So C ′ is C, and ∠AQC = 15◦.

8. [30] A knight moves on a two-dimensional grid. From any square, it can move 2 units in one axis-parallel direction, then move 1 unit in an orthogonal direction, the way a regular knight moves in agame of chess. The knight starts at the origin. As it moves, it keeps track of a number t, which isinitially 0. When the knight lands at the point (a, b), the number is changed from x to ax + b.

Show that, for any integers a and b, it is possible for the knight to land at the points (1, a) and (−1, a)with t equal to b.

Solution: For convenience, we will refer to (a, b) as [ax + b], the function it represents. This willmake it easier to follow the trajectory of t over a given sequence of moves.

Suppose we start at [x + 1] with t = a. Taking the path [x + 1] → [−x] → [x − 1] → [−x] → [x + 1]will yield t = a + 2. So we can go from t = a to t = a + 2 at [x + 1].

We can also move until we get to [−3], then go [−3]→ [x− 1] to end up with t = −4 at [x− 1]. Butgoing [x− 1]→ [3x]→ [x− 1] means we can go from t = a to t = 3a− 1 at x− 1. Since we can startwith t = −4, this means we can therefore get arbitrarily small even and odd numbers at [x− 1], hencealso [3x], hence also at [x + 1].

This implies we can get any value of t we want at [x + 1], so we can also get any value of t we want at[−x], [x−1], [−x−2], [x−3], etc., as well as [−x+2], [x+3], [−x+4], [x+5], etc. We can do a similarthing starting at [−x + 1] to get from t = a to t = a + 2, and use the [−x− 1]→ [−3x]→ [−x− 1] loopto get arbitrarily small integers of both parities. So we can get any value of t we want at all points ofthe form [±x + k] for any integer k.

9. [30] Let p(x) = anxn+an−1x

n−1 + . . .+a0 be a polynomial with complex coefficients such that ai 6= 0for all i. Prove that |r| ≤ 2 maxn−1

i=0 |ai−1ai| for all roots r of all such polynomials p. Here we let |z|

denote the absolute value of the complex number z.

Solution: If r is a root, then −anrn = an−1r

n−1 + . . . + a0. By the Triangle Inequality, |−anrn| ≤

|an−1rn−1|+ ... + |a0|. Rearranging this inequality yields |anrn| − |an−1r

n−1| − ...− |a0| ≤ 0.

Now suppose |r| = k max |ai−1ai|. Applying this over values of i ranging from m + 1 to n (assuming

m + 1 ≤ n), we get |amrm| ≤ |anrn|

kn−m . This, along with the above equation, yields:

Team Round A

|anrn| ·(

1− 1k− 1

k2− 1

k3− . . .− 1

kn

)= 0

This is only true when an = 0, r = 0, or (1− 1k −

1k2 − . . .) = 0. The first option is impossible by the

constraints in the problem. The second option implies k = 0. The third option implies that k < 2;otherwise (1− 1

k −1k2 − . . .− 1

kn ) would always remain positive. Either way, |r| ≤ 2 maxn−1i=0 |

ai−1ai|.

10. Call an 2n-digit base-10 number special if we can split its digits into two sets of size n such that thesum of the numbers in the two sets is the same. Let pn be the probability that a randomly-chosen2n-digit number is special. (We allow leading zeros in 2n-digit numbers).

(a) [20] The sequence pn converges to a constant c. Find c.

Answer: 12 We first claim that if a 2n-digit number x has at least eight 0’s and at least eight

1’s and the sum of its digits is even, then x is special.Let A be a set of eight 0’s and eight 1’s and let B be the set of all the other digits. We splitb arbitrily into two sets Y and Z of equal size. If |

∑y∈Y y −

∑z∈Z z| > 8, then we swap the

biggest element of the set with the bigger sum with the smallest element of the other set. Thistransposition always decreases the absolute value of the sum: in the worst case, a 9 from thebigger set is swapped for a 0 from the smaller set, which changes the difference by at most 18.Therefore, after a finite number of steps, we will have |

∑y∈Y y −

∑z∈Z z| ≤ 8.

Note that this absolute value is even, since the sum of all the digits is even. Without loss ofgenerality, suppose that

∑y∈Y y −

∑z∈Z z is 2k, where 0 ≤ k ≤ 4. If we add k 0’s and 8− k 1’s

to Y , and we add the other elements of A to Z, then the two sets will balance, so x is special.

(b) [45] Let qn = pn − c. There exists a unique positive constant r such that qn

rn converges to aconstant d. Find r and d.Answer: r = 1

4 , d = −1 To get the next asymptotic term after the constant term of 12 , we need

to consider what happens when the digit sum is even; we want to find the probability that sucha number isn’t balanced. We claim that the configuration that contributes the vast majority ofunbalanced numbers is when all numbers are even and the sum is 2 mod 4, or such a configurationwith all numbers increased by 1. Clearly this gives qn being asymptotic to − 1

2 ·2 ·(

12

)2n = −(

14

)n,so r = 1

4 and d = −1.To prove the claim, first note that the asymptotic probability that there are at most 4 digits thatoccur more than 10 times is asymptotically much smaller than

(12

)n, so we can assume that thereexist 5 digits that each occur at least 10 times. If any of those digits are consecutive, then thedigit sum being even implies that the number is balanced (by an argument similar to part (a)).So, we can assume that none of the numbers are consecutive. We would like to say that thisimplies that the numbers are either 0, 2, 4, 6, 8 or 1, 3, 5, 7, 9. However, we can’t quite say this yet,as we need to rule out possibilities like 0, 2, 4, 7, 9. In this case, though, we can just pair 0 and 7up with 2 and 4; by using the same argument as in part (a), except using 0 and 7 both (to get asum of 7) and 2 and 4 both (to get a sum of 6) to balance out the two sets at the end.In general, if there is ever a gap of size 3, consider the number right after it and the 3 numbersbefore it (so we have k− 4, k− 2, k, k + 3 for some k), and pair them up such that one pair has asum that’s exactly one more than the other (i.e. pair k − 4 with k + 3 and k − 2 with k). Sincewe again have pairs of numbers whose sums differ by 1, we can use the technique from part (a)of balancing out the sets at the end.So, we can assume there is no gap of size 3, which together with the condition that no two numbersare adjacent implies that the 5 digits are either 0, 2, 4, 6, 8 or 1, 3, 5, 7, 9. For the remainder of thesolution, we will deal with the 0, 2, 4, 6, 8 case, since it is symmetric with the other case under thetransformation x 7→ 9− x.If we can distribute the odd digits into two sets S1 and S2 such that (i) the differense in sums ofS1 and S2 is small; and (ii) the difference in sums of S1 and S2, plus the sum of the even digits,is divisible by 4, then the same argument as in part (a) implies that the number is good.

Team Round A

In fact, if there are any odd digits, then we can use them at the beginning to fix the parity mod4 (by adding them all in such that the sums of the two sets remain close, and then potentiallyswitching one with an even digit). Therefore, if there are any odd digits then the number is good.Also, even if there are no odd digits, if the sum of the digits is divisible by 4 then the number isgood.So, we have shown that almost all non-good numbers come from having all numbers being evenwith a digit sum that is 2 mod 4, or the analogous case under the mapping x 7→ 9 − x. Thisformalizes the claim we made in the first paragraph, so r = 1

4 and d = −1, as claimed.

Team Round A

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Team Round B

1. [10] How many ways are there to place pawns on an 8× 8 chessboard, so that there is at most 1 pawnin each horizontal row? Express your answer in the form pe1

1 · pe22 · · · , where the pi are distinct primes

and the ei are positive integers.

2. [10] In the following figure, a regular hexagon of side length 1 is attached to a semicircle of diameter1. What is the longest distance between any two points in the figure?

3. [15] Let p(x) = anxn + an−1xn−1 + . . . + a0, where each ai is either 1 or −1. Let r be a root of p. If

|r| > 158 , what is the minimum possible value of n?

4. [20] Find all 4-digit integers of the form aabb (when written in base 10) that are perfect squares.

5. [25] Compute98∑

n=1

2√

n +√

n + 2+

1√n + 1 +

√n + 2

.

6. [25] Into how many regions can a circle be cut by 10 parabolas?

7. [30] Evaluate2010∑k=1

cos2(k)

8. [30] Consider the following two-player game. Player 1 starts with a number, N . He then subtracts aproper divisor of N from N and gives the result to player 2 (a proper divisor of N is a positive divisorof N that is not equal to 1 or N). Player 2 does the same thing with the number she gets from player 1,and gives the result back to player 1. The two players continue until a player is given a prime number,at which point that player loses. For how many values of N between 2 and 100 inclusive does player 1have a winning strategy?

9. [35] Let S be the set of ordered pairs of integers (x, y) with 1 ≤ x ≤ 5 and 1 ≤ y ≤ 3. How manysubsets R of S have the property that all the points of R lie on the graph of a single cubic? A cubic isa polynomial of the form y = ax3 + bx2 + cx + d, where a, b, c, and d are real numbers (meaning thata is allowed to be 0).

10. Call an 2n-digit number special if we can split its digits into two sets of size n such that the sum ofthe numbers in the two sets is the same. Let pn be the probability that a randomly-chosen 2n-digitnumber is special (we will allow leading zeros in the number).

(a) [25] The sequence pn converges to a constant c. Find c.

(b) [30] Let qn = pn − c. There exists a unique positive constant r such that qn

rn converges to aconstant d. Find r and d.

Page 1 of 1

13thAnnual Harvard-MIT Mathematics TournamentSaturday 20 February 2010

Team Round B

1. [10] How many ways are there to place pawns on an 8× 8 chessboard, so that there is at most 1 pawnin each horizontal row? Express your answer in the form pe1

1 · pe22 · · · , where the pi are distinct primes

and the ei are positive integers.

Answer: 316 If there is at most 1 pawn in each row, then each row of the chessboard may haveeither 0 or 1 pawn somewhere in the row. There is 1 case if there are no pawns in the row. There are8 possible cases if there is 1 pawn in the row, one case for each square in the row. Hence for each row,there are 9 possible pawn arrangements. There are 8 rows, thus we have 98 = 316.

2. [10] In the following figure, a regular hexagon of side length 1 is attached to a semicircle of diameter1. What is the longest distance between any two points in the figure?

Answer: 1+√

132 Inspection shows that one point must be on the semicircle and the other must

be on the side of the hexagon directly opposite the edge with the semicircle, the bottom edge of thehexagon in the above diagram. Let O be the center of the semicircle and let M be the midpoint of thebottom edge.

We will determine the longest distance between points in the figure by comparing the lengths of allthe segments with one endpoint on the bottom edge and the other endpoint on the semicircle. Fix apoint A on the bottom edge of the hexagon. Suppose that B is chosen on the semicircle such that ABis as long as possible. Let C be the circle centered at A with radius AB. If C is not tangent to thesemicircle, then part of the semicircle is outside C, so we could pick a B′ on the semicircle such thatAB′ is longer than AB. So C must be tangent to the semicircle, and AB must pass through O.

Then OB is always 12 , no matter which A we choose on the bottom edge. All that remains is maximizing

AO. This length is the hypotenuse of a right triangle with the fixed height MO, so it is maximizedwhen AM is as large as possible - when A is an endpoint of the bottom edge. Note that MO = 2 ·

√3

2 ,and that AM can be at most 1

2 , so AO can be at most√

132 . So the maximum distance between two

points in the diagram is AO + OB = 1+√

132 .

3. [15] Let p(x) = anxn + an−1xn−1 + . . . + a0, where each ai is either 1 or −1. Let r be a root of p. If

|r| > 158 , what is the minimum possible value of n?

Answer: 4 We claim that n = 4 is the answer. First, we show that n > 3. Suppose that n ≤ 3. Letr be the root of the polynomial with |r| ≥ 15

8 . Then, by the Triangle Inequality, we have:

|anrn| =∣∣an−1r

n−1 + an−2rn−2 + . . . + a0

∣∣ ≤ ∣∣an−1rn−1∣∣+∣∣an−2r

n−2∣∣+ . . . + |a0|

|r|n ≤∣∣rn−1

∣∣+∣∣rn−2

∣∣+ . . . + |1| = |r|n − 1|r| − 1

|r|n+1 − 2|r|n + 1 ≤ 0⇒ 1 ≤ |r|n (2− |r|)The right-hand side is increasing in n, for |r| > 1, so it is bounded by |r|3(2− |r|). This expression isdecreasing in r for r ≥ 3

2 . When |r| = 158 , then the right-hand side is less than 1, which violates the

inequalities. Therefore n > 3. Now, we claim that there is a 4th degree polynomial with a root r with|r| ≥ 15

8 . Let p(x) = x4 − x3 − x2 − x − 1. Then p( 158 ) < 0 and p(2) > 2. By the Intermediate Value

Theorem, p(x) has such a root r.

Team Round B

4. [20] Find all 4-digit integers of the form aabb (when written in base 10) that are perfect squares.

Answer: 7744 Let x be an integer such that x2 is of the desired form. Then 1100a + 11b = x2.Then x2 is divisible by 11, which means x is divisible by 11. Then for some integer, y, x = 11y. Then1100a+11b = 112y2 ⇒ 100a+b = 11y2. This means that 100a+b ≡ 0 (mod 11)⇒ a+b ≡ 0 (mod 11).Because a and b must be nonzero digits, we have 2 ≤ a, b ≤ 9, so we can write b = 11− a.

Replacing b in the equation derived above, we obtain 99a + 11 = 11y2 ⇒ 9a + 1 = y2. We check thepossible values of a from 2 to 9, and only a = 7 yields a perfect square. When a = 7, b = 4, so theonly perfect square of for aabb is 7744.

5. [25] Compute98∑

n=1

2√

n +√

n + 2+

1√n + 1 +

√n + 2

.

Answer: 3√

11− 2√

2 + 19 Rationalizing the denominator of both terms in the summation yields

√n + 2−

√n+√

n + 2−√

n + 1 = 2√

n + 2−(√

n+√

n + 1). Then the sum98∑

n=1

2√

n + 2−(√

n+√

n + 1)

telescopes. All terms cancel except for −(√

1 +√

2)−√

2 + 2√

99 + 2√

100−√

99 = 3√

11− 2√

2 + 19.

6. [25] Into how many regions can a circle be cut by 10 parabolas?

Answer: 201 We will consider the general case of n parabolas, for which the answer is 2n2 + 1.

We will start with some rough intuition, then fill in the details afterwards. The intuition is that, if wemake the parabolas steep enough, we can basically treat them as two parallel lines. Furthermore, thenumber of regions is given in terms of the number of intersections of the parabolas that occur withinthe circle, since every time two parabolas cross a new region is created. Since two pairs of parallel linesintersect in 4 points, and pairs of parabolas also intersect in 4 points, as long as we can always makeall 4 points of intersection lie inside the circle, the parallel lines case is the best we can do.

In other words, the answer is the same as the answer if we were trying to add ten pairs of parallel lines.We can compute the answer for pairs of parallel lines as follows — when we add the kth set of parallellines, there are already 2k− 2 lines that the two new lines can intersect, meaning that each of the linesadds 2k − 1 new regions1. This means that we add 4k − 2 regions when adding the kth set of lines,making the answer 1+2+6+10+14+· · ·+(4n−2) = 1+2(1+3+5+7+· · ·+(2n−1)) = 1+2·n2 = 2n2+1.

Now that we have sketched out the solution, we will fill in the details more rigorously. First, if there aren parabolas inside the circle, and the they intersect in K points total, then we claim that the numberof regions the circle is divided into will be at most K + n + r + 1, where r is the number of parabolasthat intersect the circle itself in exactly four points.

We will prove this by induction. In the base case of n = 0, we are just saying that the circle itselfconsists of exactly one region.

To prove the inductive step, suppose that we have n parabolas with K points of intersection. We wantto show that if we add an additional parabola, and this parabola intersects the other parabolas in ppoints, then this new parabola adds either p + 1 or p + 2 regions to the circle, and that we get p + 2regions if and only if the parabola intersects the circle in exactly four points.

We will do this by considering how many regions the parabola cuts through, following its path fromwhen it initially enters the circle to when it exits the circle for the last time. When it initially entersthe circle, it cuts through one region, thereby increasing the number of regions by one2. Then, for eachother parabola that this parabola crosses, we cut through one additional region. It is also possible forthe parabola to leave and then re-enter the circle, which happens if and only if the parabola intersects

1This is the maximum possible number of new regions, but it’s not too hard to see that this is always attainable.2While the fact that a curve going through a region splits it into two new regions is intuitively obvious, it is actually very

difficult to prove. The proof relies on some deep results from algebraic topology and is known as the Jordan Curve Theorem.If you are interested in learning more about this, see http://en.wikipedia.org/wiki/Jordan_curve_theorem

Team Round B

the circle in four points, and also adds one additional region. Therefore, the number of regions is eitherp + 1 or p + 2, and it is p + 2 if and only if the parabola intersects the circle in four points. Thiscompletes the induction and proves the claim.

So, we are left with trying to maximize K + n + r + 1. Since a pair of parabolas intersects in atmost 4 points, and there are

(n2

)pairs of parabolas, we have K ≤ 4

(n2

)= 2n2 − 2n. Also, r ≤ n, so

K + n + r + 1 ≤ 2n2 + 1. On the other hand, as explained in the paragraphs giving the intuition, wecan attain 2n2 + 1 by making the parabolas sufficiently steep that they act like pairs of parallel lines.Therefore, the answer is 2n2 + 1, as claimed.

7. [30] Evaluate2010∑k=1

cos2(k)

Answer: 1005 + sin(4021)−sin(1)4 sin(1) We use the identity cos2(k) = 1+cos(2x)

2 . Then our expression

evalutes to 1005 + (cos(2)+...+cos(4020))2 .

To evaluate cos(2) + · · · + cos(4020), let y = cos(2) + · · · + cos(4020) ⇒ y(sin(1)) = cos(2) sin(1) +. . . + cos(4020) sin(1). Observe that for any x, cos(x) sin(1) = sin(x+1)−sin(x−1)

2 . Then y(sin(1)) =sin(3)−sin(1)

2 + sin(5)−sin(3)2 + . . . + sin 4021−sin(4019)

2 . This is clearly a telescoping sum; we get y(sin(1)) =sin(4021)−sin(1)

2 . Then we have the desired y = sin(4021)−sin(1)2 sin(1) . Then our original expression evaluates

to 1005 + sin(4021)−sin(1)4 sin(1) .

8. [30] Consider the following two-player game. Player 1 starts with a number, N . He then subtracts aproper divisor of N from N and gives the result to player 2 (a proper divisor of N is a positive divisorof N that is not equal to 1 or N). Player 2 does the same thing with the number she gets from player 1,and gives the result back to player 1. The two players continue until a player is given a prime number,at which point that player loses. For how many values of N between 2 and 100 inclusive does player 1have a winning strategy?

Answer: 47 We claim that player 1 has a winning strategy if and only if N is even and not an oddpower of 2.

First we show that if you are stuck with an odd number, then you are guaranteed to lose. Suppose youhave an odd number ab, where a and b are odd numbers, and you choose to subtract a. You pass youropponent the number a(b − 1). This cannot be a power of 2 (otherwise a is a power of 2 and hencea = 1, which is not allowed), so your opponent can find an odd proper divisor of a(b− 1) (such as a),and you will have a smaller odd number. Eventually you will get to an odd prime and lose.

Now consider even numbers that aren’t powers of 2. As with before, you can find an odd proper divisorof N and pass your opponent an odd number, so you are guaranteed to win.

Finally consider powers of 2. If you have the number N = 2k, it would be unwise to choose a properdivisor other than 2k−1; otherwise you would give your opponent an even number that isn’t a powerof 2. Therefore if k is odd, you will end up with 2 and lose. If k is even, though, your opponent willend up with 2 and you will win.

Therefore player 1 has a winning strategy for all even numbers except for odd powers of 2.

9. [35] Let S be the set of ordered pairs of integers (x, y) with 1 ≤ x ≤ 5 and 1 ≤ y ≤ 3. How manysubsets R of S have the property that all the points of R lie on the graph of a single cubic? A cubic isa polynomial of the form y = ax3 + bx2 + cx + d, where a, b, c, and d are real numbers (meaning thata is allowed to be 0).

Answer: 796 We observe that R must contain at most 1 point from each column of S, because nofunction can contain more than 1 point with the same x-coordinate. Therefore, |R| ≤ 5 (|R| denotesthe number of elements of R). Note that 4 points determine a cubic, so if R is any subset of points indistinct columns and |R| ≤ 4, then R has the desired property. There are 45 ways to choose at most 1

Team Round B

point from each column and 35 ways to choose exactly 1 point from each column. There are therefore45 − 35 = 781 subsets R of S such that |R| ≤ 4 and all points of R lie in distinct columns. As noted,these sets all automatically have the desired property.

Now we consider all sets R of size 5. As before, each point in R must come from a different column.Let us shift our origin to (3, 2), and let p be the polynomial containing all 5 points of R. ThenR = {(−2, p(−2)), (−1, p(−1)), (0, p(0)), (1, p(1)), (2, p(2))}.By the method of finite differences3, or alternately by Lagrange Interpolation4, there is a uniquepolynomial p of degree less than 5 going through 5 specified points, and this polynomial is of degreeless than 4 if and only if p(−2)− 4p(−1) + 6p(0)− 4p(1) + p(2) = 0.

Then p(−2) + p(2) + 6p(0) = 4(p(−1) + p(1)), where p(−2) + p(2) ∈ {−2− 1, 0, 1, 2}, p(−1) + p(1) ∈{−2,−1, 0, 1, 2}, and p(0) ∈ {−1, 0, 1}. We know that 6p(0) and 4(p(−1) + p(1)) are necessarily even,thus we must have p(−2) + p(2) ∈ {−2, 0, 2} in order for the equation to be satisfied.

Let (a, b, c) = (p(−2) + p(2), 6p(0), 4(p(−1) + p(1))). The possible values of (a, b, c) that are solutionsto a + b = c are then {(−2,−6,−8), (−2, 6, 4), (0, 0, 0), (2,−6,−4), (2, 6, 8)}.If (a, b, c) = (−2,−6,−8), then we need p(−2) + p(2) = −2, p(0) = −1, p(−1) + p(1) = −2. There isonly 1 possible solution to each of these equations: (p(−2), p(2)) = (−1,−1) for the first one, p(0) = −1for the second, and (p(1)) = (−1,−1) for the third. Hence there is 1 possible subset R for the case(a, b, c) = (−2,−6,−8).

If (a, b, c) = (−2, 6, 4), then there is again 1 possible solution to p(−2) + p(2) = 1. There are twosolutions to p(−1) + p(1) = 1: (p(−1), p(1)) = (0, 1), (1, 0). Also, p(0) can only be 1, so there are 2possible subsets for this case.

If (a, b, c) = (0, 0, 0), then there there are 3 possible solutions to p(−2) + p(2) = 0: (p(−2), p(2)) =(−1, 1), (0, 0), (1,−1). Similarly, there are 3 possible solutions to p(−1) + p(1) = 0. Also, p(0) can onlybe 0, so there are 9 possible subsets for this case.

If (a, b, c) = (2,−6,−4), then there is 1 possible solution to p(−2) + p(2) = 2: (p(−2), p(2)) = (1, 1).There are 2 possible solutions to p(−1) + p(1) = −1: (p(−1), p(1)) = (0,−1), (−1, 0). Also, p(0) canonly be -1, so there are 2 possible subsets for this case.

If (a, b, c) = (2, 6, 8), then there is 1 possible solution to p(−2) + p(2) = 2, as shown above. There is1 solution to p(−1) + p(1) = 2: (p(−1), p(1)) = (1, 1). Also, p(0) can only be 1, so there is 1 possiblesubset for this case.

Then there are 1 + 2 + 9 + 2 + 1 = 15 total possible subsets of size 5 that can be fit to a polynomial ofdegree less than 4. Hence there are 781 + 15 = 796 possible subsets total.

10. Call an 2n-digit number special if we can split its digits into two sets of size n such that the sum ofthe numbers in the two sets is the same. Let pn be the probability that a randomly-chosen 2n-digitnumber is special (we will allow leading zeros in the number).

(a) [25] The sequence pn converges to a constant c. Find c.

Answer: 12 We first claim that if a 2n-digit number x has at least eight 0’s and at least eight

1’s and the sum of its digits is even, then x is special.Let A be a set of eight 0’s and eight 1’s and let B be the set of all the other digits. We splitb arbitrily into two sets Y and Z of equal size. If |

∑y∈Y y −

∑z∈Z z| > 8, then we swap the

biggest element of the set with the bigger sum with the smallest element of the other set. Thistransposition always decreases the absolute value of the sum: in the worst case, a 9 from thebigger set is swapped for a 0 from the smaller set, which changes the difference by at most 18.Therefore, after a finite number of steps, we will have |

∑y∈Y y −

∑z∈Z z| ≤ 8.

Note that this absolute value is even, since the sum of all the digits is even. Without loss ofgenerality, suppose that

∑y∈Y y −

∑z∈Z z is 2k, where 0 ≤ k ≤ 4. If we add k 0’s and 8− k 1’s

to Y , and we add the other elements of A to Z, then the two sets will balance, so x is special.3See http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=1263378.4See http://en.wikipedia.org/wiki/Lagrange_polynomial.

Team Round B

(b) [30] Let qn = pn − c. There exists a unique positive constant r such that qn

rn converges to aconstant d. Find r and d.Answer: ( 1

4 ,−1) To get the next asymptotic term after the constant term of 12 , we need to

consider what happens when the digit sum is even; we want to find the probability that sucha number isn’t balanced. We claim that the configuration that contributes the vast majority ofunbalanced numbers is when all numbers are even and the sum is 2 mod 4, or such a configurationwith all numbers increased by 1. Clearly this gives qn being asymptotic to − 1

2 ·2 ·(

12

)2n = −(

14

)n,so r = 1

4 and d = −1.To prove the claim, first note that the asymptotic probability that there are at most 4 digits thatoccur more than 10 times is asymptotically much smaller than

(12

)n, so we can assume that thereexist 5 digits that each occur at least 10 times. If any of those digits are consecutive, then thedigit sum being even implies that the number is balanced (by an argument similar to part (a)).So, we can assume that none of the numbers are consecutive. We would like to say that thisimplies that the numbers are either 0, 2, 4, 6, 8 or 1, 3, 5, 7, 9. However, we can’t quite say this yet,as we need to rule out possibilities like 0, 2, 4, 7, 9. In this case, though, we can just pair 0 and 7up with 2 and 4; by using the same argument as in part (a), except using 0 and 7 both (to get asum of 7) and 2 and 4 both (to get a sum of 6) to balance out the two sets at the end.In general, if there is ever a gap of size 3, consider the number right after it and the 3 numbersbefore it (so we have k− 4, k− 2, k, k + 3 for some k), and pair them up such that one pair has asum that’s exactly one more than the other (i.e. pair k − 4 with k + 3 and k − 2 with k). Sincewe again have pairs of numbers whose sums differ by 1, we can use the technique from part (a)of balancing out the sets at the end.So, we can assume there is no gap of size 3, which together with the condition that no two numbersare adjacent implies that the 5 digits are either 0, 2, 4, 6, 8 or 1, 3, 5, 7, 9. For the remainder of thesolution, we will deal with the 0, 2, 4, 6, 8 case, since it is symmetric with the other case under thetransformation x 7→ 9− x.If we can distribute the odd digits into two sets S1 and S2 such that (i) the differense in sums ofS1 and S2 is small; and (ii) the difference in sums of S1 and S2, plus the sum of the even digits,is divisible by 4, then the same argument as in part (a) implies that the number is good.In fact, if there are any odd digits, then we can use them at the beginning to fix the parity mod4 (by adding them all in such that the sums of the two sets remain close, and then potentiallyswitching one with an even digit). Therefore, if there are any odd digits then the number is good.Also, even if there are no odd digits, if the sum of the digits is divisible by 4 then the number isgood.So, we have shown that almost all non-good numbers come from having all numbers being evenwith a digit sum that is 2 mod 4, or the analogous case under the mapping x 7→ 9 − x. Thisformalizes the claim we made in the first paragraph, so r = 1

4 and d = −1, as claimed.

Team Round B

3rdAnnual Harvard-MIT November TournamentSunday 7 November 2010

General Test

1. [2] Jacob flips five coins, exactly three of which land heads. What is the probability that the first twoare both heads?

2. [3] How many sequences a1, a2, . . . , a8 of zeroes and ones have a1a2 + a2a3 + · · ·+ a7a8 = 5?

3. [3] Triangle ABC has AB = 5, BC = 7, and CA = 8. New lines not containing but parallel to AB,BC, and CA are drawn tangent to the incircle of ABC. What is the area of the hexagon formed bythe sides of the original triangle and the newly drawn lines?

4. [4] An ant starts at the point (1, 0). Each minute, it walks from its current position to one of the fouradjacent lattice points until it reaches a point (x, y) with |x| + |y| ≥ 2. What is the probability thatthe ant ends at the point (1, 1)?

5. [5] A polynomial P is of the form ±x6 ± x5 ± x4 ± x3 ± x2 ± x ± 1. Given that P (2) = 27, what isP (3)?

6. [5] What is the sum of the positive solutions to 2x2 − xbxc = 5, where bxc is the largest integer lessthan or equal to x?

7. [6] What is the remainder when (1 + x)2010 is divided by 1 + x + x2?

8. [7] Two circles with radius one are drawn in the coordinate plane, one with center (0, 1) and the otherwith center (2, y), for some real number y between 0 and 1. A third circle is drawn so as to be tangentto both of the other two circles as well as the x axis. What is the smallest possible radius for this thirdcircle?

9. [7] What is the sum of all numbers between 0 and 511 inclusive that have an even number of 1s whenwritten in binary?

10. [8] You are given two diameters AB and CD of circle Ω with radius 1. A circle is drawn in one of thesmaller sectors formed such that it is tangent to AB at E, tangent to CD at F , and tangent to Ω atP . Lines PE and PF intersect Ω again at X and Y . What is the length of XY , given that AC = 2

3?

3rdAnnual Harvard-MIT November TournamentSunday 7 November 2010

General Test

1. [2] Jacob flips five coins, exactly three of which land heads. What is the probability that the first twoare both heads?

Answer: 310 We can associate with each sequence of coin flips a unique word where H represents

heads, and T represents tails. For example, the word HHTTH would correspond to the coin flipsequence where the first two flips were heads, the next two were tails, and the last was heads. We aregiven that exactly three of the five coin flips came up heads, so our word must be some rearrangementof HHHTT. To calculate the total number of possibilities, any rearrangement corresponds to a choiceof three spots to place the H flips, so there are

(53

)= 10 possibilities. If the first two flips are both

heads, then we can only rearrange the last three HTT flips, which corresponds to choosing one spotfor the remaining H. This can be done in

(31

)= 3 ways. Finally, the probability is the quotient of

these two, so we get the answer of 310 . Alternatively, since the total number of possiblities is small, we

can write out all rearrangements: HHHTT, HHTHT, HHTTH, HTHHT, HTHTH, HTTHH, THHHT,THHTH, THTHH, TTHHH. Of these ten, only in the first three do we flip heads the first two times,so we get the same answer of 3

10 .

2. [3] How many sequences a1, a2, . . . , a8 of zeroes and ones have a1a2 + a2a3 + · · ·+ a7a8 = 5?

Answer: 9 First, note that we have seven terms in the left hand side, and each term can be either0 or 1, so we must have five terms equal to 1 and two terms equal to 0. Thus, for n ∈ {1, 2, ..., 8}, atleast one of the an must be equal to 0. If we can find i, j ∈ {2, 3, ..., 7} such that ai = aj = 0 andi < j, then the terms ai−1ai, aiai+1, and ajaj+1 will all be equal to 0. We did not count any termtwice because i − 1 < i < j, so we would have three terms equal to 0, which cannot happen becausewe can have only two. Thus, we can find at most one n ∈ {2, 3, ..., 7} such that an = 0. We will docasework on which n in this range have an = 0.

If n ∈ {3, 4, 5, 6}, then we know that the terms an−1an = anan+1 = 0, so all other terms must be 1,so a1a2 = a2a3 = ... = an−2an−1 = 1 and an+1an+2 = ... = a7a8 = 1. Because every ai appears in oneof these equations for i 6= n, then we must have ai = 1 for all i 6= n, so we have 1 possibility for eachchoice of n and thus 4 possibilities total for this case.

If n = 2, then again we have a1a2 = a2a3 = 0, so we must have a3a4 = a4a5 = ... = a7a8 = 1, soa3 = a4 = ... = a8 = 1. However, this time a1 is not fixed, and we see that regardless of our choice ofa1 the sum will still be equal to 5. Thus, since there are 2 choices for a1, then there are 2 possibilitiestotal for this case. The case where n = 7 is analogous, with a8 having 2 possibilities, so we haveanother 2 possibilities.

Finally, if an = 1 for n ∈ {2, 3, ..., 7}, then we will have a2a3 = a3a4 = ... = a6a7 = 1. We alreadyhave five terms equal to 1, so the remaining two terms a1a2 and a7a8 must be 0. Since a2 = 1, thenwe must have a1 = 0, and since a7 = 1 then a8 = 0. Thus, there is only 1 possibility for this case.

Summing, we have 4 + 2 + 2 + 1 = 9 total sequences.

3. [3] Triangle ABC has AB = 5, BC = 7, and CA = 8. New lines not containing but parallel to AB,BC, and CA are drawn tangent to the incircle of ABC. What is the area of the hexagon formed bythe sides of the original triangle and the newly drawn lines?

Answer: 315

√3

General Test

A

B C

D E

F

GH

J

From the law of cosines we compute ]A = cos−1(

52+82−72

2(5)(8)

)= 60◦. Using brackets to denote the area

of a region, we find that

[ABC] = 12AB ·AC · sin 60◦ = 10

√3.

The radius of the incircle can be computed by the formula

r = 2[ABC]AB+BC+CA = 20

√3

20 =√

3.

Now the height from A to BC is 2[ABC]BC = 20

√3

7 . Then the height from A to DE is 20√

37 − 2r = 6

√3

7 .

Then [ADE] =(

6√

3/7

20√

3/7

)2

[ABC] = 9100 [ABC]. Here, we use the fact that 4ABC and 4ADE are

similar.

Similarly, we compute that the height from B to CA is 2[ABC]CA = 20

√3

8 = 5√

32 . Then the height from

B to HJ is 5√

32 − 2r =

√3

2 . Then [BHJ ] =( √

3/2

5√

3/2

)2

[ABC] = 125 [ABC].

Finally, we compute that the height from C to AB is 2[ABC]5 = 20

√3

5 = 4√

3. Then the height from C

to FG is 4√

3− 2r = 2√

3. Then [CFG] =(

2√

34√

3

)2

[ABC] = 14 [ABC].

Finally we can compute the area of hexagon DEFGHJ . We have

[DEFGHJ ] = [ABC]− [ADE]− [BHJ ]− [CFG] = [ABC](1− 9

100 −125 −

14

)= [ABC]

(3150

)=

10√

3(

3150

)= 31

5

√3.

4. [4] An ant starts at the point (1, 0). Each minute, it walks from its current position to one of the fouradjacent lattice points until it reaches a point (x, y) with |x| + |y| ≥ 2. What is the probability thatthe ant ends at the point (1, 1)?

Answer: 724 From the starting point of (1, 0), there is a 1

4 chance we will go directly to (1, 1), a 12

chance we will end at (2, 0) or (1,−1), and a 14 chance we will go to (0, 0). Thus, if p is the probability

that we will reach (1, 1) from (0, 0), then the desired probability is equal to 14 + 1

4p, so we need onlycalculate p. Note that we can replace the condition |x|+ |y| ≥ 2 by |x|+ |y| = 2, since in each iterationthe quantity |x| + |y| can increase by at most 1. Thus, we only have to consider the eight points(2, 0), (1, 1), (0, 2), (−1, 1), (−2, 0), (−1,−1), (0,−2), (1,−1). Let p1, p2, ..., p8 be the probability of

General Test

reaching each of these points from (0, 0), respectively. By symmetry, we see that p1 = p3 = p5 = p7

and p2 = p4 = p6 = p8. We also know that there are two paths from (0, 0) to (1, 1) and one path from(0, 0) to (2, 0), thus p2 = 2p1. Because the sum of all probabilities is 1, we have p1 + p2 + ...+ p8 = 1.Combining these equations, we see that 4p1+4p2 = 12p1 = 1, so p1 = 1

12 and p2 = 16 . Since p = p2 = 1

6 ,then the final answer is 1

4 + 14 ·

16 = 7

24

5. [5] A polynomial P is of the form ±x6 ± x5 ± x4 ± x3 ± x2 ± x ± 1. Given that P (2) = 27, what isP (3)?

Answer: 439 We use the following lemma:Lemma. The sign of ±2n ± 2n−1 ± · · · ± 2± 1 is the same as the sign of the 2n term.Proof. Without loss of generality, let 2n be positive. (We can flip all signs.) Notice that 2n ± 2n−1 ±2n−2 ± · · · 2± 1 ≥ 2n − 2n−1 − 2n−2 − · · · − 2− 1 = 1, which is positive.

We can use this lemma to uniquely determine the signs of P . Since our desired sum, 27, is positive, thecoefficient of x6 must be positive. Subtracting 64, we now have that ±25±24± . . .±2±1 = −37, so thesign of 25 must be negative. Continuing in this manner, we find that P (x) = x6−x5−x4+x3+x2−x+1,so P (3) = 36 − 35 − 34 + 33 + 32 − 3 + 1 = 439.

6. [5] What is the sum of the positive solutions to 2x2 − xbxc = 5, where bxc is the largest integer lessthan or equal to x?

Answer: 3+√

41+2√

114 We first note that bxc ≤ x, so 2x2−xbxc ≥ 2x2−x2 = x2. Since this function

is increasing on the positive reals, all solutions must be at most√

5. This gives us 3 possible values ofbxc: 0, 1, and 2.

If bxc = 0, then our equation becomes 2x2 = 5, which has positive solution x =√

52 . This number is

greater than 1, so its floor is not 0; thus, there are no solutions in this case.

If bxc = 1, then our equation becomes 2x2 − x = 5. Using the quadratic formula, we find the positivesolution x = 1+

√41

4 . Since 3 <√

41 < 7, this number is between 1 and 2, so it satisfies the equation.

If bxc = 2, then our equation becomes 2x2 − 2x = 5. We find the positive solution x = 1+√

112 . Since

3 <√

11 < 5, this number is between 2 and 3, so it satisfies the equation.

We then find that the sum of positive solutions is 1+√

414 + 1+

√11

2 = 3+√

41+2√

114 .

7. [6] What is the remainder when (1 + x)2010 is divided by 1 + x+ x2?

Answer: 1 We use polynomial congruence mod 1 + x + x2 to find the desired remainder. Sincex2 + x+ 1|x3 − 1, we have that x3 ≡ 1 (mod 1 + x+ x2). Now:

(1 + x)2010 ≡ (−x2)2010 (mod 1 + x+ x2)

≡ x4020 (mod 1 + x+ x2)

≡ (x3)1340 (mod 1 + x+ x2)

≡ 11340 (mod 1 + x+ x2)

≡ 1 (mod 1 + x+ x2)

Thus, the answer is 1.

8. [7] Two circles with radius one are drawn in the coordinate plane, one with center (0, 1) and the otherwith center (2, y), for some real number y between 0 and 1. A third circle is drawn so as to be tangentto both of the other two circles as well as the x axis. What is the smallest possible radius for this thirdcircle?

Answer: 3− 2√

2 Suppose that the smaller circle has radius r. Call the three circles (in order fromleft to right) O1, O2, and O3. The distance between the centers of O1 and O2 is 1+ r, and the distancein their y-coordinates is 1− r. Therefore, by the Pythagorean theorem, the difference in x-coordinates

General Test

is√

(1 + r)2 − (1− r)2 = 2√r, which means that O2 has a center at (2

√r, r). But O2 is also tangent

to O3, which means that the difference in x-coordinate from the right-most point of O2 to the centerof O3 is at most 1. Therefore, the center of O3 has an x-coordinate of at most 2

√r + r + 1, meaning

that 2√r + r + 1 ≤ 2. We can use the quadratic formula to see that this implies that

√r ≤√

2 − 1,so r ≤ 3 − 2

√2. We can achieve equality by placing the center of O3 at (2, r) (which in this case is

(2, 3− 2√

2)).

9. [7] What is the sum of all numbers between 0 and 511 inclusive that have an even number of 1s whenwritten in binary?

Answer: 65408 Call a digit in the binary representation of a number a bit. We claim that for anygiven i between 0 and 8, there are 128 numbers with an even number of 1s that have a 1 in the bitrepresenting 2i. To prove this, we simply make that bit a 1, then consider all possible configurationsof the other bits, excluding the last bit (or the second-last bit if our given bit is already the last bit).The last bit will then be restricted to satisfy the parity condition on the number of 1s. As there are128 possible configurations of all the bits but two, we find 128 possible numbers, proving our claim.

Therefore, each bit is present as a 1 in 128 numbers in the sum, so the bit representing 2i contributes128 · 2i to our sum. Summing over all 0 ≤ i ≤ 8, we find the answer to be 128(1 + 2 + . . . + 128) =128 · 511 = 65408.

10. [8] You are given two diameters AB and CD of circle Ω with radius 1. A circle is drawn in one of thesmaller sectors formed such that it is tangent to AB at E, tangent to CD at F , and tangent to Ω atP . Lines PE and PF intersect Ω again at X and Y . What is the length of XY , given that AC = 2

3?

Answer: 4√

23 Let O denote the center of circle Ω. We first prove that OX ⊥ AB and OY ⊥ CD.

Consider the homothety about P which maps the smaller circle to Ω. This homothety takes E to Xand also takes AB to the line tangent to circle Ω parallel to AB. Therefore, X is the midpoint of thearc AB, and so OX ⊥ AB. Similarly, OY ⊥ CD.

Let θ = ∠AOC. By the Law of Sines, we have AC = 2 sin θ2 . Thus, sin θ

2 = 13 , and cos θ2 =

√1− ( 1

3 )2 =2√

23 . Therefore,

XY = 2 sin∠XOY

2

= 2 sin(

90◦ − θ

2

)= 2 cos

θ

2

=4√

23.

General Test

3rdAnnual Harvard-MIT November TournamentSunday 7 November 2010

Theme Round

StarCraft

1. [3] 16 progamers are playing in a single elimination tournament. Each player has a different skill leveland when two play against each other the one with the higher skill level will always win. Each round,each progamer plays a match against another and the loser is eliminated. This continues until onlyone remains. How many different progamers can reach the round that has 2 players remaining?

2. [4] 16 progamers are playing in another single elimination tournament. Each round, each of theremaining progamers plays against another and the loser is eliminated. Additionally, each time aprogamer wins, he will have a ceremony to celebrate. A player’s first ceremony is ten seconds long,and afterward each ceremony is ten seconds longer than the last. What is the total length in secondsof all the ceremonies over the entire tournament?

3. [5] Dragoons take up 1 × 1 squares in the plane with sides parallel to the coordinate axes such thatthe interiors of the squares do not intersect. A dragoon can fire at another dragoon if the difference inthe x-coordinates of their centers and the difference in the y-coordinates of their centers are both atmost 6, regardless of any dragoons in between. For example, a dragoon centered at (4, 5) can fire ata dragoon centered at the origin, but a dragoon centered at (7, 0) can not. A dragoon cannot fire atitself. What is the maximum number of dragoons that can fire at a single dragoon simultaneously?

4. [5] A zerg player can produce one zergling every minute and a protoss player can produce one zealotevery 2.1 minutes. Both players begin building their respective units immediately from the beginningof the game. In a fight, a zergling army overpowers a zealot army if the ratio of zerglings to zealots ismore than 3. What is the total amount of time (in minutes) during the game such that at that timethe zergling army would overpower the zealot army?

5. [7] There are 111 StarCraft progamers. The StarCraft team SKT starts with a given set of elevenprogamers on it, and at the end of each season, it drops a progamer and adds a progamer (possiblythe same one). At the start of the second season, SKT has to field a team of five progamers to playthe opening match. How many different lineups of five players could be fielded if the order of playerson the lineup matters?

Unfair Coins

6. [4] When flipped, a coin has a probability p of landing heads. When flipped twice, it is twice as likelyto land on the same side both times as it is to land on each side once. What is the larger possible valueof p?

7. [4] George has two coins, one of which is fair and the other of which always comes up heads. Jacobtakes one of them at random and flips it twice. Given that it came up heads both times, what is theprobability that it is the coin that always comes up heads?

8. [5] Allison has a coin which comes up heads 23 of the time. She flips it 5 times. What is the probability

that she sees more heads than tails?

9. [6] Newton and Leibniz are playing a game with a coin that comes up heads with probability p. Theytake turns flipping the coin until one of them wins with Newton going first. Newton wins if he flips aheads and Leibniz wins if he flips a tails. Given that Newton and Leibniz each win the game half ofthe time, what is the probability p?

10. [7] Justine has a coin which will come up the same as the last flip 23 of the time and the other side 1

3 ofthe time. She flips it and it comes up heads. She then flips it 2010 more times. What is the probabilitythat the last flip is heads?

3rdAnnual Harvard-MIT November TournamentSunday 7 November 2010

Theme Round

StarCraft

1. [3] 16 progamers are playing in a single elimination tournament. Each player has a different skill leveland when two play against each other the one with the higher skill level will always win. Each round,each progamer plays a match against another and the loser is eliminated. This continues until onlyone remains. How many different progamers can reach the round that has 2 players remaining?

Answer: 9 Each finalist must be better than the person he beat in the semifinals, both of the peoplethey beat in the second round, and all 4 of the people any of those people beat in the first round. So,none of the 7 worst players can possibly make it to the finals. Any of the 9 best players can make itto the finals if the other 8 of the best 9 play each other in all rounds before the finals. So, exactly 9people are capable of making it to the finals.

2. [4] 16 progamers are playing in another single elimination tournament. Each round, each of theremaining progamers plays against another and the loser is eliminated. Additionally, each time aprogamer wins, he will have a ceremony to celebrate. A player’s first ceremony is ten seconds long,and afterward each ceremony is ten seconds longer than the last. What is the total length in secondsof all the ceremonies over the entire tournament?

Answer: 260 At the end of the first round, each of the 8 winners has a 10 second ceremony. Afterthe second round, the 4 winners have a 20 second ceremony. The two remaining players have 30 secondceremonies after the third round, and the winner has a 40 second ceremony after the finals. So, all ofthe ceremonies combined take 8 · 10 + 4 · 20 + 2 · 30 + 40 = 260 seconds.

3. [5] Dragoons take up 1 × 1 squares in the plane with sides parallel to the coordinate axes such thatthe interiors of the squares do not intersect. A dragoon can fire at another dragoon if the difference inthe x-coordinates of their centers and the difference in the y-coordinates of their centers are both atmost 6, regardless of any dragoons in between. For example, a dragoon centered at (4, 5) can fire ata dragoon centered at the origin, but a dragoon centered at (7, 0) can not. A dragoon cannot fire atitself. What is the maximum number of dragoons that can fire at a single dragoon simultaneously?

Answer: 168 Assign coordinates in such a way that the dragoon being fired on is centered at (0, 0).Any dragoon firing at it must have a center with x-coordinates and y-coordinates that are no smallerthan −6 and no greater than 6. That means that every dragoon firing at it must lie entirely in theregion bounded by the lines x = −6.5, x = 6.5, y = −6.5, and y = 6.5. This is a square with sides oflength 13, so there is room for exactly 169 dragoons in it. One of them is the dragoon being fired on,so there are at most 168 dragoons firing at it.

4. [5] A zerg player can produce one zergling every minute and a protoss player can produce one zealotevery 2.1 minutes. Both players begin building their respective units immediately from the beginningof the game. In a fight, a zergling army overpowers a zealot army if the ratio of zerglings to zealots ismore than 3. What is the total amount of time (in minutes) during the game such that at that timethe zergling army would overpower the zealot army?

Answer: 1.3 At the end of the first minute, the zerg player produces a zergling and has a superiorarmy for the 1.1 minutes before the protoss player produces the first zealot. At this point, the zealot isat least a match for the zerglings until the fourth is produced 4 minutes into the game. Then, the zergarmy has the advantage for the .2 minutes before a second zealot is produced. A third zealot will beproduced 6.3 minutes into the game, which will be before the zerg player accumulates the 7 zerglingsneeded to overwhelm the first 2 zealots. After this, the zerglings will never regain the advantagebecause the zerg player can never produce 3 more zerglings to counter the last zealot before anotherone is produced. So, the zerg player will have the military advantage for 1.1 + .2 = 1.3 minutes.

Theme Round

5. [7] There are 111 StarCraft progamers. The StarCraft team SKT starts with a given set of elevenprogamers on it, and at the end of each season, it drops a progamer and adds a progamer (possiblythe same one). At the start of the second season, SKT has to field a team of five progamers to playthe opening match. How many different lineups of five players could be fielded if the order of playerson the lineup matters?

Answer: 4015440 We disregard the order of the players, multiplying our answer by 5! = 120 atthe end to account for it. Clearly, SKT will be able to field at most 1 player not in the original setof eleven players. If it does not field a new player, then it has

(115

)= 462 choices. If it does field a

new player, then it has 100 choices for the new player, and(114

)= 330 choices for the 4 other players,

giving 33000 possibilities. Thus, SKT can field at most 33000 + 462 = 33462 unordered lineups, andmultiplying this by 120, we find the answer to be 4015440 .

Unfair Coins

6. [4] When flipped, a coin has a probability p of landing heads. When flipped twice, it is twice as likelyto land on the same side both times as it is to land on each side once. What is the larger possible valueof p?

Answer: 3+√

36 The probability that the coin will land on the same side twice is p2 + (1 − p)2 =

2p2 − 2p + 1. The probability that the coin will land on each side once is p(p − 1) + (p − 1)p =2p(1 − p) = 2p − 2p2. We are told that it is twice as likely to land on the same side both times, so2p2 − 2p + 1 = 2(2p − 2p2) = 4p − 4p2. Solving, we get 6p2 − 6p + 1 = 0, so p = 3±

√3

6 . The largersolution is p = 3+

√3

6 .

7. [4] George has two coins, one of which is fair and the other of which always comes up heads. Jacobtakes one of them at random and flips it twice. Given that it came up heads both times, what is theprobability that it is the coin that always comes up heads?

Answer: 45 In general, P (A|B) = P (A∩B)

P (B) , where P (A|B) is the probability of A given B and

P (A ∩ B) is the probability of A and B (See http://en.wikipedia.org/wiki/Conditional probabilityfor more information). If A is the event of selecting the “double-headed” coin and B is the event ofJacob flipping two heads, then P (A ∩ B) = (1

2 )(1), since there is a 12 chance of picking the double-

headed coin and Jacob will always flip two heads when he has it. By conditional probability, P (B) =( 12 )(1) + ( 1

2 )( 14 ) = 5

8 , so P (A|B) = 1/25/8 = 4

5 .

8. [5] Allison has a coin which comes up heads 23 of the time. She flips it 5 times. What is the probability

that she sees more heads than tails?

Answer: 6481 The probability of flipping more heads than tails is the probability of flipping 3 heads,

4 heads, or 5 heads. Since 5 flips will give n heads with probability(5n

)( 23 )n( 1

3 )5−n, our answer is(53

)( 23 )3( 1

3 )2 +(54

)( 23 )4( 1

3 )1 +(55

)( 23 )5( 1

3 )0 = 6481 .

9. [6] Newton and Leibniz are playing a game with a coin that comes up heads with probability p. Theytake turns flipping the coin until one of them wins with Newton going first. Newton wins if he flips aheads and Leibniz wins if he flips a tails. Given that Newton and Leibniz each win the game half ofthe time, what is the probability p?

Answer: 3−√

52 The probability that Newton will win on the first flip is p. The probability that

Newton will win on the third flip is (1−p)p2, since the first flip must be tails, the second must be heads,and the third flip must be heads. By the same logic, the probability Newton will win on the (2n + 1)st

flip is (1 − p)n(p)n+1. Thus, we have an infinite geometric sequence p + (1 − p)p2 + (1 − p)2p3 + . . .

which equals p1−p(1−p) . We are given that this sum must equal 1

2 , so 1− p + p2 = 2p, so p = 3−√

52 (the

other solution is greater than 1).

Theme Round

10. [7] Justine has a coin which will come up the same as the last flip 23 of the time and the other side 1

3 ofthe time. She flips it and it comes up heads. She then flips it 2010 more times. What is the probabilitythat the last flip is heads?

Answer: 32010+12·32010 Let the “value” of a flip be 1 if the flip is different from the previous flip and let

it be 0 if the flip is the same as the previous flip. The last flip will be heads if the sum of the values of

all 2010 flips is even. The probability that this will happen is1005∑i=0

(2010

2i

)(13

)2i(23

)2010−2i

.

We know that

1005∑i=0

(2010

2i

)(13

)2i(23

)2010−2i

+(

20102i + 1

)(13

)2i+1(23

)2010−(2i+1)

=

2010∑k=0

(2010

k

)(13

)k (23

)2010−k

=(

13

+23

)2010

= 1

and1005∑i=0

(2010

2i

)(13

)2i(23

)2010−2i

−(

20102i + 1

)(13

)2i+1(23

)2010−(2i+1)

=

2010∑k=0

(2010

k

)(−13

)k (23

)2010−k

=(−13

+23

)2010

=(

13

)2010

Summing these two values and dividing by 2 gives the answer 1+( 13 )2010

2 = 32010+12·32010 .

Theme Round

3rdAnnual Harvard-MIT November TournamentSunday 7 November 2010

Guts Round

1. [5] David, Delong, and Justin each showed up to a problem writing session at a random time duringthe session. If David arrived before Delong, what is the probability that he also arrived before Justin?

2. [5] A circle of radius 6 is drawn centered at the origin. How many squares of side length 1 and integercoordinate vertices intersect the interior of this circle?

3. [5] Jacob flipped a fair coin five times. In the first three flips, the coin came up heads exactly twice. Inthe last three flips, the coin also came up heads exactly twice. What is the probability that the thirdflip was heads?

4. [6] Let x be a real number. Find the maximum value of 2x(1−x).

5. [6] An icosahedron is a regular polyhedron with twenty faces, all of which are equilateral triangles. Ifan icosahedron is rotated by θ degrees around an axis that passes through two opposite vertices sothat it occupies exactly the same region of space as before, what is the smallest possible positive valueof θ?

6. [6] How many ordered pairs (S, T ) of subsets of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} are there whose union containsexactly three elements?

7. [7] Let f(x, y) = x2 + 2x + y2 + 4y. Let (x1, y1), (x2, y2), (x3, y3), and (x4, y4) be the vertices of asquare with side length one and sides parallel to the coordinate axes. What is the minimum value off(x1, y1) + f(x2, y2) + f(x3, y3) + f(x4, y4)?

8. [7] What is the sum of all four-digit numbers that are equal to the cube of the sum of their digits(leading zeros are not allowed)?

9. [7] How many functions f : {1, 2, . . . , 10} → {1, 2, . . . , 10} satisfy the property that f(i) + f(j) = 11for all values of i and j such that i+ j = 11.

10. [8] What is the smallest integer greater than 10 such that the sum of the digits in its base 17 repre-sentation is equal to the sum of the digits in its base 10 representation?

11. [8] How many nondecreasing sequences a1, a2, . . . , a10 are composed entirely of at most three dis-tinct numbers from the set {1, 2, . . . , 9} (so 1, 1, 1, 2, 2, 2, 3, 3, 3, 3 and 2, 2, 2, 2, 5, 5, 5, 5, 5, 5 are bothallowed)?

12. [8] An ant starts at the origin of a coordinate plane. Each minute, it either walks one unit to the rightor one unit up, but it will never move in the same direction more than twice in the row. In how manydifferent ways can it get to the point (5, 5)?

13. [8] How many sequences of ten binary digits are there in which neither two zeroes nor three ones everappear in a row?

14. [8] The positive integer i is chosen at random such that the probability of a positive integer k beingchosen is 3

2 times the probability of k+ 1 being chosen. What is the probability that the ith digit afterthe decimal point of the decimal expansion of 1

7 is a 2?

15. [8] Distinct points A,B,C,D are given such that triangles ABC and ABD are equilateral and bothare of side length 10. Point E lies inside triangle ABC such that EA = 8 and EB = 3, and point Flies inside triangle ABD such that FD = 8 and FB = 3. What is the area of quadrilateral AEFD?

16. [9] Triangle ABC is given in the plane. Let AD be the angle bisector of ∠BAC; let BE be the altitudefrom B to AD, and let F be the midpoint of AB. Given that AB = 28, BC = 33, CA = 37, what isthe length of EF?

Page 1 of 3

17. [9] A triangle with side lengths 5, 7, 8 is inscribed in a circle C. The diameters of C parallel to thesides of lengths 5 and 8 divide C into four sectors. What is the area of either of the two smaller ones?

18. [9] Jeff has a 50 point quiz at 11 am. He wakes up at a random time between 10 am and noon, thenarrives at class 15 minutes later. If he arrives on time, he will get a perfect score, but if he arrives morethan 30 minutes after the quiz starts, he will get a 0, but otherwise, he loses a point for each minutehe’s late (he can lose parts of one point if he arrives a nonintegral number of minutes late). What isJeff’s expected score on the quiz?

19. [11] How many 8-digit numbers begin with 1, end with 3, and have the property that each successivedigit is either one more or two more than the previous digit, considering 0 to be one more than 9?

20. [11] Given a permutation π of the set {1, 2, . . . , 10}, define a rotated cycle as a set of three integersi, j, k such that i < j < k and π(j) < π(k) < π(i). What is the total number of rotated cycles over allpermutations π of the set {1, 2, . . . , 10}?

21. [11] George, Jeff, Brian, and Travis decide to play a game of hot potato. They begin by arrangingthemselves clockwise in a circle in that order. George and Jeff both start with a hot potato. On histurn, a player gives a hot potato (if he has one) to a randomly chosen player among the other three (ifa player has two hot potatoes on his turn, he only passes one). If George goes first, and play proceedesclockwise, what is the probability that Travis has a hot potato after each player takes one turn?

22. [12] Let g1(x) = 13 (1 + x + x2 + · · · ) for all values of x for which the right hand side converges. Let

gn(x) = g1(gn−1(x)) for all integers n ≥ 2. What is the largest integer r such that gr(x) is defined forsome real number x?

23. [12] Let a1, a2, . . . be an infinite sequence of positive integers such that for integers n > 2, an =3an−1 − 2an−2. How many such sequences {an} are there such that a2010 ≤ 22012?

24. [12] Let P (x) be a polynomial of degree at most 3 such that P (x) = 11+x+x2 for x = 1, 2, 3, 4. What is

P (5)?

25. [14] Triangle ABC is given with AB = 13, BC = 14, CA = 15. Let E and F be the feet of thealtitudes from B and C, respectively. Let G be the foot of the altitude from A in triangle AFE. FindAG.

26. [14] w, x, y, z are real numbers such that

w + x+ y + z = 52w + 4x+ 8y + 16z = 7

3w + 9x+ 27y + 81z = 114w + 16x+ 64y + 256z = 1

What is the value of 5w + 25x+ 125y + 625z?

27. [14] Let f(x) = −x2 + 10x− 20. Find the sum of all 22010 solutions to f(f(. . . (x) . . .))︸ ︷︷ ︸2010fs

= 2.

28. [17] In the game of set, each card has four attributes, each of which takes on one of three values. Aset deck consists of one card for each of the 81 possible four-tuples of attributes. Given a collection of3 cards, call an attribute good for that collection if the three cards either all take on the same valueof that attribute or take on all three different values of that attribute. Call a collection of 3 cardstwo-good if exactly two attributes are good for that collection. How many two-good collections of 3cards are there? The order in which the cards appear does not matter.

29. [17] In the game of Galactic Dominion, players compete to amass cards, each of which is worth acertain number of points. Say you are playing a version of this game with only two kinds of cards,planet cards and hegemon cards. Each planet card is worth 2010 points, and each hegemon card is

Page 2 of 3

worth four points per planet card held. You start with no planet cards and no hegemon cards, and, oneach turn, starting at turn one, you take either a planet card or a hegemon card, whichever is worthmore points given the hand you currently hold. Define a sequence {an} for all positive integers n bysetting an to be 0 if on turn n you take a planet card and 1 if you take a hegemon card. What is thesmallest value of N such that the sequence aN , aN+1, . . . is necessarily periodic (meaning that there isa positive integer k such that an+k = an for all n ≥ N)?

30. [17] In the game of projective set, each card contains some nonempty subset of six distinguishabledots. A projective set deck consists of one card for each of the 63 possible nonempty subsets of dots.How many collections of five cards have an even number of each dot? The order in which the cardsappear does not matter.

31. [20] What is the perimeter of the triangle formed by the points of tangency of the incircle of a 5-7-8triangle with its sides?

32. [20] Let T be the set of numbers of the form 2a3b where a and b are integers satisfying 0 ≤ a, b ≤ 5.How many subsets S of T have the property that if n is in S then all positive integer divisors of n arein S?

33. [20] Convex quadrilateral BCDE lies in the plane. Lines EB and DC intersect at A, with AB = 2,AC = 5, AD = 200, AE = 500, and cos ∠BAC = 7

9 . What is the largest number of nonoverlappingcircles that can lie in quadrilateral BCDE such that all of them are tangent to both lines BE andCD?

34. [25] Estimate the sum of all the prime numbers less than 1, 000, 000. If the correct answer is X andyou write down A, your team will receive min

(b 25X

A c, b25AX c

)points, where bxc is the largest integer

less than or equal to x.

35. [25] A mathematician M ′ is called a descendent of mathematician M if there is a sequence of math-ematicians M = M1,M2, . . . ,Mk = M ′ such that Mi was Mi+1’s doctoral advisor for all i. Estimatethe number of descendents that the mathematician who has had the largest number of descendentshas had, according to the Mathematical Genealogy Project. Note that the Mathematical GenealogyProject has records dating back to the 1300s. If the correct answer is X and you write down A, yourteam will receive max

(25− b |X−A|

100 c, 0)

points, where bxc is the largest integer less than or equal tox.

36. [25] Paul Erdős was one of the most prolific mathematicians of all time and was renowned for hismany collaborations. The Erdős number of a mathematician is defined as follows. Erdős has an Erdősnumber of 0, a mathematician who has coauthored a paper with Erdős has an Erdős number of 1,a mathematician who has not coauthored a paper with Erdős, but has coauthored a paper with amathematician with Erdős number 1 has an Erdős number of 2, etc. If no such chain exists betweenErdős and another mathematician, that mathematician has an Erdős number of infinity. Of themathematicians with a finite Erdős number (including those who are no longer alive), what is theiraverage Erdős number according to the Erdős Number Project? If the correct answer is X and youwrite down A, your team will receive max (25− b100|X −A|c, 0) points where bxc is the largest integerless than or equal to x.

Page 3 of 3

3rdAnnual Harvard-MIT November TournamentSunday 7 November 2010

Guts Round

1. [5] David, Delong, and Justin each showed up to a problem writing session at a random time duringthe session. If David arrived before Delong, what is the probability that he also arrived before Justin?

Answer: 23 Let t1 be the time that David arrives, let t2 be the time that Delong arrives, and let t3 be

the time that Justin arrives. We can assume that all times are pairwise distinct because the probabilityof any two being equal is zero. Because the times were originally random and independent before wewere given any information, then all orders t1 < t2 < t3, t1 < t3 < t2, t3 < t1 < t2, t2 < t1 < t3,t2 < t3 < t1, t3 < t2 < t1 must all be equally likely. Since we are given that t1 < t2, then we only havethe first three cases to consider, and t1 < t3 in two cases of these three. Thus, the desired probabilityis 2

3 .

2. [5] A circle of radius 6 is drawn centered at the origin. How many squares of side length 1 and integercoordinate vertices intersect the interior of this circle?

Answer: 132 By symmetry, the answer is four times the number of squares in the first quadrant.Let’s identify each square by its coordinates at the bottom-left corner, (x, y). When x = 0, we canhave y = 0 . . . 5, so there are 6 squares. (Letting y = 6 is not allowed because that square intersectsonly the boundary of the circle.) When x = 1, how many squares are there? The equation of the circleis y =

√36− x2 =

√36− 12 is between 5 and 6, so we can again have y = 0 . . . 5. Likewise for x = 2

and x = 3. When x = 4 we have y =√

20 which is between 4 and 5, so there are 5 squares, and whenx = 5 we have y =

√11 which is between 3 and 4, so there are 4 squares. Finally, when x = 6, we have

y = 0, and no squares intersect the interior of the circle. This gives 6 + 6 + 6 + 6 + 5 + 4 = 33. Sincethis is the number in the first quadrant, we multiply by four to get 132.

3. [5] Jacob flipped a fair coin five times. In the first three flips, the coin came up heads exactly twice. Inthe last three flips, the coin also came up heads exactly twice. What is the probability that the thirdflip was heads?

Answer: 45 How many sequences of five flips satisfy the conditions, and have the third flip be heads?

We have H , so exactly one of the first two flips is heads, and exactly one of the last two flips isheads. This gives 2× 2 = 4 possibilities. How many sequences of five flips satisfy the conditions, andhave the third flip be tails? Now we have T , so the first two and the last two flips must all be heads.This gives only 1 possibility. So the probability that the third flip was heads is 4

(4+1) = 45

4. [6] Let x be a real number. Find the maximum value of 2x(1−x).

Answer: 4√

2 Consider the function 2y. This is monotonically increasing, so to maximize 2y, yousimply want to maximize y. Here, y = x(1 − x) = −x2 + x is a parabola opening downwards. Thevertex of the parabola occurs at x = (−1)/(−2) = 1/2, so the maximum value of the function is2(1/2)(1/2) = 4

√2.

5. [6] An icosahedron is a regular polyhedron with twenty faces, all of which are equilateral triangles. Ifan icosahedron is rotated by θ degrees around an axis that passes through two opposite vertices sothat it occupies exactly the same region of space as before, what is the smallest possible positive valueof θ?

Answer: 72◦ Because this polyhedron is regular, all vertices must look the same. Let’s consider justone vertex. Each triangle has a vertex angle of 60◦, so we must have fewer than 6 triangles; if we had6, there would be 360◦ at each vertex and you wouldn’t be able to “fold” the polyhedron up (that is, itwould be a flat plane). It’s easy to see that we need at least 3 triangles at each vertex, and this givesa triangular pyramid with only 4 faces. Having 4 triangles meeting at each vertex gives an octahedron(two square pyramids with the squares glued together) with 8 faces. Therefore, an icosahedron has 5triangles meeting at each vertex, so rotating by 360◦

5 = 72◦ gives another identical icosahedron.

Guts Round

Alternate solution: Euler’s formula tells us that V −E + F = 2, where an icosahedron has V vertices,E edges, and F faces. We’re told that F = 20. Each triangle has 3 edges, and every edge is common to2 triangles, so E = 3(20)

2 = 30. Additionally, each triangle has 3 vertices, so if every vertex is commonto n triangles, then V = 3(20)

n = 60n . Plugging this into the formula, we have 60

n − 30 + 20 = 2, so60n = 12 and n = 5. Again this shows that the rotation is 360◦

5 = 72◦

6. [6] How many ordered pairs (S, T ) of subsets of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} are there whose union containsexactly three elements?

Answer: 3240 Let the three elements in the union be a, b, and c. We know that a can be only inS, only in T , or both, so there are 3 possibilities for placing it. (Recall that S = {a}, T = {b, c} isdifferent from S = {b, c}, T = {a} because S and T are an ordered pair.) Likewise for b and c. Theother 7 elements are in neither S nor T , so there is only 1 possibility for placing them. This gives33 = 27 ways to pick S and T once you’ve picked the union. There are

(103

)= 120 ways to pick the

elements in the union, so we have 120× 27 = 3240 ways total.

7. [7] Let f(x, y) = x2 + 2x + y2 + 4y. Let (x1, y1), (x2, y2), (x3, y3), and (x4, y4) be the vertices of asquare with side length one and sides parallel to the coordinate axes. What is the minimum value off(x1, y1) + f(x2, y2) + f(x3, y3) + f(x4, y4)?

Answer: -18 The square’s corners must be at (x, y), (x+ 1, y), (x+ 1, y+ 1), and (x, y+ 1) for somex and y. So,

f(x1, y1) + f(x2, y2) + f(x3, y3) + f(x4, y4)

= 2(x2 + 2x) + 2((x+ 1)2 + 2(x+ 1)) + 2(y2 + 4y) + 2((y + 1)2 + 4(y + 1))

= 4x2 + 12x+ 6 + 4y2 + 20y + 10

= (2x+ 3)2 − 3 + (2y + 5)2 − 15≥ −18

This attains its minimum value of −18 when x = − 32 and y = − 5

2 .

8. [7] What is the sum of all four-digit numbers that are equal to the cube of the sum of their digits(leading zeros are not allowed)?

Answer: 10745 We want to find all integers x between 1000 and 9999 that are the cube of the sumof their digits. Of course, our search is only restricted to perfect cubes. The smallest such cube is103 = 1000 and the largest such cube is 213 = 9261. This means we only have to check 12 differentcubes, which is quite doable, but we can reduce the search even further with a little number theory.

Suppose we write our number as x = 1000a+ 100b+ 10c+d, where a, b, c, and d are the decimal digitsof x. Then we have

(a+ b+ c+ d)3 ≡ 1000a+ 100b+ 10c+ d ≡ a+ b+ c+ d (mod 9)

If we let k = a + b + c + d, then k must be a solution to the modular equation k3 ≡ k (mod 9). Aquick check of the values 0 through 8 shows that the only solutions are 0, 1, and 8.

Now, in our search, we only have to check values that are the cube of a number which is either 0, 1, or8 mod 9.

103 = 1000, but 1 + 0 + 0 + 0 6= 10.173 = 4913, and 4 + 9 + 1 + 3 = 17.183 = 5832, and 5 + 8 + 3 + 2 = 18.193 = 6859, but 6 + 8 + 5 + 9 6= 19.

So the only solutions are 4913 and 5832, which sum to 10745.

Guts Round

9. [7] How many functions f : {1, 2, . . . , 10} → {1, 2, . . . , 10} satisfy the property that f(i) + f(j) = 11for all values of i and j such that i+ j = 11.

Answer: 100000 To construct such a function f , we just need to choose a value for f(x) from{1, 2, ..., 10} for each x ∈ {1, 2, ..., 10}. But the condition that f(i) + f(j) = 11 whenever i + j = 11means that

f(10) = 11− f(1).f(9) = 11− f(2).

...f(6) = 11− f(5).

This means that once we have chosen f(1), f(2), f(3), f(4), and f(5), the five remaining values off(6), f(7), f(8), f(9), and f(10) are already determined. The answer is therefore just the numberof ways to choose these first five values. Since there are 10 possibilities for each one, we get that theanswer is 105 = 100000.

10. [8] What is the smallest integer greater than 10 such that the sum of the digits in its base 17 repre-sentation is equal to the sum of the digits in its base 10 representation?

Answer: 153 We assume that the answer is at most three digits (in base 10). Then our desirednumber can be expressed in the form abc10 = def17, where a, b, c are digits in base 10, and d, e, f aredigits in base 17. These variables then satisfy the equations

100a+ 10b+ c = 289d+ 17e+ f,

a+ b+ c = d+ e+ f.

Subtracting the second equation from the first, we obtain 99a + 9b = 288d + 16e, or 9(11a + b) =16(18d + e). From this equation, we find that 11a + b must be divisible by 16, and 18d + e must bedivisible by 9. To minimize abc, we find the minimal possible value of a: If a = 0, then the only way for11a+ b = b to be divisible by 16 is to set b = 0; however, this is disallowed by the problem condition,which stipulates that the number must be greater than 10. If we try a = 1, then we find that theonly possible value of b which lets 11a+ b = b+ 11 be divisible by 16 is b = 5. Plugging these in andsimplifying, we find that we must have 18d+ e = 9. The only possible solution to this is d = 0, e = 9.Now to satisfy a+ b+ c = d+ e+ f , we must have 1 + 5 + c = 0 + 9 + f , or c = f + 3. The minimalpossible solution to this is c = 3, f = 0. So our answer is abc = 153, which is also equal to 09017.

11. [8] How many nondecreasing sequences a1, a2, . . . , a10 are composed entirely of at most three distinctnumbers from the set {1, 2, . . . , 9} (so 1, 1, 1, 2, 2, 2, 3, 3, 3, 3 and 2, 2, 2, 2, 5, 5, 5, 5, 5, 5 are both allowed)?

Answer: 3357 From any sequence a1, a2, . . . , a10, construct a sequence b1, b2, . . . , b9, where bi countsthe number of times i occurs in the sequence. There is a correspondence from all possible sequencesb1, b2, . . . , b9 with at most 3 nonzero terms which add to 10, since any sequence of a1, a2, . . . , a10 willbe converted to this form, and from any sequence b1, b2, . . . , b9, we can construct a unique sequence ofa-s by listing i bi times (for 1 ≤ i ≤ 9) in nondecreasing order.

Our goal now is to count the number of possible sequences b1, b2, . . . , b9 meeting our conditions. Wecasework on the number of nonzero terms in the sequence:Case 1: The sequence has exactly one nonzero term.Then exactly one of b1, b2, . . . , b9 is equal to 10, and all the rest are equal to 0. This gives us 9 possiblesequences in this case.Case 2: The sequence has exactly two nonzero terms.There are

(92

)= 36 ways to choose the two terms bi, bj (i < j) which are nonzero. From here, we have

9 choices for the value of bi, namely 1 through 9 (since both bi and bj must be nonzero), and bj willbe fixed, so this case gives us 36 · 9 = 324 possible sequences.Case 3: The sequence has exactly three nonzero terms.There are

(93

)= 84 ways to choose the three terms bi, bj , bk (i < j < k) which are nonzero. Letting

Guts Round

ci = bi − 1, cj = bj − 1, ck = bk − 1, we have that ci, cj , ck are nonnegative integers which sum to 7.There are

(92

)= 36 solutions to this equation (consider placing two dividers in the nine spaces between

the ten elements), giving 84 · 36 = 3024 possbilities in this case.

We then have 9 + 324 + 3024 = 3357 possible sequences.

12. [8] An ant starts at the origin of a coordinate plane. Each minute, it either walks one unit to the rightor one unit up, but it will never move in the same direction more than twice in the row. In how manydifferent ways can it get to the point (5, 5)?

Answer: 84 We can change the ant’s sequence of moves to a sequence a1, a2, . . . , a10, with ai = 0 ifthe i-th step is up, and ai = 1 if the i-th step is right. We define a subsequence of moves ai, ai+1, . . . , aj ,(i ≤ j) as an up run if all terms of the subsequence are equal to 0, and ai−1 and aj+1 either do notexist or are not equal to 0, and define a right run similarly. In a sequence of moves, up runs and rightruns alternate, so the number of up rights can differ from the number of right runs by at most one.

Now let f(n) denote the number of sequences a1, a2, . . . , an where ai ∈ {1, 2} for 1 ≤ i ≤ n, anda1 + a2 + · · ·+ an = 5. (In essence, we are splitting the possible 5 up moves into up runs, and we aredoing the same with the right moves). We can easily compute that f(3) = 3, f(4) = 4, f(5) = 1, andf(n) = 0 otherwise.

For each possible pair of numbers of up runs and right runs, we have two choices of which type of run isfirst. Our answer is then 2(f(3)2 + f(3)f(4) + f(4)2 + f(4)f(5) + f(5)2) = 2(9 + 12 + 16 + 4 + 1) = 84.

13. [8] How many sequences of ten binary digits are there in which neither two zeroes nor three ones everappear in a row?

Answer: 28 Let an be the number of binary sequences of length n satisfying the conditions andending in 0, let bn be the number ending in 01, and let cn be the number ending in 11. From thelegal sequences of length 2 01, 11, 10, we find that a2 = b2 = c2 = 1. We now establish a recursionby building sequences of length n + 1 from sequences of length n. We can add a 0 to a sequence oflength n if and only if it ended with a 1, so an+1 = bn + cn. We can have a sequence of length n + 1ending with 01 only by adding a 1 to a sequence of length n ending in 0, so bn+1 = an. We can havea sequence of length n+ 1 ending with 11 only by adding a 1 to a sequence of length n ending in 01,so cn+1 = bn. We can now run the recursion:

n an bn cn2 1 1 13 2 1 14 2 2 15 3 2 26 4 3 27 5 4 38 7 5 49 9 7 510 12 9 7

Our answer is then 12 + 9 + 7 = 28.

14. [8] The positive integer i is chosen at random such that the probability of a positive integer k beingchosen is 3

2 times the probability of k+ 1 being chosen. What is the probability that the ith digit afterthe decimal point of the decimal expansion of 1

7 is a 2?

Answer: 108665 First we note that the probability that n is picked is 1

2 ×(

23

)n, because this is the

sequence whose terms decrease by a factor of 23 each time and whose sum is 1 (recall that probabilities

must sum to 1).

Now note that 17 = .142857142857 . . ., meaning that 2 occurs at digits 3, 9, 15, 21, etc. We can then

calculate the probability that we ever pick 2 as

Guts Round

∞∑k=0

12·(

23

)6k+3

=427

∞∑k=0

(23

)6k

=427· 1

1−(

23

)6=

427· 729

729− 64

=427· 729

665

=108665

.

15. [8] Distinct points A,B,C,D are given such that triangles ABC and ABD are equilateral and bothare of side length 10. Point E lies inside triangle ABC such that EA = 8 and EB = 3, and point Flies inside triangle ABD such that FD = 8 and FB = 3. What is the area of quadrilateral AEFD?

Answer: 91√

34 Since AEB ∼= DFB, we have ∠EBA = ∠FBD. Thus, ∠EBF = ∠EBA+∠ABF =

∠FBD+∠ABF = ∠ABD = 60◦. Since EB = BF = 3, this means that EBF is an equilateral triangleof side length 3. Now we have [AEFD] = [AEBD]− [EBF ]− [FBD] = [AEB] + [ABD]− [EBF ]−[FBD] = [ABD]− [EBF ] =

√3

4 (102 − 32) = 91√

34 .

16. [9] Triangle ABC is given in the plane. Let AD be the angle bisector of ∠BAC; let BE be the altitudefrom B to AD, and let F be the midpoint of AB. Given that AB = 28, BC = 33, CA = 37, what isthe length of EF?

Answer: 14 4ABE is a right triangle, and F is the midpoint of the hypotenuse (and therefore thecircumcenter), so EF = BF = AF = 14.

17. [9] A triangle with side lengths 5, 7, 8 is inscribed in a circle C. The diameters of C parallel to thesides of lengths 5 and 8 divide C into four sectors. What is the area of either of the two smaller ones?

Answer: 4918π Let 4PQR have sides p = 7, q = 5, r = 8. Of the four sectors determined by

the diameters of C that are parallel to PQ and PR, two have angles equal to P and the other twohave angles equal to π − P . We first find P using the law of cosines: 49 = 25 + 64 − 2(5)(8) cosPimplies cosP = 1

2 implies P = π3 . Thus the two smaller sectors will have angle π

3 . Next we find thecircumradius of 4PQR using the formula R = pqr

4[PQR] , where [PQR] is the area of 4PQR. By Heron’s

Formula we have [PQR] =√

10(5)(3)(2) = 10√

3; thus R = 5·7·84(10√

3)= 7√

3. The area of a smaller sector

is thus π/32π

(πR2

)= π

6

(7√3

)2

= 4918π

18. [9] Jeff has a 50 point quiz at 11 am. He wakes up at a random time between 10 am and noon, thenarrives at class 15 minutes later. If he arrives on time, he will get a perfect score, but if he arrives morethan 30 minutes after the quiz starts, he will get a 0, but otherwise, he loses a point for each minutehe’s late (he can lose parts of one point if he arrives a nonintegral number of minutes late). What isJeff’s expected score on the quiz?

Answer: 552 If Jeff wakes up between 10:00 and 10:45, he gets 50. If he wakes up between 10:45

and 11:15, and he wakes up k minutes after 10:45, then he gets 50− k points. Finally, if he wakes upbetween 11:15 and 12:00 he gets 0 points. So he has a 3

8 probability of 50, a 38 probability of 0, and a 1

4probability of a number chosen uniformly between 20 and 50 (for an average of 35). Thus his expectedscore is 3

8 × 50 + 14 × 35 = 75+35

4 = 1104 = 55

2 .

19. [11] How many 8-digit numbers begin with 1, end with 3, and have the property that each successivedigit is either one more or two more than the previous digit, considering 0 to be one more than 9?

Guts Round

Answer: 21 Given an 8-digit number a that satifies the conditions in the problem, let ai denotethe difference between its (i + 1)th and ith digit. Since i ∈ {1, 2} for all 1 ≤ i ≤ 7, we have 7 ≤a1 + a2 + · · · + a7 ≤ 14. The difference between the last digit and the first digit of m is 3 − 1 ≡ 2(mod 10), which means a1 + · · ·+ a7 = 12. Thus, exactly five of the ais equal to 2 and the remainingtwo equal to 1. The number of permutations of five 2s and two 1s is

(72

)= 21.

20. [11] Given a permutation π of the set {1, 2, . . . , 10}, define a rotated cycle as a set of three integersi, j, k such that i < j < k and π(j) < π(k) < π(i). What is the total number of rotated cycles over allpermutations π of the set {1, 2, . . . , 10}?

Answer: 72576000 Let us consider a triple (i, j, k) with i < j < k and determine how manypermutations rotate it. There are

(103

)choices for the values of π(i), π(j), π(k) and the choice of this

set of three determines the values of π(i), π(j), π(k). The other 7 values then have 7! ways to be arranged(any permutation of them will work), so exactly

(103

)7! permutations rotate (i, j, k). Therefore, as there

are(103

)such triples, the total number of rotated triples is

(103

)2 · 7! = 72576000.

21. [11] George, Jeff, Brian, and Travis decide to play a game of hot potato. They begin by arrangingthemselves clockwise in a circle in that order. George and Jeff both start with a hot potato. On histurn, a player gives a hot potato (if he has one) to a randomly chosen player among the other three (ifa player has two hot potatoes on his turn, he only passes one). If George goes first, and play proceedesclockwise, what is the probability that Travis has a hot potato after each player takes one turn?

Answer: 527 Notice that Travis can only have the hot potato at the end if he has two potatoes

before his turn. A little bit of casework shows that this can only happen whenCase 1 : George gives Travis his potato, while Jeff gives Brian his potato, which in then goes to Travis.The probability of this occuring is

(13

)3 = 127

Case 2 : George gives Travis his potato, while Jeff gives Travis his potato. The probability of thisoccuring is

(13

)2 = 19

Case 3 : George gives Brian his potato, Jeff gives Travis his potato, and then Brian gives Travis hispotato. The probability of this occuring is 1

27Because these events are all disjoint, the probability that Travis ends up with the hot potato is 5

27

22. [12] Let g1(x) = 13 (1 + x + x2 + · · · ) for all values of x for which the right hand side converges. Let

gn(x) = g1(gn−1(x)) for all integers n ≥ 2. What is the largest integer r such that gr(x) is defined forsome real number x?

Answer: 5 Notice that the series is geometric with ratio x, so it converges if −1 < x < 1. Alsonotice that where g1(x) is defined, it is equal to 1

3(1−x) . The image of g1(x) is then the interval ( 16 ,∞).

The image of g2(x) is simply the values of g1(x) for x in ( 16 , 1), which is the interval ( 2

5 ,∞). Similarly,the image of g3(x) is ( 5

9 ,∞), the image of g4(x) is ( 34 ,∞), and the image of g5(x) is ( 4

3 ,∞). As thisdoes not intersect the interval (−1, 1), g6(x) is not defined for any x, so the answer is 5.

23. [12] Let a1, a2, . . . be an infinite sequence of positive integers such that for integers n > 2, an =3an−1 − 2an−2. How many such sequences {an} are there such that a2010 ≤ 22012?

Answer: 36 · 22009 + 36 Consider the characteristic polynomial for the recurrence an+2 − 3an+1 +2an = 0, which is x2 − 3x + 2. The roots are at 2 and 1, so we know that numbers ai must be ofthe form ai = a2i−1 + b for integers a and b. Therefore a2010 must equal to a22009 + b, where a andb are both integers. If the expression is always positive, it is sufficient to say a1 is positive and a isnonnegative, or a+ b > 0, and a ≥ 0.For a given value of a, 1−a ≤ b ≤ 22012−a22009, so there are 22012−a22009 +a possible values of b foreach a (where the quantity is positive). a can take any value between 0 and 23, we sum over all such ain this range, to attain 9 · 22012 − (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8)22009 + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8),or 36

(22009

)+ 36, which is our answer.

24. [12] Let P (x) be a polynomial of degree at most 3 such that P (x) = 11+x+x2 for x = 1, 2, 3, 4. What is

P (5)?

Guts Round

Answer: −391 The forward difference of a polynomial P is ∆P (x) = P (x+ 1)−P (x), which is a new

polynomial with degree reduced by one. Therefore, if we apply this operation three times we’ll get aconstant function, and we can work back up to get a value of P (5). Practically, we create the followingtable of differences:

13

17

113

121−4

21−691

−8273

34273

10273−24

273

Then extend it to be the following table:

13

17

113

121

−9273−4

21−691

−8273

−22273

34273

10273

−14273−24

273−24273

So our answer is −9273 = −3

91

25. [14] Triangle ABC is given with AB = 13, BC = 14, CA = 15. Let E and F be the feet of thealtitudes from B and C, respectively. Let G be the foot of the altitude from A in triangle AFE. FindAG.

Answer: 39665 By Heron’s formula we have [ABC] =

√21(8)(7)(6) = 84. Let D be the foot of the

altitude from A to BC; then AD = 2 · 8414 = 12. Notice that because ∠BFC = ∠BEC, BFEC is

cyclic, so ∠AFE = 90 − ∠EFC = 90 − ∠EBC = ∠C. Therefore, we have 4AEF ∼ 4ABC, soAGAD = AE

AB ; 12 (BE)(AC) = 84 =⇒ BE = 56

5 =⇒ AE =√

132 −(

565

)2 =√

652−562

52 = 335 . Then

AG = AD · AEAB = 12 · 33/513 = 396

65 .

26. [14] w, x, y, z are real numbers such that

w + x+ y + z = 52w + 4x+ 8y + 16z = 7

3w + 9x+ 27y + 81z = 114w + 16x+ 64y + 256z = 1

What is the value of 5w + 25x+ 125y + 625z?

Answer: −60 We note this system of equations is equivalent to evaluating the polynomial (in a)P (a) = wa+xa2 +ya3 +za4 at 1, 2, 3, and 4. We know that P (0) = 0, P (1) = 5, P (2) = 7, P (3) = 11,P (4) = 1. The finite difference of a polynomial f is f(n + 1) − f(n), which is a polynomial withdegree one less than the degree of f . The second, third, etc finite differences come from applying thisoperation repeatedly. The fourth finite difference of this polynomial is constant because this is a fourthdegree polynomial. Repeatedly applying finite differences, we get

0 5 7 11 15 2 4 −10−3 2 −14

5 −16−21

and we see that the fourth finite difference is −21. We can extend this table, knowing that the fourthfinite difference is always -21, and we find that that P (5) = −60. The complete table is

Guts Round

0 5 7 11 1 −605 2 4 −10 −61−3 2 −14 −51

5 −16 −37−21 −21

27. [14] Let f(x) = −x2 + 10x− 20. Find the sum of all 22010 solutions to f(f(. . . (x) . . .))︸ ︷︷ ︸2010fs

= 2.

Answer: 5 · 22010 Define g(x) = f(f(. . . (x) . . .)). We calculate:

f(10−x) = −(10−x)2 + 10(10−x)−20 = −100 + 20x−x2 + 100−10x−20 = −x2 + 10x−20 = f(x).

This implies that g(10 − x) = g(x). So if g(x) = 2, then g(10 − x) = 2. Moreover, we can calculatef(5) = −25 + 50 − 20 = 5, so g(5) = 5 6= 2. Thus the possible solutions to g(x) = 2 can be groupedinto pairs, (x1, 10 − x1), (x2, 10 − x2), . . .. The sum of the members of each pair is 10, and there are22009 pairs, so the sum is

10 · 22009 = 5 · 22010.

28. [17] In the game of set, each card has four attributes, each of which takes on one of three values. Aset deck consists of one card for each of the 81 possible four-tuples of attributes. Given a collection of3 cards, call an attribute good for that collection if the three cards either all take on the same valueof that attribute or take on all three different values of that attribute. Call a collection of 3 cardstwo-good if exactly two attributes are good for that collection. How many two-good collections of 3cards are there? The order in which the cards appear does not matter.

Answer: 25272 In counting the number of sets of 3 cards, we first want to choose which of ourtwo attributes will be good and which of our two attributes will not be good. There are

(42

)= 6 such

choices.

Now consider the two attributes which are not good, attribute X and attribute Y. Since these are notgood, some value should appear exactly twice. Suppose the value a appears twice and b appears oncefor attribute X and that the value c appears twice and d appears once for attribute Y . There are threechoices for a and then two choices for b; similarly, there are three choices for c and then two choicesfor d. This gives 3 · 2 · 3 · 2 = 36 choices of a, b, c, and d.

There are two cases to consider. The first is that there are two cards which both have a and c, whilethe other card has both b and d. The second case is that only one card has both a and c, while onecard has a and d and the other has b and c.

Case 1 :

Card 1 Card 2 Card 3— Good attribute 1 —— Good attribute 2 —a a bc c d

The three cards need to be distinct. Card 3 is necessarily distinct from Card 1 and Card 2, but weneed to ensure that Card 1 and Card 2 are distinct from each other. There are 9 choices for the twogood attributes of Card 1, and then 8 choices for the two good attributes of Card 2. But we also wantto divide by 2 since we do not care about the order of Card 1 and Card 2. So there are 9·8

2 = 36 choicesfor the good attributes on Card 1 and Card 2. Then, the values of the good attributes of Card 1 andCard 2 uniquely determine the values of the good attributes of Card 3.

Case 2 :

Guts Round

Card 1 Card 2 Card 3— Good attribute 1 —— Good attribute 2 —a a bc d c

Card 1, Card 2, and Card 3 will all be distinct no matter what the values of the good attributes are,because the values of attributes X and Y are unique to each card. So there are 9 possibilities for thethe values of the good attributes on card 1, and then there are 9 more possibilities for the values ofthe good attribute on Card 2. We do not have to divide by 2 this time, since Card 1 and Card 2 havedistinct values in X and Y . So there are 92 = 81 possibilities here.

So our final answer is 6 · 62 · (36 + 81) = 25272.

29. [17] In the game of Galactic Dominion, players compete to amass cards, each of which is worth acertain number of points. Say you are playing a version of this game with only two kinds of cards,planet cards and hegemon cards. Each planet card is worth 2010 points, and each hegemon card isworth four points per planet card held. You start with no planet cards and no hegemon cards, and, oneach turn, starting at turn one, you take either a planet card or a hegemon card, whichever is worthmore points given the hand you currently hold. Define a sequence {an} for all positive integers n bysetting an to be 0 if on turn n you take a planet card and 1 if you take a hegemon card. What is thesmallest value of N such that the sequence aN , aN+1, . . . is necessarily periodic (meaning that there isa positive integer k such that an+k = an for all n ≥ N)?

Answer: 503 If you have P planets and H hegemons, buying a planet gives you 2010 + 4H pointswhile buying a hegemon gives you 4P points. Thus you buy a hegemon whenever P −H ≥ 502.5, andyou buy a planet whenever P − H ≤ 502.5. Therefore ai = 1 for 1 ≤ i ≤ 503. Starting at i = 504(at which point you have bought 503 planets) you must alternate buying planets and hegemons. Thesequence {ai}i≥503 is periodic with period 2.

30. [17] In the game of projective set, each card contains some nonempty subset of six distinguishabledots. A projective set deck consists of one card for each of the 63 possible nonempty subsets of dots.How many collections of five cards have an even number of each dot? The order in which the cardsappear does not matter.

Answer: 109368 We’ll first count sets of cards where the order does matter. Suppose we choosethe first four cards. Then there is exactly one card that can make each dot appear twice. However,this card could be empty or it could be one of the cards we’ve already chosen, so we have to subtractfor these two cases. First, there are 63 · 62 · 61 · 60 ways to choose the first four cards. Let’s now counthow many ways there are that the fifth card could be empty.

The fifth card is empty if and only if the first four cards already have an even number of each dot.Suppose we choose the first two cards. There is a possible fourth card if the third card is not either ofthe first two or the card that completes a set. If that is the case, then the fourth card is unique. Thiscomes to 63 · 62 · 60 cases.

Now consider how many ways there are for the fifth card to be a duplicate. This is just the number ofways for three cards to have an even number of each dot, then have two copies of the same card in theother two slots, one of which needs to be the fifth slot. The number of ways for three cards to havean even number of each dot is just the number of ways to choose two cards. Therefore, we’ll choosetwo cards (63 · 62 ways), choose the slot in the first four positions for the duplicate card (4 ways), andthe duplicate card, which can’t be any of the nonduplicated cards, so there are 60 choices. Therefore,there are 63 · 62 · 4 · 60 ways for the fifth card to be the same as one of the first four.

This means that the number of five card sets where the order does matter is 63 ·62 ·61 ·60−63 ·62 ·60−63 ·62 ·4 ·60, so our final answer is 63·62·61·60−63·62·60−63·62·4·60

120 = 63·62·(61−1−4)2 = 63 ·31 ·56 = 109368.

31. [20] What is the perimeter of the triangle formed by the points of tangency of the incircle of a 5-7-8triangle with its sides?

Guts Round

Answer: 9√

217 + 3 Let 4ABC be a triangle with sides a = 7, b = 5, and c = 8. Let the incircle of

4ABC be tangent to sides BC, CA, and AB at points D, E, and F . By the law of cosines (using theform cos(A) = b2+c2−a2

2bc ), we have

cos(A) =82 + 52 − 72

2(5)(8)=

12

cos(B) =82 + 72 − 52

2(7)(8)=

1114

cos(C) =52 + 72 − 82

2(7)(5)=

17

Now we observe that AEF , BDF , and CDE are all isosceles. Let us call the lengths of the legs ofthese triangles s, t, and u, respectively. Then we know that s + t = 8, t + u = 7, and u + s = 5, sos = 3, t = 5, and u = 2.

Our final observation is that an isosceles angle with legs of length l and whose non-equal angle is θhas a base of length l

√2(1− cos(θ)). This can be proven using the law of cosines or the Pythagorean

theorem.

Using this, we can calculate that

DE = 2√

2(1− cos(C))

= 2

√127

EF = 3√

2(1− cos(A))= 3

FD = 5√

2(1− cos(B))

= 5

√37,

and then

DE + EF + FD = 2

√127

+ 3 + 5

√37

= 3 + 9

√37

= 3 + 9√

217.

32. [20] Let T be the set of numbers of the form 2a3b where a and b are integers satisfying 0 ≤ a, b ≤ 5.How many subsets S of T have the property that if n is in S then all positive integer divisors of n arein S?

Answer: 924 Consider the correspondence (a, b) ↔ 2a3b for non-negative integers a and b. So wecan view T as the square of lattice points (a, b) where 0 ≤ a, b ≤ 5, and subsets of T as subsets of thissquare.

Notice then that the integer corresponding to (a1, b1) is a divisor of the integer corresponding to (a2, b2)if and only if 0 ≤ a1 ≤ a1 and 0 ≤ b1 ≤ b2. This means that subsets S ⊂ T with the desired property,

Guts Round

correspond to subsets of the square where if a point is in the set, then so are all points to the left andsouth of it.

Consider any such subset S. For each 0 ≤ x ≤ 5, let Sx be the maximum y value of any point (x, y) ∈ S,or −1 if there is no such point. We claim the values Sx uniquely characterize S. This is because eachSx characterizes the points of the form (x, y) in S. In particular, (x, z) will be in S if and only ifz ≤ Sx. If (x, z) ∈ S with z > Sx, then Sx is not the maximum value, and if (x, z) /∈ S with z ≤ Sx,then S fails to satisfy the desired property.

We now claim that Sx ≥ Sy for x < y, so the sequence S0, . . . , S1 is decreasing. This is becauseif (y, Sy) is in the set S, then so must be (x, Sy). Conversely, it is easy to see that if S0, . . . , S1 isdecreasing, then S is a set satisfying the desired property.

We now claim that decreasing sequences S0, . . . , S5 are in bijective correspondence with walks goingonly right and down from (−1, 5) to (5,−1). The sequence S0, . . . , S5 simply corresponds to the walk(−1, 5)→ (−1, S0)→ (0, S0)→ (0, S1)→ (1, S1)→ · · · → (4, S5)→ (5, S5)→ (5,−1). Geometrically,we are tracing out the outline of the set S.

The number of such walks is simply(126

), since we can view it as choosing the 6 of 12 steps at which

to move right. Thus the number of subsets S of T with the desired property is(126

)= 924.

33. [20] Convex quadrilateral BCDE lies in the plane. Lines EB and DC intersect at A, with AB = 2,AC = 5, AD = 200, AE = 500, and cos ∠BAC = 7

9 . What is the largest number of nonoverlappingcircles that can lie in quadrilateral BCDE such that all of them are tangent to both lines BE andCD?

Answer: 5 Let θ = ∠BAC, and cos θ = 79 implies cos θ2 =

√1+ 7

92 = 2

√2

3 ; sin θ2 = 1

3 ; BC =√4 + 25− 2(2)(5) 7

9 = 113 . Let O1 be the excircle of 4ABC tangent to lines AB and AC, and let r1

be its radius; let O1 be tangent to line AB at point P1. Then AP1 = AB+BC+CA2 and r1

AP1= tan θ

2 =1

2√

2=⇒ r1 = 16

3·2√

2. Let On be a circle tangent to On−1 and the lines AB and AC, and let rn

be its radius; let On be tangent to line AB at point Pn. Then OnPnAOn

= sin θ2 = 1

3 ; since 4APnOn ∼4APn−1On−1 and OnOn−1 = rn + rn−1, we have 1

3 = OnPnAOn

= rnAOn−1+On−1On

= rn3rn−1+rn+rn−1

=⇒rn = 2rn−1 = 2n−1 16

3·2√

2. We want the highest n such that On is contained inside 4ADE. Let the

incircle of 4ADE be tangent to AD at X; then the inradius of 4ADE is AXtan θ

2=

500+200− 11003

2

2√

2= 500

3·2√

2.

We want the highest n such that rn ≤ 5003·2√

2; thus 2n−1 · 16 ≤ 500 =⇒ n = 5.

34. [25] Estimate the sum of all the prime numbers less than 1, 000, 000. If the correct answer is X andyou write down A, your team will receive min

(b 25XA c, b

25AX c

)points, where bxc is the largest integer

less than or equal to x.

Answer: 37550402023 A decent approximation to the sum of all the primes can be obtained withthe following two facts. First, there are approximately n

lnn primes less than n and second, the nth

prime is approximately n lnn. We’ll approximate ln 1000000 as 15 (the actual number is 13.8), so

there are approximately 106

15 primes. Then we want∑ 106

15n=1 n lnn. If you know calculus, this can be

approximated by the integral∫ 106

151

x lnx dx, which has the antiderivative 12x

2 lnx − 14x

2, giving ananswer of around 1

2 ·1012

225 · (15 − ln 15) − 14 ·

1012

225 . Estimating ln 15 as about 3, this is approximately23·1012

900 = 2.5 ·1010 = 25, 000, 000, 000. The actual answer is 37,550,402,023, so an approximation of thisaccuracy would get 16 points. We can arrive at the same answer without calculus by approximating

lnn as∑nk=1

1k , so that our sum becomes

∑ 10615n=1

∑nk=1

nk =

∑ 10615k=1

∑ 10615n=k

nk ≈

∑ 10615k=1

12 ·

1012225 −k

2

k =12 ·

1012

225 · ln( 106

15 )− 12

∑ 10615k=1 k ≈

12 ·

1012

225 · ln( 106

15 )− 14

1012

225 ≈ 2.5 · 1010.

35. [25] A mathematician M ′ is called a descendent of mathematician M if there is a sequence of math-ematicians M = M1,M2, . . . ,Mk = M ′ such that Mi was Mi+1’s doctoral advisor for all i. Estimate

Guts Round

the number of descendents that the mathematician who has had the largest number of descendentshas had, according to the Mathematical Genealogy Project. Note that the Mathematical GenealogyProject has records dating back to the 1300s. If the correct answer is X and you write down A, yourteam will receive max

(25− b |X−A|100 c, 0

)points, where bxc is the largest integer less than or equal to

x.

Answer: 82310 First let’s estimate how many “generations” of mathematicians there have beensince 1300. If we suppose that a mathematician gets his PhD around age 30 and becomes a PhDadvisor around age 60, then we’ll get a generation length of approximately 30 years. However, not allmathematicians will train more than one PhD. Let’s say that only 40% of mathematicians train atleast 2 PhDs. Then effectively we have only 40% of the generations, or in other words each effectivegeneration takes 75 years. Then we have 22

3 branching generations. If we assume that all of these onlytrain 2 PhDs, then we get an answer of 2

223 ≈ 1625. But we can ensure that our chain has at least a

single person who trained 100 PhDs (this is approximately the largest number of advisees for a singlemathematician), allowing us to change one factor of 2 into a factor of 100. That gives us an answer of1625 · 50 = 81250, which is very close to the actual value of 82310.

36. [25] Paul Erdős was one of the most prolific mathematicians of all time and was renowned for hismany collaborations. The Erdős number of a mathematician is defined as follows. Erdős has an Erdősnumber of 0, a mathematician who has coauthored a paper with Erdős has an Erdős number of 1,a mathematician who has not coauthored a paper with Erdős, but has coauthored a paper with amathematician with Erdős number 1 has an Erdős number of 2, etc. If no such chain exists betweenErdős and another mathematician, that mathematician has an Erdős number of infinity. Of themathematicians with a finite Erdős number (including those who are no longer alive), what is theiraverage Erdős number according to the Erdős Number Project? If the correct answer is X and youwrite down A, your team will receive max (25− b100|X −A|c, 0) points where bxc is the largest integerless than or equal to x.

Answer: 4.65 We’ll suppose that each mathematician collaborates with approximately 20 people(except for Erdős himself, of course). Furthermore, if a mathematician has Erdős number k, thenwe’d expect him to be the cause of approximately 1

2kof his collaborators’ Erdős numbers. This is

because as we get to higher Erdős numbers, it is more likely that a collaborator has a lower Erdősnumber already. Therefore, we’d expect about 10 times as many people to have an Erdős number of2 than with an Erdős number of 1, then a ratio of 5, 2.5, 1.25, and so on. This tells us that moremathematicians have an Erdős number of 5 than any other number, then 4, then 6, and so on. If weuse this approximation, we have a ratio of mathematicians with Erdős number 1, 2, and so on of about1 : 10 : 50 : 125 : 156 : 97 : 30 : 4 : 0.3, which gives an average Erdős number of 4.8. This is close to theactual value of 4.65.

Guts Round

3rdAnnual Harvard-MIT November TournamentSunday 7 November 2010

Team Round

Polyhedron Hopping

1. [3] Travis is hopping around on the vertices of a cube. Each minute he hops from the vertex he’scurrently on to the other vertex of an edge that he is next to. After four minutes, what is theprobability that he is back where he started?

2. [6] In terms of k, for k > 0 how likely is he to be back where he started after 2k minutes?

3. [3] While Travis is having fun on cubes, Sherry is hopping in the same manner on an octahedron. Anoctahedron has six vertices and eight regular triangular faces. After five minutes, how likely is Sherryto be one edge away from where she started?

4. [6] In terms of k, for k > 0, how likely is it that after k minutes Sherry is at the vertex opposite thevertex where she started?

Circles in Circles

5. [4] Circle O has chord AB. A circle is tangent to O at T and tangent to AB at X such that AX = 2XB.What is AT

BT ?

6. [6] AB is a diameter of circle O. X is a point on AB such that AX = 3BX. Distinct circles ω1 andω2 are tangent to O at T1 and T2 and to AB at X. The lines T1X and T2X intersect O again at S1

and S2. What is the ratio T1T2S1S2

?

7. [7] ABC is a right triangle with ∠A = 30◦ and circumcircle O. Circles ω1, ω2, and ω3 lie outsideABC and are tangent to O at T1, T2, and T3 respectively and to AB, BC, and CA at S1, S2, and S3,respectively. Lines T1S1, T2S2, and T3S3 intersect O again at A′, B′, and C ′, respectively. What isthe ratio of the area of A′B′C ′ to the area of ABC?

Linear? What’s The Problem?

A function f(x1, x2, . . . , xn) is said to be linear in each of its variables if it is a polynomial such thatno variable appears with power higher than one in any term. For example, 1 + x + xy is linear in xand y, but 1 + x2 is not. Similarly, 2x + 3yz is linear in x, y, and z, but xyz2 is not.

8. [4] A function f(x, y) is linear in x and in y. f(x, y) = 1xy for x, y ∈ {3, 4}. What is f(5, 5)?

9. [5] A function f(x, y, z) is linear in x, y, and z such that f(x, y, z) = 1xyz for x, y, z ∈ {3, 4}. What is

f(5, 5, 5)?

10. [6] A function f(x1, x2, . . . , xn) is linear in each of the xi and f(x1, x2, . . . , xn) = 1x1x2···xn

whenxi ∈ {3, 4} for all i. In terms of n, what is f(5, 5, . . . , 5)?

3rdAnnual Harvard-MIT November TournamentSunday 7 November 2010

Team Round

Polyhedron Hopping

1. [3] Travis is hopping around on the vertices of a cube. Each minute he hops from the vertex he’scurrently on to the other vertex of an edge that he is next to. After four minutes, what is theprobability that he is back where he started?

Answer: 727 Let the cube have vertices all 0 or 1 in the x, y, z, coordinate system. Travis starts at

(0, 0, 0). If after 3 moves he is at (1, 1, 1) he cannot get back to (0, 0, 0). From any other vertex he hasa 1

3 chance of getting back on the final move. There is a 29 chance he ends up at (1, 1, 1), and thus a 7

9chance he does not end up there, and thus a 7

27 chance he ends up at (0, 0, 0).

2. [6] In terms of k, for k > 0 how likely is he to be back where he started after 2k minutes?

Answer: 14 + 3

4

(19

)k Again, Travis starts at (0, 0, 0). At each step, exactly one of the three coordi-nates will change. The parity of the sum of the three coordinates will change at each step, so after 2ksteps, the sum of the coordinates must be even. There are only four possibilites for Travis’s position:(0, 0, 0), (1, 1, 0), (1, 0, 1), and (0, 1, 1). Let pk be the probability that Travis is at (0, 0, 0) after 2ksteps. Then 1 − pk is the probability that he is on (1, 1, 0), (1, 0, 1), or (0, 1, 1). Suppose we want tocompute pk+1. There are two possibilities: we were either at (0, 0, 0) after 2k steps or not. If we were,then there is a 1

3 probability that we will return (since our (2k + 1)th step can be arbitary, but there isa 1

3 chance that we will reverse that as our (2k + 2)th step). If we were not at (0, 0, 0) after our 2kth

steps, then two of our coordinates must have been ones. There is a 23 probability that the (2k + 1)th

step will change one of those to a zero, and there is a 13 step that that the (2k + 2)th step will change

the remaining one. Hence, in this case, there is a(

23

) (13

)= 2

9 probability that Travis ends up (0, 0, 0)in this case. So we have:

pk+1 = pk

(13

)+ (1− pk)

(29

)pk+1 =

19pk +

29(

pk+1 −14

)=

19

(pk −

14

)(We get the value 1

4 either by guessing that the sequence p0, p1, p2, ... should converge to 14 or simply

by solving the equation − 19x + x = 2

9 .) This shows that p0 − 14 , p1 − 1

4 , ... is a geometric series with

ratio 19 . Since p0 − 1

4 = 1− 14 = 3

4 , we get that pk − 14 = 3

4

(19

)k, or that pk = 14 + 3

4

(19

)k.

3. [3] While Travis is having fun on cubes, Sherry is hopping in the same manner on an octahedron. Anoctahedron has six vertices and eight regular triangular faces. After five minutes, how likely is Sherryto be one edge away from where she started?

Answer: 1116 Let the starting vertex be the ’bottom’ one. Then there is a ’top’ vertex, and 4

’middle’ ones. If p(n) is the probability that Sherry is on a middle vertex after n minutes, p(0) = 0,

Team Round

p(n + 1) = (1− p(n)) + p(n) · 12 . This recurrence gives us the following equations.

p(n + 1) = 1− p(n)2

p(0) = 0p(1) = 1

p(2) =12

p(3) =34

p(4) =58

p(5) =1116

4. [6] In terms of k, for k > 0, how likely is it that after k minutes Sherry is at the vertex opposite thevertex where she started?

Answer: 16 + 1

3(−2)k Take p(n) from the last problem. By examining the last move of Sherry, the

probability that she ends up on the original vertex is equal to the probability that she ends up on thetop vertex, and both are equal to 1−p(n)

2 for n ≥ 1.

From the last problem,

p(n + 1) = 1− p(n)2

p(n + 1)− 23

= −12

(p(n)− 2

3

)and so p(n)− 2

3 is a geometric series with ratio − 12 . Since p(0) = 0, we get p(n)− 2

3 = − 23

(− 1

2

)n, orthat p(n) = 2

3 −23

(− 1

2

)n.

Now, for k ≥ 1, we have that the probability of ending up on the vertex opposite Sherry’s initial vertexafter k minutes is 1−p(k)

2 = 12 −

12

(23 −

23

(− 1

2

)k) = 16 + 1

3

(− 1

2

)k = 16 + 1

3(−2)k .

Circles in Circles

5. [4] Circle O has chord AB. A circle is tangent to O at T and tangent to AB at X such that AX = 2XB.What is AT

BT ?

Answer: 2 Let TX meet circle O again at Y . Since the homethety centered at T takes X to Yalso takes AB to the tangent line of circle O passing through Y , we have Y is the midpoint of arc AB.This means that ∠ATY = ∠Y TB. By the Angle Bisector Theorem, AT

BT = AXBX = 2.

6. [6] AB is a diameter of circle O. X is a point on AB such that AX = 3BX. Distinct circles ω1 andω2 are tangent to O at T1 and T2 and to AB at X. The lines T1X and T2X intersect O again at S1

and S2. What is the ratio T1T2S1S2

?

Answer: 35 Since the problem only deals with ratios, we can assume that the radius of O is 1. As

we have proven in Problem 5, points S1 and S2 are midpoints of arc AB. Since AB is a diameter,S1S2 is also a diameter, and thus S1S2 = 2.

Let O1, O2, and P denote the center of circles ω1, ω2, and O. Since ω1 is tangent to O, we havePO1 + O1X = 1. But O1X ⊥ AB. So 4PO1X is a right triangle, and O1X

2 + XP 2 = O1P2. Thus,

O1X2 + 1/4 = (1−O1X)2, which means O1X = 3

8 and O1P = 58 .

Since T1T2 ‖ O1O2, we have T1T2 = O1O2 · PT1PO1

= 2O1X · PT1PO1

= 2(

38

)1

5/8 = 65 . Thus T1T2

S1S2= 6/5

2 = 35 .

Team Round

7. [7] ABC is a right triangle with ∠A = 30◦ and circumcircle O. Circles ω1, ω2, and ω3 lie outsideABC and are tangent to O at T1, T2, and T3 respectively and to AB, BC, and CA at S1, S2, and S3,respectively. Lines T1S1, T2S2, and T3S3 intersect O again at A′, B′, and C ′, respectively. What isthe ratio of the area of A′B′C ′ to the area of ABC?

Answer:√

3+12 Let [PQR] denote the area of 4PQR. The key to this problem is following fact:

[PQR] = 12PQ · PR sin ∠QPR.

Assume that the radius of O is 1. Since ∠A = 30◦, we have BC = 1 and AB =√

3. So [ABC] =√

32 .

Let K denote the center of O. Notice that ∠B′KA′ = 90◦, ∠AKC ′ = 90◦, and ∠B′KA = ∠KAB =30◦. Thus, ∠B′KC ′ = ∠B′KA + ∠AKC ′ = 120◦ and consequently ∠C ′KA′ = 150◦.

Therefore, [A′B′C ′] = [A′KB′] + [B′KC ′] + [C ′KA′] = 12 + 1

2 sin 120◦+ 12 sin 150◦ = 3

4 +√

34 . This gives

the desired result that [A′B′C ′] =√

3+12 [ABC].

Linear? What’s The Problem?

A function f(x1, x2, . . . , xn) is said to be linear in each of its variables if it is a polynomial such thatno variable appears with power higher than one in any term. For example, 1 + x + xy is linear in xand y, but 1 + x2 is not. Similarly, 2x + 3yz is linear in x, y, and z, but xyz2 is not.

8. [4] A function f(x, y) is linear in x and in y. f(x, y) = 1xy for x, y ∈ {3, 4}. What is f(5, 5)?

Answer: 136 The main fact that we will use in solving this problem is that f(x+2, y)−f(x+1, y) =

f(x + 1, y) − f(x, y) whenever f is linear in x and y. Suppose that f(x, y) = axy + by + cx + d =x(ay + c) + (by + d) for some constants a, b, c, and d. Then it is easy to see that

f(x + 2, y)− f(x + 1, y) = (x + 2)(ay + c) + (by + d)− (x + 1)(ay + c)− (by + d) = ay + c

f(x + 1, y)− f(x, y) = (x + 1)(ay + c) + (by + d)− x(ay + c)− (by + d) = ay + c,

which implies that f(x + 2, y) − f(x + 1, y) = f(x + 1, y) − f(x, y). In particular, f(5, y) − f(4, y) =f(4, y) − f(3, y), so f(5, y) = 2f(4, y) − f(3, y). Similarly, f(x, 5) = 2f(x, 4) − f(x, 3). Now we seethat:

f(5, 5) = 2f(5, 4)− f(5, 3)= 2[2f(4, 4)− f(3, 4)]− [2f(4, 3)− f(3, 3)]= 4f(4, 4)− 2f(3, 4)− 2f(4, 3) + f(3, 3)

=416− 4

12+

19

=14− 1

3+

19

=19− 1

12

=136

,

so the answer is 136 .

9. [5] A function f(x, y, z) is linear in x, y, and z such that f(x, y, z) = 1xyz for x, y, z ∈ {3, 4}. What is

f(5, 5, 5)?

Answer: 1216 We use a similar method to the previous problem. Notice that f(x, y, 5) = 2f(x, y, 4)−

f(x, y, 3). Let f2 denote the function from the previous problem and f3 the function from this problem.

Team Round

Since 3f3(x, y, 3) is linear in x and y, and 3f3(x, y, 3) = 1xy for all x, y ∈ {3, 4}, the previous problem

implies that 3f3(5, 5, 3) = 136 = f2(5, 5). Similarly, 4f3(5, 5, 4) = f2(5, 5). Now we have

f3(5, 5, 5) = 2f3(5, 5, 4)− f3(5, 5, 3)

=12f2(5, 5)− 1

3f2(5, 5)

=16f2(5, 5)

=1

6 · 36

=1

216.

10. [6] A function f(x1, x2, . . . , xn) is linear in each of the xi and f(x1, x2, . . . , xn) = 1x1x2···xn

whenxi ∈ {3, 4} for all i. In terms of n, what is f(5, 5, . . . , 5)?

Answer: 16n Let fn(x1, x2, . . . , xn) denote the n-variable version of the function. We will prove

that fn(5, . . . , 5) = 16n by induction. The base case was done in the two previous problems. Suppose

we know that fn−1(5, 5, . . . , 5) = 1·6n−1 . Let g(x1, . . . , xn−1) = 3fn(x1, . . . , xn−1, 3). We have that g is

linear in x1, . . . , xn−1 and g(x1, . . . , xn−1) = 1x1···xn−1

for all x1, . . . , xn−1 ∈ {3, 4}. By the inductive

hypothesis, we have g(5, . . . , 5) = 16n−1 = fn−1(5, . . . , 5). Therefore, fn(5, . . . , 5, 3) = fn−1(5,...,5)

3 .Similarly, fn(5, . . . , 5, 4) = fn−1(5,...,5)

4 .

fn(5, 5, . . . , 5, 5) = 2fn(5, 5, . . . , 5, 4)− fn(5, 5, . . . , 5, 3)

=12fn−1(5, 5, . . . , 5)− 1

3fn−1

=16fn−1(5, 5, . . . , 5)

=1

6 · 6n−1

=16n

,

and this proves our conjecture by induction.

Team Round

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

1. Let a, b, and c be positive real numbers. Determine the largest total number of real roots that thefollowing three polynomials may have among them: ax2 + bx+ c, bx2 + cx+ a, and cx2 + ax+ b.

2. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner.At each step, he can either move one square to the right or simultaneously move one square to the leftand one square up. In how many ways can he reach the center square of the topmost row?

3. Let f : R → R be a differentiable function such that f(0) = 0, f(1) = 1, and |f ′(x)| ≤ 2 for all real

numbers x. If a and b are real numbers such that the set of possible values of

∫ 1

0

f(x) dx is the open

interval (a, b), determine b− a.

4. Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠BAC intersect line BC atD. If there exist points E and F on sides AC and BC, respectively, such that lines AD and EFare parallel and divide triangle ABC into three parts of equal area, determine the number of possibleintegral values for BC.

5. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. IfNathaniel goes first, determine the probability that he ends up winning.

6. Let a ? b = ab+ a+ b for all integers a and b. Evaluate 1 ? (2 ? (3 ? (4 ? . . . (99 ? 100) . . .))).

7. Let f : [0, 1)→ R be a function that satisfies the following condition: if

x =

∞∑n=1

an10n

= .a1a2a3 . . .

is the decimal expansion of x and there does not exist a positive integer k such that an = 9 for alln ≥ k, then

f(x) =

∞∑n=1

an102n

.

Determine f ′13

).

8. Find all integers x such that 2x2 + x− 6 is a positive integral power of a prime positive integer.

9. Let ABCDEF be a regular hexagon of area 1. Let M be the midpoint of DE. Let X be theintersection of AC and BM , let Y be the intersection of BF and AM , and let Z be the intersectionof AC and BF . If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate[BXC] + [AY F ] + [ABZ]− [MXZY ].

10. For all real numbers x, let

f(x) =1

2011√

1− x2011.

Evaluate (f(f(. . . (f(2011)) . . .)))2011, where f is applied 2010 times.

11. Evaluate

∫ ∞1

(lnx

x

)2011

dx.

12. Let f(x) = x2 + 6x+ c for all real numbers x, where c is some real number. For what values of c doesf(f(x)) have exactly 3 distinct real roots?

13. Sarah and Hagar play a game of darts. Let O0 be a circle of radius 1. On the nth turn, the playerwhose turn it is throws a dart and hits a point pn randomly selected from the points of On∈1. Theplayer then draws the largest circle that is centered at pn and contained in On∈1, and calls this circleOn. The player then colors every point that is inside On∈1 but not inside On her color. Sarah goesfirst, and the two players alternate turns. Play continues indefinitely. If Sarah’s color is red, andHagar’s color is blue, what is the expected value of the area of the set of points colored red?

14. How many polynomials P with integer coefficients and degree at most 5 satisfy 0 ≤ P (x) < 120 for allx ∈ {0, 1, 2, 3, 4, 5}?

15. Let f : [0, 1] → [0, 1] be a continuous function such that f(f(x)) = 1 for all x ∈ [0, 1]. Determine the

set of possible values of

∫ 1

0

f(x) dx.

16. Let f(x) = x2 − r2x + r3 for all real numbers x, where r2 and r3 are some real numbers. Define asequence {gn} for all nonnegative integers n by g0 = 0 and gn+1 = f(gn). Assume that {gn} satisfiesthe following three conditions: (i) g2i < g2i+1 and g2i+1 > g2i+2 for all 0 ≤ i ≤ 2011; (ii) there existsa positive integer j such that gi+1 > gi for all i > j, and (iii) {gn} is unbounded. If A is the greatestnumber such that A ≤ |r2| for any function f satisfying these properties, find A.

17. Let f : (0, 1) → (0, 1) be a differentiable function with a continuous derivative such that for everypositive integer n and odd positive integer a < 2n, there exists an odd positive integer b < 2n suchthat f

a2n

)= b

2n . Determine the set of possible values of f ′12

).

18. Let z = cos2π

2011+ i sin

2π

2011, and let

P (x) = x2008 + 3x2007 + 6x2006 + . . .2008 · 2009

2x+

2009 · 2010

2

for all complex numbers x. Evaluate P (z)P (z2)P (z3) . . . P (z2010).

19. Let

F (x) =1

(2− x− x5)2011,

and note that F may be expanded as a power series so that F (x) =

∞∑n=0

anxn. Find an ordered pair of

positive real numbers (c, d) such that limn→∞

annd

= c.

20. Let {an} and {bn} be sequences defined recursively by a0 = 2; b0 = 2, and an+1 = an√

1 + a2n + b2n−bn;

bn+1 = bn√

1 + a2n + b2n + an. Find the ternary (base 3) representation of a4 and b4.

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

Algebra & Calculus Individual Test

1. Let a, b, and c be positive real numbers. Determine the largest total number of real roots that thefollowing three polynomials may have among them: ax2 + bx+ c, bx2 + cx+ a, and cx2 + ax+ b.

Answer: 4 If all the polynomials had real roots, their discriminants would all be nonnegative: a2 ≥4bc, b2 ≥ 4ca, and c2 ≥ 4ab. Multiplying these inequalities gives (abc)2 ≥ 64(abc)2, a contradiction.Hence one of the quadratics has no real roots. The maximum of 4 real roots is attainable: for example,the values (a, b, c) = (1, 5, 6) give 〉2,〉3 as roots to x2 + 5x+ 6 and 〉1,〉 1

5 as roots to 5x2 + 6x+ 1.

2. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner.At each step, he can either move one square to the right or simultaneously move one square to the leftand one square up. In how many ways can he reach the center square of the topmost row?

Answer: 2n−1 Note that Josh must pass through the center square of each row. There are 2 waysto get from the center square of row k to the center square of row k+ 1. So there are 2n−1 ways to getto the center square of row n.

3. Let f : R → R be a differentiable function such that f(0) = 0, f(1) = 1, and |f ′(x)| ≤ 2 for all real

numbers x. If a and b are real numbers such that the set of possible values of

∫ 1

0

f(x) dx is the open

interval (a, b), determine b〉 a.

Answer: 34 Draw lines of slope ±2 passing through (0, 0) and (1, 1). These form a parallelogram with

vertices (0, 0), (.75, 1.5), (1, 1), (.25,〉.5). By the mean value theorem, no point of (x, f(x)) lies outsidethis parallelogram, but we can construct functions arbitrarily close to the top or the bottom of theparallelogram while satisfying the condition of the problem. So (b〉a) is the area of this parallelogram,which is 3

4 .

4. Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠BAC intersect line BC atD. If there exist points E and F on sides AC and BC, respectively, such that lines AD and EFare parallel and divide triangle ABC into three parts of equal area, determine the number of possibleintegral values for BC.

Answer: 13

A B

C

DE

F

7

14

Algebra & Calculus Individual Test

Note that such E,F exist if and only if[ADC]

[ADB]= 2. (1)

([ ] denotes area.) Since AD is the angle bisector, and the ratio of areas of triangles with equal heightis the ratio of their bases,

AC

AB=DC

DB=

[ADC]

[ADB].

Hence (1) is equivalent to AC = 2AB = 14. Then BC can be any length d such that the triangleinequalities are satisfied:

d+ 7 > 14

7 + 14 > d

Hence 7 < d < 21 and there are 13 possible integral values for BC.

5. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. IfNathaniel goes first, determine the probability that he ends up winning.

Answer: 511 For 1 ≤ k ≤ 6, let xk be the probability that the current player, say A, will win when

the number on the tally at the beginning of his turn is k modulo 7. The probability that the totalis l modulo 7 after his roll is 1

6 for each l 6≡ k (mod 7); in particular, there is a 16 chance he wins

immediately. The chance that A will win if he leaves l on the board after his turn is 1〉 xl. Hence for1 ≤ k ≤ 6,

xk =1

6

∑1≤l≤6, l 6=k

(1〉 xl) +1

6.

Letting s =∑6l=1 xl, this becomes xk = xk−s

6 + 1 or 5xk

6 = 〉 s6 + 1. Hence x1 = · · · = x6, and 6xk = s

for every k. Plugging this in gives 11xk

6 = 1, or xk = 611 .

Since Nathaniel cannot win on his first turn, he leaves Obediah with a number not divisible by 7.Hence Obediah’s chance of winning is 6

11 and Nathaniel’s chance of winning is 511 .

6. Let a ? b = ab+ a+ b for all integers a and b. Evaluate 1 ? (2 ? (3 ? (4 ? . . . (99 ? 100) . . .))).

Answer: 101!〉 1 We will first show that ? is both commutative and associative.

• Commutativity: a ? b = ab+ a+ b = b ? a

• Associativity: a ? (b ? c) = a(bc + b + c) + a + bc + b + c = abc + ab + ac + bc + a + b + c and(a ? b) ? c = (ab+ a+ b)c+ ab+ a+ b+ c = abc+ ab+ ac+ bc+ a+ b+ c. So a ? (b ? c) = (a ? b) ? c.

So we need only calculate ((. . . (1 ? 2) ? 3) ? 4) . . . ? 100). We will prove by induction that

((. . . (1 ? 2) ? 3) ? 4) . . . ? n) = (n+ 1)!〉 1.

• Base case (n = 2): (1 ? 2) = 2 + 1 + 2 = 5 = 3!〉 1

• Inductive step:

Suppose that(((. . . (1 ? 2) ? 3) ? 4) . . . ? n) = (n+ 1)!〉 1.

Then,

((((. . . (1 ? 2) ? 3) ? 4) . . . ? n) ? (n+ 1)) = ((n+ 1)!〉 1) ? (n+ 1)

= (n+ 1)!(n+ 1)〉 (n+ 1) + (n+ 1)!〉 1 + (n+ 1)

= (n+ 2)!〉 1

Algebra & Calculus Individual Test

Hence, ((. . . (1 ? 2) ? 3) ? 4) . . . ? n) = (n+ 1)!〉 1 for all n. For n = 100, this results to 101!〉 1.

7. Let f : [0, 1)→ R be a function that satisfies the following condition: if

x =

∞∑n=1

an10n

= .a1a2a3 . . .

is the decimal expansion of x and there does not exist a positive integer k such that an = 9 for alln ≥ k, then

f(x) =

∞∑n=1

an102n

.

Determine f ′(13

).

Answer: 0 Note that 13 =

∑∞n=1

310n .

Clearly f is an increasing function. Also for any integer n ≥ 1, we see from decimal expansions thatf( 1

3 ±1

10n )〉 f( 13 ) = ± 1

102n .

Consider h such that 10−n−1 ≤ |h| < 10−n. The two properties of f outlined above show that |f( 13 +

h)〉 f( 13 )| < 1

102n . And from | 1h | ≤ 10n+1, we get∣∣∣ f( 1

3+h)−f(13 )

h

∣∣∣ < 110n−1 . Taking n→∞ gives h→ 0

and f ′( 13 ) = limn→∞

110n−1 = 0.

8. Find all integers x such that 2x2 + x〉 6 is a positive integral power of a prime positive integer.

Answer: 〉3, 2, 5 Let f(x) = 2x2 + x 〉 6 = (2x 〉 3)(x + 2). Suppose a positive integer a divides

both 2x〉 3 and x+ 2. Then a must also divide 2(x+ 2)〉 (2x〉 3) = 7. Hence, a can either be 1 or7. As a result, 2x 〉 3 = 7n or 〉7n for some positive integer n, or either x + 2 or 2x 〉 3 is ±1. Weconsider the following cases:

• (2x〉 3) = 1. Then x = 2, which yields f(x) = 4, a prime power.

• (2x〉 3) = 〉1. Then x = 1, which yields f(x) = 〉3, not a prime power.

• (x+ 2) = 1). Then x = 〉1, which yields f(x) = 〉5 not a prime power.

• (x+ 2) = 〉1. Then x = 〉3, which yields f(x) = 9, a prime power.

• (2x〉 3) = 7. Then x = 5, which yields f(x) = 49, a prime power.

• (2x〉 3) = 〉7. Then x = 〉2, which yields f(x) = 0, not a prime power.

• (2x〉 3) = ±7n, for n ≥ 2. Then, since x+ 2 =(2x〉 3) + 7

2, we have that x+ 2 is divisible by 7

but not by 49. Hence x + 2 = ±7, yielding x = 5,〉9. The former has already been considered,while the latter yields f(x) = 147.

So x can be either -3, 2 or 5.

(Note: In the official solutions packet we did not list the answer -3. This oversight was quickly noticedon the day of the test, and only the answer 〉3, 2, 5 was marked as correct.

9. Let ABCDEF be a regular hexagon of area 1. Let M be the midpoint of DE. Let X be theintersection of AC and BM , let Y be the intersection of BF and AM , and let Z be the intersectionof AC and BF . If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate[BXC] + [AY F ] + [ABZ]〉 [MXZY ].

Answer: 0

Algebra & Calculus Individual Test

A B

C

DE

F

M

XY

Z

O

Let O be the center of the hexagon. The desired area is [ABCDEF ] 〉 [ACDM ] 〉 [BFEM ]. Notethat [ADM ] = [ADE]/2 = [ODE] = [ABC], where the last equation holds because sin 60◦ = sin 120◦.Thus, [ACDM ] = [ACD] + [ADM ] = [ACD] + [ABC] = [ABCD], but the area of ABCD is half thearea of the hexagon. Similarly, the area of [BFEM ] is half the area of the hexagon, so the answer iszero.

10. For all real numbers x, let

f(x) =1

2011√

1〉 x2011.

Evaluate (f(f(. . . (f(2011)) . . .)))2011, where f is applied 2010 times.

Answer: 20112011 Direct calculation shows that f(f(x)) =2011√

1〉 x2011〉x

and f(f(f(x))) = x.

Hence (f(f(. . . (f(x)) . . .))) = x, where f is applied 2010 times. So (f(f(. . . (f(2011)) . . .)))2011 =20112011.

11. Evaluate

∫ ∞1

(lnx

x

)2011

dx.

Answer: 2011!20102012 By the chain rule,

d

dx(lnx)n =

n lnn−1 x

x.

We calculate the definite integral using integration by parts:∫ ∞x=1

(lnx)n

x2011dx =

[(lnx)n

〉2010x2010

]x=∞x=1

〉∫ ∞x=1

n(lnx)n−1

〉2010x2011dx

But ln(1) = 0, and limx→∞(ln x)n

x2010 = 0 for all n > 0. So∫ ∞x=1

(lnx)n

x2011dx =

∫ ∞x=1

n(lnx)n−1

2010x2011dx

It follows that ∫ ∞x=1

(lnx)n

x2011dx =

n!

2010n

∫ ∞x=1

1

x2011dx =

n!

2010n+1

So the answer is 2011!20102012 .

Algebra & Calculus Individual Test

12. Let f(x) = x2 + 6x+ c for all real numbers x, where c is some real number. For what values of c doesf(f(x)) have exactly 3 distinct real roots?

Answer: 11−√13

2 Suppose f has only one distinct root r1. Then, if x1 is a root of f(f(x)), it must

be the case that f(x1) = r1. As a result, f(f(x)) would have at most two roots, thus not satisfyingthe problem condition. Hence f has two distinct roots. Let them be r1 6= r2.

Since f(f(x)) has just three distinct roots, either f(x) = r1 or f(x) = r2 has one distinct root. Assumewithout loss of generality that r1 has one distinct root. Then f(x) = x2 + 6x+ c = r1 has one root, sothat x2 + 6x + c 〉 r1 is a square polynomial. Therefore, c 〉 r1 = 9, so that r1 = c 〉 9. So c 〉 9 is aroot of f . So (c〉 9)2 + 6(c〉 9) + c = 0, yielding c2 〉 11c+ 27 = 0, or (c〉 11

2 )2 = 132 . This results to

c = 11±√13

2 .

If c = 11−√13

2 , f(x) = x2 + 6x + 11−√13

2 = (x + 7+√13

2 )(x + 5−√13

2 ). We know f(x) = −7−√13

2 has a

double root, -3. Now −5+√13

2 > −7−√13

2 so the second root is above the vertex of the parabola, and ishit twice.

If c = 11+√13

2 , f(x) = x2 + 6x + 11+√13

2 = (x + 7−√13

2 )(x + 5+√13

2 ). We know f(x) = −7+√13

2 has adouble root, -3, and this is the value of f at the vertex of the parabola, so it is its minimum value.

Since −5−√13

2 < −7+√13

2 , f(x) = −5−√13

2 has no solutions. So in this case, f has only one real root.

So the answer is c = 11−√13

2 .

Note: In the solutions packet we had both roots listed as the correct answer. We noticed this oversighton the day of the test and awarded points only for the correct answer.

13. Sarah and Hagar play a game of darts. Let O0 be a circle of radius 1. On the nth turn, the playerwhose turn it is throws a dart and hits a point pn randomly selected from the points of On−1. Theplayer then draws the largest circle that is centered at pn and contained in On−1, and calls this circleOn. The player then colors every point that is inside On−1 but not inside On her color. Sarah goesfirst, and the two players alternate turns. Play continues indefinitely. If Sarah’s color is red, andHagar’s color is blue, what is the expected value of the area of the set of points colored red?

Answer: 6π7 Let f(r) be the average area colored red on a dartboard of radius r if Sarah plays first.

Then f(r) is proportional to r2. Let f(r) = (πx)r2 for some constant x. We want to find f(1) = πx.

In the first throw, if Sarah’s dart hits a point with distance r from the center of O0, the radius of O1

will be 1〉r. The expected value of the area colored red will be (π〉π(1〉r)2)+(π(1〉r)2〉f(1〉r)) =π 〉 f(1〉 r). The value of f(1) is the average value of π 〉 f(1〉 r) over all points in O0. Using polarcoordinates, we get

f(1) =

2π∫0

1∫0

(π 〉 f(1〉 r))rdrdθ

2π∫0

1∫0

rdrdθ

πx =

1∫0

(π 〉 πx(1〉 r)2)rdr

1∫0

rdr

πx

2=

∫ 1

0

πr 〉 πxr(1〉 r)2dr

πx

2=π

2〉 πx(

1

2〉 2

3+

1

4)

πx

2=π

2〉 πx

12

Algebra & Calculus Individual Test

πx =6π

7

14. How many polynomials P with integer coefficients and degree at most 5 satisfy 0 ≤ P (x) < 120 for allx ∈ {0, 1, 2, 3, 4, 5}?

Answer: 86400000 For each nonnegative integer i, let xi = x(x〉 1) · · · (x〉 i+ 1). (Define x0 = 1.)

Lemma: Each polynomial with integer coefficients f can be uniquely written in the form

f(x) = anxn + . . .+ a1x

1 + a0x0, an 6= 0.

Proof: Induct on the degree. The base case (degree 0) is clear. If f has degree m with leading coefficientc, then by matching leading coefficients we must have m = n and an = c. By the induction hypothesis,f(x)〉 cxn can be uniquely written as an−1x

n−1(x) + . . .+ a1x1 + a0x

0.

There are 120 possible choices for a0, namely any integer in [0, 120). Once a0, . . . , ai−1 have beenchosen so 0 ≤ P (0), . . . , P (i〉 1) < 120, for some 0 ≤ i ≤ 5, then we have

P (i) = aii! + ai−1ii−1 + · · ·+ a0

so by choosing ai we can make P (i) any number congruent to ai−1ii−1 + · · ·+a0 modulo i!. Thus there

are 120i! choices for ai. Note the choice of ai does not affect the value of P (0), . . . , P (i 〉 1). Thus all

polynomials we obtain in this way are valid. The answer is

5∏i=0

120

i!= 86400000.

Note: Their is also a solution involving finite differences that is basically equivalent to this solution.One proves that for i = 0, 1, 2, 3, 4, 5 there are 5!

i! ways to pick the ith finite difference at the point 0.

15. Let f : [0, 1] → [0, 1] be a continuous function such that f(f(x)) = 1 for all x ∈ [0, 1]. Determine the

set of possible values of

∫ 1

0

f(x) dx.

Answer: ( 34 , 1] Since the maximum value of f is 1,

∫ 1

0f(x)dx ≤ 1.

By our condition f(f(x)) = 1, f is 1 at any point within the range of f . Clearly, 1 is in the range off , so f(1) = 1. Now f(x) is continuous on a closed interval so it attains a minimum value c. Since cis in the range of f , f(c) = 1.

If c = 1, f(x) = 1 for all x and∫ 1

0f(x)dx = 1.

Now assume c < 1. By the intermediate value theorem, since f is continuous it attains all valuesbetween c and 1. So for all x ≥ c, f(x) = 1. Therefore,

∫ 1

0

f(x)dx =

∫ c

0

f(x)dx+ (1〉 c).

Since f(x) ≥ c,∫ c0f(x)dx > c2, and the equality is strict because f is continuous and thus cannot be

c for all x < c and 1 at c. So

∫ 1

0

f(x)dx > c2 + (1〉 c) = (c〉 1

2)2 +

3

4≥ 3

4.

Therefore 34 <

∫ 1

0f(x)dx ≤ 1, and it is easy to show that every value in this interval can be reached.

Algebra & Calculus Individual Test

16. Let f(x) = x2 〉 r2x + r3 for all real numbers x, where r2 and r3 are some real numbers. Define asequence {gn} for all nonnegative integers n by g0 = 0 and gn+1 = f(gn). Assume that {gn} satisfiesthe following three conditions: (i) g2i < g2i+1 and g2i+1 > g2i+2 for all 0 ≤ i ≤ 2011; (ii) there existsa positive integer j such that gi+1 > gi for all i > j, and (iii) {gn} is unbounded. If A is the greatestnumber such that A ≤ |r2| for any function f satisfying these properties, find A.

Answer: 2 Consider the function f(x) 〉 x. By the constraints of the problem, f(x) 〉 x must benegative for some x, namely, for x = g2i+1, 0 ≤ i ≤ 2011. Since f(x) 〉 x is positive for x of largeabsolute value, the graph of f(x) 〉 x crosses the x-axis twice and f(x) 〉 x has two real roots, saya < b. Factoring gives f(x)〉 x = (x〉 a)(x〉 b), or f(x) = (x〉 a)(x〉 b) + x.

Now, for x < a, f(x) > x > a, while for x > b, f(x) > x > b. Let c 6= b be the number such thatf(c) = f(b) = b. Note that b is not the vertex as f(a) = a < b, so by the symmetry of quadratics, cexists and b+c

2 = r22 as the vertex of the parabola. By the same token, b+a

2 = r2+12 is the vertex of

f(x)〉 x. Hence c = a〉 1. If f(x) > b then x < c or x > b. Consider the smallest j such that gj > b.Then by the above observation, gj−1 < c. (If gi ≥ b then f(gi) ≥ gi ≥ b so by induction, gi+1 ≥ gi forall i ≥ j. Hence j > 1; in fact j ≥ 4025.) Since gj−1 = f(gj−2), the minimum value of f is less than c.The minimum value is the value of f evaluated at its vertex, b+a−12 , so

f

(b+ a〉 1

2

)< c(

b+ a〉 1

2〉 a)(

b+ a〉 1

2〉 b)

+b+ a〉 1

2< a〉 1

1〉 (b〉 a)2

4+b〉 a+ 1

2< 0

3

4<

(b〉 a)2

4〉 b〉 a

2

4 < (b〉 a〉 1)2.

Then either b 〉 a 〉 1 < 〉2 or b 〉 a 〉 1 > 2, but b > a, so the latter must hold and (b 〉 a)2 > 9.Now, the discriminant of f(x) 〉 x equals (b 〉 a)2 (the square of the difference of the two roots) and(r2 + 1)2 〉 4r3 (from the coefficients), so (r2 + 1)2 > 9 + 4r3. But r3 = g1 > g0 = 0 so |r2| > 2.

We claim that we can make |r2| arbitrarily close to 2, so that the answer is 2. First define Gi, i ≥ 0as follows. Let N ≥ 2012 be an integer. For ε > 0 let h(x) = x2 〉 2 〉 ε, gε(x) = 〉

√x+ 2 + ε and

G2N+1 = 2 + ε, and define Gi recursively by Gi = gε(Gi+1), Gi+1 = h(Gi). (These two equations areconsistent.) Note the following.

(i) G2i < G2i+1 andG2i+1 > G2i+2 for 0 ≤ i ≤ N〉1. First noteG2N = 〉√

4 + 2ε > 〉√

4 + 2ε+ ε2 =〉2 〉 ε. Let l be the negative solution to h(x) = x. Note that 〉2 〉 ε < G2N < l < 0 sinceh(G2N ) > 0 > G2N . Now gε(x) is defined as long as x ≥ 〉2〉 ε, and it sends (〉2〉 ε, l) into (l, 0)and (l, 0) into (〉2〉 ε, l). It follows that the Gi, 0 ≤ i ≤ 2N are well-defined; moreover, G2i < land G2i+1 > l for 0 ≤ i ≤ N 〉 1 by backwards induction on i, so the desired inequalities follow.

(ii) Gi is increasing for i ≥ 2N + 1. Indeed, if x ≥ 2 + ε, then x2 〉 x = x(x〉 1) > 2 + ε so h(x) > x.Hence 2 + ε = G2N+1 < G2N+2 < · · · .

(iii) Gi is unbounded. This follows since h(x)〉 x = x(x〉 2)〉 2〉 ε is increasing for x > 2 + ε, so Giincreases faster and faster for i ≥ 2N + 1.

Now define f(x) = h(x + G0) 〉 G0 = x2 + 2G0x + G20 〉 G0 〉 2 〉 ε. Note Gi+1 = h(Gi) while

gi+1 = f(gi) = h(gi + G0) 〉 G0, so by induction gi = Gi 〉 G0. Since {Gi}∞i=0 satisfies (i), (ii), and(iii), so does gi.

We claim that we can make G0 arbitrarily close to 〉1 by choosing N large enough and ε small enough;this will make r2 = 〉2G0 arbitrarily close to 2. Choosing N large corresponds to taking G0 to be alarger iterate of 2 + ε under gε(x). By continuity of this function with respect to x and ε, it suffices to

Algebra & Calculus Individual Test

take ε = 0 and show that (letting g = g0)

g(n)(2) = g(· · · g︸ ︷︷ ︸n

(2) · · · )→ 〉1 as n→∞.

But note that for 0 ≤ θ ≤ π2 ,

g(〉2 cos θ) = 〉√

2〉 2 cos θ = 〉2 sin

(θ

2

)= 2 cos

(π

2〉 θ

2

).

Hence by induction, g(n)(〉2 cos θ) = 〉2 cos(π2 〉

π4 + · · ·+ (〉1)n

(θ 〉 π

2n

)). Hence g(n)(2) = g(n−1)(〉2 cos 0)

converges to 〉2 cos(π2 〉π4 + · · · ) = 〉2 cos(π3 ) = 〉1, as needed.

17. Let f : (0, 1) → (0, 1) be a differentiable function with a continuous derivative such that for everypositive integer n and odd positive integer a < 2n, there exists an odd positive integer b < 2n suchthat f

(a2n

)= b

2n . Determine the set of possible values of f ′(12

).

Answer: {〉1, 1} The key step is to notice that for such a function f , f ′(x) 6= 0 for any x.

Assume, for sake of contradiction that there exists 0 < y < 1 such that f ′(y) = 0. Since f ′ isa continuous function, there is some small interval (c, d) containing y such that |f ′(x)| ≤ 1

2 for allx ∈ (c, d). Now there exists some n, a such that a

2n ,a+12n are both in the interval (c, d). From the

definition,f(a+1

2n )〉 f( a2n )

a+12n 〉

a2n

= 2n(b′

2n〉 b

2n) = b′ 〉 b where b, b′ are integers; one is odd, and one is

even. So b′ 〉 b is an odd integer. Since f is differentiable, by the mean value theorem there exists apoint where f ′ = b′ 〉 b. But this point is in the interval (c, d), and |b′ 〉 b| > 1

2 . This contradicts theassumption that |f ′(x)| ≤ 1

2 for all x ∈ (c, d).

Since f ′(x) 6= 0, and f ′ is a continuous function, f ′ is either always positive or always negative. So f iseither increasing or decreasing. f( 1

2 ) = 12 always. If f is increasing, it follows that f( 1

4 ) = 14 , f( 3

4 ) = 34 ,

and we can show by induction that indeed f( a2n ) = a

2n for all integers a, n. Since numbers of this formare dense in the interval (0, 1), and f is a continuous function, f(x) = x for all x.

It can be similarly shown that if f is decreasing f(x) = 1〉 x for all x. So the only possible values off ′( 1

2 ) are 〉1, 1.

Query: if the condition that the derivative is continuous were omitted, would the same result stillhold?

18. Let z = cos2π

2011+ i sin

2π

2011, and let

P (x) = x2008 + 3x2007 + 6x2006 + . . .2008 · 2009

2x+

2009 · 2010

2

for all complex numbers x. Evaluate P (z)P (z2)P (z3) . . . P (z2010).

Answer: 20112009 · (10052011 〉 10042011) Multiply P (x) by x〉 1 to get

P (x)(x〉 1) = x2009 + 2x2008 + . . .+ 2009x〉 2009 · 2010

2,

or,

P (x)(x〉 1) + 2010 · 1005 = x2009 + 2x2008 + . . .+ 2009x+ 2010.

Multiplying by x〉 1 once again:

(x〉 1)(P (x)(x〉 1) +2010 · 2011

2) = x2010 + x2009 + . . .+ x〉 2010,

= (x2010 + x2009 + . . .+ x+ 1)〉 2011.

Algebra & Calculus Individual Test

Hence,

P (x) =

(x2010 + x2009 + . . .+ x+ 1)〉 2011

x〉 1〉 2011 · 1005

x〉 1

Note that x2010+x2009+. . .+x+1 has z, z2, . . . , z2010 as roots, so they vanish at those points. Pluggingthose 2010 powers of z into the last equation, and multiplying them together, we obtain

2010∏i=1

P (zi) =(〉2011) · 1005 · (x〉 1004

1005)

(x〉 1)2.

Note that (x〉 z)(x〉 z2) . . . (x〉 z2010) = x2010 + x2009 + . . .+ 1. Using this, the product turns out tobe 20112009 · (10052011 〉 10042011).

19. Let

F (x) =1

(2〉 x〉 x5)2011,

and note that F may be expanded as a power series so that F (x) =

∞∑n=0

anxn. Find an ordered pair of

positive real numbers (c, d) such that limn→∞

annd

= c.

Answer: ( 1620112010! , 2010) First notice that all the roots of 2 〉 x 〉 x5 that are not 1 lie strictly

outside the unit circle. As such, we may write 2〉x〉x5 as 2(1〉x)(1〉 r1x)(1〉 r2x)(1〉 r3x)(1〉 r4x)

where |ri| < 1, and let 1(2−x−x5) = b0

(1−x) + b1(1−r1x) + . . .+ b4

(1−r4x) . We calculate b0 as limx→1(1−x)

(2−x−x5) =

limx→1(−1)

(−1−5x4) = 16 .

Now raise the equation above to the 2011th power.

1

(2〉 x〉 x5)2011=

(1/6

(1〉 x)+

b1(1〉 r1x)

+ . . .+b4

(1〉 r4x)

)2011

Expand the right hand side using multinomial expansion and then apply partial fractions. The resultwill be a sum of the terms (1〉 x)−k and (1〉 rix)−k, where k ≤ 2011.

Since |ri| < 1, the power series of (1〉 rix)−k will have exponentially decaying coefficients, so we onlyneed to consider the (1〉x)−k terms. The coefficient of xn in the power series of (1〉x)−k is

(n+k−1k−1

),

which is a (k 〉 1)th degree polynomial in variable n. So when we sum up all coefficients, only thepower series of (1〉 x)−2011 will have impact on the leading term n2010.

The coefficient of the (1 〉 x)−2011 term in the multinomial expansion is ( 16 )2011. The coefficient of

the xn term in the power series of (1 〉 x)−2011 is(n+20102010

)= 1

2010!n2010 + . . .. Therefore, (c, d) =

( 1620112010! , 2010).

20. Let {an} and {bn} be sequences defined recursively by a0 = 2; b0 = 2, and an+1 = an√

1 + a2n + b2n〉bn;

bn+1 = bn√

1 + a2n + b2n + an. Find the ternary (base 3) representation of a4 and b4.

Answer: 1000001100111222 and 2211100110000012

Note first that√

1 + a2n + b2n = 32n

. The proof is by induction; the base case follows trivially from what

is given. For the inductive step, note that 1+a2n+1+b2n+1 = 1+a2n(1+a2n+b2n)+b2n〉2anbn√

1 + a2n + b2n+

b2n(1 + a2n + b2n) + a2n + 2anbn√

1 + a2n + b2n = 1 + (a2n + b2n)(1 + a2n + b2n) + a2n + b2n = (1 + a2n + b2n)2.

Invoking the inductive hypothesis, we see that√

1 + a2n+1 + b2n+1 = (32n

)2 = 32n+1

, as desired.

The quickest way to finish from here is to consider a sequence of complex numbers {zn} defined byzn = an+ bni for all nonnegative integers n. It should be clear that z0 = 2 + 2i and zn+1 = zn(32

n

+ i).

Therefore, z4 = (2 + 2i)(320

+ i)(321

+ i)(322

+ i)(323

+ i). This product is difficult to evaluate in

Algebra & Calculus Individual Test

the decimal number system, but in ternary the calculation is a cinch! To speed things up, we willuse balanced ternary1, in which the three digits allowed are 〉1, 0, and 1 rather than 0, 1, and 2. Letx+ yi = (32

0

+ i)(321

+ i)(322

+ i)(323

+ i), and consider the balanced ternary representation of x andy. For all 0 ≤ j ≤ 15, let xj denote the digit in the 3j place of x, let yj denote the digit in the 3j placeof y, and let b(j) denote the number of ones in the binary representation of j. It should be clear thatxj = 〉1 if b(j) ≡ 2 (mod 4), xj = 0 if b(j) ≡ 1 (mod 2), and xj = 1 if b(j) ≡ 0 (mod 4). Similarly,yj = 〉1 if b(j) ≡ 1 (mod 4), yj = 0 if b(j) ≡ 0 (mod 2), and yj = 1 if b(j) ≡ 3 (mod 4). Convertingto ordinary ternary representation, we see that x = 2212112211220013 and y = 1100222022121203. Itremains to note that a4 = 2x〉 2y and b4 = 2x+ 2y and perform the requisite arithmetic to arrive atthe answer above.

1http://en.wikipedia.org/wiki/Balanced_ternary

Algebra & Calculus Individual Test

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

1. Let a, b, and c be positive real numbers. Determine the largest total number of real roots that thefollowing three polynomials may have among them: ax2 + bx+ c, bx2 + cx+ a, and cx2 + ax+ b.

2. A classroom has 30 students and 30 desks arranged in 5 rows of 6. If the class has 15 boys and 15girls, in how many ways can the students be placed in the chairs such that no boy is sitting in frontof, behind, or next to another boy, and no girl is sitting in front of, behind, or next to another girl?

3. Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠BAC intersect line BC atD. If there exist points E and F on sides AC and BC, respectively, such that lines AD and EFare parallel and divide triangle ABC into three parts of equal area, determine the number of possibleintegral values for BC.

4. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner.At each step, he can either move one square to the right or simultaneously move one square to the leftand one square up. In how many ways can he reach the center square of the topmost row?

5. Let a ? b = ab+ a+ b for all integers a and b. Evaluate 1 ? (2 ? (3 ? (4 ? . . . (99 ? 100) . . .))).

6. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. IfNathaniel goes first, determine the probability that he ends up winning.

7. Find all integers x such that 2x2 + x− 6 is a positive integral power of a prime positive integer.

8. Let ABCDEF be a regular hexagon of area 1. Let M be the midpoint of DE. Let X be theintersection of AC and BM , let Y be the intersection of BF and AM , and let Z be the intersectionof AC and BF . If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate[BXC] + [AY F ] + [ABZ]− [MXZY ].

9. For all real numbers x, let

f(x) =1

2011√

1− x2011.

Evaluate (f(f(. . . (f(2011)) . . .)))2011, where f is applied 2010 times.

10. The integers from 1 to n are written in increasing order from left to right on a blackboard. Davidand Goliath play the following game: starting with David, the two players alternate erasing any twoconsecutive numbers and replacing them with their sum or product. Play continues until only onenumber on the board remains. If it is odd, David wins, but if it is even, Goliath wins. Find the 2011thsmallest positive integer greater than 1 for which David can guarantee victory.

11. Let f(x) = x2 + 6x+ c for all real numbers x, where c is some real number. For what values of c doesf(f(x)) have exactly 3 distinct real roots?

12. Mike and Harry play a game on an 8× 8 board. For some positive integer k, Mike chooses k squaresand writes an M in each of them. Harry then chooses k + 1 squares and writes an H in each of them.After Harry is done, Mike wins if there is a sequence of letters forming “HMM” or “MMH,” whenread either horizontally or vertically, and Harry wins otherwise. Determine the smallest value of k forwhich Mike has a winning strategy.

13. How many polynomials P with integer coefficients and degree at most 5 satisfy 0 ≤ P (x) < 120 for allx ∈ {0, 1, 2, 3, 4, 5}?

14. The ordered pairs (2011, 2), (2010, 3), (2009, 4), . . ., (1008, 1005), (1007, 1006) are written from left toright on a blackboard. Every minute, Elizabeth selects a pair of adjacent pairs (xi, yi) and (xj , yj), with

(xi, yi) left of (xj , yj), erases them, and writes

(xiyixjyj

,xiyiyjxj

)in their place. Elizabeth continues

this process until only one ordered pair remains. How many possible ordered pairs (x, y) could appearon the blackboard after the process has come to a conclusion?

15. Let f(x) = x2 − r2x + r3 for all real numbers x, where r2 and r3 are some real numbers. Define asequence {gn} for all nonnegative integers n by g0 = 0 and gn+1 = f(gn). Assume that {gn} satisfiesthe following three conditions: (i) g2i < g2i+1 and g2i+1 > g2i+2 for all 0 ≤ i ≤ 2011; (ii) there existsa positive integer j such that gi+1 > gi for all i > j, and (iii) {gn} is unbounded. If A is the greatestnumber such that A ≤ |r2| for any function f satisfying these properties, find A.

16. Let A = {1, 2, . . . , 2011}. Find the number of functions f from A to A that satisfy f(n) ≤ n for all nin A and attain exactly 2010 distinct values.

17. Let z = cos2π

2011+ i sin

2π

2011, and let

P (x) = x2008 + 3x2007 + 6x2006 + . . .2008 · 2009

2x+

2009 · 2010

2

for all complex numbers x. Evaluate P (z)P (z2)P (z3) . . . P (z2010).

18. Let n be an odd positive integer, and suppose that n people sit on a committee that is in the processof electing a president. The members sit in a circle, and every member votes for the person either tohis/her immediate left, or to his/her immediate right. If one member wins more votes than all theother members do, he/she will be declared to be the president; otherwise, one of the the members whowon at least as many votes as all the other members did will be randomly selected to be the president.If Hermia and Lysander are two members of the committee, with Hermia sitting to Lysander’s leftand Lysander planning to vote for Hermia, determine the probability that Hermia is elected president,assuming that the other n− 1 members vote randomly.

19. Let {an} and {bn} be sequences defined recursively by a0 = 2; b0 = 2, and an+1 = an√

1 + a2n + b2n−bn;

bn+1 = bn√

1 + a2n + b2n + an. Find the ternary (base 3) representation of a4 and b4.

20. Alice and Bob play a game in which two thousand and eleven 2011×2011 grids are distributed betweenthe two of them, 1 to Bob, and the other 2010 to Alice. They go behind closed doors and fill theirgrid(s) with the numbers 1, 2, . . . , 20112 so that the numbers across rows (left-to-right) and downcolumns (top-to-bottom) are strictly increasing. No two of Alice’s grids may be filled identically. Afterthe grids are filled, Bob is allowed to look at Alice’s grids and then swap numbers on his own grid,two at a time, as long as the numbering remains legal (i.e. increasing across rows and down columns)after each swap. When he is done swapping, a grid of Alice’s is selected at random. If there exist twointegers in the same column of this grid that occur in the same row of Bob’s grid, Bob wins. Otherwise,Alice wins. If Bob selects his initial grid optimally, what is the maximum number of swaps that Bobmay need in order to guarantee victory?

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

Algebra & Combinatorics Individual Test

1. Let a, b, and c be positive real numbers. Determine the largest total number of real roots that thefollowing three polynomials may have among them: ax2 + bx+ c, bx2 + cx+ a, and cx2 + ax+ b.

Answer: 4 If all the polynomials had real roots, their discriminants would all be nonnegative: a2 ≥4bc, b2 ≥ 4ca, and c2 ≥ 4ab. Multiplying these inequalities gives (abc)2 ≥ 64(abc)2, a contradiction.Hence one of the quadratics has no real roots. The maximum of 4 real roots is attainable: for example,the values (a, b, c) = (1, 5, 6) give 〉2,〉3 as roots to x2 + 5x+ 6 and 〉1,〉 1

5 as roots to 5x2 + 6x+ 1.

2. A classroom has 30 students and 30 desks arranged in 5 rows of 6. If the class has 15 boys and 15girls, in how many ways can the students be placed in the chairs such that no boy is sitting in frontof, behind, or next to another boy, and no girl is sitting in front of, behind, or next to another girl?

Answer: 2 · 15!2 If we color the desks of the class in a checkerboard pattern, we notice that all ofone gender must go in the squares colored black, and the other gender must go in the squares coloredwhite. There are 2 ways to pick which gender goes in which color, 15! ways to put the boys into desksand 15! ways to put the girls into desks. So the number of ways is 2 · 15!2.

(There is a little ambiguity in the problem statement as to whether the 15 boys and the 15 girls aredistinguishable or not. If they are not distinguishable, the answer is clearly 2. Given the number ofcontestants who submitted the answer 2, the graders judged that there was enough ambiguity to justifyaccepting 2 as a correct answer. So both 2 and 2 · 15!2 were accepted as correct answers.)

3. Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠BAC intersect line BC atD. If there exist points E and F on sides AC and BC, respectively, such that lines AD and EFare parallel and divide triangle ABC into three parts of equal area, determine the number of possibleintegral values for BC.

Answer: 13

A B

C

DE

F

7

14

Note that such E,F exist if and only if[ADC]

[ADB]= 2. (1)

Algebra & Combinatorics Individual Test

([ ] denotes area.) Since AD is the angle bisector, and the ratio of areas of triangles with equal heightis the ratio of their bases,

AC

AB=DC

DB=

[ADC]

[ADB].

Hence (1) is equivalent to AC = 2AB = 14. Then BC can be any length d such that the triangleinequalities are satisfied:

d+ 7 > 14

7 + 14 > d

Hence 7 < d < 21 and there are 13 possible integral values for BC.

4. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner.At each step, he can either move one square to the right or simultaneously move one square to the leftand one square up. In how many ways can he reach the center square of the topmost row?

Answer: 2n−1 Note that Josh must pass through the center square of each row. There are 2 waysto get from the center square of row k to the center square of row k+ 1. So there are 2n−1 ways to getto the center square of row n.

5. Let a ? b = ab+ a+ b for all integers a and b. Evaluate 1 ? (2 ? (3 ? (4 ? . . . (99 ? 100) . . .))).

Answer: 101!〉 1 We will first show that ? is both commutative and associative.

• Commutativity: a ? b = ab+ a+ b = b ? a

• Associativity: a ? (b ? c) = a(bc + b + c) + a + bc + b + c = abc + ab + ac + bc + a + b + c and(a ? b) ? c = (ab+ a+ b)c+ ab+ a+ b+ c = abc+ ab+ ac+ bc+ a+ b+ c. So a ? (b ? c) = (a ? b) ? c.

So we need only calculate ((. . . (1 ? 2) ? 3) ? 4) . . . ? 100). We will prove by induction that

((. . . (1 ? 2) ? 3) ? 4) . . . ? n) = (n+ 1)!〉 1.

• Base case (n = 2): (1 ? 2) = 2 + 1 + 2 = 5 = 3!〉 1

• Inductive step:

Suppose that(((. . . (1 ? 2) ? 3) ? 4) . . . ? n) = (n+ 1)!〉 1.

Then,

((((. . . (1 ? 2) ? 3) ? 4) . . . ? n) ? (n+ 1)) = ((n+ 1)!〉 1) ? (n+ 1)

= (n+ 1)!(n+ 1)〉 (n+ 1) + (n+ 1)!〉 1 + (n+ 1)

= (n+ 2)!〉 1

Hence, ((. . . (1 ? 2) ? 3) ? 4) . . . ? n) = (n+ 1)!〉 1 for all n. For n = 100, this results to 101!〉 1.

6. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. IfNathaniel goes first, determine the probability that he ends up winning.

Answer: 511 For 1 ≤ k ≤ 6, let xk be the probability that the current player, say A, will win when

the number on the tally at the beginning of his turn is k modulo 7. The probability that the totalis l modulo 7 after his roll is 1

6 for each l 6≡ k (mod 7); in particular, there is a 16 chance he wins

Algebra & Combinatorics Individual Test

immediately. The chance that A will win if he leaves l on the board after his turn is 1〉 xl. Hence for1 ≤ k ≤ 6,

xk =1

6

∑1≤l≤6, l 6=k

(1〉 xl) +1

6.

Letting s =∑6l=1 xl, this becomes xk = xk−s

6 + 1 or 5xk

6 = 〉 s6 + 1. Hence x1 = · · · = x6, and 6xk = s

for every k. Plugging this in gives 11xk

6 = 1, or xk = 611 .

Since Nathaniel cannot win on his first turn, he leaves Obediah with a number not divisible by 7.Hence Obediah’s chance of winning is 6

11 and Nathaniel’s chance of winning is 511 .

7. Find all integers x such that 2x2 + x〉 6 is a positive integral power of a prime positive integer.

Answer: 〉3, 2, 5 Let f(x) = 2x2 + x 〉 6 = (2x 〉 3)(x + 2). Suppose a positive integer a divides

both 2x〉 3 and x+ 2. Then a must also divide 2(x+ 2)〉 (2x〉 3) = 7. Hence, a can either be 1 or7. As a result, 2x 〉 3 = 7n or 〉7n for some positive integer n, or either x + 2 or 2x 〉 3 is ±1. Weconsider the following cases:

• (2x〉 3) = 1. Then x = 2, which yields f(x) = 4, a prime power.

• (2x〉 3) = 〉1. Then x = 1, which yields f(x) = 〉3, not a prime power.

• (x+ 2) = 1). Then x = 〉1, which yields f(x) = 〉5 not a prime power.

• (x+ 2) = 〉1. Then x = 〉3, which yields f(x) = 9, a prime power.

• (2x〉 3) = 7. Then x = 5, which yields f(x) = 49, a prime power.

• (2x〉 3) = 〉7. Then x = 〉2, which yields f(x) = 0, not a prime power.

• (2x〉 3) = ±7n, for n ≥ 2. Then, since x+ 2 =(2x〉 3) + 7

2, we have that x+ 2 is divisible by 7

but not by 49. Hence x + 2 = ±7, yielding x = 5,〉9. The former has already been considered,while the latter yields f(x) = 147.

So x can be either -3, 2 or 5.

(Note: In the official solutions packet we did not list the answer -3. This oversight was quickly noticedon the day of the test, and only the answer 〉3, 2, 5 was marked as correct.

8. Let ABCDEF be a regular hexagon of area 1. Let M be the midpoint of DE. Let X be theintersection of AC and BM , let Y be the intersection of BF and AM , and let Z be the intersectionof AC and BF . If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate[BXC] + [AY F ] + [ABZ]〉 [MXZY ].

Answer: 0

A B

C

DE

F

M

XY

Z

O

Algebra & Combinatorics Individual Test

Let O be the center of the hexagon. The desired area is [ABCDEF ] 〉 [ACDM ] 〉 [BFEM ]. Notethat [ADM ] = [ADE]/2 = [ODE] = [ABC], where the last equation holds because sin 60◦ = sin 120◦.Thus, [ACDM ] = [ACD] + [ADM ] = [ACD] + [ABC] = [ABCD], but the area of ABCD is half thearea of the hexagon. Similarly, the area of [BFEM ] is half the area of the hexagon, so the answer iszero.

9. For all real numbers x, let

f(x) =1

2011√

1〉 x2011.

Evaluate (f(f(. . . (f(2011)) . . .)))2011, where f is applied 2010 times.

Answer: 20112011 Direct calculation shows that f(f(x)) =2011√

1〉 x2011〉x

and f(f(f(x))) = x.

Hence (f(f(. . . (f(x)) . . .))) = x, where f is applied 2010 times. So (f(f(. . . (f(2011)) . . .)))2011 =20112011.

10. The integers from 1 to n are written in increasing order from left to right on a blackboard. Davidand Goliath play the following game: starting with David, the two players alternate erasing any twoconsecutive numbers and replacing them with their sum or product. Play continues until only onenumber on the board remains. If it is odd, David wins, but if it is even, Goliath wins. Find the 2011thsmallest positive integer greater than 1 for which David can guarantee victory.

Answer: 4022 If n is odd and greater than 1, then Goliath makes the last move. No matter whattwo numbers are on the board, Goliath can combine them to make an even number. Hence Goliathhas a winning strategy in this case.

Now suppose n is even. We can replace all numbers on the board by their residues modulo 2. Initiallythe board reads 1, 0, 1, 0, . . . , 1, 0. David combines the rightmost 1 and 0 by addition to make 1, so nowthe board reads 1, 0, 1, 0, . . . , 0, 1. We call a board of this form a “good” board. When it is Goliath’sturn, and there is a good board, no matter where he moves, David can make a move to restore a goodboard. Indeed, Goliath must combine a neighboring 0 and 1; David can then combine that numberwith a neighboring 1 to make 1 and create a good board with two fewer numbers.

David can ensure a good board after his last turn. But a good board with one number is simply 1,so David wins. So David has a winning strategy if n is even. Therefore, the 2011th smallest positiveinteger greater than 1 for which David can guarantee victory is the 2011th even positive integer, whichis 4022.

11. Let f(x) = x2 + 6x+ c for all real numbers x, where c is some real number. For what values of c doesf(f(x)) have exactly 3 distinct real roots?

Answer: 11−√13

2 Suppose f has only one distinct root r1. Then, if x1 is a root of f(f(x)), it must

be the case that f(x1) = r1. As a result, f(f(x)) would have at most two roots, thus not satisfyingthe problem condition. Hence f has two distinct roots. Let them be r1 6= r2.

Since f(f(x)) has just three distinct roots, either f(x) = r1 or f(x) = r2 has one distinct root. Assumewithout loss of generality that r1 has one distinct root. Then f(x) = x2 + 6x+ c = r1 has one root, sothat x2 + 6x + c 〉 r1 is a square polynomial. Therefore, c 〉 r1 = 9, so that r1 = c 〉 9. So c 〉 9 is aroot of f . So (c〉 9)2 + 6(c〉 9) + c = 0, yielding c2 〉 11c+ 27 = 0, or (c〉 11

2 )2 = 132 . This results to

c = 11±√13

2 .

If c = 11−√13

2 , f(x) = x2 + 6x + 11−√13

2 = (x + 7+√13

2 )(x + 5−√13

2 ). We know f(x) = −7−√13

2 has a

double root, -3. Now −5+√13

2 > −7−√13

2 so the second root is above the vertex of the parabola, and ishit twice.

If c = 11+√13

2 , f(x) = x2 + 6x + 11+√13

2 = (x + 7−√13

2 )(x + 5+√13

2 ). We know f(x) = −7+√13

2 has adouble root, -3, and this is the value of f at the vertex of the parabola, so it is its minimum value.

Since −5−√13

2 < −7+√13

2 , f(x) = −5−√13

2 has no solutions. So in this case, f has only one real root.

Algebra & Combinatorics Individual Test

So the answer is c = 11−√13

2 .

Note: In the solutions packet we had both roots listed as the correct answer. We noticed this oversighton the day of the test and awarded points only for the correct answer.

12. Mike and Harry play a game on an 8× 8 board. For some positive integer k, Mike chooses k squaresand writes an M in each of them. Harry then chooses k + 1 squares and writes an H in each of them.After Harry is done, Mike wins if there is a sequence of letters forming “HMM” or “MMH,” whenread either horizontally or vertically, and Harry wins otherwise. Determine the smallest value of k forwhich Mike has a winning strategy.

Answer: 16 Suppose Mike writes k M ’s. Let a be the number of squares which, if Harry writesan H in, will yield either HMM or MMH horizontally, and let b be the number of squares which,if Harry writes an H in, will yield either HMM or MMH vertically. We will show that a ≤ k andb ≤ k. Then, it will follow that there are at most a+ b ≤ 2k squares which Harry cannot write an Hin. There will be at least 64〉 k 〉 2k = 64〉 3k squares which Harry can write in. If 64〉 3k ≥ k + 1,or k ≤ 15, then Harry wins.

We will show that a ≤ k (that b ≤ k will follow by symmetry). Suppose there are ai M ’s in row i. Ineach group of 2 or more consective M ’s, Harry cannot write H to the left or right of that group, givingat most 2 forbidden squares. Hence ai is at most the number of M ’s in row i. Summing over the rowsgives the desired result.

Mike can win by writing 16 M ’s according to the following diagram:

· · · · · · · ·· M M · · M M ·· M M · · M M ·· · · · · · · ·· · · · · · · ·· M M · · M M ·· M M · · M M ·· · · · · · · ·

13. How many polynomials P with integer coefficients and degree at most 5 satisfy 0 ≤ P (x) < 120 for allx ∈ {0, 1, 2, 3, 4, 5}?

Answer: 86400000 For each nonnegative integer i, let xi = x(x〉 1) · · · (x〉 i+ 1). (Define x0 = 1.)

Lemma: Each polynomial with integer coefficients f can be uniquely written in the form

f(x) = anxn + . . .+ a1x

1 + a0x0, an 6= 0.

Proof: Induct on the degree. The base case (degree 0) is clear. If f has degree m with leading coefficientc, then by matching leading coefficients we must have m = n and an = c. By the induction hypothesis,f(x)〉 cxn can be uniquely written as an−1x

n−1(x) + . . .+ a1x1 + a0x

0.

There are 120 possible choices for a0, namely any integer in [0, 120). Once a0, . . . , ai−1 have beenchosen so 0 ≤ P (0), . . . , P (i〉 1) < 120, for some 0 ≤ i ≤ 5, then we have

P (i) = aii! + ai−1ii−1 + · · ·+ a0

so by choosing ai we can make P (i) any number congruent to ai−1ii−1 + · · ·+a0 modulo i!. Thus there

are 120i! choices for ai. Note the choice of ai does not affect the value of P (0), . . . , P (i 〉 1). Thus all

polynomials we obtain in this way are valid. The answer is

5∏i=0

120

i!= 86400000.

Note: Their is also a solution involving finite differences that is basically equivalent to this solution.One proves that for i = 0, 1, 2, 3, 4, 5 there are 5!

i! ways to pick the ith finite difference at the point 0.

Algebra & Combinatorics Individual Test

14. The ordered pairs (2011, 2), (2010, 3), (2009, 4), . . ., (1008, 1005), (1007, 1006) are written from left toright on a blackboard. Every minute, Elizabeth selects a pair of adjacent pairs (xi, yi) and (xj , yj), with

(xi, yi) left of (xj , yj), erases them, and writes

(xiyixjyj

,xiyiyjxj

)in their place. Elizabeth continues

this process until only one ordered pair remains. How many possible ordered pairs (x, y) could appearon the blackboard after the process has come to a conclusion?

Answer: 504510 First, note that none of the numbers will ever be 0. Let ? denote the replacementoperation. For each pair on the board (xi, yi) define its primary form to be (xi, yi) and its secondaryform to be [xiyi,

xi

yi]. Note that the primary form determines the secondary form uniquely and vice

versa. In secondary form,

[a1, b1] ? [a2, b2] =

(√a1b1,

√a1b1

)?

(√a2b2,

√a2b2

)=

(a1b2,

a1b2

)= [a21, b

22].

Thus we may replace all pairs on the board by their secondary form and use the above rule for ?instead. From the above rule, we see that if the leftmost number on the board is x, then after oneminute it will be x or x2 depending on whether it was erased in the intervening step, and similarly forthe rightmost number. Let k be the number of times the leftmost pair is erased and n be the numberof times the rightmost pair is erased. Then the final pair is[

40222k

,

(1007

1006

)2n]. (2)

Any step except the last cannot involve both the leftmost and rightmost pair, so k + n ≤ 1005. Sinceevery pair must be erased at least once, k, n ≥ 1. Every pair of integers satisfying the above can occur,for example, by making 1005〉 k 〉 n moves involving only the pairs in the middle, then making k 〉 1moves involving the leftmost pair, and finally n moves involving the rightmost pair.

In light of (2), the answer is the number of possible pairs (k, n), which is

1004∑k=1

1005−k∑n=1

1 =

1004∑k=1

1005〉k =

1004∑k=1

k =1004 · 1005

2= 504510.

15. Let f(x) = x2 〉 r2x + r3 for all real numbers x, where r2 and r3 are some real numbers. Define asequence {gn} for all nonnegative integers n by g0 = 0 and gn+1 = f(gn). Assume that {gn} satisfiesthe following three conditions: (i) g2i < g2i+1 and g2i+1 > g2i+2 for all 0 ≤ i ≤ 2011; (ii) there existsa positive integer j such that gi+1 > gi for all i > j, and (iii) {gn} is unbounded. If A is the greatestnumber such that A ≤ |r2| for any function f satisfying these properties, find A.

Answer: 2 Consider the function f(x) 〉 x. By the constraints of the problem, f(x) 〉 x must benegative for some x, namely, for x = g2i+1, 0 ≤ i ≤ 2011. Since f(x) 〉 x is positive for x of largeabsolute value, the graph of f(x) 〉 x crosses the x-axis twice and f(x) 〉 x has two real roots, saya < b. Factoring gives f(x)〉 x = (x〉 a)(x〉 b), or f(x) = (x〉 a)(x〉 b) + x.

Now, for x < a, f(x) > x > a, while for x > b, f(x) > x > b. Let c 6= b be the number such thatf(c) = f(b) = b. Note that b is not the vertex as f(a) = a < b, so by the symmetry of quadratics, cexists and b+c

2 = r22 as the vertex of the parabola. By the same token, b+a

2 = r2+12 is the vertex of

f(x)〉 x. Hence c = a〉 1. If f(x) > b then x < c or x > b. Consider the smallest j such that gj > b.Then by the above observation, gj−1 < c. (If gi ≥ b then f(gi) ≥ gi ≥ b so by induction, gi+1 ≥ gi forall i ≥ j. Hence j > 1; in fact j ≥ 4025.) Since gj−1 = f(gj−2), the minimum value of f is less than c.

Algebra & Combinatorics Individual Test

The minimum value is the value of f evaluated at its vertex, b+a−12 , so

f

(b+ a〉 1

2

)< c(

b+ a〉 1

2〉 a)(

b+ a〉 1

2〉 b)

+b+ a〉 1

2< a〉 1

1〉 (b〉 a)2

4+b〉 a+ 1

2< 0

3

4<

(b〉 a)2

4〉 b〉 a

2

4 < (b〉 a〉 1)2.

Then either b 〉 a 〉 1 < 〉2 or b 〉 a 〉 1 > 2, but b > a, so the latter must hold and (b 〉 a)2 > 9.Now, the discriminant of f(x) 〉 x equals (b 〉 a)2 (the square of the difference of the two roots) and(r2 + 1)2 〉 4r3 (from the coefficients), so (r2 + 1)2 > 9 + 4r3. But r3 = g1 > g0 = 0 so |r2| > 2.

We claim that we can make |r2| arbitrarily close to 2, so that the answer is 2. First define Gi, i ≥ 0as follows. Let N ≥ 2012 be an integer. For ε > 0 let h(x) = x2 〉 2 〉 ε, gε(x) = 〉

√x+ 2 + ε and

G2N+1 = 2 + ε, and define Gi recursively by Gi = gε(Gi+1), Gi+1 = h(Gi). (These two equations areconsistent.) Note the following.

(i) G2i < G2i+1 andG2i+1 > G2i+2 for 0 ≤ i ≤ N〉1. First noteG2N = 〉√

4 + 2ε > 〉√

4 + 2ε+ ε2 =〉2 〉 ε. Let l be the negative solution to h(x) = x. Note that 〉2 〉 ε < G2N < l < 0 sinceh(G2N ) > 0 > G2N . Now gε(x) is defined as long as x ≥ 〉2〉 ε, and it sends (〉2〉 ε, l) into (l, 0)and (l, 0) into (〉2〉 ε, l). It follows that the Gi, 0 ≤ i ≤ 2N are well-defined; moreover, G2i < land G2i+1 > l for 0 ≤ i ≤ N 〉 1 by backwards induction on i, so the desired inequalities follow.

(ii) Gi is increasing for i ≥ 2N + 1. Indeed, if x ≥ 2 + ε, then x2 〉 x = x(x〉 1) > 2 + ε so h(x) > x.Hence 2 + ε = G2N+1 < G2N+2 < · · · .

(iii) Gi is unbounded. This follows since h(x)〉 x = x(x〉 2)〉 2〉 ε is increasing for x > 2 + ε, so Giincreases faster and faster for i ≥ 2N + 1.

Now define f(x) = h(x + G0) 〉 G0 = x2 + 2G0x + G20 〉 G0 〉 2 〉 ε. Note Gi+1 = h(Gi) while

gi+1 = f(gi) = h(gi + G0) 〉 G0, so by induction gi = Gi 〉 G0. Since {Gi}∞i=0 satisfies (i), (ii), and(iii), so does gi.

We claim that we can make G0 arbitrarily close to 〉1 by choosing N large enough and ε small enough;this will make r2 = 〉2G0 arbitrarily close to 2. Choosing N large corresponds to taking G0 to be alarger iterate of 2 + ε under gε(x). By continuity of this function with respect to x and ε, it suffices totake ε = 0 and show that (letting g = g0)

g(n)(2) = g(· · · g︸ ︷︷ ︸n

(2) · · · )→ 〉1 as n→∞.

But note that for 0 ≤ θ ≤ π2 ,

g(〉2 cos θ) = 〉√

2〉 2 cos θ = 〉2 sin

(θ

2

)= 2 cos

(π

2〉 θ

2

).

Hence by induction, g(n)(〉2 cos θ) = 〉2 cosπ2 〉

π4 + · · ·+ (〉1)n

θ 〉 π

2n

)). Hence g(n)(2) = g(n−1)(〉2 cos 0)

converges to 〉2 cos(π2 〉π4 + · · · ) = 〉2 cos(π3 ) = 〉1, as needed.

16. Let A = {1, 2, . . . , 2011}. Find the number of functions f from A to A that satisfy f(n) ≤ n for all nin A and attain exactly 2010 distinct values.

Answer: 22011 〉 2012 Let n be the element of A not in the range of f . Let m be the element of Athat is hit twice.

Algebra & Combinatorics Individual Test

We now sum the total number of functions over n,m. Clearly f(1) = 1, and by induction, for x ≤m, f(x) = x. Also unless n = 2011, f(2011) = 2011 because f can take no other number to 2011. Itfollows from backwards induction that for x > n, f(x) = x. Therefore n > m, and there are only n〉mvalues of f that are not fixed.

Now f(m+ 1) = m or f(m+ 1) = m+ 1. For m < k < n,given the selection of f(1), f(2), . . . , f(k〉 1),k〉 1 of the k+ 1 possible values of f(k+ 1) (1, 2, 3, . . . , k, and counting m twice) have been taken, sothere are two distinct values that f(k+ 1) can take (one of them is k+ 1, and the other is not, so theyare distinct). For f(n), when the other 2010 values of f have been assigned, there is only one missing,so f(n) is determined.

For each integer in [m,n), there are two possible values of f , so there are 2n−m−1 different functionsf for a given m,n. So our answer is

2010∑m=1

2011∑n=m+1

2n−m−1 =

2010∑m=1

2−m−12011∑

n=m+1

2n

=

2010∑m=1

2−m−1(22012 〉 2m+1)

=

2010∑m=1

22011−m 〉 1

=

(2010∑m=1

2m

)〉 2010

= 22011 〉 2012

17. Let z = cos2π

2011+ i sin

2π

2011, and let

P (x) = x2008 + 3x2007 + 6x2006 + . . .2008 · 2009

2x+

2009 · 2010

2

for all complex numbers x. Evaluate P (z)P (z2)P (z3) . . . P (z2010).

Answer: 20112009 · (10052011 〉 10042011) Multiply P (x) by x〉 1 to get

P (x)(x〉 1) = x2009 + 2x2008 + . . .+ 2009x〉 2009 · 2010

2,

or,

P (x)(x〉 1) + 2010 · 1005 = x2009 + 2x2008 + . . .+ 2009x+ 2010.

Multiplying by x〉 1 once again:

(x〉 1)(P (x)(x〉 1) +2010 · 2011

2) = x2010 + x2009 + . . .+ x〉 2010,

= (x2010 + x2009 + . . .+ x+ 1)〉 2011.

Hence,

P (x) =

(x2010 + x2009 + . . .+ x+ 1)〉 2011

x〉 1〉 2011 · 1005

x〉 1

Algebra & Combinatorics Individual Test

Note that x2010+x2009+. . .+x+1 has z, z2, . . . , z2010 as roots, so they vanish at those points. Pluggingthose 2010 powers of z into the last equation, and multiplying them together, we obtain

2010∏i=1

P (zi) =(〉2011) · 1005 · (x〉 1004

1005)

(x〉 1)2.

Note that (x〉 z)(x〉 z2) . . . (x〉 z2010) = x2010 + x2009 + . . .+ 1. Using this, the product turns out tobe 20112009 · (10052011 〉 10042011).

18. Let n be an odd positive integer, and suppose that n people sit on a committee that is in the processof electing a president. The members sit in a circle, and every member votes for the person either tohis/her immediate left, or to his/her immediate right. If one member wins more votes than all theother members do, he/she will be declared to be the president; otherwise, one of the the members whowon at least as many votes as all the other members did will be randomly selected to be the president.If Hermia and Lysander are two members of the committee, with Hermia sitting to Lysander’s leftand Lysander planning to vote for Hermia, determine the probability that Hermia is elected president,assuming that the other n〉 1 members vote randomly.

Answer: 2n−1n2n−1 Let x be the probability Hermia is elected if Lysander votes for her, and let y be the

probability that she wins if Lysander does not vote for her. We are trying to find x, and do so by firstfinding y. If Lysander votes for Hermia with probability 1

2 then the probability that Hermia is electedchairman is x

2 + y2 , but it is also 1

n by symmetry. If Lysander does not vote for Hermia, Hermia canget at most 1 vote, and then can only be elected if everyone gets one vote and she wins the tiebreaker.The probability she wins the tiebreaker is 1

n , and chasing around the circle, the probability that everyperson gets 1 vote is 1

2n−1 . (Everyone votes for the person to the left, or everyone votes for the personto the right.) Hence

y =1

n2n−1.

Then x2 + 1

n2n = 1n , so solving for x gives

x =2n 〉 1

n2n−1.

19. Let {an} and {bn} be sequences defined recursively by a0 = 2; b0 = 2, and an+1 = an√

1 + a2n + b2n〉bn;

bn+1 = bn√

1 + a2n + b2n + an. Find the ternary (base 3) representation of a4 and b4.

Answer: 1000001100111222 and 2211100110000012

Note first that√

1 + a2n + b2n = 32n

. The proof is by induction; the base case follows trivially from what

is given. For the inductive step, note that 1+a2n+1+b2n+1 = 1+a2n(1+a2n+b2n)+b2n〉2anbn√

1 + a2n + b2n+

b2n(1 + a2n + b2n) + a2n + 2anbn√

1 + a2n + b2n = 1 + (a2n + b2n)(1 + a2n + b2n) + a2n + b2n = (1 + a2n + b2n)2.

Invoking the inductive hypothesis, we see that√

1 + a2n+1 + b2n+1 = (32n

)2 = 32n+1

, as desired.

The quickest way to finish from here is to consider a sequence of complex numbers {zn} defined byzn = an+ bni for all nonnegative integers n. It should be clear that z0 = 2 + 2i and zn+1 = zn(32

n

+ i).

Therefore, z4 = (2 + 2i)(320

+ i)(321

+ i)(322

+ i)(323

+ i). This product is difficult to evaluate inthe decimal number system, but in ternary the calculation is a cinch! To speed things up, we willuse balanced ternary1, in which the three digits allowed are 〉1, 0, and 1 rather than 0, 1, and 2. Letx+ yi = (32

0

+ i)(321

+ i)(322

+ i)(323

+ i), and consider the balanced ternary representation of x andy. For all 0 ≤ j ≤ 15, let xj denote the digit in the 3j place of x, let yj denote the digit in the 3j placeof y, and let b(j) denote the number of ones in the binary representation of j. It should be clear thatxj = 〉1 if b(j) ≡ 2 (mod 4), xj = 0 if b(j) ≡ 1 (mod 2), and xj = 1 if b(j) ≡ 0 (mod 4). Similarly,yj = 〉1 if b(j) ≡ 1 (mod 4), yj = 0 if b(j) ≡ 0 (mod 2), and yj = 1 if b(j) ≡ 3 (mod 4). Convertingto ordinary ternary representation, we see that x = 2212112211220013 and y = 1100222022121203. It

1http://en.wikipedia.org/wiki/Balanced_ternary

Algebra & Combinatorics Individual Test

remains to note that a4 = 2x〉 2y and b4 = 2x+ 2y and perform the requisite arithmetic to arrive atthe answer above.

20. Alice and Bob play a game in which two thousand and eleven 2011×2011 grids are distributed betweenthe two of them, 1 to Bob, and the other 2010 to Alice. They go behind closed doors and fill theirgrid(s) with the numbers 1, 2, . . . , 20112 so that the numbers across rows (left-to-right) and downcolumns (top-to-bottom) are strictly increasing. No two of Alice’s grids may be filled identically. Afterthe grids are filled, Bob is allowed to look at Alice’s grids and then swap numbers on his own grid,two at a time, as long as the numbering remains legal (i.e. increasing across rows and down columns)after each swap. When he is done swapping, a grid of Alice’s is selected at random. If there exist twointegers in the same column of this grid that occur in the same row of Bob’s grid, Bob wins. Otherwise,Alice wins. If Bob selects his initial grid optimally, what is the maximum number of swaps that Bobmay need in order to guarantee victory?

Answer: 1Consider the grid whose entries in the jth row are, in order, 2011j〉2010, 2011j〉2009, . . . , 2011j. Callthis grid A0. For k = 1, 2 . . . , 2010, let grid Ak be the grid obtained from A0 by swapping the rightmostentry of the kth row with the leftmost entry of the k+1st row. We claim that if A ∈ {A0, A1, . . . , A2010},then given any legally numbered grid B such that A and B differ in at least one entry, there exist twointegers in the same column of B that occur in the same row of A.We first consider A0. Assume for the sake of contradiction B is a legally numbered grid distinct fromA0, such that there do not exist two integers in the same column of B that occur in the same rowof A0. Since the numbers 1, 2, . . . , 2011 occur in the same row of A0, they must all occur in differentcolumns of B. Clearly 1 is the leftmost entry in B’s first row. Let m be the smallest number that doesnot occur in the first row of B. Since each row is in order, m must be the first entry in its row. Butthen 1 and m are in the same column of B, a contradiction. It follows that the numbers 1, 2, . . . , 2011all occur in the first row of B. Proceeding by induction, 2011j〉2010, 2011j〉2009, . . . , 2011j must alloccur in the jth row of B for all 1 ≤ j ≤ 2011. Since A0 is the only legally numbered grid satsifyingthis condition, we have reached the desired contradiction.Now note that if A ∈ {A1, . . . , A2010}, there exist two integers in the same column of A0 that occur inthe same row of A. In particular, if A = Ak and 1 ≤ k ≤ 2010, then the integers 2011k 〉 2010 and2011k + 1 occur in the same column of A0 and in the same row of Ak. Therefore, it suffices to showthat for all 1 ≤ k ≤ 2010, there is no legally numbered grid B distinct from Ak and A0 such that theredo not exist two integers in the same column of B that occur in the same row of A0. Assume for thesake of contradiction that there does exist such a grid B. By the same logic as above, applied to thefirst k〉1 rows and applied backwards to the last 2010〉k〉1 rows, we see that B may only differ fromAk in the kth and k+ 1st rows. However, there are only two legally numbered grids that are identicalto Ak outside of rows k and k + 1, namely A0 and Ak. This proves the claim.It remains only to note that, by the pigeonhole principle, if one of Alice’s grids is A0, then there existsa positive integer k, 1 ≤ k ≤ 2010, such that Ak is not one of the Alice’s grids. Therefore, if Bob setshis initial grid to be A0, he will require only one swap to switch his grid to Ak after examining Alice’sgrids. If A0 is not among Alice’s grids, then if Bob sets his initial grid to be A0, he will not in factrequire any swaps at all.

Algebra & Combinatorics Individual Test

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

1. Let a, b, and c be positive real numbers. Determine the largest total number of real roots that thefollowing three polynomials may have among them: ax2 + bx+ c, bx2 + cx+ a, and cx2 + ax+ b.

2. Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠BAC intersect line BC atD. If there exist points E and F on sides AC and BC, respectively, such that lines AD and EFare parallel and divide triangle ABC into three parts of equal area, determine the number of possibleintegral values for BC.

3. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner.At each step, he can either move one square to the right or simultaneously move one square to the leftand one square up. In how many ways can he reach the center square of the topmost row?

4. Let H be a regular hexagon of side length x. Call a hexagon in the same plane a “distortion” of H ifand only if it can be obtained from H by translating each vertex of H by a distance strictly less than1. Determine the smallest value of x for which every distortion of H is necessarily convex.

5. Let a ? b = ab+ a+ b for all integers a and b. Evaluate 1 ? (2 ? (3 ? (4 ? . . . (99 ? 100) . . .))).

6. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. IfNathaniel goes first, determine the probability that he ends up winning.

7. Find all integers x such that 2x2 + x− 6 is a positive integral power of a prime positive integer.

8. Let ABCDEF be a regular hexagon of area 1. Let M be the midpoint of DE. Let X be theintersection of AC and BM , let Y be the intersection of BF and AM , and let Z be the intersectionof AC and BF . If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate[BXC] + [AY F ] + [ABZ]− [MXZY ].

9. For all real numbers x, let

f(x) =1

2011√

1− x2011.

Evaluate (f(f(. . . (f(2011)) . . .)))2011, where f is applied 2010 times.

10. Let ABCD be a square of side length 13. Let E and F be points on rays AB and AD, respectively,so that the area of square ABCD equals the area of triangle AEF . If EF intersects BC at X andBX = 6, determine DF .

11. Let f(x) = x2 + 6x+ c for all real numbers x, where c is some real number. For what values of c doesf(f(x)) have exactly 3 distinct real roots?

12. Let ABCDEF be a convex equilateral hexagon such that lines BC, AD, and EF are parallel. LetH be the orthocenter of triangle ABD. If the smallest interior angle of the hexagon is 4 degrees,determine the smallest angle of the triangle HAD in degrees.

13. How many polynomials P with integer coefficients and degree at most 5 satisfy 0 ≤ P (x) < 120 for allx ∈ {0, 1, 2, 3, 4, 5}?

14. Let ABCD be a cyclic quadrilateral, and suppose that BC = CD = 2. Let I be the incenter of triangleABD. If AI = 2 as well, find the minimum value of the length of diagonal BD.

15. Let f(x) = x2 − r2x + r3 for all real numbers x, where r2 and r3 are some real numbers. Define asequence {gn} for all nonnegative integers n by g0 = 0 and gn+1 = f(gn). Assume that {gn} satisfiesthe following three conditions: (i) g2i < g2i+1 and g2i+1 > g2i+2 for all 0 ≤ i ≤ 2011; (ii) there existsa positive integer j such that gi+1 > gi for all i > j, and (iii) {gn} is unbounded. If A is the greatestnumber such that A ≤ |r2| for any function f satisfying these properties, find A.

16. Let ABCD be a quadrilateral inscribed in the unit circle such that ∠BAD is 30 degrees. Let m denotethe minimum value of CP + PQ + CQ, where P and Q may be any points lying along rays AB andAD, respectively. Determine the maximum value of m.

17. Let z = cos2π

2011+ i sin

2π

2011, and let

P (x) = x2008 + 3x2007 + 6x2006 + . . .2008 · 2009

2x+

2009 · 2010

2

for all complex numbers x. Evaluate P (z)P (z2)P (z3) . . . P (z2010).

18. Collinear points A, B, and C are given in the Cartesian plane such that A = (a, 0) lies along thex-axis, B lies along the line y = x, C lies along the line y = 2x, and AB/BC = 2. If D = (a, a), thecircumcircle of triangle ADC intersects y = x again at E, and ray AE intersects y = 2x at F , evaluateAE/EF .

19. Let {an} and {bn} be sequences defined recursively by a0 = 2; b0 = 2, and an+1 = an√

1 + a2n + b2n−bn;

bn+1 = bn√

1 + a2n + b2n + an. Find the ternary (base 3) representation of a4 and b4.

20. Let ω1 and ω2 be two circles that intersect at points A and B. Let line l be tangent to ω1 at P and toω2 at Q so that A is closer to PQ than B. Let points R and S lie along rays PA and QA, respectively,so that PQ = AR = AS and R and S are on opposite sides of A as P and Q. Let O be the circumcenterof triangle ASR, and let C and D be the midpoints of major arcs AP and AQ, respectively. If ∠APQis 45 degrees and ∠AQP is 30 degrees, determine ∠COD in degrees.

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

Algebra & Geometry Individual Test

1. Let a, b, and c be positive real numbers. Determine the largest total number of real roots that thefollowing three polynomials may have among them: ax2 + bx+ c, bx2 + cx+ a, and cx2 + ax+ b.

Answer: 4 If all the polynomials had real roots, their discriminants would all be nonnegative: a2 ≥4bc, b2 ≥ 4ca, and c2 ≥ 4ab. Multiplying these inequalities gives (abc)2 ≥ 64(abc)2, a contradiction.Hence one of the quadratics has no real roots. The maximum of 4 real roots is attainable: for example,the values (a, b, c) = (1, 5, 6) give 〉2,〉3 as roots to x2 + 5x+ 6 and 〉1,〉 1

5 as roots to 5x2 + 6x+ 1.

2. Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠BAC intersect line BC atD. If there exist points E and F on sides AC and BC, respectively, such that lines AD and EFare parallel and divide triangle ABC into three parts of equal area, determine the number of possibleintegral values for BC.

Answer: 13

A B

C

DE

F

7

14

Note that such E,F exist if and only if[ADC]

[ADB]= 2. (1)

([ ] denotes area.) Since AD is the angle bisector, and the ratio of areas of triangles with equal heightis the ratio of their bases,

AC

AB=DC

DB=

[ADC]

[ADB].

Hence (1) is equivalent to AC = 2AB = 14. Then BC can be any length d such that the triangleinequalities are satisfied:

d+ 7 > 14

7 + 14 > d

Hence 7 < d < 21 and there are 13 possible integral values for BC.

Algebra & Geometry Individual Test

3. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner.At each step, he can either move one square to the right or simultaneously move one square to the leftand one square up. In how many ways can he reach the center square of the topmost row?

Answer: 2n−1 Note that Josh must pass through the center square of each row. There are 2 waysto get from the center square of row k to the center square of row k+ 1. So there are 2n−1 ways to getto the center square of row n.

4. Let H be a regular hexagon of side length x. Call a hexagon in the same plane a “distortion” of H ifand only if it can be obtained from H by translating each vertex of H by a distance strictly less than1. Determine the smallest value of x for which every distortion of H is necessarily convex.

Answer: 4

A1

A2

A3 A4

A5

A6X X′

Y Y ′

Let H = A1A2A3A4A5A6 be the hexagon, and for all 1 ≤ i ≤ 6, let points A′i be considered such thatAiA

′i < 1. Let H ′ = A′1A

′2A′3A′4A′5A′6, and consider all indices modulo 6. For any point P in the plane,

let D(P ) denote the unit disk {Q|PQ < 1} centered at P ; it follows that A′i ∈ D(Ai).

Let X and X ′ be points on line A1A6, and let Y and Y ′ be points on line A3A4 such that A1X =A1X

′ = A3Y = A3Y′ = 1 and X and X ′ lie on opposite sides of A1 and Y and Y ′ lie on opposite sides

of A3. If X ′ and Y ′ lie on segments A1A6 and A3A4, respectively, then segment A′1A′3 lies between the

lines XY and X ′Y ′. Note that x2 is the distance from A2 to A1A3.

Algebra & Geometry Individual Test

A1

A2

A3 A4

A5

A6X X′

Y Y ′

If x2 ≥ 2, then C(A2) cannot intersect line XY , since the distance from XY to A1A3 is 1 and the

distance from XY to A2 is at least 1. Therefore, A′1A′3 separates A′2 from the other 3 vertices of the

hexagon. By analogous reasoning applied to the other vertices, we may conclude that H ′ is convex.

If x2 < 2, then C(A2) intersects XY , so by choosing A′1 = X and A′3 = Y , we see that we may choose

A′2 on the opposite side of XY , in which case H ′ will be concave. Hence the answer is 4, as desired.

5. Let a ? b = ab+ a+ b for all integers a and b. Evaluate 1 ? (2 ? (3 ? (4 ? . . . (99 ? 100) . . .))).

Answer: 101!〉 1

We will first show that ? is both commutative and associative.

• Commutativity: a ? b = ab+ a+ b = b ? a

• Associativity: a ? (b ? c) = a(bc + b + c) + a + bc + b + c = abc + ab + ac + bc + a + b + c and(a ? b) ? c = (ab+ a+ b)c+ ab+ a+ b+ c = abc+ ab+ ac+ bc+ a+ b+ c. So a ? (b ? c) = (a ? b) ? c.

So we need only calculate ((. . . (1 ? 2) ? 3) ? 4) . . . ? 100). We will prove by induction that

((. . . (1 ? 2) ? 3) ? 4) . . . ? n) = (n+ 1)!〉 1.

• Base case (n = 2): (1 ? 2) = 2 + 1 + 2 = 5 = 3!〉 1

• Inductive step:

Suppose that(((. . . (1 ? 2) ? 3) ? 4) . . . ? n) = (n+ 1)!〉 1.

Then,

((((. . . (1 ? 2) ? 3) ? 4) . . . ? n) ? (n+ 1)) = ((n+ 1)!〉 1) ? (n+ 1)

= (n+ 1)!(n+ 1)〉 (n+ 1) + (n+ 1)!〉 1 + (n+ 1)

= (n+ 2)!〉 1

Hence, ((. . . (1 ? 2) ? 3) ? 4) . . . ? n) = (n+ 1)!〉 1 for all n. For n = 100, this results to 101!〉 1.

Algebra & Geometry Individual Test

6. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. IfNathaniel goes first, determine the probability that he ends up winning.

Answer: 511 For 1 ≤ k ≤ 6, let xk be the probability that the current player, say A, will win

when the number on the tally at the beginning of his turn is k modulo 7. The probability that thetotal is l modulo 7 after his roll is 1

6 for each l 6≡ k (mod 7); in particular, there is a 16 chance he wins

immediately. The chance that A will win if he leaves l on the board after his turn is 1〉 xl. Hence for1 ≤ k ≤ 6,

xk =1

6

∑1≤l≤6, l 6=k

(1〉 xl) +1

6.

Letting s =∑6l=1 xl, this becomes xk = xk−s

6 + 1 or 5xk

6 = 〉 s6 + 1. Hence x1 = · · · = x6, and 6xk = s

for every k. Plugging this in gives 11xk

6 = 1, or xk = 611 .

Since Nathaniel cannot win on his first turn, he leaves Obediah with a number not divisible by 7.Hence Obediah’s chance of winning is 6

11 and Nathaniel’s chance of winning is 511 .

7. Find all integers x such that 2x2 + x〉 6 is a positive integral power of a prime positive integer.

Answer: 〉3, 2, 5 Let f(x) = 2x2 + x 〉 6 = (2x 〉 3)(x + 2). Suppose a positive integer a divides

both 2x〉 3 and x+ 2. Then a must also divide 2(x+ 2)〉 (2x〉 3) = 7. Hence, a can either be 1 or7. As a result, 2x 〉 3 = 7n or 〉7n for some positive integer n, or either x + 2 or 2x 〉 3 is ±1. Weconsider the following cases:

• (2x〉 3) = 1. Then x = 2, which yields f(x) = 4, a prime power.

• (2x〉 3) = 〉1. Then x = 1, which yields f(x) = 〉3, not a prime power.

• (x+ 2) = 1). Then x = 〉1, which yields f(x) = 〉5 not a prime power.

• (x+ 2) = 〉1. Then x = 〉3, which yields f(x) = 9, a prime power.

• (2x〉 3) = 7. Then x = 5, which yields f(x) = 49, a prime power.

• (2x〉 3) = 〉7. Then x = 〉2, which yields f(x) = 0, not a prime power.

• (2x〉 3) = ±7n, for n ≥ 2. Then, since x+ 2 =(2x〉 3) + 7

2, we have that x+ 2 is divisible by 7

but not by 49. Hence x + 2 = ±7, yielding x = 5,〉9. The former has already been considered,while the latter yields f(x) = 147.

So x can be either -3, 2 or 5.

(Note: In the official solutions packet we did not list the answer -3. This oversight was quickly noticedon the day of the test, and only the answer 〉3, 2, 5 was marked as correct.

8. Let ABCDEF be a regular hexagon of area 1. Let M be the midpoint of DE. Let X be theintersection of AC and BM , let Y be the intersection of BF and AM , and let Z be the intersectionof AC and BF . If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate[BXC] + [AY F ] + [ABZ]〉 [MXZY ].

Answer: 0

Algebra & Geometry Individual Test

A B

C

DE

F

M

XY

Z

O

Let O be the center of the hexagon. The desired area is [ABCDEF ] 〉 [ACDM ] 〉 [BFEM ]. Notethat [ADM ] = [ADE]/2 = [ODE] = [ABC], where the last equation holds because sin 60◦ = sin 120◦.Thus, [ACDM ] = [ACD] + [ADM ] = [ACD] + [ABC] = [ABCD], but the area of ABCD is half thearea of the hexagon. Similarly, the area of [BFEM ] is half the area of the hexagon, so the answer iszero.

9. For all real numbers x, let

f(x) =1

2011√

1〉 x2011.

Evaluate (f(f(. . . (f(2011)) . . .)))2011, where f is applied 2010 times.

Answer: 20112011 Direct calculation shows that f(f(x)) =2011√

1〉 x2011〉x

and f(f(f(x))) = x.

Hence (f(f(. . . (f(x)) . . .))) = x, where f is applied 2010 times. So (f(f(. . . (f(2011)) . . .)))2011 =20112011.

10. Let ABCD be a square of side length 13. Let E and F be points on rays AB and AD, respectively,so that the area of square ABCD equals the area of triangle AEF . If EF intersects BC at X andBX = 6, determine DF .

Answer:√

13

A B

CD

E

F

X

Y

6

7

x

y

Algebra & Geometry Individual Test

First Solution

Let Y be the point of intersection of lines EF and CD. Note that [ABCD] = [AEF ] implies that[BEX] + [DY F ] = [CY X]. Since 4BEX ∼ 4CY X ∼ 4DY F , there exists some constant r suchthat [BEX] = r · BX2, [Y DF ] = r · CX2, and [CY X] = r · DF 2. Hence BX2 + DF 2 = CX2, soDF =

√CX2 〉BX2 =

√49〉 36 =

√13.

Second Solution

Let x = DF and y = Y D. Since 4BXE ∼ 4CXY ∼ 4DFY , we have

BE

BX=CY

CX=DY

DF=y

x.

Using BX = 6, XC = 7 and CY = 13〉 y we get BE = 6yx and 13−y

7 = yx . Solving this last equation

for y gives y = 13xx+7 . Now [ABCD] = [AEF ] gives

169 =1

2AE ·AF =

1

2

(13 +

6y

x

)(13 + x) .

169 = 6y + 13x+78y

x

13 =6x

x+ 7+ x+

78

x+ 7

0 = x2 〉 13.

Thus x =√

13.

11. Let f(x) = x2 + 6x+ c for all real numbers x, where c is some real number. For what values of c doesf(f(x)) have exactly 3 distinct real roots?

Answer: 11−√13

2 Suppose f has only one distinct root r1. Then, if x1 is a root of f(f(x)), it must

be the case that f(x1) = r1. As a result, f(f(x)) would have at most two roots, thus not satisfyingthe problem condition. Hence f has two distinct roots. Let them be r1 6= r2.

Since f(f(x)) has just three distinct roots, either f(x) = r1 or f(x) = r2 has one distinct root. Assumewithout loss of generality that r1 has one distinct root. Then f(x) = x2 + 6x+ c = r1 has one root, sothat x2 + 6x + c 〉 r1 is a square polynomial. Therefore, c 〉 r1 = 9, so that r1 = c 〉 9. So c 〉 9 is aroot of f . So (c〉 9)2 + 6(c〉 9) + c = 0, yielding c2 〉 11c+ 27 = 0, or (c〉 11

2 )2 = 132 . This results to

c = 11±√13

2 .

If c = 11−√13

2 , f(x) = x2 + 6x + 11−√13

2 = (x + 7+√13

2 )(x + 5−√13

2 ). We know f(x) = −7−√13

2 has a

double root, -3. Now −5+√13

2 > −7−√13

2 so the second root is above the vertex of the parabola, and ishit twice.

If c = 11+√13

2 , f(x) = x2 + 6x + 11+√13

2 = (x + 7−√13

2 )(x + 5+√13

2 ). We know f(x) = −7+√13

2 has adouble root, -3, and this is the value of f at the vertex of the parabola, so it is its minimum value.

Since −5−√13

2 < −7+√13

2 , f(x) = −5−√13

2 has no solutions. So in this case, f has only one real root.

So the answer is c = 11−√13

2 .

Note: In the solutions packet we had both roots listed as the correct answer. We noticed this oversighton the day of the test and awarded points only for the correct answer.

12. Let ABCDEF be a convex equilateral hexagon such that lines BC, AD, and EF are parallel. LetH be the orthocenter of triangle ABD. If the smallest interior angle of the hexagon is 4 degrees,determine the smallest angle of the triangle HAD in degrees.

Answer: 3

Algebra & Geometry Individual Test

A

B

C

D

E

FH

A′

B′

D′

A

B

C

D

E

FH

Note that ABCD and DEFA are isosceles trapezoids, so ∠BAD = ∠CDA and ∠FAD = ∠EDA. Inorder for the hexagon to be convex, the angles at B, C, E, and F have to be obtuse, so ∠A = ∠D = 4◦.Letting s be a side length of the hexagon, AD = AB cos∠BAD + BC + CD cos∠CDA = s(1 +2 cos∠BAD), so ∠BAD is uniquely determined by AD. Since the same equation holds for trapezoidDEFA, it follows that ∠BAD = ∠FAD = ∠CDA = ∠EDA = 2◦. Then ∠BCD = 180◦ 〉 2◦ = 178◦.Since 4BCD is isosceles, ∠CDB = 1◦ and ∠BDA = 1◦. (One may also note that ∠BDA = 1◦ byobserving that equal lengths AB and BC must intercept equal arcs on the circumcircle of isoscelestrapezoid ABCD).

Let A′, B′, andD′ be the feet of the perpendiculars from A, B, andD toBD, DA, and AB, respectively.Angle chasing yields

∠AHD = ∠AHB′ + ∠DHB′ = (90◦ 〉 ∠A′AB′) + (90◦ 〉 ∠D′DB′)

= ∠BDA+ ∠BAD = 1◦ + 2◦ = 3◦

∠HAD = 90◦ 〉 ∠AHB′ = 89◦

∠HDA = 90◦ 〉 ∠DHB′ = 88◦

Hence the smallest angle in 4HAD is 3◦.

It is faster, however, to draw the circumcircle of DEFA, and to note that since H is the orthocenterof triangle ABD, B is the orthocenter of triangle HAD. Then since F is the reflection of B acrossAD, quadrilateral HAFD is cyclic, so ∠AHD = ∠ADF + ∠DAF = 1◦ + 2◦ = 3◦, as desired.

Algebra & Geometry Individual Test

13. How many polynomials P with integer coefficients and degree at most 5 satisfy 0 ≤ P (x) < 120 for allx ∈ {0, 1, 2, 3, 4, 5}?

Answer: 86400000

For each nonnegative integer i, let xi = x(x〉 1) · · · (x〉 i+ 1). (Define x0 = 1.)

Lemma: Each polynomial with integer coefficients f can be uniquely written in the form

f(x) = anxn + . . .+ a1x

1 + a0x0, an 6= 0.

Proof: Induct on the degree. The base case (degree 0) is clear. If f has degree m with leading coefficientc, then by matching leading coefficients we must have m = n and an = c. By the induction hypothesis,f(x)〉 cxn can be uniquely written as an−1x

n−1(x) + . . .+ a1x1 + a0x

0.

There are 120 possible choices for a0, namely any integer in [0, 120). Once a0, . . . , ai−1 have beenchosen so 0 ≤ P (0), . . . , P (i〉 1) < 120, for some 0 ≤ i ≤ 5, then we have

P (i) = aii! + ai−1ii−1 + · · ·+ a0

so by choosing ai we can make P (i) any number congruent to ai−1ii−1 + · · ·+a0 modulo i!. Thus there

are 120i! choices for ai. Note the choice of ai does not affect the value of P (0), . . . , P (i 〉 1). Thus all

polynomials we obtain in this way are valid. The answer is

5∏i=0

120

i!= 86400000.

Note: Their is also a solution involving finite differences that is basically equivalent to this solution.One proves that for i = 0, 1, 2, 3, 4, 5 there are 5!

i! ways to pick the ith finite difference at the point 0.

14. Let ABCD be a cyclic quadrilateral, and suppose that BC = CD = 2. Let I be the incenter of triangleABD. If AI = 2 as well, find the minimum value of the length of diagonal BD.

Answer: 2√

3

A

B

C

D

I

Algebra & Geometry Individual Test

Let T be the point where the incircle intersects AD, and let r be the inradius and R be the circumradiusof 4ABD. Since BC = CD = 2, C is on the midpoint of arc BD on the opposite side of BD as A,and hence on the angle bisector of A. Thus A, I, and C are collinear. We have the following formulas:

AI =IM

sin∠IAM=

r

sin A2

BC = 2R sinA

2BD = 2R sinA

The last two equations follow from the extended law of sines on 4ABC and 4ABD, respectively.

Using AI = 2 = BC gives sin2 A2 = r

2R . However, it is well-known that R ≥ 2r with equality for an

equilateral triangle (one way to see this is the identity 1+ rR = cosA+cosB+cosD). Hence sin2 A

2 ≤14

and A2 ≤ 30◦. Then

BD = 2R

(2 sin

A

2cos

A

2

)= BC · 2 cos

A

2≥ 2

(2 ·√

3

2

)= 2√

3

with equality when 4ABD is equilateral.

Remark: Similar but perhaps simpler computations can be made by noting that if AC intersects BDat X, then AB/BX = AD/DX = 2, which follows from the exterior angle bisector theorem; if IA isthe A-excenter of triangle ABC, then AIA/XIA = 2 since it is well-known that C is the circumcenterof cyclic quadrilateral BIDIA.

15. Let f(x) = x2 〉 r2x + r3 for all real numbers x, where r2 and r3 are some real numbers. Define asequence {gn} for all nonnegative integers n by g0 = 0 and gn+1 = f(gn). Assume that {gn} satisfiesthe following three conditions: (i) g2i < g2i+1 and g2i+1 > g2i+2 for all 0 ≤ i ≤ 2011; (ii) there existsa positive integer j such that gi+1 > gi for all i > j, and (iii) {gn} is unbounded. If A is the greatestnumber such that A ≤ |r2| for any function f satisfying these properties, find A.

Answer: 2

Consider the function f(x)〉x. By the constraints of the problem, f(x)〉x must be negative for somex, namely, for x = g2i+1, 0 ≤ i ≤ 2011. Since f(x) 〉 x is positive for x of large absolute value, thegraph of f(x)〉 x crosses the x-axis twice and f(x)〉 x has two real roots, say a < b. Factoring givesf(x)〉 x = (x〉 a)(x〉 b), or f(x) = (x〉 a)(x〉 b) + x.

Now, for x < a, f(x) > x > a, while for x > b, f(x) > x > b. Let c 6= b be the number such thatf(c) = f(b) = b. Note that b is not the vertex as f(a) = a < b, so by the symmetry of quadratics, cexists and b+c

2 = r22 as the vertex of the parabola. By the same token, b+a

2 = r2+12 is the vertex of

f(x)〉 x. Hence c = a〉 1. If f(x) > b then x < c or x > b. Consider the smallest j such that gj > b.Then by the above observation, gj−1 < c. (If gi ≥ b then f(gi) ≥ gi ≥ b so by induction, gi+1 ≥ gi forall i ≥ j. Hence j > 1; in fact j ≥ 4025.) Since gj−1 = f(gj−2), the minimum value of f is less than c.The minimum value is the value of f evaluated at its vertex, b+a−12 , so

f

(b+ a〉 1

2

)< c(

b+ a〉 1

2〉 a)(

b+ a〉 1

2〉 b)

+b+ a〉 1

2< a〉 1

1〉 (b〉 a)2

4+b〉 a+ 1

2< 0

3

4<

(b〉 a)2

4〉 b〉 a

2

4 < (b〉 a〉 1)2.

Algebra & Geometry Individual Test

Then either b 〉 a 〉 1 < 〉2 or b 〉 a 〉 1 > 2, but b > a, so the latter must hold and (b 〉 a)2 > 9.Now, the discriminant of f(x) 〉 x equals (b 〉 a)2 (the square of the difference of the two roots) and(r2 + 1)2 〉 4r3 (from the coefficients), so (r2 + 1)2 > 9 + 4r3. But r3 = g1 > g0 = 0 so |r2| > 2.

We claim that we can make |r2| arbitrarily close to 2, so that the answer is 2. First define Gi, i ≥ 0as follows. Let N ≥ 2012 be an integer. For ε > 0 let h(x) = x2 〉 2 〉 ε, gε(x) = 〉

√x+ 2 + ε and

G2N+1 = 2 + ε, and define Gi recursively by Gi = gε(Gi+1), Gi+1 = h(Gi). (These two equations areconsistent.) Note the following.

(i) G2i < G2i+1 andG2i+1 > G2i+2 for 0 ≤ i ≤ N〉1. First noteG2N = 〉√

4 + 2ε > 〉√

4 + 2ε+ ε2 =〉2 〉 ε. Let l be the negative solution to h(x) = x. Note that 〉2 〉 ε < G2N < l < 0 sinceh(G2N ) > 0 > G2N . Now gε(x) is defined as long as x ≥ 〉2〉 ε, and it sends (〉2〉 ε, l) into (l, 0)and (l, 0) into (〉2〉 ε, l). It follows that the Gi, 0 ≤ i ≤ 2N are well-defined; moreover, G2i < land G2i+1 > l for 0 ≤ i ≤ N 〉 1 by backwards induction on i, so the desired inequalities follow.

(ii) Gi is increasing for i ≥ 2N + 1. Indeed, if x ≥ 2 + ε, then x2 〉 x = x(x〉 1) > 2 + ε so h(x) > x.Hence 2 + ε = G2N+1 < G2N+2 < · · · .

(iii) Gi is unbounded. This follows since h(x)〉 x = x(x〉 2)〉 2〉 ε is increasing for x > 2 + ε, so Giincreases faster and faster for i ≥ 2N + 1.

Now define f(x) = h(x + G0) 〉 G0 = x2 + 2G0x + G20 〉 G0 〉 2 〉 ε. Note Gi+1 = h(Gi) while

gi+1 = f(gi) = h(gi + G0) 〉 G0, so by induction gi = Gi 〉 G0. Since {Gi}∞i=0 satisfies (i), (ii), and(iii), so does gi.

We claim that we can make G0 arbitrarily close to 〉1 by choosing N large enough and ε small enough;this will make r2 = 〉2G0 arbitrarily close to 2. Choosing N large corresponds to taking G0 to be alarger iterate of 2 + ε under gε(x). By continuity of this function with respect to x and ε, it suffices totake ε = 0 and show that (letting g = g0)

g(n)(2) = g(· · · g︸ ︷︷ ︸n

(2) · · · )→ 〉1 as n→∞.

But note that for 0 ≤ θ ≤ π2 ,

g(〉2 cos θ) = 〉√

2〉 2 cos θ = 〉2 sin

(θ

2

)= 2 cos

(π

2〉 θ

2

).

Hence by induction, g(n)(〉2 cos θ) = 〉2 cosπ2 〉

π4 + · · ·+ (〉1)n

θ 〉 π

2n

)). Hence g(n)(2) = g(n−1)(〉2 cos 0)

converges to 〉2 cos(π2 〉π4 + · · · ) = 〉2 cos(π3 ) = 〉1, as needed.

16. Let ABCD be a quadrilateral inscribed in the unit circle such that ∠BAD is 30 degrees. Let m denotethe minimum value of CP + PQ + CQ, where P and Q may be any points lying along rays AB andAD, respectively. Determine the maximum value of m.

Answer: 2

Algebra & Geometry Individual Test

A

BC

D

E

F

G

H

P

Q

R

For a fixed quadrilateral ABCD as described, we first show that m, the minimum possible length ofCP + PQ + QC, equals the length of AC. Reflect B, C, and P across line AD to points E, F , andR, respectively, and then reflect D and F across AE to points G and H, respectively. These tworeflections combine to give a 60◦ rotation around A, so triangle ACH is equilateral. It also followsthat RH is a 60◦ rotation of PC around A, so, in particular, these segments have the same length.Because QR = QP by reflection,

CP + PQ+QC = CQ+QR+RH.

The latter is the length of a broken path CQRH from C to H, and by the “shortest path is a straightline” principle, this total length is at least as long as CH = CA. (More directly, this follows from thetriangle inequality: (CQ+QR)+RH ≥ CR+RH ≥ CH). Therefore, the lower bound m ≥ AC indeedholds. To see that this is actually an equality, note that choosing Q as the intersection of segment CHwith ray AD, and choosing P so that its reflection R is the intersection of CH with ray AE, alignspath CQRH with segment CH, thus obtaining the desired minimum m = AC.

We may conclude that the largest possible value of m is the largest possible length of AC, namely 2:the length of a diameter of the circle.

17. Let z = cos2π

2011+ i sin

2π

2011, and let

P (x) = x2008 + 3x2007 + 6x2006 + . . .2008 · 2009

2x+

2009 · 2010

2

for all complex numbers x. Evaluate P (z)P (z2)P (z3) . . . P (z2010).

Answer: 20112009 · (10052011 〉 10042011)

Multiply P (x) by x〉 1 to get

P (x)(x〉 1) = x2009 + 2x2008 + . . .+ 2009x〉 2009 · 2010

2,

or,

P (x)(x〉 1) + 2010 · 1005 = x2009 + 2x2008 + . . .+ 2009x+ 2010.

Multiplying by x〉 1 once again:

Algebra & Geometry Individual Test

(x〉 1)(P (x)(x〉 1) +2010 · 2011

2) = x2010 + x2009 + . . .+ x〉 2010,

= (x2010 + x2009 + . . .+ x+ 1)〉 2011.

Hence,

P (x) =

(x2010 + x2009 + . . .+ x+ 1)〉 2011

x〉 1〉 2011 · 1005

x〉 1

Note that x2010+x2009+. . .+x+1 has z, z2, . . . , z2010 as roots, so they vanish at those points. Pluggingthose 2010 powers of z into the last equation, and multiplying them together, we obtain

2010∏i=1

P (zi) =(〉2011) · 1005 · (x〉 1004

1005)

(x〉 1)2.

Note that (x〉 z)(x〉 z2) . . . (x〉 z2010) = x2010 + x2009 + . . .+ 1. Using this, the product turns out tobe 20112009 · (10052011 〉 10042011).

18. Collinear points A, B, and C are given in the Cartesian plane such that A = (a, 0) lies along thex-axis, B lies along the line y = x, C lies along the line y = 2x, and AB/BC = 2. If D = (a, a), thecircumcircle of triangle ADC intersects y = x again at E, and ray AE intersects y = 2x at F , evaluateAE/EF .

Answer: 7

Algebra & Geometry Individual Test

A

B

CD

E

F

O

PQ

Let points O, P , and Q be located at (0, 0), (a, 2a), and (0, 2a), respectively. Note that BC/AB = 1/2implies [OCD]/[OAD] = 1/2, so since [OPD] = [OAD], [OCD]/[OPD] = 1/2. It follows that[OCD] = [OPD]. Hence OC = CP . We may conclude that triangles OCQ and PCA are congruent,so C = (a/2, a).It follows that ∠ADC is right, so the circumcircle of triangle ADC is the midpoint of AC, whichis located at (3a/4, a/2). Let (3a/4, a/2) = H, and let E = (b, b). Then the power of the pointO with respect to the circumcircle of ADC is OD · OE = 2ab, but it may also be computed asOH2 〉HA2 = 13a/16〉 5a/16 = a/2. It follows that b = a/4, so E = (a/4, a/4).We may conclude that line AE is x + 3y = a, which intersects y = 2x at an x-coordinate of a/7.Therefore, AE/EF = (a〉 a/4)/(a/4〉 a/7) = (3a/4)/(3a/28) = 7.

Remark: The problem may be solved more quickly if one notes from the beginning that lines OA,OD, OP , and OQ form a harmonic pencil because D is the midpoint of AP and lines OQ and AP areparallel.

19. Let {an} and {bn} be sequences defined recursively by a0 = 2; b0 = 2, and an+1 = an√

1 + a2n + b2n〉bn;

bn+1 = bn√

1 + a2n + b2n + an. Find the ternary (base 3) representation of a4 and b4.

Answer: 1000001100111222 and 2211100110000012 Note first that√

1 + a2n + b2n = 32n

. Theproof is by induction; the base case follows trivially from what is given. For the inductive step, note

Algebra & Geometry Individual Test

that 1 + a2n+1 + b2n+1 = 1 + a2n(1 + a2n + b2n) + b2n 〉 2anbn√

1 + a2n + b2n + b2n(1 + a2n + b2n) + a2n +

2anbn√

1 + a2n + b2n = 1 + (a2n + b2n)(1 + a2n + b2n) + a2n + b2n = (1 + a2n + b2n)2. Invoking the inductive

hypothesis, we see that√

1 + a2n+1 + b2n+1 = (32n

)2 = 32n+1

, as desired.

The quickest way to finish from here is to consider a sequence of complex numbers {zn} defined byzn = an+ bni for all nonnegative integers n. It should be clear that z0 = 2 + 2i and zn+1 = zn(32

n

+ i).

Therefore, z4 = (2 + 2i)(320

+ i)(321

+ i)(322

+ i)(323

+ i). This product is difficult to evaluate inthe decimal number system, but in ternary the calculation is a cinch! To speed things up, we willuse balanced ternary1, in which the three digits allowed are 〉1, 0, and 1 rather than 0, 1, and 2. Letx+ yi = (32

0

+ i)(321

+ i)(322

+ i)(323

+ i), and consider the balanced ternary representation of x andy. For all 0 ≤ j ≤ 15, let xj denote the digit in the 3j place of x, let yj denote the digit in the 3j placeof y, and let b(j) denote the number of ones in the binary representation of j. It should be clear thatxj = 〉1 if b(j) ≡ 2 (mod 4), xj = 0 if b(j) ≡ 1 (mod 2), and xj = 1 if b(j) ≡ 0 (mod 4). Similarly,yj = 〉1 if b(j) ≡ 1 (mod 4), yj = 0 if b(j) ≡ 0 (mod 2), and yj = 1 if b(j) ≡ 3 (mod 4). Convertingto ordinary ternary representation, we see that x = 2212112211220013 and y = 1100222022121203. Itremains to note that a4 = 2x〉 2y and b4 = 2x+ 2y and perform the requisite arithmetic to arrive atthe answer above.

20. Let ω1 and ω2 be two circles that intersect at points A and B. Let line l be tangent to ω1 at P and toω2 at Q so that A is closer to PQ than B. Let points R and S lie along rays PA and QA, respectively,so that PQ = AR = AS and R and S are on opposite sides of A as P and Q. Let O be the circumcenterof triangle ASR, and let C and D be the midpoints of major arcs AP and AQ, respectively. If ∠APQis 45 degrees and ∠AQP is 30 degrees, determine ∠COD in degrees.

Answer: 142.5

A

B

P

Q

R

SC

D

T

T ′

We use directed angles throughout the solution.

Let T denote the point such that ∠TCD = 1/2∠APQ and ∠TDC = 1/2∠AQP . We claim that T isthe circumcenter of triangle SAR.

Since CP = CA, QP = RA, and ∠CPQ = ∠CPA + ∠APQ = ∠CPA + ∠ACP = ∠CAR, we have4CPQ ∼= 4CAR. By spiral similarity, we have 4CPA ∼ 4CQR.

Let T ′ denote the reflection of T across CD. Since ∠TCT ′ = ∠APQ = ∠ACP , we have 4TCT ′ ∼4ACP ∼ 4RCQ. Again, by spiral similarity centered at C, we have 4CTR ∼ 4CT ′Q. ButCT = CT ′, so 4CTR ∼= 4CT ′Q and TR = T ′Q. Similarly, 4DTT ′ ∼ 4DAQ, and spiral similaritycentered at D shows that 4DTA ∼= 4DT ′Q. Thus TA = T ′Q = TR.

We similarly have TA = T ′P = TS, so T is indeed the circumcenter. Therefore, we have ∠COD =∠CTD = 180◦ 〉 45◦

2 〉30◦

2 = 142.5◦.

1http://en.wikipedia.org/wiki/Balanced_ternary

Algebra & Geometry Individual Test

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

1. A classroom has 30 students and 30 desks arranged in 5 rows of 6. If the class has 15 boys and 15girls, in how many ways can the students be placed in the chairs such that no boy is sitting in frontof, behind, or next to another boy, and no girl is sitting in front of, behind, or next to another girl?

2. Let a, b, and c be positive real numbers. Determine the largest total number of real roots that thefollowing three polynomials may have among them: ax2 + bx + c, bx2 + cx + a, and cx2 + ax + b.

3. Let f : R → R be a differentiable function such that f(0) = 0, f(1) = 1, and |f ′(x)| ≤ 2 for all real

numbers x. If a and b are real numbers such that the set of possible values of

∫ 1

0

f(x) dx is the open

interval (a, b), determine b− a.

4. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner.At each step, he can either move one square to the right or simultaneously move one square to the leftand one square up. In how many ways can he reach the center square of the topmost row?

5. Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠BAC intersect line BC atD. If there exist points E and F on sides AC and BC, respectively, such that lines AD and EFare parallel and divide triangle ABC into three parts of equal area, determine the number of possibleintegral values for BC.

6. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. IfNathaniel goes first, determine the probability that he ends up winning.

7. Let ABCDEF be a regular hexagon of area 1. Let M be the midpoint of DE. Let X be theintersection of AC and BM , let Y be the intersection of BF and AM , and let Z be the intersectionof AC and BF . If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate[BXC] + [AY F ] + [ABZ]− [MXZY ].

8. Let f : [0, 1)→ R be a function that satisfies the following condition: if

x =

∞∑n=1

an10n

= .a1a2a3 . . .

is the decimal expansion of x and there does not exist a positive integer k such that an = 9 for alln ≥ k, then

f(x) =

∞∑n=1

an102n

.

Determine f ′13

).

9. The integers from 1 to n are written in increasing order from left to right on a blackboard. Davidand Goliath play the following game: starting with David, the two players alternate erasing any twoconsecutive numbers and replacing them with their sum or product. Play continues until only onenumber on the board remains. If it is odd, David wins, but if it is even, Goliath wins. Find the 2011thsmallest positive integer greater than 1 for which David can guarantee victory.

10. Evaluate

∫ ∞1

(lnx

x

)2011

dx.

11. Mike and Harry play a game on an 8× 8 board. For some positive integer k, Mike chooses k squaresand writes an M in each of them. Harry then chooses k + 1 squares and writes an H in each of them.After Harry is done, Mike wins if there is a sequence of letters forming “HMM” or “MMH,” whenread either horizontally or vertically, and Harry wins otherwise. Determine the smallest value of k forwhich Mike has a winning strategy.

12. Sarah and Hagar play a game of darts. Let O0 be a circle of radius 1. On the nth turn, the playerwhose turn it is throws a dart and hits a point pn randomly selected from the points of On∈1. Theplayer then draws the largest circle that is centered at pn and contained in On∈1, and calls this circleOn. The player then colors every point that is inside On∈1 but not inside On her color. Sarah goesfirst, and the two players alternate turns. Play continues indefinitely. If Sarah’s color is red, andHagar’s color is blue, what is the expected value of the area of the set of points colored red?

13. The ordered pairs (2011, 2), (2010, 3), (2009, 4), . . ., (1008, 1005), (1007, 1006) are written from left toright on a blackboard. Every minute, Elizabeth selects a pair of adjacent pairs (xi, yi) and (xj , yj), with

(xi, yi) left of (xj , yj), erases them, and writes

(xiyixj

yj,xiyiyjxj

)in their place. Elizabeth continues

this process until only one ordered pair remains. How many possible ordered pairs (x, y) could appearon the blackboard after the process has come to a conclusion?

14. Let f : [0, 1] → [0, 1] be a continuous function such that f(f(x)) = 1 for all x ∈ [0, 1]. Determine the

set of possible values of

∫ 1

0

f(x) dx.

15. Let A = {1, 2, . . . , 2011}. Find the number of functions f from A to A that satisfy f(n) ≤ n for all nin A and attain exactly 2010 distinct values.

16. Let f(x) = x2 − r2x + r3 for all real numbers x, where r2 and r3 are some real numbers. Define asequence {gn} for all nonnegative integers n by g0 = 0 and gn+1 = f(gn). Assume that {gn} satisfiesthe following three conditions: (i) g2i < g2i+1 and g2i+1 > g2i+2 for all 0 ≤ i ≤ 2011; (ii) there existsa positive integer j such that gi+1 > gi for all i > j, and (iii) {gn} is unbounded. If A is the greatestnumber such that A ≤ |r2| for any function f satisfying these properties, find A.

17. Let f : (0, 1) → (0, 1) be a differentiable function with a continuous derivative such that for everypositive integer n and odd positive integer a < 2n, there exists an odd positive integer b < 2n suchthat f

a2n

)= b

2n . Determine the set of possible values of f ′12

).

18. Let n be an odd positive integer, and suppose that n people sit on a committee that is in the processof electing a president. The members sit in a circle, and every member votes for the person either tohis/her immediate left, or to his/her immediate right. If one member wins more votes than all theother members do, he/she will be declared to be the president; otherwise, one of the the members whowon at least as many votes as all the other members did will be randomly selected to be the president.If Hermia and Lysander are two members of the committee, with Hermia sitting to Lysander’s leftand Lysander planning to vote for Hermia, determine the probability that Hermia is elected president,assuming that the other n− 1 members vote randomly.

19. Let

F (x) =1

(2− x− x5)2011,

and note that F may be expanded as a power series so that F (x) =

∞∑n=0

anxn. Find an ordered pair of

positive real numbers (c, d) such that limn→∞

annd

= c.

20. Alice and Bob play a game in which two thousand and eleven 2011×2011 grids are distributed betweenthe two of them, 1 to Bob, and the other 2010 to Alice. They go behind closed doors and fill theirgrid(s) with the numbers 1, 2, . . . , 20112 so that the numbers across rows (left-to-right) and downcolumns (top-to-bottom) are strictly increasing. No two of Alice’s grids may be filled identically. Afterthe grids are filled, Bob is allowed to look at Alice’s grids and then swap numbers on his own grid,two at a time, as long as the numbering remains legal (i.e. increasing across rows and down columns)after each swap. When he is done swapping, a grid of Alice’s is selected at random. If there exist twointegers in the same column of this grid that occur in the same row of Bob’s grid, Bob wins. Otherwise,Alice wins. If Bob selects his initial grid optimally, what is the maximum number of swaps that Bobmay need in order to guarantee victory?

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

Calculus & Combinatorics Individual Test

1. A classroom has 30 students and 30 desks arranged in 5 rows of 6. If the class has 15 boys and 15girls, in how many ways can the students be placed in the chairs such that no boy is sitting in frontof, behind, or next to another boy, and no girl is sitting in front of, behind, or next to another girl?

Answer: 2 · 15!2 If we color the desks of the class in a checkerboard pattern, we notice that all ofone gender must go in the squares colored black, and the other gender must go in the squares coloredwhite. There are 2 ways to pick which gender goes in which color, 15! ways to put the boys into desksand 15! ways to put the girls into desks. So the number of ways is 2 · 15!2.

(There is a little ambiguity in the problem statement as to whether the 15 boys and the 15 girls aredistinguishable or not. If they are not distinguishable, the answer is clearly 2. Given the number ofcontestants who submitted the answer 2, the graders judged that there was enough ambiguity to justifyaccepting 2 as a correct answer. So both 2 and 2 · 15!2 were accepted as correct answers.)

2. Let a, b, and c be positive real numbers. Determine the largest total number of real roots that thefollowing three polynomials may have among them: ax2 + bx+ c, bx2 + cx+ a, and cx2 + ax+ b.

Answer: 4 If all the polynomials had real roots, their discriminants would all be nonnegative: a2 ≥4bc, b2 ≥ 4ca, and c2 ≥ 4ab. Multiplying these inequalities gives (abc)2 ≥ 64(abc)2, a contradiction.Hence one of the quadratics has no real roots. The maximum of 4 real roots is attainable: for example,the values (a, b, c) = (1, 5, 6) give −2,−3 as roots to x2 + 5x+ 6 and −1,− 1

5 as roots to 5x2 + 6x+ 1.

3. Let f : R → R be a differentiable function such that f(0) = 0, f(1) = 1, and |f ′(x)| ≤ 2 for all real

numbers x. If a and b are real numbers such that the set of possible values of

∫ 1

0

f(x) dx is the open

interval (a, b), determine b− a.

Answer: 34 Draw lines of slope ±2 passing through (0, 0) and (1, 1). These form a parallelogram

with vertices (0, 0), (.75, 1.5), (1, 1), (.25,−.5). By the mean value theorem, no point of (x, f(x)) liesoutside this parallelogram, but we can construct functions arbitrarily close to the top or the bottomof the parallelogram while satisfying the condition of the problem. So (b − a) is the area of thisparallelogram, which is 3

4 .

4. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner.At each step, he can either move one square to the right or simultaneously move one square to the leftand one square up. In how many ways can he reach the center square of the topmost row?

Answer: 2n−1 Note that Josh must pass through the center square of each row. There are 2 waysto get from the center square of row k to the center square of row k+ 1. So there are 2n−1 ways to getto the center square of row n.

5. Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠BAC intersect line BC atD. If there exist points E and F on sides AC and BC, respectively, such that lines AD and EFare parallel and divide triangle ABC into three parts of equal area, determine the number of possibleintegral values for BC.

Answer: 13

Calculus & Combinatorics Individual Test

A B

C

DE

F

7

14

Note that such E,F exist if and only if[ADC]

[ADB]= 2. (1)

([ ] denotes area.) Since AD is the angle bisector, and the ratio of areas of triangles with equal heightis the ratio of their bases,

AC

AB=DC

DB=

[ADC]

[ADB].

Hence (1) is equivalent to AC = 2AB = 14. Then BC can be any length d such that the triangleinequalities are satisfied:

d+ 7 > 14

7 + 14 > d

Hence 7 < d < 21 and there are 13 possible integral values for BC.

6. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. IfNathaniel goes first, determine the probability that he ends up winning.

Answer: 511 For 1 ≤ k ≤ 6, let xk be the probability that the current player, say A, will win

when the number on the tally at the beginning of his turn is k modulo 7. The probability that thetotal is l modulo 7 after his roll is 1

6 for each l 6≡ k (mod 7); in particular, there is a 16 chance he wins

immediately. The chance that A will win if he leaves l on the board after his turn is 1− xl. Hence for1 ≤ k ≤ 6,

xk =1

6

∑1≤l≤6, l 6=k

(1− xl) +1

6.

Letting s =∑6l=1 xl, this becomes xk = xk−s

6 + 1 or 5xk

6 = − s6 + 1. Hence x1 = · · · = x6, and 6xk = s

for every k. Plugging this in gives 11xk

6 = 1, or xk = 611 .

Since Nathaniel cannot win on his first turn, he leaves Obediah with a number not divisible by 7.Hence Obediah’s chance of winning is 6

11 and Nathaniel’s chance of winning is 511 .

Calculus & Combinatorics Individual Test

7. Let ABCDEF be a regular hexagon of area 1. Let M be the midpoint of DE. Let X be theintersection of AC and BM , let Y be the intersection of BF and AM , and let Z be the intersectionof AC and BF . If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate[BXC] + [AY F ] + [ABZ]− [MXZY ].

Answer: 0

A B

C

DE

F

M

XY

Z

O

Let O be the center of the hexagon. The desired area is [ABCDEF ] − [ACDM ] − [BFEM ]. Notethat [ADM ] = [ADE]/2 = [ODE] = [ABC], where the last equation holds because sin 60◦ = sin 120◦.Thus, [ACDM ] = [ACD] + [ADM ] = [ACD] + [ABC] = [ABCD], but the area of ABCD is half thearea of the hexagon. Similarly, the area of [BFEM ] is half the area of the hexagon, so the answer iszero.

8. Let f : [0, 1)→ R be a function that satisfies the following condition: if

x =

∞∑n=1

an10n

= .a1a2a3 . . .

is the decimal expansion of x and there does not exist a positive integer k such that an = 9 for alln ≥ k, then

f(x) =

∞∑n=1

an102n

.

Determine f ′(13

).

Answer: 0 Note that 13 =

∑∞n=1

310n .

Clearly f is an increasing function. Also for any integer n ≥ 1, we see from decimal expansions thatf( 1

3 ±1

10n )− f( 13 ) = ± 1

102n .

Consider h such that 10−n−1 ≤ |h| < 10−n. The two properties of f outlined above show that |f( 13 +

h)− f( 13 )| < 1

102n . And from | 1h | ≤ 10n+1, we get∣∣∣ f( 1

3+h)−f(13 )

h

∣∣∣ < 110n−1 . Taking n→∞ gives h→ 0

and f ′( 13 ) = limn→∞

110n−1 = 0.

9. The integers from 1 to n are written in increasing order from left to right on a blackboard. Davidand Goliath play the following game: starting with David, the two players alternate erasing any twoconsecutive numbers and replacing them with their sum or product. Play continues until only onenumber on the board remains. If it is odd, David wins, but if it is even, Goliath wins. Find the 2011thsmallest positive integer greater than 1 for which David can guarantee victory.

Calculus & Combinatorics Individual Test

Answer: 4022 If n is odd and greater than 1, then Goliath makes the last move. No matter whattwo numbers are on the board, Goliath can combine them to make an even number. Hence Goliathhas a winning strategy in this case.

Now suppose n is even. We can replace all numbers on the board by their residues modulo 2. Initiallythe board reads 1, 0, 1, 0, . . . , 1, 0. David combines the rightmost 1 and 0 by addition to make 1, so nowthe board reads 1, 0, 1, 0, . . . , 0, 1. We call a board of this form a “good” board. When it is Goliath’sturn, and there is a good board, no matter where he moves, David can make a move to restore a goodboard. Indeed, Goliath must combine a neighboring 0 and 1; David can then combine that numberwith a neighboring 1 to make 1 and create a good board with two fewer numbers.

David can ensure a good board after his last turn. But a good board with one number is simply 1,so David wins. So David has a winning strategy if n is even. Therefore, the 2011th smallest positiveinteger greater than 1 for which David can guarantee victory is the 2011th even positive integer, whichis 4022.

10. Evaluate

∫ ∞1

(lnx

x

)2011

dx.

Answer: 2011!20102012 By the chain rule,

d

dx(lnx)n =

n lnn−1 x

x.

We calculate the definite integral using integration by parts:∫ ∞x=1

(lnx)n

x2011dx =

[(lnx)n

−2010x2010

]x=∞x=1

−∫ ∞x=1

n(lnx)n−1

−2010x2011dx

But ln(1) = 0, and limx→∞(ln x)n

x2010 = 0 for all n > 0. So∫ ∞x=1

(lnx)n

x2011dx =

∫ ∞x=1

n(lnx)n−1

2010x2011dx

It follows that ∫ ∞x=1

(lnx)n

x2011dx =

n!

2010n

∫ ∞x=1

1

x2011dx =

n!

2010n+1

So the answer is 2011!20102012 .

11. Mike and Harry play a game on an 8× 8 board. For some positive integer k, Mike chooses k squaresand writes an M in each of them. Harry then chooses k + 1 squares and writes an H in each of them.After Harry is done, Mike wins if there is a sequence of letters forming “HMM” or “MMH,” whenread either horizontally or vertically, and Harry wins otherwise. Determine the smallest value of k forwhich Mike has a winning strategy.

Answer: 16

Suppose Mike writes k M ’s. Let a be the number of squares which, if Harry writes an H in, will yieldeither HMM or MMH horizontally, and let b be the number of squares which, if Harry writes an Hin, will yield either HMM or MMH vertically. We will show that a ≤ k and b ≤ k. Then, it willfollow that there are at most a+ b ≤ 2k squares which Harry cannot write an H in. There will be atleast 64−k− 2k = 64− 3k squares which Harry can write in. If 64− 3k ≥ k+ 1, or k ≤ 15, then Harrywins.

We will show that a ≤ k (that b ≤ k will follow by symmetry). Suppose there are ai M ’s in row i. Ineach group of 2 or more consective M ’s, Harry cannot write H to the left or right of that group, givingat most 2 forbidden squares. Hence ai is at most the number of M ’s in row i. Summing over the rowsgives the desired result.

Mike can win by writing 16 M ’s according to the following diagram:

Calculus & Combinatorics Individual Test

· · · · · · · ·· M M · · M M ·· M M · · M M ·· · · · · · · ·· · · · · · · ·· M M · · M M ·· M M · · M M ·· · · · · · · ·

12. Sarah and Hagar play a game of darts. Let O0 be a circle of radius 1. On the nth turn, the playerwhose turn it is throws a dart and hits a point pn randomly selected from the points of On−1. Theplayer then draws the largest circle that is centered at pn and contained in On−1, and calls this circleOn. The player then colors every point that is inside On−1 but not inside On her color. Sarah goesfirst, and the two players alternate turns. Play continues indefinitely. If Sarah’s color is red, andHagar’s color is blue, what is the expected value of the area of the set of points colored red?

Answer: 6π7 Let f(r) be the average area colored red on a dartboard of radius r if Sarah plays first.

Then f(r) is proportional to r2. Let f(r) = (πx)r2 for some constant x. We want to find f(1) = πx.

In the first throw, if Sarah’s dart hits a point with distance r from the center of O0, the radius of O1

will be 1−r. The expected value of the area colored red will be (π−π(1−r)2)+(π(1−r)2−f(1−r)) =π − f(1− r). The value of f(1) is the average value of π − f(1− r) over all points in O0. Using polarcoordinates, we get

f(1) =

2π∫0

1∫0

(π − f(1− r))rdrdθ

2π∫0

1∫0

rdrdθ

πx =

1∫0

(π − πx(1− r)2)rdr

1∫0

rdr

πx

2=

∫ 1

0

πr − πxr(1− r)2dr

πx

2=π

2− πx(

1

2− 2

3+

1

4)

πx

2=π

2− πx

12

πx =6π

7

13. The ordered pairs (2011, 2), (2010, 3), (2009, 4), . . ., (1008, 1005), (1007, 1006) are written from left toright on a blackboard. Every minute, Elizabeth selects a pair of adjacent pairs (xi, yi) and (xj , yj), with

(xi, yi) left of (xj , yj), erases them, and writes

(xiyixjyj

,xiyiyjxj

)in their place. Elizabeth continues

this process until only one ordered pair remains. How many possible ordered pairs (x, y) could appearon the blackboard after the process has come to a conclusion?

Answer: 504510 First, note that none of the numbers will ever be 0. Let ? denote the replacementoperation. For each pair on the board (xi, yi) define its primary form to be (xi, yi) and its secondary

Calculus & Combinatorics Individual Test

form to be [xiyi,xi

yi]. Note that the primary form determines the secondary form uniquely and vice

versa. In secondary form,

[a1, b1] ? [a2, b2] =

(√a1b1,

√a1b1

)?

(√a2b2,

√a2b2

)=

(a1b2,

a1b2

)= [a21, b

22].

Thus we may replace all pairs on the board by their secondary form and use the above rule for ?instead. From the above rule, we see that if the leftmost number on the board is x, then after oneminute it will be x or x2 depending on whether it was erased in the intervening step, and similarly forthe rightmost number. Let k be the number of times the leftmost pair is erased and n be the numberof times the rightmost pair is erased. Then the final pair is[

40222k

,

(1007

1006

)2n]. (2)

Any step except the last cannot involve both the leftmost and rightmost pair, so k + n ≤ 1005. Sinceevery pair must be erased at least once, k, n ≥ 1. Every pair of integers satisfying the above can occur,for example, by making 1005− k − n moves involving only the pairs in the middle, then making k − 1moves involving the leftmost pair, and finally n moves involving the rightmost pair.

In light of (2), the answer is the number of possible pairs (k, n), which is

1004∑k=1

1005−k∑n=1

1 =

1004∑k=1

1005− k =

1004∑k=1

k =1004 · 1005

2= 504510.

14. Let f : [0, 1] → [0, 1] be a continuous function such that f(f(x)) = 1 for all x ∈ [0, 1]. Determine the

set of possible values of

∫ 1

0

f(x) dx.

Answer: ( 34 , 1] Since the maximum value of f is 1,

∫ 1

0f(x)dx ≤ 1.

By our condition f(f(x)) = 1, f is 1 at any point within the range of f . Clearly, 1 is in the range off , so f(1) = 1. Now f(x) is continuous on a closed interval so it attains a minimum value c. Since cis in the range of f , f(c) = 1.

If c = 1, f(x) = 1 for all x and∫ 1

0f(x)dx = 1.

Now assume c < 1. By the intermediate value theorem, since f is continuous it attains all valuesbetween c and 1. So for all x ≥ c, f(x) = 1. Therefore,

∫ 1

0

f(x)dx =

∫ c

0

f(x)dx+ (1− c).

Since f(x) ≥ c,∫ c0f(x)dx > c2, and the equality is strict because f is continuous and thus cannot be

c for all x < c and 1 at c. So

∫ 1

0

f(x)dx > c2 + (1− c) = (c− 1

2)2 +

3

4≥ 3

4.

Therefore 34 <

∫ 1

0f(x)dx ≤ 1, and it is easy to show that every value in this interval can be reached.

15. Let A = {1, 2, . . . , 2011}. Find the number of functions f from A to A that satisfy f(n) ≤ n for all nin A and attain exactly 2010 distinct values.

Answer: 22011 − 2012 Let n be the element of A not in the range of f . Let m be the element of Athat is hit twice.

Calculus & Combinatorics Individual Test

We now sum the total number of functions over n,m. Clearly f(1) = 1, and by induction, for x ≤m, f(x) = x. Also unless n = 2011, f(2011) = 2011 because f can take no other number to 2011. Itfollows from backwards induction that for x > n, f(x) = x. Therefore n > m, and there are only n−mvalues of f that are not fixed.

Now f(m+ 1) = m or f(m+ 1) = m+ 1. For m < k < n,given the selection of f(1), f(2), . . . , f(k− 1),k− 1 of the k+ 1 possible values of f(k+ 1) (1, 2, 3, . . . , k, and counting m twice) have been taken, sothere are two distinct values that f(k+ 1) can take (one of them is k+ 1, and the other is not, so theyare distinct). For f(n), when the other 2010 values of f have been assigned, there is only one missing,so f(n) is determined.

For each integer in [m,n), there are two possible values of f , so there are 2n−m−1 different functionsf for a given m,n. So our answer is

2010∑m=1

2011∑n=m+1

2n−m−1 =

2010∑m=1

2−m−12011∑

n=m+1

2n

=

2010∑m=1

2−m−1(22012 − 2m+1)

=

2010∑m=1

22011−m − 1

=

(2010∑m=1

2m

)− 2010

= 22011 − 2012

16. Let f(x) = x2 − r2x + r3 for all real numbers x, where r2 and r3 are some real numbers. Define asequence {gn} for all nonnegative integers n by g0 = 0 and gn+1 = f(gn). Assume that {gn} satisfiesthe following three conditions: (i) g2i < g2i+1 and g2i+1 > g2i+2 for all 0 ≤ i ≤ 2011; (ii) there existsa positive integer j such that gi+1 > gi for all i > j, and (iii) {gn} is unbounded. If A is the greatestnumber such that A ≤ |r2| for any function f satisfying these properties, find A.

Answer: 2

Consider the function f(x)−x. By the constraints of the problem, f(x)−x must be negative for somex, namely, for x = g2i+1, 0 ≤ i ≤ 2011. Since f(x) − x is positive for x of large absolute value, thegraph of f(x)− x crosses the x-axis twice and f(x)− x has two real roots, say a < b. Factoring givesf(x)− x = (x− a)(x− b), or f(x) = (x− a)(x− b) + x.

Now, for x < a, f(x) > x > a, while for x > b, f(x) > x > b. Let c 6= b be the number such thatf(c) = f(b) = b. Note that b is not the vertex as f(a) = a < b, so by the symmetry of quadratics, cexists and b+c

2 = r22 as the vertex of the parabola. By the same token, b+a

2 = r2+12 is the vertex of

f(x)− x. Hence c = a− 1. If f(x) > b then x < c or x > b. Consider the smallest j such that gj > b.Then by the above observation, gj−1 < c. (If gi ≥ b then f(gi) ≥ gi ≥ b so by induction, gi+1 ≥ gi forall i ≥ j. Hence j > 1; in fact j ≥ 4025.) Since gj−1 = f(gj−2), the minimum value of f is less than c.The minimum value is the value of f evaluated at its vertex, b+a−12 , so

f

(b+ a− 1

2

)< c(

b+ a− 1

2− a)(

b+ a− 1

2− b)

+b+ a− 1

2< a− 1

1− (b− a)2

4+b− a+ 1

2< 0

3

4<

(b− a)2

4− b− a

2

4 < (b− a− 1)2.

Calculus & Combinatorics Individual Test

Then either b − a − 1 < −2 or b − a − 1 > 2, but b > a, so the latter must hold and (b − a)2 > 9.Now, the discriminant of f(x) − x equals (b − a)2 (the square of the difference of the two roots) and(r2 + 1)2 − 4r3 (from the coefficients), so (r2 + 1)2 > 9 + 4r3. But r3 = g1 > g0 = 0 so |r2| > 2.

We claim that we can make |r2| arbitrarily close to 2, so that the answer is 2. First define Gi, i ≥ 0as follows. Let N ≥ 2012 be an integer. For ε > 0 let h(x) = x2 − 2 − ε, gε(x) = −

√x+ 2 + ε and

G2N+1 = 2 + ε, and define Gi recursively by Gi = gε(Gi+1), Gi+1 = h(Gi). (These two equations areconsistent.) Note the following.

(i) G2i < G2i+1 andG2i+1 > G2i+2 for 0 ≤ i ≤ N−1. First noteG2N = −√

4 + 2ε > −√

4 + 2ε+ ε2 =−2 − ε. Let l be the negative solution to h(x) = x. Note that −2 − ε < G2N < l < 0 sinceh(G2N ) > 0 > G2N . Now gε(x) is defined as long as x ≥ −2− ε, and it sends (−2− ε, l) into (l, 0)and (l, 0) into (−2− ε, l). It follows that the Gi, 0 ≤ i ≤ 2N are well-defined; moreover, G2i < land G2i+1 > l for 0 ≤ i ≤ N − 1 by backwards induction on i, so the desired inequalities follow.

(ii) Gi is increasing for i ≥ 2N + 1. Indeed, if x ≥ 2 + ε, then x2 − x = x(x− 1) > 2 + ε so h(x) > x.Hence 2 + ε = G2N+1 < G2N+2 < · · · .

(iii) Gi is unbounded. This follows since h(x)− x = x(x− 2)− 2− ε is increasing for x > 2 + ε, so Giincreases faster and faster for i ≥ 2N + 1.

Now define f(x) = h(x + G0) − G0 = x2 + 2G0x + G20 − G0 − 2 − ε. Note Gi+1 = h(Gi) while

gi+1 = f(gi) = h(gi + G0) − G0, so by induction gi = Gi − G0. Since {Gi}∞i=0 satisfies (i), (ii), and(iii), so does gi.

We claim that we can make G0 arbitrarily close to −1 by choosing N large enough and ε small enough;this will make r2 = −2G0 arbitrarily close to 2. Choosing N large corresponds to taking G0 to be alarger iterate of 2 + ε under gε(x). By continuity of this function with respect to x and ε, it suffices totake ε = 0 and show that (letting g = g0)

g(n)(2) = g(· · · g︸ ︷︷ ︸n

(2) · · · )→ −1 as n→∞.

But note that for 0 ≤ θ ≤ π2 ,

g(−2 cos θ) = −√

2− 2 cos θ = −2 sin

(θ

2

)= 2 cos

(π

2− θ

2

).

Hence by induction, g(n)(−2 cos θ) = −2 cos(π2 −

π4 + · · ·+ (−1)n

(θ − π

2n

)). Hence g(n)(2) = g(n−1)(−2 cos 0)

converges to −2 cos(π2 −π4 + · · · ) = −2 cos(π3 ) = −1, as needed.

17. Let f : (0, 1) → (0, 1) be a differentiable function with a continuous derivative such that for everypositive integer n and odd positive integer a < 2n, there exists an odd positive integer b < 2n suchthat f

(a2n

)= b

2n . Determine the set of possible values of f ′(12

).

Answer: {−1, 1} The key step is to notice that for such a function f , f ′(x) 6= 0 for any x.

Assume, for sake of contradiction that there exists 0 < y < 1 such that f ′(y) = 0. Since f ′ isa continuous function, there is some small interval (c, d) containing y such that |f ′(x)| ≤ 1

2 for allx ∈ (c, d). Now there exists some n, a such that a

2n ,a+12n are both in the interval (c, d). From the

definition,f(a+1

2n )− f( a2n )

a+12n −

a2n

= 2n(b′

2n− b

2n) = b′ − b where b, b′ are integers; one is odd, and one is

even. So b′ − b is an odd integer. Since f is differentiable, by the mean value theorem there exists apoint where f ′ = b′ − b. But this point is in the interval (c, d), and |b′ − b| > 1

2 . This contradicts theassumption that |f ′(x)| ≤ 1

2 for all x ∈ (c, d).

Since f ′(x) 6= 0, and f ′ is a continuous function, f ′ is either always positive or always negative. So f iseither increasing or decreasing. f( 1

2 ) = 12 always. If f is increasing, it follows that f( 1

4 ) = 14 , f( 3

4 ) = 34 ,

and we can show by induction that indeed f( a2n ) = a

2n for all integers a, n. Since numbers of this formare dense in the interval (0, 1), and f is a continuous function, f(x) = x for all x.

Calculus & Combinatorics Individual Test

It can be similarly shown that if f is decreasing f(x) = 1− x for all x. So the only possible values off ′( 1

2 ) are −1, 1.

Query: if the condition that the derivative is continuous were omitted, would the same result stillhold?

18. Let n be an odd positive integer, and suppose that n people sit on a committee that is in the processof electing a president. The members sit in a circle, and every member votes for the person either tohis/her immediate left, or to his/her immediate right. If one member wins more votes than all theother members do, he/she will be declared to be the president; otherwise, one of the the members whowon at least as many votes as all the other members did will be randomly selected to be the president.If Hermia and Lysander are two members of the committee, with Hermia sitting to Lysander’s leftand Lysander planning to vote for Hermia, determine the probability that Hermia is elected president,assuming that the other n− 1 members vote randomly.

Answer: 2n−1n2n−1 Let x be the probability Hermia is elected if Lysander votes for her, and let y be

the probability that she wins if Lysander does not vote for her. We are trying to find x, and do soby first finding y. If Lysander votes for Hermia with probability 1

2 then the probability that Hermiais elected chairman is x

2 + y2 , but it is also 1

n by symmetry. If Lysander does not vote for Hermia,Hermia can get at most 1 vote, and then can only be elected if everyone gets one vote and she wins thetiebreaker. The probability she wins the tiebreaker is 1

n , and chasing around the circle, the probabilitythat every person gets 1 vote is 1

2n−1 . (Everyone votes for the person to the left, or everyone votes forthe person to the right.) Hence

y =1

n2n−1.

Then x2 + 1

n2n = 1n , so solving for x gives

x =2n − 1

n2n−1.

19. Let

F (x) =1

(2− x− x5)2011,

and note that F may be expanded as a power series so that F (x) =

∞∑n=0

anxn. Find an ordered pair of

positive real numbers (c, d) such that limn→∞

annd

= c.

Answer: ( 1620112010! , 2010)

First notice that all the roots of 2 − x − x5 that are not 1 lie strictly outside the unit circle. Assuch, we may write 2− x− x5 as 2(1− x)(1− r1x)(1− r2x)(1− r3x)(1− r4x) where |ri| < 1, and let

1(2−x−x5) = b0

(1−x) + b1(1−r1x) + . . .+ b4

(1−r4x) . We calculate b0 as limx→1(1−x)

(2−x−x5) = limx→1(−1)

(−1−5x4) = 16 .

Now raise the equation above to the 2011th power.

1

(2− x− x5)2011=

(1/6

(1− x)+

b1(1− r1x)

+ . . .+b4

(1− r4x)

)2011

Expand the right hand side using multinomial expansion and then apply partial fractions. The resultwill be a sum of the terms (1− x)−k and (1− rix)−k, where k ≤ 2011.

Since |ri| < 1, the power series of (1− rix)−k will have exponentially decaying coefficients, so we onlyneed to consider the (1−x)−k terms. The coefficient of xn in the power series of (1−x)−k is

(n+k−1k−1

),

which is a (k − 1)th degree polynomial in variable n. So when we sum up all coefficients, only thepower series of (1− x)−2011 will have impact on the leading term n2010.

Calculus & Combinatorics Individual Test

The coefficient of the (1 − x)−2011 term in the multinomial expansion is ( 16 )2011. The coefficient of

the xn term in the power series of (1 − x)−2011 is(n+20102010

)= 1

2010!n2010 + . . .. Therefore, (c, d) =

( 1620112010! , 2010).

20. Alice and Bob play a game in which two thousand and eleven 2011×2011 grids are distributed betweenthe two of them, 1 to Bob, and the other 2010 to Alice. They go behind closed doors and fill theirgrid(s) with the numbers 1, 2, . . . , 20112 so that the numbers across rows (left-to-right) and downcolumns (top-to-bottom) are strictly increasing. No two of Alice’s grids may be filled identically. Afterthe grids are filled, Bob is allowed to look at Alice’s grids and then swap numbers on his own grid,two at a time, as long as the numbering remains legal (i.e. increasing across rows and down columns)after each swap. When he is done swapping, a grid of Alice’s is selected at random. If there exist twointegers in the same column of this grid that occur in the same row of Bob’s grid, Bob wins. Otherwise,Alice wins. If Bob selects his initial grid optimally, what is the maximum number of swaps that Bobmay need in order to guarantee victory?

Answer: 1

Consider the grid whose entries in the jth row are, in order, 2011j−2010, 2011j−2009, . . . , 2011j. Callthis grid A0. For k = 1, 2 . . . , 2010, let grid Ak be the grid obtained from A0 by swapping the rightmostentry of the kth row with the leftmost entry of the k+1st row. We claim that if A ∈ {A0, A1, . . . , A2010},then given any legally numbered grid B such that A and B differ in at least one entry, there exist twointegers in the same column of B that occur in the same row of A.We first consider A0. Assume for the sake of contradiction B is a legally numbered grid distinct fromA0, such that there do not exist two integers in the same column of B that occur in the same rowof A0. Since the numbers 1, 2, . . . , 2011 occur in the same row of A0, they must all occur in differentcolumns of B. Clearly 1 is the leftmost entry in B’s first row. Let m be the smallest number that doesnot occur in the first row of B. Since each row is in order, m must be the first entry in its row. Butthen 1 and m are in the same column of B, a contradiction. It follows that the numbers 1, 2, . . . , 2011all occur in the first row of B. Proceeding by induction, 2011j−2010, 2011j−2009, . . . , 2011j must alloccur in the jth row of B for all 1 ≤ j ≤ 2011. Since A0 is the only legally numbered grid satsifyingthis condition, we have reached the desired contradiction.Now note that if A ∈ {A1, . . . , A2010}, there exist two integers in the same column of A0 that occur inthe same row of A. In particular, if A = Ak and 1 ≤ k ≤ 2010, then the integers 2011k − 2010 and2011k + 1 occur in the same column of A0 and in the same row of Ak. Therefore, it suffices to showthat for all 1 ≤ k ≤ 2010, there is no legally numbered grid B distinct from Ak and A0 such that theredo not exist two integers in the same column of B that occur in the same row of A0. Assume for thesake of contradiction that there does exist such a grid B. By the same logic as above, applied to thefirst k−1 rows and applied backwards to the last 2010−k−1 rows, we see that B may only differ fromAk in the kth and k+ 1st rows. However, there are only two legally numbered grids that are identicalto Ak outside of rows k and k + 1, namely A0 and Ak. This proves the claim.It remains only to note that, by the pigeonhole principle, if one of Alice’s grids is A0, then there existsa positive integer k, 1 ≤ k ≤ 2010, such that Ak is not one of the Alice’s grids. Therefore, if Bob setshis initial grid to be A0, he will require only one swap to switch his grid to Ak after examining Alice’sgrids. If A0 is not among Alice’s grids, then if Bob sets his initial grid to be A0, he will not in factrequire any swaps at all.

Calculus & Combinatorics Individual Test

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

1. Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠BAC intersect line BC atD. If there exist points E and F on sides AC and BC, respectively, such that lines AD and EFare parallel and divide triangle ABC into three parts of equal area, determine the number of possibleintegral values for BC.

2. Let a, b, and c be positive real numbers. Determine the largest total number of real roots that thefollowing three polynomials may have among them: ax2 + bx + c, bx2 + cx + a, and cx2 + ax + b.

3. Let f : R → R be a differentiable function such that f(0) = 0, f(1) = 1, and |f ′(x)| ≤ 2 for all real

numbers x. If a and b are real numbers such that the set of possible values of

∫ 1

0

f(x) dx is the open

interval (a, b), determine b− a.

4. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner.At each step, he can either move one square to the right or simultaneously move one square to the leftand one square up. In how many ways can he reach the center square of the topmost row?

5. Let H be a regular hexagon of side length x. Call a hexagon in the same plane a “distortion” of H ifand only if it can be obtained from H by translating each vertex of H by a distance strictly less than1. Determine the smallest value of x for which every distortion of H is necessarily convex.

6. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. IfNathaniel goes first, determine the probability that he ends up winning.

7. Let ABCDEF be a regular hexagon of area 1. Let M be the midpoint of DE. Let X be theintersection of AC and BM , let Y be the intersection of BF and AM , and let Z be the intersectionof AC and BF . If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate[BXC] + [AY F ] + [ABZ]− [MXZY ].

8. Let f : [0, 1)→ R be a function that satisfies the following condition: if

x =

∞∑n=1

an10n

= .a1a2a3 . . .

is the decimal expansion of x and there does not exist a positive integer k such that an = 9 for alln ≥ k, then

f(x) =

∞∑n=1

an102n

.

Determine f ′13

).

9. Let ABCD be a square of side length 13. Let E and F be points on rays AB and AD, respectively,so that the area of square ABCD equals the area of triangle AEF . If EF intersects BC at X andBX = 6, determine DF .

10. Evaluate

∫ ∞1

(lnx

x

)2011

dx.

11. Let ABCDEF be a convex equilateral hexagon such that lines BC, AD, and EF are parallel. LetH be the orthocenter of triangle ABD. If the smallest interior angle of the hexagon is 4 degrees,determine the smallest angle of the triangle HAD in degrees.

12. Sarah and Hagar play a game of darts. Let O0 be a circle of radius 1. On the nth turn, the playerwhose turn it is throws a dart and hits a point pn randomly selected from the points of On∈1. Theplayer then draws the largest circle that is centered at pn and contained in On∈1, and calls this circleOn. The player then colors every point that is inside On∈1 but not inside On her color. Sarah goesfirst, and the two players alternate turns. Play continues indefinitely. If Sarah’s color is red, andHagar’s color is blue, what is the expected value of the area of the set of points colored red?

13. Let ABCD be a cyclic quadrilateral, and suppose that BC = CD = 2. Let I be the incenter of triangleABD. If AI = 2 as well, find the minimum value of the length of diagonal BD.

14. Let f : [0, 1] → [0, 1] be a continuous function such that f(f(x)) = 1 for all x ∈ [0, 1]. Determine the

set of possible values of

∫ 1

0

f(x) dx.

15. Let f(x) = x2 − r2x + r3 for all real numbers x, where r2 and r3 are some real numbers. Define asequence {gn} for all nonnegative integers n by g0 = 0 and gn+1 = f(gn). Assume that {gn} satisfiesthe following three conditions: (i) g2i < g2i+1 and g2i+1 > g2i+2 for all 0 ≤ i ≤ 2011; (ii) there existsa positive integer j such that gi+1 > gi for all i > j, and (iii) {gn} is unbounded. If A is the greatestnumber such that A ≤ |r2| for any function f satisfying these properties, find A.

16. Let ABCD be a quadrilateral inscribed in the unit circle such that ∠BAD is 30 degrees. Let m denotethe minimum value of CP + PQ + CQ, where P and Q may be any points lying along rays AB andAD, respectively. Determine the maximum value of m.

17. Let f : (0, 1) → (0, 1) be a differentiable function with a continuous derivative such that for everypositive integer n and odd positive integer a < 2n, there exists an odd positive integer b < 2n suchthat f

a2n

)= b

2n . Determine the set of possible values of f ′12

).

18. Collinear points A, B, and C are given in the Cartesian plane such that A = (a, 0) lies along thex-axis, B lies along the line y = x, C lies along the line y = 2x, and AB/BC = 2. If D = (a, a), thecircumcircle of triangle ADC intersects y = x again at E, and ray AE intersects y = 2x at F , evaluateAE/EF .

19. Let

F (x) =1

(2− x− x5)2011,

and note that F may be expanded as a power series so that F (x) =

∞∑n=0

anxn. Find an ordered pair of

positive real numbers (c, d) such that limn→∞

annd

= c.

20. Let ω1 and ω2 be two circles that intersect at points A and B. Let line l be tangent to ω1 at P and toω2 at Q so that A is closer to PQ than B. Let points R and S lie along rays PA and QA, respectively,so that PQ = AR = AS and R and S are on opposite sides of A as P and Q. Let O be the circumcenterof triangle ASR, and let C and D be the midpoints of major arcs AP and AQ, respectively. If ∠APQis 45 degrees and ∠AQP is 30 degrees, determine ∠COD in degrees.

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

Calculus & Geometry Individual Test

1. Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠BAC intersect line BC atD. If there exist points E and F on sides AC and BC, respectively, such that lines AD and EFare parallel and divide triangle ABC into three parts of equal area, determine the number of possibleintegral values for BC.

Answer: 13

A B

C

DE

F

7

14

Note that such E,F exist if and only if[ADC]

[ADB]= 2. (1)

([ ] denotes area.) Since AD is the angle bisector, and the ratio of areas of triangles with equal heightis the ratio of their bases,

AC

AB=DC

DB=

[ADC]

[ADB].

Hence (1) is equivalent to AC = 2AB = 14. Then BC can be any length d such that the triangleinequalities are satisfied:

d+ 7 > 14

7 + 14 > d

Hence 7 < d < 21 and there are 13 possible integral values for BC.

2. Let a, b, and c be positive real numbers. Determine the largest total number of real roots that thefollowing three polynomials may have among them: ax2 + bx+ c, bx2 + cx+ a, and cx2 + ax+ b.

Answer: 4 If all the polynomials had real roots, their discriminants would all be nonnegative: a2 ≥4bc, b2 ≥ 4ca, and c2 ≥ 4ab. Multiplying these inequalities gives (abc)2 ≥ 64(abc)2, a contradiction.Hence one of the quadratics has no real roots. The maximum of 4 real roots is attainable: for example,the values (a, b, c) = (1, 5, 6) give −2,−3 as roots to x2 + 5x+ 6 and −1,− 1

5 as roots to 5x2 + 6x+ 1.

Calculus & Geometry Individual Test

3. Let f : R → R be a differentiable function such that f(0) = 0, f(1) = 1, and |f ′(x)| ≤ 2 for all real

numbers x. If a and b are real numbers such that the set of possible values of

∫ 1

0

f(x) dx is the open

interval (a, b), determine b− a.

Answer: 34 Draw lines of slope ±2 passing through (0, 0) and (1, 1). These form a parallelogram with

vertices (0, 0), (.75, 1.5), (1, 1), (.25,−.5). By the mean value theorem, no point of (x, f(x)) lies outsidethis parallelogram, but we can construct functions arbitrarily close to the top or the bottom of theparallelogram while satisfying the condition of the problem. So (b−a) is the area of this parallelogram,which is 3

4 .

4. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner.At each step, he can either move one square to the right or simultaneously move one square to the leftand one square up. In how many ways can he reach the center square of the topmost row?

Answer: 2n−1 Note that Josh must pass through the center square of each row. There are 2 waysto get from the center square of row k to the center square of row k+ 1. So there are 2n−1 ways to getto the center square of row n.

5. Let H be a regular hexagon of side length x. Call a hexagon in the same plane a “distortion” of H ifand only if it can be obtained from H by translating each vertex of H by a distance strictly less than1. Determine the smallest value of x for which every distortion of H is necessarily convex.

Answer: 4

A1

A2

A3 A4

A5

A6X X′

Y Y ′

Let H = A1A2A3A4A5A6 be the hexagon, and for all 1 ≤ i ≤ 6, let points A′i be considered such thatAiA

′i < 1. Let H ′ = A′1A

′2A′3A′4A′5A′6, and consider all indices modulo 6. For any point P in the plane,

let D(P ) denote the unit disk {Q|PQ < 1} centered at P ; it follows that A′i ∈ D(Ai).

Let X and X ′ be points on line A1A6, and let Y and Y ′ be points on line A3A4 such that A1X =A1X

′ = A3Y = A3Y′ = 1 and X and X ′ lie on opposite sides of A1 and Y and Y ′ lie on opposite sides

of A3. If X ′ and Y ′ lie on segments A1A6 and A3A4, respectively, then segment A′1A′3 lies between the

lines XY and X ′Y ′. Note that x2 is the distance from A2 to A1A3.

Calculus & Geometry Individual Test

A1

A2

A3 A4

A5

A6X X′

Y Y ′

If x2 ≥ 2, then C(A2) cannot intersect line XY , since the distance from XY to A1A3 is 1 and the

distance from XY to A2 is at least 1. Therefore, A′1A′3 separates A′2 from the other 3 vertices of the

hexagon. By analogous reasoning applied to the other vertices, we may conclude that H ′ is convex.

If x2 < 2, then C(A2) intersects XY , so by choosing A′1 = X and A′3 = Y , we see that we may choose

A′2 on the opposite side of XY , in which case H ′ will be concave. Hence the answer is 4, as desired.

6. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. IfNathaniel goes first, determine the probability that he ends up winning.

Answer: 511 For 1 ≤ k ≤ 6, let xk be the probability that the current player, say A, will win when

the number on the tally at the beginning of his turn is k modulo 7. The probability that the totalis l modulo 7 after his roll is 1

6 for each l 6≡ k (mod 7); in particular, there is a 16 chance he wins

immediately. The chance that A will win if he leaves l on the board after his turn is 1− xl. Hence for1 ≤ k ≤ 6,

xk =1

6

∑1≤l≤6, l 6=k

(1− xl) +1

6.

Letting s =∑6l=1 xl, this becomes xk = xk−s

6 + 1 or 5xk

6 = − s6 + 1. Hence x1 = · · · = x6, and 6xk = s

for every k. Plugging this in gives 11xk

6 = 1, or xk = 611 .

Since Nathaniel cannot win on his first turn, he leaves Obediah with a number not divisible by 7.Hence Obediah’s chance of winning is 6

11 and Nathaniel’s chance of winning is 511 .

7. Let ABCDEF be a regular hexagon of area 1. Let M be the midpoint of DE. Let X be theintersection of AC and BM , let Y be the intersection of BF and AM , and let Z be the intersectionof AC and BF . If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate[BXC] + [AY F ] + [ABZ]− [MXZY ].

Answer: 0

Calculus & Geometry Individual Test

A B

C

DE

F

M

XY

Z

O

Let O be the center of the hexagon. The desired area is [ABCDEF ] − [ACDM ] − [BFEM ]. Notethat [ADM ] = [ADE]/2 = [ODE] = [ABC], where the last equation holds because sin 60◦ = sin 120◦.Thus, [ACDM ] = [ACD] + [ADM ] = [ACD] + [ABC] = [ABCD], but the area of ABCD is half thearea of the hexagon. Similarly, the area of [BFEM ] is half the area of the hexagon, so the answer iszero.

8. Let f : [0, 1)→ R be a function that satisfies the following condition: if

x =

∞∑n=1

an10n

= .a1a2a3 . . .

is the decimal expansion of x and there does not exist a positive integer k such that an = 9 for alln ≥ k, then

f(x) =

∞∑n=1

an102n

.

Determine f ′(13

).

Answer: 0 Note that 13 =

∑∞n=1

310n .

Clearly f is an increasing function. Also for any integer n ≥ 1, we see from decimal expansions thatf( 1

3 ±1

10n )− f( 13 ) = ± 1

102n .

Consider h such that 10−n−1 ≤ |h| < 10−n. The two properties of f outlined above show that |f( 13 +

h)− f( 13 )| < 1

102n . And from | 1h | ≤ 10n+1, we get∣∣∣ f( 1

3+h)−f(13 )

h

∣∣∣ < 110n−1 . Taking n→∞ gives h→ 0

and f ′( 13 ) = limn→∞

110n−1 = 0.

9. Let ABCD be a square of side length 13. Let E and F be points on rays AB and AD, respectively,so that the area of square ABCD equals the area of triangle AEF . If EF intersects BC at X andBX = 6, determine DF .

Answer:√

13

Calculus & Geometry Individual Test

A B

CD

E

F

X

Y

6

7

x

y

First Solution

Let Y be the point of intersection of lines EF and CD. Note that [ABCD] = [AEF ] implies that[BEX] + [DY F ] = [CY X]. Since 4BEX ∼ 4CY X ∼ 4DY F , there exists some constant r suchthat [BEX] = r · BX2, [Y DF ] = r · CX2, and [CY X] = r · DF 2. Hence BX2 + DF 2 = CX2, soDF =

√CX2 −BX2 =

√49− 36 =

√13.

Second Solution

Let x = DF and y = Y D. Since 4BXE ∼ 4CXY ∼ 4DFY , we have

BE

BX=CY

CX=DY

DF=y

x.

Using BX = 6, XC = 7 and CY = 13− y we get BE = 6yx and 13−y

7 = yx . Solving this last equation

for y gives y = 13xx+7 . Now [ABCD] = [AEF ] gives

169 =1

2AE ·AF =

1

2

(13 +

6y

x

)(13 + x) .

169 = 6y + 13x+78y

x

13 =6x

x+ 7+ x+

78

x+ 7

0 = x2 − 13.

Thus x =√

13.

10. Evaluate

∫ ∞1

(lnx

x

)2011

dx.

Answer: 2011!20102012 By the chain rule,

d

dx(lnx)n =

n lnn−1 x

x.

We calculate the definite integral using integration by parts:∫ ∞x=1

(lnx)n

x2011dx =

[(lnx)n

−2010x2010

]x=∞x=1

−∫ ∞x=1

n(lnx)n−1

−2010x2011dx

But ln(1) = 0, and limx→∞(ln x)n

x2010 = 0 for all n > 0. So

Calculus & Geometry Individual Test

∫ ∞x=1

(lnx)n

x2011dx =

∫ ∞x=1

n(lnx)n−1

2010x2011dx

It follows that ∫ ∞x=1

(lnx)n

x2011dx =

n!

2010n

∫ ∞x=1

1

x2011dx =

n!

2010n+1

So the answer is 2011!20102012 .

11. Let ABCDEF be a convex equilateral hexagon such that lines BC, AD, and EF are parallel. LetH be the orthocenter of triangle ABD. If the smallest interior angle of the hexagon is 4 degrees,determine the smallest angle of the triangle HAD in degrees.

Answer: 3

A

B

C

D

E

FH

A′

B′

D′

Note that ABCD and DEFA are isosceles trapezoids, so ∠BAD = ∠CDA and ∠FAD = ∠EDA. Inorder for the hexagon to be convex, the angles at B, C, E, and F have to be obtuse, so ∠A = ∠D = 4◦.Letting s be a side length of the hexagon, AD = AB cos∠BAD + BC + CD cos∠CDA = s(1 +2 cos∠BAD), so ∠BAD is uniquely determined by AD. Since the same equation holds for trapezoidDEFA, it follows that ∠BAD = ∠FAD = ∠CDA = ∠EDA = 2◦. Then ∠BCD = 180◦ − 2◦ = 178◦.Since 4BCD is isosceles, ∠CDB = 1◦ and ∠BDA = 1◦. (One may also note that ∠BDA = 1◦ byobserving that equal lengths AB and BC must intercept equal arcs on the circumcircle of isoscelestrapezoid ABCD).

Let A′, B′, andD′ be the feet of the perpendiculars from A, B, andD toBD, DA, and AB, respectively.Angle chasing yields

∠AHD = ∠AHB′ + ∠DHB′ = (90◦ − ∠A′AB′) + (90◦ − ∠D′DB′)

= ∠BDA+ ∠BAD = 1◦ + 2◦ = 3◦

∠HAD = 90◦ − ∠AHB′ = 89◦

∠HDA = 90◦ − ∠DHB′ = 88◦

Hence the smallest angle in 4HAD is 3◦.

Calculus & Geometry Individual Test

A

B

C

D

E

FH

It is faster, however, to draw the circumcircle of DEFA, and to note that since H is the orthocenterof triangle ABD, B is the orthocenter of triangle HAD. Then since F is the reflection of B acrossAD, quadrilateral HAFD is cyclic, so ∠AHD = ∠ADF + ∠DAF = 1◦ + 2◦ = 3◦, as desired.

12. Sarah and Hagar play a game of darts. Let O0 be a circle of radius 1. On the nth turn, the playerwhose turn it is throws a dart and hits a point pn randomly selected from the points of On−1. Theplayer then draws the largest circle that is centered at pn and contained in On−1, and calls this circleOn. The player then colors every point that is inside On−1 but not inside On her color. Sarah goesfirst, and the two players alternate turns. Play continues indefinitely. If Sarah’s color is red, andHagar’s color is blue, what is the expected value of the area of the set of points colored red?

Answer: 6π7 Let f(r) be the average area colored red on a dartboard of radius r if Sarah plays first.

Then f(r) is proportional to r2. Let f(r) = (πx)r2 for some constant x. We want to find f(1) = πx.

In the first throw, if Sarah’s dart hits a point with distance r from the center of O0, the radius of O1

will be 1−r. The expected value of the area colored red will be (π−π(1−r)2)+(π(1−r)2−f(1−r)) =π − f(1− r). The value of f(1) is the average value of π − f(1− r) over all points in O0. Using polarcoordinates, we get

f(1) =

2π∫0

1∫0

(π − f(1− r))rdrdθ

2π∫0

1∫0

rdrdθ

πx =

1∫0

(π − πx(1− r)2)rdr

1∫0

rdr

πx

2=

∫ 1

0

πr − πxr(1− r)2dr

πx

2=π

2− πx(

1

2− 2

3+

1

4)

πx

2=π

2− πx

12

Calculus & Geometry Individual Test

πx =6π

7.

13. Let ABCD be a cyclic quadrilateral, and suppose that BC = CD = 2. Let I be the incenter of triangleABD. If AI = 2 as well, find the minimum value of the length of diagonal BD.

Answer: 2√

3 Let T be the point where the incircle intersects AD, and let r be the inradius and Rbe the circumradius of 4ABD. Since BC = CD = 2, C is on the midpoint of arc BD on the oppositeside of BD as A, and hence on the angle bisector of A. Thus A, I, and C are collinear. We have thefollowing formulas:

AI =IM

sin∠IAM=

r

sin A2

BC = 2R sinA

2BD = 2R sinA

The last two equations follow from the extended law of sines on 4ABC and 4ABD, respectively.

Using AI = 2 = BC gives sin2 A2 = r

2R . However, it is well-known that R ≥ 2r with equality for an

equilateral triangle (one way to see this is the identity 1+ rR = cosA+cosB+cosD). Hence sin2 A

2 ≤14

and A2 ≤ 30◦. Then

BD = 2R

(2 sin

A

2cos

A

2

)= BC · 2 cos

A

2≥ 2

(2 ·√

3

2

)= 2√

3

with equality when 4ABD is equilateral.

A

B

C

D

I

Remark: Similar but perhaps simpler computations can be made by noting that if AC intersects BDat X, then AB/BX = AD/DX = 2, which follows from the exterior angle bisector theorem; if IA isthe A-excenter of triangle ABC, then AIA/XIA = 2 since it is well-known that C is the circumcenterof cyclic quadrilateral BIDIA.

Calculus & Geometry Individual Test

14. Let f : [0, 1] → [0, 1] be a continuous function such that f(f(x)) = 1 for all x ∈ [0, 1]. Determine the

set of possible values of

∫ 1

0

f(x) dx.

Answer: ( 34 , 1] Since the maximum value of f is 1,

∫ 1

0f(x)dx ≤ 1.

By our condition f(f(x)) = 1, f is 1 at any point within the range of f . Clearly, 1 is in the range off , so f(1) = 1. Now f(x) is continuous on a closed interval so it attains a minimum value c. Since cis in the range of f , f(c) = 1.

If c = 1, f(x) = 1 for all x and∫ 1

0f(x)dx = 1.

Now assume c < 1. By the intermediate value theorem, since f is continuous it attains all valuesbetween c and 1. So for all x ≥ c, f(x) = 1. Therefore,∫ 1

0

f(x)dx =

∫ c

0

f(x)dx+ (1− c).

Since f(x) ≥ c,∫ c0f(x)dx > c2, and the equality is strict because f is continuous and thus cannot be

c for all x < c and 1 at c. So∫ 1

0

f(x)dx > c2 + (1− c) = (c− 1

2)2 +

3

4≥ 3

4.

Therefore 34 <

∫ 1

0f(x)dx ≤ 1, and it is easy to show that every value in this interval can be reached.

15. Let f(x) = x2 − r2x + r3 for all real numbers x, where r2 and r3 are some real numbers. Define asequence {gn} for all nonnegative integers n by g0 = 0 and gn+1 = f(gn). Assume that {gn} satisfiesthe following three conditions: (i) g2i < g2i+1 and g2i+1 > g2i+2 for all 0 ≤ i ≤ 2011; (ii) there existsa positive integer j such that gi+1 > gi for all i > j, and (iii) {gn} is unbounded. If A is the greatestnumber such that A ≤ |r2| for any function f satisfying these properties, find A.

Answer: 2 Consider the function f(x) − x. By the constraints of the problem, f(x) − x must benegative for some x, namely, for x = g2i+1, 0 ≤ i ≤ 2011. Since f(x) − x is positive for x of largeabsolute value, the graph of f(x) − x crosses the x-axis twice and f(x) − x has two real roots, saya < b. Factoring gives f(x)− x = (x− a)(x− b), or f(x) = (x− a)(x− b) + x.

Now, for x < a, f(x) > x > a, while for x > b, f(x) > x > b. Let c 6= b be the number such thatf(c) = f(b) = b. Note that b is not the vertex as f(a) = a < b, so by the symmetry of quadratics, cexists and b+c

2 = r22 as the vertex of the parabola. By the same token, b+a

2 = r2+12 is the vertex of

f(x)− x. Hence c = a− 1. If f(x) > b then x < c or x > b. Consider the smallest j such that gj > b.Then by the above observation, gj−1 < c. (If gi ≥ b then f(gi) ≥ gi ≥ b so by induction, gi+1 ≥ gi forall i ≥ j. Hence j > 1; in fact j ≥ 4025.) Since gj−1 = f(gj−2), the minimum value of f is less than c.The minimum value is the value of f evaluated at its vertex, b+a−12 , so

f

(b+ a− 1

2

)< c(

b+ a− 1

2− a)(

b+ a− 1

2− b)

+b+ a− 1

2< a− 1

1− (b− a)2

4+b− a+ 1

2< 0

3

4<

(b− a)2

4− b− a

2

4 < (b− a− 1)2.

Then either b − a − 1 < −2 or b − a − 1 > 2, but b > a, so the latter must hold and (b − a)2 > 9.Now, the discriminant of f(x) − x equals (b − a)2 (the square of the difference of the two roots) and(r2 + 1)2 − 4r3 (from the coefficients), so (r2 + 1)2 > 9 + 4r3. But r3 = g1 > g0 = 0 so |r2| > 2.

Calculus & Geometry Individual Test

We claim that we can make |r2| arbitrarily close to 2, so that the answer is 2. First define Gi, i ≥ 0as follows. Let N ≥ 2012 be an integer. For ε > 0 let h(x) = x2 − 2 − ε, gε(x) = −

√x+ 2 + ε and

G2N+1 = 2 + ε, and define Gi recursively by Gi = gε(Gi+1), Gi+1 = h(Gi). (These two equationsare consistent.) Note the following. (i) G2i < G2i+1 and G2i+1 > G2i+2 for 0 ≤ i ≤ N − 1. Firstnote G2N = −

√4 + 2ε > −

√4 + 2ε+ ε2 = −2 − ε. Let l be the negative solution to h(x) = x. Note

that −2 − ε < G2N < l < 0 since h(G2N ) > 0 > G2N . Now gε(x) is defined as long as x ≥ −2 − ε,and it sends (−2 − ε, l) into (l, 0) and (l, 0) into (−2 − ε, l). It follows that the Gi, 0 ≤ i ≤ 2Nare well-defined; moreover, G2i < l and G2i+1 > l for 0 ≤ i ≤ N − 1 by backwards induction on i,so the desired inequalities follow. (ii) Gi is increasing for i ≥ 2N + 1. Indeed, if x ≥ 2 + ε, thenx2 − x = x(x− 1) > 2 + ε so h(x) > x. Hence 2 + ε = G2N+1 < G2N+2 < · · · . (iii) Gi is unbounded.This follows since h(x) − x = x(x − 2) − 2 − ε is increasing for x > 2 + ε, so Gi increases faster andfaster for i ≥ 2N + 1. Now define f(x) = h(x + G0) − G0 = x2 + 2G0x + G2

0 − G0 − 2 − ε. NoteGi+1 = h(Gi) while gi+1 = f(gi) = h(gi + G0) − G0, so by induction gi = Gi − G0. Since {Gi}∞i=0

satisfies (i), (ii), and (iii), so does gi.

We claim that we can make G0 arbitrarily close to −1 by choosing N large enough and ε small enough;this will make r2 = −2G0 arbitrarily close to 2. Choosing N large corresponds to taking G0 to be alarger iterate of 2 + ε under gε(x). By continuity of this function with respect to x and ε, it suffices totake ε = 0 and show that (letting g = g0)

g(n)(2) = g(· · · g︸ ︷︷ ︸n

(2) · · · )→ −1 as n→∞.

But note that for 0 ≤ θ ≤ π2 ,

g(−2 cos θ) = −√

2− 2 cos θ = −2 sin

(θ

2

)= 2 cos

(π

2− θ

2

).

Hence by induction, g(n)(−2 cos θ) = −2 cos(π2 −

π4 + · · ·+ (−1)n

(θ − π

2n

)). Hence g(n)(2) = g(n−1)(−2 cos 0)

converges to −2 cos(π2 −π4 + · · · ) = −2 cos(π3 ) = −1, as needed.

16. Let ABCD be a quadrilateral inscribed in the unit circle such that ∠BAD is 30 degrees. Let m denotethe minimum value of CP + PQ + CQ, where P and Q may be any points lying along rays AB andAD, respectively. Determine the maximum value of m.

Answer: 2

A

BC

D

E

F

G

H

P

Q

R

Calculus & Geometry Individual Test

For a fixed quadrilateral ABCD as described, we first show that m, the minimum possible length ofCP + PQ + QC, equals the length of AC. Reflect B, C, and P across line AD to points E, F , andR, respectively, and then reflect D and F across AE to points G and H, respectively. These tworeflections combine to give a 60◦ rotation around A, so triangle ACH is equilateral. It also followsthat RH is a 60◦ rotation of PC around A, so, in particular, these segments have the same length.Because QR = QP by reflection,

CP + PQ+QC = CQ+QR+RH.

The latter is the length of a broken path CQRH from C to H, and by the “shortest path is a straightline” principle, this total length is at least as long as CH = CA. (More directly, this follows from thetriangle inequality: (CQ+QR)+RH ≥ CR+RH ≥ CH). Therefore, the lower bound m ≥ AC indeedholds. To see that this is actually an equality, note that choosing Q as the intersection of segment CHwith ray AD, and choosing P so that its reflection R is the intersection of CH with ray AE, alignspath CQRH with segment CH, thus obtaining the desired minimum m = AC.

We may conclude that the largest possible value of m is the largest possible length of AC, namely 2:the length of a diameter of the circle.

17. Let f : (0, 1) → (0, 1) be a differentiable function with a continuous derivative such that for everypositive integer n and odd positive integer a < 2n, there exists an odd positive integer b < 2n suchthat f

(a2n

)= b

2n . Determine the set of possible values of f ′(12

).

Answer: {−1, 1} The key step is to notice that for such a function f , f ′(x) 6= 0 for any x.

Assume, for sake of contradiction that there exists 0 < y < 1 such that f ′(y) = 0. Since f ′ isa continuous function, there is some small interval (c, d) containing y such that |f ′(x)| ≤ 1

2 for allx ∈ (c, d). Now there exists some n, a such that a

2n ,a+12n are both in the interval (c, d). From the

definition,f(a+1

2n )− f( a2n )

a+12n −

a2n

= 2n(b′

2n− b

2n) = b′ − b where b, b′ are integers; one is odd, and one is

even. So b′ − b is an odd integer. Since f is differentiable, by the mean value theorem there exists apoint where f ′ = b′ − b. But this point is in the interval (c, d), and |b′ − b| > 1

2 . This contradicts theassumption that |f ′(x)| ≤ 1

2 for all x ∈ (c, d).

Since f ′(x) 6= 0, and f ′ is a continuous function, f ′ is either always positive or always negative. So f iseither increasing or decreasing. f( 1

2 ) = 12 always. If f is increasing, it follows that f( 1

4 ) = 14 , f( 3

4 ) = 34 ,

and we can show by induction that indeed f( a2n ) = a

2n for all integers a, n. Since numbers of this formare dense in the interval (0, 1), and f is a continuous function, f(x) = x for all x.

It can be similarly shown that if f is decreasing f(x) = 1− x for all x. So the only possible values off ′( 1

2 ) are −1, 1.

Query: if the condition that the derivative is continuous were omitted, would the same result stillhold?

18. Collinear points A, B, and C are given in the Cartesian plane such that A = (a, 0) lies along thex-axis, B lies along the line y = x, C lies along the line y = 2x, and AB/BC = 2. If D = (a, a), thecircumcircle of triangle ADC intersects y = x again at E, and ray AE intersects y = 2x at F , evaluateAE/EF .

Answer: 7

Calculus & Geometry Individual Test

A

B

CD

E

F

O

PQ

Let points O, P , and Q be located at (0, 0), (a, 2a), and (0, 2a), respectively. Note that BC/AB = 1/2implies [OCD]/[OAD] = 1/2, so since [OPD] = [OAD], [OCD]/[OPD] = 1/2. It follows that[OCD] = [OPD]. Hence OC = CP . We may conclude that triangles OCQ and PCA are congruent,so C = (a/2, a).It follows that ∠ADC is right, so the circumcircle of triangle ADC is the midpoint of AC, whichis located at (3a/4, a/2). Let (3a/4, a/2) = H, and let E = (b, b). Then the power of the pointO with respect to the circumcircle of ADC is OD · OE = 2ab, but it may also be computed asOH2 −HA2 = 13a/16− 5a/16 = a/2. It follows that b = a/4, so E = (a/4, a/4).We may conclude that line AE is x + 3y = a, which intersects y = 2x at an x-coordinate of a/7.Therefore, AE/EF = (a− a/4)/(a/4− a/7) = (3a/4)/(3a/28) = 7.

Remark: The problem may be solved more quickly if one notes from the beginning that lines OA,OD, OP , and OQ form a harmonic pencil because D is the midpoint of AP and lines OQ and AP areparallel.

Calculus & Geometry Individual Test

19. Let

F (x) =1

(2− x− x5)2011,

and note that F may be expanded as a power series so that F (x) =

∞∑n=0

anxn. Find an ordered pair of

positive real numbers (c, d) such that limn→∞

annd

= c.

Answer: ( 1620112010! , 2010) First notice that all the roots of 2 − x − x5 that are not 1 lie strictly

outside the unit circle. As such, we may write 2−x−x5 as 2(1−x)(1− r1x)(1− r2x)(1− r3x)(1− r4x)

where |ri| < 1, and let 1(2−x−x5) = b0

(1−x) + b1(1−r1x) + . . .+ b4

(1−r4x) . We calculate b0 as limx→1(1−x)

(2−x−x5) =

limx→1(−1)

(−1−5x4) = 16 .

Now raise the equation above to the 2011th power.

1

(2− x− x5)2011=

(1/6

(1− x)+

b1(1− r1x)

+ . . .+b4

(1− r4x)

)2011

Expand the right hand side using multinomial expansion and then apply partial fractions. The resultwill be a sum of the terms (1− x)−k and (1− rix)−k, where k ≤ 2011.

Since |ri| < 1, the power series of (1− rix)−k will have exponentially decaying coefficients, so we onlyneed to consider the (1−x)−k terms. The coefficient of xn in the power series of (1−x)−k is

(n+k−1k−1

),

which is a (k − 1)th degree polynomial in variable n. So when we sum up all coefficients, only thepower series of (1− x)−2011 will have impact on the leading term n2010.

The coefficient of the (1 − x)−2011 term in the multinomial expansion is ( 16 )2011. The coefficient of

the xn term in the power series of (1 − x)−2011 is(n+20102010

)= 1

2010!n2010 + . . .. Therefore, (c, d) =

( 1620112010! , 2010).

20. Let ω1 and ω2 be two circles that intersect at points A and B. Let line l be tangent to ω1 at P and toω2 at Q so that A is closer to PQ than B. Let points R and S lie along rays PA and QA, respectively,so that PQ = AR = AS and R and S are on opposite sides of A as P and Q. Let O be the circumcenterof triangle ASR, and let C and D be the midpoints of major arcs AP and AQ, respectively. If ∠APQis 45 degrees and ∠AQP is 30 degrees, determine ∠COD in degrees.

Answer: 142.5

A

B

P

Q

R

SC

D

T

T ′

We use directed angles throughout the solution.

Let T denote the point such that ∠TCD = 1/2∠APQ and ∠TDC = 1/2∠AQP . We claim that T isthe circumcenter of triangle SAR.

Calculus & Geometry Individual Test

Since CP = CA, QP = RA, and ∠CPQ = ∠CPA + ∠APQ = ∠CPA + ∠ACP = ∠CAR, we have4CPQ ∼= 4CAR. By spiral similarity, we have 4CPA ∼ 4CQR.

Let T ′ denote the reflection of T across CD. Since ∠TCT ′ = ∠APQ = ∠ACP , we have 4TCT ′ ∼4ACP ∼ 4RCQ. Again, by spiral similarity centered at C, we have 4CTR ∼ 4CT ′Q. ButCT = CT ′, so 4CTR ∼= 4CT ′Q and TR = T ′Q. Similarly, 4DTT ′ ∼ 4DAQ, and spiral similaritycentered at D shows that 4DTA ∼= 4DT ′Q. Thus TA = T ′Q = TR.

We similarly have TA = T ′P = TS, so T is indeed the circumcenter. Therefore, we have ∠COD =∠CTD = 180◦ − 45◦

2 −30◦

2 = 142.5◦.

Calculus & Geometry Individual Test

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

1. A classroom has 30 students and 30 desks arranged in 5 rows of 6. If the class has 15 boys and 15girls, in how many ways can the students be placed in the chairs such that no boy is sitting in frontof, behind, or next to another boy, and no girl is sitting in front of, behind, or next to another girl?

2. Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠BAC intersect line BC atD. If there exist points E and F on sides AC and BC, respectively, such that lines AD and EFare parallel and divide triangle ABC into three parts of equal area, determine the number of possibleintegral values for BC.

3. Let a, b, and c be positive real numbers. Determine the largest total number of real roots that thefollowing three polynomials may have among them: ax2 + bx + c, bx2 + cx + a, and cx2 + ax + b.

4. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner.At each step, he can either move one square to the right or simultaneously move one square to the leftand one square up. In how many ways can he reach the center square of the topmost row?

5. Let H be a regular hexagon of side length x. Call a hexagon in the same plane a “distortion” of H ifand only if it can be obtained from H by translating each vertex of H by a distance strictly less than1. Determine the smallest value of x for which every distortion of H is necessarily convex.

6. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. IfNathaniel goes first, determine the probability that he ends up winning.

7. Let ABCDEF be a regular hexagon of area 1. Let M be the midpoint of DE. Let X be theintersection of AC and BM , let Y be the intersection of BF and AM , and let Z be the intersectionof AC and BF . If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate[BXC] + [AY F ] + [ABZ]− [MXZY ].

8. The integers from 1 to n are written in increasing order from left to right on a blackboard. Davidand Goliath play the following game: starting with David, the two players alternate erasing any twoconsecutive numbers and replacing them with their sum or product. Play continues until only onenumber on the board remains. If it is odd, David wins, but if it is even, Goliath wins. Find the 2011thsmallest positive integer greater than 1 for which David can guarantee victory.

9. Let ABCD be a square of side length 13. Let E and F be points on rays AB and AD, respectively,so that the area of square ABCD equals the area of triangle AEF . If EF intersects BC at X andBX = 6, determine DF .

10. Mike and Harry play a game on an 8× 8 board. For some positive integer k, Mike chooses k squaresand writes an M in each of them. Harry then chooses k + 1 squares and writes an H in each of them.After Harry is done, Mike wins if there is a sequence of letters forming “HMM” or “MMH,” whenread either horizontally or vertically, and Harry wins otherwise. Determine the smallest value of k forwhich Mike has a winning strategy.

11. Let ABCDEF be a convex equilateral hexagon such that lines BC, AD, and EF are parallel. LetH be the orthocenter of triangle ABD. If the smallest interior angle of the hexagon is 4 degrees,determine the smallest angle of the triangle HAD in degrees.

12. The ordered pairs (2011, 2), (2010, 3), (2009, 4), . . ., (1008, 1005), (1007, 1006) are written from left toright on a blackboard. Every minute, Elizabeth selects a pair of adjacent pairs (xi, yi) and (xj , yj), with

(xi, yi) left of (xj , yj), erases them, and writes

(xiyixj

yj,xiyiyjxj

)in their place. Elizabeth continues

this process until only one ordered pair remains. How many possible ordered pairs (x, y) could appearon the blackboard after the process has come to a conclusion?

13. Let ABCD be a cyclic quadrilateral, and suppose that BC = CD = 2. Let I be the incenter of triangleABD. If AI = 2 as well, find the minimum value of the length of diagonal BD.

14. Let A = {1, 2, . . . , 2011}. Find the number of functions f from A to A that satisfy f(n) ≤ n for all nin A and attain exactly 2010 distinct values.

15. Let f(x) = x2 − r2x + r3 for all real numbers x, where r2 and r3 are some real numbers. Define asequence {gn} for all nonnegative integers n by g0 = 0 and gn+1 = f(gn). Assume that {gn} satisfiesthe following three conditions: (i) g2i < g2i+1 and g2i+1 > g2i+2 for all 0 ≤ i ≤ 2011; (ii) there existsa positive integer j such that gi+1 > gi for all i > j, and (iii) {gn} is unbounded. If A is the greatestnumber such that A ≤ |r2| for any function f satisfying these properties, find A.

16. Let ABCD be a quadrilateral inscribed in the unit circle such that ∠BAD is 30 degrees. Let m denotethe minimum value of CP + PQ + CQ, where P and Q may be any points lying along rays AB andAD, respectively. Determine the maximum value of m.

17. Let n be an odd positive integer, and suppose that n people sit on a committee that is in the processof electing a president. The members sit in a circle, and every member votes for the person either tohis/her immediate left, or to his/her immediate right. If one member wins more votes than all theother members do, he/she will be declared to be the president; otherwise, one of the the members whowon at least as many votes as all the other members did will be randomly selected to be the president.If Hermia and Lysander are two members of the committee, with Hermia sitting to Lysander’s leftand Lysander planning to vote for Hermia, determine the probability that Hermia is elected president,assuming that the other n− 1 members vote randomly.

18. Collinear points A, B, and C are given in the Cartesian plane such that A = (a, 0) lies along thex-axis, B lies along the line y = x, C lies along the line y = 2x, and AB/BC = 2. If D = (a, a), thecircumcircle of triangle ADC intersects y = x again at E, and ray AE intersects y = 2x at F , evaluateAE/EF .

19. Alice and Bob play a game in which two thousand and eleven 2011×2011 grids are distributed betweenthe two of them, 1 to Bob, and the other 2010 to Alice. They go behind closed doors and fill theirgrid(s) with the numbers 1, 2, . . . , 20112 so that the numbers across rows (left-to-right) and downcolumns (top-to-bottom) are strictly increasing. No two of Alice’s grids may be filled identically. Afterthe grids are filled, Bob is allowed to look at Alice’s grids and then swap numbers on his own grid,two at a time, as long as the numbering remains legal (i.e. increasing across rows and down columns)after each swap. When he is done swapping, a grid of Alice’s is selected at random. If there exist twointegers in the same column of this grid that occur in the same row of Bob’s grid, Bob wins. Otherwise,Alice wins. If Bob selects his initial grid optimally, what is the maximum number of swaps that Bobmay need in order to guarantee victory?

20. Let ω1 and ω2 be two circles that intersect at points A and B. Let line l be tangent to ω1 at P and toω2 at Q so that A is closer to PQ than B. Let points R and S lie along rays PA and QA, respectively,so that PQ = AR = AS and R and S are on opposite sides of A as P and Q. Let O be the circumcenterof triangle ASR, and let C and D be the midpoints of major arcs AP and AQ, respectively. If ∠APQis 45 degrees and ∠AQP is 30 degrees, determine ∠COD in degrees.

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

Combinatorics & Geometry Individual Test

1. A classroom has 30 students and 30 desks arranged in 5 rows of 6. If the class has 15 boys and 15girls, in how many ways can the students be placed in the chairs such that no boy is sitting in frontof, behind, or next to another boy, and no girl is sitting in front of, behind, or next to another girl?

Answer: 2 · 15!2 If we color the desks of the class in a checkerboard pattern, we notice that all ofone gender must go in the squares colored black, and the other gender must go in the squares coloredwhite. There are 2 ways to pick which gender goes in which color, 15! ways to put the boys into desksand 15! ways to put the girls into desks. So the number of ways is 2 · 15!2.

(There is a little ambiguity in the problem statement as to whether the 15 boys and the 15 girls aredistinguishable or not. If they are not distinguishable, the answer is clearly 2. Given the number ofcontestants who submitted the answer 2, the graders judged that there was enough ambiguity to justifyaccepting 2 as a correct answer. So both 2 and 2 · 15!2 were accepted as correct answers.)

2. Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠BAC intersect line BC atD. If there exist points E and F on sides AC and BC, respectively, such that lines AD and EFare parallel and divide triangle ABC into three parts of equal area, determine the number of possibleintegral values for BC.

Answer: 13

A B

C

DE

F

7

14

Note that such E,F exist if and only if[ADC]

[ADB]= 2. (1)

([ ] denotes area.) Since AD is the angle bisector, and the ratio of areas of triangles with equal heightis the ratio of their bases,

AC

AB=DC

DB=

[ADC]

[ADB].

Hence (1) is equivalent to AC = 2AB = 14. Then BC can be any length d such that the triangleinequalities are satisfied:

d+ 7 > 14

7 + 14 > d

Hence 7 < d < 21 and there are 13 possible integral values for BC.

Combinatorics & Geometry Individual Test

3. Let a, b, and c be positive real numbers. Determine the largest total number of real roots that thefollowing three polynomials may have among them: ax2 + bx+ c, bx2 + cx+ a, and cx2 + ax+ b.

Answer: 4 If all the polynomials had real roots, their discriminants would all be nonnegative: a2 ≥4bc, b2 ≥ 4ca, and c2 ≥ 4ab. Multiplying these inequalities gives (abc)2 ≥ 64(abc)2, a contradiction.Hence one of the quadratics has no real roots. The maximum of 4 real roots is attainable: for example,the values (a, b, c) = (1, 5, 6) give −2,−3 as roots to x2 + 5x+ 6 and −1,− 1

5 as roots to 5x2 + 6x+ 1.

4. Josh takes a walk on a rectangular grid of n rows and 3 columns, starting from the bottom left corner.At each step, he can either move one square to the right or simultaneously move one square to the leftand one square up. In how many ways can he reach the center square of the topmost row?

Answer: 2n−1 Note that Josh must pass through the center square of each row. There are 2 waysto get from the center square of row k to the center square of row k+ 1. So there are 2n−1 ways to getto the center square of row n.

5. Let H be a regular hexagon of side length x. Call a hexagon in the same plane a “distortion” of H ifand only if it can be obtained from H by translating each vertex of H by a distance strictly less than1. Determine the smallest value of x for which every distortion of H is necessarily convex.

Answer: 4

A1

A2

A3 A4

A5

A6X X′

Y Y ′

Let H = A1A2A3A4A5A6 be the hexagon, and for all 1 ≤ i ≤ 6, let points A′i be considered such thatAiA

′i < 1. Let H ′ = A′1A

′2A′3A′4A′5A′6, and consider all indices modulo 6. For any point P in the plane,

let D(P ) denote the unit disk {Q|PQ < 1} centered at P ; it follows that A′i ∈ D(Ai).

Let X and X ′ be points on line A1A6, and let Y and Y ′ be points on line A3A4 such that A1X =A1X

′ = A3Y = A3Y′ = 1 and X and X ′ lie on opposite sides of A1 and Y and Y ′ lie on opposite sides

of A3. If X ′ and Y ′ lie on segments A1A6 and A3A4, respectively, then segment A′1A′3 lies between the

lines XY and X ′Y ′. Note that x2 is the distance from A2 to A1A3.

Combinatorics & Geometry Individual Test

A1

A2

A3 A4

A5

A6X X′

Y Y ′

If x2 ≥ 2, then C(A2) cannot intersect line XY , since the distance from XY to A1A3 is 1 and the

distance from XY to A2 is at least 1. Therefore, A′1A′3 separates A′2 from the other 3 vertices of the

hexagon. By analogous reasoning applied to the other vertices, we may conclude that H ′ is convex.

If x2 < 2, then C(A2) intersects XY , so by choosing A′1 = X and A′3 = Y , we see that we may choose

A′2 on the opposite side of XY , in which case H ′ will be concave. Hence the answer is 4, as desired.

6. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. IfNathaniel goes first, determine the probability that he ends up winning.

Answer: 511 For 1 ≤ k ≤ 6, let xk be the probability that the current player, say A, will win when

the number on the tally at the beginning of his turn is k modulo 7. The probability that the totalis l modulo 7 after his roll is 1

6 for each l 6≡ k (mod 7); in particular, there is a 16 chance he wins

immediately. The chance that A will win if he leaves l on the board after his turn is 1− xl. Hence for1 ≤ k ≤ 6,

xk =1

6

∑1≤l≤6, l 6=k

(1− xl) +1

6.

Letting s =∑6l=1 xl, this becomes xk = xk−s

6 + 1 or 5xk

6 = − s6 + 1. Hence x1 = · · · = x6, and 6xk = s

for every k. Plugging this in gives 11xk

6 = 1, or xk = 611 .

Since Nathaniel cannot win on his first turn, he leaves Obediah with a number not divisible by 7.Hence Obediah’s chance of winning is 6

11 and Nathaniel’s chance of winning is 511 .

7. Let ABCDEF be a regular hexagon of area 1. Let M be the midpoint of DE. Let X be theintersection of AC and BM , let Y be the intersection of BF and AM , and let Z be the intersectionof AC and BF . If [P ] denotes the area of polygon P for any polygon P in the plane, evaluate[BXC] + [AY F ] + [ABZ]− [MXZY ].

Combinatorics & Geometry Individual Test

Answer: 0

A B

C

DE

F

M

XY

Z

O

Let O be the center of the hexagon. The desired area is [ABCDEF ] − [ACDM ] − [BFEM ]. Notethat [ADM ] = [ADE]/2 = [ODE] = [ABC], where the last equation holds because sin 60◦ = sin 120◦.Thus, [ACDM ] = [ACD] + [ADM ] = [ACD] + [ABC] = [ABCD], but the area of ABCD is half thearea of the hexagon. Similarly, the area of [BFEM ] is half the area of the hexagon, so the answer iszero.

8. The integers from 1 to n are written in increasing order from left to right on a blackboard. Davidand Goliath play the following game: starting with David, the two players alternate erasing any twoconsecutive numbers and replacing them with their sum or product. Play continues until only onenumber on the board remains. If it is odd, David wins, but if it is even, Goliath wins. Find the 2011thsmallest positive integer greater than 1 for which David can guarantee victory.

Answer: 4022 If n is odd and greater than 1, then Goliath makes the last move. No matter whattwo numbers are on the board, Goliath can combine them to make an even number. Hence Goliathhas a winning strategy in this case.

Now suppose n is even. We can replace all numbers on the board by their residues modulo 2. Initiallythe board reads 1, 0, 1, 0, . . . , 1, 0. David combines the rightmost 1 and 0 by addition to make 1, so nowthe board reads 1, 0, 1, 0, . . . , 0, 1. We call a board of this form a “good” board. When it is Goliath’sturn, and there is a good board, no matter where he moves, David can make a move to restore a goodboard. Indeed, Goliath must combine a neighboring 0 and 1; David can then combine that numberwith a neighboring 1 to make 1 and create a good board with two fewer numbers.

David can ensure a good board after his last turn. But a good board with one number is simply 1,so David wins. So David has a winning strategy if n is even. Therefore, the 2011th smallest positiveinteger greater than 1 for which David can guarantee victory is the 2011th even positive integer, whichis 4022.

9. Let ABCD be a square of side length 13. Let E and F be points on rays AB and AD, respectively,so that the area of square ABCD equals the area of triangle AEF . If EF intersects BC at X andBX = 6, determine DF .

Answer:√

13

Combinatorics & Geometry Individual Test

A B

CD

E

F

X

Y

6

7

x

y

First Solution

Let Y be the point of intersection of lines EF and CD. Note that [ABCD] = [AEF ] implies that[BEX] + [DY F ] = [CY X]. Since 4BEX ∼ 4CY X ∼ 4DY F , there exists some constant r suchthat [BEX] = r · BX2, [Y DF ] = r · CX2, and [CY X] = r · DF 2. Hence BX2 + DF 2 = CX2, soDF =

√CX2 −BX2 =

√49− 36 =

√13.

Second Solution

Let x = DF and y = Y D. Since 4BXE ∼ 4CXY ∼ 4DFY , we have

BE

BX=CY

CX=DY

DF=y

x.

Using BX = 6, XC = 7 and CY = 13− y we get BE = 6yx and 13−y

7 = yx . Solving this last equation

for y gives y = 13xx+7 . Now [ABCD] = [AEF ] gives

169 =1

2AE ·AF =

1

2

(13 +

6y

x

)(13 + x) .

169 = 6y + 13x+78y

x

13 =6x

x+ 7+ x+

78

x+ 7

0 = x2 − 13.

Thus x =√

13.

10. Mike and Harry play a game on an 8× 8 board. For some positive integer k, Mike chooses k squaresand writes an M in each of them. Harry then chooses k + 1 squares and writes an H in each of them.After Harry is done, Mike wins if there is a sequence of letters forming “HMM” or “MMH,” whenread either horizontally or vertically, and Harry wins otherwise. Determine the smallest value of k forwhich Mike has a winning strategy.

Answer: 16 Suppose Mike writes k M ’s. Let a be the number of squares which, if Harry writesan H in, will yield either HMM or MMH horizontally, and let b be the number of squares which,if Harry writes an H in, will yield either HMM or MMH vertically. We will show that a ≤ k andb ≤ k. Then, it will follow that there are at most a+ b ≤ 2k squares which Harry cannot write an Hin. There will be at least 64− k − 2k = 64− 3k squares which Harry can write in. If 64− 3k ≥ k + 1,or k ≤ 15, then Harry wins.

We will show that a ≤ k (that b ≤ k will follow by symmetry). Suppose there are ai M ’s in row i. Ineach group of 2 or more consective M ’s, Harry cannot write H to the left or right of that group, givingat most 2 forbidden squares. Hence ai is at most the number of M ’s in row i. Summing over the rowsgives the desired result.

Combinatorics & Geometry Individual Test

Mike can win by writing 16 M ’s according to the following diagram:

· · · · · · · ·· M M · · M M ·· M M · · M M ·· · · · · · · ·· · · · · · · ·· M M · · M M ·· M M · · M M ·· · · · · · · ·

.

11. Let ABCDEF be a convex equilateral hexagon such that lines BC, AD, and EF are parallel. LetH be the orthocenter of triangle ABD. If the smallest interior angle of the hexagon is 4 degrees,determine the smallest angle of the triangle HAD in degrees.

Answer: 3

A

B

C

D

E

FH

A′

B′

D′

Note that ABCD and DEFA are isosceles trapezoids, so ∠BAD = ∠CDA and ∠FAD = ∠EDA. Inorder for the hexagon to be convex, the angles at B, C, E, and F have to be obtuse, so ∠A = ∠D = 4◦.Letting s be a side length of the hexagon, AD = AB cos∠BAD + BC + CD cos∠CDA = s(1 +2 cos∠BAD), so ∠BAD is uniquely determined by AD. Since the same equation holds for trapezoidDEFA, it follows that ∠BAD = ∠FAD = ∠CDA = ∠EDA = 2◦. Then ∠BCD = 180◦ − 2◦ = 178◦.Since 4BCD is isosceles, ∠CDB = 1◦ and ∠BDA = 1◦. (One may also note that ∠BDA = 1◦ byobserving that equal lengths AB and BC must intercept equal arcs on the circumcircle of isoscelestrapezoid ABCD).

Let A′, B′, andD′ be the feet of the perpendiculars from A, B, andD toBD, DA, and AB, respectively.Angle chasing yields

∠AHD = ∠AHB′ + ∠DHB′ = (90◦ − ∠A′AB′) + (90◦ − ∠D′DB′)

= ∠BDA+ ∠BAD = 1◦ + 2◦ = 3◦

∠HAD = 90◦ − ∠AHB′ = 89◦

∠HDA = 90◦ − ∠DHB′ = 88◦

Hence the smallest angle in 4HAD is 3◦.

Combinatorics & Geometry Individual Test

A

B

C

D

E

FH

It is faster, however, to draw the circumcircle of DEFA, and to note that since H is the orthocenterof triangle ABD, B is the orthocenter of triangle HAD. Then since F is the reflection of B acrossAD, quadrilateral HAFD is cyclic, so ∠AHD = ∠ADF + ∠DAF = 1◦ + 2◦ = 3◦, as desired.

12. The ordered pairs (2011, 2), (2010, 3), (2009, 4), . . ., (1008, 1005), (1007, 1006) are written from left toright on a blackboard. Every minute, Elizabeth selects a pair of adjacent pairs (xi, yi) and (xj , yj), with

(xi, yi) left of (xj , yj), erases them, and writes

(xiyixjyj

,xiyiyjxj

)in their place. Elizabeth continues

this process until only one ordered pair remains. How many possible ordered pairs (x, y) could appearon the blackboard after the process has come to a conclusion?

Answer: 504510 First, note that none of the numbers will ever be 0. Let ? denote the replacementoperation. For each pair on the board (xi, yi) define its primary form to be (xi, yi) and its secondaryform to be [xiyi,

xi

yi]. Note that the primary form determines the secondary form uniquely and vice

versa. In secondary form,

[a1, b1] ? [a2, b2] =

(√a1b1,

√a1b1

)?

(√a2b2,

√a2b2

)=

(a1b2,

a1b2

)= [a21, b

22].

Thus we may replace all pairs on the board by their secondary form and use the above rule for ?instead. From the above rule, we see that if the leftmost number on the board is x, then after oneminute it will be x or x2 depending on whether it was erased in the intervening step, and similarly forthe rightmost number. Let k be the number of times the leftmost pair is erased and n be the numberof times the rightmost pair is erased. Then the final pair is[

40222k

,

(1007

1006

)2n]. (2)

Any step except the last cannot involve both the leftmost and rightmost pair, so k + n ≤ 1005. Sinceevery pair must be erased at least once, k, n ≥ 1. Every pair of integers satisfying the above can occur,for example, by making 1005− k − n moves involving only the pairs in the middle, then making k − 1moves involving the leftmost pair, and finally n moves involving the rightmost pair.

In light of (2), the answer is the number of possible pairs (k, n), which is

1004∑k=1

1005−k∑n=1

1 =

1004∑k=1

1005− k =

1004∑k=1

k =1004 · 1005

2= 504510.

Combinatorics & Geometry Individual Test

13. Let ABCD be a cyclic quadrilateral, and suppose that BC = CD = 2. Let I be the incenter of triangleABD. If AI = 2 as well, find the minimum value of the length of diagonal BD.

Answer: 2√

3 Let T be the point where the incircle intersects AD, and let r be the inradius and Rbe the circumradius of 4ABD. Since BC = CD = 2, C is on the midpoint of arc BD on the oppositeside of BD as A, and hence on the angle bisector of A. Thus A, I, and C are collinear. We have thefollowing formulas:

AI =IM

sin∠IAM=

r

sin A2

BC = 2R sinA

2BD = 2R sinA

The last two equations follow from the extended law of sines on 4ABC and 4ABD, respectively.

Using AI = 2 = BC gives sin2 A2 = r

2R . However, it is well-known that R ≥ 2r with equality for an

equilateral triangle (one way to see this is the identity 1+ rR = cosA+cosB+cosD). Hence sin2 A

2 ≤14

and A2 ≤ 30◦. Then

BD = 2R

(2 sin

A

2cos

A

2

)= BC · 2 cos

A

2≥ 2

(2 ·√

3

2

)= 2√

3

with equality when 4ABD is equilateral.

A

B

C

D

I

Remark: Similar but perhaps simpler computations can be made by noting that if AC intersects BDat X, then AB/BX = AD/DX = 2, which follows from the exterior angle bisector theorem; if IA isthe A-excenter of triangle ABC, then AIA/XIA = 2 since it is well-known that C is the circumcenterof cyclic quadrilateral BIDIA.

14. Let A = {1, 2, . . . , 2011}. Find the number of functions f from A to A that satisfy f(n) ≤ n for all nin A and attain exactly 2010 distinct values.

Answer: 22011 − 2012 Let n be the element of A not in the range of f . Let m be the element of Athat is hit twice.

We now sum the total number of functions over n,m. Clearly f(1) = 1, and by induction, for x ≤m, f(x) = x. Also unless n = 2011, f(2011) = 2011 because f can take no other number to 2011. It

Combinatorics & Geometry Individual Test

follows from backwards induction that for x > n, f(x) = x. Therefore n > m, and there are only n−mvalues of f that are not fixed.

Now f(m+ 1) = m or f(m+ 1) = m+ 1. For m < k < n,given the selection of f(1), f(2), . . . , f(k− 1),k− 1 of the k+ 1 possible values of f(k+ 1) (1, 2, 3, . . . , k, and counting m twice) have been taken, sothere are two distinct values that f(k+ 1) can take (one of them is k+ 1, and the other is not, so theyare distinct). For f(n), when the other 2010 values of f have been assigned, there is only one missing,so f(n) is determined.

For each integer in [m,n), there are two possible values of f , so there are 2n−m−1 different functionsf for a given m,n. So our answer is

2010∑m=1

2011∑n=m+1

2n−m−1 =

2010∑m=1

2−m−12011∑

n=m+1

2n

=

2010∑m=1

2−m−1(22012 − 2m+1)

=

2010∑m=1

22011−m − 1

=

(2010∑m=1

2m

)− 2010

= 22011 − 2012

15. Let f(x) = x2 − r2x + r3 for all real numbers x, where r2 and r3 are some real numbers. Define asequence {gn} for all nonnegative integers n by g0 = 0 and gn+1 = f(gn). Assume that {gn} satisfiesthe following three conditions: (i) g2i < g2i+1 and g2i+1 > g2i+2 for all 0 ≤ i ≤ 2011; (ii) there existsa positive integer j such that gi+1 > gi for all i > j, and (iii) {gn} is unbounded. If A is the greatestnumber such that A ≤ |r2| for any function f satisfying these properties, find A.

Answer: 2 Consider the function f(x) − x. By the constraints of the problem, f(x) − x must benegative for some x, namely, for x = g2i+1, 0 ≤ i ≤ 2011. Since f(x) − x is positive for x of largeabsolute value, the graph of f(x) − x crosses the x-axis twice and f(x) − x has two real roots, saya < b. Factoring gives f(x)− x = (x− a)(x− b), or f(x) = (x− a)(x− b) + x.

Now, for x < a, f(x) > x > a, while for x > b, f(x) > x > b. Let c 6= b be the number such thatf(c) = f(b) = b. Note that b is not the vertex as f(a) = a < b, so by the symmetry of quadratics, cexists and b+c

2 = r22 as the vertex of the parabola. By the same token, b+a

2 = r2+12 is the vertex of

f(x)− x. Hence c = a− 1. If f(x) > b then x < c or x > b. Consider the smallest j such that gj > b.Then by the above observation, gj−1 < c. (If gi ≥ b then f(gi) ≥ gi ≥ b so by induction, gi+1 ≥ gi forall i ≥ j. Hence j > 1; in fact j ≥ 4025.) Since gj−1 = f(gj−2), the minimum value of f is less than c.The minimum value is the value of f evaluated at its vertex, b+a−12 , so

f

(b+ a− 1

2

)< c(

b+ a− 1

2− a)(

b+ a− 1

2− b)

+b+ a− 1

2< a− 1

1− (b− a)2

4+b− a+ 1

2< 0

3

4<

(b− a)2

4− b− a

2

4 < (b− a− 1)2.

Then either b − a − 1 < −2 or b − a − 1 > 2, but b > a, so the latter must hold and (b − a)2 > 9.Now, the discriminant of f(x) − x equals (b − a)2 (the square of the difference of the two roots) and(r2 + 1)2 − 4r3 (from the coefficients), so (r2 + 1)2 > 9 + 4r3. But r3 = g1 > g0 = 0 so |r2| > 2.

Combinatorics & Geometry Individual Test

We claim that we can make |r2| arbitrarily close to 2, so that the answer is 2. First define Gi, i ≥ 0as follows. Let N ≥ 2012 be an integer. For ε > 0 let h(x) = x2 − 2 − ε, gε(x) = −

√x+ 2 + ε and

G2N+1 = 2 + ε, and define Gi recursively by Gi = gε(Gi+1), Gi+1 = h(Gi). (These two equationsare consistent.) Note the following. (i) G2i < G2i+1 and G2i+1 > G2i+2 for 0 ≤ i ≤ N − 1. Firstnote G2N = −

√4 + 2ε > −

√4 + 2ε+ ε2 = −2 − ε. Let l be the negative solution to h(x) = x. Note

that −2 − ε < G2N < l < 0 since h(G2N ) > 0 > G2N . Now gε(x) is defined as long as x ≥ −2 − ε,and it sends (−2 − ε, l) into (l, 0) and (l, 0) into (−2 − ε, l). It follows that the Gi, 0 ≤ i ≤ 2Nare well-defined; moreover, G2i < l and G2i+1 > l for 0 ≤ i ≤ N − 1 by backwards induction on i,so the desired inequalities follow. (ii) Gi is increasing for i ≥ 2N + 1. Indeed, if x ≥ 2 + ε, thenx2 − x = x(x− 1) > 2 + ε so h(x) > x. Hence 2 + ε = G2N+1 < G2N+2 < · · · . (iii) Gi is unbounded.This follows since h(x) − x = x(x − 2) − 2 − ε is increasing for x > 2 + ε, so Gi increases faster andfaster for i ≥ 2N + 1. Now define f(x) = h(x + G0) − G0 = x2 + 2G0x + G2

0 − G0 − 2 − ε. NoteGi+1 = h(Gi) while gi+1 = f(gi) = h(gi + G0) − G0, so by induction gi = Gi − G0. Since {Gi}∞i=0

satisfies (i), (ii), and (iii), so does gi.

We claim that we can make G0 arbitrarily close to −1 by choosing N large enough and ε small enough;this will make r2 = −2G0 arbitrarily close to 2. Choosing N large corresponds to taking G0 to be alarger iterate of 2 + ε under gε(x). By continuity of this function with respect to x and ε, it suffices totake ε = 0 and show that (letting g = g0)

g(n)(2) = g(· · · g︸ ︷︷ ︸n

(2) · · · )→ −1 as n→∞.

But note that for 0 ≤ θ ≤ π2 ,

g(−2 cos θ) = −√

2− 2 cos θ = −2 sin

(θ

2

)= 2 cos

(π

2− θ

2

).

Hence by induction, g(n)(−2 cos θ) = −2 cosπ2 −

π4 + · · ·+ (−1)n

θ − π

2n

)). Hence g(n)(2) = g(n−1)(−2 cos 0)

converges to −2 cos(π2 −π4 + · · · ) = −2 cos(π3 ) = −1, as needed.

16. Let ABCD be a quadrilateral inscribed in the unit circle such that ∠BAD is 30 degrees. Let m denotethe minimum value of CP + PQ + CQ, where P and Q may be any points lying along rays AB andAD, respectively. Determine the maximum value of m.

Answer: 2

A

BC

D

E

F

G

H

P

Q

R

For a fixed quadrilateral ABCD as described, we first show that m, the minimum possible length ofCP + PQ + QC, equals the length of AC. Reflect B, C, and P across line AD to points E, F , andR, respectively, and then reflect D and F across AE to points G and H, respectively. These two

Combinatorics & Geometry Individual Test

reflections combine to give a 60◦ rotation around A, so triangle ACH is equilateral. It also followsthat RH is a 60◦ rotation of PC around A, so, in particular, these segments have the same length.Because QR = QP by reflection,

CP + PQ+QC = CQ+QR+RH.

The latter is the length of a broken path CQRH from C to H, and by the “shortest path is a straightline” principle, this total length is at least as long as CH = CA. (More directly, this follows from thetriangle inequality: (CQ+QR)+RH ≥ CR+RH ≥ CH). Therefore, the lower bound m ≥ AC indeedholds. To see that this is actually an equality, note that choosing Q as the intersection of segment CHwith ray AD, and choosing P so that its reflection R is the intersection of CH with ray AE, alignspath CQRH with segment CH, thus obtaining the desired minimum m = AC.

We may conclude that the largest possible value of m is the largest possible length of AC, namely 2:the length of a diameter of the circle.

17. Let n be an odd positive integer, and suppose that n people sit on a committee that is in the processof electing a president. The members sit in a circle, and every member votes for the person either tohis/her immediate left, or to his/her immediate right. If one member wins more votes than all theother members do, he/she will be declared to be the president; otherwise, one of the the members whowon at least as many votes as all the other members did will be randomly selected to be the president.If Hermia and Lysander are two members of the committee, with Hermia sitting to Lysander’s leftand Lysander planning to vote for Hermia, determine the probability that Hermia is elected president,assuming that the other n− 1 members vote randomly.

Answer: 2n−1n2n−1 Let x be the probability Hermia is elected if Lysander votes for her, and let y be the

probability that she wins if Lysander does not vote for her. We are trying to find x, and do so by firstfinding y. If Lysander votes for Hermia with probability 1

2 then the probability that Hermia is electedchairman is x

2 + y2 , but it is also 1

n by symmetry. If Lysander does not vote for Hermia, Hermia canget at most 1 vote, and then can only be elected if everyone gets one vote and she wins the tiebreaker.The probability she wins the tiebreaker is 1

n , and chasing around the circle, the probability that everyperson gets 1 vote is 1

2n−1 . (Everyone votes for the person to the left, or everyone votes for the personto the right.) Hence

y =1

n2n−1.

Then x2 + 1

n2n = 1n , so solving for x gives

x =2n − 1

n2n−1.

18. Collinear points A, B, and C are given in the Cartesian plane such that A = (a, 0) lies along thex-axis, B lies along the line y = x, C lies along the line y = 2x, and AB/BC = 2. If D = (a, a), thecircumcircle of triangle ADC intersects y = x again at E, and ray AE intersects y = 2x at F , evaluateAE/EF .

Answer: 7

Combinatorics & Geometry Individual Test

A

B

CD

E

F

O

PQ

Let points O, P , and Q be located at (0, 0), (a, 2a), and (0, 2a), respectively. Note that BC/AB = 1/2implies [OCD]/[OAD] = 1/2, so since [OPD] = [OAD], [OCD]/[OPD] = 1/2. It follows that[OCD] = [OPD]. Hence OC = CP . We may conclude that triangles OCQ and PCA are congruent,so C = (a/2, a).It follows that ∠ADC is right, so the circumcircle of triangle ADC is the midpoint of AC, whichis located at (3a/4, a/2). Let (3a/4, a/2) = H, and let E = (b, b). Then the power of the pointO with respect to the circumcircle of ADC is OD · OE = 2ab, but it may also be computed asOH2 −HA2 = 13a/16− 5a/16 = a/2. It follows that b = a/4, so E = (a/4, a/4).We may conclude that line AE is x + 3y = a, which intersects y = 2x at an x-coordinate of a/7.Therefore, AE/EF = (a− a/4)/(a/4− a/7) = (3a/4)/(3a/28) = 7.

Remark: The problem may be solved more quickly if one notes from the beginning that lines OA,OD, OP , and OQ form a harmonic pencil because D is the midpoint of AP and lines OQ and AP areparallel.

19. Alice and Bob play a game in which two thousand and eleven 2011×2011 grids are distributed betweenthe two of them, 1 to Bob, and the other 2010 to Alice. They go behind closed doors and fill theirgrid(s) with the numbers 1, 2, . . . , 20112 so that the numbers across rows (left-to-right) and downcolumns (top-to-bottom) are strictly increasing. No two of Alice’s grids may be filled identically. Afterthe grids are filled, Bob is allowed to look at Alice’s grids and then swap numbers on his own grid,two at a time, as long as the numbering remains legal (i.e. increasing across rows and down columns)after each swap. When he is done swapping, a grid of Alice’s is selected at random. If there exist twointegers in the same column of this grid that occur in the same row of Bob’s grid, Bob wins. Otherwise,Alice wins. If Bob selects his initial grid optimally, what is the maximum number of swaps that Bobmay need in order to guarantee victory?

Answer: 1 Consider the grid whose entries in the jth row are, in order, 2011j − 2010, 2011j −2009, . . . , 2011j. Call this grid A0. For k = 1, 2 . . . , 2010, let grid Ak be the grid obtained from A0

by swapping the rightmost entry of the kth row with the leftmost entry of the k + 1st row. We claim

Combinatorics & Geometry Individual Test

that if A ∈ {A0, A1, . . . , A2010}, then given any legally numbered grid B such that A and B differ inat least one entry, there exist two integers in the same column of B that occur in the same row of A.We first consider A0. Assume for the sake of contradiction B is a legally numbered grid distinct fromA0, such that there do not exist two integers in the same column of B that occur in the same rowof A0. Since the numbers 1, 2, . . . , 2011 occur in the same row of A0, they must all occur in differentcolumns of B. Clearly 1 is the leftmost entry in B’s first row. Let m be the smallest number that doesnot occur in the first row of B. Since each row is in order, m must be the first entry in its row. Butthen 1 and m are in the same column of B, a contradiction. It follows that the numbers 1, 2, . . . , 2011all occur in the first row of B. Proceeding by induction, 2011j−2010, 2011j−2009, . . . , 2011j must alloccur in the jth row of B for all 1 ≤ j ≤ 2011. Since A0 is the only legally numbered grid satsifyingthis condition, we have reached the desired contradiction.Now note that if A ∈ {A1, . . . , A2010}, there exist two integers in the same column of A0 that occur inthe same row of A. In particular, if A = Ak and 1 ≤ k ≤ 2010, then the integers 2011k − 2010 and2011k + 1 occur in the same column of A0 and in the same row of Ak. Therefore, it suffices to showthat for all 1 ≤ k ≤ 2010, there is no legally numbered grid B distinct from Ak and A0 such that theredo not exist two integers in the same column of B that occur in the same row of A0. Assume for thesake of contradiction that there does exist such a grid B. By the same logic as above, applied to thefirst k−1 rows and applied backwards to the last 2010−k−1 rows, we see that B may only differ fromAk in the kth and k+ 1st rows. However, there are only two legally numbered grids that are identicalto Ak outside of rows k and k + 1, namely A0 and Ak. This proves the claim.It remains only to note that, by the pigeonhole principle, if one of Alice’s grids is A0, then there existsa positive integer k, 1 ≤ k ≤ 2010, such that Ak is not one of the Alice’s grids. Therefore, if Bob setshis initial grid to be A0, he will require only one swap to switch his grid to Ak after examining Alice’sgrids. If A0 is not among Alice’s grids, then if Bob sets his initial grid to be A0, he will not in factrequire any swaps at all.

20. Let ω1 and ω2 be two circles that intersect at points A and B. Let line l be tangent to ω1 at P and toω2 at Q so that A is closer to PQ than B. Let points R and S lie along rays PA and QA, respectively,so that PQ = AR = AS and R and S are on opposite sides of A as P and Q. Let O be the circumcenterof triangle ASR, and let C and D be the midpoints of major arcs AP and AQ, respectively. If ∠APQis 45 degrees and ∠AQP is 30 degrees, determine ∠COD in degrees.

Answer: 142.5

A

B

P

Q

R

SC

D

T

T ′

We use directed angles throughout the solution.

Let T denote the point such that ∠TCD = 1/2∠APQ and ∠TDC = 1/2∠AQP . We claim that T isthe circumcenter of triangle SAR.

Combinatorics & Geometry Individual Test

Since CP = CA, QP = RA, and ∠CPQ = ∠CPA + ∠APQ = ∠CPA + ∠ACP = ∠CAR, we have4CPQ ∼= 4CAR. By spiral similarity, we have 4CPA ∼ 4CQR.

Let T ′ denote the reflection of T across CD. Since ∠TCT ′ = ∠APQ = ∠ACP , we have 4TCT ′ ∼4ACP ∼ 4RCQ. Again, by spiral similarity centered at C, we have 4CTR ∼ 4CT ′Q. ButCT = CT ′, so 4CTR ∼= 4CT ′Q and TR = T ′Q. Similarly, 4DTT ′ ∼ 4DAQ, and spiral similaritycentered at D shows that 4DTA ∼= 4DT ′Q. Thus TA = T ′Q = TR.

We similarly have TA = T ′P = TS, so T is indeed the circumcenter. Therefore, we have ∠COD =∠CTD = 180◦ − 45◦

2 −30◦

2 = 142.5◦.

Combinatorics & Geometry Individual Test

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

1. [4] Let ABC be a triangle with area 1. Let points D and E lie on AB and AC, respectively, suchthat DE is parallel to BC and DE/BC = 1/3. If F is the reflection of A across DE, find the area oftriangle FBC.

2. [4] Let a ? b = sin a cos b for all real numbers a and b. If x and y are real numbers such thatx ? y 〉 y ? x = 1, what is the maximum value of x ? y + y ? x?

3. [4] Evaluate 2011× 20122012× 201320132013〉 2013× 20112011× 201220122012.

4. [4] Let p be the answer to this question. If a point is chosen uniformly at random from the squarebounded by x = 0, x = 1, y = 0, and y = 1, what is the probability that at least one of its coordinatesis greater than p?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14th HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND

5. [5] Rachelle picks a positive integer a and writes it next to itself to obtain a new positive integer b.For instance, if a = 17, then b = 1717. To her surprise, she finds that b is a multiple of a2. Find theproduct of all the possible values of b

a2 .

6. [5] Square ABCD is inscribed in circle ω with radius 10. Four additional squares are drawn inside ωbut outside ABCD such that the lengths of their diagonals are as large as possible. A sixth square isdrawn by connecting the centers of the four aforementioned small squares. Find the area of the sixthsquare.

7. [6] For any positive real numbers a and b, define a ◦ b = a+ b+ 2√ab. Find all positive real numbers

x such that x2 ◦ 9x = 121.

8. [6] Find the smallest k such that for any arrangement of 3000 checkers in a 2011× 2011 checkerboard,with at most one checker in each square, there exist k rows and k columns for which every checker iscontained in at least one of these rows or columns.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14th HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND

9. [6] Segments AA′, BB′, and CC ′, each of length 2, all intersect at a point O. If ∠AOC ′ = ∠BOA′ =∠COB′ = 60◦, find the maximum possible value of the sum of the areas of triangles AOC ′, BOA′,and COB′.

10. [6] In how many ways can one fill a 4 × 4 grid with a 0 or 1 in each square such that the sum of theentries in each row, column, and long diagonal is even?

11. [8] Rosencrantz and Guildenstern play a game in which they repeatedly flip a fair coin. Let a1 = 4,a2 = 3, and an = an−1 + an−2 for all n ≥ 3. On the nth flip, if the coin is heads, Rosencrantz paysGuildenstern an dollars, and, if the coin is tails, Guildenstern pays Rosencrantz an dollars. If playcontinues for 2010 turns, what is the probability that Rosencrantz ends up with more money than hestarted with?

12. [8] A sequence of integers {ai} is defined as follows: ai = i for all 1 ≤ i ≤ 5, and ai = a1a2 · · · ai−1 〉 1

for all i > 5. Evaluate a1a2 · · · a2011 〉2011∑i=1

a2i .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14th HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND

13. [8] Let a,b, and c be the side lengths of a triangle, and assume that a ≤ b and a ≤ c. Let x = b+c−a2 .

If r and R denote the inradius and circumradius, respectively, find the minimum value of axrR .

14. [8] Danny has a set of 15 pool balls, numbered 1, 2, . . . , 15. In how many ways can he put the balls in8 indistinguishable bins such that the sum of the numbers of the balls in each bin is 14, 15, or 16?

15. [10] Find all irrational numbers x such that x3 〉 17x and x2 + 4x are both rational numbers.

16. [10] Let R be a semicircle with diameter XY . A trapezoid ABCD in which AB is parallel to CD iscircumscribed about R such that AB contains XY . If AD = 4, CD = 5, and BC = 6, determine AB.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14th HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND

17. [10] Given positive real numbers x, y, and z that satisfy the following system of equations:

x2 + y2 + xy = 1,

y2 + z2 + yz = 4,

z2 + x2 + zx = 5,

find x+ y + z.

18. [10] In how many ways can each square of a 4× 2011 grid be colored red, blue, or yellow such that notwo squares that are diagonally adjacent are the same color?

19. [12] Find the least positive integer N with the following property: If all lattice points in [1, 3]× [1, 7]×[1, N ] are colored either black or white, then there exists a rectangular prism, whose faces are parallelto the xy, xz, and yz planes, and whose eight vertices are all colored in the same color.

20. [12] Let ABCD be a quadrilateral circumscribed about a circle with center O. Let O1, O2, O3, andO4 denote the circumcenters of 4AOB, 4BOC, 4COD, and 4DOA. If ∠A = 120◦,∠B = 80◦, and∠C = 45◦, what is the acute angle formed by the two lines passing through O1O3 and O2O4?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14th HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND

21. [12] Let ABCD be a quadrilateral inscribed in a circle with center O. Let P denote the intersectionof AC and BD. Let M and N denote the midpoints of AD and BC. If AP = 1, BP = 3, DP =

√3,

and AC is perpendicular to BD, find the area of triangle MON .

22. [12] Find the number of ordered triples (a, b, c) of pairwise distinct integers such that 〉31 ≤ a, b, c ≤ 31and a+ b+ c > 0.

23. [14] Let S be the set of points (x, y, z) in R3 such that x, y, and z are positive integers less than orequal to 100. Let f be a bijective map between S and the {1, 2, . . . , 1000000} that satisfies the followingproperty: if x1 ≤ x2, y1 ≤ y2, and z1 ≤ z2, then f(x1, y1, z1) ≤ f(x2, y2, z2). Define

Ai =

100∑j=1

100∑k=1

f(i, j, k),

Bi =

100∑j=1

100∑k=1

f(j, i, k),

and Ci =

100∑j=1

100∑k=1

f(j, k, i).

Determine the minimum value of Ai+1 〉Ai +Bj+1 〉Bj + Ck+1 〉 Ck.

24. [14] In how many ways may thirteen beads be placed on a circular necklace if each bead is either blueor yellow and no two yellow beads may be placed in adjacent positions? (Beads of the same color areconsidered to be identical, and two arrangements are considered to be the same if and only if each canbe obtained from the other by rotation).

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14th HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND

25. [14] Let n be an integer greater than 3. Let R be the set of lattice points (x, y) such that 0 ≤ x, y ≤ nand |x〉 y| ≤ 3. Let An be the number of paths from (0, 0) to (n, n) that consist only of steps of theform (x, y)→ (x, y+ 1) and (x, y)→ (x+ 1, y) and are contained entirely within R. Find the smallest

positive real number that is greater than An+1

Anfor all n.

26. [14] In how many ways can 13 bishops be placed on an 8×8 chessboard such that (i) a bishop is placedon the second square in the second row, (ii) at most one bishop is placed on each square, (iii) no bishopis placed on the same diagonal as another bishop, and (iv) every diagonal contains a bishop? (For thepurposes of this problem, consider all diagonals of the chessboard to be diagonals, not just the maindiagonals).

27. [16] Find the number of polynomials p(x) with integer coefficients satisfyingp(x) ≥ min{2x4 〉 6x2 + 1, 4〉 5x2} and p(x) ≤ max{2x4 〉 6x2 + 1, 4〉 5x2} for all x ∈ R.

28. [16] Let ABC be a triangle, and let points P and Q lie on BC such that P is closer to B than Q is.Suppose that the radii of the incircles of triangles ABP , APQ, and AQC are all equal to 1, and thatthe radii of the corresponding excircles opposite A are 3, 6, and 5, respectively. If the radius of theincircle of triangle ABC is 3

2 , find the radius of the excircle of triangle ABC opposite A.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14th HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND

29. [16] Let ABC be a triangle such that AB = AC = 182 and BC = 140. Let X1 lie on AC such thatCX1 = 130. Let the line through X1 perpendicular to BX1 at X1 meet AB at X2. Define X2, X3, . . .,as follows: for n odd and n ≥ 1, let Xn+1 be the intersection of AB with the perpendicular to Xn−1Xn

through Xn; for n even and n ≥ 2, let Xn+1 be the intersection of AC with the perpendicular toXn−1Xn through Xn. Find BX1 +X1X2 +X2X3 + . . ..

30. [16] How many ways are there to color the vertices of a 2n-gon with three colors such that no vertexhas the same color as its either of its two neighbors or the vertex directly across from it?

31. [18] Let A = {1, 2, 3, . . . , 9}. Find the number of bijective functions f : A → A for which there existsat least one i ∈ A such that

|f(i)〉 f−1(i)| > 1.

32. [18] Let p be a prime positive integer. Define a mod-p recurrence of degree n to be a sequence {ak}k≥0of numbers modulo p satisfying a relation of the form ai+n = cn−1ai+n−1 + ... + c1ai+1 + c0ai for alli ≥ 0, where c0, c1, . . . , cn−1 are integers and c0 6≡ 0 (mod p). Compute the number of distinct linearrecurrences of degree at most n in terms of p and n.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14th HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND

33. [25] Find the number of sequences consisting of 100 R’s and 2011 S’s that satisfy the property thatamong the first k letters, the number of S’s is strictly more than 20 times the number of R’s for all1 ≤ k ≤ 2111.

34. [25] Let w = w1, w2, . . . , w6 be a permutation of the integers {1, 2, . . . , 6}. If there do not exist indicesi < j < k such that wi < wj < wk or indices i < j < k < l such that wi > wj > wk > wl, then w issaid to be exquisite. Find the number of exquisite permutations.

35. [25] An independent set of a graph G is a set of vertices of G such that no two vertices among theseare connected by an edge. If G has 2000 vertices, and each vertex has degree 10, find the maximumpossible number of independent sets that G can have.

36. [25] An ordering of a set of n elements is a bijective map between the set and {1, 2, . . . , n}. Call anordering ρ of the 10 unordered pairs of distinct integers from the set {1, 2, 3, 4, 5} admissible if, for any1 ≤ a < b < c ≤ 5, either p({a, b}) < p({a, c}) < p({b, c}) or p({b, c}) < p({a, c}) < p({a, b}). Find thetotal number of admissible orderings.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

Guts Round

1. [4] Let ABC be a triangle with area 1. Let points D and E lie on AB and AC, respectively, suchthat DE is parallel to BC and DE/BC = 1/3. If F is the reflection of A across DE, find the area oftriangle FBC.

Answer: 13

Let AF intersect BC at H. Since DE/BC = 1/3 and F and A are equidistant from DE, we haveAF = 2

3AH and FH = AH 〉 AF = 13AH. Furthermore, since AF is perpendicular to DE, we have

AH and FH are the altitudes of triangles ABC and FBC respectively. Therefore the area of triangleFBC is 1

2 · FH ·BC = 12 · 13 ·AH ·BC = 1

3 .

2. [4] Let a ? b = sin a cos b for all real numbers a and b. If x and y are real numbers such thatx ? y 〉 y ? x = 1, what is the maximum value of x ? y + y ? x?

Answer: 1

We have x ? y + y ? x = sinx cos y + cosx sin y = sin(x+ y) ≤ 1. Equality is achieved when x = π2 and

y = 0. Indeed, for these values of x and y, we have x? y〉 y ?x = sinx cos y〉 cosx sin y = sin(x〉 y) =sin π

2 = 1.

3. [4] Evaluate 2011× 20122012× 201320132013〉 2013× 20112011× 201220122012.

Answer: 0

Both terms are equal to 2011× 2012× 2013× 1× 10001× 100010001.

4. [4] Let p be the answer to this question. If a point is chosen uniformly at random from the squarebounded by x = 0, x = 1, y = 0, and y = 1, what is the probability that at least one of its coordinatesis greater than p?

Answer:√5−12

The probability that a randomly chosen point has both coordinates less than p is p2, so the probabilitythat at least one of its coordinates is greater than p is 1 〉 p2. Since p is the answer to this question,

we have 1〉 p2 = p, and the only solution of p in the interval [0, 1] is√5−12 .

5. [5] Rachelle picks a positive integer a and writes it next to itself to obtain a new positive integer b.For instance, if a = 17, then b = 1717. To her surprise, she finds that b is a multiple of a2. Find theproduct of all the possible values of b

a2 .

Answer: 77

Suppose a has k digits. Then b = a(10k + 1). Thus a divides 10k + 1. Since a ≥ 10k−1, we have10k+1a ≤ 11. But since none of 2, 3, or 5 divide 10k + 1, the only possibilities are 7 and 11. These

values are obtained when a = 143 and a = 1, respectively.

6. [5] Square ABCD is inscribed in circle ω with radius 10. Four additional squares are drawn inside ωbut outside ABCD such that the lengths of their diagonals are as large as possible. A sixth square isdrawn by connecting the centers of the four aforementioned small squares. Find the area of the sixthsquare.

Answer: 144

Let DEGF denote the small square that shares a side with AB, where D and E lie on AB. Let Odenote the center of ω, K denote the midpoint of FG, and H denote the center of DEGF . The areaof the sixth square is 2 ·OH2.

