360 problems for mathematical contests [andreescu] 9739417124 (1)

'I'D ANDREESCD DORIN ANDRICA

360

Problems for

Mathematical Contests

TITU ANDREESCU DORIN ANDRICA

360 Problems

for Mathematical Contests

GIL Publishing House

© GIL Publishing House ISBN 973-9417-12-4

360 Problems for Mathematical Contests Authors: Titu Andreescu, Dorin Andrica

Copyright © 2003 by Gil. All rights reserved.

GIL Publishing House P.o. Box 44, Post Office 3, 4700, Zalau, Romania,

tel. (+40) 260/616314

fax. (+40) 260/616414

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Contents

FOREWORD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 FROM THE AUTHORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Chapter 1 . ALGEBRA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Chapter 2. NUMBER THEORY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Chapter 3. GEOMETRY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

Chapter 4. TRIGONOMETRY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

Chapter 5. MATHEMATICAL ANALYSIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

Chapter 6. COMPREHENSIVE PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

FOREWORD

I take great pleasure in recommending to all readers - Romanians or from abroad the book of professors Titu Andreescu and Dorin Andrica. This book is the fruit of a

prodigious activity of the two authors, well-known creators of mathematics questions for Olympiads and other mathematical contests. They have published innumerable original problems in various mathematical journals.

The book is organized in six chapters: algebra, number theory, geometry, trigonometry, analysis and comprehensive problems. In addition, other fields of mathematics found their place in this book, for example, combinatorial problems can be found in the last chapter, and problems involving complex numbers are included in the trigonometry section. Moreover, in all chapters of this book the serious reader can find numerous challenging inequality problems. All featured problems are interesting, with an increased level of difficulty; some of them are real gems that will give great satisfaction to any math lover attempting to solve or even extend them.

Through their outstanding work as jury members of the National Mathematical Olympiad, the Balkan Mathematics Contest (BMO) , and the International Mathematical Olympiad (IMO) , the authors also supported the excellent results of the Romanian contestants in these competitions. A great effort was given in preparing lectures for summer and winter training camps and also for creating original problems to be used in selection tests to search for truly gifted mathematics students. To support the claim that the Romanian students selected to represent the country were really the ones to deserve such honor, we note that only two mathematicians of Romanian origin, both former IMO gold-medalists, were invited recently to give conferences at the International Mathematical Congress: Dan Voiculescu (Zurich, 1994) and Daniel Tataru (Beijing, 2002) . The Romanian mathematical community unanimously recognized this outstanding activity of professors Titu Andreescu and Dorin Andrica. As a consequence, Titu Andreescu, at that time professor at Loga Academy in Timi§oara and having students on the team participating in the IMO, was appointed to serve as deputy leader of the national team. Nowadays, Titu's potential, as with other Romanians in different fields, has been fully realized in the United States, leading the USA team in the IMO, coordinating the training and selection of team contestants and serving as member of several national and regional mathematical contest juries.

One more time, I strongly express my belief that the 360 mathematics problems featured in this book will reveal the beauty of mathematics to all students and it will be a guide to their teachers and professors.

Professor loan Tomescu Department of Mathematics and Computer Science University of Bucharest Associate member of the Romanian Academy

FROM THE AUTHORS

This book is intended to help students preparing for all rounds of Mathematical Olympiads or any other significant mathematics contest. Teachers will also find this work useful in training young talented students.

Our experience as contestants was a great asset in preparing this book. To this we added our vast personal experience from the other side of the " barricade" , as creators of problems and members of numerous contest committees.

All the featured problems are supposed to be original. They are the fruit of our collaboration for the last 30 years with several elementary mathematics journals from all over the world. Many of these problems were used in contests throughout these years, from the first round to the international level. It is possible that some problems are already known, but this is not critical. The important thing is that an educated - to a certain extent - reader will find in this book problems that bring something new and will teach new ways of dealing with key mathematics concepts, a variety of methods, tactics, and strategies.

The problems are divided in chapters, although this division is not firm, for some of the problems require background in several fields of mathematics.

Besides the traditional fields: algebra, geometry, trigonometry and analysis, we devoted an entire chapter to number theory, because many contest problems require knowledge in this field.

The comprehensive problems in the last chapter are also intended to help undergraduate students participating in mathematics contests hone their problem solving skills. Students and teachers can find here ideas and questions that can be interesting topics for mathematics circles.

Due to the difficulty level of the problems contained in this book, we deemed it appropriate to give a very clear and complete presentation of all solutions. In many cases, alternative solutions are provided.

As a piece of advice to all readers, we suggest that they try to find their own solutions to the problems before reading the given ones. Many problems can be solved in multiple ways and pertain to interesting extensions.

This edition is significantly different from the 2002 Romanian edition. It features more recent problems, enhanced solutions, along with references for all published problems.

We wish to extend our gratitude to everyone who influenced in one way or another the final version of this book.

We will gladly receive any observation from the readers.

The authors

Chapter 1 ALGEBRA

PROBLEMS

1 . Let C be a set of n characters {Cl ' C2 , . . • , cn

}. We call word a string of at

most m characters, m :::; n, that does not start nor end with Cl . How many words can be formed with the characters of the set C?

2. The numbers 1, 2 , . . . , 5n are divided into two disjoint sets. Prove that these

sets contain at least n pairs (x, y

), x > y, such that the number x - y is also an

element of the set which contains the pair.

3. Let al , a2 , . . . , an be distinct numbers from the interval [a, b]

and let a be a permutation of

{I, 2, . . . , n

}.

Define the function f : [a, b]

-t [a, b]

as follows:

f(x) = { aq ( i) if x = �i' i = 1, n

x otherwIse

Prove that there is a positive integer h such that f[h](X) fafa ... af. '----v-----" htimes

x, where f[h]

4. Prove that if x, y, z are nonzero real numbers with x + y + z = 0, then x2 + y

2 y2 + Z

2 Z2 + x

2 x3

y3 Z3

--- + + = - + - + -. x + y y + Z Z + x yz zx xy

5. Let a, b, c,

d be complex numbers with a +

b + C +

d = O. Prove that

a3 + b3 + c

3 + d3 = 3(abc + bcd + cda + dab)

.

6. Let a, b, c be nonzero real numbers such that a +

b + c = 0 and a3 +

b3 + c3 =

a5 + b5 + c

5• Prove that

7. Let a, b, c,

d be integers. Prove that a +

b + c +

d divides

2(a4 + b4 + c

4 +

�) _ (a2 + b2 + c

2 + d2)2

+ 8abcd.

10 1 . ALGEBRA

8. Solve in complex numbers the equation (x +

l) (x +

2) (x +

3)2 (X +

4) (x +

5) = 360

.

9. Solve in real numbers the equation

-IX + VY + 2v'z=2 + .jU + -IV = x + y + z + u + v.

10. Find the real solutions to the equation (x + y

) 2 = (X +

1) (y -

1).

11 . Solve the equation

� x + J4X + V16X + l·· + J4nx + 3 -..;x =

1.

12 . Solve the equation

-J x + a + -J x + b + -J x + c = -J x + a + b - c,

where a, b, c are real parameters. Discuss the equation in terms of the values of the

parameters.

13. Let a and b

be distinct positive real numbers. Find all pairs of positive real numbers

(x, y

), solutions to the system of equations

14. Solve the equation

{ x4 - y4

= ax -by

x2

_ y2 = {/ a2 _

b2.

[ 25x - 2 ] = l3x +

4 4 3 '

where [a]

denotes the integer part of a real number a.

. 1+V5

15. Prove that If a � --2-' then

1. 1 . PROBLEMS 11

16. Prove that if x, y, z are real numbers such that xS + y

S + Z

S f:. 0, then the

ratio 2xyz -

(x + y + z

)

xS + y

S + Z S

equals � if and only if x + y + z = O.

