Portal through Mathematics - American Mathematical Society

Portal through Mathematics

Journey to Advanced Thinking

O.A. IvanovTranslated

by Robert G. Burns

AMS / MAA ANNELI LAX NEW MATHEMATICAL LIBRARY VOL 52

Portal throughMathematics

10.1090/nml/052

Originally published byThe Mathematical Association of America, 2017.

ISBN: 978-1-4704-4876-9LCCN: 2016960274

Copyright © 2018, held by the Amercan Mathematical SocietyPrinted in the United States of America.

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10 9 8 7 6 5 4 3 2 23 22 21 20 19 18

AMS/MAA ANNELI LAX NEWMATHEMATICAL LIBRARY

VOL 52

Portal throughMathematics

O. A. Ivanov

Council on Publications and Communications

Jennifer J. Quinn, Chair

Committee on Books

Jennifer J. Quinn, Chair

Anneli Lax New Mathematical Library Editorial Board

Karen Saxe, Editor

Timothy G. Feeman

John H. McCleary

Katharine Ott

Katherine S. Socha

James S. Tanton

Jennifer M. Wilson

ANNELI LAX NEW MATHEMATICAL LIBRARY

1. Numbers: Rational and Irrational by Ivan Niven

2. What is Calculus About? by W. W. Sawyer

3. An Introduction to Inequalities by E. F. Beckenbach and R. Bellman

4. Geometric Inequalities by N. D. Kazarinoff

5. The Contest Problem Book I Annual High School Mathematics Examinations

1950–1960. Compiled and with solutions by Charles T. Salkind

6. The Lore of Large Numbers by P. J. Davis

7. Uses of Infinity by Leo Zippin

8. Geometric Transformations I by I. M. Yaglom, translated by A. Shields

9. Continued Fractions by Carl D. Olds

10. Replaced by NML-34

11. Hungarian Problem Books I and II, Based on the Eotvos Competitions12.

}1894–1905 and 1906–1928, translated by E. Rapaport

13. Episodes from the Early History of Mathematics by A. Aaboe

14. Groups and Their Graphs by E. Grossman and W. Magnus

15. The Mathematics of Choice by Ivan Niven

16. From Pythagoras to Einstein by K. O. Friedrichs

17. The Contest Problem Book II Annual High School Mathematics Examinations

1961–1965. Compiled and with solutions by Charles T. Salkind

18. First Concepts of Topology by W. G. Chinn and N. E. Steenrod

19. Geometry Revisited by H. S. M. Coxeter and S. L. Greitzer

20. Invitation to Number Theory by Oystein Ore

21. Geometric Transformations II by I. M. Yaglom, translated by A. Shields

22. Elementary Cryptanalysis by Abraham Sinkov, revised and updated by

Todd Feil

23. Ingenuity in Mathematics by Ross Honsberger

24. Geometric Transformations III by I. M. Yaglom, translated by A. Shenitzer

25. The Contest Problem Book III Annual High School Mathematics Examina-

tions 1966–1972. Compiled and with solutions by C. T. Salkind and J. M.

Earl

26. Mathematical Methods in Science by George Polya

27. International Mathematical Olympiads—1959–1977. Compiled and with

solutions by S. L. Greitzer

28. The Mathematics of Games and Gambling, Second Edition by Edward

W. Packel

29. The Contest Problem Book IV Annual High School Mathematics Examina-

tions 1973–1982. Compiled and with solutions by R. A. Artino, A. M. Gaglione,

and N. Shell

30. The Role of Mathematics in Science by M. M. Schiffer and L. Bowden

31. International Mathematical Olympiads 1978–1985 and forty supplementary

problems. Compiled and with solutions by Murray S. Klamkin

32. Riddles of the Sphinx by Martin Gardner

33. U.S.A. Mathematical Olympiads 1972–1986. Compiled and with solutions

by Murray S. Klamkin

34. Graphs and Their Uses by Oystein Ore. Revised and updated by Robin

J. Wilson

35. Exploring Mathematics with Your Computer by Arthur Engel

36. Game Theory and Strategy by Philip D. Straffin, Jr.

37. Episodes in Nineteenth and Twentieth Century Euclidean Geometry by Ross

Honsberger

38. The Contest Problem Book V American High School Mathematics Examinations

and American Invitational Mathematics Examinations 1983–1988. Compiled

and augmented by George Berzsenyi and Stephen B. Maurer

39. Over and Over Again by Gengzhe Chang and Thomas W. Sederberg

40. The Contest Problem Book VI American High School Mathematics Examina-

tions 1989–1994. Compiled and augmented by Leo J. Schneider

41. The Geometry of Numbers by C. D. Olds, Anneli Lax, and Giuliana P. Davidoff

42. Hungarian Problem Book III, Based on the Eotvos Competitions 1929–1943,

translated by Andy Liu

43. Mathematical Miniatures by Svetoslav Savchev and Titu Andreescu

44. Geometric Transformations IV by I. M. Yaglom, translated by A. Shenitzer

45. When Life is Linear: from computer graphics to bracketology by Tim Chartier

46. The Riemann Hypothesis: A Million Dollar Problem by Roland van der Veen

and Jan van de Craats

47. Portal through Mathematics: Journey to Advanced Thinking by Oleg A. Ivanov.

Translated by Robert G. Burns.

Other titles in preparation.

MAA Service Center

P.O. Box 91112

Washington, DC 20090-1112

1-800-331-1MAA FAX: 1-240-396-5647

Contents

Foreword ix

Preface for an American Readership xi

Author’s Preface xiii

Part I Surprising and Easy 1

1 Surprising right triangles 3

2 Surprisingly short solutions of geometric problems 7

3 A natural assertion with a surprising proof 11

4 Surprising answers 17

5 A surprising connection between three sequences 23

Part II Algebra, Calculus, and Geometry: problems 27

6 Five problems and a function 29

7 Five solutions of a routine problem 33

8 Equations of the form f (x, y) = g(x, y) and their generalizations 39

9 The generalized version of Viete’s formula 49

10 Multiple roots of polynomials 55

11 Non-routine applications of the derivative 63

12 Complex numbers, polynomials, and trigonometry 71

13 Complex numbers and geometry 79

14 Areas of triangles and quadrilaterals 85

15 Constructions in solid geometry 93

16 Inequalities 101

17 Diophantine equations 111

18 Combinatorial tales 119

19 Integrals 129

vii

viii Contents

Part III Algebra, Calculus, and Geometry: theory (a little

way beyond high school mathematics) 139

20 Functional equations of elementary functions 143

21 Sequences given by recurrence relations 151

22 The “golden ratio” or solving equations of the form f (x) = x 161

23 Convex functions: inequalities and approximations 167

24 Taylor’s formula, Euler’s formula, and a combinatorial problem 177

25 Derivatives of vector-functions 187

26 Polynomials and trigonometric relations 199

27 Areas and volumes as functions of co-ordinates 207

28 Values of trigonometric functions and sequences satisfying a

certain recurrence relation 217

29 Do there exist further “numbers” beyond complex numbers? 223

Solutions of the supplementary problems 231

Index 303

Foreword

The teaching of elementary mathematics is often presented as a train track.

Students with early mastery of the mathematics “at their stop” are sent on the

next stop (bumped up a grade) without getting the chance to develop what

they know and see the landscape between stations. One of the biggest leaps

is between high school and university mathematics. What are we missing by

jumping to the next station?

Oleg Ivanov introduces us to the rich world between grade 12 plus or

minus epsilon mathematics (and here epsilon can be small or large!) and

university mathematics with a wonderful collection of mathematical tidbits

to intrigue, propel, and delight. Is there a natural way to find an explicit

formula for the nth Fibonacci number? The set of complex numbers is a field

that contains the real numbers; is there another such field? A rope is tied

around the Earth’s equator and then lengthened by 6 feet. How high can the

rope be raised off the equator to the same height all the way round? Now

suppose that the rope is pulled away from the Earth’s surface at just one

point. How high above the surface can that point of the rope be pulled?

The 29 mathematical themes presented in this text range in style and

content, background and outlook. Teachers and other life-long students of

mathematics occupy different places along the track between school and

beyond-school mathematics, and each will respond to the essays here in their

own ways. Ivanov has provided us a wide selection of deep and surprising

mathematical delights to reflect upon and savor. Wherever you are on the

track between high school mathematics and university courses, you can stop

and explore the landscape away from the track. Some things will be familiar

immediately, some will become clear a little later on. This book is a guide

to the landscape, with wonderful hikes mapped out, and promises to delight

the reader again and again.

Stop the train, pack your backpack, and follow Ivanov to the joy that

even elementary mathematics affords the explorer.

Karen Saxe for the NML Editorial Board

ix

Preface for anAmerican Readership

The purpose of this Preface is to give readers—be they teachers, students or

just someone interested in learning a little mathematics at first-year university

level (but occasionally dipping below and rising above that level)—an idea

of the level of mathematical expertise needed for a ready understanding of

each of the chapters—or “Themes”, as they are called—of the present book.

First a few words about the absolute minimum level of mathematical

skill assumed throughout. The reader should have a fairly clear idea of

the number hierarchy N ⊂ Z ⊂ Q ⊂ R ⊂ C, perhaps with the exception of

the final inclusion. The concept of a function is crucial. The reader is also

expected to know the basic facts about quadratic equations and have a good

idea of the use of mathematical induction. The basic Euclidean geometry of

triangles and circles and coordinate geometry as it applies to these figures

and parabolas is also taken for granted in not a few places. Thus possibly a

Grade 12 mathematics student and certainly a first year university calculus

student should be able to come to grips with most of the material of this

book. The more difficult or advanced Themes might be used to at least pique

the interest of high-school students, even if the techniques for answering

the questions raised are perhaps outside the high-school curriculum. Several

Themes start off gently, so that the first page or two can be read with interest

and profit by everyone: these are Themes 1, 3, 4, 12, 14, 15, 17, 18, and 22.

Robert Burns and Karen Saxe

xi

Author’s Preface

I have always been fascinated by the book “Proofs from THE BOOK” by

M. Aigner and G. M. Zigler. However, while most of the proofs that book

presents are indeed “elegant and amazing”, they tend not to be very easy; in

fact even the statements selected for proving are not so simple. The math-

ematical statements considered in the present book are, by contrast, con-

siderably more elementary. And furthermore, whereas the proofs of “Proofs

from THE BOOK” are really aimed only at professional mathematicians, the

beauty of the proofs contained in the present book can be appreciated by high

school teachers and students in schools offering more advanced mathemati-

cal instruction. It is precisely for this audience that this book is intended. It

aims at helping the teacher add variety to the lessons, and, I believe, should

bring the students to a better understanding of what mathematics is.

The book is divided into three parts. The title of Part 1, “Surprising

and Easy”, speaks for itself. Here are gathered intriguing mathematical facts

with the most striking proofs. Part 1 is by way of an introduction, where

the aim is to intrigue the reader. The other two parts are devoted to problem

solving: the problems of Part 2 involve only the standard concepts and facts

traditionally included in high school curricula, so that students may work

on these independently, while solving those of Part 3 will require a parallel

introduction to new concepts and ideas of proof. The Themes of Part 1 are

highly diverse—as indeed are those of Parts 2 and 3. Thus here the reader will

find: integer solutions of equations by geometric means (yielding surprising

answers); very short solutions of apparently difficult geometric problems,

obtained using analytic geometry; an unusual stereometric construction used

to solve a little known analogue of a well-known result of plane geometry;

an approximate solution of an unusual equation, obtained by means of the

calculus; and curious connections between the terms of three recurrence

sequences arising in the solution of a certain Theme 1 problem.

xiii

xiv Author’s Preface

Acknowledgements

The solution of mathematical problems and discussions of mathematical

topics with colleagues and students who have caught the mathematical bug,

provide a welcome sense of life’s repletion and the satisfaction it affords. I

may even say, more specifically, that without my colleagues and our mathe-

matical interchanges this book would never have been written. Thus some of

the mathematical ideas included here were topics of discussion with the panel

of judges of the Euler Olympiad, chaired by V. B. Nekrasov. It behooves me

to make special mention, however, of the role of some of my colleagues in

connection with the Themes of Part 1 of the present book.

The reduction to Pell’s equation of the equations appearing in the solution

of Problem 2 of Part 1 is due to V. M. Gol′khovoj. The fact that Problem 2

of Theme 2, although well known, would be the perfect example of the use

of algebraic methods in geometry, dawned on the author following a lecture

by R. R. Pimenov at the St. Petersburg Seminar for Mathematics Teachers.

The idea of the proof of the basic result of Theme 3 is due to A. Moshonkin.

It was B. I. Ryzhik who brought Problem 3 of Theme 4 to my attention. And

I am especially grateful to him for his constant encouragement to write this

book. Finally, it was B. M. Bekker who provided me with the first proof of

the basic result of Theme 5.

I thank all of the above from the bottom of my heart.

I am very grateful to Robert Burns, the translator of this book (and of

two others), and a friend whom I have known now for almost 20 years, for

his painstaking work producing the English versions of my books!

I am also deeply grateful to the Anneli Lax New Mathematics Library

Editorial Board and especially to its Chair, Karen Saxe, for much help in

improving the book and readying it for publication.

Oleg Ivanov, St. Petersburg, Russia

Solutions of thesupplementary problems

Theme 6. Five problems and a function

6.1. Observe first that since ax > 0, any solutions of ax = x must be positive.

Hence we may go over to the equivalent equation x ln a = ln x, or

ln a = ln xx

. We investigated the behavior of the function f (x) = ln xx

in

this section, and from our knowledge of that behavior we infer first that

the given equation has just one solution precisely if either ln a ≤ 0 or

ln a = 1e, that is, if a ∈ (0, 1] or a = e1/e, has exactly two solutions if

a ∈(1, e1/e

), and has no solutions for all other values of a. Figure 90

shows the graphs of y = ex/e and the line y = x, tangential to it at the

point (e, e). Note also that e1/e ≈ 1.44.

0 1 2 3 4

1

2

3

4

5

Figure 90

6.2. Method 1. We investigate the function f (x) = x4 2−x for x ≥ 1. Since

f ′(x) = 4x3 2−x − x4 2−x ln 2 = x3 2−x(4 − x ln 2),

this function is increasing on the interval[1, 4

ln 2

]and decreasing on[

4ln 2

,+∞). On our calculator (which is allowed!), we find that 4

ln 2≈

5.77. Hence the largest value of this function for natural values of x is

f (5) or f (6). Thus it only remains to compare these two values. Since

f (5) = 54

25 = 62532

≈ 19.5, while f (6) = 814

= 20.25, we conclude that

the largest term of the given sequence is x6.

231

232 Solutions of the supplementary problems

Method 2. Here is how best to solve this problem (or at least discuss its

solution) in a classroom of, for instance, grade tens. First one merely

calculates, that is, compiles a table of terms of the given sequence,

continuing the calculations until its behavior has become clear:

n 1 2 3 4 5 6 7

xn 0.5 4 10.125 16 19.5 20.25 18.8

It would seem that the largest term should be x6. So let’s try to prove

this directly: consider the inequality xn+1 < xn, that is, (n+1)4

2n+1 < n4

2n ,

which may be rewritten as(1 + 1

n

)4< 2. Since the left-hand expression

decreases with increasing n, once this inequality has been shown to

hold for some k, we can be sure it will then hold for all n ≥ k. From the

above table we see that this inequality holds for n = 6, so we infer that

it holds for all n ≥ 6. Hence the term x6 is the largest.

6.3. Instead of trying to minimize the given function, we minimize its loga-

rithm. Thus we want the least value of the function g(x) = x ln x. Since

g′(x) = ln x + 1, the desired least value is attained at x = 1e

≈ 0.37.

Hence the least value of the given function is e−1/e ≈ 0.69.

6.4. Consider the function f (x) = x1+1/x , whose values at natural n are

the given numbers n n√

n . As before, we consider the logarithm of this

function, that is, the function g(x) =(1 + 1

x

)ln x, for x ≥ 1. We have

g′(x) = −1

x2ln x +

(1 +

1

x

)1

x=

x + 1 − ln x

x2.

It is well known that x − 1 ≥ ln x for all x > 0, whence x + 1 − ln x >

0. Alternatively, we may differentiate. Since

(x + 1 − ln x)′ = 1 −1

x≥ 0 for x ≥ 1,

we infer that x + 1 − ln x ≥ 2 > 0 for x ≥ 1. Hence the function g is

increasing, so the function f is also increasing. We conclude, therefore,

that the numbers n n√

n, n = 1, 2, . . . are in increasing order in their

natural order.

Theme 7. Five solutions of a routine problem

7.1. Method 1. We wish to find all a for which the equation xx2+x+1

= a

has a solution. Rewrite the equation as ax2 + (a − 1)x + a = 0. When


a = 0 we get x = 0. For a �= 0 this is a quadratic equation, so has a

(real) solution if and only if its discriminant is not negative. Thus we

obtain as the condition on a that it should satisfy (a + 1)(3a − 1) ≤ 0,

which holds for −1 ≤ a ≤ 13

.

Method 2. Clearly, f (0) = 0. Assuming x �= 0, we rewrite the expres-

sion for f (x) as

f (x) =1

x + 1x

+ 1.

Putting t = x + 1x

, we get f (x) = g(t) where g(t) = 1t+1

. Since |t | ≥ 2,

we need to find the values taken by g on the half-lines (−∞,−2] and

[2,+∞). We might illustrate the situation with Figure 91:

–4 –2 2 4 6 8

–1

13

1

Figure 91

However, it is in any case clear that the set of values taken by this

function on those half-lines is the union [−1, 0) ∪(0, 1

3

].

Method 3. Let’s see where the given function is monotonic (increasing

or decreasing). Its derivative is

f ′(x) =x2 + x + 1 − x(2x + 1)

(x2 + x + 1)2=

1 − x2

(x2 + x + 1)2.

Hence the function is decreasing on each of the intervals (−∞,−1] and

[1,+∞) and increasing on [−1, 1]. Since f (−1) = −1, f (1) = 13

and

f (x) → 0 as x → ±∞, we conclude that the range of f is the interval[−1, 1

3

](Figure 92).

7.2. Method 1. We rewrite the given equation as x2 + x + 1x

+ 1x2 = a,

and investigate the function f (x) = x2 + x + 1x

+ 1x2 . In order to see

where this function is increasing and where decreasing, we examine its


–4 –2 2 4 6 8

–1

13

1

Figure 92

derivative

f ′(x) = 2x + 1 −1

x2−

2

x3=

2(x4 − 1)

x3+

x2 − 1

x2

=(x2 − 1)(2x2 + x + 2)

x3.