Let KF = x. Since KF 2 + OK2 = OF 2, we have x2 + (2x + 5√

2)2 = 102. Solving for x, we getx =√

2. Thus, we have OH = 6√

2 and 2 ·OH2 = 144.

Guts Round

7. [6] For any positive real numbers a and b, define a ◦ b = a+ b+ 2√ab. Find all positive real numbers

x such that x2 ◦ 9x = 121.

Answer: 31−3√53

2

Since a ◦ b = (√a +√b)2, we have x2 ◦ 9x = (x + 3

√x)2. Moreover, since x is positive, we have

x+ 3√x = 11, and the only possible solution is that

√x = −3+

√53

2 , so x = 31−3√53

2 .

8. [6] Find the smallest k such that for any arrangement of 3000 checkers in a 2011× 2011 checkerboard,with at most one checker in each square, there exist k rows and k columns for which every checker iscontained in at least one of these rows or columns.

Answer: 1006

If there is a chip in every square along a main diagonal, then we need at least 1006 rows and columnsto contain all these chips. We are left to show that 1006 is sufficient.

Take the 1006 rows with greatest number of chips. Assume without loss of generality they are thefirst 1006 rows. If the remaining 1005 rows contain at most 1005 chips, then we can certainly choose1006 columns that contain these chips. Otherwise, there exists a row that contains at least 2 chips, soevery row in the first 1006 rows must contain at least 2 chips. But this means that there are at least2× 1006 + 1006 = 3018 chips in total. Contradiction.

9. [6] Segments AA′, BB′, and CC ′, each of length 2, all intersect at a point O. If ∠AOC ′ = ∠BOA′ =∠COB′ = 60◦, find the maximum possible value of the sum of the areas of triangles AOC ′, BOA′,and COB′.

Answer:√

3

Extend OA to D and OC ′ to E such that AD = OA′ and C ′E = OC. Since OD = OE = 2 and∠DOE = 60◦, we have ODE is an equilateral triangle. Let F be the point on DE such that DF = OBand EF = OB′. Clearly we have 4DFA ∼= 4OBA′ and 4EFC ′ ∼= OB′C. Thus the sum of the areasof triangles AOC ′, BOA′, and COB′ is the same as the sum of the areas of triangle DFA, FEC ′,and OAC ′, which is at most the area of triangle ODE. Since ODE is an equilateral triangle with sidelength 2, its area is

√3. Equality is achieved when OC = OA′ = 0.

10. [6] In how many ways can one fill a 4 × 4 grid with a 0 or 1 in each square such that the sum of theentries in each row, column, and long diagonal is even?

Answer: 256

First we name the elements of the square as follows:

a11 a12 a13 a14a21 a22 a23 a24a31 a32 a33 a34a41 a42 a43 a44

We claim that for any given values of a11, a12, a13, a21, a22, a23, a32, and a33 (the + signs in the diagrambelow), there is a unique way to assign values to the rest of the entries such that all necessary sumsare even.

+ + + a14+ + + a24a31 + + a34a41 a42 a43 a44

Guts Round

Taking additions mod 2, we have

a14 = a11 + a12 + a13

a24 = a21 + a22 + a23

a44 = a11 + a22 + a33

a42 = a12 + a22 + a32

a43 = a13 + a23 + a33

Since the 4th column, the 4th row, and the 1st column must have entries that sum to 0, we have

a34 = a14 + a24 + a44 = a12 + a13 + a21 + a23 + a33

a41 = a42 + a43 + a44 = a11 + a12 + a13 + a23 + a32

a31 = a11 + a21 + a41 = a12 + a13 + a21 + a23 + a32

It is easy to check that the sum of entries in every row, column, and the main diagonal is even. Sincethere are 28 = 256 ways to assign the values to the initial 8 entries, there are exactly 256 ways to fillthe board.

11. [8] Rosencrantz and Guildenstern play a game in which they repeatedly flip a fair coin. Let a1 = 4,a2 = 3, and an = an−1 + an−2 for all n ≥ 3. On the nth flip, if the coin is heads, Rosencrantz paysGuildenstern an dollars, and, if the coin is tails, Guildenstern pays Rosencrantz an dollars. If playcontinues for 2010 turns, what is the probability that Rosencrantz ends up with more money than hestarted with?

Answer: 12 〉 1

21341

Since Rosencrantz and Guildenstern have an equal chance of winning each toss, both have the sameprobability of ending up with a positive amount of money. Let x denote the probability that they bothend up with zero dollars. We wish to find 1−x

2 .

We have x is equal to the probability that

s2010 := i1a1 + i2a2 + · · · i2010a2010 = 0,

where in has an equal probability of being either 1 or 〉1.

We claim that s2010 = 0 if and only if i3n = 〉i3n−1 = 〉i3n−2 for all n ≤ 670. We start with thefollowing lemma.

Lemma. We have an >

n−3∑k=1

ak for all n ≥ 4.

Proof: For the case n = 4, a4 = a3 + a2 = 2a2 + a1 > a1. In case n > 4, we have

an = an−2 + an−1 > an−2 +

n−4∑k=1

ak = an−4 +

n−3∑k=1

ak >

n−3∑k=1

ak.

It suffices to show that s3n = 0 only if i3k = 〉i3k−1 = 〉i3k−2 for all k ≤ n. The triangle inequalityimplies the following:

0 ≤∣∣∣∣∣i3n−2a3n−2 + s3n−3

∣∣〉 ∣∣i3n−1a3n−1 + i3na3n∣∣∣∣∣ ≤ |s3n| = 0

0 ≤∣∣∣∣∣i3n−1a3n−1 + s3n−3

∣∣〉 ∣∣i3n−2a3n−2 + i3na3n∣∣∣∣∣ ≤ |s3n| = 0

Guts Round

By the lemma, we have

a3n−2 +

3n−3∑k=1

ak < a3n−2 + a3n < a3n−1 + a3n

a3n−1 +

3n−3∑k=1

ak < a3n−1 + a3n−3 + a3n−4 + a3n−2 = a3n−2 + a3n

i3n = i3n−1 implies

∣∣∣∣a3n−2 +

3n−3∑k=1

ak

∣∣∣∣ <∣∣i3n−1a3n−1 + i3na3n

∣∣ and i3n = i3n−2 implies

∣∣∣∣a3n−1 +

3n−3∑k=1

ak

∣∣∣∣ < ∣∣i3n−2a3n−2+i3na3n∣∣, which are both contradictions; therefore, we must have i3n = 〉i3n−1

and i3n = 〉i3n−2.

The probability that i3n = 〉i3n−1 = 〉i3n−2 is 14 , so x =

(14

)670= 1

21340 , and 1−x2 = 1

2 〉 121341 .

12. [8] A sequence of integers {ai} is defined as follows: ai = i for all 1 ≤ i ≤ 5, and ai = a1a2 · · · ai−1 〉 1

for all i > 5. Evaluate a1a2 · · · a2011 〉2011∑i=1

a2i .

Answer: 〉1941

For all i ≥ 6, we have ai = a1a2 · · · ai−1 〉 1. So

ai+1 = a1a2 · · · ai 〉 1

= (a1a2 · · · ai−1)ai 〉 1

= (ai + 1)ai 〉 1

= a2i + ai 〉 1.

Therefore, for all i ≥ 6, we have a2i = ai+1 〉 ai + 1, and we obtain that

a1a2 · · · a2011 〉2011∑i=1

a2i

= a2012 + 1〉5∑i=1

a2i 〉2011∑i=6

a2i

= a2012 + 1〉5∑i=1

i2 〉2011∑i=6

(ai+1 〉 ai + 1)

= a2012 + 1〉 55〉 (a2012 〉 a6 + 2006)

= a6 〉 2060

= 〉1941

13. [8] Let a,b, and c be the side lengths of a triangle, and assume that a ≤ b and a ≤ c. Let x = b+c−a2 .

If r and R denote the inradius and circumradius, respectively, find the minimum value of axrR .

Answer: 3

It is well-known that both abc4R and r(a+b+c)

2 are equal to the area of triangle ABC. Thus abc4R = r(a+b+c)

2 ,and

Rr =abc

2(a+ b+ c).

Guts Round

Since a ≤ b and a ≤ c, we have a2

bc ≤ 1. We thus obtain that

ax

rR=

a(b+ c〉 a)/2abc

2(a+b+c)

=(a+ b+ c)(b+ c〉 a)

bc

=(b+ c)2 〉 a2

bc

=(b+ c)2

bc〉 a2

bc

=b

c+c

b+ 2〉 a2

bc

≥ b

c+c

b+ 2〉 1

≥ 2 + 2〉 1

= 3

Equality is achieved when a = b = c.

14. [8] Danny has a set of 15 pool balls, numbered 1, 2, . . . , 15. In how many ways can he put the balls in8 indistinguishable bins such that the sum of the numbers of the balls in each bin is 14, 15, or 16?

Answer: 122

Clearly, the balls numbered 15, 14, . . . , 9, 8 must be placed in separate bins, so we number the bins15, 14, . . . , 9, 8. Note that bins 15 and 14 may contain only one ball while all other bins must containat least two balls. We have two cases to examine.

Case 1: Only one bin contains exactly one ball. Let ai denote the number of ways to place the ballsnumbered 1, 2, . . . , i〉 1 into the bins numbered 15, 14, . . . , 15〉 i+ 1. We can place either i〉 1 or i〉 2into the bin numbered 15〉 i+ 1. If we place i〉 1 in there, then there are ai−1 ways to finish packingthe rest. If we place i〉2 in this bin, then i〉1 must be placed in the bin numbered 15〉 i+ 2, so thereare ai−2 ways to place the rest of the balls. Therefore ai = ai−1 + ai−2. Since a1 = 2 and a2 = 3, thesequence {ai} is the Fibonacci sequence, and a7 = 34.

Case 2: Both bins 14 and 15 contain only one ball. A pair of balls from 1-7 must be put together toone of the bins numbered 8 through 13. This pair has sum at most 8, so we can count for all the cases.

Balls Number of packings1, 2 161, 3 101, 4 121, 5 121, 6 101, 7 82, 3 62, 4 42, 5 42, 6 33, 4 23, 5 1

Therefore, there are 88 possibilities in this case, and the total number of possibilities is 122.

15. [10] Find all irrational numbers x such that x3 〉 17x and x2 + 4x are both rational numbers.

Answer: 〉2±√

5

Guts Round

From x2 + 4x ∈ Q, we deduce that (x + 2)2 = x2 + 4x + 4 is also rational, and hence x = 〉2 ± √y,

where y is rational. Then x3〉17x = (26〉6y)±(y〉5)√y, which forces y to be 5. Hence x = 〉2±

√5.

It is easy to check that both values satisfy the problem conditions.

16. [10] Let R be a semicircle with diameter XY . A trapezoid ABCD in which AB is parallel to CD iscircumscribed about R such that AB contains XY . If AD = 4, CD = 5, and BC = 6, determine AB.

Answer: 10

We claim that AB = AD + BC. Let O denote the center of R. Since DA and DC are both tangentto R, we have ∠ADO = ∠ODC. Since CD is parallel to AB, we also have ∠ODC = ∠DOA. Thus∠ADO = ∠DOA, and it follows that AD = AO. Similarly, we have BC = BO. Therefore, we

AB = AO +BO = AD +BC = 10.

17. [10] Given positive real numbers x, y, and z that satisfy the following system of equations:

x2 + y2 + xy = 1,

y2 + z2 + yz = 4,

z2 + x2 + zx = 5,

find x+ y + z.

Answer:√

5 + 2√

3

Let O denote the origin. Construct vectors OA, OB, and OC as follows: The lengths of OA, OB, andOC are x, y, and z, respectively, and the angle between any two vectors is 120◦. By the Law of Cosines,we have AB = 1, BC = 2, and AC =

√5. Thus ABC is a right triangle, and O is its Fermat point.

The area of triangle ABC is equal to 12 · 1 · 2 = 1. But the area is also equal to the sum of the areas of

triangles AOB, BOC, and COA, which is equal to 12 ·xy ·

√32 + 1

2 · yz ·√32 + 1

2 · zx ·√32 . We thus obtain

xy + yz + zx = 4√3. Adding the three equations given in the problem and subtracting both sides by

xy+yz+zx, we obtain x2+y2+z2 = 5〉 2√3. Therefore (x+y+z)2 = (x2+y2+z2)+2(xy+yz+zx) =

5 + 2√

3.

18. [10] In how many ways can each square of a 4× 2011 grid be colored red, blue, or yellow such that notwo squares that are diagonally adjacent are the same color?

Answer: 64 · 34020

If we first color the board in a checkerboard pattern, it is clear that the white squares are independentof the black squares in diagonal coloring, so we calculate the number of ways to color the white squaresof a 4× n board and then square it.

Let an be the number of ways to color the white squares of a 4×n board in this manner such that thetwo squares in the last column are the same color, and bn the number of ways to color it such that theyare different. We want to find their sum xn. We have a1 = 3, b1 = 6. Given any filled 4× n〉 1 gridwith the two white squares in the last column different, there is only 1 choice for the middle squarein the nth row, and two choices for the outside square, 1 choice makes them the same color, 1 makesthem different. If the two white squares are the same, there are 2 choices for the middle square andthe outer square, so 4 choices. Of these, in 2 choices, the two new squares are the same color, and inthe other 2, the two squares are different. It follows that an = 2an−1 + bn−1 and bn = 2an−1 + bn−1,so an = bn for n ≥ 2. We have xn = 8 · 3n−1 and x2011 = 8 · 32010. So the answer is 64 · 34020.

19. [12] Find the least positive integer N with the following property: If all lattice points in [1, 3]× [1, 7]×[1, N ] are colored either black or white, then there exists a rectangular prism, whose faces are parallelto the xy, xz, and yz planes, and whose eight vertices are all colored in the same color.

Answer: 127

Guts Round

First we claim that if the lattice points in [1, 3] × [1, 7] are colored either black or white, then thereexists a rectangle whose faces are parallel to the x and y axes, whose vertices are all the same color(a.k.a. monochromatic). Indeed, in every row y = i, 1 ≤ i ≤ 7, there are two lattice points with thesame color. Note there are 3 combinations of 2 columns to choose from (for the two similarly-coloredlattice points to be in), and 2 colors to choose from. By the Pigeonhole Principle, in the 2 · 3 + 1 = 7rows two rows must have a pair of similarly-colored lattice points in the same columns, i.e. there is amonochromatic rectangle.

This shows that in each cross section z = i, 1 ≤ i ≤ N there is a monochromatic rectangle. Next,note there are

(32

)(72

)possibilities for this rectangle (

(32

)ways to choose the 2 x-coordinates and

(72

)ways to choose the 2 y-coordinates), and 2 possible colors. Thus if N = 2

(32

)(72

)+ 1 = 127, then by

the Pigeonhole Principle there are two values of i such that the same-colored rectangle has the same xand y coordinates in the plane z = i, i.e. there is a monochromatic rectangular prism.

For N = 126 the assertion is not true. In each cross section z = i, we can color so that there isexactly 1 monochromatic rectangle, and in the 126 cross sections, have all 126 possible monochromaticrectangles represented. To do this, in each cross section we color so that each row has exactly 2 latticepoints of the same color, and such that 6 of the rows give all possible combinations of 2 points havingthe same color. This way, there will be exactly 1 monochromatic rectangle in each cross section; wecan obviously vary it for the different cross sections.

20. [12] Let ABCD be a quadrilateral circumscribed about a circle with center O. Let O1, O2, O3, andO4 denote the circumcenters of 4AOB, 4BOC, 4COD, and 4DOA. If ∠A = 120◦,∠B = 80◦, and∠C = 45◦, what is the acute angle formed by the two lines passing through O1O3 and O2O4?

Answer: 82.5

Lemma: Given a triangle 4ABC, let I be the incenter, IA be the excenter opposite A, and Š be thesecond intersection of AI with the circumcircle. Then Š is the center of the circle through B, I, C,and IA.

Proof. First, note

∠IBIA = ∠IBC + ∠CBIA =∠ABC

2+

180◦ 〉 ∠ABC2

= 90◦.

Similarly ∠ICIA = 90◦. Therefore BICIA is cyclic. Now note that A, I, Š , and IA are collinearbecause they are all on the angle bisector of ∠BAC. Hence

∠CIŠ = 180◦ 〉 ∠CIA = ∠CAI + ∠ACI = ∠BCŠ + ∠ICB = ∠ICŠ.

(Note ∠CAI = ∠BAŠ = ∠BCŠ since A, B, Š, and C are concyclic.) Hence ŠC = ŠI. SimilarlyŠB = ŠI. Thus Š is the center of the circle passing through B, I, and C, and therefore IA as well. �

Let BA and CD intersect at E and DA and CB intersect at F . We first show that F , O1, O, and O3

are collinear.

Let O′1 and O′3 denote the intersections of FO with the circumcircles of triangles FAB and FDC.Since O is the excenter of triangle FAB, by the lemma O′1 is the circumcenter of 4ABO; since Ois incenter of triangle FDC, by the lemma O′3 is the circumcenter of 4DOC. Hence O′1 = O1 andO′3 = O3. Thus, points F , O1, O, and O3 are collinear, and similarly, we have E, O2, O, and O3 arecollinear. Now ∠BEC = 55◦ and ∠DFC = 20◦ so considering quadrilateral EOFC, the angle betweenO1O3 and O2O4 is

∠EOF = ∠OEC + ∠OFC + ∠FCE

=∠BEC

2+∠DFC

2+ ∠FCE

= 27.5◦ + 10◦ + 45◦ = 82.5◦.

Guts Round

21. [12] Let ABCD be a quadrilateral inscribed in a circle with center O. Let P denote the intersectionof AC and BD. Let M and N denote the midpoints of AD and BC. If AP = 1, BP = 3, DP =

√3,

and AC is perpendicular to BD, find the area of triangle MON .

Answer: 34

We first prove that ONPM is a parallelogram. Note that APD and BPC are both 30◦ 〉 60◦ 〉 90◦

triangles. Let M ′ denote the intersection of MP and BC. Since ∠BPM ′ = ∠MPD = 30◦, we haveMP ⊥ BC. Since ON is the perpendicular bisector of BC, we have MP//NO. Similarly, we haveMO//NP . Thus ONPM is a parallelogram. It follows that the area of triangle MON is equal to thearea of triangle MPN , which is equal to 1

2 · 1 · 3 · sin∠MPN = 12 · 1 · 3 · sin 150◦ = 3

4 .

22. [12] Find the number of ordered triples (a, b, c) of pairwise distinct integers such that 〉31 ≤ a, b, c ≤ 31and a+ b+ c > 0.

Answer: 117690

We will find the number of such triples with a < b < c. The answer to the original problem will thenbe six times what we will get. By symmetry, the number of triples (a, b, c) with a+ b+ c > 0 is equalto the number of those with a + b + c < 0. Our main step is thus to find the number of triples withsum 0.

If b = 0, then a = 〉c, and there are 31 such triples. We will count the number of such triples withb > 0 since the number of those with b < 0 will be equal by symmetry.

For all positive n such that 1 ≤ n ≤ 15, if a = 〉2n, there are n〉 1 pairs (b, c) such that a+ b+ c = 0and b > 0, and for all positive n such that 1 ≤ n ≤ 16, if a = 〉2n+ 1, there are also n〉 1 such pairs(b, c). In total, we have 1 + 1 + 2 + 2 + 3 + 3 + ...+ 14 + 14 + 15 = 225 triples in the case b > 0 (andhence likewise for b < 0.)

In total, there are 31 + 225 + 225 = 481 triples such that a < b < c and a + b + c = 0. Since thereare

(633

)= 39711 triples (a, b, c) such that 〉31 ≤ a < b < c ≤ 31, the number of triples with the

additional restriction that a+ b+ c > 0 is 39711−4812 = 19615. So the answer to the original problem is

19615× 6 = 117690.

23. [14] Let S be the set of points (x, y, z) in R3 such that x, y, and z are positive integers less than orequal to 100. Let f be a bijective map between S and the {1, 2, . . . , 1000000} that satisfies the followingproperty: if x1 ≤ x2, y1 ≤ y2, and z1 ≤ z2, then f(x1, y1, z1) ≤ f(x2, y2, z2). Define

Ai =

100∑j=1

100∑k=1

f(i, j, k),

Bi =

100∑j=1

100∑k=1

f(j, i, k),

and Ci =

100∑j=1

100∑k=1

f(j, k, i).

Determine the minimum value of Ai+1 〉Ai +Bj+1 〉Bj + Ck+1 〉 Ck.

Answer: 30604

We examine the 6 planes, their intersections and the lines between 2 points in one of the three pairsof parallel planes. The expression is equivalent to summing differences in values along all these lines.We examine the planes intersections. There is one cube, 3 · 98 squares and 3 · 98 · 98 lines. Theminimum value of the difference along a line is 1. For a square, to minimize the differences we takefour consecutive numbers, and the minimum value is 6. To find the minimum value along a cube,we take 8 consecutive numbers. Since we are taking differences, we can add or subtract any constantto the numbers, so we assume the numbers are 1-8. Examining the cube, we see there’s 1 spotwhere the number is multiplied by -3, 3 spots where the number is multiplied by -1, 1 spot where

Guts Round

the number is multiplied by 3, and 3 spots where the number is multiplied by 1. 1 and 8 must goin the corners, and 2,3,4,5 must go in spots multiplied by -1, -1, 1,1, respectively. To minimize thedifferences we put 5 in the final spot multiplied by -1, and 4 in the spot multiplied by 1 opposite5. Then the sum of all the differences is 28, so the minimum for a cube is 28. So the answer is28 + 18(100〉 2) + 3(100〉 2)2 = 30000 + 600 + 12〉 36 + 28 = 30604. It is clear that this value can beobtained.

24. [14] In how many ways may thirteen beads be placed on a circular necklace if each bead is either blueor yellow and no two yellow beads may be placed in adjacent positions? (Beads of the same color areconsidered to be identical, and two arrangements are considered to be the same if and only if each canbe obtained from the other by rotation).

Answer: 41

Let tn be the number of arrangements of n beads in a row such that bead i and i + 1 are not bothyellow for 1 ≤ i < n. Let an and bn be the number of arrangements satisfying the additional conditionthat beads n and 1 are not both yellow, and that beads n and 1 are both yellow, respectively. Clearlytn = an + bn.

First consider tn. If bead n is blue, there are tn−1 ways to choose the remaining colors. If bead n isyellow, then bead n〉 1 must be blue, and there are tn−2 ways to choose the remaining colors. Hence

tn = tn−1 + tn−2.

Next consider bn. In this case beads 1 and n are both yellow, so beads 2 and n〉 1 must be blue; theremaining colors can be chosen in tn−4 ways. Hence

bn = tn−4.

From t0 = 1 and t1 = 2 we see that tn = Fn+2 and bn = Fn−2 and so an = Fn+2 〉 Fn−2, where Fndenotes the nth Fibonacci number.

Next, since 13 is prime, a circular arrangement corresponds to exactly 13 straight-line arrangements,except for when all beads are the same color. The all-blue’s arrangement satisfies our condition, whilethe all-yellow does not. Hence our answer is

a13 〉 1

13+ 1 =

F15 〉 F11 + 12

13=

610〉 89 + 12

13= 41.

25. [14] Let n be an integer greater than 3. Let R be the set of lattice points (x, y) such that 0 ≤ x, y ≤ nand |x〉 y| ≤ 3. Let An be the number of paths from (0, 0) to (n, n) that consist only of steps of theform (x, y)→ (x, y+ 1) and (x, y)→ (x+ 1, y) and are contained entirely within R. Find the smallest

positive real number that is greater than An+1

Anfor all n.

Answer: 2 +√

2

We first find An in terms of n. Let an be the number of ways to get to the point (n, n+ 3), and let bnbe the number of ways to get to the point (n+ 1, n+ 2). By symmetry, an is also the number of waysto get to (n+ 3, n) and bn is also the number of ways to get to the point (n+ 2, n+ 1).

a0

b0

b0

a0

a1

b1

b1

a1

Guts Round

We can easily see that a0 = 1 and b0 = 3. This also means that An = an−3 + 3bn−3 + 3bn−3 + an−3 =2an−3 + 6bn−3.

We also get the recurrence:

ai+1 = ai + bi

bi+1 = ai + 3bi

We have both 3ai+1 = 3ai + 3bi and ai+2 = ai+1 + bi+1. Subtracting these gives

ai+2 〉 3ai+1 = ai+1 〉 3ai + bi+1 〉 3bi

ai+2 〉 3ai+1 = ai+1 〉 3ai + ai

ai+2 = 4ai+1 〉 2ai

Now we can solve this recurrence using its characteristic polynomial x2 〉 4x + 2, which has roots of2 +√

2 and 2 〉√

2. We can then write ai = A(2 +√

2)i + B(2 〉√

2)i for some constants A and B.Now, a0 = a and a1 = a0 + b0 = 4. Using this, we solve for A and B to get

ai =

(1 +√

2

2

)(2 +

√2)i +

(1〉√

2

2

)(2〉

√2)i

Then,

bi = ai+1 〉 ai

=

(1 +√

2

2

)((2 +

√2)i+1 〉 (2 +

√2)i) +

(1〉√

2

2

)((2〉

√2)i+1 〉 (2〉

√2)i)

=

(1 +√

2

2

)(1 +

√2)(2 +

√2)i +

(1〉√

2

2

)(1〉

√2)(2〉

√2)i

=

(3 + 2

√2

2

)(2 +

√2)i +

(3〉 2

√2

2

)(2〉

√2)i

Therefore,

An = 2an−3 + 6bn−3 = (10 + 7√

2)(2 +√

2)n−3 + (10〉 7√

2)(2〉√

2)n−3.

We can then easily see that An < (2 +√

2)An−1. Also, since 2〉√

2 < 1, as n approaches infinity, theratio An

An−1approaches 2 +

√2. Hence the least upper bound of An

An−1is 2 +

√2.

26. [14] In how many ways can 13 bishops be placed on an 8×8 chessboard such that (i) a bishop is placedon the second square in the second row, (ii) at most one bishop is placed on each square, (iii) no bishopis placed on the same diagonal as another bishop, and (iv) every diagonal contains a bishop? (For thepurposes of this problem, consider all diagonals of the chessboard to be diagonals, not just the maindiagonals).

Answer: 1152

We color the squares of the chessboard white and black such that B2 (the second square in the secondrow) is black. Note that at most 7 bishops can go on the white squares, and if there is a bishop on b2,at most 5 more can be on the white squares. So of the other 12 bishops, 7 go on white squares and 5go on black squares.

Consider the long diagonal on the white squares, and the 6 white diagonals parallel to it. Of the 7bishops placed on the white squares, exactly one must go on each of these diagonals (this also proves

Guts Round

that at most 7 can go on the white squares). Of these diagonals there is 1 of length 8, and 2 of length2,4, and 6. There are 2 ways to place 2 bishops on the diagonals of length 2, then 2 ways to place 2bishops on the diagonals of length 4, then 2 ways to place 2 bishops on the diagonals of length 2, thenthe long diagonal bishop can go on either corner. So there are 16 ways to place 7 bishops on the whitesquares.

Now we can divide the black squares of the board into the 6 diagonals parallel to the long whitediagonal, and the long black diagonal. The bishop on b2 accounts for two of these diagonals. We areleft with a diagonal of length 3, and two diagonals of length 5,7. There are 3 ways to pick the bishopon the diagonal of length 3, 6 ways to pick two bishop for the diagonals of length 5, and 6 ways to pickthe bishop on the diagonals of length 7. So there are 72 ways to pick 5 other bishops for the blacksquares. So the answer is 72 · 16 = 1152.

27. [16] Find the number of polynomials p(x) with integer coefficients satisfyingp(x) ≥ min{2x4 〉 6x2 + 1, 4〉 5x2} and p(x) ≤ max{2x4 〉 6x2 + 1, 4〉 5x2} for all x ∈ R.

Answer: 4

We first find the intersection points of f(x) = 2x4〉6x2+1 and g(x) = 4〉5x2. If 2x4〉6x2+1 = 4〉5x2,

then 2x4 〉 x2 〉 3 = 0, so (2x2 〉 3)(x2 + 1) = 0, and x = ±√

32 . Note that this also demonstrates

that g(x) ≥ f(x) if and only if |x| ≤√

32 and that p(x) must satisfy p(x) ≤ g(x) iff |x| ≤

√32 . We

must have p(√

32 ) = 〉 7

2 , so p(x) = 〉 72 + (2x2 〉 3)q(x) for a polnomial q of degree 0, 1 or 2. We now

examine cases.

Case 1: q is constant. We have q = n2 for an integer n since 1 ≤ p(0) ≤ 4, n = 〉 3

2 or 〉 52 . Clearly

q = 〉 52 is an appropriate choice because then p = g. Let p1(x) = 〉 7

2 + 〉 32 (2x2 〉 3) = 1 〉 3x2. We

have p1(x) is ≤ g(x) and ≤ f(x) in the right places, so this function works. Thus, there are 2 solutions.

Case 2: q is linear. We have p(x) is a cubic, so p(x) 〉 g(x) is also a cubic, which means it can’t bepositive for both arbitrarily large positive values of x and arbitrarily large negative values of x. Thus,there are no solutions.

Case 3: q is quadratic. We have q(x) = ax2 + bx + c. Apply the same argument for the case when qis a constant, we have c = 〉 3

2 or 〉 52 . Since p must have integer coefficients, we have b must be an

integer. Since f(x) ≥ p(x) ≥ g(x) for large x, the leading coefficient of f must be greater than or equalto the leading coefficient of p, which must be greater than 0. Thus a = 1 or a = 1

2 . However, if a = 12 ,

then the quadratic term of p is not an integer, so a = 1.

Now if c = 〉 52 , then p(0) = 4 = g(0). But this is the maximum value of g(x), so it must be a local

maximum of p. Thus p must not have a linear term odd (otherwise the function behaves like 〉3bx+ 4around x = 0). So p(x) must be 2x4 〉 8x2 + 4. This is indeed bounded between f(x) and g(x) at allpoints.

Now suppose c = 〉 32 . We have a must equal to 1. If b 6= 0, then p will have a cubic term, which means

f(x) 〉 p(x) can’t be positive for both arbitrarily large positive x and arbitrarily large negative x, sob = 0. Therefore p(x) = 〉 7

2 + (2x2 〉 3)(x2 〉 32 ) = f(x). It is easy to check that this choice of p is

indeed bounded between f(x) and g(x).

Therefore, there are 4 solutions in total.

28. [16] Let ABC be a triangle, and let points P and Q lie on BC such that P is closer to B than Q is.Suppose that the radii of the incircles of triangles ABP , APQ, and AQC are all equal to 1, and thatthe radii of the corresponding excircles opposite A are 3, 6, and 5, respectively. If the radius of theincircle of triangle ABC is 3

2 , find the radius of the excircle of triangle ABC opposite A.

Answer: 135

Let t denote the radius of the excircle of triangle 4ABC.

Lemma: Let 4ABC be a triangle, and let r and rA be the inradius and exradius opposite A. Then

r

rA= tan

B

2tan

C

2.

Guts Round

Proof. Let I and J denote the incenter and the excenter with respect to A. Let D and E be the footof the perpendicular from I and J to BC, respectively. Then

r = ID = BI sinB

2

rA = JE = BJ sin180◦ 〉B

2= BJ cos

B

2

BI = BJ tan∠AJB = BY tanC

2.

The last equation followed from

∠AJB = 180◦ 〉 ∠ABJ 〉 ∠JAB =180◦ 〉B

2〉 A

2=C

2.

Hencer

rA=

sin B2

cos B2· BIBJ

= tanB

2· tan

C

2.

�

Noting tan ∠APB2 tan ∠APQ

2 = tan ∠AQP2 tan ∠AQC

2 = 1 and applying the lemma to 4ABC, 4ABP ,4APQ, and 4AQC give

3/2

t= tan

∠ABC2

· tan∠ACB

2

=

(tan∠ABC

2· tan

∠APB2

)·(

tan∠APQ

2· tan

∠AQP2

)·(

tan∠AQC

2· tan

∠ACB2

)=

1

3· 1

6· 1

5

Therefore, t = 135.

29. [16] Let ABC be a triangle such that AB = AC = 182 and BC = 140. Let X1 lie on AC such thatCX1 = 130. Let the line through X1 perpendicular to BX1 at X1 meet AB at X2. Define X2, X3, . . .,as follows: for n odd and n ≥ 1, let Xn+1 be the intersection of AB with the perpendicular to Xn−1Xn

through Xn; for n even and n ≥ 2, let Xn+1 be the intersection of AC with the perpendicular toXn−1Xn through Xn. Find BX1 +X1X2 +X2X3 + . . ..

Answer: 11065

Let M and N denote the perpendiculars from X1 and A to BC, respectively. Since triangle ABCis isosceles, we have M is the midpoint of BC. Moreover, since AM is parallel to X1N , we haveNC

X1C=MC

AC⇔ X1N

130=

70

182=

5

13, so NC = 50. Moreover, since X1N ⊥ BC, we find X1C = 120 by

the Pythagorean Theorem. Also, BN = BC 〉NC = 140〉 50 = 90, so by the Pythagorean Theorem,X1B = 150.

We want to compute X2X1 = X1B tan(∠ABX1). We have

tan(∠ABX1) = tan(∠ABC 〉 ∠X1BC) =1 + tan(∠ABC) tan(∠X1BC)

tan(∠ABC)〉 tan(∠X1BC)=

(125

)〉(43

)1 +

(125

) (43

)=

16156315

=16

63.

Hence X2X1 = 150 · 1663 , and by the Pythagorean Theorem again, X2B = 150 · 6563 .Next, notice that AXn

AXn+2is constant for every nonnegative integer n (where we let B = X0). Indeed,

since XnXn+1 is parallel to Xn+2Xn+3 for each n, the dilation taking Xn to Xn+2 for some n alsotakes Xk to Xk+2 for all k.

Guts Round

Since4AXn+2Xn+3 ∼ 4AXnXn+1 with ratio AXn

AXn+2for each even n, we can compute that Xn+2Xn+3

XnXn+1=

AXn+2

AXn= 1〉 150· 6563

182 for every nonnegative integer n. Notice we use all three sides of the above similartriangles.

We now split our desired sum into two geometric series, one with the even terms and one with the oddterms, to obtain

BX1 +X1X2 + . . . = (BX1 +X2X3 + . . .) + (X1X2 +X3X4 + . . .) =150

150· 6563182

+150 · 1663150· 6563182

=7963 · 150150· 6563182

=1106

5.

30. [16] How many ways are there to color the vertices of a 2n-gon with three colors such that no vertexhas the same color as its either of its two neighbors or the vertex directly across from it?

Answer: 3n + (〉2)n+1 〉 1

Let the 2n-gon have vertices A1, A2, ..., A2n, in that order. Consider the diagonals d1 = (A1, An+1),d2 = (A2, An+2), · · · , dn = (An, A2n). Suppose the three colors are red (R), green (G), and blue (B).Each diagonal can either be colored (R,G), (G,R), (G,B), (B,G), (B,R), or (R,B). We first chooseone of the six colorings for d1, which then constrains the possible colorings for d2, which constrains thepossible colorings for d3, and so on. This graph shows the possible configurations; two pairs of colorsare connected by an edge if they can be the colors for di and di+1 for any 1 ≤ i ≤ n〉 1.

(G,R)

(B,G) (R,B)

(R,G)

(G,B) (B,R)

Suppose without loss of generality that d1 is colored (R,G). (At the end, we multiply our answer by6.) Then dn must be either (R,G), (B,G), or (R,B). Now, we simply need to count the number ofpaths of length n〉 1 within this graph from (R,G) to one of these three points.

Suppose we are making a random walk of n 〉 1 steps, where at each move we pick one of the threepossible edges with probability 1

3 . We will calculate the probability that the walk ends at one of (R,G),(B,G), or (R,B).

Let ai and bi be the probability that, after i steps, we are at (R,G) and (B,G), respectively. Bysymmetry, bi is also the probability that we are at (R,B) after i steps.

Observe that after each move, the probability of arriving at either (R,G) or (G,R) will always be 13 .

Therefore, the probability of being at (G,R) after i steps is 13 〉 ai. Similarly, the probability of being

at (G,B) is 13 〉 bi and the probability of being at (B,R) is 1

3 〉 bi.

Guts Round

13 − ai

bi bi

ai

13 − bi

13 − bi

Now, for i ≥ 1 we have the recurrences

ai+1 =1

3

((1

3〉 ai

)+

(1

3〉 bi

)+

(1

3〉 bi

))=

1

3〉 1

3ai 〉

2

3bi

bi+1 =1

3

((1

3〉 ai

)+

(1

3〉 bi

)+ bi

)=

2

9〉 1

3ai

So then

ai+2 =1

3〉 1

3ai+1 〉

2

3bi+1

ai+2 =1

3〉 1

3ai+1 〉

2

3

(2

9〉 1

3ai

)ai+2 =

5

27〉 1

3ai+1 +

2

9ai(

ai+2 〉1

6

)= 〉1

3

(ai+1 〉

1

6

)+

2

9

(ai 〉

1

6

)

This recurrence has a characteristic polynomial x2 + 13x〉 2

9 , which has roots 13 and 〉 2

3 . We can writeai = 1

6 +A( 13 )i +B(〉 2

3 )i for some constants A and B for i ≥ 1. Since a1 = 0 and a2 = 13 , we can solve

for A and B and get

ai =1

6+

1

6

(1

3

)i+

1

3

(〉2

3

)i

Guts Round

The answer to the problem is then

6 · 3n−1(an−1 + 2bn−1) = 6 · 3n−1 (an−1 + 1〉 an−1 〉 3an)

= 6 · 3n−1(1〉 3an)

= 6 · 3n−1(

1〉 3

(1

6+

1

6

(1

3

)n+

1

3

(〉2

3

)n))= 6 · 3n−1

(1

2〉 1

2

(1

3

)n〉(〉2

3

)n)= 3n + (〉2)n+1 〉 1.

31. [18] Let A = {1, 2, 3, . . . , 9}. Find the number of bijective functions f : A → A for which there existsat least one i ∈ A such that

|f(i)〉 f−1(i)| > 1.

Answer: 359108

We count the complement — the number of functions f such that for all i ∈ A, |f(i)〉 f−1(i)| ≤ 1.

The condition is equivalent to |f(f(i))〉 i| ≤ 1 for all i ∈ A. If f(j) = j, the inequality is automaticallysatisfied for i = j. Otherwise, if f(f(j)) = j but f(j) = k 6= j, then we will have f(f(k)) = k, allowingthe inequality to be satisfied for i = j, k. Else, if f(f(i)) 6= i, say f(f(i)) = i + 1 and f(i) = k, thenf(f(k)) = f(i+ 1) = k + 1 or k 〉 1. Thus the function f allows us to partition the elements of A intothree groups:

(a) those such that f(i) = i,

(b) those that form pairs {i, j} such that f(i) = j and f(j) = i, and

(c) those that form quartets {i, i+ 1, j, j + 1} such that f permutes them as (i j i+ 1 j + 1) or(i j + 1 i+ 1 j), in cycle notation.

Let a be the number of elements of the second type. Note that a is even.

Case 1: There are no elements of the third type. If a = 8, there are 9 · 7 · 5 · 3 = 945 possibilities.If a = 6, there are

(93

)· 5 · 3 = 1260 possibilities. If a = 4, there are

(95

)· 3 = 378 possibilities.

If a = 2, there are(97

)= 36 possibilities. If a = 0, there is 1 possibility. In total, case 1 offers

945 + 1260 + 378 + 36 + 1 = 2620 possibilities.

Case 2: There are 4 elements of the third type. There are 21 ways to choose the quartet {i, i+1, j, j+1}.For each way, there are two ways to assign the values of the function to each element (as describedabove). For the remaining 5 elements, we divide into cases according to the value of a. If a = 4, thereare 5× 3 = 15 possibilities. If a = 2, there are

(53

)= 10 possibilities. If a = 0, there is one possibility.

In total, case 2 offers 21× 2× (15 + 10 + 1) = 1092 possibilities.

Case 3: There are 8 elements of the third type. There are 5 ways to choose the unique element notof the third type. Of the remaining eight, there are 3 ways to divide them into two quartets, and foreach quartet, there are 2 ways to assign values of f . In total, case 3 offers 5× 3× 22 = 60 possibilities.

Therefore, the number of functions f : A → A such that for at least one i ∈ A, |f(i) 〉 f−1(i)| > 1 is9!〉 2620〉 1092〉 60 = 359108.

32. [18] Let p be a prime positive integer. Define a mod-p recurrence of degree n to be a sequence {ak}k≥0of numbers modulo p satisfying a relation of the form ai+n = cn−1ai+n−1 + ... + c1ai+1 + c0ai for alli ≥ 0, where c0, c1, . . . , cn−1 are integers and c0 6≡ 0 (mod p). Compute the number of distinct linearrecurrences of degree at most n in terms of p and n.

Answer: 1〉 np−1p+1 + p2(p2n−1)(p+1)2

Guts Round

In the solution all polynomials are taken modulo p. Call a polynomial nice if it is monic with nonzeroconstant coefficient. We can associate each recurrence relation with a polynomial: associate

cnai+n + cn−1ai+n−1 + ...+ c1ai+1 + c0ai = 0

withcnx

n + cn−1xn−1 + ...+ c1x+ c0.

Let Di be the set of mod-p recurrences {ak}k≥0 where i is the least integer so that {ak}k≥0 has degreei, and let di = |Di|.Let Sn be the set of pairs ({ak}k≥0, P ) where {ak}k≥0 is a mod-p recurrence, and P is a nice polynomialassociated to a recurrence relation of degree at most n satisfied by {ak}k≥0. To find dn generally, wecount the number of elements in Sn in two ways.

One the one hand, for each sequence {ak}k≥0 in Di, there exist pn−i polynomials P such that({ak}k≥0, P ) ∈ S. Indeed, {ak}k≥0 satisfies any recurrence relation associated with a polynomialmultiple of P . When j = i there is just one nice degree j polynomial that is a multiple of P , P itself.For j > i, there are (p 〉 1)pj−i−1 nice polynomials of degree j that are multiples of P , namely QPwhere Q is a nice polynomial of degree j 〉 i. (There are p choices for the coefficients of x, . . . , xj−i−1

and p〉1 choices for the constant term.) So the number of nice polynomial multiples of degree at mostn is

1 +

n∑j=i+1

(p〉 1)pj−i−1 = 1 + (p〉 1)

(pn−i 〉 1

p〉 1

)= pn−i.

Hence

|Sn| =n∑i=0

dipn−i. (1)

On the other hand, given a monic polynomial P of degree i, there are pi recurrences {ak}k≥0 such that({ak}k≥0, P ) ∈ S, since a0, . . . , ai−1 can be chosen arbitrarily and the rest of the terms are determined.Since there are (p〉1)pi−1 nice polynomials of degree i 6= 0 (and 1 nice polynomial for i = 0), summingover i gives

|Sn| = 1 +

n∑i=1

(p〉 1)p2i−1 (2)

Now clearly d0 = 1. Setting (1) and (2) equal for n and n+ 1 give

n+1∑i=0

dipn+1−i = 1 + (p〉 1)

n+1∑i=1

p2i−1 (3)

n∑i=0

dipn−i = 1 + (p〉 1)

n∑i=1

p2i−1

=⇒n∑i=0

dipn+1−i = p+ (p〉 1)

n∑i=1

p2i. (4)

Guts Round

Subtracting (4) from (3) yields:

dn+1 = 1〉 p+ (p〉 1)

2n+1∑i=1

(〉1)i+1pi

= (p〉 1)

2n+1∑i=0

(〉1)i+1pi

= (p〉 1)2n∑i=0

p2m

= (p〉 1)2(p2n+2 〉 1

p2 〉 1

)=

(p〉 1)(p2n+2 〉 1)

p+ 1

Thus the answer is

n∑i=0

di = 1 +p〉 1

p+ 1

n∑i=1

(p2i 〉 1)

= 1 +p〉 1

p+ 1

(〉n+ p2 · p

2n 〉 1

p2 〉 1

)= 1〉 np〉 1

p+ 1+p2(p2n 〉 1)

(p+ 1)2

33. [25] Find the number of sequences consisting of 100 R’s and 2011 S’s that satisfy the property thatamong the first k letters, the number of S’s is strictly more than 20 times the number of R’s for all1 ≤ k ≤ 2111.

Answer: 112111

(2111100

)Given positive integers r and s such that s ≥ 20r, let N(s, r) denote the number of sequences of scopies of S and r copies of R such that for all 1 ≤ k ≤ r+ s〉 1, among the first k letters, the numberof S’s is strictly more than 20 times the number of R’s. We claim that

N(s, r) =s〉 20r

s+ r

(s+ r

s

).

We induct on s+ r. For the base case, note that N(20r, r) = 0 for all r and N(s, 0) = 1 for all s.

Assume the formula holds up to r+ s〉 1. We now construct a path in the Cartesian plane as follows:Let R represent moving one unit to the left and S represent moving one unit up. This induces abijection between sequences of S’s and Rs and lattice paths from the origin to the point (r, s) thatare above the line y = 20x. In order to reach the point (r, s), we must go through either (r 〉 1, s) or

(r, s〉1). Therefore, we have N(r, s) = N(s, r〉1)+N(s〉1, r) = s−20(r−1)s+r−1

(s+r−1s

)+ s−1−20r

s+r−1(s+r−1s

)=

s−20rs+r

(s+rs

).

Remark This is a special case of the ballot problem, first studied in 1887 by Joseph Bertrand, gener-alizing the Catalan numbers. A good expository article on this problem is “Four Proofs of the BallotProblem” by Marc Renault available on his website athttp://webspace.ship.edu/msrenault/ballotproblem/.

In 2009, Yufei Zhao studied a variant problem called bidirectional ballot sequences, which he used toconstruct More-Sums-Then-Differences sets in additive combinatorics. His paper is available on hiswebsite athttp://web.mit.edu/yufeiz/www/mstd construction.pdf.

Guts Round

34. [25] Let w = w1, w2, . . . , w6 be a permutation of the integers {1, 2, . . . , 6}. If there do not exist indicesi < j < k such that wi < wj < wk or indices i < j < k < l such that wi > wj > wk > wl, then w issaid to be exquisite. Find the number of exquisite permutations.

Answer: 25

Given a permutation w = w1, . . . , wn for some n, call a sequence wi1 , wi2 , . . . , wim an increasingsubsequence if i1 < · · · < im and wi1 < · · · < wim . Define decreasing subsequences similarly. Letis(w) denote the length of the longest increasing sequence and ds(w) denote the length of the longestdecreasing sequence. We wish to find the number of permutations for n = 6 such that is(w) ≤ 2 andds(w) ≤ 3. We note here that 6 = 2× 3 is not a coincidence.

Erdos and Szekeres first studied problems on the longest increasing and decreasing subsequences. In1935, they showed that for any permutation w of {1, 2, . . . , pq + 1}, either is(w) > p or ds(w) > q,which later appeared on the Russian Math Olympiad.

In 1961, Schensted proved that the bound pq + 1 is sharp, and he enumerated the number of permu-tations for n = pq such that is(w) ≤ p and ds(w) ≤ q (exquisite permutations for simplicity), with anelegant combinatorial proof based on the RSK-algorithm relating Young Tableux and permutations.

The main idea of his proof is as follows. Consider a p× q rectangle. A Young Tableau is an assignmentof 1, 2, . . . , pq, one to each unit square of the rectangle, such that every row and column is in increasingorder. There is a bijection between set of exquisite permutations and pairs of Young Tableaux. Sincethe number of ways to write 1, 2, . . . , 6 on a 2 × 3 rectangle with every row and column in increasingorder is 5, there are exactly 25 exquisite permutations.

For a thorough exposition of increasing and decreasing subsequences and a collection of interesting openquestions, see Richard Stanley’s note for his undergraduate research students at MIT, “Increasing anddecreasing subsequences and their variants” available athttp://www.math.mit.edu/∼rstan/papers/ids.pdf.

35. [25] An independent set of a graph G is a set of vertices of G such that no two vertices among theseare connected by an edge. If G has 2000 vertices, and each vertex has degree 10, find the maximumpossible number of independent sets that G can have.

Answer: 2047100

The upper bound is obtained when G is a disjoint union of bipartite graphs, each of which has 20vertices with 10 in each group such that every pair of vertices not in the same group are connected.

In 1991, during his study of the Cameron—Erdos conjecture on the number of sum-free sets, NogaAlon came across this problem on independent sets and conjectured that our construction gives thebest bound.

This problem received considerable attention due to its application to combinatorial group theory andstatistical mechanics, but no solution was found until 2009, when Yufei Zhao resolved Alon’s conjecturewith a beautiful and elementary approach.

For the reader to enjoy the full insight of Yufei’s argument, we omit the proof here and refer to hispaper “The Number of Independent Sets in a Regular Graph” available on his website athttp://web.mit.edu/yufeiz/www/indep reg.pdf.

36. [25] An ordering of a set of n elements is a bijective map between the set and {1, 2, . . . , n}. Call anordering ρ of the 10 unordered pairs of distinct integers from the set {1, 2, 3, 4, 5} admissible if, for any1 ≤ a < b < c ≤ 5, either p({a, b}) < p({a, c}) < p({b, c}) or p({b, c}) < p({a, c}) < p({a, b}). Find thetotal number of admissible orderings.

Answer: 768

This problem is a special case of the higher Bruhat order, a class of combinatorial object widelystudied for its connection to an assortment of mathematical areas such as algebraic geometry, algebraiccombinatorics, and computational geometry.

Guts Round

An admissble order in our problem—the higher Bruhat order B(5, 2)—are best viewed as the setof reduced decompositions of the permutation 4321. Loosely speaking, a reduced decomposition is asequence of adjacent transpositions that changes the permutation n, n 〉 1, . . . , 1 to 1, 2, . . . , n. Forexample, for n = 4, the sequence (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4) induces the following reduceddecomposition:

4321→ 4312→ 4132→ 1432→ 1423→ 1243→ 1234.

For each permutation above, switching the two bolded numbers yields the next permutation in thechain. For example, switching 3 and 1 in 4312 yields 4132.

Readers interested in the connection between the higher Bruhat order and reduced decompositions arereferred to Delong Meng’s paper “Reduced decompositions and permutation patterns generalized tothe higher Bruhat order” for background as well as recent development of this subject. The paper isavailable athttp://web.mit.edu/delong13/www/papers.html.

The number of reduced decompositions of n, n〉 1, . . . , 1 is equal to the number of (n〉 1)st standardYoung Tableaux of staircase shape, given by the formula(

n2

)!

1n−1 · 3n−2 · · · (2n〉 3).

When n = 5, the formula gives 768.

Young Tableaux are one of the most important tools in algebraic combinatorics, especially for problemsinvolving permutations, integer partitions, and posets. For a concise and accessible introduction to theYoung Tableaux and reduced decompositions (and much more cool combinatorial stuff!), see Section7 of Richard Stanley’s note “A Combinatorial Miscellany” athttp://math.mit.edu/∼rstan/papers/comb.pdf.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Guts Round

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

Team Round A

Warm-Up [50]

1. [20] Alice and Barbara play a game on a blackboard. At the start, zero is written on the board. Alicegoes first, and the players alternate turns. On her turn, each player replaces x – the number writtenon the board – with any real number y, subject to the constraint that 0 < y − x < 1.

(a) [10] If the first player to write a number greater than or equal to 2010 wins, determine, withproof, who has the winning strategy.

(b) [10] If the first player to write a number greater than or equal to 2010 on her 2011th turn or laterwins (if a player writes a number greater than or equal to 2010 on her 2010th turn or earlier, sheloses immediately), determine, with proof, who has the winning strategy.

2. [15] Rachel and Brian are playing a game in a grid with 1 row of 2011 squares. Initially, there is onewhite checker in each of the first two squares from the left, and one black checker in the third squarefrom the left. At each stage, Rachel can choose to either run or fight. If Rachel runs, she moves theblack checker 1 unit to the right, and Brian moves each of the white checkers one unit to the right. IfRachel chooses to fight, she pushes the checker immediately to the left of the black checker 1 unit tothe left, the black checker is moved 1 unit to the right, and Brian places a new white checker in thecell immediately to the left of the black one. The game ends when the black checker reaches the lastcell. How many different final configurations are possible?

3. [15] Let n be a positive integer, and let a1, a2, . . . , an be a set of positive integers such that a1 = 2 andam = ϕ(am+1) for all 1 ≤ m ≤ n − 1, where, for all positive integers k, ϕ(k) denotes the number ofpositive integers less than or equal to k that are relatively prime to k. Prove that an ≥ 2n−1.

Complex Numbers [35]

4. [15] Let a, b, and c be complex numbers such that |a| = |b| = |c| = |a+ b+ c| = 1. If |a− b| = |a− c|and b 6= c, prove that |a+ b||a+ c| = 2.

5. [20] Let a and b be positive real numbers. Define two sequences of real numbers {an} and {bn} for allpositive integers n by (a+ bi)n = an + bni. Prove that

|an+1|+ |bn+1||an|+ |bn|

≥ a2 + b2

a+ b

for all positive integers n.

Page 1 of 3

Coin Flipping [75]

In a one-player game, the player begins with 4m fair coins. On each of m turns, the player takes 4unused coins, flips 3 of them randomly to heads or tails, and then selects whether the 4th one is headsor tails (these four coins are then considered used). After m turns, when the sides of all 4m coins havebeen determined, if half the coins are heads and half are tails, the player wins; otherwise, the playerloses.

6. [10] Prove that whenever the player must choose the side of a coin, the optimal strategy is to chooseheads if more coins have been determined tails than heads and to choose tails if more coins have beendetermined heads than tails.

7. [15] Let T denote the number of coins determined tails and H denote the number of coins determinedheads at the end of the game. Let pm(k) be the probability that |T −H| = k after a game with 4mcoins, assuming that the player follows the optimal strategy outlined in problem 6. Clearly pm(k) = 0if k is an odd integer, so we need only consider the case when k is an even integer. (By definition,p0(0) = 1 and p0(k) = 0 for k ≥ 1). Prove that pm(0) ≥ pm+1(0) for all nonnegative integers m.

8. (a) [5] Find a1, a2, and a3, so that the following equation holds for all m ≥ 1:

pm(0) = a1pm−1(0) + a2pm−1(2) + a3pm−1(4)

(b) [5] Find b1, b2, b3, and b4, so that the following equation holds for all m ≥ 1:

pm(2) = b1pm−1(0) + b2pm−1(2) + b3pm−1(4) + b4pm−1(6)

(c) [5] Find c1, c2, c3, and c4, so that the following equation holds for all m ≥ 1 and j ≥ 2:

pm(2j) = c1pm−1(2j − 2) + c2pm−1(2j) + c3pm−1(2j + 2) + c4pm−1(2j + 4)

9. [15] We would now like to examine the behavior of pm(k) as m becomes arbitrarily large; specifi-cally, we would like to discern whether lim

m→∞pm(0) exists and, if it does, to determine its value. Let

limm→∞

pm(k) = Ak.

(a) [5] Prove that 23pm(k) ≥ pm(k + 2) for all m and k.

(b) [10] Prove that A0 exists and that A0 > 0. Feel free to assume the result of analysis that anon-increasing sequence of real numbers that is bounded below by a constant c converges to alimit that is greater than or equal to c.

10. [20] Once it has been demonstrated that limn→∞

pn(0) exists and is greater than 0, it follows that

limn→∞

pn(k) exists and is greater than 0 for all even positive integers k and that∞∑k=0

A2k = 1. It also

follows that A0 = a1A0 +a2A2 +a3A4, A2 = b1A0 + b2A2 + b3A4 + b4A6, and A2j = c1A2j−2 + c2A2j +c3A2j+2 + c4A2j+4 for all positive integers j ≥ 2, where a1, a2, a3, b1, b2, b3, b4, and c1, c2, c3, c4 are theconstants you found in problem 8. Assuming these results, determine, with proof, the value of A0.

Page 2 of 3

Geometry [90]

11. [20] Let ABC be a non-isosceles, non-right triangle, let ω be its circumcircle, and let O be its cir-cumcenter. Let M be the midpoint of segment BC. Let the circumcircle of triangle AOM intersect ωagain at D. If H is the orthocenter of triangle ABC, prove that ∠DAH = ∠MAO.

12. [70] Let ABC be a triangle, and let E and F be the feet of the altitudes from B and C, respectively.If A is not a right angle, prove that the circumcenter of triangle AEF lies on the incircle of triangleABC if and only if the incenter of triangle ABC lies on the circumcircle of triangle AEF .

Let ABC be a triangle, and let E and F be the feet of the altitudes from B and C, respectively. If Ais not a right angle, prove that the circumcenter of triangle AEF lies on the incircle of triangle ABCif and only if the incenter of triangle ABC lies on the circumcircle of triangle AEF .

Last Writes [110]

13. [30] Given positive integers a and b such that a > b, define a sequence of ordered pairs (al, bl) fornonnegative integers l by a0 = a, b0 = b, and (al+1, bl+1) = (bl, al mod bl), where, for all positiveintegers x and y, x mod y is defined to be the remainder left by x upon division by y. Define f(a, b)to be the smallest positive integer j such that bj = 0. Given a positive integer n, define g(n) to bemax1≤k≤n−1 f(n, k).

(a) [15] Given a positive integer m, what is the smallest positive integer nm such that g(nm) = m?

(b) [15] What is the second smallest?

14. [25] Given a positive integer n, a sequence of integers a1, a2, . . . , ar, where 0 ≤ ai ≤ k for all 1 ≤ i ≤ r,is said to be a “k-representation” of n if there exists an integer c such that

r∑i=1

ai =

r∑i=1

aikc−i = n.

Prove that every positive integer n has a k-representation, and that the k-representation is unique ifand only if 0 does not appear in the base-k representation of n− 1.

15. [55] Denote {1, 2, ..., n} by [n], and let S be the set of all permutations of [n]. Call a subset T of Sgood if every permutation σ in S may be written as t1t2 for elements t1 and t2 of T , where the productof two permutations is defined to be their composition. Call a subset of U of S extremely good if everypermutation σ in S may be written as s−1us for elements s of S and u of U . Let τ be the smallestvalue of |T |/|S| for all good subsets T , and let υ be the smallest value of |U |/|S| for all extremely goodsubsets U . Prove that

√υ ≥ τ .

Page 3 of 3

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

Team Round A

Warm-Up [50]

1. [20] Alice and Barbara play a game on a blackboard. At the start, zero is written on the board. Alicegoes first, and the players alternate turns. On her turn, each player replaces x – the number writtenon the board – with any real number y, subject to the constraint that 0 < y − x < 1.

(a) [10] If the first player to write a number greater than or equal to 2010 wins, determine, withproof, who has the winning strategy.

(b) [10] If the first player to write a number greater than or equal to 2010 on her 2011th turn or laterwins (if a player writes a number greater than or equal to 2010 on her 2010th turn or earlier, sheloses immediately), determine, with proof, who has the winning strategy.

Solution: Each turn of the game is equivalent to adding a number between 0 and 1 to the numberon the board.

(a) Barbara has the winning strategy: whenever Alice adds z to the number on the board, Barbaraadds 1−z. After Barbara’s ith turn, the number on the board will be i. Therefore, after Barbara’s2009th turn, Alice will be forced to write a number between 2009 and 2010, after which Barbaracan write 2010 and win the game.

(b) Alice has the winning strategy: she writes any number a on her first turn, and, after that, wheneverBarbara adds z to the number on the board, Alice adds 1− z. After Alice’s ith turn, the numberon the board will be (i− 1) + a, so after Alice’s 2010th turn, the number will be 2009 + a. SinceBarbara cannot write a number greater than or equal to 2010 on her 2010th turn, she will beforced to write a number between 2009 + a and 2010, after which Alice can write 2010 and winthe game.

2. [15] Rachel and Brian are playing a game in a grid with 1 row of 2011 squares. Initially, there is onewhite checker in each of the first two squares from the left, and one black checker in the third squarefrom the left. At each stage, Rachel can choose to either run or fight. If Rachel runs, she moves theblack checker 1 unit to the right, and Brian moves each of the white checkers one unit to the right. IfRachel chooses to fight, she pushes the checker immediately to the left of the black checker 1 unit tothe left, the black checker is moved 1 unit to the right, and Brian places a new white checker in thecell immediately to the left of the black one. The game ends when the black checker reaches the lastcell. How many different final configurations are possible?

Solution: Both operations, run and fight, move the black checker exactly one square to the right,so the game will end after exactly 2008 moves regardless of Brian’s choices. Furthermore, it is easy tosee that if we view running and fighting as operations, they commute. So the order of the moves doesnot matter, all that matters is how many times Rachel runs and how many times Rachel fights. Eachfight adds one white checker to the grid, so two games with different numbers of fights will end up indifferent final configurations. There are 2009 possible values for the number of fights, so there are 2009possible final configurations.

3. [15] Let n be a positive integer, and let a1, a2, . . . , an be a set of positive integers such that a1 = 2 andam = ϕ(am+1) for all 1 ≤ m ≤ n − 1, where, for all positive integers k, ϕ(k) denotes the number ofpositive integers less than or equal to k that are relatively prime to k. Prove that an ≥ 2n−1.

Solution: We first note that ϕ(s) < s for all positive integers s ≥ 2, so am > 2 for all m > 1.

For integers s > 2, let As be the set of all positive integers x ≤ s such that gcd(s, x) = 1. Sincegcd(s, x) = gcd(s, s − x) for all x, if a is a positive integer in As, so is s − a. Moreover, if a is in As,a and s − a are different since gcd(s, s2 ) = s

2 > 1, meaning s2 is not in As. Hence we may evenly pair

Team Round A

up the elements of As that sum to s, so ϕ(s), the number of elements of As, must be even. It followsthat am is even for all m ≤ n− 1.

If t > 2 is even, At will not contain any even number, so ϕ(t) ≤ t2 . We may conclude that am ≥ 2am−1

for all m ≤ n− 1, so an ≥ an−1 ≥ 2n−2a1 = 2n−1, as desired.

Finally, note that such a set exists for all n by letting ai = 2i for all i.

Complex Numbers [35]

4. [15] Let a, b, and c be complex numbers such that |a| = |b| = |c| = |a+ b+ c| = 1. If |a− b| = |a− c|and b 6= c, prove that |a+ b||a+ c| = 2.

Solution:

First Solution. Since |a| = 1, a cannot be 0. Let u = ba and v = c

a . Dividing the given equations by|a| = 1 gives |u| = |v| = |1+u+v| = 1 and |1−u| = |1−v|. The goal is to prove that |1+u||1+v| = 2.

By squaring |1 − u| = |1 − v|, we get (1 − u)(1− u) = (1 − v)(1− v), and thus 1 − u − ū + |u|2 =1− v − v̄ + |v|2, or u+ ū = v + v̄. This implies Re(u) = Re(v). Since u and v are on the unit circle inthe complex plane, u is equal to either v or v̄. However, b 6= c implies u 6= v, so u = v̄.

Therefore, 1 = |1+u+ ū| = |1+2 Re(u)|. Since Re(u) is real, we either have Re(u) = 0 or Re(u) = −1.The first case gives u = ±i and |1 + u||1 + v| = |1 + i||1 − i| = 2, as desired. It remains only to notethat Re(u) = −1 is in fact impossible because u is of norm 1 and u = −1 would imply u = ū = v.

Remark: by the rotational symmetry of the circle, it is acceptable to skip the first step of this solutionand assume a = 1 without loss of generality.

Second Solution. Let a, b, and c, be the vertices of a triangle inscribed in the unit circle in thecomplex plane. Since the complex coordinate of the circumcenter is 0 and the complex coordinate ofthe centroid is a+b+c

3 , it follows from well-known facts about the Euler line that the complex coordinateof the orthocenter is a+b+c. Hence the orthocenter lies on the unit circle as well. Is it not possible forthe orthocenter not to be among the three vertices of the triangle, for, if it were, two opposite angles ofthe convex cyclic quadrilateral formed by the three vertices and the orthocenter would each measuregreater than 90 degrees. It follows that the triangle is right. However, since |a− b| = |a− c|, the rightangle cannot occur at b or c, so it must occur at a, and the desired conclusion follows immediately.

5. [20] Let a and b be positive real numbers. Define two sequences of real numbers {an} and {bn} for allpositive integers n by (a+ bi)n = an + bni. Prove that

|an+1|+ |bn+1||an|+ |bn|

≥ a2 + b2

a+ b

for all positive integers n.

Solution: Let z = a+ bi. It is easy to see that what we are asked to show is equivalent to

|zn+1 + z̄n+1|+ |zn+1 − z̄n+1||zn + z̄n|+ |zn − z̄n|

≥ 2zz̄

|z + z̄|+ |z − z̄|

Cross-multiplying, we see that it suffices to show

|zn+2 + zn+1z̄ + zz̄n+1 + z̄n+2|+ |zn+2 − zn+1z̄ + zz̄n+1 − z̄n+2|

+|zn+2 + zn+1z̄ − zz̄n+1 − z̄n+2|+ |zn+2 − zn+1z̄ − zz̄n+1 + z̄n+2|

≥ 2|zn+1z̄ + zz̄n+1|+ 2|zn+1z̄ − zz̄n+1|

However, by the triangle inequality,

|zn+2 + zn+1z̄ + zz̄n+1 + z̄n+2|+ |zn+2 − zn+1z̄ − zz̄n+1 + z̄n+2| ≥ 2|zn+1z̄ + zz̄n+1|

Team Round A

and|zn+2 − zn+1z̄ + zz̄n+1 − z̄n+2|+ |zn+2 + zn+1z̄ − zz̄n+1 − z̄n+2| ≥ 2|zn+1z̄ − zz̄n+1|

This completes the proof.

Remark: more computationally intensive trigonometric solutions are also possible by reducing theproblem to maximizing and minimizing the values of the sine and cosine functions.

Coin Flipping [75]

In a one-player game, the player begins with 4m fair coins. On each of m turns, the player takes 4unused coins, flips 3 of them randomly to heads or tails, and then selects whether the 4th one is headsor tails (these four coins are then considered used). After m turns, when the sides of all 4m coins havebeen determined, if half the coins are heads and half are tails, the player wins; otherwise, the playerloses.

6. [10] Prove that whenever the player must choose the side of a coin, the optimal strategy is to chooseheads if more coins have been determined tails than heads and to choose tails if more coins have beendetermined heads than tails.

Solution: Let qn(k) be the probability that, with n turns left and |H − T | = k, the player winsplaying the optimal strategy. Note that k is always even. We prove by induction on n that with nturns left in the game, this is the optimal strategy, and that qn(k) ≥ qn(k + 2).

Base Case: n = 1, if there is one more turn in the game, then clearly if we flip 3 coins and get|H − T | ≥ 3 then our play does not matter. If we get |H − T | = 1 then the optimal strategy is to pickthe side so that |H − T | = 0. So this is the optimal strategy.

Now q1(0) = 34 , q1(2) = 1

2 , q1(4) = 18 and q1(2k) = 0 for k ≥ 3. So q1(k) ≥ q1(k + 2) for all k.

Induction Step: Assume the induction hypothesis for n − 1 turns left in the game. With n turns leftin the game, since qn−1(k) decreases when k = |H − T | increases, a larger value of |H − T | is nevermore desirable. So picking the side of the coin that minimizes |H − T | is the optimal strategy.

Now for k ≥ 2, qn(2k) = 18qn−1(2k − 4) + 3

8qn−1(2k − 2) + 38qn−1(2k) + 1

8qn−1(2k + 2), and it is clear,by the induction hypothesis that qn(2k) ≥ qn(2k+ 2) for all k ≥ 2. Similar computations make it clearthat qn(0) ≥ qn(2) ≥ qn(4). This completes the induction step.

7. [15] Let T denote the number of coins determined tails and H denote the number of coins determinedheads at the end of the game. Let pm(k) be the probability that |T −H| = k after a game with 4mcoins, assuming that the player follows the optimal strategy outlined in problem 6. Clearly pm(k) = 0if k is an odd integer, so we need only consider the case when k is an even integer. (By definition,p0(0) = 1 and p0(k) = 0 for k ≥ 1). Prove that pm(0) ≥ pm+1(0) for all nonnegative integers m.

Solution:

First solution. Using the sequences qn(k) defined above, it is clear that qn+1(0) = 34qn(0) + 1

4qn(2) ≤qn(0). So qn+1(0) ≤ qn(0). Now note that for k = 0, pn(k) = qn(k), because they are both equal tothe probability that after n turns of the optimal strategy |H − T | = 0 (this is only true for k = 0). Itfollows that pn+1(0) ≤ pn(0).

Second solution. When we play the game with n + 1 turns, there is a 34 chance that after 1 turn

|H − T | = 0 and a 14 chance that |H − T | = 2.

Now suppose we play the game, and get 2 tails and 1 heads on the first turn. Consider the followingtwo strategies for the rest of the game.

Strategy A: We pick the fourth coin to be heads, and play the optimal strategy for the other n turns.Our probability of winning is pn(0).

Strategy B: We pick the fourth coin to be heads with probability 34 and tails with probability 1

4 , thenproceed with the optimal strategy for the rest of the game. This is equivalent to throwing the first 3coins over again and applying the optimal strategy. Our probability of winning is pn+1(0).

Team Round A

By the theorem in the previous problem, Strategy A is the optimal strategy, and thus our probabilityof winning employing Strategy B does not exceed our probability of wining employing Strategy A. Sopn(0) ≥ pn+1(0).

8. (a) [5] Find a1, a2, and a3, so that the following equation holds for all m ≥ 1:

pm(0) = a1pm−1(0) + a2pm−1(2) + a3pm−1(4)

(b) [5] Find b1, b2, b3, and b4, so that the following equation holds for all m ≥ 1:

pm(2) = b1pm−1(0) + b2pm−1(2) + b3pm−1(4) + b4pm−1(6)

(c) [5] Find c1, c2, c3, and c4, so that the following equation holds for all m ≥ 1 and j ≥ 2:

pm(2j) = c1pm−1(2j − 2) + c2pm−1(2j) + c3pm−1(2j + 2) + c4pm−1(2j + 4)

Solution: We will show how to find a1 in the first equation; other coefficients can be evaluated inthe same way. a1 is the probability that, beginning with |T −H| = 0, the value of |T −H| remains 0after using the optimal strategy for one round. This only happens when the first three coins are notall heads or all tails. Therefore, a1 = 3

4 . It follows that b1 = 14 . The rest of the terms are the binomial

coefficients of the expansion of ( 1+12 )3.

(a) a1 = 34 , a2 = 1

2 , a3 = 18 .

(b) b1 = 14 , b2 = 3

8 , b3 = 38 , b4 = 1

8 .

(c) c1 = 18 , c2 = 3

8 , c3 = 38 , c4 = 1

8 .

9. [15] We would now like to examine the behavior of pm(k) as m becomes arbitrarily large; specifi-cally, we would like to discern whether lim

m→∞pm(0) exists and, if it does, to determine its value. Let

limm→∞

pm(k) = Ak.

(a) [5] Prove that 23pm(k) ≥ pm(k + 2) for all m and k.

(b) [10] Prove that A0 exists and that A0 > 0. Feel free to assume the result of analysis that anon-increasing sequence of real numbers that is bounded below by a constant c converges to alimit that is greater than or equal to c.

Solution:

(a) We proceed by induction.

Base Case: When n = 1, we get 23p1(0) = 2

3 ·34 = 1

2 >14 = p1(2). And since p1(2k) = 0 for k ≥ 2,

we get 23p1(k) ≥ p1(k + 2) for all k.

Induction Step: Assume that 23pi(k) ≥ pi(k + 2) for all i ≤ n. This clearly implies 2

3pn+1(k) ≥pn+1(k + 2) for all k ≥ 4 by the formula (c) from the previous problem.

Now, for k = 2,

2

3pn+1(2) =

2

3

(1

4pn(0) +

3

8pn(2) +

3

8pn(4) +

1

8pn(6)

)≥ 1

12pn(0) + pn+1(4)

≥ pn+1(4)

by the formula (b) from the previous problem and our induction hypothesis.

For k = 0, we write out formula (a) from the previous problem.

2

3pn+1(0) =

2

3

(3

4pn(0) +

1

2pn(2) +

1

8pn(4)

)≥ 1

4pn(0) +

3

8pn(2) +

1

2pn(4) +

1

8pn(6)

Team Round A

by our induction hypothesis. But pn+1(2) = 14pn(0) + 3

8pn(2) + 38pn(4) + 1

8pn(6), which is clearlyless than the last line above. Thus 2

3pn+1(0) ≥ pn+1(2). This completes the induction step.

Remark: This is nowhere near the strongest bound. We intentionally asked for a bound strongenough to be helpful in part b, but not strong enough to help with other problems. The strongest

bound is pn(0) ≥ 2+√5

2 pn(2) and pn(2k) ≥ (2 +√

5)pn(2k+ 2) for k ≥ 1. This is intuitive because

A0 = 2+√5

2 A2 and for k ≥ 1, A2k = (2 +√

5)A2k+2.

(b) For any n and k, pn(2k) ≤ ( 23 )kpn(0). Now

1 =

∞∑k=0

pn(2k) ≤ pn(0)

∞∑k=0

(2

3

)k= 3pn(0).

So pn(0) ≥ 13 for all n. This means pn(0) is a non-increasing sequence that is bounded below by

13 . So A0 exists and A0 ≥ 1

3 > 0.

10. [20] Once it has been demonstrated that limn→∞

pn(0) exists and is greater than 0, it follows that

limn→∞

pn(k) exists and is greater than 0 for all even positive integers k and that∞∑k=0

A2k = 1. It also

follows that A0 = a1A0 +a2A2 +a3A4, A2 = b1A0 + b2A2 + b3A4 + b4A6, and A2j = c1A2j−2 + c2A2j +c3A2j+2 + c4A2j+4 for all positive integers j ≥ 2, where a1, a2, a3, b1, b2, b3, b4, and c1, c2, c3, c4 are theconstants you found in problem 8. Assuming these results, determine, with proof, the value of A0.

Solution: By our recurrence relations,

1

4A0 =

1

2A2 +

1

8A4

5

8A2 =

1

4A0 +

3

8A4 +

1

8A6

and5

8A2k =

1

8A2k−2 +

3

8A2k+2 +

1

8A2k+4

for all k ≥ 2. So the terms A2k (k ≥ 1) satisfy a linear recurrence with characteristic equationx3 + 3x2 + 1 = 5x. The roots of the characteristic equation are 1 and −2 ±

√5, so for k ≥ 2,

A2k = c1 + c2(−2 +√

5)k + c3(−2−√

5)k. But clearly, limk→∞

A2k = 0, so c1 = c3 = 0. Thus for k ≥ 2,

we are left with A2k = c2(−2 +√

5)k = A4(−2 +√

5)k−2.

Now, since∞∑k=0

A2k = 1, we get 1 = A0 + A2 + A4

∞∑j=0

(√

5 − 2)j = A0 + A2 +A4

1− (√

5− 2)=

A0 +A2 +3 +√

5

4A4.

We now have the following equations.

1 = A0 +A2 +3 +√

5

4A4

0 = 2A0 − 4A2 −A4

0 = 2A0 − 5A2 + 3A4 +A6 = 2A0 − 5A2 + (1 +√

5)A4

We have three equations with three unknowns. Solving this linear system gives A0 =√5−12 .

Remark: The fact that A0 =√5−12 > 1

2 means that for arbitrarily large n, the player wins more oftenthan they lose, so the player is at an advantage. In addition the final answer is the reciprocal of thegolden ratio, which is simply beautiful.

The probabilistic experiment explored in this problem is a type of random walk. A random walk is aninfinite set of moves in some set (usually the Rn lattice but sometimes weird shapes or even fractals)

Team Round A

where each move is determined by a random variable. The most basic random walk is a walk on thereal line, where the walker starts at 0, and at every move has probability 1

2 of moving one unit to theleft, and probability 1

2 of moving one unit to the right.

The random walk explored in this problem is what is known as a ’biased’ random walk, where theprobabilities of movement are asymmetric, and weighted towards a certain direction or outcome. Inthis example the choice of the side of 1 out of every 4 coins biases the walk towards the origin.

An interesting inquiry for the reader to ponder is, what is the value of A0 if the game is played with 6coins flipped each turn, or 8, or 2m. If the limits A2k exist, they will satisfy a linear recurrence with

characteristic equation

(x+ 1

2

)2m−1

= xm−1, as is evident from this problem, where we solved the

m = 2 case.

We now give proofs that Ak exists for all even k, and that

∞∑k=0

A2k = 1

First note that we can use the inductive argument in problem 9a to prove the stronger bound pn(0) ≥2+√5

2 pn(2) and pn(2k) ≥ (2 +√

5)pn(2k + 2) for all k ≥ 1.

Now,

pn+2(0) +1

4pn+1(2) =

3

4pn+1(0) +

3

4pn+1(2) +

1

8pn+1(4)

=3

4pn(0) +

43

64pn(2) +

27

64pn(4) +

9

64pn(6) +

1

64pn(8)

= pn+1(0) +11

64pn(2) +

19

64pn(4) +

9

64pn(6) +

1

64pn(8)

= pn+1(0) +1

4pn(2)− 5

64pn(2) +

19

64pn(4) +

9

64pn(6) +

1

64pn(8)

≤ pn+1(0) +1

4pn(2)

So pn+1(0) + pn(2) is a non-increasing sequence that is bounded below, so it converges to a limit, callit A0 +A2. Now A2 = − lim

n→∞pn+1(0) + lim

n→∞(pn+1(0) + pn(2)) = lim

n→∞pn(2).

So limn→∞

pn(2) exists. It follows from our recursions that A2k exists for all k. Since pn(2k) ≥ 0 for all

k, it follows that all the Ai are non-negative.

Now we must show that

∞∑k=0

A2k = 1. This is not immediately clear because, while a finite sum of

limits of sequences equals the limit of the finite sum of the sequences, the sum of the limits does notnecessarily equal the limit of the sum if it is an infinite sum.

Now it follows from part 8a, that for any ε > 0 there exists N such that

N∑k=0

pn(2k) > 1− ε

2for all n.

So

N∑k=0

A2k ≥ 1− ε

2> 1− ε. Since A2k ≥ 0 for all k,

∞∑k=0

A2k ≥ 1.

Now if

∞∑k=0

A2k is greater than 1, or diverges, then there exists a pair j and ε > 0 such that

j∑k=0

A2k =

1 + ε. But,

j∑k=0

pn(2k) ≤ 1 for all n, so

j∑k=0

A2k ≤ 1, contradiction.

So

∞∑k=0

A2k = 1.

Team Round A

Geometry [90]

11. [20] Let ABC be a non-isosceles, non-right triangle, let ω be its circumcircle, and let O be its cir-cumcenter. Let M be the midpoint of segment BC. Let the circumcircle of triangle AOM intersect ωagain at D. If H is the orthocenter of triangle ABC, prove that ∠DAH = ∠MAO.

Solution:

First solution.

A

B CM

X

O

D

H

Let X be the intersection of the line tangent to ω at B with the line tangent to ω at C. Note that4OMC ∼ 4OCX since ∠OMC = ∠OCX = π

2 . Hence OMOC = OC

OX , or, equivalently, OMOA = OA

OX . BySAS similarity, it follows that 4OAM ∼ 4OXA. Therefore, ∠OAM = ∠OXA.

We claim now that ∠OAD = ∠OAX. By the similarity 4OAM ∼ OXA, we have that ∠OAX =∠OMA. Since AOMD is a cyclic quadrilateral, we have that ∠OMA = ∠ODA. Since OA = OD, wehave that ∠ODA = ∠OAD. Combining these equations tells us that ∠OAX = ∠OAD, so A, D, andX are collinear.

Finally, since both AH and OX are perpendicular to BC, it follows that AH ‖ OX, so ∠DAH =∠DXO = ∠AXO = ∠MAO, as desired.

Second solution.

Let Y be the intersection of BC with the line tangent to ω at A. Then the circumcircle of triangleAOM has diameter OY , so AD is perpendicular to OY because the radical axis of two circles isperpendicular to the line between their centers. Since Y is on the polar of A, it follows that A is onthe polar of Y , so AD ⊥ OX implies that AD is the polar of Y , i.e. AD is the symmedian from Ain triangle ABC. Hence AD and AM are isogonal. Since AH and AO are also isogonal, the desiredconclusion follows immediately.

Remark: with sufficient perserverance, angle-chasing solutions involving only the points given in thediagram may also be devised.

12. [70] Let ABC be a triangle, and let E and F be the feet of the altitudes from B and C, respectively.If A is not a right angle, prove that the circumcenter of triangle AEF lies on the incircle of triangleABC if and only if the incenter of triangle ABC lies on the circumcircle of triangle AEF .

Team Round A

Solution:

This problem is arguably the most difficult among all those appearing in the 2011 Harvard-MITMathematics Tournament. Do not feel badly if your team wasted time in a vain attempt to find asolution. It was intended by the author as a serious test for serious geometers.

Let D be the foot of the altitude from A. Let H be the orthocenter of triangle ABC. Let M be themidpoint of AH. Let I be the incenter of triangle ABC. Let ω be the incircle of triangle ABC. Let γbe the circumcircle of AEF . Let η be the nine-point circle of triangle ABC.

We first dispense with the case in which AB = AC. Since M is the circumcenter of triangle AEF , ifthe circumcenter of triangle AEF lies on the incircle of triangle ABC, then ω and η intersect in twopoints: M and D. Since ω and η are tangent by Feuerbach’s Theorem, it follows that ω and η arecoincident, in which case triangle ABC is equilateral.

A

B CD

EF

H

I

M

ω

γ

η

If the incenter of triangle ABC lies on the circumcircle of triangle AEF , then A, I, and H lie on γ.Since no circle may intersect a line in three distinct points and A is distinct from I and H, it followsthat I and H are coincident, in which case triangle ABC is equilateral. Since it is clear that if ABCis equilateral, M lies on ω and I lies on γ, this completes the proof in the case that AB = AC.

Throughout the remainder of this solution, we assume that AB 6= AC. Let Ω be the circumcircle oftriangle ABC, and let α be the A-mixtilinear incircle. Let α be tangent to AB at P , to AC at Q, andto Ω at R. Since AB 6= AC, lines PQ and BC are not parallel. Let their intersection be Z, and letthe bisectors of ∠A, ∠B, and ∠C intersect Ω at T , U , and V , respectively. Let the point diametricallyopposite T on Ω be S. Let lines RS and BC intersect at Y . We introduce a lemma here, namely thatI lies on PQ and that Y and Z are harmonic conjugates with respect to B and C.

Team Round A

A

B CD

E

F

HI

K

M

P

Q

R

S

T

U

V

Y Z

ω

γ

η

Ω

α

Let lines AR and PQ intersect at K. Since PA and QA are tangent to α at P and Q, respectively, RKis the symmedian from R in triangle PQR. By Pascal’s Theorem applied to cyclic hexagon ACV RUB,points Q, I, and P are collinear. Since AP = AQ and AI bisects ∠PAQ, I is the midpoint of PQ.Hence RI is the median from R in triangle PQR. It follows that ∠ICQ = ∠PRA = ∠IRQ and∠IBP = ∠QRA = ∠IRP . Hence quadrilaterals RBPI and RCQI are cyclic, so RI is the bisectorof ∠BRC. It follows also that ∠BIP = ∠BCI, so line PQ is tangent to the circumcircle of triangleBIC at I. Since T is the circumcenter of triangle BIC and S is diametrically opposite T on Ω, BSand CS are tangent to the circumcircle of triangle BIC at B and C, respectively. It follows that IS isthe symmedian from I in triangle BIC; since RI is the bisector of ∠BRC, R, I, and S are collinear.Hence RS is the polar of Z with respect to the circumcircle of triangle BIC, so Y and Z are harmonicconjugates with respect to B and C, as desired.

Team Round A

A

B CD

E

F

HIK

M

P

Q

R

S

T

U

V

X

Y Z

ω

γ

η

Ω

α

We now take up the main problem. We first prove that I lies on γ if M lies on ω. If M lies on ω,then M is the Feuerbach point. Let X be the center of the homothety of positive magnitude mappingω to Ω. Since the center of the homothety of positive magnitude ω to η is M and the center of thehomothety of positive magnitude mapping η to Ω is H, it follows by Monge’s Circle Theorem thatX lies on line MH. Note that R is the center of the homothety of positive magnitude mapping α toΩ. Since the center of homothety of positive magnitude mapping α to ω is A, it follows by the sametheorem that R lies on line AX, so points A, M , H, and R are collinear.

Note that triangle BRC is the reflection of triangle BHC about line BC. Since line RY bisects angleBRC, it follows that line HY bisects ∠BHC. Let the line through H parallel to PQ intersect BCat Z ′. Since ∠HBA = ∠HCA, the internal bisectors of ∠BAC and ∠BHC are parallel. Since PQ isperpendicular to AI and BZ ′ is parallel to PQ, it follows that BZ ′ is the external bisector of ∠BHC.Hence Y and Z ′ are harmonic conjugates with respect to B and C. It follows that Z and Z ′ arecoincident, and, therefore, that H and K are coincident. We may conclude that ∠AIH is right, whichimplies the desired result because γ is the circle on diameter AH. (Do think carefully about why thisargument fails if AB = AC).

Conversely, if I lies on γ, then ∠AIH is right, so points P , H, I, and Q are collinear. Since PQ isperpendicular to AI and the interal bisectors of ∠BAC and ∠BHC are parallel, it follows that PQis the external bisector of ∠BHC. Let the internal bisector of ∠BHC intersect BC at Y ′. Then Y ′

and Z are harmonic conjugates with respect to B and C, so Y and Y ′ are coincident. Let R′ be thereflection of H about line BC. Clearly R′Y is the bisector of ∠BR′C, so R′, Y , and S are collinear.Since R, Y , and S are collinear and R′ also lies on Ω, it follows that R and R′ are coincident. SinceHR′ is perpendicular to BC, it follows that points A, H, and R are collinear.

Let X ′ be the center of the homothety of positive magnitude mapping ω to Ω. Since the center of thehomothety of positive magnitude mapping ω to α is A, and the center of the homothety of positivemagnitude mapping α to Ω is R, it follows by Monge’s Circle Theorem that X ′ lies on line AR. Note

Team Round A

that H is the center of the homothety of positive magnitude mapping Ω to η. Since the center of thehomothety of positive magnitude mapping ω to η is the Feuerbach point, it follows by the same theoremthat the Feuerbach point lies on line AR as well. Since AR intersects η at M and D, the Feuerbachpoint may be either M or D. However, D may not be the Feuerbach point, else the the altitude fromA and the bisector of ∠A in triangle ABC would be coincident, contradicting the assumption thatAB 6= AC. We may conclude that M is the Feuerbach point, so, in particular, M lies on ω, as desired.This completes the proof.

Remark: more computationally intensive solutions are also possible; in particular, it may be observedthat the condition IJ = AM is equivalent to the condition MD/DA′ = IJ/JA′, where J denotesthe point where ω is tangent to BC and A′ denotes the midpoint of BC. It is left to the reader todetermine how to proceed. Let ABC be a triangle, and let E and F be the feet of the altitudes from Band C, respectively. If A is not a right angle, prove that the circumcenter of triangle AEF lies on theincircle of triangle ABC if and only if the incenter of triangle ABC lies on the circumcircle of triangleAEF .

Last Writes [110]

13. [30] Given positive integers a and b such that a > b, define a sequence of ordered pairs (al, bl) fornonnegative integers l by a0 = a, b0 = b, and (al+1, bl+1) = (bl, al mod bl), where, for all positiveintegers x and y, x mod y is defined to be the remainder left by x upon division by y. Define f(a, b)to be the smallest positive integer j such that bj = 0. Given a positive integer n, define g(n) to bemax1≤k≤n−1 f(n, k).

(a) [15] Given a positive integer m, what is the smallest positive integer nm such that g(nm) = m?

(b) [15] What is the second smallest?

Solution:

(a) The answer is Fm+1, where F1 = 1,F2 = 2, and Fi+1 = Fi + Fi−1 for all i ≥ 2.

We consider a reverse sequence as follows: starting at p0 = (k0, 0) for some positive integer k0,at each step we can take a pair pi = (ri, si) to any pair pi+1 = (si + ki+1ri, ri) = (ri+1, si+1) forsome positive integer ki+1. It is clear that any such sequence is the reverse of a legal sequence.Thus, nm is equal to the smallest possible value of the first integer in a possible pm of a reversesequence. The pair pm is uniquely determined by the choice of k0, k1, . . . , km, and lowering any kilowers the first number of pm. Thus, the minimum possible value occurs when all ki are equal to1. In this case, p0 = (1, 0), p1 = (2, 1), and pi+1 = (ri + si, ri). It is apparent that pi = (Fi+1, Fi),so the minimum is Fm+1.

(b) The answer is Lm+1, where L1 = 1,L2 = 3, and Li+1 = Li + Li−1.

The only integer n satisfying g(n) = 1 is 2, as it is the only positive integer n > 1 such that allintegers 1 ≤ i < n divide n. Thus, for m = 1, there is no such second smallest integer.

For m ≥ 2, we once again consider all possible reverse sequences. For any sequence, characterizedby k0, . . . , km, if there are at least two ki not equal to 1, then we can find a smaller possible value(of rm+1) not equal to the minimum possible by setting any one of these ki to 1. Similarly, if anyki is at least 3, we can find a smaller possible value not equal to the minimum by setting thiski to be 2. Thus, the second smallest obtainable value of rm+1 must occur when ki = 1 for all0 ≤ i ≤ m except for some kj which is equal to 2.

We now claim that the value of rm is minimized with respect to the above conditions by lettingk1 = 2, and ki = 1 for all other 0 ≤ i ≤ m. Doing so yields rm = Lm+1, with {Li} defined as in theanswer. Before we begin our proof, we first note that letting k0 = 2 instead yields rm = 2Fm+1,which is strictly larger than Lm+1 (since it is larger for m = 1, 2, andboth sequences satisfy thesame recurrence). We therefore assume that k0 = 1.

Proof of claim. Our recurrences for pairs give us that ri+1 = ki+1ri+si, si+1 = ri. Thus, we haveri+1 = ki+1ri + ri−1. Now suppose we have kj = 2 for a particular j, and for all i 6= j, 0 ≤ i ≤ m,

Team Round A

we have ki = 1. Then we have r0 = 1, r1 = 1 + k1, and for 2 ≤ i ≤ m, i 6= j, we haveri = ri−1 + ri−2. We also have rj = 2rj−1 + rj−2.

By our reasoning in part (a), ri = Fi+1 for i < j. Thus, rj = 2Fj + Fj−1 = Fj+1 + Fj = Fj+2

(where we define F0 = 1). Expressing rj−1 as Fj+1 − Fj−1, we find rj+1 = Fj+3 − Fj−1, rj+2 =Fj+4 − Fj−1, rj+3 = Fj+5 − 2Fj−1, and so on. It is easy to show by induction that for i > j, wehave ri = Fi+2 − Fi−j−1Fj−1. Thus, to minimize rm = Fm+2 − Fm−j−1Fj−1, we must maximizeFm−j−1Fj−1.

We claim that Fm−j−1Fj−1 ≤ Fm−2 for 1 ≤ j ≤ m.

Proof of claim. It is easy to see (and in fact well-known) that for all positive integers n, Fn countsthe number of distinct ways to tile a 1 × n board with 1 × 1 squares and 1 × 2 dominoes. ThusFm−j−1Fj−1 counts the number of ways to tile a 1 ×m − 2 board with 1 × 1 squares and 1 × 2dominoes such that the j − 1th and jth square are not both covered by the same domino, whichis at most the number of ways to tile a 1×m− 2 board, as desired.

Since this minimum is reached by letting j = 1, we may conclude that the answer is indeed Lm+1.

14. [25] Given a positive integer n, a sequence of integers a1, a2, . . . , ar, where 0 ≤ ai ≤ k for all 1 ≤ i ≤ r,is said to be a “k-representation” of n if there exists an integer c such that

r∑i=1

ai =

r∑i=1

aikc−i = n.

Prove that every positive integer n has a k-representation, and that the k-representation is unique ifand only if 0 does not appear in the base-k representation of n− 1.

Solution: Given a positive integer n, a sequence of integers a1, a2, . . . , ar, where 0 ≤ ai ≤ k for all1 ≤ i ≤ r, is said to be a “k-representation” of n if there exists an integer c such that

r∑i=1

ai =

r∑i=1

aikc−i = n.

Prove that every positive integer n has a k-representation, and that the k-representation is unique ifand only if 0 does not appear in the base-k representation of n− 1.

Equivalently, a k-representation is given by a sequence ap, . . . , aq, for some p < q, such that

q∑i=p

ai =

q∑i=p

aiki.

We first show existence. Let the representation of n − 1 in base k be given by∑ri=0 aik

i = n − 1.Let l =

∑ri=1 ai(k

i−1 + ki−2 + · · · + 1). Now we extend the sequence {ai} by letting ai = k − 1 if−l + 1 ≤ i ≤ −1 and a−l = k. We claim that a−l, . . . , ar−1, ar is a k-representation of n. Indeed,

r∑i=−l

aiki = (n− 1) +

( −1∑i=−l

(k − 1)ni

)+ k−l

= (n− 1) + (k − 1)k−11− k−l

1− k−1+ k−l = n

r∑i=−l

ai =

(r∑i=0

ai

)+ l(k − 1) + 1

=

(r∑i=0

ai

)+

(r∑i=1

ai(ki − 1)

)+ 1

=

(r∑i=0

aiki

)+ 1 = n

Team Round A

Now suppose there is another k-representation {bi} of n; then∑bi =

∑bik

i = n. This implies that∑ci =

∑cik

i = 0 where ci = ai − bi. The following claim specifies all possibilities of {ci}.Claim: Suppose that

∑qi=p cik

i = 0, where ci, p, q ∈ Z, p < q, and ci ∈ [−k, k]. Then the sequence {ci}must be the concatenation of subsequences of the form

±(1, 1− k, 1− k, . . . , 1− k,−k)

, possibly with 0’s in between.

Proof. If ci is not the zero sequence, then without loss of generality, we may assume cq > 0.

Since∑qi=p cik

i = 0, we have

|cqkq| = |cq−1kq−1 + . . .+ cpkp| ≤ kq + kq−1 + . . .+ kp+1 <

k

k − 1kq ≤ 2kq

so cq = 1. Now we havekq + cq−1k

q−1 + . . .+ cpkp = 0. (1)

This means (k + cq−1)kq−1 + . . .+ cpkp = 0. Hence, as above,

|(k + cq−1)kq−1| = |cq−2kq−2 + . . .+ cpkp| < 2kq−1.

Therefore, |(k + cq−1)| < 2, which means cq−1 = −k or − k + 1.

If cq−1 = −k, then cqkq + cq−1k

q−1 = 0, and we get a subsequence (1,−k).

If cq−1 = −k + 1, then cqkq + cq−1k

q−1 = kq−1. Thus

kq−1 + cq−2kq−2 + . . .+ cpk

p = 0,

which has the exact same form as (1). We can then repeat this procedure to obtain a subsequence(1, 1− k, 1− k, . . . , 1− k,−k).

Once we have such a subsequence, the terms in the sum∑qi=p cik

i = 0 corresponding to that subse-quence sum to 0, so we may remove them and apply the same argument.

We now consider the cases.

If the base k representation of n−1 contains no 0’s, then ai 6= 0 so it is impossible to have ci = ai−bi =−k. On the other hand, we know that ci is composed of subsequences of the form±(1, 1−k, 1−k, . . . , 1−k,−k). Therefore, if {ci} is not the zero sequence, then the fact that

∑ci = 0 implies that we must

have both a subsequence (1, 1−k, 1−k, . . . , 1−k,−k) and a subsequence −(1, 1−k, 1−k, . . . , 1−k,−k),meaning that there exists i for which ci = −k, contradiction.

If the base k representation of n − 1 contains a 0, then picking the largest i such that ai = 0, wecan change ai to k and ai+1 to ai+1 − 1, and append a (1,−k) to the end of the sequence. Thisyields another k-representation of n, so a k-representation of n is unique if and only if the base krepresentation of n− 1 contains no 0’s, as desired.

15. [55] Denote {1, 2, ..., n} by [n], and let S be the set of all permutations of [n]. Call a subset T of Sgood if every permutation σ in S may be written as t1t2 for elements t1 and t2 of T , where the productof two permutations is defined to be their composition. Call a subset of U of S extremely good if everypermutation σ in S may be written as s−1us for elements s of S and u of U . Let τ be the smallestvalue of |T |/|S| for all good subsets T , and let υ be the smallest value of |U |/|S| for all extremely goodsubsets U . Prove that

√υ ≥ τ .

Solution: Denote {1, 2, ..., n} by [n], and let S be the set of all permutations of [n]. Call a subsetT of S good if every permutation σ in S may be written as t1t2 for elements t1 and t2 of T , wherethe product of two permutations is defined to be their composition. Call a subset of U of S extremelygood if every permutation σ in S may be written as s−1us for elements s of S and u of U . Let τ be

Team Round A

the smallest value of |T |/|S| for all good subsets T , and let υ be the smallest value of |U |/|S| for allextremely good subsets U . Prove that

√υ ≥ τ .

Call an element t ∈ S an involution if and only if t2 is the identity permutation. We claim that theset of all involutions in S constitutes a good subset of S. The proof is simple. Let s be an arbitrarypermutation in S. Note that s may be decomposed into a product of disjoint cycles of length at least2, and suppose that there are m such cycles in its decomposition. For 1 ≤ i ≤ m, let li denote thelength of the ith cycle, so that s may be written as

(a1,1a1,2 . . . a1,l1)(a2,1a2,2 . . . a2,l2) . . . (am,1am,2 . . . am,lm)

for some pairwise distinct elements

a1,1, a1,2, . . . , a1,l1 , a2,1, a2,2, . . . , a2,l2 , . . . , am,1, am,2, . . . , am,lm

of [n]. Consider the permutations q and r defined by q(ai,j) = ai,li+1−j for all 1 ≤ j ≤ li andr(ai,1) = ai,1 and r(ai,k) = ai,li+2−k for all 2 ≤ k ≤ li, for all 1 ≤ i ≤ m, and by q(x) = r(x) = x forall x ∈ [n] otherwise. Since q, r ∈ S, q2 = r2 = 1, and rq = s, it follows that the set of all involutionsin S is indeed good, as desired.

For all integer partitions λ of n, let fλ denote the number of standard Young tableaux of shape λ. Bythe Robinson-Schensted-Knuth correspondence, the permutations of [n] are in bijection with pairs ofstandard Young tableaux of the same shape in such a way that the involutions of [n] are in bijectionwith pairs of identical standard Young tableaux. In other words, the number of permutations of [n] isequal to the number of pairs of identically shaped standard Young tableaux whose shape is a partitionof n, and the number of involutions of [n] is equal to the number of standard Young tableaux whose

shape is a partition of n. Hence n! =∑λ(fλ)2 and τ ≤

∑λfλ

n! , where both sums range over allpartitions λ of n.

For all elements u ∈ S, define the conjugacy class of u to be the set of elements that may be writtenin the form s−1us for some s ∈ S. It is easy to see that for all u, u′ ∈ S, the conjugacy classes ofu and u′ are either identical or disjoint. It follows that S may be partitioned into disjoint conjugacyclasses and that any extremely good subset of S must contain at least one element from each distinctconjugacy class. We claim that the number of distinct conjuguacy classes of S is at least the numberof integer partitions of n. It turns out that the two numbers are in fact equal, but such a result is notnecessary for the purposes of this problem. Let u be an arbitrary permutation in S. Recall that u maybe written as

(a1,1a1,2 . . . a1,l1)(a2,1a2,2 . . . a2,l2) . . . (am,1am,2 . . . am,lm)

for some pairwise distinct elements

a1,1, a1,2, . . . , a1,l1 , a2,1, a2,2, . . . , a2,l2 , . . . , am,1, am,2, . . . , am,lm

of [n]. Associate to the conjugacy class of u the partition n = lw1 + lw2 + . . .+ lwm + 1 + . . .+ 1, wherew1, w2, . . . , wm is a partition of 1, 2, . . . ,m such that w1 ≥ w2 ≥ . . . ≥ wm and the 1’s represent all thefixed points of s. Now note that if s is any permutation in S, s−1us may be written in the form

s−1(a1,1a1,2 . . . a1,l1)ss−1(a2,1a2,2 . . . a2,l2)s . . . s−1(am,1am,2 . . . am,lm)s,

which may be written as

(b1,1b1,2 . . . b1,l1)(b2,1b2,2 . . . b2,l2) . . . (bm,1bm,2 . . . bm,lm)

for some pairwise distinct elements

b1,1, b1,2, . . . , b1,l1 , b2,1, b2,2, . . . , b2,l2 , . . . , bm,1, bm,2, . . . , bm,lm

of [n] because multiplying by s−1 on the left and by s on the right is equivalent to re-indexing theletters 1, 2, . . . , n. Hence if partitions of n are associated to all conjugacy classes of S analogously, the

Team Round A

same partition that is associated to u is associated to s−1us for all s ∈ S. Since exactly one partitionis associated to each conjugacy class, it follows that the number of conjugacy classes cannot exceedthe number of partitions, as desired.

To conclude, we need only observe that υ ≥∑λ

1n! by our above claim, for then

n!√υ =

√n!∑λ

1 =

√√√√(∑λ

(fλ)2

)(∑λ

1

)≥∑λ

fλ ≥ n!τ

by the Cauchy-Schwarz inequality, and this completes the proof.

Remark: more computationally intensive solutions that do not use Young tableaux but instead calculatethe number of involutions explicitly and make use of the generating function for the number of integerpartitions are also possible.

Team Round A

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

Team Round B

Think Carefully! [55]

The problems in this section require only short answers.

1. [55] Tom, Dick, and Harry play a game in which they each pick an integer between 1 and 2011. Tompicks a number first and informs Dick and Harry of his choice. Then Dick picks a different numberand informs Harry of his choice. Finally, Harry picks a number different from both Tom’s and Dick’s.After all the picks are complete, an integer is randomly selected between 1 and 2011. The player whosenumber is closest wins 2 dollars, unless there is a tie, in which case each of the tied players wins 1dollar. If Tom knows that Dick and Harry will each play optimally and select randomly among equallyoptimal choices, there are two numbers Tom can pick to maximize his expected profit; what are they?

Complex Numbers [90]

The problems in this section require only short answers.

The norm of a complex number z = a + bi, denoted by |z|, is defined to be√a2 + b2. In the following

problems, it may be helpful to note that the norm is multiplicative and that it obeys the triangleinequality. In other words, please observe that for all complex numbers x and y, |xy| = |x||y| and|x + y| ≤ |x|+ |y|. (You may verify these facts for yourself if you like).

2. [20] Let a, b, and c be complex numbers such that |a| = |b| = |c| = |a + b + c| = 1. If |a− b| = |a− c|and b 6= c, evaluate |a + b||a + c|.

3. [30] Let x and y be complex numbers such that |x| = |y| = 1.

(a) [15] Determine the maximum value of |1 + x|+ |1 + y| − |1 + xy|.(b) [15] Determine the maximum value of |1+x|+ |1+xy|+ |1+xy2|+ . . .+ |1+xy2011|−1006|1+y|.

4. [40] Let a, b, and c be complex numbers such that |a| = |b| = |c| = 1. If

a2

bc+

b2

ca+

c2

ab= 1

as well, determine the product of all possible values of |a + b + c|.

Warm Up Your Proof Skills! [40]

The problems in this section require complete proofs.

5. (a) [20] Alice and Barbara play a game on a blackboard. At the start, zero is written on the board.The players alternate turns. On her turn, each player replaces x – the number written on theboard – with any real number y, subject to the constraint that 0 < y − x < 1. The first playerto write a number greater than or equal to 2010 wins. If Alice goes first, determine, with proof,who has the winning strategy.

(b) [20] Alice and Barbara play a game on a blackboard. At the start, zero is written on the board.The players alternate turns. On her turn, each player replaces x – the number written on theboard – with any real number y, subject to the constraint that y − x ∈ (0, 1). The first player towrite a number greater than or equal to 2010 on her 2011th turn or later wins. If a player writesa number greater than or equal to 2010 on her 2010th turn or before, she loses immediately. IfAlice goes first, determine, with proof, who has the winning strategy.

Page 1 of 2

The Euler Totient Function [40]

The problems in this section require complete proofs.

Euler’s totient function, denoted by ϕ, is a function whose domain is the set of positive integers. It isespecially important in number theory, so it is often discussed on the radio or on national TV. (Justkidding). But what is it, exactly? For all positive integers k, ϕ(k) is defined to be the number ofpositive integers less than or equal to k that are relatively prime to k. It turns out that ϕ is what youwould call a multiplicative function, which means that if a and b are relatively prime positive integers,ϕ(ab) = ϕ(a)ϕ(b). Unfortunately, the proof of this result is highly nontrivial. However, there is muchmore than that to ϕ, as you are about to discover!

6. [10] Let n be a positive integer such that n > 2. Prove that ϕ(n) is even.

7. [10] Let n be an even positive integer. Prove that ϕ(n) ≤ n2 .

8. [20] Let n be a positive integer, and let a1, a2, . . . , an be a set of positive integers such that a1 = 2 andam = ϕ(am+1) for all 1 ≤ m ≤ n− 1. Prove that an ≥ 2n∈1.

Introduction to the Symmedian [70]

The problems in this section require complete proofs.

If A, B, and C are three points in the plane that do not all lie on the same line, the symmedian fromA in triangle ABC is defined to be the reflection of the median from A in triangle ABC about thebisector of angle A. Like the ϕ function, it turns out that the symmedian satisfies some interestingproperties, too. For instance, just like how the medians from A, B, and C all intersect at the centroidof triangle ABC, the symmedians from A, B, and C all intersect at what is called (no surprises here)the symmedian point of triangle ABC. The proof of this fact is not easy, but it is unremarkable. Inthis section, you will investigate some surprising alternative constructions of the symmedian.

9. [25] Let ABC be a non-isosceles, non-right triangle, let ω be its circumcircle, and let O be its circum-center. Let M be the midpoint of segment BC. Let the tangents to ω at B and C intersect at X.Prove that ∠OAM = ∠OXA. (Hint: use SAS similarity).

10. [15] Let the circumcircle of triangle AOM intersect ω again at D. Prove that points A, D, and X arecollinear.

11. [10] Let H be the intersection of the three altitudes of triangle ABC. (This point is usually called theorthocenter). Prove that ∠DAH = ∠MAO.

12. [10] Prove that line AD is the symmedian from A in triangle ABC by showing that ∠DAB = ∠MAC.

13. [10] Prove that line AD is also the symmedian from D in triangle DBC.

Last Writes [65]

The problems in this section require complete proofs.

14. [25] Rachel and Brian are playing a game in a grid with 1 row of 2011 squares. Initially, there is onewhite checker in each of the first two squares from the left, and one black checker in the third squarefrom the left. At each stage, Rachel can choose to either run or fight. If Rachel runs, she moves theblack checker moves 1 unit to the right, and Brian moves each of the white checkers one unit to theright. If Rachel chooses to fight, she pushes the checker immediately to the left of the black checker 1unit to the left; the black checker is moved 1 unit to the right, and Brian places a new white checkerin the cell immediately to the left of the black one. The game ends when the black checker reaches thelast cell. How many different final configurations are possible?

15. [40] On Facebook, there is a group of people that satisfies the following two properties: (i) there existsa positive integers k such that any subset of 2k − 1 people in the group contains a subset of k peoplein the group who are all friends with each other, and (ii) every member of the group has 2011 friendsor fewer.

(a) [15] If k = 2, determine, with proof, the maximum number of people the group may contain.

(b) [25] If k = 776, determine, with proof, the maximum number of people the group may contain.

Page 2 of 2

14th Annual Harvard-MIT Mathematics TournamentSaturday 12 February 2011

Team Round B

Think Carefully! [55]

The problems in this section require only short answers.

1. [55] Tom, Dick, and Harry play a game in which they each pick an integer between 1 and 2011. Tompicks a number first and informs Dick and Harry of his choice. Then Dick picks a different numberand informs Harry of his choice. Finally, Harry picks a number different from both Tom’s and Dick’s.After all the picks are complete, an integer is randomly selected between 1 and 2011. The player whosenumber is closest wins 2 dollars, unless there is a tie, in which case each of the tied players wins 1dollar. If Tom knows that Dick and Harry will each play optimally and select randomly among equallyoptimal choices, there are two numbers Tom can pick to maximize his expected profit; what are they?

Answer: 503, 1509 Let x denote the number Tom chooses. By the symmetry of the problem,

picking x and picking 2012− x yield the same expected profit. If Tom picks 1006, Dick sees that if hepicks 1007, Harry’s best play is to pick 1005, and Dick will win with probability 1005

2011 , and clearly this isthe best outcome he can achieve. So Dick will pick 1007 (or 1005) and Harry will pick 1005 (or 1007),and Tom will win with probability 1

2011 . Picking the number 2 will make the probability of winning atleast 2

2011 since Dick and Harry would both be foolish to pick 1, so picking 1006 is suboptimal. It isnow clear that the answer to the problem is some pair (x, 2012− x).

By the symmetry of the problem we can assume without loss of generality that 1 ≤ x ≤ 1005. Wewill show that of these choices, x = 503 maximizes Tom’s expected profit. The trick is to examine therelationship between Tom’s choice and Dick’s choice. We claim that (a) if x > 503, Dick’s choice is2013− x and (b) if x < 503, Dick’s choice is 1341 + bx+1

3 c. Let y denote the number Dick chooses.Proof (a). Note first that if x > 503, 2013−x is the smallest value of y for which Harry always choosesa number less than x. This is obvious, for if y < 2011−x, a choice of 2012−x will give Harry a greaterexpected profit than a choice of any number less than x, and if y = 2012 − x, Harry never choosesa number between x and y since x − 1 > 2012∈x∈x

2 = 1006 − x holds for all integers x greater than503, so, by symmetry, Harry chooses a number less than x exactly half the time. The desired resultactually follows immediately because the fact that Harry always chooses a number less than x wheny = 2013− x implies that there would be no point in Dick choosing a number greater than 2013− x,so it suffices to compare Dick’s expected profit when y = 2013− x with that of all smaller values of y,which is trivial.Proof (b). This case is somewhat more difficult. First, it should be obvious that if x < 503, Harrynever chooses a number less than x. The proof is by contradiction. If y ≤ 2011 − x, Harry canobtain greater expected profit by choosing 2012 − x than by choosing any number less than x. Ify ≥ 1002 + x, Harry can obtain greater expected profit by choosing any number between x and ythan by choosing a number less than x. Hence if Harry chooses a number less than x, x and y mustsatisfy y > 2011 − x and y < 1002 + x, which implies 2x > 2009, contradiction. We next claim thatif y ≥ 1342 + bx+1

3 c, then Harry chooses a number between x and y. The proof follows from the

inequality1342+b x+1

3 c∈x2 > 2011 − (1342 + bx+1

3 c), which is equivalent to 3bx+13 c > x − 4, which is

obviously true. We also claim that if y ≤ 1340 + bx+13 c then Harry chooses a number greater than y.

The proof follows from the inequality1340+b x+1

3 c∈x2 < 2011 − (1340 + bx+1

3 c), which is equivalent to3bx+1

3 c < x + 2, which is also obviously true. Now if x 6≡ 1 (mod 3), then we have 3bx+13 c > x − 1,

so, in fact, if y = 1341 + bx+13 c, Harry still chooses a number between x and y. In this case it is clear

that such a choice of y maximizes Dick’s expected profit. It turns out that the same holds true evenif x ≡ 1 (mod 3); the computations are only slightly more involved. We omit them here because theybear tangential relation to the main proof.We may conclude that if x > 503, the optimal choice for x is 504, and if x < 503, the optimal choicefor x is 502. In the first case, y = 1509, and, in the second case, y = 1508. Since a choice of x = 503and y = 1509 clearly outperforms both of these combinations when evaluated based on Tom’s expected

Team Round B

profit, it suffices now to show that if Tom chooses 503, Dick chooses 1509. Since the computations areroutine and almost identical to those shown above, the proof is left as an exercise to the reader.Remark: For the more determined, intuition alone is sufficient to see the answer. Assuming Tom’snumber x is less than 1006, Dick will never pick a number y that makes Harry’s best move y + 1, soHarry’s best move will either be x − 1 or anything in between x and y. Clearly Tom does not wantHarry to pick x−1, and the biggest number he can pick so that Harry will have to pick in between himand Dick is 1

4 th of the way along the set of numbers, which in this case is 503. In a rigorous solution,however, the calculations must be handled with grave care.

An interesting inquiry is, what is the optimal strategy with n players?

Complex Numbers [90]

The problems in this section require only short answers.

The norm of a complex number z = a+ bi, denoted by |z|, is defined to be√a2 + b2. In the following

problems, it may be helpful to note that the norm is multiplicative and that it obeys the triangleinequality. In other words, please observe that for all complex numbers x and y, |xy| = |x||y| and|x+ y| ≤ |x|+ |y|. (You may verify these facts for yourself if you like).

2. [20] Let a, b, and c be complex numbers such that |a| = |b| = |c| = |a+ b+ c| = 1. If |a− b| = |a− c|and b 6= c, evaluate |a+ b||a+ c|.

Answer: 2 First solution.

Since |a| = 1, a cannot be 0. Let u = ba and v = c

a . Dividing the given equations by |a| = 1 gives|u| = |v| = |1 + u+ v| = 1 and |1− u| = |1− v|. The goal is to prove that |1 + u||1 + v| = 2.

By squaring |1 − u| = |1 − v|, we get (1 − u)(1− u) = (1 − v)(1− v), and thus 1 − u − ū + |u|2 =1− v − v̄ + |v|2, or u+ ū = v + v̄. This implies Re(u) = Re(v). Since u and v are on the unit circle inthe complex plane, u is equal to either v or v̄. However, b 6= c implies u 6= v, so u = v̄.

Therefore, 1 = |1+u+ ū| = |1+2 Re(u)|. Since Re(u) is real, we either have Re(u) = 0 or Re(u) = −1.The first case gives u = ±i and |1 + u||1 + v| = |1 + i||1 − i| = 2, as desired. It remains only to notethat Re(u) = −1 is in fact impossible because u is of norm 1 and u = −1 would imply u = ū = v.

Remark: by the rotational symmetry of the circle, it is acceptable to skip the first step of this solutionand assume a = 1 without loss of generality.

Second solution.

Let a, b, and c, be the vertices of a triangle inscribed in the unit circle in the complex plane. Sincethe complex coordinate of the circumcenter is 0 and the complex coordinate of the centroid is a+b+c

3 ,it follows from well-known facts about the Euler line that the complex coordinate of the orthocenteris a+ b+ c. Hence the orthocenter lies on the unit circle as well. Is it not possible for the orthocenternot to be among the three vertices of the triangle, for, if it were, two opposite angles of the convexcyclic quadrilateral formed by the three vertices and the orthocenter would each measure greater than90 degrees. It follows that the triangle is right. However, since |a− b| = |a− c|, the right angle cannotoccur at b or c, so it must occur at a, and the desired conclusion follows immediately.

3. [30] Let x and y be complex numbers such that |x| = |y| = 1.

(a) [15] Determine the maximum value of |1 + x|+ |1 + y| − |1 + xy|.(b) [15] Determine the maximum value of |1+x|+ |1+xy|+ |1+xy2|+ . . .+ |1+xy2011|−1006|1+y|.

Solution:

(a) Answer: 2√

2

(b) Answer: 2012√

2 We divide the terms into 1006 sums of the form

|1 + xy2k|+ |1 + xy2k+1| − |1 + y|.

Team Round B

For each of these, we obtain, as in part a,∣∣1 + xy2k∣∣+∣∣1 + xy2k+1

∣∣− |1 + y| ≤∣∣1 + xy2k

∣∣+∣∣xy2k+1 − y

∣∣=∣∣1 + xy2k

∣∣+ |y|∣∣1− xy2k∣∣

=∣∣1 + xy2k

∣∣+∣∣1− xy2k∣∣

≤ 2√|1− x2y4k|

Again, this is maximized when x2y4 = −1, which leaves a total sum of at most 2012√

2. If we let

x = i and y = −1, every sum of three terms will be 2√

2, for a total of 2012√

2 , as desired.

4. [40] Let a, b, and c be complex numbers such that |a| = |b| = |c| = 1. If

a2

bc+b2

ca+c2

ab= 1

as well, determine the product of all possible values of |a+ b+ c|.Answer: 2 Let s = a+ b+ c. Then

s3 = a3 + b3 + c3 + 3(a2b+ ab2 + b2c+ bc2 + c2a+ ca2) + 6abc

= abc(a2

bc+b2

ca+c2

ab+ 3(

a

b+b

a+b

c+c

b+c

a+a

c) + 6)

= abc(1 + (3(a+ b+ c)(1

a+

1

b+

1

c)− 9) + 6)

= abc(3s(1

a+

1

b+

1

c)− 2)

= abc(3ss̄− 2) (because s̄ = ā+ b̄+ c̄ = 1/a+ 1/b+ 1/c)

= abc(3|s|2 − 2)

Taking absolute values, we find |s|3 = |3|s|2 − 2|. It follows that |s| must be a positive real root ofx3 − 3x2 + 2 = 0 or x3 + 3x2 − 2 = 0. However, since the negative real roots of x3 − 3x2 + 2 = 0are exactly the additive inverses of the positive real roots of x3 − 3x2 + 2 = 0, and all three roots ofx3 − 3x2 + 2 = 0 are real (x3 − 3x2 + 2 = 0 may be factored as (x − 1)(x2 − 2x − 2) = 0, and thediscriminant of x2 − 2x− 2 is positive), the product of all possible values of |s| is (−2) · (−1)n, wheren denotes the number of negative real roots of x3 − 3x2 + 2 = 0. By Descartes’s Rule of Signs, we seethat n is odd, so the answer is 2, as desired.

Warm Up Your Proof Skills! [40]

The problems in this section require complete proofs.

5. (a) [20] Alice and Barbara play a game on a blackboard. At the start, zero is written on the board.The players alternate turns. On her turn, each player replaces x – the number written on theboard – with any real number y, subject to the constraint that 0 < y − x < 1. The first playerto write a number greater than or equal to 2010 wins. If Alice goes first, determine, with proof,who has the winning strategy.

(b) [20] Alice and Barbara play a game on a blackboard. At the start, zero is written on the board.The players alternate turns. On her turn, each player replaces x – the number written on theboard – with any real number y, subject to the constraint that y − x ∈ (0, 1). The first player towrite a number greater than or equal to 2010 on her 2011th turn or later wins. If a player writesa number greater than or equal to 2010 on her 2010th turn or before, she loses immediately. IfAlice goes first, determine, with proof, who has the winning strategy.

Solution: Each turn of the game is equivalent to adding a number between 0 and 1 to the numberon the board.

Team Round B

(a) Barbara has the winning strategy: whenever Alice adds z to the number on the board, Barbaraadds 1−z. After Barbara’s ith turn, the number on the board will be i. Therefore, after Barbara’s2009th turn, Alice will be forced to write a number between 2009 and 2010, after which Barbaracan write 2010 and win the game.

(b) Alice has the winning strategy: she writes any number a on her first turn, and, after that, wheneverBarbara adds z to the number on the board, Alice adds 1− z. After Alice’s ith turn, the numberon the board will be (i− 1) + a, so after Alice’s 2010th turn, the number will be 2009 + a. SinceBarbara cannot write a number greater than or equal to 2010 on her 2010th turn, she will beforced to write a number between 2009 + a and 2010, after which Alice can write 2010 and winthe game.

The Euler Totient Function [40]

The problems in this section require complete proofs.

Euler’s totient function, denoted by ϕ, is a function whose domain is the set of positive integers. It isespecially important in number theory, so it is often discussed on the radio or on national TV. (Justkidding). But what is it, exactly? For all positive integers k, ϕ(k) is defined to be the number ofpositive integers less than or equal to k that are relatively prime to k. It turns out that ϕ is what youwould call a multiplicative function, which means that if a and b are relatively prime positive integers,ϕ(ab) = ϕ(a)ϕ(b). Unfortunately, the proof of this result is highly nontrivial. However, there is muchmore than that to ϕ, as you are about to discover!

6. [10] Let n be a positive integer such that n > 2. Prove that ϕ(n) is even.

Solution: Let An be the set of all positive integers x ≤ n such that gcd(n, x) = 1. Since gcd(n, x) =gcd(n, n − x) for all x, if a is a positive integer in An, so is n − a. Moreover, if a is in An, a andn− a are different since gcd(n, n2 ) = n

2 > 1, meaning n2 is not in An. Hence we may evenly pair up the

elements of An that sum to n, so ϕ(n), the number of elements of An, must be even, as desired.

7. [10] Let n be an even positive integer. Prove that ϕ(n) ≤ n2 .

Solution: Again, let An be the set of all positive integers x ≤ n such that gcd(n, x) = 1. Since n iseven, no element of An may be even, and, by definition, every element of An must be at most n. Itfollows that ϕ(n), the number of elements of An, must be at most n

2 , as desired.

8. [20] Let n be a positive integer, and let a1, a2, . . . , an be a set of positive integers such that a1 = 2 andam = ϕ(am+1) for all 1 ≤ m ≤ n− 1. Prove that an ≥ 2n∈1.

Solution: From problem six, it follows that am is even for all m ≤ n − 1. From problem seven, itfollows that am ≥ 2am∈1 for all m ≤ n − 1. We may conclude that an ≥ an∈1 ≥ 2n∈2a1 = 2n∈1, asdesired.

Introduction to the Symmedian [70]

The problems in this section require complete proofs.

If A, B, and C are three points in the plane that do not all lie on the same line, the symmedian fromA in triangle ABC is defined to be the reflection of the median from A in triangle ABC about thebisector of angle A. Like the ϕ function, it turns out that the symmedian satisfies some interestingproperties, too. For instance, just like how the medians from A, B, and C all intersect at the centroidof triangle ABC, the symmedians from A, B, and C all intersect at what is called (no surprises here)the symmedian point of triangle ABC. The proof of this fact is not easy, but it is unremarkable. Inthis section, you will investigate some surprising alternative constructions of the symmedian.

9. [25] Let ABC be a non-isosceles, non-right triangle, let ω be its circumcircle, and let O be its circum-center. Let M be the midpoint of segment BC. Let the tangents to ω at B and C intersect at X.Prove that ∠OAM = ∠OXA. (Hint: use SAS similarity).

Team Round B

Solution:

A

B CM

X

O

D

H

Note that 4OMC ∼ 4OCX since ∠OMC = ∠OCX = π2 . Hence OM

OC = OCOX , or, equivalently,

OMOA = OA

OX . By SAS similarity, it follows that 4OAM ∼ 4OXA. Therefore, ∠OAM = ∠OXA.

10. [15] Let the circumcircle of triangle AOM intersect ω again at D. Prove that points A, D, and X arecollinear.

Solution: By the similarity 4OAM ∼ OXA, we have that ∠OAX = ∠OMA. Since AOMD is acyclic quadrilateral, we have that ∠OMA = ∠ODA. Since OA = OD, we have that ∠ODA = ∠OAD.Combining these equations tells us that ∠OAX = ∠OAD, so A, D, and X are collinear, as desired.

11. [10] Let H be the intersection of the three altitudes of triangle ABC. (This point is usually called theorthocenter). Prove that ∠DAH = ∠MAO.

Solution: Note that since both AH and OX are perpendicular to BC, it follows that AH ‖ OX, so∠DAH = ∠DXO = ∠AXO = ∠MAO, as desired.

12. [10] Prove that line AD is the symmedian from A in triangle ABC by showing that ∠DAB = ∠MAC.

Solution: First, we have that

∠OAC = 12 (π − ∠AOC) = π

2 − ∠ABC = ∠HAB

so, by the previous problem, we obtain

∠DAB = ∠DAH + ∠HAB = ∠MAO + ∠OAC = ∠MAC.

It follows that the lines AD and AM are reflections of each other across the bisector of ∠BAC, so ADis the symmedian from A in triangle ABC, as desired.

13. [10] Prove that line AD is also the symmedian from D in triangle DBC.

Solution: In the previous problems, we showed that the symmedian from A in triangle ABC is theline AX, where X is the intersection of the tangents to the circumcircle at B and C. Since the choiceof triangle was arbitrary, we may conclude that DX is the symmedian from D in triangle DBC. SinceA, D, and X, are collinear, this completes the proof.

Team Round B

Last Writes [65]

The problems in this section require complete proofs.

14. [25] Rachel and Brian are playing a game in a grid with 1 row of 2011 squares. Initially, there is onewhite checker in each of the first two squares from the left, and one black checker in the third squarefrom the left. At each stage, Rachel can choose to either run or fight. If Rachel runs, she moves theblack checker moves 1 unit to the right, and Brian moves each of the white checkers one unit to theright. If Rachel chooses to fight, she pushes the checker immediately to the left of the black checker 1unit to the left; the black checker is moved 1 unit to the right, and Brian places a new white checkerin the cell immediately to the left of the black one. The game ends when the black checker reaches thelast cell. How many different final configurations are possible?

Answer: 2009 Both operations, run and fight, move the black checker exactly one square to theright, so the game will end after exactly 2008 moves regardless of Brian’s choices. Furthermore, it iseasy to see that the order of the operations does not matter, so two games with the same number offights will end up in the same final configuration. Finally, note that each fight adds one white checkerto the grid, so two games with different numbers of fights will end up in different final configurations.There are 2009 possible values for the number of fights, so there are 2009 possible final configurations.

15. [40] On Facebook, there is a group of people that satisfies the following two properties: (i) there existsa positive integers k such that any subset of 2k − 1 people in the group contains a subset of k peoplein the group who are all friends with each other, and (ii) every member of the group has 2011 friendsor fewer.

(a) [15] If k = 2, determine, with proof, the maximum number of people the group may contain.

(b) [25] If k = 776, determine, with proof, the maximum number of people the group may contain.

Solution:

(a) Answer: 4024 If k = 2, then among any three people at least two of them are friends. Clearlyif we have 4024 people divided into two sets of 2012 such that everyone is friends with everyone intheir set but no one in the other set, then any triple of three people will contain two people fromthe same set, who will automatically be friends. Therefore, the group may contain 4024 people.

We now prove that 4024 is the maximum. Given any set with at least 4025 people, considerany two people who are not friends. Between them them have at most 4022 friends, so by thepigeonhole principle there exists someone who is not friends with either one. Therefore, thereexists a triple of three people who are all not friends with each other, so the group may notcontain 4025 people or more, so 4024 is the maximum, as we claimed.

(b) Answer: 4024 The answer remains the same. Considering the construction identical to thatin the above solution, we see that the group may still contain 4024 people and satisfy the desiredcriterion.

We now prove that 4024 is the maximum. Consider a group with at least 4025 people. Fromthe previous part, we know there exist three people who are not friends. Pick such a threesomeand call them X,Y , and Z. We now consider the rest of the people and successively pick pairs ofpersons Ai, Bi for 1 ≤ i ≤ 774 as follows: Once we have picked Ai, Bi, of the remaining 4019− 2ipeople, we find two who are not friends, which is always possible since 4019 − 2i > 2013, andwe name one of those people Ai+1 and the other Bi+1. Once we have done this, consider the setcontaining X, Y , and Z as well as Ai and Bi for all 1 ≤ i ≤ 774. This is a set of 2k − 1 = 1551people. Any subset where everyone is friends with each other can contain at most 1 of Ai, Bifor all i, and at most 1 of X, Y , and Z, meaning that any such subset may contain at most 775people. Hence there exists a subset containing 1551 people that does not have 776 people whoare all friends. Thus, the group may not contain 4025 people or more, so the answer is still 4024,as desired.

Team Round B

4th

Annual Harvard-MIT November TournamentSaturday 12 November 2011

General Test

1. [3] Find all ordered pairs of real numbers (x, y) such that x2y = 3 and x + xy = 4.

2. [3] Let ABC be a triangle, and let D, E, and F be the midpoints of sides BC, CA, and AB, respectively.Let the angle bisectors of ∠FDE and ∠FBD meet at P . Given that ∠BAC = 37◦ and ∠CBA = 85◦,determine the degree measure of ∠BPD.

3. [4] Alberto, Bernardo, and Carlos are collectively listening to three different songs. Each is simultane-ously listening to exactly two songs, and each song is being listened to by exactly two people. In howmany ways can this occur?

4. [4] Determine the remainder when

21·2

2 + 22·3

2 + · · · + 22011·2012

2

is divided by 7.

5. [5] Find all real values of x for which

1√

x +√

x − 2+

1√

x + 2 +√

x=

1

4.

6. [5] Five people of heights 65, 66, 67, 68, and 69 inches stand facing forwards in a line. How manyorders are there for them to line up, if no person can stand immediately before or after someone whois exactly 1 inch taller or exactly 1 inch shorter than himself?

7. [5] Determine the number of angles θ between 0 and 2π, other than integer multiples of π/2, such thatthe quantities sin θ, cos θ, and tan θ form a geometric sequence in some order.

8. [6] Find the number of integers x such that the following three conditions all hold:

• x is a multiple of 5

• 121 < x < 1331

• When x is written as an integer in base 11 with no leading 0s (i.e. no 0s at the very left), itsrightmost digit is strictly greater than its leftmost digit.

9. [7] Let P and Q be points on line l with PQ = 12. Two circles, ω and Ω, are both tangent to l at Pand are externally tangent to each other. A line through Q intersects ω at A and B, with A closer toQ than B, such that AB = 10. Similarly, another line through Q intersects Ω at C and D, with Ccloser to Q than D, such that CD = 7. Find the ratio AD/BC.

10. [8] Let r1, r2, . . . , r7 be the distinct complex roots of the polynomial P (x) = x7 − 7. Let

K =∏

1≤i<j≤7

(ri + rj),

that is, the product of all numbers of the form ri+rj , where i and j are integers for which 1 ≤ i < j ≤ 7.Determine the value of K2.

4th

Annual Harvard-MIT November TournamentSaturday 12 November 2011

General Test

1. [3] Find all ordered pairs of real numbers (x, y) such that x2y = 3 and x + xy = 4.

Answer: (1, 3), (3, 1

3) Multiplying the second equation by x gives

x2 + x2y = 4x,

and substituting our known value of x2y gives the quadratic

x2 − 4x + 3 = 0,

so x = 1 or x = 3. Hence, we obtain the solutions (x, y) = (1, 3), (3, 1/3).

2. [3] Let ABC be a triangle, and let D, E, and F be the midpoints of sides BC, CA, and AB, respectively.Let the angle bisectors of ∠FDE and ∠FBD meet at P . Given that ∠BAC = 37◦ and ∠CBA = 85◦,determine the degree measure of ∠BPD.

Answer: 61◦ Because D,E, F are midpoints, we have ABC ∼ DEF . Furthermore, we know thatFD ‖ AC and DE ‖ AB, so we have

∠BDF = ∠BCA = 180 − 37 − 85 = 58◦.

Also, ∠FDE = ∠BAC = 37◦. Hence, we have

∠BPD = 180◦ − ∠PBD − ∠PDB = 180◦ − 85◦

2−

(

37◦

2+ 58◦

)

= 61◦.

A

B CD

EF

P

58

37

85

3. [4] Alberto, Bernardo, and Carlos are collectively listening to three different songs. Each is simultane-ously listening to exactly two songs, and each song is being listened to by exactly two people. In howmany ways can this occur?

Answer: 6 We have

(

3

2

)

= 3 choices for the songs that Alberto is listening to. Then, Bernardo and

Carlos must both be listening to the third song. Thus, there are 2 choices for the song that Bernardoshares with Alberto. From here, we see that the songs that everyone is listening to are forced. Thus,there are a total of 3 × 2 = 6 ways for the three to be listening to songs.

General Test

4. [4] Determine the remainder when

21·2

2 + 22·3

2 + · · · + 22011·2012

2

is divided by 7.

Answer: 1 We have that 23 ≡ 1 (mod 7). Hence, it suffices to consider the exponents modulo3. We note that the exponents are the triangular number and upon division by 3 give the pattern ofremainders 1, 0, 0, 1, 0, 0, . . ., so what we want is

21·2

2 + · · · + 22011·2012

2 ≡ 21 + 20 + 20 + 21 + . . . + 20 + 21 (mod 7)

≡ 2010

3(21 + 20 + 20) + 21

≡ (670)(4) + 2

≡ 1.

5. [5] Find all real values of x for which

1√x +

√x − 2

+1√

x + 2 +√

x=

1

4.

Answer: 257

16We note that

1

4=

1√x +

√x − 2

+1√

x + 2 +√

x

=

√x −

√x − 2

(√

x +√

x − 2)(√

x −√

x − 2)+

√x + 2 −√

x

(√

x + 2 +√

x)(√

x + 2 −√x)

=

√x −

√x − 2

2+

√x + 2 −√

x

2

=1

2(√

x + 2 −√

x − 2),

so that2√

x + 2 − 2√

x − 2 = 1.

Squaring, we get that

8x − 8√

(x + 2)(x − 2) = 1 ⇒ 8x − 1 = 8√

(x + 2)(x − 2).

Squaring again gives64x2 − 16x + 1 = 64x2 − 256,

so we get that x = 257

16.

6. [5] Five people of heights 65, 66, 67, 68, and 69 inches stand facing forwards in a line. How manyorders are there for them to line up, if no person can stand immediately before or after someone whois exactly 1 inch taller or exactly 1 inch shorter than himself?

Answer: 14 Let the people be A,B,C,D,E so that their heights are in that order, with A thetallest and E the shortest. We will do casework based on the position of C.

• Case 1: C is in the middle. Then, B must be on one of the two ends, for two choices. This leavesonly one choice for D–the other end. Then, we know the positions of A and E since A cannotneighbor B and E cannot neighbor D. So we have 2 options for this case.

General Test

• Case 2: C is in the second or fourth spot. Then, we have two choices for the position of C.Without loss of generality, let C be in the second spot. Then, the first and third spots must be Aand E, giving us two options. This fixes the positions of B and D, so we have a total of 2× 2 = 4options for this case.

• Case 3: C is in the first or last spot. Then, we have two choices for the position of C. Withoutloss of generality, let it be in the first spot. Either A or E is in the second spot, giving us twochoices. Without loss of generality, let it be A. Then, if D is in the third spot, the positions of Band E are fixed. If E is in third spot, the positions of B and D are fixed, so we have a total of2 × 2 × (1 + 1) = 8 options for this case.

Hence, we have a total of 2 + 4 + 8 = 14 possibilities.

7. [5] Determine the number of angles θ between 0 and 2π, other than integer multiples of π/2, such thatthe quantities sin θ, cos θ, and tan θ form a geometric sequence in some order.

Answer: 4 If sin θ, cos θ, and tan θ are in a geometric progression, then the product of two mustequal the square of the third. Using this criterion, we have 3 cases.

• Case 1: sin θ · tan θ = cos2 θ. This implies that (sin2 θ) = (cos3 θ). Writing sin2 θ as 1 − cos2 θand letting cos θ = x, we have that x3 + x2 − 1 = 0. We wish to find the number of solutions ofthis where |x| ≤ 1. Clearly −1 is not a root. If −1 < x ≤ 0, we have that x2 + x3 ≤ x2 < 1 sox3 + x2 − 1 < 0 and there are no roots. If 0 < x ≤ 1, then x3 + x2 − 1 is a strictly increasingfunction. Since it has value −1 at x = 0 and value 1 and x = 1, there is exactly one root between0 and 1, non-inclusive. There are 2 values of θ such that cos θ equals this root, and thus, twosolutions in this case.

• Case 2: sin θ · cos θ = tan2 θ. This implies that cos3 θ = sin θ. To find the number of solutions inthis case, we can analyze the graphs of the functions in different ranges. Note that from θ = 0to θ = π

2, cos3 θ decreases strictly from 1 to 0 while sin θ increases strictly from 0 to 1. Hence,

there is one solution in this range. By a similar argument, a solution exists between θ = π andθ = 3π

2. In the intervals [π

2, π] and [3π

2, 2π], we have that one function is negative and the other

is positive, so there are no solutions. Thus, there are two solutions in this case.

• Case 3: cos θ · tan θ = sin2 θ. This implies that sin θ = sin2 θ, so sin θ = 0, 1. Clearly the onlysolutions of these have θ as an integer multiple of π

2. Thus, there are no pertinent solutions int

his case.

We can see that the solutions for the first two cases are mutually exclusive. Hence, there are 4 solutionsin total.

8. [6] Find the number of integers x such that the following three conditions all hold:

• x is a multiple of 5

• 121 < x < 1331

• When x is written as an integer in base 11 with no leading 0s (i.e. no 0s at the very left), itsrightmost digit is strictly greater than its leftmost digit.

Answer: 99 We will work in base 11, so let x = def11

such that d > 0. Then, based on the firsttwo conditions, we aim to find multiples of 5 between 10011 and 100011. We note that

def11

≡ 112 · d + 11 · e + f ≡ d + e + f (mod 5).

Hence, x a multiple of 5 if and only if the sum of its digits is a multiple of 5. Thus, we wish to findtriples (d, e, f) with elements in 0, 1, 2, · · · , 9, 10 such that d + e + f ≡ 0 (mod 5) and 0 < d < f .

Note that if we choose d and f such that d < f , there is exactly one value of e modulo 5 that wouldmake d + e + f ≡ 0 (mod 5). Once the this value of e is fixed, then there are two possibilities for eunless e ≡ 0 (mod 5), in which case there are three possibilities. Thus, our answer is twice the numberof ways to choose d and f such that 0 < d < f plus the number of ways to choose d and f such that

General Test

d + f ≡ 0 (mod 5) and 0 < d < f (to account for the extra choice for the value of e). Note that thenumber of ways to choose 0 < d < f is just

(

10

2

)

since any any choice of two digits yields exactly oneway to order them. The number of ways to choose d + f ≡ 0 (mod 5) and 0 < d < f can be found bylisting: (d, f) = (1, 4), (1, 9), (2, 3), (2, 8), (3, 7), (4, 6), (5, 10), (6, 9), (7, 8), for 9 such pairings.

Hence, the total is 2

(

10

2

)

+ 9 = 99 possibilities for x.

9. [7] Let P and Q be points on line l with PQ = 12. Two circles, ω and Ω, are both tangent to l at Pand are externally tangent to each other. A line through Q intersects ω at A and B, with A closer toQ than B, such that AB = 10. Similarly, another line through Q intersects Ω at C and D, with Ccloser to Q than D, such that CD = 7. Find the ratio AD/BC.

Answer: 8

9We first apply the Power of a Point theorem repeatedly. Note that QA ·QB = QP 2 =

QC · QD. Substituting in our known values, we obtain QA(QA + 10) = 122 = QC(QC + 7). Solvingthese quadratics, we get that QA = 8 and QC = 9.

We can see that AQDQ

= CQBQ

and that ∠AQD = ∠CQB, so QAD ∼ QCB . (Alternatively, goingback to the equality QA · QB = QC · QD, we realize that this is just a Power of a Point theorem onthe quadrilateral ABDC, and so this quadrilateral is cyclic. This implies that ∠ADQ = ∠ADC =∠ABC = ∠QBC.) Thus, AD

BC= AQ

QC= 8

9.

ω

Ω

A

B

C

D

P Q

8

10

9

7

12

10. [8] Let r1, r2, . . . , r7 be the distinct complex roots of the polynomial P (x) = x7 − 7. Let

K =∏

1≤i<j≤7

(ri + rj),

that is, the product of all numbers of the form ri+rj , where i and j are integers for which 1 ≤ i < j ≤ 7.Determine the value of K2.

Answer: 117649 We first note that x7 − 7 = (x − r1)(x − r2) · · · (x − r7), which implies, replacingx by −x and taking the negative of the equation, that (x + r1)(x + r2) · · · (x + r7) = x7 + 7. Also notethat the product of the ri is just the constant term, so r1r2 · · · r7 = 7.

General Test

Now, we have that

27 · 7 · K2 = (7

∏

i=1

2ri)K2

=

7∏

i=1

2ri

∏

1≤i<j≤7

(ri + rj)2

=∏

1≤i=j≤7

(ri + rj)∏

1≤i<j≤7

(ri + rj)∏

1≤j<i≤7

(ri + rj)

=∏

1≤i,j≤7

(ri + rj)

=

7∏

i=1

7∏

j=1

(ri + rj).

However, note that for any fixed i,

7∏

j=1

(ri + rj) is just the result of substuting x = ri into (x + r1)(x +

r2) · · · (x + r7). Hence,7

∏

j=1

(ri + rj) = r7

i + 7 = (r7

i − 7) + 14 = 14.

Therefore, taking the product over all i gives 147, which yields K2 = 76 = 117649.

General Test

4th

Annual Harvard-MIT November TournamentSaturday 12 November 2011

Theme Round

Fish

1. [3] Five of James’ friends are sitting around a circular table to play a game of Fish. James chooses aplace between two of his friends to pull up a chair and sit. Then, the six friends divide themselves intotwo disjoint teams, with each team consisting of three consecutive players at the table. If the order inwhich the three members of a team sit does not matter, how many possible (unordered) pairs of teamsare possible?

2. [3] In a game of Fish, R2 and R3 are each holding a positive number of cards so that they are collectivelyholding a total of 24 cards. Each player gives an integer estimate for the number of cards he is holding,such that each estimate is an integer between 80% of his actual number of cards and 120% of his actualnumber of cards, inclusive. Find the smallest possible sum of the two estimates.

3. [5] In preparation for a game of Fish, Carl must deal 48 cards to 6 players. For each card that he deals,he runs through the entirety of the following process:

1. He gives a card to a random player.

2. A player Z is randomly chosen from the set of players who have at least as many cards as everyother player (i.e. Z has the most cards or is tied for having the most cards).

3. A player D is randomly chosen from the set of players other than Z who have at most as manycards as every other player (i.e. D has the fewest cards or is tied for having the fewest cards).

4. Z gives one card to D.

He repeats steps 1-4 for each card dealt, including the last card. After all the cards have been dealt,what is the probability that each player has exactly 8 cards?

4. [6] Toward the end of a game of Fish, the 2 through 7 of spades, inclusive, remain in the hands of threedistinguishable players: DBR, RB, and DB, such that each player has at least one card. If it is knownthat DBR either has more than one card or has an even-numbered spade, or both, in how many wayscan the players’ hands be distributed?

5. [8] For any finite sequence of positive integers π, let S(π) be the number of strictly increasing sub-sequences in π with length 2 or more. For example, in the sequence π = {3, 1, 2, 4}, there are fiveincreasing sub-sequences: {3, 4}, {1, 2}, {1, 4}, {2, 4}, and {1, 2, 4}, so S(π) = 5. In an eight-playergame of Fish, Joy is dealt six cards of distinct values, which she puts in a random order π from left toright in her hand. Determine

∑

π

S(π),

where the sum is taken over all possible orders π of the card values.

Circles and Tangents

6. [3] Let ABC be an equilateral triangle with AB = 3. Circle ω with diameter 1 is drawn inside thetriangle such that it is tangent to sides AB and AC. Let P be a point on ω and Q be a point onsegment BC. Find the minimum possible length of the segment PQ.

7. [4] Let XY Z be a triangle with ∠XY Z = 40◦ and ∠Y ZX = 60◦. A circle Γ, centered at the pointI, lies inside triangle XY Z and is tangent to all three sides of the triangle. Let A be the point of

tangency of Γ with Y Z, and let ray−→XI intersect side Y Z at B. Determine the measure of ∠AIB.

8. [5] Points D,E, F lie on circle O such that the line tangent to O at D intersects ray−−→EF at P . Given

that PD = 4, PF = 2, and ∠FPD = 60◦, determine the area of circle O.

9. [6] Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. Let D be the foot of the altitudefrom A to BC. The inscribed circles of triangles ABD and ACD are tangent to AD at P and Q,respectively, and are tangent to BC at X and Y , respectively. Let PX and QY meet at Z. Determinethe area of triangle XY Z.

10. [7] Let Ω be a circle of radius 8 centered at point O, and let M be a point on Ω. Let S be the set ofpoints P such that P is contained within Ω, or such that there exists some rectangle ABCD containingP whose center is on Ω with AB = 4, BC = 5, and BC ‖ OM . Find the area of S.

4th

Annual Harvard-MIT November TournamentSaturday 12 November 2011

Theme Round

1. [3] Five of James’ friends are sitting around a circular table to play a game of Fish. James chooses aplace between two of his friends to pull up a chair and sit. Then, the six friends divide themselves intotwo disjoint teams, with each team consisting of three consecutive players at the table. If the order inwhich the three members of a team sit does not matter, how many possible (unordered) pairs of teamsare possible?

Answer: 5 Note that the team not containing James must consist of three consecutive players whoare already seated. We have 5 choices for the player sitting furthest clockwise on the team of whichJames is not a part. The choice of this player uniquely determines the teams, so we have a total of 5possible pairs.

2. [3] In a game of Fish, R2 and R3 are each holding a positive number of cards so that they are collectivelyholding a total of 24 cards. Each player gives an integer estimate for the number of cards he is holding,such that each estimate is an integer between 80% of his actual number of cards and 120% of his actualnumber of cards, inclusive. Find the smallest possible sum of the two estimates.

Answer: 20 To minimize the sum, we want each player to say an estimate as small as possible–i.e.an estimate as close to 80% of his actual number of cards as possible. We claim that the minimumpossible sum is 20.

First, this is achievable when R2 has 10 cards and estimates 8, and when R3 has 14 cards and estimates12.

Then, suppose that R2 has x cards and R3 has 24 − x. Then, the sum of their estimates is

⌈

4

5(x)

⌉

+

⌈

4

5(24 − x)

⌉

≥⌈

4

5(x) +

4

5(24 − x)

⌉

≥⌈

4

5(24)

⌉

≥ 20.

Note: We use the fact that for all real numbers a, b, ⌈a⌉ + ⌈b⌉ ≥ ⌈a + b⌉.

3. [5] In preparation for a game of Fish, Carl must deal 48 cards to 6 players. For each card that he deals,he runs through the entirety of the following process:

1. He gives a card to a random player.

2. A player Z is randomly chosen from the set of players who have at least as many cards as everyother player (i.e. Z has the most cards or is tied for having the most cards).

3. A player D is randomly chosen from the set of players other than Z who have at most as manycards as every other player (i.e. D has the fewest cards or is tied for having the fewest cards).

4. Z gives one card to D.

He repeats steps 1-4 for each card dealt, including the last card. After all the cards have been dealt,what is the probability that each player has exactly 8 cards?

Answer: 5

6After any number of cards are dealt, we see that the difference between the number of

cards that any two players hold is at most one. Thus, after the first 47 cards have been dealt, thereis only one possible distribution: there must be 5 players with 8 cards and 1 player with 7 cards. Wehave two cases:

• Carl gives the last card to the player with 7 cards. Then, this player must give a card to another,leading to a uneven distribution of cards.

• Carl gives the last card to a player already with 8 cards. Then, that player must give a card toanother; however, our criteria specify that he can only give it to the player with 7 cards, leadingto an even distribution.

Theme Round

The probability of the second case happening, as Carl deals at random, is 5

6.

4. [6] Toward the end of a game of Fish, the 2 through 7 of spades, inclusive, remain in the hands of threedistinguishable players: DBR, RB, and DB, such that each player has at least one card. If it is knownthat DBR either has more than one card or has an even-numbered spade, or both, in how many wayscan the players’ hands be distributed?

Answer: 450 First, we count the number of distributions where each player has at least 1 card.The possible distributions are:

• Case 1: 4/1/1: There are 3 choices for who gets 4 cards, 6 choices for the card that one ofthe single-card players holds, and 5 choices for the card the other single-card player holds, or3 × 6 × 5 = 90 choices.

• Case 2: 3/2/1: There are 6 choices for the single card,

(

5

2

)

= 10 choices for the pair of cards,

and 3! = 6 choices for which player gets how many cards, for a total of 6 × 10 × 6 = 360 choices.

• Case 3: 2/2/2: There are

(

6

2

)

= 15 choices for the cards DBR gets,

(

4

2

)

= 6 for the cards that

RB gets, and DB gets the remaining two cards. This gives a total of 15 × 6 = 90 choices.

Thus, we have a total of 90 + 360 + 90 = 540 ways for the cards to be distributed so that each personholds at least one.

Next, we look at the number of ways that the condition cannot be satisfied if each player has at leastone card. Then, DBR must have no more than one card, and cannot have an even spade. We only careabout cases where he has a non-zero number of cards, so he must have exactly 1 odd spade. Then, wesee that there are 25 − 2 = 30 ways to distribute the other 5 cards among RB and DB so that neitherhas 0 cards. Since there are 3 odd spades, this gives 3× 30 bad cases, so we have 540− 90 = 450 oneswhere all the problem conditions hold.

5. [8] For any finite sequence of positive integers π, let S(π) be the number of strictly increasing sub-sequences in π with length 2 or more. For example, in the sequence π = {3, 1, 2, 4}, there are fiveincreasing sub-sequences: {3, 4}, {1, 2}, {1, 4}, {2, 4}, and {1, 2, 4}, so S(π) = 5. In an eight-playergame of Fish, Joy is dealt six cards of distinct values, which she puts in a random order π from left toright in her hand. Determine

∑

π

S(π),

where the sum is taken over all possible orders π of the card values.

Answer: 8287 For each subset of Joy’s set of cards, we compute the number of orders of cards inwhich the cards in the subset are arranged in increasing order. When we sum over all subsets of Joy’scards, we will obtain the desired sum.

Consider any subset of k cards. The probability that they are arranged in increasing order is precisely1/k! (since we can form a k!-to-1 correspondence between all possible orders and orders in which thecards in our subset are in increasing order), and there are 6! = 720 total arrangements so exactly720/k! of them give an increasing subsequence in the specified cards. Now for any for k = 2, 3, 4, 5, 6,

we have

(

6

k

)

subsets of k cards, so we sum to get

∑

π

S(π) =6

∑

k=2

(

6

k

)

· 6!

k!= 8287.

6. [3] Let ABC be an equilateral triangle with AB = 3. Circle ω with diameter 1 is drawn inside thetriangle such that it is tangent to sides AB and AC. Let P be a point on ω and Q be a point onsegment BC. Find the minimum possible length of the segment PQ.

Theme Round

Answer: 3√

3−3

2Let P , Q, be the points which minimize the distance. We see that we want both

to lie on the altitude from A to BC. Hence, Q is the foot of the altitude from A to BC and AQ = 3√

3

2.

Let O, which must also lie on this line, be the center of ω, and let D be the point of tangency betweenω and AC. Then, since OD = 1

2, we have AO = 2OD = 1 because ∠OAD = 30◦, and OP = 1

2.

Consequently,

PQ = AQ − AO − OP =3√

3 − 3

2.

A

B C

O

D

P

Q

7. [4] Let XY Z be a triangle with ∠XY Z = 40◦ and ∠Y ZX = 60◦. A circle Γ, centered at the pointI, lies inside triangle XY Z and is tangent to all three sides of the triangle. Let A be the point of

tangency of Γ with Y Z, and let ray−→XI intersect side Y Z at B. Determine the measure of ∠AIB.

Answer: 10◦ Let D be the foot of the perpendicular from X to Y Z. Since I is the incenter and Athe point of tangency, IA ⊥ Y Z, so

AI ‖ XD ⇒ ∠AIB = ∠DXB.

Since I is the incenter,

∠BXZ =1

2∠Y XZ =

1

2(180◦ − 40◦ − 60◦) = 40◦.

Consequently, we get that

∠AIB = ∠DXB = ∠ZXB − ∠ZXD = 40◦ − (90◦ − 60◦) = 10◦.

X

Y ZAB

I

D

8. [5] Points D,E, F lie on circle O such that the line tangent to O at D intersects ray−−→EF at P . Given

that PD = 4, PF = 2, and ∠FPD = 60◦, determine the area of circle O.

Theme Round

Answer: 12π By the power of a point on P , we get that

16 = PD2 = (PF )(PE) = 2(PE) ⇒ PE = 8.

However, since PE = 2PD and ∠FPD = 60◦, we notice that PDE is a 30 − 60 − 90 triangle, soDE = 4

√3 and we have ED ⊥ DP . It follows that DE is a diameter of the circle, since tangents

the tangent at D must be perpendicular to the radius containing D. Hence, the area of the circle is( 1

2DE)2π = 12π.

D

E

F

P4

2

6

2√

3

9. [6] Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. Let D be the foot of the altitudefrom A to BC. The inscribed circles of triangles ABD and ACD are tangent to AD at P and Q,respectively, and are tangent to BC at X and Y , respectively. Let PX and QY meet at Z. Determinethe area of triangle XY Z.

Answer: 25

4First, note that AD = 12, BD = 5, CD = 9.

By equal tangents, we get that PD = DX, so PDX is isosceles. Because D is a right angle, we get that∠PXD = 45◦. Similarly, ∠XY Z = 45◦, so XY Z is an isosceles right triangle with hypotenuse XY .However, by tangents to the incircle, we get that XD = 1

2(12+5−13) = 2 and Y D = 1

2(12+9−15) = 3.

Hence, the area of the XYZ is 1

4(XY )2 = 1

4(2 + 3)2 = 25

4.

A

B CD

P

QZ

X Y

10. [7] Let Ω be a circle of radius 8 centered at point O, and let M be a point on Ω. Let S be the set ofpoints P such that P is contained within Ω, or such that there exists some rectangle ABCD containingP whose center is on Ω with AB = 4, BC = 5, and BC ‖ OM . Find the area of S.

Theme Round

Answer: 164 + 64π We wish to consider the union of all rectangles ABCD with AB = 4, BC = 5,and BC ‖ OM , with center X on Ω. Consider translating rectangle ABCD along the radius XO to arectangle A′B′C ′D′ now centered at O. It is now clear that that every point inside ABCD is a translateof a point in A′B′C ′D′, and furthermore, any rectangle ABCD translates along the appropriate radiusto the same rectangle A′B′C ′D′.

We see that the boundary of this region can be constructed by constructing a quarter-circle at eachvertex, then connecting these quarter-circles with tangents to form four rectangular regions. Now,splitting our region in to four quarter circles and five rectangles, we compute the desired area to be

4 · 1

4(8)2π + 2(4 · 8) + 2(5 · 8) + (4 · 5) = 164 + 64π.

O M

Theme Round

4th

Annual Harvard-MIT November TournamentSaturday 12 November 2011

Team Round

p-Polynomials

1. [2] Find the number of positive integers x less than 100 for which

3x + 5x + 7x + 11x + 13x + 17x + 19x

is prime.

2. [4] Determine the set of all real numbers p for which the polynomial Q(x) = x3 + px2− px − 1 has

three distinct real roots.

3. [6] Find the sum of the coefficients of the polynomial P (x) = x4− 29x3 + ax2 + bx + c, given that

P (5) = 11, P (11) = 17, and P (17) = 23.

4. [7] Determine the number of quadratic polynomials P (x) = p1x2 + p2x − p3, where p1, p2, p3 are not

necessarily distinct (positive) prime numbers less than 50, whose roots are distinct rational numbers.

C Coloring

5. [3] Sixteen wooden Cs are placed in a 4-by-4 grid, all with the same orientation, and each is to becolored either red or blue. A quadrant operation on the grid consists of choosing one of the fourtwo-by-two subgrids of Cs found at the corners of the grid and moving each C in the subgrid to theadjacent square in the subgrid that is 90 degrees away in the clockwise direction, without changingthe orientation of the C. Given that two colorings are the considered same if and only if one can beobtained from the other by a series of quadrant operations, determine the number of distinct coloringsof the Cs.

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

6. [5] Ten Cs are written in a row. Some Cs are upper-case and some are lower-case, and each is writtenin one of two colors, green and yellow. It is given that there is at least one lower-case C, at least onegreen C, and at least one C that is both upper-case and yellow. Furthermore, no lower-case C can befollowed by an upper-case C, and no yellow C can be followed by a green C. In how many ways canthe Cs be written?

7. [7] Julia is learning how to write the letter C. She has 6 differently-colored crayons, and wants to writeCc Cc Cc Cc Cc. In how many ways can she write the ten Cs, in such a way that each upper case Cis a different color, each lower case C is a different color, and in each pair the upper case C and lowercase C are different colors?

[GG]eometry

8. [4] Let G,A1, A2, A3, A4, B1, B2, B3, B4, B5 be ten points on a circle such that GA1A2A3A4 is a regularpentagon and GB1B2B3B4B5 is a regular hexagon, and B1 lies on minor arc GA1. Let B5B3 intersectB1A2 at G1, and let B5A3 intersect GB3 at G2. Determine the degree measure of ∠GG2G1.

9. [4] Let ABC be a triangle with AB = 9, BC = 10, and CA = 17. Let B′ be the reflection of the pointB over the line CA. Let G be the centroid of triangle ABC, and let G′ be the centroid of triangleAB′C. Determine the length of segment GG′.

10. [8] Let G1G2G3 be a triangle with G1G2 = 7, G2G3 = 13, and G3G1 = 15. Let G4 be a point

outside triangle G1G2G3 so that ray−−−→

G1G4 cuts through the interior of the triangle, G3G4 = G4G2,and ∠G3G1G4 = 30◦. Let G3G4 and G1G2 meet at G5. Determine the length of segment G2G5.

4th

Annual Harvard-MIT November TournamentSaturday 12 November 2011

Team Round

1. [2] Find the number of positive integers x less than 100 for which

3x + 5x + 7x + 11x + 13x + 17x + 19x

is prime.

Answer: 0 We claim that our integer is divisible by 3 for all positive integers x. Indeed, we have

3x + 5x + 7x + 11x + 13x + 17x + 19x ≡ (0)x + (−1)x + (1)x + (−1)x + (1)x + (−1)x + (1)x

≡ 3[(1)x + (−1)x]

≡ 0 (mod 3).

It is clear that for all x ≥ 1, our integer is strictly greater than 3, so it will always be composite,making our answer 0.

2. [4] Determine the set of all real numbers p for which the polynomial Q(x) = x3 + px2 − px − 1 hasthree distinct real roots.

Answer: p > 1 and p < −3 First, we note that

x3 + px2 − px − 1 = (x − 1)(x2 + (p + 1)x + 1).

Hence, x2 + (p + 1)x + 1 has two distinct roots. Consequently, the discriminant of this equation mustbe positive, so (p + 1)2 − 4 > 0, so either p > 1 or p < −3. However, the problem specifies that thequadratic must have distinct roots (since the original cubic has distinct roots), so to finish, we need tocheck that 1 is not a double root–we will do this by checking that 1 is not a root of x2 + (p + 1)x + 1for any value p in our range. But this is clear, since 1 + (p + 1) + 1 = 0 ⇒ p = −3, which is not in theaforementioned range. Thus, our answer is all p satisfying p > 1 or p < −3.

3. [6] Find the sum of the coefficients of the polynomial P (x) = x4 − 29x3 + ax2 + bx + c, given thatP (5) = 11, P (11) = 17, and P (17) = 23.

Answer: -3193 Define Q(x) = P (x)− x− 6 = x4 − 29x3 + ax2 + (b− 1)x + (c− 6) and notice thatQ(5) = Q(11) = Q(17) = 0. Q(x) has degree 4 and by Vieta’s Formulas the sum of its roots is 29, soits last root is 29 − 17 − 11 − 5 = −4, giving us Q(x) = (x − 5)(x − 11)(x − 17)(x + 4). This meansthat P (1) = Q(1) + 7 = (−4)(−10)(−16)(5) + 7 = −3200 + 7 = −3193.

4. [7] Determine the number of quadratic polynomials P (x) = p1x2 + p2x − p3, where p1, p2, p3 are not

necessarily distinct (positive) prime numbers less than 50, whose roots are distinct rational numbers.

Answer: 31 The existence of distinct rational roots means that the given quadratic splits into linearfactors. Then, since p1, p3 are both prime, we get that the following are the only possible factorizations:

• (p1x − p3)(x + 1) ⇒ p2 = p1 − p3

• (p1x + p3)(x − 1) ⇒ p2 = −p1 + p3

• (p1x − 1)(x + p3) ⇒ p2 = p1p3 − 1

• (p1x + 1)(x − p3) ⇒ p2 = −p1p3 + 1

In the first case, observe that since p2 + p3 = p1, we have p1 > 2, so p1 is odd and exactly one of p2, p3

is equal to 2. Thus, we get a solutions for every pair of twin primes below 50, which we enumerateto be (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), giving 12 solutions in total. Similarly, the secondcase gives p1 + p2 = p3, for another 12 solutions.

Team Round

In the third case, if p1, p3 are both odd, then p2 is even and thus equal to 2. However, this givesp1p3 = 3, which is impossible. Therefore, at least one of p1, p3 is equal to 2. If p1 = 2, we getp2 = 2p3 − 1, which we find has 4 solutions: (p2, p3) = (3, 2), (5, 3), (13, 7), (37, 19). Similarly, there arefour solutions with p3 = 2. However, we count the solution (p1, p2, p3) = (2, 3, 2) twice, so we have atotal of 7 solutions in this case. Finally, in the last case

p2 = −p1p3 + 1 < −(2)(2) + 1 < 0,

so there are no solutions. Hence, we have a total of 12 + 12 + 7 = 31 solutions.

5. [3] Sixteen wooden Cs are placed in a 4-by-4 grid, all with the same orientation, and each is to becolored either red or blue. A quadrant operation on the grid consists of choosing one of the fourtwo-by-two subgrids of Cs found at the corners of the grid and moving each C in the subgrid to theadjacent square in the subgrid that is 90 degrees away in the clockwise direction, without changingthe orientation of the C. Given that two colorings are the considered same if and only if one can beobtained from the other by a series of quadrant operations, determine the number of distinct coloringsof the Cs.

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

Answer: 1296 For each quadrant, we have three distinct cases based on the number of Cs in eachcolor:

• Case 1: all four the same color: 2 configurations (all red or all blue)

• Case 2: 3 of one color, 1 of the other: 2 configurations (three red or three blue)

• Case 3: 2 of each color: 2 configurations (red squares adjacent or opposite)

Thus, since there are 4 quadrants, there are a total of (2 + 2 + 2)4 = 1296 possible grids.

6. [5] Ten Cs are written in a row. Some Cs are upper-case and some are lower-case, and each is writtenin one of two colors, green and yellow. It is given that there is at least one lower-case C, at least onegreen C, and at least one C that is both upper-case and yellow. Furthermore, no lower-case C can befollowed by an upper-case C, and no yellow C can be followed by a green C. In how many ways canthe Cs be written?

Answer: 36 By the conditions of the problem, we must pick some point in the line where the greenCs transition to yellow, and some point where the upper-case Cs transition to lower-case. We see thatthe first transition must occur before the second, and that they cannot occur on the same C. Hence,

the answer is

(9

2

)= 36.

7. [7] Julia is learning how to write the letter C. She has 6 differently-colored crayons, and wants to writeCc Cc Cc Cc Cc. In how many ways can she write the ten Cs, in such a way that each upper case Cis a different color, each lower case C is a different color, and in each pair the upper case C and lowercase C are different colors?

Team Round

Answer: 222480 Suppose Julia writes Cc a sixth time, coloring the upper-case C with the uniquecolor different from that of the first five upper-case Cs, and doing the same with the lower-case C (note:we allow the sixth upper-case C and lower-case c to be the same color). Note that because the colorson the last Cc are forced, and any forced coloring of them is admissible, our problem is equivalent tocoloring these six pairs.

There are 6! ways for Julia to color the upper-case Cs. We have two cases for coloring the lower-caseCs:

• Case 1: the last pair of Cs use two different colors. In this case, all six lower-case Cs have adifferent color to their associated upper-case C, and in addition the six lower-case Cs all use eachcolor exactly once. In other words, we have a derangement* of the six colors, based on the colorsof the upper-case Cs. We calculate D6 = 265 ways to color the lower-case Cs here.

• Case 2: the last pair of Cs have both Cs the same color. Then, the color of the last lower-caseC is forced, and with the other five Cs we, in a similar way to before, have a derangement of theremaining five colors based on the colors of the first five lower-case Cs, so we have D5 = 44 waysto finish the coloring.

Our answer is thus 720(265 + 44) = 222480.

*A derangement is a permutation π of the set {1, 2, . . . , n} such that π(k) 6= k for all k, i.e. thereare no fixed points of the permutation. To calculate Dn, the number of derangements of an n-elementset, we can use an inclusion-exclusion argument. There are n! ways to permute the elements of the set.

Now, we subtract the number of permutations with at least one fixed point, which is

(n

1

)(n−1)! =

n!

1!,

since we choose a fixed point, then permute the other n − 1 elements. Correcting for overcounting,

we add back the number of permutations with at least two fixed points, which is

(n

2

)(n − 2)! =

n!

2!.

Continuing in this fashion by use of the principle of inclusion-exclusion, we get

Dn = n!

(1

0!− 1

1!+

1

2!+ · · · + (−1)n

n!

).

8. [4] Let G,A1, A2, A3, A4, B1, B2, B3, B4, B5 be ten points on a circle such that GA1A2A3A4 is a regularpentagon and GB1B2B3B4B5 is a regular hexagon, and B1 lies on minor arc GA1. Let B5B3 intersectB1A2 at G1, and let B5A3 intersect GB3 at G2. Determine the degree measure of ∠GG2G1.

Answer: 12◦

G

G1

G2

B1

B5

A2

A3

B3

Note that GB3 is a diameter of the circle. As a result, A2, A3 are symmetric with respect to GB3, asare B1, B5. Therefore, B1A2 and B5A3 intersect along line GB3, so in fact, B1, A2, G1, G2 are collinear.We now have

∠GG2G1 = ∠GG2B1 =ĜB1 − B̂3A2

2=

60◦ − 36◦

2= 12◦.

Team Round

9. [4] Let ABC be a triangle with AB = 9, BC = 10, and CA = 17. Let B′ be the reflection of the pointB over the line CA. Let G be the centroid of triangle ABC, and let G′ be the centroid of triangleAB′C. Determine the length of segment GG′.

Answer: 48

17

A

B

C

B′

G

G′

M

Let M be the midpoint of AC. For any triangle, we know that the centroid is located 2/3 of the wayfrom the vertex, so we have MG/MB = MG′/MB′ = 1/3, and it follows that MGG′ ∼ MBB′. Thus,GG′ = BB′/3 . However, note that BB′ is twice the altitude to AC in triangle ABC. To finish, wecalculate the area of ABC in two different ways. By Heron’s Formula, we have

[ABC] =√

18(18 − 9)(18 − 10)(18 − 17) = 36,

and we also have

[ABC] = 1

4BB′ · AC = 17

4(BB′),

from which it follows that GG′ = BB′/3 = 48/17.

10. [8] Let G1G2G3 be a triangle with G1G2 = 7, G2G3 = 13, and G3G1 = 15. Let G4 be a point

outside triangle G1G2G3 so that ray−−−→G1G4 cuts through the interior of the triangle, G3G4 = G4G2,

and ∠G3G1G4 = 30◦. Let G3G4 and G1G2 meet at G5. Determine the length of segment G2G5.

Answer: 169

23

G1

G2

G3

G4

G5

x y

713

15

We first show that quadrilateral G1G2G4G3 is cyclic. Note that by the law of cosines,

Team Round

cos ∠G2G1G3 =72 + 152 − 132

2 · 7 · 15=

1

2,

so ∠G2G1G3 = 60◦. However, we know that ∠G3G1G4 = 30◦, so G1G4 is an angle bisector. Now, let

G1G4 intersect the circumcircle of triangle G1G2G3 at X. Then, the minor arcs Ĝ2X and Ĝ3X aresubtended by the equal angles ∠G2G1X and ∠G3G1X, implying that G2X = G3X, i.e. X is on theperpendicular bisector of G2G3, l. Similarly, since G4G2 = G4G3, G4 lies on l. However, since l andG1G4 are distinct (in particular, G1 lies on G1G4 but not l), we in fact have X = G4, so G1G2G4G3

is cyclic.

We now have G5G2G4 ∼ G5G3G1 since G1G2G4G3 is cyclic. Now, we have ∠G4G3G2 = ∠G4G1G2 =30◦, and we may now compute G2G4 = G4G3 = 13/

√3. Let G5G2 = x and G5G4 = y. Now, from

G5G4G2 ∼ G5G1G3, we have:

x

y + 13/√

3=

13/√

3

15=

y

x + 7.

Equating the first and second expressions and cross-multiplying, we get

y +13√

3

3=

15√

3x

13.

Now, equating the first and third expressions and and substituting gives

(15√

3x

13− 13

√3

3

) (15√

3x

13

)= x(x + 7).

Upon dividing both sides by x, we obtain a linear equation from which we can solve to get x = 169/23.

Team Round

4th Annual Harvard-MIT November TournamentSaturday 12 November 2011

Guts Round

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4TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND

Round 1

1. [5] Determine the remainder when 1 + 2 + · · · + 2014 is divided by 2012.

2. [5] Let ABCD be a rectangle with AB = 6 and BC = 4. Let E be the point on BC with BE = 3,and let F be the point on segment AE such that F lies halfway between the segments AB and CD. IfG is the point of intersection of DF and BC, find BG.

3. [5] Let x be a real number such that 2x = 3. Determine the value of 43x+2.

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4TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND

Round 2

4. [6] Determine which of the following numbers is smallest in value: 54√

3, 144, 108√

6 − 108√

2.

5. [6] Charlie folds an 172 -inch by 11-inch piece of paper in half twice, each time along a straight line

parallel to one of the paper’s edges. What is the smallest possible perimeter of the piece after two suchfolds?

6. [6] To survive the coming Cambridge winter, Chim Tu doesn’t wear one T-shirt, but instead wears upto FOUR T-shirts, all in different colors. An outfit consists of three or more T-shirts, put on one ontop of the other in some order, such that two outfits are distinct if the sets of T-shirts used are differentor the sets of T-shirts used are the same but the order in which they are worn is different. Given thatChim Tu changes his outfit every three days, and otherwise never wears the same outfit twice, howmany days of winter can Chim Tu survive? (Needless to say, he only has four t-shirts.)

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4TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND

Round 3

7. [7] How many ordered triples of positive integers (a, b, c) are there for which a4b2c = 54000?

8. [7] Let a, b, c be not necessarily distinct integers between 1 and 2011, inclusive. Find the smallest

possible value ofab + c

a + b + c.

9. [7] Unit circle Ω has points X,Y,Z on its circumference so that XY Z is an equilateral triangle. LetW be a point other than X in the plane such that triangle WY Z is also equilateral. Determine thearea of the region inside triangle WY Z that lies outside circle Ω.

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4TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND

Round 4

10. [8] Determine the number of integers D such that whenever a and b are both real numbers with−1/4 < a, b < 1/4, then |a2 − Db2| < 1.

11. [8] For positive integers m,n, let gcd(m,n) denote the largest positive integer that is a factor of bothm and n. Compute

91∑n=1

gcd(n, 91).

12. [8] Joe has written 5 questions of different difficulties for a test with problems numbered 1 though 5.He wants to make sure that problem i is harder than problem j whenever i− j ≥ 3. In how many wayscan he order the problems for his test?

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4TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND

Round 5

13. [8] Tac is dressing his cat to go outside. He has four indistinguishable socks, four indistinguishableshoes, and 4 indistinguishable show-shoes. In a hurry, Tac randomly pulls pieces of clothing out of adoor and tries to put them on a random one of his cat’s legs; however, Tac never tries to put morethan one of each type of clothing on each leg of his cat. What is the probability that, after Tac is done,the snow-shoe on each of his cat’s legs is on top of the shoe, which is on top of the sock?

14. [8] Let AMOL be a quadrilateral with AM = 10, MO = 11, and OL = 12. Given that the perpen-dicular bisectors of sides AM and OL intersect at the midpoint of segment AO, find the length of sideLA.

15. [8] For positive integers n, let L(n) be the largest factor of n other than n itself. Determine the numberof ordered pairs of composite positive integers (m,n) for which L(m)L(n) = 80.

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4TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND

Round 6

16. [10] A small fish is holding 17 cards, labeled 1 through 17, which he shuffles into a random order.Then, he notices that although the cards are not currently sorted in ascending order, he can sort theminto ascending order by removing one card and putting it back in a different position (at the beginning,between some two cards, or at the end). In how many possible orders could his cards currently be?

17. [10] For a positive integer n, let p(n) denote the product of the positive integer factors of n. Determinethe number of factors n of 2310 for which p(n) is a perfect square.

18. [10] Consider a cube ABCDEFGH, where ABCD and EFGH are faces, and segments AE,BF,CG,DHare edges of the cube. Let P be the center of face EFGH, and let O be the center of the cube. Giventhat AG = 1, determine the area of triangle AOP .

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4TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND

Round 7

19. [10] Let ABCD be a rectangle with AB = 3 and BC = 7. Let W be a point on segment AB such thatAW = 1. Let X,Y,Z be points on segments BC,CD,DA, respectively, so that quadrilateral WXY Zis a rectangle, and BX < XC. Determine the length of segment BX.

20. [10] The UEFA Champions League playoffs is a 16-team soccer tournament in which Spanish teamsalways win against non-Spanish teams. In each of 4 rounds, each remaining team is randomly pairedagainst one other team; the winner advances to the next round, and the loser is permanently knockedout of the tournament. If 3 of the 16 teams are Spanish, what is the probability that there are 2Spanish teams in the final round?

21. [10] Let P (x) = x4 + 2x3 − 13x2 − 14x + 24 be a polynomial with roots r1, r2, r3, r4. Let Q be thequartic polynomial with roots r2

1, r22, r

23, r

24, such that the coefficient of the x4 term of Q is 1. Simplify

the quotient Q(x2)/P (x), leaving your answer in terms of x. (You may assume that x is not equal toany of r1, r2, r3, r4).

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4TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND

Round 8

22. [12] Let ABC be a triangle with AB = 23, BC = 24, and CA = 27. Let D be the point on segmentAC such that the incircles of triangles BAD and BCD are tangent. Determine the ratio CD/DA.

23. [12] Let N = 5AB37C2, where A,B,C are digits between 0 and 9, inclusive, and N is a 7-digit positiveinteger. If N is divisible by 792, determine all possible ordered triples (A,B,C).

24. [12] Three not necessarily distinct positive integers between 1 and 99, inclusive, are written in a rowon a blackboard. Then, the numbers, without including any leading zeros, are concatenated to form anew integer N . For example, if the integers written, in order, are 25, 6, and 12, then N = 25612 (andnot N = 250612). Determine the number of possible values of N .

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4TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND

Round 9

The answers to the following three problems are mutually dependent, although your answer to each willbe graded independently. Let A be the answer to problem 25, B the answer to problem 26, and C bethe answer to problem 27.

25. [12] Let XY Z be an equilateral triangle, and let K,L,M be points on sides XY, Y Z,ZX, respectively,such that XK/KY = B, Y L/LZ = 1/C, and ZM/MX = 1. Determine the ratio of the area oftriangle KLM to the area of triangle XY Z.

26. [12] Determine the positive real value of x for which√

2 + AC + 2Cx +√

AC − 2 + 2Ax =√

2(A + C)x + 2AC.

27. [12] In-Young generates a string of B zeroes and ones using the following method:

• First, she flips a fair coin. If it lands heads, her first digit will be a 0, and if it lands tails, herfirst digit will be a 1.

• For each subsequent bit, she flips an unfair coin, which lands heads with probability A. If thecoin lands heads, she writes down the number (zero or one) different from previous digit, while ifthe coin lands tails, she writes down the previous digit again.

What is the expected value of the number of zeroes in her string?

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4TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND

Round 10

28. [14] Determine the value of2011∑k=1

k − 1k!(2011 − k)!

.

29. [14] Let ABC be a triangle with AB = 4, BC = 8, and CA = 5. Let M be the midpoint of BC, andlet D be the point on the circumcircle of ABC so that segment AD intersects the interior of ABC,and ∠BAD = ∠CAM . Let AD intersect side BC at X. Compute the ratio AX/AD.

30. [14] Let S be a set of consecutive positive integers such that for any integer n in S, the sum of thedigits of n is not a multiple of 11. Determine the largest possible number of elements of S.

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4TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND

Round 11

31. [17] Each square in a 3 × 10 grid is colored black or white. Let N be the number of ways this can bedone in such a way that no five squares in an ‘X’ configuration (as shown by the black squares below)are all white or all black. Determine

√N .

32. [17] Find all real numbers x satisfying

x9 +98x6 +

2764

x3 − x +219512

= 0.

33. [17] Let ABC be a triangle with AB = 5, BC = 8, and CA = 7. Let Γ be a circle internally tangentto the circumcircle of ABC at A which is also tangent to segment BC. Γ intersects AB and AC atpoints D and E, respectively. Determine the length of segment DE.

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4TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND

Round 12

34. [20] The integer 843301 is prime. The primorial of a prime number p, denoted p#, is defined to bethe product of all prime numbers less than or equal to p. Determine the number of digits in 843301#.Your score will be

max{⌊

60(

13−

∣∣∣∣ln (A

d

)∣∣∣∣)⌋, 0

},

where A is your answer and d is the actual answer.

35. [20] Let G be the number of Google hits of “guts round” at 10:31PM on October 31, 2011. Let B bethe number of Bing hits of “guts round” at the same time. Determine B/G. Your score will be

max(

0,

⌊20

(1 − 20|a − k|

k

)⌋),

where k is the actual answer and a is your answer.

36. [20] Order any subset of the following twentieth century mathematical achievements chronologically,from earliest to most recent. If you correctly place at least six of the events in order, your score will be2(n − 5), where n is the number of events in your sequence; otherwise, your score will be zero. Note:if you order any number of events with one error, your score will be zero.

A). Axioms for Set Theory published by Zermelo

B). Category Theory introduced by Mac Lane and Eilenberg

C). Collatz Conjecture proposed

D). Erdos number defined by Goffman

E). First United States delegation sent to International Mathematical Olympiad

F). Four Color Theorem proven with computer assistance by Appel and Haken

G). Harvard-MIT Math Tournament founded

H). Hierarchy of grammars described by Chomsky

I). Hilbert Problems stated

J). Incompleteness Theorems published by Godel

K). Million dollar prize for Millennium Problems offered by Clay Mathematics Institute

L). Minimum number of shuffles needed to randomize a deck of cards established by Diaconis

M). Nash Equilibrium introduced in doctoral dissertation

N). Proof of Fermat’s Last Theorem completed by Wiles

O). Quicksort algorithm invented by Hoare

Write your answer as a list of letters, without any commas or parentheses.

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4th Annual Harvard-MIT November TournamentSaturday 12 November 2011

Guts Round

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School Team Team ID#

1. [5]

2. [5]

3. [5]

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School Team Team ID#

4. [6]

5. [6]

6. [6]

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School Team Team ID#

7. [7]

8. [7]

9. [7]

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School Team Team ID#

10. [8]

11. [8]

12. [8]

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School Team Team ID#

13. [8]

14. [8]

15. [8]

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School Team Team ID#

16. [10]

17. [10]

18. [10]

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School Team Team ID#

19. [10]

20. [10]

21. [10]

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School Team Team ID#

22. [12]

23. [12]

24. [12]

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School Team Team ID#

25. [12]

26. [12]

27. [12]

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School Team Team ID#

28. [14]

29. [14]

30. [14]

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School Team Team ID#

31. [17]

32. [17]

33. [17]

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School Team Team ID#

34. [20]

35. [20]

36. [20]

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4th

Annual Harvard-MIT November TournamentSaturday 12 November 2011

Guts Round

1. [5] Determine the remainder when 1 + 2 + · · · + 2014 is divided by 2012.

Answer: 1009 We wish to find the value of 1 + 2 + · · · + 2014 modulo 2012. We have

1 + 2 + · · · + 2014 =1

2(2014)(2015) = 1007 · 2015 ≡ 1007 · 3 = 3021 ≡ 1009 (mod 2012).

Remark: Note that, since 2 is not relatively prime to 2012, that this is not the same as

1

2(2)(3) ≡ 3 (mod 2012).

2. [5] Let ABCD be a rectangle with AB = 6 and BC = 4. Let E be the point on BC with BE = 3,and let F be the point on segment AE such that F lies halfway between the segments AB and CD. IfG is the point of intersection of DF and BC, find BG.

Answer: 1 Note that since F is a point halfway between AB and AC, the diagram must besymmetric about the line through F parallel to AB. Hence, G must be the reflection of E across themidpoint of BC. Therefore, BG = EC = 1.

3. [5] Let x be a real number such that 2x = 3. Determine the value of 43x+2.

Answer: 11664 We have

43x+2 = 43x · 42 = (22)3x · 16 = 26x · 16 = (2x)6 · 16 = 36 · 16 = 11664.

4. [6] Determine which of the following numbers is smallest in value: 54√

3, 144, 108√

6 − 108√

2.

Answer: 54√

3 We can first compare 54√

3 and 144. Note that√

3 < 2 and 144

54= 8

3> 2. Hence,

54√

3 is less. Now, we wish to compare this to 108√

6− 108√

2. This is equivalent to comparing√

3 to2(√

6−√

2). We claim that√

3 < 2(√

6−√

2). To prove this, square both sides to get 3 < 4(8− 4√

3)

or√

3 < 29

16which is true because 29

2

162 = 841

256> 3. We can reverse this sequence of squarings because,

at each step, we make sure that both our values are positive after taking the square root. Hence, 54√

3is the smallest.

5. [6] Charlie folds an 17

2-inch by 11-inch piece of paper in half twice, each time along a straight line

parallel to one of the paper’s edges. What is the smallest possible perimeter of the piece after two suchfolds?

Answer: 39

2Note that a piece of paper is folded in half, one pair of opposite sides is preserved

and the other pair is halved. Hence, the net effect on the perimeter is to decrease it by one of the sidelengths. Hence, the original perimeter is 2

(17

2

)+ 2 · 11 = 39 and by considering the cases of folding

twice along one edge or folding once along each edge, one can see that this perimeter can be decreasedby at most 11 + 17

2= 39

2. Hence, the minimal perimeter is 39

2.

6. [6] To survive the coming Cambridge winter, Chim Tu doesn’t wear one T-shirt, but instead wears upto FOUR T-shirts, all in different colors. An outfit consists of three or more T-shirts, put on one ontop of the other in some order, such that two outfits are distinct if the sets of T-shirts used are differentor the sets of T-shirts used are the same but the order in which they are worn is different. Given thatChim Tu changes his outfit every three days, and otherwise never wears the same outfit twice, howmany days of winter can Chim Tu survive? (Needless to say, he only has four t-shirts.)

Guts Round

Answer: 144 We note that there are 4 choices for Chim Tu’s innermost T-shirt, 3 choices for thenext, and 2 choices for the next. At this point, he has exactly 1 T-shirt left, and 2 choices: either heputs that one on as well or he discards it. Thus, he has a total of 4 × 3 × 2 × 2 = 48 outfits, and cansurvive for 48 × 3 = 144 days.

7. [7] How many ordered triples of positive integers (a, b, c) are there for which a4b2c = 54000?

Answer: 16 We note that 54000 = 24 ×33 ×53. Hence, we must have a = 2a13a25a3 , b = 2b13b25b3 ,c = 2c13c25c3 . We look at each prime factor individually:

• 4a1 + 2b1 + c1 = 4 gives 4 solutions: (1, 0, 0), (0, 2, 0), (0, 1, 2), (0, 0, 4)

• 4a2 + 2b2 + c2 = 3 and 4a3 + 2b3 + c3 = 3 each give 2 solutions: (0, 1, 1), (0, 1, 3).

Hence, we have a total of 4 × 2 × 2 = 16 solutions.

8. [7] Let a, b, c be not necessarily distinct integers between 1 and 2011, inclusive. Find the smallest

possible value ofab + c

a + b + c.

Answer: 2

3We have

ab + c

a + b + c=

ab − a − b

a + b + c+ 1.

We note that ab−a−ba+b+c

< 0 ⇔ (a− 1)(b− 1) < 1, which only occurs when either a = 1 or b = 1. Withoutloss of generality, let a = 1. Then, we have a value of

−1

b + c + a+ 1.

We see that this is minimized when b and c are also minimized (so b = c = 1), for a value of 2

3.

9. [7] Unit circle Ω has points X,Y,Z on its circumference so that XY Z is an equilateral triangle. LetW be a point other than X in the plane such that triangle WY Z is also equilateral. Determine thearea of the region inside triangle WY Z that lies outside circle Ω.

Answer: 3√

3−π3

Let O be the center of the circle. Then, we note that since ∠WY Z = 60◦ = ∠Y XZ,

that Y W is tangent to Ω. Similarly, WZ is tangent to Ω. Now, we note that the circular segmentcorresponding to Y Z is equal to 1

3the area of Ω less the area of triangle OY Z. Hence, our total area

is

[WY Z] − 1

3[Ω] + [Y OZ] =

3√

3

4− 1

3π +

√3

4=

3√

3 − π

3.

X

Y

Z

WO1

Ω

10. [8] Determine the number of integers D such that whenever a and b are both real numbers with−1/4 < a, b < 1/4, then |a2 − Db2| < 1.

Answer: 32 We have

−1 < a2 − Db2 < 1 ⇒ a2 − 1

b2< D <

a2 + 1

b2.

Guts Round

We havea2 − 1

b2is maximal at −15 =

.252 − 1

.252and

a2 + 1

b2is minimal at

02 + 1

.252= 16. However, since

we cannot have a, b = ±.25, checking border cases of -15 and 16 shows that both of these values arepossible for D. Hence, −15 ≤ D ≤ 16, so there are 32 possible values of D.

11. [8] For positive integers m,n, let gcd(m,n) denote the largest positive integer that is a factor of bothm and n. Compute

91∑

n=1

gcd(n, 91).

Answer: 325 Since 91 = 7 × 13, we see that the possible values of gcd(n, 91) are 1, 7, 13, 91. For1 ≤ n ≤ 91, there is only one value of n such that gcd(n, 91) = 91. Then, we see that there are 12values of n for which gcd(n, 91) = 7 (namely, multiples of 7 other than 91), 6 values of n for whichgcd(n, 91) = 13 (the multiples of 13 other than 91), and 91 − 1 − 6 − 12 = 72 values of n for whichgcd(n, 91) = 1. Hence, our answer is 1 × 91 + 12 × 7 + 6 × 13 + 72 × 1 = 325.

12. [8] Joe has written 5 questions of different difficulties for a test with problems numbered 1 though 5.He wants to make sure that problem i is harder than problem j whenever i− j ≥ 3. In how many wayscan he order the problems for his test?

Answer: 25 We will write pi > pj for integers i, j when the ith problem is harder than the jthproblem. For the problem conditions to be true, we must have p4 > p1, p5 > p2, and p5 > p1.

Then, out of 5! = 120 total orderings, we see that in half of them satisfy p4 > p1 and half satisfyp5 > p2, and that these two events occur independently. Hence, there are

(1

2

) (1

2

)(120) = 30 orderings

which satisfy the first two conditions. Then, we see that there are 4!

2!2!= 6 orderings of p1, p2, p4, p5

which work; of these, only p4 > p1 > p5 > p2 violates the condition p5 > p1. Consequently, we have5

6(30) = 25 good problem orderings.

13. [8] Tac is dressing his cat to go outside. He has four indistinguishable socks, four indistinguishableshoes, and 4 indistinguishable show-shoes. In a hurry, Tac randomly pulls pieces of clothing out of adoor and tries to put them on a random one of his cat’s legs; however, Tac never tries to put morethan one of each type of clothing on each leg of his cat. What is the probability that, after Tac is done,the snow-shoe on each of his cat’s legs is on top of the shoe, which is on top of the sock?

Answer: 1

1296On each leg, Tac’s cat will get a shoe, a sock, and a snow-shoe in a random order.

Thus, the probability that they will be put on in order for any given leg is 1

3!= 1

6. Thus, the probability

that this will occur for all 4 legs is(

1

6

)4= 1

1296.

14. [8] Let AMOL be a quadrilateral with AM = 10, MO = 11, and OL = 12. Given that the perpen-dicular bisectors of sides AM and OL intersect at the midpoint of segment AO, find the length of sideLA.

Answer:√

77 Let D be the midpoint of AM and E be the midpoint of AO. Then, we note thatADE ∼ AMO, so M is a right angle. Similarly, L is a right angle. Consequently, we get that

AO2 = OM2 + AM2 ⇒ AL =√

AO2 − OL2 =√

112 + 102 − 122 =√

77.

Guts Round

A

M

O

L

D

E

15. [8] For positive integers n, let L(n) be the largest factor of n other than n itself. Determine the numberof ordered pairs of composite positive integers (m,n) for which L(m)L(n) = 80.

Answer: 12 Let x be an integer, and let px be the smallest prime factor of x. Then, if L(a) = x,we note that we must have a = px for some prime p ≤ px. (Otherwise, if p > px, then px

px

> x. If p is

composite, then kx > x for some factor k of x.)

So we have:

• L(a) = 2, 4, 8, 10, 16, 20, 40 ⇒ 1 value for a

• L(a) = 5 ⇒ 3 values for a

Hence, we note that, since m and n are composite, we cannot have L(m) = 1 or L(n) = 1, so thepossible pairs (L(m), L(n)) are (2, 40), (4, 20), (5, 16), (8, 10) and vice-versa.

We add the number of choices for each pair, and double since m and n are interchangeable, to get2(1 × 1 + 1 × 1 + 3 × 1 + 1 × 1) = 12 possible ordered pairs (m,n).

16. [10] A small fish is holding 17 cards, labeled 1 through 17, which he shuffles into a random order.Then, he notices that although the cards are not currently sorted in ascending order, he can sort theminto ascending order by removing one card and putting it back in a different position (at the beginning,between some two cards, or at the end). In how many possible orders could his cards currently be?

Answer: 256 Instead of looking at moves which put the cards in order, we start with the cardsin order and consider possible starting positions by backtracking one move: each of 17 cards can bemoved to 16 new places. But moving card k between card k + 1 and card k + 2 is equivalent tomoving card k + 1 between card k − 1 and card k. We note that these are the only possible pairs ofmoves which produce the same result, so we have double counted 16 moves. Thus, we have a total of17 × 16 − 16 = 256 possible initial positions.

17. [10] For a positive integer n, let p(n) denote the product of the positive integer factors of n. Determinethe number of factors n of 2310 for which p(n) is a perfect square.

Answer: 27 Note that 2310 = 2 × 3 × 5 × 7 × 11. In general, we see that if n has d(n) positive

integer factors, then p(n) = nd

2 since we can pair factors (d, nd) which multiply to n. As a result, p(n)

is a square if and only if n is a square or d is a multiple of 4.

Thus, because 2310 is not divisible by the square of any prime, we claim that for integers n dividing2310, p(n) is even if and only if n is not prime. Clearly, p(n) is simply equal to n when n is prime,and p(1) = 1, so it suffices to check the case when n is composite. Suppose that n = p1p2 · · · pk, wherek > 1 and {p1, . . . , pk} is some subset of {2, 3, 5, 7, 11}. Then, we see that n has 2k factors, and that4 | 2k, so p(n) is a square.

Guts Round

Since 2310 has 25 = 32 factors, five of which are prime, 27 of them have p(n) even.

18. [10] Consider a cube ABCDEFGH, where ABCD and EFGH are faces, and segments AE,BF,CG,DHare edges of the cube. Let P be the center of face EFGH, and let O be the center of the cube. Giventhat AG = 1, determine the area of triangle AOP .

Answer:√

2

24From AG = 1, we get that AE = 1√

3and AC =

√2√3. We note that triangle AOP is

located in the plane of rectangle ACGE. Since OP ‖ CG and O is halway between AC and EG, we

get that [AOP ] = 1

8[ACGE]. Hence, [AOP ] = 1

8( 1√

3)(

√2√3) =

√2

24.

A

B

C

D

E

F

G

H

O P

1√

3

√

2√

3

19. [10] Let ABCD be a rectangle with AB = 3 and BC = 7. Let W be a point on segment AB such thatAW = 1. Let X,Y,Z be points on segments BC,CD,DA, respectively, so that quadrilateral WXY Zis a rectangle, and BX < XC. Determine the length of segment BX.

Answer: 7−√

41

2We note that

∠Y XC = 90 − ∠WXB = ∠XWB = 90 − ∠AWZ = ∠AZW

gives us that XY C ∼= ZWA and XY Z ∼ WXB. Consequently, we get that Y C = AW = 1. FromXY Z ∼ WXB, we get that

BX

BW=

CY

CX⇒ BX

2=

1

7 − BX

from which we get

BX2 − 7BX + 2 = 0 ⇒ BX =7 −

√41

2

(since we have BX < CX).

A

B C

D

W

X

Y

Z

Guts Round

20. [10] The UEFA Champions League playoffs is a 16-team soccer tournament in which Spanish teamsalways win against non-Spanish teams. In each of 4 rounds, each remaining team is randomly pairedagainst one other team; the winner advances to the next round, and the loser is permanently knockedout of the tournament. If 3 of the 16 teams are Spanish, what is the probability that there are 2Spanish teams in the final round?

Answer: 4

5We note that the probability there are not two Spanish teams in the final two is the

probability that the 3 of them have already competed against each other in previous rounds. Notethat the random pairings in each round is equivalent, by the final round, to dividing the 16 into twogroups of 8 and taking a winner from each. Now, letting the Spanish teams be A, B, and C, once wefix the group in which A is contained, the probability that B is contained in this group as well is 7/15.Likewise, the probability that C will be in the same group as A and B is now 6/14. Our answer is thus

1 −(

7

15

)(6

14

)=

4

5.

21. [10] Let P (x) = x4 + 2x3 − 13x2 − 14x + 24 be a polynomial with roots r1, r2, r3, r4. Let Q be thequartic polynomial with roots r2

1, r22, r

23, r

24, such that the coefficient of the x4 term of Q is 1. Simplify

the quotient Q(x2)/P (x), leaving your answer in terms of x. (You may assume that x is not equal toany of r1, r2, r3, r4).

Answer: x4 − 2x3 − 13x2 + 14x + 24 We note that we must have

Q(x) = (x − r21)(x − r2

2)(x − r23)(x − r2

4) ⇒ Q(x2) = (x2 − r21)(x

2 − r22)(x

2 − r23)(x

2 − r24)

. Since P (x) = (x − r1)(x − r2)(x − r3)(x − r4), we get that

Q(x2)/P (x) = (x + r1)(x + r2)(x + r3)(x + r4).

Thus, Q(x2)/P (x) = (−1)4P (−x) = P (−x), so it follows that

Q(x2)/P (x) = x4 − 2x3 − 13x2 + 14x + 24.

22. [12] Let ABC be a triangle with AB = 23, BC = 24, and CA = 27. Let D be the point on segmentAC such that the incircles of triangles BAD and BCD are tangent. Determine the ratio CD/DA.

Answer: 14

13Let X,Z,E be the points of tangency of the incircle of ABD to AB,BD,DA respec-

tively. Let Y,Z, F be the points of tangency of the incircle of CBD to CB,BD,DC respectively. Wenote that

CB + BD + DC = CY + Y B + BZ + ZD + DF + FC = 2(CY ) + 2(BY ) + 2(DF )2(24) + 2(DF )

by equal tangents, and that similarly

AB + BD + DA = 2(23) + 2(DE).

Since DE = DZ = DF by equal tangents, we can subtract the equations above to get that

CB + CD − AB − AD = 2(24) − 24(23) ⇒ CD − DA = 1.

Since we know that CD + DA = 27, we get that CD = 14, DA = 13, so the desired ratio is 14

13.

Guts Round

A

B

CDE F

X Y

Z

23. [12] Let N = 5AB37C2, where A,B,C are digits between 0 and 9, inclusive, and N is a 7-digit positiveinteger. If N is divisible by 792, determine all possible ordered triples (A,B,C).

Answer: (0, 5, 5), (4, 5, 1), (6, 4, 9) First, note that 792 = 23 × 32 × 11. So we get that

8 | N ⇒ 8 | 7C2 ⇒ 8 | 10C + 6 ⇒ C = 1, 5, 9

9 | N ⇒ 9 | 5 + A + B + 3 + 7 + C + 2 ⇒ A + B + C = 1, 10, 19

11 | N ⇒ 11 | 5 − A + B − 3 + 7 − C + 2 ⇒ −A + B − C = −11, 0

Adding the last two equations, and noting that they sum to 2B, which must be even, we get thatB = 4, 5.

Checking values of C we get possible triplets of (0, 5, 5), (4, 5, 1), and (6, 4, 9).

24. [12] Three not necessarily distinct positive integers between 1 and 99, inclusive, are written in a rowon a blackboard. Then, the numbers, without including any leading zeros, are concatenated to form anew integer N . For example, if the integers written, in order, are 25, 6, and 12, then N = 25612 (andnot N = 250612). Determine the number of possible values of N .

Answer: 825957 We will divide this into cases based on the number of digits of N .

• Case 1: 6 digits. Then each of the three numbers must have two digits, so we have 90 choices foreach. So we have a total of 903 = 729000 possibilities.

• Case 2: 5 digits. Then, exactly one of the three numbers is between 1 and 9, inclusive. Weconsider cases on the presence of 0s in N .

– No 0s. Then, we have 9 choices for each digit, for a total of 95 = 59049 choices.

– One 0. Then, the 0 can be the second, third, fourth, or fifth digit, and 9 choices for each ofthe other 4 digits. Then, we have a total of 4 × 94 = 26244 choices.

– Two 0s. Then, there must be at least one digit between them and they cannot be in the firstdigit, giving us 3 choices for the positioning of the 0s. Then, we have a total of 3 ∗ 93 = 2187choices.

So we have a total of 59049 + 26244 + 2187 = 87480 choices in this case.

• Case 3: 4 digits. Again, we casework on the presence of 0s.

– No 0s. Then, we have 94 = 6561 choices.

– One 0. Then, the 0 can go in the second, third, or fourth digit, so we have 3 × 93 = 2187choices.

So we have a total of 6561 + 2187 = 8748 choices in this case.

• Case 4: 3 digits. Then, we cannot have any 0s, so we have a total of 93 = 729 choices.

Hence, we have a total of 729000 + 87480 + 8748 + 729 = 825957 choices for N .

Guts Round

25. [12] Let XY Z be an equilateral triangle, and let K,L,M be points on sides XY, Y Z,ZX, respectively,such that XK/KY = B, Y L/LZ = 1/C, and ZM/MX = 1. Determine the ratio of the area oftriangle KLM to the area of triangle XY Z.

Answer: 1

5First, we note that

[KLM ] = [XY Z] − [XKM ] − [Y LK] − [ZML].

Then, note that

[XKM ] =XK

XY· XM

XZ· [XY Z] =

B

B + 1· 1

2· [XY Z]

[Y LK] =Y L

Y Z· Y K

Y X· [XY Z] =

1

C + 1· 1

B + 1· [XY Z]

[ZML] =ZM

ZX· ZL

ZY· [XY Z] =

1

2· 1

C + 1· [XY Z]

Consequently,

A =[KLM ]

[XY Z]

= 1 − B

B + 1· 1

2− 1

C + 1· 1

B + 1− C

C + 1· 1

2

=B + C

(B + 1)(C + 1)(2)

If we solve our system of equations for A,B,C, we get that A = 1

5.

X

Y

Z

K

L

M1

2

1

2

2

3

1

3

4

5

1

5

26. [12] Determine the positive real value of x for which

√2 + AC + 2Cx +

√AC − 2 + 2Ax =

√2(A + C)x + 2AC.

Answer: 4 Note that if we have√

a +√

b =√

a + b for non-negative reals a, b, then squaring gives

us that 2√

ab = 0, so that either a = 0 or b = 0.

Now, note that(2 + AC + 2Cx) + (AC − 2 + 2Ax) = (2(A + C)x + 2AC).

Guts Round

Consequently, either (2 + AC + 2Cx) or (AC − 2 + 2Ax) must be equal to 0. However, we observefrom the problems that both A, C, and x must be non-negative, so (2 + AC + 2Cx) > 0. As a result,we know that AC − 2 + 2Ax = 0, or that

B = x =2 − AC

2A.

If we solve our system of equations for A,B,C, we get that B = 4.

27. [12] In-Young generates a string of B zeroes and ones using the following method:

• First, she flips a fair coin. If it lands heads, her first digit will be a 0, and if it lands tails, herfirst digit will be a 1.

• For each subsequent bit, she flips an unfair coin, which lands heads with probability A. If thecoin lands heads, she writes down the number (zero or one) different from previous digit, while ifthe coin lands tails, she writes down the previous digit again.

What is the expected value of the number of zeroes in her string?

Answer: 2 Since each digit is dependent on the previous, and the first digit is random, we note thatthe probability that In Young obtains a particular string is the same probability as that she obtainsthe inverse string (i.e. that where the positions of the 0s and 1s are swapped). Consequently, we wouldexpect that half of her digits are 0s, so that

C =B

2.

If we solve our system of equations for A,B,C, we get that C = 2.

Solution of the system of equations for Problems 25, 26, 27:

Thus, we have the three equations

A =B + C

(B + 1)(C + 1)(2), B =

2 − AC

2A, C =

B

2

Plugging the last equation into the first two results in

A =3B

(B + 1)(B + 2)(2)⇒ B =

4 − AB

4A

Rearranging the second equation gives

4AB = 4 − AB ⇒ AB =4

5⇒ A =

4

5B

Then, plugging this into the first equation gives

4

5B=

3B

(B + 1)(B + 2)(2)

⇒ 15B2 = 8B2 + 24B + 16

⇒ 7B2 − 24B − 16 = 0

⇒ (7B + 4)(B − 4) = 0

Since we know that B > 0, we get that B = 4. Plugging this back in gives A = 1

5and C = 2.

Guts Round

28. [14] Determine the value of2011∑

k=1

k − 1

k!(2011 − k)!.

Answer:2009(22010)+1

2011!We note that

(2011!)2011∑

k=1

k − 1

k!(2011 − k)!=

2011∑

k=1

(2011!)(k − 1)

k!(2011 − k)!

=

2011∑

k=1

k(2011)!

k!(2011 − k)!−

2011∑

k=1

2011!

k!(2011 − k)!

=

2011∑

k=1

k

(2011

k

)−

2011∑

k=1

(2011

k

)

= (2011)(22010) − (22011 − 1)

Thus, we get an answer of(2009

(22010

)+ 1

)/(2011!).

Note: To compute the last two sums, observe that

2011∑

k=0

(2011

k

)= (1 + 1)2011 = 22011

by the Binomial Theorem, and that

2011∑

k=0

k

(2011

k

)=

1

2

(2011∑

k=0

k

(2011

k

)+

2011∑

k=0

(2011 − k)

(2011

2011 − k

))= 2011

(22010

).

29. [14] Let ABC be a triangle with AB = 4, BC = 8, and CA = 5. Let M be the midpoint of BC, andlet D be the point on the circumcircle of ABC so that segment AD intersects the interior of ABC,and ∠BAD = ∠CAM . Let AD intersect side BC at X. Compute the ratio AX/AD.

Answer: 9

41Let E be the intersection of AM with the circumcircle of ABC. We note that, by

equal angles ADC ∼ ABM , so that

AD = AC(AB

AM) =

20

AM.

Using the law of cosines on ABC, we get that

cos B =42 + 82 − 52

2(4)(8)=

55

64.

Then, using the law of cosines on ABM, we get that

AM =√

42 + 42 − 2(4)(4) cos B =3√2⇒ AD =

20√

2

3.

Applying Power of a Point on M ,

(AM)(ME) = (BM)(MC) ⇒ ME =16√

2

3⇒ AE =

41√

2

6.

Then, we note that AXB ∼ ACE, so that

AX = AB(AC

AE) =

60√

2

41⇒ AX

AD=

9

41

Guts Round

A

B C

D E

X M

30. [14] Let S be a set of consecutive positive integers such that for any integer n in S, the sum of thedigits of n is not a multiple of 11. Determine the largest possible number of elements of S.

Answer: 38 We claim that the answer is 38. This can be achieved by taking the smallest integerin the set to be 999981. Then, our sums of digits of the integers in the set are

45, . . . , 53, 45, . . . , 54, 1, . . . , 10, 2, . . . , 10,

none of which are divisible by 11.

Suppose now that we can find a larger set S: then we can then take a 39-element subset of S whichhas the same property. Note that this implies that there are consecutive integers a − 1, a, a + 1 forwhich 10b, . . . , 10b + 9 are all in S for b = a− 1, a, a + 1. Now, let 10a have sum of digits N . Then, thesums of digits of 10a + 1, 10a + 2, . . . , 10a + 9 are N + 1, N + 2, . . . , N + 9, respectively, and it followsthat n ≡ 1 (mod 11).

If the tens digit of 10a is not 9, note that 10(a + 1) + 9 has sum of digits N + 10, which is divisible by11, a contradiction. On the other hand, if the tens digit of 10a is 9, the sum of digits of 10(a − 1) isN − 1, which is also divisible by 11. Thus, S has at most 38 elements.

Motivation: We want to focus on subsets of S of the form {10a, . . . , 10a + 9}, since the sum of digitsgoes up by 1 most of the time. If the tens digit of 10a is anything other than 0 or 9, we see that S canat most contain the integers between 10a−8 and 10a+18, inclusive. However, we can attempt to make10(a− 1) + 9 have sum of digits congruent to N + 9 modulo 11, as to be able to add as many integersto the beginning as possible, which can be achieved by making 10(a − 1) + 9 end in the appropriatenumber of nines. We see that we want to take 10(a − 1) + 9 = 999999 so that the sum of digits uponadding 1 goes down by 53 ≡ 9 (mod 11), giving the example we constructed previously.

31. [17] Each square in a 3 × 10 grid is colored black or white. Let N be the number of ways this can bedone in such a way that no five squares in an ‘X’ configuration (as shown by the black squares below)are all white or all black. Determine

√N .

Guts Round

Answer: 25636 Note that we may label half of the cells in our board the number 0 and the otherhalf 1, in such a way that squares labeled 0 are adjacent only to squares labeled 1 and vice versa. Inother words, we make this labeling in a ’checkerboard’ pattern. Since cells in an ’X’ formation are alllabeled with the same number, the number of ways to color the cells labeled 0 is

√N , and the same is

true of coloring the cells labeled 1.

Let a2n be the number of ways to color the squares labeled 0 in a 3 by 2n grid without a monochromatic’X’ formation; we want to find a10. Without loss of generality, let the rightmost column of our gridhave two cells labeled 0. Let b2n be the number of such colorings on a 3 by 2n grid which do not havetwo black squares in the rightmost column and do not contain a monochromatic ’X’, which we note isalso the number of such colorings which do not have two white squares in the rightmost column.

Now, we will establish a recursion on a2n and b2n.We have two cases:

• Case 1: All three squares in the last two columns are the same color. For a2n, there are 2 ways tocolor these last three squares, and for b2n there is 1 way to color them. Then, we see that thereare b2n−2 ways to color the remaining 2n − 2 columns.

• Case 2: The last three squares are not all the same color. For a2n, there are 6 ways to color thelast three squares, and for b2n there are 5 ways to color them. Then, there are a2n−2 ways tocolor the remaining 2n − 2 columns.

Consequently, we get the recursions a2n = 6a2n−2 + 2b2n−2 and b2n = 5a2n−2 + b2n−2. From the firstequation, we get that b2n = 1

2a2n+2 − 3a2n. Plugging this in to the second equations results in the

recurion1

2a2n+2 − 3a2n = 5a2n−2 +

1

2a2n − 3a2n−2 ⇒ a2n+2 = 7a2n + 4a2n−2.

Now, we can easily see that a0 = 1 and a2 = 23 = 8, so we compute a10 = 25636.

32. [17] Find all real numbers x satisfying

x9 +9

8x6 +

27

64x3 − x +

219

512= 0.

Answer: 1

2, −1±

√13

4Note that we can re-write the given equation as

3

√x − 3

8= x3 +

3

8.

Furthermore, the functions of x on either side, we see, are inverses of each other and increasing. Let

f(x) = 3

√x − 3

8. Suppose that f(x) = y = f−1(x). Then, f(y) = x. However, if x < y, we have

f(x) > f(y), contradicting the fact that f is increasing, and similarly, if y < x, we have f(x) < f(y),again a contradiction. Therefore, if f(x) = f−1(x) and both are increasing functions in x, we requiref(x) = x. This gives the cubic

x3 − x +3

8= 0 →

(x − 1

2

)(x2 +

1

2x − 3

4

)= 0,

giving x =1

2,−1 ±

√13

4.

33. [17] Let ABC be a triangle with AB = 5, BC = 8, and CA = 7. Let Γ be a circle internally tangentto the circumcircle of ABC at A which is also tangent to segment BC. Γ intersects AB and AC atpoints D and E, respectively. Determine the length of segment DE.

Answer: 40

9

Guts Round

A

B C

DE

O

MX

Y

Z

P

First, note that a homothety h centered at A takes Γ to the circumcircle of ABC, D to B and E to C,since the two circles are tangent. As a result, we have DE ‖ BC. Now, let P be the center of Γ andO be the circumcenter of ABC: by the homothety h, we have DE/BC = AP/AO.

Let Γ be tangent to BC at X, and let ray−−→AX meet the circumcircle of ABC at Y . Note that Y is the

image of X under h. Furthermore, h takes BC to the tangent line l to the circumcircle of ABC at Y ,and since BC ‖ l, we must have that Y is the midpoint of arc B̂C. Therefore, AX bisects ∠BAC.

Now, let Z be the foot of the altitude from A to BC, and let M be the midpoint of BC, so thatOM ⊥ BC. Note that AP/AO = ZX/ZM . Now, letting BC = a = 8, CA = b = 7, and AB = c = 5,we compute

BZ = c cos B =c2 + a2 − b2

2a=

5

2

by the Law of Cosines,

BX =ac

b + c=

10

3

by the Angle Bisector Theorem, andBM = 4.

To finish,

DE =(AP )(BC)

AO=

(ZX)(BC)

ZM=

(5/6)(8)

(3/2)=

40

9.

34. [20] The integer 843301 is prime. The primorial of a prime number p, denoted p#, is defined to bethe product of all prime numbers less than or equal to p. Determine the number of digits in 843301#.Your score will be

max

{⌊60

(1

3−

∣∣∣∣ln(

A

d

)∣∣∣∣)⌋

, 0

},

where A is your answer and d is the actual answer.

Answer: 365851 Remark: 843301# − 1 is the largest known prime number of the form p# − 1,where p is prime.

35. [20] Let G be the number of Google hits of “guts round” at 10:31PM on October 31, 2011. Let B bethe number of Bing hits of “guts round” at the same time. Determine B/G. Your score will be

max

(0,

⌊20

(1 − 20|a − k|

k

)⌋),

Guts Round

where k is the actual answer and a is your answer.

Answer: .82721 The number of Google hits was 7350. The number of Bing hits was 6080. Theanswer is thus 6080/7350 = .82721.

36. [20] Order any subset of the following twentieth century mathematical achievements chronologically,from earliest to most recent. If you correctly place at least six of the events in order, your score will be2(n − 5), where n is the number of events in your sequence; otherwise, your score will be zero. Note:if you order any number of events with one error, your score will be zero.

A). Axioms for Set Theory published by Zermelo

B). Category Theory introduced by Mac Lane and Eilenberg

C). Collatz Conjecture proposed

D). Erdos number defined by Goffman

E). First United States delegation sent to International Mathematical Olympiad

F). Four Color Theorem proven with computer assistance by Appel and Haken

G). Harvard-MIT Math Tournament founded

H). Hierarchy of grammars described by Chomsky

I). Hilbert Problems stated

J). Incompleteness Theorems published by Godel

K). Million dollar prize for Millennium Problems offered by Clay Mathematics Institute

L). Minimum number of shuffles needed to randomize a deck of cards established by Diaconis

M). Nash Equilibrium introduced in doctoral dissertation

N). Proof of Fermat’s Last Theorem completed by Wiles

O). Quicksort algorithm invented by Hoare

Write your answer as a list of letters, without any commas or parentheses.

Answer: IAJCBMHODEFLNGK The dates are as follows:

A). Axioms for Set Theory published by Zermelo 1908

B). Category Theory introduced by Mac Lane and Eilenberg 1942-1945

C). Collatz Conjecture proposed 1937

D). Erdos number defined by Goffman 1969

E). First United States delegation sent to International Mathematical Olympiad 1974

F). Four Color Theorem proven with computer assistance by Appel and Haken 1976

G). Harvard-MIT Math Tournament founded 1998

H). Hierarchy of grammars described by Chomsky 1956

I). Hilbert Problems stated 1900

J). Incompleteness Theorems published by Godel 1931

K). Million dollar prize for Millennium Problems offered by Clay Mathematics Institute 2000

L). Minimum number of shuffles needed to randomize a deck of cards established by Diaconis 1992

M). Nash Equilibrium introduced in doctoral dissertation 1950

N). Proof of Fermat’s Last Theorem completed by Wiles 1994

O). Quicksort algorithm invented by Hoare 1960

so the answer is IAJCBMHODEFLNGK.

Guts Round

15th

Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Team A

1. [20] Let ABC be a triangle. Let the angle bisector of ∠A and the perpendicular bisector of BC

intersect at D. Then let E and F be points on AB and AC such that DE and DF are perpendicularto AB and AC, respectively. Prove that BE = CF .

2. [20]

(a) For what positive integers n do there exist functions f, g : {1, 2, . . . , n} → {1, 2, . . . , n} such thatfor all 1 ≤ i ≤ n, we have that exactly one of f(g(i)) = i and g(f(i)) = i holds?

(b) What if f, g must be permutations?

3. [20] Alice and Bob are playing a game of Token Tag, played on an 8×8 chessboard. At the beginningof the game, Bob places a token for each player somewhere on the board. After this, in every round,Alice moves her token, then Bob moves his token. If at any point in a round the two tokens are on thesame square, Alice immediately wins. If Alice has not won by the end of 2012 rounds, then Bob wins.

(a) Suppose that a token can legally move to any orthogonally adjacent square. Show that Bob hasa winning strategy for this game.

(b) Suppose instead that a token can legally move to any square which shares a vertex with the squareit is currently on. Show that Alice has a winning strategy for this game.

4. [20] Let ABC be a triangle with AB < AC. Let M be the midpoint of BC. Line l is drawn throughM so that it is perpendicular to AM , and intersects line AB at point X and line AC at point Y . Provethat ∠BAC = 90◦ if and only if quadrilateral XBY C is cyclic.

5. [20] Purineqa is making a pizza for Arno. There are five toppings that she can put on the pizza:mushrooms, olives, green peppers, cheese, and pepperoni. However, Arno is very picky and only likessome subset of the five toppings. Purineqa makes 5 pizzas, each with some subset of the five toppings,then for each of those Arno tells her if that pizza has any toppings he does not like. Purineqa choosesthese pizzas such that no matter which toppings Arno likes, she can then make him a sixth pizza withall the toppings he likes and no others. What are all possible combinations of the five initial pizzas forthis to be the case?

6. [30] It has recently been discovered that the right triangle with vertices (0,0), (0,2012), and (2012,0)is a giant pond home to many frogs. Frogs have the special ability that, if they are at a lattice point(x, y), they can hop to any of the three lattice points (x+1, y+1), (x−2, y+1), (x+1, y−2), assumingthe given lattice point lies in or on the boundary of the triangle.

Frog Jeff starts at the corner (0, 0), while Frog Kenny starts at the corner (0, 2012). Show that the setof points Jeff can reach is equal in size to the set of points that Kenny can reach.

7. [30] Five points are chosen on a sphere of radius 1. What is the maximum possible volume of theirconvex hull?

8. [30] For integer m,n ≥ 1, let A(m,n) denote the number of functions f : {1, 2, . . . , n} → {1, 2, . . . ,m}such that f(j) − f(i) ≥ j − i for all 1 ≤ i < j ≤ n, and let B(m,n) denote the number of functionsg : {0, 1, . . . , 2n + m} → {0, 1, . . . ,m} such that g(0) = 0, g(2n + m) = m, and |g(i)− g(i− 1)| = 1 forall 1 ≤ i ≤ 2n + m. Prove that A(m,n) = B(m,n).

For the remaining problems, let ϕ(k) denote the number of positive integers less than or

equal to k that are relatively prime to k.

9. [40] For any positive integer n, let N = φ(1) + φ(2) + ... + φ(n). Show that there exists a sequence

a1, a2, ...aN

containing exactly φ(k) instances of k for all positive integers k ≤ n such that

1a1a2

+ 1a2a3

+ ... + 1aN a1

= 1.

10. [40] For positive odd integer n, let f(n) denote the number of matrices A satisfying the followingconditions:

• A is n × n.

• Each row and column contains each of 1, 2, . . . , n exactly once in some order.

• AT = A. (That is, the element in row i and column j is equal to the one in row j and column i,for all 1 ≤ i, j ≤ n.)

Prove that f(n) ≥ n!(n−1)!ϕ(n) , where ϕ is the totient function.

15th

Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Team A

1. [20] Let ABC be a triangle with AB < AC. Let the angle bisector of ∠A and the perpendicularbisector of BC intersect at D. Then let E and F be points on AB and AC such that DE and DF areperpendicular to AB and AC, respectively. Prove that BE = CF .

Answer: see below Note that DE,DF are the distances from D to AB,AC, respectively, andbecause AD is the angle bisector of ∠BAC, we have DE = DF . Also, DB = DC because D is on theperpendicular bisector of BC. Finally, ∠DEB = ∠DFC = 90◦, so it follows that DEB ∼= DFC, andBE = CF .

2. [20] For what positive integers n do there exist functions f, g : {1, 2, . . . , n} → {1, 2, . . . , n} such thatfor each 1 ≤ i ≤ n, either f(g(i)) = i or g(f(i)) = i, but not both?

Answer: n even We claim that this is possible for all even n. First, a construction: set f(2m−1) =f(2m) = 2m− 1 and g(2m− 1) = g(2m) = 2m for m = 1, . . . , n

2 . It is easy to verify that this solutionworks.

Now, we show that this is impossible for odd n. Without loss of generality, suppose that f(g(1)) = 1and that g(1) = a ⇒ f(a) = 1. Then, we have g(f(a)) = g(a) = 1. Consequently, a 6= 1. In this case,call 1 and a a pair (we likewise regard i and j as a pair when g(f(i)) = i and f(i) = j). Now, to showthat n is even it suffices to show that all pairs are disjoint. Suppose for the sake of contradiction thatsome integer b 6= a is also in a pair with 1 (note that 1 is arbitrary). Then, we have f(g(b)) = b, g(b) = 1or g(f(b)) = b, f(b) = 1. But we already know that g(1) = a, so we must have f(g(b)) = b, g(b) = 1.But that would mean that both f(g(1)) = 1 and g(f(1)) = 1, a contradiction.

3. [20] Alice and Bob are playing a game of Token Tag, played on an 8× 8 chessboard. At the beginningof the game, Bob places a token for each player on the board. After this, in every round, Alice movesher token, then Bob moves his token. If at any point in a round the two tokens are on the same square,Alice immediately wins. If Alice has not won by the end of 2012 rounds, then Bob wins.

(a) Suppose that a token can legally move to any horizontally or vertically adjacent square. Showthat Bob has a winning strategy for this game.

(b) Suppose instead that a token can legally move to any horizontally, vertically, or diagonally adjacentsquare. Show that Alice has a winning strategy for this game.

Answer: see below For part (a), color the checkerboard in the standard way so that half of thesquares are black and the other half are white. Bob’s winning strategy is to place the two coins onthe same color, so that Alice must always move her coin on to a square with the opposite color as thesquare containing Bob’s coin.

For part (b), consider any starting configuration. By considering only the column that the tokens arein, it is easy to see that Alice can get to the same column as Bob (immediately after her move) in 7rounds. (This is just a game on a 1×8 chessboard.) Following this, Alice can stay on the same columnas Bob each turn, while getting to the same row as him. This too also takes at most 7 rounds. Thus,Alice can catch Bob in 14 < 2012 rounds from any starting position.

4. [20] Let ABC be a triangle with AB < AC. Let M be the midpoint of BC. Line l is drawn throughM so that it is perpendicular to AM , and intersects line AB at point X and line AC at point Y . Provethat ∠BAC = 90◦ if and only if quadrilateral XBY C is cyclic.

Answer: see below

Team A

A

B CM

X

Yl

p

P

First, note that XBY C cyclic is equivalent to ∡BXM = ∡ACB. However, note that ∡BXM =90◦ − ∡BAM , so XBY C cyclic is in turn equivalent to ∡BAM + ∡ACB = 90◦.

Let the line tangent to the circumcircle of △ABC at A be p, and let P be an arbitrary point on p onthe same side of AM as B. Note that ∡PAB = ∡ACB. If ∡ACB = 90◦ − ∡BAM we have l ⊥ AMand thus the circumcenter O of △ABC lies on AM . Since AB < AC, we must have O = M , and∡BAC = 90◦. Conversely, if ∡BAC = 90◦, ∡PAM = 90◦, and it follows that ∡ACB = 90◦−∡BAM .

5. [20] Purineqa is making a pizza for Arno. There are five toppings that she can put on the pizza.However, Arno is very picky and only likes some subset of the five toppings. Purineqa makes fivepizzas, each with some subset of the five toppings. For each pizza, Arno states (with either a “yes” ora “no”) if the pizza has any toppings that he does not like. Purineqa chooses these pizzas such that nomatter which toppings Arno likes, she has enough information to make him a sixth pizza with all thetoppings he likes and no others. What are all possible combinations of the five initial pizzas for thisto be the case?

Answer: see below We claim the only way for Purineqa to deduce Arno’s preferences is for eachpizza to contain exactly one topping, with no topping be repeated. It is obvious that he can deducethe toppings in this case.

We now claim that this is not possible with any other combination. Suppose that Arno tells Purineqathat he does not like any of the five pizzas. Then, Purineqa should be able to rule out at least one ofthe possibilities that Arno likes none of the toppings and that Arno likes exactly one of the toppingsT . It is clear that this is possible if and only if there is a pizza with only T on it. This is true for allfive toppings T , so we’re done.

6. [30] It has recently been discovered that the right triangle with vertices (0,0), (0,2012), and (2012,0)is a giant pond that is home to many frogs. Frogs have the special ability that, if they are at a latticepoint (x, y), they can hop to any of the three lattice points (x+1, y+1), (x−2, y+1), and (x+1, y−2),assuming the given lattice point lies in or on the boundary of the triangle.

Frog Jeff starts at the corner (0, 0), while Frog Kenny starts at the corner (0, 2012). Show that the setof points that Jeff can reach is equal in size to the set of points that Kenny can reach.

Answer: see below

Team A

(x, y) 7→ (x + y

2, y

√

3

2)

We transform the triangle as follows: map each lattice point (x, y) to the point

x(1, 0) + y(1/2,√

3/2) = (x + y/2, y√

3/2).

This transforms the right triangle into an equilateral triangle as shown above.

Now, the three allowed movements

(x, y) 7→ (x + 1, y + 1),

(x, y) 7→ (x − 2, y + 1),

(x, y) 7→ (x + 1, y − 2)

become the movements

(x, y) 7→ (x + 3/2, y +√

3/2),

(x, y) 7→ (x − 3/2, y +√

3/2),

(x, y) 7→ (x, y −√

3)

That is, each step is a movement of√

3 in any of these three directions, which are separated by 120◦

angles. The pond is now completely symmetrical with respect to 120◦ rotations, so it does not matterwhich vertex you start at. The lower left vertex corresponds to the original point (0, 0), and the topvertex corresponds to the original point (0, 2012).

7. [30] Five points are chosen on a sphere of radius 1. What is the maximum possible volume of theirconvex hull?

Answer:√

32 Let the points be A,B,C,X, Y so that X and Y are on opposite sides of the plane

defined by triangle ABC. The volume is 1/3 the product of the area of ABC and sum of the distancesfrom X and Y to the plane defined by ABC. The area of ABC is maximized when the plane containingit intersects the sphere in the largest possible cross section, and ABC is equilateral: this gives an areaof 3

√3/4. Then, the sum of the distances from X and Y to the plane of ABC is at most 2. This is

clearly obtainable when A, B, and C form an equilateral triangle circumscribed by the equator of thesphere, and X and Y are at the poles, and we get a volume of

√3/2.

8. [30] For integer n,m ≥ 1, let A(n,m) denote the number of functions f : {1, 2, . . . , n} → {1, 2, . . . ,m}such that f(j) − f(i) ≤ j − i for all 1 ≤ i < j ≤ n, and let B(n,m) denote the number of functionsg : {0, 1, . . . , 2n + m} → {0, 1, . . . ,m} such that g(0) = 0, g(2n + m) = m, and |g(i)− g(i− 1)| = 1 forall 1 ≤ i ≤ 2n + m. Prove that A(n,m) = B(n,m).

Team A

Answer: see below We first note that the condition for f is equivalent to i − f(i) ≤ j − f(j) forall 1 ≤ i < j ≤ n. Letting f ′(x) = x − f(x), we see this is equivalent to saying that f ′ is decreasing.Thus, we only need that f ′(x) ≤ f ′(x + 1); in other words, we only require the statement to be truefor j = i + 1.

Fix m,n. For any function g satisfying the conditions for B, we construct a function f satisfying theconditions for A as follows. For a given function g and 1 ≤ i ≤ 2n + m, say that g has a up step at i ifg(i)−g(i−1) = 1, and say it has a down step at i otherwise. We see that g must be composed of m+nup steps and n down steps. Let i1, i2, . . . , in be the indices for which down steps occur, in ascendingorder. Let f be the function such that f(j) = (m + 1) − g(ij) for 1 ≤ j ≤ n. By our argument in thefirst paragraph, it suffices to show that f(k)− f(k − 1) ≤ 1 for 1 < k ≤ n, or that g(ik−1)− 1 ≤ g(ik).If this were not the case for some k, then there would be at least 1 down step in between ik−1 and ik,a contradiction, so the condition indeed holds.

We now claim that this construction is a bijection. For injectivity, note that for any two distinctg, g′, there exists a k for which the values of g(ik), g′(i′k) are distinct, in which case the functionsf, f ′ must be distinct. For surjectivity, consider any suitable f . Let f ′ be the function such thatf ′(k) = (m + 1) − f(k) for all 1 ≤ k ≤ n. (The range of this function is still {1, 2, . . . ,m}.) Then, wecan find a g as follows: for each 1 ≤ k ≤ n in sequence, have g make up steps until it reaches the valuef ′(k), then take one down step. This is always possible, as f ′(k + 1) − f ′(k) ≤ 1. Thus, our claim istrue, and our proof is complete.

9. [40] For any positive integer n, let N = ϕ(1) + ϕ(2) + . . . + ϕ(n). Show that there exists a sequence

a1, a2, . . . , aN

containing exactly ϕ(k) instances of k for all positive integers k ≤ n such that

1a1a2

+ 1a2a3

+ · · · + 1aN a1

= 1.

Answer: see below We write all fractions of the form b/a, where a and b are relatively prime, and0 ≤ b ≤ a ≤ n, in ascending order. For instance, for n = 5, this is the sequence

01 , 1

5 , 14 , 1

3 , 25 , 1

2 , 35 , 2

3 , 34 , 4

5 , 11 .

This sequence is known as the Farey sequence.

Now, if we look at the the sequence of the denominators of the fractions, we see that k appears ϕ(k)times when k > 1, although 1 appears twice. Thus, there are N + 1 elements in the Farey sequence.Let the Farey sequence be

b1a1

, b2a2

, . . . , bN+1

aN+1

Now, aN+1 = 1, so the sequence a1, a2, . . . , aN contains ϕ(k) instances of k for every 1 ≤ k ≤ n. Weclaim that this sequence also satisfies

1a1a2

+ 1a2a3

+ · · · + 1aN a1

= 1.

Since a1 = aN+1 = 1, we have

1a1a2

+ 1a2a3

+ · · · + 1aN a1

= 1a1a2

+ 1a2a3

+ · · · + 1aN aN+1

.

Now, it will suffice to show that 1aiai+1

= bi+1

ai+1− bi

ai

. Once we have shown this, the above sum will

telescope to bN+1

aN+1− b1

a1= 1 − 0 = 1.

To see why 1aiai+1

= bi+1

ai+1− bi

ai

holds, we note that this is equivalent to 1 = bi+1ai − biai+1. We

can prove this fact geometrically: consider the triangle in the plane with vertices (0, 0), (ai, bi), and

Team A

(ai+1, bi+1). This triangle contains these three boundary points, but it contains no other boundary orinterior points since ai and ai+1 are relatively prime to bi and bi+1, respectively, and since no other

fraction with denominator at most n lies between bi

ai

and bi+1

ai+1. Thus, by Pick’s theorem, this triangle

has area 1/2. But the area of the triangle can also be computed as the cross product 12 (bi+1ai−biai+1);

hence bi+1ai − biai+1 = 1 and we are done.

10. [40] For positive odd integer n, let f(n) denote the number of matrices A satisfying the followingconditions:

• A is n × n.

• Each row and column contains each of 1, 2, . . . , n exactly once in some order.

• AT = A. (That is, the element in row i and column j is equal to the one in row j and column i,for all 1 ≤ i, j ≤ n.)

Prove that f(n) ≥ n!(n−1)!ϕ(n) .

Answer: see below We first note that main diagonal (the squares with row number equal tocolumn number) is a permutation of 1, 2, . . . , n. This is because each number i (1 ≤ i ≤ n) appearsan even number of times off the main diagonal, so must appear an odd number of times on the maindiagonal. Thus, we may assume that the main diagonal’s values are 1, 2, . . . , n in that order. Call anymatrix satifsying this condition and the problem conditions good. Let g(n) denote the number of good

matrices. It now remains to show that g(n) ≥ (n−1)!ϕ(n) .

Now, consider a round-robin tournament with n teams, labeled from 1 through n, with the matchesspread over n days such that on day i, all teams except team i play exactly one match (so there aren−1

2 pairings), and at the end of n days, each pair of teams has played exactly once. We consider twosuch tournaments distinct if there is some pairing of teams i, j which occurs on different days in thetournaments. We claim that the tournaments are in bijection with the good matrices.Proof of Claim: Given any good matrix A, we construct a tournament by making day k have matchesbetween team i and j for each i, j such that Ai,j = k, besides (i, j) = (k, k). Every pair will play someday, and since each column and row contains exactly one value of each number, no team will playmore than once a day. Furthermore, given two distinct good matrices, there exists a value (off themain diagonal) on which they differ; this value corresponds to the same pair playing on different dates,so the corresponding tournaments must be distinct. For the other direction, take any tournament.Make a matrix A with the main diagonal as 1, 2, . . . , n, and for each k, set Ai,j = k for each i, j suchthat teams i, j play each other on day k. This gives a good matrix. Similarly, given any two distincttournaments, there exists a team pair i, j which play each other on different days; this corresponds toa differing value on the corresponding good matrices.

It now suffices to exhibit (n−1)!ϕ(n) distinct tournaments. (It may be helpful here to think of the days in

the tournament as an unordered collection of sets of pairings, with the order implictly imposed by theteam not present in the set of pairings.) For our construction, consider a regular n-gon with centerO. Label the points as A1, A2, . . . , An as an arbitrary permutation (so there are n! possible labelings).The team k will be represented by Ak. For each k, consider the line AkO. The remaining n−1 verticescan be paired into n−1

2 groups which are perpendicular to this line; use these pairings for day k. Ofcourse, this doesn’t generate n! distinct tournaments– but how many does it make?

Consider any permutation of labels. Starting from an arbitrary point, let the points of the polygonbe Aπ(1), Aπ(2), . . . , Aπ(n) in clockwise order. Letting π(0) = π(n) and π(n + 1) = π(1), we note thatπ(i − 1) and π(i + 1) play each other on day π(i). We then see that any other permutation of labelsrepresenting the same tournament must have Aπ(i−1)Aπ(i) = Aπ(i)Aπ(i+1) for all i. Thus, if Aπ(1) isk vertices clockwise of Aπ(0), then Aπ(2) is k vertices clockwise of Aπ(1), and so on all the way upto Aπ(n−1) being k vertices clockwise of Aπ(n). This is only possible if k is relatively prime to n,sothere are ϕ(n) choices of k. There are n choices of the place to put Aπ(1), giving nϕ(n) choices ofpermutations meeting this condition. It is clear that each permutation meeting this condition providesthe same tournament, so the n! permutations can be partitioned into equivalence classes of size nϕ(n)each. Thus, there are n!

nϕ(n) distinct equivalence classes, and we are done.

Team A

15th

Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Team B

For problems 1-5, only a short answer is required.

1. [10] Triangle ABC has AB = 5, BC = 3√

2, and AC = 1. If the altitude from B to AC and theangle bisector of angle A intersect at at D, what is BD?

2. [10] You are given two line segments of length 2n for each integer 0 ≤ n ≤ 10. How many distinctnondegenerate triangles can you form with three of the segments? Two triangles are considered distinctif they aren’t congruent.

3. [10] Mac is trying to fill 2012 barrels with apple cider. He starts with 0 energy. Every minute, hemay rest, gaining 1 energy, or if he has n energy, he may expend k energy (1 ≤ k ≤ n) to fill up ton(k + 1) barrels with cider. What is the minimal number of minutes he needs to fill all the barrels?

4. [10] A restaurant has some number of seats, arranged in a line. Its customers are in parties arrangedin a queue. To seat its customers, the restaurant takes the next party in the queue and attempts tosee all of the party’s member(s) in a contiguous block of unoccupied seats. If one or more such blocksexist, then the restaurant places the party in an arbitrarily selected block; otherwise, the party leaves.

Suppose the queue has parties of sizes 6, 4, 2, 5, 3, 1 from front to back, and all seats are initially empty.What is the minimal number of seats the restaurant needs to guarantee that it will seat all of thesecustomers?

5. [10] Steph and Jeff each start with the number 4, and Travis is flipping coins. Every time he flipsa heads, Steph replaces her number x with 2x − 1, and Jeff replaces his number y with y + 8. Everytime he flips a tails, Steph replaces her number x with x+1

2, and Jeff replaces his number y with y − 3.

After some (positive) number of coin flips, Steph and Jeff miraculously end up with the same numberbelow 2012. How many times was a coin flipped?

For problems 6-10, justification is required.

6. [20] Let ABC be a triangle. Let the angle bisector of ∠A and the perpendicular bisector of BC

intersect at D. Then let E and F be points on AB and AC such that DE and DF are perpendicularto AB and AC, respectively. Prove that BE = CF .

7. [20]

(a) For what positive integers n do there exist functions f, g : {1, 2, . . . , n} → {1, 2, . . . , n} such thatfor all 1 ≤ i ≤ n, we have that exactly one of f(g(i)) = i and g(f(i)) = i holds?

(b) What if f, g must be permutations?

8. [20] Alice and Bob are playing a game of Token Tag, played on an 8×8 chessboard. At the beginningof the game, Bob places a token for each player somewhere on the board. After this, in every round,Alice moves her token, then Bob moves his token. If at any point in a round the two tokens are on thesame square, Alice immediately wins. If Alice has not won by the end of 2012 rounds, then Bob wins.

(a) Suppose that a token can legally move to any orthogonally adjacent square. Show that Bob hasa winning strategy for this game.

(b) Suppose instead that a token can legally move to any square which shares a vertex with the squareit is currently on. Show that Alice has a winning strategy for this game.

9. [20] Let ABC be a triangle with AB < AC. Let M be the midpoint of BC. Line l is drawn throughM so that it is perpendicular to AM , and intersects line AB at point X and line AC at point Y . Provethat ∠BAC = 90◦ if and only if quadrilateral XBY C is cyclic.

10. [20] Purineqa is making a pizza for Arno. There are five toppings that she can put on the pizza:mushrooms, olives, green peppers, cheese, and pepperoni. However, Arno is very picky and only likessome subset of the five toppings. Purineqa makes 5 pizzas, each with some subset of the five toppings,then for each of those Arno tells her if that pizza has any toppings he does not like. Purineqa choosesthese pizzas such that no matter which toppings Arno likes, she can then make him a sixth pizza withall the toppings he likes and no others. What are all possible combinations of the five initial pizzas forthis to be the case?

15th

Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Team B

1. [10] Triangle ABC has AB = 5, BC = 3√

2, and AC = 1. If the altitude from B to AC and the anglebisector of angle A intersect at at D, what is BD?

Answer: 5

3Let E be the foot of the perpendicular from B to line AC. By the Law of Cosines,

cos ∠BAC = 4

5, and it follows that BE = 3 and AE = 4. Now, by the Angle Bisector Theorem,

BD

BE= AB

AB+AE= 5

9, so BD = 5

3.

2. [10] You are given two line segments of length 2n for each integer 0 ≤ n ≤ 10. How many distinctnondegenerate triangles can you form with three of the segments? Two triangles are considered distinctif they are not congruent.

Answer: 55 First, observe that if we have three sticks of distinct lengths 2a < 2b < 2c, then2a + 2b < 2b+1 ≤ 2c, so we cannot form a triangle. Thus, we must have (exactly) two of our sticks thesame length, so that our triangle has side lengths 2a, 2a, 2b. This triangle is non-degenerate if and onlyif 2a+1 > 2b, and since a 6= b, this happens if and only if a > b. Clearly, there are

(

11

2

)

= 55 ways tochoose such a, b.

3. [10] Mac is trying to fill 2012 barrels with apple cider. He starts with 0 energy. Every minute, he mayrest, gaining 1 energy, or if he has n energy, he may expend k energy (0 ≤ k ≤ n) to fill up to n(k + 1)barrels with cider. What is the minimal number of minutes he needs to fill all the barrels?

Answer: 46 First, suppose that Mac fills barrels during two consecutive minutes. Let his energyimmediately before doing so be n, and the energy spent in the next two minutes be k1, k2, respectively.It is not difficult to check that he can fill at least as many barrels by spending k1 + k2 + 1 energy andresting for an additional minute before doing so, so that his starting energy is n + 1: this does notchange the total amount of time. Furthermore, this does not affect the amount of energy Mac hasremaining afterward. We may thus assume that Mac first rests for a (not necessarily fixed) period oftime, then spends one minute filling barrels, and repeats this process until all of the barrels are filled.

Next, we check that he only needs to fill barrels once. Suppose that Mac first rests for n1 minutes,then spends k1 energy, and next rests for n2 minutes and spends k2 energy. It is again not difficult tocheck that Mac can instead rest for n1 + n2 + 1 minutes and spend k1 + k2 + 1 energy, to increase thenumber of barrels filled, while not changing the amount of time nor the energy remaining. Iteratingthis operation, we can reduce our problem to the case in which Mac first rests for n minutes, thenspends n energy filling all of the barrels.

We need n(n + 1) ≥ 2012, so n ≥ 45, and Mac needs a minimum of 46 minutes.

4. [10] A restaurant has some number of seats, arranged in a line. Its customers are in parties arrangedin a queue. To seat its customers, the restaurant takes the next party in the queue and attempts toseat all of the party’s member(s) in a contiguous block of unoccupied seats. If one or more such blocksexist, then the restaurant places the party in an arbitrarily selected block; otherwise, the party leaves.

Suppose the queue has parties of sizes 6, 4, 2, 5, 3, 1 from front to back, and all seats are initially empty.What is the minimal number of seats the restaurant needs to guarantee that it will seat all of thesecustomers?

Answer: 29 First, note that if there are only 28 seats, it is possible for the seating not to bepossible, in the following way. The party of six could be seated in such a way that the remainingcontiguous regions have sizes 10 and 12. Then, the party of 4 is seated in the middle of the region ofsize 12, and the party of 2 is seated in the middle of the region of size 10, so that the remaining regionsall of size 4. Now, it is impossible to seat the party of 5. It is clear that fewer than 28 seats also makeit impossible to guarantee that everyone can be seated.

Now, given 29 seats, clearly, the first party can be seated. Afterward, 23 seats remain in at most2 contiguous regions, one of which has to have size at least 4. Next, 19 seats remain in at most 3

Team B

contiguous regions, one of which has size at least 2. 17 seats in at most 4 contiguous regions remainfor the party of 5, and one region must have size at least 5. Finally, 12 seats in at most 5 contiguousregions are available for the party of 3, and the party of 1 can take any remaining seat. Our answer istherefore 29.

Remark: We can arrive at the answer of 29 by doing the above argument in reverse.

5. [10] Steph and Jeff each start with the number 4, and Travis is flipping a coin. Every time he flipsa heads, Steph replaces her number x with 2x − 1, and Jeff replaces his number y with y + 8. Everytime he flips a tails, Steph replaces her number x with x+1

2, and Jeff replaces his number y with y − 3.

After some (positive) number of coin flips, Steph and Jeff miraculously end up with the same numberbelow 2012. How many times was the coin flipped?

Answer: 137 Suppose that a heads and b tails are flipped. Jeff’s number at the end is 4 + 8a− 3b.Note that the operations which Steph applies are inverses of each other, and as a result it is not difficultto check by induction that her final number is simply 1 + 3 · 2a−b.

We now have 3+3(a− b)+5a = 3 ·2a−b. Letting n = a− b, we see that 2n −n−1 must be divisible by5, so that a is an integer. In particular, n is a positive integer. Furthermore, we have 1+3 · 2n < 2012,so that n ≤ 9. We see that the only possibility is for n = 7 = a − b, and thus 4 + 8a − 3b = 385.Solving, we get a = 72, b = 65, so our answer is 72 + 65 = 137.

6. [20] Let ABC be a triangle with AB < AC. Let the angle bisector of ∠A and the perpendicularbisector of BC intersect at D. Then let E and F be points on AB and AC such that DE and DF areperpendicular to AB and AC, respectively. Prove that BE = CF .

Answer: see below Note that DE,DF are the distances from D to AB,AC, respectively, andbecause AD is the angle bisector of ∠BAC, we have DE = DF . Also, DB = DC because D is on theperpendicular bisector of BC. Finally, ∠DEB = ∠DFC = 90◦, so it follows that DEB ∼= DFC, andBE = CF .

7. [20] For what positive integers n do there exist functions f, g : {1, 2, . . . , n} → {1, 2, . . . , n} such thatfor each 1 ≤ i ≤ n, either f(g(i)) = i or g(f(i)) = i, but not both?

Answer: n even We claim that this is possible for all even n. First, a construction: set f(2m−1) =f(2m) = 2m− 1 and g(2m− 1) = g(2m) = 2m for m = 1, . . . , n

2. It is easy to verify that this solution

works.

Now, we show that this is impossible for odd n. Without loss of generality, suppose that f(g(1)) = 1and that g(1) = a ⇒ f(a) = 1. Then, we have g(f(a)) = g(a) = 1. Consequently, a 6= 1. In this case,call 1 and a a pair (we likewise regard i and j as a pair when g(f(i)) = i and f(i) = j). Now, to showthat n is even it suffices to show that all pairs are disjoint. Suppose for the sake of contradiction thatsome integer b 6= a is also in a pair with 1 (note that 1 is arbitrary). Then, we have f(g(b)) = b, g(b) = 1or g(f(b)) = b, f(b) = 1. But we already know that g(1) = a, so we must have f(g(b)) = b, g(b) = 1.But that would mean that both f(g(1)) = 1 and g(f(1)) = 1, a contradiction.

8. [20] Alice and Bob are playing a game of Token Tag, played on an 8× 8 chessboard. At the beginningof the game, Bob places a token for each player on the board. After this, in every round, Alice movesher token, then Bob moves his token. If at any point in a round the two tokens are on the same square,Alice immediately wins. If Alice has not won by the end of 2012 rounds, then Bob wins.

(a) Suppose that a token can legally move to any horizontally or vertically adjacent square. Showthat Bob has a winning strategy for this game.

(b) Suppose instead that a token can legally move to any horizontally, vertically, or diagonally adjacentsquare. Show that Alice has a winning strategy for this game.

Answer: see below For part (a), color the checkerboard in the standard way so that half of thesquares are black and the other half are white. Bob’s winning strategy is to place the two coins onthe same color, so that Alice must always move her coin on to a square with the opposite color as thesquare containing Bob’s coin.

Team B

For part (b), consider any starting configuration. By considering only the column that the tokens arein, it is easy to see that Alice can get to the same column as Bob (immediately after her move) in 7rounds. (This is just a game on a 1×8 chessboard.) Following this, Alice can stay on the same columnas Bob each turn, while getting to the same row as him. This too also takes at most 7 rounds. Thus,Alice can catch Bob in 14 < 2012 rounds from any starting position.

9. [20] Let ABC be a triangle with AB < AC. Let M be the midpoint of BC. Line l is drawn throughM so that it is perpendicular to AM , and intersects line AB at point X and line AC at point Y . Provethat ∠BAC = 90◦ if and only if quadrilateral XBY C is cyclic.

Answer: see below

A

B CM

X

Yl

p

P

First, note that XBY C cyclic is equivalent to ∡BXM = ∡ACB. However, note that ∡BXM =90◦ − ∡BAM , so XBY C cyclic is in turn equivalent to ∡BAM + ∡ACB = 90◦.

Let the line tangent to the circumcircle of △ABC at A be p, and let P be an arbitrary point on p onthe same side of AM as B. Note that ∡PAB = ∡ACB. If ∡ACB = 90◦ − ∡BAM we have l ⊥ AM

and thus the circumcenter O of △ABC lies on AM . Since AB < AC, we must have O = M , and∡BAC = 90◦. Conversely, if ∡BAC = 90◦, ∡PAM = 90◦, and it follows that ∡ACB = 90◦−∡BAM .

10. [20] Purineqa is making a pizza for Arno. There are five toppings that she can put on the pizza.However, Arno is very picky and only likes some subset of the five toppings. Purineqa makes fivepizzas, each with some subset of the five toppings. For each pizza, Arno states (with either a “yes” ora “no”) if the pizza has any toppings that he does not like. Purineqa chooses these pizzas such that nomatter which toppings Arno likes, she has enough information to make him a sixth pizza with all thetoppings he likes and no others. What are all possible combinations of the five initial pizzas for thisto be the case?

Answer: see below We claim the only way for Purineqa to deduce Arno’s preferences is for eachpizza to contain exactly one topping, with no topping be repeated. It is obvious that he can deducethe toppings in this case.

We now claim that this is not possible with any other combination. Suppose that Arno tells Purineqathat he does not like any of the five pizzas. Then, Purineqa should be able to rule out at least one ofthe possibilities that Arno likes none of the toppings and that Arno likes exactly one of the toppingsT . It is clear that this is possible if and only if there is a pizza with only T on it. This is true for allfive toppings T , so we’re done.

Team B

ALGEBRA TEST

This test consists of 10 short-answer problems to be solved individually in 50 minutes. Problems will beweighted with point values after the contest based on how many competitors solve each problem. There isno penalty for guessing.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted otherthan the official translation sheets. Similarly, graph paper, rulers, protractors, compasses, and other drawingaids are not permitted.

Our goal is that a closed form answer equivalent to the correct answer will be accepted. However, we donot always have the resourses to determine whether a complicated or strange answer is equivalent to ours.To assist us in awarding you all the points that you deserve, you answers should be simplified as much aspossible. Answers must be exact unless otherwise specified.

Correct mathematical notation must be used. No partial credit will be given unless otherwise specified.

If you believe the test contains an error, please submit your protest in writing to Science Center 109 duringlunchtime.

Enjoy!

15th Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Algebra Test

1. Let f be the function such that

f(x) ={

2x if x ≤ 12

2 − 2x if x > 12

What is the total length of the graph of f(f(...f︸ ︷︷ ︸2012f ′s

(x)...)) from x = 0 to x = 1?

2. You are given an unlimited supply of red, blue, and yellow cards to form a hand. Each card has a pointvalue and your score is the sum of the point values of those cards. The point values are as follows: thevalue of each red card is 1, the value of each blue card is equal to twice the number of red cards, andthe value of each yellow card is equal to three times the number of blue cards. What is the maximumscore you can get with fifteen cards?

3. Given points a and b in the plane, let a⊕b be the unique point c such that abc is an equilateral trianglewith a, b, c in the clockwise orientation.

Solve (x ⊕ (0, 0)) ⊕ (1, 1) = (1,−1) for x.

4. During the weekends, Eli delivers milk in the complex plane. On Saturday, he begins at z and deliversmilk to houses located at z3, z5, z7, . . . , z2013, in that order; on Sunday, he begins at 1 and deliversmilk to houses located at z2, z4, z6, . . . , z2012, in that order. Eli always walks directly (in a straightline) between two houses. If the distance he must travel from his starting point to the last house is√

2012 on both days, find the real part of z2.

5. Find all ordered triples (a, b, c) of positive reals that satisfy: ⌊a⌋bc = 3, a⌊b⌋c = 4, and ab⌊c⌋ = 5, where⌊x⌋ denotes the greatest integer less than or equal to x.

6. Let a0 = −2, b0 = 1, and for n ≥ 0, let

an+1 = an + bn +√

a2n + b2

n,

bn+1 = an + bn −√

a2n + b2

n.

Find a2012.

7. Let ⊗ be a binary operation that takes two positive real numbers and returns a positive real number.Suppose further that ⊗ is continuous, commutative (a ⊗ b = b ⊗ a), distributive across multiplication(a ⊗ (bc) = (a ⊗ b)(a ⊗ c)), and that 2 ⊗ 2 = 4. Solve the equation x ⊗ y = x for y in terms of x forx > 1.

8. Let x1 = y1 = x2 = y2 = 1, then for n ≥ 3 let xn = xn−1yn−2 + xn−2yn−1 and yn = yn−1yn−2 −xn−1xn−2. What are the last two digits of |x2012|?

9. How many real triples (a, b, c) are there such that the polynomial p(x) = x4 + ax3 + bx2 + ax + c hasexactly three distinct roots, which are equal to tan y, tan 2y, and tan 3y for some real y?

10. Suppose that there are 16 variables {ai,j}0≤i,j≤3, each of which may be 0 or 1. For how many settingsof the variables ai,j do there exist positive reals ci,j such that the polynomial

f(x, y) =∑

0≤i,j≤3

ai,jci,jxiyj

(x, y ∈ R) is bounded below?

15th Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Algebra Test

Name Team ID#

School Team

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Score:

15th Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Algebra Test

1. Let f be the function such that

f(x) =

{2x if x ≤ 1

22 − 2x if x > 1

2

What is the total length of the graph of f(f(...f︸ ︷︷ ︸

2012f ′s

(x)...)) from x = 0 to x = 1?

Answer:√

42012 + 1 When there are n copies of f , the graph consists of 2n segments, each of which

goes 1/2n units to the right, and alternately 1 unit up or down. So, the length is

2n

√

1 +1

22n=

√4n + 1

Taking n = 2012, the answer is√

42012 + 1

2. You are given an unlimited supply of red, blue, and yellow cards to form a hand. Each card has a pointvalue and your score is the sum of the point values of those cards. The point values are as follows: thevalue of each red card is 1, the value of each blue card is equal to twice the number of red cards, andthe value of each yellow card is equal to three times the number of blue cards. What is the maximumscore you can get with fifteen cards?

Answer: 168 If there are B blue cards, then each red card contributes 1+2B points (one for itselfand two for each blue card) and each yellow card contributes 3B points. Thus, if B > 1, it is optimalto change all red cards to yellow cards. When B = 0, the maximum number of points is 15. WhenB = 1, the number of points is always 42. When B > 1, the number of points is 3BY , where Y isthe number of yellow cards. Since B + Y = 15, the desired maximum occurs when B = 7 and Y = 8,which gives 168 points.

3. Given points a and b in the plane, let a⊕b be the unique point c such that abc is an equilateral trianglewith a, b, c in the clockwise orientation.

Solve (x ⊕ (0, 0)) ⊕ (1, 1) = (1,−1) for x.

Answer: ( 1−√

32 , 3−

√3

2 ) It is clear from the definition of ⊕ that b ⊕ (a ⊕ b) = a and if a ⊕ b = c

then b ⊕ c = a and c ⊕ a = b. Therefore x ⊕ (0, 0) = (1, 1) ⊕ (1,−1) = (1 −√

3, 0). Now this means

x = (0, 0) ⊕ (1 −√

3, 0) = (1−√

32 , 3−

√3

2 ).

4. During the weekends, Eli delivers milk in the complex plane. On Saturday, he begins at z and deliversmilk to houses located at z3, z5, z7, . . . , z2013, in that order; on Sunday, he begins at 1 and deliversmilk to houses located at z2, z4, z6, . . . , z2012, in that order. Eli always walks directly (in a straightline) between two houses. If the distance he must travel from his starting point to the last house is√

2012 on both days, find the real part of z2.

Answer: 10051006 Note that the distance between two points in the complex plane, m and n, is |m − n|.

We have that1006∑

k=1

∣∣z2k+1 − z2k−1

∣∣ =

1006∑

k=1

∣∣z2k − z2k−2

∣∣ =

√2012.

Algebra Test

However, noting that

|z| ·1006∑

k=1

∣∣z2k − z2k−2

∣∣ =

1006∑

k=1

∣∣z2k+1 − z2k−1

∣∣ ,

we must have |z| = 1. Then, since Eli travels a distance of√

2012 on each day, we have

1006∑

k=1

∣∣z2k − z2k−2

∣∣ =

∣∣z2 − 1

∣∣ ·

1006∑

k=1

∣∣z2k−2

∣∣ =

∣∣z2 − 1

∣∣ ·

1006∑

k=1

|z|2k−2

= 1006∣∣z2 − 1

∣∣ =

√2012,

so∣∣z2 − 1

∣∣ =

√2012

1006 . Since |z| = 1, we can write z = cos (θ)+ i sin (θ) and then z2 = cos (2θ)+ i sin (2θ).Hence,

∣∣z2 − 1

∣∣ =

√

(cos (2θ) − 1)2

+ sin2 (2θ) =√

2 − 2 cos (2θ) =

√2012

1006,

so 2 − 2 cos (2θ) = 21006 . The real part of z2, cos (2θ), is thus 1005

1006 .

5. Find all ordered triples (a, b, c) of positive reals that satisfy: ⌊a⌋bc = 3, a⌊b⌋c = 4, and ab⌊c⌋ = 5, where⌊x⌋ denotes the greatest integer less than or equal to x.

Answer: (√

303 ,

√304 , 2

√30

5 ), (√

303 ,

√302 ,

√305 ) Write p = abc, q = ⌊a⌋⌊b⌋⌊c⌋. Note that q is an integer.

Multiplying the three equations gives:

p =

√60

q

Substitution into the first equation,

p = 3a

⌊a⌋ < 3⌊a⌋ + 1

⌊a⌋ ≤ 6

Looking at the last equation:

p = 5c

⌊c⌋ ≥ 5⌊c⌋⌊c⌋ ≥ 5

Here we’ve used ⌊x⌋ ≤ x < ⌊x⌋ + 1, and also the apparent fact that ⌊a⌋ ≥ 1. Now:

5 ≤√

60

q≤ 6

12

5≥ q ≥ 5

3

Since q is an integer, we must have q = 2. Since q is a product of 3 positive integers, we must havethose be 1,1, and 2 in some order, so there are three cases:

Case 1: ⌊a⌋ = 2. By the equations, we’d need a = 23

√30 =

√

120/9 > 3, a contradiction, sothere are no solutions in this case.

Case 2: ⌊b⌋ = 2. We have the solution(√

30

3,

√30

2,

√30

5

)

Case 3: ⌊c⌋ = 2. We have the solution(√

30

3,

√30

4,2√

30

5

)

Algebra Test

6. Let a0 = −2, b0 = 1, and for n ≥ 0, let

an+1 = an + bn +√

a2n + b2

n,

bn+1 = an + bn −√

a2n + b2

n.

Find a2012.

Answer: 21006√

22010 + 2 − 22011 We have

an+1 + bn+1 = 2(an + bn)

an+1bn+1 = (an + bn)2 − a2n − b2

n = 2anbn

Thus,

an + bn = −2n

anbn = −2n+1

Using Viete’s formula, a2012 and b2012 are the roots of the following quadratic, and, since the squareroot is positive, a2012 is the bigger root:

x2 + 22012x − 22013

Thus,

a2012 = 21006√

22010 + 2 − 22011

7. Let ⊗ be a binary operation that takes two positive real numbers and returns a positive real number.Suppose further that ⊗ is continuous, commutative (a ⊗ b = b ⊗ a), distributive across multiplication(a ⊗ (bc) = (a ⊗ b)(a ⊗ c)), and that 2 ⊗ 2 = 4. Solve the equation x ⊗ y = x for y in terms of x forx > 1.

Answer: y =√

2 We note that (a⊗ bk) = (a⊗ b)k for all positive integers k. Then for all rational

numbers pq

we have a⊗ bp

q = (a⊗ b1

q )p = (a⊗ b)p

q . So by continuity, for all real numbers a, b, it follows

that 2a ⊗ 2b = (2⊗ 2)ab = 4ab. Therefore given positive reals x, y, we have x⊗ y = 2log2(x) ⊗ 2log

2(y) =

4log2(x) log

2(y).

If x = 4log2(x) log

2(y) = 22 log

2(x) log

2(y) then log2(x) = 2 log2(x) log2(y) and 1 = 2 log2(y) = log2(y

2).Thus y =

√2 regardless of x.

8. Let x1 = y1 = x2 = y2 = 1, then for n ≥ 3 let xn = xn−1yn−2 + xn−2yn−1 and yn = yn−1yn−2 −xn−1xn−2. What are the last two digits of |x2012|?Answer: 84 Let zn = yn + xni. Then the recursion implies that:

z1 = z2 = 1 + i,

zn = zn−1zn−2.

This implies thatzn = (z1)

Fn ,

where Fn is the nth Fibonacci number (F1 = F2 = 1). So, z2012 = (1 + i)F2012 . Notice that

(1 + i)2 = 2i.

Also notice that every third Fibonnaci number is even, and the rest are odd. So:

z2012 = (2i)F2012−1

2 (1 + i)

Algebra Test

Let m = F2012−12 . Since both real and imaginary parts of 1 + i are 1, it follows that the last two digits

of |x2012| are simply the last two digits of 2m = 2F2012−1

2 .

By the Chinese Remainder Theorem, it suffices to evaluate 2m modulo 4 and 25. Clearly, 2m is divisibleby 4. To evaluate it modulo 25, it suffices by Euler’s Totient theorem to evaluate m modulo 20.

To determine (F2012 − 1)/2 modulo 4 it suffices to determine F2012 modulo 8. The Fibonacci sequencehas period 12 modulo 8, and we find

F2012 ≡ 5 (mod 8),

m ≡ 2 (mod 4).

2 ∗ 3 ≡ 1 (mod 5), som ≡ 3F2012 − 3 (mod 5).

The Fibonacci sequence has period 20 modulo 5, and we find

m ≡ 4 (mod 5).

Combining,

m ≡ 14 (mod 20)

2m ≡ 214 = 4096 ≡ 21 (mod 25)

|x2012| ≡ 4 · 21 = 84 (mod 100).

9. How many real triples (a, b, c) are there such that the polynomial p(x) = x4 + ax3 + bx2 + ax + c hasexactly three distinct roots, which are equal to tan y, tan 2y, and tan 3y for some real y?

Answer: 18 Let p have roots r, r, s, t. Using Vieta’s on the coefficient of the cubic and linear terms,we see that 2r + s + t = r2s + r2t + 2rst. Rearranging gives 2r(1 − st) = (r2 − 1)(s + t).

If r2 − 1 = 0, then since r 6= 0, we require that 1 − st = 0 for the equation to hold. Conversely, if1 − st = 0, then since st = 1, s + t = 0 cannot hold for real s, t, we require that r2 − 1 = 0 for theequation to hold. So one valid case is where both these values are zero, so r2 = st = 1. If r = tan y(here we stipulate that 0 ≤ y < π), then either y = π

4 or y = 3π4 . In either case, the value of tan 2y

is undefined. If r = tan 2y, then we have the possible values y = π8 , 3π

8 , 5π8 , 7π

8 . In each of these cases,we must check if tan y tan 3y = 1. But this is true if y + 3y = 4y is a odd integer multiple of π

2 , whichis the case for all such values. If r = tan 3y, then we must have tan y tan 2y = 1, so that 3y is an oddinteger multiple of π

2 . But then tan 3y would be undefined, so none of these values can work.

Now, we may assume that r2 − 1 and 1− st are both nonzero. Dividing both sides by (r2 − 1)(1− st)and rearranging yields 0 = 2r

1−r2 + s+t1−st

, the tangent addition formula along with the tangent doubleangle formula. By setting r to be one of tan y, tan 2y, or tan 3y, we have one of the following:

(a) 0 = tan 2y + tan 5y

(b) 0 = tan 4y + tan 4y

(c) 0 = tan 6y + tan 3y.

We will find the number of solutions y in the interval [0, π). Case 1 yields six multiples of π7 . Case 2

yields tan 4y = 0, which we can readily check has no solutions. Case 3 yields eight multiples of π9 . In

total, we have 4 + 6 + 8 = 18 possible values of y.

10. Suppose that there are 16 variables {ai,j}0≤i,j≤3, each of which may be 0 or 1. For how many settingsof the variables ai,j do there exist positive reals ci,j such that the polynomial

f(x, y) =∑

0≤i,j≤3

ai,jci,jxiyj

Algebra Test

(x, y ∈ R) is bounded below?

Answer: 126 For some choices of the ai,j , let S = {(i, j)|ai,j = 1}, and let S′ = S ∪ {(0, 0)}. LetC(S′) denote the convex hull of S′. We claim that there exist the problem conditions are satisfied(there exist positive coefficients for the terms so that the polynomial is bounded below) if and only ifthe vertices of C(S′) all have both coordinates even.

For one direction, suppose that C(S′) has a vertex v = (i′, j′) with at least one odd coordinate; WLOG,suppose it is i′. Since v is a vertex, it maximizes some objective function ai + bj over C(S′) uniquely,and thus also over S′. Since (0, 0) ∈ S′, we must have ai′ + bj′ > 0. Now consider plugging in(x, y) = (−ta, tb) (t > 0) into f . This gives the value

f(−ta, tb) =∑

(i,j)∈S

(−1)ici,jtai+bj .

But no matter what positive ci,j we choose, this expression is not bounded below as t grows infinitely

large, as there is a −ci′,j′tai′+bj′

term, with ai′ + bj′ > 0, and all other terms have smaller powers oft. So the polynomial cannot be bounded below.

For the other direction, suppose the vertices of C(S′) all have both coordinates even. If all points in S′

are vertices of C(S′), then the polynomial is a sum of squares, so it is bounded below. Otherwise, weassume that some points in S′ are not vertices of C(S′). It suffices to consider the case where there isexactly one such point. Call this point w = (i′, j′). Let V (S′) denote the set of the vertices of C(S′),and let n = |V (S′)|. Enumerate the points of V (S′) as v1, v2, . . . , vn. Let ik, jk denote the i and jcoordinates of vk, respectively.

Since w ∈ C(S′), there exist nonnegative constants λ1, λ2, . . . , λn such that∑n

k=1 λk = 1 and∑n

k=1 λkvk =w. (Here, we are treating the ordered pairs as vectors.) Then, by weighted AM-GM, we have

n∑

k=1

λk|x|ik |y|jk ≥ |x|i′ |y|j′

.

Let c be the λ-value associated with (0, 0). Then by picking cik,jk= λk and ci′,j′ = 1, we find that

p(x, y) ≥ −c for all x, y, as desired.

We now find all possible convex hulls C(S′) (with vertices chosen from (0, 0), (0, 2), (2, 0), and (2, 2)),and for each convex hull, determine how many possible settings of ai,j give that convex hull. Thereare 8 such possible convex hulls: the point (0, 0) only, 3 lines, 3 triangles, and the square. The pointhas 2 possible choices, each line has 4 possible choices, each triangle has 16 possible choices, and thesquare has 64 possible choices, giving 2 + 3 · 4 + 3 · 16 + 64 = 126 total choices.

Algebra Test

COMBINATORICS TEST

This test consists of 10 short-answer problems to be solved individually in 50 minutes. Problems will beweighted with point values after the contest based on how many competitors solve each problem. There isno penalty for guessing.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted otherthan the official translation sheets. Similarly, graph paper, rulers, protractors, compasses, and other drawingaids are not permitted.

Our goal is that a closed form answer equivalent to the correct answer will be accepted. However, we donot always have the resourses to determine whether a complicated or strange answer is equivalent to ours.To assist us in awarding you all the points that you deserve, you answers should be simplified as much aspossible. Answers must be exact unless otherwise specified.

Correct mathematical notation must be used. No partial credit will be given unless otherwise specified.

If you believe the test contains an error, please submit your protest in writing to Science Center 109 duringlunchtime.

Enjoy!

15th Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Combinatorics Test

1. In the game of Minesweeper, a number on a square denotes the number of mines that share at leastone vertex with that square. A square with a number may not have a mine, and the blank squares areundetermined. How many ways can the mines be placed in this configuration?

2 1 2

2. Brian has a 20-sided die with faces numbered from 1 to 20, and George has three 6-sided dice with facesnumbered from 1 to 6. Brian and George simultaneously roll all their dice. What is the probabilitythat the number on Brian’s die is larger than the sum of the numbers on George’s dice?

3. In the figure below, how many ways are there to select 5 bricks, one in each row, such that any twobricks in adjacent rows are adjacent?

4. A frog is at the point (0, 0). Every second, he can jump one unit either up or right. He can only moveto points (x, y) where x and y are not both odd. How many ways can he get to the point (8, 14)?

5. Dizzy Daisy is standing on the point (0, 0) on the xy-plane and is trying to get to the point (6, 6). Shestarts facing rightward and takes a step 1 unit forward. On each subsequent second, she either takesa step 1 unit forward or turns 90 degrees counterclockwise then takes a step 1 unit forward. She maynever go on a point outside the square defined by |x| ≤ 6, |y| ≤ 6, nor may she ever go on the samepoint twice. How many different paths may Daisy take?

6. For a permutation σ of 1, 2, . . . , 7, a transposition is a swapping of two elements. (For instance, wecould apply a transposition to the permutation 3, 7, 1, 4, 5, 6, 2 and get 3, 7, 6, 4, 5, 1, 2 by swapping the1 and the 6.)

Let f(σ) be the minimum number of transpositions necessary to turn σ into the permutation 1, 2, 3, 4, 5, 6, 7.Find the sum of f(σ) over all permutations σ of 1, 2, . . . , 7.

7. You are repeatedly flipping a fair coin. What is the expected number of flips until the first time thatyour previous 2012 flips are ‘HTHT...HT’?

8. How many ways can one color the squares of a 6x6 grid red and blue such that the number of redsquares in each row and column is exactly 2?

9. A parking lot consists of 2012 parking spots equally spaced in a line, numbered 1 through 2012. Oneby one, 2012 cars park in these spots under the following procedure: the first car picks from the 2012spots uniformly randomly, and each following car picks uniformly randomly among all possible choiceswhich maximize the minimal distance from an already parked car. What is the probability that thelast car to park must choose spot 1?

10. Jacob starts with some complex number x0 other than 0 or 1. He repeatedly flips a fair coin. If thenth flip lands heads, he lets xn = 1 − xn−1, and if it lands tails he lets xn = 1

xn−1. Over all possible

choices of x0, what are all possible values of the probability that x2012 = x0?

15th Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Combinatorics Test

Name Team ID#

School Team

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Score:

15th

Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Combinatorics Test

1. In the game of Minesweeper, a number on a square denotes the number of mines that share at leastone vertex with that square. A square with a number may not have a mine, and the blank squares areundetermined. How many ways can the mines be placed in this configuration?

2 1 2

Answer: 95 Let A be the number of mines in the first two columns. Let B,C,D,E be the numberof mines in the third, fourth, fifth, and sixth columns, respectively. We need to have A + B = 2,B + C + D = 1, and D + E = 2. This can happen in three ways, which are (A,B,C,D,E) =(2, 0, 1, 0, 2), (2, 0, 0, 1, 1), (1, 1, 0, 0, 2). This gives (10)(2)(1) + (10)(3)(2) + (5)(3)(1) = 95 possibleconfigurations.

2. Brian has a 20-sided die with faces numbered from 1 to 20, and George has three 6-sided dice with facesnumbered from 1 to 6. Brian and George simultaneously roll all their dice. What is the probabilitythat the number on Brian’s die is larger than the sum of the numbers on George’s dice?

Answer: 1940 Let Brian’s roll be d and let George’s rolls be x, y, z. By pairing the situation d, x, y, z

with 21 − d, 7 − x, 7 − y, 7 − z, we see that the probability that Brian rolls higher is the same as theprobability that George rolls higher. Given any of George’s rolls x, y, z, there is exactly one numberBrian can roll which will make them tie, so the probability that they tie is 1

20 . So the probability that

Brian wins is1− 1

20

2 = 1940 .

3. In the figure below, how many ways are there to select 5 bricks, one in each row, such that any twobricks in adjacent rows are adjacent?

Answer: 61 The number of valid selections is equal to the number of paths which start at a topbrick and end at a bottom brick. We compute these by writing 1 in each of the top bricks and lettinglower bricks be the sum of the one or two bricks above them. Thus, the number inside each brick isthe number of paths from that brick to the top. The bottom row is 6, 14, 16, 15, 10, which sums to61.

1 1 1 1 1

2 2 2 2 1

2 4 4 4 3

6 8 8 7 3

6 14 16 15 10

4. A frog is at the point (0, 0). Every second, he can jump one unit either up or right. He can only moveto points (x, y) where x and y are not both odd. How many ways can he get to the point (8, 14)?

Combinatorics Test

Answer: 330 When the frog is at a point (x, y) where x and y are both even, then if that frogchooses to move right, his next move will also have to be a step right; similarly, if he moves up, hisnext move will have to be up.

If we “collapse” each double step into one step, the problem simply becomes how many ways are thereto move to the point (4, 7) using only right and up steps, with no other restrictions. That is 11 stepstotal, so the answer is

(

114

)

= 330.

5. Dizzy Daisy is standing on the point (0, 0) on the xy-plane and is trying to get to the point (6, 6). Shestarts facing rightward and takes a step 1 unit forward. On each subsequent second, she either takesa step 1 unit forward or turns 90 degrees counterclockwise then takes a step 1 unit forward. She maynever go on a point outside the square defined by |x| ≤ 6, |y| ≤ 6, nor may she ever go on the samepoint twice. How many different paths may Daisy take?

Answer: 131922 Because Daisy can only turn in one direction and never goes to the same squaretwice, we see that she must travel in an increasing spiral about the origin. Clearly, she must arrive at(6, 6) coming from below. To count her paths, it therefore suffices to consider the horizontal and verticallines along which she travels (out of 5 choices to move upward, 6 choices leftward, 6 choices downward,and 6 choices rightward). Breaking up the cases by the number of complete rotations she performs,

the total is

(

5

0

)(

6

0

)3

+

(

5

1

)(

6

1

)3

+

(

5

2

)(

6

2

)3

+

(

5

3

)(

6

3

)3

+

(

5

4

)(

6

4

)3

+

(

5

5

)(

6

5

)3

= 131922.

6. For a permutation σ of 1, 2, . . . , 7, a transposition is a swapping of two elements. (For instance, wecould apply a transposition to the permutation 3, 7, 1, 4, 5, 6, 2 and get 3, 7, 6, 4, 5, 1, 2 by swapping the1 and the 6.)

Let f(σ) be the minimum number of transpositions necessary to turn σ into the permutation 1, 2, 3, 4, 5, 6, 7.Find the sum of f(σ) over all permutations σ of 1, 2, . . . , 7.

Answer: 22212 To solve this problem, we use the idea of a cycle in a permutation. If σ is apermutation, we say that (a1 a2 · · · ak) is a cycle if σ(ai) = σ(ai+1) for 1 ≤ i ≤ k− 1 and σ(ak) = a1.Any permutation can be decomposed into disjoint cycles; for instance, the permutation 3, 7, 6, 4, 5, 1, 2,can be written as (1 3 6)(2 7)(4)(5). For a permutation σ, let g(σ) be the number of cycles in its cycledecomposition. (This includes single-element cycles.)

Claim. For any permutation σ on n elements, f(σ) = n − g(σ).

Proof. Given a cycle (a1 a2 · · · ak) (with k ≥ 2) of a permutation σ, we can turn this cycle into theidentity permutation with k − 1 transpositions; first we swap a1 and a2; that is, we replace σ with apermutation σ′ such that instead of σ(ak) = a1 and σ(a1) = a2, we have σ′(ak) = a2 and σ′(a1) = a1.Now, σ′ takes a1 to itself, so we are left with the cycle (a2 · · · an). We continue until the entire cycleis replaced by the identity, which takes k − 1 transpositions. Now, for any σ, we resolve each cycle inthis way, making a total of n − g(σ) transpositions, to turn σ into the identity permutation.

This shows that n−g(σ) transpositions; now let us that we cannot do it in less. We show that wheneverwe make a transposition, the value of n− g(σ) can never decrease by more than 1. Whenever we swaptwo elements, if they are in different cycles, then those two cycles merge into one; thus n−g(σ) actuallyincreased. If the two elements are in one cycle, then the one cycle splits into two cycles, so n − g(σ)decreased by only one, and this proves the claim.

Thus, we want to find∑

σ∈S7

(7 − g(σ)) = 7 · 7! −∑

σ∈S7

g(σ)

to evaluate the sum, we instead sum over every cycle the number of permutations it appears in. Forany 1 ≤ k ≤ 7, the number of cycles of size k is n!

(n−k)!k , and the number of permutations each such

cycle can appear in is (n − k)!. Thus we get that the answer is

7 · 7! −

7∑

k=1

7!

k= 22212.

Combinatorics Test

7. You are repeatedly flipping a fair coin. What is the expected number of flips until the first time thatyour previous 2012 flips are ‘HTHT...HT’?

Answer: (22014 − 4)/3 Let S be our string, and let f(n) be the number of binary strings of length

n which do not contain S. Let g(n) be the number of strings of length n which contain S but whoseprefix of length n − 1 does not contain S (so it contains S for the “first” time at time n).

Consider any string of length n which does not contain S and append S to it. Now, this new stringcontains S, and in fact it must contain S for the first time at either time n + 2, n + 4, ..., or n + 2012.It’s then easy to deduce the relation

f(n) = g(n + 2) + g(n + 4) + · · · + g(n + 2012)

Now, let’s translate this into a statement about probabilities. Let t be the first time our sequence ofcoin flips contains the string S. Dividing both sides by 2n, our equality becomes

P (t > n) = 4P (t = n + 2) + 16P (t = n + 4) + · · · + 22012P (t = n + 2012)

Summing this over all n from 0 to ∞, we get

∑

P (t > n) = 4 + 16 + · · · + 22012 = (22014 − 4)/3

But it is also easy to show that since t is integer-valued,∑

P (t > n) = E(t), and we are done.

8. How many ways can one color the squares of a 6x6 grid red and blue such that the number of redsquares in each row and column is exactly 2?

Answer: 67950 Assume the grid is n×n. Let f(n) denote the number of ways to color exactly twosquares in each row and column red. So f(1) = 0 and f(2) = 1. We note that coloring two squaresred in each row and column partitions the set 1, 2, . . . , n into cycles such that i is in the same cycle as,and adjacent to, j iff column i and column j have a red square in the same row. Each i is adjacent totwo other, (or the same one twice in a 2-cycle).

Now consider the cycle containing 1, and let it have size k. There are(

n2

)

ways to color two squaresred in the first column. Now we let the column that is red in the same row as the top ball in the firstcolumn, be the next number in the cycle. There are n− 1 ways to pick this column, and n− 2 ways topick the second red square in this column (unless k = 2). Then there are (n − 2)(n − 3) ways to pickthe red squares in the third column. and (n − j)(n − j + 1) ways to pick the jth ones for j ≤ k − 1.Then when we pick the kth column, the last one in the cycle, it has to be red in the same row as thesecond red square in column 1, so there are just n − k + 1 choices. Therefore if the cycle has length kthere aren(n−1)

2 × (n − 1)(n − 2) × . . . × (n − k + 1)(n − k + 2) × (n − k + 1) ways, which equals: n!(n−1)!2(n−k)!(n−k)! .

Summing over the size of the cycle containing the first column, we get

f(n) =

n∑

k=2

1

2f(n − k)

(n)!(n − 1)!

(n − k)!(n − k)!

2nf(n)

n!n!=

n∑

k=2

f(n − k)

(n − k)!(n − k)!

2nf(n)

n!n!−

2(n − 1)f(n − 1)

(n − 1)!(n − 1)!=

f(n − 2)

(n − 2)!(n − 2)!

We thus obtain the recursion:

f(n) = n(n − 1)f(n − 1) +n(n − 1)2

2f(n − 2)

Combinatorics Test

Then we get:

f(1) = 0

f(2) = 1

f(3) = 6

f(4) = 12 × 6 + 18 = 90

f(5) = 20 × 90 + 40 × 6 = 2040

f(6) = 30 × 2040 + 75 × 90 = 67950.

9. A parking lot consists of 2012 parking spots equally spaced in a line, numbered 1 through 2012. Oneby one, 2012 cars park in these spots under the following procedure: the first car picks from the 2012spots uniformly randomly, and each following car picks uniformly randomly among all possible choiceswhich maximize the minimal distance from an already parked car. What is the probability that thelast car to park must choose spot 1?

Answer: 12062300 We see that for 1 to be the last spot, 2 must be picked first (with probability

1n), after which spot n is picked. Then, cars from 3 to n − 1 will be picked until there are only gaps

of 1 or 2 remaining. At this point, each of the remaining spots (including spot 1) is picked uniformlyat random, so the probability that spot 1 is chosen last here will be the reciprocal of the number ofremaining slots.

Let f(n) denote the number of empty spots that will be left if cars park in n + 2 consecutive spotswhose ends are occupied, under the same conditions, except that the process stops when a car isforced to park immediately next to a car. We want to find the value of f(2009). Given the gapof n cars, after placing a car, there are gaps of f(⌊n−1

2 ⌋) and f(⌈n−12 ⌉) remaining. Thus, f(n) =

f(⌊n−12 ⌋) + f(⌈n−1

2 ⌉). With the base cases f(1) = 1, f(2) = 2, we can determine with induction that

f(x) =

{

x − 2n−1 + 1 if 2n ≤ x ≤ 32 · 2n − 2,

2n if 32 · 2n − 1 ≤ x ≤ 2 · 2n − 1.

Thus, f(2009) = 1024, so the total probability is 12012 · 1

1024+1 = 12062300 .

10. Jacob starts with some complex number x0 other than 0 or 1. He repeatedly flips a fair coin. If thenth flip lands heads, he lets xn = 1 − xn−1, and if it lands tails he lets xn = 1

xn−1

. Over all possible

choices of x0, what are all possible values of the probability that x2012 = x0?

Answer: 1, 22011+13·22011 Let f(x) = 1 − x, g(x) = 1

x. Then for any x, f(f(x)) = x and g(g(x)) = x.

Furthermore, f(g(x)) = 1 − 1x, g(f(g(x))) = x

x−1 , f(g(f(g(x)))) = 11−x

, g(f(g(f(g(x))))) = 1 − x =

f(x), so for all n, xn is one of x, 1x, 1 − 1

x, x

x−1 , 11−x

, 1 − x, and we can understand the coin flippingprocedure as moving either left or right with equal probability along this cycle of values.

For most x, all six of these values are distinct. In this case, suppose that we move right R times and left2012−R times between x0 and x2012. For x2012 = x0, we need to have that R−2012+R ≡ 0 (mod 6),or R ≡ 1 (mod 3). The number of possible ways to return to x0 is then a =

(

20121

)

+(

20124

)

+· · ·+(

20122011

)

.

Let b =(

20120

)

+(

20123

)

+ · · · +(

20122010

)

=(

20122

)

+(

20125

)

+ · · · +(

20122012

)

. Then we have a + 2b = 22012

and that b = (1+1)2012+(1+ω)2012+(1+ω2)2012

3 , where ω is a primitive third root of unity. It can be seenthat 1 + ω is a primitive sixth root of unity and 1 + ω2 is its inverse, so (1 + ω)2012 = (1 + ω)2 = ω,

and similarly (1 + ω2)2012 = ω2. Therefore, b = ·22012−13 , so a = 22012 − 2b = 22012+2

3 , and our desired

probability is then a22012 = 22012+2

3·22012 = 22011+13·22011 .

For some x0, however, the cycle of values can become degenerate. It could be the case that two adjacentvalues are equal. Let y be a value that is equal to an adjacent value. Then y = 1

yor y = 1 − y, which

gives y ∈ {−1, 12}. Therefore, this only occurs in the cycle of values −1, 2, 1

2 , 12 , 2,−1. In this case, note

that after 2012 steps we will always end up an even number of steps away from our starting point,and each of the numbers occupies two spaces of opposite parity, so we would need to return to our

Combinatorics Test

original location, just as if all six numbers were distinct. Therefore in this case we again have that the

probability that x2012 = x0 is 22011+13·22011 .

It is also possible that two numbers two apart on the cycle are equal. For this to be the case, let y be

the value such that f(g(y)) = y. Then 1 − 1x

= x, or x − 1 = x2, so x = 1±i√

32 . Let ζ = 1+i

√3

2 . Then

we get that the cycle of values is ζ, ζ, ζ, ζ, ζ, ζ, and since at the end we are always an even number ofspaces away from our starting location, the probability that x2012 = x0 is 1.

Finally, we need to consider the possibility that two opposite numbers are equal. In this case we havea y such that f(g(f(y))) = y, or x

x−1 = x, so x = 2. In this case we obtain the same cycle of numbers

in the case where two adjacent numbers are equal, and so we again obtain the probability 22011+13·22011 .

Therefore, the only possibilities are 1, 22011+13·22011 .

Combinatorics Test

GEOMETRY TEST

This test consists of 10 short-answer problems to be solved individually in 50 minutes. Problems will beweighted with point values after the contest based on how many competitors solve each problem. There isno penalty for guessing.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted otherthan the official translation sheets. Similarly, graph paper, rulers, protractors, compasses, and other drawingaids are not permitted.

Our goal is that a closed form answer equivalent to the correct answer will be accepted. However, we donot always have the resourses to determine whether a complicated or strange answer is equivalent to ours.To assist us in awarding you all the points that you deserve, you answers should be simplified as much aspossible. Answers must be exact unless otherwise specified.

Correct mathematical notation must be used. No partial credit will be given unless otherwise specified.

If you believe the test contains an error, please submit your protest in writing to Science Center 109 duringlunchtime.

Enjoy!

15th Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Geometry Test

1. ABC is an isosceles triangle with AB = 2 and ∡ABC = 90◦. D is the midpoint of BC and E is onAC such that the area of AEDB is twice the area of ECD. Find the length of DE.

2. Let ABC be a triangle with ∠A = 90◦, AB = 1, and AC = 2. Let ℓ be a line through A perpendicularto BC, and let the perpendicular bisectors of AB and AC meet ℓ at E and F , respectively. Find thelength of segment EF .

3. Let ABC be a triangle with incenter I. Let the circle centered at B and passing through I intersectside AB at D and let the circle centered at C passing through I intersect side AC at E. Suppose DEis the perpendicular bisector of AI. What are all possible measures of angle BAC in degrees?

4. There are circles ω1 and ω2. They intersect in two points, one of which is the point A. B lies on ω1

such that AB is tangent to ω2. The tangent to ω1 at B intersects ω2 at C and D, where D is the closerto B. AD intersects ω1 again at E. If BD = 3 and CD = 13, find EB/ED.

5. A mouse lives in a circular cage with completely reflective walls. At the edge of this cage, a smallflashlight with vertex on the circle whose beam forms an angle of 15◦ is centered at an angle of 37.5◦

away from the center. The mouse will die in the dark. What fraction of the total area of the cage cankeep the mouse alive?

37.5◦

6. Triangle ABC is an equilateral triangle with side length 1. Let X0,X1, . . . be an infinite sequence ofpoints such that the following conditions hold:

• X0 is the center of ABC

• For all i ≥ 0, X2i+1 lies on segment AB and X2i+2 lies on segment AC.• For all i ≥ 0, ∡XiXi+1Xi+2 = 90◦.• For all i ≥ 1, Xi+2 lies in triangle AXiXi+1.

Find the maximum possible value of∑∞

i=0 |XiXi+1|, where |PQ| is the length of line segment PQ.

7. Let S be the set of the points (x1, x2, . . . , x2012) in 2012-dimensional space such that |x1|+ |x2|+ · · ·+|x2012| ≤ 1. Let T be the set of points in 2012-dimensional space such that

2012maxi=1

|xi| = 2. Let p be a

randomly chosen point on T . What is the probability that the closest point in S to p is a vertex of S?

8. Hexagon ABCDEF has a circumscribed circle and an inscribed circle. If AB = 9, BC = 6, CD = 2,and EF = 4. Find {DE,FA}.

9. Let O,O1, O2, O3, O4 be points such that O1, O,O3 and O2, O,O4 are collinear in that order, OO1 =1, OO2 = 2, OO3 =

√2, OO4 = 2, and ∡O1OO2 = 45◦. Let ω1, ω2, ω3, ω4 be the circles with respective

centers O1, O2, O3, O4 that go through O. Let A be the intersection of ω1 and ω2, B be the intersectionof ω2 and ω3, C be the intersection of ω3 and ω4, and D be the intersection of ω4 and ω1, with A,B,C,Dall distinct from O. What is the largest possible area of a convex quadrilateral P1P2P3P4 such that Pi

lies on Oi and that A,B,C,D all lie on its perimeter?

10. Let C denote the set of points (x, y) ∈ R2 such that x2 + y2 ≤ 1. A sequence Ai = (xi, yi)|i ≥ 0 ofpoints in R2 is ‘centric’ if it satisfies the following properties:

• A0 = (x0, y0) = (0, 0), A1 = (x1, y1) = (1, 0).

• For all n ≥ 0, the circumcenter of triangle AnAn+1An+2 lies in C.

Let K be the maximum value of x22012 + y2

2012 over all centric sequences. Find all points (x, y) suchthat x2 + y2 = K and there exists a centric sequence such that A2012 = (x, y).

15th Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Geometry Test

Name Team ID#

School Team

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Score:

15th Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Geometry Test

1. ABC is an isosceles triangle with AB = 2 and ∡ABC = 90◦. D is the midpoint of BC and E is onAC such that the area of AEDB is twice the area of ECD. Find the length of DE.

Answer:√

173 Let F be the foot of the perpendicular from E to BC. We have [AEDB]+ [EDC] =

[ABC] = 2 ⇒ [EDC] = 23 . Since we also have [EDC] = 1

2 (EF )(DC), we get EF = FC = 43 . So

FD = 13 , and ED =

√173 by the Pythagorean Theorem.

2. Let ABC be a triangle with ∠A = 90◦, AB = 1, and AC = 2. Let ℓ be a line through A perpendicularto BC, and let the perpendicular bisectors of AB and AC meet ℓ at E and F , respectively. Find thelength of segment EF .

Answer: 3√

54 Let M,N be the midpoints of AB and AC, respectively. Then we have ∠EAB =

∠ACB and ∠EAC = ∠ABC, so AEM ∼ CBA ⇒ AE =√

54 and FAN ∼ CBA ⇒ AF =

√5.

Consequently, EF = AF − AE = 3√

54 .

3. Let ABC be a triangle with incenter I. Let the circle centered at B and passing through I intersectside AB at D and let the circle centered at C passing through I intersect side AC at E. Suppose DEis the perpendicular bisector of AI. What are all possible measures of angle BAC in degrees?

Answer: 5407 Let α = ∡BAC. DE is the perpendicular bisector of AI, so DA = DI, and

∠DIA = ∠DAI = α/2. Thus, ∠IDB = ∠DIB = α, since BD = BI. This gives ∠DBI = 180◦ − 2α,so that ∠ABC = 360◦ − 4α. Similarly, ∠ACB = 360◦ − 4α. Now, summing the angles in ABC, wefind 720◦ − 7α = 180◦, so that α = 540

7

◦.

4. There are circles ω1 and ω2. They intersect in two points, one of which is the point A. B lies on ω1

such that AB is tangent to ω2. The tangent to ω1 at B intersects ω2 at C and D, where D is the closerto B. AD intersects ω1 again at E. If BD = 3 and CD = 13, find EB/ED.

Answer: 4√

3/3

[diagram]

By power of a point, BA =√

BD · BC = 4√

3. Also, DEB ∼ DBA, so EB/ED = BA/BD = 4√

3/3.

5. A mouse lives in a circular cage with completely reflective walls. At the edge of this cage, a smallflashlight with vertex on the circle whose beam forms an angle of 15◦ is centered at an angle of 37.5◦

away from the center. The mouse will die in the dark. What fraction of the total area of the cage cankeep the mouse alive?

37.5◦

Geometry Test

Answer: 34 We claim that the lit region is the entire cage except for a circle of half the radius

of the cage in the center, along with some isolated points on the boundary of the circle and possiblyminus a set of area 0. Note that the region is the same except for a set of area 0 if we disallow thelight paths at the very edge of the beam. In that case, we can see that the lit region is an open subsetof the disk, as clearly the region after k bounces is open for each k and the union of open sets is againopen. We will then show that a dense subset of the claimed region of the cage is lit.

First, let us show that no part of the inner circle is lit. For any given light path, each chord of thecircle is the same length, and in particular the minimum distance from the center of the circle is thesame on each chord of the path. Since none of the initial chords can come closer than half the cage’sradius to the center, the circle with half the cage’s radius is indeed dark.

Now we need to show that for each open subset of the outer region, there is a light path passingthrough it, which will imply that the unlit region outside the small circle contains no open set, andthus has area 0. To do this, simply consider a light path whose angle away from the center is irrationalsuch that the distance d from the center of the cage to the midpoint of the first chord is sufficientlyclose to the distance from the center of the cage to a point in the open set we’re considering. Eachsuccessive chord of the light path can be seen as a rotation of the original one, and since at each stepit is translated by an irrational angle, we obtain a dense subset of all the possible chords. This meansthat we obtain a dense subset of the circumference of the circle of radius d centered at the center ofthe cage, and in particular a point inside the open set under consideration, as desired.

Therefore, the lit region of the cage is the area outside the concentric circle of half the radius plus orminus some regions of area 0, which tells us that 3

4 of the cage is lit.

6. Triangle ABC is an equilateral triangle with side length 1. Let X0,X1, . . . be an infinite sequence ofpoints such that the following conditions hold:

• X0 is the center of ABC

• For all i ≥ 0, X2i+1 lies on segment AB and X2i+2 lies on segment AC.

• For all i ≥ 0, ∡XiXi+1Xi+2 = 90◦.

• For all i ≥ 1, Xi+2 lies in triangle AXiXi+1.

Find the maximum possible value of∑∞

i=0 |XiXi+1|, where |PQ| is the length of line segment PQ.

Answer:√

63 Let Y be the foot of the perpendicular from A to X0X1: note that the sum we

wish to minimize is simply X0Y + Y A. However, it is not difficult to check (for example, by AM-GM) that AY + Y X0 ≥

√2AX0 =

√6/3. This may be achieved by making ∠Y X0A = 45◦, so that

∠AX1X0 = 105◦.

7. Let S be the set of the points (x1, x2, . . . , x2012) in 2012-dimensional space such that |x1|+ |x2|+ · · ·+|x2012| ≤ 1. Let T be the set of points in 2012-dimensional space such that

2012maxi=1

|xi| = 2. Let p be a

randomly chosen point on T . What is the probability that the closest point in S to p is a vertex of S?

Answer: 122011 Note that T is a hypercube in 2012-dimensional space, containing the rotated

hyperoctahedron S. Let v be a particular vertex of S, and we will consider the set of points x on Tsuch that v is the closest point to x in S. Let w be another point of S and let ℓ be the line betweenv and w. Then in order for v to be the closest point to x in S, it must also be so on the region of ℓcontained in S. This condition is then equivalent to the projection of x lying past v on the line ℓ, oralternatively that v lies in the opposite halfspace of w defined by the hyperplane perpendicular to ℓand passing through v. This can be written algebraically as (x − v) · (w − v) ≤ 0. Therefore, v is theclosest point to x if and only if (x − v) · (w − v) ≤ 0 for all w in S.

Note that these conditions do not depend on where w is on the line, so for each line intersecting Snontrivially, let us choose w such that w lies on the hyperplane H containing all the vertices of Sexcept for v and −v. We can further see that the conditions are linear in w, so them holding for all w

Geometry Test

in H is equivalent to them holding on the vertices of the region S ∩ H, which are simply the verticesof S except for v and −v. Let us now compute what these conditions look like.

Without loss of generality, let v = (1, 0, . . . , 0) and w = (0, 1, 0, . . . , 0). Then the equation is of theform (x−v) ·(−1, 1, 0, . . . , 0) ≤ 0, which we can rewrite as (x1−1, x2, x3, . . . , x2012) ·(−1, 1, 0, . . . , 0) =1 − x1 + x2 ≤ 0. For the other choices of w, we get the similar conditions that 1 − x1 + xi ≤ 0 andalso 1 − x1 − xi ≤ 0 for each i ∈ {2, . . . , 2012}. Note that if any xi ∈ {2,−2} for i 6= 1, then one ofthese conditions trivially fails, as it would require 3 − x1 ≤ 0. Therefore, the only face of T where xcan lie is the face defined by x1 = 2, which gives us the conditions that −1 + xi ≤ 0 and −1 − xi ≤ 0,so xi ∈ [−1, 1] for all i ∈ {2, . . . , 2012}. This defines a 2011-dimensional hypercube of side length 2 onthe face of T defined by x1 = 2, and we obtain similar regions on each of the other faces correspondingto the other vertices of S. Therefore, the volume of the set of x for which x is closest to a vertex is2 · 2012 · 22011 and the volume of all the choices of x is 2 · 2012 · 42011, so the desired probability is

122011 .

8. Hexagon ABCDEF has a circumscribed circle and an inscribed circle. If AB = 9, BC = 6, CD = 2,and EF = 4. Find {DE,FA}.

Answer: { 9+√

332 , 9−

√33

2 } By Brianchon’s Theorem, AD,BE,CF concur at some point P . Also, it

follows from the fact that tangents from a point to a circle have equal lengths that AB + CD + EF =BC + DE + FA. Let DE = x, so that FA = 9 − x.

Note that APB ∼ EPD, BPC ∼ FPE, and CPD ∼ APF . The second similarty gives BP/FP = 3/2,so that BP = 3y, FP = 2y for some y. From here, the first similarity gives DP = xy/3. Now, the

third similarity gives 4y = (9−x)xy

3 , so that x2 − 9x + 12 = 0. It follows that x =9 ±

√33

2, giving our

answer.

9. Let O,O1, O2, O3, O4 be points such that O1, O,O3 and O2, O,O4 are collinear in that order, OO1 =1, OO2 = 2, OO3 =

√2, OO4 = 2, and ∡O1OO2 = 45◦. Let ω1, ω2, ω3, ω4 be the circles with respective

centers O1, O2, O3, O4 that go through O. Let A be the intersection of ω1 and ω2, B be the intersectionof ω2 and ω3, C be the intersection of ω3 and ω4, and D be the intersection of ω4 and ω1, with A,B,C,Dall distinct from O. What is the largest possible area of a convex quadrilateral P1P2P3P4 such that Pi

lies on Oi and that A,B,C,D all lie on its perimeter?

Answer: 8 + 4√

2 We first maximize the area of triangle P1OP2, noting that the sum of the area ofP1OP2 and the three other analogous triangles is the area of P1P2P3P4. Note that if A 6= P1, P2, withoutloss of generality say ∠OAP1 < 90◦. Then, ∠OO1P1 = 2∠OAP1, and since ∠OAP2 = 180◦−∠OAP1 >90◦, we see that ∠OO2P2 = 2∠OAP1 as well, and it follows that OO1P1 ∼ OO2P2. This is a spiralsimilarity, so OO1O2 ∼ OP1P2, and in particular ∠P1OP2 = ∠O1OO2, which is fixed. By the sinearea formula, to maximize OP1 · OP2, which is bounded above by the diameters 2(OO1), 2(OO2). Ina similar way, we want P3, P4 to be diametrically opposite O3, O4 in their respective circles.

When we take these Pi, we indeed have A ∈ P1P2 and similarly for B,C,D, since ∠OAP1 = ∠OAP2 =90◦. To finish, the area of the quadrilateral is the sum of the areas of the four triangles, which is

1

2·√

2

2· 22 · (1 · 2 + 2 ·

√2 +

√2 · 2 + 2 · 1) = 8 + 4

√2.

10. Let C denote the set of points (x, y) ∈ R2 such that x2 + y2 ≤ 1. A sequence Ai = (xi, yi)|i ≥ 0 of

points in R2 is ‘centric’ if it satisfies the following properties:

• A0 = (x0, y0) = (0, 0), A1 = (x1, y1) = (1, 0).

• For all n ≥ 0, the circumcenter of triangle AnAn+1An+2 lies in C.

Let K be the maximum value of x22012 + y2

2012 over all centric sequences. Find all points (x, y) suchthat x2 + y2 = K and there exists a centric sequence such that A2012 = (x, y).

Geometry Test

Answer: (−1006, 1006√

3), (−1006,−1006√

3) Consider any triple of points △AnAn+1An+2 with

circumcenter Pn. By the Triangle Inequality we have AnPn ≤ AnA0+A0Pn ≤ AnA0+1. Since Pn is thecircumcenter, we have PnAn = PnAn+2. Finally we have An+2a0 ≤ PnAn+2+1 = AnPn+1 ≤ AnA0+2.

Therefore√

x2n+2 + y2

n+2 ≤√

x2n + y2

n + 2.

It is also clear that equality occurs if and only if An, A0, Pn, An+2 are all collinear and Pn lies on theunit circle.

From the above inequality it is clear that√

x22012 + y2

2012 ≤ 2012. So the maximum value of K is 20122.

Now we must find all points A2 that conforms to the conditions of the equality case. P0 must lie onthe unit circle, so it lies on the intersection of the unit circle with the perpendicular bisector of A0A1,

which is the line x = 12 . Thus P0 must be one of

(

12 ,±

√3

2

)

. From now on we assume that we take the

positive root, as the negative root just reflects all successive points about the x-axis.

If P0 =(

12 ,

√3

2

)

then A0, P0, A2 must be colinear, so A2 = (1,√

3).

Then since we must have A0, P2n, A2n, A2n+2 colinear and P2n on the unit circle, it follows that

P2n = (−1)n(

12 ,

√3

2

)

. Then by induction we have A2n = (−1)n+1(n, n√

3). To fill out the rest of the

sequence, we may take A2n+1 = (−1)n(n + 1,−n√

3) and P2n+1 = (−1)n+1(

12 ,−

√3

2

)

so that we get

the needed properties.

Therefore the answer is

A2012 ∈ {(−1006, 1006√

3), (−1006,−1006√

3)}

.

Geometry Test

15th Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Guts

1. [2] Square ABCD has side length 2, and X is a point outside the square such that AX = XB =√

2.What is the length of the longest diagonal of pentagon AXBCD?

2. [2] Let a0, a1, a2, . . . denote the sequence of real numbers such that a0 = 2 and an+1 = an

1+anfor n ≥ 0.

Compute a2012.

3. [2] Suppose x and y are real numbers such that −1 < x < y < 1. Let G be the sum of the geometricseries whose first term is x and whose ratio is y, and let G′ be the sum of the geometric series whosefirst term is y and ratio is x. If G = G′, find x + y.

4. [2] Luna has an infinite supply of red, blue, orange, and green socks. She wants to arrange 2012 socksin a line such that no red sock is adjacent to a blue sock and no orange sock is adjacent to a greensock. How many ways can she do this?

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15TH ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 11 FEBRUARY 2012 — GUTS ROUND

5. [3] Mr. Canada chooses a positive real a uniformly at random from (0, 1], chooses a positive real buniformly at random from (0, 1], and then sets c = a/(a + b). What is the probability that c liesbetween 1/4 and 3/4?

6. [3] Let rectangle ABCD have lengths AB = 20 and BC = 12. Extend ray BC to Z such that CZ = 18.Let E be the point in the interior of ABCD such that the perpendicular distance from E to AB is6 and the perpendicular distance from E to AD is 6. Let line EZ intersect AB at X and CD at Y .Find the area of quadrilateral AXY D.

7. [3] M is an 8 × 8 matrix. For 1 ≤ i ≤ 8, all entries in row i are at least i, and all entries on column iare at least i. What is the minimum possible sum of the entries of M?

8. [3] Amy and Ben need to eat 1000 total carrots and 1000 total muffins. The muffins can not be eatenuntil all the carrots are eaten. Furthermore, Amy can not eat a muffin within 5 minutes of eating acarrot and neither can Ben. If Amy eats 40 carrots per minute and 70 muffins per minute and Beneats 60 carrots per minute and 30 muffins per minute, what is the minimum number of minutes it willtake them to finish the food?

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15TH ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 11 FEBRUARY 2012 — GUTS ROUND

9. [5] Given △ABC with AB < AC, the altitude AD, angle bisector AE, and median AF are drawnfrom A, with D,E, F all lying on BC. If ∡BAD = 2∡DAE = 2∡EAF = ∡FAC, what are all possiblevalues of ∡ACB?

10. [5] Let P be a polynomial such that P (x) = P (0) + P (1)x + P (2)x2 and P (−1) = 1. Compute P (3).

11. [5] Knot is on an epic quest to save the land of Hyruler from the evil Gammadorf. To do this, he mustcollect the two pieces of the Lineforce, then go to the Temple of Lime. As shown on the figure, Knotstarts on point K, and must travel to point T , where OK = 2 and OT = 4. However, he must firstreach both solid lines in the figure below to collect the pieces of the Lineforce. What is the minimaldistance Knot must travel to do so?

K

T

15◦

30◦

15◦

O

12. [5] Knot is ready to face Gammadorf in a card game. In this game, there is a deck with twenty cardsnumbered from 1 to 20. Each player starts with a five card hand drawn from this deck. In each round,Gammadorf plays a card in his hand, then Knot plays a card in his hand. Whoever played a cardwith greater value gets a point. At the end of five rounds, the player with the most points wins. IfGammadorf starts with a hand of 1, 5, 10, 15, 20, how many five-card hands of the fifteen remainingcards can Knot draw which always let Knot win (assuming he plays optimally)?

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15TH ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 11 FEBRUARY 2012 — GUTS ROUND

13. [7] Niffy’s favorite number is a positive integer, and Stebbysaurus is trying to guess what it is. Niffytells her that when expressed in decimal without any leading zeros, her favorite number satisfies thefollowing:

• Adding 1 to the number results in an integer divisible by 210.

• The sum of the digits of the number is twice its number of digits.

• The number has no more than 12 digits.

• The number alternates in even and odd digits.

Given this information, what are all possible values of Niffy’s favorite number?

14. [7] Let triangle ABC have AB = 5, BC = 6, and AC = 7, with circumcenter O. Extend ray AB topoint D such that BD = 5, and extend ray BC to point E such that OD = OE. Find CE.

15. [7] Let f(x) = x2 + ax + b and g(x) = x2 + cx + d be two distinct real polynomials such that thex-coordinate of the vertex of f is a root of g, the x-coordinate of the vertex of g is a root of f and bothf and g have the same minimum value. If the graphs of the two polynomials intersect at the point(2012,−2012), what is the value of a + c?

16. [7] Let A, B, C, and D be points randomly selected independently and uniformly within the unitsquare. What is the probability that the six lines AB, AC, AD, BC, BD, and CD all have positiveslope?

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15TH ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 11 FEBRUARY 2012 — GUTS ROUND

17. [11] Mark and William are playing a game. Two walls are placed 1 meter apart, with Mark andWilliam each starting an orb at one of the walls. Simultaneously, they release their orbs directlytoward the other. Both orbs are enchanted such that, upon colliding with each other, they instantlyreverse direction and go at double their previous speed. Furthermore, Mark has enchanted his orb sothat when it collides with a wall it instantly reverses direction and goes at double its previous speed(William’s reverses direction at the same speed). Initially, Mark’s orb is moving at 1

1000 meters/s, andWilliam’s orb is moving at 1 meter/s. Mark wins when his orb passes the halfway point between thetwo walls. How fast, in meters/s, is his orb going when this first happens?

18. [11] Let x and y be positive real numbers such that x2 + y2 = 1 and (3x − 4x3)(3y − 4y3) = − 12 .

Compute x + y.

19. [11] Given that P is a real polynomial of degree at most 2012 such that P (n) = 2n for n = 1, 2, . . . , 2012,what choice(s) of P (0) produce the minimal possible value of P (0)2 + P (2013)2?

20. [11] Let n be the maximum number of bishops that can be placed on the squares of a 6× 6 chessboardsuch that no two bishops are attacking each other. Let k be the number of ways to put n bishops onan 6 × 6 chessboard such that no two bishops are attacking each other. Find n + k. (Two bishops areconsidered to be attacking each other if they lie on the same diagonal. Equivalently, if we label thesquares with coordinates (x, y), with 1 ≤ x, y ≤ 6, then the bishops on (a, b) and (c, d) are attackingeach other if and only if |a − c| = |b − d|.). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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15TH ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 11 FEBRUARY 2012 — GUTS ROUND

21. [13] Let N be a three-digit integer such that the difference between any two positive integer factorsof N is divisible by 3. Let d(N) denote the number of positive integers which divide N . Find themaximum possible value of N · d(N).

22. [13] For each positive integer n, there is a circle around the origin with radius n. Rainbow Dash startsoff somewhere on the plane, but not on a circle. She takes off in some direction in a straight path. Shemoves

√5

5 units before crossing a circle, then√

5 units, then 3√

55 units. What distance will she travel

before she crosses another circle?

23. [13] Points X and Y are inside a unit square. The score of a vertex of the square is the minimumdistance from that vertex to X or Y . What is the minimum possible sum of the scores of the verticesof the square?

24. [13] Franklin has four bags, numbered 1 through 4. Initially, the first bag contains fifteen balls,numbered 1 through 15, and the other bags are empty. Franklin randomly pulls a pair of balls out ofthe first bag, throws away the ball with the lower number, and moves the ball with the higher numberinto the second bag. He does this until there is only one ball left in the first bag. He then repeats thisprocess in the second and third bag until there is exactly one ball in each bag. What is the probabilitythat ball 14 is in one of the bags at the end?

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15TH ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 11 FEBRUARY 2012 — GUTS ROUND

25. [17] FemtoPravis is walking on an 8×8 chessboard that wraps around at its edges (so squares on the leftedge of the chessboard are adjacent to squares on the right edge, and similarly for the top and bottomedges). Each femtosecond, FemtoPravis moves in one of the four diagonal directions uniformly atrandom. After 2012 femtoseconds, what is the probability that FemtoPravis is at his original location?

26. [17] Suppose ABC is a triangle with circumcenter O and orthocenter H such that A,B,C,O, and Hare all on distinct points with integer coordinates. What is the second smallest possible value of thecircumradius of ABC?

27. [17] Let S be the set {1, 2, . . . , 2012}. A perfectutation is a bijective function h from S to itself suchthat there exists an a ∈ S such that h(a) 6= a, and that for any pair of integers a ∈ S and b ∈ S suchthat h(a) 6= a, h(b) 6= b, there exists a positive integer k such that hk(a) = b. Let n be the number ofordered pairs of perfectutations (f, g) such that f(g(i)) = g(f(i)) for all i ∈ S, but f 6= g. Find theremainder when n is divided by 2011.

28. [17] Alice is sitting in a teacup ride with infinitely many layers of spinning disks. The largest disk hasradius 5. Each succeeding disk has its center attached to a point on the circumference of the previousdisk and has a radius equal to 2/3 of the previous disk. Each disk spins around its center (relative tothe disk it is attached to) at a rate of π/6 radians per second. Initially, at t = 0, the centers of thedisks are aligned on a single line, going outward. Alice is sitting at the limit point of all these disks.After 12 seconds, what is the length of the trajectory that Alice has traced out?

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15TH ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 11 FEBRUARY 2012 — GUTS ROUND

29. [19] Consider the cube whose vertices are the eight points (x, y, z) for which each of x, y, and z iseither 0 or 1. How many ways are there to color its vertices black or white such that, for any vertex,if all of its neighbors are the same color then it is also that color? Two vertices are neighbors if theyare the two endpoints of some edge of the cube.

30. [19] You have a twig of length 1. You repeatedly do the following: select two points on the twigindependently and uniformly at random, make cuts on these two points, and keep only the largestpiece. After 2012 repetitions, what is the expected length of the remaining piece?

31. [19] Let S7 denote all the permutations of 1, 2, . . . , 7. For any π ∈ S7, let f(π) be the smallest positiveinteger i such that π(1), π(2), . . . , π(i) is a permutation of 1, 2, . . . , i. Compute

∑π∈S7

f(π).

32. [19] Let S be a set of size 3. How many collections T of subsets of S have the property that for anytwo subsets U ∈ T and V ∈ T , both U ∩ V and U ∪ V are in T?

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15TH ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 11 FEBRUARY 2012 — GUTS ROUND

33. [23] Compute the decimal expansion of√

π. Your score will be min(23, k), where k is the number ofconsecutive correct digits immediately following the decimal point in your answer.

For each of the remaining three problems, it is difficult to obtain an exact answer. Instead,give an interval [L,U ], where L and U are positive real numbers written in decimal. If[L,U ] contains the answer to the problem, you will receive credit based off of how close Land U are. Otherwise, you will receive no credit.

34. [23] Let Q be the product of the sizes of all the non-empty subsets of {1, 2, . . . , 2012}, and let M =log2(log2(Q)). Give lower and upper bounds L and U for M . If 0 < L ≤ M ≤ U , then your score willbe min(23,

⌊23

3(U−L)

⌋). Otherwise, your score will be 0.

35. [23] Let N be the number of distinct roots of∏2012

k=1 (xk − 1). Give lower and upper bounds L and U

on N . If 0 < L ≤ N ≤ U , then your score will be⌊

23(U/L)1.7

⌋. Otherwise, your score will be 0.

36. [23] Maria is hopping up a flight of stairs with 100 steps. At every hop, she advances some integernumber of steps. Each hop she makes has fewer steps. However, the positive difference between thelength of consecutive hops decreases. Let P be the number of distinct ways she can hop up the stairs.

Find lower and upper bounds L and U for P . If 0 < L ≤ P ≤ U , your score will be⌊

23√U/L

⌋.

Otherwise, your score will be 0.

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School Team Team ID#

1. [2]

2. [2]

3. [2]

4. [2]

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School Team Team ID#

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School Team Team ID#

9. [5]

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12. [5]

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School Team Team ID#

13. [7]

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16. [7]

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School Team Team ID#

17. [11]

18. [11]

19. [11]

20. [11]

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School Team Team ID#

21. [13]

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24. [13]

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School Team Team ID#

25. [17]

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28. [17]

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School Team Team ID#

29. [19]

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32. [19]

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School Team Team ID#

33. [23]

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15th

Annual Harvard-MIT Mathematics TournamentSaturday 11 February 2012

Guts

1. [2] Square ABCD has side length 2, and X is a point outside the square such that AX = XB =√

2.What is the length of the longest diagonal of pentagon AXBCD?

Answer:√

10 Since AX = XB =√

2 and AB = 2, we have ∠AXB = 90◦. Hence, the distancefrom X to AB is 1 and the distance from X to CD is 3. By inspection, the largest diagonals are thusBX = CX =

√32 + 12.

2. [2] Let a0, a1, a2, . . . denote the sequence of real numbers such that a0 = 2 and an+1 = an

1+an

for n ≥ 0.Compute a2012.

Answer: 24025 Calculating out the first few terms, note that they follow the pattern an = 2

2n+1 .

Plugging this back into the recursion shows that it indeed works.

3. [2] Suppose x and y are real numbers such that −1 < x < y < 1. Let G be the sum of the geometricseries whose first term is x and whose ratio is y, and let G′ be the sum of the geometric series whosefirst term is y and ratio is x. If G = G′, find x + y.

Answer: 1 We note that G = x/(1− y) and G′ = y/(1− x). Setting them equal gives x/(1− y) =y/(1 − x) ⇒ x2 − x = y2 − x ⇒ (x + y − 1)(x − y) = 0, so we get that x + y − 1 = 0 ⇒ x + y = 1.

4. [2] Luna has an infinite supply of red, blue, orange, and green socks. She wants to arrange 2012 socksin a line such that no red sock is adjacent to a blue sock and no orange sock is adjacent to a greensock. How many ways can she do this?

Answer: 4 · 32011 Luna has 4 choices for the first sock. After that, she has 3 choices for each of2011 remaining socks for a total of 4 · 32011.

5. [3] Mr. Canada chooses a positive real a uniformly at random from (0, 1], chooses a positive real buniformly at random from (0, 1], and then sets c = a/(a + b). What is the probability that c liesbetween 1/4 and 3/4?

Answer: 2/3 From c ≥ 1/4 we get

a

a + b≥ 1

4⇐⇒ b ≤ 3a

and similarly c ≤ 3/4 givesa

a + b≤ 3

4⇐⇒ a ≤ 3b.

Choosing a and b randomly from [0, 1] is equivalent to choosing a single point uniformly and randomlyfrom the unit square, with a on the horizontal axis and b on the vertical axis:

a = 3b

b = 3a

To find the probability that b ≤ 3a and a ≤ 3b, we need to find the area of the shaded region of thesquare. The area of each of the triangles on the side is (1/2)(1)(1/3) = 1/6, and so the area of theshaded region is 1 − 2(1/6) = 2/3.

Guts

6. [3] Let rectangle ABCD have lengths AB = 20 and BC = 12. Extend ray BC to Z such that CZ = 18.Let E be the point in the interior of ABCD such that the perpendicular distance from E to AB is6 and the perpendicular distance from E to AD is 6. Let line EZ intersect AB at X and CD at Y .Find the area of quadrilateral AXY D.

Answer: 72 Draw the line parallel to AD through E, intersecting AB at F and CD at G. Itis clear that XFE and Y GE are congruent, so the area of AXY D is equal to that of AFGD. ButAFGD is simply a 12 by 6 rectangle, so the answer must be 72. (Note: It is also possible to directlycompute the values of AX and DY , then use the formula for the area of a trapezoid.)

7. [3] M is an 8 × 8 matrix. For 1 ≤ i ≤ 8, all entries in row i are at least i, and all entries on column iare at least i. What is the minimum possible sum of the entries of M?

Answer: 372 Let sn be the minimum possible sum for an n by n matrix. Then, we note thatincreasing it by adding row n + 1 and column n + 1 gives 2n + 1 additional entries, each of which hasminimal size at least n + 1. Consequently, we obtain sn+1 = sn + (2n + 1)(n + 1) = sn + 2n2 + 3n + 1.Since s0 = 0, we get that s8 = 2(72 + . . . + 02) + 3(7 + . . . + 0) + 8 = 372.

8. [3] Amy and Ben need to eat 1000 total carrots and 1000 total muffins. The muffins can not be eatenuntil all the carrots are eaten. Furthermore, Amy can not eat a muffin within 5 minutes of eating acarrot and neither can Ben. If Amy eats 40 carrots per minute and 70 muffins per minute and Beneats 60 carrots per minute and 30 muffins per minute, what is the minimum number of minutes it willtake them to finish the food?

Answer: 23.5 or 47/2 Amy and Ben will continuously eat carrots, then stop (not necessarily at

the same time), and continuously eat muffins until no food is left. Suppose that Amy and Ben finisheating the carrots in T1 minutes and the muffins T2 minutes later; we wish to find the minimum valueof T1 + T2. Furthermore, suppose Amy finishes eating the carrots at time a1, and Ben does so at timeb1, so that T1 = max(a1, b1).

First, suppose that a1 ≤ b1, and let b1 − a1 = c. We have 40(T1 − c)+60T1 = 1000, so T1 is minimizedwhen c = 0. Also, 30(T2 − 5)+70(T2 −max(5− c, 0)) = 1000. Wee see that T1 +T2 is minimized whenc = 5, and T1 + T2 = 23.5. In a similar way, we see that when b1 ≤ a1, T1 + T2 > 23.5, so our answeris 23.5.

9. [5] Given △ABC with AB < AC, the altitude AD, angle bisector AE, and median AF are drawnfrom A, with D,E, F all lying on BC. If ∡BAD = 2∡DAE = 2∡EAF = ∡FAC, what are all possiblevalues of ∡ACB?

Answer: 30◦ or π/6 radians Let H and O be the orthocenter and circumcenter of ABC, respctively:

it is well-known (and not difficult to check) that ∡BAH = ∡CAO. However, note that ∡BAH =∡BAD = ∡CAF , so ∡CAF = ∡CAO, that is, O lies on median AF , and since AB < AC, it followsthat F = O. Therefore, ∡BAC = 90◦.

Now, we compute ∡ACB = ∡BAD = 26∡BAC = 30◦.

10. [5] Let P be a polynomial such that P (x) = P (0) + P (1)x + P (2)x2 and P (−1) = 1. Compute P (3).

Answer: 5 Plugging in x = −1, 1, 2 results in the trio of equations 1 = P (−1) = P (0)−P (1)+P (2),P (1) = P (0) + P (1) + P (2) ⇒ P (1) + P (2) = 0, and P (2) = P (0) + 2P (1) + 4P (2). Solving theseas a system of equations in P (0), P (1), P (2) gives P (0) = −1, P (1) = −1, P (2) = 1. Consequently,P (x) = x2 − x − 1 ⇒ P (3) = 5.

11. [5] Knot is on an epic quest to save the land of Hyruler from the evil Gammadorf. To do this, he mustcollect the two pieces of the Lineforce, then go to the Temple of Lime. As shown on the figure, Knotstarts on point K, and must travel to point T , where OK = 2 and OT = 4. However, he must firstreach both solid lines in the figure below to collect the pieces of the Lineforce. What is the minimaldistance Knot must travel to do so?

Guts

K

T

15◦

30◦

15◦

O

Answer: 2√

5 Let l1 and l2 be the lines as labeled in the above diagram. First, suppose Knot

visits l1 first, at point P1, then l2, at point P2. Let K ′ be the reflection of K over l1, and let T ′ be thereflection of T over l2. The length of Knot’s path is at least

KP1 + P1P2 + P2T = K ′P1 + P1P2 + P2T′ ≥ K ′T ′

by the Triangle Inequality (This bound can be achieved by taking P1, P2 to be the intersections ofK ′T ′ with l1, l2, respectively.) Also, note that ∡K ′OT ′ = 90◦, so that K ′T ′ = 2

√5.

Now, suppose Knot instead visits l2 first, at point Q2, then l1, at point Q1. Letting K ′′ be the reflectionof K over l2 and T ′′ be the reflection of T over l1, by similar logic to before the length of his path isat least the length of K ′′T ′′. However, by inspection K ′′T ′′ > K ′T ′, so our answer is 2

√5.

12. [5] Knot is ready to face Gammadorf in a card game. In this game, there is a deck with twenty cardsnumbered from 1 to 20. Each player starts with a five card hand drawn from this deck. In each round,Gammadorf plays a card in his hand, then Knot plays a card in his hand. Whoever played a cardwith greater value gets a point. At the end of five rounds, the player with the most points wins. IfGammadorf starts with a hand of 1, 5, 10, 15, 20, how many five-card hands of the fifteen remainingcards can Knot draw which always let Knot win (assuming he plays optimally)?

Answer: 2982 Knot can only lose if all of his cards are lower than 10; if not he can win by playingthe lowest card that beats Gammadorf’s card, or if this is not possible, his lowest card, each turn.There are

(

75

)

= 21 losing hands, so he has(

155

)

−(

75

)

possible winning hands.

13. [7] Niffy’s favorite number is a positive integer, and Stebbysaurus is trying to guess what it is. Niffytells her that when expressed in decimal without any leading zeros, her favorite number satisfies thefollowing:

• Adding 1 to the number results in an integer divisible by 210.

• The sum of the digits of the number is twice its number of digits.

• The number has no more than 12 digits.

• The number alternates in even and odd digits.

Given this information, what are all possible values of Niffy’s favorite number?

Answer: 1010309 Note that Niffy’s favorite number must end in 9, since adding 1 makes itdivisible by 10. Also, the sum of the digits of Niffy’s favorite number must be even (because it is equalto twice the number of digits) and congruent to 2 modulo 3 (because adding 1 gives a multiple of 3).Furthermore, the sum of digits can be at most 24, because there at most 12 digits in Niffy’s favoritenumber, and must be at least 9, because the last digit is 9. This gives the possible sums of digits 14and 20. However, if the sum of the digits of the integer is 20, there are 10 digits, exactly 5 of whichare odd, giving an odd sum of digits, which is impossible. Thus, Niffy’s favorite number is a 7 digitnumber with sum of digits 14.

Guts

The integers which we seek must be of the form ABCDEF9, where A,C,E are odd, B,D,F are even,and A+B+C +D+E +F = 5. Now, note that {A,C,E} = {1, 1, 1} or {1, 1, 3}, and these correspondto {B,D,F} = {0, 0, 2} and {0, 0, 0}, respectively. It suffices to determine which of these six integersare congruent to −1 (mod 7), and we see that Niffy’s favorite number must be 1010309.

14. [7] Let triangle ABC have AB = 5, BC = 6, and AC = 7, with circumcenter O. Extend ray AB topoint D such that BD = 5, and extend ray BC to point E such that OD = OE. Find CE.

Answer:√

59 − 3 Because OD = OE, D and E have equal power with respect to the circle, so

(EC)(EB) = (DB)(DA) = 50. Letting EC = x, we have x(x + 6) = 50, and taking the positive rootgives x =

√59 − 3.

15. [7] Let f(x) = x2 + ax + b and g(x) = x2 + cx + d be two distinct real polynomials such that thex-coordinate of the vertex of f is a root of g, the x-coordinate of the vertex of g is a root of f and bothf and g have the same minimum value. If the graphs of the two polynomials intersect at the point(2012,−2012), what is the value of a + c?

Answer: −8048 It is clear, by symmetry, that 2012 is the equidistant from the vertices of thetwo quadratics. Then it is clear that reflecting f about the line x = 2012 yields g and vice versa.Thus the average of each pair of roots is 2012. Thus the sum of the four roots of f and g is 8048, soa + c = −8048.

16. [7] Let A, B, C, and D be points randomly selected independently and uniformly within the unitsquare. What is the probability that the six lines AB, AC, AD, BC, BD, and CD all have positiveslope?

Answer: 124 Consider the sets of x-coordinates and y-coordinates of the points. In order to

make 6 lines of positive slope, we must have smallest x-coordinate must be paired with the smallesty-coordinate, the second smallest together, and so forth. If we fix the order of the x-coordinates, theprobability that the corresponding y-coordinates are in the same order is 1/24.

17. [11] Mark and William are playing a game. Two walls are placed 1 meter apart, with Mark andWilliam each starting an orb at one of the walls. Simultaneously, they release their orbs directlytoward the other. Both orbs are enchanted such that, upon colliding with each other, they instantlyreverse direction and go at double their previous speed. Furthermore, Mark has enchanted his orb sothat when it collides with a wall it instantly reverses direction and goes at double its previous speed(William’s reverses direction at the same speed). Initially, Mark’s orb is moving at 1

1000 meters/s, andWilliam’s orb is moving at 1 meter/s. Mark wins when his orb passes the halfway point between thetwo walls. How fast, in meters/s, is his orb going when this first happens?

Answer: 217/125 If the two orbs leave their respective walls at the same time, then they will return

to their walls at the same time (because colliding affects both their speeds). After returning to thewall n times, Mark’s orb will travel at 4n

1000 meter/s and William’s will travel at 2n meter/s. Mark

wins when his orb is traveling faster at n = 10. 410

1000 = 217

125

18. [11] Let x and y be positive real numbers such that x2 + y2 = 1 and (3x − 4x3)(3y − 4y3) = − 12 .

Compute x + y.

Answer:√

62 Solution 1: Let x = cos(θ) and y = sin(θ). Then, by the triple angle formulae, we

have that 3x − 4x3 = − cos(3θ) and 3y − 4y3 = sin(3θ), so − sin(3θ) cos(3θ) = − 12 . We can write

this as 2 sin(3θ) cos(3θ) = sin(6θ) = 1, so θ = 16 sin−1(1) = π

12 . Thus, x + y = cos(

π12

)

+ sin(

π12

)

=√6+

√2

4 +√

6−√

24 =

√6

2 .

Solution 2: Expanding gives 9xy +16x3y3 − 12xy3 − 12x3y = 9(xy)+16(xy)3 − 12(xy)(x2 + y2) = − 12 ,

and since x2 + y2 = 1, this is −3(xy) + 16(xy)3 = − 12 , giving xy = − 1

2 , 14 . However, since x and y are

positive reals, we must have xy = 14 . Then, x + y =

√

x2 + y2 + 2xy =√

1 + 2 · 14 =

√

32 =

√6

2 .

Guts

19. [11] Given that P is a real polynomial of degree at most 2012 such that P (n) = 2n for n = 1, 2, . . . , 2012,what choice(s) of P (0) produce the minimal possible value of P (0)2 + P (2013)2?

Answer: 1 − 22012 Define ∆1(n) = P (n + 1) − P (n) and ∆i(n) = ∆i−1(n + 1) − ∆i−1(n) for

i > 1. Since P (n) has degree at most 2012, we know that ∆2012(n) is constant. Computing, weobtain ∆1(0) = 2 − P (0) and ∆i(0) = 2i−1 for 1 < i ≤ 2012. We see that continuing on gives∆2012(0) = ∆2012(1) = P (0) and ∆i(2012 − i) = 22013−i for 1 ≤ i ≤ 2011. Then, P (2013) =P (2012) + ∆1(2012) = . . . = P (2012) + ∆1(2011) + . . . + ∆2012(0) = P (0) + 22013 − 2. Now, we wantto minimize the value of P (0)2 + P (2013)2 = 2P (0)2 + 2P (0)(22013 − 2) + (22013 − 2)2, but this occurssimply when P (0) = − 1

2 (22013 − 2) = 1 − 22012.

20. [11] Let n be the maximum number of bishops that can be placed on the squares of a 6× 6 chessboardsuch that no two bishops are attacking each other. Let k be the number of ways to put n bishops onan 6 × 6 chessboard such that no two bishops are attacking each other. Find n + k. (Two bishops areconsidered to be attacking each other if they lie on the same diagonal. Equivalently, if we label thesquares with coordinates (x, y), with 1 ≤ x, y ≤ 6, then the bishops on (a, b) and (c, d) are attackingeach other if and only if |a − c| = |b − d|.)Answer: 74 Color the square with coordinates (i, j) black if i + j is odd and white otherwise,for all 1 ≤ i, j ≤ 6. Looking at the black squares only, we note that there are six distinct diagonalswhich run upward and to the right, but that two of them consist only of a corner square; we cannotsimultaneously place bishops on both of these corner squares. Consequently, we can place at most fivebishops on black squares. (This can be achieved by placing bishops on (1, 2), (1, 4), (6, 1), (6, 3), (6, 5).)If there are five bishops on black squares, there must be exactly one bishop on one of the two blackcorner squares, (6, 1) and (1, 6): suppose without loss of generality that we place a bishop on (1, 6).Then, exactly one of (3, 6) and (1, 4) must also contain a bishop, and there are 2 ways to place twobishops on the four remaining black squares that are not yet under attack. Thus, we have a total of2 · 2 · 2 possible placements on black squares.

Similarly, there are at most 5 bishops which can be placed on white squares and 23 ways to place them,so that n = 10 and k = 26. Finally, n + k = 10 + 26 = 74.

21. [13] Let N be a three-digit integer such that the difference between any two positive integer factorsof N is divisible by 3. Let d(N) denote the number of positive integers which divide N . Find themaximum possible value of N · d(N).

Answer: 5586 We first note that all the prime factors of n must be 1 modulo 3 (and thus 1 modulo6). The smallest primes with this property are 7, 13, 19, . . .. Since 74 = 2401 > 1000, the number canhave at most 3 prime factors (including repeats). Since 7 ·13 ·19 = 1729 > 1000, the most factors N canhave is 6. Consider the number 72 ·19 = 931, which has 6 factors. For this choice of N , N ·d(N) = 5586.For another N to do better, it must have at least 6 factors, for otherwise, N · d(N) < 1000 · 5 = 5000.It is easy to verify that 72 · 19 is the greatest number with 6 prime factors satisfying our conditions, sothe answer must be 5586.

22. [13] For each positive integer n, there is a circle around the origin with radius n. Rainbow Dash startsoff somewhere on the plane, but not on a circle. She takes off in some direction in a straight path. She

moves√

55 units before crossing a circle, then

√5 units, then 3

√5

5 units. What distance will she travelbefore she crosses another circle?

Answer: 2√

170−9√

55 Note that the distance from Rainbow Dash’s starting point to the first place in

which she hits a circle is irrelevant, except in checking that this distance is small enough that she doesnot hit another circle beforehand. It will be clear at the end that our configuration does not allow this(by the Triangle Inequality). Let O be the origin, and let Rainbow Dash’s first three meeting points

be A,B,C so that AB =√

5 and BC = 3√

55 .

Consider the lengths of OA,OB,OC. First, note that if OA = OC = n (i.e. A and C lie on the samecircle), then we need OB = n−1, but since she only crosses the circle containing B once, it follows thatthe circle passing through B is tangent to AC, which is impossible since AB 6= AC. If OA = OB = n,

Guts

note that OC = n + 1. Dropping a perpendicular from O to AB, we see that by the PythagoreanTheorem,

n2 − 5

4= (n + 1)2 − 121

20,

from which we get that n is not an integer. Similarly, when OB = OC = n, we have OA = n + 1, andn is not an integer.

Therefore, either OA = n + 2, OB = n + 1, OC = n or OA = n,OB = n + 1, OC = n + 2. In the firstcase, by Stewart’s Theorem,

24√

5

5+ (n + 1)2 · 8

√5

5= n2 ·

√5 + (n + 2)2 · 3

√5

5.

This gives a negative value of n, so the configuration is impossible. In the final case, we have, againby Stewart’s Theorem,

24√

5

5+ (n + 1)2 · 8

√5

5= (n + 2)2 ·

√5 + n2 · 3

√5

5.

Solving gives n = 3, so OA = 3, OB = 4, OC = 5.

Next, we compute, by the Law of Cosines, cos ∠OAB = − 13√

5, so that sin∠OAB = 2

√11

3√

5. Let the

projection from O to line AC be P ; we get that OP = 2√

11√5

. Rainbow Dash will next hit the circle of

radius 6 at D. Our answer is now CD = PD − PC = 2√

1705 − 9

√5

5 by the Pythagorean Theorem.

23. [13] Points X and Y are inside a unit square. The score of a vertex of the square is the minimumdistance from that vertex to X or Y . What is the minimum possible sum of the scores of the verticesof the square?

Answer:√

6+√

22 Let the square be ABCD. First, suppose that all four vertices are closer to X than

Y . Then, by the triangle inequality, the sum of the scores is AX + BX + CX + DX ≥ AB + CD = 2.Similarly, suppose exactly two vertices are closer to X than Y . Here, we have two distinct cases: thevertices closer to X are either adjacent or opposite. Again, by the Triangle Inequality, it follows thatthe sum of the scores of the vertices is at least 2.

On the other hand, suppose that A is closer to X and B,C,D are closer to Y . We wish to computethe minimum value of AX + BY + CY + DY , but note that we can make X = A to simply minimizeBY + CY + DY . We now want Y to be the Fermat point of triangle BCD, so that ∡BY C =∡CY D = ∡DY B = 120◦. Note that by symmetry, we must have ∡BCY = ∡DCY = 45◦, so∡CBY = ∡CDY = 15◦.

And now we use the law of sines: BY = DY = sin 45◦

sin 120◦and CY = sin 15◦

sin 120◦. Now, we have BY + CY +

DY =√

2+√

62 , which is less than 2, so this is our answer.

24. [13] Franklin has four bags, numbered 1 through 4. Initially, the first bag contains fifteen balls,numbered 1 through 15, and the other bags are empty. Franklin randomly pulls a pair of balls out ofthe first bag, throws away the ball with the lower number, and moves the ball with the higher numberinto the second bag. He does this until there is only one ball left in the first bag. He then repeats thisprocess in the second and third bag until there is exactly one ball in each bag. What is the probabilitythat ball 14 is in one of the bags at the end?

Answer: 23 Pretend there is a 16th ball numbered 16. This process is equivalent to randomly

drawing a tournament bracket for the 16 balls, and playing a tournament where the higher ranked ballalways wins. The probability that a ball is left in a bag at the end is the probability that it loses toball 16. Of the three balls 14, 15, 16, there is a 1

3 chance 14 plays 15 first, a 13 chance 14 plays 16

first, and a 13 chance 15 plays 16 first. In the first case, 14 does not lose to 16, and instead loses to 15;

otherwise 14 loses to 16, and ends up in a bag. So the answer is 23 .

Guts

25. [17] FemtoPravis is walking on an 8×8 chessboard that wraps around at its edges (so squares on the leftedge of the chessboard are adjacent to squares on the right edge, and similarly for the top and bottomedges). Each femtosecond, FemtoPravis moves in one of the four diagonal directions uniformly atrandom. After 2012 femtoseconds, what is the probability that FemtoPravis is at his original location?

Answer:(

1+21005

21007

)2

We note the probability that he ends up in the same row is equal to the

probability that he ends up in the same column by symmetry. Clearly these are independent, so wecalculate the probability that he ends up in the same row.

Now we number the rows 0 − 7 where 0 and 7 are adjacent. Suppose he starts at row 0. After twomore turns, the probability he is in row 2 (or row 6) is 1

4 , and the probability he is in row 0 again is 12 .

Let an, bn, cn and dn denote the probability he is in row 0,2,4,6 respectively after 2n moves.

We have a0 = 1, and for n ≥ 0 we have the following equations:

an+1 =1

2an +

1

4bn +

1

4dn

bn+1 =1

2bn +

1

4an +

1

4cn

cn+1 =1

2cn +

1

4bn +

1

4dn

dn+1 =1

2dn +

1

4an +

1

4cn

From which we get the following equations:

an + cn =1

2

xn = an − cn =1

2(an−1 − cn−1) =

xn−1

2

So

a1006 + c1006 =1

2

x0 = 1, x1006 =1

21006

a1006 =1 + 21005

21007

And thus the answer is(

1+21005

21007

)2

.

26. [17] Suppose ABC is a triangle with circumcenter O and orthocenter H such that A,B,C,O, and Hare all on distinct points with integer coordinates. What is the second smallest possible value of thecircumradius of ABC?

Answer:√

10 Assume without loss of generality that the circumcenter is at the origin. By wellknown properties of the Euler line, the centroid G is such that O, G, and H are collinear, with G inbetween O and H, such that GH = 2GO. Thus, since G = 1

3 (A + B + C), and we are assuming O isthe origin, we have H = A + B + C. This means that as long as A, B, and C are integer points, Hwill be as well.

However, since H needs to be distinct from A, B, and C, we must have △ABC not be a right triangle,since in right triangles, the orthocenter is the vertex where the right angle is.

Now, if a circle centered at the origin has any integer points, it will have at least four integer points.(If it has a point of the form (a, 0), then it will also have (−a, 0), (0, a), and (0,−a). If it has a pointof the form (a, b), with a, b 6= 0, it will have each point of the form (±a,±b).) But in any of these caseswhere there are only four points, any triangle which can be made from those points is a right triangle.

Guts

Thus we need the circumcircle to contain at least eight lattice points. The smallest radius this occursat is

√12 + 22 =

√5, which contains the eight points (±1,±2) and (±2,±1). We get at least one valid

triangle with this circumradius:

A = (−1, 2), B = (1, 2), C = (2, 1).

The next valid circumradius is√

12 + 32 =√

10 which has the valid triangle

A = (−1, 3), B = (1, 3), C = (3, 1).

27. [17] Let S be the set {1, 2, . . . , 2012}. A perfectutation is a bijective function h from S to itself suchthat there exists an a ∈ S such that h(a) 6= a, and that for any pair of integers a ∈ S and b ∈ S suchthat h(a) 6= a, h(b) 6= b, there exists a positive integer k such that hk(a) = b. Let n be the number ofordered pairs of perfectutations (f, g) such that f(g(i)) = g(f(i)) for all i ∈ S, but f 6= g. Find theremainder when n is divided by 2011.

Answer: 2 Note that both f and g, when written in cycle notation, must contain exactly one cyclethat contains more than 1 element. Assume f has k fixed points, and that the other 2012−k elementsform a cycle, (of which there are (2011 − k)! ways).

Then note that if f fixes a then f(g(a)) = g(f(a)) = g(a) implies f fixes g(a) So g must send fixedpoints of f to fixed points of f . It must, therefore send non-fixed points to non-fixed points. Thispartitions S into two sets, at least one of which must be fixed by g, since g is a perfectutation.

If g fixes all of the the non-fixed points of f , then, since any function commutes with the identity, g fixessome m of the fixed points and cycles the rest in (k−m−1)! ways. So there are

∑k−2m=0

(

km

)

(k−m−1)!

choices, which is∑k−2

m=0k!

(k−m)m! .

If g fixes all of the fixed points of f , then order the non-fixed points of f a1, a2, . . . , a2012−k such thatf(ai) = ai+1. If g(ai) = aj then f(g(ai)) = aj+1 thus g(ai+1) = aj+1. Therefore the choice of g(a1)uniquely determines g(ai) for the rest of the i, and g(am) = am+j−i. But g has to be a perfectutation,so g cycles through all the non-fixed points of f , which happens if and only if j − i is relatively primeto 2012 − k. So there are φ(2012 − k) choices.

Therefore for any f there are∑k−2

m=0k!

(k−m)m! +φ(2012−k) choices of g, but one of them will be g = f ,

which we cannot have by the problem statement. So there are −1 +∑k−2

m=0k!

(k−m)m! + φ(2012 − k)options.

Now note that a permutation can not fix all but one element. So n =∑2010

k=0

(

2012k

)

(2011 − k)!(−1 +∑k−2

m=0k!

(k−m)m! + φ(2012 − k))

Modulo 2011 (which is prime), note that all terms in the summand except the one where k = 1 vanish.Thus, n ≡ (2010)!(−1 + (−1)) ≡ 2 (mod 2011) by Wilson’s Theorem.

28. [17] Alice is sitting in a teacup ride with infinitely many layers of spinning disks. The largest disk hasradius 5. Each succeeding disk has its center attached to a point on the circumference of the previousdisk and has a radius equal to 2/3 of the previous disk. Each disk spins around its center (relative tothe disk it is attached to) at a rate of π/6 radians per second. Initially, at t = 0, the centers of thedisks are aligned on a single line, going outward. Alice is sitting at the limit point of all these disks.After 12 seconds, what is the length of the trajectory that Alice has traced out?

Answer: 18π Suppose the center of the largest teacup is at the origin in the complex plane, and

let z = 23eπit/6. The center of the second disk is at 5eπit/6 at time t; that is, 15

2 z. Then the center ofthe third disk relative to the center of the second disk is at 15

2 z2, and so on. Summing up a geometricseries, we get that Alice’s position is

15

2(z + z2 + z3 + · · · ) =

15

2(1 + z2 + z3 + · · · ) − 15

2

=15

2

(

1

1 − z

)

− 15

2.

Guts

Now, after 12 seconds, z has made a full circle in the complex plane centered at 0 and of radius 2/3.Thus 1 − z is a circle centered at 1 of radius 2/3.

So 1 − z traces a circle, and now we need to find the path that 11−z traces. In the complex plane,

taking the reciprocal corresponds to a reflection about the real axis followed by a geometric inversionabout the unit circle centered at 0. It is well known that geometric inversion maps circles not passingthrough the center of the inversion to circles.

Now, the circle traced by 1 − z contains the points 1 − 2/3 = 1/3, and 1 + 2/3 = 5/3. Therefore thecircle 1

1−z contains the points 3 and 3/5, with the center lying halfway between. So the radius of thecircle is

1

2

(

3 − 3

5

)

=6

5

and so the perimeter is 2π(6/5) = 12π/5. Scaling by 15/2 gives an answer of

15

2

(

12π

5

)

= 18π.

29. [19] Consider the cube whose vertices are the eight points (x, y, z) for which each of x, y, and z iseither 0 or 1. How many ways are there to color its vertices black or white such that, for any vertex,if all of its neighbors are the same color then it is also that color? Two vertices are neighbors if theyare the two endpoints of some edge of the cube.

Answer: 118 Divide the 8 vertices of the cube into two sets A and B such that each set contains4 vertices, any two of which are diagonally adjacent across a face of the cube. We do casework basedon the number of vertices of each color in set A.

• Case 1: 4 black. Then all the vertices in B must be black, for 1 possible coloring.

• Case 2: 3 black, 1 white. Then there are 4 ways to assign the white vertex. The vertex in Bsurrounded by the black vertices must also be black. Meanwhile, the three remaining vertices inB may be any configuration except all black, for a total of 4(23 − 1) = 28 possible colorings.

• Case 3: 2 black, 2 white. Then, there are 6 ways to assign the 2 white vertices. The 4 vertices of Bcannot all be the same color. Additionally, we cannot have 3 black vertices of B surround a whitevertex of A with the other vertex of B white, and vice-versa, so we have a total of 6(24−2−4) = 60possible colorings.

• Case 4: 1 black, 3 white. As in case 2, there are 28 possible colorings.

• Case 5: 5 white. As in case 1, there is 1 possible coloring.

So there is a total of 1 + 28 + 60 + 28 + 1 = 118 possible colorings.

30. [19] You have a twig of length 1. You repeatedly do the following: select two points on the twigindependently and uniformly at random, make cuts on these two points, and keep only the largestpiece. After 2012 repetitions, what is the expected length of the remaining piece?

Answer: (11/18)2012 First let p(x) be the probability density of x being the longest length.

Let an be the expected length after n cuts. an =∫ 1

0p(x) · (xan−1)dx = an−1

∫ 1

0xp(x)dx = a1an−1.

It follows that an = an1 , so our answer is (a1)

2012. We now calculate a1.

Let P (z) be the probability that the longest section is ≤ z. Clearly P (z) = 0 for z ≤ 13 .

To simulate making two cuts we pick two random numbers x, y from [0, 1], and assume without lossof generality that x ≤ y. Then picking two such points is equivalent to picking a point in the topleft triangle half of the unit square. This figure has area 1

2 so our P (z) will be double the area wherex ≤ z, y ≥ 1 − z and y − x ≤ z. For 1

3 ≤ z ≤ 12 the probability is double the area bounded by

x = z, 1 − z = y, y − x = z. This is 2( 12 (3z − 1)2) = (3z − 1)2.

Guts

For 12 ≤ z ≤ 1 this value is double the hexagon bounded by x = 0, y = 1 − z, y = x, x = z, y = 1,

y = x+z. The complement of this set, however, is three triangles of area (1−z)2

2 , so P (z) = 1−3(1−z)2

for 12 ≤ z ≤ 1.

Now note that P ′(z) = p(z). Therefore by integration by parts a1 =∫ 1

0zp(z)dz =

∫ 1

0zP ′(z)dz =

[10zP (z) −∫ 1

0P (z)dz. This equals

1 −∫ 1

2

1

3

(3z − 1)2dz −∫ 1

1

2

1 − 3(1 − z)2dz

= 1 − [1

2

1

3

(3z − 1)3

9− 1

2+ [11

2

(z − 1)3

.

= 1 − 1

72− 1

2+

1

8=

1

2+

1

9

=11

18

So the answer is ( 1118 )2012.

31. [19] Let S7 denote all the permutations of 1, 2, . . . , 7. For any π ∈ S7, let f(π) be the smallest positiveinteger i such that π(1), π(2), . . . , π(i) is a permutation of 1, 2, . . . , i. Compute

∑

π∈S7f(π).

Answer: 29093 Extend the definition of f to apply for any permutation of 1, 2, . . . , n, for anypositive integer n. For positive integer n, let g(n) denote the number of permutations π of 1, 2, . . . , nsuch that f(π) = n. We have g(1) = 1. For fixed n, k (with k ≤ n), the number of permutationsπ of 1, 2, . . . , n such that f(π) = k is g(k)(n − k)!. This gives us the recursive formula g(n) =

n!−∑n−1k=1 g(k)(n−k)!. Using this formula, we find that the first 7 values of g are 1, 1, 3, 13, 71, 461, 3447.

Our sum is then equal to∑7

k=1 k · g(k)(7 − k)!. Using our computed values of g, we get that the sumevaluates to 29093.

32. [19] Let S be a set of size 3. How many collections T of subsets of S have the property that for anytwo subsets U ∈ T and V ∈ T , both U ∩ V and U ∪ V are in T?

Answer: 74 Let us consider the collections T grouped based on the size of the set X =⋃

U∈T U ,which we can see also must be in T as long as T contains at least one set. This leads us to countthe number of collections on a set of size at most 3 satisfying the desired property with the additionalproperty that the entire set must be in the collection. Let Cn denote that number of such collectionson a set of size n. Our answer will then be 1 +

(

30

)

C0 +(

31

)

C1 +(

32

)

C2 +(

33

)

C3, with the additional 1coming from the empty collection.

Now for such a collection T on a set of n elements, consider the set I =⋂

U∈T U . Suppose this set hassize k. Then removing all these elements from consideration gives us another such collection on a setof size n − k, but now containing the empty set. We can see that for each particular choice of I, thisgives a bijection to the collections on the set S to the collections on the set S − I. This leads us toconsider the further restricted collections that must contain both the entire set and the empty set.

It turns out that such restricted collections are a well-studied class of objects called topological spaces.Let Tn be the number of topological spaces on n elements. Our argument before shows that Cn =∑n

k=0

(

nk

)

Tk. It is relatively straightforward to see that T0 = 1, T1 = 1, and T2 = 4. For a set of size3, there are the following spaces. The number of symmetric versions is shown in parentheses.

• ∅, {a, b, c} (1)

• ∅, {a, b}, {a, b, c} (3)

• ∅, {a}, {a, b, c} (3)

• ∅, {a}, {a, b}, {a, b, c} (6)

• ∅, {a}, {b, c}, {a, b, c} (3)

Guts

• ∅, {a}, {a, b}, {a, c}, {a, b, c} (3)

• ∅, {a}, {b}, {a, b}.{a, b, c} (3)

• ∅, {a}, {b}, {a, b}, {a, c}, {a, b, c} (6)

• ∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} (1)

which gives T3 = 29. Tracing back our reductions, we have that C0 =(

00

)

T0 = 1, C1 =(

10

)

T0 +(

11

)

T1 =

2, C2 =(

20

)

T0 +(

21

)

T1 +(

22

)

T2 = 7, C3 =(

30

)

T0 +(

31

)

T1 +(

32

)

T2 +(

33

)

T3 = 45, and then our answer is

1 +(

30

)

C0 +(

31

)

C1 +(

32

)

C2 +(

33

)

C3 = 1 + 1 + 6 + 21 + 45 = 74.

33. [23] Compute the decimal expansion of√

π. Your score will be min(23, k), where k is the number ofconsecutive correct digits immediately following the decimal point in your answer.

Answer: 1.77245385090551602729816... For this problem, it is useful to know the following squareroot algorithm that allows for digit-by-digit extraction of

√x and gives one decimal place of

√x for

each two decimal places of x. We will illustrate how to extract the second digit after the decimal pointof

√π, knowing that π = 3.1415 · · · and

√π = 1.7 · · · .

Let d be the next decimal digit. Then d should be the largest digit such that (1.7+0.01d)2 < π, whichin this case we will treat as (1.7+0.01d)2 < 3.1415. Expanding this, we get 2.89+0.034d+0.0001d2 <3.1415, from which we get the value of d to be approximately ⌊ 3.1415−2.89

0.034 ⌋ = ⌊ 0.25150.034 ⌋ = 7, since the

0.0001d2 term is negligible. Indeed, 7 is the largest such digit, and so d = 7 is the second digit of√

π.Because we are constantly subtracting the square of our extracted answer so far, we can record thedifference in a manner similar to long division, which yields a quick method of extracting square rootsby hand.

34. [23] Let Q be the product of the sizes of all the non-empty subsets of {1, 2, . . . , 2012}, and let M =log2(log2(Q)). Give lower and upper bounds L and U for M . If 0 < L ≤ M ≤ U , then your score will

be min(23,⌊

233(U−L)

⌋

). Otherwise, your score will be 0.

Answer: 2015.318180... In this solution, all logarithms will be taken in base 2. It is clear that

log(Q) =∑2012

k=1

(

2012k

)

log(k). By paring k with 2012 − k, we get∑2011

k=1 0.5 ∗ log(k(2012 − k))(

2012k

)

+

log(2012), which is between 0.5 ∗ log(2012)∑2012

k=0

(

2012k

)

and log(2012)∑2012

k=0

(

2012k

)

; i.e., the answer isbetween log(2012)22011 and log(2012)22012. Thus log(log(Q)) is between 2011 + log(log(2012)) and2012 + log(log(2012)). Also 3 < log(log(2012)) < 4. So we get 2014 < M < 2016.