17. Solve in real numbers the equation 1 vx;:-=-r +

2 .JX2 -

4 + . . . + n

JX n - n

2 = "2(xl + X2 + . . . + xn

).

18. Find the real solutions to the system of equations 1 1 1 -+ - =

9

x y

1 1 1 1 - + - 1 + - 1 + - - 18 (?Ix �)( ?Ix)( �) -

19. Solve in real numbers the system of equations

y2 + u2 + v

2 + w

2 = 4x - 1

x2 + u2 + v

2 + w

2 = 4y - 1 x2 + y

2 + v

2 + w

2 = 4u - 1

x2 + y2 + u2 + w

2 = 4v - 1

x2 + y2 + u

2 + v2 = 4w - 1

20. Let aI , a2 , a s, a4, a5 be real numbers such that al + a2 + a s + a4 + a5 = 0 and max

lai - aj

l < 1. Prove that a

i + a

� + a

� + a

� + a

� < 10. l�i<j�5 - -

21 . Let a, b, c be positive real numbers. Prove that

1 1 1 1 1 1 - + - + - > -- + -- + --2a

2b 2c - a +

b b + c c + a

22. Let a, b, c be real numbers such that the sum of any two of them is not equal

to zero. Prove that

23. Let a, b, c be real numbers such that a

bc = 1 . Prove that at most two of the

numbers

are greater than 1 .

1 2a - -b '

1 2b- -c '

1 2c - -

a

12 1 . ALGEBRA

24. Let a, b, c,

d be real numbers. Prove that

2 2 � 2 1

min (a -b b

- c c - a-d - a

) < -, , , - 4'

25. Let aI , a2 , ' . . , an be numbers in the interval (0,1)

and let k ;::: 2 be an integer

Find the maximum value of the expression n L V'

ai(1 - ai+1

),

i=l

26. Let m and n be positive integers. Prove that xmn -

1 xn -

l --- > --

m - x for any positive real number x.

27. Prove that m! ;::: (n!) [�]

for all positive integers m and n.

28. Prove that 1 1 1 �2 1

+ - + - + ' ' ' + - > n - -v'2 v'3 \Iii n + 1

for any integer n ;::: 2 .

29. Prove that

n (1-1/ vn) + 1 > 1 + � + � + . . . + .!. > n (\In +

1-1)

2 3

n for any positive integer n.

( ) na1 a2 ' " an 30. Let aI , a2 , . . . , an E 0,1

and let tn = . Prove that a1 + a2 + . . . + an

n L loga. tn ;:::

(n -

l)n.

i=l

31. Prove that between n and 3n there is at least a perfect cube for any intege

n ;::: 10.

32. Compute the sum

33. Compute the sums:

n "'" 1c(Ic+l) S

n = �(-1)

2 k.

k=l

a) Sn = � (k + l)l(k + 2) (�); b)

Tn = � (k + l) (k + 2) (k + 3) (�).

1 .1 . PROBLEMS

34. Show that for any positive integer n the number

Cn; 1) 22n + Cn: 1) 22n-2 .3 + . . . + Cn

2: 1) 3n

is the sum of two consecutive perfect squares.

35. Evaluate the sums:

36. Prove that

12 (�) + 32 (�) +

52 (�) + . . . = n(n + 1)2n-3

for all integers n � 3.

37. Prove that 2n L

[log2 k] = (n - 2)2n + n + 2 k=l

for all positive integers n.

38. Let Xn = 22n

+ 1, n = 1 , 2, 3, . . . Prove that


1 2 22 2n -l 1 - + - + - + . . . + -- <

Xl X2 X3 Xn 3

39. Let f : C -t C be a function such that f(z)f(iz) = Z2 for all z E C. Prove that

f (z) + f ( -z) = 0

for all z E C

13

40. Consider a function f : (0, 00

) -t � and a real number a >

0 such that

f(a) = 1. Prove that if

f(x)f(y) + f (�) f (�) = 2f(xy) for all x, y E (0, 00

),

then f is a constant function.

41. Find with proof if the function t: � -t [- 1, 1] , f(x) = sin[x] is periodical.

n 42. For all i, j = 1, n define S(i, j) =

L ki+i. Evaluate the determinant � = k=l

IS(i , j ) l ·

14

43. Let

1. ALGEBRA

{ ai if i = j

xii = 0 if i i j

, i + j i

2n + 1 bi if i +

j = 2n + 1

where ai , bi are real numbers. Evaluate the determinant

�2n =

IXii

l·

44. a) Compute the determinant

x y z v y x v z z v x y v z y x

b) Prove that if the numbers abed, bade, cdab

, deba are divisible by a prime p,

then at least one of the numbers

a + b

+ e + d, a +

b - e -

d, a -

b + e -

d, a -

b - e +

d,

is divisible by p.

45. Consider the quadratic polynomials tl (x)

= x2 + PI X + qr

and t2 (x)

= x2 + P2X + q

�, where PI , P2, ql , q2 are real numbers.

Prove that if polynomials tl and t2 have zeros of the same nature, then the polynomial

has real zeros.

46. Let a, b, e be real numbers with a > 0 such that the quadratic polynomial

T (x)

= ax2

+ bex +

b S + e

S -4abe

has nonreal zeros. Prove that exactly one of the polynomials

TI (x)

= ax2 + bx + e and

T2 (x)

ax2 + ex +

b has only positive values.

47. Consider the polynomials with complex coefficients P(x)

= xn + alxn-l + . . . + an and Q(x)

= xn + bl xn-l + . . . +

bn

having zeros Xl , X2 , . . . , Xn and xi, x�, . . . , x

; respectively.

Prove that if al + a s + a5 + . . . and a2 + a4 + a6 + . . . are real numbers, then bl +

b2 + . . . +

bn is also a real number.

1 . 1 . BROBLEMS

48. Let P(x)

be a polynomial of degree n. If k P(k)

= -k- for k

= 0 , 1 , . . . , n + 1

evaluate P(m), where m > n.

49. Find all polynomials P(x)

with integral coefficients such that

for all real numbers x.

15

50. Consider the polynomials Pi, i = 1, 2, . . . , n with degrees at least 1. Prove that if the polynomial

P(x)

= P1(Xn+1

) + XP2

(xn+1

) + . . . + xn-1pn

(xn+1

),

is divisible by xn +xn-1 + . . · + x + 1, then all polynomials Pi (x) , i = 1, n, are divisible by x - 1.

51 . Let P be a prime number and let P(x)

= aoxn + a1xn-1 + . . . + an

be a polynomial with integral coefficients such that an =1= 0 (mod p). Prove that

if there are n + 1 integers 0'1 , 0'2 , • • • , an+ 1 such that P ( ar)

== 0 (mod p)

for all r = 1, 2, . . . , n + 1, then there exist i , j with i i- j such that ai == aj (mod p

).

52. Determine all polynomials P

with real coefficients such that pn(x)

= P(xn)

for all real numbers x, where n > 1 is a given integer.

53. Let P(x)

= aoxn + a1xn-1 + . . . + an, an i- 0,

be a polynomial with complex coefficients such that there is an integer m with

Prove that the polynomial P

has at least a zero with the absolute value less than 1 .

54. Find all polynomials P

of degree n having only real zeros Xl , X2 , . . . , Xn such that n 1 n2 � P(x) - Xi

= XP I(

X) '

for all nonzero real numbers x.

55. Consider the polynomial with real coefficients P(x)

= aoxn + a1xn-1 + . . . + an ,

16 1 . ALGEBRA

and an f:.

O. Prove that if the equation P(x) = 0 has all of its roots real and distinct, then the

equation x2 PII(x) + 3xP' (x) + P(x) = 0

has the same property.

56. Let R�? and Rf�) be the sets of polynomials with real coefficients having no multiple zeros and having multiple zeros of order n respectively. Prove that if P(x) E R�? and P(Q(x)) E R�? , then Q' (x) E R�]

-1).

