Miscellaneous Problems and Essays Ross Honsberger

AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS VOL 19

In Pólya’s Footsteps

Miscellaneous Problems and Essays

Ross Honsberger

In P6lya's Footsteps


The inspiration for mfulY of these problems cLune from the OlympiadComer of Crux MathenulticorlUl1, now Crux Mathel1UlticorU111 }vithMathenultical MaYhe111, published by the Canadian Mathematical Societywith support from the University of Calgary, Memorial University ofNewfoundland, and the lJniversity of Ottawa. Full attribution ell1 befound with each problem.

@1997 byThe Mathematical Association ofAmerica (lncorporatedj

Library (?f Congress Catalog Card Number 97-70506

Complete Set ISBN 0-88385-300-0Vol. 19 ISBN 0-88385-326-4

Printed in the United States ofAmerica

Current printing (last digit):10 9 8 7 6 5 4 3 2 1

The Do/ciani Mathenlatical EX/Jositions

NUMBER NINETEEN



Ross HonsbergerUniversity) o.f Waterloo

Puhlished and Distrihuted hyTHE yLt\THEMATICAL ASSOCL\TIO~ OF AMERICA

10.1090/dol/019

THEDOLCIANI MATHEMATICAL EXPOSITIONS

Published byTHE MATHEMATICAL ASSOCIATION OF AMERICA

Committee on PublicationsJAMES W. DANIEL, Chair

Dolciani Mathematical Expositions Editorial BoardBRUCE P. PALKA, EditorCHRISTINE W. AYOUB

EDWARD J. BARBEAU

IRL C. BIVENS

The DOLCJANI MATHEMATICAL EXPOSITIONS series of the MathematicalAssociation of Alnerica was established through a generous gift to the Association fromMary P. Dolciani, Professor of Mathematics at Hunter College of the City Universityof New York. In making the gift, Professor Dolciani, herself an exceptionally talentedand successful expositor of mathematics, had the purpose of furthering the ideal ofexcellence in mathematical exposition.

The Association, for its part, was delighted to accept the gracious gesture initiating the revolving fund for this series from one who has served the Association withdistinction, both as a member of the Comlnittee on Publications and as a member ofthe Board of Governors. It was with genuine pleasure that the Board chose to namethe series in her honor.

The books in the series are selected for their lucid expository style and stimulatingmathematical content. Typically, they contain an alnple supply of exercises, manywith accompanying solutions. They are intended to be sufficiently elementary for theundergraduate and even the mathematically inclined high-school student to understandand enjoy, but also to be interesting and sometimes challenging to the more advancedmathematician.

1. Mathematical Gelns, Ross Honsberger2. Mathematical Gelns 11, Ross Honsberger3. Mathematical Morsels, Ross Honsberger4. Mathematical Plums, Ross Honsberger (ed.)5. Great Moments in Mathematics (Before 1650), Howard Eves6. Maxima and Minima without Calculus, Ivan Niven7. Great Moments in Mathematics (A.fter 1650), Howard Eves8. Map Coloring, Polyhedra, and the Four-Color Prohlem, David Barnette9. Mathematical Gems Ill, Ross Honsberger

10. More Mathematical Morsels, Ross Honsberger11. Old and New Unsolved Problems in Plane Geolnetry and Numher Theory, Victor

Klee and Stan Wagon12. Problems .lor Mathematicians, Young and Old, Paul R. Halmos13. Excursions in Calculus: An Intefplay oj'the Continuous and the Discrete, Robert

M. Young14. The Wohascum County Prohlenl Book, George T. Gilbert, Mark Krusemeyer, and

Loren C. Larson15. Lion Hunting and Other Mathenlatical Pursuits: A Collection 0.( Mathematics,

Verse, and Stories by Ralph P Boas, if:, edited by Gerald L. Alexanderson andDale H. Mugler

16. Linear Algebra Prohlem Book, Paul R. Halmos17. From ErdtJs to Kiev: Proble/ns 0.( Olympiad Caliber, Ross Honsberger18. Which Way Did the Bicycle Go? ... and Other Intriguing Mathenlatical A{vster

ies, Joseph D. E. Konhauser, Dan Velleman, and Stan Wagon19. In P6~ya 's Footsteps: Miscellaneous Problems and Essays, Ross Honsberger

Preface

Just as a recording of a Mozart concerto makes no pretense of teaching oneto compose music~ these mathematical perfonnances are not motivated bya desire to teach mathematics; they are offered solely .for your enjoyment.

There is no denying that a certain degree of concentration is required forthe appreciation of their beautiful ideas~ but it is hoped that a leisurely paceand generous explanations \vill make them a pleasure to read. The technicaldemands are very modest; a high school graduate should be well equippedto handle many of the topics and a university undergraduate in mathematicsought to be perfectly comfortable throughout.

I hope you will find something exciting in each of these topics-a surprising result~ an intriguing approach~ a stroke of ingenuity-and that youwill approach them as entertainment. You are certainly not required to attempt these problems before going through the solutions~ but if you are ableto give them a little thought first~ I'm sure you will find them all the moreexciting.