35. [23] Let N be the number of distinct roots of∏2012

k=1 (xk − 1). Give lower and upper bounds L and U

on N . If 0 < L ≤ N ≤ U , then your score will be⌊

23(U/L)1.7

⌋

. Otherwise, your score will be 0.

Answer: 1231288 For x to be such a number is equivalent to x being an kth root of unity for somek up to 2012. For each k, there are ϕ(k) primitive kth roots of unity, so the total number of roots is∑2012

k=1 ϕ(k).

We will give a good approximation of this number using well known facts about the Möbius function,

defined by µ(n) =

{

0 if n is not squarefree

(−1)r if n has r distinct prime factors. It turns out that if f(n) =

∑

d|n g(d),

then g(n) =∑

d|n µ(d)f(nd ). Using this fact, since n =

∑

d|n ϕ(d), we have that ϕ(n) =∑

d|n µ(d)nd .

Now we have reduced the problem to estimating∑2012

k=1

∑

d|k µ(d)kd . Let a = k

d , so we obtain∑2012

k=1

∑

d|k aµ(d).

Guts

We can interchange the order of summation by writing

2012∑

d=1

⌊ 2012

d⌋

∑

a=1

aµ(d) ≈2012∑

d=1

µ(d)1

2

(

⌊2012

d⌋)2

≈2012∑

d=1

µ(d)20122

2d2

=20122

2

2012∑

d=1

µ(d)

d2

≈ 20122

2

∞∑

d=1

µ(d)

d2

The Möbius function also satisfies the property that∑

d|n µ(d) =

{

1 if n = 1

0 otherwise, which can be seen

as a special case of the theorem above (letting f(n) = 1, g(n) =

{

1 if n = 1

0 otherwise). We can then see that

(

∑∞d=1

µ(d)d2

)

(∑∞

c=11c2

)

= 112 = 1, so

∑∞d=1

µ(d)d2 = 6

π2 . Therefore, we have∑2012

k=1 ϕ(k) ≈ 3π2 · 20122 =

1230488.266... 2012 is large enough that all of our approximations are pretty accurate and we shouldbe comfortable perturbing this estimate by a small factor to give bounding values.

36. [23] Maria is hopping up a flight of stairs with 100 steps. At every hop, she advances some integernumber of steps. Each hop she makes has fewer steps. However, the positive difference between thelength of consecutive hops decreases. Let P be the number of distinct ways she can hop up the stairs.

Find lower and upper bounds L and U for P . If 0 < L ≤ P ≤ U , your score will be

⌊

23√U/L

⌋

.

Otherwise, your score will be 0.

Answer: 6922 Consider the sequence of hops backwards. It is an increasing sequence wherethe first finite differences are increasing, so all the second finite differences are all positive integers.Furthermore, given positive integers a, e0 (representing the initial value and initial first finite difference),and a sequence of positive integers e1, e2, . . . , ek, representing the second finite differences, we obtain asequence of k+2 integers satisfying our constraints where the i-th term is equal to a+

∑i−2j=0(i−j−1)ej .

The sum of these values is equal to (k +2)a+∑k

j=0(j+1)(j+2)

2 ek−j . The number of sequences of length

k+2 with sum 100 is then the number of positive integer solutions to (k+2)a+∑k

j=0(j+1)(j+2)

2 ek−j =100 in a, e0, . . . , ek.

We can now approximate a count of the total number of solutions by caseworking over the possiblevalues of k. First, we must consider all sequences of length 1 or 2; it easy to see that there are 1 and49 of these, respectively. Otherwise, we want to consider the following equations:

x1 + 3x2 + 3x3 = 100,

x1 + 4x2 + 3x3 + 6x4 = 100,

x1 + 5x2 + 3x3 + 6x4 + 10x5 = 100,

x1 + 6x2 + 3x3 + 6x4 + 10x5 + 15x6 = 100,

x1 + 7x2 + 3x3 + 6x4 + 10x5 + 15x6 + 21x7 = 100,

x1 + 8x2 + 3x3 + 6x4 + 10x5 + 15x6 + 21x7 + 28x8 = 100.

(Any larger values of k will yield equations with no solutions, as the sum of the coefficients willbe greater than 100 in all of these.) We can compute the number of solutions to the last equationeasily: there are 8. For the remaining 5 equations, we can use the following observation to estimatethe number of solutions. Suppose we wanted to count the number of positive integer solutions to

Guts

∑ki=1 cixi = n, where ci = 1. This is equivalent to finding the number of nonnegative integer solutions

to∑k

i=1 cixi = n − ∑ki=1 ci, which is also equivalent to finding the number of nonnegative integer

solutions to∑k

i=2 cixi ≤ n − ∑ki=1 ci. Let n′ = n − ∑k

i=1 ci. Each xi can take values from 0 to n′

ci

,

giving about (n′)k−1

Q

k

i=2ci

choices. Of course, not all of these choices work; we try to estimate the probability

that one does. We can think of a random selection of xi as approximated by choosing Xi uniformlyfrom [0, 1], then setting xi = Xi · n′

ci

. The condition would then be∑k

i=2 Xk ≤ 1. The probability

of this occuring is 1(k−1)! , so an approximate number of solutions would be (n′)k−1

(k−1)!Q

k

i=2ci

. We can use

these values to get a lower bound of around 4000, and we can also use n instead of n′ for a reasonableupper bound of 16000.

Guts

GENERAL TEST

This test consists of 10 short-answer problems to be solved individually in 50 minutes. Problems are unequallyweighted with point values as shown in brackets on the answer form. The maximum possible score is 50 points.There is no point penalty for guessing, though in the case of a tie it is slightly more advantageous not to answerthan to answer incorrectly.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted. Similarly,graph paper, rulers, protractors, compasses, and other drawing aids are not permitted.

Answers should be simplified as much as is reasonably possible and must be exact unless otherwise specified. Ra-tional numbers should be written in lowest terms, although denominators of irrationals need not be rationalized.An nth root should be simplified so that the radicand is not divisible by the nth power of any prime.

Correct mathematical notation must be used. No partial credit will be given unless otherwise specified.

If you believe the test contains an error, please submit your protest in writing to the Science Center Lobby duringlunchtime.

Enjoy!

HMMT November 2012Saturday 10 November 2012

General Test

1. [3] What is the sum of all of the distinct prime factors of 253 − 272?

2. [3] Let Q(x) = x2 + 2x + 3, and suppose that P (x) is a polynomial such that

P (Q(x)) = x6 + 6x5 + 18x4 + 32x3 + 35x2 + 22x + 8.

Compute P (2).

3. [3] ABCD is a rectangle with AB = 20 and BC = 3. A circle with radius 5, centered at the midpointof DC, meets the rectangle at four points: W , X, Y , and Z. Find the area of quadrilateral WXY Z.

4. [4] If you roll four fair 6-sided dice, what is the probability that at least three of them will show thesame value?

5. [4] How many ways are there to arrange three indistinguishable rooks on a 6× 6 board such that notwo rooks are attacking each other? (Two rooks are attacking each other if and only if they are in thesame row or the same column.)

6. [5] ABCD is a parallelogram satisfying AB = 7, BC = 2, and ∠DAB = 120◦. ParallelogramECFA is contained in ABCD and is similar to it. Find the ratio of the area of ECFA to the area ofABCD.

7. [6] Find the number of ordered 2012-tuples of integers (x1, x2, . . . , x2012), with each integer between0 and 2011 inclusive, such that the sum x1 + 2x2 + 3x3 + · · · + 2012x2012 is divisible by 2012.

8. [7] Let n be the 200th smallest positive real solution to the equation x− π2 = tanx. Find the greatest

integer that does not exceed n2 .

9. [7] Consider triangle ABC where BC = 7, CA = 8, and AB = 9. D and E are the midpoints of BCand CA, respectively, and AD and BE meet at G. The reflection of G across D is G′, and G′E meetsCG at P . Find the length PG.

10. [8] Let α and β be reals. Find the least possible value of

(2 cos α + 5 sin β − 8)2 + (2 sin α + 5 cos β − 15)2.

HMMT November 2012Saturday 10 November 2012

General Test

Name Team ID#

School Team

1. [3]

2. [3]

3. [3]

4. [4]

5. [4]

6. [5]

7. [6]

8. [7]

9. [7]

10. [8]

Score:

HMMT November 2012Saturday 10 November 2012

General Test

1. [3] What is the sum of all of the distinct prime factors of 253 − 272?

Answer: 28 We note that 253 − 272 = 56 − 36 = (53 − 33)(53 + 33) = (5 − 3)(52 + 5 · 3 + 32)(5 +3)(52 − 5 · 3 + 32) = 2 · 72 · 23 · 19, so the sum of the distinct prime factors is 2 + 7 + 19 = 28.

2. [3] Let Q(x) = x2 + 2x + 3, and suppose that P (x) is a polynomial such that

P (Q(x)) = x6 + 6x5 + 18x4 + 32x3 + 35x2 + 22x + 8.

Compute P (2).

Answer: 2 Note that Q(−1) = 2. Therefore, P (2) = P (Q(−1)) = 1−6+18−32+35−22+8 = 2.

3. [3] ABCD is a rectangle with AB = 20 and BC = 3. A circle with radius 5, centered at the midpointof DC, meets the rectangle at four points: W , X, Y , and Z. Find the area of quadrilateral WXY Z.

Answer: 27 Suppose that X and Y are located on AB with X closer to A than B. Let O be thecenter of the circle, and let P be the midpoint of AB. We have OP ⊥ AB so OPX and OPY areright triangles with right angles at P . Because OX = OY = 5 and OP = 3, we have XP = PY = 4by the Pythagorean theorem. Now, WXY Z is a trapezoid with WZ = WO + OZ = 5 + 5 = 10 ,XY = XP + PY = 8, and height 3, so its area is

(

10+82

)

× 3 = 27.

4. [4] If you roll four fair 6-sided dice, what is the probability that at least three of them will show thesame value?

Answer: 772 We have two cases: either three of the dice show one value and the last shows a

different value, or all four dice show the same value. In the first case, there are six choices for the valueof the dice which are the same and

(

43

)

choice for which dice show that value. Then there are 5 choices

for the last die. In total, there are 6(

43

)

5 = 120 possibilities. For the second case, there are 6 valuesthat the last die can show. Consequently, the overall probability is, 120+6

64 = 12664 = 7

72 .

5. [4] How many ways are there to arrange three indistinguishable rooks on a 6× 6 board such that notwo rooks are attacking each other? (Two rooks are attacking each other if and only if they are in thesame row or the same column.)

Answer: 2400 There are 6 × 6 = 36 possible places to place the first rook. Since it cannot be inthe same row or column as the first, the second rook has 5× 5 = 25 possible places, and similarly, thethird rook has 4× 4 = 16 possible places. However, the rooks are indistinguishable, so there are 3! = 6ways to reorder them. Therefore, the number of arrangements is 36×25×16

6 = 2400.

6. [5] ABCD is a parallelogram satisfying AB = 7, BC = 2, and ∠DAB = 120◦. Parallelogram ECFA

is contained in ABCD and is similar to it. Find the ratio of the area of ECFA to the area of ABCD.

Answer: 3967 First, note that BD is the long diagonal of ABCD, and AC is the long diagonal of

ECFA. Because the ratio of the areas of similar figures is equal to the square of the ratio of their side

lengths, we know that the ratio of the area of ECFA to the area of ABCD is equal to the ratio AC2

BD2 .

Using law of cosines on triangle ABD, we have BD2 = AD2 + AB2 − 2(AD)(AB) cos(120◦) = 22 +72 − 2(2)(7)(− 1

2 ) = 67.

Using law of cosines on triangle ABC, we have AC2 = AB2 +BC2 − 2(AB)(BC) cos(60◦) = 72 +22 −2(7)(2)(1

2 ) = 39.

Finally, AC2

BD2 = 3967 .

General Test

7. [6] Find the number of ordered 2012-tuples of integers (x1, x2, . . . , x2012), with each integer between0 and 2011 inclusive, such that the sum x1 + 2x2 + 3x3 + · · · + 2012x2012 is divisible by 2012.

Answer: 20122011 We claim that for any choice of x2, x3, ..., x2012, there is exactly one possiblevalue of x1 satisfying the condition. We have x1 + 2x2 + ... + 2012x2012 ≡ 0 (mod 2012) or x1 ≡−(2x2 + ... + 2012x2012) (mod 2012). Indeed, we see that the right hand side is always an integerbetween 0 and 2011, so x1 must equal this number.

Now, there are 2012 choices for each of the 2011 variables x2, ..., x2012, and each of the 20122011 possiblecombinations gives exactly one valid solution, so the total number of 2012-tuples is 20122011.

8. [7] Let n be the 200th smallest positive real solution to the equation x− π

2 = tanx. Find the greatestinteger that does not exceed n

2 .

Answer: 314 Drawing the graphs of the functions y = x− π

2 and y = tanx, we may observe that the

graphs intersect exactly once in each of the intervals(

(2k−1)π2 ,

(2k+1)π2

)

for each k = 1, 2, · · · . Hence, the

200th intersection has x in the range (399π

2 , 401π

2 ). At this intersection, y = x− π

2 is large, and thus, theintersection will be slightly less than 401π

2 . We have that ⌊ 401π

4 ⌋ = ⌊100π+ π

4 ⌋ = ⌊314.16+ π

4 ⌋ = 314.

9. [7] Consider triangle ABC where BC = 7, CA = 8, and AB = 9. D and E are the midpoints of BC

and CA, respectively, and AD and BE meet at G. The reflection of G across D is G′, and G′E meetsCG at P . Find the length PG.

Answer:√

1459 Observe that since G′ is a reflection and GD = 1

2AG, we have AG = GG′ and

therefore, P is the centroid of triangle ACG′. Thus, extending CG to hit AB at F , PG = 13CG =

29CF = 2

9

√

2(82+72)−92

4 =√

1459 by the formula for the length of a median.

10. [8] Let α and β be reals. Find the least possible value of

(2 cos α + 5 sin β − 8)2 + (2 sin α + 5 cos β − 15)2.

Answer: 100 Let the vector −→v = (2 cos α, 2 sin α) and −→w = (5 sin β, 5 cos β). The locus of ends ofvectors expressible in the form −→v + −→w are the points which are five units away from a point on thecircle of radius two about the origin. The expression that we desire to minimize is the square of thedistance from this point to X = (8, 15). Thus, the closest distance from such a point to X is whenthe point is 7 units away from the origin along the segment from the origin to X. Thus, since X is 17units away from the origin, the minimum is 102 = 100.

General Test

THEME ROUND

This test consists of 10 short-answer problems to be solved individually in 50 minutes. Problems are unequallyweighted with point values as shown in brackets on the answer form. The maximum possible score is 50 points.There is no point penalty for guessing, though in the case of a tie it is slightly more advantageous not to answerthan to answer incorrectly.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted. Similarly,graph paper, rulers, protractors, compasses, and other drawing aids are not permitted.

Answers should be simplified as much as is reasonably possible and must be exact unless otherwise specified. Ra-tional numbers should be written in lowest terms, although denominators of irrationals need not be rationalized.An nth root should be simplified so that the radicand is not divisible by the nth power of any prime.

Correct mathematical notation must be used. No partial credit will be given unless otherwise specified.

If you believe the test contains an error, please submit your protest in writing to the Science Center Lobby duringlunchtime.

Enjoy!

HMMT November 2012Saturday 10 November 2012

Theme Round

Power Towers

In a tower of exponents, the exponents are computed from the top down. For example, 223= 28 =

256.

1. [3] If 444= 128

√222n

, find n.

2. [4] If xx = 201220122013, find x.

3. [4] Find the smallest positive integer n such that 22··2︸︷︷︸n

> 3333

. (The notation 22··2︸︷︷︸n

is used to denote

a power tower with n 2’s. For example, 22··2︸︷︷︸n

with n = 4 would equal 2222

.)

4. [7] Find the sum of all real solutions for x to the equation (x2 + 2x + 3)(x2+2x+3)(x2+2x+3)

= 2012.

5. [7] Given any positive integer, we can write the integer in base 12 and add together the digits of its

base 12 representation. We perform this operation on the number 765432

1

repeatedly until a single base12 digit remains. Find this digit.

Rock-paper-scissors

In the game of rock-paper-scissors, two players each choose one of rock, paper, or scissors to play.Rock beats scissors, scissors beats paper, and paper beats rock. If the players play the same thing, thematch is considered a draw.

6. [3] A rectangular piece of paper with vertices ABCD is being cut by a pair of scissors. The pair ofscissors starts at vertex A, and then cuts along the angle bisector of DAB until it reaches another edgeof the paper. One of the two resulting pieces of paper has 4 times the area of the other piece. Whatis the ratio of the longer side of the original paper to the shorter side?

7. [4] The game of rock-scissors is played just like rock-paper-scissors, except that neither player isallowed to play paper. You play against a poorly-designed computer program that plays rock with50% probability and scissors with 50% probability. If you play optimally against the computer, findthe probability that after 8 games you have won at least 4.

8. [4] In the game of rock-paper-scissors-lizard-Spock, rock defeats scissors and lizard, paper defeatsrock and Spock, scissors defeats paper and lizard, lizard defeats paper and Spock, and Spock defeatsrock and scissors, as shown in the below diagram. As before, if two players choose the same move,then there is a draw. If three people each play a game of rock-paper-scissors-lizard-Spock at the sametime by choosing one of the five moves at random, what is the probability that one player beats theother two?

Lizard

Spock

Scissors

Paper

Rock

9. [6] 64 people are in a single elimination rock-paper-scissors tournament, which consists of a 6-roundknockout bracket. Each person has a different rock-paper-scissors skill level, and in any game, theperson with the higher skill level will always win. For how many players P is it possible that P winsthe first four rounds that he plays?

(A 6-round knockout bracket is a tournament which works as follows:

(a) In the first round, all 64 competitors are paired into 32 groups, and the two people in each groupplay each other. The winners advance to the second round, and the losers are eliminated.

(b) In the second round, the remaining 32 players are paired into 16 groups. Again, the winner ofeach group proceeds to the next round, while the loser is eliminated.

(c) Each round proceeds in a similar way, eliminating half of the remaining players. After the sixthround, only one player will not have been eliminated. That player is declared the champion.)

10. [8] In a game of rock-paper-scissors with n people, the following rules are used to determine achampion:

(a) In a round, each person who has not been eliminated randomly chooses one of rock, paper, orscissors to play.

(b) If at least one person plays rock, at least one person plays paper, and at least one person playsscissors, then the round is declared a tie and no one is eliminated. If everyone makes the samemove, then the round is also declared a tie.

(c) If exactly two moves are represented, then everyone who made the losing move is eliminated fromplaying in all further rounds (for example, in a game with 8 people, if 5 people play rock and 3people play scissors, then the 3 who played scissors are eliminated).

(d) The rounds continue until only one person has not been eliminated. That person is declared thechampion and the game ends.

If a game begins with 4 people, what is the expected value of the number of rounds required for achampion to be determined?

HMMT November 2012Saturday 10 November 2012

Theme Round

Name Team ID#

School Team

1. [3]

2. [4]

3. [4]

4. [7]

5. [7]

6. [3]

7. [4]

8. [4]

9. [6]

10. [8]

Score:

HMMT November 2012Saturday 10 November 2012

Theme Round

1. [3] If 444

=128√

222n

, find n.

Answer: 4 We rewrite the left hand side as

(22)44

= 22·44

= 229

,

and the right hand side as

(222n

)1

128 = 22(2n−7)

.

Equating, we find 2n − 7 = 9, yielding n = 4.

2. [4] If xx = 201220122013

, find x.

Answer: 20122012 We have

201220122013

= 20122012·20122012

= (20122012)20122012

.

Thus, x = 20122012.

3. [4] Find the smallest positive integer n such that 22·

·

2

︸︷︷︸

n

> 3333

. (The notation 22·

·

2

︸︷︷︸

n

is used to denote

a power tower with n 2’s. For example, 22·

·

2

︸︷︷︸

n

with n = 4 would equal 2222

.)

Answer: 6 Clearly, n ≥ 5. When we take n = 5, we have

22222

= 2216

< 3327

= 3333

.

On the other hand, when n = 6, we have

222222

= 2265536

= 4265535

> 4427

> 3327

= 3333

.

Our answer is thus n = 6.

4. [7] Find the sum of all real solutions for x to the equation (x2 + 2x + 3)(x2+2x+3)(x

2+2x+3)

= 2012.

Answer: −2 When y = x2 +2x+3, note that there is a unique real number y such that yyy

= 2012

because yyy

is increasing in y. The sum of the real distinct solutions of the equation x2 +2x+3 = y is−2 by Vieta’s Formulae as long as 22 + 4(y − 3) > 0, which is equivalent to y > 2. This is easily seento be the case; therefore, our answer is −2.

5. [7] Given any positive integer, we can write the integer in base 12 and add together the digits of its

base 12 representation. We perform this operation on the number 765432

1

repeatedly until a single base12 digit remains. Find this digit.

Answer: 4 For a positive integer n, let s(n) be the sum of digits when n is expressed in base 12. Weclaim that s(n) ≡ n (mod 11) for all positive integers n. Indeed, if n = dk12k + dk−112k−1 + · · · + d0

with each di an integer between 0 and 11, inclusive, because 12 ≡ 1 (mod 11), reducing modulo 11

gives exactly s(n). Thus, our answer is congruent to N = 765432

1

modulo 11, and furthermore must bea one-digit integer in base 12; these two conditions uniquely determine the answer.

By Fermat’s Little Theorem, 710 ≡ 1 (mod 11), and also observe that 654321

≡ 6 (mod 10) because6 ∼= 0 (mod 2) and 6 ∼= 1 (mod 5). Thus, N ≡ 76 ≡ 3432 ≡ 22 ≡ 4 (mod 11), which is our answer.(Additionally, we note that this process of writing the number in base twelve and summing the digitsmust eventually terminate because the value decreases after each step.)

Theme Round

6. [3] A rectangular piece of paper with vertices ABCD is being cut by a pair of scissors. The pair ofscissors starts at vertex A, and then cuts along the angle bisector of DAB until it reaches another edgeof the paper. One of the two resulting pieces of paper has 4 times the area of the other piece. Whatis the ratio of the longer side of the original paper to the shorter side?

Answer: 52 Without loss of generality, let AB > AD, and let x = AD, y = AB. Let the cut along

the angle bisector of ∠DAB meet CD at E. Note that ADE is a 45-45-90 triangle, so DE = AD = x,

and EC = y−x. Now, [ADE] = x2

2 , and [AECB] = x(y− x2 ) = 4[ADE]. Equating and dividing both

sides by x, we find that 2x = y − x2 , so y/x = 5

2 .

7. [4] The game of rock-scissors is played just like rock-paper-scissors, except that neither player isallowed to play paper. You play against a poorly-designed computer program that plays rock with50% probability and scissors with 50% probability. If you play optimally against the computer, findthe probability that after 8 games you have won at least 4.

Answer: 163256 Since rock will always win against scissors, the optimum strategy is for you to always

play rock; then, you win a game if and only if the computer plays scissors. Let pn be the probabilitythat the computer plays scissors n times; we want p0 + p1 + p2 + p3 + p4. Note that by symmetry,pn = p8−n for n = 0, 1, . . . , 8, and because p0+p1+· · ·+p8 = 1, p0+· · ·+p3 = p5+· · ·+p8 = (1−p4)/2.Our answer will thus be (1 + p4)/2.

If the computer is to play scissors exactly 4 times, there are(84

)ways in which it can do so, compared

to 28 possible combinations of eight plays. Thus, p4 =(84

)/28 = 35/128. Our answer is thus

1+ 35128

2 =163256 .

8. [4] In the game of rock-paper-scissors-lizard-Spock, rock defeats scissors and lizard, paper defeatsrock and Spock, scissors defeats paper and lizard, lizard defeats paper and Spock, and Spock defeatsrock and scissors, as shown in the below diagram. As before, if two players choose the same move,then there is a draw. If three people each play a game of rock-paper-scissors-lizard-Spock at the sametime by choosing one of the five moves at random, what is the probability that one player beats theother two?

Lizard

Spock

Scissors

Paper

Rock

Answer: 1225 Let the three players be A,B,C. Our answer will simply be the sum of the probability

that A beats both B and C, the probability that B beats both C and A, and the probability that Cbeats A and B, because these events are all mutually exclusive. By symmetry, these three probabilitiesare the same, so we only need to compute the probability that A beats both B and C. Given A’splay, the probability that B’s play loses to that of A is 2/5, and similarly for C. Thus, our answer is3 ·

(25

)·(

25

)= 12

25 .

9. [6] 64 people are in a single elimination rock-paper-scissors tournament, which consists of a 6-roundknockout bracket. Each person has a different rock-paper-scissors skill level, and in any game, theperson with the higher skill level will always win. For how many players P is it possible that P winsthe first four rounds that he plays?

(A 6-round knockout bracket is a tournament which works as follows:

Theme Round

(a) In the first round, all 64 competitors are paired into 32 groups, and the two people in each groupplay each other. The winners advance to the second round, and the losers are eliminated.

(b) In the second round, the remaining 32 players are paired into 16 groups. Again, the winner ofeach group proceeds to the next round, while the loser is eliminated.

(c) Each round proceeds in a similar way, eliminating half of the remaining players. After the sixthround, only one player will not have been eliminated. That player is declared the champion.)

Answer: 49 Note that a sub-bracket, that is, a subset of games of the tournament that themselvesconstitute a bracket, is always won by the person with the highest skill level. Therefore, a person winsher first four rounds if and only if she has the highest skill level among the people in her 16-personsub-bracket. This is possible for all but the people with the 16 − 1 = 15 lowest skill levels, so ouranswer is 64 − 15 = 49.

10. [8] In a game of rock-paper-scissors with n people, the following rules are used to determine achampion:

(a) In a round, each person who has not been eliminated randomly chooses one of rock, paper, orscissors to play.

(b) If at least one person plays rock, at least one person plays paper, and at least one person playsscissors, then the round is declared a tie and no one is eliminated. If everyone makes the samemove, then the round is also declared a tie.

(c) If exactly two moves are represented, then everyone who made the losing move is eliminated fromplaying in all further rounds (for example, in a game with 8 people, if 5 people play rock and 3people play scissors, then the 3 who played scissors are eliminated).

(d) The rounds continue until only one person has not been eliminated. That person is declared thechampion and the game ends.

If a game begins with 4 people, what is the expected value of the number of rounds required for achampion to be determined?

Answer: 4514 For each positive integer n, let En denote the expected number of rounds required to

determine a winner among n people. Clearly, E1 = 0. When n = 2, on the first move, there is a 13

probability that there is a tie, and a 23 probability that a winner is determined. In the first case, the

expected number of additional rounds needed is exactly E2; in the second, it is E1. Therefore, we getthe relation

E2 =1

3(E2 + 1) +

2

3(E1 + 1),

from which it follows that E2 = 32 .

Next, if n = 3, with probability 19 there is only one distinct play among the three players, and with

probability 627 = 2

9 all three players make different plays. In both of these cases, no players areeliminated. In all remaining situations, which occur with total probability 2

3 , two players make oneplay and the third makes a distinct play; with probability 1

3 two players are eliminated and withprobability 1

3 one player is eliminated. This gives the relation

E3 =1

3(E3 + 1) +

1

3(E2 + 1) +

1

3(E1 + 1),

from which we find that E3 = 94 .

Finally, suppose n = 4. With probability 127 , all four players make the same play, and with probability

3·6·281 = 4

9 , two players make one play, and the other two players make the other two plays; in both casesno players are eliminated, with total probability 1

27 + 49 = 13

27 over the two cases. With probability6·481 = 8

27 , three players make one play and the fourth makes another; thus, there is a probability of427 for exactly one player being eliminated and a probability of 4

27 of three players being eliminated.

Theme Round

Then, there is a remaining probability of 6·381 = 2

9 , two players make one play and the other two playersmake another. Similar analysis from before yields

E4 =13

27(E4 + 1) +

4

27(E3 + 1) +

2

9(E2 + 1) +

4

27(E1 + 1),

so it follows that E4 = 4514 .

Theme Round

HMMT November 2012Saturday 10 November 2012

Team Round

Divisors

In this section, the word divisor is used to refer to a positive divisor of an integer; that is, a divisor ofa positive integer n is a positive integer d such that n

dis an integer.

1. [3] Find the number of integers between 1 and 200 inclusive whose distinct prime divisors sum to 16.(For example, the sum of the distinct prime divisors of 12 is 2 + 3 = 5.)

2. [5] Find the number of ordered triples of divisors (d1, d2, d3) of 360 such that d1d2d3 is also a divisorof 360.

3. [6] Find the largest integer less than 2012 all of whose divisors have at most two 1’s in their binaryrepresentations.

Permutations

A permutation π is defined as a function from a set of integers to itself that rearranges the elements ofthe set. For example, a possible permutation of the numbers from 1 through 4 is the function π givenby π(1) = 2, π(2) = 4, π(3) = 3, π(4) = 1.

4. [3] Let π be a permutation of the numbers from 2 through 2012. Find the largest possible value oflog2 π(2) · log3 π(3) · · · log2012 π(2012).

5. [4] Let π be a randomly chosen permutation of the numbers from 1 through 2012. Find the probabilitythat π(π(2012)) = 2012.

6. [6] Let π be a permutation of the numbers from 1 through 2012. What is the maximum possiblenumber of integers n with 1 ≤ n ≤ 2011 such that π(n) divides π(n + 1)?

7. [8] Let A1A2 . . . A100 be the vertices of a regular 100-gon. Let π be a randomly chosen permutation ofthe numbers from 1 through 100. The segments Aπ(1)Aπ(2), Aπ(2)Aπ(3), . . . , Aπ(99)Aπ(100), Aπ(100)Aπ(1)

are drawn. Find the expected number of pairs of line segments that intersect at a point in the interiorof the 100-gon.

Circumcircles

The circumcircle of a triangle is the circle passing through all three vertices of the triangle.

8. [4] ABC is a triangle with AB = 15, BC = 14, and CA = 13. The altitude from A to BC is extendedto meet the circumcircle of ABC at D. Find AD.

9. [5] Triangle ABC satisfies ∠B > ∠C. Let M be the midpoint of BC, and let the perpendicularbisector of BC meet the circumcircle of △ABC at a point D such that points A, D, C, and B appearon the circle in that order. Given that ∠ADM = 68◦ and ∠DAC = 64◦, find ∠B.

10. [6] Triangle ABC has AB = 4, BC = 5, and CA = 6. Points A′, B′, C ′ are such that B′C ′ is tangentto the circumcircle of △ABC at A, C ′A′ is tangent to the circumcircle at B, and A′B′ is tangent tothe circumcircle at C. Find the length B′C ′.

HMMT November 2012Saturday 10 November 2012

Team Round

1. [3] Find the number of integers between 1 and 200 inclusive whose distinct prime divisors sum to 16.(For example, the sum of the distinct prime divisors of 12 is 2 + 3 = 5.)

Answer: 6 The primes less than 16 are 2, 3, 5, 7, 11, and 13. We can write 16 as the sum of suchprimes in three different ways and find the integers less than 200 with those prime factors:

• 13 + 3: 3 · 13 = 39 and 32 · 13 = 117.

• 11 + 5: 5 · 11 = 55

• 11 + 3 + 2: 2 · 3 · 11 = 66, 22 · 3 · 11 = 132, and 2 · 32 · 11 = 198.

There are therefore 6 numbers less than 200 whose prime divisors sum to 16.

2. [5] Find the number of ordered triples of divisors (d1, d2, d3) of 360 such that d1d2d3 is also a divisorof 360.

Answer: 800 Since 360 = 23 · 32 · 5, the only possible prime divisors of di are 2, 3, and 5, so wecan write di = 2ai · 3bi · 5ci , for nonnegative integers ai, bi, and ci. Then, d1d2d3|360 if and only if thefollowing three inequalities hold.

a1 + a2 + a3 ≤ 3

b1 + b2 + b3 ≤ 2

c1 + c2 + c3 ≤ 1

Now, one can count that there are 20 assignments of ai that satisfy the first inequality, 10 assignmentsof bi that satisfy the second inequality, and 4 assignments of ci that satisfy the third inequality, for atotal of 800 ordered triples (d1, d2, d3).

(Alternatively, instead of counting, it is possible to show that the number of nonnegative-integer triples(a1, a2, a3) satisfying a1 + a2 + a3 ≤ n equals

(

n+33

)

, since this is equal to the number of nonnegative-integer quadruplets (a1, a2, a3, a4) satisfying a1 + a2 + a3 + a4 = n.)

3. [6] Find the largest integer less than 2012 all of whose divisors have at most two 1’s in their binaryrepresentations.

Answer: 1536 Call a number good if all of its positive divisors have at most two 1’s in their binaryrepresentations. Then, if p is an odd prime divisor of a good number, p must be of the form 2k + 1.The only such primes less than 2012 are 3, 5, 17, and 257, so the only possible prime divisors of n are2, 3, 5, 17, and 257.

Next, note that since (2i + 1)(2j + 1) = 2i+j + 2i + 2j + 1, if either i or j is greater than 1, thenthere will be at least 3 1’s in the binary representation of (2i + 1)(2j + 1), so (2i + 1)(2j + 1) cannotdivide a good number. On the other hand, if i = j = 1, then (21 + 1)(21 + 1) = 9 = 23 + 1, so 9 isa good number and can divide a good number. Finally, note that since multiplication by 2 in binaryjust appends additional 0s, so if n is a good number, then 2n is also a good number.

It therefore follows that any good number less than 2012 must be of the form c · 2k, where c belongs to{1, 3, 5, 9, 17, 257} (and moreover, all such numbers are good). It is then straightforward to check thatthe largest such number is 1536 = 3 · 29.

4. [3] Let π be a permutation of the numbers from 2 through 2012. Find the largest possible value oflog2 π(2) · log3 π(3) · · · log2012 π(2012).

Team Round

Answer: 1 Note that

2012∏

i=2

logi π(i) =

2012∏

i=2

log π(i)

log i

=

∏2012i=2 log π(i)∏2012

i=2 log i

= 1,

where the last equality holds since π is a permutation of the numbers 2 through 2012.

5. [4] Let π be a randomly chosen permutation of the numbers from 1 through 2012. Find the probabilitythat π(π(2012)) = 2012.

Answer: 11006 There are two possibilities: either π(2012) = 2012 or π(2012) = i and π(i) = 2012 for

i 6= 2012. The first case occurs with probability 2011!/2012! = 1/2012, since any permutation on theremaining 2011 elements is possible. Similarly, for any fixed i, the second case occurs with probability2010!/2012! = 1/(2011 ·2012), since any permutation on the remaining 2010 elements is possible. Sincethere are 2011 possible values for i, and since our two possibilities are disjoint, the overall probabilitythat π(π(2012)) = 2012 equals

1

2012+ (2011)

1

2011 · 2012=

1

1006.

6. [6] Let π be a permutation of the numbers from 1 through 2012. What is the maximum possiblenumber of integers n with 1 ≤ n ≤ 2011 such that π(n) divides π(n + 1)?

Answer: 1006 Since any proper divisor of n must be less than or equal to n/2, none of the numbersgreater than 1006 can divide any other number less than or equal to 2012. Since there are at most1006 values of n for which π(n) ≤ 1006, this means that there can be at most 1006 values of n forwhich π(n) divides π(n + 1).

On the other hand, there exists a permutation for which π(n) divides π(n + 1) for exactly 1006 valuesof n, namely the permutation:

(1, 2, 22, 23, . . . , 210, 3, 2 · 3, 22 · 3, 23 · 3, . . . , 29 · 3, 5, . . . )

Formally, for each odd number ℓ ≤ 2012, we construct the sequence ℓ, 2ℓ, 4ℓ, . . . , 2kℓ, where k is thelargest integer such that 2kℓ ≤ 2012. We then concatenate all of these sequences to form a permutationof the numbers 1 through ℓ (note that no number occurs in more than one sequence). It follows that ifπ(n) ≤ 1006, then π(n+1) will equal 2π(n), and therefore π(n) will divide π(n+1) for all 1006 valuesof n satisfying 1 ≤ π(n) ≤ 1006.

7. [8] Let A1A2 . . . A100 be the vertices of a regular 100-gon. Let π be a randomly chosen permutation ofthe numbers from 1 through 100. The segments Aπ(1)Aπ(2), Aπ(2)Aπ(3), . . . , Aπ(99)Aπ(100), Aπ(100)Aπ(1)

are drawn. Find the expected number of pairs of line segments that intersect at a point in the interiorof the 100-gon.

Answer: 48503 By linearity of expectation, the expected number of total intersections is equal to

the sum of the probabilities that any given intersection will occur.

Let us compute the probability pi,j that Aπ(i)Aπ(i+1) intersects Aπ(j)Aπ(j+1) (where 1 ≤ i, j ≤ 100,i 6= j, and indices are taken modulo 100). Note first that if j = i + 1, then these two segments sharevertex π(i + 1) and therefore will not intersect in the interior of the 100-gon; similarly, if i = j + 1,these two segments will also not intersect. On the other hand, if π(i), π(i + 1), π(j), and π(j + 1) areall distinct, then there is a 1/3 chance that Aπ(i)Aπ(i+1) intersects Aπ(j)Aπ(j+1) ; in any set of four

Team Round

points that form a convex quadrilateral, exactly one of the three ways of pairing the points into twopairs (two pairs of opposite sides and the two diagonals) forms two segments that intersect inside thequadrilateral (namely, the two diagonals).

Now, there are 100 ways to choose a value for i, and 97 ways to choose a value for j which is not i,i + 1, or i − 1, there are 9700 ordered pairs (i, j) where pi,j = 1/3. Since each pair is counted twice(once as (i, j) and once as (j, i)), there are 9700/2 = 4850 distinct possible intersections, each of whichoccurs with probability 1/3, so the expected number of intersections is equal to 4850/3.

8. [4] ABC is a triangle with AB = 15, BC = 14, and CA = 13. The altitude from A to BC is extendedto meet the circumcircle of ABC at D. Find AD.

Answer: 634 Let the altitude from A to BC meet BC at E. The altitude AE has length 12; one

way to see this is that it splits the triangle ABC into a 9 − 12 − 15 right triangle and a 5 − 12 − 13right triangle; from this, we also know that BE = 9 and CE = 5.

Now, by Power of a Point, AE · DE = BE · CE, so DE = (BE · CE)/AE = (9 · 5)/(12) = 15/4. Itthen follows that AD = AE + DE = 63/4.

9. [5] Triangle ABC satisfies ∠B > ∠C. Let M be the midpoint of BC, and let the perpendicularbisector of BC meet the circumcircle of △ABC at a point D such that points A, D, C, and B appearon the circle in that order. Given that ∠ADM = 68◦ and ∠DAC = 64◦, find ∠B.

Answer: 86◦ Extend DM to hit the circumcircle at E. Then, note that since ADEB is a cyclicquadrilateral, ∠ABE = 180◦ − ∠ADE = 180◦ − ∠ADM = 180◦ − 68◦ = 112◦.

We also have that ∠MEC = ∠DEC = ∠DAC = 64◦. But now, since M is the midpoint of BCand since EM ⊥ BC, triangle BEC is isosceles. This implies that ∠BEM = ∠MEC = 64◦, and∠MBE = 90◦ − ∠MEB = 26◦. It follows that ∠B = ∠ABE − ∠MBE = 112◦ − 26◦ = 86◦.

10. [6] Triangle ABC has AB = 4, BC = 5, and CA = 6. Points A′, B′, C ′ are such that B′C ′ is tangentto the circumcircle of △ABC at A, C ′A′ is tangent to the circumcircle at B, and A′B′ is tangent tothe circumcircle at C. Find the length B′C ′.

Answer: 803 Note that by equal tangents, B′A = B′C, C ′A = C ′B, and A′B = A′C. Moreover,

since the line segments A′B′, B′C ′, and C ′A′ are tangent to the circumcircle of ABC at C, A, andB respectively, we have that ∠A′BC = ∠A′CB = ∠A, ∠B′AC = ∠B′CA = ∠B, and ∠C ′BA =∠C ′AB = ∠C. By drawing the altitudes of the isosceles triangles BC ′A and AC ′B, we therefore havethat C ′A = 2/ cos C and B′A = 3/ cos B.

Now, by the Law of Cosines, we have that

cos B =a2 + c2 − b2

2ac=

25 + 16 − 36

2(5)(4)=

1

8

cos C =a2 + b2 − c2

2ab=

25 + 36 − 16

2(5)(6)=

3

4.

Therefore,

B′C ′ = C ′A + B′A = 2

(

4

3

)

+ 3(8) =80

3.

Team Round

HMMT November 2012Saturday 10 November 2012

Guts Round

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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

1. [5] Find the number of prime numbers less than 30.

2. [5] Albert is very hungry, and goes to his favorite burger shop to buy a meal, which consists of aburger, a side, and a drink. Given that there are 5 different types of burgers offered, 3 different typesof sides, and 12 different types of drinks, find the number of meals Albert could get.

3. [5] Find the area of the region between two concentric circles that have radii 100 and 99.

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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

4. [6] ABCD is a concave quadrilateral such that ∠CBA = 50◦, ∠BAD = 80◦, ∠ADC = 30◦, andCB = CD. Find ∠CBD.

5. [6] a and b are complex numbers such that 2a + 3b = 10 and 4a2 + 9b2 = 20. Find ab.

6. [6] Given the following formulas:

1 + 2 + · · · + n =n(n + 1)

2

12 + 22 + · · · + n2 =n(n + 1)(2n + 1)

6

13 + 23 + · · · + n3 =

(

n(n + 1)

2

)2

,

find(13 + 3 · 12 + 3 · 1) + (23 + 3 · 22 + 3 · 2) + · · · + (993 + 3 · 992 + 3 · 99).

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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

7. [7] Consider the sequence given by a0 = 1, a1 = 1 + 3, a2 = 1 + 3 + 32, a3 = 1 + 3 + 32 + 33, . . .. Findthe number of terms among a0, a1, a2, . . . , a2012 that are divisible by 7.

8. [7] Three cards are drawn from the top of a shuffled standard 52-card deck. Find the probability thatthey are all of different suits. (A suit is either spades, clubs, hearts, or diamonds.)

9. [7] How many sets consist of distinct composite numbers that add up to 23?

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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

10. [8] Consider the sequence defined by a0 = 0, a1 = 1, and an = 2013an−1 + 2012an−2 + n for n ≥ 2.Find the remainder when a2012 is divided by 2012.

11. [8] Find the smallest positive integer n such that the number of zeroes that n! ends with is a positivemultiple of 5.

12. [8] Three circles of radius 1 are drawn, whose centers form the vertices of an equilateral triangle ofside length 1. Find the area of the region common to at least two of the circles.

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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

13. [9] Triangle ABC has a perimeter of 18. ∠A = 60◦, and BC = 7. Find the area of the triangle.

14. [9] Mark would like to place the numbers 1 to 7 on a circle such that the sequences along both arcsgoing from 1 to 7 are increasing. For example, one arc could be 1, 2, 4, 7 and the other could be 1,3, 5, 6, 7. In how many distinct ways can Mark place the numbers? Two arrangements are distinct ifand only if one cannot be rotated to match the other.

15. [9] Find the area of the region in the xy-plane consisting of all points (a, b) such that the quadraticax2 + 2(a + b − 7)x + 2b = 0 has fewer than two real solutions for x.

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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

16. [10] Yuhang is making a bracelet for his one true love using beads. How many distinct bracelets canbe made from 2 red beads, 2 green beads, and 2 blue beads, if two bracelets are distinct if and only ifone cannot be made into the other through rotations and reflections?

17. [10] Given that x = ln 30, y = ln 360, and z = ln 270, and that there are rational numbers p, q, r suchthat ln 5400 = px + qy + rz, find the ordered triple (p, q, r).

18. [10] Let △ABC be a triangle with AC = 1 and ∠ABC obtuse. Let D and E be points on AC suchthat ∠DBC = ∠ABE = 90◦. If AD = DE = EC, find AB + AC.

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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

19. [11] Find the number of triples of nonnegative integers (x, y, z) such that 15x + 21y + 35z = 525.

20. [11] An elementary school teacher is taking a class of 20 students on a field trip. To make sureher students don’t get lost, she uses the buddy system—using the complete class roster, she pairs thestudents into 10 pairs and leaves if no person reports that someone from his or her pair is missing. Inparticular, the teacher will not notice if both students from a pair go missing. Suppose each studentindependently gets lost with probability 1

10. Given that the teacher leaves, what is the probability that

no student got lost?

21. [11] ABC is a triangle with AB = 7, BC = 10, and CA = 13. Point D lies on segment BC such thatDC = 2BD. Point E lies on segment AD such that AE = 4ED. Point F lies on segment AC suchthat FC = 5AF . Find the area of △EFC.

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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

22. [12] Alice generates an infinite decimal by rolling a fair 6-sided die with the numbers 1, 2, 3, 5, 7,and 9 infinitely many times, and appending the resulting numbers after a decimal point. What is theexpected value of her number?

23. [12] ABC is a triangle with AB = 4, BC = 6, and CA = 7. Let Ω be the circumcircle of ABC; thetangent to Ω passing through A meets the extension of side BC at P . Find the length PB.

24. [12] 12 children sit around a circle. Find the number of ways we can distribute 4 pieces of candyamong them such that each child gets at most one piece of candy and no two adjacent children bothget a piece of candy.

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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

25. [13] Triangle ABC satisfies AB = 8, BC = 9, and CA = 10. Evaluatesin2 B + sin2 C − sin2 A

sin B · sin C.

26. [13] x1, x2, x3, . . . is a sequence of real numbers satisfying x1 = 1, x2 = 2, and xn+1 = 2xn−xn−1 +2n

for n ≥ 2. Find x2012.

27. [13] Find the number of ordered triples of positive integers (a, b, c) satisfying a2−b2+ac−bc = 2012.

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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

28. [15] Find the set of all possible values that can be attained by the expressionab + b2

a2 + b2, where a and

b are positive real numbers. Express your answer in interval notation.

29. [15] Find the sum of the real values of x satisfying (x + 1)(2x + 1)(3x + 1)(4x + 1) = 16x4.

30. [15] A monkey forms a string of letters by repeatedly choosing one of the letters a, b, or c to type atrandom. Find the probability that he first types the string aaa before he first types the string abc.

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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

31. [17] Let ABCD be a regular tetrahedron with AB = 1, and let P be the center of face BCD. LetM be the midpoint of segment AP . Ray BM meets face ACD at Q. Determine the area of triangleQDC.

32. [17] Define f(n) to be the remainder when nnn

is divided by 23 for each positive integer n. Find thesmallest positive integer k such that f(n + k) = f(n) for all positive integers n.

33. [17] You are playing pool on a 1× 1 table, and the cue ball is at the bottom left corner of the square.Find the smallest angle larger than 45◦ at which you could shoot the ball such that the ball bouncesagainst walls exactly 2012 times before arriving at a vertex of the square.

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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

34. [20] For a positive integer n, let τ(n) be the number of divisors of n. Determine∑2012

n=1τ(n). Your

score will be max

{

0,

⌊

20(

1 − |S−k|S

)3⌋}

, where k is your answer and S is the actual answer.

35. [20] A regular 2012-gon has a circumcircle with radius 1. Compute the area of the 2012-gon. Your

score will be min(20, ⌊k2

5⌋), where k is the number of consecutive correct digits immediately following

the decimal point of your answer.

36. [20] Let π1 and π2 be permutations of the numbers from 1 through 7. Call π1 superior to π2 if thesum of all i such that π1(i) > π2(i) exceeds the sum of all i such that π2(i) > π1(i). Write down apermutation of the integers from 1 through 7. Let N be the total number of answers submitted forthis problem, and let n be the number of submitted answers your answer is superior to. Your scorewill be ⌈20 n

N⌉.

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HMMT November 2012Saturday 10 November 2012

Guts Round

1. [5]

Answer: 10 The prime numbers under 30 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. There are 10 intotal.

2. [5]

Answer: 180 Albert’s choice of burgers, sides, and drinks are independent events. Therefore,we can use the Multiplication Principle of Counting Independent Events to find the total number ofdifferent meals that Albert can get is:

5 × 3 × 12 = 180

3. [5]

Answer: 199π The area of a circle of radius 100 is 1002π, and the area of a circle of radius 99 is 992π.Therefore, the area of the region between them is (1002 − 992)π = (100 + 99)(100 − 99)π = 199π.

4. [6]

Answer: 10 ∠CBD = ∠CDB because BC = CD. Notice that ∠BCD = 80 + 50 + 30 = 160, so∠CBD = ∠CDB = 10.

5. [6]

Answer: 203

20 = 4a2 + 9b2

20 + 12ab = 4a2 + 12ab + 9b2

20 + 12ab = (2a + 3b)2

20 + 12ab = 100

12ab = 80

ab =20

3

6. [6]

Answer: 25502400

(13 + 3 · 12 + 3 · 1) + (23 + 3 · 22 + 3 · 2) + · · · + (993 + 3 · 992 + 3 · 99)

= (13 + 3 · 12 + 3 · 1 + 1) + (23 + 3 · 22 + 3 · 2 + 1) + · · · + (993 + 3 · 992 + 3 · 99 + 1) − 99

= (1 + 1)3 + (2 + 1)3 + ...(99 + 1)3 − 99

= 13 + (1 + 1)3 + (2 + 1)3 + ...(99 + 1)3 − 100

=

(

100(100 + 1)

2

)2

− 100

= 25502400

Guts Round

7. [7]

Answer: 335

an = 1 + 3 + 32 + ... + 3n

3an = 3 + 32 + 33 + ... + 3n+1

3an = 3n+1 − 1 + Sn

an =3n+1 − 1

2

Because gcd(2, 7) = 1, to demonstrate that 7|an = 3n+1−12 is equivalent to showing that 3n+1 − 1 ≡ 0

mod 7

However 3n+1 − 1 = 0 mod 7 =⇒ n + 1 ≡ 0 mod 6, hence the values of n ≤ 2012 are 5, 11, 17, ...,2009, which is in total

2009 + 1

6= 335

possible terms.

8. [7]

Answer: 169425 There are 13 cards of each suit. The probability that the second card has a different

suit than the first is 3·1352−1 . The probability that the third card has a different suit than the first and

second card is 2·1352−2

Since the two events need to both occur, we use the multiplicative rule:

3 · 13

52 − 1· 2 · 13

52 − 2

=169

425

9. [7]

Answer: 4 Because 23 is odd, we must have an odd number of odd numbers in our set. Since thesmallest odd composite number is 9, we cannot have more than 2 odd numbers, as otherwise the sumwould be at least 27. Therefore, the set has exactly one odd number.

The only odd composite numbers less than 23 are 9, 15, and 21. If we include 21, then the rest of theset must include composite numbers that add up to 2, which is impossible. If we include 15, then therest of the set must include distinct even composite numbers that add up to 8. The only possibility isthe set {8}. If we include 9, the rest of the set must contain distinct even composite numbers that addto 14. The only possibilities are {14}, {4, 10}, and {6, 8}. We have exhausted all cases, so there are atotal of 4 sets.

10. [8]

Answer: 1006 Taking both sides modulo 2012, we see that an ≡ an−1 + n (mod 2012). Therefore,

a2012 ≡ a2011 + 2012 ≡ a2010 + 2011 + 2012 ≡ ... ≡ 1 + 2 + ... + 2012 ≡ (2012)(2013)2 ≡ (1006)(2013) ≡

(1006)(1) ≡ 1006 (mod 2012).

Guts Round

11. [8]

Answer: 45 The number of zeroes that n! ends with is the largest power of 10 dividing n!. Theexponent of 5 dividing n! exceeds the exponent of 2 dividing n!, so we simply seek the exponent of 5dividing n!.

For a number less than 125, this exponent is just the number of multiples of 5, but not 25, less than nplus twice the number of multiples of 25 less than n.

Counting up, we see that 24! ends with 4 zeroes while 25! ends with 6 zeroes, so n! cannot end with 5zeroes. Continuing to count up, we see that the smallest n such that n! ends with 10 zeroes is 45.

12. [8]

Answer: 2π−√

32 Let A△ be the equilateral triangle. Let A1 be the area of the region outside of

the equilateral triangle but inside the second and third circles. Define A2, A3 analogously. We haveA1 = A2 = A3 = Ak =

(

12 · π3

− 12 · sin 120

2

)

=4π − 3

√3

12,

and

A△ =12 · sin 60

2=

√3

4.

Thus, the total area is

A1 + A2 + A3 + A△ = 3 · 4π − 3√

3

12+

√3

4=

2π −√

3

2.

13. [9]

Answer: 6√

3 Let AB be x. Then, AC = 18 − 7 − x = 11 − x. Using the law of cosines gives:

x2 + (11 − x)2 − 2x(11 − x) cos 60◦ = 72.

3x2 − 33x + 72 = 0.

x = 3 or 8.

Therefore, AB = 8 and AC = 3 or AB = 3 and AC = 8. In both cases, the area of the triangle is:

12 · 8 · 3 sin 60◦ = 6

√3 .

14. [9]

Answer: 32 First, we can fix the position of the 1. Then, by the condition that the numbers areincreasing along each arc from 1, we know that the 2 must be adjacent to the 1; so we have two optionsfor its placement. Similarly, we have two options for placing each of 3,4,5,6 in that order. Finally, the7 must go in the remaining space, for a total of 25 = 32 orderings.

15. [9]

Answer: 49π To find the region in question, we want to find (a, b) such that the discriminant ofthe quadratic is not positive. In other words, we want

4(a + b − 7)2 − 4(a)(2b) ≤ 0 ⇔ a2 + b2 − 7a − 7b + 49 ≤ 0 ⇔ (a − 7)2 + (b − 7)2 ≤ 49,

which is a circle of radius 7 centered at (7, 7) and hence has area 49π.

16. [10]

Answer: 11 We divide this problem into cases based on the relative position of the two red beads:

Guts Round

• They are adjacent. Then, there are 4 possible placements of the green and blue beads: GGBB,GBBG, GBGB, BGGB.

• They are 1 bead apart. Then, there are two choices for the bead between then and 2 choices forthe other bead of that color, for a total of 4.

• They are opposite. Then, there are three choices for the placement of the green beads.

This gives a total of 11 arrangements.

17. [10]

Answer:(

54 , 1

2 , 14

)

We can rewrite the equation in terms of ln 2, ln 3, ln 5, to get

3 ln 2 + 3 ln 3 + 2 ln 5 = ln 5400 = px + qy + rz = (p + 3q + r) ln 2 + (p + 2q + 3r) ln 3 + (p + q + r) ln 5.

Consequently, since p, q, r are rational we want to solve the system of equations p + 3q + r = 3, p +2q + 3r = 3, p + q + r = 2, which results in the ordered triple

(

54 , 1

2 , 14

)

.

18. [10]

Answer: 1 +√

33 DBC is a right triangle with hypotenuse DC. Since DE = EC,E is the midpoint

of this right triangle’s hypotenuse, and it follows that E is the circumcenter of the triangle. It followsthat BE = DE = CE, as these are all radii of the same circle. A similar argument shows thatBD = DE = AD.

Thus, BD = DE = DE, and triangle BDE is equilateral. So, ∠DBE = ∠BED = ∠EDB = 60◦. Wehave ∠BEC = 180◦−∠BED = 120◦. Because BE = CE, triangle BEC is isosceles and ∠ECB = 30◦.Therefore, DBC is a right triangle with ∠DBC = 90◦, ∠BCD = 30◦, and ∠CDB = 60◦. This

means that CD = 2√3BC. Combined iwth CD = 2

3 , we have BC =√

33 . Similarly, AB =

√3

3 , so

AB + AC = 1 +√

33 .

19. [11]

Answer: 21 First, note that 525 = 3 × 7 × 5 × 5. Then, taking the equation modulo 7 gives that7 | x; let x = 7x′ for some nonnegative integer x′. Similarly, we can write y = 5y′ and z = 3z′ forsome nonnegative integers y′, z′. Then, after substitution and division of both sides by 105, the givenequation is equivalent to x′ + y′ + z′ = 5. This is the same as the problem of placing 2 dividers among

5 balls, so is

(

7

2

)

= 21.

20. [11]

Answer: 8110

8210 The teacher will leave if the students from each pair are either both present or both

not present; the probability that both are present is81

100and the probability that neither are present

is1

100. If the teacher leaves, then the probability that both students in any given pair did not get lost

is81100

81100 + 1

100

=81

82. Since there are ten pairs, the overall probability is

(

81

82

)10

=8110

8210.

21. [11]

Answer: 80√

39 By shared bases, we know that

[EFC] =

(

5

6

)

[AEC] =

(

5

6

)(

4

5

)

[ADC] =

(

5

6

)(

4

5

) (

2

3

)

[ABC].

By Heron’s formula, we find that [ABC] =√

(15)(8)(2)(5) = 20√

3, so [AEC] = 80√

39

Guts Round

22. [12]

Answer: 12 Let Xn be the nth number rolled. The number formed, 0.X1X2 · · · , is simply

∑∞n=1

Xn

10n .

By linearity of expectation, the expected value is∑∞

n=1 E(

Xn

10n

)

=∑∞

n=1E(Xn)10n . However, the rolls are

independent: for all n, E(Xn) = 16 (1+2+3+5+7+9) = 9

2 . So, our answer is 92 ·

∑∞n=1

110n = 9

2 · 19 = 12 .

23. [12]

Answer: 3211 Since AP is a tangent to Ω, we know that ∠PAB = ∠PCA, so △PAB ∼ △PCA, so

we get thatPB

PA=

4

7=

PA

PB + 6.

Solving, we get that 7PB = 4PA, so

4(PB + 6) = 7PA =49

4PB ⇒ 33

4PB = 24 ⇒ PB =

32

11.

24. [12]

Answer: 105 Since 4 pieces of candy are distributed, there must be exactly 8 children who do notreceive any candy; since no two consecutive children do receive candy, the 8 who do not must consistof 4 groups of consecutive children. We divide into cases based on the sizes of these groups:

• {5, 1, 1, 1}: there are 12 places to begin the group of 5 children who do not receive any candy

• {4, 2, 1, 1}: there are 12 places to begin the group of 4 children who do not receive candy andthen 3 choices for the group of 2 children which does not receive candy, for a total of 36 choices

• {3, 3, 1, 1}: these 8 children can either be bunched in the order 3,3,1,1, or in the order 3,1,3,1; thefirst has 12 positions in which to begin the first group of 3 non-candy receiving children and thesecond has 6 possibilities (due to symmetry), for a total of 18

• {3, 2, 2, 1}: there are 12 places to begin the group of 3 children who do not receive candy andthen 3 choices for the group of 1 child which does not receive candy, for a total of 36 choices

• {2, 2, 2, 2}: there are 12/4 = 3 ways in which this can occur

This gives a total of 12 + 36 + 18 + 36 + 3 = 105

25. [13]

Answer: 8380 Using the Law of Sines, we have

sin2 B + sin2 C − sin2 A

sinB sinC=

sinB

sin C+

sinC

sin B− sin A

sinB

sin A

sin C=

AC

AB+

AB

AC− BC

AC

BC

AB=

83

80.

26. [13]

Answer: 22013 − 6036 Let yn = xn − 2n+1. Note that

xn+1 = 2xn − xn−1 + 2n ⇔ yn+1 = 2yn − yn−1 ⇔ yn+1 − yn = yn − yn−1.

Using the values for x1, x2, we get that y1 = −3 and y2 = −6, so yn+1 − yn = (−6) − (−3) = −3. Byinduction, yn = −3n. Then, we get that xn = 2n+1 − 3n, so x2012 = 22013 − 6036.

Guts Round

27. [13]

Answer: 1755 We write this as (a− b)(a + b) + (a− b)(c) = (a− b)(a + b + c) = 2012. Since a, b, care positive integers, a− b < a + b + c. So, we have three possibilities: a− b = 1 and a + b + c = 2012,a − b = 2 and a + b + c = 1006, and a − b = 4 and a + b + c = 503.

The first solution gives a = b + 1 and c = 2011 − 2b, so b can range from 1 through 1005, whichdetermines a and c completely.

Similarly, the second solution gives a = b + 2 and c = 1004 − 2b, so b can range from 1 through 501.

Finally, the third solution gives a = b + 4 and c = 499 − 2b, so b can range from 1 through 249.

Hence, the total number of solutions is 1005 + 501 + 249 = 1755.

28. [15]

Answer: (0, 1+√

22 ] Suppose that k = ab+b2

a2+b2 for some positive real a, b. We claim that k lies in

(0, 1+√

22 ]. Let x = a

b . We have that ab+b2

a2+b2 =a

b+1

( a

b )2+1

= x+1x2+1 . Thus, x + 1 = k(x2 + 1), so the

quadratic kt2 − t + k − 1 = 0 has a positive real root. Thus, its discriminant must be nonnegative, so

12 ≥ 4(k − 1)(k) =⇒ (2k − 1)2 ≤ 2, which implies 1−√

22 ≤ k ≤ 1+

√2

2 . Since x > 0, we also have

k > 0, so we know that k must lie in (0, 1+√

22 ].

Now, take any k in the interval (0, 1+√

22 ]. We thus know that 12 ≥ 4k(k − 1), so the quadratic

kt2 − t+k− 1 = 0 has a positive solution,1+

√1−4k(k−1)

2k . Call this solution x. Then k(x2 +1) = x+1,

so x+1x2+1 = k. If we set a = x and b = 1, we get that ab+b2

a2+b2 = k. Thus, the set of all attainable values

of ab+b2

a2+b2 is the interval (0, 1+√

22 ].

29. [15]

Answer: − 25+5√

178 First, note that (x+1)(2x+1)(3x+1)(4x+1) = ((x+1)(4x+1))((2x+1)(3x+

1)) = (4x2 + 5x + 1)(6x2 + 5x + 1) = (5x2 + 5x + 1 − x2)(5x2 + 5x + 1 + x2) = (5x2 + 5x + 1)2 − x4.

Therefore, the equation is equivalent to (5x2 + 5x + 1)2 = 17x4, or 5x2 + 5x + 1 = ±√

17x2.

If 5x2 +5x+1 =√

17x2, then (5−√

17)x2 +5x+1 = 0. The discriminant of this is 25− 4(5−√

17) =

5 + 4√

17, so in this case, there are two real roots and they sum to − 55−

√17

= − 25+5√

178 .

If 5x2 +5x+1 = −√

17x2, then (5+√

17)x2 +5x+1 = 0. The discriminant of this is 25−4(5+√

17) =5 − 4

√17. This is less than zero, so there are no real solutions in this case. Therefore, the sum of all

real solutions to the equation is − 25+5√

178 .

30. [15]

Answer: 37 It suffices to assume that the monkey starts all over as soon as he has typed a string

that ends in no prefix of either abc or aaa. For instance, if the monkey gets to abb we can throw theseout because there’s no way to finish one of those strings from this without starting all over.

Now, we draw the tree of all possible intermediate stages under this assumption; there are not manypossibilities. The paths from the root ”a” are:

a- aa- aaa

a- aa- aab- aabc

a- ab- abc

The first and last possibilities have probability 1/27 each and the middle one has probability 1/81, so

in total the probability of getting the first before the second or the third is 1/271/27+1/81+1/27 = 3

7 .

Guts Round

31. [17]

Answer: 3√

320 We place the points in the coordinate plane. We let A = (0, 0,

√6

3 ), B = (0,√

33 , 0),

C = (− 12 ,−

√3

6 , 0), and D = (12 ,

√3

6 , 0). The point P is the origin, while M is (0, 0,√

66 ). The line

through B and M is the line x = 0, y =√

33 − z

√2. The plane through A, C, and D has equation

z = 2√

2y +√

23 . The coordinates of Q are the coordinates of the intersection of this line and this

plane. Equating the equations and solving for y and z, we see that y = − 15√

3and z =

√6

5 , so the

coordinates of Q are (0,− 15√

3,√

65 ).

Let N be the midpoint of CD, which has coordinates (0,−√

36 , 0). By the distance formula, QN = 3

√3

10 .

Thus, the area of QCD is QN ·CD2 = 3

√3

20 .

32. [17]

Answer: 2530 Note that φ(23) = 22 and φ(22) = 10, so if lcm(23, 22, 10) = 2530|k then f(n+k) ≡f(n) (mod 23) is always true. We show that this is necessary as well.

Choosing n ≡ 0 (mod 23), we see that k ≡ 0 (mod 23). Thus n + k ≡ n (mod 23) always, and we canmove to the exponent by choosing n to be a generator modulo 23:

(n + k)n+k ≡ nn (mod 22)

The choice of n here is independent of the choice (mod 23) since 22 and 23 are coprime. Thus wemust have again that 22|k, by choosing n ≡ 0 (mod 22). But then n + k ≡ n (mod 11) always, andwe can go to the exponent modulo φ(11) = 10 by choosing n a generator modulo 11:

n + k ≡ n (mod 10)

From here it follows that 10|k as well. Thus 2530|k and 2530 is the minimum positive integer desired.

33. [17]

Answer: tan−1 ( 10091005 ) As per usual with reflection problems instead of bouncing off the sides of

a 1 × 1 square we imagine the ball to travel in a straight line from origin in an infinite grid of 1 × 1squares, ”bouncing” every time it meets a line x = m or y = n. Let the lattice point it first meetsafter leaving the origin be (a, b), so that b > a. Note that a and b are coprime, otherwise the ball willreach a vertex before the 2012th bounce. We wish to minimize the slope of the line to this point fromorigin, which is b/a.

Now, the number of bounces up to this point is a − 1 + b − 1 = a + b − 2, so the given statement isjust a + b = 2014. To minimize b/a with a and b relatively prime, we must have a = 1005, b = 1009,so that the angle is tan−1 ( 1009

1005 ).

34. [20]

Answer: 15612 Note that the number of integers between 1 and 2012 that have n as a divi-sor is ⌊ 2012

n ⌋. Therefore, if we sum over the possible divisors, we see that the sum is equivalent to∑2012

d=1 ⌊ 2012d ⌋. This can be approximated by

∑2012d=1

2012d = 2012

∑2012d=1

1d ≈ 2012 ln(2012). As it turns

out, 2012 ln(2012) ≈ 15300, which is worth 18 points. Using the very rough approximation ln(2012) ≈ 7still gives 14 points.

35. [20]

Answer: 3.1415875473 The answer is 1006 sin π1006 . Approximating directly by π = 3.1415 . . . is

worth only 3 points.

Using the third-degree Taylor polynomial for sin we can approximate sinx ≈ x − x3

6 . This gives ananswer of 3.1415875473 worth full points. If during the calculation we use the approximation π3 ≈ 30,this gives an answer worth 9 points.

Guts Round

36. [20]

Answer: x Submit solution

Guts Round

TEAM ROUND

This test consists of ten problems to be solved by a team in one hour. The problems are unequally weighted

with point values as shown in brackets. They are not necessarily in order of di�culty, though harder problems are

generally worth more points. We do not expect most teams to get through all the problems.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted. Similarly,

graph paper, rulers, protractors, compasses, and other drawing aids are not permitted. If there is a chalkboard in

the room, you may write on it during the round only, not before it starts.

Numerical answers should, where applicable, be simplified as much as reasonably possible and must be exact unless

otherwise specified. Correct mathematical notation must be used. Your proctor cannot assist you in interpreting

or solving problems but has our phone number and may help with any administrative di�culties.

All problems require full written proof/justification. Even if the answer is numerical, you

must prove your result.

If you believe the test contains an error, please submit your protest in writing to Lobby 10 during lunchtime.

Enjoy!

HMMT 2013Saturday 16 February 2013

Team Round

Organization

Team Team ID#

1. [10]

HMMT 2013Saturday 16 February 2013

Team Round

Organization

Team Team ID#

2. [15]

HMMT 2013Saturday 16 February 2013

Team Round

Organization

Team Team ID#

3. [15]

HMMT 2013Saturday 16 February 2013

Team Round

Organization

Team Team ID#

4. [20]

HMMT 2013Saturday 16 February 2013

Team Round

Organization

Team Team ID#

5. [25]

HMMT 2013Saturday 16 February 2013

Team Round

Organization

Team Team ID#

6. [25]

HMMT 2013Saturday 16 February 2013

Team Round

Organization

Team Team ID#

7. [30]

HMMT 2013Saturday 16 February 2013

Team Round

Organization

Team Team ID#

8. [35]

HMMT 2013Saturday 16 February 2013

Team Round

Organization

Team Team ID#

9. [35]

HMMT 2013Saturday 16 February 2013

Team Round

Organization

Team Team ID#

10. [40]

HMMT 2013Saturday 16 February 2013

Team Round

1. [10] Let a and b be real numbers such thatab

a2 + b2=

14. Find all possible values of

|a2 � b2|a2 + b2

.

2. [15] A cafe has 3 tables and 5 individual counter seats. People enter in groups of size between 1 and4, inclusive, and groups never share a table. A group of more than 1 will always try to sit at a table,but will sit in counter seats if no tables are available. Conversely, a group of 1 will always try to sit atthe counter first. One morning, M groups consisting of a total of N people enter and sit down. Then,a single person walks in, and realizes that all the tables and counter seats are occupied by some personor group. What is the minimum possible value of M + N?

3. [15] Let ABC be a triangle with circumcenter O such that AC = 7. Suppose that the circumcircleof AOC is tangent to BC at C and intersects the line AB at A and F . Let FO intersect BC at E.Compute BE.

4. [20] Let a1, a2, a3, a4, a5 be real numbers whose sum is 20. Determine with proof the smallest possiblevalue of

X

1i<j5

bai + ajc.

5. [25] Thaddeus is given a 2013 ⇥ 2013 array of integers each between 1 and 2013, inclusive. He isallowed two operations:1. Choose a row, and subtract 1 from each entry.2. Chooses a column, and add 1 to each entry.He would like to get an array where all integers are divisible by 2013. On how many arrays is thispossible?

6. [25] Let triangle ABC satisfy 2BC = AB + AC and have incenter I and circumcircle !. Let D bethe intersection of AI and ! (with A,D distinct). Prove that I is the midpoint of AD.

7. [30] There are n children and n toys such that each child has a strict preference ordering on the toys.We want to distribute the toys: say a distribution A dominates a distribution B 6= A if in A, eachchild receives at least as preferable of an toy as in B. Prove that if some distribution is not dominatedby any other, then at least one child gets his/her favorite toy in that distribution.

8. [35] Let points A and B be on circle ! centered at O. Suppose that !A and !B are circles notcontaining O which are internally tangent to ! at A and B, respectively. Let !A and !B intersect atC and D such that D is inside triangle ABC. Suppose that line BC meets ! again at E and let lineEA intersect !A at F . If FC ? CD, prove that O, C, and D are collinear.

9. [35] Let m be an odd positive integer greater than 1. Let Sm be the set of all non-negative integersless than m which are of the form x + y, where xy � 1 is divisible by m. Let f(m) be the number ofelements of Sm.

(a) Prove that f(mn) = f(m)f(n) if m,n are relatively prime odd integers greater than 1.(b) Find a closed form for f(pk), where k > 0 is an integer and p is an odd prime.

10. [40] Chim Tu has a large rectangular table. On it, there are finitely many pieces of paper with non-overlapping interiors, each one in the shape of a convex polygon. At each step, Chim Tu is allowed toslide one piece of paper in a straight line such that its interior does not touch any other piece of paperduring the slide. Can Chim Tu always slide all the pieces of paper o↵ the table in finitely many steps?

HMMT 2013Saturday 16 February 2013

Team Round

1. [10] Let a and b be real numbers such thatab

a2 + b2=

14. Find all possible values of

|a2 � b2|a2 + b2

.

2. [15] A cafe has 3 tables and 5 individual counter seats. People enter in groups of size between 1 and4, inclusive, and groups never share a table. A group of more than 1 will always try to sit at a table,but will sit in counter seats if no tables are available. Conversely, a group of 1 will always try to sit atthe counter first. One morning, M groups consisting of a total of N people enter and sit down. Then,a single person walks in, and realizes that all the tables and counter seats are occupied by some personor group. What is the minimum possible value of M + N?

3. [15] Let ABC be a triangle with circumcenter O such that AC = 7. Suppose that the circumcircleof AOC is tangent to BC at C and intersects the line AB at A and F . Let FO intersect BC at E.Compute BE.

4. [20] Let a1, a2, a3, a4, a5 be real numbers whose sum is 20. Determine with proof the smallest possiblevalue of

X

1i<j5

bai + ajc.

5. [25] Thaddeus is given a 2013 ⇥ 2013 array of integers each between 1 and 2013, inclusive. He isallowed two operations:1. Choose a row, and subtract 1 from each entry.2. Chooses a column, and add 1 to each entry.He would like to get an array where all integers are divisible by 2013. On how many arrays is thispossible?

6. [25] Let triangle ABC satisfy 2BC = AB + AC and have incenter I and circumcircle !. Let D bethe intersection of AI and ! (with A,D distinct). Prove that I is the midpoint of AD.

7. [30] There are n children and n toys such that each child has a strict preference ordering on the toys.We want to distribute the toys: say a distribution A dominates a distribution B 6= A if in A, eachchild receives at least as preferable of an toy as in B. Prove that if some distribution is not dominatedby any other, then at least one child gets his/her favorite toy in that distribution.

8. [35] Let points A and B be on circle ! centered at O. Suppose that !A and !B are circles notcontaining O which are internally tangent to ! at A and B, respectively. Let !A and !B intersect atC and D such that D is inside triangle ABC. Suppose that line BC meets ! again at E and let lineEA intersect !A at F . If FC ? CD, prove that O, C, and D are collinear.

9. [35] Let m be an odd positive integer greater than 1. Let Sm be the set of all non-negative integersless than m which are of the form x + y, where xy � 1 is divisible by m. Let f(m) be the number ofelements of Sm.

(a) Prove that f(mn) = f(m)f(n) if m,n are relatively prime odd integers greater than 1.(b) Find a closed form for f(pk), where k > 0 is an integer and p is an odd prime.

10. [40] Chim Tu has a large rectangular table. On it, there are finitely many pieces of paper with non-overlapping interiors, each one in the shape of a convex polygon. At each step, Chim Tu is allowed toslide one piece of paper in a straight line such that its interior does not touch any other piece of paperduring the slide. Can Chim Tu always slide all the pieces of paper o↵ the table in finitely many steps?

HMMT 2013Saturday 16 February 2013

Team Round

1. [10] Let a and b be real numbers such thatab

a2 + b2=

14. Find all possible values of

|a2 � b2|a2 + b2

.

2. [15] A cafe has 3 tables and 5 individual counter seats. People enter in groups of size between 1 and4, inclusive, and groups never share a table. A group of more than 1 will always try to sit at a table,but will sit in counter seats if no tables are available. Conversely, a group of 1 will always try to sit atthe counter first. One morning, M groups consisting of a total of N people enter and sit down. Then,a single person walks in, and realizes that all the tables and counter seats are occupied by some personor group. What is the minimum possible value of M + N?

3. [15] Let ABC be a triangle with circumcenter O such that AC = 7. Suppose that the circumcircleof AOC is tangent to BC at C and intersects the line AB at A and F . Let FO intersect BC at E.Compute BE.

4. [20] Let a1, a2, a3, a4, a5 be real numbers whose sum is 20. Determine with proof the smallest possiblevalue of

X

1i<j5

bai + ajc.

5. [25] Thaddeus is given a 2013 ⇥ 2013 array of integers each between 1 and 2013, inclusive. He isallowed two operations:1. Choose a row, and subtract 1 from each entry.2. Chooses a column, and add 1 to each entry.He would like to get an array where all integers are divisible by 2013. On how many arrays is thispossible?

6. [25] Let triangle ABC satisfy 2BC = AB + AC and have incenter I and circumcircle !. Let D bethe intersection of AI and ! (with A,D distinct). Prove that I is the midpoint of AD.

7. [30] There are n children and n toys such that each child has a strict preference ordering on the toys.We want to distribute the toys: say a distribution A dominates a distribution B 6= A if in A, eachchild receives at least as preferable of an toy as in B. Prove that if some distribution is not dominatedby any other, then at least one child gets his/her favorite toy in that distribution.

8. [35] Let points A and B be on circle ! centered at O. Suppose that !A and !B are circles notcontaining O which are internally tangent to ! at A and B, respectively. Let !A and !B intersect atC and D such that D is inside triangle ABC. Suppose that line BC meets ! again at E and let lineEA intersect !A at F . If FC ? CD, prove that O, C, and D are collinear.

9. [35] Let m be an odd positive integer greater than 1. Let Sm be the set of all non-negative integersless than m which are of the form x + y, where xy � 1 is divisible by m. Let f(m) be the number ofelements of Sm.

(a) Prove that f(mn) = f(m)f(n) if m,n are relatively prime odd integers greater than 1.(b) Find a closed form for f(pk), where k > 0 is an integer and p is an odd prime.

10. [40] Chim Tu has a large rectangular table. On it, there are finitely many pieces of paper with non-overlapping interiors, each one in the shape of a convex polygon. At each step, Chim Tu is allowed toslide one piece of paper in a straight line such that its interior does not touch any other piece of paperduring the slide. Can Chim Tu always slide all the pieces of paper o↵ the table in finitely many steps?

HMMT 2013Saturday 16 February 2013

Team Round

1. [10] Let a and b be real numbers such thatab

a2 + b2=

14. Find all possible values of

|a2 � b2|a2 + b2

.

2. [15] A cafe has 3 tables and 5 individual counter seats. People enter in groups of size between 1 and4, inclusive, and groups never share a table. A group of more than 1 will always try to sit at a table,but will sit in counter seats if no tables are available. Conversely, a group of 1 will always try to sit atthe counter first. One morning, M groups consisting of a total of N people enter and sit down. Then,a single person walks in, and realizes that all the tables and counter seats are occupied by some personor group. What is the minimum possible value of M + N?

3. [15] Let ABC be a triangle with circumcenter O such that AC = 7. Suppose that the circumcircleof AOC is tangent to BC at C and intersects the line AB at A and F . Let FO intersect BC at E.Compute BE.

4. [20] Let a1, a2, a3, a4, a5 be real numbers whose sum is 20. Determine with proof the smallest possiblevalue of

X

1i<j5

bai + ajc.

5. [25] Thaddeus is given a 2013 ⇥ 2013 array of integers each between 1 and 2013, inclusive. He isallowed two operations:1. Choose a row, and subtract 1 from each entry.2. Chooses a column, and add 1 to each entry.He would like to get an array where all integers are divisible by 2013. On how many arrays is thispossible?

6. [25] Let triangle ABC satisfy 2BC = AB + AC and have incenter I and circumcircle !. Let D bethe intersection of AI and ! (with A,D distinct). Prove that I is the midpoint of AD.

7. [30] There are n children and n toys such that each child has a strict preference ordering on the toys.We want to distribute the toys: say a distribution A dominates a distribution B 6= A if in A, eachchild receives at least as preferable of an toy as in B. Prove that if some distribution is not dominatedby any other, then at least one child gets his/her favorite toy in that distribution.

8. [35] Let points A and B be on circle ! centered at O. Suppose that !A and !B are circles notcontaining O which are internally tangent to ! at A and B, respectively. Let !A and !B intersect atC and D such that D is inside triangle ABC. Suppose that line BC meets ! again at E and let lineEA intersect !A at F . If FC ? CD, prove that O, C, and D are collinear.

9. [35] Let m be an odd positive integer greater than 1. Let Sm be the set of all non-negative integersless than m which are of the form x + y, where xy � 1 is divisible by m. Let f(m) be the number ofelements of Sm.

(a) Prove that f(mn) = f(m)f(n) if m,n are relatively prime odd integers greater than 1.(b) Find a closed form for f(pk), where k > 0 is an integer and p is an odd prime.

10. [40] Chim Tu has a large rectangular table. On it, there are finitely many pieces of paper with non-overlapping interiors, each one in the shape of a convex polygon. At each step, Chim Tu is allowed toslide one piece of paper in a straight line such that its interior does not touch any other piece of paperduring the slide. Can Chim Tu always slide all the pieces of paper o↵ the table in finitely many steps?

HMMT 2013Saturday 16 February 2013

Team Round

1. [10] Let a and b be real numbers such thatab

a2 + b2=

14. Find all possible values of

|a2 � b2|a2 + b2

.

2. [15] A cafe has 3 tables and 5 individual counter seats. People enter in groups of size between 1 and4, inclusive, and groups never share a table. A group of more than 1 will always try to sit at a table,but will sit in counter seats if no tables are available. Conversely, a group of 1 will always try to sit atthe counter first. One morning, M groups consisting of a total of N people enter and sit down. Then,a single person walks in, and realizes that all the tables and counter seats are occupied by some personor group. What is the minimum possible value of M + N?

3. [15] Let ABC be a triangle with circumcenter O such that AC = 7. Suppose that the circumcircleof AOC is tangent to BC at C and intersects the line AB at A and F . Let FO intersect BC at E.Compute BE.

4. [20] Let a1, a2, a3, a4, a5 be real numbers whose sum is 20. Determine with proof the smallest possiblevalue of

X

1i<j5

bai + ajc.

5. [25] Thaddeus is given a 2013 ⇥ 2013 array of integers each between 1 and 2013, inclusive. He isallowed two operations:1. Choose a row, and subtract 1 from each entry.2. Chooses a column, and add 1 to each entry.He would like to get an array where all integers are divisible by 2013. On how many arrays is thispossible?

6. [25] Let triangle ABC satisfy 2BC = AB + AC and have incenter I and circumcircle !. Let D bethe intersection of AI and ! (with A,D distinct). Prove that I is the midpoint of AD.

7. [30] There are n children and n toys such that each child has a strict preference ordering on the toys.We want to distribute the toys: say a distribution A dominates a distribution B 6= A if in A, eachchild receives at least as preferable of an toy as in B. Prove that if some distribution is not dominatedby any other, then at least one child gets his/her favorite toy in that distribution.

8. [35] Let points A and B be on circle ! centered at O. Suppose that !A and !B are circles notcontaining O which are internally tangent to ! at A and B, respectively. Let !A and !B intersect atC and D such that D is inside triangle ABC. Suppose that line BC meets ! again at E and let lineEA intersect !A at F . If FC ? CD, prove that O, C, and D are collinear.

9. [35] Let m be an odd positive integer greater than 1. Let Sm be the set of all non-negative integersless than m which are of the form x + y, where xy � 1 is divisible by m. Let f(m) be the number ofelements of Sm.

(a) Prove that f(mn) = f(m)f(n) if m,n are relatively prime odd integers greater than 1.(b) Find a closed form for f(pk), where k > 0 is an integer and p is an odd prime.

10. [40] Chim Tu has a large rectangular table. On it, there are finitely many pieces of paper with non-overlapping interiors, each one in the shape of a convex polygon. At each step, Chim Tu is allowed toslide one piece of paper in a straight line such that its interior does not touch any other piece of paperduring the slide. Can Chim Tu always slide all the pieces of paper o↵ the table in finitely many steps?

HMMT 2013Saturday 16 February 2013

Team Round

1. [10] Let a and b be real numbers such thatab

a2 + b2=

14. Find all possible values of

|a2 � b2|a2 + b2

.

2. [15] A cafe has 3 tables and 5 individual counter seats. People enter in groups of size between 1 and4, inclusive, and groups never share a table. A group of more than 1 will always try to sit at a table,but will sit in counter seats if no tables are available. Conversely, a group of 1 will always try to sit atthe counter first. One morning, M groups consisting of a total of N people enter and sit down. Then,a single person walks in, and realizes that all the tables and counter seats are occupied by some personor group. What is the minimum possible value of M + N?

3. [15] Let ABC be a triangle with circumcenter O such that AC = 7. Suppose that the circumcircleof AOC is tangent to BC at C and intersects the line AB at A and F . Let FO intersect BC at E.Compute BE.

4. [20] Let a1, a2, a3, a4, a5 be real numbers whose sum is 20. Determine with proof the smallest possiblevalue of

X

1i<j5

bai + ajc.

5. [25] Thaddeus is given a 2013 ⇥ 2013 array of integers each between 1 and 2013, inclusive. He isallowed two operations:1. Choose a row, and subtract 1 from each entry.2. Chooses a column, and add 1 to each entry.He would like to get an array where all integers are divisible by 2013. On how many arrays is thispossible?

6. [25] Let triangle ABC satisfy 2BC = AB + AC and have incenter I and circumcircle !. Let D bethe intersection of AI and ! (with A,D distinct). Prove that I is the midpoint of AD.

7. [30] There are n children and n toys such that each child has a strict preference ordering on the toys.We want to distribute the toys: say a distribution A dominates a distribution B 6= A if in A, eachchild receives at least as preferable of an toy as in B. Prove that if some distribution is not dominatedby any other, then at least one child gets his/her favorite toy in that distribution.

8. [35] Let points A and B be on circle ! centered at O. Suppose that !A and !B are circles notcontaining O which are internally tangent to ! at A and B, respectively. Let !A and !B intersect atC and D such that D is inside triangle ABC. Suppose that line BC meets ! again at E and let lineEA intersect !A at F . If FC ? CD, prove that O, C, and D are collinear.

9. [35] Let m be an odd positive integer greater than 1. Let Sm be the set of all non-negative integersless than m which are of the form x + y, where xy � 1 is divisible by m. Let f(m) be the number ofelements of Sm.

(a) Prove that f(mn) = f(m)f(n) if m,n are relatively prime odd integers greater than 1.(b) Find a closed form for f(pk), where k > 0 is an integer and p is an odd prime.

10. [40] Chim Tu has a large rectangular table. On it, there are finitely many pieces of paper with non-overlapping interiors, each one in the shape of a convex polygon. At each step, Chim Tu is allowed toslide one piece of paper in a straight line such that its interior does not touch any other piece of paperduring the slide. Can Chim Tu always slide all the pieces of paper o↵ the table in finitely many steps?

HMMT 2013Saturday 16 February 2013

Team Round

1. [10] Let a and b be real numbers such thatab

a2 + b2=

14. Find all possible values of

|a2 � b2|a2 + b2

.

2. [15] A cafe has 3 tables and 5 individual counter seats. People enter in groups of size between 1 and4, inclusive, and groups never share a table. A group of more than 1 will always try to sit at a table,but will sit in counter seats if no tables are available. Conversely, a group of 1 will always try to sit atthe counter first. One morning, M groups consisting of a total of N people enter and sit down. Then,a single person walks in, and realizes that all the tables and counter seats are occupied by some personor group. What is the minimum possible value of M + N?

3. [15] Let ABC be a triangle with circumcenter O such that AC = 7. Suppose that the circumcircleof AOC is tangent to BC at C and intersects the line AB at A and F . Let FO intersect BC at E.Compute BE.

4. [20] Let a1, a2, a3, a4, a5 be real numbers whose sum is 20. Determine with proof the smallest possiblevalue of

X

1i<j5

bai + ajc.

5. [25] Thaddeus is given a 2013 ⇥ 2013 array of integers each between 1 and 2013, inclusive. He isallowed two operations:1. Choose a row, and subtract 1 from each entry.2. Chooses a column, and add 1 to each entry.He would like to get an array where all integers are divisible by 2013. On how many arrays is thispossible?

6. [25] Let triangle ABC satisfy 2BC = AB + AC and have incenter I and circumcircle !. Let D bethe intersection of AI and ! (with A,D distinct). Prove that I is the midpoint of AD.

7. [30] There are n children and n toys such that each child has a strict preference ordering on the toys.We want to distribute the toys: say a distribution A dominates a distribution B 6= A if in A, eachchild receives at least as preferable of an toy as in B. Prove that if some distribution is not dominatedby any other, then at least one child gets his/her favorite toy in that distribution.

8. [35] Let points A and B be on circle ! centered at O. Suppose that !A and !B are circles notcontaining O which are internally tangent to ! at A and B, respectively. Let !A and !B intersect atC and D such that D is inside triangle ABC. Suppose that line BC meets ! again at E and let lineEA intersect !A at F . If FC ? CD, prove that O, C, and D are collinear.

9. [35] Let m be an odd positive integer greater than 1. Let Sm be the set of all non-negative integersless than m which are of the form x + y, where xy � 1 is divisible by m. Let f(m) be the number ofelements of Sm.

(a) Prove that f(mn) = f(m)f(n) if m,n are relatively prime odd integers greater than 1.(b) Find a closed form for f(pk), where k > 0 is an integer and p is an odd prime.

10. [40] Chim Tu has a large rectangular table. On it, there are finitely many pieces of paper with non-overlapping interiors, each one in the shape of a convex polygon. At each step, Chim Tu is allowed toslide one piece of paper in a straight line such that its interior does not touch any other piece of paperduring the slide. Can Chim Tu always slide all the pieces of paper o↵ the table in finitely many steps?

HMMT 2013Saturday 16 February 2013

Team Round

1. [10] Let a and b be real numbers such thatab

a2 + b2=

14. Find all possible values of

|a2 � b2|a2 + b2

.

2. [15] A cafe has 3 tables and 5 individual counter seats. People enter in groups of size between 1 and4, inclusive, and groups never share a table. A group of more than 1 will always try to sit at a table,but will sit in counter seats if no tables are available. Conversely, a group of 1 will always try to sit atthe counter first. One morning, M groups consisting of a total of N people enter and sit down. Then,a single person walks in, and realizes that all the tables and counter seats are occupied by some personor group. What is the minimum possible value of M + N?

3. [15] Let ABC be a triangle with circumcenter O such that AC = 7. Suppose that the circumcircleof AOC is tangent to BC at C and intersects the line AB at A and F . Let FO intersect BC at E.Compute BE.

4. [20] Let a1, a2, a3, a4, a5 be real numbers whose sum is 20. Determine with proof the smallest possiblevalue of

X

1i<j5

bai + ajc.

5. [25] Thaddeus is given a 2013 ⇥ 2013 array of integers each between 1 and 2013, inclusive. He isallowed two operations:1. Choose a row, and subtract 1 from each entry.2. Chooses a column, and add 1 to each entry.He would like to get an array where all integers are divisible by 2013. On how many arrays is thispossible?

6. [25] Let triangle ABC satisfy 2BC = AB + AC and have incenter I and circumcircle !. Let D bethe intersection of AI and ! (with A,D distinct). Prove that I is the midpoint of AD.

7. [30] There are n children and n toys such that each child has a strict preference ordering on the toys.We want to distribute the toys: say a distribution A dominates a distribution B 6= A if in A, eachchild receives at least as preferable of an toy as in B. Prove that if some distribution is not dominatedby any other, then at least one child gets his/her favorite toy in that distribution.

8. [35] Let points A and B be on circle ! centered at O. Suppose that !A and !B are circles notcontaining O which are internally tangent to ! at A and B, respectively. Let !A and !B intersect atC and D such that D is inside triangle ABC. Suppose that line BC meets ! again at E and let lineEA intersect !A at F . If FC ? CD, prove that O, C, and D are collinear.

9. [35] Let m be an odd positive integer greater than 1. Let Sm be the set of all non-negative integersless than m which are of the form x + y, where xy � 1 is divisible by m. Let f(m) be the number ofelements of Sm.

(a) Prove that f(mn) = f(m)f(n) if m,n are relatively prime odd integers greater than 1.(b) Find a closed form for f(pk), where k > 0 is an integer and p is an odd prime.

10. [40] Chim Tu has a large rectangular table. On it, there are finitely many pieces of paper with non-overlapping interiors, each one in the shape of a convex polygon. At each step, Chim Tu is allowed toslide one piece of paper in a straight line such that its interior does not touch any other piece of paperduring the slide. Can Chim Tu always slide all the pieces of paper o↵ the table in finitely many steps?

HMMT 2013Saturday 16 February 2013

Team Round

1. [10] Let a and b be real numbers such thatab

a2 + b2=

1

4. Find all possible values of

|a2 − b2|a2 + b2

.

Answer:√

32 The hypothesis statement is equivalent to

a2 + b2 = 4ab

1 : (a + b)2 = 6ab

2 : (a − b)2 = 2ab

Multiplying equations 1 and 2,(a2 − b2)2 = 12(ab)2

|a2 − b2| = ±√

12ab

Since the left hand side and ab are both positive,

|a2 − b2| =√

12ab

|a2 − b2|a2 + b2

=

√12ab

a2 + b2=

√12

4=

√3

2

(It is clear that such a and b exist: for example, we can take a = 1 and solve for b by way of thequadratic formula.)

2. [15] A cafe has 3 tables and 5 individual counter seats. People enter in groups of size between 1 and4, inclusive, and groups never share a table. A group of more than 1 will always try to sit at a table,but will sit in counter seats if no tables are available. Conversely, a group of 1 will always try to sit atthe counter first. One morning, M groups consisting of a total of N people enter and sit down. Then,a single person walks in, and realizes that all the tables and counter seats are occupied by some personor group. What is the minimum possible value of M + N?

Answer: 16 We first show that M + N ≥ 16. Consider the point right before the last table isoccupied. We have two cases: first suppose there exists at least one open counter seat. Then, everytable must contribute at least 3 to the value of M + N , because no groups of 1 will have taken a tablewith one of the counter seats open. By the end, the counter must contribute at least 5 + 2 = 7 toM + N , as there must be at least two groups sitting at the counter. It follows that M + N ≥ 16. Forthe second case, assume the counter is full right before the last table is taken. Then, everybody sittingat the counter must have entered as a singleton, since they entered when a table was still available.Consequently, the counter must contribute 10 to M +N , and each table contributes at least 2, so onceagain M + N ≥ 16.

Now, M +N = 16 is achievable with eight groups of one, who first fill the counter seats, then the threetables. Thus, our answer is 16.

3. [15] Let ABC be a triangle with circumcenter O such that AC = 7. Suppose that the circumcircleof AOC is tangent to BC at C and intersects the line AB at A and F . Let FO intersect BC at E.Compute BE.

Answer: EB = 72 O is the circumcenter of △ABC =⇒ AO = CO =⇒ ∠OCA = ∠OAC.

Because AC is an inscribed arc of circumcircle △AOC, ∠OCA = ∠OFA. Furthermore BC is tangentto circumcircle △AOC, so ∠OAC = ∠OCB. However, again using the fact that O is the circumcenterof △ABC, ∠OCB = ∠OBC.

We now have that CO bisects ∠ACB, so it follows that triangle CA = CB. Also, by AA similaritywe have EOB ∼ EBF . Thus, EB2 = EO · EF = EC2 by the similarity and power of a point, soEB = BC/2 = AC/2 = 7/2.

Team Round

4. [20] Let a1, a2, a3, a4, a5 be real numbers whose sum is 20. Determine with proof the smallest possiblevalue of

∑

1≤i<j≤5

⌊ai + aj⌋.

Answer: 72 We claim that the minimum is 72. This can be achieved by taking a1 = a2 = a3 =a4 = 0.4 and a5 = 18.4. To prove that this is optimal, note that

∑

1≤i<j≤5

⌊ai + aj⌋ =∑

1≤i<j≤5

(ai + aj) − {ai + aj} = 80 −∑

1≤i<j≤5

{ai + aj},

so it suffices to maximize

∑

1≤i<j≤5

{ai + aj} =5∑

i=1

{ai + ai+2} +5∑

i=1

{ai + ai+1},

where a6 = a1 and a7 = a2, Taking each sum modulo 1, it is clear that both are integers. Thus, theabove sum is at most 2 ·4 = 8, and our original expression is at least 80−8 = 72, completing the proof.

5. [25] Thaddeus is given a 2013 × 2013 array of integers each between 1 and 2013, inclusive. He isallowed two operations:

1. Choose a row, and subtract 1 from each entry.

2. Chooses a column, and add 1 to each entry.

He would like to get an array where all integers are divisible by 2013. On how many arrays is thispossible?

Answer: 20134025 We claim that the set of grids on which it is possible to obtain an array of allzeroes (mod 2013) is indexed by ordered 4025-tuples of residues (mod 2013), corresponding to thestarting entries in the first row and first column of the grid, giving the answer of 20134025. To do this,we show that given after fixing all of the entries in the first row and column, there is a unique startinggrid which can become an array of all zeroes after applying the appropriate operations.

Let ai,j be the entry in the i-th row and the j-th column. Suppose there is a sequence of operationsgiving all zeroes in the array; let ri be the number of times we operate on row i, and let cj be thenumber of times we operate on column j. It is enough to take all of these values to be residues modulo2013. Clearly, ai,j + ri + cj ≡ 0 (mod 2013) for each i, j. In particular, r1 + c1 ≡ a1,1. Now, for eachi, j, we have

ai,j ≡ −ri − cj

≡ (ai,1 + c1) + (a1,j + r1)

≡ ai,1 + a1,j − a1,1,

which is fixed. Thus, there rest of the entries in the grid are forced.

Conversely, if we set ai,j to be the appropriate representative of the residue class of ai,1 + a1,j − a1,1

modulo 2013, we may take ri ≡ −ai,1 (mod 2013), and cj ≡ a1,1 − a1,j (mod 2013) for each i, j. It isclear that ai,j + ri + cj ≡ 0 (mod 2013) for each i, j, so we’re done.

6. [25] Let triangle ABC satisfy 2BC = AB + AC and have incenter I and circumcircle ω. Let D bethe intersection of AI and ω (with A,D distinct). Prove that I is the midpoint of AD.

Answer: N/A Since AD is an angle bisector, D is the midpoint of arc BC opposite A on ω. It is

well-known that B, I, and C lie on a circle centered at D. Thus BD = DC = DI. Applying Ptolemy’stheorem to cyclic quadrilateral ABDC, we get

AB · DC + AC · BD = AD · BC = AD · (AB + AC)/2

Team Round

Using BD = DC we have immediately that AD = 2BD = 2DI so I is the midpoint of AD as desired.

Solution 2: Let P and Q be the midpoints of AB and AC, and take the point E on segment BC suchthat BE = BP . Note that CE = AB − BE = AB − BP = AB+AC

2 − AB2 = AC

2 = CQ, so triangles

BPE and CQE are isosceles. In addition, BEEC = AB/2

AC/2 = ABAC , so by the angle bisector theorem, AE

bisects ∠CAB, whence E must lie on the bisector of ∠A.

Since triangles BPE and CQE are isosceles, the bisectors of angles B and C are the perpendicu-lar bisectors of segments PE and EQ, respectively. Thus, the circumcenter of △PQE is I, so theperpendicular bisector of PQ meets the bisector of ∠A at I.

Furthermore, since ∠DAB = ∠CAD, arcs B̂D and D̂C have the same measure, so BD = DC, whencethe perpendicular bisector of BC meets the bisector of ∠A at D. A homothety centered at A withfactor 1/2 maps BC to PQ, and so maps D to I. Thus, D is the midpoint of AI.

Solution 3: Let a = BC, b = CA, and c = AB, let r and R denote the lengths of the inradius andcircumradius of △ABC, respectively, let E be the intersection of segments AD and BC, and let O bethe circumcenter of △ABC. I is the midpoint of chord AD if and only if OI ⊥ AD, which is trueif and only if OA2 = OI2 + AI2. By Euler’s distance formula, OI2 = R(R − 2r), and by Stewart’s

theorem and the angle bisector theorem we can find AI2 =(

b+ca+b+c

)2

bc(1 −(

ab+c

)2

) = bc3 . Thus, it

remains to show that 6Rr = bc.

Now, we use the combine the well-known formulas abc4K = R and K = rs to get abc = 4Rrs, where K is

the area of △ABC and s is its semiperimeter. We have bc = Rr 4sa = Rr 2a+2(b+c)

a = Rr 2a+2(2a)a = 6Rr,

as desired.

7. [30] There are n children and n toys such that each child has a strict preference ordering on the toys.We want to distribute the toys: say a distribution A dominates a distribution B 6= A if in A, eachchild receives at least as preferable of an toy as in B. Prove that if some distribution is not dominatedby any other, then at least one child gets his/her favorite toy in that distribution.

Answer: N/A Suppose we have a distribution A assigning each child Ci, i = 1, 2, . . . , n, toy Ti,

such that no child Ci gets their top preference T ′i 6= Ti. Then, pick an arbitrary child C1 and construct

the sequence of children Ci1 , Ci2 , Ci3 , . . . where i1 = 1 and Cik+1was assigned the favorite toy T ′

ikof

the last child Cik. Eventually, some Cik

= Ci1 ; at this point, just pass the toys around this cycleso that each of these children gets their favorite toy. Clearly the resulting distribution dominates theoriginal, so we’re done.

8. [35] Let points A and B be on circle ω centered at O. Suppose that ωA and ωB are circles notcontaining O which are internally tangent to ω at A and B, respectively. Let ωA and ωB intersect atC and D such that D is inside triangle ABC. Suppose that line BC meets ω again at E and let lineEA intersect ωA at F . If FC ⊥ CD, prove that O, C, and D are collinear.

Answer: N/A Let H = CA ∩ ω, and G = BH ∩ ωB . There are homotheties centered at A and

B taking ωA → ω and ωB → ω that take A : F 7→ E, A : C 7→ H, B : C 7→ E and B : G 7→ H. Inparticular CF ||EH||CG, so C,F,G are collinear, lying on a line perpendicular to DC.

Because of the right angles at C, DF and DG are diameters of ωA, ωB , respectively. Also, we have thatthe ratio of the sizes of ωA and ωB , under the two homotheties above, is CF/EH ·EH/CG = CF/CG.Therefore, DF/DG = CF/CG, but then △DCF and △DCG are both right triangles which share oneside and have hypotenuse and other side in proportion; it is obvious now that the two circles ωA andωB are congruent.

Therefore, O has the same distance to A and B, and so the same distances to the centers of the twocircles as well. As a result, O lies on the radical axis CD as desired.

9. [35] Let m be an odd positive integer greater than 1. Let Sm be the set of all non-negative integersless than m which are of the form x + y, where xy − 1 is divisible by m. Let f(m) be the number ofelements of Sm.

Team Round

(a) Prove that f(mn) = f(m)f(n) if m,n are relatively prime odd integers greater than 1.

(b) Find a closed form for f(pk), where k > 0 is an integer and p is an odd prime.

Answer: N/A For a positive integer n, let Z/nZ denote the set of residues modulo n and (Z/nZ)∗

denote the set of residues modulo n that are relatively prime to n. Then, rephrased, Sm is the set ofresidues modulo m of the form x + x−1, where x ∈ (Z/mZ)∗.

For part (a), suppose a = x + x−1 ∈ Sm and b = y + y−1 ∈ Sn. By the Chinese RemainderTheorem, there exists a residue z ∈ (Z/mnZ)∗ such that m|(x − z) and n|(y − z), and thus z + z−1 ≡x+x−1 (mod m) and z +z−1 ≡ y +y−1 (mod n). Therefore, all f(m)f(n) residues modulo mn whichresult from applying the Chinese Remainder Theorem to an element each of Sm and Sn are in Smn.Conversely, given z + z−1 ∈ (Z/mnZ)∗, taking z + z−1 modulo m and n gives elements of Sm and Sn,so indeed f(mn) = f(m)f(n).

We now proceed to part (b). For each x ∈ (Z/pkZ)∗, denote q(x) to be the largest non-negative integer

i ≤ k such that pi divides x2 − 1 (this is clearly well-defined). For a given x, let g(x) be the numberof y ∈ (Z/pk

Z)∗ such that x + x−1 ≡ y + y−1 (mod pk). Note that this condition is equivalent to(x − y)(xy − 1) ≡ 0 (mod pk).

First, consider the case in which q(x) ≥ k/2, in which case we have p⌈k/2⌉|(x − 1)(x + 1), and becausep is odd, x ≡ ±1 (mod p⌈k/2⌉). Thus, either (x − 1)2 ≡ 0 (mod pk) or (x + 1)2 ≡ 0 (mod pk), and itfollows that x+x−1 ≡ ±2 (mod pk) (clearly 2 and −2 are distinct). Conversely, x+x−1 ≡ ±2 implies(x ± 1) ≡ 0 (mod p⌈k/2⌉), which in turn implies q(x) ≥ k/2. It is now clear that there are exactly twoelements of Sm corresponding to residues of the form x + x−1 with q(x) ≥ k/2, and all other elementsof Sm come from x with q(x) < k/2.

Fix x with q(x) < k/2; we will compute g(x). Suppose (x − y)(xy − 1) ≡ 0 (mod pk), and sayx − y, xy − 1 have j, j′ factors of p, respectively. If j ≤ j′, note that x2 − 1 ≡ xy − 1 ≡ 0 (mod pj), soj ≤ q(x). Similarly, if j′ ≤ j, x ≡ y (mod pj′

), so x2 − 1 ≡ xy − 1 ≡ 0 (mod pj′

), and so j′ ≤ q(x). Itfollows that min(j, j′) ≤ q(x), and thus max(j, j′) ≥ k − q(x).

Suppose pk−q(x)|(x − y). Then, xy − 1 ≡ x2 − 1 ≡ 0 (mod pq(x)) because q(x) < k/2, so any y withpk−q(x)|(x− y) satisfies (x− y)(xy− 1) ≡ 0 (mod pk). Now, suppose pk−q(x)|(xy − 1), that is, y ≡ x−1

(mod pk−q(x)). Then, xy ≡ 1 (mod pq(x)), and since x2 ≡ 1 (mod pq(x)), we have x ≡ y (mod pq(x)),so again we have (x−y)(xy−1) ≡ 0 (mod pk). It follows that the set of y satisfying (x−y)(xy−1) ≡ 0(mod pk) is exactly the set of y with y ≡ x (mod pk−q(x)) or y ≡ x−1 (mod pk−q(x)). x, x−1 are distinctresidues modulo pk−q(x), because x2−1 has fewer than k/2 factors of p, so it follows that g(x) = 2pq(x).In particular, note that these values are distinct for different values of q(x) < k/2, so x+x−1 ≡ y+y−1

implies q(x) = q(y) or q(x), q(y) ≥ k/2.

For each integer i with 0 ≤ i < k/2, we need to compute the number of x ∈ (Z/pkZ) with q(x) = i.

Clearly, this is the number of x with pi|x2 − 1 minus the number of x with pi+1|x2 − 1. When i = 0,the number of x with pi|x2 − 1 is clearly pk−1(p− 1), and when i > 0, this number is 2pk−i, as we havex ≡ ±1 (mod pi).

We can now count the number of elements of Spk using casework on the value of q(x) where we takex to be such that x + x−1 is a particular element of Sm. Applying the results from the previous twoparagraphs, our answer is

2 +pk−1(p − 3)

2+

⌈k/2⌉−1∑

i=1

2pk−i − 2pk−i−1

2pi,

where the summand 2 comes from ±2 ∈ Spk , corresponding to all x with q(x) > k/2, the next summandcomes from those x with q(x) = 0, and each additional summand comes from those x with q(x) = i inthe relevant range. We may evaluate the last sum as a geometric series, to obtain the final closed formanswer of

2 +pk−1(p − 3)

2+

pk−1 − p1+(−1)k

2

p + 1.

Team Round

10. [40] Chim Tu has a large rectangular table. On it, there are finitely many pieces of paper with non-overlapping interiors, each one in the shape of a convex polygon. At each step, Chim Tu is allowed toslide one piece of paper in a straight line such that its interior does not touch any other piece of paperduring the slide. Can Chim Tu always slide all the pieces of paper off the table in finitely many steps?

Answer: N/A Let the pieces of paper be P1, P2, . . . , Pn in the Cartesian plane. It suffices to show

that for any constant distance D, they can be slid so that each pairwise distance is at least D. Then,we can apply this using D equal to the diameter of the rectangle, sliding all but at most one of thepieces of paper off the table, and then slide this last one off arbitrarily.

We show in particular that there is always a polygon Pi which can be slid arbitrarily far to the right(i.e. in the positive x-direction). For each Pi let Bi be a bottommost (i.e. lowest y-coordinate) point onthe boundary of Pi. Define a point Q to be exposed if the ray starting at Q in the positive x directionmeets the interior of no piece of paper.

Consider the set of all exposed Bi; this set is nonempty because certainly the bottommost of all theBi is exposed. Of this set, let Bk be the exposed Bi with maximal y-coordinate, and if there are morethan one such, choose the one with maximal x-coordinate. We claim that the corresponding Pk canbe slid arbitrarily far to the right.

Suppose for the sake of contradiction that there is some polygon blocking this path. To be precise,if Ak is the highest point of Pk, then the region R formed by the right-side boundary of Pk and therays pointing in the positive x direction from Ak and Bk, must contain interior point(s) of some set ofpolygon(s) Pj in its interior. All of their bottommost points Bj must lie in R, since none of them canhave boundary intersecting the ray from Bk, by the construction of Bk.

Because Bk was chosen to be rightmost out of all the exposed Bi with that y-coordinate, it must bethat all of the Bj corresponding to the blocking Pj have larger y-coordinate. Now, choose of these theone with smallest y-coordinate - it must be exposed, and it has strictly higher y-coordinate than Bk,contradiction. It follows that the interior of R intersects no pieces of paper.

Now, for a fixed D such that D is at least the largest distance between any two points of two polygons,we can shift this exposed piece of paper nD to the right, the next one of the remaining pieces (n−1)D,and so on, so that the pairwise distances between pieces of paper, even when projected onto the x-axis,are at least D each. We’re done.

Team Round

ALGEBRA TEST

This test consists of 10 short-answer problems to be solved individually in 50 minutes. Problems will be weightedwith point values after the contest based on how many competitors solve each problem. There is no penalty forguessing.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted. Similarly,graph paper, rulers, protractors, compasses, and other drawing aids are not permitted.

Answers should be simplified as much as is reasonably possible and must be exact unless otherwise specified. Ra-tional numbers should be written in lowest terms, although denominators of irrationals need not be rationalized.An nth root should be simplified so that the radicand is not divisible by the nth power of any prime.

Correct mathematical notation must be used. No partial credit will be given unless otherwise specified.

If you believe the test contains an error, please submit your protest in writing to Lobby 10 during lunchtime.

Enjoy!

HMMT 2013Saturday 16 February 2013

Algebra Test

1. Let x and y be real numbers with x > y such that x2y2 +x2 + y2 +2xy = 40 and xy +x+ y = 8. Findthe value of x.

2. Let {an}n≥1 be an arithmetic sequence and {gn}n≥1 be a geometric sequence such that the first fourterms of {an + gn} are 0, 0, 1, and 0, in that order. What is the 10th term of {an + gn}?

3. Let S be the set of integers of the form 2x + 2y + 2z, where x, y, z are pairwise distinct non-negativeintegers. Determine the 100th smallest element of S.

4. Determine all real values of A for which there exist distinct complex numbers x1, x2 such that thefollowing three equations hold:

x1(x1 + 1) = A

x2(x2 + 1) = A

x41 + 3x3

1 + 5x1 = x42 + 3x3

2 + 5x2.

5. Let a and b be real numbers, and let r, s, and t be the roots of f(x) = x3 + ax2 + bx − 1. Also,g(x) = x3 + mx2 + nx + p has roots r2, s2, and t2. If g(−1) = −5, find the maximum possible valueof b.

6. Find the number of integers n such that

1 +⌊

100n

101

⌋=

⌈99n

100

⌉.

7. Compute∞∑

a1=0

∞∑a2=0

· · ·∞∑

a7=0

a1 + a2 + · · · + a7

3a1+a2+···+a7.

8. Let x, y be complex numbers such that x2+y2

x+y = 4 and x4+y4

x3+y3 = 2. Find all possible values of x6+y6

x5+y5 .

9. Let z be a non-real complex number with z23 = 1. Compute

22∑k=0

11 + zk + z2k

.

10. Let N be a positive integer whose decimal representation contains 11235 as a contiguous substring,and let k be a positive integer such that 10k > N . Find the minimum possible value of

10k − 1gcd(N, 10k − 1)

.

HMMT 2013Saturday 16 February 2013

Algebra Test

PUT LABEL HERE

Name Team ID#

Organization Team

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Score:

HMMT 2013Saturday 16 February 2013

Algebra Test

1. Let x and y be real numbers with x > y such that x2y2 +x2 + y2 +2xy = 40 and xy +x+ y = 8. Findthe value of x.

Answer: 3 +√

7 We have (xy)2 + (x + y)2 = 40 and xy + (x + y) = 8. Squaring the second

equation and subtracting the first gives xy(x + y) = 12 so xy, x + y are the roots of the quadratica2−8a+12 = 0. It follows that {xy, x+y} = {2, 6}. If x+y = 2 and xy = 6, then x, y are the roots ofthe quadratic b2−2b+6 = 0, which are non-real, so in fact x+y = 6 and xy = 2, and x, y are the roots

of the quadratic b2 − 6b + 2 = 0. Because x > y, we take the larger root, which is 6+√

282 = 3 +

√7.

2. Let {an}n≥1 be an arithmetic sequence and {gn}n≥1 be a geometric sequence such that the first fourterms of {an + gn} are 0, 0, 1, and 0, in that order. What is the 10th term of {an + gn}?Answer: −54 Let the terms of the geometric sequence be a, ra, r2a, r3a. Then, the terms of

the arithmetic sequence are −a,−ra,−r2a + 1,−r3a. However, if the first two terms of this sequenceare −a,−ra, the next two terms must also be (−2r + 1)a, (−3r + 2)a. It is clear that a 6= 0 becausea3 +g3 6= 0, so −r3 = −3r+2 ⇒ r = 1 or −2. However, we see from the arithmetic sequence that r = 1is impossible, so r = −2. Finally, by considering a3, we see that −4a + 1 = 5a, so a = 1/9. We alsosee that an = (3n− 4)a and gn = (−2)n−1a, so our answer is a10 + g10 = (26− 512)a = −486a = −54.

3. Let S be the set of integers of the form 2x + 2y + 2z, where x, y, z are pairwise distinct non-negativeintegers. Determine the 100th smallest element of S.

Answer: 577 S is the set of positive integers with exactly three ones in its binary representation.

The number of such integers with at most d total bits is(

d3

)

, and noting that(

93

)

= 84 and(

103

)

= 120,we want the 16th smallest integer of the form 29 + 2x + 2y, where y < x < 9. Ignoring the 29 term,

there are(

d′

2

)

positive integers of the form 2x + 2y with at most d′ total bits. Because(

62

)

= 15, ouranswer is 29 + 26 + 20 = 577. (By a bit, we mean a digit in base 2.)

4. Determine all real values of A for which there exist distinct complex numbers x1, x2 such that thefollowing three equations hold:

x1(x1 + 1) = A

x2(x2 + 1) = A

x41 + 3x3

1 + 5x1 = x42 + 3x3

2 + 5x2.

Answer: −7 Applying polynomial division,

x41 + 3x3

1 + 5x1 = (x21 + x1 − A)(x2

1 + 2x1 + (A − 2)) + (A + 7)x1 + A(A − 2)

= (A + 7)x1 + A(A − 2).

Thus, in order for the last equation to hold, we need (A+7)x1 = (A+7)x2, from which it follows thatA = −7. These steps are reversible, so A = −7 indeed satisfies the needed condition.

5. Let a and b be real numbers, and let r, s, and t be the roots of f(x) = x3 + ax2 + bx − 1. Also,g(x) = x3 + mx2 + nx + p has roots r2, s2, and t2. If g(−1) = −5, find the maximum possible valueof b.

Answer: 1 +√

5 By Vieta’s Formulae, m = −(r2 + s2 + t2) = −a2 + 2b, n = r2s2 + s2t2 + t2r2 =

b2 + 2a, and p = −1. Therefore, g(−1) = −1 − a2 + 2b − b2 − 2a − 1 = −5 ⇔ (a + 1)2 + (b − 1)2 = 5.This is an equation of a circle, so b reaches its maximum when a + 1 = 0 ⇒ a = −1. When a = −1,b = 1 ±

√5, so the maximum is 1 +

√5.

Algebra Test

6. Find the number of integers n such that

1 +

⌊

100n

101

⌋

=

⌈

99n

100

⌉

.

Answer: 10100 Consider f(n) = ⌈ 99n100 ⌉ − ⌊ 100n

101 ⌋. Note that f(n + 10100) = ⌈ 99n100 + 99 · 101⌉ −

⌊ 100n101 + 1002⌋ = f(n) + 99 · 101− 1002 = f(n)− 1. Thus, for each residue class r modulo 10100, there

is exactly one value of n for which f(n) = 1 and n ≡ r (mod 10100). It follows immediately that theanswer is 10100.

7. Compute∞∑

a1=0

∞∑

a2=0

· · ·∞∑

a7=0

a1 + a2 + · · · + a7

3a1+a2+···+a7

.

Answer: 15309/256 Note that, since this is symmetric in a1 through a7,

∞∑

a1=0

∞∑

a2=0

· · ·∞∑

a7=0

a1 + a2 + · · · + a7

3a1+a2+···+a7

= 7

∞∑

a1=0

∞∑

a2=0

· · ·∞∑

a7=0

a1

3a1+a2+···+a7

= 7

( ∞∑

a1=0

a1

3a1

) ( ∞∑

a=0

1

3a

)6

.

If S =∑

a3a , then 3S−S =

∑

13a = 3/2, so S = 3/4. It follows that the answer equals 7· 34 ·

(

32

)6= 15309

256 .

Alternatively, let f(z) =∑∞

a1=0

∑∞a2=0 · · ·

∑∞a7=0 za1+a2+···+a7 . Note that we can rewrite f(z) =

(∑∞

a=0 za)7 = 1(1−z)7 . Furthermore, note that zf ′(z) =

∑∞a1=0

∑∞a2=0 · · ·

∑∞a7=0(a1 + a2 + · · · +

a7)za1+a2+···+a7 , so the sum in question is simply f ′(1/3)

3 . Since f ′(x) = 7(1−z)8 , it follows that the sum

is equal to 7·37

28 = 15309256 .

8. Let x, y be complex numbers such that x2+y2

x+y = 4 and x4+y4

x3+y3 = 2. Find all possible values of x6+y6

x5+y5 .

Answer: 10 ± 2√

17 Let A = 1x + 1

y and let B = xy + y

x . Then

B

A=

x2 + y2

x + y= 4,

so B = 4A. Next, note that

B2 − 2 =x4 + y4

x2y2and AB − A =

x3 + y3

x2y2,

so

B2 − 2

AB − A= 2.

Substituting B = 4A and simplifying, we find that 4A2 +A−1 = 0, so A = −1±√

178 . Finally, note that

64A3 − 12A = B3 − 3B =x6 + y6

x3y3and 16A3 − 4A2 − A = A(B2 − 2) − (AB − A) =

x5 + y5

x3y3,

so

Algebra Test

x6 + y6

x5 + y5=

64A2 − 12

16A2 − 4A − 1=

4 − 16A

3 − 8A,

where the last inequality follows from the fact that 4A2 = 1−A. If A = −1+√

178 , then this value equals

10 + 2√

17. Similarly, if A = −1−√

178 , then this value equals 10 − 2

√17.

(It is not hard to see that these values are achievable by noting that with the values of A and B wecan solve for x + y and xy, and thus for x and y.)

9. Let z be a non-real complex number with z23 = 1. Compute

22∑

k=0

1

1 + zk + z2k.

Answer: 46/3 First solution: Note that

22∑

k=0

1

1 + zk + z2k=

1

3+

22∑

k=1

1 − zk

1 − z3k=

1

3+

22∑

k=1

1 − (z24)k

1 − z3k=

1

3+

22∑

k=1

7∑

ℓ=0

z3kℓ.

3 and 23 are prime, so every non-zero residue modulo 23 appears in an exponent in the last sum exactly7 times, and the summand 1 appears 22 times. Because the sum of the 23rd roots of unity is zero, ouranswer is 1

3 + (22 − 7) = 463 .

Second solution: For an alternate approach, we first prove the following identity for an arbitrarycomplex number a:

22∑

k=0

1

a − zk=

23a22

a23 − 1.

To see this, let f(x) = x23 − 1 = (x − 1)(x − z)(x − z2) . . . (x − z22). Note that the sum in question is

merely f ′(a)f(a) , from which the identity follows.

Now, returning to our orignal sum, let ω 6= 1 satisfy ω3 = 1. Then

22∑

k=0

1

1 + zk + z2k=

1

ω2 − ω

22∑

k=0

1

ω − zk− 1

ω2 − zk

=1

ω2 − ω

(

22∑

k=0

1

ω − zk−

22∑

k=0

1

ω2 − zk

)

=1

ω2 − ω

(

23ω22

ω23 − 1− 23ω44

ω46 − 1

)

=23

ω2 − ω

(

ω

ω2 − 1− ω2

ω − 1

)

=23

ω2 − ω

(ω2 − ω) − (ω − ω2)

2 − ω − ω2

=46

3.

10. Let N be a positive integer whose decimal representation contains 11235 as a contiguous substring,and let k be a positive integer such that 10k > N . Find the minimum possible value of

Algebra Test

10k − 1

gcd(N, 10k − 1).

Answer: 89 Set m = 10k−1gcd(N,10k−1)

. Then, in lowest terms, N10k−1

= am for some integer a. On the

other hand, the decimal expansion of N10k−1

simply consists of the decimal expansion of N , possiblywith some padded zeros, repeating. Since N contains 11235 as a contiguous substring, the decimalrepresentation of a

m must as well.

Conversely, if m is relatively prime to 10 and if there exists an a such that the decimal representation

of am contains the substring 11235, we claim that m is an attainable value for 10k−1

gcd(N,10k−1). To see

this, note that since m is relatively prime to 10, there exists a value of k such that m divides 10k − 1(for example, k = φ(m)). Letting ms = 10k − 1 and N = as, it follows that a

m = asms = N

10k−1. Since

the decimal expansion of this fraction contains the substring 11235, it follows that N must also, andtherefore m is an attainable value.

We are therefore looking for a fraction am which contains the substring 11235 in its decimal expansion.

Since 1, 1, 2, 3, and 5 are the first five Fibonacci numbers, it makes sense to look at the value of theinfinite series

∞∑

i=1

Fi

10i.

A simple generating function argument shows that∑∞

i=1 Fixi = x

1−x−x2 , so substituting x = 1/10leads us to the fraction 10/89 (which indeed begins 0.11235 . . . ).

How do we know no smaller values of m are possible? Well, if a′/m′ contains the substring 11235somewhere in its infinitely repeating decimal expansion, then note that there is an i such that thedecimal expansion of the fractional part of 10i(a′/m′) begins with 0.11235 . . . . We can therefore,without loss of generality, assume that the decimal representation of a′/m′ begins 0.11235 . . . . Butsince the decimal representation of 10/89 begins 0.11235 . . . , it follows that

∣

∣

∣

∣

10

89− a′

m′

∣

∣

∣

∣

≤ 10−5.

On the other hand, this absolute difference, if non-zero, is at least 189m′

. If m′ < 89, this is at least1

892 > 10−5, and therefore no smaller values of m′ are possible.

Algebra Test

COMBINATORICS TEST

This test consists of 10 short-answer problems to be solved individually in 50 minutes. Problems will be weightedwith point values after the contest based on how many competitors solve each problem. There is no penalty forguessing.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted. Similarly,graph paper, rulers, protractors, compasses, and other drawing aids are not permitted.

Answers should be simplified as much as is reasonably possible and must be exact unless otherwise specified. Ra-tional numbers should be written in lowest terms, although denominators of irrationals need not be rationalized.An nth root should be simplified so that the radicand is not divisible by the nth power of any prime.

Correct mathematical notation must be used. No partial credit will be given unless otherwise specified.

If you believe the test contains an error, please submit your protest in writing to Lobby 10 during lunchtime.

Enjoy!

HMMT 2013Saturday 16 February 2013

Combinatorics Test

1. A standard 52-card deck contains cards of 4 suits and 13 numbers, with exactly one card for eachpairing of suit and number. If Maya draws two cards with replacement from this deck, what is theprobability that the two cards have the same suit or have the same number, but not both?

2. If Alex does not sing on Saturday, then she has a 70% chance of singing on Sunday; however, to resther voice, she never sings on both days. If Alex has a 50% chance of singing on Sunday, find theprobability that she sings on Saturday.

3. On a game show, Merble will be presented with a series of 2013 marbles, each of which is either red orblue on the outside. Each time he sees a marble, he can either keep it or pass, but cannot return to aprevious marble; he receives 3 points for keeping a red marble, loses 2 points for keeping a blue marble,and gains 0 points for passing. All distributions of colors are equally likely and Merble can only seethe color of his current marble. If his goal is to end with exactly one point and he plays optimally,what is the probability that he fails?

4. How many orderings (a1, . . . , a8) of (1, 2, . . . , 8) exist such that a1−a2 +a3−a4 +a5−a6 +a7−a8 = 0?

5. At a certain chocolate company, each bar is 1 unit long. To make the bars more interesting, thecompany has decided to combine dark and white chocolate pieces. The process starts with two bars,one completely dark and one completely white. At each step of the process, a new number p is chosenuniformly at random between 0 and 1. Each of the two bars is cut p units from the left, and the pieceson the left are switched: each is grafted onto the opposite bar where the other piece of length p waspreviously attached. For example, the bars might look like this after the first step:

0 p 1

Each step after the first operates on the bars resulting from the previous step. After a total of100 steps, what is the probability that on each bar, the chocolate 1/3 units from the left is the sametype of chocolate as that 2/3 units from the left?

6. Values a1, . . . , a2013 are chosen independently and at random from the set {1, . . . , 2013}. What isexpected number of distinct values in the set {a1, . . . , a2013}?

7. A single-elimination ping-pong tournament has 22013 players, seeded in order of ability. If the playerwith seed x plays the player with seed y, then it is possible for x to win if and only if x ≤ y + 3. Forhow many players P it is possible for P to win? (In each round of a single elimination tournament, theremaining players are randomly paired up; each player plays against the other player in his pair, withthe winner from each pair progressing to the next round and the loser eliminated. This is repeateduntil there is only one player remaining.)

8. It is known that exactly one of the three (distinguishable) musketeers stole the truffles. Each musketeermakes one statement, in which he either claims that one of the three is guilty, or claims that one of thethree is innocent. It is possible for two or more of the musketeers to make the same statement. Afterhearing their claims, and knowing that exactly one musketeer lied, the inspector is able to deduce whostole the truffles. How many ordered triplets of statements could have been made?

9. Given a permutation σ of {1, 2, . . . , 2013}, let f(σ) to be the number of fixed points of σ – that is,the number of k ∈ {1, 2, . . . , 2013} such that σ(k) = k. If S is the set of all possible permutations σ,compute ∑

σ∈S

f(σ)4.

(Here, a permutation σ is a bijective mapping from {1, 2, . . . , 2013} to {1, 2, . . . , 2013}.)10. Rosencrantz and Guildenstern each start with $2013 and are flipping a fair coin. When the coin

comes up heads Rosencrantz pays Guildenstern $1 and when the coin comes up tails Guildenstern paysRosencrantz $1. Let f(n) be the number of dollars Rosencrantz is ahead of his starting amount aftern flips. Compute the expected value of max{f(0), f(1), f(2), . . . , f(2013)}.

HMMT 2013Saturday 16 February 2013

Combinatorics Test

PUT LABEL HERE

Name Team ID#

Organization Team

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Score:

HMMT 2013Saturday 16 February 2013

Combinatorics Test

1. A standard 52-card deck contains cards of 4 suits and 13 numbers, with exactly one card for eachpairing of suit and number. If Maya draws two cards with replacement from this deck, what is theprobability that the two cards have the same suit or have the same number, but not both?

Answer: 1552 After drawing the first card, there are 12 other cards from the same suit and 3 other

cards with the same number, so the probability is 12+352 .

2. If Alex does not sing on Saturday, then she has a 70% chance of singing on Sunday; however, to resther voice, she never sings on both days. If Alex has a 50% chance of singing on Sunday, find theprobability that she sings on Saturday.

Answer: 27 Let p be the probability that Alex sings on Saturday. Then, the probability that she

sings on Sunday is .7(1 − p); setting this equal to .5 gives p = 27 .

3. On a game show, Merble will be presented with a series of 2013 marbles, each of which is either red orblue on the outside. Each time he sees a marble, he can either keep it or pass, but cannot return to aprevious marble; he receives 3 points for keeping a red marble, loses 2 points for keeping a blue marble,and gains 0 points for passing. All distributions of colors are equally likely and Merble can only seethe color of his current marble. If his goal is to end with exactly one point and he plays optimally,what is the probability that he fails?

Answer: 122012 First, we note that if all the marbles are red or all are blue, then it is impossible

for Merble to win; we claim that he can guarantee himself a win in every other case. In particular, hisstrategy should be to keep the first red and first blue marble that he encounters, and to ignore all theothers. Consequently, the probability that he cannot win is 2

22013 = 122012 .

4. How many orderings (a1, . . . , a8) of (1, 2, . . . , 8) exist such that a1−a2+a3−a4+a5−a6+a7−a8 = 0?

Answer: 4608 We can divide the numbers up based on whether they have a + or - beforethem. Both the numbers following +’s and -’s must add up to 18. Without loss of generality, we canassume the +’s contain the number 1 (and add a factor of 2 at the end to account for this). Thepossible 4-element sets containing a 1 which add to 18 are {1, 2, 7, 8}, {1, 3, 6, 8}, {1, 4, 5, 8}, {1, 4, 6, 7}.Additionally, there are 4! ways to order the numbers following a + and 4! ways to order the numbersfollowing a −. Thus the total number of possibilities is 4 × 2 × 4! × 4! = 4608.

5. At a certain chocolate company, each bar is 1 unit long. To make the bars more interesting, thecompany has decided to combine dark and white chocolate pieces. The process starts with two bars,one completely dark and one completely white. At each step of the process, a new number p is chosenuniformly at random between 0 and 1. Each of the two bars is cut p units from the left, and the pieceson the left are switched: each is grafted onto the opposite bar where the other piece of length p waspreviously attached. For example, the bars might look like this after the first step:

0 p 1

Combinatorics Test

Each step after the first operates on the bars resulting from the previous step. After a total of100 steps, what is the probability that on each bar, the chocolate 1/3 units from the left is the sametype of chocolate as that 2/3 units from the left?

Answer: 12

[

(

13

)100+ 1

]

If the values of p chosen are p1, . . . , p100, then note that the color of a

bar changes at each value of pi. Consequently, we want to find the probability that exactly an evennumber of pi are in (1

3 , 23 ). Summing, this is equal to

(

100

0

)(

1

3

)0 (

2

3

)100

+

(

100

2

) (

1

3

)2 (

2

3

)98

+ . . .

(

100

100

)(

1

3

)100 (

2

3

)0

.

To compute, we note that this is equal to

1

2

[

(

2

3+

1

3

)100

+

(

2

3−

1

3

)100]

after expanding using the binomial theorem, since any terms with odd exponents are cancelled outbetween the two terms.

6. Values a1, . . . , a2013 are chosen independently and at random from the set {1, . . . , 2013}. What isexpected number of distinct values in the set {a1, . . . , a2013}?

Answer: 20132013−20122013

20132012 For each n ∈ {1, 2, . . . , 2013}, let Xn = 1 if n appears in {a1, a2, . . . , a2013}

and 0 otherwise. Defined this way, E[Xn] is the probability that n appears in {a1, a2, . . . , a2013}. Sinceeach ai (1 ≤ i ≤ 2013) is not n with probability 2012/2013, the probability that n is none of the

ai’s is(

20122013

)2013, so E[Xn], the probability that n is one of the ai’s, is 1 −

(

20122013

)2013. The ex-

pected number of distinct values in {a1, . . . , a2013} is the expected number of n ∈ {1, 2, . . . , 2013}such that Xn = 1, that is, the expected value of X1 + X2 + · · · + X2013. By linearity of expectation,

E[X1 + X2 + · · · + X2013] = E[X1] + E[X2] + · · · + E[Xn] = 2013(

1 −(

20122013

)2013)

= 20132013−20122013

20132012 .

7. A single-elimination ping-pong tournament has 22013 players, seeded in order of ability. If the playerwith seed x plays the player with seed y, then it is possible for x to win if and only if x ≤ y + 3. Forhow many players P it is possible for P to win? (In each round of a single elimination tournament, theremaining players are randomly paired up; each player plays against the other player in his pair, withthe winner from each pair progressing to the next round and the loser eliminated. This is repeateduntil there is only one player remaining.)

Answer: 6038 We calculate the highest seed n that can win. Below, we say that a player xvicariously defeats a player y if x defeats y directly or indirectly through some chain (i.e. x defeats x1,who defeated x2, . . ., who defeated xn, who defeated y for some players x1, . . . , xn).

We first consider the highest seeds that are capable of making the semifinals. The eventual winnermust be able to beat two of these players and thus must be able to beat the second best player in thesemifinals. The seed of the player who vicariously beats the 1-seed is maximized if 1 loses to 4 in thefirst round, 4 to 7 in the second round, etc. Therefore 3 · 2011 + 1 = 6034 is the maximum value ofthe highest seed in the semifinals. If 1, and 2 are in different quarters of the draw, then by a similarargument 6035 is the largest possible value of the second best player in the semis, and thus 6038 isthe highest that can win. If 1 and 2 are in the same quarter, then in one round the highest remainingseed will not be able to go up by 3, when the player who has vicariously beaten 1 plays the player whovicariously beat 2, so 3 · 2011− 1 = 6032 is the highest player the semifinalist from that quarter couldbe. But then the eventual winner still must be seeded at most 6 above this player, and thus 6038 isstill the upper bound.

Therefore 6038 is the worst seed that could possibly win, and can do so if 6034, 6035, 6036, 6038 allmake the semis, which is possible (it is not difficult to construct such a tournament). Then, note thatany player x with a lower seed can also win for some tournament – in particular, it suffices to take

Combinatorics Test

the tournament where it is possible for player 6038 to win and switch the positions of 6038 and x.Consequently, there are 6038 players for whom it is possible to win under some tournament.

8. It is known that exactly one of the three (distinguishable) musketeers stole the truffles. Each musketeermakes one statement, in which he either claims that one of the three is guilty, or claims that one of thethree is innocent. It is possible for two or more of the musketeers to make the same statement. Afterhearing their claims, and knowing that exactly one musketeer lied, the inspector is able to deduce whostole the truffles. How many ordered triplets of statements could have been made?

Answer: 99 We divide into cases, based on the number of distinct people that statements aremade about.

• The statements are made about 3 distinct people. Then, since exactly one person is guilty, andbecause exactly one of the three lied, there are either zero statements of guilt or two statements ofguilt possible; in either case, it is impossible by symmetry to determine who is guilty or innocent.

• The statements are made about 2 distinct people or 1 distinct person. Then, either at least twoof the statements are the same, or all are different.

– If two statements are the same, then those two statements must be true because only onemusketeer lied. Consequently, the lone statement must be false. If all the statements areabout the same person, there there must be 2 guilty claims and 1 innocence claim (otherwisewe would not know which of the other two people was guilty). Then, there are 3 choices forwho the statement is about and 3 choices for who makes the innocence claim, for a 3 · 3 = 9triplets of statements. Meanwhile, if the statements are about two different people, then thisis doable unless both of the distinct statements imply guilt for the person concerned (i.e.where there are two guilty accusations against one person and one claim of innocence againstanother). Consequently, there are 3 sets of statements that can be made, 3 · 2 = 6 ways todetermine who they are made about, and 3 ways to determine who makes which statement,for a total of 3 · 6 · 3 = 54 triplets in this case.

– If all the statements are different, then they must be about two different people. Here, theremust be one person, who we will call A, who has both a claim of innocence and an accusationof guilt against him. The last statement must concern another person, B. If the statementaccuses B of being guilty, then we can deduce that he is the guilty one. On the other hand, ifthe statement claims that B is innocent, either of the other two musketeers could be guilty.Consequently, there are 3 · 2 = 6 ways to choose A and B, and 3! = 6 ways to choose whomakes which statement, for a total of 6 · 6 = 36 triplets of statements.

In total, we have 9 + 54 + 36 = 99 possible triplets of statements.

9. Given a permutation σ of {1, 2, . . . , 2013}, let f(σ) to be the number of fixed points of σ – that is,the number of k ∈ {1, 2, . . . , 2013} such that σ(k) = k. If S is the set of all possible permutations σ,compute

∑

σ∈S

f(σ)4.

(Here, a permutation σ is a bijective mapping from {1, 2, . . . , 2013} to {1, 2, . . . , 2013}.)

Answer: 15(2013!) First, note that

∑

σ∈S

f(σ)4 =∑

σ∈S

∑

1≤a1,a2,a3,a4≤2013

g(σ, a1, a2, a3, a4),

where g(σ, a1, a2, a3, a4) = 1 if all ai are fixed points of σ and 0 otherwise. (The ai’s need not bedistinct.) Switching the order of summation, we find that the desired sum is

∑

1≤a1,a2,a3,a4≤2013

∑

σ∈S

g(σ, a1, a2, a3, a4).

Combinatorics Test

Note that the inner sum is equal to the number of permutations on {1, 2, . . . , 2013} that fix a1, a2, a3,and a4. This depends on the number of distinct values the ais take. If they take on exactly k distinctvalues, then the inner sum will evaluate to (2013 − k)!, because σ can be any permutation of theremaining 2013 − k elements. (For example, if a1 = a2 but a1, a3, and a4 are distinct, the inner sumis 2010! because σ can be any permutation that fixes a1, a3, and a4.)

Now, suppose we are given which of the ai are equal (for example, we could be given a1 = a2 buta1, a3, a4 mutually distinct, as per the above example). Assuming there are k distinct values amongthe ai, there are 2013(2013 − 1) · · · (2013 − k + 1) ways to choose the ai. At this point, there are(2013 − k)! ways to choose σ on the remaining (2013 − k) values such that it fixes the ai, for a totalof 2013! choices for (σ, a1, a2, a3, a4) such that g(σ, a1, a2, a3, a4) = 1 and the ai satisfy the correctequality relations.

Thus the answer is 2013! times the number of ways to choose equivalence classes on the ai, so theproblem reduces to finding the number of ways to partition 4 elements into nonempty sets. Thisprocess can be accelerated by doing casework based on the number of sets:

(a) One set must contain all four elements, only one possibility. (i.e. all the ais are equal)

(b) Either one set contains 3 elements and the other contains the fourth (4 possibilities) or one setcontains 2 elements and the other contains the other two (3 possibilities). (i.e. there are twodistinct values of ai)

(c) One set contains two elements, the other two each contain one. There are(

42

)

= 6 ways to choosethe two elements in the set with two elements, and this uniquely determines the partition. (i.e.there are three distinct values of ai)

(d) All sets contain one element, in which case there is only one possibility. (i.e. all the ai are distinct)

Thus the number of ways to construct such a partition is 1 + 4 + 3 + 6 + 1 = 15, and our answer is15 · 2013!.

10. Rosencrantz and Guildenstern each start with $2013 and are flipping a fair coin. When the coincomes up heads Rosencrantz pays Guildenstern $1 and when the coin comes up tails Guildenstern paysRosencrantz $1. Let f(n) be the number of dollars Rosencrantz is ahead of his starting amount aftern flips. Compute the expected value of max{f(0), f(1), f(2), . . . , f(2013)}.

Answer: −12 +

(1007)(2013

1006)22012 We want to calculate Γ =

∞∑

i=0

i · P (max profit=i), where we consider

the maximum profit Rosencrantz has at any point over the first 2013 coin flips. By summation by

parts this is equal to2013∑

a=1

P (max profit ≥ a).

Let pa be the probability that Rosencrantz’ max profit is at least a and let En,a be the set of sequencesof flips such that Rosencrantz first reaches a profit of a on exactly the nth flip. Let Q+

a , Q−a , Qa be the

sets such that after all 2013 flips Rosencrantz’ final profit is (respectively) greater than, less than, orequal to a.

Then,

Γ =

2013∑

a=1

P (max profit ≥a)

=

2013∑

a=1

2013∑

n=a

P (En,a)

=2013∑

a=1

2013∑

n=a

P (En,a ∩ Q+a ) + P (En,a ∩ Q−

a ) + P (En,a ∩ Qa).

By symmetry, P (En,a ∩ Q+a ) = P (En,a ∩ Q−

a ) because, for any sequence in En,a ∩ Q+a , we can reverse

all the flips after the nth flip to get a sequence in En,a ∩ Q−a , and vice-versa.

Combinatorics Test

Furthermore,

2013∑

n=a

P (En,a ∩ Q+a ) = P (Q+

a ) and

2013∑

n=a

P (En,a ∩ Qa) = P (Qa).

So we have2013∑

a=1

P (max profit ≥a) =

2013∑

a=1

(

P (Qa) + 2P (Q+a )

)

.

Since by symmetry P (Qa) = P (Q−a) and we have an odd number of flips, we have

2013∑

a=1

P (Qa) =1

2.

Also P (Q+a ) =

1

22013

2013∑

k=⌈ 2014+a

2 ⌉

(

2013

k

)

.

So the rest is just computation. We have:

Γ =1

2+

1

22012

2013∑

a=1

2013∑

k=⌈ 2014+a

2 ⌉

(

2013

k

)

=1

2+

1

22012

2013∑

k=1008

2k−2014∑

a=1

(

2013

k

)

=1

2+

1

22012

2013∑

k=1008

(

2013

k

)

(k + k − 2013 − 1)

=1

2+

1

22012

2013∑

k=1008

2013

(

2012

k − 1

)

− 2013

(

2012

k

)

−

(

2013

k

)

=1

2+

2013(

20121007

)

− 22012 +(

20131007

)

22012

=−1

2+

(1007)(

20131006

)

22012.

So the answer is −12 +

(1007)(2013

1006)22012 (for reference, approximately 35.3).

Combinatorics Test

GEOMETRY TEST

This test consists of 10 short-answer problems to be solved individually in 50 minutes. Problems will be weightedwith point values after the contest based on how many competitors solve each problem. There is no penalty forguessing.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted. Similarly,graph paper, rulers, protractors, compasses, and other drawing aids are not permitted.

Answers should be simplified as much as is reasonably possible and must be exact unless otherwise specified. Ra-tional numbers should be written in lowest terms, although denominators of irrationals need not be rationalized.An nth root should be simplified so that the radicand is not divisible by the nth power of any prime.

Correct mathematical notation must be used. No partial credit will be given unless otherwise specified.

If you believe the test contains an error, please submit your protest in writing to Lobby 10 during lunchtime.

Enjoy!

HMMT 2013Saturday 16 February 2013

Geometry Test

1. Jarris the triangle is playing in the (x, y) plane. Let his maximum y coordinate be k. Given that hehas side lengths 6, 8, and 10 and that no part of him is below the x-axis, find the minimum possiblevalue of k.

2. Let ABCD be an isosceles trapezoid such that AD = BC, AB = 3, and CD = 8. Let E be a point inthe plane such that BC = EC and AE ⊥ EC. Compute AE.

3. Let A1A2A3A4A5A6 be a convex hexagon such that AiAi+2 ‖ Ai+3Ai+5 for i = 1, 2, 3 (we takeAi+6 = Ai for each i). Segment AiAi+2 intersects segment Ai+1Ai+3 at Bi, for 1 ≤ i ≤ 6, asshown. Furthermore, suppose that △A1A3A5

∼= △A4A6A2. Given that [A1B5B6] = 1, [A2B6B1] =4, and [A3B1B2] = 9 (by [XY Z] we mean the area of △XY Z), determine the area of hexagonB1B2B3B4B5B6.

A1

A2

A3

A4

A5

A6

B1

B2B3

B4

B5B6

(not to scale)

14

9

?

4. Let ω1 and ω2 be circles with centers O1 and O2, respectively, and radii r1 and r2, respectively. Supposethat O2 is on ω1. Let A be one of the intersections of ω1 and ω2, and B be one of the two intersectionsof line O1O2 with ω2. If AB = O1A, find all possible values of r1

r2.

5. In triangle ABC, ∠A = 45◦ and M is the midpoint of BC. AM intersects the circumcircle of ABCfor the second time at D, and AM = 2MD. Find cos ∠AOD, where O is the circumcenter of ABC.

6. Let ABCD be a quadrilateral such that ∠ABC = ∠CDA = 90◦, and BC = 7. Let E and F be onBD such that AE and CF are perpendicular to BD. Suppose that BE = 3. Determine the productof the smallest and largest possible lengths of DF .

7. Let ABC be an obtuse triangle with circumcenter O such that ∠ABC = 15◦ and ∠BAC > 90◦.Suppose that AO meets BC at D, and that OD2 + OC · DC = OC2. Find ∠C.

8. Let ABCD be a convex quadrilateral. Extend line CD past D to meet line AB at P and extend lineCB past B to meet line AD at Q. Suppose that line AC bisects ∠BAD. If AD = 7

4 , AP = 212 , and

AB = 1411 , compute AQ.

9. Pentagon ABCDE is given with the following conditions:

(a) ∠CBD + ∠DAE = ∠BAD = 45◦, ∠BCD + ∠DEA = 300◦

(b) BADA = 2

√2

3 , CD = 7√

53 , and DE = 15

√2

4

(c) AD2 · BC = AB · AE · BD

Compute BD.

10. Triangle ABC is inscribed in a circle ω. Let the bisector of angle A meet ω at D and BC at E. Letthe reflections of A across D and C be D′ and C ′, respectively. Suppose that ∠A = 60◦, AB = 3, andAE = 4. If the tangent to ω at A meets line BC at P , and the circumcircle of APD′ meets line BCat F (other than P ), compute FC ′.

HMMT 2013Saturday 16 February 2013

Geometry Test

PUT LABEL HERE

Name Team ID#

Organization Team

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Score:

HMMT 2013Saturday 16 February 2013

Geometry Test

1. Jarris the triangle is playing in the (x, y) plane. Let his maximum y coordinate be k. Given that hehas side lengths 6, 8, and 10 and that no part of him is below the x-axis, find the minimum possiblevalue of k.

Answer: 245 By playing around, we find that Jarris should have his hypotenuse flat on the x-

axis. The desired minimum value of k is then the length of the altitude to the hypotenuse. Thus, bycomputing the area of the triangle in two ways, 1

2 · 10 · k = 12 · 6 · 8 and so k = 24

5 .

2. Let ABCD be an isosceles trapezoid such that AD = BC, AB = 3, and CD = 8. Let E be a point inthe plane such that BC = EC and AE ⊥ EC. Compute AE.

Answer: 2√

6 Let r = BC = EC = AD. △ACE has right angle at E, so by the PythagoreanTheorem,

AE2 = AC2 − CE2 = AC2 − r2

Let the height of △ACD at A intersect DC at F . Once again, by the Pythagorean Theorem,

AC2 = FC2 + AF 2 =

(

8 − 3

2+ 3

)2

+ AD2 − DF 2 =

(

11

2

)2

+ r2 −(

5

2

)2

Plugging into the first equation,

AE2 =

(

11

2

)2

+ r2 −(

5

2

)2

− r2,

so AE = 2√

6.

3. Let A1A2A3A4A5A6 be a convex hexagon such that AiAi+2 ‖ Ai+3Ai+5 for i = 1, 2, 3 (we takeAi+6 = Ai for each i). Segment AiAi+2 intersects segment Ai+1Ai+3 at Bi, for 1 ≤ i ≤ 6, asshown. Furthermore, suppose that △A1A3A5

∼= △A4A6A2. Given that [A1B5B6] = 1, [A2B6B1] =4, and [A3B1B2] = 9 (by [XY Z] we mean the area of △XY Z), determine the area of hexagonB1B2B3B4B5B6.

A1

A2

A3

A4

A5

A6

B1

B2B3

B4

B5B6

(not to scale)

1

4

9

?

Answer: 22 Because B6A3B3A6 and B1A4B4A1 are parallelograms, B6A3 = A6B3 and A1B1 =A4B4. By the congruence of the large triangles A1A3A5 and A2A4A6, A1A3 = A4A6. Thus, B6A3 +A1B1 − A1A3 = A6B3 + A4B4 − A4A6, so B6B1 = B3B4. Similarly, opposite sides of hexagonB1B2B3B4B5B6 are equal, and implying that the triangles opposite each other on the outside of thishexagon are congruent.

Furthermore, by definition B5B6 ‖ A3A5, B3B4 ‖ A1A3, B6B1 ‖ A4A6 andB1B2 ‖ A1A5. Let the areaof triangle A1A3A5 and triangle A2A4A6 be k2. Then, by similar triangles,

Geometry Test

√

1

k2=

A1B6

A1A3√

4

k2=

B6B1

A4A6=

B1B6

A1A3√

9

k2=

A3B1

A1A3

Summing yields 6/k = 1, so k2 = 36. To finish, the area of B1B2B3B4B5B6 is equivalent to the areaof the triangle A1A3A5 minus the areas of the smaller triangles provided in the hypothesis. Thus, ouranswer is 36 − 1 − 4 − 9 = 22.

4. Let ω1 and ω2 be circles with centers O1 and O2, respectively, and radii r1 and r2, respectively. Supposethat O2 is on ω1. Let A be one of the intersections of ω1 and ω2, and B be one of the two intersectionsof line O1O2 with ω2. If AB = O1A, find all possible values of r1

r2

.

Answer: −1+√

52 , 1+

√5

2 There are two configurations to this problem, namely, B in between the

segment O1O2 and B on the ray O1O2 passing through the side of O2 Case 1: Let us only considerthe triangle ABO2. AB = AO1 = O1O2 = r1 because of the hypothesis and AO1 and O1O2 are radiiof w1. O2B = O2A = r2 because they are both radii of w2.

Then by the isosceles triangles, ∠AO1B = ∠ABO1 = ∠ABO2 = ∠O2AB. Thus can establish that△ABO1 ∼ △O2AB.

Thus,r2

r1=

r1

r2 − r1

r21 − r2

2 + r1r2 = 0

By straightforward quadratic equation computation and discarding the negative solution,

r1

r2=

−1 +√

5

2

Case 2: Similar to case 1, let us only consider the triangle ABO1. AB = AO1 = O1O2 = r1 becauseof the hypothesis and AO1 and O1O2 are radii of w1. O2B = O2A = r2 because they are both radii ofw2.

Then by the isosceles triangles, ∠AO1B = ∠ABO1 = ∠ABO2 = ∠O2AB. Thus can establish that△ABO1 ∼ △O2AB.

Now,r2

r1=

r1

r2 + r1

r21 − r2

2 − r1r2 = 0

By straightforward quadratic equation computation and discarding the negative solution,

r1

r2=

1 +√

5

2

5. In triangle ABC, ∠A = 45◦ and M is the midpoint of BC. AM intersects the circumcircle of ABCfor the second time at D, and AM = 2MD. Find cos ∠AOD, where O is the circumcenter of ABC.

Answer: − 18 ∠BAC = 45◦, so ∠BOC = 90◦. If the radius of the circumcircle is r, BC =

√2r,

and BM = CM =√

22 r. By power of a point, BM ·CM = AM ·DM , so AM = r and DM = 1

2r, andAD = 3

2r. Using the law of cosines on triangle AOD gives cos ∠AOD = − 18 .

Geometry Test

6. Let ABCD be a quadrilateral such that ∠ABC = ∠CDA = 90◦, and BC = 7. Let E and F be onBD such that AE and CF are perpendicular to BD. Suppose that BE = 3. Determine the productof the smallest and largest possible lengths of DF .

Answer: 9 By inscribed angles, ∠CDB = ∠CAB, and ∠ABD = ∠ACD. By definition,∠AEB = ∠CDA = ∠ABC = ∠CFA. Thus, △ABE ∼ △ADC and △CDF ∼ △CAB. This showsthat

BE

AB=

CD

CAand

DF

CD=

AB

BD

Based on the previous two equations, it is sufficient to conclude that 3 = EB = FD. Thus, FD mustequal to 3, and the product of its largest and smallest length is 9.

7. Let ABC be an obtuse triangle with circumcenter O such that ∠ABC = 15◦ and ∠BAC > 90◦.Suppose that AO meets BC at D, and that OD2 + OC · DC = OC2. Find ∠C.

Answer: 35 Let the radius of the circumcircle of △ABC be r.

OD2 + OC · CD = OC2

OC · CD = OC2 − OD2

OC · CD = (OC + OD)(OC − OD)

OC · CD = (r + OD)(r − OD)

By the power of the point at D,OC · CD = BD · DC

r = BD

Then, △OBD and △OAB and △AOC are isosceles triangles. Let ∠DOB = α. ∠BAO = 90 − α2 . In

△ABD, 15 + 90 − α2 = α. This means that α = 70. Furthermore, ∠ACB intercepts minor arc AB,

thus ∠ACB = ∠AOB2 = 70

2 = 35

8. Let ABCD be a convex quadrilateral. Extend line CD past D to meet line AB at P and extend lineCB past B to meet line AD at Q. Suppose that line AC bisects ∠BAD. If AD = 7

4 , AP = 212 , and

AB = 1411 , compute AQ.

Answer: 4213 We prove the more general statement 1

AB+ 1

AP= 1

AD+ 1

AQ, from which the answer

easily follows.

Denote ∠BAC = ∠CAD = γ, ∠BCA = α, ∠ACD = β. Then we have that by the law of sines,ACAB

+ ACAP

= sin(γ+α)sin(α) + sin(γ−β)

sin(β) = sin(γ−α)sin(α) + sin(γ+β)

sin(β) = ACAD

+ ACAQ

where we have simply used the sine

addition formula for the middle step.

Dividing the whole equation by AC gives the desired formula, from which we compute AQ = (1114 +

221 − 4

7 )−1 = 4213 .

9. Pentagon ABCDE is given with the following conditions:

(a) ∠CBD + ∠DAE = ∠BAD = 45◦, ∠BCD + ∠DEA = 300◦

(b) BADA

= 2√

23 , CD = 7

√5

3 , and DE = 15√

24

(c) AD2 · BC = AB · AE · BD

Compute BD.

Answer:√

39 As a preliminary, we may compute that by the law of cosines, the ratio ADBD

= 3√5.

Now, construct the point P in triangle ABD such that △APB ∼ △AED. Observe that APAD

=AE·ABAD·AD

= BCBD

(where we have used first the similarity and then condition 3). Furthermore, ∠CBD =∠DAB − ∠DAE = ∠DAB − ∠PAB = ∠PAD so by SAS, we have that △CBD ∼ △PAD.

Geometry Test

Therefore, by the similar triangles, we may compute PB = DE · ABAD

= 5 and PD = CD · ADBD

= 7.Furthermore, ∠BPD = 360 − ∠BPA − ∠DPA = 360 − ∠BCD − ∠DEA = 60 and therefore, by thelaw of cosines, we have that BD =

√39.

10. Triangle ABC is inscribed in a circle ω. Let the bisector of angle A meet ω at D and BC at E. Letthe reflections of A across D and C be D′ and C ′, respectively. Suppose that ∠A = 60◦, AB = 3, andAE = 4. If the tangent to ω at A meets line BC at P , and the circumcircle of APD′ meets line BCat F (other than P ), compute FC ′.

Answer: 2√

13 − 6√

3 First observe that by angle chasing, ∠PAE = 180 − 12∠BAC − ∠ABC =

∠AEP , so by the cyclic quadrilateral APD′F , ∠EFD′ = ∠PAE = ∠PEA = ∠D′EF . Thus, ED′Fis isosceles.

Define B′ to be the reflection of A about B, and observe that B′C ′||EF and B′D′C ′ is isosceles. Itfollows that B′EFC ′ is an isosceles trapezoid, so FC ′ = B′E, which by the law of cosines, is equal to√

AB′2 + AE2 − 2AB · AE cos 30 = 2√

13 − 6√

3.

Geometry Test

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT 2013, 16 FEBRUARY 2013 — GUTS ROUND

Organization Team Team ID#

1. [4] Arpon chooses a positive real number k. For each positive integer n, he places a marker at the point(n, nk) in the (x, y) plane. Suppose that two markers whose x coordinates differ by 4 have distance31. What is the distance between the markers at (7, 7k) and (19, 19k)?

2. [4] The real numbers x, y, z satisfy 0 ≤ x ≤ y ≤ z ≤ 4. If their squares form an arithmetic progressionwith common difference 2, determine the minimum possible value of |x〉 y|+ |y 〉 z|.

3. [4] Find the rightmost non-zero digit of the expansion of (20)(13!).

4. [4] Spencer is making burritos, each of which consists of one wrap and one filling. He has enoughfilling for up to four beef burritos and three chicken burritos. However, he only has five wraps for theburritos; in how many orders can he make exactly five burritos?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT 2013, 16 FEBRUARY 2013 — GUTS ROUND

Organization Team Team ID#

5. [5] Rahul has ten cards face-down, which consist of five distinct pairs of matching cards. During eachmove of his game, Rahul chooses one card to turn face-up, looks at it, and then chooses another toturn face-up and looks at it. If the two face-up cards match, the game ends. If not, Rahul flips bothcards face-down and keeps repeating this process. Initially, Rahul doesn’t know which cards are which.Assuming that he has perfect memory, find the smallest number of moves after which he can guaranteethat the game has ended.

6. [5] Let R be the region in the Cartesian plane of points (x, y) satisfying x ≥ 0, y ≥ 0, and x + y +bxc+ byc ≤ 5. Determine the area of R.

7. [5] Find the number of positive divisors d of 15! = 15 · 14 · · · · · 2 · 1 such that gcd(d, 60) = 5.

8. [5] In a game, there are three indistinguishable boxes; one box contains two red balls, one contains twoblue balls, and the last contains one ball of each color. To play, Raj first predicts whether he will drawtwo balls of the same color or two of different colors. Then, he picks a box, draws a ball at random,looks at the color, and replaces the ball in the same box. Finally, he repeats this; however, the boxesare not shuffled between draws, so he can determine whether he wants to draw again from the samebox. Raj wins if he predicts correctly; if he plays optimally, what is the probability that he will win?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT 2013, 16 FEBRUARY 2013 — GUTS ROUND

Organization Team Team ID#

9. [6] I have 8 unit cubes of different colors, which I want to glue together into a 2 × 2 × 2 cube. Howmany distinct 2× 2× 2 cubes can I make? Rotations of the same cube are not considered distinct, butreflections are.

10. [6] Wesyu is a farmer, and she’s building a cao (a relative of the cow) pasture. She starts with atriangle A0A1A2 where angle A0 is 90◦, angle A1 is 60◦, and A0A1 is 1. She then extends the pasture.First, she extends A2A0 to A3 such that A3A0 = 1

2A2A0 and the new pasture is triangle A1A2A3.Next, she extends A3A1 to A4 such that A4A1 = 1

6A3A1. She continues, each time extending AnAn→2to An+1 such that An+1An→2 = 1

2n→2AnAn→2. What is the smallest K such that her pasture neverexceeds an area of K?

11. [6] Compute the prime factorization of 1007021035035021007001. (You should write your answer in theform pe11 pe22 . . . pekk , where p1, . . . , pk are distinct prime numbers and e1, . . . , ek are positive integers.)

12. [6] For how many integers 1 ≤ k ≤ 2013 does the decimal representation of kk end with a 1?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT 2013, 16 FEBRUARY 2013 — GUTS ROUND

Organization Team Team ID#

13. [8] Find the smallest positive integer n such that 5n+1+2n+1

5n+2n > 4.99.

14. [8] Consider triangle ABC with ∠A = 2∠B. The angle bisectors from A and C intersect at D, andthe angle bisector from C intersects AB at E. If DE

DC = 13 , compute AB

AC .

15. [8] Tim and Allen are playing a match of tenus. In a match of tenus, the two players play a seriesof games, each of which is won by one of the two players. The match ends when one player has wonexactly two more games than the other player, at which point the player who has won more gameswins the match. In odd-numbered games, Tim wins with probability 3/4, and in the even-numberedgames, Allen wins with probability 3/4. What is the expected number of games in a match?

16. [8] The walls of a room are in the shape of a triangle ABC with ∠ABC = 90◦, ∠BAC = 60◦, andAB = 6. Chong stands at the midpoint of BC and rolls a ball toward AB. Suppose that the ballbounces off AB, then AC, then returns exactly to Chong. Find the length of the path of the ball.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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HMMT 2013, 16 FEBRUARY 2013 — GUTS ROUND

Organization Team Team ID#

17. [11] The lines y = x, y = 2x, and y = 3x are the three medians of a triangle with perimeter 1. Findthe length of the longest side of the triangle.

18. [11] Define the sequence of positive integers {an} as follows. Let a1 = 1, a2 = 3, and for each n > 2,let an be the result of expressing an→1 in base n〉1, then reading the resulting numeral in base n, thenadding 2 (in base n). For example, a2 = 310 = 112, so a3 = 113 + 23 = 610. Express a2013 in base ten.

19. [11] An isosceles trapezoid ABCD with bases AB and CD has AB = 13, CD = 17, and height 3. LetE be the intersection of AC and BD. Circles Ω and ω are circumscribed about triangles ABE andCDE. Compute the sum of the radii of Ω and ω.

20. [11] The polynomial f(x) = x3 〉 3x2 〉 4x + 4 has three real roots r1, r2, and r3. Let g(x) =x3 + ax2 + bx + c be the polynomial which has roots s1, s2, and s3, where s1 = r1 + r2z + r3z

2,

s2 = r1z + r2z2 + r3, s3 = r1z

2 + r2 + r3z, and z = →1+i√3

2 . Find the real part of the sum of thecoefficients of g(x).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT 2013, 16 FEBRUARY 2013 — GUTS ROUND

Organization Team Team ID#

21. [14] Find the number of positive integers j ≤ 32013 such that

j =

m∑k=0

((〉1)

k · 3ak

)for some strictly increasing sequence of nonnegative integers {ak}. For example, we may write 3 = 31

and 55 = 30 〉 33 + 34, but 4 cannot be written in this form.

22. [14] Sherry and Val are playing a game. Sherry has a deck containing 2011 red cards and 2012 blackcards, shuffled randomly. Sherry flips these cards over one at a time, and before she flips each cardover, Val guesses whether it is red or black. If Val guesses correctly, she wins 1 dollar; otherwise, sheloses 1 dollar. In addition, Val must guess red exactly 2011 times. If Val plays optimally, what is herexpected profit from this game?

23. [14] Let ABCD be a parallelogram with AB = 8, AD = 11, and ∠BAD = 60◦. Let X be on segmentCD with CX/XD = 1/3 and Y be on segment AD with AY/Y D = 1/2. Let Z be on segment ABsuch that AX, BY , and DZ are concurrent. Determine the area of triangle XY Z.

24. [14] Given a point p and a line segment l, let d(p, l) be the distance between them. Let A, B, and Cbe points in the plane such that AB = 6, BC = 8, AC = 10. What is the area of the region in the(x, y)-plane formed by the ordered pairs (x, y) such that there exists a point P inside triangle ABCwith d(P,AB) + x = d(P,BC) + y = d(P,AC)?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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HMMT 2013, 16 FEBRUARY 2013 — GUTS ROUND

Organization Team Team ID#

25. [17] The sequence (zn) of complex numbers satisfies the following properties:

• z1 and z2 are not real.

• zn+2 = z2n+1zn for all integers n ≥ 1.

• zn+3

z2nis real for all integers n ≥ 1.

•∣∣∣∣z3z4∣∣∣∣ =

∣∣∣∣z4z5∣∣∣∣ = 2.

Find the product of all possible values of z1.

26. [17] Triangle ABC has perimeter 1. Its three altitudes form the side lengths of a triangle. Find theset of all possible values of min(AB,BC,CA).

27. [17] Let W be the hypercube {(x1, x2, x3, x4) | 0 ≤ x1, x2, x3, x4 ≤ 1}. The intersection of W and ahyperplane parallel to x1 + x2 + x3 + x4 = 0 is a non-degenerate 3-dimensional polyhedron. What isthe maximum number of faces of this polyhedron?

28. [17] Let z0 +z1 +z2 + · · · be an infinite complex geometric series such that z0 = 1 and z2013 = 120132013 .

Find the sum of all possible sums of this series.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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HMMT 2013, 16 FEBRUARY 2013 — GUTS ROUND

Organization Team Team ID#

29. [20] Let A1, A2, ..., Am be finite sets of size 2012 and let B1, B2, ..., Bm be finite sets of size 2013 suchthat Ai ∩Bj = ∅ if and only if i = j. Find the maximum value of m.

30. [20] How many positive integers k are there such that

k

2013(a + b) = lcm(a, b)

has a solution in positive integers (a, b)?

31. [20] Let ABCD be a quadrilateral inscribed in a unit circle with center O. Suppose that ∠AOB =∠COD = 135◦, BC = 1. Let B′ and C ′ be the reflections of A across BO and CO respectively. LetH1 and H2 be the orthocenters of AB′C ′ and BCD, respectively. If M is the midpoint of OH1, andO′ is the reflection of O about the midpoint of MH2, compute OO′.

32. [20] For an even positive integer n Kevin has a tape of length 4n with marks at 〉2n,〉2n+1, . . . , 2n〉1, 2n. He then randomly picks n points in the set 〉n,〉n + 1,〉n + 2, . . . , n〉 1, n, and places a stoneon each of these points. We call a stone ‘stuck’ if it is on 2n or 〉2n, or either all the points to theright, or all the points to the left, all contain stones. Then, every minute, Kevin shifts the unstuckstones in the following manner:

• He picks an unstuck stone uniformly at random and then flips a fair coin.

• If the coin came up heads, he then moves that stone and every stone in the largest contiguous setcontaining that stone one point to the left. If the coin came up tails, he moves every stone in that setone point right instead.

• He repeats until all the stones are stuck.

Let pk be the probability that at the end of the process there are exactly k stones in the right half.Evaluate

pn→1 〉 pn→2 + pn→3 〉 . . . + p3 〉 p2 + p1pn→1 + pn→2 + pn→3 + . . . + p3 + p2 + p1

in terms of n.

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HMMT 2013, 16 FEBRUARY 2013 — GUTS ROUND

Organization Team Team ID#

33. [25] Compute the value of 125 + 224 + 323 + . . .+ 242 + 251. If your answer is A and the correct answer

is C, then your score on this problem will be⌊25 min

((AC

)2,(CA

)2)⌋.

34. [25] For how many unordered sets {a, b, c, d} of positive integers, none of which exceed 168, do thereexist integers w, x, y, z such that (〉1)wa + (〉1)xb + (〉1)yc + (〉1)zd = 168? If your answer is A and

the correct answer is C, then your score on this problem will be b25e→3|C−A|

C c.

35. [25] Let P be the number to partition 2013 into an ordered tuple of prime numbers? What islog2(P )? If your answer is A and the correct answer is C, then your score on this problem will be⌊1252

(min

(CA , A

C

)〉 3

5

)⌋or zero, whichever is larger.

36. [24] (Mathematicians A to Z) Below are the names of 26 mathematicians, one for each letter ofthe alphabet. Your answer to this question should be a subset of {A,B, · · · , Z}, where each letterrepresents the corresponding mathematician. If two mathematicians in your subset have birthdatesthat are within 20 years of each other, then your score is 0. Otherwise, your score is max(3(k 〉 3), 0)where k is the number of elements in your subset.

Niels Abel Isaac Netwon

Étienne Bézout Nicole OresmeAugustin-Louis Cauchy Blaise PascalRené Descartes Daniel QuillenLeonhard Euler Bernhard RiemannPierre Fatou Jean-Pierre SerreAlexander Grothendieck Alan TuringDavid Hilbert Stanislaw UlamKenkichi Iwasawa John VennCarl Jacobi Andrew WilesAndrey Kolmogorov Leonardo XimenesJoseph-Louis Lagrange Shing-Tung YauJohn Milnor Ernst Zermelo

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HMMT 2013Saturday 16 February 2013

Guts Round

1. [4] Arpon chooses a positive real number k. For each positive integer n, he places a marker at the point(n, nk) in the (x, y) plane. Suppose that two markers whose x coordinates differ by 4 have distance31. What is the distance between the markers at (7, 7k) and (19, 19k)?

Answer: 93 The difference of the x-coordinates of the markers is 12 = 3 · 4. Thus, by similartriangles (where we draw right triangles whose legs are parallel to the axes and whose hypotenuses liealong the line y = kx), the distance between the markers is 3 · 31 = 93.

2. [4] The real numbers x, y, z satisfy 0 ≤ x ≤ y ≤ z ≤ 4. If their squares form an arithmetic progressionwith common difference 2, determine the minimum possible value of |x − y| + |y − z|.

Answer: 4 − 2√

3 Clearly |x−y|+ |y− z| = z−x = z2−x2

z+x= 4

z+x, which is minimized when z = 4

and x =√

12. Thus, our answer is 4 −√

12 = 4 − 2√

3.

3. [4] Find the rightmost non-zero digit of the expansion of (20)(13!).

Answer: 6 We can rewrite this as (10 ∗ 2)(13 ∗ 12 ∗ 11 ∗ 10 ∗ 9 ∗ 8 ∗ 7 ∗ 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1) =(103)(2 ∗ 13 ∗ 12 ∗ 11 ∗ 9 ∗ 8 ∗ 7 ∗ 6 ∗ 4 ∗ 3); multiplying together the units digits for the terms not equalto 10 reveals that the rightmost non-zero digit is 6.

4. [4] Spencer is making burritos, each of which consists of one wrap and one filling. He has enoughfilling for up to four beef burritos and three chicken burritos. However, he only has five wraps for theburritos; in how many orders can he make exactly five burritos?

Answer: 25 Spencer’s burrito-making can include either 3, 2, or 1 chicken burrito; consequently,

he has(

53

)

+(

52

)

+(

51

)

= 25 orders in which he can make burritos.

5. [5] Rahul has ten cards face-down, which consist of five distinct pairs of matching cards. During eachmove of his game, Rahul chooses one card to turn face-up, looks at it, and then chooses another toturn face-up and looks at it. If the two face-up cards match, the game ends. If not, Rahul flips bothcards face-down and keeps repeating this process. Initially, Rahul doesn’t know which cards are which.Assuming that he has perfect memory, find the smallest number of moves after which he can guaranteethat the game has ended.

Answer: 4 Label the 10 cards a1, a2, ..., a5, b1, b2, ..., b5 such that ai and bi match for 1 ≤ i ≤ 5.

First, we’ll show that Rahul cannot always end the game in less than 4 moves, in particular, whenhe turns up his fifth card (during the third move), it is possible that the card he flips over is not onewhich he has yet encountered; consequently, he will not guarantee being able to match it, so he cannotguarantee that the game can end in three moves.

However, Rahul can always end the game in 4 moves. To do this, he can turn over 6 distinct cards inhis first 3 moves. If we consider the 5 sets of cards {a1, b1}, {a2, b2}, {a3, b3}, {a4, b4}, {a5, b5}, then bythe pigeonhole principle, at least 2 of the 6 revealed cards must be from the same set. Rahul can thenturn over those 2 cards on the fourth move, ending the game.

6. [5] Let R be the region in the Cartesian plane of points (x, y) satisfying x ≥ 0, y ≥ 0, and x + y +⌊x⌋ + ⌊y⌋ ≤ 5. Determine the area of R.

Answer: 92 We claim that a point in the first quadrant satisfies the desired property if the point

is below the line x + y = 3 and does not satisfy the desired property if it is above the line.

To see this, for a point inside the region, x+ y < 3 and ⌊x⌋+ ⌊y⌋ ≤ x+ y < 3 However, ⌊x⌋+ ⌊y⌋ mustequal to an integer. Thus, ⌊x⌋ + ⌊y⌋ ≤ 2. Adding these two equations, x + y + ⌊x⌋ + ⌊y⌋ < 5, whichsatisfies the desired property. Conversely, for a point outside the region, ⌊x⌋+⌊y⌋+{x}+{y} = x+y > 3However, {x} + {y} < 2. Thus, ⌊x⌋ + ⌊y⌋ > 1, so ⌊x⌋ + ⌊y⌋ ≥ 2, implying that x + y + ⌊x⌋ + ⌊y⌋ > 5.

Guts Round

To finish, R is the region bounded by the x-axis, the y-axis, and the line x + y = 3 is a right trianglewhose legs have length 3. Consequently, R has area 9

2 .

7. [5] Find the number of positive divisors d of 15! = 15 · 14 · · · · · 2 · 1 such that gcd(d, 60) = 5.

Answer: 36 Since gcd(d, 60) = 5, we know that d = 5id′ for some integer i > 0 and some integerd′ which is relatively prime to 60. Consequently, d′ is a divisor of (15!)/5; eliminating common factorswith 60 gives that d′ is a factor of (72)(11)(13), which has (2 + 1)(1 + 1)(1 + 1) = 12 factors. Finally,i can be 1, 2, or 3, so there are a total of 3 · 12 = 36 possibilities.

8. [5] In a game, there are three indistinguishable boxes; one box contains two red balls, one contains twoblue balls, and the last contains one ball of each color. To play, Raj first predicts whether he will drawtwo balls of the same color or two of different colors. Then, he picks a box, draws a ball at random,looks at the color, and replaces the ball in the same box. Finally, he repeats this; however, the boxesare not shuffled between draws, so he can determine whether he wants to draw again from the samebox. Raj wins if he predicts correctly; if he plays optimally, what is the probability that he will win?

Answer: 56 Call the box with two red balls box 1, the box with one of each color box 2, and the

box with two blue balls box 3. Without loss of generality, assume that the first ball that Bob draws isred. If Bob picked box 1, then he would have picked a red ball with probability 1, and if Bob pickedbox 2, then he would have picked a red ball with probability 1

2 . Therefore, the probability that hepicked box 1 is 1

1+ 12

= 23 , and the probability that he picked box 2 is 1

3 . We will now consider both

possible predictions and find which one gives a better probability of winning, assuming optimal play.

If Bob predicts that he will draw two balls of the same color, then there are two possible plays: hedraws from the same box, or he draws from a different box. If he draws from the same box, then in the23 chance that he originally picked box 1, he will always win, and in the 1

3 chance that he picked box2, he will win with probability 1

2 , for a total probability of 23 + 1

3 · 12 = 5

6 . If he draws from a differentbox, then if he originally picked box 1, he will win with probability 1

4 and if he originally picked box2, he will win with probability 1

2 , for a total probability of 23 · 1

4 + 13 · 1

2 = 13 .

If Bob predicts that he will draw two balls of different colors, then we can consider the same two possibleplays. Using similar calculations, if he draws from the same box, then he will win with probability 1

6 ,and if he draws from a different box, then he will win with probability 2

3 . Looking at all cases, Bob’sbest play is to predict that he will draw two balls of the same color and then draw the second ball fromthe same box, with a winning probability of 5

6 .

9. [6] I have 8 unit cubes of different colors, which I want to glue together into a 2 × 2 × 2 cube. Howmany distinct 2× 2× 2 cubes can I make? Rotations of the same cube are not considered distinct, butreflections are.

Answer: 1680 Our goal is to first pin down the cube, so it can’t rotate. Without loss of generality,suppose one of the unit cubes is purple, and let the purple cube be in the top left front position. Now,look at the three positions that share a face with the purple cube. There are

(

73

)

ways to pick the threecubes that fill those positions and two ways to position them that are rotationally distinct. Now, we’vetaken care of any possible rotations, so there are simply 4! ways to position the final four cubes. Thus,our answer is

(

73

)

· 2 · 4! = 1680 ways.

10. [6] Wesyu is a farmer, and she’s building a cao (a relative of the cow) pasture. She starts with atriangle A0A1A2 where angle A0 is 90◦, angle A1 is 60◦, and A0A1 is 1. She then extends the pasture.First, she extends A2A0 to A3 such that A3A0 = 1

2A2A0 and the new pasture is triangle A1A2A3.Next, she extends A3A1 to A4 such that A4A1 = 1

6A3A1. She continues, each time extending AnAn−2

to An+1 such that An+1An−2 = 12n−2AnAn−2. What is the smallest K such that her pasture never

exceeds an area of K?

Answer:√

3 First, note that for any i, after performing the operation on triangle AiAi+1Ai+2,the resulting pasture is triangle Ai+1Ai+2Ai+3. Let Ki be the area of triangle AiAi+1Ai+2. From

Guts Round

An+1An−2 = 12n−2AnAn−2 and AnAn+1 = AnAn−2+An−2An+1, we have AnAn+1 = (1+ 1

2n−2 )AnAn−2.We also know that the area of a triangle is half the product of its base and height, so if we let the base oftriangle An−2An−1An be AnAn−2, its area is Kn−2 = 1

2hAnAn−2. The area of triangle An−1AnAn+1

is Kn−1 = 12hAnAn+1. The h’s are equal because the distance from An−1 to the base does not change.

We now have Kn−1

Kn−2= AnAn+1

AnAn−2= 1 + 1

2n−2 = 2n−12n−2 . Therefore, K1

K0= 3

2 , K2

K0= K2

K1

K1

K0= 7

6 · 32 = 7

4 ,K3

K0= K3

K2

K2

K0= 15

14 · 74 = 15

8 . We see the pattern Kn

K0= 2n+1−1

2n , which can be easily proven by induction.

As n approaches infinity, Kn

K0grows arbitrarily close to 2, so the smallest K such that the pasture never

exceeds an area of K is 2K0 =√

3.

11. [6] Compute the prime factorization of 1007021035035021007001. (You should write your answer in theform pe1

1 pe2

2 . . . pek

k , where p1, . . . , pk are distinct prime numbers and e1, . . . , ek are positive integers.)

Answer: 77 · 117 · 137 The number in question is

7∑

i=0

(

7

i

)

1000i = (1000 + 1)7 = 10017 = 77 · 117 · 137.

12. [6] For how many integers 1 ≤ k ≤ 2013 does the decimal representation of kk end with a 1?

Answer: 202 We claim that this is only possible if k has a units digit of 1. Clearly, it is true inthese cases. Additionally, kk cannot have a units digit of 1 when k has a units digit of 2, 4, 5, 6, or 8. Ifk has a units digit of 3 or 7, then kk has a units digit of 1 if and only if 4|k, a contradiction. Similarly,if k has a units digit of 9, then kk has a units digit of 1 if and only if 2|k, also a contradiction. Sincethere are 202 integers between 1 and 2013, inclusive, with a units digit of 1, there are 202 such k whichfulfill our criterion.

13. [8] Find the smallest positive integer n such that 5n+1+2n+1

5n+2n > 4.99.

Answer: 7 Writing 5n+1 = 5 ·5n and 2n+1 = 2 ·2n and cross-multiplying yields 0.01 ·5n > 2.99 ·2n,and re-arranging yields (2.5)n > 299. A straightforward calculation shows that the smallest n for whichthis is true is n = 7.

14. [8] Consider triangle ABC with ∠A = 2∠B. The angle bisectors from A and C intersect at D, andthe angle bisector from C intersects AB at E. If DE

DC= 1

3 , compute ABAC

.

Answer: 79 Let AE = x and BE = y. Using angle-bisector theorem on △ACE we have

x : DE = AC : DC, so AC = 3x. Using some angle chasing, it is simple to see that ∠ADE = ∠AED,so AD = AE = x. Then, note that △CDA ∼ △CEB, so y : (DC+DE) = x : DC, so y : x = 1+ 1

3 = 43 ,

so AB = x + 43x = 7

3x. Thus the desired answer is AB : AC = 73x : 3x = 7

9 .

15. [8] Tim and Allen are playing a match of tenus. In a match of tenus, the two players play a seriesof games, each of which is won by one of the two players. The match ends when one player has wonexactly two more games than the other player, at which point the player who has won more gameswins the match. In odd-numbered games, Tim wins with probability 3/4, and in the even-numberedgames, Allen wins with probability 3/4. What is the expected number of games in a match?

Answer: 163 Let the answer be E. If Tim wins the first game and Allen wins the second game

or vice versa, which occurs with probability (3/4)2 + (1/4)2 = 5/8, the expected number of additionalgames is just E, so the expected total number of games is E +2. If, on the other hand, one of Tim andAllen wins both of the first two games, with probability 1 − (5/8) = 3/8, there are exactly 2 games inthe match. It follows that

E =3

8· 2 +

5

8· (E + 2),

and solving gives E = 163 .

Guts Round

16. [8] The walls of a room are in the shape of a triangle ABC with ∠ABC = 90◦, ∠BAC = 60◦, andAB = 6. Chong stands at the midpoint of BC and rolls a ball toward AB. Suppose that the ballbounces off AB, then AC, then returns exactly to Chong. Find the length of the path of the ball.

Answer: 3√

21 Let C ′ be the reflection of C across AB and B′ be the reflection of B across

AC ′; note that B′, A,C are collinear by angle chasing. The image of the path under these reflectionsis just the line segment MM ′, where M is the midpoint of BC and M ′ is the midpoint of B′C ′, soour answer is just the length of MM ′. Applying the Law of Cosines to triangle M ′C ′M , we haveMM ′2 = 27 + 243 − 2 · 3

√3 · 9

√3 · 1

2 = 189, so MM ′ = 3√

21.

17. [11] The lines y = x, y = 2x, and y = 3x are the three medians of a triangle with perimeter 1. Findthe length of the longest side of the triangle.

Answer:√

582+

√34+

√58

The three medians of a triangle contain its vertices, so the three vertices of

the triangle are (a, a), (b, 2b) and (c, 3c) for some a, b, and c. Then, the midpoint of (a, a) and (b, 2b),which is (a+b

2 , a+2b2 ), must lie along the line y = 3x. Therefore,

a + 2b

2= 3 · a + b

2,

a + 2b = 3a + 3b,

−2a = b.

Similarly, the midpoint of (b, 2b) and (c, 3c), which is ( b+c2 , 2b+3c

2 ), must lie along the line y = x.Therefore,

2b + 3c

2=

b + c

2,

2b + 3c = b + c,

b = −2c,

c = −1

2b = a.

From this, three points can be represented as (a, a), (−2a,−4a), and (a, 3a). Using the distanceformula, the three side lengths of the triangle are 2|a|,

√34|a|, and

√58|a|. Since the perimeter of the

triangle is 1, we find that |a| = 12+

√34+

√58

and therefore the longest side length is√

582+

√34+

√58

.

18. [11] Define the sequence of positive integers {an} as follows. Let a1 = 1, a2 = 3, and for each n > 2,let an be the result of expressing an−1 in base n − 1, then reading the resulting numeral in base n,then adding 2 (in base n). For example, a2 = 310 = 112, so a3 = 113 + 23 = 610. Express a2013 in baseten.

Answer: 23097 We claim that for nonnegative integers m and for 0 ≤ n < 3 · 2m, a3·2m+n =(3 · 2m + n)(m + 2) + 2n. We will prove this by induction; the base case for a3 = 6 (when m = 0,n = 0) is given in the problem statement. Now, suppose that this is true for some pair m and n. Wewill divide this into two cases:

• Case 1: n < 3 · 2m − 1. Then, we want to prove that this is true for m and n + 1. In particular,writing a3·2m+n in base 3 · 2m + n results in the digits m + 2 and 2n. Consequently, reading it inbase 3·2m+n+1 gives a3·2m+n+1 = 2+(3·2m+n+1)(m+2)+2n = (2·2m+n+1)(m+2)+2(n+1),as desired.

• Case 2: n = 3 · 2m − 1. Then, we want to prove that this is true for m + 1 and 0. Similarlyto the previous case, we get that a3·2m+n+1 = a3·2m+1 = 2 + (3 · 2m + n + 1)(m + 2) + 2n =2 + (3 · 2m+1)(m + 2) + 2(3 · 2m − 1) = (3 · 2m+1 + 0)((m + 1) + 2) + 2(0), as desired.

In both cases, we have proved our claim.

Guts Round

19. [11] An isosceles trapezoid ABCD with bases AB and CD has AB = 13, CD = 17, and height 3. LetE be the intersection of AC and BD. Circles Ω and ω are circumscribed about triangles ABE andCDE. Compute the sum of the radii of Ω and ω.

Answer: 39 Let Ω have center O and radius R and let ω have center P and radius M . Let Qbe the intersection of AB and OE. Note that OE is the perpendicular bisector of AB because thetrapezoid is isosceles. Also, we see OE is the circumradius of Ω. On the other hand, we know bysimilarity of △AEB and △CED that QE = 13

13+17 · 3 = 1330 · 3. And, because BQ = 13/2 and is

perpendicular to OQ, we can apply the Pythagorean theorem to △OQB to see OQ =

√

R2 −(

132

)2.

Since OE = OQ + QE, R = 1330 · 3 +

√

R2 −(

132

)2. Solving this equation for R yields R = 13

30 · 39.

Since by similarity M = 1713R, we know R + M = 30

13R, so R + M = 39.

20. [11] The polynomial f(x) = x3 − 3x2 − 4x + 4 has three real roots r1, r2, and r3. Let g(x) =x3 + ax2 + bx + c be the polynomial which has roots s1, s2, and s3, where s1 = r1 + r2z + r3z

2,

s2 = r1z + r2z2 + r3, s3 = r1z

2 + r2 + r3z, and z = −1+i√

32 . Find the real part of the sum of the

coefficients of g(x).

Answer: −26 Note that z = e2π

3i = cos 2π

3 + i sin 2π3 , so that z3 = 1 and z2 + z + 1 = 0. Also,

s2 = s1z and s3 = s1z2.

Then, the sum of the coefficients of g(x) is g(1) = (1−s1)(1−s2)(1−s3) = (1−s1)(1−s1z)(1−s1z2) =

1 − (1 + z + z2)s1 + (z + z2 + z3)s21 − z3s3

1 = 1 − s31.

Meanwhile, s31 = (r1 + r2z + r3z

2)3 = r31 + r3

2 + r33 + 3r2

1r2z + 3r21r3z

2 + 3r22r3z + 3r2

2r1z2 + 3r2

3r1z +3r2

3r2z2 + 6r1r2r3.

Since the real parts of both z and z2 are − 12 , and since all of r1, r2, and r3 are real, the real part of s3

1 isr31+r3

2+r33− 3

2 (r21r2+· · ·+r2

3r2)+6r1r2r3 = (r1+r2+r3)3− 9

2 (r1+r2+r3)(r1r2+r2r3+r3r1)+272 r1r2r3 =

33 − 92 · 3 · −4 + 27

2 · −4 = 27.

Therefore, the answer is 1 − 27 = −26.

21. [14] Find the number of positive integers j ≤ 32013 such that

j =m

∑

k=0

(

(−1)k · 3ak

)

for some strictly increasing sequence of nonnegative integers {ak}. For example, we may write 3 = 31

and 55 = 30 − 33 + 34, but 4 cannot be written in this form.

Answer: 22013 Clearly m must be even, or the sum would be negative. Furthermore, if am ≤ 2013,

the sum cannot exceed 32013 since j = 3am +∑m−1

k=0

(

(−1)k · 3ak

)

≤ 3am . Likewise, if am > 2013, then

the sum necessarily exceeds 32013, which is not hard to see by applying the Triangle Inequality andsumming a geometric series. Hence, the elements of {ak} can be any subset of {0, 1, . . . , 2013} with anodd number of elements. Since the number of even-sized subsets is equal to the number of odd-sized

elements, there are 22014

2 = 22013 such subsets.

Now, it suffices to show that given such an {ak}, the value of j can only be obtained in this way. Supposefor the the sake of contradiction that there exist two such sequences {ak}0≤k≤ma

and {bk}0≤k≤mb

which produce the same value of j for j positive or negative, where we choose {ak}, {bk} such thatmin(ma,mb) is as small as possible. Then, we note that since 3a0 + 3a1 + . . . + 3(ama

−1) ≤ 30 + 31 +

. . . + 3(ama−1) < 2(3(ama−1)), we have that

∑ma

k=0

(

(−1)k · 3ak

)

> 3(ama−1). Similarly, we get that

3(amb−1) ≥ ∑mb

k=0

(

(−1)k · 3ak

)

> 3(mb−1); for the two to be equal, we must have ma = mb. However,

this means that the sequences obtained by removing amaand amb

from {ak} {bk} have smaller maximumvalue but still produce the same alternating sum, contradicting our original assumption.

Guts Round

22. [14] Sherry and Val are playing a game. Sherry has a deck containing 2011 red cards and 2012 blackcards, shuffled randomly. Sherry flips these cards over one at a time, and before she flips each cardover, Val guesses whether it is red or black. If Val guesses correctly, she wins 1 dollar; otherwise, sheloses 1 dollar. In addition, Val must guess red exactly 2011 times. If Val plays optimally, what is herexpected profit from this game?

Answer: 14023 We will prove by induction on r + b that the expected profit for guessing if there

are r red cards, b black cards, and where g guesses must be red, is equal to (b − r) + 2(r−b)(r+b) g. It is

not difficult to check that this holds in the cases (r, b, g) = (1, 0, 0), (0, 1, 0), (1, 0, 1), (0, 1, 1). Then,suppose that this is true as long as the number of cards is strictly less than r + b; we will prove thatit also holds true when there are r red and b blue cards.

Let f(r, b, g) be her expected profit under these conditions. If she guesses red, her expected profit is

r

r + b(1 + f(r − 1, b, g − 1)) +

b

r + b(−1 + f(r, b − 1, g − 1)) = (b − r) +

2(r − b)

(r + b)g.

Similarly, if she guesses black, her expected profit is

r

r + b(−1 + f(r − 1, b, g)) +

b

r + b(1 + f(r, b − 1, g)) = (b − r) +

2(r − b)

(r + b)g.

Plugging in the our starting values gives an expected profit of 14023 .

23. [14] Let ABCD be a parallelogram with AB = 8, AD = 11, and ∠BAD = 60◦. Let X be on segmentCD with CX/XD = 1/3 and Y be on segment AD with AY/Y D = 1/2. Let Z be on segment ABsuch that AX, BY , and DZ are concurrent. Determine the area of triangle XY Z.

Answer: 19√

32 Let AX and BD meet at P . We have DP/PB = DX/AB = 3/4. Now, applying

Ceva’s Theorem in triangle ABD, we see that

AZ

ZB=

DP

PB· AY

Y D=

3

4· 1

2=

3

8.

Now,[AY Z]

[ABCD]=

[AY Z]

2[ABD]=

1

2· 1

3· 3

11=

1

22,

and similarly[DY X]

[ABCD]=

1

2· 2

3· 3

4=

1

4.

Also,[XCBZ]

[ABCD]=

1

2

(

1

4+

8

11

)

=43

88.

The area of XY Z is the rest of the fraction of the area of ABCD not covered by the three abovepolygons, which by a straightforward calculation 19/88 the area of ABCD, so our answer is

8 · 11 · sin 60◦ · 19

88=

19√

3

2.

24. [14] Given a point p and a line segment l, let d(p, l) be the distance between them. Let A, B, and Cbe points in the plane such that AB = 6, BC = 8, AC = 10. What is the area of the region in the(x, y)-plane formed by the ordered pairs (x, y) such that there exists a point P inside triangle ABCwith d(P,AB) + x = d(P,BC) + y = d(P,AC)?

Answer: 2885 . Place ABC in the coordinate plane so that A = (0, 6), B = (0, 0), C = (8, 0).

Consider a point P = (a, b) inside triangle ABC. Clearly, d(P,AB) = a, d(P,BC) = b. Now, we see

Guts Round

that the area of triangle ABC is 6·82 = 24, but may also be computed by summing the areas of triangles

PAB,PBC,PCA. The area of triangle PAB is 6·a2 = 3a, and similarly the area of triangle PBC is

4b. Thus, it follows easily that d(P,CA) = 24−3a−4b5 . Now, we have

(x, y) =

(

24

5− 8

5a − 4

bb,

24

5− 3

5a − 9

5b

)

.

The desired region is the set of (x, y) obtained by those (a, b) subject to the constraints a ≥ 0, b ≥0, 6a + 8b ≤ 48.

Consequently, our region is the triangle whose vertices are obtained by evaluating (x, y) at the vertices(a, b) of the triangle. To see this, let f(a, b) output the corresponding (x, y) according to the above.Then, we can write every point P in ABC as P = m(0, 0) + n(0, 6) + p(8, 0) for some m + n + p = 1.Then, f(P ) = mf(0, 0) + nf(0, 6) + pf(8, 0) = m( 24

5 , 245 ) + n(−8, 0) + p(0,−6), so f(P ) ranges over

the triangle with those three vertices.

Therefore, we need the area of the triangle with vertices(

245 , 24

5

)

, (0,−6), (−8, 0), which is easilycomputed (for example, using determinants) to be 288

5 .

25. [17] The sequence (zn) of complex numbers satisfies the following properties:

• z1 and z2 are not real.

• zn+2 = z2n+1zn for all integers n ≥ 1.

• zn+3

z2n

is real for all integers n ≥ 1.

•∣

∣

∣

∣

z3

z4

∣

∣

∣

∣

=

∣

∣

∣

∣

z4

z5

∣

∣

∣

∣

= 2.

Find the product of all possible values of z1.

Answer: 65536 All complex numbers can be expressed as r(cos θ + i sin θ) = reiθ. Let zn berneiθn .

zn+3

z2n

=z2n+2zn+1

z2n

=z5n+1z

2n

z2n

= z5n+1 is real for all n ≥ 1, so θn =

πkn

5for all n ≥ 2, where kn is an

integer. θ1 + 2θ2 = θ3, so we may write θ1 =πk1

5with k1 an integer.

r3

r4=

r4

r5⇒ r5 =

r24

r3= r2

4r3, so r3 = 1.r3

r4= 2 ⇒ r4 =

1

2, r4 = r2

3r2 ⇒ r2 =1

2, and r3 = r2

2r1 ⇒ r1 = 4.

Therefore, the possible values of z1 are the nonreal roots of the equation x10−410 = 0, and the product

of the eight possible values is410

42= 48 = 65536. For these values of z1, it is not difficult to construct

a sequence which works, by choosing z2 nonreal so that |z2| = 12 .

26. [17] Triangle ABC has perimeter 1. Its three altitudes form the side lengths of a triangle. Find theset of all possible values of min(AB,BC,CA).

Answer: ( 3−√

54 , 1

3 ] Let a, b, c denote the side lengths BC,CA, and AB, respectively. Without

loss of generality, assume a ≤ b ≤ c; we are looking for the possible range of a.

First, note that the maximum possible value of a is 13 , which occurs when ABC is equilateral. It

remains to find a lower bound for a.

Now rewrite c = xa and b = ya, where we have x ≥ y ≥ 1. Note that for a non-equilateral triangle,x > 1. The triangle inequality gives us a + b > c, or equivalently, y > x − 1. If we let K be thearea, the condition for the altitudes gives us 2K

c+ 2K

b> 2K

a, or equivalently, 1

b> 1

a− 1

c, which after

some manipulation yields y < xx−1 . Putting these conditions together yields x − 1 < x

x−1 , and after

rearranging and solving a quadratic, we get x < 3+√

52 .

Guts Round

We now use the condition a(1 + x + y) = 1, and to find a lower bound for a, we need an upper boundfor 1 + x + y. We know that 1 + x + y < 1 + x + x

x−1 = x − 1 + 1x−1 + 3.

Now let f(x) = x − 1 + 1x−1 + 3. If 1 < x < 2, then 1 + x + y ≤ 1 + 2x < 5. But for x ≥ 2, we see

that f(x) attains a minimum of 5 at x = 2 and continues to strictly increase after that point. Since

x < 3+√

52 , we have f(x) < f

(

3+√

52

)

= 3 +√

5 > 5, so this is a better upper bound than the case for

which 1 < x < 2. Therefore, a >(

13+

√5

)

= 3−√

54 .

For any a such that√

5 − 2 ≥ a > 3−√

54 , we can let b = 1+

√5

2 a and c = 1 − a − b. For any otherpossible a, we can let b = c = 1−a

2 . The triangle inequality and the altitude condition can both beverified algebraically.

We now conclude that the set of all possible a is 3−√

54 < a ≤ 1

3 .

27. [17] Let W be the hypercube {(x1, x2, x3, x4) | 0 ≤ x1, x2, x3, x4 ≤ 1}. The intersection of W and ahyperplane parallel to x1 + x2 + x3 + x4 = 0 is a non-degenerate 3-dimensional polyhedron. What isthe maximum number of faces of this polyhedron?

Answer: 8 The number of faces in the polyhedron is equal to the number of distinct cells (3-dimensional faces) of the hypercube whose interior the hyperplane intersects. However, it is possibleto arrange the hyperplane such that it intersects all 8 cells. Namely, x1 + x2 + x3 + x4 = 3

2 intersectsall 8 cells because it passes through (0, 1

2 , 12 , 1

2 ) (which is on the cell x1 = 0), (1, 16 , 1

6 , 16 ) (which is on

the cell x1 = 1), and the points of intersection with the other 6 cells can be found by permuting thesequadruples.

28. [17] Let z0 +z1 +z2 + · · · be an infinite complex geometric series such that z0 = 1 and z2013 = 120132013 .

Find the sum of all possible sums of this series.

Answer: 20132014

20132013−1 Clearly, the possible common ratios are the 2013 roots r1, r2, . . . , r2013 of the

equation r2013 = 120132013 . We want the sum of the values of xn = 1

1−rn, so we consider the polynomial

whose roots are x1, x2, . . . , x2013. It is easy to see that (1− 1xn

)2013 = 120132013 , so it follows that the xn

are the roots of the polynomial equation 120132013 x2013 − (x− 1)2013 = 0. The leading coefficient of this

polynomial is 120132013 − 1, and it follows easily from the Binomial Theorem that the next coefficient is

2013, so our answer is, by Vieta’s Formulae,

− 20131

20132013 − 1=

20132014

20132013 − 1.

29. [20] Let A1, A2, ..., Am be finite sets of size 2012 and let B1, B2, ..., Bm be finite sets of size 2013 suchthat Ai ∩ Bj = ∅ if and only if i = j. Find the maximum value of m.

Answer:(

40252012

)

In general, we will show that if each of the sets Ai contain a elements and if each

of the sets Bj contain b elements, then the maximum value for m is(

a+ba

)

.

Let U denote the union of all the sets Ai and Bj and let |U | = n. Consider the n! orderings of theelements of U . Note that for any specific ordering, there is at most one value of i such that all theelements in Ai come before all the elements in Bi in this ordering; this follows since Aj shares at leastone element with Bi and Bj shares at least one element with Ai for any other j 6= i.

On the other hand, the number of ways to permute the (a + b) elements in Ai ∪ Bi so that all theelements in Ai come first is equal to a!b!. Therefore, the number of permutations of U where all theelements in Ai come before all the elements in Bi is equal to:

n! · a!b!

(a + b)!=

n!(

a+ba

)

Summing over all m values of i, the total number of orderings where, for some i, the elements in Ai

come before Bi is equal to

Guts Round

n!m(

a+ba

)

But there are at most u! such orderings, since there are u! total orderings, so it follows that m ≤(

a+ba

)

.Equality is attained by taking U to be a set containing (a + b) elements, letting Ai range over alla-element subsets of U , and letting Bi = U \ Ai for each i.

30. [20] How many positive integers k are there such that

k

2013(a + b) = lcm(a, b)

has a solution in positive integers (a, b)?

Answer: 1006 First, we can let h = gcd(a, b) so that (a, b) = (hA, hB) where gcd(A,B) = 1.

Making these subtitutions yields k2013 (hA+hB) = hAB, so k = 2013AB

A+B. Because A and B are relatively

prime, A + B shares no common factors with neither A nor B, so in order to have k be an integer,A + B must divide 2013, and since A and B are positive, A + B > 1.

We first show that for different possible values of A + B, the values of k generated are distinct. Inparticular, we need to show that 2013AB

A+B6= 2013A′B′

A′+B′ whenever A + B 6= A′ + B′. Assume that such anequality exists, and cross-multiplying yields AB(A′ +B′) = A′B′(A+B). Since AB is relatively primeto A + B, we must have A + B divide A′ + B′. With a similar argument, we can show that A′ + B′

must divide A + B, so A + B = A′ + B′.

Now, we need to show that for the same denominator A + B, the values of k generated are alsodistinct for some relatively prime non-ordered pair (A,B). Let n = A + B = C + D. Assumethat 2013AB

n= 2013CD

n, or equivalently, A(n − A) = C(n − C). After some rearrangement, we have

(C + A)(C − A) = n(C − A) This implies that either C = A or C = n − A = B. But in either case,(C,D) is some permutation of (A,B).

Our answer can therefore be obtained by summing up the totients of the factors of 2013 (excluding 1)and dividing by 2 since (A,B) and (B,A) correspond to the same k value, so our answer is 2013−1

2 =1006.

Remark: It can be proven that the sum of the totients of all the factors of any positive integer Nequals N , but in this case, the sum of the totients can be computed by hand.

31. [20] Let ABCD be a quadrilateral inscribed in a unit circle with center O. Suppose that ∠AOB =∠COD = 135◦, BC = 1. Let B′ and C ′ be the reflections of A across BO and CO respectively. LetH1 and H2 be the orthocenters of AB′C ′ and BCD, respectively. If M is the midpoint of OH1, andO′ is the reflection of O about the midpoint of MH2, compute OO′.

Answer: 14 (8 −

√6 − 3

√2) Put the diagram on the complex plane with O at the origin and A at

1. Let B have coordinate b and C have coordinate c. We obtain easily that B′ is b2, C ′ is c2, and Dis bc. Therefore, H1 is 1 + b2 + c2 and H2 is b + c + bc (we have used the fact that for triangles on theunit circle, their orthocenter is the sum of the vertices). Finally, we have that M is 1

2 (1 + b2 + c2), sothe reflection of O about the midpoint of MH2 is 1

2 (1 + b2 + c2 + 2b + 2c + 2bc) = 12 (b + c + 1)2, so we

just seek 12 |b + c + 1|2. But we know that b = cis135◦ and c = cis195◦, so we obtain that this value is

14 (8 −

√6 − 3

√2).

32. [20] For an even integer positive integer n Kevin has a tape of length 4n with marks at −2n,−2n +1, . . . , 2n − 1, 2n. He then randomly picks n points in the set −n,−n + 1,−n + 2, . . . , n − 1, n, andplaces a stone on each of these points. We call a stone ‘stuck’ if it is on 2n or −2n, or either all thepoints to the right, or all the points to the left, all contain stones. Then, every minute, Kevin shiftsthe unstuck stones in the following manner:

• He picks an unstuck stone uniformly at random and then flips a fair coin.

Guts Round

• If the coin came up heads, he then moves that stone and every stone in the largest contiguous setcontaining that stone one point to the left. If the coin came up tails, he moves every stone in that setone point right instead.

• He repeats until all the stones are stuck.

Let pk be the probability that at the end of the process there are exactly k stones in the right half.Evaluate

pn−1 − pn−2 + pn−3 − . . . + p3 − p2 + p1

pn−1 + pn−2 + pn−3 + . . . + p3 + p2 + p1

in terms of n.

Answer: 1n−1 After we have selected the positions of the initial n stones, we number their

positions: a1 < a2 < . . . < an. The conditions on how we move the stones imply that the expectedvalue of (ai − aj) after t minutes is still equal to ai − aj . In addition, if bi is the final position of theith stone, E(bi+1 − bi) = E(ai+1 − ai). But this quantity is also equal to (3n + 2) · pi + 1 · (1 − pi).

Now, let’s calculate the expected value of ai+1 − ai. This is the sum over g = ai+1 − ai, and j, thenumber of spaces before ai of g ·

(

ji−1

)(

2n−j−gn−i+1

)

, so we get

1(

2n+1n

)

∑

g

g ·∑

j

(

j

i − 1

)(

2n − j − g

n − i − 1

)

But∑

j

(

ji−1

)(

2n−j−gn−i−1

)

is just(

2n−g+1n−1

)

. Therefore the expected value of ai+1 − ai is independent of i,

so pi is constant for all i 6= 0, n. It follows that the answer is 1n−1 .

33. [25] Compute the value of 125 +224 +323 + . . .+242 +251. If your answer is A and the correct answer

is C, then your score on this problem will be⌊

25min(

(

AC

)2,(

CA

)2)⌋

.

Answer: 66071772829247409 The sum is extremely unimodal, so we want to approximate it usingits largest term. Taking logs of each term, we see that the max occurs when (26 − n) log n peaks, andtaking derivatives gives

x + x log x = 26

From here it’s easy to see that the answer is around 10, and slightly less (it’s actually about 8.3, butin any case it’s hard to find powers of anything except 10). Thus the largest term will be somethinglike 1016, which is already an order of magnitude within the desired answer 6.6 × 1016.

To do better we’d really need to understand the behavior of the function x26−x, but what approximatelyhappens is that only the four or five largest terms in the sum are of any substantial size; thus it isreasonable here to pick some constant from 4 to 20 to multiply our guess 1016; any guess between4.0 × 1016 and 2.0 × 1017 is reasonable.

34. [25] For how many unordered sets {a, b, c, d} of positive integers, none of which exceed 168, do thereexist integers w, x, y, z such that (−1)wa + (−1)xb + (−1)yc + (−1)zd = 168? If your answer is A and

the correct answer is C, then your score on this problem will be ⌊25e−3|C−A|

C ⌋.Answer: 761474 As an approximation, we assume a, b, c, d are ordered to begin with (so we

have to divide by 24 later) and add to 168 with a unique choice of signs; then, it suffices to counte + f + g + h = 168 with each e, f, g, h in [−168, 168] and then divide by 24 (we drop the conditionthat none of them can be zero because it shouldn’t affect the answer that much).

One way to do this is generating functions. We want the coefficient of t168 in the generating function

(t−168 + t−167 + . . . + t167 + t168)4 = (t169 − t−168)4/(t − 1)4

Clearing the negative powers, it suffices to find the coefficient of t840 in

(t337 − 1)4/(t − 1)4 = (1 − 4t337 + 6t674 − . . .) 1(t−1)4 .

To do this we expand the bottom as a power series in t:

Guts Round

1(t−1)4 =

∑

n≥0

(

n+33

)

tn

It remains to calculate(

840+33

)

− 4 ·(

840−337+33

)

+ 6 ·(

840−674+33

)

. This is almost exactly equal to16 (8433 − 4 · 5063 + 6 · 1693) ≈ 1.83 × 107.

Dividing by 24, we arrive at an estimation 762500. Even if we use a bad approximation 16·24 (8503 − 4 ·

5003 + 6 · 1503) we get approximately 933000, which is fairly close to the answer.

35. [25] Let P be the number to partition 2013 into an ordered tuple of prime numbers? What islog2(P )? If your answer is A and the correct answer is C, then your score on this problem will be⌊

1252

(

min(

CA

, AC

)

− 35

)⌋

or zero, whichever is larger.

Answer: 614.519... We use the following facts and heuristics.

(1) The ordered partitions of n into any positive integers (not just primes) is 2n−1. This can be guessedby checking small cases and finding a pattern, and is not difficult to prove.

(2) The partitions of 2013n

into any positive integers equals the partitions of 2013 into integers from theset {n, 2n, 3n, · · · }.(3) The small numbers matter more when considering partitions.

(4) The set of primes {2, 3, 5, 7, · · · } is close in size (near the small numbers) to {3, 6, 9, · · · } or{2, 4, 6, · · · }.(5) The prime numbers get very sparse compared to the above two sets in the larger numbers.

Thus, using these heuristics, the number of partitions of 2013 into primes is approximately 22013

3−1

or 22013

2−1, which, taking logarithms, give 670 and 1005.5, respectively. By (5), we should estimate

something that is slightly less than these numbers.

36. [24] (Mathematicians A to Z) Below are the names of 26 mathematicians, one for each letter ofthe alphabet. Your answer to this question should be a subset of {A,B, · · · , Z}, where each letterrepresents the corresponding mathematician. If two mathematicians in your subset have birthdatesthat are within 20 years of each other, then your score is 0. Otherwise, your score is max(3(k − 3), 0)where k is the number of elements in your subset.

Niels Abel Isaac Netwon

Étienne Bézout Nicole OresmeAugustin-Louis Cauchy Blaise PascalRené Descartes Daniel QuillenLeonhard Euler Bernhard RiemannPierre Fatou Jean-Pierre SerreAlexander Grothendieck Alan TuringDavid Hilbert Stanislaw UlamKenkichi Iwasawa John VennCarl Jacobi Andrew WilesAndrey Kolmogorov Leonardo XimenesJoseph-Louis Lagrange Shing-Tung YauJohn Milnor Ernst Zermelo

Answer: {O, D, P, E, B, C, R, H, K, S, Y} A knowledgeable math historian might come up with

this 11-element subset, earning 24 points: {O, D, P, E, B, C, R, H, K, S, Y}.

Guts Round

HMMT November 2013Saturday 9 November 2013

General Test

1. [2] What is the smallest non-square positive integer that is the product of four prime numbers (notnecessarily distinct)?

2. [3] Plot points A,B,C at coordinates (0, 0), (0, 1), and (1, 1) in the plane, respectively. Let S denotethe union of the two line segments AB and BC. Let X1 be the area swept out when Bobby rotatesS counterclockwise 45 degrees about point A. Let X2 be the area swept out when Calvin rotates Sclockwise 45 degrees about point A. Find X1+X2

2 .

3. [4] A 24-hour digital clock shows times h : m : s, where h, m, and s are integers with 0 ≤ h ≤ 23,0 ≤ m ≤ 59, and 0 ≤ s ≤ 59. How many times h : m : s satisfy h+m = s?

4. [4] A 50-card deck consists of 4 cards labeled “i” for i = 1, 2, . . . , 12 and 2 cards labeled “13”. If Bobrandomly chooses 2 cards from the deck without replacement, what is the probability that his 2 cardshave the same label?

5. [5] Let ABC be an isosceles triangle with AB = AC. Let D and E be the midpoints of segments

AB and AC, respectively. Suppose that there exists a point F on ray−−→DE outside of ABC such that

triangle BFA is similar to triangle ABC. Compute ABBC .

6. [5] Find the number of positive integer divisors of 12! that leave a remainder of 1 when divided by 3.

7. [6] Find the largest real number λ such that a2 + b2 + c2 + d2 ≥ ab + λbc + cd for all real numbersa, b, c, d.

8. [6] How many of the first 1000 positive integers can be written as the sum of finitely many distinctnumbers from the sequence 30, 31, 32, . . .?

9. [7] Let ABC be a triangle and D a point on BC such that AB =√

2, AC =√

3, ∠BAD = 30◦, and∠CAD = 45◦. Find AD.

10. [8] How many functions f : {1, 2, . . . , 2013} → {1, 2, . . . , 2013} satisfy f(j) < f(i)+j−i for all integersi, j such that 1 ≤ i < j ≤ 2013?

HMMT November 2013Saturday 9 November 2013

General Test

1. [2] What is the smallest non-square positive integer that is the product of four prime numbers (notnecessarily distinct)?

Answer: 24 The smallest two integers that are the product of four primes are 24 = 16 and23 · 3 = 24. Since 16 is a perfect square and 24 is not, the answer is 24.

2. [3] Plot points A,B,C at coordinates (0, 0), (0, 1), and (1, 1) in the plane, respectively. Let S denotethe union of the two line segments AB and BC. Let X1 be the area swept out when Bobby rotatesS counterclockwise 45 degrees about point A. Let X2 be the area swept out when Calvin rotates Sclockwise 45 degrees about point A. Find X1+X2

2 .

Answer: π4 It’s easy to see X1 = X2. Simple cutting and pasting shows that X1 equals the area

of 18 of a circle with radius AC =

√2, so X1+X2

2 = X1 = 18π(√

2)2 = π4 .

3. [4] A 24-hour digital clock shows times h : m : s, where h, m, and s are integers with 0 ≤ h ≤ 23,0 ≤ m ≤ 59, and 0 ≤ s ≤ 59. How many times h : m : s satisfy h+m = s?

Answer: 1164 We are solving h+m = s in 0 ≤ s ≤ 59, 0 ≤ m ≤ 59, and 0 ≤ h ≤ 23. If s ≥ 24,each h corresponds to exactly 1 solution, so we get 24(59−23) = 24(36) in this case. If s ≤ 23, we wantthe number of nonnegative integer solutions to h + m ≤ 23, which by lattice point counting (or ballsand urns) is

23+2

2

)= (23 + 2)(23 + 1)/2 = 25 · 12. Thus our total is 12(72 + 25) = 12(100− 3) = 1164.

4. [4] A 50-card deck consists of 4 cards labeled “i” for i = 1, 2, . . . , 12 and 2 cards labeled “13”. If Bobrandomly chooses 2 cards from the deck without replacement, what is the probability that his 2 cardshave the same label?

Answer: 731225 All pairs of distinct cards (where we distinguish cards even with the same label) are

equally likely. There are22

)+ 12

42

)= 73 pairs of cards with the same label and

502

)= 100 · 494 = 1225

pairs of cards overall, so the desired probability is 731225 .

5. [5] Let ABC be an isosceles triangle with AB = AC. Let D and E be the midpoints of segments

AB and AC, respectively. Suppose that there exists a point F on ray−−→DE outside of ABC such that

triangle BFA is similar to triangle ABC. Compute ABBC .

Answer:√

2 Let α = ∠ABC = ∠ACB, AB = 2x, and BC = 2y, so AD = DB = AE = EC = xand DE = y. Since 4BFA ∼ 4ABC and BA = AC, we in fact have 4BFA ∼= 4ABC, soBF = BA = 2x, FA = 2y, and ∠DAF = α. But DE ‖ BC yields ∠ADF = ∠ABC = α as well,whence 4FAD ∼ 4ABC gives 2y

x = FAAD = AB

BC = 2x2y =⇒ AB

BC = xy =√

2.

6. [5] Find the number of positive integer divisors of 12! that leave a remainder of 1 when divided by 3.

Answer: 66 First we factor 12! = 210355271111, and note that 2, 5, 11 ≡ −1 (mod 3) while 7 ≡ 1(mod 3). The desired divisors are precisely 2a5b7c11d with 0 ≤ a ≤ 10, 0 ≤ b ≤ 2, 0 ≤ c ≤ 1, 0 ≤ d ≤ 1,and a+ b+ d even. But then for any choice of a, b, exactly one d ∈ {0, 1} makes a+ b+ d even, so wehave exactly one 1 (mod 3)-divisor for every triple (a, b, c) satisfying the inequality constraints. Thisgives a total of (10 + 1)(2 + 1)(1 + 1) = 66.

7. [6] Find the largest real number λ such that a2 + b2 + c2 + d2 ≥ ab + λbc + cd for all real numbersa, b, c, d.

Answer: 32 Let f(a, b, c, d) = (a2 +b2 +c2 +d2)− (ab+λbc+cd). For fixed (b, c, d), f is minimized

at a = b2 , and for fixed (a, b, c), f is minimized at d = c

2 , so simply we want the largest λ such that

f( b2 , b, c,c2 ) = 3

4 (b2 +c2)−λbc is always nonnegative. By AM-GM, this holds if and only if λ ≤ 2 34 = 3

2 .

General Test

8. [6] How many of the first 1000 positive integers can be written as the sum of finitely many distinctnumbers from the sequence 30, 31, 32, . . .?

Answer: 105 We want to find which integers have only 0’s and 1’s in their base 3 representation.Note that 100010 = 11010013. We can construct a bijection from all such numbers to the binary strings,by mapping x3 ↔ x2. Since 11010012 = 10510, we conclude that the answer is 105.

9. [7] Let ABC be a triangle and D a point on BC such that AB =√

2, AC =√

3, ∠BAD = 30◦, and∠CAD = 45◦. Find AD.

Answer:√62 OR

√3√2

Note that [BAD] + [CAD] = [ABC]. If α1 = ∠BAD, α2 = ∠CAD, then

we deduce sin(α1+α2)AD = sinα1

AC + sinα2

AB upon division by AB ·AC ·AD. Now

AD =sin(30◦ + 45◦)sin 30◦√

3+ sin 45◦√

2

.

But sin(30◦ + 45◦) = sin 30◦ cos 45◦ + sin 45◦ cos 30◦ = sin 30◦ 1√2

+ sin 45◦√32 =

√62 ( sin 30◦√

3+ sin 45◦√

2), so

our answer is√62 .

10. [8] How many functions f : {1, 2, . . . , 2013} → {1, 2, . . . , 2013} satisfy f(j) < f(i)+j−i for all integersi, j such that 1 ≤ i < j ≤ 2013?

Answer:40252013

)Note that the given condition is equivalent to f(j) − j < f(i) − i for all

1 ≤ i < j ≤ 2013. Let g(i) = f(i) − i, so that the condition becomes g(j) < g(i) for i < j and1 − i ≤ g(i) ≤ 2013 − i. However, since g is decreasing, we see by induction that g(i + 1) is inthe desired range so long as g(i) is in the desired range. Hence, it suffices to choose 2013 values forg(1), . . . , g(2013) in decreasing order from [−2012, 2012], for a total of

40252013

)possible functions.

General Test

HMMT November 2013Saturday 9 November 2013

Theme Round

Traveling

1. [2] Two cars are driving directly towards each other such that one is twice as fast as the other. Thedistance between their starting points is 4 miles. When the two cars meet, how many miles is thefaster car from its starting point?

2. [4] You are standing at a pole and a snail is moving directly away from the pole at 1 cm/s. When thesnail is 1 meter away, you start “Round 1”. In Round n (n ≥ 1), you move directly toward the snailat n+ 1 cm/s. When you reach the snail, you immediately turn around and move back to the startingpole at n+ 1 cm/s. When you reach the pole, you immediately turn around and Round n+ 1 begins.

At the start of Round 100, how many meters away is the snail?

3. [5] Let ABC be a triangle with AB = 5, BC = 4, and CA = 3. Initially, there is an ant at eachvertex. The ants start walking at a rate of 1 unit per second, in the direction A → B → C → A (so

the ant starting at A moves along ray−−→AB, etc.). For a positive real number t less than 3, let A(t) be

the area of the triangle whose vertices are the positions of the ants after t seconds have elapsed. Forwhat positive real number t less than 3 is A(t) minimized?

4. [7] There are 2 runners on the perimeter of a regular hexagon, initially located at adjacent vertices.Every second, each of the runners independently moves either one vertex to the left, with probability12 , or one vertex to the right, also with probability 1

2 . Find the probability that after a 2013 secondrun (in which the runners switch vertices 2013 times each), the runners end up at adjacent verticesonce again.

5. [7] Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Company XYZ wants to locate theirbase at the point P in the plane minimizing the total distance to their workers, who are located atvertices A, B, and C. There are 1, 5, and 4 workers at A, B, and C, respectively. Find the minimumpossible total distance Company XYZ’s workers have to travel to get to P .

Bases

Many of you may be familiar with the decimal (or base 10) system. For example, when we say 201310,we really mean 2·103+0·102+1·101+3·100. Similarly, there is the binary (base 2) system. For example,111110111012 = 1 ·210 +1 ·29 +1 ·28 +1 ·27 +1 ·26 +0 ·25 +1 ·24 +1 ·23 +1 ·22 +0 ·21 +1 ·20 = 201310.

In general, if we are given a string (anan∈1 . . . a0)b in base b (the subscript b means that we are in baseb), then it is equal to

∑ni=0 aib

i.

It turns out that for every positive integer b > 1, every positive integer k has a unique base b repre-sentation. That is, for every positive integer k, there exists a unique n and digits 0 ≤ a0, . . . , an < bsuch that (anan∈1 . . . a0)b = k.

We can adapt this to bases b < −1. It actually turns out that if b < −1, every nonzero integer has aunique base b representation. That is, for every nonzero integer k, there exists a unique n and digits0 ≤ a0, . . . , an < |b| such that (anan∈1 . . . a0)b = k. The next five problems involve base −4.

Note: Unless otherwise stated, express your answers in base 10.

6. [2] Evaluate 1201201∈4.

7. [3] Express −2013 in base −4.

8. [5] Let b(n) be the number of digits in the base −4 representation of n. Evaluate∑2013

i=1 b(i).

9. [7] Let N be the largest positive integer that can be expressed as a 2013-digit base −4 number. Whatis the remainder when N is divided by 210?

10. [8] Find the sum of all positive integers n such that there exists an integer b with |b| 6= 4 such thatthe base −4 representation of n is the same as the base b representation of n.

HMMT November 2013Saturday 9 November 2013

Theme Round

1. [2] Two cars are driving directly towards each other such that one is twice as fast as the other. Thedistance between their starting points is 4 miles. When the two cars meet, how many miles is thefaster car from its starting point?

Answer: 83 Note that the faster car traveled twice the distance of the slower car, and together,

the two cars traveled the total distance between the starting points, which is 4 miles. Let the distancethat the faster car traveled be x. Then, x + x

2 = 4 =⇒ x = 83 . Thus, the faster car traveled 8

3 milesfrom the starting point.

2. [4] You are standing at a pole and a snail is moving directly away from the pole at 1 cm/s. When thesnail is 1 meter away, you start “Round 1”. In Round n (n ≥ 1), you move directly toward the snailat n+ 1 cm/s. When you reach the snail, you immediately turn around and move back to the startingpole at n+ 1 cm/s. When you reach the pole, you immediately turn around and Round n+ 1 begins.

At the start of Round 100, how many meters away is the snail?

Answer: 5050 Suppose the snail is xn meters away at the start of round n, so x1 = 1, and the runner

takes 100xn

(n+1)−1 = 100xn

n seconds to catch up to the snail. But the runner takes the same amount of time

to run back to the start, so during round n, the snail moves a distance of xn+1〉xn = 200xn

n1

100 = 2xn

n .

Finally, we have x100 = 10199 x99 = 101

9910098 x98 = · · · = 101!/2!

99! x1 = 5050.

3. [5] Let ABC be a triangle with AB = 5, BC = 4, and CA = 3. Initially, there is an ant at eachvertex. The ants start walking at a rate of 1 unit per second, in the direction A → B → C → A (so

the ant starting at A moves along ray〉〉→AB, etc.). For a positive real number t less than 3, let A(t) be

the area of the triangle whose vertices are the positions of the ants after t seconds have elapsed. Forwhat positive real number t less than 3 is A(t) minimized?

Answer: 4724 We instead maximize the area of the remaining triangles. This area (using 1

2xy sin θ)

is 12 (t)(5〉 t) 3

5 + 12 (t)(3〉 t) 4

5 + 12 (t)(4〉 t)1 = 1

10 (〉12t2 +47t), which has a maximum at t = 4724 ∈ (0, 3).

4. [7] There are 2 runners on the perimeter of a regular hexagon, initially located at adjacent vertices.Every second, each of the runners independently moves either one vertex to the left, with probability12 , or one vertex to the right, also with probability 1

2 . Find the probability that after a 2013 secondrun (in which the runners switch vertices 2013 times each), the runners end up at adjacent verticesonce again.

Answer: 23 + 1

3 ( 14 )2013 OR 24027+1

3·24026 OR 23 + 1

3 ( 12 )4026 OR 2

3 + 13 ( 1

64 )671 Label the runners A and

B and arbitrarily fix an orientation of the hexagon. Let pt(i) be the probability that A is i (mod 6)vertices to the right of B at time t, so without loss of generality p0(1) = 1 and p0(2) = · · · = p0(6) = 0.Then for t > 0, pt(i) = 1

4pt−1(i〉 2) + 12pt−1(i) + 1

4pt−1(i+ 2).

In particular, pt(2) = pt(4) = pt(6) = 0 for all t, so we may restrict our attention to pt(1), pt(3), pt(5).Thus pt(1) + pt(3) + pt(5) = 1 for all t ≥ 0, and we deduce pt(i) = 1

4 + 14pt−1(i) for i = 1, 3, 5.

Finally, let f(t) = pt(1) + pt(5) denote the probability that A,B are 1 vertex apart at time t, sof(t) = 1

2 + 14f(t〉 1) =⇒ f(t)〉 2

3 = 14 [f(t〉 1)〉 2

3 ], and we conclude that f(2013) = 23 + 1

3 ( 14 )2013.

5. [7] Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Company XYZ wants to locate theirbase at the point P in the plane minimizing the total distance to their workers, who are located atvertices A, B, and C. There are 1, 5, and 4 workers at A, B, and C, respectively. Find the minimumpossible total distance Company XYZ’s workers have to travel to get to P .

Answer: 69 We want to minimize 1 · PA + 5 · PB + 4 · PC. By the triangle inequality, (PA +PB) + 4(PB+PC) ≥ AB+ 4BC = 13 + 56 = 69, with equality precisely when P = [AB]∩ [BC] = B.

Theme Round

6. [2] Evaluate 1201201−4.

Answer: 2017 The answer is 1+2(〉4)2 +(〉4)3 +2(〉4)5 +(〉4)6 = 1〉2 ·42 +2 ·45 = 2049〉32 =2017.

7. [3] Express 〉2013 in base 〉4.

Answer: 200203 〉2013 ≡ 3 (mod 4), so the last digit is 3; now −2013−3−4 = 504 ≡ 0, so the next

digit (to the left) is 0; then 504−0−4 = 〉126 ≡ 2; −126−2

−4 = 32 ≡ 0; 32−0−4 = 〉8 ≡ 0; −8−0

−4 = 2.

Thus 〉201310 = 200203−4.

8. [5] Let b(n) be the number of digits in the base 〉4 representation of n. Evaluate∑2013

i=1 b(i).

Answer: 12345 We have the following:

• b(n) = 1 for n between 1 and 3.

• b(n) = 3 for n between 42 〉 3 · 4 = 4 and 3 · 42 + 3 = 51. (Since a · 42 〉 b · 4 + c takes on 3 · 4 · 4distinct values over 1 ≤ a ≤ 3, 0 ≤ b ≤ 3, 0 ≤ c ≤ 3, with minimum 4 and maximum 51.)

• b(n) = 5 for n between 44 〉 3 · 43 〉 3 · 4 = 52 and 3 · 44 + 3 · 42 + 3 = 819.

• b(n) = 7 for n between 46 〉 3 · 45 〉 3 · 43 〉 3 · 41 = 820 and 3 · 46 + 3 · 44 + 3 · 42 + 3 > 2013.

Thus2013∑i=1

b(i) = 7(2013)〉 2(819 + 51 + 3) = 14091〉 2(873) = 14091〉 1746 = 12345.

9. [7] Let N be the largest positive integer that can be expressed as a 2013-digit base 〉4 number. Whatis the remainder when N is divided by 210?

Answer: 51 The largest is∑1006

i=0 3 · 42i = 3 161007−116−1 = 161007−1

5 .

This is 1 (mod 2), 0 (mod 3), 3 · 1007 ≡ 21 ≡ 1 (mod 5), and 3(21007〉 1) ≡ 3(28〉 1) ≡ 3(22〉 1) ≡ 2(mod 7), so we need 1 (mod 10) and 9 (mod 21), which is 9 + 2 · 21 = 51 (mod 210).

10. [8] Find the sum of all positive integers n such that there exists an integer b with |b| 6= 4 such thatthe base 〉4 representation of n is the same as the base b representation of n.

Answer: 1026 All 1 digit numbers, 0, 1, 2, 3, are solutions when, say, b = 5. (Of course, d ∈{0, 1, 2, 3} works for any base b of absolute value greater than d but not equal to 4.)

Consider now positive integers n = (ad . . . a1a0)4 with more than one digit, so d ≥ 1, ad 6= 0, and0 ≤ ak ≤ 3 for k = 0, 1, . . . , d. Then n has the same representation in base b if and only if |b| > max akand

∑dk=0 ak(〉4)k =

∑dk=0 akb

k, or equivalently,∑d

k=0 ak(bk 〉 (〉4)k) = 0.

First we prove that b ≤ 3. Indeed, if b ≥ 4, then b 6= 4 =⇒ b ≥ 5, so bk 〉 (〉4)k is positive for all

k ≥ 1 (and zero for k = 0). But then∑d

k=0 ak(bk 〉 (〉4)k) ≥ ad(bd 〉 (〉4)d) must be positive, andcannot vanish.

Next, we show b ≥ 2. Assume otherwise for the sake of contradiction; b cannot be 0,±1 (thesebases don’t make sense in general) or 〉4, so we may label two distinct negative integers 〉r,〉s with

r 〉 1 ≥ s ≥ 2 such that {r, s} = {4,〉b}, s > max ak, and∑d

k=0 ak((〉r)k 〉 (〉s)k) = 0, which,combined with the fact that rk 〉 sk ≥ 0 (equality only at k = 0), yields

rd 〉 sd ≤ ad(rd 〉 sd) =

d−1∑k=0

(〉1)d−1−kak(rk 〉 sk)

≤d−1∑k=0

(s〉 1)(rk 〉 sk) = (s〉 1)rd 〉 1

r 〉 1〉 (sd 〉 1).

Hence rd 〉 1 ≤ (s〉 1) rd−1r−1 < (r 〉 1) rd−1

r−1 = rd 〉 1, which is absurd.

Theme Round

Thus b ≥ 2, and since b ≤ 3 we must either have b = 2 or b = 3. In particular, all ak must be at mostb〉 1. We now rewrite our condition as

ad(4d 〉 (〉b)d) =

d−1∑k=0

(〉1)d−1−kak(4k 〉 (〉b)k).

Since 4k 〉 (〉b)k ≥ 0 for k ≥ 0, with equality only at k = 0, we deduce

ad(4d 〉 (〉b)d) ≤∑

k≡d−1 (mod 2)

(b〉 1)(4k 〉 (〉b)k).

If d〉 1 is even (d is odd), this gives

ad(4d + bd) ≤ (b〉 1)4d+1 〉 40

42 〉 1〉 (b〉 1)

bd+1 〉 b0

b2 〉 1,

so 4d < (b〉 1) 4d+1

15 =⇒ b > 1 + 154 , which is impossible.

Thus d〉 1 is odd (d is even), and we get

ad(4d 〉 bd) ≤ (b〉 1)4d+1 〉 41

42 〉 1+ (b〉 1)

bd+1 〉 b1

b2 〉 1⇐⇒ bd 〉 1

4d 〉 1≥ad 〉 4

15 (b〉 1)

ad + bb+1

.

If b = 2, then ad = 1, so 12d+1

= 2d−14d−1

≥ 1125 , which is clearly impossible (d ≥ 2).

If b = 3 and ad = 2, then 9d/2−116d/2−1

≤ 815 . Since d is even, it’s easy to check this holds only for d/2 = 1,

with equality, so ak = b 〉 1 if k ≡ d 〉 1 (mod 2). Thus (ad, . . . , a0) = (2, 2, a0), yielding solutions(22x)3 (which do work; note that the last digit doesn’t matter).

Otherwise, if b = 3 and ad = 14, then 9d/2−116d/2−1

≤ 415 . It’s easy to check d/2 ∈ {1, 2}.

If d/2 = 1, we’re solving 16a2 〉 4a1 + a0 = 9a2 + 3a1 + a0 ⇐⇒ a2 = a1. We thus obtain the workingsolution (11x)3. (Note that 110 = 1

2220 in bases 〉4, 3.)

If d/2 = 2, we want 256a4〉64a3+16a2〉4a1+a0 = 81a4+27a3+9a2+3a1+a0, or 175 = 91a3〉7a2+7a1,which simplifies to 25 = 13a3 〉 a2 + a1. This gives the working solutions (1210x)3, (1221x)3. (Notethat 12100 = 1102 and 12210 = 1102 + 110 in bases 〉4, 3.)

The list of all nontrivial (≥ 2-digit) solutions (in base 〉4 and b) is then 11x, 22x, 1210x, 1221x, whereb = 3 and x ∈ {0, 1, 2}. In base 10, they are 12 + x, 2 · 12 + x, 122 + x, 122 + 12 + x, with sum3(2 · 122 + 4 · 12) + 4(0 + 1 + 2) = 1020.

Finally, we need to include the trivial solutions n = 1, 2, 3, for a total sum of 1026.

Theme Round

HMMT November 2013Saturday 9 November 2013

Team Round

Expected Value

1. [3] Tim the Beaver can make three different types of geometrical figures: squares, regular hexagons,and regular octagons. Tim makes a random sequence F0, F1, F2, F3, . . . of figures as follows:

• F0 is a square.

• For every positive integer i, Fi is randomly chosen to be one of the 2 figures distinct from Fi−1(each chosen with equal probability 1

2 ).

• Tim takes 4 seconds to make squares, 6 to make hexagons, and 8 to make octagons. He makesone figure after another, with no breaks in between.

Suppose that exactly 17 seconds after he starts making F0, Tim is making a figure with n sides. Whatis the expected value of n?

2. [4] Gary plays the following game with a fair n-sided die whose faces are labeled with the positiveintegers between 1 and n, inclusive: if n = 1, he stops; otherwise he rolls the die, and starts over witha k-sided die, where k is the number his n-sided die lands on. (In particular, if he gets k = 1, he willstop rolling the die.) If he starts out with a 6-sided die, what is the expected number of rolls he makes?

3. [6] The digits 1, 2, 3, 4, 5, 6 are randomly chosen (without replacement) to form the three-digit numbersM = ABC and N = DEF . For example, we could have M = 413 and N = 256. Find the expectedvalue of M ·N .

Power of a Point

4. [4] Consider triangle ABC with side lengths AB = 4, BC = 7, and AC = 8. Let M be the midpointof segment AB, and let N be the point on the interior of segment AC that also lies on the circumcircleof triangle MBC. Compute BN .

5. [4] In triangle ABC, ∠BAC = 60◦. Let ω be a circle tangent to segment AB at point D and segmentAC at point E. Suppose ω intersects segment BC at points F and G such that F lies in between Band G. Given that AD = FG = 4 and BF = 1

2 , find the length of CG.

6. [6] Points A,B,C lie on a circle ω such that BC is a diameter. AB is extended past B to point B′

and AC is extended past C to point C ′ such that line B′C ′ is parallel to BC and tangent to ω at pointD. If B′D = 4 and C ′D = 6, compute BC.

7. [7] In equilateral triangle ABC, a circle ω is drawn such that it is tangent to all three sides of thetriangle. A line is drawn from A to point D on segment BC such that AD intersects ω at points Eand F . If EF = 4 and AB = 8, determine |AE 〉 FD|.

Periodicity

8. [2] Define the sequence {xi}i≥0 by x0 = x1 = x2 = 1 and xk = xk−1+xk−2+1xk−3

for k > 2. Find x2013.

9. [7] For an integer n ≥ 0, let f(n) be the smallest possible value of |x+ y|, where x and y are integerssuch that 3x〉 2y = n. Evaluate f(0) + f(1) + f(2) + · · ·+ f(2013).

10. [7] Let ω = cos 2π727 + i sin 2π

727 . The imaginary part of the complex number

13∏k=8

(1 + ω3k−1

+ ω2·3k−1)

is equal to sinα for some angle α between 〉π2 and π2 , inclusive. Find α.

HMMT November 2013Saturday 9 November 2013

Team Round

1. [3] Tim the Beaver can make three different types of geometrical figures: squares, regular hexagons,and regular octagons. Tim makes a random sequence F0, F1, F2, F3, . . . of figures as follows:

• F0 is a square.

• For every positive integer i, Fi is randomly chosen to be one of the 2 figures distinct from Fi−1(each chosen with equal probability 1

2 ).

• Tim takes 4 seconds to make squares, 6 to make hexagons, and 8 to make octagons. He makesone figure after another, with no breaks in between.

Suppose that exactly 17 seconds after he starts making F0, Tim is making a figure with n sides. Whatis the expected value of n?

Answer: 7 We write Fi = n as shorthand for “the ith figure is an n-sided polygon.”

If F1 = 8, the F2 = 6 or F2 = 4. If F2 = 6, Tim is making a 6-gon at time 13 (probability contribution1/4). If F2 = 4, F3 = 6 or F3 = 8 will take the time 13 mark (1/8 contribution each).

If F1 = 6, F2 = 8 or F2 = 4. If F2 = 8, it takes the 13 mark (1/4 contribution). If F2 = 4, F3 = 6 orF3 = 8 will take the 13 mark (1/8 contribution each).

Thus, the expected value of the number of sides at time 13 is 0(4)+( 14 + 1

8 + 18 )(6)+( 1

8 + 14 + 1

8 )(8) = 7.

2. [4] Gary plays the following game with a fair n-sided die whose faces are labeled with the positiveintegers between 1 and n, inclusive: if n = 1, he stops; otherwise he rolls the die, and starts over with ak-sided die, where k is the number his n-sided die lands on. (In particular, if he gets k = 1, he will stoprolling the die.) If he starts out with a 6-sided die, what is the expected number of rolls he makes?

Answer: 19760 If we let an be the expected number of rolls starting with an n-sided die, we

see immediately that a1 = 0, and an = 1 + 1n

∑ni=1 ai for n > 1. Thus a2 = 2, and for n ≥ 3,

an = 1 + 1nan + n−1

n (an−1 〉 1), or an = an−1 + 1n−1 . Thus an = 1 +

∑n−1i=1

1i for n ≥ 2, so a6 =

1 + 60+30+20+15+1260 = 197

60 .

3. [6] The digits 1, 2, 3, 4, 5, 6 are randomly chosen (without replacement) to form the three-digit numbersM = ABC and N = DEF . For example, we could have M = 413 and N = 256. Find the expectedvalue of M ·N .

Answer: 143745 By linearity of expectation and symmetry,

E[MN ] = E[(100A+ 10B + C)(100D + 10E + F )] = 1112 · E[AD].

Since

E[AD] =(1 + 2 + 3 + 4 + 5 + 6)2 〉 (12 + 22 + 32 + 42 + 52 + 62)

6 · 5=

350

30,

our answer is 111 · 35 · 37 = 111 · 1295 = 143745.

4. [4] Consider triangle ABC with side lengths AB = 4, BC = 7, and AC = 8. Let M be the midpointof segment AB, and let N be the point on the interior of segment AC that also lies on the circumcircleof triangle MBC. Compute BN .

Answer:√2104 OR

√105

2√2

Let ∠BAC = θ. Then, cos θ = 42+82−722·4·8 . Since AM = 4

2 = 2, and power

of a point gives AM · AB = AN · AC, we have AN = 2·48 = 1, so NC = 8〉 1 = 7. Law of cosines on

triangle BAN gives

BN2 = 42 + 12 〉 2 · 4 · 1 · 42 + 82 〉 72

2 · 4 · 8= 17〉 16 + 15

8= 15〉 15

8=

105

8,

so BN =√2104 .

Team Round

5. [4] In triangle ABC, ∠BAC = 60◦. Let ω be a circle tangent to segment AB at point D and segmentAC at point E. Suppose ω intersects segment BC at points F and G such that F lies in between Band G. Given that AD = FG = 4 and BF = 1

2 , find the length of CG.

Answer: 165 Let x = CG. First, by power of a point, BD =

√BF (BF + FG) = 3

2 , and

CE =√x(x+ 4). By the law of cosines, we have

(x+9

2)2 =

(11

2

)2

+ (4 +√x(x+ 4))2 〉 11

2(4 +

√x(x+ 4)),

which rearranges to 2(5x 〉 4) = 5√x(x+ 4). Squaring and noting x > 4

5 gives (5x 〉 16)(15x 〉 4) =0 =⇒ x = 16

5 .

6. [6] Points A,B,C lie on a circle ω such that BC is a diameter. AB is extended past B to point B′

and AC is extended past C to point C ′ such that line B′C ′ is parallel to BC and tangent to ω at pointD. If B′D = 4 and C ′D = 6, compute BC.

Answer: 245 Let x = AB and y = AC, and define t > 0 such that BB′ = tx and CC ′ = ty.

Then 10 = B′C ′ = (1 + t)√x2 + y2, 42 = t(1 + t)x2, and 62 = t(1 + t)y2 (by power of a point), so

52 = 42 + 62 = t(1 + t)(x2 + y2) gives 1325 = 52

102 = t(1+t)(1+t)2 = t

1+t =⇒ t = 1312 . Hence BC =

√x2 + y2 =

101+t = 10

25/12 = 245 .

7. [7] In equilateral triangle ABC, a circle ω is drawn such that it is tangent to all three sides of thetriangle. A line is drawn from A to point D on segment BC such that AD intersects ω at points Eand F . If EF = 4 and AB = 8, determine |AE 〉 FD|.

Answer: 4√5

OR 4√5

5 Without loss of generality, A,E, F,D lie in that order. Let x = AE, y = DF .

By power of a point, x(x+ 4) = 42 =⇒ x = 2√

5〉 2, and y(y + 4) = (x+ 4 + y)2 〉 (4√

3)2 =⇒ y =48−(x+4)2

2(x+2) = 12−(1+√5)2√

5. It readily follows that x〉 y = 4√

5= 4√5

5 .

8. [2] Define the sequence {xi}i≥0 by x0 = x1 = x2 = 1 and xk = xk−1+xk−2+1xk−3

for k > 2. Find x2013.

Answer: 9 We have x3 = 1+1+11 = 3, x4 = 3+1+1

1 = 5, x5 = 5+3+11 = 9, x6 = 9+5+1

3 = 5. Bythe symmetry of our recurrence (or just further computation—it doesn’t matter much), x7 = 3 andx8 = x9 = x10 = 1, so our sequence has period 8. Thus x2013 = x13 = x5 = 9.

9. [7] For an integer n ≥ 0, let f(n) be the smallest possible value of |x+ y|, where x and y are integerssuch that 3x〉 2y = n. Evaluate f(0) + f(1) + f(2) + · · ·+ f(2013).

Answer: 2416 First, we can use 3x 〉 2y = n to get x = n+2y3 . Thus |x + y| = |n+5y

3 |. Givena certain n, the only restriction on y is that 3 | n + 2y ⇐⇒ 3 | n + 5y. Hence the set of possiblex+ y equals the set of integers of the form n+5y

3 , which in turn equals the set of integers congruent to3−1n ≡ 2n (mod 5). (Prove this!)

Thus f(n) = |x + y| is minimized when x + y equals the least absolute remainder (2n)5 when 2n isdivided by 5, i.e. the number between 〉2 and 2 (inclusive) congruent to 2n modulo 5. We immediatelyfind f(n) = f(n + 5m) for all integers m, and the following initial values of f : f(0) = |(0)5| = 0;f(1) = |(2)5| = 2; f(2) = |(4)5| = 1; f(3) = |(6)5| = 1; and f(4) = |(8)5| = 2.

Since 2013 = 403·5〉2, it follows that f(0)+f(1)+· · ·+f(2013) = 403[f(0)+f(1)+· · ·+f(4)]〉f(2014) =403 · 6〉 2 = 2416.

10. [7] Let ω = cos 2π727 + i sin 2π

727 . The imaginary part of the complex number

13∏k=8

(1 + ω3k−1

+ ω2·3k−1)

Team Round

is equal to sinα for some angle α between 〉π2 and π2 , inclusive. Find α.

Answer: 12π727 Note that 727 = 36 〉 2. Our product telescopes to 1−ω313

1−ω37= 1−ω12

1−ω6 = 1 + ω6, which

has imaginary part sin 12π727 , giving α = 12π

727 .

Team Round

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HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

1. [5] Evaluate 2 + 5 + 8 + · · ·+ 101.

2. [5] Two fair six-sided dice are rolled. What is the probability that their sum is at least 10?

3. [5] A square is inscribed in a circle of radius 1. Find the perimeter of the square.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

4. [6] Find the minimum possible value of (x2 + 6x+ 2)2 over all real numbers x.

5. [6] How many positive integers less than 100 are relatively prime to 200? (Two numbers are relativelyprime if their greatest common factor is 1.)

6. [6] A right triangle has area 5 and a hypotenuse of length 5. Find its perimeter.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

7. [7] Marty and three other people took a math test. Everyone got a non-negative integer score. Theaverage score was 20. Marty was told the average score and concluded that everyone else scored belowaverage. What was the minimum possible score Marty could have gotten in order to definitively reachthis conclusion?

8. [7] Evaluate the expression1

2−1

2−1

2− · · ·1

2−1

2

,

where the digit 2 appears 2013 times.

9. [7] Find the remainder when 12 + 32 + 52 + · · ·+ 992 is divided by 1000.

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HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

10. [8] How many pairs of real numbers (x, y) satisfy the equation

y4 − y2 = xy3 − xy = x3y − xy = x4 − x2 = 0?

11. [8] David has a unit triangular array of 10 points, 4 on each side. A looping path is a sequenceA1, A2, . . . , A10 containing each of the 10 points exactly once, such that Ai and Ai+1 are adjacent(exactly 1 unit apart) for i = 1, 2, . . . , 10. (Here A11 = A1.) Find the number of looping paths in thisarray.

12. [8] Given that 622 + 1222 = 18728, find positive integers (n,m) such that n2 +m2 = 9364.

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HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

13. [9] Let S = {1, 2, . . . , 2013}. Find the number of ordered triples (A,B,C) of subsets of S such thatA ⊆ B and A ∪B ∪ C = S.

14. [9] Find all triples of positive integers (x, y, z) such that x2 + y − z = 100 and x+ y2 − z = 124.

15. [9] Find all real numbers x between 0 and 360 such that√

3 cos 10◦ = cos 40◦ + sinx◦.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

16. [10] A bug is on one exterior vertex of solid S, a 3 × 3 × 3 cube that has its center 1 × 1 × 1 cuberemoved, and wishes to travel to the opposite exterior vertex. Let O denote the outer surface of S(formed by the surface of the 3× 3× 3 cube). Let L(S) denote the length of the shortest path throughS. (Note that such a path cannot pass through the missing center cube, which is empty space.) Let

L(O) denote the length of the shortest path through O. What is the ratio L(S)L(O)?

17. [10] Find the sum of 1n over all positive integers n with the property that the decimal representation

of 1n terminates.

18. [10] The rightmost nonzero digit in the decimal expansion of 101! is the same as the rightmost nonzerodigit of n!, where n is an integer greater than 101. Find the smallest possible value of n.

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HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

19. [11] Let p, q, r, s be distinct primes such that pq − rs is divisible by 30. Find the minimum possiblevalue of p+ q + r + s.

20. [11] There exist unique nonnegative integers A,B between 0 and 9, inclusive, such that

(1001 ·A+ 110 ·B)2 = 57, 108, 249.

Find 10 ·A+B.

21. [11] Suppose A, B, C, and D are four circles of radius r > 0 centered about the points (0, r), (r, 0),(0,−r), and (−r, 0) in the plane. Let O be a circle centered at (0, 0) with radius 2r. In terms of r,what is the area of the union of circles A, B, C, and D subtracted by the area of circle O that is notcontained in the union of A, B, C, and D?

(The union of two or more regions in the plane is the set of points lying in at least one of the regions.)

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HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

22. [12] Let S be a subset of {1, 2, 3, . . . , 12} such that it is impossible to partition S into k disjointsubsets, each of whose elements sum to the same value, for any integer k ≥ 2. Find the maximumpossible sum of the elements of S.

23. [12] The number 989 · 1001 · 1007 + 320 can be written as the product of three distinct primes p, q, rwith p < q < r. Find (p, q, r).

24. [12] Find the number of subsets S of {1, 2, . . . 6} satisfying the following conditions:

• S is non-empty.

• No subset of S has the property that the sum of its elements is 10.

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HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

25. [13] Let a, b be positive reals with a > b > 12a. Place two squares of side lengths a, b next to each

other, such that the larger square has lower left corner at (0, 0) and the smaller square has lower leftcorner at (a, 0). Draw the line passing through (0, a) and (a + b, 0). The region in the two squareslying above the line has area 2013. If (a, b) is the unique pair maximizing a+ b, compute a

b .

26. [13] Trapezoid ABCD is inscribed in the parabola y = x2 such that A = (a, a2), B = (b, b2),C = (−b, b2), and D = (−a, a2) for some positive reals a, b with a > b. If AD +BC = AB + CD, andAB = 3

4 , what is a?

27. [13] Find all triples of real numbers (a, b, c) such that a2 + 2b2 − 2bc = 16 and 2ab− c2 = 16.

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HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

28. [15] Triangle ABC has AB = 4, BC = 3, and a right angle at B. Circles ω1 and ω2 of equal radii aredrawn such that ω1 is tangent to AB and AC, ω2 is tangent to BC and AC, and ω1 is tangent to ω2.Find the radius of ω1.

29. [15] Let 4XY Z be a right triangle with ∠XY Z = 90◦. Suppose there exists an infinite sequence ofequilateral triangles X0Y0T0, X1Y1T1, . . . such that X0 = X,Y0 = Y , Xi lies on the segment XZ for alli ≥ 0, Yi lies on the segment Y Z for all i ≥ 0, XiYi is perpendicular to Y Z for all i ≥ 0, Ti and Y areseparated by line XZ for all i ≥ 0, and Xi lies on segment Yi→1Ti→1 for i ≥ 1.

Let P denote the union of the equilateral triangles. If the area of P is equal to the area of XY Z, findXYY Z .

30. [15] Find the number of ordered triples of integers (a, b, c) with 1 ≤ a, b, c ≤ 100 and a2b+ b2c+ c2a =ab2 + bc2 + ca2.

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HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

31. [17] Chords AB and CD of circle ω intersect at E such that AE = 8, BE = 2, CD = 10, and∠AEC = 90◦. Let R be a rectangle inside ω with sides parallel to AB and CD, such that no point inthe interior of R lies on AB, CD, or the boundary of ω. What is the maximum possible area of R?

32. [17] Suppose that x and y are chosen randomly and uniformly from (0, 1). What is the probability

that⌊√

xy

⌋is even? Hint:

∑∞n=1

1n2 = π2

6 .

33. [17] On each side of a 6 by 8 rectangle, construct an equilateral triangle with that side as one edgesuch that the interior of the triangle intersects the interior of the rectangle. What is the total area ofall regions that are contained in exactly 3 of the 4 equilateral triangles?

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HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

34. [20] Find the number of positive integers less than 1000000 that are divisible by some perfect cubegreater than 1. Your score will be max

{0, b20− 200|1− k

S |c}

, where k is your answer and S is theactual answer.

35. [20] Consider the following 4 by 4 grid with one corner square removed:

You may start at any square in this grid and at each move, you may either stop or travel to an adjacentsquare (sharing a side, not just a corner) that you have not already visited (the square you start at isautomatically marked as visited). Determine the distinct number of paths you can take. Your scorewill be max

{0, b20− 200|1− k

S |c}

, where k is your answer and S is the actual answer.

36. [20] Pick a subset of at least four of the following seven numbers, order them from least to greatest,and write down their labels (corresponding letters from A through G) in that order: (A) π; (B)√

2 +√

3; (C)√

10; (D) 355113 ; (E) 16 tan→1 1

5 − 4 tan→1 1240 ; (F) ln(23); and (G) 2

√e. If the ordering of

the numbers you picked is correct and you picked at least 4 numbers, then your score for this problemwill be (N − 2)(N − 3), where N is the size of your subset; otherwise, your score is 0.

HMMT November 2013Saturday 9 November 2013

Guts Round

1. [5] Evaluate 2 + 5 + 8 + · · ·+ 101.

Answer: 1751 There are 1023 = 34 terms with average 2+101

2 , so their sum is 17 · 103 = 1751.

2. [5] Two fair six-sided dice are rolled. What is the probability that their sum is at least 10?

Answer: 16 There are 3, 2, 1 outcomes with sum 10, 11, 12, so the probability is 3+2+1

62 = 16 .

3. [5] A square is inscribed in a circle of radius 1. Find the perimeter of the square.

Answer: 4√

2 OR 8√2

The square has diagonal length 2, so side length√

2 and perimeter 4√

2.

4. [6] Find the minimum possible value of (x2 + 6x+ 2)2 over all real numbers x.

Answer: 0 This is ((x+ 3)2 − 7)2 ≥ 0, with equality at x+ 3 = ±√

7.

5. [6] How many positive integers less than 100 are relatively prime to 200? (Two numbers are relativelyprime if their greatest common factor is 1.)

Answer: 40 1 ≤ n < 100 is relatively prime to 200 if and only if it’s relatively prime to 100(200, 100 have the same prime factors). Thus our answer is φ(100) = 100 1

245 = 40.

6. [6] A right triangle has area 5 and a hypotenuse of length 5. Find its perimeter.

Answer: 5 + 3√

5 If x, y denote the legs, then xy = 10 and x2 + y2 = 25, so x+ y +√x2 + y2 =√

(x2 + y2) + 2xy + 5 =√

45 + 5 = 5 + 3√

5.

7. [7] Marty and three other people took a math test. Everyone got a non-negative integer score. Theaverage score was 20. Marty was told the average score and concluded that everyone else scored belowaverage. What was the minimum possible score Marty could have gotten in order to definitively reachthis conclusion?

Answer: 61 Suppose for the sake of contradiction Marty obtained a score of 60 or lower. Sincethe mean is 20, the total score of the 4 test takers must be 80. Then there exists the possibility of 2students getting 0, and the last student getting a score of 20 or higher. If so, Marty could not haveconcluded with certainty that everyone else scored below average.

With a score of 61, any of the other three students must have scored points lower or equal to 19 points.Thus Marty is able to conclude that everyone else scored below average.

8. [7] Evaluate the expression1

2−1

2−1

2− · · ·1

2−1

2

,

where the digit 2 appears 2013 times.

Answer: 20132014 Let f(n) denote the corresponding expression with the digit 2 appearing exactly n

times. Then f(1) = 12 and for n > 1, f(n) = 1

2−f(n−1) . By induction using the identity 12−N−1

N

= NN+1 ,

f(n) = nn+1 for all n ≥ 1, so f(2013) = 2013

2014 .

Guts Round

9. [7] Find the remainder when 12 + 32 + 52 + · · ·+ 992 is divided by 1000.

Answer: 650 We have S =∑49i=0(2i + 1)2 =

∑49i=0 4i2 + 4i + 1 = 4 · 49·50·996 + 4 · 49·502 + 50 ≡

700 + 900 + 50 (mod 1000) ≡ 650 (mod 1000).

10. [8] How many pairs of real numbers (x, y) satisfy the equation

y4 − y2 = xy3 − xy = x3y − xy = x4 − x2 = 0?

Answer: 9 We can see that if they solve the first and fourth equations, they are automaticallysolutions to the second and third equations. Hence, the solutions are just the 32 = 9 points where x, ycan be any of −1, 0, 1.

11. [8] David has a unit triangular array of 10 points, 4 on each side. A looping path is a sequenceA1, A2, . . . , A10 containing each of the 10 points exactly once, such that Ai and Ai+1 are adjacent(exactly 1 unit apart) for i = 1, 2, . . . , 10. (Here A11 = A1.) Find the number of looping paths in thisarray.

Answer: 60 There are 10 · 2 times as many loop sequences as loops. To count the number ofloops, first focus on the three corners of the array: their edges are uniquely determined. It’s now easyto see there are 3 loops (they form “V -shapes”), so the answer is 10 · 2 · 3 = 60.

12. [8] Given that 622 + 1222 = 18728, find positive integers (n,m) such that n2 +m2 = 9364.

Answer: (30, 92) OR (92, 30) If a2 + b2 = 2c, then (a+b2 )2 + (a−b2 )2 = 2a2+2b2

4 = a2+b2

2 = c. Thus,

n = 62+1222 = 92 and m = 122−62

2 = 30 works.

13. [9] Let S = {1, 2, . . . , 2013}. Find the number of ordered triples (A,B,C) of subsets of S such thatA ⊆ B and A ∪B ∪ C = S.

Answer: 52013 OR 125671 Let n = 2013. Each of the n elements can be independently placed in 5spots: there are 23 − 1 choices with element x in at least one set, and we subtract the 21 choices withelement x in set A but not B. Specifying where the elements go uniquely determines A,B,C, so thereare 5n = 52013 ordered triples.

14. [9] Find all triples of positive integers (x, y, z) such that x2 + y − z = 100 and x+ y2 − z = 124.

Answer: (12, 13, 57) Cancel z to get 24 = (y − x)(y + x − 1). Since x, y are positive, we have

y + x− 1 ≥ 1 + 1− 1 > 0, so 0 < y − x < y + x− 1. But y − x and y + x− 1 have opposite parity, so(y − x, y + x− 1) ∈ {(1, 24), (3, 8)} yields (y, x) ∈ {(13, 12), (6, 3)}.Finally, 0 < z = x2 + y − 100 forces (x, y, z) = (12, 13, 57).

15. [9] Find all real numbers x between 0 and 360 such that√

3 cos 10◦ = cos 40◦ + sinx◦.

Answer: 70, 110 (need both, but order doesn’t matter) Note that√

3 = 2 cos 30◦, so sinx◦ =

cos 20◦ =⇒ x ∈ {70, 110}.

16. [10] A bug is on one exterior vertex of solid S, a 3 × 3 × 3 cube that has its center 1 × 1 × 1 cuberemoved, and wishes to travel to the opposite exterior vertex. Let O denote the outer surface of S(formed by the surface of the 3× 3× 3 cube). Let L(S) denote the length of the shortest path throughS. (Note that such a path cannot pass through the missing center cube, which is empty space.) Let

L(O) denote the length of the shortest path through O. What is the ratio L(S)L(O)?

Answer:√29

3√5

OR√14515 By (*), the shortest route in O has length 2

√1.52 + 32 = 3

√5. By (**),

the shortest route overall (in S) has length 2√

1.52 + 12 + 22 =√

32 + 22 + 42 =√

29. Therefore the

desired ratio is√29

3√5

=√14515 .

Guts Round

(*) Suppose we’re trying to get from (0, 0, 0) to (3, 3, 3) through O. Then one minimal-length paththrough O is (0, 0, 0)→ (1.5, 0, 3)→ (3, 3, 3).

(**) Suppose we’re trying to get from (0, 0, 0) to (3, 3, 3) through S. Then the inner hole is [1, 2] ×[1, 2]× [1, 2], and one minimal-length path is (0, 0, 0)→ (1.5, 1, 2)→ (3, 3, 3).

To justify (*), note that we must at some point hit one of the three faces incident to (3, 3, 3), andtherefore one of the edges of those faces. Without loss of generality, the first of these edges (whichmust lie on a face incident to (0, 0, 0)) is {(t, 0, 3) : 0 ≤ t ≤ 3}. Then the shortest path goes directlyfrom the origin to the edge, and then directly to (3, 3, 3); t = 1.5 minimizes the resulting distance. (Onemay either appeal to the classic geometric “unfolding” argument, or just direct algebraic minimization.)

To justify (**), consider the portion of S “visible” from (0, 0, 0). It sees 3 mutually adjacent faces ofthe center cube “hole” (when looking inside the solid) and its 6 edges. (3, 3, 3) can also see these 6edges. The shortest path through S must be a straight line from the start vertex to some point P onthe surface of the center cube (+), and it’s easy to see, using the triangle inequality, that this point Pmust be on one of the 6 edges. Without loss of generality, it’s on the edge {(t, 1, 2) : 1 ≤ t ≤ 2}. Theremaining path is a straight line to (3, 3, 3). Then once again, t = 1.5 minimizes the distance. (Andonce again, one may either appeal to the classic geometric “unfolding” argument—though this timeit’s a little trickier—or just direct algebraic minimization.)

Comment. (+) can be proven in many ways. The rough physical intuition is that “a fully stretchedrubber band” with ends at (0, 0, 0) and (3, 3, 3) must not have any “wiggle room” (and so touches theinner cube). Perhaps more rigorously, we can try shortening a path that does not hit the center cube,for instance by “projecting the path down” towards the closest edge of the inner cube.

17. [10] Find the sum of 1n over all positive integers n with the property that the decimal representation

of 1n terminates.

Answer: 52 The decimal representation of 1

n terminates if and only if n = 2i5j for some nonnegative

integers i, j, so our desired sum is∑i≥0

∑j≥0

2−i5−j =∑i≥0

2−i∑j≥0

5−j = (1− 2−1)−1(1− 5−1)−1 =2

1

5

4=

5

2.

18. [10] The rightmost nonzero digit in the decimal expansion of 101! is the same as the rightmost nonzerodigit of n!, where n is an integer greater than 101. Find the smallest possible value of n.

Answer: 103 101! has more factors of 2 than 5, so its rightmost nonzero digit is one of 2, 4, 6, 8.Notice that if the rightmost nonzero digit of 101! is 2k (1 ≤ k ≤ 4), then 102! has rightmost nonzerodigit 102(2k) ≡ 4k (mod 10), and 103! has rightmost nonzero digit 103(4k) ≡ 2k (mod 10). Hencen = 103.

19. [11] Let p, q, r, s be distinct primes such that pq − rs is divisible by 30. Find the minimum possiblevalue of p+ q + r + s.

Answer: 54 The key is to realize none of the primes can be 2, 3, or 5, or else we would haveto use one of them twice. Hence p, q, r, s must lie among 7, 11, 13, 17, 19, 23, 29, . . .. These optionsgive remainders of 1 (mod 2) (obviously), 1,−1, 1,−1, 1,−1,−1, . . . modulo 3, and 2, 1, 3, 2, 4, 3, 4, . . .modulo 5. We automatically have 2 | pq − rs, and we have 3 | pq − rs if and only if pqrs ≡ (pq)2 ≡ 1(mod 3), i.e. there are an even number of −1 (mod 3)’s among p, q, r, s.

If {p, q, r, s} = {7, 11, 13, 17}, then we cannot have 5 | pq − rs, or else 12 ≡ pqrs ≡ (pq)2 (mod 5) is aquadratic residue. Our next smallest choice (in terms of p+ q + r + s) is {7, 11, 17, 19}, which works:7 · 17− 11 · 19 ≡ 22 − 4 ≡ 0 (mod 5). This gives an answer of 7 + 17 + 11 + 19 = 54.

20. [11] There exist unique nonnegative integers A,B between 0 and 9, inclusive, such that

(1001 ·A+ 110 ·B)2 = 57, 108, 249.

Find 10 ·A+B.

Guts Round

Answer: 75 We only need to bound for AB00; in other words, AB2 ≤ 5710 but (AB+1)2 ≥ 5710.A quick check gives AB = 75. (Lots of ways to get this...)

21. [11] Suppose A, B, C, and D are four circles of radius r > 0 centered about the points (0, r), (r, 0),(0,−r), and (−r, 0) in the plane. Let O be a circle centered at (0, 0) with radius 2r. In terms of r,what is the area of the union of circles A, B, C, and D subtracted by the area of circle O that is notcontained in the union of A, B, C, and D?

(The union of two or more regions in the plane is the set of points lying in at least one of the regions.)

Answer: 8r2 Solution 1. Let U denote the union of the four circles, so we seek

U − ([O]− U) = 2U − [O] = 2[(2r)2 + 4 · 1

2πr2]− π(2r)2 = 8r2.

(Here we decompose U into the square S with vertices at (±r,±r) and the four semicircular regions ofradius r bordering the four sides of U .)

Solution 2. There are three different kinds of regions: let x be an area of a small circle that does notcontain the two intersections with the other two small circles, y be an area of intersection of two smallcircles, and z be one of those four areas that is inside the big circle but outside all of the small circles.

Then the key observation is y = z. Indeed, adopting the union U notation from the previous solution,we have 4z = [O] − U = π(2r)2 − U , and by the inclusion-exclusion principle, 4y = [A] + [B] + [C] +[D] − U = 4π2 − U , so y = z. Now U = (4x + 4y) − (4z) = 4x. But the area of each x is simply 2r2

by moving the curved outward parts to fit into the curved inward parts to get a r × 2r rectangle. Sothe answer is 8r2.

22. [12] Let S be a subset of {1, 2, 3, . . . , 12} such that it is impossible to partition S into k disjointsubsets, each of whose elements sum to the same value, for any integer k ≥ 2. Find the maximumpossible sum of the elements of S.

Answer: 77 We note that the maximum possible sum is 78 (the entire set). However, this couldbe partitioned into 2 subsets with sum 39: {1, 2, 3, 10, 11, 12} and {4, 5, 6, 7, 8, 9}. The next largestpossible sum is 77 (the entire set except 1). If k ≥ 2 subsets each had equal sum, then they wouldhave to be 7 subsets with sum 11 each or 11 subsets with sum 7 each. However, the subset containing12 will have sum greater than 11; hence there is no way to partition the subset {2, . . . , 12} into equalsubsets.

23. [12] The number 989 · 1001 · 1007 + 320 can be written as the product of three distinct primes p, q, rwith p < q < r. Find (p, q, r).

Answer: (991, 997, 1009) Let f(x) = x(x − 12)(x + 6) + 320 = x3 − 6x2 − 72x + 320, so that

f(1001) = 989 · 1001 · 1007 + 320. But f(4) = 4(−8)(10) + 320 = 0, so f(x) = (x− 4)(x2 − 2x− 80) =(x− 4)(x− 10)(x+ 8).

Thus f(1001) = 991 · 997 · 1009, as desired.

24. [12] Find the number of subsets S of {1, 2, . . . 6} satisfying the following conditions:

• S is non-empty.

• No subset of S has the property that the sum of its elements is 10.

Answer: 34 We do casework based on the largest element of S. Call a set n-free if none of itssubsets have elements summing to n.

Case 1: The largest element of S is 6. Then 4 /∈ S. If 5 /∈ S, then we wish to find all 4-free subsetsof {1, 2, 3} (note that 1 + 2 + 3 = 6 < 10). We just cannot include both 1, 3, so we have 2(22 − 1) = 6choices here.

If 5 ∈ S, then we want 4, 5-free subsets of {1, 2, 3}. The only 4-but-not-5-free subset is {2, 3}, so wehave 6− 1 choices here, for a case total of 6 + 5 = 11.

Guts Round

Case 2: The largest element of S is 5. We seek 5, 10-free subsets of {1, 2, 3, 4}. We just cannothave both 1, 4 or both 2, 3 (note that getting 10 requires the whole set), so we have (22 − 1)2 = 9subsets in this case.

Case 3: The largest element of S is at most 4. (So we want a 4-free subset of {1, 2, 3, 4}.) Theonly way to sum to 10 with 1, 2, 3, 4 is by using all the terms, so we simply discount the empty set and{1, 2, 3, 4}, for a total of 24 − 2 = 14 subsets.

In conclusion, the total number of subsets is 11 + 9 + 14 = 34.

25. [13] Let a, b be positive reals with a > b > 12a. Place two squares of side lengths a, b next to each

other, such that the larger square has lower left corner at (0, 0) and the smaller square has lower leftcorner at (a, 0). Draw the line passing through (0, a) and (a + b, 0). The region in the two squareslying above the line has area 2013. If (a, b) is the unique pair maximizing a+ b, compute a

b .

Answer: 53 Let t = a

b ∈ (1, 2); we will rewrite the sum a + b as a function of t. The area

condition easily translates to a2−ab+2b2

2 = 2013, or b2(t2 − t + 2) = 4026 ⇐⇒ b =√

4026t2−t+2 . Thus

a+ b is a function f(t) = (1 + t)√

4026t2−t+2 of t, and our answer is simply the value of t maximizing f ,

or equivalently g(t) = f2

4026 = (1+t)2

t2−t+2 , over the interval (1, 2). (In general, such maximizers/maximumsneed not exist, but we shall prove there’s a unique maximum here.)

We claim that λ = 167 is the maximum of (1+t)2

t2−t+2 . Indeed,

λ− g(t) =(λ− 1)t2 − (λ+ 2)t+ (2λ− 1)

t2 − t+ 2

=1

7

9t2 − 30t+ 25

t2 − t+ 2=

1

7

(3t− 5)2

(t− 12 )2 + 7

4

≥ 0

for all reals t (not just t ∈ (1, 2)), with equality at t = 53 ∈ (1, 2).

Comment. To motivate the choice of λ, suppose λ were the maximum of f , attained at t0 ∈ (1, 2);then h(t) = λ(t2 − t+ 2)− (t+ 1)2 is quadratic and nonnegative on (1, 2), but zero at t = t0. If g is anontrivial quadratic (nonzero leading coefficient), then t0 must be a double root, so g has determinant0. Of course, g could also be constant or linear over (1, 2), but we can easily rule out both of thesepossibilities.

Alternatively, we can simply take a derivative of f to find critical points.

26. [13] Trapezoid ABCD is inscribed in the parabola y = x2 such that A = (a, a2), B = (b, b2),C = (−b, b2), and D = (−a, a2) for some positive reals a, b with a > b. If AD +BC = AB + CD, andAB = 3

4 , what is a?

Answer: 2740 Let t = 3

4 , so 2a + 2b = 2√

(a− b)2 + (a2 − b2)2 = 2t gives a = b = t and

t2 = (a− b)2[1 + (a+ b)2] = (a− b)2[1 + t2]. Thus a =t+ t√

1+t2

2 =34+

35

2 = 2740 .

27. [13] Find all triples of real numbers (a, b, c) such that a2 + 2b2 − 2bc = 16 and 2ab− c2 = 16.

Answer: (4, 4, 4), (−4,−4,−4) (need both, but order doesn’t matter) a2 + 2b2 − 2bc and 2ab− c2

are both homogeneous degree 2 polynomials in a, b, c, so we focus on the homogeneous equation a2 +2b2−2bc = 2ab− c2, or (a− b)2 + (b− c)2 = 0. So a = b = c, and a2 = 2ab− c2 = 16 gives the solutions(4, 4, 4) and (−4,−4,−4).

28. [15] Triangle ABC has AB = 4, BC = 3, and a right angle at B. Circles ω1 and ω2 of equal radii aredrawn such that ω1 is tangent to AB and AC, ω2 is tangent to BC and AC, and ω1 is tangent to ω2.Find the radius of ω1.

Answer: 57 Solution 1. Denote by r the common radius of ω1, ω2, and let O1, O2 be the centers

of ω1 and ω2 respectively. Suppose ωi hits AC at Bi for i = 1, 2, so that O1O2 = B1B2 = 2r.

Guts Round

Extend angle bisector AO1 to hit BC at P . By the angle bisector theorem and triangle similarity4AB1O1 ∼ 4ABP , we deduce r

AB1= BP

AB = 34+5 . Similarly, r

CB2= 4

3+5 , so

5 = AC = AB1 +B1B2 +B2C = 3r + 2r + 2r = 7r,

or r = 57 .

Solution 2. Use the same notation as in the previous solution, and let α = 12∠A. By constructing

right triangles with hypotenuses AO1, O1O2, and O2C and legs parallel to AB and BC, we obtain

4 = AB = r cotα+ 2r cos∠A+ r.

But cotα = 1+cos 2αsin 2α =

1+ 45

35

= 3 and cos∠A = 45 , so the above equation simplifies to

4 = r(3 +8

5+ 1) =

28r

5,

or r = 57 .

29. [15] Let 4XY Z be a right triangle with ∠XY Z = 90◦. Suppose there exists an infinite sequence ofequilateral triangles X0Y0T0, X1Y1T1, . . . such that X0 = X,Y0 = Y , Xi lies on the segment XZ for alli ≥ 0, Yi lies on the segment Y Z for all i ≥ 0, XiYi is perpendicular to Y Z for all i ≥ 0, Ti and Y areseparated by line XZ for all i ≥ 0, and Xi lies on segment Yi−1Ti−1 for i ≥ 1.

Let P denote the union of the equilateral triangles. If the area of P is equal to the area of XY Z, findXYY Z .

Answer: 1 For any region R, let [R] denote its area.

Let a = XY , b = Y Z, ra = X1Y1. Then [P] = [XY T0](1 + r2 + r4 + · · · ), [XY Z] = [XY Y1X1](1 +r2 + r4 + · · · ), Y Y1 = ra

√3, and b = ra

√3(1 + r + r2 + · · · ) (although we can also get this by similar

triangles).

Hence a2√3

4 = 12 (ra+ a)(ra

√3), or 2r(r + 1) = 1 =⇒ r =

√3−12 . Thus XY

Y Z = ab = 1−r

r√3

= 1.

30. [15] Find the number of ordered triples of integers (a, b, c) with 1 ≤ a, b, c ≤ 100 and a2b+ b2c+ c2a =ab2 + bc2 + ca2.

Answer: 29800 This factors as (a− b)(b− c)(c− a) = 0. By the inclusion-exclusion principle, weget 3 · 1002 − 3 · 100 + 100 = 29800.

31. [17] Chords AB and CD of circle ω intersect at E such that AE = 8, BE = 2, CD = 10, and∠AEC = 90◦. Let R be a rectangle inside ω with sides parallel to AB and CD, such that no point inthe interior of R lies on AB, CD, or the boundary of ω. What is the maximum possible area of R?

Answer: 26 + 6√

17 By power of a point, (CE)(ED) = (AE)(EB) = 16, and CE+ED = CD =10. Thus CE,ED are 2, 8. Without loss of generality, assume CE = 8 and DE = 2.

Assume our circle is centered at the origin, with points A = (−3, 5), B = (−3,−5), C = (5,−3),D = (−5,−3), and the equation of the circle is x2 + y2 = 34. Clearly the largest possible rectanglemust lie in the first quadrant, and if we let (x, y) be the upper-right corner of the rectangle, then the

area of the rectangle is (x+ 3)(y+ 3) = 9 + 6(x+ y) +xy ≤ 9 + 12√

x2+y2

2 + x2+y2

2 = 26 + 6√

17, where

equality holds if and only if x = y =√

17.

32. [17] Suppose that x and y are chosen randomly and uniformly from (0, 1). What is the probability

that⌊√

xy

⌋is even? Hint:

∑∞n=1

1n2 = π2

6 .

Answer: 1− π2

24 OR 24−π2

24 Note that for every positive integer n, the probability that b√

xy c = n

is just the area of the triangle formed between (0, 0), (1, 1n2 ), (1, 1

(n+1)2 ), which is just 12

(1n2 − 1

(n+1)2

).

Guts Round

Thus the probability that b√

xy c is odd is

∞∑k=1

1

2

(1

(2k − 1)2− 1

(2k)2

)=

1

2

∞∑k=1

(1

(2k − 1)2+

1

(2k)2

)−∞∑k=1

1

(2k)2

=1

2

∞∑n=1

1

n2− 1

4

∞∑k=1

1

k2

=π2

12− π2

24=π2

24.

Thus our answer is just 1− π2

24 .

33. [17] On each side of a 6 by 8 rectangle, construct an equilateral triangle with that side as one edgesuch that the interior of the triangle intersects the interior of the rectangle. What is the total area ofall regions that are contained in exactly 3 of the 4 equilateral triangles?

Answer: 96√3−154√3

OR 288−154√3

3 OR 96− 154√3

OR 96− 154√3

3 Let the rectangle be ABCD with

AB = 8 and BC = 6. Let the four equilateral triangles be ABP1, BCP2, CDP3, and DAP4 (forconvenience, call them the P1-, P2-, P3-, P4- triangles). Let W = AP1 ∩DP3, X = AP1 ∩DP4, andY = DP4 ∩ CP2. Reflect X,Y over the line P2P4 (the line halfway between AB and DC) to pointsX ′, Y ′.

First we analyze the basic configuration of the diagram. Since AB = 8 < 2 ·6√32 , the P2-, P4- triangles

intersect. Furthermore, AP1 ⊥ BP2, so if T = BP2 ∩ AP1, then BP2 = 6 < 4√

3 = BT . Therefore P2

lies inside triangle P1BA, and by symmetry, also triangle P3DC.

It follows that the area we wish to compute is the union of two (congruent) concave hexagons, one ofwhich is WXY P2Y

′X ′. (The other is its reflection over Y Y ′, the mid-line of AD and BC.) So we seek

2[WXY P2Y′X ′] = 2([WXP4X

′]− [P2Y P4Y′]).

It’s easy to see that [WXP4X′] = 1

3 [ADP4] = 1362√3

4 = 3√

3, since WXP4X′ and its reflections over

lines DWX ′ and AWX partition 4ADP4.

It remains to consider P2Y P4Y′, a rhombus with (perpendicular) diagonals P2P4 and Y Y ′. If O denotes

the intersection of these two diagonals (also the center of ABCD), thenOP2 is P2B√32 −

12AB = 3

√3−4,

the difference between the lengths of the P2-altitude in 4CBP2 and the distance between the parallellines Y Y ′, CB. Easy angle chasing gives OY = OP2√

3, so

[P2Y P4Y′] = 4 · OP2 ·OY

2=

2√3OP 2

2 =2√3

(3√

3− 4)2 =86− 48

√3√

3,

and our desired area is

2[WXP4X′]− 2[P2Y P4Y

′] = 6√

3− 172− 96√

3√3

=96√

3− 154√3

,

or 288−154√3

3 .

34. [20] Find the number of positive integers less than 1000000 that are divisible by some perfect cubegreater than 1. Your score will be max

{0, b20− 200|1− k

S |c}

, where k is your answer and S is theactual answer.

Answer: 168089 Using the following code, we get the answer (denoted by the variable ans):

ans = 0

for n in xrange(1,1000000):

Guts Round

divisible_by_cube = True

for i in xrange(2,101):

if n%(i*i*i)==0:

divisible_by_cube = False

break

if divisible_by_cube: ans = ans + 1

print ans

This gives the output

168089

Alternatively, let N = 1000000 and denote by P the set of primes. Then by PIE, the number ofn ∈ (0, N) divisible by a nontrivial cube, or equivalently, by p3 for some p ∈ P , is∑

p∈PbN − 1

p3c −

∑p<q∈P

bN − 1

p3q3c ± · · · ,

which deviates from∑p∈P

N − 1

p3−∑

p<q∈P

N − 1

p3q3± · · · = (N − 1)(1−

∏p∈P

(1− p−3))

by at most the sum of

• N1/3 supt∈R|t− btc| = N1/3, for terms b N−1p31···p3r

c with p1 · · · pr ≤ (N − 1)1/3, and

• (N − 1)∑k>(N−1)1/3 k

−3 < (N − 1)[(N − 1)−1 +∫∞(N−1)1/3 x

−3 dx] = 1 + (N − 1) (N−1)−2/3

2 =

O(N1/3), for the remaining terms.

So we are really interested in 106 − 106∏p∈P (1 − p−3) (which, for completeness, is 168092.627 . . .).

There are a few simple ways to approximate this:

• We can use a partial product of∏p∈P (1 − p−3). Using just 1 − 2−3 = 0.875 gives an answer of

125000 (this is also just the number of x ≤ N divisible by 23 = 8), (1 − 2−3)(1 − 3−3) ≈ 0.843gives 157000 (around the number of x divisible by 23 or 33), etc. This will give a lower bound,of course, so we can guess a bit higher. For instance, while 157000 gives a score of around 7,rounding up to 160000 gives ≈ 10.

• We can note that∏p∈P (1− p−3) = ζ(3)−1 is the inverse of 1 + 2−3 + 3−3 + · · · . This is a bit less

efficient, but successive partial sums (starting with 1 + 2−3) give around 111000, 139000, 150000,157000, etc. Again, this gives a lower bound, so we can guess a little higher.

• We can optimize the previous approach with integral approximation after the rth term: ζ(3) isthe sum of 1+2−3 + · · ·+r−3 plus something between

∫∞r+1

x−3 dx = 12 (r+1)−2 and

∫∞rx−3 dx =

12r−2. Then starting with r = 1, we get intervals of around (111000, 334000), (152000, 200000),

(161000, 179000), (165000, 173000), etc. Then we can take something like the average of the twoendpoints as our guess; such a strategy gets a score of around 10 for r = 2 already, and ≈ 17 forr = 3.

35. [20] Consider the following 4 by 4 grid with one corner square removed:

You may start at any square in this grid and at each move, you may either stop or travel to an adjacentsquare (sharing a side, not just a corner) that you have not already visited (the square you start at isautomatically marked as visited). Determine the distinct number of paths you can take. Your scorewill be max

{0, b20− 200|1− k

S |c}

, where k is your answer and S is the actual answer.

Answer: 14007

Guts Round

36. [20] Pick a subset of at least four of the following seven numbers, order them from least to greatest,and write down their labels (corresponding letters from A through G) in that order: (A) π; (B)√

2 +√

3; (C)√

10; (D) 355113 ; (E) 16 tan−1 1

5 − 4 tan−1 1240 ; (F) ln(23); and (G) 2

√e. If the ordering of

the numbers you picked is correct and you picked at least 4 numbers, then your score for this problemwill be (N − 2)(N − 3), where N is the size of your subset; otherwise, your score is 0.

Answer: F,G,A,D,E,B,C OR F < G < A < D < E < B < C OR C > B > E > D > A > G > F

We have ln(23) < 2√e < π < 355

113 < 16 tan−1 15 − 4 tan−1 1

240 <√

2 +√

3 <√

10.

Guts Round

HMMT 2013Saturday 16 February 2013

HMIC

1. [5] Let S be a set of size n, and k be a positive integer. For each 1 ≤ i ≤ kn, there is a subset Si ⊂ Ssuch that |Si| = 2. Furthermore, for each e ∈ S, there are exactly 2k values of i such that e ∈ Si.Show that it is possible to choose one element from Si for each 1 ≤ i ≤ kn such that every element ofS is chosen exactly k times.

2. [7] Find all functions f : R → R such that, for all real numbers x, y,

(x− y)(f(x)− f(y)) = f(x− f(y))f(f(x)− y).

3. [8] Triangle ABC is inscribed in a circle ω such that ∠A = 60◦ and ∠B = 75◦. Let the bisector ofangle A meet BC and ω at E and D, respectively. Let the reflections of A across D and C be D′ andC ′, respectively. If the tangent to ω at A meets line BC at P , and the circumcircle of APD′ meetsline AC at F 6= A, prove that the circumcircle of C ′FE is tangent to BC at E.

4. [10] A subset U ⊂ R is open if for any x ∈ U , there exist real numbers a, b such that x ∈ (a, b) ⊂ U .Suppose S ⊂ R has the property that any open set intersecting (0, 1) also intersects S. Let T be acountable collection of open sets containing S. Prove that the intersection of all of the sets of T is nota countable subset of R.

(A set Γ is countable if there exists a bijective function f : Γ → Z.)

5. [12]

(a) Given a finite set X of points in the plane, let fX(n) be the largest possible area of a polygon withat most n vertices, all of which are points of X. Prove that if m,n are integers with m ≥ n > 2,then fX(m) + fX(n) ≥ fX(m+ 1) + fX(n− 1).

(b) Let P0 be a 1-by-2 rectangle (including its interior), and inductively define the polygon Pi to bethe result of folding Pi−1 over some line that cuts Pi−1 into two connected parts. The diameter

of a polygon Pi is the maximum distance between two points of Pi. Determine the smallestpossible diameter of P2013.

(In other words, given a polygon Pi−1, a fold of Pi−1 consists of a line l dividing Pi−1 into twoconnected parts A and B, and the folded polygon Pi = A ∪ Bl, where Bl is the reflection of Bover the line l.)

HMMT 2013Saturday 16 February 2013

HMIC

1. [5] Let S be a set of size n, and k be a positive integer. For each 1 ≤ i ≤ kn, there is a subset Si ⊂ Ssuch that |Si| = 2. Furthermore, for each e ∈ S, there are exactly 2k values of i such that e ∈ Si.Show that it is possible to choose one element from Si for each 1 ≤ i ≤ kn such that every element ofS is chosen exactly k times.

Answer: N/A Consider the undirected graph G = (S, E) where the elements of S are the vertices,

and for each 1 ≤ i ≤ kn, there is an edge between the two elements of Si. (Note that there might bemultiedges if two subsets are the same, but there are no self-loops.) Consider any connected componentC of G, which must be a 2k-regular graph, and because 2k is even, C has an Eulerian circuit. Pickan orientation of the circuit, and hence a direction for each edge in C. Then, for each i such that theedge corresponding to Si is in C, pick the element that is pointed to by that edge. Since the circuitgoes into each vertex of C k times, each element in C is picked exactly k times as desired. Repeatingfor each connected component finishes the problem.

2. [7] Find all functions f : R → R such that, for all real numbers x, y,

(x− y)(f(x)− f(y)) = f(x− f(y))f(f(x)− y).

Answer: f(x) = 0, f(x) = x First, suppose that f(0) 6= 0. Then, by letting x = y = 0, we get that

either f(f(0)) or f(−f(0)) is zero. The former gives a contradiction by plugging in x = f(0), y = 0into the original equation. Thus, f(−f(0)) = 0.

Now plugging in x = 0, y = −f(0) gives that f(2f(0)) = f(0). Finally, we note that by symmetry ofthe left hand side of the functional equation, f(x − f(y))f(f(x) − y) = f(y − f(x))f(f(y) − x), andletting y = 0 and x = 2f(0) we get f(f(0))f(f(0)) = f(−f(0))f(−f(0)) = 0, giving and so f(f(0) = 0,again a contradiction. Therefore, f(0) = 0.

Setting y = 0 yields xf(x) = f(x)f(f(x)), so if f(x) 6= 0, then f(f(x)) = x. On the other hand, settingy = f(x) yields (x−f(x))(f(x)−f(f(x)) = f(x−f(f(x)))f(0), so in fact f(f(x)) = f(x) → f(x) = x.Thus, for all x, f(x) = 0 or f(x) = x.

Clearly, f ≡ 0 satisfies the original equation, so suppose there exists a non-zero x with f(x) = x. Then,for all y 6= x, if f(y) 6= 0, the original equation yields x(x − y) = xf(x − y). Thus, f(x − y) = x − y.Similarly, if f(y) = y, we find (x− y)2 = f(x− y)2, so f(x− y) = x− y. It follows that f must be theidentity, which indeed is a solution to the original equation. The proof is complete.

3. [8] Triangle ABC is inscribed in a circle ω such that ∠A = 60◦ and ∠B = 75◦. Let the bisector ofangle A meet BC and ω at E and D, respectively. Let the reflections of A across D and C be D′ andC ′, respectively. If the tangent to ω at A meets line BC at P , and the circumcircle of APD′ meetsline AC at F 6= A, prove that the circumcircle of C ′FE is tangent to BC at E.

Answer: N/A We will show that CE2 = (CF )(CC ′). By a simple computation using the given

angles, one may find that this is equivalent to CF = AC −AB, or AF = 2AC −AB.

We compute AF by trigonometry. Assume for simplicity that AC = 12 , so AD′ = 2AD = 2AC = 1

because △ACD is isosceles by angle chasing. We first compute PD′ by the law of cosines on triangleAPD′, which yields PD′2 = AP 2+AD′2−2(AP )(AD′) cos 75 = AP 2+1−2AP cos 75. We may easily

compute AP by the law of sines in triangle BAP to be 1√2. Thus, PD′2 = 1

2 + 1−√2√6−

√2

4 = 4−√3

2 .

By the law of sines within the circumcircle of APD′F , we have FD′

sin 30 = PD′

sin 75 . From this, we find

that FD′ = PD′ sin 30sin 75 = 4−

√3√

2+√6. Thus, we have by the law of cosines on triangle D′AF that AF 2 +

AD′2 − 2(AF )(AD′) cos 30 = AF 2 + 1 −√3AF = FD′2 = 4−

√3

4+2√3. Finally, solving for AF yields

AF =√3±(2

√3−3)

2 from which we indeed get AF = 3−√3

2 which equals 2AC −AB.

HMIC

4. [10] A subset U ⊂ R is open if for any x ∈ U , there exist real numbers a, b such that x ∈ (a, b) ⊂ U .Suppose S ⊂ R has the property that any open set intersecting (0, 1) also intersects S. Let T be acountable collection of open sets containing S. Prove that the intersection of all of the sets of T is nota countable subset of R.

(A set Γ is countable if there exists a bijective function f : Γ → Z.)

Answer: N/A If S is uncountable then we’re done, so assume S is countable. We may also assume

that the supersets are a chain V1 ⊃ V2 ⊃ V3 ⊃ · · · by taking intersections.

We will use the following fact from point set topology:

If K1 ⊃ K2 ⊃ · · · is a sequence of nonempty compact sets, then their intersection is nonempty.

Now, we construct an uncountable family of such sequences K1 ⊃ K2 ⊃ · · · such that Ki is a nontrivial(i.e. [a, b] with b > a) closed interval contained in Vi, and any two such sequences have disjointintersection. The construction proceeds as follows. At each step, we choose out of countably manyoptions one closed interval Ki+1 ⊂ Ki.

To choose K1, we claim that there exist countably many disjoint closed intervals in V1. To do this,choose an open interval in V1 of length at most 1/2, and take some closed interval inside it. In theremainder of (0, 1), which has measure at least 1/2 and where V1 is still dense, choose another openinterval of length at most 1/4, and a closed interval inside it. This process can be repeated indefinitelyto find a countably infinite family of nontrivial disjoint closed intervals in V1. Choose one of them tobe K1.

Now, inductively apply this construction on Ki to find countably many choices for Ki+1, all disjoint.As a result, we get a total of N

N, i.e. uncountably many, choices of sequences Ki satisfying thegiven properties. Any pair of sequences eventually have disjoint Ki after some point, so no twointersections will intersect. Thus we can choose a point out of the intersection of each Ki sequence tofind uncountably many points in the intersection of all the Vi, as desired.

5. [12]

(a) Given a finite set X of points in the plane, let fX(n) be the largest possible area of a polygon withat most n vertices, all of which are points of X. Prove that if m,n are integers with m ≥ n > 2,then fX(m) + fX(n) ≥ fX(m+ 1) + fX(n− 1).

(b) Let P0 be a 1-by-2 rectangle (including its interior), and inductively define the polygon Pi to bethe result of folding Pi−1 over some line that cuts Pi−1 into two connected parts. The diameterof a polygon Pi is the maximum distance between two points of Pi. Determine the smallestpossible diameter of P2013.

(In other words, given a polygon Pi−1, a fold of Pi−1 consists of a line l dividing Pi−1 into twoconnected parts A and B, and the folded polygon Pi = A ∪ Bl, where Bl is the reflection of Bover the line l.)

Answer: N/A

Note from http://artofproblemsolving.com/community/c129h529051p3039219. Minor slip: InSection 1.3, Case 2 (for #5), how are B′ and A′ defined exactly? Right now it’s not clear to me thatthey have m + 1 and n − 1 vertices, respectively, since it’s not always possible to find Bi, Bi′ withf(i′) − f(i) = i′ − i, is it? (The simplest counterexample is when m = 3, n = 5, and the Aj , Bk arepairwise distinct.)

[b]Edit:[/b] I think it should be f(i′) − f(i) = i′ − i + 1 (which is not hard to find using a discretecontinuity argument) and then B′ = B1 . . . BiAf(i)+1 . . . Af(i′)Bi′+1 . . . Bm (A′ the complement of B′),in which case everything seems to work out. (We are assuming at most two points lie between Bk, Bk+1

for all k; otherwise we can just use Case 1.) Thanks to Tim for clarifying most of this! I probablyshould have tried the m = 3, n = 5 case a little harder before posting this, since it generalizes fairlyeasily.

HMIC

1 CONVEXITY OF THE LARGEST AREA OF A POLYGON FUNCTION

1 Convexity of the largest area of a polygon function

(a)

Let V be a finite set of points in the plane. Let fV be a function that takes integers ≥ 3 as input, andoutputs a polygon of largest area with vertices in V .

Lemma 1.1. [fV (a+ 1)] + [fV (a− 1)] ≤ 2[fV (a)], where [P ] represents the area of polygon P .

Proof. Let the vertices of fV (a + 1) be 1a+1, 2a+1 . . . , (a + 1)a+1, and let the vertices of fv(a − 1) be1a−1, 2a−1, . . . , (a− 1)a−1.

Lemma 1.2. The collection of vertices

W = (1a−1, 2a−1, . . . (a− 1)a−1, 1a+1, . . . , (a+ 1)a+1)

. It’s clear that W ⊂ V . are in convex configuration.

Proof. Suppose that the vertex sx ∈ W (for some integers x ∈ {a − 1, a + 1}, s) is inside the convexhull of W . Then there exists a vertex k on the convex hull of W such that replacing vertex sx in fV (x)with k increases the area of the polygon fV (x). However, fV (x) is defined to be the polygon with xsides with maximum area. Contradiction. This proves Lemma 2.

Therefore the vertices in W are in convex configuration. Label them with 1, 2, . . . 2a in clockwise order,starting from an arbitrary vertex. Let (2a) denote the vertex labeled (2a). (The vertex (1) will beshorthanded as 1 when there is no ambiguity).

Lemma 1.3. We have the inequality

[fV (a+ 1)] + [fV (a− 1)] ≤ [P1357 . . . (2a− 1)] + [P2468 . . . (2a)].

(Throughout, the symbol P denotes an arbitrary polygon.)

Proof. It suffices to show that

2[P1234 . . . (2a)]− [fV (a+ 1)]− [fV (a− 1)]

≥ 2[P1234 . . . (2a)]− [P1357 . . . (2a− 1)]− [P2468 . . . (2a)].

We will do this using the next few lemmas.

Lemma 1.4. Let x and y be integers ≤ 2a. If ray−−−−−−−→(x)(x+ 1) intersects (or is parallel to) ray

−−−−−−→(y)(y − 1).

Then [P(x)(x+1)(x+2) . . . (y− 1)(y)] ≥ [△(x)(x+1)(x+2)]+ [△(x+1)(x+2)(x+3)]+ . . .+△[(y−2)(y − 1)(y)] (where (2a+ k) is equivalent to (k) for all k).

Proof. [P(x)(x+1)(x+2) . . . (y−1)(y)] = [△(x)(x+1)(x+2)]+[△(x)(x+2)(x+3)]+. . .+[△(x)(y−1)(y)].

Suppose ray−−−−−−−→(x)(x+ 1) intersects ray

−−−−−−→(y)(y − 1) at point D, the convexity of W tells us that all vertices

in between x and y exclusive going clockwise from x are contained inside triangle (x)D(y). It followsthat for any x′ between x and y exclusive going clockwise from x, that (x) is farther from line (x′)(x′ + 1)than (x′ − 1). Therefore

[△(x)(x′)(x′ + 1)] ≤ [△(x′ − 1)(x′)(x′ + 1)].

Summing these relations over all x in between x and y exclusive going clockwise from x gives thestatement of Lemma 4, as desired.

An identical argument works if ray−−−−−−−→(x)(x+ 1) and ray

−−−−−−→(y)(y − 1) are parallel, except instead of D, all

the points between x and y are bounded by the two rays and the line (x)(y).

This proves Lemma 4.

HMIC

2 FOLDING A PAPER

Let z and s be integers such that z ∈ {a − 1, a + 1} and both vertices sz and (s + 1)z are adjacentvertices of fV (z). u and v are positive integers defined such that sz has label (u) and sz+1 has label(v).

Lemma 1.5. Rays−−−−−−−→(u)(u+ 1) and

−−−−−−→(v)(v − 1) intersect or are parallel to each other.

Proof. Since (u), (u + 1), (v − 1), (v) are in convex formation, then either−−−−−−−→(u)(u+ 1) and

−−−−−−→(v)(v − 1)

intersect or are parallel to each other, or else−−−−−−−→(u+ 1)(u) and

−−−−−−→(v − 1)(v) intersect. If we suppose that

the lemma statement does not hold for the sake of contradiction, then it follows that rays−−−−−−−→(u+ 1)(u) and−−−−−−→

(v − 1)(v) must intersect. Let Q be the collection of vertices between v and u in W , going clockwise,

inclusive. Since Q is in convex formation (it’s a subset of W ), and−−−−−−−→(u+ 1)(u) and

−−−−−−→(v + 1)(v) intersect,

it follows that u + 1 is farther than u from any line joining two vertices of Q. Therefore removing sz(which is the same vertex as (u)) from fV (z) and replacing it with vertex (u+ 1) increases the area offV (z), contradicting the assertion that fV (z) was the polygon of z vertices with maximal area on theset V . Therefore Lemma 5 is proven.

Lemma 5 allows us to apply Lemma 4 to the polygons formed by taking (P1234 . . . (2a)) and subtractingout fV (a+1)], as well as the polygons formed by taking (P1234 . . . (2a)) and subtracting out fV (a−1).Adding the results tells us that

2[P1234 . . . (2a)]− [fV (a+ 1)]− [fV (a− 1)]

≥ 2[P1234 . . . (2a)]− [P1357 . . . (2a− 1)]− [P2468 . . . (2a)],

which is sufficient to prove Lemma 3.

Since (P135 . . . (2a − 1)]), (P246 . . . (2a)) ≤ fV (a) since (1), (2), . . . (2a) ∈ V . Using this fact withLemma 3, we have proven Lemma 1 as desired.

Lemma 1.6. Any function g from integers to real numbers that satisfies g(x+ 1) + g(x− 1) ≤ 2g(x)satisfies the inequality g(x)+g(y) ≤ g(x+1)+g(y−1) for y−1 ≥ x+1, and is thus a convex function.

Proof. We proceed by induction on y − x. When y − x = 2, the problem is trivial.

For y−x > 2, note that g(x)+g(x+2)+g(y)+g(y−2) ≤ 2g(x+1)+2g(y−1), but the inductive statementshows that g(x+2)+ g(y− 2) ≥ g(x+ 1)+ g(y− 1). It follows that g(x) + g(y) ≤ g(x+ 1)+ g(y− 1),which completes our induction as desired.

It follows from the preceding lemma that fV (a) is a convex function, as desired.

2 Folding a paper

Let the smallest diameter be M . We claim that M is√2

21006 . Let the P0 be a 1 by 2 rectangle. For0 ≤ i ≤ 2012, polygon Pi+1 is polygon Pi folded across line li. Line li splits Pi into P ′

i and P ′′i , and

P ′′i is reflected to P ′′′

i , and Pi+1 = P ′i ∪ P ′′′

i .

Claim 1. For 1 ≤ i ≤ 2013, Pi can be expressed as the union of no more than 2i convex polygons,and the average number of sides of these polygons is not more than 4.

Proof. We proceed by induction. P0 clearly satisfies the inductive hypothesis. Suppose that Pi is theunion of convex polygons A1, A2, . . . An. Folding A1, A2, . . . An across li and taking the union of theresulting polygons yields Pi+1.

Let S(P ) be the number of sides of polygon P .

Lemma. A convex polygon Z is folded once. Z ′ and Z ′′′ are convex, and S(Z ′) + S(Z ′′′) ≤ S(Z) + 4.

HMIC

2 FOLDING A PAPER

Proof. Any line l through Z splits Z into two convex polygons, and four new vertices are formed by l.The number of vertices in a convex polygon equals the number of sides, proving our lemma.

Thus Pi+1 can be expressed as the union of no more than 2i+1 convex polygons, and the averagenumber of sides of these convex polygons does not exceed 4. Claim 1 is proven.

Claim 2. For a quadrilateral of area A, the diameter has length less than√2A.

Proof. Without loss of generality, let the fixed area be 1. The diagonal is not longer than the diameterof the quadrilateral. Half of the product of the diagonals is not less than the area, and Claim 2follows.

Claim 1 tells us that P2013 the union of no more than 22013 convex polygons, whose average numberof sides is no more than 4. Part (a) of the problem tells us that there is some quadrilateral with area≥ 1

22012 , with vertices at the vertices of the final folded figure. Claim 2 tells us that the diameter of

this quadrilateral, and thus of P2013, is ≥√2

21006 .

HMIC

TEAM

This test consists of ten problems to be solved by a team in one hour. The problems are unequally weighted

with point values as shown in brackets. They are not necessarily in order of di�culty, though harder problems are

generally worth more points. We do not expect most teams to get through all the problems.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted. Similarly,

graph paper, rulers, protractors, compasses, and other drawing aids are not permitted. If there is a chalkboard in

the room, you may write on it during the round only, not before it starts.

Numerical answers should, where applicable, be simplified as much as reasonably possible and must be exact unless

otherwise specified. Correct mathematical notation must be used. Your proctor cannot assist you in interpreting

or solving problems but has our phone number and may help with any administrative di�culties.

All problems require full written proof/justification. Even if the answer is numerical, you

must prove your result.

If you believe the test contains an error, please submit your protest in writing to the Science Center Lobby during

lunchtime.

Enjoy!

HMMT 2014Saturday 22 February 2014

Team

1. [10] Let ! be a circle, and let A and B be two points in its interior. Prove that there exists a circlepassing through A and B that is contained in the interior of !.

2. [15] Let a1, a2, . . . be an infinite sequence of integers such that ai divides ai+1 for all i � 1, and letbi be the remainder when ai is divided by 210. What is the maximal number of distinct terms in thesequence b1, b2, . . .?

3. [15] There are n girls G1, . . . , Gn and n boys B1, . . . , Bn. A pair (Gi, Bj) is called suitable if andonly if girl Gi is willing to marry boy Bj . Given that there is exactly one way to pair each girl with adistinct boy that she is willing to marry, what is the maximal possible number of suitable pairs?

4. [20] Compute100X

k=0

�2100

250 + 2k

⌫.

(Here, if x is a real number, then bxc denotes the largest integer less than or equal to x.)

5. [25] Prove that there exists a nonzero complex number c and a real number d such that

����

����1

1 + z + z2

���������

1

1 + z + z2� c

����

���� = d

for all z with |z| = 1 and 1 + z + z2 6= 0. (Here, |z| denotes the absolute value of the complex numberz, so that |a+ bi| =

pa2 + b2 for real numbers a, b.)

6. [25] Let n be a positive integer. A sequence (a0, . . . , an) of integers is acceptable if it satisfies thefollowing conditions:

(a) 0 = |a0| < |a1| < · · · < |an�1| < |an|.(b) The sets {|a1 � a0|, |a2 � a1|, . . . , |an�1 � an�2|, |an � an�1|} and {1, 3, 9, . . . , 3n�1} are equal.

Prove that the number of acceptable sequences of integers is (n+ 1)!.

7. [30] Find the maximum possible number of diagonals of equal length in a convex hexagon.

8. [35] Let ABC be an acute triangle with circumcenter O such that AB = 4, AC = 5, and BC = 6. LetD be the foot of the altitude from A to BC, and E be the intersection of AO with BC. Suppose thatX is on BC between D and E such that there is a point Y on AD satisfying XY k AO and Y O ? AX.Determine the length of BX.

9. [35] For integers m,n � 1, let A(n,m) be the number of sequences (a1, · · · , anm) of integers satisfyingthe following two properties:

(a) Each integer k with 1 k n occurs exactly m times in the sequence (a1, · · · , anm).

(b) If i, j, and k are integers such that 1 i nm and 1 j k n, then j occurs in the sequence(a1, · · · , ai) at least as many times as k does.

For example, if n = 2 and m = 5, a possible sequence is (a1, · · · , a10) = (1, 1, 2, 1, 2, 2, 1, 2, 1, 2). Onthe other hand, the sequence (a1, · · · , a10) = (1, 2, 1, 2, 2, 1, 1, 1, 2, 2) does not satisfy property (2) fori = 5, j = 1, and k = 2.

Prove that A(n,m) = A(m,n).

10. [40] Fix a positive real number c > 1 and positive integer n. Initially, a blackboard contains thenumbers 1, c, . . . , cn�1. Every minute, Bob chooses two numbers a, b on the board and replaces themwith ca+ c2b. Prove that after n� 1 minutes, the blackboard contains a single number no less than

✓cn/L � 1

c1/L � 1

◆L

,

where � = 1+p5

2 and L = 1 + log�(c).

HMMT 2014Saturday 22 February 2014

Team

Organization

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1. [10]

HMMT 2014Saturday 22 February 2014

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2. [15]

HMMT 2014Saturday 22 February 2014

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3. [15]

HMMT 2014Saturday 22 February 2014

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4. [20]

HMMT 2014Saturday 22 February 2014

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5. [25]

HMMT 2014Saturday 22 February 2014

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6. [25]

HMMT 2014Saturday 22 February 2014

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7. [30]

HMMT 2014Saturday 22 February 2014

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8. [35]

HMMT 2014Saturday 22 February 2014

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9. [35]

HMMT 2014Saturday 22 February 2014

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Organization

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10. [40]

HMMT 2014Saturday 22 February 2014

Team

1. [10] Let ω be a circle, and let A and B be two points in its interior. Prove that there exists a circlepassing through A and B that is contained in the interior of ω.

Answer: N/A WLOG, suppose OA ≥ OB. Let ω′ be the circle of radius OA centered at O. We

have that B lies inside ω′. Thus, it is possible to scale ω′ down about the point A to get a circle ω′′

passing through both A and B. Since ω′′ lies inside ω′ and ω′ lies inside ω, ω′′ lies inside ω.

Alternative solution 1 : WLOG, suppose OA ≥ OB. Since OA ≥ OB, the perpendicular bisector ofAB intersects segment OA at some point C. We claim that the circle ω′ passing through A and Band centered at C lies entirely in ω. Let x = OA and y = AC = BC. Note that y is the length of theradius of ω′. By definition, any point P contained in ω′ is of distance at most y from C. Applying thetriangle inequality to OCP , we see that OP ≤ OC + CP ≤ (x − y) + y = x, so P lies in ω. Since Pwas arbitrary, it follows that ω′ lies entirely in ω.

Alternative solution 2 : Draw line AB, and let it intersect ω at A′ and B′, where A and A′ are on thesame side of B. Choose X inside the segment AB so that A′X/AX = B′X/BX; such a point existsby the intermediate value theorem. Notice that X is the center of a dilation taking A′B′ to AB - thesame dilation carries ω to ω′ which goes through A and B. Since ω′ is ω dilated with respect to apoint in its interior, it’s clear that ω′ must be contained entirely within ω, and so we are done.

2. [15] Let a1, a2, . . . be an infinite sequence of integers such that ai divides ai+1 for all i ≥ 1, and letbi be the remainder when ai is divided by 210. What is the maximal number of distinct terms in thesequence b1, b2, . . .?

Answer: 127 It is clear that the sequence {ai} will be a concatenation of sequences of the

form {vi}N0

i=1, {wi · p1}N1

i=1, {xi · p1p2}N2

i=1, {yi · p1p2p3}N3

i=1, and {zi · p1p2p3p4}N4

i=1, for some permutation(p1, p2, p3, p4) of (2, 3, 5, 7) and some sequences of integers {vi}.{wi}.{xi}.{yi}.{zi}, each coprime with210.

In {vi}N0

i=1, there are a maximum of φ(210) distinct terms mod 210. In {wi · p1}N1

i=1, there are a

maximum of φ( 210p1

) distinct terms mod 210. In {xi ·p1p2}N2

i=1, there are a maximum of φ( 210p1p2

) distinct

terms mod 210. In {yi · p1p2p3}N3

i=1, there are a maximum of φ( 210p1p2p3

) distinct terms mod 210. In

{zi · p1p2p3p4}N4

i=1, there can only be one distinct term mod 210.

Therefore we wish to maximize φ(210) + φ( 210p1

) + φ( 210p1p2

) + φ( 210p1p2p3

) + 1 over all permutations

(p1, p2, p3, p4) of (2, 3, 5, 7). It’s easy to see that the maximum occurs when we take p1 = 2, p2 =3, p3 = 5, p4 = 7 for an answer of φ(210)+φ(105)+φ(35)+φ(7)+1 = 127. This upper bound is clearlyattainable by having the vi’s cycle through the φ(210) integers less than 210 coprime to 210, the wi’scycle through the φ( 210

p1

) integers less than 210p1

coprime to 210p1

, etc.

3. [15] There are n girls G1, . . . , Gn and n boys B1, . . . , Bn. A pair (Gi, Bj) is called suitable if andonly if girl Gi is willing to marry boy Bj . Given that there is exactly one way to pair each girl with adistinct boy that she is willing to marry, what is the maximal possible number of suitable pairs?

Answer:n(n+1)

2 We represent the problem as a graph with vertices G1, . . . , Gn, B1, . . . , Bn such

that there is an edge between vertices Gi and Bj if and only if (Gi, Bj) is suitable, so we want tomaximize the number of edges while having a unique matching.

We claim the answer is n(n+1)2 . First, note that this can be achieved by having an edge between Gi

and Bj for all pairs j ≤ i, because the only possible matching in this case is pairing Gi with Bi for alli. To prove that this is maximal, we first assume without loss of generality that our unique matchingconsists of pairing Gi with Bi for all i, which takes n edges. Now, note that for any i, j, at most oneof the two edges GiBj and GjBi can be added, because if both were added, we could pair Gi with Bj

and Gj with Bi instead to get another valid matching. Therefore, we may add at most(

n2

)

·1 = n(n−1)2

edges, so the maximal number of edges is n + n(n−1)2 = n(n+1)

2 as desired.

4. [20] Compute100∑

k=0

⌊

2100

250 + 2k

⌋

.

(Here, if x is a real number, then ⌊x⌋ denotes the largest integer less than or equal to x.)

Answer: 101 · 249 − 50 Let ak = 2100

250+2k . Notice that, for k = 0, 1, . . . , 49,

ak + a100−k =2100

250 + 2k+

2100

250 + 2100−k=

2100

250 + 2k+

250+k

2k + 250= 250

It is clear that for k = 0, 1, . . . , 49, ak, a100−k /∈ Z, so ⌊ak⌋+ ⌊a100−k⌋ = 250 − 1 (since the sum of floorsis an integer less than ak + a100−k but greater than ak − 1 + a100−k − 1). Thus,

100∑

k=0

⌊ak⌋ = 50 · (250 − 1) + 249 = 101 · 249 − 50

5. [25] Prove that there exists a nonzero complex number c and a real number d such that∣

∣

∣

∣

∣

∣

∣

∣

1

1 + z + z2

∣

∣

∣

∣

−∣

∣

∣

∣

1

1 + z + z2− c

∣

∣

∣

∣

∣

∣

∣

∣

= d

for all z with |z| = 1 and 1 + z + z2 6= 0. (Here, |z| denotes the absolute value of the complex numberz, so that |a + bi| =

√a2 + b2 for real numbers a, b.)

Answer: 43 Let f(z) =

∣

∣

∣

11+z+z2

∣

∣

∣. Parametrize z = eit = cos t + i sin t and let g(t) = f(eit),

0 ≤ t < 2π. Writing out 11+z+z2 in terms of t and simplifying, we find that g(t) = cos t−i sin t

1+2 cos t . Lettingx(t) = ℜ(g(t)) and y(t) = ℑ(g(t)) (the real and imaginary parts of g(t), respectively), what we wish toprove is equivalent to showing that {(x(t), y(t)) | 0 ≤ t < 2π} is a hyperbola with one focus at (0, 0).However

9

(

x(t) − 2

3

)2

− 3y(t)2 = 1

holds for all t, so from this equation we find that the locus of points (x(t), y(t)) is a hyperbola, withcenter ( 2

3 , 0) and focal length 23 , so the foci are at (0, 0) and (4

3 , 0). Hence c = 43 .

6. [25] Let n be a positive integer. A sequence (a0, . . . , an) of integers is acceptable if it satisfies thefollowing conditions:

(a) 0 = |a0| < |a1| < · · · < |an−1| < |an|.(b) The sets {|a1 − a0|, |a2 − a1|, . . . , |an−1 − an−2|, |an − an−1|} and {1, 3, 9, . . . , 3n−1} are equal.

Prove that the number of acceptable sequences of integers is (n + 1)!.

Answer: N/A We actually prove a more general result via strong induction on n.

First, we state the more general result we wish to prove.

For n > 0, define a great sequence to be a sequence of integers (a0, . . . , an) such that1. 0 = |a0| < |a1| < · · · < |an−1| < |an|2. Let (b1, b2, . . . , bn) be a set of positive integers such that 3bi ≤ bi+1 for all i. The sets {|a1−a0|, |a2−a1|, . . . , |an−1 − an−2|, |an − an−1|} and {b1, b2, . . . , bn} are equal.

Then, the number of great sequences is (n + 1)!.

If we prove this statement, then we can just consider the specific case of b1 = 1, 3bi = bi+1 to solveour problem.

Before we proceed, we will prove a lemma.Lemma: Let (b1, b2, . . . , bn) be a set of positive integers such that 3bi ≤ bi+1 for all i. Then bi >

2i−1∑

k=1

bk.

Proof: bi

2 = bi

3 + bi

9 + · · · ≥i−1∑

k=1

bk. Equality only occurs when the sequence bi is infinite, which is not

the case, so the inequality holds.

We can now proceed with the induction. The base case is obvious. Assume it is true up to n = j − 1.Next, consider some permutation of the set B = {b1, b2, b3, · · · , bj}. Denote it as C = {c1, c2, . . . , cj}.Find the element ck = bj . For the first k − 1 elements of C, we can put them in order and applythe inductive hypothesis. The number of great sequences such that the set of differences {|a1 −a0|, . . . , |ak−1 − ak−2|} is equal to {c1, c2, . . . , ck−1} is k!.

Then, ak can be either ak−1 + ck or ak−1 − ck. This is because, by the lemma, ck > 2k−1∑

m=1cm =

2k−1∑

m=1|am −am−1| ≥ 2|ak−1 −ak−2 +ak−2 −ak−3 + · · ·+a1 −a0| = 2|ak−1|. It is easy to check that for

either possible value of ak, |ak| > |ak−1|. After that, there is only one possible value for ak+1, . . . , aj

because only one of ai ± ci+1 will satisfy |ai| > |ai−1|.There are

(

j−1k−1

)

possible ways to choose c1, c2, . . . , ck−1 from B. Given those elements of C, there arek! ways to make a great sequence (a0, a1, . . . , ak−1). Then, there are 2 possible values for ak. Afterthat, there are (j − k)! ways to order the remaining elements of C, and for each such ordering, thereis exactly 1 possible great sequence (a0, a1, . . . , aj).

Now, counting up all the possible ways to do this over all values of k, we get that the number of great

sequences is equal toj

∑

k=1

(

j−1k−1

)

k!2(j −k)! =j

∑

k=1

2(j − 1)!(k) = 2(j − 1)! j(j+1)2 = (j +1)!. The induction

is complete, and this finishes the proof.

Alternate solution: Another method of performing the induction is noting that any acceptable se-quence (a0, . . . , an) can be matched with n + 2 acceptable sequences of length n + 2 because we cantake (3a0, . . . , 3an) and add an element with a difference of 1 in any of n + 2 positions.

7. [30] Find the maximum possible number of diagonals of equal length in a convex hexagon.

Answer: 7 First, we will prove that 7 is possible. Consider the following hexagon ABCDEF

whose vertices are located at A(0, 0), B( 12 , 1−

√3

2 ), C( 12 ,

√3

2 ), D(0, 1), E(− 12 ,

√3

2 ), F (− 12 , 1−

√3

2 ). Onecan easily verify that all diagonals but BE and CF have length 1.

Now suppose that there are at least 8 diagonals in a certain convex hexagon ABCDEF whose lengthsare equal. There must be a diagonal such that, with this diagonal taken out, the other 8 have equallength. There are two cases.

Case I. The diagonal is one of AC,BD,CE,DF,EA,FB. WLOG, assume it is AC. We have EC =EB = FB = FC. Thus, B and C are both on the perpendicular bisector of EF . Since ABCDEF isconvex, both B and C must be on the same side of line EF , but this is impossible as one of B or C,must be contained in triangle CEF . Contradiction.

Case II: The diagonal is one of AD,BE,CF . WLOG, assume it is AD. Again, we have EC = EB =FB = FC. By the above reasoning, this is a contradiction.

Thus, 7 is the maximum number of possible diagonals.

8. [35] Let ABC be an acute triangle with circumcenter O such that AB = 4, AC = 5, and BC = 6. LetD be the foot of the altitude from A to BC, and E be the intersection of AO with BC. Suppose thatX is on BC between D and E such that there is a point Y on AD satisfying XY ‖ AO and Y O ⊥ AX.Determine the length of BX.

Answer: 96/41 Let AX intersect the circumcircle of △ABC again at K. Let OY intersect AK

and BC at T and L, respectively. We have ∠LOA = ∠OY X = ∠TDX = ∠LAK, so AL is tangentto the circumcircle. Furthermore, OL ⊥ AK, so △ALK is isosceles with AL = AK, so AK is alsotangent to the circumcircle. Since BC and the tangents to the circumcircle at A and K all intersectat the same point L, CL is a symmedian of △ACK. Then AK is a symmedian of △ABC. Then we

can use BXXC = (AB)2

(AC)2 to compute BX = 9641 .

9. [35] For integers m,n ≥ 1, let A(n,m) be the number of sequences (a1, · · · , anm) of integers satisfyingthe following two properties:

(a) Each integer k with 1 ≤ k ≤ n occurs exactly m times in the sequence (a1, · · · , anm).

(b) If i, j, and k are integers such that 1 ≤ i ≤ nm and 1 ≤ j ≤ k ≤ n, then j occurs in the sequence(a1, · · · , ai) at least as many times as k does.

For example, if n = 2 and m = 5, a possible sequence is (a1, · · · , a10) = (1, 1, 2, 1, 2, 2, 1, 2, 1, 2). Onthe other hand, the sequence (a1, · · · , a10) = (1, 2, 1, 2, 2, 1, 1, 1, 2, 2) does not satisfy property (2) fori = 5, j = 1, and k = 2.

Prove that A(n,m) = A(m,n).

Answer: N/A Solution 1: We show that A(n,m) is equal to the the number of standard Young

tableaux with n rows and m columns (i.e. fillings of an n × m matrix with the numbers 1, 2, ..., nmso that numbers are increasing in each row and column). Consider the procedure where every timea k appears in the sequences, you add a number to the leftmost empty spot of the k-th row. Doingthis procedure will result in a valid standard Young tableau. The entries are increasing along everyrow because new elements are added from left to right. The elements are also increasing along everycolumn. This is because the condition about the sequences implies that there will always be at leastas many elements in row i as there are in row j for i < j. At the end of this procedure, the YoungTableaux has been filled because each of the n numbers have been added m times.

Now, consider a n × m standard Young tableau. If the number p is in row k, you add a k to thesequence. This will produce a valid sequence. To see this, suppose that p appears in the entry (x, y).Then all of the entries (q, y) where q < x have already been added to the sequence because they mustcontain entries less than p. Thus, the numbers 1 through x − 1 have all already been added at least ytimes Then, when we process p, we are adding the y-th x, which is valid. At the end of this procedure,n numbers have been added m times to the sequence.

The two procedures given above are inverses of each other. Thus, A(n,m) is equal to the number ofn × m standard Young tableaux. For every n × m Tableaux, we can transpose it to form an m × ntableau. The number of such tableaux is A(m,n). Thus, A(n,m) = A(m,n).

Solution 2: We can also form a direct bijection to show A(n,m) = A(m,n), as follows. Supposethat a = (a1, . . . , amn) is a sequence satisfying properties 1 and 2. We will define a sequence f(a) =(b1, . . . , bnm) satisfying the same properties 1 and 2, but with m and n switched.

The bijection f is simple: just define bi, for 1 ≤ i ≤ nm, to be equal to the number of j with1 ≤ j ≤ i such that aj = ai. In other words, to obtain f(a) from a, replace the kth occurrence ofeach number with the number k. For example, if n = 2 and m = 5, then f(1, 1, 2, 1, 2, 2, 1, 2, 1, 2) =(1, 2, 1, 3, 2, 3, 4, 4, 5, 5).

First, it is clear that f(a) satisfies property 1 with m and n switched. Indeed, it follows directly fromthe definition of f that for each pair (k1, k2) with 1 ≤ k1 ≤ n and 1 ≤ k2 ≤ m, there is exactly oneindex i for which (ai, bi) = (k1, k2). This implies the desired result.

Second, we show that f(a) satisfies property 2 with m and n switched. For this, suppose that i, j, kare integers with 1 ≤ i ≤ nm and 1 ≤ j ≤ k ≤ m. Then, the number of times that k appears in thesequence (b1, . . . , bi) is exactly equal to the number of integers ℓ with 1 ≤ ℓ ≤ n such that ℓ appears atleast k times in the sequence (a1, . . . , ai). Similarly, the number of times that j appears in the sequence(b1, . . . , bi) is exactly equal to the number of integers ℓ with 1 ≤ j ≤ n such that ℓ appears at least j

times in the sequence (a1, . . . , ai). In particular, the number j appears in the sequence (b1, . . . , bi) atleast as many times as k does.

Now, we show that f is a bijection. We claim that f is actually an involution; that is, f(f(a)) = a.(Note in particular that this implies that f is a bijection, and its inverse is itself.)

Fix an index i with 1 ≤ i ≤ mn. Let ℓ = ai and k = bi; then, k is the number of times that the term ℓappears in the sequence (a1, . . . , ai). It is enough to show that ℓ is the number of times that the term kappears in the sequence (b1, . . . , bi). First, note that by definition of f , the number of times k appearsin the sequence (b1, . . . , bi) is equal to the number of integers j such that the sequence (a1, . . . , ai)contains the number j at least k times.

Now, ℓ occurs exactly k times in the sequence (a1, . . . , ai). So by property 2, each j with j ≤ ℓ appearsat least k times in the sequence (a1, . . . , ai). Furthermore, each j with j > ℓ appears at most k − 1times in the sequence (a1, . . . , ai−1) and thus appears at most k−1 times in the sequence (a1, . . . , ai) aswell (because ai = ℓ). Therefore, the number of integers j such that the sequence (a1, . . . , ai) containsthe number j at least k times is exactly equal to ℓ. This shows the desired result.

Note: The bijections given in solutions 1 and 2 can actually be shown to be the same.

10. [40] Fix a positive real number c > 1 and positive integer n. Initially, a blackboard contains thenumbers 1, c, . . . , cn−1. Every minute, Bob chooses two numbers a, b on the board and replaces themwith ca + c2b. Prove that after n − 1 minutes, the blackboard contains a single number no less than

(

cn/L − 1

c1/L − 1

)L

,

where φ = 1+√

52 and L = 1 + logφ(c).

Answer: N/A By a simple reverse induction, we can show that at any instant, any number on

the board takes the form∑

α∈A crαcα for a certain A ⊆ {0, . . . , n− 1} for non-negative integer weightsrα satisfying

∑

α∈A φ−rα = 1, where these subsets A partition {0, . . . , n − 1}, and after each minute

the number of parts decreases by 1. (Note that 1 = φ−1 + φ−2, and when we “merge” two of thesesets A,B corresponding to a, b → ca + c2b, we add 1 to all the weights of A, and 2 to all the weightsof B). Therefore, the final number takes the form

∑n−1i=0 ctici for positive integer weights t0, . . . , tn−1

satisfying∑n−1

i=0 φ−ti = 1.

Let M = logφ(c) > 0 (so L = 1 + M). Then c = φM , so by Holder’s inequality,

(

n−1∑

i=0

ctici

)1 (

n−1∑

i=0

φ−ti

)M

≥(

n−1∑

i=0

[ci(cφ−M )ti ]1/L

)L

=

(

cn/L − 1

c1/L − 1

)L

,

as desired.

Comment: Note that the same method applies for any choice of initial numbers x1, x2, . . . , xn (thechoice c0, c1, . . . , cn−1 simply gives a relatively clean closed form). The invariant

∑

φ−ti = 1 wasinspired by the corresponding invariant

∑

2−ti = 1 for ISL 2007 A5, where the relevant operation is{a, b} 7→ {c1a+c1b} (specifically, for c = 2, but this is not important) rather than {a, b} 7→ {c1a+c2b}.Alternate solution: We have cxL + c2yL ≥ (x + y)L for any nonnegative reals x, y, since (cxL +c2yL)(c−1/(L−1) + c−2/(L−1))L−1 ≥ (x + y)L by Holder’s inequality, c1/(L−1) = c1/ logφ(c) = φ, and ofcourse φ−1 + φ−2 = 1.

In particular, (ca+c2b)1/L ≥ a1/L+b1/L. Thus∑n−1

i=0 (ci)1/L ≤ N1/L, where N is the final (nonnegative)

number on the board. Hence N1/L ≥ cn/L−1c1/L−1

.

ALGEBRA

This test consists of 10 short-answer problems to be solved individually in 50 minutes. Problems will be weighted

with point values after the contest based on how many competitors solve each problem. There is no penalty for

guessing.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted. Similarly,

graph paper, rulers, protractors, compasses, and other drawing aids are not permitted.

Answers should be simplified as much as is reasonably possible and must be exact unless otherwise specified. Ra-

tional numbers should be written in lowest terms, although denominators of irrationals need not be rationalized.

An nth root should be simplified so that the radicand is not divisible by the nth power of any prime.

Correct mathematical notation must be used. No partial credit will be given unless otherwise specified.

If you believe the test contains an error, please submit your protest in writing to the Science Center Lobby during

lunchtime.

Enjoy!

HMMT 2014Saturday 22 February 2014

Algebra

1. Given that x and y are nonzero real numbers such that x + 1y

= 10 and y + 1x

= 512 , find all possible

values of x.

2. Find the integer closest to1

4p54 + 1� 4

p54 � 1

.

3. Let

A =1

6

�(log2(3))

3 � (log2(6))3 � (log2(12))

3 + (log2(24))3�.

Compute 2A.

4. Let b and c be real numbers, and define the polynomial P (x) = x

2 + bx+ c. Suppose that P (P (1)) =P (P (2)) = 0, and that P (1) 6= P (2). Find P (0).

5. Find the sum of all real numbers x such that 5x4 � 10x3 + 10x2 � 5x� 11 = 0.

6. Given that w and z are complex numbers such that |w + z| = 1 and |w2 + z

2| = 14, find the smallestpossible value of |w3 + z

3|. Here, | · | denotes the absolute value of a complex number, given by|a+ bi| =

pa

2 + b

2 whenever a and b are real numbers.

7. Find the largest real number c such that

101X

i=1

x

2i

� cM

2

whenever x1, . . . , x101 are real numbers such that x1+· · ·+x101 = 0 andM is the median of x1, . . . , x101.

8. Find all real numbers k such that r4 + kr

3 + r

2 + 4kr + 16 = 0 is true for exactly one real number r.

9. Given that a, b, and c are complex numbers satisfying

a

2 + ab+ b

2 = 1 + i

b

2 + bc+ c

2 = �2c

2 + ca+ a

2 = 1,

compute (ab+ bc+ ca)2. (Here, i =p�1.)

10. For an integer n, let f9(n) denote the number of positive integers d 9 dividing n. Suppose that m isa positive integer and b1, b2, . . . , bm are real numbers such that f9(n) =

Pm

j=1 bjf9(n� j) for all n > m.Find the smallest possible value of m.

HMMT 2014Saturday 22 February 2014

Algebra

PUT LABEL HERE

Name Team ID#

Organization Team

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Score:

HMMT 2014Saturday 22 February 2014

Algebra

1. Given that x and y are nonzero real numbers such that x + 1y = 10 and y + 1

x = 512 , find all possible

values of x.

Answer: 4, 6 OR 6, 4 Let z = 1y . Then x + z = 10 and 1

x + 1z = 5

12 . Since 1x + 1

z = x+zxz = 10

xz , we

have xz = 24. Thus, x(10 − x) = 24, so x2 − 10x + 24 = (x − 6)(x − 4) = 0, whence x = 6 or x = 4.

Alternate solution: Clearing denominators gives xy + 1 = 10y and yx + 1 = 512x, so x = 24y. Thus we

want to find all real (nonzero) x such that x2

24 + 1 = 512x (and for any such x, y = x/24 will satisfy the

original system of equations). This factors as (x − 4)(x − 6) = 0, so precisely x = 4, 6 work.

2. Find the integer closest to1

4√

54 + 1 − 4√

54 − 1.

Answer: 250 Let x = (54 + 1)1/4 and y = (54 − 1)1/4. Note that x and y are both approximately5. We have

1

x − y=

(x + y)(x2 + y2)

(x − y)(x + y)(x2 + y2)=

(x + y)(x2 + y2)

x4 − y4

=(x + y)(x2 + y2)

2≈ (5 + 5)(52 + 52)

2= 250.

Note: To justify the ≈, note that 1 = x4 − 54 implies

0 < x − 5 =1

(x + 5)(x2 + 52)<

1

(5 + 5)(52 + 52)=

1

500,

and similarly 1 = 54 − y4 implies

0 < 5 − y =1

(5 + y)(52 + y2)<

1

(4 + 4)(42 + 42)=

1

256.

Similarly,

0 < x2 − 52 =1

x2 + 52<

1

2 · 52=

1

50

and

0 < 52 − y2 =1

52 + y2<

1

52 + 4.52<

1

45.

Now

|x + y − 10| = |(x − 5) − (5 − y)| < max(|x − 5|, |5 − y|) <1

256,

and similarly |x2 + y2 − 2 · 52| < 145 . It’s easy to check that (10 − 1/256)(50 − 1/45) > 499.5 and

(10 + 1/256)(50 + 1/45) < 500.5, so we’re done.

3. Let

A =1

6

(

(log2(3))3 − (log2(6))3 − (log2(12))3 + (log2(24))3)

.

Compute 2A.

Answer: 72 Let a = log2(3), so 2a = 3 and A = 16 [a3 − (a + 1)3 − (a + 2)3 + (a + 3)3]. But

(x + 1)3 − x3 = 3x2 + 3x + 1, so A = 16 [3(a + 2)2 + 3(a + 2)− 3a2 − 3a] = 1

2 [4a + 4 + 2] = 2a + 3. Thus2A = (2a)2(23) = 9 · 8 = 72.

4. Let b and c be real numbers, and define the polynomial P (x) = x2 + bx + c. Suppose that P (P (1)) =P (P (2)) = 0, and that P (1) 6= P (2). Find P (0).

Answer: − 32 OR −1.5 OR −1 1

2 Since P (P (1)) = P (P (2)) = 0, but P (1) 6= P (2), it follows that

P (1) = 1 + b + c and P (2) = 4 + 2b + c are the distinct roots of the polynomial P (x). Thus, P (x)factors:

P (x) = x2 + bx + c

= (x − (1 + b + c))(x − (4 + 2b + c))

= x2 − (5 + 3b + 2c)x + (1 + b + c)(4 + 2b + c).

It follows that −(5 + 3b + 2c) = b, and that c = (1 + b + c)(4 + 2b + c). From the first equation, wefind 2b + c = −5/2. Plugging in c = −5/2 − 2b into the second equation yields

−5/2 − 2b = (1 + (−5/2) − b)(4 + (−5/2)).

Solving this equation yields b = − 12 , so c = −5/2 − 2b = − 3

2 .

5. Find the sum of all real numbers x such that 5x4 − 10x3 + 10x2 − 5x − 11 = 0.

Answer: 1 Rearrange the equation to x5 + (1 − x)5 − 12 = 0. It’s easy to see this has two realroots, and that r is a root if and only if 1 − r is a root, so the answer must be 1.

Alternate solution: Note that 5x4−10x3 +10x2−5x−11 = 5x(x−1)(x2−x+1)−11 = 5u(u+1)−11,where u = x2 − x. Of course, x2 − x = u has real roots if and only if u ≥ − 1

4 , and distinct real roots

if and only if u > − 14 . But the roots of 5u(u + 1) − 11 are

−5±√

52+4(5)(11)

2·5 = −5±7√

510 , one of which is

greater than − 14 and the other less than − 1

4 . For the larger root, x2 − x = u has exactly two distinctreal roots, which sum up to 1 by Vieta’s.

6. Given that w and z are complex numbers such that |w + z| = 1 and |w2 + z2| = 14, find the smallestpossible value of |w3 + z3|. Here, | · | denotes the absolute value of a complex number, given by|a + bi| =

√a2 + b2 whenever a and b are real numbers.

Answer: 412 OR 20.5 OR 20 1

2 We can rewrite |w3 +z3| = |w+z||w2−wz +z2| = |w2−wz +z2| =

|32 (w2 + z2) − 12 (w + z)2|.

By the triangle inequality, |32 (w2 +z2)− 12 (w+z)2 + 1

2 (w+z)2| ≤ |32 (w2 +z2)− 12 (w+z)2|+ | 12 (w+z)2|.

By rearranging and simplifying, we get |w3 + z3| = |32 (w2 + z2)− 12 (w + z)2| ≥ 3

2 |w2 + z2|− 12 |w + z|2 =

32 (14) − 1

2 = 412 .

To achieve 41/2, it suffices to take w, z satisfying w + z = 1 and w2 + z2 = 14.

7. Find the largest real number c such that

101∑

i=1

x2i ≥ cM2

whenever x1, . . . , x101 are real numbers such that x1 + · · · + x101 = 0 and M is the median ofx1, . . . , x101.

Answer: 515150 OR 103.02 OR 103 1

50 Suppose without loss of generality that x1 ≤ · · · ≤ x101 and

M = x51 ≥ 0.

Note that f(t) = t2 is a convex function over the reals, so we may “smooth” to the case x1 = · · · =x50 ≤ x51 = · · · = x101 (the x51 = · · · is why we needed to assume x51 ≥ 0). Indeed, by Jensen’sinequality, the map x1, x2, . . . , x50 → x1+···+x50

50 , . . . , x1+···+x50

50 will decrease or fix the LHS, whilepreserving the ordering condition and the zero-sum condition.

Similarly, we may without loss of generality replace x51, . . . , x101 with their average (which will decreaseor fix the LHS, but also either fix or increase the RHS). But this simplified problem has x1 = · · · =

x50 = −51r and x51 = · · · = x101 = 50r for some r ≥ 0, and by homogeneity, C works if and only if

C ≤ 50(51)2+51(50)2

502 = 51(101)50 = 5151

50 .

Comment: One may also use the Cauchy-Schwarz inequality or the QM-AM inequality instead ofJensen’s inequality.

Comment: For this particular problem, there is another solution using the identity 101∑

x2i−(

∑

xi)2 =

∑

(xj − xi)2. Indeed, we may set u = x51 − (x1 + · · · + x50)/50 and v = (x52 + · · · + x101)/50 − x51,

and use the fact that (u − v)2 ≤ u2 + v2.

8. Find all real numbers k such that r4 + kr3 + r2 + 4kr + 16 = 0 is true for exactly one real number r.

Answer: ± 94 (OR 9

4 ,− 94 OR − 9

4 , 94 ) OR ±2 1

4 OR ±2.25 Any real quartic has an even number of

real roots with multiplicity, so there exists real r such that x4 + kx3 + x2 + 4kx + 16 either takes theform (x + r)4 (clearly impossible) or (x + r)2(x2 + ax + b) for some real a, b with a2 < 4b. Clearly

r 6= 0, so b = 16r2 and 4k = 4(k) yields 32

r + ar2 = 4(2r + a) =⇒ a(r2 − 4) = 8 r2−4r . Yet a 6= 8

r (or elsea2 = 4b), so r2 = 4, and 1 = r2 + 2ra + 16

r2 =⇒ a = −72r . Thus k = 2r − 7

2r = ± 94 (since r = ±2).

It is easy to check that k = 94 works, since x4 +(9/4)x3 +x2 +4(9/4)x+16 = 1

4 (x+2)2(4x2 −7x+16).The polynomial given by k = − 9

4 is just 14 (−x + 2)2(4x2 + 7x + 16).

Alternate solution: x4 +kx3 +x2 +4kx+16 = (x2 + k2x+4)2 +(1−8− k2

4 )x2, so for some ǫ ∈ {−1, 1},2x2+(k−ǫ

√k2 + 28)x+8 has a single real root and thus takes the form 2(x+r)2 (using the same notation

as above). But then (k − ǫ√

k2 + 28)2 = 4(2)(8) = 82, so we conclude that (k ± 8)2 = (ǫ√

k2 + 28)2

and k = ±(4 − 74 ) = ± 9

4 .

9. Given that a, b, and c are complex numbers satisfying

a2 + ab + b2 = 1 + i

b2 + bc + c2 = −2

c2 + ca + a2 = 1,

compute (ab + bc + ca)2. (Here, i =√−1.)

Answer: −11−4i3 OR − 11+4i

3 More generally, suppose a2 + ab + b2 = z, b2 + bc + c2 = x,

c2 + ca + a2 = y for some complex numbers a, b, c, x, y, z.

We show that

f(a, b, c, x, y, z) = (1

2(ab + bc + ca) sin 120◦)2 − (

1

4)2[(x + y + z)2 − 2(x2 + y2 + z2)]

holds in general. Plugging in x = −2, y = 1, z = 1 + i will then yield the desired answer,

(ab + bc + ca)2 =16

3

1

16[(x + y + z)2 − 2(x2 + y2 + z2)]

=i2 − 2(4 + 1 + (1 + i)2)

3=

−1 − 2(5 + 2i)

3=

−11 − 4i

3.

Solution 1: Plug in x = b2 + bc + c2, etc. to get a polynomial g in a, b, c (that agrees with f for everyvalid choice of a, b, c, x, y, z). It suffices to show that g(a, b, c) = 0 for all positive reals a, b, c, as thenthe polynomial g will be identically 0.

But this is easy: by the law of cosines, we get a geometrical configuration with a point P inside atriangle ABC with PA = a, PB = b, PC = c, ∠PAB = ∠PBC = ∠PCA = 120◦, x = BC2,

y = CA2, z = AB2. By Heron’s formula, we have

(1

2(ab + bc + ca) sin 120◦)2 = [ABC]2

=(√

x +√

y +√

z)∏

cyc(√

x +√

y −√z)

24

=1

16[(√

x +√

y)2 − z][(√

x −√y)2 − z]

=1

16[(x − y)2 + z2 − 2z(x + y)]

= (1

4)2[(x + y + z)2 − 2(x2 + y2 + z2)],

as desired.

Solution 2: Let s = a + b + c. We have x − y = b2 + bc − ca − a2 = (b − a)s and cyclic, sox + as = y + bs = z + cs (they all equal

∑

a2 +∑

bc). Now add all equations to get

x + y + z = 2∑

a2 +∑

bc = s2 +1

2

∑

(b − c)2;

multiplying both sides by 4s2 yields 4s2(x + y + z) = 4s4 + 2∑

(z − y)2, so [2s2 − (x + y + z)]2 =(x + y + z)2 − 2

∑

(z − y)2 = 6∑

yz − 3∑

x2. But 2s2 − (x + y + z) = s2 − 12

∑

(b − c)2 = (a + b +c)2 − 1

2

∑

(b − c)2 = 3(ab + bc + ca), so

9(ab + bc + ca)2 = 6∑

yz − 3∑

x2 = 3[(x + y + z)2 − 2(x2 + y2 + z2)],

which easily rearranges to the desired.

Comment: Solution 2 can be done with less cleverness. Let u = x + as = y + bs = z + cs, so a = u−xs ,

etc. Then s =∑

u−xs , or s2 = 3u− s(x + y + z). But we get another equation in s, u by just plugging

in directly to a2 + ab + b2 = z (and after everything is in terms of s, we can finish without too muchtrouble).

10. For an integer n, let f9(n) denote the number of positive integers d ≤ 9 dividing n. Suppose that m isa positive integer and b1, b2, . . . , bm are real numbers such that f9(n) =

∑mj=1 bjf9(n−j) for all n > m.

Find the smallest possible value of m.

Answer: 28 Let M = 9. Consider the generating function

F (x) =∑

n≥1

fM (n)xn =

M∑

d=1

∑

k≥1

xdk =

M∑

d=1

xd

1 − xd.

Observe that fM (n) = fM (n + M !) for all n ≥ 1 (in fact, all n ≤ 0 as well). Thus fM (n) satisfies adegree m linear recurrence if and only if it eventually satisfies a degree m linear recurrence. But thelatter occurs if and only if P (x)F (x) is a polynomial for some degree m polynomial P (x). (Why?)

Suppose P (x)F (x) = Q(x) is a polynomial for some polynomial P of degree m. We show that xs − 1 |P (x) for s = 1, 2, . . . ,M , or equivalently that P (ω) = 0 for all primitive sth roots of unity 1 ≤ s ≤ M).Fix a primitive sth root of unity ω, and define a function

Fω(z) = (1 − ω−1z)∑

s∤d≤M

zd

1 − zd+

∑

s|d≤M

zd

1 + (ω−1z) + · · · + (ω−1z)d−1

for all z where all denominators are nonzero (in particular, this includes z = ω).

Yet Fω(z)−F (z)(1−ω−1z) = 0 for all complex z such that z1, z2, . . . , zM 6= 1, so P (z)Fω(z)−Q(z)(1−ω−1z) = 0 holds for all such z as well. In particular, the rational function P (x)Fω(x)−Q(x)(1−ω−1x)

has infinitely many roots, so must be identically zero once we clear denominators. But no denominatorvanishes at x = ω, so we may plug in x = ω to the polynomial identity and then divide out by theoriginal (nonzero) denominators to get 0 = P (ω)Fω(ω) − Q(ω)(1 − ω−1ω) = P (ω)Fω(ω). However,

Fω(ω) =∑

s|d≤M

ωd

1 + (ω−1ω) + · · · + (ω−1ω)d−1=

∑

s|d≤M

1

d

is a positive integer multiple of 1/d, and therefore nonzero. Thus P (ω) = 0, as desired.

Conversely, if xs − 1 | P (x) for s = 1, 2, . . . ,M , then P (x) will clearly suffice. So we just want thedegree of the least common multiple of the xs − 1 for s = 1, 2, . . . ,M , or just the number of roots ofunity of order at most M , which is

∑Ms=1 φ(s) = 1 + 1 + 2 + 2 + 4 + 2 + 6 + 4 + 6 = 28.

Comment: Only at the beginning do we treat F (x) strictly as a formal power series; later once we get

the rational function representation∑6

d=1xd

1−xd , we can work with polynomial identities in general anddon’t have to worry about convergence issues for |x| ≥ 1.

COMBINATORICS

This test consists of 10 short-answer problems to be solved individually in 50 minutes. Problems will be weighted

with point values after the contest based on how many competitors solve each problem. There is no penalty for

guessing.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted. Similarly,

graph paper, rulers, protractors, compasses, and other drawing aids are not permitted.

Answers should be simplified as much as is reasonably possible and must be exact unless otherwise specified. Ra-

tional numbers should be written in lowest terms, although denominators of irrationals need not be rationalized.

An nth root should be simplified so that the radicand is not divisible by the nth power of any prime.

Correct mathematical notation must be used. No partial credit will be given unless otherwise specified.

If you believe the test contains an error, please submit your protest in writing to the Science Center Lobby during

lunchtime.

Enjoy!

HMMT 2014Saturday 22 February 2014

Combinatorics

1. There are 100 students who want to sign up for the class Introduction to Acting. There are threeclass sections for Introduction to Acting, each of which will fit exactly 20 students. The 100 students,including Alex and Zhu, are put in a lottery, and 60 of them are randomly selected to fill up the classes.What is the probability that Alex and Zhu end up getting into the same section for the class?

2. There are 10 people who want to choose a committee of 5 people among them. They do this by firstelecting a set of 1, 2, 3, or 4 committee leaders, who then choose among the remaining people tocomplete the 5-person committee. In how many ways can the committee be formed, assuming thatpeople are distinguishable? (Two committees that have the same members but di↵erent sets of leadersare considered to be distinct.)

3. Bob writes a random string of 5 letters, where each letter is either A, B, C, or D. The letter in eachposition is independently chosen, and each of the letters A,B,C,D is chosen with equal probability.Given that there are at least two A’s in the string, find the probability that there are at least threeA’s in the string.

4. Find the number of triples of sets (A,B,C) such that:

(a) A,B,C ✓ {1, 2, 3, . . . , 8}.(b) |A \B| = |B \ C| = |C \A| = 2.

(c) |A| = |B| = |C| = 4.

Here, |S| denotes the number of elements in the set S.

5. Eli, Joy, Paul, and Sam want to form a company; the company will have 16 shares to split among the4 people. The following constraints are imposed:

• Every person must get a positive integer number of shares, and all 16 shares must be given out.

• No one person can have more shares than the other three people combined.

Assuming that shares are indistinguishable, but people are distinguishable, in how many ways can theshares be given out?

6. We have a calculator with two buttons that displays an integer x. Pressing the first button replacesx by bx

2 c, and pressing the second button replaces x by 4x + 1. Initially, the calculator displays 0.How many integers less than or equal to 2014 can be achieved through a sequence of arbitrary buttonpresses? (It is permitted for the number displayed to exceed 2014 during the sequence. Here, bycdenotes the greatest integer less than or equal to the real number y.)

7. Six distinguishable players are participating in a tennis tournament. Each player plays one match oftennis against every other player. There are no ties in this tournament—each tennis match resultsin a win for one player and a loss for the other. Suppose that whenever A and B are players in thetournament such that A wins strictly more matches than B over the course of the tournament, it is alsotrue that A wins the match against B in the tournament. In how many ways could the tournamenthave gone?

8. The integers 1, 2, . . . , 64 are written in the squares of a 8⇥8 chess board, such that for each 1 i < 64,the numbers i and i+ 1 are in squares that share an edge. What is the largest possible sum that canappear along one of the diagonals?

9. There is a heads up coin on every integer of the number line. Lucky is initially standing on the zeropoint of the number line facing in the positive direction. Lucky performs the following procedure: helooks at the coin (or lack thereof) underneath him, and then,

• If the coin is heads up, Lucky flips it to tails up, turns around, and steps forward a distance ofone unit.

• If the coin is tails up, Lucky picks up the coin and steps forward a distance of one unit facing thesame direction.

• If there is no coin, Lucky places a coin heads up underneath him and steps forward a distance ofone unit facing the same direction.

He repeats this procedure until there are 20 coins anywhere that are tails up. How many times hasLucky performed the procedure when the process stops?

10. An up-right path from (a, b) 2 R2 to (c, d) 2 R2 is a finite sequence (x1, y1), . . . , (xk

, y

k

) of pointsin R2 such that (a, b) = (x1, y1), (c, d) = (x

k

, y

k

), and for each 1 i < k we have that either(x

i+1, yi+1) = (xi

+1, yi

) or (xi+1, yi+1) = (x

i

, y

i

+1). Two up-right paths are said to intersect if theyshare any point.

Find the number of pairs (A,B) where A is an up-right path from (0, 0) to (4, 4), B is an up-right pathfrom (2, 0) to (6, 4), and A and B do not intersect.

HMMT 2014Saturday 22 February 2014

Combinatorics

PUT LABEL HERE

Name Team ID#

Organization Team

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Score:

HMMT 2014Saturday 22 February 2014

Combinatorics

1. There are 100 students who want to sign up for the class Introduction to Acting. There are threeclass sections for Introduction to Acting, each of which will fit exactly 20 students. The 100 students,including Alex and Zhu, are put in a lottery, and 60 of them are randomly selected to fill up the classes.What is the probability that Alex and Zhu end up getting into the same section for the class?

Answer: 19/165 There is a 60100 = 3

5 chance that Alex is in the class. If Alex is in the class, the

probability that Zhu is in his section is 1999 . So the answer is 3

5 · 1999 = 19

165 .

2. There are 10 people who want to choose a committee of 5 people among them. They do this by firstelecting a set of 1, 2, 3, or 4 committee leaders, who then choose among the remaining people tocomplete the 5-person committee. In how many ways can the committee be formed, assuming thatpeople are distinguishable? (Two committees that have the same members but different sets of leadersare considered to be distinct.)

Answer: 7560 There are(

105

)

ways to choose the 5-person committee. After choosing the

committee, there are 25 − 2 = 30 ways to choose the leaders. So the answer is 30 ·(

105

)

= 7560.

3. Bob writes a random string of 5 letters, where each letter is either A, B, C, or D. The letter in eachposition is independently chosen, and each of the letters A,B,C,D is chosen with equal probability.Given that there are at least two A’s in the string, find the probability that there are at least threeA’s in the string.

Answer: 53188 There are

(

52

)

33 = 270 strings with 2 A’s. There are(

53

)

32 = 90 strings with 3 A’s.

There are(

54

)

31 = 15 strings with 4 A’s. There is(

55

)

30 = 1 string with 5 A’s.

The desired probability is 90+15+1270+90+15+1 = 53

188 .

4. Find the number of triples of sets (A,B,C) such that:

(a) A,B,C ⊆ {1, 2, 3, . . . , 8}.

(b) |A ∩ B| = |B ∩ C| = |C ∩ A| = 2.

(c) |A| = |B| = |C| = 4.

Here, |S| denotes the number of elements in the set S.

Answer: 45360 We consider the sets drawn in a Venn diagram.

A

B C

R1

R2 R3

R4

R5R6

R7

Note that each element that is in at least one of the subsets lies in these seven possible spaces. Wesplit by casework, with the cases based on N = |R7| = |A ∩ B ∩ C|.

Case 1: N = 2

Because we are given that |R4|+ N = |R5|+ N = |R6|+ N = 2, we must have |R4| = |R5| = |R6| = 0.But we also know that |R1|+ |R5|+ |R6|+ N = 4, so |R1| = 2. Similarly, |R2| = |R3| = 2. Since theseregions are distinguishable, we multiply through and obtain

(

82

)(

62

)(

42

)(

22

)

= 2520 ways.

Case 2: N = 1

In this case, we can immediately deduce |R4| = |R5| = |R6| = 1. From this, it follows that |R1| =4 − 1 − 1 − 1 = 1, and similarly, |R2| = |R3| = 1. All seven regions each contain one integer, so thereare a total of (8)(7) . . . (2) = 40320 ways.

Case 3: N = 0

Because |R4| + N = |R5| + N = |R6| + N = 2, we must have |R4| = |R5| = |R6| = 2. Since|R1|+ |R5|+ |R6|+ N = 4, we immediately see that |R1| = 0. Similarly, |R2| = |R3| = 0. The numberof ways to fill R4, R5, R6 is

(

82

)(

62

)(

42

)

= 2520.

This clearly exhausts all the possibilities, so adding gives us 40320 + 2520 + 2520 = 45360 ways.

5. Eli, Joy, Paul, and Sam want to form a company; the company will have 16 shares to split among the4 people. The following constraints are imposed:

• Every person must get a positive integer number of shares, and all 16 shares must be given out.