57. Let P(x) be a polynomial with real coefficients of degree at least 2. Prove that if there is a real number a such that

P(a)plI(a) > (P' (a))2 ,

then P has at least two nonreal zeros.

58. Consider the equation

aoxn + a1Xn-1 + . . . + an = 0

with real coefficients ai . Prove that if the equation has all of its roots real, then (n - l)ar 2:: 2naOa2 . Is the reciprocal true?


X4 - (2m + 1)x3 + (m - 1)x2 + (2m2 + l)x + m = 0,

where m is a real parameter .


x2n + a1x2n-1 + . . . + a2n_2X2 - 2nx + 1 = 0,

if all of its roots are real.

SOLUTIONS

1 . Let Nk be the number of words having exactly k characters from the set C, 1 :s; k :s; m. Clearly,

N1 = n - 1. The number that we seek is

N1 +

N2 + . . . +

Nm·

Let Ak = {I, 2, . . . , k } , 1 :s; k :s; m. We need to find out the number of functions

f : Ak ---+ A, k = 2, n with the properties

f(l) f:. a1 and f(k)

f:. a1

For f(l) and f(k) there are n - 1 possibilities of choosing a character from C2 , • . • , Cn and for f(i), 1 < i < k there are n such possibilities. Therefore the number of strings f(l)f(2) . . . f(k - l)f(k) is

Nk = (n - 1)2nk-2

It follows that N1 +

N2 + ... +

Nm = (n - 1) + (n - 1)2nO + (n - 1)2n1 + . . . + (n - 1)2nm-2 =

(Dorin Andrica)

2. Suppose, for the sake of contradiction, that there are two sets A and B such that Au B =

{I, 2, . . . , 5n}, An B = 0 and the sets contain together less than n pairs

(x, y), x > y, with the desired property.

Let k be a given number, k = 1 , n. If k and 2k are in the same set - A or B -the same can be said about the difference 2k - k = k. The same argument is applied for 4k and 2k. Consider the case when k and 4k are elements of A and 2k is an element of B . If 3k is an element of A, then 4k - 3k = k E A, so let 3k E B. Now if 5k E A, then 5k - 4k = k E A and if 5k E B, then 5k - 3k = 2k E B; so among the numbers k , 2k, 3k, 4k, 5k there is at least a pair with the desired property. Because k = 1, 2, . . . , n, it follows that there are at least n pairs with the desired property. (D

orin Andrica, Revista Matematica Timi§oara (RMT), No. 2(1978) , pp. 75,

Problem 3698)

3. Note that (1)

17

18 1. ALGEBRA

and furthermore (2)

for all integers ml , m2 2:: 1 . Suppose that for all integers k 2:: 1 we have

I[k](x) f=. x.

Because there are n ! permutations, it follows that for k > n ! there are distinct positive integers nl > n2 such that

(3)

Let h = nl - n2 > 0 and observe that for all k the functions I[k

] are injective,

since numbers ai , i

= 1, n are distinct. From relation (3) we derive that

or

l[n2+

h] (x)

= l[n2] (x), x E

[a, b],

(10 l[n2+

h-l]) (x)

= (10 l[n2-1

]) (x), X E

[a, b].

Because I

is injective, we obtain l[n2+

h-l] (x)

= l[n2-1

] (x), x E

[a, b]

and in the same manner

or

l[h+l](X)

= I(x), x E

[a, b]

l[h](

X)

= x, x E [a, b].

Alternat

ive so

lution. Let

Sn be the symmetric group of order n and

Hn the cyclic

subgroup generated by a. It is clear that Hn is a finite group and therefore there is

integer h such that a[h]

is identical permutation. Notice that

I[k] (x)

= { aq[kJ (i) if x = �i ' i

= 1, n x otherwIse

Then I[h](

x)

= x and the solution is complete. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2

(1978)

, pp. 53, Problem 3540)

4. Because x + y + z = 0, we obtain

x + y = -z, y + z = -x, z + x = -y,

or, by squaring and rearranging,

x2

+ y2 = Z2 _ 2xy, y

2 + Z

2 = x2

_ 2yz, Z2

+ x2 = y

2 - 2zx.

The given equality is equivalent to Z2 - 2xy x

2 - 2y Z y

2 - 2zx x3 y3 Z3

--- + + = - + - + -, -z -x -y yz zx xy

1.2. SOLUTIONS

and consequently to (Xy yz zx) X3 y3 Z3

-(x + y + z

)+ 2 - +-+ - = - +- + - .

z x Y yz zx xy

The last equality is equivalent to

2(X2y2 + y

2z2 + Z

2X2 ) = x

4 + y

4 + Z

4.

On the other side, from x + y + z = 0

we obtain (x + y + Z

)2 = 0

or

x2 + y

2 + Z

2 = -2( xy + y z + zx

).

Squaring yields

X4 + y

4 + Z

4 + 2

(X2y2 + y

2z2 + Z

2X2 ) =

4(X2y2 + y

2z2 + Z

2X2 ) + 8xyz

(x + y + z

)

or

as desired.

19

( Titu

Andreescu, Revista Matematica Timi§oara (RMT) , No.

3(1971), pp. 25,

Problem 483

; Gazeta Matematica (GM-B) , No. 12(1977)

, pp. 501, Problem

6090)

5. We assume that numbers a, b, c,

d are different from zero. Consider the equation

x4 - (2: a) x3 + (2: ab) x2 - (2: abc) x + abed =

0

with roots a, b, c,

d. Substituting x with a,

b, e and

d and simplifying by a,

b, c,

d f:. 0,

after summing up we obtain

Because 2: a = 0, it follows that

If one of the numbers is zero, say a, then b+ c +

d= 0,

or b + c = -

d. It is left to prove that

b3 + c3 + d3 =

3bcd. Now

b3 + c3 + d3 =

b3 + c3 - (b + c)3 = -3bc(b + c) =

3bcd

as desired. (Dorin Andri

ca, Revista Matematica Timi§oara (RMT) , No. 1-2(1979)

, pp. 47,

Problem 3803)

6. Because a + b + c =

0, we obtain

a3 +

b3 + c3 =

3abc and a

5 + b5 + c

5 = 5abc(a2 + b2 + c2 + ab + bc + ca)

20

or

1. ALGEBRA

The given relation becomes 3abc =

5abc(a2 + b2 + c

2 + a

b + bc + ca

)

3

a2 + b2 + c

2 + a

b + bc + ca = 5'

since a, b, c are nonzero numbers. It follows that

1 3

-[(a +

b + C

)2 + a

2 + b2 + c

2] = -2 5

and, using again the relation a + b + c =

0, we obtain

a2 + b2 + c

2 = � 5 ' as desired. ( Ti

tu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2

(1977)

, pp. 59,

Problem 3016)

7. Consider the equation

x4 - (L: a) x3 + (L: ab) x2 - (L: abc) x + abed =

0,

with roots a, b, e,

d. Substituting x with a,

b, e and

d, respectively, we obtain after

summation that L: a4 + (L: a

b) L: a2 +4abed

is divisible by L: a. Taking into account that

we deduce that

is divisible by L: a. Hence

2(a4 + b4 + e

4 + d4)

-(a2 + b2 + e

2 + d2 )2

+ 8abed

is divisible by a + b + e +

d, as desired. (D

orin Andrica)

8. The equation is equivalent to (x2 + 6x +

5) (x2 + 6x +

8) (x2 + 6x +

9) = 360

.

Setting x2 + 6x = y yields

(y +

5) (y +

8) (y +

9) = 360

,

1.2 . SOLUTIONS 21 or y3 + 22y2 + 157y = 0, with solutions

Yl = 0, Y2 = -11 + 6i, Y3 = -11 -

6i.

Turning back to the substitution, we obtain a first equation, X2 + 6x = 0, with solutions Xl =

0, X2 = -

6.