It is a pleasure to acknowledge the great debt this volume owes the undergraduate problems journal Crux Mathematicarum, now Crux Mathematicarunl vvith Mathematical Mayhem, published by the Canadian MathematicalSociety with support from the University of Calgary~ Memorial Universityof Newfoundland~ and the University of Ottawa. For interesting elementaryproblems~ this publication is in a class by itself. I came across the great majority of the problems discussed in the present volume in the Olympiad Comercolumns of Crux Mathenlaticorum. Sometimes a solution given in the presentcollection has been based on a solution published in this column~ but unlessotherwise acknowledged~ the solutions here are based on my own work (withone exception) and in all cases the responsibility for shortcomings in presentation is mine alone. The exception is the essay The Infinite Checkerboard~

vii

viii In P61ya's Footsteps

which was written probably twenty years ago, and with my deepest apologies~

I cannot recall where I encountered the topic. The sections may be read inany order. No attempt has been made to bring together all the problems on aspecific subject; in fact, this has been avoided in the hope of encouraging afeeling of spontaneity throughout the work.

I would like to extend my wannest thanks to chainnan Bruce Palka of theDolciani Subcommittee and to the members Christine Ayoub and Irl Bivensfor their generous reception of this manuscript and their perceptive revie\vs.Finally, it has again been my great good fortune to have had Beverly Ruediand Elaine Pedreira see this work through publication; their unfailing genialityand excellence in all phases of this trying process are deeply appreciated.

CONTENTS

Preface viiFour Engaging Problems 1A Problem from the 1991 Asian Pacific Olympiad 6Four Problems from the First Round of the 1988 Spanish Olympiad 9Problem K797 from Kvant 14An Unused Problem from the 1990 International Olympiad 16A Problem from the 1990 Nordic Olympiad 19Three Problems from the 1991 AIME 21An Elementary Inequality 26Six Geometry Problems 29Two Problems from the 1989 Swedish Olympiad 36Two Problems from the 1989 Austrian-Polish Mathematics Competition .42Two Problems from the 1990 Australian Olympiad 45Problem 1367 from Crux Mathematicorum 48Three Problems from Japan 51Two Problems from the 1990 Canadian Olympiad 59A Problem from the 1989 U. S. A. Olympiad 65A Problem on Seating Rearrangements 67Three Problems from the 1980 and 1981 Chinese New Year's Contests .73A Problem in Arithmetic 78A Checkerboard Problem 81Two Problems from the 1990 Asian Pacific Olympiad 85Four Problems from the 1989 AIME 89Five Unused Problems from the 1989 International Olympiad 98Four Geometry Problems 111Five Problems from the 1980 All-Union Russian Olympiad 121The Fundamental Theorem of 3-Bar Motion 129

ix

x In Polya's Footsteps

Three Problems from the 1989 Austrian Olympiad 132Three Problems from the Tournament of the Towns Competitions 138Problem 1506 from Crux Mathematicorum 143Three Unused Problems from the 1987 International Olympiad 145Two Problems from the 1981 Leningrad High School Olympiad 151Four Problems from the Pi Mu Epsilon Journal-Fall 1992 156An Elegant Solution to Morsel 26 165Two Euclidean Problems from The Netherlands 168Two Problems from the 1989 Singapore Mathematical

Society Interschool Competitions 174Problem M1046 from Kvant (1987) 177Two Theorems on Convex Figures 179The Infinite Checkerboard 184Two Problems from the 1986 Swedish Mathematical Competition 188A Brilliant 1-1 Correspondence 192The Steiner-Lehmus Problem Revisited 194Two Problems from the 1987 Bulgarian Olympiad 197A Problem from the 1987 Hungarian National Olympiad 202A Problem from the 1987 Canadian Olympiad 206Problem 1123 from Crux Mathematicorum 208A Problem from the 1987 AIME 211A Generalization of Old Morsel 3 213Two Problems from the 1991 Canadian Olympiad 216An Old Chestnut 221A Combinatorial Discontinuity 223A Surprising Theorem of Kummer 229A Combinatorial Problem in Solid Geometry 234Two Problems from the 1989 Indian Olympiad 239A Gem from Combinatorics 243Two Problems from the 1989 Asian Pacific Olympiad 247A Selection of Joseph Liouville's Amazing Identities Concerning

the Arithmetic Functions a(n), T(n), ¢(n), Jl(n), A(n) 251A Problem from the 1988 Austrian-Polish Mathematics Competition 267An Excursion into the Complex Plane 271Two Problems from the 1990 International Olympiad 280Exercises 289Solutions to the Exercises 293

Praise God from Whomall blessings flow.

Exercises

These exercises are generally much easier than most of the problems we havebeen considering and they are included just for your amusement. They arequestions, some slightly revised, that occurred on various Grade 13 ProblemsPapers of Ontario, a defunct examination which used to be taken by candidateswriting for university scholarships involving mathematics.

1. (1960). If the lengths of the sides of a triangle are in the ratios 3 : 7 : 8,show that its angles are in arithmetic progression.

2. (1947). A candidate at an examination writes four papers. If the maximumnumber of marks obtainable on each paper is m, show that the number ofways of obtaining a total of 2m marks is !(m + 1) (2m2 + 4m + 3).

3. (1945). Find the coefficient of x 6 yzt5 in the expansion of

4. (1950). Prove that on the axis of any parabola there is a certain point Kwhich has the property that, if a chord PQ of the parabola is drawn throughit, then

1 1PK2 + QK2

is constant for all positions of the chord.

5. (1949). Find the roots of

17x4 + 36x3- 14x2

- 4x + 1 == 0,

given that they are in harmonic progression.