• No one person can have more shares than the other three people combined.

Assuming that shares are indistinguishable, but people are distinguishable, in how many ways can theshares be given out?

Answer: 315 We are finding the number of integer solutions to a + b + c + d = 16 with 1 ≤a, b, c, d ≤ 8. We count the number of solutions to a+b+c+d = 16 over positive integers, and subtractthe number of solutions in which at least one variable is larger than 8. If at least one variable is largerthan 8, exactly one of the variables is larger than 8. We have 4 choices for this variable. The numberof solutions to a + b + c + d = 16 over positive integers, where a > 8, is just the number of solutionsto a′ + b + c + d = 8 over positive integers, since we can substitute a′ = a − 8. Thus, by the stars andbars formula (the number of positive integer solutions to x1 + · · · + xm = n is

(

n−1m−1

)

), the answer is(

16−14−1

)

−(

41

)(

(16−8)−14−1

)

= 35 · 13 − 4 · 35 = 315.

6. We have a calculator with two buttons that displays an integer x. Pressing the first button replacesx by ⌊x

2 ⌋, and pressing the second button replaces x by 4x + 1. Initially, the calculator displays 0.How many integers less than or equal to 2014 can be achieved through a sequence of arbitrary buttonpresses? (It is permitted for the number displayed to exceed 2014 during the sequence. Here, ⌊y⌋denotes the greatest integer less than or equal to the real number y.)

Answer: 233 We consider the integers from this process written in binary. The first operationtruncates the rightmost digit, while the second operation appends 01 to the right.

We cannot have a number with a substring 11. For simplicity, call a string valid if it has no consecutive1′s. Note that any number generated by this process is valid, as truncating the rightmost digit andappending 01 to the right of the digits clearly preserve validity.

Since we can effectively append a zero by applying the second operation and then the first operation,we see that we can achieve all valid strings.

Note that 2014 has eleven digits when written in binary, and any valid binary string with eleven digitsis at most 10111111111 = 1535. Therefore, our problem reduces to finding the number of eleven-digitvalid strings. Let Fn denote the number of valid strings of length n. For any valid string of length n, wecan create a valid string of length n+1 by appending a 0, or we can create a valid string of length n+2by appending 01. This process is clearly reversible, so our recursion is given by Fn = Fn−1+Fn−2, withF1 = 2, F2 = 3. This yields a sequence of Fibonacci numbers starting from 2, and some computationshows that our answer is F11 = 233.

7. Six distinguishable players are participating in a tennis tournament. Each player plays one match oftennis against every other player. There are no ties in this tournament—each tennis match resultsin a win for one player and a loss for the other. Suppose that whenever A and B are players in thetournament such that A wins strictly more matches than B over the course of the tournament, it is alsotrue that A wins the match against B in the tournament. In how many ways could the tournamenthave gone?

Answer: 2048 We first group the players by wins, so let G1 be the set of all players with the mostwins, G2 be the set of all players with the second most wins, ..., Gn be the set of all players with theleast wins. By the condition in the problem, everyone in group Gi must beat everyone in group Gj forall i < j. Now, consider the mini-tournament consisting of the matches among players inside a singlegroup Gi. Each must have the same number of wins, say xi. But the total number of games is

(

|Gi|2

)

and each game corresponds to exactly one win, so we must have(

|Gi|2

)

= |Gi|xi =⇒ |Gi| = 2xi + 1.Therefore, the number of players in each Gi is odd.

We now have∑

|Gi| = 6 and all |Gi| are odd, so we can now do casework on the possibilities.

Case 1: Gi’s have sizes 5 and 1. In this case, there are 2 ways to permute the groups (i.e. either|G1| = 5, |G2| = 1 or |G1| = 1, |G2| = 5). There are 6 ways to distribute the players into the twogroups. There are 24 possible mini-tournaments in the group of size 5; to prove this, we label theplayers p1, . . . , p5 and note that each player has 2 wins. Without loss of generality, let p1 beat p2

and p3, and also without loss of generality let p2 beat p3. It’s easy to verify that there are 2 possiblemini-tournaments, depending on whether p4 beats p5 or p5 beats p4. Since there are

(

42

)

· 2 = 12 waysto pick the two players p1 defeats and choose which one beats the other, there are indeed 12 · 2 = 24tournaments. Then the total number of possible tournaments in this case is 2 · 6 · 24 = 288.

Case 2: The sizes are 3, 3. In this case, there are(

63

)

= 20 ways to distribute the players into thegroups, and 2 possible mini-tournaments in either group, so the total here is 20 · 2 · 2 = 80.

Case 3: The sizes are 3, 1, 1, 1. In this case, there are 4 ways to permute the groups,(

63

)

· 6 = 120ways to distribute the players into groups, and 2 possible mini-tournaments in the group of size 3, fora total of 4 · 120 · 2 = 960.

Case 4: The sizes are 1, 1, 1, 1, 1, 1. There are 720 ways to distribute the players into groups.

The final answer is 288 + 80 + 960 + 720 = 2048.

8. The integers 1, 2, . . . , 64 are written in the squares of a 8×8 chess board, such that for each 1 ≤ i < 64,the numbers i and i + 1 are in squares that share an edge. What is the largest possible sum that canappear along one of the diagonals?

Answer: 432 Our answer is 26 + 52 + 54 + 56 + 58 + 60 + 62 + 64.One possible configuration:

WLOG, we seek to maximize the sum of the numbers on the main diagonal (top left to bottom right).If we color the squares in a checker-board pattern and use the fact that a and a + 1 lie on differentcolored squares, we notice that all numbers appearing on the main diagonal must be of the same parity.

Consider the smallest value m on the main diagonal. All numbers from 1 to m− 1 must lie on one sideof the diagonal since the main diagonal disconnects the board into two regions, and by assumption, all

26 25 24 23 18 17 8 727 52 53 22 19 16 9 628 51 54 21 20 15 10 529 50 55 56 57 14 11 430 49 44 43 58 13 12 331 48 45 42 59 60 61 232 47 46 41 40 39 62 133 34 35 36 37 38 63 64

numbers less than m cannot lie on the main diagonal. Therefore, m ≤ 29 (one more than the seventhtriangular number) But if m = 29, then the sum of the numbers on the main diagonal is at most29 + 51 + 53 + 55 + 57 + 59 + 61 + 63 = 428, as these numbers must be odd. Similarly, m = 27 is alsonot optimal.

This leaves m = 28 as a possibility. But if this were the case, the only way it beats our answer is if wehave 28+52+54+ ...+64, which would require 52, 54, ..., 64 to appear sequentially along the diagonal,forcing 28 to be in one of the corners.

Now label the squares (row, column) with (1, 1) being the top left and (8, 8) being the bottom right.Assume WLOG that 28 occupies (1, 1). Since 62 and 64 are in (7, 7) and (8, 8), respectively, we musthave 63 in (7, 8) or (8, 7), and WLOG, assume it’s in (8, 7). Since 61 is next to 60, it is not difficultto see that (7, 8) must be occupied by 1 (all numbers a between 2 and 60 must have a − 1 and a + 1as neighbors) . Since 1 is above the main diagonal, all numbers from 1 to 27 must also be above themain diagonal. Since there are 28 squares above the main diagonal, there is exactly one number abovethe main diagonal greater than 28.

Notice that 61 must occupy (7, 6) or (6, 7). If it occupies (7, 6), then we are stuck at (8, 6), since itmust contain a number between 2 and 59, which is impossible. Therefore, 61 must occupy (6, 7), andno more numbers greater than 28 can be above the main diagonal. This forces 59, 57, 55, and 53 tooccupy (6, 5), (5, 4), (4, 3), (3, 2), respectively. But we see that 27 occupies (1, 2) and 29 occupies (2, 1),leaving nowhere for 51.

This is a contradiction, so our answer is therefore optimal.

Alternate solution: Another method of proving that m ≤ 26 is to note that each side of the diagonalhas 28 squares, 16 of which are one color and 12 of which are the other color. As the path has toalternate colors, one can make at most 13 + 12 = 25 steps before moving on the diagonal.

9. There is a heads up coin on every integer of the number line. Lucky is initially standing on the zeropoint of the number line facing in the positive direction. Lucky performs the following procedure: helooks at the coin (or lack thereof) underneath him, and then,

• If the coin is heads up, Lucky flips it to tails up, turns around, and steps forward a distance ofone unit.

• If the coin is tails up, Lucky picks up the coin and steps forward a distance of one unit facing thesame direction.

• If there is no coin, Lucky places a coin heads up underneath him and steps forward a distance ofone unit facing the same direction.

He repeats this procedure until there are 20 coins anywhere that are tails up. How many times hasLucky performed the procedure when the process stops?

Answer: 6098 We keep track of the following quantities: Let N be the sum of 2k, where k rangesover all nonnegative integers such that position −1 − k on the number line contains a tails-up coin.Let M be the sum of 2k, where k ranges over all nonnegative integers such that position k contains atails-up coin.

We also make the following definitions: A ”right event” is the event that Lucky crosses from thenegative integers on the number line to the non-negative integers. A ”left event” is the event thatLucky crosses from the non-negative integers on the number line to the negative integers.

We now make the following claims:

(a) Every time a right event or left event occurs, every point on the number line contains a coin.

(b) Suppose that n is a positive integer. When the nth left event occurs, the value of M is equal ton. When the nth right event occurs, the value of N is equal to n.

(c) For a nonzero integer n, denote by ν2(n) the largest integer k such that 2k divides n. The numberof steps that elapse between the (n−1)st right event and the nth left event is equal to 2ν2(n)+1.The number of steps that elapse between the nth left event and the nth right event is also equalto 2ν2(n) + 1. (If n − 1 = 0, then the “(n − 1)st right event” refers to the beginning of thesimulation.)

(d) The man stops as soon as the 1023rd right event occurs. (Note that 1023 = 210 − 1.)

In other words, Lucky is keeping track of two numbers M and N , which are obtained by interpretingthe coins on the number line as binary strings, and alternately incrementing each of them by one. Wewill prove claim 2; the other claims follow from very similar reasoning and their proofs will be omitted.

Clearly, left and right events alternate. That is, a left event occurs, then a right event, then a leftevent, and so on. So it’s enough to prove that, between each right event and the following left event,the value of M is incremented by 1, and that between each left event and the following right event, thevalue of N is incremented by 1. We will show the first statement; the second follows from symmetry.

Suppose that a right event has just occurred. Then, by claim 1, every space on the number linecotnains a coin. So, there is some nonnegative integer ℓ for which positions 0, . . . , ℓ− 1 on the numberline contain a tails up coin, and position ℓ contains a heads up coin. Since Lucky is standing at position0 facing rightward, the following sequence of steps will occur:

(a) Lucky will take ℓ steps to the right, eventually reaching position ℓ. During this process, he willpick up the coins at positions 0, . . . , ℓ − 1.

(b) Then, Lucky turn the coin at position ℓ to a tails up coin and turn around.

(c) Finally, Lucky will take ℓ + 1 steps to the left, eventually reaching position −1 (at which point aleft event occurs). During this process, he will place a heads up coin at positions 0, . . . , ℓ − 1.

During this sequence, the tails up coins at positions 0, . . . , ℓ− 1 have been changed to heads up coins,and the heads up coin at position ℓ has been changed to a tails up coin. So the value of M has beenincremented by

2ℓ −

ℓ−1∑

i=0

2i = 1

as desired.

Now, it remains to compute the answer to the question. By claims 3 and 4, the total number of stepstaken by the simulation is

2

1023∑

n=1

(2ν2(n) + 1).

This can be rewritten as

4

1023∑

n=1

ν2(n) + 2 · 1023 = 4ν2(1023!) + 2046.

We can compute ν2(1023!) = 1013 using Legendre’s formula for the highest power of 2 dividing afactorial. This results in the final answer 6098.

10. An up-right path from (a, b) ∈ R2 to (c, d) ∈ R

2 is a finite sequence (x1, y1), . . . , (xk, yk) of pointsin R

2 such that (a, b) = (x1, y1), (c, d) = (xk, yk), and for each 1 ≤ i < k we have that either(xi+1, yi+1) = (xi + 1, yi) or (xi+1, yi+1) = (xi, yi + 1). Two up-right paths are said to intersect if theyshare any point.

Find the number of pairs (A,B) where A is an up-right path from (0, 0) to (4, 4), B is an up-right pathfrom (2, 0) to (6, 4), and A and B do not intersect.

Answer: 1750 The number of up-right paths from (0, 0) to (4, 4) is(

84

)

because any such up-right path is identical to a sequence of 4 U’s and 4 R’s, where U corresponds to a step upwards andR corresponds to a step rightwards. Therefore, the total number of pairs of (possibly intersecting)

up-right paths from (0, 0) to (4, 4) and (2, 0) to (6, 4) is(

84

)2.

We will now count the number of intersecting pairs of up-right paths and subtract it to get the answer.Consider an up-right path A from (0, 0) to (4, 4) and an up-right path B from (2, 0) to (6, 4). If theyintersect, take the point (x, y) where they first meet each other, and switch the parts of the paths after(x, y) to make an up-right path A′ from (0, 0) to (6, 4) and an up-right path B′ from (2, 0) to (4, 4).

Conversely, given an up-right path A′ from (0, 0) to (6, 4) and an up-right path B′ from (2, 0) to (4, 4),they must intersect somewhere, so we can again take their first intersection point and switch the endsto get the original up-right path A from (0, 0) to (4, 4) and up-right path B from (2, 0) to (6, 4), whereA and B intersect.

Consequently, the number of intersecting pairs of up-right paths is exactly equal to the number of pairsof up-right paths from (0, 0) to (6, 4) and (2, 0) to (4, 4), which is

(

104

)(

64

)

. The number of pairs that

do not intersect is therefore(

84

)2−

(

104

)(

64

)

= 4900 − 3150 = 1750.

GEOMETRY

This test consists of 10 short-answer problems to be solved individually in 50 minutes. Problems will be weighted

with point values after the contest based on how many competitors solve each problem. There is no penalty for

guessing.

No translators, books, notes, slide rules, calculators, abaci, or other computational aids are permitted. Similarly,

graph paper, rulers, protractors, compasses, and other drawing aids are not permitted.

Answers should be simplified as much as is reasonably possible and must be exact unless otherwise specified. Ra-

tional numbers should be written in lowest terms, although denominators of irrationals need not be rationalized.

An nth root should be simplified so that the radicand is not divisible by the nth power of any prime.

Correct mathematical notation must be used. No partial credit will be given unless otherwise specified.

If you believe the test contains an error, please submit your protest in writing to the Science Center Lobby during

lunchtime.

Enjoy!

HMMT 2014Saturday 22 February 2014

Geometry

1. Let O1 and O2 be concentric circles with radii 4 and 6, respectively. A chord AB is drawn in O1 with

length 2. Extend AB to intersect O2 in points C and D. Find CD.

2. Point P and line ` are such that the distance from P to ` is 12. Given that T is a point on ` such that

PT = 13, find the radius of the circle passing through P and tangent to ` at T .

3. ABC is a triangle such that BC = 10, CA = 12. Let M be the midpoint of side AC. Given that BMis parallel to the external bisector of \A, find area of triangle ABC. (Lines AB and AC form two

angles, one of which is \BAC. The external bisector of \A is the line that bisects the other angle.)

4. In quadrilateral ABCD, \DAC = 98

�,\DBC = 82

�,\BCD = 70

�, and BC = AD. Find \ACD.

AB

C D

98�82�

5. Let C be a circle in the xy plane with radius 1 and center (0, 0, 0), and let P be a point in space

with coordinates (3, 4, 8). Find the largest possible radius of a sphere that is contained entirely in the

slanted cone with base C and vertex P .

6. In quadrilateral ABCD, we have AB = 5, BC = 6, CD = 5, DA = 4, and \ABC = 90

�. Let AC and

BD meet at E. Compute

BEED .

7. Triangle ABC has sides AB = 14, BC = 13, and CA = 15. It is inscribed in circle �, which has center

O. Let M be the midpoint of AB, let B0be the point on � diametrically opposite B, and let X be the

intersection of AO and MB0. Find the length of AX.

8. Let ABC be a triangle with sides AB = 6, BC = 10, and CA = 8. Let M and N be the midpoints of

BA and BC, respectively. Choose the point Y on ray CM so that the circumcircle of triangle AMYis tangent to AN . Find the area of triangle NAY .

9. Two circles are said to be orthogonal if they intersect in two points, and their tangents at either point of

intersection are perpendicular. Two circles !1 and !2 with radii 10 and 13, respectively, are externally

tangent at point P . Another circle !3 with radius 2

p2 passes through P and is orthogonal to both !1

and !2. A fourth circle !4, orthogonal to !3, is externally tangent to !1 and !2. Compute the radius

of !4.

10. Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. Let � be the circumcircle of ABC, let

O be its circumcenter, and let M be the midpoint of minor arc

dBC. Circle !1 is internally tangent to

� at A, and circle !2, centered at M , is externally tangent to !1 at a point T . Ray AT meets segment

BC at point S, such that BS � CS = 4/15. Find the radius of !2.

HMMT 2014Saturday 22 February 2014

Geometry

PUT LABEL HERE

Name Team ID#

Organization Team

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Score:

HMMT 2014Saturday 22 February 2014

Geometry

1. Let O1 and O2 be concentric circles with radii 4 and 6, respectively. A chord AB is drawn in O1 withlength 2. Extend AB to intersect O2 in points C and D. Find CD.

Answer: 2√

21 Let O be the common center of O1 and O2, and let M be the midpoint of AB.

Then OM ⊥ AB, so by the Pythagorean Theorem, OM =√

42 − 12 =√

15. Thus CD = 2CM =2√

62 − 15 = 2√

21.

2. Point P and line ℓ are such that the distance from P to ℓ is 12. Given that T is a point on ℓ such thatPT = 13, find the radius of the circle passing through P and tangent to ℓ at T .

Answer: 169/24 Let O be the center of the given circle, Q be the foot of the altitude from P to

ℓ, and M be the midpoint of PT . Then since OM ⊥ PT and ∠OTP = ∠TPQ, ∆OMP ∼ ∆TQP .

Thus the OP = TP · PMPQ = 13 · 13/2

12 = 16924

3. ABC is a triangle such that BC = 10, CA = 12. Let M be the midpoint of side AC. Given that BMis parallel to the external bisector of ∠A, find area of triangle ABC. (Lines AB and AC form twoangles, one of which is ∠BAC. The external bisector of ∠A is the line that bisects the other angle.)

Answer: 8√

14 Since BM is parallel to the external bisector of ∠A = ∠BAM , it is perpendicular

to the angle bisector of ∠BAM . Thus BA = BM = 12BC = 6. By Heron’s formula, the area of

∆ABC is therefore√

(14)(8)(4)(2) = 8√

14.

4. In quadrilateral ABCD, ∠DAC = 98◦,∠DBC = 82◦,∠BCD = 70◦, and BC = AD. Find ∠ACD.

AB

C D

98◦

82◦

Answer: 28 Let B′ be the reflection of B across CD. Note that AD = BC, and ∠DAC+∠CB′D =180◦, so ACB′D is a cyclic trapezoid. Thus, ACB′D is an isosceles trapezoid, so ∠ACB′ = 98◦. Notethat ∠DCB′ = ∠BCD = 70◦, so ∠ACD = ∠ACB′ − ∠DCB′ = 98◦ − 70◦ = 28◦.

5. Let C be a circle in the xy plane with radius 1 and center (0, 0, 0), and let P be a point in spacewith coordinates (3, 4, 8). Find the largest possible radius of a sphere that is contained entirely in theslanted cone with base C and vertex P .

Answer: 3 −√

5 Consider the plane passing through P that is perpendicular to the plane of thecircle. The intersection of the plane with the cone and sphere is a cross section consisting of a circleinscribed in a triangle with a vertex P . By symmetry, this circle is a great circle of the sphere, andhence has the same radius. The other two vertices of the triangle are the points of intersection betweenthe plane and the unit circle, so the other two vertices are (3

5 , 45 , 0), (− 3

5 ,− 45 , 0).

Using the formula A = rs and using the distance formula to find the side lengths, we find thatr = 2A

2s = 2∗82+10+4

√5

= 3 −√

5.

6. In quadrilateral ABCD, we have AB = 5, BC = 6, CD = 5,DA = 4, and ∠ABC = 90◦. Let AC andBD meet at E. Compute BE

ED .

Answer:√

3 We find that AC =√

61, and applying the law of cosines to triangle ACD tells us that

∠ADC = 120. Then BEED is the ratio of the areas of triangles ABC and ADC, which is (5)(6)

(4)(5)√

3

2

=√

3.

7. Triangle ABC has sides AB = 14, BC = 13, and CA = 15. It is inscribed in circle Γ, which has centerO. Let M be the midpoint of AB, let B′ be the point on Γ diametrically opposite B, and let X be theintersection of AO and MB′. Find the length of AX.

Answer: 65/12 Since B′B is a diameter, ∠B′AB = 90◦, so B′A ‖ OM , so OMB′A = BM

BA = 12 . Thus

AXXO = B′A

OM = 2, so AX = 23R, where R = abc

4A = (13)(14)(15)4(84) = 65

8 is the circumradius of ABC. Putting

it all together gives AX = 6512 .

8. Let ABC be a triangle with sides AB = 6, BC = 10, and CA = 8. Let M and N be the midpoints ofBA and BC, respectively. Choose the point Y on ray CM so that the circumcircle of triangle AMYis tangent to AN . Find the area of triangle NAY .

Answer: 600/73 Let G = AN ∩CM be the centroid of ABC. Then GA = 23GN = 10

3 and GM =

13CM = 1

3

√82 + 32 =

√733 . By power of a point, (GM)(GY ) = GA2, so GY = GA2

GY = (10/3)2√

73

3

= 1003√

73.

Thus

[NAY ] = [GAM ] · [GAY ]

[GAM ]· [NAY ]

[GAY ]

=1

6[ABC] · GY

GM· NA

GA

= 4 · 100

73· 3

2=

600

73

9. Two circles are said to be orthogonal if they intersect in two points, and their tangents at either point ofintersection are perpendicular. Two circles ω1 and ω2 with radii 10 and 13, respectively, are externallytangent at point P . Another circle ω3 with radius 2

√2 passes through P and is orthogonal to both ω1

and ω2. A fourth circle ω4, orthogonal to ω3, is externally tangent to ω1 and ω2. Compute the radiusof ω4.

Answer: 9261 Let ωi have center Oi and radius ri. Since ω3 is orthogonal to ω1, ω2, ω4, it has equal

power r23 to each of them. Thus O3 is the radical center of ω1, ω2, ω4, which is equidistant to the three

sides of △O1O2O4 and therefore its incenter.

For distinct i, j ∈ {1, 2, 4}, ωi ∩ ωj lies on the circles with diameters O3Oi and O3Oj , and hence ω3

itself. It follows that ω3 is the incircle of △O1O2O4, so 8 = r23 = r1r2r4

r1+r2+r4

= 130r4

23+r4

=⇒ r4 = 9261 .

Comment: The condition P ∈ ω3 is unnecessary.

10. Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. Let Γ be the circumcircle of ABC, letO be its circumcenter, and let M be the midpoint of minor arc B̂C. Circle ω1 is internally tangent toΓ at A, and circle ω2, centered at M , is externally tangent to ω1 at a point T . Ray AT meets segmentBC at point S, such that BS − CS = 4/15. Find the radius of ω2.

Answer: 1235/108 Let N be the midpoint of BC. Notice that BS − CS = 415 means that

NS = 215 . let lines MN and AS meet at P , and let D be the foot of the altitude from A to BC.

Then BD = 5 and AD = 12, so DN = 2 and DS = 3215 . Thus NP = AD SN

SD = 12 2/1532/15 = 3

4 . Now

OB = R = abc4A = (13)(14)(15)

4(84) = 658 , so ON =

√OB2 − BN2 =

√(658

)2 − 72 = 338 . Thus OP = 27

8

and PM = OM − OP = 194 . By Monge’s theorem, the exsimilicenter of ω1 and Γ (which is A), the

insimilicenter of ω1 and ω2 (which is T ), and the insimilicenter of ω2 and Γ (call this P ′) are collinear.But notice that this means P ′ = OM ∩ AT = P . From this we get

radius of ω2

R=

MP

OP=

38

27.

Thus the radius of ω2 is 658 · 38

27 = 1235108 .

GUTS

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

1. [4] Compute the prime factorization of 159999.

2. [4] Let x1, . . . , x100 be defined so that for each i, xi is a (uniformly) random integer between 1 and 6inclusive. Find the expected number of integers in the set {x1, x1 + x2, . . . , x1 + x2 + . . .+ x100} thatare multiples of 6.

3. [4] Let ABCDEF be a regular hexagon. Let P be the circle inscribed in 4BDF . Find the ratio ofthe area of circle P to the area of rectangle ABDE.

4. [4] Let D be the set of divisors of 100. Let Z be the set of integers between 1 and 100, inclusive. Markchooses an element d of D and an element z of Z uniformly at random. What is the probability thatd divides z?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

5. [5] If four fair six-sided dice are rolled, what is the probability that the lowest number appearing onany die is exactly 3?

6. [5] Find all integers n for whichn3 + 8

n2 � 4is an integer.

7. [5] The Evil League of Evil is plotting to poison the city’s water supply. They plan to set out fromtheir headquarters at (5, 1) and put poison in two pipes, one along the line y = x and one along theline x = 7. However, they need to get the job done quickly before Captain Hammer catches them.What’s the shortest distance they can travel to visit both pipes and then return to their headquarters?

8. [5] The numbers 20, 21, · · · , 215, 216 = 65536 are written on a blackboard. You repeatedly take twonumbers on the blackboard, subtract one from the other, erase them both, and write the result of thesubtraction on the blackboard. What is the largest possible number that can remain on the blackboardwhen there is only one number left?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

9. [6] Compute the side length of the largest cube contained in the region

{(x, y, z) : x2 + y2 + z2 25 and x � 0}

of three-dimensional space.

10. [6] Find the number of nonempty sets F of subsets of the set {1, . . . , 2014} such that:

(a) For any subsets S1, S2 2 F , S1 \ S2 2 F .

(b) If S 2 F , T ✓ {1, . . . , 2014}, and S ✓ T , then T 2 F .

11. [6] Two fair octahedral dice, each with the numbers 1 through 8 on their faces, are rolled. Let N bethe remainder when the product of the numbers showing on the two dice is divided by 8. Find theexpected value of N .

12. [6] Find a nonzero monic polynomial P (x) with integer coe�cients and minimal degree such thatP (1� 3

p2 + 3

p4) = 0. (A polynomial is called monic if its leading coe�cient is 1.)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

13. [8] An auditorium has two rows of seats, with 50 seats in each row. 100 indistinguishable people sitin the seats one at a time, subject to the condition that each person, except for the first person to sitin each row, must sit to the left or right of an occupied seat, and no two people can sit in the sameseat. In how many ways can this process occur?

14. [8] Let ABCD be a trapezoid with AB k CD and \D = 90�. Suppose that there is a point E on CDsuch that AE = BE and that triangles AED and CEB are similar, but not congruent. Given thatCDAB = 2014, find BC

AD .

15. [8] Given a regular pentagon of area 1, a pivot line is a line not passing through any of the pentagon’svertices such that there are 3 vertices of the pentagon on one side of the line and 2 on the other. Apivot point is a point inside the pentagon with only finitely many non-pivot lines passing through it.Find the area of the region of pivot points.

16. [8] Suppose that x and y are positive real numbers such that x2 � xy + 2y2 = 8. Find the maximumpossible value of x2 + xy + 2y2.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

17. [11] Let f : N ! N be a function satisfying the following conditions:

(a) f(1) = 1.

(b) f(a) f(b) whenever a and b are positive integers with a b.

(c) f(2a) = f(a) + 1 for all positive integers a.

How many possible values can the 2014-tuple (f(1), f(2), ..., f(2014)) take?

18. [11] Find the number of ordered quadruples of positive integers (a, b, c, d) such that a, b, c, and d areall (not necessarily distinct) factors of 30 and abcd > 900.

19. [11] Let ABCD be a trapezoid with AB k CD. The bisectors of \CDA and \DAB meet at E, thebisectors of \ABC and \BCD meet at F , the bisectors of \BCD and \CDA meet at G, and thebisectors of \DAB and \ABC meet at H. Quadrilaterals EABF and EDCF have areas 24 and 36,respectively, and triangle ABH has area 25. Find the area of triangle CDG.

20. [11] A deck of 8056 cards has 2014 ranks numbered 1–2014. Each rank has four suits—hearts,diamonds, clubs, and spades. Each card has a rank and a suit, and no two cards have the same rankand the same suit. How many subsets of the set of cards in this deck have cards from an odd numberof distinct ranks?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

21. [14] Compute the number of ordered quintuples of nonnegative integers (a1, a2, a3, a4, a5) such that0 a1, a2, a3, a4, a5 7 and 5 divides 2a1 + 2a2 + 2a3 + 2a4 + 2a5 .

22. [14] Let ! be a circle, and let ABCD be a quadrilateral inscribed in !. Suppose that BD and ACintersect at a point E. The tangent to ! at B meets line AC at a point F , so that C lies between Eand F . Given that AE = 6, EC = 4, BE = 2, and BF = 12, find DA.

23. [14] Let S = {�100,�99,�98, . . . , 99, 100}. Choose a 50-element subset T of S at random. Find theexpected number of elements of the set {|x| : x 2 T}.

24. [14] Let A = {a1, a2, . . . , a7} be a set of distinct positive integers such that the mean of the elementsof any nonempty subset of A is an integer. Find the smallest possible value of the sum of the elementsin A.

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HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

25. [17] Let ABC be an equilateral triangle of side length 6 inscribed in a circle !. Let A1, A2 be thepoints (distinct from A) where the lines through A passing through the two trisection points of BCmeet !. Define B1, B2, C1, C2 similarly. Given that A1, A2, B1, B2, C1, C2 appear on ! in that order,find the area of hexagon A1A2B1B2C1C2.

26. [17] For 1 j 2014, define

bj = j20142014Y

i=1,i 6=j

(i2014 � j2014)

where the product is over all i 2 {1, . . . , 2014} except i = j. Evaluate

1

b1+

1

b2+ · · ·+ 1

b2014.

27. [17] Suppose that (a1, . . . , a20) and (b1, . . . , b20) are two sequences of integers such that the sequence(a1, . . . , a20, b1, . . . , b20) contains each of the numbers 1, . . . , 40 exactly once. What is the maximumpossible value of the sum

20X

i=1

20X

j=1

min(ai, bj)?

28. [17] Let f(n) and g(n) be polynomials of degree 2014 such that f(n) + (�1)ng(n) = 2n for n =1, 2, . . . , 4030. Find the coe�cient of x2014 in g(x).

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HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

29. [20] Natalie has a copy of the unit interval [0, 1] that is colored white. She also has a black marker,and she colors the interval in the following manner: at each step, she selects a value x 2 [0, 1] uniformlyat random, and

(a) If x 12 she colors the interval [x, x+ 1

2 ] with her marker.

(b) If x > 12 she colors the intervals [x, 1] and [0, x� 1

2 ] with her marker.

What is the expected value of the number of steps Natalie will need to color the entire interval black?

30. [20] Let ABC be a triangle with circumcenter O, incenter I, \B = 45�, and OI k BC. Find cos\C.

31. [20] Compute1007X

k=1

✓cos

✓⇡k

1007

◆◆2014

.

32. [20] Find all ordered pairs (a, b) of complex numbers with a2+b2 6= 0, a+ 10ba2+b2 = 5, and b+ 10a

a2+b2 = 4.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

33. [25] An up-right path from (a, b) 2 R2 to (c, d) 2 R2 is a finite sequence (x1, y1), . . . , (xk, yk) ofpoints in R2 such that (a, b) = (x1, y1), (c, d) = (xk, yk), and for each 1 i < k we have that either(xi+1, yi+1) = (xi + 1, yi) or (xi+1, yi+1) = (xi, yi + 1).

Let S be the set of all up-right paths from (�400,�400) to (400, 400). What fraction of the paths inS do not contain any point (x, y) such that |x|, |y| 10? Express your answer as a decimal numberbetween 0 and 1.

If C is the actual answer to this question and A is your answer, then your score on this problem isdmax{25(1� 10|C �A|), 0}e.

34. [25] Consider a number line, with a lily pad placed at each integer point. A frog is standing at thelily pad at the point 0 on the number line, and wants to reach the lily pad at the point 2014 on thenumber line. If the frog stands at the point n on the number line, it can jump directly to either pointn+2 or point n+3 on the number line. Each of the lily pads at the points 1, · · · , 2013 on the numberline has, independently and with probability 1/2, a snake. Let p be the probability that the frog canmake some sequence of jumps to reach the lily pad at the point 2014 on the number line, without everlanding on a lily pad containing a snake. What is p1/2014? Express your answer as a decimal number.

If C is the actual answer to this question and A is your answer, then your score on this problem isdmax{25(1� 20|C �A|), 0}e.

35. [25] How many times does the letter “e” occur in all problem statements in this year’s HMMT Februarycompetition?

If C is the actual answer to this question and A is your answer, then your score on this problem isdmax{25(1� | log2(C/A)|), 0}e.

36. [25] We have two concentric circles C1 and C2 with radii 1 and 2, respectively. A random chord of C2

is chosen. What is the probability that it intersects C1?

Your answer to this problem must be expressed in the form mn , where m and n are positive integers. If

your answer is in this form, your score for this problem will be b 25·XY c, where X is the total number of

teams who submit the answer mn (including your own team), and Y is the total number of teams who

submit a valid answer. Otherwise, your score is 0. (Your answer is not graded based on correctness,whether your fraction is in lowest terms, whether it is at most 1, etc.)

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HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

1. [4]

2. [4]

3. [4]

4. [4]

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HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

5. [5]

6. [5]

7. [5]

8. [5]

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HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

9. [6]

10. [6]

11. [6]

12. [6]

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HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

13. [8]

14. [8]

15. [8]

16. [8]

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HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

17. [11]

18. [11]

19. [11]

20. [11]

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HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

21. [14]

22. [14]

23. [14]

24. [14]

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HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

25. [17]

26. [17]

27. [17]

28. [17]

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HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

29. [20]

30. [20]

31. [20]

32. [20]

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HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND

Organization Team Team ID#

33. [25]

34. [25]

35. [25]

36. [25]

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HMMT 2014Saturday 22 February 2014

Guts

1. [4] Compute the prime factorization of 159999.

Answer: 3 · 7 · 19 · 401 We have 159999 = 160000 − 1 = 204 − 1 = (20 − 1)(20 + 1)(202 + 1) =3 · 7 · 19 · 401.

2. [4] Let x1, . . . , x100 be defined so that for each i, xi is a (uniformly) random integer between 1 and 6inclusive. Find the expected number of integers in the set {x1, x1 + x2, . . . , x1 + x2 + . . .+ x100} thatare multiples of 6.

Answer: 503 Note that for any i, the probability that x1 + x2 + . . . + xi is a multiple of 6 is 1

6

because exactly 1 value out of 6 possible values of xi works. Because these 100 events are independent,the expected value is 100 · 16 = 50

3 .

3. [4] Let ABCDEF be a regular hexagon. Let P be the circle inscribed in 4BDF . Find the ratio ofthe area of circle P to the area of rectangle ABDE.

Answer: π√3

12 Let the side length of the hexagon be s. The length of BD is s√

3, so the area

of rectangle ABDE is s2√

3. Equilateral triangle BDF has side length s√

3. The inradius of anequilateral triangle is

√3/6 times the length of its side, and so has length s

2 . Thus, the area of circle

P is πs2

4 , so the ratio is πs2/4

s2√3

= π√3

12 .

4. [4] Let D be the set of divisors of 100. Let Z be the set of integers between 1 and 100, inclusive. Markchooses an element d of D and an element z of Z uniformly at random. What is the probability thatd divides z?

Answer: 217900 As 100 = 22 · 52, there are 3 · 3 = 9 divisors of 100, so there are 900 possible pairs

of d and z that can be chosen.

If d is chosen, then there are 100d possible values of z such that d divides z, so the total number of valid

pairs of d and z is∑d|100

100d =

∑d|100 d = (1 + 2 + 22)(1 + 5 + 52) = 7 · 31 = 217. The answer is

therefore 217900 .

5. [5] If four fair six-sided dice are rolled, what is the probability that the lowest number appearing onany die is exactly 3?

Answer: 175/1296 The probability that all the die rolls are at least 3 is 46

4. The probability

they are all at least 4 is 36

4. The probability of being in the former category but not the latter is thus

46

4 − 36

4= 256−81

1296 = 1751296 .

6. [5] Find all integers n for whichn3 + 8

n2 − 4is an integer.

Answer: 0, 1, 3, 4, 6 We have n3+8n2−4 = (n+2)(n2−2n+4)

(n+2)(n−2) = n2−2n+4n−2 for all n 6= −2. Then

n2−2n+4n−2 = n + 4

n−2 , which is an integer if and only if 4n−2 is an integer. This happens when

n − 2 = −4,−2,−1, 1, 2, 4, corresponding to n = −2, 0, 1, 3, 4, 6, but we have n 6= −2 so the answersare 0, 1, 3, 4, 6.

7. [5] The Evil League of Evil is plotting to poison the city’s water supply. They plan to set out fromtheir headquarters at (5, 1) and put poison in two pipes, one along the line y = x and one along the linex = 7. However, they need to get the job done quickly before Captain Hammer catches them. What’sthe shortest distance they can travel to visit both pipes and then return to their headquarters?

Answer: 4√

5 After they go to y = x, we reflect the remainder of their path in y = x, along

with the second pipe and their headquarters. Now, they must go from (5, 1) to y = 7 crossing y = x,

Guts

and then go to (1, 5). When they reach y = 7, we reflect the remainder of their path again, so nowtheir reflected headquarters is at (1, 9). Thus, they just go from (5, 1) to (1, 9) in some path thatinevitably crosses y = x and y = 7. The shortest path they can take is a straight line with length√

42 + 82 = 4√

5.

Comment. These ideas can be used to prove that the orthic triangle of an acute triangle has thesmallest possible perimeter of all inscribed triangles.

Also, see if you can find an alternative solution using Minkowski’s inequality!

8. [5] The numbers 20, 21, · · · , 215, 216 = 65536 are written on a blackboard. You repeatedly take twonumbers on the blackboard, subtract one from the other, erase them both, and write the result of thesubtraction on the blackboard. What is the largest possible number that can remain on the blackboardwhen there is only one number left?

Answer: 131069 If we reverse the order of the numbers in the final subtraction we perform,then the final number will be negated. Thus, the possible final numbers come in pairs with oppositesigns. Therefore, the largest possible number is the negative of the smallest possible number. To getthe smallest possible number, clearly we can take the smallest number originally on the board andsubtract all of the other numbers from it (you can make this rigorous pretty easily if needed), so the

smallest possible number is 1−∑16k=1 2k = 1−131070 = −131069, and thus the largest possible number

is 131069.

9. [6] Compute the side length of the largest cube contained in the region

{(x, y, z) : x2 + y2 + z2 ≤ 25 and x ≥ 0}

of three-dimensional space.

Answer: 5√6

3 The given region is a hemisphere, so the largest cube that can fit inside it has one

face centered at the origin and the four vertices of the opposite face on the spherical surface. Let theside length of this cube be s. Then, the radius of the circle is the hypotenuse of a triangle with side

lengths s and√22 s. So, by the Pythagorean Theorem, the radius equals

√62 s. Since the radius of the

hemisphere is 5, the side length of the cube is 5√6

3 .

10. [6] Find the number of nonempty sets F of subsets of the set {1, . . . , 2014} such that:

(a) For any subsets S1, S2 ∈ F , S1 ∩ S2 ∈ F .

(b) If S ∈ F , T ⊆ {1, . . . , 2014}, and S ⊆ T , then T ∈ F .

Answer: 22014 For a subset S of {1, . . . , 2014}, let FS be the set of all sets T such that S ⊆ T ⊆{1, . . . , 2014}. It can be checked that the sets FS satisfy the conditions 1 and 2. We claim that theFS are the only sets of subsets of {1, . . . , 2014} satisfying the conditions 1 and 2. (Thus, the answer isthe number of subsets S of {1, . . . , 2014}, which is 22014.)

Suppose that F satisfies the conditions 1 and 2, and let S be the intersection of all the sets of F . Weclaim that F = FS . First, by definition of S, all elements T ∈ F are supersets of S, so F ⊆ FS . Onthe other hand, by iterating condition 1, it follows that S is an element of F , so by condition 2 any setT with S ⊆ T ⊆ {1, . . . , 2014} is an element of F . So F ⊇ FS . Thus F = FS .

11. [6] Two fair octahedral dice, each with the numbers 1 through 8 on their faces, are rolled. Let N bethe remainder when the product of the numbers showing on the two dice is divided by 8. Find theexpected value of N .

Answer: 114 If the first die is odd, which has 1

2 probability, then N can be any of 0, 1, 2, 3,

4, 5, 6, 7 with equal probability, because multiplying each element of {0, . . . , 7} with an odd numberand taking modulo 8 results in the same numbers, as all odd numbers are relatively prime to 8. Theexpected value in this case is 3.5.

Guts

If the first die is even but not a multiple of 4, which has 14 probability, then using similar reasoning,

N can be any of 0, 2, 4, 6 with equal probability, so the expected value is 3.

If the first die is 4, which has 18 probability, then N can by any of 0, 4 with equal probability, so the

expected value is 2.

Finally, if the first die is 8, which has 18 probability, then N = 0. The total expected value is 1

2 (3.5) +14 (3) + 1

8 (2) + 18 (0) = 11

4 .

12. [6] Find a nonzero monic polynomial P (x) with integer coefficients and minimal degree such thatP (1− 3

√2 + 3√

4) = 0. (A polynomial is called monic if its leading coefficient is 1.)

Answer: x3 − 3x2 + 9x− 9 Note that (1− 3√

2 + 3√

4)(1 + 3√

2) = 3, so 1− 3√

2 + 3√

4 = 31+ 3√2

.

Now, if f(x) = x3−2, we have f( 3√

2) = 0, so if we let g(x) = f(x−1) = (x−1)3−2 = x3−3x2+3x−3,then g(1+ 3

√2) = f( 3

√2) = 0. Finally, we let h(x) = g( 3

x ) = 27x3− 27

x2 + 9x−3 so h( 3

1+ 3√2) = g(1+ 3

√2) = 0.

To make this a monic polynomial, we multiply h(x) by −x3

3 to get x3 − 3x2 + 9x− 9.

13. [8] An auditorium has two rows of seats, with 50 seats in each row. 100 indistinguishable people sitin the seats one at a time, subject to the condition that each person, except for the first person to sitin each row, must sit to the left or right of an occupied seat, and no two people can sit in the sameseat. In how many ways can this process occur?

Answer:(10050

)298 First, note that there are 249 ways a single row can be filled, because each of

the 49 people after the first in a row must sit to the left or to the right of the current group of peoplein the row, so there are 2 possibilities for each of these 49 people.

Now, there are(10050

)ways to choose the order in which people are added to the rows, and 249 ways to

fill up each row separately, for a total of(10050

)298 ways to fill up the auditorium.

14. [8] Let ABCD be a trapezoid with AB ‖ CD and ∠D = 90◦. Suppose that there is a point E on CDsuch that AE = BE and that triangles AED and CEB are similar, but not congruent. Given thatCDAB = 2014, find BC

AD .

Answer:√

4027 Let M be the midpoint of AB. Let AM = MB = ED = a, ME = AD = b,and AE = BE = c. Since 4BEC ∼ 4DEA, but 4BEC is not congruent to 4DAE, we must have4BEC ∼ 4DEA. Thus, BC/BE = AD/DE = b/a, so BC = bc/a, and CE/EB = AE/ED = c/a,

so EC = c2/a. We are given that CD/AB =c2

a +a

2a = c2

2a2 + 12 = 2014 ⇒ c2

a2 = 4027. Thus,

BC/AD = bc/ab = c/a =

√4027.

15. [8] Given a regular pentagon of area 1, a pivot line is a line not passing through any of the pentagon’svertices such that there are 3 vertices of the pentagon on one side of the line and 2 on the other. Apivot point is a point inside the pentagon with only finitely many non-pivot lines passing through it.Find the area of the region of pivot points.

Answer: 12 (7− 3

√5) Let the pentagon be labeled ABCDE. First, no pivot point can be on

the same side of AC as vertex B. Any such point P has the infinite set of non-pivot lines within thehourglass shape formed by the acute angles between lines PA and PC. Similar logic can be appliedto points on the same side of BD as C, and so on. The set of pivot points is thus a small pentagonwith sides on AC,BD,CE,DA,EB. The side ratio of this small pentagon to the large pentagon is

(2 cos(72◦))2 =3−√

5

2,

so the area of the small pentagon is (3−√

5

2

)2

=1

2(7− 3

√5).

Guts

16. [8] Suppose that x and y are positive real numbers such that x2 − xy + 2y2 = 8. Find the maximumpossible value of x2 + xy + 2y2.

Answer: 72+32√2

7 Let u = x2 + 2y2. By AM-GM, u ≥√

8xy, so xy ≤ u√8

. If we let xy = ku

where k ≤ 1√8

, then we have

u(1− k) = 8

u(1 + k) = x2 + xy + 2y2

that is, u(1 + k) = 8 · 1 + k

1− k. It is not hard to see that the maximum value of this expression occurs at

k =1√8

, so the maximum value is 8 ·1 + 1√

8

1− 1√8

=72 + 32

√2

7.

17. [11] Let f : N→ N be a function satisfying the following conditions:

(a) f(1) = 1.

(b) f(a) ≤ f(b) whenever a and b are positive integers with a ≤ b.(c) f(2a) = f(a) + 1 for all positive integers a.

How many possible values can the 2014-tuple (f(1), f(2), ..., f(2014)) take?

Answer: 1007 Note that f(2014) = f(1007)+1, so there must be exactly one index 1008 ≤ i ≤ 2014such that f(i) = f(i − 1) + 1, and for all 1008 ≤ j ≤ 2014, j 6= i we must have f(j) = f(j − 1). Wefirst claim that each value of i corresponds to exactly one 2014-tuple (f(1), . . . , f(2014)). To provethis, note that f(1024) = 11, so each i uniquely determines the values of f(1007), . . . , f(2014). Thenall of f(1), . . . , f(1006) can be uniquely determined from these values because for any 1 ≤ k ≤ 1006,there exists a unique n such that 1007 ≤ k · 2n ≤ 2014. It’s also clear that these values satisfy thecondition that f is nondecreasing, so we have a correspondence from each 1008 ≤ i ≤ 2014 to a unique2014-tuple.

Also, given any valid 2014-tuple (f(1), . . . , f(2014)), we know that f(1), . . . , f(1006) can be uniquelydetermined by f(1007), . . . , f(2014), which yields some 1008 ≤ i ≤ 2014 where f(i) = f(i− 1) + 1, sowe actually have a bijection between possible values of i and 2014-tuples. Therefore, the total numberof possible 2014-tuples is 1007.

18. [11] Find the number of ordered quadruples of positive integers (a, b, c, d) such that a, b, c, and d areall (not necessarily distinct) factors of 30 and abcd > 900.

Answer: 1940 Since abcd > 900 ⇐⇒ 30a

30b

30c

30d < 900, and there are

(42

)3solutions to

abcd = 223252, the answer is 12 (84 −

(42

)3) = 1940 by symmetry.

19. [11] Let ABCD be a trapezoid with AB ‖ CD. The bisectors of ∠CDA and ∠DAB meet at E, thebisectors of ∠ABC and ∠BCD meet at F , the bisectors of ∠BCD and ∠CDA meet at G, and thebisectors of ∠DAB and ∠ABC meet at H. Quadrilaterals EABF and EDCF have areas 24 and 36,respectively, and triangle ABH has area 25. Find the area of triangle CDG.

Answer: 2567 Let M,N be the midpoints of AD,BC respectively. Since AE and DE are bisectors

of supplementary angles, the triangle AED is right with right angle E. Then EM is the median ofa right triangle from the right angle, so triangles EMA and EMD are isosceles with vertex M . Butthen ∠MEA = ∠EAM = ∠EAB, so EM ‖ AB. Similarly, FN ‖ BA. Thus, both E and F are onthe midline of this trapezoid. Let the length of EF be x. Triangle EFH has area 1 and is similar totriangle ABH, which has area 25, so AB = 5x. Then, letting the heights of trapezoids EABF andEDCF be h (they are equal since EF is on the midline), the area of trapezoid EABF is 6xh

2 = 24. So

the area of trapezoid EDCF is 36 = 9xh2 . Thus DC = 8x. Then, triangle GEF is similar to and has

164 times the area of triangle CDG. So the area of triangle CDG is 64

63 times the area of quadrilateralEDCF , or 256

7 .

Guts

20. [11] A deck of 8056 cards has 2014 ranks numbered 1–2014. Each rank has four suits—hearts,diamonds, clubs, and spades. Each card has a rank and a suit, and no two cards have the same rankand the same suit. How many subsets of the set of cards in this deck have cards from an odd numberof distinct ranks?

Answer: 12 (162014 − 142014) There are

(2014k

)ways to pick k ranks, and 15 ways to pick the suits

in each rank (because there are 16 subsets of suits, and we must exclude the empty one). We thereforewant to evaluate the sum

(20141

)151 +

(20143

)153 + · · ·+

(20142013

)152013.

Note that (1 + 15)2014 = 1 +(20141

)151 +

(20142

)152 + . . . +

(20142013

)152013 + 152014 and (1 − 15)2014 =

1 −(20141

)151 +

(20142

)152 − . . . −

(20142013

)152013 + 152014, so our sum is simply (1+15)2014−(1−15)2014

2 =12 (162014 − 142014).

21. [14] Compute the number of ordered quintuples of nonnegative integers (a1, a2, a3, a4, a5) such that0 ≤ a1, a2, a3, a4, a5 ≤ 7 and 5 divides 2a1 + 2a2 + 2a3 + 2a4 + 2a5 .

Answer: 6528 Let f(n) denote the number of n-tuples (a1, . . . , an) such that 0 ≤ a1, . . . , an ≤ 7and 5 | 2a1 + . . .+2an . To compute f(n+1) from f(n), we note that given any n-tuple (a1, . . . , an) suchthat 0 ≤ a1, . . . , an ≤ 7 and 5 - 2a1 + . . .+ 2an , there are exactly two possible values for an+1 such that0 ≤ an+1 ≤ 7 and 5 | 2a1 + . . .+2an+1 , because 2n ≡ 1, 2, 4, 3, 1, 2, 4, 3 (mod 5) for n = 0, 1, 2, 3, 4, 5, 6, 7respectively.

Also, given any valid (n + 1)-tuple (a1, . . . , an+1), we can remove an+1 to get an n-tuple (a1, . . . , an)such that 0 ≤ a1, . . . , an ≤ 7 and 5 - 2a1 + . . .+ 2an , so these are in bijection. There are a total of 8n

n-tuples, f(n) of which satisfy 5 | 2a1 + . . .+ 2an , so there are 8n − f(n) for which 5 - 2a1 + . . .+ 2an .Therefore, f(n+ 1) = 2(8n − f(n)).

We now have f(1) = 0, f(2) = 2(8 − 0) = 16, f(3) = 2(64 − 16) = 96, f(4) = 2(512 − 96) = 832,f(5) = 2(4096− 832) = 6528.

22. [14] Let ω be a circle, and let ABCD be a quadrilateral inscribed in ω. Suppose that BD and ACintersect at a point E. The tangent to ω at B meets line AC at a point F , so that C lies between Eand F . Given that AE = 6, EC = 4, BE = 2, and BF = 12, find DA.

Answer: 2√

42 By power of a point, we have ED ·EB = EA ·EC, whence ED = 12. Additionally,

by power of a point, we have 144 = FB2 = FC ·FA = FC(FC+ 10), so FC = 8. Note that ∠FBC =∠FAB and ∠CFB = ∠AFB, so 4FBC ∼ 4FAB. Thus, AB/BC = FA/FB = 18/12 = 3/2, soAB = 3k and BC = 2k for some k. Since 4BEC ∼ 4AED, we have AD/BC = AE/BE = 3, soAD = 3BC = 6k. By Stewart’s theorem on 4EBF , we have

(4)(8)(12) + (2k)2(12) = (2)2(8) + (12)2(4) =⇒ 8 + k2 = 8/12 + 12,

whence k2 = 14/3. Thus,

DA = 6k = 6√

14/3 = 6

√42

3= 2√

42.

23. [14] Let S = {−100,−99,−98, . . . , 99, 100}. Choose a 50-element subset T of S at random. Find theexpected number of elements of the set {|x| : x ∈ T}.

Answer: 8825201 Let us solve a more generalized version of the problem: Let S be a set with 2n+ 1

elements, and partition S into sets A0, A1, . . . , An such that |A0| = 1 and |A1| = |A2| = · · · = |An| = 2.(In this problem, we have A0 = {0} and Ak = {k,−k} for k = 1, 2, . . . , 100.) Let T be a randomlychosen m-element subset of S. What is the expected number of Ak’s that have a representative in T?

For k = 0, 1, . . . , n, let wk = 1 if T ∩ Ak 6= ∅ and 0 otherwise, so that the number of Ak’s that havea representative in T is equal to

∑nk=0 wk. It follows that the expected number of Ak’s that have a

representative in T is equal to

E[w0 + w1 + · · ·+ wn] = E[w0] + E[w1] + · · ·+ E[wn] = E[w0] + nE[w1],

Guts

since E[w1] = E[w2] = · · · = E[wn] by symmetry.

Now E[w0] is equal to the probability that T ∩A0 6= ∅, that is, the probability that the single elementof A0 is in T , which is T |/|S| = m/(2n+1). Similarly, E[w1] is the probability that T ∩A1 6= ∅, that is,the probability that at least one of the two elements of A1 is in T . Since there are

(2n−1m

)m-element

subsets of S that exclude both elements of A1, and there are(2n+1m

)m-element subsets of S in total,

we have that

E[w1] = 1−(2n−1m

)(2n+1m

) = 1− (2n−m)(2n−m+ 1)

2n(2n+ 1).

Putting this together, we find that the expected number of Ak’s that have a representative in T is

m

2n+ 1+ n− (2n−m+ 1)(2n−m)

2(2n+ 1).

In this particular problem, we have n = 100 and m = 50, so substituting these values gives our answer

of8825

201.

24. [14] Let A = {a1, a2, . . . , a7} be a set of distinct positive integers such that the mean of the elementsof any nonempty subset of A is an integer. Find the smallest possible value of the sum of the elementsin A.

Answer: 1267 For 2 ≤ i ≤ 6, we claim that a1 ≡ . . . ≡ a7 (mod i). This is because if we considerany i− 1 of the 7 numbers, the other 7− (i− 1) = 8− i of them must all be equal modulo i, becausewe want the sum of all subsets of size i to be a multiple of i. However, 8 − i ≥ 2, and this argumentapplies to any 8− i of the 7 integers, so in fact all of them must be equal modulo i.

We now have that all of the integers are equivalent modulo all of 2, . . . , 6, so they are equivalent modulo60, their least common multiple. Therefore, if the smallest integer is k, then the other 6 integers mustbe at least k + 60, k + 60 · 2, . . . , k + 60 · 6. This means the sum is 7k + 60 · 21 ≥ 7 + 60 · 21 = 1267.1267 is achievable with {1, 1 + 60, . . . , 1 + 60 · 6}, so it is the answer.

25. [17] Let ABC be an equilateral triangle of side length 6 inscribed in a circle ω. Let A1, A2 be thepoints (distinct from A) where the lines through A passing through the two trisection points of BCmeet ω. Define B1, B2, C1, C2 similarly. Given that A1, A2, B1, B2, C1, C2 appear on ω in that order,find the area of hexagon A1A2B1B2C1C2.

Answer: 846√3

49 Let A′ be the point on BC such that 2BA′ = A′C. By law of cosines on triangle

AA′B, we find that AA′ = 2√

7. By power of a point, A′A1 = 2∗42√7

= 4√7. Using side length ratios,

A1A2 = 2AA1

AA′ = 22√7+ 4√

7

2√7

= 187 .

Now our hexagon can be broken down into equilateral triangle A1B1C1 and three copies of triangleA1C1C2. Since our hexagon has rotational symmetry, ∠C2 = 120, and using law of cosines on thistriangle with side lengths 18

7 and 6, a little algebra yields A1C2 = 307 (this is a 3-5-7 triangle with an

angle 120).

The area of the hexagon is therefore 62√3

4 + 3 ∗ 12187

307

√32 = 846

√3

49

26. [17] For 1 ≤ j ≤ 2014, define

bj = j20142014∏

i=1,i6=j

(i2014 − j2014)

where the product is over all i ∈ {1, . . . , 2014} except i = j. Evaluate

1

b1+

1

b2+ · · ·+ 1

b2014.

Guts

Answer: 12014!2014 We perform Lagrange interpolation on the polynomial P (x) = 1 through the

points 12014, 22014, . . . , 20142014. We have

1 = P (x) =

2014∑j=1

∏2014i=1,i6=j(x− i2014)∏2014

i=1,i6=j(j2014 − i2014)

.

Thus,

1 = P (0) =

2014∑j=1

((−1)2013) 2014!2014

j2014

(−1)2013∏2014i=1,i6=j(i

2014 − j2014),

which equals

2014!20142014∑j=1

1

j2014∏2014i=1,i6=j(i

2014 − j2014)= 2014!2014

(1

b1+

1

b2+ · · ·+ 1

b2014

),

so the desired sum is 12014!2014 .

27. [17] Suppose that (a1, . . . , a20) and (b1, . . . , b20) are two sequences of integers such that the sequence(a1, . . . , a20, b1, . . . , b20) contains each of the numbers 1, . . . , 40 exactly once. What is the maximumpossible value of the sum

20∑i=1

20∑j=1

min(ai, bj)?

Answer: 5530 Let xk, for 1 ≤ k ≤ 40, be the number of integers i with 1 ≤ i ≤ 20 such thatai ≥ k. Let yk, for 1 ≤ k ≤ 40, be the number of integers j with 1 ≤ j ≤ 20 such that bj ≥ k. Itfollows from the problem statement that xk +yk is the number of elements of the set {1, . . . , 40} whichare greater than or equal to 40, which is just 41− k.

Note that if 1 ≤ i, j ≤ 20, and 1 ≤ k ≤ 40, then min(ai, bj) ≥ k if and only if ai ≥ k and bj ≥ k. Sofor a fixed k with 1 ≤ k ≤ 40, the number of pairs (i, j) with 1 ≤ i, j ≤ 20 such that min(ai, bj) ≥ k isequal to xkyk. So we can rewrite

20∑i=1

20∑j=1

min(ai, bj) =

40∑k=1

xkyk.

Since xk + yk = 41− k for 1 ≤ k ≤ 40, we have

xkyk ≤⌊

41− k2

⌋⌈41− k

2

⌉by a convexity argument. So

20∑i=1

20∑j=1

min(ai, bj) ≤40∑k=1

⌊41− k

2

⌋⌈41− k

2

⌉= 5530.

Equality holds when (a1, . . . , a20) = (2, 4, . . . , 38, 40) and (b1, . . . , b20) = (1, 3, . . . , 37, 39).

28. [17] Let f(n) and g(n) be polynomials of degree 2014 such that f(n) + (−1)ng(n) = 2n for n =1, 2, . . . , 4030. Find the coefficient of x2014 in g(x).

Answer: 32014

22014·2014! Define the polynomial functions h1 and h2 by h1(x) = f(2x) + g(2x) and

h2(x) = f(2x−1)−g(2x−1). Then, the problem conditions tell us that h1(x) = 22x and h2(x) = 22x−1

for x = 1, 2, . . . , 2015.

Guts

By the Lagrange interpolation formula, the polynomial h1 is given by

h1(x) =

2015∑i=1

22i2015∏j=1i 6=j

x− ji− j

.

So the coefficient of x2014 in h1(x) is

2015∑i=1

22i2015∏j=1i 6=j

1

i− j=

1

2014!

2015∑i=1

22i(−1)2015−i(

2014

i− 1

)=

4 · 32014

2014!

where the last equality follows from the binomial theorem. By a similar argument, the coefficient of

x2014 in h2(x) is 2·320142014! .

We can write g(x) = 12 (h1(x/2)− h2((x+ 1)/2)). So, the coefficient of x2014 in g(x) is

1

2

(4 · 32014

22014 · 2014!− 2 · 32014

22014 · 2014!

)=

32014

22014 · 2014!.

29. [20] Natalie has a copy of the unit interval [0, 1] that is colored white. She also has a black marker,and she colors the interval in the following manner: at each step, she selects a value x ∈ [0, 1] uniformlyat random, and

(a) If x ≤ 12 she colors the interval [x, x+ 1

2 ] with her marker.

(b) If x > 12 she colors the intervals [x, 1] and [0, x− 1

2 ] with her marker.

What is the expected value of the number of steps Natalie will need to color the entire interval black?

Answer: 5 The first choice always wipes out half the interval. So we calculate the expected valueof the amount of time needed to wipe out the other half.

Solution 1 (non-calculus):

We assume the interval has 2n points and we start with the last n colored black. We let f(k) be theexpected value of the number of turns we need if there are k white points left. So we must calculatef(n).

We observe that

f(k) = 1 +(n− k + 1) · 0 + (n− k + 1) · f(k) + 2

∑k−1i=1 f(i)

2n

f(k)n+ k − 1

2n= 1 +

∑k−1i=1 f(i)

n

f(k + 1)n+ k

2n= 1 +

∑ki=1 f(i)

n

f(k + 1) = f(k)n+ k + 1

n+ k

f(k) = f(1)n+ k

n+ 1

And note that f(1) = 2 so f(n) = 4nn+1 and limn→∞ f(n) = 4.

Therefore adding the first turn, the expected value is 5.

Solution 2 (calculus):

We let f(x) be the expected value with length x uncolored. Like above, limx→0 f(x) = 2.

Guts

Similarly we have the recursion

f(x) = 1 + (1

2− x)f(x) + 2

∫ x

0

f(y)dy

f ′(x) = 0 +1

2f ′(x)− f(x)− xf ′(x) + 2f(x)

f ′(x)

f(x)=

1

x+ 12

And solving yields f(x) = c( 12 + x) and since limx→0 f(x) = 2, c = 4. So f(x) = 2 + 4x and f( 1

2 ) = 4.

Therefore adding the first turn, our expected value is 5.

30. [20] Let ABC be a triangle with circumcenter O, incenter I, ∠B = 45◦, and OI ‖ BC. Find cos∠C.

Answer: 1−√22 Let M be the midpoint of BC, and D the foot of the perpendicular of I with BC.

Because OI||BC, we have OM = ID. Since ∠BOC = 2∠A, the length of OM is OA cos∠BOM =OA cosA = R cosA, and the length of ID is r, where R and r are the circumradius and inradius of4ABC, respectively.

Thus, r = R cosA, so 1 + cosA = (R+ r)/R. By Carnot’s theorem, (R+ r)/R = cosA+ cosB+ cosC,

so we have cosB + cosC = 1. Since cosB =√22 , we have cosC = 1−

√22 .

31. [20] Compute1007∑k=1

(cos

(πk

1007

))2014

.

Answer:2014(1+(2013

1007))22014 Let ω = e

2πi2014 . We have ω2014 = 1. Note that cos( πk

1007 ) = 12 (ωk + ω−k).

Our desired expression is

1

22014

1007∑k=1

(ωk + ω−k)2014

Using binomial expansion and switching the order of the resulting summation, this is equal to

1

22014

2014∑j=0

(2014

j

) 1007∑k=1

(ω2014−2j)k

.

Note that unless ω2014−2j = 1, the summand

1007∑k=1

(ω2014−2j)k

is the sum of roots of unity spaced evenly around the unit circle in the complex plane (in particularthe 1007th, 19th, and 53rd roots of unity), so it is zero. Thus, we must only sum over those j for whichω2014−2j = 1, which holds for j = 0, 1007, 2014. This yields the answer

1

22014

(1007 + 1007

(2014

1007

)+ 1007

)=

2014(1 +

(20131007

))22014

.

32. [20] Find all ordered pairs (a, b) of complex numbers with a2+b2 6= 0, a+ 10ba2+b2 = 5, and b+ 10a

a2+b2 = 4.

Answer: (1, 2), (4, 2), ( 52 , 2±

32 i) Solution 1. First, it is easy to see that ab 6= 0. Thus, we can

write5− ab

=4− ba

=10

a2 + b2.

Guts

Then, we have10

a2 + b2=

4a− aba2

=5b− abb2

=4a+ 5b− 2ab

a2 + b2.

Therefore, 4a + 5b − 2ab = 10, so (2a − 5)(b − 2) = 0. Now we just plug back in and get the foursolutions: (1, 2), (4, 2), ( 5

2 , 2±32 i). It’s not hard to check that they all work.

Solution 2. The first equation plus i times the second yields 5 + 4i = a+ bi+ 10(b+ai)a2+b2 = a+ bi− 10i

a+bi ,

which is equivalent to a+ bi = (5±3)+4i2 by the quadratic formula.

Similarly, the second equation plus i times the first yields 4 + 5i = b + ai − 10ib+ai , which is equivalent

to b+ ai = 4+(5±3)i2 .

Letting ε1, ε2 ∈ {−1, 1} be the signs in a+bi and b+ai, we get (a, b) = 12 (a+bi, b+ai)− 1

2 i(b+ai, a+bi) =

( 10+(ε1+ε2)34 , 8+(ε2−ε1)3i

4 ).

Comment. Many alternative approaches are possible. For instance, 5−ab = 4−b

a =⇒ b − 2 =

ε√

(a− 1)(a− 4) for some ε ∈ {−1, 1}, and substituting in and expanding gives 0 = (−2a2+5a)ε√

(a− 1)(a− 4).

More symmetrically, we may write a = λ(4− b), b = λ(5− a) to get (a, b) = λ1−λ2 (4− 5λ, 5− 4λ), and

then plug into a2 + b2 = 10λ to get 0 = 10(λ4 + 1)− 41(λ3 +λ) + 60λ2 = (λ− 2)(2λ− 1)(5λ2− 8λ+ 5).

33. [25] An up-right path from (a, b) ∈ R2 to (c, d) ∈ R2 is a finite sequence (x1, y1), . . . , (xk, yk) ofpoints in R2 such that (a, b) = (x1, y1), (c, d) = (xk, yk), and for each 1 ≤ i < k we have that either(xi+1, yi+1) = (xi + 1, yi) or (xi+1, yi+1) = (xi, yi + 1).

Let S be the set of all up-right paths from (−400,−400) to (400, 400). What fraction of the paths inS do not contain any point (x, y) such that |x|, |y| ≤ 10? Express your answer as a decimal numberbetween 0 and 1.

If C is the actual answer to this question and A is your answer, then your score on this problem isdmax{25(1− 10|C −A|), 0}e.

Answer: 0.2937156494680644 . . . Note that any up-right path must pass through exactly one pointof the form (n,−n) (i.e. a point on the upper-left to lower-right diagonal), and the number of such

paths is(

800400−n

)2because there are

(800

400−n)

up-right paths from (−400,−400) to (n,−n) and another(800

400−n)

from (n,−n) to (400, 400). An up-right path contains a point (x, y) with |x|, |y| ≤ 10 if andonly if −10 ≤ n ≤ 10, so the probability that this happens is∑10

n=−10(

800400−n

)2∑400n=−400

(800

400−n)2 =

∑10n=−10

(800

400−n)2(

1600800

)To estimate this, recall that if we normalize

(800n

)to be a probability density function, then it will

be approximately normal with mean 400 and variance 800 · 14 = 200. If this is squared, then it isproportional to a normal distribution with half the variance and the same mean, because the probability

density function of a normal distribution is proportional to e−(x−µ)2

2σ2 , where µ is the mean and σ2 isthe variance.

Therefore, the(800n

)2probability density function is roughly proportional to a normal distribution with

mean 400 and variance 100, or standard deviation 10. So∑10n=−10

(800

400−n)2

represents roughly onestandard deviation. Recall that approximately 68 percent of a normal distribution lies within onestandard deviation of the mean (look up the 68-95-99.7 rule to read more), so a good guess would bearound .32. This guess can be improved by noting that we’re actually summing 21 values instead of20, so you’d have approximately .68 · 2120 ≈ .71 of the normal distribution, giving an answer of .29.

34. [25] Consider a number line, with a lily pad placed at each integer point. A frog is standing at thelily pad at the point 0 on the number line, and wants to reach the lily pad at the point 2014 on thenumber line. If the frog stands at the point n on the number line, it can jump directly to either point

Guts

n+ 2 or point n+ 3 on the number line. Each of the lily pads at the points 1, · · · , 2013 on the numberline has, independently and with probability 1/2, a snake. Let p be the probability that the frog canmake some sequence of jumps to reach the lily pad at the point 2014 on the number line, without everlanding on a lily pad containing a snake. What is p1/2014? Express your answer as a decimal number.

If C is the actual answer to this question and A is your answer, then your score on this problem isdmax{25(1− 20|C −A|), 0}e.Answer: 0.9102805441016536

First, we establish a rough upper bound for the probability p. Let q be the probability that the frogcan reach the lily pad at the point 2014 on the number line if it is allowed to jump from a point non the number line to the point n + 1, in addition to the points n + 2 and n + 3. Clearly, p ≤ q.Furthermore, p is approximated by q; it should be easy to convince one’s self that jumps from a pointn to the point n+ 1 are only useful for reaching the lily pad at point 2014 in very few situations.

Now we compute q. We note that, if the frog can jump from points n to points n+ 1, n+ 2, and n+ 3,then it can reach the lily pad at the point 2014 on the number line if and only if each snake-free lilypad is at most 3 units away from the closest snake-free lily pad on the left.

Define the sequence {am}∞m=1 by a0 = 1, a1 = 1, a2 = 2, and am+3 = am+2 + am+1 + am for m ≥ 0.Then, it can be shown by induction that am is the number of possible arrangements of snakes on lilypads at points 1, · · · ,m− 1 so that the frog can make some sequence of jumps (of size 1, 2, or 3) fromthe lily pad at point 0 to the lily pad at point m without landing on a lily pad containing a snake. Itfollows that q = a2014/2

2013. So

p1/2014 ≈ q1/2014 = (a2014)1/2014/22013/2014 ≈ (a2014)1/2014/2.

Analyzing the recurrence relation am+3 = am+2 +am+1 +am yields that (a2014)1/2014 is approximatelyequal to the largest real root r of the characteristic polynomial equation r3 − r2 − r − 1 = 0. So toroughly approximate p, it suffices to find the largest real root of this equation.

For this, we apply Newton’s method, or one of many other methods for computing the roots of apolynomial. With an initial guess of 2, one iteration of Newton’s method yields r ≈ 13/7, so p ≈ r/2 ≈13/14 ≈ 0.928571. A second iteration yields r ≈ 1777/966, so p ≈ r/2 ≈ 1777/1932 ≈ 0.919772. (Itturns out that the value of r is 1.839286 . . ., yielding p ≈ r/2 = 0.919643 . . ..)

Using tools from probability theory, we can get an even better estimate for p. We model the problemusing a discrete-time Markov chain. The state of the Markov chain at time n, for n = 0, 1, . . . , 2013,indicates which of the lily pads at positions n − 2, n − 1, n are reachable by the frog. It is clear thatthe state of the Markov chain at time n only depends (randomly) on its state at time n− 1. There are23 = 8 possible states for this Markov chain, because each of the lily pads at positions n−2, n−1, n canbe either reachable or unreachable by the frog. Number each state using the number 1+d2 +2d1 +4d0,where di is 1 if the lily pad at point n − i is reachable, and 0 otherwise. So, for example, at timen = 0, the lily pad at point n is reachable (d0 = 1) whereas the lily pads at points n− 1 and n− 2 areunreachable (d1 = d2 = 0), so the Markov chain is in state number 1 + d2 + 2d1 + 4d0 = 5.

The transition matrix M for the Markov chain can now be computed directly from the conditions ofthe problem. It is equal to

M :=

1 0 0 0 12 0 0 0

0 0 0 0 12 0 0 0

0 1 0 0 0 12 0 0

0 0 0 0 0 12 0 0

0 0 12 0 0 0 1

2 00 0 1

2 0 0 0 12 0

0 0 0 12 0 0 0 1

20 0 0 1

2 0 0 0 12

.

(The verification of this transition matrix is left as an exercise for the reader.) So the state vector vfor the Markov chain at time 2013 is

v := M2014[0, 1, 0, 0, 0, 0, 0, 0]t.

Guts

Now, the lily pad at point 2014 is reachable by the frog if and only if the Markov chain is in state3, 4, 5, 6, 7, or 8 at time 2013. This happens with probability

p = [0, 0, 1, 1, 1, 1, 1, 1] v.

By expanding [0, 1, 0, 0, 0, 0, 0, 0]t in an eigenbasis for M , we find that p1/2014 is approximately equalto the second-largest real eigenvalue of the matrix M . The characteristic polynomial of M is

det(λI −M) = −λ3

8+

3λ4

8+λ6

4− 3λ7

2+ λ8,

so its eigenvalues are the roots of this polynomial. The largest real root of this characteristic polynomialis λ = 1, and the second-largest real root is 0.9105247383471604 . . . (which can be found, again, usingNewton’s method, after factoring out (λ− 1)λ3 from the polynomial), which is a good approximationfor p.

35. [25] How many times does the letter “e” occur in all problem statements in this year’s HMMT Februarycompetition?

If C is the actual answer to this question and A is your answer, then your score on this problem isdmax{25(1− | log2(C/A)|), 0}e.Answer: 1661 It is possible to arrive at a good estimate using Fermi estimation. See http:

//en.wikipedia.org/wiki/Fermi_problem for more details.

For example, there are 76 problems on the HMMT this year. You might guess that the average numberof words in a problem is approximately 40, and the average number of of letters in a word is about 5.The frequency of the letter ”e” in the English language is about 10%, resulting in an estimate of

76 · 40 · 5 · 0.1 = 1520.

This is remarkably close to the actual answer.

36. [25] We have two concentric circles C1 and C2 with radii 1 and 2, respectively. A random chord of C2

is chosen. What is the probability that it intersects C1?

Your answer to this problem must be expressed in the form mn , where m and n are positive integers. If

your answer is in this form, your score for this problem will be b 25·XY c, where X is the total number ofteams who submit the answer m

n (including your own team), and Y is the total number of teams whosubmit a valid answer. Otherwise, your score is 0. (Your answer is not graded based on correctness,whether your fraction is in lowest terms, whether it is at most 1, etc.)

Answer: N/A The question given at the beginning of the problem statement is a famous problem

in probability theory widely known as Bertrand’s paradox. Depending on the interpretation of thephrase “random chord,” there are at least three different possible answers to this question:

• If the random chord is chosen by choosing two (uniform) random endpoints on circle C2 andtaking the chord joining them, the answer to the question is 1/3.

• If the random chord is chosen by choosing a (uniformly) random point P the interior of C2 (otherthan the center) and taking the chord with midpoint P , the answer to the question becomes 1/4.

• If the random chord is chosen by choosing a (uniformly) random diameter d of C, choosing apoint P on d, and taking the chord passing through P and perpendicular to d, the answer tothe question becomes 1/2. (This is also the answer resulting from taking a uniformly randomhorizontal chord of C2.)

You can read more about Bertrand’s paradox online at http://en.wikipedia.org/wiki/Bertrand_

paradox_(probability). We expect that many of the valid submissions to this problem will be equalto 1/2, 1/3, or 1/4.

However, your score on this problem is not based on correctness, but is rather proportional to thenumber of teams who wrote the same answer as you! Thus, this becomes a problem of finding what

Guts

is known in game theory as the “focal point,” or “Schelling point.” You can read more about focalpoints at http://en.wikipedia.org/wiki/Focal_point_(game_theory) or in economist ThomasSchelling’s book The Strategy Of Conflict.

Guts

HMMT November 2014Saturday 15 November 2014

General Test

1. Two circles ω and γ have radii 3 and 4 respectively, and their centers are 10 units apart. Let x be theshortest possible distance between a point on ω and a point on γ, and let y be the longest possibledistance between a point on ω and a point on γ. Find the product xy.

2. Let ABC be a triangle with ∠B = 90◦. Given that there exists a point D on AC such that AD = DC

and BD = BC, compute the value of the ratio ABBC

.

3. Compute the greatest common divisor of 48 − 1 and 812 − 1.

4. In rectangle ABCD with area 1, point M is selected on AB and points X, Y are selected on CD suchthat AX < AY . Suppose that AM = BM . Given that the area of triangle MXY is 1

2014, compute

the area of trapezoid AXY B.

5. Mark and William are playing a game with a stored value. On his turn, a player may either multiplythe stored value by 2 and add 1 or he may multiply the stored value by 4 and add 3. The first playerto make the stored value exceed 2100 wins. The stored value starts at 1 and Mark goes first. Assumingboth players play optimally, what is the maximum number of times that William can make a move?

(By optimal play, we mean that on any turn the player selects the move which leads to the best possibleoutcome given that the opponent is also playing optimally. If both moves lead to the same outcome,the player selects one of them arbitrarily.)

6. Let ABC be a triangle with AB = 5, AC = 4, BC = 6. The angle bisector of C intersects side AB atX. Points M and N are drawn on sides BC and AC, respectively, such that XM ‖ AC and XN ‖ BC.Compute the length MN .

7. Consider the set of 5-tuples of positive integers at most 5. We say the tuple (a1, a2, a3, a4, a5) is perfectif for any distinct indices i, j, k, the three numbers ai, aj , ak do not form an arithmetic progression(in any order). Find the number of perfect 5-tuples.

8. Let a, b, c, x be reals with (a+ b)(b+ c)(c+ a) 6= 0 that satisfy

a2

a+ b=

a2

a+ c+ 20,

b2

b+ c=

b2

b+ a+ 14, and

c2

c+ a=

c2

c+ b+ x.

Compute x.

9. For any positive integers a and b, define a⊕ b to be the result when adding a to b in binary (base 2),neglecting any carry-overs. For example, 20⊕ 14 = 101002 ⊕ 11102 = 110102 = 26. (The operation ⊕is called the exclusive or.) Compute the sum

22014

−1∑

k=0

(

k ⊕

⌊

k

2

⌋)

.

Here ⌊x⌋ is the greatest integer not exceeding x.

10. Suppose that m and n are integers with 1 ≤ m ≤ 49 and n ≥ 0 such that m divides nn+1 + 1. Whatis the number of possible values of m?

HMMT November 2014Saturday 15 November 2014

General Test

1. Two circles ω and γ have radii 3 and 4 respectively, and their centers are 10 units apart. Let x be theshortest possible distance between a point on ω and a point on γ, and let y be the longest possibledistance between a point on ω and a point on γ. Find the product xy.

Answer: 51 Let ℓ be the line connecting the centers of ω and γ. Let A and B be the intersectionsof ℓ with ω, and let C and D be the intersections of ℓ with γ, so that A, B, C, and D are collinear,in that order. The shortest distance between a point on ω and a point on γ is BC = 3. The longestdistance is AD = 3 + 10 + 4 = 17. The product is 51.

2. Let ABC be a triangle with ∠B = 90◦. Given that there exists a point D on AC such that AD = DCand BD = BC, compute the value of the ratio AB

BC.

Answer:√3 D is the circumcenter of ABC because it is the midpoint of the hypotenuse.

Therefore, DB = DA = DC because they are all radii of the circumcircle, so DBC is an equilateral

triangle, and ∠C = 60◦. This means that ABC is a 30◦ − 60◦ − 90◦ triangle, withAB

BC=

√3 .

3. Compute the greatest common divisor of 48 − 1 and 812 − 1.

Answer: 15 Let d = gcd(a, b) for some a, b ∈ Z+.

Then, we can write d = ax− by, where x, y ∈ Z+, and

2a − 1 | 2ax − 1 (1)

2b − 1 | 2by − 1 (2)

Multiplying the right-hand side of (2) by 2d, we get,

2b − 1 | 2ax − 2d

Thus, gcd(2a − 1, 2b − 1) = 2d − 1 = 2gcd(a,b) − 1.

Using a = 16 and b = 36, we get

gcd(216 − 1, 236 − 1) = 2gcd(16,36) − 1 = 24 − 1 = 15

4. In rectangle ABCD with area 1, point M is selected on AB and points X, Y are selected on CD suchthat AX < AY . Suppose that AM = BM . Given that the area of triangle MXY is 1

2014 , computethe area of trapezoid AXY B.

Answer: 12 + 1

2014 OR 5041007 Notice that [AMX] + [BYM ] = 1

2 [ABCD] = 12 . Thus,

[AXY B] = [AMX] + [BYM ] + [MXY ] =1

2+

1

2014=

504

1007

5. Mark and William are playing a game with a stored value. On his turn, a player may either multiplythe stored value by 2 and add 1 or he may multiply the stored value by 4 and add 3. The first playerto make the stored value exceed 2100 wins. The stored value starts at 1 and Mark goes first. Assumingboth players play optimally, what is the maximum number of times that William can make a move?

(By optimal play, we mean that on any turn the player selects the move which leads to the best possibleoutcome given that the opponent is also playing optimally. If both moves lead to the same outcome,the player selects one of them arbitrarily.)

Answer: 33 We will work in the binary system in this solution.

General Test

Let multiplying the stored value by 2 and adding 1 be Move A and multiplying the stored value by 4and adding 3 be Move B. Let the stored value be S. Then, Move A affixes one 1 to S, while Move Baffixes two 1s. The goal is to have greater than or equal to 101 1s. If any player makes the numberof 1s in S congruent to 2 mod 3, then no matter what the other player does, he will lose, since thenumber of 1s in S reaches 101 or 102 only from 99 ≡ 0 (mod 3) or 100 ≡ 1 (mod 3).

Mark’s winning strategy: Do Move A. In the succeeding moves, if William does Move B, then Markdoes Move A, and vice versa, which in total, affixes three 1s to S. This ensures that William alwaystakes his turn while the number of 1s in S is congruent to 2 mod 3. Note that Mark has to follow thisstrategy because once he does not, then William can follow the same strategy and make Mark lose, acontradiction to the required optimal play. Since S starts out with one 1, this process gives William amaximum of 33 moves.

6. Let ABC be a triangle with AB = 5, AC = 4, BC = 6. The angle bisector of C intersects side AB atX. Points M and N are drawn on sides BC and AC, respectively, such that XM ‖ AC and XN ‖ BC.Compute the length MN .

Answer: 3√145 By Stewart’s Theorem on the angle bisector,

CX2 = AC ·BC

(

1− AB

AC +BC

2)

Thus,

CX2 = 4 · 6(

1− 5

10

2)

= 18

Since XM ‖ AC and XN ‖ BC, we produce equal angles. So, by similar triangles, XM = XN =4·610 = 12

5 .Moreover, triangles MCX and NCX are congruent isosceles triangles with vertices M and N , respec-tively. Since CX is an angle bisector, then CX and MN are perpendicular bisectors of each other.Therefore,

MN2 = 4(XN2 − (CX/2)2) = 4 ·(12

5

)2

− 18 =126

25

and

MN =3√14

5

7. Consider the set of 5-tuples of positive integers at most 5. We say the tuple (a1, a2, a3, a4, a5) is perfectif for any distinct indices i, j, k, the three numbers ai, aj , ak do not form an arithmetic progression(in any order). Find the number of perfect 5-tuples.

Answer: 780 There are two situations.

1. The multiset is aabbc; the only condition here is c 6= 12 (a + b), for (

(53

)− |S|) ·

(31

)= 18 such

triples, where S is the set of unordered triples (a, b, c) which do not satisfy the condition, and S ={(1, 2, 3), (2, 3, 4), (3, 4, 5), (1, 3, 5)}. Each one gives 5!

2!2! = 30 orderings, so 18 · 30 = 540 in this case.

2. There are four distinct elements in the tuple. Then, the elements must be {1, 2, 4, 5}. All of themwork, for an additional 4 · 60 = 240.

Therefore, there are 540 + 240 = 780 such tuples.

8. Let a, b, c, x be reals with (a+ b)(b+ c)(c+ a) 6= 0 that satisfy

a2

a+ b=

a2

a+ c+ 20,

b2

b+ c=

b2

b+ a+ 14, and

c2

c+ a=

c2

c+ b+ x.

Compute x.

General Test

Answer: −34 Note that

a2

a+ b+

b2

b+ c+

c2

c+ a− a2

c+ a− b2

a+ b− c2

b+ c=

a2 − b2

a+ b+

b2 − c2

b+ c+

c2 − a2

c+ a

= (a− b) + (b− c) + (c− a)

= 0.

Thus, when we sum up all the given equations, we get that20 + 14 + x = 0. Therefore, x = −34.

9. For any positive integers a and b, define a⊕ b to be the result when adding a to b in binary (base 2),neglecting any carry-overs. For example, 20⊕ 14 = 101002 ⊕ 11102 = 110102 = 26. (The operation ⊕is called the exclusive or.) Compute the sum

22014−1∑

k=0

(

k ⊕⌊k

2

⌋)

.

Here ⌊x⌋ is the greatest integer not exceeding x.

Answer: 22013(22014 − 1) OR 24027 − 22013 Let k = a2013a2012...a0 in base 2. Then⌊k2

⌋=

0a2013...a1 in base 2. So the leftmost digit of k ⊕⌊k2

⌋is 1 if and only if a2013 = 1, and the nth digit

from the right is 1 if and only if an 6= an−1 (1 ≤ n ≤ 2013).

In either case, the probability of each digit being 1 is 12 . Therefore, the sum of all such numbers is

1

2· 22014 · 11 . . . 112

︸ ︷︷ ︸

2014 digits

= 22013(22014 − 1).

10. Suppose that m and n are integers with 1 ≤ m ≤ 49 and n ≥ 0 such that m divides nn+1 + 1. Whatis the number of possible values of m?

Answer: 29 If n is even, n+ 1 | nn+1 + 1, so we can cover all odd m.

If m is even and m | nn+1 +1, then n must be odd, so n+1 is even, and m cannot be divisible by 4 orany prime congruent to 3 (mod 4). Conversely, if m/2 has all factors 1 (mod 4), then by CRT thereexists N ≡ 1 (mod 4) such that m | N2 + 1 | NN+1 + 1 (note (N + 1)/2 is odd).

So the only bad numbers take the form 2k, where 1 ≤ k ≤ 24 is divisible by at least one of2, 3, 7, 11, 19, 23, 31, . . .. We count k = 2, 4, . . . , 24 (there are 12 numbers here), k = 3, 9, 15, 21 (anotherfour), k = 7, 11, 19, 23 (another four), giving a final answer of 49− 12− 4− 4 = 29.

General Test

HMMT November 2014Saturday 15 November 2014

Theme Round

Townspeople and Goons

In the city of Lincoln, there is an empty jail, at least two townspeople and at least one goon. A gameproceeds over several days, starting with morning.

• Each morning, one randomly selected unjailed person is placed in jail. If at this point all goonsare jailed, and at least one townsperson remains, then the townspeople win. If at this point alltownspeople are jailed and at least one goon remains, then the goons win.

• Each evening, if there is at least one goon and at least one townsperson not in jail, then onerandomly selected townsperson is jailed. If at this point there are at least as many goons remainingas townspeople remaining, then the goons win.

The game ends immediately after any group wins.

1. Find the probability that the townspeople win if there are initially two townspeople and one goon.

2. Find the smallest positive integer n such that, if there are initially 2n townspeople and 1 goon, thenthe probability the townspeople win is greater than 50%.

3. Find the smallest positive integer n such that, if there are initially n + 1 townspeople and n goons,then the probability the townspeople win is less than 1%.

4. Suppose there are initially 1001 townspeople and two goons. What is the probability that, when thegame ends, there are exactly 1000 people in jail?

5. Suppose that there are initially eight townspeople and one goon. One of the eight townspeople isnamed Jester. If Jester is sent to jail during some morning, then the game ends immediately in hissole victory. (However, the Jester does not win if he is sent to jail during some night.)

Find the probability that only the Jester wins.

Parabolas

A parabola is the set of points in the plane equidistant from a point F (its focus) and a line ℓ (itsdirectrix).

ℓ

F

Parabolas can be represented as polynomial equations of degree two, such as y = x2.

6. Let P1, P2, P3 be pairwise distinct parabolas in the plane. Find the maximum possible number ofintersections between two or more of the Pi. In other words, find the maximum number of points thatcan lie on two or more of the parabolas P1, P2, P3.

7. Let P be a parabola with focus F and directrix ℓ. A line through F intersects P at two points A andB. Let D and C be the feet of the altitudes from A and B onto ℓ, respectively. Given that AB = 20and CD = 14, compute the area of ABCD.

8. Consider the parabola consisting of the points (x, y) in the real plane satisfying

(y + x) = (y − x)2 + 3(y − x) + 3.

Find the minimum possible value of y.

9. In equilateral triangle ABC with side length 2, let the parabola with focus A and directrix BC intersectsides AB and AC at A1 and A2, respectively. Similarly, let the parabola with focus B and directrixCA intersect sides BC and BA at B1 and B2, respectively. Finally, let the parabola with focus C anddirectrix AB intersect sides CA and CB at C1 and C2, respectively.

Find the perimeter of the triangle formed by lines A1A2, B1B2, C1C2.

10. Let z be a complex number and k a positive integer such that zk is a positive real number other than1. Let f(n) denote the real part of the complex number zn. Assume the parabola p(n) = an2 + bn+ c

intersects f(n) four times, at n = 0, 1, 2, 3. Assuming the smallest possible value of k, find the largestpossible value of a.

HMMT November 2014Saturday 15 November 2014

Theme Round

1. Find the probability that the townspeople win if there are initially two townspeople and one goon.

Answer: 1

3The goon is chosen on the first turn with probability 1

3, and this is necessary and

sufficient for the townspeople to win.

2. Find the smallest positive integer n such that, if there are initially 2n townspeople and 1 goon, thenthe probability the townspeople win is greater than 50%.

Answer: 3 We instead consider the probability the goon wins. The game clearly must last n days.The probability the goon is not sent to jail on any of these n days is then

2n

2n+ 1· 2n− 2

2n− 1· · · · · 2

3.

If n = 2 then the probability the goon wins is 4

5· 23= 8

15> 1

2, but when n = 3 we have 6

7· 8

15= 16

35< 1

2,

so the answer is n = 3.

Alternatively, the let pn be the probability that 2n townspeople triumph against 1 goon. There is a1

2n+1chance that the goon is jailed during the first morning and the townspeople win. Otherwise, the

goon eliminates one townperson during the night. We thus have 2n − 2 townspeople and 1 goon left,so the probability that the town wins is pn−1. We obtain the recursion

pn =1

2n+ 1+

2n

2n+ 1pn−1.

By the previous question, we have the initial condition p1 = 1

3. We find that p2 = 7

15< 1

2and

p3 = 19

35> 1

2, yielding n = 3 as above.

3. Find the smallest positive integer n such that, if there are initially n + 1 townspeople and n goons,then the probability the townspeople win is less than 1%.

Answer: 6 By a similar inductive argument, the probability for a given n is

pn =n!

(2n+ 1)!!.

Clearly this is decreasing in n. It is easy to see that

p5 =1 · 2 · 3 · 4 · 53 · 5 · 7 · 9 · 11 =

8

693> 0.01

and

p6 =6

13p5 =

48

693 · 13 < 0.01.

Hence the answer is n = 6. Heuristically, pn+1 ≈ 1

2pn for each n, so arriving at these estimates for the

correct answer of n is not difficult.

4. Suppose there are initially 1001 townspeople and two goons. What is the probability that, when thegame ends, there are exactly 1000 people in jail?

Answer: 3

1003By considering the parity of the number of people in jail, we see that this situation

arises if and only if the goons win after the 500th night. That means that at this point we must haveexactly one townsperson and two goons remaining. In other words, this situation arises if and only ifno goon is ever sent to jail. The probability that this occurs is

1001

1003· 999

1001· 997999

· . . . 35=

3

1003.

Theme Round

5. Suppose that there are initially eight townspeople and one goon. One of the eight townspeople isnamed Jester. If Jester is sent to jail during some morning, then the game ends immediately in hissole victory. (However, the Jester does not win if he is sent to jail during some night.)

Find the probability that only the Jester wins.

Answer: 1

3Let an denote the answer when there are 2n− 1 regular townies, one Jester, and one

goon. It is not hard to see that a1 = 1

3. Moreover, we have a recursion

an =1

2n+ 1· 1 + 1

2n+ 1· 0 + 2n− 1

2n+ 1

(

1

2n− 1· 0 + 2n− 2

2n− 1· an−1

)

.

The recursion follows from the following consideration: during the day, there is a 1

2n+1chance the Jester

is sent to jail and a 1

2n+1chance the goon is sent to jail, at which point the game ends. Otherwise, there

is a 1

2n−1chance that the Jester is selected to be jailed from among the townies during the evening. If

none of these events occur, then we arrive at the situation of an−1.

Since a1 = 1

3, we find that an = 1

3for all values of n. This gives the answer.

6. Let P1, P2, P3 be pairwise distinct parabolas in the plane. Find the maximum possible number ofintersections between two or more of the Pi. In other words, find the maximum number of points thatcan lie on two or more of the parabolas P1, P2, P3.

Answer: 12 Note that two distinct parabolas intersect in at most 4 points, which is not difficultto see by drawing examples.1 Given three parabolas, each pair intersects in at most 4 points, for atmost 4 · 3 = 12 points of intersection in total. It is easy to draw an example achieving this maximum,for example, by slanting the parabolas at different angles.

7. Let P be a parabola with focus F and directrix ℓ. A line through F intersects P at two points A andB. Let D and C be the feet of the altitudes from A and B onto ℓ, respectively. Given that AB = 20and CD = 14, compute the area of ABCD.

Answer: 140 Observe that AD + BC = AF + FB = 20, and that ABCD is a trapezoid withheight BC = 14. Hence the answer is 1

2(AD +BC)(14) = 140.

8. Consider the parabola consisting of the points (x, y) in the real plane satisfying

(y + x) = (y − x)2 + 3(y − x) + 3.

Find the minimum possible value of y.

Answer: − 1

2Let w = y − x. Adding w to both sides and dividing by two gives

y =w2 + 4w + 3

2=

(w + 2)2 − 1

2,

which is minimized when w = −2. This yields y = − 1

2.

9. In equilateral triangle ABC with side length 2, let the parabola with focus A and directrix BC intersectsides AB and AC at A1 and A2, respectively. Similarly, let the parabola with focus B and directrixCA intersect sides BC and BA at B1 and B2, respectively. Finally, let the parabola with focus C anddirectrix AB intersect sides CA and CB at C1 and C2, respectively.

Find the perimeter of the triangle formed by lines A1A2, B1B2, C1C2.

Answer: 66− 36√3 Since everything is equilateral it’s easy to find the side length of the wanted

triangle. By symmetry, it’s just AA1 + 2A1B2 = 3AA1 − AB. Using the definition of a parabola,

AA1 =√3

2A1B so some calculation gives a side length of 2(11− 6

√3), thus the perimeter claimed.

1This can be seen rigorously by showing that five points in the plane determine at most one quadratic equation ax2 + by2 +

cxy + dx+ ey + f = 0, up to scaling.

Theme Round

10. Let z be a complex number and k a positive integer such that zk is a positive real number other than1. Let f(n) denote the real part of the complex number zn. Assume the parabola p(n) = an2 + bn+ c

intersects f(n) four times, at n = 0, 1, 2, 3. Assuming the smallest possible value of k, find the largestpossible value of a.

Answer: 1

3Let r = |z|, θ = arg z, and C = ℜz

|z| = cos θ = cos 2πj

kfor some j with gcd(j, k) = 1. The

condition of the four consecutive points lying on a parabola is equivalent to having the finite difference

f(3)− 3f(2) + 3f(1)− f(0) = 0.

This implies

f(3)− f(0) = 3 [f(2)− f(1)]

⇐⇒ r3 cos(3θ)− 1 = 3(

r2 cos(2θ)− r cos(θ))

⇐⇒ r3(4C3 − 3C)− 1 = 3(r2(2C2 − 1)− rC).

Now we simply test the first few possible values of k.

k = 1 implies C = 1, which gives r3 − 1 = 3(r2 − r) =⇒ (r − 1)3 = 0 =⇒ r = 1. This is not allowedsince r = 1 implies a periodic function.

k = 2 implies C = −1, which gives −r3 − 1 = 3r2 + r =⇒ (r+1)3 = 0, again not allowed since r > 0.

k = 3 implies C = − 1

2. This gives r3 − 1 = −3

2(r2 − r) =⇒ (r− 1)(r+ 1

2)(r+ 2) = 0. These roots are

either negative or 1, again not allowed.

k = 4 implies C = 0. This gives −1 = −3r2 =⇒ r = ± 1√3. r = 1√

3is allowed, so this will generate

our answer.

Again by finite differences (or by any other method of interpolating with a quadratic), we get 2a =f(0) + f(2)− 2f(1) = 2

3, so a = 1

3.

Theme Round

HMMT November 2014Saturday 15 November 2014

Team Round

Mins and Maxes

1. [3] What is the smallest positive integer n which cannot be written in any of the following forms?

• n = 1 + 2 + · · ·+ k for a positive integer k.

• n = pk for a prime number p and integer k.

• n = p+ 1 for a prime number p.

2. [5] Let f(x) = x2 + 6x + 7. Determine the smallest possible value of f(f(f(f(x)))) over all realnumbers x.

3. [5] The side lengths of a triangle are distinct positive integers. One of the side lengths is a multipleof 42, and another is a multiple of 72. What is the minimum possible length of the third side?

Enumeration

4. [3] How many ways are there to color the vertices of a triangle red, green, blue, or yellow such thatno two vertices have the same color? Rotations and reflections are considered distinct.

5. [5] Let A,B,C,D,E be five points on a circle; some segments are drawn between the points so thateach of the

(

5

2

)

= 10 pairs of points is connected by either zero or one segments. Determine the numberof sets of segments that can be drawn such that:

• It is possible to travel from any of the five points to any other of the five points along drawnsegments.

• It is possible to divide the five points into two nonempty sets S and T such that each segmenthas one endpoint in S and the other endpoint in T .

6. [6] Find the number of strictly increasing sequences of nonnegative integers with the following prop-erties:

• The first term is 0 and the last term is 12. In particular, the sequence has at least two terms.

• Among any two consecutive terms, exactly one of them is even.

7. [7] Sammy has a wooden board, shaped as a rectangle with length 22014 and height 32014. The boardis divided into a grid of unit squares. A termite starts at either the left or bottom edge of the rectangle,and walks along the gridlines by moving either to the right or upwards, until it reaches an edge oppositethe one from which the termite started. Depicted below are two possible paths of the termite.

The termite’s path dissects the board into two parts. Sammy is surprised to find that he can stillarrange the pieces to form a new rectangle not congruent to the original rectangle. This rectangle hasperimeter P . How many possible values of P are there?

Hexagons

8. [3] Let H be a regular hexagon with side length one. Peter picks a point P uniformly and at randomwithin H, then draws the largest circle with center P that is contained in H. What is this probabilitythat the radius of this circle is less than 1

2?

9. [5] How many lines pass through exactly two points in the following hexagonal grid?

10. [8] Let ABCDEF be a convex hexagon with the following properties.

(a) AC and AE trisect ∠BAF .

(b) BE ‖ CD and CF ‖ DE.

(c) AB = 2AC = 4AE = 8AF .

Suppose that quadrilaterals ACDE and ADEF have area 2014 and 1400, respectively. Find the areaof quadrilateral ABCD.

HMMT November 2014Saturday 15 November 2014

Team Round

1. [3] What is the smallest positive integer n which cannot be written in any of the following forms?

• n = 1 + 2 + · · ·+ k for a positive integer k.

• n = pk for a prime number p and integer k.

• n = p+ 1 for a prime number p.

Answer: 22 Consider 1, 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19 are in the form pk. So we are left with6, 10, 12, 14, 15, 18, 20, 21, 22, ...

Next, 6, 12, 14, 18, 20 are in the form p+ 1, so we are left with 10, 15, 21, 22, ...

Finally, 10, 15, 21 are in the form n = 1 + 2 + · · ·+ k, so we are left with 22, ...

Since 22 = 2 · 11 is not a prime power, 22− 1 = 21 is not prime, and 1 + 2 + · · ·+ 6 = 21 < 22 < 28 =1 + 2 + · · ·+ 7, 22 is the smallest number not in the three forms, as desired.

2. [5] Let f(x) = x2 + 6x + 7. Determine the smallest possible value of f(f(f(f(x)))) over all realnumbers x.

Answer: 23 Consider that f(x) = x2 + 6x+ 7 = (x+ 3)2 − 2. So f(x) ≥ −2 for real numbers x.Also, f is increasing on the interval [−3,∞). Therefore

f(f(x)) ≥ f(−2) = −1,

f(f(f(x))) ≥ f(−1) = 2,

andf(f(f(f(x)))) ≥ f(2) = 23.

Thus, the minimum value of f(f(f(f(x)))) is 23 and equality is obtained when x = −3.

3. [5] The side lengths of a triangle are distinct positive integers. One of the side lengths is a multipleof 42, and another is a multiple of 72. What is the minimum possible length of the third side?

Answer: 7 Suppose that two of the side lengths are 42a and 72b, for some positive integers aand b. Let c be the third side length. We know that 42a is not equal to 72b, since the side lengths aredistinct. Also, 6|42a− 72b. Therefore, by the triangle inequality, we get c > |42a− 72b| ≥ 6 and thusc ≥ 7. Hence, the minimum length of the third side is 7 and equality is obtained when a = 7 and b = 4.

4. [3] How many ways are there to color the vertices of a triangle red, green, blue, or yellow such thatno two vertices have the same color? Rotations and reflections are considered distinct.

Answer: 24 There are 4 ways to color the first vertex, then 3 ways to color the second vertex tobe distinct from the first, and finally 2 ways to color the third vertex to be distinct from the earliertwo vertices. Multiplying gives 24 ways.

5. [5] Let A,B,C,D,E be five points on a circle; some segments are drawn between the points so thateach of the

(

5

2

)

= 10 pairs of points is connected by either zero or one segments. Determine the numberof sets of segments that can be drawn such that:

• It is possible to travel from any of the five points to any other of the five points along drawnsegments.

• It is possible to divide the five points into two nonempty sets S and T such that each segmenthas one endpoint in S and the other endpoint in T .

Team Round

Answer: 195 First we show that we can divide the five points into sets S and T according to thesecond condition in only one way. Assume that we can divide the five points into S ∪ T and S′ ∪ T ′.Then, let A = S′ ∩ S,B = S′ ∩ T,C = T ′ ∩ S, and D = T ′ ∩ T . Since S, T and S′, T ′ partition the setof five points, A,B,C,D also partition the set of five points.

Now, according to the second condition, there can only be segments between S and T and between S′

and T ′. Therefore, the only possible segments are between points in A and D, or between points in Band C. Since, according to the first condition, the points are all connected via segments, it must bethat A = D = ∅ or B = C = ∅. If A = D = ∅, then it follows that S′ = T and T ′ = S. Otherwise, ifB = C = ∅, then S′ = S and T ′ = T . In either case, S, T and S′, T ′ are the same partition of the fivepoints, as desired.

We now determine the possible sets of segments with regard to the sets S and T .

Case 1 : the two sets contain 4 points and 1 point. Then, there are(

5

1

)

= 5 ways to partition the pointsin this manner. Moreover, the 1 point (in its own set) must be connected to each of the other 4 points,and these are the only possible segments. Therefore, there is only 1 possible set of segments, which,combining with the 5 ways of choosing the sets, gives 5 possible sets of segments.

Case 2 : the two sets contain 3 points and 2 points. Then, there are(

5

2

)

= 10 ways to partition thepoints in this manner. Let S be the set containing 3 points and T the set containing 2 points. Weconsider the possible degrees of the points in T .

• If both points have degree 3, then each point must connect to all points in S, and the five pointsare connected via segments. So the number of possible sets of segments is 1.

• If the points have degree 3 and 2. Then, we can swap the points in 2 ways, and, for the pointwith degree 2, we can choose the elements of S it connects to in

(

3

2

)

= 3 ways. In each case, thefive points are guaranteed to be connected via segments. Hence 6 ways.

• If the points have degree 3 and 1. Similarly, we can swap the points in 2 ways and connect thepoint with degree 1 to the elements of S in

(

3

1

)

= 3 ways. Since all five points are connected inall cases, we have 6 ways.

• If both points have degree 2. Then, in order for the five points to be connected, the two pointsmust connect to a common element of S. Call this common element A. Then, for the other twoelements of S, each one must be connected to exactly one element of T . We can choose A in 3ways, and swap the correspondence between the other two elements of S with the elements of Tin 2 ways. Hence 6 ways.

• If the points have degree 2 and 1. Then, in order to cover S, the point with degree 2 must connectto 2 points in S, and the point with degree 1 to the remaining point in S. But then, the fivepoints will not be connected via segments, an impossibility.

• If both points have degree 1. Then, similar to the previous case, it is impossible to cover all the3 points in S with only 2 segments, a contradiction.

Combining the subcases, we have 1+6+6+6 = 19 possible sets of segments with regard to a partition.With 10 possible partitions, we have a total of 19 · 10 = 190 possible sets of segments.

Finally, combining this number with the 5 possibilities from case 1, we have a total of 5 + 190 = 195possibilities, as desired.

6. [6] Find the number of strictly increasing sequences of nonnegative integers with the following prop-erties:

• The first term is 0 and the last term is 12. In particular, the sequence has at least two terms.

• Among any two consecutive terms, exactly one of them is even.

Answer: 144 For a natural number n, let An be a set containing all sequences which satisfy theproblem conditions but which 12 is replaced by n. Also, let an be the size of An.

Team Round

We first consider a1 and a2. We get a1 = 1, as the only sequence satisfying the problem conditions is0, 1. We also get a2 = 1, as the only possible sequence is 0, 1, 2.

Next, we show that an+2 = an+1 + an for all natural number n. We consider the second-to-last termsof each sequence in An+2.

Case 1. The second-to-last term is n+1. When we leave out the last term, the remaining sequence willstill satisfy the problem conditions, and hence is in An+1. Conversely, for a sequence in An+1, we couldadd n+ 2 at the end of that sequence, and since n+ 1 and n+ 2 have different parities, the resultingsequence will be in An+2. Therefore, there is a one-to-one correspondence between the sequences inthis case and the sequences in An+1. So the number of sequences in this case is an+1.

Case 2. The second-to-last term is less than or equal n. But n and n + 2 have the same parity, sothe second-to-last term cannot exceed n − 1. When we substitute the last term (n + 2) with n, theresulting sequence will satisfy the problem conditions and will be in An. Conversely, for a sequence inAn, we could substitute its last term n, with n+2. As n and n+2 have the same parity, the resultingsequence will be in An. Hence, in this case, the number of sequences is an.

Now, since an+2 = an+1 + an for all natural numbers n, we can recursively compute that the numberof all possible sequences having their last terms as 12 is a12 = 144. Note that the resulting sequence(an) is none other than the Fibonacci numbers.

7. [7] Sammy has a wooden board, shaped as a rectangle with length 22014 and height 32014. The boardis divided into a grid of unit squares. A termite starts at either the left or bottom edge of the rectangle,and walks along the gridlines by moving either to the right or upwards, until it reaches an edge oppositethe one from which the termite started. Depicted below are two possible paths of the termite.

The termite’s path dissects the board into two parts. Sammy is surprised to find that he can stillarrange the pieces to form a new rectangle not congruent to the original rectangle. This rectangle hasperimeter P . How many possible values of P are there?

Answer: 4 Let R be the original rectangle and R′ the new rectangle which is different from R.We see that the perimeter of R′ depends on the possibilities for the side lengths of R′.

We will prove that the dividing line must have the following characterization: starting from the lowerleft corner of R, walk to the right by distance a, then walk up by distance b, for some positive numbera and b, and repeat the two steps until one reaches the upper right corner of R, with the condition thatthe last step is a walk to the right. (The directions stated here depends on the orientation of R, butwe can always orient R so as to fit the description.) Let there be n+ 1 walks to the right and n walksto the top, then we have that this division would rearrange a rectangle of dimension (n+1)a×nb intoa rectangle of dimension na× (n+ 1)b.

Let us first assume the above. Now, according to the problem, it suffices to find n, a, b such that(n+ 1)a = 22014, nb = 32014 or (n+ 1)a = 32014, nb = 22014. This means that n+ 1 and n are a powerof 3 and a power of 2, whose exponents do not exceed 2014. This corresponds to finding nonnegativeintegers k, l ≤ 2014 such that |2k − 3l| = 1. The only possible pairs of (2k, 3l) are (2, 1), (2, 3), (3, 4)and (8, 9). So there are 4 possible configurations of R′.

Now, we prove our claim. For completeness, we will actually prove the claim more generally for anycut, not just ones that move right and up (hence the length of the solution which follows, but only theabove two paragraphs are relevant for the purposes of finding the answer).

Team Round

First we show that the dividing boundary between the two pieces must meet the boundary of R attwo points, each being on opposite sides of R as the other. To see why, consider that otherwise, therewould be two consecutive sides of R which belong to the same piece. Then, the smallest rectanglecontaining such a configuration must have each side being as large as each of the two sides, and thusit is R. Since this piece is also part of R′, R′ must contain R, but their areas are equal, so R′ = R, acontradiction.

Now, let the dividing boundary go from the top side to the bottom side of R, and call the right piece”piece 1” and the left piece ”piece 2.” We orient R′ in such a way that piece 1 is fixed and piece 2 ismoved from the original position in some way to create R′. We will show that piece 2 must be movedby translation by some vector v, tv. Otherwise, piece 2 is affected by tv as a well as a rotation by 90◦

or 180◦. We show that these cases are impossible.

First, consider the case where there is a 90◦ rotation. Let the distance from the top side to the bottomside of R be x. Then, the two pieces are contained between a pair of horizontal lines which are ofdistance x apart from one another. If piece 2 is rotated by 90◦, then these horizontal lines become apair of vertical lines which are of distance x apart from one another. So R′ is contained within a unionof regions between a pair of horizontal lines and a pair of vertical lines.

Now, we show that R′ must be contained within only one of these regions. Consider if there existspoints (x1, y1) and (x2, y2) in R′ such that (x1, y1) is not in the horizontal region (so that y1 is outof range) and (x2, y2) is not in the vertical region (so that x2 is out of range). Then, it follows that(x2, y1) is also in the rectangle R′. But (x2, y1) cannot be contained in either region, since both of itsx and y coordinates are out of range, a contradiction.

So let us assume, without loss of generality, that R′ is contained in the vertical region (the one whichcontains piece 2). Then, the horizontal side of R′ cannot have length greater than x, the width of theregion. However, piece 2 is contained in the region, and its width is exactly x. Therefore, the width ofR′ must be exactly x, rendering it to be the same shape as R, a contradiction.

Next, we show that the case with a 180◦ rotation is also impossible. We modify our considerationsfrom the previous case by considering a half-region of the region between a pair of horizontal lines(which are still are of distance x apart), which we define as a part of the region on the right or onthe left of a certain vertical line. Then, piece 2 is contained within a certain half-region going to theright and piece 1 is contained within a certain half-region going to the left. Now, in R′, since piece 2 isrotated by 180◦, we would have both half-regions going to the left, and R′ is contained within a unionof them.

Now, consider the ”end” of each half-region (the part of the boundary that is vertical). The ends ofboth half-regions must be contained in R′, since they are part of piece 1 and piece 2. Consider a vectorthat maps the end of one half-region to the other. If the vector is horizontal, then the union of theregions have vertical distance x. Similarly to the previous case, we deduce that the vertical side of Rmust be of length no more than x, and so must be exactly x, but then R′ = R, a contradiction.

Now, if the vector has both nonzero horizontal and vertical components, then the parallelogram gen-erated by the locus of the end of a half-region being translated by the vector to the end of the otherhalf-region must be contained within R′ (since R′ is convex). However, the parallelogram is not con-tained within the union of the two half-regions, a contradiction.

Finally, if the vector is vertical, then the two half-regions must be on top of one another, and so willhave no region in common. Then, since R′ is a rectangle, the intersection of R′ with each half-regionwill also be a rectangle. So pieces 1 and 2 must be rectangles. But then a rotation of 180◦ would mappiece 2 to itself. So this reduces to a case of pure translation.

We now consider the translation tv by a vector v on piece 2. Since R′ must contain the ends of thehalf-regions (which retain their original orientations), the vertical side of R′ must be at least of lengthx. But R′ 6= R, so the vertical side of R′ has length strictly greater than x. This implies that thehorizontal side of R′ must be strictly shorter than that of R, since they have equal area. However,the horizontal side of R′ is at least as long as the horizontal distance between the ends of the twohalf-regions, so the ends of the two half-regions must have moved closer to one another horizontally.

Team Round

This implies that the vector v has a positive x component. Also, v cannot be entirely horizontal,because there is no more space for piece 2 to move into. So v has a nonzero y component. Withoutloss of generality, let us assume that v has a positive y component.

Before we continue further, let us label the vertices of R as A,B,C,D, going in counter-clockwisedirection, with the left side of R being AB and the right side of R being CD. So AB is in piece 2and CD is in piece 1. Call the translated piece 2 that is part of the rectangle R′ piece 2′, with thecorresponding points A′ and B′.

Now, consider the half-regions of piece 1 and piece 2. They are half-regions with the end AB goingto the right and the one with the end CD going to the left. So, in R′, the half-regions are with endA′B′ going to the right and with end CD going to the left, and R′ is contained within the union ofthese two. Now, there cannot be a point in R′ that is to the left of A′B′, since the smallest rectanglecontaining that point and A′ would not be contained in the union of the two half-regions. Similarly,there cannot be a point in R′ that is to the right of CD for the same reason. These restrictions implythat R′ must be contained within the union of the two half-regions that lie horizontally between A′B′

and CD. However, since the smallest rectangle containing A′ and C is precisely this region, R′ mustbe this region. Let A′B′ intersect BC at L and let CD intersect the line passing through A′ which isparallel to AD at M . We have R′ = A′LCM .

Now, consider the segments BL and LB′. We know that BL is a boundary of piece 2. Also, since LB′

is a boundary of R′ and it is below piece 2′, it cannot be a boundary of piece 2′. Therefore, it must bea boundary of piece 1. Since LB′ is not a boundary of R but is a boundary of piece 1, it must be partof the dividing boundary between piece 1 and 2, and so must also be a boundary of piece 2.

We now prove that: from a sequence of B,B′, B′′, ..., each being translated from the preceding one byv, one of them must eventually lie on AD. Also, if L,L′, L′′, ... is also a sequence of L being successivelytranslated by v, then, using Bi and Li to designate the ith term of each sequence: BiLi and LiBi+1

must be part of the boundary of piece 2 for all i ≤ N − 1, where BN is the last point, the one that lieson AD.

We have already proved the assertion for i = 1. We now set out to prove, by induction, that BiLi andLiBi+1 must be part of the boundary of piece 2 for all i such that Bi+1 is still within R or on the edgeof R.

Consider, by induction hypothesis, that Bi−1Li−1 and Li−1Bi are parts of the boundary of piece 2.Then, by mapping, BiLi and LiBi+1 must be parts of the boundary of piece 2′. Since the dividingboundary of R go from top to bottom, LiBi+1 cannot be on the rightmost edge of R, which is alsothe rightmost edge of R′. This means that LiBi+1 is not a boundary of R′. So we have that BiLi andLiBi+1 are not boundaries of R′ but are boundaries of piece 2′, so they must be boundaries of piece 1.Now, since they are not boundaries of R either but are boundaries of piece 1, they must be boundariesof piece 2, as desired.

Now we are left to show that one of Bi must lie exactly on AD. Let Bi be the last term of the sequencethat is contained within R or its boundary. Then, by the previous result, Bi−1Li−1 and Li−1Bi areboundaries of piece 2. Then, by mapping, BiLi is a boundary of piece 2′. Since BiLi is on or belowAD, it cannot be on the boundary of R′, but since it is a boundary of piece 2′, it must also a boundaryof piece 1. Now, if Bi is not on AD, then BiLi will not be a boundary of R, but since it is a boundaryof piece 1, it must also be a boundary of piece 2. By mapping, this implies that Bi+1Li+1 must be aboundary of piece 2′. However, Bi+1Li+1 is not contained within R or its boundary, and so cannot beon piece 1’s boundary. Therefore, since Bi+1Li+1 is a boundary of piece 2′ but not of piece 1, it mustbe a boundary of R′. This means that Bi+1Li+1 is on the upper edge of R′. Mapping back, we getthat BiLi must be on the upper edge of R, a contradiction to the assumption that Bi is not on AD(the upper edge of R). So Bi is on AD, as desired.

Let BN be the term of the sequence that is on AD. We have now shown that

L1B2, B2L2, . . . , BN−1LN−1, LN−1BN

completely defines the dividing line between piece 1 and piece 2. Moreover, B1 = B, defining thestarting point of the dividing line. We now add the final description: LN = D. To see why, note that

Team Round

LN must be on the upper edge of R, as it is in the same horizontal level as BN . However, since LN−1

is one of furthest points to the right of piece 2, by mapping, LN must be one of the furthest points tothe right of piece 2′, and so must be on the rightmost edge of piece 2′, which is the rightmost edge ofR′ and of R. Therefore, LN is on the upper edge and the rightmost edge of R, and so it must be D,as desired.

8. [3] Let H be a regular hexagon with side length one. Peter picks a point P uniformly and at randomwithin H, then draws the largest circle with center P that is contained in H. What is this probabilitythat the radius of this circle is less than 1

2?

Answer: 2√3−1

3We first cut the regular hexagon H by segments connecting its center to each

vertex into six different equilateral triangles with side lengths 1. Therefore, each point inside H iscontained in some equilateral triangle. We first see that for each point inside an equilateral triangle,the radius of the largest circle with center P which is contained in H equals the shortest distance fromP to the nearest side of the hexagon, which is also a side of the triangle in which it is contained.

Consider that the height of each triangle is√3

2. Therefore, the region inside the triangle containing all

points with distance more than 1

2to the side of the hexagon is an equilateral triangle with a height of

√3−1

2. Consequently, the area inside the triangle containing all points with distance less than 1

2to the

side of the hexagon has area√3

4

(

1−(√

3−1√3

)2)

=√3

4·(

2√3−1

3

)

. This is of the ratio 2√3−1

3to the

area of the triangle, which is√3

4. Since all triangles are identical and the point P is picked uniformly

within H, the probability that the radius of the largest circle with center P which is contained in H is

less than 1

2is 2

√3−1

3, as desired.

9. [5] How many lines pass through exactly two points in the following hexagonal grid?

Answer: 60 First solution. From a total of 19 points, there are(

19

2

)

= 171 ways to choose twopoints. We consider lines that pass through more than 2 points.

• There are 6 + 6 + 3 = 15 lines that pass through exactly three points. These are: the six sidesof the largest hexagon, three lines through the center (perpendicular to the sides of the largesthexagon), and the other six lines perpendiculars to the sides of the largest hexagon.

• There are 6 lines that pass through exactly four points. (They are parallel to the sides of thelargest hexagon.)

• There are 3 lines that pass through exactly five points. (They all pass through the center.)

For each n = 3, 4, 5, a line that passes through n points will be counted(

n

2

)

times, and so the corre-sponding amount will have to be subtracted. Hence the answer is

171−

(

3

2

)

· 15−

(

4

2

)

· 6−

(

5

2

)

· 3 = 171− 45− 36− 30 = 60.

Second solution. We divide the points into 4 groups as follows.

• Group 1 consists of the center point.

• Group 2 consists of the 6 points surrounding the center.

• Group 3 consists of the 6 vertices of the largest hexagon.

Team Round

• Group 4 consists of the 6 midpoints of the sides of the largest hexagon.

We wish to count the number of lines that pass through exactly 2 points. Consider: all lines connectingpoints in group 1 and 2, 1 and 3, and 1 and 4 pass through more than 2 points. So it is sufficient torestrict our attention to group 2, 3 and 4.

• For lines connecting group 2 and 2, the only possibilities are those that the two endpoints are 120degrees apart with respect to the center, so 6 possibilities.

• For lines connecting group 3 and 3, it is impossible.

• For lines connecting group 4 and 4, the two endpoints must be 60 degrees apart with respect tothe center, so 6 possibilities.

• For lines connecting group 3 and 2. For each point in group 3, the only possible points in group2 are those that are 120 degrees apart from the point in group 3. So 2 · 6 = 12 possibilities.

• For lines connecting group 4 and 2, the endpoints must be 150 degrees apart with respect to thecenter, so 2 · 6 = 12 possibilities.

• For lines connecting group 4 and 3. For each point in group 4, any point in group 3 works exceptthose that are on the side on the largest hexagon of which the point in group 4 is the midpoint.Hence 4 · 6 = 24 possibilities.

Therefore, the number of lines passing through 2 points is 6 + 6 + 12 + 12 + 24 = 60, as desired.

10. [8] Let ABCDEF be a convex hexagon with the following properties.

(a) AC and AE trisect ∠BAF .

(b) BE ‖ CD and CF ‖ DE.

(c) AB = 2AC = 4AE = 8AF .

Suppose that quadrilaterals ACDE and ADEF have area 2014 and 1400, respectively. Find the areaof quadrilateral ABCD.

Answer: 7295 From conditions (a) and (c), we know that triangles AFE, AEC and ACBare similar to one another, each being twice as large as the preceding one in each dimension. LetAE ∩ FC = P and AC ∩ EB = Q. Then, since the quadrilaterals AFEC and AECB are similar toone another, we have AP : PE = AQ : QC. Therefore, PQ ‖ EC.

Let PC∩QE = T . We know by condition (b) that BE ‖ CD and CF ‖ DE. Therefore, triangles PQTand ECD have their three sides parallel to one another, and so must be similar. From this we deducethat the three lines joining the corresponding vertices of the two triangles must meet at a point, i.e.,that PE, TD,QC are concurrent. Since PE and QC intersect at A, the points A, T,D are collinear.Now, because TCDE is a parallelogram, TD bisects EC. Therefore, since A, T,D are collinear, ADalso bisects EC. So the triangles ADE and ACD have equal area.

Now, since the area of quadrilateral ACDE is 2014, the area of triangle ADE is 2014/2 = 1007.And since the area of quadrilateral ADEF is 1400, the area of triangle AFE is 1400 − 1007 = 393.Therefore, the area of quadrilateral ABCD is 16 · 393 + 1007 = 7295, as desired.

Team Round

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HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND

Organization Team Team ID#

1. [5] Solve for x in the equation 20 · 14 + x = 20 + 14 · x.

2. [5] Find the area of a triangle with side lengths 14, 48, and 50.

3. [5] Victoria wants to order at least 550 donuts from Dunkin’ Donuts for the HMMT 2014 Novembercontest. However, donuts only come in multiples of twelve. Assuming every twelve donuts cost $7.49,what is the minimum amount Victoria needs to pay, in dollars? (Because HMMT is affiliated withMIT, the purchase is tax exempt. Moreover, because of the size of the order, there is no delivery fee.)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND

Organization Team Team ID#

4. [6] How many two-digit prime numbers have the property that both digits are also primes?

5. [6] Suppose that x, y, z are real numbers such that

x = y + z + 2, y = z + x+ 1, and z = x+ y + 4.

Compute x+ y + z.

6. [6] In the octagon COMPUTER exhibited below, all interior angles are either 90◦ or 270◦ and wehave CO = OM = MP = PU = UT = TE = 1.

C

O M

P U

T E

R

D

Point D (not to scale in the diagram) is selected on segment RE so that polygons COMPUTED andCDR have the same area. Find DR.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND

Organization Team Team ID#

7. [7] Let ABCD be a quadrilateral inscribed in a circle with diameter AD. If AB = 5, AC = 6, andBD = 7, find CD.

8. [7] Find the number of digits in the decimal representation of 241.

9. [7] Let f be a function from the nonnegative integers to the positive reals such that f(x+y) = f(x)·f(y)holds for all nonnegative integers x and y. If f(19) = 524288k, find f(4) in terms of k.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND

Organization Team Team ID#

10. [8] Let ABC be a triangle with CA = CB = 5 and AB = 8. A circle ω is drawn such that the interiorof triangle ABC is completely contained in the interior of ω. Find the smallest possible area of ω.

11. [8] How many integers n in the set {4, 9, 14, 19, . . . , 2014} have the property that the sum of thedecimal digits of n is even?

12. [8] Sindy writes down the positive integers less than 200 in increasing order, but skips the multiplesof 10. She then alternately places + and − signs before each of the integers, yielding an expression+1− 2+ 3− 4+ 5− 6+ 7− 8+ 9− 11+ 12− · · · − 199. What is the value of the resulting expression?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND

Organization Team Team ID#

13. [9] Let ABC be a triangle with AB = AC = 25

14BC. Let M denote the midpoint of BC and let X

and Y denote the projections of M onto AB and AC, respectively. If the areas of triangle ABC andquadrilateral AXMY are both positive integers, find the minimum possible sum of these areas.

14. [9] How many ways can the eight vertices of a three-dimensional cube be colored red and blue suchthat no two points connected by an edge are both red? Rotations and reflections of a given coloringare considered distinct.

15. [9] Carl is on a vertex of a regular pentagon. Every minute, he randomly selects an adjacent vertex(each with probability 1

2) and walks along the edge to it. What is the probability that after 10 minutes,

he ends up where he had started?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND

Organization Team Team ID#

16. [10] A particular coin has a 1

3chance of landing on heads (H), 1

3chance of landing on tails (T), and

1

3chance of landing vertically in the middle (M). When continuously flipping this coin, what is the

probability of observing the continuous sequence HMMT before HMT?

17. [10] Let ABC be a triangle with AB = AC = 5 and BC = 6. Denote by ω the circumcircle of ABC.We draw a circle Ω which is externally tangent to ω as well as to the lines AB and AC (such a circleis called an A-mixtilinear excircle). Find the radius of Ω.

18. [10] For any positive integer x, define Accident(x) to be the set of ordered pairs (s, t) with s ∈{0, 2, 4, 5, 7, 9, 11} and t ∈ {1, 3, 6, 8, 10} such that x + s − t is divisible by 12. For any nonnegativeinteger i, let ai denote the number of x ∈ {0, 1, . . . , 11} for which |Accident(x)| = i. Find

a20 + a21 + a22 + a23 + a24 + a25.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND

Organization Team Team ID#

19. [11] Let a sequence {an}∞n=0 be defined by a0 =√2, a1 = 2, and an+1 = ana

2n−1 for n ≥ 1. The

sequence of remainders when a0, a1, a2, · · · are divided by 2014 is eventually periodic with some minimalperiod p (meaning that am = am+p for all sufficiently large integers m, and p is the smallest suchpositive integer). Find p.

20. [11] Determine the number of sequences of sets S1, S2, . . . , S999 such that

S1 ⊆ S2 ⊆ · · · ⊆ S999 ⊆ {1, 2, . . . , 999}.

Here A ⊆ B means that all elements of A are also elements of B.

21. [11] If you flip a fair coin 1000 times, what is the expected value of the product of the number ofheads and the number of tails?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND

Organization Team Team ID#

22. [12] Evaluate the infinite sum∞∑

n=2

log2

(

1− 1

n

1− 1

n+1

)

.

23. [12] Seven little children sit in a circle. The teacher distributes pieces of candy to the children in sucha way that the following conditions hold.

• Every little child gets at least one piece of candy.

• No two little children have the same number of pieces of candy.

• The numbers of candy pieces given to any two adjacent little children have a common factor otherthan 1.

• There is no prime dividing every little child’s number of candy pieces.

What is the smallest number of pieces of candy that the teacher must have ready for the little children?

24. [12] Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. We construct isosceles righttriangle ACD with ∠ADC = 90◦, where D,B are on the same side of line AC, and let lines AD andCB meet at F . Similarly, we construct isosceles right triangle BCE with ∠BEC = 90◦, where E,Aare on the same side of line BC, and let lines BE and CA meet at G. Find cos∠AGF .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND

Organization Team Team ID#

25. [13] What is the smallest positive integer n which cannot be written in any of the following forms?

• n = 1 + 2 + · · ·+ k for a positive integer k.

• n = pk for a prime number p and integer k.

• n = p+ 1 for a prime number p.

• n = pq for some distinct prime numbers p and q

26. [13] Consider a permutation (a1, a2, a3, a4, a5) of {1, 2, 3, 4, 5}. We say the tuple (a1, a2, a3, a4, a5) isflawless if for all 1 ≤ i < j < k ≤ 5, the sequence (ai, aj , ak) is not an arithmetic progression (in thatorder). Find the number of flawless 5-tuples.

27. [13] In triangle ABC, let the parabola with focus A and directrix BC intersect sides AB and AC atA1 and A2, respectively. Similarly, let the parabola with focus B and directrix CA intersect sides BCand BA at B1 and B2, respectively. Finally, let the parabola with focus C and directrix AB intersectsides CA and CB at C1 and C2, respectively.

If triangle ABC has sides of length 5, 12, and 13, find the area of the triangle determined by linesA1C2, B1A2 and C1B2.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND

Organization Team Team ID#

28. [15] Let x be a complex number such that x+ x−1 is a root of the polynomial p(t) = t3 + t2 − 2t− 1.Find all possible values of x7 + x−7.

29. [15] Let ω be a fixed circle with radius 1, and let BC be a fixed chord of ω such that BC = 1. Thelocus of the incenter of ABC as A varies along the circumference of ω bounds a region R in the plane.Find the area of R.

30. [15] Suppose we keep rolling a fair 2014-sided die (whose faces are labelled 1, 2, . . . , 2014) until weobtain a value less than or equal to the previous roll. Let E be the expected number of times we rollthe die. Find the nearest integer to 100E.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND

Organization Team Team ID#

31. [17] Flat Albert and his buddy Mike are watching the game on Sunday afternoon. Albert is drinkinglemonade from a two-dimensional cup which is an isosceles triangle whose height and base measure 9cmand 6cm; the opening of the cup corresponds to the base, which points upwards. Every minute afterthe game begins, the following takes place: if n minutes have elapsed, Albert stirs his drink vigorouslyand takes a sip of height 1

n2 cm. Shortly afterwards, while Albert is busy watching the game, Mike addscranberry juice to the cup until it’s once again full in an attempt to create Mike’s cranberry lemonade.Albert takes sips precisely every minute, and his first sip is exactly one minute after the game begins.

After an infinite amount of time, let A denote the amount of cranberry juice that has been poured (insquare centimeters). Find the integer nearest 27

π2A.

32. [17] Let f(x) = x2 − 2, and let fn denote the function f applied n times. Compute the remainderwhen f24(18) is divided by 89.

33. [17] How many ways can you remove one tile from a 2014 × 2014 grid such that the resulting figurecan be tiled by 1× 3 and 3× 1 rectangles?

Warning: The next set of three problems will consist of estimation problems.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND

Organization Team Team ID#

34. [20] Let M denote the number of positive integers which divide 2014!, and let N be the integer closestto ln(M). Estimate the value of N . If your answer is a positive integer A, your score on this problemwill be the larger of 0 and

⌊

20− 1

8|A−N |

⌋

. Otherwise, your score will be zero.

35. [20] Ten points are equally spaced on a circle. A graph is a set of segments (possibly empty) drawnbetween pairs of points, so that every two points are joined by either zero or one segments. Two graphsare considered the same if we can obtain one from the other by rearranging the points.

Let N denote the number of graphs with the property that for any two points, there exists a path fromone to the other among the segments of the graph. Estimate the value of N . If your answer is a positiveinteger A, your score on this problem will be the larger of 0 and ⌊20− 5 |ln(A/N)|⌋. Otherwise, yourscore will be zero.

36. [20] Pick a subset of at least four of the following geometric theorems, order them from earliest tolatest by publication date, and write down their labels (a single capital letter) in that order. If atheorem was discovered multiple times, use the publication date corresponding to the geometer forwhich the theorem is named.

C. (Ceva) Three cevians AD, BE, CF of a triangle ABC are concurrent if and only if BDDC

CEEA

AFFB

= 1.

E. (Euler) In a triangle ABC with incenter I and circumcenter O, we have IO2 = R(R−2r), wherer is the inradius and R is the circumradius of ABC.

H. (Heron) The area of a triangle ABC is√

s(s− a)(s− b)(s− c), where s = 1

2(a+ b+ c).

M. (Menelaus) IfD, E, F lie on lines BC, CA, AB, then they are collinear if and only if BDDC

CEEA

AFFB

=−1, where the ratios are directed.

P. (Pascal) Intersections of opposite sides of cyclic hexagons are collinear.

S. (Stewart) Let ABC be a triangle and D a point on BC. Set m = BD, n = CD, d = AD. Thenman+ dad = bmb+ cnc.

V. (Varignon) The midpoints of the sides of any quadrilateral are the vertices of a parallelogram.

If your answer is a list of 4 ≤ N ≤ 7 labels in a correct order, your score will be (N − 2)(N − 3).Otherwise, your score will be zero.

HMMT November 2014Saturday 15 November 2014

Guts Round

1. [5] Solve for x in the equation 20 · 14 + x = 20 + 14 · x.Answer: 20 By inspection, 20 + 14 · 20 = 20 · 14 + 20. Alternatively, one can simply computex = 20·14−20

14−1 = 20.

2. [5] Find the area of a triangle with side lengths 14, 48, and 50.

Answer: 336 Note that this is a multiple of the 7-24-25 right triangle. The area is therefore

14(48)

2= 336.

3. [5] Victoria wants to order at least 550 donuts from Dunkin’ Donuts for the HMMT 2014 Novembercontest. However, donuts only come in multiples of twelve. Assuming every twelve donuts cost $7.49,what is the minimum amount Victoria needs to pay, in dollars? (Because HMMT is affiliated withMIT, the purchase is tax exempt. Moreover, because of the size of the order, there is no delivery fee.)

Answer: 344.54 The smallest multiple of 12 larger than 550 is 552 = 12 · 46. So the answer is46 · $7.49. To make the multiplication easier, we can write this as 46 · ($7.5− $0.01) = $345− $0.46 =$344.54.

Note: this is the actual cost of donuts at the 2014 HMMT November contest.

4. [6] How many two-digit prime numbers have the property that both digits are also primes?

Answer: 4 When considering the 16 two-digit numbers with 2, 3, 5, and 7 as digits, we find thatonly 23, 37, 53, and 73 have this property.

5. [6] Suppose that x, y, z are real numbers such that

x = y + z + 2, y = z + x+ 1, and z = x+ y + 4.

Compute x+ y + z.

Answer: −7 Adding all three equations gives

x+ y + z = 2(x+ y + z) + 7,

from which we find that x+ y + z = −7.

6. [6] In the octagon COMPUTER exhibited below, all interior angles are either 90◦ or 270◦ and wehave CO = OM = MP = PU = UT = TE = 1.

C

O M

P U

T E

R

D

Point D (not to scale in the diagram) is selected on segment RE so that polygons COMPUTED andCDR have the same area. Find DR.

Answer: 2 The area of the octagon COMPUTER is equal to 6. So, the area of CDR must be3. So, we have the equation 1

2 ∗ CD ∗DR = [CDR] = 3. And from CD = 3, we have DR = 2.

Guts Round

7. [7] Let ABCD be a quadrilateral inscribed in a circle with diameter AD. If AB = 5, AC = 6, andBD = 7, find CD.

Answer:√38 We have AD2 = AB2+BD2 = AC2+CD2, so CD =

√AB2 +BD2 −AC2 =

√38.

8. [7] Find the number of digits in the decimal representation of 241.

Answer: 13 Noticing that 210 = 1024 ≈ 1000 allows for a good estimate. Alternatively, thenumber of decimal digits of n is given by ⌊log10(n)⌋ + 1. Using log10(2) ≈ 0.31 also gives the correctanswer. The exact value of 241 is 2199023255552.

9. [7] Let f be a function from the nonnegative integers to the positive reals such that f(x+y) = f(x)·f(y)holds for all nonnegative integers x and y. If f(19) = 524288k, find f(4) in terms of k.

Answer: 16k4/19 The given condition implies f(mn) = f(m)n, so

f(4)19 = f(4 · 19) = f(19 · 4) = f(19)4

and it follows that f(4) = 16k4/19.

10. [8] Let ABC be a triangle with CA = CB = 5 and AB = 8. A circle ω is drawn such that the interiorof triangle ABC is completely contained in the interior of ω. Find the smallest possible area of ω.

Answer: 16π We need to contain the interior of AB, so the diameter is at least 8. This bound is

sharp because the circle with diameter AB contains all of ABC. Hence the minimal area is 16π.

11. [8] How many integers n in the set {4, 9, 14, 19, . . . , 2014} have the property that the sum of thedecimal digits of n is even?

Answer: 201

We know that 2014 does not qualify the property. So, we’ll consider {4, 9, 14, ..., 2009} instead. Now,we partition this set into 2 sets: {4, 14, 24, ..., 2004} and {9, 19, 29, ..., 2009}.For each so the first and second set are basically x4 and x9, where x = 0, 1, 2, ..., 200, respectively. Andwe know that for each value of x, x must be either even or odd, which makes exactly one of {x4, x9}has even sum of decimal digits. Therefore, there are in total of 201 such numbers.

12. [8] Sindy writes down the positive integers less than 200 in increasing order, but skips the multiplesof 10. She then alternately places + and − signs before each of the integers, yielding an expression+1− 2+ 3− 4+ 5− 6+ 7− 8+ 9− 11+ 12− · · · − 199. What is the value of the resulting expression?

Answer: −100 Group the numbers into (1− 2+ 3− 4+ ...+18− 19) + (21− 22+ ...+38− 39) +... + (181 − 182 + ... + 198 − 199). We can easily show that each group is equal to −10, and so theanswer is −100.

13. [9] Let ABC be a triangle with AB = AC = 2514BC. Let M denote the midpoint of BC and let X

and Y denote the projections of M onto AB and AC, respectively. If the areas of triangle ABC andquadrilateral AXMY are both positive integers, find the minimum possible sum of these areas.

Answer: 1201 By similar triangles, one can show that [AXMY ] = 2 · [AMX] = ( 2425 )2 ·2[ABM ] =

( 2425 )2 · [ABC]. Thus the answer is 252 + 242 = 1201.

14. [9] How many ways can the eight vertices of a three-dimensional cube be colored red and blue suchthat no two points connected by an edge are both red? Rotations and reflections of a given coloringare considered distinct.

Answer: 35 We do casework on R, the number of red vertices. Let the cube be calledABCDEFGH, with opposite faces ABCD and EFGH, such that A is directly above E.

• R = 0: There is one such coloring, which has only blue vertices.

Guts Round

• R = 1: There are 8 ways to choose the red vertex, and all other vertices must be blue. There are8 colorings in this case.

• R = 2: Any pair not an edge works, so the answer is(

82

)

− 12 = 16.

• R = 3: Each face ABCD and EFGH has at most two red spots. Assume WLOG ABCD hasexactly two and EFGH has exactly one (multiply by 2 at the end). There are two ways to pickthose in ABCD (two opposite corners), and two ways after that to pick EFGH. Hence the grandtotal for this subcase is 2 · 2 · 2 = 8.

• R = 4: There are only two ways to do this.

Hence, the sum is 35.

15. [9] Carl is on a vertex of a regular pentagon. Every minute, he randomly selects an adjacent vertex(each with probability 1

2 ) and walks along the edge to it. What is the probability that after 10 minutes,he ends up where he had started?

Answer: 127512 Let A denote a clockwise move and B denote a counterclockwise move. We want to

have some combination of 10 A’s and B’s, with the number of A’s and the number of B’s differing bya multiple of 5. We have

(

100

)

+(

105

)

+(

1010

)

= 254. Hence the answer is 254210 = 127

512 .

16. [10] A particular coin has a 13 chance of landing on heads (H), 1

3 chance of landing on tails (T), and13 chance of landing vertically in the middle (M). When continuously flipping this coin, what is theprobability of observing the continuous sequence HMMT before HMT?

Answer: 14 For a string of coin flips S, let PS denote the probability of flipping HMMT before

HMT if S is the starting sequence of flips. We know that the desired probability, p, is 13PH+ 1

3PM+ 13PT .

Now, using conditional probability, we find that

PH =1

3PHH +

1

3PHM +

1

3PHT

=1

3PH +

1

3PHM +

1

3PT .

We similarly find that

PM = PT =1

3PH +

1

3PM +

1

3PT

PHM =1

3PHMM +

1

3PH

PHMM =1

3+

1

3PM +

1

3PH .

Solving gives PH = PM = PT = 14 . Thus, p = 1

4 .

17. [10] Let ABC be a triangle with AB = AC = 5 and BC = 6. Denote by ω the circumcircle of ABC.We draw a circle Ω which is externally tangent to ω as well as to the lines AB and AC (such a circleis called an A-mixtilinear excircle). Find the radius of Ω.

Answer: 758 Let M be the midpoint of BC. Let D be the point diametrically opposite A on the

circumcircle, and let the A-mixtilinear excircle be tangent to lines AB and AC at X and Y . Let O bethe center of the A-mixtilinear excircle.

Notice that △AOX ∼ △ABM . If we let x be the desired radius, we have

x+AD

x=

5

3.

We can compute AD5 = 5

4 since △ADB ∼ △ABM , we derive AD = 254 . From here it follows that

x = 758 .

Guts Round

18. [10] For any positive integer x, define Accident(x) to be the set of ordered pairs (s, t) with s ∈{0, 2, 4, 5, 7, 9, 11} and t ∈ {1, 3, 6, 8, 10} such that x + s − t is divisible by 12. For any nonnegativeinteger i, let ai denote the number of x ∈ {0, 1, . . . , 11} for which |Accident(x)| = i. Find

a20 + a21 + a22 + a23 + a24 + a25.

Answer: 26 Modulo twelve, the first set turns out to be {−1 · 7, 0 · 7, . . . , 5 · 7} and the second setturns out to be be {6 · 7, . . . , 10 · 7}. We can eliminate the factor of 7 and shift to reduce the problemto s ∈ {0, 1, . . . , 6} and t ∈ {7, . . . , 11}. With this we can easily compute (a0, a1, a2, a3, a4, a5) =(1, 2, 2, 2, 2, 3). Therefore, the answer is 26.

It turns out that the choice of the sets correspond to the twelve keys of a piano in an octave. Thefirst set gives the locations of the white keys, while the second set gives the location of the black keys.Steps of seven then correspond to perfect fifths. See, for example:

• http://en.wikipedia.org/wiki/Music_and_mathematics

• http://en.wikipedia.org/wiki/Guerino_Mazzola

19. [11] Let a sequence {an}∞n=0 be defined by a0 =√2, a1 = 2, and an+1 = ana

2n−1 for n ≥ 1. The

sequence of remainders when a0, a1, a2, · · · are divided by 2014 is eventually periodic with some minimalperiod p (meaning that am = am+p for all sufficiently large integers m, and p is the smallest suchpositive integer). Find p.

Answer: 12 Let an = 2bn , so notice b1 = 1, b2 = 2, and bn+1 = bn + 2bn−1 for n ≥ 1, so by

inspection bn = 2n−1 for all n; thus an = 22n−1

. 2014 = 2 · 19 · 53 so we just want to find the lcm ofthe eventual periods of 2n mod ord19(2) and ord53(2). These orders divide 18 and 52 respectively, andwe can manually check ord9(2) = 6 and ord13(2) = 12. The lcm of these is 12, so to show the answer

is 12, it suffices to show that 13|ord53(2). This is true since 24 6≡ 1 (mod 53). So, the answer is 12 .

20. [11] Determine the number of sequences of sets S1, S2, . . . , S999 such that

S1 ⊆ S2 ⊆ · · · ⊆ S999 ⊆ {1, 2, . . . , 999}.

Here A ⊆ B means that all elements of A are also elements of B.

Answer: 102997 OR 1000999 The idea is to look at each element individually, rather than eachsubset. For each k ∈ {1, 2, ..., 999}, there are 1000 choices for the first subset in the chain that containsk. This count includes the possibility that k doesn’t appear in any of the subsets. If Si is the firstsubset containing k, for some i ∈ {1, 2, ..., 999}, then k is also in Sj , for all i < j ≤ 999. As a result,picking the first subset that contains k uniquely determines the appearance of k in all the subsets. Itfollows that there are 1000999 such subset chains.

21. [11] If you flip a fair coin 1000 times, what is the expected value of the product of the number ofheads and the number of tails?

Answer: 249750 We solve the problem for n coins. We want to find

E(n) =

n∑

k=0

1

2n

(

n

k

)

k(n− k).

We present three methods for evaluating this sum.

Method 1: Discard the terms k = 0, k = n. Since(

nk

)

k(n − k) = n(n − 1)(

n−2k−1

)

by the factorialdefinition, we may rewrite the sum as

E(n) =n(n− 1)

2n·n−1∑

k=1

(

n− 2

k − 1

)

.

Guts Round

But clearly∑n−1

k=1

(

n−2k−1

)

= 2n−2, so the answer is n(n−1)4 .

Method 2: Let E[·] denote expected value. Using linearity of expectation, we want to find the expectedvalue of

E[X(n−X)] = nE[X]− E[X2]

where X is the number of heads. Moreover, we have

Var(X) = E[X2]− E[X]2.

The variance of each individual coin flip is 14 , so Var(X) = n

4 . Hence E[X2] = 14n

2 + n4 . Consequently

E[X(n−X)] = n · n2−(

1

4n2 +

n

4

)

=n(n− 1)

4.

Method 3: Differentiating the binomial theorem, we obtain

∂

∂x

∂

∂y(x+ y)n =

n∑

k=0

∂

∂x

∂

∂y

(

n

k

)

xkyn−k =n∑

k=0

(

n

k

)

k(n− k)xk−1yn−k−1.

We also know that∂

∂x

∂

∂y(x+ y)n = n(n− 1)(x+ y)n−2.

Plugging in x = y = 1, we find that E(n) = n(n−1)4 .

22. [12] Evaluate the infinite sum∞∑

n=2

log2

(

1− 1n

1− 1n+1

)

.

Answer: −1 Using the identity log2(

ab

)

= log2 a− log2 b, the sum becomes

∞∑

n=2

log2

(

n− 1

n

)

−∞∑

n=2

log2

(

n

n+ 1

)

.

Most of the terms cancel out, except the log2(

12

)

term from the first sum. Therefore, the answer is−1.

23. [12] Seven little children sit in a circle. The teacher distributes pieces of candy to the children in sucha way that the following conditions hold.

• Every little child gets at least one piece of candy.

• No two little children have the same number of pieces of candy.

• The numbers of candy pieces given to any two adjacent little children have a common factor otherthan 1.

• There is no prime dividing every little child’s number of candy pieces.

What is the smallest number of pieces of candy that the teacher must have ready for the little children?

Answer: 44 An optimal arrangement is 2-6-3-9-12-4-8.

Note that at least two prime factors must appear. In addition, any prime factor that appears mustappear in at least two non-prime powers unless it is not used as a common factor between any twoadjacent little children. Thus with the distinctness condition we easily see that, if we are to beat 44,5 and 7 cannot be included. More comparison shows that 12 or something higher cannot be avoided,so this is optimal.

Guts Round

24. [12] Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. We construct isosceles righttriangle ACD with ∠ADC = 90◦, where D,B are on the same side of line AC, and let lines AD andCB meet at F . Similarly, we construct isosceles right triangle BCE with ∠BEC = 90◦, where E,Aare on the same side of line BC, and let lines BE and CA meet at G. Find cos∠AGF .

Answer: − 513 We see that ∠GAF = ∠GBF = 45◦, hence quadrilateral GFBA is cyclic.

Consequently ∠AGF + ∠FBA = 180◦. So cos∠AGF = − cos∠FBA. One can check directly thatcos∠CBA = 5

13 (say, by the Law of Cosines).

25. [13] What is the smallest positive integer n which cannot be written in any of the following forms?

• n = 1 + 2 + · · ·+ k for a positive integer k.

• n = pk for a prime number p and integer k.

• n = p+ 1 for a prime number p.

• n = pq for some distinct prime numbers p and q

Answer: 40 The first numbers which are neither of the form pk nor pq are 12, 18, 20, 24, 28, 30, 36, 40, . . . .Of these 12, 18, 20, 24, 30 are of the form p+ 1 and 28, 36 are triangular. Hence the answer is 40.

26. [13] Consider a permutation (a1, a2, a3, a4, a5) of {1, 2, 3, 4, 5}. We say the tuple (a1, a2, a3, a4, a5) isflawless if for all 1 ≤ i < j < k ≤ 5, the sequence (ai, aj , ak) is not an arithmetic progression (in thatorder). Find the number of flawless 5-tuples.

Answer: 20 We do casework on the position of 3.

• If a1 = 3, then the condition is that 4 must appear after 5 and 2 must appear after 1. It is easyto check there are six ways to do this.

• If a2 = 3, then there are no solutions; since there must be an index i ≥ 3 with ai = 6− a1.

• If a3 = 3, then 3 we must have {{a1, a2}, {a4, a5}} = {{1, 5}, {2, 4}}. It’s easy to see there are23 = 8 such assignments.

• The case a4 = 3 is the same as a2 = 3, for zero solutions.

• The case a5 = 3 is the same as a1 = 3, for six solutions.

Hence, the total is 6 + 8 + 6 = 20.

27. [13] In triangle ABC, let the parabola with focus A and directrix BC intersect sides AB and AC atA1 and A2, respectively. Similarly, let the parabola with focus B and directrix CA intersect sides BCand BA at B1 and B2, respectively. Finally, let the parabola with focus C and directrix AB intersectsides CA and CB at C1 and C2, respectively.

If triangle ABC has sides of length 5, 12, and 13, find the area of the triangle determined by linesA1C2, B1A2 and C1B2.

Answer: 67283375 By the definition of a parabola, we get AA1 = A1B sinB and similarly for the

other points. So AB2

AB = AC1

AC , giving B2C1 ‖ BC, and similarly for the other sides. So DEF (WLOG,in that order) is similar to ABC. It suffices to scale after finding the length of EF , which is

B2C1 −B2F − EC1

The parallel lines also give us B2A1F ∼ BAC and so forth, so expanding out the ratios from thesesimilarities in terms of sines eventually gives

EF

BC=

2∏

cyc sinA+∑

cyc sinA sinB − 1∏

cyc(1 + sinA)

Plugging in, squaring the result, and multiplying by KABC = 30 gives the answer.

Guts Round

28. [15] Let x be a complex number such that x+ x−1 is a root of the polynomial p(t) = t3 + t2 − 2t− 1.Find all possible values of x7 + x−7.

Answer: 2 Since x+ x−1 is a root,

0 =(

x+ x−1)3

+(

x+ x−1)2 − 2

(

x+ x−1)

− 1

= x3 + x−3 + 3x+ 3x−1 + x2 + 2 + x−2 − 2x− 2x−1 − 1

= x3 + x−3 + x2 + x−2 + x+ x−1 + 1

= x−3(

1 + x+ x2 + · · ·+ x6)

.

Since x 6= 0, the above equality holds only if x is a primitive seventh root of unity, i.e. x7 = 1 andx 6= 1. Therefore, the only possible value of x7 + x−7 is 1 + 1 = 2.

29. [15] Let ω be a fixed circle with radius 1, and let BC be a fixed chord of ω such that BC = 1. Thelocus of the incenter of ABC as A varies along the circumference of ω bounds a region R in the plane.Find the area of R.

Answer: π(

3−√3

3

)

− 1 We will make use of the following lemmas.

Lemma 1: If ABC is a triangle with incenter I, then ∠BIC = 90 + A2 .

Proof: Consider triangle BIC. Since I is the intersection of the angle bisectors, ∠IBC = B2 and

∠ICB = C2 . It follows that ∠BIC = 180− B

2 − C2 = 90 + A

2 .

Lemma 2: If A is on major arc BC, then the circumcenter of △BIC is the midpoint of minor arcBC, and vice-versa.

Proof: Let M be the midpoint of minor arc BC. It suffices to show that ∠BMC + 2∠BIC = 360◦,since BM = MC. This follows from Lemma 1 and the fact that ∠BMC = 180− ∠A. The other caseis similar.

Let O be the center of ω. Since BC has the same length as a radius, △OBC is equilateral. We nowbreak the problem into cases depending on the location of A.

Case 1: If A is on major arc BC, then ∠A = 30◦ by inscribed angles. If M is the midpoint of minorarc BC, then ∠BMC = 150◦. Therefore, if I is the incenter of △ABC, then I traces out a circularsegment bounded by BC with central angle 150◦, on the same side of BC as A.

Case 2: A similar analysis shows that I traces out a circular segment bounded by BC with centralangle 30◦, on the other side of BC.

The area of a circular segment of angle θ (in radians) is given by 12θR

2− 12R

2 sin θ, where R is the radiusof the circular segment. By the Law of Cosines, since BC = 1, we also have that 2R2 − 2R2 cos θ = 1.Computation now gives the desired answer.

30. [15] Suppose we keep rolling a fair 2014-sided die (whose faces are labelled 1, 2, . . . , 2014) until weobtain a value less than or equal to the previous roll. Let E be the expected number of times we rollthe die. Find the nearest integer to 100E.

Answer: 272 Let n = 2014. Let pk denote the probability the sequence has length at least k. Weobserve that

pk =

(

nk

)

nk

since every sequence of k rolls can be sorted in exactly one way. Now the answer is

∑

k≥0

pk =

(

1 +1

n

)n

.

As n → ∞, this approaches e. Indeed, one can check from here that the answer is 272.

Guts Round

31. [17] Flat Albert and his buddy Mike are watching the game on Sunday afternoon. Albert is drinkinglemonade from a two-dimensional cup which is an isosceles triangle whose height and base measure 9cmand 6cm; the opening of the cup corresponds to the base, which points upwards. Every minute afterthe game begins, the following takes place: if n minutes have elapsed, Albert stirs his drink vigorouslyand takes a sip of height 1

n2 cm. Shortly afterwards, while Albert is busy watching the game, Mike addscranberry juice to the cup until it’s once again full in an attempt to create Mike’s cranberry lemonade.Albert takes sips precisely every minute, and his first sip is exactly one minute after the game begins.

After an infinite amount of time, let A denote the amount of cranberry juice that has been poured (insquare centimeters). Find the integer nearest 27

π2A.

Answer: 26 Let A0 = 12 (6)(9) = 27 denote the area of Albert’s cup; since area varies as the square

of length, at time n Mike adds

A

(

1−(

1− 1

9n2

)2)

whence in all, he adds

A0

∞∑

n=1

(

2

9n2− 1

81n4

)

=2A0ζ(2)

9− A0ζ(4)

81= 6ζ(2)− 1

3ζ(4)

where ζ is the Riemann zeta function. Since ζ(2) = π2

6 and ζ(4) = π4

90 , we find that A = π2 − π4

270 , so27Aπ2 = 27− π2

10 , which gives an answer 26.

Note that while the value of ζ(2) is needed to reasonable precision, we only need the fact that 0.5 <9π2 ζ(4) < 1.5 in order to obtain a sufficiently accurate approximation. This is not hard to obtainbecause the terms of the expansion ζ(4) decrease rapidly.

32. [17] Let f(x) = x2 − 2, and let fn denote the function f applied n times. Compute the remainderwhen f24(18) is divided by 89.

Answer: 47 Let Ln denote the Lucas numbers given by L0 = 2, L1 = 1, and Ln+2 = Ln+1 + Ln.Note that L2

n − 2 = L2n when n is even (one can show this by induction, or explicitly using Ln =

( 1+√5

2 )n + ( 1−√5

2 )n). So, f24(L6) = L3·225 .

Now note that since 89 ≡ 4 (mod 5), we have 5p−1

2 ≡ 1 (mod 89) so L89 = ( 1+√5

2 )p + ( 1−√5

2 )p ≡ L1

(mod 89) and similarly L90 ≡ L2, so the sequence Ln (mod 89) is periodic with period 88. (Alter-natively, reason by analog of Fermat’s little theorem, since we can substitute an integer residue for√5.)

We have 3 · 225 ≡ 3 · 25 ≡ 8 (mod 11) and ≡ 0 (mod 8), so L3·225 ≡ L8 (mod 89). Computing L8 = 47gives the answer.

33. [17] How many ways can you remove one tile from a 2014 × 2014 grid such that the resulting figurecan be tiled by 1× 3 and 3× 1 rectangles?

Answer: 451584 Number the rows and columns of the grid from 0 to 2013, thereby assigning anordered pair to each tile. We claim that a tile (i, j) may be selected if and only if i ≡ j ≡ 0 (mod 3);call such a square good.

First, let us show that this condition is sufficient. Observe that any such square s is the corner of acanonical 4× 4 square S whose vertices are all good. Then the sides of S partition the board into ninedistinct regions. It’s easy to see that all of them can be suitably tiled.

Guts Round

Now we show that only good squares can be removed. Let ω be a non-real cube root of unity. In thetile with coordinates (i, j), place the complex number ωi+j . Note that any 1 × 3 or 3 × 1 rectangleplaced on the grid must cover three squares with sum 1 + ω + ω2 = 0. Now, note that the sum of thenumbers on the whole 2014× 2014 grid, including the removed tile, is

2013∑

k=0

2013∑

l=0

ωk+l =

(

2013∑

k=0

ωk

)2

which can be simplified to 1 using the identity 1+ω+ω2 = 0. Therefore, it is necessary that i+ j ≡ 0(mod 3). By placing the complex number ωi−j instead of ωi+j , the same calculations show that i−j ≡ 0(mod 3) is necessary. This can only occur if i ≡ j ≡ 0 (mod 3).

Hence the answer is exactly the set of good squares, of which there are 6722 = 451584.

34. [20] Let M denote the number of positive integers which divide 2014!, and let N be the integer closestto ln(M). Estimate the value of N . If your answer is a positive integer A, your score on this problemwill be the larger of 0 and

⌊

20− 18 |A−N |

⌋

. Otherwise, your score will be zero.

Answer: 439 Combining Legendre’s Formula and the standard prime approximations, the answeris

∏

p

(

1 +2014− sp(2014)

p− 1

)

where sp(n) denotes the sum of the base p-digits of n.

Estimate ln 1000 ≈ 8, and ln 2014 ≈ 9. Using the Prime Number Theorem or otherwise, one mightestimate about 150 primes less than 1007 and 100 primes between 1008 and 2014. Each prime between1008 and 2014 contributes exactly ln 2. For the other 150 primes we estimate ln 2014/p as theircontribution, which gives

∑

p<1000(ln 2014 − ln p). Estimating the average ln p for p < 1000 to beln 1000−1 ≈ 7 (i.e. an average prime less than 1000 might be around 1000/e), this becomes 150·2 = 300.So these wildly vague estimates give 300 + 150 ln 2 ≈ 400, which is not far from the actual answer.

The following program in Common Lisp then gives the precise answer of 438.50943.

;;;; First, generate a list of all the primes

(defconstant +MAXP+ 2500)

(defun is-prime (p)

(loop for k from 2 to (isqrt p) never (zerop (mod p k))))

(defparameter *primes* (loop for p from 2 to +MAXP+

if (is-prime p) collect p))

;;;; Define NT functions

Guts Round

(defconstant +MAXDIGITS+ 15)

(defun base-p-digit (p i n)

(mod (truncate n (expt p i)) p))

(defun sum-base-p-digit (p n)

(loop for i from 0 to +MAXDIGITS+ sum (base-p-digit p i n)))

(defun vp-n-factorial (p n)

(/ (- n (sum-base-p-digit p n)) (1- p)))

;;;; Compute product

(princ (loop for p in *primes*

sum (log (1+ (vp-n-factorial p 2014)))))

35. [20] Ten points are equally spaced on a circle. A graph is a set of segments (possibly empty) drawnbetween pairs of points, so that every two points are joined by either zero or one segments. Two graphsare considered the same if we can obtain one from the other by rearranging the points.

Let N denote the number of graphs with the property that for any two points, there exists a path fromone to the other among the segments of the graph. Estimate the value of N . If your answer is a positiveinteger A, your score on this problem will be the larger of 0 and ⌊20− 5 |ln(A/N)|⌋. Otherwise, yourscore will be zero.

Answer: 11716571 The question asks for the number of isomorphism classes of connected graphson 10 vertices. This is enumerated in http://oeis.org/A001349; the answer is 11716571.

In fact, of the 245 ≈ 3.51 · 1013 ≈ 3 · 1013 graphs on 10 labelled vertices, virtually all (about 3.45 · 1013)are connected. You might guess this by noticing that an “average” graph has 22.5 edges, which is fairlydense (and virtually all graphs with many edges are connected). Moreover, a “typical” isomorphismclass contains 10! ≈ 3 · 106 elements, one for each permutation of the vertices. So estimating the

quotient 3·10133·106 = 107 gives a very close estimate.

36. [20] Pick a subset of at least four of the following geometric theorems, order them from earliest tolatest by publication date, and write down their labels (a single capital letter) in that order. If atheorem was discovered multiple times, use the publication date corresponding to the geometer forwhich the theorem is named.

C. (Ceva) Three cevians AD, BE, CF of a triangle ABC are concurrent if and only if BDDC

CEEA

AFFB = 1.

E. (Euler) In a triangle ABC with incenter I and circumcenter O, we have IO2 = R(R−2r), wherer is the inradius and R is the circumradius of ABC.

H. (Heron) The area of a triangle ABC is√

s(s− a)(s− b)(s− c), where s = 12 (a+ b+ c).

M. (Menelaus) IfD, E, F lie on lines BC, CA, AB, then they are collinear if and only if BDDC

CEEA

AFFB =

−1, where the ratios are directed.

P. (Pascal) Intersections of opposite sides of cyclic hexagons are collinear.

S. (Stewart) Let ABC be a triangle and D a point on BC. Set m = BD, n = CD, d = AD. Thenman+ dad = bmb+ cnc.

V. (Varignon) The midpoints of the sides of any quadrilateral are the vertices of a parallelogram.

If your answer is a list of 4 ≤ N ≤ 7 labels in a correct order, your score will be (N − 2)(N − 3).Otherwise, your score will be zero.

Answer: HMPCVSE The publication dates were as follows.

• Heron: 60 AD, in his book Metrica.

• Menelaus: We could not find the exact date the theorem was published in his book Spherics, butbecause Menelaus lived from 70AD to around 130AD, this is the correct placement.

• Pascal: 1640 AD, when he was just 17 years old. He wrote of the theorem in a note one yearbefore that.

Guts Round

• Ceva: 1678 AD, in his work De lineis rectis. But it was already known at least as early as the11th century.

• Varignon: 1731 AD.

• Stewart: 1746 AD.

• Euler: 1764 AD, despite already being published in 1746.

Guts Round

HMMT 2014Saturday 22 February 2014

HMIC

1. [20] Consider a regular n-gon with n > 3, and call a line acceptable if it passes through the interiorof this n-gon. Draw m different acceptable lines, so that the n-gon is divided into several smallerpolygons.

(a) Prove that there exists an m, depending only on n, such that any collection of m acceptable linesresults in one of the smaller polygons having 3 or 4 sides.

(b) Find the smallest possible m which guarantees that at least one of the smaller polygons will have3 or 4 sides.

2. [25] 2014 triangles have non-overlapping interiors contained in a circle of radius 1. What is the largestpossible value of the sum of their areas?

3. [30] Fix positive integers m and n. Suppose that a1, a2, . . . , am are reals, and that pairwise distinctvectors v1, . . . , vm ∈ R

n satisfy ∑

j 6=i

aj

vj − vi

‖vj − vi‖3= 0

for i = 1, 2, . . . ,m.

Prove that ∑

1≤i<j≤m

aiaj

‖vj − vi‖= 0.

4. [35] Let ω be a root of unity and f be a polynomial with integer coefficients. Show that if |f(ω)| = 1,then f(ω) is also a root of unity.

5. [40] Let n be a positive integer, and let A and B be n × n matrices with complex entries such thatA2 = B2. Show that there exists an n × n invertible matrix S with complex entries that satisfiesS(AB − BA) = (BA − AB)S.

HMMT 2014Saturday 22 February 2014

HMIC

1. [20] Consider a regular n-gon with n > 3, and call a line acceptable if it passes through the interiorof this n-gon. Draw m different acceptable lines, so that the n-gon is divided into several smallerpolygons.

(a) Prove that there exists an m, depending only on n, such that any collection of m acceptable linesresults in one of the smaller polygons having 3 or 4 sides.

(b) Find the smallest possible m which guarantees that at least one of the smaller polygons will have3 or 4 sides.

Answer: N/A We will prove that if m ≥ n− 4, then there is guaranteed to be a smaller polygon

with 3 or 4 sides, while if m ≤ n − 5, there might not be a polygon with 3 or 4 sides. This will solveboth parts of the problem.

Given a configuration of lines, let P1, . . . , Pk be all of the resulting smaller polygons. Let E(Pi) bethe number of edges in polygon Pi, and let E = E(P1) + · · · + E(Pk). First, note that whenever anew polygon is formed, it must have been because a larger polygon was split into two smaller polygonsby a line passing through it. When this happens, k increases by 1 and E increases by at most 4 (itmight be less than 4 if the line passes through vertices of the larger polygon). Therefore, if adding anacceptable line increases the number of polygons by a, then E increases by at most 4a.

Now, assume m ≥ n − 4. At the beginning, we have k = 1 and E = n. If the number of polygons atthe end is p + 1, then E ≤ n + 4p, so the average number of edges per polygon is less than or equalto n+4p

1+p. Now, note that each acceptable line introduces at least one new polygon, so p ≥ m ≥ n− 4.

Also, note that as p increases, n+4p1+p

strictly decreases, so it is maximized at p = n− 4, where n+4p1+p

=5n−16n−3 = 5− 1

n−3 < 5. Therefore, the average number of edges per polygon is less than 5, so there mustexist a polygon with either 3 or 4 edges, as desired.

We will now show that if m ≤ n − 5, then we can draw m acceptable lines in a regular n-gonA1A2 . . . An such that there are no polygons with 3 or 4 sides. Let M1 be the midpoint of A3A4,M2 be the midpoint of A4A5, ..., Mn−5 be the midpoint of An−3An−4. Let the m acceptable lines beA1M1, A1M2, . . . , A1Mm. We can see that all resulting polygons have 5 sides or more, so we are done.

2. [25] 2014 triangles have non-overlapping interiors contained in a circle of radius 1. What is the largestpossible value of the sum of their areas?

Answer: N/A This problem turned out to be much trickier than we expected. We have yet to

see a complete solution, but let us know if you find one!

Comment. We apologize for the oversight on our part. Our test-solvers essentially all misread theproblem to contain the additional assumption that all triangle vertices lie on the circumference (whichis not the case).

3. [30] Fix positive integers m and n. Suppose that a1, a2, . . . , am are reals, and that pairwise distinctvectors v1, . . . , vm ∈ Rn satisfy

∑

j 6=i

ajvj − vi

‖vj − vi‖3= 0

for i = 1, 2, . . . ,m.

Prove that∑

1≤i<j≤m

aiaj‖vj − vi‖

= 0.

HMIC

Answer: N/A Since vi · (vj − vi) + vj · (vi − vj) = −‖vj − vi‖2 for any 1 ≤ i < j ≤ m, we have

0 =m∑

i=1

aivi · 0 =m∑

i=1

aivi∑

j 6=i

ajvj − vi

‖vj − vi‖3= −

∑

1≤i<j≤m

aiaj‖vj − vi‖

,

as desired.

Alternative Solution. Fix a1, . . . , am, and define f : (Rn)m → R by

f(x1, . . . , xm) =∑

1≤i<j≤m

aiaj‖xj − xi‖

.

Now if we view f(x1, . . . , xm) individually as a function in xk (k fixed), then the problem condition fori = k says precisely that the gradient of f with respect to xk (but of course, multiplied by ak) evaluatedat (v1, . . . , vm) is 0.

Finally, the multivariate chain rule shows that F (t) := f(tv1, . . . , tvm) = 1tf(v1, . . . , vm) must satisfy

F ′(1) = 0. Yet F ′(1) = − 112 f(v1, . . . , vm), so we have f(v1, . . . , vm) = 0, as desired.

Comment. This is (sort of) an energy minimization problem from physics (more precisely, minimizingelectrostatic or gravitational potential).

Comment. It would be interesting to determine all differentiable A : R+ → R such that the problemstill holds when the terms are replaced by

vj−vi

‖vj−vi‖A′(‖vj−vi‖) and A(‖vj−vi‖). (Here A(t) = 1/t, which

has a nice physical interpretation, but that’s not important.) The obvious examples are A(t) = cts

for constant c and real s (where both solutions easily carry through). But there may be others whereinstead of defining F (t) through the scaling transformation (v1, . . . , vm) → (tv1, . . . , tvm), there’s asubtler translation.

4. [35] Let ω be a root of unity and f be a polynomial with integer coefficients. Show that if |f(ω)| = 1,then f(ω) is also a root of unity.

Answer: N/A Suppose ω is a primitive nth root, so that Φn(x) is the minimal polynomial of ω

(over Q). Thus f(ω)f(ω−1) = 1 implies Φn(x) | f(x)f(x−1)− 1 (divisibility as Laurent polynomials).Hence |f(ωk)| = 1 whenever gcd(k, n) = 1, and

∏

gcd(k,n)=1(t − f(ωk)) is a monic integer-coefficient

polynomial with all roots within the unit disk, so its roots must all be roots of unity (since 0 is clearlynot a root), and we’re done.

Comment. The result used at the end (a monic integer-coefficient polynomial with all roots withinthe unit disk has every root either 0 or a root of unity) is classical. See for instance the discussion hereor here. This result is sometimes called Kronecker’s theorem, and appears, for instance, in Problems

from the Book.

Comment. Note that the proof works as long as f ∈ Q[x] and f(ω) is an algebraic integer. Once is it

known that f(ω) is a root of unity, more can be said about the order m of f(ω). If ω is a primitive nthroot of unity, then the conjugates of f(ω) lie among the ≤ φ(n) numbers in {f(ωk) : gcd(k, n) = 1}.In particular, φ(m) ≤ φ(n), which (e.g. through a naive estimate using p − 1 ≥ √

p for p ≥ 3 to get

φ(m) ≥√

m/2) shows that m ≤ 2φ(n)2 (for instance), so there exists g ∈ Q[x] with g(ω) a root ofunity of maximal order N , which then must generate the unit group of Q(ω) (or else we could find alarger order). In particular, since ω lies in this group, n | N . But φ(N) ≤ φ(n) by our earlier argument,so either N = n or N = 2n if n is odd. Either way, it’s clear that g(ω) = ±ωk for some k, so ouroriginal f(ω) must also be ±ωj for some j.

Alternative Solution. We follow the paper here, but fill in some of the details to make the ideasas accessible as possible. We will use the fact that if A is a finitely generated abelian group, thenany subgroup B is also finitely generated, and furthermore, rank(B) ≤ rank(A). (For instance, seeTheorem 14.6.5 in Artin’s Algebra, 3rd ed. for a proof of finite generation, which for our particularcase of modules over a PID (namely, Z) can be done in parallel to the rank(B) ≤ rank(A) proof—theproof is very similar to Gaussian elimination.)

HMIC

Let K = Q(ω), V the set of algebraic integers in K of modulus 1 (which must be units of K), U ⊇ Vthe set of units in K, R = U ∩R the set of real units in K (which, importantly, is also the set of unitsin K ∩ R), and L = K ∩ R denote the maximal real subfield of K.

The key is that rank(V ) = rank(U) − rank(R). To prove this, we note that for u ∈ U , u ∈ U aswell (since K = K), so uu = |u|2 ∈ R and u/u ∈ V , which gives u2 = (uu)(u/u) ∈ RV . ThusS := {u2 : u ∈ U} ⊆ RV ⊆ U . But U is a finitely generated abelian group by Dirichlet’s unit theorem,so S,RV are as well, and furthermore rank(S) ≤ rank(RV ) ≤ rank(U). Now take multiplicativelyindependent u1, . . . , urank(U) ∈ U ; then clearly u2

1, . . . , u2rank(U) ∈ S are multiplicatively independent as

well, so rank(U) ≤ rank(S), whence rank(RV ) = rank(U). Of course, R ∩ V = {±1}, so rank(RV ) =rank(R) + rank(V ) − rank(R ∩ V ) = rank(R) + rank(V ) (the proof here for Z-modules is essentiallythe same as that for Q-vector spaces), finishing the proof.

V contains only roots of unity if and only if rank(V ) = 0 (no free elements), so we must proverank(U) = rank(R). Let r, s denote the number of real and complex embeddings of a number field, so[K : Q] = r(K) + 2s(K), [L : Q] = r(L) + 2s(L), and thus r(K) + 2s(K) = [K : L](r(L) + 2s(L)). ByDirichlet’s unit theorem, rank(U) = r(K) + s(K)− 1 and rank(R) = r(L) + s(L)− 1.

If K = L (equivalently, K is totally real), we’re trivially done. Otherwise, if [K : L] > 1, it’s easyto check that we have rank(U) = rank(R) if and only if [K : L] = 2 and r(K) = s(L) = 0 (K istotally imaginary and L is totally real)—then r(L) = s(K) automatically holds. This is precisely thecondition that K is a CM-field.

(CM-field K just means K is totally imaginary (no real embeddings), and K is a quadratic extensionof K ∩ R.)

In particular, the cyclotomic field K = Q(ω) = Q[ω] is a totally imaginary quadratic extension of thetotally real field Q(ω+ω−1) = Q[ω+ω−1]. Indeed, it is easy to check that the latter is simply K ∩R,and K = Q(ω + ω−1, ω − ω−1 = ±

√

(ω + ω−1)2 − 4). So we’re done.

5. [40] Let n be a positive integer, and let A and B be n × n matrices with complex entries such thatA2 = B2. Show that there exists an n × n invertible matrix S with complex entries that satisfiesS(AB −BA) = (BA−AB)S.

Answer: N/A Let X = A+B and Y = A−B, so XY = BA−AB and Y X = AB −BA. Note

that XY = −Y X.

It suffices (actually is equivalent) to show that AB −BA and BA−AB have the same Jordan forms.In other words, we need to show that for any complex number λ, the Jordan λ-block decompositionsof AB −BA and BA−AB are the same.

However, the λ-decomposition of a matrix M is uniquely determined by the infinite (eventually con-stant) sequence (dimker(M −λI)1, dimker(M −λI)2, . . .), so we only have to show that dimker(XY −λI)k = dimker(Y X − λI)k for all positive integers k and complex numbers λ.

For λ = 0, this follows trivially from XY = −Y X: we have (XY )kv = 0 if and only if (Y X)kv = 0.

Now suppose λ 6= 0, and define the (matrix) polynomial p(T ) = (T − λI)k. Let V1 = {v ∈ Cn :p(XY )v = 0}, and V2 = {v ∈ Cn : p(Y X)v = 0}. Note that if v ∈ V1, then 0 = [Y p(XY )]v =[p(Y X)Y ]v = p(Y X)[Y v], so Y v ∈ V2. Furthermore, if v 6= 0, then Y v 6= 0, or else p(XY )v =p(0)v = (−λI)kv 6= 0 (since (XY )jv = 0 for all j ≥ 1), contradiction. Viewing Y (more precisely,v → Y v) as a linear map from V1 to V2, we have dimV1 = dimY V1 + dimkerY = dimY V1 ≤ dimV2

(alternatively, if v1, . . . , vr is a basis of V1, then Y v1, . . . , Y vr are linearly independent in V2). Bysymmetry, dimV1 = dimV2, and we’re done.

Comment. If A,B are additionally real, then we can also choose S to be real (not just complex).Indeed, note that if S = P + Qi for real P,Q, then since A,B are real, it suffices to find c ∈ R suchthat det(P + cQ) 6= 0. However, det(P + xQ) is a polynomial in x that is not identically zero (it isnonzero for x = i), so it has finitely many (complex, and thus real) roots. Thus the desired c clearlyexists.

HMIC

Comment. Note that A(AB − BA) = (BA − AB)A, since A2B = B3 = BA2. (The same holds forA → B.) Thus the problem is easy unless all linear combinations of A,B are noninvertible. However,it seems difficult to finish without analyzing generalized eigenvectors as above.

Comment. See this paper by Ross Lippert and Gilbert Strang for a general analysis of the Jordanforms of MN and NM .

HMIC

HMMT February 2015Saturday 21 February 2015

Team

For any geometry problem (e.g. Problem 2, 4, or 5), your team’s solution should include diagrams asappropriate (sufficiently large, in-scale, clearly labeled, etc.). Failure to meet any of these requirementswill result in a 2-point automatic deduction. Thanks in advance for your effort and understanding,and we hope you enjoy the problems!

1. [5] The complex numbers x, y, z satisfy

xyz = −4

(x+ 1)(y + 1)(z + 1) = 7

(x+ 2)(y + 2)(z + 2) = −3.

Find, with proof, the value of (x+ 3)(y + 3)(z + 3).

2. [10] Let P be a (non-self-intersecting) polygon in the plane. Let C1, . . . , Cn be circles in the planewhose interiors cover the interior of P . For 1 ≤ i ≤ n, let ri be the radius of Ci. Prove that there is asingle circle of radius r1 + · · ·+ rn whose interior covers the interior of P .

3. [15] Let z = a+bi be a complex number with integer real and imaginary parts a, b ∈ Z, where i =√−1

(i.e. z is a Gaussian integer). If p is an odd prime number, show that the real part of zp − z is aninteger divisible by p.

4. [15] (Convex) quadrilateral ABCD with BC = CD is inscribed in circle Ω; the diagonals of ABCD

meet at X. Suppose AD < AB, the circumcircle of triangle BCX intersects segment AB at a point

Y 6= B, and ray−−→CY meets Ω again at a point Z 6= C. Prove that ray

−−→DY bisects angle ZDB.

(We have only included the conditions AD < AB and that Z lies on ray−−→CY for everyone’s convenience.

With slight modifications, the problem holds in general. But we will only grade your team’s solutionin this special case.)

5. [20] For a convex quadrilateral P , let D denote the sum of the lengths of its diagonals and let S

denote its perimeter. Determine, with proof, all possible values of SD.

6. [30] $indy has $100 in pennies (worth $0.01 each), nickels (worth $0.05 each), dimes (worth $0.10each), and quarters (worth $0.25 each). Prove that she can split her coins into two piles, each withtotal value exactly $50.

7. [35] Let f : [0, 1] → C be a nonconstant complex-valued function on the real interval [0, 1]. Prove thatthere exists ǫ > 0 (possibly depending on f) such that for any polynomial P with complex coefficients,there exists a complex number z with |z| ≤ 1 such that |f(|z|)− P (z)| ≥ ǫ.

8. [40] Let π be a permutation of {1, 2, . . . , 2015}. With proof, determine the maximum possible numberof ordered pairs (i, j) ∈ {1, 2, . . . , 2015}2 with i < j such that π(i) · π(j) > i · j.

9. [40] Let z = e2πi

101 and let ω = e2πi

10 . Prove that

9∏

a=0

100∏

b=0

100∏

c=0

(ωa + zb + zc)

is an integer and find (with proof) its remainder upon division by 101.

10. [40] The sequences of real numbers {ai}∞i=1 and {bi}∞i=1 satisfy an+1 = (an−1 − 1)(bn + 1) andbn+1 = anbn−1 − 1 for n ≥ 2, with a1 = a2 = 2015 and b1 = b2 = 2013. Evaluate, with proof, theinfinite sum

∞∑

n=1

bn

(

1

an+1

− 1

an+3

)

.

HMMT February 2015Saturday 21 February 2015

Team

1. [5] The complex numbers x, y, z satisfy

xyz = −4

(x+ 1)(y + 1)(z + 1) = 7

(x+ 2)(y + 2)(z + 2) = −3.

Find, with proof, the value of (x+ 3)(y + 3)(z + 3).

Answer: −28 Solution 1. Consider the cubic polynomial f(t) = (x + t)(y + t)(z + t). By thetheory of finite differences, f(3) − 3f(2) + 3f(1) − f(0) = 3! = 6, since f is monic. Thus f(3) =6 + 3f(2)− 3f(1) + f(0) = 6 + 3(−3)− 3(7) + (−4) = −28.

Solution 2. Alternatively, note that the system of equations is a (triangular) linear system in w := xyz,v := xy + yz + zx, and u := x + y + z. The unique solution (u, v, w) to this system is

(

− 272 , 47

2 ,−4)

.Plugging in yields

(x+ 3)(y + 3)(z + 3) = w + 3v + 9u+ 27

= −4 + 3 · 472

+ 9 ·(

−27

2

)

+ 27

= −28.

Remark. Since f(0) < 0, f(1) > 0, f(2) < 0, f(+∞) > 0, the intermediate value theorem tells us theroots −x,−y,−z of f are real numbers in (0, 1), (1, 2), and (2,+∞), in some order. With a little morecalculation, one finds that x, y, z are the three distinct zeroes of the polynomial X3+ 27

2 X2+ 472 X +4.

The three zeroes are approximately −11.484,−1.825, and −0.191.

2. [10] Let P be a (non-self-intersecting) polygon in the plane. Let C1, . . . , Cn be circles in the planewhose interiors cover the interior of P . For 1 ≤ i ≤ n, let ri be the radius of Ci. Prove that there is asingle circle of radius r1 + · · ·+ rn whose interior covers the interior of P .

Answer: N/A If n = 1, we are done. Suppose n > 1. Since P is connected, there must be a

point x on the plane which lies in the interiors of two circles, say Ci, Cj . Let Oi, Oj , respectively, bethe centers of Ci, Cj . Since OiOj < ri + rj , we can choose O to be a point on segment OiOj such thatOiO ≤ rj and OjO ≤ ri. Replace the two circles Ci and Cj with the circle C centered at O of radiusri + rj . Note that C covers both Ci and Cj . Induct to finish.

3. [15] Let z = a+bi be a complex number with integer real and imaginary parts a, b ∈ Z, where i =√−1

(i.e. z is a Gaussian integer). If p is an odd prime number, show that the real part of zp − z is aninteger divisible by p.

Answer: N/A Solution 1. We directly compute/expand

Re(zp − z) = Re ((a+ bi)p − (a+ bi))

=

[

ap −(

p

2

)

ap−2b2 +

(

p

4

)

ap−4b4 − · · ·]

− a.

Since(

pi

)

is divisible by p for all i = 2, 4, 6, . . . (since 1 ≤ i ≤ p− 1), we have

[ap −(

p

2

)

ap−2b2 +

(

p

4

)

ap−4b4 − · · · ]− a ≡ ap − a ≡ 0 (mod p)

by Fermat’s little theorem. Thus p divides the real part of zp − z.

Team

Remark. The fact that(

pi

)

is divisible by p (for 1 ≤ i ≤ p − 1) is essentially equivalent to thepolynomial congruence (rX + sY )p ≡ rXp + sY p (mod p) (here the coefficients are taken modulo p),a fundamental result often called that “Frobenius endomorphism”.

Solution 2. From the Frobenius endomorphism,

zp = (a+ bi)p ≡ ap + (bi)p = ap ± bpi ≡ a± bi (mod p · Z[i]),

where we’re using congruence of Gaussian integers (so that u ≡ v (mod p) if and only if u−vp is a

Gaussian integer). This is equivalent to the simultaneous congruence of the real and imaginary partsmodulo p, so the real part of zp is congruent to a, the real part of z. So indeed p divides the real partof zp − z.

4. [15] (Convex) quadrilateral ABCD with BC = CD is inscribed in circle Ω; the diagonals of ABCDmeet at X. Suppose AD < AB, the circumcircle of triangle BCX intersects segment AB at a point

Y 6= B, and ray−−→CY meets Ω again at a point Z 6= C. Prove that ray

−−→DY bisects angle ZDB.

(We have only included the conditions AD < AB and that Z lies on ray−−→CY for everyone’s convenience.

With slight modifications, the problem holds in general. But we will only grade your team’s solutionin this special case.)

Answer: N/A This is mostly just angle chasing. The conditions AD < AB (or ∠ABD < ∠ADB)

and the assumption Z ∈ −−→CY are not crucial, as long as we’re careful with configurations (for example,

DY may only be an external angle bisector of ∠ZDB in some cases), 1 but it’s the easiest to visualize.

In this case Y and Z lie between A and B, on the respective segment/arc. We’ll prove Y is the

incenter of △ZDB; it will follow that ray−−→DY indeed (internally) bisects ∠ZDB. It suffices to prove

the following two facts:

• BY is the internal angle bisector of ∠DBZ.2

• ZY is the internal angle bisector of ∠BZD, since CB = CD. Indeed (more explicitly), arcs BCand CD are equal, so ∠BZC = ∠CZD, i.e. Y Z bisects ∠BZD.

5. [20] For a convex quadrilateral P , let D denote the sum of the lengths of its diagonals and let Sdenote its perimeter. Determine, with proof, all possible values of S

D .

Answer: 1 < SD < 2 Suppose we have a convex quadrilateral ABCD with diagonals AC and BD

intersecting at E (convexity is equivalent to having E on the interiors of segments AC and BD).

To prove the lower bound, note that by the triangle inequality, AB+BC > AC and AD+DC > AC,so S = AB +BC +AD +DC > 2AC. Similarly, S > 2BD, so 2S > 2AC + 2BD = 2D gies S > D.

To prove the upper bound, note that again by the triangle inequality, AE+EB > AB, CE+BE >BC, AE + ED > AD, CE + ED > CD. Adding these yields

2(AE + EC +BE + ED) > AB +BC +AD + CD = S.

Now since ABCD is convex, E is inside the quadrilateral, so AE + EC = AC and BE + ED = BD.Thus 2(AC +BD) = D > S.

To achieve every real value in this range, first consider a square ABCD. This has SD =

√2. Suppose

now that we have a rectangle with AB = CD = 1 and BC = AD = x, where 0 < x ≤ 1. As x

1The cleanest way to see this is via directed angles, or more rigorously by checking that the problem is equivalent to afour-variable polynomial identity in complex numbers.

2This is true in general; it doesn’t require CB = CD. It’s part of the spiral similarity configuration centered at B : Y X → ZA

and B : ZY → AX, due to Y Z ∩AX = C and B = (CY X) ∩ (CZA).More explicitly, this follows from the angle chase

∠DBA = ∠XBY = ∠XCY = ∠ACZ = ∠ABZ.

Team

approaches 0 (i.e. our rectangle gets thinner), SD gets arbitrarily close to 1, so by intermediate

value theorem, we hit every value SD ∈ (1,

√2].

To achieve the other values, we let AB = BC = CD = DA = 1 and let θ = m∠ABE vary from 45◦

down to 0◦ (i.e. a rhombus that gets thinner). This means AC = 2 sin θ and BD = 2 cos θ. Wehave S = 4 and D = 2(sin θ + cos θ). When θ = 45◦, S

D =√2, and when θ = 0◦, S

D = 2. Thus by the

intermediate value theorem, we are able to choose θ to obtain any value in the range [√2, 2).

Putting this construction together with the strict upper and lower bounds, we find that all possiblevalues of S

D are all real values in the open interval (1, 2).

6. [30] $indy has $100 in pennies (worth $0.01 each), nickels (worth $0.05 each), dimes (worth $0.10each), and quarters (worth $0.25 each). Prove that she can split her coins into two piles, each withtotal value exactly $50.

Answer: N/A Solution 1. First, observe that if there are pennies in the mix, there must be a

multiple of 5 pennies (since 5, 10, 25 ≡ 0 (mod 5)), so we can treat each group of 5 pennies as nickelsand reduce the problem to the case of no pennies, treated below.

Indeed, suppose there are no pennies and that we are solving the problem with only quarters, dimes,and nickels. So we have 25q + 10d+ 5n = 10000 (or equivalently, 5q + 2d+ n = 2000), where q is thenumber of quarters, d is the number of dimes, and n is the number of nickels. Notice that if q ≥ 200,d ≥ 500, or n ≥ 1000, then we can just take a subset of 200 quarters, 500 dimes, or 1000 nickels,respectively, and make that our first pile and the remaining coins our second pile.

Thus, we can assume that q ≤ 199, d ≤ 499, n ≤ 999. Now if there are only quarters and dimes, wehave 5q + 2d < 1000 + 1000 = 2000, so there must be nickels in the mix. By similar reasoning, theremust be dimes and quarters in the mix.

So to make our first pile, we throw all the quarters in that pile. The total money in the pile is atmost $49.75, so now,

• if possible, we keep adding in dimes until we get an amount equal to or greater than$50. If it is $50, then we are done. Otherwise, it must be $50.05 since the total money valuebefore the last dime is either $49.90 or $49.95. In that case, we take out the last dime and replaceit with a nickel, giving us $50.

• if not possible (i.e. we run out of dimes before reaching $50), then all the quarters anddimes have been used up, so we only have nickels left. Of course, our money value now has ahundredth’s place of 5 or 0 and is less than $50, so we simply keep adding nickels until we aredone.

Solution 2. First, observe that if there are pennies in the mix, there must be a multiple of 5 pennies(since 5, 10, 25 ≡ 0 (mod 5)), so we can treat each group of 5 pennies as nickels and reduce the problemto the case of no pennies, treated below. Similarly, we can treat each group of 2 nickels as dimes, andreduce the problem to the case of at most one nickel. We have two cases:

Case 1. There are no nickels. Then, either the dimes alone total to at least $50, in which case $indycan form a pile with $50 worth of dimes and a pile with the rest of the money, or the quarters alonetotal to at least $50, in which case $indy can form a pile with $50 worth of quarters and a pile withthe rest of the money.

Case 2. There is exactly one nickel. By examining the value of the money modulo $0.25, we find thatthere must be at least two dimes. Then, we treat the nickel and two dimes as a quarter, and reducethe problem to the previously solved case.

Remark. There are probably many other solutions/algorithms for this specific problem. For example,many teams did casework based on the parities of q, d, n, after “getting rid of pennies”.

Remark. It would certainly be interesting to investigate the natural generalizations of this problem(see e.g. the knapsack problem and partition problem). For instance, it’s true that given a set of

Team

positive integers ranging from 1 to n with sum no less than 2n!, there exists a subset of them summingup to exactly n!. (See here and here for discussion on two AoPS blogs.)

Remark. The problem author (Carl Lian) says the problem comes from algebraic geometry!

7. [35] Let f : [0, 1] → C be a nonconstant complex-valued function on the real interval [0, 1]. Prove thatthere exists ǫ > 0 (possibly depending on f) such that for any polynomial P with complex coefficients,there exists a complex number z with |z| ≤ 1 such that |f(|z|)− P (z)| ≥ ǫ.

Answer: N/A Here’s a brief summary of how we’ll proceed: Use a suitably-centered high-degree

roots of unity filter, or alternatively integrate over suitably-centered circles (“continuous roots of unityfilter”).

We claim we can choose ǫ(f) = max(|f(x1)−f(x2)|)/2. Fix P and suppose for the sake of contradictionthat for all z with |z| ≤ 1 it is the case that |f(z)− P (z)| < ǫ. We can write z = reiθ so that

|f(r)− P (reiθ)| < ǫ.

Let p be a prime larger than deg(P ) and set θ = 0, 2π/p, 4π/p, . . . in the above. Averaging, and usingtriangle inequality,

|f(r)− 1

p

p−1∑

k=0

P (re2πik/p)| < ǫ.

The sum inside the absolute value is a roots of unity filter; since p > deg(P ) it is simply the constantterm of P - denote this value by p0. Thus for all r,

|f(r)− p0| < ǫ.

Setting r = x1, x2 and applying triangle inequality gives the desired contradiction.

Remark. This is more or less a concrete instance of Runge’s approximation theorem from complexanalysis. However, do not be intimidated by the fancy names here: in fact, the key ingredient behindthis more general theorem (namely, or similar3) uses the same core idea of the “continuous roots ofunity filter”.

8. [40] Let π be a permutation of {1, 2, . . . , 2015}. With proof, determine the maximum possible numberof ordered pairs (i, j) ∈ {1, 2, . . . , 2015}2 with i < j such that π(i) · π(j) > i · j.

Answer:(

20142

)

Let n = 2015. The only information we will need about n is that n > 5.4 For the

construction, take π to be the n-cycle defined by

π(k) =

{

k + 1 if 1 ≤ k ≤ n− 1

1 if k = n.

Then π(i) > i for 1 ≤ i ≤ n− 1. So π(i)π(j) > ij for at least(

n−12

)

pairs i < j.

For convenience let zi = π(i)i , so that we are trying to maximize the number of pairs (i, j), i < j

with zizj > 1. Notice that over any cycle c = (i1 i2 · · · ik) in the cycle decomposition of π we have∏

i∈c zi = 1. In particular, multiplying over all such cycles gives∏n

i=1 zi = 1.

Construct a graph G on vertex set V = [n] such that there is an edge between i and j whenever

π(i)π(j) > ij. For any cycle C = (v1, v2, . . . , vk) in this graph we get 1 <∏k

i=1 zvizvi+1

=∏

v∈C z2v .So, we get in particular that G is non-Hamiltonian.

By the contrapositive of Ore’s theorem,5 there are two distinct indices u, v ∈ [n] such that d(u)+d(v) ≤n− 1. The number of edges is then at most

(

n−22

)

+ (n− 1) =(

n−12

)

+ 1.

3which generally underlies a great deal of complex analysis4If you want, you can try running a computer check for n ≤ 5. One of the problem czars believes the answer is still

(

4

2

)

= 6(compare with the Remark at the end for a looser general problem) based on computer program, but didn’t check carefully.

5see also this StackExchange thread

Team

If equality is to hold, we need G \ {u, v} to be a complete graph. We also need d(u) + d(v) = n − 1,with uv not an edge. This implies d(u), d(v) ≥ 1. Since n > 5, the pigeonhole principle gives that atleast one of u, v has degree at least 3. WLOG d(u) ≥ 3.

Let w be a neighbor of v and let a, b, c be neighbors of u; WLOG w 6= a, b. Since G \ {u, v, w} is acomplete graph, we can pick a Hamiltonian path in G \ {u, v, w} with endpoints a, b. Connecting u tothe ends of this path forms an (n− 2)-cycle C.

This gives us∏

x∈C z2x > 1. But we also have zvzw > 1, so 1 =∏n

i=i zi > 1, contradiction.

So,(

n−12

)

+ 1 cannot be attained, and(

n−12

)

is indeed the maximum number of pairs possible.

Remark. Let’s think about the more general problem where we forget about the permutation structure(and only remember

∏

zi = 1 and zi > 0). If n = 4 it’s easy to show by hand that(

n−12

)

is still the

best possible. If n = 5, then we can get 1 +(

n−12

)

= 7: consider exponentiating the zero-sum set{t, t, t2− t, t2− t,−2t2} for some sufficiently small t. A computer check would easily determine whethera construction exists in the original permutation setting. (If the following code is correct, the answeris ‘no’, i.e. 6 is the best: see http://codepad.org/Efsv8Suy.)

9. [40] Let z = e2πi

101 and let ω = e2πi

10 . Prove that

9∏

a=0

100∏

b=0

100∏

c=0

(ωa + zb + zc)

is an integer and find (with proof) its remainder upon division by 101.

Answer: 13 Solution 1. Let p = 101 and r = 10. Note that p ∤ r.

In the sequel, we will repeatedly use the polynomial identities∏

k (mod p)(x − zk) = xp − 1, and∏

j (mod r)(x− ωj) = xr − 1.

The product is an integer by standard symmetric sum theory (concrete precursor to Galois theory,though one doesn’t need the full language). More precisely, it expands into an integer-coefficientpolynomial symmetric in the ωj (for the r residues j (mod r)), and also symmetric in the zk (for thep residues k (mod p)). So by the fundamental theorem of symmetric sums, it can be written as aninteger-coefficient polynomial in the symmetric sums of the ωj together with the symmetric sums of thezk. But these symmetric sums are integers themselves, so the original expression is indeed an integer.

To actually compute the remainder modulo p, note6 by the Frobenius endomorphism (u+v)p ≡ up+vp

(mod p) that

∏

a,b,c

(ωa + zb + zc) =∏

a,b

(−1)p((−ωa − zb)p − 1)

≡∏

a,b

(ωpa + zpb − (−1)p) =∏

a,b

(ωpa + 2) (mod p),

where we use p ≡ 1 (mod 2) in the last step. This simplifies further to

∏

a

(−1)p(−2− ωpa)p =∏

a

(−1)p(−2− ωa)p

= (−1)pr[(−2)r − 1]p ≡ (−1)pr[(−2)r − 1] (mod p),

where we’ve used the fact that {pa (mod r)} = {a (mod r)} (since p is invertible mod r), as well asFermat’s little theorem on (−2)r − 1.

Finally, we plug in the specific numbers:

(−1)101·10[(−2)10 − 1] = 1023 ≡ 13 (mod 101).

6Here we are using congruence of algebraic integers modulo p, so that α ≡ β (mod p) if and only if α−β

pis an algebraic

integer. In particular, for integers u, v, u−vp

is an algebraic integer if and only if u−vp

is an integer.

Team

Remark. By repeatedly getting rid of “z” terms and applying the Frobenius map, one can show that∏

(ωa + zb1 + · · · + zbm) is an integer congruent to something like mr + 1 modulo p (as long as p isodd, etc.).

Solution 2 (sketch). Here we sketch a better solution found by several teams, but missed by theauthor (primarily due to blindness from having first found the first solution). The proof of the firstpart (that the product is an integer) is the same, so we only sketch different proofs of the second part(computation of the remainder mod p).

For example, we can use the algebraic integer formulation from the previous solution. Indeed, notethat D :=

∏

a,b,c(ωa + zb + zc)−∏

a,b,c(ωa + 1b + 1c) is an algebraic integer divisible by z − 1. But D

is also a difference of integers, hence an integer itself. The only way Dz−1 can be an algebraic integer is

if p | D (Why?), so it simply remains to compute the remainder when the integer∏

a,b,c(ωa + 1b + 1c)

is divided by p, which is quite easy.

Alternatively (as some contestants/teams found), we can get rid of ω first (rather than z as in theprevious solution), so (up to sign) we want to evaluate

∏

b,c(1 + (zb + zc)r) (mod p). But we have

a polynomial identity∏

b,c(1 + (T b + T c)r) ∈ Φp(T ) · Z[x] + M for some constant integer M , where

Φp(T ) = T p−1 + · · · + T + 1 denotes the pth cyclotomic polynomial. But note that Φp(z) = 0 iscongruent to Φp(1) = p modulo p, so

∏

b,c

(1 + (zb + zc)r)−∏

b,c

(1 + (1b + 1c)r)

is an integer divisible by p. The rest is easy.

Remark. Without using the Frobenius map, this second solution gives another proof that∏

(ωa +zb1 + · · ·+ zbm) is an integer congruent to something like mr + 1 modulo p (as long as p is odd, etc.).Can you reconcile this difference (e.g. is the use of Frobenius in the first solution equivalent to a stepof the second solution)?

Remark. The central idea behind the second solution is that if f(z) is an integer for some polynomialf ∈ Z[x], then it’s congruent to (the integer) f(1) modulo p. Can you generalize this, say when z isnot necessarily a root of unity?

10. [40] The sequences of real numbers {ai}∞i=1 and {bi}∞i=1 satisfy an+1 = (an−1 − 1)(bn + 1) andbn+1 = anbn−1 − 1 for n ≥ 2, with a1 = a2 = 2015 and b1 = b2 = 2013. Evaluate, with proof, theinfinite sum

∞∑

n=1

bn

(

1

an+1− 1

an+3

)

.

Answer: 1 + 12014·2015 OR 4058211

4058210 First note that an and bn are weakly increasing and tend to

infinity. In particular, an, bn /∈ {0,−1, 1} for all n.

For n ≥ 1, we have an+3 = (an+1 − 1)(bn+2 + 1) = (an+1 − 1)(an+1bn), so

bnan+3

=1

an+1(an+1 − 1)=

1

an+1 − 1− 1

an+1.

Therefore,

∞∑

n=1

bnan+1

− bnan+3

=

∞∑

n=1

bnan+1

−(

1

an+1 − 1− 1

an+1

)

=

∞∑

n=1

bn + 1

an+1− 1

an+1 − 1.

Team

Furthermore, bn + 1 = an+1

an−1−1 for n ≥ 2. So the sum over n ≥ 2 is

∞∑

n=2

(

1

an−1 − 1− 1

an+1 − 1

)

= limN→∞

N∑

n=2

(

1

an−1 − 1− 1

an+1 − 1

)

=1

a1 − 1+

1

a2 − 1− lim

N→∞

(

1

aN − 1+

1

aN+1 − 1

)

=1

a1 − 1+

1

a2 − 1.

Hence the final answer is(

b1 + 1

a2− 1

a2 − 1

)

+

(

1

a1 − 1+

1

a2 − 1

)

.

Cancelling the common terms and putting in our starting values, this equals

2014

2015+

1

2014= 1− 1

2015+

1

2014= 1 +

1

2014 · 2015 .

Team

HMMT February 2015Saturday 21 February 2015

Algebra

1. Let Q be a polynomialQ(x) = a0 + a1x+ · · ·+ anx

n,

where a0, . . . , an are nonnegative integers. Given that Q(1) = 4 and Q(5) = 152, find Q(6).

2. The fraction 1

2015has a unique “(restricted) partial fraction decomposition” of the form

1

2015=

a

5+

b

13+

c

31,

where a, b, c are integers with 0 ≤ a < 5 and 0 ≤ b < 13. Find a+ b.

3. Let p be a real number and c 6= 0 an integer such that

c− 0.1 < xp

(

1− (1 + x)10

1 + (1 + x)10

)

< c+ 0.1

for all (positive) real numbers x with 0 < x < 10−100. (The exact value 10−100 is not important. Youcould replace it with any “sufficiently small number”.)

Find the ordered pair (p, c).

4. Compute the number of sequences of integers (a1, . . . , a200) such that the following conditions hold.

• 0 ≤ a1 < a2 < · · · < a200 ≤ 202.

• There exists a positive integer N with the following property: for every index i ∈ {1, . . . , 200}there exists an index j ∈ {1, . . . , 200} such that ai + aj −N is divisible by 203.

5. Let a, b, c be positive real numbers such that a+b+c = 10 and ab+bc+ca = 25. Letm = min{ab, bc, ca}.Find the largest possible value of m.

6. Let a, b, c, d, e be nonnegative integers such that 625a + 250b + 100c + 40d + 16e = 153. What is themaximum possible value of a+ b+ c+ d+ e?

7. Suppose (a1, a2, a3, a4) is a 4-term sequence of real numbers satisfying the following two conditions:

• a3 = a2 + a1 and a4 = a3 + a2;

• there exist real numbers a, b, c such that

an2 + bn+ c = cos(an)

for all n ∈ {1, 2, 3, 4}.

Compute the maximum possible value of

cos(a1)− cos(a4)

over all such sequences (a1, a2, a3, a4).

8. Find the number of ordered pairs of integers (a, b) ∈ {1, 2, . . . , 35}2 (not necessarily distinct) such thatax+ b is a “quadratic residue modulo x2 + 1 and 35”, i.e. there exists a polynomial f(x) with integercoefficients such that either of the following equivalent conditions holds:

• there exist polynomials P,Q with integer coefficients such that f(x)2 − (ax+ b) = (x2 +1)P (x)+35Q(x);

• or more conceptually, the remainder when (the polynomial) f(x)2 − (ax + b) is divided by (thepolynomial) x2 + 1 is a polynomial with (integer) coefficients all divisible by 35.

9. Let N = 302015. Find the number of ordered 4-tuples of integers (A,B,C,D) ∈ {1, 2, . . . , N}4 (notnecessarily distinct) such that for every integer n, An3 +Bn2 + 2Cn+D is divisible by N .

10. Find all ordered 4-tuples of integers (a, b, c, d) (not necessarily distinct) satisfying the following systemof equations:

a2 − b2 − c2 − d2 = c− b− 2

2ab = a− d− 32

2ac = 28− a− d

2ad = b+ c+ 31.

HMMT February 2015Saturday 21 February 2015

Algebra

1. Let Q be a polynomialQ(x) = a0 + a1x+ · · ·+ anx

n,

where a0, . . . , an are nonnegative integers. Given that Q(1) = 4 and Q(5) = 152, find Q(6).

Answer: 254 Since each ai is a nonnegative integer, 152 = Q(5) ≡ a0 (mod 5) and Q(1) = 4 =⇒ai ≤ 4 for each i. Thus, a0 = 2. Also, since 54 > 152 = Q(5), a4, a5, . . . , an = 0.

Now we simply need to solve the system of equations

5a1 + 52a22 + 53a33 = 150

a1 + a2 + a3 = 2

to get

a2 + 6a3 = 7.

Since a2 and a3 are nonnegative integers, a2 = 1, a3 = 1, and a1 = 0. Therefore, Q(6) = 63 +62 +2 =254.

2. The fraction 12015 has a unique “(restricted) partial fraction decomposition” of the form

1

2015=

a

5+

b

13+

c

31,

where a, b, c are integers with 0 ≤ a < 5 and 0 ≤ b < 13. Find a+ b.

Answer: 14 This is equivalent to 1 = 13 · 31a+5 · 31b+5 · 13c.1 Taking modulo 5 gives 1 ≡ 3 · 1a(mod 5), so a ≡ 2 (mod 5). Taking modulo 13 gives 1 ≡ 5 · 5b = 25b ≡ −b (mod 13), so b ≡ 12(mod 13). The size constraints on a, b give a = 2, b = 12, so a+ b = 14.

Remark. This problem illustrates the analogy between polynomials and integers, with prime powers(here 51, 131, 311) taking the role of powers of irreducible polynomials (such as (x − 1)1 or (x2 + 1)3,when working with polynomials over the real numbers).

Remark. The “partial fraction decomposition” needs to be restricted since it’s only unique “modulo1”. Abstractly, the abelian group (or Z-module) Q/Z has a “prime power direct sum decomposition”(more or less equivalent to Bezout’s identity, or the Chinese remainder theorem), but Q itself (as anabelian group under addition) does not.

You may wonder whether there’s a similar “prime power decomposition” of Q that accounts not justfor addition, but also for multiplication (i.e. the full ring structure of the rationals). In some sense,the “adeles/ideles” serve this purpose, but it’s not as clean as the partial fraction decomposition (foradditive structure alone)—in fact, the subtlety of adeles/ideles reflects much of the difficulty in numbertheory!

3. Let p be a real number and c 6= 0 an integer such that

c− 0.1 < xp

(

1− (1 + x)10

1 + (1 + x)10

)

< c+ 0.1

for all (positive) real numbers x with 0 < x < 10−100. (The exact value 10−100 is not important. Youcould replace it with any “sufficiently small number”.)

Find the ordered pair (p, c).

1Note that this does actually have integer solutions by Bezout’s identity, as gcd(13 · 31, 5 · 31, 5 · 13) = 1.

Algebra

Answer: (−1,−5) This is essentially a problem about limits, but phrased concretely in terms of

“small numbers” (like 0.1 and 10−100).

We are essentially studying the rational function f(x) := 1−(1+x)10

1+(1+x)10 = −10x+O(x2)2+O(x) , where the “big-O”

notation simply make precise the notion of “error terms”.2

Intuitively, f(x) ≈ −10x2 = −5x for “small nonzero x”. (We could easily make this more precise if

we wanted to, by specifying the error terms more carefully, but it’s not so important.) So g(x) :=xpf(x) ≈ −5xp+1 for “small nonzero x”.

• If p + 1 > 0, g will approach 0 (“get very small”) as x approaches 0 (often denoted x → 0), sothere’s no way it can stay above the lower bound c− 0.1 for all small nonzero x.

• If p+ 1 < 0, g will approach −∞ (“get very large in the negative direction”) as x → 0, so there’sno way it can stay below the upper bound c+ 0.1 for all small nonzero x.

• If p + 1 = 0, g ≈ −5 becomes approximately constant as x → 0. Since c is an integer, we musthave c = −5 (as −5 is the only integer within 0.1 of −5).

Remark. Why does (p, c) = (−1,−5) actually satisfy the inequality? This is where the 10−100 kicksin: for such small values of x, the “error” |g(x) − (−5)| of the approximation g ≈ −5 does actuallylie within the permitted threshold of ±0.1. (You can easily work out the details yourself, if you’reinterested. It’s something you might want to work out once or twice in your life, but rational functionsare “well-behaved” enough that we can usually rely on our intuition in these kinds of scenarios.)

4. Compute the number of sequences of integers (a1, . . . , a200) such that the following conditions hold.

• 0 ≤ a1 < a2 < · · · < a200 ≤ 202.

• There exists a positive integer N with the following property: for every index i ∈ {1, . . . , 200}there exists an index j ∈ {1, . . . , 200} such that ai + aj −N is divisible by 203.

Answer: 20503 Let m := 203 be an integer not divisible by 3. We’ll show the answer for generalsuch m is m⌈m−1

2 ⌉.Let x, y, z be the three excluded residues. Then N works if and only if {x, y, z} ≡ {N−x,N−y,N−z}(mod m). Since x, y, z (mod m) has opposite orientation as N − x,N − y,N − z (mod m), this isequivalent to x, y, z forming an arithmetic progression (in some order) modulo m centered at one ofx, y, z (or algebraically, one of N ≡ 2x ≡ y + z, N ≡ 2y ≡ z + x, N ≡ 2z ≡ x+ y holds, respectively).

Since 3 ∤ m, it’s impossible for more than one of these congruences to hold (or else x, y, z would have tobe equally spaced modulo m, i.e. x−y ≡ y−z ≡ z−x). So the number of distinct 3-sets correspondingto arithmetic progressions is m⌈m−1

2 ⌉ (choose a center and a difference, noting that ±d give the samearithmetic progression). Since our specific m = 203 is odd this gives mm−1

2 = 203 · 101 = 20503.

Remark. This problem is a discrete analog of certain so-called Frieze patterns. (See also Chapter 6,Exercise 5.8 of Artin’s Algebra textbook.)

5. Let a, b, c be positive real numbers such that a+b+c = 10 and ab+bc+ca = 25. Letm = min{ab, bc, ca}.Find the largest possible value of m.

Answer: 259 Without loss of generality, we assume that c ≥ b ≥ a. We see that 3c ≥ a+b+c = 10.

Therefore, c ≥ 103 .

2For instance, the O(x2) refers to a function bounded by C|x|2 for some positive constant C, whenever x is close enough to0 (and as the 10−100 suggests, that’s all we care about).

Algebra

Since

0 ≤ (a− b)2

= (a+ b)2 − 4ab

= (10− c)2 − 4 (25− c(a+ b))

= (10− c)2 − 4 (25− c(10− c))

= c(20− 3c),

we obtain c ≤ 203 . Considerm = min{ab, bc, ca} = ab, as bc ≥ ca ≥ ab. We compute ab = 25−c(a+b) =

25− c(10− c) = (c− 5)2. Since 103 ≤ c ≤ 20

3 , we get that ab ≤ 259 . Therefore, m ≤ 25

9 in all cases andthe equality can be obtained when (a, b, c) = ( 53 ,

53 ,

203 ).

6. Let a, b, c, d, e be nonnegative integers such that 625a + 250b + 100c + 40d + 16e = 153. What is themaximum possible value of a+ b+ c+ d+ e?

Answer: 153 The intuition is that as much should be in e as possible. But divisibility obstructionslike 16 ∤ 153 are in our way. However, the way the coefficients 54 > 53 · 2 > · · · are set up, we can atleast easily avoid having a, b, c, d too large (speifically, ≥ 2). This is formalized below.

First, we observe that (a1, a2, a3, a4, a5) = (5, 1, 0, 0, 0) is a solution. Then given a solution, replacing(ai, ai+1) with (ai − 2, ai+1 + 5), where 1 ≤ i ≤ 4, also yields a solution. Given a solution, it turns outall solutions can be achieved by some combination of these swaps (or inverses of these swaps).

Thus, to optimize the sum, we want (a, b, c, d) ∈ {0, 1}4, since in this situation, there would be no wayto make swaps to increase the sum. So the sequence of swaps looks like (5, 1, 0, 0, 0) → (1, 11, 0, 0, 0) →(1, 1, 25, 0, 0) → (1, 1, 1, 60, 0) → (1, 1, 1, 0, 150), yielding a sum of 1 + 1 + 1 + 0 + 150 = 153.

Why is this optimal? Suppose (a, b, c, d, e) maximizes a + b + c + d + e. Then a, b, c, d ≤ 1, or elsewe could use a replacement (ai, ai+1) → (ai − 2, ai+1 + 5) to strictly increase the sum. But modulo 2forces a odd, so a = 1. Subtracting off and continuing in this manner3 shows that we must have b = 1,then c = 1, then d = 0, and finally e = 150.

Remark. The answer is coincidentally obtained by dropping the exponent of 153 into the one’s place.

7. Suppose (a1, a2, a3, a4) is a 4-term sequence of real numbers satisfying the following two conditions:

• a3 = a2 + a1 and a4 = a3 + a2;

• there exist real numbers a, b, c such that

an2 + bn+ c = cos(an)

for all n ∈ {1, 2, 3, 4}.

Compute the maximum possible value of

cos(a1)− cos(a4)

over all such sequences (a1, a2, a3, a4).

Answer: −9 + 3√13 Let f(n) = cos an and m = 1. The second (“quadratic interpolation”)

condition on f(m), f(m + 1), f(m + 2), f(m + 3) is equivalent to having a vanishing third finitedifference

f(m+ 3)− 3f(m+ 2) + 3f(m+ 1)− f(m) = 0.

3This is analogous to the “number theoretic” proof of the uniqueness of the base 2 expansion of a nonnegative integer.

Algebra

This is equivalent to

f(m+ 3)− f(m) = 3 [f(m+ 2)− f(m+ 1)]

⇐⇒ cos(am+3)− cos(am) = 3 (cos(am+2)− cos(am+1))

= −6 sin

(

am+2 + am+1

2

)

sin

(

am+2 − am+1

2

)

= −6 sin(am+3

2

)

sin(am

2

)

.

Set x = sin(am+3

2

)

and y = sin(

am

2

)

. Then the above rearranges to

(1− 2x2)− (1− 2y2) = −6xy ⇐⇒ x2 − y2 = 3xy.

Solving gives y = x−3±√13

2 . The expression we are trying to maximize is 2(x2− y2) = 6xy, so we want

x, y to have the same sign; thus y = x−3+√13

2 .

Then |y| ≤ |x|, so since |x|, |y| ≤ 1, to maximize 6xy we can simply set x = 1, for a maximal value of

6 · −3+√13

2 = −9 + 3√13.

8. Find the number of ordered pairs of integers (a, b) ∈ {1, 2, . . . , 35}2 (not necessarily distinct) such thatax+ b is a “quadratic residue modulo x2 + 1 and 35”, i.e. there exists a polynomial f(x) with integercoefficients such that either of the following equivalent conditions holds:

• there exist polynomials P,Q with integer coefficients such that f(x)2 − (ax+ b) = (x2 +1)P (x)+35Q(x);

• or more conceptually, the remainder when (the polynomial) f(x)2 − (ax + b) is divided by (thepolynomial) x2 + 1 is a polynomial with (integer) coefficients all divisible by 35.

Answer: 225 By the Chinese remainder theorem, we want the product of the answers modulo 5and modulo 7 (i.e. when 35 is replaced by 5 and 7, respectively).

First we do the modulo 7 case. Since x2 +1 is irreducible modulo 7 (or more conceptually, in F7[x]),exactly half of the nonzero residues modulo x2 + 1 and 7 (or just modulo x2 + 1 if we’re working in

F7[x]) are quadratic residues, i.e. our answer is 1 + 72−12 = 25 (where we add back one for the zero

polynomial).

Now we do the modulo 5 case. Since x2+1 factors as (x+2)(x−2) modulo 5 (or more conceptually,in F5[x]), by the polynomial Chinese remainder theorem modulo x2 + 1 (working in F5[x]), we wantthe product of the number of polynomial quadratic residues modulo x± 2. By centering/evaluatingpolynomials at ∓2 accordingly, the polynomial squares modulo these linear polynomials are just thosereducing to integer squares modulo 5.4 So we have an answer of (1 + 5−1

2 )2 = 9 in this case.

Our final answer is thus 25 · 9 = 225.

Remark. This problem illustrates the analogy between integers and polynomials (specifically here,polynomials over the finite field of integers modulo 5 or 7), with x2 + 1 (mod 7) or x ± 2 (mod 5)taking the role of a prime number. Indeed, just as in the integer case, we expect exactly half of the(coprime) residues to be (coprime, esp. nonzero) quadratic residues.

9. Let N = 302015. Find the number of ordered 4-tuples of integers (A,B,C,D) ∈ {1, 2, . . . , N}4 (notnecessarily distinct) such that for every integer n, An3 +Bn2 + 2Cn+D is divisible by N .

Answer: 24 Note that n0 =(

n0

)

, n1 =(

n1

)

, n2 = 2(

n2

)

+(

n1

)

, n3 = 6(

n3

)

+6(

n2

)

+(

n1

)

(generally seehttp://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind). Thus the polynomialrewrites as

6A

(

n

3

)

+ (6A+ 2B)

(

n

2

)

+ (A+B + 2C)

(

n

1

)

+D

(

n

0

)

,

4This is more explicit than necessary. By the same reasoning as in the previous paragraph, we can abstractly count 1+ 51−1

2

quadratic residues modulo x±2 (irreducible polynomials in F5[x]) each (and then multiply/square to get the answer for x2+1).

Algebra

which by the classification of integer-valued polynomials is divisible by N always if and only if 6A, 6A+2B,A+B + 2C,D are always divisible by N .

We can eliminate B and (trivially) D from the system: it’s equivalent to the system 6A ≡ 0 (mod N),4A− 4C ≡ 0 (mod N), B ≡ −A− 2C (mod N), D ≡ 0 (mod N). So we want 12 times the number of(A,C) with A ≡ 0 (mod N/6), C ≡ A (mod N/4). So there are N/(N/6) = 6 choices for A, and thengiven such a choice of A there are N/(N/4) = 4 choices for C. So we have 6 ·4 ·12 = 24 solutions total.

10. Find all ordered 4-tuples of integers (a, b, c, d) (not necessarily distinct) satisfying the following systemof equations:

a2 − b2 − c2 − d2 = c− b− 2

2ab = a− d− 32

2ac = 28− a− d

2ad = b+ c+ 31.

Answer: (5,−3, 2, 3) We first give two systematic solutions using standard manipulations and

divisibility conditions (with some casework), and then a third solution using quaternionic numbertheory (not very practical, so mostly for your cultural benefit).

Solution 1. Subtract the second equation from the third to get a(c − b + 1) = 30. Add the secondand third to get 2a(b+ c) = −4− 2d. Substitute into the fourth to get

2a(2ad− 31) = −4− 2d ⇐⇒ a(31− 2ad) = 2 + d ⇐⇒ d =31a− 2

2a2 + 1,

which in particular gives a 6≡ 1 (mod 3). Then plugging in a factor of 30 for a gives us the system ofequations b + c = 2ad − 31 and c − b + 1 = 30/a in b, c. Here, observe that b + c is odd, so c − b + 1is even. Thus a must be odd (and from earlier a 6≡ 1 (mod 3)), so a ∈ {−1,±3, 5,±15}. Manuallychecking these, we see that the only possibilities we need to check are (a, d) = (5, 3), (−1,−11), (−3,−5),corresponding to (b, c) = (−3, 2), (11,−20), (5,−6). Then check the three candidates against firstcondition a2 − b2 − c2 − d2 = c− b− 2 to find our only solution (a, b, c, d) = (5,−3, 2, 3).

Solution 2. Here’s an alternative casework solution. From 2ad = b + c + 31, we have that b + c isodd. So, b and c has different parity. Thus, b2 + c2 ≡ 1 (mod 4). Plugging this into the first equation,we get that a and d also have the same parity.

So, a2 − b2 − c2 − d2 ≡ −1 (mod 4). Thus, c− b− 2 ≡ −1 (mod 4). So, c ≡ b+ 1 (mod 4).

From taking modulo a in the second and third equation, we have a | d+ 32 and a | 28− d. So, a | 60.Now, if a is even, let a = 2k and d = 2m. Plugging this in the second and third equation, we get2kc = 14− k −m and 2kb = k −m− 16. So, k(c− b) = 15− k.

We can see that k 6= 0. Therefore, c− b = 15−kk

= 15k− 1.

But c− b ≡ 1 (mod 4). So, 15k− 1 ≡ 1 (mod 4), or 15

k≡ 2 (mod 4) which leads to a contradiction.

So, a is odd. And we have a | 60. So, a | 15. This gives us 8 easy possibilities to check...

Solution 3. The left hand sides clue us in to the fact that this problem is secretly about quaternions.Indeed, we see that letting z = a+ bi+ cj + dk gives

(z − i+ j)z = −2− 32i+ 28j + 31k.

Taking norms gives N(z−i+j)N(z) = 22+322+282+312 = 2773 = 47 ·59. By the triangle inequality,N(z), N(z − i+ j) aren’t too far apart, so they must be 47, 59 (in some order).

Thus z, z − i + j are Hurwitz primes.5 We rely on the following foundational lemma in quaternionnumber theory:

5For the purposes of quaternion number theory, it’s simpler to work in the the Hurwitz quaternions H = 〈i, j, k, 1+i+j+k

2〉Z,

which has a left- (or right-) division algorithm, left- (resp. right-) Euclidean algorithm, is a left- (resp. right-) principal idealdomain, etc. There’s no corresponding division algorithms when we’re working with the Lipschitz quaternions, i.e. those withinteger coordinates.

Algebra

Lemma. Let p ∈ Z be an integer prime, and A a Hurwitz quaternion. If p | N(A), then the HA+Hp(a left ideal, hence principal) has all element norms divisible by p, hence is nontrivial. (So it’s eitherHp or of the form HP for some Hurwitz prime P .)

In our case, it will suffice to apply the lemma for A = −2−32i+28j+31k at primes p = 47 and q = 59to get factorizations (unique up to suitable left/right unit multiplication) A = QP and A = P ′Q′

(respectively), with P, P ′ Hurwitz primes of norm p, and Q,Q′ Hurwitz primes of norm q. Indeed,these factorizations come from HA+Hp = HP and HA+Hq = HQ′.

We compute by the Euclidean algorithm:

HA+H(47) = H(−2− 32i+ 28j + 31k) +H(47)

= H(−2 + 15i− 19j − 16k) +H(47)

= [H(47 · 18) +H(47)(−2− 15i+ 19j + 16k)]−2 + 15i− 19j − 16k

47 · 18= [H18 +H(−2 + 3i+ j − 2k)]

−2 + 15i− 19j − 16k

18

= H(−2 + 3i+ j − 2k)−2 + 15i− 19j − 16k

18

= H−54− 90i+ 54j − 36k

18= H(−3− 5i+ 3j − 2k).

Thus6 there’s a unit7 ǫ such that P = ǫ(−3− 5i+ 3j − 2k).

Similarly, to get P ′, we compute

AH+ 47H = (−2− 32i+ 28j + 31k)H+ 47H

= (−2 + 15i− 19j − 16k)H+ 47H

=−2 + 15i− 19j − 16k

47 · 18 [(47 · 18)H+ 47(−2− 15i+ 19j + 16k)H]

=−2 + 15i− 19j − 16k

18[18H+ (−2 + 3i+ j − 2k)H]

=−2 + 15i− 19j − 16k

18(−2 + 3i+ j − 2k)H

=−54 + 18i+ 18j + 108k

18H

= (−3 + i+ j + 6k)H,

so there’s a unit ǫ′ with P ′ = (−3 + i+ j + 6k)ǫ′.

Finally, we have either z = ǫ(−3− 5i+ 3j − 2k) for some ǫ, or z − i+ j = (−3 + i+ j + 6k)ǫ′ for someǫ′. Checking the 24 + 24 cases (many of which don’t have integer coefficients, and can be ruled outimmediately) gives z = iP = 5− 3i+ 2j + 3k as the only possibility.

Remark. We have presented the most conceptual proof possible. It’s also possible to directly computebased on the norms only, and do some casework. For example, since 47 ≡ 3 (mod 4), it’s easy tocheck the only ways to write it as a sum of four squares are (±5)2 + (±3)2 + (±3)2 + (±2)2 and(±3)2 + (±1)2 + (±1)2 + (±6)2.

Remark. For a systematic treatment of quaternions (including the number theory used above), onegood resource is On Quaternions and Octonions: Their Geometry, Arithmetic, and Symmetry by JohnH. Conway and Derek A. Smith. A more focused treatment is the expository paper Factorization of

Hurwitz Quaternions by Boyd Coan and Cherng-tiao Perng.

For an example of interesting research in this rather exotic area, see the Metacommutation of Hurwitz

primes paper by Henry Cohn and Abhinav Kumar.

6Hidden computations: we’ve used 47 · 18 = 846 = 22 + 152 + 192 + 162, and 18 = N(−2 + 3i+ j − 2k).7i.e. one of ±1,±i,±j,±k,

±1±i±j±k

2

Algebra

HMMT February 2015Saturday 21 February 2015

Combinatorics

1. Evan’s analog clock displays the time 12:13; the number of seconds is not shown. After 10 secondselapse, it is still 12:13. What is the expected number of seconds until 12:14?

2. Victor has a drawer with 6 socks of 3 different types: 2 complex socks, 2 synthetic socks, and 2trigonometric socks. He repeatedly draws 2 socks at a time from the drawer at random, and stops ifthe socks are of the same type. However, Victor is “synthetic-complex type-blind”, so he also stops ifhe sees a synthetic and a complex sock.

What is the probability that Victor stops with 2 socks of the same type? Assume Victor returns bothsocks to the drawer after each step.

3. Starting with the number 0, Casey performs an infinite sequence of moves as follows: he chooses anumber from {1, 2} at random (each with probability 1

2) and adds it to the current number. Let pm

be the probability that Casey ever reaches the number m. Find p20 − p15.

4. Alice Czarina is bored and is playing a game with a pile of rocks. The pile initially contains 2015 rocks.At each round, if the pile has N rocks, she removes k of them, where 1 ≤ k ≤ N , with each possiblek having equal probability. Alice Czarina continues until there are no more rocks in the pile. Let p bethe probability that the number of rocks left in the pile after each round is a multiple of 5. If p is ofthe form 5a · 31b · c

d, where a, b are integers and c, d are positive integers relatively prime to 5 · 31, find

a+ b.

5. For positive integers x, let g(x) be the number of blocks of consecutive 1’s in the binary expansion ofx. For example, g(19) = 2 because 19 = 100112 has a block of one 1 at the beginning and a blockof two 1’s at the end, and g(7) = 1 because 7 = 1112 only has a single block of three 1’s. Computeg(1) + g(2) + g(3) + · · ·+ g(256).

6. Count the number of functions f : Z → {‘green’, ‘blue’} such that f(x) = f(x + 22) for all integers x

and there does not exist an integer y with f(y) = f(y + 2) = ‘green’.

7. 2015 people sit down at a restaurant. Each person orders a soup with probability 1

2. Independently,

each person orders a salad with probability 1

2. What is the probability that the number of people who

ordered a soup is exactly one more than the number of people who ordered a salad?

8. Let S be the set of all 3-digit numbers with all digits in the set {1, 2, 3, 4, 5, 6, 7} (so in particular, allthree digits are nonzero). For how many elements abc of S is it true that at least one of the (notnecessarily distinct) “digit cycles”

abc, bca, cab

is divisible by 7? (Here, abc denotes the number whose base 10 digits are a, b, and c in that order.)

9. Calvin has a bag containing 50 red balls, 50 blue balls, and 30 yellow balls. Given that after pullingout 65 balls at random (without replacement), he has pulled out 5 more red balls than blue balls, whatis the probability that the next ball he pulls out is red?

10. A group of friends, numbered 1, 2, 3, . . . , 16, take turns picking random numbers. Person 1 picks anumber uniformly (at random) in [0, 1], then person 2 picks a number uniformly (at random) in [0, 2],and so on, with person k picking a number uniformly (at random) in [0, k]. What is the probabilitythat the 16 numbers picked are strictly increasing?

HMMT February 2015Saturday 21 February 2015

Combinatorics

1. Evan’s analog clock displays the time 12:13; the number of seconds is not shown. After 10 secondselapse, it is still 12:13. What is the expected number of seconds until 12:14?

Answer: 25 At first, the time is uniformly distributed between 12:13:00 and 12:13:50. After 10seconds, the time is uniformly distributed between 12:13:10 and 12:14:00. Thus, it takes on average 25seconds to reach 12:14(:00).

2. Victor has a drawer with 6 socks of 3 different types: 2 complex socks, 2 synthetic socks, and 2trigonometric socks. He repeatedly draws 2 socks at a time from the drawer at random, and stops ifthe socks are of the same type. However, Victor is “synthetic-complex type-blind”, so he also stops ifhe sees a synthetic and a complex sock.

What is the probability that Victor stops with 2 socks of the same type? Assume Victor returns bothsocks to the drawer after each step.

Answer: 37 Let the socks be C1, C2, S1, S2, T1, T2, where C, S and T stand for complex, synthetic

and trigonometric respectively. The possible stopping points consist of three pairs of socks of the sametype plus four different complex-synthetic (C-S) pairs, for a total of 7. So the answer is 3

7 .

3. Starting with the number 0, Casey performs an infinite sequence of moves as follows: he chooses anumber from {1, 2} at random (each with probability 1

2 ) and adds it to the current number. Let pmbe the probability that Casey ever reaches the number m. Find p20 − p15.

Answer: 11220 We note that the only way n does not appear in the sequence is if n−1 and then n+1

appears. Hence, we have p0 = 1, and pn = 1− 12pn−1 for n > 0. This gives pn − 2

3 = − 12

(

pn−1 −23

)

,so that

pn =2

3+

1

3·

(

−1

2

)n

,

so p20 − p15 is just1− (−2)5

3 · 220=

11

220.

4. Alice Czarina is bored and is playing a game with a pile of rocks. The pile initially contains 2015 rocks.At each round, if the pile has N rocks, she removes k of them, where 1 ≤ k ≤ N , with each possiblek having equal probability. Alice Czarina continues until there are no more rocks in the pile. Let p bethe probability that the number of rocks left in the pile after each round is a multiple of 5. If p is ofthe form 5a · 31b · c

d , where a, b are integers and c, d are positive integers relatively prime to 5 · 31, finda+ b.

Answer: −501 We claim that p = 15

610

1115

1620 · · ·

20062010

20112015 . Let pn be the probability that, starting

with n rocks, the number of rocks left after each round is a multiple of 5. Indeed, using recursions wehave

p5k =p5k−5 + p5k−10 + · · ·+ p5 + p0

5k

for k ≥ 1. For k ≥ 2 we replace k with k − 1, giving us

p5k−5 =p5k−10 + p5k−15 + · · ·+ p5 + p0

5k − 5

=⇒ (5k − 5)p5k−5 = p5k−10 + p5k−15 + · · ·+ p5 + p0

Substituting this back into the first equation, we have

5kp5k = p5k−5 + (p5k−10 + p5k−15 + · · ·+ p5 + p0) = p5k−5 + (5k − 5)p5k−5,

Combinatorics

which gives p5k = 5k−45k p5k−5. Using this equation repeatedly along with the fact that p0 = 1 proves

the claim.

Now, the power of 5 in the denominator is v5(2015!) = 403 + 80 + 16 + 3 = 502, and 5 does not divideany term in the numerator. Hence a = −502. (The sum counts multiples of 5 plus multiples of 52 plusmultiples of 53 and so on; a multiple of 5n but not 5n+1 is counted exactly n times, as desired.)

Noting that 2015 = 31 ·65, we found that the numbers divisible by 31 in the numerator are those of theform 31 + 155k where 0 ≤ k ≤ 12, including 312 = 961; in the denominator they are of the form 155kwhere 1 ≤ k ≤ 13. Hence b = (13 + 1) − 13 = 1 where the extra 1 comes from 312 in the numerator.Thus a+ b = −501.

5. For positive integers x, let g(x) be the number of blocks of consecutive 1’s in the binary expansion ofx. For example, g(19) = 2 because 19 = 100112 has a block of one 1 at the beginning and a blockof two 1’s at the end, and g(7) = 1 because 7 = 1112 only has a single block of three 1’s. Computeg(1) + g(2) + g(3) + · · ·+ g(256).

Answer: 577 Solution 1. We prove that g(1)+ g(2)+ · · ·+ g(2n) = 1+2n−2(n+1) for all n ≥ 1,giving an answer of 1 + 26 · 9 = 577. First note that g(2n) = 1, and that we can view 0, 1, . . . , 2n − 1as n-digit binary sequences by appending leading zeros as necessary. (Then g(0) = 0.)

Then for 0 ≤ x ≤ 2n−1, x and 2n−x are complementary n-digit binary sequences (of 0’s and 1’s), withx’s strings of 1’s (0’s) corresponding to 2n−x’s strings of 0’s (resp. 1’s). It follows that g(x)+g(2n−x)is simply 1 more than the number of digit changes in x (or 2n − x), i.e. the total number of 01 and10 occurrences in x. Finally, because exactly half of all n-digit binary sequences have 0, 1 or 1, 0 atpositions k, k + 1 (for 1 ≤ k ≤ n− 1 fixed), we conclude that the average value of g(x) + g(2n − x) is1 + n−1

2 = n+12 , and thus that the total value of g(x) is 1

2 · 2n · n+12 = (n+ 1)2n−2, as desired.

Solution 2. We prove that g(1)+ g(2)+ · · ·+ g(2n − 1) = 2n−2(n+1). Identify each block of 1’s withits rightmost 1. Then it suffices to count the number of these “rightmost 1’s.” For each 1 ≤ k ≤ n− 1,the kth digit from the left is a rightmost 1 if and only if the k and k+1 digits from the left are 1 and 0respectively. Thus there are 2n−2 possible numbers. For the rightmost digit, it is a rightmost 1 if andonly if it is a 1, so there are 2n−1 possible numbers. Sum up we have: (n−1)2n−2+2n−1 = 2n−2(n+1),as desired.

Remark. We can also solve this problem using recursion or generating functions.

6. Count the number of functions f : Z → {‘green’, ‘blue’} such that f(x) = f(x + 22) for all integers xand there does not exist an integer y with f(y) = f(y + 2) = ‘green’.

Answer: 39601 It is clear that f is determined by f(0), . . . , f(21). The colors of the 11 evenintegers are independent of those of the odd integers because evens and odds are never exactly 2 apart.

First, we count the number of ways to “color” the even integers. f(0) can either be ‘green’ or ‘blue’.If f(0) is ‘green’, then f(2) = f(20) = ‘blue’. A valid coloring of the 8 other even integers correspondsbijectively to a string of 8 bits such that no two consecutive bits are 1. In general, the number of suchlength n strings is well known to be Fn+2 (indexed according to F0 = 0, F1 = 1, Fn+2 = Fn+1 + Fn),which can be proven by recursion. Therefore, the number of colorings of even integers in this case isF10 = 55.

If f(0) is ‘blue’, then a valid coloring of the 10 other even integers corresponds bijectively to a stringas above, of 10 bits. The number of colorings for this case is F12 = 144. The total number of coloringsof even integers is 55 + 144 = 199.

Using the same reasoning for coloring the odd integers, we see that the number of colorings of all ofthe integers is 1992 = 39601.

7. 2015 people sit down at a restaurant. Each person orders a soup with probability 12 . Independently,

each person orders a salad with probability 12 . What is the probability that the number of people who

ordered a soup is exactly one more than the number of people who ordered a salad?

Combinatorics

Answer:(40302016)24030 OR

(40302014)24030 OR

(40322016)−2(40302015)24031 Solution 1. Note that total soups = total salads + 1

is equivalent to total soups + total not-salads = 2016. So there are precisely(

2015+20152016

)

possibilities,

each occurring with probability (1/2)2015+2015. Thus our answer is(40302016)24030 .

Solution 2. To count the number of possibilities, we can directly evaluate the sum∑2014

i=0

(

2015i

)(

2015i+1

)

.

One way is to note(

2015i+1

)

=(

20152014−i

)

, and finish with Vandermonde’s identity:∑2014

i=0

(

2015i

)(

20152014−i

)

=(

2015+20152014

)

=(

40302014

)

(which also equals(

40302016

)

).

(We could have also used(

2015i

)

=(

20152015−i

)

to get∑2014

i=0

(

20152015−i

)(

2015i+1

)

=(

2015+20152016

)

directly, which iscloser in the spirit of the previous solution.)

Solution 3 (sketch). It’s also possible to get a handle on∑2014

i=0

(

2015i

)(

2015i+1

)

by squaring Pascal’s

identity(

2015i

)

+(

2015i+1

)

=(

2016i+1

)

and summing over 0 ≤ i ≤ 2014. This gives an answer of(40322016)−2(40302015)

24031 ,

which can be simplified by noting(

40322016

)

= 40322016

(

40312015

)

, and then applying Pascal’s identity.

8. Let S be the set of all 3-digit numbers with all digits in the set {1, 2, 3, 4, 5, 6, 7} (so in particular, allthree digits are nonzero). For how many elements abc of S is it true that at least one of the (notnecessarily distinct) “digit cycles”

abc, bca, cab

is divisible by 7? (Here, abc denotes the number whose base 10 digits are a, b, and c in that order.)

Answer: 127 Since the value of each digit is restricted to {1, 2, . . . , 7}, there is exactly one digitrepresentative of each residue class modulo 7.

Note that 7 | abc if and only if 100a+10b+ c ≡ 0 (mod 7) or equivalently 2a+3b+ c ≡ 0. So we wantthe number of triples of residues (a, b, c) such that at least one of 2a + 3b + c ≡ 0, 2b + 3c + a ≡ 0,2c+ 3a+ b ≡ 0 holds.

Let the solution sets of these three equations be S1, S2, S3 respectively, so by PIE and cyclic symmetrywe want to find 3|S1| − 3|S1 ∩ S2|+ |S1 ∩ S2 ∩ S3|.

Clearly |S1| = 72, since for each of a and b there is a unique c that satisfies the equation.

For S1∩S2, we may eliminate a to get the system 0 ≡ 2(2b+3c)− (3b+c) = b+5c (and a ≡ −2b−3c),which has 7 solutions (one for each choice of c).

For S1 ∩ S2 ∩ S3 ⊆ S1 ∩ S2, we have from the previous paragraph that b ≡ −5c and a ≡ 10c− 3c ≡ 0.By cyclic symmetry, b, c ≡ 0 as well, so there’s exactly 1 solution in this case.

Thus the answer is 3 · 72 − 3 · 7 + 1 = 127.

9. Calvin has a bag containing 50 red balls, 50 blue balls, and 30 yellow balls. Given that after pullingout 65 balls at random (without replacement), he has pulled out 5 more red balls than blue balls, whatis the probability that the next ball he pulls out is red?

Answer: 926 Solution 1. The only information this gives us about the number of yellow balls

left is that it is even. A bijection shows that the probability that there are k yellow balls left is equalto the probability that there are 30− k yellow balls left (flip the colors of the red and blue balls, andthen switch the 65 balls that have been picked with the 65 balls that have not been picked). So theexpected number of yellow balls left is 15. Therefore the expected number of red balls left is 22.5. Sothe answer is 22.5

65 = 45130 = 9

26 .

Solution 2. Let w(b) =(

50b

)(

50r=b+5

)(

3060−2b

)

be the number of possibilities in which b blue balls havebeen drawn (precisely 15 ≤ b ≤ 30 are possible). For fixed b, the probability of drawing red next is

50−r50+50+30−65 = 45−b

65 . So we want to evaluate

∑30b=15 w(b)

45−b65

∑30b=15 w(b)

.

Combinatorics

Note the symmetry of weights:

w(45− b) =

(

50

45− b

)(

50

50− b

)(

30

2b− 30

)

=

(

50

b+ 5

)(

50

b

)(

30

60− 2b

)

,

so the 45−b65 averages out with 45−(45−b)

65 to give a final answer of 45/265 = 9

26 .

Remark. If one looks closely enough, the two approaches are not so different. The second solutionmay be more conceptually/symmetrically phrased in terms of the number of yellow balls.

10. A group of friends, numbered 1, 2, 3, . . . , 16, take turns picking random numbers. Person 1 picks anumber uniformly (at random) in [0, 1], then person 2 picks a number uniformly (at random) in [0, 2],and so on, with person k picking a number uniformly (at random) in [0, k]. What is the probabilitythat the 16 numbers picked are strictly increasing?

Answer: 1715

16!2 Solution 1 (intuitive sketch). If person i picks ai, this is basically a continuous

version of Catalan paths (always y ≤ x) from (0, 0) to (17, 17), with “up-right corners” at the (i, ai).

A cyclic shifts argument shows that “ 117” of the increasing sequences (x1, . . . , x16) in [0, 17]16 work

(i.e. have xi ∈ [0, i] for all i), so contribute volume1 117

1716

16! . Explicitly, the cyclic shift we’re using is

TC : (x1, . . . , x16) 7→ (x2 − x1, . . . , x16 − x1, C − x1)

for C = 17 (though it’s the same for any C > 0), which sends increasing sequences in [0, C]16 toincreasing sequences in [0, C]16. The “ 1

17” essentially follows from the fact that T has period 17, andalmost every2 T -orbit (of 17 (increasing) sequences) contains exactly 1 working sequence.3

But to be more rigorous, we still need some more justification.4

The volume contribution of permitted sequences (i.e. ai ∈ [0, i] for all i; those under consideration inthe first place) (a1, . . . , a16) ∈ [0, 17]16 is 16!, so based on the previous paragraph, our final probability

is 1715

16!2 .

Solution 2. Here we present a discrete version of the previous solution.

To do this, we consider several related events.

Let X be a 16-tuple chosen uniformly and randomly from [0, 17]16 (used to define events A,B,C). LetZ be a 16-tuple chosen uniformly and randomly from {1, 2, . . . , 17}16 (used to define event D).

• A is the event that X’s coordinates are ordered ascending;

• B is the event that X lies in the “box” [0, 1]× · · · × [0, 16];

• C is the event that whenX’s coordinates are sorted ascending to form Y (e.g. ifX = (1, 3.2, 3, 2, 5, 6, . . . , 16)then Y = (1, 2, 3, 3.2, 5, 6, . . . , 16)), Y lies in the box;

• D is the event that when Z’s coordinates are sorted ascending to form W , W lies in the afore-mentioned box. When Z satisfies this condition, Z is known as a parking function.

1If we wanted to be completely rigorous, we’d want to also show that the set of such sequences is measurable. However, oursets here are simple enough that this is not a major concern.

2i.e. “outside a set of measure 0”3Try to visualize this yourself! Compare the “exactly 1 of 17” with the discrete solutions below.4We’re using a symmetry/“bijection-of-integrals” argument here (to equate 17 integrals with sum 17

16

16!), so we need to be a

little careful with the bijections. In particular, we must make sure the cyclic shift

T : (x1, . . . , x16) 7→ (x2 − x1, . . . , x16 − x1, 17− x1)

is “measure-preserving” on R16. (An example of a non-measure-preserving bijection is t 7→ t2 on [0, 1].) The standard way

to check this is by taking a Jacobian (determinant); more conceptually, since T 17 = Id, and T is affine (so the Jacobian isconstant), we should be able to avoid direct computation by using the chain rule. Alternatively, more “combinatorially”, itshould also be possible to do so by properly discretizing our space.

Combinatorics

We want to find P (A|B) because given that X is in B, X has a uniform distribution in the box, justas in the problem. Now note

P (A|B) =P (A ∩B)

P (B)=

P (A ∩B)

P (A)

P (A)

P (B)= P (B|A)

P (A)

P (B).

C is invariant with respect to permutations, so 116! = P (A|C) = P (A∩C)

P (C) = P (A∩B)P (C) . Since P (A) = 1

16! ,

we have P (B|A) = P (A∩B)P (A) = P (C).

Furthermore, P (C) = P (D) because C only depends on the ceilings of the coordinates. So P (A|B) =

P (C)P (A)P (B) = P (D)P (A)

P (B) . (*)

Given a 16-tuple Z from {1, 2, . . . , 17}16, let Z + n (for integers n) be the 16-tuple formed by addingn to each coordinate and then reducing modulo 17 so that each coordinate lies in [1, 17].

Key claim (discrete analog of cyclic shifts argument). Exactly one of Z,Z + 1, . . . , Z + 16 is aparking function.

First, assuming this claim, it easily follows that P (D) = 117 . Substituting P (A) = 1

16! , P (B) = 16!1716

into (*) gives P (A|B) = 1715

16!2 . It now remains to prove the claim.

Proof. Consider the following process.

Begin with 17 parking spots around a circle, labelled 1 to 17 clockwise and all unoccupied. There are16 cars, 1 to 16, and they park one at a time, from 1 to 16. The ith car tries to park in the spot givenby the ith coordinate of Z. If this spot is occupied, that car parks in the closest unoccupied spot inthe clockwise direction. Because there are only 16 cars, each car will be able to park, and exactly onespot will be left.

Suppose that number 17 is left. For any integer n (1 ≤ n ≤ 16), the n cars that ended up parking inspots 1 through n must have corresponded to coordinates at most n. (If not, then the closest spot inthe clockwise direction would have to be before spot 17 and greater than n, a contradiction.) It followsthat the nth lowest coordinate is at most n and that when Z is sorted, it lies in the box.

Suppose now that D is true. For any integer n (1 ≤ n ≤ 16), the nth lowest coordinate is at most n,so there are (at least) n cars whose corresponding coordinates are at most n. At least one of these carsdoes not park in spots 1 through n− 1. Consider the first car to do so. It either parked in spot n, orskipped over it because spot n was occupied. Therefore spot n is occupied at the end. This is true forall n not equal to 17, so spot 17 is left.

It follows that Z is a parking function if and only if spot 17 is left. The same is true for Z+1 (assumingthat the process uses Z + 1 instead of Z), etc.

Observe that the process for Z + 1 is exactly that of Z, rotated by 1 spot clockwise. In particular, itsempty spot is one more than that of Z, (where 1 is one more than 17.) It follows that exactly one ofZ,Z + 1, . . . , Z + 16 leaves the spot 17, and that exactly one of these is a parking function. �

Solution 3. Suppose that person i picks a number in the interval [bi − 1, bi] where bi ≤ i. Then wehave the condition: b1 ≤ b2 ≤ · · · ≤ b16. Let ci be the number of bj ’s such that bj = i. Then, for eachadmissible sequence b1, b2, . . . , b16, there is the probability

1c1!c2!···c16!

that the problem condition holds,

since if ci numbers are picked uniformly and randomly in the interval [i− 1, i], then there is 1ci!

chanceof them being in an increasing order. Thus the answer we are looking for is

1

16!

∑

bi≤ib1≤···≤b16

1

c1!c2! · · · c16!=

1

16!2

∑

bi≤ib1≤···≤b16

(

c1 + · · ·+ c16c1, c2, . . . , c16

)

.

Thus it suffices to prove that∑

bi≤ib1≤···≤b16

(

c1 + · · ·+ c16c1, c2, . . . , c16

)

= 1715.

Combinatorics

The left hand side counts the number of 16-tuple such that the nth smallest entry is less than or equalto n. In other words, this counts the number of parking functions of length 16.5 Since the numberof parking functions of length n is 1

n+1 · (n + 1)n = (n + 1)n−1 (as proven for n = 16 in the previoussolution), we obtain the desired result.

5See http://www-math.mit.edu/~rstan/transparencies/parking.pdf.

Combinatorics

HMMT February 2015Saturday 21 February 2015

Geometry

1. Let R be the rectangle in the Cartesian plane with vertices at (0, 0), (2, 0), (2, 1), and (0, 1). R can bedivided into two unit squares, as shown.

Pro selects a point P uniformly at random in the interior of R. Find the probability that the linethrough P with slope 1

2 will pass through both unit squares.

2. Let ABC be a triangle with orthocenter H; suppose that AB = 13, BC = 14, CA = 15. Let GA bethe centroid of triangle HBC, and define GB , GC similarly. Determine the area of triangle GAGBGC .

3. Let ABCD be a quadrilateral with ∠BAD = ∠ABC = 90◦, and suppose AB = BC = 1, AD = 2.The circumcircle of ABC meets AD and BD at points E and F , respectively. If lines AF and CD

meet at K, compute EK.

4. Let ABCD be a cyclic quadrilateral with AB = 3, BC = 2, CD = 2, DA = 4. Let lines perpendicularto BC from B and C meet AD at B′ and C ′, respectively. Let lines perpendicular to AD from A and

D meet BC at A′ and D′, respectively. Compute the ratio [BCC′B′][DAA′D′] , where [̟] denotes the area of

figure ̟.

5. Let I be the set of points (x, y) in the Cartesian plane such that

x >

(

y4

9+ 2015

)1/4

Let f(r) denote the area of the intersection of I and the disk x2 + y2 ≤ r2 of radius r > 0 centered atthe origin (0, 0). Determine the minimum possible real number L such that f(r) < Lr2 for all r > 0.

6. In triangle ABC, AB = 2, AC = 1 +√5, and ∠CAB = 54◦. Suppose D lies on the extension of AC

through C such that CD =√5 − 1. If M is the midpoint of BD, determine the measure of ∠ACM ,

in degrees.

7. Let ABCDE be a square pyramid of height 12 with square base ABCD of side length AB = 12 (so

E is the vertex of the pyramid, and the foot of the altitude from E to ABCD is the center of squareABCD). The faces ADE and CDE meet at an acute angle of measure α (so that 0◦ < α < 90◦).Find tanα.

8. Let S be the set of discs D contained completely in the set {(x, y) : y < 0} (the region below thex-axis) and centered (at some point) on the curve y = x2 − 3

4 . What is the area of the union of theelements of S?

9. Let ABCD be a regular tetrahedron with side length 1. Let X be the point in triangle BCD suchthat [XBC] = 2[XBD] = 4[XCD], where [̟] denotes the area of figure ̟. Let Y lie on segment AXsuch that 2AY = Y X. Let M be the midpoint of BD. Let Z be a point on segment AM such thatthe lines Y Z and BC intersect at some point. Find AZ

ZM .

10. Let G be the set of all points (x, y) in the Cartesian plane such that 0 ≤ y ≤ 8 and

(x− 3)2 + 31 = (y − 4)2 + 8√

y(8− y).

There exists a unique line ℓ of negative slope tangent to G and passing through the point (0, 4).Suppose ℓ is tangent to G at a unique point P . Find the coordinates (α, β) of P .

HMMT February 2015Saturday 21 February 2015

Geometry

1. Let R be the rectangle in the Cartesian plane with vertices at (0, 0), (2, 0), (2, 1), and (0, 1). R can bedivided into two unit squares, as shown.

Pro selects a point P uniformly at random in the interior of R. Find the probability that the linethrough P with slope 1

2 will pass through both unit squares.

Answer: 34

Precisely the middle two (of four) regions satisfiy the problem conditions, and it’s easy to compute the(fraction of) areas as 3

4 .

2. Let ABC be a triangle with orthocenter H; suppose that AB = 13, BC = 14, CA = 15. Let GA bethe centroid of triangle HBC, and define GB , GC similarly. Determine the area of triangle GAGBGC .

Answer: 28/3 Let D,E, F be the midpoints of BC, CA, and AB, respectively. Then GAGBGC

is the DEF about H with a ratio of 23 , and DEF is the dilation of ABC about H with a ratio of − 1

2 ,

so GAGBGC is the dilation of ABC about H with ratio − 13 . Thus [GAGBGC ] =

[ABC]9 . By Heron’s

formula, the area of ABC is√21 · 8 · 7 · 6 = 84, so the area of GAGBGC is [ABC]/9 = 84/9 = 28/3.

3. Let ABCD be a quadrilateral with ∠BAD = ∠ABC = 90◦, and suppose AB = BC = 1, AD = 2.The circumcircle of ABC meets AD and BD at points E and F , respectively. If lines AF and CDmeet at K, compute EK.

Answer:√22 Assign coordinates such that B is the origin, A is (0, 1), and C is (1, 0). Clearly, E

is the point (1, 1). Since the circumcenter of ABC is ( 12 ,12 ), the equation of the circumcircle of ABC is

(x− 12 )

2+(y− 12 )

2 = 12 . Since line BD is given by x = 2y, we find that F is at ( 65 ,

35 ). The intersection

of AF with CD is therefore at ( 32 ,12 ), so K is the midpoint of CD. As a result, EK =

√22 .

This is in fact a special case of APMO 2013, Problem 5, when the quadrilateral is a square.

4. Let ABCD be a cyclic quadrilateral with AB = 3, BC = 2, CD = 2, DA = 4. Let lines perpendicularto BC from B and C meet AD at B′ and C ′, respectively. Let lines perpendicular to AD from A and

D meet BC at A′ and D′, respectively. Compute the ratio [BCC′B′][DAA′D′] , where [̟] denotes the area of

figure ̟.

Answer: 3776 To get a handle on the heights CB′, etc. perpendicular to BC and AD, let

X = BC ∩AD, which lies on ray−−→BC and

−−→AD since ÂB > ĈD (as chords AB > CD).

By similar triangles we have equality of ratios XC : XD : 2 = (XD + 4) : (XC + 2) : 3, so we have asystem of linear equations: 3XC = 2XD+8 and 3XD = 2XC +4, so 9XC = 6XD+24 = 4XC +32

gives XC = 325 and XD = 84/5

3 = 285 .

It’s easy to compute the trapezoid area ratio [BC′B′C][AA′D′D] =

BC(CB′+BC′)AD(AD′+DA′) = BC

ADXC+XBXA+XD (where we have

similar right triangles due to the common angle at X). This is just BCAD

XC+BC/2XD+AD/2 = 2

432/5+128/5+2 = 37

76 .

Geometry

5. Let I be the set of points (x, y) in the Cartesian plane such that

x >

(y4

9+ 2015

)1/4

Let f(r) denote the area of the intersection of I and the disk x2 + y2 ≤ r2 of radius r > 0 centered atthe origin (0, 0). Determine the minimum possible real number L such that f(r) < Lr2 for all r > 0.

Answer: π3 Let B(P, r) be the (closed) disc centered at P with radius r. Note that for all

(x, y) ∈ I, x > 0, and x >(

y4

9 + 2015)1/4

> |y|√3. Let I ′ = {(x, y) : x

√3 > |y|}. Then I ⊆ I ′ and the

intersection of I ′ with B((0, 0), r) is π3 r

2, so the f(r) = the area of I ∩B((0, 0), r) is also less than π3 r

2.Thus L = π

3 works.

On the other hand, if x > |y|√3+7, then x > |y|√

3+7 >

((|y|9

)4

+ 74)1/4

>(

y4

9 + 2015)1/4

, which means

that if I ′′ = {(x, y) : (x−7)√3 > |y|}, then I ′′ ⊆ I ′. However, for r > 7, the area of I ′′∩B((7, 0), r−7)

is π3 (r − 7)2, and I ′′ ⊆ I, B((7, 0), r − 7) ⊆ B((0, 0), r), which means that f(r) > π

3 (r − 7)2 for allr > 7, from which it is not hard to see that L = π

3 is the minimum possible L.

Remark: The lines y = ±√3x are actually asymptotes for the graph of 9x4 − y4 = 2015. The bulk of

the problem generalizes to the curve |√3x|α − |y|α = C (for a positive real α > 0 and any real C); the

case α = 0 is the most familiar case of a hyperbola.

6. In triangle ABC, AB = 2, AC = 1 +√5, and ∠CAB = 54◦. Suppose D lies on the extension of AC

through C such that CD =√5 − 1. If M is the midpoint of BD, determine the measure of ∠ACM ,

in degrees.

Answer: 63 Let E be the midpoint of AD. EC =√5 + 1 −

√5 = 1, and EM = 1 by similar

triangles (ABD ∼ EMD). △ECM is isosceles, with m∠CEM = 54◦. Thus m∠ACM = m∠ECM =180−54

2 = 63◦.

7. Let ABCDE be a square pyramid of height 12 with square base ABCD of side length AB = 12 (so

E is the vertex of the pyramid, and the foot of the altitude from E to ABCD is the center of squareABCD). The faces ADE and CDE meet at an acute angle of measure α (so that 0◦ < α < 90◦).Find tanα.

Answer: 17144 Let X be the projection of A onto DE. Let b = AB = 12.

The key fact in this computation is that if Y is the projection of A onto face CDE, then the projectionof Y onto line DE coincides with the projection of A onto line DE (i.e. X as defined above). Wecompute AY = b√

b2+1by looking at the angle formed by the faces and the square base (via 1/2-b/2-

√b2 + 1/2 right triangle). Now we compute AX = 2[AED]/ED = b

√b2+1/2√

2b2+1/2.

But α = ∠AXY , so from (b2 + 1)2 − (√2b2 + 1)2 = (b2)2, it easily follows that tanα =

√2b2+1b2 = 17

144 .

8. Let S be the set of discs D contained completely in the set {(x, y) : y < 0} (the region below thex-axis) and centered (at some point) on the curve y = x2 − 3

4 . What is the area of the union of theelements of S?

Answer: 2π3 +

√34 Solution 1. An arbitrary point (x0, y0) is contained in S if and only if there

exists some (x, y) on the curve (x, x2 − 34 ) such that (x− x0)

2 + (y− y0)2 < y2, since the radius of the

circle is at most the distance from (x, y) to the x-axis. Some manipulation yields x2 − 2y0(x2 − 3

4 ) −2xx0 + x2

0 + y20 < 0.

Observe that (x0, y0) ∈ S if and only if the optimal choice for x that minimizes the expression satisfiesthe inequality. The minimum is achieved for x = x0

1−2y0

. After substituting and simplifying, we obtain

y0(−x2

0

1−2y0

+x20+y20+

32y0) < 0. Since y0 < 0 and 1−2y0 > 0, we find that we need −2x2

0−2y20+32−2y0 >

0 ⇐⇒ 1 > x0 + (y0 +12 )

2.

Geometry

S is therefore the intersection of the lower half-plane and a circle centered at (0,− 12 ) of radius 1. This

is a circle of sector angle 4π/3 and an isosceles triangle with vertex angle 2π/3. The sum of these areas

is 2π3 +

√34 .

Solution 2. Let O = (0,− 12 ) and ℓ = {y = −1} be the focus and directrix of the given parabola. Let

ℓ′ denote the x-axis. Note that a point P ′ is in S iff there exists a point P on the parabola in the lowerhalf-plane for which d(P, P ′) < d(P, ℓ′). However, for all such P , d(P, ℓ′) = 1 − d(P, ℓ) = 1 − d(P,O),which means that P ′ is in S iff there exists a P on the parabola for which d(P ′, P ) + d(P,O) < 1. Itis not hard to see that this is precisely the intersection of the unit circle centered at O and the lowerhalf-plane, so now we can proceed as in Solution 1.

9. Let ABCD be a regular tetrahedron with side length 1. Let X be the point in triangle BCD suchthat [XBC] = 2[XBD] = 4[XCD], where [̟] denotes the area of figure ̟. Let Y lie on segment AXsuch that 2AY = Y X. Let M be the midpoint of BD. Let Z be a point on segment AM such thatthe lines Y Z and BC intersect at some point. Find AZ

ZM .

Answer: 47 We apply three-dimensional barycentric coordinates with reference tetrahedron

ABCD. The given conditions imply that

X = (0 : 1 : 2 : 4)

Y = (14 : 1 : 2 : 4)

M = (0 : 1 : 0 : 1)

Z = (t : 1 : 0 : 1)

for some real number t. Normalizing, we obtain Y =(1421 ,

121 ,

221 ,

421

)and Z =

(t

t+2 ,1

t+2 , 0,1

t+2

). If

Y Z intersects line BC then there exist parameters α + β = 1 such that αY + βZ has zero A and Dcoordinates, meaning

14

21α+

t

t+ 2β = 0

4

21α+

1

t+ 2β = 0

α+ β = 1.

Adding twice the second equation to the first gives 2221α+ β = 0, so α = −22, β = 21, and thus t = 7

2 .

It follows that Z = (7 : 2 : 0 : 2), and AZZM = 2+2

7 = 47 .

10. Let G be the set of all points (x, y) in the Cartesian plane such that 0 ≤ y ≤ 8 and

(x− 3)2 + 31 = (y − 4)2 + 8√

y(8− y).

There exists a unique line ℓ of negative slope tangent to G and passing through the point (0, 4).Suppose ℓ is tangent to G at a unique point P . Find the coordinates (α, β) of P .

Answer: ( 125 , 85 ) Let G be G restricted to the strip of plane 0 ≤ y ≤ 4 (we only care about this

region since ℓ has negative slope going down from (0, 4)). By completing the square, the originalequation rearranges to (x− 3)2 + (

√y(8− y)− 4)2 = 1. One could finish the problem in a completely

standard way via the single-variable parameterization (x,√

y(8− y)) = (3 + cos t, 4 + sin t) on theappropriate interval of t—just take derivatives with respect to t to find slopes (the computationswould probably not be too bad)—but we will present a slightly cleaner solution.

Consider the bijective plane transformation Φ : (x, y) 7→ (x,√

y(8− y)), with inverse Φ−1 : (x, y) 7→(x, 4−

√16− y2). In general, Φ maps curves as follows: Φ({(x, y) : f(x, y) = c}) = {Φ(x, y) : f(x, y) =

c} = {(x′, y′) : f(Φ−1(x′, y′)) = c}.Our line ℓ has the form y − 4 = −mx for some m > 0. We have Φ(G) = {(x− 3)2 + (y − 4)2 = 1 : 0 ≤y ≤ 4} and Φ({4− y = mx : 0 ≤ y ≤ 4}) = {

√16− y2 = mx : 0 ≤ y ≤ 4}. Since ℓ is unique, m must

Geometry

also be. But it’s easy to see that m = 1 gives a tangency point, so if our original tangency point was(u, v), then our new tangency point is (u,

√v(8− v)) = 4

5 (3, 4) =(125 , 16

5

), and so (u, v) =

(125 , 8

5

).

Remark. To see what G looks like, see Wolfram Alpha using the plotting/graphing commands.

Geometry

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HMMT FEBRUARY 2015, 21 FEBRUARY 2015 — GUTS ROUND

Organization Team Team ID#

1. [4] Let R be the rectangle in the Cartesian plane with vertices at (0, 0), (2, 0), (2, 1), and (0, 1). R canbe divided into two unit squares, as shown.

The resulting figure has 7 segments of unit length, connecting neighboring lattice points (those lyingon or inside R). Compute the number of paths from (0, 1) (the upper left corner) to (2, 0) (the lowerright corner) along these 7 segments, where each segment can be used at most once.

2. [4] Let ABCDE be a convex pentagon such that ∠ABC = ∠ACD = ∠ADE = 90◦ and AB = BC =CD = DE = 1. Compute AE.

3. [4] Find the number of pairs of union/intersection operations (�1,�2) ∈ {∪,∩}2 satisfying the fol-lowing condition: for any sets S, T , function f : S → T , and subsets X,Y, Z of S, we have equality ofsets

f(X)�1(f(Y )�2f(Z)) = f (X�1(Y�2Z)) ,

where f(X) denotes the image of X: the set {f(x) : x ∈ X}, which is a subset of T . The images f(Y )(of Y ) and f(Z) (of Z) are similarly defined.

4. [4] Consider the function z(x, y) describing the paraboloid

z = (2x− y)2 − 2y2 − 3y.

Archimedes and Brahmagupta are playing a game. Archimedes first chooses x. Afterwards, Brah-magupta chooses y. Archimedes wishes to minimize z while Brahmagupta wishes to maximize z.Assuming that Brahmagupta will play optimally, what value of x should Archimedes choose?

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HMMT FEBRUARY 2015, 21 FEBRUARY 2015 — GUTS ROUND

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5. [5] LetH be the unit hypercube of dimension 4 with a vertex at (x, y, z, w) for each choice of x, y, z, w ∈{0, 1}. (Note that H has 24 = 16 vertices.) A bug starts at the vertex (0, 0, 0, 0). In how many wayscan the bug move to (1, 1, 1, 1) (the opposite corner of H) by taking exactly 4 steps along the edges ofH?

6. [5] Let D be a regular ten-sided polygon with edges of length 1. A triangle T is defined by choosingthree vertices of D and connecting them with edges. How many different (non-congruent) triangles Tcan be formed?

7. [5] Let C be a cube of side length 2. We color each of the faces of C blue, then subdivide it into23 = 8 unit cubes. We then randomly rearrange these cubes (possibly with rotation) to form a new3-dimensional cube.

What is the probability that its exterior is still completely blue?

8. [5] Evaluatesin(arcsin(0.4) + arcsin(0.5)) · sin(arcsin(0.5)− arcsin(0.4)),

where for x ∈ [−1, 1], arcsin(x) denotes the unique real number y ∈ [−π, π] such that sin(y) = x.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Organization Team Team ID#

9. [6] Let a, b, c be integers. Define f(x) = ax2 + bx + c. Suppose there exist pairwise distinct integersu, v, w such that f(u) = 0, f(v) = 0, and f(w) = 2. Find the maximum possible value of thediscriminant b2 − 4ac of f .

10. [6] Let b(x) = x2 + x + 1. The polynomial x2015 + x2014 + · · · + x + 1 has a unique “base b(x)”representation

x2015 + x2014 + · · ·+ x+ 1 =

N∑

k=0

ak(x)b(x)k,

where

• N is a nonnegative integer;

• each “digit” ak(x) (for 0 ≤ k ≤ N) is either the zero polynomial (i.e. ak(x) = 0) or a nonzeropolynomial of degree less than deg b = 2; and

• the “leading digit aN (x)” is nonzero (i.e. not the zero polynomial).

Find aN (0) (the “leading digit evaluated at 0”).

11. [6] Find

∞∑

k=0

1 +√

20000004k

2

,

where ⌊x⌋ denotes the largest integer less than or equal to x.

12. [6] For integers a, b, c, d, let f(a, b, c, d) denote the number of ordered pairs of integers (x, y) ∈{1, 2, 3, 4, 5}2 such that ax + by and cx + dy are both divisible by 5. Find the sum of all possiblevalues of f(a, b, c, d).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT FEBRUARY 2015, 21 FEBRUARY 2015 — GUTS ROUND

Organization Team Team ID#

13. [8] Let P (x) = x3 + ax2 + bx + 2015 be a polynomial all of whose roots are integers. Given thatP (x) ≥ 0 for all x ≥ 0, find the sum of all possible values of P (−1).

14. [8] Find the smallest integer n ≥ 5 for which there exists a set of n distinct pairs (x1, y1), . . . , (xn, yn)of positive integers with 1 ≤ xi, yi ≤ 4 for i = 1, 2, . . . , n, such that for any indices r, s ∈ {1, 2, . . . , n}(not necessarily distinct), there exists an index t ∈ {1, 2, . . . , n} such that 4 divides xr + xs − xt andyr + ys − yt.

15. [8] Find the maximum possible value of H · M · M · T over all ordered triples (H,M, T ) of integerssuch that H ·M ·M · T = H +M +M + T .

16. [8] Determine the number of unordered triples of distinct points in the 4 × 4 × 4 lattice grid{0, 1, 2, 3}3 that are collinear in R3 (i.e. there exists a line passing through the three points).

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HMMT FEBRUARY 2015, 21 FEBRUARY 2015 — GUTS ROUND

Organization Team Team ID#

17. [11] Find the least positive integer N > 1 satisfying the following two properties:

• There exists a positive integer a such that N = a(2a− 1).

• The sum 1 + 2 + · · ·+ (N − 1) is divisible by k for every integer 1 ≤ k ≤ 10.

18. [11] Let f : Z → Z be a function such that for any integers x, y, we have

f(x2 − 3y2) + f(x2 + y2) = 2(x+ y)f(x− y).

Suppose that f(n) > 0 for all n > 0 and that f(2015) · f(2016) is a perfect square. Find the minimumpossible value of f(1) + f(2).

19. [11] Find the smallest positive integer n such that the polynomial (x+ 1)n − 1 is “divisible by x2 + 1modulo 3”, or more precisely, either of the following equivalent conditions holds:

• there exist polynomials P,Q with integer coefficients such that (x+1)n−1 = (x2+1)P (x)+3Q(x);

• or more conceptually, the remainder when (the polynomial) (x + 1)n − 1 is divided by (thepolynomial) x2 + 1 is a polynomial with (integer) coefficients all divisible by 3.

20. [11] What is the largest real number θ less than π (i.e. θ < π) such that

10∏

k=0

cos(2kθ) 6= 0

and10∏

k=0

(

1 +1

cos(2kθ)

)

= 1?

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HMMT FEBRUARY 2015, 21 FEBRUARY 2015 — GUTS ROUND

Organization Team Team ID#

21. [14] Define a sequence ai,j of integers such that a1,n = nn for n ≥ 1 and ai,j = ai−1,j + ai−1,j+1 forall i, j ≥ 1. Find the last (decimal) digit of a128,1.

22. [14] Let A1,A2, . . .,A2015 be distinct points on the unit circle with center O. For every two distinctintegers i, j, let Pij be the midpoint of Ai and Aj . Find the smallest possible value of

∑

1≤i<j≤2015

OP 2ij .

23. [14] Let S = {1, 2, 4, 8, 16, 32, 64, 128, 256}. A subset P of S is called squarely if it is nonempty andthe sum of its elements is a perfect square. A squarely set Q is called super squarely if it is not a propersubset of any squarely set. Find the number of super squarely sets.

(A set A is said to be a proper subset of a set B if A is a subset of B and A 6= B.)

24. [14] ABCD is a cyclic quadrilateral with sides AB = 10, BC = 8, CD = 25, and DA = 12. A circleω is tangent to segments DA, AB, and BC. Find the radius of ω.

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HMMT FEBRUARY 2015, 21 FEBRUARY 2015 — GUTS ROUND

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25. [17] Let r1, . . . , rn be the distinct real zeroes of the equation

x8 − 14x4 − 8x3 − x2 + 1 = 0.

Evaluate r21 + · · ·+ r2n.

26. [17] Let a =√17 and b = i

√19, where i =

√−1. Find the maximum possible value of the ratio

|a− z|/|b− z| over all complex numbers z of magnitude 1 (i.e. over the unit circle |z| = 1).

27. [17] Let a, b be integers chosen independently and uniformly at random from the set {0, 1, 2, . . . , 80}.Compute the expected value of the remainder when the binomial coefficient

(

ab

)

= a!b!(a−b)! is divided

by 3. (Here(

00

)

= 1 and(

ab

)

= 0 whenever a < b.)

28. [17] Let w, x, y, and z be positive real numbers such that

0 6= cosw cosx cos y cos z

2π = w + x+ y + z

3 tanw = k(1 + secw)

4 tanx = k(1 + secx)

5 tan y = k(1 + sec y)

6 tan z = k(1 + sec z).

(Here sec t denotes 1cos t when cos t 6= 0.) Find k.

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Organization Team Team ID#

29. [20] Let ABC be a triangle whose incircle has center I and is tangent to BC, CA, AB, at D, E, F .

Denote by X the midpoint of major arc B̂AC of the circumcircle of ABC. Suppose P is a point online XI such that DP ⊥ EF .

Given that AB = 14, AC = 15, and BC = 13, compute DP .

30. [20] Find the sum of squares of all distinct complex numbers x satisfying the equation

0 = 4x10 − 7x9 + 5x8 − 8x7 + 12x6 − 12x5 + 12x4 − 8x3 + 5x2 − 7x+ 4.

31. [20] Define a power cycle to be a set S consisting of the nonnegative integer powers of an integer a,i.e. S = {1, a, a2, . . . } for some integer a. What is the minimum number of power cycles required suchthat given any odd integer n, there exists some integer k in one of the power cycles such that n ≡ k(mod 1024)?

32. [20] A wealthy king has his blacksmith fashion him a large cup, whose inside is a cone of height 9inches and base diameter 6 inches (that is, the opening at the top of the cup is 6 inches in diameter).At one of his many feasts, he orders the mug to be filled to the brim with cranberry juice.

For each positive integer n, the king stirs his drink vigorously and takes a sip such that the heightof fluid left in his cup after the sip goes down by 1

n2 inches. Shortly afterwards, while the king isdistracted, the court jester adds pure Soylent to the cup until it’s once again full. The king takes sipsprecisely every minute, and his first sip is exactly one minute after the feast begins.

As time progresses, the amount of juice consumed by the king (in cubic inches) approaches a numberr. Find r.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT FEBRUARY 2015, 21 FEBRUARY 2015 — GUTS ROUND

Organization Team Team ID#

33. [25] Let N denote the sum of the decimal digits of(

1000100

)

. Estimate the value of N . If your answer isa positive integer A written fully in decimal notation (for example, 521495223), your score will be the

greatest integer not exceeding 25 · (0.99)|A−N |. Otherwise, your score will be zero.

34. [25] For an integer n, let f(n) denote the number of pairs (x, y) of integers such that x2+xy+y2 = n.Compute the sum

106∑

n=1

nf(n).

Write your answer in the form a · 10b, where b is an integer and 1 ≤ a < 10 is a decimal number.

If your answer is written in this form, your score will be max{0, 25 − ⌊100|log10(A/N)|⌋)}, whereN = a · 10b is your answer to this problem and A is the actual answer. Otherwise, your score will bezero.

35. [25] Let P denote the set of all subsets of {1, . . . , 23}. A subset S ⊆ P is called good if whenever A,Bare sets in S, the set (A \ B) ∪ (B \ A) is also in S. (Here, A \ B denotes the set of all elements in Athat are not in B, and B \ A denotes the set of all elements in B that are not in A.) What fractionof the good subsets of P have between 2015 and 3015 elements, inclusive?

If your answer is a decimal number or a fraction (of the form m/n, where m and n are positive integers),then your score on this problem will be equal to max{0, 25− ⌊1000|A−N |⌋}, where N is your answerand A is the actual answer. Otherwise, your score will be zero.

36. [25] A prime number p is twin if at least one of p+2 or p− 2 is prime and sexy if at least one of p+6and p− 6 is prime.

How many sexy twin primes (i.e. primes that are both twin and sexy) are there less than 109? Expressyour answer as a positive integer N in decimal notation; for example, 521495223. If your answer isin this form, your score for this problem will be max

{

0, 25− ⌊ 110000 |A−N |

⌋

}, where A is the actualanswer to this problem. Otherwise, your score will be zero.

HMMT February 2015Saturday 21 February 2015

Guts

1. [4] Let R be the rectangle in the Cartesian plane with vertices at (0, 0), (2, 0), (2, 1), and (0, 1). R canbe divided into two unit squares, as shown.

The resulting figure has 7 segments of unit length, connecting neighboring lattice points (those lyingon or inside R). Compute the number of paths from (0, 1) (the upper left corner) to (2, 0) (the lowerright corner) along these 7 segments, where each segment can be used at most once.

Answer: 4 Just count them directly. If the first step is to the right, there are 2 paths. If the firststep is downwards (so the next step must be to the right), there are again 2 paths. This gives a totalof 4 paths.

2. [4] Let ABCDE be a convex pentagon such that ∠ABC = ∠ACD = ∠ADE = 90◦ and AB = BC =CD = DE = 1. Compute AE.

Answer: 2 By Pythagoras,

AE2 = AD2 + 1 = AC2 + 2 = AB2 + 3 = 4,

so AE = 2.

3. [4] Find the number of pairs of union/intersection operations (�1,�2) ∈ {∪,∩}2 satisfying the fol-lowing condition: for any sets S, T , function f : S → T , and subsets X,Y, Z of S, we have equality ofsets

f(X)�1(f(Y )�2f(Z)) = f (X�1(Y�2Z)) ,

where f(X) denotes the image of X: the set {f(x) : x ∈ X}, which is a subset of T . The images f(Y )(of Y ) and f(Z) (of Z) are similarly defined.

Answer: 1 If and only if�1 = �2 = ∪. See http://math.stackexchange.com/questions/359693/overview-of-basic-

4. [4] Consider the function z(x, y) describing the paraboloid

z = (2x− y)2 − 2y2 − 3y.

Archimedes and Brahmagupta are playing a game. Archimedes first chooses x. Afterwards, Brah-magupta chooses y. Archimedes wishes to minimize z while Brahmagupta wishes to maximize z.Assuming that Brahmagupta will play optimally, what value of x should Archimedes choose?

Answer: − 38 Viewing x as a constant and completing the square, we find that

z = 4x2 − 4xy + y2 − 2y2 − 3y

= −y2 − (4x+ 3)y + 4x2

= −(

y +4x+ 3

2

)2

+

(

4x+ 3

2

)2

+ 4x2.

Bhramagupta wishes to maximize z, so regardless of the value of x, he will pick y = − 4x+32 . The

expression for z then simplifies to

z = 8x2 + 6x+9

4.

Guts

Archimedes knows this and will therefore pick x to minimize the above expression. By completing thesquare, we find that x = − 3

8 minimizes z.

Alternatively, note that z is convex in x and concave in y, so we can use the minimax theorem to switchthe order of moves. If Archimedes goes second, he will set x = y

2 to minimize z, so Brahmagupta willmaximize −2y2 − 3y by setting y = − 3

4 . Thus Archimedes should pick x = − 38 , as above.

5. [5] LetH be the unit hypercube of dimension 4 with a vertex at (x, y, z, w) for each choice of x, y, z, w ∈{0, 1}. (Note that H has 24 = 16 vertices.) A bug starts at the vertex (0, 0, 0, 0). In how many wayscan the bug move to (1, 1, 1, 1) (the opposite corner of H) by taking exactly 4 steps along the edges ofH?

Answer: 24 You may think of this as sequentially adding 1 to each coordinate of (0, 0, 0, 0). Thereare 4 ways to choose the first coordinate, 3 ways to choose the second, and 2 ways to choose the third.The product is 24.

6. [5] Let D be a regular ten-sided polygon with edges of length 1. A triangle T is defined by choosingthree vertices of D and connecting them with edges. How many different (non-congruent) triangles Tcan be formed?

Answer: 8 The problem is equivalent to finding the number of ways to partition 10 into a sum ofthree (unordered) positive integers. These can be computed by hand to be (1, 1, 8), (1, 2, 7), (1, 3, 6),(1, 4, 5), (2, 2, 6), (2, 3, 5), (2, 4, 4), (3, 3, 4).

7. [5] Let C be a cube of side length 2. We color each of the faces of C blue, then subdivide it into23 = 8 unit cubes. We then randomly rearrange these cubes (possibly with rotation) to form a new3-dimensional cube.

What is the probability that its exterior is still completely blue?

Answer: 1224 or 1

88 or 116777216 Each vertex of the original cube must end up as a vertex of the

new cube in order for all the old blue faces to show. There are 8 such vertices, each corresponding toone unit cube, and each has a probability 1

8 of being oriented with the old outer vertex as a vertex ofthe new length-2 cube. Multiplying gives the answer.

8. [5] Evaluatesin(arcsin(0.4) + arcsin(0.5)) · sin(arcsin(0.5)− arcsin(0.4)),

where for x ∈ [−1, 1], arcsin(x) denotes the unique real number y ∈ [−π, π] such that sin(y) = x.

Answer: 0.09 OR 9100 Use the difference of squares identity1 sin(a−b) sin(a+b) = sin(a)2−sin(b)2

to get 0.52 − 0.42 = 0.32 = 0.09 = 9100 .

9. [6] Let a, b, c be integers. Define f(x) = ax2 + bx + c. Suppose there exist pairwise distinct integersu, v, w such that f(u) = 0, f(v) = 0, and f(w) = 2. Find the maximum possible value of thediscriminant b2 − 4ac of f .

Answer: 16 By the factor theorem, f(x) = a(x − u)(x − v), so the constraints essentially boildown to 2 = f(w) = a(w − u)(w − v). (It’s not so important that u 6= v; we merely specified it for ashorter problem statement.)

We want to maximize the discriminant b2−4ac = a2[(u+v)2−4uv] = a2(u−v)2 = a2[(w−v)−(w−u)]2.Clearly a | 2. If a > 0, then (w − u)(w − v) = 2/a > 0 means the difference |u − v| is less than 2/a,whereas if a < 0, since at least one of |w − u| and |w − v| equals 1, the difference |u − v| of factors isgreater than 2/|a|.So the optimal choice occurs either for a = −1 and |u− v| = 3, or a = −2 and |u− v| = 2. The latterwins, giving a discriminant of (−2)2 · 22 = 16.

1proven most easily with complex numbers or the product-to-sum identity on sin(a − b) sin(a + b) (followed by the doubleangle formula for cosine)

Guts

10. [6] Let b(x) = x2 + x + 1. The polynomial x2015 + x2014 + · · · + x + 1 has a unique “base b(x)”representation

x2015 + x2014 + · · ·+ x+ 1 =

N∑

k=0

ak(x)b(x)k,

where

• N is a nonnegative integer;

• each “digit” ak(x) (for 0 ≤ k ≤ N) is either the zero polynomial (i.e. ak(x) = 0) or a nonzeropolynomial of degree less than deg b = 2; and

• the “leading digit aN (x)” is nonzero (i.e. not the zero polynomial).

Find aN (0) (the “leading digit evaluated at 0”).

Answer: −1006 Comparing degrees easily gives N = 1007. By ignoring terms of degree at most2013, we see

aN (x)(x2 + x+ 1)1007 ∈ x2015 + x2014 +O(x2013).

Write aN (x) = ux+ v, so

aN (x)(x2 + x+ 1)1007 ∈ (ux+ v)(x2014 + 1007x2013 +O(x2012))

⊆ ux2015 + (v + 1007u)x2014 +O(x2013).

Finally, matching terms gives u = 1 and v + 1007u = 1, so v = 1− 1007 = −1006.

Remark. This problem illustrates the analogy between integers and polynomials, with the nonconstant(degree ≥ 1) polynomial b(x) = x2 + x+ 1 taking the role of a positive integer base b > 1.

11. [6] Find

∞∑

k=0

1 +√

20000004k

2

,

where ⌊x⌋ denotes the largest integer less than or equal to x.

Answer: 1414 The kth floor (for k ≥ 0) counts the number of positive integer solutions to4k(2x− 1)2 ≤ 2 · 106. So summing over all k, we want the number of integer solutions to 4k(2x− 1)2 ≤2 · 106 with k ≥ 0 and x ≥ 1. But each positive integer can be uniquely represented as a power of 2times an odd (positive) integer, so there are simply ⌊103

√2⌋ = 1414 solutions.

12. [6] For integers a, b, c, d, let f(a, b, c, d) denote the number of ordered pairs of integers (x, y) ∈{1, 2, 3, 4, 5}2 such that ax + by and cx + dy are both divisible by 5. Find the sum of all possiblevalues of f(a, b, c, d).

Answer: 31 Standard linear algebra over the field F5 (the integers modulo 5). The dimension ofthe solution set is at least 0 and at most 2, and any intermediate value can also be attained. So theanswer is 1 + 5 + 52 = 31.

This also can be easily reformulated in more concrete equation/congruence-solving terms, especiallysince there are few variables/equations.

13. [8] Let P (x) = x3 + ax2 + bx + 2015 be a polynomial all of whose roots are integers. Given thatP (x) ≥ 0 for all x ≥ 0, find the sum of all possible values of P (−1).

Answer: 9496 Since all the roots of P (x) are integers, we can factor it as P (x) = (x−r)(x−s)(x−t)for integers r, s, t. By Viete’s formula, the product of the roots is rst = −2015, so we need three integersto multiply to −2015.

P (x) cannot have two distinct positive roots u, v since otherwise, P (x) would be negative at least insome infinitesimal region x < u or x > v, or P (x) < 0 for u < x < v. Thus, in order to have two

Guts

positive roots, we must have a double root. Since 2015 = 5× 13× 31, the only positive double root isa perfect square factor of 2015, which is at x = 1, giving us a possibility of P (x) = (x− 1)2(x+2015).

Now we can consider when P (x) only has negative roots. The possible unordered triplets are (−1,−1,−2015), (−1,−5,−403)(−1,−31,−65), (−5,−13,−31), which yield the polynomials(x+1)2(x+2015), (x+1)(x+5)(x+403), (x+1)(x+13)(x+155), (x+1)(x+31)(x+65), (x+5)(x+13)(x+ 31), respectively.

Noticing that P (−1) = 0 for four of these polynomials, we see that the nonzero values are P (−1) =(−1− 1)2(2014), (5− 1)(13− 1)(31− 1), which sum to 8056 + 1440 = 9496.

14. [8] Find the smallest integer n ≥ 5 for which there exists a set of n distinct pairs (x1, y1), . . . , (xn, yn)of positive integers with 1 ≤ xi, yi ≤ 4 for i = 1, 2, . . . , n, such that for any indices r, s ∈ {1, 2, . . . , n}(not necessarily distinct), there exists an index t ∈ {1, 2, . . . , n} such that 4 divides xr + xs − xt andyr + ys − yt.

Answer: 8 In other words, we have a set S of n pairs in (Z/4Z)2 closed under addition. Since1 + 1 + 1 + 1 ≡ 0 (mod 4) and 1 + 1 + 1 ≡ −1 (mod 4), (0, 0) ∈ S and S is closed under (additive)inverses. Thus S forms a group under addition (a subgroup of (Z/4Z)2). By Lagrange’s theorem(from basic group theory), n | 42, so n ≥ 8. To achieve this bound, one possible construction is{1, 2, 3, 4} × {2, 4}.Remark. In fact, S is a finite abelian group. Such groups have a very clean classification; this isclarified by the fact that abelian groups are the same as modules over Z, the ring of integers.

15. [8] Find the maximum possible value of H · M · M · T over all ordered triples (H,M, T ) of integerssuch that H ·M ·M · T = H +M +M + T .

Answer: 8 If any of H,M, T are zero, the product is 0. We can do better (examples below), sowe may now restrict attention to the case when H,M, T 6= 0.

WhenM ∈ {−2,−1, 1, 2}, a little casework gives all the possible (H,M, T ) = (2, 1, 4), (4, 1, 2), (−1,−2, 1), (1,−2,−1).

• If M = −2, i.e. H − 4+ T = 4HT , then −15 = (4H − 1)(4T − 1), so 4H − 1 ∈ {±1,±3,±5,±15}(only−1,+3,−5,+15 are possible) corresponding to 4T−1 ∈ {∓15,∓5,∓3,∓1} (only +15,−5,+3,−1are possible). But H,T are nonzero, we can only have 4H − 1 ∈ {+3,−5}, yielding (−1,−2, 1)and (1,−2,−1).

• If M = +2, i.e. H + 4 + T = 4HT , then 17 = (4H − 1)(4T − 1), so 4H − 1 ∈ {±1,±17} (only−1,−17 are possible) corresponding to 4T −1 ∈ {±17,±1} (only −17,−1 are possible). But H,Tare nonzero, so there are no possibilities here.

• If M = −1, i.e. H − 2 + T = HT , then −1 = (H − 1)(T − 1), so we have H − 1 ∈ {±1} andT − 1 ∈ {∓1}, neither of which is possible (as H,T 6= 0).

• If M = +1, i.e. H + 2 + T = HT , then 3 = (H − 1)(T − 1), so we have H − 1 ∈ {±1,±3}. SinceH,T 6= 0, H − 1 ∈ {+1,+3}, yielding (2, 1, 4) and (4, 1, 2).

Now suppose there is such a triple (H,M, T ) for |M | ≥ 3. The equation in the problem gives (M2H −1)(M2T − 1) = 2M3 + 1. Note that since H,T 6= 0, |2M3 + 1| = |M2H − 1| · |M2T − 1| ≥ min(M2 −1,M2 + 1)2 = M4 − 2M2 + 1 > 2|M |3 + 1 gives a contradiction.

16. [8] Determine the number of unordered triples of distinct points in the 4 × 4 × 4 lattice grid{0, 1, 2, 3}3 that are collinear in R3 (i.e. there exists a line passing through the three points).

Answer: 376 Define a main plane to be one of the xy, yz, zx planes. Define a space diagonal tobe a set of collinear points not parallel to a main plane. We classify the lines as follows:

(a) Lines parallel to two axes (i.e. orthogonal to a main plane). Notice that given a planeof the form v = k, where v ∈ {x, y, z}, k ∈ {0, 1, 2, 3}, there are 8 such lines, four in one directionand four in a perpendicular direction. There are 4× 3 = 12 such planes. However, each line liesin two of these (v, k) planes, so there are 8×4×3

2 = 48 such lines. Each of these lines has 4 points,so there are 4 possible ways to choose 3 collinear points, giving 4× 48 = 192 triplets.

Guts

(b) Diagonal lines containing four points parallel to some main plane. Consider a plane ofthe form (v, k), as defined above. These each have 2 diagonals that contain 4 collinear points.Each of these diagonals uniquely determines v, k so these diagonals are each counted once. Thereare 12 possible (v, k) pairs, yielding 12× 2× 4 = 96 triplets.

(c) Diagonal lines containing three points parallel to some main plane. Again, consider aplane (v, k). By inspection, there are four such lines and one way to choose the triplet of pointsfor each of these lines. This yields 4× 12 = 48 triplets.

(d) Main diagonals. There are four main diagonals, each with 4 collinear points, yielding 4×4 = 16triplets.

(e) Space diagonals containing three points. Choose one of the points in the set {1, 2}3 to bethe midpoint of the line. Since these 8 possibilities are symmetric, say we take the point (1, 1, 1).There are four space diagonals passing through this point, but one is a main diagonal. So each ofthe 8 points has 3 such diagonals with 3 points each, yielding 8× 3 = 24 ways.

Adding all these yields 192 + 96 + 48 + 16 + 24 = 376.

17. [11] Find the least positive integer N > 1 satisfying the following two properties:

• There exists a positive integer a such that N = a(2a− 1).

• The sum 1 + 2 + · · ·+ (N − 1) is divisible by k for every integer 1 ≤ k ≤ 10.

Answer: 2016 The second condition implies that 16 divides a(2a− 1)(2a2 − a− 1), which showsthat a ≡ 0 or 1 modulo 16. The case a = 1 would contradict the triviality-avoiding condition N > 1.a cannot be 16, because 7 does not divide a(2a− 1)(2a2 − a− 1). a cannot be 17, because 9 does notdivide a(2a − 1)(2a2 − a − 1). It can be directly verified that a = 32 is the smallest positive integerfor which 1 + 2 + · · ·+ (N − 1) = 24 · 32 · 5 · 7 · 13 · 31 which is divisible by 1, 2, . . . , 10. For this a, wecompute N = 32(2 · 32− 1) = 2016.

18. [11] Let f : Z → Z be a function such that for any integers x, y, we have

f(x2 − 3y2) + f(x2 + y2) = 2(x+ y)f(x− y).

Suppose that f(n) > 0 for all n > 0 and that f(2015) · f(2016) is a perfect square. Find the minimumpossible value of f(1) + f(2).

Answer: 246 Plugging in −y in place of y in the equation and comparing the result with theoriginal equation gives

(x− y)f(x+ y) = (x+ y)f(x− y)

This shows that whenever a, b ∈ Z− {0} with a ≡ b(mod 2), we have

f(a)

a=

f(b)

b

which implies that there are constants α = f(1) ∈ Z>0, β = f(2) ∈ Z>0 for which f satisfies theequation (∗):

f(n) =

{

n · α when 2 ∤ nn2 · β when 2|n

Therefore, f(2015)f(2016) = 2015α · 1008β = 24 · 32 · 5 · 7 · 13 · 31αβ, so αβ = 5 · 7 · 13 · 31 · t2 forsome t ∈ Z>0. We claim that (α, β, t) = (5 · 31, 7 · 13, 1) is a triple which gives the minimum α+ β. Inparticular, we claim α+ β ≥ 246.

Consider the case t ≥ 2 first. We have, by AM-GM, α + β ≥ 2 ·√αβ ≥ 4 ·

√14105 > 246. Suppose

t = 1. We have α · β = 5 · 7 · 13 · 31. Because (α + β)2 − (α − β)2 = 4αβ is fixed, we want to have αas close as β as possible. This happens when one of α, β is 5 · 31 and the other is 7 · 13. In this case,α+ β = 91 + 155 = 246.

Guts

Finally, we note that the equality f(1) + f(2) = 246 can be attained. Consider f : Z → Z such thatf(n) = 91n for every odd n ∈ Z and f(n) = 155

2 n for every even n ∈ Z. It can be verified that fsatisfies the condition in the problem and f(1) + f(2) = 246 as claimed.

19. [11] Find the smallest positive integer n such that the polynomial (x+ 1)n − 1 is “divisible by x2 + 1modulo 3”, or more precisely, either of the following equivalent conditions holds:

• there exist polynomials P,Q with integer coefficients such that (x+1)n−1 = (x2+1)P (x)+3Q(x);

• or more conceptually, the remainder when (the polynomial) (x + 1)n − 1 is divided by (thepolynomial) x2 + 1 is a polynomial with (integer) coefficients all divisible by 3.

Answer: 8 Solution 1. We have (x+ 1)2 = x2 + 2x+ 1 ≡ 2x, (x+ 1)4 ≡ (2x)2 ≡ −4 ≡ −1, and(x + 1)8 ≡ (−1)2 = 1. So the order n divides 8, as x + 1 and x2 + 1 are relatively prime polynomialsmodulo 3 (or more conceptually, in F3[x]), but cannot be smaller by our computations of the 2nd and4th powers.

Remark. This problem illustrates the analogy between integers and polynomials (specifically here,polynomials over the finite field of integers modulo 3), with x2 + 1 (mod 3) taking the role of a

prime number. Indeed, we can prove analogously to Fermat’s little theorem that f(x)3deg(x2+1) ≡ f(x)

(mod x2 + 1, 3) for any polynomial f(x): try to work out the proof yourself! (This kind of problem,with 3 replaced by any prime p and x2 + 1 by any irreducible polynomial modulo p, is closely relatedto the theory of (extensions of) finite fields.)

Solution 2. Here’s a solution avoiding the terminology of groups and fields. Let R(x) = (x+1)n−1−P (x)(x2 + 1) = 3Q(x) be the remainder (here P is the quotient), whose coefficients (upon expansion)must all be divisible by 3. Now consider R(i). The coefficients of the even and odd powers of x aredivisible by 3, so R(i) must have real and imaginary parts divisible by 3. Notice that (1 + i)2 = 2i,(1 + i)4 = −4, (1 + i)6 = −8i, (1 + i)8 = 16, and that (1 + i)8 − 1 = 15, which has real and imaginaryparts divisible by 3. Since any even power of (1+ i) (for n ≤ 6) yields a purely real or purely imaginarynumber with a coefficient not divisible by 3, multiplying it by 1 + i will yield an imaginary part notdivisible by 3. To see that n = 8 works, notice that, when taken modulo 3,

(x+ 1)8 − 1 = x8 + 8x7 + 28x6 + 56x5 + 70x4 + 56x3 + 28x2 + 8x

≡ x8 − x7 + x6 − x5 + x4 − x3 + x2 − x

= (x2 + 1)(x6 − x5 + x2 − x) (mod 3),

as desired.

20. [11] What is the largest real number θ less than π (i.e. θ < π) such that

10∏

k=0

cos(2kθ) 6= 0

and10∏

k=0

(

1 +1

cos(2kθ)

)

= 1?

Answer: 2046π2047 For equality to hold, note that θ cannot be an integer multiple of π (or else sin = 0

and cos = ±1).

Let z = eiθ/2 6= ±1. Then in terms of complex numbers, we want

10∏

k=0

(1 +2

z2k+1 + z−2k+1 ) =

10∏

k=0

(z2k

+ z−2k)2

z2k+1 + z−2k+1 ,

Guts

which partially telescopes to

z + z−1

z211 + z−211

10∏

k=0

(z2k

+ z−2k).

Using a classical telescoping argument (or looking at binary representation; if you wish we may notethat z − z−1 6= 0, so the ultimate telescoping identity holds2), this simplifies to

z + z−1

z211 + z−211

z211 − z−211

z − z−1=

tan(210θ)

tan(θ/2).

Since tanx is injective modulo π (i.e. π-periodic and injective on any given period), θ works if andonly if θ

2 + ℓπ = 1024θ for some integer ℓ, so θ = 2ℓπ2047 . The largest value for ℓ such that θ < π is at

ℓ = 1023, which gives θ = 2046π2047 .3

Remark. It’s also possible to do this without complex numbers, but it’s less systematic. The steps

are the same, though, first note that 1 + sec 2kθ = 1+cos 2kθcos 2kθ

= 2 cos2 2k−1θcos 2kθ

using the identity cos 2x =2 cos2 x− 1 (what does this correspond to in complex numbers?). hen we telescope using the identity2 cosx = sin 2x

sin x (again, what does this correspond to in complex numbers?).

21. [14] Define a sequence ai,j of integers such that a1,n = nn for n ≥ 1 and ai,j = ai−1,j + ai−1,j+1 forall i, j ≥ 1. Find the last (decimal) digit of a128,1.

Answer: 4 By applying the recursion multiple times, we find that a1,1 = 1, a2,n = nn+(n+1)n+1,and a3,n = nn + 2(n + 1)n+1 + (n + 2)n+2. At this point, we can conjecture and prove by inductionthat

am,n =

m−1∑

k=0

(

m− 1

k

)

(n+ k)n+k =∑

k≥0

(

m− 1

k

)

(n+ k)n+k.

(The second expression is convenient for dealing with boundary cases. The induction relies on(

m0

)

=(

m−10

)

on the k = 0 boundary, as well as(

mk

)

=(

m−1k

)

+(

m−1k−1

)

for k ≥ 1.) We fix m = 128. Note that(

127k

)

≡ 1 (mod 2) for all 1 ≤ k ≤ 127 and(

127k

)

≡ 0 (mod 5) for 3 ≤ k ≤ 124, by Lucas’ theorem onbinomial coefficients. Therefore, we find that

a128,1 =

127∑

k=0

(

127

k

)

(k + 1)k+1 ≡127∑

k=0

(k + 1)k+1 ≡ 0 (mod 2)

and

a128,1 ≡∑

k∈[0,2]∪[125,127]

(

127

k

)

(k + 1)k+1 ≡ 4 (mod 5).

Therefore, a128,1 ≡ 4 (mod 10).

22. [14] Let A1,A2, . . .,A2015 be distinct points on the unit circle with center O. For every two distinctintegers i, j, let Pij be the midpoint of Ai and Aj . Find the smallest possible value of

∑

1≤i<j≤2015

OP 2ij .

Answer: 2015·20134 OR 4056195

4 Use vectors.∑ |ai + aj |2/4 =

∑

(2 + 2ai · aj)/4 = 12

(

20152

)

+14 (|∑

ai|2 −∑ |ai|2) ≥ 2015 · 20144 − 2015

4 = 2015·20134 , with equality if and only if

∑

ai = 0, whichoccurs for instance for a regular 2015-gon.

2The identity still holds even if z2k

− z−2k

= 0 for some k ≥ 1 used in the telescoping argument: why?3This indeed works, since

∏10

k=0cos(2kθ) 6= 0: why?

Guts

23. [14] Let S = {1, 2, 4, 8, 16, 32, 64, 128, 256}. A subset P of S is called squarely if it is nonempty andthe sum of its elements is a perfect square. A squarely set Q is called super squarely if it is not a propersubset of any squarely set. Find the number of super squarely sets.

(A set A is said to be a proper subset of a set B if A is a subset of B and A 6= B.)

Answer: 5 Clearly we may biject squarely sets with binary representations of perfect squaresbetween 1 and 20 + · · · + 28 = 29 − 1 = 511, so there are 22 squarely sets, corresponding to n2 forn = 1, 2, . . . , 22. For convenience, we say N is (super) squarely if and only if the set corresponding toN is (super) squarely.

The general strategy is to rule out lots of squares at time, by searching for squares with few missingdigits (and ideally most 1’s consecutive, for simplicity). We can restrict ourselves (for now) to odds;(2k)2 is just k2 with two additional zeros at the end. 1, 9, 25, 49, 81 are ineffective, but 121 = 27 − 7 =26 + 25 + 24 + 23 + 20 immediately rules out all odd squares up to 92, as they must be 1 (mod 8).

Fortunately, 222 = 4 · 112 is in our range (i.e. less than 512), ruling out all even squares up to 202 aswell.

This leaves us with 112, 132, 152, 172, 192, 212, 222, with binary representations 001111001, 010101001,011100001, 100100001, 101101001 (kills 172), 110111001 (kills 132), 111100100 (kills nothing by parity).Thus 112, 152, 192, 212, 222 are the only super squarely numbers, for a total of 5.

24. [14] ABCD is a cyclic quadrilateral with sides AB = 10, BC = 8, CD = 25, and DA = 12. A circleω is tangent to segments DA, AB, and BC. Find the radius of ω.

Answer:√

12097 OR

√84637 Denote E an intersection point of AD and BC. Let x = EA and

y = EB. Because ABCD is a cyclic quadrilateral, △EAB is similar to △ECD. Therefore, y+8x = 25

10and x+12

y = 2510 . We get x = 128

21 and y = 15221 . Note that ω is the E-excircle of △EAB, so we may

finish by standard calculations.

Indeed, first we compute the semiperimeter s = EA+AB+BE2 = x+y+10

2 = 353 . Now the radius of ω is

(by Heron’s formula for area)

rE =[EAB]

s−AB=

√

s(s− x)(s− y)

s− 10=

√

1209

7=

√8463

7.

25. [17] Let r1, . . . , rn be the distinct real zeroes of the equation

x8 − 14x4 − 8x3 − x2 + 1 = 0.

Evaluate r21 + · · ·+ r2n.

Answer: 8 Observe that

x8 − 14x4 − 8x3 − x2 + 1 = (x8 + 2x4 + 1)− (16x4 + 8x3 + x2)

= (x4 + 4x2 + x+ 1)(x4 − 4x2 − x+ 1).

The polynomial x4 + 4x2 + x + 1 = x4 + 154 x2 + (x2 + 1)2 has no real roots. On the other hand,

let P (x) = x4 − 4x2 − x + 1. Observe that P (−∞) = +∞ > 0, P (−1) = −1 < 0, P (0) = 1 > 0,P (1) = −3 < 0, P (+∞) = +∞ > 0, so by the intermediate value theorem, P (x) = 0 has four distinctreal roots, which are precisely the real roots of the original degree 8 equation. By Vieta’s formula onP (x),

r21 + r22 + r23 + r24 = (r1 + r2 + r3 + r4)2 − 2 · (

∑

i<j

rirj)

= 02 − 2(−4) = 8.

Guts

26. [17] Let a =√17 and b = i

√19, where i =

√−1. Find the maximum possible value of the ratio

|a− z|/|b− z| over all complex numbers z of magnitude 1 (i.e. over the unit circle |z| = 1).

Answer: 43 Let |a− z|/|b− z| = k. We wish to determine the minimum and maximum value of k.

Squaring and expansion give:

|a− z|2 = |b− z|2 · k2

|a|2 − 2a · z + 1 = (|b|2 − 2b · z + 1)k2

|a|2 + 1− (|b|2 + 1)k2 = 2(a− bk2) · z,

where · is a dot product of complex numbers, i.e., the dot product of vectors corresponding to thecomplex numbers in the complex plane. Now, since z has modulus 1 but can assume any direction,the only constraint on the value of k is

||a|2 + 1− (|b|2 + 1)k2| ≤ 2|a− bk2|.

Squaring again and completing the square, the inequality reduces to:

(|a|2 − 1)2 + (|b|2 − 1)2k4 + 2(4a · b− (|a|2 + 1)(|b|2 + 1))k2 ≤ 0

((|a|2 − 1)− (|b|2 − 1)k2)2 − 4|a− b|2k2 ≤ 0

|(|a|2 − 1)− (|b|2 − 1)k2| ≤ 2|a− b|k.

At this stage all the relevant expressions are constant real numbers. Denote, for simplicity, A =|a|2−1, B = |b|2−1, and C = |a− b|. Then, we are looking for k such that |A−Bk2| ≤ 2Ck. If B = 0,then k ≥ | A

2C |, so the minimum value is | A2C | and the maximum value is +∞. Otherwise, consider

C2 +AB = (|a|2 − 2a · b+ |b|2) + (|a|2 − 1)(|b|2 − 1)

= |ab|2 − 2a · b+ 1

= |ab|2 − 2ℜ(ab) + 1

= |ab− 1|2

So let D = |ab−1| =√C2 +AB. We may assume B > 0 (the another case is analogous: just substitute

A,B with −A,−B). Then, k is determined by the following inequalities:

Bk2 + 2Ck −A ≥ 0

Bk2 − 2Ck −A ≤ 0

The first inequality gives k ≤ −C−DB or k ≥ −C+D

B , and the second gives C−DB ≤ k ≤ C+D

B . Combining,

this gives |C−DB | ≤ k ≤ |C+D

B |, as claimed.

To summarize the general answer, let A = |a|2 − 1, B = |b|2 − 1, C = |a − b|, D = |ab − 1|. Then, if|b| = 1, min is | A

2C | and max is +∞; otherwise, min is |C−DB | and max is |C+D

B |.In the special case a =

√17 and b =

√19i, we have A = 16, B = 18, C =

∣

∣

√17−

√19i∣

∣ =√36 = 6,

and D =√17 · 19 + 1 = 18. Thus the answer is C+D

B = 6+1818 = 4

3 .

27. [17] Let a, b be integers chosen independently and uniformly at random from the set {0, 1, 2, . . . , 80}.Compute the expected value of the remainder when the binomial coefficient

(

ab

)

= a!b!(a−b)! is divided

by 3. (Here(

00

)

= 1 and(

ab

)

= 0 whenever a < b.)

Answer: 18166561 By Lucas’ Theorem we’re looking at

4∏

i=1

(

aibi

)

Guts

where the ai and bi are the digits of a and b in base 3. If any ai < bi, then the product is zero modulo3.

Otherwise, the potential residues are(

20

)

= 1,(

21

)

= 2,(

22

)

= 1,(

10

)

= 1,(

11

)

= 1,(

00

)

= 1.

So each term in the product has a 13 chance of being zero; given that everything is nonzero, each term

has a 16 chance of being 2 and a 5

6 chance of being 1. The probability that an even number of termsare 1 given that none are zero is then given by the roots of unity filter

( 56 + 16 · (1))4 + ( 56 + 1

6 · (−1))4

2=

81 + 16

162=

97

162.

Thus the expected value is(

2

3

)4(

2− 97

162

)

=1816

6561.

28. [17] Let w, x, y, and z be positive real numbers such that

0 6= cosw cosx cos y cos z

2π = w + x+ y + z

3 tanw = k(1 + secw)

4 tanx = k(1 + secx)

5 tan y = k(1 + sec y)

6 tan z = k(1 + sec z).

(Here sec t denotes 1cos t when cos t 6= 0.) Find k.

Answer:√19 From the identity tan u

2 = sinu1+cosu , the conditions work out to 3 tan w

2 = 4 tan x2 =

5 tan y2 = 6 tan z

2 = k. Let a = tan w2 , b = tan x

2 , c = tan y2 , and d = tan z

2 . Using the identity

tan(M +N) = tanM+tanN1−tanM tanN , we obtain

tan

(

w + x

2+

y + z

2

)

=tan

(

w+x2

)

+ tan(

y+z2

)

1− tan(

w+x2

)

tan(

y+z2

)

=a+b1−ab +

c+d1−cd

1−(

a+b1−ab

)(

c+d1−cd

)

=a+ b+ c+ d− abc− abd− bcd− acd

1 + abcd− ab− ac− ad− bc− bd− cd.

Because x+ y + z + w = π, we get that tan(

x+y+z+w2

)

= 0 and thus a+b+c+d = abc+abd+bcd+acd.Substituting a, b, c, d corresponding to the variable k, we obtain that k3 − 19k = 0. Therefore, k canbe only 0,

√19,−

√19. However, k = 0 is impossible as w, x, y, z will all be 0. Also, k = −

√19 is

impossible as w, x, y, z will exceed π. Therefore, k =√19.

29. [20] Let ABC be a triangle whose incircle has center I and is tangent to BC, CA, AB, at D, E, F .

Denote by X the midpoint of major arc B̂AC of the circumcircle of ABC. Suppose P is a point online XI such that DP ⊥ EF .

Given that AB = 14, AC = 15, and BC = 13, compute DP .

Answer: 4√5

5 Let H be the orthocenter of triangle DEF . We claim that P is the midpoint of DH.

Indeed, consider an inversion at the incicrle of ABC, denoting the inverse of a point with an asterik.It maps ABC to the nine-point circle of △DEF . According to ∠IAX = 90◦, we have ∠A∗X∗I = 90◦.Hence line XI passes through the point diametrically opposite to A∗, which is the midpoint of DH,as claimed.

Guts

The rest is a straightforward computation. The inradius of △ABC is r = 4. The length of EF is givenby EF = 2AF ·r

AI = 16√5. Then,

DP 2 =

(

1

2DH

)2

=1

4

(

4r2 − EF 2)

= 42 − 64

5=

16

5.

Hence DP = 4√5

5 .

Remark. This is also not too bad of a coordinate bash.

30. [20] Find the sum of squares of all distinct complex numbers x satisfying the equation

0 = 4x10 − 7x9 + 5x8 − 8x7 + 12x6 − 12x5 + 12x4 − 8x3 + 5x2 − 7x+ 4.

Answer: − 716 For convenience denote the polynomial by P (x). Notice 4 + 8 = 7 + 5 = 12 and

that the consecutive terms 12x6 − 12x5 + 12x4 are the leading terms of 12Φ14(x), which is suggestive.Indeed, consider ω a primitive 14-th root of unity; since ω7 = −1, we have 4ω10 = −4ω3, −7ω9 = 7ω2,and so on, so that

P (ω) = 12(ω6 − ω5 + · · ·+ 1) = 12Φ14(ω) = 0.

Dividing, we findP (x) = Φ14(x)(4x

4 − 3x3 − 2x2 − 3x+ 4).

This second polynomial is symmetric; since 0 is clearly not a root, we have

4x4 − 3x3 − 2x2 − 3x+ 4 = 0 ⇐⇒ 4(x+1

x)2 − 3(x+

1

x)− 10 = 0.

Setting y = x + 1/x and solving the quadratic gives y = 2 and y = −5/4 as solutions; replacingy with x + 1/x and solving the two resulting quadratics give the double root x = 1 and the roots(−5 ± i

√39)/8 respectively. Together with the primitive fourteenth roots of unity, these are all the

roots of our polynomial.

Explicitly, the roots are

eπi/7, e3πi/7, e5πi/7, e9πi/7, e11πi/7, e13πi/7, 1, (−5± i√39)/8.

The sum of squares of the roots of unity (including 1) is just 0 by symmetry (or a number of other

methods). The sum of the squares of the final conjugate pair is 2(52−39)82 = − 14

32 = − 716 .

31. [20] Define a power cycle to be a set S consisting of the nonnegative integer powers of an integer a,i.e. S = {1, a, a2, . . . } for some integer a. What is the minimum number of power cycles required suchthat given any odd integer n, there exists some integer k in one of the power cycles such that n ≡ k(mod 1024)?

Answer: 10 Solution 1. Partition the odd residues mod 1024 into 10 classes:

• Class 1: 1 (mod 4).

• Class n (2 ≤ n ≤ 9): 2n − 1 (mod 2n+1).

• Class 10: −1 (mod 1024).

Let Sa be the power cycle generated by a. If a is in class 1, all of Sa is in class 1. If a is in class n(2 ≤ n ≤ 9), then Sa is in the union of class n and the residues 1 (mod 2n+1). If a is in class 10, thenSa is in the union of class n and the residues 1 (mod 1024). Therefore, Sa cannot contain two of thefollowing residues: 5, 22 − 1, 23 − 1, . . . 210 − 1, and that at least 10 cycles are needed.

Note that 5128−1 = (5−1)(5+1)(52+1) · · · (564+1) has exactly 9 factors of 2 in its prime factorization,while 5256 − 1 = (5128 − 1)(5128 + 1) is divisible by 1024 so the order of 5 modulo 1024, the smallest

Guts

positive power of 5 that is congruent to 1, is 256. Observe that among 50, 51, . . . 5255, the ratio betweenany two is a positive power of 5 smaller than 5256, so the ratio is not congruent to 1 and any two termsare not congruent mod 1024. In addition, all terms are in class 1, and class 1 has 256 members, so S5

contains members congruent to each element of class 1.

Similarly, let 2 ≤ n ≤ 9. Then the order of a, where a = 2n − 1, is 210−n. The 29−n termsa1, a3, . . . a2

10−n−1 are pairwise not congruent and all in class n. Class n only has 29−n members,so Sa contains members congruent to each element of class n.

Finally, S−1 contains members congruent to the element of class 10.

The cycles S5, S−1, and 8 cycles Sa cover all the residues mod 1024, so the answer is 10.

Solution 2. Lemma. Given a positive integer n ≥ 3, there exists an odd integer x such that theorder of x modulo 2n is 2n−2.

Proof. We apply induction on n. The base cases of n = 3, 4 are clearly true with x = 3, so suppose wehave the statement holds for n− 1 and we wish to show it for n where n ≥ 5. Suppose no such integerx exists, so we have x2n−3 ≡ 1 (mod 2n) for all odd x. But then remark that (x2n−4 −1)(x2n−4

+1) ≡ 0

(mod 2n). As n− 4 ≥ 1 we have x2n−4

+ 1 ≡ 2 (mod 4) as all squares are 1 (mod 4), so it follows for

the above relation to be true we require 2n−1 divides x2n−4 − 1 for all odd x. However, by taking x tohave order 2n−3 modulo 2n−1 (which exists by the inductive hypothesis) we get a contradiction so weare done. �

Now, let x have order 28 modulo 210. Remark that if for some integer k we had xk ≡ −1 (mod 210), wewould have x2k ≡ 1 (mod 210) so 27 | k. In that case k is even so as x is odd we have xk ≡ 1 (mod 4)but −1 + 1024m is never 1 (mod 4) for any integer m so it follows −1 is not equal to xk modulo 1024for any integer k. Thus it follows that when we let S to be the set of powers of x, then no two elementsin S and −S are congruent modulo 1024. As the order of x is 256 and there are 512 possible oddremainders upon dividing by 1024, it immediately follows that every integer x is congruent modulo1024 to ±xk for some 1 ≤ k ≤ 256 and some choice of sign.

Now, it is easy so see that x,−x,−x2,−x4, . . . ,−x27 ,−x28 generating 10 power cycles works (xa for

any a is in the first, and then −x2ka for odd a is in the power cycle generated by −x2k). To showthat we cannot do any better suppose there exist 9 power cycles which include all odd integers modulo

1024. Then remark that −x2k is congruent to a number in a power cycle modulo 1024 if and only if

the power cycle is generated by an integer congruent to x−2k·a (mod 1024) where a is an odd integer.It follows that 9 integers must be congruent to −xa1 ,−x2a2 , . . . ,−x256a9 (mod 1024) for some oddintegers a1, a2, . . . , a9. But then 3 is clearly not in any of the power cycles, contradiction so it followswe must have at least 10 power cycles so we are done.

32. [20] A wealthy king has his blacksmith fashion him a large cup, whose inside is a cone of height 9inches and base diameter 6 inches (that is, the opening at the top of the cup is 6 inches in diameter).At one of his many feasts, he orders the mug to be filled to the brim with cranberry juice.

For each positive integer n, the king stirs his drink vigorously and takes a sip such that the heightof fluid left in his cup after the sip goes down by 1

n2 inches. Shortly afterwards, while the king isdistracted, the court jester adds pure Soylent to the cup until it’s once again full. The king takes sipsprecisely every minute, and his first sip is exactly one minute after the feast begins.

As time progresses, the amount of juice consumed by the king (in cubic inches) approaches a numberr. Find r.

Answer: 216π3−2187√3

8π2 First, we find the total amount of juice consumed. We can simply subtract

the amount of juice remaining at infinity from the initial amount of juice in the cup, which of courseis simply the volume of the cup; we’ll denote this value by V .

Since volume in the cup varies as the cube of height, the amount of juice remaining in the cup after mminutes is

V ·m∏

n=1

(

9− 1n2

9

)3

= V ·(

m∏

n=1

(

1− 1

9n2

)

)3

.

Guts

We can now factor the term inside the product to find

V

(

m∏

n=1

(3n+ 1)(3n− 1)

9n2

)3

= V

(

(3m+ 1)!

33m(m!)3

)3

.

If remains to evaluate the limit of this expression as m goes to infinity.

However, by Stirling’s approximation, we have

limm→∞

(3m+ 1)!

33m(m!)3= lim

m→∞

( 3n+1e )3n+1 ·

√

2π(3n+ 1)

( 3ne )3n√

(2πn)3

= limm→∞

(3n+ 1)√3

2πne

(

3n+ 1

3n

)3n

=3√3

2π.

Therefore the total amount of juice the king consumes is

V − V

(

3√3

2π

)3

=

(

32 · π · 93

)

(

8π3 − 81√3

8π3

)

=216π3 − 2187

√3

8π2.

Remark. We present another way to calculate the limit at m → ∞ of f(m) = (3m+1)!33m(m!)3 . We have

f(m+ 1) =(3m+ 4)!

33m+3(m+ 1)!3= f(m)

(m+ 23 )(m+ 4

3 )

(m+ 1)2,

whence we can write

f(m) =cΓ(m+ 2

3 )Γ(m+ 43 )

Γ(m+ 1)2

for some constant c. We can find c by equating the expressions at m = 0; we have

1 = f(0) =cΓ( 23 )Γ(

43 )

Γ(1)2,

so that c = Γ(1)2/Γ( 23 )Γ(43 ).

Of course, Γ(1) = 0! = 1. We can evaluate the other product as follows:

Γ

(

2

3

)

Γ

(

4

3

)

=1

3Γ

(

2

3

)

Γ

(

1

3

)

=1

3· π

sinπ/3=

2π

3√3.

Here the first step follows from Γ(n + 1) = nΓ(n), while the second follows from Euler’s reflectionformula. Thus c = 3

√3/2π. We can now compute

limm→∞

f(m) = limm→∞

cΓ(m+ 23 )Γ(m+ 4

3 )

Γ(m+ 1)2=

3√3

2πlim

m→∞

Γ(m+ 23 )Γ(m+ 4

3 )

Γ(m+ 1)2.

Since limn→∞ Γ(n+ α)/[Γ(n)nα] = 1, this final limit is 1 and f(m) → 3√3/2π as m → ∞.

33. [25] Let N denote the sum of the decimal digits of(

1000100

)

. Estimate the value of N . If your answer isa positive integer A written fully in decimal notation (for example, 521495223), your score will be the

greatest integer not exceeding 25 · (0.99)|A−N |. Otherwise, your score will be zero.

Answer: 621 http://www.wolframalpha.com/input/?i=sum+of+digits+of+nCr(1000,100)

To see this, one can estimate there are about 150 digits, and we expect the digits to be roughly random,for 150 · 4.5 ≈ 675, which is already very close to the actual answer. The actual number of digits is140, and here 140 · 4.5 = 630 is within 9 of the actual answer.

Guts

34. [25] For an integer n, let f(n) denote the number of pairs (x, y) of integers such that x2+xy+y2 = n.Compute the sum

106∑

n=1

nf(n).

Write your answer in the form a · 10b, where b is an integer and 1 ≤ a < 10 is a decimal number.

If your answer is written in this form, your score will be max{0, 25 − ⌊100|log10(A/N)|⌋)}, whereN = a · 10b is your answer to this problem and A is the actual answer. Otherwise, your score will bezero.

Answer: 1.813759629294 · 1012 Rewrite the sum as

∑

x2+xy+y2≤106

(x2 + xy + y2),

where the sum is over all pairs (x, y) of integers with x2 + xy + y2 ≤ 106. We can find a crude upperbound for this sum by noting that

x2 + xy + y2 =3

4x2 +

(x

2+ y)2

≥ 3

4x2,

so each term of this sum has |x| ≤ 2√3103. Similarly, |y| ≤ 2√

3103. Therefore, the number of terms in

the sum is at most(

4√3103 + 1

)2

≈ 106.

(We are throwing away “small” factors like 163 in the approximation.) Furthermore, each term in the

sum is at most 106, so the total sum is less than about 1012. The answer 1 · 1012 would unfortunatelystill get a score of 0.

For a better answer, we can approximate the sum by an integral:

∑

x2+xy+y2≤106

(x2 + xy + y2) ≈∫∫

x2+xy+y2≤106(x2 + xy + y2) dy dx.

Performing the change of variables (u, v) =(√

32 x, 1

2x+ y)

and then switching to polar coordinates

(r, θ) = (√u2 + v2, tan−1(v/u)) yields

∫∫

x2+xy+y2≤106(x2 + xy + y2) dy dx =

2√3

∫∫

u2+v2≤106(u2 + v2) dv du

=2√3

∫ 2π

0

∫ 103

0

r3 dr dθ

=4π√3

∫ 103

0

r3 dr

=π√3· 1012.

This is approximately 1.8138 · 1012, which is much closer to the actual answer. (An answer of 1.8 · 1012is good enough for full credit.)

The answer can also be computed exactly by the Common Lisp code:

(defconstant +MAX+ 1e6)

(defvar +lower+ -2000)

(defvar +upper+ 2000)

(princ

Guts

(loop for x from +lower+ to +upper+ sum

(loop for y from +lower+ to +upper+

sum

(let ((S (+ (* x x) (* x y) (* y y))))

(if (and (<= S +MAX+) (> S 0)) S 0)))))

35. [25] Let P denote the set of all subsets of {1, . . . , 23}. A subset S ⊆ P is called good if whenever A,Bare sets in S, the set (A \ B) ∪ (B \ A) is also in S. (Here, A \ B denotes the set of all elements in Athat are not in B, and B \ A denotes the set of all elements in B that are not in A.) What fractionof the good subsets of P have between 2015 and 3015 elements, inclusive?

If your answer is a decimal number or a fraction (of the form m/n, where m and n are positive integers),then your score on this problem will be equal to max{0, 25− ⌊1000|A−N |⌋}, where N is your answerand A is the actual answer. Otherwise, your score will be zero.

Answer: 1883918387767004194221830714712250060123547691684840486192422095701784512492731212 ≈ 0.3950203047068107 Let n = 23, and ℓ =

⌊n/2⌋ = 11.

We use the well-known rephrasing of the symmetric difference ((A \B)∪ (B \A)) in terms of additionmodulo 2 of “indicators/characteristic vectors”. So we simply want the number

(

nℓ

)

2of dimension ℓ

subspaces of the F := F2-vector space V := Fn2 . Indeed, good subsets of 2d elements simply correspond

to dimension d subspaces (in particular, good subsets can only have sizes equal to powers of |F | = 2,and 2ℓ is the only power between 2015 and 3015, inclusive).

To do this, it’s easier to first count the number of (ordered) tuples of ℓ linearly independent elements ofV , and divide (to get the subspace count) by the number of (ordered) tuples of ℓ linearly independentelements of any ℓ-dimensional subspace of V (a well-defined number independent of the choice ofsubspace).

In general, if we want to count tuples of m linearly independent elements in an n-dimensional space(with n ≥ m), just note that we are building on top of (0) (the zero-dimensional subspace), and oncewe’ve chosen r ≤ m elements (with 0 ≤ r ≤ m − 1), there are 2n − 2r elements linearly independentto the previous r elements (which span a subspace of dimension r, hence of 2r “bad” elements). Thusthe number of m-dimensional subspaces of an n-dimensional space is

(

n

m

)

2

:=(2n − 20)(2n − 21) · · · (2n − 2m−1)

(2m − 20)(2m − 21) · · · (2m − 2m−1),

a “Gaussian binomial coefficient.”

We want to estimate(

nℓ

)

2∑n

m=0

(

nm

)

2

=1

2

(

2311

)

2∑11

m=0

(

23m

)

2

.

To do this, note that(

nm

)

2=(

nn−m

)

2, so we may restrict our attention to the lower half. Intuitively, the

Gaussian binomial coefficients should decay exponentially (or similarly quickly) away from the center;indeed, if m ≤ n/2, then

(

n

m− 1

)

2

/

(

n

m

)

2

=(2m − 20) · 2m−1

(2n − 2m−1)≈ 22m−1−n.

So in fact, the decay is super-exponential, starting (for n = 23 odd and m ≤ ℓ = 11) at an ≈ 14 rate.

So most of the terms (past the first 2 to 4, say) are negligible in our estimation. If we use the first twoterms, we get an approximation of 1

2 · 11+ 1

4

= 25 = 0.4, which is enough for 20 points. (Including the

next term gives an approximation of 3281 ≈ 0.3950617, which is good enough to get full credit.)

To compute the exact answer, we used the following python3 code:

Guts

from functools import lru_cache

from fractions import Fraction

@lru_cache(maxsize=None)

def gauss_binom(n, k, e):

if k < 0 or k > n:

return 0

if k == 0 or k == n:

return 1

return e ** k * gauss_binom(n - 1, k, e) + \

gauss_binom(n - 1, k - 1, e)

N = 23

K = 11

good = gauss_binom(N, K, 2)

total = sum(gauss_binom(N, i, 2) for i in range(N + 1))

print(Fraction(good, total))

print(float(good/total))

36. [25] A prime number p is twin if at least one of p+2 or p− 2 is prime and sexy if at least one of p+6and p− 6 is prime.

How many sexy twin primes (i.e. primes that are both twin and sexy) are there less than 109? Expressyour answer as a positive integer N in decimal notation; for example, 521495223. If your answer isin this form, your score for this problem will be max

{

0, 25− ⌊ 110000 |A−N |

⌋

}, where A is the actualanswer to this problem. Otherwise, your score will be zero.

Answer: 1462105 The Hardy-Littlewood conjecture states that given a set A of integers, thenumber of integers x such that x+ a is a prime for all a ∈ A is

x

(lnx)|A|

∏

p

1− w(p;A)p

(1− 1p )

k(1 + o(1))

where w(p;A) is the number of distinct residues of A modulo p and the o(1) term goes to 0 as x goes toinfinity. Note that for the 4 tuples of the form (0,±2,±6), w(p;A) = 3, and using the approximation1−k/p

(1−1/p)k≈ 1−

(

k2

)

/p2 ≈ (1− 1p2 )

(k2), we have

∏

p>3

1− kp

(1− 1p )

k≈(

6

π2

)(k2)·(

4

3

)(k2)(9

8

)(k2)≈(

9

10

)(k2)

Applying this for the four sets A = (0,±2,±6), x = 109 (and approximating lnx = 20 and just takingthe p = 2 and p = 3 terms, we get the approximate answer

4 · 109

2031− 1

2

( 12 )3

1− 23

( 13 )3

(

9

10

)3

= 1640250

One improvement we can make is to remove the double-counted tuples, in particular, integers x suchthat x, x+ 6, x− 6, and one of x± 2 is prime. Again by the Hardy-Littlewood conjecture, the numberof such x is approximately (using the same approximations)

2 · 109

2041− 1

2

( 12 )4

1− 23

( 13 )4

(

9

10

)6

≈ 90000

Subtracting gives an estimate of about 1550000. Note that this is still an overestimate, as ln 109 is

actually about 20.7 and 1−k/p(1−1/p)k

< (1− 1p2 )

(k2).

Here is the C++ code that we used to generate the answer:

Guts

#include<iostream>

#include<cstring> // memset

using namespace std;

const int MAXN = 1e9;

bool is_prime[MAXN + 6];

int main(){

// Sieve of Eratosthenes

memset(is_prime, true, sizeof(is_prime));

is_prime[0] = is_prime[1] = false;

for (int i=2; i<MAXN + 6; i++){

if (is_prime[i]){

for (int j=2 * i; j < MAXN + 6; j += i){

is_prime[j] = false;

}

}

}

// Count twin sexy primes.

int ans = 1; // 5 is the only twin sexy prime < 6.

for (int i=6; i<MAXN; i++){

if (is_prime[i]

&& (is_prime[i-6] || is_prime[i+6])

&& (is_prime[i-2] || is_prime[i+2])) {

ans++;

}

}

cout << ans << endl;

return 0;

}

Guts

HMMT November 2015November 14, 2015

Individual

1. Find the number of triples (a, b, c) of positive integers such that a + ab + abc = 11.

2. Let a and b be real numbers randomly (and independently) chosen from the range [0, 1]. Find theprobability that a, b and 1 form the side lengths of an obtuse triangle.

3. Neo has an infinite supply of red pills and blue pills. When he takes a red pill, his weight will double,and when he takes a blue pill, he will lose one pound. If Neo originally weighs one pound, what is theminimum number of pills he must take to make his weight 2015 pounds?

4. Chords AB and CD of a circle are perpendicular and intersect at a point P . If AP = 6, BP = 12, andCD = 22, find the area of the circle.

5. Let S be a subset of the set {1, 2, 3, . . . , 2015} such that for any two elements a, b ∈ S, the differencea− b does not divide the sum a + b. Find the maximum possible size of S.

6. Consider all functions f : Z→ Z satisfying

f(f(x) + 2x + 20) = 15.

Call an integer n good if f(n) can take any integer value. In other words, if we fix n, for any integerm, there exists a function f such that f(n) = m. Find the sum of all good integers x.

7. Let 4ABC be a right triangle with right angle C. Let I be the incenter of ABC, and let M lie on ACand N on BC, respectively, such that M, I,N are collinear and MN is parallel to AB. If AB = 36and the perimeter of CMN is 48, find the area of ABC.

8. Let ABCD be a quadrilateral with an inscribed circle ω that has center I. If IA = 5, IB = 7, IC =4, ID = 9, find the value of AB

CD .

9. Rosencrantz plays n ≤ 2015 games of question, and ends up with a win rate(

i.e. # of games won# of games played

)of

k. Guildenstern has also played several games, and has a win rate less than k. He realizes that if,after playing some more games, his win rate becomes higher than k, then there must have been somepoint in time when Rosencrantz and Guildenstern had the exact same win-rate. Find the product ofall possible values of k.

10. Let N be the number of functions f from {1, 2, . . . , 101} → {1, 2, . . . , 101} such that f101(1) = 2. Findthe remainder when N is divided by 103.

HMMT November 2015November 14, 2015

Individual

1. Find the number of triples (a, b, c) of positive integers such that a+ ab+ abc = 11.

Proposed by: Yang Liu

Answer: 3

We can write a+ ab+ abc = a(1 + b+ bc). Since 11 is prime, a = 11 or a = 1. But since b, c are bothpositive integers, we cannot have a = 11, and so a = 1. Then 1 + b + bc = 11 =⇒ b + bc = 10 =⇒b(c + 1) = 10, and since c is a positive integer, only b = 1, 2, 5 are possible. This gives the 3 triples(a, b, c) = (1, 1, 9), (1, 2, 4), (1, 5, 1).

2. Let a and b be real numbers randomly (and independently) chosen from the range [0, 1]. Find theprobability that a, b and 1 form the side lengths of an obtuse triangle.

Proposed by: Alexander Katz

Answer: π→24

We require a + b > 1 and a2 + b2 < 1. Geometrically, this is the area enclosed in the quarter-circlecentered at the origin with radius 1, not including the area enclosed by a + b < 1 (an isosceles righttriangle with side length 1). As a result, our desired probability is π→2

4 .

3. Neo has an infinite supply of red pills and blue pills. When he takes a red pill, his weight will double,and when he takes a blue pill, he will lose one pound. If Neo originally weighs one pound, what is theminimum number of pills he must take to make his weight 2015 pounds?

Proposed by: Alexander Katz

Answer: 13

Suppose instead Neo started at a weight of 2015 pounds, instead had green pills, which halve hisweight, and purple pills, which increase his weight by a pound, and he wished to reduce his weight toone pound. It is clear that, if Neo were able to find such a sequence of pills in the case where he goesfrom 2015 pounds to 1 pound, he can perform the sequence in reverse (replacing green pills with redpills and purple pills with blue pills) to achieve the desired weight, so this problem is equivalent to theoriginal.

Suppose at some point, Neo were to take two purple pills followed by a green pill; this changes hisweight from 2k to k + 1. However, the same effect could be achieved using less pills by first taking agreen pill and then taking a purple pill, so the optimal sequence will never contain consecutive purplepills. As a result, there is only one optimal sequence for Neo if he is trying to lose weight: take a purplepill when his weight is odd, and a green pill when his weight is even. His weight thus becomes

2015→ 2016→ 1008→ 504→ 252→ 126→ 63

→ 64→ 32→ 16→ 8→ 4→ 2→ 1

which requires a total of 13 pills. Reversing this sequence solves the original problem directly.

4. Chords AB and CD of a circle are perpendicular and intersect at a point P . If AP = 6, BP = 12, andCD = 22, find the area of the circle.

Proposed by: Sam Korsky

Answer: 130π

Let O be the center of the circle and let M be the midpoint of segment AB and let N be the midpointof segment CD. Since quadrilateral OMPN is a rectangle we have that ON = MP = AM −AP = 3so

OC =√ON2 +NC2 =

√9 + 121 =

√130

Hence the desired area is 130π .

5. Let S be a subset of the set {1, 2, 3, . . . , 2015} such that for any two elements a, b ∈ S, the differencea− b does not divide the sum a+ b. Find the maximum possible size of S.

Proposed by: Sam Korsky

Answer: 672

From each of the sets {1, 2, 3}, {4, 5, 6}, {7, 8, 9}, . . . at most 1 element can be in S. This leads to an

upper bound of⌈20153

⌉= 672 which we can obtain with the set {1, 4, 7, . . . , 2014}.

6. Consider all functions f : Z→ Z satisfying

f(f(x) + 2x+ 20) = 15.

Call an integer n good if f(n) can take any integer value. In other words, if we fix n, for any integerm, there exists a function f such that f(n) = m. Find the sum of all good integers x.

Proposed by: Yang Liu

Answer: -35

For almost all integers x, f(x) 6= −x− 20. If f(x) = −x− 20, then

f(−x− 20 + 2x+ 20) = 15 =⇒ −x− 20 = 15 =⇒ x = −35.

Now it suffices to prove that the f(−35) can take any value.

f(−35) = 15 in the function f(x) ≡ 15. Otherwise, set f(−35) = c, and f(x) = 15 for all other x. Itis easy to check that these functions all work.

7. Let 4ABC be a right triangle with right angle C. Let I be the incenter of ABC, and let M lie on ACand N on BC, respectively, such that M, I,N are collinear and MN is parallel to AB. If AB = 36and the perimeter of CMN is 48, find the area of ABC.

Proposed by: Alexander Katz

Answer: 252

Note that ∠MIA = ∠BAI = ∠CAI, so MI = MA. Similarly, NI = NB. As a result, CM +MN + NC = CM + MI + NI + NC = CM + MA + NB + NC = AC + BC = 48. Furthermore,AC2+BC2 = 362. As a result, we have AC2+2AC ·BC+BC2 = 482, so 2AC ·BC = 482−362 = 12·84,and so AC·BC

2 = 3 · 84 = 252 .

8. Let ABCD be a quadrilateral with an inscribed circle ω that has center I. If IA = 5, IB = 7, IC =4, ID = 9, find the value of AB

CD .

Proposed by: Sam Korsky

Answer: 3536

The I-altitudes of triangles AIB and CID are both equal to the radius of ω, hence have equal length.

Therefore [AIB][CID] = AB

CD . Also note that [AIB] = IA · IB · sinAIB and [CID] = IC · ID · sinCID,

but since lines IA, IB, IC, ID bisect angles ∠DAB,∠ABC,∠BCD,∠CDA respectively we have that∠AIB+∠CID = (180◦−∠IAB−∠IBA) + (180◦−∠ICD−∠IDC) = 180◦. So, sinAIB = sinCID.

Therefore [AIB][CID] = IA·IB

IC·ID . Hence

AB

CD=IA · IBIC · ID

=35

36.

9. Rosencrantz plays n ≤ 2015 games of question, and ends up with a win rate(

i.e. # of games won# of games played

)of

k. Guildenstern has also played several games, and has a win rate less than k. He realizes that if,after playing some more games, his win rate becomes higher than k, then there must have been somepoint in time when Rosencrantz and Guildenstern had the exact same win-rate. Find the product ofall possible values of k.

Proposed by: Alexander Katz

Answer: 12015

Write k = mn , for relatively prime integers m,n. For the property not to hold, there must exist integers

a and b for whicha

b<m

n<a+ 1

b+ 1

(i.e. at some point, Guildenstern must ”jump over” k with a single win)

⇐⇒ an+ n−m > bm > an

hence there must exist a multiple of m strictly between an and an+ n−m.

If n−m = 1, then the property holds as there is no integer between an and an+n−m = an+1. We nowshow that if n−m 6= 1, then the property does not hold. By Bzout’s Theorem, as n and m are relativelyprime, there exist a and x such that an = mx−1, where 0 < a < m. Then an+n−m ≥ an+2 = mx+1,so b = x satisfies the conditions. As a result, the only possible k are those in the form n

n+1 .

We know that Rosencrantz played at most 2015 games, so the largest non-perfect winrate he could

possibly have is 20142015 . Therefore, k ∈ { 12 ,

23 , . . . ,

20142015}, the product of which is

1

2015.

10. Let N be the number of functions f from {1, 2, . . . , 101} → {1, 2, . . . , 101} such that f101(1) = 2. Findthe remainder when N is divided by 103.

Proposed by: Yang Liu

Answer: 43

For convenience, let n = 101. Compute the number of functions such that fn(1) = 1. Since n is aprime, there are 2 cases: the order of 1 is either 1 or n. The first case gives nn→1 functions, and thesecond case gives (n− 1)! functions. By symmetry, the number of ways for fn(1) = 2 is

1

n− 1· (nn − nn→1 − (n− 1)!) = nn→1 − (n− 2)!.

Plugging in n = 101, we need to find

101100 − 99! ≡ (−2)→2 − 101!

6

= 1/4− 1/6 = 1/12 = 43 (mod 103).

HMMT November 2015November 14, 2015

Theme round

1. Consider a 1×1 grid of squares. Let A;B;C;D be the vertices of this square, and let E be the midpointof segment CD. Furthermore, let F be the point on segment BC satisfying BF = 2CF , and let P bethe intersection of lines AF and BE. Find AP

PF .

2. Consider a 2× 2 grid of squares. David writes a positive integer in each of the squares. Next to eachrow, he writes the product of the numbers in the row, and next to each column, he writes the productof the numbers in each column. If the sum of the eight numbers he writes down is 2015, what is theminimum possible sum of the four numbers he writes in the grid?

3. Consider a 3 × 3 grid of squares. A circle is inscribed in the lower left corner, the middle square ofthe top row, and the rightmost square of the middle row, and a circle O with radius r is drawn suchthat O is externally tangent to each of the three inscribed circles. If the side length of each square is1, compute r.

4. Consider a 4× 4 grid of squares. Aziraphale and Crowley play a game on this grid, alternating turns,with Aziraphale going first. On Aziraphales turn, he may color any uncolored square red, and onCrowleys turn, he may color any uncolored square blue. The game ends when all the squares arecolored, and Aziraphales score is the area of the largest closed region that is entirely red. If Aziraphalewishes to maximize his score, Crowley wishes to minimize it, and both players play optimally, whatwill Aziraphales score be?

5. Consider a 5 × 5 grid of squares. Vladimir colors some of these squares red, such that the centers ofany four red squares do not form an axis-parallel rectangle (i.e. a rectangle whose sides are parallel tothose of the squares). What is the maximum number of squares he could have colored red?

6. Consider a 6 × 6 grid of squares. Edmond chooses four of these squares uniformly at random. Whatis the probability that the centers of these four squares form a square?

7. Consider a 7 × 7 grid of squares. Let f : {1; 2; 3; 4; 5; 6; 7} → {1; 2; 3; 4; 5; 6; 7} be a function; in otherwords, f(1); f(2); : : : ; f(7) are each (not necessarily distinct) integers from 1 to 7. In the top row ofthe grid, the numbers from 1 to 7 are written in order; in every other square, f(x) is written where x isthe number above the square. How many functions have the property that the bottom row is identicalto the top row, and no other row is identical to the top row?

8. Consider an 8× 8 grid of squares. A rook is placed in the lower left corner, and every minute it movesto a square in the same row or column with equal probability (the rook must move; i.e. it cannot stayin the same square). What is the expected number of minutes until the rook reaches the upper rightcorner?

9. Consider a 9× 9 grid of squares. Haruki fills each square in this grid with an integer between 1 and 9,inclusive. The grid is called a super-sudoku if each of the following three conditions hold:

• Each column in the grid contains each of the numbers 1; 2; 3; 4; 5; 6; 7; 8; 9 exactly once.

• Each row in the grid contains each of the numbers 1; 2; 3; 4; 5; 6; 7; 8; 9 exactly once.

• Each 3× 3 subsquare in the grid contains each of the numbers 1; 2; 3; 4; 5; 6; 7; 8; 9 exactly once.

How many possible super-sudoku grids are there?

10. Consider a 10 × 10 grid of squares. One day, Daniel drops a burrito in the top left square, where awingless pigeon happens to be looking for food. Every minute, if the pigeon and the burrito are inthe same square, the pigeon will eat 10% of the burrito’s original size and accidentally throw it intoa random square (possibly the one it is already in). Otherwise, the pigeon will move to an adjacentsquare, decreasing the distance between it and the burrito. What is the expected number of minutesbefore the pigeon has eaten the entire burrito?

HMMT November 2015November 14, 2015

Theme round

1. Consider a 1×1 grid of squares. Let A,B,C,D be the vertices of this square, and let E be the midpointof segment CD. Furthermore, let F be the point on segment BC satisfying BF = 2CF , and let P bethe intersection of lines AF and BE. Find AP

PF .

Proposed by: Sam Korsky

Answer: 3

Let line BE hit line DA at Q. It’s clear that triangles AQP and FBP are similar so

AP

PF=

AQ

BF=

2AD23BC

= 3

2. Consider a 2× 2 grid of squares. David writes a positive integer in each of the squares. Next to eachrow, he writes the product of the numbers in the row, and next to each column, he writes the productof the numbers in each column. If the sum of the eight numbers he writes down is 2015, what is theminimum possible sum of the four numbers he writes in the grid?

Proposed by: Alexander Katz

Answer: 88

Let the four numbers be a, b, c, d, so that the other four numbers are ab, ad, bc, bd. The sum of theseeight numbers is a + b + c + d + ab + ad + bc + bd = (a + c) + (b + d) + (a + c)(b + d) = 2015, andso (a + c + 1)(b + d + 1) = 2016. Since we seek to minimize a + b + c + d, we need to find the twofactors of 2016 that are closest to each other, which is easily calculated to be 42 ·48 = 2016; this makesa + b + c + d = 88 .

3. Consider a 3 × 3 grid of squares. A circle is inscribed in the lower left corner, the middle square ofthe top row, and the rightmost square of the middle row, and a circle O with radius r is drawn suchthat O is externally tangent to each of the three inscribed circles. If the side length of each square is1, compute r.

Proposed by: Sam Korsky

Answer: 5√2−36

Let A be the center of the square in the lower left corner, let B be the center of the square in themiddle of the top row, and let C be the center of the rightmost square in the middle row. It’s clearthat O is the circumcenter of triangle ABC - hence, the desired radius is merely the circumradius oftriangle ABC minus 1

2 . Now note that by the Pythagorean theorem, BC =√

2 and AB = AC =√

5 so

we easily find that the altitude from A in triangle ABC has length 3√2

2 . Therefore the area of triangleABC is 3

2 . Hence the circumradius of triangle ABC is given by

BC · CA ·AB

4 · 32=

5√

2

6

and so the answer is 5√2

6 〉12 =

5√

2〉 3

6.

4. Consider a 4× 4 grid of squares. Aziraphale and Crowley play a game on this grid, alternating turns,with Aziraphale going first. On Aziraphales turn, he may color any uncolored square red, and onCrowleys turn, he may color any uncolored square blue. The game ends when all the squares arecolored, and Aziraphales score is the area of the largest closed region that is entirely red. If Aziraphalewishes to maximize his score, Crowley wishes to minimize it, and both players play optimally, whatwill Aziraphales score be?

Proposed by: Alexander Katz

Answer: 6

We claim that the answer is 6.

On Aziraphale’s first two turns, it is always possible for him to take 2 adjacent squares from thecentral four; without loss of generality, suppose they are the squares at (1, 1) and (1, 2). If allowed,Aziraphale’s next turn will be to take one of the remaining squares in the center, at which point therewill be seven squares adjacent to a red square, and so Aziraphale can guarantee at least two moreadjacent red squares. After that, since the number of blue squares is always at most the number ofred squares, Aziraphale can guarantee another adjacent red square, making his score at least 6.

If, however, Crowley does not allow Aziraphale to attain another central red square – i.e. coloring theother two central squares blue – then Aziraphale will continue to take squares from the second row,WLOG (1, 3). If Aziraphale is also allowed to take (1, 0), he will clearly attain at least 6 adjacent redsquares as each red square in this row has two adjacent squares to it, and otherwise (if Crowley takes(1, 0)), Aziraphale will take (0, 1) and guarantee a score of at least 4 + 4

2 = 6 as there are 4 uncoloredsquares adjacent to a red one.

Therefore, the end score will be at least 6. We now show that this is the best possible for Aziraphale;i.e. Crowley can always limit the score to 6. Crowley can play by the following strategy: if Aziraphalecolors a square in the second row, Crowley will color the square below it, if Aziraphale colors a squarein the third row, Crowley will color the square above it. Otherwise, if Aziraphale colors a square inthe first or fourth rows, Crowley will color an arbitrary square in the same row. It is clear that the two”halves” of the board cannot be connected by red squares, and so the largest contiguous red regionwill occur entirely in one half of the grid, but then the maximum score is 4 + 4

2 = 6.

The optimal score is thus both at least 6 and at most 6, so it must be 6 as desired.

5. Consider a 5 × 5 grid of squares. Vladimir colors some of these squares red, such that the centers ofany four red squares do not form an axis-parallel rectangle (i.e. a rectangle whose sides are parallel tothose of the squares). What is the maximum number of squares he could have colored red?

Proposed by: Sam Korsky

Answer: 12

We claim that the answer is 12. We first show that if 13 squares are colored red, then some four forman axis-parallel rectangle. Note that we can swap both columns and rows without affecting whetherfour squares form a rectangle, so we may assume without loss of generality that the top row has themost red squares colored; suppose it has k squares colored. We may further suppose that, without lossof generality, these k red squares are the first k squares in the top row from the left.

Consider the k × 5 rectangle formed by the first k columns. In this rectangle, no more than 1 squareper row can be red (excluding the top one), so there are a maximum of k+4 squares colored red. In theremaining (5〉k)×5 rectangle, at most 4(5〉k) squares are colored red (as the top row of this rectanglehas no red squares), so there are a maximum of (k + 4) + 4(5〉 k) = 24〉 3k squares colored red in the5× 5 grid. By assumption, at least 13 squares are colored red, so we have 13 ≤ 24〉 3k ⇐⇒ k ≤ 3.

Hence there are at most 3 red squares in any row. As there are at least 13 squares colored red, thisimplies that at least 3 rows have 3 red squares colored. Consider the 3× 5 rectangle formed by thesethree rows. Suppose without loss of generality that the leftmost three squares in the top row arecolored red, which forces the rightmost three squares in the second row to be colored red. But then,by the Pigeonhole Principle, some 2 of the 3 leftmost squares or some 2 of the 3 rightmost squares inthe bottom row will be colored red, leading to an axis-parallel rectangle – a contradiction.

Hence there are most 12 squares colored red. It remains to show that there exists some coloring whereexactly 12 squares are colored red, one example of which is illustrated below:

R R R RR RR RR RR R

The maximum number of red squares, therefore, is 12 .

6. Consider a 6 × 6 grid of squares. Edmond chooses four of these squares uniformly at random. Whatis the probability that the centers of these four squares form a square?

Proposed by: Alexander Katz

Answer: 1561 OR 105

(364 )

Firstly, there are364

)possible combinations of points. Call a square proper if its sides are parallel to

the coordinate axes and improper otherwise. Note that every improper square can be inscribed in aunique proper square. Hence, an n × n proper square represents a total of n squares: 1 proper andn〉 1 improper.

There are thus a total of

6∑i=1

i(6〉 i)2 =

6∑i=1

(i3 〉 12i2 + 36i)

=

6∑i=1

i3 〉 12

6∑i=1

i2 + 36∑

i = 16i

= 441〉 12(91) + 36(21)

= 441〉 1092 + 756

= 105

squares on the grid. Our desired probability is thus105364

) =1

561.

7. Consider a 7 × 7 grid of squares. Let f : {1, 2, 3, 4, 5, 6, 7} → {1, 2, 3, 4, 5, 6, 7} be a function; in otherwords, f(1), f(2), . . . , f(7) are each (not necessarily distinct) integers from 1 to 7. In the top row ofthe grid, the numbers from 1 to 7 are written in order; in every other square, f(x) is written where x isthe number above the square. How many functions have the property that the bottom row is identicalto the top row, and no other row is identical to the top row?