We see that f (x) increases on each of the intervals [−1, 0) and [1,+∞),

and decreases on (−∞,−1] and (0, 1]. Furthermore, f (−1) = 0,

f (1) = 4 and f (x) → +∞ as x → ∞ and as x → 0. Thus the graph

of the function has the following form (Figure 93):

–2 –1 1 2

2

4

6

8

Figure 93

Hence the equation f (x) = a has two solutions precisely for those a in

the interval (0, 4).

Method 2. Putting t = x + 1x

changes the equation to t2 + t = a + 2,

and once again, since∣∣x + 1

x

∣∣ ≥ 2, we need to examine this equation

only on the union of the intervals (−∞,−2] and [2,+∞). The graph

of the binomial t2 + t is shown in Figure 94.

Hence for a + 2 > 6, that is, for each a > 4, the quadratic equation

t2 + t = a + 2 has two solutions t1 and t2, one of which is less than

−2 and the other greater than 2. It follows that in this case the original

equation will have four solutions. If a = 4 then t1 < −2, and t2 = 2,


–3 –2 –1 1 2 3

2

4

6

8

Figure 94

in which case the original equation has two negative solutions as well

as the solution x = 1. If 0 < a < 4 then once again we have t1 < −2,

but now 0 < t2 < 2, so that in this case the original equation has two

solutions.

In this last approach, we might instead have analyzed the situations

t1 ≤ −2 and t2 ≥ 2 by algebraic means, but shall not pursue this further.

7.3. We give several solutions of Part a) of the problem.

Method 1. We examine the graph of y =√

x + 3 to see how the number

of its points of intersection with the line y = 1 + ax varies with the slope

a of that line. Every such line passes through the point (0, 1), which

lies below the “half-parabola” y =√

x + 3. Figure 95 presents a sketch

of that half-parabola together with the line y = 1 + x3

(which passes

through the points (−3, 0) and (0, 1)) and two other straight lines. One

of the latter two lines has slope greater than 13

while the other has

negative slope.

–3 –2 2 4 6

1

3

Figure 95

Clearly, for 0 < a ≤ 13

the given equation will have two solutions while

for every other value of the parameter a it will have only one solution.


Method 2. We rewrite the given equation as√

x + 3 − 1

x= a,

and sketch the graph of the function f (x) =√

x + 3 − 1

x. Its derivative

is

f ′(x) =x

2√

x+3−

√x + 3 + 1

x2

=x − 2(x + 3) + 2

√x + 3

2x2√

x + 3=

2√

x + 3 − x − 6

2x2√

x + 3.

On squaring both sides of the inequality 2√

x + 3 ≤ x + 6, we get the

inequality x2 + 8x + 24 ≥ 0, which is easily verified as valid for all x.

Hence f ′(x) < 0 everywhere on the domain of the function f , which

therefore decreases on that domain, that is, on the intervals [−3, 0)

and (0,+∞). It is easy to see that f (x) → 0 as x → +∞. All that

remains for us to be able to sketch the graph is to note that f (−3) = 13

.

From the following sketch it is clear that the equation in question has

two solutions for 0 < a ≤ 13

and just one for every other value of the

parameter a (Figure 96).

–3 2 4 6

1

3

Figure 96

Method 3. Setting t =√

x + 3 yields the equation at2 − t + 1 − 3a =0. The number of non-negative solutions of this equation will then be

the same as the number of solutions of the original equation. If a = 0

then t = 1. Assuming a �= 0, we rewrite the equation in the form

t2 −t

a+

1 − 3a

a= 0.

By Viete’s formula, if the constant term here is negative, then this

quadratic equation will have roots of opposite sign, and therefore just

one positive root. Thus if a < 0 or a > 13

there will be just one positive


root. If a = 13

then the equation becomes t2 − 3t = 0, whence t = 0 or

3. Now suppose 0 < a < 13

. The discriminant of the equation is

1

a2−

4(1 − 3a)

a=

12a2 − 4a + 1

a2,

which is positive for all a, so the equation has two roots for all a in the

interval of present interest. Since for these a their product is positive

and their sum is 1a

> 0, both roots must be positive.

Method 4. Again we make the change of variable t =√

x + 3, but this

time rewrite the resulting equation in the form t−1t2−3

= a and investigate

the behavior of the function f (t) = t−1t2−3

for t ≥ 0. Since

f ′(t) = −t2 − 2t + 3

(t2 − 3)2< 0

for all t , this function is decreasing on each of the intervals [0,√

3 )

and (√

3 ,+∞). We also have f (t) → 0 as t → +∞ and f (0) = 13

.

Putting all this together we obtain the following sketch of the graph,

from which the solution of the problem can be read off (Figure 97).

1 2 3 4 5

–2

–1

1

2

3

4

Figure 97

We now see how each of the above four approaches works for Part b).

Method 1. The diagram is significantly different since now the point

through which all the relevant lines—those with equation of the

form y = ax + 2—pass, namely (0, 2), lies above the half-parabola

y =√

x + 3. Hence not all such lines meet the half-parabola, and

two of them are tangential to it. Let’s find those values of a for

which the line y = ax + 2 is tangential to the graph of y =√

x + 3.

The simplest way to do this is to see for which a the equation

(x + 3) = (ax + 2)2 has just one solution. The discriminant of this

equation is (4a − 1)2 − 4a2, which vanishes when 4a − 1 = 2a, that

is, a = 12

, or when 4a − 1 = −2a, that is, a = 16

. In Figure 98 the two


dashed lines are the tangent lines y = x2

+ 2 and y = x6

+ 2, while the

solid line represents the straight line y = 2x3

+ 2 passing through the

points (−3, 0) and (0, 2).

–3 –2 2 4 6 8

1

3

Figure 98

From this diagram we can read off the answer: the given equation has

no solutions for 16

< a < 12

, exactly one solution for a = 16, a = 1

2, and

also for a ≤ 0 and a > 23, and two solutions for 1

2< a ≤ 2

3.

Method 2. We examine the behavior of the function f (x) =√x + 3 − 2

x. Its derivative is

f ′(x) =x

2√

x+3−

√x + 3 + 2

x2

=x − 2(x + 3) + 4

√x + 3

2x2√

x + 3=

4√

x + 3 − x − 6

2x2√

x + 3.

Thus this time we need to solve the inequality 4√

x + 3 ≤ x + 6.

Squaring both sides yields the inequality x2 − 4x − 12 ≥ 0, whence

x ∈ [−3,−2] ∪ [6,+∞). Hence the given function is decreasing on

each of the intervals [−3,−2] and [6,+∞) and increasing on each of

[−2, 0) and (0, 6]. Clearly, f (x) → 0 as x → +∞ and f (x) → ∞ as

x → 0. We tabulate the values of the function at the key values of x:

x −3 −2 6

f (x) 23

12

16

Here, then, is a sketch of the graph of the function (Figure 99), from

which the solution can be read off:

Method 3. The discriminant of the equation t2 − ta

+ 2−3aa

= 0 is12a2−8a+1

a2 , so this equation has solutions if and only if a ≤ 16

or a ≥ 12

.


–3 2 4 6

3

Figure 99

The remaining argumentation and calculations are similar to those used

above in connection with the solution of Part a) by this method.

Method 4. We set t =√

x + 3 and investigate the function f (t) = t−2t2−3

for t ≥ 0. Since

f ′(t) = −(t − 1)(t − 3)

(t2 − 3)2,

this function is decreasing on the intervals [0, 1] and [3,+∞), increas-

ing on [1,√

3 ) and (√

3 , 2], and furthermore f (t) → 0 as t → +∞,

f (0) = 23

, f (1) = 12

and f (3) = 16

. Hence the graph is as shown in

Figure 100, whence the answer.

1 2 3 4 5

–2

–1

1

2

3

4

Figure 100

7.4. Method 1. The graph of y =√

4 − x2 is the upper semicircle of radius 2

with center at the origin of coordinates. The set defined by the equation

y = |x − 2a| − 3a is the absolute value function translated through the

vector (2a,−3a), so its graph looks like a big open “V”, or a checkmark,

with its “corner”, that is, vertex, at the point P (2a,−3a). The vertex P

lies on the straight line y = − 32x for all values of a. Figure 101 shows

the situation when the right-hand side of the “V” passes through the

right-most point of the semicircle.


–3 –2 –1 1 2 3

–2

–1

1

2

3

Figure 101

The value of the parameter a in this situation is obtained by solving the

equation |2 − 2a| − 3a = 0. Observe first that a ≥ 0. We have either

2 − 2a = 3a, yielding a = 25

, or 2a − 2 = 3a, yielding a = −2, which

we can ignore. For a > 25

the graph of y = |x − 2a| − 3a will lie below

that depicted in Figure 101, so the original equation will have either one

solution of none. It is easy to check that in fact for a ∈ ( 25, 2] there is

just one solution and for a > 2 none.

As a decreases, starting from the value a = 25, the vertex P of the big

“V” moves up the line y = − 23x and the situation becomes that shown

in Figure 102a.

–3 –2 –1 1 2 3

–1

1

2

3

–3 –2 –1 1 2 3

–1

1

2

3

(a) (b)

Figure 102

Now suppose that the point P = (2a,−3a) is on the semicircle (as in

Figure 102b). For this to be the case we need 4a2 + 9a2 = 4, whence

a = − 2√13

. Since the slope of the tangent line to the semicircle at P is23

, our “V” lies above that tangent line, so that, apart from P , the “V”

has no other points of intersection with the semicircle. It follows that for

a < − 2√13

the original equation has no solutions and has exactly two

solutions for a ∈(− 2√

13, 2

5

].


Method 2. Here is a sketch of the set of points (x, y) satisfying the

equation√

4 − x2 = |x − 2y| − 3y, that is, of its graph (Figure 103).

–3 –2 –1 1 2 3

–1

1

2

3

Figure 103

Its left-most point is (−2, 2), its lowest point(− 4√

13,− 2√

13

), and its

right-most point(2, 2

5

). Since for each value of the parameter a the

number of solutions of the original equation is equal to the number

of points of intersection of the above graph with the line y = a, we

immediately infer that the original equation has just one solution when

a = − 2√13

or a ∈(

25, 2

], two solutions when a ∈

(− 2√

13, 2

5

], and none

for all other values of a.

But of course we have yet to explain how the above sketch was arrived

at!

Suppose first that x − 2y ≥ 0. In this case the equation whose graph

we seek to construct simplifies to√

4 − x2 = x − 5y or y = 15

(x −√4 − x2 ). Since we are in the case y ≤ x

2, it follows that

15

(x −

√4 − x2

)≤

x

2, or − 3x ≤ 2

√4 − x2 .

The latter inequality is obviously true for x ∈ [0, 2]. Suppose x ≤ 0.

Squaring the second inequality above, we obtain 9x2 ≤ 16 − 4x2,

or x2 ≤ 1613

, whence x ∈[− 4√

13, 0

]. We conclude that y = 1

5(x −

√4 − x2 ) for x ∈

[− 4√

13, 2

].

In the case 2y ≥ x, the function we are investigating becomes y =−x −

√4 − x2. The inequality −2x − 2

√4 − x2 ≥ x is equivalent to

−3x ≥ 2√

4 − x2, which is easily shown to hold precisely when x ∈[−2,− 4√

13

].

Thus we have so far shown that

y =

⎧⎨⎩

−x −√

4 − x2 for x ∈[−2,− 4√

13

],

15

(x −√

4 − x2 ) for x ∈[− 4√

13, 2

].


Consider first the function f (x) = −x −√

4 − x2 . Its derivative is

f ′(x) = −1 +x

√4 − x2

.

Since the relevant values of x are negative, it follows that f ′(x) < 0 for

those x. Hence for x ∈[−2,− 4√

13

]the function we are investigating is

decreasing.

Now consider the function f (x) = 15

(x −√

4 − x2 ). Differentiating,

we obtain

f ′(x) =1

5

(1 +

x√

4 − x2

).

Examining this to see where f ′(x) ≥ 0, we obtain the inequality√4 − x2 ≥ −x, whence x ≥ −

√2 . Since − 4√

13> −2, the function

in question is increasing for x ∈[− 4√

13, 2

].

The final data needed to obtain the above sketch of the graph of our

function are its values at x = ±2 and − 4√13

(see the table).

x −2 − 4√13

2

f (x) 2 − 2√13

25

Theme 8. On equations of the form f (x, y) = g(x, y)and their generalizations

8.1. Solution of Exercise 1. The straight line y = kx + b avoids the parabola

y = x2 precisely if the equation x2 − kx − b = 0 has no (real) solution,

which is the case precisely when its discriminant is negative: k2 + 4b <

0. Similarly, the line y = kx + b avoids the parabola y = −x2 + 6x − 8

precisely when the discriminant of the equation x2 + (k − 6)x + b +8 = 0 is negative, that is, (k − 6)2 − 4b − 32 < 0. Hence the straight

line y = kx + b avoids both of the given parabolas precisely if k and

b satisfy the system of inequalities k2 − 12k + 4 < 4b < −k2. This

system has solutions if and only if the inequality k2 − 12k + 4 < −k2

is solvable, that is, if and only if k2 − 6k + 2 < 0 is solvable—which it

is. It is satisfied by k = 1, for instance. With this value of k, the above

system of inequalities becomes −7 < 4b < −1, which has b = −1 as

a solution, for example. Hence the straight line y = x − 1 is just one

example of a line meeting neither of the two given parabolas.


Solution of Exercise 2. There’s not much to prove here, of course. We

have

M(x0, y0) ∈ A ∪ B ⇐⇒ M ∈ A

or M ∈ B ⇐⇒ f (x0, y0) = 0

or g(x0, y0) = 0 ⇐⇒ f (x0, y0)g(x0, y0) = 0.

8.2. Suppose that q(x, y) = 0 is a quartic curve that is the union of two

closed curves, one of which is entirely contained in the interior of the

other, together with further points. Let K be one of these additional

points and L a point different from K in the (or a) region bounded by

the inner of the two curves. Let ax + by + c = 0 be an equation of the

line containing the points K and L, and consider the system{

q(x, y) = 0,

ax + by + c = 0.

On the one hand, solving this system reduces to solving a quartic equa-

tion in one variable, and therefore has at most four solutions, while on

the other hand the line KL meets the closed curves contained one in the

other in at least four points (see Figure 104), so has at least five points

in common with the given quartic curve. This contradiction establishes

the assertion of the problem.

Figure 104

8.3. The first question needing to be settled here is the following one: what

conditions on the coefficients of an equation of the form

a11x2 + 2a12xy + a22y

2 + 2a13x + 2a23y + a33 = 0,

ensure that it is the equation of a circle? Since a circle is defined by

an equation of the form a(x − x0)2 + a(y − y0)2 = d where ad > 0,

it is necessary that a12 = 0 and a11 = a22. We need to see how these

conditions can be satisfied by a linear combination of the two given

equations. We can ensure that there is no term in xy in such a linear


combination by adding b12 times the first to −a12 times the second. The

coefficients of x2 and y2 in the resulting equation are then respectively

a11b12 − a12b11 and a22b12 − a12b22. Hence there exists a linear combi-

nation of the two given equations that is the equation of a circle only if

a11b12 − a12b11 = a22b12 − a12b22, or (a11 − a22)b12 = (b11 − b22)a12,

or

a11 − a22

a12

=b11 − b22

b12

.

This condition has a transparent geometric meaning, about which, how-

ever, the author will say nothing further except to note that it was this

that suggested the next problem to him.

8.4. If we choose our coordinate system with axes coinciding with the angle

bisectors, then the two angles (each considered as a pair of rays) are

contained in the respective straight-line pairs given by equations of the

form (x − a)2 = k1y2 and x2 = k2(y − b)2, or

x2 − k1y2 − 2ax + a2 = 0 and x2 − k2y

2 + 2k2by − k2b2 = 0.

Multiplying the first of these equations by 1 + k2 and the second by

−(1 + k1) and adding, we obtain a quadratic equation in x and y with

the sum of its degree-two terms equal to (k2 − k1)(x2 + y2). We have

thus obtained the equation of a circle passing through the four points of

intersection of the arms of the respective angles (Figure 105).

Figure 105

Of course, this problem also has a geometrical solution. Find it!

Theme 9. The generalized version of Viete’sformula

9.1. a) From x1 + x2 + x3 = 0 it follows that

x21 + x2

2 + x23 = −2(x1x2 + x2x3 + x3x1) = 6.


b) From x1 + x2 + x3 = 0 it follows that

x31 + x3

2 + x33 = 3x1x2x3 = −3.

9.2. Set x1 = cos 2π9

, x2 = cos 8π9

and x3 = cos 14π9

. Since cos 2π3

=cos 8π

3= cos 14π

3= − 1

2and cos 3t = 4 cos3 t − 3 cos t , it follows that

x1, x2 and x3 are the roots of the cubic equation 4x3 − 3x + 12

= 0.

Hence we deduce, as in the solution of the preceding problem, that

x31 + x3

2 + x33 = 3x1x2x3 = − 3

8.

Of course, this problem partakes of the subject-matter of Theme 12,

where the idea behind it is developed much further.

9.3. The given condition on a , b, and c implies that (a + b + c)(ab + bc +ac) = abc, whence

0 = a2b + abc + a2c + ab2 + b2c + abc + abc + bc2 + ac2 − abc

= ab(a + c) + ac(a + c) + b2(a + c) + bc(a + c)

= (a + c)(b2 + ab + bc + ac) = (a + c)(a + b)(b + c),

so that at least one of the final three factors must be zero.

Here is a different solution using Viete’s formula. Consider the

polynomial

p(x) = (x − a)(x − b)(x − c) = x3 − Ax2 + Bx − C.

The assumption that (a + b + c)(ab + bc + ac) = abc then becomes

C = AB. Hence

p(x) = x3 − Ax2 + Bx − AB = x2(x − A) + B(x − A)

= (x − A)(x2 + B),

and we see that A is actually a root of the polynomial. Hence we must

have a + b + c = a or a + b + c = b or a + b + c = c, so that some

two of the given numbers add to zero.

In conclusion we note that this problem is often formulated instead as

follows. Prove that

if1

a+

1

b+

1

c=

1

a + b + c, then

1

a5+

1

b5+

1

c5=

1

(a + b + c)5.

9.4. Let x1 < x2 < x3 be the roots of the given polynomial. If they are in

arithmetical progression, then x1 + x2 + x3 = 3x2, whence, by Viete’s


theorem, x2 = − a3. Thus − a

3is a root of the equation, that is,

−a3

27+

a3

9−

ab

3+ c =

2a3

27−

ab

3+ c = 0,

whence 2a3 + 27c = 9ab.