289

290 In P61ya's Footsteps

6. (1936). A and B are arbitrary points inside a given circle K. Constructa circle through A and B to touch K.

7. (1938). Reduce

to its simplest surd fonn.

8. (1937). A quadratic expression in x is posItIve except for the range- 2 :S x :S 1, and a second quadratic expression is negative except for therange -1 :S x :S 4. Each expression has the value 60 when :r == 3. For whatother value of x are these expressions equal?

9. (1940). Find the equation of the locus of a point P which moves so thatthe tangents from P to the circle x 2 + y2 == a2 cut off a segment of length2a on the line x == a.

10. (1937). Show that for any positive integer p and any integer s > 1 thereare p consecutive odd numbers whose sum is pS, and find the first of thesenumbers.

11. (1945). Find a polynomial f(x) of degree five such that f(x) - 1 IS

divisible by (x - 1)3 and f (x) is itself divisible by x3 .

12. (1958). A unifonn cylindrical tub can be filled with water from the coldwater tap in 14 minutes, and drained completely through a hole in the bottomin 21 minutes. When both hot and cold water taps are turned on and the plugremoved from the hole, the tub fills in 12.6 minutes. How long would it takethe hot water tap to fill the tub by itself?

13. (1934). A, B, C, and D are arbitrary fixed points in 3-space and P isa variable point. Show that P A 2 + P B 2 + PC2 + P D 2 is a minimum for Pat the midpoint of the segment which joins the midpoints of .<4C and BD.

14. (1951). Show that the equation

1 1 1 1--+--==-+-x+2 y+2 2 z+2

is not satisfied by any set of positive integers x, y, Z, in which :r ~ 4. Hencefind all solutions (x. y ~ z) in positive integers.

Exercises 291

15. (a) (1956). If an ellipse JI rolls without slipping on an identical fixedellipse E, starting with them touching at a pair of vertices, prove that thelocus of a focus of l\I is a circle with center at a focus of E.

(b) (1956). If a parabola 1.\1 rolls without slipping on an identical fixedparabola P, starting with them touching at their vertices, prove that the locusof the focus of ~~1 is the directrix of P. (More interestingly, the locus of thevertex is a cissoid of Diocles, a curve famous for providing a solution to theancient problem of doubling the cube.)

16. (1934). Prove that

• • 0 • 0 • 0 sin2 50 0

SIn 10 + SIn 3 + SIn 5 + ... + SIn 99 == .sin 10

17. (1942). Of the 9! numbers formed by pennuting the digits 1, 2, 3, 4, 5,6, 7, 8, 9, how many are between 678000000 and 859000000?

18. (a) (1959). Show that the segment of a tangent to a hyperbola whichis intercepted between the asymptotes is bisected by the point of contact andthat, with the asymptotes, such a segment fonns a triangle of constant area.

(b) (1954). A straight line cuts a hyperbola at the points Q and Randits asymptotes at P and S. Prove the midpoint of QR is also the midpoint ofPS.

19. (1961). Given that one side of a triangle is twice as long as another side,show that the angle opposite the longer of these sides is greater than twicethe angle opposite the lesser side.

20. (1935). If II, 12 , 13 are the centers of the escribed circles of 6ABC,prove that the area of

abc6111213 == -.

2r

21. (1943). A triangle ABC is such that 3AB == 2AC. Also, a point D onBC is such that BD == 2DC and AD == BC. Show that

tan~LADB= f5.2 V19

22. (1950). Let x be a number between -1 and 1. Noting that n 2

n(n + 1) - n, find a fonnula for the sum of the infinite series

292 In P6lya's Footsteps

23. (1942). If m == esc () - sin () and n == sec () - cos (), show that

m 2/ 3 + n 2/3 == (mn)-2/3.

24. (1961). If 8 n denotes the sum of the first n positive integers, find thesum of the infinite series

81 82 83 8 ns== -+-+-+ ... +--+ ....1 2 4 2n - 1

25. (1941). Find the smallest positive value of

1 1 1 1 1 14tan- - - tan- - + tan- -5 70 99

Solutions to the Exercises

1. (1960). The key to this problem is the recognition that the angles of a

triangle are in arithmetic progression if and only if one of them is 60°:if an angle is 60°, the amount a second angle is less than 60° isprecisely the amount the third angle must exceed 60° in order tomake the sum 180° (in general, three numbers are in arithmeticprogression if and only if one of them is equal to their average).

In a scalene triangle, then, the 60° angle must be the mid-sized angle andtherefore be opposite the side of intermediate length. Thus if the sides are oflengths 3k, 7k and 8k, we need only show that the angle x opposite the sideof length 7k is 60°. But this is immediate from the law of cosines:

49k2 == 9k2 + 64k2- 2 . 3k . 8k cos x~

73k2- 49k2 24 1

cos x == 48k2 48 2

2. (1947). The exponents in the generating function

f (x) == 1 + x + x 2 + x 3 + ... + x'm

are the possible scores on a paper. We may consider scores of a, b, C, and d onthe four papers to be registered in the product f (;r)4 by the term x a+b+c+d

:

f(X)4 == (1 + x + x2 + + x m)(l + x + x 2 + + xTn)x

(1 + x + x 2 + + x r7l )(l +:T + x 2 + + x 17l)

== ... + (xa)(xb)(xc)(.r d) + ...