Proposed by: Alexander Katz

Answer: 1470

Consider the directed graph with 1, 2, 3, 4, 5, 6, 7 as vertices, and there is an edge from i to j if andonly if f(i) = j. Since the bottom row is equivalent to the top one, we have f6(x) = x. Therefore, thegraph must decompose into cycles of length 6, 3, 2, or 1. Furthermore, since no other row is equivalentto the top one, the least common multiple of the cycle lengths must be 6. The only partitions of 7satisfying these constraints are 7 = 6 + 1, 7 = 3 + 2 + 2, and 7 = 3 + 2 + 1 + 1.

If we have a cycle of length 6 and a cycle of length 1, there are 7 ways to choose which six vertices willbe in the cycle of length 6, and there are 5! = 120 ways to determine the values of f within this cycle(to see this, pick an arbitrary vertex in the cycle: the edge from it can connect to any of the remaining5 vertices, which can connect to any of the remaining 4 vertices, etc.). Hence, there are 7 · 120 = 840possible functions f in this case.

If we have a cycle of length 3 and two cycles of length 2, there are(72)(

52)

2 = 105 possible ways to assignwhich vertices will belong to which cycle (we divide by two to avoid double-counting the cycles of

length 2). As before, there are 2! · 1! · 1! = 2 assignments of f within the cycles, so there are a total of210 possible functions f in this case.

Finally, if we have a cycle of length 3, a cycle of length 2, and two cycles of length 1, there are73

)42

)= 210 possible ways to assign the cycles, and 2! · 1! · 0! · 0! = 2 ways to arrange the edges within

the cycles, so there are a total of 420 possible functions f in this case.

Hence, there are a total of 840 + 210 + 420 = 1470 possible f .

8. Consider an 8× 8 grid of squares. A rook is placed in the lower left corner, and every minute it movesto a square in the same row or column with equal probability (the rook must move; i.e. it cannot stayin the same square). What is the expected number of minutes until the rook reaches the upper rightcorner?

Proposed by: Alexander Katz

Answer: 70

Let the expected number of minutes it will take the rook to reach the upper right corner from the topor right edges be Ee, and let the expected number of minutes it will take the rook to reach the upperright corner from any other square be Ec. Note that this is justified because the expected time fromany square on the top or right edges is the same, as is the expected time from any other square (thisis because swapping any two rows or columns doesn’t affect the movement of the rook). This gives ustwo linear equations:

Ec =2

14(Ee + 1) +

12

14(Ec + 1)

Ee =1

14(1) +

6

14(Ee + 1) +

7

14(Ec + 1)

which gives the solution Ee = 63, Ec = 70 .

9. Consider a 9× 9 grid of squares. Haruki fills each square in this grid with an integer between 1 and 9,inclusive. The grid is called a super-sudoku if each of the following three conditions hold:

• Each column in the grid contains each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 exactly once.

• Each row in the grid contains each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 exactly once.

• Each 3× 3 subsquare in the grid contains each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 exactly once.

How many possible super-sudoku grids are there?

Proposed by: Alexander Katz

Answer: 0

Without loss of generality, suppose that the top left corner contains a 1, and examine the top left 3×4:

1 x x xx x x *x x x *

There cannot be another 1 in any of the cells marked with an x, but the 3×3 on the right must containa 1, so one of the cells marked with a * must be a 1. Similarly, looking at the top left 4× 3:

1 x xx x xx x xx * *

One of the cells marked with a * must also contain a 1. But then the 3 × 3 square diagonally belowthe top left one:

1 x x xx x x *x x x *x * * ?

must contain multiple 1s, which is a contradiction. Hence no such supersudokus exist.

10. Consider a 10 × 10 grid of squares. One day, Daniel drops a burrito in the top left square, where awingless pigeon happens to be looking for food. Every minute, if the pigeon and the burrito are inthe same square, the pigeon will eat 10% of the burrito’s original size and accidentally throw it intoa random square (possibly the one it is already in). Otherwise, the pigeon will move to an adjacentsquare, decreasing the distance between it and the burrito. What is the expected number of minutesbefore the pigeon has eaten the entire burrito?

Proposed by: Sam Korsky

Answer: 71.8

Label the squares using coordinates, letting the top left corner be (0, 0). The burrito will end up in 10(not necessarily different) squares. Call them p1 = (x1, y1) = (0, 0), p2 = (x2, y2), . . . , p10 = (x10, y10).p2 through p10 are uniformly distributed throughout the square. Let di = |xi+1 〉 xi|+ |yi+1 〉 yi|, thetaxicab distance between pi and pi+1.

After 1 minute, the pigeon will eat 10% of the burrito. Note that if, after eating the burrito, the pigeonthrows it to a square taxicab distance d from the square it’s currently in, it will take exactly d minutesfor it to reach that square, regardless of the path it takes, and another minute for it to eat 10% of theburrito.

Hence, the expected number of minutes it takes for the pigeon to eat the whole burrito is

1 + E

(9∑

i=1

(di + 1)

)= 1 + E

(9∑

i=1

1 + |xi+1 〉 xi|+ |yi+1 〉 yi|

)

= 10 + 2 · E

(9∑

i=1

|xi+1 〉 xi|

)

= 10 + 2 ·

(E(|x2|) + E(

9∑i=2

|xi+1 〉 xi|)

)= 10 + 2 · (E(|x2|) + 8 · E(|xi+1 〉 xi|))

= 10 + 2 ·

(4.5 + 8 · 1

100·

9∑k=1

k(20〉 2k)

)= 10 + 2 · (4.5 + 8 · 3.3)

= 71.8

HMMT November 2015November 14, 2015

Team

1. [3] Triangle ABC is isosceles, and ∠ABC = x◦. If the sum of the possible measures of ∠BAC is 240◦,find x.

2. [3] Bassanio has three red coins, four yellow coins, and five blue coins. At any point, he may giveShylock any two coins of different colors in exchange for one coin of the other color; for example, hemay give Shylock one red coin and one blue coin, and receive one yellow coin in return. Bassaniowishes to end with coins that are all the same color, and he wishes to do this while having as manycoins as possible. How many coins will he end up with, and what color will they be?

3. [3] Let bxc denote the largest integer less than or equal to x, and let fxg denote the fractional part ofx. For example, bπc = 3, and fπg = 0.14159 . . . , while b100c = 100 and f100g = 0. If n is the largest

solution to the equation bncn = 20152016 , compute fng.

4. [5] Call a set of positive integers good if there is a partition of it into two sets S and T , such that theredo not exist three elements a, b, c 2 S such that ab = c and such that there do not exist three elementsa, b, c 2 T such that ab = c (a and b need not be distinct). Find the smallest positive integer n suchthat the set f2, 3, 4, . . . , ng is not good.

5. [5] Kelvin the Frog is trying to hop across a river. The river has 10 lilypads on it, and he must hopon them in a specific order (the order is unknown to Kelvin). If Kelvin hops to the wrong lilypad atany point, he will be thrown back to the wrong side of the river and will have to start over. AssumingKelvin is infinitely intelligent, what is the minimum number of hops he will need to guarantee reachingthe other side?

6. [5] Marcus and four of his relatives are at a party. Each pair of the five people are either friends orenemies. For any two enemies, there is no person that they are both friends with. In how many waysis this possible?

7. [6] Let ABCD be a convex quadrilateral whose diagonals AC and BD meet at P . Let the area oftriangle APB be 24 and let the area of triangle CPD be 25. What is the minimum possible area ofquadrilateral ABCD?

8. [6] Find any quadruple of positive integers (a, b, c, d) satisfying a3 + b4 + c5 = d11 and abc < 105.

9. [7] A graph consists of 6 vertices. For each pair of vertices, a coin is flipped, and an edge connectingthe two vertices is drawn if and only if the coin shows heads. Such a graph is good if, starting fromany vertex V connected to at least one other vertex, it is possible to draw a path starting and endingat V that traverses each edge exactly once. What is the probability that the graph is good?

10. [7] A number n is bad if there exists some integer c for which xx ≡ c (mod n) has no integer solutionsfor x. Find the number of bad integers between 2 and 42 inclusive.

HMMT November 2015November 14, 2015

Team

1. [3] Triangle ABC is isosceles, and ∠ABC = x◦. If the sum of the possible measures of ∠BAC is 240◦,find x.

Proposed by: Alexander Katz

Answer: 20

There are three possible triangles: either ∠ABC = ∠BCA, in which case ∠BAC = 180−2x, ∠ABC =∠BAC, in which case ∠BAC = x, or ∠BAC = ∠BCA, in which case ∠BAC = 180−x

2 . These sum to540−3x

2 , so we have 540−3x2 = 240 =⇒ x = 20 .

2. [3] Bassanio has three red coins, four yellow coins, and five blue coins. At any point, he may giveShylock any two coins of different colors in exchange for one coin of the other color; for example, hemay give Shylock one red coin and one blue coin, and receive one yellow coin in return. Bassaniowishes to end with coins that are all the same color, and he wishes to do this while having as manycoins as possible. How many coins will he end up with, and what color will they be?

Proposed by: Yang Liu

Answer: 7 yellow coins

Let r, y, b denote the numbers of red, yellow, and blue coins respectively. Note that each of the threepossible exchanges do not change the parities of y − r, b − y, or b − r, and eventually one of thesedifferences becomes zero. Since b− r is the only one of these differences that is originally even, it mustbe the one that becomes zero, and so Bassanio will end with some number of yellow coins. Furthermore,Bassanio loses a coin in each exchange, and he requires at least five exchanges to rid himself of theblue coins, so he will have at most 12− 5 = 7 yellow coins at the end of his trading.

It remains to construct a sequence of trades that result in seven yellow coins. First, Bassanio willexchange one yellow and one blue coin for one red coin, leaving him with four red coins, three yellowcoins, and four blue coins. He then converts the red and blue coins into yellow coins, resulting in

7 yellow coins , as desired.

3. [3] Let bxc denote the largest integer less than or equal to x, and let {x} denote the fractional part ofx. For example, bπc = 3, and {π} = 0.14159 . . . , while b100c = 100 and {100} = 0. If n is the largest

solution to the equation bncn = 20152016 , compute {n}.

Proposed by: Alexander Katz

Answer: 20142015

Note that n = bnc+ {n}, so

bncn

=bnc

bnc+ {n}

=2015

2016=⇒ 2016bnc = 2015bnc+ 2015{n}

=⇒ bnc = 2015{n}

Hence, n = bnc + {n} = 20162015bnc, and so n is maximized when bnc is also maximized. As bnc is an

integer, and {n} < 1, the maximum possible value of bnc is 2014. Therefore, {n} = bnc2015 =

2014

2015.

4. [5] Call a set of positive integers good if there is a partition of it into two sets S and T , such that theredo not exist three elements a, b, c ∈ S such that ab = c and such that there do not exist three elementsa, b, c ∈ T such that ab = c (a and b need not be distinct). Find the smallest positive integer n suchthat the set {2, 3, 4, . . . , n} is not good.

Proposed by: Sam Korsky

Answer: 65536

First, we claim that the set {2, 4, 8, 256, 65536} is not good. Assume the contrary and say 2 ∈ S. Thensince 22 = 4, we have 4 ∈ T . And since 44 = 256, we have 256 ∈ S. Then since 2562 = 65536, we have65536 ∈ T . Now, note that we cannot place 8 in either S or T , contradiction.

Hence n ≤ 65536. And the partition S = {2, 3} ∪ {256, 257, . . . , 65535} and T = {4, 5, . . . , 255} shows

that n ≥ 65536. Therefore n = 65536 .

5. [5] Kelvin the Frog is trying to hop across a river. The river has 10 lilypads on it, and he must hopon them in a specific order (the order is unknown to Kelvin). If Kelvin hops to the wrong lilypad atany point, he will be thrown back to the wrong side of the river and will have to start over. AssumingKelvin is infinitely intelligent, what is the minimum number of hops he will need to guarantee reachingthe other side?

Proposed by: Alexander Katz

Answer: 176

Kelvin needs (at most) i(10−i) hops to determine the ith lilypad he should jump to, then an additional

11 hops to actually get across the river. Thus he requires∑10

i=1 i(10− i)+11 = 176 hops to guaranteesuccess.

6. [5] Marcus and four of his relatives are at a party. Each pair of the five people are either friends orenemies. For any two enemies, there is no person that they are both friends with. In how many waysis this possible?

Proposed by: Alexander Katz

Answer: 52

Denote friendship between two people a and b by a ∼ b. Then, assuming everyone is friends withthemselves, the following conditions are satisfied:

• a ∼ a• If a ∼ b, then b ∼ a• If a ∼ b and b ∼ c, then a ∼ c

Thus we can separate the five people into a few groups (possibly one group), such that people arefriends within each group, but two people are enemies when they are in different groups. Here comesthe calculation. Since the number of group(s) can be 1, 2, 3, 4, or 5, we calculate for each of thosecases. When there’s only one group, then we only have 1 possibility that we have a group of 5, andthe total number of friendship assignments in this case is

55

)= 1; when there are two groups, we

have 5 = 1 + 4 = 2 + 3 are all possible numbers of the two groups, with a total of51

)+52

)= 15

choices; when there are three groups, then we have 5 = 1 + 1 + 3 = 1 + 2 + 2, with53

)+

(51)(

52)

2 = 25possibilities; when there are four of them, then we have 5 = 1 + 1 + 1 + 2 be its only possibility, with52

)= 10 separations; when there are 5 groups, obviously we have 1 possibility. Hence, we have a total

of 1 + 15 + 25 + 10 + 1 = 52 possibilities.

Alternatively, we can also solve the problem recursively. Let Bn be the number of friendship graphswith n people, and consider an arbitrary group. If this group has size k, then there are

nk

)possible such

groups, and Bn−k friendship graphs on the remaining n− k people. Therefore, we have the recursion

Bn =

n∑k=0

(n

k

)Bn−k

with the initial condition B1 = 1. Calculating routinely gives B5 = 52 as before.

7. [6] Let ABCD be a convex quadrilateral whose diagonals AC and BD meet at P . Let the area oftriangle APB be 24 and let the area of triangle CPD be 25. What is the minimum possible area ofquadrilateral ABCD?

Proposed by: Sam Korsky

Answer: 49 + 20√

6

Note that ∠APB = 180◦ − ∠BPC = ∠CPD = 180◦ − ∠DPA so 4[BPC][DPA] = (PB · PC ·sinBPC)(PD ·PA·sinDPA) = (PA·PB ·sinAPB)(PC ·PD ·sinCPD) = 4[APB][CPD] = 2400 =⇒[BPC][DPA] = 600. Hence by AM-GM we have that

[BPC] + [DPA] ≥ 2√

[BPC][DPA] = 20√

6

so the minimum area of quadrilateral ABCD is 49 + 20√

6 .

8. [6] Find any quadruple of positive integers (a, b, c, d) satisfying a3 + b4 + c5 = d11 and abc < 105.

Proposed by: Sam Korsky

Answer: (a, b, c, d) = (128, 32, 16, 4) or (a, b, c, d) = (160, 16, 8, 4)

It’s easy to guess that there are solutions such that a, b, c, d are in the form of nx, where n is a rathersmall number. After a few attempts, we can see that we obtain simple equations when n = 2 or n = 3:for n = 2, the equation becomes in the form of 2t + 2t + 2t+1 = 2t+2 for some non-negative integert; for n = 3, the equation becomes in the form of 3t + 3t + 3t = 3t+1 for some non-negative integert. In the first case, we hope that t is a multiple of two of 3, 4, 5, that t + 1 is a multiple of the lastone, and that t + 2 is a multiple of 11. Therefore, t ≡ 15, 20, 24 (mod 60) and t ≡ 9 (mod 11). It’seasy to check that the only solution that satisfies the given inequality is the solution with t = 20, and(a, b, c, d) = (128, 32, 16, 4). In the case where n = 3, we must have that t is a multiple of 60, whichobviously doesn’t satisfy the inequality restriction. Remark: By programming, we find that the onlytwo solutions are (a, b, c, d) = (128, 32, 16, 4) and (a, b, c, d) = (160, 16, 8, 4), with the the former beingthe intended solution.

9. [7] A graph consists of 6 vertices. For each pair of vertices, a coin is flipped, and an edge connectingthe two vertices is drawn if and only if the coin shows heads. Such a graph is good if, starting fromany vertex V connected to at least one other vertex, it is possible to draw a path starting and endingat V that traverses each edge exactly once. What is the probability that the graph is good?

Proposed by: Sam Korsky

Answer: 50716384 or 210−10

215 or 29−5214

First, we find the probability that all vertices have even degree. Arbitrarily number the vertices 1, 2,3, 4, 5, 6. Flip the coin for all the edges out of vertex 1; this vertex ends up with even degree withprobability 1

2 . Next we flip for all the remaining edges out of vertex 2; regardless of previous edges,vertex 2 ends up with even degree with probability 1

2 , and so on through vertex 5. Finally, if vertices1 through 5 all have even degree, vertex 6 must also have even degree. So all vertices have even degreewith probability 1

25 = 132 . There are

62

)= 15 edges total, so there are 215 total possible graphs, of

which 210 have all vertices with even degree. Observe that exactly 10 of these latter graphs are notgood, namely, the 1

2

63

)graphs composed of two separate triangles. So 210− 10 of our graphs are good,

and the probability that a graph is good is 210−10215 .

10. [7] A number n is bad if there exists some integer c for which xx ≡ c (mod n) has no integer solutionsfor x. Find the number of bad integers between 2 and 42 inclusive.

Proposed by: Sam Korsky

Answer: 25

Call a number good if it is not bad . We claim all good numbers are products of distinct primes, noneof which are equivalent to 1 modulo another.

We first show that all such numbers are good . Consider n = p1p2 . . . pk, and let x be a number satisfyingx ≡ c (mod p1p2 . . . pk) and x ≡ 1 (mod (p1−1)(p2−1) . . . (pk−1)). Since, by assumption, p1p2 . . . pkand (p1 − 1)(p2 − 1) . . . (pk − 1) are relatively prime, such an x must exist by CRT. Then xx ≡ c1 = c(mod n), for any c, as desired.

We now show that all other numbers are bad . Suppose that there exist some p1, p2 | n such thatgcd(p1, p2−1) 6= 1 (which must hold for some two primes by assumption), and hence gcd(p1, p2−1) = p1.Consider some c for which p1c is not a p1th power modulo p2, which must exist as p1c can take anyvalue modulo p2 (as p1, p2 are relatively prime). We then claim that xx ≡ p1c (mod n) is not solvable.

Since p1p2 | n, we have xx ≡ p1c (mod p1p2), hence p1 | x. But then xx ≡ p1c is a p1th power modulop2 as p1 | x, contradicting our choice of c. As a result, all such numbers are bad .

Finally, it is easy to see that n is bad if it is not squarefree. If p1 divides n twice, then letting c = p1makes the given equivalence unsolvable.

Hence, there are 16 numbers (13 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41; and 3 semiprimes:

3 · 5 = 15, 3 · 11 = 33, 5 · 7 = 35) that are good , which means that 41− 16 = 25 numbers are bad .

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HMMT November 2015, November 14, 2015 — GUTS ROUND

Organization Team Team ID#

1. [5] Farmer Yang has a 2015 × 2015 square grid of corn plants. One day, the plant in the very centerof the grid becomes diseased. Every day, every plant adjacent to a diseased plant becomes diseased.After how many days will all of Yangs corn plants be diseased?

2. [5] The three sides of a right triangle form a geometric sequence. Determine the ratio of the length ofthe hypotenuse to the length of the shorter leg.

3. [5] A parallelogram has 2 sides of length 20 and 15. Given that its area is a positive integer, find theminimum possible area of the parallelogram.

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HMMT November 2015, November 14, 2015 — GUTS ROUND

Organization Team Team ID#

4. [6] Eric is taking a biology class. His problem sets are worth 100 points in total, his three midterms areworth 100 points each, and his final is worth 300 points. If he gets a perfect score on his problem setsand scores 60%, 70%, and 80% on his midterms respectively, what is the minimum possible percentagehe can get on his final to ensure a passing grade? (Eric passes if and only if his overall percentage isat least 70%).

5. [6] James writes down three integers. Alex picks some two of those integers, takes the average of them,and adds the result to the third integer. If the possible final results Alex could get are 42, 13, and 37,what are the three integers James originally chose?

6. [6] Let AB be a segment of length 2 with midpoint M . Consider the circle with center O and radiusr that is externally tangent to the circles with diameters AM and BM and internally tangent to thecircle with diameter AB. Determine the value of r.

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HMMT November 2015, November 14, 2015 — GUTS ROUND

Organization Team Team ID#

7. [7] Let n be the smallest positive integer with exactly 2015 positive factors. What is the sum ofthe (not necessarily distinct) prime factors of n? For example, the sum of the prime factors of 72 is2 + 2 + 2 + 3 + 3 = 14.

8. [7] For how many pairs of nonzero integers (c, d) with 〉2015 ≤ c, d ≤ 2015 do the equations cx = dand dx = c both have an integer solution?

9. [7] Find the smallest positive integer n such that there exists a complex number z, with positive realand imaginary part, satisfying zn = (z)n.

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HMMT November 2015, November 14, 2015 — GUTS ROUND

Organization Team Team ID#

10. [8] Call a string of letters S an almost palindrome if S and the reverse of S differ in exactly two places.

Find the number of ways to order the letters in HMMTTHEMETEAM to get an almost palindrome.

11. [8] Find all integers n, not necessarily positive, for which there exist positive integers a, b, c satisfyingan + bn = cn.

12. [8] Let a and b be positive real numbers. Determine the minimum possible value of√a2 + b2 +

√(a〉 1)2 + b2 +

√a2 + (b〉 1)2 +

√(a〉 1)2 + (b〉 1)2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2015, November 14, 2015 — GUTS ROUND

Organization Team Team ID#

13. [9] Consider a 4 × 4 grid of squares, each of which are originally colored red. Every minute, Piet canjump on one of the squares, changing the color of it and any adjacent squares (two squares are adjacentif they share a side) to blue. What is the minimum number of minutes it will take Piet to change theentire grid to blue?

14. [9] Let ABC be an acute triangle with orthocenter H. Let D,E be the feet of the A,B-altitudesrespectively. Given that AH = 20 and HD = 15 and BE = 56, find the length of BH.

15. [9] Find the smallest positive integer b such that 1111b (1111 in base b) is a perfect square. If no suchb exists, write ”No solution”.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2015, November 14, 2015 — GUTS ROUND

Organization Team Team ID#

16. [10] For how many triples (x, y, z) of integers between 〉10 and 10 inclusive do there exist reals a, b, cthat satisfy

ab = x

ac = y

bc = z?

17. [10] Unit squares ABCD and EFGH have centers O1 and O2 respectively, and are originally situatedsuch that B and E are at the same position and C and H are at the same position. The squares thenrotate clockwise about their centers at the rate of one revolution per hour. After 5 minutes, what isthe area of the intersection of the two squares?

18. [10] A function f satisfies, for all nonnegative integers x and y:

• f(0, x) = f(x, 0) = x

• If x ≥ y ≥ 0, f(x, y) = f(x〉 y, y) + 1

• If y ≥ x ≥ 0, f(x, y) = f(x, y 〉 x) + 1

Find the maximum value of f over 0 ≤ x, y ≤ 100.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2015, November 14, 2015 — GUTS ROUND

Organization Team Team ID#

19. [11] Each cell of a 2× 5 grid of unit squares is to be colored white or black. Compute the number ofsuch colorings for which no 2× 2 square is a single color.

20. [11] Let n be a three-digit integer with nonzero digits, not all of which are the same. Define f(n) to bethe greatest common divisor of the six integers formed by any permutation of ns digits. For example,f(123) = 3, because gcd(123, 132, 213, 231, 312, 321) = 3. Let the maximum possible value of f(n) bek. Find the sum of all n for which f(n) = k.

21. [11] Consider a 2 × 2 grid of squares. Each of the squares will be colored with one of 10 colors, andtwo colorings are considered equivalent if one can be rotated to form the other. How many distinctcolorings are there?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2015, November 14, 2015 — GUTS ROUND

Organization Team Team ID#

22. [12] Find all the roots of the polynomial x5 〉 5x4 + 11x3 〉 13x2 + 9x〉 3.

23. [12] Compute the smallest positive integer n for which

0 < 4√n〉 b 4

√nc < 1

2015.

24. [12] Three ants begin on three different vertices of a tetrahedron. Every second, they choose one of thethree edges connecting to the vertex they are on with equal probability and travel to the other vertexon that edge. They all stop when any two ants reach the same vertex at the same time. What is theprobability that all three ants are at the same vertex when they stop?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2015, November 14, 2015 — GUTS ROUND

Organization Team Team ID#

25. [13] Let ABC be a triangle that satisfies AB = 13, BC = 14, AC = 15. Given a point P in the plane,let PA, PB , PC be the reflections of A,B,C across P . Call P good if the circumcircle of PAPBPC

intersects the circumcircle of ABC at exactly 1 point. The locus of good points P encloses a region S.Find the area of S.

26. [13] Let f : R+ → R be a continuous function satisfying f(xy) = f(x) + f(y) + 1 for all positive realsx, y. If f(2) = 0, compute f(2015).

27. [13] Let ABCD be a quadrilateral with A = (3, 4), B = (9,〉40), C = (〉5,〉12), D = (〉7, 24). Let Pbe a point in the plane (not necessarily inside the quadrilateral). Find the minimum possible value ofAP + BP + CP + DP .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2015, November 14, 2015 — GUTS ROUND

Organization Team Team ID#

28. [15] Find the shortest distance between the lines x+22 = y−1

3 = z1 and x−3

−1 = y1 = z+1

2

29. [15] Find the largest real number k such that there exists a sequence of positive reals {ai} for which∑∞n=1 an converges but

∑∞n=1

√an

nk does not.

30. [15] Find the largest integer n such that the following holds: there exists a set of n points in the planesuch that, for any choice of three of them, some two are unit distance apart.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2015, November 14, 2015 — GUTS ROUND

Organization Team Team ID#

31. [17] Two random points are chosen on a segment and the segment is divided at each of these twopoints. Of the three segments obtained, find the probability that the largest segment is more thanthree times longer than the smallest segment.

32. [17] Find the sum of all positive integers n ≤ 2015 that can be expressed in the form dx2 e + y + xy,where x and y are positive integers.

33. [17] How many ways are there to place four points in the plane such that the set of pairwise distancesbetween the points consists of exactly 2 elements? (Two configurations are the same if one can beobtained from the other via rotation and scaling.)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2015, November 14, 2015 — GUTS ROUND

Organization Team Team ID#

34. [20] Let n be the second smallest integer that can be written as the sum of two positive cubes in twodifferent ways. Compute n. If your guess is a, you will receive max

25〉 5 ·max

an ,

na

), 0)

points,rounded up.

35. [20] Let n be the smallest positive integer such that any positive integer can be expressed as the sumof n integer 2015th powers. Find n. If your answer is a, your score will be max(20 〉 1

5 | log10an |, 0),

rounded up.

36. [20] Consider the following seven false conjectures with absurdly high counterexamples. Pick anysubset of them, and list their labels in order of their smallest counterexample (the smallest n for whichthe conjecture is false) from smallest to largest. For example, if you believe that the below list isalready ordered by counterexample size, you should write ”PECRSGA”.

• P. (Polya’s conjecture) For any integer n, at least half of the natural numbers below n have anodd number of prime factors.

• E. (Euler’s conjecture) There is no perfect cube n that can be written as the sum of threepositive cubes.

• C. (Cyclotomic) The polynomial with minimal degree whose roots are the primitive nth rootsof unity has all coefficients equal to -1, 0, or 1.

• R. (Prime race) For any integer n, there are more primes below n equal to 2 (mod 3) than thereare equal to 1 (mod 3).

• S. (Seventeen conjecture) For any integer n, n17 + 9 and (n + 1)17 + 9 are relatively prime.

• G. (Goldbach’s (other) conjecture) Any odd composite integer n can be written as the sumof a prime and twice a square.

• A. (Average square) Let a1 = 1 and ak+1 =1+a2

1+a22+...+a2

k

k . Then an is an integer for any n.

If your answer is a list of 4 ≤ n ≤ 7 labels in the correct order, your score will be (n 〉 2)(n 〉 3).Otherwise, it will be 0.

HMMT November 2015November 14, 2015

Guts round

1. [5] Farmer Yang has a 2015 × 2015 square grid of corn plants. One day, the plant in the very centerof the grid becomes diseased. Every day, every plant adjacent to a diseased plant becomes diseased.After how many days will all of Yangs corn plants be diseased?

Proposed by: Alexander Katz

Answer: 2014

After k minutes, the diseased plants are the ones with taxicab distance at most k from the center. Theplants on the corner are the farthest from the center and have taxicab distance 2014 from the center,so all the plants will be diseased after 2014 minutes.

2. [5] The three sides of a right triangle form a geometric sequence. Determine the ratio of the length ofthe hypotenuse to the length of the shorter leg.

Proposed by: Alexander Katz

Answer: 1+√5

2

Let the shorter leg have length `, and the common ratio of the geometric sequence be r > 1. Then thelength of the other leg is `r, and the length of the hypotenuse is `r2. Hence,

`2 + (`r)2 = (`r2)2

=⇒ `2(r2 + 1) = `2r4

=⇒ r2 + 1 = r4

Hence, r4 − r2 − 1 = 0, and therefore r2 = 1±√5

2 . As r > 1, we have r2 =1 +√

5

2, completing the

problem as the ratio of the hypotenuse to the shorter side is `r2

` = r2.

3. [5] A parallelogram has 2 sides of length 20 and 15. Given that its area is a positive integer, find theminimum possible area of the parallelogram.

Proposed by: Yang Liu

Answer: 1

The area of the parallelogram can be made arbitrarily small, so the smallest positive integer area is 1.

4. [6] Eric is taking a biology class. His problem sets are worth 100 points in total, his three midterms areworth 100 points each, and his final is worth 300 points. If he gets a perfect score on his problem setsand scores 60%, 70%, and 80% on his midterms respectively, what is the minimum possible percentagehe can get on his final to ensure a passing grade? (Eric passes if and only if his overall percentage isat least 70%).

Proposed by: Alexander Katz

Answer: 60%

We see there are a total of 100 + 3× 100 + 300 = 700 points, and he needs 70%× 700 = 490 of them.He has 100 + 60 + 70 + 80 = 310 points before the final, so he needs 180 points out of 300 on the final,which is 60%.

5. [6] James writes down three integers. Alex picks some two of those integers, takes the average of them,and adds the result to the third integer. If the possible final results Alex could get are 42, 13, and 37,what are the three integers James originally chose?

Proposed by: Alexander Katz

Answer: −20, 28, 38

Let x, y, z be the integers. We have

x+ y

2+ z = 42

y + z

2+ x = 13

x+ z

2+ y = 37

Adding these three equations yields 2(x+ y + z) = 92, so x+y2 + z = 23 + z

2 = 42 so z = 38. Similarly,x = −20 and y = 28.

6. [6] Let AB be a segment of length 2 with midpoint M . Consider the circle with center O and radiusr that is externally tangent to the circles with diameters AM and BM and internally tangent to thecircle with diameter AB. Determine the value of r.

Proposed by: Sam Korsky

Answer: 13

Let X be the midpoint of segment AM . Note that OM ⊥ MX and that MX = 12 and OX = 1

2 + rand OM = 1− r. Therefore by the Pythagorean theorem, we have

OM2 +MX2 = OX2 =⇒ (1− r)2 +1

22=

(1

2+ r

)2

which we can easily solve to find that r =1

3.

7. [7] Let n be the smallest positive integer with exactly 2015 positive factors. What is the sum ofthe (not necessarily distinct) prime factors of n? For example, the sum of the prime factors of 72 is2 + 2 + 2 + 3 + 3 = 14.

Proposed by: Alexander Katz

Answer: 116

Note that 2015 = 5 × 13 × 31 and that N = 230 · 312 · 54 has exactly 2015 positive factors. We claimthis is the smallest such integer. Note that N < 266.

If n has 3 distinct prime factors, it must be of the form p30q12r4 for some primes p, q, r, so n ≥ 230·312·54.

If n has 2 distinct prime factors, it must be of the form peqf > 2e+f where (e + 1)(f + 1) = 2015. Itis easy to see that this means e+ f > 66 so n > 266 > N .

If n has only 1 prime factor, we have n ≥ 22014 > N .

So N is the smallest such integer, and the sum of its prime factors is 2 · 30 + 3 · 12 + 5 · 4 = 116.

8. [7] For how many pairs of nonzero integers (c, d) with −2015 ≤ c, d ≤ 2015 do the equations cx = dand dx = c both have an integer solution?

Proposed by: Yang Liu

Answer: 8060

We need both c/d and d/c to be integers, which is equivalent to |c| = |d|, or d = ±c. So there are 4030ways to pick c and 2 ways to pick d, for a total of 8060 pairs.

9. [7] Find the smallest positive integer n such that there exists a complex number z, with positive realand imaginary part, satisfying zn = (z)n.

Proposed by: Alexander Katz

Since |z| = |z| we may divide by |z| and assume that |z| = 1. Then z = 1z , so we are looking for the

smallest positive integer n such that there is a 2nth root of unity in the first quadrant. Clearly thereis a sixth root of unity in the first quadrant but no fourth or second roots of unity, so n = 3 is thesmallest.

10. [8] Call a string of letters S an almost palindrome if S and the reverse of S differ in exactly two places.

Find the number of ways to order the letters in HMMTTHEMETEAM to get an almost palindrome.

Proposed by: Yang Liu

Answer: 2160

Note that T,E,A are used an odd number of times. Therefore, one must go in the middle spot andthe other pair must match up. There are are 3 · 2( 6!

2! ) = 2160 ways to fill in the first six spots with theletters T,H,E,M,M and a pair of different letters. The factor of 3 accounts for which letter goes inthe middle.

11. [8] Find all integers n, not necessarily positive, for which there exist positive integers a, b, c satisfyingan + bn = cn.

Proposed by: Rikhav Shah

Answer: ±1,±2

By Fermat’s Last Theorem, we know n < 3. Suppose n ≤ −3. Then an + bn = cn =⇒ (bc)−n +(ac)−n = (ab)−n, but since −n ≥ 3, this is also impossible by Fermat’s Last Theorem. As a result,|n| < 3.

Furthermore, n 6= 0, as a0 + b0 = c0 =⇒ 1 + 1 = 1, which is false. We now just need to findconstructions for n = −2,−1, 1, 2. When n = 1, (a, b, c) = (1, 2, 3) suffices, and when n = 2, (a, b, c) =(3, 4, 5) works nicely. When n = −1, (a, b, c) = (6, 3, 2) works, and when n = −2, (a, b, c) = (20, 15, 12)

is one example. Therefore, the working values are n = ±1,±2 .

12. [8] Let a and b be positive real numbers. Determine the minimum possible value of√a2 + b2 +

√(a− 1)2 + b2 +

√a2 + (b− 1)2 +

√(a− 1)2 + (b− 1)2

Proposed by: Alexander Katz

Answer: 2√

2

Let ABCD be a square with A = (0, 0), B = (1, 0), C = (1, 1), D = (0, 1), and P be a point in the sameplane as ABCD. Then the desired expression is equivalent to AP +BP +CP +DP . By the triangleinequality, AP +CP ≥ AC and BP +DP ≥ BD, so the minimum possible value is AC +BD = 2

√2.

This is achievable when a = b = 12 , so we are done.

13. [9] Consider a 4 × 4 grid of squares, each of which are originally colored red. Every minute, Piet canjump on one of the squares, changing the color of it and any adjacent squares (two squares are adjacentif they share a side) to blue. What is the minimum number of minutes it will take Piet to change theentire grid to blue?

Proposed by: Alexander Katz

Answer: 4

Piet can change the colors of at most 5 squares per minute, so as there are 16 squares, it will takehim at least four minutes to change the colors of every square. Some experimentation yields that it isindeed possible to make the entire grid blue after 4 minutes; one example is shown below:

XX

XX

Here, jumping on the squares marked with an X provides the desired all-blue grid.

14. [9] Let ABC be an acute triangle with orthocenter H. Let D,E be the feet of the A,B-altitudesrespectively. Given that AH = 20 and HD = 15 and BE = 56, find the length of BH.

Proposed by: Sam Korsky

Answer: 50

Let x be the length of BH. Note that quadrilateral ABDE is cyclic, so by Power of a Point, x(56−x) =20 · 15 = 300. Solving for x, we get x = 50 or 6. We must have BH > HD so x = 50 is the correctlength.

15. [9] Find the smallest positive integer b such that 1111b (1111 in base b) is a perfect square. If no suchb exists, write ”No solution”.

Proposed by: Alexander Katz

Answer: 7

We have 1111b = b3 + b2 + b+ 1 = (b2 + 1)(b+ 1). Note that gcd(b2 + 1, b+ 1) = gcd(b2 + 1− (b+ 1)(b−1), b+ 1) = gcd(2, b+ 1), which is either 1 or 2. If the gcd is 1, then there is no solution as this impliesb2 + 1 is a perfect square, which is impossible for positive b. Hence the gcd is 2, and b2 + 1, b+ 1 areboth twice perfect squares.

Let b+ 1 = 2a2. Then b2 + 1 = (2a2 − 1)2 + 1 = 4a4 − 4a2 + 2 = 2(2a4 − 2a2 + 1), so 2a4 − 2a2 + 1 =(a2 − 1)2 + (a2)2 must be a perfect square. This first occurs when a2 − 1 = 3, a2 = 4 =⇒ a = 2, and

thus b = 7 . Indeed, 11117 = 202.

16. [10] For how many triples (x, y, z) of integers between −10 and 10 inclusive do there exist reals a, b, cthat satisfy

ab = x

ac = y

bc = z?

Proposed by: Yang Liu

Answer: 4061

If none are of x, y, z are zero, then there are 4 · 103 = 4000 ways, since xyz must be positive. Indeed,(abc)2 = xyz. So an even number of them are negative, and the ways to choose an even number of 3variables to be negative is 4 ways. If one of x, y, z is 0, then one of a, b, c is zero at least. So at leasttwo of x, y, z must be 0. If all 3 are zero, this gives 1 more solution. If exactly 2 are negative, thenthis gives 3 · 20 more solutions. This comes from choosing one of x, y, z to be nonzero, and choosingits value in 20 ways.

Our final answer is 4000 + 60 + 1 = 4061.

17. [10] Unit squares ABCD and EFGH have centers O1 and O2 respectively, and are originally situatedsuch that B and E are at the same position and C and H are at the same position. The squares thenrotate clockwise about their centers at the rate of one revolution per hour. After 5 minutes, what isthe area of the intersection of the two squares?

Proposed by: Alexander Katz

Answer: 2−√3

4

Note that AE = BF = CG = DH = 1 at all times. Suppose that the squares have rotated θ radians.Then ∠O1O2H = π

4 − θ = ∠O1DH, so ∠HDC = π4 − ∠O1DH = θ. Let P be the intersection

of AB and EH and Q be the intersection of BC and GH. Then PH ‖ BQ and HQ ‖ PB, and∠PHG = π

2 , so PBQH - our desired intersection - is a rectangle. We have BQ = 1−QC = 1− sin θand HQ = 1− cos θ, so our desired area is (1− cos θ)(1− sin θ). After 5 minutes, we have θ = 2π

12 = π6 ,

so our answer is2−√

3

4.

18. [10] A function f satisfies, for all nonnegative integers x and y:

• f(0, x) = f(x, 0) = x

• If x ≥ y ≥ 0, f(x, y) = f(x− y, y) + 1

• If y ≥ x ≥ 0, f(x, y) = f(x, y − x) + 1

Find the maximum value of f over 0 ≤ x, y ≤ 100.

Proposed by: Alexander Katz

Answer: 101

Firstly, f(100, 100) = 101.

To see this is maximal, note that f(x, y) ≤ max{x, y}+ 1, say by induction on x+ y.

19. [11] Each cell of a 2× 5 grid of unit squares is to be colored white or black. Compute the number ofsuch colorings for which no 2× 2 square is a single color.

Proposed by: Alexander Katz

Answer: 634

Let an denote the number of ways to color a 2 × n grid subject only to the given constraint, and bndenote the number of ways to color a 2 × n grid subject to the given constraint, but with the addedrestriction that the first column cannot be colored black-black.

Consider the first column of a 2 × n grid that is not subject to the additional constraint. It canbe colored black-white or white-black, in which case the leftmost 2x2 square is guaranteed not to bemonochromatic, and so the remaining 2× (n−1) subgrid can be colored in an−1 ways. Otherwise, it iscolored white-white or black-black; WLOG, assume that it’s colored black-black. Then the remaining2×(n−1) subgrid is subject to both constraints, so there are bn−1 ways to color the remaining subgrid.Hence an = 2an−1 + 2bn−1.

Now consider the first column of a 2 × n grid that is subject to the additional constraint. The firstcolumn cannot be colored black-black, and if it is colored white-black or black-white, there are an−1ways to color the remaining subgrid by similar logic to the previous case. If it is colored white-white,then there are bn−1 ways to color the remaining subgrid, again by similar logic to the previous case.Hence bn = 2an−1 + bn−1.

Therefore, we have bn = 2an−1 + 12 (an − 2an−1), and so an = 2an−1 + 2bn−1 = 2an−1 + 2(2an−2 +

12 (an−1 − 2an−2)) = 3an−1 + 2an−2. Finally, we have a0 = 1 (as the only possibility is to, well, do

nothing) and a1 = 4 (as any 2× 1 coloring is admissible), so a2 = 14, a3 = 50, a4 = 178, a5 = 634 .

20. [11] Let n be a three-digit integer with nonzero digits, not all of which are the same. Define f(n) to bethe greatest common divisor of the six integers formed by any permutation of ns digits. For example,f(123) = 3, because gcd(123, 132, 213, 231, 312, 321) = 3. Let the maximum possible value of f(n) bek. Find the sum of all n for which f(n) = k.

Proposed by: Alexander Katz

Answer: 5994

Let n = abc, and assume without loss of generality that a ≥ b ≥ c. We have k | 100a + 10b + c andk | 100a+ 10c+ b, so k | 9(b− c). Analogously, k | 9(a− c) and k | 9(a− b). Note that if 9 | n, then 9also divides any permutation of ns digits, so 9 | f(n) as well; ergo, f(n) ≥ 9, implying that k ≥ 9. If kis not a multiple of 3, then we have k | c− a =⇒ k ≤ c− a < 9, contradiction, so 3 | k.

Let x = min(a − b, b − c, a − c). If x = 1, then we have k | 9, implying k = 9 - irrelevant to ourinvestigation. So we can assume x ≥ 2. Note also that x ≤ 4, as 2x ≤ (a− b) + (b− c) = a− c ≤ 9− 1,and if x = 4 we have n = 951 =⇒ f(n) = 3. If x = 3, then since 3 | k | 100a+10b+c =⇒ 3 | a+b+c,we have a ≡ b ≡ c (mod 3) (e.g. if b−c = 3, then b ≡ c (mod 3), so a ≡ b ≡ c (mod 3) - the other casesare analogous). This gives us the possibilites n = 147, 258, 369, which give f(n) = 3, 3, 9 respectively.

Hence we can conclude that x = 2; therefore k | 18. We know also that k ≥ 9, so either k = 9 ork = 18. If k = 18, then all the digits of n must be even, and n must be a multiple of 9; it is clear thatthese are sufficient criteria. As n’s digits are all even, the sum of them is also even, and hence theirsum is 18. Since a ≥ b ≥ c, we have a+ b+ c = 18 ≤ 3a =⇒ a ≥ 6, but if a = 6 then a = b = c = 6,contradicting the problem statement. Thus a = 8, and this gives us the solutions n = 882, 864 alongwith their permutations.

It remains to calculate the sum of the permutations of these solutions. In the n = 882 case, each digitis either 8, 8, or 2 (one time each), and in the n = 864 case, each digit is either 8, 6, or 4 (twice each).

Hence the desired sum is 111(8 + 8 + 2) + 111(8 · 2 + 6 · 2 + 4 · 2) = 111(54) = 5994 .

21. [11] Consider a 2 × 2 grid of squares. Each of the squares will be colored with one of 10 colors, andtwo colorings are considered equivalent if one can be rotated to form the other. How many distinctcolorings are there?

Proposed by: Sam Korsky

Answer: 2530

This solution will be presented in the general case with n colors. Our problem asks for n = 10.

We isolate three cases:

Case 1: Every unit square has the same color

In this case there are clearly n ways to color the square.

Case 2: Two non-adjacent squares are the same color, and the other two squares are also the samecolor (but not all four squares are the same color).

In this case there are clearlyn2

)= n(n−1)

2 ways to color the square.

Case 3: Every other case

Since without the ”rotation” condition there would be n4 colorings, we have that in this case by

complementary counting there are n4−n(n−1)−n4 ways to color the square.

Therefore the answer is

n+n2 − n

2+n4 − n2

4=n4 + n2 + 2n

4= 2530

22. [12] Find all the roots of the polynomial x5 − 5x4 + 11x3 − 13x2 + 9x− 3.

Proposed by: Sam Korsky

Answer: 1, 3+√3i

2 , 1−√3i

2 , 3−√3i

2 , 1+√3i

2

The x5−5x4 at the beginning of the polynomial motivates us to write it as (x−1)5+x3−3x2+4x−2 andagain the presence of the x3−3x2 motivates writing the polynomial in the form (x−1)5+(x−1)3+(x−1).Let a and b be the roots of the polynomial x2 + x+ 1. It’s clear that the roots of our polynomial aregiven by 1 and the roots of the polynomials (x−1)2 = a and (x−1)2 = b. The quadratic formula shows

that WLOG a = −1+√3i

2 and b = −1−√3i

2 so we find that either x − 1 = ± 1+√3i

2 or x − 1 = ± 1−√3i

2 .

Hence our roots are 1,3 +√

3i

2,

1−√

3i

2,

3−√

3i

2,

1 +√

3i

2.

23. [12] Compute the smallest positive integer n for which

0 < 4√n− b 4

√nc < 1

2015.

Proposed by: Alexander Katz

Answer: 4097

Let n = a4 + b where a, b are integers and 0 < b < 4a3 + 6a2 + 4a+ 1. Then

4√n− b 4

√nc < 1

20154√a4 + b− a < 1

20154√a4 + b < a+

1

2015

a4 + b <

(a+

1

2015

)4

a4 + b < a4 +4a3

2015+

6a2

20152+

4a

20153+

1

20154

To minimize n = a4 + b, we clearly should minimize b, which occurs at b = 1. Then

1 <4a3

2015+

6a2

20152+

4a

20153+

1

20154.

If a = 7, then 6a2

20152 ,4a

20153 ,1

20154 <1

2015 , so 4a3

2015 + 6a2

20152 + 4a20153 + 1

20154 <4·73+32015 < 1, so a ≥ 8. When

a = 8, we have 4a3

2015 = 20482015 > 1, so a = 8 is the minimum.

Hence, the minimum n is 84 + 1 = 4097 .

24. [12] Three ants begin on three different vertices of a tetrahedron. Every second, they choose one of thethree edges connecting to the vertex they are on with equal probability and travel to the other vertexon that edge. They all stop when any two ants reach the same vertex at the same time. What is theprobability that all three ants are at the same vertex when they stop?

Proposed by: Anna Ellison

Answer: 116

At every second, each ant can travel to any of the three vertices they are not currently on. Given that,at one second, the three ants are on different vertices, the probability of them all going to the samevertex is 1

27 and the probability of them all going to different vertices is 1127 , so the probability of the

three ants all meeting for the first time on the nth step is ( 1127 )n−1× 1

27 . Then the probability the three

ants all meet at the same time is∑∞i=0 ( 11

27 )i × 127 =

127

1− 1127

= 116 .

25. [13] Let ABC be a triangle that satisfies AB = 13, BC = 14, AC = 15. Given a point P in the plane,let PA, PB , PC be the reflections of A,B,C across P . Call P good if the circumcircle of PAPBPCintersects the circumcircle of ABC at exactly 1 point. The locus of good points P encloses a region S.Find the area of S.

Proposed by: Yang Liu

Answer: 422564 π

By the properties of reflection, the circumradius of PAPBPC equals the circumradius of ABC. There-fore, the circumcircle of PAPBPC must be externally tangent to the circumcircle of ABC. Now it’seasy to see that the midpoint of the 2 centers of ABC and PAPBPC lies on the circumcircle of ABC.So the locus of P is simply the circumcircle of ABC.

Since [ABC] = abc4R , we find the circumradius is R = 13·14·15

84·4 = 658 , so the enclosed region has area

422564 π.

26. [13] Let f : R+ → R be a continuous function satisfying f(xy) = f(x) + f(y) + 1 for all positive realsx, y. If f(2) = 0, compute f(2015).

Proposed by: Alexander Katz

Answer: log2 2015− 1

Let g(x) = f(x) + 1. Substituting g into the functional equation, we get that

g(xy)− 1 = g(x)− 1 + g(y)− 1 + 1

g(xy) = g(x) + g(y).

Also, g(2) = 1. Now substitute x = ex′, y = ey

′, which is possible because x, y ∈ R+. Then set

h(x) = g(ex). This gives us that

g(ex′+y′) = g(ex

′) + g(ey

′) =⇒ h(x′ + y′) = h(x′) + h(y′)

for al x′, y′ ∈ R. Also h is continuous. Therefore, by Cauchy’s functional equation, h(x) = cx for areal number c. Going all the way back to g, we can get that g(x) = c log x. Since g(2) = 1, c = 1

log 2 .

Therefore, g(2015) = c log 2015 = log 2015log 2 = log2 2015.

Finally, f(2015) = g(2015)− 1 = log2 2015− 1.

27. [13] Let ABCD be a quadrilateral with A = (3, 4), B = (9,−40), C = (−5,−12), D = (−7, 24). Let Pbe a point in the plane (not necessarily inside the quadrilateral). Find the minimum possible value ofAP +BP + CP +DP .

Proposed by: Alexander Katz

Answer: 16√

17 + 8√

5

By the triangle inequality, AP +CP ≥ AC and BP +DP ≥ BD. So P should be on AC and BD; i.e.it should be the intersection of the two diagonals. Then AP +BP +CP +DP = AC +BD, which iseasily computed to be 16

√17 + 8

√5 by the Pythagorean theorem.

Note that we require the intersection of the diagonals to actually exist for this proof to work, butABCD is convex and this is not an issue.

28. [15] Find the shortest distance between the lines x+22 = y−1

3 = z1 and x−3

−1 = y1 = z+1

2

Proposed by: Sam Korsky

Answer: 5√3

3

First we find the direction of a line perpendicular to both of these lines. By taking the cross product(2, 3, 1)× (−1, 1, 2) = (5,−5, 5) we find that the plane x− y + z + 3 = 0 contains the first line and isparallel to the second. Now we take a point on the second line, say the point (3, 0,−1) and find the

distance between this point and the plane. This comes out to |3−0+(−1)+3|√12+12+12

= 5√3

=5√

3

3.

29. [15] Find the largest real number k such that there exists a sequence of positive reals {ai} for which∑∞n=1 an converges but

∑∞n=1

√annk does not.

Proposed by: Alexander Katz

Answer: 12

For k > 12 , I claim that the second sequence must converge. The proof is as follows: by the Cauchy-

Schwarz inequality, 0@∑n≥1

√annk

1A2

≤

0@∑n≥1

an

1A0@∑n≥1

1

n2k

1ASince for k > 1

2 ,∑n≥1

1n2k converges, the right hand side converges. Therefore, the left hand side must

also converge.

For k ≤ 12 , the following construction surprisingly works: an = 1

n log2 n. It can be easily verified that∑

n≥1 an converges, while ∑n≥1

√an

n12

=∑n≥1

1

n log n

does not converge.

30. [15] Find the largest integer n such that the following holds: there exists a set of n points in the planesuch that, for any choice of three of them, some two are unit distance apart.

Proposed by: Alexander Katz

Answer: 7

We can obtain n = 7 in the following way: Consider a rhombus ABCD made up of two equilateraltriangles of side length 1, where ∠DAB = 60◦. Rotate the rhombus clockwise about A to obtain a newrhombus AB′C ′D′ such that DD′ = 1. Then one can verify that the seven points A,B,C,D,B′, C ′, D′

satisfy the problem condition.

To prove that n = 8 points is unobtainable, one interprets the problem in terms of graph theory.Consider a graph on 8 vertices, with an edge drawn between two vertices if and only if the vertices areat distance 1 apart. Assume for the sake of contradiction that this graph has no three points, no twoof which are at distance 1 apart (in terms of graph theory, this means the graph has no independentset of size 3).

First, note that this graph cannot contain a complete graph of size 4 (it’s clear that there can’t existfour points in the plane with any two having the same pairwise distance).

I claim that every vertex has degree 4. It is easy to see that if a vertex has degree 5 or higher, then thereexists an independent set of size 3 among its neighbors, contradiction (one can see this by drawingthe 5 neighbors on a circle of radius 1 centered at our initial vertex and considering their pairwisedistances). Moreover, if a vertex has degree 3 or lower then there are at least four vertices that arenot at distance 1 from that vertex, and since not all four of these vertices can be at distance 1 fromone another, there exists an independent set of of size 3, contradiction.

Now, we consider the complement of our graph. Every vertex of this new graph has degree 3 and by ourobservations, contains no independent set of size 4. Moreover, by assumption this graph contains notriangle (a complete graph on three vertices). But we can check by hand that there are only six distinctgraphs on eight vertices with each vertex having degree 3 (up to isomorphism), and five of these graphscontain a triangle, and the remaining graph contains an independent set of size 4, contradiction!

Hence the answer is n = 7

31. [17] Two random points are chosen on a segment and the segment is divided at each of these twopoints. Of the three segments obtained, find the probability that the largest segment is more thanthree times longer than the smallest segment.

Proposed by: Sam Korsky

Answer: 2735

We interpret the problem with geometric probability. Let the three segments have lengths x, y, 1−x−yand assume WLOG that x ≥ y ≥ 1 − x − y. The every possible (x, y) can be found in the triangledetermined by the points

13 ,

13

), ( 1

2 ,12 ), (1, 0) in R2, which has area 1

12 . The line x = 3(1 − x − y)

intersects the lines x = y and y = 1 − x − y at the points37 ,

37

)and

35 ,

15

)Hence x ≤ 3(1 − x − y)

if (x, y) is in the triangle determined by points13 ,

13

),37 ,

37

),35 ,

15

)which by shoelace has area 2

105 .Hence the desired probability is given by

112 −

2105

112

=27

35

32. [17] Find the sum of all positive integers n ≤ 2015 that can be expressed in the form dx2 e + y + xy,where x and y are positive integers.

Proposed by: Calvin Deng

Answer: 2029906

Lemma: n is expressible as dx2 e+ y + xy iff 2n+ 1 is not a Fermat Prime.

Proof: Suppose n is expressible. If x = 2k, then 2n + 1 = (2k + 1)(2y + 1), and if x = 2k − 1, thenn = k(2y + 1). Thus, if 2n + 1 isn’t prime, we can factor 2n + 1 as the product of two odd integers2x + 1, 2y + 1 both greater than 1, resulting in positive integer values for x and y. Also, if n has anodd factor greater than 1, then we factor out its largest odd factor as 2y + 1, giving a positive integervalue for x and y. Thus n is expressible iff 2n + 1 is not prime or n is not a power of 2. That leavesonly the n such that 2n+ 1 is a prime one more than a power of two. These are well-known, and arecalled the Fermat primes.

It’s a well-known fact that the only Fermat primes ≤ 2015 are 3, 5, 17, 257, which correspond ton = 1, 2, 8, 128. Thus the sum of all expressible numbers is 2015·2016

2 − (1 + 2 + 8 + 128) = 2029906.

33. [17] How many ways are there to place four points in the plane such that the set of pairwise distancesbetween the points consists of exactly 2 elements? (Two configurations are the same if one can beobtained from the other via rotation and scaling.)

Proposed by: Alexander Katz

Answer: 6

Let A,B,C,D be the four points. There are 6 pairwise distances, so at least three of them must beequal.

Case 1: There is no equilateral triangle. Then WLOG we have AB = BC = CD = 1.

• Subcase 1.1: AD = 1 as well. Then AC = BD 6= 1, so ABCD is a square.

• Subcase 1.2: AD 6= 1. Then AC = BD = AD, so A,B,C,D are four points of a regular pentagon.

Case 2: There is an equilateral triangle, say ABC, of side length 1.

• Subcase 2.1: There are no more pairs of distance 1. Then D must be the center of the triangle.

• Subcase 2.2: There is one more pair of distance 1, say AD. Then D can be either of the twointersections of the unit circle centered at A with the perpendicular bisector of BC. This givesus 2 kites.

• Subcase 2.3: Both AD = BD = 1. Then ABCD is a rhombus with a 60◦ angle.

This gives us 6 configurations total.

34. [20] Let n be the second smallest integer that can be written as the sum of two positive cubes in twodifferent ways. Compute n. If your guess is a, you will receive max

25− 5 ·max

an ,

na

), 0)

points,rounded up.

Proposed by: Alexander Katz

Answer: 4104

A computer search yields that the second smallest number is 4104. Indeed, 4104 = 93 + 153 = 23 + 163

35. [20] Let n be the smallest positive integer such that any positive integer can be expressed as the sumof n integer 2015th powers. Find n. If your answer is a, your score will be max(20 − 1

5 | log10an |, 0),

rounded up.

Proposed by: Alexander Katz

Answer: 22015 +⌊

32

)2015⌋− 2

In general, if k ≤ 471600000, then any integer can be expressed as the sum of 2k +⌊

32

)k⌋− 2 integer

kth powers. This bound is optimal.

The problem asking for the minimum number of k-th powers needed to add to any positive integer iscalled Waring’s problem.

36. [20] Consider the following seven false conjectures with absurdly high counterexamples. Pick anysubset of them, and list their labels in order of their smallest counterexample (the smallest n for whichthe conjecture is false) from smallest to largest. For example, if you believe that the below list isalready ordered by counterexample size, you should write ”PECRSGA”.

• P. (Polya’s conjecture) For any integer n, at least half of the natural numbers below n have anodd number of prime factors.

• E. (Euler’s conjecture) There is no perfect cube n that can be written as the sum of threepositive cubes.

• C. (Cyclotomic) The polynomial with minimal degree whose roots are the primitive nth rootsof unity has all coefficients equal to -1, 0, or 1.

• R. (Prime race) For any integer n, there are more primes below n equal to 2 (mod 3) than thereare equal to 1 (mod 3).

• S. (Seventeen conjecture) For any integer n, n17 + 9 and (n+ 1)17 + 9 are relatively prime.

• G. (Goldbach’s (other) conjecture) Any odd composite integer n can be written as the sumof a prime and twice a square.

• A. (Average square) Let a1 = 1 and ak+1 =1+a21+a

22+...+a

2k

k . Then an is an integer for any n.

If your answer is a list of 4 ≤ n ≤ 7 labels in the correct order, your score will be (n − 2)(n − 3).Otherwise, it will be 0.

Proposed by: Alexander Katz

Answer: ACGPRES

The smallest counterexamples are:

• Polya’s conjecture: 906,150,257

• Euler’s sum of powers: 31,858,749,840,007,945,920,321

• Cyclotomic polynomials: 105

• Prime race: 23,338,590,792

• Seventeen conjecture: 8,424,432,925,592,889,329,288,197,322,308,900,672,459,420,460,792,433

• Goldbach’s other conjecture: 5777

• Average square: 44

HMMT February 2015Saturday 21 February 2015

HMICTesting window: April 1–5

1. [20] Let S be the set of positive integers n such that the inequality

φ(n) · τ(n) ≥

√

n3

3

holds, where φ(n) is the number of positive integers k ≤ n that are relatively prime to n, and τ(n) isthe number of positive divisors of n. Prove that S is finite.

2. [25] Let m,n be positive integers with m ≥ n. Let S be the set of pairs (a, b) of relatively primepositive integers such that a, b ≤ m and a+ b > m.

For each pair (a, b) ∈ S, consider the nonnegative integer solution (u, v) to the equation au − bv = nchosen with v ≥ 0 minimal, and let I(a, b) denote the (open) interval (v/a, u/b).

Prove that I(a, b) ⊆ (0, 1) for every (a, b) ∈ S, and that any fixed irrational number α ∈ (0, 1) lies inI(a, b) for exactly n distinct pairs (a, b) ∈ S.

3. [30] Let M be a 2014×2014 invertible matrix, and let F(M) denote the set of matrices whose rows area permutation of the rows of M . Find the number of matrices F ∈ F(M) such that det(M + F ) 6= 0.

4. [35] Prove that there exists a positive integer N such that for any positive integer n ≥ N , there areat least 2015 non-empty subsets S of {n2 + 1, n2 + 2, . . . , n2 + 3n} with the property that the productof the elements of S is a perfect square.

5. [40] Let ω = e2πi/5 be a primitive fifth root of unity. Prove that there do not exist integers a, b, c, d, kwith k > 1 such that

(a+ bω + cω2 + dω3)k = 1 + ω.

HMMT February 2015Saturday 21 February 2015

HMICTesting window: April 1–5

1. [20] Let S be the set of positive integers n such that the inequality

φ(n) · τ(n) ≥√n3

3

holds, where φ(n) is the number of positive integers k ≤ n that are relatively prime to n, and τ(n) isthe number of positive divisors of n. Prove that S is finite.

Answer: N/A

Proof. Let S be the set of all positive integers n such that

φ(n) · τ(n) ≥√n3

3.

Define a function Φ on all positive integers n by

Φ(n) =φ(n)2 · τ(n)2

n3.

An important observation is that Φ has the property that for every relatively prime positive integersm,n, we have Φ(mn) = Φ(m)Φ(n).

Define another function ψ on all ordered-pairs (a, p) of positive integer a and prime number p as follows:

ψ(a, p) :=(a+ 1)2(1− 1/p)2

pa.

If we express n in its canonical form as n =∏ki=1 p

aii , then we have

Φ(n) =

k∏i=1

ψ(ai, pi).

Therefore, S is actually the set of all n =∏ki=1 p

aii such that

k∏i=1

ψ(ai, pi) ≥1

3.

It is straightforward to establish the following: for every prime p and positive integer a,

• if p ≥ 11, then ψ(a, p) < 13 ;

• if p = 5 and a ≥ 2, then ψ(a, p) < 13 ;

• if p = 3 and a ≥ 3, then ψ(a, p) < 13 ;

• if p = 2 and a ≥ 5, then ψ(a, p) < 13 ;

• ψ(a, p) is always less than 1√3unless (a, p) = (1, 3) where ψ(1, 3) = 16

27 .

The data above shows that in the case n = pa, in which there is only one prime dividing n, there areat most 8 possible n in S: 21, 22, 23, 24, 31, 32, 51, 71. If n is divisible by at least two distinct primes,

HMICTesting window: April 1–5

then one of them must be 31 and 31 fully divides n (that is, 32 - n). In the latter case n = 3 ·n0. Write

n0 =∏lj=1 q

bjj . In order for Φ(n) ≥ 1

3 , we require

l∏j=1

ψ(bj , qj) ≥9

16.

This is only possible when (b, q) = (2, 2), where n = 12. Also, note that when n = 1, Φ(1) = 1, so1 ∈ S. Hence, there are at most 10 (a finite number of) possible values of n in S.

Remark 0.1. This problem was proposed by Pakawut Jiradilok.

Remark 0.2. By explicit calculations, there are in fact exactly ten elements in S: 1, 2, 3, 4, 5, 7, 8, 9, 12, 16.

Remark 0.3. To only get a finite bound, one does not need to be nearly as careful as we are in theabove proof.

2. [25] Let m,n be positive integers with m ≥ n. Let S be the set of pairs (a, b) of relatively primepositive integers such that a, b ≤ m and a+ b > m.

For each pair (a, b) ∈ S, consider the nonnegative integer solution (u, v) to the equation au − bv = nchosen with v ≥ 0 minimal, and let I(a, b) denote the (open) interval (v/a, u/b).

Prove that I(a, b) ⊆ (0, 1) for every (a, b) ∈ S, and that any fixed irrational number α ∈ (0, 1) lies inI(a, b) for exactly n distinct pairs (a, b) ∈ S.

Answer: N/A

Proof. The fact that I(a, b) ⊆ (0, 1) follows from the small-ness of n ≤ m < a+b: the smallest solution

(u, v) has 0 ≤ v ≤ a− 1, so u = n+bva < (a+b)+b(a−1)

a = b+ 1 forces 1 ≤ u ≤ b.For the main part of the problem, it suffices (actually, is equivalent) to show that

(i) 0, 1 each appear exactly n times as endpoints of intervals I(a, b); and

(ii) each reduced rational p/q ∈ (0, 1) appears an equal (possibly zero, if q is large) number of timesas left and right endpoints of intervals I(a, b).

We prove these separately as follows:

(i) 0 is a (left) endpoint precisely when v/a = 0, or equivalently (a, b) ∈ S has a | n (look at v moda). Since n ≤ m, we get φ(d) good pairs (d, b) ∈ S for fixed d | n. Indeed, for any coprime residueclass modulo d, there’s exactly one representative b with b ≤ m < b + d. (Conceptually, it maybe enlightening to think of S as part of the Euclidean algorithm tree generated by (1, 1).)

Thus 0 occurs as an endpoint∑d|n φ(d) = n times. Similarly, 1 is an endpoint precisely when

u = b, or equivalently (a, b) ∈ S has b | n (look at u mod b). By the same reasoning as before, weget n right endpoint occurrences of 1.

(ii) Fix p/q ∈ (0, 1) with p, q coprime, so q ≥ 2. We want to show that v1/a1 = p/q occurs for thesame number of (a1, b1) ∈ S as u2/b2 = p/q occurs for (a2, b2) ∈ S. (In fact, we will show that thenumber of occurrences of the former in (x, b1) ∈ S equals the number of the latter in (a2, x) ∈ S.)

The former occurs precisely when 1 ≤ a1 ≤ m, q | a1, and (given the first two conditions)v1 ≡ p(a1/q) (mod a1). Given the first two conditions, the third is equivalent to −nb−11 ≡p(a1/q) (mod a1), or b1(a1/q) ≡ −np−1 (mod a1). This is equivalent to having (a1/q) | n andb1 ≡ (−n/(a1/q))p−1 (mod q).

Similarly, the latter occurs precisely when 1 ≤ b2 ≤ m, q | b2, (b2/q) | n, and a2 ≡ (n/(b2/q))p−1

(mod q).

As hinted at before, if we consider the occurrences for a fixed value x = a1 = b2, then thenumber of permitted residue classes b1 (mod a1 = x) (in the former) is the same as the number ofpermitted residue classes a2 (mod b2 = x) (in the latter): y (mod x) is permitted in the former ifand only if −y (mod x) is permitted in the latter; note that (y, x) = 1 if and only if (−y, x) = 1.

HMICTesting window: April 1–5

Remark 0.4. This problem was proposed by Victor Wang.

Remark 0.5. This problem was inspired by ISL 2013 N7. Note that the n = 1 case gives the famousFarey fractions sequence (for fractions of denominator at most m).

3. [30] Let M be a 2014×2014 invertible matrix, and let F(M) denote the set of matrices whose rows area permutation of the rows of M . Find the number of matrices F ∈ F(M) such that det(M +F ) 6= 0.

Answer: 2013!!2

Proof. Number the rows of M as r1, r2, . . . , r2014, and remark that by invertibility, these rows arelinearly independent. We can write det(M + F ) as the sum of the determinants of the 22014 matriceswhich can be obtained by replacing some set of the rows of M with the corresponding rows of F (thisfollows from distributivity and expansion by minors) - we’ll call these the insertion matrices of M . Ofcourse, the rows of F are themselves some permutation of the rows of F .

Now, all insertion matrices in which some row appears at least twice have determinant 0 (by lineardependence of rows), so the determinant of M + F is the sum of the determinants of the insertionmatrices whose rows are a permutation of the ri. Let the i-th row of F be rσ(i) and partition σ intodisjoint cycles; we claim det(M + F ) = 0 iff σ contains an even cycle.

Indeed, if we replace row ri in M , in order for the rows of M + F to remain a permutation of the riwe must replace row rσ(i) as well, and so on. In other words, we either replace none of the rows in anyparticular cycle, or we replace all of them. Yet if any cycle is even, the determinants of any insertionmatrix in which that cycle is replaced and the same insertion matrix, except not replacing that cycle,are the same in magnitude and opposite in sign (this follows from the fact that a row transpositionflips the sign of the determinant, and we must perform an odd number of these). Thus det(M + F )vanishes if σ contains any even cycles. On the other hand, if σ consists of only odd cycles, everynon-zero insertion matrix has determinant det(M) 6= 0 by analogous reasoning (this time, the evennumber of row transpositions preserves sign). It remains only to count the number of elements of S2014

which can be decomposed into exclusively odd cycles.

We begin by writing a recursion for the general problem. Let xn denote the number of permutationson [n] which can be decomposed into exclusively odd cycles. By casework on the cycle containing 1,we find the recursion

xn =∑i

(n− 1

2i− 2

)(2i− 2)!xn−2i+1

where x0 = x1 = x2 = 1.

The intended solution from this point was combinatorial in nature, similar to the one from the fourthcomment under Evan Chen’s “Recursion” blog post (from June 2012). The following much nicer finishis due to Lawrence Sun:

Set yn = xn/n! and rewrite the recursion as

nyn = yn−1 + yn−3 + . . .

Shifting, we find in addition that(n− 2)yn−2 = yn−3 + . . .

Subtracting,nyn = yn−1 + (n− 2)yn−2

for all n. It now follows from a straightforward induction that y2k = [(2k − 1)!!]2/(2k)! and y2k+1 =[(2k + 1)!!(2k − 1)!!]/(2k + 1)!. In particular,

x2014 = 2014!y2014 = (2013!!)2

as desired.

HMICTesting window: April 1–5

Here is an alternative approach: define the power series F (z) in the variable z by

F (z) =

∞∑n=0

xnzn

n!.

The recurrence relation is equivalent to the differential equation

F ′(z) =

( ∞∑n=0

z2n

)F (z) =

1

1− z2F (z).

Solving this equation using the initial condition F (0) = 1 gives

F (z) =

√1 + z

1− z= (1 + z)(1− z2)−1/2.

By the binomial theorem, we have

F (z) = (1 + z)∞∑n=0

(−1/2

n

)(−1)nz2n,

so

x2014 = 2014!

(−1/2

1007

)(−1)1007 = (2013!!)2

as desired.

Remark 0.6. This problem was proposed by Dhroova Aiylam.

Remark 0.7. We can write the matrices F ∈ F(M) as F = PM , where P ranges over all 2014× 2014permutation matrices. Then, we have F + M = (P + 12014×2014)M , which is invertible if and only ifP + 12014×2014 is. Using this, we can give an alternative proof that the answer to the question is equalto the number of permutations of {1, . . . , 2014} with cycles of odd length.

4. [35] Prove that there exists a positive integer N such that for any positive integer n ≥ N , there areat least 2015 non-empty subsets S of {n2 + 1, n2 + 2, . . . , n2 + 3n} with the property that the productof the elements of S is a perfect square.

Answer: N/A

Lemma 0.1. Let n, k be integers with 1 ≤ k ≤ n. Then(n

k

)≤(enk

)kwhere e ≈ 2.71828 . . . is Euler’s number.

Proof. We have (n

k

)=n(n− 1) · · · (n− k + 1)

k!

≤ nk

k!

=(nk

)k kkk!

≤(nk

)k ∞∑m=0

km

m!

=(enk

)kHMIC

Testing window: April 1–5

where the last step follows from the series expansion

ex =

∞∑m=0

xm

m!

applied to x = k.

In the solution below, all logarithms are to the base e. If f and g are two functions of the variable n,then write f(n) ∼ g(n) to denote that

limn→∞

f(n)

g(n)= 1.

Proof of original problem. Let U be the set of elements of {n2 + 1, . . . , n2 + 3n} with no prime factorgreater than 3n.1 Let s be the number of primes p ≤ 3n. By the pigeonhole principle, we can find

distinct subsets X1, . . . , Xr ⊂ U with r ≥ 2|U|

2s such that every prime p ≤ 3n either divides all theproducts ∏

x∈X1

x, . . . ,∏x∈Xr

x

an even number of times or divides all of them an odd number of times. Define the subsets Y1, . . . , Yr ⊂U by

Yi := X1∆Xi = (X1 −Xi) ∪ (Xi −X1)

for 1 ≤ i ≤ r. Then by construction, the sets Y1, . . . , Yr are distinct, and they have the property thatevery prime p ≤ 3n divides each of the products∏

y∈Y1

y, . . . ,∏y∈Yr

y

an even number of times. Since Y1, . . . , Yr ⊂ U , the products∏y∈Y1

y, . . . ,∏y∈Yr

y

do not have any prime factor greater than 3n, so they are perfect squares.

We have found r distinct subsets of {n2+1, . . . , n2+3n} whose products are perfect squares. Therefore,

it remains only to show that r ≥ 2016 for n large enough. Since r ≥ 2|U|

2s , it suffices to show that|U | ≥ s+ 11 for n large enough.

Let T be the set of prime numbers p > 3n that divide an element of {n2 + 1, . . . , n2 + 3n}. Everyelement of T evidently divides the binomial coefficient(

n2 + 3n

3n

)=

(n2 + 1) · · · (n2 + 3n)

(3n)!,

so the product of the elements of T dividesn2+3n

3n

)as well. We then have the chain of inequalities

(3n)|T | ≤∏p∈T

p

≤(n2 + 3n

3n

)≤(e(n+ 3)

3

)3n

1The reason for choosing the threshold 3n will become clear.

HMICTesting window: April 1–5

where the last inequality follows from the lemma. Therefore,

|T | ≤log

((e(n+3)

3

)3n)log(3n)

= 3n1 + log(n+ 3)− log 3

log 3 + log n.

It follows that

|U | ≥ 3n− |T |

≥ 3n

(1− 1 + log(n+ 3)− log 3

log 3 + log n

)∼ (2 log 3− 1)

3n

log(3n).

On the other hand, by the prime number theorem we have

s ∼ 3n

log(3n).

Since 2 log 3− 1 > 1, we conclude that |U | ≥ s+ 11 for n large enough, as desired.

Remark 0.8. This problem was proposed by Dhroova Aiylam. Thanks to Mitchell Lee for writing thesolution, and also working out Remark 0.10.

Remark 0.9. In fact, the above proof gives a lower bound for r of the form 2(2 log 3−2+o(1)) 3nlog(3n) where

o(1) denotes a term that approaches 0 as n→∞.

Remark 0.10. In fact, for any real number ε > 0 there is an integer Nε such that the set(n2 + 1, . . . , n2 +

&(√3

2e3/2 + ε

)n

log n

')

contains a nonempty subset whose product is a square for any n > Nε. The proof of this is similar tothe above solution with 3n replaced by (√

3

2e3/2 + ε

)n

log n

and a more careful lower bound for ∏p∈T

p.

Remark 0.11. Compare with the following problem (addressing higher powers): http://www.aops.

com/community/c6h495623_perfect_powers_in_intervals.

5. [40] Let ω = e2πi/5 be a primitive fifth root of unity. Prove that there do not exist integers a, b, c, d, kwith k > 1 such that

(a+ bω + cω2 + dω3)k = 1 + ω.

Answer: N/A

Proof. Let ζ = ω = e2πi/5 be a primitive fifth root of unity. If x = a + bζ + cζ2 + dζ3 (a, b, c, d ∈ Z),

then from ζ + ζ−1 =√5−12 and ζ2 + ζ−2 = ζ3 + ζ−3 = −

√5+12 , we deduce |x|2 = (a2 + b2 + c2 + d2) +

√5−12 (ab+ bc+ cd)−

√5+12 (ac+ bd+ ad) ∈ Z[φ], where φ = 1+

√5

2 .

HMICTesting window: April 1–5

But |x|2k = |1 + ζ|2 = 2 + φ−1 = φ2, so φ2/k = |x|2 ∈ Z[φ]. It follows that |x|2 has norm φ2/kφ2/k

in Z[φ], and so must be a positive unit. Yet by the lemma below, the positive units are precisely theinteger powers of φ, so k > 1 forces k = 2.

However, there must exist f, g ∈ Z[t] such that f(t)k = 1 + t + g(t)(t4 + t3 + t2 + t + 1) (if f(ζ) = x,then the minimal polynomial of ζ over Q must divide f(t)k − 1 − t), whence f(1)k ≡ 2 (mod 5), so2 - k, contradiction.

(Alternatively, k = 2 would force a2 + b2 + c2 + d2 − ab − bc − cd = 0, which can only occur ifa = b = c = d = 0.2)

Lemma 0.2. The positive units of Z[φ] are precisely the integer powers of φ. (In other words,3 φ isthe fundamental unit of Q(

√5).)

Proof. To prove this, one can either use Pell equation theory, or reason more directly as follows: aunit m + nφ > 1 must be at least φ. (Sketch: unit is equivalent to norm ±1, and so its conjugatem+nφ = ±1

m+nφ lies in (−1, 1), so n√

5 = n(φ−φ) > 0 forces n positive, and then −1 < m+nφ forces

m > −nφ− 1 = nφ − 1 > −1. Hence m+ nφ ≥ 0 + 1φ = φ, as desired.) If it’s not a positive power of

φ, then for some ` > 0, m+nφφ` is a unit lying in (1, φ), contradiction.

Remark 0.12. This problem was proposed by Carl Lian.

Remark 0.13. The “morally correct” (purely algebraic) way to think about about the above solutionis via the tower of (number field) extensions

Q(ζ)/Q(√

5)/Q.

Instead of thinking in terms of complex absolute value |·|, it is better to reason in terms of the normNQ(ζ)/Q(

√5)(·) : Q(ζ) → Q(

√5) (which happens to be the same as looking at the square of complex

absolute value, in our case), which maps algebraic integers to algebraic integers; the norm of xk = 1+ζover Q(

√5) is φ2, and φ is the fundamental unit of Q(

√5). Then one proceeds as in the above solution.

2On a silly note, this inspired HMMT November 2013 General Problem 7.3since Z[φ] is the ring of integers of Q(

√5)

HMICTesting window: April 1–5

HMMT February 2016February 20, 2016

Team

1. [25] Let a and b be integers (not necessarily positive). Prove that a3 + 5b3 6= 2016.

2. [25] For positive integers n, let cn be the smallest positive integer for which ncn − 1 is divisible by 210,if such a positive integer exists, and cn = 0 otherwise. What is c1 + c2 + · · ·+ c210?

3. [30] Let ABC be an acute triangle with incenter I and circumcenter O. Assume that ∠OIA = 90◦.Given that AI = 97 and BC = 144, compute the area of 4ABC.

4. [30] Let n > 1 be an odd integer. On an n × n chessboard the center square and four corners aredeleted. We wish to group the remaining n2−5 squares into 1

2 (n2−5) pairs, such that the two squaresin each pair intersect at exactly one point (i.e. they are diagonally adjacent, sharing a single corner).

For which odd integers n > 1 is this possible?

5. [35] Find all prime numbers p such that y2 = x3 + 4x has exactly p solutions in integers modulo p.

In other words, determine all prime numbers p with the following property: there exist exactly pordered pairs of integers (x, y) such that x, y ∈ {0, 1, . . . , p− 1} and

p divides y2 − x3 − 4x.

6. [35] A nonempty set S is called well-filled if for every m ∈ S, there are fewer than 12m elements of S

which are less than m. Determine the number of well-filled subsets of {1, 2, . . . , 42}.

7. [40] Let q(x) = q1(x) = 2x2 + 2x− 1, and let qn(x) = q(qn→1(x)) for n > 1. How many negative realroots does q2016(x) have?

8. [40] Compute ∫ π

0

2 sin θ + 3 cos θ − 3

13 cos θ − 5dθ.

9. [40] Fix positive integers r > s, and let F be an infinite family of sets, each of size r, no two of whichshare fewer than s elements. Prove that there exists a set of size r − 1 that shares at least s elementswith each set in F .

10. [50] Let ABC be a triangle with incenter I whose incircle is tangent to BC, CA, AB at D, E, F .Point P lies on EF such that DP ⊥ EF . Ray BP meets AC at Y and ray CP meets AB at Z. PointQ is selected on the circumcircle of 4AY Z so that AQ ⊥ BC.

Prove that P , I, Q are collinear.

HMMT February 2016February 20, 2016

Team

1. [25] Let a and b be integers (not necessarily positive). Prove that a3 + 5b3 6= 2016.

Proposed by: Evan Chen

Since cubes are 0 or ±1 modulo 9, by inspection we see that we must have a3 ≡ b3 ≡ 0 (mod 3) forthis to be possible. Thus a, b are divisible by 3. But then we get 33 | 2016, which is a contradiction.

One can also solve the problem in the same manner by taking modulo 7 exist, since all cubes are 0 or±1 modulo 7. The proof can be copied literally, noting that 7 | 2016 but 73 - 2016.

2. [25] For positive integers n, let cn be the smallest positive integer for which ncn − 1 is divisible by 210,if such a positive integer exists, and cn = 0 otherwise. What is c1 + c2 + · · ·+ c210?

Proposed by: Joy Zheng

Answer: 329

In order for cn 6= 0, we must have gcd(n, 210) = 1, so we need only consider such n. The numberncn − 1 is divisible by 210 iff it is divisible by each of 2, 3, 5, and 7, and we can consider the orderof n modulo each modulus separately; cn will simply be the LCM of these orders. We can ignore themodulus 2 because order is always 1. For the other moduli, the sets of orders are

a ∈ {1, 2} mod 3

b ∈ {1, 2, 4, 4} mod 5

c ∈ {1, 2, 3, 3, 6, 6} mod 7.

By the Chinese Remainder Theorem, each triplet of choices from these three multisets occurs forexactly one n in the range {1, 2, . . . , 210}, so the answer we seek is the sum of lcm(a, b, c) over a, b, cin the Cartesian product of these multisets. For a = 1 this table of LCMs is as follows:

1 2 3 3 6 61 1 2 3 3 6 62 2 2 6 6 6 64 4 4 12 12 12 124 4 4 12 12 12 12

which has a sum of 21 + 56 + 28 + 56 = 161. The table for a = 2 is identical except for the top row,where 1, 3, 3 are replaced by 2, 6, 6, and thus has a total sum of 7 more, or 168. So our answer is161 + 168 = 329 .

This can also be computed by counting how many times each LCM occurs:

• 12 appears 16 times when b = 4 and c ∈ {3, 6}, for a contribution of 12× 16 = 192;

• 6 appears 14 times, 8 times when c = 6 and b ≤ 2 and 6 times when c = 3 and (a, b) ∈{(1, 2), (2, 1), (2, 2)}, for a contribution of 6× 14 = 84;

• 4 appears 8 times when b = 4 and a, c ∈ {1, 2}, for a contribution of 4× 8 = 32;

• 3 appears 2 times when c = 3 and a = b = 1, for a contribution of 3× 2 = 6;

• 2 appears 7 times when a, b, c ∈ {1, 2} and (a, b, c) 6= (1, 1, 1), for a contribution of 2× 7 = 14;

• 1 appears 1 time when a = b = c = 1, for a contribution of 1× 1 = 1.

The result is again 192 + 84 + 32 + 6 + 14 + 1 = 329.

3. [30] Let ABC be an acute triangle with incenter I and circumcenter O. Assume that ∠OIA = 90◦.Given that AI = 97 and BC = 144, compute the area of 4ABC.

Proposed by: Evan Chen

Answer: 14040

We present five different solutions and outline a sixth and seventh one. In what follows, let a = BC,b = CA, c = AB as usual, and denote by r and R the inradius and circumradius. Let s = 1

2 (a+ b+ c).

In the first five solutions we will only prove that

∠AIO = 90◦ =⇒ b+ c = 2a.

Let us see how this solves the problem. This lemma implies that s = 216. If we let E be the foot of Ion AB, then AE = s−BC = 72, consequently the inradius is r =

√972 − 722 = 65. Finally, the area

is sr = 216 · 65 = 14040 .

First Solution. Since OI ⊥ DA, AI = DI. Now, it is a well-known fact that DI = DB = DC (this isoccasionally called “Fact 5”). Then by Ptolemy’s Theorem,

DB ·AC +DC ·AB = DA ·BC =⇒ AC +AB = 2BC.

Second Solution. As before note that I is the midpoint of AD. Let M and N be the midpoints of ABand AC, and let the reflection of M across BI be P ; thus BM = BP . Also, MI = PI, but we knowMI = NI as I lies on the circumcircle of triangle AMN . Consequently, we get PI = NI; moreoverby angle chasing we have

∠INC = ∠AMI = 180◦ − ∠BPI = ∠IPC.

Thus triangles INC and PIC are congruent (CI is a bisector) so we deduce PC = NC. Thus,

BC = BP + PC = BM + CN =1

2(AB +AC).

Third Solution. We appeal to Euler’s Theorem, which states that IO2 = R(R− 2r).

Thus by the Pythagorean Theorem on 4AIO (or by Power of a Point) we may write

(s− a)2 + r2 = AI2 = R2 − IO2 = 2Rr =abc

2s

with the same notations as before. Thus, we derive that

abc = 2s((s− a)2 + r2

)= 2(s− a) (s(s− a) + (s− b)(s− c))

=1

2(s− a)

((b+ c)2 − a2 + a2 − (b− c)2

)= 2bc(s− a).

From this we deduce that 2a = b+ c, and we can proceed as in the previous solution.

Fourth Solution. From Fact 5 again (DB = DI = DC), drop perpendicular from I to AB at E; callM the midpoint of BC. Then, by AAS congruency on AIE and CDM , we immediately get thatCM = AE. As AE = 1

2 (AB +AC −BC), this gives the desired conclusion.

Fifth Solution. This solution avoids angle-chasing and using the fact that BI and CI are angle-bisectors. Recall the perpendicularity lemma, where

WX ⊥ Y Z ⇐⇒ WY 2 −WZ2 = XY 2 −XZ2.

Let B′ be on the extension of ray CA such that AB′ = AB. Of course, as in the proof of the anglebisector theorem, BB′ ‖ AI, meaning that BB′ ⊥ IO. Let I ′ be the reflection of I across A; of course, I ′

is then the incenter of triangle AB′C ′. Now, we have B′I2−BI2 = B′O2−BO2 by the perpendicularityand by power of a point B′O2−BO2 = B′A ·B′C. Moreover BI2 +B′I2 = BI2 +BI ′2 = 2BA2 +2AI2

by the median formula. Subtracting, we get BI2 = AI2 + 12 (AB)(AB − AC). We have a similar

expression for CI, and subtracting the two results in BI2 − CI2 = 12 (AB2 −AC2). Finally,

BI2 − CI2 =1

4[(BC +AB −AC)2 − (BC −AB +AC)2]

from which again, the result 2BC = AB +AC follows.

Sixth Solution, outline. Use complex numbers, setting I = ab+bc+ca, A = −a2, etc. on the unit circle(scale the picture to fit in a unit circle; we calculate scaling factor later). Set a = 1, and let u = b+ cand v = bc. Write every condition in terms of u and v, and the area in terms of u and v too. There

should be two equations relating u and v: 2u+ v+ 1 = 0 and u2 = 13097

2v from the right angle and the

144 to 97 ratio, respectively. The square area can be computed in terms of u and v, because the areaitself is antisymmetric so squaring it suffices. Use the first condition to homogenize (not coincidentallythe factor (1− b2)(1− c2) = (1 + bc)2− (b+ c)2 = (1 + v)2−u2 from the area homogenizes perfectly. . .because AB ·AC = AI ·AIA, where IA is the A-excenter, and of course the way the problem is set upAIA = 3AI.), and then we find the area of the scaled down version. To find the scaling factor simplydetermine |b− c| by squaring it, writing in terms again of u and v, and comparing this to the value of144.

Seventh Solution, outline. Trigonometric solutions are also possible. One can you write everything interms of the angles and solve the equations; for instance, the ∠AIO = 90◦ condition can be rewritten

as 12 cos B→C2 = 2 sin B

2 sin C2 and the 97 to 144 ratio condition can be rewritten as

2 sin B2 2 sin C

2

sinA = 97144 .

The first equation implies sin A2 = 2 sin B

2 sin C2 , which we can plug into the second equation to get

cos A2 .

4. [30] Let n > 1 be an odd integer. On an n × n chessboard the center square and four corners aredeleted. We wish to group the remaining n2−5 squares into 1

2 (n2−5) pairs, such that the two squaresin each pair intersect at exactly one point (i.e. they are diagonally adjacent, sharing a single corner).

For which odd integers n > 1 is this possible?

Proposed by: Evan Chen

Answer: 3,5

Constructions for n = 3 and n = 5 are easy. For n > 5, color the odd rows black and the even rowswhite. If the squares can be paired in the way desired, each pair we choose must have one black celland one white cell, so the numbers of black cells and white cells are the same. The number of blackcells is n+1

2 n − 4 or n+12 n − 5 depending on whether the removed center cell is in an odd row. The

number of white cells is n→12 n or n→1

2 n− 1. But(n+ 1

2n− 5

)− n− 1

2n = n− 5

so for n > 5 this pairing is impossible. Thus the answer is n = 3 and n = 5.

5. [35] Find all prime numbers p such that y2 = x3 + 4x has exactly p solutions in integers modulo p.

In other words, determine all prime numbers p with the following property: there exist exactly pordered pairs of integers (x, y) such that x, y ∈ {0, 1, . . . , p− 1} and

p divides y2 − x3 − 4x.

Proposed by: Evan Chen

Answer: p = 2 and p ≡ 3 (mod 4)

Clearly p = 2 works with solutions (0, 0) and (1, 1) and not (0, 1) or (1, 0).

If p ≡ 3 (mod 4) then −1 is not a quadratic residue, so for x3 + 4x 6= 0, exactly one of x3 + 4xand −x3 − 4x is a square and gives two solutions (for positive and negative y), so there’s exactly twosolutions for each such pair {x,−x}. If x is such that x3 + 4x = 0, there’s exactly one solution.

If p ≡ 1 (mod 4), let i be a square root of −1 (mod p). The right hand side factors as x(x+2i)(x−2i).For x = 0, 2i,−2i this is zero, there is one choice of y, namely zero. Otherwise, the right hand side isnonzero. For any fixed x, there are either 0 or 2 choices for y. Replacing x by −x negates the righthand side, again producing two choices for y since −1 is a quadratic residue. So the total number ofsolutions (x, y) is 3 (mod 4), and thus there cannot be exactly p solutions.

Remark: This is a conductor 36 elliptic curve with complex multiplication, and the exact formula forthe number of solutions is given in http://www.mathcs.emory.edu/~ono/publications-cv/pdfs/

026.pdf.

6. [35] A nonempty set S is called well-filled if for every m ∈ S, there are fewer than 12m elements of S

which are less than m. Determine the number of well-filled subsets of {1, 2, . . . , 42}.Proposed by: Casey Fu

Answer:(4321

)− 1

Let an be the number of well-filled subsets whose maximum element is n (setting a0 = 1). Then it’seasy to see that

a2k+1 = a2k + a2k→1 + · · ·+ a0

a2k+2 = (a2k+1 − Ck) + a2k + · · ·+ a0.

where Ck is the number of well-filled subsets of size k + 1 with maximal element 2k + 1.

We proceed to compute Ck. One can think of such a subset as a sequence of numbers 1 ≤ s1 < · · · <sk+1 ≤ 2k + 1 such that si ≥ 2i − 1 for every 1 ≤ i ≤ k + 1. Equivalently, letting si = i + 1 + ti it’sthe number of sequences 0 ≤ t1 ≤ · · · ≤ tk+1 ≤ k + 1 such that ti ≥ i for every i. This gives the list ofx-coordinates of steps up in a Catalan path from (0, 0) to (k + 1, k + 1), so

Ck =1

k + 2

(2(k + 1)

(k + 1)

)is equal to the (k + 1)th Catalan number.

From this we can solve the above recursion to derive that

an =

(n

b(n− 1)/2c

).

Consequently, for even n,

a0 + · · ·+ an = an+1 =

(n+ 1

bn/2c

).

Putting n = 42 gives the answer, after subtracting off the empty set (counted in a0).

7. [40] Let q(x) = q1(x) = 2x2 + 2x− 1, and let qn(x) = q(qn→1(x)) for n > 1. How many negative realroots does q2016(x) have?

Proposed by: Ernest Chiu

Answer: 22017+13

Define g(x) = 2x2 − 1, so that q(x) = − 12 + g

(x+ 1

2

). Thus

qN (x) = 0 ⇐⇒ 1

2= gN

(x+

1

2

)where N = 2016.

But, viewed as function g : [−1, 1]→ [−1, 1] we have that g(x) = cos(2 arccos(x)). Thus, the equationqN (x) = 0 is equivalent to

cos

(22016 arccos

(x+

1

2

))=

1

2.

Thus, the solutions for x are

x = −1

2+ cos

(π/3 + 2πn

22016

)n = 0, 1, . . . , 22016 − 1.

So, the roots are negative for the values of n such that

1

3π <

π/3 + 2πn

22016<

5

3π

which is to say1

6(22016 − 1) < n <

1

6(5 · 22016 − 1).

The number of values of n that fall in this range is 16 (5 ·22016−2)− 1

6 (22016 +2)+1 = 16 (4 ·22016 +2) =

13 (22017 + 1).

8. [40] Compute ∫ π

0

2 sin θ + 3 cos θ − 3

13 cos θ − 5dθ.

Proposed by: Carl Lian

Answer: 3π13 −

413 log 3

2

We have ∫ π

0

2 sin θ + 3 cos θ − 3

13 cos θ − 5dθ = 2

∫ π/2

0

2 sin 2x+ 3 cos 2x− 3

13 cos 2x− 5dx

= 2

∫ π/2

0

4 sinx cosx− 6 sin2 x

8 cos2 x− 18 sin2 xdx

= 2

∫ π/2

0

sinx(2 cosx− 3 sinx)

(2 cosx+ 3 sinx)(2 cosx− 3 sinx)dx

= 2

∫ π/2

0

sinx

2 cosx+ 3 sinx.

To compute the above integral we want to write sinx as a linear combination of the denominator andits derivative:

2

∫ π/2

0

sinx

2 cosx+ 3 sinx= 2

∫ π/2

0

− 113 [−3(2 cosx+ 3 sinx) + 2(3 cosx− 2 sinx)]

2 cosx+ 3 sinx

= − 2

13

[∫ π/2

0

(−3) + 2

∫ π

0

−2 sinx+ 3 cosx

2 cosx+ 3 sinx

]

= − 2

13

[−3π

2+ 2 log(3 sinx+ 2 cosx) |π/20

]= − 2

13

[−3π

2+ 2 log

3

2

]=

3π

13− 4

13log

3

2.

9. [40] Fix positive integers r > s, and let F be an infinite family of sets, each of size r, no two of whichshare fewer than s elements. Prove that there exists a set of size r − 1 that shares at least s elementswith each set in F .

Proposed by: Victor Wang

This is a generalization of 2002 ISL C5.

Solution 1. Say a set S s-meets F if it shares at least s elements with each set in F . Suppose nosuch set of size (at most) r − 1 exists. (Each S ∈ F s-meets F by the problem hypothesis.)

Let T be a maximal set such that T ⊆ S for infinitely many S ∈ F , which form F ′ ⊆ F (such T exists,since the empty set works). Clearly |T | < r, so by assumption, T does not s-meet F , and there existsU ∈ F with |U ∩T | ≤ s− 1. But U s-meets F ′, so by pigeonhole, there must exist u ∈ U \T belongingto infinitely many S ∈ F ′, contradicting the maximality of T .

Comment. Let X be an infinite set, and a1, . . . , a2r→2→s elements not in X. Then F = {B ∪ {x} :B ⊆ {a1, . . . , a2r→2→s}, |B| = r − 1, x ∈ X} shows we cannot replace r − 1 with any smaller number.

Solution 2. We can also use a more indirect approach (where the use of contradiction is actuallyessential).

Fix S ∈ F and a ∈ S. By assumption, S \ {a} does not s-meet F , so there exists S′ ∈ F such that S′

contains at most s − 1 elements of S \ {a}, whence S ∩ S′ is an s-set containing a. We will derive acontradiction from the following lemma:

Lemma. Let F,G be families of r-sets such that any f ∈ F and g ∈ G share at least s elements. Thenthere exists a finite set H such that for any f ∈ F and g ∈ G, |f ∩ g ∩H| ≥ s.Proof. Suppose not, and take a counterexample with r + s minimal; then F,G must be infinite andr > s > 0.

Take arbitrary f0 ∈ F and g0 ∈ G; then the finite set X = f0∪g0 meets F,G. For every subset Y ⊆ X,let FY = {S ∈ F : S ∩X = Y }; analogously define GY . Then the FY , GY partition F,G, respectively.For any FY and y ∈ Y , define FY (y) = {S \ {y} : S ∈ FY }.Now fix subsets Y, Z ⊆ X. If one of FY , GZ is empty, define HY,Z = ∅.Otherwise, if Y, Z are disjoint, take arbitrary y ∈ Y , z ∈ Z. By the minimality assumption, thereexists finite HY,Z such that for any f ∈ FY (y) and g ∈ GZ(z), |f ∩ g ∩HY,Z | ≥ s.If Y,Z share an element a, and s = 1, take HY,Z = {a}. Otherwise, if s ≥ 2, we find again byminimality a finite HY,Z(a) such that for f ∈ FY (a) and g ∈ GZ(a), |f ∩ g ∩HY,Z | ≥ s− 1; then takeHY,Z = HY,Z(a) ∪ {a}.Finally, we see that H =

⋃Y,Z⊆X HY,Z shares at least s elements with each f ∩ g (by construction),

contradicting our assumption.

10. [50] Let ABC be a triangle with incenter I whose incircle is tangent to BC, CA, AB at D, E, F .Point P lies on EF such that DP ⊥ EF . Ray BP meets AC at Y and ray CP meets AB at Z. PointQ is selected on the circumcircle of 4AY Z so that AQ ⊥ BC.

Prove that P , I, Q are collinear.

Proposed by: Evan Chen

A

B C

I

D

E

FP

Y

Z

X

Q

The proof proceeds through a series of seven lemmas.

Lemma 1. Lines DP and EF are the internal and external angle bisectors of ∠BPC.

Proof. Since DEF the cevian triangle of ABC with respect to its Gregonne point, we have that

−1 =(EF ∩BC,D;B,C

).

Then since ∠DPF = 90◦ we see P is on the Apollonian circle of BC through D. So the conclusionfollows.

Lemma 2. Triangles BPF and CEP are similar.

Proof. Invoking the angle bisector theorem with the previous lemma gives

BP

BF=BP

BD=CP

CD=CP

CE.

But ∠BFP = ∠CEP , so 4BFP ∼ 4CEP .

Lemma 3. Quadrilateral BZY C is cyclic; in particular, line Y Z is the antiparallel of line BC through∠BAC.

Proof. Remark that ∠Y BZ = ∠PBF = ∠ECP = ∠Y CZ.

Lemma 4. The circumcircles of triangles AY Z, AEF , ABC are concurrent at a point X such that4XBF ∼ 4XCE.

Proof. Note that line EF is the angle bisector of ∠BPZ = ∠CPY . Thus

ZF

FB=ZP

PB=Y P

PC=Y E

EC.

Then, if we let X be the Miquel point of quadrilateral ZY CB, it follows that the spiral similaritymapping segment BZ to segment CY maps E to F ; therefore the circumcircle of 4AEF must passthrough X too.

Lemma 5. Ray XP bisects ∠FXE.

Proof. The assertion amounts toXF

XE=BF

EC=FP

PE.

The first equality follows from the spiral similarity 4BFX ∼ 4CEX, while the second is from4BFP ∼ 4CEP . So the proof is complete by the converse of angle bisector theorem.

Lemma 6. Points X, P , I are collinear.

Proof. On one hand, ∠FXI = ∠FAI = 12∠A. On the other hand, ∠FXP = 1

2∠FXE = 12∠A. Hence,

X, Y , I collinear.

Lemma 7. Points X, Q, I are collinear.

Proof. On one hand, ∠AXQ = 90◦, because we established earlier that line Y Z was antiparallel toline BC through ∠A, hence AQ ⊥ BC means exactly that ∠AZQ = AY Q = 90◦. On the other hand,∠AXI = 90◦ according to the fact that X lies on the circle with diameter AI. This completes theproof of the lemma.

Finally, combining the final two lemmas solves the problem.

HMMT February 2016February 20, 2016

Algebra

1. Let z be a complex number such that |z| = 1 and |z − 1.45| = 1.05. Compute the real part of z.

2. For which integers n ∈ {1, 2, . . . , 15} is nn + 1 a prime number?

3. Let A denote the set of all integers n such that 1 ≤ n ≤ 10000, and moreover the sum of the decimaldigits of n is 2. Find the sum of the squares of the elements of A.

4. Determine the remainder when2015∑i=0

⌊2i

25

⌋is divided by 100, where bxc denotes the largest integer not greater than x.

5. An infinite sequence of real numbers a1, a2, . . . satisfies the recurrence

an+3 = an+2 − 2an+1 + an

for every positive integer n. Given that a1 = a3 = 1 and a98 = a99, compute a1 + a2 + · · ·+ a100.

6. Call a positive integer N ≥ 2 “special” if for every k such that 2 ≤ k ≤ N , N can be expressed as asum of k positive integers that are relatively prime to N (although not necessarily relatively prime toeach other). How many special integers are there less than 100?

7. Determine the smallest positive integer n ≥ 3 for which

A ≡ 210n (mod 2170)

where A denotes the result when the numbers 210, 220, . . . , 210n are written in decimal notation andconcatenated (for example, if n = 2 we have A = 10241048576).

8. Define φ!(n) as the product of all positive integers less than or equal to n and relatively prime to n.Compute the number of integers 2 ≤ n ≤ 50 such that n divides φ!(n) + 1.

9. For any positive integer n, Sn be the set of all permutations of {1, 2, 3, . . . , n}. For each permutationπ ∈ Sn, let f(π) be the number of ordered pairs (j, k) for which π(j) > π(k) and 1 ≤ j < k ≤ n.Further define g(π) to be the number of positive integers k ≤ n such that π(k) ≡ k ± 1 (mod n).Compute ∑

π2S999

(−1)f(π)+g(π).

10. Let a, b and c be positive real numbers such that

a2 + ab+ b2 = 9

b2 + bc+ c2 = 52

c2 + ca+ a2 = 49.

Compute the value of49b2 − 33bc+ 9c2

a2.

HMMT February 2016February 20, 2016

Algebra

1. Let z be a complex number such that |z| = 1 and |z − 1.45| = 1.05. Compute the real part of z.

Proposed by: Evan Chen

Answer: 2029

From the problem, let A denote the point z on the unit circle, B denote the point 1.45 on the real axis,and O the origin. Let AH be the height of the triangle OAH and H lies on the segment OB. The realpart of z is OH. Now we have OA = 1, OB = 1.45, and AB = 1.05. Thus

OH = OA cos∠AOB = cos∠AOB =12 + 1.452 − 1.052

2 · 1 · 1.45=

20

29.

2. For which integers n ∈ {1, 2, . . . , 15} is nn + 1 a prime number?

Proposed by: Evan Chen

Answer: 1,2,4

n = 1 works. If n has an odd prime factor, you can factor, and this is simulated also by n = 8:

a2k+1 + 1 = (a+ 1)(

2k∑i=0

(−a)i)

with both parts larger than one when a > 1 and k > 0. So it remains to check 2 and 4, which work.Thus the answers are 1,2,4.

3. Let A denote the set of all integers n such that 1 ≤ n ≤ 10000, and moreover the sum of the decimaldigits of n is 2. Find the sum of the squares of the elements of A.

Proposed by: Evan Chen

Answer: 7294927

From the given conditions, we want to calculate

3∑i=0

3∑j=i

(10i + 10j)2.

By observing the formula, we notice that each term is an exponent of 10. 106 shows up 7 times, 105

shows up 2 times, 104 shows up 9 times, 103 shows up 4 times, 102 shows up 9 times, 10 shows 2 times,1 shows up 7 times. Thus the answer is 7294927.

4. Determine the remainder when2015∑i=0

⌊2i

25

⌋is divided by 100, where bxc denotes the largest integer not greater than x.

Proposed by: Alexander Katz

Answer: 14

Let ri denote the remainder when 2i is divided by 25. Note that because 2φ(25) ≡ 220 ≡ 1 (mod 25),r is periodic with length 20. In addition, we find that 20 is the order of 2 mod 25. Since 2i is never amultiple of 5, all possible integers from 1 to 24 are represented by r1, r2, ..., r20 with the exceptions of5, 10 ,15, and 20. Hence,

∑20i=1 ri =

∑24i=1 i− (5 + 10 + 15 + 20) = 250.

We also have

2015∑i=0

⌊2i

25

⌋=

2015∑i=0

2i − ri25

=

2015∑i=0

2i

25−

2015∑i=0

ri25

=22016 − 1

25−

1999∑i=0

ri25−

15∑i=0

ri25

=22016 − 1

25− 100

(250

25

)−

15∑i=0

ri25

≡ 22016 − 1

25−

15∑i=0

ri25

(mod 100)

We can calculate∑15i=0 ri = 185, so

2015∑i=0

⌊2i

25

⌋≡ 22016 − 186

25(mod 100)

Now 2φ(625) ≡ 2500 ≡ 1 (mod 625), so 22016 ≡ 216 ≡ 536 (mod 625). Hence 22016 − 186 ≡ 350(mod 625), and 22016 − 186 ≡ 2 (mod 4). This implies that 22016 − 186 ≡ 350 (mod 2500), and so22016→186

25 ≡ 14 (mod 100).

5. An infinite sequence of real numbers a1, a2, . . . satisfies the recurrence

an+3 = an+2 − 2an+1 + an

for every positive integer n. Given that a1 = a3 = 1 and a98 = a99, compute a1 + a2 + · · ·+ a100.

Proposed by: Evan Chen

Answer: 3

A quick telescope gives that a1 + · · ·+ an = 2a1 + a3 + an→1 − an→2 for all n ≥ 3:

n∑k=1

ak = a1 + a2 + a3 +

n→3∑k=1

(ak − 2ak+1 + 2ak+2)

= a1 + a2 + a3 +

n→3∑k=1

ak − 2

n→2∑k=2

ak +

n→1∑k=3

ak

= 2a1 + a3 − an→2 + an→1.

Putting n = 100 gives the answer.

One actual value of a2 which yields the sequence is a2 = 742745601954597303450449 .

6. Call a positive integer N ≥ 2 “special” if for every k such that 2 ≤ k ≤ N , N can be expressed as asum of k positive integers that are relatively prime to N (although not necessarily relatively prime toeach other). How many special integers are there less than 100?

Proposed by: Casey Fu

Answer: 50

We claim that all odd numbers are special, and the only special even number is 2. For any even N > 2,the numbers relatively prime to N must be odd. When we consider k = 3, we see that N can’t beexpressed as a sum of 3 odd numbers.

Now suppose that N is odd, and we look at the binary decomposition of N , so write N = 2a1 + 2a2 +... + 2aj as a sum of distinct powers of 2. Note that all these numbers only have factors of 2 and aretherefore relatively prime to N . We see that j < log2N + 1.

We claim that for any k ≥ j, we can write N as a sum of k powers of 2. Suppose that we have Nwritten as N = 2a1 + 2a2 + ...+ 2ak . Suppose we have at least one of these powers of 2 even, say 2a1 .We can then write N = 2a1→1 + 2a1→1 + 2a2 + ... + 2ak , which is k + 1 powers of 2. The only waythis process cannot be carried out is if we write N as a sum of ones, which corresponds to k = N .Therefore, this gives us all k > log2N .

Now we consider the case k = 2. Let 2a be the largest power of 2 such that 2a < N . We can writeN = 2a + (N − 2a). Note that since 2a and N are relatively prime, so are N − 2a and N . Note thata < log2N . Now similar to the previous argument, we can write 2a as a sum of k powers of 2 for1 < k < 2a, and since 2a > N

2 , we can achieve all k such that 2 ≤ k < N2 + 1.

Putting these together, we see that since N2 + 1 > log2N for N ≥ 3, we can achieve all k from 2

through N , where N is odd.

7. Determine the smallest positive integer n ≥ 3 for which

A ≡ 210n (mod 2170)

where A denotes the result when the numbers 210, 220, . . . , 210n are written in decimal notation andconcatenated (for example, if n = 2 we have A = 10241048576).

Proposed by: Evan Chen

Answer: 14

Note that210n = 1024n = 1.024n × 103n.

So 210n has roughly 3n+ 1 digits for relatively small n’s. (Actually we have that for 0 < x < 1,

(1 + x)2 = 1 + 2x+ x2 < 1 + 3x.

Therefore, 1.0242 < 1.032 < 1.09, 1.092 < 1.27, 1.272 < 1.81 < 2, and 22 = 4, so 1.02416 < 4. Thusthe conclusion holds for n ≤ 16.)

For any positive integer n ≤ 16,

A =

n∑i=1

210i × 10∑nj=i+1(3j+1).

LetAi = 210i × 10

∑nj=i+1(3j+1)

for 1 ≤ i ≤ n, then we know that

A− 210n =

n→1∑i=1

Ai

andAi = 210i+

∑nj=i+1(3j+1) × 5

∑nj=i+1(3j+1) = 2ui × 5vi

where ui = 10i+∑nj=i+1(3j + 1), vi =

∑nj=i+1(3j + 1). We have that

ui − ui→1 = 10− (3i+ 1) = 3(3− i).

Thus, for 1 ≤ i ≤ n − 1, ui is minimized when i = 1 or i = n − 1, with u1 = 3n2+5n+122 and

un→1 = 13n− 9. When n = 5,

A− 210n = A1 +A2 +A3 +A4 = 210 × 1046 + 220 × 1039 + 230 × 1029 + 240 × 1016

is at most divisible by 257 instead of 2170. For all other n’s, we have that u1 6= un→1, so we shouldhave that both 170 ≤ u1 and 170 ≤ un→1. Therefore, since 170 ≤ un→1, we have that 14 ≤ n. We can

see that u1 > 170 and 14 < 16 in this case. Therefore, the minimum of n is 14 .

8. Define φ!(n) as the product of all positive integers less than or equal to n and relatively prime to n.Compute the number of integers 2 ≤ n ≤ 50 such that n divides φ!(n) + 1.

Proposed by: Alexander Katz

Answer: 30

Note that, if k is relatively prime to n, there exists a unique 0 < k→1 < n such that kk→1 ≡ 1 (mod n).Hence, if k2 6≡ 1 (mod n), we can pair k with its inverse to get a product of 1.

If k2 ≡ 1 (mod n), then (n − k)2 ≡ 1 (mod n) as well, and k(n − k) ≡ −k2 ≡ −1 (mod n). Hencethese k can be paired up as well, giving products of -1. When n 6= 2, there is no k such that k2 ≡ 1(mod n) and k ≡ n− k (mod n), so the total product (mod n) is (−1)

m2 , where m is the number of

k such that k2 ≡ 1 (mod n).

For prime p and positive integer i, the number of solutions to k2 ≡ 1 (mod pi) is 2 if p is odd, 4 ifp = 2 and i ≥ 3, and 2 if p = i = 2. So, by Chinese remainder theorem, if we want the product to be−1, we need n = pk, 2pk, or 4. We can also manually check the n = 2 case to work.

Counting the number of integers in the allowed range that are of one of these forms (or, easier, doingcomplementary counting), we get an answer of 30.

(Note that this complicated argument basically reduces to wanting a primitive root.)

9. For any positive integer n, Sn be the set of all permutations of {1, 2, 3, . . . , n}. For each permutationπ ∈ Sn, let f(π) be the number of ordered pairs (j, k) for which π(j) > π(k) and 1 ≤ j < k ≤ n.Further define g(π) to be the number of positive integers k ≤ n such that π(k) ≡ k ± 1 (mod n).Compute ∑

π2S999

(−1)f(π)+g(π).

Proposed by: Ritesh Ragavender

Answer: 995× 2998

Define an n × n matrix An(x) with entries ai,j = x if i ≡ j ± 1 (mod n) and 1 otherwise. Let

F (x) =∑π2Sn(−1)f(π)xg(π) (here (−1)f(π) gives the sign

∏ π(u)→π(v)u→v of the permutation π). Note by

construction that F (x) = det(An(x)).

We find that the eigenvalues of An(x) are 2x+ n− 2 (eigenvector of all ones) and (x− 1)(ωj + ω→1j ),

where ωj = e2πjin , for 1 ≤ j ≤ n− 1. Since the determinant is the product of the eigenvalues,

F (x) = (2x+ n− 2)2n→1(x− 1)n→1n→1∏k=1

cos

(2πk

n

).

Evaluate the product and plug in x = −1 to finish. (As an aside, this approach also tells us that thesum is 0 whenever n is a multiple of 4.)

10. Let a, b and c be positive real numbers such that

a2 + ab+ b2 = 9

b2 + bc+ c2 = 52

c2 + ca+ a2 = 49.

Compute the value of49b2 − 33bc+ 9c2

a2.

Proposed by: Alexander Katz

Answer: 52

Consider a triangle ABC with Fermat point P such that AP = a,BP = b, CP = c. Then

AB2 = AP 2 +BP 2 − 2AP ·BP cos(120◦)

by the Law of Cosines, which becomes

AB2 = a2 + ab+ b2

and hence AB = 3. Similarly, BC =√

52 and AC = 7.

Furthermore, we have

BC2 = 52 = AB2 +BC2 − 2AB ·BC cos∠BAC

= 32 + 72 − 2 · 3 · 7 cos∠BAC

= 58− 42 cos∠BAC

And so cos∠BAC = 17 .

Invert about A with arbitrary radius r. Let B′, P ′, C ′ be the images of B,P,C respectively. Since∠APB = ∠AB′P ′ = 120◦ and ∠APC = ∠AC ′P ′ = 120◦, we note that ∠B′P ′C ′ = 120◦ − ∠BAC,and so

cos∠B′P ′C ′ = cos(120◦ − ∠BAC)

= cos 120◦ cos∠BAC − sin 120◦ sin∠BAC

= −1

2

(1

7

)+

√3

2

(4√

3

7

)

=11

14

Furthermore, using the well-known result

B′C ′ =r2BC

AB ·ACfor an inversion about A, we have

B′P ′ =BPr2

AB ·AP

=br2

a · 3

=br2

3a

and similarly P ′C ′ = cr2

7a , B′C ′ = r2√52

21 . Applying the Law of Cosines to B′P ′C ′ gives us

B′C ′2 = B′P ′2 + P ′C ′2 − 2B′P ′ · P ′C ′ cos(120◦ − ∠BAC)

=⇒ 52r4

212=b2r4

9a2+c2r4

49a2− 11bcr4

147a2

=⇒ 52

212=

b2

9a2+

c2

49a2− 11bc

147a2

=⇒ 52

212=

49b2 − 33bc+ 9c2

212a2

and so 49b2→33bc+9c2

a2 = 52 .

Motivation: the desired sum looks suspiciously like the result of some Law of Cosines, so we should trybuilding a triangle with sides 7b

a and 3ca . Getting the − 33bc

a term is then a matter of setting cos θ = 1114 .

Now there are two possible leaps: noticing that cos θ = cos(120 − ∠BAC), or realizing that it’s prettydifficult to contrive a side of 7b

a – but it’s much easier to contrive a side of b3a . Either way leads to the

natural inversion idea, and the rest is a matter of computation.

HMMT February 2016February 20, 2016

Combinatorics

1. For positive integers n, let Sn be the set of integers x such that n distinct lines, no three concurrent,can divide a plane into x regions (for example, S2 = {3, 4}, because the plane is divided into 3 regionsif the two lines are parallel, and 4 regions otherwise). What is the minimum i such that Si contains atleast 4 elements?

2. Starting with an empty string, we create a string by repeatedly appending one of the letters H, M ,T with probabilities 1

4 , 12 , 1

4 , respectively, until the letter M appears twice consecutively. What is theexpected value of the length of the resulting string?

3. Find the number of ordered pairs of integers (a, b) such that a, b are divisors of 720 but ab is not.

4. Let R be the rectangle in the Cartesian plane with vertices at (0, 0), (2, 0), (2, 1), and (0, 1). R can bedivided into two unit squares, as shown; the resulting figure has seven edges.

How many subsets of these seven edges form a connected figure?

5. Let a, b, c, d, e, f be integers selected from the set {1, 2, . . . , 100}, uniformly and at random withreplacement. Set

M = a+ 2b+ 4c+ 8d+ 16e+ 32f.

What is the expected value of the remainder when M is divided by 64?

6. Define the sequence a1, a2 . . . as follows: a1 = 1 and for every n ≥ 2,

an =

{n− 2 if an−1 = 0

an−1 − 1 if an−1 6= 0

A non-negative integer d is said to be jet-lagged if there are non-negative integers r, s and a positiveinteger n such that d = r + s and that an+r = an + s. How many integers in {1, 2, . . . , 2016} arejet-lagged?

7. Kelvin the Frog has a pair of standard fair 8-sided dice (each labelled from 1 to 8). Alex the sketchyKat also has a pair of fair 8-sided dice, but whose faces are labelled differently (the integers on eachAlex’s dice need not be distinct). To Alex’s dismay, when both Kelvin and Alex roll their dice, theprobability that they get any given sum is equal!

Suppose that Alex’s two dice have a and b total dots on them, respectively. Assuming that a 6= b, findall possible values of min{a, b}.

8. Let X be the collection of all functions f : {0, 1, . . . , 2016} → {0, 1, . . . , 2016}. Compute the numberof functions f ∈ X such that

maxg∈X

(min

0≤i≤2016

(max(f(i), g(i))

)− max

0≤i≤2016

(min(f(i), g(i))

))= 2015.

9. Let V = {1, . . . , 8}. How many permutations σ : V → V are automorphisms of some tree?

(A graph consists of a some set of vertices and some edges between pairs of distinct vertices. It isconnected if every two vertices in it are connected by some path of one or more edges. A tree G onV is a connected graph with vertex set V and exactly |V | − 1 edges, and an automorphism of G is apermutation σ : V → V such that vertices i, j ∈ V are connected by an edge if and only if σ(i) andσ(j) are.)

10. Kristoff is planning to transport a number of indivisible ice blocks with positive integer weights fromthe north mountain to Arendelle. He knows that when he reaches Arendelle, Princess Anna and QueenElsa will name an ordered pair (p, q) of nonnegative integers satisfying p+q ≤ 2016. Kristoff must thengive Princess Anna exactly p kilograms of ice. Afterward, he must give Queen Elsa exactly q kilogramsof ice.

What is the minimum number of blocks of ice Kristoff must carry to guarantee that he can alwaysmeet Anna and Elsa’s demands, regardless of which p and q are chosen?

HMMT February 2016February 20, 2016

Combinatorics

1. For positive integers n, let Sn be the set of integers x such that n distinct lines, no three concurrent,can divide a plane into x regions (for example, S2 = {3, 4}, because the plane is divided into 3 regionsif the two lines are parallel, and 4 regions otherwise). What is the minimum i such that Si contains atleast 4 elements?

Proposed by:

Answer: 4

For S3, either all three lines are parallel (4 regions), exactly two are parallel (6 regions), or none areparallel (6 or seven regions, depending on whether they all meet at one point), so |S3| = 3. Then, forS4, either all lines are parallel (5 regions), exactly three are parallel (8 regions), there are two sets ofparallel pairs (9 regions), exactly two are parallel (9 or 10 regions), or none are parallel (8, 9, 10, or 11regions), so |S4| = 4.

2. Starting with an empty string, we create a string by repeatedly appending one of the letters H, M ,T with probabilities 1

4 , 12 , 1

4 , respectively, until the letter M appears twice consecutively. What is theexpected value of the length of the resulting string?

Proposed by:

Answer: 6

Let E be the expected value of the resulting string. Starting from the empty string,

• We have a 12 chance of not selecting the letter M ; from here the length of the resulting string is

1 + E.

• We have a 14 chance of selecting the letter M followed by a letter other than M , which gives a

string of length 2 + E.

• We have a 14 chance of selecting M twice, for a string of length 2.

Thus, E = 12 (1 + E) + 1

4 (2 + E) + 14 (2). Solving gives E = 6.

3. Find the number of ordered pairs of integers (a, b) such that a, b are divisors of 720 but ab is not.

Proposed by: Casey Fu

Answer: 2520

First consider the case a, b > 0. We have 720 = 24 ·32 ·5, so the number of divisors of 720 is 5∗3∗2 = 30.We consider the number of ways to select an ordered pair (a, b) such that a, b, ab all divide 720. Usingthe balls and urns method on each of the prime factors, we find the number of ways to distribute thefactors of 2 across a and b is

(62

), the factors of 3 is

(42

), the factors of 5 is

(32

). So the total number of

ways to select (a, b) with a, b, ab all dividing 720 is 15 ∗ 6 ∗ 3 = 270. The number of ways to select any(a, b) with a and b dividing 720 is 30 ∗ 30 = 900, so there are 900 〉 270 = 630 ways to select a and bsuch that a, b divide 720 but ab doesn’t.

Now, each a, b > 0 corresponds to four solutions (±a,±b) giving the final answer of 2520. (Note thatab 6= 0.)

4. Let R be the rectangle in the Cartesian plane with vertices at (0, 0), (2, 0), (2, 1), and (0, 1). R can bedivided into two unit squares, as shown; the resulting figure has seven edges.

How many subsets of these seven edges form a connected figure?

Proposed by: Joy Zheng

Answer: 81

We break this into cases. First, if the middle edge is not included, then there are 6 ∗ 5 = 30 ways tochoose two distinct points for the figure to begin and end at. We could also allow the figure to includeall or none of the six remaining edges, for a total of 32 connected figures not including the middle edge.Now let’s assume we are including the middle edge. Of the three edges to the left of the middle edge,there are 7 possible subsets we can include (8 total subsets, but we subtract off the subset consistingof only the edge parallel to the middle edge since it’s not connected). Similarly, of the three edges tothe right of the middle edge, there are 7 possible subsets we can include. In total, there are 49 possibleconnected figures that include the middle edge. Therefore, there are 32 + 49 = 81 possible connectedfigures.

5. Let a, b, c, d, e, f be integers selected from the set {1, 2, . . . , 100}, uniformly and at random withreplacement. Set

M = a+ 2b+ 4c+ 8d+ 16e+ 32f.

What is the expected value of the remainder when M is divided by 64?

Proposed by: Evan Chen

Answer: 632

Consider M in binary. Assume we start with M = 0, then add a to M , then add 2b to M , thenadd 4c to M , and so on. After the first addition, the first bit (defined as the rightmost bit) of M istoggled with probability 1

2 . After the second addition, the second bit of M is toggled with probability12 . After the third addition, the third bit is toggled with probability 1

2 , and so on for the remainingthree additions. As such, the six bits of M are each toggled with probability 1

2 - specifically, the kth

bit is toggled with probability 12 at the kth addition, and is never toggled afterwards. Therefore, each

residue from 0 to 63 has probability 164 of occurring, so they are all equally likely. The expected value

is then just 632 .

6. Define the sequence a1, a2 . . . as follows: a1 = 1 and for every n ≥ 2,

an =

{n〉 2 if an−1 = 0

an−1 〉 1 if an−1 6= 0

A non-negative integer d is said to be jet-lagged if there are non-negative integers r, s and a positiveinteger n such that d = r + s and that an+r = an + s. How many integers in {1, 2, . . . , 2016} arejet-lagged?

Proposed by: Pakawut Jiradilok

Let N = n+r, and M = n. Then r = N〉M , and s = aN〉aM , and d = r+s = (aN +N)〉(aM +M).So we are trying to find the number of possible values of (aN + N) 〉 (aM + M), subject to N ≥ Mand aN ≥ aM .

Divide the ai into the following “blocks”:

• a1 = 1, a2 = 0,

• a3 = 1, a4 = 0,

• a5 = 3, a6 = 2, a7 = 1, a8 = 0,

• a9 = 7, a10 = 6, . . . , a16 = 0,

and so on. The kth block contains ai for 2k−1 < i ≤ 2k. It’s easy to see by induction that a2k = 0 andthus a2k+1 = 2k 〉 1 for all k ≥ 1. Within each block, the value an + n is constant, and for the kthblock (k ≥ 1) it equals 2k. Therefore, d = (aN +N)〉 (aM +M) is the difference of two powers of 2,say 2n 〉 2m. For any n ≥ 1, it is clear there exists an N such that aN + N = 2n (consider the nth

block). We can guarantee aN ≥ aM by setting M = 2m. Therefore, we are searching for the numberof integers between 1 and 2016 that can be written as 2n〉 2m with n ≥ m ≥ 1. The pairs (n,m) withn > m ≥ 1 and n ≤ 10 all satisfy 1 ≤ 2n 〉 2m ≤ 2016 (45 possibilities). In the case that n = 11, wehave that 2n 〉 2m ≤ 2016 so 2m ≥ 32, so m ≥ 5 (6 possibilities). There are therefore 45 + 6 = 51jetlagged numbers between 1 and 2016.

7. Kelvin the Frog has a pair of standard fair 8-sided dice (each labelled from 1 to 8). Alex the sketchyKat also has a pair of fair 8-sided dice, but whose faces are labelled differently (the integers on eachAlex’s dice need not be distinct). To Alex’s dismay, when both Kelvin and Alex roll their dice, theprobability that they get any given sum is equal!

Suppose that Alex’s two dice have a and b total dots on them, respectively. Assuming that a 6= b, findall possible values of min{a, b}.Proposed by: Alexander Katz

Answer: 24, 28, 32

Ed. note: I’m probably horribly abusing notation

Define the generating function of an event A as the polynomial

g(A, x) =∑

pixi

where pi denotes the probability that i occurs during event A. We note that the generating is multi-plicative; i.e.

g(A AND B, x) = g(A)g(B) =∑

piqjxi+j

where qj denotes the probability that j occurs during event B.

In our case, events A and B are the rolling of the first and second dice, respectively, so the generatingfunctions are the same:

g(die, x) =1

8x1 +

1

8x2 +

1

8x3 +

1

8x4 +

1

8x5 +

1

8x6 +

1

8x7 +

1

8x8

and so

g(both dice rolled, x) = g(die, x)2 =1

64(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8)2

where the coefficient of xi denotes the probability of rolling a sum of i.

We wish to find two alternate dice, C and D, satisfying the following conditions:

• C and D are both 8-sided dice; i.e. the sum of the coefficients of g(C, x) and g(D,x) are both 8(or g(C, 1) = g(D, 1) = 8).

• The faces of C and D are all labeled with a positive integer; i.e. the powers of each term of g(C, x)and g(D,x) are positive integer (or g(C, 0) = g(D, 0) = 0).

• The probability of rolling any given sum upon rolling C and D is equal to the probability of rollingany given sum upon rolling A and B; i.e. g(C, x)g(D,x) = g(A, x)g(B, x).

Because the dice are “fair” – i.e. the probability of rolling any face is 18 – we can multiply g(A, x), g(B, x), g(C, x)

and g(D,x) by 8 to get integer polynomials; as this does not affect any of the conditions, we can as-sume g(C, x) and g(D,x) are integer polynomials multiplying to (x1 + x2 + . . . + x8)2 (and subjectto the other two conditions as well). Since Z is a UFD (i.e. integer polynomials can be expressedas the product of integer polynomials in exactly one way, up to order and scaling by a constant), allfactors of g(C, x) and g(D,x) must also be factors of x1 + x2 + . . . + x8. Hence it is useful to factorx1 + x2 + . . .+ x8 = x(x+ 1)(x2 + 1)(x4 + 1).

We thus have g(C, x)g(D,x) = x2(x+ 1)2(x2 + 1)2(x4 + 1)2. We know that g(C, 0) = g(D, 0) = 0, sox | g(C, x), g(D,x). It remains to distribute the remaining term (x+1)2(x2 +1)2(x4 +1)2; we can vieweach of these 6 factors as being “assigned” to either C or D. Note that since g(C, 1) = g(D, 1) = 8,and each of the factors x + 1, x2 + 1, x4 + 1 evaluates to 2 when x = 1, exactly three factors must beassigned to C and exactly three to D. Finally, assigning x + 1, x2 + 1, and x4 + 1 to C results in thestandard die, with a = b = 28.. This gives us the three cases (and their permutations):

• g(C, x) = x(x + 1)2(x2 + 1), g(D,x) = x(x2 + 1)(x4 + 1)2. In this case we get g(C, x) = x5 +2x4 + 2x3 + 2x2 + x and g(D,x) = x11 + x9 + 2x7 + 2x5 + x3 + x, so the “smaller” die has faces5, 4, 4, 3, 3, 2, 2, and 1 which sum to 24.

• g(C, x) = x(x + 1)(x2 + 1)2, g(D,x) = x(x + 1)(x4 + 1)2. In this case we have g(C, x) =x6 + x5 + 2x4 + 2x3 + x2 + x and g(D,x) = x10 + x9 + 2x6 + 2x5 + x2 + x, so the “smaller” diehas faces 6, 5, 4, 4, 3, 3, 2 and 1 which sum to 28.

• g(C, x) = x(x2 + 1)2(x4 + 1), g(D,x) = x(x + 1)2(x4 + 1). In this case we have g(C, x) =x9 + 2x7 + 2x5 + 2x3 + x and g(D,x) = x7 + 2x6 + x5 + x3 + 2x2 + x, so the “smaller die” hasfaces 7, 6, 6, 5, 3, 2, 2, 1 which sum to 32.

Therefore, min{a, b} is equal to 24, 28, or 32 .

8. Let X be the collection of all functions f : {0, 1, . . . , 2016} → {0, 1, . . . , 2016}. Compute the numberof functions f ∈ X such that

maxg∈X

(min

0≤i≤2016

(max(f(i), g(i))

)〉 max

0≤i≤2016

(min(f(i), g(i))

))= 2015.

Proposed by:

Answer: 2 · (32017 〉 22017)

For each f, g ∈ X, we define

d(f, g) := min0≤i≤2016

(max(f(i), g(i))

)〉 max

0≤i≤2016

(min(f(i), g(i))

)Thus we desire maxg∈X d(f, g) = 2015.

First, we count the number of functions f ∈ X such that

9g : mini

max{f(i), g(i)} ≥ 2015 and 9g : mini

max{f(i), g(i)} = 0.

That means for every value of i, either f(i) = 0 (then we pick g(i) = 2015) or f(i) ≥ 2015 (then wepick g(i) = 0). So there are A = 32017 functions in this case.

Similarly, the number of functions such that

9g : mini

max{f(i), g(i)} = 2016 and 9g : mini

max{f(i), g(i)} ≤ 1

is also B = 32017.

Finally, the number of functions such that

9g : mini

max{f(i), g(i)} = 2016 and 9g : mini

max{f(i), g(i)} = 0

is C = 22017.

Now A+B〉C counts the number of functions with maxg∈X d(f, g) ≥ 2015 and C counts the numberof functions with maxg∈X d(f, g) ≥ 2016, so the answer is A+B 〉 2C = 2 · (32017 〉 22017).

9. Let V = {1, . . . , 8}. How many permutations σ : V → V are automorphisms of some tree?

(A graph consists of a some set of vertices and some edges between pairs of distinct vertices. It isconnected if every two vertices in it are connected by some path of one or more edges. A tree G onV is a connected graph with vertex set V and exactly |V | 〉 1 edges, and an automorphism of G is apermutation σ : V → V such that vertices i, j ∈ V are connected by an edge if and only if σ(i) andσ(j) are.)

Proposed by: Mitchell Lee

Answer: 30212

We decompose into cycle types of σ. Note that within each cycle, all vertices have the same degree;also note that the tree has total degree 14 across its vertices (by all its seven edges).

For any permutation that has a 1 in its cycle type (i.e it has a fixed point), let 1 ≤ a ≤ 8 be a fixedpoint. Consider the tree that consists of the seven edges from a to the seven other vertices - thispermutation (with a as a fixed point) is an automorphism of this tree.

For any permutation that has cycle type 2 + 6, let a and b be the two elements in the 2-cycle. If the6-cycle consists of c, d, e, f, g, h in that order, consider the tree with edges between a and b, c, e, g andbetween b and d, f, h. It’s easy to see σ is an automorphism of this tree.

For any permutation that has cycle type 2+2+4, let a and b be the two elements of the first two-cycle.Let the other two cycle consist of c and d, and the four cycle be e, f, g, h in that order. Then considerthe tree with edges between a and b, a and c, b and d, a and e, b and f , a and g, b and h. It’s easy tosee σ is an automorphism of this tree.

For any permutation that has cycle type 2 + 3 + 3, let a and b be the vertices in the 2-cycle. One of aand b must be connected to a vertex distinct from a, b (follows from connectedness), so there must bean edge between a vertex in the 2-cycle and a vertex in a 3-cycle. Repeatedly applying σ to this edgeleads to a cycle of length 4 in the tree, which is impossible (a tree has no cycles). Therefore, thesepermutations cannot be automorphisms of any tree.

For any permutation that has cycle type 3 + 5, similarly, there must be an edge between a vertex inthe 3-cycle and a vertex in the 5-cycle. Repeatedly applying σ to this edge once again leads to a cyclein the tree, which is not possible. So these permutations cannot be automorphisms of any tree.

The only remaining possible cycle types of σ are 4 + 4 and 8. In the first case, if we let x and y bethe degrees of the vertices in each of the cycles, then 4x+ 4y = 14, which is impossible for integer x, y.In the second case, if we let x be the degree of the vertices in the 8-cycle, then 8x = 14, which is notpossible either.

So we are looking for the number of permutations whose cycle type is not 2 + 2 + 3, 8, 4 + 4, 3 + 5. Thenumber of permutations with cycle type 2+2+3 is

(82

)12

(63

)(2!)2 = 1120, with cycle type 8 is 7! = 5040,

with cycle type 4 + 4 is 12

(84

)(3!)2 = 1260, with cycle type 3 + 5 is

(83

)(2!)(4!) = 2688. Therefore,

by complementary counting, the number of permutations that ARE automorphisms of some tree is8!〉 1120〉 1260〉 2688〉 5040 = 30212.

10. Kristoff is planning to transport a number of indivisible ice blocks with positive integer weights fromthe north mountain to Arendelle. He knows that when he reaches Arendelle, Princess Anna and QueenElsa will name an ordered pair (p, q) of nonnegative integers satisfying p+q ≤ 2016. Kristoff must thengive Princess Anna exactly p kilograms of ice. Afterward, he must give Queen Elsa exactly q kilogramsof ice.

What is the minimum number of blocks of ice Kristoff must carry to guarantee that he can alwaysmeet Anna and Elsa’s demands, regardless of which p and q are chosen?

Proposed by: Pakawut Jiradilok

Answer: 18

The answer is 18.

First, we will show that Kristoff must carry at least 18 ice blocks. Let

0 < x1 ≤ x2 ≤ · · · ≤ xn

be the weights of ice blocks he carries which satisfy the condition that for any p, q ∈ Z≥0 such thatp+ q ≤ 2016, there are disjoint subsets I, J of {1, . . . , n} such that

∑α∈I xα = p and

∑α∈J xα = q.

Claim: For any i, if x1 + · · ·+ xi ≤ 2014, then

xi+1 ≤⌊x1 + · · ·+ xi

2

⌋+ 1.

Proof. Suppose to the contrary that xi+1 ≥ bx1+···+xi

2 c+2. Consider when Anna and Elsa both demandbx1+···+xi

2 c+ 1 kilograms of ice (which is possible as 2×(bx1+···+xi

2 c+ 1)≤ x1 + · · ·+ xi + 2 ≤ 2016).

Kristoff cannot give any ice xj with j ≥ i + 1 (which is too heavy), so he has to use from x1, . . . , xi.Since he is always able to satisfy Anna’s and Elsa’s demands, x1 + · · ·+ xi ≥ 2×

(bx1+···+xi

2 c+ 1)≥

x1 + · · ·+ xi + 1. A contradiction.

It is easy to see x1 = 1, so by hand we compute obtain the inequalities x2 ≤ 1, x3 ≤ 2, x4 ≤ 3, x5 ≤ 4,x6 ≤ 6, x7 ≤ 9, x8 ≤ 14, x9 ≤ 21, x10 ≤ 31, x11 ≤ 47, x12 ≤ 70, x13 ≤ 105, x14 ≤ 158, x15 ≤ 237,x16 ≤ 355, x17 ≤ 533, x18 ≤ 799. And we know n ≥ 18; otherwise the sum x1 + · · · + xn would notreach 2016.

Now we will prove that n = 18 works. Consider the 18 numbers named above, say a1 = 1, a2 = 1,a3 = 2, a4 = 3, . . . , a18 = 799. We claim that with a1, . . . , ak, for any p, q ∈ Z≥0 such thatp + q ≤ a1 + · · · + ak, there are two disjoint subsets I, J of {1, . . . , k} such that

∑α∈I xα = p and∑

α∈J xα = q. We prove this by induction on k. It is clear for small k = 1, 2, 3. Now suppose this istrue for a certain k, and we add in ak+1. When Kristoff meets Anna first and she demands p kilogramsof ice, there are two cases.

Case I: if p ≥ ak+1, then Kristoff gives the ak+1 block to Anna first, then he consider p′ = p 〉 ak+1

and the same unknown q. Now p′ + q ≤ a1 + · · · + ak and he has a1, . . . , ak, so by induction he cansuccessfully complete his task.

Case II: if p < ak+1, regardless of the value of q, he uses the same strategy as if p+q ≤ a1+ · · ·+ak andhe uses ice from a1, . . . , ak without touching ak+1. Then, when he meets Elsa, if q ≤ a1 + · · ·+ ak 〉 p,he is safe. If q ≥ a1 + · · ·+ak〉p+ 1, we know q〉ak+1 ≥ a1 + · · ·+ak〉p+ 1〉

(ba1+···+ak2 c+ 1

)≥ 0.

So he can give the ak+1 to Elsa first then do as if q′ = q 〉 ak+1 is the new demand by Elsa. He cannow supply the ice to Elsa because p+ q′ ≤ a1 + · · ·+ ak. Thus, we finish our induction.

Therefore, Kristoff can carry those 18 blocks of ice and be certain that for any p+ q ≤ a1 + · · ·+ a18 =2396, there are two disjoint subsets I, J ⊆ {1, . . . , 18} such that

∑α∈I aα = p and

∑α∈J aα = q. In

other words, he can deliver the amount of ice both Anna and Elsa demand.

HMMT February 2016February 20, 2016

Geometry

1. Dodecagon QWARTZSPHINX has all side lengths equal to 2, is not self-intersecting (in particular,the twelve vertices are all distinct), and moreover each interior angle is either 90◦ or 270◦. What areall possible values of the area of 4SIX?

2. Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Let H be the orthocenter of ABC. Findthe distance between the circumcenters of triangles AHB and AHC.

3. In the below picture, T is an equilateral triangle with a side length of 5 and ω is a circle with a radiusof 2. The triangle and the circle have the same center. Let X be the area of the shaded region, andlet Y be the area of the starred region. What is X − Y ?

T

ω

4. Let ABC be a triangle with AB = 3, AC = 8, BC = 7 and let M and N be the midpoints of AB andAC, respectively. Point T is selected on side BC so that AT = TC. The circumcircles of trianglesBAT , MAN intersect at D. Compute DC.

5. Nine pairwise noncongruent circles are drawn in the plane such that any two circles intersect twice.For each pair of circles, we draw the line through these two points, for a total of

92

)= 36 lines. Assume

that all 36 lines drawn are distinct. What is the maximum possible number of points which lie on atleast two of the drawn lines?

6. Let ABC be a triangle with incenter I, incircle γ and circumcircle Γ. Let M , N , P be the midpointsof sides BC, CA, AB and let E, F be the tangency points of γ with CA and AB, respectively. Let U ,V be the intersections of line EF with line MN and line MP , respectively, and let X be the midpoint

of arc B̂AC of Γ. Given that AB = 5, AC = 8, and ∠A = 60◦, compute the area of triangle XUV .

7. Let S = {(x, y)|x, y ∈ Z, 0 ≤ x, y,≤ 2016}. Given points A = (x1, y1), B = (x2, y2) in S, define

d2017(A,B) = (x1 − x2)2 + (y1 − y2)2 (mod 2017).

The points A = (5, 5), B = (2, 6), C = (7, 11) all lie in S. There is also a point O ∈ S that satisfies

d2017(O,A) = d2017(O,B) = d2017(O,C).

Find d2017(O,A).

8. For i = 0, 1, . . . , 5 let li be the ray on the Cartesian plane starting at the origin, an angle θ = iπ3counterclockwise from the positive x-axis. For each i, point Pi is chosen uniformly at random fromthe intersection of li with the unit disk. Consider the convex hull of the points Pi, which will (withprobability 1) be a convex polygon with n vertices for some n. What is the expected value of n?

9. In cyclic quadrilateral ABCD with AB = AD = 49 and AC = 73, let I and J denote the incenters oftriangles ABD and CBD. If diagonal BD bisects IJ , find the length of IJ .

10. The incircle of a triangle ABC is tangent to BC at D. Let H and Γ denote the orthocenter andcircumcircle of 4ABC. The B-mixtilinear incircle, centered at OB , is tangent to lines BA and BCand internally tangent to Γ. The C-mixtilinear incircle, centered at OC , is defined similarly. Supposethat DH ⊥ OBOC , AB =

√3 and AC = 2. Find BC.

HMMT February 2016February 20, 2016

Geometry

1. Dodecagon QWARTZSPHINX has all side lengths equal to 2, is not self-intersecting (in particular,the twelve vertices are all distinct), and moreover each interior angle is either 90◦ or 270◦. What areall possible values of the area of 4SIX?

Proposed by: Evan Chen

Answer: 2, 6

The dodecagon has to be a ”plus shape” of area 20, then just try the three non-congruent possibilities.

2. Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Let H be the orthocenter of ABC. Findthe distance between the circumcenters of triangles AHB and AHC.

Proposed by: Evan Chen

Answer: 14

Let HB be the reflection of H over AC and let HC be the reflection of H over AB. The reflec-tions of H over AB,AC lie on the circumcircle of triangle ABC. Since the circumcenters of trianglesAHCB,AHBC are both O, the circumcenters of AHB,AHC are reflections of O over AB,AC respec-tively. Moreover, the lines from O to the circumcenters in question are the perpendicular bisectors ofAB and AC. Now we see that the distance between the two circumcenters is simply twice the lengthof the midline of triangle ABC that is parallel to BC, meaning the distance is 2( 1

2BC) = 14.

3. In the below picture, T is an equilateral triangle with a side length of 5 and ω is a circle with a radiusof 2. The triangle and the circle have the same center. Let X be the area of the shaded region, andlet Y be the area of the starred region. What is X 〉 Y ?

T

ω

Proposed by:

Answer: 25√3

12 〉4π3

Let K(P ) denote the area of P . Note that K(T )〉K(ω) = 3(X 〉 Y ), which gives our answer.

4. Let ABC be a triangle with AB = 3, AC = 8, BC = 7 and let M and N be the midpoints of AB andAC, respectively. Point T is selected on side BC so that AT = TC. The circumcircles of trianglesBAT , MAN intersect at D. Compute DC.

Proposed by: Evan Chen

Answer: 7√3

3

We note thatD is the circumcenter O of ABC, since 2∠C = ∠ATB = ∠AOB. So we are merely lookingfor the circumradius of triangle ABC. By Heron’s Formula, the area of the triangle is

√9 · 6 · 1 · 2 =

6√

3, so using the formula abc4R = K, we get an answer of 3·8·7

4·6√3

= 7√3

3 . Alternatively, one can compute

the circumradius using trigonometric methods or the fact that ∠A = 60◦.

5. Nine pairwise noncongruent circles are drawn in the plane such that any two circles intersect twice.For each pair of circles, we draw the line through these two points, for a total of

(92

)= 36 lines. Assume

that all 36 lines drawn are distinct. What is the maximum possible number of points which lie on atleast two of the drawn lines?

Proposed by: Evan Chen

Answer: 462

The lines in question are the radical axes of the 9 circles. Three circles with noncollinear centers havea radical center where their three pairwise radical axes concur, but all other intersections between twoof the

(92

)lines can be made to be distinct. So the answer is((9

2

)2

)〉 2

(9

3

)= 462

by just counting pairs of lines, and then subtracting off double counts due to radical centers (eachcounted three times).

6. Let ABC be a triangle with incenter I, incircle γ and circumcircle Γ. Let M , N , P be the midpointsof sides BC, CA, AB and let E, F be the tangency points of γ with CA and AB, respectively. Let U ,V be the intersections of line EF with line MN and line MP , respectively, and let X be the midpoint

of arc B̂AC of Γ. Given that AB = 5, AC = 8, and ∠A = 60◦, compute the area of triangle XUV .

Proposed by: Evan Chen

Answer: 21√3

8

Let segments AI and EF meet at K. Extending AK to meet the circumcircle again at Y , we see thatX and Y are diametrically opposite, and it follows that AX and EF are parallel. Therefore the heightfrom X to UV is merely AK. Observe that AE = AF , so 4AEF is equilateral; since MN,MP areparallel to AF,AE respectively, it follows that 4MV U,4UEN,4FPV are equilateral as well. ThenMV = MP 〉PV = 1

2AC 〉FP = 12AC 〉AF +AP = 1

2AC 〉AF + 12AB = 1

2BC, since E,F are thetangency points of the incircle. Since 4MV U is equilateral, we have UV = MU = MV = 1

2BC.

Now we can compute BC = 7, whence UV = 72 and

AK =AB +AC 〉BC

2· cos 30◦ =

3√

3

2.

Hence, the answer is 21√3

8 .

7. Let S = {(x, y)|x, y ∈ Z, 0 ≤ x, y,≤ 2016}. Given points A = (x1, y1), B = (x2, y2) in S, define

d2017(A,B) = (x1 〉 x2)2 + (y1 〉 y2)2 (mod 2017).

The points A = (5, 5), B = (2, 6), C = (7, 11) all lie in S. There is also a point O ∈ S that satisfies

d2017(O,A) = d2017(O,B) = d2017(O,C).

Find d2017(O,A).

Proposed by: Yang Liu

Answer: 1021

Note that the triangle is a right triangle with right angle at A. Therefore, R2 = (7→2)2+(11→6)24 = 25

2 =(25)(2→1) ≡ 1021 (mod 2017). (An equivalent approach works for general triangles; the fact that thetriangle is right simply makes the circumradius slightly easier to compute.)

8. For i = 0, 1, . . . , 5 let li be the ray on the Cartesian plane starting at the origin, an angle θ = iπ3counterclockwise from the positive x-axis. For each i, point Pi is chosen uniformly at random fromthe intersection of li with the unit disk. Consider the convex hull of the points Pi, which will (withprobability 1) be a convex polygon with n vertices for some n. What is the expected value of n?

Proposed by:

Answer: 2 + 4 ln(2)

A vertex Pi is part of the convex hull if and only if it is not contained in the triangle formed bythe origin and the two adjacent vertices. Let the probability that a given vertex is contained inthe aforementioned triangle be p. By linearity of expectation, our answer is simply 6(1 〉 p). Say|P0| = a, |P2| = b. Stewart’s Theorem and the Law of Cosines give that p is equal to the probability

that |P1| <√ab〉 aba2+b2+ab(a+b)2 = ab

a+b ; alternatively this is easy to derive using coordinate methods.

The corresponding double integral evaluates to p = 23 (1〉 ln(2)), thus telling us our answer.

9. In cyclic quadrilateral ABCD with AB = AD = 49 and AC = 73, let I and J denote the incenters oftriangles ABD and CBD. If diagonal BD bisects IJ , find the length of IJ .

Proposed by: Evan Chen

Answer: 285

√69

Let O be circumcenter, R the circumradius and r the common inradius. We have IO2 = JO2 =R(R〉 2r) by a result of Euler; denote x for the common value of IO and JO. Additionally, we knowAJ = AB = AD = 49 (angle chase to find that ∠BJA = ∠JBA). Since A is the midpoint of the arc

B̂D not containing C, both J and A lie on the angle bisector of angle ∠BCD, so C, J,A are collinear.So by Power of a Point we have

R2 〉 x2 = 2Rr = AJ · JC = 49 · 24.

Next, observe that the angle bisector of angle BAD contains both I and O, so A, I,O are collinear.Let M be the midpoint of IJ , lying on BD. Let K be the intersection of IO and BD. Observing thatthe right triangles 4IMO and 4IKM are similar, we find IM2 = IK · IO = rx, so IJ2 = 4rx. Nowapply Stewart’s Theorem to 4AOJ to derive

R (x(R〉 x) + 4rx) = 492x+ x2(R〉 x).

Eliminating the common factor of x and rearranging gives

492 〉 (R〉 x)2 = 4Rr = 48 · 49

so R〉 x = 7. Hence R+ x = 49·247 = 168, and thus 2R = 175, 2x = 161. Thus r = 49·24

175 = 16825 .

Finally, IJ = 2√rx = 2

√84·161

25 = 28√69

5 .

10. The incircle of a triangle ABC is tangent to BC at D. Let H and Γ denote the orthocenter andcircumcircle of 4ABC. The B-mixtilinear incircle, centered at OB , is tangent to lines BA and BCand internally tangent to Γ. The C-mixtilinear incircle, centered at OC , is defined similarly. Supposethat DH ⊥ OBOC , AB =

√3 and AC = 2. Find BC.

Proposed by: Evan Chen

Answer:√

13 (7 + 2

√13)

Let the B-mixtilinear incircle ωB touch Γ at TB , BA at B1 and BC at B2. Define TC ∈ Γ, C1 ∈ CB,C2 ∈ CA, and ωC similarly. Call I the incenter of triangle ABC, and γ the incircle.

We first identify two points on the radical axis of the B and C mixtilinear incircles:

• The midpoint M of arc BC of the circumcircle of ABC. This follows from the fact that M , B1,TB are collinear with

MB2 = MC2 = MB1 ·MTB

and similarly for C.

• The midpoint N of ID. To see this, first recall that I is the midpoint of segments B1B2 andC1C2. From this, we can see that the radical axis of ωB and γ contains N (since it is the linethrough the midpoints of the common external tangents of ωB , γ). A similar argument for Cshows that the midpoint of ID is actually the radical center of the ωB , ωC , γ.

Now consider a homothety with ratio 2 at I. It sends line MN to the line through D and the A-excenterIA (since M is the midpoint of IIA, by “Fact 5”). Since DH was supposed to be parallel to line MN ,it follows that line DH passes through IA; however a homothety at D implies that this occurs only ifH is the midpoint of the A-altitude.

Let a = BC, b = CA = 2 and c = AB =√

3. So, we have to just find the value of a such that theorthocenter of ABC lies on the midpoint of the A-altitude. This is a direct computation with the Lawof Cosines, but a more elegant solution is possible using the fact that H has barycentric coordinates(SBSC : SCSA : SASB), where SA = 1

2 (b2 + c2 〉 a2) and so on. Indeed, as H is on the A-midline wededuce directly that

SBSC = SA(SB + SC) = a2SA =⇒ 1

4(a2 〉 1)(a2 + 1) =

1

2a2(7〉 a2).

Solving as a quadratic in a2 and taking the square roots gives

3a4 〉 14a2 〉 1 = 0 =⇒ a =

√1

3(7 + 2

√13)

as desired. 1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2016, February 20, 2016 — GUTS ROUND

Organization Team Team ID#

1. [5] Let x and y be complex numbers such that x+ y =√

20 and x2 + y2 = 15. Compute |x〉 y|.

2. [5] Sherry is waiting for a train. Every minute, there is a 75% chance that a train will arrive. However,she is engrossed in her game of sudoku, so even if a train arrives she has a 75% chance of not noticingit (and hence missing the train). What is the probability that Sherry catches the train in the next fiveminutes?

3. [5] Let PROBLEMZ be a regular octagon inscribed in a circle of unit radius. Diagonals MR, OZmeet at I. Compute LI.

4. [5] Consider a three-person game involving the following three types of fair six-sided dice.

• Dice of type A have faces labelled 2, 2, 4, 4, 9, 9.

• Dice of type B have faces labelled 1, 1, 6, 6, 8, 8.

• Dice of type C have faces labelled 3, 3, 5, 5, 7, 7.

All three players simultaneously choose a die (more than one person can choose the same type of die,and the players don’t know one another’s choices) and roll it. Then the score of a player P is thenumber of players whose roll is less than P ’s roll (and hence is either 0, 1, or 2). Assuming all threeplayers play optimally, what is the expected score of a particular player?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2016, February 20, 2016 — GUTS ROUND

Organization Team Team ID#

5. [6] Patrick and Anderson are having a snowball fight. Patrick throws a snowball at Anderson which isshaped like a sphere with a radius of 10 centimeters. Anderson catches the snowball and uses the snowfrom the snowball to construct snowballs with radii of 4 centimeters. Given that the total volume ofthe snowballs that Anderson constructs cannot exceed the volume of the snowball that Patrick threw,how many snowballs can Anderson construct?

6. [6] Consider a 2 × n grid of points and a path consisting of 2n 〉 1 straight line segments connectingall these 2n points, starting from the bottom left corner and ending at the upper right corner. Sucha path is called efficient if each point is only passed through once and no two line segments intersect.How many efficient paths are there when n = 2016?

7. [6] A contest has six problems worth seven points each. On any given problem, a contestant can scoreeither 0, 1, or 7 points. How many possible total scores can a contestant achieve over all six problems?

8. [6] For each positive integer n and non-negative integer k, define W (n, k) recursively by

W (n, k) =

{nn k = 0

W (W (n, k 〉 1), k 〉 1) k > 0.

Find the last three digits in the decimal representation of W (555, 2).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2016, February 20, 2016 — GUTS ROUND

Organization Team Team ID#

9. [7] Victor has a drawer with two red socks, two green socks, two blue socks, two magenta socks, twolavender socks, two neon socks, two mauve socks, two wisteria socks, and 2000 copper socks, for a totalof 2016 socks. He repeatedly draws two socks at a time from the drawer at random, and stops if thesocks are of the same color. However, Victor is red-green colorblind, so he also stops if he sees a redand green sock.

What is the probability that Victor stops with two socks of the same color? Assume Victor returnsboth socks to the drawer at each step.

10. [7] Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Let O be the circumcenter of ABC.Find the distance between the circumcenters of triangles AOB and AOC.

11. [7] Define φ!(n) as the product of all positive integers less than or equal to n and relatively prime ton. Compute the remainder when ∑

2≤n≤50gcd(n,50)=1

φ!(n)

is divided by 50.

12. [7] Let R be the rectangle in the Cartesian plane with vertices at (0, 0), (2, 0), (2, 1), and (0, 1). R canbe divided into two unit squares, as shown; the resulting figure has seven edges.

Compute the number of ways to choose one or more of the seven edges such that the resulting figureis traceable without lifting a pencil. (Rotations and reflections are considered distinct.)

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HMMT February 2016, February 20, 2016 — GUTS ROUND

Organization Team Team ID#

13. [9] A right triangle has side lengths a, b, and√

2016 in some order, where a and b are positive integers.Determine the smallest possible perimeter of the triangle.

14. [9] Let ABC be a triangle such that AB = 13, BC = 14, CA = 15 and let E, F be the feet of thealtitudes from B and C, respectively. Let the circumcircle of triangle AEF be ω. We draw three lines,tangent to the circumcircle of triangle AEF at A, E, and F . Compute the area of the triangle thesethree lines determine.

15. [9] Compute tan(π7

)tan

(2π7

)tan

(3π7

).

16. [9] Determine the number of integers 2 ≤ n ≤ 2016 such that nn 〉 1 is divisible by 2, 3, 5, 7.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2016, February 20, 2016 — GUTS ROUND

Organization Team Team ID#

17. [11] Compute the sum of all integers 1 ≤ a ≤ 10 with the following property: there exist integers pand q such that p, q, p2 + a and q2 + a are all distinct prime numbers.

18. [11] Alice and Bob play a game on a circle with 8 marked points. Alice places an apple beneath one ofthe points, then picks five of the other seven points and reveals that none of them are hiding the apple.Bob then drops a bomb on any of the points, and destroys the apple if he drops the bomb either onthe point containing the apple or on an adjacent point. Bob wins if he destroys the apple, and Alicewins if he fails. If both players play optimally, what is the probability that Bob destroys the apple?

19. [11] Let

A = limn→∞

2016∑i=0

(〉1)i ·(ni

)(ni+2

)(ni+1

)2Find the largest integer less than or equal to 1

A .

The following decimal approximation might be useful: 0.6931 < ln(2) < 0.6932, where ln denotes thenatural logarithm function.

20. [11] Let ABC be a triangle with AB = 13, AC = 14, and BC = 15. Let G be the point on AC suchthat the reflection of BG over the angle bisector of ∠B passes through the midpoint of AC. Let Y bethe midpoint of GC and X be a point on segment AG such that AX

XG = 3. Construct F and H on ABand BC, respectively, such that FX k BG k HY . If AH and CF concur at Z and W is on AC suchthat WZ k BG, find WZ.

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HMMT February 2016, February 20, 2016 — GUTS ROUND

Organization Team Team ID#

21. [12] Tim starts with a number n, then repeatedly flips a fair coin. If it lands heads he subtracts 1from his number and if it lands tails he subtracts 2. Let En be the expected number of flips Tim doesbefore his number is zero or negative. Find the pair (a, b) such that

limn→∞

(En 〉 an〉 b) = 0.

22. [12] On the Cartesian plane R2, a circle is said to be nice if its center is at the origin (0, 0) and it passesthrough at least one lattice point (i.e. a point with integer coordinates). Define the points A = (20, 15)and B = (20, 16). How many nice circles intersect the open segment AB?

For reference, the numbers 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683,691 are the only prime numbers between 600 and 700.

23. [12] Let t = 2016 and p = ln 2. Evaluate in closed form the sum

∞∑k=1

(1〉

k−1∑n=0

e−ttn

n!

)(1〉 p)k−1 p.

24. [12] Let ∆A1B1C be a triangle with ∠A1B1C = 90◦ and CA1

CB1=√

5 + 2. For any i ≥ 2, define Ai tobe the point on the line A1C such that AiBi−1 ⊥ A1C and define Bi to be the point on the line B1Csuch that AiBi ⊥ B1C. Let Γ1 be the incircle of ∆A1B1C and for i ≥ 2, Γi be the circle tangent toΓi−1, A1C,B1C which is smaller than Γi−1.

How many integers k are there such that the line A1B2016 intersects Γk?

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HMMT February 2016, February 20, 2016 — GUTS ROUND

Organization Team Team ID#

25. [14] A particular coin can land on heads (H), on tails (T), or in the middle (M), each with proba-bility 1

3 . Find the expected number of flips necessary to observe the contiguous sequence HMMTH-MMT...HMMT, where the sequence HMMT is repeated 2016 times.

26. [14] For positive integers a, b, a "" b is defined as follows: a "" 1 = a, and a "" b = aa↑↑(b−1) if b > 1.

Find the smallest positive integer n for which there exists a positive integer a such that a "" 6 6≡ a "" 7mod n.

27. [14] Find the smallest possible area of an ellipse passing through (2, 0), (0, 3), (0, 7), and (6, 0).

28. [14] Among citizens of Cambridge there exist 8 different types of blood antigens. In a crowded lecturehall are 256 students, each of whom has a blood type corresponding to a distinct subset of the antigens;the remaining of the antigens are foreign to them.

Quito the Mosquito flies around the lecture hall, picks a subset of the students uniformly at random,and bites the chosen students in a random order. After biting a student, Quito stores a bit of anyantigens that student had. A student bitten while Quito had k blood antigen foreign to him/her willsuffer for k hours. What is the expected total suffering of all 256 students, in hours?

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HMMT February 2016, February 20, 2016 — GUTS ROUND

Organization Team Team ID#

29. [16] Katherine has a piece of string that is 2016 millimeters long. She cuts the string at a locationchosen uniformly at random, and takes the left half. She continues this process until the remainingstring is less than one millimeter long. What is the expected number of cuts that she makes?

30. [16] Determine the number of triples 0 ≤ k,m, n ≤ 100 of integers such that

2mn〉 2nm = 2k.

31. [16] For a positive integer n, denote by τ(n) the number of positive integer divisors of n, and denoteby φ(n) the number of positive integers that are less than or equal to n and relatively prime to n.Call a positive integer n good if ϕ(n) + 4τ(n) = n. For example, the number 44 is good becauseϕ(44) + 4τ(44) = 44.

Find the sum of all good positive integers n.

32. [16] How many equilateral hexagons of side length√

13 have one vertex at (0, 0) and the other fivevertices at lattice points?

(A lattice point is a point whose Cartesian coordinates are both integers. A hexagon may be concavebut not self-intersecting.)

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HMMT February 2016, February 20, 2016 — GUTS ROUND

Organization Team Team ID#

33. [20] (Lucas Numbers) The Lucas numbers are defined by L0 = 2, L1 = 1, and Ln+2 = Ln+1 + Lnfor every n ≥ 0. There are N integers 1 ≤ n ≤ 2016 such that Ln contains the digit 1. Estimate N .

An estimate of E earns b20〉 2|N 〉 E|c or 0 points, whichever is greater.

34. [20] (Caos) A cao [sic] has 6 legs, 3 on each side. A walking pattern for the cao is defined as anordered sequence of raising and lowering each of the legs exactly once (altogether 12 actions), startingand ending with all legs on the ground. The pattern is safe if at any point, he has at least 3 legs onthe ground and not all three legs are on the same side. Estimate N , the number of safe patterns.

An estimate of E > 0 earns⌊20 min(N/E,E/N)4

⌋points.

35. [20] (Maximal Determinant) In a 17 × 17 matrix M , all entries are ±1. The maximum possiblevalue of |detM | is N . Estimate N .

An estimate of E > 0 earns⌊20 min(N/E,E/N)2

⌋points.

36. [20] (Self-Isogonal Cubics) Let ABC be a triangle with AB = 2, AC = 3, BC = 4. The isogonalconjugate of a point P , denoted P ∗, is the point obtained by intersecting the reflection of lines PA,PB, PC across the angle bisectors of ∠A, ∠B, and ∠C, respectively.

Given a point Q, let K(Q) denote the unique cubic plane curve which passes through all points P suchthat line PP ∗ contains Q. Consider:

(a) the M’Cay cubic K(O), where O is the circumcenter of 4ABC,

(b) the Thomson cubic K(G), where G is the centroid of 4ABC,

(c) the Napoleon-Feurerbach cubic K(N), where N is the nine-point center of 4ABC,

(d) the Darboux cubic K(L), where L is the de Longchamps point (the reflection of the orthocenteracross point O),

(e) the Neuberg cubic K(X30), where X30 is the point at infinity along line OG,

(f) the nine-point circle of 4ABC,

(g) the incircle of 4ABC, and

(h) the circumcircle of 4ABC.

Estimate N , the number of points lying on at least two of these eight curves. An estimate of E earns⌊20 · 2−|N−E|/6

⌋points.

HMMT February 2016February 20, 2016

Guts

1. [5] Let x and y be complex numbers such that x+ y =√

20 and x2 + y2 = 15. Compute |x− y|.Proposed by: Evan Chen

Answer:√

10

We have (x− y)2 + (x+ y)2 = 2(x2 + y2), so (x− y)2 = 10, hence |x− y| =√

10.

2. [5] Sherry is waiting for a train. Every minute, there is a 75% chance that a train will arrive. However,she is engrossed in her game of sudoku, so even if a train arrives she has a 75% chance of not noticingit (and hence missing the train). What is the probability that Sherry catches the train in the next fiveminutes?

Proposed by:

Answer: 1−(

1316

)5During any given minute, the probability that Sherry doesn’t catch the train is 1

4 +(

34

)2= 13

16 . Thedesired probability is thus one minus the probability that she doesn’t catch the train for the next five

minutes: 1−(

1316

)5.

3. [5] Let PROBLEMZ be a regular octagon inscribed in a circle of unit radius. Diagonals MR, OZmeet at I. Compute LI.

Proposed by: Evan Chen

Answer:√

2

If W is the center of the circle then I is the incenter of 4RWZ. Moreover, PRIZ is a rhombus. Itfollows that PI is twice the inradius of a 1-1-

√2 triangle, hence the answer of 2−

√2. So LI =

√2.

Alternatively, one can show (note, really) that the triangle OIL is isosceles.

4. [5] Consider a three-person game involving the following three types of fair six-sided dice.

• Dice of type A have faces labelled 2, 2, 4, 4, 9, 9.

• Dice of type B have faces labelled 1, 1, 6, 6, 8, 8.

• Dice of type C have faces labelled 3, 3, 5, 5, 7, 7.

All three players simultaneously choose a die (more than one person can choose the same type of die,and the players don’t know one another’s choices) and roll it. Then the score of a player P is thenumber of players whose roll is less than P ’s roll (and hence is either 0, 1, or 2). Assuming all threeplayers play optimally, what is the expected score of a particular player?

Proposed by:

Answer: 89

Short version: third player doesn’t matter; against 1 opponent, by symmetry, you’d both play the samestrategy. Type A beats B, B beats C, and C beats A all with probability 5/9. It can be determinedthat choosing each die with probability 1/3 is the best strategy. Then, whatever you pick, there is a1/3 of dominating, a 1/3 chance of getting dominated, and a 1/3 chance of picking the same die (whichgives a 1/3 · 2/3 + 1/3 · 1/3 = 1/3 chance of rolling a higher number). Fix your selection; then theexpected payout is then 1/3 · 5/9 + 1/3 · 4/9 + 1/3 · 1/3 = 1/3 + 1/9 = 4/9. Against 2 players, yourEV is just E(p1) + E(p2) = 2E(p1) = 8/9

5. [6] Patrick and Anderson are having a snowball fight. Patrick throws a snowball at Anderson which isshaped like a sphere with a radius of 10 centimeters. Anderson catches the snowball and uses the snowfrom the snowball to construct snowballs with radii of 4 centimeters. Given that the total volume ofthe snowballs that Anderson constructs cannot exceed the volume of the snowball that Patrick threw,how many snowballs can Anderson construct?

Proposed by:

Answer: 15 ⌊(10

4

)3⌋

=

⌊125

8

⌋= 15.

6. [6] Consider a 2 × n grid of points and a path consisting of 2n − 1 straight line segments connectingall these 2n points, starting from the bottom left corner and ending at the upper right corner. Sucha path is called efficient if each point is only passed through once and no two line segments intersect.How many efficient paths are there when n = 2016?

Proposed by: Casey Fu

Answer:

(4030

2015

)The general answer is

(2(n− 1)

n− 1

): Simply note that the points in each column must be taken in order,

and anything satisfying this avoids intersections, so just choose the steps during which to be in thefirst column.

7. [6] A contest has six problems worth seven points each. On any given problem, a contestant can scoreeither 0, 1, or 7 points. How many possible total scores can a contestant achieve over all six problems?

Proposed by: Evan Chen

Answer: 28

For 0 ≤ k ≤ 6, to obtain a score that is k (mod 6) exactly k problems must get a score of 1. Theremaining 6− k problems can generate any multiple of 7 from 0 to 7(6− k), of which there are 7− k.

So the total number of possible scores is6∑k=0

(7− k) = 28.

8. [6] For each positive integer n and non-negative integer k, define W (n, k) recursively by

W (n, k) =

{nn k = 0

W (W (n, k − 1), k − 1) k > 0.

Find the last three digits in the decimal representation of W (555, 2).

Proposed by:

Answer: 875

For any n, we have

W (n, 1) = W (W (n, 0), 0) = (nn)nn

= nnn+1

.

Thus,

W (555, 1) = 555555556

.

Let N = W (555, 1) for brevity, and note that N ≡ 0 (mod 125), and N ≡ 3 (mod 8). Then,

W (555, 2) = W (N, 1) = NNN+1

is 0 (mod 125) and 3 (mod 8).

From this we can conclude (by the Chinese Remainder Theorem) that the answer is 875.

9. [7] Victor has a drawer with two red socks, two green socks, two blue socks, two magenta socks, twolavender socks, two neon socks, two mauve socks, two wisteria socks, and 2000 copper socks, for a totalof 2016 socks. He repeatedly draws two socks at a time from the drawer at random, and stops if thesocks are of the same color. However, Victor is red-green colorblind, so he also stops if he sees a redand green sock.

What is the probability that Victor stops with two socks of the same color? Assume Victor returnsboth socks to the drawer at each step.

Proposed by: Evan Chen

Answer: 19990081999012

There are

(2000

2

)+ 8

(2

2

)= 1999008 ways to get socks which are matching colors, and four extra

ways to get a red-green pair, hence the answer.

10. [7] Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Let O be the circumcenter of ABC.Find the distance between the circumcenters of triangles AOB and AOC.

Proposed by: Evan Chen

Answer: 916

Let S, T be the intersections of the tangents to the circumcircle of ABC at A,C and at A,B respectively.Note that ASCO is cyclic with diameter SO, so the circumcenter of AOC is the midpoint of OS, andsimilarly for the other side. So the length we want is 1

2ST . The circumradius R of ABC can be

computed by Heron’s formula and K = abc4R , giving R = 65

8 . A few applications of the Pythagorean

theorem and similar triangles gives AT = 656 , AS = 39

2 , so the answer is91

6.

11. [7] Define φ!(n) as the product of all positive integers less than or equal to n and relatively prime ton. Compute the remainder when ∑

2≤n≤50gcd(n,50)=1

φ!(n)

is divided by 50.

Proposed by: Evan Chen

Answer: 12

First, φ!(n) is even for all odd n, so it vanishes modulo 2.

To compute the remainder modulo 25, we first evaluate φ!(3) + φ!(7) + φ!(9) ≡ 2 + 5 · 4 + 5 · 3 ≡ 12(mod 25). Now, for n ≥ 11 the contribution modulo 25 vanishes as long as 5 - n.

We conclude the answer is 12.

12. [7] Let R be the rectangle in the Cartesian plane with vertices at (0, 0), (2, 0), (2, 1), and (0, 1). R canbe divided into two unit squares, as shown; the resulting figure has seven edges.

Compute the number of ways to choose one or more of the seven edges such that the resulting figureis traceable without lifting a pencil. (Rotations and reflections are considered distinct.)

Proposed by: Joy Zheng

Answer: 61

We have two cases, depending on whether we choose the middle edge. If so, then either all the remainingedges are either to the left of or to the right of this edge, or there are edges on both sides, or neither;in the first two cases there are 6 ways each, in the third there are 16 + 1 = 17 ways, and in the lastthere is 1 way. Meanwhile, if we do not choose the middle edge, then we have to choose a beginningand endpoint, plus the case where we have a loop, for a total of 6 · 5 + 1 = 31 cases. This gives a totalof 6 + 6 + 17 + 1 + 31 = 61 possible cases.

13. [9] A right triangle has side lengths a, b, and√

2016 in some order, where a and b are positive integers.Determine the smallest possible perimeter of the triangle.

Proposed by: Evan Chen

Answer: 48 +√

2016

There are no integer solutions to a2 + b2 = 2016 due to the presence of the prime 7 on the right-handside (by Fermat’ s Christmas Theorem). Assuming a < b, the minimal solution (a, b) = (3, 45) whichgives the answer above.

14. [9] Let ABC be a triangle such that AB = 13, BC = 14, CA = 15 and let E, F be the feet of thealtitudes from B and C, respectively. Let the circumcircle of triangle AEF be ω. We draw three lines,tangent to the circumcircle of triangle AEF at A, E, and F . Compute the area of the triangle thesethree lines determine.

Proposed by: Christopher Shao

Answer: 4625

Note that AEF ∼ ABC. Let the vertices of the triangle whose area we wish to compute be P,Q,R,opposite A,E, F respectively. Since H,O are isogonal conjugates, line AH passes through the circum-center of AEF , so QR ‖ BC.

Let M be the midpoint of BC. We claim that M = P . This can be seen by angle chasing at E,F tofind that ∠PFB = ∠ABC, ∠PEC = ∠ACB, and noting that M is the circumcenter of BFEC. So,the height from P to QR is the height from A to BC, and thus if K is the area of ABC, the area wewant is QR

BCK.

Heron’s formula gives K = 84, and similar triangles QAF,MBF and RAE,MCE give QA = BC2

tanBtanA ,

RA = BC2

tanCtanA , so that QR

BC = tanB+tanC2 tanA = tanB tanC−1

2 = 1110 ,

since the height from A to BC is 12. So our answer is462

5.

15. [9] Compute tan(π7

)tan

(2π7

)tan

(3π7

).

Proposed by: Alexander Katz

Answer:√

7

Consider the polynomial P (z) = z7 − 1. Let z = eix = cosx+ i sinx. Then

z7 − 1 =

(cos7 x−

(7

2

)cos5 x sin2 x+

(7

4

)cos3 x sin4 x−

(7

6

)cosx sin6 x− 1

)+ i

(− sin7 x+

(7

2

)sin5 x cos2 x−

(7

4

)sin3 x cos4 x+

(7

6

)sinx cos 6x

)Consider the real part of this equation. We may simplify it to 64 cos7 x − . . . − 1, where the middleterms are irrelevant. The roots of P are x = 2π

7 ,4π7 , . . ., so

∏7k=1 cos

(2πk

7

)= 1

64 . But

7∏k=1

cos

(2πk

7

)=

(3∏k=1

cos

(kπ

7

))2

so∏3k=1 cos

(kπ7

)= 1

8 .

Now consider the imaginary part of this equation. We may simplify it to −64 sin11 x + . . . + 7 sinx,where again the middle terms are irrelevant. We can factor out sinx to get −64 sin10 x+ . . .+ 7, andthis polynomial has roots x = 2π

7 , . . . ,12π7 (but not 0). Hence

∏6k=1 sin

(2πk

7

)= − 7

64 . But, like before,we have

6∏k=1

sin

(2πk

7

)= −

(3∏k=1

sin

(2πk

7

))2

hence∏3k=1 sin

(kπ7

)=√

78 . As a result, our final answer is

√7

818

=√

7 .

16. [9] Determine the number of integers 2 ≤ n ≤ 2016 such that nn − 1 is divisible by 2, 3, 5, 7.

Proposed by: Evan Chen

Answer: 9

Only n ≡ 1 (mod 210) work. Proof: we require gcd(n, 210) = 1. Note that ∀p ≤ 7 the order of n(mod p) divides p− 1, hence is relatively prime to any p ≤ 7. So nn ≡ 1 (mod p) ⇐⇒ n ≡ 1 (mod p)for each of these p.

17. [11] Compute the sum of all integers 1 ≤ a ≤ 10 with the following property: there exist integers pand q such that p, q, p2 + a and q2 + a are all distinct prime numbers.

Proposed by: Evan Chen

Answer: 20

Odd a fail for parity reasons and a ≡ 2 (mod 3) fail for mod 3 reasons. This leaves a ∈ {4, 6, 10}. Itis easy to construct p and q for each of these, take (p, q) = (3, 5), (5, 11), (3, 7), respectively.

18. [11] Alice and Bob play a game on a circle with 8 marked points. Alice places an apple beneath one ofthe points, then picks five of the other seven points and reveals that none of them are hiding the apple.Bob then drops a bomb on any of the points, and destroys the apple if he drops the bomb either onthe point containing the apple or on an adjacent point. Bob wins if he destroys the apple, and Alicewins if he fails. If both players play optimally, what is the probability that Bob destroys the apple?

Proposed by:

Answer: 12

Let the points be 0, . . . , 7 (mod 8), and view Alice’s reveal as revealing the three possible locations ofthe apple. If Alice always picks 0, 2, 4 and puts the apple randomly at 0 or 4, by symmetry Bob cannotachieve more than 1

2 . Here’s a proof that 12 is always possible.

Among the three revealed indices a, b, c, positioned on a circle, two must (in the direction in whichthey’re adjacent) have distance at least 3, so without loss of generality the three are 0, b, c where1 ≤ b < c ≤ 5. Modulo reflection and rotation, the cases are: (0, 1, 2): Bob places at 1 and wins.(0, 1, 3): Bob places at 1 half the time and 3 half the time, so wherever the apple is Bob wins withprobability 1

2 . (0, 1, 4): Bob places at 1 or 4, same as above. (0, 2, 4): Bob places at 1 or 3, same asabove. (0, 2, 5): Bob places at 1 or 5, same as above.

These cover all cases, so we’re done.

19. [11] Let

A = limn→∞

2016∑i=0

(−1)i ·(ni

)(ni+2

)(ni+1

)2Find the largest integer less than or equal to 1

A .

The following decimal approximation might be useful: 0.6931 < ln(2) < 0.6932, where ln denotes thenatural logarithm function.

Proposed by: Pakawut Jiradilok

Answer: 1

Note2016∑i=0

(−1)i ·(ni

)(ni+2

)(ni+1

)2 =

2016∑i=0

(−1)i · (i+ 1)(n− i− 1)

(i+ 2)(n− i),

So

limn→∞

2016∑i=0

(−1)i ·(ni

)(ni+2

)(ni+1

)2 =

2016∑i=0

(−1)i · (i+ 1)

(i+ 2)= 1−

2016∑i=2

(−1)i

i≈ ln(2).

Then 1A ≈

1ln(2) ≈ 1.44, so the answer is 1.

20. [11] Let ABC be a triangle with AB = 13, AC = 14, and BC = 15. Let G be the point on AC suchthat the reflection of BG over the angle bisector of ∠B passes through the midpoint of AC. Let Y bethe midpoint of GC and X be a point on segment AG such that AX

XG = 3. Construct F and H on ABand BC, respectively, such that FX ‖ BG ‖ HY . If AH and CF concur at Z and W is on AC suchthat WZ ‖ BG, find WZ.

Proposed by: Ritesh Ragavender

Answer: 1170√

371379

Observe that BG is the B-symmedian, and thus AGGC = c2

a2 . Stewart’s theorem gives us

BG =

√2a2c2b

b(a2 + c2)− a2b2c2

a2 + c2=

ac

a2 + c2

√2(a2 + c2)− b2 =

390√

37

197.

Then by similar triangles,

ZW = HYZA

HA= BG

Y C

GC

ZA

HA= BG

1

2

6

7=

1170√

37

1379

where ZAHA is found with mass points or Ceva.

21. [12] Tim starts with a number n, then repeatedly flips a fair coin. If it lands heads he subtracts 1from his number and if it lands tails he subtracts 2. Let En be the expected number of flips Tim doesbefore his number is zero or negative. Find the pair (a, b) such that

limn→∞

(En − an− b) = 0.

Proposed by:

Answer: ( 23 ,

29 )

We have the recurrence En = 12 (En−1 + 1) + 1

2 (En−2 + 1), or En = 1 + 12 (En−1 + En−2), for n ≥ 2.

Let Fn = En − 23n. By directly plugging this into the recurrence for En, we get the recurrence Fn =

12 (Fn−1 + Fn−1). The roots of the characteristic polynomial of this recurrence are 1 and − 1

2 , so Fn =A + B(− 1

2 )n for some A and B depending on the initial conditions. But clearly we have E0 = 0 andE1 = 1 so F0 = 0 and F1 = 1

3 so A = 29 and B = − 2

9 .

Hence, En = 23n+ 2

9 −29 (− 1

2 )n, so limn→∞(En − 23n−

29 ) = 0. Hence ( 2

3 ,29 ) is the desired pair.

22. [12] On the Cartesian plane R2, a circle is said to be nice if its center is at the origin (0, 0) and it passesthrough at least one lattice point (i.e. a point with integer coordinates). Define the points A = (20, 15)and B = (20, 16). How many nice circles intersect the open segment AB?

For reference, the numbers 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683,691 are the only prime numbers between 600 and 700.

Proposed by:

Answer: 10

The square of the radius of a nice circle is the sum of the square of two integers.

The nice circle of radius r intersects (the open segment) AB if and only if a point on AB is a distancer from the origin. AB consists of the points (20, t) where t ranges over (15, 16). The distance from theorigin is

√202 + t2 =

√400 + t2. As t ranges over (15, 16),

√400 + t2 ranges over (

√625,√

656), so thenice circle of radius r intersects AB if and only if 625 < r2 < 656.

The possible values of r2 are those in this range that are the sum of two perfect squares, and each suchvalue corresponds to a unique nice circle. By Fermat’s Christmas theorem, an integer is the sum oftwo squares if an only if in its prime factorization, each prime that is 3 mod 4 appears with an evenexponent (possibly 0.) In addition, since squares are 0, 1, or 4 mod 8, we can quickly eliminate integersthat are 3, 6, or 7 mod 8.

Now I will list all the integers that aren’t 3, 6, or 7 mod 8 in the range and either supply the bad primefactor or write ”nice” with the prime factorization.

626: nice (2 · 313)

628: nice (22 · 157)

629: nice (17 · 37)

632: 79

633: 3

634: nice (2 · 317)

636: 3

637: nice (72 · 13)

640: nice (27 · 5)

641: nice (641)

642: 3

644: 7

645: 3

648: nice (23 · 34)

649: 11

650: nice (2 · 52 · 13)

652: 163

653: nice (653). There are 10 nice circles that intersect AB.

23. [12] Let t = 2016 and p = ln 2. Evaluate in closed form the sum

∞∑k=1

(1−

k−1∑n=0

e−ttn

n!

)(1− p)k−1

p.

Proposed by: Aaron Landesman

Answer: 1−(

12

)2016

Let q = 1− p. Then

∞∑k=1

(1−

k−1∑n=0

e−ttn

n!

)qk−1p =

∞∑k=1

qk−1p−∞∑k=1

k−1∑n=0

e−ttn

n!qk−1p

= 1−∞∑k=1

k−1∑n=0

e−ttn

n!qk−1p

= 1−∞∑n=0

∞∑k=n+1

e−ttn

n!qk−1p

= 1−∞∑n=0

e−ttn

n!qn

= 1−∞∑n=0

e−t(qt)n

n!= 1− e−teqt = 1− e−pt.

Thus the answer is 1−(

12

)2016.

24. [12] Let ∆A1B1C be a triangle with ∠A1B1C = 90◦ and CA1

CB1=√

5 + 2. For any i ≥ 2, define Ai tobe the point on the line A1C such that AiBi−1 ⊥ A1C and define Bi to be the point on the line B1Csuch that AiBi ⊥ B1C. Let Γ1 be the incircle of ∆A1B1C and for i ≥ 2, Γi be the circle tangent toΓi−1, A1C,B1C which is smaller than Γi−1.

How many integers k are there such that the line A1B2016 intersects Γk?

Proposed by:

Answer: 4030

We claim that Γ2 is the incircle of 4B1A2C. This is because 4B1A2C is similar to A1B1C withdilation factor

√5 − 2, and by simple trigonometry, one can prove that Γ2 is similar to Γ1 with the

same dilation factor. By similarities, we can see that for every k, the incircle of 4AkBkC is Γ2k−1,and the incircle of 4BkAk+1C is Γ2k. Therefore, A1B2016 intersects all Γ1, . . . ,Γ4030 but not Γk forany k ≥ 4031.

25. [14] A particular coin can land on heads (H), on tails (T), or in the middle (M), each with proba-bility 1

3 . Find the expected number of flips necessary to observe the contiguous sequence HMMTH-MMT...HMMT, where the sequence HMMT is repeated 2016 times.

Proposed by: Ritesh Ragavender

Answer: 38068−8180

Let E0 be the expected number of flips needed. Let E1 be the expected number more of flips neededif the first flip landed on H. Let E2 be the expected number more if the first two landed on HM. Ingeneral, let Ek be the expected number more of flips needed if the first k flips landed on the first kvalues of the sequence HMMTHMMT...HMMT.

We have

Ei =

{1 + 1

3Ei+1 + 13E1 + 1

3E0 i 6≡ 0 (mod 4)1 + 1

3Ei+1 + 23E0 i ≡ 0 (mod 4)

Using this relation for i = 0 gives us E1 = E0 − 3. Let Fi = 13iEi. By simple algebraic manipulations

we have

Fi+1 − Fi =

{− 2

3i+1 · E0 i 6≡ 0 (mod 4)− 1

3i − 23i+1 · E0 i ≡ 0 (mod 4)

We clearly have F2016·4 = 0 and F0 = E0. So adding up the above relations for i = 0 to i = 2016 · 4− 1gives

−E0 = −2E0

2016·4∑i=1

1

3i−

2015∑k=0

1

34k

= E0

(1

32016·4 − 1

)−

1− 132016·4

8081

so E0 = 38068−8180 .

26. [14] For positive integers a, b, a ↑↑ b is defined as follows: a ↑↑ 1 = a, and a ↑↑ b = aa↑↑(b−1) if b > 1.

Find the smallest positive integer n for which there exists a positive integer a such that a ↑↑ 6 6≡ a ↑↑ 7mod n.

Proposed by: Sammy Luo

Answer: 283

We see that the smallest such n must be a prime power, because if two numbers are distinct mod n,they must be distinct mod at least one of the prime powers that divide n. For k ≥ 2, if a ↑↑ k anda ↑↑ (k + 1) are distinct mod pr, then a ↑↑ (k − 1) and a ↑↑ k must be distinct mod φ(pr). In fact they

need to be distinct mod φ(pr)2 if p = 2 and r ≥ 3 because then there are no primitive roots mod pr.

Using this, for 1 ≤ k ≤ 5 we find the smallest prime p such that there exists a such that a ↑↑ k anda ↑↑ (k + 1) are distinct mod p. The list is: 3, 5, 11, 23, 47. We can easily check that the next largestprime for k = 5 is 139, and also any prime power other than 121 for which a ↑↑ 5 and a ↑↑ 6 aredistinct is also larger than 139.

Now if a ↑↑ 6 and a ↑↑ 7 are distinct mod p, then p− 1 must be a multiple of 47 or something that iseither 121or at least 139. It is easy to see that 283 is the smallest prime that satisfies this.

If n is a prime power less than 283 such that a ↑↑ 6 and a ↑↑ 7 are distinct mod n, then the prime canbe at most 13 and clearly this doesn’t work because φ(pr) = pr−1(p− 1).

To show that 283 works, choose a so that a is a primitive root mod 283, 47, 23, 11, 5 and 3. This ispossible by the Chinese Remainder theorem, and it is easy to see that this a works by induction.

27. [14] Find the smallest possible area of an ellipse passing through (2, 0), (0, 3), (0, 7), and (6, 0).

Proposed by: Calvin Deng

Answer: 56π√

39

Let Γ be an ellipse passing through A = (2, 0), B = (0, 3), C = (0, 7), D = (6, 0), and let P = (0, 0) bethe intersection of AD and BC. Area of Γ

Area of ABCD is unchanged under an affine transformation, so we just

have to minimize this quantity over situations where Γ is a circle and PAPD = 1

3 and PBBC = 3

7 . In fact,

we may assume that PA =√

7, PB = 3, PC = 7, PD = 3√

7. If ∠P = θ, then we can compute lengthsto get

r =Area of Γ

Area of ABCD= π

32− 20√

7 cos θ + 21 cos2 θ

9√

7 · sin3 θ

Let x = cos θ. Then if we treat r as a function of x,

0 =r′

r=

3x

1− x2+

42x− 20√

7

32− 20x√

7 + 21x2

which means that 21x3 − 40x√

7 + 138x− 20√

7 = 0. Letting y = x√

7 gives

0 = 3y3 − 40y2 + 138y − 140 = (y − 2)(3y2 − 34y + 70)

The other quadratic has roots that are greater than√

7, which means that the minimum ratio isattained when cos θ = x = y√

7= 2√

7. Plugging that back in gives that the optimum Area of Γ

Area of ABCD is

28π√

381 , so putting this back into the original configuration gives Area of Γ ≥ 56π

√3

9 . If you want tocheck on Geogebra, this minimum occurs when the center of Γ is ( 8

3 ,73 ).

28. [14] Among citizens of Cambridge there exist 8 different types of blood antigens. In a crowded lecturehall are 256 students, each of whom has a blood type corresponding to a distinct subset of the antigens;the remaining of the antigens are foreign to them.

Quito the Mosquito flies around the lecture hall, picks a subset of the students uniformly at random,and bites the chosen students in a random order. After biting a student, Quito stores a bit of anyantigens that student had. A student bitten while Quito had k blood antigen foreign to him/her willsuffer for k hours. What is the expected total suffering of all 256 students, in hours?

Proposed by: Sammy Luo

Answer: 2135−2128+12119·129

Let n = 8.

First, consider any given student S and an antigen a foreign to him/her. Assuming S has been bitten,we claim the probability S will suffer due to a is

1− 22n−1+1 − 1

22n−1(2n−1 + 1).

Indeed, let N = 2n−1 denote the number of students with a. So considering just these students andsumming over the number bitten, we obtain a probability

1

2N

N∑t=0

(N

t

)(N

t

)t

t+ 1=

1

2N2NN − 2N + 1

N + 1.

We now use linearity over all pairs (S, a) of students S and antigens a foreign to them. Noting thateach student is bitten with probability 1

2 , and retaining the notation N = 2n−1, we get

1

2

n∑k=0

[(n

k

)· k(

2NN − 2N + 1

2N (N + 1)

)]=nN(2NN − 2N + 1)

2N+1(N + 1).

Finally, setting n = 8 = 23 and N = 2n−1 = 27 = 128, we get the claimed answer.

29. [16] Katherine has a piece of string that is 2016 millimeters long. She cuts the string at a locationchosen uniformly at random, and takes the left half. She continues this process until the remainingstring is less than one millimeter long. What is the expected number of cuts that she makes?

Proposed by:

Answer: 1 + log(2016)

Letting f(x) be the expected number of cuts if the initial length of the string is x, we get the integralequation f(x) = 1 + 1

x

∫ x1f(y)dy. Letting g(x) =

∫ x1f(y)dy, we get dg/dx = 1 + 1

xg(x). Usingintegrating factors, we see that this has as its solution g(x) = x log(x), and thus f(x) = 1 + log(x).

30. [16] Determine the number of triples 0 ≤ k,m, n ≤ 100 of integers such that

2mn− 2nm = 2k.

Proposed by: Casey Fu

Answer: 22

First consider when n ≥ m, so let n = m + d where d ≥ 0. Then we have 2m(m + d − 2dm) =2m(m(1− 2d) + d), which is non-positive unless m = 0. So our first set of solutions is m = 0, n = 2j .

Now, we can assume that m > n, so let m = n + d where d > 0. Rewrite 2mn − 2nm = 2n+dn −2n(n+ d) = 2n((2d − 1)n− d). In order for this to be a power of 2, (2d − 1)n− d must be a power of2. This implies that for some j, 2j ≡ −d (mod 2d − 1). But notice that the powers of 2 (mod 2d − 1)are 1, 2, 4, . . . , 2d−1 (2d ≡ 1 so the cycle repeats).

In order for the residues to match, we need 2j + d = c(2d− 1), where 0 ≤ j ≤ d− 1 and c ≥ 1. In orderfor this to be true, we must have 2d−1 + d ≥ 2d − 1 ⇐⇒ d + 1 ≥ 2d−1. This inequality is only truefor d = 1, 2, 3. We plug each of these into the original expression (2d − 1)n− d.

For d = 1: n− 1 is a power of 2. This yields the set of solutions (2j + 2, 2j + 1) for j ≥ 0.

For d = 2: 3n − 2 is a power of 2. Note that powers of 2 are −2 (mod 3) if and only if it is an even

power, so n = 22j+23 . This yields the solution set ( 22j+8

3 , 22j+23 ), j ≥ 0.

For d = 3: 7n− 3 is a power of 2. Powers of 2 have a period of 3 when taken (mod 7), so inspection

tells us 7n− 3 = 23j+2, yielding the solution set (23j+2+247 , 23j+2+3

7 ), j ≥ 0.

Therefore, all the solutions are of the form

(m,n) = (0, 2j), (2j + 2, 2j + 1)

(22j + 8

3,

22j + 2

3), (

23j+2 + 24

7,

23j+2 + 3

7)

for j ≥ 0.

Restricting this family to m,n ≤ 100 gives 7 + 7 + 5 + 3 = 22.

31. [16] For a positive integer n, denote by τ(n) the number of positive integer divisors of n, and denoteby φ(n) the number of positive integers that are less than or equal to n and relatively prime to n.Call a positive integer n good if ϕ(n) + 4τ(n) = n. For example, the number 44 is good becauseϕ(44) + 4τ(44) = 44.

Find the sum of all good positive integers n.

Proposed by: Lawrence Sun

Answer: 172

We claim that 44, 56, 72 are the only good numbers. It is easy to check that these numbers work.

Now we prove none others work. First, remark that as n = 1, 2 fail so we have ϕ(n) is even, thus n iseven. This gives us ϕ(n) ≤ n/2. Now remark that τ(n) < 2

√n, so it follows we need n/2 + 8

√n >

n =⇒ n ≤ 256. This gives us a preliminary bound. Note that in addition we have 8τ(n) > n.

Now, it is easy to see that powers of 2 fail. Thus let n = 2apb1 where p1 is an odd prime. From8τ(n) > n we get 8(a+ 1)(b+ 1) > 2apb1 ≥ 2a3b from which we get that (a, b) is one of

(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1).

Remark that p1 ≤ b

√8(a+1)(b+1)

2a . From this we can perform some casework:

• If a = 1, b = 1 then p1 − 1 + 16 = 2p1 but then p = 15, absurd.

• If a = 1, b = 2 then we have p1 ≤ 5 which is obviously impossible.

• If a = 1, b = 3 then p1 ≤ 4 which is impossible.

• If a = 2, b = 1 then p1 ≤ 12 and it is easy to check that p1 = 11 and thus n = 44 is the onlysolution.

• If a = 2, b = 2 then p1 ≤ 4 which is impossible.

• If a = 3, b = 1 then p1 ≤ 8 and only p1 = 7 or n = 56 works.

• If a = 3, b = 2 then p1 ≤ 3 and p1 = 3, n = 72 works.

• If a = 4, b = 1 then p1 ≤ 1 which is absurd.

Now suppose n is the product of 3 distinct primes, so n = 2apb1pc2 so we have 8(a+ 1)(b+ 1)(c+ 1) >

2a3b5c then we must have (a, b, c) equal to one of

(1, 1, 1), (1, 2, 1), (2, 1, 1), (3, 1, 1).

Again, we can do some casework:

• If a = b = c = 1 then 8τ(n) = 64 > 2p1p2 but then p1 = 3, p2 = 5 or p1 = 3, p2 = 7 is forcedneither of which work.

• If a = 1, b = 2, c = 1 then 8τ(n) = 96 > 2p21p2 but then p1 = 3, p2 = 5 is forced which does not

work.

• If a = 2, b = 1, c = 1 then 8τ(n) = 96 > 4p1p2 forces p1 = 3, p2 = 5 or p1 = 3, p2 = 7 neither ofwhich work.

• If a = 3, b = 1, c = 1 then 8τ(n) = 108 > 8p1p2 which has no solutions for p1, p2.

Finally, take the case where n is the product of at least 4 distinct primes. But then n ≥ 2 ·3 ·5 ·7 = 210and as 2 · 3 · 5 · 11 > 256, it suffices to check only the case of 210. But 210 clearly fails, so it followsthat 44, 56, 72 are the only good numbers so we are done.

32. [16] How many equilateral hexagons of side length√

13 have one vertex at (0, 0) and the other fivevertices at lattice points?

(A lattice point is a point whose Cartesian coordinates are both integers. A hexagon may be concavebut not self-intersecting.)

Proposed by: Casey Fu

Answer: 216

We perform casework on the point three vertices away from (0, 0). By inspection, that point can be(±8,±3), (±7,±2), (±4,±3), (±3,±2), (±2,±1) or their reflections across the line y = x. The casesare as follows:

If the third vertex is at any of (±8,±3) or (±3,±8), then there are 7 possible hexagons. There are 8points of this form, contributing 56 hexagons.

If the third vertex is at any of (±7,±2) or (±2,±7), there are 6 possible hexagons, contributing 48hexagons.

If the third vertex is at any of (±4,±3) or (±3,±4), there are again 6 possible hexagons, contributing48 more hexagons.

If the third vertex is at any of (±3,±2) or (±2,±3), then there are again 6 possible hexagons, con-tributing 48 more hexagons.

Finally, if the third vertex is at any of (±2,±1), then there are 2 possible hexagons only, contributing16 hexagons.

Adding up, we get our answer of 216 .

33. [20] (Lucas Numbers) The Lucas numbers are defined by L0 = 2, L1 = 1, and Ln+2 = Ln+1 + Lnfor every n ≥ 0. There are N integers 1 ≤ n ≤ 2016 such that Ln contains the digit 1. Estimate N .

An estimate of E earns b20− 2|N − E|c or 0 points, whichever is greater.

Proposed by: Evan Chen

Answer: 1984

lucas_ones n = length . filter (elem ’1’) $ take (n + 1) lucas_strs

where

lucas = 2 : 1 : zipWith (+) lucas (tail lucas)

lucas_strs = map show lucas

main = putStrLn . show $ lucas_ones 2016

34. [20] (Caos) A cao [sic] has 6 legs, 3 on each side. A walking pattern for the cao is defined as anordered sequence of raising and lowering each of the legs exactly once (altogether 12 actions), startingand ending with all legs on the ground. The pattern is safe if at any point, he has at least 3 legs onthe ground and not all three legs are on the same side. Estimate N , the number of safe patterns.

An estimate of E > 0 earns⌊20 min(N/E,E/N)4

⌋points.

Proposed by:

Answer: 1416528

# 1 = on ground, 0 = raised, 2 = back on ground

cache = {}

def pangzi(legs):

if legs == (2,2,2,2,2,2): return 1

elif legs.count(0) > 3: return 0

elif legs[0] + legs[1] + legs[2] == 0: return 0

elif legs[3] + legs[4] + legs[5] == 0: return 0

elif cache.has_key(legs): return cache[legs]

cache[legs] = 0

for i in xrange(6): # raise a leg

if legs[i] == 1:

new = list(legs)

new[i] = 0

cache[legs] += pangzi(tuple(new))

elif legs[i] == 0: # lower a leg

new = list(legs)

new[i] = 2

cache[legs] += pangzi(tuple(new))

return cache[legs]

print pangzi((1,1,1,1,1,1))

35. [20] (Maximal Determinant) In a 17 × 17 matrix M , all entries are ±1. The maximum possiblevalue of |detM | is N . Estimate N .

An estimate of E > 0 earns⌊20 min(N/E,E/N)2

⌋points.

Proposed by: Evan Chen

Answer: 327680 · 216

This is Hadamard’s maximal determinant problem. There’s an upper bound of n12n which empirically

seems to give reasonably good estimates, but in fact this is open for general n.

36. [20] (Self-Isogonal Cubics) Let ABC be a triangle with AB = 2, AC = 3, BC = 4. The isogonalconjugate of a point P , denoted P ∗, is the point obtained by intersecting the reflection of lines PA,PB, PC across the angle bisectors of ∠A, ∠B, and ∠C, respectively.

Given a point Q, let K(Q) denote the unique cubic plane curve which passes through all points P suchthat line PP ∗ contains Q. Consider:

(a) the M’Cay cubic K(O), where O is the circumcenter of 4ABC,

(b) the Thomson cubic K(G), where G is the centroid of 4ABC,

(c) the Napoleon-Feurerbach cubic K(N), where N is the nine-point center of 4ABC,

(d) the Darboux cubic K(L), where L is the de Longchamps point (the reflection of the orthocenteracross point O),

(e) the Neuberg cubic K(X30), where X30 is the point at infinity along line OG,

(f) the nine-point circle of 4ABC,

(g) the incircle of 4ABC, and

(h) the circumcircle of 4ABC.

Estimate N , the number of points lying on at least two of these eight curves. An estimate of E earns⌊20 · 2−|N−E|/6

⌋points.

Proposed by: Evan Chen

Answer: 49

The first main insight is that all the cubics pass through the points A, B, C, H (orthocenter), O,and the incenter and three excenters. Since two cubics intersect in at most nine points, this is all theintersections of a cubic with a cubic.

On the other hand, it is easy to see that among intersections of circles with circles, there are exactly3 points; the incircle is tangent to the nine-point circle at the Feurerbach point while being containedcompletely in the circumcircle; on the other hand for this obtuse triangle the nine-point circle and thecircumcircle intersect exactly twice.

All computations up until now are exact, so it remains to estimate:

• Intersection of the circumcircle with cubics. Each cubic intersects the circumcircle at an evennumber of points, and moreover we already know that A, B, C are among these, so the numberof additional intersections contributed is either 1 or 3; it is the former only for the Neuberg cubicwhich has a “loop”. Hence the actual answer in this case is 1+3+3+3+3 = 13 (but an estimateof 3 · 5 = 15 is very reasonable).

• Intersection of the incircle with cubics. Since ∠A is large the incircle is small, but on the otherhand we know I lies on each cubic. Hence it’s very likely that each cubic intersects the incircletwice (once “coming in” and once “coming out”). This is the case, giving 2 · 5 = 10 new points.

• Intersection of the nine-point with cubics. We guess this is close to the 10 points of the incircle,as we know the nine-point circle and the incircle are tangent to each other. In fact, the exactcount is 14 points; just two additional branches appear.

In total, N = 9 + 3 + 13 + 10 + 14 = 49.

B C

A

O

I

H

N

Fe

IA

IB

IC

HMMT November 2016November 12, 2016

General

1. If a and b satisfy the equations a + 1b = 4 and 1

a + b = 1615 , determine the product of all possible values

of ab.

2. I have five different pairs of socks. Every day for five days, I pick two socks at random withoutreplacement to wear for the day. Find the probability that I wear matching socks on both the thirdday and the fifth day.

3. Let V be a rectangular prism with integer side lengths. The largest face has area 240 and the smallestface has area 48. A third face has area x, where x is not equal to 48 or 240. What is the sum of allpossible values of x?

4. A rectangular pool table has vertices at (0, 0) (12, 0) (0, 10), and (12, 10). There are pockets only inthe four corners. A ball is hit from (0, 0) along the line y = x and bounces off several walls beforeeventually entering a pocket. Find the number of walls that the ball bounces off of before entering apocket.

5. Let the sequence {ai}∞i=0 be defined by a0 = 12 and an = 1 + (an∈1 − 1)2. Find the product

∞∏i=0

ai = a0a1a2 . . .

6. The numbers 1, 2 · · · 11 are arranged in a line from left to right in a random order. It is observed thatthe middle number is larger than exactly one number to its left. Find the probability that it is largerthan exactly one number to its right.

7. Let ABC be a triangle with AB = 13, BC = 14, CA = 15. The altitude from A intersects BC at D.Let ω1 and ω2 be the incircles of ABD and ACD, and let the common external tangent of ω1 and ω2

(other than BC) intersect AD at E. Compute the length of AE.

8. Let S = {1, 2, . . . 2016}, and let f be a randomly chosen bijection from S to itself. Let n be the smallestpositive integer such that f (n)(1) = 1, where f (i)(x) = f(f (i∈1)(x)). What is the expected value of n?

9. Let the sequence ai be defined as ai+1 = 2ai . Find the number of integers 1 ≤ n ≤ 1000 such that ifa0 = n, then 100 divides a1000 − a1.

10. Quadrilateral ABCD satisfies AB = 8, BC = 5, CD = 17, DA = 10. Let E be the intersection of ACand BD. Suppose BE : ED = 1 : 2. Find the area of ABCD.

HMMT November 2016November 12, 2016

General

1. If a and b satisfy the equations a+ 1b = 4 and 1

a + b = 1615 , determine the product of all possible values

of ab.

Proposed by: Eshaan Nichani

Answer: 1

We multiply a + 1b = 4 and 1

a + b = 1615 to get 2 + ab + 1

ab = 6415 ⇐⇒ (ab)2 − 34

15 (ab) + 1 = 0. Since3414

)2−4 ·1 ·1 > 0, there are two roots ab, so the product of both possible values of ab is 1 by Vieta’s.

2. I have five different pairs of socks. Every day for five days, I pick two socks at random withoutreplacement to wear for the day. Find the probability that I wear matching socks on both the thirdday and the fifth day.

Proposed by: Kevin Yang

Answer: 163

I get a matching pair on the third day with probability 19 because there is a 1

9 probability of thesecond sock matching the first. Given that I already removed a matching pair of the third day, I get a

matching pair on the fifth day with probability 17 . We multiply these probabilities to get

1

63.

3. Let V be a rectangular prism with integer side lengths. The largest face has area 240 and the smallestface has area 48. A third face has area x, where x is not equal to 48 or 240. What is the sum of allpossible values of x?

Proposed by: Eshaan Nichani

Answer: 260

Let the length, width, and height of the prism be s1, s2, s3. WIthout loss of generality, assume thats1 ≤ s2 ≤ s3. Then, we have that s1s2 = 48 and s2s3 = 240. Noting that s1 ≤ s2, we must have(s1, s2) = (1, 48), (2, 24), (3, 16), (4, 12), (6, 8). We must also have s2s3 = 240 and s2 ≤ s3, and theonly possibilities for (s1, s2) that yield integral s3 that satisfy these conditions are (4, 12), which givess3 = 20, and (6, 8), which gives s3 = 30. Thus, the only valid (s1, s2, s3) are (4, 12, 20) and (6, 8, 30).It follows that the only possible areas of the third face are 4(20) = 80 and 6(30) = 180, so the desired

answer is 80 + 180 = 260 .

4. A rectangular pool table has vertices at (0, 0) (12, 0) (0, 10), and (12, 10). There are pockets only inthe four corners. A ball is hit from (0, 0) along the line y = x and bounces off several walls beforeeventually entering a pocket. Find the number of walls that the ball bounces off of before entering apocket.

Proposed by: Allen Liu

Answer: 9

Consider the tiling of the plane with the 12 × 10 rectangle to form a grid. Then the reflection of theball off a wall is equivalent to traveling along the straight line y = x into another 12 × 10 rectangle.Hence we want to find the number of walls of the grid that the line y = x hits before it reaches thefirst corner it hits, (60, 60).

The line y = x hits each of the horizontal lines y = 10, 20, 30, 40, 50 and each of the vertical linesx = 12, 24, 36, 48. This gives a total of 9 walls hit before entering a pocket.

5. Let the sequence {ai}∞i=0 be defined by a0 = 12 and an = 1 + (an∈1 − 1)2. Find the product

∞∏i=0

ai = a0a1a2 . . .

Proposed by: Henrik Boecken

Answer: 23

Let {bi}∞i=0 be defined by bn = an − 1 and note that bn = b2n∈1. The infinite product is then

(1 + b0)(1 + b20)(1 + b40) . . . (1 + b2k

0 ) . . .

By the polynomial identity

(1 + x)(1 + x2)(1 + x4) . . . (1 + x2k

) · · · = 1 + x+ x2 + x3 + · · · = 1

1− x

Our desired product is then simply

1

1− (a0 − 1)=

2

3

6. The numbers 1, 2 · · · 11 are arranged in a line from left to right in a random order. It is observed thatthe middle number is larger than exactly one number to its left. Find the probability that it is largerthan exactly one number to its right.

Proposed by: Allen Liu

Answer: 1033

Suppose the middle number is k. Then there are k − 1 ways to pick the number smaller than k toits left and

11∈k

4

)ways to pick the 4 numbers larger than k to its right. Hence there is a total of∑7

k=2(k − 1) ·11∈k

4

)ways for there to be exactly one number smaller than k to its left. We calculate

this total:

7∑k=2

(k − 1) ·(

11− k4

)=

9∑j=4

j∑i=4

(i

4

)

=

9∑j=4

(j + 1

5

)

=

(11

6

).

The only way k can be larger than exactly one number to its right is if k = 3. Then the probability of

this happening is2·(8

4)(11

6 )=

10

33.

7. Let ABC be a triangle with AB = 13, BC = 14, CA = 15. The altitude from A intersects BC at D.Let ω1 and ω2 be the incircles of ABD and ACD, and let the common external tangent of ω1 and ω2

(other than BC) intersect AD at E. Compute the length of AE.

Proposed by: Eshaan Nichani

Answer: 7

Solution 1:

Let I1, I2 be the centers of ω1, ω2, respectively, X1, X2 be the tangency points of ω1, ω2 with BC,respectively, and Y1, Y2 be the tangency points of ω1, ω2 with AD, respectively. Let the two commonexternal tangents of ω1, ω2 meet at P . Note that line I1I2 also passes through P .

By Heron’s formula, the area of triangle ABC is 84. Thus, 12AD ·BC = 84, and so AD = 12. By the

Pythagorean Theorem on right triangles ABD and ACD, BD = 5 and CD = 9.

The inradius of ABD, rABD, is [ABD]sABD

, where [ABD] is the area of ABD and sABD is its semiperimeter.ABD is a 5-12-13 right triangle, so [ABD] = 30 and sABD = 15. Thus, rABD = 2. Similarly, we get

that ACD’s inradius is rACD = 3. I1Y1DX1 is a square, so X1D = I1X1 = rABD = 2, and similarlyX2D = 3. X1 and X2 are on opposite sides of D, so X1X2 = X1D +X2D = 5.

Since P lies on lines I1I2, X1X2, and I1X1, I2X2 are parallel, triangles PX1I1 and PX2I2 are similar.Thus, X1I1

X2I2= 2

3 = PX1

PX2= PX1

PX1+X1X2= PX1

PX1+5 . Solving gives PX1 = 10. Letting ∠I1PX1 = θ, since

I1X1P is a right angle, we have tan θ = X1I1X1P1

= 15 . D and E lie on different common external tangents,

so PI1 bisects ∠EPD, and thus ∠EPD = 2θ. Thus, tan∠EPD = tan 2θ = 2 tan θ1∈tan2 θ = 5

12 .

ED is perpendicular to BC, so triangle EDP is a right triangle with right angle at D. Thus, 512 =

tan∠EPD = EDPD . PD = PX1 +X1D = 12, so ED = 5. AD = 12, so it follows that AE = 7 .

Solution 2:

Lemma: Let Ω1,Ω2 be two non-intersecting circles. Let `A, `B be their common external tangents, and`C be one of their common internal tangents. Ω1 intersects `A, `B at points A1, B1, respectively, andΩ2 intersects `A, `B at points A2, B2. If `C intersects `A at X, and `B at Y , then XY = A1A2 = B1B2.

Proof: Let `C intersect Ω1,Ω2 at C1, C2, respectively. Then, examining the tangents to Ω1,Ω2 fromX,Y , we have A1X = C1X,A2X = C2X,B1Y = C1Y,B2Y = C2Y . Thus, 2A1A2 = 2B1B2 =A1A2 +B1B2 = A1X +A2X +B1Y +B2Y = C1X +C2X +C1Y +C2Y = 2XY , and the conclusionfollows.

Using the notation from above, we apply the lemma to circles ω1, ω2, and conclude that ED = X1X2.Then, we proceed as above to compute X1X2 = 5 = ED. Thus, AD = 7 .

8. Let S = {1, 2, . . . 2016}, and let f be a randomly chosen bijection from S to itself. Let n be the smallestpositive integer such that f (n)(1) = 1, where f (i)(x) = f(f (i∈1)(x)). What is the expected value of n?

Proposed by: Eshaan Nichani

Answer: 20172

Say that n = k. Then 1, f(1), f2(1), . . . , f (k∈1)(1) are all distinct, which means there are 2015 ·2014 · · · (2016− k + 1) ways to assign these values. There is 1 possible value of fk(1), and (2016− k)!ways to assign the image of the 2016− k remaining values. Thus the probability that n = k is 1

2016 .

Therefore the expected value of n is 12016 (1 + 2 + · · ·+ 2016) = 2017

2

9. Let the sequence ai be defined as ai+1 = 2ai . Find the number of integers 1 ≤ n ≤ 1000 such that ifa0 = n, then 100 divides a1000 − a1.

Proposed by: Allen Liu

Answer: 50

We claim that a1000 is constant mod 100.

a997 is divisible by 2. This means that a998 is divisible by 4. Thus a999 is constant mod 5. Since itis also divisible by 4, it is constant mod 20. Thus a1000 is constant mod 25, since φ(25) = 20. Sincea1000 is also divisible by 4, it is constant mod 100.

We know that a1000 is divisible by 4, and let it be congruent to k mod 25.

Then 2n is divisible by 4 (n ≥ 2) and 2n ≡ k mod 25 We can also show that 2 is a primitive root mod 25,

so there is one unique value of n mod 20. It suffices to show this value isn’t 1. But 220mod4 ≡ 216mod20

mod 25, so n ≡ 16 mod 20. Thus there are 1000/20 = 50 values of n.

10. Quadrilateral ABCD satisfies AB = 8, BC = 5, CD = 17, DA = 10. Let E be the intersection of ACand BD. Suppose BE : ED = 1 : 2. Find the area of ABCD.

Proposed by: Brice Huang

Answer: 60

Since BE : ED = 1 : 2, we have [ABC] : [ACD] = 1 : 2.

Suppose we cut off triangle ACD, reflect it across the perpendicular bisector of AC, and re-attach itas triangle A′C ′D′ (so A′ = C,C ′ = A).

Triangles ABC and C ′A′D′ have vertex A = C ′ and bases BC and A′D′ . Their areas and bases areboth in the ratio 1 : 2. Thus in fact BC and A′D′ are collinear.

Hence the union of ABC and C ′A′D′ is the 8− 15− 17 triangle ABD′, which has area 12 · 8 · 15 = 60.

HMMT November 2016November 12, 2016

Theme Round

1. DeAndre Jordan shoots free throws that are worth 1 point each. He makes 40% of his shots. If hetakes two shots find the probability that he scores at least 1 point.

2. Point P1 is located 600 miles West of point P2. At 7:00 AM a car departs from P1 and drives East ata speed of 50 miles per hour. At 8:00 AM another car departs from P2 and drives West at a constantspeed of x miles per hour. If the cars meet each other exactly halfway between P1 and P2, what is thevalue of x?

3. The three points A, B, C form a triangle. AB = 4, BC = 5 AC = 6. Let the angle bisector of ∠Aintersect side BC at D. Let the foot of the perpendicular from B to the angle bisector of ∠A be E.Let the line through E parallel to AC meet BC at F . Compute DF .

4. A positive integer is written on each corner of a square such that numbers on opposite vertices arerelatively prime while numbers on adjacent vertices are not relatively prime. What is the smallestpossible value of the sum of these 4 numbers?

5. Steph Curry is playing the following game and he wins if he has exactly 5 points at some time. Flip afair coin. If heads, shoot a 3-point shot which is worth 3 points. If tails, shoot a free throw which isworth 1 point. He makes 1

2 of his 3-point shots and all of his free throws. Find the probability he willwin the game. (Note he keeps flipping the coin until he has exactly 5 or goes over 5 points)

6. Let P1, P2, . . . , P6 be points in the complex plane, which are also roots of the equation x6+6x3�216 = 0.Given that P1P2P3P4P5P6 is a convex hexagon, determine the area of this hexagon.

7. Seven lattice points form a convex heptagon with all sides having distinct lengths. Find the minimumpossible value of the sum of the squares of the sides of the heptagon.

8. Let P1P2 . . . P8 be a convex octagon. An integer i is chosen uniformly at random from 1 to 7, inclusive.For each vertex of the octagon, the line between that vertex and the vertex i vertices to the right ispainted red. What is the expected number times two red lines intersect at a point that is not one ofthe vertices, given that no three diagonals are concurrent?

9. The vertices of a regular nonagon are colored such that 1) adjacent vertices are different colors and 2)if 3 vertices form an equilateral triangle, they are all different colors.

Let m be the minimum number of colors needed for a valid coloring, and n be the total number ofcolorings using m colors. Determine mn. (Assume each vertex is distinguishable.)

10. We have 10 points on a line A1, A2 · · ·A10 in that order. Initially there are n chips on point A1. Nowwe are allowed to perform two types of moves. Take two chips on Ai, remove them and place onechip on Ai+1, or take two chips on Ai+1, remove them, and place a chip on Ai+2 and Ai . Find theminimum possible value of n such that it is possible to get a chip on A10 through a sequence of moves.

HMMT November 2016November 12, 2016

Theme Round

1. DeAndre Jordan shoots free throws that are worth 1 point each. He makes 40% of his shots. If hetakes two shots find the probability that he scores at least 1 point.

Proposed by: Allen Liu

Answer: 1625

We want to find the probability of making at least one shot. The probability he makes no shots is35

)2, so the probability of making at least one is 1−

35

)2=

16

25.

2. Point P1 is located 600 miles West of point P2. At 7:00 AM a car departs from P1 and drives East ata speed of 50 miles per hour. At 8:00 AM another car departs from P2 and drives West at a constantspeed of x miles per hour. If the cars meet each other exactly halfway between P1 and P2, what is thevalue of x?

Proposed by: Eshaan Nichani

Answer: 60

Each car meets having traveled 300 miles. Therefore the first car traveled for 300/50 = 6 hours, and

so the second car traveled for 5 hours. The second car must have traveled 300/5 = 60 miles per hour.

3. The three points A, B, C form a triangle. AB = 4, BC = 5 AC = 6. Let the angle bisector of ∠Aintersect side BC at D. Let the foot of the perpendicular from B to the angle bisector of ∠A be E.Let the line through E parallel to AC meet BC at F . Compute DF .

Proposed by: Allen Liu

Answer: 1/2

Since AD bisects ∠A, by the angle bisector theorem ABBD = AC

CD , so BD = 2 and CD = 3. Extend BEto hit AC at X. Since AE is the perpendicular bisector of BX, AX = 4. Since B,E,X are collinear,applying Menelaus’ Theorem to the triangle ADC, we have

AE

ED· DBBC· CXXA

= 1

This implies that AEED = 5, and since EF ‖ AC, DF

DC = DEDA , so DF = DC

6 = 12 .

4. A positive integer is written on each corner of a square such that numbers on opposite vertices arerelatively prime while numbers on adjacent vertices are not relatively prime. What is the smallestpossible value of the sum of these 4 numbers?

Proposed by: Eshaan Nichani

Answer: 60

Two opposite vertices are relatively prime, but they both share a factor with their common neighbor.So that common neighbor must have two prime factors. So each of the 4 numbers has two primefactors, which are not shared with the opposite vertex. Moreover, it suffices to choose the vertices tobe the numbers ab, bc, cd, da for some prime numbers a, b, c, d. It’s clear that we should choose them tobe the smallest primes 2, 3, 5, 7 in some order. The order that minimizes the sum of all of the numbersgives 14, 10, 15, 21 for a sum of 60.

5. Steph Curry is playing the following game and he wins if he has exactly 5 points at some time. Flip afair coin. If heads, shoot a 3-point shot which is worth 3 points. If tails, shoot a free throw which isworth 1 point. He makes 1

2 of his 3-point shots and all of his free throws. Find the probability he willwin the game. (Note he keeps flipping the coin until he has exactly 5 or goes over 5 points)

Proposed by: Allen Liu

Answer: 140243

If he misses the shot, then the state of the game is the same as before he flipped the coin. Since theprobability of making a free throw is 1

2 and the probability of making a 3-point shot is 14 . Therefore,

given that he earns some point, the probability it is a 3-point shot is1

3. The possible ways of earning

points are 11111,113, 131, and 311, which have probabilities 32243 , 4

27 , 427 , and 4

27 , which sum to140

243.

6. Let P1, P2, . . . , P6 be points in the complex plane, which are also roots of the equation x6+6x3−216 = 0.Given that P1P2P3P4P5P6 is a convex hexagon, determine the area of this hexagon.

Proposed by: Eshaan Nichani

Answer: 9√

3

Factor x6 + 6x3 − 216 = (x3 − 12)(x3 + 18). This gives us 6 points equally spaced in terms of theirangles from the origin, alternating in magnitude between 3

√12 and 3

√18. This means our hexagon is

composed of 6 triangles, each with sides of length 3√

12 and 3√

18 and with a 60 degree angle in between

them. This yields the area of each triangle as 3√3

2 , so the total area of the hexagon is 9√

3.

7. Seven lattice points form a convex heptagon with all sides having distinct lengths. Find the minimumpossible value of the sum of the squares of the sides of the heptagon.

Proposed by: Allen Liu

Answer: 42

Consider the vectors corresponding to the sides of the heptagon, and call them [xi, yi] for i between1 and 7. Then since

∑xi =

∑yi = 0, and a2 has the same parity as a, we have that

∑x2i + y2i

must be an even number. A side length of a lattice valued polygon must be expressible as√a2 + b2,

so the smallest possible values are√

1,√

2,√

4,√

5,√

8,√

9,√

10. However, using the seven smallestlengths violates the parity constraint. If we try

√13, we indeed can get a heptagon with lengths√

1,√

2,√

4,√

5,√

8,√

9,√

13. One example is the heptagon (0, 0), (3, 0), (5, 1), (6, 2), (3, 4), (2, 4), (0, 2),

and its sum of squares of side lengths is 1 + 2 + 4 + 5 + 8 + 9 + 13 = 42 .

8. Let P1P2 . . . P8 be a convex octagon. An integer i is chosen uniformly at random from 1 to 7, inclusive.For each vertex of the octagon, the line between that vertex and the vertex i vertices to the right ispainted red. What is the expected number times two red lines intersect at a point that is not one ofthe vertices, given that no three diagonals are concurrent?

Proposed by: Eshaan Nichani

Answer: 547

If i = 1 or i = 7, there are 0 intersections. If i = 2 or i = 6 there are 8. If i = 3 or i = 5 there are16 intersections. When i = 4 there are 6 intersections (since the only lines drawn are the four longdiagonals).

Thus the final answer is 8+16+6+16+87 =

54

7

9. The vertices of a regular nonagon are colored such that 1) adjacent vertices are different colors and 2)if 3 vertices form an equilateral triangle, they are all different colors.

Let m be the minimum number of colors needed for a valid coloring, and n be the total number ofcolorings using m colors. Determine mn. (Assume each vertex is distinguishable.)

Proposed by: Eshaan Nichani

Answer: 54

It’s clear that m is more than 2 since it’s impossible to alternate the color of the vertices withouthaving two of the same color adjacent (since the graph is not bipartite). However, it’s possible to use3 colors. Number the vertices 1 through 9 in order and let the colors be A,B,C. Coloring the verticesin the order BCBCACABA gives a configuration that works, so m is 3. To determine n, we canpartition the nonagon into three equilateral triangles. Vertices 1, 4, 7 must be different colors, whichwe can choose in 3! = 6 ways. Suppose WLOG that they’re A,B,C respectively. Then we look atvertices 2, 5, 8. Vertex 2 can be colored B or C. If 2 is B, then vertex 8 must be A, and vertex 5 mustbe C. In this case there are two ways to color the remaining vertices 3, 6, 9. Otherwise, if vertex 2 isC, then vertex 5 must be A, and vertex 8 must be B. This gives us only 1 possible coloring for theremaining three vertices. So n is 6(2 + 1) = 18. So our answer is mn = 54.

10. We have 10 points on a line A1, A2 · · ·A10 in that order. Initially there are n chips on point A1. Nowwe are allowed to perform two types of moves. Take two chips on Ai, remove them and place onechip on Ai+1, or take two chips on Ai+1, remove them, and place a chip on Ai+2 and Ai . Find theminimum possible value of n such that it is possible to get a chip on A10 through a sequence of moves.

Proposed by: Allen Liu

Answer: 46

We claim that n = 46 is the minimum possible value of n. As having extra chips cannot hurt, it isalways better to perform the second operation than the first operation, except on point A1. Assign thevalue of a chip on point Ai to be i. Then the total value of the chips initially is n. Furthermore, bothtypes of operations keep the total values of the chips the same, as 2 · 1 = 2 and i+ (i+ 2) = 2 · (i+ 1).

When n = 46, we claim that any sequence of these moves will eventually lead to a chip reaching A10.If, for the sake of contradiction, that there was a way to get stuck with no chip having reached A10,then there could only be chips on A1 through A9, and furthermore at most one chip on each. The totalvalue of these chips is at most 45, which is less than the original value of chips 46.

However, if n ≤ 45, we claim that it is impossible to get one chip to A10. To get a chip to A10,an operation must have been used on each of A1 through A9 at least once. Consider the last timethe operation was used on Ak for 2 ≤ k ≤ 9. After this operation, there must be a chip on Ak−1.Additionally, since no chip is ever moved past Ak again, there is no point to perform any operationson any chips left of Ak, which means that a chip will remain on Ak−1 until the end. Therefore, if thereis a way to get a chip to A10, there must also be a way to get a chip to A10 and also A1 through A8,which means that the original value of the chips must have been already 1 + 2 + · · ·+ 8 + 10 = 46.

HMMT November 2016November 12, 2016

Team

1. [3] Two circles centered at O1 and O2 have radii 2 and 3 and are externally tangent at P . The commonexternal tangent of the two circles intersects the line O1O2 at Q. What is the length of PQ?

2. [3] What is the smallest possible perimeter of a triangle whose side lengths are all squares of distinctpositive integers?

3. [3] Complex number ω satisfies ω5 = 2. Find the sum of all possible values of

ω4 + ω3 + ω2 + ω + 1.

4. [5] Meghal is playing a game with 2016 rounds 1, 2, · · · , 2016. In round n, two rectangular double-sidedmirrors are arranged such that they share a common edge and the angle between the faces is 2π

n+2 .Meghal shoots a laser at these mirrors and her score for the round is the number of points on the twomirrors at which the laser beam touches a mirror. What is the maximum possible score Meghal couldhave after she finishes the game?

5. [5] Allen and Brian are playing a game in which they roll a 6-sided die until one of them wins. Allenwins if two consecutive rolls are equal and at most 3. Brian wins if two consecutive rolls add up to 7and the latter is at most 3. What is the probability that Allen wins?

6. [5] Let ABC be a triangle with AB = 5, BC = 6, and AC = 7. Let its orthocenter be H and the feetof the altitudes from A,B,C to the opposite sides be D,E, F respectively. Let the line DF intersectthe circumcircle of AHF again at X. Find the length of EX.

7. [6] Rachel has two indistinguishable tokens, and places them on the first and second square of a 1× 6grid of squares, She can move the pieces in two ways:

• If a token has free square in front of it, then she can move this token one square to the right

• If the square immediately to the right of a token is occupied by the other token, then shecan“leapfrog” the first token; she moves the first token two squares to the right, over the othertoken, so that it is on the square immediately to the right of the other token.

If a token reaches the 6th square, then it cannot move forward any more, and Rachel must move theother one until it reaches the 5th square. How many different sequences of moves for the tokens canRachel make so that the two tokens end up on the 5th square and the 6th square?

8. [6] Alex has an 20 × 16 grid of lightbulbs, initially all off. He has 36 switches, one for each row andcolumn. Flipping the switch for the ith row will toggle the state of each lightbulb in the ith row (sothat if it were on before, it would be off, and vice versa). Similarly, the switch for the jth column willtoggle the state of each bulb in the jth column. Alex makes some (possibly empty) sequence of switchflips, resulting in some configuration of the lightbulbs and their states. How many distinct possibleconfigurations of lightbulbs can Alex achieve with such a sequence? Two configurations are distinct ifthere exists a lightbulb that is on in one configuration and off in another.

9. [7] A cylinder with radius 15 and height 16 is inscribed in a sphere. Three congruent smaller spheresof radius x are externally tangent to the base of the cylinder, externally tangent to each other, andinternally tangent to the large sphere. What is the value of x?

10. [7] Determine the largest integer n such that there exist monic quadratic polynomials p1(x), p2(x),p3(x) with integer coefficients so that for all integers i ∈ [1, n] there exists some j ∈ [1, 3] and m ∈ Zsuch that pj(m) = i.

HMMT November 2016November 12, 2016

Team

1. [3] Two circles centered at O1 and O2 have radii 2 and 3 and are externally tangent at P . The commonexternal tangent of the two circles intersects the line O1O2 at Q. What is the length of PQ?

Proposed by: Eshaan Nichani

Answer: 12

Let the common external tangent intersect the circles centered at O1, O2 at X, Y respectively. ThenO2QO1Q

= OYOX = 3

2 , so O1O2

O1Q= O2Q→O1Q

O1Q= 1

2 . Since O1O2 = 2 + 3 = 5, O1Q = 10 and hencePQ = O1Q + O1P = 12.

2. [3] What is the smallest possible perimeter of a triangle whose side lengths are all squares of distinctpositive integers?

Proposed by: Eshaan Nichani

Answer: 77

There exist a triangle with side lengths 42, 52, 62, which has perimeter 77. If the sides have lengthsa2, b2, c2 with 0 < a < b < c, then a2 + b2 > c2 by the triangle inequality. Therefore (b − 1)2 + b2 ≥a2 + b2 > c2 ≥ (b + 1)2. Solving this inequality gives b > 4. If b ≥ 6, then a2 + b2 + c2 ≥ 62 + 72 > 77.

If b = 5, then c ≥ 7 is impossible, while c = 6 forces a = 4, which gives a perimeter of 77 .

3. [3] Complex number ω satisfies ω5 = 2. Find the sum of all possible values of

ω4 + ω3 + ω2 + ω + 1.

Proposed by: Henrik Boecken

Answer: 5

The value of ω4 + ω3 + ω2 + ω + 1 =ω5 − 1

ω − 1=

1

ω − 1. The sum of these values is therefore the sum of

1

ω − 1over the five roots ω. Substituting z = ω − 1, we have that (z + 1)5 = 2, so z5 + 5z4 + 10z3 +

10z2 + 5z − 1 = 0. The sum of the reciprocals of the roots of this equation is − 5

−1= 5 by Vieta’s.

4. [5] Meghal is playing a game with 2016 rounds 1, 2, · · · , 2016. In round n, two rectangular double-sidedmirrors are arranged such that they share a common edge and the angle between the faces is 2π

n+2 .Meghal shoots a laser at these mirrors and her score for the round is the number of points on the twomirrors at which the laser beam touches a mirror. What is the maximum possible score Meghal couldhave after she finishes the game?

Proposed by: Rachel Zhang

Answer: 1019088

Let points O,A1, A2 lie in a plane such that ∠A1OA2 = 2πn+2 . We represent the mirrors as line segments

extending between O and A1, and O and A2. Also let points A3, A4, · · · , An+2 lie in the plane suchthat Ai+1 is the reflection of Ai→1 over OAi.

If Meghal shoots a laser along line l such that the first point of contact with a mirror is along OA2, thenext point of contact, if it exists, is the point on OA1 that is a reflection of the intersection of l withOA3. If we continue this logic, we find that the maximum score for round n is equal to the maximumnumber of intersection points between l and OAi for some i. We do casework on whether n is even orodd. If n is even, there are at most n+2

2 spokes such that l can hit OAi, and if n is odd, there are at mostn+32 such spokes. Then we must sum 2+2+3+3+ · · ·+1009+1009 = 1009 ·1010−1−1 = 1019088 .

5. [5] Allen and Brian are playing a game in which they roll a 6-sided die until one of them wins. Allenwins if two consecutive rolls are equal and at most 3. Brian wins if two consecutive rolls add up to 7and the latter is at most 3. What is the probability that Allen wins?

Proposed by: Eshaan Nichani

Answer: 5/12

Note that at any point in the game after the first roll, the probability that Allen wins depends only onthe most recent roll, and not on any rolls before that one. So we may define p as the probability thatAllen wins at any point in the game, given that the last roll was a 1, 2, or 3, and q as the probabilitythat he wins given that the last roll was a 4, 5, or 6.

Suppose at some point, the last roll was r1 ∈ {1, 2, 3}, and the next roll is r2 ∈ {1, 2, 3, 4, 5, 6}. By thedefinition of p, Allen wins with probability p. Furthermore, if r2 = r1, which happens with probability16 , Allen wins. If r2 ∈ {1, 2, 3} but r2 6= r1, which happens with probability 2

6 , neither Allen nor Brianwins, so they continue playing the game, now where the last roll was r2. In this case, Allen wins withprobability p. If r2 ∈ {4, 5, 6}, which happens with probability 3

6 , neither Allen nor Brian wins, so theycontinue playing, now where the last roll was r2. In this case, Allen wins with probability q. Hence,the probability that Allen wins in this case can be expressed as 1

6 + 26p + 3

6q, and thus

p =1

6+

2

6p +

3

6q

By a similar analysis for q, we find that

q =1

6· 0 +

2

6p +

3

6q

Solving, we get p = 12 and q = 1

3 . Allen wins with probability p = 12 if the first roll is 1, 2, or 3, and he

wins with probability q = 13 if the first roll is 4, 5, or 6. We conclude that the overall probability that

he wins the game is 12p + 1

2q =5

12.

6. [5] Let ABC be a triangle with AB = 5, BC = 6, and AC = 7. Let its orthocenter be H and the feetof the altitudes from A,B,C to the opposite sides be D,E, F respectively. Let the line DF intersectthe circumcircle of AHF again at X. Find the length of EX.

Proposed by: Allen Liu

Answer: 19049

Since ∠AFH = ∠AEH = 90◦, E is on the circumcircle of AHF . So ∠XEH = ∠HFD = ∠HBD,which implies that XE ‖ BD. Hence EX

BD = EYY B . Let DF and BE intersect at Y . Note that

∠EDY = 180◦ −∠BDF −∠CDE = 180◦ − 2∠A, and ∠BDY = ∠A. Applying the sine rule to EYDand BYD, we get

EY

Y B=

ED

BD· sin∠EDY

sin∠BDY=

ED

BD· sin 2∠A

sin∠A=

ED

BD· 2 cos∠A

Next, letting x = CD and y = AE, by Pythagoras we have

AB2 − (6− x)2 = AD2 = AC2 − x2

BC2 − (7− y)2 = BE2 = BA2 − y2

Solving, we get x = 5, y = 197 . Drop the perpendicular from E to DC at Z. Then ED cos∠A =

ED cos∠EDZ = DZ. But AD ‖ EZ, so DZ = AEAC ·DC = 95

49 . Therefore

EX =EY

Y B·BD = 2ED cos∠A = 2DZ =

190

49

7. [6] Rachel has two indistinguishable tokens, and places them on the first and second square of a 1× 6grid of squares, She can move the pieces in two ways:

• If a token has free square in front of it, then she can move this token one square to the right

• If the square immediately to the right of a token is occupied by the other token, then shecan“leapfrog” the first token; she moves the first token two squares to the right, over the othertoken, so that it is on the square immediately to the right of the other token.

If a token reaches the 6th square, then it cannot move forward any more, and Rachel must move theother one until it reaches the 5th square. How many different sequences of moves for the tokens canRachel make so that the two tokens end up on the 5th square and the 6th square?

Proposed by: Christopher Shao

Answer: 42

We put a marker on (i, j) when a token is on ith and jth square and i > j. When the token infront/behind moves one step forward to a blank square, move the marker rightward/upward one unitcorrespondingly. When a ”leapfrog” happens, the marker moves from (x− 1, x) to (x, x + 1). We cantranslate this movement into: 1. move the marker upward to (x, x); 2. move the marker rightwardto (x, x + 1). Thus, we set up a lattice path way from (2, 1) to (6, 5) staying under y = x. This is abijection since every intersection of the path way and y = x indicates a ”leapfrog”. According to thedefinition of Catalan Number, the answer is the number of such lattice path ways, which is C5 = 42.

8. [6] Alex has an 20 × 16 grid of lightbulbs, initially all off. He has 36 switches, one for each row andcolumn. Flipping the switch for the ith row will toggle the state of each lightbulb in the ith row (sothat if it were on before, it would be off, and vice versa). Similarly, the switch for the jth column willtoggle the state of each bulb in the jth column. Alex makes some (possibly empty) sequence of switchflips, resulting in some configuration of the lightbulbs and their states. How many distinct possibleconfigurations of lightbulbs can Alex achieve with such a sequence? Two configurations are distinct ifthere exists a lightbulb that is on in one configuration and off in another.

Proposed by: Christopher Shao

Answer: 235

The switch flip operations are commutative, so for any given sequence of switch flips S, we get the sameconfiguration regardless of the order we do them in. We can arrange the switch flips so that all of theflips of the same switch happen consecutively. Furthermore, two consecutive flips of the same switchleave the configuration unchanged, so we can remove them, resulting in a sequence of switch flips S′

that involves flipping a switch for a row or column at most once that achieves the same configurationas S. The order of the flips in S′ also doesn’t matter, so we can treat S′ as a set of switches that areflipped to produce the same configuration as S.

The desired number is then equal to the number of distinct configurations that can be obtained byflipping exactly the switches in some subset S′ of the set of all of the switches. We claim that if S1

and S2 are distinct sets of switches that result in the same configuration of lights, then S1 and S2

are complements. Indeed, without loss of generality, suppose that the first row’s switch is in S1 andthat it isn’t in S2. In order to have the same configuration of lights in the first row, we must havethat every column switch is in S1 if and only if it isn’t in S2. Applying the same argument to thefirst column yields that every row switch is in S1 if and only if it isn’t in S2, and the claim follows.Thus, for every set of switches, there is exactly one other set that attains the same configuration as it,namely its complement. There are 2m+n sets of switches possible, and so the total number of possibleconfigurations is 2m+n/2 = 2m+n→1.

9. [7] A cylinder with radius 15 and height 16 is inscribed in a sphere. Three congruent smaller spheresof radius x are externally tangent to the base of the cylinder, externally tangent to each other, andinternally tangent to the large sphere. What is the value of x?

Proposed by: Eshaan Nichani

Answer: 15√37→754

Let O be the center of the large sphere, and let O1, O2, O3 be the centers of the small spheres. ConsiderG, the center of equilateral 4O1O2O3. Then if the radii of the small spheres are r, we have thatOG = 8+r and O1O2 = O2O3 = O3O1 = 2r, implying that O1G = 2r√

3. Then OO1 =

√OG2 + OO2

1 =√(8 + r)2 + 4

3r2. Now draw the array OO1, and suppose it intersects the large sphere again at P .

Then P is the point of tangency between the large sphere and the small sphere with center O1, so

OP =√

152 + 82 = 17 = OO1 + O1P =√

(8 + r)2 + 43r

2 + r. We rearrange this to be

17− r =

√(8 + r)2 +

4

3r2

⇐⇒ 289− 34r + r2 =7

3r2 + 16r + 64

⇐⇒ 4

3r2 + 50r − 225 = 0

=⇒ r =−50±

√502 + 4 · 43 · 225

2 · 43

=15√

37− 75

4.

10. [7] Determine the largest integer n such that there exist monic quadratic polynomials p1(x), p2(x),p3(x) with integer coefficients so that for all integers i ∈ [1, n] there exists some j ∈ [1, 3] and m ∈ Zsuch that pj(m) = i.

Proposed by: Eshaan Nichani

Answer: 9

The construction for n = 9 can be achieved with the polynomials x2 + x + 1, x2 + x + 2, and x2 + 5.

First we consider what kinds of polynomials we can have. Let p(x) = (x + h)2 + k. h is either aninteger or half an integer. Let k = 0. If h is an integer then p(x) hits the perfect squares 0, 1, 4, 9, etc.If h is half an integer, then let k = 1/4. Then p(x) hits the product of two consecutive integers, i.e. 0,2, 6, 12, etc.

Assume there is a construction for n = 10. In both of the cases above, the most a polynomial can hitout of 10 is 4, in the 0, 1, 4, 9 case. Thus p1 must hit 1, 2, 5, 10, and p2 and p3 hit 3 integers each, outof 3, 4, 6, 7, 8, 9. The only ways we can hit 3 out of 7 consecutive integers is with the sequences 0, 2, 6or 0, 1, 4. The only way a 0, 2, 6 works is if it hits 3, 5, and 9, which doesn’t work since 5 was hit by p2.Otherwise, p2 is 0, 1, 4, which doesn’t work as p2 hits 3, 4, and 7, and p3 must hit 6, 8, and 9, whichis impossible. Thus no construction for n = 10 exists.

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HMMT November 2016, November 12, 2016 — GUTS ROUND

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1. [5] If five fair coins are flipped simultaneously, what is the probability that at least three of them showheads?

2. [5] How many perfect squares divide 1010?

3. [5] Evaluate 2016!2

2015!2017! . Here n! denotes 1× 2× · · · × n.

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HMMT November 2016, November 12, 2016 — GUTS ROUND

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4. [6] A square can be divided into four congruent figures as shown:

For how many n with 1 ≤ n ≤ 100 can a unit square be divided into n congruent figures?

5. [6] If x+ 2y − 3z = 7 and 2x− y + 2z = 6, determine 8x+ y.

6. [6] Let ABCD be a rectangle, and let E and F be points on segment AB such that AE = EF = FB.If CE intersects the line AD at P , and PF intersects BC at Q, determine the ratio of BQ to CQ.

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HMMT November 2016, November 12, 2016 — GUTS ROUND

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7. [7] What is the minimum value of the product

6∏i=1

ai − ai+1

ai+2 − ai+3

given that (a1, a2, a3, a4, a5, a6) is a permutation of (1, 2, 3, 4, 5, 6)? (note a7 = a1, a8 = a2 · · · )

8. [7] Danielle picks a positive integer 1 ≤ n ≤ 2016 uniformly at random. What is the probability thatgcd(n, 2015) = 1?

9. [7] How many 3-element subsets of the set {1, 2, 3, . . . , 19} have sum of elements divisible by 4?

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HMMT November 2016, November 12, 2016 — GUTS ROUND

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10. [8] Michael is playing basketball. He makes 10% of his shots, and gets the ball back after 90% of hismissed shots. If he does not get the ball back he stops playing. What is the probability that Michaeleventually makes a shot?

11. [8] How many subsets S of the set {1, 2, . . . , 10} satisfy the property that, for all i ∈ [1, 9], either i ori+ 1 (or both) is in S?

12. [8] A positive integer ABC, where A, B, C are digits, satisfies

ABC = BC −A

Find ABC.

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HMMT November 2016, November 12, 2016 — GUTS ROUND

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13. [9] How many functions f : {0, 1}3 → {0, 1} satisfy the property that, for all ordered triples (a1, a2, a3)and (b1, b2, b3) such that ai ≥ bi for all i, f(a1, a2, a3) ≥ f(b1, b2, b3)?

14. [9] The very hungry caterpillar lives on the number line. For each non-zero integer i, a fruit sits on thepoint with coordinate i. The caterpillar moves back and forth; whenever he reaches a point with food,he eats the food, increasing his weight by one pound, and turns around. The caterpillar moves at aspeed of 2−w units per day, where w is his weight. If the caterpillar starts off at the origin, weighingzero pounds, and initially moves in the positive x direction, after how many days will he weigh 10pounds?

15. [9] Let ABCD be an isosceles trapezoid with parallel bases AB = 1 and CD = 2 and height 1. Findthe area of the region containing all points inside ABCD whose projections onto the four sides of thetrapezoid lie on the segments formed by AB,BC,CD and DA.

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HMMT November 2016, November 12, 2016 — GUTS ROUND

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16. [10] Create a cube C1 with edge length 1. Take the centers of the faces and connect them to form anoctahedron O1. Take the centers of the octahedron’s faces and connect them to form a new cube C2.Continue this process infinitely. Find the sum of all the surface areas of the cubes and octahedrons.

17. [10] Let p(x) = x2 − x+ 1. Let α be a root of p(p(p(p(x))). Find the value of

(p(α)− 1)p(α)p(p(α))p(p(p(α))

18. [10] An 8 by 8 grid of numbers obeys the following pattern:

1) The first row and first column consist of all 1s.

2) The entry in the ith row and jth column equals the sum of the numbers in the (i − 1) by (j − 1)sub-grid with row less than i and column less than j.

What is the number in the 8th row and 8th column?

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HMMT November 2016, November 12, 2016 — GUTS ROUND

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19. [11] Let S be the set of all positive integers whose prime factorizations only contain powers of theprimes 2 and 2017 (1, powers of 2, and powers of 2017 are thus contained in S). Compute

∑s∈S

1s .

20. [11] Let V be the volume enclosed by the graph

x2016 + y2016 + z2 = 2016

Find V rounded to the nearest multiple of ten.

21. [11] Zlatan has 2017 socks of various colours. He wants to proudly display one sock of each of thecolours, and he counts that there are N ways to select socks from his collection for display. Given thisinformation, what is the maximum value of N?

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22. [12] Let the function f : Z→ Z take only integer inputs and have integer outputs. For any integers xand y, f satisfies

f(x) + f(y) = f(x+ 1) + f(y − 1)

If f(2016) = 6102 and f(6102) = 2016, what is f(1)?

23. [12] Let d be a randomly chosen divisor of 2016. Find the expected value of

d2

d2 + 2016

24. [12] Consider an infinite grid of equilateral triangles. Each edge (that is, each side of a small triangle)is colored one of N colors. The coloring is done in such a way that any path between any two non-adjecent vertices consists of edges with at least two different colors. What is the smallest possible valueof N?

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HMMT November 2016, November 12, 2016 — GUTS ROUND

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25. [13] Chris and Paul each rent a different room of a hotel from rooms 1 − 60. However, the hotelmanager mistakes them for one person and gives ”Chris Paul” a room with Chris’s and Paul’s roomconcatenated. For example, if Chris had 15 and Paul had 9, ”Chris Paul” has 159. If there are 360rooms in the hotel, what is the probability that ”Chris Paul” has a valid room?

26. [13] Find the number of ways to choose two nonempty subsets X and Y of {1, 2, . . . , 2001}, such that|Y | = 1001 and the smallest element of Y is equal to the largest element of X.

27. [13] Let r1, r2, r3, r4 be the four roots of the polynomial x4 − 4x3 + 8x2 − 7x+ 3. Find the value of

r21r22 + r23 + r24

+r22

r21 + r23 + r24+

r23r21 + r22 + r24

+r24

r21 + r22 + r23

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28. [15] The numbers 1−10 are written in a circle randomly. Find the expected number of numbers whichare at least 2 larger than an adjacent number.

29. [15] We want to design a new chess piece, the American, with the property that (i) the American cannever attack itself, and (ii) if an American A1 attacks another American A2, then A2 also attacks A1.Let m be the number of squares that an American attacks when placed in the top left corner of an 8 by8 chessboard. Let n be the maximal number of Americans that can be placed on the 8 by 8 chessboardsuch that no Americans attack each other, if one American must be in the top left corner. Find thelargest possible value of mn.

30. [15] On the blackboard, Amy writes 2017 in base-a to get 133201a. Betsy notices she can erase a digitfrom Amy’s number and change the base to base-b such that the value of the the number remains thesame. Catherine then notices she can erase a digit from Betsy’s number and change the base to base-csuch that the value still remains the same. Compute, in decimal, a+ b+ c.

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31. [17] Define a number to be an anti-palindrome if, when written in base 3 as anan−1...a0, then ai+an−i =2 for any 0 ≤ i ≤ n. Find the number of anti-palindromes less than 312 such that no two consecutivedigits in base 3 are equal.

32. [17] Let Ck,n denote the number of paths on the Cartesian plane along which you can travel from(0, 0) to (k, n), given the following rules: 1) You can only travel directly upward or directly rightward2) You can only change direction at lattice points 3) Each horizontal segment in the path must be atmost 99 units long.

Find∞∑j=0

C100j+19,17

33. [17] Camille the snail lives on the surface of a regular dodecahedron. Right now he is on vertex P1

of the face with vertices P1, P2, P3, P4, P5. This face has a perimeter of 5. Camille wants to get tothe point on the dodecahedron farthest away from P1. To do so, he must travel along the surface adistance at least L. What is L2?

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34. [20] Find the sum of the ages of everyone who wrote a problem for this year’s HMMT Novembercontest. If your answer is X and the actual value is Y , your score will be max(0, 20− |X − Y |)

35. [20] Find the total number of occurrences of the digits 0, 1 . . . , 9 in the entire guts round. If your

answer is X and the actual value is Y , your score will be max(0, 20− |X−Y |2 )

36. [20] Find the number of positive integers less than 1000000 which are less than or equal to thesum of their proper divisors. If your answer is X and the actual value is Y , your score will bemax(0, 20− 80|1− X

Y |) rounded to the nearest integer.

HMMT November 2016November 12, 2016

Guts

1. [5] If five fair coins are flipped simultaneously, what is the probability that at least three of them showheads?

Proposed by: Allen Liu

Answer: 12

The coin either lands heads-up at least three times or lands heads-down at least three times. Thesescenarios are symmetric so the probably is just 1

2 .

Alternatively, we can explicitly compute the probability, which is just(53)+(5

4)+(55)

25 = 1632 = 1

2

2. [5] How many perfect squares divide 1010?

Proposed by: Evan Chen

Answer: 36

A perfect square s divides 1010 if and only if s = 2a · 5b where a, b ∈ {0, 2, 4, 6, 8, 10}. There are 36choices, giving 36 different s’s.

3. [5] Evaluate 2016!2

2015!2017! . Here n! denotes 1× 2× · · · × n.

Proposed by: Evan Chen

Answer: 20162017

2016!2

2015!2017! = 2016!2015!

2016!2017! = 2016

11

2017 = 20162017

4. [6] A square can be divided into four congruent figures as shown:

For how many n with 1 ≤ n ≤ 100 can a unit square be divided into n congruent figures?

Proposed by: Kevin Sun

Answer: 100

We can divide the square into congruent rectangles for all n, so the answer is 100 .

5. [6] If x+ 2y − 3z = 7 and 2x− y + 2z = 6, determine 8x+ y.

Proposed by: Eshaan Nichani

Answer: 32

8x+ y = 2(x+ 2y − 3z) + 3(2x− y + 2x) = 2(7) + 3(6) = 32

6. [6] Let ABCD be a rectangle, and let E and F be points on segment AB such that AE = EF = FB.If CE intersects the line AD at P , and PF intersects BC at Q, determine the ratio of BQ to CQ.

Proposed by: Eshaan Nichani

Answer: 1/3

Because4PAE ∼ 4PDC and AE : DC = 1 : 3, we have that PA : PD = 1 : 3 =⇒ PA : AB = PA :BC = 1 : 2. Also, by similar triangles 4PAF ∼ 4QBF , since AF : BF = 2 : 1, PA : BQ = 2 : 1.

Then BQ = 12PA = 1

2 ·12BC = 1

4BC. Then BQ : CQ =1

3.

7. [7] What is the minimum value of the product

6∏i=1

ai − ai+1

ai+2 − ai+3

given that (a1, a2, a3, a4, a5, a6) is a permutation of (1, 2, 3, 4, 5, 6)? (note a7 = a1, a8 = a2 · · · )

Proposed by: Kevin Sun

Answer: 1

The product always evaluates to 1 .

8. [7] Danielle picks a positive integer 1 ≤ n ≤ 2016 uniformly at random. What is the probability thatgcd(n, 2015) = 1?

Proposed by: Evan Chen

Answer: 14412016

We split the interval [1, 2016] into [1, 2015] and 2016. The number of integers in [1, 2015] that arerelatively prime to 2015 is φ(2015) = 4

5 ·1213 ·

3031 · 2015 = 1440. Also, 2016 is relatively prime to 2015, so

there are a total of 1441 numbers in [1, 2016] that are relatively prime to 2015. Then the probability

of picking a number relatively prime to 2015 is1441

2016.

9. [7] How many 3-element subsets of the set {1, 2, 3, . . . , 19} have sum of elements divisible by 4?

Proposed by: Sam Korsky

Answer: 244

Consider the elements of the sets mod 4. Then we would need to have sets of the form {0, 0, 0},{0, 2, 2}, {0, 1, 3}, {1, 1, 2}, or {2, 3, 3}. In the set {1, 2, . . . , 19} there four elements divisible by 4 and5 elements congruent to each of 1, 2, 3 mod 4. Hence the desired number is given by(

4

3

)+

(4

1

)(5

2

)+

(4

1

)(5

1

)(5

1

)+

(5

2

)(5

1

)+

(5

1

)(5

2

)= 244

10. [8] Michael is playing basketball. He makes 10% of his shots, and gets the ball back after 90% of hismissed shots. If he does not get the ball back he stops playing. What is the probability that Michaeleventually makes a shot?

Proposed by: Eshaan Nichani

Answer: 10/19

We find the probability Michael never makes a shot. We do casework on the number of shots Michaeltakes. He takes only one shot with probability 9

10 ·110 (he misses with probability 9

10 and does not getthe ball back with probability 1

10 ). Similarly, he takes two shots with probability 910 ·

910 ·

910 ·

110 , three

shots with probability

910

)5 110 , and so on. So we want to sum

∑∞i=1

910

)2i−1 · 110 =

110 ·

910

1− 81100

= 919 . Then

the probability Michael makes a shot is 1− 919 =

10

19.

11. [8] How many subsets S of the set {1, 2, . . . , 10} satisfy the property that, for all i ∈ [1, 9], either i ori+ 1 (or both) is in S?

Proposed by: Eshaan Nichani

Answer: 144

We do casework on the number of i’s not in S. Notice that these i’s that are not in S cannot beconsecutive, otherwise there exists an index i such that both i and i + 1 are both not in S. Hence if

there are k i’s not in S, we want to arrange k black balls and 10−k white balls such that no two blackballs are consecutive. Take out k− 1 white balls to insert back between black balls later, then we wantto arrange k black balls and 11− 2k white balls arbitrarily, which can be done in

11−k

k

)ways. Hence

we want to find the sum110

)+101

)+92

)+83

)+74

)+65

), which is equal to 144 ways.

12. [8] A positive integer ABC, where A, B, C are digits, satisfies

ABC = BC −A

Find ABC.

Proposed by: Henrik Boecken

The equation is equivalent to 100A+10B+C = BC−A. Suppose A = 0, so that we get 10B+C = BC .Reducing mod B, we find that C must be divisible by B. C 6= 0, since otherwise 10B = 1, contradiction,so C ≥ B. Thus 10B + C ≥ BB for digits B,C. For B ≥ 4, we have 100 > 10B + C ≥ BB > 100,a contradiction, so B = 1, 2, 3. We can easily test that these do not yield solutions, so there are nosolutions when A = 0.

Thus A ≥ 1, and so 100 ≤ 100A + 10B + C ≤ 1000, and thus 100 ≤ BC − A ≤ 1000. 1 ≤ A ≤ 10,so we have 101 ≤ BC ≤ 1010. We can test that the only pairs (B,C) that satisfy this condition are(2, 7), (2, 8), (2, 9), (3, 5), (3, 6), (4, 4), (5, 3), (6, 3), (7, 3), (8, 3), (9, 3). Of these pairs, only (2, 7) yields asolution to the original equation, namely A = 1, B = 2, C = 7. Thus ABC = 127.

13. [9] How many functions f : {0, 1}3 → {0, 1} satisfy the property that, for all ordered triples (a1, a2, a3)and (b1, b2, b3) such that ai ≥ bi for all i, f(a1, a2, a3) ≥ f(b1, b2, b3)?

Proposed by: Eshaan Nichani

Answer: 20

Consider the unit cube with vertices {0, 1}3. Let O = (0, 0, 0), A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1),D = (0, 1, 1), E = (1, 0, 1), F = (1, 1, 0), and P = (1, 1, 1). We want to find a function f on thesevertices such that f(1, y, z) ≥ f(0, y, z) (and symmetric representations). For instance, if f(A) = 1,then f(E) = f(F ) = f(P ) = 1 as well, and if f(D) = 1, then f(P ) = 1 as well.

We group the vertices into four levels: L0 = {O}, L1 = {A,B,C}, L2 = {D,E, F}, and L3 = {P}.We do casework on the lowest level of a 1 in a function.

• If the 1 is in L0, then f maps everything to 1, for a total of 1 way.

• If the 1 is in L1, then f(O) = 0. If there are 3 1’s in L1, then everything but O must be mappedto 1, for 1 way. If there are 2 1’s in L1, then f(L2) = f(L3) = 1, and there are 3 ways tochoose the 2 1’s in L1, for a total of 3 ways. If there is one 1, then WLOG f(A) = 1. Thenf(E) = f(F ) = f(P ) = 1, and f(D) equals either 0 or 1. There are 3 · 2 = 6 ways to do this. Intotal, there are 1 + 3 + 6 = 10 ways for the lowest 1 to be in L1.

• If the lowest 1 is in L2, then f(O) = f(L1) = 0. If there are 3 1’s in L2, there is one way to makef . If there are 2 1’s, then we can pick the 2 1’s in 3 ways. Finally, if there is one 1, then we pickthis 1 in 3 ways. There are 1 + 3 + 3 = 7 ways.

• The lowest 1 is in L3. There is 1 way.

• There are no 1’s. Then f sends everything to 0. There is 1 way.

In total, there are 1 + 10 + 7 + 1 + 1 = 20 total f ’s.

14. [9] The very hungry caterpillar lives on the number line. For each non-zero integer i, a fruit sits on thepoint with coordinate i. The caterpillar moves back and forth; whenever he reaches a point with food,he eats the food, increasing his weight by one pound, and turns around. The caterpillar moves at aspeed of 2−w units per day, where w is his weight. If the caterpillar starts off at the origin, weighingzero pounds, and initially moves in the positive x direction, after how many days will he weigh 10pounds?

Proposed by: Eshaan Nichani

Answer: 9217

On the nth straight path, the caterpillar travels n units before hitting food and his weight is n − 1.Then his speed is 21−n. Then right before he turns around for the nth time, he has traveled a total timeof∑n

i=1i

21−i = 12

∑ni=1 i · 2i. We want to know how many days the catepillar moves before his weight

is 10, so we want to take n = 10 so that his last straight path was taken at weight 9. Hence we want to

evaluate S = 12

∑10i=1 i ·2i. Note that 2S = 1

2

∑11i=2(i−1) ·2i, so S = 2S−S = 1

2

(11 · 211 −

∑10i=1 2i

)=

12

10 · 211 − 211 + 2

)= 9217 .

15. [9] Let ABCD be an isosceles trapezoid with parallel bases AB = 1 and CD = 2 and height 1. Findthe area of the region containing all points inside ABCD whose projections onto the four sides of thetrapezoid lie on the segments formed by AB,BC,CD and DA.

Proposed by: Allen Liu

Answer: 58

Let E, F , be the projections of A, B on CD. A point whose projections lie on the sides must becontained in the square ABFE. Furthermore, the point must lie under the perpendicular to AD at Aand the perpendicular to BC at B, which have slopes 1

2 and − 12 . The area of the desired pentagon is

1− 14 −

18 =

5

8.

16. [10] Create a cube C1 with edge length 1. Take the centers of the faces and connect them to form anoctahedron O1. Take the centers of the octahedron’s faces and connect them to form a new cube C2.Continue this process infinitely. Find the sum of all the surface areas of the cubes and octahedrons.

Proposed by: Shyam Narayanan

Answer: 54+9√3

8

The lengths of the second cube are one-third of the lengths of the first cube, so the surface areadecreases by a factor of one-ninth. Since the first cube has surface area 6 and the first octahedron has

surface area√

3, the total area is (6 +√

3) ·(

1 +1

9+

1

92+ · · ·

)=

54 + 9√

3

8

17. [10] Let p(x) = x2 − x+ 1. Let α be a root of p(p(p(p(x))). Find the value of

(p(α)− 1)p(α)p(p(α))p(p(p(α))

Proposed by: Henrik Boecken

Answer: −1

Since (x− 1)x = p(x)− 1, we can set

(p(α)− 1)p(α)p(p(α))p(p(p(α)) = (p(p(α))− 1)p(p(α))p(p(p(α))

= (p(p(p(α)))− 1)p(p(p(α))

= p(p(p(p(α))))− 1

= −1

18. [10] An 8 by 8 grid of numbers obeys the following pattern:

1) The first row and first column consist of all 1s.

2) The entry in the ith row and jth column equals the sum of the numbers in the (i − 1) by (j − 1)sub-grid with row less than i and column less than j.

What is the number in the 8th row and 8th column?

Proposed by: Eshaan Nichani

Answer: 2508

Let xi,j be the number in the ith row and the jth column. Then if i, j ≥ 2, xi+1,j+1−xi+1,j−xi,j+1+xi,jonly counts the term xi,j since every other term is added and subtracted the same number of times.Thus xi+1,j+1 = xi+1,j + xi,j+1 when i, j ≥ 2. Also, x2,i = xi,2 = i so xi+1,j+1 = xi+1,j + xi,j+1 holdsfor all i, j ≥ 1 except for (i, j) = (1, 1) where x2,2 is one less than expected. This means that xi,j isthe number ways of travelling from (1, 1) to (i, j), minus the number of ways of travelling from (2, 2)

to (i, j), which is

(14

7

)−(

12

6

)= 2508 .

19. [11] Let S be the set of all positive integers whose prime factorizations only contain powers of theprimes 2 and 2017 (1, powers of 2, and powers of 2017 are thus contained in S). Compute

∑s∈S

1s .

Proposed by: Daniel Qu

Answer: 20171008

Since every s can be written as 2i ·2017j for non-negative integers i and j, the given sum can be writtenas (

∑∞i=0

12i )(

∑∞j=0

12017j ). We can easily find the sum of these geometric series since they both have

common ratio of magnitude less than 1, giving us ( 11− 1

2

) · 11− 1

2017

) = 21 ·

20172016 = 2017

1008 .

20. [11] Let V be the volume enclosed by the graph

x2016 + y2016 + z2 = 2016

Find V rounded to the nearest multiple of ten.

Proposed by: Henrik Boecken

Answer: 360

Let R be the region in question. Then we have

[−1, 1]2 × [−√

2014,√

2014] ⊂ R ⊂ [− 2016√

2016,2016√

2016]2 × [√

2016,√

2016]

We find some bounds: we have √2016 <

√2025 = 45.

By concavity of√·, we have the bound

√2014 ≤ 11

89

√1936 +

78

89

√2025 = 45− 11

89.

Finally, if we let 2016√

2016 = 1 + ε, then (1 + ε)2016 = 2016, so(2016

5

)ε5 < 2016⇒ ε <

5

√120

2015 · 2014 · 2013 · 2012<

5

√120

20004=

5

√7500

1015=

5√

7500

1000<

6

1000.

Therefore, the volume of R is lower-bounded by

22 · 2(

45− 11

89

)= 360− 88

89> 355

and upper-bounded by

22(

1 +0.8

100

)2

· 2(45) = 360

(1 +

6

1000

)2

< 365.

Thus, R rounded to the nearest ten is 360.

21. [11] Zlatan has 2017 socks of various colours. He wants to proudly display one sock of each of thecolours, and he counts that there are N ways to select socks from his collection for display. Given thisinformation, what is the maximum value of N?

Proposed by: Saranesh Prembabu

Answer: 3671 · 4Say that there are k sock types labeled 1, 2, . . . , k, and ai socks of type i. The problem asks to maximize∏k

i=1 ai subject to∑k

i=1 ai = 2017, over all k and all sequences of positive integers a1, . . . , ak.

The optimal (a1, . . . , ak) cannot have any ai = 1 for any i, because if there exists ai = 1 we can deletethis ai and add 1 to any aj (j 6= i) to increase the product while keeping the sum constant.

There exists an optimal (a1, . . . , ak) without any ai ≥ 4 because if there exists ai ≥ 4 we can replacethis ai with ai − 2 and 2, which nonstrictly increases the product while keeping the sum constant.

Therefore, there exists an optimal (a1, . . . , ak) whose terms are all 2 or 3. The optimal (a1, . . . , ak)cannot have more than two 2s, because we can replace three 2s with two 3s, which increases the sum

by a factor of 32

23 = 98 while keeping the sum constant.

It follows that we want to partition 2017 into 671 3s and two 2s, for a product of 3671 · 4.

22. [12] Let the function f : Z→ Z take only integer inputs and have integer outputs. For any integers xand y, f satisfies

f(x) + f(y) = f(x+ 1) + f(y − 1)

If f(2016) = 6102 and f(6102) = 2016, what is f(1)?

Proposed by: Henrik Boecken

Answer: 8117

We havef(x+ 1) = f(x) + f(y)− f(y − 1)

If y is fixed, we havef(x+ 1) = f(x) + constant

implying f is linear. Using our two points, then, we get f(x) = 8118− x, so f(1) = 8117

23. [12] Let d be a randomly chosen divisor of 2016. Find the expected value of

d2

d2 + 2016

Proposed by: Henrik Boecken

Answer: 12

Let ab = 2016. Then

a2

a2 + 2016+

b2

b2 + 2016=

a2

a2 + 2016+

( 2016a )2

( 2016a )2 + 2016

=a2

a2 + 2016+

2016

a2 + 2016= 1

Thus, every divisor d pairs up with 2016d to get 1, so our desired expected value is

1

2.

24. [12] Consider an infinite grid of equilateral triangles. Each edge (that is, each side of a small triangle)is colored one of N colors. The coloring is done in such a way that any path between any two non-adjecent vertices consists of edges with at least two different colors. What is the smallest possible valueof N?

Proposed by: Henrik Boecken

Answer: 6

Note that the condition is equivalent to having no edges of the same color sharing a vertex by justconsidering paths of length two. Consider a hexagon made out of six triangles. Six edges meet at thecenter, so N ≥ 6. To prove N = 6, simply use two colors for each of the three possible directions of anedge, and color edges of the same orientation alternatingly with different colors.

25. [13] Chris and Paul each rent a different room of a hotel from rooms 1 − 60. However, the hotelmanager mistakes them for one person and gives ”Chris Paul” a room with Chris’s and Paul’s roomconcatenated. For example, if Chris had 15 and Paul had 9, ”Chris Paul” has 159. If there are 360rooms in the hotel, what is the probability that ”Chris Paul” has a valid room?

Proposed by: Meghal Gupta

Answer: 1531180

There are 60 · 59 = 3540 total possible outcomes, and we need to count the number of these whichconcatenate into a number at most 60. Of these, 9 ·8 result from both Chris and Paul getting one-digitroom numbers. If Chris gets a two-digit number, then he must get a number at most 35 and Paulshould get a one-digit room number, giving (35− 9) · 9 possibilties. If Chris gets a one-digit number, itmust be 1, 2, or 3. If Chris gets 1, 2 or 3, Paul can get any two-digit number from 10 to 60 to guaranteea valid room, giving 51 ·3 outcomes. The total number of correct outcomes is 72+51∗3+26∗9 = 459,

so the desired probability is153

1180.

26. [13] Find the number of ways to choose two nonempty subsets X and Y of {1, 2, . . . , 2001}, such that|Y | = 1001 and the smallest element of Y is equal to the largest element of X.

Proposed by: Allen Liu

Answer: 22000

We claim that there is a bijection between pairs (X,Y ) and sets S with at least 1001 elements. To getS from X and Y , take S = X ∪ Y , which contains Y and thus has at least 1001 elements. To form(X,Y ) from S, make Y the largest 1001 elements of S, and make X everything except the largest 1000elements of S. Therefore we need to count the number of subsets of {1, 2, . . . , 2001} with at least 1001elements. For every subset of {1, 2, . . . , 2001}, either it or its complement has at least 1001 elements,

so number of possible subsets is 12 · 2

2001 = 22000 .

27. [13] Let r1, r2, r3, r4 be the four roots of the polynomial x4 − 4x3 + 8x2 − 7x+ 3. Find the value of

r21r22 + r23 + r24

+r22

r21 + r23 + r24+

r23r21 + r22 + r24

+r24

r21 + r22 + r23

Proposed by: Henrik Boecken

Answer: −4

Add 1 to each fraction to get

r21 + r22 + r23 + r24r22 + r23 + r24

+r21 + r22 + r23 + r24r21 + r23 + r24

+r21 + r22 + r23 + r24r21 + r22 + r24

+r21 + r22 + r23 + r24r21 + r22 + r23

This seems like a difficult problem until one realizes that

r21 + r22 + r23 + r24 = (r1 + r2 + r3 + r4)2 − 2(r1r2 + r1r3 + r1r4 + r2r3 + r2r4 + r3r4) = 42 − 2 · 8 = 0

Thus, our current expression is 0. Noting that we added 4, the original value had to be −4 .

28. [15] The numbers 1−10 are written in a circle randomly. Find the expected number of numbers whichare at least 2 larger than an adjacent number.

Proposed by: Shyam Narayanan

Answer: 173

For 1 ≤ i ≤ 10, let Xi be the random variable that is 1 if the i in the circle is at least 2 larger thanone of its neighbors, and 0 otherwise. The random variable representing number of numbers that areat least 2 larger than one of their neighbors is then just X1 + X2 + · · · + X10. The expected valueE[X1 +X2 + · · ·+X10] is equal to E[X1] + E[X2] + · · ·+ E[X10] by the linearity of expectation, so itsuffices to compute E[Xi] for all 1 ≤ i ≤ 10.

By the definition of expected value, E[Xi] = 1·P (the i is at least 2 larger than one of its neighbors)+0·P (it is not at least 2 larger than either of its neighbors) = P (the i is at least 2 larger than one of its neighbors) =1−P (the i is at most 1 larger than both of its neighbors). For the last probability, i’s neighbors mustbe drawn from the set {max(1, i − 1),max(1, i − 1) + 1, . . . , 10}, excluding i itself. This set has

10 − max(1, i − 1) elements, so there are a total of10−max(1,i−1)

2

)sets of two neighbors for i that

satisfy the condition, out of a total of92

)possible sets of two neighbors from all of the numbers that

are not i. The last probability is then(10−max(1,i−1)

2 )(92)

, so E[Xi] = 1− (10−max(1,i−1)2 )

(92)

.

The final sum we wish to calculate then becomes (1− (92)

(92)

)+(1− (92)

(92)

)+(1− (82)

(92)

)+(1− (72)

(92)

)+· · ·+(1− (12)

(92)

) =

0 + 0 + (1− 2836 ) + (1− 21

36 ) + · · ·+ (1− 0) =17

3.

29. [15] We want to design a new chess piece, the American, with the property that (i) the American cannever attack itself, and (ii) if an American A1 attacks another American A2, then A2 also attacks A1.Let m be the number of squares that an American attacks when placed in the top left corner of an 8 by8 chessboard. Let n be the maximal number of Americans that can be placed on the 8 by 8 chessboardsuch that no Americans attack each other, if one American must be in the top left corner. Find thelargest possible value of mn.

Proposed by: Kevin Yang

Answer: 1024

Since one of the Americans must be in the top left corner, that eliminates m squares from considerationfor placing additional Americans. So m+ n is at most 64, which implies mn can be at most 1024. Toachieve 1024, we can color a chessboard the normal way, and say that an American attacks all squaresof the opposite color. Then the American in the top left corner attacks the 32 squares of the oppositecolor, and placing all Americans on the squares of the same color as the top-left corner guarantees noAmericans attack each other.

30. [15] On the blackboard, Amy writes 2017 in base-a to get 133201a. Betsy notices she can erase a digitfrom Amy’s number and change the base to base-b such that the value of the the number remains thesame. Catherine then notices she can erase a digit from Betsy’s number and change the base to base-csuch that the value still remains the same. Compute, in decimal, a+ b+ c.

Proposed by: Daniel Qu

Answer: 22

2017 = 1332014 = 132016 = 120112

31. [17] Define a number to be an anti-palindrome if, when written in base 3 as anan−1...a0, then ai+an−i =2 for any 0 ≤ i ≤ n. Find the number of anti-palindromes less than 312 such that no two consecutivedigits in base 3 are equal.

Proposed by: Shyam Narayanan

Answer: 126

Note once the middle digit/pair of digits is determined, it suffices to choose the digits in the left halfof the number and ensure no pair of consecutive digits are equal. For a number with an even numberof digits, the middle pair is 02 or 20 while for a number with an odd number of digits, the middle digitis 1. We can now count recursively.

Let an be the number of ways to choose n digits no two of which are consecutive and equal such thatthe leading digit is nonzero and the ending digit is 1. Let bn be the number ways to do the same suchthat the ending digit is 0 or 2.

Note an = bn−1. Also bn = bn−1 + 2an−1.

Solving for the terms of the sequence, they are a1 = 1, a2 = b1 = 1, a3 = b2 = 3, a4 = b3 = 5, a5 = b4 =11, a6 = b5 = 21, b6 = 43. Therefore, there are 43 twelve-digit numbers satisfying the condition, 21eleven-digit numbers, 21 ten-digit numbers.... and 1 one-digit number. The sum of these values givesus a final answer of 126

32. [17] Let Ck,n denote the number of paths on the Cartesian plane along which you can travel from(0, 0) to (k, n), given the following rules: 1) You can only travel directly upward or directly rightward2) You can only change direction at lattice points 3) Each horizontal segment in the path must be atmost 99 units long.

Find∞∑j=0

C100j+19,17

Proposed by: Saranesh Prembabu

Answer: 10017

If we are traveling from (0, 0) to (n, 17), we first travel x0 rightwards, then up one, then x1 rightwards,then up one, ..., until we finally travel x17 rightwards. x0, ..., x17 are all at most 99 by our constraint,but can equal 0. Given that x0, ..., x16 are fixed, there is exactly one way to choose x17 so thatx0 + ... + x17 is congruent to 19 mod 100. Then, this means that the sum equals the total number ofways to choose x0, ..., x16, which equals 10017 = 1034.

33. [17] Camille the snail lives on the surface of a regular dodecahedron. Right now he is on vertex P1

of the face with vertices P1, P2, P3, P4, P5. This face has a perimeter of 5. Camille wants to get tothe point on the dodecahedron farthest away from P1. To do so, he must travel along the surface adistance at least L. What is L2?

Proposed by: Saranesh Prembabu

Answer: 17+7√5

2

Consider the net of the dodecahedron. It suffices to look at three pentagons ABCDE, EDFGH, andGFIJK, where AJ = L. This can be found by the law of cosines on triangle AEJ . We have AE = 1,

EJ = tan 72◦, and ∠AEJ = 162◦. Thus L2 = 1 + tan2 72◦ + 2 · tan 72◦ · cos 18◦ = 17+7√5

2

34. [20] Find the sum of the ages of everyone who wrote a problem for this year’s HMMT Novembercontest. If your answer is X and the actual value is Y , your score will be max(0, 20− |X − Y |)Proposed by: Allen Liu

Answer: 258

There was one problem for which I could not determine author information, so I set the author asone of the problem czars at random. Then, I ran the following command on a folder containing TeXsolutions files to all four contests:

evan@ArchMega ~/Downloads/November

$ grep --no-filename "Proposed by: " *.tex | sort | uniq -c

15 {\em Proposed by: Allen Liu }

1 {\em Proposed by: Brice Huang }

2 {\em Proposed by: Christopher Shao }

2 {\em Proposed by: Daniel Qu }

21 {\em Proposed by: Eshaan Nichani }

3 {\em Proposed by: Evan Chen }

9 {\em Proposed by: Henrik Boecken }

2 {\em Proposed by: Kevin Sun }

2 {\em Proposed by: Kevin Yang }

1 {\em Proposed by: Meghal Gupta }

1 {\em Proposed by: Rachel Zhang }

1 {\em Proposed by: Sam Korsky }

3 {\em Proposed by: Saranesh Prembabu }

3 {\em Proposed by: Shyam Narayanan }

This gave the counts of problem proposals; there were 14 distinct authors of problems for November2016. Summing their ages (concealed for privacy) gives 258.

35. [20] Find the total number of occurrences of the digits 0, 1 . . . , 9 in the entire guts round. If your

answer is X and the actual value is Y , your score will be max(0, 20− |X−Y |2 )

Proposed by: Allen Liu

Answer: 559

To compute the answer, I extracted the flat text from the PDF file and ran word-count against the listof digit matches.

evan@ArchMega ~/Downloads/November

$ pdftotext HMMTNovember2016GutsTest.pdf guts-test-text.txt

evan@ArchMega ~/Downloads/November

$ cat guts-test-text.txt | egrep "[0-9]" --only-matching | wc -l

559

36. [20] Find the number of positive integers less than 1000000 which are less than or equal to thesum of their proper divisors. If your answer is X and the actual value is Y , your score will bemax(0, 20− 80|1− X

Y |) rounded to the nearest integer.

Proposed by: Allen Liu

Answer: 247548

N = 1000000

s = [0] * N

ans = 0

for i in range(1, N):

if i <= s[i]:

ans += 1

for j in range(i + i, N, i):

s[j] += i

print(ans)

HMMT Invitational Competition

April 2016

1. [5] Theseus starts at the point (0, 0) in the plane. If Theseus is standing at the point (x, y) in theplane, he can step one unit to the north to point (x, y + 1), one unit to the west to point (x − 1, y),one unit to the south to point (x, y − 1), or one unit to the east to point (x + 1, y). After a sequenceof more than two such moves, starting with a step one unit to the south (to point (0,−1)), Theseusfinds himself back at the point (0, 0). He never visited any point other than (0, 0) more than once, andnever visited the point (0, 0) except at the start and end of this sequence of moves.

Let X be the number of times that Theseus took a step one unit to the north, and then a step one unitto the west immediately afterward. Let Y be the number of times that Theseus took a step one unitto the west, and then a step one unit to the north immediately afterward. Prove that |X − Y | = 1.

2. [7] Let ABC be an acute triangle with circumcenter O, orthocenter H, and circumcircle Ω. Let M bethe midpoint of AH and N the midpoint of BH. Assume the points M , N , O, H are distinct and lieon a circle ω. Prove that the circles ω and Ω are internally tangent to each other.

3. [8] Denote by N the positive integers. Let f : N→ N be a function such that, for any w, x, y, z ∈ N,

f(f(f(z)))f(wxf(yf(z))) = z2f(xf(y))f(w).

Show that f(n!) ≥ n! for every positive integer n.

4. [10] Let P be an odd-degree integer-coefficient polynomial. Suppose that xP (x) = yP (y) for infinitelymany pairs x, y of integers with x 6= y. Prove that the equation P (x) = 0 has an integer root.

5. [12] Let S = {a1, . . . , an} be a finite set of positive integers of size n ≥ 1, and let T be the set of allpositive integers that can be expressed as sums of perfect powers (including 1) of distinct numbers inS, meaning

T =

{n∑

i=1

aeii | e1, e2, . . . , en ≥ 0

}.

Show that there is a positive integer N (only depending on n) such that T contains no arithmeticprogression of length N .

HMMT Invitational CompetitionApril 2016

1. [5] Theseus starts at the point (0, 0) in the plane. If Theseus is standing at the point (x, y) in theplane, he can step one unit to the north to point (x, y + 1), one unit to the west to point (x − 1, y),one unit to the south to point (x, y − 1), or one unit to the east to point (x+ 1, y). After a sequenceof more than two such moves, starting with a step one unit to the south (to point (0,−1)), Theseusfinds himself back at the point (0, 0). He never visited any point other than (0, 0) more than once, andnever visited the point (0, 0) except at the start and end of this sequence of moves.

Let X be the number of times that Theseus took a step one unit to the north, and then a step one unitto the west immediately afterward. Let Y be the number of times that Theseus took a step one unitto the west, and then a step one unit to the north immediately afterward. Prove that |X − Y | = 1.

Proposed by: Mitchell Lee

The path Theseus traces out is a closed, non-self-intersecting path in the plane. Since each move isalong a line segment, the path forms the boundary of a polygon in the plane. WLOG this polygon istraversed counterclockwise (i.e. with the interior of the polygon always on Theseus’s left); the argumentin the clockwise case is analogous, with a sign flip.

Let i = 0 correspond to the direction east, i = 1 correspond to north, and so forth. For 0 ≤ i ≤ 3and j = i ± 1 let Xij be the number of times Theseus switches from going in direction i to going indirection j (where indices are taken mod 4). So, X = X1,2, Y = X2,1. We have the following relationsamong the Xij :

The sum n =∑Xi,j is the number of vertices of the polygon. The sum of the interior angles of the

polygon is thus π(n−2). Each counterclockwise turn contributes π2 , and each clockwise turn contributes

3π2 to this sum. The number of the former is counted by

∑Xi,i+1, and the latter by

∑Xi+1,i. So,

π(n−2) = π2

∑Xi,i+1+ 3π

2

∑Xi+1,i = πn+ π

2

∑(Xi+1,i−Xi,i+1), which gives

∑(Xi,i+1−Xi+1,i) = 4.

Finally, observe by starting from the middle of any edge that since we start and end in the samedirection, the number of times we turn into direction i must equal the number of times we turn out ofit. So Xi,i+1 + Xi,i∈1 = Xi+1,i + Xi∈1,i, which rearranges to give Xi,i+1 −Xi+1,i = Xi∈1,i −Xi∈1,i.So 4 =

∑(Xi,i+1 −Xi+1,i) = 4(X1,2 −X2,1) = 4(X − Y ).

So X − Y = 1 as desired.

2. [7] Let ABC be an acute triangle with circumcenter O, orthocenter H, and circumcircle Ω. Let M bethe midpoint of AH and N the midpoint of BH. Assume the points M , N , O, H are distinct and lieon a circle ω. Prove that the circles ω and Ω are internally tangent to each other.

Proposed by: Evan Chen

Let R be the circumradius of 4ABC. Recall that the circumcircle of 4HBC has radius equal to R.By a homothety at H with factor 2 at H, it follows that 4HMN has circumradius R/2. Thus ω hascircumradius R/2, and since O lies on ω the conclusion follows.

Remark. This problem is a slight modification of a proposal by Dhroova Aiylam.

3. [8] Denote by N the positive integers. Let f : N→ N be a function such that, for any w, x, y, z ∈ N,

f(f(f(z)))f(wxf(yf(z))) = z2f(xf(y))f(w).

Show that f(n!) ≥ n! for every positive integer n.

Proposed by: Pakawut Jiradilok

If f(z1) = f(z2), then plugging in (w, x, y, z1) and (w, x, y, z2) yields z1 = z2. Thus, f is injective.Substitution of (z, w) with (1, f(f(1))) in the main equation yields

f(f(f(1))xf(yf(1))

)= f(xf(y))

Because f is injective, we have

f(yf(1)) · f(f(1)) = f(y),∀y ∈ N (∗)

We can easily show by induction that

f(y) = f(yf(1)n)f(f(1))n,∀n ∈ N

Thus, f(f(1))n|f(y),∀n ∈ N which implies that f(f(1)) = 1. Using this equality in (∗) yields f(1) = 1.Substitution of (y, z) with (1, 1) in the main equation yields f(xy) = f(x)f(y).

So,

f(n!) = f( n∏i=1

i)

=

n∏i=1

f(i).

By injectivity, each f(i) in this product is a distinct positive integer, so their product is at least∏ni=1 i = n!, as desired.

Remark 1. The equation condition of f in the problem is actually equivalent to the following twoconditions combined:

(C1) f(xy) = f(x)f(y),∀x, y ∈ N, and

(C2) f(f(f(x))) = x,∀x ∈ N

4. [10] Let P be an odd-degree integer-coefficient polynomial. Suppose that xP (x) = yP (y) for infinitelymany pairs x, y of integers with x 6= y. Prove that the equation P (x) = 0 has an integer root.

Proposed by: Victor Wang

Proof. Let n be the (odd) degree of P . Suppose, for contradiction, that P has no integer roots, andlet Q(x) = xP (x) = anx

n+1 + an∈1xn + · · · + a0x

1. WLOG an > 0, so there exists M > 0 such thatQ(M) < Q(M + 1) < · · · and Q(−M) < Q(−M − 1) < · · · (as n+ 1 is even).

There exist finitely many solutions (t, y) for any fixed t, so infinitely many solutions (x, y) with |x|, |y| ≥M . By definition of M , there WLOG exist infinitely many solutions with |x| ≥ M and y ≥ −x > 0(otherwise replace P (t) with P (−t)).Equivalently, there exist infinitely many (t, k) with t ≥M and k ≥ 0 such that

0 = Q(t+ k)−Q(−t) = an[(t+ k)n+1 − (−t)n+1] + · · ·+ a0[(t+ k)1 − (−t)1].

If k 6= 0, then Q(x + k) − Q(−x) has constant term Q(k) − Q(0) = kP (k) 6= 0, and if k = 0, thenQ(x + k) − Q(−x) = 2(a0x

1 + a2x3 + · · · + an∈1x

n) has x1 coefficient nonzero (otherwise P (0) = 0),so no fixed k yields infinitely many solutions.

Hence for any N ≥ 0, there exist k = kN ≥ N and t = tN ≥M such that 0 = Q(t+ k)−Q(−t). Butthen the triangle inequality yields

an[(t+ k)n+1 − (−t)n+1]︸ ︷︷ ︸=∑n−1

i=0 ∈ai[(t+k)i+1∈(∈t)i]

≤ 2(t+ k)n(|an∈1|+ · · ·+ |a0|). (1)

But (t+ k)n+1 − tn+1 ≥ (t+ k)n+1 − t · (t+ k)n = k · (t+ k)n,1 so ankN ≤ 2(|an∈1|+ · · ·+ |a0|) for allN . Since kN ≥ N , this gives a contradiction for sufficiently large N .

1However, the mean value theorem estimate (t + k)n+1 − tn+1 ≥ k · (n + 1)tn will not suffice a priori, if k happens to bemuch larger than t. Of course, it is easy to prove that k cannot be too much larger than t (even before finishing the problem);the execution would just take slightly longer.

Remark 2. Another way to finish after the triangle inequality estimate (1) is to observe that (t +

k)n+1 − (−t)n+1 = (t + k)n+1 − tn+1 ≥ tn · k + kn+1 (by binomial expansion), sokN (tnN+knN )(tN+kN )n (which

by Holder’s inequality is at least kN2n−1 ≥ N

2n−1 ) is bounded above (independently of N), contradiction.(Here we use Holder’s inequality to deduce (1 + 1)n∈1(an + bn) ≥ (a+ b)n for positive reals a, b.)

Remark 3. The statement is vacuous for even-degree polynomials. The case of degree 3 polynomialswas given as ISL 2002 A3.

5. [12] Let S = {a1, . . . , an} be a finite set of positive integers of size n ≥ 1, and let T be the set of allpositive integers that can be expressed as sums of perfect powers (including 1) of distinct numbers inS, meaning

T =

{n∑i=1

aeii | e1, e2, . . . , en ≥ 0

}.

Show that there is a positive integer N (only depending on n) such that T contains no arithmeticprogression of length N .

Proposed by: Yang Liu

Answer: N/A

In general we can assume that each ai > 1, since replacing ai = 1 by some large integer a creates a setT containing the original T as a subset (by setting ei = 0).

We proceed by induction on n. For the base case n = 1, an arithmetic progression of length at least 3would give ae11 + ae31 = 2ae21 where e3 > e2 > e1. But ae31 ≥ 2ae21 , so this is impossible. Thus our resultholds for n = 1.

Assume the result is true for some n− 1, and let An∈1 be a number such that the longest progressionwhen |S| = n− 1 has length less than An∈1. Let M be a large integer that we will choose later. Takean arithmetic progression of length 2M + 1, calling the terms b1, b2, . . . , b2M+1. Note that b2M+1 ≤2bM+1, since b is a sequence of positive integers. For each term from bM+1 to b2M+1 assign to it themaximum power that is part of the sum. Call this value c(bi). More explicitly, if bi =

∑nj=1 a

ejj , then

c(bi) = max(ae11 , ae22 , . . . , a

enn ).

Since bM+1

n ≤ c(bi) ≤ b2M+1 ≤ 2bM+1 for M + 1 ≤ i ≤ 2M + 1, there are at most∑nj=1(logai 2n+ 1) ≤

n(log2 2n+ 1) different values of c(bi). By Van der Waerden’s Theorem, there exists a value of M suchthat coloring an arithmetic progression of length M with n(log2 2n+ 1) colors yields a monochromaticarithmetic progression of length An∈1. In particular, we can take M = W (An∈1, n(log2 2n + 1)),where W (n, k) denotes the Van der Waerden number. So, we color bM+1, . . . , b2M+1 by their c(bi).Subtracting the common perfect power from each term of the monochromatic arithmetic progressionobtained gives an arithmetic progression of An∈1 integers expressible as the sum of perfect powersof distinct numbers in S \ aj . By the inductive hypothesis, this is a contradiction. So no arithmeticprogression of length 2M + 1 can be contained in T , and we can take An = 2M + 1.

By induction, we thus have such an An for all n.

February 2017February 18, 2017

Team

1. [15] Let P (x), Q(x) be nonconstant polynomials with real number coefficients. Prove that if

bP (y)c = bQ(y)c

for all real numbers y, then P (x) = Q(x) for all real numbers x.

2. [25] Does there exist a two-variable polynomial P (x, y) with real number coefficients such that P (x, y)is positive exactly when x and y are both positive?

3. [30] A polyhedron has 7n faces. Show that there exist n+ 1 of the polyhedron’s faces that all have thesame number of edges.

4. [35] Let w = w1w2 . . . wn be a word. Define a substring of w to be a word of the form wiwi+1 . . . wj→1wj ,for some pair of positive integers 1 ≤ i ≤ j ≤ n. Show that w has at most n distinct palindromicsubstrings.

For example, aaaaa has 5 distinct palindromic substrings, and abcata has 5 (a, b, c, t, ata).

5. [35] Let ABC be an acute triangle. The altitudes BE and CF intersect at the orthocenter H, andpoint O denotes the circumcenter. Point P is chosen so that ∠APH = ∠OPE = 90◦, and point Q ischosen so that ∠AQH = ∠OQF = 90◦. Lines EP and FQ meet at point T . Prove that points A, T,Oare collinear.

6. [40] Let r be a positive integer. Show that if a graph G has no cycles of length at most 2r, then it hasat most |V |2016 cycles of length exactly 2016r, where |V | denotes the number of vertices in the graphG.

7. [45] Let p be a prime. A complete residue class modulo p is a set containing at least one elementequivalent to k (mod p) for all k.

(a) (20) Show that there exists an n such that the nth row of Pascal’s triangle forms a completeresidue class modulo p.

(b) (25) Show that there exists an n ≤ p2 such that the nth row of Pascal’s triangle forms a completeresidue class modulo p.

8. [45] Does there exist an irrational number α > 1 such that

bαnc ≡ 0 (mod 2017)

for all integers n ≥ 1?

9. [65] Let n be a positive odd integer greater than 2, and consider a regular n-gon G in the plane centeredat the origin. Let a subpolygon G′ be a polygon with at least 3 vertices whose vertex set is a subset ofthat of G. Say G′ is well-centered if its centroid is the origin. Also, say G′ is decomposable if its vertexset can be written as the disjoint union of regular polygons with at least 3 vertices. Show that allwell-centered subpolygons are decomposable if and only if n has at most two distinct prime divisors.

10. [65] Let LBC be a fixed triangle with LB = LC, and let A be a variable point on arc LB of itscircumcircle. Let I be the incenter of 4ABC and AK the altitude from A. The circumcircle of 4IKLintersects lines KA and BC again at U 6= K and V 6= K. Finally, let T be the projection of I onto lineUV . Prove that the line through T and the midpoint of IK passes through a fixed point as A varies.

February 2017February 18, 2017

Team

1. [15] Let P (x), Q(x) be nonconstant polynomials with real number coefficients. Prove that if

bP (y)c = bQ(y)c

for all real numbers y, then P (x) = Q(x) for all real numbers x.

Proposed by: Alexander Katz

Answer:

By the condition, we know that |P (x) − Q(x)| ≤ 1 for all x. This can only hold if P (x) − Q(x) is aconstant polynomial. Now take a constant c such that P (x) = Q(x) + c. Without loss of generality,we can assume that c ≥ 0. Assume that c > 0. By continuity, if degP = degQ > 0, we can select aninteger r and a real number x0 such that Q(x0) + c = r. Then bP (x0)c = bQ(x0) + cc = r. On theother hand, bQ(x0)c = br − cc < r as r was an integer. This is a contradiction. Therefore, c = 0 asdesired.

2. [25] Does there exist a two-variable polynomial P (x, y) with real number coefficients such that P (x, y)is positive exactly when x and y are both positive?

Proposed by: Alexander Katz

Answer: No

For any ε and positive x, P (x, ε) > 0 and P (x,−ε) ≤ 0. Thus by continuity/IVT, P (x, 0) = 0 for allpositive x. Similarly P (0, y) = 0 for all positive y. This implies xy | P (x, y), and so we can writeP (x, y) = xyQ(x, y). But then this same logic holds for Q, and this cannot continue infinitely unlessP is identically 0 – in which case the conditions do not hold. So no such polynomial exists.

3. [30] A polyhedron has 7n faces. Show that there exist n+ 1 of the polyhedron’s faces that all have thesame number of edges.

Proposed by: Alexander Katz

Let V,E, and F denote the number of vertices, edges, and faces respectively. Let ak denote the numberof faces with k sides, and let M be the maximum number of sides any face has.

Suppose that ak ≤ n for all k and that M > 8. Note that each edge is part of exactly two faces, andeach vertex is part of at least three faces. It follows that

M∑k=3

ak = F

M∑k=3

kak2

= E

M∑k=3

kak3≥ V

and in particular

M∑k=3

ak

(1− k

2+k

3

)≥ F − E + V = 2

by Euler’s formula. But on the other hand, ak ≤ n by assumption, so

2 ≤M∑k=3

ak

(1− k

6

)

≤M∑k=3

n

(1− k

6

)

=

8∑k=3

n

(1− k

6

)+

M∑k=9

n

(1− k

6

)≤ 1

2n− 1

2n(M − 8)

where the last step follows from the fact that 1− k6 ≤ −

12 for k ≥ 9. Thus

2 ≤ 9

2n− 1

2nM =⇒ M ≤

92n− 2

12n

< 9

contradicting the fact that M > 8. It follows that M ≤ 8, and as each face has at least 3 edges, theresult follows directly from Pigeonhole.

4. [35] Let w = w1w2 . . . wn be a word. Define a substring of w to be a word of the form wiwi+1 . . . wj→1wj ,for some pair of positive integers 1 ≤ i ≤ j ≤ n. Show that w has at most n distinct palindromicsubstrings.

For example, aaaaa has 5 distinct palindromic substrings, and abcata has 5 (a, b, c, t, ata).

Proposed by: Yang Liu

For each palindrome substring appearing in w, consider only the leftmost position in which is appears.I claim that now, no two substrings share the same right endpoint. If some two do, then you can reflectthe smaller one about the center of the larger one to move the smaller one left.

5. [35] Let ABC be an acute triangle. The altitudes BE and CF intersect at the orthocenter H, andpoint O denotes the circumcenter. Point P is chosen so that ∠APH = ∠OPE = 90◦, and point Q ischosen so that ∠AQH = ∠OQF = 90◦. Lines EP and FQ meet at point T . Prove that points A, T,Oare collinear.

Proposed by: Evan Chen

Observe that T is the radical center of the circles with diameter OE, OF , AH. So T lies on the radicalaxis of (OE), (OF ) which is the altitude from O to EF , hence passing through A.

So ATO are collinear, done.

6. [40] Let r be a positive integer. Show that if a graph G has no cycles of length at most 2r, then it hasat most |V |2016 cycles of length exactly 2016r, where |V | denotes the number of vertices in the graphG.

Proposed by: Yang Liu

The key idea is that there is at most 1 path of length r between any pair of vertices, or else you geta cycle of length ≤ 2r. Now, start at any vertex (|V | choices) and walk 2015 times. There’s at most|V |2016 ways to do this by the previous argument. Now you have to go from the end to the start, andthere’s only one way to do this. So we’re done.

7. [45] Let p be a prime. A complete residue class modulo p is a set containing at least one elementequivalent to k (mod p) for all k.

(a) (20) Show that there exists an n such that the nth row of Pascal’s triangle forms a completeresidue class modulo p.

(b) (25) Show that there exists an n ≤ p2 such that the nth row of Pascal’s triangle forms a completeresidue class modulo p.

Proposed by: Alexander Katz

We use the following theorem of Lucas:

Theorem. Given a prime p and nonnegative integers a, b written in base p as a = anan→1 . . . a0p and

b = bnbn→1 . . . b0p respectively, where 0 ≤ ai, bi ≤ p− 1 for 0 ≤ i ≤ n, we have(a

b

)=

n∏i=0

(aibi

)(mod p).

Now, let n = (p − 1) × p + (p − 2) = p2 − 2. For k = pq + r with 0 ≤ q, r ≤ p − 1, applying Lucas’stheorem gives (

n

k

)≡(p− 1

q

)(p− 2

r

)(mod p).

Note that (p− 1

q

)=

q∏i=1

p− ii≡ (−1)q (mod p),

and (p− 2

r

)=

r∏i=1

p− 1− ii

≡ (−1)r(r + 1)!

r!= (−1)r(r + 1) (mod p).

So for 2 ≤ i ≤ p we can take k = (p + 1)(i − 1) and obtainnk

)≡ i (mod p), while for i = 1 we can

take k = 0. Thus this row satisfies the desired property.

8. [45] Does there exist an irrational number α > 1 such that

bαnc ≡ 0 (mod 2017)

for all integers n ≥ 1?

Proposed by: Sam Korsky

Answer: Yes

Let α > 1 and 0 < β < 1 be the roots of x2 − 4035x+ 2017. Then note that bαnc = αn + βn − 1. Letxn = αn +βn for all nonnegative integers n. It’s easy to verify that xn = 4035xn→1−2017xn→2 ≡ xn→1(mod 2017) so since x1 = 4035 ≡ 1 (mod 2017) we have that xn ≡ 1 (mod 2017) for all n. Thus αsatisfies the problem.

9. [65] Let n be a positive odd integer greater than 2, and consider a regular n-gon G in the plane centeredat the origin. Let a subpolygon G′ be a polygon with at least 3 vertices whose vertex set is a subset ofthat of G. Say G′ is well-centered if its centroid is the origin. Also, say G′ is decomposable if its vertexset can be written as the disjoint union of regular polygons with at least 3 vertices. Show that allwell-centered subpolygons are decomposable if and only if n has at most two distinct prime divisors.

Proposed by: Yang Liu

⇒, i.e. n has ≥ 3 prime divisors: Let n =∏peii . Note it suffices to only consider regular pi-gons.

Label the vertices of the n-gon 0, 1, . . . , n− 1. Let S = {xnp1: 0 ≤ x ≤ p1 − 1}, and let Sj = S + jn

p3for

0 ≤ j ≤ p3 − 2. (S + a = {s+ a : s ∈ S}.) Then let Sp3→1 = {xnp2: 0 ≤ x ≤ p2 − 1}+ (p3→1)n

p3. Finally,

let S′ = {xnp3: 0 ≤ x ≤ p3 − 1}. Then I claim(

p3→1Gi=0

Si

)\S′

is well-centered but not decomposable. Well-centered follows from the construction: I only added andsubtracted off regular polygons. To show that its decomposable, consider n

p1. Clearly this is in the set,

but isn’t in S′. I claim that np1

isn’t in any more regular pi-gons. For i ≥ 4, this means that np1

+ npi

isin some set. But this is a contradiction, as we can easily check that all points we added in are multiplesof peii , while n

piisn’t.

For i = 1, note that 0 was removed by S′. For i = 2, note that the only multiples of pe33 that are in

some Sj are 0, np1, . . . , (p1→1)n

p1. In particular, n

p1+ n

p2isn’t in any Sj . So it suffices to consider the case

i = 3, but it is easy to show that np1

+ (p3→1)np3

isn’t in any Si. So we’re done.

⇐, i.e. n has ≤ 2 prime divisors: This part seems to require knowledge of cyclotomic polynomials.These will easily give a solution in the case n = pa. Now, instead turn to the case n = paqb. The nextlemma is the key ingredient to the solution.

Lemma: Every well-centered subpolygon can be gotten by adding in and subtracting off regularpolygons.

Note that this is weaker than the problem claim, as the problem claims that adding in polygons isenough.

Proof. It is easy to verify that φn(x) = (xn→1)(xnpq →1)

(xnp →1)(x

nq →1)

. Therefore, it suffices to check that there exist

integer polynomials c(x), d(x) such that

xn − 1

xnp − 1

· c(x) +xn − 1

xnq − 1

· d(x) =(xn − 1)(x

npq − 1)

(xnp − 1)(x

nq − 1)

.

Rearranging means that we want

(xnq − 1) · c(x) + (x

np − 1) · d(x) = x

npq − 1.

But now, since gcd(n/p, n/q) = n/pq, there exist positive integers s, t such that snq −

tnp = n

pq . Now

choose c(x) = xsnq →1

xnq →1

, d(x) = xsnq →x

npq

xnp →1

to finish.

Now we can finish combinatorially. Say we need subtraction, and at some point we subtract off a p-gon.All the points in the p-gon must have been added at some point. If any of them was added from ap-gon, we could just cancel both p-gons. If they all came from a q-gon, then the sum of those p q-gonswould be a pq-gon, which could have been instead written as the sum of q p-gons. So we don’t needsubtraction either way. This completes the proof.

10. [65] Let LBC be a fixed triangle with LB = LC, and let A be a variable point on arc LB of itscircumcircle. Let I be the incenter of 4ABC and AK the altitude from A. The circumcircle of 4IKLintersects lines KA and BC again at U 6= K and V 6= K. Finally, let T be the projection of I onto lineUV . Prove that the line through T and the midpoint of IK passes through a fixed point as A varies.

Proposed by: Sam Korsky

Answer:

Let M be the midpoint of arc BC not containing L and let D be the point where the incircle of triangleABC touches BC. Also let N be the projection from I to AK. We claim that M is the desired fixedpoint.

By Simson’s Theorem on triangle KUV and point I we have that points T,D,N are collinear andsince quadrilateral NKDI is a rectangle we have that line DN passes through the midpoint of IK.Thus it suffices to show that M lies on line DN .

Now, let Ia, Ib, Ic be the A,B,C-excenters of triangle ABC respectively. Then I is the orthocenter oftriangle IaIbIc and ABC is the Cevian triangle of I with respect to triangle IaIbIc. It’s also well-knownthat M is the midpoint of IIa.

Let D′ be the reflection of I over BC and let N ′ be the reflection of I over AK. Clearly K is themidpoint of D′N ′. If we could prove that Ia, D

′,K,N ′ were collinear then by taking a homothetycentered at I with ratio 1

2 we would have that points M,D,N were collinear as desired. Thus itsuffices to show that points Ia, D

′,K are collinear.

Let lines BC and IbIc intersect at R and let lines AI and BC intersect at S. Then it’s well-known that(Ib, Ic;A,R) is harmonic and projecting from C we have that (Ia, I;S,A) is harmonic. But KS ⊥ KAwhich means that KS bisects angle ∠IKIa. But it’s clear by the definition of D′ that KS bisectsangle ∠IKD′ which implies that points Ia,K,D

′ are collinear as desired. This completes the proof.

February 2017February 18, 2017

Algebra and Number Theory

1. Let Q(x) = a0 + a1x + · · · + anxn be a polynomial with integer coefficients, and 0 ≤ ai < 3 for all

0 ≤ i ≤ n.Given that Q(

√3) = 20 + 17

√3, compute Q(2).

2. Find the value of

∑1≤a<b<c

1

2a3b5c

(i.e. the sum of 12a3b5c

over all triples of positive integers (a, b, c) satisfying a < b < c)

3. Let f : R→ R be a function satisfying f(x)f(y) = f(x− y). Find all possible values of f(2017).

4. Find all pairs (a, b) of positive integers such that a2017 + b is a multiple of ab.

5. Kelvin the Frog was bored in math class one day, so he wrote all ordered triples (a, b, c) of positiveintegers such that abc = 2310 on a sheet of paper. Find the sum of all the integers he wrote down. Inother words, compute ∑

abc=2310a,b,c∈N

(a+ b+ c),

where N denotes the positive integers.

6. A polynomial P of degree 2015 satisfies the equation P (n) = 1n2 for n = 1, 2, . . . , 2016. Find

b2017P (2017)c.

7. Determine the largest real number c such that for any 2017 real numbers x1, x2, . . . , x2017, the inequality

2016∑i=1

xi(xi + xi+1) ≥ c · x22017

holds.

8. Consider all ordered pairs of integers (a, b) such that 1 ≤ a ≤ b ≤ 100 and

(a+ b)(a+ b+ 1)

ab

is an integer.

Among these pairs, find the one with largest value of b. If multiple pairs have this maximal value ofb, choose the one with largest a. For example choose (3, 85) over (2, 85) over (4, 84). Note that youranswer should be an ordered pair.

9. The Fibonacci sequence is defined as follows: F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for all integersn ≥ 2. Find the smallest positive integer m such that Fm ≡ 0 (mod 127) and Fm+1 ≡ 1 (mod 127).

10. Let N denote the natural numbers. Compute the number of functions f : N→ {0, 1, . . . , 16} such that

f(x+ 17) = f(x) and f(x2) ≡ f(x)2 + 15 (mod 17)

for all integers x ≥ 1.

February 2017February 18, 2017

Algebra and Number Theory

1. Let Q(x) = a0 + a1x + · · · + anxn be a polynomial with integer coefficients, and 0 ≤ ai < 3 for all

0 ≤ i ≤ n.Given that Q(

√3) = 20 + 17

√3, compute Q(2).

Proposed by: Yang Liu

Answer: 86

One can evaluate

Q(√

3) = (a0 + 3a2 + 32a4 + . . . ) + (a1 + 3a3 + 32a5 + . . . )√

3.

Therefore, we have that

(a0 + 3a2 + 32a4 + . . . ) = 20 and (a1 + 3a3 + 32a5 + . . . ) = 17.

This corresponds to the base-3 expansions of 20 and 17. This gives us that Q(x) = 2+2x+2x3+2x4+x5,so Q(2) = 86.

2. Find the value of

∑1≤a<b<c

1

2a3b5c

(i.e. the sum of 12a3b5c

over all triples of positive integers (a, b, c) satisfying a < b < c)

Proposed by: Alexander Katz

Answer: 1/1624

Let x = b− a and y = c− b so that b = a+ x and c = a+ x+ y. Then

2a3b5c = 2a3a+x5a+x+y = 30a15x5y

and a, x, y are any positive integers. Thus

∑1≤a≤b<c

1

2a3b5c=

∑1≤a,x,y

1

30a15x5y

=∑1≤a

1

30a

∑1≤x

1

15x

∑1≤y

1

5y

=1

29· 1

14· 1

4

=1

1624

3. Let f : R→ R be a function satisfying f(x)f(y) = f(x− y). Find all possible values of f(2017).

Proposed by: Alexander Katz

Let P (x, y) be the given assertion. From P (0, 0) we get f(0)2 = f(0) =⇒ f(0) = 0, 1.

From P (x, x) we get f(x)2 = f(0). Thus, if f(0) = 0, we have f(x) = 0 for all x, which satisfies thegiven constraints. Thus f(2017) = 0 is one possibility.

Now suppose f(0) = 1. We then have P (0, y) =⇒ f(−y) = f(y), so that P (x,−y) =⇒ f(x)f(y) =f(x − y) = f(x)f(−y) = f(x + y). Thus f(x − y) = f(x + y), and in particular f(0) = f

x2 −

x2

)=

fx2 + x

2

)= f(x). It follows that f(x) = 1 for all x, which also satisfies all given constraints.

Thus the two possibilities are f(2017) = 0, 1 .

4. Find all pairs (a, b) of positive integers such that a2017 + b is a multiple of ab.

Proposed by: Yang Liu

Answer: (1, 1) and (2, 22017).

We want ab|a2017 + b. This gives that a|b. Therefore, we can set b = b2017a. Substituting this givesb2017a

2|a2017 + b2017a, so b2017a|a2016 + b2017. Once again, we get a|b2017, so we can set b2017 = b2016a.Continuing this way, if we have bi+1a|ai + bi+1, then a|bi+1, so we can set bi+1 = bia and derivebia|ai−1 + bi. Continuing down to i = 1, we would have b = b1a

2017 so ab1|1 + b1. If a ≥ 3, thenab1 > 1 + b1 for all b1 ≥ 1, so we need either a = 1 or a = 2. If a = 1, then b|b + 1, so b = 1. Thisgives the pair (1, 1). If a = 2, we need 2b|b + 22017. Therefore, we get b|22017, so we can write b = 2k

for 0 ≤ k ≤ 2017. Then we need 2k+1|2k + 22017. As k ≤ 2017, we need 2|1 + 22017−k. This can onlyhappen is k = 2017. This gives the pair (2, 22017).

5. Kelvin the Frog was bored in math class one day, so he wrote all ordered triples (a, b, c) of positiveintegers such that abc = 2310 on a sheet of paper. Find the sum of all the integers he wrote down. Inother words, compute ∑

abc=2310a,b,c∈N

(a+ b+ c),

where N denotes the positive integers.

Proposed by: Yang Liu

Answer: 49140

Note that 2310 = 2 · 3 · 5 · 7 · 11. The given sum clearly equals 3∑abc=2310 a by symmetry. The inner

sum can be rewritten as ∑a|2310

a · τ(

2310

a

),

as for any fixed a, there are τ2310a

)choices for the integers b, c.

Now consider the function f(n) =∑a|n a · τ

na

). Therefore, f = n ∗ τ , where n denotes the function

g(n) = n and ∗ denotes Dirichlet convolution. As both n and τ are multiplicative, f is also multiplica-tive.

It is easy to compute that f(p) = p+ 2 for primes p. Therefore, our final answer is 3(2 + 2)(3 + 2)(5 +2)(7 + 2)(11 + 2) = 49140.

6. A polynomial P of degree 2015 satisfies the equation P (n) = 1n2 for n = 1, 2, . . . , 2016. Find

b2017P (2017)c.Proposed by: Alexander Katz

Answer: −9

Let Q(x) = x2P (x)− 1. Then Q(n) = n2P (n)− 1 = 0 for n = 1, 2, . . . , 2016, and Q has degree 2017.Thus we may write

Q(x) = x2P (x)− 1 = (x− 1)(x− 2) . . . (x− 2016)L(x)

where L(x) is some linear polynomial. Then Q(0) = −1 = (−1)(−2) . . . (−2016)L(0), so L(0) = − 12016! .

Now note that

Q′(x) = x2P ′(x) + 2xP (x)

=

2016∑i=1

(x− 1) . . . (x− (i− 1))(x− (i+ 1)) . . . (x− 2016)L(x) + (x− 1)(x− 2) . . . (x− 2016)L′(x)

Thus

Q′(0) = 0 = L(0)

(2016!

−1+

2016!

−2+ . . .+

2016!

−2016

)+ 2016!L′(0)

whence L′(0) = L(0)11 + 1

2 + . . .+ 12016

)= −H2016

2016! , where Hn denotes the nth harmonic number.

As a result, we have L(x) = −H2016x+12016! . Then

Q(2017) = 20172P (2017)− 1 = 2016!

(−2017H2016 + 1

2016!

)which is −2017H2016 − 1. Thus

P (2017) =−H2016

2017.

From which we get 2017P (2017) = −H2016. It remains to approximate H2016. We alter the well knownapproximation

Hn ≈∫ n

1

1

xdx = log x

to

Hn ≈ 1 +1

2+

∫ n

3

1

xdx = 1 +

1

2+ log(2016)− log(3) ≈ log(2016) +

1

2

so that it suffices to lower bound log(2016). Note that e3 ≈ 20, which is close enough for our purposes.Then e6 ≈ 400 =⇒ e7 ≈ 1080, and e3 ≈ 20 < 25 =⇒ e0.6 << 2 =⇒ e7.6 < 2016, so thatlog(2016) > 7.6. It follows that H2016 ≈ log(2016) + 0.5 = 7.6 + 0.5 > 8 (of course these are looseestimates, but more than good enough for our purposes). Thus −9 < 2017P (2017) < −8, making our

answer −9 .

Alternatively, a well-read contestant might know that Hn ≈ log n + γ, where γ ≈ .577 is the Euler-Mascheroni constant. The above solution essentially approximates γ as 0.5 which is good enough forour purposes.

7. Determine the largest real number c such that for any 2017 real numbers x1, x2, . . . , x2017, the inequality

2016∑i=1

xi(xi + xi+1) ≥ c · x22017

holds.

Proposed by: Pakawut Jiradilok

Answer: − 10082017

Let n = 2016. Define a sequence of real numbers {pk} by p1 = 0, and for all k ≥ 1,

pk+1 =1

4(1− pk).

Note that, for every i ≥ 1,

(1− pi) · x2i + xixi+1 + pi+1x2i+1 =

(xi

2√pi+1

+√pi+1xi+1

)2

≥ 0.

Summing from i = 1 to n gives

n∑i=1

xi(xi + xi+1) ≥ −pn+1x2n+1.

One can show by induction that pk = k−12k . Therefore, our answer is −p2017 = − 1008

2017 .

8. Consider all ordered pairs of integers (a, b) such that 1 ≤ a ≤ b ≤ 100 and

(a+ b)(a+ b+ 1)

ab

is an integer.

Among these pairs, find the one with largest value of b. If multiple pairs have this maximal value ofb, choose the one with largest a. For example choose (3, 85) over (2, 85) over (4, 84). Note that youranswer should be an ordered pair.

Proposed by: Alexander Katz

Answer: (35,90)

Firstly note that (a+b)(a+b+1)ab = 2+ a2+b2+a+b

ab . Let c be this fraction so that (a+b)(a+b+1) = ab(c+2)for some integers a, b, c. Suppose (a, b) with a ≥ b is a solution for some c. Consider the quadratic

x2 − (bc− 1)x+ b2 + b = 0

It has one root a, and the other root is therefore bc − a − 1. Furthermore the other root can also be

expressed as b2+ba ≤ b2+b

b+1 = b, so that 0 < bc− a− 1 ≤ b. In particular, (b, bc− a− 1) is a solution aswell.

Thus all solutions (a, b) reduce to a solution where a = b, at which point c = 2 + 2a . Since a, c are

positive integers we thus have a = 1, 2, and so c = 3, 4 .

Through this jumping process, we iteratively find the solutions for c = 3:

(2, 2)→ (2, 3)→ (3, 6)→ (6, 14)→ (14, 35)→ (35, 90)

and c = 4:

(1, 2)→ (2, 6)→ (6, 21)→ (21, 77)

so that the desired pair is (35, 90) .

9. The Fibonacci sequence is defined as follows: F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for all integersn ≥ 2. Find the smallest positive integer m such that Fm ≡ 0 (mod 127) and Fm+1 ≡ 1 (mod 127).

Proposed by: Sam Korsky

Answer: 256

First, note that 5 is not a quadratic residue modulo 127. We are looking for the period of the Fibonaccinumbers mod 127. Let p = 127. We work in Fp2 for the remainder of this proof. Let α and β be the

roots of x2 − x − 1. Then we know that Fn = αn−βn

α−β . Note that since x → xp is an automorphismand since automorphisms cycle the roots of a polynomial we have that αp = β and βp = α. ThenFp = αp−βp

α−β = −1 and Fp+1 = αβ−βαα−β = 0 and similarly we obtain F2p+1 = 1 and F2p+2 = 0. Thus

since 2p + 2 is a power of 2 and since the period does not divide p + 1, we must have the answer is2p+ 2 = 256 .

10. Let N denote the natural numbers. Compute the number of functions f : N→ {0, 1, . . . , 16} such that

f(x+ 17) = f(x) and f(x2) ≡ f(x)2 + 15 (mod 17)

for all integers x ≥ 1.

Proposed by: Yang Liu

Answer: 12066

By plugging in x = 0, we get that f(0) can be either −1, 2. As f(0) is unrelated to all other values,we need to remember to multiply our answer by 2 at the end. Similarly, f(1) = −1 or 2.

Consider the graph x→ x2. It is a binary tree rooted at −1, and there is an edge −1→ 1, and a loop1 → 1. Our first case is f(1) = −1. Note that if x, y satisfy x2 = y, then f(y) 6= 1. Otherwise, wewould have f(x)2 = 3 (mod 17), a contradiction as 3 is a nonresidue. So only the 8 leaves can takethe value 1. This contributes 28.

For f(1) = 2, we can once again propagate down the tree. While it looks like we have 2 choices at eachnode (for the square roots), this is wrong, as if f(x) = −2 and y2 = x, then f(y) = 0 is forced.

Given this intuition, let an denote the answer for a binary tree of height n where the top is either −2or 2. Therefore, a1 = 2, a2 = 5. You can show the recurrence an = a2n−1 + 22

n−4. This is because if thetop is 2, then we get a contribution of a2n−1. If the top is −2, then both entries below it must be 0.After that, you can show that each of the remaining 2n − 4 vertices can be either of 2 possible squareroots. Therefore, we get the recurrence as claimed. One can compute that a4 = 5777, so we get thefinal answer 2(256 + 5777) = 12066.

February 2017February 18, 2017

Combinatorics

1. Kelvin the Frog is going to roll three fair ten-sided dice with faces labelled 0, 1, 2, . . . , 9. First he rollstwo dice, and finds the sum of the two rolls. Then he rolls the third die. What is the probability thatthe sum of the first two rolls equals the third roll?

2. How many ways are there to insert +’s between the digits of 111111111111111 (fifteen 1’s) so that theresult will be a multiple of 30?

3. There are 2017 jars in a row on a table, initially empty. Each day, a nice man picks ten consecutivejars and deposits one coin in each of the ten jars. Later, Kelvin the Frog comes back to see that N ofthe jars all contain the same positive integer number of coins (i.e. there is an integer d > 0 such thatN of the jars have exactly d coins). What is the maximum possible value of N?

4. Sam spends his days walking around the following 2× 2 grid of squares.

1 24 3

Say that two squares are adjacent if they share a side. He starts at the square labeled 1 and everysecond walks to an adjacent square. How many paths can Sam take so that the sum of the numberson every square he visits in his path is equal to 20 (not counting the square he started on)?

5. Kelvin the Frog likes numbers whose digits strictly decrease, but numbers that violate this conditionin at most one place are good enough. In other words, if di denotes the ith digit, then di ≤ di+1 forat most one value of i. For example, Kelvin likes the numbers 43210, 132, and 3, but not the numbers1337 and 123. How many 5-digit numbers does Kelvin like?

6. Emily starts with an empty bucket. Every second, she either adds a stone to the bucket or removes astone from the bucket, each with probability 1

2 . If she wants to remove a stone from the bucket andthe bucket is currently empty, she merely does nothing for that second (still with probability 1

2 ). Whatis the probability that after 2017 seconds her bucket contains exactly 1337 stones?

7. There are 2017 frogs and 2017 toads in a room. Each frog is friends with exactly 2 distinct toads. LetN be the number of ways to pair every frog with a toad who is its friend, so that no toad is pairedwith more than one frog. Let D be the number of distinct possible values of N , and let S be the sumof all possible values of N . Find the ordered pair (D,S).

8. Kelvin and 15 other frogs are in a meeting, for a total of 16 frogs. During the meeting, each pair ofdistinct frogs becomes friends with probability 1

2 . Kelvin thinks the situation after the meeting is coolif for each of the 16 frogs, the number of friends they made during the meeting is a multiple of 4. Saythat the probability of the situation being cool can be expressed in the form a

b , where a and b arerelatively prime. Find a.

9. Let m be a positive integer, and let T denote the set of all subsets of {1, 2, . . . ,m}. Call a subset Sof T δ-good if for all s1, s2 ∈ S, s1 6= s2, |∆(s1, s2)| ≥ δm, where ∆ denotes symmetric difference (thesymmetric difference of two sets is the set of elements that is in exactly one of the two sets). Find thelargest possible integer s such that there exists an integer m and a 1024

2047 -good set of size s.

10. Compute the number of possible words w = w1w2 . . . w100 satisfying:

• w has exactly 50 A’s and 50 B’s (and no other letters).

• For i = 1, 2, . . . , 100, the number of A’s among w1, w2, . . . , wi is at most the number of B’s amongw1, w2, . . . , wi.

• For all i = 44, 45, . . . , 57, if wi is an B, then wi+1 must be an B.

February 2017February 18, 2017

Combinatorics

1. Kelvin the Frog is going to roll three fair ten-sided dice with faces labelled 0, 1, 2, . . . , 9. First he rollstwo dice, and finds the sum of the two rolls. Then he rolls the third die. What is the probability thatthe sum of the first two rolls equals the third roll?

Proposed by: Yang Liu

Answer: 11200

First, there are 103 = 1000 triples (a, b, c). Now, we should count how many of these triples satisfya+ b = c. If c = 0, we get 1 triple (0, 0, 0). If c = 1, we get two triples (1, 0, 1) and (0, 1, 1). Continuing,this gives that the total number of triples is 1 + 2 + · · · + 10 = 55. Therefore, our final answer is55

1000 = 11200 .

2. How many ways are there to insert +’s between the digits of 111111111111111 (fifteen 1’s) so that theresult will be a multiple of 30?

Proposed by: Yang Liu

Answer: 2002

Note that because there are 15 1’s, no matter how we insert +’s, the result will always be a multipleof 3. Therefore, it suffices to consider adding +’s to get a multiple of 10. By looking at the unitsdigit, we need the number of summands to be a multiple of 10. Because there are only 15 digits inour number, we have to have exactly 10 summands. Therefore, we need to insert 9 +’s in 14 possiblepositions, giving an answer of

149

)= 2002.

3. There are 2017 jars in a row on a table, initially empty. Each day, a nice man picks ten consecutivejars and deposits one coin in each of the ten jars. Later, Kelvin the Frog comes back to see that N ofthe jars all contain the same positive integer number of coins (i.e. there is an integer d > 0 such thatN of the jars have exactly d coins). What is the maximum possible value of N?

Proposed by: Kevin Sun

Answer: 2014

Label the jars 1, 2, . . . , 2017. I claim that the answer is 2014. To show this, we need both a constructionand an upper bound. For the construction, for 1 ≤ i ≤ 201, put a coin in the jars 10i + 1, 10i +2, . . . , 10i+ 10. After this, each of the jars 1, 2, . . . , 2010 has exactly one coin. Now, put a coin in eachof the jars 2008, 2009, . . . , 2017. Now, the jars 1, 2, . . . , 2007, 2011, 2012, . . . , 2017 all have exactly onecoin. This gives a construction for N = 2014 (where d = 1).

Now, we show that this is optimal. Let c1, c2, . . . , c2017 denote the number of coins in each of the jars.For 1 ≤ j ≤ 10, define

sj = cj + cj+10 + cj+20 + . . . .

Note that throughout the process, s1 = s2 = · · · = sj . It is also easy to check that the sums s1, s2, . . . , s7each involve 202 jars, while the sums s8, s9, s10 each involve 201 jars.

Call a jar good if it has exactly d coins. If there are at least 2015 good jars, then one can check that it isforced that at least one of s1, s2, . . . , s7 only involves good jars, and similarly, at least one of s8, s9, s10only involves good jars. But this would mean that 202d = 201d as all si are equal, contradiction.

4. Sam spends his days walking around the following 2× 2 grid of squares.

1 24 3

Say that two squares are adjacent if they share a side. He starts at the square labeled 1 and everysecond walks to an adjacent square. How many paths can Sam take so that the sum of the numberson every square he visits in his path is equal to 20 (not counting the square he started on)?

Proposed by: Sam Korsky

Answer: 167

Note that on the first step, Sam can either step on 2 or 4. On the second step, Sam can either stepon 1 or 3, regardless of whether he is on 2 or 4. Now, for example, say that Sam takes 8 steps. Histotal sum will be 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 2a, where a is the number of times that he decides tostep on the larger number of his two choices. Solving gives a = 4. As he took 8 steps, this gives him84

)= 70 ways in this case.

We can follow a similar approach by doing casework on the number of steps he takes. I will simply listthem out here for brevity. For 8 steps, we get

84

)= 70. For 9 steps, we get

93

)= 84. For 12 steps, we

get a contribution on121

)= 12. For 13 steps, we get a contribution of

130

)= 1. Therefore, the final

answer is 70 + 84 + 12 + 1 = 167.

5. Kelvin the Frog likes numbers whose digits strictly decrease, but numbers that violate this conditionin at most one place are good enough. In other words, if di denotes the ith digit, then di ≤ di+1 forat most one value of i. For example, Kelvin likes the numbers 43210, 132, and 3, but not the numbers1337 and 123. How many 5-digit numbers does Kelvin like?

Proposed by: Alexander Katz

Answer: 14034

Suppose first that no digit violates the constraint; i.e. the digits are in strictly decreasing order. Thereare

105

)ways to choose the digits of the number, and each set of digits can be arranged in exactly one

way, so there are105

)such numbers.

We now perform casework on which digit violates the constraint. If it is the final digit, the first fourdigits must be arranged in decreasing order, which there are

104

)ways to do. The final digit can then

be any digit, but we have overcounted the ones in which the number is in fully decreasing order (thiscan happen, for example, if the first 4 digits we chose were 5, 4, 3, and 2 – a last digit of 1 was alreadycounted in the first case). Therefore, there are

104

)101

)− 252 new numbers in this case.

If the offending digit is second from the right, the first 3 digits must be decreasing, as must the last2 digits. There are

103

)102

)ways to do this. As before, we overcount the case where the second digit

from the right is not actually an offender, so we again overcount the case where all 5 digits decrease.Hence there are

103

)102

)− 252 new numbers in this case.

The case where the third digit is the offender is identical to the previous case, so there are another103

)102

)− 252 numbers to account for. The final case is when the second digit is the offending digit,

in which case there are104

)ways to choose the final 4 digits, but only 9 ways to choose the opening

digit (as 0 cannot be a leading digit). Accounting for the usual overcounting, our final answer is

252 +

[(10

4

)(10

1

)− 252

]+ 2

[(10

3

)(10

2

)− 252

]+

[(10

4

)· 9− 252

]which is easily calculated as 14034 .

6. Emily starts with an empty bucket. Every second, she either adds a stone to the bucket or removes astone from the bucket, each with probability 1

2 . If she wants to remove a stone from the bucket andthe bucket is currently empty, she merely does nothing for that second (still with probability 1

2 ). Whatis the probability that after 2017 seconds her bucket contains exactly 1337 stones?

Proposed by: Sam Korsky

Answer:(2017

340 )22017

Replace 2017 with n and 1337 with k and denote the general answer by f(n, k). I claim that f(n, k) =( n

bn−k2 c)2n . We proceed by induction on n.

The claim is obviously true for n = 0 since f(0, 0) = 1. Moreover, we have that f(n, 0) = 12f(n −

1, 0) + 12f(n− 1, 1) and f(n, k) = 1

2f(n− 1, k− 1) + 12f(n− 1, k+ 1) for k > 0 so the inductive step is

immediate by Pascal’s identity. This concludes the proof.

7. There are 2017 frogs and 2017 toads in a room. Each frog is friends with exactly 2 distinct toads. LetN be the number of ways to pair every frog with a toad who is its friend, so that no toad is pairedwith more than one frog. Let D be the number of distinct possible values of N , and let S be the sumof all possible values of N . Find the ordered pair (D,S).

Proposed by: Yang Liu

Answer: (1009, 21009 − 2)

I claim that N can equal 0 or 2i for 1 ≤ i ≤ 1008. We prove this now. Note that the average number offriends a toad has is also 2. If there is a toad with 0 friends, then clearly N = 0. If a toad has 1 friend,then it must be paired with its only friend, so we have reduced to a smaller case. Otherwise, all toadsand frogs have exactly degree 2, so the graph is a union of cycles. Each cycle can be paired off in ex-actly two ways. The number of cycles can range anywhere from 1 to 1008, and this completes the proof.

To construct all N = 21, 22, . . . , 21008, we can simply let our graph be a union of i cycles, which wouldhave 2i matchings. Clearly we can choose any i = 1, 2, . . . , 1008.

Therefore, D = 1009 and S = 21 + 22 + · · ·+ 21008 = 21009 − 2.

8. Kelvin and 15 other frogs are in a meeting, for a total of 16 frogs. During the meeting, each pair ofdistinct frogs becomes friends with probability 1

2 . Kelvin thinks the situation after the meeting is coolif for each of the 16 frogs, the number of friends they made during the meeting is a multiple of 4. Saythat the probability of the situation being cool can be expressed in the form a

b , where a and b arerelatively prime. Find a.

Proposed by: Yang Liu

Consider the multivariate polynomial ∏1≤i<j≤16

(1 + xixj)

We’re going to filter this by summing over all 416 16-tuples (x1, x2, . . . , x16) such that xj = ±1,±i.Most of these evaluate to 0 because i2 = (−i)2 = −1, and 1 · −1 = −1. If you do this filtering, you getthe following 4 cases:

Case 1: Neither of i or −i appears. Then the only cases we get are when all the xj are 1, or they’reall −1. Total is 2121. (120 =

162

).)

Case 2: i appears, but −i does not. Then all the remaining xj must be all 1 or all −1. This contributesa sum of (1 + i)15 · 2105 + (1− i)15 · 2105 = 2113. i can be at any position, so we get 16 · 2113.

Case 3: −i appears, but i does not. Same contribution as above. 16 · 2113.

Case 4: Both i and −i appear. Then all the rest of the xj must be all 1 or all −1. This contributes asum of 2 · (1 + i(−i)) · (1 + i)14 · (1− i)14 · 291 = 2107. i and −i can appear in 16 · 15 places, so we get240 · 2107.

So the final answer is this divided a factor for our filter. (416 = 232.) So our final answer is 289+16·282+240·2752120 =

1167241 .

Therefore, the answer is 1167.

9. Let m be a positive integer, and let T denote the set of all subsets of {1, 2, . . . ,m}. Call a subset Sof T δ-good if for all s1, s2 ∈ S, s1 6= s2, |∆(s1, s2)| ≥ δm, where ∆ denotes symmetric difference (thesymmetric difference of two sets is the set of elements that is in exactly one of the two sets). Find thelargest possible integer s such that there exists an integer m and a 1024

2047 -good set of size s.

Proposed by: Yang Liu

Answer: 2048

Let n = |S|. Let the sets in S be s1, s2, . . . , sn. We bound the sum∑

1≤i<j≤n |∆(si, sj)| in two ways.On one hand, by the condition we have the obvious bound∑

1≤i<j≤n

|∆(si, sj)| ≥(n

2

)δm.

On the other hand, for 1 ≤ i ≤ m, let ti = |{1 ≤ j ≤ n : i ∈ sj}|. Then it is clear that

∑1≤i<j≤n

|∆(si, sj)| =m∑

k=1

tk(n− tk) ≤ n2

4m

by AM-GM. Therefore, we get the bound(n

2

)δm ≤ n2

4m⇒ n ≤ 2δ

2δ − 1= 2048.

To give a construction with n = 2048, take m = 2047. For the rest of this construction, we will beinterpreting the integers 1, 2, . . . ,m as 11-digit integers in binary. Given this interpretation, define adot product x � y of two positive integers 0 ≤ x, y ≤ m the following way. If x = (x1x2 . . . x11)2, y =(y1y2 . . . y11)2 in binary, then

x� y =∑

xiyi (mod 2).

Now we can define the sets s1, s2, . . . , s2048. Define

si = {1 ≤ j ≤ m : (i− 1)� j = 1.}.

A computation shows that this construction works.

Some notes: here is the motivation behind the construction. We are treating the integers 0, 1, . . . ,m asthe vector space V = F11

2 , and the sets si correspond to linear functionals fi : V → F2. In particular,the function fi : V → F2 is simply defined as fi(x) = (i − 1) � x, which one can easily check to belinear. This construction corresponds to Hadamard matrices of size 211.

10. Compute the number of possible words w = w1w2 . . . w100 satisfying:

• w has exactly 50 A’s and 50 B’s (and no other letters).

• For i = 1, 2, . . . , 100, the number of A’s among w1, w2, . . . , wi is at most the number of B’s amongw1, w2, . . . , wi.

• For all i = 44, 45, . . . , 57, if wi is an B, then wi+1 must be an B.

Proposed by: Allen Liu

Call the last property in the problem statement P (i, j) where in the statement i = 44, j = 57. We showthat the number of words satisfying the first two conditions and P (m,m+ k) is the same independentof m (assuming k is fixed). It suffices to show that the number of words satisfying P (m− 1,m+ k− 1)is the same as the number of words satisfying P (m,m + k). Construct a bijection as follows: for aword satisfying P (m− 1,m+ k− 1), if it satisfies P (m,m+ k), leave it as is. Otherwise, the characterin position m + k must be B and the character in position m + k + 1 must be A. In this case, movethese two characters to positions m − 1,m (shifting all other characters back). It is not difficult toverify that this is indeed a bijection.

Thus, the problem is now equivalent to computing the number of words satisfying the first two condi-tions and P (1, 14). However, this condition simply means that the first 15 characters must be B. Nowwe are essentially counting the number of paths from (15, 0) to (50, 50) that don’t go above y = x.There is a bijection between paths from (15, 0) to (50, 50) that do cross y = x and paths from (15, 0)to (49, 51) (using the standard reflection argument). Thus the answer is

8535

)−8534

)

February 2017February 18, 2017

Geometry

1. Let A,B,C,D be four points on a circle in that order. Also, AB = 3, BC = 5, CD = 6, and DA = 4.Let diagonals AC and BD intersect at P . Compute AP

CP .

2. Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. Let ` be a line passing through twosides of triangle ABC. Line ` cuts triangle ABC into two figures, a triangle and a quadrilateral, thathave equal perimeter. What is the maximum possible area of the triangle?

3. Let S be a set of 2017 distinct points in the plane. Let R be the radius of the smallest circle containingall points in S on either the interior or boundary. Also, let D be the longest distance between two ofthe points in S. Let a, b are real numbers such that a ≤ D

R ≤ b for all possible sets S, where a is aslarge as possible and b is as small as possible. Find the pair (a, b).

4. Let ABCD be a convex quadrilateral with AB = 5, BC = 6, CD = 7, and DA = 8. Let M,P,N,Q bethe midpoints of sides AB,BC,CD,DA respectively. Compute MN2 − PQ2.

5. Let ABCD be a quadrilateral with an inscribed circle ω and let P be the intersection of its diagonalsAC and BD. Let R1, R2, R3, R4 be the circumradii of triangles APB, BPC, CPD, DPA respectively.If R1 = 31 and R2 = 24 and R3 = 12, find R4.

6. In convex quadrilateral ABCD we have AB = 15, BC = 16, CD = 12, DA = 25, and BD = 20. LetM and γ denote the circumcenter and circumcircle of 4ABD. Line CB meets γ again at F , line AFmeets MC at G, and line GD meets γ again at E. Determine the area of pentagon ABCDE.

7. Let ω and Γ by circles such that ω is internally tangent to Γ at a point P . Let AB be a chord of Γtangent to ω at a point Q. Let R 6= P be the second intersection of line PQ with Γ. If the radius of Γis 17, the radius of ω is 7, and AQ

BQ = 3, find the circumradius of triangle AQR.

8. Let ABC be a triangle with circumradius R = 17 and inradius r = 7. Find the maximum possiblevalue of sin A

2 .

9. Let ABC be a triangle, and let BCDE,CAFG,ABHI be squares that do not overlap the trianglewith centers X,Y, Z respectively. Given that AX = 6, BY = 7, and CZ = 8, find the area of triangleXY Z.

10. Let ABCD be a quadrilateral with an inscribed circle ω. Let I be the center of ω let IA = 12,IB = 16, IC = 14, and ID = 11. Let M be the midpoint of segment AC. Compute IM

IN , where N isthe midpoint of segment BD.

February 2017February 18, 2017

Geometry

1. Let A,B,C,D be four points on a circle in that order. Also, AB = 3, BC = 5, CD = 6, and DA = 4.Let diagonals AC and BD intersect at P . Compute AP

CP .

Proposed by: Sam Korsky

Answer: 25

Note that 4APB ∼ 4DPC so APAB = DP

CD . Similarly, 4BPC ∼ 4APD so CPBC = DP

DA . Dividing thesetwo equations yields

AP

CP=AB ·DABC · CD

=2

5

2. Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. Let ` be a line passing through twosides of triangle ABC. Line ` cuts triangle ABC into two figures, a triangle and a quadrilateral, thathave equal perimeter. What is the maximum possible area of the triangle?

Proposed by: Sam Korsky

Answer: 132326

There are three cases: ` intersects AB,AC, ` intersects AB,BC, and ` intersects AC,BC. These casesare essentially identical, so let ` intersect segment AB at M and segment AC at N .

Then the condition is equivalent to

AM +MN +AN = MB +BC + CN +MN

AM +AN = MB + CN + 15

but AN + CN = 14 and AM +BM = 13, so that

BM + CN = 27−AM −AN = AM −AN − 15

implying that AM +AN = 21.

Now let ∠BAC = θ for convenience, so that

[AMN ] =1

2AM ·AN · sin θ

which is maximized when AM = AN = 212 . Further we can easily calculate sin θ = 12

13 (e.g. by LOC);note that this is why the area is maximized in this case (we want to maximize sin θ, which is equivalentto maximizing θ, so θ should be opposite the largest side). Our answer is thus

1

2· 21

2· 21

2· 12

13=

1323

26

Alternatively we could also calculate

[AMN ] = [ABC] · AMAB

· ANAC

= 84 ·212

13·

212

14

which gives the same answer.

3. Let S be a set of 2017 distinct points in the plane. Let R be the radius of the smallest circle containingall points in S on either the interior or boundary. Also, let D be the longest distance between two ofthe points in S. Let a, b are real numbers such that a ≤ D

R ≤ b for all possible sets S, where a is aslarge as possible and b is as small as possible. Find the pair (a, b).

Proposed by: Yang Liu

Answer: (√

3, 2)

It is easy to verify that the smallest circle enclosing all the points will either have some 2 points in Sas its diameter, or will be the circumcircle of some 3 points in S who form an acute triangle.

Now, clearly DR ≤ 2. Indeed consider the two farthest pair of points S1, S2. Then D = |S1S2| ≤ 2R,

as both points S1, S2 are inside a circle of radius R. We can achieve this upper bound by taking S tohave essentially only 2 points, and the remaining 2015 points in S are at the same place as these 2 points.

For the other direction, I claim DR ≥

√3. Recall that the smallest circle is either the circumcircle of

3 points, or has some 2 points as the diameter. In the latter case, say the diameter is S1S2. ThenD ≥ |S1S2| = 2R, so D

R ≥ 2 in that case. Now say the points S1, S2, S3 are the circumcircle. WLOG,say that S1S2 is the longest side of the triangle. As remarked above, we can assume this triangle isacute. Therefore, π

3 ≤ ∠S1S3S2 ≤ π2 . By the Law of Sines we have that

D ≥ |S1S2| = 2R sin∠S1S3S2 ≥ 2R sinπ

3= R√

3.

This completes the proof. To achieve equality, we can take S to have 3 points in the shape of anequilateral triangle.

4. Let ABCD be a convex quadrilateral with AB = 5, BC = 6, CD = 7, and DA = 8. Let M,P,N,Q bethe midpoints of sides AB,BC,CD,DA respectively. Compute MN2 − PQ2.

Proposed by: Sam Korsky

Answer: 13

Draw in the diagonals of the quad and use the median formula three times to get MN2 in terms of thediagonals. Do the same for PQ2 and subtract, the diagonal length terms disappear and the answer is

BC2 +DA2 −AB2 − CD2

2= 13

5. Let ABCD be a quadrilateral with an inscribed circle ω and let P be the intersection of its diagonalsAC and BD. Let R1, R2, R3, R4 be the circumradii of triangles APB, BPC, CPD, DPA respectively.If R1 = 31 and R2 = 24 and R3 = 12, find R4.

Proposed by: Sam Korsky

Answer: 19

Note that ∠APB = 180◦ − ∠BPC = ∠CPD = 180◦ − ∠DPA so sinAPB = sinBPC = sinCPD =sinDPA. Now let ω touch sides AB,BC,CD,DA at E,F,G,H respectively. Then AB + CD =AE +BF + CG+DH = BC +DA so

AB

sinAPB+

CD

sinCPD=

BC

sinBPC+

DA

sinDPA

and by the Extended Law of Sines this implies

2R1 + 2R3 = 2R2 + 2R4

which immediately yields R4 = R1 +R3 −R2 = 19 .

6. In convex quadrilateral ABCD we have AB = 15, BC = 16, CD = 12, DA = 25, and BD = 20. LetM and γ denote the circumcenter and circumcircle of 4ABD. Line CB meets γ again at F , line AFmeets MC at G, and line GD meets γ again at E. Determine the area of pentagon ABCDE.

Proposed by: Evan Chen

Answer: 396

Note that ∠ADB = ∠DCB = 90◦ and BC ‖ AD. Now by Pascal theorem on DDEBFA implies thatB, M , E are collinear. So [ADE] = [ABD] = 150 and [BCD] = 96, so the total area is 396.

7. Let ω and Γ by circles such that ω is internally tangent to Γ at a point P . Let AB be a chord of Γtangent to ω at a point Q. Let R 6= P be the second intersection of line PQ with Γ. If the radius of Γis 17, the radius of ω is 7, and AQ

BQ = 3, find the circumradius of triangle AQR.

Proposed by: Sam Korsky

Answer:√

170

Let r denote the circumradius of triangle AQR. By Archimedes Lemma, R is the midpoint of arc ABof Γ. Therefore ∠RAQ = ∠RPB = ∠RPA so 4RAQ ∼ 4RPA. By looking at the similarity ratiobetween the two triangles we have

r

17=AQ

AP

Now, let AP intersect ω again at X 6= P . By homothety we have XQ ‖ AR so

AX

AP= 1− PQ

PR= 1− 7

17=

10

17

But we also knowAX ·AP = AQ2

so10

17AP 2 = AQ2

Thusr

17=AQ

AP=

√10

17

so we compute r =√

170 as desired.

8. Let ABC be a triangle with circumradius R = 17 and inradius r = 7. Find the maximum possiblevalue of sin A

2 .

Proposed by: Sam Korsky

Answer: 17+√51

34

Letting I and O denote the incenter and circumcenter of triangle ABC we have by the triangle in-equality that

AO ≤ AI +OI =⇒ R ≤ r

sin A2

+√R(R− 2r)

and by plugging in our values for r and R we get

sinA

2≤ 17 +

√51

34

as desired. Equality holds when ABC is isosceles and I lies between A and O.

9. Let ABC be a triangle, and let BCDE,CAFG,ABHI be squares that do not overlap the trianglewith centers X,Y, Z respectively. Given that AX = 6, BY = 7, and CZ = 8, find the area of triangleXY Z.

Proposed by: Sam Korsky

Answer: 21√15

4

By the degenerate case of Von Aubel’s Theorem we have that Y Z = AX = 6 and ZX = BY = 7 and

XY = CZ = 8 so it suffices to find the area of a 6− 7− 8 triangle which is given by21√

15

4.

To prove that AX = Y Z, note that by LoC we get

Y X2 =b2

2+c2

2+ bc sin∠A

and

AX2 = b2 +a2

2− ab(cos∠C − sin∠C)

= c2 +a2

2− ac(cos∠B − sin∠B)

=b2 + c2 + a(b sin∠C + c sin∠B)

2

=b2

2+c2

2+ ah

where h is the length of the A-altitude of triangle ABC. In these calculations we used the well-knownfact that b cos∠C + c cos∠B = a which can be easily seen by drawing in the A-altitude. Then sincebc sin∠A and ah both equal twice the area of triangle ABC, we are done.

10. Let ABCD be a quadrilateral with an inscribed circle ω. Let I be the center of ω let IA = 12,IB = 16, IC = 14, and ID = 11. Let M be the midpoint of segment AC. Compute IM

IN , where N isthe midpoint of segment BD.

Proposed by: Sam Korsky

Answer: 2122

Let points W,X, Y, Z be the tangency points between ω and lines AB,BC,CD,DA respectively. Nowinvert about ω. Then A′, B′, C ′, D′ are the midpoints of segments ZW,WX,XY, Y Z respectively.Thus by Varignon’s Theorem A′B′C ′D′ is a parallelogram. Then the midpoints of segments A′C ′ andB′D′ coincide at a point P . Note that figure IA′PC ′ is similar to figure ICMA with similitude ratior2

IA·IC where r is the radius of ω. Similarly figure IB′PD′ is similar to figure IDMB with similitude

ratio r2

IB·ID . Therefore

IP =r2

IA · IC· IM =

r2

IB · ID· IN

which yields

IM

IN=IA · ICIB · ID

=12 · 14

16 · 11=

21

22

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

February 2017, February 18, 2017 — GUTS ROUND

Organization Team Team ID#

1. [4] A random number generator will always output 7. Sam uses this random number generator once.What is the expected value of the output?

2. [4] Let A,B,C,D,E, F be 6 points on a circle in that order. Let X be the intersection of AD and BE,Y is the intersection of AD and CF , and Z is the intersection of CF and BE. X lies on segments BZand AY and Y lies on segment CZ. Given that AX = 3, BX = 2, CY = 4, DY = 10, EZ = 16, andFZ = 12, find the perimeter of triangle XY Z.

3. [4] Find the number of pairs of integers (x, y) such that x2 + 2y2 < 25.

4. [4] Find the number of ordered triples of nonnegative integers (a, b, c) that satisfy

(ab+ 1)(bc+ 1)(ca+ 1) = 84.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

February 2017, February 18, 2017 — GUTS ROUND

Organization Team Team ID#

5. [6] Find the number of ordered triples of positive integers (a, b, c) such that

6a+ 10b+ 15c = 3000.

6. [6] Let ABCD be a convex quadrilateral with AC = 7 and BD = 17. Let M,P,N,Q be the midpointsof sides AB,BC,CD,DA respectively. Compute MN2 + PQ2

7. [6] An ordered pair of sets (A,B) is good if A is not a subset of B and B is not a subset of A. Howmany ordered pairs of subsets of {1, 2, . . . , 2017} are good?

8. [6] You have 128 teams in a single elimination tournament. The Engineers and the Crimson are two ofthese teams. Each of the 128 teams in the tournament is equally strong, so during each match, eachteam has an equal probability of winning.

Now, the 128 teams are randomly put into the bracket.

What is the probability that the Engineers play the Crimson sometime during the tournament?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

February 2017, February 18, 2017 — GUTS ROUND

Organization Team Team ID#

9. [7] Jeffrey writes the numbers 1 and 100000000 = 108 on the blackboard. Every minute, if x, y are onthe board, Jeffrey replaces them with

x+ y

2and 2

(1

x+

1

y

)−1.

After 2017 minutes the two numbers are a and b. Find min(a, b) to the nearest integer.

10. [7] Let ABC be a triangle in the plane with AB = 13, BC = 14, AC = 15. Let Mn denote the smallest

possible value of (APn +BPn + CPn)1n over all points P in the plane. Find limn→∞Mn.

11. [7] Consider the graph in 3-space of

0 = xyz(x+ y)(y + z)(z + x)(x〉 y)(y 〉 z)(z 〉 x).

This graph divides 3-space into N connected regions. What is N?

12. [7] In a certain college containing 1000 students, students may choose to major in exactly one of math,computer science, finance, or English. The diversity ratio d(s) of a student s is the defined as numberof students in a different major from s divided by the number of students in the same major as s(including s). The diversity D of the college is the sum of all the diversity ratios d(s).

Determine all possible values of D.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

February 2017, February 18, 2017 — GUTS ROUND

Organization Team Team ID#

13. [9] The game of Penta is played with teams of five players each, and there are five roles the players canplay. Each of the five players chooses two of five roles they wish to play. If each player chooses theirroles randomly, what is the probability that each role will have exactly two players?

14. [9] Mrs. Toad has a class of 2017 students, with unhappiness levels 1, 2, . . . , 2017 respectively. Todayin class, there is a group project and Mrs. Toad wants to split the class in exactly 15 groups. Theunhappiness level of a group is the average unhappiness of its members, and the unhappiness of theclass is the sum of the unhappiness of all 15 groups. What’s the minimum unhappiness of the classMrs. Toad can achieve by splitting the class into 15 groups?

15. [9] Start by writing the integers 1, 2, 4, 6 on the blackboard. At each step, write the smallest positiveinteger n that satisfies both of the following properties on the board.

• n is larger than any integer on the board currently.

• n cannot be written as the sum of 2 distinct integers on the board.

Find the 100-th integer that you write on the board. Recall that at the beginning, there are already 4integers on the board.

16. [9] Let a and b be complex numbers satisfying the two equations

a3 〉 3ab2 = 36

b3 〉 3ba2 = 28i.

Let M be the maximum possible magnitude of a. Find all a such that |a| = M .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

February 2017, February 18, 2017 — GUTS ROUND

Organization Team Team ID#

17. [10] Sean is a biologist, and is looking at a string of length 66 composed of the letters A, T,C,G. Asubstring of a string is a contiguous sequence of letters in the string. For example, the string AGTC has10 substrings: A,G, T,C,AG,GT, TC,AGT,GTC,AGTC. What is the maximum number of distinctsubstrings of the string Sean is looking at?

18. [10] Let ABCD be a quadrilateral with side lengths AB = 2, BC = 3, CD = 5, and DA = 4. Whatis the maximum possible radius of a circle inscribed in quadrilateral ABCD?

19. [10] Find (in terms of n ≥ 1) the number of terms with odd coefficients after expanding the product:∏1≤i<j≤n

(xi + xj)

e.g., for n = 3 the expanded product is given by x21x2 + x21x3 + x22x3 + x22x1 + x23x1 + x23x2 + 2x1x2x3and so the answer would be 6.