Suppose now that 2a3 + 27c = 9ab, or, equivalently, c = ab3

− 2a3

27. We

shall then have

x3 + ax2 + bx + c = x3 + ax2 + bx +ab

3−

2a3

27

= x3 +a3

27+ ax2 −

a3

9+ bx +

ab

3

=(x +

a

3

)(x2 −

ax

3+

a2

9

)

+ a(x +

a

3

) (x −

a

3

)+ b

(x +

a

3

)

=(x +

a

3

)(x2 +

2ax

3+ b −

2a2

9

).

Hence the number x1 = − a3

is one of the roots of the given equation. The

sum of the roots of the quadratic trinomial factor of the final expression

above is − 2a3

= 2x1, so if these roots are real and distinct then the three

roots of the original cubic will be in arithmetic progression. Hence

the necessary further condition on a, b and c for those roots to be in

arithmetic progression is that the quadratic trinomial

x2 +2ax

3+ b −

2a2

9

have two distinct real roots, or, equivalently, that its discriminant be

positive, which will be the case precisely if a2 > 3b.

Theme 10. Multiple roots of polynomials

10.1. Since the equation x3 + 3x2 − 4x + 2 = 1 − 7x may be rewritten

as (x + 1)3 = 0, it has just one solution. Hence the straight line

y = 1 − 7x has exactly one point of intersection with the graph of the

polynomial p(x) = x3 + 3x2 − 4x + 2. Furthermore, since the equa-

tion p(x) = 1 − 7x has the number −1 as a root of multiplicity three,

that straight line must be tangential to the graph of the cubic. It is easy

to see that this is the only line satisfying the conditions of the problem.


It is worthwhile mentioning by the way that the point of intersection

of the graph and the line is a center of symmetry of that graph.

10.2. a) One such straight line is easy to find. For, since

x4 + 2x3 + x2 + 2x + 1 = x2(x + 1)2 + 2x + 1,

it follows that the line y = 2x + 1 is tangential to the graph of the

given polynomial function at the points with abscissas 0 and −1. It

seems intuitively clear (especially in view of the result of Problem

4) that this is the only such line. But let’s prove it rigorously. Thus

let y = ℓ1(x) and y = ℓ2(x) be equations of two lines each of which

is tangential to the graph of the given polynomial function p(x) of

degree 4 at two distinct points. Then

p(x) − ℓ1(x) = a(x − t1)2(x − t2)2 = a(q1(x)

)2,

p(x) − ℓ2(x) = a(x − u1)2(x − u2)2 = a(q2(x)

)2,

whence

ℓ2(x) − ℓ1(x) = a(q2(x) − q1(x)

)(q2(x) + q1(x)

),

which is impossible since the polynomial on the right-hand side of

this equation has degree at least 2 while that on the left-hand side

has degree at most 1.

b) If x1 and x2 are the abscissas of two points where the straight line

y = kx + d is tangential to the graph of the given polynomial, then

x1 and x2 must each be roots of multiplicity two of the equation

x4 + ax2 + bx + c = kx + d, that is, roots of multiplicity two of

the polynomial x4 + ax2 + (b − k)x + c − d. Since the coefficient

of x3 of this polynomial is zero, we have x1 + x2 = 0, whence

x4 + ax2 + (b − k)x + c − d = (x − x1)2(x + x1)2

= (x2 − x21 )2 = x4 − 2x2

1x2 + x41 .

Hence a = −2x21 , b − k = 0 and c − d = x4

1 , whence k = b and

d = c − x41 = c − a2

4. We conclude that the desired “double tan-

gent line” has equation y = bx + c − a2

4.

10.3. Clearly, we can position a circle of any radius so that it is tangential

to any given parabola at two distinct points. All one needs to do is to

place the circle inside the parabola and let it descend till they touch.


Obviously, in this final position the center of the circle will lie on the

axis of symmetry of the parabola (Figure 106).

Figure 106

However, this is not what the problem is about. It asks rather whether

it is possible for just one of the two points of intersection not to be a

point of tangency.

Consider the parabola y = x2 and a circle (x − x0)2 + (y − y0)2 = r2.

The abscissas of the points of intersection of these two curves are the

solutions of the equation

(x − x0)2 + (x2 − y0)2 = r2,

an equation of degree 4. We are assuming that this equation has just

two roots x1 and x2, where x1 is the abscissa of a point of tangency of

the two curves and therefore a multiple root. If in fact x1 is a double

root, then x2 will also have to be double root, and the second point

of intersection of the parabola and circle will then also be a point of

tangency. However, is it not at least theoretically possible that the first

root x1 has multiplicity three?

Let’s suppose that the point A(1, 1) is one of the points of intersection

of the parabola y = x2 with some circle of radius r . The tangent line

to the parabola at that point has equation y = 2x − 1. Let P denote

the center of the circle. Since this straight line is also tangential to the

circle, the vector AP is perpendicular to the that line, and therefore

parallel to the vector a(−2, 1). Hence the point P has coordinates of

the form (1 − 2t, 1 + t), whence r2 = 5t2. Hence the abscissas of the

points of intersection of parabola and circle are roots of the equation

(x − 1 + 2t)2 +(x2 − 1 − t

)2 = 5t2.

By construction, for every t the number x = 1 is a multiple root of

this equation. Our idea is to choose the value of t so that it is a triple


root. We now carry out the algebraic manipulation needed to exhibit

(x − 1)2 as a factor of the difference of the two sides of this equation:

f (x) = (x − 1 + 2t)2 +(x2 − 1 − t

)2 − 5t2

= (x − 1)2 + 4t(x − 1) +(x2 − 1

)2 − 2t(x2 − 1)

= (x − 1)(x − 1 + 4t + (x − 1)(x + 1)2 − 2t(x + 1)

)

= (x − 1)(x − 1 + (x − 1)(x + 1)2 − 2t(x − 1)

)

= (x − 1)2((x + 1)2 + 1 − 2t

).

Hence we see that x = 1 is a root of multiplicity greater than two if

and only if (x + 1)2 + 1 − 2t = 0 when x = 1, which occurs at the

value t = 52. Since for this value of t we have

f (x) = (x − 1)3(x + 3),

the second point of intersection of the parabola y = x2 and the circle

(x + 4)2 +(y − 7

2

)2 = 1254

is the point B(−3, 9). Figure 107 shows

the parabola and the circle we have just constructed.

–10 –5 –3 5–2

2

4

6

9

Figure 107

Exercise. Verify that the graph of the function f (x) is as sketched in

Figure 108.

–2 –1 1 2

–20

10

Figure 108


Theme 11. Non-routine applications of thederivative

11.1. Since

∣∣f ′(y)∣∣=

∣∣∣∣ limx→y

f (x) − f (y)

x − y

∣∣∣∣= limx→y

∣∣∣∣f (x) − f (y)

x − y

∣∣∣∣ ≤ limx→y

|x − y|=0,

we infer that f ′(y) = 0 for all y ∈ R, so that f (x) is constant.

Here is a proof avoiding use of the derivative. Consider any numbers

a < b and subdivide the interval [a, b] into n equal subintervals by

means of points x0 = a < x1 < . . . < xn = b. By assumption, we have

|f (xk+1) − f (xk)| ≤ (xk+1 − xk)2 = (b−a)2

n2 , whence

|f (b) − f (a)| =

∣∣∣∣∣

n−1∑

k=0

(f (xk+1) − f (xk)

)∣∣∣∣∣

≤n−1∑

k=0

∣∣(f (xk+1) − f (xk))∣∣ =

(b − a)2

n.

Since the number n was arbitrary, and (b−a)2

n→ 0 as n → ∞, we

conclude that f (b) = f (a).

11.2. We may assume that a > b > 0. Rewriting the desired pair of inequal-

ities in the form

√ab

<

ab

− 1

ln ab

<

ab

+ 1

2

and setting t = ab

> 1, we obtain

√t <

t − 1

ln t<

t + 1

2, or

2(t − 1)

t + 1< ln t <

√t −

1√

t.

The following diagram shows what these inequalities look like geo-

metrically (Figure 109):

Now let’s prove them. Since

(ln t −

2(t − 1)

t + 1

)′=

1

t−

4

(t + 1)2=

(t − 1)2

t(t + 1)2≥ 0 ,

this difference is increasing. Since it is equal to zero at t = 1, it will

therefore be positive for t > 1, whence

ln t >2(t − 1)

t + 1.


1 2 3 4 5

1

Figure 109

For the other difference we have(√

t −1

√t

− ln t

)′=

1

2√

t+

1

2√

t3−

1

t

=t + 1 −

√t

2√

t3=

(√

t − 1)2

2√

t3≥ 0,

showing that this function is also increasing. Since it is zero at t = 1,

it will be positive for all t > 1, whence

√t −

1√

t> ln t.

Remark. The inequality of this problem has an interesting interpreta-

tion. We first need to rewrite it as

2

a + b<

ln a − ln b

a − b<

1√

ab.

By the Mean-Value Theorem the quotient ln a−ln ba−b

is equal to the deriva-

tive of the logarithm function at some point c of the interval (b, a).

Since (ln x)′ = 1x

, we infer that

2

a + b<

1

c<

1√

ab, or

√ab < c <

a + b

2.

Thus it turns out that the “intermediate point” c lies between the

geometric and arithmetic means of the endpoints of the interval [b, a].

11.3. Since 4ab ≤ (a + b)2 while 2(a2 + b2) ≥ (a + b)2, neither of these

inequalities is any use in helping us to determine which of the num-

bers 8ab(a2 + b2) and (a + b)4 is larger. However, one can establish

8ab(a2 + b2) ≤ (a + b)4 by direct means (see the remark at the end of

the present solution).


We shall prove that the inequality in question holds for all a, b ≥ 0

if and only if k ≥ n. Set u = (a + b)2 and v = 2ab. Then a2 + b2 =u − v and u ≥ 2v. The inequality in question then becomes

vk(u − v)n ≤(u

2

)k+n

, or(v

u

)k (1 −

v

u

)n

≤1

2k+n.

We now set t = vu

≤ 12, and examine the behavior of the function

f (t) = tk(1 − t)n on the interval[0, 1

2

]. We have

f ′(t) = ktk−1(1 − t)n − ntk(1 − t)n−1 = tk−1(1 − t)n−1(k − (k + n)t).

If k ≥ n then kk+n

≥ 12, so f (t) is increasing on the interval

[0, 1

2

],

whence f (t) ≤ f(

12

)= 1

2k+n . On the other hand, if k < n then kk+n

<12, whence we infer that the function f attains its largest value at some

interior point of the interval, which will therefore be greater than f(

12

).

Remark. Here is a different proof of the fact that if k ≥ n then the

inequality in question holds for all a, b ≥ 0. Set ℓ = k − n and rewrite

that inequality in the form

(ab)n(a2 + b2)n(ab)ℓ ≤(a + b)4n

8n·

(a + b)2ℓ

4ℓ.

It now only remains to observe that ab ≤ (a+b)2

4and ab(a2 + b2) ≤

(a+b)4

8, the latter inequality holding since

(a + b)4 − 8ab(a2 + b2) = (a − b)4 ≥ 0.

11.4. Suppose Mike turns off the road after walking along it for x km. Thus

the time he spends walking along the road is x5

hours. From that point

on he walks in a straight line through the fields to the pond, a distance

of√

4 + (2 − x)2 =√

x2 − 4x + 8 km. Hence that part of his journey

takes√

x2−4x+83

hours. Thus the time he takes to walk the whole route

is

t(x) =x

5+

√x2 − 4x + 8

3.

The derivative of this function is

t ′(x) =1

5+

x − 2

3√

x2 − 4x + 8.

Note that t ′(0) = 15

− 1

3√

2< 0, while t ′(2) = 1

5> 0. Hence the least

value of the function t(x) occurs at some point of the interval (0, 2). In

order to find that point we solve the inequality t ′(x) ≤ 0 for x ∈ [0, 2].


We have, in succession,

1

5≤

2 − x

3√

x2 − 4x + 8,3√

x2 − 4x + 8 ≤ 5(2 − x),4x2 − 16x + 7 ≥0,

whence we get x ∈[0, 1

2

], and conclude that Mike should walk along

the road for only a half-kilometer before turning off and heading

directly for the pond.

In the second case, we need to minimize the function

t(x) =x

5+

√x2 − 4x + 20

3,

whose derivative is

t ′(x) =1

5+

x − 2

3√

x2 − 4x + 20.

Since t ′(0) = 15

− 1

3√

5> 0, we might suspect that in this case the

function is increasing on the whole interval [0, 2]. And in fact on

solving the inequality

1

5+

x − 2

3√

x2 − 4x + 20≥ 0,

we arrive at the inequality x2 − 4x − 5 ≤ 0, valid for all x ∈ [0, 2].

The following diagram shows the graphs of the function t(x) in each

of the two cases (Figure 110):

0.5 1.0 1.5 2.0

0.960.981.001.021.041.06

0.5 1.0 1.5 2.0

1.55

1.60

1.65

1.70

(a) (b)

Figure 110

In fact, it is more appropriate to solve this problem once and for all in

the general case. Thus let u and v denote the respective speeds with

which Mike walks along the road and through the fields. And denote

by a the distance along the road to the point from which the direct path

to the pond is at 90◦ to the road, and b the distance from this point to


the pond. We then have

t(x) =x

u+

√x2 − 2ax + a2 + b2

v, x ∈ [0, a],

so that

t ′(x) =1

u+

x − a

v√

x2 − 2ax + a2 + b2.

We look for a relation between the various parameters of the problem

ensuring that the function t(x) attains its least value somewhere on

the interval (0, a). To this end, we solve the inequality t ′(x) ≤ 0. After

some routine algebraic manipulation, one arrives at the inequality

x2 − 2ax + a2 −b2v2

u2 − v2≥ 0.

The left-hand quadratic trinomial is negative at x = a (we are assuming

u > v). Hence in order for there to be a value of x in the interval (0, a)

for which this inequality is valid, the trinomial should be positive at

x = 0. Thus we require that a2 > b2v2

u2−v2 , or au >√

a2 + b2 v.

We conclude that if au >√

a2 + b2 v then Mike should say on the

road for the distance a − bv√u2−v2

, while if au ≤√

a2 + b2 v then he

should head straight for the pond from his starting point in the village.

11.5. This problem is, in fact, a routine exercise on the derivative. The author

included it in this section for the following reason. In his experience,

the standard reaction of students to the problem is to start “guessing”;

instead of beginning the honest labor of calculation, they try to guess

the answer based on their idea of “geometrical intuition”. Thus this

application of the derivative is, after all, non-routine, but only in the

sense of helping to overcome the students’ routine reaction to it.

We shall for convenience assume the side of the base of the pyramid

to have length 2. An apothem (the altitude of a side face measured

from the apex, see Figure 90a of the pyramid will form a side of an

isosceles triangle with base of length 2 and vertical angle of cosine 2326

(namely, the vertical triangle though the middle of the pyramid with

its base parallel to two sides of the pyramid’s base, shown in Figure

111b. Hence the length d of an apothem is d = 2

√133

(use the Law of

Cosines). A cross-section of the pyramid by a plane passing through

an edge of the square base will have the form of a trapezoid with one

base of length 2 (Figure 112a).


d d d

2

(a) (b)

Figure 111

Let the length of the other base be 2x. We infer the height h of such

a trapezoid by applying the Law of Cosines to the triangle with sides

of lengths d and dx (Figure 112b) meeting at the apex at an angle of

cosine 2326

(and with base the middle line of the trapezoid, equal in

length to its height), obtaining

h =2

√3

√13x2 − 23x + 13 .

Hence the area of the cross-section is 2√3

(x + 1)√

13x2 − 23x + 13 .

Define f (x) = (x + 1)√

13x2 − 23x + 13 . With a view to finding the

h

d

h

2

dx

(a) (b)

Figure 112


largest value of this function, we calculate its derivative:

f ′(x) =√

13x2 − 23x + 13 +(x + 1)(26x − 23)

2√

13x2 − 23x + 13

=52x2 − 43x + 3

2√

13x2 − 23x + 13.

We see that the derivative vanishes at x1 = 113

and x2 = 34

, with the

original function having a local maximum at x1 and a local minimum

at x2. Hence the largest of the relevant values of f (x) is either f (x1) or

f (1). Since f (x1) > f (0) =√

13 > 2√

3 = f (1), we conclude that

its largest value occurs at x1.

Thus the cross-section of the type in question with largest area is that

cutting each of two of the lateral edges of the pyramid in the ratio

1 : 12, measured from the apex.

Theme 12. Complex numbers, polynomials, andtrigonometry

12.1. Since the nth roots of unity other than 1 are precisely the roots of the

polynomial xn−1 + xn−2 + · · · + 1, we have

(x − z1)(x − z2) · · · (x − zn−1) = xn−1 + xn−2 + · · · + 1.

Substituting x = 1 in this equation, we obtain

(1 − z1)(1 − z2) · · · (1 − zn−1) = n.

12.2. The roots of the polynomial q(x) are the numbers cos α ± i sin α, so it

is enough to check that these are also roots of the polynomial pn(x).

Substituting these values for x in that polynomial and appealing to de

Moivre’s theorem, we obtain

(cos nα ± i sin nα) sin α − (cos α ± i sin α) sin nα + sin(n − 1)α

= cos nα sin α − cos α sin nα + sin(n − 1)α

= − sin(n − 1)α + sin(n − 1)α = 0,

as required.

12.3. a) Let ε be a cube root of unity other than 1. We look for all n

for which ε is a root of the polynomial (t + 1)n − tn − 1. Since

ε2 + ε + 1 = 0, we have ε + 1 = −ε2, so that

(ε + 1)n − εn − 1 = (−1)nε2n − εn − 1.


We now examine the six cases corresponding to the six possible

remainders obtained on dividing n by 6:

If n = 6k then (−1)nε2n − εn − 1 = 1 − 1 − 1 �= 0.

If n = 6k + 1 then (−1)nε2n − εn − 1 = −ε2 − ε − 1 = 0.

If n = 6k + 2 then (−1)nε2n − εn − 1 = ε − ε2 − 1 �= 0.

If n = 6k + 3 then (−1)nε2n − εn − 1 = −1 − 1 − 1 �= 0.

If n = 6k + 4 then (−1)nε2n − εn − 1 = ε2 − ε − 1 �= 0.

If n = 6k + 5 then (−1)nε2n − εn − 1 = −ε − ε2 − 1 = 0.