293


From this viewpoint, the number of ways of accumulating a total score of 2mis the number of ways of generating the tenn x 2m in this product, makin~

the required number simply the coefficient of x 2m in f(x)4. Letting [xn]g(x ~

denote the coefficient of xn in the function g(x), we want to detennine thtnumber [x 2m ]f(x)4.

Now

f(X)4 == (1 + x + x 2 + x 3 + ... + Xm )4

== (1_Xrn+1)4

I-x

== (1 - x m +1 )4(1 - x)-4,

and so the required number is

== [x2m ](1- Xm +1)4(1_ x)-4,

== [x 2m ](1 - 4xm +1 + 6x2m+2 - ... )(1 - x)-4

== [x2m ](1 - 4xm +1)(1 - x)-4

(higher powers of x m +1 do not contribute to the coefficient of x 2m )

== [x2m ](1 _ 4xm +1 ) L: (4 +: -l)x i

i~O

= [x2m](1 - 4xm +1)~C~ 3) Xi

= Cm3+ 3)_4(m;2)== (2m + 3)(2m + 2)(2m + 1) _ 4. (m + 2)(m + l)m

6 62(m+1) 2== (4m2 + 8m + 3 - 2m - 4m)

6m+1

== -3-(2m2 + 4m + 3),

as required.

3. (1945). Since the term x 6 y zt5 contains only a single y, in multiplyingout the product (x + y + z+ t) lO(X + z) (x 2 + y2), from the factor (x 2 + y2) itmust be the x 2 that is taken toward the formation of a term in x 6 yzt5 . Thus

[X6yzt5](X + Y + z + t)10(X + z)(x2 + y2)

==[X4yzt5](x + Y + z + t)10(X + z)


==[x4yzf5J(x + Y + z + t)lO. 1-~ + [X4yzt5](X + Y + z + t)lO. Z

==[X3yzt5](X + Y + z + t)10 + [x4yf5](~r + y + z + t)lO

= C30) G) G) G) + C40) G) G)10 ·9·8 10 . 9 ·8· 7

3.2 ·7·6·1+ 4.3.2 ·6·1

==5040 + 1260

==6300.

295

4. (1950). Let the parabola be y2 == x, let P( a2 ~ a) be any point on it andK(k,O) be a point on its axis (figure 189). Then

1 1PK2 (a 2 -k)2+ a2'

The slopem of PK ism == a/(a2- k), implying a2 - k == aim, and

therefore

1

PK2

y

1

2Y =x

--0+-------+-----.;.----- X

FIGURE 189


Let P K extend into chord PQ and that Q is (b 2. - b). Then

.. bslope QK == 'm == k _ b2 .

implying k - b2 == ~, giving

Then

1

QK2

1 1,)

m-

Now

a+b 1slope PQ == m == 2 b2 == --b·a - a-

implying

a2 + b2

N== (a-b)2

a2b

2[1 + (a ~ b)2 ]

Also m == al(a2- k) implies a2

- k == aim, giving

2 a 2k == a - - == a - a(a - b) == abom

Thus we want to show that there is a value of ab which makes j\;'" a constant.One way this could happen is if

a2 + b2

----== 1(a - b)2 + 1 '

making

1 1N == a2 b2 == k 2 ~

a constant for all chords through K. In this case, we would have

giving ab == ~ and N == -&- == 4. Thus N is constant for all chords throughthe point K (~ , 0), and, after observing that the focus of our parabola is (±. 0),the required point for the parabola y2 == 4p:r: is K(2p, 0).

Solutions to the Exercises 297

5. (1949). Since the roots are in hannonic progression, by definition theirreciprocals are in arithmetic progression. Thus, let y == ~ and let the fourvalues of y be a - 3d, a - d, a + d, and a. + 3d. Then x == 1. and these valuesy

of yare the roots of the equation

17 36 14 4- + - - - - - + 1 == O.y4 y3 y2 Y

that is,

y4 _ 4y3 - 14y2 + 36y + 17 == O.

Thus

the sum of the roots == 4a == 4, and a == 1,

and the roots are 1 - 3d, 1 - d, 1+d, and 1+3d. Also the sum of the productsof the roots taken two at a time is

(1 - 4d + 3d2) + (1 - 2d - 3d2

) + (1 - 9d2) + (1 - d2

)

+(1 + 2d - 3d2) + (1 + 4d + 3d2

) == -14,

giving

-10d2 + 6 == -14,

d2 == 2,

d == ±J2.

Both these values of d give the same set of roots,

y == 1 ± 3J2 and 1 ± J2.

Hence the required roots x are the reciprocals

1 1. _-_1_±_3_J2_2 and _ 1 ± J2.1 ± 3J2' 1 ± J2' I.e., 17

6. (1936). Let the desired circle touch K at T and let the common tangentat T meet the chord XY, through A and B, at the exterior point P (figure190). Then

PT2 == P X . PY == P A . P B,

implying

PX PBPA PY


T

BA

K

I----~----~~r__-~p

FIGURE 190

Subtracting 1 from each side yields

PX-PA

PA

PB-PY

PY

that is,

XA BYAP yp'

and we have

XA APYB pya

That is to say, P can be found on XY extended as the point which dividesAY externally in the known ratio XA: YB.

Thus, after first locating P, draw the tangent to K from P to get T andthen draw the circle through A, B, and T.