The equation X2 + 6x = -11 + 6i is equivalent to (x + 3)2 = -2 + 6i. Setting X + 3 = u + iv, u, v E lR, we obtain the system { U2 -v

2 = -2

2uv = 6

It follows that (u2 + V2)2 = (u2 -V2)2 + (2UV)2 = 4 + 36 = 40. Therefore { U2 - v

2 = -2 u2 +V2 = 2v'W

and u2 = v'W

- 1, v2 = y'lO + 1, yielding the solutions

X3,4 = -3

± V V16 - 1 ± iV V16 + 1

where the signs + and - correspond. The equation X2 + 6x = -11 - 6i

can be solved in a similar way and it has the solutions

X5 = -3 + J y'lO - 1 -

iV y'lO + 1, X6 = -3 - V.Jill - 1 + iV y'lO + l .

( Titu


3(1972), pp. 2

6,

Problem 1255)

or

9. The equation is equivalent to

X -..;x + Y - VY + z -2

vz-=-2 + u - Vu + v - ..;v = 0,

(vx -D2 + (v'Y -D2 + (vz-=-2_1)2 + + (.ru-D\ (VV-D2

=0

Because x, y, Z, U, v are real numbers, it follows that

Hence

..;x = VY = Vu =

..;v = � and vz-=-2 = 1.

1 X = Y = U = v = 4' Z =

3.

( Titu

Andreescu, Revista Matematica Timi§oara (RMT) , No. 2

(1974)

, pp. 47,

Problem 2002; Gazeta Matematica (GM-B) , No. 1

0(1974)

, pp. 560

, Problem 14536)

22

or

1. ALGEBRA

10. Setting X = x + 1 and Y = y - 1 yields

(X + y)2 = Xy �[X2 + y2 + (X + y)2] =

0. 2

Hence X = Y = 0, so the solution is x = -1 and y = 1 . ( Ti

tu Andreescu, Revista Matematidl Timi§oara (RMT) , No. 1

(1977)

, pp. 40,

Problem 2811)

1 1. The equation is equivalent to

� x + V 4x + V 16x + V/ . . + v' 4nx + 3 = ..jii

+ 1

Squaring the equation yields

V4X + V16X + V/ . . + v'4nx + 3 =

2..jii + 1

Squaring again implies

V16X + V/ . . +

v'4nx +

3 = 4..jii

+ 1

Continuing this procedure yields 4nx +

3 = 4nx + 2 · 2n

..;x + 1

1 and 2 . 2n Vx = 2. Hence x = 4n

' ( Ti

tu Andreescu, Revista Matematidl Timi§oara (RMT) , No.

4-5(1972), pp.

43,

Problem 1385)

12 . We distinguish two cases: 1) b

= c. The equation becomes Jx + a +

Jx +

b = Jx + a,

so x;:= -b.

2) b f:.

c. The equation is equivalent to Jx +

b+ Jx + c =

Jx + a +

b - c -

Jx + a,

or b- c

b- c J

x +b-Jx + c

- Jx + a +

b- c +

Jx + a

'

so Jx +

b -Jx + c =

Jx + a +

b - c +

Jx + a.

(1)

(2)

1 .2. SOLUTIONS

Summing up relations (1)

and (2) we obtain v' x +

b = v' x + a +

b - c,

and then a = c. To conclude, we have found that (i) If

b = c, then the equation has the solution x = -

b.

(ii) If b f=. c and a f=. c, there is no solution. (iii) If

b f=. c and a = c, then x = -a is the only solution.

23

( Titu


(1978)

, pp. 26,

Problem 3017)

13. Because a and b

are distinct numbers, x and y are distinct as well. The second equation could be written as

and the system could be solved in terms of a and b. We have

a2b2

= b2y2 + 2

by(x4 _ y4

) + (X4 _ y

4)2

a2x2 = b2x2 + X

2 (X2 _ y

2 )3 .

Subtracting the first equation from the second yields

which reduces to

Solving the quadratic equation in b

yields b = y3 +

3x2y (and a = x3 +

3xy2)

or b = y3 - x

2y (and a = x3 - xy

2). The second alternative is not possible because

a = x(x2 _y

2 ) and b = y

(y2 _X

2) cannot be both positive. It follows that a = x3+3xy2

and b = 3x2y + y3 . Hence a +

b = (x + y

)3 and a -b = (x - y

)3 . The system now becomes

x + y =�a +

b

x - y ={!a -

b

and its unique solution is x = (�

a + b + �a -

b)/2, y =

(�a +

b -�a -

b)/2. ( Ti

tu A ndreescu, Korean Mathematics Com petitions, 2

001)

13x +

4

14. Let 3 = y, Y E Z. It follows that 3y -

4

x = 13

24 1. ALGEBRA

and the equation is equivalent to

or

[ � (3Y : 4) - 2 ]

= Y,

[75Y - 126] = 5

2 Y

Using that for any real number a, [a] ::; a <

[a] + 1, we obtain

75y - 12

6 1

Y ::; 52

< y + ,

126 178

or 126 ::; 2

3y <

178, so 23 ::; y < 23·

Note that y E Z, therefore y = 6

or y = 7, thus

14 17

Xl = 13 and X2 = 13 '

are the desired solutions. ( Titu


3(1972), pp. 2

5,

Problem 155

2)

1+V5

. 1 15. From a � -- 2- we obtam a2 - a -

I � 0, or a � a + 1. We have

Hence

That is because

and

[1 + na2 ] 1 = - + na - a,

a a

O::;a < 1.

[ 1+ [1: ana2 ] ]

= [ 1 + � : na - a ]

= [( 1 + � - a) � + n] = n, n � O.

( 1 ) 1 1 a 1 + - - a -::; (a - a)- =

1 - - < 1

a a a a

(1 + � - a) � > � > O.

a a a2

( Titu


(1978)

, pp. 45,

Problem 3479)

16. First we consider the case when X + y + z = O. Then x3 + y3 + Z3 =

3xyz and

the ratio equals �, as desired. 3

Conversely, if

then

and so

Using the formula

1.2. SOLUTIONS

2xy z -(x + y + z) 2 x3 + y3 + Z3 - "3'

2(x3 + y3 + Z3 -3xyz) + 3(x + y + z) = O.

x3 + y3 + Z3 -3xyz = (x + y + z)(x2 + y2 + Z2 -xy -yz -zx), we obtain by factorization that

and so (x + y + z)[(x -y)2 + (y -Z)2 + (z -X)2 + 3] = O.

Because the second factor is positive, it follows that x + y + z = 0, as desired.

25

( Titu

Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1 973), pp. 30,

Problem 1513) 17. We write the equation as

Xl -2JXI -1 + X2 -2· 2VX2 -22 + . . . + Xn -2nvxn -n2 = 0,

or (JXI -1 -1)2 + ( VX2 -22 -2) 2 + ... + (V Xn -n2 _ n) 2 = O.

Because the numbers Xi, i = 1, n are real, it follows that

( Titu

Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1977), pp. 14,

Problem 2243) 18. Using the identity

(a + b + c)3 = a3 + b3 + c3 + 3(a + b) (b + c)(c + a) we obtain

(� + � +1) 3 = � + � +1+3 (� +�) (1+ _1_) (1+ _1 _) = {IX flY X Y {IX flY {IX flY = 9+ 1 + 54 = 64

Hence

26 1. ALGEBRA

and so 1 1 - + - =

3. fIX flY

The system is now reduced to

1 �+�=9

x y

1 1

fIX + flY = 3,

which is a symmetric system, having the solution 1 1

x = '8' y = 1 and x = 1, Y = '8' ( Ti

tu Andreescu, Revista Matematica Timi§oara (RMT) , No.

4-5(1972) , pp.

43,

Problem 1386)

19. By summing up the equations of the system we obtain

It follows that

(4x2 -4x + 1

) + (4y2 -4y + 1

) + (4u2 -4u + 1

)+

+(4v2 -4v + 1

) + (4w2 -4w + 1

) = O.