We conclude that the number ε is a root of the polynomial

(t + 1)n − tn − 1 if and only if n = 6k ± 1. For these n this poly-

nomial is divisible by (t − ε1)(t − ε2) = t2 + t + 1, whence

(x

y+ 1

)n

−xn

yn− 1 = q

(x

y

)(x2

y2+

x

y+ 1

),

where q(t) is a polynomial of degree n − 3. Multiplying this equa-

tion throughout by yn, we see that the given polynomial is divisible

by x2 + xy + y2 if and only if n = 6k ± 1.

b) The solution of Part a) shows us that the given polynomial is divis-ible by x2 + xy + y2. It is obvious that it is also divisible by x + y.Using this information, we factor the given polynomial by meansof the following algebraic manipulations:

(x + y)7 − x7 − y7 = 7x6y+21x5y2+35x4y3+35x3y4+21x2y5+7xy6

= 7xy(x5 + 3x4y + 5x3y2 + 5x2y3 + 3xy4 + y5)

= 7xy(x5 + x4y + 2x4y + 2x3y2 + 3x3y2

+ 3x2y3 + 2x2y3 + 2xy4 + xy4 + y5)

= 7xy(x + y)(x4 + 2x3y + 3x2y2 + 2xy3 + y4)

= 7xy(x+y)(x2(x2+xy+y2)+xy(x2+xy + y2)

+ y2(x2 + xy + y2))

= 7xy(x + y)(x2 + xy + y2)2.

A different solution. Consider the polynomial p(t) = (t + 1)7 −t7 − 1. Among its roots are, obviously, 0 and −1, and also, as

we know from Part a), ε1 and ε2 where ε2i + εi + 1 = 0. Since

p′(t) = 7(t + 1)6 − 7t6, we have

p′(εi) = 7(εi + 1)6 − 7ε6i = (−ε2

i )6 − 1 = 0,


so that each of the numbers ε1 and ε2 is actually a multiple root

of the polynomial p(t). Hence p(t) is divisible by the polynomial

t(t + 1)(t2 + t + 1)2. However, since p(t) has degree 6, we must in

fact have

p(t) = at(t + 1)(t2 + t + 1)2.

Putting t = 1 yields 126 = 18a, whence a = 7. Setting t = xy

, we

get the equation

(x

y+ 1

)7

−x7

y7− 1 = 7

x

y

(x

y+ 1

)(x2

y2+

x

y+ 1

)2

,

and multiplying throughout by y7, we get the desired factorization.

Theme 13. Complex numbers and geometry

13.1. a) The solution of this problem derives directly from the result of

Problem 12.1. Write εk = cos 2πk5

+ i sin 2πk5

, k = 1, 2, 3, 4; these are

just the 5th roots of unity other than 1. We may take the corresponding

points of the complex plane to be vertices of the pentagon in question,

with the vertex A1 lying on the real axis (that is, corresponding to 1).

Then

|A1A2|2|A1A3|2 = |1 − ε1|2|1 − ε2|2

= (1 − ε1)(1 − ε1)(1 − ε2)(1 − ε2)

= (1 − ε1)(1 − ε4)(1 − ε2)(1 − ε3) = 5

by the result of Problem 12.1.

b) Once Part a) has been solved, it is clear how to generalize

it. Consider a regular (2n + 1)-gon inscribed in the unit circle,

with vertices A1, . . . , A2n+1 at the (2n + 1)th roots of unity. Then

|A1A2| |A1A3| . . . |A1An+1| =√

2n + 1 .

13.2. Choose a coordinate system with origin at the center of the heptagon

and x-axis an arbitrarily chosen straight line through that center. For

convenience we may assume that the radius of the circumcircle of the

heptagon is 1. The coordinates of the kth vertex Ak of the heptagon may

then be taken to be(cos

(α + 2πk

7

), sin

(α + 2πk

7

)), k = 0, 1, . . . , 6, for

some angle α. The sum of the squares of the distances of the vertices


from the line—that is, from the x-axis—is then

6∑

k=0

sin2(α + 2πk

7

)= 7

2− 1

2

6∑

k=0

cos(2α + 4πk

7

).

We shall show that the sum of cosines in the right-hand side of this

equation is equal to zero, whence it follows that the sum of the squares

of the distances in question is independent of α,, and therefore of the

orientation of the straight line relative to the heptagon.

Write u = cos 2α + i sin 2α and v = cos 2π7

+ i sin 2π7

. Since v7 =1 it follows that u + uv + . . . + uv6 = u(v7−1)

v−1= 0. It now only

remains to observe that the real part of this sum is equal to the sum∑6k=0 cos

(2α + 4πk

7

).

13.3. This problem may be solved, of course, by means of analytic geometry

or vector algebra. It might also serve to illustrate the use of geometric

transformations. However, as we shall see, the simplest approach of

all is just to calculate with complex numbers.

Suppose the vertices of the square ABCD correspond to complex

numbers u, v, w and t , and those of the square A1B1C1D1 to complex

numbers u1, v1, w1 and t1. We may assume that the vertices of the

two squares are read off counterclockwise. Then the line segment

AD, for instance, is obtained from the segment AB by means of a

counterclockwise rotation of the latter through 90◦, whence i(v − u) =t − u. Similarly, we have i(v1 − u1) = t1 − u1. Let A2, B2, C2 and D2

denote the midpoints of the respective segments AA1, BB1, CC1 and

DD1. The complex numbers corresponding to these points are given

by

u2 =u + u1

2, v2 =

v + v1

2, w2 =

w + w1

2, t2 =

t + t1

2.

Since

i(v2 − u2) = i

(v − u

2+

v1 − u1

2

)=

t − u

2+

t1 − u1

2= t2 − u2,

we see that if we rotate the line segment A2B2 through 90◦ about its

endpoint A2 it moves into the position of A2D2. Hence the segments

A2B2 and A2D2 are equal in length and perpendicular. The perpendic-

ularity and equality in length of the other adjacent pairs of sides of the

quadrilateral A2B2C2D2 are established similarly.

13.4. Let the given parallelogram be ABCD, and consider the squares

erected on the sides AB, BC and AD (as in Figure 113). We shall


prove that the line segments KL and KM (in the notation of Fig-

ure 113) are equal in length and perpendicular. This time we shall

identify planar vectors with complex numbers. Let u denote the com-

plex number corresponding to the vector AB and by v the complex

number corresponding to the vector AD. Since the products ±iu and

±iv determine vectors perpendicular to AB and AD and of the same

respective lengths, they correspond to sides of the squares erected on

AB and AD. Hence the vector AK is determined by the number u−iu2

A B

CD

K

M

Lu

vviv

-iu

-iv

Figure 113

and AM by v+iv2

, and therefore the vector KM by

z =v + iv

2−

u − iu

2=

1

2

(v − u + i(u + v)

).

One shows similarly that the vector KL is defined by the number

w =v − iv

2+

u + iu

2=

1

2

(u + v + i(u − v)

).

Then since

iw =1

2

(i(u + v) − (u − v)

)= z,

we infer that the vector KM can be obtained from the vector KL by

rotating the latter through 90◦, as we wished to prove.


13.5. We have

n∑

k=1

|z − zk|2 =n∑

k=1

(z − zk)(z − zk)

= n|z|2 − z

n∑

k=1

zk − z

n∑

k=1

zk +n∑

k=1

|zk|2 = n(|z|2 + 1

).

13.6. a) The given equality 1a

+ 1b

+ 1c

= 0 implies 1a

+ 1

b+ 1

c= 0, or

a|a|2 + b

|b|2 + c|c|2 = 0. Multiplying the later equality by a suitable

positive real number, we get an equation of the form αa + βb +γ c = 0 with α, β, γ > 0 and α + β + γ = 1. Interpreted geomet-

rically, this means that the origin of coordinates lies in the triangle

with the points a, b, c as vertices.

Here is the proof of this fact. Let αa + βb + γ c = 0 where

α, β, γ > 0 and α + β + γ = 1. We rewrite the equation αa +βb + γ c = 0 in the form

(1 − γ )

(α

1 − γa +

β

1 − γb

)+ γ c = 0.

Since α1−γ

+ β

1−γ= α+β

1−γ= 1, we infer that d lies on the line seg-

ment with endpoints a and b (Figure 114).

c

b

a

d

O

Figure 114

Then since (1 − γ )d + γ c = 0, the origin lies on the line segment

with endpoints d and c, and therefore in the triangle with vertices

a, b and c. Write d = α1−γ

a + β

1−γb.

b) If z0 is a solution of the given equation, then, writing ai = ci − z0,

we have 1a1

+ 1a2

+ 1a3

= 0. By the result of Part a) above, the origin

lies in the triangle with vertices a1, a2, a3. Hence z0 lies in the

triangle with vertices c1, c2, c3.


c) Let p(x) = (x − c1)(x − c2)(x − c3) where |ci | ≤ 1. If p(x) has no

repeated root then the equation p′(x) = 0 has the same solutions as

p′(x)

p(x)=

1

x − c1

+1

x − c2

+1

x − c3

= 0.

By Part b) above this implies that the roots all lie in the triangle

with vertices at points in the unit circle, so that they must lie in that

circle. What if p(x) has a repeated root?

13.7. Let ε be a complex number satisfying ε2 + ε + 1 = 0. We choose our

coordinate system so that the center of the given equilateral triangle

ABC is at the origin. If the vertex A then corresponds to the complex

number u, the other two vertices will correspond to the numbers εu

and ε2u. Thus the problem comes down to showing that the inequality

|z − u| ≤ |z − εu| + |z − ε2u| holds for every complex number z. And

in fact we have

|z − u| = |(ε + ε2)(z − u)| = |ε2z − εu + εz − ε2u|

≤ |ε2z − εu| + |εz − ε2u| = |z − ε2u| + |z − εu|.

Theme 14. Areas of triangles and quadrilaterals

14.1. a) Since the diagonal AC divides our quadrilateral ABCD into tri-

angles of equal area, their heights h1 and h2 as measured from this

diagonal as base (see Figure 115a) must be equal. It follows that

the point P of intersection of the diagonals must be the midpoint of

the diagonal BD. For similar reasons, the point P must also be the

midpoint of the diagonal AC. Since the point of intersection of the

diagonals is the midpoint of both, it follows that the quadrilateral is a

parallelogram.

A

B

C

D

Ph1

h2

A B

CD

L

K

h1 h2

(a) (b)

Figure 115


b) By assumption, the quadrilaterals ADKL and KCBL are of equal

area (see Figure 115b). Since, obviously, SDKL = SKCL (S denoting,

as before, area), it follows that SADL = SLCB . Since the bases of the

latter two triangles are equal in length, their heights h1 and h2 must also

be equal, so that the line segments AB and CD are parallel. Similarly,

we have AD ‖ BC. Hence the given quadrilateral is a parallelogram.

14.2. Denote by a, b and c the lengths of the sides of the original triangle,

and by α, β and γ the respective angles opposite those sides. Let

A1, B1 and C1 be the centers of the equilateral triangles erected on

the sides BC, AC and AB respectively, of the original triangle so

that, for instance, the points A and A1 lie on opposite sides of the

line segment BC. These are the vertices of the “external” Napoleonic

triangle (Figure 116). On the other hand, we have the points A2, B2

A

B

C

B1

C1

A1

A2

B2C2

d1

d2

α

Figure 116

and C2, also centers of equilateral triangles erected on the sides BC,

AC and AB respectively, but so that, for instance, the points A and A2

lie on the same side of the line segment BC; these are the vertices of

the “internal” Napoleonic triangle.

In Theme 13 it was proved that the triangle A1B1C1 is equilateral. It

can be proved similarly that the triangle A2B2C2 is also equilateral,

and this we shall assume done. Let d1 and d2 be the lengths of a side of

the triangles A1B1C1 and A2B2C2 respectively. We get an expression

for the length d1 by considering the triangle AB1C1. Since the angle

between the sides AB1 and AC1 is equal to α + π3

, AB1 = c√3, and

AC1 = b√3

, we have

d21 =

b2

3+

c2

3−

2bc

3cos

(α +

π

3

).


Similarly, we have

d22 =

b2

3+

c2

3−

2bc

3cos

(α −

π

3

).

Hence

SA1B1C1− SA2B2C2

=√

3

4(d2

1 − d22 )

=bc

2√

3

(cos

(α −

π

3

)− cos

(α +

π

3

))

=bc√

3sin α sin

π

3

=bc

2sin α = SABC .

14.3. The assumption is that AB = BC = CD = 2. For each possible fixed

disposition of B, C and D, the triangle ABD has largest area when

its sides AB and BD are perpendicular to each other (Figure 117).

Similarly, for each given (possible) disposition of the points A, B

A D

CB

P

Figure 117

and C, the area of the triangle ACD will be largest when AC ⊥ CD.

Hence the quadrilateral ABCD will have greatest area when the points

B and C lie on the circle with diameter AD. The center P of this circle

is the midpoint of the line segment AD. For this quadrilateral we

have PA = PB = PC = PD and, since AB = BC = CD = 2, the

triangles APB, PBC and PCD are congruent. Hence the angle at the

vertex P of each of these isosceles triangles is equal to 60◦, so that in

fact they are equilateral. Thus the greatest area of such a quadrilateral

is 3√

3.


Theme 15. Constructions in solid geometry

15.1. By means of a sufficient number of diagonals, subdivide each face of

the given polyhedron into triangles. Then for each face the outwards-

directed vector n perpendicular to that face and of length equal to its

area, will be the sum of vectors with the same direction as n and with

lengths equal to the areas of the triangles of the subdivision of that

face. It follows that we may assume without loss of generality that

all of the faces of the polyhedron are triangular. This assumed, we

choose an arbitrary point inside the polyhedron and join it by means

of straight line segments to the polyhedron’s vertices. Each face will

then form the base of a tetrahedron with apex at the chosen interior

point of the polyhedron. The polyhedron is thus subdivided into N ,

say, tetrahedrons. Let ai , bi , ci and di be vectors perpendicular to

the faces of the ith such tetrahedron (i = 1, . . . , N), of lengths equal

to the areas of the corresponding faces, and directed outwards from

the tetrahedron, and furthermore with di perpendicular to the base of

the ith tetrahedron. Thus di is perpendicular to the ith face of the

given polyhedron, and∑n

i=1 di is the sum of vectors perpendicular

to the faces of the polyhedron that we wish to show is zero. Since

ai + bi + ci + di = 0 (already proven), it follows that

0 =N∑

i=1

(ai + bi + ci + di) =N∑

i=1

di +N∑

i=1

(ai + bi + ci).

Hence the assertion we wish to prove will follow from the equation∑Ni=1(ai + bi + ci) = 0, which we shall now prove. Each lateral face

(that is, other than the base) of a tetrahedron of the subdivision of the

polyhedron is a face of exactly two such tetrahedrons. Hence the set

{ai, bi, ci} of all vectors corresponding to these faces consists of pairs

of oppositely directed vectors of equal length, which cancel in the sum.

15.2. The idea of the solution consists in examining the possible plane

projections of a regular tetrahedron. If the projection is a triangle then it

coincides with the projection of a face of the tetrahedron, so the largest

area of such a projection is the area of a face, namely a2√

34

where a is the

length of an edge of our tetrahedron. If the projection is a quadrilateral,

then its area S is 12d1d2 sin α, where d1 and d2 are the lengths of its

diagonals and α is the angle between them. Here the diagonals are the

projections of two opposite edges of the tetrahedron, whence it is clear

that S ≤ a2

2. It is also clear that this bound is attained precisely when


the plane of projection is parallel to two of the tetrahedron’s skew

(or, equivalently, opposite) edges. Since√

34

< 12

we conclude that the

greatest area a projection can have is a2

2.

15.3. There are two different ways for the four edges of length 1 to be

arranged in the tetrahedron. If three of these edges are the sides of a

face, then that face will be an equilateral triangle, and so of area√

34

.

Since the height of the tetrahedron measured with this face as base is

at most 1, the maximal volume of such a tetrahedron is√

312

.

The other possibility is that the four edges of unit length form a closed,

connected, four-segment, broken line (the 1-skeleton of the tetrahedron

with one pair of opposite edges removed). Consider the parallelepiped

having the edges of the tetrahedron as diagonals of its faces. Those

four faces of the parallelepiped having an edge of the aforementioned

broken line segment as a diagonal must be rectangles, whence it fol-

lows, in particular, that there is an edge of the parallelepiped—namely,

a common edge of two such (adjacent) rectangular faces—that is an

altitude. Let x be the length of such an edge of the parallelepiped.

The two other faces of the parallelepiped must be rhombi with side of

length√

1 − x2 . The area of these faces will be greatest when they

are actually squares, of area 1 − x2. Since the volume v of the tetra-

hedron is a third of that of this rectangular parallelepiped, we have

v = 13x(1 − x2). It remains to find the largest value of this function

on the interval [0, 1]. Differentiating, we find that this value is attained

at x = 1√3

, whence we infer that the greatest volume such a tetrahe-

dron can have is 2

9√

3. It now only remains to observe that

√3

12> 2

9√

3,

whence we conclude that the tetrahedron of the required sort with

greatest volume is that with three edges forming an equilateral triangle

of side 1 and with another edge of length 1 perpendicular to this face.

15.4. Let n4 ⊥ (ABC) with |n4 | = SABC , n1 ⊥ (ABD) with |n1 | = SABD ,

n2 ⊥ (BCD) with |n2 | = SBCD , n3 ⊥ (ACD) with |n3 | = SACD . Let

αAD , αBD and αCD be the sizes of the dihedral angles at the indicated

edges. Observe that then the angle between the vectors n1 and n2, for

instance, is π − αAB . Hence n1 · n2 = −SABDSBCD cos αBD . Since

n24 = (n1 + n2 + n3)2 = n2

1 + n22 + n2

3 + 2n1 ·n2 + 2n2 ·n3 + 2n1 ·n3,

it follows that

S2ABC = S2

ABD + S2BCD + S2

ACD − 2SABDSBCD cos αBD

− 2SBCDSACD cos αCD − 2SABDSACD cos αAD.


Thus the statement of Problem 6 may be considered as a sort of “three-

dimensional Pythagoras’ theorem”, and the above formula the analo-

gous “three-dimensional law of cosines”.