7. (1938). Observe that 6- 2J5 == (J5 -1)2 and 9 - 4J5 == (J5 - 2)2.We have

V3-J5 J2-V7-3J5J2 + V7 - 3J5 J2 - V7 - 3J5V6 - 2J5 - V36 - 16J5

3J5 - 5

J5 - 1- 2(J5 - 2)3J5 -5


3- V5:3)b -.5

3 - V5 3)b + .5

3V5 - ·5 . 3)b + .5

-:l)b vb-- --

20 5

299

8. (1937). For some value a, the first expression must be a(x - l)(x + 2),and for some value b, the second expression must be b(x - 4)(x + 1). Since,for x == 3, we get

a(2)(5) == 60 == b(-1)(4)~

it follows that a == 6 and b == -1.5. Hence the expressions are equal when

6(x - l)(x + 2) == -15(x - 4)(x + 1),

2(x - l)(:r + 2) == -5(x - 4)(x~ + 1),

7x 2- 13x - 24 == 0

(x - 3)(7x + 8) == O.

giving x == 3 and -~.

9. (1940). Let Q(a. k) and R(a. k + 2a) be any two points on x == a whichare a distance 2a apart and let the tangents from Q and R to the given circlemeet at P(x,y) (figure 191). We want the locus of P.

A non-vertical straight line through Q has equation

y - k == m(x - a)~ i.e., mx - y + k - am == 0, 'm finite.

To be a tangent to the given circle, we must have

k-am--;=::::;:===== == ±a.J'm2 + 1

k - a:rn == ±ay/m2 + 1.

k 2- 2am.k + a2 m 2 == a2 m 2 + a2

,

k 2 - a2

--- == m..2ak

implying the equation of the tangent is

k2 -a2 k2 -02

---.T - Y + k - == 02ak' 2k

300

FIGURE 191

In P61ya's Footsteps

x=ay

R (a, k + 2a)

2a

Q (a, k)

---+-----I-----+----xo

2 2 2X + Y = a

(k2 - a2)x - 2aky + 2ak2 - ak2 + a3 == 0

(k 2- a2)x - 2aky + ak2 + a3 == o.

Similarly, the tangent through R has equation

(1)

[(k + 2a)2 - a2]x - 2a(k + 2a)y + a(k + 2a)2 + a3 == O. (2)

Thus the equation of the locus in question is obtained by eliminating k fromequations (1) and (2).

Subtracting (1) from (2) gives

which, divided by 4a, yields

(k + a)x - ay + ak + a2 == 0,

giving

a(y - x - a)k(x+a)==a(y-x-a) and k== .

x+a


Substituting in (1), we get

[a2(y-x-a)2 _ 2] _ . a(y-x-a) a2 (y-x-a)2 3_

( )2 a x 2ay + a ( )2 + a - 0,x+a x+a x+a

which, after considerable but straightforward simplification, reduces to

Thus the locus consists of all points on the parabola having vertex at (a, 0)

and the x-axis for its axis except for the points given by x == ±a, whichrequire special interpretation to be admitted and are better rejected altogether.

10. (1937). (a) If p is odd, say 2n + 1, then we wish to solve the followingequation for a:

(a- 2n) +... + (a -4) + (a- 2) +a+ (a+ 2) + (a+4) +... + (a+2n) == pS

that is, (2n + l)a == pS, and since (2n + 1) == p, we may take a == ps-l,

which is odd, and begin at the number

a - 2n == ps-l - (p - 1) == ps-1 - P + 1.

(b) If p is even, say 2n, we similarly wish to solve

(a - 2n) + ... + (a - 4) + (a - 2) + a + (a + 2)+(a + 4)

+ ... + (a + 2n - 2) == pS

2na - 2n == pS

2n(a - 1) == p(a - 1) == pS

a==ps-l+1,

which is odd, and again having the series stat1 at the same number pS -1 - p+ 1:

a - 2n == a - p == ps-l - P + 1.

11. (1945). Let

f(x) == (x - 1)3(ax2 + bx + c) + 1

== (x3 - 3x2 + 3x - 1)(ax2 + bx + c) + 1.


Since f (x) is divisible by x3 , the terms in x2 and :r, as well as the absoluteterm, must vanish. Thus

the absolute term == -c + 1 == 0, giving c == 1.

the coefficient of x == 3c - b == 3 - b == 0, giving b == 3.

the coefficient of x 2 == -3 + 3b - a == 0, giving a == 6.

and making

f(x) == (x 3- 3x2 + 3x - 1)(6x2 + 3x + 1) + 1

== 6x5- 15x4 + 10x3

.

12. (1958). Let V denote the volume of the tub and let the hot water tap byitself take x minutes to fill the tub. Thus the hot water tap supplies water atthe rate of ¥- units per minute, the cold water tap supplies water at the rateof ~ units per minute, while the hole in the bottom drains water at the rateof ~ units per minute. Thus

V V V \tl-; + 14 - 21 == 21.6'

1 1 1 1 1 7;; == 21.6 + 21 - 14 == 126 (10 + 6 - 9) == 126'

giving

126 .x == - == 18 mInutes.