Due to the fact that x, y, u, v, w are real numbers, we obtain 1

x = y = u = v = w = -2 (Dorin Andrica, Gazeta Matematica (GM-B) , No.

8(1977) , pp.

321, Problem

1678

2)

20. From the triangle inequality we deduce

Hence

So

l ai - aj l ::; lai - ai+l l + . . . + laj-l - ajl ::;

< (J' -i)

max la· - a"1 < J" -i

1 < _ i

< J" < _ 5

- lSi<jS5 � J - ,

I: (ai - aj)2::; I: (i -j)2

= lSi<jS5 lSi<jS5

5 5 4 � a� - 2 � a"a" <

50

� l � � J-i=l i,i=l i>Fi

1.2. SOLUTIONS

and consequently

5 t, a� - (t, a;r � 50

5 5

Note that L

ai = 0 and so L

ar

::; 10, as claimed. i=l i=l

27

( Titu Andreescu, Romanian Mathematica Olympiad - second round 1979; Revista

Matematica Timi§oara (RMT), No. 1-2(1980), pp. 61, Problem 4094)

21 . The inequality (a +

b)2 �

4ab

yields 1 1 4 - + - > --

a b- a +

b

1 1 4 1 1 4 and similarly - + - > -- - + - > -- . , ' b e -

b+ e ' e a - e + a

Summing up these inequalities yields 1 1 1 1 1 1 2a

+ 2b + 2e �

a + b + b

+ e + e + a'

as desired. (Dorin Andrica, Gazeta Matematica (GM-B), No. 8(1977) , Problem 5966)

22. Using the identities

a5 + b5 + e

5 = (a +

b + e

)5 -5(a +

b) (b + e

) (e + a)(a2 + b2 + e2 + ab + be + 00)

and

we obtain

or

a5 + b5 + e

5 -(a +

b + e

)5 5 2 2 2

3 b3 c3 ( b )3 = -3

(a +

b + e + a

b + be + ca)

a + + - a + + e

It suffices now to prove that 5 10 3 (a2 + b2 + e2 + a

b + be + 00) � 9 (a + b + e)2

3(a2 + b2 + e

2 + a

b + be + ca) � 2

(a2 + b2 + e

2 + 2a

b + 2

be + 2ca).

The last inequality is equivalent to

a2 + b2 + e

2 � a

b + be + 00,

which is clearly true. ( Titu

Andreescu, Revista Matematica Timi§oara (RMT), No. 1 (1981) , pp. 49,

Problem 4295; Gazeta Matematica (GM-B) , No. 6(1980), pp. 280, Problem 0-148; No. 11 (1982) , pp. 422, Problem 19450)

28 1. ALGEBRA

23. Assume by contradiction that all numbers 2a -�, 2b -�, 2c -� are greater b c a than 1. Then

and (2a-D (2b-D (2C-D > 1

(2a-D + (2b-D + (2C-D >3. From the relation (1) and using abc = 1 we obtain

3>2(a+b+c)-G+�+D. On the other hand, relation (2) gives

2(a+b+c)-G+�+D >3 which is a contradiction.

The proof is complete.

(1)

(2)

(3)

( Titu

Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1986) , pp. 72,

Problem 5982)

24. Assume by contradiction that all numbers are greater than 1/4. Then

hence

2 2 �J2 2 1 1 1 1 a -b + b -c + c -a-+ d -a > -+ -+ -+ -4 4 4 4

0> G-ar + G-br + G-cr + G-dr This is a contradiction so the claim holds. ( Ti

tu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 (1985) , pp. 59,

Problem 5479)

25. Setting ai = sin2 ai for i

= 1, n, where aI , a2 , . . . , an are real numbers, the expression becomes

n

E = 2: \!sin2 ai cos2 ai+b an+l = al· i=1

Using the AM-GM inequality yields

1 k -2: bf ;::: b1 b2 • • . bk, bi > 0, i = U

. k i=1

For b1 = sinf ai , b2 = cosf ai+l and b3 = b4 = . . . = bk = k� we obtain {/2 1 ( . 2 2

k -2) 1 \I' 2 2 k sm ai + cos ai+l + -2- ;::: 2¥ sm ai cos ai+l·

1.2. SOLUTIONS

Summing up these relations for k

= 1, 2, . . . , n yields

and so

� ( n(k - 2

) ) > _l_E

k n + 2 - /0 -2 , 2-/0 -

n · 2/ok2 n · 21-

� E< ---- 2 2

n

{14 Hence the maximum value of

E is ;; and it is reached if and only if

1 al = a2 = . . . = an = 2"'

29

(Dorin Andrica, Revista Matematidl Timi§oara (RMT) , No. 1

(1978)

, pp. 63, Problem 3266)

26. Because x and m are positive, we have to prove that

x(xmn - 1

) - m

(xn - 1

) � 0,

or (xn -

l) [(xn)m-Ix +

(xn)m-

2x + . . . + x - m

] � 0.

Define E(x)

= (xn)m-I x +

(xn)m-2x + . . . + x - m and note that if x � 1 , then

xn � 1 and E(x)

� 0, so the inequality holds. In the other case, when x < 1, we have xn <

1 and

E(x) < ° and again the inequality holds, as claimed. ( Ti


(1978)

, pp. 45, Problem 3480)

27. For m :::; n the inequality is clearly true, so consider m > n and define p = [:]. This implies that m = pn + q with q E

{O, 1, . . . , n -

I} and the inequality

can be written as (pn + q)! �

(n!)P.

We have (pn + q)! � (pn)! =

= (1· 2 . . . n

) (n +

1) . . .

(2n) ... ((P - l

)n + 1

) . . . (pn) �

(n!)P,

and we are done. ( Ti


(1977)

, pp. 61, Problem 3034)

28. We will use the inequality

Xml + x2

m + . . . + x

m > _1_ (Xl + X2 + . . . + X

)m n - nm-l n ,

which holds for all positive real numbers XI , X2 , . . . , Xn and all m E (-00 , 0] U [1 ,00

).

30 1. ALGEBRA

Now set Xl = 1 , X2 = 2, . . . , Xn = n and m = _.!.. We obtain n

1 1 1 1 1 [n (n + 1) ] -n �2

1 + - + - + · · · + - > -- = n - -v'2 y'3 \Iii n-�-l 2 n + 1 '

as desired. ( Titu

Andreescu, Revista Matematica Timi§oara (RMT) , No. 2 (1974) , pp. 52,

Problem 2035)

or

29. From the AM-GM inequality we deduce

.!. t i � 1 � n IT i � 1 = \in + 1 , n i=l � i=l �

1 n 1 1 + -2: -;- � \in + 1 , n i=l �

and so 1 + ! + ! + . . . + .!. > n

( \in + 1 - 1) 2 3 n '

as desired. Ob h h · 1· · . b h b

i + 1 . -1 -serve t at t e mequa Ity IS stnct ecause t e num ers -. -, � = , n, are

� distinct.

or

In order to prove the first inequality we apply the AM-GM inequality in the form

Therefore

1 2 n -1

1 + - + - + · · · + -- if! 1 2 3 n > n _ = _.

n n \Iii

n (1 - _1_) + 1

> 1 + � + . . . + .!.. \Iii 2 n (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 62,

Problem 3037)

30. Because the numbers al , a2 , . . . , an are positive, from the AM-GM inequality al + a2 + . . . + an ------- � -tlala2 . . . an n

we deduce that tn ::; (ala2 . . . an

) n;;:l . Using that numbers ai are less than 1 we obtain

n - 1 logai tn � -- logai (al a2 . . . an

).

n

1 . 2. SOLUTIONS

Summing up these inequalities yields n 1 n 2: loga; tn 2:: n - 2: loga; (a1 a2 . . . an

) =

i=1 n i=1 n - 1

= -n- [n + (logal a2 + loga2 a1

) + . . . +

(logal an + logan a1

) + . . . +

+(logan an-1 + logan_l an

) ]·

1 Note that a + - > 2 for all a > 0, so a -

� n - 1 L.J logai tn 2:: -- [n + 2(n - 1) + 2 (n - 2) + . . . + 2]

= (n - l)n, i=1 n

as claimed.