15.5. a) Consider a tetrahedron ABCD with the angles of the faces around

its apex D all right angles. For the sake of brevity, we write a = AD,

b = BD and c = CD. We look for solutions for a, b and c of the

system of equations

⎧⎪⎪⎨⎪⎪⎩

ab = 4,

bc = 6,

ac = 12,

since with such values of a, b and c we shall have SABD = 2,

SBCD = 3, and SACD = 6. Solving the above system we first see that

(abc)2 = 288 whence abc = 12√

2, and thence a = 2√

2, b =√

2

and c = 3√

2 . It remains to observe that then, by the formula of

Problem 6, S2ABC = 4 + 9 + 36 = 49, whence SABC = 7.

b) Consider the right triangle ABC with sides of lengths AC = 3,

BC = 4 and AB = 5. Its area is 6, and the radius of its incircle is

1. Let O be the center of the incircle. Since SACO = 32

, SBCO = 2

and SABO = 52

, if the side faces of the tetrahedron ABCD are

inclined to the base at an angle of 60◦, then their areas will be 3, 4

and 5.

Theme 16. Inequalities

16.1. a) The least value of the given function is 3 since

x +4

x2=

x

2+

x

2+

4

x2≥ 3

√x

2·x

2·

4

x2= 3,

with equality at x = 2.

b) The least value of the given function is 3 since

x2 +2

x= x2 +

1

x+

1

x≥ 3

√x2 ·

1

x·

1

x= 3,

with equality at x = 1. Note that the substitution t = 2x

transforms

this problem into the preceding one.


16.2. Suppose (x, y, z) is any triple satisfying x2 + 2y2 + 3z2 = 6. Then by

the Cauchy-Schwarz inequality

|x + 2y + 3z| = |1 · x +√

2 ·√

2 y +√

3 ·√

3 y|

≤√

1 + 2 + 3√

x2 + 2y2 + 3z2 = 6.

Hence the given system has solutions only if |a| ≤ 6. Consider first the

case a = 6. Then for every solution of the system, the above inequality

is an equality, which occurs in the Cauchy-Schwarz inequality if and

only if the two n-tuples involved (here triples) are proportional. Hence

the triple (x,√

2 y,√

3 z) is proportional to the triple (1,√

2 ,√

3 ),

that is, x = t , y = t and z = t . The first equation x2 + 2y2 + 3z2 = 6

then yields t2 = 1, whence, in the present case, t = 1, and we infer

that in the case a = 6 the only solution is (x, y, z) = (1, 1, 1). In the

case a = −6 the only solution is (−1,−1,−1). For each a < 6 there

are in fact infinitely many solutions. This is easiest seen geometrically.

The first equation of the given system has graph an ellipsoid centered

at the origin, while the second is the equation of a plane parametrized

by a. The two planes corresponding to a = ±6 are tangent to the

ellipsoid and sandwich it between them. Intermediate planes—that is,

with |a| < 6—meet the ellipsoid in elliptical cross-sections.

16.3. a) The given inequality is valid since

2(ak+n+bk+n

)−(ak + bk

) (an + bn

)= ak+n−akbn + bk+n−anbk

= an(ak−bk

)−bn

(ak−bk

)

=(ak−bk

) (an−bn

)≥ 0,

regardless of whether a ≥ b or a ≤ b.

b) Since k + n = ℓ + m, k ≤ n and ℓ ≤ m, we have m − k = n − ℓ >

0. Write p = m − k. Then

(ak + bk

) (an + bn

)−

(aℓ + bℓ

) (am + bm

)

=(akbn − ambℓ

)−

(aℓbm − anbk

)

= akbℓ(bn−ℓ − am−k

)− aℓbk

(bm−k − an−ℓ

)

=(bp − ap

) (akbℓ − aℓbk

)

= akbk(bp − ap

) (bℓ−k − aℓ−k

)≥ 0


regardless of whether a ≥ b or a ≤ b, in view of the fact that p > 0

and ℓ − k > 0.

16.4. a) Of course, this is immediate from the inequality between the arith-

metic and geometric means. However, the latter inequality follows

from this one, and in fact this represents one of the standard ways

of establishing it. For this reason we prove it instead by induc-

tion on the number n of factors. Thus suppose inductively that

x1x2 · · · xn = 1 implies x1 + x2 + · · · + xn > n provided these n

(positive) numbers are not all equal. Consider n + 1 positive num-

bers x1, . . . , xn+1 such that x1x2 · · · xnxn+1 = 1. If they are all equal

then they are all 1, and their sum is n + 1. Suppose on the other

hand that some are less than 1, some greater. Without loss of

generality we may assume (by re-indexing the numbers if nec-

essary) that xn < 1 < xn+1. Then (xn+1 − 1)(1 − xn) > 0, whence

xn + xn+1 > xnxn+1 + 1. Hence

x1 + x2 +· · ·+ xn + xn+1 > x1 +· · ·+ xn−1 + xnxn+1 + 1 ≥ n + 1

by the inductive assumption, since x1x2 · · · xn−1(xnxn+1) = 1.

b) The given inequality is a direct consequence of the result of Part a)

since the product of the terms in the left-hand sum is equal to 1.

16.5. Denote the lengths of the sides of the triangle by a, b and c, and its

area by S. Since

ha =2S

a, hb =

2S

b, hc =

2S

cand r =

2S

a + b + c,

the inequality ha + hb + hc ≥ 9r may be expressed in the form 1a

+1b

+ 1c

≥ 9a+b+c

or, equivalently, as

(a + b + c)

(1

a+

1

b+

1

c

)≥ 9.

We shall now give four proofs of this inequality. Once rewritten in the

form

31a

+ 1b

+ 1c

≤a + b + c

3,

we see that it is essentially just the inequality between the harmonic

and arithmetic means of three positive numbers. The second proof

uses the inequality between the arithmetic and geometric means of


three numbers:

(a + b + c)

(1

a+

1

b+

1

c

)≥ 3

3√

abc · 31

3√

abc= 9.

For the third, we have

(a + b + c)

(1

a+

1

b+

1

c

)= 3 +

a

b+

b

a+

b

c+

c

b+

a

c+

c

a

≥ 3 + 2 + 2 + 2 = 9.

The final proof is via the Cauchy-Schwarz inequality, and this one is,

in the author’s opinion, the most natural:

(a + b + c)

(1

a+

1

b+

1

c

)

≥(√

a ·1

√a

+√

b ·1

√b

+√

c ·1

√c

)2

= 9.

16.6. We take the equation of the plane to be ax + by + cz = 1. Setting

y = z = 0, we get x = 1a

, so the point of intersection A of the plane

with the x-axis has coordinates(

1a, 0, 0

). Similarly, the plane meets the

other two coordinate axes in the points B(0, 1

b, 0

)and C

(0, 0, 1

c

). The

volume V of the tetrahedron ABCD is 16abc

. Since the point M(4, 2, 1)

lies on the plane, we have 4a + 2b + c = 1. By the inequality between

the arithmetic and geometric means of three numbers, we have

1 = 4a + 2b + c ≥ 33√

8abc = 63√

abc,

whence 13√

abc≥ 6. Hence 1

abc≥ 63 and then V ≥ 62 = 36. Equality

holds when 4a = 2b = c = 13

, that is, when a = 112

, b = 16

and c = 13

.

This answer has an interesting geometric interpretation. Since the ver-

tices of the face ABC of the tetrahedron are the points A(12, 0.0),

B(0, 6, 0) and C(0, 0, 3), the point M is the point of intersection of the

medians of that face, that is, the centroid or center of mass.

16.7. a) Squaring both sides and cancelling like terms, we get the inequality

x1y1 + x2y2 + · · · + xnyn

≤√

x21 + x2

2 + · · · + x2n

√y2

1 + y22 + · · · + y2

n ,

which is a consequence of the Cauchy-Schwarz inequality. It is

worth remarking that in terms of vectors x(x1, x2, . . . , xn) and

y(y1, y2, . . . , yn) in the (Euclidean) space Rn the given inequality


may be rewritten in the form |x + y| ≤ |x| + | y|, standing revealed

as the familiar triangle inequality.

b) Consider plane vectors ai(xi, yi), i = 1, 2, . . . .k. From Part a), just

solved, we know that |a + b| ≤ |a| + |b|. Hence

|a1 + a2 + · · · + ak| ≤ |a1 + a2 + · · · + ak−1| + |ak|

≤ |a1 + a2 + · · · + ak−2| + |ak−1| + |ak|

≤ · · · ≤ |a1| + |a2| + · · · + |ak|.

c) The general inequality embracing those of both Parts a) and b) is

|a1 + a2 + · · · + ak| ≤ |a1| + |a2| + · · · + |ak|,

true for any vectors in Euclidean Rn.

Theme 17. Diophantine equations

17.1. a) Rewriting the given equation as (x − y)(x + y) = 2012 and setting

a = x − y and b = x + y, we get the equation ab = 2012. hence

at least one of the integers a, b must be even. However, the system

{x − y = a,

x + y = b

has integer solutions only if a and b have the same parity. Hence they

must both be even. Writing a = 2k and b = 2n, we obtain the equa-

tion kn = 503. Then since the number 503 is prime, we conclude

that the integer solutions of the given equation are just (1, 503),

(503, 1), (−1,−503) and (−503,−1). Thus the given equation has

exactly four solutions in integers.

b) If ab = 2014 then one of a and b must be even and the other odd.

But then the system

{x − y = a,

x + y = b

has no solutions in integers. Hence the given equation has no integer

solutions.

c) In the notation of Part a), we get kn = 504 = 23 · 32 · 7. The number

504 therefore has (3 + 1)(2 + 1)(1 + 1) = 24 natural divisors and


hence 48 integer divisors. Hence the given equation has exactly 48

integer solutions.

17.2. a) Write the integer x in the form x = 2011k + d, where d may be

any of 0, 1, . . . 2010. Since[

x2010

]=

[x

2011

]+ 1 = k + 1, we must

have k + 1 ≤ x2010

< 2011, whence

2010k + 2010 ≤ 2011k + d < 2010k + 4020,

and therefore 2010 ≤ k + d < 4020, or 2010 ≤ k + d ≤ 4019.

Hence for each of the 2011 possible values of d there are 2010

values of the number k for which the number x = 2011k + d is a

solution of the given equation. Hence that equation has altogether

2010 · 2011 solutions.

b) The largest value of k is k = 4019, with d = 0. Hence the largest

solution of the given equation is the number x = 4019 · 2011 =8 082 209.

17.3. Two easy solutions of the given equation are the pairs (0, 0) and (2, 2).

Thus in searching for further solutions we may suppose y ≥ 3. Since

y3 + 1 = (y + 1)(y2 − y + 1) we must have y + 1 = 3a and y2 − y +1 = 3b where 2 ≤ a < b. Since 32a − 3a+1 + 3 = 3b, it follows that

3 = 3b + 3a+1 − 32a , which is impossible since the right-hand side is

divisible by 9 and the left-hand side is not so divisible.

17.4. Easy integer solutions of the equation 3x + 1 = 2y are the pairs (0, 1)

and (1, 2). Since the remainders on dividing a power of 3 by 8 are just

1 or 3, no number of the form 3x + 1 is divisible by 8, whence y ≤ 2.

Hence there are no further integer solutions of the given equation.

17.5. One should not be too surprised at seeing such a problem appear in

connection with the theme “Diophantine equations”. As you will now

see, the trigonometry is a red herring. Since the values of the sine and

cosine functions lie in the interval [−1, 1], the given equation holds if

and only if{

sin 1992π2

x= 1,

cos x = 1or

{sin 1992π2

x= −1,

cos x = −1.

Consider the second case: sin 1992π2

x= −1 implies 1992π2

x= −π

2+

2πk, k ∈ Z, and cos x = −1 implies x = π + 2πn, n ∈ Z. Hence

1992π2 =π2

2(2n + 1)(4k − 1), or 3984 = (2n + 1)(4k − 1),

which is impossible since 3984 is even.


In the first case, one obtains similarly the equation n(4k + 1) = 1992

where n and k are integers. Thus 4k + 1 is a divisor of 1992 with

remainder 1 on division by 4. Since 1992 = 8 · 3 · 83, one calculates

that the only such divisors of 1992 are 1, 249, −3 and −83. The cor-

responding values of n are then n = 1992, 8,−664,−24 respectively.

Hence x = 3984π, 16π,−1328π,−48π .

17.6. Suppose the number m has (nonzero) remainder d on being divided by

17. From the table referred to in Problem 9, we see that whatever the

value of d there is an n for which the remainder on dividing 3n by 17

is 17 − d, so that 3n + m is divisible by 17.

Theme 18. Combinatorial tales

18.1. Of course, there are just two “tetrahedral dice”—a “right-handed” one

and a “left-handed” one, so to speak. To see this, place the tetrahedron

on the table with the face having four pips against the table top, and

rotate it about its vertical axis until the face with three pips is facing

you. The remaining faces have either one or two pips, and these can

be assigned in two ways, namely, with the face having one pip to the

right or to the left.

18.2. The following table, represents a chessboard with each white square

inscribed with the number of shortest paths by which a bishop can

reach that square starting from any of the four white squares of the first

row. These entries should be calculated starting from the first row and

proceeding upwards.

8 35 89 103 69

7 35 54 49 20

6 10 25 29 20

5 10 15 14 6

4 3 7 8 6

3 3 4 4 2

2 1 2 2 2

1 1 1 1 1

a b c d e f g h

Thus in each white square of the first row (squares b1, d1, f1 and h1)

we have, appropriately, the number 1. There is only one shortest way

of getting to the square a2, namely directly from b1, so that square

also contains the number 1. However, there are two shortest ways of


getting to the square c2, namely directly either from square b1 or

d1. It follows that this table closely resembles Pascal’s triangle, the

difference being due to the fact that the white squares of the extreme

columns a and h have only one lower adjacent white square. Thus

the number of shortest paths to square a8, for example is equal to the

number of such paths to the square b7. Once the table is complete,

it only remains to total the numbers in the top row, yielding 296

paths.

18.3. We are interested in tickets with just three of the six winning numbers

marked. These three numbers may be any of the six winning numbers,

so the number of possibilities for them is(

6

3

)= 20. The three remaining

numbers must come from the 49 − 6 = 43 non-winning numbers, so

can be chosen in(

43

3

)= 12341 ways. Hence the number of tickets with

exactly three winning numbers marked is 20 · 12341 = 246820.

18.4. Suppose x1, x2, . . . , xk are non-negative integers constituting a solu-

tion of the equation x1 + x2 + · · · + xk = n. The numbers yi = xi + 1

will then be natural. The latter numbers constitute a solution of

the equation (y1 − 1) + (y2 − 1) + · · · + (yk − 1) = n, that is, of the

equation y1 + y2 + . . . + yk = n + k. Thus in view of the formula for

the number of solutions of the given equation in natural numbers (see

Problem 3), the number of solutions in non-negative integers is(n+k−1

k−1

).

18.5. In the partial solution of Problem 10 in the main text the case where

three edges of the tetrahedron are of length 20 inches and three of

length 33 inches was completely settled: if edges of the same length

are edges of a single face, then such a tetrahedron exists, but there is no

such tetrahedron in which three edges of the same length form an open,

continuous (triple-segmented) broken line—the only other possibility.

We now consider the remaining cases, in order of increasing com-

plexity. Of course, from six edges of either length one can construct a

(regular) tetrahedron. And a tetrahedron can have five edges of length

33 inches and just one of length 20 inches. However, it is not so easy

to see if there is one with the lengths the other way around: five edges

of length 20 inches and one of length 33 inches. To prove that this is

possible, we imagine two equilateral triangles of side 20 inches with

an edge in common, situated in the plane. The distance between their

opposing (that is, unattached) vertices is then 20√

3, which, as was

noted in the partial solution of Problem 10, exceeds 33. These two

triangles can, therefore, be bent up about their common edge until the

distance between their free vertices is 33 inches.


It is easy to show similarly that there is a tetrahedron with two

opposite edges of length 20 inches and the other four of length 33

inches. But is there a tetrahedron with two opposite edges of length 33

inches and the rest of length 20 inches? Imagine laid out in the plane

two isosceles triangles having a common base of length 33 inches

and side edges of length 20 inches. The tetrahedron in question will

exist if and only if the distance between the opposing vertices of the

triangles exceeds 33 inches. We carry out the calculation in general

terms. As before, let a and b be positive numbers with a > b, and

let a now assume the role of the number 33 and b that of 20 in the

preceding picture. The tetrahedron in question will then exist precisely

if 2

√b2 − a2

4=

√4b2 − a2 > a. On solving this equation, we obtain

ab

<√

2 . In our case, we have ab

= 3320

> 32

>√

2, so there is no such

tetrahedron.

There are two more cases. Does there exist a tetrahedron with two

adjacent edges of length 20 and the rest of length 33? Once again we

carry out the argument in general terms. Thus, imagine laid out in the

plane an equilateral triangle of side a with an edge in common with

an isosceles triangle (lying outside it) with its equal sides of length

b, where, as before, b < a. A tetrahedron with two adjacent edges of

length b and the rest of length a will exist if and only if the distance

between opposing vertices of this figure is greater than a, that is, if the

inequality

a√

3

2+

√b2 −

a2

4> a

holds. Solving, we obtain the condition ab

< 1+√

3√2

≈ 1.93. In our con-

crete case, we have ab

= 3320

, so such a tetrahedron does exist.

We turn finally to the case where there are two adjacent edges of

the greater length and the rest of the shorter length. Thus imagine laid

out in the plane an equilateral triangle with side of length b, having an

edge in common with an isosceles triangle with equal sides of length

a, and lying inside that triangle. A tetrahedron with two adjacent sides

of length a and the rest of length b will exist if and only if the distance

between the free vertices of these two triangles is less than b, that is,

if the inequality

√a2 −

b2

4−

b√

3

2< b,


holds. Solving, we obtain the condition ab

< 1+√

3√2

, as in the preceding

case. Hence such a tetrahedron exists.

Thus the final answer to Problem 10 is: there are nine such distinct

tetrahedrons. In conclusion, we remark that it follows from our inves-

tigation of this problem, that the number of distinct tetrahedrons with

edges of lengths a or b depends on which of the indicated intervals

the ratio ab

falls into (Figure 118). The greater this ratio the fewer

1 2 3 2

1+ 5

2

1+ 3

2

Figure 118

tetrahedrons there are.

Theme 19. Integrals

19.1. a) The number ln n is equal to the area under the graph of the

function y = 1x

and above the interval [1, n]:∫ n

11x

dx = ln x∣∣n1

=ln n − ln 1 = ln n. We estimate this area using “lower” rectangles

and then “upper” rectangles, as shown in Figure 119.

1 2 3 4 5

0.2

0.4

0.6

0.8

1.0

1 2 3 4 5

0.2

0.4

0.6

0.8

1.0

(a) (b)

Figure 119

Consider the rectangles with bases the intervals [k − 1, k], k =2, 3, . . . , n, and heights 1

k(the case n = 5 is shown in Figure 119a).