7

13. (1934). Let the midpoints of AC and BD be X and Y and the midpointof XY be Z (figure 192). Our solution is based on the theorem that the sumof the squares of two sides of a triangle is equal to twice the square of the

c

x

z

B "-------~-------...;;;:- DY

A

FIGURE 192


x

FIGURE 193

p

y

303

median to the third side increased by one-half the square of the third side.When the point P is at Z, we have both

and therefore

ZA2 + ZB2 + ZC2 + ZD2

1== 2ZX2 + 2Zy2 + -(AC2 + BD2)

2

=4ZX2 + ~(AC2 + BD2) (recall Z bisects XY)2

= Xy2 + ~(AC2 + BD2).2

For P in general position, we have similarly that

PA2 + PB2+ PC2 + PD2 == (PA2 + PC2) + (PB2 + PD2)

== 2PX2 + ~AC2 + 2py2 + ~BD22 2

= 2(PX2 + py2) + ~(AC2 + BD2).2

In order to show a minimum occurs when P is at Z, then, we need to showthat

But, in 6.PXY (figure 193), PZ is a median and we have

PX2 + py2 == 2PZ2 + ~Xy2 > ~XY222'

from which (1) follows immediately, completing the solution.

(1)


14. (1951). If X 2: 4, then

1 1 1 1 1 1- + -- == -- + -- < - + --.2 z+2 x+2 y+2-6 y+2

giving

1 1 1 1 1---==-<-----26 3-y+2 z+2·

But y 2: 1, implying y + 2 2: 3, and Y~2 ~ ~. Hence

1 1 1-- - -- <-~

y+2 z+2 3'

a contradiction, and it follows that x < 4.

Now, if both x and yare greater than 1, then each of x~2 and Y~2 is

less than or equal to ~, making it impossible for x~2 + Y~2 to exceed ~,

which it must do in order to be equal to ~ + Z~2. Hence either one or both ofx and y must equal 1. Also, the equation is symmetrical in x and y, implyinga limit of 3 on the value of y as well as on x, and that the values of x and ycan be interchanged for the same value of z.

For x == 1 and y == 1, we get z == 4, and the solution (x, y, z) == (1~ 1,4);also

x == 1, Y == 2, gives (1,2,10);

x == 1, Y == 3, gives (1,3,28);

and interchanging x and y we also have the additional two solutions

(2, 1, 4) and (3, 1, 28) .

15. (1956). (a) In all positions, ]vI is the reflection of E in the commontangent at their point of contact P. Therefore the focal radii to P in AI arerespectively the same lengths as those to P in E. However, in figure 194,the reflector property of an ellipse gives Lx == Ly, and the reflection of Ein the tangent gives Lx == Lz. Thus y == z and APA' is straight. Hence atevery position AA' == rl + r2 == 2a, the length of the major axis, implyingthe locus of A' is the circle A (2a ). Similarly B' traces the circle B (2a ).

(b) This is similar to part (a). The solution is based on the followingproperty:

the perpendicular from the focus of a parabola to a tangent meetsthe tangent at the point where it crosses the tangent at the vertex.

The details are left as an additional exercise.


FIGURE 194

B M

start

305

16. (1934). Clearing of fractions, we need to show that

sin 1° sin 1° + sin 1° sin 3C + sin 1° sin 5° + ... + sin 1° sin 99° == sin2 50° .

But this is immediate from sin A sin B == - ~ [cos(A + B) - cos(A - B)J:

1the left side == - - [(cos 2° - cos 0°) + (cos 40

- cos 2°)2

+ (cos 6° - cos4°) + ... + (cos 100° - cos 98°)J

= - ~ (cos 1000 - cos 00 )2

= -~ [(1 - 2sin2 500) - 1]

2

== sin2 50°.

17. (1942). The initial digits of the integers in question can be classifiednicely into 16 types in each of which the remaining digits may be appendedin any order:

678 ,679 ,68 ,69 ,7 ,

81 ,82 ,83 ,84 ,

giving 6! ways of putting in the remaining 6 digits,giving 6! ways of putting in the remaining 6 digits,giving 7! ways of putting in the remaining 7 digits,giving 7! ways of putting in the remaining 7 digits,giving 8! ways of putting in the remaining 8 digits,giving 7! ways of putting in the remaining 7 digits,giving 7! ways of putting in the remaining 7 digits,giving 7! ways of putting in the remaining 7 digits,giving 7! ways of putting in the remaining 7 digits,

306 . In P6lya's Footsteps

851 , giving 6! ways of putting in the remaining 6 digits,852 , giving 6! ways of putting in the remaining 6 digits,853 , giving 6! ways of putting in the remaining 6 digits,854 , giving 6! ways of putting in the remaining 6 digits,(no integer can begin 855 ... since there is only one 5)856 , giving 6! ways of putting in the remaining 6 digits,857 , giving 6! ways of putting in the remaining 6 digits.

Hence the required total is

8 . 6! + 6 . 7! + 8! == 6!(8 + 42 + 56)

== 720 . 106 == 76320.

18. (a) (1959). If P(s, t) is a point on the hyperbola b2 x 2 - a2 y2 == a2 b2,

the equation of the tangent QPR is

Solving with the equation of the asymptotes, b2 x 2- a2 y2 == 0, we have

y

Q

--------tf-----w----+-+------ x

~ bx±ay=O /

FIGURE 195


therefore

b2 (a2b2 + a2t y )2 _ 2,2 = 0

b2 s a y .