31

(Dorin Andrica, Revista Matematica Timi§oara

(RMT

) , No. 2(1977) , pp. 62,

Problem 3038)

31 . We begin with the following lemma. Lemma.

Let a >

b be two pos

itive

integers suc

h that

� - � > 1. Then

between num

bers a an

d b there

is at

least a per

fect cu

be. P

roof. Suppose, for the sake of contradiction, that there is no perfect cube between

a and b. Then there is an integer c such that

c3 � b

< a � (c + 1) 3 .

This means c � � < � � c + 1,

so

which is false. 0 Now we can easily check that for n = 10, 11 , 12, 13, 14, 15 the statement holds. If n 2:: 16, then

or

Hence

3 1 1 n > (2, 5) =

(1 , 4 - 1)3 > (�_ 1) 3 '

aC 1 y n > �_(

{I3n - -rn > 1 , and using the above lemma the problem is solved.

( Titu

Andreescu, Revista Matematica Timi§oara

(RMT

), No. 1-2(1990) , pp. 59,

Problem 4080)

32 1. ALGEBRA

k(k + 1) . . 32. Note that the number 2 IS odd for k = 4p + 1 or k = 4p + 2 and IS

even for k = 4p + 3 or k = 4p, where p is a positive integer. We have the following cases: i) if n = 4m, then

n m-l "" k(k+l) "" Sn= L...,.(-1) 2 k= L..J

(-4p -

1-4p -

2+4p +

3+4p +

4)=4m.

k=1 p=O

ii) if n = 4m + 1, then

Sn = 4m - (4m + 1) = -1.

iii) if n = 4m + 2, then

Sn = 4m - (4m + 1) -

(4m + 2) = - (4m + 3) .

iv) if n = 4m + 3 then

Sn = 4m - (4m + 1) -

(4m + 2) + (4m + 3) = O.

Hence { 4m if n=4m -1 if n=4

m + 1 Sn= �(4m +

3) if n=4m+2 if n = 4m +

3.

(Dorin Andrica, Revista Matematidl Timi§oara (RMT) , No. 1 (1981) , pp. 50,

Problem 4303)

33. a) Summing up the identities

(n + 2) (n + 2)(n + 1) (n) k - (n + 2 - k)(n + 1 - k) k for k = 0 to k = n yields

Sn= (n + 2)1(n + 1)

(�(n;2) - (�:�) - (�:�)) = 1 [2n+2 ( 2) 1] _ 2n+2 - (n + 3) (n + 2)(n + 1) - n +

- - (n + 2)(n + 1) . b) Summing up the identities

(n + 3) (n + 3)(n + 2)(n + 1) (n) k - (n + 3 -k)(n + 2 - k)(n + 1 - k) k

for k = 0 to k = n yields

1'. _ 1 n - (n + 3)(n + 2)(n + 1)

1 .2. SOLUTIONS . (� (n + 3) _ (n + 3) _ (n + 3) _ (n + 3) ) = k=O k n + 1 n + 2 n + 3

= 1 (2n+3 _

�(n2 + 3n + 2)) = (n + 3) (n + 2) (n + 1) 2 _ 2n+4 - (n2 + 3n + 2) - 2(n + 3) (n + 2) (n + 1)

.

33

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2 (1975) , pp. 43,

Problem 2116)

34. Let Sn be the number in the statement. It is not difficult to see that

Sn = � [(2 + v'3t<+l + (2 - v'3tH] .

The required property says: there exists k > 0 such that Sn = (k - 1)2 + k2 , or, equi val entl y,

2k2 - 2k + 1 - Sn = O.

The discriminant of this equation is Ll = 4(2Sn - 1) , and, after usual computa-tions, we obtain

� = ( (1 + v3) 2n+l ; (1 - v3) 2nH ) 2 Solving the equation, we find

k = 2n+1 + (1 + v'.3) 2n+l

+ (1 - v'3) 2n+l

2n+2 Therefore, it is sufficient to prove that k is an integer . Let us denote

Em

(1 + .J3) m

+ (1 - .J3) m, where m is a positive integer. Clearly, Em is an integer for

all m. We will prove that 2[ �]

divides Em, m = 1 , 2, 3, . . . Moreover, the numbers E

m satisfy the relation Em = 2

Em-1 +

2Em-2•

The property now follows by induction. (Dorin Andrica, Romanian IMO Selection Test, 1999)

35. Differentiating the identity

yields

where

sin nx = sinn X ((�) cotn-1 X - (�) cotn-3 x + (�)

cotn-5 X - • • • ) n cosnx = n sinn-1 x cos xP(cot x) - sinn X�PI (cot X) ,

sm X

34

and

1. ALGEBRA

For x = � we obtain

Because

7f (0.) n

n cosn "4 = 2 (nP(l)

- 2P' (

1) ).

nP(l)

= n(�)

- n(;)

+ n(�)

- . . .

-2P' (1)

= -2(n - 1

) (�) + 2

(n -

3) (;) - 2

(n - 5)

(�) + . . . ,

we have

nP(1)- 2

P' (1)= -

[(n - 2

) (�) -(n -

6) (;) +(n - 1

0) (�) - . . .

J =

= -n ( (�)

-(;)

+ (�)

- . . . ) + 28n·

To conclude, use that

hence n (0.) n ( n 7f n 7f) S

= cos - + sin-n 2 4 4 (Dorin Andri

ca, Revista Matematica Timi§oara (RMT) , No. 2(1977)

, pp. Problem

3200)

36. Differentiating with respect to x the identities (x +

l)n = (�) + (�) X + . . . + (�)

xn

and (x -

l)n = (�)

xn -(

n � 1

) xn-1 + . . . +

(-1

)n (�) yields

and

n(x _ 1

)n-1 = n(�)

xn-1 -(n - 1

) (

n � 1

)xn-

2 + . . . +

(_1

)n-1 (�). MUltiplying by x gives

nx(x + 1

)n-1 = (�)

x + 2 (;)

X2 + . . . + n

(�) xn

and

nx(x _

1)n-1 = n(�)

x" -(n - 1)

(

n � l) x"-

' + . . . +

(_1

),,-1

(�)x.

and

1 .2. SOLUTIONS

Differentiating again we obtain

n(x +

l)n-l + n(n -

l)x(x + 1

)n-2 = (�) + 22 (�) x + . . . + n2 (�)

xn-1

n(x -

l)n-l + n (n -l)x(x - 1

)n-2 = n2 (�)

xn-1 - (n - 1

)2 (n � 1) xn-2+

+ . . . +(_W

-l (�) Setting x = 1 yields

12 (�) + 22 (�) + . . . + n2 (�) = n (n + 1)2n-2

Summing up the last two identities gives Sn = 1

2 (�) + 32 (�) + . . . = n (n + 1)2n-

a,

as desired.

35

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1

(1978)

, pp. 90,

Problem 3438)

37. Note that 2n n_1 2

i+1 _1 2: [log2 k]

= 2: 2: [log2 k] + [log2 2

n] , k=l i=O k=2i

and [log2 k]

= i

for 2i ::;

k < 2i+1 .

Hence

as claimed.

2n n-1 2:�Og2

k] = 2: i . 2i + [log2

2n] = (n -2)2n + n +

2

k=l i=O

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2

(1981), pp.

63,

Problem 4585

; Gazeta Matematica (GM-B) , No. 2-3(1982), pp.