Since these rectangles all lie below the graph of the function in

question, the sum of their areas, namely∑n

k=21k, will be less than

the area under the graph, that is, less than ln n. To get the upper

estimate for ln n, one takes the rectangles on the intervals [k, k + 1],

k = 1, 2, . . . , n − 1, and heights 1k. Their union contains the area


under the graph above the interval [1, n] (see Figure 119b), so the

sum∑n−1

k=11k

is greater than ln n.

b) The proof is similar to that of Part a):

1√

n+

1√

n + 1+ · · · +

1√

n2 − 1>

∫ n2

n

dx√

x= 2(n −

√n ).

19.2. The proof is similar to that of Part a) of the preceding problem:

1k + 2k + · · · + nk <

∫ n+1

0

xk dx =(n + 1)k+1

k + 1.

19.3. Under the given assumption,

∫ 1

0

(ax2 + bx + c) dx =a

3+

b

2+ c =

2a + 3b + 6c

6= 0,

so the given quadratic function can be neither everywhere positive nor

everywhere negative on the interval (0, 1), so must have a root in that

interval.

Although this solution may seem appealing, it does not really get

to the heart of the matter. Let’s try a different approach. Write p(x) =ax2 + bx + c and consider the sum

p(0) + 4p(

12

)+ p(1) = c + 4

(a

4+

b

2+ c

)+ a + b + c

= 2a + 3b + 6c = 0.

It follows that the numbers p(0), p(

12

)and p(1) cannot all have the

same sign, so the function p(x) must have opposite signs at certain

points of the interval [0, 1] and therefore vanish at some interior point

of that interval.

The connection between these two solutions is revealed in Simpson’s

formula—see Problem 19.7.

19.4. On changing a product of three cosines to a sum of cosines in the

standard manner, we obtain

cos k1x cos k2x cos k3x =1

4

(cos(k1 + k2 + k3)x + cos(k1 + k2 − k3)x

+ cos(k1 − k2 + k3)x + cos(k1 − k2 − k3)x).

In the first of the given integrals we have k1 = 1, k2 = 7 and k3 = 9.

Since these numbers are all odd, none of the corresponding numbers


k1 ± k2 ± k3 can be zero, and the integral of each of the above sum-

mands will in this case be zero. Thus the answer is 0.

Since 2 + 7 − 9 = 0, the second of the given integrals will be equal

to∫ 2π

014dx = π

2.

19.5. We know that the function F (t) =∫ t

0f (x) dx is an antiderivative of

the function f (t), and that every antiderivative differs from this one by

a constant. The function F (t) is 2π -periodic if F (t + 2π ) = F (t), or

∫ t+2π

0

f (x) dx −∫ t

0

f (x) dx =∫ t+2π

t

f (x) dx =∫ 2π

0

f (x) dx = 0.

19.6. The ordinates (y-coordinates) of the points of intersection of the

straight line and parabola are given by yi = ax2i + bxi + c, i = 1, 2.

The area in question is equal to the difference between the area of the

trapezoid with bases y1 and y2, which is

y1 + y2

2(x2 − x1) =

a(x21 + x2

2 ) + b(x1 + x2) + 2c

2(x2 − x1) ,

and the area under the graph of the given quadratic function above the

interval [x1, x2], which is

a(x32 − x3

1 )

3+

b(x22 − x2

1 )

2+ c(x2 − x1).

Hence the desired area is

a(x22 + x2

1 )(x2 − x1)

2−

a(x32 − x3

1 )

3

=a(x2 − x1)

6(3x2

1 + 3x22 − 2x2

1 − 2x1x2 − 2x22 )

=a(x2 − x1)

6(x2

1 − 2x1x2 + x22 ) =

a(x2 − x1)3

6.

19.7. The inequality in question follows from Schwarz’s inequality, since

(∫ b

a

f (x) dx

)2

=(∫ b

a

1 · f (x) dx

)2

≤∫ b

a

1 dx ·∫ b

a

f 2(x) dx = (b − a)

∫ b

a

f 2(x) dx.

19.8. Consider first the special case a = 0 and b = 1. Since the equality in

question is supposed to hold for any polynomial of degree less than or

equal to 3, it should be valid in particular for the particular functions

f (x) = 1, f (x) = x and f (x) = x2. Replacing f in the given formula


by each of these in turn, we obtain the system

⎧⎪⎪⎨⎪⎪⎩

1 = 16

(A + B + C),

12

= 16

(B2

+ C),

13

= 16

(B4

+ C),

or, after simplifying,

⎧⎪⎪⎨⎪⎪⎩

A + B + C = 6,

B + 2C = 6,

B + 4C = 8,

which has the unique solution A = C = 1, B = 4.

We have to show that these values of A, B, and C work for any

polynomial f (x) of degree at most 3 and any a and b, that is, that

∫ b

a

f (x) dx = b−a6

(f (a) + 4f

(a+b

2

)+ f (b)

).

is valid generally. To begin with, we verify this formula for the func-

tions f1(x) = 1, f2(x) = x, f3(x) = x2 and f4(x) = x3. The following

table gives the integrals of these functions:

n 1 2 3 4∫ b

afn(x) dx b − a b2−a2

2b3−a3

3b4−a4

4

We now calculate the right-hand side of the above equation for each

of the fi in turn. Thus for f1 we have

b−a6

(f1(a) + 4f1

(a+b

2

)+ f1(b)

)= b−a

6· 6 = b − a.

For f2 we have

b−a6

(f2(a) + 4f2

(a+b

2

)+ f2(b)

)= b−a

6·(a + 4 · a+b

2+ b

)

= b−a6

· 3(a + b) = b2−a2

2.

For f3 we have

b−a6

(f3(a) + 4f3

(a+b

2

)+ f3(b)

)= b−a

6·(a2 + 4 · (a+b)2

4+ b2

)

= b−a6

· 2(a2 + ab + b2) = b3−a3

3.


And, finally,

b−a6

(f4(a) + 4f4

(a+b

2

)+ f4(b)

)

= b−a6

·(a3 + (a+b)3

2+ b3

)

= b−a6

· a+b2

· (2a2 − 2ab + 2b2 + a2 + 2ab + b2)

= b2−a2

12· 3(a2 + b2) = b4−a4

4.

It now only remains to observe that every polynomial f (x) of degree

3 or less can be written in the form c1f1(x) + c2f2(x) + c3f3(x) +c4f4(x). And then

∫ b

a

f (x) dx =∫ b

a

(4∑

i=1

cifi(x)

)dx

= ci

4∑

i=1

∫ b

a

fi(x) dx

= ci

4∑

i=1

b−a6

(fi(a) + 4fi

(a+b

2

)+ fi(b)

)

= b−a6

(f (a) + 4f

(a+b

2

)+ f (b)

).

19.9. Both the sphere and the truncated cone are solids of revolution. How-

ever, let’s instead integrate over areas of cross-sections. The area S(x)

of a cross-section of the sphere by a plane depends quadratically on

the distance x of the plane from the sphere’s center. Hence we may use

Simpson’s rule to calculate the integral∫ r

−rS(x) dx. In this case we

have a = −r , b = r , whence a+b2

= 0. Since planes at a distance r from

the center of the sphere are tangential to it, we have S(−r) = S(r) = 0.

Hence by Simpson’s rule 2r6

· 4πr2 = 43πr3.

Let r and R be the radii of the top and bottom of the truncated

cone, and h its height. The radius of a cross-section S(x) by a plane

parallel to the bases is a linear function of the distance x of the plane

from one or the other base. Hence the area of such a cross-section

is a quadratic function of that distance. The volume of the truncated

cone is∫ h

0S(x) dx. Since S(0) = πR2, S(h) = πr2, and the area of

the cross-section an equal distance from the bases is equal to π (R+r)2

4,

we conclude from Simpson’s rule that the desired volume is

h

6π(R2 + (R + r)2 + r2

)=

πh

3

(R2 + Rr + r2

).


Theme 20. Functional equations ofelementary functions

20.1. Writing f (0) = b and g(x) = f (x) − f (0), we have

g

(x + y

2

)= f

(x + y

2

)− f (0) =

f (x) + f (y)

2− f (0)

=f (x) − f (0) + f (y) − f (0)

2

=g(x) + g(y)

2,

so g(x) satisfies the same relation but with g(0) = 0. Write a = g(1).

For all natural numbers n we have g(n) = g(n+1)+g(n−1)

2, whence

g(n + 1) = 2g(n) − g(n − 1). Setting n = 1 in this, we get g(2) = 2a.

Setting n = 2, we get g(3) = 2g(2) − g(1) = 4a − a = 3a. Arguing

inductively, we obtain more generally g(n) = an for all natural n.

Putting y = −x in the relation for g yields g(−x) = −g(x), and, argu-

ing inductively as before, we ultimately get g(k) = ak for all inte-

gers k. Putting y = 0 in the relation for g yields g(

x2

)= g(x)

2. Hence

g(

k2

)= a · k

2for all k ∈ Z, whence, by induction,

g

(k

2n

)= a ·

k

2n

for all integers k and non-negative integers n. Since every real number

x is the limit of a sequence (rn) of numbers of the form k2n (its binary

expansion), it follows that

g(x) = limn→∞

g(rn) = limn→∞

arn = ax,

whence g(x) = ax for all x ∈ R. Hence f (x) = ax + b, a “linear func-

tion in the high-school sense”.

20.2. In the solution of Problem 2 it was first shown that if the function

satisfying the conditions of the problem is not identically zero, then it

must be positive. It therefore makes sense to introduce the new function

g(x) = ln f (x), and then

g(x + y) = ln f (x + y) = ln(f (x)f (y)

)

= ln f (x) + ln f (y) = g(x) + g(y).

whence g(x) = αx and therefore f (x) = eαx = ax .


20.3. Putting y = 0, we get f (0) = f (x) + f (0), so that f (x) = 0 for all

x ∈ R.

20.4. Write g(x) = f (ex). Then

g(x + y) = f (ex+y) = f (exey) = f (ex) + f (ey) = g(x) + g(y),

whence g(x) = αx. Thus f (ex) = αx, whence f (x) = α ln x, and, if

α �= 0, then f (x) = loga x for an appropriate number a.

20.5. Putting x = y = 0 in the given relation, we get f (0) = 2f (0)

1−f 2(0). Since

1 − f 2(0) �= 2, it follows that f (0) = 0. Putting y = −x then yields

f (−x) = −f (x). We cannot have |f (x)| = 1 for any x, since oth-

erwise f would be undefined at 2x. Hence by the assumption that

f is continuous and since f (0) = 0, the range of f (x) must be

contained in the interval (−1, 1). Suppose f (a) �= 0 for some num-

ber a. Then |f (2a)| = 2|f (a)|1−f 2(a)

≥ 2|f (a)|. Arguing inductively, we

get |f (2na)| ≥ 2n|f (a)| for all natural numbers n, contradicting the

boundedness of the function just established.

20.6. Setting k = 1, we obtain S1 · Sn = Sn, so that provided neither pro-

gression is just the zero sequence, we must have S1 = 1. Let d be the

common difference. Then S4 = S22 is equivalent to 6d + 4 = (d + 2)2,

whence d2 = 2d, and either d = 0 or d = 2. If d = 0 then Sn = n and

indeed SnSk = nk = Snk . If d = 2 then, as is well known, Sn = n2 (the

sum of the first n odd numbers), and SnSk = n2k2 = Snk . We conclude

that the given relation is satisfied by just three arithmetic progressions,

two of them the constant sequences consisting of all zeroes or all ones,

and the third the sequence of odd numbers.

Theme 21. Sequences given byrecurrence relations

21.1. Let b be the general term of the constant sequence satisfying the

given relation; thus b = qb + d. Write yn = xn − b. Substituting xn =yn + b and xn+1 = yn+1 + b in the given recurrence relation yields

yn+1 + b = q(yn + b) + d, or yn+1 = qyn,

since b = qb + d. Hence (yn) is a geometrical progression. Hence

if a is the first term of that progression, then yn = aqn−1, and xn =aqn−1 + b.


21.2. By definition, xn = xn−1+xn−2

2. The characteristic equation of this recur-

rence relation is 2t2 − t − 1 = 0, and since the roots of this equation

are 1 and − 12

, we infer that xn = a + b(− 1

2

)n. From the assumptions

x0 = 0 and x1 = 1, it follows that a = −b = 23

. Hence

xn =2

3+

(−1)n−1

3 · 2n−1,

and the desired point is 23.

21.3. Answer: xn = c1Fn + c2Fn+1. Since the Fibonacci sequence (with

x0 = 0, x1 = 1) and its translate (x0 = 1, x1 = 1) satisfy the given

recurrence relation, the one shown here will also satisfy it. Let the first

two terms of any such sequence be x0 = a and x1 = b. Then a = c2

and b = c1 + c2, whence xn = (b − a)Fn + aFn+1.

21.4. a) Answer: any sequence with nth term of the form xn = an2 + bn +c. One might try informed guesswork to get this. Thus one might

first observe that the characteristic equation t3 − 3t2 + 3t − 1 = 0

of the given relation has t = 1 as a root of multiplicity three. The

sequences satisfying the recurrence relation xn+2 = 2xn+1 − xn,

whose characteristic equation has the root t = 1 of multiplicity

two, are, as we know (see Theorem 21.5), just the arithmetic pro-

gressions, that is, with nth term xn = an + b. Hence it is natural to

suspect that the sequence (xn) = (n2) might be one satisfying the

given relation. This is easy to check out, since, on the one hand,

(n + 3)2 = n2 + 6n + 9, and, on the other,

3(n + 2)2 − 3(n + 1)2 + n2 = 3n2 + 12n + 12

− 3n2 − 6n − 3 + n2 = n2 + 6n + 9.

However, there is another approach using an idea analogous to that

of the proof of Theorem 21.5. We set yn = xn+1 − xn, and note that

the given recurrence relation may be rewritten as

xn+3 − xn+2 = 2(xn+2 − xn+1) − (xn+1 − xn),

or yn+2 = 2yn+1 − yn. We infer that the sequence (yn) is an (arbi-

trary) arithmetic progression. Since xn − x0 =∑n−1

i=0 yi , and is

therefore equal to the sum of the first n terms of an arithmetic

progression, it will have quadratic form.

b) An example of a recurrence relation with the desired property

is xn+2 = 3xn − 2xn−1. The way to come up with this is to


write down a cubic polynomial having 1 as a root of multiplic-

ity two—for example, (t − 1)2(t + 2) = t3 − 3t + 2—, and then

write down the recurrence relation having this as its characteristic

equation.

21.5. By contrast with the examples of Theme 21, here the desired recurrence

relation is not linear. A sequence (xn) is a geometric progression pre-

cisely ifxn+1

xn= xn

xn−1, or xn+1 = x2

n

xn−1. Of course, we are here including

constant sequences as geometric.

21.6. To solve this we use the solution of Problem 3 of Theme 21, where

it was shown that the Fibonacci number Fn+1 gives the number of

ways of tiling a 2 × n strip with 2 × 1 dominoes. Any two such tilings

differ only in the different arrangements of “vertical” tiles and pairs

of “horizontal tiles” (with one tile above the other in each pair). Let k

be the number of such horizontal pairs in a tiling. Then since 2k ≤ n,

we have k ≤ [n/2]. For each such k we need to count the number

of ways of choosing k 2 × 2 blocks in the 2 × n strip. For this we

argue as follows. For any particular such choice, imagine that the

vertical dominoes are black and that each 2 × 2 block consisting of

two horizontal dominoes has been replaced by a single vertical white

domino. Thus we now have a tiling of a 2 × (n − k) strip by dominoes

placed vertically, of which k are white and the rest black. The number

of such tilings is obviously(n−k

k

). Since each such tiling corresponds

to precisely one choice of k 2 × 2 blocks from the original 2 × n

strip, the number of the latter choices is likewise(n−k

k

). Summing(

n−k

k

)over k = 0, 1, . . . , [n/2], then gives the total number of tilings

of the 2 × n strip, which we know from the solution to Problem 3 to

be Fn+1.

Theme 22. The “golden ratio” or solvingequations of the form f (x) = x

22.1. Consider the function f (x) = x+2x+1

= 1 + 1x+1

on the interval [1,+∞).

It is, obviously, everywhere decreasing on that interval and

f ′(x) = − 1(x+1)2 > −1. The equation f (x) = x has the unique solu-

tion c =√

2 . Since x1 = 1 <√

2 , it follows from Theorem 22.1

that the sequence x1, x3, x5, . . . is increasing and the sequence

x2, x4, x6, . . . decreasing—both converging to√

2.


It is worth mentioning in this connection that this sequence consists

of the successive finite continued fractions of the number√

2, since if

we define sn by

sn = 1 +1

2 +1

2 +1

2 + · · ·

(n − 1 fraction bars),

then

sn+1 = 1 +1

2 +1

2 +1

2 + · · ·

= 1 +1

1 + sn

=sn + 2

sn + 1,

and also s1 = 1.

22.2. a) An example of such a function is f (x) = 12

(x +

√x2 + 4

), whose

graph is sketched in Figure 120. This example shows that there is

Figure 120

no direct analogue of Theorem 22.1 for increasing functions.

b) That the equation f (x) = x has exactly one solution follows from

the fact that the function x − f (x) is increasing with slope at

least 1 − q > 0 at every point of its graph. For its derivative is

(x − f (x))′ = 1 − f ′(x) > 1 − q > 0 by assumption and it is not

difficult to prove (via the Mean-Value Theorem) that if a function

g defined on R has positive derivative bounded away from zero

then limx→±∞ g(x) = ±∞, so its graph must cross the x-axis, and

moreover at just one point.

22.3. Let f (c) = c. By the assumptions of the theorem c is then the only

solution of the equation f (x) = x.


a) In the case x1 = c, the sequence defined by the recurrence rela-

tion xn+1 = f (xn) is constant. In the case x1 < c we have, by

assumption, that x2 = f (x1) > x1, and then in turn x2 = f (x1) <

f (c) = c. Thus x1 < x2 < c. The same argument shows that if

xn < c then xn < xn+1 < c, so by induction we conclude that if

x1 < c then the sequence (xn) is increasing and bounded above by

the number c. In the case x1 > c, we have x2 = f (x1) < x1 and

x2 = f (x1) > f (c) = c, and we conclude similarly that in this case

the sequence (xn) is decreasing and bounded below by c.

b) As shown above, in the case x1 < c the sequence (xn) is increasing

and bounded above. Hence by Weierstrass’s theorem it converges.

Let a denote its limit. Proceeding to the limit in the equation xn+1 =f (xn), we obtain f (a) = a. However, since by the assumptions of

the problem the equation x = f (x) has just one solution, we must

have a = c, that is, xn → c. The argument in the case x1 > c is

completely analogous.