(a

2b

2 + a2ty ) _

b b2s - ±ay~

ab2 + aty = ±bsy~

ab2 = y(±bs - at),

giving

ab2 ab2

y - or- bs - at -bs - at'

Hence the ordinate of the midpoint of QR is

1 (ab2

ab2

)y = 2 bs - at - bs + at

= ab2

(bS + at - bs + at)2 b2 s2 - a2 t 2

a2 b2 t

307

b2s2 - a2t 2 .

But (s, t) satisfies the equation of the hyperbola, giving b2 s2 - a2 t2 = a2 b2,

and implying Y = t.Substituting the values of y in the equation of the asymptotes, similar

calculations give Q and R to have the coordinates

Q ( a2b , ab

2) and R ( a

2b , -ab

2) .

bs - at bs - at bs + at bs + at

Accordingly, it is easily checked that the abscissa of the midpoint of Q Ris t, making the midpoint P, as required. Also, the area of 6QOR is easilyfound to be the constant ab for all values of sand t.

(b) (1954). Solving the equations b2x 2 - a2y2 = a2b2 and y = mx + k

for the coordinates of Q and R (figure 196), we have

b2x 2 - a2(m2x 2 + 2mkx + k 2) = a2b2,

(b2 - a2m 2)x2- 2a2mkx - a2 (k 2 + b2

) = 0, (1)

giving the x-coordinate of the midpoint At of QR to be

1 ( 2a2mk) a

2mk

2 b2 - a 2 m 2 = b2 - a2 m 2 .

308

FIGURE 196


y

-----t----~---_+___+----x

o

s

Solving the equations b2 x 2 - a2 y2 == 0 and y == mx + k for the midpointof P S clearly does not alter the coefficients of x 2 and x in the equationcorresponding to (1) (only the absolute term is changed), and so the samex-coordinate is obtained for the midpoint. Similarly for the y-coordinates.

19. (1961). In figure 197, we would like to show that ¢ > O. By the law ofsines, we have

x 2x ( x)sin () == sin 2¢ == sin ¢ cos ¢ ,

implyingsin () == sin ¢ cos ¢.

x

FIGURE 197


Now 0 > O~ implying cos 0 < 1, and therefore

sin e < sin o.

309

But because 20 < 180 0 (in the triangle)~ then (J) < 90° ~ and since the sine

function is increasing in the range (0°. 900) ~ sin e< sin dJ implies () < ¢.

20. (1935). Recall the standard relations

1. AL == s~ the semiperimeter of triangle ABC~2. the area ~ of triangle ABC is given by TS and by J s(s - a)(s - b)(s - c)~

3. (from similar triangles~ see the figure)

r

s-a

givingrs ~

rl == -- == --~s-a s-a

and similarly for T2 and r3.

From figure 198, we have that

1~ABII == -CTI

2

FIGURE 198


and hence, adding, we get ABI1C == ~rl (b + c); similarly for .4BC12 andACB13 , and we have altogether that

~111213 == ABI1C + ABC12 + ACB13 - 2~ABC

= Dr1(b+C)] + [~r2(a+c)] + Dr3(a+b)] -2~1

= 2h(2s - a) + r2(2s - b) + r3(2s - c)] - 2~

= ~ {rds + (s - a)] + r2 [s + (s - b)] + r3 [s + (s - c)J} - 2~1

== - [s(rl + r2 + r3) + 3rsl - 2rs (recall rl(s - a) == rs ==~, etc.)2 -1

== 2s(r1 + r2 + r3 - r).

Thus we want to show that

1 abc-s(r1 +r2 +r3 -r)==-2 '2 r

that is,

or equivalently,

~ ~ ~ ~ abc--+--+----==-8-a s-b s-c s ~'

or finally,~2 ~2 ~2 ~2

-- + -- + -- - - == abc.s-a s-b s-c s

Recalling that ~ == vis(s - a)(s - b)(s - c), the left side is

s(s - b)(s - e) + s(s - a)(s - e) + s(s - a)(s - b)

- (s-a)(s-b)(s-c)

== (s3 - bs2 - es2 + bcs) + (s3 - as 2- cs2 + acs)

+ (s3 - as2- bs2 + abs)

- (s3 - bs2 - cs2 + bcs - as2 + abs + acs - abc)

== 2s3 - s2(2a + 2b + 2c - a - b - c)

+ s(be + ae + ab - be - ae - ab) + abe

== 2s3 - s2(2s) + abe

== abc.


A

B 2y D y c

FIGURE 199

21. (1943). In LABD (figure 199), the semiperimeter s = x + ~y, andtherefore

(x+~)(x-~)

(x + ~y)(~y - x)

(s - a)(s - b)s(s - d)

1tan 2LADB =

245 y2 - x 2

Now in LABD, the law of cosines gives

implying

/ADB = 13y2 - 4x2

COSL 2 21 y

similarly, from LADC we get

and9x2 - 10y2

cosLADC = 2.6y

But cos LADE = - cos LADC (supplementary angles), and so

13y2 - 4x2 9x2 - 10y212y 2 6y2

13y2 - 4x2 = 18x2 - 20y2,

312

giving

Hence


33y 2 == 22x2 ~

3y2 == 2x2.

22. (1950).

1tan -LADB ==

2

6y2 - y2 (525y2 - 6y2 == V19'

== (1· 2 - 1) + (2·3 - 2)x + (3·4 - 3)x2

+ ... + [n(n + 1) - n]xn-

1 + ....