83, Problem 1

9113)

38. Let Yn = 22n - 1 for all positive integers n. Then

� _ _ 2_

1 2 (22n )2

- 2 . 22n

+ 1

Yn Yn+1 22n - 1 22n+1 - 1

(22n - 1

) (22n+1 - 1

)

and therefore

_ (22n _ 1)2

_

22n - 1 _

1 _ 1 - (

22n _ 1) (22n +1 - 1

) - (22n

)2 - 1 -

22n + 1

-xn

1 1 2 = - - -

36 1. ALGEBRA

Xn Yn Yn+l Summing up these relations yields

1 2 22 2n-1 1 2n 1 - + - + - + . . . +-- = - - -- < Xl X2 X3 Xn YI Yn+l YI

for all positive integers n, as desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2

(1980)

, pp. 67, Problem

4135)

39. Substituting z with iz in the relation

J(z)J(i

z)

= Z2

yields J(iz)J(

-z)

= _Z2.

Summing up gives J(iz) (J(

z) + J( -z

) ) = 0,

so J(i

z)

= 0

or J(z) + J( -z

) = o.

From the relation J(z)J(i

z)

= Z2 we deduce that

J(z)

= 0

if and only if z = O. Hence if z -:f.

0, then

J(iz)

-:f. 0

and so J(z) + J( -z

) =

0 and, if z =

0, then J(

z) + J( -z

) = 2

J(0) =

O. Clearly,

J(z) + J( -z

) = 0

for all numbers z E C, as desired. R

emark. A function

J : C -+ C satisfying the relation

J(z)J(i

z)

= Z2 is

J(z)

= (-� + i�) z. ( Ti


(1976), pp.

56,

Problem 2583)

40. Setting X = Y = 1 yields J2 (1

) + J2 (

a) = 2 J (1 )

and .(J(l)

- 1)2 = 0

so J(l)

= 1. Substituting Y = 1 gives

or

J(x)J(l)

+ J (�) J(a) = 2

J(x)

a Take now Y = - and observe that X

J(x)J (�) + J (�) J(x) = 2

J(a).

1 .2. SOLUTIONS 37

Consequently, f(

x) f (;) = 1,

therefore f2 (

x) = 1, x > o.

Now set x = y = 0, that gives 12 (,ji)

+ 12 (�) = 2/(t)

and because the left-hand side is positive, it follows that f

is positive and f (

x) = 1

for all x. Then f

is a constant function, as claimed. ( Titu

Andreescu, Revista Matematica Timi§oara

(RMT

), No. 12

(1977)

, pp. 45,

Problem 2849

; Gazeta Matematica (GM-B), No. 1

0(1980)

, pp. 439

, Problem 18455)

41 . The function is not periodical. Suppose, by way of contradiction, that there is a number

T > 0

such that f(

x + T)

= f(x)

or sin[x +

T] = sin

[x], for all x E JR.

It follows that [x +

T] -

[x] = 2k

(x)7r, X E 1R,

where k : JR -+ Z is a function. Because 7r is irrational, we deduce that k(x) = 0

for all x E JR and therefore

[x] = [x +

T] for all x E JR

which is false, since the greatest integer function is not periodical. (Dorin Andri

ca, Revista Matematica Timi§oara (RMT

) , No. 1

(1978)

, pp. 89,

Pro blem 3430)

42. Considering the determinant

1 2 3

n 8=

12 22 32

n2

In 2n 3n nn

we have 1 12 13

In

� = IS(i

,j ) 1

= 8·

21 22 23 2n = 82

,

n1 n2

n3 nn

because the second determinant is obtained from 8

by interchanging rows and columns.

38 1 . ALGEBRA

On the other hand,

1 1 1 1

8 = n! 1 2 3 n

1n-1 2n-1 3n-1 nn-l

(here we used the known result on Vandermonde determinants) . Therefore

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1 (1982), pp. 52,

Problem 3862)

43. The determinant is

al 0 0 0 0 0 a2 0 0

b2

0 0 a3 b3 0 �

2n = 0 0

b2n-2 a2n-2 0

0 b2n-l 0 0 a2n-l b

2n 0 0 0 0

Expanding along the first and then the last row we obtain

which gives n �

2n = II (

aka2n-k+l -bkb2n-k+

d

k=l

b1 0 0

0 0 a2n

(Dorin Andri

ca, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 90, Problem 3201; Gazeta Matematica (GM-B) , No. 8 (1977) , pp. 325, Problem 16808)

44. a) Adding the last three columns to the first one yields that x + y + z + v divides the determinant.

Adding the first and second columns and subtracting the last two columns implies that x + y - z - v divides the determinant.

Analogously we can check that x-y+z-v and x-y-z+v divide the determinant, and taking into account that it has degree

4 in each of the variables, the determinant

equals ..\

(x + y + z + v) (x + y - z - v) (x - y + z - v

) (x - y - z + v

),

where ..\ is a constant.

1 . 2. SOLUTIONS

Because the coefficient of X4

is equal to 1, we have A = 1 and so

x y z w y x v z z v x y v z y x

= (x + y + z + v

) (x + y - z - v

) (x - y + z - v

) (x - y - z + v

)

b) As shown above, we have

ab

e d

39

6..=

ba

d e

e d

ab

de

b a

= (a +

b + e +

d) (a +

b - e -

d) (a -

b + e -

d) (a -

b - e + d)

On the other hand, mUltiplying the first column by 1000

, the second by 100,

the third by 10

and adding all these to the fourth, we obtain on the last column the numbers a

bed, bade, e

dab, debao Because all those numbers are divisible by the

prime number p, it follows that p divides 6..

and therefore p divides at least one of the numbers a +

b + e +

d, a +

b - e -

d, a -

b + e -

d, a -

b - e +

d. ( Ti

tu Andreescu

)

45. Because the quadratic polynomials h(x)

and t2(X)

have zeros of the same nature, it follows that their discriminants have the same sign, hence

Consequently, (P1P2 +

4q1 q2

)2 -4(P1 q2 + P2Q 1

)2 2:: O.

Note now that the left-hand side of the inequality is the discriminant of the quadratic polynomial t and the conclusion follows. ( Ti


(1978)

, pp. 63,

Problem 3267; Gazeta Matematica (GM-B) , No.

5(1979)

, pp. 191, Problem 1

7740)

46. Because the quadratic polynomial T

has nonreal zeros, the discriminant 6.. = b2e2 -4a(b3 + e3 -

4abe)

is negative. Observe that 6..

= (b2

-4ae) (e2 -4ab)

< 0,

where 6..1 =

b2 -4ae and

6..2 = e

2 -4ab

are the discriminants of the quadratic polynomials

T1 and

T2• Hence exactly one of the numbers

6..1 and

6..2 is negative and

since a > 0, the conclusion follows. ( Ti


(1977)

, pp. 40,

Problem 2810)

40 1 . ALGEBRA

47. Observe that al +a2 + . . · +an and al -a2 + . . . + (_l)n-Ian are real numbers, that is

P(l) and

P( -1

) are real numbers. Hence

and

P(l) = P(l)

and P(-l)

= P(-l)

Because P(x)

= (x - Xl

) . . . (X - xn

), the relations

(1)

become (1 - xI

) . . .

(1 - xn

) = (1 -

xI) . . .

(1 -

xn)

(1 + xI

) . . .

(1 + xn

) = (1 +

xI) . . .

(1 + xn )

Multiplying these relations yields (1 - x

�) . . .

(1 - x

�) = (1 -

x�) . . .

(1 -

x�),

or Q(l)

= Q(l)

. Therefore bl +

b2 + . . . +

bn is a real number.

(1)

( Titu


(1977)

, pp. 47,

Pro blem 2864)

48. Because P(O)

= 0, there is a polynomial

Q with

P(x)

= xQ(x). Then

1 Q(k) = k

+ l '

k = 1, n.

Define H(

x)

= (x +

l)Q (x) - 1. It is clear that

degH(

x)

= n and H(k)

= 0

for all

k = 1, n, hence

H(x)

= (x +

l)Q (x) - 1 = ao

(x -

l) (x - 2

) . . .

(x - n

)

Setting x = m, m > n in relation (1)

yields Q (m)

= ao (m - 1

) (m - 2

) . . .