22.4. Let’s consider the general case of a function of the form f (x) =√a + x and x1 =

√a , where a is any positive number. We consider

the function f (x) only for x in the interval [0,+∞). This function

is then easily seen to satisfy the conditions of Problem 22.3, whence

we infer that the sequence (xn) defined by xn+1 = f (xn), x1 =√

a,

converges to the solution of the equation√

a + x = x, that is, to the

number 1+√

1+4a

2. Hence, in particular: a) xn → 1+

√5

2; b) yn → 3.

22.5. 1) By the Mean-Value Theorem and the given assumptions, we have

for all numbers a, b that

|f (b) − f (a)| = |f ′(c)| |b − a| ≤ q |b − a|.a) Hence |xn+1 − xn| = |f (xn) − f (xn−1)| ≤ q |xn − xn−1|.b) The desired inequality follows from the chain of inequalities

|xn+1−xn| ≤ q |xn−xn−1| ≤ q2|xn−1−xn−2| ≤· · ·≤ qn|x1−x0|.c) We have

|xn+k−xn|=|xn+k−xn+k−1 + xn+k−1−xn+k−2+· · ·+xn+1 − xn|

≤ |xn+k−xn!+!k−1| + |xn+k−1 − xn+k−!2| + · · · + |xn+1−xn|

≤(qn+k−1+qn+k−2+· · ·+qn

)|x1−x0|

≤ (qn+qn+1+· · · )|x1−x0|

=qn

1 − q|x1 − x0|.


2) From the inequalities of Part c) above (for all n, k) it follows that

the sequence (xn) is a so-called Cauchy sequence, that is—roughly

speaking—one whose terms get arbitrarily close to one another the

further along in the sequence one goes. It is a fundamental result of

mathematical analysis that every Cauchy sequence is convergent.

Thus suppose xn → c. Then proceeding to the limit in the equation

xn+1 = f (xn), we get c = f (c).

22.6. Before having your students work on this problem by themselves, try

to pique their interest—for instance, by showing them some pictures.

Figure 121a shows the behavior of the first seven terms of the sequence

of the given sort with first term chosen to be x1 = 2, and the same for

for the graph in Figure 121b except now with first term x1 = 5. It

seems pretty clear from these graphs that both sequences converge

to π . Figure 122 shows the sequence satisfying the given recurrence

1 2 3 4 5 6 7

2

1 2 3 4 5 6 7

4

(a) (b)

Figure 121

relation with first term x1 = −1. Here it would seem that the sequence

converges to −π . Thus although it might in principle be thought that

1 2 3 4 5 6 7

Figure 122

all solutions of the equation sin x = 0—all numbers of the form kπ—

should figure as limits of of sequences satisfying the given relation,


these examples seem to indicate that for some reason the sequences in

question “refuse” to tend to 0, for example.

It is easy to see that the function f (x) = x + sin x is increas-

ing on the whole real line and moreover that f (x) > x for all

x ∈ (0, π ). Hence 0 = f (0) < f (x1) < f (π ) = π (where we are tak-

ing x1 ∈ (0, π )), that is, 0 < x2 < π . Note also that f (x1) > x1, so

that x2 > x1. It follows by induction that for any x1 ∈ (0, π ) the

resulting sequence (xn) is strictly increasing, positive, and bounded

above by π . Then since its limit must be a number of the form

kπ , that limit must in fact be π . Now let’s consider what hap-

pens if, for instance, x1 ∈ (−π, 0). Since sin x < 0 on this interval,

we have f (x) < x, whence x2 = f (x1) < x1. An induction shows

that in this case we obtain a decreasing sequence tending to −π .

The general answer is as follows: if x1 ∈ (2πk, π + 2πk), then the

sequence (xn) is increasing and converges to π + 2πk, while if x1 ∈(π + 2πk, 2π + 2πk), the sequence (xn) is decreasing and tends to

π + 2πk.

A question worth asking the students is the following one: “How

can the conditions of this problem be changed so that the analo-

gous sequences converge to numbers of the form 2πk?” Answer:

Consider the sequences satisfying the recurrence relation xn+1 =xn − sin xn.

Theme 23. Convex functions: inequalitiesand approximations

23.1. Since f ′(0) = n, the straight line with equation t = 1 + nx is tangen-

tial to the graph of y = (1 + x)n at the point (0, 1). Since the function

f (x) is convex, its graph lies above its tangent line, whence Bernoulli’s

inequality (1 + x)n ≥ 1 + nx for all x > −1.

a) The above argument goes through for the more general function

f (x) = (1 + x)α with α > 1. Hence (1 + x)α ≥ 1 + αx for all x ≥−1 provided α > 1. Figure 123a gives the geometric picture for

α = 43

.

b) For 0 < α < 1 the function f (x) = (1 + x)α is concave on the

interval (−1,∞), so that for all x ≥ −1 the inequality (1 + x)α ≤1 + αx holds. (In Figure 123b, α = 3

4.)


(a) (b)

Figure 123

23.2. The standard special case of Jensen’s inequality for the function f (x) =1x

, convex on the interval (0,+∞), has the form

n

x1 + x2 + · · · + xn

≤1x1

+ 1x2

+ · · · + 1xn

n.

This may be rewritten as the inequality

n1x1

+ 1x2

+ · · · + 1xn

≤x1 + x2 + · · · + xn

n

between the harmonic and arithmetic means of n positive numbers.

However, in the author’s view this inequality is more naturally proved

by rewriting it in the form

(x1 + x2 + · · · + xn)

(1

x1

+1

x2

+ · · · +1

xn

)≥ n2

and appealing to the Cauchy-Schwarz inequality. Or one may establish

it as follows. Consider the inequality

x1 + x2 + · · · + xn

n≥ n

√x1x2 · · · xn ,

between the arithmetic and geometric means of n positive numbers,

and the same inequality, but this time applied to the reciprocals of

those numbers:

1x1

+ 1x2

+ · · · + 1xn

n≥

1

n√

x1x2 · · · xn

.

It is then immediate from these two inequalities that

n1x1

+ 1x2

+ · · · + 1xn

≤ n√

x1x2 · · · xn ≤x1 + x2 + · · · + xn

n.


23.3. We raise both sides of the inequality

(x

p

1 + xp

2 + · · · + xpn

n

) 1p

≤(

xq

1 + xq

2 + · · · + xqn

n

) 1q

,

to the power p and rewrite the resulting inequality in terms of new

variables ti defined by xq

i = ti , i = 1, 2, . . . , n. Then since xi = t1/q

i ,

we have xp

i = tp/q

i , and the above inequality takes the form

tp

q

1 + tp

q

2 + · · · + tp

q

n

n≤

(t1 + t2 + · · · + tn

n

) p

q

.

It now only remains to observe that this inequality is Jensen’s inequality

as determined by the concave function f (x) = xp/q .

23.4. a) Since (sin x)′′ = − sin x < 0 for x ∈(0, π

2

], the sine function is

concave on that interval, so that its graph lies above every chord with

end-points having abscissas in the interval [0, π ]—a fact familiar

from sketches of the graph of y = sin x! In particular, the graph

of the sine function lies above the chord joining the points (0, 0)

and (π2, 1). Since this chord has equation y = 2x

πit follows that

sin x > 2xπ

for all x ∈(0, π

2

).

b) The idea is similar to that of the proof of the item a). The graph

of the sine function lies above the chord joining the points (0, 0)

and(

π6, 1

2

). Since this chord has equation y = 3x

πit follows that

sin x > 3xπ

for all x ∈(0, π

6

).

23.5. We proved in Part b) of the preceding problem that sin x > 3xπ

for

all x ∈(0, π

6

). Thus, in particular, sin π

30> 3

π· π

30= 1

10. Hence b =

a2 sin π

30

< 5a.

aa a a

a

b

b

b

Figure 124


Here is another—geometric—solution. Lay off five copies of the given

triangle in the plane in such a way that neighboring copies have a side

of length b in common (as in Figure 124).

Since 5 · 12◦ = 60◦, the distance between the furthermost endpoints

of the bases of the first and fifth of these triangles is equal to b, which

is therefore less than the length of the broken line made up of the bases

of the five triangles, namely 5a.

Theme 25. Derivatives of vector-functions

25.1. Exercise1is just amatterofdifferentiating. InExample1,wehave f ′(t)=(a1, b1, c1). In Example 2, f ′(t) = (−aω sin ωt, aω cos ωt, 0), whence

| f ′(t)| =√

a2ω2 sin2 ωt + a2ω2 cos2 ωt = aω. In Example 3 we have

f ′(t) = (−aω sin ωt, aω cos ωt, b), whence | f ′(t)| =√

a2ω2 + b2.

Furthermore, the angle between the velocity vector and the z-axis has

cosine b√a2ω2+b2

and is therefore constant.

Exercise 2 may be done componentwise using the known rules for

differentiating an ordinary real-valued function, or directly from the

definition as the limit of a difference quotient, in which case the proofs

resemble the ones for real-valued functions. We give both proofs.

1) We have

(f (t) + g(t)

)′ = limh→0

1

h

[( f (t + h) + g(t + h)) − ( f (t) + g(t))

]

= limh→0

1

h

[( f (t + h) − f (t)) + (g(t + h) − g(t))

]

= limh→0

( f (t + h) − f (t)

h

)+ lim

h→0

( g(t + h) − g(t)

h

)

= f ′(t) + g′(t).

And here is the componentwise argument. Since, relative to

some Cartesian coordinate system, we have f (t) + g(t) =(f1(t) +

g1(t), f2(t) + g2(t), f3(t) + g3(t)), it follows that

(f (t) + g(t)

)′ =(f ′

1(t) + g′1(t), f ′

2(t) + g′2(t), f ′

3(t) + g′3(t)

)

=(f ′

1(t), f ′2(t), f ′

3(t))+

(g′

1(t), g′2(t), g′

3(t))

= f ′(t) + g′(t).


2) We have

(ϕ(t) f (t)

)′ = limh→0

1

h

[ϕ(t + h) f (t + h) − ϕ(t) f (t)

]

= limh→0

1

h

[ϕ(t + h) f (t + h) + ϕ(t) f (t + h)

− ϕ(t) f (t + h) − ϕ(t) f (t)]

= limh→0

1

h

[(ϕ(t + h) − ϕ(t)) f (t + h) + ϕ(t)( f (t + h) − f (t))

]

= limh→0

[ϕ(t + h) − ϕ(t)

hf (t + h)

]+ lim

h→0

[ϕ(t)

f (t + h) − f (t)

h

]

= ϕ′(t) f (t) + ϕ(t) f ′(t).

For the componentwise argument, note first that

ϕ(t) f (t) =(ϕ(t)f1(t), ϕ(t)f2(t), ϕ(t)f3(t)

).

Hence

(ϕ(t) f (t)

)′ =(ϕ′(t)f1(t) + ϕ(t)f ′

1(t), ϕ′(t)f2(t) + ϕ(t)f ′2(t),

ϕ′(t)f3(t) + ϕ(t)f ′3(t)

)

= ϕ′(t)(f1(t), f2(t), f3(t)

)+ ϕ(t)

(f ′

1(t), f ′2(t), f ′

3(t))

= ϕ′(t) f (t) + ϕ(t) f ′(t).

3) The proof that(

f (t) · g(t))′ = f ′(t) · g(t) + f (t) · g′(t) is similar

to the preceding one. One inserts f (t) · g(t + h) − f (t) · g(t + h)

into the numerator of the difference quotient and breaks the limit

up into two limits (and uses the fact that the dot product distributes

over vector addition).

For the componentwise argument, note first that

f (t) · g(t) = f1(t)g1(t) + f2(t)g2(t) + f3(t)g3(t).

Hence

(f (t) · g(t)

)′ = f ′1(t)g1(t) + f1(t)g′

1(t) + f ′2(t)g2(t) + f2(t)g′

2(t)

+ f ′3(t)g3(t) + f3(t)g′

3(t)

= f ′(t) · g(t) + f (t) · g′(t).


4) If f (t) is a constant vector on an interval of values of t , then

limh→0

f (t + h) − f (t)

h= lim

h→0

0

h= 0.

For the converse we use the representation of f (t) in terms

of components relative to some coordinate system, say f (t) =(f1(t), f2(t), f3(t)

). Then if f ′(t) = 0 for all t in an interval,

it follows that f ′i (t) = 0 (i = 1, 2, 3) on that interval, whence

fi(t) = const (i = 1, 2, 3) (the proof of which requires the Mean-

Value Theorem!). Hence f (t) = const .

25.2. Consider a point-particle moving in the plane of the triangle ABC,

and passing through the point M in question at time t0. Write Mt

for the position of the particle at time t . We define vector-functions

f (t), g(t) and h(t) by f (t) = MtA, g(t) = MtB and h(t) = MtC,

and write ϕ(t) = | f (t)| + |g(t)| + |h(t)|. By assumption, the func-

tion ϕ(t) attains its least value at the time t = t0, so we must have

ϕ′(t0) = 0. Since g(t) = MtB = MtA + AB = f (t) + AB, it follows

that g′(t) = f ′(t), and, similarly, h′(t) = f ′(t). Let v denote the value

of the derivative of these vector-functions at time t = t0, and by e1,

e2 and e3 the unit vectors parallel to the vectors MA, MB and MC

respectively. Then

0 = ϕ′(t0) = v · e1 + v · e2 + v · e3 = v · (e1 + e2 + e3).

Since the value of the velocity vector v at t0 may clearly be arranged

to be nonzero and not perpendicular to e1 + e2 + e3, we infer from

the fact that v · (e1 + e2 + e3) = 0 that e1 + e2 + e3 = 0. However,

the only way three planar unit vectors can sum to zero is if the angle

between each two of them is 120◦.

However, this argument is defective. It nowhere uses the assumption

in the problem that the triangle ABC be acute-angled. What has been

overlooked in the above argument is that if the point M should coincide

with one of the vertices of the triangle, then the function ϕ would

not be differentiable at t = t0 and Fermat’s result that the derivative

vanishes at a point where a function has a local extreme value would

be inapplicable.

Thus we need to complete the argument. It is not difficult to show

that if the triangle ABC is acute-angled, then the point M for which

the sum of its distances from the vertices is least cannot coincide with

a vertex.


For the sake of completeness, we give the answer also in the case

where the triangle is not acute-angled. If its angles are still all less than

120◦, then the answer is as in the acute-angled case. If, on the other

hand, one of the angles is greater than or equal to 120◦, then the vertex

where the triangle has that angle is the point such that the sum of the

distances from it to the three vertices is least.

25.3. From f ′(t) = f (t) × h and a basic property of the cross product, it

follows that f ′(t) ⊥ f (t) and f ′(t) ⊥ h. Hence f ′(t) · f (t) = 0 and

f ′(t) · h = 0, whence | f (t)| = const and f (t) · h = const . Hence a

point-particle with motion given by f (t) is restricted to a sphere and

also to a plane, whence it follows that it must move in a circle. Further-

more, in view of the constancy of the angle between the vectors f (t)

and h (taken together with the given equation for f ′(t)), the quantity

| f ′(t)| is also constant. Hence if we introduce a rectangular Carte-

sian coordinate system with the z-axis parallel to the vector h, then in

terms of components referred to that frame, f (t) will have the form

(a cos ωt, a sin ωt, b).

25.4. We introduce a rectangular Cartesian coordinate system with z-axis

parallel to the vector H . Then from the solution of the preceding prob-

lem we have that the derivative f ′(t) of the vector-function f (t) giving

the motion of the electron, has components (a cos ωt, a sin ωt, b) rel-

ative to that frame. Hence in terms of its component functions, the

vector-function f (t) will have the form

(x0 − a

ωsin ωt, y0 + a

ωcos ωt, z0 + bt

),

showing that the electron does indeed move along a helix.

Theme 26. Polynomials andtrigonometric relations

26.1. In solving Problem 2 of this Theme, we showed that the numbers

x1 = cos 2π5

and x2 = cos 4π5

are both roots of multiplicity two of the

polynomial 16x5 − 20x3 + 5x − 1, the fifth of whose roots is 1. Hence

by Viete’s theorem we have x21x2

2 = 116

. Then since x1 > 0 and x2 < 0,

we conclude that x1x2 = − 14

.

26.2. It follows from the proof of Problem 3 that the square of the product in

question is equal to 2n + 1. Since all factors in the product are positive,

we conclude that it must be equal to√

2n + 1 .


26.3. Since Tn(x) = 2xTn−1(x) − Tn−2(x), increasing n by 1 results in the

multiplication of the leading coefficient by the factor 2. Then since

T1(x) = x, the leading coefficient of the polynomial Tn(x) must be

2n−1.

26.4. (1) Since Tn(cos t) = cos nt , we have Tn(xk) = cos πk = (−1)k .

(2) if x ∈ (xk+1, xk) then x = cos t for some t ∈(

kπn

, (k+1)π

n

).

Since Tn(cos t) = cos nt , we have T ′n(cos t) sin t = n sin nt . Obviously

sin t > 0 on these intervals. And since kπ < nt < (k + 1)π , the func-

tion sin nt does not change sign on the interval(

kπn

, (k+1)π

n

). Hence the

derivative T ′n(x) preserves its sign on each of the intervals (xk+1, xk)

and Tn(x) is therefore monotonic on each of them. Since it is cer-

tainly continuous, Tn(x) is therefore monotonic on each closed interval

[xk+1, xk].

26.5. 1) Here one needs to “face the music” and verify directly that Tn(x) sat-

isfies the given differential equation—by means of differentiation and

algebraic manipulation—on each of the intervals (−∞,−1), (−1, 1)

and (1,+∞) separately. On the first and third of these one can use

the explicit form of Tn(x) given by Lemma 26.3, and on the interval

(−1, 1) the definition of the Chebyshev polynomials.

2) On making the substitution x = cos t , t ∈ [0, π ], in the given inte-

gral, we obtain the integral∫ π

0cos kx cos nx dx, and showing that this

vanishes is standard:

∫ π

0

cos kx cos nx dx =1

2

∫ π

0

(cos(k + n)x + cos(k − n)x

)dx

=1

2

(sin(k + n)x

k + n+

sin(k − n)x

k − n

)∣∣∣∣π

0

= 0.

26.6. Since the left-hand side of the given identity is a polynomial, it is

enough to prove that it holds on the interval [−1, 1]. Thus it suffices

to verify the identity

T 2n (cos t) − (cos2 t − 1)U 2

n−1(cos t) = 1.

Here the left-hand side is

cos2 nt +(sin t Un−1(cos t)

)2 = cos2 nt + sin2 nt = 1,

completing the verification.