Now, the order of these tenns may be rearranged provided the series is absolutely convergent for -1 < x < 1. This is the case, but at this point let ussimply assume it. Then

S == [1 ·2 + 2 . 3x + 3 . 4x2 + ... + n(n + l)xn-

1 + ...]

- (1 + 2x + 3x2 + ... + nxn-

1 + .. '),

which is the derivative of (1 +2x +3x2 +... +nxn - 1 +...) minus the seriesitself. Since 1 + 2x + 3x2 + ... + nxn - 1 + ... == (1- x)-2, we have

s = d(l - x)-2 _ (1 _ x)-2dx

2 1(l-x)3 (l-x)2

2-(I-x)(1 - x)3

l+x(1 - x)3 .


To justify our assumption concerning convergence, we need to check thisresult. We have

[xn-1](1 + x)(l - ~r)-3 == [xn-1](1 + x) L (i ~ 2)X i

i~O

(n + l)n + n(n - 1)2

as desired.

23. (1942). We have

1m == esc () - sin () == -:--() - sin ()

SIn1 - sin2 ()

sin ()and

1n == sec () - cos () == --() - cos ()

cos1 - cos2 () sin2 ()

cos () cos () .

Therefore mn == sin () cos () .We want to show that

(

C~S2 ()) 2/3 + (sin2()) 2/3

sIn () cos ()

that is,

(1 ) 2/3

sin () cos ()

COS4 / 3 () sin4 / 3 () 1--- + == ------sin2/ 3 () cos2/ 3 () sin2/ 3 () cos2 / 3 ()'

or

But, since

COS4 / 3 () cos2 / 3 () + sin4 / 3 () sin2 / 3 ()

sin2 / 3 () cos2 / 3 ()

1

sin2 / 3 () coS2 / 3 () .

COS4 / 3 () COS2/ 3 () + sin4

/3

() sin2

/3

() == cos2 () + sin2 () == 1,

this is clearly so, and because all the steps in our derivation are reversible,the desired conclusion follows.

314 In Polya's Footsteps

n(n + 1)24. (1961). Sn == , and therefore

2

S == " n(n + 1)~ 2nn2::1

evaluated at x == ~,

t - 1a x - 2'

1at x == 2'

1at x == 2'

1at x == 2'

d 2 3 4== - (x + 2x + 3x + ...)dx

d [ 2 2]== - x (1 + 2x + 3x + ...)dx

d[x2(1 - X)-2]

dx== [2x(1 - X)-2 + x 2(-2)(1 - x)-3( -1)]

(1) -2 1 ( 1) -3

==1· 2 +4. 2 2

==4+4

== 8.

25. (1941). Let tan- 1 .g. == x, tan- 1 /0 == y, and tan- 1 gIg == z. Then wewant the smallest value of 4x - y + z == 4x - (y - z).

Now,

2 tan 2x 2 ( 1 ~t::n~ x )tan4x == 21- tan 2x 1 (2 tan x ) 2

- 1 - tan2 x

4 tanx(l - tan2 x)

== (1-tan2 x)2-4tan2 x

4(.g.)(1 - <15)(1 - is)2 - 4(25)

4(5)(25 - 1)(25 - 1)2 - 4(25)

480 120- -476 119


Also,

tan y - tanz _tan(y - z) = 1 + tanytanz

29 1== 6931 == 239

99 -70

70 . 99 + 1

Therefore

tan4x - tan(y - z)tan(4x - y + z) = tan[4x - (y - z)] = 1 + tan 4x tan(y _ z)

~ - 2k 120 . 239 - 119== 120 1 119.239+1201 + 119 . 239

== 28561 == 1.28561

Hence the desired

1r4x - y + z == -.

· 4

AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS

In Pólya’s FootstepsRoss Honsberger was born in Toronto, Canada, in 1929 and attended the University of Toronto. After more than a decade of teaching mathe matics in Toronto, he took advantage of a sab batical leave to continue his studies at the University of Waterloo, Canada. He joined its faculty in 1964 in the Department of Combina torics and Optimization, and has been there ever since. He has published a number of best selling books with the Mathematical Association of America, including Episodes in Nineteenth and Twentieth Century Euclidean Geometry, and From Erdos to Kiev. In Pólya’s Footsteps is his eighth

book published in the Dolciani Mathematical Exposition Series.

The study of mathematics is often undertaken with an air of such seriousness that it doesn’t always seem to be much fun at the time. However, it is quite amazing how many surprising results and brilliant arguments one is in a posi-tion to enjoy with just a high school background. This is a book of miscellaneous delights, presented not in an attempt to instruct but as a harvest of rewards that are due good high school students and, of course, those more advanced—their teachers, and everyone in the university mathematics community. Admittedly, they take a little concentration, but the price is a bargain for such gems.

A half dozen essays are sprinkled among some hundred problems, most of which are the easier problems that have appeared on various national and inter-national olympiads. Many subjects are represented—combinatorics, geometry, number theory, algebra, probability. The sections may be read in any order. The book concludes with twenty-fi ve exercises and their detailed solutions.

It is hoped that something to delight will be found in every section—a surprising result, an intriguing approach, a stroke of ingenuity—and that the leisurely pace and generous explanations will make them a pleasure to read.

The inspiration for many of the problems came from the Olympiad Corner of Crux Mathematicorum, published by the Canadian Mathematical Society.

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Miscellaneous Problems and Essays Ross Honsberger

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