20. [10] For positive integers a and N , let r(a,N) ∈ {0, 1, . . . , N 〉 1} denote the remainder of a whendivided by N . Determine the number of positive integers n ≤ 1000000 for which

r(n, 1000) > r(n, 1001).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

February 2017, February 18, 2017 — GUTS ROUND

Organization Team Team ID#

21. [12] Let P and A denote the perimeter and area respectively of a right triangle with relatively prime

integer side-lengths. Find the largest possible integral value of P 2

A

22. [12] Kelvin the Frog and 10 of his relatives are at a party. Every pair of frogs is either friendly orunfriendly. When 3 pairwise friendly frogs meet up, they will gossip about one another and end up ina fight (but stay friendly anyway). When 3 pairwise unfriendly frogs meet up, they will also end upin a fight. In all other cases, common ground is found and there is no fight. If all

113

)triples of frogs

meet up exactly once, what is the minimum possible number of fights?

23. [12] Five points are chosen uniformly at random on a segment of length 1. What is the expecteddistance between the closest pair of points?

24. [12] At a recent math contest, Evan was asked to find 22016 (mod p) for a given prime number p with100 < p < 500. Evan has forgotten what the prime p was, but still remembers how he solved it:

• Evan first tried taking 2016 modulo p〉 1, but got a value e larger than 100.

• However, Evan noted that e〉 12 (p〉 1) = 21, and then realized the answer was 〉221 (mod p).

What was the prime p?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

February 2017, February 18, 2017 — GUTS ROUND

Organization Team Team ID#

25. [15] Find all real numbers x satisfying the equation x3 〉 8 = 16 3√x+ 1.

26. [15] Kelvin the Frog is hopping on a number line (extending to infinity in both directions). Kelvinstarts at 0. Every minute, he has a 1

3 chance of moving 1 unit left, a 13 chance of moving 1 unit right

and 13 chance of getting eaten. Find the expected number of times Kelvin returns to 0 (not including

the start) before getting eaten.

27. [15] Find the smallest possible value of x + y where x, y ≥ 1 and x and y are integers that satisfyx2 〉 29y2 = 1

28. [15] Let . . . , a−1, a0, a1, a2, . . . be a sequence of positive integers satisfying the following relations:an = 0 for n < 0, a0 = 1, and for n ≥ 1,

an = an−1 + 2(n〉 1)an−2 + 9(n〉 1)(n〉 2)an−3 + 8(n〉 1)(n〉 2)(n〉 3)an−4.

Compute ∑n≥0

10nann!

.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

February 2017, February 18, 2017 — GUTS ROUND

Organization Team Team ID#

29. [17] Yang has the sequence of integers 1, 2, . . . , 2017. He makes 2016 swaps in order, where a swapchanges the positions of two integers in the sequence. His goal is to end with 2, 3, . . . , 2017, 1. Howmany different sequences of swaps can Yang do to achieve his goal?

30. [17] Consider an equilateral triangular grid G with 20 points on a side, where each row consists ofpoints spaced 1 unit apart. More specifically, there is a single point in the first row, two points in thesecond row, . . . , and 20 points in the last row, for a total of 210 points. Let S be a closed non-self-intersecting polygon which has 210 vertices, using each point in G exactly once. Find the sum of allpossible values of the area of S.

31. [17] A baseball league has 6 teams. To decide the schedule for the league, for each pair of teams, acoin is flipped. If it lands head, they will play a game this season, in which one team wins and oneteam loses. If it lands tails, they don’t play a game this season. Define the imbalance of this scheduleto be the minimum number of teams that will end up undefeated, i.e. lose 0 games. Find the expectedvalue of the imbalance in this league.

32. [17] Let a, b, c be non-negative real numbers such that ab+ bc+ ca = 3. Suppose that

a3b+ b3c+ c3a+ 2abc(a+ b+ c) =9

2.

What is the minimum possible value of ab3 + bc3 + ca3?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

February 2017, February 18, 2017 — GUTS ROUND

Organization Team Team ID#

33. [20] Welcome to the USAYNO, where each question has a yes/no answer. Choose any subset of thefollowing six problems to answer. If you answer n problems and get them all correct, you will receivemax(0, (n〉 1)(n〉 2)) points. If any of them are wrong, you will receive 0 points.

Your answer should be a six-character string containing ’Y’ (for yes), ’N’ (for no), or ’B’ (for blank).For instance if you think 1, 2, and 6 are ’yes’ and 3 and 4 are ’no’, you would answer YYNNBY (andreceive 12 points if all five answers are correct, 0 points if any are wrong).

(a) a, b, c, d, A,B,C, and D are positive real numbers such that ab >

AB and c

d >CD . Is it necessarily

true that a+cb+d >

A+CB+D?

(b) Do there exist irrational numbers α and β such that the sequence bαc+ bβc, b2αc+ b2βc, b3αc+b3βc, . . . is arithmetic?

(c) For any set of primes P, let SP denote the set of integers whose prime divisors all lie in P. Forinstance S{2,3} = {2a3b | a, b ≥ 0} = {1, 2, 3, 4, 6, 8, 9, 12, . . .}. Does there exist a finite set ofprimes P and integer polynomials P and Q such that gcd(P (x), Q(y)) ∈ SP for all x, y?

(d) A function f is called P-recursive if there exists a positive integer m and real polynomialsp0(n), p1(n), . . . , pm(n) satisfying

pm(n)f(n+m) = pm−1(n)f(n+m〉 1) + . . .+ p0(n)f(n)

for all n. Does there exist a P-recursive function f satisfying limn→∞f(n)

n√

2= 1?

(e) Does there exist a nonpolynomial function f : Z→ Z such that a〉 b divides f(a)〉 f(b) for allintegers a 6= b?

(f) Do there exist periodic functions f, g : R→ R such that f(x) + g(x) = x for all x?

34. [20]

(a) Can 1000 queens be placed on a 2017 × 2017 chessboard such that every square is attacked bysome queen? A square is attacked by a queen if it lies in the same row, column, or diagonal asthe queen.

(b) A 2017× 2017 grid of squares originally contains a 0 in each square. At any step, Kelvin the Frogchooses two adjacent squares (two squares are adjacent if they share a side) and increments thenumbers in both of them by 1. Can Kelvin make every square contain a different power of 2?

(c) A tournament consists of single games between every pair of players, where each game has awinner and loser with no ties. A set of people is dominated if there exists a player who beats allof them. Does there exist a tournament in which every set of 2017 people is dominated?

(d) Every cell of a 19 × 19 grid is colored either red, yellow, green, or blue. Does there necessarilyexist a rectangle whose sides are parallel to the grid, all of whose vertices are the same color?

(e) Does there exist a c ∈ R+ such that max(|A ·A|, |A+A|) ≥ c|A| log2 |A| for all finite sets A ⊂ Z?

(f) Can the set {1, 2, . . . , 1093} be partitioned into 7 subsets such that each subset is sum-free (i.e.no subset contains a, b, c with a+ b = c)?

35. [20]

(a) Does there exist a finite set of points, not all collinear, such that a line between any two pointsin the set passes through a third point in the set?

(b) Let ABC be a triangle and P be a point. The isogonal conjugate of P is the intersection of thereflection of line AP over the A-angle bisector, the reflection of line BP over the B-angle bisector,and the reflection of line CP over the C-angle bisector. Clearly the incenter is its own isogonalconjugate. Does there exist another point that is its own isogonal conjugate?

(c) Let F be a convex figure in a plane, and let P be the largest pentagon that can be inscribed inF . Is it necessarily true that the area of P is at least 3

4 the area of F?

(d) Is it possible to cut an equilateral triangle into 2017 pieces, and rearrange the pieces into a square?

(e) Let ABC be an acute triangle and P be a point in its interior. Let D,E, F lie on BC,CA,AB re-spectively so that PD bisects ∠BPC, PE bisects ∠CPA, and PF bisects ∠APB. Is it necessarilytrue that AP +BP + CP ≥ 2(PD + PE + PF )?

(f) Let P2018 be the surface area of the 2018-dimensional unit sphere, and let P2017 be the surfacearea of the 2017-dimensional unit sphere. Is P2018 > P2017?

36. [20]

(a) Does∑p−1

i=11i ≡ 0 (mod p2) for all odd prime numbers p? (Note that 1

i denotes the number suchthat i · 1i ≡ 1 (mod p2))

(b) Do there exist 2017 positive perfect cubes that sum to a perfect cube?

(c) Does there exist a right triangle with rational side lengths and area 5?

(d) A magic square is a 3× 3 grid of numbers, all of whose rows, columns, and major diagonals sumto the same value. Does there exist a magic square whose entries are all prime numbers?

(e) Is∏

pp2+1p2−1 = 22+1

22−1 ·32+132−1 ·

52+152−1 ·

72+172−1 · . . . a rational number?

(f) Do there exist an infinite number of pairs of distinct integers (a, b) such that a and b have thesame set of prime divisors, and a+ 1 and b+ 1 also have the same set of prime divisors?

February 2017February 18, 2017

Guts

1. [4] A random number generator will always output 7. Sam uses this random number generator once.What is the expected value of the output?

Proposed by: Sam Korsky

Answer: 7

The only output is 7, so the expected value is 7.

2. [4] Let A,B,C,D,E, F be 6 points on a circle in that order. Let X be the intersection of AD and BE,Y is the intersection of AD and CF , and Z is the intersection of CF and BE. X lies on segments BZand AY and Y lies on segment CZ. Given that AX = 3, BX = 2, CY = 4, DY = 10, EZ = 16, andFZ = 12, find the perimeter of triangle XY Z.

Proposed by: Sam Korsky

Answer: 776

Let XY = z, Y Z = x, and ZX = y. By Power of a Point, we have that

3(z + 10) = 2(y + 16), 4(x+ 12) = 10(z + 3), and 12(x+ 4) = 16(y + 2).

Solving this system gives XY = 113 and Y Z = 14

3 and ZX = 92 . Therefore, our answer if XY + Y Z +

ZX = 776 .

3. [4] Find the number of pairs of integers (x, y) such that x2 + 2y2 < 25.

Proposed by: Allen Liu

Answer: 55

We do casework on y.

If y = 0, we have x2 < 25, so we get 9 values of x. If y = ±1, then x2 < 23, so we still have 9 valuesof x. If y = ±2, we have x2 < 17, so we have 9 values of x. If y = ±3, we have x2 < 7, we get 5 valuesof x.

Therefore, the final answer is 9 + 2(9 + 9 + 5) = 55.

4. [4] Find the number of ordered triples of nonnegative integers (a, b, c) that satisfy

(ab+ 1)(bc+ 1)(ca+ 1) = 84.

Proposed by: Evan Chen

Answer: 12

The solutions are (0, 1, 83) and (1, 2, 3) up to permutation. First, we do the case where at least oneof a, b, c is 0. WLOG, say a = 0. Then we have 1 + bc = 84 =⇒ bc = 83. As 83 is prime, the onlysolution is (0, 1, 83) up to permutation.

Otherwise, we claim that at least one of a, b, c is equal to 1. Otherwise, all are at least 2, so (1 +ab)(1 + bc)(1 + ac) ≥ 53 > 84. So WLOG, set a = 1. We now need (b + 1)(c + 1)(bc + 1) = 84.Now, WLOG, say b ≤ c. If b = 1, then (c + 1)2 = 42, which has no solution. If b ≥ 3, then(b + 1)(c + 1)(bc + 1) ≥ 42 · 10 = 160 > 84. So we need b = 2. Then we need (c + 1)(2c + 1) = 21.Solving this gives c = 3, for the solution (1, 2, 3).

Therefore, the answer is 6 + 6 = 12.

5. [6] Find the number of ordered triples of positive integers (a, b, c) such that

6a+ 10b+ 15c = 3000.

Proposed by: Yang Liu

Answer: 4851

Note that 6a must be a multiple of 5, so a must be a multiple of 5. Similarly, b must be a multiple of3, and c must be a multiple of 2.

Set a = 5A, b = 3B, c = 2C. Then the equation reduces to A + B + C = 100. This has992

)= 4851

solutions.

6. [6] Let ABCD be a convex quadrilateral with AC = 7 and BD = 17. Let M,P,N,Q be the midpointsof sides AB,BC,CD,DA respectively. Compute MN2 + PQ2

Proposed by: Sam Korsky

Answer: 169

MPNQ is a parallelogram whose side lengths are 3.5 and 8.5 so the sum of squares of its diagonals is72+172

2 = 169

7. [6] An ordered pair of sets (A,B) is good if A is not a subset of B and B is not a subset of A. Howmany ordered pairs of subsets of {1, 2, . . . , 2017} are good?

Proposed by: Alexander Katz

Answer: 42017 〉 2 · 32017 + 22017

Firstly, there are 42017 possible pairs of subsets, as each of the 2017 elements can be in neither subset,in A only, in B only, or in both.

Now let us count the number of pairs of subsets for which A is a subset of B. Under these conditions,each of the 2017 elements could be in neither subset, in B only, or in both A and B. So there are 32017

such pairs.

By symmetry, there are also 32017 pairs of subsets where B is a subset of A. But this overcounts thepairs in which A is a subset of B and B is a subset of A, i.e. A = B. There are 22017 such subsets.

Thus, in total, there are 42017 〉 2 · 32017 + 22017 good pairs of subsets.

8. [6] You have 128 teams in a single elimination tournament. The Engineers and the Crimson are two ofthese teams. Each of the 128 teams in the tournament is equally strong, so during each match, eachteam has an equal probability of winning.

Now, the 128 teams are randomly put into the bracket.

What is the probability that the Engineers play the Crimson sometime during the tournament?

Proposed by: Yang Liu

Answer: 164

There are1282

)= 127 · 64 pairs of teams. In each tournament, 127 of these pairs play.

By symmetry, the answer is 127127·64 = 1

64 .

9. [7] Jeffrey writes the numbers 1 and 100000000 = 108 on the blackboard. Every minute, if x, y are onthe board, Jeffrey replaces them with

x+ y

2and 2

(1

x+

1

y

)−1.

After 2017 minutes the two numbers are a and b. Find min(a, b) to the nearest integer.

Proposed by: Yang Liu

Answer: 10000

Note that the product of the integers on the board is a constant. Indeed, we have that

x+ y

2· 2(

1

x+

1

y

)−1= xy.

Therefore, we expect that the answer to the problem is approximately√

1 · 108 = 104.

To be more rigorous, we have to show that the process indeed converges quickly enough. To show this,we bound the difference between the integers on the board at time i. Say that at time i, the integerson the board are ai < bi. Note that

di+1 = bi+1 〉 ai+1 =ai + bi

2〉 2

(1

ai+

1

bi

)−1=

(ai 〉 bi)2

2(ai + bi)

<bi 〉 ai

2=di2.

The inequality at the end follows from that obvious fact that bi〉ai < bi +ai. Therefore, di+1 ≤ di2 , so

d2017 <108

22017 , which is extremely small. So the difference is essentially 0 at time 2017, which completesthe argument.

10. [7] Let ABC be a triangle in the plane with AB = 13, BC = 14, AC = 15. Let Mn denote the smallest

possible value of (APn +BPn + CPn)1n over all points P in the plane. Find limn→∞Mn.

Proposed by: Yang Liu

Answer: 65/8

Let R denote the circumradius of triangle ABC. As ABC is an acute triangle, it isn’t hard to checkthat for any point P , we have either AP ≥ R,BP ≥ R, or CP ≥ R. Also, note that if we chooseP = O (the circumcenter) then (APn +BPn + CPn) = 3 ·Rn. Therefore, we have the inequality

R ≤ minP∈R2

(APn +BPn + CPn)1n ≤ (3Rn)

1n = R · 3 1

n .

Taking n→∞ yieldsR ≤ lim

n→∞Mn ≤ R

(as limn→∞ 31n = 1), so the answer is R = 65

8 .

11. [7] Consider the graph in 3-space of

0 = xyz(x+ y)(y + z)(z + x)(x〉 y)(y 〉 z)(z 〉 x).

This graph divides 3-space into N connected regions. What is N?

Proposed by: Evan Chen

Answer: 48

Note that reflecting for each choice of sign for x, y, z, we get new regions. Therefore, we can restrict tothe case where x, y, z > 0. In this case, the sign of the expression only depends on (x〉y)(y〉z)(z〉x).It is easy to see that for this expression, every one of the 3! = 6 orderings for {x, y, z} contributes aregion.

Therefore, our answer is 23 · 3! = 48.

12. [7] In a certain college containing 1000 students, students may choose to major in exactly one of math,computer science, finance, or English. The diversity ratio d(s) of a student s is the defined as numberof students in a different major from s divided by the number of students in the same major as s(including s). The diversity D of the college is the sum of all the diversity ratios d(s).

Determine all possible values of D.

Proposed by: Evan Chen

Answer: {0, 1000, 2000, 3000}

It is easy to check that if n majors are present, the diversity is 1000(n 〉 1). Therefore, taking n =1, 2, 3, 4 gives us all possible answers.

13. [9] The game of Penta is played with teams of five players each, and there are five roles the players canplay. Each of the five players chooses two of five roles they wish to play. If each player chooses theirroles randomly, what is the probability that each role will have exactly two players?

Proposed by: Alexander Katz

Consider a graph with five vertices corresponding to the roles, and draw an edge between two verticesif a player picks both roles. Thus there are exactly 5 edges in the graph, and we want to find theprobability that each vertex has degree 2. In particular, we want to find the probability that the graphis composed entirely of cycles.

Thus there are two cases. The first case is when the graph is itself a 5-cycle. There are 4! ways tochoose such a directed cycle (pick an arbitrary vertex A and consider a vertex it connects to, etc.),and thus 4!

2 = 12 ways for the undirected graph to be a 5-cycle. Now, there are 5! ways to assign theedges in this cycle to people, giving a total contribution of 12 · 5!.

The second case is when the graph is composed of a 2-cycle and a 3-cycle, which only requires choosingthe two vertices to be the 2-cycle, and so there are

52

)= 10 ways. To assign the players to edges, there

are52

)= 10 ways to assign the players to the 2-cycle. For the 3-cycle, any of the 3! = 6 permutations

of the remaining 3 players work. The total contribution is 10 · 10 · 6.

Therefore, out answer is

12 · 120 + 10 · 10 · 6105

=51

2500.

14. [9] Mrs. Toad has a class of 2017 students, with unhappiness levels 1, 2, . . . , 2017 respectively. Todayin class, there is a group project and Mrs. Toad wants to split the class in exactly 15 groups. Theunhappiness level of a group is the average unhappiness of its members, and the unhappiness of theclass is the sum of the unhappiness of all 15 groups. What’s the minimum unhappiness of the classMrs. Toad can achieve by splitting the class into 15 groups?

Proposed by: Yang Liu

One can show that the optimal configuration is {1}, {2}, . . . , {14}, {15, . . . , 2017}. This would give usan answer of 1 + 2 + · · ·+ 14 + 15+2017

2 = 105 + 1016 = 1121.

15. [9] Start by writing the integers 1, 2, 4, 6 on the blackboard. At each step, write the smallest positiveinteger n that satisfies both of the following properties on the board.

• n is larger than any integer on the board currently.

• n cannot be written as the sum of 2 distinct integers on the board.

Find the 100-th integer that you write on the board. Recall that at the beginning, there are already 4integers on the board.

Proposed by: Yang Liu

Answer: 388

The sequence goes1, 2, 4, 6, 9, 12, 17, 20, 25, . . . .

Common differences are 5, 3, 5, 3, 5, 3, . . . , starting from 12. Therefore, the answer is 12 + 47× 8 = 388.

16. [9] Let a and b be complex numbers satisfying the two equations

a3 〉 3ab2 = 36

b3 〉 3ba2 = 28i.

Let M be the maximum possible magnitude of a. Find all a such that |a| = M .

Proposed by: Alexander Katz

Answer: 3,〉 32 + 3i

√3

2 ,〉 32 〉

3i√3

2

Notice that

(a〉 bi)3 = a3 〉 3a2bi〉 3ab2 + b3i

= (a3 〉 3ab2) + (b3 〉 3ba2)i

= 36 + i(28i)

= 8

so that a〉 bi = 2 + i. Additionally

(a+ bi)3 = a3 + 3a2bi〉 3ab2 〉 b3i= (a3 〉 3ab2)〉 (b3 〉 3ba2)i

= 36〉 i(28i)

= 64

It follows that a〉 bi = 2ω and a+ bi = 4ω′ where ω, ω′ are third roots of unity. So a = ω+ 2ω′. Fromthe triangle inequality |a| ≤ |ω| + |2ω′| = 3, with equality when ω and ω′ point in the same direction(and thus ω = ω′). It follows that a = 3, 3ω, 3ω2, and so

a = 3,〉3

2+

3i√

3

2,〉3

2〉 3i√

3

2

17. [10] Sean is a biologist, and is looking at a string of length 66 composed of the letters A, T,C,G. Asubstring of a string is a contiguous sequence of letters in the string. For example, the string AGTC has10 substrings: A,G, T,C,AG,GT, TC,AGT,GTC,AGTC. What is the maximum number of distinctsubstrings of the string Sean is looking at?

Proposed by: Yang Liu

Answer: 2100

Let’s consider the number of distinct substrings of length `. On one hand, there are obviously at most4` distinct substrings. On the other hand, there are 67〉 ` substrings of length ` in a length 66 string.Therefore, the number of distinct substrings is at most

66∑`=1

min(4`, 67〉 `) = 2100.

To show that this bound is achievable, one can do a construction using deBrujin sequences that wewon’t elaborate on here.

18. [10] Let ABCD be a quadrilateral with side lengths AB = 2, BC = 3, CD = 5, and DA = 4. Whatis the maximum possible radius of a circle inscribed in quadrilateral ABCD?

Proposed by: Sam Korsky

Answer: 2√307

Let the tangent lengths be a, b, c, d so that

a+ b = 2

b+ c = 3

c+ d = 5

d+ a = 4

Then b = 2〉 a and c = 1 + a and d = 4〉 a. The radius of the inscribed circle of quadrilateral ABCDis given by √

abc+ abd+ acd+ bcd

a+ b+ c+ d=

√〉7a2 + 16a+ 8

7

This is clearly maximized when a = 87 which leads to a radius of

√12049 =

2√

30

7.

19. [10] Find (in terms of n ≥ 1) the number of terms with odd coefficients after expanding the product:∏1≤i<j≤n

(xi + xj)

e.g., for n = 3 the expanded product is given by x21x2 + x21x3 + x22x3 + x22x1 + x23x1 + x23x2 + 2x1x2x3and so the answer would be 6.

Proposed by: Sam Korsky

Answer: n!

Note that if we take (mod 2), we get that∏1≤i<j≤n

(xi + xj) ≡∏

1≤i<j≤n

(xj 〉 xi) = det(M),

where M is the matrix with Mij = xj−1i . This is called a Vandermonde determinant. Expanding thisdeterminant using the formula

det(M) =∑σ

n∏i=1

xi−1σ(i),

where the sum if over all n! permutations σ, gives the result.

20. [10] For positive integers a and N , let r(a,N) ∈ {0, 1, . . . , N 〉 1} denote the remainder of a whendivided by N . Determine the number of positive integers n ≤ 1000000 for which

r(n, 1000) > r(n, 1001).

Proposed by: Pakawut Jiradilok

Answer: 499500

Note that 0 ≤ r(n, 1000) ≤ 999 and 0 ≤ r(n, 1001) ≤ 1000. Consider the10002

)= 499500 ways to

choose pairs (i, j) such that i > j.By the Chinese Remainder Theorem, there is exactly one n such that1 ≤ n ≤ 1000 ·1001 such that n ≡ i (mod 1000) and n ≡ j (mod 1001). Finally, it is easy to check thatnone of the n in the range 1000001 to 1001000 satisfy the condition, so the answer is exactly 499500.

21. [12] Let P and A denote the perimeter and area respectively of a right triangle with relatively prime

integer side-lengths. Find the largest possible integral value of P 2

A

Proposed by: Sam Korsky

Assume WLOG that the side lengths of the triangle are pairwise coprime. Then they can be writtenas m2 〉 n2, 2mn,m2 + n2 for some coprime integers m and n where m > n and mn is even. Then weobtain

P 2

A=

4m(m+ n)

n(m〉 n)

But n,m〉n,m,m+n are all pairwise coprime so for this to be an integer we need n(m〉n)|4 and by

checking each case we find that (m,n) = (5, 4) yields the maximum ratio of 45 .

22. [12] Kelvin the Frog and 10 of his relatives are at a party. Every pair of frogs is either friendly orunfriendly. When 3 pairwise friendly frogs meet up, they will gossip about one another and end up ina fight (but stay friendly anyway). When 3 pairwise unfriendly frogs meet up, they will also end upin a fight. In all other cases, common ground is found and there is no fight. If all

113

)triples of frogs

meet up exactly once, what is the minimum possible number of fights?

Proposed by: Alexander Katz

Answer: 28

Consider a graph G with 11 vertices – one for each of the frogs at the party – where two vertices areconnected by an edge if and only if they are friendly. Denote by d(v) the number of edges emanatingfrom v; i.e. the number of friends frog v has. Note that d(1) + d(2) + . . .+ d(11) = 2e, where e is thenumber of edges in this graph.

Focus on a single vertex v, and choose two other vertices u,w such that uv is an edge but wv is not.There are then d(v) choices for u and 10〉 d(v) choices for w, so there are d(v)(10〉 d(v)) sets of threefrogs that include v and do not result in a fight. Each set, however, is counted twice – if uw is an edge,then we count this set both when we focus on v and when we focus on w, and otherwise we count itwhen we focus on v and when we focus on u. As such, there are a total of

1

2

∑v

d(v)(10〉 d(v))

sets of 3 frogs that do not result in a fight.

Note that d(v)+10−d(v)2 = 5 ≥

√d(v)(10〉 d(v)) =⇒ d(v)(10〉 d(v)) ≤ 25 by AM-GM. Thus there are

a maximum of1

2

∑v

d(v)(10〉 d(v)) ≤ 1

2(25 · 11) =

275

2

sets of three frogs that do not result in a fight; since this number must be an integer, there are amaximum of 137 such sets. As there are a total of

113

)= 165 sets of 3 frogs, this results in a minimum

165〉 137 = 28 number of fights.

It remains to show that such an arrangement can be constructed. Set d(1) = d(2) = . . . = d(10) = 5and d(11) = 4. Arrange these in a circle, and connect each to the nearest two clockwise neighbors; thisgives each vertex 4 edges. To get the final edge for the first ten vertices, connect 1 to 10, 2 to 9, 3 to8, 4 to 7, and 5 to 6. Thus 28 is constructable, and is thus the true minimum.

23. [12] Five points are chosen uniformly at random on a segment of length 1. What is the expecteddistance between the closest pair of points?

Proposed by: Meghal Gupta

Answer: 124

Choose five points arbitrarily at a1, a2, a3, a4, a5 in increasing order. Then the intervals (a2〉x, a2), (a3〉x, a3), (a4 〉 x, a4), (a5 〉 x, a5) must all be unoccupied. The probability that this happens is the same

as doing the process in reverse: first defining these intervals, then choosing five random points none ofwhich lie in the four intervals. This transformed process clearly has a (1〉 4x)5 probability of success.It follows that the desired probability is

∫ 1/4

0

(1〉 4x)5dx =1

24

24. [12] At a recent math contest, Evan was asked to find 22016 (mod p) for a given prime number p with100 < p < 500. Evan has forgotten what the prime p was, but still remembers how he solved it:

• Evan first tried taking 2016 modulo p〉 1, but got a value e larger than 100.

• However, Evan noted that e〉 12 (p〉 1) = 21, and then realized the answer was 〉221 (mod p).

What was the prime p?

Proposed by: Evan Chen

Answer: 211

Answer is p = 211. Let p = 2d+ 1, 50 < d < 250. The information in the problem boils down to

2016 = d+ 21 (mod 2d).

From this we can at least read off d | 1995.

Now factor 1995 = 3 · 5 · 7 · 19. The values of d in this interval are 57, 95, 105, 133. The prime valuesof 2d+ 1 are then 191 and 211. Of these, we take 211 since (2/191) = +1 while (2/211) = 〉1.

Also, this is (almost) a true story: the contest in question was the PUMaC 2016 Live Round.

25. [15] Find all real numbers x satisfying the equation x3 〉 8 = 16 3√x+ 1.

Proposed by: Alexander Katz

Answer: 〉2, 1±√

5

Let f(x) = x3−88 . Then f−1(x) = 3

√8x+ 8 = 2 3

√x+ 1, and so the given equation is equivalent to

f(x) = f−1(x). This implies f(f(x)) = x. However, as f is monotonically increasing, this implies that

f(x) = x. As a result, we have x3−88 = x =⇒ x3 〉 8x〉 8 = 0 =⇒ (x+ 2)(x2 〉 2x〉 4) = 0, and so

x = 〉2, 1±√

5.

26. [15] Kelvin the Frog is hopping on a number line (extending to infinity in both directions). Kelvinstarts at 0. Every minute, he has a 1

3 chance of moving 1 unit left, a 13 chance of moving 1 unit right

and 13 chance of getting eaten. Find the expected number of times Kelvin returns to 0 (not including

the start) before getting eaten.

Proposed by: Allen Liu

First we compute probability that the mouse returns to 0 before being eaten. Then probability thatit is at 0 in 2n minutes without being eaten is given by 1

32n

2nn

). Therefore, the overall expectation is

given by ∑n≥1

(2n

n

)9−n = 〉1 +

∑n≥0

(2n

n

)9−n

= 〉1 +1√

1〉 4/9= 〉1 +

3√5

=3√

5〉 5

5

where we use the well known fact that∑n≥0

(2n

n

)xn =

1√1〉 4x

.

for x = 19 .

27. [15] Find the smallest possible value of x + y where x, y ≥ 1 and x and y are integers that satisfyx2 〉 29y2 = 1

Proposed by: Sam Korsky

Answer: 11621

Continued fraction convergents to√

29 are 5, 112 ,163 ,

275 ,

7013 and you get 702 〉 29 · 132 = 〉1 so since

(70 + 13√

29)2 = 9801 + 1820√

29 the answer is 9801 + 1820 = 11621

28. [15] Let . . . , a−1, a0, a1, a2, . . . be a sequence of positive integers satisfying the following relations:an = 0 for n < 0, a0 = 1, and for n ≥ 1,

an = an−1 + 2(n〉 1)an−2 + 9(n〉 1)(n〉 2)an−3 + 8(n〉 1)(n〉 2)(n〉 3)an−4.

Compute ∑n≥0

10nann!

.

Proposed by: Yang Liu

Answer: e23110.

Let y =∑n≥0

xnann! . Then y′ = (1 + 2x+ 9x2 + 8x3)y by definition. So y = C exp(x+ x2 + 3x3 + 2x4).

Take x = 0 to get C = 1. Take x = 10 to get the answer.

29. [17] Yang has the sequence of integers 1, 2, . . . , 2017. He makes 2016 swaps in order, where a swapchanges the positions of two integers in the sequence. His goal is to end with 2, 3, . . . , 2017, 1. Howmany different sequences of swaps can Yang do to achieve his goal?

Proposed by: Yang Liu

Answer: 20172015

Let n = 2017. The problem is asking to write a cycle permutation of n integers as the product of n〉 1transpositions. Say that the transpositions Yang uses are (ai, bi) (i.e. swapping the ai-th integer inthe sequence with the bi-th integer in the sequence). Draw the graph with edges (ai, bi). One canshow that the result is a cycle if and only if the resulting graph is acyclic, so it must be a tree. Thereare nn−2 trees by Cayley’s formula, and for each tree, it can be made in (n 〉 1)! ways (any orderingof the edges). So the total number of ways to end with a cycle is nn−2 · (n 〉 1)!. By symmetry, eachcycle can be made in the same number of ways, so in particular the cycle 2, 3, . . . , n, 1 can be made innn−2·(n−1)!

(n−1)! = nn−2 ways.

30. [17] Consider an equilateral triangular grid G with 20 points on a side, where each row consists ofpoints spaced 1 unit apart. More specifically, there is a single point in the first row, two points in thesecond row, . . . , and 20 points in the last row, for a total of 210 points. Let S be a closed non-self-intersecting polygon which has 210 vertices, using each point in G exactly once. Find the sum of allpossible values of the area of S.

Proposed by: Sammy Luo

Answer: 52√

3

Imagine deforming the triangle lattice such that now it looks like a lattice of 45-45-90 right triangleswith legs of length 1. Note that by doing this, the area has multiplied by 2√

3, so we need to readjust

out answer on the isosceles triangle lattice by a factor of√32 at the end. By Pick’s Theorem, the area

in the new lattice is given by I + P2 〉 1 = 0 + 105〉 1 = 104. Therefore, the answer is 104 ·

√32 = 52

√3.

31. [17] A baseball league has 6 teams. To decide the schedule for the league, for each pair of teams, acoin is flipped. If it lands head, they will play a game this season, in which one team wins and oneteam loses. If it lands tails, they don’t play a game this season. Define the imbalance of this schedule

to be the minimum number of teams that will end up undefeated, i.e. lose 0 games. Find the expectedvalue of the imbalance in this league.

Proposed by: Yang Liu

Let n denote the number of teams.

Lemma: Given a connected graph G, the imbalance of G is 1 iff G is a tree. Let’s just talk in terms ofdirected graphs and indegree/outdegree.

Proof. If there is a cycle, direct the cycle such that it is a directed cycle. Then from this cycle, pointall remaining edges outwards. If G is a tree, induct on the size. Take any leaf. If it wins its game, itis undefeated. Otherwise, it must lose to its neighbor. Then induct on the tree resulting after deletingthe leaf.

Now the finish is a simple counting argument using expected values. Using Cayley’s formula, for eachsubset of vertices, we compute the probability that it is a maximal connected component and is a tree.This ends up being

2−(n2)

n∑i=1

(n

i

)· ii−2 · 2(n−i

2 ).

This evaluates to 5055215 for n = 6.

32. [17] Let a, b, c be non-negative real numbers such that ab+ bc+ ca = 3. Suppose that

a3b+ b3c+ c3a+ 2abc(a+ b+ c) =9

2.

What is the minimum possible value of ab3 + bc3 + ca3?

Proposed by: Pakawut Jiradilok

Answer: 18

Expanding the inequality∑cycab(b+ c〉 2a)2 ≥ 0 gives

∑cyc

ab3)

+ 4∑

cyc

a3b)〉 4∑

cyc

a2b2)〉 abc(a+ b+ c) ≥ 0

Using∑cyca3b)

+ 2abc(a+ b+ c) = 92 in the inequality above yields

∑cyc

ab3)〉 4(ab+ bc+ ca)2 ≥

∑cyc

ab3)〉 4∑

cyc

a2b2)〉 9abc(a+ b+ c) ≥ 〉18

Since ab+ bc+ ca = 3, we have∑cycab3 ≥ 18 as desired.

The equality occurs when (a, b, c) ∼cyc

(√

32 ,√

6, 0).

33. [20] Welcome to the USAYNO, where each question has a yes/no answer. Choose any subset of thefollowing six problems to answer. If you answer n problems and get them all correct, you will receivemax(0, (n〉 1)(n〉 2)) points. If any of them are wrong, you will receive 0 points.

Your answer should be a six-character string containing ’Y’ (for yes), ’N’ (for no), or ’B’ (for blank).For instance if you think 1, 2, and 6 are ’yes’ and 3 and 4 are ’no’, you would answer YYNNBY (andreceive 12 points if all five answers are correct, 0 points if any are wrong).

(a) a, b, c, d, A,B,C, and D are positive real numbers such that ab >

AB and c

d >CD . Is it necessarily

true that a+cb+d >

A+CB+D?

(b) Do there exist irrational numbers α and β such that the sequence bαc+ bβc, b2αc+ b2βc, b3αc+b3βc, . . . is arithmetic?

(c) For any set of primes P, let SP denote the set of integers whose prime divisors all lie in P. Forinstance S{2,3} = {2a3b | a, b ≥ 0} = {1, 2, 3, 4, 6, 8, 9, 12, . . .}. Does there exist a finite set ofprimes P and integer polynomials P and Q such that gcd(P (x), Q(y)) ∈ SP for all x, y?

(d) A function f is called P-recursive if there exists a positive integer m and real polynomialsp0(n), p1(n), . . . , pm(n) satisfying

pm(n)f(n+m) = pm−1(n)f(n+m〉 1) + . . .+ p0(n)f(n)

for all n. Does there exist a P-recursive function f satisfying limn→∞f(n)

n√

2= 1?

(e) Does there exist a nonpolynomial function f : Z→ Z such that a〉 b divides f(a)〉 f(b) for allintegers a 6= b?

(f) Do there exist periodic functions f, g : R→ R such that f(x) + g(x) = x for all x?

Proposed by: Alexander Katz

Answer: NNNYYY

34. [20]

(a) Can 1000 queens be placed on a 2017 × 2017 chessboard such that every square is attacked bysome queen? A square is attacked by a queen if it lies in the same row, column, or diagonal asthe queen.

(b) A 2017× 2017 grid of squares originally contains a 0 in each square. At any step, Kelvin the Frogchooses two adjacent squares (two squares are adjacent if they share a side) and increments thenumbers in both of them by 1. Can Kelvin make every square contain a different power of 2?

(c) A tournament consists of single games between every pair of players, where each game has awinner and loser with no ties. A set of people is dominated if there exists a player who beats allof them. Does there exist a tournament in which every set of 2017 people is dominated?

(d) Every cell of a 19 × 19 grid is colored either red, yellow, green, or blue. Does there necessarilyexist a rectangle whose sides are parallel to the grid, all of whose vertices are the same color?

(e) Does there exist a c ∈ R+ such that max(|A ·A|, |A+A|) ≥ c|A| log2 |A| for all finite sets A ⊂ Z?

(f) Can the set {1, 2, . . . , 1093} be partitioned into 7 subsets such that each subset is sum-free (i.e.no subset contains a, b, c with a+ b = c)?

Proposed by: Alexander Katz

Answer: NNYYYY

35. [20]

(a) Does there exist a finite set of points, not all collinear, such that a line between any two pointsin the set passes through a third point in the set?

(b) Let ABC be a triangle and P be a point. The isogonal conjugate of P is the intersection of thereflection of line AP over the A-angle bisector, the reflection of line BP over the B-angle bisector,and the reflection of line CP over the C-angle bisector. Clearly the incenter is its own isogonalconjugate. Does there exist another point that is its own isogonal conjugate?

(c) Let F be a convex figure in a plane, and let P be the largest pentagon that can be inscribed inF . Is it necessarily true that the area of P is at least 3

4 the area of F?

(d) Is it possible to cut an equilateral triangle into 2017 pieces, and rearrange the pieces into a square?

(e) Let ABC be an acute triangle and P be a point in its interior. Let D,E, F lie on BC,CA,AB re-spectively so that PD bisects ∠BPC, PE bisects ∠CPA, and PF bisects ∠APB. Is it necessarilytrue that AP +BP + CP ≥ 2(PD + PE + PF )?

(f) Let P2018 be the surface area of the 2018-dimensional unit sphere, and let P2017 be the surfacearea of the 2017-dimensional unit sphere. Is P2018 > P2017?

Proposed by: Alexander Katz

Answer: NYYYYN

36. [20]

(a) Does∑p−1i=1

1i ≡ 0 (mod p2) for all odd prime numbers p? (Note that 1

i denotes the number suchthat i · 1i ≡ 1 (mod p2))

(b) Do there exist 2017 positive perfect cubes that sum to a perfect cube?

(c) Does there exist a right triangle with rational side lengths and area 5?

(d) A magic square is a 3× 3 grid of numbers, all of whose rows, columns, and major diagonals sumto the same value. Does there exist a magic square whose entries are all prime numbers?

(e) Is∏pp2+1p2−1 = 22+1

22−1 ·32+132−1 ·

52+152−1 ·

72+172−1 · . . . a rational number?

(f) Do there exist an infinite number of pairs of distinct integers (a, b) such that a and b have thesame set of prime divisors, and a+ 1 and b+ 1 also have the same set of prime divisors?

Proposed by: Alexander Katz

Answer: NYYYYY

HMMT November 2017November 11, 2017

General Round

1. Find the sum of all positive integers whose largest proper divisor is 55. (A proper divisor of n is adivisor that is strictly less than n.)

2. Determine the sum of all distinct real values of x such that

||| · · · ||x|+ x| · · · |+ x|+ x| = 1,

where there are 2017 x’s in the equation.

3. Find the number of integers n with 1 ≤ n ≤ 2017 so that (n − 2)(n − 0)(n − 1)(n − 7) is an integermultiple of 1001.

4. Triangle ABC has AB = 10, BC = 17, and CA = 21. Point P lies on the circle with diameter AB.What is the greatest possible area of APC?

5. Given that a, b, c are integers with abc = 60, and that complex number ω 6= 1 satisfies ω3 = 1, find theminimum possible value of

∣∣a+ bω + cω2∣∣.

6. A positive integer n is magical if ⌊√⌈√n⌉⌋

=

⌈√⌊√n⌋⌉,

where b·c and d·e represent the floor and ceiling function respectively. Find the number of magicalintegers between 1 and 10,000, inclusive.

7. Reimu has a wooden cube. In each step, she creates a new polyhedron from the previous one by cuttingoff a pyramid from each vertex of the polyhedron along a plane through the trisection point on eachadjacent edge that is closer to the vertex. For example, the polyhedron after the first step has sixoctagonal faces and eight equilateral triangular faces. How many faces are on the polyhedron after thefifth step?

8. Marisa has a collection of 28− 1 = 255 distinct nonempty subsets of {1, 2, 3, 4, 5, 6, 7, 8}. For each stepshe takes two subsets chosen uniformly at random from the collection, and replaces them with eithertheir union or their intersection, chosen randomly with equal probability. (The collection is allowed tocontain repeated sets.) She repeats this process 28 − 2 = 254 times until there is only one set left inthe collection. What is the expected size of this set?

9. Find the minimum possible value of

√58− 42x+

√149− 140

√1− x2

where −1 ≤ x ≤ 1.

10. Five equally skilled tennis players named Allen, Bob, Catheryn, David, and Evan play in a roundrobin tournament, such that each pair of people play exactly once, and there are no ties. In each ofthe ten games, the two players both have a 50% chance of winning, and the results of the games areindependent. Compute the probability that there exist four distinct players P1, P2, P3, P4 such that Pi

beats Pi+1 for i = 1, 2, 3, 4. (We denote P5 = P1).

HMMT November 2017November 11, 2017

General Round

1. Find the sum of all positive integers whose largest proper divisor is 55. (A proper divisor of n is adivisor that is strictly less than n.)

Proposed by: Michael Tang

Answer: 550

The largest proper divisor of an integer n is np , where p is the smallest prime divisor of n. So n = 55p

for some prime p. Since 55 = 5 · 11, we must have p ≤ 5, so p = 2, 3, 5 gives all solutions. The sum ofthese solutions is 55(2 + 3 + 5) = 550.

2. Determine the sum of all distinct real values of x such that

||| · · · ||x|+ x| · · · |+ x|+ x| = 1,

where there are 2017 x’s in the equation.

Proposed by: Yuan Yao

Answer: − 20162017

Note that |x+ |x|| = 2x when x is nonnegative, and is equal to 0 otherwise. Thus, when there are 2017x’s, the expression equals 2017x when x ≥ 0 and −x otherwise, so the two solutions to the equationare x = −1 and 1

2017 , and their sum is − 20162017 .

3. Find the number of integers n with 1 ≤ n ≤ 2017 so that (n − 2)(n − 0)(n − 1)(n − 7) is an integermultiple of 1001.

Proposed by: Dhruv Rohatgi

Answer: 99

Note that 1001 = 7 · 11 · 13, so the stated product must be a multiple of 7, as well as a multiple of11, as well as a multiple of 13. There are 4 possible residues of n modulo 11 for which the productis a multiple of 11; similarly, there are 4 possible residues of n modulo 13 for which the product is amultiple of 13. However, there are only 3 possible residues of n modulo 7 for which the product is amultiple of 7.

Consider each of these 4 · 4 · 3 = 48 possible triples of remainders. By the Chinese Remainder Theoremthere is exactly one value of n with 1 ≤ n ≤ 1001 achieving those remainders, and exactly one valueof n with 16 ≤ n ≤ 1016 achieving those remainders. Similarly, there is exactly one value of n with1017 ≤ n ≤ 2017 with those same remainders. Hence there are 96 values of n with 16 ≤ n ≤ 2017 suchthat (n− 2)(n− 0)(n− 1)(n− 7) is a multiple of 1001.

It remains to check n ∈ {1, 2, 3, . . . , 15}. Since the product must be a multiple of 7, we can narrowthe set to {1, 2, 7, 8, 9, 14}. The first 3 values work trivially, since the product is 0. It can be easilychecked that none of the remaining values of n yield a product which is a multiple of 11. Hence, thefinal answer is 96 + 3 = 99.

4. Triangle ABC has AB = 10, BC = 17, and CA = 21. Point P lies on the circle with diameter AB.What is the greatest possible area of APC?

Proposed by: Michael Tang

Answer: 1892

To maximize [APC], point P should be the farthest point on the circle from AC. Let M be themidpoint of AB and Q be the projection of M onto AC. Then PQ = PM + MQ = 1

2AB + 12hB ,

where hB is the length of the altitude from B to AC. By Heron’s formula, one finds that the area ofABC is

√24 · 14 · 7 · 3 = 84, so hB = 2·84

AC = 8. Then PQ = 12 (10 + 8) = 9, so the area of APC is

12 · 21 · 9 = 189

2 .

5. Given that a, b, c are integers with abc = 60, and that complex number ω 6= 1 satisfies ω3 = 1, find theminimum possible value of

∣∣a+ bω + cω2∣∣.

Proposed by: Ashwin Sah

Answer:√

3

Since ω3 = 1, and ω 6= 1, ω is a third root of unity. For any complex number z, |z|2 = z · z̄. Lettingz = a+ bω + cω2, we find that z̄ = a+ cω + bω2, and

|z|2 = a2 + abω + acω2 + abω2 + b2 + bcω + acω + bcω2 + c2

= (a2 + b2 + c2) + (ab+ bc+ ca)(ω) + (ab+ bc+ ca)(ω2)

= (a2 + b2 + c2)− (ab+ bc+ ca)

=1

2((a− b)2 + (b− c)2 + (c− a)2),

where we have used the fact that ω3 = 1 and that ω + ω2 = −1. This quantity is minimized whena, b, and c are as close to each other as possible, making a = 3, b = 4, c = 5 the optimal choice, giving|z|2 = 3. (A smaller value of |z| requires two of a, b, c to be equal and the third differing from them byat most 2, which is impossible.) So |z|min =

√3.

6. A positive integer n is magical if ⌊√⌈√n⌉⌋

=

⌈√⌊√n⌋⌉,

where b·c and d·e represent the floor and ceiling function respectively. Find the number of magicalintegers between 1 and 10,000, inclusive.

Proposed by: Yuan Yao

Answer: 1330

First of all, we have b√nc = d

√ne when n is a perfect square and b

√nc = d

√ne − 1 otherwise.

Therefore, in the first case, the original equation holds if and only if√n is a perfect square itself, i.e.,

n is a fourth power. In the second case, we need m = b√nc to satisfy the equation b

√m+ 1c = d

√me,

which happens if and only if either m or m + 1 is a perfect square k2. Therefore, n is magical if andonly if (k2−1)2 < n < (k2 +1)2 for some (positive) integer k. There are (k2 +1)2− (k2−1)2 = 4k2−1integers in this range. The range in the problem statement includes k = 1, 2, . . . , 9 and the interval(992, 1002], so the total number of magical numbers is

4(12 + 22 + · · ·+ 92)− 9 + (1002 − 992) = 4 · 9 · (9 + 1) · (18 + 1)

6+ 190 = 1330.

7. Reimu has a wooden cube. In each step, she creates a new polyhedron from the previous one by cuttingoff a pyramid from each vertex of the polyhedron along a plane through the trisection point on eachadjacent edge that is closer to the vertex. For example, the polyhedron after the first step has sixoctagonal faces and eight equilateral triangular faces. How many faces are on the polyhedron after thefifth step?

Proposed by: Qi Qi

Answer: 974

Notice that the number of vertices and edges triple with each step. We always have 3 edges meetingat one vertex, and slicing off a pyramid doesn’t change this (we make new vertices from which oneedge from the previous step and two of the pyramid edges emanate). So at each step we replace thesliced-off vertex with three new vertices, and to each edge we create four new “half-edges” (two fromthe pyramid at each endpoint), which is equivalent to tripling the number of vertices and edges. Then,by Euler’s Theorem the number of faces is E − V + 2 = 12 · 35 − 8 · 35 + 2 = 974.

8. Marisa has a collection of 28− 1 = 255 distinct nonempty subsets of {1, 2, 3, 4, 5, 6, 7, 8}. For each stepshe takes two subsets chosen uniformly at random from the collection, and replaces them with eithertheir union or their intersection, chosen randomly with equal probability. (The collection is allowed tocontain repeated sets.) She repeats this process 28 − 2 = 254 times until there is only one set left inthe collection. What is the expected size of this set?

Proposed by: Yuan Yao

Answer: 1024255

It suffices to compute the probability of each number appearing in the final subset. For any giveninteger n ∈ [1, 8], there are 27 = 128 subsets with n and 27 − 1 = 127 without. When we focus ononly this element, each operation is equivalent to taking two random sets and discarding one of themrandomly. Therefore there is a 128

255 probability that n is in the final subset, and the expected value ofits size is 8 · 128255 = 1024

255 .

Alternatively, since |A|+ |B| = |A∪B|+ |A∩B|, the expected value of the average size of all remainingsubsets at a given step is constant, so the answer is simply the average size of all 255 subsets, which is8·128255 = 1024

255 .

9. Find the minimum possible value of

√58− 42x+

√149− 140

√1− x2

where −1 ≤ x ≤ 1.

Proposed by: Serina Hu

Answer:√

109

Substitute x = cos θ and√

1− x2 = sin θ, and notice that 58 = 32 + 72, 42 = 2 · 3 · 7, 149 = 72 + 102,and 140 = 2 · 7 · 10. Therefore the first term is an application of Law of Cosines on a triangle thathas two sides 3 and 7 with an angle measuring θ between them to find the length of the third side;similarly, the second is for a triangle with two sides 7 and 10 that have an angle measuring 90 − θbetween them. ”Gluing” these two triangles together along their sides of length 7 so that the mergedtriangles form a right angle, we see that the minimum length of the sum of their third sides occurswhen the glued triangles form a right triangle. This right triangle has legs of length 3 and 10, so itshypotenuse has length

√109.

10. Five equally skilled tennis players named Allen, Bob, Catheryn, David, and Evan play in a roundrobin tournament, such that each pair of people play exactly once, and there are no ties. In each ofthe ten games, the two players both have a 50% chance of winning, and the results of the games areindependent. Compute the probability that there exist four distinct players P1, P2, P3, P4 such that Pi

beats Pi+1 for i = 1, 2, 3, 4. (We denote P5 = P1).

Proposed by: Steven Hao

Answer: 4964

We make the following claim: if there is a 5-cycle (a directed cycle involving 5 players) in the tourna-ment, then there is a 4-cycle.

Proof: Assume that A beats B, B beats C, C beats D, D beats E and E beats A. If A beats Cthen A,C,D,E forms a 4-cycle, and similar if B beats D, C beats E, and so on. However, if all fivereversed matches occur, then A,D,B,C is a 4-cycle.

Therefore, if there are no 4-cycles, then there can be only 3-cycles or no cycles at all.

Case 1: There is a 3-cycle. Assume that A beats B, B beats C, and C beats A. (There are(53

)= 10

ways to choose the cycle and 2 ways to orient the cycle.) Then D either beats all three or is beaten byall three, because otherwise there exists two people X and Y in these three people such that X beatsY , and D beats Y but is beaten by X, and then X,D, Y, Z will form a 4-cycle (Z is the remaining

person of the three). The same goes for E. If D and E both beat all three or are beaten by all three,then there is no restriction on the match between D and E. However, if D beats all three and E losesto all three, then E cannot beat D because otherwise E,D,A,B forms a 4-cycle. This means thatA,B,C is the only 3-cycle in the tournament, and once the cycle is chosen there are 2 · 2 + 2 · 1 = 6ways to choose the results of remaining matches, for 10 · 2 · 6 = 120 ways in total.

Case 2: There are no cycles. This means that the tournament is a complete ordering (the person witha higher rank always beats the person with a lower rank). There are 5! = 120 ways in this case as well.

Therefore, the probability of not having a 4-cycle is 120+120210 = 15

64 , and thus the answer is 1− 1564 = 49

64 .

HMMT November 2017November 11, 2017

Theme Round

Based on a true story.

1. Two ordered pairs (a, b) and (c, d), where a, b, c, d are real numbers, form a basis of the coordinateplane if ad 6= bc. Determine the number of ordered quadruples (a, b, c, d) of integers between 1 and 3inclusive for which (a, b) and (c, d) form a basis for the coordinate plane.

2. Horizontal parallel segments AB = 10 and CD = 15 are the bases of trapezoid ABCD. Circle γ ofradius 6 has center within the trapezoid and is tangent to sides AB, BC, and DA. If side CD cutsout an arc of γ measuring 120◦, find the area of ABCD.

3. Emilia wishes to create a basic solution with 7% hydroxide (OH) ions. She has three solutions ofdifferent bases available: 10% rubidium hydroxide (Rb(OH)), 8% cesium hydroxide (Cs(OH)), and 5%francium hydroxide (Fr(OH)). (The Rb(OH) solution has both 10% Rb ions and 10% OH ions, andsimilar for the other solutions.) Since francium is highly radioactive, its concentration in the finalsolution should not exceed 2%. What is the highest possible concentration of rubidium in her solution?

4. Mary has a sequence m2,m3,m4, . . . , such that for each b ≥ 2, mb is the least positive integer m forwhich none of the base-b logarithms logb(m), logb(m + 1), . . . , logb(m + 2017) are integers. Find thelargest number in her sequence.

5. Each of the integers 1, 2, . . . , 729 is written in its base-3 representation without leading zeroes. Thenumbers are then joined together in that order to form a continuous string of digits: 12101112202122 . . . .How many times in this string does the substring 012 appear?

6. Rthea, a distant planet, is home to creatures whose DNA consists of two (distinguishable) strands ofbases with a fixed orientation. Each base is one of the letters H, M, N, T, and each strand consists ofa sequence of five bases, thus forming five pairs. Due to the chemical properties of the bases, each pairmust consist of distinct bases. Also, the bases H and M cannot appear next to each other on the samestrand; the same is true for N and T. How many possible DNA sequences are there on Rthea?

7. On a blackboard a stranger writes the values of s7(n)2 for n = 0, 1, . . . , 720 〉 1, where s7(n) denotesthe sum of digits of n in base 7. Compute the average value of all the numbers on the board.

8. Undecillion years ago in a galaxy far, far away, there were four space stations in the three-dimensionalspace, each pair spaced 1 light year away from each other. Admiral Ackbar wanted to establish a basesomewhere in space such that the sum of squares of the distances from the base to each of the stationsdoes not exceed 15 square light years. (The sizes of the space stations and the base are negligible.)Determine the volume, in cubic light years, of the set of all possible locations for the Admiral’s base.

9. New this year at HMNT: the exciting game of RNG baseball ! In RNG baseball, a team of infinitelymany people play on a square field, with a base at each vertex; in particular, one of the bases is calledthe home base. Every turn, a new player stands at home base and chooses a number n uniformly atrandom from {0, 1, 2, 3, 4}. Then, the following occurs:

• If n > 0, then the player and everyone else currently on the field moves (counterclockwise) aroundthe square by n bases. However, if in doing so a player returns to or moves past the home base,he/she leaves the field immediately and the team scores one point.

• If n = 0 (a strikeout), then the game ends immediately; the team does not score any more points.

What is the expected number of points that a given team will score in this game?

10. Denote φ = 1+√5

2 and consider the set of all finite binary strings without leading zeroes. Each string Shas a “base-φ” value p(S). For example, p(1101) = φ3 +φ2 + 1. For any positive integer n, let f(n) be

the number of such strings S that satisfy p(S) = φ48n→1φ48→1 . The sequence of fractions f(n+1)

f(n) approaches

a real number c as n goes to infinity. Determine the value of c.

HMMT November 2017November 11, 2017

Theme Round

1. Two ordered pairs (a, b) and (c, d), where a, b, c, d are real numbers, form a basis of the coordinateplane if ad 6= bc. Determine the number of ordered quadruples (a, b, c, d) of integers between 1 and 3inclusive for which (a, b) and (c, d) form a basis for the coordinate plane.

Proposed by: Ashwin Sah

Answer: 66

Any pair of distinct points will form a basis except when (a, b) and (c, d) are both from {(1, 1), (2, 2), (3, 3)},so the answer is 9 · 8〉 3 · 2 = 66.

2. Horizontal parallel segments AB = 10 and CD = 15 are the bases of trapezoid ABCD. Circle γ ofradius 6 has center within the trapezoid and is tangent to sides AB, BC, and DA. If side CD cutsout an arc of γ measuring 120◦, find the area of ABCD.

Proposed by: Michael Tang

Answer: 2252

Suppose that the center of the circle is O and the circle intersects CD at X and Y . Since ∠XOY = 120◦

and triangle XOY is isosceles, the distance from O to XY is 6 · sin(30◦) = 3. On the other hand, thedistance from O to AB is 6 as the circle is tangent to AB, and O is between AB and CD, so the height

of the trapezoid is 6 + 3 = 9 and its area is 9·(10+15)2 = 225

2 .

3. Emilia wishes to create a basic solution with 7% hydroxide (OH) ions. She has three solutions ofdifferent bases available: 10% rubidium hydroxide (Rb(OH)), 8% cesium hydroxide (Cs(OH)), and 5%francium hydroxide (Fr(OH)). (The Rb(OH) solution has both 10% Rb ions and 10% OH ions, andsimilar for the other solutions.) Since francium is highly radioactive, its concentration in the finalsolution should not exceed 2%. What is the highest possible concentration of rubidium in her solution?

Proposed by: Yuan Yao

Answer: 1%

Suppose that Emilia uses R liters of Rb(OH), C liters of Cs(OH), and F liters of Fr(OH), then we have

10% ·R+ 8% · C + 5% · FR+ C + F

= 7% and5% · F

R+ C + F≤ 2%.

The equations simplify to 3R+ C = 2F and 3F ≤ 2R+ 2C, which gives

9R+ 3C

2≤ 2R+ 2C ⇒ 5R ≤ C.

Therefore the concentration of rubidium is maximized when 5R = C, so F = 4R, and the concentrationof rubidium is

10% ·RR+ C + F

= 1%.

4. Mary has a sequence m2,m3,m4, . . . , such that for each b ≥ 2, mb is the least positive integer m forwhich none of the base-b logarithms logb(m), logb(m + 1), . . . , logb(m + 2017) are integers. Find thelargest number in her sequence.

Proposed by: Michael Tang

Answer: 2188

It is not difficult to see that for all of the logarithms to be non-integers, they must lie strictly betweenn and n+ 1 for some integer n. Therefore, we require bn+1 〉 bn > 2018, and so mb = bn + 1 where nis the smallest integer that satisfies the inequality. In particular, this means that bn 〉 bn−1 ≤ 2018.

Note that m2 = 211 + 1 = 2049 (since 212 〉 211 = 2048 > 2018) and m3 = 37 + 1 = 2188 (since38 〉 37 = 4374 > 2018). we now show that 2188 is the maximum possible value for mb.

If n = 0, then mb = 1 + 1 = 2.

If n = 1, then b〉 1 ≤ 2018 and thus mb = b+ 1 ≤ 2020.

If n = 2, then b2 〉 b ≤ 2018, which gives b ≤ 45, and thus mb = b2 + 1 ≤ 2018 + b+ 1 ≤ 2065.

If n = 3, then b3 〉 b2 ≤ 2018, which gives b ≤ 12, and thus mb = b3 + 1 ≤ 123 + 1 = 1729.

If n = 4, then b4 〉 b3 ≤ 2018, which gives b ≤ 6, and thus mb = b4 + 1 ≤ 64 + 1 = 1297.

It then remains to check the value of m4 and m5. Indeed, m4 = 45 + 1 = 1025 and m5 = 54 + 1 = 626,so no values of mb exceeds 2188.

5. Each of the integers 1, 2, . . . , 729 is written in its base-3 representation without leading zeroes. Thenumbers are then joined together in that order to form a continuous string of digits: 12101112202122 . . . .How many times in this string does the substring 012 appear?

Proposed by: Michael Tang

Answer: 148

Ignore 729 = 36 = 10000003 since it will not contribute to a 012 substring. Break into cases on how012 appears: (i) when an individual integer contains the string 012; (ii) when 01 are the last two digitsof an integer and 2 is the first digit of the next integer; and (iii) when 0 is the last digit of an integerand 12 are the first two digits of the next integer.

For case (i), we want to find the total number of appearances of the string 012 in 1, 2, 3, . . . , N . Sinceeach number has at most six digits, 012 appears at most once per number. If such a number has ddigits, 4 ≤ d ≤ 6, then there are d 〉 3 possible positions for the substring 012 and 2 · 3d−4 possiblechoices for the remaining digits (since the leftmost digit must be nonzero). Thus there are

6∑d=4

(d〉 3) ·2 · 3d−4

)= 1 · 2 + 2 · 6 + 3 · 18 = 68

appearances of 012 in case (i).

For case (ii), we have an integer n for which n ends in 01 and n + 1 starts with 2. Then n must alsostart with 2. Hence it suffices to count the number of integers 1 ≤ n < N which start with 2 and endwith 01 in base three. If n has d digits, 3 ≤ d ≤ 6, then there are 3d−3 possibilities for the middledigits, so we have

6∑d=3

3d−3 = 1 + 3 + 9 + 27 = 40

appearances of 012 in case (ii).

For case (ii), we have an integer n for which n ends in 0 and n + 1 starts with 12. Then n must alsostart with 12, so we want to count the number of 1 ≤ n < N starting with 12 and ending in 0. Likecase (ii), there are also 40 appearances of 012 in case (iii).

In total, the substring 012 appears 68 + 40 + 40 = 148 times.

6. Rthea, a distant planet, is home to creatures whose DNA consists of two (distinguishable) strands ofbases with a fixed orientation. Each base is one of the letters H, M, N, T, and each strand consists ofa sequence of five bases, thus forming five pairs. Due to the chemical properties of the bases, each pairmust consist of distinct bases. Also, the bases H and M cannot appear next to each other on the samestrand; the same is true for N and T. How many possible DNA sequences are there on Rthea?

Proposed by: Yuan Yao

Answer: 12 · 74 or 28812

There are 4 · 3 = 12 ways to choose the first base pairs, and regardless of which base pair it is, thereare 3 possibilities for the next base on one strand and 3 possibilities for the next base on the other

strand. Among these possibilities, exactly 2 of them have identical bases forming a base pair (usingone of the base not in the previous base pair if the previous pair is H-M or N-T, or one of the base inthe previous pair otherwise), which is not allowed. Therefore there are 3 · 3 〉 2 = 7 ways to chooseeach of the following base pairs. Thus in total there are 12 · 74 = 28812 possible DNA (which is alsothe maximum number of species).

7. On a blackboard a stranger writes the values of s7(n)2 for n = 0, 1, . . . , 720 〉 1, where s7(n) denotesthe sum of digits of n in base 7. Compute the average value of all the numbers on the board.

Proposed by: Ashwin Sah

Answer: 3680

Solution 1: We solve for 0 to bn 〉 1 and sb(n)2 (i.e. base b).

Let n = d1 . . . dn in base b, where there may be leading zeros. Then sb(n) = d1 + · · ·+ dn, regardlessof the leading zeros.

E[sd(n)2

]= E

[(d1 + · · ·+ dn)2

]=∑

1≤i≤n

E[d2i]

+ 2∑

1≤i<j≤n

E[didj ],

and now notice that we can treat choosing n uniformly as choosing the di uniformly independentlyfrom {0, . . . , b〉 1}. Thus this simplifies to

E[sd(n)2

]= nE

[d21]

+ n(n〉 1)E [d1]2.

Now

E[d21]

=02 + · · ·+ (b〉 1)2

b=

(b〉 1)(2b〉 1)

6,

E[d1] =0 + · · ·+ (b〉 1)

b=b〉 1

2,

so the answer is

n · (b〉 1)(2b〉 1)

6+ n(n〉 1) ·

(b〉 1

2

)2

.

Plugging in b = 7, n = 20 yields the result.

Solution 2: There are two theorems we will cite regarding variance and expected value. The first isthat, for any variable X,

V ar(X) = E[X2]〉 E[X]2.

The second is that, for two independent variables X and Y,

V ar(X + Y ) = V ar(X) + V ar(Y ).

Let X be the sum of all of the digits. We want to find E[X2]. The expected of a single digit is17 (0+1+2+3+4+5+6) = 3. Thus, the expected value of the sum of the digits is E[X] = 20×3 = 60, soE[X]2 = 3600. The variance of a single digit is 1

7 [(0〉3)2+(1〉3)2+...+(6〉3)2] = 9+4+1+0+1+4+97 = 4.

Since the digits are independent, their variances add by the second theorem above. Therefore, thevariance of the sum of all of the digits is V ar(X) = 20 × 4 = 80. Finally, using the first theorem, wehave E[X2] = E[X]2 + V ar(X) = 3680.

8. Undecillion years ago in a galaxy far, far away, there were four space stations in the three-dimensionalspace, each pair spaced 1 light year away from each other. Admiral Ackbar wanted to establish a basesomewhere in space such that the sum of squares of the distances from the base to each of the stationsdoes not exceed 15 square light years. (The sizes of the space stations and the base are negligible.)Determine the volume, in cubic light years, of the set of all possible locations for the Admiral’s base.

Proposed by: Yuan Yao

Answer: 27√6

8 π

Solution 1: Set up a coordinate system where the coordinates of the stations are ( 12√2, 12√2, 12√2),

(〉 12√2,〉 1

2√2, 12√2), ( 1

2√2,〉 1

2√2,〉 1

2√2), and (〉 1

2√2, 12√2,〉 1

2√2). The sum of squares of the distances

is then ∑sym

2(x〉 1

2√

2)2 + 2(x+

1

2√

2)2 = 4(x2 + y2 + z2) +

3

2= 4r2 +

3

2,

where r is the distance from the center of the tetrahedron. It follows from 4r2 + 32 ≤ 15 that r ≤ 3

√6

4 ,

so the set is a ball with radius R = 3√6

4 , and the volume is 4π3 R

3 = 27√6

8 π.

Solution 2: Let P be the location of the base; S1, S2, S3, S4 be the stations; and G be the center ofthe tetrahedron. We have:

4∑i=1

PS2i =

4∑i=1

~PSi · ~PSi

4∑i=1

PS2i =

4∑i=1

(~PG+ ~GSi

)·(~PG+ ~GSi

)4∑i=1

PS2i =

4∑i=1

(~PG · ~PG+ 2 ~PG · ~GSi + ~GSi · ~GSi

)Since GS1 = GS2 = GS3 = GS4, we have:

4∑i=1

PS2i = 4PG2 + 4GS2

1 + 2

4∑i=1

(~PG · ~GSi

)4∑i=1

PS2i = 4PG2 + 4GS2

1 + 2 ~PG ·

(4∑i=1

~GSi

)Since G is the center of the tetrahedron, ~GS1 + ~GS2 + ~GS3 + ~GS4 = ~0. Thus:

4∑i=1

PS2i = 4PG2 + 4GS2

1

Since GS21 = 3

8 , we have PG2 ≤ 278 . Thus, the locus of all good points is a ball centered at G with

radius r =√

278 . Then, the volume is V = 4

3πr3 = 27

√6π

8 .

9. New this year at HMNT: the exciting game of RNG baseball ! In RNG baseball, a team of infinitelymany people play on a square field, with a base at each vertex; in particular, one of the bases is calledthe home base. Every turn, a new player stands at home base and chooses a number n uniformly atrandom from {0, 1, 2, 3, 4}. Then, the following occurs:

• If n > 0, then the player and everyone else currently on the field moves (counterclockwise) aroundthe square by n bases. However, if in doing so a player returns to or moves past the home base,he/she leaves the field immediately and the team scores one point.

• If n = 0 (a strikeout), then the game ends immediately; the team does not score any more points.

What is the expected number of points that a given team will score in this game?

Proposed by: Yuan Yao

Answer: 409125

For i = 0, 1, 2, 3, let Pi be the probability that a player on the i-th base scores a point before strikeout(with zeroth base being the home base). We have the following equations:

P0 =1

5(P1 + P2 + P3 + 1)

P1 =1

5(P2 + P3 + 1 + 1)

P2 =1

5(P3 + 1 + 1 + 1)

P3 =1

5(1 + 1 + 1 + 1)

Solving the system of equations gives P3 = 45 , P2 = 19

25 , P1 = 89125 , P0 = 409

625 , so the probability that abatter scores a point himself is 409

625 , given that he is able to enter the game before the game is over.Since the probability that the nth player will be able to stand on the home base is ( 4

5 )n−1 (none of theprevious n〉1 players receive a strikeout), the expected value is 409

625 (1+ 45 +( 4

5 )2+· · · ) = 409625 ·

11− 4

5

= 409125 .

10. Denote φ = 1+√5

2 and consider the set of all finite binary strings without leading zeroes. Each string Shas a “base-φ” value p(S). For example, p(1101) = φ3 +φ2 + 1. For any positive integer n, let f(n) be

the number of such strings S that satisfy p(S) = φ48n−1φ48−1 . The sequence of fractions f(n+1)

f(n) approaches

a real number c as n goes to infinity. Determine the value of c.

Proposed by: Ashwin Sah

Answer: 25+3√69

2

We write everything in base φ. Notice that

φ48n 〉 1

φ48 〉 1= 10 . . . 010 . . . 01 . . . 10 . . . 01,

where there are n 〉 1 blocks of 47 zeros each. We can prove that every valid base-φ representationcomes from replacing a consecutive string 100 with a 011 repeatedly. Using this, we can easily classifywhat base-φ representations are counted by f(n).

Notice that 10000000 = 01100000 = 01011000 = 01010110 and similar, so that in each block of zeroswe can choose how many times to perform a replacement. It turns out that we can do anywhere from0 to 24 such replacements, but that if we choose to do 24 then the next block cannot have chosen 0replacements. (An analogy with lower numbers is 10001000 = 01101000 = 01100110 = 01011110, withthe first block ”replaced twice,” which was only allowed since the second block had ”replaced once,”opening up the slot which was filled by the last 1 in the final replacement 011).

Thus we have a bijection from f(n) to sequences in {0, . . . , 24}n−1 such that (a) the sequence does notend in 24 and (b) the sequence never has a 24 followed by a 0.

We let an denote the number of length-n sequences starting with a 0, bn for the number of suchsequences starting with any of 1 to 23, and cn for the number of such sequences starting with 24. Weknow a1 = 1, b1 = 23, c0 = 0 and that f(n) = an−1 + bn−1 + cn−1.

Now,

an = an−1 + bn−1 + cn−1

bn = 23(an−1 + bn−1 + cn−1)

cn = bn−1 + cn−1

so bn = 23an for all n. Substituting gives an = 24an−1 + cn−1, cn = 23an−1 + cn−1. Solving forcn = an+1 〉 24an−1 and plugging in gives

an+1 〉 24an = an 〉 an−1,

which gives a characteristic polynomial of λ2〉 25λ+ 1 = 0. We easily find that an grows as λn (whereλ is the larger solution to the quadratic equation) and thus bn, cn do as well, implying that f(n) growsas λn, where

λ =25 +

√252 〉 4

2=

25 + 3√

69

2,

which is our answer.

HMMT November 2017November 11, 2017

Team Round

1. [15] A positive integer k is called powerful if there are distinct positive integers p, q, r, s, t such thatp2, q3, r5, s7, t11 all divide k. Find the smallest powerful integer.

2. [20] How many sequences of integers (a1, . . . , a7) are there for which 〉1 ≤ ai ≤ 1 for every i, and

a1a2 + a2a3 + a3a4 + a4a5 + a5a6 + a6a7 = 4?

3. [25] Michael writes down all the integers between 1 and N inclusive on a piece of paper and discoversthat exactly 40% of them have leftmost digit 1. Given that N > 2017, find the smallest possible valueof N .

4. [30] An equiangular hexagon has side lengths 1, 1, a, 1, 1, a in that order. Given that there exists acircle that intersects the hexagon at 12 distinct points, we have M < a < N for some real numbers Mand N . Determine the minimum possible value of the ratio N

M .

5. [35] Ashwin the frog is traveling on the xy-plane in a series of 22017 〉 1 steps, starting at the origin.At the nth step, if n is odd, then Ashwin jumps one unit to the right. If n is even, then Ashwin jumpsm units up, where m is the greatest integer such that 2m divides n. If Ashwin begins at the origin,what is the area of the polygon bounded by Ashwin’s path, the line x = 22016, and the x-axis?

6. [40] Consider five-dimensional Cartesian space

R5 = {(x1, x2, x3, x4, x5) | xi ∈ R},

and consider the hyperplanes with the following equations:

• xi = xj for every 1 ≤ i < j ≤ 5;

• x1 + x2 + x3 + x4 + x5 = 〉1;

• x1 + x2 + x3 + x4 + x5 = 0;

• x1 + x2 + x3 + x4 + x5 = 1.

Into how many regions do these hyperplanes divide R5?

7. [50] There are 12 students in a classroom; 6 of them are Democrats and 6 of them are Republicans.Every hour the students are randomly separated into four groups of three for political debates. Ifa group contains students from both parties, the minority in the group will change his/her politicalalignment to that of the majority at the end of the debate. What is the expected amount of timeneeded for all 12 students to have the same political alignment, in hours?

8. [55] Find the number of quadruples (a, b, c, d) of integers with absolute value at most 5 such that

(a2 + b2 + c2 + d2)2 = (a+ b+ c+ d)(a〉 b+ c〉 d)((a〉 c)2 + (b〉 d)2).

9. [60] Let A,B,C,D be points chosen on a circle, in that order. Line BD is reflected over lines ABand DA to obtain lines `1 and `2 respectively. If lines `1, `2, and AC meet at a common point and ifAB = 4, BC = 3, CD = 2, compute the length DA.

10. [70] Yannick has a bicycle lock with a 4-digit passcode whose digits are between 0 and 9 inclusive.(Leading zeroes are allowed.) The dials on the lock is currently set at 0000. To unlock the lock, everysecond he picks a contiguous set of dials, and increases or decreases all of them by one, until the dialsare set to the passcode. For example, after the first second the dials could be set to 1100, 0010, or9999, but not 0909 or 0190. (The digits on each dial are cyclic, so increasing 9 gives 0, and decreasing0 gives 9.) Let the complexity of a passcode be the minimum number of seconds he needs to unlockthe lock. What is the maximum possible complexity of a passcode, and how many passcodes have thismaximum complexity? Express the two answers as an ordered pair.

HMMT November 2017November 11, 2017

Team Round

1. [15] A positive integer k is called powerful if there are distinct positive integers p, q, r, s, t such thatp2, q3, r5, s7, t11 all divide k. Find the smallest powerful integer.

Proposed by: Yuan Yao

Answer: 1024

First of all, 1024 is powerful because it can be divided evenly by 162 = 256, 83 = 512, 45 = 1024, 27 =128, and 111 = 1.

Now we show that 1024 is the smallest powerful number. Since s 6= t, at least one of them is at least 2.If t ≥ 2 or s ≥ 3, then we need the number to be divisible by at least 211 = 2048 or 37 = 2187, whichboth exceed 1024, so we must have s = 2 and t = 1. If r = 3, then the number must be divisible by35 = 243 and 27 = 128, which means that the number is at least 243 · 128 > 1024, so r ≥ 4, and thenumber is at least 45 = 1024. Therefore the smallest powerful number is indeed 1024.

2. [20] How many sequences of integers (a1, . . . , a7) are there for which 〉1 ≤ ai ≤ 1 for every i, and

a1a2 + a2a3 + a3a4 + a4a5 + a5a6 + a6a7 = 4?

Proposed by: mendel keller

Answer: 38

For i = 1, 2, . . . , 6, let bi = aiai+1. From the problem condition each of b1, b2, . . . , b6 can only be 〉1, 0,or 1. Since the sum of these six numbers is 4, either there are five 1s and a 〉1 or there are four 1s andtwo 0s.

In the first case, there are 6 ways to choose i such that bi = 〉1. Once that is fixed, determining thevalue of a1 (one of 1 and 〉1) will determine the value of all the remaining ai’s, so there are 6 · 2 = 12possible ways in this case.

In the second case, since if one of b2, b3, b4, b5 is zero, then one of the adjacent term to this zero termmust also be zero. Therefore the two zeroes must be next to each other or be b1 and b6.

If b1 = b2 = 0, then a2 must be zero. a1’s value doesn’t matter, and a3, a4, . . . , a7 must have the samesign. The same goes for b5 = b6 = 0, giving 3 · 2 · 2 = 12 possibilities in these two cases.

If bi = bi+1 = 0 for i = 2, 3, 4, then ai+1 must be zero. Moreover, a1, a2, . . . , ai must have the samesign, and so do ai+2, . . . , a7. this gives 2 · 2 · 3 = 12 possibilities in these three cases.

If b1 = b6 = 0, then a1 = a7 = 0. Also, a2, a3, . . . , a6 must have the same sign so there are 2 possibilities.

Combining these cases gives 12 + 12 + 12 + 2 = 38 possible sequences in total.

3. [25] Michael writes down all the integers between 1 and N inclusive on a piece of paper and discoversthat exactly 40% of them have leftmost digit 1. Given that N > 2017, find the smallest possible valueof N .

Proposed by: Michael Tang

Answer: 1,481,480

Let d be the number of digits of N . Suppose that N does not itself have leftmost digit 1. Then thenumber of integers 1, 2, . . . , N which have leftmost digit 1 is

1 + 10 + 102 + . . .+ 10d→1 =10d 〉 1

9,

so we must have 10d→19 = 2N

5 , or 5(10d〉 1) = 18N . But the left-hand side is odd, so this is impossible.

Thus N must have leftmost digit 1. In this case, the number of integers 1, 2, . . . , N which have leftmostdigit 1 is

1 + 10 + 102 + . . .+ 10d→2 + (N 〉 10d→1 + 1)

=10d→1 〉 1

9+N 〉 10d→1 + 1

= N 〉 8

(10d→1 〉 1

9

).

Therefore we need N 〉 8(

10d−1→19

)= 2

5N , or N = 403

(10d−1→1

9

). Then, 10d−1→1

9 must be divisible by

3. The base-ten representation of 10d−1→19 has d〉 1 ones, so d〉 1 must be divisible by 3. Both d = 1

and d = 4 make N less than 2017, but d = 7 gives the answer N = 403 (111111) = 1481480 .

4. [30] An equiangular hexagon has side lengths 1, 1, a, 1, 1, a in that order. Given that there exists acircle that intersects the hexagon at 12 distinct points, we have M < a < N for some real numbers Mand N . Determine the minimum possible value of the ratio N

M .

Proposed by: Yuan Yao

Answer: 3√3→1 OR 3

√3+32

We claim that the greatest possible value of M is√

3〉 1, whereas the least possible value of N is 3.

To begin, note that the condition requires the circle to intersect each side of the hexagon at two pointson its interior. This implies that the center must be inside the hexagon as its projection onto all six sidesmust be on their interior. Suppose that the hexagon is ABCDEF , with AB = BC = DE = EF = 1,CD = FA = a, and the center O.

When a ≤√

3〉 1, we note that the distance from O to CD (which is√32 ) is greater than or equal to

the distance from O to B or E (which is a+12 ). However, for the circle to intersect all six sides at two

points each, the distance from the center of the circle to CD and to FA must be strictly less than thatfrom the center to B and to E, because otherwise any circle that intersects CD and FA at two pointseach must include B or E on its boundary or interior, which will not satisfy the condition. WLOGassume that the center of the circle is closer to FA than to CD, including equality (in other words,the center is on the same side of BE as FA, possibly on BE itself), then note that the parabola withfoci B and E and common directrix CD intersects on point O, which means that there does not exista point in the hexagon on the same side of BE as FA that lies on the same side of both parabola asCD. This means that the center of the circle cannot be chosen.

When a =√

3〉1+ε for some very small real number ε > 0, the circle with center O and radius r =√32

intersects sides AB,BC,DE,EF at two points each and is tangent to CD and FA on their interior.Therefore, there exists a real number ε′ > 0 such that the circle with center O and radius r′ = r + ε′

satisfy the requirement.

When a ≥ 3, we note that the projection of BF onto BC has length | 12 〉a2 | ≥ 1, which means that the

projection of F onto side BC is not on its interior, and the same goes for side EF onto BC. However,for a circle to intersect both BC and EF at two points, the projection of center of the circle onto thetwo sides must be on their interior, which cannot happen in this case.

When a = 3〉 ε for some very small real number ε > 0, a circle with center O and radius r =√34 (a+ 1)

intersects AF and CD at two points each and is tangent to all four other sides on their interior.Therefore, there exists a real number ε′ > 0 such that the circle with center O and radius r′ = r + ε′

satisfy the requirement.

With M ≤√

3〉 1 and N ≥ 3, we have NM ≥

3√3→1 = 3

√3+32 , which is our answer.

5. [35] Ashwin the frog is traveling on the xy-plane in a series of 22017 〉 1 steps, starting at the origin.At the nth step, if n is odd, then Ashwin jumps one unit to the right. If n is even, then Ashwin jumpsm units up, where m is the greatest integer such that 2m divides n. If Ashwin begins at the origin,what is the area of the polygon bounded by Ashwin’s path, the line x = 22016, and the x-axis?

Proposed by: Nikhil Reddy

Answer: 22015 · (22017 〉 2018)

Notice that since v2(x) = v2(22017 〉 x), the path divides the rectangle bonded by the coordinate axesand the two lines passing through Ashwin’s final location parallel to the axes. The answer is thereforehalf of the product of the coordinates of Ashwin’s final coordinates. The x-coordinate is the number ofodd number steps (which is 22016). The y-coordinate is the number of total powers of 2 in (22017〉 1)!.The final answer is therefore 22015 · (22017 〉 2018).

6. [40] Consider five-dimensional Cartesian space

R5 = {(x1, x2, x3, x4, x5) | xi ∈ R},

and consider the hyperplanes with the following equations:

• xi = xj for every 1 ≤ i < j ≤ 5;

• x1 + x2 + x3 + x4 + x5 = 〉1;

• x1 + x2 + x3 + x4 + x5 = 0;

• x1 + x2 + x3 + x4 + x5 = 1.

Into how many regions do these hyperplanes divide R5?

Proposed by: Mehtaab Sawhney

Answer: 480

(Joint with Junyao Peng)

Note that given a set of plane equations Pi(x1, x2, x3, x4, x5) = 0, for i = 1, 2, . . . , n, each regionthat the planes separate the space into correspond to a n-tuple of 〉1 and 1, representing the sign ofP1, P2, . . . Pn for all points in that region.

Therefore, the first set of planes separate the space into 5! = 120 regions, with each region representingan ordering of the five coordinates by numerical size. Moreover, the next three planes are parallel toeach other and perpendicular to all planes in the first set, so these three planes separate each regioninto 4. Therefore, a total of 4 · 120 = 480 regions is created.

7. [50] There are 12 students in a classroom; 6 of them are Democrats and 6 of them are Republicans.Every hour the students are randomly separated into four groups of three for political debates. Ifa group contains students from both parties, the minority in the group will change his/her politicalalignment to that of the majority at the end of the debate. What is the expected amount of timeneeded for all 12 students to have the same political alignment, in hours?

Proposed by: Yuan Yao

Answer: 34154

When the party distribution is 6 〉 6, the situation can change (to 3 〉 9) only when a group of threecontains three people from the same party, and the remaining three are distributed evenly across theother three groups (to be converted).

To compute the probability, we assume that the groups and the members of the group are ordered (sothere are 12! ways of grouping). There are 2 ways to choose the party, 4 ways to choose the group,6 · 5 · 4 ways to choose the three members of the group, 9 · 6 · 3 ways to place the other three membersof the party, and 6! ways to fill in the members of the other party. The probability is then

2 · 4 · 6 · 5 · 4 · 9 · 6 · 3 · 6!

12!=

2 · 4 · 6 · 5 · 4 · 9 · 6 · 312 · 11 · 10 · 9 · 8 · 7

=18

77.

This means that the shift in distribution will happen in 7718 hours on average.

When the distribution is 3 〉 9, the situation can change (to 0 〉 12) only when the three members ofthe minority party are all in different groups. Using the similar method as above, there are 12 · 9 · 6ways to place the three members and 9! ways to place the rest, so the probability is

12 · 9 · 6 · 9!

12!=

12 · 9 · 612 · 11 · 10

=27

55.

This means that the shift in distribution will happen in 5527 hours on average.

By linearity of expectation, we can add up the two results and get that the expected value is 7718 + 55

27 =34155 hours.

8. [55] Find the number of quadruples (a, b, c, d) of integers with absolute value at most 5 such that

(a2 + b2 + c2 + d2)2 = (a+ b+ c+ d)(a〉 b+ c〉 d)((a〉 c)2 + (b〉 d)2).

Proposed by: Mehtaab Sawhney

Answer: 49

Let x = a+ c, y = a〉 c, w = b+ d, and z = b〉 d. Then

(w2 + x2 + y2 + z2)2 = 4(x2 〉 w2)(y2 + z2)

and since |x2 + w2| ≥ |x2 〉 w2| it follows that w = 0 or y = z = 0. Now y = z = 0 impliesa = b = c = d = 0. Now w = 0 gives b = 〉d. Then for equality to hold x2 = y2 +z2. This is equivalentto ac = b2, which includes the previous case. It suffices to count the number of triples (a, b, c) thatsatisfy the equation.

When b = 0, either a or c is zero, which gives 11 + 11〉 1 = 21 triples.

When b = ±1, we have |a| = |c| = 1 and a, c have the same sign, for 2 · 2 = 4 triples.

When b = ±2, we have (a, c) = (1, 4), (2, 2), (4, 1) or their negatives, for 2 · 6 = 12 triples.

When b = ±3,±4,±5, we have |a| = |b| = |c| and a, c have the same sign, for 6 · 2 = 12 triples.

So in total there are 21 + 4 + 12 + 12 = 49 solutions.

9. [60] Let A,B,C,D be points chosen on a circle, in that order. Line BD is reflected over lines ABand DA to obtain lines `1 and `2 respectively. If lines `1, `2, and AC meet at a common point and ifAB = 4, BC = 3, CD = 2, compute the length DA.

Proposed by: Ashwin Sah

Answer:√

21

Let the common point be E. Then since lines BE and BD are symmetric about line BA, BA isan exterior bisector of ∠DBE, and similarly DA is also an exterior bisector of ∠BDE. ThereforeA is the E-excenter of triangle BDE and thus lie on the interior bisector of ∠BED. Since C lieson line AE, and by the fact that A,B,C,D are concyclic, we get that ∠ABC + ∠ADC = 180◦,which implies ∠DBC + ∠BDC = 1

2 (∠DBE + ∠BDE), so C is the incenter of triangle BDE. Thus∠ABC = ∠CDA = π

2 , and thus DA2 = AC2 〉 CD2 = AB2 +BC2 〉 CD2 = 32 + 42 〉 22 = 21. The

length of DA is then√

21.

10. [70] Yannick has a bicycle lock with a 4-digit passcode whose digits are between 0 and 9 inclusive.(Leading zeroes are allowed.) The dials on the lock is currently set at 0000. To unlock the lock, everysecond he picks a contiguous set of dials, and increases or decreases all of them by one, until the dialsare set to the passcode. For example, after the first second the dials could be set to 1100, 0010, or9999, but not 0909 or 0190. (The digits on each dial are cyclic, so increasing 9 gives 0, and decreasing0 gives 9.) Let the complexity of a passcode be the minimum number of seconds he needs to unlockthe lock. What is the maximum possible complexity of a passcode, and how many passcodes have thismaximum complexity? Express the two answers as an ordered pair.

Proposed by: Yuan Yao

Answer: (12, 2)

To simplify the solution, we instead consider the equivalent problem of reducing a passcode to 0000using the given move.

Given a passcode a1a2a3a4, define a differential of the passcode to be a quintuple (d1, d2, d3, d4, d5)such that di ≡ ai 〉 ai→1 (mod 10) for i = 1, 2, 3, 4, 5, where we define a0 = a5 = 0.

Claim 1: For any passcode, there exists a differential that satisfies the following two conditions:

• d1 + d2 + d3 + d4 + d5 = 0;

• The range (difference between the maximum and minimum) of these five numbers is at most 10.

Proof: We first see that the differential defined by di = ai 〉 ai→1 satisfy the first condition since thesum of the five numbers is a5 〉 a0 = 0. Suppose that a differential satisfying the first condition has arange greater than 10, then we take one of the largest number dm and one of the smallest number dn(where dm 〉 dn > 10), and replace the former by dm 〉 10 and the latter by dn + 10. This will eitherreduce the range or reduce the number of maximal and minimal numbers, so the process will terminateafter finitely many iterations. Thus we can find a differential that satisfies both conditions.

(Note: we call such a differential a standard differential from now on, although it is important toremember that there may be more than one standard differential for one passcode. As a corollary, allof the numbers in a standard differential must be in the range [〉9, 9], as a number greater than 9 willbe more than 10 away from a negative number, and similar for a number smaller than 〉9.)

Given a passcode, we define the magnitude of one of its standard differentials to be the sum of all thepositive values in the differential (which is also the absolute value of the sum all the negative values).

Claim 2: The magnitude of the a passcode’s standard differential is equal to the passcode’s complexity.

Proof: Obviously 0000 is the only passcode with complexity 0 whose standard differential has mag-nitude 0. Suppose that the magnitude of a standard differential is M , then it suffices show that themagnitude can be reduced to 0 in M moves, and that the magnitude can only decrease by at most 1with each move.

The first part can be shown via the following algorithm. When the magnitude is not zero, there mustbe a positive number di and a negative number dj . WLOG assume that i < j, then after taking thedials ai, ai+1, . . . , aj→1 and decreases them all by 1, di decreases by 1 and dj increases by 1. This willdecrease the magnitude by 1 (and the differential remains standard), so by repeating this process Mtimes we can bring the magnitude down to 0.

For the second part, we assume WLOG that we take the dials ai, ai+1, . . . , aj→1 and decrease them allby 1, and then di is replaced by d′i = di 〉 1 and d′j = dj + 1. If the differential remains standard, thenthe magnitude decreases by 1 (when di > 0 and dj < 0), remains the same (when either di ≤ 0 anddj < 0 or di > 0 and dj ≥ 0), or increases by 1 (when di ≤ 0 and dj ≥ 0).

In the latter two cases, it is possible that the differential is no longer standard.

If the magnitude previously remained the same (WLOG suppose that di > 0 and dj ≥ 0), then thereexists a negative dk that is minimal such that dj 〉 dk = 10 and now d′j is the unique maximum.Replacing dk by dk + 10 = dj and d′j by d′j 〉 10 = dk + 1 will reduce the magnitude by 1, and the newdifferential will be standard because the unique maximum d′j is no longer present and the minimum isnow either dk or dk + 1. This means that the magnitude decreases by at most 1.

If the magnitude previously increased by 1 (when di ≤ 0 and dj ≥ 0), then there exists either a negativedk that is (previously) minimal such that dj 〉 dk = 10, or a positive dl that is (previously) maximalsuch that dl 〉 di = 10, or both. By similar logic as the previous case, replacing dk by dk + 10 and d′jby d′j 〉 10, or replacing dl by dl〉 10 and d′i by d′i + 10 (or both, if both dk and dl exist). will decreasethe magnitude by 1, and ensure that the new differential is standard. The replacement will decreasethe current magnitude by at most 2, this means that the original magnitude decreases by at most 1 intotal.

These considerations finishes the second part and therefore the proof.

With this claim, we also see that the magnitudes of all possible standard differentials of a given passcodeare the same, so the choice of the differential is irrelevant.

We can now proceed to find the maximum possible complexity. Suppose that there are m positivenumbers and n negative numbers in the differential, and suppose that the maximum and the minimumare x and 〉y respectively. Since the sum of all positive numbers is at most mx and the absolute valueof the sum of all negative numbers is at most ny, the complexity is at most C = min(mx, ny). Itsuffices to maximize C under the condition that m + n ≤ 5 and x + y = x 〉 (〉y) ≤ 10. It is notdifficult to see (via casework) that the maximal C is 12, achieved by m = 2, n = 3, x = 6, y = 4 orm = 3, n = 2, x = 4, y = 6. In the first case, the digits must increase from 0 by 6 twice and decreasesby 4 three times (and reduced modulo 10), which gives the passcode 6284; in the second case the digitsincrease by 4 three times and decreases by 6 twice instead, which gives the passcode 4826. Since allinequalities are tight, these two passcodes are the only ones that has the maximal complexity of 12.

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HMMT November 2017, November 11, 2017 — GUTS ROUND

Organization Team Team ID#

1. [2] Suppose x is a rational number such that x√

2 is also rational. Find x.

2. [2] Regular octagon CHILDREN has area 1. Find the area of pentagon CHILD.

3. [2] The length of a rectangle is three times its width. Given that its perimeter and area are bothnumerically equal to k > 0, find k.

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HMMT November 2017, November 11, 2017 — GUTS ROUND

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4. [3] Alec wishes to construct a string of 6 letters using the letters A, C, G, and N, such that:

• The first three letters are pairwise distinct, and so are the last three letters;

• The first, second, fourth, and fifth letters are pairwise distinct.

In how many ways can he construct the string?

5. [3] Define a sequence {an} by a1 = 1 and an = (an∈1)! + 1 for every n > 1. Find the least n for whichan > 1010.

6. [3] Lunasa, Merlin, and Lyrica each have a distinct hat. Every day, two of these three people, selectedrandomly, switch their hats. What is the probability that, after 2017 days, every person has their ownhat back?

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HMMT November 2017, November 11, 2017 — GUTS ROUND

Organization Team Team ID#

7. [3] Compute

1002 + 992 〉 982 〉 972 + 962 + 952 〉 942 〉 932 + . . . + 42 + 32 〉 22 〉 12.

8. [3] Regular octagon CHILDREN has area 1. Find the area of quadrilateral LINE.

9. [3] A malfunctioning digital clock shows the time 9 : 57 AM; however, the correct time is 10 : 10 AM.There are two buttons on the clock, one of which increases the time displayed by 9 minutes, and anotherwhich decreases the time by 20 minutes. What is the minimum number of button presses necessary tocorrectly set the clock to the correct time?

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HMMT November 2017, November 11, 2017 — GUTS ROUND

Organization Team Team ID#

10. [4] Computex

wif w 6= 0 and

x + 6y 〉 3z

〉3x + 4w=〉2y + z

x〉 w=

2

3.

11. [4] Find the sum of all real numbers x for which

bb· · · bbbxc+ xc+ xc · · · c+ xc = 2017 and {{· · · {{{x}+ x}+ x} · · · }+ x} =1

2017

where there are 2017 x’s in both equations. (bxc is the integer part of x, and {x} is the fractional partof x.) Express your sum as a mixed number.

12. [4] Trapezoid ABCD, with bases AB and CD, has side lengths AB = 28, BC = 13, CD = 14, andDA = 15. Let diagonals AC and BD intersect at P , and let E and F be the midpoints of AP andBP , respectively. Find the area of quadrilateral CDEF .

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HMMT November 2017, November 11, 2017 — GUTS ROUND

Organization Team Team ID#

13. [5] Fisica and Ritmo discovered a piece of Notalium shaped like a rectangular box, and wanted tofind its volume. To do so, Fisica measured its three dimensions using a ruler with infinite precision,multiplied the results and rounded the product to the nearest cubic centimeter, getting a result of Vcubic centimeters. Ritmo, on the other hand, measured each dimension to the nearest centimeter andmultiplied the rounded measurements, getting a result of 2017 cubic centimeters. Find the positivedifference between the least and greatest possible positive values for V .

14. [5] Points A, B, C, and D lie on a line in that order such thatAB

BC=

DA

CD. If AC = 3 and BD = 4,

find AD.

15. [5] On a 3 × 3 chessboard, each square contains a knight with 12 probability. What is the probability

that there are two knights that can attack each other? (In chess, a knight can attack any piece whichis two squares away from it in a particular direction and one square away in a perpendicular direction.)

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HMMT November 2017, November 11, 2017 — GUTS ROUND

Organization Team Team ID#

16. [7] A repunit is a positive integer, all of whose digits are 1s. Let a1 < a2 < a3 < . . . be a list of all thepositive integers that can be expressed as the sum of distinct repunits. Compute a111.

17. [7] A standard deck of 54 playing cards (with four cards of each of thirteen ranks, as well as twoJokers) is shuffled randomly. Cards are drawn one at a time until the first queen is reached. What isthe probability that the next card is also a queen?

18. [7] Mr. Taf takes his 12 students on a road trip. Since it takes two hours to walk from the school tothe destination, he plans to use his car to expedite the journey. His car can take at most 4 students ata time, and travels 15 times as fast as traveling on foot. If they plan their trip optimally, what is theshortest amount of time it takes for them to all reach the destination, in minutes?

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HMMT November 2017, November 11, 2017 — GUTS ROUND

Organization Team Team ID#

19. [9] The skeletal structure of coronene, a hydrocarbon with the chemical formula C24H12, is shownbelow.

Each line segment between two atoms is at least a single bond. However, since each carbon (C) requiresexactly four bonds connected to it and each hydrogen (H) requires exactly one bond, some of the linesegments are actually double bonds. How many arrangements of single/double bonds are there suchthat the above requirements are satisfied? (Rotations and reflections of the same arrangement areconsidered distinct.)

20. [9] Rebecca has four resistors, each with resistance 1 ohm. Every minute, she chooses any two resistorswith resistance of a and b ohms respectively, and combine them into one by one of the followingmethods:

• Connect them in series, which produces a resistor with resistance of a + b ohms;

• Connect them in parallel, which produces a resistor with resistance of aba+b ohms;

• Short-circuit one of the two resistors, which produces a resistor with resistance of either a or bohms.

Suppose that after three minutes, Rebecca has a single resistor with resistance R ohms. How manypossible values are there for R?

21. [9] A box contains three balls, each of a different color. Every minute, Randall randomly draws a ballfrom the box, notes its color, and then returns it to the box. Consider the following two conditions:

(1) Some ball has been drawn at least three times (not necessarily consecutively).

(2) Every ball has been drawn at least once.

What is the probability that condition (1) is met before condition (2)?

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HMMT November 2017, November 11, 2017 — GUTS ROUND

Organization Team Team ID#

22. [12] A sequence of positive integers a1, a2, . . . , a2017 has the property that for all integers m where1 ≤ m ≤ 2017, 3(

∑mi=1 ai)

2 =∑m

i=1 a3i . Compute a1337.

23. [12] A string of digits is defined to be similar to another string of digits if it can be obtained byreversing some contiguous substring of the original string. For example, the strings 101 and 110 aresimilar, but the strings 3443 and 4334 are not. (Note that a string is always similar to itself.) Considerthe string of digits

S = 01234567890123456789012345678901234567890123456789,

consisting of the digits from 0 to 9 repeated five times. How many distinct strings are similar to S?

24. [12] Triangle ABC has side lengths AB = 15, BC = 18, CA = 20. Extend CA and CB to points Dand E respectively such that DA = AB = BE. Line AB intersects the circumcircle of CDE at P andQ. Find the length of PQ.

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HMMT November 2017, November 11, 2017 — GUTS ROUND

Organization Team Team ID#

Déjà vu?

25. [15] Fisica and Ritmo discovered a piece of Notalium shaped like a rectangular box, and wanted tofind its volume. To do so, Fisica measured its three dimensions using a ruler with infinite precision,multiplied the results and rounded the product to the nearest cubic centimeter, getting a result of2017 cubic centimeters. Ritmo, on the other hand, measured each dimension to the nearest centimeterand multiplied the rounded measurements, getting a result of V cubic centimeters. Find the positivedifference between the least and greatest possible positive values for V .

26. [15] Points A,B,C,D lie on a circle in that order such thatAB

BC=

DA

CD. If AC = 3 and BD = BC = 4,

find AD.

27. [15] On a 3 × 3 chessboard, each square contains a Chinese knight with 12 probability. What is the

probability that there are two Chinese knights that can attack each other? (In Chinese chess, a Chineseknight can attack any piece which is two squares away from it in a particular direction and one squareaway in a perpendicular direction, under the condition that there is no other piece immediately adjacentto it in the first direction.)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2017, November 11, 2017 — GUTS ROUND

Organization Team Team ID#

28. [19] Compute the number of functions f : {1, 2, . . . , 9} → {1, 2, . . . , 9} which satisfy f(f(f(f(f(x))))) =x for each x ∈ {1, 2, . . . , 9}.

29. [19] Consider a sequence xn such that x1 = x2 = 1, x3 =2

3. Suppose that xn =

x2n∈1xn∈2

2x2n∈2 〉 xn∈1xn∈3

for all n ≥ 4. Find the least n such that xn ≤1

106.

30. [19] Given complex number z, define sequence z0, z1, z2, . . . as z0 = z and zn+1 = 2z2n + 2zn for n ≥ 0.Given that z10 = 2017, find the minimum possible value of |z|.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2017, November 11, 2017 — GUTS ROUND

Organization Team Team ID#

31. [24] In unit square ABCD, points E,F,G are chosen on side BC,CD,DA respectively such that AE

is perpendicular to EF and EF is perpendicular to FG. Given that GA =404

1331, find all possible

values of the length of BE.

32. [24] Let P be a polynomial with integer coefficients such that P (0) + P (90) = 2018. Find the leastpossible value for |P (20) + P (70)|.

33. [24] Tetrahedron ABCD with volume 1 is inscribed in circumsphere ω such that AB = AC = AD = 2and BC · CD ·DB = 16. Find the radius of ω.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2017, November 11, 2017 — GUTS ROUND

Organization Team Team ID#

Déjà vu??

34. [30] The skeletal structure of circumcircumcircumcoronene, a hydrocarbon with the chemical formulaC150H30, is shown below.

Each line segment between two atoms is at least a single bond. However, since each carbon (C) requiresexactly four bonds connected to it and each hydrogen (H) requires exactly one bond, some of the linesegments are actually double bonds. How many arrangements of single/double bonds are there suchthat the above requirements are satisfied? (Rotations and reflections of the same arrangement areconsidered distinct.)

If the correct answer is C and your answer is A, you get max(⌊

30(

1〉∣∣∣loglog2 C

AC

∣∣∣)⌋ , 0) points.

35. [30] Rebecca has twenty-four resistors, each with resistance 1 ohm. Every minute, she chooses anytwo resistors with resistance of a and b ohms respectively, and combine them into one by one of thefollowing methods:

• Connect them in series, which produces a resistor with resistance of a + b ohms;

• Connect them in parallel, which produces a resistor with resistance of aba+b ohms;

• Short-circuit one of the two resistors, which produces a resistor with resistance of either a or bohms.

Suppose that after twenty-three minutes, Rebecca has a single resistor with resistance R ohms. Howmany possible values are there for R?

If the correct answer is C and your answer is A, you get max(⌊

30(

1〉∣∣∣loglog2 C

AC

∣∣∣)⌋ , 0) points.

36. [30] A box contains twelve balls, each of a different color. Every minute, Randall randomly draws aball from the box, notes its color, and then returns it to the box. Consider the following two conditions:

(1) Some ball has been drawn at least twelve times (not necessarily consecutively).

(2) Every ball has been drawn at least once.

What is the probability that condition (1) is met before condition (2)?

If the correct answer is C and your answer is A, you get max(⌊

30(1〉 1

2 |log2 A〉 log2 C|)⌋

, 0)

points.

HMMT November 2017November 11, 2017

Guts Round

1. [2] Suppose x is a rational number such that x√

2 is also rational. Find x.

Proposed by: Michael Kural

Answer: 0

Suppose x 6= 0. Then x√2

x =√

2 is the quotient of two nonzero rationals, and so is rational. However,

it is well-known that√

2 is irrational. Therefore any solution x must satisfy x = 0. We can see that 0is rational and 0

√2 = 0 is rational, so the answer is indeed x = 0 .

2. [2] Regular octagon CHILDREN has area 1. Find the area of pentagon CHILD.

Proposed by: Yuan Yao

Answer: 12

The pentagon CHILD is congruent to the pentagon NERDC, as their corresponding angles and sidesare congruent. Moreover, the two pentagons together compose the entire octagon, so each pentagonmust have area one-half of the area of the octagon, or 1

2 .

3. [2] The length of a rectangle is three times its width. Given that its perimeter and area are bothnumerically equal to k > 0, find k.

Proposed by: Yuan Yao

Answer: 643

Let a be the width of the rectangle. Then the length of the rectangle is 3a, so the perimeter is2(a+ 3a) = 8a, and the area is 3a2. Since the length is numerically equal to the width, we know that

8a = 3a2 = k.

Because k > 0, the rectangle is non-degenerate. It follows that 8 = 3a, so a = 83 . Therefore, k =

64

3.

4. [3] Alec wishes to construct a string of 6 letters using the letters A, C, G, and N, such that:

• The first three letters are pairwise distinct, and so are the last three letters;

• The first, second, fourth, and fifth letters are pairwise distinct.

In how many ways can he construct the string?

Proposed by: Yuan Yao

There are 4! = 24 ways to decide the first, second, fourth, and fifth letters because these letters canbe selected sequentially without replacement from the four possible letters. Once these four lettersare selected, there are 2 ways to select the third letter because two distinct letters have already beenselected for the first and second letters, leaving two possibilities. The same analysis applies to the sixthletter. Thus, there are 24 · 22 = 96 total ways to construct the string.

5. [3] Define a sequence {an} by a1 = 1 and an = (an−1)! + 1 for every n > 1. Find the least n for whichan > 1010.

Proposed by: Michael Tang

Answer: 6

We have a2 = 2, a3 = 3, a4 = 7, a5 = 7! + 1 = 5041, and a6 = 5041! + 1. But

5041! + 1� 5041 · 5040 · 5039 > 1010.

Hence, the answer is 6 .

6. [3] Lunasa, Merlin, and Lyrica each have a distinct hat. Every day, two of these three people, selectedrandomly, switch their hats. What is the probability that, after 2017 days, every person has their ownhat back?

Proposed by: Yuan Yao

Answer: 0

Imagine that the three hats are the vertices of an equilateral triangle. Then each day the exchange isequivalent to reflecting the triangle along one of its three symmetry axes, which changes the orientationof the triangle (from clockwise to counterclockwise or vice versa). Thus, an even number of suchexchanges must be performed if the orientation is to be preserved. Since the triangle is reflected 2017times, it is impossible for the final triangle to have the same orientation as the original triangle, so thedesired probability is 0 .

7. [3] Compute

1002 + 992 − 982 − 972 + 962 + 952 − 942 − 932 + . . .+ 42 + 32 − 22 − 12.

Proposed by: Michael Tang

Answer: 10100

Note that (n+ 3)2 − (n+ 2)2 − (n+ 1)2 + n2 = 4 for every n. Therefore, adding 02 to the end of thegiven sum and applying this identity for every four consecutive terms after 1002, we see that the givensum is equivalent to 1002 + 25 · 4 = 10100 .

Alternatively, we can apply the difference-of-squares factorization to rewrite 1002 − 982 = (100 −98)(100 + 98) = 2(100 + 98), 992 − 972 = (99− 97)(99 + 97) = 2(99 + 97), etc. Thus, the given sum isequivalent to 2(100 + 99 + · · ·+ 2 + 1) = 2 · 100·1012 = 10100.

8. [3] Regular octagon CHILDREN has area 1. Determine the area of quadrilateral LINE.

Proposed by: Yuan Yao

Answer: 12

Suppose that the side length CH =√

2a, then the area of the octagon is ((2 +√

2)a)2 − 4 · 12a2 =

(4 + 4√

2)a2, and the area of LINE is (√

2a)((2 +√

2)a) = (2 + 2√

2)a2, which is exactly one-half ofthe area of the octagon. Therefore the area of LINE is 1

2 .

9. [3] A malfunctioning digital clock shows the time 9 : 57 AM; however, the correct time is 10 : 10 AM.There are two buttons on the clock, one of which increases the time displayed by 9 minutes, and anotherwhich decreases the time by 20 minutes. What is the minimum number of button presses necessary tocorrectly set the clock to the correct time?

Proposed by: Kevin Sun

Answer: 24

We need to increase the time by 13 minutes. If we click the 9 minute button a times and the 20minute button b times, then we must have 9a− 20b = 13. Note that if this equation is satisfied, thenb increases as a increases, so it suffices to minimize a. This means that a must end in a 7. However,since 63− 20b = 13 has no integer solution, the next smallest possible value of a is 17, which gives thesolution (a, b) = (17, 7), or 24 button presses.

10. [4] Computex

wif w 6= 0 and

x+ 6y − 3z

−3x+ 4w=−2y + z

x− w=

2

3.

Proposed by: Angela Deng

Answer: 23

We have x+ 6y − 3z = 23 (−3x+ 4w) and −2y + z = 2

3 (x− w), so

x

w=

(x+ 6y − 3z) + 3(−2y + z)

(−3x+ 4w) + 3(x− w)=

23 (−3x+ 4w) + 3 · 23 (x− w)

(−3x+ 4w) + 3(x− w)=

23 [(−3x+ 4w) + 3(x− w)]

(−3x+ 4w) + 3(x− w)=

2

3

.

11. [4] Find the sum of all real numbers x for which

bb· · · bbbxc+ xc+ xc · · · c+ xc = 2017 and {{· · · {{{x}+ x}+ x} · · · }+ x} =1

2017

where there are 2017 x’s in both equations. (bxc is the integer part of x, and {x} is the fractional partof x.) Express your sum as a mixed number.

Proposed by: Yuan Yao

Answer: 3025 12017 or 6101426

2017

The two equations are equivalent to 2017bxc = 2017 and {2017x} = 12017 , respectively. The first

equation reduces to bxc = 1, so we must have x = 1 + r for some real r satisfying 0 ≤ r < 1. From thesecond equation, we deduce that {2017x} = {2017 + 2017r} = {2017r} = 1

2017 , so 2017r = n + 12017 ,

where n is an integer. Dividing both sides of this equation by 2017 yields r = n2017 + 1

20172 , wheren = 0, 1, 2, . . . , 2016 so that we have 0 ≤ r < 1. Thus, we have x = 1 + r = 1 + n

2017 + 120172

for n = 0, 1, 2, . . . , 2016. The sum of these solutions is 2017 · 1 + 2016·20172 · 1

2017 + 2017 · 120172 =

2017 + 20162 + 1

2017 = 30251

2017.

12. [4] Trapezoid ABCD, with bases AB and CD, has side lengths AB = 28, BC = 13, CD = 14, andDA = 15. Let diagonals AC and BD intersect at P , and let E and F be the midpoints of AP andBP , respectively. Find the area of quadrilateral CDEF .

Proposed by: Christopher Shao

Answer: 112

Note that EF is a midline of triangle APB, so EF is parallel to AB and EF = 12AB = 14 = CD. We

also have that EF is parallel to CD, and so CDEF is a parallelogram. From this, we have EP = PCas well, so CE

CA = 23 . It follows that the height from C to EF is 2

3 of the height from C to AB. We cancalculate that the height from C to AB is 12, so the height from C to EF is 8. Therefore CDEF is aparallelogram with base 14 and height 8, and its area is 14 · 8 = 112.

13. [5] Fisica and Ritmo discovered a piece of Notalium shaped like a rectangular box, and wanted tofind its volume. To do so, Fisica measured its three dimensions using a ruler with infinite precision,multiplied the results and rounded the product to the nearest cubic centimeter, getting a result of Vcubic centimeters. Ritmo, on the other hand, measured each dimension to the nearest centimeter andmultiplied the rounded measurements, getting a result of 2017 cubic centimeters. Find the positivedifference between the least and greatest possible positive values for V .

Proposed by: Yuan Yao

Answer: 4035

The only possible way for Ritmo to get 2017 cubic centimeters is to have his measurements rounded to1, 1, 2017 centimeters respectively. Therefore the largest value of V is achieved when the dimensionsare (1.5− ε)(1.5− ε)(2017.5− ε) = 4539.375− ε′ for some very small positive real ε, ε′, and the smallestvalue of V is achieved when the dimensions are (0.5 + ε)(0.5 + ε)(2016.5 + ε) = 504.125 + ε′ for somevery small positive real ε, ε′. Therefore the positive difference is 4539− 504 = 4035.

14. [5] Points A, B, C, and D lie on a line in that order such thatAB

BC=DA

CD. If AC = 3 and BD = 4,

find AD.

Proposed by: Yuan Yao

Answer: 6

Let BC = x, then the equation becomes 3−xx = 7−x

4−x . This simplifies to a quadratic equation withsolutions x = 1 and x = 6. Since x < 3, we have x = 1 and AD = 7− x = 6.

15. [5] On a 3 × 3 chessboard, each square contains a knight with 12 probability. What is the probability

that there are two knights that can attack each other? (In chess, a knight can attack any piece whichis two squares away from it in a particular direction and one square away in a perpendicular direction.)

Proposed by: Yuan Yao

Answer: 209256

Notice that a knight on the center square cannot attack any other square on the chessboard, so whetherit contains a knight or not is irrelevant.

For ease of reference, we label the other eight squares as follows:

0 5 23 X 76 1 4

Notice that a knight in square i attacks both square i+ 1 and i−1 (where square numbers are reducedmodulo 8). We now consider the number of ways such that no two knights attack each other.

• 0 knights: 1 way.

• 1 knights: 8 ways.

• 2 knights:(82

)− 8 = 20 ways.

• 3 knights: 8 + 8 = 16 ways, where the two 8s represent the number of ways such that the“distances” between the knights (index-wise) are 2, 2, 4 and 2, 3, 3 respectively.

• 4 knights: 2 ways.

Therefore, out of 28 = 256 ways, 1 + 8 + 20 + 16 + 2 = 47 of them doesn’t have a pair of attackingknights. Thus the answer is 256−47

256 = 209256 .

16. [7] A repunit is a positive integer, all of whose digits are 1s. Let a1 < a2 < a3 < . . . be a list of all thepositive integers that can be expressed as the sum of distinct repunits. Compute a111.

Proposed by: Michael Tang

Answer: 1223456

Let {rn}n≥0 be the repunits (so r0 = 1, r1 = 11, and so on). We see that for any n, there is

rn−1 + rn−2 + · · ·+ r0 <rn10

+rn100

+ · · · < rn9< rn,

so rn is only needed when all possible combinations of the first n repunits are exhausted (after 2n

terms), which shows that there is a bijection between the sequence {an} and the binary numbers.In particular, if k = 2n1 + 2n2 + · · · + 2ns for distinct ni’s, then ak = rn1

+ rn2+ · · · + rns

. Since111 = 11011112 = 20 + 21 + 22 + 23 + 25 + 26, we have

a111 = r0 + r1 + r2 + r3 + r5 + r6 = 1223456.

17. [7] A standard deck of 54 playing cards (with four cards of each of thirteen ranks, as well as twoJokers) is shuffled randomly. Cards are drawn one at a time until the first queen is reached. What isthe probability that the next card is also a queen?

Proposed by: Serina Hu

Answer: 227

Since the four queens are equivalent, we can compute the probability that a specific queen, say thequeen of hearts, is right after the first queen. Remove the queen of hearts; then for every ordering ofthe 53 other cards, there are 54 locations for the queen of hearts, and exactly one of those is after thefirst queen. Therefore the probability that the queen of hearts immediately follows the first queen is154 , and the probability any queen follows the first queen is 1

54 · 4 = 227 .

18. [7] Mr. Taf takes his 12 students on a road trip. Since it takes two hours to walk from the school tothe destination, he plans to use his car to expedite the journey. His car can take at most 4 students ata time, and travels 15 times as fast as traveling on foot. If they plan their trip optimally, what is theshortest amount of time it takes for them to all reach the destination, in minutes?

Proposed by: Yuan Yao

Answer: 30.4 OR 1525

A way to plan the trip is to have Mr. Taf drive 4 students to the 80% mark, then drive back to the10% mark to pick up another 4 students to the 90% mark, and finally drive back to the 20% mark topick up the last 4 students to the destination. All students will reach the destination at the same time,and Mr. Taf would have driven for (0.8 + 0.7 + 0.8 + 0.7 + 0.8) · 12015 = 30.4 minutes.

Now we show that 30.4 minutes is necessary. First of all, for a trip to be optimal, Mr. Taf must notcarry students when he was driving away from the destination, and all student not on the car mustkeep walking towards the destination. Suppose that among all the students, the student that walkedfor the longest walked for 15m minutes, where 0 ≤ m ≤ 8, then he spent 120−15m

15 = 8 −m minuteson the car, so it took them exactly 14m+ 8 minutes to get to the destination. Moreover, all studentsmust have spent at least 12(8 − m) minutes on the car total, and Mr. Taf would need to spend atleast 3(8−m) = 24− 3m minutes driving students towards the destination. Since it takes Mr. Taf 8minutes to drive the entire trip, he would need to drive 3(8 −m) − 8 = 16 − 3m minutes away fromthe destination, so Mr. Fat drove for at least 40 − 6m minutes. From this we derive the inequality40− 6m ≥ 14m+ 8, which comes out to m ≥ 1.6, so the journey is at least 14(1.6) + 8 = 30.4 minutes.

19. [9] The skeletal structure of coronene, a hydrocarbon with the chemical formula C24H12, is shownbelow.

Each line segment between two atoms is at least a single bond. However, since each carbon (C) requiresexactly four bonds connected to it and each hydrogen (H) requires exactly one bond, some of the linesegments are actually double bonds. How many arrangements of single/double bonds are there suchthat the above requirements are satisfied?

Proposed by: Yuan Yao

Answer: 20

Note that each carbon needs exactly one double bond. Label the six carbons in the center 1, 2, 3, 4, 5, 6clockwise. We consider how these six carbons are double-bonded. If a carbon in the center is notdouble-bonded to another carbon in the center, it must double-bond to the corresponding carbon onthe outer ring. This will result in the outer ring broken up into (some number of) strings insteadof a loop, which means that there will be at most one way to pair off the outer carbons throughdouble-bonds. (In fact, as we will demonstrate later, there will be exactly one way.)

Now we consider how many double bonds are on the center ring.

• 3 bonds. There are 2 ways to pair of the six carbons, and 2 ways to pair of the outer ring as well,for 4 ways in total.

• 2 bonds. Then either two adjacent carbons (6 ways) or two diametrically opposite carbons (3ways) are not double-bonded, and in the former case the outer ring will be broken up into two“strands” with 2 and 14 carbons each, while in the latter case it will be broken up into two strandsboth with 8 carbons each, and each produce one valid way of double-bonding, for 9 ways in total.

• 1 bond. There are 6 ways to choose the two double-bonded center carbon, and the outer ringwill be broken up into four strands with 2, 2, 2, 8 carbons each, which gives one valid way ofdouble-bonding, for 6 ways in total.

• 0 bonds. Then the outer ring is broken up into six strands of 2 carbons each, giving 1 way.

Therefore, the number of possible arrangements is 4 + 9 + 6 + 1 = 20.

Note: each arrangement of single/double bonds is also called a resonance structure of coronene.

20. [9] Rebecca has four resistors, each with resistance 1 ohm. Every minute, she chooses any two resistorswith resistance of a and b ohms respectively, and combine them into one by one of the followingmethods:

• Connect them in series, which produces a resistor with resistance of a+ b ohms;

• Connect them in parallel, which produces a resistor with resistance of aba+b ohms;

• Short-circuit one of the two resistors, which produces a resistor with resistance of either a or bohms.

Suppose that after three minutes, Rebecca has a single resistor with resistance R ohms. How manypossible values are there for R?

Proposed by: Yuan Yao

Answer: 15

Let Rn be the set of all possible resistances using exactly n 1-ohm circuit segments (without shorting

any of them), then we get Rn =n−1⋃i=1

({a+ b | a ∈ Ri, b ∈ Rn−i} ∪ { ab

a+b | a ∈ Ri, b ∈ Rn−i})

, starting

with R1 = {1}, we get:

R2 = {1

2, 2}

R3 = {1

3,

2

3,

3

2, 3}

R4 = {1

4,

2

5,

3

5,

3

4, 1,

4

3,

5

3,

5

2, 4}

Their union is the set of all possible effective resistances we can get, which contains 2 + 4 + 9 = 15values. (Note that R1 ⊂ R4 and the sets R2, R3, R4 are disjoint.)

21. [9] A box contains three balls, each of a different color. Every minute, Randall randomly draws a ballfrom the box, notes its color, and then returns it to the box. Consider the following two conditions:

(1) Some ball has been drawn at least three times (not necessarily consecutively).

(2) Every ball has been drawn at least once.

What is the probability that condition (1) is met before condition (2)?

Proposed by: Yuan Yao

Answer: 1327

At any time, we describe the current state by the number of times each ball is drawn, sorted innonincreasing order. For example, if the red ball has been drawn twice and green ball once, then thestate would be (2, 1, 0). Given state S, let PS be the probability that the state was achieved at somepoint of time before one of the two conditions are satisfied. Starting with P(0,0,0) = 1, we compute:

P(1,0,0) = 1;

P(1,1,0) =2

3, P(2,0,0) =

1

3;

P(1,1,1) =1

3P(1,1,0) =

2

9, P(2,1,0) =

2

3P(1,1,0) +

2

3P(2,0,0) =

2

3, P(3,0,0) =

1

3P(2,0,0) =

1

9;

P(2,1,1) = P(2,2,0) = P(3,1,0) =1

3P(2,1,0) =

2

9;

P(2,2,1) =1

3P(2,2,0) =

2

27, P(3,2,0) =

2

3P(2,2,0) =

4

27.

Therefore, the probability that the first condition is satisfied first is P(3,0,0) + P(3,1,0) + P(3,2,0) =19 + 2

9 + 427 = 13

27 .

22. [12] A sequence of positive integers a1, a2, . . . , a2017 has the property that for all integers m where1 ≤ m ≤ 2017, 3(

∑mi=1 ai)

2 =∑m

i=1 a3i . Compute a1337.

Proposed by: Serina Hu

Answer: 4011

I claim that ai = 3i for all i. We can conjecture that the sequence should just be the positive multiplesof three because the natural numbers satisfy the property that the square of their sum is the sum oftheir cubes, and prove this by induction. At i = 1, we have that 3a2i = a3i , so ai = 3. Now assumingthis holds for i = m, we see that

3

(m+1∑i=1

ai

)2

= 3

(am+1 +

m∑i=1

ai

)2

= 3a2m+1 +

m∑i=1

a3i + 6am+1

m∑i=1

ai

= 3a2m+1 +

m∑i=1

a3i + 6am+1 · 3(m(m+ 1)

2

)

=

m+1∑i=1

a3i .

Therefore,

a3m+1 = 3a2m+1 + am+1(9m2 + 9m)

0 = a2m+1 − 3am+1 − (9m2 + 9m)

0 = (am+1 − (3m+ 3))(am+1 + 3m)

and because the sequence is positive, am+1 = 3m + 3, which completes the induction. Then a1337 =1337 · 3 = 4011.

23. [12] A string of digits is defined to be similar to another string of digits if it can be obtained byreversing some contiguous substring of the original string. For example, the strings 101 and 110 aresimilar, but the strings 3443 and 4334 are not. (Note that a string is always similar to itself.) Considerthe string of digits

S = 01234567890123456789012345678901234567890123456789,

consisting of the digits from 0 to 9 repeated five times. How many distinct strings are similar to S?

Proposed by: Kevin Sun

Answer: 1126

We first count the number of substrings that one could pick to reverse to yield a new substring. If weinsert two dividers into the sequence of 50 digits, each arrangement of 2 dividers among the 52 totalobjects specifies a substring that is contained between the two dividers, for a total of

(522

)substrings.

Next, we account for overcounting. Every substring of length 0 or 1 will give the identity string whenreversed, so we are overcounting here by 51 + 50− 1 = 100 substrings. Next, for any longer substrings that starts and ends with the same digit, removing the digit from both ends results in a substrings′, such that reversing s would give the same string as reversing s′. Therefore, we are overcounting by10 ·

(52

)substrings. Our total number of strings similar to S is therefore

(522

)− 100− 10 ·

(52

)= 1126 .

24. [12] Triangle ABC has side lengths AB = 15, BC = 18, CA = 20. Extend CA and CB to points Dand E respectively such that DA = AB = BE. Line AB intersects the circumcircle of CDE at P andQ. Find the length of PQ.

Proposed by: Yuan Yao

Answer: 37

WLOG suppose that P is closer to A than to B. Let DA = AB = BE = c = 15, BC = a = 18, CA =b = 20, PA = x, and QB = y. By Power of a Point on B and A, we get ac = (x+c)y and bc = (y+c)x,respectively. Subtracting the two equations gives cy − cx = ac − bc ⇒ y − x = a − b. Substitutingy = x+ a− b into the first equation gives ac = (x+ c)(x+ a− b) = x2 + (a− b+ c)x+ ac− bc, which

is a quadratic with unique positive solution x =(b−a−c)+

√(a−b+c)2+4bc

2 . Thus,

PQ = x+y+c = (y−x)+2x+c = (a−b+c)+(b−a−c)+√

(a− b+ c)2 + 4bc =√

132 + 4 · 20 · 15 = 37.

25. [15] Fisica and Ritmo discovered a piece of Notalium shaped like a rectangular box, and wanted tofind its volume. To do so, Fisica measured its three dimensions using a ruler with infinite precision,multiplied the results and rounded the product to the nearest cubic centimeter, getting a result of2017 cubic centimeters. Ritmo, on the other hand, measured each dimension to the nearest centimeterand multiplied the rounded measurements, getting a result of V cubic centimeters. Find the positivedifference between the least and greatest possible positive values for V .

Proposed by: Yuan Yao

Answer: 7174

It is not difficult to see that the maximum possible value of V can be achieved when the dimensionsare (0.5 + ε)× (0.5 + ε)× (8070− ε′) = 2017.5− ε′′ for some very small reals ε, ε′, ε′′ > 0, which whenmeasured by Ritmo, gives V = 1 · 1 · 8070 = 8070. Similarly, the minimum possible positive value ofV can be achieved when the dimensions are (1.5 − ε) × (1.5 − ε) × ( 8066

9 + ε′) = 2016.5 + ε′′ for somevery small reals ε, ε′, ε′′ > 0, which when measured by Ritmo, gives V = 1 · 1 · 896 = 896. Therefore,the difference between the maximum and minimum is 8070− 896 = 7174.

26. [15] Points A,B,C,D lie on a circle in that order such thatAB

BC=DA

CD. If AC = 3 and BD = BC = 4,

find AD.

Proposed by: Yuan Yao

Answer: 32

By Ptolemy’s theorem, we have AB · CD + BC · DA = AC · BD = 3 · 4 = 12. Since the conditionimplies AB · CD = BC ·DA, we have DA = 6

BC = 32 .

27. [15] On a 3 × 3 chessboard, each square contains a Chinese knight with 12 probability. What is the

probability that there are two Chinese knights that can attack each other? (In Chinese chess, a Chineseknight can attack any piece which is two squares away from it in a particular direction and one squareaway in a perpendicular direction, under the condition that there is no other piece immediately adjacentto it in the first direction.)

Proposed by: Yuan Yao

Answer: 79256

Suppose the 3×3 square is

A B CD E FG H I

We count the number of ways a board could have two knights

attack each other using PIE. First notice that in any setup with two knights attack each other, thecenter square must be empty. Also, for any pair of knights that attack each other, one must be in acorner, and the other at the center of a nonadjacent side. There are 8 · 25 ways for one pair of knightsto attack each other. Next, we count the number of ways two pairs of knights attack each other: up tosymmetry, there are four cases: knights at A, B, G, H, and D and E empty; knights at A, H, F, andB, D, E empty; knights at A, B, H, I, and D, E, F empty; and knights at A, C, H, and D, E, F empty.Four each of these cases, there are four symmetries, so there are a total of 4 · (23 + 23 + 22 + 23) waysto have two pairs of knights attack each other. Next, there’s only one way for three pairs of knightsto attack each other, discounting symmetry: A, B, G, H, I have knights, and D, E, F empty. Thenthere are 4 · 2 · 2 ways for three knights to attack. Finally, there is only one way for four knights toattack: knights at A, B, C, G, H, I and empty squares at D, E, F, for a total of 2 ways after countingsymmetries.

Applying PIE, we get that the total number of boards with at least one pair of knights attacking eachother is

8 · 25 − 4 · (23 + 23 + 22 + 23) + 4 · 2 · 2− 2 = 158.

Then the probability the 3× 3 board has a pair of knights attacking each other is 15829 =

79

256.

28. [19] Compute the number of functions f : {1, 2, . . . , 9} → {1, 2, . . . , 9} which satisfy f(f(f(f(f(x))))) =x for each x ∈ {1, 2, . . . , 9}.Proposed by: Evan Chen

Answer: 3025

All cycles lengths in the permutation must divide 5, which is a prime number. Either f(x) = x for all x,or there exists exactly one permutation cycle of length 5. In the latter case, there are

(95

)ways to choose

which numbers are in the cycle and 4! ways to create the cycle. The answer is thus 1 +(95

)· 4! = 3025.

29. [19] Consider a sequence xn such that x1 = x2 = 1, x3 = 23 . Suppose that xn =

x2n−1xn−2

2x2n−2 − xn−1xn−3for all n ≥ 4. Find the least n such that xn ≤ 1

106 .

Proposed by: Mehtaab Sawhney

Answer: 13

The recursion simplifies to xn−1

xn+ xn−3

xn−2= 2xn−2

xn−1. So if we set yn = xn−1

xnfor n ≥ 2 then we have

yn − yn−1 = yn−1 − yn−2 for n ≥ 3, which means that {yn} is an arithmetic sequence. From the

starting values we have y2 = 1, y3 = 32 , so yn = n

2 for all n. (This means that xn = 2n−1

n! .) Sincex1

xn= y2y3 · · · yn, it suffices to find the minimal n such that the RHS is at least 106. Note that

y2y3 · · · y12 = 1 · (1.5 · 2 · 2.5 · 3 · 3.5) · (4 · 4.5 · 5 · 5.5 · 6) < 2.55 · 55 = 12.55 < 2002 · 12.5 = 500000 < 106,

while

y2y3 · · · y13 = 1 · (1.5 · 2 · 2.5 · 3) · (3.5 · 4 · 4.5) · (5 · 5.5 · 6 · 6.5) > 20 · 60 · 900 = 1080000 > 106,

so the answer is 13.

30. [19] Given complex number z, define sequence z0, z1, z2, . . . as z0 = z and zn+1 = 2z2n + 2zn for n ≥ 0.Given that z10 = 2017, find the minimum possible value of |z|.Proposed by: Yuan Yao

Answer:1024√4035−1

2

Define wn = zn + 12 , so zn = wn − 1

2 , and the original equation becomes

wn+1 −1

2= 2(wn −

1

2)2 + 2(wn −

1

2) = 2w2

n −1

2,

which reduces to wn+1 = 2w2n. it is not difficult to show that

z10 +1

2= 2017 +

1

2=

4035

2= w10 = 21023w1024

0 ,

and thus w0 =1024√4035

2 ω1024, where ω1024 is one of the 1024th roots of unity. Since |w0| =1024√4035

2 > 12 ,

to minimize the magnitude of z = w0 − 12 , we need ω1024 = −1, which gives |z| =

1024√4035−12 .

31. [24] In unit square ABCD, points E,F,G are chosen on side BC,CD,DA respectively such that AEis perpendicular to EF and EF is perpendicular to FG. Given that GA = 404

1331 , find all possible valuesof the length of BE.

Proposed by: Yuan Yao

Answer: 911

Let BE = x, then since triangles ABE,ECF, FDG are all similar, we have CE = 1 − x,CF =x(1− x), FD = 1− x(1− x), DG = x− x2(1− x), GA = 1− x+ x2(1− x) = (1− x)(x2 + 1), thereforeit remains to solve the equation

(1− x)(x2 + 1) =404

1331.

We first seek rational solutions x = pq for relatively prime positive integers p, q. Therefore we have

(q−p)(p2+q2)q3 = 404

1331 . Since both q−p and p2+q2 are relatively prime to q3, we have q3 = 1331⇒ q = 11,

so (11 − p)(p2 + 121) = 404 = 22 · 101, and it is not difficult to see that p = 9 is the only integralsolution. We can therefore rewrite the original equation as

(x− 9

11)(x2 − 2

11x+

103

121) = 0.

It is not difficult to check that the quadratic factor has no zeroes, therefore BE = x = 911 is the only

solution.

32. [24] Let P be a polynomial with integer coefficients such that P (0) + P (90) = 2018. Find the leastpossible value for |P (20) + P (70)|.Proposed by: Michael Tang

Answer: 782

First, note that P (x) = x2−3041 satisfy the condition and gives |P (70)+P (20)| = |4900+400−6082| =782. To show that 782 is the minimum, we show 2800 | P (90) − P (70) − P (20) + P (0) for every P ,since −782 is the only number in the range [−782, 782] that is congruent to 2018 modulo 2800.

Proof: It suffices to show that 2800 | 90n − 70n − 20n + 0n for every n ≥ 0 (having 00 = 1).

Let Q(n) = 90n − 70n − 20n + 0n, then we note that Q(0) = Q(1) = 0, Q(2) = 2800, and Q(3) = (93 −73−23) ·103 = 378000 = 135 ·2800. For n ≥ 4, we note that 400 divides 104, and 90n + 0n ≡ 70n + 20n

(mod 7). Therefore 2800 | Q(n) for all n.

33. [24] Tetrahedron ABCD with volume 1 is inscribed in circumsphere ω such that AB = AC = AD = 2and BC · CD ·DB = 16. Find the radius of ω.

Proposed by: Caleb He

Answer: 53

Let X be the foot of the perpendicular from A to ∆BCD. Since AB = AC = AD, it follows that X isthe circumcenter of ∆BCD. Denote XB = XC = XD = r. By the Pythagorean Theorem on ∆ABX,we have AX =

√4− r2. Now, from the extended law of sines on ∆BCD, we have the well-known

identityBC · CD ·DB

4r= [BCD],

where [BCD] denotes the area of ∆BCD. However, we have

V =1

3AX · [BCD],

where V is the volume of ABCD, which yields the expression

[BCD] =3√

4− r2.

Now, given that BC · CD ·DB = 16, we have

4

r=

3√4− r2

.

Solving, we get r = 85 . Now, let O be the center of ω. Since OB = OC = OD, it follows that the foot

of the perpendicular from O to ∆BCD must also be the circumcenter of ∆BCD, which is X. Thus,A,X,O are collinear. Let R be the radius of ω. Then we have

R = OA

= OX +XA

=√R2 − r2 +

√4− r2

=

√R2 − 64

25+

6

5.

Solving, we get R = 53 . (Note: solving for R from OA = OX −XA gives a negative value for R.)

34. [30] The skeletal structure of circumcircumcircumcoronene, a hydrocarbon with the chemical formulaC150H30, is shown below.

Each line segment between two atoms is at least a single bond. However, since each carbon (C) requiresexactly four bonds connected to it and each hydrogen (H) requires exactly one bond, some of the linesegments are actually double bonds. How many arrangements of single/double bonds are there suchthat the above requirements are satisfied? If the correct answer is C and your answer is A, you get

max(⌊

30(

1−∣∣∣loglog2 C

AC

∣∣∣)⌋ , 0) points.

Proposed by: Yuan Yao

Answer: 267227532

The problem is equivalent to the one in OEIS A008793, a.k.a. ”number of ways to tile hexagon of edgen with diamonds of side 1.” Notice that there is a bjiection between such a tiling and the number ofways to stack some unit cubes alongside a corner of an n× n× n box (see the Art of Problem Solvinglogo as an example, also known as 3-dimensional Young diagrams), where this problem n = 5. It isknown that there are

(2nn

)= 252 ways to stack one layer (since each way correspond a way to walk from

a corner of a 5 by 5 grid to the opposite one), so 2525

5! ≈ 8× 109 gives a somewhat loose upper bound(generate five layers and sort them by size, and hope that it will be a valid stack in general). Thisresult can be improved by dividing out a reasonable constant factor after considering the probabilitythat sorting by size indeed gives a valid stack (for example, it would be fair to guess that there is abouta 1

4! chance that the first row of each layer will be in the right order, given that each row has a smallrole in determining the final size of the layer; dividing 24 from the previous result gives a very closeguess). In general, a guess anywhere between 108 and 109 is a fair guess.

35. [30] Rebecca has twenty-four resistors, each with resistance 1 ohm. Every minute, she chooses anytwo resistors with resistance of a and b ohms respectively, and combine them into one by one of thefollowing methods:

• Connect them in series, which produces a resistor with resistance of a+ b ohms;

• Connect them in parallel, which produces a resistor with resistance of aba+b ohms;

• Short-circuit one of the two resistors, which produces a resistor with resistance of either a or bohms.

Suppose that after twenty-three minutes, Rebecca has a single resistor with resistance R ohms. Howmany possible values are there for R? If the correct answer is C and your answer is A, you get

max(⌊

30(

1−∣∣∣loglog2 C

AC

∣∣∣)⌋ , 0) points.

Proposed by: Yuan Yao

Answer: 1015080877

This is the same problem as in OEIS A153588. It is helpful to see (or guess) that neither the numeratoror the denominator of the final resistance exceed the (n+ 1)-th Fibonacci number, which in this caseis F25 = 75025, using concepts on the line of continued fractions. So 750252 ≈ 5.6 × 109 is an upperbound for the total number, which is already close to the final answer. Multiplying by some constantfactor to remove non-reduced fractions (such as 3

4 to deal with parity) will improve this result.

36. [30] A box contains twelve balls, each of a different color. Every minute, Randall randomly draws aball from the box, notes its color, and then returns it to the box. Consider the following two conditions:

(1) Some ball has been drawn at least twelve times (not necessarily consecutively).

(2) Every ball has been drawn at least once.

What is the probability that condition (1) is met before condition (2)? If the correct answer is C andyour answer is A, you get max

(⌊30(1− 1

2 |log2 C − log2A|)⌋, 0)

points.

Proposed by: Yuan Yao

Answer: 0.02236412255 . . .

Below is a python implementation to compute the probability, using the same method as the solutionto the easier version (with three balls).

from fractions import Fraction

N = 12

probs = [{} for i in range((N-1)*(N-1)+2)]

prob1 = Fraction()

prob2 = Fraction()

init = tuple(0 for i in range(N))

probs[0][init] = Fraction(1,1)

for i in range((N-1)*(N-1)+1):

for t in probs[i]:

for j in range(N):

val = probs[i][t] * Fraction(1,N)

l = list(t)

l[j] += 1

l.sort()

l = tuple(l)

if (l[-1] == N):

prob1 = prob1 + val

elif (l[0] == 1):

prob2 = prob2 + val

else:

probs[i+1][l] = probs[i+1].setdefault(l, Fraction()) + val

print(prob1)

Intuitively the probability should be quite small, since the distribution tends towards the secondcondition instead of the first. Indeed, the exact fraction is p = M

N , where

M =663659309086473387879121984765654681548533307869748367531

919050571107782711246694886954585701687513519369602069583,

N =2967517762021717138065641019865112420616209349876886946382

1672067789922444492392280614561539198623553884143178743808.

Note: This is a simplified variant of the Bingo Paradox, which is a phenomenon where horizontalbingos are significantly more frequent than vertical bingos. For more information, see https://www.

maa.org/sites/default/files/pdf/Mathhorizons/pdfs/The_Bingo_Paradox_MH_Sept17.pdf.

HMIC 2017April 10, 2017

HMIC

1. [6] Kevin and Yang are playing a game. Yang has 2017 +20172

)cards with their front sides face down

on the table. The cards are constructed as follows:

• For each 1 ≤ n ≤ 2017, there is a blue card with n written on the back, and a fraction an

bnwritten

on the front, where gcd(an, bn) = 1 and an, bn > 0.

• For each 1 ≤ i < j ≤ 2017, there is a red card with (i, j) written on the back, and a fractionai+aj

bi+bj

written on the front.

It is given no two cards have equal fractions. In a turn Kevin can pick any two cards and Yang tellsKevin which card has the larger fraction on the front. Show that, in fewer than 10000 turns, Kevincan determine which red card has the largest fraction out of all of the red cards.

2. [7] Let S = {1, 2 . . . , n} for some positive integer n, and let A be an n-by-n matrix having as entriesonly ones and zeros. Define an infinite sequence {xi}i≥0 to be strange if:

• xi ∈ S for all i,

• axkxk+1= 1 for all k, where aij denotes the element in the ith row and jth column of A.

Prove that the set of strange sequences is empty if and only if A is nilpotent, i.e. Am = 0 for someinteger m.

3. [9] Let v1, v2, . . . , vm be vectors in Rn, such that each has a strictly positive first coordinate. Considerthe following process. Start with the zero vector w = (0, 0, . . . , 0) ∈ Rn. Every round, choose an i suchthat 1 ≤ i ≤ m and w · vi ≤ 0, and then replace w with w + vi.

Show that there exists a constant C such that regardless of your choice of i at each step, the processis guaranteed to terminate in C rounds. The constant C may depend on the vectors v1, . . . , vm.

4. [9] Let G be a weighted bipartite graph A ∪ B, with |A| = |B| = n. In other words, each edge inthe graph is assigned a positive integer value, called its weight. Also, define the weight of a perfectmatching in G to be the sum of the weights of the edges in the matching.

Let G′ be the graph with vertex set A ∪ B, and contains the edge e if and only if e is part of someminimum weight perfect matching in G.

Show that all perfect matchings in G′ have the same weight.

5. [11] Let S be the set {−1, 1}n, that is, n-tuples such that each coordinate is either −1 or 1. For

s = (s1, s2, . . . , sn), t = (t1, t2, . . . , tn) ∈ {−1, 1}n,

define s� t = (s1t1, s2t2, . . . , sntn).

Let c be a positive constant, and let f : S → {−1, 1} be a function such that there are at least(1 − c) · 22n pairs (s, t) with s, t ∈ S such that f(s � t) = f(s)f(t). Show that there exists a functionf ′ such that f ′(s� t) = f ′(s)f ′(t) for all s, t ∈ S and f(s) = f ′(s) for at least (1− 10c) · 2n values ofs ∈ S.

HMIC 2017April 10, 2017

HMIC

1. [6] Kevin and Yang are playing a game. Yang has 2017 +(20172

)cards with their front sides face down

on the table. The cards are constructed as follows:

• For each 1 ≤ n ≤ 2017, there is a blue card with n written on the back, and a fraction an

bnwritten

on the front, where gcd(an, bn) = 1 and an, bn > 0.

• For each 1 ≤ i < j ≤ 2017, there is a red card with (i, j) written on the back, and a fractionai+aj

bi+bj

written on the front.

It is given no two cards have equal fractions. In a turn Kevin can pick any two cards and Yang tellsKevin which card has the larger fraction on the front. Show that, in fewer than 10000 turns, Kevincan determine which red card has the largest fraction out of all of the red cards.

Proposed by: Evan Chen

We will provide an algorithm to determine which red card has the highest value in 2016 + 2015 turns.

We start with a lemma: If the blue card with the largest fraction is labeled k and the red card withthe largest fraction is labeled (i, j), then i = k or j = k.

To prove this lemma, we assume for contradiction that i 6= k and j 6= k. Assume without loss of gener-

ality that ai

bi>

aj

bj. Then ai+ak

bi+bk− ai+aj

bi+bj=

(ai+ak)(bi+bj)−(ai+aj)(bi+bk)(bi+bk)(bi+bj)

=akbi+aibj+akbj−aibk−ajbi−ajbk

(bi+bk)(bi+bj).

However, we know since ak

bk> ai

bi>

aj

bj, we have that akbi − aibk > 0, akbj − ajbk > 0, aibj − ajbi > 0,

so ai+ak

bi+bk− ai+aj

bi+bj> 0.

Now, we can first find the blue card with the maximum fraction in 2016 turns. Afterwards, we knowk, so we only need to consider pairs (i, j) with i = k or j = k. There are only 2016 of these pairs, sowe can find the maximum of these in 2015 turns. By the lemma, the maximum of the cards with i = kor j = k is the same as the global maximum of all the red cards.

Therefore we have provided and algorithm to determine the (i, j) with the maximum fraction in 2016+2015 turns.

2. [7] Let S = {1, 2 . . . , n} for some positive integer n, and let A be an n-by-n matrix having as entriesonly ones and zeros. Define an infinite sequence {xi}i≥0 to be strange if:

• xi ∈ S for all i,

• axkxk+1= 1 for all k, where aij denotes the element in the ith row and jth column of A.

Prove that the set of strange sequences is empty if and only if A is nilpotent, i.e. Am = 0 for someinteger m.

Proposed by: Henrik Boecken

Answer:

Consider the directed graph G on n labeled vertices whose adjacency matrix is A. Then, observe thata strange sequence is simply an infinite path on G. Since powers of the adjacency matrix count paths,if A is nilpotent, there exists no infinite path. If A is not nilpotent, a cycle must exist, so we’re done.

3. [9] Let v1, v2, . . . , vm be vectors in Rn, such that each has a strictly positive first coordinate. Considerthe following process. Start with the zero vector w = (0, 0, . . . , 0) ∈ Rn. Every round, choose an i suchthat 1 ≤ i ≤ m and w · vi ≤ 0, and then replace w with w + vi.

Show that there exists a constant C such that regardless of your choice of i at each step, the processis guaranteed to terminate in C rounds. The constant C may depend on the vectors v1, . . . , vm.

Proposed by: Allen Liu

Let w0 = 0, let wr be the vector w after r rounds. Also, let e1 denote the vector (1, 0, . . . , 0). Notethat

w2r+1 = (wr + vi)

2 = w2r + v2i + 2wr · vi ≤ w2

r + v2i ,

as we have by the condition that wr · vi ≤ 0. Therefore, we have by induction that |wr| ≤M√r, where

M = maxi |vi|. Also, we have that

wr+1 · e1 = wr · e1 + vi · e1,

so wr · e1 ≥ mr, where m = mini(vi · e1) > 0.

Now using the obvious inequality

M√r ≥ |wr| ≥ wr · e1 ≥ mr,

we get that r ≤ M2

m2 , as desired.

4. [9] Let G be a weighted bipartite graph A ∪ B, with |A| = |B| = n. In other words, each edge inthe graph is assigned a positive integer value, called its weight. Also, define the weight of a perfectmatching in G to be the sum of the weights of the edges in the matching.

Let G′ be the graph with vertex set A ∪ B, and contains the edge e if and only if e is part of someminimum weight perfect matching in G.

Show that all perfect matchings in G′ have the same weight.

Proposed by: Yang Liu

Let m denote the minimum weight of a matching in G. Let G′′ be the (multi)graph formed by takingthe union of all minimum weight perfect matchings, but keeping edges multiple times. Note that G′′ isregular. Now, assume that some perfect matching in G′ (equivalently G′′) has weight M > m. Deletethis matching from G′′, and call the resulting graph H. H is still regular, so it can be decomposedinto a union of perfect matchings. Now using pigeonhole directly gives us a contradiction.

To show that all regular bipartite graphs can be decomposed into a union of perfect matchings, useHall’s marriage lemma to take out one perfect matching and use induction.

5. [11] Let S be the set {−1, 1}n, that is, n-tuples such that each coordinate is either −1 or 1. For

s = (s1, s2, . . . , sn), t = (t1, t2, . . . , tn) ∈ {−1, 1}n,

define s� t = (s1t1, s2t2, . . . , sntn).

Let c be a positive constant, and let f : S → {−1, 1} be a function such that there are at least(1 − c) · 22n pairs (s, t) with s, t ∈ S such that f(s � t) = f(s)f(t). Show that there exists a functionf ′ such that f ′(s� t) = f ′(s)f ′(t) for all s, t ∈ S and f(s) = f ′(s) for at least (1− 10c) · 2n values ofs ∈ S.Proposed by: Yang Liu

We use finite fourier analysis, which we explain the setup for below. For a subset T ⊆ S, define thefunction χT : S → {−1, 1} to satisfy χT (s) =

∏t∈T st. Note that χT (s � t) = χT (s)χT (t) for all

s, t ∈ S. Therefore, we should show that there exists a subset T such that taking f ′ = χT satisfiesthe constraint. This subset T also has to exist, as the χT are all the multiplicative functions fromS → {−1, 1}. Also, for two functions f, g : S → R define

〈f, g〉 =1

2n

∑s∈S

f(s)g(s).

Now we claim a few facts below that are easy to verify. One, for any subsets T1, T2 ∈ S we have that

〈χT1, χT2〉 =

{0 if T1 6= T2

1 if T1 = T2.

We will refer to this claim as orthogonality. Also, note that for any function f : S → R, we can write

f =∑T⊆S

〈f, χT 〉χT .

This is called expressing f by its Fourier basis. By the given condition, we have that∑s,t∈S

f(s� t)f(s)f(t) = 1− 2c.

Expanding the left-hand side in terms of the Fourier basis, we get that

1−2c =1

22n

∑s,t∈S

f(s�t)f(s)f(t) =1

22n

∑T1,T2,T3⊆S

∑s,t∈S〈f, χT1

〉〈f, χT2〉〈f, χT3

〉χT1(s)χT1

(t)χT2(s)χT3

(t)

=∑

T1,T2,T3⊆S

〈f, χT1〉〈f, χT2〉〈f, χT3〉〈χT1 , χT2〉〈χT1 , χT3〉 =∑T⊆S

〈f, χT 〉3

by orthogonality. Also note by orthogonality that

1 = 〈f, f〉 =

⟨∑T1⊆S

〈f, T1〉χT1(s),

∑T2⊆S

〈f, T2〉χT2(s)

⟩=∑T⊆S

〈f, χT 〉2.

Finally, from above we have that

1− 2c =∑T⊆S

〈f, χT 〉3 ≤(

maxT⊆S〈f, χT 〉

)∑T⊆S

〈f, χT 〉2 = maxT⊆S〈f, χT 〉.

This last equation implies that some χT and f agree on (1− c) · 2n values, as desired.

HMMT February 2018February 10, 2018

Team Round

1. [20] In an n× n square array of 1× 1 cells, at least one cell is colored pink. Show that you can alwaysdivide the square into rectangles along cell borders such that each rectangle contains exactly one pinkcell.

2. [25] Is the number (1 +

1

2

)(1 +

1

4

)(1 +

1

6

). . .

(1 +

1

2018

)greater than, less than, or equal to 50?

3. [30] Michelle has a word with 2n letters, where a word can consist of letters from any alphabet. Michelleperforms a switcheroo on the word as follows: for each k = 0, 1, . . . , n − 1, she switches the first 2k

letters of the word with the next 2k letters of the word. For example, for n = 3, Michelle changes

ABCDEFGH → BACDEFGH → CDBAEFGH → EFGHCDBA

in one switcheroo.

In terms of n, what is the minimum positive integer m such that after Michelle performs the switcheroooperation m times on any word of length 2n, she will receive her original word?

4. [30] In acute triangle ABC, let D,E, and F be the feet of the altitudes from A,B, and C respectively,and let L,M, and N be the midpoints of BC,CA, and AB, respectively. Lines DE and NL intersectat X, lines DF and LM intersect at Y , and lines XY and BC intersect at Z. Find ZB

ZC in terms ofAB, AC, and BC.

5. [30] Is it possible for the projection of the set of points (x, y, z) with 0 ≤ x, y, z ≤ 1 onto sometwo-dimensional plane to be a simple convex pentagon?

6. [35] Let n ≥ 2 be a positive integer. A subset of positive integers S is said to be comprehensive iffor every integer 0 ≤ x < n, there is a subset of S whose sum has remainder x when divided by n.Note that the empty set has sum 0. Show that if a set S is comprehensive, then there is some (notnecessarily proper) subset of S with at most n− 1 elements which is also comprehensive.

7. [50] Let [n] denote the set of integers {1, 2, . . . , n}. We randomly choose a function f : [n] → [n],out of the nn possible functions. We also choose an integer a uniformly at random from [n]. Findthe probability that there exist positive integers b, c ≥ 1 such that f b(1) = a and f c(a) = 1. (fk(x)denotes the result of applying f to x k times).

8. [60] Allen plays a game on a tree with 2n vertices, each of whose vertices can be red or blue. Initially,all of the vertices of the tree are colored red. In one move, Allen is allowed to take two vertices of thesame color which are connected by an edge and change both of them to the opposite color. He wins ifat any time, all of the vertices of the tree are colored blue.

(a) (20) Show that Allen can win if and only if the vertices can be split up into two groups V1 andV2 of size n, such that each edge in the tree has one endpoint in V1 and one endpoint in V2.

(b) (40) Let V1 = {a1, . . . , an} and V2 = {b1, . . . , bn} from part (a). Let M be the minimum over allpermutations σ of {1, . . . , n} of the quantity

n∑i=1

d(ai, bσ(i)),

where d(v, w) denotes the number of edges along the shortest path between vertices v and w inthe tree.

Show that if Allen can win, then the minimum number of moves that it can take for Allen to winis equal to M .

(A graph consists of a set of vertices and some edges between distinct pairs of vertices. It is connectedif every pair of vertices are connected by some path of one or more edges. A tree is a graph which isconnected, in which the number of edges is one less than the number of vertices.)

9. [60] Evan has a simple graph with v vertices and e edges. Show that he can delete at least e∈v+12 edges

so that each vertex still has at least half of its original degree.

10. [60] Let n and m be positive integers which are at most 1010. Let R be the rectangle with cornersat (0, 0), (n, 0), (n,m), (0,m) in the coordinate plane. A simple non-self-intersecting quadrilateral withvertices at integer coordinates is called far-reaching if each of its vertices lie on or inside R, but each sideof R contains at least one vertex of the quadrilateral. Show that there is a far-reaching quadrilateralwith area at most 106.

(A side of a rectangle includes the two endpoints.)

HMMT February 2018February 10, 2018

Team Round

1. [20] In an n× n square array of 1× 1 cells, at least one cell is colored pink. Show that you can alwaysdivide the square into rectangles along cell borders such that each rectangle contains exactly one pinkcell.

Proposed by: Kevin Sun

We claim that the statement is true for arbitrary rectangles. We proceed by induction on the numberof marked cells. Our base case is k = 1 marked cell, in which case the original rectangle works.

To prove it for k marked cells, we split the rectangle into two smaller rectangles, both of which containsat least one marked cell. By induction, we can divide the two smaller rectangles into rectangles withexactly one marked cell. Combining these two sets of rectangles gives a way to divide our originalrectangle into rectangles with exactly one marked cell, completing the induction.

2. [25] Is the number (1 +

1

2

)(1 +

1

4

)(1 +

1

6

). . .

(1 +

1

2018

)greater than, less than, or equal to 50?

Proposed by: Henrik Boecken

Call the expression S. Note that(1 +

1

2

)(1 +

1

4

)(1 +

1

6

). . .

(1 +

1

2018

)<

(1 +

1

1

)(1 +

1

3

)(1 +

1

5

). . .

(1 +

1

2017

)Multiplying these two products together, we get(

1 +1

1

)(1 +

1

2

)(1 +

1

3

). . .

(1 +

1

2018

)=

2

1· 3

2· 4

3. . .

2019

2018= 2019

This shows thatS2 < 2019 =⇒ S <

√2019 < 50

as desired.

3. [30] Michelle has a word with 2n letters, where a word can consist of letters from any alphabet. Michelleperforms a switcheroo on the word as follows: for each k = 0, 1, . . . , n − 1, she switches the first 2k

letters of the word with the next 2k letters of the word. For example, for n = 3, Michelle changes

ABCDEFGH → BACDEFGH → CDBAEFGH → EFGHCDBA

in one switcheroo.

In terms of n, what is the minimum positive integer m such that after Michelle performs the switcheroooperation m times on any word of length 2n, she will receive her original word?

Proposed by: Mehtaab Sawhney

Let m(n) denote the number of switcheroos needed to take a word of length 2n back to itself. Considera word of length 2n for some n > 1. After 2 switcheroos, one has separately performed a switcheroo onthe first half of the word and on the second half of the word, while returning the (jumbled) first halfof the word to the beginning and the (jumbled) second half of the word to the end.

After 2 ·m(n−1) switcheroos, one has performed a switcheroo on each half of the word m(n−1) timeswhile returning the halves to their proper order. Therefore, the word is in its proper order. However,it is never in its proper order before this, either because the second half precedes the first half (i.e.after an odd number of switcheroos) or because the halves are still jumbled (because each half has hadfewer than m(n− 1) switcheroos performed on it).

It follows that m(n) = 2m(n− 1) for all n > 1. We can easily see that m(1) = 2, and a straightforwardproof by induction shows that m = 2n.

4. [30] In acute triangle ABC, let D,E, and F be the feet of the altitudes from A,B, and C respectively,and let L,M, and N be the midpoints of BC,CA, and AB, respectively. Lines DE and NL intersectat X, lines DF and LM intersect at Y , and lines XY and BC intersect at Z. Find ZB

ZC in terms ofAB, AC, and BC.

Proposed by: Faraz Masroor

Because NL||AC we have triangles DXL and DEC are similar. From angle chasing, we also havethat triangle DEC is similar to triangle ABC. We have ∠XNA = 180◦ −∠XNB = 180◦ −∠LNB =180−CAB = ∠LMA. In addition, we have NX

NA = XD·XEXL·NA = AB

BCXELC

NMNA = AB

BCEDDC

BCAB = ED

DC = ABAC =

MLMA . These two statements mean that triangles ANX and AML are similar, and ∠XAB = ∠XAN =∠LAM = ∠LAC. Similarly, ∠XAY = ∠LAC, making A,X, and Y collinear, with ∠Y AB = ∠XAB =∠LAC; ie. line AXY is a symmedian of triangle ABC.

Then ZBZC = AB

ACsin∠ZABsin∠ZAC = AB

ACsin∠LACsin∠LAB , by the ratio lemma. But using the ratio lemma, 1 = LB

LC =ABAC

sin∠LABsin∠LAC , so sin∠LAC

sin∠LAB = ABAC , so ZB

ZC = AB2

AC2 .

5. [30] Is it possible for the projection of the set of points (x, y, z) with 0 ≤ x, y, z ≤ 1 onto sometwo-dimensional plane to be a simple convex pentagon?

Proposed by: Yuan Yao

It is not possible. Consider P , the projection of (12 ,

12 ,

12 ) onto the plane. Since for any point (x, y, z)

in the cube, (1 − x, 1 − y, 1 − z) is also in the cube, and the midpoint of their projections will be theprojection of their midpoint, which is P , the projection of the cube onto this plane will be a centrallysymmetric region around P , and thus cannot be a pentagon.

6. [35] Let n ≥ 2 be a positive integer. A subset of positive integers S is said to be comprehensive iffor every integer 0 ≤ x < n, there is a subset of S whose sum has remainder x when divided by n.Note that the empty set has sum 0. Show that if a set S is comprehensive, then there is some (notnecessarily proper) subset of S with at most n− 1 elements which is also comprehensive.

Proposed by: Allen Liu

We will show that if |S| ≥ n, we can remove one element from S and still have a comprehensive set.Doing this repeatedly will always allow us to find a comprehensive subset of size at most n− 1.

Write S = {s1, s2, . . . , sk} for some k ≥ n. Now start with the empty set and add in the elements siin order. During this process, we will keep track of all possible remainders of sums of any subset.

If T is the set of current remainders at any time, and we add an element si, the set of remainderswill be T ∪ {t + si|t ∈ T}. In particular, the set of remainders only depends on the previous set ofremainders and the element we add in.

At the beginning of our process, the set of possible remainders is {0} for the empty set. Since weassumed that S is comprehensive, the final set is {0, 1, . . . , n}. The number of elements changes from 1to n− 1. However, since we added k ≥ n elements, at least one element did not change the size of ourremainder set. This implies that adding this element did not contribute to making any new remaindersand S is still comprehensive without this element, proving our claim.

7. [50] Let [n] denote the set of integers {1, 2, . . . , n}. We randomly choose a function f : [n] → [n],out of the nn possible functions. We also choose an integer a uniformly at random from [n]. Findthe probability that there exist positive integers b, c ≥ 1 such that f b(1) = a and f c(a) = 1. (fk(x)denotes the result of applying f to x k times).

Proposed by: Allen Liu

Answer: 1n

Given a function f , define N(f) to be the number of numbers that are in the same cycle as 1 (including 1itself), if there is one, and zero if there is no such cycle. The problem is equivalent to finding E(N(f))/n.Note that

P (N(f) = k) =n− 1

n· n− 2

n· · · · · n− k + 1

n· 1

n,

and it suffices to computen∑k=1

Pk where Pk = knP (N(f) = k). Observe that

Pn =

(n− 1

n· n− 2

n· · · · · 3

n· 2

n· 1

n

)· nn

· 1n

Pn→1 =

(n− 1

n· n− 2

n· · · · · 3

n· 2

n

)· n− 1

n· 1n

⇒ Pn + Pn→1 =

(n− 1

n· n− 2

n· · · · · 3

n· 2

n

)· 1n

Pn→2 =

(n− 1

n· n− 2

n· · · · · 3

n

)· n− 2

n· 1n

⇒ Pn + Pn→1 + Pn→2 =

(n− 1

n· n− 2

n· · · · · 3

n

)· 1n

· · ·

⇒n∑k=1

Pk = 1 · 1n

Therefore the answer is 1n .

8. [60] Allen plays a game on a tree with 2n vertices, each of whose vertices can be red or blue. Initially,all of the vertices of the tree are colored red. In one move, Allen is allowed to take two vertices of thesame color which are connected by an edge and change both of them to the opposite color. He wins ifat any time, all of the vertices of the tree are colored blue.

(a) (20) Show that Allen can win if and only if the vertices can be split up into two groups V1 andV2 of size n, such that each edge in the tree has one endpoint in V1 and one endpoint in V2.

(b) (40) Let V1 = {a1, . . . , an} and V2 = {b1, . . . , bn} from part (a). Let M be the minimum over allpermutations σ of {1, . . . , n} of the quantity

n∑i=1

d(ai, bσ(i)),

where d(v, w) denotes the number of edges along the shortest path between vertices v and w inthe tree.

Show that if Allen can win, then the minimum number of moves that it can take for Allen to winis equal to M .

(A graph consists of a set of vertices and some edges between distinct pairs of vertices. It is connectedif every pair of vertices are connected by some path of one or more edges. A tree is a graph which isconnected, in which the number of edges is one less than the number of vertices.)

Proposed by: Kevin Sun

Part (a): First we show that if we can’t split the vertices in the desired way then Allen cannot win. Todo so, observe that there is a unique way to split the vertices into two groups so that all edges crossbetween the two groups, since trees are bipartite. Each of Allen’s moves either adds one more bluevertex to each group or removes a blue vertex from each group, so the difference in the number of bluevertices between the two groups is invariant. Both groups initially have no blue vertices, so Allen canonly possibly arrive at a state where both groups have all blue vertices if both groups were initially ofequal size.

To show that Allen can win when the splitting is possible, we induct on n. It’s trivial for n = 1. Atree on 2n vertices has 2n− 1 edges, so the total degree for each of the two n-vertex groups is 2n− 1.Therefore, at least one vertex in each group is a leaf, say L1 in one group and L2 in the other. Allencan then flip all the vertices from L1 to L2 since there is a path of even length (in terms of numberof vertices) from one to the other, including flipping L1 and L2, and then flip back all of the middlevertices in the path, with the end result being that L1 and L2 are blue while all other vertices in thetree remain red. This lets us remove L1 and L2 from consideration and apply the inductive hypothesison the remaining graph.

Part (b): For each edge e in the graph, we can determine a minimum number of times Allen mustperform an operation on the endpoints of that edge (henceforth, ”flip this edge”), as follows. Observethat removing e disconnects the graph into two components. Each component is still a tree whichcan be split into two groups as before, though not necessarily with an equal number of vertices. Inparticular, suppose that the two groups differ by d vertices. Flipping any edge other than e does notchange the difference in number of blue vertices between the two groups of the component. So e mustflip at least d times if Allen is to win. Let me be the minimum number of times that edge e must flip,as determined in this way, and let M ′ be the total sum of all of the me. And let T ′ be the original treewith every edge duplicated me times (or removed if me = 0).

Observe that any permutation σ that gives us n paths must have total length at least M ′ for the samereasons as in the above paragraph, so M ′ ≤M .

In fact, we will show that M ′ = M . To show that M ′ = M , it suffices to show that M ′ ≥M , which wewill show by finding a partitioning of T ′ into n odd-length paths for which every vertex is the endpointof exactly one path. This will show that M ′ is the sum of path lengths for some permutation, meaningit is at least the sum of path lengths for the optimal permutation. We do this by strong induction.The base case is trivial, and if any me is 0 then we’re immediately done because we can split the T ′

into two smaller trees. Otherwise we repeat the argument from part (a) where we use a path betweentwo leaves, since each group of n vertices must contain a leaf, thereby reducing the number of verticesin our tree by 2 and allowing us to use our inductive hypothesis.

It now suffices to show that Allen can win the game in M ′ moves, but this is exactly the same inductionas in the previous paragraph. If any me is 0 then we’re immediately done by splitting T ′ into twosmaller trees, otherwise we use the path between the two leaves to use our inductive hypothesis.

9. [60] Evan has a simple graph with v vertices and e edges. Show that he can delete at least e→v+12 edges

so that each vertex still has at least half of its original degree.

Proposed by: Allen Liu

Fix v. We use strong induction on the number of edges e. If e ≤ v − 1, the result trivially holds byremoving 0 edges. Now take e > v − 1 and assume the result has been shown for all smaller values ofe. Consider a graph G with v vertices and e edges.

Suppose G contains a cycle C of even length 2k, where vertices (but not edges) may be repeated inthe cycle. Let G′ be the subgraph of G with the edges of C removed. Then G′ has v vertices ande − 2k edges. By the inductive hypothesis, it is possible to remove e→2k→v+1

2 edges from G′ so thateach vertex still has at least half its original degree. In the original graph G, remove these same edges,

and also remove every other edge of C (so, if the vertices of C are v1, · · · , v2k in order, we remove theedges between v2i→1 and v2i for 1 ≤ i ≤ k). In total, we have removed e→2k→v+1

2 + k = e→v+12 edges.

Furthermore, all vertices in G still have at least half their original degrees, as desired.

The remaining case to consider is if G has no cycles of even length. Then no two cycles in G can haveany vertices or edges in common. Suppose the contrary; then two odd cycles overlap, so their union isconnected and has an even number of edges. This union has an Eulerian tour, which is a cycle withan even number of edges, contradicting our assumption.

The number of edges in G is at most v + c− 1, where c is the number of cycles. So, we must removeat least e→v+1

2 = c2 edges from G. But we can remove c edges from G, one from each cycle. No vertex

has its degree decreased by more than 1, and each vertex whose degree is decreased is in a cycle andso has degree at least 2. Therefore each vertex still has at least half of its original degree, and we haveremoved at least e→v+1

2 edges, as desired.

Thus our claim holds for a graph with e edges, and thus by induction holds for any number of edges,as needed.

10. [60] Let n and m be positive integers which are at most 1010. Let R be the rectangle with cornersat (0, 0), (n, 0), (n,m), (0,m) in the coordinate plane. A simple non-self-intersecting quadrilateral withvertices at integer coordinates is called far-reaching if each of its vertices lie on or inside R, but each sideof R contains at least one vertex of the quadrilateral. Show that there is a far-reaching quadrilateralwith area at most 106.

(A side of a rectangle includes the two endpoints.)

Proposed by: Kevin Sun

Let g = gcd(n,m), with n = g · a and m = g · b. Note that the number of points on the diagonal of Rconnecting (0, 0) and (n,m) is g + 1. We construct two far-reaching quadrilaterals and show that atleast one of them has small area.

For our first quadrilateral, let (x1, y1) and (x2, y2) be the points with the shortest nonzero distancesto the diagonal between (0, 0) and (n,m) which lie above and below the diagonal, respectively. Nowconsider the quadrilateral with vertices (0, 0), (x1, y1), (n,m), (x2, y2). Note that the only latticepoints which can lie on or inside this quadrilateral are (x1, y1), (x2, y2), and points on the diagonal ofR, as otherwise we could find a closer point to the diagonal than (x1, y1) and (x2, y2). Thus by Pick’sTheorem, the area of this quadrilateral is at most g.

For our second quadrilateral, we will take as our vertices the points (0, 0), (n − 1,m), (a, b), and(n,m − 1). This is a concave quadrilateral which can be split into two triangle of areas a

2 and b2 , so

the area of this quadrilateral is at most 12 (a+ b).

We have therefore shown that there is always a far-reaching quadrilateral with area at most min(g, 12 (a+b)). Since g · 12 (a+ b) = 1

2 (n+m) ≤ 1010, we have that min(g, 12 (a+ b)) ≤ 105, so we can always finda far-reaching quadrilateral with area at most 105 as desired.

HMMT February 2018February 10, 2018

Algebra and Number Theory

1. For some real number c, the graphs of the equation y = |x 〉 20| + |x + 18| and the line y = x + cintersect at exactly one point. What is c?

2. Compute the positive real number x satisfying

x(2x6) = 3.

3. There are two prime numbers p so that 5p can be expressed in the form⌊n2

5

⌋for some positive integer

n. What is the sum of these two prime numbers?

4. Distinct prime numbers p, q, r satisfy the equation

2pqr + 50pq = 7pqr + 55pr = 8pqr + 12qr = A

for some positive integer A. What is A?

5. Let ω1, ω2, . . . , ω100 be the roots of x101−1x−1 (in some order). Consider the set

S = {ω11 , ω

22 , ω

33 , . . . , ω

100100}.

Let M be the maximum possible number of unique values in S, and let N be the minimum possiblenumber of unique values in S. Find M 〉N .

6. Let α, β, and γ be three real numbers. Suppose that

cosα+ cosβ + cos γ = 1

sinα+ sinβ + sin γ = 1.

Find the smallest possible value of cosα.

7. Rachel has the number 1000 in her hands. When she puts the number x in her left pocket, the numberchanges to x+ 1. When she puts the number x in her right pocket, the number changes to x−1. Eachminute, she flips a fair coin. If it lands heads, she puts the number into her left pocket, and if itlands tails, she puts it into her right pocket. She then takes the new number out of her pocket. If theexpected value of the number in Rachel’s hands after eight minutes is E, then compute b E10c.

8. For how many pairs of sequences of nonnegative integers (b1, b2, . . . , b2018) and (c1, c2, . . . , c2018) doesthere exist a sequence of nonnegative integers (a0, . . . , a2018) with the following properties:

• For 0 ≤ i ≤ 2018, ai < 22018;

• For 1 ≤ i ≤ 2018, bi = ai−1 + ai and ci = ai−1|ai;

where | denotes the bitwise or operation?

(The bitwise or of two nonnegative integers x = · · ·x3x2x1x0 and y = · · · y3y2y1y0 expressed in binaryis defined as x|y = · · · z3z2z1z0, where zi = 1 if at least one of xi and yi is 1, and 0 otherwise.)

9. Assume the quartic x4 〉 ax3 + bx2 〉 ax + d = 0 has four real roots 12 ≤ x1, x2, x3, x4 ≤ 2. Find the

maximum possible value of (x1+x2)(x1+x3)x4

(x4+x2)(x4+x3)x1(over all valid choices of a, b, d).

10. Let S be a randomly chosen 6-element subset of the set {0, 1, 2, . . . , n}. Consider the polynomialP (x) =

∑i∈S x

i. Let Xn be the probability that P (x) is divisible by some nonconstant polynomialQ(x) of degree at most 3 with integer coefficients satisfying Q(0) 6= 0. Find the limit of Xn as n goesto infinity.

HMMT February 2018February 10, 2018

Algebra and Number Theory

1. For some real number c, the graphs of the equation y = |x 〉 20| + |x + 18| and the line y = x + cintersect at exactly one point. What is c?

Proposed by: Henrik Boecken

Answer: 18

We want to know the value of c so that the graph |x〉20|+ |x+18|〉x = c has one solution. The graphof the function |x〉20|+ |x+ 18|〉x consists of an infinite section of slope 〉3 for x ∈ (〉∞,〉18], thena finite section of slope 〉1 for x ∈ [〉18, 20], then an infinite section of slope 1 for x ∈ [20,∞). Noticethat this graph is strictly decreasing on (〉∞, 20] and strictly increasing on [20,∞). Therefore anyhorizontal line will intersect this graph 0 or 2 times, except the one that passes through the “vertex”(20, |20〉 20|+ |20 + 18| 〉 20) = (20, 18), giving a value of c = 18.

2. Compute the positive real number x satisfying

x(2x6) = 3.

Proposed by: Henrik Boecken

Answer: 6√

3

Let t = x6, so x2t = 3. Taking this to the third power gives x6t = 27, or equivalently tt = 33. We cansee by inspection that t = 3, and this is the only solution as for t > 1, the function tt is monotonicallyincreasing, and if 0 < t < 1, tt < 1. Solving for x gives x6 = 3, or x = 6

√3.

3. There are two prime numbers p so that 5p can be expressed in the form⌊n2

5

⌋for some positive integer

n. What is the sum of these two prime numbers?

Proposed by: Kevin Sun

Answer: 52

Note that the remainder when n2 is divided by 5 must be 0, 1, or 4. Then we have that 25p = n2

or 25p = n2 〉 1 or 25p = n2 〉 4. In the first case there are no solutions. In the second case, if25p = (n 〉 1)(n + 1), then we must have n 〉 1 = 25 or n + 1 = 25 as n 〉 1 and n + 1 cannotboth be divisible by 5, and also cannot both have a factor besides 25. Similarly, in the third case,25p = (n〉 2)(n+ 2), so we must have n〉 2 = 25 or n+ 2 = 25.

Therefore the n we have to check are 23, 24, 26, 27. These give values of p = 21, p = 23, p = 27, andp = 29, of which only 23 and 29 are prime, so the answer is 23 + 29 = 52.

4. Distinct prime numbers p, q, r satisfy the equation

2pqr + 50pq = 7pqr + 55pr = 8pqr + 12qr = A

for some positive integer A. What is A?

Proposed by: Kevin Sun

Answer: 1980

Note that A is a multiple of p, q, and r, so K = Apqr is an integer. Dividing through, we have that

K = 8 +12

p= 7 +

55

q= 2 +

50

r.

Then p ∈ {2, 3}, q ∈ {5, 11}, and r ∈ {2, 5}. These values give K ∈ {14, 12}, K ∈ {18, 12}, and K ∈{27, 12}, giving K = 12 and (p, q, r) = (3, 11, 5). We can then compute A = pqr·K = 3·11·5·12 = 1980.

5. Let ω1, ω2, . . . , ω100 be the roots of x101−1x−1 (in some order). Consider the set

S = {ω11 , ω

22 , ω

33 , . . . , ω

100100}.

Let M be the maximum possible number of unique values in S, and let N be the minimum possiblenumber of unique values in S. Find M 〉N .

Proposed by: Henrik Boecken

Answer: 98

Throughout this solution, assume we’re working modulo 101.

First, N = 1. Let ω be a primitive 101st root of unity. We then let ωn = ω1/n, which we can dobecause 101 is prime, so 1/n exists for all nonzero n and 1/n = 1/m =⇒ m = n. Thus the setcontains only one distinct element, ω.

M = 100 is impossible. Fix ζ, a primitive 101st root of unity, and let ωn = ζπ(n) for each n. Supposethat there are 100 distinct such nπ(n) exponents; then π permutes the set {1, 2, · · · , 100}. Fix g, aprimitive root of 101; write n = gen and π(n) = gτ(en). Then {en} = {0, 1, 2, . . . , 100} and τ is a

permutation of this set, as is en+τ(en). However, this is impossible:∑100n=1 en+τ(en) = 5050+5050 ≡

5050 (mod 100), which is a contradiction. Thus there cannot be 100 distinct exponents.

M = 99 is possible. Again, let ζ be a primitive root of unity and let ωn = ζ1/(n+1), except whenn = 100, in which case let ω100 be the last possible root. Notice that n

n+1 = mm+1 if and only if n = m,

so this will produce 99 different elements in the set.

Thus M 〉N = 99〉 1 = 98.

6. Let α, β, and γ be three real numbers. Suppose that

cosα+ cosβ + cos γ = 1

sinα+ sinβ + sin γ = 1.

Find the smallest possible value of cosα.

Proposed by: Henrik Boecken

Answer: −1−√7

4

Let a = cosα+ i sinα, b = cosβ + i sinβ, and c = cos γ + i sin γ.

We then havea+ b+ c = 1 + i

where a, b, c are complex numbers on the unit circle. Now, to minimize cosα = Re[a], consider atriangle with vertices a, 1 + i, and the origin. We want a as far away from 1 + i as possible whilemaintaining a nonnegative imaginary part. This is achieved when b and c have the same argument, so|b + c| = |1 + i 〉 a| = 2. Now a, 0, and 1 + i form a 1 〉 2 〉

√2 triangle. The value of cosα is now

the cosine of the angle between the 1 and√

2 sides plus the π4 angle from 1 + i. Call the first angle δ.

Then

cos δ =12 + (

√2)2 〉 22

2 · 1 ·√

2

=〉1

2√

2

and

cosα = cos(π

4+ δ)

= cosπ

4cos δ 〉 sin

π

4sin δ

=

√2

2· 〉1

2√

2〉√

2

2·√

7

2√

2

=〉1〉

√7

4

7. Rachel has the number 1000 in her hands. When she puts the number x in her left pocket, the numberchanges to x+ 1. When she puts the number x in her right pocket, the number changes to x−1. Eachminute, she flips a fair coin. If it lands heads, she puts the number into her left pocket, and if itlands tails, she puts it into her right pocket. She then takes the new number out of her pocket. If theexpected value of the number in Rachel’s hands after eight minutes is E, then compute b E10c.Proposed by: Kevin Sun

Answer: 13

Call a real number very large if x ∈ [1000, 1008], very small if x ∈ [0, 11000 ], and medium-sized if

x ∈ [ 18 , 8]. Every number Rachel is ever holding after at most 8 steps will fall under one of thesecategories. Therefore the main contribution to E will come from the probability that Rachel is holdinga number at least 1000 at the end.

Note that if her number ever becomes medium-sized, it will never become very large or very smallagain. Therefore the only way her number ends up above 1000 is if the sequence of moves consists ofx → x + 1 moves and consecutive pairs of x → x−1 moves. Out of the 256 possible move sequences,the number of ways for the number to stay above 1000 is the number of ways of partitioning 8 into anordered sum of 1 and 2, or the ninth Fibonacci number F9 = 34.

Therefore34

256· 1000 ≤ E ≤ 34

256· 1000 + 8,

where34

256· 1000 ≈ 132.8. Furthermore, the extra contribution will certainly not exceed 7, so we get

that b E10c = 13.

(The actual value of E is

1538545594943410132524842390483285519695831541468827074238984121209064525621

11415831910281261197289931074429782903650103348754306523894286954489856000,

which is approximately equal to 134.77297. We can see that the extra contribution is about 2 and isvery insignificant.)

8. For how many pairs of sequences of nonnegative integers (b1, b2, . . . , b2018) and (c1, c2, . . . , c2018) doesthere exist a sequence of nonnegative integers (a0, . . . , a2018) with the following properties:

• For 0 ≤ i ≤ 2018, ai < 22018;

• For 1 ≤ i ≤ 2018, bi = ai−1 + ai and ci = ai−1|ai;

where | denotes the bitwise or operation?

(The bitwise or of two nonnegative integers x = · · ·x3x2x1x0 and y = · · · y3y2y1y0 expressed in binaryis defined as x|y = · · · z3z2z1z0, where zi = 1 if at least one of xi and yi is 1, and 0 otherwise.)

Proposed by: Kevin Sun

Answer: (22019 〉 1)2018

Define the bitwise and of two nonnegative integers x = · · ·x3x2x1x0 and y = · · · y3y2y1y0 expressed inbinary to be x&y = · · · z3z2z1z0, where zi = 1 if both xi and yi are 1, and 0 otherwise.

Now, we can prove that from the definitions of | and & that x + y = (x|y) + (x&y). Therefore itsuffices to count pairs of sequences (c1, c2, . . . , c2018) and (d1, d2, . . . , d2018) such that ci = ai−1|ai anddi = ai−1&ai for 0 ≤ ai < 22018.

Since both |,& are bitwise operations, it suffices to count the number of sequences {ci} and {di}restricting each ai to {0, 2k} for each k ∈ [0, 2017] and multiply these counts together. Each sequence(a0, . . . , a2018) leads to a unique {ci} and {di} except for the sequences (2k, 0, 2k, 0, . . . , 2k) and thesequences (0, 2k, 0, 2k, . . . , 0), which lead to the same {ci} and {di}.Therefore for each k, there are 22019〉 1 ways to determine the k-th bits of each ci and di. Multiplyingthis over all k gives a final count of (22019 〉 1)2018.

9. Assume the quartic x4 〉 ax3 + bx2 〉 ax + d = 0 has four real roots 12 ≤ x1, x2, x3, x4 ≤ 2. Find the

maximum possible value of (x1+x2)(x1+x3)x4

(x4+x2)(x4+x3)x1(over all valid choices of a, b, d).

Proposed by: Allen Liu

Answer: 54

We can rewrite the expression as

x24x21· (x1 + x1)(x1 + x2)(x1 + x3)(x1 + x4)

(x4 + x1)(x4 + x2)(x4 + x3)(x4 + x4)

x24x21· f(〉x1)

f(〉x4)

where f(x) is the quartic. We attempt to find a simple expression for f(〉x1). We know that

f(〉x1)〉 f(x1) = 2a · x31 + 2a · x1

Since x1 is a root, we havef(〉x1) = 2a · x31 + 2a · x1

Plugging this into our previous expression:

x24x21· x

31 + x1x34 + x4

x1 + 1x1

x4 + 1x4

The expression x + 1x is maximized at x = 2, 12 and minimized at x = 1. We can therefore maximize

the numerator with x1 = 2 and minimize the denominator with x4 = 1 to achieve the answer of 54 . It

can be confirmed that such an answer can be achieved such as with x2 = x3 =√10−13 .

10. Let S be a randomly chosen 6-element subset of the set {0, 1, 2, . . . , n}. Consider the polynomialP (x) =

∑i∈S x

i. Let Xn be the probability that P (x) is divisible by some nonconstant polynomialQ(x) of degree at most 3 with integer coefficients satisfying Q(0) 6= 0. Find the limit of Xn as n goesto infinity.

Proposed by: Allen Liu

Answer: 1001520736

We begin with the following claims:

Claim 1: There are finitely many Q(x) that divide some P (x) of the given form.

Proof: First of all the leading coefficient of Q must be 1, because if Q divides P then P/Q must haveinteger coefficients too. Note that if S = {s1, s2, s3, s4, s5, s6} with elements in increasing order, then

|P (x)| ≥ |xs6 | 〉 |xs5 | 〉 |xs4 | 〉 · · · 〉 |xs1 | = |x|s6 〉 |x|s5 〉 |x|s4 〉 · · · 〉 |x|s1 ,

so all the roots of P must have magnitude less than 2, and so do all the roots of Q. Therefore, allthe symmetric expressions involving the roots of Q are also bounded, so by Vieta’s Theorem all thecoefficients of Q of a given degree are bounded, and the number of such Q is therefore finite.

Claim 2: If Q has a nonzero root that does not have magnitude 1, then the probability that it dividesa randomly chosen P vanishes as n goes to infinity.

Proof: WLOG suppose that Q has a root r with |r| > 1 (similar argument will apply for |r| < 1).Then from the bound given in the proof of Claim 1, it is not difficult to see that s6 〉 s5 is boundedsince

|P (r)| > |r|s6 〉 5|r|s5 > |r|s6−s5 〉 5,

which approaches infinity as s6〉s5 goes to infinity. By similar argument we can show that s5〉s4, s4〉s3, . . . are all bounded. Therefore, the probability of choosing the correct coefficients is bounded aboveby the product of five fixed numbers divided by n5/5!, which vanishes as n goes to infinity.

From the claims above, we see that we only need to consider polynomials with roots of magnitude1, since the sum of all other possibilities vanishes as n goes to infinity. Moreover, this implies thatwe only need to consider roots of unity. Since Q has degree at most 3, the only possible roots are

〉1,±i, −1±i√3

2 , 1±i√3

2 , corresponding to x+ 1, x2 + 1, x2 + x+ 1, x2 〉 x+ 1 (note that eighth root ofunity is impossible because x4 + 1 cannot be factored in the rationals).

Now we compute the probability of P (r) = 0 for each possible root r. Since the value of xs cycles withs, and we only care about n→∞, we may even assume that the exponents are chosen independentlyat random, with repetition allowed.

Case 1: When r = 〉1, the number of odd exponents need to be equal to the number of even exponents,

which happens with probability(63)26 = 5

16 .

Case 2: When r = ±i, the number of exponents that are 0 modulo 4 need to be equal to thosethat are 2 modulo 4, and same for 1 modulo 4 and 3 modulo 4, which happens with probability(60)26 ·

(00)(

63)

26 +(62)26 ·

(21)(

42)

26 +(64)26 ·

(42)(

21)

26 +(66)26 ·

(63)(

00)

26 = 25256 . Note that Case 1 and Case 2 have no overlaps,

since the former requires 3 even exponents, and the latter requires 0, 2, 4, or 6 even exponents.

Case 3: When r = −1±i√3

2 , the number of exponents that are 0, 1, 2 modulo 3 need to be equal to

each other, so the probability is( 62,2,2)36 = 10

81 .

Case 4: When r = 1±i√3

2 , then if ni is the number of exponents that are i modulo 6 (i = 0, 1, 2, 3, 4, 5),then n0 〉 n3 = n2 〉 n5 = n4 〉 n1 = k for some k. Since 3k ≡ n0 + n1 + · · · + n5 = 6 ≡ 0 (mod 2),k must be one of 〉2, 0, 2. When k = 0, we have n0 + n2 + n4 = n1 + n3 + n5, which is the same asCase 1. When k = 2, we have n0 = n2 = n4 = 2, which is covered in Case 3, and similar for k = 〉2.Therefore we do not need to consider this case.

Now we deal with over-counting. Since Case 1 and 2 deal with the exponents modulo 4 and Case 3 dealwith exponents modulo 3, the probabilities are independent from each other. So by complementarycounting, we compute the final probability as

1〉 (1〉 5

16〉 25

256)(1〉 10

81) = 1〉 151

256· 71

81=

10015

20736.

HMMT February 2018February 10, 2018

Combinatorics

1. Consider a 2 × 3 grid where each entry is one of 0, 1, and 2. For how many such grids is the sum ofthe numbers in every row and in every column a multiple of 3? One valid grid is shown below.[

1 2 02 1 0

]2. Let a and b be five-digit palindromes (without leading zeroes) such that a < b and there are no other

five-digit palindromes strictly between a and b. What are all possible values of b− a? (A number is apalindrome if it reads the same forwards and backwards in base 10.)

3. A 4× 4 window is made out of 16 square windowpanes. How many ways are there to stain each of thewindowpanes, red, pink, or magenta, such that each windowpane is the same color as exactly two ofits neighbors? Two different windowpanes are neighbors if they share a side.

4. How many ways are there for Nick to travel from (0, 0) to (16, 16) in the coordinate plane by movingone unit in the positive x or y direction at a time, such that Nick changes direction an odd number oftimes?

5. A bag contains nine blue marbles, ten ugly marbles, and one special marble. Ryan picks marblesrandomly from this bag with replacement until he draws the special marble. He notices that none ofthe marbles he drew were ugly. Given this information, what is the expected value of the number oftotal marbles he drew?

6. Sarah stands at (0, 0) and Rachel stands at (6, 8) in the Euclidean plane. Sarah can only move 1 unitin the positive x or y direction, and Rachel can only move 1 unit in the negative x or y direction. Eachsecond, Sarah and Rachel see each other, independently pick a direction to move at the same time,and move to their new position. Sarah catches Rachel if Sarah and Rachel are ever at the same point.Rachel wins if she is able to get to (0, 0) without being caught; otherwise, Sarah wins. Given that bothof them play optimally to maximize their probability of winning, what is the probability that Rachelwins?

7. A tourist is learning an incorrect way to sort a permutation (p1, . . . , pn) of the integers (1, . . . , n). Wedefine a fix on two adjacent elements pi and pi+1, to be an operation which swaps the two elementsif pi > pi+1, and does nothing otherwise. The tourist performs n − 1 rounds of fixes, numbereda = 1, 2, . . . , n − 1. In round a of fixes, the tourist fixes pa and pa+1, then pa+1 and pa+2, and so on,

up to pn∈1 and pn. In this process, there are (n− 1) + (n− 2) + · · ·+ 1 = n(n∈1)2 total fixes performed.

How many permutations of (1, . . . , 2018) can the tourist start with to obtain (1, . . . , 2018) after per-forming these steps?

8. A permutation of {1, 2, . . . , 7} is chosen uniformly at random. A partition of the permutation intocontiguous blocks is correct if, when each block is sorted independently, the entire permutation becomessorted. For example, the permutation (3, 4, 2, 1, 6, 5, 7) can be partitioned correctly into the blocks[3, 4, 2, 1] and [6, 5, 7], since when these blocks are sorted, the permutation becomes (1, 2, 3, 4, 5, 6, 7).

Find the expected value of the maximum number of blocks into which the permutation can be parti-tioned correctly.

9. How many ordered sequences of 36 digits have the property that summing the digits to get a numberand taking the last digit of the sum results in a digit which is not in our original sequence? (Digitsrange from 0 to 9.)

10. Lily has a 300× 300 grid of squares. She now removes 100× 100 squares from each of the four cornersand colors each of the remaining 50000 squares black and white. Given that no 2× 2 square is coloredin a checkerboard pattern, find the maximum possible number of (unordered) pairs of squares suchthat one is black, one is white and the squares share an edge.

HMMT February 2018February 10, 2018

Combinatorics

1. Consider a 2 × 3 grid where each entry is one of 0, 1, and 2. For how many such grids is the sum ofthe numbers in every row and in every column a multiple of 3? One valid grid is shown below.[

1 2 02 1 0

]Proposed by: Henrik Boecken

Answer: 9

Any two elements in the same row fix the rest of the grid, so 32 = 9.

2. Let a and b be five-digit palindromes (without leading zeroes) such that a < b and there are no otherfive-digit palindromes strictly between a and b. What are all possible values of b− a? (A number is apalindrome if it reads the same forwards and backwards in base 10.)

Proposed by: Kevin Sun

Answer: 100, 110, 11

Let xyzyx be the digits of the palindrome a. There are three cases. If z < 9, then the next palindromegreater than xyzyx is xy(z + 1)yx, which differs by 100. If z = 9 but y < 9, then the next palindrome upis x(y + 1)0(y + 1)x, which differs from xy9yx by 110. Finally, if y = z = 9, then the next palindromeafter x999x is (x+ 1)000(x+ 1), which gives a difference of 11. Thus, the possible differences are11, 100, 110.

3. A 4× 4 window is made out of 16 square windowpanes. How many ways are there to stain each of thewindowpanes, red, pink, or magenta, such that each windowpane is the same color as exactly two ofits neighbors? Two different windowpanes are neighbors if they share a side.

Proposed by: Kevin Sun

Answer: 24

For the purpose of explaining this solution, let’s label the squares as

11 12 13 1421 22 23 2431 32 33 3441 42 43 44

Note that since the corner squares 11, 14, 41, 44 each only have two neighbors, each corner square isthe same color as both of its neighbors (for example, 11, 12, and 21 are the same color, 31, 41, and 42are the same color, etc.). This corner square constraint heavily limits the possible colorings. We willnow use casework.

Case 1: Suppose two corner squares on the same side (without loss of generality, let them be 11 and14) have the same color (without loss of generality, red). Then 21, 11, 12, 13, 14, 24 are all red, and 12has two red neighbors (11 and 13) so its third neighbor ( 22) is a color different from red (withoutloss of generality, magenta). But 22 has two red neighbors (12 and 21), so its other two neighbors (23and 32)must be magenta. Applying the same logic symmetrically, we find that all four interior squares(22, 23, 32, 33) have the same color. Furthermore, 21 has one magenta neighbor 22, so 31 must be red.Symmetrically, 34 is red, and by the corner square constraint we have that all the exterior squares are

the same color. Thus in general, this case is equivalent to a window taking the following form (withdistinct colors A and B)

A A A AA B B AA B B AA A A A

The number of choices of A and B is 3 · 2 = 6.

Case 2: No two corner squares on the same side have the same color.

Then from the corner square constraint 12 has neighbor 11 of the same color and neighbor 13 of adifferent color, so its neighbor 22 must be the same color as 12. Therefore, this case is equivalent tocoloring each quadrant entirely in one color such two quadrants sharing a side have different colors.(A quadrant refers to the four squares on one vertical half and one horizontal half, e.g. 13, 14, 23, 24).

If only two colors are used, the window will take the form (with distinct colors A and B):

A A B BA A B BB B A AB B A A

Again there are 3 · 2 = 6 ways to chose A and B.

If all three colors are used, the window will take the form (with distinct colors A, B and C)

A A B BA A B BC C A AC C A A

or

A A B BA A B BB B C CB B C C

There are 3 · 2 · 1 = 6 ways to select colors for each of these forms.

Therefore, there are 6 colorings in Case 1 and 6 + 6 + 6 in Case 2, for a total of 24 colorings.

4. How many ways are there for Nick to travel from (0, 0) to (16, 16) in the coordinate plane by movingone unit in the positive x or y direction at a time, such that Nick changes direction an odd number oftimes?

Proposed by: Huaiyu Wu

Answer: 2 ·(

30

15

)= 310235040

This condition is equivalent to the first and last step being in different directions, as if you switchdirections an odd number of times, you must end in a different direction than you started. If the firststep is in the x direction and the last step is in the y direction, it suffices to count the number of paths

from (1, 0) to (16, 15), of which there are

(30

15

). Similarly, in the other case, it suffices to count the

number of paths from (0, 1) to (15, 16), of which there are also

(30

15

). Therefore the total number of

paths is 2 ·(

30

15

).

5. A bag contains nine blue marbles, ten ugly marbles, and one special marble. Ryan picks marblesrandomly from this bag with replacement until he draws the special marble. He notices that none ofthe marbles he drew were ugly. Given this information, what is the expected value of the number oftotal marbles he drew?

Proposed by: Kevin Sun

Answer: 2011

The probability of drawing k marbles is the probability of drawing k − 1 blue marbles and then the

special marble, which is pk =(

920

)k−1 × 120 . The probability of drawing no ugly marbles is therefore∑∞

k=1 pk = 111 .

Then given that no ugly marbles were drawn, the probability that k marbles were drawn is 11pk. Theexpected number of marbles Ryan drew is

∞∑k=1

k(11pk) =11

20

∞∑k=1

k

(9

20

)k−1

=11

20× 400

121=

20

11.

(To compute the sum in the last step, let S =∑∞

k=1 k(

920

)k−1and note that 9

20S = S−∑∞

k=1

(920

)k−1=

S − 2011 ).

6. Sarah stands at (0, 0) and Rachel stands at (6, 8) in the Euclidean plane. Sarah can only move 1 unitin the positive x or y direction, and Rachel can only move 1 unit in the negative x or y direction. Eachsecond, Sarah and Rachel see each other, independently pick a direction to move at the same time,and move to their new position. Sarah catches Rachel if Sarah and Rachel are ever at the same point.Rachel wins if she is able to get to (0, 0) without being caught; otherwise, Sarah wins. Given that bothof them play optimally to maximize their probability of winning, what is the probability that Rachelwins?

Proposed by: Rachel Zhang

Answer: 6364

We make the following claim: In a game with n × m grid where n ≤ m and n ≡ m (mod 2), theprobability that Sarah wins is 1

2n under optimal play.

Proof: We induct on n. First consider the base case n = 0. In this case Rachel is confined on a line,so Sarah is guaranteed to win.

We then consider the case where n = m (a square grid). If Rachel and Sarah move in parallel directionsat first, then Rachel can win if she keep moving in this direction, since Sarah will not be able to catchRachel no matter what. Otherwise, the problem is reduced to a (n− 1)× (n− 1) grid. Therefore, theoptimal strategy for both players is to choose a direction completely randomly, since any bias can beabused by the other player. So the reduction happens with probability 1

2 , and by induction hypothesisSarah will with probability 1

2n−1 , so on a n× n grid Sarah wins with probability 12n .

Now we use induction to show that when n < m, both player will move in the longer (m) directionuntil they are at corners of a square grid (in which case Sarah wins with probability 1

2n . If Sarahmoves in the n direction and Rachel moves in the m (or n) direction, then Rachel can just move inthe n direction until she reaches the other side of the grid and Sarah will not be able to catch her.If Rachel moves in the n direction and Sarah moves in the m direction, then the problem is reducedto a (n− 1)× (m− 1) grid, which means that Sarah’s winning probability is now doubled to 1

2n−1 byinduction hypothesis. Therefore it is suboptimal for either player to move in the shorter (n) direction.This shows that the game will be reduced to n× n with optimal play, and thus the claim is proved.

From the claim, we can conclude that the probability that Rachel wins is 1− 126 = 63

64 .

7. A tourist is learning an incorrect way to sort a permutation (p1, . . . , pn) of the integers (1, . . . , n). Wedefine a fix on two adjacent elements pi and pi+1, to be an operation which swaps the two elementsif pi > pi+1, and does nothing otherwise. The tourist performs n − 1 rounds of fixes, numbereda = 1, 2, . . . , n − 1. In round a of fixes, the tourist fixes pa and pa+1, then pa+1 and pa+2, and so on,

up to pn−1 and pn. In this process, there are (n− 1) + (n− 2) + · · ·+ 1 = n(n−1)2 total fixes performed.

How many permutations of (1, . . . , 2018) can the tourist start with to obtain (1, . . . , 2018) after per-forming these steps?

Proposed by: Kevin Sun

Answer: 1009! · 1010!

Note that the given algorithm is very similar to the well-known Bubble Sort algorithm for sorting anarray. The exception is that in the i-th round through the array, the first i− 1 pairs are not checked.

We claim a necessary and sufficient condition for the array to be sorted after the tourist’s process is: forall i, after i rounds, the numbers 1, · · · , i are in the correct position. Firstly, this is necessary becausethese indices of the array are not touched in future rounds - so if a number was incorrect, then it wouldstay incorrect. On the other hand, suppose this condition holds. Then, we can ”add” the additionalfixes during each round (of the first i− 1 pairs during the i-th round) to make the process identical tobubble sort. The tourist’s final result won’t change because by our assumption these swaps won’t doanything. However, this process is now identical to bubble sort, so the resulting array will be sorted.Thus, our condition is sufficient.

Now, there are two positions the 1 can be in (p1, p2). There are three positions the 2 can be in(p1, · · · , p4 except for the position of 1). Similarly, for 1 ≤ i ≤ 1009 there are 2i − (i − 1) = i + 1positions i can be in, and after that the remaining 1009 numbers can be arranged arbitrarily. Thus,the answer is 1010! · 1009!.

8. A permutation of {1, 2, . . . , 7} is chosen uniformly at random. A partition of the permutation intocontiguous blocks is correct if, when each block is sorted independently, the entire permutation becomessorted. For example, the permutation (3, 4, 2, 1, 6, 5, 7) can be partitioned correctly into the blocks[3, 4, 2, 1] and [6, 5, 7], since when these blocks are sorted, the permutation becomes (1, 2, 3, 4, 5, 6, 7).

Find the expected value of the maximum number of blocks into which the permutation can be parti-tioned correctly.

Proposed by: Mehtaab Sawhney

Answer: 151105

Let σ be a permutation on {1, . . . , n}. Call m ∈ {1, . . . , n} a breakpoint of σ if {σ(1), . . . , σ(m)} ={1, . . . ,m}. Notice that the maximum partition is into k blocks, where k is the number of breakpoints:if our breakpoints are m1, . . . ,mk, then we take {1, . . . ,m1}, {m1 +1, . . . ,m2}, . . . , {mk−1 +1, . . . ,mk}as our contiguous blocks.

Now we just want to findE[k] = E[X1 + · · ·+Xn],

where Xi = 1 if i is a breakpoint, and Xi = 0 otherwise. We use linearity of expectation and noticethat

E[Xi] =i!(n− i)!

n!,

since this is the probability that the first i numbers are just 1, . . . , i in some order. Thus,

E[k] =

n∑i=1

i!(n− i)!n!

=

n∑i=1

(n

i

)−1.

We can compute for n = 7 that the answer is151

105.

9. How many ordered sequences of 36 digits have the property that summing the digits to get a numberand taking the last digit of the sum results in a digit which is not in our original sequence? (Digitsrange from 0 to 9.)

Proposed by: Kevin Sun

Answer: 936 + 4

We will solve this problem for 36 replaced by n. We use [n] to denote {1, 2, . . . , n} and σs to denotethe last digit of the sum of the digits of s.

Let D be the set of all sequences of n digits and let Si be the set of digit sequences s such that si = σs,

the ith digit of s. The quantity we are asked to compute is equal to

∣∣∣∣∣D \n⋃

i=1

Si

∣∣∣∣∣. We use the principle

of inclusion-exclusion to compute this:∣∣∣∣∣D \n⋃

i=1

Si

∣∣∣∣∣ =∑J⊆[n]

(−1)|J|

∣∣∣∣∣∣⋂j∈J

Sj

∣∣∣∣∣∣Note that a digit sequence is in Si if and only if the n− 1 digits which are not i sum to a multiple of10. This gives that |Si| = 10 · 10n−2 = 10n−1 as there are 10 ways to pick the ith digit and 10n−2 waysto pick the other digits.

Similarly, given a subset J ⊆ [n], we can perform a similar analysis. If a string s is in⋂j∈J

Sj , we must

have that sj = σs for all j ∈ J . There are 10 ways to pick σs, which determines sj for all j ∈ J . Fromthere, there are 10(n−|J|)−1 ways to pick the remaining digits as if we fix all but one, the last digit isuniquely determined. This gives 10n−|J| choices.

However, this breaks down when |J | = n, as not all choices of σs lead to any valid solutions. When|J | = n, J = [n] and we require that the last digit of nσs is σs, which happens for gcd(n− 1, 10) valuesof σs.

We now compare our expression from the principle of inclusion-exclusion to the binomial expansion of(10− 1)n. By the binomial theorem,

9n = (10− 1)n =∑J⊆[n]

(−1)|J|10n−|J|.

These agree on every term except for the term where J = [n]. In this case, we need to add an extra(−1)n gcd(n− 1, 10) and subtract (−1)n.

Thus our final value for

∣∣∣∣∣D \n⋃

i=1

Si

∣∣∣∣∣ is 9n + (−1)n(gcd(n− 1, 10)− 1), which is 936 + 4 for n = 36.

10. Lily has a 300× 300 grid of squares. She now removes 100× 100 squares from each of the four cornersand colors each of the remaining 50000 squares black and white. Given that no 2× 2 square is coloredin a checkerboard pattern, find the maximum possible number of (unordered) pairs of squares suchthat one is black, one is white and the squares share an edge.

Proposed by: Allen Liu

Answer: 49998

First we show an upper bound. Define a grid point as a vertex of one of the squares in the figure.Construct a graph as follows. Place a vertex at each grid point and draw an edge between two adjacentpoints if that edge forms a black-white boundary. The condition of there being no 2× 2 checkerboardis equivalent to no vertex having degree more than 2. There are 1012 + 4 · 992 = 49405 vertices thatare allowed to have degree 2 and 12 · 99 = 1188 vertices (on the boundary) that can have degree 1.

This gives us an upper bound of 49999 edges. We will show that exactly this many edges is impossible.Assume for the sake of contradiction that we have a configuration achieving exactly this many edges.

Consider pairing up the degree 1 vertices so that those on a horizontal edge pair with the other vertexin the same column and those on a vertical edge pair with the other vertex in the same row. If wecombine the pairs into one vertex, the resulting graph must have all vertices with degree exactly 2.This means the graph must be a union of disjoint cycles. However all cycles must have even lengthand there are an odd number of total vertices so this is impossible. Thus we have an upper bound of49998.

We now describe the construction. The top row alternates black and white. The next 99 rows alternatebetween all black and all white. Lets say the second row from the top is all white. The 101st rowalternates black and white for the first 100 squares, is all black for the next 100 and alternates betweenwhite and black for the last 100 squares. The next 98 rows alternate between all black and all white(the 102nd row is all white). Finally, the bottom 101 rows are a mirror of the top 101 rows with thecolors reversed. We easily verify that this achieves the desired. We illustrate the construction for 300replaced by 12.

HMMT February 2018February 10, 2018

Geometry

1. Triangle GRT has GR = 5, RT = 12, and GT = 13. The perpendicular bisector of GT intersects theextension of GR at O. Find TO.

Proposed by: Henrik Boecken

Answer: 16910

First, note that TO = GO as O lies on the perpendicular bisector of GT . Then if M is the midpoint

of GT , we have that 4GRT ∼ 4GMO, so we can compute TO = GO = GM · GTGR

=13

2· 13

5=

169

10.

2. Points A,B,C,D are chosen in the plane such that segments AB, BC, CD, DA have lengths 2, 7, 5,12, respectively. Let m be the minimum possible value of the length of segment AC and let M be themaximum possible value of the length of segment AC. What is the ordered pair (m,M)?

Proposed by: Kevin Sun

Answer: (7, 9)

By the triangle inequality on triangle ACD, AC + CD ≥ AD, or AC ≥ 7. The minimum of 7 canbe achieved when A, C, D lie on a line in that order. By the triangle inequality on triangle ABC,AB + BC ≥ AC, or AC ≤ 9. The maximum of 9 can be achieved when A, B, C lie on a line in thatorder. This gives the answer (7, 9).

3. How many noncongruent triangles are there with one side of length 20, one side of length 17, and one60◦ angle?

Proposed by: Dai Yang

Answer: 2

There are 3 possible vertices that can have an angle of 60◦, we will name them. Call the vertex wherethe sides of length 20 and 17 meet α, denote the vertex where 17 doesn’t meet 20 by β, and the finalvertex, which meets 20 but not 17, we denote by γ.

The law of cosines states that if we have a triangle, then we have the equation c2 = a2 + b2− 2ab cosCwhere C is the angle between a and b. But cos 60◦ = 1

2 so this becomes c2 = a2 + b2 − ab. Wethen try satisfying this equation for the 3 possible vertices and find that, for α the equation readsc2 = 400 + 289 − 340 = 349 so that c =

√349. For β we find that 400 = 289 + b2 − 17b or rather

b2 − 17b− 111 = 0 this is a quadratic, solving we find that it has two roots b = 17±√289+4442 , but since√

733 > 17 only one of these roots is positive. We can also see that this isn’t congruent to the othertriangle we had, as for both the triangles the shortest side has length 17, and so if they were congruentthe lengths of all sides would need to be equal, but 18 <

√349 < 19 and since 232 < 733 clearly

17±√289+4442 > 17+23

2 = 20 and so the triangles aren’t congruent. If we try applying the law of cosinesto γ however, we get the equation 289 = a2 + 400 − 20a which we can rewrite as a2 − 20a + 111 = 0which has no real solutions, as the discriminant 400 − 4 ∗ 111 = −44 is negative. Thus, γ cannot be60◦, and there are exactly two non congruent triangles with side lengths 20 and 17 with an angle being60◦.

4. A paper equilateral triangle of side length 2 on a table has vertices labeled A, B, C. Let M be thepoint on the sheet of paper halfway between A and C. Over time, point M is lifted upwards, foldingthe triangle along segment BM , while A, B, and C remain on the table. This continues until A andC touch. Find the maximum volume of tetrahedron ABCM at any time during this process.

Proposed by: Dhruv Rohatgi

Answer:

√3

6

View triangle ABM as a base of this tetrahedron. Then relative to triangle ABM , triangle CBMrotates around segment BM on a hinge. Therefore the volume is maximized when C is farthest fromtriangle ABM , which is when triangles ABM and CBM are perpendicular. The volume in this case

can be calculated using the formula for the volume of a tetrahedron as1

6· 1 · 1 ·

√3 =

√3

6.

5. In the quadrilateral MARE inscribed in a unit circle ω, AM is a diameter of ω, and E lies on theangle bisector of ∠RAM . Given that triangles RAM and REM have the same area, find the area ofquadrilateral MARE.

Proposed by: Yuan Yao

Answer: 8√2

9

Since AE bisects ∠RAM , we have RE = EM , and E,A lie on different sides of RM . Since AM is adiameter, ∠ARM = 90◦. If the midpoint of RM is N , then from [RAM ] = [REM ] and ∠ARM = 90◦,we find AR = NE. Note that O, the center of ω, N , and E are collinear, and by similarity of trianglesNOM and RAM , ON = 1

2AR = 12NE. Therefore, ON = 1

3 and NE = 23 . By the Pythagorean

theorem on triangle RAM , RM = 4√2

3 , Therefore, the area of MARE is 2 · 12 ·4√2

3 ·23 = 8

√2

9 .

6. Let ABC be an equilateral triangle of side length 1. For a real number 0 < x < 0.5, let A1 and A2

be the points on side BC such that A1B = A2C = x, and let TA = 4AA1A2. Construct trianglesTB = 4BB1B2 and TC = 4CC1C2 similarly.

There exist positive rational numbers b, c such that the region of points inside all three trianglesTA, TB , TC is a hexagon with area

8x2 − bx+ c

(2− x)(x+ 1)·√

3

4.

Find (b, c).

Proposed by: Kevin Sun

Answer: (8, 2)

Solution 1:

Notice that the given expression is defined and continuous not only on 0 < x < 0.5, but also on0 ≤ x ≤ 0.5. Let f(x) be the function representing the area of the (possibly degenerate) hexagon forx ∈ [0, 0.5]. Since f(x) is equal to the given expression over (0, 0.5), we can conclude that f(0) andf(0.5) will also be equal to the expression when x = 0 and x = 0.5 respectively. (In other words, f(x)

is equal to the expression over [0, 0.5].) In each of the cases, we can compute easily that f(0) =√34 and

f(0.5) = 0, so by plugging them in, we get c2·1 = 1 and 2→b/2+c

(3/2)·(3/2) = 0, which gives b = 8 and c = 2.

Solution 2:

Let P = AA1∩CC2, Q = AA2∩BB1, R = BB1∩CC2. These three points are the points on the hexagonfarthest away from A. For reasons of symmetry, the area of the hexagon (call it H for convenience) is:

[H] = [ABC]− 3[BPRQC].

Also, [BPC] = [BQC] by symmetry, so:

[BPRQC] = [BPC] + [BQC]− [BRC]

[BPRQC] = 2[BPC]− [BRC].

From this, one can see that all we need to do is calculate the A-level of the points P and R in barycentriccoordinates. Ultimately, the A-level of P is x

x+1 , and the A-level of R is x2→x . From this, straightforward

calculation shows that:

[H] =8x2 − 8x+ 2

(2− x)(x+ 1)·√

3

4,

thus giving us the answer (b, c) = (8, 2).

7. Triangle ABC has sidelengths AB = 14, AC = 13, and BC = 15. Point D is chosen in the interiorof AB and point E is selected uniformly at random from AD. Point F is then defined to be theintersection point of the perpendicular to AB at E and the union of segments AC and BC. Supposethat D is chosen such that the expected value of the length of EF is maximized. Find AD.

Proposed by: Gabriel Mintzer

Answer:√

70

Let G be the intersection of the altitude to AB at point D with AC ∪ BC. We first note that the

maximal expected value is obtained when DG = [ADGC]AD , where [P ] denotes the area of polygon P .

Note that if DG were not equal to this value, we could move D either closer or further from A andincrease the value of the fraction, which is the expected value of EF . We note that this equality canonly occur if D is on the side of the altitude to AB nearest point B. Multiplying both sides of thisequation by AD yields AD ·DG = [ADGC], which can be interpreted as meaning that the area of therectangle with base AD and height DG must have area equal to that of quadrilateral ADGC. We cannow solve this problem with algebra.

Let x = BD. We first compute the area of the rectangle with base AD and height DG. We have thatAD = AB −BD = 14− x. By decomposing the 13− 14− 15 triangle into a 5− 12− 13 triangle anda 9− 12− 15 triangle, and using a similarity argument, we find that DG = 4

3x. Thus, the area of thisrectangle is 4

3x(14− x) = 563 x−

43x

2.

We next compute the area of quadrilateral ADGC. We note that [ADGC] = [ABC] − [BDG]. Wehave that [ABC] = 1

2 (12)(14) = 84. We have BD = x and DG = 43x, so [BDG] = 1

2 (x)43x)

= 23x

2.Therefore, we have [ADGC] = [ABC]− [BDG] = 84− 2

3x2.

Equating these two areas, we have

56

3x− 4

3x2 = 84− 2

3x2,

or, simplifying,x2 − 28x+ 126 = 0.

Solving yields x = 14±√

70, but 14 +√

70 exceeds AB, so we discard it as an extraneous root. Thus,BD = 14−

√70 and

AD = AB −BD = 14− (14−√

70) =√

70 .

Remark: if the altitude to point C meets AB at point H, then the general answer to this problemis√AH ·AB. This result can be derived by considering the effects of dilation in the AB direction

and dilation in the CH direction then performing dilations such that ∠C is right and carrying out thecalculation described above while considering congruent triangles.

8. Let ABC be an equilateral triangle with side length 8. Let X be on side AB so that AX = 5 andY be on side AC so that AY = 3. Let Z be on side BC so that AZ,BY,CX are concurrent. LetZX,ZY intersect the circumcircle of AXY again at P,Q respectively. Let XQ and Y P intersect atK. Compute KX ·KQ.

Proposed by: Allen Liu

Answer: 304

Let BY and CX meet at O. O is on the circumcircle of AXY , since 4AXC ∼= 4CY B.

We claim that KA and KO are tangent to the circumcircle of AXY . Let XY and BC meet at L.Then, LBZC is harmonic. A perspectivity at X gives AY OP is harmonic. Similarly, a perspectivityat Y gives AXOQ is harmonic. Thus, K is the pole of chord AO.

Now we compute. Denote r as the radius and θ as ∠AXO. Then,

r =XY√

3=

√52 + 32 − 3 · 5√

3=

√19

3;

sin θ = sin 60◦ · ACXC

=

√3

2· 8√

52 + 82 − 5 · 8=

4

7

√3;

KX ·KQ = KA2 = (r · tan θ)2 = (

√19

3· 4√

3)2 = 304.

9. Po picks 100 points P1, P2, . . . , P100 on a circle independently and uniformly at random. He then drawsthe line segments connecting P1P2, P2P3, . . . , P100P1. When all of the line segments are drawn, thecircle is divided into a number of regions. Find the expected number of regions that have all sidesbounded by straight lines.

Proposed by: Allen Liu

Answer: 48533

If the 100 segments do not intersect on the interior, then the circle will be cut into 101 regions. ByEuler’s formula, each additional intersection cuts two edges into two each, and adds one more vertex,so since V − E + F is constant, there will be one more region as well. It then suffices to computethe expected number of intersections, where two segments that share a vertex are not counted asintersections.

We use linearity of expectation to compute this value. It suffices to compute the expected number ofsegments that each segment intersects. Consider one such segment P1P2. It cannot possibly intersecta segment that shares an endpoint, so that leaves 97 possible other segments. Again, by linearity ofexpection, it suffices to compute the probability that P1P2 intersects PiPi+1. However, since each ofthe points was chosen uniformly at random, this is equal to the probability that AC intersects BD,where A,B,C,D are chosen uniformly at random from the circle. Since this probability is 1/3, eachsegment intersects with 97

3 segments on average.

Now, we can sum over all segments and divide by two to get (100 · 973 )/2 = 48503 intersections, since

each intersection is counted twice. Accounting for the fact that there are 101 regions to begin with,and exactly 100 of them have an arc on the boundary, we get 4850

3 + 101− 100 = 48533 as the answer.

10. Let ABC be a triangle such that AB = 6, BC = 5, AC = 7. Let the tangents to the circumcircle ofABC at B and C meet at X. Let Z be a point on the circumcircle of ABC. Let Y be the foot ofthe perpendicular from X to CZ. Let K be the intersection of the circumcircle of BCY with line AB.Given that Y is on the interior of segment CZ and Y Z = 3CY , compute AK.

Proposed by: Allen Liu

Answer: 14710

Let ω1 denote the circumcircle of ABC and ω2 denote the circle centered at X through B and C. Letω2 intersect AB,AC again at B′, C ′. The (signed) power of Y with respect to ω1 is −CY · Y Z. Thepower of Y with respect to ω2 is XY 2−CX2 = −CY 2. Thus the ratio of the powers of Y with respectto the two circles is 3 : 1. The circumcircle of BCY passes through the intersection points of ω1 and ω2

(B and C) and thus contains exactly the set of points such that the ratio of their powers with respectto ω1 and ω2 is 3 : 1 (this fact can be verified in a variety of ways). We conclude that K must be the

point on line AB such that KB′

KA = 3. It now suffices to compute AB′. Note AB′ = AC·B′C′

BC by similartriangles. Also an angle chase gives that B′XC ′ are collinear. We compute

BX =BC

2 cosA=

5

2 · 57=

7

2

and thus B′C ′ = 7 so AB′ = 495 and AK = 3

2AB′ = 147

10 .

HMMT February 2018February 10, 2018

Geometry

1. Triangle GRT has GR = 5, RT = 12, and GT = 13. The perpendicular bisector of GT intersects theextension of GR at O. Find TO.

2. Points A,B,C,D are chosen in the plane such that segments AB, BC, CD, DA have lengths 2, 7, 5,12, respectively. Let m be the minimum possible value of the length of segment AC and let M be themaximum possible value of the length of segment AC. What is the ordered pair (m,M)?

3. How many noncongruent triangles are there with one side of length 20, one side of length 17, and one60◦ angle?

4. A paper equilateral triangle of side length 2 on a table has vertices labeled A, B, C. Let M be thepoint on the sheet of paper halfway between A and C. Over time, point M is lifted upwards, foldingthe triangle along segment BM , while A, B, and C remain on the table. This continues until A andC touch. Find the maximum volume of tetrahedron ABCM at any time during this process.

5. In the quadrilateral MARE inscribed in a unit circle ω, AM is a diameter of ω, and E lies on theangle bisector of ∠RAM . Given that triangles RAM and REM have the same area, find the area ofquadrilateral MARE.

6. Let ABC be an equilateral triangle of side length 1. For a real number 0 < x < 0.5, let A1 and A2

be the points on side BC such that A1B = A2C = x, and let TA = 4AA1A2. Construct trianglesTB = 4BB1B2 and TC = 4CC1C2 similarly.

There exist positive rational numbers b, c such that the region of points inside all three trianglesTA, TB , TC is a hexagon with area

8x2 − bx + c

(2− x)(x + 1)·√

3

4.

Find (b, c).

7. Triangle ABC has sidelengths AB = 14, AC = 13, and BC = 15. Point D is chosen in the interiorof AB and point E is selected uniformly at random from AD. Point F is then defined to be theintersection point of the perpendicular to AB at E and the union of segments AC and BC. Supposethat D is chosen such that the expected value of the length of EF is maximized. Find AD.

8. Let ABC be an equilateral triangle with side length 8. Let X be on side AB so that AX = 5 andY be on side AC so that AY = 3. Let Z be on side BC so that AZ,BY,CX are concurrent. LetZX,ZY intersect the circumcircle of AXY again at P,Q respectively. Let XQ and Y P intersect atK. Compute KX ·KQ.

9. Po picks 100 points P1, P2, . . . , P100 on a circle independently and uniformly at random. He then drawsthe line segments connecting P1P2, P2P3, . . . , P100P1. When all of the line segments are drawn, thecircle is divided into a number of regions. Find the expected number of regions that have all sidesbounded by straight lines.

10. Let ABC be a triangle such that AB = 6, BC = 5, AC = 7. Let the tangents to the circumcircle ofABC at B and C meet at X. Let Z be a point on the circumcircle of ABC. Let Y be the foot ofthe perpendicular from X to CZ. Let K be the intersection of the circumcircle of BCY with line AB.Given that Y is on the interior of segment CZ and Y Z = 3CY , compute AK.

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HMMT February 2018, February 10, 2018 — GUTS ROUND

Organization Team Team ID#

1. [4] A square can be divided into four congruent figures as shown:

If each of the congruent figures has area 1, what is the area of the square?

2. [4] John has a 1 liter bottle of pure orange juice. He pours half of the contents of the bottle intoa vat, fills the bottle with water, and mixes thoroughly. He then repeats this process 9 more times.Afterwards, he pours the remaining contents of the bottle into the vat. What fraction of the liquid inthe vat is now water?

3. [4] Allen and Yang want to share the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. How many ways are there tosplit all ten numbers among Allen and Yang so that each person gets at least one number, and eitherAllen’s numbers or Yang’s numbers sum to an even number?

4. [4] Find the sum of the digits of 11 · 101 · 111 · 110011.

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HMMT February 2018, February 10, 2018 — GUTS ROUND

Organization Team Team ID#

5. [6] Randall proposes a new temperature system called Felsius temperature with the following conversionbetween Felsius ◦E, Celsius ◦C, and Fahrenheit ◦F :

◦E =7×◦C

5+ 16 =

7×◦F − 80

9.

For example, 0 ◦C = 16 ◦E. Let x, y, z be real numbers such that x ◦C = x ◦E, y ◦E = y ◦F, z ◦C =z ◦F . Find x + y + z.

6. [6] A bug is on a corner of a cube. A healthy path for the bug is a path along the edges of the cubethat starts and ends where the bug is located, uses no edge multiple times, and uses at most two ofthe edges adjacent to any particular face. Find the number of healthy paths.

7. [6] A triple of integers (a, b, c) satisfies a + bc = 2017 and b + ca = 8. Find all possible values of c.

8. [6] Suppose a real number x > 1 satisfies

log2(log4 x) + log4(log16 x) + log16(log2 x) = 0.

Computelog2(log16 x) + log16(log4 x) + log4(log2 x).

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HMMT February 2018, February 10, 2018 — GUTS ROUND

Organization Team Team ID#

9. [7] In a game, N people are in a room. Each of them simultaneously writes down an integer between0 and 100 inclusive. A person wins the game if their number is exactly two-thirds of the average ofall the numbers written down. There can be multiple winners or no winners in this game. Let m bethe maximum possible number such that it is possible to win the game by writing down m. Find thesmallest possible value of N for which it is possible to win the game by writing down m in a room ofN people.

10. [7] Let a positive integer n be called a cubic square if there exist positive integers a, b with n =gcd(a2, b3). Count the number of cubic squares between 1 and 100 inclusive.

11. [7] FInd the value of60∑k=1

k∑n=1

n2

61− 2n.

12. [7] 4PNR has side lengths PN = 20, NR = 18, and PR = 19. Consider a point A on PN . 4NRAis rotated about R to 4N ′RA′ so that R, N ′, and P lie on the same line and AA′ is perpendicular toPR. Find PA

AN .

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HMMT February 2018, February 10, 2018 — GUTS ROUND

Organization Team Team ID#

13. [9] Suppose 4ABC has lengths AB = 5, BC = 8, and CA = 7, and let ω be the circumcircle of4ABC. Let X be the second intersection of the external angle bisector of ∠B with ω, and let Y bethe foot of the perpendicular from X to BC. Find the length of Y C.

14. [9] Given that x is a positive real, find the maximum possible value of

sin(

tan→1(x

9

)− tan→1

( x

16

)).

15. [9] Michael picks a random subset of the complex numbers {1, ω, ω2, . . . , ω2017} where ω is a primitive2018th root of unity and all subsets are equally likely to be chosen. If the sum of the elements in hissubset is S, what is the expected value of |S|2? (The sum of the elements of the empty set is 0.)

16. [9] Solve for x:xbxbxbxbxcccc = 122.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2018, February 10, 2018 — GUTS ROUND

Organization Team Team ID#

17. [10] Compute the value ofcos 30.5◦ + cos 31.5◦ + ... + cos 44.5◦

sin 30.5◦ + sin 31.5◦ + ... + sin 44.5◦.

18. [10] Compute the number of integers n ∈ {1, 2, . . . , 300} such that n is the product of two distinctprimes, and is also the length of the longest leg of some nondegenerate right triangle with integer sidelengths.

19. [10] Suppose there are 100 cookies arranged in a circle, and 53 of them are chocolate chip, with theremainder being oatmeal. Pearl wants to choose a contiguous subsegment of exactly 67 cookies andwants this subsegment to have exactly k chocolate chip cookies. Find the sum of the k for which Pearlis guaranteed to succeed regardless of how the cookies are arranged.

20. [10] Triangle 4ABC has AB = 21, BC = 55, and CA = 56. There are two points P in the plane of4ABC for which ∠BAP = ∠CAP and ∠BPC = 90◦. Find the distance between them.

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HMMT February 2018, February 10, 2018 — GUTS ROUND

Organization Team Team ID#

21. [12] You are the first lucky player to play in a slightly modified episode of Deal or No Deal! Initially,there are sixteen cases marked 1 through 16. The dollar amounts in the cases are the powers of 2 from21 = 2 to 216 = 65536, in some random order. The game has eight turns. In each turn, you choose acase and claim it, without opening it. Afterwards, a random remaining case is opened and revealed toyou, then removed from the game.

At the end of the game, all eight of your cases are revealed and you win all of the money inside them.

However, the hosts do not realize you have X-ray vision and can see the amount of money inside eachcase! What is the expected amount of money you will make, given that you play optimally?

22. [12] How many graphs are there on 10 vertices labeled 1, 2, . . . , 10 such that there are exactly 23 edgesand no triangles?

23. [12] Kevin starts with the vectors (1, 0) and (0, 1) and at each time step, he replaces one of the vectorswith their sum. Find the cotangent of the minimum possible angle between the vectors after 8 timesteps.

24. [12] Find the largest positive integer n for which there exist n finite sets X1, X2, . . . , Xn with theproperty that for every 1 ≤ a < b < c ≤ n, the equation

|Xa ∪Xb ∪Xc| =l√

abcm

holds.

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HMMT February 2018, February 10, 2018 — GUTS ROUND

Organization Team Team ID#

25. [15] Fran writes the numbers 1, 2, 3, . . . , 20 on a chalkboard. Then she erases all the numbers bymaking a series of moves; in each move, she chooses a number n uniformly at random from the setof all numbers still on the chalkboard, and then erases all of the divisors of n that are still on thechalkboard (including n itself). What is the expected number of moves that Fran must make to eraseall the numbers?

26. [15] Let ABC be a triangle with ∠A = 18◦,∠B = 36◦. Let M be the midpoint of AB, D a point onray CM such that AB = AD; E a point on ray BC such that AB = BE, and F a point on ray ACsuch that AB = AF . Find ∠FDE.

27. [15] There are 2018 frogs in a pool and there is 1 frog on the shore. In each time-step thereafter, onerandom frog moves position. If it was in the pool, it jumps to the shore, and vice versa. Find theexpected number of time-steps before all frogs are in the pool for the first time.

28. [15] Arnold and Kevin are playing a game in which Kevin picks an integer 1 ≤ m ≤ 1001, and Arnoldis trying to guess it. On each turn, Arnold first pays Kevin 1 dollar in order to guess a number k ofArnold’s choice. If m ≥ k, the game ends and he pays Kevin an additional m − k dollars (possiblyzero). Otherwise, Arnold pays Kevin an additional 10 dollars and continues guessing.

Which number should Arnold guess first to ensure that his worst-case payment is minimized?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2018, February 10, 2018 — GUTS ROUND

Organization Team Team ID#

29. [17] Let a, b, c be positive integers. All the roots of each of the quadratics

ax2 + bx + c, ax2 + bx− c, ax2 − bx + c, ax2 − bx− c

are integers. Over all triples (a, b, c), find the triple with the third smallest value of a + b + c.

30. [17] Find the number of unordered pairs {a, b}, where a, b ∈ {0, 1, 2, . . . , 108} such that 109 dividesa3 + b3 − ab.

31. [17] In triangle ABC, AB = 6, BC = 7 and CA = 8. Let D, E, F be the midpoints of sides BC,AC, AB, respectively. Also let OA, OB , OC be the circumcenters of triangles AFD, BDE, and CEF ,respectively. Find the area of triangle OAOBOC .

32. [17] How many 48-tuples of positive integers (a1, a2, . . . , a48) between 0 and 100 inclusive have theproperty that for all 1 ≤ i < j ≤ 48, ai 6∈ {aj , aj + 1}?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2018, February 10, 2018 — GUTS ROUND

Organization Team Team ID#

33. [20] 679 contestants participated in HMMT February 2017. Let N be the number of these contestantswho performed at or above the median score in at least one of the three individual tests. Estimate N .

An estimate of E earns⌊20− jE→N j2

⌋or 0 points, whichever is greater.

34. [20] The integers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 are written on a blackboard. Each day, a teacher chooses oneof the integers uniformly at random and decreases it by 1. Let X be the expected value of the numberof days which elapse before there are no longer positive integers on the board. Estimate X.

An estimate of E earns b20 · 2→jX→Ej/8c points.

35. [20] In a wooden block shaped like a cube, all the vertices and edge midpoints are marked. The cubeis cut along all possible planes that pass through at least four marked points. Let N be the numberof pieces the cube is cut into. Estimate N .

An estimate of E > 0 earns b20 min(N/E,E/N)c points.

36. [20] In the game of Connect Four, there are seven vertical columns which have spaces for six tokens.These form a 7 × 6 grid of spaces. Two players White and Black move alternately. A player takes aturn by picking a column which is not already full and dropping a token of their color into the lowestunoccupied space in that column. The game ends when there are four consecutive tokens of the samecolor in a line, either horizontally, vertically, or diagonally. The player who has four tokens in a row oftheir color wins.

Assume two players play this game randomly. Each player, on their turn, picks a random columnwhich is not full and drops a token of their color into that column. This happens until one player winsor all of the columns are filled. Let P be the probability that all of the columns are filled without anyplayer obtaining four tokens in a row of their color. Estimate P .

An estimate of E > 0 earns b20 min(P/E,E/P )c points.

HMMT February 2018February 10, 2018

Guts Round

1. [4] A square can be divided into four congruent figures as shown:

If each of the congruent figures has area 1, what is the area of the square?

Proposed by: Kevin Sun

Answer: 4

There are four congruent figures with area 1, so the area of the square is 4.

2. [4] John has a 1 liter bottle of pure orange juice. He pours half of the contents of the bottle intoa vat, fills the bottle with water, and mixes thoroughly. He then repeats this process 9 more times.Afterwards, he pours the remaining contents of the bottle into the vat. What fraction of the liquid inthe vat is now water?

Proposed by: Farrell Eldrian Wu

Answer: 56

All the liquid was poured out eventually. 5 liters of water was poured in, and he started with 1 liter oforange juice, so the fraction is 5

1+5 = 56 .

3. [4] Allen and Yang want to share the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. How many ways are there tosplit all ten numbers among Allen and Yang so that each person gets at least one number, and eitherAllen’s numbers or Yang’s numbers sum to an even number?

Proposed by: Kevin Sun

Answer: 1022

Since the sum of all of the numbers is odd, exactly one of Allen’s sum and Yang’s sum must be odd.Therefore any way of splitting the numbers up where each person receives at least one number is valid,so the answer is 210 − 2 = 1022.

4. [4] Find the sum of the digits of 11 · 101 · 111 · 110011.

Proposed by: Evan Chen

Answer: 48

There is no regrouping, so the answer is 2 · 2 · 3 · 4 = 48. The actual product is 13566666531.

5. [6] Randall proposes a new temperature system called Felsius temperature with the following conversionbetween Felsius ◦E, Celsius ◦C, and Fahrenheit ◦F :

◦E =7×◦C

5+ 16 =

7×◦F − 80

9.

For example, 0 ◦C = 16 ◦E. Let x, y, z be real numbers such that x ◦C = x ◦E, y ◦E = y ◦F, z ◦C =z ◦F . Find x+ y + z.

Proposed by: Yuan Yao

Answer: −120

Notice that (5k) ◦C = (7k + 16) ◦E = (9k + 32) ◦F , so Felsius is an exact average of Celsius andFahrenheit at the same temperature. Therefore we conclude that x = y = z, and it is not difficult tocompute that they are all equal to −40.

6. [6] A bug is on a corner of a cube. A healthy path for the bug is a path along the edges of the cubethat starts and ends where the bug is located, uses no edge multiple times, and uses at most two ofthe edges adjacent to any particular face. Find the number of healthy paths.

Proposed by: Dhruv Rohatgi

Answer: 6

There are 6 symmetric ways to choose the first two edges on the path. After these are chosen, allsubsequent edges are determined, until the starting corner is reached once again.

7. [6] A triple of integers (a, b, c) satisfies a+ bc = 2017 and b+ ca = 8. Find all possible values of c.

Proposed by: Ashwin Sah

Answer: −6, 0, 2, 8

Add and subtract the two equations to find

(b+ a)(c+ 1) = 8 + 2017,

(b− a)(c− 1) = 2017− 8.

We see that c is even and then that every integer c with c + 1|2025, c − 1|2009 works. We factor andsolve.

The full solutions are (2017, 8, 0), (−667, 1342, 2), (−59,−346,−6), (−31, 256, 8).

8. [6] Suppose a real number x > 1 satisfies

log2(log4 x) + log4(log16 x) + log16(log2 x) = 0.

Computelog2(log16 x) + log16(log4 x) + log4(log2 x).

Proposed by: Michael Tang

Answer: − 14

Let A and B be these sums, respectively. Then

B −A = log2

(log16 x

log4 x

)+ log4

(log2 x

log16 x

)+ log16

(log4 x

log2 x

)= log2(log16 4) + log4(log2 16) + log16(log4 2)

= log2

(1

2

)+ log4 4 + log16

(1

2

)= (−1) + 1 +

(−1

4

)= −1

4.

Since A = 0, we have the answer B = − 14 .

9. [7] In a game, N people are in a room. Each of them simultaneously writes down an integer between0 and 100 inclusive. A person wins the game if their number is exactly two-thirds of the average ofall the numbers written down. There can be multiple winners or no winners in this game. Let m bethe maximum possible number such that it is possible to win the game by writing down m. Find thesmallest possible value of N for which it is possible to win the game by writing down m in a room ofN people.

Proposed by: Kevin Sun

Answer: 34

Since the average of the numbers is at most 100, the winning number is an integer which is at mosttwo-thirds of 100, or at most 66. This is achieved in a room with 34 people, in which 33 people pick100 and one person picks 66, so the average number is 99.

Furthermore, this cannot happen with less than 34 people. If the winning number is 66 and there areN people, the sum of the numbers must be 99. then we must have that 99N ≤ 66 + 100(N − 1), whichreduces to N ≥ 34.

10. [7] Let a positive integer n be called a cubic square if there exist positive integers a, b with n =gcd(a2, b3). Count the number of cubic squares between 1 and 100 inclusive.

Proposed by: Ashwin Sah

Answer: 13

This is easily equivalent to vp(n) 6≡ 1, 5 (mod 6) for all primes p. We just count: p ≥ 11 =⇒ vp(n) = 1is clear, so we only look at the prime factorizations with primes from {2, 3, 5, 7}. This is easy to compute:we obtain 13.

11. [7] FInd the value of60∑k=1

k∑n=1

n2

61− 2n.

Proposed by: Henrik Boecken

Answer: −18910

Change the order of summation and simplify the inner sum:

60∑k=1

k∑n=1

n2

61− 2n=

60∑n=1

60∑k=n

n2

61− 2n

=

60∑n=1

n2(61− n)

61− 2n

Then, we rearrange the sum to add the terms corresponding to n and 61− n:

60∑n=1

n2(61− n)

61− 2n=

30∑n=1

(n2(61− n)

61− 2n+

(61− n)2(61− (61− n))

61− 2(61− n)

)

=

30∑n=1

n2(61− n)− n(61− n)2

61− 2n

=

30∑n=1

n(61− n)(n− (61− n))

61− 2n

=

30∑n=1

−n(61− n)

=

30∑n=1

n2 − 61n

Finally, using the formulas for the sum of the first k squares and sum of the first k positive integers,we conclude that this last sum is

30(31)(61)

6− 61

30(31)

2= −18910

So, the original sum evaluates to −18910.

12. [7] 4PNR has side lengths PN = 20, NR = 18, and PR = 19. Consider a point A on PN . 4NRAis rotated about R to 4N ′RA′ so that R, N ′, and P lie on the same line and AA′ is perpendicular toPR. Find PA

AN .

Proposed by: Henrik Boecken

Answer: 1918

Denote the intersection of PR and AA′ be D. Note RA′ = RA, so D, being the altitude of an isoscelestriangle, is the midpoint of AA′. Thus,

∠ARD = ∠A′RD = ∠NRA

so RA is the angle bisector of PNR through R. By the angle bisector theorem, we have PAAN = PR

RN = 1918

13. [9] Suppose 4ABC has lengths AB = 5, BC = 8, and CA = 7, and let ω be the circumcircle of4ABC. Let X be the second intersection of the external angle bisector of ∠B with ω, and let Y bethe foot of the perpendicular from X to BC. Find the length of Y C.

Proposed by: Caleb He

Answer: 132

Extend ray−−→AB to a point D, Since BX is an angle bisector, we have ∠XBC = ∠XBD = 180◦ −

∠XBA = ∠XCA, so XC = XA by the inscribed angle theorem. Now, construct a point E on BCso that CE = AB. Since ∠BAX ∼= ∠BCX, we have 4BAX ∼= 4ECX by SAS congruence. Thus,XB = XE, so Y bisects segment BE. Since BE = BC−EC = 8−5 = 3, we have Y C = EC+Y E =5 + 1

2 · 3 = 132 .

(Archimedes Broken Chord Thoerem).

14. [9] Given that x is a positive real, find the maximum possible value of

sin(

tan−1(x

9

)− tan−1

( x16

)).

Proposed by: Yuan Yao

Answer: 725

Consider a right triangle AOC with right angle at O, AO = 16 and CO = x. Moreover, let Bbe on AO such that BO = 9. Then tan−1 x

9 = ∠CBO and tan−1 x16 = ∠CAO, so their difference

is equal to ∠ACB. Note that the locus of all possible points C given the value of ∠ACB is partof a circle that passes through A and B, and if we want to maximize this angle then we need tomake this circle as small as possible. This happens when OC is tangent to the circumcircle of ABC, soOC2 = OA·OB = 144 = 122, thus x = 12, and it suffices to compute sin(α−β) where sinα = cosβ = 4

5and cosα = sinβ = 3

5 . By angle subtraction formula we get sin(α− β) = (45 )2 − ( 3

5 )2 = 725 .

15. [9] Michael picks a random subset of the complex numbers {1, ω, ω2, . . . , ω2017} where ω is a primitive2018th root of unity and all subsets are equally likely to be chosen. If the sum of the elements in hissubset is S, what is the expected value of |S|2? (The sum of the elements of the empty set is 0.)

Proposed by: Nikhil Reddy

Answer: 10092

Consider a and −a of the set of complex numbers. If x is the sum of some subset of the other complexnumbers, then expected magnitude squared of the sum including a and −a is

(x+ a)(x+ a) + xx+ xx+ (x− a)(x− a)

4

xx+aa

2

xx+1

2

By repeating this process on the remaining 2016 elements of the set, we can obtain a factor of 12 every

time. In total, the answer is1009

2

16. [9] Solve for x:xbxbxbxbxcccc = 122.

Proposed by: John Michael Wu

Answer: 12241

This problem can be done without needless casework.

(For negative values of x, the left hand side will be negative, so we only need to consider positive valuesof x.)

The key observation is that for x ∈ [2, 3), 122 is an extremely large value for the expression. Indeed,we observe that:

bxc = 2

bxbxcc ≤ 2(3)− 1 = 5

bxbxbxccc ≤ 3(5)− 1 = 14

bxbxbxbxcccc ≤ 3(14)− 1 = 41

xbxbxbxbxcccc < 3(41) = 123

So the expression can only be as large as 122 if ALL of those equalities hold (the the fourth line equaling40 isn’t good enough), and x = 122

41 . Note that this value is extremely close to 3. We may check thatthis value of x indeed works. Note that the expression is strictly increasing in x, so x = 122

41 is the onlyvalue that works.

17. [10] Compute the value ofcos 30.5◦ + cos 31.5◦ + ...+ cos 44.5◦

sin 30.5◦ + sin 31.5◦ + ...+ sin 44.5◦.

Proposed by: Sujay Kazi

Answer: (√

2− 1)(√

3 +√

2) = 2−√

2−√

3 +√

6

Consider a 360-sided regular polygon with side length 1, rotated so that its sides are at half-degreeinclinations (that is, its sides all have inclinations of 0.5◦, 1.5◦, 2.5◦, and so on. Go to the bottom pointon this polygon and then move clockwise, numbering the sides 1, 2, 3, ..., 360 as you go. Then, take thesection of 15 sides from side 31 to side 45. These sides have inclinations of 30.5◦, 31.5◦, 32.5◦, and so on,up to 44.5◦. Therefore, over this section, the horizontal and vertical displacements are, respectively:

H = cos 30.5◦ + cos 31.5◦ + ...+ cos 44.5◦

V = sin 30.5◦ + sin 31.5◦ + ...+ sin 44.5◦

However, we can also see that, letting R be the circumradius of this polygon:

H = R(sin 45◦ − sin 30◦)

V = R[(1− cos 45◦)− (1− cos 30◦)]

From these, we can easily compute that our desired answer is HV = (

√2−1)(

√3+√

2) = 2−√

2−√

3+√

6.

18. [10] Compute the number of integers n ∈ {1, 2, . . . , 300} such that n is the product of two distinctprimes, and is also the length of the longest leg of some nondegenerate right triangle with integer sidelengths.

Proposed by:

Answer: 13

Let n = p · q for primes p < q. If n is the second largest side of a right triangle there exist integers c, asuch that a < pq and (pq)2 = c2 − a2 = (c− a)(c+ a). Since c− a < c+ a there are three cases for thevalues of c− a, c+ a, and in each case we determine when a < pq.

(a) c− a = 1 and c+ a = p2q2: Then a = p2q2−12 > pq, so there are no solutions.

(b) c− a = p and c+ a = pq2: Then a = pq2−p2 > pq.

(c) c−a = p2 and c+a = q2. Then a = q2−p2

2 which we require to be less than pq. This is equivalentto

q2 − p2

2< pq

q2 < 2pq + p2

2q2 < (q + p)2

√2q < q + p

(√

2− 1)q < p < q

So the problem is equivalent to finding the number of distinct prime pairs (p, q) such that pq < 300

and (√

2− 1)q < p < q. There are 13 such pairs:

{(3, 5), (3, 7), (5, 7), (5, 11), (7, 11), (7, 13), (11, 13), (11, 17), (11, 19), (11, 23), (13, 17), (13, 19), (13, 23)}

and 13 · 23 = 299 which is the biggest such pair.

The most interesting borderline case are (3, 7) : 37 ≈ .42 >

√2−1, which leads to the (20, 21, 29) triangle,

(5, 13) : 513 ≈ .385 <

√2− 1, which leads to the (65, 72, 97) triangle, and (7, 17) : 7

17 ≈ .411 <√

2− 1which leads to the (119, 120, 169) right triangle.

19. [10] Suppose there are 100 cookies arranged in a circle, and 53 of them are chocolate chip, with theremainder being oatmeal. Pearl wants to choose a contiguous subsegment of exactly 67 cookies andwants this subsegment to have exactly k chocolate chip cookies. Find the sum of the k for which Pearlis guaranteed to succeed regardless of how the cookies are arranged.

Proposed by: Alexander Wei

Answer: 71

We claim that the only values of k are 35 and 36.

WLOG assume that the cookies are labelled 0 through 99 around the circle. Consider the follow-ing arrangement: cookies 0 through 17, 34 through 50, and 67 through 84 are chocolate chip, andthe remaining are oatmeal. (The cookies form six alternating blocks around the circle of length18, 16, 17, 16, 18, 15.) Consider the block of 33 cookies that are not chosen. It is not difficult to see thatsince the sum of the lengths of each two adjacent block is always at least 33 and at most 34, this blockof unchosen cookies always contains at least one complete block of cookies of the same type (and noother cookies of this type). So this block contains 17 or 18 or 33− 16 = 17 or 33− 15 = 18 chocolatechip cookies. Therefore, the block of 67 chosen cookies can only have 53 − 17 = 36 or 53 − 18 = 35chocolate chip cookies.

Now we show that 35 and 36 can always be obtained. Consider all possible ways to choose 67 cookies:cookies 0 through 66, 1 through 67, ..., 99 through 65. It is not difficult to see that the number ofchocolate chip cookies in the block changes by at most 1 as we advance from one way to the next.

Moreover, each cookie will be chosen 67 times, so on average there will be 67·53100 = 35.51 chocolate chip

cookies in each block. Since not all blocks are below average and not all blocks are above average, theremust be a point where a block below average transition into a block above average. The difference ofthese two blocks is at most 1, so one must be 35 and one must be 36.

Therefore, the sum of all possible values of k is 35 + 36 = 71.

20. [10] Triangle 4ABC has AB = 21, BC = 55, and CA = 56. There are two points P in the plane of4ABC for which ∠BAP = ∠CAP and ∠BPC = 90◦. Find the distance between them.

Proposed by: Michael Tang

Answer: 52

√409

Let P1 and P2 be the two possible points P , with AP1 < AP2. Both lie on the ∠A-bisector and thecircle γ with diameter BC. Let D be the point where the ∠A-bisector intersects BC, let M be themidpoint of BC, and let X be the foot of the perpendicular from M onto the ∠A-bisector.

Since we know the radius of γ, to compute P1P2 it suffices to compute MX. By the angle bisectortheorem we find BD = 15 and DC = 40, so Stewart’s theorem gives

15 · 40 · 55 + 55 ·AD2 = 212 · 40 + 562 · 15 =⇒ AD = 24.

Then cos∠ADB = −212+152+242

2·15·24 = 12 , so ∠ADB = ∠MDX = 60◦. Since DM = BM − BD =

552 − 15 = 25

2 , we get MX = DM sin∠MDX = 25√3

4 . Hence

P1P2 = 2

vuut(55

2

)2

−

(25√

3

4

)2

=5√

409

2.

21. [12] You are the first lucky player to play in a slightly modified episode of Deal or No Deal! Initially,there are sixteen cases marked 1 through 16. The dollar amounts in the cases are the powers of 2 from21 = 2 to 216 = 65536, in some random order. The game has eight turns. In each turn, you choose acase and claim it, without opening it. Afterwards, a random remaining case is opened and revealed toyou, then removed from the game.

At the end of the game, all eight of your cases are revealed and you win all of the money inside them.

However, the hosts do not realize you have X-ray vision and can see the amount of money inside eachcase! What is the expected amount of money you will make, given that you play optimally?

Proposed by: Kevin Sun

Answer:7 · 218 + 4

15(or

1835012

15)

Firstly, note that it is always optimal for you to take the case with the largest amount of money.To prove this rigorously, consider a strategy where you don’t - then change the first move where youdeviate to taking the maximal case. This can only increase your return.

We calculate the probability f(n, k) that, if there are n cases numbered 1, · · · , n in increasing orderof value, that you will take case k in the course of play. We claim that f(n, k) = k−1

n−1 and prove thisby induction on n/2 (n always even). The base case n = 2 is true because you will always take case2 and leave case 1. Then, for the general n, you will always take case n (so f(n, n) = 1). Afterward,one case at random will be removed. When calculating f(n, k) there is a n−1−k

n−1 probability a casenumbered greater than k is removed, which inductively gives a probability f(n − 2, k). Also, thereis a k−1

n−1 probability a case numbered less than k is removed, which inductively gives a probability

f(n− 2, k − 1). We can compute

f(n, k) =n− 1− kn− 1

f(n− 2, k) +k − 1

n− 1f(n− 2, k − 1)

=n− 1− kn− 1

· k − 1

n− 3+k − 1

n− 1

k − 2

n− 3

=k − 1

(n− 1)(n− 3)(n− 1− k + k − 2)

=k − 1

n− 1

as desired.

Finally, we must find∑16

i=1 f(16, i)2i−1. Using standard procedures, we get

16∑i=1

f(16, i)2i−1 =

15∑i=1

i

152i

=

15∑i=1

i

15(217 − 2i+1)

=1

15(15)(217)− 1

15

(15∑i=1

2i+1

)

= 217 − 1

15(217 − 4)

=14 · 217 + 4

15

22. [12] How many graphs are there on 10 vertices labeled 1, 2, . . . , 10 such that there are exactly 23 edgesand no triangles?

Proposed by: Allen Liu

Answer: 42840

Note that the sum of the degrees of the graph is 23 · 2 = 46, so at least one vertex has degree 5 ormore. We casework on the maximal degree n.

Case 1: n ≥ 7, then none of the n neighbors can have an edge between each other, for(n2

)edges

unusable, and the vertex with maximal degree cannot connect to the 9 − n other vertices. Then wehave

(n2

)+ 9− n >

(102

)− 23 = 22 when n ≥ 7, so there cannot be any graph in this case.

Case 2: n = 6. WLOG suppose that 1 is connected to 2, 3, 4, 5, 6, 7, then none of 2, 3, 4, 5, 6, 7 canconnect to each other.

Case 2.1: There is at least one edge between 8, 9, 10, then each of 2, 3, 4, 5, 6, 7 can connect to at mosttwo of 8, 9, 10, for at most 6 · 2 +

(32

)= 15 additional edges. Along with the 6 original edges, it is not

enough to each 23 edges.

Case 2.2: There are no edges between 8, 9, 10, then this graph is a bipartite graph between 1, 8, 9, 10and 2, 3, 4, 5, 6, 7. There can be at most 4 · 6 = 24 edges in this graph, so exactly one edge is removedfrom this graph. There are

(104

)· 24 = 5040 possible graphs in this case.

Case 3: n = 5. WLOG suppose that 1 is connected to 2, 3, 4, 5, 6, then none of 2, 3, 4, 5, 6 can connectto each other.

Case 3.1: There is at least one edge between 7, 8, 9, 10. Then each of 2, 3, 4, 5, 6 can connect to at mostthree of 7, 8, 9, 10, for 5 · 3 = 15 edges. In this case at least three of 7, 8, 9, 10 must not be connectedto each other, so there can be at most three edges, for 5 + 15 + 3 = 23 edges at most. However, thisrequires the three disconnected vertices of 7, 8, 9, 10 to be connected to all of 2, 3, 4, 5, 6 and the other

vertex of 7, 8, 9, 10, causing them to have degree 6. We can therefore ignore this case. (The case where2, 3, 4, 5, 6 can connect to two or less of 7, 8, 9, 10 can be easily ruled out.)

Case 3.2: There are no edges between 7, 8, 9, 10, then this graph is a bipartite graph between 1, 7, 8, 9, 10and 2, 3, 4, 5, 6. This is a K5,5 with two edges removed, which accounts for

(105

)/2 ·

(252

)= 126 · 300 =

37800 graphs.

It is not difficult to see that Case 2.2 and Case 3.2 are disjoint (by considering max degree), so thereare 5040 + 37800 = 42840 graphs in total.

23. [12] Kevin starts with the vectors (1, 0) and (0, 1) and at each time step, he replaces one of the vectorswith their sum. Find the cotangent of the minimum possible angle between the vectors after 8 timesteps.

Proposed by: Allen Liu

Answer: 987

Say that the vectors Kevin has at some step are (a, b) and (c, d). Notice that regardless of which vectorhe replaces with (a+ c, b+ d), the area of the triangle with vertices (0, 0), (a, b), and (c, d) is preservedwith the new coordinates. We can see this geometrically: the parallelogram with vertices (0, 0), (a, b),(c, d), and (a + c, b + d) can be cut in half by looking at the triangle formed by any 3 of the vertices,which include the original triangle, and both possible triangles that might arise in the next step.

Because the area is preserved, the minimum possible angle then arises when the two vectors, our sides,are as long as possible. This occurs when we alternate which vector is getting replaced for the sum.Given two vectors (a, b) and (c, d), with

√a2 + b2 >

√c2 + d2, we would rather replace (c, d) than

(a, b), and (a + c, b + d) has a larger norm than (a, b). Then at the nth step, Kevin has the vectors(Fn, Fn−1) and (Fn+1, Fn), where F0 = 0 and F1 = 1. The tangent of the angle between them is thetangent of the difference of the angles they make with the x-axis, which is just their slope. We canthen compute the cotangent as∣∣∣∣∣1 + Fn−1

Fn· Fn

Fn+1

Fn

Fn+1− Fn−1

Fn

∣∣∣∣∣ =

∣∣∣∣Fn(Fn+1 + Fn−1)

F 2n − Fn−1Fn+1

∣∣∣∣ .We can show (by induction) that Fn(Fn+1 + Fn−1) = F2n and F 2

n − Fn−1Fn+1 = (−1)n+1. Thus atthe 8th step, the cotangent of the angle is F16 = 987.

24. [12] Find the largest positive integer n for which there exist n finite sets X1, X2, . . . , Xn with theproperty that for every 1 ≤ a < b < c ≤ n, the equation

|Xa ∪Xb ∪Xc| =l√

abcm

holds.

Proposed by: Pakawut Jiradilok

Answer: 4

First, we construct an example for N = 4. Let X1, X2, X3, X4 be pairwise disjoint sets such thatX1 = ∅, |X2| = 1, |X3| = 2, and |X4| = 2. It is straightforward to verify the condition.

We claim that there are no five sets X1, X2, . . . , X5 for which #(Xa ∪Xb ∪Xc) =l√

abcm, for 1 ≤ a <

b < c ≤ 5. Note that showing the non-existence of five such sets implies that there are no n sets withthe desired property for n ≥ 5 as well.

Suppose, for sake of contradiction, that there are such X1, . . . , X5. Then, note that |X1∪X2∪X4| = 3,|X1 ∪X2 ∪X5| = 4, and |X2 ∪X4 ∪X5| = 7. Note that

|X1 ∪X2 ∪X4|+ |X1 ∪X2 ∪X5| = |X2 ∪X4 ∪X5|.

For any sets A,B,C,D, we have the following two inequalities:

|A ∪B ∪ C|+ |A ∪B ∪D| ≥ |A ∪B ∪ C ∪D| ≥ |B ∪ C ∪D|.

For A = X1, B = X2, C = X4, and D = X5 in the situation above, we conclude that the equalitiesmust both hold in both inequalities. The first equality shows that X1 ∪X2 = ∅, and therefore bothX1 and X2 are empty.

Now observe that |X1 ∪X4 ∪X5| = 5 6= 7 = |X2 ∪X4 ∪X5|. This gives a contradiction.

25. [15] Fran writes the numbers 1, 2, 3, . . . , 20 on a chalkboard. Then she erases all the numbers bymaking a series of moves; in each move, she chooses a number n uniformly at random from the setof all numbers still on the chalkboard, and then erases all of the divisors of n that are still on thechalkboard (including n itself). What is the expected number of moves that Fran must make to eraseall the numbers?

Proposed by: Michael Tang

Answer: 13110

For each n, 1 ≤ n ≤ 20, consider the first time that Fran chooses one of the multiples of n. It is inthis move that n is erased, and all the multiples of n at most 20 are equally likely to be chosen forthis move. Hence this is the only move in which Fran could possibly choose n; since there are b20/ncmultiples of n at most 20, this means that the probability that n is ever chosen is 1/b20/nc. Thereforethe expected number of moves is

E =

20∑n=1

1

b20/nc

=1

20+

1

10+

1

6+

1

5+

1

4+

1

3+ 4

(1

2

)+ 10 (1) =

131

10.

(This sum is easier to compute than it may seem, if one notes that 1/20 + 1/5 + 1/4 = 1/2 and1/6 + 1/3 = 1/2).

26. [15] Let ABC be a triangle with ∠A = 18◦,∠B = 36◦. Let M be the midpoint of AB, D a point onray CM such that AB = AD; E a point on ray BC such that AB = BE, and F a point on ray ACsuch that AB = AF . Find ∠FDE.

Proposed by: Faraz Masroor

Answer: 27

Let ∠ABD = ∠ADB = x, and ∠DAB = 180 − 2x. In triangle ACD, by the law of sines, CD =AD

sin∠ACM ·sin 198− 2x, and by the law of sines in triangle BCD, CD = BDsin∠BCM ·sinx+ 36. Combining

the two, we have 2 cosx = BDAD = sin 198−2x

sin x+36 ·sin∠BCMsin∠ACM . But by the ratio lemma, 1 = MB

MA = CBCA

sin∠BCMsin∠ACM ,

meaning that sin∠BCMsin∠ACM = CA

CB = sin 36sin 18 = 2 cos 18. Plugging this in and simplifying, we have 2 cosx =

sin 198−2xsin x+36 · 2 cos 18 = cos 108−2x

cos 54−x · 2 cos 18, so that cos xcos 18 = cos 108−2x

cos 54−x . We see that x = 36◦ is a solutionto this equation, and by carefully making rough sketches of both functions, we can convince ourselvesthat this is the only solution where x is between 0 and 90 degrees. Therefore ∠ABD = ∠ADB = 36,∠DAB = 108. Simple angle chasing yields ∠AEB = 72,∠ECA = 54,∠EAC = 54,∠EAB = 72,making D,A, and E collinear, and so ∠BDE = 36. And because AF = AB = AD, ∠FDB =1/2∠FAB = 9, so ∠FDE = 36− 9 = 27.

27. [15] There are 2018 frogs in a pool and there is 1 frog on the shore. In each time-step thereafter, onerandom frog moves position. If it was in the pool, it jumps to the shore, and vice versa. Find theexpected number of time-steps before all frogs are in the pool for the first time.

Proposed by: Dhruv Rohatgi

Answer: 22018 − 1

Consider the general case of n frogs. Let Ei be the expected time for all frogs to enter the pool wheni frogs are on the shore and n− i frogs are in the pool. We have E0 = 0, En = 1 + En−1, and

Ei =i

nEi−1 +

n− in

Ei+1 + 1

for 0 < i < n. Define fi so that

Ei =fi

(n− 1)(n− 2) · · · (i)+ Ei−1.

Then by plugging this equation into the first equation, we can show that

fi = n(n− 1) · · · (i+ 1) + (n− i)fi+1.

Furthermore, we know that fn = 1. Therefore

f1 =

n∑i=1

n!

i!

(n− 1)!

(n− i)!

= (n− 1)!

n∑i=1

(n

i

)= (n− 1)!(2n − 1).

Therefore

E1 =(n− 1)!(2n − 1)

(n− 1)!+ E0 = 2n − 1.

Plugging in n = 2018 yields E1 = 22018 − 1.

28. [15] Arnold and Kevin are playing a game in which Kevin picks an integer 1 ≤ m ≤ 1001, and Arnoldis trying to guess it. On each turn, Arnold first pays Kevin 1 dollar in order to guess a number k ofArnold’s choice. If m ≥ k, the game ends and he pays Kevin an additional m − k dollars (possiblyzero). Otherwise, Arnold pays Kevin an additional 10 dollars and continues guessing.

Which number should Arnold guess first to ensure that his worst-case payment is minimized?

Proposed by:

Answer: 859

We let f(n) denote the smallest amount we can guarantee to pay at most if Arnold’s first choice is n.For each k < n, if Arnold’s first choice is k + 1, in both worst case scenarios, he could end up payingeither n− k or 11 + f(k). It is then clear that f(n) = mink+1<n max{n− k, 11 + f(k)}.Now clearly f(k) is a non-decreasing function of k, and n − k is a strictly decreasing function of k.Therefore if there exists k such that n− k = 11 + f(k), we have f(n) = n− k = 11 + f(k) with pickingk + 1 as an optimal play (and picking K + 1 also optimal iff K ≥ k and f(K) = f(k).

Now note that f(k) = k for k ≤ 12 (but f(13) = 12 though it’s not relevant to the solution). Leta1 = 11. Now recursively define ai such that ai − ai−1 = 11 + f(ai−1). Thus f(ai) = ai − ai−1 withthe optimal move to pick ai−1 + 1.

a1 = 11

a2 − 11 = 11 + 11 : a2 = 33, f(a2) = 22

a3 − 33 = 11 + f(33) : a3 = 66, f(a3) = 33

It is clear by induction that ai is 11 times the ith triangular number. 1001 is 11× 91 = 14×132 , so the

optimal strategy is to pick 1 more than 11× 12×132 = 858. So the answer is 859.

29. [17] Let a, b, c be positive integers. All the roots of each of the quadratics

ax2 + bx+ c, ax2 + bx− c, ax2 − bx+ c, ax2 − bx− c

are integers. Over all triples (a, b, c), find the triple with the third smallest value of a+ b+ c.

Proposed by: Henrik Boecken

Answer: (1, 10, 24)

The quadratic formula yields that the answers to these four quadratics are ±b±√b2±4ac2a . Given that

all eight of these expressions are integers, we can add or subtract appropriate pairs to get that ba

and√b2±4ac

a are integers. Let b′ = ba and c′ = 4c

a . We can rewrite the expressions to get that b′

and√b′2 ± c′ are positive integers, which also tells us that c′ is a positive integer. Let b′2 + c′ = n2,

b′2 − c′ = m2.

Notice that a + b + c = a(1 + b′ + c′

4 ), so to find the third smallest value of a + b + c, we first findsmall solutions to (b′, c′). To do this, we find triples (m, b′, n) such that m2, b′2, n2 form an arithmeticsequence. Because odd squares are 1 mod 4 and even squares are 0 mod 4, if any of these three termsis odd, then all three terms must be odd. By dividing these terms by the largest possible power of 2then applying the same logic, we can extend our result to conclude that v2(m) = v2(b′) = v2(n). Thus,we only need to look at (m, b′, n) all odd, then multiply them by powers of 2 to get even solutions.

We then plug in b′ = 3, 5, 7, 9, and find that out of these options, only (n, b′,m) = (1, 5, 7) works,giving (b′, c′) = (5, 24), a+ b+ c = 12a. Multiplying by 2 yields that (n, b′,m) = (2, 10, 14) also works,giving (b′, c′) = (10, 96), a + b + c = 35a. For 11 ≤ b ≤ 17, we can check that m = b + 2 fails to give

an integer n. For 11 ≤ b ≤ 17, m 6= b + 2, a + b + c = a(1 + b′ + c′

4 ) ≥ a(1 + 11 + 152−1124 ) = 38a,

the smallest possible value of which is greater than 12a with a = 1, 12a with a = 2, and 35a witha = 1. Thus, it cannot correspond to the solution with the third smallest a + b + c. For b ≥ 19,

a + b + c = a(1 + b′ + c′

4 ) ≥ a(1 + 19 + 212+192

4 ) = 40a, which, similar as before, can’t correspond tothe solution with the third smallest a+ b+ c.

Thus the smallest solution is (a, b′, c′) = (1, 5, 24), (a, b, c) = (1, 5, 6), the second smallest solution is(a, b′, c′) = (2, 5, 24), (a, b, c) = (2, 10, 12), and the third smallest solution that the problem asks for is(a, b′, c′) = (1, 10, 96), (a, b, c) = (1, 10, 24).

30. [17] Find the number of unordered pairs {a, b}, where a, b ∈ {0, 1, 2, . . . , 108} such that 109 dividesa3 + b3 − ab.Proposed by: Henrik Boecken

Answer: 54

We start with the equationa3 + b3 ≡ ab (mod 109).

If either a or b are 0, then we get a3 ≡ 0, implying that both are 0. Thus, {0, 0} is a pair. For the restof the problem, let’s assume that neither a nor b are 0. Multiplying both sides by a−1b−2 yields

(ab−1)2 + a−1b ≡ b−1

from which we make the substitutiona = xy−1

b = y−1

to get the equationy ≡ x2 + x−1.

Plugging this value back into (a, b), we get that all solutions must be of the form

(a, b) = ((x+ x−2)−1, (x2 + x−1)−1),

where 1 ≤ x ≤ 108. It now suffices to find all nonzero unordered pairs {m,n} of the form {x+x−2, x2+x−1}, where 1 ≤ x ≤ 108. There are four values of x for which x+x−2 ≡ x2 +x−1, and of these values,three of them give x+ x−2 ≡ 0. This is because we can re-arrange the equation at hand to get

x4 − x3 + x− 1 ≡ 0,

which factors into(x− 1)(x3 + 1) ≡ 0.

If x = 1, then {m,n} = {2, 2}, and if x3 + 1 ≡ 0 (which has three solutions: 46, 64 and 108), then

{m,n} = {x−1(x3 + 1), x−2(x3 + 1)} = {0, 0}.

Therefore, we keep x = 1 and discard x = 46, 64, 108. Of the remaining 104 values of x, m 6= n, andneither are 0. We have to worry about collisions between distinct values of x. There are two ways acollision can occur: if there exists x 6= y such that

(x+ x−2, x2 + x−1) = (y + y−2, y2 + y−1),

or if there exists x 6= y such that

(x+ x−2, x2 + x−1) = (y2 + y−1, y + y−2).

The first case cannot occur: if x+x−2 ≡ y+ y−2, we have that x2 +x−1 = x(x+x−2) 6= y(x+x−2) =y(y + y−2) = y2 + y−1. As a consequence of this, the second case only occurs if y = x−1. Therefore,the remaining 104 values of x can be partitioned into 52 pairs of (x, x−1), which ends up producing 52distinct unordered pairs {m,n}. Adding this to the x = 1 case and {0, 0}, we get a total of

52 + 1 + 1 = 54

unordered pairs.

31. [17] In triangle ABC, AB = 6, BC = 7 and CA = 8. Let D, E, F be the midpoints of sides BC,AC, AB, respectively. Also let OA, OB , OC be the circumcenters of triangles AFD, BDE, and CEF ,respectively. Find the area of triangle OAOBOC .

Proposed by: Henrik Boecken

Let AB = z,BC = x,CA = y. LetX,Y, Z,O,N be the circumcenter of AEF,BFD,CDE,ABC,DEFrespectively. Note thatN is the nine-point center ofABC, andX,Y, Z are the midpoints ofOA,OB,OCrespectively, and thus XY Z is the image of homothety of ABC with center O and ratio 1

2 , so this tri-angle has side lengths x

2 ,y2 ,

z2 . Since NX perpendicularly bisects EF , which is parallel to BC and thus

Y Z, we see that N is the orthocenter of XY Z. Moreover, O1 lies on Y N and O1X is perpendicularto XY .

To compute the area of O1O2O3, it suffices to compute [NO1O2] + [NO2O3] + [NO3O1]. Note thatO1X is parallel to NO2, and O2Y is parallel to XN , so [NO1O2] = [NXO2] = [NXY ]. Similarly theother two triangles have equal area as [NY Z] and [NZX] respectively, so the desired area is simplythe area of [XY Z], which is

1

4

√(x+ y + z)(x+ y − z)(x− y + z)(−x+ y + z)

4=

√21 · 9 · 5 · 7

16=

21√

15

16.

32. [17] How many 48-tuples of positive integers (a1, a2, . . . , a48) between 0 and 100 inclusive have theproperty that for all 1 ≤ i < j ≤ 48, ai 6∈ {aj , aj + 1}?Proposed by: Mehtaab Sawhney

Answer: 5448

(With Ashwin Sah) The key idea is write the elements of the sequence in increasing order. These setsare in bijection with solutions to d1 + . . . + dk = 48 and a1 + . . . + ak+1 = 53 with di ≥ 1, ai ≥ 1 for

2 ≤ I ≤ k, and a1, ak+1 ≥ 0. Notice that there are(54k

)solutions to the second equation and then

there are 48!d1!···dk!

solutions for each {di} set. Then this gives that the answer is∑1≤k≤48

(54

k

) ∑d1+...+dk=48

48!Qki=1 di!

= 48![x48]∑

1≤k≤48

(ex − 1)k(

54

k

)

= 48![x48]∑

0≤k≤54

(ex − 1)k(

54

k

)= 4854

= 5448.

33. [20] 679 contestants participated in HMMT February 2017. Let N be the number of these contestantswho performed at or above the median score in at least one of the three individual tests. Estimate N .

An estimate of E earns⌊20− |E−N |2

⌋or 0 points, whichever is greater.

Proposed by: Henrik Boecken

Answer: 516

Out of the 679 total contestants at HMMT February 2017, 188 contestants scored at least the medianon all three tests, 159 contestants scored at least the median on two tests, and 169 contestants scoredat least the median on one test, giving a total of 516 contestants

34. [20] The integers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 are written on a blackboard. Each day, a teacher chooses oneof the integers uniformly at random and decreases it by 1. Let X be the expected value of the numberof days which elapse before there are no longer positive integers on the board. Estimate X.

An estimate of E earns b20 · 2−|X−E|/8c points.

Proposed by: Mehtaab Sawhney

Answer: 120.75280458176904

296737619803103602904351875450233789256821656791141519377689665997643364314731919528626892568856634269487507484041490597091266476395426712013/2457397331936622422709580324717167259850640860619179389144469725473122744052437980010340128296796160000000000000000000000000000000000000000

35. [20] In a wooden block shaped like a cube, all the vertices and edge midpoints are marked. The cubeis cut along all possible planes that pass through at least four marked points. Let N be the numberof pieces the cube is cut into. Estimate N .

An estimate of E > 0 earns b20 min(N/E,E/N)c points.

Proposed by: Yuan Yao

Answer: 15600

36. [20] In the game of Connect Four, there are seven vertical columns which have spaces for six tokens.These form a 7 × 6 grid of spaces. Two players White and Black move alternately. A player takes aturn by picking a column which is not already full and dropping a token of their color into the lowestunoccupied space in that column. The game ends when there are four consecutive tokens of the samecolor in a line, either horizontally, vertically, or diagonally. The player who has four tokens in a row oftheir color wins.

Assume two players play this game randomly. Each player, on their turn, picks a random columnwhich is not full and drops a token of their color into that column. This happens until one player winsor all of the columns are filled. Let P be the probability that all of the columns are filled without anyplayer obtaining four tokens in a row of their color. Estimate P .

An estimate of E > 0 earns b20 min(P/E,E/P )c points.

Proposed by: Allen Liu

Answer: 0.0025632817

HMMT November 2018November 10, 2018

General Round

1. What is the largest factor of 130000 that does not contain the digit 0 or 5?

2. Twenty-seven players are randomly split into three teams of nine. Given that Zack is on a differentteam from Mihir and Mihir is on a different team from Andrew, what is the probability that Zack andAndrew are on the same team?

3. A square in the xy-plane has area A, and three of its vertices have x-coordinates 2, 0, and 18 in someorder. Find the sum of all possible values of A.

4. Find the number of eight-digit positive integers that are multiples of 9 and have all distinct digits.

5. Compute the smallest positive integer n for which√100 +

√n +

√100−

√n

is an integer.

6. Call a polygon normal if it can be inscribed in a unit circle. How many non-congruent normal polygonsare there such that the square of each side length is a positive integer?

7. Anders is solving a math problem, and he encounters the expression√

15!. He attempts to simplifythis radical by expressing it as a

√b where a and b are positive integers. The sum of all possible distinct

values of ab can be expressed in the form q · 15! for some rational number q. Find q.

8. Equilateral triangle ABC has circumcircle Ω. Points D and E are chosen on minor arcs AB and ACof Ω respectively such that BC = DE. Given that triangle ABE has area 3 and triangle ACD hasarea 4, find the area of triangle ABC.

9. 20 players are playing in a Super Smash Bros. Melee tournament. They are ranked 1− 20, and playern will always beat player m if n < m. Out of all possible tournaments where each player plays 18distinct other players exactly once, one is chosen uniformly at random. Find the expected number ofpairs of players that win the same number of games.

10. Real numbers x, y, and z are chosen from the interval [−1, 1] independently and uniformly at random.What is the probability that

|x|+ |y|+ |z|+ |x + y + z| = |x + y|+ |y + z|+ |z + x|?

HMMT November 2018November 10, 2018

General Round

1. What is the largest factor of 130000 that does not contain the digit 0 or 5?

Proposed by: Farrell Eldrian Wu

Answer: 26

If the number is a multiple of 5, then its units digit will be either 0 or 5. Hence, the largest suchnumber must have no factors of 5.

We have 130000 = 24 · 54 · 13. Removing every factor of 5, we get that our number must be a factor of24 · 13 = 208.

If our number contains a factor of 13, we cancel the factor of 2 from 208, 104, and 52 until we get 26.Otherwise, the largest number we can have is 24 = 16. We conclude that the answer is 26.

2. Twenty-seven players are randomly split into three teams of nine. Given that Zack is on a differentteam from Mihir and Mihir is on a different team from Andrew, what is the probability that Zack andAndrew are on the same team?

Proposed by: Yuan Yao

Answer: 817

Once we have assigned Zack and Mihir teams, there are 8 spots for more players on Zack’s team and9 for more players on the third team. Andrew is equally likely to occupy any of these spots, so ouranswer is 8

17 .

3. A square in the xy-plane has area A, and three of its vertices have x-coordinates 2, 0, and 18 in someorder. Find the sum of all possible values of A.

Proposed by: Yuan Yao

Answer: 1168

More generally, suppose three vertices of the square lie on lines y = y1, y = y2, y = y3. One of thesevertices must be adjacent to two others. If that vertex is on y = y1 and the other two are on y = y2and y = y3, then we can use the Pythagorean theorem to get that the square of the side length is(y2 − y1)2 + (y3 − y1)2.

For (y1, y2, y3) = (2, 0, 18), the possibilities are 22 + 162, 22 + 182, 162 + 182, so the sum is 2(22 + 162 +182) = 2(4 + 256 + 324) = 1168.

4. Find the number of eight-digit positive integers that are multiples of 9 and have all distinct digits.

Proposed by: James Lin

Answer: 181440

Note that 0 + 1 + · · ·+ 9 = 45. Consider the two unused digits, which must then add up to 9. If it’s 0and 9, there are 8 · 7! ways to finish; otherwise, each of the other four pairs give 7 · 7! ways to finish,since 0 cannot be the first digit. This gives a total of 36 · 7! = 181440.

5. Compute the smallest positive integer n for which√100 +

√n +

√100−

√n

is an integer.

Proposed by: Michael Tang

Answer: 6156

The number√

100 +√n+

√100−

√n is a positive integer if and only if its square is a perfect square.

We have(√100 +

√n +

√100−

√n

)2

= (100 +√n) + (100−

√n) + 2

√(100 +

√n)(100−

√n)

= 200 + 2√

10000− n.

To minimize n, we should maximize the value of this expression, given that it is a perfect square. Forthis expression to be a perfect square,

√10000− n must be an integer. Then 200 + 2

√10000− n is

even, and it is less than 200 + 2√

10000 = 400 = 202. Therefore, the greatest possible perfect squarevalue of 200 + 2

√10000− n is 182 = 324. Solving

200 + 2√

10000− n = 324

for n gives the answer, n = 6156 .

6. Call a polygon normal if it can be inscribed in a unit circle. How many non-congruent normal polygonsare there such that the square of each side length is a positive integer?

Proposed by: Yuan Yao

Answer: 14

The side lengths of the polygon can only be from the set {1,√

2,√

3, 2}, which take up 60◦, 90◦, 120◦, 180◦

of the circle respectively. By working modulo 60 degrees we see that√

2 must be used an even numberof times. We now proceed to casework on the longest side of the polygon.

Case 1: If the longest side has length 2, then the remaining sides must contribute the remaining 180degrees. There are 3 possibilities: (1, 1, 1, 2), (1,

√3, 2), (

√2,√

2, 2).

Case 2: If the longest side has length√

3, then it takes up either 120◦ or 240◦ of the circle. Inthe former case we have 6 possibilities: (1, 1, 1, 1,

√3), (1,

√2,√

2,√

3), (√

2, 1,√

2,√

3), (1, 1,√

3,√

3),(1,√

3, 1,√

3), (√

3,√

3,√

3). In the latter case there is only 1 possibility: (1, 1,√

3).

Case 3: If the longest side has length√

2, then it shows up either twice or four times. In the former casewe have 2 possibilities: (1, 1, 1,

√2,√

2), (1, 1,√

2, 1,√

2). In the latter case there is only 1 possibility:(√

2,√

2,√

2,√

2).

Case 4: If all sides have length 1, then there is 1 possibility: (1, 1, 1, 1, 1, 1).

Adding up all cases, we have 3 + 6 + 1 + 2 + 1 + 1 = 14 polygons.

7. Anders is solving a math problem, and he encounters the expression√

15!. He attempts to simplifythis radical by expressing it as a

√b where a and b are positive integers. The sum of all possible distinct

values of ab can be expressed in the form q · 15! for some rational number q. Find q.

Proposed by: Nikhil Reddy

Answer: 4

Note that 15! = 211 ·36 ·53 ·72 ·111 ·131. The possible a are thus precisely the factors of 25 ·33 ·51 ·71 =30240. Since ab

15! = aba2b = 1

a , we have

q =1

15!

∑a,b:

a√

b=√

15!

ab

=∑

a|30420

ab

15!

=∑

a|30420

1

a

=

(1 +

1

2+

1

4+

1

8+

1

16+

1

32

)(1 +

1

3+

1

9+

1

27

)(1 +

1

5

)(1 +

1

7

)=

(63

32

)(40

27

)(6

5

)(8

7

)= 4.

8. Equilateral triangle ABC has circumcircle Ω. Points D and E are chosen on minor arcs AB and ACof Ω respectively such that BC = DE. Given that triangle ABE has area 3 and triangle ACD hasarea 4, find the area of triangle ABC.

Proposed by: Yuan Yao

Answer: 377

A rotation by 120◦ about the center of the circle will take ABE to BCD, so BCD has area 3. LetAD = x,BD = y, and observe that ∠ADC = ∠CDB = 60◦. By Ptolemy’s Theorem, CD = x + y.We have

4 = [ACD] =1

2AD · CD · sin 60◦ =

√3

4x(x + y)

3 = [BCD] =1

2BD · CD · sin 60◦ =

√3

4y(x + y)

By dividing these equations find x : y = 4 : 3. Let x = 4t, y = 3t. Substitute this into the first equation

to get 1 =√34 · 7t

2. By the Law of Cosines,

AB2 = x2 + xy + y2 = 37t2.

The area of ABC is thenAB2

√3

4=

37

7.

9. 20 players are playing in a Super Smash Bros. Melee tournament. They are ranked 1− 20, and playern will always beat player m if n < m. Out of all possible tournaments where each player plays 18distinct other players exactly once, one is chosen uniformly at random. Find the expected number ofpairs of players that win the same number of games.

Proposed by: Anders Olsen

Answer: 4

Consider instead the complement of the tournament: The 10 possible matches that are not played. Inorder for each player to play 18 games in the tournament, each must appear once in these 10 unplayedmatches. Players n and n+1 will win the same number of games if, in the matching, they are matchedwith each other, or n plays a player a > n + 1 and n + 1 plays a player b < n. (Note no other pairs ofplayers can possibly win the same number of games.) The first happens with probability 1

19 (as there

are 19 players for player n to be paired with), and the second happens with probability (n−1)(20−n−1)19·17 .

By linearity of expectation, the expected number of pairs of players winning the same number of gamesis the sum of these probabilities. We compute

19∑n=1

(1

19+

(n− 1)(20− n− 1)

323

)=

18∑n=0

(1

19+

n(18− n)

323

)= 1 +

193

)323

= 4.

10. Real numbers x, y, and z are chosen from the interval [−1, 1] independently and uniformly at random.What is the probability that

|x|+ |y|+ |z|+ |x + y + z| = |x + y|+ |y + z|+ |z + x|?

Proposed by: Yuan Yao

Answer: 38

We assume that x, y, z are all nonzero, since the other case contributes zero to the total probabilty.

If x, y, z are all positive or all negative then the equation is obviously true. Otherwise, since flippingthe signs of all three variables or permuting them does not change the equality, we assume WLOGthat x, y > 0 and z < 0.

If x + y + z > 0, then the LHS of the original equation becomes x + y − z + x + y = z = 2x + 2y, andthe RHS becomes x + y + |x + z|+ |y + z|, so we need |x + z|+ |y + z| = x + y. But this is impossiblewhen −x−y < z < 0, since the equality is achieved only at the endpoints and all the values in betweenmake the LHS smaller than the RHS. (This can be verified via simple casework.)

If x+ y + z < 0, then x+ z, y + z < 0 as well, so the LHS of the original equation becomes x+ y− z−x− y − z = −2z and the RHS becomes x + y − x− z − y − z = −2z. In this case, the equality holdstrue.

Thus, we seek the volume of all points (x, y, z) ∈ [0, 1]3 that satisfy x + y − z < 0 (we flip the signof z here for convenience). The equation x + y − z = 0 represents a plane through the vertices(1, 0, 1), (0, 0, 0), (0, 1, 1), and the desired region is the triangular pyramid, above the plane inside theunit cube, which has vertices (1, 0, 1), (0, 0, 0), (0, 1, 1), (0, 0, 1). This pyramid has volume 1

6 .

So the total volume of all points in [−1, 1]3 that satisfy the equation is 2 · 1 + 6 · 16 = 3, out of 23 = 8,so the probability is 3

8 .

Note: A more compact way to express the equality condition is that the equation holds true if andonly if xyz(x + y + z) ≥ 0.

HMMT November 2018November 10, 2018

Theme Round

1. Square CASH and regular pentagon MONEY are both inscribed in a circle. Given that they do notshare a vertex, how many intersections do these two polygons have?

2. Consider the addition problem:

C A S H+ M EO S I D E

where each letter represents a base-ten digit, and C,M,O 6= 0. (Distinct letters are allowed to representthe same digit) How many ways are there to assign values to the letters so that the addition problemis true?

3. HOW , BOW , and DAH are equilateral triangles in a plane such that WO = 7 and AH = 2. Giventhat D,A,B are collinear in that order, find the length of BA.

4. I have two cents and Bill has n cents. Bill wants to buy some pencils, which come in two differentpackages. One package of pencils costs 6 cents for 7 pencils, and the other package of pencils costs adime for a dozen pencils (i.e. 10 cents for 12 pencils). Bill notes that he can spend all n of his centson some combination of pencil packages to get P pencils. However, if I give my two cents to Bill, hethen notes that he can instead spend all n + 2 of his cents on some combination of pencil packages toget fewer than P pencils. What is the smallest value of n for which this is possible?

Note: Both times Bill must spend all of his cents on pencil packages, i.e. have zero cents after eitherpurchase.

5. Lil Wayne, the rain god, determines the weather. If Lil Wayne makes it rain on any given day, theprobability that he makes it rain the next day is 75%. If Lil Wayne doesn’t make it rain on one day,the probability that he makes it rain the next day is 25%. He decides not to make it rain today. Findthe smallest positive integer n such that the probability that Lil Wayne makes it rain n days fromtoday is greater than 49.9%.

6. Farmer James invents a new currency, such that for every positive integer n ≤ 6, there exists an n-coinworth n! cents. Furthermore, he has exactly n copies of each n-coin. An integer k is said to be niceif Farmer James can make k cents using at least one copy of each type of coin. How many positiveintegers less than 2018 are nice?

7. Ben “One Hunna Dolla” Franklin is flying a kite KITE such that IE is the perpendicular bisector ofKT . Let IE meet KT at R. The midpoints of KI, IT, TE,EK are A,N,M,D, respectively. Giventhat [MAKE] = 18, IT = 10, [RAIN ] = 4, find [DIME].

Note: [X] denotes the area of the figure X.

8. Crisp All, a basketball player, is dropping dimes and nickels on a number line. Crisp drops a dime onevery positive multiple of 10, and a nickel on every multiple of 5 that is not a multiple of 10. Crispthen starts at 0. Every second, he has a 2

3 chance of jumping from his current location x to x+ 3, anda 1

3 chance of jumping from his current location x to x + 7. When Crisp jumps on either a dime or anickel, he stops jumping. What is the probability that Crisp stops on a dime?

9. Circle ω1 of radius 1 and circle ω2 of radius 2 are concentric. Godzilla inscribes square CASH in ω1

and regular pentagon MONEY in ω2. It then writes down all 20 (not necessarily distinct) distancesbetween a vertex of CASH and a vertex of MONEY and multiplies them all together. What is themaximum possible value of his result?

10. One million bucks (i.e. one million male deer) are in different cells of a 1000× 1000 grid. The left andright edges of the grid are then glued together, and the top and bottom edges of the grid are gluedtogether, so that the grid forms a doughnut-shaped torus. Furthermore, some of the bucks are honestbucks, who always tell the truth, and the remaining bucks are dishonest bucks, who never tell the truth.Each of the million bucks claims that “at most one of my neighboring bucks is an honest buck.” Apair of neighboring bucks is said to be buckaroo if exactly one of them is an honest buck. What is theminimum possible number of buckaroo pairs in the grid?

Note: Two bucks are considered to be neighboring if their cells (x1, y1) and (x2, y2) satisfy either:x1 = x2 and y1 − y2 ≡ ±1 (mod 1000), or x1 − x2 ≡ ±1 (mod 1000) and y1 = y2.

HMMT November 2018November 10, 2018

Theme Round

1. Square CASH and regular pentagon MONEY are both inscribed in a circle. Given that they do notshare a vertex, how many intersections do these two polygons have?

Proposed by: Yuan Yao

Answer: 8

Pentagon MONEY divides the circumference into 5 circular arcs, and each vertex of CASH liesin a different arc. Then each side of CASH will intersect two sides of MONEY , for a total of 8intersections.

2. Consider the addition problem:

C A S H+ M EO S I D E

where each letter represents a base-ten digit, and C,M,O 6= 0. (Distinct letters are allowed to representthe same digit) How many ways are there to assign values to the letters so that the addition problemis true?

Proposed by: James Lin

Answer: 0

Clearly, CASH and ME cannot add up to 11000 or more, so O = 1 and S = 0. By examining theunits digit, we find that H = 0. Then CASH +ME < 9900 + 99 < 10000, so there are no solutions.

3. HOW , BOW , and DAH are equilateral triangles in a plane such that WO = 7 and AH = 2. Giventhat D,A,B are collinear in that order, find the length of BA.

Proposed by: James Lin

Answer: 11

Note that H 6= B since otherwise DAB is an equilateral triangle. Let M be the midpoint of DA, soHB = 7

√3 and HM =

√3, and ∠HMB = 90◦. By the Pythagorean theorem,

BM =

√(7√

3)2 − (√

3)2 = 12.

Then BA = BM −AM = 11.

4. I have two cents and Bill has n cents. Bill wants to buy some pencils, which come in two differentpackages. One package of pencils costs 6 cents for 7 pencils, and the other package of pencils costs adime for a dozen pencils (i.e. 10 cents for 12 pencils). Bill notes that he can spend all n of his centson some combination of pencil packages to get P pencils. However, if I give my two cents to Bill, hethen notes that he can instead spend all n+ 2 of his cents on some combination of pencil packages toget fewer than P pencils. What is the smallest value of n for which this is possible?

Note: Both times Bill must spend all of his cents on pencil packages, i.e. have zero cents after eitherpurchase.

Proposed by: James Lin

Answer: 100

Suppose that Bill buys a packages of 7 and b packages of 12 in the first scenario and c packages of 7and d packages of 12 in the second scenario. Then we have the following system:

6a+ 10b = n

6c+ 10d = n+ 2

7a+ 12b > 7c+ 12d.

Since the packages of 12 give more pencils per cent, we must have b > d. Subtract the first twoequations and divide by 2 to get

3(c− a)− 5(b− d) = 1.

Note that the last inequality is 12(b − d) > 7(c − a). The minimal solutions to the equation withb− d > 0 are

(c− a, b− d) = (2, 1), (7, 4), (12, 7), (17, 10).

(17, 10) is the first pair for which 12(b− d) > 7(c− a). Hence b ≥ 10 so n ≥ 100. We can easily verifythat (a, b, c, d, n) = (0, 10, 17, 0, 100) satisfies the system of equations.

5. Lil Wayne, the rain god, determines the weather. If Lil Wayne makes it rain on any given day, theprobability that he makes it rain the next day is 75%. If Lil Wayne doesn’t make it rain on one day,the probability that he makes it rain the next day is 25%. He decides not to make it rain today. Findthe smallest positive integer n such that the probability that Lil Wayne makes it rain n days fromtoday is greater than 49.9%.

Proposed by: Anders Olsen

Answer: 9

Let pn denote the probability that Lil Wayne makes it rain n days from today. We have p0 = 0 and

pn+1 =3

4pn +

1

4(1− pn) =

1

4+

1

2pn.

This can be written as

pn+1 −1

2=

1

2

(pn −

1

2

)and we can check that the solution of this recurrence is

pn =1

2− 1

2n+1

We want 12n+1 <

11000 , which first occurs when n = 9.

6. Farmer James invents a new currency, such that for every positive integer n ≤ 6, there exists an n-coinworth n! cents. Furthermore, he has exactly n copies of each n-coin. An integer k is said to be niceif Farmer James can make k cents using at least one copy of each type of coin. How many positiveintegers less than 2018 are nice?

Proposed by: Nikhil Reddy

Answer: 210

We use the factorial base, where we denote

(dn . . . d1)∗ = dn × n! + · · ·+ d1 × 1!

The representation of 201810 is 244002∗ and the representation of 72010 is 100000∗. The largest nicenumber less than 244002∗ is 243321∗. Notice that for the digit di of a nice number, we can vary itsvalue from 1 to i, while for a generic number in the factorial base, di→1 can vary from 0 to i − 1.Hence we can map nice numbers to all numbers by truncating the last digit and reducing each previousdigit by 1, and likewise reverse the procedure by increasing all digits by 1 and adding 1 at the end.Furthermore, this procedure preserves the ordering of numbers. Applying this procedure to 243321∗gives 13221∗. We count from 0∗ to 13221∗ (since the first nice number is 1∗), to get an answer of

13221∗ + 1 = 210.

7. Ben “One Hunna Dolla” Franklin is flying a kite KITE such that IE is the perpendicular bisector ofKT . Let IE meet KT at R. The midpoints of KI, IT, TE,EK are A,N,M,D, respectively. Giventhat [MAKE] = 18, IT = 10, [RAIN ] = 4, find [DIME].

Note: [X] denotes the area of the figure X.

Proposed by: Michael Ren

Answer: 16

Let [KIR] = [RIT ] = a and [KER] = [TER] = b. We will relate all areas to a and b. First,

[RAIN ] = [RAI] + [INR] =1

2a+

1

2a = a.

Next, we break up [MAKE] = [MAD] + [AKD] + [DEM ]. We have

[MAD] =AD ·DM

2=

1

2· IE

2· KT

2=

[KITE]

4=a+ b

2

[AKD] =[KIE]

4=a+ b

4

[DEM ] =[KTE]

4=b

2.

After adding these we get [MAKE] = 3a+5b4 . We want to find

[DIME] = 2[IME] = [ITE] = a+ b =4

5

(3a+ 5b

4

)+

2

5a =

4

5· 18 +

2

5· 4 = 16.

8. Crisp All, a basketball player, is dropping dimes and nickels on a number line. Crisp drops a dime onevery positive multiple of 10, and a nickel on every multiple of 5 that is not a multiple of 10. Crispthen starts at 0. Every second, he has a 2

3 chance of jumping from his current location x to x+ 3, anda 1

3 chance of jumping from his current location x to x+ 7. When Crisp jumps on either a dime or anickel, he stops jumping. What is the probability that Crisp stops on a dime?

Proposed by: James Lin

Answer: 2031

Let “a 3” mean a move in which Crisp moves from x to x+ 3, and “a 7” mean a move in which Crispmoves from x to x+ 7. Note that Crisp stops precisely the first time his number of 3’s and number of7’s differs by a multiple of 5, and that he’ll stop on a dime if they differ by 0, and stop on a nickel ifthey differ by 5. This fact will be used without justification.

We split into two cases:

(a) Crisp begins with a 3. Rather than consider the integer Crisp is on, we’ll count the difference, n,between his number of 3’s and his number of 7’s. Each 3 increases n by 1, and each 7 decreases nby 1. Currently, n is 1. The probability of stopping on a dime, then, is the probability n reaches 0before n reaches 5, where n starts at 1. Let ai be the probability n reaches 0 first, given a currentposition of i, for i = 1, 2, 3, 4. We desire a1. We have the system of linear equations

a1 =2

3a2 +

1

3· 1

a2 =2

3a3 +

1

3a1

a3 =2

3a4 +

1

3a2

a4 =2

3· 0 +

1

3a3

From which we determine that a1 = 1531 .

(b) Crisp begins with a 7. Now, let m be the difference between his number of 7’s and his number of3’s. Let bi denote his probability of stopping on a dime, given his current position of m = i. Wedesire b1. We have the system of linear equations

b1 =1

3b2 +

2

3· 1

b2 =1

3b3 +

2

3b1

b3 =1

3b4 +

2

3b2

b4 =1

3· 0 +

2

3b3

From which we determine that b1 = 3031 .

We conclude that the answer is 23a1 + 1

3b1 = 23 ·

1531 + 1

3 ·3031 = 20

31 .

9. Circle ω1 of radius 1 and circle ω2 of radius 2 are concentric. Godzilla inscribes square CASH in ω1

and regular pentagon MONEY in ω2. It then writes down all 20 (not necessarily distinct) distancesbetween a vertex of CASH and a vertex of MONEY and multiplies them all together. What is themaximum possible value of his result?

Proposed by: Michael Ren

Answer: 1048577 or 220 + 1

We represent the vertices with complex numbers. Place the vertices of CASH at 1, i,−1,−i and thevertices of MONEY at 2α, 2αω, 2αω2, 2αω2, 2αω3, 2αω4 with |α| = 1 and ω = e

2πi5 . We have that the

product of distances from a point z to the vertices of CASH is |(z− 1)(z− i)(z+ 1)(z+ i)| = |z4− 1|,so we want to maximize |(16α4 − 1)(16α4ω4 − 1)(16α4ω3 − 1)(16α4ω2 − 1)(16α4ω − 1)|, which justcomes out to be |220α20 − 1|. By the triangle inequality, this is at most 220 + 1, and it is clear thatsome α makes equality hold.

10. One million bucks (i.e. one million male deer) are in different cells of a 1000× 1000 grid. The left andright edges of the grid are then glued together, and the top and bottom edges of the grid are gluedtogether, so that the grid forms a doughnut-shaped torus. Furthermore, some of the bucks are honestbucks, who always tell the truth, and the remaining bucks are dishonest bucks, who never tell the truth.Each of the million bucks claims that “at most one of my neighboring bucks is an honest buck.” Apair of neighboring bucks is said to be buckaroo if exactly one of them is an honest buck. What is theminimum possible number of buckaroo pairs in the grid?

Note: Two bucks are considered to be neighboring if their cells (x1, y1) and (x2, y2) satisfy either:x1 = x2 and y1 − y2 ≡ ±1 (mod 1000), or x1 − x2 ≡ ±1 (mod 1000) and y1 = y2.

Proposed by: James Lin

Answer: 1200000

Note that each honest buck has at most one honest neighbor, and each dishonest buck has at leasttwo honest neighbors. The connected components of honest bucks are singles and pairs. Then if thereare K honest bucks and B buckaroo pairs, we get B ≥ 3K. From the dishonest buck condition we getB ≥ 2(1000000 −K), so we conclude that B ≥ 1200000. To find equality, partition the grid into fivedifferent parts with side

√5, and put honest bucks on every cell in two of the parts.

HMMT November 2018November 10, 2018

Team Round

1. [15] Four standard six-sided dice are rolled. Find the probability that, for each pair of dice, the productof the two numbers rolled on those dice is a multiple of 4.

2. [20] Alice starts with the number 0. She can apply 100 operations on her number. In each operation,she can either add 1 to her number, or square her number. After applying all operations, her score isthe minimum distance from her number to any perfect square. What is the maximum score she canattain?

3. [25] For how many positive integers n ≤ 100 is it true that 10n has exactly three times as many positivedivisors as n has?

4. [30] Let a and b be real numbers greater than 1 such that ab = 100. The maximum possible value of

a(log10 b)2 can be written in the form 10x for some real number x. Find x.

5. [35] Find the sum of all positive integers n such that 1 + 2 + · · ·+ n divides

15[(n + 1)2 + (n + 2)2 + · · ·+ (2n)2].

6. [45] Triangle 4PQR, with PQ = PR = 5 and QR = 6, is inscribed in circle ω. Compute the radiusof the circle with center on QR which is tangent to both ω and PQ.

7. [50] A 5 × 5 grid of squares is filled with integers. Call a rectangle corner-odd if its sides are gridlines and the sum of the integers in its four corners is an odd number. What is the maximum possiblenumber of corner-odd rectangles within the grid?

Note: A rectangle must have four distinct corners to be considered corner-odd ; i.e. no 1× k rectanglecan be corner-odd for any positive integer k.

8. [55] Tessa has a unit cube, on which each vertex is labeled by a distinct integer between 1 and 8inclusive. She also has a deck of 8 cards, 4 of which are black and 4 of which are white. At each stepshe draws a card from the deck, and

• if the card is black, she simultaneously replaces the number on each vertex by the sum of thethree numbers on vertices that are distance 1 away from this vertex;

• if the card is white, she simultaneously replaces the number on each vertex by the sum of thethree numbers on vertices that are distance

√2 away from this vertex.

When Tessa finishes drawing all cards of the deck, what is the maximum possible value of a numberthat is on the cube?

9. [60] Let A,B,C be points in that order along a line, such that AB = 20 and BC = 18. Let ω be a circleof nonzero radius centered at B, and let `1 and `2 be tangents to ω through A and C, respectively.Let K be the intersection of `1 and `2. Let X lie on segment KA and Y lie on segment KC such thatXY ‖ BC and XY is tangent to ω. What is the largest possible integer length for XY ?

10. [65] David and Evan are playing a game. Evan thinks of a positive integer N between 1 and 59,inclusive, and David tries to guess it. Each time David makes a guess, Evan will tell him whether theguess is greater than, equal to, or less than N . David wants to devise a strategy that will guaranteethat he knows N in five guesses. In David’s strategy, each guess will be determined only by Evan’sresponses to any previous guesses (the first guess will always be the same), and David will only guessa number which satisfies each of Evan’s responses. How many such strategies are there?

Note: David need not guess N within his five guesses; he just needs to know what N is after fiveguesses.

HMMT November 2018November 10, 2018

Team Round

1. [15] Four standard six-sided dice are rolled. Find the probability that, for each pair of dice, the productof the two numbers rolled on those dice is a multiple of 4.

Proposed by: Michael Tang

Answer: 31432

If any two of the dice show an odd number, then this is impossible, so at most one of the dice can showan odd number. We take two cases:

Case 1: If exactly one of the dice shows an odd number, then all three other dice must show a multiple

of 4, which can only be the number 4. The probability that this occurs is 4 · 12 ·16

)3= 1

108 .

Case 2: If all of the dice show even numbers, then the condition is satisfied. The probability that this

occurs is12

)4= 1

16 .

The total probability is 1108 + 1

16 = 31432 .

2. [20] Alice starts with the number 0. She can apply 100 operations on her number. In each operation,she can either add 1 to her number, or square her number. After applying all operations, her score isthe minimum distance from her number to any perfect square. What is the maximum score she canattain?

Proposed by: Dhruv Rohatgi

Answer: 94

Note that after applying the squaring operation, Alice’s number will be a perfect square, so she canmaximize her score by having a large number of adding operations at the end. However, her scoresneeds to be large enough that the many additions to not bring her close to a larger square. Hencethe strategy is as follows: 2 additions to get to 2, 4 consecutive squares to get to 65536, and 94 moreadditions for a score of 94.

3. [25] For how many positive integers n ≤ 100 is it true that 10n has exactly three times as many positivedivisors as n has?

Proposed by: James Lin

Answer: 28

Let n = 2a5bc, where 2, 5 - c. Then, the ratio of the number of divisors of 10n to the number of divisorsof n is a+2

a+1b+2b+1 = 3. Solving for b, we find that b = 1→a

2a+1 . This forces (a, b) = (0, 1), (1, 0). Therefore,the answers are of the form 2k and 5k whenever gcd(k, 10) = 1. There are 50 positive numbers of theform 2k and 20 positive numbers of the form 5k less than or equal to 100. Of those 70 numbers, only12 ·

45 have k relatively prime to 10, so the answer is 70 · 12 ·

45 = 28 .

4. [30] Let a and b be real numbers greater than 1 such that ab = 100. The maximum possible value of

a(log10 b)2 can be written in the form 10x for some real number x. Find x.

Proposed by: James Lin

Answer: 3227

Let p = log10 a, q = log10 b. Since a, b > 1, p and q are positive. The condition ab = 100 translates top+ q = 2. We wish to maximize

x = log10 a(log10 b)2 = (log10 a)(log10 b)

2 = pq2.

By AM-GM,27

4pq2 ≤

(p+

q

2+q

2

)3= 8.

Hence pq2 ≤ 3227 with equality when p = 2

3 , q = 43 .

5. [35] Find the sum of all positive integers n such that 1 + 2 + · · ·+ n divides

15[(n+ 1)2 + (n+ 2)2 + · · ·+ (2n)2].

Proposed by: Michael Ren

Answer: 64

We can compute that 1 + 2 + · · ·+n = n(n+1)2 and (n+ 1)2 + (n+ 2)2 + · · ·+ (2n)2 = 2n(2n+1)(4n+1)

6 −n(n+1)(2n+1)

6 = n(2n+1)(7n+1)6 , so we need 15(2n+1)(7n+1)

3(n+1) = 5(2n+1)(7n+1)n+1 to be an integer. The remain-

der when (2n+1)(7n+1) is divided by (n+1) is 6, so after long division we need 30n+1 to be an integer.

The solutions are one less than a divisor of 30 so the answer is

1 + 2 + 4 + 5 + 9 + 14 + 29 = 64.

6. [45] Triangle 4PQR, with PQ = PR = 5 and QR = 6, is inscribed in circle ω. Compute the radiusof the circle with center on QR which is tangent to both ω and PQ.

Proposed by: Michael Tang

Answer: 209

Solution 1: Denote the second circle by γ. Let T and r be the center and radius of γ, respectively, andlet X and H be the tangency points of γ with ω and PQ, respectively. Let O be the center of ω, andlet M be the midpoint of QR. Note that QM = MR = 1

2QR = 3, so 4PMQ and 4PMR are 3−4−5triangles. Since 4QHT ∼ 4QMP and HT = r, we get QT = 5

4r. Then TM = QM −QT = 3− 54r.

By the extended law of sines, the circumradius of 4PQR is OP = PR2 sin∠PQR = 5

2(4/5) = 258 , so

OM = MP − OP = 4 − 258 = 7

8 . Also, we have OT = OX − XT = 258 − r. Therefore, by the

Pythagorean theorem, 3− 5

4r)2

+78

)2=258 − r

)2.

This simplifies to 916r

2 − 54r = 0, so r = 5

4 ·169 = 20

9 .

P

Q RT

O

M

H

X

Solution 2: Following the notation of the previous solution, we compute the power of T with respect toω in two ways. One of these is −QT ·TR = − 5

4r6− 5

4r). The other way is −(OX−OT )(OX+OT ) =

−r( 254 − r). Equating these two yields r = 20

9 .

7. [50] A 5 × 5 grid of squares is filled with integers. Call a rectangle corner-odd if its sides are gridlines and the sum of the integers in its four corners is an odd number. What is the maximum possiblenumber of corner-odd rectangles within the grid?

Note: A rectangle must have four distinct corners to be considered corner-odd ; i.e. no 1× k rectanglecan be corner-odd for any positive integer k.

Proposed by: Henrik Boecken

Answer: 60

Consider any two rows and the five numbers obtained by adding the two numbers which share a givencolumn. Suppose a of these are odd and b of these are even. The number of corner-odd rectangles withtheir sides contained in these two rows is ab. Since a+ b = 5, we have ab ≤ 6. Therefore every pair ofrows contains at most 6 corner-odd rectangles.

There are52

)= 10 pairs of rows, so there are at most 60 corner-odd rectangles. Equality holds when

we place 1 along one diagonal and 0 everywhere else.

8. [55] Tessa has a unit cube, on which each vertex is labeled by a distinct integer between 1 and 8inclusive. She also has a deck of 8 cards, 4 of which are black and 4 of which are white. At each stepshe draws a card from the deck, and

• if the card is black, she simultaneously replaces the number on each vertex by the sum of thethree numbers on vertices that are distance 1 away from this vertex;

• if the card is white, she simultaneously replaces the number on each vertex by the sum of thethree numbers on vertices that are distance

√2 away from this vertex.

When Tessa finishes drawing all cards of the deck, what is the maximum possible value of a numberthat is on the cube?

Proposed by: Yuan Yao

Answer: 42648

The order of the deck does not matter as black cards and white cards commute, therefore we can assumethat the cards are alternating black and white, and only worry about the arrangement of the numbers.After each pair of black and white cards, each number is replaced by the sum of two times the edgeneighbors and three times the diagonally opposite number. We can compute that after four pairs ofoperations, the number at vertex V will be 1641v+1640(d1 +d2 +d3), where v is the number originallyat v and d1, d2, d3 are the numbers at diagonally adjacent vertices. Set v = 8 and d1, d2, d3 = 5, 6, 7 inany order to obtain the maximum number 42648.

9. [60] Let A,B,C be points in that order along a line, such that AB = 20 and BC = 18. Let ω be a circleof nonzero radius centered at B, and let `1 and `2 be tangents to ω through A and C, respectively.Let K be the intersection of `1 and `2. Let X lie on segment KA and Y lie on segment KC such thatXY ‖ BC and XY is tangent to ω. What is the largest possible integer length for XY ?

Proposed by: James Lin

Answer: 35

Note that B is the K-excenter of KXY , so XB is the angle bisector of ∠AKY . As AB and XY areparallel, ∠XAB + 2∠AXB = 180◦, so ∠XBA = 180◦ − ∠AXB − ∠XAB. This means that AXB isisosceles with AX = AB = 20. Similarly, Y C = BC = 18. As KXY is similar to KAC, we have thatKXKY = XA

Y C = 2018 . Let KA = 20x,KC = 18x, so the Triangle Inequality applied to triangle KAC gives

KA < KC +AC =⇒ 20x < 18x+ 38 =⇒ x < 19. Then, XY = AC · KXKA = 38 · x→1

x = 38− 38x < 36,

so the maximum possible integer length of XY is 35. The optimal configuration is achieved when theradius of ω becomes arbitrarily small and `1 and `2 are on opposite sides of AC.

10. [65] David and Evan are playing a game. Evan thinks of a positive integer N between 1 and 59,inclusive, and David tries to guess it. Each time David makes a guess, Evan will tell him whether theguess is greater than, equal to, or less than N . David wants to devise a strategy that will guarantee

that he knows N in five guesses. In David’s strategy, each guess will be determined only by Evan’sresponses to any previous guesses (the first guess will always be the same), and David will only guessa number which satisfies each of Evan’s responses. How many such strategies are there?

Note: David need not guess N within his five guesses; he just needs to know what N is after fiveguesses.

Proposed by: Anders Olsen

Answer: 36440

We can represent each strategy as a binary tree labeled with the integers from 1 to 59, where Davidstarts at the root and moves to the right child if he is too low and to the left child if he is too high.Our tree must have at most 6 layers as David must guess at most 5 times. Once David has been toldthat he guessed correctly or if the node he is at has no children, he will be sure of Evan’s number.Consider the unique strategy for David when 59 is replaced with 63. This is a tree where every node inthe first 5 layers has two children, and it can only be labeled in one way such that the strategy satisfiesthe given conditions. In order to get a valid strategy for 59, we only need to delete 4 of the verticesfrom this tree and relabel the vertices as necessary. Conversely, every valid strategy tree for 59 can becompleted to the strategy tree for 63. If we delete a parent we must also delete its children. Thus,we can just count the number of ways to delete four nodes from the tree for 63 so that if a parent isdeleted then so are its children. We cannot delete a node in the fourth layer, as that means we deleteat least 1 + 2 + 4 = 7 nodes. If we delete a node in the fifth layer, then we delete its two children aswell, so in total we delete three nodes. There are now two cases: if we delete all four nodes from thesixth layer or if we delete one node in the fifth layer along with its children and another node in thesixth layer. There are

324

)ways to pick 4 from the sixth layer and 16 · 30 to pick one from the fifth

layer along with its children and another node that is from the sixth layer, for a total of 36440.

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HMMT November 2018, November 10, 2018 — GUTS ROUND

Organization Team Team ID#

1. [5] A positive integer is called primer if it has a prime number of distinct prime factors. Find thesmallest primer number.

2. [5] Pascal has a triangle. In the nth row, there are n + 1 numbers an,0, an,1, an,2, . . . , an,n wherean,0 = an,n = 1. For all 1 ≤ k ≤ n− 1, an,k = an−1,k − an−1,k−1. What is the sum of all numbers inthe 2018th row?

3. [5] An n×m maze is an n×m grid in which each cell is one of two things: a wall, or a blank. A mazeis solvable if there exists a sequence of adjacent blank cells from the top left cell to the bottom rightcell going through no walls. (In particular, the top left and bottom right cells must both be blank.)Determine the number of solvable 2× 2 mazes.

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HMMT November 2018, November 10, 2018 — GUTS ROUND

Organization Team Team ID#

4. [6] Let a, b, c, n be positive real numbers such that a+ba = 3, b+c

b = 4, and c+ac = n. Find n.

5. [6] Jerry has ten distinguishable coins, each of which currently has heads facing up. He chooses onecoin and flips it over, so it now has tails facing up. Then he picks another coin (possibly the same oneas before) and flips it over. How many configurations of heads and tails are possible after these twoflips?

6. [6] An equilateral hexagon with side length 1 has interior angles 90◦, 120◦, 150◦, 90◦, 120◦, 150◦ in thatorder. Find its area.

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HMMT November 2018, November 10, 2018 — GUTS ROUND

Organization Team Team ID#

7. [7] At Easter-Egg Academy, each student has two eyes, each of which can be eggshell, cream, orcornsilk. It is known that 30% of the students have at least one eggshell eye, 40% of the students haveat least one cream eye, and 50% of the students have at least one cornsilk eye. What percentage ofthe students at Easter-Egg Academy have two eyes of the same color?

8. [7] Pentagon JAMES is such that AM = SJ and the internal angles satisfy ∠J = ∠A = ∠E = 90◦,and ∠M = ∠S. Given that there exists a diagonal of JAMES that bisects its area, find the ratio ofthe shortest side of JAMES to the longest side of JAMES.

9. [7] Farmer James has some strange animals. His hens have 2 heads and 8 legs, his peacocks have 3heads and 9 legs, and his zombie hens have 6 heads and 12 legs. Farmer James counts 800 heads and2018 legs on his farm. What is the number of animals that Farmer James has on his farm?

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HMMT November 2018, November 10, 2018 — GUTS ROUND

Organization Team Team ID#

10. [8] Abbot writes the letter A on the board. Every minute, he replaces every occurrence of A with ABand every occurrence of B with BA, hence creating a string that is twice as long. After 10 minutes,there are 210 = 1024 letters on the board. How many adjacent pairs are the same letter?

11. [8] Let 4ABC be an acute triangle, with M being the midpoint of BC, such that AM = BC. LetD and E be the intersection of the internal angle bisectors of ∠AMB and ∠AMC with AB and AC,respectively. Find the ratio of the area of 4DME to the area of 4ABC.

12. [8] Consider an unusual biased coin, with probability p of landing heads, probability q ≤ p of landingtails, and probability 1

6 of landing on its side (i.e. on neither face). It is known that if this coin isflipped twice, the likelihood that both flips will have the same result is 1

2 . Find p.

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HMMT November 2018, November 10, 2018 — GUTS ROUND

Organization Team Team ID#

13. [9] Find the smallest positive integer n for which

1!2! · · · (n− 1)! > n!2.

14. [9] Call a triangle nice if the plane can be tiled using congruent copies of this triangle so that any twotriangles that share an edge (or part of an edge) are reflections of each other via the shared edge. Howmany dissimilar nice triangles are there?

15. [9] On a computer screen is the single character a. The computer has two keys: c (copy) and p (paste),which may be pressed in any sequence.

Pressing p increases the number of a’s on screen by the number that were there the last time c waspressed. c doesn’t change the number of a’s on screen. Determine the fewest number of keystrokesrequired to attain at least 2018 a’s on screen. (Note: pressing p before the first press of c doesnothing).

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HMMT November 2018, November 10, 2018 — GUTS ROUND

Organization Team Team ID#

16. [10] A positive integer is called primer if it has a prime number of distinct prime factors. A positiveinteger is called primest if it has a primer number of distinct primer factors. Find the smallest primestnumber.

17. [10] Pascal has a triangle. In the nth row, there are n + 1 numbers an,0, an,1, an,2, . . . , an,n wherean,0 = an,n = 1. For all 1 ≤ k ≤ n − 1, an,k = an−1,k − an−1,k−1. What is the sum of the absolutevalues of all numbers in the 2018th row?

18. [10] An n×m maze is an n×m grid in which each cell is one of two things: a wall, or a blank. A mazeis solvable if there exists a sequence of adjacent blank cells from the top left cell to the bottom rightcell going through no walls. (In particular, the top left and bottom right cells must both be blank.)Determine the number of solvable 2× 5 mazes.

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HMMT November 2018, November 10, 2018 — GUTS ROUND

Organization Team Team ID#

19. [11] Let A be the number of unordered pairs of ordered pairs of integers between 1 and 6 inclusive,and let B be the number of ordered pairs of unordered pairs of integers between 1 and 6 inclusive.(Repetitions are allowed in both ordered and unordered pairs.) Find A−B.

20. [11] Let z be a complex number. In the complex plane, the distance from z to 1 is 2, and the distancefrom z2 to 1 is 6. What is the real part of z?

21. [11] A function f : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} is said to be nasty if there do not exist distincta, b ∈ {1, 2, 3, 4, 5} satisfying f(a) = b and f(b) = a. How many nasty functions are there?

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HMMT November 2018, November 10, 2018 — GUTS ROUND

Organization Team Team ID#

22. [12] In a square of side length 4, a point on the interior of the square is randomly chosen and a circleof radius 1 is drawn centered at the point. What is the probability that the circle intersects the squareexactly twice?

23. [12] Let S be a subset with four elements chosen from {1, 2, . . . , 10}. Michael notes that there is a wayto label the vertices of a square with elements from S such that no two vertices have the same label,and the labels adjacent to any side of the square differ by at least 4. How many possibilities are therefor the subset S?

24. [12] Let ABCD be a convex quadrilateral so that all of its sides and diagonals have integer lengths.Given that ∠ABC = ∠ADC = 90◦, AB = BD, and CD = 41, find the length of BC.

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HMMT November 2018, November 10, 2018 — GUTS ROUND

Organization Team Team ID#

25. [13] Let a0, a1, . . . and b0, b1, . . . be geometric sequences with common ratios ra and rb, respectively,such that

∞∑i=0

ai =

∞∑i=0

bi = 1 and

( ∞∑i=0

a2i

)( ∞∑i=0

b2i

)=

∞∑i=0

aibi.

Find the smallest real number c such that a0 < c must be true.

26. [13] Points E,F,G,H are chosen on segments AB,BC,CD,DA, respectively, of square ABCD. Giventhat segment EG has length 7, segment FH has length 8, and that EG and FH intersect inside ABCDat an acute angle of 30◦, then compute the area of square ABCD.

27. [13] At lunch, Abby, Bart, Carl, Dana, and Evan share a pizza divided radially into 16 slices. Eachone takes takes one slice of pizza uniformly at random, leaving 11 slices. The remaining slices of pizzaform “sectors” broken up by the taken slices, e.g. if they take five consecutive slices then there is onesector, but if none of them take adjacent slices then there will be five sectors. What is the expectednumber of sectors formed?

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HMMT November 2018, November 10, 2018 — GUTS ROUND

Organization Team Team ID#

28. [15] What is the 3-digit number formed by the 9998th through 10000th digits after the decimal pointin the decimal expansion of 1

998?

Note: Make sure your answer has exactly three digits, so please include any leading zeroes if necessary.

29. [15] An isosceles right triangle ABC has area 1. Points D,E, F are chosen on BC,CA,AB respectivelysuch that DEF is also an isosceles right triangle. Find the smallest possible area of DEF .

30. [15] Let n be a positive integer. Let there be Pn ways for Pretty Penny to make exactly n dollars outof quarters, dimes, nickels, and pennies. Also, let there be Bn ways for Beautiful Bill to make exactlyn dollars out of one dollar bills, quarters, dimes, and nickels. As n goes to infinity, the sequence offractions Pn

Bnapproaches a real number c. Find c.

Note: Assume both Pretty Penny and Beautiful Bill each have an unlimited number of each type ofcoin. Pennies, nickels, dimes, quarters, and dollar bills are worth 1, 5, 10, 25, 100 cents respectively.

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HMMT November 2018, November 10, 2018 — GUTS ROUND

Organization Team Team ID#

31. [17] David and Evan each repeatedly flip a fair coin. David will stop when he flips a tail, and Evanwill stop once he flips 2 consecutive tails. Find the probability that David flips more total heads thanEvan.

32. [17] Over all real numbers x and y, find the minimum possible value of

(xy)2 + (x + 7)2 + (2y + 7)2.

33. [17] Let ABC be a triangle with AB = 20, BC = 10, CA = 15. Let I be the incenter of ABC, and letBI meet AC at E and CI meet AB at F . Suppose that the circumcircles of BIF and CIE meet at apoint D different from I. Find the length of the tangent from A to the circumcircle of DEF .

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HMMT November 2018, November 10, 2018 — GUTS ROUND

Organization Team Team ID#

34. [20] A positive integer is called primer if it has a prime number of distinct prime factors. A positiveinteger is called primest if it has a primer number of distinct primer factors. A positive integer iscalled prime-minister if it has a primest number of distinct primest factors. Let N be the smallestprime-minister number. Estimate N .

An estimate of E > 0 earns b20 minNE , E

N

)c points.

35. [20] Pascal has a triangle. In the nth row, there are n + 1 numbers an,0, an,1, an,2, . . . , an,n wherean,0 = an,n = 1. For all 1 ≤ k ≤ n− 1, an,k = an−1,k − an−1,k−1. Let N be the value of the sum

2018∑k=0

|a2018,k|2018k

) .

Estimate N .

An estimate of E > 0 earns b20 · 2−|N−E|/70c points.

36. [20] An n×m maze is an n×m grid in which each cell is one of two things: a wall, or a blank. A mazeis solvable if there exists a sequence of adjacent blank cells from the top left cell to the bottom rightcell going through no walls. (In particular, the top left and bottom right cells must both be blank.)Let N be the number of solvable 5× 5 mazes. Estimate N .

An estimate of E > 0 earns b20 minNE , E

N

)2c points.

HMMT November 2018November 10, 2018

Guts Round

1. [5] A positive integer is called primer if it has a prime number of distinct prime factors. Find thesmallest primer number.

Proposed by: Yuan Yao

Answer: 6

A primer number must have at least two distinct prime factors, and 6 will work.

2. [5] Pascal has a triangle. In the nth row, there are n + 1 numbers an,0, an,1, an,2, . . . , an,n wherean,0 = an,n = 1. For all 1 k n � 1, an,k = an�1,k � an�1,k�1. What is the sum of all numbers inthe 2018th row?

Proposed by: Michael Ren

Answer: 2

In general, the sum of the numbers on the nth row will be

nX

k=0

an,k = an,0 +n�1X

k=1

(an�1,k � an�1,k�1) + an,n = an,0 + (an�1,n�1 � an�1,0) + an,n = 2.

3. [5] An n⇥m maze is an n⇥m grid in which each cell is one of two things: a wall, or a blank. A mazeis solvable if there exists a sequence of adjacent blank cells from the top left cell to the bottom rightcell going through no walls. (In particular, the top left and bottom right cells must both be blank.)Determine the number of solvable 2⇥ 2 mazes.

Proposed by: James Lin

Answer: 3

We must have both top-left and bottom-right cells blank, and we cannot have both top-right andbottom-left cells with walls. As long as those conditions are satisfied, the maze is solvable, so theanswer is 3.

4. [6] Let a, b, c, n be positive real numbers such that a+ba = 3, b+c

b = 4, and c+ac = n. Find n.

Proposed by: Evan Chen

Answer:

76

We have

1 =b

a· cb· ac= (3� 1)(4� 1)(n� 1).

Solving for n yields n = 76 .

5. [6] Jerry has ten distinguishable coins, each of which currently has heads facing up. He chooses onecoin and flips it over, so it now has tails facing up. Then he picks another coin (possibly the same oneas before) and flips it over. How many configurations of heads and tails are possible after these twoflips?

Proposed by: Andrew Gu

Answer: 46

We have two cases:

Case 1: Jerry picks the same coin twice. Then, the first time he flips the coin, it becomes tails, andthen the second time, it becomes heads again, giving us the original state of all heads.

Case 2: Jerry picks two di↵erent coins. In this case, there are two coins with tails face up, and therest are heads face up. There are

�102

�= 10·9

2 = 45 ways to pick which two coins have tails.

Adding up the possibilities from both cases, we have a total of 1 + 45 = 46 possible configurations.

6. [6] An equilateral hexagon with side length 1 has interior angles 90�, 120�, 150�, 90�, 120�, 150� in thatorder. Find its area.

Proposed by: Yuan Yao

Answer:

3+p3

2

The area of this hexagon is the area of a 32 ⇥

⇣1 +

p32

⌘rectangle (with the 90� angles of the hexagon

at opposite vertices) minus the area of an equilateral triangle with side length 1. Then this is

6 + 3p3

4�

p3

4=

3 +p3

2.

7. [7] At Easter-Egg Academy, each student has two eyes, each of which can be eggshell, cream, orcornsilk. It is known that 30% of the students have at least one eggshell eye, 40% of the students haveat least one cream eye, and 50% of the students have at least one cornsilk eye. What percentage ofthe students at Easter-Egg Academy have two eyes of the same color?

Proposed by: Yuan Yao

Answer: 80%

For the purposes of this solution, we abbreviate “eggshell” by “egg”, and “cornsilk” by “corn”. Weknow that there are only six combinations of eye color possible: egg-cream, egg-corn, egg-egg, cream-corn, cream-cream, corn-corn. If we let the proportions for each of these be represented by a, b, c, d,e, and f respectively, we have the following four equalities:

a+ b+ c = .3

a+ d+ e = .4

b+ d+ f = .5

a+ b+ c+ d+ e+ f = 1

where the first three equalities come from the given conditions. Adding the first three equations andsubtracting the fourth, we obtain that

a+ b+ d = .2

which is the proportion of people with di↵erent colored eyes. The proportion of people with the sameeye color is thus 1� .2 = .8.

8. [7] Pentagon JAMES is such that AM = SJ and the internal angles satisfy \J = \A = \E = 90�,and \M = \S. Given that there exists a diagonal of JAMES that bisects its area, find the ratio ofthe shortest side of JAMES to the longest side of JAMES.

Proposed by: James Lin

Answer:

14

Since \J = \A = 90� and AM = JS, JAMS must be a rectangle. In addition, \M + \S = 270�,so \M = \S = 135�. Therefore, \ESM = \EMS = 45�, which means MES is an isosceles righttriangle. Note that AME and JSE are congruent, which means that [JAES] = [JAE] + [JSE] =[JAE] + [AME] > [AME], so AE cannot be our diagonal. Similarly, JE cannot be our diagonal.Diagonals SA and JM bisect rectangle JAMS, so they also cannot bisect the pentagon. Thus,the only diagonal that can bisect [JAMES] is MS, which implies [JAMS] = [MES]. We know[JAMS] = JA ·AM and [MES] = ME·ES

2 , and ME = ES = JAp2, which implies

JA ·AM =JA2

4=) AM

JA=

1

4

Finally, EM and MS are both 1p2the length of SM = JA. This means that AM is our shortest side

and JA is our longest side, so 14 is our answer.

9. [7] Farmer James has some strange animals. His hens have 2 heads and 8 legs, his peacocks have 3heads and 9 legs, and his zombie hens have 6 heads and 12 legs. Farmer James counts 800 heads and2018 legs on his farm. What is the number of animals that Farmer James has on his farm?

Proposed by: James Lin

Answer: 203

Note that each animal has 6 more legs than heads. Thus, if there are n animals, then there are 6nmore legs than heads in total. There are 2018� 800 = 1218 more legs than heads in total, so there are12186 = 203 animals.

10. [8] Abbot writes the letter A on the board. Every minute, he replaces every occurrence of A with ABand every occurrence of B with BA, hence creating a string that is twice as long. After 10 minutes,there are 210 = 1024 letters on the board. How many adjacent pairs are the same letter?

Proposed by: Yuan Yao

Answer: 341

Let an denote the number of adjacent pairs of letters that are the same after n minutes, and bn thenumber of adjacent pairs that are di↵erent.

Lemma 1. an = bn�1 for all n � 0.

Proof. Any adjacent pair of identical letters XX at stage n either came from the same letter of stagen � 1 (W ! XX), or two adjacent letters of stage n � 1 (VW ! MXXN). Because A ! AB andB ! BA, they cannot have come from the same letter.

If they came from a pair of adjacent letters, then observing what each adjacent pair of letters resultsin in the next minute,

AA ! ABAB

AB ! ABBA

BA ! BAAB

BB ! BABA

we see that our adjacent pair VW must have been AB or BA. The number of such pairs is preciselybn�1.

From the relation an + bn = 2n � 1 for all n � 0, we obtain the recurrence relation

an = 2n�1 � 1� an�1

from which we obtain values a0 = 0, a1 = 0, a2 = 1, a3 = 2, a4 = 5, a5 = 10, a6 = 21, a7 = 42,a8 = 85, a9 = 170, and a10 = 341 .

11. [8] Let 4ABC be an acute triangle, with M being the midpoint of BC, such that AM = BC. LetD and E be the intersection of the internal angle bisectors of \AMB and \AMC with AB and AC,respectively. Find the ratio of the area of 4DME to the area of 4ABC.

Proposed by: Handong Wang

Answer:

29

Let [XY Z] denote the area of 4XY Z.

Solution 1: Let AM = `, let DE = d, and let the midpoint of DE be F . Since ADAB = AE

AC = 23

by the angle bisector theorem, F lies on AM and 4ADE is similar to 4ABC. Note that \DME is

formed by angle bisectors of \AMB and \AMC, which add up to 180�. Thus \DME is right, soboth 4DMF and 4EMF are isosceles. This implies that FM = d

2 . Applying the similarity between4ADE and 4ABC, we get d

` = ADAB = 2

3 . Since AF = 2FM , [ADE] = 2[DME]. Finally, since

[ADE][ABC] =

49 ,

[DME][ABC] = 1

2 · 49 =

2

9.

Solution 2: We compute the ratio [DME][ABC] by finding 1 � [ADE]

[ABC] �[DBM ][ABC] � [EMC]

[ABC] . Since ADDB =

AEEC = AM

12BC

= 2 by the angle bisector theorem, we see that [ADE][ABC] =

�23

� �23

�= 4

9 . Also, since

BM = CM = 12BC, [DBM ]

[ABC] =�13

� �12

�= 1

6 , and[EMC][ABC] =

�12

� �13

�= 1

6 . Thus, [DME][ABC] = 1 � [ADE]

[ABC] �

[DBM ][ABC] � [EMC]

[ABC] = 1� 49 � 1

6 � 16 =

2

9.

12. [8] Consider an unusual biased coin, with probability p of landing heads, probability q p of landingtails, and probability 1

6 of landing on its side (i.e. on neither face). It is known that if this coin isflipped twice, the likelihood that both flips will have the same result is 1

2 . Find p.

Proposed by: Brian Reinhart

Answer:

23

The probability that both flips are the same is p2 + q2 + 136 . For this to be 1

2 , we must have

p2 + q2 +1

36= p2 +

✓5

6� p

◆2

+1

36=

1

2

Using the quadratic formula, p = 23 or 1

6 . Since p � q, we have that p = 23 .

13. [9] Find the smallest positive integer n for which

1!2! · · · (n� 1)! > n!2.

Proposed by: Andrew Gu

Answer: 8

Dividing both sides by n!2, we obtain

1!2!...(n� 3)!(n� 2)!(n� 1)!

[n(n� 1)!] [n(n� 1)(n� 2)!]> 1

1!2!...(n� 3)!

n2(n� 1)> 1

1!2!...(n� 3)! > n2(n� 1)

Factorials are small at first, so we can rule out some small cases: when n = 6, the left hand side is1!2!3! = 12, which is much smaller than 62 · 5. (Similar calculations show that n = 1 through n = 5 donot work. either.)

Setting n = 7, the left-hand side is 288, which is still smaller than 72 · 6. However, n = 8 gives34560 > 448, so 8 is the smallest integer for which the inequality holds.

14. [9] Call a triangle nice if the plane can be tiled using congruent copies of this triangle so that any twotriangles that share an edge (or part of an edge) are reflections of each other via the shared edge. Howmany dissimilar nice triangles are there?

Proposed by: Yuan Yao

Answer: 4

The triangles are 60-60-60, 45-45-90, 30-60-90, and 30-30-120.

We make two observations.

• By reflecting ”around” the same point, any angle of the triangle must be an integer divisor of360�.

• if any angle is an odd divisor of 360�, i.e equals 360k for odd k, then the two adjacent sides must

be equal.

We do casework on the largest angle.

• 60�. We are forced into a 60-60-60 triangle, which works.

• 72�. By observation 2, this triangle’s other two angles are 54�. This is not an integer divisor of360�.

• 90�. The second largest angle is at least 45�. If it is 45�, it is the valid 90-45-45 triangle. If it

is 360�

7 , the triangle is invalid by observation 2. If it is 60�, it is the valid 90-60-30 triangle. Ifit is 72�, the triangle is invalid by observation 2.

• 120�. By observation 2, the other angles are 30�, meaning it is the valid 120-30-30 triangle.

The conclusion follows.

15. [9] On a computer screen is the single character a. The computer has two keys: c (copy) and p (paste),which may be pressed in any sequence.

Pressing p increases the number of a’s on screen by the number that were there the last time c waspressed. c doesn’t change the number of a’s on screen. Determine the fewest number of keystrokesrequired to attain at least 2018 a’s on screen. (Note: pressing p before the first press of c doesnothing).

Proposed by: John Michael Wu

Answer: 21

The first keystroke must be c and the last keystroke must be p. If there are k c’s pressed in total, letni denote one more than the number of p’s pressed immediately following the i’th c, for 1 i k.

Then, we have that the total number of keystrokes is

s :=kX

i=1

ni

and the total number of a’s is

r :=kY

i=1

ni

We desire to minimize s with the constraint that r � 2018. We claim that the minimum possible s iss = 21.

This value of s is achieved by k = 7 and n1 = n2 = n3 = n4 = n5 = n6 = n7 = 3, so it remains toshow that s = 20 is not possible.

Suppose it were for some k and ni. By the AM-GM inequality,

✓n1 + n2 + · · ·+ nk

k

◆� k

pn1n2 · · ·nk

implying that

2018 n1n2 · · ·nk

✓n1 + n2 + · · ·+ nk

k

◆k

=

✓20

k

◆k

which is satisfied by no positive integers k. More rigorously, the function f(x) = x1x is well known to

have a maximum at x = e. Making the substitution u = 20k , we obtain

✓20

k

◆k

= u20u

=⇣u

1u

⌘20

which is maximized by setting u = e. However, e20e ⇡ 1568.05, meaning that s = 20 is not possible.

16. [10] A positive integer is called primer if it has a prime number of distinct prime factors. A positiveinteger is called primest if it has a primer number of distinct primer factors. Find the smallest primestnumber.

Proposed by: Yuan Yao

Answer: 72

We claim the answer is 72, as it has 6 primer factors: 6, 12, 24, 18, 36, 72, and 6 is a primer.

We now prove that there is no smaller primer. Suppose there were a smaller primer r < 72. We docasework on the number of distinct prime factors of r.

• r has � 4 distinct prime factors. Then r � 2 · 3 · 5 · 7 = 210, which is larger than 72.

• r has 3 distinct prime factors. If each of these factors has multiplicity 1, i.e r = pqs for distinctprimes p, q, s, then r has precisely 4 primer factors: pq, qs, sp, pqs, and 4 is not a primer. Thus,r contains at least one factor of multiplicity at least 2. If r is p2qs for distinct primes p, q, s,then r has 7 distinct primer factors: pq, qs, sp, pqs, p2q, sp2, p2qs, and 7 is not a primer. Thus, ifr = paqbsc, a+ b+ c � 5, and r � 23 · 3 · 5 = 120, which is � 72.

• r has 2 distinct prime factors. If r = paqb, for distinct primts p, q, then r’s primer factors areprecisely its divisors of the form piqj , where 1 i a and 1 j b, meaning that it has abprimer factors. Thus, ab is a primer, meaning that ab � 6. Thus r � 23 · 32 = 72, where the otherpossibilities can be ruled out through easy casework.

• r has 1 distinct prime factor. Then it doesn’t have any primer factors, and thus cannot possiblyhave a primer number of them.

We conclude that 72 is the smallest primer number.

17. [10] Pascal has a triangle. In the nth row, there are n + 1 numbers an,0, an,1, an,2, . . . , an,n wherean,0 = an,n = 1. For all 1 k n � 1, an,k = an�1,k � an�1,k�1. What is the sum of the absolutevalues of all numbers in the 2018th row?

Proposed by: Michael Ren

Answer:

22018+23

Let sn be the sum of the absolute values of numbers in the nth row. For odd n, we have thatan,1, . . . , an,n�1 alternate in sign as �,+,�,+, . . . ,+, with the last term being an,n�1 = 1. For even

n, we have that an,1, . . . , an,n�2 alternate in sign as �,+,�,+, . . . ,+, and an,n�1 = 0. These facts canbe proven by induction. Thus, sn = 1� an,1 + an,2 � · · ·+(�1)n�1an,n�1 +1. Applying the recursion,for n > 0 this becomes sn = 1 � (an�1,1 � an�1,0) + (an�1,2 � an�1,1) � · · · + (�1)n�1(an�1,n�1 �an�1,n�2)+ 1 = 2(1� an�1,1 + an�1,2 � · · ·+(�1)n�2an�1,n�2 +1)� 1+ (�1)n�1. In other words, if nis even then sn = 2sn�1 � 2 and if n is odd then sn = 2sn�1. This means that s2n = 4s2n�2 � 2. Since2018 is even, we can write s2018 = 4s2016 � 2 = 22018 � 22017 � 22015 � · · · � 2. Applying the formulafor the sum of a geometric series, we get s2018 = 22018 � 22019�2

4�1 = 22018+23 .

18. [10] An n⇥m maze is an n⇥m grid in which each cell is one of two things: a wall, or a blank. A mazeis solvable if there exists a sequence of adjacent blank cells from the top left cell to the bottom rightcell going through no walls. (In particular, the top left and bottom right cells must both be blank.)Determine the number of solvable 2⇥ 5 mazes.

Proposed by: John Michael Wu

Answer: 49

Solution 1: Replace 5 by an arbitrary n. Label the cells of the maze by (x, y) where 1 x n and1 y 2. Let an denote the number of solvable 2⇥ n mazes, and let bn denote the number of 2⇥ nmazes where there exists a sequence of adjacent blank cells from the (1, 1) to (n, 1). We observe therelations

an = 2an�1 + bn�2

bn = 2bn�1 + an�2.

The first relation follows from dividing into cases depending on if (n�1, 2 is blank. If this cell is blank,then the maze is solvable if and only if there is a path to (n� 1, 2) and the cell (n, 2) is blank. The cell(n, 1) is arbitrary so we get 2an�1. If (n� 1, 2) is a wall, then the maze is solvable if and only if thereis a path to (n� 2, 1) and each of the cells (n� 1, 1), (n, 1), (n, 2) are blank. This gives the term bn�2.

The second relation follows similarly dividing into cases based on whether the cell (n � 1, 1) is blankor not.

The base cases are a1 = 1, a2 = 3, b1 = 2, b2 = 4. We thus obtain:

n an bn1 1 22 3 43 8 94 20 215 49 50

The answer is then 49.

Solution 2: Call a maze reachable if there exists a sequence of adjacent blank cells from (1, 1) to anycell in column n. Let xn denote the number of reachable 2 ⇥ n mazes where the rightmost columnhas a blank in the top cell and a wall in the bottom cell, let yn denote the number of reachable 2⇥ nmazes where the rightmost column has a wall in the top cell and a blank in the bottom cell, and letzn denote the number of reachable 2⇥ n mazes where the rightmost column has a blank in both of itscells. Then, observe the relations

xn = zn�1 + xn�1

yn = zn�1 + yn�1

zn = zn�1 + xn�1 + yn�1,

as well as base cases x1 = 1, y1 = 0, z1 = 1. We thus obtain:

n xn yn zn1 1 0 12 2 1 23 4 3 54 9 8 125 21 20 29

Note that the answer is y5 + z5, which is 20 + 29 = 49.

19. [11] Let A be the number of unordered pairs of ordered pairs of integers between 1 and 6 inclusive,and let B be the number of ordered pairs of unordered pairs of integers between 1 and 6 inclusive.(Repetitions are allowed in both ordered and unordered pairs.) Find A�B.

Proposed by: Yuan Yao

Answer: 225

There are 6 · 6 ordered pairs of integers between 1 and 6 inclusive and 21 unordered pairs of integers(�62

�= 15 di↵erent pairs and 6 doubles). Then, A =

�362

�+ 36 = 666 and B = 21 · 21 = 441. Therefore

A�B = 225.

For general n, there are n2 ordered pairs of integers and n(n+1)2 unordered pairs of integers. Then

A = n2(n2+1)2 and B = n2(n+1)2

4 so

A�B =n2(2(n2 + 1)� (n+ 1)2)

4=

✓n(n� 1)

2

◆2

.

20. [11] Let z be a complex number. In the complex plane, the distance from z to 1 is 2, and the distancefrom z2 to 1 is 6. What is the real part of z?

Proposed by: James Lin

Answer:

54

Note that we must have |z� 1| = 2 and |z2 � 1| = 6, so |z+1| = |z2�1||z�1| = 3. Thus, the distance from z

to 1 in the complex plane is 2 and the distance from z to �1 in the complex plane is 3. Thus, z, 1,�1form a triangle with side lengths 2, 3, 3. The area of a triangle with sides 2, 2, 3 can be computed to

be 3p7

4 by standard techniques, so the length of the altitude from z to the real axis is 3p7

4 · 22 = 3

p7

4 .

The distance between 1 and the foot from z to the real axis is

r22 �

⇣3p7

4

⌘2= 1

4 by the Pythagorean

Theorem. It is clear that z has positive imaginary part as the distance from z to �1 is greater thanthe distance from z to 1, so the distance from 0 to the foot from z to the real axis is 1 + 1

4 = 54 . This

is exactly the real part of z that we are trying to compute.

21. [11] A function f : {1, 2, 3, 4, 5} ! {1, 2, 3, 4, 5} is said to be nasty if there do not exist distincta, b 2 {1, 2, 3, 4, 5} satisfying f(a) = b and f(b) = a. How many nasty functions are there?

Proposed by: Michael Ren

Answer: 1950

We use complementary counting. There are 55 = 3125 total functions. If there is at least one pair ofnumbers which map to each other, there are

�52

�= 10 ways to choose the pair and 53 = 125 ways to

assign the other values of the function for a total of 1250. But we overcount each time there are two suchpairs, so we must subtract o↵ 5 · 3 · 5 = 75, where there are 5 options for which number is not in a pair,3 options for how the other four numbers are paired up, and 5 options for where the function outputswhen the unpaired number is inputted. This results in a final answer of 3125� (1250� 75) = 1950.

22. [12] In a square of side length 4, a point on the interior of the square is randomly chosen and a circleof radius 1 is drawn centered at the point. What is the probability that the circle intersects the squareexactly twice?

Proposed by: Jason Lu

Answer:

⇡+816

Consider the two intersection points of the circle and the square, which are either on the same sideof the square or adjacent sides of the square. In order for the circle to intersect a side of the squaretwice, it must be at distance at most 1 from that side and at least 1 from all other sides. The regionof points where the center could be forms a 2⇥ 1 rectangle.

In the other case, a square intersects a pair of adjacent sides once each if it it at distance at most onefrom the corner, so that the circle contains the corner. The region of points where the center could beis a quarter-circle of radius 1.

The total area of the regions where the center could be is ⇡ + 8, so the probability is ⇡+816 .

23. [12] Let S be a subset with four elements chosen from {1, 2, . . . , 10}. Michael notes that there is a wayto label the vertices of a square with elements from S such that no two vertices have the same label,and the labels adjacent to any side of the square di↵er by at least 4. How many possibilities are therefor the subset S?

Proposed by: James Lin

Answer: 36

Let the four numbers be a, b, c, d around the square. Assume without loss of generality that a is thelargest number, so that a > b and a > d. Note that c cannot be simultaneously smaller than one ofb, d and larger than the other because, e.g. if b > c > d, then a > b > c > d and a � d+ 12. Hence cis either smaller than b and d or larger than b and d.

Case 1: c is smaller than b and d. Then we have a� c � 8, but when a� c = 8, we have b = c+4 = d,so we need a� c = 9, giving the only set {1, 5, 6, 10}.Case 2: c is larger than b and d. Since a > c and b, d are both at most c�4, the range of possible valuesfor c is {6, 7, 8, 9}. When c = 9, 8, 7, 6, there are 1, 2, 3, 4 choices for a respectively and

�52

�,�42

�,�32

�,�22

�

for b and d respectively (remember that order of b and d does not matter). So there are 1 · 10+ 2 · 6+3 · 3 + 4 · 1 = 35 sets in this case.

Therefore we have 1 + 35 = 36 possible sets in total.

24. [12] Let ABCD be a convex quadrilateral so that all of its sides and diagonals have integer lengths.Given that \ABC = \ADC = 90�, AB = BD, and CD = 41, find the length of BC.

Proposed by: Anders Olsen

Answer: 580

Let the midpoint of AC be O which is the center of the circumcircle of ABCD. ADC is a right trianglewith a leg of length 41, and 412 = AC2 �AD2 = (AC �AD)(AC +AD). As AC,AD are integers and41 is prime, we must have AC = 840, AD = 841. Let M be the midpoint of AD. 4AOM ⇠ 4ACD, soBM = BO +OM = 841/2 + 41/2 = 441. Then AB =

p4202 + 4412 = 609 (this is a 20-21-29 triangle

scaled up by a factor of 21). Finally, BC2 = AC2 �AB2 so BC =p8412 � 6092 = 580.

25. [13] Let a0, a1, . . . and b0, b1, . . . be geometric sequences with common ratios ra and rb, respectively,such that

1X

i=0

ai =1X

i=0

bi = 1 and

1X

i=0

a2i

! 1X

i=0

b2i

!=

1X

i=0

aibi.

Find the smallest real number c such that a0 < c must be true.

Proposed by: Handong Wang

Answer:

43

Let a0 = a and b0 = b. FromP1

i=0 ai =a0

1�ra

= 1 we have a0 = 1� ra and similarly b0 = 1� rb. This

meansP1

i=0 a2i = a2

01�r2

a

= a2

(1�ra

)(1+ra

) =a2

a(2�a) =a

2�a , soP1

i=0 a2i

P1i=0 b

2i =

P1i=0 aibi yields

a

2� a· b

2� b=

ab

1� (1� a)(1� b).

Since the numerators are equal, the denominators must be equal, which when expanded gives 2ab �3a� 3b+ 4 = 0, which is equivalent to (2a� 3)(2b� 3) = 1. But note that 0 < a, b < 2 since we needthe sequences to converge (or |ra|, |rb| < 1), so then �3 < 2b� 3 < 1, and thus 2a� 3 > 1 (impossible)or 2a� 3 < � 1

3 . Hence a < 43 , with equality when b approaches 0.

26. [13] Points E,F,G,H are chosen on segments AB,BC,CD,DA, respectively, of square ABCD. Giventhat segment EG has length 7, segment FH has length 8, and that EG and FH intersect inside ABCDat an acute angle of 30�, then compute the area of square ABCD.

Proposed by: Kevin Sun

Answer:

78419

Rotate EG by 90� about the center of the square to E0G0 with E0 2 AD and G0 2 BC. Now E0G0

and FH intersect at an angle of 60�. Then consider the translation which takes E0 to H and G0 to I.Triangle FHI has FH = 8, HI = 7 and \FHI = 60�. Furthermore, the height of this triangle is theside length of the square. Using the Law of Cosines,

FI =p72 � 7 · 8 + 82 =

p57.

By computing the area of FHI in two ways, if h is the height then

1

2⇥

p57⇥ h =

1

2⇥

p3

2⇥ 7⇥ 8.

Then h = 28p19

and the area of the square is h2 = 78419 .

27. [13] At lunch, Abby, Bart, Carl, Dana, and Evan share a pizza divided radially into 16 slices. Eachone takes takes one slice of pizza uniformly at random, leaving 11 slices. The remaining slices of pizzaform “sectors” broken up by the taken slices, e.g. if they take five consecutive slices then there is onesector, but if none of them take adjacent slices then there will be five sectors. What is the expectednumber of sectors formed?

Proposed by: Andrew Gu

Answer:

113

Consider the more general case where there are N slices and M > 0 slices are taken. Let S denote thenumber of adjacent pairs of slices of pizza which still remain. There are N �M slices and a sector ofk slices contributes k � 1 pairs to S. Hence the number of sectors is N � M � S. We compute theexpected value of S by looking at each adjacent pair in the original pizza:

E(S) = N

�N�2M

��NM

� = N(N �M)(N �M � 1)

N(N � 1)=

(N �M)(N �M � 1)

N � 1

The expected number of sectors is then

N �M � (N �M)(N �M � 1)

N � 1=

(N �M)M

N � 1.

For N = 16,M = 5 this yields 113 .

28. [15] What is the 3-digit number formed by the 9998th through 10000th digits after the decimal pointin the decimal expansion of 1

998?

Note: Make sure your answer has exactly three digits, so please include any leading zeroes if necessary.

Proposed by: Sujay Kazi

Answer: 042

Note that 1998 + 1

2 = 250499 repeats every 498 digits because 499 is prime, so 1

998 does as well (after thefirst 498 block). Now we need to find 38th to 40th digits. We expand this as a geometric series

1

998=

11000

1� 21000

= .001 + .001⇥ .002 + .001⇥ .0022 + · · · .

The contribution to the 36th through 39th digits is 4096, the 39th through 42nd digits is 8192, and 41st

through 45th digits is 16384. We add these together:

4 0 9 68 1 9 2

1 6 8 3 44 1 0 4 2 0 · · ·

The remaining terms decrease too fast to have e↵ect on the digits we are looking at, so the 38th to40th digits are 042.

29. [15] An isosceles right triangle ABC has area 1. Points D,E, F are chosen on BC,CA,AB respectivelysuch that DEF is also an isosceles right triangle. Find the smallest possible area of DEF .

Proposed by: Yuan Yao

Answer:

15

Without loss of generality, suppose that AB is the hypotenuse.

If F is the right angle, then F must be the midpoint of AB. To prove this, let X and Y be the feetfrom F to BC and AC. Since \XFY = \DFE = 90�, we have \XFD = \Y FE so

XF = DF cos\XFD = EF cos\Y FE = Y F.

Hence F is equidistant from AC and BC so it is the midpoint of AB. Then the minimum area isachieved by minimizing DF ; this occurs when DF is perpendicular to BC. The triangle DEF thenbecomes the medial triangle of ABC, so its area is 1

4 .

If F is not the right angle, without loss of generality, let the right angle be D. Place this triangle inthe complex plane such that C is the origin, B =

p2, and A =

p2i.

Now since D is on the real axis and E is on the imaginary axis, D = x and E = yi, and we can obtainF by a 90 degree counterclockwise rotation of D around E: this evaluates to F = y + (x+ y)i. For Fto be on AB, the only constraint is to have y + (x+ y) =

p2 =) x =

p2� 2y.

To minimize the area, we minimize

DE2

2=

x2 + y2

2=

(p2� 2y)2 + y2

2=

5y2 � 4p2y + 2

2

which has a minimum of 15 at y = 2

p2

5 . Since this is between 0 andp2, this is indeed a valid

configuration.

Finally, we take the smallest area of 15 .

30. [15] Let n be a positive integer. Let there be Pn ways for Pretty Penny to make exactly n dollars outof quarters, dimes, nickels, and pennies. Also, let there be Bn ways for Beautiful Bill to make exactlyn dollars out of one dollar bills, quarters, dimes, and nickels. As n goes to infinity, the sequence offractions P

n

Bn

approaches a real number c. Find c.