26.7. 1) Since Un−1(x) = 1n

T ′n(x), we have

Un−1(cos t) sin t =1

nT ′

n(cos t) sin t

= −1

n

(Tn(cos t)

)′ = −1

n(cos nt)′ = sin nt.

2) Making the substitution x = cos t , t ∈ [0, π ], in the given integral,

we get the integral

∫ π

0

Uk(cos t)Un(cos t) sin2 t dt

=∫ π

0

sin(k + 1)t sin(n + 1)t dt

=1

2

∫ π

0

(cos(k − n)t

− cos(k + n + 2)t)dt = 0.

Theme 27. Areas and volumes as functionsof co-ordinates

27.1. let A(x1, y1) be any point on the line ℓ. The point B(x2, y2) with

coordinates x2 = x1 − b and y2 = y1 + a is then also on the line ℓ

since

ax2 + by2 + c = a(x1 − b) + b(y1 + a) + c = ax1 + by1 + c = 0.

Consider the triangle ABM . The altitude of this triangle dropped from

the vertex M is the desired distance d from M to the line ℓ. The

formula for the Euclidean distance between two points gives |AB| =√a2 + b2 , whence

SABM =1

2

√a2 + b2 · d(M, ℓ).

On the other hand, by the formula (2) this area is equal to

1

2|(x0 − x1)(y2 − y1) − (y0 − y1)(x2 − x1)| =

1

2|a(x0 − x1) + b(y0 − y1)|

=1

2|ax0 + by0 − (ax1 + by1)|

=1

2|ax0 + by0 + c|.


Hence

1

2

√a2 + b2 · d(M, ℓ) =

1

2|ax0 + by0 + c|,

whence the desired formula.

27.2. Since any polygon can be dissected into triangles, it suffices to prove

the formula (6) in the case that M is a triangle, with vertices A,B,C,

say. Since the area of a triangle is unchanged by translations we may

without loss of generality assume that the vertex C is the origin.

This assumed, consider the other vertices A(x1, y1), B(x2, y2) and their

images A1(a1x1 + b1y1, a2x1 + b2y1), B1(a1x2 + b1y2, a2x2 + b2y2)

under the mapping �. (Note that of course � fixes C.) Then since

(a1x1 + b1y1)(a2x2 + b2y2) − (a2x1 + b2y1)(a1x2 + b1y2)

= a1a2x1x2 + a1b2x1y2 + a2b1x2y1 + b1b2y1y2a1a2x1x2

− a2b1x1y2 − a1b2x2y1 − b1b2y1y2

= (a1b2 − a2b1)x1y2 − (a1b2 − a2b1)x2y1

= (a1b2 − a2b1)(x1y2 − x2y1),

the desired formula follows from the formula (1).

27.3. Under the translation taking the vertex D to the origin, the other

three vertices are sent to the points A1(x1 − x4, y1 − y4, z1 − z4),

B1(x2 − x4, y2 − y4, z2 − z4) and C1(x3 − x4, y3 − y4, z3 − z4), and

the volume of the original tetrahedron ABCD is the same as that of

the tetrahedron OA1B1C1. Consider the parallelepiped determined by

the edges OA1, OB1 and OC1. The area of the triangle OA1B1 is then

half that of the parallelogram defined by the edges OA1 and OB1, and

since the height of the parallelepiped as measured from the paralel-

logram containing the triangle OA1B1 is the height of the tetrahedron

as measured from that face as base, we obtain (as in the solution of

Problem 1 of Theme 15)

VABCD =1

3·

1

2V0 =

1

6V0,

where V0 is the volume of the parallelepiped. Hence from the for-

mula (3) we get the following formula for the volume of the given


tetrahedron:

VABCD =1

6

∣∣∣∣∣∣

∣∣∣∣∣∣

x1 − x4 y1 − y4 z1 − z4

x2 − x4 y2 − y4 z2 − z4

x3 − x4 y3 − y4 z3 − z4

∣∣∣∣∣∣

∣∣∣∣∣∣.

27.4. Consider the vectors a(a1, a2) and b(b1, b2). Part a) is immediate from

the formula (5) with one pair of coordinates set equal to zero in that

formula, that is, from the formula

S(a, b) = a1b2 − a2b1

for the oriented area of the parallelogram.

For the proof of Part b) we need the additional vector c(c1, c2). Note

that S(d, d) = 0 for every vector d. The given system is xa + yb = c.

Hence

S(c, b) = S(xa + yb, b) = xS(a, b) + yS(b, b) = xS(a, b),

whence we get, via the formula (5), the desired formula for x. The

formula for y is obtained similarly.

Theme 28. Values of trigonometric functionsand sequences satisfying a certain recurrencerelation

28.1. We write xk = cos 2kπ2n+1

, as in Problem 9. We wish to prove that x1 �= xk

for all k = 2, 3, . . . , n. If x1 = xk , then

2kπ

2n + 1= ±

2π

2n + 1+ 2πm, m ∈ Z.

Simplifying, we get

2k−1 = ±1 + m(2n + 1), or 2k−1 ± 1 = m(2n + 1),

which is impossible since the number 2k−1 ± 1 is positive and less than

2n + 1.

28.2. We know that the first terms of periodic sequences of the sort in

question must lie in the interval [−1, 1], so we may write x1 = cos α,

and then xn+1 = cos 2nα. Hence x1 = xn+1 if and only if 2nα = ±α +2πk, whence

α =2πk

2n ± 1, and x1 = cos

2πk

2n ± 1.


28.3. Consider the interval (u, v) with u = arccos β and v = arccos α. Thus

cos x ∈ (α, β) for all x ∈ (u + 2πm, v + 2πm), m ∈ Z. Choose the

natural number n large enough for 2π2n+1

< v − u to hold, and choose

the natural number m so that 2π2n+1

< u + 2πm. The common difference

of the arithmetic progression with kth term sk = 2πk2n+1

, is then less

than the length of the interval (u + 2πm, v + 2πm), so there exists

a natural number k such that u + 2πm < 2πk2n+1

< v + 2πm. Thus by

construction the number x = cos 2πk2n+1

is the first term of a periodic

sequence of the sort in question and lies in the interval (α, β).

28.4. Admittedly, the conditions of this problem are such that it looks rather

formidable. However, someone who has read the earlier Themes atten-

tively may guess from the shape of the graph that Chebyshev poly-

nomials (see Theme 26) are involved. Writing f (x) = 2x2 − 1 (the

Chebyshev polynomial T2(x)), we define

fn(x) = f (f (· · · (f︸︷︷︸n times

(x) · · · ).

The polynomial fn (the Chebyshev polynomial T2n(x)) has degree 2n,

and the polynomial given in the problem is just f3(x). Hence the abscis-

sas of the points of intersection of the graph of f3(x) with the line y = x

are the solutions of the equation f3(x) = x. Let a be any of these solu-

tions, and consider the sequence (xn) defined by the recurrence relation

xn+1 = 2x2n − 1 = f (xn) together with the specification x1 = a. Then

x2 = f (x1), x3 = f (x2) = f (f (x1)), and x4 = f3(x1) = f3(a) = a, so

that x4 = x1. Clearly, f (x) = x implies f3(x) = x, so the solutions of

the equation f3(x) = x include the numbers 1 and − 12, the first terms

of the constant sequences of the sort in question. For all sequences

with first term a solution of this equation other than 1 and − 12, we will

have x2 �= x1. But is it possible that x3 = x1? This would imply that

f (f (x1)) = x1, whence x1 = f (f (f (x1))) = f (x1), an impossibility.

We conclude that the roots other than 1 and − 12

are first terms of

sequences of the sort in question of period 3.

28.5. In the notation of the preceding solution, we have fn(x) = T2n(x),

the 2nth Chebyshev polynomial of the first kind. By Lemma 26.5,

on each subinterval [xk+1, xk] of the interval [−1, 1], where xk =cos πk

2n , k = 0, 1, 2, . . . , 2n, the function fn(x) is monotonic. Further-

more the image under fn of each of these subintervals is [−1, 1].

Hence the equation fn(x) = x has at least one solution in each of these


subintervals. Since the number of such subintervals is 2n, which is also

the degree of fn(x), the equation fn(x) = x has precisely 2n real roots.

By the solution of the previous problem the roots of the equation

fn(x) = x are first terms of sequences of the required sort with xn+1 =xn. Hence the period d of such a sequence must be a divisor of n.

28.6. From the solution of the previous problem and the inclusion-exclusion

principle (see Problem 2 of Theme 24) it follows that the number of

solutions yielding sequences of period exactly n is

2n −∑

ki |n

2ki +∑

ki>kj |n

2(ki ,kj ) −∑

...

2(ki ,kj ,kl ) + · · · ,

where ki, kj , . . . are the positive divisors of n, and (ki, kj , . . .) denotes

the greatest common divisor of the numbers between the parentheses.

Theme 29. Do there exist further “numbers”beyond complex numbers?

29.1. a) We have

uu = (a + bi + cj + dk)(a − bi − cj − dk)

= a2 + b2 + c2 + d2 + a(bi + cj + dk) − a(bi + cj + dk)

− bcij − bcji − bdik − bdki − cdjk − cdkj

= a2 + b2 + c2 + d2 = |u|2.

b) Unquestionably, all one needs do here is calculate carefully. Since

we shall be needing the formula for the product of two quaternions a

little further on, let’s first do that calculation. If

u = a + bi + cj + dk and v = A = Bi + Cj + Dk,

then

uv = aA − bB − cC − dD + (aB + bA + cD − dC)i

+ (aC + cA + dB − bD)j + (aD + dA + bC − cB)k.

Then in order to get the formula for the product vu in the other order, all

we have to do is switch the big and small letters in the above formula.


Thus

vu = aA − bB − cC − dD + (aB + bA + dC − cD)i

+ (aC + cA + bD − dB)j + (aD + dA + cB − bC)k.

Then in order to get the formula for the product v u, one needs to

introduce minus signs in front of each of the letters b, B, c, C, d,D.

This yields

v u = aA − bB − cC − dD + (−aB − bA + dC − cD)i

+ (−aC − cA + bD − dB)j + (−aD − dA + cB − bC)k.

It then only remains to observe that

uv = aA − bB − cC − dD − (aB + bA + cD − dC)i

− (aC + cA + dB − bD)j − (aD + dA + bC − cB)k = v u.

c) Using the equalities established in the previous two solutions

together with the associativity of the multiplication of quaternions,

we get

|uv|2 = (uv)(uv) = (uv)(v u) = (u(vv))u = (uu)|v|2 = |u|2|v|2.

d) From the above formula for the product of two quaternions it follows

that

|uv|2 = (aA − bB − cC − dD)2 + (aB + bA + cD − dC)2

+ (aC + cA + dB − bD)2 + (aD + dA + bC − cB)2,

and since

|u|2|v|2 = (a2 + b2 + c2 + d2)2(A2 + B2 + C2 + D2)2,

we infer via the result of Part c) the identity

(a2 + b2 + c2 + d2)2(A2 + B2 + C2 + D2)2

= (aA − bB − cC − dD)2 + (aB + bA + cD − dC)2

+ (aC + cA + dB − bD)2 + (aD + dA + bC − cB)2.

29.2. Writing u1 = b1i + c1j + d1k and u2 = b2i + c2j + d2k, we have

u1u2 = −(b1b2 + c1c2 + d1d2) + (b1c2 − b2c1)i

+(a2c1 − a1c2)j + (a1b2 − a2b1)k.


The very notation i, j and k suggests that we should look for the

desired geometric interpretation in the vector space of space vectors,

that is, consider u1 and u2 as vectors u1(a1, b1, c1) and u2(a2, b2, c2).

The quantity b1b2 + c1c2 + d1d2 is then the dot product u1 · u2 of the

vectors u1 and u2, while the vector

(b1c2 − b2c1, a2c1 − a1c2, a1b2 − a2b1)

is one that has appeared several times before, for instance in Theme 27.

It is perpendicular to both of the vectors u1 and u2 and has magnitude

equal to the area of the parallelogram determined by those vectors. Of

course, this is none other than the cross product u1 × u2. We conclude

that if we identify the purely imaginary quaternions with the vectors

of Euclidean R3, then we shall have u1u2 = −u1 · u2 + u1 × u2.

Index

acceleration, 189

centripetal, 193

area

of the faces of a tetrahedron, 11

of planar regions, 129

of polygons, 215

of quadrilaterals, 66, 85, 262

of a trapezoid, 85, 131

of triangles, 85, 207, 208, 214, 263

signed, 214

Bernoulli’s inequality, 176, 288

Binet’s formula, 154

binomial

coefficients, 119, 160, 178

relations between, 120

theorem, 178

Cauchy sequence, 286

Cauchy’s mean-value theorem, 20, 180

Cauchy-Schwarz inequality, 105, 267

Ceva’s theorem, 88

Chebyshev polynomials

of the first kind, 202, 299

of the second kind, 204

properties of, 202, 203

circumcircle, 102

combinatorics, 119, 155, 184

complex numbers, 72, 73, 76, 79, 158, 224,

258

congruence of numbers with respect to a

modulus, 114

continued fraction(s), 163, 284

convergent sequences, 23, 143, 162, 163,

164, 174, 175, 281

Cramer’s rule, 216

curvature, 196

cyclic quadrilateral, 66

de Moivre’s theorem, 73, 79, 200, 256

determinant, 213

derivative, 30, 33, 53, 55, 57, 63, 66, 178,

233, 251

definition of, 188, 250

of vector-functions, 187

diophantine equations, 111, 271

distance from a point to a line, 215

division with remainder, 112, 115, 257, 272

divisors, 111

equation

of a circle, 7, 41, 43

of an ellipse, 34, 43

of a parabola, 9, 39

of a straight line, 8, 41

of a union of sets, 45, 243

set given by, 39, 243

Euler’s formula for complex exponents, 183

exponent rule, 143, 183

Fermat’s optical principle, 192

Fibonacci numbers, 153, 160, 162, 284

relations between, 155

field, 224

axioms of a, 223

first nontrivial limit, 19

functional equation, 143, 146

of exponential functions, 144

of linear functions, 143

of power functions, 144

functions

continuous, 143

contracting, 166, 286

convex, 167

differentiable, 146, 164

Fundamental Theorem

of Algebra, 224

of Calculus, 129, 134, 137

golden ratio, 161

graph of a function, 30, 33, 53, 55, 231, 233

helix, 188, 197

Heron’s formula, 102, 207

incircle, 89, 102

inclusion-exclusion formula for sets, 185

303

304 Index

inequalities

between arithmetic and geometric means

(a.m.–g.m. inequality), 101, 102, 103,

269

generalized version, 172

between arithmetic and harmonic means,

108, 269, 288

between arithmetic and quadratic means,

105

geometric interpretation of, 132

integral, 20, 129

application of, 131, 276

evaluation of, 130, 135, 138

Jensen’s inequality, 170

Law of Cosines, 66, 68, 208

generalization of, 266

length of a circle, 17

logarithm, 29, 250

means of two numbers, 101, 108, 167,

251

mean-value theorem, 165, 251, 286

Minkowski’s inequality, 106

multiplicity of a root, 57

Napoleon’s problem, 80

Newton’s tangent method, 24, 174

number e, 29, 186

optical property of ellipses, 191

orthogonal projection, 11, 211

parallelepiped, 93

Pell’s equation, 4, 205

periodic sequence(s), 219

permutation, 184

polynomials, 55

divisibility of, 71

factorizing of, 66, 226, 256

multiple roots of, 49, 57, 178, 247

roots of, 49, 63, 66, 199, 201, 243

prime number(s), 115, 116

probability, 123, 124, 184

pyramid, 254

Pythagoras’ theorem, 3, 17


Pythagorean triple(s), 3

quaternions, 229, 300

radius of curvature, 196

rational approximations, 143, 186

recurrence relation, 23, 151, 153, 162, 164,

202, 218

characteristic equation of, 156

Rolle’s theorem, 20, 169

roots of unity. 73, 74, 79, 81, 256, 258

rotation, 79, 80, 259

Schwarz’s inequality, 137, 278

signed area, 214

Simpson’s rule, 138, 278, 280

Snell’s law of refraction, 192

solid of revolution, 134

speed, 108, 189

of light in a medium, 192

tangent line, 24, 57, 60, 169, 191, 231, 247

Taylor polynomial, 179

Taylor series

for the exponential function, 181

for the sine and cosine functions, 181

Taylor’s theorem, 19, 180

for polynomials, 177

tetrahedron, 11, 93, 125

tiling, 153, 155, 284

tower of Hanoi, 151

triangle inequality, 11, 290


trigonometric relations, 73, 76, 199, 201, 258

values of trigonometric functions, 73, 217

vector-functions, 187

coordinates of, 187

vectors, 12, 79, 259, 265

cross product of, 96, 98, 195, 212, 301

non-coplanar, 15

velocity, 188

Viete’s formula, 49, 200, 201, 244

volume

of a cone, 135

of a doughnut, 135

of a parallelepiped, 93, 98, 212

of a solid of revolution, 134

of a tetrahedron, 93, 99, 266

Weierstrass’ theorem on continuous

functions, 225

Young’s inequality, 132

NML/52

For additional information

and updates on this book, visit

www.ams.org/bookpages/nml-52

AMS / MAA ANNELI LAX NEW MATHEMATICAL LIBRARY

Portal through Mathematics is a collection of puzzles and problems mostly

on topics related to secondary mathematics.

The problems and topics are fresh and interesting and frequently

surprising. One example: the puzzle that asks how much length must be

added to a belt around the Earth’s equator to raise it one foot has probably

achieved old chestnut status. Ivanov, after explaining the surprising answer

to this question, goes a step further and asks, if you grabbed that too long

belt at some point and raised it as high as possible, how high would that

be? The answer to that is more surprising than the classic puzzle’s answer.

The book is organized into twenty-nine themes, each a topic from algebra,

geometry or calculus and each launched from an opening puzzle or

problem. There are excursions into number theory, solid geometry, physics

and combinatorics. Always there is an emphasis on surprise and delight.

And every theme begins at a level approachable with minimal background

requirements. With well over 250 puzzles and problems, there is some-

thing here sure to appeal to everyone.

Portal through Mathematics will be useful for prospective secondary

teachers of mathematics and may be used (as a supplementary resource)

in university courses in algebra, geometry, calculus, and discrete math-

ematics. It can be also used for professional development for teachers

looking for inspiration. However, the intended audience is much broader.

Every fan of mathematics will fi nd enjoyment in it.

Portal through Mathematics - American Mathematical Society

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Transcript of Portal through Mathematics - American Mathematical Society