Note: Assume both Pretty Penny and Beautiful Bill each have an unlimited number of each type ofcoin. Pennies, nickels, dimes, quarters, and dollar bills are worth 1, 5, 10, 25, 100 cents respectively.

Proposed by: James Lin

Answer: 20

Let dx be the number ways to make exactly x cents using only dimes and nickels. It is easy to see thatwhen x is a multiple of 5,

dx =j x

10

k+ 1.

Now, let cx be the number of ways to make exactly x cents using only quarters, dimes and nickels.Again, it is easy to see that when x is a multiple of 5,

cx = cx�25 + dx.

(We can either use 1 or more quarters, which corresponds to the cx�25 term, or we can use 0 quarters,which corresponds to the dx term.) Combining these two equations, we see that cx can be approximatedby a polynomial of degree 2. (In fact, we get five di↵erent approximations of cx, depending on thevalue of x (mod 25), but they all only di↵er by a constant, which will not a↵ect the limit case.) Wealso see that

Bn = c100n + c100(n�1) + . . .+ c0

andPn = c100n + c100n�5 + . . .+ c0.

Suppose a is the value such that limn!1cn

an2 = 1. Then

limn!1

Bn

Pn= lim

n!1

Pbn/100ck=0 a(100k)2Pbn/5c

k=0 a(5k)2= lim

n!1

400 · n100 (

n100 + 1)(2 · n

100 + 1)n5 (

n5 + 1)(2 · n

5 + 1)= 20.

31. [17] David and Evan each repeatedly flip a fair coin. David will stop when he flips a tail, and Evanwill stop once he flips 2 consecutive tails. Find the probability that David flips more total heads thanEvan.

Proposed by: Anders Olsen

Answer:

15

Solution 1: We can find the values of the functions D(h) and E(h), the probabilities that David andEvan, respectively, flip exactly h heads. It is easy to see that D(h) = 2�h�1. In order to find E(h), wenote that each sequence must end with the flips HTT (unless Evan flips only 2 heads). We disregardthese flips for now. Then there are h prior places we can include an extra tails in the sequence, onebetween each pair of heads. There is a 2�h+1 probability of this happening with no extra tails, h2�h

probability with 1 extra tail,�h2

�2�h�1 probability with 2 extra tails, and so on. This sum is

2�h+1hX

n=0

2�n

✓h

n

◆= 2

✓3

4

◆h

.

We divide by 8 to account for the probability of getting HTT to finish our sequence to get that

E(h) =3h

4h+1.

Our answer is1X

n=0

E(n)

1X

m=n+1

D(m)

!=

1X

n=0

3n

8n+1=

1

5.

Solution 2: Since we only care about the number of heads, we think of this as a “survival” gamewhere they flip a single head each round, such that David has a 1

2 chance of flipping another head andEvan has a 3

4 chance of flipping another head. (If they don’t get to flip another head, they lose.) Davidwins if and only if when at least one of David and Evan loses, David does not lose but Evan loses. Theprobability that at least one of them lose each round is 1� 1

2 · 34 = 5

8 , and David wins this round withprobability 1

2 · 14 = 1

8 , so the overall probability is 15 .

32. [17] Over all real numbers x and y, find the minimum possible value of

(xy)2 + (x+ 7)2 + (2y + 7)2.

Proposed by: James Lin

Answer: 45

Solution 1: Rewrite the given expression as (x2 +4)(1+ y2) + 14(x+2y) + 94. By Cauchy-Schwartz,this is at least (x + 2y)2 + 14(x + 2y) + 94 = (x + 2y + 7)2 + 45. The minimum is 45, attained whenxy = 2, x+ 2y = �7.

Solution 2: Let z = 2y, s = x+ z, p = xz. We seek to minimize

⇣xz2

⌘2+ (x+ 7)2 + (z + 7)2 =

p2

4+ (x2 + z2) + 14(x+ z) + 98

=p2

4+ s2 � 2p+ 14s+ 98

=⇣p2� 2⌘2

+ (s+ 7)2 + 45

� 45.

Equality holds when s = �7, p = 4. Since s2 � 4p, this system has a real solution for x and z.

33. [17] Let ABC be a triangle with AB = 20, BC = 10, CA = 15. Let I be the incenter of ABC, and letBI meet AC at E and CI meet AB at F . Suppose that the circumcircles of BIF and CIE meet at apoint D di↵erent from I. Find the length of the tangent from A to the circumcircle of DEF .

Proposed by: Michael Ren

Answer: 2p30

Solution 1: Let O = AI \ (AEF ). We claim that O is the circumcenter of DEF . Indeed, note that\EDF = \ECI + \FBI = \B+\C

2 = \EOF2 , and OE = OF , so the claim is proven.

Now note that the circumcircle ofDEF passes through the incenter of AEF , so power of A with respectto (DEF ) is

pAE ·AF . We can compute that AE = 10, AF = 12 by the angle bisector theorem, so

the length of the tangent isp10 · 12 = 2

p30.

Solution 2: Let E0 be the reflection of E across AI. Note that E0 lies on AB. We claim that E0

lies on the circumcircle of DEF , which will imply that the power of A with respect to (DEF ) isAE0 ·AF = AE ·AF and we proceed as in Solution 1. We can easily compute

\EE0F = 90� +\A2

and

\EDF = \EDI + \IDF = \ECI + \IBF = 90� � \A2

.

Therefore \EDF + \EE0F = 180�, so E0 lies on the circumcircle of DEF as desired.

34. [20] A positive integer is called primer if it has a prime number of distinct prime factors. A positiveinteger is called primest if it has a primer number of distinct primer factors. A positive integer iscalled prime-minister if it has a primest number of distinct primest factors. Let N be the smallestprime-minister number. Estimate N .

An estimate of E > 0 earns b20min�NE , E

N

�c points.

Proposed by: Yuan Yao

Answer: 24 · 33 · 53 · 7 = 378000

One heuristic for estimating the answer is that numbers of the form pqrs for primes p, q, r, s withp 6= r, q 6= s are primest. Thus, primest numbers are not very rare, so we can expect the answer to berelatively small with only a few distinct prime factors.

from operator import *

primes = []

primers = []

primests = []

for i in range(2,3000):

prime_factors = 0

primer_factors = 0

temp = i

for x in primes:

if x > temp:

break

if temp % x == 0:

prime_factors += 1

while temp % x == 0:

temp = temp // x

if (prime_factors == 0):

primes.append(i)

continue

elif (prime_factors in primes):

primers.append(i)

for x in primers:

if i % x == 0:

primer_factors += 1

if (primer_factors in primers):

primests.append(i)

def product(L):

ans = 1

for i in L:

ans*= i

return ans

def sum_prime_product(L, curr = ()):

if (L == ()):

#print(curr)

if len(curr) in primes:

return product(curr)

return 0

return sum_prime_product(L[1:], curr + (L[0],)) + sum_prime_product(L[1:], curr)

def count_primests(L, curr = ()):

if (L == ()):

if (sum_prime_product(curr) in primers):

return 1

return 0

ans = 0

for i in range(0,L[0]+1):

ans += count_primests(L[1:], curr+(i,))

return ans

def compute(L):

ans = 1

for i in range(len(L)):

ans *= (primes[i]**L[i])

return ans

def find_best(M, best = 2**20 * 3**5, curr = ()):

num = compute(curr)

if (num > best):

return False

if (count_primests(curr) in primests):

print(num, curr)

return num

for i in range(1,M):

result = find_best(M, best, curr + (i,))

if (result == False):

break

elif (result < best):

best = result

return best

print("Answer:", find_best(30))

35. [20] Pascal has a triangle. In the nth row, there are n + 1 numbers an,0, an,1, an,2, . . . , an,n wherean,0 = an,n = 1. For all 1 k n� 1, an,k = an�1,k � an�1,k�1. Let N be the value of the sum

2018X

k=0

|a2018,k|�2018k

� .

Estimate N .

An estimate of E > 0 earns b20 · 2�|N�E|/70c points.

Proposed by: Michael Ren

Answer: 780.9280674537

A good estimate for this question is to use the fact that

2018X

k=0

|a2018,k| =22018 + 2

3,

the answer to Guts 17. This suggests that each |a2018,k| is roughly 13 of its corresponding entry

✓2018

k

◆

in the usual Pascal’s triangle, as the sum of the terms in the 2018th row of Pascal’s triangle is 22018.This then gives an estimate of 2018

3 , which earns 6 points. Code for computing answer in Python 3:

import math

lists=[[1]]

for i in range(2018):

newlist=[]

for j in range(i):

newlist.append(lists[-1][j+1]-lists[-1][j])

lists.append([1]+newlist+[1])

big=math.factorial(2018)

sum=0

for i in range(2019):

sum+=abs(lists[-1][i])/(big//math.factorial(i)//math.factorial(2018-i))

print(sum)

36. [20] An n⇥m maze is an n⇥m grid in which each cell is one of two things: a wall, or a blank. A mazeis solvable if there exists a sequence of adjacent blank cells from the top left cell to the bottom rightcell going through no walls. (In particular, the top left and bottom right cells must both be blank.)Let N be the number of solvable 5⇥ 5 mazes. Estimate N .

An estimate of E > 0 earns b20min�NE , E

N

�2c points.

Proposed by: John Michael Wu

Answer: 1225194

The following code solves the problem in Python 3.

# dfs that returns all paths with no adjacent vertices other than those consecutive in the path

def dfs(graph,start,end,path):

if start==end:

return [path]

paths=[]

for child in graph[start]:

skip=False

if child in path:

continue

for vert in graph[child]:

if vert in path[:-1]:

skip=True

break

if not skip:

paths=paths+dfs(graph,child,end,path[:]+[child])

return paths

# construct graph representing 5x5 grid

graph={}

for a in range(5):

for b in range(5):

graph[(a,b)]=[]

for a in range(4):

for b in range(5):

graph[(a,b)].append((a+1,b));graph[(a+1,b)].append((a,b))

graph[(b,a)].append((b,a+1));graph[(b,a+1)].append((b,a))

paths=dfs(graph,(0,0),(4,4),[(0,0)])

paths.sort(key=len)

intpaths=[0]*len(paths)

# convert paths to 25-bit binary integers

for i in range(len(paths)):

for j in paths[i]:

intpaths[i]+=2**(5*j[0]+j[1])

mazes=0

for j in range(2**23):

k=2*j

# disregard cases that are common and never valid

if k&8912896==8912896 or k&34==34 or k&4472832==4472832 or k&1092==1092:

continue

for path in intpaths:

# cehck if case has empty spaces along whole path

if path&k==0:

mazes+=1

break

print(mazes)

Alternatively, the following code solves the problem in Java SE 8.

import java.util.*;

public class guts36 {

static int M = 5;

static int N = 5;

static long[] pow2 = new long[M * N];

static int[][] dir = new int[][] {new int[] {0, 1}, new int[] {1, 0}, new int[] {0, -1}, new int[] {-1, 0}};

public static void main(String[] args) {

pow2[0] = 1;

for (int i = 1; i < pow2.length; i++) {

pow2[i] = pow2[i - 1] * 2;

}

boolean[][] grid = new boolean[M][N];

grid[0][0] = true;

grid[M - 1][N - 1] = true;

int ans = 0;

for (long c = 0; c < pow2[M * N - 2]; c++) {

long d = c;

for (int b = 0; b < M * N - 2; b++) {

int i = (b + 1) / N;

int j = (b + 1) % N;

grid[i][j] = ((d & 1) > 0);

d >>= 1;

}

if (check(grid)) {

ans++;

}

}

System.out.println("answer: " + ans);

}

static int[] add(int[] a, int[] b) {

return new int[] {a[0] + b[0], a[1] + b[1]};

}

static boolean get(boolean[][] g, int[] a) {

return g[a[0]][a[1]];

}

static void set(boolean[][] g, int[] a, boolean v) {

g[a[0]][a[1]] = v;

}

static boolean valid(int[] a) {

return (a[0] >= 0) && (a[1] >= 0) && (a[0] < M) && (a[1] < N);

}

static boolean check(boolean[][] grid) {

Stack<int[]> q = new Stack<int[]>();

q.add(new int[] {0, 0});

boolean[][] reached = new boolean[M][N];

reached[0][0] = true;

while (!q.isEmpty()) {

int[] a = q.pop();

for (int[] d: dir) {

int[] b = add(a, d);

if (valid(b) && get(grid, b) && !get(reached, b)) {

if (b[0] == M - 1 && b[1] == N - 1) {

return true;

}

set(reached, b, true);

q.add(b);

}

}

}

return false;

}

}

HMIC 2018April 09, 2018

HMIC 2018

1. [6] Let m > 1 be a fixed positive integer. For a nonempty string of base-ten digits S, let c(S) be thenumber of ways to split S into contiguous nonempty strings of digits such that the base-ten numberrepresented by each string is divisible by m. These strings are allowed to have leading zeroes.

In terms of m, what are the possible values that c(S) can take?

For example, if m = 2, then c(1234) = 2 as the splits 1234 and 12|34 are valid, while the other sixsplits are invalid.

2. [7] Consider a finite set of points T ∈ Rn contained in the n-dimensional unit ball centered at theorigin, and let X be the convex hull of T . Prove that for all positive integers k and all points x ∈ X,there exist points t1, t2, . . . , tk ∈ T , not necessarily distinct, such that their centroid

t1 + t2 + · · ·+ tkk

has Euclidean distance at most1√k

from x.

(The n-dimensional unit ball centered at the origin is the set of points in Rn with Euclidean distanceat most 1 from the origin. The convex hull of a set of points T ∈ Rn is the smallest set of points Xcontaining T such that each line segment between two points in X lies completely inside X.)

3. [8] A polygon in the plane (with no self-intersections) is called equitable if every line passing throughthe origin divides the polygon into two (possibly disconnected) regions of equal area.

Does there exist an equitable polygon which is not centrally symmetric about the origin?

(A polygon is centrally symmetric about the origin if a 180-degree rotation about the origin sends thepolygon to itself.)

4. [10] Find all functions f : R+ → R+ such that

f(x + f(y + xy)) = (y + 1)f(x + 1)− 1

for all x, y ∈ R+.

(R+ denotes the set of positive real numbers.)

5. [11] Let G be an undirected simple graph. Let f(G) be the number of ways to orient all of the edgesof G in one of the two possible directions so that the resulting directed graph has no directed cycles.Show that f(G) is a multiple of 3 if and only if G has a cycle of odd length.

HMIC 2018April 09, 2018

HMIC 2018

1. [6] Let m > 1 be a fixed positive integer. For a nonempty string of base-ten digits S, let c(S) be thenumber of ways to split S into contiguous nonempty strings of digits such that the base-ten numberrepresented by each string is divisible by m. These strings are allowed to have leading zeroes.

In terms of m, what are the possible values that c(S) can take?

For example, if m = 2, then c(1234) = 2 as the splits 1234 and 12|34 are valid, while the other sixsplits are invalid.

Proposed by: Kevin Sun

Answer: 0 and 2n for all nonnegative integer n

First, we note that c(1) = 0 and c(00 . . . 0) = 2n→1 if there are n zeroes in the string. Now we showthat these are the only possibilities. Note that a split can be added if and only if the string before thissplit (ignoring all other splits) represent a multiple of m (if there is a split before it, then removing thedigits before this preceding split is equivalent to subtracting the removed number times a power of 10,which will also be a multiple of m, so the remaining number between the two splits remain a multipleof m). Thus, whether we can add a split or not depends only on the string itself and no other splits,so c(S) is 0 if the number is not divisible by m and a power of two otherwise.

2. [7] Consider a finite set of points T ∈ Rn contained in the n-dimensional unit ball centered at theorigin, and let X be the convex hull of T . Prove that for all positive integers k and all points x ∈ X,there exist points t1, t2, . . . , tk ∈ T , not necessarily distinct, such that their centroid

t1 + t2 + · · ·+ tkk

has Euclidean distance at most1√k

from x.

(The n-dimensional unit ball centered at the origin is the set of points in Rn with Euclidean distanceat most 1 from the origin. The convex hull of a set of points T ∈ Rn is the smallest set of points Xcontaining T such that each line segment between two points in X lies completely inside X.)

Proposed by: Henrik Boecken

By the definition of convex hull, we can write x =∑m

i=1 λizi, where each zi ∈ T , each λi ≥ 0 and∑mi=1 λi = 1. Consider then a random variable Z that takes on value zi with probability λi. We have

E[Z] = x. Let Z̄ = 1k

∑ki=1 Zi, where each Zi is an independent copy of Z. Then we wish to compute

Var[Z̄] =1

k2

k∑i=1

Var[Zi]

Finally, we haveVar[Zi] = E[||Zi 〉 x||2] = E[||Zi||2]〉 x2 ≤ E[||Zi||2] ≤ 1

The second equality follows from the identity Var[X] = E[X2]〉 E[X]2. Now, we know that

E[||x〉 1

k

k∑i=1

Zi||2] = Var[Z̄] ≤ 1

k

Thus, there must exist some realization of xi of the Zi such that

||x〉 1

k

k∑i=1

xi||2 ≤1

k

and we are done.

3. [8] A polygon in the plane (with no self-intersections) is called equitable if every line passing throughthe origin divides the polygon into two (possibly disconnected) regions of equal area.

Does there exist an equitable polygon which is not centrally symmetric about the origin?

(A polygon is centrally symmetric about the origin if a 180-degree rotation about the origin sends thepolygon to itself.)

Proposed by: Kevin Sun

Consider the polygon with vertices

A(1, 0), B(0, 1), C(0, 5), D(5, 0), E(7, 0),

F (0,〉7), G(〉7, 0),H(〉√

84, 0), I(0,√

84), J(0, 6),

K(〉6, 0), L(〉5, 0),M(0,〉5), N(0,〉1).

Notice that its intersection with each of the four quadrants are all trapezoids with angles 45◦, 45◦, 135◦, 135◦,

and the trapezoids in quadrant 1 and 3 both have area 52→122 = 72→52

2 = 12, and the trapezoids in

quadrant 2 and 4 both have area 72→122 = 84→62

2 = 24, so by similar triangles, we can show that thispolygon is indeed equitable. It is also apparent that this polygon is not centrally symmetric.

4. [10] Find all functions f : R+ → R+ such that

f(x+ f(y + xy)) = (y + 1)f(x+ 1)〉 1

for all x, y ∈ R+.

(R+ denotes the set of positive real numbers.)

Proposed by: Ashwin Sah

Let P (x, y) denote the assertion that

f(x+ f(y + xy)) = (y + 1)f(x+ 1)〉 1.

Claim 1. f is injective. Proof: If f(a) = f(b) then P(x, a

x+1

), P(x, b

x+1

)yields a = b, since

f(x+ 1) ∈ R+ so in particular is nonzero. ◦ Now P(x, 1

f(x+1)

)yields

f

(x+ f

(x+ 1

f(x+ 1)

))= f(x+ 1)

hence by injectivity

x+ f

(x+ 1

f(x+ 1)

)= x+ 1

so that

f

(x+ 1

f(x+ 1)

)= 1.

By injectivity, this equals some constant c, so that

x+ 1

f(x+ 1)= c

for all x ∈ R+. Now letting x, y > 1 in P (x, y) automatically yields

x

c+y + xy

c2= (y + 1)

(x+ 1

c

)〉 1,

which immediately yields c = 1 if we take x, y large. Finally, we have f(x+1) = x+18x ∈ R+. Finally,

P(x, y

x+1

)yields

f(x+ f(y)) =

(y

x+ 1+ 1

)f(x+ 1)〉 1 = x+ y,

so that fixing y and letting x > 1 yields

x+ f(y) = x+ y,

so that f(y) = y. This was for arbitrary positive y, so that f(x) = x8x ∈ R+, which clearly works.

5. [11] Let G be an undirected simple graph. Let f(G) be the number of ways to orient all of the edgesof G in one of the two possible directions so that the resulting directed graph has no directed cycles.Show that f(G) is a multiple of 3 if and only if G has a cycle of odd length.

Proposed by: Yang Liu

Let fG(q) be the number of ways to color G with q colors. This is the chromatic polynomial of G,and turns out to be polynomial in q. Indeed, choose an edge e and let Gne be the graph G with eremoved and let G/e be graph G with the vertices on either side of e merged together (multiple edgesare removed). It is not hard to see that

fG(q) = fGne(q)〉 fG/e(q)

and that for an empty graph En with n vertices, we have fEn(q) = qn. Induction finishes.

Now we can also show that the desired number of ways to direct G to avoid directed cycles is cG =(〉1)nfG(〉1) where n is the number of vertices of G. The way to do it is to show that

cG = cGne + cG/e,

and cEn = 1. Thus (〉1)ncG will satisfy the desired recurrence, after multiplying the above by (〉1)n,and satisfies the correct initial conditions.

Now all we need to know is that fG(q) has integer coefficients, so that (〉1)ncG ≡ fG(〉1) ≡ fG(2)(mod 3), and fG(2) = 0 when G is not bipartite and is 2c(G) where c(G) is the number of connectedcomponents of G otherwise. The result follows.

HMMT February 2019February 16, 2019

Team Round

1. [20] Let ABCD be a parallelogram. Points X and Y lie on segments AB and AD respectively, andAC intersects XY at point Z. Prove that

AB

AX+AD

AY=AC

AZ.

2. [20] Let N = {1, 2, 3, . . .} be the set of all positive integers, and let f be a bijection from N to N. Mustthere exist some positive integer n such that (f(1), f(2), . . . , f(n)) is a permutation of (1, 2, . . . , n)?

3. [25] For any angle 0 < θ < π/2, show that

0 < sin θ + cos θ + tan θ + cot θ − sec θ − csc θ < 1.

4. [35] Find all positive integers n for which there do not exist n consecutive composite positive integersless than n!.

5. [40] Find all positive integers n such that the unit segments of an n × n grid of unit squares can bepartitioned into groups of three such that the segments of each group share a common vertex.

6. [45] Scalene triangle ABC satisfies ∠A = 60◦. Let the circumcenter of ABC be O, the orthocenter beH, and the incenter be I. Let D,T be the points where line BC intersects the internal and externalangle bisectors of ∠A, respectively. Choose point X on the circumcircle of 4IHO such that HX ‖ AI.Prove that OD ⊥ TX.

7. [50] A convex polygon on the plane is called wide if the projection of the polygon onto any line in thesame plane is a segment with length at least 1. Prove that a circle of radius 1

3 can be placed completelyinside any wide polygon.

8. [50] Can the set of lattice points {(x, y)|x, y ∈ Z, 1 ≤ x, y ≤ 252, x 6= y} be colored using 10 distinctcolors such that for all a 6= b, b 6= c, the colors of (a, b) and (b, c) are distinct?

9. [55] Let p > 2 be a prime number. Fp[x] is defined as the set of all polynomials in x with coefficientsin Fp (the integers modulo p with usual addition and subtraction), so that two polynomials are equalif and only if the coefficients of xk are equal in Fp for each nonnegative integer k. For example,(x+ 2)(2x+ 3) = 2x2 + 2x+ 1 in F5[x] because the corresponding coefficients are equal modulo 5.

Let f, g ∈ Fp[x]. The pair (f, g) is called compositional if

f(g(x)) ≡ xp2

− x

in Fp[x]. Find, with proof, the number of compositional pairs (in terms of p).

10. [60] Prove that for all positive integers n, all complex roots r of the polynomial

P (x) = (2n)x2n + (2n− 1)x2n→1 + · · ·+ (n+ 1)xn+1 + nxn + (n+ 1)xn→1 + · · ·+ (2n− 1)x+ 2n

lie on the unit circle (i.e. |r| = 1).

HMMT February 2019February 16, 2019

Team Round

1. [20] Let ABCD be a parallelogram. Points X and Y lie on segments AB and AD respectively, andAC intersects XY at point Z. Prove that

AB

AX+AD

AY=AC

AZ.

Proposed by: Yuan Yao

Solution 1. (Similar Triangles)

Let X ′ and Y ′ lie on segments AB and AD respectively such that ZX ′ ‖ AD and ZY ′ ‖ AB. We notethat triangles AXY and Y ′Y Z are similar, and that triangles AY ′Z and ADC are similar. Thus, wehave

AC

AZ=AD

AY ′and

AY ′

AY=XZ

XY.

This means thatAD

AY=AD

AY ′· AY

′

AY=XZ

XY· ACAZ

,

and similarly,AB

AX=ZY

XY· ACAZ

.

Therefore we haveAB

AX+AD

AY=

(XZ

XY+ZY

XY

)· ACAZ

=AC

AZ,

as desired.

Solution 2. (Affine Transformations)

We recall that affine transformations preserve both parallel lines and ratios between distances ofcollinear points. It thus suffices to show the desired result when ABCD is a square. This can bedone in a variety of ways. For instance, a coordinate bash can be applied by setting A to be the origin.Let the length of the square be 1 and set X and Y as (a, 0) and (0, b) respectively, so the line XY hasequation bx+ ay = ab. Then, we note that Z is the point ( ab

a+b ,aba+b ), so

AB

AX+AD

AY=

1

a+

1

b=a+ b

ab=AC

AZ.

2. [20] Let N = {1, 2, 3, . . .} be the set of all positive integers, and let f be a bijection from N to N. Mustthere exist some positive integer n such that (f(1), f(2), . . . , f(n)) is a permutation of (1, 2, . . . , n)?

Proposed by: Michael Tang

Answer: No

Consider the bijection f defined by

(f(1), f(2), f(3), f(4), . . .) = (2, 4, 6, 1, 8, 3, 10, 5, 12, . . .),

which alternates between even and odd numbers after the second entry. (More formally, we definef(n) = 2n for n = 1, 2, f(n) = n+ 3 for odd n ≥ 3 and f(n) = n− 3 for even n ≥ 4.) No such n canexist for this f as the largest number among f(1), f(2), . . . , f(n) is more than n for all n: for k ≥ 2,the maximum of the first 2k − 1 or 2k values is achieved by f(2k − 1) = 2k + 2. (Checking n = 1 andn = 2 is trivial.)

3. [25] For any angle 0 < θ < π/2, show that

0 < sin θ + cos θ + tan θ + cot θ − sec θ − csc θ < 1.

Proposed by: Yuan Yao

We use the following geometric construction, which follows from the geometric definition of the trigono-metric functions: Let Z be a point on the unit circle in the coordinate plane with origin O. Let X1, Y1 bethe projections of Z onto the x- and y-axis respectively, and let X2, Y2 lie on x- and y-axis respectivelysuch that X2Y2 is tangent to the unit circle at Z. Then we have

OZ = X1Y1 = 1, X1Z = sin θ, Y1Z = cos θ,X2Z = tan θ, Y2Z = cot θ,OX2 = sec θ,OY2 = csc θ.

It then suffices to show that 0 < X2Y2 − X1X2 − Y1Y2 < 1 = X1Y1. The left inequality is truebecause X1X2 and Y1Y2 are the projections of ZX2 and ZY2 onto x- and y-axis respectively. The rightinequality is true because X1X2 +X1Y1 + Y1Y2 > X2Y2 by triangle inequality. Therefore we are done.

4. [35] Find all positive integers n for which there do not exist n consecutive composite positive integersless than n!.

Proposed by: Brian Reinhart

Answer: 1, 2, 3, 4

Solution 1. First, note that clearly there are no composite positive integers less than 2!, and no 3consecutive composite positive integers less than 3!. The only composite integers less than 4! are

4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22,

and it is easy to see that there are no 4 consecutive composite positive integers among them. Therefore,all n ≤ 4 works.

Define M = lcm(1, 2, . . . , n + 1). To see that there are no other such positive integers, we first showthat for all n ≥ 5, n! > M . Let k = blog2(n+ 1)c. Note that v2(M) = k, while

v2((n+ 1)!) =

k∑i=1

⌊n+ 1

2i

⌋≥

k∑i=1

(n+ 1

2i− 1

)=

(n+ 1− n+ 1

2k

)− k ≥ (n+ 1− 2)− k = n− k− 1.

This means that at least (n− k− 1)− k = n− 2k− 1 powers of 2 are lost when going from (n+ 1)! toM . Since M |(n+ 1)!, when n− 2k − 1 ≥ k + 1 ⇐⇒ n ≥ 3k + 2, we have

M ≤ (n+ 1)!

2k+1≤ (n+ 1)!

2(n+ 1)< n!,

as desired. Since n ≥ 2k − 1, we can rule out all k such that 2k ≥ 3k + 3, which happens when k ≥ 4or n ≥ 15. Moreover, when k = 3, we may also rule out all n ≥ 3k + 2 = 11.

We thus need only check values of n between 5 and 10:

n = 5: n! = 120,M = 60;

n = 6: n! = 720,M = 420;

n = 7: n! = 5040,M = 840;

n ∈ {8, 9, 10}: n! ≥ 40320,M ≤ 27720.

In all cases, n! > M , as desired.

To finish, note that M − 2,M − 3, . . . ,M − (n + 1) are all composite (divisible by 2, 3, . . . , n + 1respectively), which gives the desired n consecutive numbers. Therefore, all integers n ≥ 5 do notsatisfy the problem condition, and we are done.

Solution 2. Here is a different way to show that constructions exist for n ≥ 5. Note that when n+ 1is not prime, the numbers n!− 2, n!− 3, . . . , n!− (n+ 1) are all composite (the first n− 1 are clearly

composite, the last one is composite because n + 1 | n! and n! > 2(n + 1)). Otherwise, if n = p − 1for prime p ≥ 7, then the numbers (n − 1)!, (n − 1)! − 1, (n − 1)! − 2, . . . , (n − 1)! − (n − 1) are allcomposite (the first one and the last n− 2 are clearly composite since (n− 1)! > 2(n− 1), the secondone is composite since p | (p− 2)!− 1 = (n− 1)!− 1 by Wilson’s theorem).

5. [40] Find all positive integers n such that the unit segments of an n × n grid of unit squares can bepartitioned into groups of three such that the segments of each group share a common vertex.

Proposed by: Yuan Yao

Answer: n ≡ 0, 2 (mod 6)

We first prove that n ≡ 0, 2 (mod 6) is necessary for there to be such a partitioning. We break thisdown into proving that n has to be even and that n ≡ 0, 2 (mod 3).

The only way a segment on a side of the square can be part of such a T-shape is as one of the twoconsecutive segments along the longer side of the T-shape, so they must come in pairs and therefore,the length of each side has to be even. On the other hand, the total number of segments, which is2n(n+ 1), has to be a multiple of three as each T-shape consists of three segments, hence either n orn+ 1 is a multiple of 3, implying that n ≡ 0, 2 (mod 3).

We can then show that these two conditions is sufficient by showing that n = 2 and n = 6 works andn = k+ 6 works whenever n = k works. The construction for n = 2 is simple; just put a T-shape withthe longer side on each of the four sides. For n = 6 and to go from n = k to n = k + 6, consider thefollowing diagram:

There are two main parts – the cycle of stacks of T’s in all four orientation (see the red, blue, yellow,and green stacks), and the border (seen here by the cyan, brown, and black T-shapes). The case n = 6can be considered as a special case where the middle square is a single point.

6. [45] Scalene triangle ABC satisfies ∠A = 60◦. Let the circumcenter of ABC be O, the orthocenter beH, and the incenter be I. Let D,T be the points where line BC intersects the internal and external

angle bisectors of ∠A, respectively. Choose point X on the circumcircle of 4IHO such that HX ‖ AI.Prove that OD ⊥ TX.Proposed by: Wanlin Li

Let IA denote the A-excenter. Because ∠A = 60◦, AI is the perpendicular bisector of OH andB,H,O,C all lie on the circle with diameter IIA. We are given that X is on this circle as well, andsince HI = OI, XIOIA is also an isosceles trapezoid. But IIA is a diameter, so this means X mustbe diametrically opposite O on (BOC) and is actually the intersection of the tangents to (ABC) fromB and C.

Now (T,D;B,C) = −1, so T is on the polar of D with respect to (ABC). BC is the polar of X andD lies on BC, so X must also lie on the polar of D. Therefore TX is the polar of D with respect to(ABC), and OD ⊥ TX as desired.

7. [50] A convex polygon on the plane is called wide if the projection of the polygon onto any line in thesame plane is a segment with length at least 1. Prove that a circle of radius 1

3 can be placed completelyinside any wide polygon.

Proposed by: Shengtong Zhang

Solution 1.

Lemma. for any polygon including its boundary, there exists a largest circle contained inside it.

Proof. Its easy to see that for any circle inside the polygon, it can be increased in size until it is tangentto at least three sides of the polygon. Then for any three sides of the polygon, there is only one circletangent to all three, so there are only finitely many possibilities. Therefore there exists a largest one.

Alternatively, one can show that the space of valid (x, y, r) such that the circle with center (x, y) andradius r is compact, e.g. by showing the complement is open and that the complement is open. Thenthe map (x, y, r)→ r is continuous and therefore has a maximum.

Now, take the largest circle. It clearly must be tangent to three sides. If the circle lies inside thetriangle made by the three lines, we can expand the polygon to that triangle and solve it for thetriangle instead. Otherwise, we have the following diagram:

Here the circle is an excircle of the triangle ABC made by the lines AB,AD, and BE. (Note that ADand AB dont have to be consecutive sides of the polygon, but the ones in between dont really matter.)

Then since the circle is an excircle, we can consider a homothety at C with power 1 + ε, which sendsthe circle to a slightly larger circle which does not touch line AB. If this homothety causes the circleto leave the polygon for small enough ε, it must be because the circle was initially tangent to anotherline `, for which it would be an incircle of the triangle made by ` and lines AD, BE, bringing us backto the first case.

Thus we can reduce to a case where we have a triangle with each height at least 1, and we want toshow the inradius is at least 1/3. Let K be the area of the triangle, so the heights 2K

a ,2Kb ,

2Kc are all

at least 1. Then the inradius r satisfies

r =K

s=

2K

a+ b+ c≥ 2K

2K + 2K + 2K=

1

3,

as desired.

Solution 2. Consider the center of mass G. We will use the notion of support lines for convex shapes.(Support lines are the lines that touches the shape but does not cut through it.) If a circle centered atG with radius 1/3 cannot be contained inside the polygon, then there exist a point P on the boundarythat GP < 1/3. Let `1 be the support line passing through p, `2 be the line parallel to `1 and passingthrough G, and `3 be the other support line that is parallel to `1, touching the polygon at P ′. Suppose`2 intersects the polygon at A and B. Extend P ′A and P ′B, intersecting `1 at A′ and B′. Then, if weconsider the two parts of the polygon that `2 divides the polygon into, we have:

• the part of the polygon that contains P ′ contains the triangle P ′AB;

• the part of the polygon that contains P is contained in the quadrilateral AA′B′B.

Then we conclude that the center of mass G′ of the triangle P ′A′B′ lies between `2 and `1, which byassumption is less than 1

3 away from `1. However, because the height from P ′ to `1 is at least 1, thedistance from G′ to `1 is at least 1

3 , so we have a contradiction. Therefore no such P exists and thecircle can be placed inside the polygon.

8. [50] Can the set of lattice points {(x, y)|x, y ∈ Z, 1 ≤ x, y ≤ 252, x 6= y} be colored using 10 distinctcolors such that for all a 6= b, b 6= c, the colors of (a, b) and (b, c) are distinct?

Proposed by: Franklyn Wang

Answer: Yes

Associate to each number from 1 to 252 a distinct 5-element subset of S = {1, 2, . . . , 10}. Then assignto (a, b) an element of S that is in the subset associated to a but not in that associated to b. It’s notdifficult to see that this numerical assignment is a valid coloring: the color assigned to (a, b) is not inb, while the color assigned to (b, c) is in b, so they must be distinct.

9. [55] Let p > 2 be a prime number. Fp[x] is defined as the set of all polynomials in x with coefficientsin Fp (the integers modulo p with usual addition and subtraction), so that two polynomials are equalif and only if the coefficients of xk are equal in Fp for each nonnegative integer k. For example,(x+ 2)(2x+ 3) = 2x2 + 2x+ 1 in F5[x] because the corresponding coefficients are equal modulo 5.

Let f, g ∈ Fp[x]. The pair (f, g) is called compositional if

f(g(x)) ≡ xp2

− x

in Fp[x]. Find, with proof, the number of compositional pairs (in terms of p).

Proposed by: Ashwin Sah

Answer: 4p(p− 1)

Solution 1. First, notice that (deg f)(deg g) = p2 and both polynomials are clearly nonconstant.Therefore there are three possibilities for the ordered pair (deg f, deg g), which are (1, p2), (p2, 1), and(p, p).

In the subsequent parts of the solution, equalities are modulo p. If f(x) = ax + b, a 6= 0 is linear,

then it is invertible so then g is uniquely determined as g(x) = f−1(f(g(x))) = xp2−x−ba . Similarly,

if g(x) = cx + d, c 6= 0 (mod p) is linear then f is uniquely determined as f(x) = f(g(g−1(x))) =x−dc

)p2 − x−dc ). In each case there are p(p− 1) compositional pairs.

The last case is deg f = deg g = p. We take the derivative of both sides (we use the formal derivativeDxf(x) =

∑n≥1 nfnx

n−1, which satisfies the usual chain and product rules but can be used onarbitrary polynomials, including those in Fp[x]).

Thusf ′(g(x))g′(x) = p2xp

2−1 − 1 = −1,

using that p = 0 in Fp. Now g′(x) and f ′(g(x)) must both be constant polynomials. Since g isnonconstant, this means that f ′(x) is also a constant polynomial. We must be careful here, as unlikein R, nonlinear polynomials can have constant derivatives. From the formula of derivative, we see thath′(x) = 0 as a polynomial exactly when h(x) is a linear combination of 1, xp, x2p, . . . (remember thatp = 0). Thus f ′, g′ both being constant and f, g being of degree p tells us

f(x) = axp + bx+ c, g(x) = dxp + ex+ f

where a, b, c, d, e, f are some elements of Fp. Now we must have

a(dxp + ex+ f)p + b(dxp + ex+ f) + c = xp2

− x

over Fp[x]. We use the fact that (x+y)p = xp+yp as polynomials in Fp, since the binomial coefficientspj

)≡ 0 (mod p) for 1 ≤ j ≤ p− 1. This implies (x+ y+ z)p = xp + yp + zp. Therefore we can expand

the previous equation as

a(dpxp2

+ epxp + fp) + b(dxp + ex+ f) + c = xp2

− x.

Equating coefficients, we see that

adp = 1,

aep + bd = 0,

be = −1,

afp + bf + c = 0.

The first and third equations imply that a, d, b, e are nonzero (mod p) and a = d−p, b = −e−1. Thenaep + bd = 0 gives

d−pep − e−1d = 0

or ep+1 = dp+1. Recalling that ep−1 = dp−1 = 1 in (mod p), this tells us d2 = e2 so d = ±e.Furthermore, any choice of such (d, e) give unique (a, b) which satisfy the first three equations. Finally,once we have determined a, b, d, e, any choice of f gives a unique valid choice of c.

Thus we have p − 1 choices for d, two choices for e after choosing d (n.b. for p = 2 there is only onechoice for e, so the assumption p > 2 is used here), and then p choices for f , for a total of 2p(p − 1)compositional pairs in this case.

Finally, adding the number of compositional pairs from all three cases, we obtain 4p(p−1) compositionalpairs in total.

Solution 2. The key step is obtaining

f(x) = axp + bx+ c, g(x) = dxp + ex+ f

in the case where deg f = deg g = p. We present an alternative method of obtaining this, with the restof the solution being the same as the first solution. Let

f(x) = fpxp + fp−1x

p−1 + · · ·+ f0

g(x) = gpxp + gp−1x

p−1 + · · ·+ g0

where fp, gp are nonzero. Like before, we have g(x)p = g(xp) in Fp[x], so

xp2

− x = fpg(xp) + fp−1g(x)p−1 + · · ·+ f0.

Consider the maximal k < p for which fk 6= 0. (It is not hard to see that in fact k ≥ 1, as fpg(xp) + f0cannot be xp

2 − x.) First assume that k > 1. We look at the xkp−1 coefficient, which is affected onlyby the fkg(x)k term. By expanding, the coefficient is kfkg

k−1p gp−1. Therefore gp−1 = 0. Then we look

at the xkp−2 coefficient, then the xkp−3 coefficient, etc. down to the xkp−p+1 coefficient to concludethat gp−1 = gp−2 = · · · = g1 = 0. However, then the x coefficient of f(g(x)) is zero, contradiction.

Therefore we must have k = 1, so f is of the form axp + bx + c. Using the same method as we usedwhen k > 1, we get gp−1 = gp−2 = · · · g2 = 0, though the xkp−p+1 coefficient is now the x coefficientwhich we want to be nonzero. Hence we do not obtain g1 = 0 anymore and we find that g is of theform dxp + ex+ f .

10. [60] Prove that for all positive integers n, all complex roots r of the polynomial

P (x) = (2n)x2n + (2n− 1)x2n−1 + · · ·+ (n+ 1)xn+1 + nxn + (n+ 1)xn−1 + · · ·+ (2n− 1)x+ 2n

lie on the unit circle (i.e. |r| = 1).

Proposed by: Faraz Masroor

Note that neither 0 nor 1 are roots of the polynomial. Consider the function

Q(x) = P (x)/xn = (2n)xn+(2n)x−n+(2n−1)xn−1 +(2n−1)x−n+1 + · · ·+(n+1)x1 +(n+1)x−1 +n.

All 2n of the complex roots of P (x) will be roots of Q(x).

If |x| = 1, then x = eiθ, and

Q(x) = (2n)(xn + x−n) + (2n− 1)(xn−1 + x−n+1) + · · ·+ (n+ 1)(x+ x−1) + n

= (2n)(einθ + e−inθ) + (2n− 1)(ei(n−1)θ + e−i(n−1)θ) + · · ·+ (n+ 1)(eiθ + e−iθ) + n

= (2n)(2 cos(nθ)) + (2n− 1)(2 cos((n− 1)θ)) + · · ·+ (n+ 1)(cos(θ)) + n,

which is real. Thus on the unit circle, we have Q(x) is real, and we want to show it has 2n roots there.

Rewrite

P (x) = (2n)x2n + (2n− 1)x2n−1 + · · ·+ (n+ 1)xn+1 + nxn + (n+ 1)xn−1 + · · ·+ 2n

= (2n)(x2n + x2n−1 + · · ·+ 1)

− (x2n−1 + 2x2n−2 + · · ·+ (n− 1)xn + nxn−1 + (n− 1)xn−2 + · · ·+ 2x2 + x)

= 2nx2n+1 − 1

x− 1− x(x2n−2 + 2x2n−3 + · · ·+ (n− 1)xn + nxn−1 + (n− 1)xn−2 + · · ·+ 2x+ 1)

= 2nx2n+1 − 1

x− 1− x(xn−1 + xn−2 + · · ·+ x+ 1)2

= 2nx2n+1 − 1

x− 1− x

(xn − 1

x− 1

)2

,

and thus

Q(x) =2n

xnx2n+1 − 1

x− 1− x

xn

(xn − 1

x− 1

)2

.

Consider the roots of unity rj = ei2π2n j , for j = 0 to 2n− 1. There are 2n such roots of unity: they all

have r2nj = 1, and they alternate between those which satisfy rnj = 1 or rnj = −1. At those x = rj , if

rnj = 1 but x 6= 1, then

Q(x) =2n

xnx2n+1 − 1

x− 1− x

xn

(xn − 1

x− 1

)2

= 2nx1 − 1

x− 1− x

(1− 1

x− 1

)2

= 2n > 0.

At x = 1, we can easily see Q(1) > 0.

If rnj = −1, then

Q(x) =2n

xnx2n+1 − 1

x− 1− x

xn

(xn − 1

x− 1

)2

= −2nx1 − 1

x− 1+ x

(−1− 1

x− 1

)2

= −2n+4x

(x− 1)2

= −2n+4

x− 2 + 1/x

= −2n+4

2 cos( 2π2nj)− 2

< −2n− 4 < 0

since the denominator of this second term is strictly negative (j 6= 0).

Thus at each of the 2n-roots of unity, Q(x) alternates in sign, and because Q(x) is real and continuouson the unit circle, it has at least one root between every pair of consecutive roots of unity. Since thereare 2n of these pairs, and we know that Q(x) has exactly 2n roots (by the Fundamental Theorem ofAlgebra), we have found all of Q’s roots, and therefore those of P .

HMMT February 2019February 16, 2019

Algebra and Number Theory

1. What is the smallest positive integer that cannot be written as the sum of two nonnegative palin-dromic integers? (An integer is palindromic if the sequence of decimal digits are the same when readbackwards.)

2. Let N = 2(22) and x be a real number such that N(NN) = 2(2x). Find x.

3. Let x and y be positive real numbers. Define a = 1+ xy and b = 1+ y

x . If a2 +b2 = 15, compute a3 +b3.

4. Let N be the set of positive integers, and let f : N→ N be a function satisfying

• f(1) = 1;

• for n ∈ N, f(2n) = 2f(n) and f(2n + 1) = 2f(n)〉 1.

Determine the sum of all positive integer solutions to f(x) = 19 that do not exceed 2019.

5. Let a1, a2, . . . be an arithmetic sequence and b1, b2, . . . be a geometric sequence. Suppose that a1b1 = 20,a2b2 = 19, and a3b3 = 14. Find the greatest possible value of a4b4.

6. For positive reals p and q, define the remainder when p is divided by q as the smallest nonnegativereal r such that p−r

q is an integer. For an ordered pair (a, b) of positive integers, let r1 and r2 be the

remainder when a√

2 + b√

3 is divided by√

2 and√

3 respectively. Find the number of pairs (a, b) suchthat a, b ≤ 20 and r1 + r2 =

√2.

7. Find the value of∞∑a=1

∞∑b=1

∞∑c=1

ab(3a + c)

4a+b+c(a + b)(b + c)(c + a).

8. There is a unique function f : N→ R such that f(1) > 0 and such that∑d|n

f(d)f(nd

)= 1

for all n ≥ 1. What is f(20182019)?

9. Tessa the hyper-ant has a 2019-dimensional hypercube. For a real number k, she calls a placement ofnonzero real numbers on the 22019 vertices of the hypercube k-harmonic if for any vertex, the sum ofall 2019 numbers that are edge-adjacent to this vertex is equal to k times the number on this vertex.Let S be the set of all possible values of k such that there exists a k-harmonic placement. Find

∑k∈S|k|.

10. The sequence of integers {ai}∞i=0 satisfies a0 = 3, a1 = 4, and

an+2 = an+1an +

⌈√a2n+1 〉 1

√a2n 〉 1

⌉for n ≥ 0. Evaluate the sum

∞∑n=0

(an+3

an+2〉 an+2

an+

an+1

an+3〉 an

an+1

).

HMMT February 2019February 16, 2019

Algebra and Number Theory

1. What is the smallest positive integer that cannot be written as the sum of two nonnegative palin-dromic integers? (An integer is palindromic if the sequence of decimal digits are the same when readbackwards.)

Proposed by: Yuan Yao

Answer: 21

We need to first prove that every positive integer N less than 21 can be written as sum of twononnegative palindromic integers. If N is in the interval [1, 9], then it can be written as 0 + N . If Nis in the interval [10, 18], it can be written as 9 + (N − 9). In addition, 19 and 20 can be written as11 + 8 and 11 + 9, respectively.

Second, we need to show that 21 cannot be expressed in such a way. Lets suppose 21 = a + b witha ≤ b. It follows that b has to be at least 11. Since b ≤ 21, the only way for b to be palindromic isthat b = 11. However, this leads to a = 21 − b = 10, which is not a palindrome. Therefore, 21 is thesmallest number that satisfy the problem condition.

2. Let N = 2(22) and x be a real number such that N(NN) = 2(2x). Find x.

Proposed by: Yuan Yao

Answer: 66

We compute

N(NN) = 161616

= 24·24·24

= 2226+2

= 2266

,

so x = 66.

3. Let x and y be positive real numbers. Define a = 1+ xy and b = 1+ y

x . If a2 +b2 = 15, compute a3 +b3.

Proposed by: Michael Tang

Answer: 50

Note that a− 1 = xy and b− 1 = y

x are reciprocals. That is,

(a− 1)(b− 1) = 1 =⇒ ab− a− b+ 1 = 1 =⇒ ab = a+ b.

Let t = ab = a+ b. Then we can write

a2 + b2 = (a+ b)2 − 2ab = t2 − 2t,

so t2 − 2t = 15, which factors as (t − 5)(t + 3) = 0. Since a, b > 0, we must have t = 5. Then, wecompute

a3 + b3 = (a+ b)3 − 3ab(a+ b) = 53 − 3 · 52 = 50.

4. Let N be the set of positive integers, and let f : N→ N be a function satisfying

• f(1) = 1;

• for n ∈ N, f(2n) = 2f(n) and f(2n+ 1) = 2f(n)− 1.

Determine the sum of all positive integer solutions to f(x) = 19 that do not exceed 2019.

Proposed by: Yuan Yao

Answer: 1889

For n = 2a0 +2a1 + · · ·+2ak where a0 > a1 > · · · > ak, we can show that f(n) = 2a0−2a1−· · ·−2ak =2a0+1 − n by induction: the base case f(1) = 1 clearly holds; for the inductive step, when n is even

we note that f(n) = 2f(n2 ) = 2(2a0 − n

2 ) = 2a0+1 − n as desired, and when n is odd we also havef(n) = 2f(n−1

2 )− 1 = 2(2a0 − n−12 )− 1 = 2a0+1 − n, again as desired.

Since 19 = f(n) ≤ 2a0 ≤ n, we have a0 ≥ 5 and n = 2a0+1 − 19 ≤ 2019 gives a0 ≤ 9. So the answer is9∑

a=5(2a+1 − 19) = (211 − 26)− 19 · 5 = 1889.

5. Let a1, a2, . . . be an arithmetic sequence and b1, b2, . . . be a geometric sequence. Suppose that a1b1 = 20,a2b2 = 19, and a3b3 = 14. Find the greatest possible value of a4b4.

Proposed by: Michael Tang

Answer: 374

We present two solutions: the first more algebraic and computational, the second more conceptual.

Solution 1. Let {an} have common difference d and {bn} have common ratio d; for brevity, let a1 = aand b1 = b. Then we have the equations ab = 20, (a+ d)br = 19, and (a+ 2d)br2 = 14, and we wantto maximize (a+ 3d)br3.

The equation (a + d)br = 19 expands as abr + dbr = 19, or 20r + bdr = 19 since ab = 20. Similarly,(20 + 2bd)r2 = 14, or 10r2 + bdr2 = 7. Multiplying the first equation by r and subtracting the second,we get

10r2 = 19r − 7 =⇒ (5r − 7)(2r − 1) = 0,

so either r = 75 or r = 1

2 .

For each value of r, we have bd = 19−20rr = 19

r − 20, so

(a+ 3d)br3 = (20 + 3bd)r3 =

(57

r− 40

)r3 = r2(57− 40r).

The greater value of this expression is 374 , achieved when r = 1

2 .

Solution 2. The key is to find a (linear) recurrence relation that the sequence cn = anbn satisfies.Some knowledge of theory helps here: cn is of the form snrn + trn for some constants r, s, t, so {cn}satisfies a linear recurrence relation with characteristic polynomial (x− r)2 = x2 − 2rx+ r2. That is,

cn = 2rcn−1 − r2cn−2

for some constant r.

Taking n = 3, we get 14 = 2r · 19 − r2 · 20, which factors as (5r − 7)(2r − 1) = 0, so either r = 75 or

r = 12 . Then

c4 = 2rc3 − r2c2 = 28r − 19r2.

This expression is maximized at r = 1419 , and strictly decreases on either side. Since 1

2 is closer to 1419

than 75 , we should choose r = 1

2 , giving the answer c4 = 14− 194 = 37

4 .

6. For positive reals p and q, define the remainder when p is divided by q as the smallest nonnegativereal r such that p−r

q is an integer. For an ordered pair (a, b) of positive integers, let r1 and r2 be the

remainder when a√

2 + b√

3 is divided by√

2 and√

3 respectively. Find the number of pairs (a, b) suchthat a, b ≤ 20 and r1 + r2 =

√2.

Proposed by: Yuan Yao

Answer: 16

The remainder when we divide a√

2 + b√

3 by√

2 is defined to be the smallest non-negative real r1such that a

√2+b√3−r1√

2is integral. As x√

2is integral iff x is an integral multiple of

√2, it follows that

r1 = b√

3− c√

2, for some integer c. Furthermore given any real r such that a√2+b√3−r√

2is integral, we

may add or subtract√

2 to r and the fraction remains an integer. Thus, the smallest non-negative realr1 such that the fraction is an integer must satisfy 0 ≤ r1 <

√2.

Similarly, we find r2 = a√

2− d√

3 for some integer d and 0 ≤ r2 <√

3. Since r1 + r2 =√

2, then

(a− c)√

2 + (b− d)√

3 =√

2 ⇐⇒ a− c = 1 and b− d = 0.

Finally, substituting in c = a− 1 and d = b plugging back into our bounds for r1 and r2, we get{0 ≤ b

√3− (a− 1)

√2 <√

2

0 ≤ a√

2− b√

3 <√

3

or 8>><>>:(a− 1)

√2 ≤ b

√3

b√

3 < a√

2

b√

3 ≤ a√

2

a√

2 < (b+ 1)√

3

Note that b√

3 < a√

2 =⇒ b√

3 ≤ a√

2 and

(a− 1)√

2 ≤ b√

3 =⇒ a√

2 ≤ b√

3 +√

2 < b√

3 +√

3 = (b+ 1)√

3

so the last two inequalities are redundant. We are left with

(a− 1)√

2 ≤ b√

3 < a√

2.

Since the non-negative number line is partitioned by intervals of the form[(a− 1)

√2, a√

2)

for positiveintegers a, for any positive integer b, we can find a positive integer a that satisfies the inequalities. Asclearly a > b, it remains to find the number of b such that a ≤ 20. This is bounded by

b√

3 < a√

2 ≤ 20√

2 ⇐⇒ b <20√

2√3

=⇒ b ≤ 16

so there are 16 values of b and thus 16 ordered pairs of positive integers (a, b) that satisfy the problem.

7. Find the value of∞∑a=1

∞∑b=1

∞∑c=1

ab(3a+ c)

4a+b+c(a+ b)(b+ c)(c+ a).

Proposed by: Andrew Gu

Answer: 154

Let S denote the given sum. By summing over all six permutations of the variables a, b, c we obtain

6S =

∞∑a=1

∞∑b=1

∞∑c=1

3(a2b+ a2c+ b2a+ b2c+ c2a+ c2b) + 6abc

4a+b+c(a+ b)(b+ c)(c+ a)

=

∞∑a=1

∞∑b=1

∞∑c=1

3

4a+b+c

= 3

( ∞∑a=1

1

4a

)( ∞∑b=1

1

4b

)( ∞∑c=1

1

4c

)

= 3

(1

3

)3

=1

9.

Hence S = 154 .

8. There is a unique function f : N→ R such that f(1) > 0 and such that∑d|n

f(d)f(nd

)= 1

for all n ≥ 1. What is f(20182019)?

Proposed by: Ashwin Sah

Answer:(40382019)

2

28076 OR (4038!)2

(2019!)4·28076

Fix any prime p, and let an = f(pn) for n ≥ 0. Notice that using the relation for pn, we obtain

n∑i=0

aian−i = 1,

which means that if we let g(x) =∑

n≥0 anxn, then g(x)2 = 1 + x + x2 + · · · = 1

1−x as a generating

function. Thus g(x) = (1− x)−12 , and this is well-known to have generating function with coefficients

an =(2n

n )4n . One way to see this is using the Taylor series and then reorganizing terms; it is also

intimately related to the generating function for the Catalan numbers. In particular, an is independentof our choice of p.

Now if we define f0(n) =∏

p|n avp(n), then we see that f = f0 on the prime powers.

If we define the Dirichlet convolution of two functions χ1, χ2 : N→ R as χ3 such that

χ3(n) =∑d|n

χ1(d)χ2

(nd

),

then it is well-known that multiplicative functions (χ(m)χ(n) = χ(mn) if gcd(m,n), so e.g. φ(n), theEuler totient function) convolve to a multiplicative function.

In particular, f0 is a multiplicative function by definition (it is equivalent to only define it at primepowers then multiply), so the convolution of f0 with itself is multiplicative. By definition of an, theconvolution of f0 with itself equals 1 at all prime powers. Thus by multiplicativity, it equals theconstant function 1 everywhere.

Two final things to note: f0(1) = a0 = 1 > 0, and f satisfying the conditions in the problem statementis indeed unique (proceed by induction on n that f(n) is determined uniquely and that the resultingalgorithm for computing f gives a well-defined function). Therefore f0, satisfying those same conditions,must equal f .

At last, we have

f(p2019) = f0(p2019) =

40382019

)42019

so

f(20182019) = f(22019)f(10092019) =

40382019

)244038

=

40382019

)228076

.

9. Tessa the hyper-ant has a 2019-dimensional hypercube. For a real number k, she calls a placement ofnonzero real numbers on the 22019 vertices of the hypercube k-harmonic if for any vertex, the sum ofall 2019 numbers that are edge-adjacent to this vertex is equal to k times the number on this vertex.Let S be the set of all possible values of k such that there exists a k-harmonic placement. Find

∑k∈S|k|.

Proposed by: Yuan Yao

Answer: 2040200

By adding up all the equations on each vertex, we get 2019S = kS where S is the sum of all entries, sok = 2019 unless S = 0. In the latter case, by adding up all the equations on a half of the cube, we get

2018S − S = kS where S is the sum of all entries on that half of the cube, so k = 2017 unless S = 0.In the latter case (the sum of all entries of any half is zero), by adding up all the equations on a half ofthe half-cube, we get 2017S − 2S = kS, so k = 2015 unless S = 0. We continue this chain of caseworkuntil we get that the sum of every two vertices connected by unit segments is zero, in which case wehave k = −2019. This means that k can take any odd value between −2019 and 2019 inclusive, so thesum of absolute values is 2 · 10102 = 2040200.

To achieve these values, suppose that the vertices of the hypercube are {0, 1}2019 and that the label of(x1, x2, . . . , x2019) is ax1

1 ax22 . . . ax2019

2019 for constants a1, a2, . . . , a2019 ∈ {−1, 1}, then it is not difficult tosee that this labeling is (a1 + a2 + · · ·+ a2019)-harmonic for any choice of ai’s, so this can achieve allodd values between −2019 and 2019 inclusive.

10. The sequence of integers {ai}∞i=0 satisfies a0 = 3, a1 = 4, and

an+2 = an+1an +

⌈√a2n+1 − 1

√a2n − 1

⌉for n ≥ 0. Evaluate the sum

∞∑n=0

(an+3

an+2− an+2

an+an+1

an+3− anan+1

).

Proposed by: Ernest Chiu

Answer:14

69

The key idea is to note that an+1an +√a2n+1 − 1

√a2n − 1 is the larger zero of the quadratic

fn(x) = x2 − (2an+1an)x+ a2n + a2n+1 − 1.

Since an+2 is the smallest integer greater than or equal to this root, it follows that a2n + a2n+1 +a2n+2 − 2anan+1an+2 − 1 is some small nonnegative integer. For these particular initial conditions

(a0 = 3, a1 = 4, a2 = 12+d√

120e = 23), this integer is (32+42)+(232−2·3·4·23)−1 = 25−23−1 = 1.

We now use induction to prove both

an+3 = 2an+2an+1 − an and a2n + a2n+1 + a2n+2 − 2anan+1an+2 − 1 = 1

for n ≥ 0. The base case is not difficult to check: a3 = 4 · 23 + d√

7920e = 181 = 2 · 4 · 23 − 3, andthe other equation has been checked above. Since the quadratic equation fn+1(x) = 1 has a solutionan by induction hypothesis. Then, using Vieta’s theorem, 2an+1an+2 − an is also a solution. Then,the two roots of fn+1(x) = 0 must be in strictly between an and 2an+1an+2 − an, so we have thatan+3 ≤ 2an+1an+2 − an since an+3 is the ceiling of the larger root. In fact, since an+1an+2 is muchlarger than an for n ≥ 1 (it is not difficult to see that an grows faster than exponential), meaningthat the two roots of fn+1 are more than 1 away from the minimum, and f(2an+1an+2 − an) = 1,we have f(2an+1an+2 − an − 1) < 0, which mean that we must have an+3 = 2an+1an+2 − an, whichsimultaneously proves both statements due to Vieta jumping.

To finish, note that the above recurrence gives

an+3

an+2− an+2

an= 2an+1 −

anan+2

− 2an+1 +an−1an

= − anan+2

+an−1an

,

which telescopes with the other two terms. (Convergence can be shown since the ratio of adjacent termsis bounded above by 1/3. In fact it goes to zero rapidly.) The only leftover terms after telescoping area3a2− a2a0

= −a0a2

+a−1a0

= − 3

23+

1

3=

14

69, giving the answer. (Here, we use the backwards recurrence

an−1 = 2anan+1 − an+2 to find a−1 = 2 · 3 · 4− 23 = 1.)

HMMT February 2019February 16, 2019

Combinatorics

1. How many distinct permutations of the letters of the word REDDER are there that do not contain apalindromic substring of length at least two? (A substring is a contiguous block of letters that is partof the string. A string is palindromic if it is the same when read backwards.)

2. Your math friend Steven rolls five fair icosahedral dice (each of which is labelled 1, 2, . . . , 20 on itssides). He conceals the results but tells you that at least half of the rolls are 20. Suspicious, youexamine the first two dice and find that they show 20 and 19 in that order. Assuming that Steven istruthful, what is the probability that all three remaining concealed dice show 20?

3. Reimu and Sanae play a game using 4 fair coins. Initially both sides of each coin are white. Startingwith Reimu, they take turns to color one of the white sides either red or green. After all sides arecolored, the 4 coins are tossed. If there are more red sides showing up, then Reimu wins, and if thereare more green sides showing up, then Sanae wins. However, if there is an equal number of red sidesand green sides, then neither of them wins. Given that both of them play optimally to maximize theprobability of winning, what is the probability that Reimu wins?

4. Yannick is playing a game with 100 rounds, starting with 1 coin. During each round, there is a n%chance that he gains an extra coin, where n is the number of coins he has at the beginning of theround. What is the expected number of coins he will have at the end of the game?

5. Contessa is taking a random lattice walk in the plane, starting at (1, 1). (In a random lattice walk,one moves up, down, left, or right 1 unit with equal probability at each step.) If she lands on a pointof the form (6m, 6n) for m,n ∈ Z, she ascends to heaven, but if she lands on a point of the form(6m+3, 6n+3) for m,n ∈ Z, she descends to hell. What is the probability that she ascends to heaven?

6. A point P lies at the center of square ABCD. A sequence of points {Pn} is determined by P0 = P ,and given point Pi, point Pi+1 is obtained by reflecting Pi over one of the four lines AB,BC,CD,DA,chosen uniformly at random and independently for each i. What is the probability that P8 = P?

7. In an election for the Peer Pressure High School student council president, there are 2019 voters andtwo candidates Alice and Celia (who are voters themselves). At the beginning, Alice and Celia bothvote for themselves, and Alice’s boyfriend Bob votes for Alice as well. Then one by one, each of theremaining 2016 voters votes for a candidate randomly, with probabilities proportional to the currentnumber of the respective candidate’s votes. For example, the first undecided voter David has a 2

3probability of voting for Alice and a 1

3 probability of voting for Celia.

What is the probability that Alice wins the election (by having more votes than Celia)?

8. For a positive integer N , we color the positive divisors of N (including 1 and N) with four colors. Acoloring is called multichromatic if whenever a, b and gcd(a, b) are pairwise distinct divisors of N , thenthey have pairwise distinct colors. What is the maximum possible number of multichromatic coloringsa positive integer can have if it is not the power of any prime?

9. How many ways can one fill a 3× 3 square grid with nonnegative integers such that no nonzero integerappears more than once in the same row or column and the sum of the numbers in every row andcolumn equals 7?

10. Fred the Four-Dimensional Fluffy Sheep is walking in 4-dimensional space. He starts at the origin.Each minute, he walks from his current position (a1, a2, a3, a4) to some position (x1, x2, x3, x4) withinteger coordinates satisfying

(x1−a1)2+(x2−a2)2+(x3−a3)2+(x4−a4)2 = 4 and |(x1 + x2 + x3 + x4)− (a1 + a2 + a3 + a4)| = 2.

In how many ways can Fred reach (10, 10, 10, 10) after exactly 40 minutes, if he is allowed to passthrough this point during his walk?

HMMT February 2019February 16, 2019

Combinatorics

1. How many distinct permutations of the letters of the word REDDER are there that do not contain apalindromic substring of length at least two? (A substring is a contiguous block of letters that is partof the string. A string is palindromic if it is the same when read backwards.)

Proposed by: Yuan Yao

Answer: 6

If two identical letters are adjacent or have a single letter in between, there is clearly a palindromicsubstring of length (respectively) two or three. So there cannot be any such substrings.

Say we have a permutation of the word REDDER without any palindromic substrings. Let us call thefirst letter X. The second letter has to be different, let us call it Y. The third letter cannot be X or Y,let it be Z. Again, the fourth letter cannot be Y or Z, and we only have 3 letters to choose from, soit has to be X. Continuing analogously, the fifth letter has to be Y, and the sixth letter has to be Z.So any word satisfying the problem statement has to be of the form XYZXYZ. It is easy to check thatsuch a word indeed does not have any palindromic substrings. X, Y, Z can be any permutation of R,E, D, giving a total of 6 possibilities.

2. Your math friend Steven rolls five fair icosahedral dice (each of which is labelled 1, 2, . . . , 20 on itssides). He conceals the results but tells you that at least half of the rolls are 20. Suspicious, youexamine the first two dice and find that they show 20 and 19 in that order. Assuming that Steven istruthful, what is the probability that all three remaining concealed dice show 20?

Proposed by: Evan Chen

Answer: 158

The given information is equivalent to the first two dice being 20 and 19 and there being at least two20’s among the last three dice. Thus, we need to find the probability that given at least two of the lastthree dice are 20’s, all three are. Since there is only one way to get all three 20’s and 3 · 19 = 57 waysto get exactly two 20’s, the probability is 1

1+57 = 158 .

3. Reimu and Sanae play a game using 4 fair coins. Initially both sides of each coin are white. Startingwith Reimu, they take turns to color one of the white sides either red or green. After all sides arecolored, the 4 coins are tossed. If there are more red sides showing up, then Reimu wins, and if thereare more green sides showing up, then Sanae wins. However, if there is an equal number of red sidesand green sides, then neither of them wins. Given that both of them play optimally to maximize theprobability of winning, what is the probability that Reimu wins?

Proposed by: Yuan Yao

Answer: 516

Clearly Reimu will always color a side red and Sanae will always color a side green, because theirsituation is never worse off when a side of a coin changes to their own color. Since the number ofred-only coins is always equal to the number of green-only coins, no matter how Reimu and Sanaecolor the coins, they will have an equal probability of winning by symmetry, so instead they willcooperate to make sure that the probability of a tie is minimized, which is when all 4 coins havedifferent colors on both sides (which can easily be achieved by Reimu coloring one side of a new coinred and Sanae immediately coloring the opposite side green). Therefore, the probability of Reimu

winning is(43)+(4

4)24 = 5

16 .

4. Yannick is playing a game with 100 rounds, starting with 1 coin. During each round, there is a n%chance that he gains an extra coin, where n is the number of coins he has at the beginning of theround. What is the expected number of coins he will have at the end of the game?

Proposed by: Yuan Yao

Answer: 1.01100

Let Xi be the random variable which is the number of coins at the end of round i. Say that X0 = 1for convenience. Fix i > 0 and some positive integer x. Conditioning on the event Xi∈1 = x, there areonly two cases with positive probability. In particular,

Pr[Xi = x+ 1 | Xi∈1 = x] =x

100

andPr[Xi = x | Xi∈1 = x] = 1〉 x

100.

Therefore

E[Xi] =∑x>0

x · Pr[Xi = x]

=∑x>0

x ·((

1〉 x

100

)Pr[Xi∈1 = x] +

x〉 1

100Pr[Xi∈1 = x〉 1]

)=∑x>0

xPr[Xi∈1 = x]〉 1

100

∑x>0

xPr[Xi∈1 = x〉 1]

+1

100

∑x>0

x2 Pr[Xi∈1 = x〉 1]〉 1

100

∑x>0

x2 Pr[Xi = x]

=99

100E[Xi∈1]〉 1

100+

1

50E[Xi∈1] +

1

100

=101

100E[Xi∈1].

(A different way to understand this is that no matter how many coins Yannick has currently (as longas he does not have more than 100 coins, which is guaranteed in this problem), the expected numberof coins after one round is always 1.01 times the current number of coins, so the expected value ismultiplied by 1.01 each round.)

Therefore

E[X100] =

(101

100

)100

E[X0] = 1.01100.

5. Contessa is taking a random lattice walk in the plane, starting at (1, 1). (In a random lattice walk,one moves up, down, left, or right 1 unit with equal probability at each step.) If she lands on a pointof the form (6m, 6n) for m,n ∈ Z, she ascends to heaven, but if she lands on a point of the form(6m+3, 6n+3) for m,n ∈ Z, she descends to hell. What is the probability that she ascends to heaven?

Proposed by: John Michael Wu

Answer: 1322

Let P (m,n) be the probability that she ascends to heaven from point (m,n). Then P (6m, 6n) = 1 andP (6m+ 3, 6n+ 3) = 0 for all integers m,n ∈ Z. At all other points,

4P (m,n) = P (m〉 1, n) + P (m+ 1, n) + P (m,n〉 1) + P (m,n+ 1) (1)

This gives an infinite system of equations. However, we can apply symmetry arguments to cut downthe number of variables to something more manageable. We have P (m,n) = P (m + 6a, n + 6b) fora, b ∈ Z, and P (m,n) = P (n,m), and P (m,n) = P (〉m,n), and P (m,n) = 1〉 P (m+ 3, n+ 3) (sinceany path from the latter point to heaven corresponds with a path from the former point to hell, andvice versa).

Thus for example we have

P (1, 2) = P (〉1,〉2) = 1〉 P (2, 1) = 1〉 P (1, 2),

so P (1, 2) = 1/2.

Applying Equation (1) to points (1, 1), (0, 1), and (0, 2), and using the above symmetries, we get theequations

4P (1, 1) = 2P (0, 1) + 1,

4P (0, 1) = P (0, 2) + 2P (1, 1) + 1,

4P (0, 2) = P (0, 1) + 3/2.

Solving yields P (1, 1) = 13/22.

6. A point P lies at the center of square ABCD. A sequence of points {Pn} is determined by P0 = P ,and given point Pi, point Pi+1 is obtained by reflecting Pi over one of the four lines AB,BC,CD,DA,chosen uniformly at random and independently for each i. What is the probability that P8 = P?

Proposed by: Yuan Yao

Answer: 122516384

Solution 1. WLOG, AB and CD are horizontal line segments and BC and DA are vertical. Thenobserve that we can consider the reflections over vertical lines separately from those over horizontallines, as each reflection over a vertical line moves Pi horizontally to point Pi+1, and vice versa. Nowconsider only the reflections over horizontal segments AB and CD. Note that it is impossible for P8

to be in the same location vertical location as P if there are an odd number of these reflections. Thenwe consider the reflections in pairs: let w denote reflecting twice over AB, let x denote reflecting overAB and then CD, let y denote reflecting over CD and then AB, and let z denote reflecting twice overCD. Note that both w and z preserve the position of our point. Also note that in order to end at thesame vertical location as P, we must have an equal number of x’s and y’s. Now we count the numberof sequences of length at most 4 with this property:

• Case 1: Length 0

There is just the empty sequence here, so 1.

• Case 2: Length 1

There are just the sequences w and z, so 2.

• Case 3: Length 2

We may either have an x and a y or two characters that are either w or z. There are 2 sequencesof the former type and 4 of the latter, for 6 total.

• Case 4: Length 3

There are 12 sequences with an x, a y, and either a w or a z, and 8 sequences of only w’s and z’s,for 12 total.

• Case 5: Length 4

There are 6 sequences of 2 x’s and 2 y’s, 48 with one of each and two terms that are either w orz, and 16 of just w’s and z’s, for a total of 70.

Now let the number of such sequences of length k be ak (so a3 = 20). Note that these counts work alsoif we consider only reflections over vertical line segments BC and AD. Now to finish, we only need tocount the number of ways to combine 2 valid sequences of total length 4. This is

4∑i=0

aia4∈i

(8

2i

),

as there are ai sequences of reflections over AB and CD, a4∈i sequences of reflections over BC andAD such that there are 8 total reflections, and

82i

)ways to choose which of the 8 reflections will be

over AB or CD. We compute that this sum is 1 ·70 ·1 + 2 ·20 ·28 + 6 ·6 ·70 + 20 ·2 ·28 + 70 ·1 ·1 = 4900total sequences of reflections that place P8 at P. There are of course 48 = 65536 total sequences of 8reflections, each chosen uniformly at random, so our answer is 4900

65536 = 122516384 .

Solution 2. Suppose that P0 is the origin and the four lines are x = ±0.5 and y = ±0.5. We considera permutation of the lattice points on the coordinate plane, where all points with even x-coordinatesare reflected across the y-axis and all points with even y-coordinates are reflected across the x-axis, sothat the x- and y-coordinates are both rearranged in the following order:

. . . , 4,〉3, 2,〉1, 0, 1,〉2, 3,〉4, . . . .

It is not difficult to see that a reflection across one of the lines corresponds to changing one of thecoordinates from one number to either the previous number of the next number. Therefore, afterthe permutation, the question is equivalent to asking for the number of lattice walks of length 8 thatreturns to the origin. For such a lattice walk to return to origin, there needs to be the same numberof up and down moves, and the same number of left and right moves. This condition is equivalentto having four moves that are left or up (LU), and four moves that are right or up (RU). Moreover,knowing whether a move is LU and whether it is RU uniquely determines what the move is, so it

suffices to designate four LU moves and four RU moves, giving84

)2= 4900 possible walks. Hence the

probability is 490048 = 1225

16384 .

7. In an election for the Peer Pressure High School student council president, there are 2019 voters andtwo candidates Alice and Celia (who are voters themselves). At the beginning, Alice and Celia bothvote for themselves, and Alice’s boyfriend Bob votes for Alice as well. Then one by one, each of theremaining 2016 voters votes for a candidate randomly, with probabilities proportional to the currentnumber of the respective candidate’s votes. For example, the first undecided voter David has a 2

3probability of voting for Alice and a 1

3 probability of voting for Celia.

What is the probability that Alice wins the election (by having more votes than Celia)?

Proposed by: Yuan Yao

Answer: 15132017

Let Pn(m) be the probability that after n voters have voted, Alice gets m votes. We show by inductionthat for n ≥ 3, the ratio Pn(2) : Pn(3) : · · · : Pn(n 〉 1) is equal to 1 : 2 : · · · : (n 〉 2). We take abase case of n = 3, for which the claim is obvious. Then suppose the claim holds for n = k. ThenPk(m) = 2m∈2

(k∈1)(k∈2) . Then

Pk+1(i) =k 〉 ik

Pk(i) +i〉 1

kPk(i〉 1) =

(k 〉 i)(2i〉 2) + (i〉 1)(2i〉 4)

k(k 〉 1)(k 〉 2)=

2i〉 2

k(k 〉 1).

Also, we can check Pk+1(2) = 2k(k∈1) and Pk+1(k) = 2

k , so indeed the claim holds for n = k + 1, and

thus by induction our claim holds for all n ≥ 3. The probability that Ceila wins the election is then

1009∑m=2

P2019(m)

2018∑m=2

P2019(m)

=1008 · (1 + 1008)/2

2017 · (1 + 2017)/2=

504

2017,

and thus the probability that Alice wins is 15132017 .

8. For a positive integer N , we color the positive divisors of N (including 1 and N) with four colors. Acoloring is called multichromatic if whenever a, b and gcd(a, b) are pairwise distinct divisors of N , thenthey have pairwise distinct colors. What is the maximum possible number of multichromatic coloringsa positive integer can have if it is not the power of any prime?

Proposed by: Evan Chen

Answer: 192

First, we show that N cannot have three distinct prime divisors. For the sake of contradiction, supposepqr|N for three distinct primes p, q, r. Then by the problem statement, (p, q, 1), (p, r, 1), and (q, r, 1)have three distinct colors, so (p, q, r, 1) has four distinct colors. In addition, (pq, r, 1), (pq, pr, p), and(pq, qr, q) have three distinct colors, so (pq, p, q, r, 1) has five distinct colors, contradicting the fact thatthere are only four possible colors.

Similarly, if p3q|N for some distinct primes p and q, then (p, q, 1), (p2, q, 1), (p3, q, 1), (p2, pq, p),(p3, pq, p), and (p3, p2q, p2) are all triples with distinct colors, so (1, q, p, p2, p3) must have five distinctcolors, which is again a contradiction. In addition, if p2q2|N for some distinct primes p and q, then(p, q, 1), (p2, q2, 1), (p2, q, 1), and (p, q2, 1) are all triples with pairwise distinct colors, so (1, p, q, p2, q2)must have five distinct colors, another contradiction.

We are therefore left with two possibilities:

• Case 1: N = pq

In this case, the only triple of factors that must have pairwise distinct colors is (p, q, 1). We have4 · 3 · 2 = 24 choices for these three, and 4 choices for pq itself, giving 4 · 24 = 96 multichromaticcolorings.

• Case 2: N = p2q

In this case, the triples of pairwise distinctly colored factors are (p, q, 1), (p2, q, 1), and (p2, pq, p).From this, we see that (1, p, q, p2) must have four distinct colors, and the color of pq must bedistinct from p and p2. There are 4 · 3 · 2 · 1 = 24 ways to assign the four distinct colors, 2 ways toassign the color of pq after that, and 4 ways to color p2q after that, giving a total of 24 ·2 ·4 = 192monochromatic colorings.

Therefore, there can be at most 192 multichromatic colorings.

9. How many ways can one fill a 3× 3 square grid with nonnegative integers such that no nonzero integerappears more than once in the same row or column and the sum of the numbers in every row andcolumn equals 7?

Proposed by: Sam Korsky

Answer: 216

In what ways could we potentially fill a single row? The only possibilities are if it contains the numbers(0, 0, 7) or (0, 1, 6) or (0, 2, 5) or (0, 3, 4) or (1, 2, 4). Notice that if we write these numbers in binary,in any choices for how to fill the row, there will be exactly one number with a 1 in its rightmost digit,exactly one number with a 1 in the second digit from the right, and exactly exactly one number with a1 in the third digit from the right. Thus, consider the following operation: start with every unit squarefilled with the number 0. Add 1 to three unit squares, no two in the same row or column. Then add 2to three unit squares, no two in the same row or column. Finally, add 4 to three unit squares, no twoin the same row or column. There are clearly 63 = 216 ways to perform this operation and every suchoperation results in a unique, suitably filled-in 3 by 3 square. Hence the answer is 216.

10. Fred the Four-Dimensional Fluffy Sheep is walking in 4-dimensional space. He starts at the origin.Each minute, he walks from his current position (a1, a2, a3, a4) to some position (x1, x2, x3, x4) withinteger coordinates satisfying

(x1〉a1)2+(x2〉a2)2+(x3〉a3)2+(x4〉a4)2 = 4 and |(x1 + x2 + x3 + x4)〉 (a1 + a2 + a3 + a4)| = 2.

In how many ways can Fred reach (10, 10, 10, 10) after exactly 40 minutes, if he is allowed to passthrough this point during his walk?

Proposed by: Brice Huang

Answer:4010

)4020

)3

The possible moves correspond to the vectors ±h2, 0, 0, 0i, ±h1, 1, 1,〉1i, and their permutations. It’snot hard to see that these vectors form the vertices of a 4-dimensional hypercube, which motivates thechange of coordinates

(x1, x2, x3, x4)⇒(x1 + x2 + x3 + x4

2,x1 + x2 〉 x3 〉 x4

2,x1 〉 x2 + x3 〉 x4

2,x1 〉 x2 〉 x3 + x4

2

).

Under this change of coordinates, Fred must travel from (0, 0, 0, 0) to (20, 0, 0, 0), and the possiblemoves correspond to the sixteen vectors h±1,±1,±1,±1i. The new x1-coordinate must increase 30times and decrease 10 times during Fred’s walk, while the other coordinates must increase 20 times

and decrease 20 times. Therefore, there are4010

)4020

)3possible walks.

HMMT February 2019February 16, 2019

Geometry

1. Let d be a real number such that every non-degenerate quadrilateral has at least two interior angleswith measure less than d degrees. What is the minimum possible value for d?

Proposed by: James Lin

Answer: 120

The sum of the internal angles of a quadrilateral triangle is 360◦. To find the minimum d, we notethe limiting case where three of the angles have measure d and the remaining angle has measureapproaching zero. Hence, d ≥ 360◦/3 = 120. It is not difficult to see that for any 0 < α < 120, aquadrilateral of which three angles have measure α degrees and fourth angle has measure (360 − 3α)degrees can be constructed.

2. In rectangle ABCD, points E and F lie on sides AB and CD respectively such that both AF and CEare perpendicular to diagonal BD. Given that BF and DE separate ABCD into three polygons withequal area, and that EF = 1, find the length of BD.

Proposed by: Yuan Yao

Answer:√

3

Observe that AECF is a parallelogram. The equal area condition gives that BE = DF = 13AB.

Let CE ∩ BD = X, then EXCX = BE

CD = 13 , so that BX2 = EX · CX = 3EX2 ⇒ BX =

√3EX ⇒

∠EBX = 30◦. Now, CE = 2BE = CF , so CEF is an equilateral triangle and CD = 32CF = 3

2 .

Hence, BD = 2√3· 32 =

√3.

3. Let AB be a line segment with length 2, and S be the set of points P on the plane such that thereexists point X on segment AB with AX = 2PX. Find the area of S.

Proposed by: Yuan Yao

Answer:√

3 + 2π3

Observe that for any X on segment AB, the locus of all points P such that AX = 2PX is a circlecentered at X with radius 1

2AX. Note that the point P on this circle where PA forms the largest anglewith AB is where PA is tangent to the circle at P , such that ∠PAB = arcsin(1/2) = 30◦. Therefore,if we let Q and Q′ be the tangent points of the tangents from A to the circle centered at B (call itω) with radius 1

2AB, we have that S comprises the two 30-60-90 triangles AQB andAQ′B, each with

area 12

√3 and the 240◦ sector of ω bounded by BQ and BQ′ with area 2

3π. Therefore the total area

is√

3 + 2π3 .

4. Convex hexagon ABCDEF is drawn in the plane such that ACDF and ABDE are parallelogramswith area 168. AC and BD intersect at G. Given that the area of AGB is 10 more than the area ofCGB, find the smallest possible area of hexagon ABCDEF .

Proposed by: Andrew Lin

Answer: 196

Since ACDF and ABDE have area 168, triangles ABD and ACD (which are each half a parallelogram)both have area 84. Thus, B and C are the same height away from AD, and since ABCDEF is convex,B and C are on the same side of AD. Thus, BC is parallel to AD, and ABCD is a trapezoid.In particular, we have that the area of ABG equals the area of CDG. Letting this quantity bex, we have that the area of BCG is x − 10, and the area of ADG is 84 − x. Then notice that[ABG][CBG] = AG

GC = [ADG][CDG] . This means that x

x→10 = 84→xx . Simplifying, we have x2 − 47x + 420 = 0; this

has solutions x = 12 and x = 35. The area of ABCDEF is twice the area of trapezoid ABCD, or2[x + (x − 10) + (84 − x) + x] = 4x + 148; choosing x = 12, we get that the smallest possible area is48 + 148 = 196.

5. Isosceles triangle ABC with AB = AC is inscribed in a unit circle Ω with center O. Point D is thereflection of C across AB. Given that DO =

√3, find the area of triangle ABC.

Proposed by: Lillian Zhang

Answer:√2+12 OR

√2→12

Solution 1. Observe that

∠DBO = ∠DBA+ ∠ABO = ∠CBA+ ∠BAO =1

2(∠CBA+ ∠BCA) +

1

2(∠BAC) =

1

2(180◦) = 90◦.

Thus BC = BD =√

2 by the Pythagorean Theorem on 4DBO. Then ∠BOC = 90◦, and the distance

from O to BC is√22 . Depending on whether A is on the same side of BC as O, the height from A to

BC is either 1 +√22 or 1−

√22 , so the area is (

√2 · (1±

√22 ))/2 =

√2±12 .

Solution 2. One can observe that ∠DBA = ∠CBA = ∠ACB by property of reflection and ABCbeing isosceles, hence DB is tangent to Ω and Power of a Point (and reflection property) gives BC =BD =

√OD2 −OB2 =

√2. Proceed as in Solution 1.

Note. It was intended, but not specified in the problem statement that triangle ABC is acute, so weaccepted either of the two possible answers.

6. Six unit disks C1, C2, C3, C4, C5, C6 are in the plane such that they don’t intersect each other and Ciis tangent to Ci+1 for 1 ≤ i ≤ 6 (where C7 = C1). Let C be the smallest circle that contains all sixdisks. Let r be the smallest possible radius of C, and R the largest possible radius. Find R− r.Proposed by: Daniel Liu

Answer:√

3− 1

The minimal configuration occurs when the six circles are placed with their centers at the vertices ofa regular hexagon of side length 2. This gives a radius of 3.

The maximal configuration occurs when four of the circles are placed at the vertices of a square of sidelength 2. Letting these circles be C1, C3, C4, C6 in order, we place the last two so that C2 is tangent toC1 and C3 and C5 is tangent to C4 and C6. (Imagine pulling apart the last two circles on the plane;this is the configuration you end up with.) The resulting radius is 2 +

√3, so the answer is

√3− 1.

Now we present the proofs for these configurations being optimal. First, we rephrase the problem: givenan equilateral hexagon of side length 2, let r be the minimum radius of a circle completely containingthe vertices of the hexagon. Find the difference between the minimum and maximum values in r.(Technically this r is off by one from the actual problem, but since we want R − r in the actualproblem, this difference doesn’t matter.)

Proof of minimality. We claim the minimal configuration stated above cannot be covered by a cir-cle with radius r < 2. If r < 2 and all six vertices O1, O2, . . . , O6 are in the circle, then we havethat ∠O1OO2 > 60◦ since O1O2 is the largest side of the triangle O1OO2, and similar for other an-gles ∠O2OO3,∠O3OO4, . . . , but we cannot have six angles greater than 60◦ into 360◦, contradiction.Therefore r ≥ 2.

Proof of maximality. Let ABCDEF be the hexagon, and choose the covering circle to be centered atO, the midpoint of AD, and radius

√3 + 1. We claim the other vertices are inside this covering circle.

First, we will show the claim for B. Let M be the midpoint of AC. Since ABC is isosceles and AM ≥ 1,we must have BM ≤

√4− 1 =

√3. Furthermore, MO is a midline of ACD, so MO = CD

2 = 1. Thus

by the triangle inequality, OB ≤MB +OM =√

3 + 1, proving the claim. A similar argument provesthe claim for C,E, F . Finally, an analogous argument to above shows if we define P as the midpointof BE, then AP ≤

√3 + 1 and DP ≤

√3 + 1, so by triangle inequality AD ≤ 2(

√3 + 1). Hence

OA = OD ≤√

3 + 1, proving the claim for A and D. Thus the covering circle contains all six verticesof ABCDEF .

7. Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Let H be the orthocenter of ABC. Findthe radius of the circle with nonzero radius tangent to the circumcircles of AHB,BHC,CHA.

Proposed by: Michael Ren

Answer: 654

Solution 1. We claim that the circle in question is the circumcircle of the anticomplementary triangleof ABC, the triangle for which ABC is the medial triangle.

Let A′B′C ′ be the anticomplementary triangle of ABC, such that A is the midpoint of B′C ′, B is themidpoint of A′C ′, and C is the midpoint of A′B′. Denote by ω the circumcircle of A′B′C ′. Denote byωA the circumcircle of BHC, and similarly define ωB , ωC .

Since ∠BA′C = ∠BAC = 180◦ − ∠BHC, we have that ωA passes through A′. Thus, ωA can beredefined as the circumcircle of A′BC. Since triangle A′B′C ′ is triangle A′BC dilated by a factor of 2from point A′, ω is ωA dilated by a factor of 2 from point A′. Thus, circles ω and ωA are tangent atA′.

By a similar logic, ω is also tangent to ωB and ωC . Therefore, the circumcircle of the anticomplementarytriangle of ABC is indeed the circle that the question is asking for.

Using the formula R = abc4A , we can find that the circumradius of triangle ABC is 65

8 . The circumradiusof the anticomplementary triangle is double of that, so the answer is 65

4 .

Solution 2. It is well-known that the circumcircle of AHB is the reflection of the circumcircle ofABC over AB. In particular, the circumcircle of AHB has radius equal to the circumradius R = 65

8 .Similarly, the circumcircles of BHC and CHA have radii R. Since H lies on all three circles (in thequestion), the circle centered at H with radius 2R = 65

4 is tangent to each circle at the antipode of Hin that circle.

8. In triangle ABC with AB < AC, let H be the orthocenter and O be the circumcenter. Given that themidpoint of OH lies on BC, BC = 1, and the perimeter of ABC is 6, find the area of ABC.

Proposed by: Andrew Lin

Answer: 67

Solution 1. Let A′B′C ′ be the medial triangle of ABC, where A′ is the midpoint of BC and soon. Notice that the midpoint of OH, which is the nine-point-center N of triangle ABC, is also thecircumcircle of A′B′C ′ (since the midpoints of the sides of ABC are on the nine-point circle). Thus, ifN is on BC, then NA′ is parallel to B′C ′, so by similarity, we also know that OA is parallel to BC.

Next, AB < AC, so B is on the minor arc AC. This means that ∠OAC = ∠OCA = ∠C, so ∠AOC =180− 2∠C. This gives us the other two angles of the triangle in terms of angle C: ∠B = 90 +∠C and∠A = 90− 2∠C. To find the area, we now need to find the height of the triangle from A to BC, andthis is easiest by finding the circumradius R of the triangle.

We do this by the Extended Law of Sines. Letting AC = x and AB = 5− x,

1

sin(90− 2C)=

x

sin(90 + C)=

5− xsinC

= 2R,

which can be simplified to1

cos 2C=

x

cosC=

5− xsinC

= 2R.

This means that1

cos 2C=

(x) + (5− x)

(cosC) + (sinC)=

5

cosC + sinC

and the rest is an easy computation:

cosC + sinC = 5 cos 2C = 5(cos2 C − sin2 C)

1

5= cosC − sinC

Squaring both sides,1

25= cos2 C − 2 sinC cosC + sin2 C = 1− sin 2C

so sin 2C = 2425 , implying that cos 2C = 7

25 . Therefore, since 1cos 2C = 2R from above, R = 25

14 . Finally,viewing triangle ABC with BC = 1 as the base, the height is√

R2 −(BC

2

)2

=12

7

by the Pythagorean Theorem, yielding an area of 12 · 1 ·

127 = 6

7 .

Solution 2. The midpoint of OH is the nine-point center N . We are given N lies on BC, and we alsoknow N lies on the perpendicular bisector of EF , where E is the midpoint of AC and F is the midpointof AB. The main observation is that N is equidistant from M and F , where M is the midpoint of BC.

Translating this into coordinates, we pick B(−0.5, 0) and C(0.5, 0), and arbitrarily set A(a, b) where(without loss of generality) b > 0. We get E(a+0.5

2 , b2 ), F (a→0.52 , b2 ), M(0, 0). Thus N must havex-coordinate equal to the average of those of E and F , or a

2 . Since N lies on BC, we have N(a2 , 0).

Since MN = EN , we have a2

4 = 116 + b2

4 . Thus a2 = b2 + 14 . The other equation is AB + AC = 5,

which is just √(a+ 0.5)2 + b2 +

√(a− 0.5)2 + b2 = 5.

This is equivalent to √2a2 + a+

√2a2 − a = 5√

2a2 + a = 5−√

2a2 − a

2a2 + a = 25− 10√

2a2 − a+ 2a2 − a

25− 2a = 10√

2a2 − a

625− 100a+ 4a2 = 200a2 − 100a

196a2 = 625

Thus a2 = 625196 , so b2 = 576

196 . Thus b = 2414 = 12

7 , so [ABC] = b2 = 6

7 .

9. In a rectangular box ABCDEFGH with edge lengths AB = AD = 6 and AE = 49, a plane slicesthrough point A and intersects edges BF,FG,GH,HD at points P,Q,R, S respectively. Given thatAP = AS and PQ = QR = RS, find the area of pentagon APQRS.

Proposed by: Yuan Yao

Answer: 141√11

2

Let AD be the positive x-axis, AB be the positive y-axis, and AE be the positive z-axis, with A theorigin. The plane, which passes through the origin, has equation k1x+k2y = z for some undeterminedparameters k1, k2. Because AP = AS and AB = AD, we get PB = SD, so P and S have the samez-coordinate. But P (0, 6, 6k2) and S(6, 0, 6k1), so k1 = k2 = k for some k. Then Q and R both havez-coordinate 49, so Q( 49

k − 6, 6, 49) and R(6, 49k − 6, 49). The equation QR2 = RS2 then gives(49

k− 6

)2

+ (49− 6k)2 = 2

(12− 49

k− 12

)2

.

This is equivalent to(49− 6k)2(k2 + 1) = 2(49− 12k)2,

which factors as(k − 7)(36k3 − 336k2 − 203k + 343) = 0.

This gives k = 7 as a root. Note that for Q and R to actually lie on FG and GH respectively, wemust have 49

6 ≥ k ≥ 4912 . Via some estimation, one can show that the cubic factor has no roots in this

range (for example, it’s easy to see that when k = 1 and k = 33636 = 28

3 , the cubic is negative, and italso remains negative between the two values), so we must have k = 7.

Now consider projecting APQRS onto plane ABCD. The projection is ABCD save for a triangleQ′CR′ with side length 12 − 49

k = 5. Thus the projection has area 36 − 252 = 47

2 . Since the areaof the projection equals [APQRS] · cos θ, where θ is the (smaller) angle between planes APQRSand ABCD, and since the planes have normal vectors (k, k,−1) and (0, 0, 1) respectively, we get

cos θ = (k,k,→1)·(0,0,1)√k2+k2+1

= 1√2k2+1

= 1√99

and so

[APQRS] =47√

99

2=

141√

11

2.

10. In triangle ABC, AB = 13, BC = 14, CA = 15. Squares ABB1A2, BCC1B2, CAA1C2 are constructedoutside the triangle. Squares A1A2A3A4, B1B2B3B4, C1C2C3C4 are constructed outside the hexagonA1A2B1B2C1C2. Squares A3B4B5A6, B3C4C5B6, C3A4A5C6 are constructed outside the hexagonA4A3B4B3C4C3. Find the area of the hexagon A5A6B5B6C5C6.

Proposed by: Yuan Yao

Answer: 19444

Solution 1.

We can use complex numbers to find synthetic observations. Let A = a,B = b, C = c. Notice that B2 isa rotation by−90◦ (counter-clockwise) of C about B, and similarly C1 is a rotation by 90◦ of B about C.Since rotation by 90◦ corresponds to multiplication by i, we have B2 = (c− b) · (−i) + b = b(1 + i)− ciand C1 = (b − c) · i + c = bi + c(1 − i). Similarly, we get C2 = c(1 + i) − ai, A1 = ci + a(1 − i),A2 = a(1 + i) − bi, B1 = ai + b(1 − i). Repeating the same trick on B1B2B3B4 et. al, we getC4 = −a + b(−1 + i) + c(3 − i), C3 = a(−1 − i) − b + c(3 + i), A4 = −b + c(−1 + i) + a(3 − i),A3 = b(−1− i)− c+ a(3 + i), B4 = −c+ a(−1 + i) + b(3− i), B3 = c(−1− i)− a+ b(3 + i). Finally,repeating the same trick on the outermost squares, we get B6 = −a + b(3 + 5i) + c(−3 − 3i), C5 =−a+ b(−3 + 3i) + c(3− 5i), C6 = −b+ c(3 + 5i) + a(−3− 3i), A5 = −b+ c(−3 + 3i) + a(3− 5i), A6 =−c+ a(3 + 5i) + b(−3− 3i), B5 = −c+ a(−3 + 3i) + b(3− 5i).

From here, we observe the following synthetic observations.

S1. B2C1C4B3, C2A1A4C3, A2B1B4A3 are trapezoids with bases of lengthsBC, 4BC; AC, 4AC; AB, 4ABand heights ha, hb, hc respectively (where ha is the length of the altitude from A to BC, and like-wise for hb, hc)

S2. If we extend B5B4 and B6B3 to intersect at B7, then B7B4B3∼= BB1B2 ∼ B7B5B6 with scale

factor 1:5. Likewise when we replace all B’s with A’s or C’s.

Proof of S1. Observe C1 −B2 = c− b and C4 −B3 = 4(c− b), hence B2C1‖B3C4 and B3C4 = 4B2C1.Furthermore, since translation preserves properties of trapezoids, we can translate B2C1C4B3 suchthat B2 coincides with A. Being a translation of a − B2, we see that B3 maps to B′3 = 2b − c andC4 maps to C ′4 = −2b + 3c. Both 2b − c and −2b + 3c lie on the line determined by b and c (since−2 + 3 = 2 − 1 = 1), so the altitude from A to BC is also the altitude from A to B′3C

′4. Thus ha

equals the length of the altitude from B2 to B3C4, which is the height of the trapezoid B2C1C4B3.This proves S1 for B2C1C4B3; the other trapezoids follow similarly.

Proof of S2. Notice a translation of −a + 2b − c maps B1 to B4, B2 to B3, and B to a pointB8 = −a + 3b − c. This means B8B3B4

∼= BB1B2. We can also verify that 4B8 + B6 = 5B3 and4B8 + B5 = 5B4, showing that B8B5B6 is a dilation of B8B4B3 with scale factor 5. We also get B8

lies on B3B6 and B5B4, so B8 = B7. This proves S2 for B3B4B5B6, and similar arguments prove thelikewise part.

Now we are ready to attack the final computation. By S2, [B3B4B5B6] + [BB1B2] = [B7B5B6] =[BB1B2]. But by the 1

2ac sinB formula, [BB1B2] = [ABC] (since ∠B1BB2 = 180◦ −∠ABC). Hence,

[B3B4B5B6] + [BB1B2] = 25[ABC]. Similarly, [C3C4C5C6] + [CC1C2] = 25[ABC] and [A3A4A5A6] +[AA1A2] = 25[ABC]. Finally, the formula for area of a trapezoid shows [B2C1C4B3] = 5BC

2 · ha =5[ABC], and similarly the other small trapezoids have area 5[ABC]. The trapezoids thus contributearea (75 + 3 · 5) = 90[ABC]. Finally, ABC contributes area [ABC] = 84.

By S1, the outside squares have side lengths 4BC, 4CA, 4AB, so the sum of areas of the outside squaresis 16(AB2 +AC2 +BC2). Furthermore, a Law of Cosines computation shows A1A

22 = AB2 +AC2 +

2 · AB · AC · cos∠BAC = 2AB2 + 2AC2 − BC2, and similarly B1B22 = 2AB2 + 2BC2 − AC2 and

C1C22 = 2BC2 +2AC2−AB2. Thus the sum of the areas of A1A2A3A4 et. al is 3(AB2 +AC2 +BC2).

Finally, the small squares have area add up to AB2 +AC2 +BC2. Aggregating all contributions fromtrapezoids, squares, and triangle, we get

[A5A6B5B6C5C6] = 91[ABC] + 20(AB2 +AC2 +BC2) = 7644 + 11800 = 19444.

Solution 2. Let a = BC, b = CA, c = AB. We can prove S1 and S2 using some trigonometry instead.

Proof of S1. The altitude from B3 to B2C1 has length B2B3 sin∠BB2B1 = B1B2 sin∠BB2B1 =BB1 sin∠B1BB2 = AB sin∠ABC = ha using Law of Sines. Similarly, we find the altitude fromC4 to B2C1 equals ha, thus proving B2C1C4B3 is a trapezoid. Using B1B2 =

√2a2 + 2c2 − b2 from

end of Solution 1, we get the length of the projection of B2B3 onto B3C4 is B2B3 cosBB2B1 =(2a2+2c2→b2)+a2→c2

2a = 3a2+c2→b22a , and similarly the projection of C1C4 onto B3C4 has length 3a2+b2→c2

2a .

It follows that B3C4 = 3a2+c2→b22a + a + 3a2+b2→c2

2a = 4a, proving S1 for B2C1C4B3; the other casesfollow similarly.

Proof of S2. Define B8 to be the image of B under the translation taking B1B2 to B4B3. We claim B8

lies on B3B6. Indeed, B8B4B3∼= BB1B2, so ∠B8B3B4 = ∠BB2B1 = 180◦ − ∠B3B2C1 = ∠B2B3C4.

Thus ∠B8B3C4 = ∠B4B3B2 = 90◦. But ∠B6B3C4 = 90◦, hence B8, B3, B6 are collinear. Similarly wecan prove B5B4 passes through B8, so B8 = B7. Finally, B7B3

B7B6= B7B4

B7B5= 1

5 (using B3B6 = 4a,B4B5 =4c,B7B3 = a,B7B4 = c) shows B7B4B3 ∼ B7B5B6 with scale factor 1:5, as desired. The likewise partfollows similarly.

HMMT February 2019February 16, 2019

Geometry

1. Let d be a real number such that every non-degenerate quadrilateral has at least two interior angleswith measure less than d degrees. What is the minimum possible value for d?

2. In rectangle ABCD, points E and F lie on sides AB and CD respectively such that both AF and CEare perpendicular to diagonal BD. Given that BF and DE separate ABCD into three polygons withequal area, and that EF = 1, find the length of BD.

3. Let AB be a line segment with length 2, and S be the set of points P on the plane such that thereexists point X on segment AB with AX = 2PX. Find the area of S.

4. Convex hexagon ABCDEF is drawn in the plane such that ACDF and ABDE are parallelogramswith area 168. AC and BD intersect at G. Given that the area of AGB is 10 more than the area ofCGB, find the smallest possible area of hexagon ABCDEF .

5. Isosceles triangle ABC with AB = AC is inscribed in a unit circle Ω with center O. Point D is thereflection of C across AB. Given that DO =

√3, find the area of triangle ABC.

6. Six unit disks C1, C2, C3, C4, C5, C6 are in the plane such that they don’t intersect each other and Ci

is tangent to Ci+1 for 1 ≤ i ≤ 6 (where C7 = C1). Let C be the smallest circle that contains all sixdisks. Let r be the smallest possible radius of C, and R the largest possible radius. Find R− r.

7. Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Let H be the orthocenter of ABC. Findthe radius of the circle with nonzero radius tangent to the circumcircles of AHB,BHC,CHA.

8. In triangle ABC with AB < AC, let H be the orthocenter and O be the circumcenter. Given that themidpoint of OH lies on BC, BC = 1, and the perimeter of ABC is 6, find the area of ABC.

9. In a rectangular box ABCDEFGH with edge lengths AB = AD = 6 and AE = 49, a plane slicesthrough point A and intersects edges BF,FG,GH,HD at points P,Q,R, S respectively. Given thatAP = AS and PQ = QR = RS, find the area of pentagon APQRS.

10. In triangle ABC, AB = 13, BC = 14, CA = 15. Squares ABB1A2, BCC1B2, CAA1C2 are constructedoutside the triangle. Squares A1A2A3A4, B1B2B3B4, C1C2C3C4 are constructed outside the hexagonA1A2B1B2C1C2. Squares A3B4B5A6, B3C4C5B6, C3A4A5C6 are constructed outside the hexagonA4A3B4B3C4C3. Find the area of the hexagon A5A6B5B6C5C6.

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HMMT February 2019, February 16, 2019 — GUTS ROUND

Organization Team Team ID#

There are nine sets of four problems each. The point values of the problems in each set, in order, are 3, 4,5, 7, 9, 12, 15, 20, 25. Good luck!

Also, a heads up: the problems in the last set (problems 33 to 36) of this guts round will NOT give possiblepartial credits as per tradition, but the set is nevertheless different from a normal guts set. In light of thischange, you may want to reserve more time for the last set than usual.

1. [3] Find the sum of all real solutions to x2 + cosx = 2019.

2. [3] There are 100 people in a room with ages 1, 2, . . . , 100. A pair of people is called cute if each ofthem is at least seven years older than half the age of the other person in the pair. At most how manypairwise disjoint cute pairs can be formed in this room?

3. [3] Let S(x) denote the sum of the digits of a positive integer x. Find the maximum possible value ofS(x+ 2019)− S(x).

4. [3] Tessa has a figure created by adding a semicircle of radius 1 on each side of an equilateral trianglewith side length 2, with semicircles oriented outwards. She then marks two points on the boundary ofthe figure. What is the greatest possible distance between the two points?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2019, February 16, 2019 — GUTS ROUND

Organization Team Team ID#

5. [4] Call a positive integer n weird if n does not divide (n−2)!. Determine the number of weird numbersbetween 2 and 100 inclusive.

6. [4] The pairwise products ab, bc, cd, and da of positive integers a, b, c, and d are 64, 88, 120, and 165 insome order. Find a+ b+ c+ d.

7. [4] For any real number α, define

sign(α) =

+1 if α > 0,

0 if α = 0,

−1 if α < 0.

How many triples (x, y, z) ∈ R3 satisfy the following system of equations

x = 2018− 2019 · sign(y + z),

y = 2018− 2019 · sign(z + x),

z = 2018− 2019 · sign(x+ y)?

8. [4] A regular hexagon PROFIT has area 1. Every minute, greedy George places the largest possibleequilateral triangle that does not overlap with other already-placed triangles in the hexagon, with tiesbroken arbitrarily. How many triangles would George need to cover at least 90% of the hexagon’s area?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2019, February 16, 2019 — GUTS ROUND

Organization Team Team ID#

9. [5] Define P = {S,T} and let P be the set of all proper subsets of P . (A proper subset is a subset thatis not the set itself.) How many ordered pairs (S, T ) of proper subsets of P are there such that

(a) S is not a proper subset of T and T is not a proper subset of S; and

(b) for any sets S ∈ S and T ∈ T , S is not a proper subset of T and T is not a proper subset of S?

10. [5] Let

A = (1 + 2√

2 + 3√

3 + 6√

6)(2 + 6√

2 +√

3 + 3√

6)(3 +√

2 + 6√

3 + 2√

6)(6 + 3√

2 + 2√

3 +√

6),

B = (1 + 3√

2 + 2√

3 + 6√

6)(2 +√

2 + 6√

3 + 3√

6)(3 + 6√

2 +√

3 + 2√

6)(6 + 2√

2 + 3√

3 +√

6).

Compute the value of A/B.

11. [5] In the Year 0 of Cambridge there is one squirrel and one rabbit. Both animals multiply in numbersquickly. In particular, if there are m squirrels and n rabbits in Year k, then there will be 2m + 2019squirrels and 4n− 2 rabbits in Year k + 1. What is the first year in which there will be strictly morerabbits than squirrels?

12. [5] Bob is coloring lattice points in the coordinate plane. Find the number of ways Bob can color fivepoints in {(x, y) | 1 ≤ x, y ≤ 5} blue such that the distance between any two blue points is not aninteger.

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HMMT February 2019, February 16, 2019 — GUTS ROUND

Organization Team Team ID#

13. [7] Reimu has 2019 coins C0, C1, . . . , C2018, one of which is fake, though they look identical to eachother (so each of them is equally likely to be fake). She has a machine that takes any two coins andpicks one that is not fake. If both coins are not fake, the machine picks one uniformly at random. Foreach i = 1, 2, . . . , 1009, she puts C0 and Ci into the machine once, and machine picks Ci. What is theprobability that C0 is fake?

14. [7] Let ABC be a triangle where AB = 9, BC = 10, CA = 17. Let Ω be its circumcircle, and letA1, B1, C1 be the diametrically opposite points from A,B,C, respectively, on Ω. Find the area of theconvex hexagon with the vertices A,B,C,A1, B1, C1.

15. [7] Five people are at a party. Each pair of them are friends, enemies, or frenemies (which is equivalentto being both friends and enemies). It is known that given any three people A,B,C:

• If A and B are friends and B and C are friends, then A and C are friends;

• If A and B are enemies and B and C are enemies, then A and C are friends;

• If A and B are friends and B and C are enemies, then A and C are enemies.

How many possible relationship configurations are there among the five people?

16. [7] Let R be the set of real numbers. Let f : R→ R be a function such that for all real numbers x andy, we have

f(x2) + f(y2) = f(x+ y)2 − 2xy.

Let S =2019∑

n=−2019f(n). Determine the number of possible values of S.

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HMMT February 2019, February 16, 2019 — GUTS ROUND

Organization Team Team ID#

17. [9] Let ABC be a triangle with AB = 3, BC = 4, and CA = 5. Let A1, A2 be points on side BC,B1, B2 be points on side CA, and C1, C2 be points on side AB. Suppose that there exists a point Psuch that PA1A2, PB1B2, and PC1C2 are congruent equilateral triangles. Find the area of convexhexagon A1A2B1B2C1C2.

18. [9] 2019 points are chosen independently and uniformly at random on the interval [0, 1]. Tairitsupicks 1000 of them randomly and colors them black, leaving the remaining ones white. Hikari thencomputes the sum of the positions of the leftmost white point and the rightmost black point. What isthe probability that this sum is at most 1?

19. [9] Complex numbers a, b, c form an equilateral triangle with side length 18 in the complex plane. If|a+ b+ c| = 36, find |bc+ ca+ ab|.

20. [9] On floor 0 of a weird-looking building, you enter an elevator that only has one button. You pressthe button twice and end up on floor 1. Thereafter, every time you press the button, you go up byone floor with probability X

Y , where X is your current floor, and Y is the total number of times youhave pressed the button thus far (not including the current one); otherwise, the elevator does nothing.Between the third and the 100th press inclusive, what is the expected number of pairs of consecutivepresses that both take you up a floor?

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HMMT February 2019, February 16, 2019 — GUTS ROUND

Organization Team Team ID#

21. [12] A regular hexagon ABCDEF has side length 1 and center O. Parabolas P1, P2, . . . , P6 areconstructed with common focus O and directrices AB,BC,CD,DE,EF, FA respectively. Let χ bethe set of all distinct points on the plane that lie on at least two of the six parabolas. Compute∑

X∈χ|OX|.

(Recall that the focus is the point and the directrix is the line such that the parabola is the locus ofpoints that are equidistant from the focus and the directrix.)

22. [12] Determine the number of subsets S of {1, 2, . . . , 1000} that satisfy the following conditions:

• S has 19 elements, and

• the sum of the elements in any non-empty subset of S is not divisible by 20.

23. [12] Find the smallest positive integer n such that

222···

2︸ ︷︷ ︸n 2’s

> ((· · · ((100!)!)! · · · )!)!︸ ︷︷ ︸100 factorials

.

24. [12] Let S be the set of all positive factors of 6000. What is the probability of a random quadruple(a, b, c, d) ∈ S4 satisfies

lcm(gcd(a, b), gcd(c, d)) = gcd(lcm(a, b), lcm(c, d))?

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HMMT February 2019, February 16, 2019 — GUTS ROUND

Organization Team Team ID#

25. [15] A 5 by 5 grid of unit squares is partitioned into 5 pairwise incongruent rectangles with sides lyingon the gridlines. Find the maximum possible value of the product of their areas.

26. [15] Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Let IA, IB , IC be the A,B,C excentersof this triangle, and let O be the circumcenter of the triangle. Let γA, γB , γC be the correspondingexcircles and ω be the circumcircle. X is one of the intersections between γA and ω. Likewise, Y is anintersection of γB and ω, and Z is an intersection of γC and ω. Compute

cos∠OXIA + cos∠OY IB + cos∠OZIC .

27. [15] Consider the eighth-sphere {(x, y, z) | x, y, z ≥ 0, x2 + y2 + z2 = 1}. What is the area of itsprojection onto the plane x+ y + z = 1?

28. [15] How many positive integers 2 ≤ a ≤ 101 have the property that there exists a positive integer Nfor which the last two digits in the decimal representation of a2

n

is the same for all n ≥ N?

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HMMT February 2019, February 16, 2019 — GUTS ROUND

Organization Team Team ID#

29. [20] Yannick picks a number N randomly from the set of positive integers such that the probabilitythat n is selected is 2−n for each positive integer n. He then puts N identical slips of paper numbered1 through N into a hat and gives the hat to Annie. Annie does not know the value of N, but she drawsone of the slips uniformly at random and discovers that it is the number 2. What is the expected valueof N given Annie’s information?

30. [20] Three points are chosen inside a unit cube uniformly and independently at random. What is theprobability that there exists a cube with side length 1

2 and edges parallel to those of the unit cube thatcontains all three points?

31. [20] Let ABC be a triangle with AB = 6, AC = 7, BC = 8. Let I be the incenter of ABC. Points Zand Y lie on the interior of segments AB and AC respectively such that Y Z is tangent to the incircle.Given point P such that

∠ZPC = ∠Y PB = 90◦,

find the length of IP .

32. [20] For positive integers a and b such that a is coprime to b, define ordb(a) as the least positive integerk such that b | ak − 1, and define ϕ(a) to be the number of positive integers less than or equal to awhich are coprime to a. Find the least positive integer n such that

ordn(m) <ϕ(n)

10

for all positive integers m coprime to n.

Warning: The next (and final) set contains problems in different formats.

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HMMT February 2019, February 16, 2019 — GUTS ROUND

Organization Team Team ID#

Welcome to Ultra Relay! This is an event where teams of twenty-five students race to solve a set oftwenty-five problems and ultimately find the answers to the four \anchor" problems. Many of the problemswill depend on answers to other problems, hence the team members will need to relay the answers from oneperson/problem to the next, along the network shown below. (The anchor problems are indicated by a box.)Only the answers to the four anchor problems will be graded.

H1 H2

H3 H4 H5

H6 H7

M1 M2 M3 M4

M5 M6 M7

M8 M9

M10 M11 M12 M13

T1 T2 T3

T4

T5

What? You don’t have twenty-five people on your team, you say? That’s unfortunate. Anyway, here are allthe problems. (You may work on them together.) Have fun!

For ease of reference, the label of each problem also denotes the answer to that problem.

H1. Let r = H1 be the answer to this problem. Given that r is a nonzero real number, what is the valueof r4 + 4r3 + 6r2 + 4r?

H2. Given two distinct points A, B and line ` that is not perpendicular to AB, what is the maximumpossible number of points P on ` such that ABP is an isosceles triangle?

H3. Let A = H1, B = H6 + 1. A real number x is chosen randomly and uniformly in the interval [A,B].Find the probability that x2 > x3 > x.

H4. Let A = d1/H3e, B = dH5/2e. How many ways are there to partition the set {1, 2, . . . , A + B} intotwo sets U and V with size A and B respectively such that the probability that a number chosen fromU uniformly at random is greater than a number chosen from V uniformly at random is exactly 1

2?

H5. Let A = H2, B = H7. Two circles with radii A and B respectively are given in the plane. If the lengthof their common external tangent is twice the length of their common internal tangent (where bothtangents are considered as segments with endpoints being the points of tangency), find the distancebetween the two centers.

H6. How many ways are there to arrange the numbers 21, 22, 33, 35 in a row such that any two adjacentnumbers are relatively prime?

H7. How many pairs of integers (x, y) are there such that |x2 − 2y2| ≤ 1 and |3x− 4y| ≤ 1?

M1. Let S = M10. Determine the number of ordered triples (a, b, c) of nonnegative integers such thata+ 2b+ 4c = S.

M2. Let S = bM5c. Two integersm and n are chosen between 1 and S inclusive uniformly and independentlyat random. What is the probability that mn = nm?

M3. Let S = dM7e. In right triangle ABC, ∠C = 90◦, AC = 27, BC = 36. A circle with radius S istangent to both AC and BC and intersects AB at X and Y . Find the length of XY .

M4. Let S = M13 + 5. Compute the product of all positive divisors of S.

M5. Let A =√M1, B = dM11e. Given complex numbers x and y such that x+ 1

y = A, 1x +y = B, compute

the value of xy + 1xy .

M6. Let A = b1/M2c, B = bM23 /100c. Let P and Q both be quadratic polynomials. Given that the real

roots of P (Q(x)) = 0 are 0, A,B,C in some order, find the sum of all possible values of C.

M7. Let A = dlog2M4e, B = M12 + 1. A 5-term sequence of positive reals satisfy that the first three termsand the last three terms both form an arithmetic sequence and the middle three terms form a geometricsequence. If the first term is A and the fifth term is B, determine the third term of the sequence.

M8. Let A = bM25 c, B = bM2

6 c. A regular A-gon, a regular B-gon, and a circle are given in the plane.What is the greatest possible number of regions that these shapes divide the plane into?

M9. Let A and B be the unit digits of d7M6e and b6M7c respectively. When all the positive integers notcontaining digit A or B are written in increasing order, what is the 2019th number in the list?

M10. What is the smallest positive integer with remainder 2, 3, 4 when divided by 3, 5, 7 respectively?

M11. An equiangular hexagon has side lengths 1, 2, 3, 4, 5, 6 in some order. Find the nonnegative differencebetween the largest and the smallest possible area of this hexagon.

M12. Determine the second smallest positive integer n such that n3 + n2 + n+ 1 is a perfect square.

M13. Given that A,B are nonzero base-10 digits such that A ·AB +B = BB, find AB.

T1. Let S, P,A,C,E be (not necessarily distinct) decimal digits where E 6= 0. Given that N =√ESCAPE

is a positive integer, find the minimum possible value of N .

T2. Let X = bT1/8c, Y = T3 − 1, Z = T4 − 2. A point P lies inside the triangle ABC such that PA =X,PB = Y, PC = Z. Find the largest possible area of the triangle.

T3. How many ways can one tile a 2× 8 board with 1× 1 and 2× 2 tiles? Rotations and reflections of thesame configuration are considered distinct.

T4. Let S = T5. Given real numbers a, b, c such that a2 + b2 + c2 + (a + b + c)2 = S, find the maximumpossible value of (a+ b)(b+ c)(c+ a).

T5. A regular tetrahedron has volume 8. What is the volume of the set of all the points in the space (notnecessarily inside the tetrahedron) that are closer to the center of the tetrahedron than any of the fourvertices?

33. [25] Determine the value of H4.

34. [25] Determine the value of M8.

35. [25] Determine the value of M9.

36. [25] Determine the value of T2.

HMMT February 2019February 16, 2019

Guts Round

1. [3] Find the sum of all real solutions to x2 + cosx = 2019.

Proposed by: Evan Chen

Answer: 0

The left-hand side is an even function, hence for each x that solves the equation, −x will also be asolution. Pairing the solutions up in this way, we get that the sum must be 0.

2. [3] There are 100 people in a room with ages 1, 2, . . . , 100. A pair of people is called cute if each ofthem is at least seven years older than half the age of the other person in the pair. At most how manypairwise disjoint cute pairs can be formed in this room?

Proposed by: Yuan Yao

Answer: 43

For a cute pair (a, b) we would have

a ≥ b

2+ 7, b ≥ a

2+ 7.

Solving the system, we get that a and b must both be at least 14. However 14 could only be pairedwith itself or a smaller number; therefore, only people with age 15 or above can be paired with someoneof different age. Pairing consecutive numbers (15, 16), (17, 18), . . . , (99, 100) works, giving 100−14

2 = 43pairs.

3. [3] Let S(x) denote the sum of the digits of a positive integer x. Find the maximum possible value ofS(x+ 2019)− S(x).

Proposed by: Alec Sun

Answer: 12

We note that S(a+ b) ≤ S(a) + S(b) for all positive a and b, since carrying over will only decrease thesum of digits. (A bit more rigorously, one can show that S(x + a · 10b) − S(x) ≤ a for 0 ≤ a ≤ 9.)Hence we have S(x+ 2019)− S(x) ≤ S(2019) = 12, and equality can be achieved with x = 100000 forexample.

4. [3] Tessa has a figure created by adding a semicircle of radius 1 on each side of an equilateral trianglewith side length 2, with semicircles oriented outwards. She then marks two points on the boundary ofthe figure. What is the greatest possible distance between the two points?

Proposed by: Yuan Yao

Answer: 3

Note that both points must be in different semicircles to reach the maximum distance. Let these pointsbe M and N , and O1 and O2 be the centers of the two semicircles where they lie respectively. Then

MN ≤MO1 +O1O2 +O2N.

Note that the the right side will always be equal to 3 (MO1 = O2N = 1 from the radius condition,and O1O2 = 1 from being a midline of the equliateral triangle), hence MN can be at most 3. Finally,if the four points are collinear (when M and N are defined as the intersection of line O1O2 with thetwo semicircles), then equality will hold. Therefore, the greatest possible distance between M and Nis 3.

5. [4] Call a positive integer n weird if n does not divide (n−2)!. Determine the number of weird numbersbetween 2 and 100 inclusive.

Proposed by: Yuan Yao

Answer: 26

We claim that all the weird numbers are all the prime numbers and 4. Since no numbers between 1and p− 2 divide prime p, (p− 2)! will not be divisible by p. We also have 2! = 2 not being a multipleof 4.

Now we show that all other numbers are not weird. If n = pq where p 6= q and p, q ≥ 2, then sincep and q both appear in 1, 2, . . . , n − 2 and are distinct, we have pq | (n − 2)!. This leaves the onlycase of n = p2 for prime p ≥ 3. In this case, we can note that p and 2p are both less than p2 − 2, so2p2 | (n− 2)! and we are similarly done.

Since there are 25 prime numbers not exceeding 100, there are 25 + 1 = 26 weird numbers.

6. [4] The pairwise products ab, bc, cd, and da of positive integers a, b, c, and d are 64, 88, 120, and 165 insome order. Find a+ b+ c+ d.

Proposed by: Anders Olsen

Answer: 42

The sum ab+ bc+ cd+ da = (a+ c)(b+ d) = 437 = 19 · 23, so {a+ c, b+ d} = {19, 23} as having eitherpair sum to 1 is impossible. Then the sum of all 4 is 19 + 23 = 42. (In fact, it is not difficult to see thatthe only possible solutions are (a, b, c, d) = (8, 8, 11, 15) or its cyclic permutations and reflections.)

7. [4] For any real number α, define

sign(α) =

+1 if α > 0,

0 if α = 0,

−1 if α < 0.

How many triples (x, y, z) ∈ R3 satisfy the following system of equations

x = 2018− 2019 · sign(y + z),

y = 2018− 2019 · sign(z + x),

z = 2018− 2019 · sign(x+ y)?

Proposed by: Pakawut Jiradilok

Answer: 3

Since sign(x + y) can take one of 3 values, z can be one of 3 values: 4037, 2018, or −1. The sameis true of x and y. However, this shows that x + y cannot be 0, so z can only be 4037 or −1. Thesame is true of x and y. Now note that, if any two of x, y, z are −1, then the third one must be 4037.Furthermore, if any one of x, y, z is 4037, then the other two must be −1. Thus, the only possibilityis to have exactly two of x, y, z be −1 and the third one be 4037. This means that the only remainingtriples are (−1,−1, 4037) and its permutations. These all work, so there are exactly 3 ordered triples.

8. [4] A regular hexagon PROFIT has area 1. Every minute, greedy George places the largest possibleequilateral triangle that does not overlap with other already-placed triangles in the hexagon, with tiesbroken arbitrarily. How many triangles would George need to cover at least 90% of the hexagon’s area?

Proposed by: Yuan Yao

Answer: 46

It’s not difficult to see that the first triangle must connect three non-adjacent vertices (e.g. POI),which covers area 1

2 , and leaves three 30-30-120 triangles of area 16 each. Then, the next three triangles

cover 13 of the respective small triangle they are in, and leave six 30-30-120 triangles of area 1

18 each.

This process continues, doubling the number of 30-30-120 triangles each round and the area of eachtriangle is divided by 3 each round. After 1 + 3 + 6 + 12 + 24 = 46 triangles, the remaining area is3·246·34 = 48

486 = 881 < 0.1, and the last triangle removed triangle has area 1

486 , so this is the minimumnumber necessary.

9. [5] Define P = {S,T} and let P be the set of all proper subsets of P . (A proper subset is a subset thatis not the set itself.) How many ordered pairs (S, T ) of proper subsets of P are there such that

(a) S is not a proper subset of T and T is not a proper subset of S; and

(b) for any sets S ∈ S and T ∈ T , S is not a proper subset of T and T is not a proper subset of S?

Proposed by: Yuan Yao

Answer: 7

For ease of notation, we let 0 = ∅, 1 = {S}, 2 = {T}. Then both S and T are proper subsets of{0, 1, 2}. We consider the following cases:

Case 1. If S = ∅, then S is a proper subset of anyset except the empty set, so we must have T = ∅.

Case 2. If S = {0}, then T cannot be empty, nor can it contain either 1 or 2, so we must haveT = {0}. This also implies that if S contains another element, then there would be no choice of Tbecause {0} would be a proper subset.

Case 3. If S = {1}, then T cannot contain 0, and cannot contain both 1 and 2 (or it becomes a propersuperset of S), so it can only be {1} or {2}, and both work. The similar apply when S = {2}.Case 4. If S = {1, 2}, then since T cannot contain 0, it must contain both 1 and 2 (or it becomes aproper subset of S), so T = {1, 2}.Hence, all the possibilities are

(S, T ) = (∅,∅), ({0}, {0}), ({1}, {1}), ({1}, {2}), ({2}, {1}), ({2}, {2}), ({1, 2}, {1, 2}),

for 7 possible pairs in total.

10. [5] Let

A = (1 + 2√

2 + 3√

3 + 6√

6)(2 + 6√

2 +√

3 + 3√

6)(3 +√

2 + 6√

3 + 2√

6)(6 + 3√

2 + 2√

3 +√

6),

B = (1 + 3√

2 + 2√

3 + 6√

6)(2 +√

2 + 6√

3 + 3√

6)(3 + 6√

2 +√

3 + 2√

6)(6 + 2√

2 + 3√

3 +√

6).

Compute the value of A/B.

Proposed by: Yuan Yao

Answer: 1

Note that

A = ((1 + 2√

2)(1 + 3√

3))((2 +√

3)(1 + 3√

2))((3 +√

2)(1 + 2√

3))((3 +√

3)(2 +√

2)),

B = ((1 + 3√

2)(1 + 2√

3))((2 +√

2)(1 + 3√

3))((3 +√

3)(1 + 2√

2))((2 +√

3)(3 +√

2)).

It is not difficult to check that they have the exact same set of factors, so A = B and thus the ratio is1.

11. [5] In the Year 0 of Cambridge there is one squirrel and one rabbit. Both animals multiply in numbersquickly. In particular, if there are m squirrels and n rabbits in Year k, then there will be 2m + 2019squirrels and 4n− 2 rabbits in Year k + 1. What is the first year in which there will be strictly morerabbits than squirrels?

Proposed by: Yuan Yao

Answer: 13

In year k, the number of squirrels is

2(2(· · · (2 · 1 + 2019) + 2019) + · · · ) + 2019 = 2k + 2019 · (2k−1 + 2k−2 + · · ·+ 1) = 2020 · 2k − 2019

and the number of rabbits is

4(4(· · · (4 · 1− 2)− 2)− · · · )− 2 = 4k − 2 · (4k−1 + 4k−2 + · · ·+ 1) =4k + 2

3.

For the number of rabbits to exceed that of squirrels, we need

4k + 2 > 6060 · 2k − 6057⇔ 2k > 6059.

Since 213 > 6059 > 212, k = 13 is the first year for which there are more rabbits than squirrels.

12. [5] Bob is coloring lattice points in the coordinate plane. Find the number of ways Bob can color fivepoints in {(x, y) | 1 ≤ x, y ≤ 5} blue such that the distance between any two blue points is not aninteger.

Proposed by: Michael Ren

Answer: 80

We can see that no two blue points can have the same x or y coordinate. The blue points then mustmake a permutation of 1, 2, 3, 4, 5 that avoid the pattern of 3-4-5 triangles. It is not hard to usecomplementary counting to get the answer from here.

There are 8 possible pairs of points that are a distance of 5 apart while not being in the same rowor column (i.e. a pair that is in the 3-4-5 position). If such a pair of points is included in the choiceof five points, then there are 3! = 6 possibilities for the remaining three points, yielding 8 × 6 = 48configurations that have violations. However, we now need to consider overcounting.

The only way to have more than one violation in one configuration is to have a corner point and thentwo points adjacent to the opposite corner, e.g. (1, 1), (4, 5), (5, 4). In each such case, there are exactly2! = 2 possibilities for the other two points, and there are exactly two violations so there are a total of2× 4 = 8 configurations that are double-counted.

Therefore, there are 48−8 = 40 permutations that violate the no-integer-condition, leaving 120−40 =80 good configurations.

13. [7] Reimu has 2019 coins C0, C1, . . . , C2018, one of which is fake, though they look identical to eachother (so each of them is equally likely to be fake). She has a machine that takes any two coins andpicks one that is not fake. If both coins are not fake, the machine picks one uniformly at random. Foreach i = 1, 2, . . . , 1009, she puts C0 and Ci into the machine once, and machine picks Ci. What is theprobability that C0 is fake?

Proposed by: Yuan Yao

Answer: 21009

21009+1009

Let E denote the event that C0 is fake, and let F denote the event that the machine picks Ci over C0 for

all i = 1, 2, ...1009. By the definition of conditional probability, P (E|F ) = P (E∩F )P (F ) . Since E implies F,

P (E∩F ) = P (E) = 12019 . Now we want to compute P (F ). If C0 is fake, F is guaranteed to happen. If Ci

is fake for some 1 ≤ i ≤ 1009, then F is impossible. Finally, if Ci is fake for some 1010 ≤ i ≤ 2018, thenF occurs with probability 2−1009, since there is a 1

2 probability for each machine decision. Therefore,

P (F ) = 12019 · 1 + 1009

2019 · 0 + 10092019 · 2

−1009 = 21009+10092019·21009 . Therefore, P (E|F ) = 21009

21009+1009 .

14. [7] Let ABC be a triangle where AB = 9, BC = 10, CA = 17. Let Ω be its circumcircle, and letA1, B1, C1 be the diametrically opposite points from A,B,C, respectively, on Ω. Find the area of theconvex hexagon with the vertices A,B,C,A1, B1, C1.

Proposed by: Yuan Yao

Answer: 11554

We first compute the circumradius of ABC: Since cosA = 92−172−1022·9·17 = − 15

17 , we have sinA = 817 and

R = a2 sinA = 170

16 . Moreover, we get that the area of triangle ABC is 12bc sinA = 36.

Note that triangle ABC is obtuse, The area of the hexagon is equal to twice the area of triangle ABC(which is really [ABC] + [A1B1C1]) plus the area of rectangle ACA1C1. The dimensions of ACA1C1

are AC = 17 and A1C =√

(2R)2 −AC2 = 514 , so the area of the hexagon is 36 · 2 + 17 · 514 = 1155

4 .

15. [7] Five people are at a party. Each pair of them are friends, enemies, or frenemies (which is equivalentto being both friends and enemies). It is known that given any three people A,B,C:

• If A and B are friends and B and C are friends, then A and C are friends;

• If A and B are enemies and B and C are enemies, then A and C are friends;

• If A and B are friends and B and C are enemies, then A and C are enemies.

How many possible relationship configurations are there among the five people?

Proposed by: Yuan Yao

Answer: 17

If A and B are frenemies, then regardless of whether another person C is friends or enemies with A,C will have to be frenemies with B and vice versa. Therefore, if there is one pair of frenemies then allof them are frenemies with each other, and there is only one possibility.

If there are no frenemies, then one can always separate the five people into two possibly “factions”(one of which may be empty) such that two people are friends if and only if they belong to the samefaction. Since the factions are unordered, there are 25/2 = 16 ways to assign the “alignments” thateach gives a unique configuration of relations. So in total there are 16 + 1 = 17 possibilities.

16. [7] Let R be the set of real numbers. Let f : R→ R be a function such that for all real numbers x andy, we have

f(x2) + f(y2) = f(x+ y)2 − 2xy.

Let S =2019∑

n=−2019f(n). Determine the number of possible values of S.

Proposed by: Yuan Yao

Answer: 2039191 OR(20202

)+ 1

Letting y = −x givesf(x2) + f(x2) = f(0)2 + 2x2

for all x. When x = 0 the equation above gives f(0) = 0 or f(0) = 2.

If f(0) = 2, then f(x) = x+ 2 for all nonegative x, so the LHS becomes x2 + y2 + 4, and RHS becomesx2 + y2 + 4x+ 4y + 4 for all x+ y ≥ 0, which cannot be equal to LHS if x+ y > 0.

If f(0) = 0 then f(x) = x for all nonnegative x. Moreover, letting y = 0 gives

f(x2) = f(x)2 ⇒ f(x) = ±x

for all x. Since negative values are never used as inputs on the LHS and the output on the RHS isalways squared, we may conclude that for all negative x, f(x) = x and f(x) = −x are both possible(and the values are independent). Therefore, the value of S can be written as

S = f(0) + (f(1) + f(−1)) + (f(2) + f(−2)) + · · ·+ (f(2019) + f(−2019)) = 2

2019∑i=1

iδi

for δ1, δ2, . . . , δ2019 ∈ {0, 1}. It is not difficult to see that S2 can take any integer value between 0 and

2020·20192 = 2039190 inclusive, so there are 2039191 possible values of S.

17. [9] Let ABC be a triangle with AB = 3, BC = 4, and CA = 5. Let A1, A2 be points on side BC,B1, B2 be points on side CA, and C1, C2 be points on side AB. Suppose that there exists a point Psuch that PA1A2, PB1B2, and PC1C2 are congruent equilateral triangles. Find the area of convexhexagon A1A2B1B2C1C2.

Proposed by: Michael Ren

Answer: 12+22√3

15

Since P is the shared vertex between the three equilateral triangles, we note that P is the incenterof ABC since it is equidistant to all three sides. Since the area is 6 and the semiperimeter is also 6,we can calculate the inradius, i.e. the altitude, as 1, which in turn implies that the side length of theequilateral triangle is 2√

3. Furthermore, since the incenter is the intersection of angle bisectors, it is

easy to see that AB2 = AC1, BC2 = BA1, and CA2 = CB1. Using the fact that the altitudes fromP to AB and CB form a square with the sides, we use the side lengths of the equilateral triangle tocompute that AB2 = AC1 = 2− 1√

3, BA1 = BC2 = 1− 1√

3, and CB1 = CA2 = 3− 1√

3. We have that

the area of the hexagon is therefore

6−

(1

2

(2− 1√

3

)2

· 4

5+

1

2

(1− 1√

3

)2

+1

2

(3− 1√

3

)2

· 3

5

)=

12 + 22√

3

15.

18. [9] 2019 points are chosen independently and uniformly at random on the interval [0, 1]. Tairitsupicks 1000 of them randomly and colors them black, leaving the remaining ones white. Hikari thencomputes the sum of the positions of the leftmost white point and the rightmost black point. What isthe probability that this sum is at most 1?

Proposed by: Yuan Yao

Answer: 10192019

Note that each point is chosen uniformly and independently from 0 to 1, so we can apply symmetry.Given any coloring, suppose that we flip all the positions of the black points: then the problem becomescomputing the probability that the leftmost white point is to the left of the leftmost black point, whichis a necessary and sufficient condition for the sum of the original leftmost white point and the originalrightmost black point being at most 1. This condition, however, is equivalent to the leftmost point ofall 2019 points being white. Since there are 1019 white points and 1000 black points and each point isequally likely to be the leftmost, this happens with probability 1019

2019 .

19. [9] Complex numbers a, b, c form an equilateral triangle with side length 18 in the complex plane. If|a+ b+ c| = 36, find |bc+ ca+ ab|.Proposed by: Henrik Boecken

Answer: 432

Using basic properties of vectors, we see that the complex number d = a+b+c3 is the center of the

triangle. From the given, |a+ b+ c| = 36 =⇒ |d| = 12. Then, let a′ = a− d, b′ = b− d, and c′ = c− d.Due to symmetry, |a′ + b′ + c′| = 0 and |b′c′ + c′a′ + a′b′| = 0.

Finally, we compute

|bc+ ca+ ab| = |(b′ + d)(c′ + d) + (c′ + d)(a′ + d) + (a′ + d)(b′ + d)|= |b′c′ + c′a′ + a′b′ + 2d(a′ + b′ + c′) + 3d2|= |3d2| = 3 · 122 = 432.

20. [9] On floor 0 of a weird-looking building, you enter an elevator that only has one button. You pressthe button twice and end up on floor 1. Thereafter, every time you press the button, you go up byone floor with probability X

Y , where X is your current floor, and Y is the total number of times youhave pressed the button thus far (not including the current one); otherwise, the elevator does nothing.

Between the third and the 100th press inclusive, what is the expected number of pairs of consecutivepresses that both take you up a floor?

Proposed by: Kevin Yang

Answer: 973

By induction, we can determine that after n total button presses, your current floor is uniformlydistributed from 1 to n − 1: the base case n = 2 is trivial to check, and for the n + 1th press, theprobability that you are now on floor i is 1

n−1 (1− in ) + 1

n−1 ( i−1n ) = 1n for i = 1, 2, . . . , n, finishing the

inductive step.

Hence, the probability that the (n+ 1)-th and (n+ 2)-th press both take you up a floor is

1

n− 1

n−1∑i=1

i

n· i+ 1

n+ 1=

∑n−1i=1 i

2 + i

(n− 1)n(n+ 1)=

(n−1)n(2n−1)6 + n(n−1)

2

(n− 1)n(n+ 1)=

1

3.

Since there are 100− 3 = 97 possible pairs of consecutive presses, the expected value is 973 .

21. [12] A regular hexagon ABCDEF has side length 1 and center O. Parabolas P1, P2, . . . , P6 areconstructed with common focus O and directrices AB,BC,CD,DE,EF, FA respectively. Let χ bethe set of all distinct points on the plane that lie on at least two of the six parabolas. Compute∑

X∈χ|OX|.

(Recall that the focus is the point and the directrix is the line such that the parabola is the locus ofpoints that are equidistant from the focus and the directrix.)

Proposed by: Yuan Yao

Answer: 35√

3

Recall the focus and the directrix are such that the parabola is the locus of points equidistant fromthe focus and the directrix. We will consider pairs of parabolas and find their points of intersections(we label counterclockwise):

(1): P1 ∩ P2, two parabolas with directrices adjacent edges on the hexagon (sharing vertex A). Theintersection inside the hexagon can be found by using similar triangles: by symmetry this X must lieon OA and must have that its distance from AB and FA are equal to |OX| = x, which is to say

sin 60◦ =

√3

2=

x

|OA| − x=

x

1− x=⇒ x = 2

√3− 3.

By symmetry also, the second intersection point, outside the hexagon, must lie on OD. Furthermore,X must have that its distance AB and FA are equal to |OX|. Then again by similar triangles

sin 60◦ =

√3

2=

x

|OA|+ x=

x

1 + x=⇒ x = 2

√3 + 3.

(2): P1 ∩ P3, two parabolas with directrices edges one apart on the hexagon, say AB and CD. Theintersection inside the hexagon is clearly immediately the circumcenter of triangle BOC (equidistancecondition), which gives

x =

√3

3.

Again by symmetry the X outside the hexagon must lie on the lie through O and the midpoint of EF ;then one can either observe immediately that x =

√3 or set up

sin 30◦ =1

2=

x

x+√

3=⇒ x =

√3

where we notice√

3 is the distance from O to the intersection of AB with the line through O and themidpoint of BC.

(3): P1 ∩ P4, two parabolas with directrices edges opposite on the hexagon, say AB and DE. Clearlythe two intersection points are both inside the hexagon and must lie on CF , which gives

x =

√3

2.

These together give that the sum desired is

6(2√

3− 3) + 6(2√

3 + 3) + 6

(√3

3

)+ 6(√

3) + 6

(√3

2

)= 35

√3.

22. [12] Determine the number of subsets S of {1, 2, . . . , 1000} that satisfy the following conditions:

• S has 19 elements, and

• the sum of the elements in any non-empty subset of S is not divisible by 20.

Proposed by: Alec Sun

Answer: 8 ·(5019

)First we prove that each subset must consist of elements that have the same residue mod 20. Let asubset consist of elements a1, . . . , a19, and consider two lists of partial sums

a1, a1 + a2, a1 + a2 + a3, . . . , a1 + a2 + · · ·+ a19

a2, a1 + a2, a1 + a2 + a3, . . . , a1 + a2 + · · ·+ a19.

The residues mod 20 of the partial sums in each list must be pairwise distinct, otherwise subtractingthe sum with less terms from the sum with more terms yields a subset whose sum of elements is 0(mod 20). Since the residues must also be nonzero, each list forms a complete nonzero residue classmod 20. Since the latter 18 sums in the two lists are identical, a1 ≡ a2 (mod 20). By symmetricarguments, ai ≡ aj (mod 20) for any i, j.

Furthermore this residue 1 ≤ r ≤ 20 must be relatively prime to 20, because if d = gcd(r, 20) > 1then any 20/d elements of the subset will sum to a multiple of 20. Hence there are ϕ(20) = 8 possibleresidues. Since there are 50 elements in each residue class, the answer is

(5019

). We can see that any

such subset whose elements are a relatively prime residue r (mod 20) works because the sum of any1 ≤ k ≤ 19 elements will be kr 6= 0 (mod 20).

23. [12] Find the smallest positive integer n such that

222···

2︸ ︷︷ ︸n 2’s

> ((· · · ((100!)!)! · · · )!)!︸ ︷︷ ︸100 factorials

.

Proposed by: Zack Chroman

Answer: 104

Note that 2222

> 1002. We claim that a > b2 =⇒ 2a > (b!)2, for b > 2. This is because

2a > b2b ⇐⇒ a > 2b log2(b),

and log2(b) < b2/2 for b > 2. Then since bb > b! this bound works. Then

(222...2

)︸ ︷︷ ︸m 2’s

> ((((100!)!)!)! . . . )2︸ ︷︷ ︸m−4 factorials

for all m ≥ 4 by induction. So n = 104 works. The lower bound follows from the fact that n! > 2n forn > 3, and since 100 > 22

2

, we have

(((100!)!)!)! . . . )︸ ︷︷ ︸100 factorials

> 22...2

100︸ ︷︷ ︸100 2’s

> 22...2︸ ︷︷ ︸

103 2’s

.

24. [12] Let S be the set of all positive factors of 6000. What is the probability of a random quadruple(a, b, c, d) ∈ S4 satisfies

lcm(gcd(a, b), gcd(c, d)) = gcd(lcm(a, b), lcm(c, d))?

Proposed by: Yuan Yao

Answer: 41512

For each prime factor, let the greatest power that divides a, b, c, d be p, q, r, s. WLOG assume thatp ≤ q and r ≤ s, and further WLOG assume that p ≤ r. Then we need r = min(q, s). If q = r thenwe have p ≤ q = r ≤ s, and if r = s then we have p ≤ r = s ≤ q, and in either case the conditionreduces to the two “medians” among p, q, r, s are equal. (It is not difficult to see that this condition isalso sufficient.)

Now we compute the number of quadruples (p, q, r, s) of integers between 0 and n inclusive that satisfythe above condition. If there are three distinct numbers then there are

(n+13

)ways to choose the three

numbers and 4!/2 = 12 ways to assign them (it must be a 1-2-1 split). If there are two distinct numbersthen there are

(n+12

)ways to choose the numbers and 4 + 4 = 8 ways to assign them (it must be a 3-1

or a 1-3 split). If there is one distinct number then there are n+ 1 ways to assign. Together we have

12

(n+ 1

3

)+ 8

(n+ 1

2

)+ (n+ 1) = 2(n+ 1)n(n− 1) + 4(n+ 1)n+ (n+ 1) = (n+ 1)(2n(n+ 1) + 1)

possible quadruples. So if we choose a random quadruple then the probability that it satisfies the

condition is (n+1)(2n(n+1)+1)(n+1)4 = 2n(n+1)+1

(n+1)3 .

Since 6000 = 24 · 53 · 31 and the power of different primes are independent, we plug in n = 4, 3, 1 to getthe overall probability to be

41

125· 25

64· 5

8=

41

512.

25. [15] A 5 by 5 grid of unit squares is partitioned into 5 pairwise incongruent rectangles with sides lyingon the gridlines. Find the maximum possible value of the product of their areas.

Proposed by: Yuan Yao

Answer: 2304

The greatest possible value for the product is 3 · 4 · 4 · 6 · 8 = 2304, achieved when the rectangles are3 × 1, 1 × 4, 2 × 2, 2 × 3, 4 × 2. To see that this is possible, orient these rectangles so that the firstnumber is the horizontal dimension and the second number is the vertical dimension. Then, place thebottom-left corners of these rectangles at (2, 4), (4, 0), (2, 2), (0, 2), (0, 0) respectively on the grid.

We will now prove that no larger product can be achieved. Suppose that there is at least one rectangleof area at most 2. Then the product is at most 2·5.754 = 2·33.06252 < 2·1100 = 2200 by AM-GM. Nowsuppose that there is at least one rectangle of area at least 9. Then the product is at most 9 ·44 = 2304by AM-GM. (Neither of these is tight, since you cannot have non-integer areas, nor can you have fourrectangles all of area 4.)

Now consider the last possibility that is not covered by any of the above: that there are no rectangles ofsize at most 2 and no rectangles of area at least 9. There can be at most one rectangle of area 3, 5, 6, 8each, at most two rectangles of area 4, and no rectangles of area 7. The only way to achieve a sum of25 with these constraints is 3, 4, 4, 6, 8, which produces a product of 2304. We have shown through theearlier cases that a larger product cannot be achieved, so this is indeed the maximum.

26. [15] Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Let IA, IB , IC be the A,B,C excentersof this triangle, and let O be the circumcenter of the triangle. Let γA, γB , γC be the correspondingexcircles and ω be the circumcircle. X is one of the intersections between γA and ω. Likewise, Y is anintersection of γB and ω, and Z is an intersection of γC and ω. Compute

cos∠OXIA + cos∠OY IB + cos∠OZIC .

Proposed by: Andrew Gu

Answer: − 4965

Let rA, rB , rC be the exradii. Using OX = R,XIA = rA, OIA =√R(R+ 2rA) (Euler’s theorem for

excircles), and the Law of Cosines, we obtain

cos∠OXIA =R2 + r2A −R(R+ 2rA)

2RrA=rA2R− 1.

Therefore it suffices to compute rA+rB+rC2R − 3. Since

rA + rB + rC − r = 2K

(1

−a+ b+ c+

1

a− b+ c+

1

a+ b− c− 1

a+ b+ c

)= 2K

8abc

(4K)2=abc

K= 4R

where K = [ABC], this desired quantity the same as r2R − 1. For this triangle, r = 4 and R = 65

8 , sothe answer is 4

65/4 − 1 = − 4965 .

27. [15] Consider the eighth-sphere {(x, y, z) | x, y, z ≥ 0, x2 + y2 + z2 = 1}. What is the area of itsprojection onto the plane x+ y + z = 1?

Proposed by: Yuan Yao

Answer: π√3

4

Consider the three flat faces of the eighth-ball. Each of these is a quarter-circle of radius 1, so each hasarea π

4 . Furthermore, the projections of these faces cover the desired area without overlap. To find theprojection factor one can find the cosine of the angle θ between the planes, which is the same as theangle between their normal vectors. Using the dot product formula for the cosine of the angle between

two vectors, cos θ = (1,0,0)·(1,1,1)|(1,0,0)||(1,1,1)| = 1√

3. Therefore, each area is multiplied by 1√

3by the projection,

so the area of the projection is 3 · π4 ·1√3

= π√3

4 .

28. [15] How many positive integers 2 ≤ a ≤ 101 have the property that there exists a positive integer Nfor which the last two digits in the decimal representation of a2

n

is the same for all n ≥ N?

Proposed by: Pakawut Jiradilok

Answer: 36

Solution 1. It suffices to consider the remainder mod 100. We start with the four numbers that havethe same last two digits when squared: 0, 1, 25, 76.

We can now go backwards, repeatedly solving equations of the form x2 ≡ n (mod 100) where n is anumber that already satisfies the condition.

0 and 25 together gives all multiples of 5, for 20 numbers in total.

1 gives 1, 49, 51, 99, and 49 then gives 7, 43, 57, 93. Similarly 76 gives 24, 26, 74, 76, and 24 then gives18, 32, 68, 82, for 16 numbers in total.

Hence there are 20 + 16 = 36 such numbers in total.

Solution 2. An equivalent formulation of the problem is to ask for how many elements of Z100 themap x 7→ x2 reaches a fixed point. We may separately solve this modulo 4 and modulo 25.

Modulo 4, it is easy to see that all four elements work.

Modulo 25, all multiples of 5 will work, of which there are 5. For the remaining 25 elements that arecoprime to 5, we may use the existence of a primitive root to equivalently ask for how many elementsof Z20 the map y 7→ 2y reaches a fixed point. The only fixed point is 0, so the only valid choices arethe multiples of 5 again. There are 5 + 4 = 9 solutions here.

Finally, the number of solutions modulo 100 is 4× 9 = 36.

29. [20] Yannick picks a number N randomly from the set of positive integers such that the probabilitythat n is selected is 2−n for each positive integer n. He then puts N identical slips of paper numbered1 through N into a hat and gives the hat to Annie. Annie does not know the value of N, but she drawsone of the slips uniformly at random and discovers that it is the number 2. What is the expected valueof N given Annie’s information?

Proposed by: Yuan Yao

Answer: 12 ln 2−1

Let S denote the value drawn from the hat. The probability that 2 is picked is 1n if n ≥ 2 and 0 if

n = 1. Thus, the total probability X that 2 is picked is

P (S = 2) =

∞∑k=2

2−k

k.

By the definition of conditional probability, P (N = n|S = 2) = P (N=n,S=2)P (S=2) = 2−n/n

X if n ≥ 2 and 0 if

n = 1. Thus the conditional expectation of N is

E[N |S = 2] =

∞∑n=1

n · P (N = n|S = 2) =

∞∑n=2

n · 2−n/n

X=

1

X

∞∑n=2

2−n =1

2X.

It remains to compute X. Note that∑∞k=0 x

k = 11−x for |x| < 1. Integrating both sides with respect

to x yields∞∑k=1

xk

k= − ln(1− x) + C

for some constant C, and plugging in x = 0 shows that C = 0. Plugging in x = 12 shows that∑∞

k=12−k

k = ln 2. Note that X is exactly this summation but without the first term. Thus, X = ln 2− 12 ,

so 12X = 1

2 ln 2−1 .

30. [20] Three points are chosen inside a unit cube uniformly and independently at random. What is theprobability that there exists a cube with side length 1

2 and edges parallel to those of the unit cube thatcontains all three points?

Proposed by: Yuan Yao

Answer: 18

Let the unit cube be placed on a xyz-coordinate system, with edges parallel to the x, y, z axes. Supposethe three points are labeled A,B,C. If there exists a cube with side length 1

2 and edges parallel tothe edges of the unit cube that contain all three points, then there must exist a segment of length 1

2that contains all three projections of A,B,C onto the x-axis. The same is true for the y- and z- axes.Likewise, if there exists segments of length 1

2 that contains each of the projections of A,B,C onto thex,y, and z axes, then there must exist a unit cube of side length 1

2 that contains A,B,C. It is easyto see that the projection of a point onto the x-axis is uniform across a segment of length 1, and thateach of the dimensions are independent. The problem is therefore equivalent to finding the cube of theprobability that a segment of length 1

2 can cover three points chosen randomly on a segment of length1.

Note that selecting three numbers p < q < r uniformly and independently at random from 0 to 1 splitsthe number line into four intervals. That is, we can equivalently sample four positive numbers a, b, c, d

uniformly satisfying a+b+c+d = 1 (here, we set a = p, b = q−p, c = r−q, d = 1−r). The probabilitythat the points p, q, r all lie on a segment of length 1

2 is the probability that r − q ≤ 12 , or b + c ≤ 1

2 .Since a+d and b+c are symmetric, we have that this probability is 1

2 and our final answer is (12 )3 = 1

8 .

31. [20] Let ABC be a triangle with AB = 6, AC = 7, BC = 8. Let I be the incenter of ABC. Points Zand Y lie on the interior of segments AB and AC respectively such that Y Z is tangent to the incircle.Given point P such that

∠ZPC = ∠Y PB = 90◦,

find the length of IP .

Proposed by: Zack Chroman

Answer:√302

Solution 1. Let PU,PV tangent from P to the incircle. We will invoke the dual of the DesarguesInvolution Theorem, which states the following:

Given a point P in the plane and four lines `1, `2, `3, `4, consider the set of conics tangent to all fourlines. Then we define a function on the pencil of lines through P by mapping one tangent from Pto each conic to the other. This map is well defined and is a projective involution, and in particularmaps PA→ PD,PB → PE,PC → PF , where ABCDEF is the complete quadrilateral given by thepairwise intersections of `1, `2, `3, `4. �

An overview of the projective background behind the (Dual) Desargues Involution Theorem can befound here: https://www.scribd.com/document/384321704/Desargues-Involution-Theorem, anda proof can be found at https://www2.washjeff.edu/users/mwoltermann/Dorrie/63.pdf.

Now, we apply this to the point P and the lines AB,AC,BC, Y Z, to get that the pairs

(PU,PV ), (PY, PB), (PZ, PC)

are swapped by some involution. But we know that the involution on lines through P which rotatesby 90◦ swaps the latter two pairs, thus it must also swap the first one and ∠UPV = 90. It followsby equal tangents that IUPV is a square, thus IP = r

√2 where r is the inradius of ABC. Since

r = 2Ka+b+c = 21

√15/221 =

√152 , we have IP =

√302 .

Solution 2. Let H be the orthocenter of ABC.

Lemma. HI2 = 2r2 − 4R2 cos(A) cos(B) cos(C), where r is the inradius and R is the circumradius.

Proof. This follows from barycentric coordinates or the general result that for a point X in the plane,

aXA2 + bXB2 + cXC2 = (a+ b+ c)XI2 + aAI2 + bBI2 + cCI2,

which itself is a fact about vectors that follows from barycentric coordinates. This can also be computeddirectly using trigonometry.

Let E = BH ∩ AC,F = CH ∩ AB, then note that B,P,E, Y are concyclic on the circle of diameterBY , and C,P, F, Z are concyclic on the circle of diameter CZ. Let Q be the second intersection ofthese circles. Since BCY Z is a tangential quadrilateral, the midpoints of BY and CZ are collinearwith I (this is known as Newton’s theorem), which implies that IP = IQ by symmetry. Note that asBH ·HE = CH ·HF , H lies on the radical axis of the two circles, which is PQ. Thus, if IP = IQ = x,BH · HE is the power of H with respect to the circle centered at I with radius x, which impliesBH ·HE = x2 −HI2.As with the first solution, we claim that x = r

√2, which by the lemma is equivalent to BH ·HE =

4R2 cos(A) cos(B) cos(C). Then note that

BH ·HE = BH · CH cos(A) = (2R cos(B))(2R cos(C)) cos(A),

so our claim holds and we finish as with the first solution.

Note. Under the assumption that the problem is well-posed (the answer does not depend on the choiceof Y,Z, or P ), then here is an alternative method to obtain IP = r

√2 by making convenient choices.

Let U be the point where Y Z is tangent to the incircle, and choose U so that IU ‖ BC (and thereforeY Z ⊥ BC). Note that Y Z ∩BC is a valid choice for P , so assume that P is the foot from U to BC.If D is the point where BC is tangent to the incircle, then IUPD is a square so IP = r

√2. (This

disregards the condition that Y and Z are in the interior of segments AC and AB, but there is noreason to expect that this condition is important.)

32. [20] For positive integers a and b such that a is coprime to b, define ordb(a) as the least positive integerk such that b | ak − 1, and define ϕ(a) to be the number of positive integers less than or equal to awhich are coprime to a. Find the least positive integer n such that

ordn(m) <ϕ(n)

10

for all positive integers m coprime to n.

Proposed by: Andrew Gu

Answer: 240

The maximum order of an element modulo n is the Carmichael function, denoted λ(n). The followingproperties of the Carmichael function are established:

• For primes p > 2 and positive integers k, λ(pk) = (p− 1)pk−1.

• For a positive integer k,

λ(2k) =

{2k−2 if k ≥ 3

2k−1 if k ≤ 2.

• For a positive integer n with prime factorization n =∏pkii ,

λ(n) = lcm(λ(pk11 ), λ(pk22 ), . . .

)Meanwhile, for n =

∏pkii , we have ϕ(n) =

∏(pi − 1)pki−1i . Hence the intuition is roughly that the

(pi − 1)pki−1i terms must share divisors in order to reach a high value of ϕ(n)λ(n) .

We will now show that n ≥ 240 by doing casework on the prime divisors of z = ϕ(n)λ(n) . Suppose p | z

and p > 2. This requires two terms among λ(pk11 ), λ(pk22 ), . . . to be multiples of p because λ(n) is thelcm of the terms whereas the product of these numbers has the same number of factors of p as ϕ(n)(note that this does not hold for p = 2 because λ(2k) 6= 2k−1 in general). These correspond to eitherp2 | n or q | n with q ≡ 1 (mod p). Therefore

n ≥ max(p2(2p+ 1), (2p+ 1)(4p+ 1))

because the smallest primes congruent to 1 (mod p) are at least 2p+ 1 and 4p+ 1. For p ≥ 5 this givesn > 240, so we may assume p ≤ 3.

First we address the case p = 3. This means that two numbers among 9, 7, 13, 19, 31, 37, . . . divide n.As 7× 37 > 240, we discard primes greater than 31. Of the remaining numbers, we have

λ(9) = 6, λ(7) = 6, λ(13) = 12, λ(19) = 18, λ(31) = 30.

No candidate value of n is the product of just two of these numbers as the gcd of any two of theassociated λ values is at most 6. Furthermore, multiplying by just 2 will not affect ϕ(n) or λ(n), sowe must multiply at least two of these numbers by a number greater than 2. Throwing out numbersgreater than 240, this leaves only 3×9×7, which does not work. (A close candidate is 3×7×13 = 273,for which ϕ(n) = 144, λ(n) = 12.)

The remaining case is when the only prime divisors of ϕ(n)λ(n) are 2. It is not hard to see that λ(n) ≥ 4

when n - 24 (and when n | 24 it’s clear that φ(n) ≤ 8, so we do not need to consider them). Whenλ(n) = 4, we need ϕ(n) ≥ 4 · 24 = 64 and v2(n) ≤ 4, so the smallest such integer is n = 24 · 3 · 5 = 240,

which we can check does indeed satisfy ϕ(n)λ(n) > 10. It is not difficult to check that higher values of λ(n)

will not yield any n below 240, so 240 is indeed the smallest possible n.

Note: The sequence ϕ(n)λ(n) is given by A034380 in the OEIS.

H1. Let r = H1 be the answer to this problem. Given that r is a nonzero real number, what is the valueof r4 + 4r3 + 6r2 + 4r?

Answer: −1

Since H1 is the answer, we know r4 + 4r3 + 6r2 + 4r = r ⇒ (r + 1)4 = r + 1. Either r + 1 = 0, or(r + 1)3 = 1⇒ r = 0. Since r is nonzero, r = −1.

H2. Given two distinct points A, B and line ` that is not perpendicular to AB, what is the maximumpossible number of points P on ` such that ABP is an isosceles triangle?

Answer: 5

In an isosceles triangle, one vertex lies on the perpendicular bisector of the opposite side. Thus, eitherP is the intersection of AB and `, or P lies on the circle centered at A with radius AB, or P lies onthe circle centered at B with radius AB. Each circle-line intersection has at most two solutions, andthe line-line intersection has at most one, giving 5. This can be easily constructed by taking any AB,and taking ` that isn’t a diameter but intersects both relevant circles twice.

H3. Let A = H1, B = H6 + 1. A real number x is chosen randomly and uniformly in the interval [A,B].Find the probability that x2 > x3 > x.

Answer: 14

A = −1, B = 3. For x3 > x, either x > 1 or −1 < x < 0. However, for x > 1, x2 < x3, so there are nosolutions. −1 < x < 0 also satisfies x2 > x3, so our answer is 1/4.

H4. Let A = d1/H3e, B = dH5/2e. How many ways are there to partition the set {1, 2, . . . , A + B} intotwo sets U and V with size A and B respectively such that the probability that a number chosen fromU uniformly at random is greater than a number chosen from V uniformly at random is exactly 1

2?

Answer: 24

A = 4, B = 7. There are 28 total ways of choosing an element from U and V , so there must be 14ways where U ’s is larger. If we relabel the elements to be 0, 1, · · · , 10, then element i is greater thanexactly i elements in the set. However, we overcount other elements in U , so the four elements inU = {a, b, c, d} must satisfy

(a− 0) + (b− 1) + (c− 2) + (d− 3) = 14⇒ a+ b+ c+ d = 20.

To remove the uniqueness condition, we subtract 1 from b, 2 from c, and 3 from d, so we wish to findsolutions a ≤ b ≤ c ≤ d ≤ 7 to a+ b+ c+ d = 14. From here, we do casework. If a = 0, b = 0, 1, 2, 3, 4give 1, 1, 2, 2, 3 solutions, respectively. If a = 1, b = 1, 2, 3, 4 give 2, 2, 3, 1 solutions, respectively. Ifa = 2, b = 2, 3, 4 give 3, 2, 1 solutions, respectively. If a = 3, the only solution is 3, 3, 4, 4. Thus, theanswer is (1 + 1 + 2 + 2 + 3) + (2 + 2 + 3 + 1) + (3 + 2 + 1) + 1 = 24.

H5. Let A = H2, B = H7. Two circles with radii A and B respectively are given in the plane. If the lengthof their common external tangent is twice the length of their common internal tangent (where bothtangents are considered as segments with endpoints being the points of tangency), find the distancebetween the two centers.

Answer: 2√4293

Let the distance between the centers be d. The length of the common external tangent is E =√d2 − (7− 5)2 =

√d2 − 4, and the length of the internal tangent is I = 12

5

√( 512d)2 − 52. Solving the

equation E = 2I gives d = 2√4293 (≈ 13.8).

H6. How many ways are there to arrange the numbers 21, 22, 33, 35 in a row such that any two adjacentnumbers are relatively prime?

Answer: 2

21 cannot be adjacent to 33 or 35, so it must be on one end bordering 22. 33 cannot be adjacent to 21or 22, so it must be on the other end bording 35. Thus, there are only 2 orderings: 21, 22, 35, 33, and33, 35, 22, 21.

H7. How many pairs of integers (x, y) are there such that |x2 − 2y2| ≤ 1 and |3x− 4y| ≤ 1?

Answer: 7

Note that if (x, y) is a solution, so is (−x,−y). Thus, we consider x ≥ 0.

When x ≡ 0 (mod 4), y = 3x/4 by inequality 2. Inequality 1 gives |x2/9| ≤ 1, so x ≤ 3, so x = 0.

When x ≡ 1 (mod 4), y = (3x + 1)/4 by inequality 2. Beyond x = 1, 2y2 − x2 > 1, so there are nomore solutions.

When x ≡ 2 (mod 4), there are no solutions for y.

When x ≡ 3 (mod 4), y = (3x − 1)/4 by inequality 2. Beyond x = 7, 2y2 − x2 > 1, so there are nomore solutions.

Thus, the solutions are (0, 0), (1, 1), (3, 2), (7, 5), and the negations of the latter three, giving 7 solutions.

M1. Let S = M10. Determine the number of ordered triples (a, b, c) of nonnegative integers such thata+ 2b+ 4c = S.

Answer: 196

S = 53. Firstly, the number of solutions is the same as the number of solutions to a + 2b + 4c = 52,since 2b, 4c are both even. Then, a + 2b = 2x has x + 1 solutions in nonnegative integers, so we wishto find 27 + 25 + · · ·+ 1. This is the sum of the first 14 odd numbers, which is 142 = 196.

M2. Let S = bM5c. Two integersm and n are chosen between 1 and S inclusive uniformly and independentlyat random. What is the probability that mn = nm?

Answer: 772

S = 12. The solutions are (x, x) for all x, and (2, 4), (4, 2). Thus, there are S + 2 = 14 solutions outof S2 = 196 possibilities, so the answer is 14/144 = 7/72.

M3. Let S = dM7e. In right triangle ABC, ∠C = 90◦, AC = 27, BC = 36. A circle with radius S istangent to both AC and BC and intersects AB at X and Y . Find the length of XY .

Answer: 16√

3

S = 14. We first note that the distance from the center of the circle to AB (which has length45 by Pythagorean theorem) is 27·36−27·14−36·14

45 = 2, so the length of the chord XY is equal to

2√

142 − 22 = 16√

3.

M4. Let S = M13 + 5. Compute the product of all positive divisors of S.

Answer: 810000

S = 30 = 2 · 3 · 5. The divisors of S are 1, 2, 3, 5, 6, 10, 15, 30. Each prime factor appears 4 times, sothe product is 243454 = 304 = 810000.

M5. Let A =√M1, B = dM11e. Given complex numbers x and y such that x+ 1

y = A, 1x +y = B, compute

the value of xy + 1xy .

Answer: 12

A = 14, B = 1. Multiplying the two given equations gives xy + 1/(xy) + 2 = 14, so the answer is14− 2 = 12.

M6. Let A = b1/M2c, B = bM23 /100c. Let P and Q both be quadratic polynomials. Given that the real

roots of P (Q(x)) = 0 are 0, A,B,C in some order, find the sum of all possible values of C.

Answer: 17

A = 10, B = 7. Let the roots of P (x) = 0 be p, q. Then, the roots of P (Q(x)) = 0 are when Q(x) = por Q(x) = q. If these are r, s, t, u in order, then note that (r+s)/2 = (t+u)/2 must both be the centerof the parabola Q. Thus, 0, 10, 7, C must divide into two pairs with equal sum. When we consider thethree possible groupings, we get that the possible values are 10+7−0 = 17, 10+0−7 = 3, 7+0−10 = −3.Therefore the sum is 17 + (−3) + 3 = 17.

M7. Let A = dlog2M4e, B = M12 + 1. A 5-term sequence of positive reals satisfy that the first three termsand the last three terms both form an arithmetic sequence and the middle three terms form a geometricsequence. If the first term is A and the fifth term is B, determine the third term of the sequence.

Answer: 403

A = 20, B = 8. If the middle term is x, then (x + 20)/2, x, (x + 8)/2 forms a geometric series.Thismeans that (x+20)/2 · (x+8)/2 = x2, which upon solving gives x = 40/3 or x = −4 (which we discardbecause x > 0).

M8. Let A = bM25 c, B = bM2

6 c. A regular A-gon, a regular B-gon, and a circle are given in the plane.What is the greatest possible number of regions that these shapes divide the plane into?

Answer: 1156

A = 144, B = 289. First, note that with only the circle, there are 2 regions. If the three shapes nevercoincide at a point, then each intersection adds precisely one region. Optimistically, we wish to havethe maximal number of intersections where all intersections have both shapes. The maximum numberof intersections between the 289-gon and the circle is 578, since each side can only intersect the circletwice. Similarly, the 144-gon and the circle add at most 288. Finally, each side of the 144-gon canonly intersect the 289-gon twice, so this adds another 288. This maximum can be achieved when allthree shapes have the same circumcenter and circumradius, and are rotated slightly. The answer is2 + 598 + 288 + 288 = 1156.

M9. Let A and B be the unit digits of d7M6e and b6M7c respectively. When all the positive integers notcontaining digit A or B are written in increasing order, what is the 2019th number in the list?

Answer: 3743

A = 9, B = 0. First, there are 8 numbers with 1 digit, 64 with two digits, and 512 with three digits.This leaves 2019−512−64−8 = 1435 of four-digit numbers we have to go through, starting with 1111.Since we don’t have two digits 0 and 9, we are basically counting in base 8, where the digits 01234567in base 8 are actually 12345678, and 111110 = 00008. Converting 1435 to base 8, we get 26338, andmapping this back, we get 3744. However, we must remember that 1111 is 08 under our mapping, sothe 1435th four-digit number is actually 3743.

M10. What is the smallest positive integer with remainder 2, 3, 4 when divided by 3, 5, 7 respectively?

Answer: 53

We note that if we double the number then it leaves a remainder of 1 when divided by all of 3, 5, and7. The smallest even number satisfying this is 3 · 5 · 7 + 1 = 106, so the smallest possible number is106/2 = 53.

M11. An equiangular hexagon has side lengths 1, 2, 3, 4, 5, 6 in some order. Find the nonnegative differencebetween the largest and the smallest possible area of this hexagon.

Answer:√32

Extending three sides of the equiangular hexagon gives an equilateral triangle. Thus, if the sidesare a, b, c, d, e, f , in order, then a + b + c = a + f + e ⇒ b + c = e + f ⇒ f − c = b − e. By asymmetric argument, we see that d−a = f−c = b−e holds, which means that they must be separatedinto three groups of two with equal differences. If the grouping is (1, 2), (3, 4), (5, 6), then we have1, 4, 5, 2, 3, 6 around the hexagon. If the grouping is (1, 4), (2, 5), (3, 6), then we get 1, 5, 3, 4, 2, 6 as theother possibility. Finally, we can use our equilateral triangle trick to find the areas. For the first, weget a big triangle of side 1 + 4 + 5 = 10, and must subtract smaller triangles of sides 1, 5, 3. This gives(100− 1− 25− 9)

√3/4 = 65

√3/4. For the other, we get (81− 1− 9− 4)

√3/4 = 67

√3/4. The positive

difference between these is√

3/2.

M12. Determine the second smallest positive integer n such that n3 + n2 + n+ 1 is a perfect square.

Answer: 7

n3 + n2 + n + 1 = (n + 1)(n2 + 1). Note that gcd(n2 + 1, n + 1) = gcd(2, n + 1) = 1 or 2, and sincen2 + 1 is not a perfect square for n ≥ 1, we must have n2 + 1 = 2p2 and n+ 1 = 2q2 for some integersp and q. The first equation is a variant of Pell’s equation, which (either by brute-forcing small casesor using the known recurrence) gives solutions (n, p) = (1, 1), (7, 5), . . . . Incidentally, both smallestsolutions n = 1 and n = 7 allows an integer solution to the second equation, so n = 7 is the secondsmallest integer that satisfy the condition.

M13. Given that A,B are nonzero base-10 digits such that A ·AB +B = BB, find AB.

Answer: 25

We know A · AB ends in 0. Since neither is 0, they must be 2, 5 in some order. We easily find thatA = 2, B = 5 works while the opposite doesn’t, so AB = 25.

T1. Let S, P,A,C,E be (not necessarily distinct) decimal digits where E 6= 0. Given that N =√ESCAPE

is a positive integer, find the minimum possible value of N .

Answer: 319

Since E 6= 0, the 6-digit number ESCAPE is at least 105, so N ≥ 317. If N were 317 or 318, the lastdigit of N2 would not match the first digit of N2, which contradicts the condition. However, N = 319will work, since the first and last digit of N2 are both 1.

T2. Let X = bT1/8c, Y = T3 − 1, Z = T4 − 2. A point P lies inside the triangle ABC such that PA =X,PB = Y, PC = Z. Find the largest possible area of the triangle.

Answer: 1344

X = 39, Y = 33, Z = 25. Fix some position for P , A, and B, and we shall find the optimal positionfor C. Letting AB be the base of the triangle, we wish to maximize the height. The legal positionsfor C are a subset of the circle with center P and radius PC, so the height is maximized when PCis orthogonal to AB. Symmetrically, we deduce that P is the orthocenter of ABC when the area ismaximized (moreover, P must be inside the triangle). If ray BP intersects AC at E, then since AEBis similar to PEC, we have

√392 − x233 + x

=

√252 − x2x

⇒ x2(x+ 33)2 = (392 − x2)(252 − x2)⇒ 66x3 + 3235x2 − 950625 = 0.

The LHS factors to (x− 15)(66x2 + 4225x+ 63375), meaning that x = 15 is the only positive solution,giving AE = 36, BE = 20, and therefore the maximum area of triangle ABC is (33 + 15)(36 + 20)/2 =1344.

T3. How many ways can one tile a 2× 8 board with 1× 1 and 2× 2 tiles? Rotations and reflections of thesame configuration are considered distinct.

Answer: 34

Let f(n) denote the number of ways to fill a 2×n board. One can fill the leftmost column with two 1×1tiles, leaving f(n−1) ways, or one can fill the leftmost two columns with one 2×2 tile, leaving f(n−2)ways. Therefore, f(n) = f(n − 1) + f(n − 2). One can also directly verify that f(0) = f(1) = 1.Therefore, f(n) = Fn+1, where Fn is the nth Fibonacci number. Easy calculation shows that thedesired quantity is f(8) = F9 = 34.

T4. Let S = T5. Given real numbers a, b, c such that a2 + b2 + c2 + (a + b + c)2 = S, find the maximumpossible value of (a+ b)(b+ c)(c+ a).

Answer: 27

Notice that S = 27 = a2 + b2 + c2 + (a + b + c)2 = (a + b)2 + (b + c)2 + (c + a)2. By AM-GM,S3 ≥ ((a + b)(b + c)(c + a))2/3 with equality if and only if a + b = b + c = c + a, i.e. a = b = c. Thus,

the maximum possible value is(S3

)3/2= 27, achieved at a = b = c = 3

2 .

T5. A regular tetrahedron has volume 8. What is the volume of the set of all the points in the space (notnecessarily inside the tetrahedron) that are closer to the center of the tetrahedron than any of the fourvertices?

Answer: 27

Let h denote the height of the tetrahedron. The center of the tetrahedron is a distance h4 from each

face. Therefore, the perpendicular bisector plane of the segment connecting the center to a vertex liesa distance 3

8h away from both the vertex and the center. Symmetrical considerations with the otherthree vertices will thus show that the desired region is also a regular tetrahedron, with the center ofthe original tetrahedron a distance 3

8h away from each face. Based on the distance from the center to

a face, one can see that the scale factor of this tetrahedron is 3h8 : h

4 = 3 : 2 relative to the original

tetrahedron, so its volume is 8 ·(32

)3= 27.

33. [25] Determine the value of H4.

Proposed by: Yuan Yao

Answer: 24

34. [25] Determine the value of M8.

Proposed by: Yuan Yao

Answer: 1156

35. [25] Determine the value of M9.

Proposed by: Yuan Yao

Answer: 3743

36. [25] Determine the value of T2.

Proposed by: Yuan Yao

Answer: 1344

HMMT November 2019November 9, 2019

General Round

1. Dylan has a 100×100 square, and wants to cut it into pieces of area at least 1. Each cut must be a straightline (not a line segment) and must intersect the interior of the square. What is the largest number of cutshe can make?

2. Meghana writes two (not necessarily distinct) primes q and r in base 10 next to each other on a blackboard,resulting in the concatenation of q and r (for example, if q = 13 and r = 5, the number on the blackboardis now 135). She notices that three more than the resulting number is the square of a prime p. Find allpossible values of p.

3. Katie has a fair 2019-sided die with sides labeled 1, 2, . . . , 2019. After each roll, she replaces her n-sideddie with an (n+1)-sided die having the n sides of her previous die and an additional side with the numbershe just rolled. What is the probability that Katie’s 2019th roll is a 2019?

4. In 4ABC, AB = 2019, BC = 2020, and CA = 2021. Yannick draws three regular n-gons in the plane of4ABC so that each n-gon shares a side with a distinct side of 4ABC and no two of the n-gons overlap.What is the maximum possible value of n?

5. Let a, b, c be positive real numbers such that a ≤ b ≤ c ≤ 2a. Find the maximum possible value of

b

a+

c

b+

a

c.

6. Find all ordered pairs (a, b) of positive integers such that 2a+ 1 divides 3b− 1 and 2b+ 1 divides 3a− 1.

7. In Middle-Earth, nine cities form a 3 by 3 grid. The top left city is the capital of Gondor and the bottomright city is the capital of Mordor. How many ways can the remaining cities be divided among the twonations such that all cities in a country can be reached from its capital via the grid-lines without passingthrough a city of the other country?

8. Compute the number of ordered pairs of integers (x, y) such that x2 + y2 < 2019 and

x2 +min(x, y) = y2 +max(x, y).

9. Let ABCD be an isosceles trapezoid with AD = BC = 255 and AB = 128. Let M be the midpoint ofCD and let N be the foot of the perpendicular from A to CD. If ∠MBC = 90◦, compute tan∠NBM .

10. An up-right path between two lattice points P and Q is a path from P to Q that takes steps of 1 uniteither up or to the right. A lattice point (x, y) with 0 ≤ x, y ≤ 5 is chosen uniformly at random. Computethe expected number of up-right paths from (0, 0) to (5, 5) not passing through (x, y).

HMMT November 2019November 9, 2019

General Round

1. Dylan has a 100×100 square, and wants to cut it into pieces of area at least 1. Each cut must be a straightline (not a line segment) and must intersect the interior of the square. What is the largest number of cutshe can make?

Proposed by: Carl Schildkraut

Answer: 9999

Since each piece has area at least 1 and the original square has area 10000, Dylan can end up with atmost 10000 pieces. There is initially 1 piece, so the number of pieces can increase by at most 9999. Eachcut increases the number of pieces by at least 1, so Dylan can make at most 9999 cuts. Notice that thisis achievable if Dylan makes 9999 vertical cuts spaced at increments of 1

100 units.

2. Meghana writes two (not necessarily distinct) primes q and r in base 10 next to each other on a blackboard,resulting in the concatenation of q and r (for example, if q = 13 and r = 5, the number on the blackboardis now 135). She notices that three more than the resulting number is the square of a prime p. Find allpossible values of p.

Proposed by: Carl Schildkraut

Answer: 5

Trying p = 2, we see that p2 − 3 = 1 is not the concatenation of two primes, so p must be odd. Thenp2 − 3 is even. Since r is prime and determines the units digit of the concatenation of q and r, r mustbe 2. Then p2 will have units digit 5, which means that p will have units digit 5. Since p is prime, wefind that p can only be 5, and in this case, p2 − 3 = 22 allows us to set q = r = 2 to satisfy the problemstatement. So there is a valid solution when p = 5, and this is the only possibility.

3. Katie has a fair 2019-sided die with sides labeled 1, 2, . . . , 2019. After each roll, she replaces her n-sideddie with an (n+1)-sided die having the n sides of her previous die and an additional side with the numbershe just rolled. What is the probability that Katie’s 2019th roll is a 2019?

Proposed by: Freddie Zhao

Answer:1

2019

Since Katie’s original die is fair, the problem is perfectly symmetric. So on the 2019th roll, each numberis equally probable as any other. Therefore, the probability of rolling a 2019 is just 1

2019 .

4. In 4ABC, AB = 2019, BC = 2020, and CA = 2021. Yannick draws three regular n-gons in the plane of4ABC so that each n-gon shares a side with a distinct side of 4ABC and no two of the n-gons overlap.What is the maximum possible value of n?

Proposed by: Carl Schildkraut

Answer: 11

If any n-gon is drawn on the same side of one side of 4ABC as 4ABC itself, it will necessarily overlapwith another triangle whenever n > 3. Thus either n = 3 or the triangles are all outside ABC. Theinterior angle of a regular n-gon is 180◦ · n−2

n , so we require

360◦ · n− 2

n+ max(∠A,∠B,∠C) < 360◦.

As 4ABC is almost equilateral (in fact the largest angle is less than 60.1◦), each angle is approximately60◦, so we require

360 · n− 2

n< 300 =⇒ n < 12.

Hence the answer is n = 11.

5. Let a, b, c be positive real numbers such that a ≤ b ≤ c ≤ 2a. Find the maximum possible value of

b

a+

c

b+

a

c.

Proposed by: Carl Schildkraut

Answer:7

2

Fix the values of b, c. By inspecting the graph of

f(x) =b

x+

x

c,

we see that on any interval the graph attains its maximum at an endpoint. This argument applies whenwe fix any two variables, so it suffices to check boundary cases in which b = a or b = c, and c = b orc = 2a. All pairs of these conditions determine the ratio between a, b, c, except b = c and c = b, in whichcase the boundary condition on a tells us that a = b or 2a = b = c. In summary, these cases are

(a, b, c) ∈ {(a, a, a), (a, a, 2a), (a, 2a, 2a)}.

The largest value achieved from any of these three is 72 .

6. Find all ordered pairs (a, b) of positive integers such that 2a + 1 divides 3b− 1 and 2b + 1 divides 3a− 1.

Proposed by: Milan Haiman

Answer: (2, 2), (12, 17), (17, 12)

This is equivalent to the existence of nonnegative integers c and d such that 3b − 1 = c(2a + 1) and3a− 1 = d(2b + 1). Then

cd =(3b− 1)(3a− 1)

(2a + 1)(2b + 1)=

3a− 1

2a + 1· 3b− 1

2b + 1<

3

2· 3

2= 2.25.

Neither c nor d can equal 0 since that would give a = 13 or b = 1

3 , so cd ≤ 2.25 implies (c, d) ∈{(1, 1), (2, 1), (1, 2)}. Substituting (c, d) back in gives three systems of equations and the three solutions:(2, 2), (12, 17), (17, 12).

7. In Middle-Earth, nine cities form a 3 by 3 grid. The top left city is the capital of Gondor and the bottomright city is the capital of Mordor. How many ways can the remaining cities be divided among the twonations such that all cities in a country can be reached from its capital via the grid-lines without passingthrough a city of the other country?

Proposed by: Shengtong Zhang

Answer: 30

For convenience, we will center the grid on the origin of the coordinate plane and align the outer cornersof the grid with the points (±1,±1), so that (−1, 1) is the capital of Gondor and (1,−1) is the capital ofMordor.

We will use casework on which nation the city at (0, 0) is part of. Assume that is belongs to Gondor.Then consider the sequence of cities at (1, 0), (1, 1), (0, 1). If one of these belongs to Mordor, then all of theprevious cities belong to Mordor, since Mordor must be connected. So we have 4 choices for which citiesbelong to Mordor. Note that this also makes all the other cities in the sequence connected to Gondor.Similarly, we have 4 (independent) choices for the sequence of cities (0,−1), (−1− 1), (−1, 0). All of thesechoices keep (0, 0) connected to Gondor except the choice that assigns all cities in both sequences toMordor. Putting this together, the answer is 2(4 · 4− 1) = 30.

8. Compute the number of ordered pairs of integers (x, y) such that x2 + y2 < 2019 and

x2 + min(x, y) = y2 + max(x, y).

2

Proposed by: Milan Haiman

Answer: 127

We havex2 − y2 = max(x, y)−min(x, y) = |x− y|

Now if x 6= y, we can divide by x− y to obtain x + y = ±1. Thus x = y or x + y = ±1.

If x = y, we see that 2019 > x2 + y2 = 2x2, so we see that −31 ≤ x ≤ 31. There are 63 ordered pairs inthis case.

In the second case, note that |x| ≥ |y| since x2 − y2 = |x− y| ≥ 0. Since x + y = ±1, we cannot havexy > 0, so either x ≥ 0, y ≤ 0, or x ≤ 0, y ≥ 0. In the first case, x+ y = 1; in the second case, x+ y = −1.Thus, the solutions for (x, y) are of the form (k, 1 − k) or (−k, k − 1) for some k > 0. In either case, wemust have k2 + (k − 1)2 < 2019, which holds true for any 1 ≤ k ≤ 32 but fails for k = 33. There are atotal of 32 · 2 = 64 solutions in this case.

In summary, there are a total of 63 + 64 = 127 integer solutions to the equation x2 + min (x, y) =y2 + max (x, y) with x2 + y2 < 2019.

9. Let ABCD be an isosceles trapezoid with AD = BC = 255 and AB = 128. Let M be the midpoint ofCD and let N be the foot of the perpendicular from A to CD. If ∠MBC = 90◦, compute tan∠NBM .

Proposed by: Milan Haiman

Answer:120

353

Construct P , the reflection of A over CD. Note that P , M , and B are collinear. As ∠PNC = ∠PBC =90◦, PNBC is cyclic. Thus, ∠NBM = ∠NCP , so our desired tangent is tan∠ACN = AN

CN . Note thatNM = 1

2AB = 64. Since 4AND ∼ 4MAD,

255

64 + ND=

ND

255.

Solving, we find ND = 225, which gives AN = 120. Then we calculate ANCN = 120

128+225 = 120353 .

10. An up-right path between two lattice points P and Q is a path from P to Q that takes steps of 1 uniteither up or to the right. A lattice point (x, y) with 0 ≤ x, y ≤ 5 is chosen uniformly at random. Computethe expected number of up-right paths from (0, 0) to (5, 5) not passing through (x, y).

Proposed by: Mehtaab Sawnhey

Answer: 175

For a lattice point (x, y), let F (x, y) denote the number of up-right paths from (0, 0) to (5, 5) that don’tpass through (x, y), and let

S =∑

0≤x≤5

∑0≤y≤5

F (x, y).

Our answer is S36 , as there are 36 lattice points (x, y) with 0 ≤ x, y ≤ 5.

Notice that the number of up-right paths from (0, 0) to (5, 5) is105

)= 252 because each path consists of

10 steps, of which we can choose 5 to be to the right. Each of these paths passes through 11 lattice points(x, y) with 0 ≤ x, y ≤ 5, so each path contributes 36−11 = 25 to the quantity we are counting in S. ThenS = 25 · 252, so our answer is 25·252

36 = 175.

3

HMMT November 2019November 9, 2019

Theme Round

1. For breakfast, Mihir always eats a bowl of Lucky Charms cereal, which consists of oat pieces and marsh-mallow pieces. He defines the luckiness of a bowl of cereal to be the ratio of the number of marshmallowpieces to the total number of pieces. One day, Mihir notices that his breakfast cereal has exactly 90 oatpieces and 9 marshmallow pieces, and exclaims, “This is such an unlucky bowl!” How many marshmallowpieces does Mihir need to add to his bowl to double its luckiness?

2. Sandy likes to eat waffles for breakfast. To make them, she centers a circle of waffle batter of radius3cm at the origin of the coordinate plane and her waffle iron imprints non-overlapping unit-square holescentered at each lattice point. How many of these holes are contained entirely within the area of thewaffle?

3. For breakfast, Milan is eating a piece of toast shaped like an equilateral triangle. On the piece of toastrests a single sesame seed that is one inch away from one side, two inches away from another side, and fourinches away from the third side. He places a circular piece of cheese on top of the toast that is tangent toeach side of the triangle. What is the area of this piece of cheese?

4. To celebrate 2019, Faraz gets four sandwiches shaped in the digits 2, 0, 1, and 9 at lunch. However, thefour digits get reordered (but not flipped or rotated) on his plate and he notices that they form a 4-digitmultiple of 7. What is the greatest possible number that could have been formed?

5. Alison is eating 2401 grains of rice for lunch. She eats the rice in a very peculiar manner: every step, ifshe has only one grain of rice remaining, she eats it. Otherwise, she finds the smallest positive integerd > 1 for which she can group the rice into equal groups of size d with none left over. She then groupsthe rice into groups of size d, eats one grain from each group, and puts the rice back into a single pile.How many steps does it take her to finish all her rice?

6. Wendy eats sushi for lunch. She wants to eat six pieces of sushi arranged in a 2× 3 rectangular grid, butsushi is sticky, and Wendy can only eat a piece if it is adjacent to (not counting diagonally) at most twoother pieces. In how many orders can Wendy eat the six pieces of sushi, assuming that the pieces of sushiare distinguishable?

7. Carl only eats food in the shape of equilateral pentagons. Unfortunately, for dinner he receives a piece ofsteak in the shape of an equilateral triangle. So that he can eat it, he cuts off two corners with straightcuts to form an equilateral pentagon. The set of possible perimeters of the pentagon he obtains is exactlythe interval [a, b), where a and b are positive real numbers. Compute a

b .

8. Omkar, Krit1, Krit2, and Krit3 are sharing x > 0 pints of soup for dinner. Omkar always takes 1 pint ofsoup (unless the amount left is less than one pint, in which case he simply takes all the remaining soup).Krit1 always takes 1

6 of what is left, Krit2 always takes 15 of what is left, and Krit3 always takes 1

4 of whatis left. They take soup in the order of Omkar, Krit1, Krit2, Krit3, and then cycle through this order untilno soup remains. Find all x for which everyone gets the same amount of soup.

9. For dinner, Priya is eating grilled pineapple spears. Each spear is in the shape of the quadrilateral PINE,with PI = 6cm, IN = 15cm, NE = 6cm, EP = 25cm, and ∠NEP + ∠EPI = 60◦. What is the area ofeach spear, in cm2?

10. For dessert, Melinda eats a spherical scoop of ice cream with diameter 2 inches. She prefers to eat herice cream in cube-like shapes, however. She has a special machine which, given a sphere placed in space,cuts it through the planes x = n, y = n, and z = n for every integer n (not necessarily positive). Melindacenters the scoop of ice cream uniformly at random inside the cube 0 ≤ x, y, z ≤ 1, and then cuts it intopieces using her machine. What is the expected number of pieces she cuts the ice cream into?

HMMT November 2019November 9, 2019

Theme Round

Illustrations by Hanna Yang

1. For breakfast, Mihir always eats a bowl of Lucky Charms cereal, which consists of oat pieces and marsh-mallow pieces. He defines the luckiness of a bowl of cereal to be the ratio of the number of marshmallowpieces to the total number of pieces. One day, Mihir notices that his breakfast cereal has exactly 90 oatpieces and 9 marshmallow pieces, and exclaims, “This is such an unlucky bowl!” How many marshmallowpieces does Mihir need to add to his bowl to double its luckiness?

Proposed by: Krit BoonsirisethAnswer: 11Let x be the number of marshmallows to add. We are given that

2 · 9

99=

9 + x

99 + x.

Rearanging this gives2(99 + x) = 11(9 + x).

Thus 9x = 99 and x = 11.

2. Sandy likes to eat waffles for breakfast. To make them, she centers a circle of waffle batter of radius3cm at the origin of the coordinate plane and her waffle iron imprints non-overlapping unit-square holescentered at each lattice point. How many of these holes are contained entirely within the area of thewaffle?

Proposed by: Carl SchildkrautAnswer: 21First, note that each divet must have its sides parallel to the coordinate axes; if the divet centeredat the lattice point (a, b) does not have this orientation, then it contains the point (a + 1/2, b) in itsinterior, so it necessarily overlaps with the divet centered at (a+ 1, b).

If we restrict our attention to one quadrant, we see geometrically that the divets centered at(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), and (2, 1) are completely contained in the waffle, andno others are. We can make this more rigorous by considering the set of points (x, y) such thatx2 + y2 < 9. We count 1 divet centered at the origin, 8 divets centered on the axes that are notcentered at the origin, and 12 divets not centered on the axes, for a total of 21 divets.

3. For breakfast, Milan is eating a piece of toast shaped like an equilateral triangle. On the piece of toastrests a single sesame seed that is one inch away from one side, two inches away from another side, and fourinches away from the third side. He places a circular piece of cheese on top of the toast that is tangent toeach side of the triangle. What is the area of this piece of cheese?

Proposed by: Carl Schildkraut

Answer:49π

9Suppose the toast has side length s. If we draw the three line segments from the sesame seed to thethree vertices of the triangle, we partition the triangle into three smaller triangles, with areas s

2 , s, and2s, so the entire piece of toast has area 7s

2 . Suppose the cheese has radius r. We similarly see that the

toast has area 3rs2 . Equating these, we see that r = 7

3 , so the area of the cheese is π73

)2= 49π

9 .

4. To celebrate 2019, Faraz gets four sandwiches shaped in the digits 2, 0, 1, and 9 at lunch. However, thefour digits get reordered (but not flipped or rotated) on his plate and he notices that they form a 4-digitmultiple of 7. What is the greatest possible number that could have been formed?

Proposed by: Milan HaimanAnswer: 1092Note that 2 and 9 are equivalent mod 7. So we will replace the 9 with a 2 for now. Since 7 is a divisorof 21, a four digit multiple of 7 consisting of 2, 0, 1, and 2 cannot have a 2 followed by a 1 (otherwisewe could subtract a multiple of 21 to obtain a number of the form 2 · 10k). Thus our number eitherstarts with a 1 or has a 0 followed by a 1. We can check that 2201 and 2012 are not divisible by 7.Thus our number starts with a 1. Checking 1220, 1202, and 1022 gives that 1022 is the only possibility.So the answer is 1092.

5. Alison is eating 2401 grains of rice for lunch. She eats the rice in a very peculiar manner: every step, ifshe has only one grain of rice remaining, she eats it. Otherwise, she finds the smallest positive integerd > 1 for which she can group the rice into equal groups of size d with none left over. She then groupsthe rice into groups of size d, eats one grain from each group, and puts the rice back into a single pile.How many steps does it take her to finish all her rice?

Proposed by: Carl SchilkrautAnswer: 17Note that 2401 = 74. Also, note that the operation is equivalent to replacing n grains of rice withn · p→1p grains of rice, where p is the smallest prime factor of n.

Now, suppose that at some moment Alison has 7k grains of rice. After each of the next foursteps, she will have 6 · 7k→1, 3 · 7k→1, 2 · 7k→1, and 7k→1 grains of rice, respectively. Thus, it takesher 4 steps to decrease the number of grains of rice by a factor of 7 given that she starts at a power of 7.

Thus, it will take 4 · 4 = 16 steps to reduce everything to 70 = 1 grain of rice, after which itwill take one step to eat it. Thus, it takes a total of 17 steps for Alison to eat all of the rice.

6. Wendy eats sushi for lunch. She wants to eat six pieces of sushi arranged in a 2× 3 rectangular grid, butsushi is sticky, and Wendy can only eat a piece if it is adjacent to (not counting diagonally) at most twoother pieces. In how many orders can Wendy eat the six pieces of sushi, assuming that the pieces of sushiare distinguishable?

Proposed by: Milan HaimanAnswer: 360Call the sushi pieces A,B,C in the top row and D,E, F in the bottom row of the grid. Note thatWendy must first eat either A,C,D, or F. Due to the symmetry of the grid, all of these choices areequivalent. Without loss of generality, suppose Wendy eats piece A.

Now, note that Wendy cannot eat piece E, but can eat all other pieces. If Wendy eats pieceB,D, or F, then in the resulting configuration, all pieces of sushi are adjacent to at most 2 pieces, soshe will have 4! ways to eat the sushi. Thus, the total number of possibilities in this case is 4·3·4! = 288.

If Wendy eats A and then C, then Wendy will only have 3 choices for her next piece of sushi,after which she will have 3! ways to eat the remaining 3 pieces of sushi. Thus, the total number ofpossibilities in this case is 4 · 1 · 3 · 3! = 72.Thus, the total number of ways for Wendy to eat the sushi is 288 + 72 = 360.

7. Carl only eats food in the shape of equilateral pentagons. Unfortunately, for dinner he receives a piece ofsteak in the shape of an equilateral triangle. So that he can eat it, he cuts off two corners with straightcuts to form an equilateral pentagon. The set of possible perimeters of the pentagon he obtains is exactlythe interval [a, b), where a and b are positive real numbers. Compute a

b .

2

Proposed by: Krit Boonsiriseth and Milan Haiman

Answer: 4√

3〉 6

Assume that the triangle has side length 1. We will show the pentagon side length x is in [2√

3〉 3, 12 ).Call the triangle ABC and let corners B,C be cut. Choose P on AB, Q,R on BC, and S on AC suchthat APQRS is equilateral. If x ≥ 1

2 then Q is to the right of R, causing self-intersection. Also thedistance from P to BC is at most x, so

x = PQ ≥ PB sin 60◦ = (1〉 x) ·√

3

2.

Solving gives (2 +√

3)x ≥√

3, or x ≥√

3(2〉√

3) = 2√

3〉3. Finally, these are attainable if we chooseP such that AP = x, then Q such that PQ = x, and so on. Therefore a

b = 4√

3〉 6.

8. Omkar, Krit1, Krit2, and Krit3 are sharing x > 0 pints of soup for dinner. Omkar always takes 1 pint ofsoup (unless the amount left is less than one pint, in which case he simply takes all the remaining soup).Krit1 always takes 1

6 of what is left, Krit2 always takes 15 of what is left, and Krit3 always takes 1

4 of whatis left. They take soup in the order of Omkar, Krit1, Krit2, Krit3, and then cycle through this order untilno soup remains. Find all x for which everyone gets the same amount of soup.

Proposed by: Krit Boonsiriseth and Milan Haiman

Answer:49

3The main observation is that if x > 1 pints of soup are left, then in one round, Omkar gets 1 and eachKritn gets x→1

6 , with x→12 soup left. Thus it is evident that each Kritn gets the same amount of soup,

which means it suffices to find x for which Omkar gets x4 .

Omkar gets 1 for each cycle and then all the remaining soup when there is less than one pint remaining.The amount of soup becomes (after each cycle)

x→ x〉 1

2→ x〉 3

4→ · · · → x+ 1

2n〉 1,

so if n is the number of cycles, then Omkar’s soup is n+ x+12n 〉 1. Setting this equal to x

4 , we obtain

x =n+ 1/2n 〉 1

1/4〉 1/2n=

(n〉 1)2n + 1

2n→2 〉 1.

This immediately implies n > 2. On the other hand, we necessarily have 0 ≤ x+12n 〉 1 ≤ 1, so

2n ≤ x+ 1 ≤ 2n+1. But

x+ 1 =(n〉 1)2n + 2n→2

2n→2 〉 1≤ (n〉 1)2n + 2n

2n→3= 8n,

So 2n ≤ 8n =⇒ n ≤ 5. Testing n = 3, 4, 5:

• For n = 3 we get x = 17 which is greater than 24.

• For n = 4 we get x = 493 which works.

• For n = 5 we get x = 1297 which is less than 25.

We see that only n = 4 and x = 493 works.

9. For dinner, Priya is eating grilled pineapple spears. Each spear is in the shape of the quadrilateral PINE,with PI = 6cm, IN = 15cm, NE = 6cm, EP = 25cm, and ∠NEP + ∠EPI = 60◦. What is the area ofeach spear, in cm2?

3

Proposed by: Milan Haiman

Answer:100√

3

3We consider a configuration composed of 2 more quadrilaterals congruent to PINE. Let them beP ′I ′N ′E′, with E′ = P and N ′ = I, and P ′′I ′′N ′′E′′ with P ′′ = E,E′′ = P ′, N ′′ = I ′, and I ′′ = N .Notice that this forms an equilateral triangle of side length 25 since ∠PP ′P ′′ = ∠PP ′′P ′ = ∠P ′PP ′′ =60◦. Also, we see that the inner triangle NN ′N ′′ forms an equilateral triangle of side length 15 sinceall the side lengths are equal. So the area inside the big equilateral triangle and outside the small one

is 625√3

4 〉 225√3

4 = 100√

3. Since there are two other congruent quadrilaterals to PINE, we have that

the area of one of them is 100√3

3 .

10. For dessert, Melinda eats a spherical scoop of ice cream with diameter 2 inches. She prefers to eat herice cream in cube-like shapes, however. She has a special machine which, given a sphere placed in space,cuts it through the planes x = n, y = n, and z = n for every integer n (not necessarily positive). Melindacenters the scoop of ice cream uniformly at random inside the cube 0 ≤ x, y, z ≤ 1, and then cuts it intopieces using her machine. What is the expected number of pieces she cuts the ice cream into?

Proposed by: Carl Schildkraut

Answer: 7 +13π

3

Note that if we consider the division of R3 into unit cubes by the given planes, we only need to computethe sum of the probabilities that the ice cream scoop intersects each cube. There are three types ofcubes that can be intersected:

• The cube 0 ≤ x, y, z ≤ 1 in which the center lies, as well as the 6 face-adjacent cubes are alwaysintersected, for a total of 7.

• The cubes edge-adjacent to the center cube are intersected if the center of the ice cream lieswithin 1 unit of the connecting edge, which happens with probability π

4 . There are 12 suchcubes, for a total of 3π.

• The cubes corner-adjacent to the center cube are intersected if the center of the ice cream lieswithin 1 unit of the connecting corner, which happens with probability π

6 . There are 8 such cubes,for a total of 4π

3 .

Adding these all up gives our answer of 7 + 13π3 .

An alternate solution is possible:

We compute the number of regions into which a convex region S in R3 is divided by planes:Let a be the number of planes intersecting S. Let b be the number of lines (intersections of two planes)passing through S. Let c be the number of points (intersections of three planes) lying inside S. Then Sis divided into a+ b+ c+ 1 regions. Then the computation for the problem is fairly straight forward.Note that the only planes, lines, and points that can intersect the ice cream scoop I are the faces,edges, and vertices of the cube 0 ≤ x, y, z ≤ 1. The computation is essentially the same as in thefirst solution. The scoop intersects each of the 6 faces with probability 1, each of the 12 edges withprobability π

4 , and each of the 8 vertices with probability π6 , for a total expected number of regions

1 + 6 + 3π + 4π3 = 7 + 13π

3 .

4

HMMT November 2019November 9, 2019

Team Round

1. [20] Each person in Cambridge drinks a (possibly different) 12 ounce mixture of water and apple juice,where each drink has a positive amount of both liquids. Marc McGovern, the mayor of Cambridge, drinks16 of the total amount of water drunk and 1

8 of the total amount of apple juice drunk. How many peopleare in Cambridge?

2. [20] 2019 students are voting on the distribution of N items. For each item, each student submits a voteon who should receive that item, and the person with the most votes receives the item (in case of a tie,no one gets the item). Suppose that no student votes for the same person twice. Compute the maximumpossible number of items one student can receive, over all possible values of N and all possible ways ofvoting.

3. [30] The coefficients of the polynomial P (x) are nonnegative integers, each less than 100. Given thatP (10) = 331633 and P (−10) = 273373, compute P (1).

4. [35] Two players play a game, starting with a pile of N tokens. On each player’s turn, they must remove2n tokens from the pile for some nonnegative integer n. If a player cannot make a move, they lose. Forhow many N between 1 and 2019 (inclusive) does the first player have a winning strategy?

5. [40] Compute the sum of all positive real numbers x ≤ 5 satisfying

x =dx2e+ dxe · bxcdxe+ bxc

.

6. [45] Let ABCD be an isosceles trapezoid with AB = 1, BC = DA = 5, CD = 7. Let P be the intersectionof diagonals AC and BD, and let Q be the foot of the altitude from D to BC. Let PQ intersect AB atR. Compute sin∠RPD.

7. [55] Consider sequences a of the form a = (a1, a2, . . . , a20) such that each term ai is either 0 or 1. Foreach such sequence a, we can produce a sequence b = (b1, b2, . . . , b20), where

bi =

ai + ai+1 i = 1

ai→1 + ai + ai+1 1 < i < 20

ai→1 + ai i = 20.

How many sequences b are there that can be produced by more than one distinct sequence a?

8. [60] In 4ABC, the external angle bisector of ∠BAC intersects line BC at D. E is a point on ray−→AC

such that ∠BDE = 2∠ADB. If AB = 10, AC = 12, and CE = 33, compute DBDE .

9. [65] Will stands at a point P on the edge of a circular room with perfectly reflective walls. He shinestwo laser pointers into the room, forming angles of n◦ and (n + 1)◦ with the tangent at P , where n is apositive integer less than 90. The lasers reflect off of the walls, illuminating the points they hit on thewalls, until they reach P again. (P is also illuminated at the end.) What is the minimum possible numberof illuminated points on the walls of the room?

10. [70] A convex 2019-gon A1A2 . . . A2019 is cut into smaller pieces along its 2019 diagonals of the formAiAi+3 for 1 ≤ i ≤ 2019, where A2020 = A1, A2021 = A2, and A2022 = A3. What is the least possiblenumber of resulting pieces?

HMMT November 2019November 9, 2019

Team Round

1. [20] Each person in Cambridge drinks a (possibly different) 12 ounce mixture of water and apple juice,where each drink has a positive amount of both liquids. Marc McGovern, the mayor of Cambridge, drinks16 of the total amount of water drunk and 1

8 of the total amount of apple juice drunk. How many peopleare in Cambridge?

Proposed by: Alec Sun

Answer: 7

The total amount of liquid drunk must be more than 6 times of the amount that Marc drinks but lessthan 8 times of the amount that Marc drinks. Therefore, the number of people in Cambridge is more than6 but less than 8, so it must be 7.

2. [20] 2019 students are voting on the distribution of N items. For each item, each student submits a voteon who should receive that item, and the person with the most votes receives the item (in case of a tie,no one gets the item). Suppose that no student votes for the same person twice. Compute the maximumpossible number of items one student can receive, over all possible values of N and all possible ways ofvoting.

Proposed by: Milan Haiman

Answer: 1009

To get an item, a student must receive at least 2 votes on that item. Since each student receives at most2019 votes, the number of items one student can receive does not exceed 2019

2 = 1009.5. So, the answer isat most 1009. This occurs when N = 2018 and item i was voted to student 1, 1, 2, 3, . . . , 2018 by student2i−1, 2i−2, . . . , 2019, 1, . . . , 2i−2 respectively for i = 1, 2, . . . , 2018. Thus, the maximum possible numberof items one student can receive is 1009.

3. [30] The coefficients of the polynomial P (x) are nonnegative integers, each less than 100. Given thatP (10) = 331633 and P (−10) = 273373, compute P (1).

Proposed by: Carl Joshua Quines

Answer: 100

LetP (x) = a0 + a1x + a2x

2 + . . .

Then1

2(P (10) + P (−10)) = a0 + 100a2 + . . .

and1

2(P (10)− P (−10)) = 10a1 + 1000a3 + . . .

Since all the coefficients are nonnegative integers, these expressions give us each of the coefficients by justtaking two digits in succession. Thus we have a0 = 3, a1 = 13, a2 = 25, a3 = 29, a4 = 30 and an = 0 forn > 4. Thus

P (1) = a0 + a1 + a2 + · · · = 100.

4. [35] Two players play a game, starting with a pile of N tokens. On each player’s turn, they must remove2n tokens from the pile for some nonnegative integer n. If a player cannot make a move, they lose. Forhow many N between 1 and 2019 (inclusive) does the first player have a winning strategy?

Proposed by: Milan Haiman

Answer: 1346

The first player has a winning strategy if and only if N is not a multiple of 3. We show this by inductionon N . If N = 0, then the first player loses.

If N is a multiple of 3, then N − 2n is never a multiple of 3 for any n, so the second player has a winningstrategy.

If N is not a multiple of 3, the first player can remove either 1 or 2 coins to get the number of coins inthe pile down to a multiple of 3, so the first player will always win.

5. [40] Compute the sum of all positive real numbers x ≤ 5 satisfying

x =dx2e+ dxe · bxcdxe+ bxc

.

Proposed by: Milan Haiman

Answer: 85

Note that all integer x work. If x is not an integer then suppose n < x < n+ 1. Then x = n+ k2n+1 , where

n is an integer and 1 ≤ k ≤ 2n is also an integer, since the denominator of the fraction on the right handside is 2n + 1. We now show that all x of this form work.

Note that

x2 = n2 +2nk

2n + 1+

(k

2n + 1

)2

= n2 + k − k

2n + 1+

(k

2n + 1

)2

.

For k2n+1 between 0 and 1, − k

2n+1 +(

k2n+1

)2is between − 1

4 and 0, so we have n2 + k− 1 < x2 ≤ n2 + k,

and dx2e = n2 + k.

Then,dx2e+ dxe · bxcdxe+ bxc

=n2 + k + n · (n + 1)

2n + 1= n +

k

2n + 1= x,

so all x of this form work.

Now, note that the 2n solutions in the interval (n, n + 1), together with the solution n + 1, form anarithmetic progression with 2n + 1 terms and average value n + n+1

2n+1 . Thus, the sum of the solutions in

the interval (n, n + 1] is 2n2 + 2n + 1 = n2 + (n + 1)2. Summing this for n from 0 to 4, we get that theanswer is

02 + 2(12 + 22 + 32 + 42) + 52 = 85.

6. [45] Let ABCD be an isosceles trapezoid with AB = 1, BC = DA = 5, CD = 7. Let P be the intersectionof diagonals AC and BD, and let Q be the foot of the altitude from D to BC. Let PQ intersect AB atR. Compute sin∠RPD.

Proposed by: Milan Haiman

Answer:4

5

Let M be the foot of the altitude from B to CD. Then 2CM + AB = CD =⇒ CM = 3. ThenDM = 4 and by the Pythagorean theorem, BM = 4. Thus BMD is a right isosceles triangle i.e.∠BDM = ∠PDC = π

4 . Similarly, ∠PCD = π4 . Thus ∠DPC = π

2 , which means quadrilateral PQDC iscyclic. Now, sin∠RPD = sin∠DCQ = sin∠MCB = 4

5 .

An alternate solution is also possible:

Note that AC ⊥ BD since AB2 + CD2 = 12 + 72 = 52 + 52 = BC2 + DA2. Thus P is the foot of thealtitude from D to AC. Since D is on the circumcircle of 4ABC, line PQR is the Simson line of D. ThusR is the foot from D to AB. Then from quadrilateral RAPD being cyclic we have ∠RPD = ∠RAD. Sosin∠RPD = 4

5 .

7. [55] Consider sequences a of the form a = (a1, a2, . . . , a20) such that each term ai is either 0 or 1. Foreach such sequence a, we can produce a sequence b = (b1, b2, . . . , b20), where

bi =

ai + ai+1 i = 1

ai→1 + ai + ai+1 1 < i < 20

ai→1 + ai i = 20.

2

How many sequences b are there that can be produced by more than one distinct sequence a?

Proposed by: Benjamin Qi

Answer: 64

Let the two sequences be b and b̂. Then, observe that given a, if b1 = b̂1 and b2 = b̂2, then b = b̂ (since a

will uniquely determine the remaining elements in b and b̂). Thus, b and b̂ must start with (1, 0, . . . ) and(0, 1, . . . ), respectively (without loss of generality).

Note that a3 is either 1 (in which case b3 = b̂3 = 0) or 2 (in which case b3 = b̂3 = 1). Moreover, b4, b5must be the same as b1, b2 (and the same for b̂) for the sequences to generate the same a3, a4. We canthen pick a6, a9, . . . .

Observe, that the last elements also have to be (. . . , 1, 0) for b and (. . . , 0, 1) for b̂. Thus, the answer isnonzero only for sequence lengths of 2 (mod 3), in which case, our answer is 2k, where the length is 3k+2(since we have two choices for every third element).

Here, since N = 20 = 3(6) + 2, the answer is 26 = 64.

8. [60] In 4ABC, the external angle bisector of ∠BAC intersects line BC at D. E is a point on ray−→AC

such that ∠BDE = 2∠ADB. If AB = 10, AC = 12, and CE = 33, compute DBDE .

Proposed by: Milan Haiman

Answer:2

3

Let F be a point on ray−→CA such that ∠ADF = ∠ADB. 4ADF and 4ADB are congruent, so AF = 10

and DF = DB. So, CF = CA + AF = 22. Since ∠FDC = 2∠ADB = ∠EDC, by the angle bisectortheorem we compute DF

DE = CFCE = 22

33 = 23 .

9. [65] Will stands at a point P on the edge of a circular room with perfectly reflective walls. He shinestwo laser pointers into the room, forming angles of n◦ and (n + 1)◦ with the tangent at P , where n is apositive integer less than 90. The lasers reflect off of the walls, illuminating the points they hit on thewalls, until they reach P again. (P is also illuminated at the end.) What is the minimum possible numberof illuminated points on the walls of the room?

Proposed by: Handong Wang

Answer: 28

Note that we want the path drawn out by the lasers to come back to P in as few steps as possible. Observethat if a laser is fired with an angle of n degrees from the tangent, then the number of points it createson the circle is 180

gcd(180,n) . (Consider the regular polygon created by linking all the points that show up on

the circle–if the center of the circle is O, and the vertices are numbered V1, V2, . . . , Vk, the angle ∠V1OV2

is equal to 2 gcd(180, n), so there are a total of 3602 gcd(180,n) sides).

Now, we consider the case with both n and n + 1. Note that we wish to minimize the value 180gcd(180,n) +

180gcd(180,n+1) , or maximize both gcd(180, n) and gcd(180, n+ 1). Note that since n and n+ 1 are relatively

prime and 180 = (4)(9)(5), the expression is maximized when gcd(180, n) = 20 and gcd(180, n + 1) = 9(or vice versa). This occurs when n = 80. Plugging this into our expression, we have that the number ofpoints that show up from the laser fired at 80 degrees is 180

20 = 9 and the number of points that appearfrom the laser fired at 81 degrees is 180

9 = 20. However, since both have a point that shows up at P (andno other overlapping points since gcd(9, 20) = 1), we see that the answer is 20 + 9− 1 = 28.

3

10. [70] A convex 2019-gon A1A2 . . . A2019 is cut into smaller pieces along its 2019 diagonals of the formAiAi+3 for 1 ≤ i ≤ 2019, where A2020 = A1, A2021 = A2, and A2022 = A3. What is the least possiblenumber of resulting pieces?

Proposed by: Krit Boonsiriseth

Answer: 5049

Each time we draw in a diagonal, we create one new region, plus one new region for each intersection onthat diagonal. So, the number of regions will be

1 + (number of diagonals) + (number of intersections),

where (number of intersections) counts an intersection of three diagonals twice. Since no four diagonalscan pass through a point, the only nonconstant term in our expression is the last one. To minimize thisterm, we want to maximize the number of triples of diagonals passing through the same point. Considerthe set S of triples of diagonals AnAn+3 that intersect at a single point. Each triple in S must come fromthree consecutive diagonals, and two different triples can only have one diagonal in common, so S has atmost

⌊20192

⌋= 1009 triples. Hence the number of resulting pieces is at least

1 + (2019) + (2 · 2019− 1009) = 5049.

To show that 5049 is attainable, we use the following construction. Let B1 . . . B1010 be a regular 1010-

gon, and let `n denote the external angle bisector of ∠Bn→1BnBn+1. Let A1 =←−−−−−−→B1009B1010 ∩

←−−→B1B2,

A2018 =←−−−−−−→B1008B1009 ∩

←−−−−→B1010B1, A2019 = `1 ∩ `1009, and for n = 1, . . . , 1008, define A2n = `n+1 ∩

←−−−−→Bn→1Bn

and A2n+1 = `n ∩←−−−−−−→Bn+1Bn+2. It follows that, for all n = 0, . . . , 1008, A2n→1A2n+2, A2nA2n+3, and

A2n+1A2n+4 intersect at Bn+1.

4

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HMMT November 2019, November 9, 2019 — GUTS ROUND

Organization Team Team ID#

1. [5] A polynomial P with integer coefficients is called tricky if it has 4 as a root.

A polynomial is called teeny if it has degree at most 1 and integer coefficients between −7 and 7, inclusive.

How many nonzero tricky teeny polynomials are there?

2. [5] You are trying to cross a 6 foot wide river. You can jump at most 4 feet, but you have one stone youcan throw into the river; after it is placed, you may jump to that stone and, if possible, from there to theother side of the river. However, you are not very accurate and the stone ends up landing uniformly atrandom in the river. What is the probability that you can get across?

3. [5] For how many positive integers a does the polynomial

x2 − ax+ a

have an integer root?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2019, November 9, 2019 — GUTS ROUND

Organization Team Team ID#

4. [6] In 2019, a team, including professor Andrew Sutherland of MIT, found three cubes of integers whichsum to 42:

42 = (−8053873881207597 )3 + (80435758145817515)3 + (12602123297335631)3

One of the digits, labeled by an underscore, is missing. What is that digit?

5. [6] A point P is chosen uniformly at random inside a square of side length 2. If P1, P2, P3, and P4 are thereflections of P over each of the four sides of the square, find the expected value of the area of quadrilateralP1P2P3P4.

6. [6] Compute the sum of all positive integers n < 2048 such that n has an even number of 1’s in its binaryrepresentation.

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HMMT November 2019, November 9, 2019 — GUTS ROUND

Organization Team Team ID#

7. [7] Let S be the set of all nondegenerate triangles formed from the vertices of a regular octagon with sidelength 1. Find the ratio of the largest area of any triangle in S to the smallest area of any triangle in S.

8. [7] There are 36 students at the Multiples Obfuscation Program, including a singleton, a pair of identicaltwins, a set of identical triplets, a set of identical quadruplets, and so on, up to a set of identical octuplets.Two students look the same if and only if they are from the same identical multiple. Nithya the teachingassistant encounters a random student in the morning and a random student in the afternoon (bothchosen uniformly and independently), and the two look the same. What is the probability that they areactually the same person?

9. [7] Let p be a real number between 0 and 1. Jocelin has a coin that lands heads with probability p andtails with probability 1 − p; she also has a number written on a blackboard. Each minute, she flips thecoin, and if it lands heads, she replaces the number x on the blackboard with 3x+ 1; if it lands tails shereplaces it with x/2. Given that there are constants a, b such that the expected value of the value writtenon the blackboard after t minutes can be written as at+ b for all positive integers t, compute p.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2019, November 9, 2019 — GUTS ROUND

Organization Team Team ID#

10. [8] Let ABCD be a square of side length 5, and let E be the midpoint of side AB. Let P and Q be thefeet of perpendiculars from B and D to CE, respectively, and let R be the foot of the perpendicular fromA to DQ. The segments CE,BP,DQ, and AR partition ABCD into five regions. What is the medianof the areas of these five regions?

11. [8] Let a, b, c, d be real numbers such that

min(20x+ 19, 19x+ 20) = (ax+ b)− |cx+ d|

for all real numbers x. Find ab+ cd.

12. [8] Four players stand at distinct vertices of a square. They each independently choose a vertex of thesquare (which might be the vertex they are standing on). Then, they each, at the same time, beginrunning in a straight line to their chosen vertex at 10mph, stopping when they reach the vertex. If at anytime two players, whether moving or not, occupy the same space (whether a vertex or a point inside thesquare), they collide and fall over. How many different ways are there for the players to choose verticesto go to so that none of them fall over?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2019, November 9, 2019 — GUTS ROUND

Organization Team Team ID#

13. [9] In4ABC, the incircle centered at I touches sides AB and BC at X and Y , respectively. Additionally,the area of quadrilateral BXIY is 2

5 of the area of ABC. Let p be the smallest possible perimeter of a4ABC that meets these conditions and has integer side lengths. Find the smallest possible area of sucha triangle with perimeter p.

14. [9] Compute the sum of all positive integers n for which

9√n+ 4

√n+ 2− 3

√n+ 16

is an integer.

15. [9] Let a, b, c be positive integers such that

a

77+

b

91+

c

143= 1.

What is the smallest possible value of a+ b+ c?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT November 2019, November 9, 2019 — GUTS ROUND

Organization Team Team ID#

16. [10] Equilateral 4ABC has side length 6. Let ω be the circle through A and B such that CA and CB areboth tangent to ω. A point D on ω satisfies CD = 4. Let E be the intersection of line CD with segmentAB. What is the length of segment DE?

17. [10] Kelvin the frog lives in a pond with an infinite number of lily pads, numbered 0, 1, 2, 3, and so forth.Kelvin starts on lily pad 0 and jumps from pad to pad in the following manner: when on lily pad i, hewill jump to lily pad (i+ k) with probability 1

2kfor k > 0. What is the probability that Kelvin lands on

lily pad 2019 at some point in his journey?

18. [10] The polynomial x3 − 3x2 + 1 has three real roots r1, r2, and r3. Compute

3√

3r1 − 2 + 3√

3r2 − 2 + 3√

3r3 − 2.

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HMMT November 2019, November 9, 2019 — GUTS ROUND

Organization Team Team ID#

19. [11] Let ABC be a triangle with AB = 5, BC = 8, CA = 11. The incircle ω and A-excircle1 Γ arecentered at I1 and I2, respectively, and are tangent to BC at D1 and D2, respectively. Find the ratio ofthe area of 4AI1D1 to the area of 4AI2D2.

20. [11] Consider an equilateral triangle T of side length 12. Matthew cuts T into N smaller equilateraltriangles, each of which has side length 1, 3, or 8. Compute the minimum possible value of N .

21. [11] A positive integer n is infallible if it is possible to select n vertices of a regular 100-gon so that theyform a convex, non-self-intersecting n-gon having all equal angles. Find the sum of all infallible integersn between 3 and 100, inclusive.

1The A-excircle of triangle ABC is the unique circle lying outside the triangle that is tangent to segment BC and theextensions of sides AB and AC.

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HMMT November 2019, November 9, 2019 — GUTS ROUND

Organization Team Team ID#

22. [12] Let f(n) be the number of distinct digits of n when written in base 10. Compute the sum of f(n) asn ranges over all positive 2019-digit integers.

23. [12] For a positive integer n, let, τ(n) be the number of positive integer divisors of n. How many integers1 ≤ n ≤ 50 are there such that τ(τ(n)) is odd?

24. [12] Let P be a point inside regular pentagon ABCDE such that ∠PAB = 48◦ and ∠PDC = 42◦. Find∠BPC, in degrees.

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HMMT November 2019, November 9, 2019 — GUTS ROUND

Organization Team Team ID#

25. [13] In acute 4ABC with centroid G, AB = 22 and AC = 19. Let E and F be the feet of the altitudesfrom B and C to AC and AB respectively. Let G′ be the reflection of G over BC. If E, F , G, and G′ lieon a circle, compute BC.

26. [13] Dan is walking down the left side of a street in New York City and must cross to the right side atone of 10 crosswalks he will pass. Each time he arrives at a crosswalk, however, he must wait t seconds,where t is selected uniformly at random from the real interval [0, 60] (t can be different at differentcrosswalks). Because the wait time is conveniently displayed on the signal across the street, Dan employsthe following strategy: if the wait time when he arrives at the crosswalk is no more than k seconds, hecrosses. Otherwise, he immediately moves on to the next crosswalk. If he arrives at the last crosswalkand has not crossed yet, then he crosses regardless of the wait time. Find the value of k which minimizeshis expected wait time.

27. [13] For a given positive integer n, we define ϕ(n) to be the number of positive integers less than or equalto n which share no common prime factors with n. Find all positive integers n for which

ϕ(2019n) = ϕ(n2).

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HMMT November 2019, November 9, 2019 — GUTS ROUND

Organization Team Team ID#

28. [15] A palindrome is a string that does not change when its characters are written in reverse order. LetS be a 40-digit string consisting only of 0’s and 1’s, chosen uniformly at random out of all such strings.Let E be the expected number of nonempty contiguous substrings of S which are palindromes. Computethe value of bEc.

29. [15] In isosceles 4ABC, AB = AC and P is a point on side BC. If ∠BAP = 2∠CAP , BP =√

3, andCP = 1, compute AP .

30. [15] A function f : Z→ Z satisfies: f(0) = 0 and∣∣f((n+ 1)2k)− f(n2k)∣∣ ≤ 1

for all integers k ≥ 0 and n. What is the maximum possible value of f(2019)?

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HMMT November 2019, November 9, 2019 — GUTS ROUND

Organization Team Team ID#

31. [17] James is standing at the point (0, 1) on the coordinate plane and wants to eat a hamburger. Foreach integer n ≥ 0, the point (n, 0) has a hamburger with n patties. There is also a wall at y = 2.1 whichJames cannot cross. In each move, James can go either up, right, or down 1 unit as long as he does notcross the wall or visit a point he has already visited.

Every second, James chooses a valid move uniformly at random, until he reaches a point with a hamburger.Then he eats the hamburger and stops moving. Find the expected number of patties that James eats onhis burger.

32. [17] A sequence of real numbers a0, a1, . . . , a9 with a0 = 0, a1 = 1, and a2 > 0 satisfies

an+2anan→1 = an+2 + an + an→1

for all 1 ≤ n ≤ 7, but cannot be extended to a10. In other words, no values of a10 ∈ R satisfy

a10a8a7 = a10 + a8 + a7.

Compute the smallest possible value of a2.

33. [17] A circle Γ with center O has radius 1. Consider pairs (A,B) of points so that A is inside the circleand B is on its boundary. The circumcircle Ω of OAB intersects Γ again at C 6= B, and line AC intersectsΓ again at X 6= C. The pair (A,B) is called techy if line OX is tangent to Ω. Find the area of the regionof points A so that there exists a B for which (A,B) is techy.

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HMMT November 2019, November 9, 2019 — GUTS ROUND

Organization Team Team ID#

34. [20] A polynomial P with integer coefficients is called tricky if it has 4 as a root.

A polynomial is called k-tiny if it has degree at most 7 and integer coefficients between −k and k, inclusive.

A polynomial is called nearly tricky if it is the sum of a tricky polynomial and a 1-tiny polynomial.

Let N be the number of nearly tricky 7-tiny polynomials. Estimate N .

An estimate of E will earn⌊20 min

(NE ,

EN

)4⌋points.

35. [20] You are trying to cross a 400 foot wide river. You can jump at most 4 feet, but you have many stonesyou can throw into the river. You will stop throwing stones and cross the river once you have placedenough stones to be able to do so. You can throw straight, but you can’t judge distance very well, soeach stone ends up being placed uniformly at random along the width of the river. Estimate the expectednumber N of stones you must throw before you can get across the river.

An estimate of E will earn⌊20 min

(NE ,

EN

)3⌋points.

36. [20] Let N be the number of sequences of positive integers (a1, a2, a3, . . . , a15) for which the polynomials

x2 − aix+ ai+1

each have an integer root for every 1 ≤ i ≤ 15, setting a16 = a1. Estimate N .

An estimate of E will earn⌊20 min

(NE ,

EN

)2⌋points.

HMMT November 2019November 9, 2019

Guts Round

1. [5] A polynomial P with integer coefficients is called tricky if it has 4 as a root.

A polynomial is called teeny if it has degree at most 1 and integer coefficients between −7 and 7,inclusive.

How many nonzero tricky teeny polynomials are there?

Proposed by: Carl Schildkraut

Answer: 2

If a degree 0 polynomial has 4 as a root, then it must be the constant zero polynomial. Thus, we willonly consider polynomials of degree 1.

If P has degree 1, integer coefficients, and 4 as a root, then it must be of the form P (x) = a(x− 4) =ax − 4a for some nonzero integer a. Since all integer coefficients are between −7 and 7, inclusive, werequire −7 ≤ 4a ≤ 7, which gives us a = −1, 1. Note that for both values, the coefficient of x is alsobetween −7 and 7, so there are 2 tricky teeny polynomials.

2. [5] You are trying to cross a 6 foot wide river. You can jump at most 4 feet, but you have one stoneyou can throw into the river; after it is placed, you may jump to that stone and, if possible, fromthere to the other side of the river. However, you are not very accurate and the stone ends up landinguniformly at random in the river. What is the probability that you can get across?

Proposed by: Carl Schildkraut and Milan Haiman

Answer:1

3

To be able to cross, the stone must land between 2 and 4 feet from the river bank you are standingon. Therefore the probability is 2

6 = 13 .

3. [5] For how many positive integers a does the polynomial

x2 − ax+ a

have an integer root?

Proposed by: Krit Boonsiriseth

Answer: 1

Let r, s be the roots of x2 − ax+ a = 0. By Vieta’s, we have r + s = rs = a. Note that if one root isan integer, then both roots must be integers, as they sum to an integer a. Then,

rs− (r + s) + 1 = 1 =⇒ (r − 1)(s− 1) = 1.

Because we require r, s to be both integers, we have r − 1 = s − 1 = ±1, which yields r = s = 0, 2.If r = 0 or s = 0, then a = 0, but we want a to be a positive integer. Therefore, our only possibilityis when r = s = 2, which yields a = 4, so there is exactly 1 value of a (namely, a = 4) such thatx2 − ax− a has an integer root.

4. [6] In 2019, a team, including professor Andrew Sutherland of MIT, found three cubes of integerswhich sum to 42:

42 = (−8053873881207597 )3 + (80435758145817515)3 + (12602123297335631)3

One of the digits, labeled by an underscore, is missing. What is that digit?

Proposed by: Andrew Gu

Answer: 4

Let the missing digit be x. Then, taking the equation modulo 10, we see that 2 ≡ −x3 + 53 + 13. Thissimplifies to x3 ≡ 4 (mod 10), which gives a unique solution of x = 4.

5. [6] A point P is chosen uniformly at random inside a square of side length 2. If P1, P2, P3, and P4

are the reflections of P over each of the four sides of the square, find the expected value of the area ofquadrilateral P1P2P3P4.

Proposed by: Carl Schildkraut

Answer: 8

Let ABCD denote the square defined in the problem. We see that if P1 is the reflection of P over AB,then the area of P1AB is the same as the area of PAB. Furthermore, if P4 is the reflection of P overDA, P1, A, and P4 are collinear, as A is the midpoint of P1P4. Repeating this argument for the otherpoints gives us that the desired area is

[P1AB]+[P2BC]+[P3CD]+[P4DA]+[ABCD] = [PAB]+[PBC]+[PCD]+[PDA]+[ABCD] = 2[ABCD] = 8.

6. [6] Compute the sum of all positive integers n < 2048 such that n has an even number of 1’s in itsbinary representation.

Proposed by: Milan Haiman

Answer: 1048064

Note that the positive integers less than 2047 are those with at most 11 binary digits. Consider thecontribution from any one of those digits. If we set that digit to 1, then the remaining 10 digits canbe set in 29 = 512 ways so that the number of 1’s is even. Therefore the answer is

512(20 + · · ·+ 210) = 512 · 2047 = 1048064.

7. [7] Let S be the set of all nondegenerate triangles formed from the vertices of a regular octagon withside length 1. Find the ratio of the largest area of any triangle in S to the smallest area of any trianglein S.

Proposed by: Carl Schildkraut

Answer: 3 + 2√

2

By a smoothing argument, the largest triangle is that where the sides span 3, 3, and 2 sides of theoctagon respectively (i.e. it has angles 45◦, 67.5◦, and 67.5◦), and the smallest triangle is that formedby three adjacent vertices of the octagon. Scaling so that the circumradius of the octagon is 1, ouranswer is

sin (90◦) + 2 sin (135◦)

2 sin (45◦)− sin (90◦)=

1 +√

2√2− 1

= 3 + 2√

2,

where the numerator is derived from splitting the large triangle by the circumradii, and the denominatoris derived from adding the areas of the two triangles formed by the circumradii, then subtracting thearea not in the small triangle.

8. [7] There are 36 students at the Multiples Obfuscation Program, including a singleton, a pair ofidentical twins, a set of identical triplets, a set of identical quadruplets, and so on, up to a set ofidentical octuplets. Two students look the same if and only if they are from the same identicalmultiple. Nithya the teaching assistant encounters a random student in the morning and a randomstudent in the afternoon (both chosen uniformly and independently), and the two look the same. Whatis the probability that they are actually the same person?

Proposed by: Yuan Yao

Answer:3

17

Let X and Y be the students Nithya encounters during the day. The number of pairs (X,Y ) for whichX and Y look the same is 1 · 1 + 2 · 2 + . . .+ 8 · 8 = 204, and these pairs include all the ones in whichX and Y are identical. As X and Y are chosen uniformly and independently, all 204 pairs are equallylikely to be chosen, thus the problem reduces to choosing one of the 36 pairs in 204, the probabilityfor which is 3

17 .

9. [7] Let p be a real number between 0 and 1. Jocelin has a coin that lands heads with probability p andtails with probability 1− p; she also has a number written on a blackboard. Each minute, she flips thecoin, and if it lands heads, she replaces the number x on the blackboard with 3x + 1; if it lands tailsshe replaces it with x/2. Given that there are constants a, b such that the expected value of the valuewritten on the blackboard after t minutes can be written as at+ b for all positive integers t, computep.

Proposed by: Carl Schildkraut

Answer:1

5

If the blackboard has the value x written on it, then the expected value of the value after one flip is

f(x) = p(3x− 1) + (1− p)x/2.

Because this expression is linear, we can say the same even if we only know the blackboard’s initialexpected value is x. Therefore, if the blackboard value is x0 at time 0, then after t minutes, theexpected blackboard value is f t(x0). We are given that x0, f(x0), f2(x0), . . . is an arithmetic sequence,so for there to be a constant difference, we must have f(x) = x+ c for some c.

This only occurs when 3p+ (1− p)/2 = 1, so p = 1/5.

10. [8] Let ABCD be a square of side length 5, and let E be the midpoint of side AB. Let P and Q be thefeet of perpendiculars from B and D to CE, respectively, and let R be the foot of the perpendicularfrom A to DQ. The segments CE,BP,DQ, and AR partition ABCD into five regions. What is themedian of the areas of these five regions?

Proposed by: Carl Schildkraut

Answer: 5

We have DQ ⊥ CE and AR ⊥ DQ, so AR||CE. Thus, we can show that 4ARD ∼= 4DQC ∼= 4CPB,so the median of the areas of the five regions is equal to the area of one of the three triangles listedabove.

Now, note that 4EBC ∼ 4BPC, so BPBC = EB

EC = 1√5. This means that BP =

√5, so CP = 2

√5.

Therefore, the area of 4BPC, the median area, is 5.

11. [8] Let a, b, c, d be real numbers such that

min(20x+ 19, 19x+ 20) = (ax+ b)− |cx+ d|

for all real numbers x. Find ab+ cd.

Proposed by: Yuan Yao

Answer: 380

In general, min(p, q) = p+q2 −

∣∣p−q2

∣∣. Letting p = 20x + 19 and q = 19x + 20 gives a = b = 19.5 andc = d = ±0.5. Then the answer is 19.52 − 0.52 = 19 · 20 = 380.

12. [8] Four players stand at distinct vertices of a square. They each independently choose a vertex of thesquare (which might be the vertex they are standing on). Then, they each, at the same time, beginrunning in a straight line to their chosen vertex at 10mph, stopping when they reach the vertex. Ifat any time two players, whether moving or not, occupy the same space (whether a vertex or a pointinside the square), they collide and fall over. How many different ways are there for the players tochoose vertices to go to so that none of them fall over?

Proposed by: Carl Schildkraut

Answer: 11

Observe that no two players can choose the same vertex, and no two players can choose each othersvertices. Thus, if two players choose their own vertices, then the remaining two also must choose their

own vertices (because they cant choose each others vertices), thus all 4 players must choose their ownvertices. There is 1 way to choose the vertices in this case.

Name the players top left, top right, bottom left, and bottom right, based on their initial positions.Assume exactly one player (without loss of generality, say the top left) chooses their own vertex. Then,the remaining 3 players have to form a triangle (recall no two player can choose each others vertices).There are 4 ways to choose which player chooses their own vertex, and 2 ways to choose which directionthe players move in the triangle, thus there are 8 ways to choose the vertices in this case.

Lastly, assume no one chooses their own vertex. We will first prove that no player can choose thevertex across them. Assume the contrary, without loss of generality, let the top left player chooses thebottom right vertex. Then, neither of the bottom left and the top right players can choose the othersvertex, because they would meet the top left player at the center of the square. As they cant choosebottom right (it is chosen by the top left player), and cant choose their own vertex (by assumption),they both have to choose the top left vertex, which is an immediate contradiction.

Now, the top left player has to choose either the top right vertex or the bottom left. Without lossof generality, let the player choose the top right vertex. Then, the top right player has to choose thebottom right vertex (as they can neither go across nor back to top left), the bottom right player has tochoose the bottom left vertex, and the bottom left player has to choose the top left vertex, and all thechoices are determined by the first players choice. There are 2 ways to choose where the first playergoes, thus there are 2 ways to choose the vertices in this case.

In total, there are 1 + 8 + 2 = 11 ways to choose the vertices.

13. [9] In 4ABC, the incircle centered at I touches sides AB and BC at X and Y , respectively. Addition-ally, the area of quadrilateral BXIY is 2

5 of the area of ABC. Let p be the smallest possible perimeterof a 4ABC that meets these conditions and has integer side lengths. Find the smallest possible areaof such a triangle with perimeter p.

Proposed by: Joey Heerens

Answer: 2√

5

Note that ∠BXI = ∠BY I = 90, which means that AB and BC are tangent to the incircle of ABC

at X and Y respectively. So BX = BY = AB+BC−AC2 , which means that 2

5 = [BXIY ][ABC] = AB+BC−AC

AB+BC+AC .

The smallest perimeter is achieved when AB = AC = 3 and BC = 4. The area of this triangle ABCis 2√

5.

14. [9] Compute the sum of all positive integers n for which

9√n+ 4

√n+ 2− 3

√n+ 16

is an integer.

Proposed by: Milan Haiman

Answer: 18

For the expression to be an integer at least one of n and n+ 2 must be a perfect square. We also notethat at most one of n and n+ 2 can be a square, so exactly one of them is a square.

Case 1: n is a perfect square. By our previous observation, it must be that 4√n+ 2 = 3

√n+ 16 ⇒

n = 16.

Case 2: n + 2 is a perfect square. By our previous observation, it must be that 9√n = 3

√n+ 16 ⇒

n = 2.

Consequently, the answer is 16 + 2 = 18.

15. [9] Let a, b, c be positive integers such that

a

77+

b

91+

c

143= 1.

What is the smallest possible value of a+ b+ c?

Proposed by: James Lin

Answer: 79

We need 13a + 11b + 7c = 1001, which implies 13(a + b + c − 77) = 2b + 6c. Then 2b + 6c must bedivisible by both 2 and 13, so it is minimized at 26 (e.g. with b = 10, c = 1). This gives a+ b+ c = 79.

16. [10] Equilateral 4ABC has side length 6. Let ω be the circle through A and B such that CA and CBare both tangent to ω. A point D on ω satisfies CD = 4. Let E be the intersection of line CD withsegment AB. What is the length of segment DE?

Proposed by: Benjamin Qi

Answer:20

13

Let F be the second intersection of line CD with ω. By power of a point, we have CF = 9, so DF = 5.

This means that [ADB][AFB] = DE

EF = DE5−DE . Now, note that triangle CAD is similar to triangle CFA,

so FAAD = CA

CD = 32 . Likewise, FB

BD = CBCD = 3

2 . Also, note that ∠ADB = 180 − ∠DAB − ∠DBA =

180 − ∠CAB = 120, and ∠AFB = 180 − ∠ADB = 60. This means that [ADB][AFB] = AD·BD·sin 120

FA·FB·sin 60 = 49 .

Therefore, we have that DE5−DE = 4

9 . Solving yields DE = 2013 .

17. [10] Kelvin the frog lives in a pond with an infinite number of lily pads, numbered 0, 1, 2, 3, and soforth. Kelvin starts on lily pad 0 and jumps from pad to pad in the following manner: when on lilypad i, he will jump to lily pad (i + k) with probability 1

2kfor k > 0. What is the probability that

Kelvin lands on lily pad 2019 at some point in his journey?

Proposed by: Nikhil Reddy

Answer:1

2

Suppose we combine all of the lily pads with numbers greater than 2019 into one lily pad labeled ∞.Also, let Kelvin stop once he reaches one of these lily pads.

Now at every leap, Kelvin has an equal chance of landing on 2019 as landing on ∞. Furthermore,Kelvin is guaranteed to reach 2019 or ∞ within 2020 leaps. Therefore the chance of landing on 2019is the same as missing it, so our answer is just 1

2 .

18. [10] The polynomial x3 − 3x2 + 1 has three real roots r1, r2, and r3. Compute

3√

3r1 − 2 + 3√

3r2 − 2 + 3√

3r3 − 2.

Proposed by: Milan Haiman

Answer: 0

Let r be a root of the given polynomial. Then

r3 − 3r2 + 1 = 0 =⇒ r3 − 3r2 + 3r − 1 = 3r − 2 =⇒ r − 1 = 3√

3r − 2.

Now by Vieta the desired value is r1 + r2 + r3 − 3 = 3− 3 = 0.

19. [11] Let ABC be a triangle with AB = 5, BC = 8, CA = 11. The incircle ω and A-excircle1 Γ arecentered at I1 and I2, respectively, and are tangent to BC at D1 and D2, respectively. Find the ratioof the area of 4AI1D1 to the area of 4AI2D2.

Proposed by: Carl Schildkraut

Answer:1

9

Let D′1 and D′2 be the points diametrically opposite D1 and D2 on the incircle and A-excircle, respec-tively. As Ix is the midpoint of Dx and D′x, we have

[AI1D1]

[AI2D2]=

[AD1D′1]

[AD2D′2].

Now, 4AD1D′1 and 4AD2D

′2 are homothetic with ratio r

rA= s−a

s , where r is the inradius, rA is theA-exradius, and s is the semiperimeter. Our answer is thus(

s− as

)2

=

(4

12

)=

1

9.

20. [11] Consider an equilateral triangle T of side length 12. Matthew cuts T into N smaller equilateraltriangles, each of which has side length 1, 3, or 8. Compute the minimum possible value of N .

Proposed by: Matthew Cho

Answer: 16

Matthew can cut T into 16 equilateral triangles with side length 3. If he instead included a triangleof side 8, then let him include a triangles of side length 3. He must include 122 − 82 − 32a = 80− 9atriangles of side length 1. Thus a ≤ 8, giving that he includes at least

(80− 9a) + (a) + 1 = 81− 8a ≥ 17

total triangles, so 16 is minimal.

21. [11] A positive integer n is infallible if it is possible to select n vertices of a regular 100-gon so thatthey form a convex, non-self-intersecting n-gon having all equal angles. Find the sum of all infallibleintegers n between 3 and 100, inclusive.

Proposed by: Benjamin Qi

Answer: 262

Suppose A1A2 . . . An is an equiangular n-gon formed from the vertices of a regular 100-gon. Note thatthe angle ∠A1A2A3 is determined only by the number of vertices of the 100-gon between A1 and A3.Thus in order for A1A2 . . . An to be equiangular, we require exactly that A1, A3, . . . are equally spacedand A2, A4, . . . are equally spaced. If n is odd, then all the vertices must be equally spaced, meaningn | 100. If n is even, we only need to be able to make a regular

(n2

)-gon from the vertices of a 100-gon,

which we can do if n | 200. Thus the possible values of n are 4, 5, 8, 10, 20, 25, 40, 50, and 100, for atotal of 262.

22. [12] Let f(n) be the number of distinct digits of n when written in base 10. Compute the sum of f(n)as n ranges over all positive 2019-digit integers.

Proposed by: Milan Haiman

Answer: 9(102019 − 92019

)Write

f(n) = f0(n) + · · ·+ f9(n),

where fd(n) = 1 if n contains the digit d and 0 otherwise. The sum of fd(n) over all 2019-digit positiveintegers n is just the number of 2019-digit positive integers that contain the digit d. For 1 ≤ d ≤ 9,∑

n

fd(n) = 9 · 102018 − 8 · 92018.

Also, ∑n

f0(n) = 9 · 102018 − 92019.

Summing over all possible values of d, we compute

∑n

f(n) =

9∑d=0

∑n

fd(n) = 9(9 · 102018 − 8 · 92018) + 9 · 102018 − 92019 = 9(102019 − 92019

).

23. [12] For a positive integer n, let, τ(n) be the number of positive integer divisors of n. How manyintegers 1 ≤ n ≤ 50 are there such that τ(τ(n)) is odd?

Proposed by: Kevin Liu

Answer: 17

Note that τ(n) is odd if and only if n is a perfect square. Thus, it suffices to find the number of integersn in the given range such that τ(n) = k2 for some positive integer k.

If k = 1, then we obtain n = 1 as our only solution. If k = 2, we see that n is either in the form pqor p3, where p and q are distinct primes. The first subcase gives 8 + 4 + 1 = 13 solutions, while thesecond subcase gives 2 solutions. k = 3 implies that n is a perfect square, and it is easy to see thatonly 62 = 36 works. Finally, k ≥ 4 implies that n is greater than 50, so we’ve exhausted all possiblecases. Our final answer is 1 + 13 + 2 + 1 = 17.

24. [12] Let P be a point inside regular pentagon ABCDE such that ∠PAB = 48◦ and ∠PDC = 42◦.Find ∠BPC, in degrees.

Proposed by: Dylan Liu

Answer: 84◦

Since a regular pentagon has interior angles 108◦, we can compute ∠PDE = 66◦, ∠PAE = 60◦, and∠APD = 360◦−∠AED−∠PDE−∠PAE = 126◦. Now observe that drawing PE divides quadrilateralPAED into equilateral triangle PAE and isosceles triangle PED, where ∠DPE = ∠EDP = 66◦. Thatis, we get PA = PE = s, where s is the side length of the pentagon.

Now triangles PAB and PED are congruent (with angles 48◦−66◦−66◦), so PD = PB and ∠PDC =∠PBC = 42◦. This means that triangles PDC and PBC are congruent (side-angle-side), so ∠BPC =∠DPC.

Finally, we compute ∠BPC+∠DPC = 2∠BPC = 360◦−∠APB −∠EPA−∠DPE = 168◦, meaning∠BPC = 84◦.

25. [13] In acute 4ABC with centroid G, AB = 22 and AC = 19. Let E and F be the feet of the altitudesfrom B and C to AC and AB respectively. Let G′ be the reflection of G over BC. If E, F , G, and G′

lie on a circle, compute BC.

Proposed by: Milan Haiman

Answer: 13

Note that B,C,E, F lie on a circle. Moreover, since BC bisects GG′, the center of the circle that goes

through E,F,G,G′

must lie on BC. Therefore, B,C,E, F,G,G′

lie on a circle. Specifically, the centerof this circle is M , the midpoint of BC, as ME = MF because M is the center of the circumcircleof BCEF . So we have GM = BC

2 , which gives AM = 3BC2 . Then, by Apollonius’s theorem, we have

AB2 +AC2 = 2(AM2 +BM2). Thus 845 = 5BC2 and BC = 13.

26. [13] Dan is walking down the left side of a street in New York City and must cross to the right sideat one of 10 crosswalks he will pass. Each time he arrives at a crosswalk, however, he must wait tseconds, where t is selected uniformly at random from the real interval [0, 60] (t can be different atdifferent crosswalks). Because the wait time is conveniently displayed on the signal across the street,Dan employs the following strategy: if the wait time when he arrives at the crosswalk is no more thank seconds, he crosses. Otherwise, he immediately moves on to the next crosswalk. If he arrives at thelast crosswalk and has not crossed yet, then he crosses regardless of the wait time. Find the value ofk which minimizes his expected wait time.

Answer: 60

(1−

(1

10

) 19

)

With probability(1− k

60

)9, Dan reaches the last crosswalk without crossing at any previous site, in

which case the expected value of his wait time is 30 seconds. Otherwise, with probability 1−(1− k

60

)9,

Dan crosses at an earlier crosswalk, in which case the expected value of his wait time is k2 . We want

to find the k that minimizes

30

(1− k

60

)9

+k

2

(1−

(1− k

60

)9)

= 30−(

30− k

2

)(1−

(1− k

60

)9)

Letting a = 1− k60 , we can use weighted AM-GM:

9110

(a(1− a9

)) 910 =

(9a9) 1

10(1− a9

) 910 ≤ 9

10

where equality occurs when 9a9 = 1 − a9, or a =(

110

) 19 , meaning that k = 60

(1−

(110

) 19

). Because

our original expression can be written as

30− 30a(1− a9),

the minimum occurs at the same value, k = 60(

1−(

110

) 19

).

27. [13] For a given positive integer n, we define ϕ(n) to be the number of positive integers less than orequal to n which share no common prime factors with n. Find all positive integers n for which

ϕ(2019n) = ϕ(n2).

Proposed by: Carl Schildkraut

Answer: 1346, 2016, 2019

Let p1, p2, . . . , pk be the prime divisors of n. Then it is known that ϕ(n) = n · p1−1p1. . . pk−1pk

. As n2 and

n has the same set of prime divisors, it also holds that ϕ(n2) = n2 · p1−1p1. . . pk−1pk

. We will examine theequality in four cases.

• gcd(n, 2019) = 1 In this case, 2019 · n has also 3 and 673 as prime divisors, thus ϕ(2019 · n) =2019 · n · p1−1p1

. . . pk−1pk· 23 ·

672673 , and the equality implies n = 1342, however gcd(1342, 3) 6= 1,

contradiction. Thus, there is no answer in this case.

• gcd(n, 2019) = 3 In this case, 2019 · n has also 673 as a prime divisor, thus ϕ(2019 · n) =2019 ·n · p1−1p1

. . . pk−1pk· 672673 , and the equality implies n = 2016, which satisfies the equation. Thus,

the only answer in this case is n = 2016.

• gcd(n, 2019) = 673 In this case, 2019 · n has also 3 as a prime divisor, thus ϕ(2019 · n) =2019 · n · p1−1p1

. . . pk−1pk· 23 , and the equality implies n = 1346, which satisfies the equation. Thus,

the only answer in this case is n = 1346.

• gcd(n, 2019) = 2019 In this case, 2019 · n has the same set of prime divisors, thus ϕ(2019 · n) =2019 ·n · p1−1p1

. . . pk−1pk, and the equality implies n = 2019, which satisfies the equation. Thus, the

only answer in this case is n = 2019.

Thus, all the answers are n = 1346, 2016, 2019.

28. [15] A palindrome is a string that does not change when its characters are written in reverse order.Let S be a 40-digit string consisting only of 0’s and 1’s, chosen uniformly at random out of all suchstrings. Let E be the expected number of nonempty contiguous substrings of S which are palindromes.Compute the value of bEc.

Proposed by: Benjamin Qi

Answer: 113

Note that S has 41 − n contiguous substrings of length n, so we see that the expected number ofpalindromic substrings of length n is just (41− n) · 2−bn/2c. By linearity of expectation, E is just thesum of this over all n from 1 to 40. However, it is much easier to just compute

∞∑n=1

(41− n) · 2−bn/2c.

The only difference here is that we have added some insignificant negative terms in the cases wheren > 41, so E is in fact slightly greater than this value (in fact, the difference between E and this sumis 7

1048576 ). To make our infinite sum easier to compute, we can remove the floor function by pairingup consecutive terms. Then our sum becomes

40 +

∞∑n=1

81− 4n

2n,

which is just 40 + 81 − 8 = 113. E is only slightly larger than this value, so our final answer isbEc = 113 .

29. [15] In isosceles 4ABC, AB = AC and P is a point on side BC. If ∠BAP = 2∠CAP , BP =√

3,and CP = 1, compute AP .

Proposed by: Milan Haiman

Answer:√

2

Let ∠CAP = α, By the Law of Sines,√3

sin 2α = 1sinα which rearranges to cosα =

√32 ⇒ α = π

6 . This

implies that ∠BAC = π2 . By the Pythagorean Theorem, 2AB2 = (

√3 + 1)2, so AB2 = 2 +

√3.

Applying Stewart’s Theorem, it follows that AP 2 = (√3+1)(2+

√3)√

3+1−√

3⇒ AP =√

2.

30. [15] A function f : Z→ Z satisfies: f(0) = 0 and∣∣f((n+ 1)2k)− f(n2k)∣∣ ≤ 1

for all integers k ≥ 0 and n. What is the maximum possible value of f(2019)?

Proposed by: Krit Boonsiriseth

Answer: 4

Consider a graph on Z with an edge between (n + 1)2k and n2k for all integers k ≥ 0 and n. Eachvertex m is given the value f(m). The inequality |f((n + 1)2k) − f(n2k)| ≤ 1 means that any twoadjacent vertices of this graph must have values which differ by at most 1. Then it follows that for allm,

f(m) ≤ number of edges in shortest path from 0 to m

because if we follow a path from 0 to m, along each edge the value increases by at most 1. Conversely,if we define f(m) to be the number of edges in the shortest path between 0 and m, then this is a validfunction because for any two adjacent vertices, the lengths of their respective shortest paths to 0 differby at most 1. Hence it suffices to compute the distance from 0 to 2019 in the graph.

There exists a path with 4 edges, given by

0→ 2048→ 2016→ 2018→ 2019.

Suppose there existed a path with three edges. In each step, the number changes by a power of 2, sowe have 2019 = ±2k1 ± 2k2 ± 2k3 for some nonnegative integers k1, k2, k3 and choice of signs. Since2019 is odd, we must have 20 somewhere. Then we have ±2k1 ± 2k2 ∈ {2018, 2020}. Without loss of

generality assume that k1 ≥ k2. Then we can write this as ±2k2(2k1k2 ± 1) ∈ {2018, 2020}. It is easyto check that k1 = k2 is impossible, so the factorization 2k2(2k1k2 ± 1) is a product of a power of twoand an odd number. Now compute 2018 = 2× 1009 and 2020 = 4× 505. Neither of the odd parts areof the form 2k1−k2 ± 1, so there is no path of three steps.

We conclude that the maximum value of f(2019) is 4.

31. [17] James is standing at the point (0, 1) on the coordinate plane and wants to eat a hamburger. Foreach integer n ≥ 0, the point (n, 0) has a hamburger with n patties. There is also a wall at y = 2.1which James cannot cross. In each move, James can go either up, right, or down 1 unit as long as hedoes not cross the wall or visit a point he has already visited.

Every second, James chooses a valid move uniformly at random, until he reaches a point with ahamburger. Then he eats the hamburger and stops moving. Find the expected number of patties thatJames eats on his burger.

Proposed by: Joey Heerens

Answer:7

3

Note that we desire to compute the number of times James moves to the right before moving down tothe line y = 0. Note also that we can describe James’s current state based on whether his y-coordinateis 0 or 1 and whether or not the other vertically adjacent point has been visited. Let E(1, N) be theexpected number of times James will go right before stopping if he starts at a point with y-coordinate 1and the other available point with the same x-coordinate has not been visited. Define E(1, Y ), E(2, N),and E(2, Y ) similarly. Then we can construct equations relating the four variables:

E(1, N) =1

3E(2, Y ) +

1

3(E(1, N) + 1),

as James can either go up, right, or down with probability 1/3 each if he starts in the state (1, N).Similarly, we have

E(2, N) =1

2E(1, Y ) +

1

2(E(2, N) + 1), E(1, Y ) =

1

2(E(1, N) + 1),

and E(2, Y ) = E(2, N)+1. Solving these equations, we get E(1, N) = 73 , which is our answer, as James

starts in that state having gone left 0 times.

32. [17] A sequence of real numbers a0, a1, . . . , a9 with a0 = 0, a1 = 1, and a2 > 0 satisfies

an+2anan−1 = an+2 + an + an−1

for all 1 ≤ n ≤ 7, but cannot be extended to a10. In other words, no values of a10 ∈ R satisfy

a10a8a7 = a10 + a8 + a7.

Compute the smallest possible value of a2.

Proposed by: Dylan Liu

Answer:√

2− 1

Say a2 = a. Then using the recursion equation, we have a3 = −1, a4 = a+1a−1 , a5 = −a+1

a+1 , a6 = − 1a ,

a7 = − 2aa2−1 , and a8 = 1.

Now we have a10a8a7 = a10 + a8 + a7. No value of a10 can satisfy this equation iff a8a7 = 1 anda8 + a7 6= 0. Since a8 is 1, we want 1 = a7 = − 2a

a2−1 , which gives a2 + 2a − 1 = 0. The only positive

root of this equation is√

2− 1.

This problem can also be solved by a tangent substitution. Write an = tanαn. The given conditionbecomes

αn+2 + αn + αn−1 = 0.

We are given α0 = 0, α1 = π/4, and α2 ∈ (0, π/2). Using this, we can recursively compute α3, α4, . . .in terms of α2 until we get to α10 = 3π

4 − 2α2. For a10 not to exist, we need α10 ≡ π/2 mod π. The

only possible value of α2 ∈ (0, π/2) is α2 = π/8, which gives a2 = tanπ/8 =√

2− 1.

33. [17] A circle Γ with center O has radius 1. Consider pairs (A,B) of points so that A is inside thecircle and B is on its boundary. The circumcircle Ω of OAB intersects Γ again at C 6= B, and line ACintersects Γ again at X 6= C. The pair (A,B) is called techy if line OX is tangent to Ω. Find the areaof the region of points A so that there exists a B for which (A,B) is techy.

Proposed by: Carl Schildkraut and Milan Haiman

Answer:3π

4

We claim that (A,B) is techy if and only if OA = AB.

Note that OX is tangent to the circle (OBC) if and only if OX is perpendicular to the angle bisectorof ∠BOC, since OB = OC. Thus (A,B) is techy if and only if OX is parallel to BC. Now sinceOC = OX,

OX ‖ BC ⇐⇒ ∠BCA = ∠OXA ⇐⇒ ∠BCA = ∠ACO ⇐⇒ OA = AB.

From the claim, the desired region of points A is an annulus between the circles centered at O withradii 1

2 and 1. So the answer is 3π4 .

34. [20] A polynomial P with integer coefficients is called tricky if it has 4 as a root.

A polynomial is called k-tiny if it has degree at most 7 and integer coefficients between −k and k,inclusive.

A polynomial is called nearly tricky if it is the sum of a tricky polynomial and a 1-tiny polynomial.

Let N be the number of nearly tricky 7-tiny polynomials. Estimate N .

An estimate of E will earn⌊20 min

(NE ,

EN

)4⌋points.

Proposed by: Carl Schildkraut

Answer: 64912347

A tricky 7-tiny polynomial takes the form

(c6x6 + . . .+ c1x+ c0)(x− 4).

For each fixed value of k, ck − 4ck+1 should lie in [−7, 7], so if we fix ck, there are around 15/4 waysof choosing ck+1. Therefore if we pick c0, . . . , c6 in this order, there should be around (15/4)7 tricky7-tiny polynomials.

A 1-tiny polynomial takes the form ε6x7 + · · ·+ ε1x+ ε0 with εi ∈ {−1, 0,+1}, so there are 38 1-tiny

polynomials.

A nearly tricky 7-tiny polynomial P takes the form Q+T where Q is roughly a tricky 7-tiny polynomial,and T is 1-tiny. Furthermore, there is a unique decomposition Q + T because T (4) = P (4) and eachinteger n can be written in the form

∑εk4k in at most one way. Therefore the number of nearly tricky

7-tiny is around (15/4)7 · 38 ≈ 68420920, which is worth 16 points.

The exact answer can be found by setting up recurrences. Let t(d, `) be the number of polynomials ofdegree at most i of the form

(`xd−1 + cd−2xd−2 + · · ·+ c0)(x− 4) + (εd−1x

d−1 + · · ·+ ε1x+ ε0).

which has integer coefficients between −7 and 7 except the leading term `xd. It follows that t(0, 0) =1, t(0, k) = 0 for all k 6= 0, and t(d+ 1, `) can be computed as follows: for each value of cd−1, there are

t(d, cd−1) ways to pick cd−2, . . . , c0, εd−1, . . . , ε0, and exactly w(cd−1 − 4`) ways of picking εd, wherew(k) = min(9− |k|, 3) for |k| ≤ 8 and 0 otherwise. Therefore setting c = cd−1 − 4` we have

t(d+ 1, `) =

8∑c=−8

t(d, c+ 4`)w(c).

The number of nearly tricky 7-tiny polynomials is simply t(8, 0), which can be computed to be 64912347using the following C code.

int w(int a){

if(a < -9 || a > 9) return 0;

else if(a == -8 || a == 8) return 1;

else if(a == -7 || a == 7) return 2;

else return 3;

}

int main()

{

int m=8,n=7,r=4,d,l,c,c4l;

int mid = 2 + n/r;

int b = 2*mid+1;

long int t[500][500];

for(l=0; l<b; l++){

t[0][l] = (l == mid) ? 1 : 0;

}

for(d=0; d<m+1; d++){

for(l=0; l<b; l++){

t[d+1][l] = 0;

for(c=-8; c<9; c++){

c4l = c + 4*(l-mid) + mid;

t[d+1][l] += (c4l >= 0 && c4l <= 2*mid) ? t[d][c4l]*w(c) : 0;

}

}

}

printf("%ld",t[8][mid]);

}

35. [20] You are trying to cross a 400 foot wide river. You can jump at most 4 feet, but you have manystones you can throw into the river. You will stop throwing stones and cross the river once you haveplaced enough stones to be able to do so. You can throw straight, but you can’t judge distance verywell, so each stone ends up being placed uniformly at random along the width of the river. Estimatethe expected number N of stones you must throw before you can get across the river.

An estimate of E will earn⌊20 min

(NE ,

EN

)3⌋points.

Proposed by: Carl Schildkraut and Milan Haiman

Answer: 712.811

If we divide the river into 100 4-foot sections, then to be able to cross we need to get at least one stoneinto each section. On average, this takes

100

100+

100

99+ · · ·+ 100

1≈ 100 ln 100

stone throws (it takes 100100−k moves on average to get a stone into a new section if k sections already

have a stone). So the answer is at least 100 ln 100 ≈ 450.

On the other hand, if we divide the river into 200 2-foot sections, then once we have a stone in eachsection we are guaranteed to be able to cross. By a similar argument, we obtain that the answer is atmost 200 ln 200 ≈ 1050.

Estimates near these bounds earn about 5 to 7 points. An estimate in between can earn close to 20points.

To compute the answer (almost) exactly, we use the following argument.

Scale the problem so the river is of size 1, and the jumps are of size 0.01. Suppose that after n throws,the stones thrown are located at positions 0 < x1 < x2 < · · · < xn < 1. Let x0 = 0, xn+1 = 1, r = 0.01.Define P (n) to be the probability that you still cannot cross the river after n throws. In other words,there exists i such that xi+1 − xi > r. Then our answer is

∑∞n=0 P (n).

By PIE we can write

P (n) =

∞∑i=1

(−1)i−1(n+ 1

i

)max(1− ir, 0)n.

based on which intervals xi+1− xi have length greater than r. Now we switch the order of summation:

∞∑n=0

P (n) =

∞∑n=0

∞∑i=1

(−1)i−1(n+ 1

i

)max(1− ir, 0)n =

∞∑i=1

(−1)i−1∞∑n=0

(n+ 1

i

)max(1− ir, 0)n.

Let x = max(1− ir, 0). Then

∞∑n=0

(n+ 1

i

)xn = xi−1

∞∑j=0

(i+ j

i

)xj =

xi−1

(1− x)i+1.

Thus, our answer isb1/rc∑i=1

(−1)i−1(1− ir)i−1

(ir)i+1≈ 712.811,

where the last approximation uses the C++ code below.

#include <bits/stdc++.h>

using namespace std;

typedef long double ld;

int main() {

ld sum = 0, r = 0.01;

for (int i = 1; ; ++i) {

ld x = 1-r*i; if (x <= 0) break;

ld ex = pow(x/(1-x),i-1)/(1-x)/(1-x);

if (i&1) sum += ex;

else sum -= ex;

}

cout << fixed << setprecision(8) << sum;

}

36. [20] Let N be the number of sequences of positive integers (a1, a2, a3, . . . , a15) for which the polynomials

x2 − aix+ ai+1

each have an integer root for every 1 ≤ i ≤ 15, setting a16 = a1. Estimate N .

An estimate of E will earn⌊20 min

(NE ,

EN

)2⌋points.

Proposed by: Krit Boonsiriseth

Answer: 1409

We note that ai+1 = x(ai − x) for some positive integer x, so ai+1 ≥ ai − 1. So, the only way ai candecrease is decreasing by 1.

As it cannot decrease that quickly, we will make the assumption that if ai ≥ 10, ai+1 = ai − 1, asotherwise it will increase at least above 16 at which point it will take many moves to go back downbelow 10. Write that a→ b if b is a possible value of ai+1 given a = ai. We have

5→ 6, 6→ 5, 8, 9, 7→ 6, 8→ 7, 9→ 8,

and in addition by going to 10 and above, 7 can go to 9 in 2 or 4 steps, 8 can in 4, 7, 8 steps, and9 can in 6, 10, 12 steps. We see from this that the vast majority of sequences should pass through 8.By looking at cycles from 8, we can determine exactly when a sequence can start at 8 and return to8 (there is one way in 3 steps, two in 4 steps, etc.), and from there we can generate a list of types ofsequences by when 8s occur. By dividing by the number of 8s and multiplying by 15, we can get thenumber of sequences that include 8, which gives us an estimate of 1235, giving us 15 points. As wenote that this is a lower estimate, we may round up slightly to get better results.

To find the exact answer, we will first show that no element larger than 32 can occur in the sequence.Reorder the sequence to make a1 maximal; we have

ai+1 ≥ ai − 1 =⇒ a15 ≥ a1 − 14.

Also, since a1 > a15, a1 ≥ 2a15 − 4, giving

a1 − 14 ≤ a1 + 4

2=⇒ a1 ≤ 32.

We then construct the following Python code:

def p36(max_val,length):

L=[[i] for i in range(1,max_val+1)]

for j in range(length):

newL=[]

for k in L:

poss=[x*(k[-1]-x) for x in range(1,k[-1]//2+1)]

for t in poss:

if 1<=t<=max_val:

newL.append(k+[t])

L=newL

return len(L)

print(p36(32,15))

This gives the exact answer of 1409.

HMIC 2019March 30, 2019 – April 5, 2019

1. [5] Let ABC be an acute scalene triangle with incenter I. Show that the circumcircle of BIC intersectsthe Euler line of ABC in two distinct points.

(Recall that the Euler line of a scalene triangle is the line that passes through its circumcenter, centroid,orthocenter, and the nine-point center.)

2. [7] Annie has a permutation (a1, a2, . . . , a2019) of S = {1, 2, . . . , 2019}, and Yannick wants to guess herpermutation. With each guess Yannick gives Annie an n-tuple (y1, y2, . . . , y2019) of integers in S, andthen Annie gives the number of indices i ∈ S such that ai = yi.

(a) Show that Yannick can always guess Annie’s permutation with at most 1200000 guesses.

(b) Show that Yannick can always guess Annie’s permutation with at most 24000 guesses.

3. [8] Do there exist four points Pi = (xi, yi) ∈ R2 (1 ≤ i ≤ 4) on the plane such that:

• for all i = 1, 2, 3, 4, the inequality x4i + y4i ≤ x3i + y3i holds, and

• for all i 6= j, the distance between Pi and Pj is greater than 1?

4. [10] A cactus is a finite simple connected graph where no two cycles share an edge. Show that in anonempty cactus, there must exist a vertex which is part of at most one cycle.

5. [12] Let p = 2017 be a prime and Fp be the integers modulo p. A function f : Z→ Fp is called good ifthere is α ∈ Fp with α 6≡ 0 (mod p) such that

f(x)f(y) = f(x+ y) + αyf(x〉 y) (mod p)

for all x, y ∈ Z. How many good functions are there that are periodic with minimal period 2016?

HMIC 2019March 30, 2019 – April 5, 2019

1. [5] Let ABC be an acute scalene triangle with incenter I. Show that the circumcircle of BIC intersectsthe Euler line of ABC in two distinct points.

(Recall that the Euler line of a scalene triangle is the line that passes through its circumcenter, centroid,orthocenter, and the nine-point center.)

Proposed by: Andrew Gu

Let O and H be the circumcenter and orthocenter of ABC. Recall that

∠BOC = 2∠A,

∠BHC = 180◦ − ∠A,

∠BIC = 90◦ +1

2∠A.

As ABC is acute, A, I,O,H all lie on the same side of BC.

• If ∠A > 60◦, then ∠BOC > ∠BIC, so O lies inside (BIC).

• If ∠A < 60◦, then ∠BHC > ∠BIC, so H lies inside (BIC).

• If ∠A = 60◦, then O and H are on (BIC).

In all cases, two intersections are guaranteed.

Note: For an obtuse triangle, the statement is not necessarily true.

2. [7] Annie has a permutation (a1, a2, . . . , a2019) of S = {1, 2, . . . , 2019}, and Yannick wants to guess herpermutation. With each guess Yannick gives Annie an n-tuple (y1, y2, . . . , y2019) of integers in S, andthen Annie gives the number of indices i ∈ S such that ai = yi.

(a) Show that Yannick can always guess Annie’s permutation with at most 1200000 guesses.

(b) Show that Yannick can always guess Annie’s permutation with at most 24000 guesses.

Proposed by: Yuan Yao

Part (a) uses the idea that for x 6= y, the guess (x, y, y, ..., y) returns 0 if the first number is y, 2 if the firstnumber is x, and 1 if the number is something else. If he tests the pairs (x, y) = (1, 2), . . . , (2017, 2018),he will get a 0 or 2 (and therefore find out what the first number is) or get only 1 (so the first numberis 2019). In general, with n numbers unknown, Yannick needs

⌊n2

⌋guesses to find out the next number

in the permutation. Using this strategy, he can guess the permutation within at most

2019∑n=1

⌊n2

⌋+ 1 = 1009 · 1010 + 1 = 1019091

guesses.

Part (b) uses the idea that after the first number has been determined and n numbers remain undeter-mined, it takes dlog2 ne guesses to determine the index of another number, by doing a binary searchto halve the candidate positions each time (using the first number as a filler). Using this Yannick canguess the permutation within

1009 +

2018∑n=1

dlog2 ne+ 1 = 1009 + 1 · 1 + 2 · 2 + 4 · 3 + · · ·+ 512 · 10 + 994 · 11 + 1 = 21161

guesses.

3. [8] Do there exist four points Pi = (xi, yi) ∈ R2 (1 ≤ i ≤ 4) on the plane such that:

• for all i = 1, 2, 3, 4, the inequality x4i + y4i ≤ x3i + y3i holds, and

• for all i 6= j, the distance between Pi and Pj is greater than 1?

Proposed by: Pakawut Jiradilok

In fact, there are! One might think that the region

{(x, y) ∈ R2 | x4 + y4 ≤ x3 + y3}

is inside the ball defined by x2 + y2 ≤ x+ y, which is a ball of radius 1/√

2. It turns out that it is notthe case.

We claim that for all ε > 0 small enough, we can choose (0, 0), (1, 1), (1− ε2,−2ε), and (−2ε, 1− ε2).First we check the condition x4i + y4i ≤ x3i + y3i . This is obviously satisfied for the first two points, andfor the other two points this translates to

(1− ε2)4 + (−2ε)4 ≤ (1− ε2)3 + (−2ε)3

(2ε)3(1 + 2ε) ≤ (1− ε2)3(ε2)

8ε(1 + 2ε) ≤ (1− ε2)3.

This is satisfied for small ε. Next we check that the distances are greater than 1. Clearly the thirdand fourth points are at distance greater than 1 from (1, 1) as they have negative y and x coordinatesrespectively. They are also at distance greater than 1 from each other for small ε as their distancetends to

√2 as ε→ 0. Also, (0, 0) and (1, 1) are at distance greater than 1 as well. The only remaining

distances to check are the distance from the first point to the third and fourth points (which are thesame distance). This distance is√

(1− ε2)2 + (−2ε)2 =√

(1 + ε2)2 = 1 + ε2 > 0

so all conditions are satisfied (for sufficiently small ε).

For example, one may choose ε = 1/20 to obtain the four points (0, 0), (1, 1), (− 110 ,

399400 ) and ( 399

400 ,−110 ).

4. [10] A cactus is a finite simple connected graph where no two cycles share an edge. Show that in anonempty cactus, there must exist a vertex which is part of at most one cycle.

Proposed by: Kevin Yang

Let C be the original cactus. For every cycle in C, arbitrarily remove one of its edges, yielding a newgraph T . Observe that since the cycles are edge-disjoint, we removed exactly one edge from every cycle,meaning the graph stays connected. However, there are no longer any cycles, so T is a tree.

Now consider any leaf v of T (note that any nonempty tree must have a leaf). If v is the only vertexin the graph we’re trivially done, since then v was the only vertex in C. Otherwise v has degree 1.If v was originally a leaf of C, we’re done. If not, observe that in the process of turning C into T , avertex’s degree cannot decrease by more than half, because for every cycle that a vertex is part of in C,it gains a degree of 2, but can only lose 1 degree from an edge of that cycle being removed. Therefore,the original degree of v in C was at most 2, meaning it could have been part of at most 1 cycle, asdesired.

Remark. The problem originally did not state that the graph is finite. We apologize for the omission.

5. [12] Let p = 2017 be a prime and Fp be the integers modulo p. A function f : Z→ Fp is called good ifthere is α ∈ Fp with α 6≡ 0 (mod p) such that

f(x)f(y) = f(x+ y) + αyf(x− y) (mod p)

for all x, y ∈ Z. How many good functions are there that are periodic with minimal period 2016?

Proposed by: Ashwin Sah

Answer: 1327392

We will classify all good functions with parameter α ∈ Fp\{0}. We will use = to denote equalitymodulo p when appropriate. Let the given statement be P (x, y). Then P (x, 0) gives f(x)f(0) = 2f(x)for all x ∈ Z. f cannot be the zero function because it does not have minimal period 2016. Thereforef(0) = 2.

P (x, 1) gives f(x + 1) − f(1)f(x) + αf(x − 1) = 0. This means that f satisfies a recurrence of depthtwo.

We can interpret f as a function f : Z → Fp2 , embedding the codomain Fp inside of Fp2 . Let τ, ρ bethe unique roots of t2 − f(1)t + α = 0 in Fp2 , so that τ + ρ = f(1), τρ = α. Since α is nonzero, soare τ and ρ. Notice that f(0) = τ0 + ρ0 and f(1) = τ1 + ρ1. By induction up and down, we see thatf(n) = τn + ρn for all n ∈ Z.

We can also easily check that given any value of f(1) and α with α 6≡ 0 (mod p), we get a unique suchfunction. Thus there are p(p− 1) total such functions, one corresponding to each quadratic t2− at+ b,where b 6= 0. If τ, ρ are the roots of the quadratic, the associated function is f(n) = τn + ρn.

Now if f is q-periodic then f(n) = f(n+ q) for all n ∈ Z, so τn + ρn = τn+q + ρn+q for all n ∈ Z so

τn(1− τ q) = ρn(ρq − 1).

We have two cases.

Case 1. τ 6= ρ.

Then n = 0 and n = 1 above give 1 = τ q = ρq, and this is furthermore sufficient to be q-periodic. Thusthe period of f is the minimal q such that τ q = ρq = 1, so the lcm of the orders of τ, ρ as elements of Fp2 .

Case 2. τ = ρ.

Then we see that τ = ρ = f(1)2 , so they are in Fp. This automatically means that f(n) = 2

(f(1)2

)nfor

all n ∈ Z, where f(1) 6= 0 is forced (as τρ = α). This has period dividing p− 1. More specifically, the

period is the order of f(1)2 as an element of Fp.

In the first case, note that the orders of τ, ρ dividing p− 1 implies that τ, ρ are actually in Fp, so thatwe can restrict to counting recurrences f(n) = an + bn where a, b are nonzero elements of Fp.The unordered pair {a, b} determines a + b = f(1) and ab = α, so it determines a unique f by ourearlier observation. We need to count the number of such pairs with lcm(ordp(a), ordp(b)) = p− 1.

Recall that for d | p− 1, there are φ(d) elements of order d in F×p . If a = b we need ordp(a) = p− 1, sothere are φ(p−1) functions. If S is the number of ordered pairs (a, b) with lcm(ordp(a), ordp(b)) = p−1,

then we see that the desired count is φ(p−1)+S2 (as each unordered pair {a, b} is counted twice as and

ordered pair except for those with a = b).

Now let n = p−1. We are counting pairs with the lcm equal to n. The number of pairs with respectiveorders d1, d2 is φ(d1)φ(d2) hence our desired sum is

∑d1,d2|n,lcm(d1,d2)=n

φ(d1)φ(d2). Now notice thatφ is a multiplicative function, and the condition on the sum splits among the prime powers dividing n,so we can choose how d1, d2 behave on the prime factors of n independently. Therefore if we define

χ(n) =∑

d1,d2|n,lcm(d1,d2)=n

φ(d1)φ(d2),

then χ is multiplicative. Finally, if q is prime and e ≥ 1 we have

χ(qe) =∑

d1,d2|qe,lcm(d1,d2)=qe

φ(d1)φ(d2)

= 2φ(qe)(φ(1) + φ(q) + · · ·+ φ(qe−1)) + φ(qe)2

= 2qe−1(q − 1)qe−1 + q2e−2(q − 1)2

= q2e−2(q2 − 1) = q2e(

1− 1

q2

).

This implies that χ(n) = n2∏q|n

(1− 1

q2

)by multiplicativity. Finally, our original answer is

φ(p− 1) + (p− 1)2∏q|p−1

(1− 1

q2

)2

,

which we now compute for p = 2017. We have p−1 = 2016 = 25 ·32 ·7. Thus φ(p−1) = 24 ·2 ·3 ·6 = 576

and∏q|2016

(1− 1

q2

)= 3

4 ·89 ·

4849 = 32

49 . Thus the number of functions is

576 + 3249 · 20162

2=

576 + 215 · 34

2= 288 + 214 · 34 = 1327392 .

HMMT February 2020February 15, 2020

Team Round

1. [20] Let n be a positive integer. Define a sequence by a0 = 1, a2i+1 = ai, and a2i+2 = ai + ai+1 foreach i ≥ 0. Determine, with proof, the value of a0 + a1 + a2 + · · ·+ a2n∈1.

2. [25] Let n be a fixed positive integer. An n-staircase is a polyomino with n(n+1)2 cells arranged in the

shape of a staircase, with arbitrary size. Here are two examples of 5-staircases:

Prove that an n-staircase can be dissected into strictly smaller n-staircases.

3. [25] Let ABC be a triangle inscribed in a circle ω and ` be the tangent to ω at A. The line through Bparallel to AC meets ` at P , and the line through C parallel to AB meets ` at Q. The circumcirclesof ABP and ACQ meet at S 6= A. Show that AS bisects BC.

4. [35] Alan draws a convex 2020-gon A = A1A2 · · ·A2020 with vertices in clockwise order and chooses2020 angles θ1, θ2, . . . , θ2020 ∈ (0, π) in radians with sum 1010π. He then constructs isosceles triangles4AiBiAi+1 on the exterior of A with BiAi = BiAi+1 and ∠AiBiAi+1 = θi. (Here, A2021 = A1.)Finally, he erases A and the point B1. He then tells Jason the angles θ1, θ2, . . . , θ2020 he chose. Showthat Jason can determine where B1 was from the remaining 2019 points, i.e. show that B1 is uniquelydetermined by the information Jason has.

5. [40] Let a0, b0, c0, a, b, c be integers such that gcd(a0, b0, c0) = gcd(a, b, c) = 1. Prove that there existsa positive integer n and integers a1, a2, . . . , an = a, b1, b2, . . . , bn = b, c1, c2, . . . , cn = c such that for all1 ≤ i ≤ n, ai∈1ai + bi∈1bi + ci∈1ci = 1.

6. [40] Let n > 1 be a positive integer and S be a collection of 12

(2nn

)distinct n-element subsets of

{1, 2, . . . , 2n}. Show that there exists A,B ∈ S such that |A ∩B| ≤ 1.

7. [50] Positive real numbers x and y satisfy∣∣∣∣∣∣∣∣∣· · · ∣∣∣∣∣|x| − y∣∣− x∣∣∣ · · · − y∣∣∣∣− x

∣∣∣∣∣ =

∣∣∣∣∣∣∣∣∣· · · ∣∣∣∣∣|y| − x∣∣− y∣∣∣ · · · − x∣∣∣∣− y

∣∣∣∣∣where there are 2019 absolute value signs | · | on each side. Determine, with proof, all possible valuesof x

y .

8. [50] Let ABC be a scalene triangle with angle bisectors AD, BE, and CF so that D, E, and F lie onsegments BC, CA, and AB respectively. Let M and N be the midpoints of BC and EF respectively.Prove that line AN and the line through M parallel to AD intersect on the circumcircle of ABC ifand only if DE = DF .

9. [55] Let p > 5 be a prime number. Show that there exists a prime number q < p and a positive integern such that p divides n2 − q.

10. [60] Let n be a fixed positive integer, and choose n positive integers a1, . . . , an. Given a permutationπ on the first n positive integers, let Sπ = {i | ai

π(i) is an integer}. Let N denote the number of distinct

sets Sπ as π ranges over all such permutations. Determine, in terms of n, the maximum value of Nover all possible values of a1, . . . , an.

HMMT February 2020February 15, 2020

Team Round

1. [20] Let n be a positive integer. Define a sequence by a0 = 1, a2i+1 = ai, and a2i+2 = ai + ai+1 foreach i ≥ 0. Determine, with proof, the value of a0 + a1 + a2 + · · ·+ a2n−1.

Proposed by: Kevin Ren

Answer: 3n+12

Solution 1: Note that a2n−1 = 1 for all n by repeatedly applying a2i+1 = ai. Now let bn =a0 + a1 + a2 + · · ·+ a2n−1. Applying the given recursion to every term of bn except a0 gives

bn =a0 + a1 + a2 + a3 + · · ·+ a2n−1=a0 + a2 + a4 + · · ·+ a2n−2 + a1 + a3 + · · ·+ a2n−1=a0 + (a0 + a1) + (a1 + a2) + (a2 + a3) + · · ·+ (a2n−1−2 + a2n−1−1)

+ a0 + a1 + a2 + · · ·+ a2n−1−1=3a0 + 3a1 + 3a2 + · · ·+ 3a2n−1−2 + 3a2n−1−1 − a2n−1−1=3bn−1 − 1.

Now we easily obtain bn = 3n+12 by induction.

Solution 2: Define a binary string to be good if it is the null string or of the form 101010 . . . 10. Let cnbe the number of good subsequences of n when written in binary form. We see c0 = 1 and c2n+1 = cnbecause the trailing 1 in 2n + 1 cannot be part of a good subsequence. Furthermore, c2n+2 − cn+1

equals the number of good subsequences of 2n+ 2 that use the trailing 0 in 2n+ 2. We will show thatthis number is exactly cn.

Let s be a good subsequence of 2n + 2 that contains the trailing 0. If s uses the last 1, remove boththe last 1 and the trailing 0 from s; the result s′ will be a good subsequence of n. If s does not use thelast 1, consider the sequence s′ where the trailing 0 in 2n + 2 is replaced by the last 0 in n (which isat the same position as the last 1 in 2n+ 2.) The map s 7→ s′ can be seen to be a bijection, and thusc2n+2 = cn + cn+1.

Now it is clear that an = cn for all n. Consider choosing each binary string between 0 and 2n − 1with equal probability. The probability that a given subsequence of length 2k is good is 1

22k. There

are(n2k

)subsequences of length 2k, so by linearity of expectation, the total expected number of good

subsequences isbn/2c∑k=0

(n2k

)22k

=(1 + 1/2)n + (1− 1/2)n

2=

3n + 1

2n+1.

This is equal to the average of a0, . . . , a2n−1, therefore the sum a0 + · · ·+ a2n−1 is 3n+12 .

2. [25] Let n be a fixed positive integer. An n-staircase is a polyomino with n(n+1)2 cells arranged in the

shape of a staircase, with arbitrary size. Here are two examples of 5-staircases:

Prove that an n-staircase can be dissected into strictly smaller n-staircases.

Proposed by: James Lin

Solution 1: Viewing the problem in reverse, it is equivalent to show that we can use multiple n-staircases to make a single, larger n-staircase, because that larger n-staircase is made up of strictlysmaller n-staircases, and is the example we need.

For the construction, we first attach two n-staircases of the same size together to make an n× (n+ 1)rectangle. Then, we arrange n(n+1) of these rectangles in a (n+1)×n grid, giving an n(n+1)×n(n+1)

size square. Finally, we can use n(n+1)2 of these squares to create a larger n-staircase of n2(n + 1)2

smaller staircases, so we are done.

Solution 2: An alternative construction using only 2n + 2 staircases was submitted by team YeahKnights A. We provide a diagram for n = 5 and allow the interested reader to fill in the details.

3. [25] Let ABC be a triangle inscribed in a circle ω and ` be the tangent to ω at A. The line through Bparallel to AC meets ` at P , and the line through C parallel to AB meets ` at Q. The circumcirclesof ABP and ACQ meet at S 6= A. Show that AS bisects BC.

Proposed by: Andrew Gu

Solution 1: In directed angles, we have

]CBP = ]BCA = ]BAP,

so BC is tangent to the circumcircle of ABP . Likewise, BC is tangent to the circumcircle of ACQ. LetM be the midpoint of BC. Then M has equal power MB2 = MC2 with respect to the circumcirclesof ABP and ACQ, so the radical axis AS passes through M .

A

B C

P

Q

S

M

A′

Solution 2: Since]CBP = ]BCA = ]BAP = ]CQP,

quadrilateral BCQP is cyclic. Then AS, BP , and CQ concur at a point A′. Since A′B ‖ AC andA′C ‖ AB, quadrilateral ABA′C is a parallelogram so line ASA′ bisects BC.

4. [35] Alan draws a convex 2020-gon A = A1A2 · · ·A2020 with vertices in clockwise order and chooses2020 angles θ1, θ2, . . . , θ2020 ∈ (0, π) in radians with sum 1010π. He then constructs isosceles triangles4AiBiAi+1 on the exterior of A with BiAi = BiAi+1 and ∠AiBiAi+1 = θi. (Here, A2021 = A1.)Finally, he erases A and the point B1. He then tells Jason the angles θ1, θ2, . . . , θ2020 he chose. Showthat Jason can determine where B1 was from the remaining 2019 points, i.e. show that B1 is uniquelydetermined by the information Jason has.

Proposed by: Andrew Gu, Colin Tang

Solution 1: For each i, let τi be the transformation of the plane which is rotation by θi counterclockwiseabout Bi. Recall that a composition of rotations is a rotation or translation, and that the angles ofrotation add. Consider the composition τ2020◦τ2019◦· · ·◦τ1, with total rotation angle 1010π. This mustbe a translation because 1010π = 505(2π). Also note that the composition sends A1 to itself becauseτi(Ai) = Ai+1. Therefore it is the identity. Now Jason can identify the map τ1 as τ−12 ◦ τ

−13 ◦ · · · ◦ τ

−12020,

and B1 is the unique fixed point of this map.

Solution 2: Fix an arbitrary coordinate system. For 1 ≤ k ≤ 2020, let ak, bk be the complex numberscorresponding to Ak, Bk. The given condition translates to

eiθk(bk − ak) = (bk − ak+1). (1 ≤ k ≤ 2020)

In other words(eiθk − 1)bk = eiθkak − ak+1,

or(e−i(θk−1+···+θ1) − e−i(θk+···+θ1))bk = e−i(θk−1+···+θ1)ak − e−i(θk+···+θ1)ak+1.

Summing over all k, and using the fact that

e−i(θ1+···+θ2020) = 1,

we see that the right hand side cancels to 0, thus

2020∑k=1

(e−i(θk−1+···+θ1) − e−i(θk+···+θ1))bk = 0.

Jason knows b2, . . . , b2020 and all the θi, so the equation above is a linear equation in b1. We finish bynoting that the coefficient of b1 is 1− e−iθ1 which is non-zero, as θ1 ∈ (0, π). Thus Jason can solve forb1 uniquely.

Solution 3: Let A1A2 · · ·A2020 and Ã1Ã2 · · · Ã2020 be two 2020-gons that satisfy the conditions inthe problem statement, and let Bk, B̃k be the points Alan would construct with respect to these twopolygons. It suffices to show that if Bk = B̃k for k = 2, 3, . . . , 2020, then B1 = B̃1.

For 2 ≤ k ≤ 2020, we note that

AkBk = Ak+1Bk, ÃkBk = Ãk+1Bk

Furthermore, we have the equality of directed angles ∠AkBkAk+1 = ∠ÃkBkÃk+1 = θk, therefore∠AkBkÃk = ∠Ak+1BkÃk+1. This implies the congruence 4AkBkÃk ∼= 4Ak+1BkÃk+1.

The congruence shows that AkÃk = Ak+1Ãk+1; furthermore, the angle from the directed segment−−−→AkÃk to

−−−−−−−→Ak+1Ãk+1 is θk counterclockwise. This holds for k = 2, 3, . . . , 2020; we conclude that

A1Ã1 = A2Ã2, and the angle from the directed segments−−−→A1Ã1 to

−−−→A2Ã2 is

−2020∑k=2

θk = θ1 − 1010π = θ1

counterclockwise.

Finally we observe that A1B1 = A2B1, and the angle from the directed segment−−−→A1B1 to

−−−→A2B1 is

θ1 counterclockwise. This implies ∠B1A1Ã1 = ∠B1A2Ã2, so 4A1B1Ã1∼= 4A2B1Ã2. Thus Ã1B1 =

Ã2B1, and the angle from−−−→Ã1B1 to

−−−→Ã2B1 is θ1 counterclockwise. We conclude that B1 = B̃1.

5. [40] Let a0, b0, c0, a, b, c be integers such that gcd(a0, b0, c0) = gcd(a, b, c) = 1. Prove that there existsa positive integer n and integers a1, a2, . . . , an = a, b1, b2, . . . , bn = b, c1, c2, . . . , cn = c such that for all1 ≤ i ≤ n, ai−1ai + bi−1bi + ci−1ci = 1.

Proposed by: Michael Ren

Solution: The problem statement is equivalent to showing that we can find a sequence of vectors,each with 3 integer components, such that the first vector is (a0, b0, c0), the last vector is (a, b, c), andevery pair of adjacent vectors has dot product equal to 1.

We will show that any vector (a, b, c) can be sent to (1, 0, 0). This is sufficient, because given vectors(a0, b0, c0) and (a, b, c), we take the sequence from (a0, b0, c0) to (1, 0, 0) and then add the reverse ofthe sequence from (a, b, c) to (1, 0, 0).

First, suppose that some two of a, b, c are relatively prime. Here we will suppose that a and b arerelatively prime; the other cases are similar. If neither of a or b is 0, then by Bezout’s identity, thereexist p, q such that |p|+ |q| < |a|+ |b| and ap+bq = 1, so we can send (a, b, c) to (p, q, 0). (Finding suchnumbers can be done using the extended Euclidean algorithm.) Clearly p and q must also be relativelyprime, so we can apply Bezout’s identity repeatedly until we eventually have (1, 0, 0), (−1, 0, 0), (0, 1, 0),or (0,−1, 0). Now, starting from (0,−1, 0), we can do (0,−1, 0) → (1,−1, 0) → (1, 0, 0), and we cando something similar to convert (−1, 0, 0) to (0, 1, 0).

Now suppose that no two of a, b, c are relatively prime. Let f = gcd(a, b). We claim that we canfind x, y, z such that axy + bx + cz = 1. Notice that this is the same as (ay + b)x + cz = 1. Since

gcd(a, b, c) = 1, there exists y such that gcd(ay + b, c) = 1. Then by Bezout’s identity, there exist x, zsuch that (ay + b)x + cz = 1. Therefore, we can send (a, b, c) to (xy, x, z). Clearly x and z must berelatively prime, so we have reduced to the case above, and we can apply the process described abovefor that case.

At the end of this process, we will have (1, 0, 0), (0, 1, 0), or (0, 0, 1). The second of these can beconverted into (1, 0, 0) by doing (0, 1, 0)→ (1, 1, 0)→ (1, 0, 0), and a similar sequence shows the samefor the third. Therefore, (a, b, c) can be sent to (1, 0, 0).

6. [40] Let n > 1 be a positive integer and S be a collection of 12

(2nn

)distinct n-element subsets of

{1, 2, . . . , 2n}. Show that there exists A,B ∈ S such that |A ∩B| ≤ 1.

Proposed by: Michael Ren

Solution 1: Assume for the sake of contradiction that there exist no such A, B. Pair up each subsetwith its complement, like so:

{1, 2, 3, . . . , n} ↔ {n+ 1, n+ 2, . . . , 2n}{1, 2, 3, . . . , n− 1, n+ 1} ↔ {n, n+ 2, . . . , 2n}{1, 2, 3, . . . , n− 1, n+ 2} ↔ {n, n+ 1, n+ 3, . . . , 2n}

...

Note that for each pair, we can have at most one of the two in S. Since S has 12 of the total number of

subsets with size n, it must be that we have exactly one element from each pair in S. For any s0 ∈ S,none of the subsets that share exactly one element with s0 can be in S, so their complements must bein S. This means that every subset with n − 1 shared elements with s0 must be in S. Without lossof generality, assume {1, 2, 3, . . . , n} ∈ S. Then, {2, 3, 4, . . . , n + 1} ∈ S, so {3, 4, 5, . . . , n + 2} ∈ S.Continuing on in this manner, we eventually reach {n+ 1, n+ 2, . . . , 2n} ∈ S, contradiction.

Solution 2: Let [2n] = {1, 2, . . . , 2n}. Consider the following cycle of 2n − 1 sets such that any twoadjacent sets have an intersection of size 1:

{1, 2, 3, . . . , n},{1, n+ 1, n+ 2, . . . , 2n− 1},{1, 2n, 2, 3, . . . , n− 1},...

{1, n+ 2, n+ 3, . . . , 2n}.

If S contains two adjacent elements of the cycle then we are done. For each permutation σ of [2n], wecan consider the cycle Cσ of sets obtained after applying σ, i.e. {σ(1), σ(2), σ(3), . . . , σ(n)} and so on.In total, each subset of [2n] with size n appears (2n− 1)(n!)2 times across all the cycles Cσ, so∑

σ

|Cσ ∩ S| = |S|(2n− 1)(n!)2 =2n− 1

2(2n)!

where the sum is over all (2n)! permutations of [2n]. This means that on average across possiblecycles, 2n−1

2 of its elements are in S. Thus, if we select a cycle Cσ uniformly at random, with positiveprobability we will have |Cσ ∩ S| ≥ n, so two adjacent elements in this cycle will be in S. Therefore,there must exist some two subsets in S that share at most one element.

This proof will work under the weaker condition |S| > n−12n−1

(2nn

).

Remark. A family of sets such that |A ∩ B| ≥ t for every pair of distinct sets A, B is calledt-intersecting. Ahlswede and Khachatrian solved the problem of determining the largest k-uniform t-intersecting family. See “Katona’s Intersection Theorem: Four Proofs” or “The Complete Intersection

Theorem for Systems of Finite Sets” for exact results.

7. [50] Positive real numbers x and y satisfy∣∣∣∣∣∣∣∣∣· · · ∣∣∣∣∣|x| − y∣∣− x∣∣∣ · · · − y∣∣∣∣− x

∣∣∣∣∣ =

∣∣∣∣∣∣∣∣∣· · · ∣∣∣∣∣|y| − x∣∣− y∣∣∣ · · · − x∣∣∣∣− y

∣∣∣∣∣where there are 2019 absolute value signs | · | on each side. Determine, with proof, all possible valuesof x

y .

Proposed by: Krit Boonsiriseth

Answer: 13 , 1, 3

Solution: Clearly x = y works. Else WLOG x < y, define d = y − x, and define f(z) :=∣∣|z − y| − x∣∣

so our expression reduces tof1009(x) =

∣∣f1009(0)− y∣∣ .

Now note that for z ∈ [0, y], f(z) can be written as

f(z) =

(d− z, 0 ≤ z ≤ dz − d, d < z ≤ y

Hence f(f(z)) = f(d− z) = z for all z ∈ [0, d]. Therefore∣∣f1009(0)− y∣∣ = |f(0)− y| = x.

If x > d then f1009(x) < x which is impossible (if f1009(x) ≤ d then the conclusion trivially holds, andif f1009(x) > d we must have f1009(x) = x− 1009d < x). Therefore x ≤ d, so f1009(x) = f(x) = d− xand we must have d− x = x. Hence y = 3x which is easily seen to work. To summarize, the possiblevalues of x

y are 13 , 1, 3.

8. [50] Let ABC be a scalene triangle with angle bisectors AD, BE, and CF so that D, E, and F lie onsegments BC, CA, and AB respectively. Let M and N be the midpoints of BC and EF respectively.Prove that line AN and the line through M parallel to AD intersect on the circumcircle of ABC ifand only if DE = DF .

Proposed by: Michael Ren

Solution 1:

A

B CD

EF

X

Y

Z

M

N

P

LA

Let X, Y be on AB, AC such that CX ‖ BE and BY ‖ CF . Then BX = BC = CY . Let Z be the

midpoint of XY . Then−−→MZ = 1

2 (−−→BX +

−−→CY ), which bisects the angle between BX and CY because

they have the same length. Therefore MZ ‖ AD. Furthermore, by similar triangles we have

AE ·AX = AB ·AC = AF ·AY.

This rearranges to AEAF = AY

AX , so EF ‖ XY . Therefore Z is the intersection of the lines in the problemstatement. Then

sin∠BZXsin∠CZY

=BX sin∠ZBX

XZ

CY sin∠ZCYY Z

= 1

iff Z ∈ (ABC), so XY is the external angle bisector of ∠BZC iff Z ∈ (ABC). Thus if P = AD∩XY ,P ∈ (ABC) if Z ∈ (ABC). Additionally the spiral similarity from BX to CY gives LAZ ⊥ XY whereLA is the midpoint of arc BAC, so if P ∈ (ABC) then Z must be on (ABC) because ∠LAZP = 90◦.Therefore Z ∈ (ABC) iff P ∈ (ABC).

From the previous length computation, we know that an inversion at A with radius√AB ·AC com-

posed with reflection about AD will send X and Y to E and F . We have P ∈ (ABC) iff its imageunder the inversion is D, but since P was defined as AD ∩XY this is true iff (AEDF ) is cyclic. SinceABC is scalene and AE 6= AF, this is true iff DE = DF.

Solution 2:

A

BC

D

EF

MA

LA

M

N

T

X

Y Y ′

X ′

Let LA be the midpoint of arc BAC and let MA be diametrically opposite LA. Let EF, ALA, and BCmeet at T so ∠DAT = 90◦; note that DE = DF iff DN ⊥ EF, which is equivalent to (TAND) beingcyclic. Let AN ∩ (ABC) = X and XM ∩ (ABC) = Y, and let Y ′ be the reflection of Y over LAMA

with similarly X ′ the reflection of X over LAMA. We wish to show N ∈ (TAD) iff XY ‖ AMA.

We claim AY ′ ‖ EF. By projecting −1 = (B,C;M,∞)X= (B,C;Y,X ′) and reflecting over LAMA, we

find (X,Y ′;B,C) = −1. Then projecting through A gives (N,AY ′ ∩ EF ;F,E) = −1, and since N isthe midpoint of EF we find AY ′ ‖ EF.Now (TAND) cyclic iff ]DAN = ]DTN, and ]DTN = ]Y Y ′A by the parallel lines. But we have]DAN = ]MAAX, so arcs MAX and Y A are equal iff (TAND) is cyclic. Thus XY ‖ AMA iffDE = DF as desired.

9. [55] Let p > 5 be a prime number. Show that there exists a prime number q < p and a positive integern such that p divides n2 − q.Proposed by: Andrew Gu

Solution 1: Note that the condition p | n2 − q just means that q is a quadratic residue modulo p, or

that the Legendre symbol(qp

)is 1. We use these standard facts about the Legendre symbol:

• If p ≡ ±1 (mod 8), then(

2p

)= 1.

• For an odd prime p, (−1

p

)=

(−1 if p ≡ 3 (mod 4)

+1 if p ≡ 1 (mod 4).

• Quadratic reciprocity: for distinct odd primes p and q,(p

q

)(q

p

)=

(−1 if p ≡ q ≡ 3 (mod 4)

+1 otherwise.

If p is a Fermat prime or Mersenne prime, then p is congruent to 1 or 7 modulo 8 respectively, sincep > 5. In that case q = 2 works. Otherwise assume p is not a Fermat prime or Mersenne prime, sothat p− 1 and p+ 1 are not powers of 2.

If p ≡ 1 (mod 4), then let q be an odd prime divisor of p−1, so that p ≡ 1 (mod q). Then by quadratic

reciprocity(qp

)=(pq

)= 1.

If p ≡ 3 (mod 4), then let q be an odd prime divisor of p + 1, so that p ≡ −1 (mod q). Either q ≡ 1

(mod 4) so that(qp

)=(pq

)= 1 or q ≡ 3 (mod 4) so that

(qp

)= −

(pq

)= −

(−1q

)= 1.

Solution 2: (Ankan Bhattacharya) We assume the same standard facts about quadratic residues asthe previous solution.

If p ≡ 1 (mod 4), then since p > 5, there exists an odd prime divisor q of p− 4, which gives(q

p

)=

(p

q

)=

(4

q

)= 1.

If p ≡ 7 (mod 8), then we can take q = 2.

If p ≡ 3 (mod 8), then by Legendre’s three square theorem there exist odd a, b, c satisfying p =a2 + b2 + c2. Since p > 3, these are not all equal and we may assume without loss of generality thatb 6= c. Then p− a2 = b2 + c2 has a prime divisor q ≡ 1 (mod 4), which gives(

q

p

)=

(p

q

)=

(a2

q

)= 1.

Remark. For an odd prime number p, let l(p) be the least prime number which is a quadratic residuemodulo p and h(−p) be the class number of the quadratic field Q[

√−p]. In the paper “The Least Prime

Quadratic Residue and the Class Number” by Chowla, Cowles, and Cowles, the following results wereproven:

• If p > 5 and p ≡ 5 (mod 8), then l(p) <√p.

• If p > 3, p ≡ 3 (mod 8), and h(−p) > 1, then l(p) <√p/3.

• If p > 3, p ≡ 3 (mod 8), and h(−p) = 1, then l(p) = p+14 .

The proofs of the second and third results require knowledge of binary quadratic forms.

10. [60] Let n be a fixed positive integer, and choose n positive integers a1, . . . , an. Given a permutationπ on the first n positive integers, let Sπ = {i | ai

π(i) is an integer}. Let N denote the number of distinct

sets Sπ as π ranges over all such permutations. Determine, in terms of n, the maximum value of Nover all possible values of a1, . . . , an.

Proposed by: James Lin

Answer: 2n − nSolution: The answer is 2n − n.

Let D = (dij) be the matrix where dij is 1 if i is a divisor of aj and 0 otherwise. For a subset S of [n],let DS be the matrix obtained from D by flipping (0↔ 1) every entry dij where j /∈ S. Observe thatS = Sπ if and only if (DS)π(i)i = 1 for all i.

To show that N ≤ 2n − n we consider two cases. If all the rows of D are distinct, then there existn different possibilities for S that set a row equal to zero. In this case, there is clearly no π so thatSπ = S. Thus there are at most 2n − n possible Sπ. Otherwise, if two rows in D are the same, thenchoose an S0 such that DS0 has two zero rows. Then, the n+ 1 sets S that are at most “one elementaway” from S0 are such that DS only has one column with nonzero entries in those two rows. Thismakes it impossible for Sπ = S as well, so N ≤ 2n − n− 1.

Now we construct N = 2n − n by setting aj = j. By Hall’s marriage theorem, it suffices to prove thefollowing:

Assuming that DS has no completely-zero rows, given a set I = {i1, i2, . . . , ik} there exist at least kvalues of j so that there exists an i ∈ I so that (DS)ij = 1. Call such j admissible.

Without loss of generality assume i1 < i2 < · · · < ik.

Note that if {dij | i ∈ I} = {0, 1}, then j is admissible. Therefore the k− 1 numbers i1, i2, . . . , ik−1 areadmissible, since for α < k, iα divides iα but ik does not. So we only need to find one more admissiblej. Assume that ik is not admissible; now it must be the case that all the iα are divisors of ik.

At this point we note that the k = 1 case is easy, since no row of DS is zero. Moreover, if k = 2,{(DS)i1i1 , (DS)i2i1} = {0, 1}, so in the row with the zero there must be 1 somewhere, yielding a secondadmissible column.

In the case where k ≥ 3, note that i1 ≤ ik/3. Therefore ik − i1 /∈ I, but i1 divides ik − i1 and ik doesnot. Thus we have found the last admissible column. Having exhausted all cases, we are done.

HMMT February 2020February 15, 2020

Algebra and Number Theory

1. Let P (x) = x3 + x2 − r2x− 2020 be a polynomial with roots r, s, t. What is P (1)?

2. Find the unique pair of positive integers (a, b) with a < b for which

2020− aa

· 2020− bb

= 2.

3. Let a = 256. Find the unique real number x > a2 such that

loga loga loga x = loga2 loga2 loga2 x.

4. For positive integers n and k, let f(n, k) be the number of distinct prime divisors of n that are at leastk. For example, f(90, 3) = 2, since the only prime factors of 90 that are at least 3 are 3 and 5. Findthe closest integer to

∞∑n=1

∞∑k=1

f(n, k)

3n+k∈7 .

5. A positive integer N is piquant if there exists a positive integer m such that if ni denotes the numberof digits in mi (in base 10), then n1 + n2 + · · ·+ n10 = N . Let pM denote the fraction of the first Mpositive integers that are piquant. Find lim

M→∞pM .

6. A polynomial P (x) is a base-n polynomial if it is of the form adxd + ad∈1x

d∈1 + · · ·+ a1x+ a0, whereeach ai is an integer between 0 and n− 1 inclusive and ad > 0. Find the largest positive integer n suchthat for any real number c, there exists at most one base-n polynomial P (x) for which P (

√2+√

3) = c.

7. Find the sum of all positive integers n for which

15 · n!2 + 1

2n− 3

is an integer.

8. Let P (x) be the unique polynomial of degree at most 2020 satisfying P (k2) = k for k = 0, 1, 2, . . . , 2020.Compute P (20212).

9. Let P (x) = x2020 + x+ 2, which has 2020 distinct roots. Let Q(x) be the monic polynomial of degree20202

)whose roots are the pairwise products of the roots of P (x). Let α satisfy P (α) = 4. Compute

the sum of all possible values of Q(α2)2.

10. We define F101[x] as the set of all polynomials in x with coefficients in F101 (the integers modulo 101with usual addition and subtraction), so that two polynomials are equal if and only if the coefficientsof xk are equal in F101 for each nonnegative integer k. For example, (x+3)(100x+5) = 100x2+2x+15in F101[x] because the corresponding coefficients are equal modulo 101.

We say that f(x) ∈ F101[x] is lucky if it has degree at most 1000 and there exist g(x), h(x) ∈ F101[x]such that

f(x) = g(x)(x1001 − 1) + h(x)101 − h(x)

in F101[x]. Find the number of lucky polynomials.

HMMT February 2020February 15, 2020

Algebra and Number Theory

1. Let P (x) = x3 + x2 − r2x− 2020 be a polynomial with roots r, s, t. What is P (1)?

Proposed by: James Lin

Answer: −4038

Solution 1: Plugging in x = r gives r2 = 2020. This means P (1) = 2− r2 − 2020 = −4038.

Solution 2: Vieta’s formulas give the following equations:

r + s+ t = −1

rs+ st+ tr = −r2

rst = 2020.

The second equation is (r+t)(r+s) = 0. Without loss of generality, let r+t = 0. Then s = r+s+t = −1.Finally r2 = rst = 2020, so P (1) = −4038.

2. Find the unique pair of positive integers (a, b) with a < b for which

2020− aa

· 2020− bb

= 2.

Proposed by: James Lin

Answer: (505, 1212)

Solution 1: If either a or b is larger than 2020, then both must be for the product to be positive.However, the resulting product would be less than 1, so this case is impossible. Now, we see that( 2020−a

a , 2020−bb ) must be in the form (xy ,

2yx ), in some order, for relatively prime positive integers x and

y. Then 2020a = x+y

y and 2020b = x+2y

x , so x+ y and x+ 2y are relatively prime factors of 2020. Since

x + y < x + 2y < 2(x + y), the only possibility is x + y = 4, x + 2y = 5. Thus, (x, y) = (3, 1), and( 2020−a

a , 2020−bb ) = (3, 23 ) because a < b. Solving gives (a, b) = (505, 1212).

Solution 2: We rearrange to find that (a+2020)(b+2020) = 2 ·20202. Note that a+2020 and b+2020are both less than 4040 < 1012, so they must both be divisible by 101. Hence, we divide out a factorof 1012 and solve the equivalent problem of (a′ + 20)(b′ + 20) = 800, where a′ = a

101 and b′ = b101 .

Because each factor must be less than 4040101 = 40, we get that (a′ + 20, b′ + 20) = (25, 32), which yields

(a, b) = (505, 1212).

3. Let a = 256. Find the unique real number x > a2 such that

loga loga loga x = loga2 loga2 loga2 x.

Proposed by: James Lin

Answer: 232

Solution: Let y = loga x so loga loga y = loga2 loga212y. Setting z = loga y, we find loga z =

loga2( 12z−

116 ), or z2− 1

2z+ 116 = 0. Thus, we have z = 1

4 , so we can backsolve to get y = 4 and x = 232.

4. For positive integers n and k, let f(n, k) be the number of distinct prime divisors of n that are at leastk. For example, f(90, 3) = 2, since the only prime factors of 90 that are at least 3 are 3 and 5. Findthe closest integer to

∞∑n=1

∞∑k=1

f(n, k)

3n+k−7 .

Proposed by: Daniel Zhu

Answer: 167

Solution: A prime p is counted in f(n, k) if p | n and k ≤ p. Thus, for a given prime p, the totalcontribution from p in the sum is

37∞∑

m=1

p∑k=1

1

3pm+k= 37

∑i≥p+1

1

3i=

37−p

2.

Therefore, if we consider p ∈ {2, 3, 5, 7, . . .} we get

∞∑n=1

∞∑k=1

f(n, k)

3n+k−7 =35

2+

34

2+

32

2+

30

2+ ε = 167 + ε,

where ε <∑∞

i=1137−i

2 = 1108 �

12 . The closest integer to the sum is 167.

5. A positive integer N is piquant if there exists a positive integer m such that if ni denotes the numberof digits in mi (in base 10), then n1 + n2 + · · ·+ n10 = N . Let pM denote the fraction of the first Mpositive integers that are piquant. Find lim

M→∞pM .

Proposed by: James Lin

Answer: 3255

Solution: For notation, let ni(m) denote the number of digits of mi and N(m) = n1(m) + n2(m) +· · · + n10(m). Observe that ni(10m) = ni(m) + i so N(10m) = N(m) + 55. We will determine, fork →∞, how many of the integers from N(10k) to N(10k+1)− 1, inclusive, are piquant.

Increment m by 1 from 10k to 10k+1. The number of digits of mi increases by one if mi < 10h ≤(m + 1)i, or m < 10

hi ≤ m + 1 for some integer h. This means that, as we increment m by 1,

the sum n1 + n2 + · · · + n10 increases when m “jumps over” 10hi for i ≤ 10. Furthermore, when

m is big enough, all “jumps” are distinguishable, i.e. there does not exist two h1

i16= h2

i2such that

m < 10h1/i1 < 10h2/i2 ≤ m+ 1.

Thus, for large k, the number of times n1(m) + n2(m) + · · ·+ n10(m) increases as m increments by 1

from 10k to 10k+1 is the number of different 10hi in the range (10k, 10k+1]. If we take the fractional

part of the exponent, this is equivalent to the number of distinct fractions 0 < ji ≤ 1 where 1 ≤ i ≤ 10.

The number of such fractions with denominator i is ϕ(i), so the total number of such fractions isϕ(1) + ϕ(2) + · · ·+ ϕ(10) = 32.

We have shown that for sufficiently large k, N(10k+1) − N(10k) = 55 and exactly 32 integers in therange [N(10k), N(10k+1)) are piquant. This implies that limM→∞ pM = 32

55 .

6. A polynomial P (x) is a base-n polynomial if it is of the form adxd + ad−1x

d−1 + · · ·+ a1x+ a0, whereeach ai is an integer between 0 and n− 1 inclusive and ad > 0. Find the largest positive integer n suchthat for any real number c, there exists at most one base-n polynomial P (x) for which P (

√2+√

3) = c.

Proposed by: James Lin

Answer: 9

Solution: It is equivalent to determine the largest n such that we cannot find two distinct base-n polynomials P1 and P2 such that P1(

√2 +√

3) = P2(√

2 +√

3). The difference of two base-npolynomials is a polynomial with integer coefficients whose absolute values are less than n, and allsuch polynomials are the difference of two base-n polynomials. We compute the minimal polynomialof x =

√2 +√

3 first: since x2 = 5 + 2√

6, we have (x2 − 5)2 = 24 so x4 − 10x2 + 1 = 0. Therefore√2 +√

3 is a root of (x2 + 1)(x4 − 10x2 + 1) = x6 − 9x4 − 9x2 + 1. The coefficients of this polynomialhave magnitude at most 9, so n < 10.

In the other direction, observe that (√

2 +√

3)k is of the form a + b√

6 for integers a and b if k iseven, and a

√2 + b

√3 if k is odd. As no integer linear combination of the first expression can equal the

second, we can treat these cases separately. Suppose Q(x) = cdx2d + cd−1x

2d−2 + · · · + c0 is an evenpolynomial with |ci| < 9 for all i and cd 6= 0. Let y = (

√2 +√

3)2 = 5 + 2√

6 and observe that y > 9.Then

|cdyd| ≥ yd

>8

y − 1(yd − 1)

= 8yd−1 + 8yd−2 + · · ·+ 8y + 8

≥ |cd−1yd−1 + cd−2yd−2 + · · ·+ c0|.

Therefore Q(√

2+√

3) = cdyd + cd−1y

d−1 + · · ·+ c0 6= 0, so no two distinct base-9 polynomials coincideat x =

√2 +√

3.

The same logic applies for the odd polynomial case after dividing out a factor of√

2 +√

3, so n = 9works.

7. Find the sum of all positive integers n for which

15 · n!2 + 1

2n− 3

is an integer.

Proposed by: Andrew Gu

Answer: 90

Solution: It is clear that n = 1 and n = 2 work so assume that n > 2. If 2n − 3 is composite thenits smallest prime factor is at most 2n−3

2 < n so will be coprime to 15 · n!2 + 1. Therefore assume that2n− 3 = p is prime. We can rewrite the numerator as

(−1)n · 15 ·(

1 · 2 · · · p+ 3

2

)·(p− 3

2· p− 1

2· · · (p− 1)

)+ 1 (mod p).

By Wilson’s Theorem, (p− 1)! ≡ −1 (mod p), so the expression simplifies to

(−1)n+1 · 15 · p− 3

2· p− 1

2· p+ 1

2· p+ 3

2+ 1 ≡ (−1)n+1 · 135

16+ 1 (mod p).

If p ≡ 3 (mod 4), then we have

135 + 16

16≡ 151

16≡ 0 (mod p).

If p ≡ 1 (mod 4), then we have

135− 16

16≡ 119

16≡ 0 (mod p).

So p must be a prime divisor of 151 or 119, which means that p ∈ {7, 17, 151}. All of these numberswork aside from 7 (because 7 ≡ 3 (mod 4)) and the corresponding values of n are 10 and 77. The sumof the solutions is then 1 + 2 + 10 + 77 = 90.

8. Let P (x) be the unique polynomial of degree at most 2020 satisfying P (k2) = k for k = 0, 1, 2, . . . , 2020.Compute P (20212).

Proposed by: Milan Haiman

Answer: 2021−40402020

)Solution 1: Since P (0) = 0, we see that P has no constant term. Let Q(x) = P (x2)−x

x be a polynomialwith degree at most 4039. From the given values of P , we see that Q(k) = 0 and Q(−k) = −2 fork = 1, 2, 3, . . . , 2020.

Now, consider the polynomial R(x) = Q(x + 1) − Q(x), which has degree at most 4038. Then R hasroots −2020, −2019, . . . , −2, 1, 2, . . . , 2019, so

R(x) = a(x+ 2020) · · · (x+ 2)(x− 1) · · · (x− 2019)

for some real number a. Using R(0) +R(−1) = Q(1)−Q(−1) = 2 yields a = − 12020!2019! , so

Q(2021) = R(2020) +Q(2020) = −4040 · · · 2022 · 2019 · · · 12020!2019!

+ 0 = − 1

2021

(4040

2020

).

It follows that P (20212) = 2021Q(2021) + 2021 = 2021−40402020

).

Solution 2: By Lagrange interpolation,

P (x) =

2020∑k=0

k∏

0≤j≤2020j 6=k

(x− j2)

(k2 − j2)=

2020∑k=0

2(−1)kk

(2020− k)!(2020 + k)!(x− k2)

2020∏j=0

(x− j2).

Therefore, by applying Pascal’s identity multiple times, we get that

P (20212) =

2020∑k=0

4042!(−1)kk

(2021− k)!(2021 + k)!

=

2020∑k=0

(4042

2021− k

)(−1)kk

= 2021−

(2021∑k=0

(−1)k+1k

((4041

2021− k

)+

(4041

2020− k

)))

= 2021−

(2021∑k=1

(−1)k+1

(4041

2021− k

))

= 2021−

(2021∑k=1

(−1)k+1

((4040

2021− k

)+

(4040

2020− k

)))

= 2021−(

4040

2020

).

9. Let P (x) = x2020 + x+ 2, which has 2020 distinct roots. Let Q(x) be the monic polynomial of degree20202

)whose roots are the pairwise products of the roots of P (x). Let α satisfy P (α) = 4. Compute

the sum of all possible values of Q(α2)2.

Proposed by: Milan Haiman

Answer: 2020 · 22019

Solution: Let P (x) have degree n = 2020 with roots r1, . . . , rn. Let R(x) =∏

i(x− r2i ). Then

∏i

rni P

(x

ri

)=∏i

∏j

(x− rirj) = Q(x)2R(x).

Using R(x2) = (−1)nP (x)P (−x) and Vieta, we obtain

P (x)P (−x)Q(x2)2 = P (0)n∏i

P

(x2

ri

).

Plugging in x = α, we use the facts that P (α) = 4, P (−α) = 4− 2α, and also

P

(α2

ri

)=α4040

r2020i

+α2

ri+ 2 = − (α− 2)2

ri + 2+α2

ri+ 2 =

2(−α− ri)2

ri(ri + 2).

This will give us

P (α)P (−α)Q(α2)2 = 2n∏i

2(−α− ri)2

ri(ri + 2)= 2n · 2nP (−α)2

P (0)P (−2).

Therefore,

Q(α2)2 =4nP (−α)

P (0)P (−2)P (α)

=4n(4− 2α)

2 · 22020 · 4

=24041(2− α)

22023

= 22018(2− α).

We can check that P (x) − 4 has no double roots (e.g. by checking that it shares no roots with itsderivative), which means that all possible α are distinct. Therefore, adding over all α gives 2020 ·22019,because the sum of the roots of P (x)− 4 is 0.

10. We define F101[x] as the set of all polynomials in x with coefficients in F101 (the integers modulo 101with usual addition and subtraction), so that two polynomials are equal if and only if the coefficientsof xk are equal in F101 for each nonnegative integer k. For example, (x+3)(100x+5) = 100x2+2x+15in F101[x] because the corresponding coefficients are equal modulo 101.

We say that f(x) ∈ F101[x] is lucky if it has degree at most 1000 and there exist g(x), h(x) ∈ F101[x]such that

f(x) = g(x)(x1001 − 1) + h(x)101 − h(x)

in F101[x]. Find the number of lucky polynomials.

Proposed by: Michael Ren

Answer: 101954

Solution 1: Let p = 101, m = 1001, and work in the ring R := Fp[x]/(xm − 1). We want to find thenumber of elements a of this ring that are of the form xp − x. We first solve this question for a fieldextension Fpd of Fp. Note that (x+n)p−(x+n) = xp−x for any n ∈ Fp, and the polynomial tp− t = bhas at most p solutions in Fpd for any b ∈ Fpd . Combining these implies that tp − t = b always haseither p or 0 solutions in Fpd , so there are pd−1 elements of Fpd expressible in the form xp − x. Now,note that we may factor R into a product of field extensions of Fp, each corresponding to an irreduciblefactor of xm − 1 in Fp, as the polynomial xm − 1 has no double roots in Fp as p - m. By the ChineseRemainder Theorem, we may multiply the number of lucky polynomials for each of the field extensionsto find the final answer. A field extension of degree d will yield pd−1 lucky polynomials. Thus, thefinal answer is pm−q, where q is the number of fields in the factorization of R into fields.

To do determine q, we first factor

xm − 1 =∏k|m

Φk(x)

in Z[x] where Φk(x) are the cyclotomic polynomials. Then we compute the number of irreducible

divisors of the cyclotomic polynomial Φk(x) in Fp[x]. We claim that this is equal to ϕ(k)ordk(p)

. Indeed,

note that given a root ω of Φk in the algebraic closure of Fp, the roots of its minimal polynomial are

ω, ωp, ωp2

, . . ., and this will cycle after the numerator repeats modulo k, from which it follows that

the degree of the minimal polynomial of ω is ordk(p). Thus, Φk(x) factors into ϕ(k)ordk(p)

irreducible

polynomials.

It remains to compute orders. We have that

ord7(101) = ord7(3) = 6,

ord11(101) = ord11(2) = 10,

ord13(101) = ord13(10) = 6.

Thus,

ord77(101) = 30,

ord91(101) = 6,

ord143(101) = 30,

ord1001(101) = 30,

ord1(101) = 1.

The number of factors of x1001 − 1 in F101[x] is thus

1

1+

6

6+

10

10+

12

6+

60

30+

72

6+

120

30+

720

30= 1 + 1 + 1 + 2 + 2 + 12 + 4 + 24 = 47,

so the total number of lucky polynomials is 1011001−47 = 101954.

Solution 2: As in the previous solution, we work in the ring R = Fp/(xm − 1), which we can treat as

the set of polynomials in Fp[x] with degree less than m. The problem is asking us for the number ofelements of the map α : h 7→ hp − h in R. Note that this map is linear because (a + b)p = ap + bp inany field where p = 0 (which R is an example of). Hence it suffices to determine the size of the kernelof α.

We can directly compute that if

h(x) = am−1xm−1 + am−2x

m−2 + · · ·+ a1x+ a0,

thenh(x)p = am−1x

p(m−1) + am−2xp(m−2) + · · ·+ a1x

p + a0,

where exponents are taken modulo m. Therefore h is in the kernel if and only if ak = apk for all kwhere indices are taken modulo m. Letting q denote the number of orbits of the map x 7→ px in Z/mZ,

the size of the kernel is then pq so the size of the image is pm−q. It remains to compute q, which willend up being the same computation as in the previous solution.

HMMT February 2020February 15, 2020

Combinatorics

1. How many ways can the vertices of a cube be colored red or blue so that the color of each vertex isthe color of the majority of the three vertices adjacent to it?

2. How many positive integers at most 420 leave different remainders when divided by each of 5, 6, and7?

3. Each unit square of a 4× 4 square grid is colored either red, green, or blue. Over all possible coloringsof the grid, what is the maximum possible number of L-trominos that contain exactly one square ofeach color? (L-trominos are made up of three unit squares sharing a corner, as shown below.)

4. Given an 8 × 8 checkerboard with alternating white and black squares, how many ways are there tochoose four black squares and four white squares so that no two of the eight chosen squares are in thesame row or column?

5. Let S be a set of intervals defined recursively as follows:

• Initially, [1, 1000] is the only interval in S.

• If l 6= r and [l, r] ∈ S, then both[l,⌊l+r2

⌋],[⌊

l+r2

⌋+ 1, r

]∈ S.

(Note that S can contain intervals such as [1, 1], which contain a single integer.) An integer i is chosenuniformly at random from the range [1, 1000]. What is the expected number of intervals in S whichcontain i?

6. Alice writes 1001 letters on a blackboard, each one chosen independently and uniformly at randomfrom the set S = {a, b, c}. A move consists of erasing two distinct letters from the board and replacingthem with the third letter in S. What is the probability that Alice can perform a sequence of moveswhich results in one letter remaining on the blackboard?

7. Anne-Marie has a deck of 16 cards, each with a distinct positive factor of 2002 written on it. Sheshuffles the deck and begins to draw cards from the deck without replacement. She stops when thereexists a nonempty subset of the cards in her hand whose numbers multiply to a perfect square. Whatis the expected number of cards in her hand when she stops?

8. Let Γ1 and Γ2 be concentric circles with radii 1 and 2, respectively. Four points are chosen on thecircumference of Γ2 independently and uniformly at random, and are then connected to form a convexquadrilateral. What is the probability that the perimeter of this quadrilateral intersects Γ1?

9. Farmer James wishes to cover a circle with circumference 10π with six different types of colored arcs.Each type of arc has radius 5, has length either π or 2π, and is colored either red, green, or blue. Hehas an unlimited number of each of the six arc types. He wishes to completely cover his circle withoutoverlap, subject to the following conditions:

• Any two adjacent arcs are of different colors.

• Any three adjacent arcs where the middle arc has length π are of three different colors.

Find the number of distinct ways Farmer James can cover his circle. Here, two coverings are equivalentif and only if they are rotations of one another. In particular, two colorings are considered distinct ifthey are reflections of one another, but not rotations of one another.

10. Max repeatedly throws a fair coin in a hurricane. For each throw, there is a 4% chance that the coingets blown away. He records the number of heads H and the number of tails T before the coin is lost.(If the coin is blown away on a toss, no result is recorded for that toss.) What is the expected value of|H − T |?

HMMT February 2020February 15, 2020

Combinatorics

1. How many ways can the vertices of a cube be colored red or blue so that the color of each vertex isthe color of the majority of the three vertices adjacent to it?

Proposed by: Milan Haiman

Answer: 8

Solution: If all vertices of the cube are of the same color, then there are 2 ways. Otherwise, look ata red vertex. Since it must have at least 2 red neighbors, there is a face of the cube containing 3 redvertices. The last vertex on this face must also be red. Similarly, all vertices on the opposite face mustbe blue. Thus, all vertices on one face of the cube are red while the others are blue. Since a cube has6 faces, the answer is 2 + 6 = 8.

2. How many positive integers at most 420 leave different remainders when divided by each of 5, 6, and7?

Proposed by: Milan Haiman

Answer: 250

Solution: Note that 210 = 5 · 6 · 7 and 5, 6, 7 are pairwise relatively prime. So, by the ChineseRemainder Theorem, we can just consider the remainders n leaves when divided by each of 5, 6, 7.To construct an n that leaves distinct remainders, first choose its remainder modulo 5, then modulo6, then modulo 7. We have 5 = 6 〉 1 = 7 〉 2 choices for each remainder. Finally, we multiply by 2because 420 = 2 · 210. The answer is 2 · 53 = 250.

3. Each unit square of a 4× 4 square grid is colored either red, green, or blue. Over all possible coloringsof the grid, what is the maximum possible number of L-trominos that contain exactly one square ofeach color? (L-trominos are made up of three unit squares sharing a corner, as shown below.)

Proposed by: Andrew Lin

Answer: 18

Solution: Notice that in each 2× 2 square contained in the grid, we can form 4 L-trominoes. By thepigeonhole principle, some color appears twice among the four squares, and there are two trominoeswhich contain both. Therefore each 2× 2 square contains at most 2 L-trominoes with distinct colors.Equality is achieved by coloring a square (x, y) red if x + y is even, green if x is odd and y is even,and blue if x is even and y is odd. Since there are nine 2 × 2 squares in our 4× 4 grid, the answer is9× 2 = 18.

4. Given an 8 × 8 checkerboard with alternating white and black squares, how many ways are there tochoose four black squares and four white squares so that no two of the eight chosen squares are in thesame row or column?

Proposed by: James Lin

Answer: 20736

Solution: Number both the rows and the columns from 1 to 8, and say that black squares are the oneswhere the rows and columns have the same parity. We will use, e.g. “even rows” to refer to rows 2, 4,6, 8. Choosing 8 squares all in different rows and columns is equivalent to matching rows to columns.

For each of the 8 rows, we first decide whether they will be matched with a column of the same parityas itself (resulting in a black square) or with one of a different parity (resulting in a white square).Since we want to choose 4 squares of each color, the 4 rows matched to same-parity columns must

contain 2 even rows and 2 odd rows. There are(42

)2= 62 ways to choose 2 odd rows and 2 even rows

to match with same-parity columns.

After choosing the above, we have fixed which 4 rows should be matched with odd columns (while theother 4 should be matched with even columns). Then there are (4!)2 = 242 ways to assign the columnsto the rows, so the answer is (6 · 24)2 = 1442 = 20736.

5. Let S be a set of intervals defined recursively as follows:

• Initially, [1, 1000] is the only interval in S.

• If l 6= r and [l, r] ∈ S, then both[l,⌊l+r2

⌋],[⌊

l+r2

⌋+ 1, r

]∈ S.

(Note that S can contain intervals such as [1, 1], which contain a single integer.) An integer i is chosenuniformly at random from the range [1, 1000]. What is the expected number of intervals in S whichcontain i?

Proposed by: Benjamin Qi

Answer: 10.976

Solution: The answer is given by computing the sum of the lengths of all intervals in S and dividingthis value by 1000, where the length of an interval [i, j] is given by j 〉 i + 1. An interval may becategorized based on how many times [1, 1000] must be split to attain it. An interval that is derivedfrom splitting [1, 1000] k times will be called a k-split.

The only 0-split is [1, 1000], with a total length of 1000. The 1-splits are [1, 500] and [501, 1000], witha total length of 1000. As long as none of the k-splits have length 1, the (k + 1)-splits will have thesame total length. Since the length of the intervals is reduced by half each time (rounded down), wefind that the sum of the lengths of the k-splits is 1000 for 0 ≤ k ≤ 9.

Note that the 9-splits consist of 210 〉 1000 intervals of length 1 and 1000 〉 29 intervals of length 2.Then the 10-splits consist of 2(1000 〉 29) intervals of length 1, with total length 2(1000 〉 29). Thetotal interval length across all splits is equal to 12(1000)〉 210, so our answer is

12〉 210

1000= 10.976.

6. Alice writes 1001 letters on a blackboard, each one chosen independently and uniformly at randomfrom the set S = {a, b, c}. A move consists of erasing two distinct letters from the board and replacingthem with the third letter in S. What is the probability that Alice can perform a sequence of moveswhich results in one letter remaining on the blackboard?

Proposed by: Daniel Zhu

Answer: 3−3−999

4

Solution: Let na, nb, and nc be the number of a’s, b’s, and c’s on the board, respectively The keyobservation is that each move always changes the parity of all three of na, nb, and nc. Since the finalconfiguration must have na, nb, and nc equal to 1, 0, 0 in some order, Alice cannot leave one letter onthe board if na, nb, and nc start with the same parity (because then they will always have the sameparity). Alice also cannot leave one letter on the board if all the letters are initially the same (becauseshe will have no moves to make).

We claim that in all other cases, Alice can make a sequence of moves leaving one letter on the board.The proof is inductive: the base cases na + nb + nc ≤ 2 are easy to verify, as the possible tuplesare (1, 0, 0), (1, 1, 0), and permutations. If na + nb + nc ≥ 3, assume without loss of generality thatna ≥ nb ≥ nc. Then nb ≥ 1 (because otherwise all the letters are a) and na ≥ 2 (because otherwise(na, nb, nc) = (1, 1, 1), which all have the same parity). Then Alice will replace a and b by c, reducingto a smaller case.

We begin by computing the probability that na, nb, and nc start with the same parity. Suppose mletters are chosen at random in the same way (so that we are in the case m = 1001). Let xm be theprobability that na, nb, and nc all have the same parity. We have the recurrence xm+1 = 1

3 (1 〉 xm)because when when choosing the (m + 1)th letter, the ni can only attain the same parity if theydid not before, and the appropriate letter is drawn. Clearly x0 = 1, which enables us to computexm = 1

4 (1 + 3 · (〉3)−m). Then x1001 is the probability that na, nb, and nc have the same parity.

The probability that all the letters are initially the same is 3−1000, as this occurs exactly when all thesubsequent letters match the first. Thus our final answer is

1〉 3−1000 〉 1

4(1 + 3 · (〉3)−1001) =

3

4〉 1

4 · 3999.

7. Anne-Marie has a deck of 16 cards, each with a distinct positive factor of 2002 written on it. Sheshuffles the deck and begins to draw cards from the deck without replacement. She stops when thereexists a nonempty subset of the cards in her hand whose numbers multiply to a perfect square. Whatis the expected number of cards in her hand when she stops?

Proposed by: Michael Ren

Answer: 837208

Solution: Note that 2002 = 2 · 7 · 11 · 13, so that each positive factor of 2002 is included on exactlyone card. Each card can identified simply by whether or not it is divisible by each of the 4 primes, andwe can uniquely achieve all of the 24 possibilities. Also, when considering the product of the values onmany cards, we only care about the values of the exponents in the prime factorization modulo 2, aswe have a perfect square exactly when each exponent is even.

Now suppose Anne-Marie has already drawn k cards. Then there are 2k possible subsets of cards fromthose she has already drawn. Note that if any two of these subsets have products with the same fourexponents modulo 2, then taking the symmetric difference yields a subset of cards in her hand whereall four exponents are 0 (mod 2), which would cause her to stop. Now when she draws the (k + 1)thcard, she achieves a perfect square subset exactly when the the exponents modulo 2 match those froma subset of the cards she already has. Thus if she has already drawn k cards, she will not stop if shedraws one of 16〉 2k cards that don’t match a subset she already has.

Let pk be the probability that Anne-Marie draws at least k cards. We have the recurrence

pk+2 =16〉 2k

16〉 kpk+1

because in order to draw k + 2 cards, the (k + 1)th card, which is drawn from the remaining 16 〉 kcards, must not be one of the 16〉 2k cards that match a subset of Anne-Marie’s first k cards. We nowcompute

p1 = 1,

p2 =15

16,

p3 =14

15p2 =

7

8,

p4 =12

14p3 =

3

4,

p5 =8

13p4 =

6

13,

p6 = 0.

The expected number of cards that Anne-Marie draws is

p1 + p2 + p3 + p4 + p5 = 1 +15

16+

7

8+

3

4+

6

13=

837

208.

8. Let Γ1 and Γ2 be concentric circles with radii 1 and 2, respectively. Four points are chosen on thecircumference of Γ2 independently and uniformly at random, and are then connected to form a convexquadrilateral. What is the probability that the perimeter of this quadrilateral intersects Γ1?

Proposed by: Yuan Yao

Answer: 2227

Solution:

R1

Define a triplet as three points on Γ2 that form the vertices of an equilateral triangle. Note that dueto the radii being 1 and 2, the sides of a triplet are all tangent to Γ1. Rather than choosing four pointson Γ2 uniformly at random, we will choose four triplets of Γ2 uniformly at random and then choosea random point from each triplet. (This results in the same distribution.) Assume without loss ofgenerality that the first step creates 12 distinct points, as this occurs with probability 1.

In the set of twelve points, a segment between two of those points does not intersect Γ1 if and only ifthey are at most three vertices apart. (In the diagram shown above, the segments connecting R1 tothe other red vertices are tangent to Γ1, so the segments connecting R1 to the six closer vertices do notintersect Γ1.) There are two possibilities for the perimeter of the convex quadrilateral to not intersectΓ1: either the convex quadrilateral contains Γ1 or is disjoint from it.

Case 1: The quadrilateral contains Γ1.

Each of the four segments of the quadrilateral passes at most three vertices, so the only possibility isthat every third vertex is chosen. This is shown by the dashed quadrilateral in the diagram, and thereare 3 such quadrilaterals.

Case 2: The quadrilateral does not contain Γ1.

In this case, all of the chosen vertices are at most three apart. This is only possible if we choose fourconsecutive vertices, which is shown by the dotted quadrilateral in the diagram. There are 12 suchquadrilaterals.

Regardless of how the triplets are chosen, there are 81 ways to pick four points and 12 + 3 = 15 ofthese choices result in a quadrilateral whose perimeter does not intersect Γ1. The desired probabilityis 1〉 5

27 = 2227 .

Remark. The problem can easily be generalized for a larger number of vertices, where Γ1 and Γ2 arethe inscribed and circumscribed circles of a regular n-gon and n + 1 points are chosen uniformly atrandom on Γ2. The probability that the perimeter of the convex (n+ 1)-gon formed by those verticesintersects Γ1 is 1〉 n+2

nn .

9. Farmer James wishes to cover a circle with circumference 10π with six different types of colored arcs.Each type of arc has radius 5, has length either π or 2π, and is colored either red, green, or blue. Hehas an unlimited number of each of the six arc types. He wishes to completely cover his circle withoutoverlap, subject to the following conditions:

• Any two adjacent arcs are of different colors.

• Any three adjacent arcs where the middle arc has length π are of three different colors.

Find the number of distinct ways Farmer James can cover his circle. Here, two coverings are equivalentif and only if they are rotations of one another. In particular, two colorings are considered distinct ifthey are reflections of one another, but not rotations of one another.

Proposed by: James Lin

Answer: 93

Solution: Fix an orientation of the circle, and observe that the the problem is equivalent to findingthe number of ways to color ten equal arcs of the circle such that each arc is one of three differentcolors, and any two arcs which are separated by exactly one arc are of different colors. We can considerevery other arc, so we are trying to color just five arcs so that no two adjacent arcs are of the samecolor. This is independent from the coloring of the other five arcs.

Let ai be the number of ways to color i arcs in three colors so that no two adjacent arcs are the samecolor. Note that a1 = 3 and a2 = 6. We claim that ai + ai+1 = 3 · 2i for i ≥ 2. To prove this, observethat ai counts the number of ways to color i+ 1 points in a line so that no two adjacent points are thesame color, and the first and (i+ 1)th points are the same color. Meanwhile, ai+1 counts the numberof ways to color i+ 1 points in a line so that no two adjacent points are the same color, and the firstand (i+ 1)th points are different colors. Then ai + ai+1 is the number of ways to color i+ 1 points ina line so that no two adjacent points are the same color. There are clearly 3 · 2i ways to do this, as wepick the colors from left to right, with 3 choices for the first color and 2 for the rest. We then computea3 = 6, a4 = 18, a5 = 30. Then we can color the whole original circle by picking one of the 30 possiblecolorings for each of the two sets of 5 alternating arcs, for 302 = 900 total.

Now, we must consider the rotational symmetry. If a configuration has no rotational symmetry, thenwe have counted it 10 times. If a configuration has 180◦ rotational symmetry, then we have countedit 5 times. This occurs exactly when we have picked the same coloring from our 30 for both choices,and in exactly one particular orientation, so there are 30 such cases. Having 72◦ or 36◦ rotational

symmetry is impossible, as arcs with exactly one arc between them must be different colors. Thenafter we correct for overcounting our answer is

900〉 30

10+

30

5= 93.

10. Max repeatedly throws a fair coin in a hurricane. For each throw, there is a 4% chance that the coingets blown away. He records the number of heads H and the number of tails T before the coin is lost.(If the coin is blown away on a toss, no result is recorded for that toss.) What is the expected value of|H 〉 T |?Proposed by: Krit Boonsiriseth

Answer: 247

Solution 1: (Dilhan Salgado) In all solutions, p = 125 will denote the probability that the coin is

blown away. Let D = |H 〉 T |. Note that if D 6= 0, the expected value of D is not changed by acoin flip, whereas if D = 0, the expected value of D increases by 1. Therefore E(D) can be computedas the sum over all n of the probability that the nth coin flip occurs when D = 0. This only occurswhen n = 2k + 1 is odd, where the probability that the first n coin flips occur is (1 〉 p)2k+1 and the

probability that D = 0 after the first n〉 1 flips is(2k

k )4k

. Therefore

E(D) = (1〉 p)∞∑k=0

(1〉 p

2

)2k (2k

k

)=

1〉 p√1〉 (1〉 p)2

using the generating function∞∑k=0

(2k

k

)xk =

1√1〉 4x

.

Plugging in p = 125 yields E(D) = 24

7 .

Solution 2: For each n > 0, the probability that Max made n successful throws (not counting thelast throw) is p(1〉 p)n.

Claim: Assuming Max made n > 1 throws, the expected value of |H 〉 T | is given by

b(n−1)/2c∏k=1

2k + 1

2k.

Proof. If n is odd then the expected value for n + 1 will be equal to that for n; since |H 〉 T | will benonzero, it will be equally likely to increase or decrease after the coin is flipped. Therefore, it sufficesto compute the expected value for the n odd case. This is∑(n−1)/2

i=0

(ni

)· (n〉 2i)

2n−1= n〉

∑(n−1)/2i=0

(ni

)· 2i

2n−1

= n ·

(1〉

2 ·∑(n−3)/2

i=0

(n−1i

)2n−1

)

= n ·

(n−1

(n−1)/2)

2n−1

=n!

(n〉 1)!!2

=n!!

(n〉 1)!!

=

(n−1)/2∏k=1

2k + 1

2k,

as desired.

Using the claim, we have

E(|H 〉 T |) = p

∞∑n=1

(1〉 p)nb(n−1)/2c∏

k=1

2k + 1

2k

= p(1〉 p)(2〉 p)

∞∑m=0

((1〉 p)2m

m∏k=1

2k + 1

2k

)= p(1〉 p)(2〉 p)

(1〉 (1〉 p)2

)−3/2=

1〉 p√p(2〉 p)

.

Plugging in p = 125 gives

E(|H 〉 T |) =24

25· 5 · 5

7=

24

7.

Solution 3: Let En be the expected value of |H 〉 T + n|. By symmetry, E−n = En for all n.Considering what happens in the next throw gives

2En = (1〉 p)En−1 + (1〉 p)En+1 + 2pn

for all n > 0. Now let α =1−√

p(2−p)1−p < 1 be the smaller root of (1〉 p)x2 〉 2x+ (1〉 p) = 0. From

∞∑n=1

2αnEn =

∞∑n=1

αn ((1〉 p)En−1 + (1〉 p)En+1 + 2pn)

= α(1〉 p)E0 + α2(1〉 p)E1 +

∞∑n=1

2pnαn +

∞∑n=2

((1〉 p)αn−1 + (1〉 p)αn+1)

)En

= α(1〉 p)E0 + (2α〉 (1〉 p))E1 +

∞∑n=1

2pnαn +

∞∑n=2

2αnEn,

we have

(1〉 p)E1 〉 α(1〉 p)E0 =

∞∑n=1

2pnαn =2pα

(1〉 α)2.

As E0 = (1〉 p)E1, this gives

E0 (1〉 α(1〉 p)) =2pα

(1〉 α)2.

Plugging in p = 125 and α = 3

4 gives E0 = 247 .

HMMT February 2020February 15, 2020

Geometry

1. Let DIAL, FOR, and FRIEND be regular polygons in the plane. If ID = 1, find the product of allpossible areas of OLA.

2. Let ABC be a triangle with AB = 5, AC = 8, and ∠BAC = 60◦. Let UVWXY Z be a regularhexagon that is inscribed inside ABC such that U and V lie on side BA, W and X lie on side AC,and Z lies on side CB. What is the side length of hexagon UVWXY Z?

3. Consider the L-shaped tromino below with 3 attached unit squares. It is cut into exactly two pieces ofequal area by a line segment whose endpoints lie on the perimeter of the tromino. What is the longestpossible length of the line segment?

4. Let ABCD be a rectangle and E be a point on segment AD. We are given that quadrilateral BCDEhas an inscribed circle ω1 that is tangent to BE at T . If the incircle ω2 of ABE is also tangent to BEat T , then find the ratio of the radius of ω1 to the radius of ω2.

5. Let ABCDEF be a regular hexagon with side length 2. A circle with radius 3 and center at A isdrawn. Find the area inside quadrilateral BCDE but outside the circle.

6. Let ABC be a triangle with AB = 5, BC = 6, CA = 7. Let D be a point on ray AB beyond B suchthat BD = 7, E be a point on ray BC beyond C such that CE = 5, and F be a point on ray CAbeyond A such that AF = 6. Compute the area of the circumcircle of DEF .

7. Let Γ be a circle, and ω1 and ω2 be two non-intersecting circles inside Γ that are internally tangent toΓ at X1 and X2, respectively. Let one of the common internal tangents of ω1 and ω2 touch ω1 and ω2

at T1 and T2, respectively, while intersecting Γ at two points A and B. Given that 2X1T1 = X2T2 andthat ω1, ω2, and Γ have radii 2, 3, and 12, respectively, compute the length of AB.

8. Let ABC be an acute triangle with circumcircle Γ. Let the internal angle bisector of ∠BAC intersectBC and Γ at E and N , respectively. Let A′ be the antipode of A on Γ and let V be the point whereAA′ intersects BC. Given that EV = 6, V A′ = 7, and A′N = 9, compute the radius of Γ.

9. Circles ωa, ωb, ωc have centers A,B,C, respectively and are pairwise externally tangent at pointsD,E, F (with D ∈ BC,E ∈ CA,F ∈ AB). Lines BE and CF meet at T . Given that ωa has ra-dius 341, there exists a line ` tangent to all three circles, and there exists a circle of radius 49 tangentto all three circles, compute the distance from T to `.

10. Let Γ be a circle of radius 1 centered at O. A circle Ω is said to be friendly if there exist distinct circlesω1, ω2, . . ., ω2020, such that for all 1 ≤ i ≤ 2020, ωi is tangent to Γ, Ω, and ωi+1. (Here, ω2021 = ω1.)For each point P in the plane, let f(P ) denote the sum of the areas of all friendly circles centered atP . If A and B are points such that OA = 1

2 and OB = 13 , determine f(A)− f(B).

HMMT February 2020February 15, 2020

Geometry

1. Let DIAL, FOR, and FRIEND be regular polygons in the plane. If ID = 1, find the product of allpossible areas of OLA.

Proposed by: Andrew Gu

Answer: 132

Solution: Focusing on FRIEND and FOR first, observe that either DIO is an equilateral triangleor O is the midpoint of ID. Next, OLA is always an isosceles triangle with base LA = 1. The possible

distances of O from LA are 1 and 1±√32 as the distance from O to ID in the equilateral triangle case

is√32 .

F

R

I

E

N

D

A1

L1

O1O2

A2

L2

The three possibilities are shown in the diagram as shaded triangles 4O1L1A1, 4O2L2A2, and4O1L2A2.

The product of all possible areas is thus

1 ·(

1−√32

)·(

1 +√32

)23

=1

25=

1

32.

2. Let ABC be a triangle with AB = 5, AC = 8, and ∠BAC = 60◦. Let UVWXY Z be a regularhexagon that is inscribed inside ABC such that U and V lie on side BA, W and X lie on side AC,and Z lies on side CB. What is the side length of hexagon UVWXY Z?

Proposed by: Ryan Kim

Answer: 4021

Solution: Let the side length of UVWXY Z be s. We have WZ = 2s and WZ ‖ AB by properties ofregular hexagons. Thus, triangles WCZ and ACB are similar. AWV is an equilateral triangle, so wehave AW = s. Thus, using similar triangles, we have

WC

WZ=AC

AB=⇒ 8− s

2s=

8

5,

so 5(8− s) = 8(2s) =⇒ s = 4021 .

A B

C

UV

W

X Y

Z

3. Consider the L-shaped tromino below with 3 attached unit squares. It is cut into exactly two pieces ofequal area by a line segment whose endpoints lie on the perimeter of the tromino. What is the longestpossible length of the line segment?

Proposed by: James Lin

Answer: 52

Solution: Let the line segment have endpoints A and B. Without loss of generality, let A lie belowthe lines x+ y =

√3 (as this will cause B to be above the line x+ y =

√3) and y = x (we can reflect

about y = x to get the rest of the cases):

A

B

Now, note that as A ranges from (0, 0) to (1.5, 0), B will range from (1, 1) to (1, 2) to (0, 2), as indicatedby the red line segments. Note that these line segments are contained in a rectangle bounded by x = 0,y = 0, x = 1.5, and y = 2, and so the longest line segment in this case has length

√22 + 1.52 = 5

2 .

As for the rest of the cases, as A = (x, 0) ranges from (1.5, 0) to (√

3, 0), B will be the point (0, 3x ), so

it suffices to maximize√x2 + 9

x2 given 1.5 ≤ x ≤√

3. Note that the further away x2 is from 3, the

larger x2 + 9x2 gets, and so the maximum is achieved when x = 1.5, which gives us the same length as

before.

Thus, the maximum length is 52 .

4. Let ABCD be a rectangle and E be a point on segment AD. We are given that quadrilateral BCDEhas an inscribed circle ω1 that is tangent to BE at T . If the incircle ω2 of ABE is also tangent to BEat T , then find the ratio of the radius of ω1 to the radius of ω2.

Proposed by: James Lin

Answer: 3+√5

2

Solution: Let ω1 be tangent to AD, BC at R, S and ω2 be tangent to AD, AB at X, Y . LetAX = AY = r, EX = ET = ER = a, BY = BT = BS = b. Then noting that RS ‖ CD, we seethat ABSR is a rectangle, so r + 2a = b. Therefore AE = a + r, AB = b + r = 2(a + r), and so

BE = (a+ r)√

5. On the other hand, BE = b+ a = r+ 3a. This implies that a = 1+√5

2 r. The desired

ratio is then RS2AY = AB

2r = a+rr = 3+

√5

2 .

A B

CD

E

X

Y

T

R S

5. Let ABCDEF be a regular hexagon with side length 2. A circle with radius 3 and center at A isdrawn. Find the area inside quadrilateral BCDE but outside the circle.

Proposed by: Carl Joshua Quines

Answer: 4√

3− 32π

Solution: Rotate the region 6 times about A to form a bigger hexagon with a circular hole. The largerhexagon has side length 4 and area 24

√3, so the area of the region is 1

6 (24√

3− 9π) = 4√

3− 32π.

A

B C

D

EF

6. Let ABC be a triangle with AB = 5, BC = 6, CA = 7. Let D be a point on ray AB beyond B suchthat BD = 7, E be a point on ray BC beyond C such that CE = 5, and F be a point on ray CAbeyond A such that AF = 6. Compute the area of the circumcircle of DEF .

Proposed by: James Lin

Answer: 2513 π

Solution 1: Let I be the incenter of ABC. We claim that I is the circumcenter of DEF .

A

BC

D

E

F

X

Y

ZI

To prove this, let the incircle touch AB, BC, and AC at X, Y , and Z, respectively. Noting thatXB = BY = 2, Y C = CZ = 4, and ZA = AX = 3, we see that XD = Y E = ZF = 9. Thus, sinceIX = IY = IZ = r (where r is the inradius) and ∠IXD = ∠IY E = ∠IZF = 90◦, we have threecongruent right triangles, and so ID = IE = IF , as desired.

Let s = 5+6+72 = 9 be the semiperimeter. By Heron’s formula, [ABC] =

√9(9− 5)(9− 6)(9− 7) =

6√

6, so r = [ABC]s = 2

√6

3 . Then the area of the circumcircle of DEF is

ID2π = (IX2 +XD2)π = (r2 + s2)π =251

3π.

Solution 2: Let D′ be a point on ray CB beyond B such that BD′ = 7, and similarly define E′, F ′.Noting that DA = E′A and AF = AF ′, we see that DE′F ′F is cyclic by power of a point. Similarly,EF ′D′D and FD′E′E are cyclic. Now, note that the radical axes for the three circles circumscribingthese quadrilaterals are the sides of ABC, which are not concurrent. Therefore, DD′FF ′EE′ is cyclic.We can deduce that the circumcenter of this circle is I in two ways: either by calculating that themidpoint of D′E coincides with the foot from I to BC, or by noticing that the perpendicular bisectorof FF ′ is AI. The area can then be calculated the same way as the previous solution.

A

B C

D

E

F

D′

E′

F ′

I

Remark. The circumcircle of DEF is the Conway circle of ABC.

7. Let Γ be a circle, and ω1 and ω2 be two non-intersecting circles inside Γ that are internally tangent toΓ at X1 and X2, respectively. Let one of the common internal tangents of ω1 and ω2 touch ω1 and ω2

at T1 and T2, respectively, while intersecting Γ at two points A and B. Given that 2X1T1 = X2T2 andthat ω1, ω2, and Γ have radii 2, 3, and 12, respectively, compute the length of AB.

Proposed by: James Lin

Answer: 96√10

13

Solution 1: Let ω1, ω2, Γ have centers O1, O2, O and radii r1, r2, R respectively. Let d be thedistance from O to AB (signed so that it is positive if O and O1 are on the same side of AB).

A BT1

T2 X1

X2

OO1

O2

M

Note that

OOi = R− ri,

cos∠T1O1O =O1T1 −OM

OO1=r1 − dR− r1

,

cos∠T2O2O =O2T2 +OM

OO1=r2 + d

R− r2.

Then

X1T1 = r1√

2− 2 cos∠X1O1T1

= ri√

2 + 2 cos∠T1O1O

= r1

√2 + 2

r1 − dR− r1

= r1

√2R− dR− r1

.

Likewise,

X2T2 = r2

√2R+ d

R− r2.

From 2X1T1 = X2T2 we have

8r21

(R− dR− r1

)= 4X1T

21 = X2T

22 = 2r22

(R+ d

R− r2

).

Plugging in r1 = 2, r2 = 3, R = 12 and solving yields d = 3613 . Hence AB = 2

√R2 − d2 = 96

√10

13 .

Solution 2: We borrow the notation from the previous solution. Let X1T1 and X2T2 intersect Γ againat M1 and M2. Note that, if we orient AB to be horizontal, then the circles ω1 and ω2 are on oppositesides of AB. In addition, for i ∈ {1, 2} there exist homotheties centered at Xi with ratio R

riwhich send

ωi to Γ. Since T1 and T2 are points of tangencies and thus top/bottom points, we see that M1 and M2

are the top and bottom points of Γ, and so M1M2 is a diameter perpendicular to AB.

M1

M2

A BT1

T2 X1

X2

M

Now, note that through power of a point and the aforementioned homotheties,

P (M1, ω1) = M1T1 ·M1X1 = X1T21

(R

r1

)(R

r1− 1

)= 30X1T

21 ,

and similarly P (M2, ω2) = 12X2T22 . (Here P is the power of a point with respect to a circle). Then

P (M1, ω1)

P (M2, ω2)=

30X1T21

12X2T 22

=30

12(2)2=

5

8.

Let M be the midpoint of AB, and suppose M1M = R + d (here d may be negative). Noting thatM1 and M2 are arc bisectors, we have ∠AX1M1 = ∠T1AM1, so 4M1AT1 ∼ 4M1X1A, meaningthat M1A

2 = M1T1 ·M1X1 = P (M1, ω1). Similarly, 4M2AT2 ∼ 4M2X2A, so M2A2 = P (M2, ω2).

Therefore,

P (M1, ω1)

P (M2, ω2)=M1A

2

M2A2=

(R2 − d2) + (R+ d)2

(R2 − d2) + (R− d)2=

2R2 + 2Rd

2R2 − 2Rd=R+ d

R− d=

5

8,

giving d = − 313R. Finally, we compute AB = 2R

√1−

313

)2= 8R

√10

13 = 96√10

13 .

8. Let ABC be an acute triangle with circumcircle Γ. Let the internal angle bisector of ∠BAC intersectBC and Γ at E and N , respectively. Let A′ be the antipode of A on Γ and let V be the point whereAA′ intersects BC. Given that EV = 6, V A′ = 7, and A′N = 9, compute the radius of Γ.

Proposed by: James Lin

Answer: 152

Solution 1: Let Ha be the foot of the altitude from A to BC. Since AE bisects ∠HaAV , by the anglebisector theorem AHa

HaE= AV

V E . Note that 4AHaE ∼ 4ANA′ are similar right triangles, so ANNA′ = AHa

HaE.

Let R be the radius of Γ. We know that AA′ = 2R, so AN =√AA′2 −NA′2 =

√4R2 − 81 and

AV = AA′ − V A′ = 2R− 7. Therefore√

4R2 − 81

9=

AN

NA′=AHa

HaE=AV

V E=

2R− 7

6.

The resulting quadratic equation is

0 = 9(2R− 7)2 − 4(4R2 − 81) = 20R2 − 252R+ 765 = (2R− 15)(10R− 51).

We are given that ABC is acute so V A′ < R. Therefore R = 152 .

A

B C

A′

V

N

EHa

K

Solution 2: Let Ψ denote inversion about A with radius√AB ·AC composed with reflection about

AE. Note that Ψ swaps the pairs {B,C}, {E,N}, and {Ha, A′}. Let K = Ψ(V ), which is also the

second intersection of AHa with Γ. Since AE bisects ∠KAA′, we have NK = NA′ = 9. By theinversion distance formula,

NK =AB ·AC · V EAE ·AV

=AE ·AN · V EAE ·AV

=AN · V EAV

.

This leads to the same equation as the previous solution.

9. Circles ωa, ωb, ωc have centers A,B,C, respectively and are pairwise externally tangent at pointsD,E, F (with D ∈ BC,E ∈ CA,F ∈ AB). Lines BE and CF meet at T . Given that ωa has ra-dius 341, there exists a line ` tangent to all three circles, and there exists a circle of radius 49 tangentto all three circles, compute the distance from T to `.

Proposed by: Andrew Gu

Answer: 294

Solution 1: We will use the following notation: let ω be the circle of radius 49 tangent to each ofωa, ωb, ωc. Let ωa, ωb, ωc have radii ra, rb, rc respectively. Let γ be the incircle of ABC, with centerI and radius r. Note that DEF is the intouch triangle of ABC and γ is orthogonal to ωa, ωb, ωc

(i.e. ID, IE, IF are the common internal tangents). Since AD,BE,CF are concurrent at T , we haveK = AB ∩DE satisfies (A,B;F,K) = −1, so K is the external center of homothety of ωa and ωb. Inparticular, K lies on `. Similarly, BC ∩EF also lies on `, so ` is the polar of T to γ. Hence IT ⊥ ` soif L is the foot from I to `, we have IT · IL = r2.

An inversion about γ preserves ωa, ωb, ωc and sends ` to the circle with diameter IT . Since inversionpreserves tangency, the circle with diameter IT must be ω. Therefore IT = 98 by the condition of theproblem statement. Letting a, b, c be the radii of ωa, ωb, ωc respectively and invoking Heron’s formulaas well as A = rs for triangle ABC, we see that γ has radius

r =

√rarbrc

ra + rb + rc.

We will compute this quantity using Descartes’ theorem. Note that there are two circles tangent to ωa,ωb, ωc, one with radius IT/2 and one with radius ∞. By Descartes’ circle theorem, we have (whereka := 1/a is the curvature)

ka + kb + kc + 2√kakb + kbkc + kakc =

1

IT/2

andka + kb + kc − 2

√kakb + kbkc + kcka = 0,

which implies √ra + rb + rcrarbrc

=√kakb + kbkc + kcka =

1

2IT.

Therefore r = 2IT , which means IL = r2

IT = 4IT and TL = 3IT = 294.

I

T

A

B

C

D

E

F

L K

Solution 2: Using the same notation as the previous solution, note that the point T can be expressedwith un-normalized barycentric coordinates(

1

ra:

1

rb:

1

rc

)with respect to ABC because T is the Gergonne point of triangle ABC. The distance from T to ` canbe expressed as a weighted average of the distances from each of the points A, B, C, which is

1/ra1/ra + 1/rb + 1/rc

· ra +1/rb

1/ra + 1/rb + 1/rc· rb +

1/rc1/ra + 1/rb + 1/rc

· rc =3

1/ra + 1/rb + 1/rc.

Note that there are two circles tangent to ωa, ωb, ωc, one with radius 49 and one with radius ∞. ByDescartes’ circle theorem, we have (where ka := 1/ra is the curvature)

ka + kb + kc + 2√kakb + kbkc + kakc =

1

49

andka + kb + kc − 2

√kakb + kbkc + kcka = 0,

so ka + kb + kc = 1/98. The distance from T to ` is then 3 · 98 = 294.

Solution 3: As in the first solution, we deduce that IT is a diameter of ω, with T being the point onω closest to `. By Steiner’s porism, we can hold ω and ` fixed while making ωc into a line parallel to`, resulting in the following figure:

I

T

A B

DE

F

L

Let R be the common radius of ωa and ωb and r be the radius of ω. Notice that ABDE is a rectan-gle with center T , so R = AE = 2IT = 4r. The distance from T to ` is IL−IT = 2R−2r = 6r = 294.

10. Let Γ be a circle of radius 1 centered at O. A circle Ω is said to be friendly if there exist distinct circlesω1, ω2, . . ., ω2020, such that for all 1 ≤ i ≤ 2020, ωi is tangent to Γ, Ω, and ωi+1. (Here, ω2021 = ω1.)For each point P in the plane, let f(P ) denote the sum of the areas of all friendly circles centered atP . If A and B are points such that OA = 1

2 and OB = 13 , determine f(A)− f(B).

Proposed by: Michael Ren

Answer: 10009 π

Solution: Let P satisfy OP = x. (For now, we focus on f(P ) and ignore the A and B from theproblem statement.) The key idea is that if we invert at some point along OP such that the imagesof Γ and Ω are concentric, then ωi still exist. Suppose that this inversion fixes Γ and takes Ω to Ω′ ofradius r (and X to X ′ in general). If the inversion is centered at a point Q along ray OP such thatOQ = d, then the radius of inversion is

√d2 − 1. Let the diameter of Ω meet OQ at A and B with A

closer to Q than B. Then, (AB;PP∞) = −1 inverts to (A′B′;P ′Q) = −1, where P∞ is the point at

infinity along line OP , so P ′ is the inverse of Q in Ω′. We can compute OP ′ = r2

d so P ′Q = d − r2

d

and PQ = d2→1d→ r2d

. Thus, we get the equation d2→1d→ r2d

+ x = d, which rearranges to 1→r2d2→r2 d = x, or

d2 − x→1(1− r2)d− r2 = 0. Now, we note that the radius of Ω is

1

2AB =

1

2

(d2 − 1

d− r− d2 − 1

d+ r

)=r(d2 − 1)

d2 − r2= r

(1 +

r2 − 1

d2 − r2

)= r

(1− x

d

).

The quadratic formula gives us that d =(1→r2)±

√r4→(2→4x2)r2+1

2x , so xd = − 1→r2±

√r4→(2→4x2)r2+1

2r2 ,which means that the radius of Ω is

r2 + 1±√r4 − (2− 4x2)r2 + 1

2r=r + 1

r ±√r2 + 1

r2 − 2 + 4x2

2.

Note that if r gives a valid chain of 2020 circles, so will 1r by homothety/inversion. Thus, we can think

of each pair of r, 1r as giving rise to two possible values of the radius of Ω, which arer+ 1

r±√

r2+ 1r2→1

2 .This means that the pairs have the same sum of radii as the circles centered at O, and the productof the radii is 1 − x2. (A simpler way to see this is to note that inversion at P with radius

√1− x2

swaps the two circles.) From this, it follows that the difference between the sum of the areas for each

pair is 2π

122 −

132

)= 5

18π. There are ϕ(2020)2 = 400 such pairs, which can be explicitly computed as

1→sin πk2020

1+sin πk2020

,1+sin πk

2020

1→sin πk2020

for positive integers k < 1010 relatively prime to 2020. Thus, the answer is 10009 π.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2020, February 15, 2020 — GUTS ROUND

Organization Team Team ID#

1. [4] Submit an integer x as your answer to this problem. The number of points you receive will bemax(0, 8〉 |8x〉 100|). (Non-integer answers will be given 0 points.)

2. [4] Let ABC be a triangle and ω be its circumcircle. The point M is the midpoint of arc BC notcontaining A on ω and D is chosen so that DM is tangent to ω and is on the same side of AM as C. Itis given that AM = AC and ∠DMC = 38◦. Find the measure of angle ∠ACB.

3. [4] Let ABC be a triangle and D, E, and F be the midpoints of sides BC, CA, and AB respectively.What is the maximum number of circles which pass through at least 3 of these 6 points?

4. [4] Compute the value of√

1053 〉 1043, given that it is a positive integer.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2020, February 15, 2020 — GUTS ROUND

Organization Team Team ID#

5. [5] Alice, Bob, and Charlie roll a 4, 5, and 6-sided die, respectively. What is the probability that a numbercomes up exactly twice out of the three rolls?

6. [5] Two sides of a regular n-gon are extended to meet at a 28◦ angle. What is the smallest possible valuefor n?

7. [5] Ana and Banana are rolling a standard six-sided die. Ana rolls the die twice, obtaining a1 and a2,then Banana rolls the die twice, obtaining b1 and b2. After Ana’s two rolls but before Banana’s two rolls,they compute the probability p that a1b1 + a2b2 will be a multiple of 6. What is the probability thatp = 1

6?

8. [5] Tessa picks three real numbers x, y, z and computes the values of the eight expressions of the form±x ± y ± z. She notices that the eight values are all distinct, so she writes the expressions down inincreasing order. For example, if x = 2, y = 3, z = 4, then the order she writes them down is

〉x〉 y 〉 z, +x〉 y 〉 z, 〉x+ y 〉 z, 〉x〉 y + z, +x+ y 〉 z, +x〉 y + z, 〉x+ y + z, +x+ y + z.

How many possible orders are there?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2020, February 15, 2020 — GUTS ROUND

Organization Team Team ID#

9. [6] Let P (x) be the monic polynomial with rational coefficients of minimal degree such that 1√2,

1√3, 1√

4, . . . , 1√

1000are roots of P . What is the sum of the coefficients of P?

10. [6] Jarris is a weighted tetrahedral die with faces F1, F2, F3, F4. He tosses himself onto a table, so thatthe probability he lands on a given face is proportional to the area of that face (i.e. the probability he

lands on face Fi is [Fi][F1]+[F2]+[F3]+[F4]

where [K] is the area of K). Let k be the maximum distance any

part of Jarris is from the table after he rolls himself. Given that Jarris has an inscribed sphere of radius3 and circumscribed sphere of radius 10, find the minimum possible value of the expected value of k.

11. [6] Find the number of ordered pairs of positive integers (x, y) with x, y ≤ 2020 such that 3x2+10xy+3y2

is the power of some prime.

12. [6] An 11× 11 grid is labeled with consecutive rows 0, 1, 2, . . . , 10 and columns 0, 1, 2, . . . , 10 so that it isfilled with integers from 1 to 210, inclusive, and the sum of all of the numbers in row n and in column nare both divisible by 2n. Find the number of possible distinct grids.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2020, February 15, 2020 — GUTS ROUND

Organization Team Team ID#

13. [8] Let 4ABC be a triangle with AB = 7, BC = 1, and CA = 4√

3. The angle trisectors of C intersectAB at D and E, and lines AC and BC intersect the circumcircle of4CDE again at X and Y , respectively.Find the length of XY .

14. [8] Let ϕ(n) denote the number of positive integers less than or equal to n which are relatively prime ton. Let S be the set of positive integers n such that 2n

'(n) is an integer. Compute the sum

∑n∈S

1

n.

15. [8] You have six blocks in a row, labeled 1 through 6, each with weight 1. Call two blocks x ≤ y connectedwhen, for all x ≤ z ≤ y, block z has not been removed. While there is still at least one block remaining,you choose a remaining block uniformly at random and remove it. The cost of this operation is the sumof the weights of the blocks that are connected to the block being removed, including itself. Compute theexpected total cost of removing all the blocks.

16. [8] Determine all triplets of real numbers (x, y, z) satisfying the system of equations

x2y + y2z = 1040

x2z + z2y = 260

(x〉 y)(y 〉 z)(z 〉 x) = 〉540.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2020, February 15, 2020 — GUTS ROUND

Organization Team Team ID#

17. [10] Let ABC be a triangle with incircle tangent to the perpendicular bisector of BC. If BC = AE = 20,where E is the point where the A-excircle touches BC, then compute the area of 4ABC.

18. [10] A vertex-induced subgraph is a subset of the vertices of a graph together with any edges whoseendpoints are both in this subset.

An undirected graph contains 10 nodes and m edges, with no loops or multiple edges. What is theminimum possible value of m such that this graph must contain a nonempty vertex-induced subgraphwhere all vertices have degree at least 5?

19. [10] The Fibonacci numbers are defined by F0 = 0, F1 = 1, and Fn = Fn−1 +Fn−2 for n ≥ 2. There existunique positive integers n1, n2, n3, n4, n5, n6 such that

100∑i1=0

100∑i2=0

100∑i3=0

100∑i4=0

100∑i5=0

Fi1+i2+i3+i4+i5 = Fn1 〉 5Fn2 + 10Fn3 〉 10Fn4 + 5Fn5 〉 Fn6 .

Find n1 + n2 + n3 + n4 + n5 + n6.

20. [10] There exist several solutions to the equation

1 +sinx

sin 4x=

sin 3x

sin 2x,

where x is expressed in degrees and 0◦ < x < 180◦. Find the sum of all such solutions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2020, February 15, 2020 — GUTS ROUND

Organization Team Team ID#

21. [12] We call a positive integer t good if there is a sequence a0, a1, . . . of positive integers satisfyinga0 = 15, a1 = t, and

an−1an+1 = (an 〉 1)(an + 1)

for all positive integers n. Find the sum of all good numbers.

22. [12] Let A be a set of integers such that for each integer m, there exists an integer a ∈ A and positiveinteger n such that an ≡ m (mod 100). What is the smallest possible value of |A|?

23. [12] A function f : A→ A is called idempotent if f(f(x)) = f(x) for all x ∈ A. Let In be the number ofidempotent functions from {1, 2, . . . , n} to itself. Compute

∞∑n=1

Inn!.

24. [12] In 4ABC, ω is the circumcircle, I is the incenter and IA is the A-excenter. Let M be the midpoint

of arc B̂AC on ω, and suppose that X, Y are the projections of I onto MIA and IA onto MI, respectively.If 4XY IA is an equilateral triangle with side length 1, compute the area of 4ABC.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2020, February 15, 2020 — GUTS ROUND

Organization Team Team ID#

25. [15] Let S be the set of 34 points in four-dimensional space where each coordinate is in {〉1, 0, 1}. LetN be the number of sequences of points P1, P2, . . . , P2020 in S such that PiPi+1 = 2 for all 1 ≤ i ≤ 2020and P1 = (0, 0, 0, 0). (Here P2021 = P1.) Find the largest integer n such that 2n divides N .

26. [15] Let ABCD be a cyclic quadrilateral, and let segments AC and BD intersect at E. Let W and Ybe the feet of the altitudes from E to sides DA and BC, respectively, and let X and Z be the midpointsof sides AB and CD, respectively. Given that the area of AED is 9, the area of BEC is 25, and∠EBC 〉 ∠ECB = 30◦, then compute the area of WXY Z.

27. [15] Let {ai}i≥0 be a sequence of real numbers defined by

an+1 = a2n 〉1

22020·2n−1

for n ≥ 0. Determine the largest value for a0 such that {ai}i≥0 is bounded.

28. [15] Let 4ABC be a triangle inscribed in a unit circle with center O. Let I be the incenter of 4ABC,and let D be the intersection of BC and the angle bisector of ∠BAC. Suppose that the circumcircle of4ADO intersects BC again at a point E such that E lies on IO. If cosA = 12

13 , find the area of 4ABC.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT February 2020, February 15, 2020 — GUTS ROUND

Organization Team Team ID#

29. [18] Let ABCD be a tetrahedron such that its circumscribed sphere of radius R and its inscribed sphereof radius r are concentric. Given that AB = AC = 1 ≤ BC and R = 4r, find BC2.

30. [18] Let S = {(x, y) | x > 0, y > 0, x+ y < 200, and x, y ∈ Z}. Find the number of parabolas P withvertex V that satisfy the following conditions:

• P goes through both (100, 100) and at least one point in S,

• V has integer coordinates, and

• P is tangent to the line x+ y = 0 at V .

31. [18] Anastasia is taking a walk in the plane, starting from (1, 0). Each second, if she is at (x, y), shemoves to one of the points (x 〉 1, y), (x + 1, y), (x, y 〉 1), and (x, y + 1), each with 1

4 probability. Shestops as soon as she hits a point of the form (k, k). What is the probability that k is divisible by 3 whenshe stops?

32. [18] Find the smallest real constant α such that for all positive integers n and real numbers 0 = y0 <y1 < · · · < yn, the following inequality holds:

α

n∑k=1

(k + 1)3/2√y2k 〉 y2k−1

≥n∑

k=1

k2 + 3k + 3

yk.

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HMMT February 2020, February 15, 2020 — GUTS ROUND

Organization Team Team ID#

33. [22] Estimate

N =

∞∏n=1

nn−1.25

.

An estimate of E > 0 will receive b22 min(N/E,E/N)c points.

34. [22] For odd primes p, let f(p) denote the smallest positive integer a for which there does not exist aninteger n satisfying p | n2 〉 a. Estimate N , the sum of f(p)2 over the first 105 odd primes p.

An estimate of E > 0 will receive b22 min(N/E,E/N)3c points.

35. [22] A collection S of 10000 points is formed by picking each point uniformly at random inside a circleof radius 1. Let N be the expected number of points of S which are vertices of the convex hull of the S.(The convex hull is the smallest convex polygon containing every point of S.) Estimate N .

An estimate of E > 0 will earn max(b22〉 |E 〉N |c, 0) points.

36. [22] A snake of length k is an animal which occupies an ordered k-tuple (s1, . . . , sk) of cells in a n × ngrid of square unit cells. These cells must be pairwise distinct, and si and si+1 must share a side fori = 1, . . . , k〉 1. If the snake is currently occupying (s1, . . . , sk) and s is an unoccupied cell sharing a sidewith s1, the snake can move to occupy (s, s1, . . . , sk−1) instead.

Initially, a snake of length 4 is in the grid {1, 2, . . . , 30}2 occupying the positions (1, 1), (1, 2), (1, 3), (1, 4)with (1, 1) as its head. The snake repeatedly makes a move uniformly at random among moves it canlegally make. Estimate N , the expected number of moves the snake makes before it has no legal movesremaining.

An estimate of E > 0 will earn b22 min(N/E,E/N)4c points.

HMMT February 2020February 15, 2020

Guts Round

1. [4] Submit an integer x as your answer to this problem. The number of points you receive will bemax(0, 8〉 |8x〉 100|). (Non-integer answers will be given 0 points.)

Proposed by: Andrew Gu, Andrew Lin

Answer: 12 or 13

Solution: We want to minimize |8x〉100|, so x should equal either the floor or the ceiling of 1008 = 12.5.

Note that no other answers receive any points, while both 12 and 13 receive 4 points.

2. [4] Let ABC be a triangle and ω be its circumcircle. The point M is the midpoint of arc BC notcontaining A on ω and D is chosen so that DM is tangent to ω and is on the same side of AM as C.It is given that AM = AC and ∠DMC = 38◦. Find the measure of angle ∠ACB.

Proposed by: Joseph Heerens

Answer: 33◦

Solution: By inscribed angles, we know that ∠BAC = 38◦·2 = 76◦ which means that ∠C = 104◦〉∠B.Since AM = AC, we have ∠ACM = ∠AMC = 90◦ 〉 ∠MAC

2 = 71◦. Once again by inscribed angles,this means that ∠B = 71◦ which gives ∠C = 33◦.

A

B C

DM

3. [4] Let ABC be a triangle and D, E, and F be the midpoints of sides BC, CA, and AB respectively.What is the maximum number of circles which pass through at least 3 of these 6 points?

Proposed by: Andrew Gu

Answer: 17

Solution: All(63

)= 20 triples of points can produce distinct circles aside from the case where the

three points are collinear (BDC, CEA, AFB).

4. [4] Compute the value of√

1053 〉 1043, given that it is a positive integer.

Proposed by: Andrew Gu

Answer: 181

Solution 1: First compute 1053 〉 1043 = 1052 + 105 · 104 + 1042 = 3 · 105 · 104 + 1 = 32761. Notethat 1802 = 32400, so 1812 = 1802 + 2 · 180 + 1 = 32761 as desired.

Solution 2: We have 1053 〉 1043 = 1052 + 105 · 104 + 1042. Thus

104√

3 <√

1053 〉 1043 < 105√

3.

Now estimating gives the bounds 180 < 104√

3 and 105√

3 < 182. So the answer is 181.

5. [5] Alice, Bob, and Charlie roll a 4, 5, and 6-sided die, respectively. What is the probability that anumber comes up exactly twice out of the three rolls?

Proposed by: Andrew Lin

Answer: 1330

Solution 1: There are 4 · 5 · 6 = 120 different ways that the dice can come up. The common numbercan be any of 1, 2, 3, 4, or 5: there are 3 + 4 + 5 = 12 ways for it to be each of 1, 2, 3, or 4, because wepick one of the three people’s rolls to disagree, and there are 3, 4, and 5 ways that roll can come up(for Alice, Bob, and Charlie respectively). Finally, there are 4 ways for Bob and Charlie to both rolla 5 and Alice to roll any number. Thus there are 52 different ways to satisfy the problem condition,and our answer is 52

120 = 1330 .

Solution 2: If Bob rolls the same as Alice, Charlie must roll a different number. Otherwise Charliemust roll the same as either Alice or Bob. So the answer is

1

5· 5

6+

4

5· 2

6=

13

30.

Solution 3: By complementary counting, the answer is

1〉 1

5· 1

6〉 4

5· 4

6=

13

30.

The first and second products correspond to rolling the same number three times and rolling threedistinct numbers, respectively.

6. [5] Two sides of a regular n-gon are extended to meet at a 28◦ angle. What is the smallest possiblevalue for n?

Proposed by: James Lin

Answer: 45

Solution: We note that if we inscribe the n-gon in a circle, then according to the inscribed angletheorem, the angle between two sides is 1

2 times some x〉 y, where x and y are integer multiples of thearc measure of one side of the n-gon. Thus, the angle is equal to 1

2 times an integer multiple of 360n , so

12 · k ·

360n = 28 for some integer k. Simplifying gives 7n = 45k, and since all k are clearly attainable,

the smallest possible value of n is 45.

7. [5] Ana and Banana are rolling a standard six-sided die. Ana rolls the die twice, obtaining a1 and a2,then Banana rolls the die twice, obtaining b1 and b2. After Ana’s two rolls but before Banana’s tworolls, they compute the probability p that a1b1 + a2b2 will be a multiple of 6. What is the probabilitythat p = 1

6?

Proposed by: James Lin

Answer: 23

Solution: If either a1 or a2 is relatively prime to 6, then p = 16 . If one of them is a multiple of 2

but not 6, while the other is a multiple of 3 but not 6, we also have p = 16 . In other words, p = 1

6

if gcd(a1, a2) is coprime to 6, and otherwise p 6= 16 . The probability that p = 1

6 is (32−1)(22−1)62 = 2

3

where q2−1q2 corresponds to the probability that at least one of a1 and a2 is not divisible by q for q = 2, 3.

8. [5] Tessa picks three real numbers x, y, z and computes the values of the eight expressions of the form±x ± y ± z. She notices that the eight values are all distinct, so she writes the expressions down inincreasing order. For example, if x = 2, y = 3, z = 4, then the order she writes them down is

〉x〉 y 〉 z, +x〉 y 〉 z, 〉x+ y 〉 z, 〉x〉 y + z, +x+ y 〉 z, +x〉 y + z, 〉x+ y + z, +x+ y + z.

How many possible orders are there?

Proposed by: Yuan Yao

Answer: 96

Solution: There are 23 = 8 ways to choose the sign for each of x, y, and z. Furthermore, we can order|x|, |y|, and |z| in 3! = 6 different ways. Now assume without loss of generality that 0 < x < y < z.Then there are only two possible orders depending on the sign of x+ y 〉 z:

〉x〉 y 〉 z, +x〉 y 〉 z, 〉x+ y 〉 z, 〉x〉 y + z, x+ y 〉 z, x〉 y + z, 〉x+ y + z, x+ y + z

〉x〉 y 〉 z, +x〉 y 〉 z, 〉x+ y 〉 z, x+ y 〉 z, 〉x〉 y + z, x〉 y + z, 〉x+ y + z, x+ y + z

Thus, the answer is 8 · 6 · 2 = 96.

9. [6] Let P (x) be the monic polynomial with rational coefficients of minimal degree such that 1√2,

1√3, 1√

4, . . . , 1√

1000are roots of P . What is the sum of the coefficients of P?

Proposed by: Carl Joshua Quines

Answer: 116000

Solution: For irrational 1√r, 〉 1√

rmust also be a root of P . Therefore

P (x) =

(x2 〉 1

2

) (x2 〉 1

3

)· · ·(x2 〉 1

1000

)(x+ 1

2

) (x+ 1

3

)· · ·(x+ 1

31

) .

We get the sum of the coefficients of P by setting x = 1, so we use telescoping to get

P (1) =12 ·

23 · · ·

9991000

32 ·

43 · · ·

3231

=1

16000.

10. [6] Jarris is a weighted tetrahedral die with faces F1, F2, F3, F4. He tosses himself onto a table, so thatthe probability he lands on a given face is proportional to the area of that face (i.e. the probability

he lands on face Fi is [Fi][F1]+[F2]+[F3]+[F4]

where [K] is the area of K). Let k be the maximum distance

any part of Jarris is from the table after he rolls himself. Given that Jarris has an inscribed sphere ofradius 3 and circumscribed sphere of radius 10, find the minimum possible value of the expected valueof k.

Proposed by: James Lin

Answer: 12

Solution: Since the maximum distance to the table is just the height, the expected value is equal to4∑

i=1hi[Fi]

4∑i=1

[Fi]

. Let V be the volume of Jarris. Recall that V = 13hi[Fi] for any i, but also V = r

3

(4∑i=1

[Fi]

)where r is the inradius (by decomposing into four tetrahedra with a vertex at the incenter). Therefore

4∑i=1

hi[Fi]

4∑i=1

[Fi]

=12V

3V/r= 4r = 12.

11. [6] Find the number of ordered pairs of positive integers (x, y) with x, y ≤ 2020 such that 3x2 +10xy+3y2 is the power of some prime.

Proposed by: James Lin

Answer: 29

Solution: We can factor as (3x+ y)(x+ 3y). If x ≥ y, we need 3x+yx+3y ∈ {1, 2} to be an integer. So we

get the case where x = y, in which we need both to be a power of 2, or the case x = 5y, in which case weneed y to be a power of 2. This gives us 11+9+9 = 29 solutions, where we account for y = 5x as well.

12. [6] An 11 × 11 grid is labeled with consecutive rows 0, 1, 2, . . . , 10 and columns 0, 1, 2, . . . , 10 so thatit is filled with integers from 1 to 210, inclusive, and the sum of all of the numbers in row n and incolumn n are both divisible by 2n. Find the number of possible distinct grids.

Proposed by: Joseph Heerens

Answer: 21100

Solution: We begin by filling the 10 by 10 grid formed by rows and columns 1 through 10 with anyvalues, which we can do in (210)100 = 21000 ways. Then in column 0, there is at most 1 way to fill in thesquare in row 10, 2 ways for the square in row 9, down to 210 ways in row 0. Similarly, there is 1 way tofill in the square in row 0 and column 10, 2 ways to fill in the square in row 0 and column 9, etc. Overall,the number of ways to fill out the squares in row or column 0 is 21 ·22 ·23 · · · 29 ·210 ·29 ·28 · · · 21 = 2100,so the number of possible distinct grids 21000 · 2100 = 21100.

13. [8] Let 4ABC be a triangle with AB = 7, BC = 1, and CA = 4√

3. The angle trisectors of Cintersect AB at D and E, and lines AC and BC intersect the circumcircle of 4CDE again at X andY , respectively. Find the length of XY .

Proposed by: Hahn Lheem

Answer: 11265

Solution: Let O be the cirumcenter of 4CDE. Observe that 4ABC ∼ 4XY C. Moreover, 4ABCis a right triangle because 12 + (4

√3)2 = 72, so the length XY is just equal to 2r, where r is the

radius of the circumcircle of 4CDE. Since D and E are on the angle trisectors of angle C, we see

that 4ODE, 4XDO, and 4Y EO are equilateral. The length of the altitude from C to AB is 4√3

7 .

The distance from C to XY is XYAB ·

4√3

7 = 2r7 ·

4√3

7 , while the distance between lines XY and AB isr√3

2 . Hence we have

4√

3

7=

2r

7· 4√

3

7+r√

3

2.

Solving for r gives that r = 5665 , so XY = 112

65 .

A B

C

D E

X YO

14. [8] Let ϕ(n) denote the number of positive integers less than or equal to n which are relatively primeto n. Let S be the set of positive integers n such that 2n

ϕ(n) is an integer. Compute the sum

∑n∈S

1

n.

Proposed by: Joseph Heerens, James Lin

Answer: 103

Solution: Let Tn be the set of prime factors of n. Then

2n

φ(n)= 2

∏p∈T

p

p〉 1.

We can check that this is an integer for the following possible sets:

∅, {2}, {3}, {2, 3}, {2, 5}, {2, 3, 7}.

For each set T , the sum of the reciprocals of the positive integers having that set of prime factors is

∏p∈T

( ∞∑m=1

1

pm

)=∏p∈T

1

p〉 1.

Therefore the desired sum is

1 + 1 +1

2+

1

2+

1

4+

1

12=

10

3.

15. [8] You have six blocks in a row, labeled 1 through 6, each with weight 1. Call two blocks x ≤ yconnected when, for all x ≤ z ≤ y, block z has not been removed. While there is still at least one blockremaining, you choose a remaining block uniformly at random and remove it. The cost of this operationis the sum of the weights of the blocks that are connected to the block being removed, including itself.Compute the expected total cost of removing all the blocks.

Proposed by: Benjamin Qi

Answer: 16310

Solution: Note that the total cost is the total number of ordered pairs (x, y) with 1 ≤ x, y ≤ 6 suchthat x and y are connected right before x gets removed.

The probability that blocks x and y are connected just before block x is removed is simply 1|x−y|+1 , since

all of the |x〉 y|+ 1 relevant blocks are equally likely to be removed first. Summing over 1 ≤ x, y ≤ 6,combining terms with the same value of |x〉 y|, we get

2

6+

4

5+

6

4+

8

3+

10

2+ 6 =

163

10.

16. [8] Determine all triplets of real numbers (x, y, z) satisfying the system of equations

x2y + y2z = 1040

x2z + z2y = 260

(x〉 y)(y 〉 z)(z 〉 x) = 〉540.

Proposed by: Krit Boonsiriseth

Answer: (16, 4, 1), (1, 16, 4)

Solution: Call the three equations (1), (2), (3). (1)/(2) gives y = 4z. (3) + (1)〉 (2) gives

(y2 〉 z2)x = 15z2x = 240,

so z2x = 16. Therefore

z(x+ 2z)2 = x2z + z2y + 4z2x =81

5

z(x〉 2z)2 = x2z + z2y 〉 4z2x =49

5

so∣∣∣x+2zx−2z

∣∣∣ = 97 . Thus either x = 16z or x = z

4 .

If x = 16z, then (1) becomes 1024z3 + 16z3 = 1040, so (x, y, z) = (16, 4, 1).

If x = z4 , then (1) becomes 1

4z3 + 16z3 = 1040, so (x, y, z) = (1, 16, 4).

17. [10] Let ABC be a triangle with incircle tangent to the perpendicular bisector of BC. If BC = AE =20, where E is the point where the A-excircle touches BC, then compute the area of 4ABC.

Proposed by: Tristan Shin

Answer: 100√

2

Solution: Let the incircle and BC touch at D, the incircle and perpendicular bisector touch at X, Ybe the point opposite D on the incircle, and M be the midpoint of BC. Recall that A, Y , and E arecollinear by homothety at A. Additionally, we have MD = MX = ME so ∠DXY = ∠DXE = 90◦.Therefore E, X, and Y are collinear. Since MX ⊥ BC, we have ∠AEB = 45◦. The area of ABC is

1

2BC ·AE · sin∠AEB = 100

√2.

MD E

I

Y

B C

A

X

18. [10] A vertex-induced subgraph is a subset of the vertices of a graph together with any edges whoseendpoints are both in this subset.

An undirected graph contains 10 nodes and m edges, with no loops or multiple edges. What is theminimum possible value of m such that this graph must contain a nonempty vertex-induced subgraphwhere all vertices have degree at least 5?

Proposed by: Benjamin Qi

Answer: 31

Solution: Suppose that we want to find the vertex-induced subgraph of maximum size where eachvertex has degree at least 5. To do so, we start with the entire graph and repeatedly remove any vertexwith degree less than 5.

If there are vertices left after this process terminates, then the subgraph induced by these verticesmust have all degrees at least 5. Conversely, if there is a vertex-induced subgraph where all degrees areat least 5, then none of these vertices can be removed during the removing process. Thus, there arevertices remaining after this process if and only if such a vertex-induced subgraph exists.

If the process ends with an empty graph, the largest possible number of edges are removed when thefirst 5 removed vertices all have 4 edges at the time of removal, and the last 5 vertices are all con-nected to each other, resulting in 5×4+4+3+2+1+0 = 30 removed edges. The answer is 30+1 = 31.

19. [10] The Fibonacci numbers are defined by F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2. Thereexist unique positive integers n1, n2, n3, n4, n5, n6 such that

100∑i1=0

100∑i2=0

100∑i3=0

100∑i4=0

100∑i5=0

Fi1+i2+i3+i4+i5 = Fn1〉 5Fn2

+ 10Fn3〉 10Fn4

+ 5Fn5〉 Fn6

.

Find n1 + n2 + n3 + n4 + n5 + n6.

Proposed by: Andrew Gu

Answer: 1545

Solution: We make use of the identity

∑̀i=0

Fi = F`+2 〉 1,

(easily proven by induction) which implies

∑̀i=k

Fi = F`+2 〉 Fk+1.

Applying this several times yields

100∑i1=0

100∑i2=0

100∑i3=0

100∑i4=0

100∑i5=0

Fi1+i2+i3+i4+i5

=

100∑i1=0

100∑i2=0

100∑i3=0

100∑i4=0

(Fi1+i2+i3+i4+102 〉 Fi1+i2+i3+i4+1)

=

100∑i1=0

100∑i2=0

100∑i3=0

(Fi1+i2+i3+204 〉 2Fi1+i2+i3+103 + Fi1+i2+i3+2)

=100∑i1=0

100∑i2=0

(Fi1+i2+306 〉 3Fi1+i2+205 + 3Fi1+i2+104 〉 Fi1+i2+3)

=

100∑i1=0

(Fi1+408 〉 4Fi1+307 + 6Fi1+206 〉 4Fi1+105 + Fi1+4)

=F510 〉 5F409 + 10F308 〉 10F207 + 5F106 〉 F5.

This representation is unique because the Fibonacci terms grow exponentially quickly, so e.g. the F510

term dominates, forcing n1 = 510 and similarly for the other terms. The final answer is

510 + 409 + 308 + 207 + 106 + 5 = 1545.

20. [10] There exist several solutions to the equation

1 +sinx

sin 4x=

sin 3x

sin 2x,

where x is expressed in degrees and 0◦ < x < 180◦. Find the sum of all such solutions.

Proposed by: Benjamin Qi

Answer: 320◦

Solution: We first apply sum-to-product and product-to-sum:

sin 4x+ sinx

sin 4x=

sin 3x

sin 2x

2 sin(2.5x) cos(1.5x) sin(2x) = sin(4x) sin(3x)

Factoring out sin(2x) = 0,sin(2.5x) cos(1.5x) = cos(2x) sin(3x)

Factoring out cos(1.5x) = 0 (which gives us 60◦ as a solution),

sin(2.5x) = 2 cos(2x) sin(1.5x)

sin(2.5x) = sin(3.5x)〉 sin(0.5x)

Convert into complex numbers, we get

(x3.5 〉 x−3.5)〉 (x0.5 〉 x−0.5) = (x2.5 〉 x−2.5)

x7 〉 x6 〉 x4 + x3 + x〉 1 = 0

(x〉 1)(x6 〉 x3 + 1) = 0

We recognize the latter expression as x9+1x3+1 , giving us

x = 0◦, 20◦, 100◦, 140◦, 220◦, 260◦, 340◦.

The sum of the solutions is20◦ + 60◦ + 100◦ + 140◦ = 320◦.

21. [12] We call a positive integer t good if there is a sequence a0, a1, . . . of positive integers satisfyinga0 = 15, a1 = t, and

an−1an+1 = (an 〉 1)(an + 1)

for all positive integers n. Find the sum of all good numbers.

Proposed by: Krit Boonsiriseth

Answer: 296

Solution: By the condition of the problem statement, we have

a2n 〉 an−1an+1 = 1 = a2n−1 〉 an−2an.

This is equivalent toan−2 + anan−1

=an−1 + an+1

an.

Let k = a0+a2a1

. Then we have

an−1 + an+1

an=an−2 + anan−1

=an−3 + an−1

an−2= · · · = a0 + a2

a1= k.

Therefore we have an+1 = kan 〉 an−1 for all n ≥ 1. We know that k is a positive rational numberbecause a0, a1, and a2 are all positive integers. We claim that k must be an integer. Suppose thatk = p

q with gcd(p, q) = 1. Since kan = an−1 + an+1 is always an integer for n ≥ 1, we must have q | anfor all n ≥ 1. This contradicts a22 〉 a1a3 = 1. Conversely, if k is an integer, inductively all ai areintegers.

Now we compute a2 = t2−115 , so k = t2+224

15t is an integer. Therefore 15k 〉 t = 224t is an integer.

Combining with the condition that a2 is an integer limits the possible values of t to 1, 4, 14, 16, 56,224. The values t < 15 all lead to an = 0 for some n whereas t > 15 leads to a good sequence. Thesum of the solutions is

16 + 56 + 224 = 296.

22. [12] Let A be a set of integers such that for each integer m, there exists an integer a ∈ A and positiveinteger n such that an ≡ m (mod 100). What is the smallest possible value of |A|?Proposed by: Daniel Zhu

Answer: 41

Solution: Work in R = Z/100Z ∼= Z/4Z× Z/25Z.

Call an element r ∈ R type (s, t) if s = ν2(r) ≤ 2 and t = ν5(r) ≤ 2. Also, define an element r ∈ R tobe coprime if it is of type (0, 0), powerful if it is of types (0, 2), (2, 0), or (2, 2), and marginal otherwise.

Then, note that if if r ∈ R is marginal, then any power of r is powerful. Therefore all marginal elementsmust be in A.

We claim that all powerful elements are the cube of some marginal element. To show this take apowerful element r. In modulo 4 or 25, if r is a unit, then since 3 is coprime to both the sizes of(Z/4Z)× and (Z/25Z)×, it is the cube of some element. Otherwise, if r is zero then it is the cube of 2or 5, respectively (since this case happens at least once this means that the constructed cube root ismarginal).

We now claim that 4 additional elements are needed to generate the coprime elements. To see this,note that R× ∼= Z/2Z×Z/20Z since there are primitive roots mod 4 and 25. Under this isomorphism,one can show that (1, 1), (1, 2), (1, 4), and (0, 1) generate anything, and that no element in R× hasmore than one of these as a multiple.

To wrap up, note that there are 100 〉 (20 + 1)(2 + 1) = 37 marginal elements, so 41 elements areneeded in total.

23. [12] A function f : A→ A is called idempotent if f(f(x)) = f(x) for all x ∈ A. Let In be the numberof idempotent functions from {1, 2, . . . , n} to itself. Compute

∞∑n=1

Inn!.

Proposed by: Carl Schildkraut

Answer: ee 〉 1

Solution: Let Ak,n denote the number of idempotent functions on a set of size n with k fixed points.We have the formula

Ak,n =

(n

k

)kn−k

for 1 ≤ k ≤ n because there are(nk

)ways to choose the fixed points and all n〉 k remaining elements

must map to fixed points, which can happen in kn−k ways. Hence

∞∑n=1

Inn!

=

∞∑n=1

n∑k=1

Ak,nn!

=

∞∑n=1

n∑k=1

kn−k

k!(n〉 k)!

=

∞∑k=1

1

k!

∞∑n=k

kn−k

(n〉 k)!

=

∞∑k=1

1

k!

∞∑n=0

kn

n!

=

∞∑k=1

1

k!ek

= ee 〉 1.

24. [12] In 4ABC, ω is the circumcircle, I is the incenter and IA is the A-excenter. Let M be the

midpoint of arc B̂AC on ω, and suppose that X, Y are the projections of I onto MIA and IA ontoMI, respectively. If 4XY IA is an equilateral triangle with side length 1, compute the area of 4ABC.

Proposed by: Michael Diao

Answer:√67

Solution 1: Using Fact 5, we know that IIA intersects the circle (ABC) at MA, which is the centerof (IIABCXY ). Let R be the radius of the latter circle. We have R = 1√

3.

We have ∠AIM = ∠Y IIA = ∠Y IX = π3 . Also, ∠IIAM = ∠IMIA by calculating the angles from the

equilateral triangle. Using 90-60-30 triangles, we have:

AI =1

2MI =

1

2IIA = R

AM =

√3

2MI =

√3R

MM2A = AM2 +AM2

A = 7R2

Now, let J and N be the feet of the altitudes from A and B respectively on MMA. Note that as M isan arc midpoint of BC, N is actually the midpoint of BC.

MAJ =AM2

A

MMA=

4√7R

MAN =BM2

A

MMA=

1√7R

Thus JN = 3√7R. Also, we have,

BN2 = MAN ·MN =6

7R2

Now, [ABC] = 12JN ·BC = JN ·BN = 3

√6

7 R2 =√67 .

Solution 2: By Fact 5, we construct the diagram first with 4XY IA as the reference triangle.

MA

IA

X

Y

I

M

A

B CD

Let MA be the circumcenter of4XY IA and let Ω be the circumcircle, which has circumradius R = 1√3.

Then by Fact 5, MA is the midpoint of minor arc dBC, and B,C ∈ Ω. Now we show the followingresult:

Claim. b+ c = 2a.

Proof. Letting D be the intersection of IIA with BC, we have MAD = 12R. Then by the Shooting

Lemma,MAA ·MAD = R2 =⇒ MAA = 2R.

On the other hand, by Ptolemy,

MAB ·AC +MAC ·AB = MAA ·BC =⇒ R · (b+ c) = 2R · a,

whence the result follows.

By the triangle area formula, we have

[ABC] = sr =1

2(a+ b+ c)r =

3

2ar.

Therefore, we are left to compute a and r.

Since MA is the antipode of M on (ABC) we get MB,MC are tangent to Ω; in particular, MBMACis a kite with MB = MC and MAB = MAC. Then by Power of a Point, we get

MB2 = MC2 = MI ·MY = 2R · 3R = 2,

where MI = IIA = 2R and IY = R.

Since ID = DMA, we know that r is the length of the projection from MA to BC. Finally, given ourabove information, we can compute

r =BM2

A

MAM=

BM2A√

BM2A +MB2

=1√21

a = 2rBMA

MB=

2√

2

7.

The area of ABC is then3

2

(2√

2√7

)(1√21

)=

√6

7.

25. [15] Let S be the set of 34 points in four-dimensional space where each coordinate is in {〉1, 0, 1}. LetN be the number of sequences of points P1, P2, . . . , P2020 in S such that PiPi+1 = 2 for all 1 ≤ i ≤ 2020and P1 = (0, 0, 0, 0). (Here P2021 = P1.) Find the largest integer n such that 2n divides N .

Proposed by: James Lin

Answer: 4041

Solution: From (0, 0, 0, 0) we have to go to (±1,±1,±1,±1), and from (1, 1, 1, 1) (or any of the othersimilar points), we have to go to (0, 0, 0, 0) or (〉1, 1, 1, 1) and its cyclic shifts. If ai is the numberof ways to go from (1, 1, 1, 1) to point of the form (±1,±1,±1,±1) in i steps, then we need to findν2(16a2018). To find a recurrence relation for ai, note that to get to some point in (±1,±1,±1,±1), wemust either come from a previous point of the form (±1,±1,±1,±1) or the point (0, 0, 0, 0). In orderto go to one point of the form (±1,±1,±1,±1) through (0, 0, 0, 0) from the point (±1,±1,±1,±1),we have one way of going to the origin and 16 ways to pick which point we go to after the origin.

Additionally, if the previous point we visit is another point of the form (±1,±1,±1,±1) then we have4 possible directions to go in. Therefore the recurrence relation for ai is ai = 4ai−1 + 16ai−2. Solvingthe linear recurrence yields

ai =1√5

(2 + 2√

5)i 〉 1√5

(2〉 2√

5)i = 4iFi+1,

so it suffices to find ν2(F2019). We have Fn ≡ 0, 1, 1, 2, 3, 1 (mod 4) for n ≡ 0, 1, 2, 3, 4, 5 (mod 6), soν2(F2019) = 1, and the answer is 4 + 2 · 2018 + 1 = 4041.

26. [15] Let ABCD be a cyclic quadrilateral, and let segments AC and BD intersect at E. Let W andY be the feet of the altitudes from E to sides DA and BC, respectively, and let X and Z be themidpoints of sides AB and CD, respectively. Given that the area of AED is 9, the area of BEC is 25,and ∠EBC 〉 ∠ECB = 30◦, then compute the area of WXY Z.

Proposed by: James Lin

Answer: 17 + 152

√3

Solution: Reflect E across DA to EW , and across BC to EY . As ABCD is cyclic, 4AED and4BEC are similar. Thus EWAED and EBEY C are similar too.

Now since W is the midpoint of EWE, X is the midpoint of AB, Y is the midpoint of EEY , and Z isthe midpoint of DC, we have that WXY Z is similar to EWAED and EBEY C.

AB

C

D

E

W

X

Y

Z

EW

EY

From the given conditions, we have EW : EY = 3 : 5 and ∠WEY = 150◦. Suppose EW = 3x and

EY = 5x. Then by the law of cosines, we have WY =√

34 + 15√

3x.

Thus, EWE : WY = 6 :√

34 + 15√

3. So by the similarity ratio,

[WXY Z] = [EWAED]

(√34 + 15

√3

6

)2

= 2 · 9 ·

(34 + 15

√3

36

)= 17 +

15

2

√3.

27. [15] Let {ai}i≥0 be a sequence of real numbers defined by

an+1 = a2n 〉1

22020·2n−1

for n ≥ 0. Determine the largest value for a0 such that {ai}i≥0 is bounded.

Proposed by: Joshua Lee

Answer: 1 + 122020

Solution: Let a0 = 1√22020

(t+ 1

t

), with t ≥ 1. (If a0 <

1√22018 then no real t exists, but we ignore

these values because a0 is smaller.) Then, we can prove by induction that

an =1

√22020·2n

(t2

n

+1

t2n

).

For this to be bounded, it is easy to see that we just need

t2n

√22020·2n =

(t

√22020

)2n

to be bounded, since the second term approaches 0. We see that this is is equivalent to t ≤ 22020/2,which means

a0 ≤1

√22020

(√

22020

+

(1√2

)2020)

= 1 +1

22020.

28. [15] Let 4ABC be a triangle inscribed in a unit circle with center O. Let I be the incenter of 4ABC,and let D be the intersection of BC and the angle bisector of ∠BAC. Suppose that the circumcircleof 4ADO intersects BC again at a point E such that E lies on IO. If cosA = 12

13 , find the area of4ABC.

Proposed by: Michael Diao

Answer: 15169

Solution: Consider the following lemma:

Lemma. AD ⊥ EO.

Proof. By the Shooting Lemma, the reflection of the midpoint M of arc BC not containing A overBC lies on (ADO). Hence

]ADE + ]DEO = ]MDC + ]DM ′O = ]MDC + ]M ′MD = 90◦.

This is enough to imply AD ⊥ EO.

A

B C

IO

M

M ′

DE

Thus I is the foot from O onto AD. Now

AI2 + IO2 = AO2.

By Euler’s formula, (r

sin A2

)2

+R2 〉 2Rr = R2.

Hence

r = 2R sin2 A

2.

Thens = a+

r

tan A2

= a+R sinA = 3R sinA

and

[ABC] = rs =

(2R sin2 A

2

)(3R sinA) .

Since R = 1, we get[ABC] = 3 (1〉 cosA) sinA.

Plugging in sinA = 513 and cosA = 12

13 , we get

[ABC] = 3 · 1

13· 5

13=

15

169.

Remark. On the contest, this problem stated that 4ABC is an acute triangle and that sinA = 513 ,

rather than cosA = 1213 . This is erroneous because there is no acute triangle satisfying the conditions

of the problem statement (the given diagram is not accurate). We apologize for the mistake.

29. [18] Let ABCD be a tetrahedron such that its circumscribed sphere of radius R and its inscribedsphere of radius r are concentric. Given that AB = AC = 1 ≤ BC and R = 4r, find BC2.

Proposed by: Michael Ren

Answer: 1 +√

715

Solution: Let O be the common center of the two spheres. Projecting O onto each face of thetetrahedron will divide it into three isosceles triangles. Unfolding the tetrahedron into its net, thereflection of any of these triangles about a side of the tetrahedron will coincide with another one ofthese triangles. Using this property, we can see that each of the faces is broken up into the same threetriangles. It follows that the tetrahedron is isosceles, i.e. AB = CD, AC = BD, and AD = BC.

Let P be the projection of O onto ABC and x = BC. By the Pythagorean Theorem on triangle POA,P has distance

√R2 〉 r2 = r

√15 from A, B, and C. Using the area-circumcenter formula, we compute

[ABC] =AB ·AC ·BC

4PA=

x

4r√

15.

However, by breaking up the volume of the tetrahedron into the four tetrahedra OABC, OABD,OACD, OBCD, we can write [ABC] = V

43 r

, where V = [ABCD]. Comparing these two expressions

for [ABC], we get x = 3√

15V .

Using the formula for the volume of an isosceles tetrahedron (or some manual calculations), we can

compute V = x2√

172 (2〉 x2). Substituting into the previous equation (and taking the solution which

is ≥ 1), we eventually get x2 = 1 +√

715 .

30. [18] Let S = {(x, y) | x > 0, y > 0, x+ y < 200, and x, y ∈ Z}. Find the number of parabolas P withvertex V that satisfy the following conditions:

• P goes through both (100, 100) and at least one point in S,

• V has integer coordinates, and

• P is tangent to the line x+ y = 0 at V .

Proposed by: James Lin

Answer: 264

Solution: We perform the linear transformation (x, y)→ (x〉 y, x+ y), which has the reverse trans-formation (a, b) →

(a+b2 , b−a2

). Then the equivalent problem has a parabola has a vertical axis of

symmetry, goes through A = (0, 200), a point B = (u, v) in

S′ = {(x, y) | x+ y > 0, x > y, y < 200, x, y ∈ Z, and x ≡ y mod 2},

and a new vertex W = (w, 0) on y = 0 with w even. Then(1〉 u

w

)2= v

200 . The only way the RHS

can be the square of a rational number is if uw = v′

10 where v = 2(10〉 v′)2. Since v is even, we can findconditions so that u,w are both even:

v′ ∈ {1, 3, 7, 9} =⇒ (2v′) | u, 20 | wv′ ∈ {2, 4, 6, 8} =⇒ v′ | u, 10 | w

v′ = 5 =⇒ 2 | u, 4 | w

It follows that any parabola that goes through v′ ∈ {3, 7, 9} has a point with v′ = 1, and any parabolathat goes through v′ ∈ {4, 6, 8} has a point with v′ = 2. We then count the following parabolas:

• The number of parabolas going through (2k, 162), where k is a nonzero integer with |2k| < 162.

• The number of parabolas going through (2k, 128) not already counted, where k is a nonzero integerwith |2k| < 128. (Note that this passes through (k, 162).)

• The number of parabolas going through (2k, 50) not already counted, where k is a nonzero integerwith |2k| < 50. (Note that this passes through

(2k5 , 162

), and any overlap must have been counted

in the first case.)

The number of solutions is then

2

(80 +

1

2· 64 +

4

5· 25

)= 264.

31. [18] Anastasia is taking a walk in the plane, starting from (1, 0). Each second, if she is at (x, y), shemoves to one of the points (x〉 1, y), (x+ 1, y), (x, y〉 1), and (x, y+ 1), each with 1

4 probability. Shestops as soon as she hits a point of the form (k, k). What is the probability that k is divisible by 3when she stops?

Proposed by: Michael Ren

Answer: 3−√3

3 or 1〉 1√3

Solution: The key idea is to consider (a+ b, a〉 b), where (a, b) is where Anastasia walks on. Then,the first and second coordinates are independent random walks starting at 1, and we want to find theprobability that the first is divisible by 3 when the second reaches 0 for the first time. Let Cn be thenth Catalan number. The probability that the second random walk first reaches 0 after 2n〉 1 steps is

Cn−1

22n−1 , and the probability that the first is divisible by 3 after 2n〉 1 steps is 122n−1

∑i≡n mod 3

(2n−1i

)(by letting i be the number of 〉1 steps). We then need to compute

∞∑n=1

(Cn−142n−1

∑i≡n mod 3

(2n〉 1

i

)).

By a standard root of unity filter, ∑i≡n mod 3

(2n〉 1

i

)=

4n + 2

6.

Letting

P (x) =2

1 +√

1〉 4x=

∞∑n=0

Cnxn

be the generating function for the Catalan numbers, we find that the answer is

1

6P

(1

4

)+

1

12P

(1

16

)=

1

3+

1

12· 2

1 +√

34

=3〉√

3

3.

32. [18] Find the smallest real constant α such that for all positive integers n and real numbers 0 = y0 <y1 < · · · < yn, the following inequality holds:

α

n∑k=1

(k + 1)3/2√y2k 〉 y2k−1

≥n∑k=1

k2 + 3k + 3

yk.

Proposed by: Andrew Gu

Answer: 16√2

9

Solution: We first prove the following lemma:

Lemma. For positive reals a, b, c, d, the inequality

a3/2

c1/2+b3/2

d1/2≥ (a+ b)3/2

(c+ d)1/2

holds.

Proof. Apply Hölder’s inequality in the form(a3/2

c1/2+b3/2

d1/2

)2

(c+ d) ≥ (a+ b)3.

For k ≥ 2, applying the lemma to a = (k 〉 1)2, b = 8k + 8, c = y2k−1, d = y2k 〉 y2k−1 yields

(k 〉 1)3

yk−1+

(8k + 8)3/2√y2k 〉 y2k−1

≥ (k + 3)3

yk.

We also have the equality(8 · 1 + 8)3/2√

y21 〉 y20=

(1 + 3)3

y1.

Summing the inequality from k = 2 to k = n with the equality yields

n∑k=1

(8k + 8)3/2√y2k 〉 y2k−1

≥n∑k=1

(k + 3)3 〉 k3

yk+n3

yn≥

n∑k=1

9(k2 + 3k + 3)

yk.

Hence the inequality holds for α = 16√2

9 . In the reverse direction, this is sharp when yn = n(n+1)(n+

2)(n+ 3) (so that yk−1 = k−1k+3yk for k = 2, . . . , n) and n→∞.

33. [22] Estimate

N =

∞∏n=1

nn−1.25

.

An estimate of E > 0 will receive b22 min(N/E,E/N)c points.

Proposed by: Sujay Kazi

Answer: ≈ 8282580

Solution: We approximate

lnN =

∞∑n=1

lnn

n5/4

with an integral as ∫ ∞1

lnx

x5/4dx =

(〉4x−1/4 lnx〉 16x−1/4

) ∣∣∣∣∣∞

1

= 16.

Therefore e16 is a good approximation. We can estimate e16 by repeated squaring:

e ≈ 2.72

e2 ≈ 7.4

e4 ≈ 55

e8 ≈ 3000

e16 ≈ 9000000.

The true value of e16 is around 8886111, which is reasonably close to the value of N . Both e16 and9000000 would be worth 20 points.

34. [22] For odd primes p, let f(p) denote the smallest positive integer a for which there does not exist aninteger n satisfying p | n2 〉 a. Estimate N , the sum of f(p)2 over the first 105 odd primes p.

An estimate of E > 0 will receive b22 min(N/E,E/N)3c points.

Proposed by: Michael Ren

Answer: 2266067

Solution: Note that the smallest quadratic nonresidue a is always a prime, because if a = bc withb, c > 1 then one of b and c is also a quadratic nonresidue. We apply the following heuristic: if p1,p2, . . . are the primes in increasing order, then given a “uniform random prime” q, the values of(p1q

),(p2q

), . . . are independent and are 1 with probability 1

2 and 〉1 with probability 12 .

Of course, there is no such thing as a uniform random prime. More rigorously, for any n, the joint dis-

tributions of(p1q

), . . . ,

(pnq

)where q is a uniform random prime less than N converges in distribution

to n independent coin flips between 1 and 〉1 as N → ∞. For ease of explanation, we won’t adoptthis more formal view, but it is possible to make the following argument rigorous by looking at primesq < N and sending N → ∞. Given any n, the residue of q mod n is uniform over the ϕ(n) residuesmod n that are relatively prime to n. By quadratic reciprocity, conditioned on either q ≡ 1 (mod 4) or

q ≡ 3 (mod 4), exactly half of the nonzero residues mod pn satisfy(pnq

)= 1 and exactly half satisfy(

pnq

)= 〉1 for odd pn (the case of pn = 2 is slightly different and one must look mod 8, but the result

is the same). The residue of q mod 8, p2, p3, . . . , pn are independent as these are pairwise relativelyprime, yielding our heuristic.

Thus, we may model our problem of finding the smallest quadratic nonresidue with the followingprocess: independent fair coins are flipped for each prime, and we take the smallest prime that flipped

heads. We can estimate the expected value of f(p)2 as∑∞n=1

p2n2n . Looking at the first few terms gives

us22

2+

32

4+

52

8+

72

16+

112

32+

132

64+

172

128+

192

256+

232

512+

292

1024≈ 22.

The terms after this decay rapidly, so a good approximation is E = 22 ·105, good enough for 20 points.The more inaccurate E = 20 · 105 earns 15 points.

This Python code computes the exact answer:

def smallest_nqr(p):

for a in range(1,p):

if pow(a,(p-1)//2,p)==p-1:

return a

import sympy

print(sum([smallest_nqr(p)**2 for p in sympy.ntheory.primerange(3,sympy.prime(10**5+2))]))

Remark. In 1961, Erdős showed that as N → ∞, the average value of f(p) over odd primes p < Nwill converge to

∑∞n=1

pn2n ≈ 3.675.

35. [22] A collection S of 10000 points is formed by picking each point uniformly at random inside a circleof radius 1. Let N be the expected number of points of S which are vertices of the convex hull of theS. (The convex hull is the smallest convex polygon containing every point of S.) Estimate N .

An estimate of E > 0 will earn max(b22〉 |E 〉N |c, 0) points.

Proposed by: Shengtong Zhang

Answer: ≈ 72.8

Solution: Here is C++ code by Benjamin Qi to estimate the answer via simulation. It is known thatthe expected number of vertices of the convex hull of n points chosen uniformly at random inside acircle is O(n1/3). See “On the Expected Complexity of Random Convex Hulls” by Har-Peled.

36. [22] A snake of length k is an animal which occupies an ordered k-tuple (s1, . . . , sk) of cells in a n× ngrid of square unit cells. These cells must be pairwise distinct, and si and si+1 must share a side fori = 1, . . . , k 〉 1. If the snake is currently occupying (s1, . . . , sk) and s is an unoccupied cell sharing aside with s1, the snake can move to occupy (s, s1, . . . , sk−1) instead.

Initially, a snake of length 4 is in the grid {1, 2, . . . , 30}2 occupying the positions (1, 1), (1, 2), (1, 3), (1, 4)with (1, 1) as its head. The snake repeatedly makes a move uniformly at random among moves it canlegally make. Estimate N , the expected number of moves the snake makes before it has no legal movesremaining.

An estimate of E > 0 will earn b22 min(N/E,E/N)4c points.

Proposed by: Andrew Gu

Answer: ≈ 4571.8706930

Solution: Let n = 30. The snake can get stuck in only 8 positions, while the total number of positions

is about n2 × 4× 3× 3 = 36n2. We can estimate the answer as 36n2

8 = 4050, which is good enough for13 points.

Let’s try to compute the answer as precisely as possible. For each head position (a, b) and tail orien-tation c ∈ [0, 36), let x = 36(na+ b) + c be an integer denoting the current state of the snake. Let Exby the expected number of moves the snake makes if it starts at state x. If from state x the snake cantransition to any of states y1, y2, . . . , yk, then add an equation of the form

Ex 〉1

k

k∑i=1

Eyi = 1.

Otherwise, if there are no transitions out of state x then set Ex = 0.

It suffices to solve a system of 36n2 linear equations for E0, E1, . . . , E36n2−1. Then the answer willequal Ei, where i corresponds to the state described in the problem statement. Naively, using Gaussianelimination would require about (36n2)3 ≈ 3.4 · 1013 operations, which is too slow. Also, it will requiretoo much memory to store (36n2)2 real numbers at once.

We can use the observation that initially, the maximum difference between any two indices withinthe same equation is at most ≈ 72n, so Gaussian elimination only needs to perform approximately(36n2) · (72n)2 ≈ 1.5 · 1011 operations. Furthermore, we’ll only need to store ≈ (36n2) · (72n) realnumbers at a time. Benjamin Qi’s solution ends up finishing in less than two minutes for n = 30 (C++code).

Here are some more examples of problems in which Gaussian elimination can be sped up by exploitingsome sort of special structure:

• https://codeforces.com/contest/963/problem/E

• https://codeforces.com/gym/102501/problem/E

HMMO 2020November 14–21, 2020

General Round

1. In the Cartesian plane, a line segment with midpoint (2020, 11) has one endpoint at (a, 0) and theother endpoint on the line y = x. Compute a.

2. Let T be a trapezoid with two right angles and side lengths 4, 4, 5, and√

17. Two line segments aredrawn, connecting the midpoints of opposite sides of T and dividing T into 4 regions. If the differencebetween the areas of the largest and smallest of these regions is d, compute 240d.

3. Jody has 6 distinguishable balls and 6 distinguishable sticks, all of the same length. How many ways arethere to use the sticks to connect the balls so that two disjoint non-interlocking triangles are formed?Consider rotations and reflections of the same arrangement to be indistinguishable.

4. Nine fair coins are flipped independently and placed in the cells of a 3 by 3 square grid. Let p be theprobability that no row has all its coins showing heads and no column has all its coins showing tails.If p = a

b for relatively prime positive integers a and b, compute 100a + b.

5. Compute the sum of all positive integers a ≤ 26 for which there exist integers b and c such thata + 23b + 15c− 2 and 2a + 5b + 14c− 8 are both multiples of 26.

6. A sphere is centered at a point with integer coordinates and passes through the three points (2, 0, 0),(0, 4, 0), (0, 0, 6), but not the origin (0, 0, 0). If r is the smallest possible radius of the sphere, computer2.

7. In triangle ABC with AB = 8 and AC = 10, the incenter I is reflected across side AB to point X andacross side AC to point Y . Given that segment XY bisects AI, compute BC2. (The incenter I is thecenter of the inscribed circle of triangle ABC.)

8. A bar of chocolate is made of 10 distinguishable triangles as shown below:

How many ways are there to divide the bar, along the edges of the triangles, into two or more contiguouspieces?

9. In the Cartesian plane, a perfectly reflective semicircular room is bounded by the upper half of theunit circle centered at (0, 0) and the line segment from (−1, 0) to (1, 0). David stands at the point(−1, 0) and shines a flashlight into the room at an angle of 46◦ above the horizontal. How many timesdoes the light beam reflect off the walls before coming back to David at (−1, 0) for the first time?

10. A sequence of positive integers a1, a2, a3, . . . satisfies

an+1 = n⌊ann

⌋+ 1

for all positive integers n. If a30 = 30, how many possible values can a1 take? (For a real number x,bxc denotes the largest integer that is not greater than x.)

HMMO 2020November 14–21, 2020

General Round

1. In the Cartesian plane, a line segment with midpoint (2020, 11) has one endpoint at (a, 0) and theother endpoint on the line y = x. Compute a.

Proposed by: Lingyi Qiu

Answer: 4018

Solution: Let the other endpoint be (t, t). The midpoint of (a, 0) and (t, t) is(a+t2 , t

2

). So, we know

that a+t2 = 2020 and t

2 = 11. The second equation yields t = 22. Substituting this into the first yieldsa = 2 · 2020− 22 = 4018.

2. Let T be a trapezoid with two right angles and side lengths 4, 4, 5, and√

17. Two line segments aredrawn, connecting the midpoints of opposite sides of T and dividing T into 4 regions. If the differencebetween the areas of the largest and smallest of these regions is d, compute 240d.

Proposed by: Shengtong Zhang

Answer: 120

Solution:

A B

CD

By checking all the possibilities, one can show that T has height 4 and base lengths 4 and 5. Orient Tso that the shorter base is on the top.

Then, the length of the cut parallel to the bases is 4+52 = 9

2 . Thus, the top two pieces are trapezoidswith height 2 and base lengths 2 and 9

4 , while the bottom two pieces are trapezoids with height 2 andbase lengths 9

4 and 52 . Thus, using the area formula for a trapezoid, the difference between the largest

and smallest areas is

d =

(52 + 9

4 −94 − 2

)· 2

2=

1

2.

3. Jody has 6 distinguishable balls and 6 distinguishable sticks, all of the same length. How many ways arethere to use the sticks to connect the balls so that two disjoint non-interlocking triangles are formed?Consider rotations and reflections of the same arrangement to be indistinguishable.

Proposed by: Daniel Zhu

Answer: 7200

Solution 1: For two disjoint triangles to be formed, three of the balls must be connected into a triangleby three of the sticks, and the three remaining balls must be connected by the three remaining sticks.

There are(63

)ways to pick the 3 balls for the first triangle. Note that once we choose the 3 balls for

the first triangle, the remaining 3 balls must form the vertices of the second triangle.

Now that we have determined the vertices of each triangle, we can assign the 6 sticks to the 6 totaledges in the two triangles. Because any ordering of the 6 sticks works, there are 6! = 720 total waysto assign the sticks as edges.

Finally, because the order of the two triangles doesn’t matter (i.e. our initial choice of 3 balls couldhave been used for the second triangle), we must divide by 2 to correct for overcounting. Hence thefinal answer is

(63

)· 6!/2 = 7200.

Solution 2: First, we ignore all the symmetries in the problem. There are then 6! ways to arrangethe balls and 6! ways to arrange the sticks. However, each triangle can be rotated or reflected, so wehave overcounted by a factor of 62. Moreover, the triangles can be swapped, so we must also divide by2. Thus the answer is

(6!)2

62 · 2=

1202

2= 7200.

4. Nine fair coins are flipped independently and placed in the cells of a 3 by 3 square grid. Let p be theprobability that no row has all its coins showing heads and no column has all its coins showing tails.If p = a

b for relatively prime positive integers a and b, compute 100a+ b.

Proposed by: Daniel Zhu

Answer: 8956

Solution: Consider the probability of the complement. It is impossible for some row to have all headsand some column to have tails, since every row intersects every column. Let q be the probability thatsome row has all heads. By symmetry, q is also the probability that some column has all tails. We canthen conclude that p = 1− 2q.

The probability that a given row does not have all heads is 78 . So, the probability that none of the

three rows have all heads is(78

)3, implying that q = 1− 343

512 = 169512 . Thus p = 1− 169

256 = 87256 .

5. Compute the sum of all positive integers a ≤ 26 for which there exist integers b and c such thata+ 23b+ 15c− 2 and 2a+ 5b+ 14c− 8 are both multiples of 26.

Proposed by: David Vulakh

Answer: 31

Solution: Assume b and c exist. Considering the two values modulo 13, we find{a+ 10b+ 2c ≡ 2 (mod 13)

2a+ 5b+ c ≡ 8 (mod 13).

Subtracting twice the second equation from the first, we get −3a ≡ −14 (mod 13). So, we have a ≡ 9(mod 13). Therefore we must either have a = 9 or a = 22.

Moreover, both a = 9 and a = 22 yield solutions with b = 0 and c = 3, 16, depending on the value ofa. Thus the answer is 9 + 22 = 31.

6. A sphere is centered at a point with integer coordinates and passes through the three points (2, 0, 0),(0, 4, 0), (0, 0, 6), but not the origin (0, 0, 0). If r is the smallest possible radius of the sphere, computer2.

Proposed by: James Lin

Answer: 51

Solution: Let (x, y, z) be the center of the sphere. By the given condition, we have

(x− 2)2 + y2 + z2 = x2 + (y − 4)2 + z2 = x2 + y2 + (z − 6)2.

Subtracting x2 + y2 + z2 yields

x2 − (x− 2)2 = y2 − (y − 4)2 = z2 − (z − 6)2,

or4(x− 1) = 8(y − 2) = 12(z − 3).

Therefore (x− 1, y − 2, z − 3) must be (6t, 3t, 2t) for some integer t. Checking small values of t yieldsthat the possibilities for (x, y, z) closest to (2, 0, 0) are (−5,−1, 1), (1, 2, 3), and (7, 5, 5). The secondyields a sphere that passes through the origin and is thus forbidden. The other two yield r2 = 51 andr2 = 75, so 51 is the answer.

7. In triangle ABC with AB = 8 and AC = 10, the incenter I is reflected across side AB to point X andacross side AC to point Y . Given that segment XY bisects AI, compute BC2. (The incenter I is thecenter of the inscribed circle of triangle ABC.)

Proposed by: Carl Schildkraut

Answer: 84

Solution 1:

A

E F

I

X Y

P

Q

Let E,F be the tangency points of the incircle to sides AC,AB, respectively. Due to symmetry aroundline AI, AXIY is a rhombus. Therefore

∠XAI = 2∠EAI = 2(90◦ − ∠EIA) = 180◦ − 2∠XAI,

which implies that 60◦ = ∠XAI = 2∠EAI = ∠BAC. By the law of cosines,

BC2 = 82 + 102 − 2 · 8 · 10 · cos 60◦ = 84.

Solution 2: Define points as above and additionally let P and Q be the intersections of AI with EFand XY , respectively. Since IX = 2IE and IY = 2IF , 4IEF ∼ 4IXY with ratio 2, implying thatIP = 1

2IQ = 14IA.

Let θ = ∠EAI = ∠IEP . Then IPIA = IP

IEIEIA = sin2 θ, implying that sin θ = 1/2 and θ = 30◦. From

here, proceed as in solution 1.

8. A bar of chocolate is made of 10 distinguishable triangles as shown below:

How many ways are there to divide the bar, along the edges of the triangles, into two or more contiguouspieces?

Proposed by: Steven Noah Raphael

Answer: 1689

Solution: Every way to divide the bar can be described as a nonempty set of edges to break, withthe condition that every endpoint of a broken edge is either on the boundary of the bar or connects toanother broken edge.

Let the center edge have endpoints X and Y . We do casework on whether the center edge is broken.

If the center edge is broken, then we just need some other edge connecting to X to be broken, andsome other edge connecting to Y to be broken. We have 25 choices for the edges connecting to X,of which 1 fails. Similarly, we have 25 − 1 valid choices for the edges connecting to Y . This yields(25 − 1)2 = 961 possibilities.

If the center edge is not broken, then the only forbidden arrangements are those with exactly onebroken edge at X or those with exactly one broken edge at Y . Looking at just the edges connectingto X, we have 5 cases with exactly one broken edge. Thus, there are 25 − 5 = 27 ways to break theedges connecting to X. Similarly there are 27 valid choices for the edges connecting to Y . This yields272 − 1 = 728 cases, once we subtract the situation where no edges are broken.

The final answer is 961 + 728 = 1689.

9. In the Cartesian plane, a perfectly reflective semicircular room is bounded by the upper half of theunit circle centered at (0, 0) and the line segment from (−1, 0) to (1, 0). David stands at the point(−1, 0) and shines a flashlight into the room at an angle of 46◦ above the horizontal. How many timesdoes the light beam reflect off the walls before coming back to David at (−1, 0) for the first time?

Proposed by: Kevin Tong

Answer: 65

Solution: Note that when the beam reflects off the x-axis, we can reflect the entire room across thex-axis instead. Therefore, the number of times the beam reflects off a circular wall in our semicircularroom is equal to the number of times the beam reflects off a circular wall in a room bounded by theunit circle centered at (0, 0). Furthermore, the number of times the beam reflects off the x-axis wall inour semicircular room is equal to the number of times the beam crosses the x-axis in the room boundedby the unit circle. We will count each of these separately.

We first find the number of times the beam reflects off a circular wall. Note that the path of thebeam is made up of a series of chords of equal length within the unit circle, each chord connectingthe points from two consecutive reflections. Through simple angle chasing, we find that the anglesubtended by each chord is 180− 2 · 46 = 88◦. Therefore, the nth point of reflection in the unit circleis (− cos(88n), sin(88n)). The beam returns to (−1, 0) when

88n ≡ 0 (mod 360) ⇐⇒ 11n ≡ 0 (mod 45)→ n = 45

but since we’re looking for the number of time the beam is reflected before it comes back to David, weonly count 45− 1 = 44 of these reflections.

Next, we consider the number of times the beam is reflected off the x-axis. This is simply the numberof times the beam crosses the x-axis in the unit circle room before returning to David, which happensevery 180◦ around the circle. Thus, we have 88·45

180 −1 = 21 reflections off the x-axis, where we subtract1 to remove the instance when the beam returns to (−1, 0). Thus, the total number of reflections is44 + 21 = 65.

10. A sequence of positive integers a1, a2, a3, . . . satisfies

an+1 = n⌊ann

⌋+ 1

for all positive integers n. If a30 = 30, how many possible values can a1 take? (For a real number x,bxc denotes the largest integer that is not greater than x.)

Proposed by: Carl Joshua Quines

Answer: 274

Solution: It is straightforward to show that if a1 = 1, then an = n for all n. Since an+1 is anincreasing function in an, it follows that the set of possible a1 is of the form {1, 2, . . . ,m} for some m,which will be the answer to the problem.

Consider the sequence bn = an+1 − 1, which has the recurrence

bn+1 = n

⌊bn + 1

n

⌋.

It has the property that bn is divisible by n. Rearranging the recurrence, we see that

bn+1

n+ 1≤ bn + 1

n+ 1<

bn+1

n+ 1+ 1,

and as the bi are integers, we get bn+1 − 1 ≤ bn < bn+1 + n. For n ≥ 2, this means that the largestpossible value of bn (call this b∗n) is the smallest multiple of n which is at least bn+1. Also, sinceb1 = b0 + 1, we find b∗0 = b∗1 − 1, meaning that the largest value for a1 is b∗1, and thus the answer is b∗1.

We have now derived a procedure for deriving b∗1 from b∗29 = 29. To speed up the computation, letcn = b∗n/n. Then, since

b∗n = n

⌈b∗n+1

n

⌉,

we find

cn =

⌈n+ 1

ncn+1

⌉= cn+1 +

⌈cn+1

n

⌉.

We now start from c29 = 1 and wish to find c1.

Applying the recurrence, we find c28 = 2, c27 = 3, and so on until we reach c15 = 15. Then, dcn+1/nebecomes greater than 1 and we find c14 = 17, c13 = 19, and so on until c11 = 23. The rest can be donemanually, with c10 = 26, c9 = 29, c8 = 33, c7 = 38, c6 = 45, c5 = 54, c4 = 68, c3 = 91, c2 = 137, andc1 = 274. The last few steps may be easier to perform by converting back into the b∗n.

HMMO 2020November 14–21, 2020

Theme Round

1. Chelsea goes to La Verde’s at MIT and buys 100 coconuts, each weighing 4 pounds, and 100 honeydews, eachweighing 5 pounds. She wants to distribute them among n bags, so that each bag contains at most 13 pounds offruit. What is the minimum n for which this is possible?

2. In the future, MIT has attracted so many students that its buildings have become skyscrapers. Ben and Jerrydecide to go ziplining together. Ben starts at the top of the Green Building, and ziplines to the bottom of the StataCenter. After waiting a seconds, Jerry starts at the top of the Stata Center, and ziplines to the bottom of the GreenBuilding. The Green Building is 160 meters tall, the Stata Center is 90 meters tall, and the two buildings are 120meters apart. Furthermore, both zipline at 10 meters per second. Given that Ben and Jerry meet at the pointwhere the two ziplines cross, compute 100a.

3. Harvard has recently built a new house for its students consisting of n levels, where the kth level from the top canbe modeled as a 1-meter-tall cylinder with radius k meters. Given that the area of all the lateral surfaces (i.e. thesurfaces of the external vertical walls) of the building is 35 percent of the total surface area of the building (includingthe bottom), compute n.

4. Points G and N are chosen on the interiors of sides ED and DO of unit square DOME, so that pentagon GNOME

has only two distinct side lengths. The sum of all possible areas of quadrilateral NOME can be expressed as a−b√c

d ,where a, b, c, d are positive integers such that gcd(a, b, d) = 1 and c is square-free (i.e. no perfect square greater than1 divides c). Compute 1000a + 100b + 10c + d.

5. The classrooms at MIT are each identified with a positive integer (with no leading zeroes). One day, as PresidentReif walks down the Infinite Corridor, he notices that a digit zero on a room sign has fallen off. Let N be theoriginal number of the room, and let M be the room number as shown on the sign. The smallest interval containingall possible values of M

N can be expressed as [ab ,cd ) where a, b, c, d are positive integers with gcd(a, b) = gcd(c, d) = 1.

Compute 1000a + 100b + 10c + d.

6. The elevator buttons in Harvard’s Science Center form a 3× 2 grid of identical buttons, and each button lights upwhen pressed. One day, a student is in the elevator when all the other lights in the elevator malfunction, so thatonly the buttons which are lit can be seen, but one cannot see which floors they correspond to. Given that at leastone of the buttons is lit, how many distinct arrangements can the student observe? (For example, if only one buttonis lit, then the student will observe the same arrangement regardless of which button it is.)

7. While waiting for their food at a restaurant in Harvard Square, Ana and Banana draw 3 squares �1,�2,�3 on oneof their napkins. Starting with Ana, they take turns filling in the squares with integers from the set {1, 2, 3, 4, 5}such that no integer is used more than once. Ana’s goal is to minimize the minimum value M that the polynomiala1x

2 + a2x + a3 attains over all real x, where a1, a2, a3 are the integers written in �1,�2,�3 respectively. Bananaaims to maximize M . Assuming both play optimally, compute the final value of 100a1 + 10a2 + a3.

8. After viewing the John Harvard statue, a group of tourists decides to estimate the distances of nearby locationson a map by drawing a circle, centered at the statue, of radius

√n inches for each integer 2020 ≤ n ≤ 10000, so

that they draw 7981 circles altogether. Given that, on the map, the Johnston Gate is 10-inch line segment which isentirely contained between the smallest and the largest circles, what is the minimum number of points on this linesegment which lie on one of the drawn circles? (The endpoint of a segment is considered to be on the segment.)

9. While waiting for their next class on Killian Court, Alesha and Belinda both write the same sequence S on a piece ofpaper, where S is a 2020-term strictly increasing geometric sequence with an integer common ratio r. Every second,Alesha erases the two smallest terms on her paper and replaces them with their geometric mean, while Belindaerases the two largest terms in her paper and replaces them with their geometric mean. They continue this processuntil Alesha is left with a single value A and Belinda is left with a single value B. Let r0 be the minimal value of rsuch that A

B is an integer. If d is the number of positive factors of r0, what is the closest integer to log2 d?

10. Sean enters a classroom in the Memorial Hall and sees a 1 followed by 2020 0’s on the blackboard. As he is earlyfor class, he decides to go through the digits from right to left and independently erase the nth digit from the leftwith probability n−1

n . (In particular, the 1 is never erased.) Compute the expected value of the number formedfrom the remaining digits when viewed as a base-3 number. (For example, if the remaining number on the board is1000, then its value is 27.)

HMMO 2020November 14–21, 2020

Theme Round

1. Chelsea goes to La Verde’s at MIT and buys 100 coconuts, each weighing 4 pounds, and 100 honeydews,each weighing 5 pounds. She wants to distribute them among n bags, so that each bag contains atmost 13 pounds of fruit. What is the minimum n for which this is possible?

Proposed by: Daniel Zhu

Answer: 75

Solution: The answer is n = 75, given by 50 bags containing one honeydew and two coconuts (13pounds), and 25 bags containing two honeydews (10 pounds).

To show that this is optimal, assign each coconut 1 point and each honeydew 2 points, so that 300points worth of fruit are bought in total. Then, we claim that each bag can contain at most 4 pointsof fruit, thus requiring n ≥ 300/4 = 75. To see this, note that each bag containing greater than 4points must contain either five coconuts (20 pounds), three coconuts and a honeydew (17 pounds), onecoconut and two honeydews (14 pounds), or three honeydews (15 pounds).

2. In the future, MIT has attracted so many students that its buildings have become skyscrapers. Benand Jerry decide to go ziplining together. Ben starts at the top of the Green Building, and ziplines tothe bottom of the Stata Center. After waiting a seconds, Jerry starts at the top of the Stata Center,and ziplines to the bottom of the Green Building. The Green Building is 160 meters tall, the StataCenter is 90 meters tall, and the two buildings are 120 meters apart. Furthermore, both zipline at 10meters per second. Given that Ben and Jerry meet at the point where the two ziplines cross, compute100a.

Proposed by: Esha Bhatia

Answer: 740

Solution: Define the following lengths:

Green

Statax y

z

u

v

Note that due to all the 3-4-5 triangles, we find xz = z

y = 43 , so 120 = x+ y = 25

12z. Then,

u =5

3x =

20

9z =

16

15120 = 128,

while

v =5

4y =

15

16z =

9

20120 = 54.

Thus u− v = 74, implying that a = 7.4.

3. Harvard has recently built a new house for its students consisting of n levels, where the kth level fromthe top can be modeled as a 1-meter-tall cylinder with radius k meters. Given that the area of all thelateral surfaces (i.e. the surfaces of the external vertical walls) of the building is 35 percent of the totalsurface area of the building (including the bottom), compute n.

Proposed by: Daniel Zhu

Answer: 13

Solution: The kth layer contributes a lateral surface area of 2kπ, so the total lateral surface area is

2(1 + 2 + · · ·+ n)π = n(n+ 1)π.

On the other hand, the vertical surface area is 2n2π (No need to sum layers, just look at the buildingfrom above and from below). Therefore,

n+ 1 =7

20(3n+ 1),

and n = 13.

4. Points G and N are chosen on the interiors of sides ED and DO of unit square DOME, so thatpentagon GNOME has only two distinct side lengths. The sum of all possible areas of quadrilateral

NOME can be expressed as a−b√c

d , where a, b, c, d are positive integers such that gcd(a, b, d) = 1 andc is square-free (i.e. no perfect square greater than 1 divides c). Compute 1000a+ 100b+ 10c+ d.

Proposed by: Andrew Lin

Answer: 10324

Solution:

D O

ME

G

N x

x

Since MO = ME = 1, but ON and GE are both less than 1, we must have either ON = NG = GE = x(call this case 1) or ON = GE = x,NG = 1 (call this case 2).

Either way, the area of NOME (a trapezoid) is 1+x2 , and triangle NGT is a 45-45-90 triangle. In case

1, we have 1 = ON +NT = x(

1 +√22

), so x = 2−

√2 and the area of the trapezoid is 3−

√2

2 . In case

2, we have 1 = ON +NT = x+√22 , which yields an area of 4−

√2

4 as x = 2−√2

2 . The sum of these two

answers is 10−3√2

4 .

5. The classrooms at MIT are each identified with a positive integer (with no leading zeroes). One day, asPresident Reif walks down the Infinite Corridor, he notices that a digit zero on a room sign has fallenoff. Let N be the original number of the room, and let M be the room number as shown on the sign.

The smallest interval containing all possible values of MN can be expressed as [ab ,

cd ) where a, b, c, d are

positive integers with gcd(a, b) = gcd(c, d) = 1. Compute 1000a+ 100b+ 10c+ d.

Proposed by: Andrew Lin

Answer: 2031

Solution: Let A represent the portion of N to the right of the deleted zero, and B represent the restof N . For example, if the unique zero in N = 12034 is removed, then A = 34 and B = 12000. Then,MN = A+B/10

A+B = 1− 910

BN .

The maximum value for B/N is 1, which is achieved when A = 0. Also, if the 0 removed is in the 10k’splace (k = 2 in the example above), we find that A < 10k and B ≥ 10k+1, meaning that A/B < 1/10and thus B/N > 10/11. Also, B/N can get arbitrarily close to 10/11 via a number like 1099 . . . 9.

Therefore the fraction MN achieves a minimum at 1

10 and always stays below 211 , though it can get

arbitrarily close. The desired interval is then [ 110 ,

211 ).

6. The elevator buttons in Harvard’s Science Center form a 3 × 2 grid of identical buttons, and eachbutton lights up when pressed. One day, a student is in the elevator when all the other lights in theelevator malfunction, so that only the buttons which are lit can be seen, but one cannot see whichfloors they correspond to. Given that at least one of the buttons is lit, how many distinct arrangementscan the student observe? (For example, if only one button is lit, then the student will observe the samearrangement regardless of which button it is.)

Proposed by: Sheldon Kieren Tan

Answer: 44

Solution 1: We first note that there are 26− 1 = 63 possibilities for lights in total. We now count thenumber of duplicates we need to subtract by casework on the number of buttons lit. To do this, we docasework on the size of the minimal “bounding box” of the lights:

• If the bounding box is 1× 1, the only arrangement up to translation is a solitary light, which canbe translated 6 ways. This means we must subtract 5.

• If the bounding box is 2× 1, there is 1 arrangement and 4 translations, so we must subtract 3.

• If the bounding box is 1× 2, there is 1 arrangement and 3 translations, so we must subtract 2.

• If the bounding box is 3× 1, there are 2 arrangements and 2 translations, so we must subtract 2.

• If the bounding box is 2 × 2, there are 2 arrangements with 2 lights, 4 with 3 lights, and 1 with4 lights—7 in total. Since there are two translations, we must subtract 7.

The final answer is 63− 5− 3− 2− 2− 7 = 44.

Solution 2: We may also count duplicates by doing casework on buttons lit:

• 1 buttons lit: There are 6 arrangements but all are the same, so we need to subtract 5 duplicatesin this case.

• 2 buttons lit: There are 4 indistinguishable ways for the buttons to be vertically adjacent, 3 tobe horizontally adjacent, 2 ways for the buttons to be diagonally adjacent for each of 2 directionsof diagonals, and 2 for when the lights are in the same vertical line but not adjacent. Since weneed to count each of these cases only once, the number of duplicates we need to subtract is 3 (2vertically adjacent), 2 (2 horizontally adjacent), 2 × 1 (2 diagonally adjacent), and 1 (2 in samevertical line but not adjacent) for a total of 8 duplicates.

• 3 buttons lit: There are 2 indistinguishable ways for all the buttons in a column to be lit and 2ways for the buttons to be lit in the shape of an L, given the rotation of the L. Thus, the numberof duplicates we need to subtract is 1 (1 column), 1×4 (rotations of L), for a total of 5 duplicates.

• 4 buttons lit: There are 2 indistinguishable ways for the lights to be arranged in a square (andno other duplicates), so we need to subtract 1 duplicate in this case.

• When there are 5 or 6 buttons lit, all of the arrangements of lights are distinct, so we do notsubtract any duplicates for these cases.

Thus, the total number of arrangements is 64− (1 + 5 + 8 + 5 + 1) = 44.

7. While waiting for their food at a restaurant in Harvard Square, Ana and Banana draw 3 squares�1,�2,�3 on one of their napkins. Starting with Ana, they take turns filling in the squares withintegers from the set {1, 2, 3, 4, 5} such that no integer is used more than once. Ana’s goal is tominimize the minimum value M that the polynomial a1x

2 + a2x + a3 attains over all real x, wherea1, a2, a3 are the integers written in �1,�2,�3 respectively. Banana aims to maximize M . Assumingboth play optimally, compute the final value of 100a1 + 10a2 + a3.

Proposed by: Sheldon Kieren Tan

Answer: 451

Solution: Relabel a1, a2, a3 as a, b, c. This is minimized at x = −b2a , so M = c− b2

4a .

If in the end a = 5 or b ∈ {1, 2}, then b2

4a ≤ 1 and M ≥ 0. The only way for Ana to block this is to setb = 5, which will be optimal if we show that it allows Ana to force M < 0, which we will now do. Atthis point, Banana has two choices:

• If Banana fixes a value of a, Ana’s best move is to pick c = 1, or c = 2 if it has not already beenused. The latter case yields M < −1, while the optimal move in the latter case (a = 4) yieldsM = 1− 25

16 > −1.

• If Banana fixes a value of c, then if that a value is not 1 Ana can put a = 1, yielding M ≤ 4− 254 <

−1. On the other hand, if Banana fixes c = 1 then Ana’s best move is to put a = 2, yieldingM = 1− 25

8 < −1.

Thus Banana’s best move is to set a = 4, eliciting a response of c = 1. Since 1− 2516 < 0, this validates

our earlier claim that b = 5 was the best first move.

8. After viewing the John Harvard statue, a group of tourists decides to estimate the distances of nearbylocations on a map by drawing a circle, centered at the statue, of radius

√n inches for each integer

2020 ≤ n ≤ 10000, so that they draw 7981 circles altogether. Given that, on the map, the JohnstonGate is 10-inch line segment which is entirely contained between the smallest and the largest circles,what is the minimum number of points on this line segment which lie on one of the drawn circles?(The endpoint of a segment is considered to be on the segment.)

Proposed by: Daniel Zhu

Answer: 49

Solution: Consider a coordinate system on any line ` where 0 is placed at the foot from (0, 0)to `. Then, by the Pythagorean theorem, a point (x, y) on ` is assigned a coordinate u for whichx2 + y2 = u2 + a for some fixed a (dependent only on `). Consider this assignment of coordinates forour segment.

First, suppose that along the line segment u never changes sign; without loss of generality, assumeit is positive. Then, if u0 is the minimum value of u, the length of the interval covered by u2 is(u0 + 10)2 − u20 = 100 + 20u0 ≥ 100, meaning that at least 100 points lie on the given circles.

Now suppose that u is positive on a length of k and negative on a length of 10 − k. Then, it mustintersect the circles at least bk2c + b(10 − k)2c points, which can be achieved for any k by settinga = 2020 + ε for very small ε.

To minimize this quantity note that k2 +(10−k)2 ≥ 50, so bk2c+b(10−k)2c > k2 +(10−k)2−2 ≥ 48,proving the bound. For a construction, set k = 4.99999.

9. While waiting for their next class on Killian Court, Alesha and Belinda both write the same sequenceS on a piece of paper, where S is a 2020-term strictly increasing geometric sequence with an integercommon ratio r. Every second, Alesha erases the two smallest terms on her paper and replaces themwith their geometric mean, while Belinda erases the two largest terms in her paper and replaces themwith their geometric mean. They continue this process until Alesha is left with a single value A andBelinda is left with a single value B. Let r0 be the minimal value of r such that A

B is an integer. If dis the number of positive factors of r0, what is the closest integer to log2 d?

Proposed by: Hahn Lheem

Answer: 2018

Solution: Because we only care about when the ratio of A to B is an integer, the value of the firstterm in S does not matter. Let the initial term in S be 1. Then, we can write S as 1, r, r2, . . . , r2019.Because all terms are in terms of r, we can write A = ra and B = rb. We will now solve for a and b.

Observe that the geometric mean of two terms rm and rn is simply rm+n

2 , or r raised to the arithmeticmean of m and n. Thus, to solve for a, we can simply consider the sequence 0, 1, 2, . . . , 2019, whichcomes from the exponents of the terms in S, and repeatedly replace the smallest two terms with theirarithmetic mean. Likewise, to solve for b, we can consider the same sequence 0, 1, 2, . . . , 2019 andrepeatedly replace the largest two terms with their arithmetic mean.

We begin by computing a. If we start with the sequence 0, 1, . . . , 2019 and repeatedly take the arith-metic mean of the two smallest terms, the final value will be

a =

0+12

+2

2 +3

2 + · · ·+ 2019

2=

2019∑k=1

k

22020−k.

Then, we can compute

2a =

2019∑k=1

k

22019−k

=⇒ a = 2a− a =

2019∑k=1

k

22019−k−

2019∑k=1

k

22020−k

=

2019∑k=1

k

22019−k−

2018∑k=0

k + 1

22019−k

= 2019−2019∑j=1

1

2j

= 2019−(

1− 1

22019

)= 2018 +

1

22019.

Likewise, or by symmetry, we can find b = 1− 122019 .

Since we want AB = ra

rb= ra−b to be a positive integer, and a − b =

2018 + 1

22019

)−1− 1

22019

)=

2017 + 122018 , r must be a perfect (22018)th power. Because r > 1, the minimal possible value is

r = 222018

. Thus, d = 22018 + 1, and so log2 d is clearly closest to 2018.

10. Sean enters a classroom in the Memorial Hall and sees a 1 followed by 2020 0’s on the blackboard.As he is early for class, he decides to go through the digits from right to left and independently erasethe nth digit from the left with probability n−1

n . (In particular, the 1 is never erased.) Compute theexpected value of the number formed from the remaining digits when viewed as a base-3 number. (Forexample, if the remaining number on the board is 1000, then its value is 27.)

Proposed by: Vincent Bian

Answer: 681751

Solution: Suppose Sean instead follows this equivalent procedure: he starts with M = 10 . . . 0, onthe board, as before. Instead of erasing digits, he starts writing a new number on the board. He goesthrough the digits of M one by one from left to right, and independently copies the nth digit fromthe left with probability 1

n . Now, let an be the expected value of Sean’s new number after he has gonethrough the first n digits of M. Note that the answer to this problem will be the expected value ofa2021, since M has 2021 digits.

Note that a1 = 1, since the probability that Sean copies the first digit is 1.

For n > 1, note that an is 3an−1 with probability 1n , and is an−1 with probability n−1

n . Thus,

E[an] =1

nE[3an−1] +

n− 1

nE[an−1] =

n+ 2

nE[an−1].

Therefore,

E[a2021] =4

2· 5

3· · · 2023

2021=

2022 · 2023

2 · 3= 337 · 2023 = 681751.

HMMO 2020November 14–21, 2020

Team Round

1. [20] For how many positive integers n ≤ 1000 does the equation in real numbers

xbxc = n

have a positive solution for x? (For a real number x, bxc denotes the largest integer that is not greaterthan x.)

2. [25] How many ways are there to arrange the numbers {1, 2, 3, 4, 5, 6, 7, 8} in a circle so that every twoadjacent elements are relatively prime? Consider rotations and reflections of the same arrangement tobe indistinguishable.

3. [30] Let A be the area of the largest semicircle that can be inscribed in a quarter-circle of radius 1.Compute 120A

π .

4. [35] Marisa has two identical cubical dice labeled with the numbers {1, 2, 3, 4, 5, 6}. However, the twodice are not fair, meaning that they can land on each face with different probability. Marisa rollsthe two dice and calculates their sum. Given that the sum is 2 with probability 0.04, and 12 withprobability 0.01, the maximum possible probability of the sum being 7 is p. Compute b100pc.

5. [40] For each positive integer n, let an be the smallest nonnegative integer such that there is only onepositive integer at most n that is relatively prime to all of n, n + 1, . . . , n + an. If n < 100, computethe largest possible value of n− an.

6. [40] Regular hexagon P1P2P3P4P5P6 has side length 2. For 1 ≤ i ≤ 6, let Ci be a unit circle centeredat Pi and `i be one of the internal common tangents of Ci and Ci+2, where C7 = C1 and C8 = C2.Assume that the lines {`1, `2, `3, `4, `5, `6} bound a regular hexagon. The area of this hexagon can beexpressed as

√ab , where a and b are relatively prime positive integers. Compute 100a+ b.

7. [45] Roger the ant is traveling on a coordinate plane, starting at (0, 0). Every second, he movesfrom one lattice point to a different lattice point at distance 1, chosen with equal probability. He willcontinue to move until he reaches some point P for which he could have reached P more quickly hadhe taken a different route. For example, if he goes from (0, 0) to (1, 0) to (1, 1) to (1, 2) to (0, 2), hestops at (0, 2) because he could have gone from (0, 0) to (0, 1) to (0, 2) in only 2 seconds. The expectednumber of steps Roger takes before he stops can be expressed as a

b , where a and b are relatively primepositive integers. Compute 100a+ b.

8. [50] Altitudes BE and CF of acute triangle ABC intersect at H. Suppose that the altitudes of triangleEHF concur on line BC. If AB = 3 and AC = 4, then BC2 = a

b , where a and b are relatively primepositive integers. Compute 100a+ b.

9. [55] Alice and Bob take turns removing balls from a bag containing 10 black balls and 10 white balls,with Alice going first. Alice always removes a black ball if there is one, while Bob removes one ofthe remaining balls uniformly at random. Once all balls have been removed, the expected number ofblack balls which Bob has can be expressed as a

b , where a and b are relatively prime positive integers.Compute 100a+ b.

10. [60] Let x and y be non-negative real numbers that sum to 1. Compute the number of ordered pairs(a, b) with a, b ∈ {0, 1, 2, 3, 4} such that the expression xayb + yaxb has maximum value 21∈a∈b.

HMMO 2020November 14, 2020

Team Round

1. [20] For how many positive integers n ≤ 1000 does the equation in real numbers

xbxc = n

have a positive solution for x? (For a real number x, bxc denotes the largest integer that is not greaterthan x.)

Proposed by: John Michael Wu

Answer: 412

Solution: If bxc = 0, then xbxc = 1 = 1b1c, so we can safely ignore this case, as it does not introducenew solutions.

If bxc = k for some k > 0, x 7→ xbxc = xk is a continuous and increasing function on the interval[k, k + 1). Therefore, the xbxc can take on any value in [kk, (k + 1)k) when bxc = k. Because 54 <1000 < 55, it suffices to count the number of integers in the interval [kk, (k + 1)k) for 1 ≤ k ≤ 4.

Thus, the number of valid n is simply

4∑k=1

(k + 1)k − kk = (2− 1) + (9− 4) + (64− 27) + (625− 256) = 412.

2. [25] How many ways are there to arrange the numbers {1, 2, 3, 4, 5, 6, 7, 8} in a circle so that every twoadjacent elements are relatively prime? Consider rotations and reflections of the same arrangement tobe indistinguishable.

Proposed by: Daniel Zhu

Answer: 36

Solution: Note that 6 can only be adjacent to 1, 5, and 7, so there are32

)= 3 ways to pick its

neighbors. Since each of 1, 5, and 7 is relatively prime to every number in {1, 2, 3, 4, 5, 6, 7, 8} but itself(and hence can have arbitrary neighbors), without loss of generality suppose we have picked 1 and 5as neighbors of 6. Observe that fixing the positions of 1, 5, and 6 eliminates the indistinguishability ofrotations and reflections.

Now, we have to consecutively arrange {2, 3, 4, 7, 8} so that no two of 2, 4, and 8 are adjacent. Thereare 3! · 2! = 12 ways of doing so, so the final answer is 3 · 12 = 36.

3. [30] Let A be the area of the largest semicircle that can be inscribed in a quarter-circle of radius 1.Compute 120A

π .

Proposed by: Akash Das

Answer: 20

Solution:

P

X

Y

M

NO

The optimal configuration is when the two ends X and Y of the semicircle lie on the arc of the quartercircle. Let O and P be the centers of the quarter circle and semicircle, respectively. Also, let M andN be the points where the semicircle is tangent to the radii of the quartercircle.

Let r be the radius of the semicircle. Since PM = PN , PMON is a square and OP =√

2r. Bythe Pythagorean theorem on triangle OPX, 1 = 2r2 + r2, so r = 1/

√3. The area of the semicircle is

therefore π213 = π

6 .

4. [35] Marisa has two identical cubical dice labeled with the numbers {1, 2, 3, 4, 5, 6}. However, the twodice are not fair, meaning that they can land on each face with different probability. Marisa rollsthe two dice and calculates their sum. Given that the sum is 2 with probability 0.04, and 12 withprobability 0.01, the maximum possible probability of the sum being 7 is p. Compute b100pc.Proposed by: Shengtong Zhang

Answer: 28

Solution: Let pi be the probability that the dice lands on the number i. The problem gives thatp21 = 0.04, p26 = 0.01, so we have

p1 = 0.2, p6 = 0.1, p2 + p3 + p4 + p5 = 0.7.

We are asked to maximize

2(p1p6 + p2p5 + p3p4) = 2(0.02 + p2p5 + p3p4).

Let x = p2 + p5 and y = p3 + p4. Then by AM-GM, p2p5 ≤ x2

4 , p3p4 ≤ y2

4 . Also,

x2 + y2

4≤ x2 + 2xy + y2

4=

(x + y)2

4=

0.72

4= 0.1225.

Hence,2(p1p6 + p2p5 + p3p4) ≤ 2(0.02 + 0.1225) = 0.285,

where equality holds if p2 = p5 = 0.35, p3 = p4 = 0.

Thus, we conclude that p = 0.285 and b100pc = 28.

5. [40] For each positive integer n, let an be the smallest nonnegative integer such that there is only onepositive integer at most n that is relatively prime to all of n, n + 1, . . . , n + an. If n < 100, computethe largest possible value of n− an.

Proposed by: Hahn Lheem

Answer: 16

Solution: Note that 1 is relatively prime to all positive integers. Therefore, the definition of an canequivalently be stated as: “an is the smallest nonnegative integer such that for all integers x, 2 ≤ x ≤ n,x shares a prime factor with at least one of n, n + 1, . . . n + an.”

The condition is equivalent the statement that the integers from n to n+ an must include multiples ofall primes less than n. Therefore, if p is the largest prime satisfying p < n, then n + an ≥ 2p.

We now claim that an = 2p − n works for all n > 11. For all primes q at most an + 1, it is apparentthat n, n+ 1, . . . , n+ an indeed contains a multiple of q. For primes an + 1 < q ≤ p, we then find that2q ≤ n + an. To finish, we claim that 2q ≥ n, which would be implied by 2(an + 2) ≥ n ⇐⇒ p ≥3n/4− 1. This is indeed true for all 11 < n < 100.

We therefore wish to maximize n − an = n − (2p − n) = 2(n − p). Therefore, the answer is twice thelargest difference between two primes less than 100. This difference is 8 (from 89 to 97), so the answeris 16. Since this is greater than 11, we have not lost anything by ignoring the smaller cases.

6. [40] Regular hexagon P1P2P3P4P5P6 has side length 2. For 1 ≤ i ≤ 6, let Ci be a unit circle centeredat Pi and `i be one of the internal common tangents of Ci and Ci+2, where C7 = C1 and C8 = C2.Assume that the lines {`1, `2, `3, `4, `5, `6} bound a regular hexagon. The area of this hexagon can beexpressed as

√ab , where a and b are relatively prime positive integers. Compute 100a + b.

Proposed by: Daniel Zhu

Answer: 1603

Solution: The only way for the lines `i to bound a regular hexagon H is if they are rotationallysymmetric around the center O of the original hexagon. (A quick way to see this is to note that theangle between the two internal common tangents of Ci and Ci+2 cannot be a multiple of 60◦.) Thusall we need to do is to compute h, the distance from the center O to the sides of H, because then wecan compute the side length of H as 2√

3h and thus its area as

6

√3

4

(2h√

3

)2

= 2√

3h2.

P1

P2

P3

T1

T3

O

M

Q

C1

C3

Without loss of generality, let’s only consider `1. Let M be the midpoint of P1P3 and let T1 and T3 bethe tangency points between `1 and C1 and C3, respectively. Without loss of generality, assume T1 iscloser to O than T3. Finally, let Q be the projection of O onto `1, so that h = OQ.

Now, note that ∠OMQ = 90◦ − ∠T1MP1 = ∠MP1T1, so 4OMQ ∼ 4MP1T1. Therefore, sinceOM = OP2/2 = 1, we find

h = OQ =OQ

OM=

T1M

MP1=

√MP 2

1 − P1T 21

MP1=

√2

3,

since MP1 = P1P3/2 =√

3. Thus the final area is 232√

3 =√

16/3.

7. [45] Roger the ant is traveling on a coordinate plane, starting at (0, 0). Every second, he movesfrom one lattice point to a different lattice point at distance 1, chosen with equal probability. He willcontinue to move until he reaches some point P for which he could have reached P more quickly hadhe taken a different route. For example, if he goes from (0, 0) to (1, 0) to (1, 1) to (1, 2) to (0, 2), hestops at (0, 2) because he could have gone from (0, 0) to (0, 1) to (0, 2) in only 2 seconds. The expectednumber of steps Roger takes before he stops can be expressed as a

b , where a and b are relatively primepositive integers. Compute 100a + b.

Proposed by: Carl Schildkraut

Answer: 1103

Solution: Roger is guaranteed to be able to take at least one step. Suppose he takes that step in adirection u. Let e1 be the expectation of the number of additional steps Roger will be able to takeafter that first move. Notice that Roger is again guaranteed to be able to make a move, and that threetypes of steps are possible:

(1) With probability 14 , Roger takes a step in the direction −u and his path ends.

(2) With probability 14 , Roger again takes a step in the direction u, after which he is expected to take

another e1 steps.

(3) With probability 12 , Roger takes a step in a direction w perpendicular to u, after which he is

expected to take some other number e2 of additional steps.

If Roger makes a move of type (3), he is again guaranteed to be able to take a step. Here are theoptions:

(1) With probability 12 , Roger takes a step in one of the directions −u and −w and his path ends.

(2) With probability 12 , Roger takes a step in one of the directions u and w, after which he is expected

to take an additional e2 steps.

Using these rules, we can set up two simple linear equations to solve the problem.

e2 =1

2e2 + 1 =⇒ e2 = 2

e1 =1

2e2 +

1

4e1 + 1 =

1

4e1 + 2 =⇒ e1 =

8

3

Since Roger takes one step before his expectation is e1, the answer is 113 .

8. [50] Altitudes BE and CF of acute triangle ABC intersect at H. Suppose that the altitudes of triangleEHF concur on line BC. If AB = 3 and AC = 4, then BC2 = a

b , where a and b are relatively primepositive integers. Compute 100a + b.

Proposed by: Jeffrey Lu

Answer: 33725

Solution:

A

B C

E

F

H

P

Let P be the orthocenter of 4EHF . Then EH ⊥ FP and EH ⊥ AC, so FP is parallel to AC.Similarly, EP is parallel to AB. Using similar triangles gives

1 =BP

BC+

CP

BC=

AE

AC+

AF

AB=

AB cosA

AC+

AC cosA

AB,

so cosA = 1225 . Then by the law of cosines, BC2 = 32 + 42 − 2(3)(4)( 12

25 ) = 33725 .

9. [55] Alice and Bob take turns removing balls from a bag containing 10 black balls and 10 white balls,with Alice going first. Alice always removes a black ball if there is one, while Bob removes one ofthe remaining balls uniformly at random. Once all balls have been removed, the expected number ofblack balls which Bob has can be expressed as a

b , where a and b are relatively prime positive integers.Compute 100a + b.

Proposed by: Benjamin Qi

Answer: 4519

Solution: Suppose a is the number of black balls and b is the number of white balls, and let Ea,bdenote the expected number of black balls Bob has once all the balls are removed with Alice goingfirst. Then we want to find E10,10. It is evident that if E0,b = 0. Also, since Bob chooses a black ballwith probability a−1

a+b−1 , if a > 0 we have

Ea,b =a− 1

a + b− 1(Ea−2,b + 1) +

b

a + b− 1Ea−1,b−1

=(a− 1)(Ea−2,b + 1) + bEa−1,b−1

a + b− 1

We claim that Ea,b = a(a−1)2(a+b−1) , which will yield an answer of 45

19 . To prove this, we use induction. In

the base case of a = 0 we find a(a−1)2(a+b−1) = 0, as desired. Also, for a > 0 we have that by the inductive

hypothesis

Ea,b =(a− 1)((a− 2)(a− 3) + 2(a + b− 3)) + b(a− 1)(a− 2)

2(a + b− 1)(a + b− 3)

=(a− 1)(a− 2)(a + b− 3) + 2(a− 1)(a + b− 3)

2(a + b− 1)(a + b− 3)

=a(a− 1)

2(a + b− 1),

as desired.

10. [60] Let x and y be non-negative real numbers that sum to 1. Compute the number of ordered pairs(a, b) with a, b ∈ {0, 1, 2, 3, 4} such that the expression xayb + yaxb has maximum value 21−a−b.

Proposed by: Shengtong Zhang

Answer: 17

Solution: Let f(x, y) = xayb + yaxb. Observe that 21−a−b is merely the value of f( 12 ,

12 ), so this value

is always achievable.

We claim (call this result (>)) that if (a, b) satisfies the condition, so does (a + 1, b + 1). To seethis, observe that if f(x, y) ≤ 21−a−b, then multiplying by the inequality xy ≤ 1

4 yields xa+1yb+1 +ya+1xb+1 ≤ 2−1−a−b, as desired.

For the rest of the solution, without loss of generality we consider the a ≥ b case. If a = b = 0, thenf(x, y) = 2, so (0, 0) works. If a = 1 and b = 0, then f(x, y) = x + y = 1, so (1, 0) works. For a ≥ 2,(a, 0) fails since f(1, 0) = 1 > 21−a.

If a = 3 and b = 1, f(x, y) = xy(x2+y2) = xy(1−2xy), which is maximized at xy = 14 ⇐⇒ x = y = 1

2 ,so (3, 1) works. However, if a = 4 and b = 1, f(x, y) = xy(x3 + y3) = xy((x + y)3 − 3xy(x + y)) =xy(1− 3xy), which is maximized at xy = 1

6 . Thus (4, 1) does not work.

From these results and (>), we are able to deduce all the pairs that do work (↙ represents those pairsthat work by (>)):

0 1 2 3 4

4

3

2

1

0 3 3 7 7 7

3 ↙ ↙ 3 7

7 ↙ ↙ ↙ ↙

7 3 ↙ ↙ ↙

7 7 ↙ ↙ ↙

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMO 2020, November 14–21, 2020 — GUTS ROUND

Organization Team Team ID#

1. [5] Two hexagons are attached to form a new polygon P . Compute the minimum number of sides thatP can have.

2. [5] Let a be a positive integer such that 2a has units digit 4. What is the sum of the possible units digitsof 3a?

3. [5] How many six-digit multiples of 27 have only 3, 6, or 9 as their digits?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMO 2020, November 14–21, 2020 — GUTS ROUND

Organization Team Team ID#

4. [6] Ainsley and Buddy play a game where they repeatedly roll a standard fair six-sided die. Ainsley winsif two multiples of 3 in a row are rolled before a non-multiple of 3 followed by a multiple of 3, and Buddywins otherwise. If the probability that Ainsley wins is a

b for relatively prime positive integers a and b,compute 100a+ b.

5. [6] The points (0, 0), (1, 2), (2, 1), (2, 2) in the plane are colored red while the points (1, 0), (2, 0), (0, 1), (0, 2)are colored blue. Four segments are drawn such that each one connects a red point to a blue point andeach colored point is the endpoint of some segment. The smallest possible sum of the lengths of thesegments can be expressed as a+

√b, where a, b are positive integers. Compute 100a+ b.

6. [6] If x, y, z are real numbers such that xy = 6, x− z = 2, and x+ y + z = 9, compute xy −

zx −

z2

xy .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMO 2020, November 14–21, 2020 — GUTS ROUND

Organization Team Team ID#

7. [7] Compute the maximum number of sides of a polygon that is the cross-section of a regular hexagonalprism.

8. [7] A small village has n people. During their yearly elections, groups of three people come up to a stageand vote for someone in the village to be the new leader. After every possible group of three people hasvoted for someone, the person with the most votes wins.

This year, it turned out that everyone in the village had the exact same number of votes! If 10 ≤ n ≤ 100,what is the number of possible values of n?

9. [7] A fair coin is flipped eight times in a row. Let p be the probability that there is exactly one pairof consecutive flips that are both heads and exactly one pair of consecutive flips that are both tails. Ifp = a

b , where a, b are relatively prime positive integers, compute 100a+ b.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMO 2020, November 14–21, 2020 — GUTS ROUND

Organization Team Team ID#

10. [8] The number 3003 is the only number known to appear eight times in Pascal’s triangle, at positions(3003

1

),

(3003

3002

),

(a

2

),

(a

a− 2

),

(15

b

),

(15

15− b

),

(14

6

),

(14

8

).

Compute a+ b(15− b).

11. [8] Two diameters and one radius are drawn in a circle of radius 1, dividing the circle into 5 sectors. Thelargest possible area of the smallest sector can be expressed as a

bπ, where a, b are relatively prime positiveintegers. Compute 100a+ b.

12. [8] In a single-elimination tournament consisting of 29 = 512 teams, there is a strict ordering on the skilllevels of the teams, but Joy does not know that ordering. The teams are randomly put into a bracket andthey play out the tournament, with the better team always beating the worse team. Joy is then given theresults of all 511 matches and must create a list of teams such that she can guarantee that the third-bestteam is on the list. What is the minimum possible length of Joy’s list?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMO 2020, November 14–21, 2020 — GUTS ROUND

Organization Team Team ID#

13. [9] Wendy is playing darts with a circular dartboard of radius 20. Whenever she throws a dart, it landsuniformly at random on the dartboard. At the start of her game, there are 2020 darts placed randomlyon the board. Every turn, she takes the dart farthest from the center, and throws it at the board again.What is the expected number of darts she has to throw before all the darts are within 10 units of thecenter?

14. [9] A point (x, y) is selected uniformly at random from the unit square S = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.If the probability that (3x+2y, x+4y) is in S is a

b , where a, b are relatively prime positive integers, compute100a+ b.

15. [9] For a real number r, the quadratics x2 + (r − 1)x + 6 and x2 + (2r + 1)x + 22 have a common realroot. The sum of the possible values of r can be expressed as a

b , where a, b are relatively prime positiveintegers. Compute 100a+ b.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMO 2020, November 14–21, 2020 — GUTS ROUND

Organization Team Team ID#

16. [10] Three players play tic-tac-toe together. In other words, the three players take turns placing an ”A”,”B”, and ”C”, respectively, in one of the free spots of a 3 × 3 grid, and the first player to have three oftheir label in a row, column, or diagonal wins. How many possible final boards are there where the playerwho goes third wins the game? (Rotations and reflections are considered different boards, but the orderof placement does not matter.)

17. [10] Let N>1 denote the set of positive integers greater than 1. Let f : N>1 → N>1 be a function suchthat f(mn) = f(m)f(n) for all m,n ∈ N>1. If f(101!) = 101!, compute the number of possible values off(2020 · 2021).

18. [10] Suppose Harvard Yard is a 17×17 square. There are 14 dorms located on the perimeter of the Yard.If s is the minimum distance between two dorms, the maximum possible value of s can be expressed asa−√b where a, b are positive integers. Compute 100a+ b.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMO 2020, November 14–21, 2020 — GUTS ROUND

Organization Team Team ID#

19. [11] Three distinct vertices of a regular 2020-gon are chosen uniformly at random. The probability thatthe triangle they form is isosceles can be expressed as a

b , where a and b are relatively prime positiveintegers. Compute 100a+ b.

20. [11] Let ω1 be a circle of radius 5, and let ω2 be a circle of radius 2 whose center lies on ω1. Let the twocircles intersect at A and B, and let the tangents to ω2 at A and B intersect at P . If the area of 4ABPcan be expressed as a

√b

c , where b is square-free and a, c are relatively prime positive integers, compute100a+ 10b+ c.

21. [11] Let f(n) be the number of distinct prime divisors of n less than 6. Compute

2020∑n=1

f(n)2.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMO 2020, November 14–21, 2020 — GUTS ROUND

Organization Team Team ID#

22. [12] In triangle ABC, AB = 32, AC = 35, and BC = x. What is the smallest positive integer x suchthat 1 + cos2A, cos2B, and cos2 C form the sides of a non-degenerate triangle?

23. [12] Two points are chosen inside the square {(x, y) | 0 ≤ x, y ≤ 1} uniformly at random, and a unitsquare is drawn centered at each point with edges parallel to the coordinate axes. The expected areaof the union of the two squares can be expressed as a

b , where a, b are relatively prime positive integers.Compute 100a+ b.

24. [12] Compute the number of positive integers less than 10! which can be expressed as the sum of at most4 (not necessarily distinct) factorials.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMO 2020, November 14–21, 2020 — GUTS ROUND

Organization Team Team ID#

25. [13] Let a1, a2, a3, . . . be a sequence of positive integers where a1 =∑100

i=0 i! and ai+ai+1 is an odd perfectsquare for all i ≥ 1. Compute the smallest possible value of a1000.

26. [13] Two players play a game where they are each given 10 indistinguishable units that must be distributedacross three locations. (Units cannot be split.) At each location, a player wins at that location if thenumber of units they placed there is at least 2 more than the units of the other player. If both playersdistribute their units randomly (i.e. there is an equal probability of them distributing their units for anyattainable distribution across the 3 locations), the probability that at least one location is won by one ofthe players can be expressed as a

b , where a, b are relatively prime positive integers. Compute 100a+ b.

27. [13] In 4ABC, D and E are the midpoints of BC and CA, respectively. AD and BE intersect at G.Given that GECD is cyclic, AB = 41, and AC = 31, compute BC.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMO 2020, November 14–21, 2020 — GUTS ROUND

Organization Team Team ID#

28. [15] Bernie has 2020 marbles and 2020 bags labeled B1, . . . , B2020 in which he randomly distributes themarbles (each marble is placed in a random bag independently). If E the expected number of integers1 ≤ i ≤ 2020 such that Bi has at least i marbles, compute the closest integer to 1000E.

29. [15] In acute triangle ABC, let H be the orthocenter and D the foot of the altitude from A. Thecircumcircle of triangle BHC intersects AC at E 6= C, and AB at F 6= B. If BD = 3, CD = 7, andAHHD = 5

7 , the area of triangle AEF can be expressed as ab , where a, b are relatively prime positive integers.

Compute 100a+ b.

30. [15] Let a1, a2, a3, . . . be a sequence of positive real numbers that satisfies

∞∑n=k

(n

k

)an =

1

5k,

for all positive integers k. The value of a1 − a2 + a3 − a4 + · · · can be expressed as ab , where a, b are

relatively prime positive integers. Compute 100a+ b.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMO 2020, November 14–21, 2020 — GUTS ROUND

Organization Team Team ID#

31. [17] For some positive real α, the set S of positive real numbers x with {x} > αx consists of the union ofseveral intervals, with total length 20.2. The value of α can be expressed as a

b , where a, b are relativelyprime positive integers. Compute 100a+ b. (Here, {x} = x− bxc is the fractional part of x.)

32. [17] The numbers 1, 2, . . . , 10 are written in a circle. There are four people, and each person randomlyselects five consecutive integers (e.g. 1, 2, 3, 4, 5, or 8, 9, 10, 1, 2). If the probability that there exists somenumber that was not selected by any of the four people is p, compute 10000p.

33. [17] In quadrilateral ABCD, there exists a point E on segment AD such that AEED = 1

9 and ∠BEC is aright angle. Additionally, the area of triangle CED is 27 times more than the area of triangle AEB. If∠EBC = ∠EAB, ∠ECB = ∠EDC, and BC = 6, compute the value of AD2.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMO 2020, November 14–21, 2020 — GUTS ROUND

Organization Team Team ID#

34. [20] Let a be the proportion of teams that correctly answered problem 1 on the Guts round. EstimateA = b10000ac. An estimate of E earns max(0, b20− |A−E|/20c) points. If you have forgotten, question1 was the following:

Two hexagons are attached to form a new polygon P . What is the minimum number of sides that P canhave?

35. [20] Estimate A, the number of times an 8-digit number appears in Pascal’s triangle. An estimate of Eearns max(0, b20− |A− E|/200c) points.

36. [20] Let pi be the ith prime. Let

f(x) =

50∑i=1

pixi−1 = 2 + 3x+ · · ·+ 229x49.

If a is the unique positive real number with f(a) = 100, estimate A = b100000ac. An estimate of E willearn max(0, b20− |A− E|/250c) points.

HMMO 2020November 14–21, 2020

Guts Round

1. [5] Two hexagons are attached to form a new polygon P . Compute the minimum number of sides thatP can have.

Proposed by: Daniel Zhu

Answer: 3

Solution: A triangle can be split into two hexagons; in other words, two concave hexagons can beattached to form a triangle, like the following:

2. [5] Let a be a positive integer such that 2a has units digit 4. What is the sum of the possible unitsdigits of 3a?

Proposed by: Daniel Zhu

Answer: 7

Solution: If 2a has last digit 4, then the last digit of a is either 2 or 7. In the former case, 3a has lastdigit 6, and in the latter case, 3a has last digit 1. This gives a final answer of 6 + 1 = 7.

3. [5] How many six-digit multiples of 27 have only 3, 6, or 9 as their digits?

Proposed by: Daniel Zhu

Answer: 51

Solution: Divide by 3. We now want to count the number of six-digit multiples of 9 that only have 1,2, or 3 as their digits. Due to the divisibility rule for 9, we only need to consider when the digit sumis a multiple of 9. Note that 3 · 6 = 18 is the maximum digit sum.

If the sum is 18, the only case is 333333.

Otherwise, the digit sum is 9. The possibilities here, up to ordering of the digits, are 111222 and111123. The first has

(63

)= 20 cases, while the second has 6 · 5 = 30. Thus the final answer is

1 + 20 + 30 = 51.

4. [6] Ainsley and Buddy play a game where they repeatedly roll a standard fair six-sided die. Ainsleywins if two multiples of 3 in a row are rolled before a non-multiple of 3 followed by a multiple of 3, andBuddy wins otherwise. If the probability that Ainsley wins is a

b for relatively prime positive integersa and b, compute 100a+ b.

Proposed by: Dora Woodruff

Answer: 109

Solution: We let X be the event of a multiple of 3 being rolled and Y be the event of a nonmultipleof 3 being rolled. In order for Ainsley to win, she needs event X to happen consecutively; meanwhile,Buddy just needs Y then X to occur. Thus, if Y occurs in the first two rolls, Buddy will be guaranteedto win, since the next time X happens, it will have been preceded by an X. Thus, the probability of Awinning is equivalent to the probability of X happening in each of the first two times, or (1/3)2 = 1/9.

5. [6] The points (0, 0), (1, 2), (2, 1), (2, 2) in the plane are colored red while the points (1, 0), (2, 0), (0, 1), (0, 2)are colored blue. Four segments are drawn such that each one connects a red point to a blue point andeach colored point is the endpoint of some segment. The smallest possible sum of the lengths of thesegments can be expressed as a+

√b, where a, b are positive integers. Compute 100a+ b.

Proposed by: Andrew Gu

Answer: 305

Solution:

If (2, 2) is connected to (0, 1) or (1, 0), then the other 6 points can be connected with segments of totallength 3, which is minimal. This leads to a total length of 3 +

√5.

On the other hand, if (2, 2) is connected to (0, 2) or (0, 2), then connecting the other points with seg-ments of total length 2 is impossible, so the minimal length is at least 2 + 2 +

√2 = 4 +

√2 > 3 +

√5.

6. [6] If x, y, z are real numbers such that xy = 6, x− z = 2, and x+ y + z = 9, compute xy − z

x − z2

xy .

Proposed by: Ragulan Sivakumar

Answer: 2

Solution: Let k = xy − z

x − z2

xy = x2−yz−z2xy . We have

k + 1 =x2 + xy − yz − z2

xy=x2 − xz + xy − yz + zx− z2

xy=

(x+ y + z)(x− z)xy

=9 · 2

6= 3,

so k = 2.

7. [7] Compute the maximum number of sides of a polygon that is the cross-section of a regular hexagonalprism.

Proposed by: Hahn Lheem

Answer: 8

Solution: Note that since there are 8 faces to a regular hexagonal prism and a cross-section may onlyintersect a face once, the upper bound for our answer is 8.

Indeed, we can construct a cross-section of the prism with 8 sides. Let ABCDEF and A′B′C ′D′E′F ′

be the two bases of the prism, with A being directly over A′. Choose points P and Q on line segmentsAB and BC, respectively, and choose points P ′ and Q′ on segments D′E′ and E′F ′, respectively, suchthat PQ ‖ P ′Q′. Then, the cross-section of the prism from the plane that goes through P , Q, P ′, andQ′ forms a polygon with 8 sides.

8. [7] A small village has n people. During their yearly elections, groups of three people come up to astage and vote for someone in the village to be the new leader. After every possible group of threepeople has voted for someone, the person with the most votes wins.

This year, it turned out that everyone in the village had the exact same number of votes! If 10 ≤ n ≤100, what is the number of possible values of n?

Proposed by: Vincent Bian

Answer: 61

Solution: The problem asks for the number of n that divide(n3

), which happens exactly when

(n−1)(n−2)2·3 is an integer. Regardless of the parity of n, (n − 1)(n − 2) is always divisible by 2. Also,

(n− 1)(n− 2) is divisible by 3 if and only if n is not a multiple of 3. Of the 91 values from 10 to 100,30 are divisible by 3, so our answer is 61.

9. [7] A fair coin is flipped eight times in a row. Let p be the probability that there is exactly one pairof consecutive flips that are both heads and exactly one pair of consecutive flips that are both tails. Ifp = a

b , where a, b are relatively prime positive integers, compute 100a+ b.

Proposed by: Yannick Yao

Answer: 1028

Solution: Separate the sequence of coin flips into alternating blocks of heads and tails. Of the blocksof heads, exactly one block has length 2, and all other blocks have length 1. The same statementapplies to blocks of tails. Thus, if there are k blocks in total, there are k − 2 blocks of length 1 and 2blocks of length 2, leading to k + 2 coins in total. We conclude that k = 6, meaning that there are 3blocks of heads and 3 blocks of tails.

The blocks of heads must have lengths 1, 1, 2 in some order, and likewise for tails. There are 32 = 9ways to choose these two orders, and 2 ways to assemble these blocks into a sequence, depending onwhether the first coin flipped is heads or tails. Thus the final probability is 18/28 = 9/128.

10. [8] The number 3003 is the only number known to appear eight times in Pascal’s triangle, at positions(3003

1

),

(3003

3002

),

(a

2

),

(a

a− 2

),

(15

b

),

(15

15− b

),

(14

6

),

(14

8

).

Compute a+ b(15− b).Proposed by: Carl Joshua Quines

Answer: 128

//////////////////////////////////////////////////////////////////////////////////// (C) 2012, Michail Vidiassov, John C. Bowman, Alexander Grahn//// asylabels.js//// version 20120912////////////////////////////////////////////////////////////////////////////////////// 3D JavaScript to be used with media9.sty (option `add3Djscript') for// Asymptote generated PRC files//// adds billboard behaviour to text labels in Asymptote PRC files so that// they always face the camera under 3D rotation.////// This work may be distributed and/or modified under the// conditions of the LaTeX Project Public License.// // The latest version of this license is in// http://mirrors.ctan.org/macros/latex/base/lppl.txt// // This work has the LPPL maintenance status `maintained'.// // The Current Maintainer of this work is A. Grahn.//////////////////////////////////////////////////////////////////////////////////var bbnodes=new Array(); // billboard meshesvar bbtrans=new Array(); // billboard transformsfunction fulltransform(mesh) { var t=new Matrix4x4(mesh.transform); if(mesh.parent.name != "") { var parentTransform=fulltransform(mesh.parent); t.multiplyInPlace(parentTransform); return t; } else return t; } // find all text labels in the scene and determine pivoting pointsvar nodes=scene.nodes;var nodescount=nodes.count;var third=1.0/3.0;for(var i=0; i < nodescount; i++) { var node=nodes.getByIndex(i); var name=node.name; var end=name.lastIndexOf(".")-1; if(end > 0) { if(name.charAt(end) == "\001") { var start=name.lastIndexOf("-")+1; if(end > start) { node.name=name.substr(0,start-1); var nodeMatrix=fulltransform(node.parent); var c=nodeMatrix.translation; // position var d=Math.pow(Math.abs(nodeMatrix.determinant),third); // scale bbnodes.push(node); bbtrans.push(Matrix4x4().scale(d,d,d).translate(c).multiply(nodeMatrix.inverse)); } } }}var camera=scene.cameras.getByIndex(0); var zero=new Vector3(0,0,0);var bbcount=bbnodes.length;// event handler to maintain camera-facing text labelsbillboardHandler=new RenderEventHandler();billboardHandler.onEvent=function(event){ var T=new Matrix4x4(); T.setView(zero,camera.position.subtract(camera.targetPosition), camera.up.subtract(camera.position)); for(var j=0; j < bbcount; j++) bbnodes[j].transform.set(T.multiply(bbtrans[j])); runtime.refresh(); }runtime.addEventHandler(billboardHandler);runtime.refresh();

//////////////////////////////////////////////////////////////////////////////////// (C) 2012--today, Alexander Grahn//// 3Dmenu.js//// version 20140923////////////////////////////////////////////////////////////////////////////////////// 3D JavaScript used by media9.sty//// Extended functionality of the (right click) context menu of 3D annotations.//// 1.) Adds the following items to the 3D context menu://// * `Generate Default View'//// Finds good default camera settings, returned as options for use with// the \includemedia command.//// * `Get Current View'//// Determines camera, cross section and part settings of the current view,// returned as `VIEW' section that can be copied into a views file of// additional views. The views file is inserted using the `3Dviews' option// of \includemedia.//// * `Cross Section'//// Toggle switch to add or remove a cross section into or from the current// view. The cross section can be moved in the x, y, z directions using x,// y, z and X, Y, Z keys on the keyboard, be tilted against and spun// around the upright Z axis using the Up/Down and Left/Right arrow keys// and caled using the s and S keys.//// 2.) Enables manipulation of position and orientation of indiviual parts and// groups of parts in the 3D scene. Parts which have been selected with the// mouse can be scaled moved around and rotated like the cross section as// described above. To spin the parts around their local up-axis, keep// Control key pressed while using the Up/Down and Left/Right arrow keys.//// This work may be distributed and/or modified under the// conditions of the LaTeX Project Public License.// // The latest version of this license is in// http://mirrors.ctan.org/macros/latex/base/lppl.txt// // This work has the LPPL maintenance status `maintained'.// // The Current Maintainer of this work is A. Grahn.//// The code borrows heavily from Bernd Gaertners `Miniball' software,// originally written in C++, for computing the smallest enclosing ball of a// set of points; see: http://www.inf.ethz.ch/personal/gaertner/miniball.html////////////////////////////////////////////////////////////////////////////////////host.console.show();//constructor for doubly linked listfunction List(){ this.first_node=null; this.last_node=new Node(undefined);}List.prototype.push_back=function(x){ var new_node=new Node(x); if(this.first_node==null){ this.first_node=new_node; new_node.prev=null; }else{ new_node.prev=this.last_node.prev; new_node.prev.next=new_node; } new_node.next=this.last_node; this.last_node.prev=new_node;};List.prototype.move_to_front=function(it){ var node=it.get(); if(node.next!=null && node.prev!=null){ node.next.prev=node.prev; node.prev.next=node.next; node.prev=null; node.next=this.first_node; this.first_node.prev=node; this.first_node=node; }};List.prototype.begin=function(){ var i=new Iterator(); i.target=this.first_node; return(i);};List.prototype.end=function(){ var i=new Iterator(); i.target=this.last_node; return(i);};function Iterator(it){ if( it!=undefined ){ this.target=it.target; }else { this.target=null; }}Iterator.prototype.set=function(it){this.target=it.target;};Iterator.prototype.get=function(){return(this.target);};Iterator.prototype.deref=function(){return(this.target.data);};Iterator.prototype.incr=function(){ if(this.target.next!=null) this.target=this.target.next;};//constructor for node objects that populate the linked listfunction Node(x){ this.prev=null; this.next=null; this.data=x;}function sqr(r){return(r*r);}//helper function//Miniball algorithm by B. Gaertnerfunction Basis(){ this.m=0; this.q0=new Array(3); this.z=new Array(4); this.f=new Array(4); this.v=new Array(new Array(3), new Array(3), new Array(3), new Array(3)); this.a=new Array(new Array(3), new Array(3), new Array(3), new Array(3)); this.c=new Array(new Array(3), new Array(3), new Array(3), new Array(3)); this.sqr_r=new Array(4); this.current_c=this.c[0]; this.current_sqr_r=0; this.reset();}Basis.prototype.center=function(){return(this.current_c);};Basis.prototype.size=function(){return(this.m);};Basis.prototype.pop=function(){--this.m;};Basis.prototype.excess=function(p){ var e=-this.current_sqr_r; for(var k=0;k<3;++k){ e+=sqr(p[k]-this.current_c[k]); } return(e);};Basis.prototype.reset=function(){ this.m=0; for(var j=0;j<3;++j){ this.c[0][j]=0; } this.current_c=this.c[0]; this.current_sqr_r=-1;};Basis.prototype.push=function(p){ var i, j; var eps=1e-32; if(this.m==0){ for(i=0;i<3;++i){ this.q0[i]=p[i]; } for(i=0;i<3;++i){ this.c[0][i]=this.q0[i]; } this.sqr_r[0]=0; }else { for(i=0;i<3;++i){ this.v[this.m][i]=p[i]-this.q0[i]; } for(i=1;i<this.m;++i){ this.a[this.m][i]=0; for(j=0;j<3;++j){ this.a[this.m][i]+=this.v[i][j]*this.v[this.m][j]; } this.a[this.m][i]*=(2/this.z[i]); } for(i=1;i<this.m;++i){ for(j=0;j<3;++j){ this.v[this.m][j]-=this.a[this.m][i]*this.v[i][j]; } } this.z[this.m]=0; for(j=0;j<3;++j){ this.z[this.m]+=sqr(this.v[this.m][j]); } this.z[this.m]*=2; if(this.z[this.m]<eps*this.current_sqr_r) return(false); var e=-this.sqr_r[this.m-1]; for(i=0;i<3;++i){ e+=sqr(p[i]-this.c[this.m-1][i]); } this.f[this.m]=e/this.z[this.m]; for(i=0;i<3;++i){ this.c[this.m][i]=this.c[this.m-1][i]+this.f[this.m]*this.v[this.m][i]; } this.sqr_r[this.m]=this.sqr_r[this.m-1]+e*this.f[this.m]/2; } this.current_c=this.c[this.m]; this.current_sqr_r=this.sqr_r[this.m]; ++this.m; return(true);};function Miniball(){ this.L=new List(); this.B=new Basis(); this.support_end=new Iterator();}Miniball.prototype.mtf_mb=function(it){ var i=new Iterator(it); this.support_end.set(this.L.begin()); if((this.B.size())==4) return; for(var k=new Iterator(this.L.begin());k.get()!=i.get();){ var j=new Iterator(k); k.incr(); if(this.B.excess(j.deref()) > 0){ if(this.B.push(j.deref())){ this.mtf_mb(j); this.B.pop(); if(this.support_end.get()==j.get()) this.support_end.incr(); this.L.move_to_front(j); } } }};Miniball.prototype.check_in=function(b){ this.L.push_back(b);};Miniball.prototype.build=function(){ this.B.reset(); this.support_end.set(this.L.begin()); this.mtf_mb(this.L.end());};Miniball.prototype.center=function(){ return(this.B.center());};Miniball.prototype.radius=function(){ return(Math.sqrt(this.B.current_sqr_r));};//functions called by menu itemsfunction calc3Dopts () { //create Miniball object var mb=new Miniball(); //auxiliary vector var corner=new Vector3(); //iterate over all visible mesh nodes in the scene for(i=0;i<scene.meshes.count;i++){ var mesh=scene.meshes.getByIndex(i); if(!mesh.visible) continue; //local to parent transformation matrix var trans=mesh.transform; //build local to world transformation matrix by recursively //multiplying the parent's transf. matrix on the right var parent=mesh.parent; while(parent.transform){ trans=trans.multiply(parent.transform); parent=parent.parent; } //get the bbox of the mesh (local coordinates) var bbox=mesh.computeBoundingBox(); //transform the local bounding box corner coordinates to //world coordinates for bounding sphere determination //BBox.min corner.set(bbox.min); corner.set(trans.transformPosition(corner)); mb.check_in(new Array(corner.x, corner.y, corner.z)); //BBox.max corner.set(bbox.max); corner.set(trans.transformPosition(corner)); mb.check_in(new Array(corner.x, corner.y, corner.z)); //remaining six BBox corners corner.set(bbox.min.x, bbox.max.y, bbox.max.z); corner.set(trans.transformPosition(corner)); mb.check_in(new Array(corner.x, corner.y, corner.z)); corner.set(bbox.min.x, bbox.min.y, bbox.max.z); corner.set(trans.transformPosition(corner)); mb.check_in(new Array(corner.x, corner.y, corner.z)); corner.set(bbox.min.x, bbox.max.y, bbox.min.z); corner.set(trans.transformPosition(corner)); mb.check_in(new Array(corner.x, corner.y, corner.z)); corner.set(bbox.max.x, bbox.min.y, bbox.min.z); corner.set(trans.transformPosition(corner)); mb.check_in(new Array(corner.x, corner.y, corner.z)); corner.set(bbox.max.x, bbox.min.y, bbox.max.z); corner.set(trans.transformPosition(corner)); mb.check_in(new Array(corner.x, corner.y, corner.z)); corner.set(bbox.max.x, bbox.max.y, bbox.min.z); corner.set(trans.transformPosition(corner)); mb.check_in(new Array(corner.x, corner.y, corner.z)); } //compute the smallest enclosing bounding sphere mb.build(); // //current camera settings // var camera=scene.cameras.getByIndex(0); var res=''; //initialize result string //aperture angle of the virtual camera (perspective projection) *or* //orthographic scale (orthographic projection) if(camera.projectionType==camera.TYPE_PERSPECTIVE){ var aac=camera.fov*180/Math.PI; if(host.util.printf('%.4f', aac)!=30) res+=host.util.printf('\n3Daac=%s,', aac); }else{ camera.viewPlaneSize=2.*mb.radius(); res+=host.util.printf('\n3Dortho=%s,', 1./camera.viewPlaneSize); } //camera roll var roll = camera.roll*180/Math.PI; if(host.util.printf('%.4f', roll)!=0) res+=host.util.printf('\n3Droll=%s,',roll); //target to camera vector var c2c=new Vector3(); c2c.set(camera.position); c2c.subtractInPlace(camera.targetPosition); c2c.normalize(); if(!(c2c.x==0 && c2c.y==-1 && c2c.z==0)) res+=host.util.printf('\n3Dc2c=%s %s %s,', c2c.x, c2c.y, c2c.z); // //new camera settings // //bounding sphere centre --> new camera target var coo=new Vector3(); coo.set((mb.center())[0], (mb.center())[1], (mb.center())[2]); if(coo.length) res+=host.util.printf('\n3Dcoo=%s %s %s,', coo.x, coo.y, coo.z); //radius of orbit if(camera.projectionType==camera.TYPE_PERSPECTIVE){ var roo=mb.radius()/ Math.sin(aac * Math.PI/ 360.); }else{ //orthographic projection var roo=mb.radius(); } res+=host.util.printf('\n3Droo=%s,', roo); //update camera settings in the viewer var currol=camera.roll; camera.targetPosition.set(coo); camera.position.set(coo.add(c2c.scale(roo))); camera.roll=currol; //determine background colour rgb=scene.background.getColor(); if(!(rgb.r==1 && rgb.g==1 && rgb.b==1)) res+=host.util.printf('\n3Dbg=%s %s %s,', rgb.r, rgb.g, rgb.b); //determine lighting scheme switch(scene.lightScheme){ case scene.LIGHT_MODE_FILE: curlights='Artwork';break; case scene.LIGHT_MODE_NONE: curlights='None';break; case scene.LIGHT_MODE_WHITE: curlights='White';break; case scene.LIGHT_MODE_DAY: curlights='Day';break; case scene.LIGHT_MODE_NIGHT: curlights='Night';break; case scene.LIGHT_MODE_BRIGHT: curlights='Hard';break; case scene.LIGHT_MODE_RGB: curlights='Primary';break; case scene.LIGHT_MODE_BLUE: curlights='Blue';break; case scene.LIGHT_MODE_RED: curlights='Red';break; case scene.LIGHT_MODE_CUBE: curlights='Cube';break; case scene.LIGHT_MODE_CAD: curlights='CAD';break; case scene.LIGHT_MODE_HEADLAMP: curlights='Headlamp';break; } if(curlights!='Artwork') res+=host.util.printf('\n3Dlights=%s,', curlights); //determine global render mode switch(scene.renderMode){ case scene.RENDER_MODE_BOUNDING_BOX: currender='BoundingBox';break; case scene.RENDER_MODE_TRANSPARENT_BOUNDING_BOX: currender='TransparentBoundingBox';break; case scene.RENDER_MODE_TRANSPARENT_BOUNDING_BOX_OUTLINE: currender='TransparentBoundingBoxOutline';break; case scene.RENDER_MODE_VERTICES: currender='Vertices';break; case scene.RENDER_MODE_SHADED_VERTICES: currender='ShadedVertices';break; case scene.RENDER_MODE_WIREFRAME: currender='Wireframe';break; case scene.RENDER_MODE_SHADED_WIREFRAME: currender='ShadedWireframe';break; case scene.RENDER_MODE_SOLID: currender='Solid';break; case scene.RENDER_MODE_TRANSPARENT: currender='Transparent';break; case scene.RENDER_MODE_SOLID_WIREFRAME: currender='SolidWireframe';break; case scene.RENDER_MODE_TRANSPARENT_WIREFRAME: currender='TransparentWireframe';break; case scene.RENDER_MODE_ILLUSTRATION: currender='Illustration';break; case scene.RENDER_MODE_SOLID_OUTLINE: currender='SolidOutline';break; case scene.RENDER_MODE_SHADED_ILLUSTRATION: currender='ShadedIllustration';break; case scene.RENDER_MODE_HIDDEN_WIREFRAME: currender='HiddenWireframe';break; } if(currender!='Solid') res+=host.util.printf('\n3Drender=%s,', currender); //write result string to the console host.console.show();// host.console.clear(); host.console.println('%%\n%% Copy and paste the following text to the\n'+ '%% option list of \\includemedia!\n%%' + res + '\n');}function get3Dview () { var camera=scene.cameras.getByIndex(0); var coo=camera.targetPosition; var c2c=camera.position.subtract(coo); var roo=c2c.length; c2c.normalize(); var res='VIEW%=insert optional name here\n'; if(!(coo.x==0 && coo.y==0 && coo.z==0)) res+=host.util.printf(' COO=%s %s %s\n', coo.x, coo.y, coo.z); if(!(c2c.x==0 && c2c.y==-1 && c2c.z==0)) res+=host.util.printf(' C2C=%s %s %s\n', c2c.x, c2c.y, c2c.z); if(roo > 1e-9) res+=host.util.printf(' ROO=%s\n', roo); var roll = camera.roll*180/Math.PI; if(host.util.printf('%.4f', roll)!=0) res+=host.util.printf(' ROLL=%s\n', roll); if(camera.projectionType==camera.TYPE_PERSPECTIVE){ var aac=camera.fov * 180/Math.PI; if(host.util.printf('%.4f', aac)!=30) res+=host.util.printf(' AAC=%s\n', aac); }else{ if(host.util.printf('%.4f', camera.viewPlaneSize)!=1) res+=host.util.printf(' ORTHO=%s\n', 1./camera.viewPlaneSize); } rgb=scene.background.getColor(); if(!(rgb.r==1 && rgb.g==1 && rgb.b==1)) res+=host.util.printf(' BGCOLOR=%s %s %s\n', rgb.r, rgb.g, rgb.b); switch(scene.lightScheme){ case scene.LIGHT_MODE_FILE: curlights='Artwork';break; case scene.LIGHT_MODE_NONE: curlights='None';break; case scene.LIGHT_MODE_WHITE: curlights='White';break; case scene.LIGHT_MODE_DAY: curlights='Day';break; case scene.LIGHT_MODE_NIGHT: curlights='Night';break; case scene.LIGHT_MODE_BRIGHT: curlights='Hard';break; case scene.LIGHT_MODE_RGB: curlights='Primary';break; case scene.LIGHT_MODE_BLUE: curlights='Blue';break; case scene.LIGHT_MODE_RED: curlights='Red';break; case scene.LIGHT_MODE_CUBE: curlights='Cube';break; case scene.LIGHT_MODE_CAD: curlights='CAD';break; case scene.LIGHT_MODE_HEADLAMP: curlights='Headlamp';break; } if(curlights!='Artwork') res+=' LIGHTS='+curlights+'\n'; switch(scene.renderMode){ case scene.RENDER_MODE_BOUNDING_BOX: defaultrender='BoundingBox';break; case scene.RENDER_MODE_TRANSPARENT_BOUNDING_BOX: defaultrender='TransparentBoundingBox';break; case scene.RENDER_MODE_TRANSPARENT_BOUNDING_BOX_OUTLINE: defaultrender='TransparentBoundingBoxOutline';break; case scene.RENDER_MODE_VERTICES: defaultrender='Vertices';break; case scene.RENDER_MODE_SHADED_VERTICES: defaultrender='ShadedVertices';break; case scene.RENDER_MODE_WIREFRAME: defaultrender='Wireframe';break; case scene.RENDER_MODE_SHADED_WIREFRAME: defaultrender='ShadedWireframe';break; case scene.RENDER_MODE_SOLID: defaultrender='Solid';break; case scene.RENDER_MODE_TRANSPARENT: defaultrender='Transparent';break; case scene.RENDER_MODE_SOLID_WIREFRAME: defaultrender='SolidWireframe';break; case scene.RENDER_MODE_TRANSPARENT_WIREFRAME: defaultrender='TransparentWireframe';break; case scene.RENDER_MODE_ILLUSTRATION: defaultrender='Illustration';break; case scene.RENDER_MODE_SOLID_OUTLINE: defaultrender='SolidOutline';break; case scene.RENDER_MODE_SHADED_ILLUSTRATION: defaultrender='ShadedIllustration';break; case scene.RENDER_MODE_HIDDEN_WIREFRAME: defaultrender='HiddenWireframe';break; } if(defaultrender!='Solid') res+=' RENDERMODE='+defaultrender+'\n'; //detect existing Clipping Plane (3D Cross Section) var clip=null; if( clip=scene.nodes.getByName('$$$$$$')|| clip=scene.nodes.getByName('Clipping Plane') ); for(var i=0;i<scene.nodes.count;i++){ var nd=scene.nodes.getByIndex(i); if(nd==clip||nd.name=='') continue; var ndUTFName=''; for (var j=0; j<nd.name.length; j++) { var theUnicode = nd.name.charCodeAt(j).toString(16); while (theUnicode.length<4) theUnicode = '0' + theUnicode; ndUTFName += theUnicode; } var end=nd.name.lastIndexOf('.'); if(end>0) var ndUserName=nd.name.substr(0,end); else var ndUserName=nd.name; respart=' PART='+ndUserName+'\n'; respart+=' UTF16NAME='+ndUTFName+'\n'; defaultvals=true; if(!nd.visible){ respart+=' VISIBLE=false\n'; defaultvals=false; } if(nd.opacity<1.0){ respart+=' OPACITY='+nd.opacity+'\n'; defaultvals=false; } if(nd.constructor.name=='Mesh'){ currender=defaultrender; switch(nd.renderMode){ case scene.RENDER_MODE_BOUNDING_BOX: currender='BoundingBox';break; case scene.RENDER_MODE_TRANSPARENT_BOUNDING_BOX: currender='TransparentBoundingBox';break; case scene.RENDER_MODE_TRANSPARENT_BOUNDING_BOX_OUTLINE: currender='TransparentBoundingBoxOutline';break; case scene.RENDER_MODE_VERTICES: currender='Vertices';break; case scene.RENDER_MODE_SHADED_VERTICES: currender='ShadedVertices';break; case scene.RENDER_MODE_WIREFRAME: currender='Wireframe';break; case scene.RENDER_MODE_SHADED_WIREFRAME: currender='ShadedWireframe';break; case scene.RENDER_MODE_SOLID: currender='Solid';break; case scene.RENDER_MODE_TRANSPARENT: currender='Transparent';break; case scene.RENDER_MODE_SOLID_WIREFRAME: currender='SolidWireframe';break; case scene.RENDER_MODE_TRANSPARENT_WIREFRAME: currender='TransparentWireframe';break; case scene.RENDER_MODE_ILLUSTRATION: currender='Illustration';break; case scene.RENDER_MODE_SOLID_OUTLINE: currender='SolidOutline';break; case scene.RENDER_MODE_SHADED_ILLUSTRATION: currender='ShadedIllustration';break; case scene.RENDER_MODE_HIDDEN_WIREFRAME: currender='HiddenWireframe';break; //case scene.RENDER_MODE_DEFAULT: // currender='Default';break; } if(currender!=defaultrender){ respart+=' RENDERMODE='+currender+'\n'; defaultvals=false; } } if(origtrans[nd.name]&&!nd.transform.isEqual(origtrans[nd.name])){ var lvec=nd.transform.transformDirection(new Vector3(1,0,0)); var uvec=nd.transform.transformDirection(new Vector3(0,1,0)); var vvec=nd.transform.transformDirection(new Vector3(0,0,1)); respart+=' TRANSFORM=' +lvec.x+' '+lvec.y+' '+lvec.z+' ' +uvec.x+' '+uvec.y+' '+uvec.z+' ' +vvec.x+' '+vvec.y+' '+vvec.z+' ' +nd.transform.translation.x+' ' +nd.transform.translation.y+' ' +nd.transform.translation.z+'\n'; defaultvals=false; } respart+=' END\n'; if(!defaultvals) res+=respart; } if(clip){ var centre=clip.transform.translation; var normal=clip.transform.transformDirection(new Vector3(0,0,1)); res+=' CROSSSECT\n'; if(!(centre.x==0 && centre.y==0 && centre.z==0)) res+=host.util.printf( ' CENTER=%s %s %s\n', centre.x, centre.y, centre.z); if(!(normal.x==1 && normal.y==0 && normal.z==0)) res+=host.util.printf( ' NORMAL=%s %s %s\n', normal.x, normal.y, normal.z); res+=host.util.printf( ' VISIBLE=%s\n', clip.visible); res+=host.util.printf( ' PLANECOLOR=%s %s %s\n', clip.material.emissiveColor.r, clip.material.emissiveColor.g, clip.material.emissiveColor.b); res+=host.util.printf( ' OPACITY=%s\n', clip.opacity); res+=host.util.printf( ' INTERSECTIONCOLOR=%s %s %s\n', clip.wireframeColor.r, clip.wireframeColor.g, clip.wireframeColor.b); res+=' END\n';// for(var propt in clip){// console.println(propt+':'+clip[propt]);// } } res+='END\n'; host.console.show();// host.console.clear(); host.console.println('%%\n%% Add the following VIEW section to a file of\n'+ '%% predefined views (See option "3Dviews"!).\n%%\n' + '%% The view may be given a name after VIEW=...\n' + '%% (Remove \'%\' in front of \'=\'.)\n%%'); host.console.println(res + '\n');}//add items to 3D context menuruntime.addCustomMenuItem("dfltview", "Generate Default View", "default", 0);runtime.addCustomMenuItem("currview", "Get Current View", "default", 0);runtime.addCustomMenuItem("csection", "Cross Section", "checked", 0);//menu event handlersmenuEventHandler = new MenuEventHandler();menuEventHandler.onEvent = function(e) { switch(e.menuItemName){ case "dfltview": calc3Dopts(); break; case "currview": get3Dview(); break; case "csection": addremoveClipPlane(e.menuItemChecked); break; }};runtime.addEventHandler(menuEventHandler);//global variable taking reference to currently selected node;var target=null;selectionEventHandler=new SelectionEventHandler();selectionEventHandler.onEvent=function(e){ if(e.selected&&e.node.name!=''){ target=e.node; }else{ target=null; }}runtime.addEventHandler(selectionEventHandler);cameraEventHandler=new CameraEventHandler();cameraEventHandler.onEvent=function(e){ var clip=null; runtime.removeCustomMenuItem("csection"); runtime.addCustomMenuItem("csection", "Cross Section", "checked", 0); if(clip=scene.nodes.getByName('$$$$$$')|| //predefined scene.nodes.getByName('Clipping Plane')){ //added via context menu runtime.removeCustomMenuItem("csection"); runtime.addCustomMenuItem("csection", "Cross Section", "checked", 1); } if(clip){//plane in predefined views must be rotated by 90 deg around normal clip.transform.rotateAboutLineInPlace( Math.PI/2,clip.transform.translation, clip.transform.transformDirection(new Vector3(0,0,1)) ); } for(var i=0; i<rot4x4.length; i++){rot4x4[i].setIdentity()} target=null;}runtime.addEventHandler(cameraEventHandler);var rot4x4=new Array(); //keeps track of spin and tilt axes transformations//key event handler for scaling moving, spinning and tilting objectskeyEventHandler=new KeyEventHandler();keyEventHandler.onEvent=function(e){ var backtrans=new Matrix4x4(); var trgt=null; if(target) { trgt=target; var backtrans=new Matrix4x4(); var trans=trgt.transform; var parent=trgt.parent; while(parent.transform){ //build local to world transformation matrix trans.multiplyInPlace(parent.transform); //also build world to local back-transformation matrix backtrans.multiplyInPlace(parent.transform.inverse.transpose); parent=parent.parent; } backtrans.transposeInPlace(); }else{ if( trgt=scene.nodes.getByName('$$$$$$')|| trgt=scene.nodes.getByName('Clipping Plane') ) var trans=trgt.transform; } if(!trgt) return; var tname=trgt.name; if(typeof(rot4x4[tname])=='undefined') rot4x4[tname]=new Matrix4x4(); if(target) var tiltAxis=rot4x4[tname].transformDirection(new Vector3(0,1,0)); else var tiltAxis=trans.transformDirection(new Vector3(0,1,0)); var spinAxis=rot4x4[tname].transformDirection(new Vector3(0,0,1)); //get the centre of the mesh if(target&&trgt.constructor.name=='Mesh'){ var centre=trans.transformPosition(trgt.computeBoundingBox().center); }else{ //part group (Node3 parent node, clipping plane) var centre=new Vector3(trans.translation); } switch(e.characterCode){ case 30://tilt up rot4x4[tname].rotateAboutLineInPlace( -Math.PI/900,rot4x4[tname].translation,tiltAxis); trans.rotateAboutLineInPlace(-Math.PI/900,centre,tiltAxis); break; case 31://tilt down rot4x4[tname].rotateAboutLineInPlace( Math.PI/900,rot4x4[tname].translation,tiltAxis); trans.rotateAboutLineInPlace(Math.PI/900,centre,tiltAxis); break; case 28://spin right if(e.ctrlKeyDown&&target){ trans.rotateAboutLineInPlace(-Math.PI/900,centre,spinAxis); }else{ rot4x4[tname].rotateAboutLineInPlace( -Math.PI/900,rot4x4[tname].translation,new Vector3(0,0,1)); trans.rotateAboutLineInPlace(-Math.PI/900,centre,new Vector3(0,0,1)); } break; case 29://spin left if(e.ctrlKeyDown&&target){ trans.rotateAboutLineInPlace(Math.PI/900,centre,spinAxis); }else{ rot4x4[tname].rotateAboutLineInPlace( Math.PI/900,rot4x4[tname].translation,new Vector3(0,0,1)); trans.rotateAboutLineInPlace(Math.PI/900,centre,new Vector3(0,0,1)); } break; case 120: //x translateTarget(trans, new Vector3(1,0,0), e); break; case 121: //y translateTarget(trans, new Vector3(0,1,0), e); break; case 122: //z translateTarget(trans, new Vector3(0,0,1), e); break; case 88: //shift + x translateTarget(trans, new Vector3(-1,0,0), e); break; case 89: //shift + y translateTarget(trans, new Vector3(0,-1,0), e); break; case 90: //shift + z translateTarget(trans, new Vector3(0,0,-1), e); break; case 115: //s trans.translateInPlace(centre.scale(-1)); trans.scaleInPlace(1.01); trans.translateInPlace(centre.scale(1)); break; case 83: //shift + s trans.translateInPlace(centre.scale(-1)); trans.scaleInPlace(1/1.01); trans.translateInPlace(centre.scale(1)); break; } trans.multiplyInPlace(backtrans);}runtime.addEventHandler(keyEventHandler);//translates object by amount calculated from Canvas sizefunction translateTarget(t, d, e){ var cam=scene.cameras.getByIndex(0); if(cam.projectionType==cam.TYPE_PERSPECTIVE){ var scale=Math.tan(cam.fov/2) *cam.targetPosition.subtract(cam.position).length /Math.min(e.canvasPixelWidth,e.canvasPixelHeight); }else{ var scale=cam.viewPlaneSize/2 /Math.min(e.canvasPixelWidth,e.canvasPixelHeight); } t.translateInPlace(d.scale(scale));}function addremoveClipPlane(chk) { var curTrans=getCurTrans(); var clip=scene.createClippingPlane(); if(chk){ //add Clipping Plane and place its center either into the camera target //position or into the centre of the currently selected mesh node var centre=new Vector3(); if(target){ var trans=target.transform; var parent=target.parent; while(parent.transform){ trans=trans.multiply(parent.transform); parent=parent.parent; } if(target.constructor.name=='Mesh'){ var centre=trans.transformPosition(target.computeBoundingBox().center); }else{ var centre=new Vector3(trans.translation); } target=null; }else{ centre.set(scene.cameras.getByIndex(0).targetPosition); } clip.transform.setView( new Vector3(0,0,0), new Vector3(1,0,0), new Vector3(0,1,0)); clip.transform.translateInPlace(centre); }else{ if( scene.nodes.getByName('$$$$$$')|| scene.nodes.getByName('Clipping Plane') ){ clip.remove();clip=null; } } restoreTrans(curTrans); return clip;}//function to store current transformation matrix of all nodes in the scenefunction getCurTrans() { var tA=new Array(); for(var i=0; i<scene.nodes.count; i++){ var nd=scene.nodes.getByIndex(i); if(nd.name=='') continue; tA[nd.name]=new Matrix4x4(nd.transform); } return tA;}//function to restore transformation matrices given as argfunction restoreTrans(tA) { for(var i=0; i<scene.nodes.count; i++){ var nd=scene.nodes.getByIndex(i); if(tA[nd.name]) nd.transform.set(tA[nd.name]); }}//store original transformation matrix of all mesh nodes in the scenevar origtrans=getCurTrans();//set initial state of "Cross Section" menu entrycameraEventHandler.onEvent(1);//host.console.clear();

Solution: We first solve for a. Note that 3003 = 3 ·7 ·11 ·13. We have 3003 =(a2

)= a(a−1)

2 ≈ a2

2 . This

means we can estimate a ≈√

3003 · 2, so a is a little less than 80. Furthermore, 11 | 2 ·3003 = a(a−1),meaning one of a or a−1 must be divisible by 11. Thus, either a = 77 or a = 78. Conveniently, 13 | 78,so we get a = 78 and we can verify that

(782

)= 3003.

We solve for b < 15 − b satisfying(15b

)= 3003. Because

(15b

)= 15!

b!(15−b)! is divisible by 11, we must

have b ≥ 5. But we’re given(146

)= 3003, so b < 6. We conclude that b = 5, and it follows that

a+ b(15− b) = 128.

11. [8] Two diameters and one radius are drawn in a circle of radius 1, dividing the circle into 5 sectors.The largest possible area of the smallest sector can be expressed as a

bπ, where a, b are relatively primepositive integers. Compute 100a+ b.

Proposed by: William Qian

Answer: 106

Solution: Let the two diameters split the circle into four sectors of areas A, B, A, and B, whereA+B = π

2 . Without loss of generality, let A ≤ B.

If our radius cuts into a sector of area A, the area of the smallest sector will be of the form min(x,A−x).Note that min(A− x, x) ≤ A

2 ≤ π8 .

If our radius cuts into a sector of area B, then the area of the smallest sector will be of the formmin(A, x,B−x) ≤ min(A, B2 ) = min(A, π4 − A

2 ). This equals A if A ≤ π6 and it equals π

4 − A2 if A ≥ π

6 .This implies that the area of the smallest sector is maximized when A = π

6 , and we get an area of π6 .

12. [8] In a single-elimination tournament consisting of 29 = 512 teams, there is a strict ordering on theskill levels of the teams, but Joy does not know that ordering. The teams are randomly put into abracket and they play out the tournament, with the better team always beating the worse team. Joyis then given the results of all 511 matches and must create a list of teams such that she can guaranteethat the third-best team is on the list. What is the minimum possible length of Joy’s list?

Proposed by: Cory Hixson

Answer: 45

Solution: The best team must win the tournament. The second-best team has to be one of the 9teams that the first best team beat; call these teams marginal. The third best team must have lost toeither the best or the second-best team, so it must either be marginal or have lost to a marginal team.Since there is exactly one marginal team that won k games for each integer 0 ≤ k ≤ 8, we can thenconclude that there are 1 + 2 + · · ·+ 9 = 45 teams that are either marginal or lost to a marginal team.Moreover, it is not hard to construct a scenario in which the third-best team is any of these 45 teams,so we cannot do better.

13. [9] Wendy is playing darts with a circular dartboard of radius 20. Whenever she throws a dart, itlands uniformly at random on the dartboard. At the start of her game, there are 2020 darts placedrandomly on the board. Every turn, she takes the dart farthest from the center, and throws it at theboard again. What is the expected number of darts she has to throw before all the darts are within10 units of the center?

Proposed by: Vincent Bian

Answer: 6060

Solution: Consider an individual dart. There is a 14 probability it is already within 10 units of the

center. If not, for every throw there is a 14 probability it is not thrown again. Thus, if E is the expected

value of times it is thrown, we find E = 1 + 34E =⇒ E = 4.

As a result, the expected number of times each dart is thrown is 34 · 4 = 3. By linearity of expectation,

the answer is 2020 · 3 = 6060.

14. [9] A point (x, y) is selected uniformly at random from the unit square S = {(x, y) | 0 ≤ x ≤ 1, 0 ≤y ≤ 1}. If the probability that (3x + 2y, x + 4y) is in S is a

b , where a, b are relatively prime positiveintegers, compute 100a+ b.

Proposed by: Christopher Xu

Answer: 820

Solution:

Under the transformation (x, y) 7→ (3x + 2y, x + 4y), S is mapped to a parallelogram with vertices(0, 0), (3, 1), (5, 5), and (2, 4). Using the shoelace formula, the area of this parallelogram is 10.

The intersection of the image parallelogram and S is the quadrilateral with vertices (0, 0),(1, 13), (1, 1),

and(12 , 1). To get this quadrilateral, we take away a right triangle with legs 1 and 1

2 and a right trianglewith legs 1 and 1

3 from the unit square. So the quadrilateral has area 1 − 12 · 12 − 1

2 · 13 = 712 . Then

the fraction of the image parallelogram that lies within S is712

10 = 7120 , which is the probability that a

point stays in S after the mapping.

15. [9] For a real number r, the quadratics x2 + (r− 1)x+ 6 and x2 + (2r+ 1)x+ 22 have a common realroot. The sum of the possible values of r can be expressed as a

b , where a, b are relatively prime positiveintegers. Compute 100a+ b.

Proposed by: Andrew Yao

Answer: 405

Solution: Let the common root be s. Then,

s2 + (r − 1)s+ 6 = s2 + (2r + 1)s+ 22,

and s = − 16r+2 . Substituting this into s2 + (r − 1)s+ 6 = 0 yields

256

(r + 2)2− 16(r − 1)

r + 2+ 6 = 0.

After multiplying both sides by (r + 2)2, the equation becomes

256− 16(r − 1)(r + 2) + 6(r + 2)2 = 0,

which simplifies into5r2 − 4r − 156 = 0.

Thus, by Vieta’s Formulas, the sum of the possible values of r is 45 .

16. [10] Three players play tic-tac-toe together. In other words, the three players take turns placing an“A”, “B”, and “C”, respectively, in one of the free spots of a 3 × 3 grid, and the first player to havethree of their label in a row, column, or diagonal wins. How many possible final boards are there wherethe player who goes third wins the game? (Rotations and reflections are considered different boards,but the order of placement does not matter.)

Proposed by: Andrew Lin

Answer: 148

Solution: In all winning cases for the third player, every spot in the grid must be filled. There aretwo ways that player C wins along a diagonal, and six ways that player C wins along a row or col-umn. In the former case, any arrangement of the As and Bs is a valid board, since every other row,column, and diagonal is blocked. So there are

(63

)= 20 different finishing boards each for this case.

However, in the latter case, we must make sure players A and B do not complete a row or columnof their own, so only 20−2 = 18 of the finishing boards are valid. The final answer is 2·20+6·18 = 148.

17. [10] Let N>1 denote the set of positive integers greater than 1. Let f : N>1 → N>1 be a function suchthat f(mn) = f(m)f(n) for all m,n ∈ N>1. If f(101!) = 101!, compute the number of possible valuesof f(2020 · 2021).

Proposed by: Sheldon Kieren Tan

Answer: 66

Solution: For a prime p and positive integer n, we let vp(n) denote the largest nonnegative integer ksuch that pk | n. Note that f is determined by its action on primes. Since f(101!) = 101!, by countingprime factors, f must permute the set of prime factors of 101!; moreover, if p and q are prime factors of101! and f(p) = q, we must have vp(101!) = vq(101!). This clearly gives f(2) = 2, f(5) = 5, so it sufficesto find the number of possible values for f(43 · 47 · 101). (We can factor 2021 = 452 − 22 = 43 · 47.)

There are 4 primes with vp(101!) = 2 (namely, 37, 41, 43, 47), so there are 6 possible values forf(43 · 47). Moreover, there are 11 primes with vp(101!) = 1 (namely, 53, 59, 61, 67, 71, 73, 79, 83, 89,97, 101). Hence there are 66 possible values altogether.

18. [10] Suppose Harvard Yard is a 17 × 17 square. There are 14 dorms located on the perimeter of theYard. If s is the minimum distance between two dorms, the maximum possible value of s can beexpressed as a−

√b where a, b are positive integers. Compute 100a+ b.

Proposed by: Hahn Lheem

Answer: 602

Solution: If two neighboring dorms are separated by a distance of more than s, we can move themslightly closer together and adjust the other dorms, increasing s. Therefore, in an optimal arrangement,the dorms form an equilateral 14-gon with side length s.

By scaling, the problem is now equivalent to finding the smallest a such that there exist 14 verticeson the boundary of an a× a square that form an equilateral 14-gon with side length 1. Such a 14-gonmust be centrally symmetric, yielding the following picture:

x1

y1

x2

y2

1 1 1

We know that x21 + y21 = x22 + y22 = 1 and x1 + x2 + 3 = y1 + y2 + 2 = a. Moreover, if these equationsare satisfied, then such a 14-gon exists. We now consider the vectors ~v1 = (x1, y1) and ~v2 = (x2, y2).These unit vectors are in the first quadrant and add to (a− 3, a− 2), which lies on the line y = x+ 1.

Since ~v1 and ~v2 must lie on the first quadrant, from the above diagram we deduce that the minimum

value of a occurs when one of ~v1, ~v2 is (0, 1), meaning that (a− 3, a− 2) = (√22 ,√22 + 1). This means

that a =√22 + 3, so the maximum possible value of s is

17√22 + 3

= 17 · 3−√22

17/2= 6−

√2.

19. [11] Three distinct vertices of a regular 2020-gon are chosen uniformly at random. The probability thatthe triangle they form is isosceles can be expressed as a

b , where a and b are relatively prime positiveintegers. Compute 100a+ b.

Proposed by: Hahn Lheem

Answer: 773

Solution: The number of isosceles triangles that share vertices with the 2020-gon is 2020 · 1009,since there are 2020 ways to choose the apex of the triangle and then 1009 ways to choose the othertwo vertices. (Since 2020 is not divisible by 3, there are no equilateral triangles, so no triangle isovercounted.)

Therefore, the probability is

2020 · 1009(20203

) =2020 · 2018/2

2020 · 2019 · 2018/6=

3

2019=

1

673.

20. [11] Let ω1 be a circle of radius 5, and let ω2 be a circle of radius 2 whose center lies on ω1. Let thetwo circles intersect at A and B, and let the tangents to ω2 at A and B intersect at P . If the area of

4ABP can be expressed as a√b

c , where b is square-free and a, c are relatively prime positive integers,compute 100a+ 10b+ c.

Proposed by: Hahn Lheem

Answer: 19285

Solution:

O1O2

A

B

P

Let O1 and O2 be the centers of ω1 and ω2, respectively. Because

∠O2AP + ∠O2BP = 90◦ + 90◦ = 180◦,

quadrilateral O2APB is cyclic. But O2, A, and B lie on ω1, so P lies on ω1 and O2P is a diameter ofω1.

From the Pythagorean theorem on triangle PAO2, we can calculate AP = 4√

6, so sin∠AO2P = 2√6

5and cos∠AO2P = 1

5 . Because 4AO2P and 4BO2P are congruent, we have

sin∠APB = sin 2∠AO2P = 2 sin∠AO2P cos∠AO2P =4√

6

25,

implying that

[APB] =PA · PB

2sin∠APB =

192√

6

25.

21. [11] Let f(n) be the number of distinct prime divisors of n less than 6. Compute

2020∑n=1

f(n)2.

Proposed by: Milan Haiman

Answer: 3431

Solution: Define

1a|n =

{1 a | n0 otherwise

Then

f(n)2 = (12|n + 13|n + 15|n)2

= 12|n + 13|n + 15|n + 2(12|n13|n + 12|n15|n + 13|n15|n)

= 12|n + 13|n + 15|n + 2(16|n + 110|n + 115|n).

So summing f(n)2 over integers 1 ≤ n ≤ 2020 is the same as summing 1 for each time n is divisible by2, 3, or 5, and additionally summing 2 for each time n is divisible by 6, 10, or 15.

2020∑n=1

f(n)2 =

⌊2020

2

⌋+

⌊2020

3

⌋+

⌊2020

5

⌋+ 2

(⌊2020

6

⌋+

⌊2020

10

⌋+

⌊2020

15

⌋)= 1010 + 673 + 404 + 2(336 + 202 + 134) = 3431.

22. [12] In triangle ABC, AB = 32, AC = 35, and BC = x. What is the smallest positive integer x suchthat 1 + cos2A, cos2B, and cos2 C form the sides of a non-degenerate triangle?

Proposed by: Hahn Lheem

Answer: 48

Solution: By the triangle inequality, we wish cos2B+ cos2 C > 1 + cos2A. The other two inequalitiesare always satisfied, since 1 + cos2A ≥ 1 ≥ cos2B, cos2 C. Rewrite the above as

2− sin2B − sin2 C > 2− sin2A,

so it is equivalent to sin2B+sin2 C < sin2A. By the law of sines, sinA : sinB : sinC = BC : AC : AB.Therefore,

sin2B + sin2 C < sin2A ⇐⇒ CA2 +AB2 < x2.

Since CA2 +AB2 = 2249, the smallest possible value of x such that x2 > 2249 is 48.

23. [12] Two points are chosen inside the square {(x, y) | 0 ≤ x, y ≤ 1} uniformly at random, and a unitsquare is drawn centered at each point with edges parallel to the coordinate axes. The expected areaof the union of the two squares can be expressed as a

b , where a, b are relatively prime positive integers.Compute 100a+ b.

Proposed by: Yannick Yao

Answer: 1409

Solution:

A

B

1−∆y

1−∆x

Let ∆x and ∆y be the positive differences between the x coordinates and y coordinates of the centersof the squares, respectively. Then, the length of the intersection of the squares along the x dimensionis 1−∆x, and likewise the length along the y dimension is 1−∆y. In order to find the expectation of∆x and ∆y, we can find the volume of the set of points (a, b, c) such that 0 ≤ a, b ≤ 1 and c ≤ |a− b|.This set is composed of the two pyramids of volume 1

6 shown below:

Since the expected distance between two points on a unit interval is therefore 13 , we have that

E[1 − ∆x] = E[1 − ∆y] = 23 . The expectation of the product of independent variables equals the

product of their expectations, so the expected area of intersection is 49 and the expected area of union

is 2− 49 = 14

9 .

24. [12] Compute the number of positive integers less than 10! which can be expressed as the sum of atmost 4 (not necessarily distinct) factorials.

Proposed by: Sheldon Kieren Tan

Answer: 648

Solution: Since 0! = 1! = 1, we ignore any possible 0!’s in our sums.

Call a sum of factorials reduced if for all positive integers k, the term k! appears at most k times. Itis straightforward to show that every positive integer can be written uniquely as a reduced sum offactorials. Moreover, by repeatedly replacing k + 1 occurrences of k! with (k + 1)!, every non-reducedsum of factorials is equal to a reduced sum with strictly fewer terms, implying that the aforementionedreduced sum associated to a positive integer n in fact uses the minimum number of factorials necessary.

It suffices to compute the number of nonempty reduced sums involving {1!, 2!, . . . , 9!} with at most 4terms. By stars and bars, the total number of such sums, ignoring the reduced condition, is

(139

)= 714.

The sums that are not reduced must either contain two copies of 1!, three copies of 2!, or four copiesof 3!. Note that at most one of these conditions is true, so we can count them separately. If kterms are fixed, there are

(13−k

9

)ways to choose the rest of the terms, meaning that we must subtract(

119

)+(109

)+(99

)= 66. Our final answer is 714− 66 = 648.

25. [13] Let a1, a2, a3, . . . be a sequence of positive integers where a1 =∑100i=0 i! and ai + ai+1 is an odd

perfect square for all i ≥ 1. Compute the smallest possible value of a1000.

Proposed by: Sheldon Kieren Tan

Answer: 7

Solution: Note that a1 ≡ 1 + 1 + 2 + 6 ≡ 2 (mod 8). Since a1 + a2 must be an odd perfect square,we must have a1 + a2 ≡ 1 (mod 8) =⇒ a2 ≡ 7 (mod 8). Similarly, since a2 + a3 is an odd perfect

square, we must have a3 ≡ 2 (mod 8). We can continue this to get a2k−1 ≡ 2 (mod 8) and a2k ≡ 7(mod 8), so in particular, we have a1000 ≡ 7 (mod 8), so a1000 ≥ 7.

Now, note that we can find some large enough odd perfect square t2 such that t2 − a1 ≥ 23. Leta2 = t2− a1. Since a2 ≡ 7 (mod 8), we can let a2− 7 = 8k for some integer k ≥ 2. Now, since we have(2k + 1)2 − (2k − 1)2 = 8k, if we let a3 = (2k − 1)2 − 7, then

a2 + a3 = a2 + ((2k − 1)2 − 7) = (2k − 1)2 + (a2 − 7) = (2k − 1)2 + 8k = (2k + 1)2,

which is an odd perfect square. Now, we can let a4 = 7 and we will get a3 + a4 = (2k − 1)2. Fromhere, we can let 2 = a5 = a7 = a9 = · · · and 7 = a4 = a6 = a8 = · · · , which tells us that the leastpossible value for a1000 is 7.

26. [13] Two players play a game where they are each given 10 indistinguishable units that must bedistributed across three locations. (Units cannot be split.) At each location, a player wins at thatlocation if the number of units they placed there is at least 2 more than the units of the other player.If both players distribute their units randomly (i.e. there is an equal probability of them distributingtheir units for any attainable distribution across the 3 locations), the probability that at least onelocation is won by one of the players can be expressed as a

b , where a, b are relatively prime positiveintegers. Compute 100a+ b.

Proposed by: Sheldon Kieren Tan

Answer: 1011

Solution: By stars and bars, the total number of distributions is(122

)2= 662. If no locations are won,

either both distributions are identical or the difference between the two is (1, 0,−1), in some order.The first case has 66 possibilities. If the difference is (1, 0,−1), we can construct all such possiblitiesby choosing nonnegative integers a, b, c that sum to 9, and having the two players choose (a + 1, b, c)and (a, b, c + 1). This can be done in

(112

)= 55 ways. In total, the second case has 6 · 55 = 5 · 66

possibilities.

Thus the probability that no locations are won is 6·66662 = 1

11 , meaning that the answer is 1011 .

27. [13] In 4ABC, D and E are the midpoints of BC and CA, respectively. AD and BE intersect at G.Given that GECD is cyclic, AB = 41, and AC = 31, compute BC.

Proposed by: Sheldon Kieren Tan

Answer: 49

Solution:

A

B CD

EG

By Power of a Point,23AD

2 = AD ·AG = AE ·AC = 12 · 312

so AD2 = 34 · 312. The median length formula yields

AD2 = 14 (2AB2 + 2AC2 −BC2),

whenceBC =

√2AB2 + 2AC2 − 4AD2 =

√2 · 412 + 2 · 312 − 3 · 312 = 49.

28. [15] Bernie has 2020 marbles and 2020 bags labeled B1, . . . , B2020 in which he randomly distributesthe marbles (each marble is placed in a random bag independently). If E the expected number ofintegers 1 ≤ i ≤ 2020 such that Bi has at least i marbles, compute the closest integer to 1000E.

Proposed by: Benjamin Kang

Answer: 1000

Solution: Let pi be the probability that a bag has i marbles. Then, by linearity of expectation, wefind

E = (p1 + p2 + · · · ) + (p2 + p3 + · · · ) + · · · = p1 + 2p2 + 3p3 + · · · .This is precisely the expected value of the number of marbles in a bag. By symmetry, this is 1.

29. [15] In acute triangle ABC, let H be the orthocenter and D the foot of the altitude from A. Thecircumcircle of triangle BHC intersects AC at E 6= C, and AB at F 6= B. If BD = 3, CD = 7,and AH

HD = 57 , the area of triangle AEF can be expressed as a

b , where a, b are relatively prime positiveintegers. Compute 100a+ b.

Proposed by: Andrew Yao

Answer: 12017

Solution:

A

B CD

H

P

Q

E

F

Let AH intersect the circumcircle of 4ABC again at P , and the circumcircle of 4BHC again atQ. Because ∠BHC = 180 − ∠A = ∠BPC, P is the reflection of H over D. Thus, we know thatPD = HD. From power of a point and AD = 12HD

7 ,

BD · CD = AD · PD =12HD2

7.

From this, HD = 72 and AH = 5

2 . Furthermore, because 4BHC is the reflection of 4BPC overBC, the circumcircle of 4BHC is the reflection of the circumcircle of 4ABC over BC. Then, AQ =2AD = 12. Applying Power of a Point,

AC ·AE = AB ·AF = AH ·AQ = 30.

We can compute AC =√

85 and AB = 3√

5, which means that AE = 6√85

17 and AF = 2√

5. Also,

[ABC] = BC·AD2 = 30. Therefore,

[AEF ] =AE ·AFAC ·AB · [ABC] =

4

17· 30 =

120

17.

30. [15] Let a1, a2, a3, . . . be a sequence of positive real numbers that satisfies

∞∑n=k

(n

k

)an =

1

5k,

for all positive integers k. The value of a1 − a2 + a3 − a4 + · · · can be expressed as ab , where a, b are

relatively prime positive integers. Compute 100a+ b.

Proposed by: Akash Das

Answer: 542

Solution: Let Sk = 15k

. In order to get the coefficient of a2 to be −1, we need to have S1 − 3S3.This subtraction makes the coefficient of a3 become −6. Therefore, we need to add 7S3 to make thecoefficient of a4 equal to 1. The coefficient of a4 in S1 − 3S3 + 7S5 is 14, so we must subtract 15S4.We can continue to pattern to get that we want to compute S1 − 3S2 + 7S3 − 15S4 + 31S5 − · · · . Toprove that this alternating sum equals a1 − a2 + a3 − a4 + · · · , it suffices to show

n∑i=1

(−(−2)i + (−1)i)

(n

i

)= (−1)i+1.

To see this is true, note that the left hand side equals −(1 − 2)i + (1 − 1)i = (−1)i+1 by binomialexpansion. (We may rearrange the sums since the positivity of the ai’s guarantee absolute convergence.)Now, all that is left to do is to compute

∞∑i=1

(2i − 1)(−1)i−1

5i=

∞∑i=1

(−1)i

5i−∞∑i=1

(−2)i

5i=

−15

1− −15−

−25

1− −25=

5

42.

31. [17] For some positive real α, the set S of positive real numbers x with {x} > αx consists of theunion of several intervals, with total length 20.2. The value of α can be expressed as a

b , where a, b arerelatively prime positive integers. Compute 100a+ b. (Here, {x} = x−bxc is the fractional part of x.)

Proposed by: Daniel Zhu

Answer: 4633

Solution: If we note that x = {x}+ bxc, then we can rewrite our given inequality as {x} > α1−αbxc.

However, since {x} < 1, we know that we must have α1−αbxc < {x} < 1, so each interval is of the form

(n + α1−αn, n + 1) for some integer n, which has length 1−(n+1)α

1−α . If we let k be the smallest integer

such that 1−(k+1)α1−α < 0, then the total length of all our intervals is the sum

k−1∑n=0

1− (n+ 1)α

1− α =k − k(k+1)

2 α

1− α .

If we set this to 20.2, we can solve for α to get

α =k − 20.2

k(k+1)2 − 20.2

.

Since we defined k to be the smallest integer such that 1− (k+ 1)α < 0, we know that k is the largestinteger such that kα < 1. If we plug in our value for α, we get that this is equivalent to

k2 − 20.2kk(k+1)

2 − 20.2< 1 =⇒ k < 40.4.

Thus, we have k = 40, and plugging this in for our formula for α gives us

α =40− 20.2

40·412 − 20.2

=33

1333.

32. [17] The numbers 1, 2, . . . , 10 are written in a circle. There are four people, and each person randomlyselects five consecutive integers (e.g. 1, 2, 3, 4, 5, or 8, 9, 10, 1, 2). If the probability that there existssome number that was not selected by any of the four people is p, compute 10000p.

Proposed by: Hahn Lheem

Answer: 3690

Solution: The unselected numbers must be consecutive. Suppose that {1, 2, . . . , k} are the unselectednumbers for some k.

In this case, 1 cannot be selected, so there are 5 possible sets of consecutive numbers the people couldhave chosen. This leads to 54 possibilities. Moreover, 10 must be selected, so we must subtract 44

possibilities where neither 1 nor 10 are selected.

Therefore, accounting for the rotation of the unselected numbers, we find p = 10(54−44)104 = 3690

10000 .

33. [17] In quadrilateral ABCD, there exists a point E on segment AD such that AEED = 1

9 and ∠BEC isa right angle. Additionally, the area of triangle CED is 27 times more than the area of triangle AEB.If ∠EBC = ∠EAB, ∠ECB = ∠EDC, and BC = 6, compute the value of AD2.

Proposed by: Akash Das

Answer: 320

Solution:

A

B

C

DE

F

Extend sides AB and CD to intersect at point F . The angle conditions yield 4BEC ∼ 4AFD, so∠AFD = 90◦. Therefore, since ∠BFC and ∠BEC are both right angles, quadrilateral EBFC is cyclicand

∠EFC = ∠EBC = 90◦ − ∠ECB = 90◦ − ∠EDF,

implying that EF ⊥ AD.

Since AFD is a right triangle, we have ( FAFD )2 = AEED = 1

9 , so FAFD = 1

3 . Therefore EBEC = 1

3 . Since thearea of CED is 27 times more than the area of AEB, ED = 9 · EA, and EC = 3 · EB, we get that∠DEC = ∠AEB = 45◦. Since BECF is cyclic, we obtain ∠FBC = ∠FCB = 45◦, so FB = FC.

Since BC = 6, we get FB = FC = 3√

2. From 4EAB ∼ 4EFC we find AB = 13FC =

√2, so

FA = 4√

2. Similarly, FD = 12√

2. It follows that AD2 = FA2 + FD2 = 320.

34. [20] Let a be the proportion of teams that correctly answered problem 1 on the Guts round. EstimateA = b10000ac. An estimate of E earns max(0, b20−|A−E|/20c) points. If you have forgotten, question1 was the following:

Two hexagons are attached to form a new polygon P . What is the minimum number of sides that Pcan have?

Proposed by: Daniel Zhu

Answer: 2539

Solution: 689 teams participated in the guts round. Of these,

• 175 teams submitted 3, the correct answer;• 196 teams submitted 4;• 156 teams submitted 10 (the correct answer if the hexagons had to be regular);• 64 teams submitted 6 (the correct answer if one of the hexagons had to be regular);• 19 teams submitted 8 (the correct answer if the hexagons had to be convex);• 17 teams submitted 11;• 13 teams submitted other incorrect answers;• 49 teams did not submit an answer.

35. [20] Estimate A, the number of times an 8-digit number appears in Pascal’s triangle. An estimate ofE earns max(0, b20− |A− E|/200c) points.

Proposed by: Daniel Zhu

Answer: 180020660

Solution: We can obtain a good estimate by only counting terms of the form(a1

),(a2

),(aa−1), and(

aa−2). The last two cases are symmetric to the first two, so we will only consider the first two and

multiply by 2 at the end.

Since(a1

)= a, there are 90000000 values of a for which

(a1

)has eight digits. Moreover, since

(a2

)≈ a2/2,

the values of a for which(a2

)has eight digits vary from about

√2 · 107 to

√2 · 108, leading to about

104√

2(1− 10−1/2) ≈ 14000 · 0.69 = 9660 values for a.

Therefore, these terms yield an estimate of 180019320, good enough for 13 points. Of course, onewould expect this to be an underestimate, and even rounding up to 180020000 would give 16 points.

36. [20] Let pi be the ith prime. Let

f(x) =

50∑i=1

pixi−1 = 2 + 3x+ · · ·+ 229x49.

If a is the unique positive real number with f(a) = 100, estimate A = b100000ac. An estimate of Ewill earn max(0, b20− |A− E|/250c) points.

Proposed by: Daniel Zhu

Answer: 83601

Solution: Note f(x) is increasing. Since f(0) = 2 and f(1) ≈ 50000, we have 0 < a < 1.

Since we know that p50 = 229, we can crudely bound

f(x) .∞∑i=1

5ixi−1 =5

(1− x)2.

Setting this equal to 100 yields x = 1− 20−1/2 ≈ 0.78, so this is a good lower bound for a, though justoutside the window to receive points.

A better estimate can be obtained by noting that since p25 = 100, it is more accurate to write

f(x) .∞∑i=1

4ixi−1 =4

(1− x)2,

which yields a = 0.8, good enough for 5 points.

However, we can do better. If we know that a ≈ 0.8, the “most significant terms” will occur at thei where pi/pi+1 ≈ 0.8. The first few primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, so this transitionoccurs roughly at p8 = 19. Thus, it is more accurate to approximate f(x) =

∑∞i=1

198 ix

i−1, so

a = 1−√

19/800 ≈ 1− 40−1/2 ≈ 0.85, good enough for 14 points.

Repeating this process again with the new estimate for a reveals that p9 = 23 may have been a betterchoice, which yield a = 1−

√23/900 ≈ 1−

√0.0256 = 0.84. This is good enough for 18 points.

HMIC 2020March 28, 2020 – April 3, 2020

1. [6] Sir Alex is coaching a soccer team of n players of distinct heights. He wants to line themup so that for each player P , the total number of players that are either to the left of Pand taller than P or to the right of P and shorter than P is even. In terms of n, how manypossible orders are there?

2. [7] Some bishops and knights are placed on an infinite chessboard, where each square hasside length 1 unit. Suppose that the following conditions hold:

• For each bishop, there exists a knight on the same diagonal as that bishop (there maybe another piece between the bishop and the knight).

• For each knight, there exists a bishop that is exactly√

5 units away from it.

• If any piece is removed from the board, then at least one of the above conditions is nolonger satisfied.

If n is the total number of pieces on the board, find all possible values of n.

3. [9] Let P1P2P3P4 be a tetrahedron in R3 and let O be a point equidistant from each of itsvertices. Suppose there exists a point H such that for each i, the line PiH is perpendicular tothe plane through the other three vertices. Line P1H intersects the plane through P2, P3, P4

at A, and contains a point B 6= P1 such that OP1 = OB. Show that HB = 3HA.

4. [9] Let Ck = 1k+1

(2kk

)denote the kth Catalan number and p be an odd prime. Prove that

exactly half of the numbers in the set{p−1∑k=1

Cknk

∣∣∣∣∣n ∈ {1, 2, . . . , p− 1}

}

are divisible by p.

5. [11] A triangle and a circle are in the same plane. Show that the area of the intersection ofthe triangle and the circle is at most one third of the area of the triangle plus one half of thearea of the circle.

Time limit: 4 hours

HMIC 2020 SolutionsApril 21, 2020

1. [6] Sir Alex is coaching a soccer team of n players of distinct heights. He wants to line them up sothat for each player P , the total number of players that are either to the left of P and taller than Por to the right of P and shorter than P is even. In terms of n, how many possible orders are there?

Proposed by: Michael Ren

Answer:⌊n2

⌋!⌈n2

⌉!

Solution: We want to determine the number of permutations π of 1, . . . , n such that every numberis involved in an even number of inversions. The key observation is that the number of inversions k isinvolved in has the same parity as π(k) − k. This is since π(k) − k can be interpreted as the numberof i with π(i) < π(k) minus the number of i with i < k, meaning that i is counted with a weight of±1 ≡ 1 (mod 2) if and only if i is in an inversion with k.

From this, it follows that such permutations are the ones that permute the even and odd numbers, sothe answer is

⌊n2

⌋!⌈n2

⌉!.

2. [7] Some bishops and knights are placed on an infinite chessboard, where each square has side length1 unit. Suppose that the following conditions hold:

• For each bishop, there exists a knight on the same diagonal as that bishop (there may be anotherpiece between the bishop and the knight).

• For each knight, there exists a bishop that is exactly√

5 units away from it.

• If any piece is removed from the board, then at least one of the two conditions above is no longersatisfied.

If n is the total number of pieces on the board, find all possible values of n.

Proposed by: Sheldon Kieren Tan

Answer: 4 | n

Solution: Color the chessboard with the usual chessboard coloring. Note that bishops can only attackknights on the same colored squares, while knights can only attack bishops on different colored squares.Let BB ,WB , BN ,WN denote the number of bishops on black colored squares, bishops on white coloredsquares, knights on black colored squares and knights on white colored squares respectively.

Since removing a knight on a white colored square will cause a leave a bishop on a white colored squarewith no knight to attack, WN ≤ WB . Similarly, BN ≤ BB , BB ≤ WN , WB ≤ BN . Combining theinequalities, we find that WN = WB = BN = BB . Hence 4 | n.

To construct n = 4, we can place bishops at (1, 0), (2, 2) and knights at (0, 1), (3, 1). We can easilyextend this to any multiple of 4 by placing this configuration as many times as required across thegrid, making sure that we avoid the (finitely many) diagonals that existing bishops occupy.

(A harder variant of the problem involves changing the third condition from ”If any piece” to ”If anyproper subset of pieces”. The proof remains unchanged but the construction is more tedious as thepieces must be ”connected” without accidentally attacking one another. The question can also bephrased in a non-chess way to obscure the motivation behind the coloring.)

3. [9] Let P1P2P3P4 be a tetrahedron in R3 and let O be a point equidistant from each of its vertices.Suppose there exists a point H such that for each i, the line PiH is perpendicular to the plane throughthe other three vertices. Line P1H intersects the plane through P2, P3, P4 at A, and contains a pointB 6= P1 such that OP1 = OB. Show that HB = 3HA.

Proposed by: Michael Ren

Solution 1: Note that A is the orthocenter of triangle P2P3P4 by projecting the givens about H.

Next, let P2A intersect P3P4 at C and (P2P3P4) at D. Then AD = 2AC.

Note that H is the orthocenter of triangle P1P2C, with A being the foot of the altitude from P1. ThusAP2 ·AC = AH ·AP1.

Finally, by Power of a Point at A,

AP1 ·AB = AP2 ·AD = 2AP2 ·AC = 2AH ·AP1.

Thus AB = 2AH which gives HB = 3HA, as desired.

Solution 2: We use vectors with H as the origin. Then Pi · (Pj − Pk) = 0 from the definition of H.Thus Pi · Pj = Pi · Pk for any distinct i, j, k. In particular, the quantity Pi · Pj is a constant over alldistinct choices of i, j. Let this value be D. Let

O′ =1

2(P1 + P2 + P3 + P4).

We claim O′ = O. To see this, note that

(O′ − P1) · (O′ − P1) =1

4(P2 + P3 + P4 − P1) · (P2 + P3 + P4 − P1)

will be symmetric after expanding. Thus O′ is equidistant from the Pi’s, which means O′ = O.

Now, the projection of O onto P1H is O·P1

P1·P1P1, so

B = P1

((2O − P1) · P1

P1 · P1

)= P1

(3D

P1 · P1

).

Meanwhile, A is the projection of P2 on P1H, so

A = P1

(P1 · P2

P1 · P1

)= P1

(D

P1 · P1

).

Thus HB = 3HA, as desired.

4. [9] Let Ck = 1k+1

(2kk

)denote the kth Catalan number and p be an odd prime. Prove that exactly half

of the numbers in the set {p∈1∑k=1

Cknk

∣∣∣∣∣n ∈ {1, 2, . . . , p− 1}

}are divisible by p.

Proposed by: Tristan Shin

Solution 1: We work in Fp[X].

We claim that

(1− 4X)p+12 − 1 + 2Xp + 2X

p∈1∑k=0

CkXk = 0

The solution follows from this claim, as

p∈1∑k=1

Cknk ≡ 0 (mod p) ⇐⇒ (1− 4n)

p+12 − 1 + 2np + 2n ≡ 0 (mod p).

Since np ≡ n (mod p), this is equivalent to n ≡ 14 (mod p) or 1−4n being a non-zero quadratic residue

in Fp. But we must omit the solution of n = 0, so we get 1 + (p∈12 − 1) = p∈12 values of n that work.

Now we prove the claim. Observe the following facts:

• Cp∈1 = 12p∈1

(2p∈1p

)= −1 by Lucas’ theorem.

• For p∈12 < k < p− 1,

νp(Ck) = νp((2k)!)− νp((k + 1)!)− νp(k!) = 1

since k + 1 < p ≤ 2k, so Ck = 0.

We show that the coefficient of Xk in the LHS is 0 for k = 0, . . . , p. This is obvious for k > p+12 and

k = 0 by the above facts, as the coefficient is 2 + 2Cp∈1 for k = p, 2Ck for p+12 < k < p, and 0 for

k = 0.

Now for 1 ≤ k ≤ p+12 , the coefficient is

( p+12k

)(−4)k + 2Ck∈1 = 0. Write(p+1

2

k

)=

p+12 ·

p∈12 ·

p∈32 · · ·

p∈2k+32

k!

=12 · (−

12 ) · (− 3

2 ) · · · (− 2k∈32 )

k!

=(−1)k∈1

2k· 1 · 3 · · · (2k − 3)

k!

=(−1)k∈1

2k· (2k − 2)!

2k∈1(k − 1)!k!

= −2Ck∈1(−4)k

so the coefficient is 0.

Solution 2: We present an alternate proof of the key claim. Use the same starting facts as before.

Let Q(X) = −2 +

p−12∑

k=0

CkXk = −2 +Xp∈1 +

p∈1∑k=0

CkXk.

Square Q and multiply by X to get

XQ2 = 4X − 4X

p−12∑

k=0

CkXk +X

p−12∑

k=0

CkXk

2

= −4X − 4XQ+X

p∈2∑k=0

k∑j=0

CjCk∈jXk + C2

p−12

Xp

= −4X − 4XQ+

p∈2∑k=0

Ck+1Xk+1 + 4Xp

= −4X − 4XQ+Q+ 1−Xp∈1 + 4Xp

= (1− 4X)Q− (1− 4X)(Xp∈1 − 1).

It follows that

0 = 4X · (XQ2 − (1− 4X)Q+ (1− 4X) · (Xp∈1 − 1))

= 4X2Q2 − 4X(1− 4X)Q+ 4(1− 4X) · (Xp −X)

= 4X2Q2 − 4X(1− 4X)Q+ (1− 4X)2 − (1− 4X)p+1

= (2XQ+ (1− 4X)p+12 − (1− 4X)) · (2XQ− (1− 4X)

p+12 − (1− 4X)).

Evaluating the second factor at 0 gives −2, so it is not the zero polynomial. Thus the first factor isthe zero polynomial, from which the claim follows.

Solution 3: We prove the following generalization: Let q be a power of p. Then the polynomialq∈1∑k=1

CkXk has q+1

2 roots in Fq and p+12 roots in Fp. It once again suffices to prove the key claim, just

with p replaced by q.

Work in Fq[[X]], the ring of formal power series over Fq. Then

∞∑k=0

(2k

k

)Xk = (1− 4X)∈

12

= (1− 4X)q−12 ((1− 4X)q)∈

12

= (1− 4X)q−12 (1− 4Xq)∈

12

= (1− 4X)q−12

∞∑k=0

(2k

k

)Xqk.

Taking both sides mod Xq,q∈1∑k=0

(2k

k

)Xk = (1− 4X)

q−12 .

Then using(2qq

)= 2, we have

√1− 4X = (1− 4X) · (1− 4X)∈

12 ≡ (1− 4X) · ((1− 4X)

q−12 + 2Xq) (mod Xq+1)

so

X

∞∑k=0

CkXk =

1−√

1− 4X

2≡ 1− (1− 4X)

q+12 − 2Xq

2(mod Xq+1)

from which the claim follows.

Alternatively, one can also finish by integrating (1− 4X)p−12 , noting that the “constant of integration”

is no longer a constant but rather a polynomial of the form a+ bXp.

5. [11] A triangle and a circle are in the same plane. Show that the area of the intersection of the triangleand the circle is at most one third of the area of the triangle plus one half of the area of the circle.

Proposed by: Krit Boonsiriseth

Solution 1: (Ankan Bhattacharya1) Let 4 denote the triangle, ◦ denote the circle, and 4′ denote thereflection of the triangle in the center of the circle. Letting overline denote complement,

[4∩ ◦] ≤ 1

3[4] +

1

2[◦]

⇐⇒ [4∩4′ ∩ ◦] + [4∩4′ ∩ ◦] ≤ 1

3[4] +

1

2

[[◦ ∩ 4 ∩4′] + 2[◦ ∩ 4 ∩4′] + [◦ ∩ 4 ∩4′]

]⇐⇒ 1

2[4∩4′ ∩ ◦] ≤ 1

3[4] +

1

2[◦ ∩ 4 ∩4′]

⇐⇒ [4∩4′ ∩ ◦] ≤ 2

3[4] + [◦ ∩ 4 ∩4′].

Since 4∩4′ ∩ ◦ is centrally symmetric, it’s enough to show the following well-known lemma.

Lemma. A centrally symmetric region R of a triangle can have area at most 23 that of the triangle.

Let the triangle be ABC with medial triangle DEF . We take two cases.

1https://artofproblemsolving.com/community/c6h2052367p14590886

• Case 1. The center of symmetry is in one of the outer triangles, say 4AEF . Then the maximalpossible R is a parallelogram with one vertex A and two other vertices on AB and AC. Byenlargening the parallelogram, we may assume its fourth vertex lies on BC. If this vertex dividesBC into an a : b ratio with a+ b = 1, then the fraction of the area taken up by the parallelogramis 1− a2 − b2 ≤ 1

2 .

• Case 2. The center of symmetry is in 4DEF . Then the maximal possible R is a centrallysymmetric hexagon with two vertices on each side. Then there are three little triangles similar to4ABC on the hexagon; by equal lengths, the similarity ratios a, b, c sum to 1. Then the fractionof the area taken up by the hexagon is 1− a2 − b2 − c2 ≤ 2

3 . �

We are done.

Solution 2: It is also possible to approach this as an optimization problem. Fix a triangle 4 = ABCon the plane, vary the circle � = �(O, r), and consider the objective function

f(O, r) := [4∩�]− 1

2[�].

By a compactness argument we can show that f reaches a maximum, and at that maximum @f@r = 0

and ∇Of = ~0 must simultaneously hold.

• We have

∂

∂rf(O, r) =

∂

∂r[4∩�]− 1

2

∂

∂r[�]

= r ·((total angle subtended by AB ∩ �, BC ∩ �, CA ∩ � at point O)− π

),

therefore

∂

∂rf(O, r) = 0 ⇐⇒ the total angle subtended by the sides of 4 inside � is π. (1)

• Let ~na, ~nb, ~nc denote the normal vectors of BC,CA and AB. Define ka to be the length of BC∩�,and define kb, kc similarly.

∇Of(O, r) = ∇O[4∩�]− 1

2∇O[�]

= (ka~na + kb~nb + kc~nc)−~0

It is easy to see that a~na + b~nb + c~nc = ~0 is the only linear relation between ~na, ~nb, ~nc, so

∇Of(O, r) = ~0 ⇐⇒ ka : kb : kc = a : b : c (2)

Now suppose that �(O, r) is chosen so that @f@r = 0 and ∇Of = ~0, so the angle condition in (1) holds,

and (ka, kb, kc) = k(a, b, c) for some k. There are two possible cases:

• Case 1. All vertices of 4 do not lie inside �. In this case � intersects 4 at six points. The linesjoining O and these six points divide 4∩� into three sectors and three triangles. By the anglecondition in (1), the three sectors have total area 1

2 [�]. The three triangles have total area k[4],and it suffices to show that k 6 1

3 .

As the total angle of the triangles at O is π, we may join two copies of each of the three trianglesto form a cyclic hexagon. Therefore,

2k · [4] = 2 · (total area of three triangles)

= area of cyclic hexagon with side lengths ka, kb, kc, ka, kb, kc

> area of hexagon with side lengths ka, kb, kc, ka, kb, kc

created by six copies of a triangle with side lengths ka, kb, kc

= 6k2 · [4],

and our conclusion readily follows.

• Case 2. A vertex of 4 lies in the interior of �. WLOG let that vertex be A. Let � intersect ABat X, AC at Y , and BC at D and E. From (2), AX : AY : DE = AB : AC : BC, XY ‖ BCand XY = DE, therefore �XY ED is a rectangle. From (1), ∠XOD + ∠Y OE = π, so in fact�XY ED is a square. Let x be the side length of �XYDE, and let h be the height of the altitudefrom A to BC. Clearly h > x. As A lies inside � = (XYDE), ∠BAC > 3π

4 , so a > 3h > 3x.Now we are done because k = x

a <13 . �

HMMT Spring 2021March 06, 2021

Team Round

1. [40] Let a and b be positive integers with a > b. Suppose that√√a+√b+

√√a−√b

is a integer.

(a) Must√a be an integer?

(b) Must√b be an integer?

2. [50] Let ABC be a right triangle with ∠A = 90◦. A circle ω centered on BC is tangent to AB at Dand AC at E. Let F and G be the intersections of ω and BC so that F lies between B and G. If linesDG and EF intersect at X, show that AX = AD.

3. [50] Let m be a positive integer. Show that there exists a positive integer n such that each of the2m+ 1 integers

2n −m, 2n − (m− 1), . . . , 2n + (m− 1), 2n +m

is positive and composite.

4. [60] Let k and n be positive integers and let

S = {(a1, . . . , ak) ∈ Zk | 0 ≤ ak ≤ · · · ≤ a1 ≤ n, a1 + · · ·+ ak = k}.

Determine, with proof, the value of ∑(a1,...,ak)∈S

(n

a1

)(a1a2

)· · ·(ak−1ak

)

in terms of k and n, where the sum is over all k-tuples (a1, . . . , ak) in S.

5. [60] A convex polyhedron has n faces that are all congruent triangles with angles 36◦, 72◦, and 72◦.Determine, with proof, the maximum possible value of n.

6. [70] Let f(x) = x2 + x+ 1. Determine, with proof, all positive integers n such that f(k) divides f(n)whenever k is a positive divisor of n.

7. [70] In triangleABC, letM be the midpoint ofBC andD be a point on segmentAM . Distinct points Y

and Z are chosen on rays−→CA and

−−→BA, respectively, such that ∠DY C = ∠DCB and ∠DBC = ∠DZB.

Prove that the circumcircle of 4DY Z is tangent to the circumcircle of 4DBC.

8. [80] For each positive real number α, define

bαNc := {bαmc | m ∈ N}.

Let n be a positive integer. A set S ⊆ {1, 2, . . . , n} has the property that: for each real β > 0,

if S ⊆ bβNc , then {1, 2, . . . , n} ⊆ bβNc .

Determine, with proof, the smallest possible size of S.

9. [90] Let scalene triangle ABC have circumcenter O and incenter I. Its incircle ω is tangent to sidesBC, CA, and AB at D, E, and F , respectively. Let P be the foot of the altitude from D to EF , andlet line DP intersect ω again at Q 6= D. The line OI intersects the altitude from A to BC at T . Giventhat OI ‖ BC, show that PQ = PT .

10. [100] Let n > 1 be a positive integer. Each unit square in an n × n grid of squares is colored eitherblack or white, such that the following conditions hold:

• Any two black squares can be connected by a sequence of black squares where every two consec-utive squares in the sequence share an edge;

• Any two white squares can be connected by a sequence of white squares where every two consec-utive squares in the sequence share an edge;

• Any 2× 2 subgrid contains at least one square of each color.

Determine, with proof, the maximum possible difference between the number of black squares andwhite squares in this grid (in terms of n).

HMMT Spring 2021March 06, 2021

Team Round

1. [40] Let a and b be positive integers with a > b. Suppose that√√a+√b+

√√a−√b

is a integer.

(a) Must√a be an integer?

(b) Must√b be an integer?

Proposed by: Daniel Zhu

Answer: (a) Yes (b) No

Solution 1: Let r =

√√a+√b and s =

√√a−√b. We know r2 + s2 = 2

√a and r2 − s2 = 2

√b.

If r + s is an integer k, then

√a =

r2 + s2

2=

(r + s)2 + (r − s)2

4=k2 + 4b/k2

4,

which is rational. Recall that since a is a positive integer,√a is either an integer or an irrational

number. (Proof: if√a = p/q for relatively prime positive integers p, q, then p2/q2 is an integer, which

implies q = 1.) Thus the answer to part (a) is yes.

The answer to part (b) is no because

(2±√

2)2 = 6±√

32,

meaning that setting a = 36 and b = 32 is a counterexample.

Solution 2: Second solution for part (a): Squaring

√√a+√b+

√√a−√b, we see that 2

√a+2√a− b

is an integer. Hence√a− b = m −

√a for some rational number m. Squaring both sides of this, we

see that a− b = m2 − 2m√a+ a, so

√a = m2+b

2m , a rational number. As in the first solution, it followsthat

√a is an integer.

2. [50] Let ABC be a right triangle with ∠A = 90◦. A circle ω centered on BC is tangent to AB at Dand AC at E. Let F and G be the intersections of ω and BC so that F lies between B and G. If linesDG and EF intersect at X, show that AX = AD.

Proposed by: Krit Boonsiriseth

Solution 1: In all solutions, let O be the center of ω. Then ∠DOE = 90◦, so ∠DFE = 45◦, so

∠DXE = 135◦. Let Γ be the circle centered at A with radius AD = AE, and let X ′ =−−→AX ∩ Γ. Then

∠DXE = ∠DX ′E = 135◦, so X = X ′.

Solution 2: Let N be the midpoint of arc FDEG. Note that DOEA is a square. Also, DXEN isa parallelogram; one way to see this is that by considering inscribed angles, ∠NEX = ∠NEF = 45◦,∠NDX = ∠NDG = 45◦, and ∠END = 135◦. This means that 4AXD ∼= 4ONE, because AD =OE, XD = NE, and ∠ADX = ∠OEN by considering parallel lines. So AX = ON = OE = AD.

Solution 3: Let Y = DF ∩ EG. Since GX ⊥ FY and FX ⊥ GY , X is the orthocenter of 4FGY .Since (AODE) passes through the midpoint of FG, and the feet of altitudes from F and G, (AODE)is the nine-point circle of 4FGY . Since AO is the diameter of (AODE), it follows that A is the

midpoint of XY , so A is the center of (DXEY ), so AX = AD.

Solution 4: Let Z = DE∩FG, possibly at infinity. Then A and X are on the polar of Z with respectto ω, so AX ⊥ BC. Let J = AX ∩ BC. Then (XJFD) is cyclic, so ∠ADX = ∠DFJ = ∠DXA, soAX = AD.

Solution 5: We use complex numbers, with (FDEG) being the unit circle, and d = −1, e = −i. AsFG is a diameter of the unit circle, we have g = −f . We have a = −1− i, from either intersecting thetangents to the unit circle at D and E or noting that ADOE is a square.

Now, intersecting the chords DG and EF , we obtain

x =dg(e+ f)− ef(d+ g)

dg − ef=f(i+ f) + if(1− f)

f + if=f − i− i− if

1 + i= −1− i+ f · 1− i

1 + i.

So,

|x− a| =∣∣∣∣f · 1− i

1 + i

∣∣∣∣ = 1 ·√

2√2

= 1.

So AX and AD have the same length (1 unit), as desired.

3. [50] Let m be a positive integer. Show that there exists a positive integer n such that each of the2m+ 1 integers

2n −m, 2n − (m− 1), . . . , 2n + (m− 1), 2n +m

is positive and composite.

Proposed by: Michael Ren

Solution: Let P be the set of prime divisors of the 2m+ 1 numbers

2m+1 −m, 2m+1 −m+ 1, . . . , 2m+1 +m.

We claim thatn = m+ 1 +

∏p∈P

(p− 1)

works. To check this, let k be any integer with |k| ≤ m. We can take some prime q|2m+1 + k, as2m+1 + k ≥ 2m+1 −m ≥ 3. Let

∏p∈P (p− 1) = (q − 1)αq. Then, applying Fermat’s little theorem, we

have2n ≡ 2m+1 · (2q−1)αq ≡ 2m+1 ≡ −k (mod q)

Thus, 2n + k is divisible by q and is bigger than q, so it is positive and composite. This works for eachof the required k, so we are done.

4. [60] Let k and n be positive integers and let

S = {(a1, . . . , ak) ∈ Zk | 0 ≤ ak ≤ · · · ≤ a1 ≤ n, a1 + · · ·+ ak = k}.

Determine, with proof, the value of ∑(a1,...,ak)∈S

(n

a1

)(a1a2

)· · ·(ak−1ak

)

in terms of k and n, where the sum is over all k-tuples (a1, . . . , ak) in S.

Proposed by: Milan Haiman

Answer:(k+n−1

k

)=(k+n−1n−1

)Solution 1: Let

T = {(b1, . . . , bn) | 0 ≤ b1, . . . , bn ≤ k, b1 + · · ·+ bn = k}.

The sum in question counts |T |, by letting ai be the number of bj that are at least i. By stars and

bars, |T | =(k+n−1

k

).

One way to think about T is as follows. Suppose we wish to choose k squares in a grid of squares withk rows and n columns, such that each square not in the bottom row has a square below it. If we dividethe grid into columns and let bj be the number of chosen squares in the jth column then we get thatT is in bijection with valid ways to choose our k squares.

On the other hand, if we divide the grid into rows, and let ai be the number of chosen squares in theith row (counting up from the bottom), then we obtain the sum in the problem. This is because wehave

(na1

)choices for the squares in the first row, and

(ai−1

ai

)choices for the squares in the ith row,

given the squares in the row below, for each i = 2, . . . , k.

Solution 2: Define

Fk(x, y) =∑

n,a1,...,ak

(n

a1

)· · ·(ak−1ak

)xnya1+···+ak ,

where the sum is over all nonegative integers n, a1, . . . , ak (the nonzero terms have n ≥ a1 ≥ · · · ≥ ak).Note that we are looking for the coefficient of xnyk in Fk(x, y). By first summing over ak on the insideand using the Binomial theorem, we obtain

Fk(x, y) =∑

n,a1,...,ak−1

(n

a1

)· · ·(ak−2ak−1

)xnya1+···+ak−1(1 + y)ak−1 .

Now, we repeat this by summing over ak−1, then ak−2, and so on. We obtain

Fk(x, y) =∑n

xn(1 + y + · · ·+ yk)n.

So the answer is just the coefficient yk in (1 + y + · · ·+ yk)n. By stars and bars, this is(k+n−1

k

).

Although we didn’t need it, another way to write Fk(x, y) is

Fk(x, y) =1

1− x− xy − · · · − xyk.

Solution 3: Let

S(n, k, k′) = {(a1, . . . , ak) | 0 ≤ ak ≤ · · · ≤ a1 ≤ n, a1 + · · ·+ ak = k′},

and note that S(n, k, k) is the set S in the problem.

Define

f(n, k, k′) =∑

(a1,...,ak)∈S(n,k,k′)

(n

a1

)(a1a2

)· · ·(ak−1ak

).

Now, consider (a1, . . . , ak) ∈ S(n, k, k′). We have

iai ≤ a1 + · · ·+ ak = k′.

So ai ≤ k′

i . In particular, if k > k′, then we have ai = 0 for each k′ < i ≤ k. This means thatf(n, k, k′) = f(n, k′, k′) if k > k′.

Now, let f(n, k) = f(n, k, k) be the answer. Splitting the sum based on the possible values of a1 gives

f(n, k) =∑a1

(n

a1

)f(a1, k − 1, k − a1).

For k > 0, we must have a1 ≥ 1, which means

f(a1, k − 1, k − a1) = f(a1, k − a1, k − a1) = f(a1, k − a1).

So, we obtain the recurrence

f(n, k) =∑a1

(n

a1

)f(a1, k − a1).

Now we claim f(n, k) =(k+n−1

k

). We proceed by induction, with our base case being k = 1. The claim

is easy to verify for k = 1.

In the inductive step, we obtain

f(n, k) =∑a1

(n

a1

)(k − 1

k − a1

)=

(k + n− 1

k

),

where we applied Vandermonde’s identity in the last equality.

5. [60] A convex polyhedron has n faces that are all congruent triangles with angles 36◦, 72◦, and 72◦.Determine, with proof, the maximum possible value of n.

Proposed by: Handong Wang

Answer: 36

Solution: Consider such a polyhedron with V vertices, E edges, and F = n faces. By Euler’s formulawe have V + F = E + 2.

Next, note that the number of pairs of incident faces and edges is both 2E and 3F , so 2E = 3F .

Now, since our polyhedron is convex, the sum of the degree measures at each vertex is strictly less than360 = 36 · 10. As all angle measures of the faces of our polyhedron are divisible by 36, the maximumdegree measure at a given vertex is 36 · 9 = 324. On the other hand, the total degree measure at allvertices is the total degree measure over all faces, which is 180F . Thus we have 180F ≤ 324V , or10F ≤ 18V .

Putting our three conditions together, we have

10F ≤ 18V = 18(E + 2− F ) = 9(2E) + 36− 18F = 9(3F ) + 36− 18F = 9F + 36.

Thus F ≤ 36.

F = 36 is attainable by taking a 9-gon antiprism with a 9-gon pyramid attached on the top and thebottom. Thus the answer is 36.

6. [70] Let f(x) = x2 + x+ 1. Determine, with proof, all positive integers n such that f(k) divides f(n)whenever k is a positive divisor of n.

Proposed by: Milan Haiman

Answer: n can be 1, a prime that is 1 mod 3, or the square of any prime except 3.

Solution: The answer is n can be 1, a prime that is 1 mod 3, or the square of any prime except 3. Itis easy to verify that all of these work.

First note that n must be 1 mod 3 since 1 divides n implies f(1) divides f(n).

Next, suppose for sake of contradiction that n = ab, with a > b > 1. We are given that f(a) dividesf(n), which means f(a) divides f(n)− f(a). We can write this as

a2 + a+ 1 | n2 + n− a2 − a = (n− a)(n+ a+ 1).

Since we are working mod a2 + a+ 1, we can replace a+ 1 with −a2, so we have

a2 + a+ 1 | (n− a)(n− a2) = a2(b− 1)(b− a).

However, a2 + a+ 1 cannot share any factors with a, and 0 < |(b− 1)(b− a)| < a2 + a+ 1, which is acontradiction.

7. [70] In triangleABC, letM be the midpoint ofBC andD be a point on segmentAM . Distinct points Y

and Z are chosen on rays−→CA and

−−→BA, respectively, such that ∠DY C = ∠DCB and ∠DBC = ∠DZB.

Prove that the circumcircle of 4DY Z is tangent to the circumcircle of 4DBC.

Proposed by: Joseph Heerens

Solution 1: We first note that the circumcircles of DBZ and Y DC are tangent to BC from our anglecriteria. By power of a point, we obtain that M lies on the radical axis of the two circles and clearlyD does as well. Therefore, we find that A lies on the radical axis so AY ·AC = AB ·AZ implying thatBY ZC is a cyclic quadrilateral.

Next, by Reim’s Theorem on (BY ZC) and (DY Z), we get that (DY Z) intersects AB,AC at B′, C ′

where BC,B′C ′ are parallel. Then a negative homothety maps B to B′ and C to C ′, so (DBC) getsmapped to (DB′C ′), and we have tangent circles.

Solution 2: Let (DY Z) intersect AB and AC at B′ and C ′, respectively. We see that ]Y C ′B′ =]Y ZB′ = ]Y ZB = ]Y CB. Thus, BC ‖ B′C ′. This means that there exists a negative homothetytaking B to B′ and C to C ′ which will map (DBC) to (DB′C ′) which is also (DY Z).

8. [80] For each positive real number α, define

bαNc := {bαmc | m ∈ N}.

Let n be a positive integer. A set S ⊆ {1, 2, . . . , n} has the property that: for each real β > 0,

if S ⊆ bβNc , then {1, 2, . . . , n} ⊆ bβNc .

Determine, with proof, the smallest possible size of S.

Proposed by: Krit Boonsiriseth

Answer: bn/2c+ 1

Solution: For each k ∈ {dn/2e, . . . , n}, picking β = 1 + 1/k gives

bβNc ∩ [n] = [n] \ {k}

so S must contain k.

Now we show that S = {dn/2e, . . . , n} works; this set S has bn/2c+ 1 elements.

Suppose β satisfy S ⊆ bβNc, and suppose for the sake of contradiction that [n] 6⊂ bβNc. Since we mayincrease β by a small amount ε without affecting bβNc ∩ [n], we may assume β is irrational. Let αsatisfy 1/α+ 1/β = 1. By Beatty’s Theorem, bαNc and bβNc are complement sets in N.

Let m be the maximal element of [n] that is not in bβNc. Then m = bkαc for some integer k. Considerm′ = b2kαc ∈ {2m, 2m+ 1}, which must be an element of bαNc. Clearly, m′ > m, and since m < n/2,m′ 6 n, so m′ is also an element of [n] that is not in bβNc. This contradicts the maximality of m, andwe are done.

9. [90] Let scalene triangle ABC have circumcenter O and incenter I. Its incircle ω is tangent to sidesBC, CA, and AB at D, E, and F , respectively. Let P be the foot of the altitude from D to EF , andlet line DP intersect ω again at Q 6= D. The line OI intersects the altitude from A to BC at T . Giventhat OI ‖ BC, show that PQ = PT .

Proposed by: Carl Schildkraut, Milan Haiman

Solution: Let H be the orthocenter of ∆DEF . We first claim that O, I,H are collinear. We presenttwo proofs.

Proof 1. Invert about ω. Circle (ABC) inverts to a circle with center on OI, but A,B,C invert to themidpoints of EF,FD,DE, respectively, so the nine-point center of ∆DEF is on OI. As this center isthe midpoint of IH, we get that H, I,O are collinear. �

Proof 2. Let QA, QB , QC be the second intersections of the D−, E−, and F− altitudes, respectively,in ∆DEF with ω. We claim ∆QAQBQC is homothetic with ∆ABC. Indeed, as QB is the reflectionof H over DF and QC is the reflection of H over DE, DQB = DQC , so the perpendicular bisector ofQBQC is line ID. As ID ⊥ BC, QBQC ||BC, whence the homothety follows. This homothety takesthe incircle to the circumcircle, so it is centered on line OI. However, it also takes the incenter H ofQAQBQC to the incenter I of ABC, so it is centered on line IH. So, O, I,H are collinear. �

As P is the midpoint of QH, it suffices to show that P is on the circle with diameter QH, or that∠QTH = 90◦. As AT ⊥ TH = IO, it suffices to show that Q is on line AT . We also present twoproofs of this.

Proof 1. Let D′ be the antipode of D, and let AD′ intersect BC at X. As X is the A-extouch point,the midpoint M of DX is also the midpoint of BC. We have

OM

MX=IDDX2

=DD′

DX

and ∠OMX = ∠D′DX = 90◦, so D′, O,X are collinear, so D′ is on line AO. As QD′||EF , AQ andAD′ are isogonal in ∠BAC, so AQ and AO are isogonal, which means Q is on the A-altitude, finishingthe proof. �

Proof 2. Let Γ denote the circumcircle of ∆ABC, and let M be the midpoint of arc BC on Γ notcontaining A.

Lemma. The intersection T ′ of MD and the A-altitude to BC is on the line through I parallel toBC.

Proof. Let D′ = MA ∩BC. As ∠D′BM = ∠CBM = ∠CAM = ∠MAB, 4D′BM ∼ 4BAM , and

MI2 = MB2 = MD′ ·MA.

Since AT ′‖ID, we haveMT ′

MD=MA

MI=

MI

MD′,

so IT ′||DD′, finishing the proof. �

By the above lemma, T is on MD. Consider a homothety centered at T that takes D to M . It takesω to a circle centered on line IT that is tangent to Γ at M ; since O is on line IT this circle must beΓ itself. So, T is the exsimilicenter of Γ and ω. By Proof 2 above, T is the center of the homothetywhich sends QAQBQC to ABC, so T , Q = QA, and A are collinear, finishing the proof. �

10. [100] Let n > 1 be a positive integer. Each unit square in an n × n grid of squares is colored eitherblack or white, such that the following conditions hold:

• Any two black squares can be connected by a sequence of black squares where every two consec-utive squares in the sequence share an edge;

• Any two white squares can be connected by a sequence of white squares where every two consec-utive squares in the sequence share an edge;

• Any 2× 2 subgrid contains at least one square of each color.

Determine, with proof, the maximum possible difference between the number of black squares andwhite squares in this grid (in terms of n).

Proposed by: Yuan Yao

Answer: 2n+ 1 if n is odd, 2n− 2 if n is even.

Solution: The first two conditions also imply that there can be no 2×2 checkerboards, so the boundarybetween black squares and white squares is either a lattice path or cycle (if one color encloses the other).Therefore, the set of squares of each color is the interior of a lattice polygon of genus 0 or 1. (In thelatter case, the genus-1 color uses all squares on the outer boundary, and the opposite color must begenus-0.)

The third condition requires that the perimeter of each color passes through all (n−1)2 interior latticepoints, or else there will be a monochromatic 2×2 subgrid. Hence, by Pick’s Theorem, the area of onecolor is at least (n−1)2/2−1 = (n2−2n−1)/2, and the difference is at most n2−(n2−2n−1) = 2n+1.

For even n, the number of interior lattice points is odd so there is no cycle that only uses them. (Inparticular, this means that both colors are genus-0.) It is impossible for the perimeter to only gothrough one boundary point either, so we need to add at least three more boundary points, whichmeans that we lose 2(3/2) = 3 from the bound for odd n.

Here is one possible set of constructions. Throughout, we’ll label the squares as (x, y), for 1 ≤ x, y ≤ n:

• For n = 2, we color (2, 2) black and the others white.

• For odd values of n, we create a comb shape using black squares. Specifically, the base of thecomb will consist of the squares (i, 2), for i = 2, 3, . . . , n−1. The teeth of the comb will be (2k, j),for k = 1, 2, . . . , n−12 , and j = 3, 4, . . . , n− 1.

• For even values of n > 2, we make a modified comb shape. The base of the comb will be (i, 2)for i = 2, 3, . . . , n, and the teeth will be (2k, j) for k = 1, 2, . . . , n2 − 1 and j = 3, 4, . . . , n − 1.Furthermore, we add the square (n, 3), and the squares (n− 1, 2k + 3) for k = 1, 2, . . . , n2 − 2.

HMMT Spring 2021March 06, 2021

Algebra and Number Theory Round

1. Compute the sum of all positive integers n for which the expression

n+ 7√n− 1

is an integer.

2. Compute the number of ordered pairs of integers (a, b), with 2 ≤ a, b ≤ 2021, that satisfy the equation

alogb(a−4) = bloga(ba−3).

3. Among all polynomials P (x) with integer coefficients for which P (−10) = 145 and P (9) = 164, computethe smallest possible value of |P (0)|.

4. Suppose that P (x, y, z) is a homogeneous degree 4 polynomial in three variables such that P (a, b, c) =P (b, c, a) and P (a, a, b) = 0 for all real a, b, and c. If P (1, 2, 3) = 1, compute P (2, 4, 8).

Note: P (x, y, z) is a homogeneous degree 4 polynomial if it satisfies P (ka, kb, kc) = k4P (a, b, c) for allreal k, a, b, c.

5. Let n be the product of the first 10 primes, and let

S =∑xyjn

ϕ(x) · y,

where ϕ(x) denotes the number of positive integers less than or equal to x that are relatively prime tox, and the sum is taken over ordered pairs (x, y) of positive integers for which xy divides n. ComputeSn .

6. Suppose that m and n are positive integers with m < n such that the interval [m,n) contains moremultiples of 2021 than multiples of 2000. Compute the maximum possible value of n−m.

7. Suppose that x, y, and z are complex numbers of equal magnitude that satisfy

x+ y + z = −√

3

2− i√

5

andxyz =

√3 + i

√5.

If x = x1 + ix2, y = y1 + iy2, and z = z1 + iz2 for real x1, x2, y1, y2, z1, and z2, then

(x1x2 + y1y2 + z1z2)2

can be written as ab for relatively prime positive integers a and b. Compute 100a+ b.

8. For positive integers a and b, let M(a, b) = lcm(a,b)gcd(a,b) , and for each positive integer n ≥ 2, define

xn = M(1,M(2,M(3, . . . ,M(n− 2,M(n− 1, n)) . . . ))).

Compute the number of positive integers n such that 2 ≤ n ≤ 2021 and 5x2n + 5x2n+1 = 26xnxn+1.

9. Let f be a monic cubic polynomial satisfying f(x) + f(−x) = 0 for all real numbers x. For all realnumbers y, define g(y) to be the number of distinct real solutions x to the equation f(f(x)) = y.Suppose that the set of possible values of g(y) over all real numbers y is exactly {1, 5, 9}. Computethe sum of all possible values of f(10).

10. Let S be a set of positive integers satisfying the following two conditions:

• For each positive integer n, at least one of n, 2n, . . . , 100n is in S.

• If a1, a2, b1, b2 are positive integers such that gcd(a1a2, b1b2) = 1 and a1b1, a2b2 ∈ S, thena2b1, a1b2 ∈ S.

Suppose that S has natural density r. Compute the minimum possible value of b105rc.Note: S has natural density r if 1

n |S ∩ {1, ..., n}| approaches r as n approaches ∞.

HMMT Spring 2021March 06, 2021

Algebra and Number Theory Round

1. Compute the sum of all positive integers n for which the expression

n+ 7√n− 1

is an integer.

Proposed by: Ryan Kim

Answer: 89

Solution: We know√n− 1 must be a positive integer, because the numerator is a positive integer,

and the square root of an integer cannot be a non-integer rational. From this,

n+ 7√n− 1

=√n− 1 +

8√n− 1

is a positive integer, so we√n− 1 must be a positive integer that divides 8. There are 4 such positive

integers: 1, 2, 4, 8, which give n = 2, 5, 17, 65, so the answer is 89.

2. Compute the number of ordered pairs of integers (a, b), with 2 ≤ a, b ≤ 2021, that satisfy the equation

alogb(a−4) = bloga(ba−3).

Proposed by: Vincent Bian

Answer: 43

Solution: Taking loga of both sides and simplifying tives

−4 logb a = (loga b)2 − 3 loga b.

Plugging in x = loga b and using logb a = 1loga b

gives

x3 − 3x2 + 4 = 0.

We can factor the polynomial as (x − 2)(x − 2)(x + 1), meaning b = a2 or b = a−1. The secondcase is impossible since both a and b are positive integers. So, we need only count the number of1 < a, b ≤ 2021 for which b = a2, which is b

√2021c − 1 = 43.

3. Among all polynomials P (x) with integer coefficients for which P (−10) = 145 and P (9) = 164, computethe smallest possible value of |P (0)|.Proposed by: Carl Schildkraut

Answer: 25

Solution: Since a− b|P (a)− P (b) for any integer polynomial P and integers a and b, we require that10|P (0)− P (−10) and 9|P (0)− P (9). So, we are looking for an integer a near 0 for which

a ≡ 5 mod 10, a ≡ 2 mod 9.

The smallest such positive integer is 65, and the smallest such negative integer is −25. This is achiev-able, for example, if P (x) = 2x2 + 3x− 25, so our answer is 25.

4. Suppose that P (x, y, z) is a homogeneous degree 4 polynomial in three variables such that P (a, b, c) =P (b, c, a) and P (a, a, b) = 0 for all real a, b, and c. If P (1, 2, 3) = 1, compute P (2, 4, 8).

Note: P (x, y, z) is a homogeneous degree 4 polynomial if it satisfies P (ka, kb, kc) = k4P (a, b, c) for allreal k, a, b, c.

Proposed by: Milan Haiman

Answer: 56

Solution: Since P (a, a, b) = 0, (x−y) is a factor of P , which means (y−z) and (z−x) are also factorsby the symmetry of the polynomial. So,

P (x, y, z)

(x− y)(y − z)(z − x)

is a symmetric homogeneous degree 1 polynomial, so it must be k(x+ y + z) for some real k. So, theanswer is

P (2, 4, 8)

P (1, 2, 3)=

(2 + 4 + 8)(2− 4)(4− 8)(8− 2)

(1 + 2 + 3)(1− 2)(2− 3)(3− 1)= 56.

5. Let n be the product of the first 10 primes, and let

S =∑xy|n

ϕ(x) · y,

where ϕ(x) denotes the number of positive integers less than or equal to x that are relatively prime tox, and the sum is taken over ordered pairs (x, y) of positive integers for which xy divides n. ComputeSn .

Proposed by: Hahn Lheem

Answer: 1024

Solution 1: We see that, for any positive integer n,

S =∑xy|n

ϕ(x) · y =∑x|n

ϕ(x)

∑y|nx

y

=∑x|n

ϕ(x)σ(nx

).

Since ϕ and σ are both weakly multiplicative (if x and y are relatively prime, then ϕ(xy) = ϕ(x)ϕ(y)and σ(xy) = σ(x)σ(y)), we may break this up as∏

p

(ϕ(p) + σ(p)),

where the product is over all primes that divide n. This is simply 210n, giving an answer of 210 = 1024.

Solution 2: We recall that ∑d|n

ϕ(d) = n.

So, we may break up the sum as

S =∑xy|n

ϕ(x) · y =∑y|n

y∑x|ny

ϕ(x) =∑y|n

y

(n

y

),

so S is simply n times the number of divisors of n. This number is 210 = 1024.

Solution 3: When constructing a term in the sum, for each prime p dividing n, we can choose toinclude p in x, or in y, or in neither. This gives a factor of p − 1, p, or 1, respectively. Thus we canfactor the sum as

S =∏p|n

(p− 1 + p+ 1) =∏p|n

2p = 210n.

So the answer is 1024.

6. Suppose that m and n are positive integers with m < n such that the interval [m,n) contains moremultiples of 2021 than multiples of 2000. Compute the maximum possible value of n−m.

Proposed by: Carl Schildkraut

Answer: 191999

Solution: Let a = 2021 and b = 2000. It is clear that we may increase y − x unless both x − 1 andy + 1 are multiples of b, so we may assume that our interval is of length b(k + 1)− 1, where there arek multiples of b in our interval. There are at least k + 1 multiples of a, and so it is of length at leastak + 1. We thus have that

ak + 1 ≤ b(k + 1)− 1 =⇒ (a− b)k ≤ b− 2 =⇒ k ≤⌊b− 2

a− b

⌋.

So, the highest possible value of k is 95, and this is achievable by the Chinese remainder theorem,giving us an answer of 191999.

7. Suppose that x, y, and z are complex numbers of equal magnitude that satisfy

x+ y + z = −√

3

2− i√

5

andxyz =

√3 + i

√5.

If x = x1 + ix2, y = y1 + iy2, and z = z1 + iz2 for real x1, x2, y1, y2, z1, and z2, then

(x1x2 + y1y2 + z1z2)2

can be written as ab for relatively prime positive integers a and b. Compute 100a+ b.

Proposed by: Akash Das

Answer: 1516

Solution: From the conditions, it is clear that a, b, c all have magnitude√

2. Conjugating the first

equation gives 2(ab+bc+caabc ) = −√32 + i

√5, which means ab + bc + ca = (−

√34 + i

√52 )(√

3 + i√

5) =−13+i

√15

4 . Then,

a1a2 + b1b2 + c1c2 =1

2Im(a2 + b2 + c2)

=1

2Im((a+ b+ c)2)− Im(ab+ bc+ ca)

=

√15

4,

so the answer is 1516.

Remark:

{a, b, c} =

{−√

3− i√

5

2,−3√

3− i√

5

4,

3√

3− i√

5

4

}

8. For positive integers a and b, let M(a, b) = lcm(a,b)gcd(a,b) , and for each positive integer n ≥ 2, define

xn = M(1,M(2,M(3, . . . ,M(n− 2,M(n− 1, n)) . . . ))).

Compute the number of positive integers n such that 2 ≤ n ≤ 2021 and 5x2n + 5x2n+1 = 26xnxn+1.

Proposed by: Hahn Lheem

Answer: 20

Solution: The desired condition is that xn = 5xn+1 or xn+1 = 5xn.

Note that for any prime p, we have νp(M(a, b)) = |νp(a)− νp(b)|. Furthermore, νp(M(a, b)) ≡ νp(a) +νp(b) mod 2. So, we have that

νp(xn) ≡ νp(1) + νp(2) + · · ·+ νp(n) mod 2.

Subtracting gives that νp(xn+1)−νp(xn) ≡ νp(n+1) mod 2. In particular, for p 6= 5, νp(n+1) must be

even, and ν5(n+1) must be odd. So n+1 must be a 5 times a perfect square. There are⌊√

20215

⌋= 20

such values of n in the interval [2, 2021].

Now we show that it is sufficient for n + 1 to be 5 times a perfect square. The main claim is that ifB > 0 and a sequence a1, a2, . . . , aB of nonnegative real numbers satisfies an ≤ B +

∑i<n ai for all

1 ≤ n ≤ N , then ∣∣∣∣∣a1 −∣∣∣∣a2 − ∣∣∣· · · − ∣∣aN−1 − aN ∣∣∣∣∣ · · · ∣∣∣∣

∣∣∣∣∣ ≤ B.This can be proved by a straightforward induction on N . We then apply this claim, with B = 1, tothe sequence ai = νp(i); it is easy to verify that this sequence satisfies the condition. This gives

νp(xn) =

∣∣∣∣∣νp(1)−∣∣∣∣νp(2)−

∣∣∣· · · − ∣∣νp(n− 1)− νp(n)∣∣∣∣∣ · · · ∣∣∣∣

∣∣∣∣∣ ≤ 1,

so νp(xn) must be equal to(νp(1) + · · · + νp(n)

)mod 2. Now suppose n + 1 = 5k2 for some k; then

νp(n + 1) ≡ 0 mod 2 for p 6= 5 and ν5(n + 1) ≡ 1 mod 2. Therefore νp(xn+1) = νp(xn) for p 6= 5, andν5(xn+1) = (ν5(xn) + 1) mod 2, and this implies xn+1/xn ∈ {1/5, 5} as we wanted.

9. Let f be a monic cubic polynomial satisfying f(x) + f(−x) = 0 for all real numbers x. For all realnumbers y, define g(y) to be the number of distinct real solutions x to the equation f(f(x)) = y.Suppose that the set of possible values of g(y) over all real numbers y is exactly {1, 5, 9}. Computethe sum of all possible values of f(10).

Proposed by: Sujay Kazi

Answer: 970

Solution: We claim that we must have f(x) = x3−3x. First, note that the condition f(x)+f(−x) = 0implies that f is odd. Combined with f being monic, we know that f(x) = x3 + ax for some realnumber a. Note that a must be negative; otherwise f(x) and f(f(x)) would both be increasing and 1would be the only possible value of g(y).

Now, consider the condition that the set of possible values of g(y) is {1, 5, 9}. The fact that we canhave g(y) = 9 means that some horizontal line crosses the graph of f(f(x)) 9 times. Since f(f(x)) hasdegree 9, this means that its graph will have 4 local maxima and 4 local minima.

Now, suppose we start at some value of y such that g(y) = 9, and slowly increase y. At some point,the value of g(y) will decrease. This happens when y is equal to a local maximum of f. Since g(y)

must jump from 9 down to 5, all four local maxima must have the same value. Similarly, all fourlocal minima must also have the same value. Since f is odd, it suffices to just consider the four localmaxima.

The local maximum of f(x) occurs when 3x2+a = 0. For convenience, let a = −3b2, so f(x) = x3−3b2x.Then, the local maximum is at x = −b, and has a value of f(−b) = 2b3.

We consider the local maxima of f(f(x)) next. They occur either when x = −b (meaning f(x) is ata local maximum) or f(x) = −b. If f(x) = −b, then f(f(x)) = f(−b) = 2b3. Thus, we must havef(f(−b)) = f(2b3) = 2b3.

This yields the equationf(2b3) = 8b9 − 3b2 · 2b3 = 2b3

which factors as 2b3(b2 − 1

) (2b2 + 1

)2. The only possible value of b2 is 1. Thus, f(x) = x3 − 3x, and

our answer is 103 − 3 · 10 = 970.

10. Let S be a set of positive integers satisfying the following two conditions:

• For each positive integer n, at least one of n, 2n, . . . , 100n is in S.

• If a1, a2, b1, b2 are positive integers such that gcd(a1a2, b1b2) = 1 and a1b1, a2b2 ∈ S, thena2b1, a1b2 ∈ S.

Suppose that S has natural density r. Compute the minimum possible value of b105rc.Note: S has natural density r if 1

n |S ∩ {1, ..., n}| approaches r as n approaches ∞.

Proposed by: Milan Haiman

Answer: 396

Solution: The optimal value of r is 1252 . This is attained by letting S be the set of integers n for

which ν2(n) ≡ 4 mod 5 and ν3(n) ≡ 1 mod 2.

Let S be a set of positive integers satisfying the two conditions. For each prime p, let Ap = {νp(n) :n ∈ S}. We claim that in fact S is precisely the set of positive integers n for which νp(n) ∈ Ap foreach prime p.

Let p be prime and suppose that a1pe1 , a2p

e2 ∈ S, with p - a1, a2. Then, setting b1 = pe1 and b2 = pe2

in the second condition gives that a1pe2 ∈ S as well. So, if we have an integer n for which νp(n) ∈ Ap

for each prime p, we can start with any element n′ of S and apply this step for each prime divisor ofn and n′ to obtain n ∈ S.

Now we deal with the first condition. Let n be any positive integer. We will compute the least positiveinteger m such that mn ∈ S. By the above result, we can work with each prime separately. For a givenprime p, let ep be the least element of Ap with ep ≥ νp(n). Then we must have νp(m) ≥ ep − νp(n),and equality for all primes p is sufficient. So, if the elements of Ap are cp,1 < cp,2 < cp,3 < cp,4 < . . . ,then

cp = max(cp,1, cp,2 − cp,1 − 1, cp,3 − cp,2 − 1, cp,4 − cp,3 − 1, . . . )

is the worst case value for νp(m).

We conclude two things from this. First, we must have∏p p

cp ≤ 100 by condition 1, and in fact thisis sufficient. Second, since we only care about cp and would like to minimize r, the optimal choice forAp is an arithmetic progression with first term cp and common difference cp + 1. So we assume thateach Ap is of this form.

Let t =∏p p

cp . We now compute r. Note that S is the set of integers n such that for each prime p,

n ≡ apk(cp+1)−1 mod pk(cp+1)

for some positive integers a, k with a < p. This means that each prime p contributes a factor of

p− 1

pcp+1+

p− 1

p2cp+2+

p− 1

p3cp+3+ · · · = p− 1

pcp+1 − 1=

1

1 + p+ · · ·+ pcp

to the density of S. Multiplying over all primes p gives r = 1σ(t) , where σ(t) is the sum of divisors of t.

So, it suffices to maximize σ(t) for t ≤ 100. By inspection, t = 96 is optimal, giving r = 1252 .

HMMT Spring 2021March 06, 2021

Combinatorics Round

1. Leo the fox has a 5 by 5 checkerboard grid with alternating red and black squares. He fills in thegrid with the numbers 1, 2, 3, . . . , 25 such that any two consecutive numbers are in adjacent squares(sharing a side) and each number is used exactly once. He then computes the sum of the numbers inthe 13 squares that are the same color as the center square. Compute the maximum possible sum Leocan obtain.

2. Ava and Tiffany participate in a knockout tournament consisting of a total of 32 players. In eachof 5 rounds, the remaining players are paired uniformly at random. In each pair, both players areequally likely to win, and the loser is knocked out of the tournament. The probability that Ava andTiffany play each other during the tournament is a

b , where a and b are relatively prime positive integers.Compute 100a + b.

3. Let N be a positive integer. Brothers Michael and Kylo each select a positive integer less than orequal to N , independently and uniformly at random. Let pN denote the probability that the productof these two integers has a units digit of 0. The maximum possible value of pN over all possible choicesof N can be written as a

b , where a and b are relatively prime positive integers. Compute 100a + b.

4. Let S = {1, 2, . . . , 9}. Compute the number of functions f : S → S such that, for all s ∈ S, f(f(f(s))) =s and f(s)− s is not divisible by 3.

5. Teresa the bunny has a fair 8-sided die. Seven of its sides have fixed labels 1, 2, . . . , 7, and the labelon the eighth side can be changed and begins as 1. She rolls it several times, until each of 1, 2, . . . , 7appears at least once. After each roll, if k is the smallest positive integer that she has not rolled sofar, she relabels the eighth side with k. The probability that 7 is the last number she rolls is a

b , wherea and b are relatively prime positive integers. Compute 100a + b.

6. A light pulse starts at a corner of a reflective square. It bounces around inside the square, reflectingoff of the square’s perimeter n times before ending in a different corner. The path of the light pulse,when traced, divides the square into exactly 2021 regions. Compute the smallest possible value of n.

7. Let S = {1, 2, . . . , 2021}, and let F denote the set of functions f : S → S. For a function f ∈ F , let

Tf ={f2021(s) : s ∈ S

},

where f2021(s) denotes f(f(· · · (f(s)) · · · )) with 2021 copies of f . Compute the remainder when∑f∈F

|Tf |

is divided by the prime 2017, where the sum is over all functions f in F .

8. Compute the number of ways to fill each cell in a 8× 8 square grid with one of the letters H,M, or Tsuch that every 2× 2 square in the grid contains the letters H,M,M, T in some order.

9. An up-right path between two lattice points P and Q is a path from P to Q that takes steps of length1 unit either up or to the right.

How many up-right paths from (0, 0) to (7, 7), when drawn in the plane with the line y = x − 2.021,enclose exactly one bounded region below that line?

10. Jude repeatedly flips a coin. If he has already flipped n heads, the coin lands heads with probability1

n+2 and tails with probability n+1n+2 . If Jude continues flipping forever, let p be the probability that he

flips 3 heads in a row at some point. Compute b180pc .

HMMT Spring 2021March 06, 2021

Combinatorics Round

1. Leo the fox has a 5 by 5 checkerboard grid with alternating red and black squares. He fills in thegrid with the numbers 1, 2, 3, . . . , 25 such that any two consecutive numbers are in adjacent squares(sharing a side) and each number is used exactly once. He then computes the sum of the numbers inthe 13 squares that are the same color as the center square. Compute the maximum possible sum Leocan obtain.

Proposed by: Milan Haiman

Answer: 169

Solution: Since consecutive numbers are in adjacent squares and the grid squares alternate in color,consecutive numbers must be in squares of opposite colors. Then the odd numbers 1, 3, 5, . . . , 25 allshare the same color while the even numbers 2, 4, . . . , 24 all share the opposite color. Since we have13 odd numbers and 12 even numbers, the odd numbers must correspond to the color in the centersquare, so Leo’s sum is always 1 + 3 + 5 + · · ·+ 25 = 169.

2. Ava and Tiffany participate in a knockout tournament consisting of a total of 32 players. In eachof 5 rounds, the remaining players are paired uniformly at random. In each pair, both players areequally likely to win, and the loser is knocked out of the tournament. The probability that Ava andTiffany play each other during the tournament is a

b , where a and b are relatively prime positive integers.Compute 100a+ b.

Proposed by: Sheldon Kieren Tan

Answer: 116

Solution: Each match eliminates exactly one player, so exactly 32− 1 = 31 matches are played, eachof which consists of a different pair of players. Among the

322

)= 32·31

2 = 496 pairs of players, each pairis equally likely to play each other at some point during the tournament. Therefore, the probabilitythat Ava and Tiffany form one of the 31 pairs of players that play each other is 31

496 = 116 , giving an

answer of 100 · 1 + 16 = 116.

3. Let N be a positive integer. Brothers Michael and Kylo each select a positive integer less than orequal to N , independently and uniformly at random. Let pN denote the probability that the productof these two integers has a units digit of 0. The maximum possible value of pN over all possible choicesof N can be written as a

b , where a and b are relatively prime positive integers. Compute 100a+ b.

Proposed by: James Lin

Answer: 2800

Solution: For k ∈ {2, 5, 10}, let qk = bN/kcN be the probability that an integer chosen uniformly at

random from [N ] is a multiple of k. Clearly, qk ≤ 1k , with equality iff k divides N .

The product of p1, p2 ∈ [N ] can be a multiple of 10 in two ways:

• one of them is a multiple of 10; this happens with probability q10(2− q10);

• one of them is a multiple of 2 (but not 5) and the other is a multiple of 5 (but not 2); this happenswith probability 2(q2 − q10)(q5 − q10).

This gives

pN = q10 · (2− q10) + 2(q2 − q10)(q5 − q10)

≤ q10 · (2− q10) + 2

(1

2− q10

)(1

5− q10

)=

1

5(1 + 3q10 + 5q210)

≤ 1

5

(1 +

3

10+

5

100

)=

27

100,

and equality holds iff N is a multiple of 10.

4. Let S = {1, 2, . . . , 9}. Compute the number of functions f : S → S such that, for all s ∈ S, f(f(f(s))) =s and f(s)− s is not divisible by 3.

Proposed by: James Lin

Answer: 288

Solution: Since f(f(f(s))) = s for all s ∈ S, each cycle in the cycle decomposition of f must havelength 1 or 3. Also, since f(s) 6≡ s mod 3 for all s ∈ S, each cycle cannot contain two elements a, b suchthat a = b mod 3. Hence each cycle has exactly three elements, one from each of residue classes mod3.

In particular, 1, 4, 7 belong to distinct cycles. There are 6 · 3 ways to choose two other numbers in thecycle containing 1. Then, there are 4 · 2 ways to choose two other numbers in the cycle containing 4.Finally, there are 2 · 1 ways to choose two other numbers in the cycle containing 7. Hence the desirednumber of functions f is 6 · 3 · 4 · 2 · 2 · 1 = 288.

5. Teresa the bunny has a fair 8-sided die. Seven of its sides have fixed labels 1, 2, . . . , 7, and the labelon the eighth side can be changed and begins as 1. She rolls it several times, until each of 1, 2, . . . , 7appears at least once. After each roll, if k is the smallest positive integer that she has not rolled sofar, she relabels the eighth side with k. The probability that 7 is the last number she rolls is a

b , wherea and b are relatively prime positive integers. Compute 100a+ b.

Proposed by: Milan Haiman

Answer: 104

Solution 1: Let n = 7 and p = 14 .

Let qk be the probability that n is the last number rolled, if k numbers less than n have already beenrolled. We want q0 and we know qn−1 = 1.

We have the relation

qk = (1− p) k

n− 1qk +

[1− (1− p)k + 1

n− 1

]qk+1.

This rearranges to [1− (1− p) k

n− 1

]qk =

[1− (1− p)k + 1

n− 1

]qk+1.

This means that the expression on the LHS does not depend on k, so

[1− 0] · q0 = [1− (1− p)] · qn−1 = p.

Solution 2: For a given sequence of Teresa’s rolls, let xi be the ith distinct number rolled. We wantto compute the probability that x7 = 7.

For a given index i, we say that xi is correct if xi is the least positive integer not in {x1, . . . , xi−1}.Note that the probability of a given sequence x1, . . . , x7 depends only on the number of correct xi,since the probability of rolling the correct number on a given roll is higher by a factor of 2.

Now, suppose x7 = 7. Consider x′i = xi−1 + 1 for 1 < i ≤ 7, and x′1 = 1. Note that this operation onsequences x1, . . . , x7 pairs sequences ending in 7 with sequences starting with 1. Additionally, we havethat x7 and x′1 are both correct, and that x′i is correct if and only if xi−1 is correct. Thus x1, . . . , x7and x′1, . . . , x

′7 have the same probability.

So, we conclude that the probability of x7 = 7 is the same as the probability of x1 = 1. But this isjust 1

4 .

6. A light pulse starts at a corner of a reflective square. It bounces around inside the square, reflectingoff of the square’s perimeter n times before ending in a different corner. The path of the light pulse,when traced, divides the square into exactly 2021 regions. Compute the smallest possible value of n.

Proposed by: Krit Boonsiriseth

Answer: 129

Solution: The main claim is that if the light pulse reflects vertically (on the left/right edges) a times

and horizontally b times, then gcd(a+1, b+1) = 1, and the number of regions is (a+2)(b+2)2 . This claim

can be conjectured by looking at small values of a and b; we give a full proof at the end.

Assuming the claim, we are trying to find the least possible value of a+b when (a+2)(b+2) = 2·2021 =2 · 43 · 47. This happens when (a + 2, b + 2) = (47, 86), which also satisfies gcd(a + 1, b + 1) = 1, andgives a+ b = 47 + 86− 4 = 129.

We now prove the claim. Imagine that at each reflection, it is the square that gets reflected instead.Then the path p of the light pulse becomes a straight segment s from (0, 0) to (a + 1, b + 1) of slope+m = a+1

b+1 .

• The square starts as 1 region; the light pulse hitting a corner at the end creates 1 more region.

• Each reflection of the light pulse creates a region. These correspond to intersections of s with aline x = n or y = n for x ∈ [a], y ∈ [b]. There are a+ b such intersections.

• Each self-intersection of p creates a region. An intersection on p corresponds to two on s, andeach intersection of s happens with a line of slope −m passing through an even integral point,i.e. a line of the form (b + 1)x + (a + 1)y = 2k. The open segment s intersects these lines fork ∈ [ab + a + b]. However, the a + b intersections that happens on a gridline x ∈ Z or y ∈ Z donot count, so here we have an additional ab/2 regions.

Therefore, the total number of regions is

2 + a+ b+ab

2=

(a+ 2)(b+ 2)

2.

7. Let S = {1, 2, . . . , 2021}, and let F denote the set of functions f : S → S. For a function f ∈ F , let

Tf ={f2021(s) : s ∈ S

},

where f2021(s) denotes f(f(· · · (f(s)) · · · )) with 2021 copies of f . Compute the remainder when∑f∈F

|Tf |

is divided by the prime 2017, where the sum is over all functions f in F .

Proposed by: Milan Haiman

Answer: 255

Solution: The key idea is that t ∈ Tf if and only if fk(t) = t for some k > 0. To see this, let s ∈ Sand consider

s, f(s), f(f(s)), . . . , f2021(s).

This sequence has 2022 terms that are all in S, so we must have a repeat. Suppose fm(s) = fn(s) with0 ≤ n < m ≤ 2021. Then f2021(s) = f2021+m−n(s). In particular, for t = f2021(s), we have fk(t) = twith k = m− n. On the other hand, if fk(t) = t, then letting s = f2021k−2021(t) gives f2021(s) = t.

We will compute the number of f for which fk(1) = 1 for some k, and then multiply by 2021. We dothis by casework on the minimum possible value of k.

Given k, we just need to choose distinct values in {2, . . . , 2021} for each of f1(1), f2(1), . . . , fk−1(1).We have 2020!

(2021−k)! ways to do this. For each of the 2021 − k other values with f not yet determined,

we can do anything we want, giving 20212021−k choices.

So, ∑f∈F

|Tf | = 2021

2021∑k=1

2020!

(2021− k)!· 20212021−k.

Taking this mod 2017, all terms with k > 4 reduce to 0, and 20212021−k reduces to 45−k for k ≤ 4. Weare thus left with∑

f∈F

|Tf | ≡ 4[44 + 3 · 43 + 3 · 2 · 42 + 3 · 2 · 1 · 41

]≡ 255 (mod 2017).

8. Compute the number of ways to fill each cell in a 8× 8 square grid with one of the letters H,M, or Tsuch that every 2× 2 square in the grid contains the letters H,M,M, T in some order.

Proposed by: Sheldon Kieren Tan

Answer: 1076

Solution: We solve the problem for general n × n boards where n even. Let the cell in the i-th rowand j-th column be ai,j .

Claim: In any valid configuration, either the rows (or columns) alternate between (· · · , H,M,H,M, · · · )and (· · · , T,M, T,M, · · · ) or (· · · ,M,M,M,M, · · · ) and (· · · , H, T,H, T, · · · ).Proof: First note that all configurations which follow the above criteria are valid.

If the rows alternate as above we are done. Else there exists one of the below configurations in one ofthe rows, from which we can deduce the rest of the 3 columns as follows:

(ai,j−1, ai,j , ai,j+1) (ai+1,j−1, ai+1,j , ai+1,j+1) (ai+2,j−1, ai+2,j , ai+2,j+1)

(H,M, T ) (T,M,H) (H,M, T )(T,M,H) (H,M, T ) (T,M,H)(H,T,M) (M,M,H) (H,T,M)(M,T,H) (H,M,M) (M,T,H)(T,H,M) (M,M,T ) (T,H,M)(M,H, T ) (T,M,M) (M,H, T )(T,M,M) (M,H, T ) (T,M,M)(M,M,T ) (T,H,M) (M,M,T )(H,M,M) (M,T,H) (H,M,M)(M,M,H) (H,T,M) (M,M,H)

It can be noted that the configurations alternate as we move down/up the columns, implying that the3 columns consist of alternating letters (or (M,M, · · · )). We can now check that all columns obey theabove form, and in particular, must alternate as stated in the claim.

It now suffices to count the number of cases. When the rows alternate between (· · · , H,M,H,M, · · · )and (· · · , T,M, T,M, · · · ), there are 2 ways to choose which one occupies the odd-numbered rows,and 2n ways to alternate between the 2 letters in each row. When the rows alternate between(· · · , H, T,H, T, · · · ) and (· · · ,M,M,M,M, · · · ), there are 2 ways to choose which occupies the odd-numbered rows, and 2

n2 ways to alternate between the 2 letters in the rows. The number of cases for

columns is the same.

Finally, if both the rows and columns alternate as above, it suffices to fix the first 2 rows (then therest of the board is uniquely determined by extending the columns). There are 2× 22 = 8 ways to dothis if the rows are (· · · , H,M,H,M, · · · ) and (· · · , T,M, T,M, · · · ), and 2× 2 = 4 ways to do this ifthe rows are (· · · ,M,M,M,M, · · · ) and (· · · , H, T,H, T, · · · ).Hence the total number of configurations is 2(2n+1 + 2

n2 +1)− 12 = 2n+2 + 2

n2 +2 − 12.

9. An up-right path between two lattice points P and Q is a path from P to Q that takes steps of length1 unit either up or to the right.

How many up-right paths from (0, 0) to (7, 7), when drawn in the plane with the line y = x − 2.021,enclose exactly one bounded region below that line?

Proposed by: Anders Olsen

Answer: 637

Solution: We will make use of a sort of bijection which is typically used to prove the closed formfor the Catalan numbers. 1 We will count these paths with complementary counting. Since both thestarting and ending points are above the line x− 2.021, any path which traverses below this line (andhence includes a point on the line y = x− 3) will enclose at least one region. In any such path, we canreflect the portion of the path after the first visit to the line y = x−3 over that line to get a path from(0, 0) to (10, 4). This process is reversible for any path to (10, 4), so the number of paths enclosing atleast one region is

144

).

More difficult is to count the paths that enclose at least two regions. For any such path, considerthe first and final times it intersects the line y = x − 3. Since at least two regions are enclosed, theremust be some point on the intermediate portion of the path on the line y = x− 2. Then we can reflectonly this portion of the path over the line y = x− 3 to get a new path containing a point on the liney = x − 4. We can then do a similar reflection starting from the first such point to get a path from(0, 0) to (11, 3). This process is reversible, so the number of paths which enclose at least two regions is143

). Then the desired answer is just

144

)−143

)= 637.

10. Jude repeatedly flips a coin. If he has already flipped n heads, the coin lands heads with probability1

n+2 and tails with probability n+1n+2 . If Jude continues flipping forever, let p be the probability that he

flips 3 heads in a row at some point. Compute b180pc .Proposed by: Anders Olsen

Answer: 47

Solution: Let pn be the probability that the nth head is flipped after a tail and Jude has yet to flip3 heads consecutively to this point. For example, p2 = 2

3 , as it is impossible for 3 heads to be flippedconsecutively and the second head comes after a tail exactly when the first flip after the first head is a

1See https://en.wikipedia.org/wiki/Catalan_number%23Second_proof for useful pictures

tail, which happens with probability 23 . Similarly, p3 = 3

4 . We now establish a recursion between valuesof pn:

pn =n

n+ 1pn−1 +

1

n+ 1pn−2.

The first term comes from when the previous head had tails both before and after, and the secondterm comes from when the previous 2 heads were consecutive. Of course there cannot be other terms,as this would imply that 3 heads were flipped consecutively. This enables us to easily compute thenext few terms: 11

15 ,5372 ,

103140 , and so on. Notably, the differences between consecutive terms (starting

from p3−p2) are 224 ,−

2120 ,

2720 ,−

25040 , and so on. This leads us to guess that pn = 2

∑n+1i=0

(−1)ii! , which

indeed satisfies the given recurrence relation. Then

limn→∞

pn = 2

∞∑i=0

(−1)i

i!=

2

e.

But since the probability that the nth head comes after a tail approaches 1 as n increases, this limit isthe same as the limit of the probability that the first n heads do not include 3 that came consecutively.Then this limit is just the probability that we never flip 3 consecutive heads. Then the desiredprobability is just p = 1 − 2

e . We are asked to compute b180pc. This is the floor of 180 − 360e . To

compute 360/e, note that we can just truncate the infinite sum

360

e=

∞∑n=0

360(−1)n

n!,

as it converges rather quickly. The first several terms are 360− 360 + 180− 60 + 15− 3 + 12 , and the

rest are insignificant. This sums to about 132.5, giving an answer of b180− 132.5c = 47.

HMMT Spring 2021March 06, 2021

Geometry Round

1. A circle contains the points (0, 11) and (0,−11) on its circumference and contains all points (x, y) withx2 + y2 < 1 in its interior. Compute the largest possible radius of the circle.

2. Let X0 be the interior of a triangle with side lengths 3, 4, and 5. For all positive integers n, define Xn

to be the set of points within 1 unit of some point in Xn→1. The area of the region outside X20 butinside X21 can be written as aπ + b, for integers a and b. Compute 100a+ b.

3. Triangle ABC has a right angle at C, and D is the foot of the altitude from C to AB. Points L,M , and N are the midpoints of segments AD, DC, and CA, respectively. If CL = 7 and BM = 12,compute BN2.

4. Let ABCD be a trapezoid with AB ‖ CD, AB = 5, BC = 9, CD = 10, and DA = 7. Lines BCand DA intersect at point E. Let M be the midpoint of CD, and let N be the intersection of thecircumcircles of4BMC and4DMA (other than M). If EN2 = a

b for relatively prime positive integersa and b, compute 100a+ b.

5. Let AEF be a triangle with EF = 20 and AE = AF = 21. Let B and D be points chosen on segmentsAE and AF , respectively, such that BD is parallel to EF . Point C is chosen in the interior of triangle

AEF such that ABCD is cyclic. If BC = 3 and CD = 4, then the ratio of areas [ABCD][AEF ] can be

written as ab for relatively prime positive integers a, b. Compute 100a+ b.

6. In triangle ABC, let M be the midpoint of BC, H be the orthocenter, and O be the circumcenter.Let N be the reflection of M over H. Suppose that OA = ON = 11 and OH = 7. Compute BC2.

7. Let O and A be two points in the plane with OA = 30, and let Γ be a circle with center O and radiusr. Suppose that there exist two points B and C on Γ with ∠ABC = 90◦ and AB = BC. Computethe minimum possible value of brc.

8. Two circles with radii 71 and 100 are externally tangent. Compute the largest possible area of a righttriangle whose vertices are each on at least one of the circles.

9. Let ABCD be a trapezoid with AB ‖ CD and AD = BD. Let M be the midpoint of AB, and letP 6= C be the second intersection of the circumcircle of 4BCD and the diagonal AC. Suppose thatBC = 27, CD = 25, and AP = 10. If MP = a

b for relatively prime positive integers a and b, compute100a+ b.

10. Acute triangle ABC has circumcircle Γ. Let M be the midpoint of BC. Points P and Q lie on Γ sothat ∠APM = 90◦ and Q 6= A lies on line AM . Segments PQ and BC intersect at S. Suppose that

BS = 1, CS = 3, PQ = 8√

737 , and the radius of Γ is r. If the sum of all possible values of r2 can be

expressed as ab for relatively prime positive integers a and b, compute 100a+ b.

HMMT Spring 2021March 06, 2021

Geometry Round

1. A circle contains the points (0, 11) and (0,−11) on its circumference and contains all points (x, y) withx2 + y2 < 1 in its interior. Compute the largest possible radius of the circle.

Proposed by: Carl Schildkraut

Answer: 61

Solution: Such a circle will be centered at (t, 0) for some t; without loss of generality, let t > 0. Ourconditions are that

t2 + 112 = r2

andr ≥ t+ 1.

So, t2 ≤ (r − 1)2, which means

(r − 1)2 + 112 ≥ r2 =⇒ 122 ≥ 2r,

so our answer is 61 (r = 61 is attainable with t = 60).

2. Let X0 be the interior of a triangle with side lengths 3, 4, and 5. For all positive integers n, define Xn

to be the set of points within 1 unit of some point in Xn→1. The area of the region outside X20 butinside X21 can be written as aπ + b, for integers a and b. Compute 100a+ b.

Proposed by: Hahn Lheem

Answer: 4112

Solution:

Xn is the set of points within n units of some point in X0. The diagram above shows X0, X1, X2, andX3. As seen above it can be verified that Xn is the union of

• X0,

• three rectangles of height n with the sides of X0 as bases, and

• three sectors of radius n centered at the vertices and joining the rectangles

Therefore the total area of Xn is

[X0] + n · perimeter(X0) + n2π.

Since Xn→1 is contained entirely within Xn, the area within Xn but not within Xn→1 is

perimeter(X0) + (2n− 1)π.

Since X0 is a (3, 4, 5) triangle, and n = 21, this is 12 + 41π.

3. Triangle ABC has a right angle at C, and D is the foot of the altitude from C to AB. Points L,M , and N are the midpoints of segments AD, DC, and CA, respectively. If CL = 7 and BM = 12,compute BN2.

Proposed by: Hahn Lheem

Answer: 193

Solution: Note that CL, BM , and BN are corresponding segments in the similar triangles 4ACD ∼4CBD ∼ 4ABC. So, we have

CL : BM : BN = AD : CD : AC.

Since AD2 + CD2 = AC2, we also have CL2 +BM2 = BN2, giving an answer of 49 + 144 = 193.

4. Let ABCD be a trapezoid with AB ‖ CD, AB = 5, BC = 9, CD = 10, and DA = 7. Lines BCand DA intersect at point E. Let M be the midpoint of CD, and let N be the intersection of thecircumcircles of4BMC and4DMA (other than M). If EN2 = a

b for relatively prime positive integersa and b, compute 100a+ b.

Proposed by: Milan Haiman

Answer: 90011

Solution: From 4EAB ∼ 4EDC with length ratio 1 : 2, we have EA = 7 and EB = 9. This meansthat A,B,M are the midpoints of the sides of 4ECD. Let N ′ be the circumcenter of 4ECD. SinceN ′ is on the perpendicular bisectors of EC and CD, we have ∠N ′MC = ∠N ′BC = 90◦. Thus N ′ ison the circumcircle of 4BMC. Similarly, N ′ is on the circumcircle of 4DMA. As N ′ 6= M , we musthave N ′ = N . So it suffices to compute R2, where R is the circumradius of 4ECD.

We can compute K = [4ECD] to be 21√

11 from Heron’s formula, giving

R =10 · 14 · 18

4K=

30√11.

So R2 = 90011 , and the final answer is 90011.

5. Let AEF be a triangle with EF = 20 and AE = AF = 21. Let B and D be points chosen on segmentsAE and AF , respectively, such that BD is parallel to EF . Point C is chosen in the interior of triangle

AEF such that ABCD is cyclic. If BC = 3 and CD = 4, then the ratio of areas [ABCD][AEF ] can be

written as ab for relatively prime positive integers a, b. Compute 100a+ b.

Proposed by: Akash Das

Answer: 5300

Solution 1: Rotate 4ABC around A to 4AB′C ′, such that B′ is on segment AF . Note that asBD ‖ EF , AB = AD. From this, AB′ = AB = AD, and B′ = D. Note that

∠ADC ′ = ∠ABC = 180− ∠ADC,

because ABCD is cyclic. Therefore, C,D, and C ′ are collinear. Also, AC ′ = AC, and

∠CAC ′ = ∠DAC + ∠C ′AD = ∠DAC + ∠CAB = ∠EAF.

Thus, since AE = AF , 4ACC ′ ∼ 4AEF . Now, we have

[ACC ′] = [ACD] + [ADC ′] = [ACD] + [ABC] = [ABCD].

But, [ACC ′] = CC′2

EF 2 · [AEF ], and we know that CC ′ = CD +DC ′ = 4 + 3 = 7. Thus,

[ABCD]

[AEF ]=

[ACC ′]

[AEF ]=

72

202=

49

400.

The answer is 100 · 49 + 400 = 5300.

Solution 2:

A

E F

C

B D

Since BD is parallel to EF and AE = AF , we have AB = AD. Since ABCD is cyclic, ∠ABC +∠ADC = 180◦. Thus we can glue 4ABC and 4ADC as shown in the diagram above to create atriangle that is similar to 4AEF and has the same area as ABCD. The base of this triangle haslength BC + CD = 3 + 4 = 7, so the desired ratio is

72

202=

49

400.

6. In triangle ABC, let M be the midpoint of BC, H be the orthocenter, and O be the circumcenter.Let N be the reflection of M over H. Suppose that OA = ON = 11 and OH = 7. Compute BC2.

Proposed by: Milan Haiman

Answer: 288

Solution: Let ω be the circumcircle of 4ABC. Note that because ON = OA, N is on ω. Let P bethe reflection of H over M . Then, P is also on ω. If Q is the midpoint of NP , note that because

NH = HM = MP,

Q is also the midpoint of HM . Since OQ ⊥ NP , we know that OQ ⊥ HM . As Q is also the midpointof HM ,

OM = OH = 7.

With this,

BM =√OB2 −BM2 = 6

√2,

and BC = 2BM = 12√

2. Therefore, BC2 = 288.

7. Let O and A be two points in the plane with OA = 30, and let Γ be a circle with center O and radiusr. Suppose that there exist two points B and C on Γ with ∠ABC = 90◦ and AB = BC. Computethe minimum possible value of brc.Proposed by: Hahn Lheem, Milan Haiman

Answer: 12

Solution: Let f1 denote a 45◦ counterclockwise rotation about point A followed by a dilation centeredA with scale factor 1/

√2. Similarly, let f2 denote a 45◦ clockwise rotation about point A followed

by a dilation centered A with scale factor 1/√

2. For any point B in the plane, there exists a pointC on Γ such that ∠ABC = 90◦ and AB = BC if and only if B lies on f1(Γ) or f2(Γ). Thus, suchpoints B and C on Γ exist if and only if Γ intersects f1(Γ) or f2(Γ). So, the minimum possible valueof r occurs when Γ is tangent to f1(Γ) and f2(Γ). This happens when r/

√2 + r = 30/

√2, i.e., when

r = 30√2+1

= 30√

2− 30. Therefore, the minimum possible value of brc is b30√

2− 30c = 12.

8. Two circles with radii 71 and 100 are externally tangent. Compute the largest possible area of a righttriangle whose vertices are each on at least one of the circles.

Proposed by: David Vulakh

Answer: 24200

Solution:

A

B

C

O

In general, let the radii of the circles be r < R, and let O be the center of the larger circle. If bothendpoints of the hypotenuse are on the same circle, the largest area occurs when the hypotenuse is adiameter of the larger circle, with [ABC] = R2.

If the endpoints of the hypotenuse are on different circles (as in the diagram above), then the distancefrom O to AB is half the distance from C to AB. Thus

[ABC] = 2[AOB] = AO ·OB sin∠AOB.

AO·OB and sin∠AOB are simultaneously maximized when AO·OB = (2r+R)·R and m∠AOB = 90◦,so the answer is R2 + 2Rr = 24200.

9. Let ABCD be a trapezoid with AB ‖ CD and AD = BD. Let M be the midpoint of AB, and letP 6= C be the second intersection of the circumcircle of 4BCD and the diagonal AC. Suppose thatBC = 27, CD = 25, and AP = 10. If MP = a

b for relatively prime positive integers a and b, compute100a+ b.

Proposed by: Krit Boonsiriseth

Answer: 2705

Solution 1: As ∠PBD = ∠PCD = ∠PAB, DB is tangent to (ABP ). As DA = DB, DA is alsotangent to (ABP ). Let CB intersect (ABP ) again at X 6= B; it follows that XD is the X-symmedianof 4AXB. As ∠AXC = ∠DAB = ∠ADC, X also lies on (ACD). Therefore ∠PXB = ∠PAB =∠PCD = ∠AXD, so XP is the median of 4AXB, i.e. XP passes through M .

Now we have 4MPA ∼ 4MBX and 4ABX ∼ 4BCD. Therefore

MP

AP=MB

XB=

AB

2XB=

BC

2DC=⇒ MP =

BC ·AP2DC

=27 · 10

2 · 25=

27

5.

Solution 2: Let E be the point such that �BDCE is a parallelogram. Then �ADCE is an isoscelestrapezoid, therefore ∠PAB = ∠CAB = ∠BED = ∠CDE; this angle is also equal to ∠PBD = ∠PCD.Now, ∠ABP = ∠ABD − ∠PBD = ∠BEC − ∠BED = ∠DEC, therefore 4PAB ∼ 4CDE. LetF be the midpoint of DE, which lies on BC because �BDCE is a parallelogram. It follows that(4PAB,M) ∼ (4CDE,F ), therefore

MP

AP=FC

DC=

BC

2DC,

and again this gives MP = 275 .

10. Acute triangle ABC has circumcircle Γ. Let M be the midpoint of BC. Points P and Q lie on Γ sothat ∠APM = 90◦ and Q 6= A lies on line AM . Segments PQ and BC intersect at S. Suppose that

BS = 1, CS = 3, PQ = 8√

737 , and the radius of Γ is r. If the sum of all possible values of r2 can be

expressed as ab for relatively prime positive integers a and b, compute 100a+ b.

Proposed by: Jeffrey Lu

Answer: 3703

Solution: Let A′ be the A-antipode in Γ, let O be the center of Γ, and let T := AA′ ∩BC. Note thatA′ lies on line PM . The key observation is that T is the reflection of S about M ; this follows by theButterfly Theorem on chords PA′ and AQ.

Let θ := ∠AMP and x = OT = OS. Observe that cos θ = PMAM = PQ

AA′ = PQ2r . We find the area of

4AMA′ in two ways. First, we have

2[AMA′] = AM ·MA · sin θ= AM · MB·MC

PM · sin θ= 4 tan θ

= 8r√

37448 −

14r2 .

On the other hand,

2[AMA′] = MT ·AA′ · sin∠OTM

= 2r√

1− 1x2 .

Setting the two expressions equal and squaring yields 3728 −

4r2 = 1 − 1

x2 . By Power of a Point,3 = BS · SC = r2 − x2, so x2 = r2 − 3. Substituting and solving the resulting quadratic in r2 givesr2 = 16

3 and r2 = 7. Thus ab = 37

3 , so 100a+ b = 3703.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT Spring 2021, March 06, 2021 — GUTS ROUND

Organization Team Team ID#

1. [8] Amelia wrote down a sequence of consecutive positive integers, erased one integer, and scrambled therest, leaving the sequence below. What integer did she erase?

6, 12, 1, 3, 11, 10, 8, 15, 13, 9, 7, 4, 14, 5, 2

2. [8] Suppose there exists a convex n-gon such that each of its angle measures, in degrees, is an odd primenumber. Compute the difference between the largest and smallest possible values of n.

3. [8] A semicircle with radius 2021 has diameter AB and center O. Points C and D lie on the semicirclesuch that ∠AOC < ∠AOD = 90◦. A circle of radius r is inscribed in the sector bounded by OA and OCand is tangent to the semicircle at E. If CD = CE, compute brc.

4. [8] In a 3 by 3 grid of unit squares, an up-right path is a path from the bottom left corner to the top rightcorner that travels only up and right in steps of 1 unit. For such a path p, let Ap denote the number ofunit squares under the path p. Compute the sum of Ap over all up-right paths p.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT Spring 2021, March 06, 2021 — GUTS ROUND

Organization Team Team ID#

5. [9] Let m,n > 2 be integers. One of the angles of a regular n-gon is dissected into m angles of equal sizeby (m−1) rays. If each of these rays intersects the polygon again at one of its vertices, we say n is m-cut.Compute the smallest positive integer n that is both 3-cut and 4-cut.

6. [9] In a group of 50 children, each of the children in the group have all of their siblings in the group. Eachchild with no older siblings announces how many siblings they have; however, each child with an oldersibling is too embarrassed, and says they have 0 siblings.

If the average of the numbers everyone says is 1225 , compute the number of different sets of siblings

represented in the group.

7. [9] Milan has a bag of 2020 red balls and 2021 green balls. He repeatedly draws 2 balls out of the baguniformly at random. If they are the same color, he changes them both to the opposite color and returnsthem to the bag. If they are different colors, he discards them. Eventually the bag has 1 ball left. Let pbe the probability that it is green. Compute b2021pc.

8. [9] Compute the product of all positive integers b ≥ 2 for which the base b number 111111b has exactly bdistinct prime divisors.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT Spring 2021, March 06, 2021 — GUTS ROUND

Organization Team Team ID#

9. [10] Let AD, BE, and CF be segments sharing a common midpoint, with AB < AE and BC < BF .Suppose that each pair of segments forms a 60◦ angle, and that AD = 7, BE = 10, and CF = 18. Let Kdenote the sum of the areas of the six triangles 4ABC, 4BCD, 4CDE, 4DEF , 4EFA, and 4FAB.Compute K

√3.

10. [10] Let a1, a2, . . . , an be a sequence of distinct positive integers such that a1 + a2 + · · ·+ an = 2021 anda1a2 · · · an is maximized. If M = a1a2 · · · an, compute the largest positive integer k such that 2k |M .

11. [10] For each positive integer 1 ≤ m ≤ 10, Krit chooses an integer 0 ≤ am < m uniformly at random.Let p be the probability that there exists an integer n for which n ≡ am (mod m) for all m. If p can bewritten as a

b for relatively prime positive integers a and b, compute 100a+ b.

12. [10] Compute the number of labelings f : {0, 1}3 → {0, 1, . . . , 7} of the vertices of the unit cube such that

|f(vi)− f(vj)| ≥ d(vi, vj)2

for all vertices vi, vj of the unit cube, where d(vi, vj) denotes the Euclidean distance between vi and vj .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT Spring 2021, March 06, 2021 — GUTS ROUND

Organization Team Team ID#

13. [11] A tournament among 2021 ranked teams is played over 2020 rounds. In each round, two teams areselected uniformly at random among all remaining teams to play against each other. The better rankedteam always wins, and the worse ranked team is eliminated. Let p be the probability that the second bestranked team is eliminated in the last round. Compute b2021pc.

14. [11] In triangle ABC, ∠A = 2∠C. Suppose that AC = 6, BC = 8, and AB =√a− b, where a and b are

positive integers. Compute 100a+ b.

15. [11] Two circles Γ1 and Γ2 of radius 1 and 2, respectively, are centered at the origin. A particle is placedat (2, 0) and is shot towards Γ1. When it reaches Γ1, it bounces off the circumference and heads backtowards Γ2. The particle continues bouncing off the two circles in this fashion.

If the particle is shot at an acute angle θ above the x-axis, it will bounce 11 times before returning to(2, 0) for the first time. If cot θ = a−

√b for positive integers a and b, compute 100a+ b.

16. [11] Let f : Z2 → Z be a function such that, for all positive integers a and b,

f(a, b) =

b if a > b

f(2a, b) if a ≤ b and f(2a, b) < a

f(2a, b)− a otherwise.

Compute f(1000, 32021).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT Spring 2021, March 06, 2021 — GUTS ROUND

Organization Team Team ID#

17. [12] Let k be the answer to this problem. The probability that an integer chosen uniformly at randomfrom {1, 2, . . . , k} is a multiple of 11 can be written as a

b for relatively prime positive integers a and b.Compute 100a+ b.

18. [12] Triangle ABC has side lengths AB = 19, BC = 20, and CA = 21. Points X and Y are selected onsides AB and AC, respectively, such that AY = XY and XY is tangent to the incircle of 4ABC. If thelength of segment AX can be written as a

b , where a and b are relatively prime positive integers, compute100a+ b.

19. [12] Almondine has a bag with N balls, each of which is red, white, or blue. If Almondine picks threeballs from the bag without replacement, the probability that she picks one ball of each color is larger than23 percent. Compute the largest possible value of

⌊N3

⌋.

20. [12] Let f(x) = x3 − 3x. Compute the number of positive divisors of⌊f

(f

(f

(f

(f

(f

(f

(f

(5

2

))))))))⌋,

where f is applied 8 times.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT Spring 2021, March 06, 2021 — GUTS ROUND

Organization Team Team ID#

21. [14] Bob knows that Alice has 2021 secret positive integers x1, . . . , x2021 that are pairwise relatively prime.Bob would like to figure out Alice’s integers. He is allowed to choose a set S ⊆ {1, 2, . . . , 2021} and askher for the product of xi over i ∈ S. Alice must answer each of Bob’s queries truthfully, and Bob may useAlice’s previous answers to decide his next query. Compute the minimum number of queries Bob needsto guarantee that he can figure out each of Alice’s integers.

22. [14] Let E be a three-dimensional ellipsoid. For a plane p, let E(p) be the projection of E onto the planep. The minimum and maximum areas of E(p) are 9π and 25π, and there exists a p where E(p) is a circleof area 16π. If V is the volume of E, compute V/π.

23. [14] Let f : N → N be a strictly increasing function such that f(1) = 1 and f(2n)f(2n + 1) = 9f(n)2 +3f(n) for all n ∈ N. Compute f(137).

24. [14] Let P be a point selected uniformly at random in the cube [0, 1]3. The plane parallel to x+y+z = 0passing through P intersects the cube in a two-dimensional region R. Let t be the expected value of theperimeter of R. If t2 can be written as a

b , where a and b are relatively prime positive integers, compute100a+ b.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT Spring 2021, March 06, 2021 — GUTS ROUND

Organization Team Team ID#

25. [16] Let n be a positive integer. Claudio has n cards, each labeled with a different number from 1 to n.He takes a subset of these cards, and multiplies together the numbers on the cards. He remarks that,given any positive integer m, it is possible to select some subset of the cards so that the difference betweentheir product and m is divisible by 100. Compute the smallest possible value of n.

26. [16] Let triangle ABC have incircle ω, which touches BC, CA, and AB at D, E, and F , respectively.Then, let ω1 and ω2 be circles tangent to AD and internally tangent to ω at E and F , respectively. LetP be the intersection of line EF and the line passing through the centers of ω1 and ω2. If ω1 and ω2 haveradii 5 and 6, respectively, compute PE · PF .

27. [16] Let P be the set of points{(x, y) | 0 ≤ x, y ≤ 25, x, y ∈ Z},

and let T be the set of triangles formed by picking three distinct points in P (rotations, reflections, andtranslations count as distinct triangles). Compute the number of triangles in T that have area larger than300.

28. [16] Caroline starts with the number 1, and every second she flips a fair coin; if it lands heads, she adds1 to her number, and if it lands tails she multiplies her number by 2. Compute the expected number ofseconds it takes for her number to become a multiple of 2021.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT Spring 2021, March 06, 2021 — GUTS ROUND

Organization Team Team ID#

29. [18] Compute the number of complex numbers z with |z| = 1 that satisfy

1 + z5 + z10 + z15 + z18 + z21 + z24 + z27 = 0.

30. [18] Let f(n) be the largest prime factor of n2 + 1. Compute the least positive integer n such thatf(f(n)) = n.

31. [18] Roger initially has 20 socks in a drawer, each of which is either white or black. He chooses a sockuniformly at random from the drawer and throws it away. He repeats this action until there are equalnumbers of white and black socks remaining.

Suppose that the probability he stops before all socks are gone is p. If the sum of all distinct possiblevalues of p over all initial combinations of socks is a

b for relatively prime positive integers a and b, compute100a+ b.

32. [18] Let acute triangle ABC have circumcenter O, and let M be the midpoint of BC. Let P be theunique point such that ∠BAP = ∠CAM , ∠CAP = ∠BAM , and ∠APO = 90◦. If AO = 53, OM = 28,and AM = 75, compute the perimeter of 4BPC.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

HMMT Spring 2021, March 06, 2021 — GUTS ROUND

Organization Team Team ID#

33. [20] After the Guts round ends, HMMT organizers will collect all answers submitted to all 66 questions(including this one) during the individual rounds and the guts round. Estimate N , the smallest positiveinteger that no one will have submitted at any point during the tournament.

An estimate of E will receive max (0, 24− 4|E −N |) points.

34. [20] Let f(n) be the largest prime factor of n. Estimate

N =

⌊104 ·

∑106

n=2 f(n2 − 1

)∑106

n=2 f(n)

⌋.

An estimate of E will receive max

(0,

⌊20− 20

(|E−N |103

)1/3⌋)points.

35. [20] Geoff walks on the number line for 40 minutes, starting at the point 0. On the nth minute, he flipsa fair coin. If it comes up heads he walks 1

n in the positive direction and if it comes up tails he walks1n in the negative direction. Let p be the probability that he never leaves the interval [−2, 2]. EstimateN = b104pc.

An estimate of E will receive max

(0,

⌊20− 20

(|E−N |160

)1/3⌋)points.

36. [20] A set of 6 distinct lattice points is chosen uniformly at random from the set {1, 2, 3, 4, 5, 6}2. Let Abe the expected area of the convex hull of these 6 points. Estimate N = b104Ac.

An estimate of E will receive max

(0,

⌊20− 20

(|E−N |104

)1/3⌋)points.

HMMT Spring 2021March 06, 2021

Guts Round

1. [8] Amelia wrote down a sequence of consecutive positive integers, erased one integer, and scrambledthe rest, leaving the sequence below. What integer did she erase?

6, 12, 1, 3, 11, 10, 8, 15, 13, 9, 7, 4, 14, 5, 2

Proposed by: Andrew Gu

Answer: 16

Solution: The sequence of positive integers exactly contains every integer between 1 and 15, inclusive.16 is the only positive integer that could be added to this sequence such that the resulting sequencecould be reordered to make a sequence of consecutive positive integers. Therefore, Amelia must haveerased the integer 16.

2. [8] Suppose there exists a convex n-gon such that each of its angle measures, in degrees, is an oddprime number. Compute the difference between the largest and smallest possible values of n.

Proposed by: Andrew Gu

Answer: 356

Solution: We can’t have n = 3 since the sum of the angles must be 180◦ but the sum of threeodd numbers is odd. On the other hand, for n = 4 we can take a quadrilateral with angle measures83◦, 83◦, 97◦, 97◦.

The largest possible value of n is 360. For larger n we can’t even have all angles have integer measure,and 179 happens to be prime.

So, the answer is 360− 4 = 356.

3. [8] A semicircle with radius 2021 has diameter AB and center O. Points C and D lie on the semicirclesuch that ∠AOC < ∠AOD = 90◦. A circle of radius r is inscribed in the sector bounded by OA andOC and is tangent to the semicircle at E. If CD = CE, compute brc.Proposed by: Hahn Lheem

Answer: 673

Solution: We are givenm∠EOC = m∠COD

andm∠AOC +m∠COD = 2m∠EOC +m∠COD = 90◦.

So m∠EOC = 30◦ and m∠AOC = 60◦. Letting the radius of the semicircle be R, we have

(R− r) sin∠AOC = r ⇒ r =1

3R,

so

brc =

⌊2021

3

⌋= 673.

4. [8] In a 3 by 3 grid of unit squares, an up-right path is a path from the bottom left corner to the topright corner that travels only up and right in steps of 1 unit. For such a path p, let Ap denote thenumber of unit squares under the path p. Compute the sum of Ap over all up-right paths p.

Proposed by: Freddie Zhao

Answer: 90

Solution: Each path consists of 3 steps up and 3 steps to the right, so there are(63

)= 20 total paths.

Consider the sum of the areas of the regions above all of these paths. By symmetry, this is the sameas the answer to the problem. For any path, the sum of the areas of the regions above and below it is32 = 9, so the sum of the areas of the regions above and below all paths is 9 · 20 = 180. Therefore, ourfinal answer is 1

2 · 180 = 90.

5. [9] Let m,n > 2 be integers. One of the angles of a regular n-gon is dissected into m angles of equalsize by (m− 1) rays. If each of these rays intersects the polygon again at one of its vertices, we say nis m-cut. Compute the smallest positive integer n that is both 3-cut and 4-cut.

Proposed by: Carl Schildkraut

Answer: 14

Solution: For the sake of simplicity, inscribe the regular polygon in a circle. Note that each interiorangle of the regular n−gon will subtend n−2 of the n arcs on the circle. Thus, if we dissect an interiorangle into m equal angles, then each must be represented by a total of n−2

m arcs. However, since eachof the rays also passes through another vertex of the polygon, that means n−2

m is an integer and thusour desired criteria is that m divides n− 2.

That means we want the smallest integer n > 2 such that n − 2 is divisible by 3 and 4 which is just12 + 2 = 14.

6. [9] In a group of 50 children, each of the children in the group have all of their siblings in the group.Each child with no older siblings announces how many siblings they have; however, each child with anolder sibling is too embarrassed, and says they have 0 siblings.

If the average of the numbers everyone says is 1225 , compute the number of different sets of siblings

represented in the group.

Proposed by: Vincent Bian

Answer: 26

Solution: For i ≥ 1, let ai be the number of families that have i members in the group. Then, amongeach family with i children in the group, the oldest child will say i− 1, and the rest will say 0. Thus,the sum of all the numbers said will be a2 + 2a3 + 3a4 + 4a5 + · · · = 50× 12

25 = 24.

Also because there are 50 children total, we know that a1 + 2a2 + 3a3 + · · · = 50. We can subtractthese two equations to get a1 + a2 + a3 + · · · = 50− 24 = 26.

7. [9] Milan has a bag of 2020 red balls and 2021 green balls. He repeatedly draws 2 balls out of thebag uniformly at random. If they are the same color, he changes them both to the opposite color andreturns them to the bag. If they are different colors, he discards them. Eventually the bag has 1 ballleft. Let p be the probability that it is green. Compute b2021pc.Proposed by: Milan Haiman

Answer: 2021

Solution: The difference between the number of green balls and red balls in the bag is always 1modulo 4. Thus the last ball must be green and p = 1.

8. [9] Compute the product of all positive integers b ≥ 2 for which the base b number 111111b has exactlyb distinct prime divisors.

Proposed by: Esha Bhatia

Answer: 24

Solution: Notice that this value, in base b, is

b6 − 1

b− 1= (b+ 1)(b2 − b+ 1)(b2 + b+ 1).

This means that, if b satisfies the problem condition, (b+ 1)(b2 − b+ 1)(b2 + b+ 1) > p1 . . . pb, wherepi is the ith smallest prime.

We claim that, if b ≥ 7, then p1 . . . pb > (b + 1)(b2 − b + 1)(b2 + b + 1). This is true for b = 7 bycalculation, and can be proven for larger b by induction and the estimate pi ≥ i.All we have to do is to check b ∈ 2, 3, 4, 5, 6. Notice that for b = 6, the primes cannot include 2, 3

and hence we want 66−15 to be divisible product of 6 primes the smallest of which is 5. However,

5 · 7 · · · · 17 > 66−15 , and by checking we rule out 5 too. All that is left is {2, 3, 4}, all of which work,

giving us an answer of 24.

9. [10] Let AD, BE, and CF be segments sharing a common midpoint, with AB < AE and BC < BF .Suppose that each pair of segments forms a 60◦ angle, and that AD = 7, BE = 10, and CF = 18. LetK denote the sum of the areas of the six triangles 4ABC, 4BCD, 4CDE, 4DEF , 4EFA, and4FAB. Compute K

√3.

Proposed by: Milan Haiman

Answer: 141

Solution: Let M be the common midpoint, and let x = 7, y = 10, z = 18. One can verify that hexagonABCDEF is convex. We have

[ABC] = [ABM ]+[BCM ]− [ACM ] =1

2·√

3

2· x2· y2

+1

2·√

3

2· y2· z2− 1

2·√

3

2· x2· z2

=

√3(xy + yz − zx)

16.

Summing similar expressions for all 6 triangles, we have

K =

√3(2xy + 2yz + 2zx)

16.

Substituting x, y, z gives K = 47√

3, for an answer of 141.

Remark: As long as hexagon ABCDEF is convex, K is the area of this hexagon.

10. [10] Let a1, a2, . . . , an be a sequence of distinct positive integers such that a1+a2+ · · ·+an = 2021 anda1a2 · · · an is maximized. If M = a1a2 · · · an, compute the largest positive integer k such that 2k |M .

Proposed by: Sheldon Kieren Tan

Answer: 62

Solution: We claim that the optimal set is {2, 3, · · · , 64} \ {58}. We first show that any optimal setis either of the form {b, b+ 1, b+ 2, . . . , d} or {b, b+ 1, . . . , d} \ {c}, for some b < c < d.

Without loss of generality, assume that the sequence a1 < a2 < · · · < an has the maximum product.Suppose aj+1 > aj + 2. Then, increasing aj by 1 and decreasing aj+1 by 1 will increase the product

M, contradicting the assumption that the sequence has the optimal product. Thus, any ”gaps” in theai can only have size 1.

Now, we show that there can only be one such gap. Suppose aj+1 = aj + 2, and ak+1 = ak + 2, forj < k. Then, we can increase aj by 1 and decrease ai+1 by 1 to increase the total product. Thus, thereis at most one gap, and the sequence ai is of one of the forms described before.

We now show that either b = 2 or b = 3. Consider any set of the form {b, b + 1, b + 2, . . . , d} or{b, b + 1, . . . , d} \ {c}. If b = 1, then we can remove b and increase d by 1 to increase the product. Ifb > 4, then we can remove b and replace it with 2 and b − 2 to increase the product. Thus, we haveb = 2, 3, or 4.

Suppose b = 4. If the next element is 5, we can replace it with a 2 and a 3 to increase the product, andif the next element is 6, we can replace it with a 1, 2, and 3 without making the product any smaller.Thus, we can assume that either b = 2 or b = 3.

The nearest triangular number to 2021 is 2016 = 1+2+· · ·+64. Using this, we can compute that if b = 2,our set must be {2, 3, · · · , 64} \ {58}, leading to a product of 64!

58 . If b = 3, our set is {3, · · · , 64} \ {56},leading to a product of 64!

2·56 .

Thus, the maximum product is 64!58 . We now compute the highest power of 2 that divides this expres-

sion. 64! includes 32 elements that contribute at least one power of 2, 16 that contribute at least twopowers of 2, and so on until the one element that contributes at least six powers of 2. This means thehighest power of 2 that divides 64! is 32 + 16 + · · · + 2 + 1 = 63. Finally, dividing by 58 removes oneof these powers of 2, making the answer 62.

11. [10] For each positive integer 1 ≤ m ≤ 10, Krit chooses an integer 0 ≤ am < m uniformly at random.Let p be the probability that there exists an integer n for which n ≡ am (mod m) for all m. If p canbe written as a

b for relatively prime positive integers a and b, compute 100a+ b.

Proposed by: Daniel Zhu

Answer: 1540

Solution: Tuples of valid am correspond with residues mod lcm(1, 2, . . . , 10), so the answer is

lcm(1, 2, . . . , 10)

10!=

23 · 32 · 5 · 728 · 34 · 52 · 7

=1

1440.

12. [10] Compute the number of labelings f : {0, 1}3 → {0, 1, . . . , 7} of the vertices of the unit cube suchthat

|f(vi)− f(vj)| ≥ d(vi, vj)2

for all vertices vi, vj of the unit cube, where d(vi, vj) denotes the Euclidean distance between vi andvj .

Proposed by: Krit Boonsiriseth

Answer: 144

Solution: Let B = {0, 1}3, let E = {(x, y, z) ∈ B : x + y + z is even}, and let O = {(x, y, z) ∈ B :x + y + z is odd}. As all pairs of vertices within E (and within O) are

√2 apart, is easy to see that

{f(E), f(O)} = {{0, 2, 4, 6}, {1, 3, 5, 7}}.

• There are two ways to choose f(E) and f(O); from now on WLOG assume f(E) = {0, 2, 4, 6}.• There are 4! ways to assign the four labels to the four vertices in E.

• The vertex opposite the vertex labeled 0 is in O, and it must be labeled 3, 5, or 7. It is easyto check that for each possible label of this vertex, there is exactly one way to label the threeremaining vertices.

Therefore the total number of labelings is 2 · 4! · 3 = 144.

13. [11] A tournament among 2021 ranked teams is played over 2020 rounds. In each round, two teamsare selected uniformly at random among all remaining teams to play against each other. The betterranked team always wins, and the worse ranked team is eliminated. Let p be the probability that thesecond best ranked team is eliminated in the last round. Compute b2021pc.Proposed by: Milan Haiman

Answer: 674

Solution: In any given round, the second-best team is only eliminated if it plays against the best team.If there are k teams left and the second-best team has not been eliminated, the second-best team playsthe best team with probability 1

(k2)

, so the second-best team survives the round with probability

1− 1(k2

) = 1− 2

k(k − 1)=k2 − k − 2

k(k − 1)=

(k + 1)(k − 2)

k(k − 1).

So, the probability that the second-best team survives every round before the last round is

2021∏k=3

(k + 1)(k − 2)

k(k − 1),

which telescopes to

2022!3! ·

2019!0!

2021!2! ·

2020!1!

=2022! · 2019!

2021! · 2020!· 2! · 1!

3! · 0!=

2022

2020· 1

3=

337

1010= p.

So,

b2021pc =

⌊2021 · 337

1010

⌋=

⌊337 · 2 + 337 · 1

1010

⌋= 337 · 2 = 674.

14. [11] In triangle ABC, ∠A = 2∠C. Suppose that AC = 6, BC = 8, and AB =√a− b, where a and b

are positive integers. Compute 100a+ b.

Proposed by: Freddie Zhao, Milan Haiman

Answer: 7303

Solution: Let x = AB, and ∠C = θ, then ∠A = 2θ and ∠B = 180− 3θ.

Extend ray BA to D so that AD = AC. We know that ∠CAD = 180−2θ, and since4ADC is isosceles,it follows that ∠ADC = ∠ACD = θ, and so ∠DCB = 2θ = ∠BAC, meaning that 4BAC ∼ 4BCD.

Therefore, we havex+ 6

8=

8

x=⇒ x(x+ 6) = 82

Since x > 0, we have x = −3 +√

73. So 100a+ b = 7303.

15. [11] Two circles Γ1 and Γ2 of radius 1 and 2, respectively, are centered at the origin. A particle isplaced at (2, 0) and is shot towards Γ1. When it reaches Γ1, it bounces off the circumference and headsback towards Γ2. The particle continues bouncing off the two circles in this fashion.

If the particle is shot at an acute angle θ above the x-axis, it will bounce 11 times before returning to(2, 0) for the first time. If cot θ = a−

√b for positive integers a and b, compute 100a+ b.

Proposed by: Hahn Lheem

Answer: 403

Solution: By symmetry, the particle must bounce off of Γ2 at points that make angles of 60◦, 120◦,180◦, 240◦, and 300◦ with the positive x-axis. Similarly, the particle must bounce off of Γ1 at pointsthat make angles of 30◦, 90◦, 150◦, 210◦, 270◦, and 330◦ with the positive x-axis.

In particular, the first point that the ball touches on Γ1 is (cos 30◦, sin 30◦). So,

cot θ =2− cos 30◦

sin 30◦= 4−

√3.

16. [11] Let f : Z2 → Z be a function such that, for all positive integers a and b,

f(a, b) =

b if a > b

f(2a, b) if a ≤ b and f(2a, b) < a

f(2a, b)− a otherwise.

Compute f(1000, 32021).

Proposed by: Akash Das

Answer: 203

Solution: Note that f(a, b) is the remainder of b when divided by a. If a > b then f(a, b) is exactly bmod a. If instead a ≤ b, our ”algorithm” doubles our a by n times until we have a × 2n > b. At thispoint, we subtract a2n−1 from f(a · 2n, b) and iterate back down until we get a > b − a · k > 0 andf(a, b) = b− a · k for some positive integer k. This expression is equivalent to b− a · k mod a, or b moda.

Thus, we want to compute 32021 mod 1000. This is equal to 3 mod 8 and 78 mod 125. By CRT, thisimplies that the answer is 203.

17. [12] Let k be the answer to this problem. The probability that an integer chosen uniformly at randomfrom {1, 2, . . . , k} is a multiple of 11 can be written as a

b for relatively prime positive integers a and b.Compute 100a+ b.

Proposed by: Carl Schildkraut, Hahn Lheem

Answer: 1000

Solution: We write k = 11q + r for integers q, r with 0 ≤ r < 11. There are q multiples of 11 from 1to k, inclusive, so our probability is a

b = q11q+r . Let d = gcd(q, r) = gcd(q, 11q+r), so that the fraction

q/d(11q+r)/d is how we would write q

11q+r in simplified form. Since we require that a and b be relatively

prime, we find a = qd and b = 11q+r

d .

Plugging these into the equation k = 100a+b, we find 11q+r = 100 qd + 11q+r

d , or d(11q+r) = 111q+r.Since d divides r and r ≤ 10, we have d ≤ 10.

If we test the case d = 10, our equation becomes q = 9r. Since r = 10 is the only valid value that is amultiple of d, we get q = 90 and k = 1000. 10 is, in fact, the gcd of q and r, so we have found that

k = 1000 satisfies the problem. Testing other values of d does not produce a valid answer.

18. [12] Triangle ABC has side lengths AB = 19, BC = 20, and CA = 21. Points X and Y are selectedon sides AB and AC, respectively, such that AY = XY and XY is tangent to the incircle of 4ABC.If the length of segment AX can be written as a

b , where a and b are relatively prime positive integers,compute 100a+ b.

Proposed by: Hahn Lheem

Answer: 6710

Solution: Note that the incircle of 4ABC is the A-excenter of 4AXY . Let r be the radius of thiscircle. We can compute the area of 4AXY in two ways:

KAXY =1

2·AX ·AY sinA

= r · (AX +AY −XY )/2

=⇒ AY =r

sinA

We also know that

KABC =1

2· 19 · 21 sinA

= r · (19 + 20 + 21)/2

=⇒ r

sinA=

19 · 21

60=

133

20

so AY = 133/20.

Let the incircle of 4ABC be tangent to AB and AC at D and E, respectively. We know thatAX +AY +XY = AD +AE = 19 + 21− 20, so AX = 20− 133

10 = 6710 .

19. [12] Almondine has a bag with N balls, each of which is red, white, or blue. If Almondine picks threeballs from the bag without replacement, the probability that she picks one ball of each color is largerthan 23 percent. Compute the largest possible value of

⌊N3

⌋.

Proposed by: James Lin

Answer: 29

Solution: If k = bN3 c, then the maximum possible probability is 6k3

(3k)(3k−1)(3k−2) . with equality when

there are k balls of each of the three colors. Going from 3k → 3k + 1 replaces k3k−2 →

k+13k+1 , which is

smaller, and going from 3k + 1→ 3k + 2 replaces k3k−1 →

k+13k+2 , which is again smaller. For this to be

larger than 23100 , we find we need 0 > 7k2 − 207k + 46, and so k = 29 is the maximal value.

20. [12] Let f(x) = x3 − 3x. Compute the number of positive divisors of⌊f

(f

(f

(f

(f

(f

(f

(f

(5

2

))))))))⌋,

where f is applied 8 times.

Proposed by: Akash Das

Answer: 6562

Solution: Note that f(y + 1y ) = (y + 1

y )3 − 3(y + 1y ) = y3 + 1

y3 . Thus, f(2 + 12 ) = 23 + 1

23 , and in

general fk(2 + 1

2

)= 23

k

+ 1

23k, where f is applied k times. It follows that we just need to find the

number of divisors of⌊23

8

+ 1238

⌋= 23

8

, which is just 38 + 1 = 6562.

21. [14] Bob knows that Alice has 2021 secret positive integers x1, . . . , x2021 that are pairwise relativelyprime. Bob would like to figure out Alice’s integers. He is allowed to choose a set S ⊆ {1, 2, . . . , 2021}and ask her for the product of xi over i ∈ S. Alice must answer each of Bob’s queries truthfully, andBob may use Alice’s previous answers to decide his next query. Compute the minimum number ofqueries Bob needs to guarantee that he can figure out each of Alice’s integers.

Proposed by: David Vulakh

Answer: 11

Solution: In general, Bob can find the values of all n integers asking only blog2 nc+ 1 queries.

For each of Alice’s numbers xi, let Qi be the set of queries S such that i ∈ S. Notice that all Qi mustbe nonempty and distinct. If there exists an empty Qi, Bob has asked no queries that include xi andhas no information about its value. If there exist i, j, i 6= j such that Qi = Qj , xi and xj could beinterchanged without the answer to any query changing, so there does not exist a unique sequence ofnumbers described by the answers to Bob’s queries (Alice can make her numbers distinct).

From the above, blog2 nc+1 is a lower bound on the number of queries, because the number of distinctnonempty subsets of {1, . . . , n} is 2n − 1.

If Bob asks any set of queries such that all Qi are nonempty and disjoint, he can uniquely determineAlice’s numbers. In particular, since the values x1, . . . , x2021 are relatively prime, each prime factorof xi occurs in the answer to query Sj iff j ∈ Q(i) (and that prime factor will occur in each answerexactly to the power with which it appears in the factorization of xi). Since all Q(i) are unique, all xican therefore be uniquely recovered by computing the product of the prime powers that occur exactlyin the answers to queries Q(i).

It is possible for Bob to ask blog2 nc+1 queries so that each i is contained in a unique nonempty subsetof them. One possible construction is to include the index i in the jth query iff the 2i−1-value bit isset in the binary representation of j. So the answer is blog2 2021c+ 1 = 11.

22. [14] Let E be a three-dimensional ellipsoid. For a plane p, let E(p) be the projection of E onto theplane p. The minimum and maximum areas of E(p) are 9π and 25π, and there exists a p where E(p)is a circle of area 16π. If V is the volume of E, compute V/π.

Proposed by: Daniel Zhu

Answer: 75

Solution: Let the three radii of E be a < b < c. We know that ab = 9 and bc = 25.

Consider the plane p where projection E(p) has area 9π. Fixing p, rotate E on the axis passing throughthe radius with length b until E(p) has area 25π. The projection onto p will be an ellipse with radii band r, where r increases monotonically from a to c.

By Intermediate Value Theorem, there must exist a circular projection with radius b. As the area ofthis projection is 16π, b = 4. Thus,

V =4

3π · abc =

4

3· 225

4π = 75π.

23. [14] Let f : N → N be a strictly increasing function such that f(1) = 1 and f(2n)f(2n + 1) =9f(n)2 + 3f(n) for all n ∈ N. Compute f(137).

Proposed by: Sheldon Kieren Tan

Answer: 2215.

Solution: Plugging in n = 1 gives f(2)f(3) = 12, therefore (f(2), f(3)) = (2, 6) or (3, 4). However,the former implies

f(4)f(5) ≥ (6 + 1)(6 + 2) > 42 = 9 · 22 + 3 · 2,

which is impossible; therefore f(2) = 3 and f(3) = 4. We now show by induction with step size 2 thatf(2n) = 3f(n) and f(2n+ 1) = 3f(n) + 1 for all n; the base case n = 1 has already been proven.

Assume the statement is true for n < 2k. Applying the given and the inductive hypothesis, we have

f(4k)f(4k + 1) = (3f(2k))(3f(2k) + 1) = (9f(k))(9f(k) + 1)

f(4k + 2)f(4k + 3) = (3f(2k + 1))(3f(2k + 1) + 1) = (9f(k) + 3)(9f(k) + 4)

Let x = f(4k + 1). Since f is strictly increasing, this implies x ≥√f(4k)f(4k + 1) > 9f(k) and

x ≤√f(4k + 2)f(4k + 3) − 1 < 9f(k) + 3. So x = 9f(k) + 1 or x = 9f(k) + 2. Since 9f(k) + 2 does

not divide 9f(k)(9f(k) + 1), we must have f(4k + 1) = x = 9f(k) + 1 and f(4k) = 9f(k). A similarargument shows that f(4k+2) = 9f(k)+3 and f(4k+3) = 9f(k)+4, and this completes the inductivestep.

Now it is a straightforward induction to show that f is the function that takes a number’s binary digitsand treats it as base 3. Since 137 = 100010012 in binary, f(137) = 100010013 = 37 + 33 + 1 = 2215.

Remark: 137 = 20214.

24. [14] Let P be a point selected uniformly at random in the cube [0, 1]3. The plane parallel to x+y+z = 0passing through P intersects the cube in a two-dimensional region R. Let t be the expected value ofthe perimeter of R. If t2 can be written as a

b , where a and b are relatively prime positive integers,compute 100a+ b.

Proposed by: Michael Ren

Answer: 12108

Solution: We can divide the cube into 3 regions based on the value of x + y + z which defines theplane: x + y + z < 1, 1 ≤ x + y + z ≤ 2, and x + y + z > 2. The two regions on the ends createtetrahedra, each of which has volume 1/6. The middle region is a triangular antiprism with volume2/3.

If our point P lies in the middle region, we can see that we will always get the same value 3√

2 for theperimeter of R.Now let us compute the expected perimeter given that we pick a point P in the first region x+y+z < 1.If x + y + z = a, then the perimeter of R will just be 3

√2a, so it is sufficient to find the expected

value of a. a is bounded between 0 and 1, and forms a continuous probability distribution with valueproportional to a2, so we can see with a bit of calculus that its expected value is 3/4.

The region x+y+z > 2 is identical to the region x+y+z < 1, so we get the same expected perimeter.Thus we have a 2/3 of a guaranteed 3

√2 perimeter, and a 1/3 of having an expected 9

4

√2 perimeter,

which gives an expected perimeter of 23 · 3√

2 + 13 ·

94

√2 = 11

√2

4 . The square of this is 1218 , giving an

extraction of 12108.

25. [16] Let n be a positive integer. Claudio has n cards, each labeled with a different number from 1 ton. He takes a subset of these cards, and multiplies together the numbers on the cards. He remarks

that, given any positive integer m, it is possible to select some subset of the cards so that the differencebetween their product and m is divisible by 100. Compute the smallest possible value of n.

Proposed by: Carl Schildkraut

Answer: 17

Solution: We require that n ≥ 15 so that the product can be divisible by 25 without being even. Inaddition, for any n > 15, if we can acquire all residues relatively prime to 100, we may multiply themby some product of {1, 2, 4, 5, 15} to achieve all residues modulo 100, so it suffices to acquire only thoseresidues. For n = 15, we have the numbers {3, 7, 9, 11, 13} to work with (as 1 is superfluous); thesegive only 25 = 32 distinct products, so they cannot be sufficient. So, we must have n ≥ 17, whence wehave the numbers {3, 7, 9, 11, 13, 17}. These generators are in fact sufficient. The following calculationsare motivated by knowledge of factorizations of some small numbers, as well as careful considerationof which sets of numbers we have and haven’t used. It is also possible to simply write out a table ofwhich residues relatively prime to 100 are included once each number is added, which likely involvesfewer calculations.

First, consider the set {3, 11, 13, 17}. This set generates, among other numbers, those in {1, 11, 21, 31, 51, 61}.Since {7, 9} generates {1, 7, 9, 63}, which spans every residue class mod 10 relatively prime to 10, weneed only worry about

{41, 71, 81, 91} × {1, 7, 9, 63}.

Since 41 can be generated as 3 · 7 · 13 · 17 and 91 can be generated as 7 · 13, we need not worry aboutthese times 1 and 9, and we may verify

41 · 7 ≡ 87 ≡ 11 · 17, 91 · 63 ≡ 33 ≡ 3 · 11.

and91 · 7 ≡ 37 ≡ 3 · 9 · 11 · 13 · 17,

using the method we used to generate 49 earlier. So, we only need to worry about

{71, 81} × {1, 7, 9, 63}.

We calculate71 ≡ 7 · 9 · 17, 71 · 9 ≡ 39 ≡ 3 · 13, 71 · 63 ≡ 73 ≡ 3 · 7 · 13,

each of which doesn’t use 11, allowing us to get all of

{71, 81} × {1, 9, 63},

so we are only missing 71 · 7 ≡ 97 and 81 · 7 ≡ 67. We find

97 ≡ 3 · 9 · 11

and67 ≡ 3 · 9 · 13 · 17,

so all numbers are achievable and we are done.

26. [16] Let triangle ABC have incircle ω, which touches BC, CA, and AB at D, E, and F , respectively.Then, let ω1 and ω2 be circles tangent to AD and internally tangent to ω at E and F , respectively.Let P be the intersection of line EF and the line passing through the centers of ω1 and ω2. If ω1 andω2 have radii 5 and 6, respectively, compute PE · PF .

Proposed by: Akash Das

Answer: 3600

Solution: Let the centers of ω1 and ω2 be O1 and O2. Let DE intersect ω1 again at Q, and let DFintersect ω2 again at R. Note that since ω1 and ω2 must be tangent to AD at the same point (by equaltangents), so AD must be the radical axis of ω1 and ω2, so RQEF is cyclic. Thus, we have

∠O1QR = ∠EQR− ∠O1QE = 180◦ − ∠EFD − ∠O1EQ = 90◦

Thus, we have QR is tangent to ω1, and similarly it must be tangent to ω2 as well.

Now, note that by Monge’s theorem on ω, ω1, and ω2, we have that P must be the intersection of theexternal tangents of ω1 and ω2. Since RQ is an external tangent, we have P , Q, and R are collinear.Thus, by power of a point, we have PE · PF = PR · PQ. Note that PR = 10

√30 and PQ = 12

√30.

Thus, we have PE · PF = 3600.

27. [16] Let P be the set of points

{(x, y) | 0 ≤ x, y ≤ 25, x, y ∈ Z},

and let T be the set of triangles formed by picking three distinct points in P (rotations, reflections,and translations count as distinct triangles). Compute the number of triangles in T that have arealarger than 300.

Proposed by: Andrew Lin, Haneul Shin

Answer: 436

Solution: Lemma: The area of any triangle inscribed in an a by b rectangle is at most ab2 . (Any

triangle’s area can be increased by moving one of its sides to a side of the rectangle). Given this,because any triangle in T is inscribed in a 25 × 25 square, we know that the largest possible area of

a triangle is 252

2 , and any triangle which does not use the full range of x or y-values will have area nomore than 25·24

2 = 300.

There are 4 · 25 = 100 triangles of maximal area: pick a side of the square and pick one of the 26vertices on the other side of our region; each triangle with three vertices at the corners of the squareis double-counted once. To get areas between 25·24

2 and 25·252 , we need to pick a vertex of the square

((0, 0) without loss of generality), as well as (25, y) and (x, 25). By Shoelace, this has area 252−xy2 , and

since x and y must both be integers, there are d(n) ways to get an area of 252−n2 in this configuration,

where d(n) denotes the number of divisors of n.

Since we can pick any of the four vertices to be our corner, there are then 4d(n) triangles of area 252−n2

for 1 ≤ n ≤ 25. So, we compute the answer to be

|P | = 100 + 4(d(1) + . . .+ d(24))

= 4∑k≤24

⌊24

k

⌋= 100 + 4(24 + 12 + 8 + 6 + 4 + 4 + 3 + 3 + 2 · 4 + 1 · 12)

= 436.

28. [16] Caroline starts with the number 1, and every second she flips a fair coin; if it lands heads, sheadds 1 to her number, and if it lands tails she multiplies her number by 2. Compute the expectednumber of seconds it takes for her number to become a multiple of 2021.

Proposed by: Carl Schildkraut, Krit Boonsiriseth

Answer: 4040

Solution: Consider this as a Markov chain on Z/2021Z. This Markov chain is aperiodic (since 0 cango to 0) and any number can be reached from any other number (by adding 1), so it has a uniquestationary distribution π, which is uniform (since the uniform distribution is stationary).

It is a well-known theorem on Markov chains that the expected return time from a state i back to i isequal to the inverse of the probability πi of i in the stationary distribution. (One way to see this is totake a length n→∞ random walk on this chain, and note that i occurs roughly πi of the time.) Sincethe probability of 0 is 1

2021 , the expected return time from 0 to 0 is 2021.

After the first step (from 0), we are at 1 with probability 1/2 and 0 with probability 1/2, so the numberof turns it takes to get from 1 to 0 on expectation is 2 · 2021− 2 = 4040.

29. [18] Compute the number of complex numbers z with |z| = 1 that satisfy

1 + z5 + z10 + z15 + z18 + z21 + z24 + z27 = 0.

Proposed by: Daniel Zhu

Answer: 11

Solution: Let the polynomial be f(z). One can observe that

f(z) =1− z15

1− z5+ z15

1− z15

1− z3=

1− z20

1− z5+ z18

1− z12

1− z3,

so all primitive 15th roots of unity are roots, along with −1 and ±i.To show that there are no more, we can try to find gcd(f(z), f(1/z)). One can show that there exista, b so that zaf(z)− zbf(1/z) can be either of these four polynomials:

(1 + z5 + z10)(1− z32), (1 + z5 + z10 + z15)(1− z30),

(1 + z3 + z6 + z9 + z12)(z32 − 1), (1 + z3 + z6 + z9)(z30 − 1).

Thus any unit circle root of f(z) must divide the four polynomials (1 − z15)(1 − z32)/(1 − z5),(1 − z20)(1 − z30)/(1 − z5), (1 − z15)(1 − z32)/(1 − z3), (1 − z12)(1 − z30)/(1 − z3). This impliesthat z must be a primitive kth root of unity, where k ∈ {1, 2, 4, 15}. The case k = 1 is clearly extrane-ous, so we are done.

30. [18] Let f(n) be the largest prime factor of n2 + 1. Compute the least positive integer n such thatf(f(n)) = n.

Proposed by: Milan Haiman

Answer: 89

Solution: Suppose f(f(n)) = n, and let m = f(n). Note that we have mn | m2 + n2 + 1.

First we find all pairs of positive integers that satisfy this condition, using Vieta root jumping.

Suppose m2 + n2 + 1 = kmn, for some positive integer k. Considering this as a quadratic in m, letthe other root (besides m) be m′. We have m′ + m = kn, so m′ is an integer. Also, mm′ = n2 + 1.So if m > n then m′ ≤ n. So if we have a solution (m,n) we can find a smaller solution (n,m′). Inparticular, it suffices to find all small solutions to describe all solutions. A minimal solution must havem = n, which gives only m = n = 1. We have that k = 3.

Now the recurrence a0 = a1 = 1, an + an+2 = 3an+1 describes all solutions with consecutive terms. Infact this recurrence gives precisely other Fibonacci number: 1, 1, 2, 5, 13, 34, 89, 233, . . .

Checking these terms gives an answer of 89.

31. [18] Roger initially has 20 socks in a drawer, each of which is either white or black. He chooses a sockuniformly at random from the drawer and throws it away. He repeats this action until there are equalnumbers of white and black socks remaining.

Suppose that the probability he stops before all socks are gone is p. If the sum of all distinct possiblevalues of p over all initial combinations of socks is a

b for relatively prime positive integers a and b,compute 100a+ b.

Proposed by: Michael Diao

Answer: 20738

Solution: Let bi and wi be the number of black and white socks left after i socks have been thrownout. In particular, b0 + w0 = 20.

The key observation is that the ratio ri = bibi+wi

is a martingale (the expected value of ri+1 given ri isjust ri).

Suppose WLOG that b0 < w0 (we will deal with the case b0 = w0 later). Say that we stop at i if bi = 0or bi = wi. Then the expected value of ri when we stop is

1

2· p+ 0 · (1− p) =

b0b0 + w0

This rearranges to p = 2b0b0+w0

.

Meanwhile, if b0 = w0 = 10, we can reduce to the case b1 = 9 < 10 = w1. Hence

10∑b0=0

p =

(9∑b0

2b020

)+

18

19=

9

2+

18

19=

207

38.

32. [18] Let acute triangle ABC have circumcenter O, and let M be the midpoint of BC. Let P be theunique point such that ∠BAP = ∠CAM , ∠CAP = ∠BAM , and ∠APO = 90◦. If AO = 53, OM = 28,and AM = 75, compute the perimeter of 4BPC.

Proposed by: Jeffrey Lu

Answer: 192

Solution: The point P has many well-known properties, including the property that ∠BAP = ∠ACPand ∠CAP = ∠BAP . We prove this for completeness.

Invert at A with radius√AB ·AC and reflect about the A-angle bisector. Let P ′ be the image of P .

The angle conditions translate to

• P ′ lies on line AM

• P ′ lies on the line parallel to BC that passes through the reflection of A about BC (since P lieson the circle with diameter AO)

In other words, P ′ is the reflection of A about M . Then BP ′ ‖ AC and CP ′ ‖ AB, so the circumcirclesof 4ABP and 4ACP are tangent to AC and AB, respectively. This gives the desired result. �

Extend BP and CP to meet the circumcircle of 4ABC again at B′ and C ′, respectively. Then∠C ′BA = ∠ACP = ∠BAP , so BC ′ ‖ AP . Similarly, CB′ ‖ AP , so BCB′C ′ is an isosceles trapezoid.In particular, this means B′P = CP , so BP + PC = BB′. Now observe that ∠ABP = ∠CAP =∠BAM , so if AM meets the circumcircle of 4ABC again at A′, then AA′ = BB′. Thus the perimeterof 4BPC is BP + PC +BC = BB′ +BC = AA′ +BC.

Now we compute. We have

BC = 2√AO2 −OM2 = 2

√81 · 25 = 90

and Power of a Point gives

MA′ =BM2

AM=

452

75= 27.

Thus AA′ +BC = 75 + 27 + 90 = 192.

33. [20] After the Guts round ends, HMMT organizers will collect all answers submitted to all 66 questions(including this one) during the individual rounds and the guts round. Estimate N , the smallest positiveinteger that no one will have submitted at any point during the tournament.

An estimate of E will receive max (0, 24− 4|E −N |) points.

Proposed by: Carl Schildkraut

Answer: 139

Solution: The correct answer was 139.

Remark: Until the end of the Guts round, no team had submitted 71 as the answer to any question.One team, however, submitted 71 as their answer to this question, increasing the answer up to 139.

34. [20] Let f(n) be the largest prime factor of n. Estimate

N =

⌊104 ·

∑106

n=2 f(n2 − 1

)∑106

n=2 f(n)

⌋.

An estimate of E will receive max

(0,

⌊20− 20

(|E−N |103

)1/3⌋)points.

Proposed by: Carl Schildkraut

Answer: 18215

Solution: We remark thatf(n2 − 1) = max(f(n− 1), f(n+ 1)).

Let X be a random variable that evaluates to f(n) for a randomly chosen 2 ≤ n ≤ 106; we essentiallywant to estimate

E[max(X1, X2)]

E[X3]

where Xi denotes a variable with distribution identical to X (this is assuming that the largest primefactors of n− 1 and n+ 1 are roughly independent).

A crude estimate can be compiled by approximating that f(n) is roughly 106 whenever n is prime and0 otherwise. Since a number in this interval should be prime with ”probability” 1

ln 106 , we may replaceeach Xi with a Bernoulli random variable that is 1 with probability 1

ln 106 ∼114 and 0 otherwise. This

gives us an estimate of1 · 2·14−1142

114

=27

14.

However, this estimate has one notable flaw: n−1 and n+1 are more likely to share the same primalitythan arbitrarily chosen numbers, since they share the same parity. So, if we restrict our sums to onlyconsidering f(n) for odd numbers, we essentially replace each Xi with a Bernoulli random variablewith expectation 1/7, giving us an estimate of 13

7 , good for 5 points.

This estimate can be substantially improved if we consider other possible factors, which increases thecorrelation between f(n− 1) and f(n+ 1) and thus decreases one’s estimate. The correct value of Nis 18215.

35. [20] Geoff walks on the number line for 40 minutes, starting at the point 0. On the nth minute, heflips a fair coin. If it comes up heads he walks 1

n in the positive direction and if it comes up tails hewalks 1

n in the negative direction. Let p be the probability that he never leaves the interval [−2, 2].Estimate N = b104pc.

An estimate of E will receive max

(0,

⌊20− 20

(|E−N |160

)1/3⌋)points.

Proposed by: Michael Ren

Answer: 8101

Solution: To estimate it by hand, we’ll do casework on the most likely ways that Geoff will go past+2, and double the answer. If Geoff starts with one of the three sequences below, he will be past 2 orvery close to 2:

(+,+,+,+), (+,+,+,−,+,+), (+,+,−,+,+,+).

The probability of one of these happening is 116 + 2

64 = 332 . This gives an estimate of p = 3

16 , whichgives E = 8125 and earns 9 points.

We can justify throwing out other starting sequences as follows. For example, suppose we start with(+,+,−,−). At this point we are at 11

12 . The variance of the rest of our random walk is

40∑n=5

1

n2<π2

6− 1− 1

4− 1

9− 1

16< 0.25.

So, the standard deviation of the rest of our walk is bounded by 0.5, which is much less than the 1312

Geoff needs to go to get to +2. One can use similar estimates for other sequences to justify them asnegligible. Furthermore, we can even use similar estimates to justify that if Geoff get close enough to+2, he is very likely to escape the interval [−2, 2].

The exact value for p is 0.8101502670..., giving N = 8101.

36. [20] A set of 6 distinct lattice points is chosen uniformly at random from the set {1, 2, 3, 4, 5, 6}2. LetA be the expected area of the convex hull of these 6 points. Estimate N = b104Ac.

An estimate of E will receive max

(0,

⌊20− 20

(|E−N |104

)1/3⌋)points.

Proposed by: Milan Haiman

Answer: 104552

Solution: The main tools we will use are linearity of expectation and Pick’s theorem. Note that theresulting polygon is a lattice polygon, and this the expected area A satisfies

A = I +B

2− 1,

where I is the expected number of interior points and B is the expected number of boundary points.We may now use linearity of expectation to write this as

A = −1 +∑

p∈{1,2,...,6}2E[Xp],

where Xp is 1 if the point is inside the polygon, 1/2 if the point is on the boundary, and 0 otherwise.Letting f(p) = E[Xp], we may write this by symmetry as

A = −1 + 4f(1, 1) + 8f(1, 2) + 8f(1, 3) + 4f(2, 2) + 8f(2, 3) + 4f(3, 3).

There are many ways to continue the estimation from here; we outline one approach. Since X(1,1) is1/2 if and only if (1, 1) is one of the selected points (and 0 otherwise), we see

f(1, 1) =1

12.

On the other hand, we may estimate that a central point is exceedingly likely to be within the polygon,and guess f(3, 3) ≈ 1. We may also estimate f(1, y) for y ∈ {2, 3}; such a point is on the boundary ifand only if (1, y) is selected or (1, z) is selected for some z < y and for some z > y. The first eventhappens with probability 1/6, and the second event happens with some smaller probability that canbe estimated by choosing the 6 points independently (without worrying about them being distinct);this works out to give the slight overestimate

f(1, 2), f(1, 3) ≈ 1

8.

From here, it is not so clear how to estimate f(2, 2) and f(2, 3), but one way is to make f(x, y)somewhat linear in each component; this works out to give

f(2, 2) ≈ 1

4, f(2, 3) ≈ 1

2.

(In actuality the estimates we’d get would be slightly higher, but each of our estimates for f(x, y) upuntil this point have been slight overestimates.) Summing these up gives us an estimate of A ≈ 31

3 orE = 103333, which earns 10 points. The actual value of A is 10.4552776..., and so N = 104552.

HMIC 2021April 24, 2021 – May 2, 2021

1. [5] 2021 people are sitting around a circular table. In one move, you may swap the positionsof two people sitting next to each other. Determine the minimum number of moves necessaryto make each person end up 1000 positions to the left of their original position.

2. [7] Let n be a positive integer. Alice writes n real numbers a1, a2, . . . , an in a line (in thatorder). Every move, she picks one number and replaces it with the average of itself and itsneighbors (an is not a neighbor of a1, nor vice versa). A number changes sign if it changesfrom being nonnegative to negative or vice versa. In terms of n, determine the maximumnumber of times that a1 can change sign, across all possible values of a1, a2, . . . , an and allpossible sequences of moves Alice may make.

3. [8] Let A be a set of n ≥ 2 positive integers, and let f(x) =∑

a∈A xa. Prove that there exists

a complex number z with |z| = 1 and |f(z)| =√n− 2.

4. [10] Let A1A2A3A4, B1B2B3B4, and C1C2C3C4 be three regular tetrahedra in 3-dimensionalspace, no two of which are congruent. Suppose that, for each i ∈ {1, 2, 3, 4}, Ci is the midpointof the line segment AiBi. Determine whether the four lines A1B1, A2B2, A3B3, and A4B4

must concur.

5. [12] In an n×n square grid, n squares are marked so that every rectangle composed of exactlyn grid squares contains at least one marked square. Determine all possible values of n.

Time limit: 4 hours

HMIC 2021April 24, 2021 – May 2, 2021

1. [5] 2021 people are sitting around a circular table. In one move, you may swap the positions of twopeople sitting next to each other. Determine the minimum number of moves necessary to make eachperson end up 1000 positions to the left of their original position.

Proposed by: Milan Haiman

Answer: 1021000

Solution 1: We claim that the answer is 1000·1021 = 1021000. To see how we can obtain this, label thepeople around the table s1, s2, ..., s2021. We then start with s1 and swap them 1000 positions to the left,then we take s2 and swap them 1000 positions to the left, and so on until we swap s1021 1000 positionsto the left. We see that after swapping s1 1000 positions to the left, we have s1022, s1023, ..., s2021still in order followed by s2, s3, ..., s1021, s1. Further, we can show that after moving sk to the left1000 times, sk is to the left of s1022 with the order s1022, s1023, ..., s2021, sk+1, sk+2, ..., s1021, s1, s2, ..., skfor 1 ≤ k ≤ 1021. As we never swap any sk for 1 ≤ k ≤ 1021 other than moving it 1000 times tothe left, when the operation is done, s1, s2, ..., s1021 are all shifted 1000 places to the left followed bys1022, s1023, ..., s2021. Therefore, every person is shifted 1000 to the left and this is done in a total of1000 · 1021 moves.

Now, to prove that this is a sufficient lower bound, we note that if a total of M moves are made, thensome student will move one seat left M times and one student will move right M times. In order tomove every student 1000 seats to the left, we can either have them make at least 1000 total leftwardmovements or 1021 total rightward movements. This means that there can be at most b M

1000c students

who get 1000 seats to the left via a net leftward movement and at most b M1021c students who get 1000

seats to the left via a net rightward movement. Thus, we see that⌊M

1000

⌋+

⌊M

1021

⌋≥ 2021 =⇒ M ≥ 1021 · 1000,

as desired.

Solution 2: After getting the same construction as above, we can let the students be labeled si for1 ≤ i ≤ 2021, and say that in order for each student to end up 1000 spaces to the left, they are movedleft ai times and right bi times. Then the condition is trivially equivalent to ai− bi = 1000 mod 2021.Further, as each move will move one person right and one person left, we see that

∑2021i=1 ai and

∑2021i=1 bi

count the same event. Thus,2021∑i=1

ai =

2021∑i=1

bi.

Now, let ai − bi = 2021ci + 1000 for integer ci and all 1 ≤ i ≤ 2021. We see that

0 =

2021∑i=1

ai −2021∑i=1

bi =

2021∑i=1

(ai − bi) =

2021∑i=1

(1000 + 2021ci) = 1000 · 2021 + 2021

2021∑i=1

ci.

Therefore, we find that2021∑i=1

ci = −1000.

We then can use the fact that ai + bi ≥ |ai − bi| = |2021ci + 1000| as both ai and bi are nonnegative.

Further, the total number of moves is equal to both∑2021

i=1 ai and∑2021

i=1 bi so we can then find that

2021∑i=1

ai + bi2

≥2021∑i=1

|ai − bi|2

=

2021∑i=1

|2021ci + 1000|2

.

Now, let∑

ci≥0 ci = S and∑

ci<0 ci = −S − 1000. Then, we find that

2021∑i=1

|2021ci + 1000|2

=∑ci≥0

2021ci + 1000

2+∑ci<0

−2021ci − 1000

2

=2021(2S + 1000)

2+ 500

−2021 + 2∑ci≥0

1

.

Combining all of this, we find that

2021∑i=1

ai + bi2

≥2021∑i=1

|ai − bi|2

≥ 2021(2S + 1000)

2+ 500(−2021 + 2

∑ci≥0

1) = 2021S + 1000 ·∑ci≥0

1.

To finish this, we know that if∑

ci<0 ci = −S − 1000, then there are at most −S − 1000 values ofci < 0, as the ci with smallest absolute value is −1. Thus, we find that

2021S + 1000 ·∑ci≥0

1 ≥ 2021S + 1000(2021− (S + 1000)) = 1021S + 1000 · 1021.

As S ≥ 0, we find that this makes the total number of moves,∑2021

i=1ai+bi

2 at least 1000 · 1021. There-fore, we have our desired lower bound and the result follows.

2. [7] Let n be a positive integer. Alice writes n real numbers a1, a2, . . . , an in a line (in that order).Every move, she picks one number and replaces it with the average of itself and its neighbors (an isnot a neighbor of a1, nor vice versa). A number changes sign if it changes from being nonnegative tonegative or vice versa. In terms of n, determine the maximum number of times that a1 can changesign, across all possible values of a1, a2, . . . , an and all possible sequences of moves Alice may make.

Proposed by: Zhao Yu Ma

Answer: n− 1

Solution: The maximum number is n − 1. We first prove the upper bound. For simplicity, colorall negative numbers red, and all non-negative numbers blue. Let X be the number of color changesamong adjacent elements (i.e. pairs of adjacent elements with different colors). It is clear that thefollowing two statements are true:

(1) When a1 changes sign, X decreases by 1. If a1 changes from negative (red) to non-negative (blue),a2 must have been non-negative (blue), so the first two colors changed from RB to BB. The sameapplies when a1 changes from non-negative to negative.

(2) X cannot increase after a move. Suppose Alice picks ai for her move where 1 ≤ i ≤ n. If ai doesnot change sign, then X clearly remains the same. Else, if ai changes sign and i = 1 or n, then Xdecreases by 1 (from (1)). Finally, if ai changes sign and i 6= 1, n, we have two cases:

Case 1: ai−1, ai+1 are of the same color. If they are both negative (red), then if ai changes color, itmust be from non-negative to negative (blue to red). Thus, the colors change from RBR to RRR andX decreases by 2. The same holds if both ai−1, ai+1 are non-negative.

Case 2: ai−1, ai+1 are of of different colors. No matter what the color of ai is, there is exactly onecolor change among the three numbers, so X will remain the same.

Now, since the initial value of X is at most n− 1, it can decrease by 1 at most n− 1 times. Hence, a1can change signs at most n− 1 times.

Now we prove the lower bound by constructing such a sequence inductively. Specifically, we induct onthe following statement:

For every n ≥ 2, there exists a sequence a1, a2, · · · , an such that by picking

a1, a2, a1, a3, a2, a1, · · · , an−1, an−2, · · · , a2, a1

in that order, a1 changes sign n− 1 times.

When n = 2, we can let a1 change sign once by starting with the sequence (1,−3), and picking a1 toobtain (−1,−3), which satisfies the conditions in our statement.

Suppose we have proven the statement for n−1. For n, let a1, a2, · · · , an−1 be as defined in our construc-tion for n−1 (we shall fix the value of an later). After executing the steps a1, a2, a1, a3, a2, a1, · · · , an−2,an−3, · · · , a2, a1, a1 would have changed sign n− 2 times.

It now remains to pick an−1, an−2, · · · , a2, a1 in order so that a1 changes sign one more time. This isalways possible as long as an is sufficiently large in magnitude and of the opposite sign as a1. Sincethe value of an has remained unchanged since the start (as we have not picked an at all), it suffices tolet an be a number satisfying the above conditions at the start.

This completes our induction, and we conclude that the maximum number of times that a1 can changesign is n− 1.

3. [8] Let A be a set of n ≥ 2 positive integers, and let f(x) =∑

a∈A xa. Prove that there exists a

complex number z with |z| = 1 and |f(z)| =√n− 2.

Proposed by: Milan Haiman

Solution 1: Let A consist of the numbers a1 < a2 < · · · < an. Let d = an− a1. Then, note that if wechoose z uniformly at random from the set X = {x : xd = −1}, then we have

f(z) = (za1 + zan) +

n−1∑i=2

zai =

n−1∑i=2

zai

Then, note that

|f(z)|2 = f(z)f(z̄) =

(n−1∑i=2

zai

)(n−1∑i=2

z̄ai

)= n− 2 +

∑i6=j,2≤i,j≤n−1

zai−aj

However, since 0 < |ai − aj | < d, for all 2 ≤ i, j ≤ n− 1 with i 6= j, we know that the average value ofzai−aj over all z ∈ X is just 0. Thus, the average value of |f(z)|2 is n−2, and thus, there exists z1 andz2 on the unit circle such that |f(z1)|2 ≤ n − 2 ≤ |f(z2)|2, and thus by Intermediate Value Theorem,there exists z on the unit circle such that |f(z)|2 = n− 2, so we are done.

Solution 2: Note that we have

|f(x)|2 = f(x)f(x̄) =

(∑a∈A

xa

)(∑b∈A

x̄b

)= n+

∑a 6=b

xa−b

Let t = 1+maxa6=b,a,b∈A v2(a−b). Let z be chosen uniformly at random from X = {x : x2t

= 1, x2t−1

=−1}, the set of all primitive 2tth roots of unity. Note that for all distinct a, b ∈ A, we know that theexpected value of za−b is 1 if v2(t) ≥ t, −1 if v2(a− b) = t− 1, and 0 if v2(a− b) < t− 1. However, we

know that there exists distinct â, b̂ ∈ A such that v2(â− b̂) = t− 1, so the expected values of zâ−b̂ and

zb̂−â are both −1. Also, we know that the expected value of every other term in this sum∑

a 6=b xa−b

is less than or equal to 0. Thus, the expected value of |f(z)|2 is at most n − 2. Additionally, since|f(1)|2 = n2, and then finish with Intermediate Value Theorem.

4. [10] Let A1A2A3A4, B1B2B3B4, and C1C2C3C4 be three regular tetrahedra in 3-dimensional space,no two of which are congruent. Suppose that, for each i ∈ {1, 2, 3, 4}, Ci is the midpoint of the linesegment AiBi. Determine whether the four lines A1B1, A2B2, A3B3, and A4B4 must concur.

Proposed by: Daniel Zhu

Answer: Yes

Solution 1: Let Pi(t) be lines in space so that Pi(0) = Ai, Pi(1) = Ci, and Pi(2) = Bi. Then observethat PiP

2j are quadratics in t. The difference between any of these two is zero at t = 0, 1, 2, so it must

be identically zero. Therefore, we find that P1P2P3P4 is a regular tetrahedron or a point for all t.

Now consider the signed volume [P1P2P3P4]. It can be written as a determinant of a matrix whoseentries are constant or linear polynomials in t, so it must be V (t) for some polynomial V that is atmost cubic. We claim that it actually is cubic. One way to see this is that since PiP

2j is nonconstant,

it must be Θ(t2) for large t, which implies that |V (t)| = Θ(t3).

However, every cubic polynomial has a real root. A nondegenerate regular tetrahedron cannot have zerovolume, so we conclude that P1, P2, P3, P4 are equal for some t. This yields the desired concurrency.

Solution 2: We will prove that the three tetrahedra are homothetic, which will prove the desiredconcurrency. Now translate so that all the tetrahedra are centered at the origin. This preservesmidpoints and homothetic-ness, so it suffices to solve the problem in this case.

Let M be the matrix that takes Ai to Bi. Since the two tetrahedra are regular, we find that M isconformal and thus must be rT for some T ∈ SO(3) and r ∈ R. Similarly, 1

2 (I +M) must be s′U forsome U ∈ SO(3). Making the substitution s = 2s′, we must solve I + rT = sU .

The remainder of the solution is a straightforward computation by eigenvalues. We know that theeigenvalues of rT must have magnitude |r|. Also, the eigenvalues of I + rT , which are just one plusthe eigenvalues of rT , must have magnitude |s|. Thus, the eigenvalues z of M must satisfy a systemof the form |z| = |r| and |1 + z| = |s|.If the tetrahedra are not homothetic, then M cannot be a multiple of the identity, meaning that Mmust have more than one distinct eigenvalue (everything in SO(3) is normal and thus unitarily di-agonalizable). However, by, say, geometry, that system can only have two solutions only if they arenonreal and conjugates of each other. But every 3× 3 matrix has a real eigenvalue, contradiction.

5. [12] In an n× n square grid, n squares are marked so that every rectangle composed of exactly n gridsquares contains at least one marked square. Determine all possible values of n.

Proposed by: Krit Boonsiriseth

Answer: n = 1, p, or p2 for some prime p.

Solution: In this solution, we will reference cells by row and column, measured from left to right andtop to bottom. The cell at (0, 0) is then in the top-left corner; the cell at (1, 3) is in the fourth cell ofthe second row.

(0, 0)

(1, 3)

We begin by constructing solutions for the possible values of n.

For n = 1, we can mark the sole cell.

For n = p, where p is a prime, we need only ensure that each 1 × n and n × 1 rectangle contains amark. This is the same as ensuring that all rows and columns contain a mark, which can be done, forexample, by marking all cells (k, k) for integers k ∈ [0, n).

For n = p2, there must be a mark in each row and column and also in each p× p rectangle. To achievethis, we can mark all cells of the form (j + kp, k + jp), for integers j, k ∈ [0, p). An illustration forn = 9 follows:

Note that there are exactly p2 = n such cells, and that row r contains a cell in the column⌊r

p

⌋+ (r mod p) · p

Likewise, there is a marked cell in each column. Now, we will show that all p× p rectangles contain amarked cell. Consider the rectangle R with upper-left corner at (xp+ y, z) for integers x, y ∈ [0, p), z ∈[0, n). Then, if z > x+ (y − 1)p, R contains the cell(

xp+

⌈z − xp

⌉, x+

⌈z − xp

⌉p

)and otherwise it contains the cell(

(x+ 1)p+

⌊z − xp

⌋, x+

⌊z − xp

⌋p+ 1

).

Since z ≤ n− p, this cell is always a valid marked cell. This concludes the proof that the constructionis correct.

Now, we prove that no other values of n are satisfiable. First, we again note that every row andcolumn must have exactly one marked cell, as there are n of each and they are disjoint. Let n = abfor 1 < a, b < n and partition the n× n grid into a b× a grid of a× b rectangles. In particular, let Rij

be the rectangle that has upper-left corner at (bi, aj) and lower-right corner at (bi+ b− 1, aj + a− 1)for i ∈ [0, a), j ∈ [0, b). Then each Rij must contain exactly one marked cell, since there are also n ofthem and they are disjoint.

Consider the set of rectangles Ri0. There are exactly a marked cells among these rectangles — one ineach of the columns from 0 to a− 1, and also one in each of the a rectangles Ri0. Therefore, there issome permutation X of the integers [0, a) so that Ri0 has a marked cell in column Xi. Now, we willprove the claim that if a rectangle Rij contains a marked cell in column c of the grid, c ≡ Xi (mod a).We prove this by induction on j, having already demonstrated the case j = 0. Suppose the claim holdsfor some j; we will show it must also hold for j′ = j + 1. Consider k satisfying Xk = 0. We knowthat the unique marked cell in rectangle Rkj is in column aj. Therefore, Rkj′ must contain a mark incolumn aj + a, as otherwise the a× b rectangle with top-left corner at (bk, aj + 1) has no mark. Next,consider k′ satisfying Xk = 1. The unique marked cell in rectangle Rk′j is in column aj + 1. Also, therectangle Rk′j′ cannot contain a marked cell in column aj + a, since there is already a mark in thatcolumn. So Rk′j′ must contain a marked cell in column aj + a + 1, as otherwise the a × b rectanglewith upper-left corner at (bk, aj + 2) has no mark. Proceeding in order of increasing Xi in this wayforces the claim to hold for k′ as well, completing the inductive proof.

Likewise, there is a permutation Y of the integers [0, b) so that, if Rij contains a marked cell in row rof the grid, r ≡ Yj (mod b).

We will now show that either X and Y are both increasing or both decreasing. Suppose not, so thatfor some i, j, Xi > Xi+1 and Yj < Yj+1 (or the same with X,Y swapped, which is handled identically).Then, consider the rectangles Rij , R(i+1)j , Ri(j+1), and R(i+1)(j+1). An example for n = 6 follows:

Taking an a×b rectangle with upper-left corner directly below the marked cell in Rij , we find that it liesentirely inside these four rectangles and yet does not contain any of their marked cells. In particular,we can note that, since Y is increasing at this point, the horizontal distance between the marks inthe lower left and upper right is strictly greater than a. Also, since X is decreasing at this point, thevertical distance between the marks in the top left and lower right is strictly greater than b. This mensthat an a× b rectangle with no mark exists, a contradiction. Therefore, either X is increasing or Y isdecreasing.

Suppose, without loss of generality, that both are increasing (noting that the increasing and decreasingcases are simply 90◦ rotations of each other). Then the marked cell in row 1 must be a + 1. But ifn is divisible by two distinct primes or the cube of a prime, there is no unique value for a, which is acontradiction. Therefore, n must be 1, a prime, or the square of a prime.