(m - n

) + 1

. m + 1

On the other hand, setting x = -1 in the same relation implies ( _l)n+1

ao = (n +

I) !

Therefore

and then

Q(m)

=

(_l)n+l (m -

l) (m - 2

) . . .

(m - n

)

+ _1_ (

n + l) ! (

m + 1)

m + 1

P(m)

=

(_l)n+lm

(m - 1

) . . .

(m - n

)

+�

. (n +

l) ! (m + 1

) m + 1

(1)

(Dorin Andrica, Gazeta Matematica (GM-B) , No.

8(1977)

, pp. 329, Problem

16833

; Revista Matematica Timi§oara (RMT) , No. 1-2(1980)

, p. 67, Problem

4133)

49. We are looking for a polynomial with integral coefficients P(x)

= aoxn + alxn-l + . . . + an, ao

-:j:. O.

1 .2. SOLUTIONS 41

We have

P' (x) = naOxn-l + (n -

l)alXn-

2 + . . . + an-l

and by identifying the coefficient of x(n-l)n in the relation P(P' (x) ) = P' (P(x)) , we obtain

or aonn-l = 1. Hence 1

a - --0- nn-l

and since ao is an integer, we deduce that n = 1 and ao = 1. Then P(x) = x + aI , P' (x) =

1 and P(P' (x)) = P' (P(x)) yields 1 + al = 1 or al = O.

Therefore P(x) = x is the only polynomial with the desired property. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1-2 (1979), Problem

3902)

50. Let (h, O2 , . . . ,

On be the roots of the equation

xn + Xn-l + . . . + x + 1 = O.

They are all distinct and Of+l = 1,

i = 1, n.

Because P(x) is divisible by xn + xn-l + . . . + x + 1, it follows that P(O

i)

= 0, i = 1, n, hence

Pl(l)

+ OlP2

(I) + . . . +

Or-lpn(l)

= 0

PI (1) + 02P2

(1) + . . . +

O�-lPn(l )

= 0

The above system of equations has the determinant

1 01

on-l 1

V = 1

O2

O�-l

1 On

on-l n

Because all of the numbers 01 , 02 , . . . ,

On are distinct, it follows that V -:f. 0 and

so the system has only the trivial solution PI (1) = P2(I)

= . . . = Pn(l)

= O. This is just another way of saying that x -

I divides Pi

(X)

for all i

= 1 , n. (Dorin Andri

ca, Revista Matematica Timi§oara (RMT) , No. 2 (1977) , pp. 75, Problem 3120; Gazeta Matematica (GM-B) , No. 8(1977) , pp. 329, Problem 16834)

42 1 . ALGEBRA

51. Consider the determinant

n+1 v = = an

II (ak - al

)

"', 1=1 "' > 1

Multiplying the second row by an-I , the third by an-2 , . . . , the last by ao and adding all to the first yields

P(ad P(

a2) P(

an+d

al a2 an+1 n+1 v = a

� a� 2

an+1 = an II (

ak - at)

"' , 1=1 "' >1

a?

a�

a�+1

On the other hand, P(ar)

== 0 (mod p), for all r = 1, n + 1 and an ;j. 0 (mod p

)

implies II (

ak - al)

== 0 (mod p). Therefore there are at least two numbers

1 :Sl<k�n+1 ai , aj , i =I j such that ai - aj == 0 (mod p

) and so ai == aj (mod p

), as desired. (D

orin Andrica, Gazeta Matematica (GM-B) , No. 8(1977) , pp. 329, Problem

16835)

52. Let m = degP(x)

and let P(x)

= aoxm +

Q(x), ao =I 0

If follows that degQ(x)

= r � m - 1 . From pn

(x)

= P(xn)

we obtain

a�xmn +

(�)a�-lx

m(n-l)Q (x

) + . . . +

Qn(x)

= aoxmn +

Q(xn),

or a�x

mn + R(x)

= aoxmn +

S(x),

where degR(x)

= m(n - 1) + r and degS(x)

= nr.

Because ao =I 0, it follows that ao = 1 if n is even and ao = 1 or ao = -1 if n is odd. Moreover, d

egR(x)

= degS(x)

or m(n - 1) + r = nr

and so (n - l) (m - r) = O.

1 .2. SOLUTIONS 43

This is impossible if Q(x)

-:f. 0, because n > 1 and m > r, therefore Q(x)

= 0 and the polynomials are P(

x)

= xm, for n even and

P(x)

= ±xm, for n odd.

Alternat

ive so

lution. Let

degP(x)

= m and let P(x)

= aoxm + alxm-1 + . . . + am.

If P(x)

= xkQ(x)

with k

a positive integer, then

xknQn (x

) = xkn

Q(xn)

or Qn(x

) = Q(xn)

Note that Q

satisfies the same condition as P. Assume that

P(O) -:f. O.

Setting x = 0 in the initial condition yields a�

= am' Then am = 1 if n is even and am =

±1 if n is odd. Differentiating the relations implies

npn-1(x)P' (

x)

= nP' (

xn)xn-l .

Setting now x = 0 gives P' (O)

= 0 and so am-l = O.

Differentiating again in relation (1) yields analogously am-2 = 0 and then

The polynomials are

am-3 = am-4 = . . . = ao = O.

P(x)

= xm, if n is even and

P(x)

= ±xm, if n is odd.

(1)

( Titu

Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1-2 (1979) , pp. 59,

Problem 3884)

and

53. From the relations between the zeros and the coefficients we obtain am

= (_l)m L

XIX2 . . ' Xm ao

It follows that

and by applying the triangle's inequality for complex numbers, we deduce that L 1 (n) I

Xll lx21· · .

Ixn-m

l > m

.

Consider Xo = min{lxl l ,

IX21, · · . ,

Ixn-m

l }. Then

so Xo < 1, as claimed.

44 1. ALGEBRA

( Titu


Problem 3531)

or

54. We have n PI (x) n2 L

P(x) - X · = -;-. i=l �

Integrating the equation yields n L

ln IP(x) - xi i = n2 In Clx l , C > 0 i=l

n IT

2 2 In IP(x) - xi i = In Cn Ixln

i=l Hence

or

Ill. (F(x) -Xi) I = klx ln' , k > 0

IP(P(x)) 1 = klx ln2.

Eliminating the modules gives 2

P(P(x) ) = AXn , A E JR.

Therefore P (x) = axn with a E R ( Titu


Problem 2863; Gazeta Matematica (GM-B) , No. 1 (1977) , pp. 22, Problem 17034)

55. Define Q(x) = xP(x) . Because an -:f. 0, the polynomial Q has distinct real zeros, so the polynomial QI has distinct real zeros as well.

Consider H(x) = XQI (X) . Again, we deduce that HI has distinct real zeros, and since

HI (x) = x2 PII (x) + 3xPI (x) + P(x)

the conclusion follows. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp. 52,

Problem 3530)

56. Let m = degP(x) and let

P(x) = ao(x - xI ) (x - X2) ' " (x - xm) .

Because P(x) E R[x] , Xl , X2 , . . . , Xm are distinct zeros. Now

P(Q(x) ) = (Q (X) - xI ) (Q(X) - X2) . . . (Q(X) - Xm)

1 .2. SOLUTIONS 45

has a multiple zero a of order k. Since P(Q(

a) )

= 0, we have m

ao II(Q(

a)

- Xi)

= 0, i=l

and so there is an integer p, 1 � p � m, such that Q(a) - xp = 0. Observe that

Q(a) - xp -:f. 0, for all j -:f. p, otherwise Xj = xp, which is false. Hence

Q(x)

- Xj has the mUltiple zero a of order k and so

Q' (x)

= (Q(

x) - xp

)' = Q' (

x)

has a multiple zero of order k - 1. This concludes the proof. (D

orin Andri

ca, Romanian Mathematical Olympiad - final round, 1978; Revista Matematica. Timi§oara (RMT), No. 2(1978) , pp. 67, Problem 3614)

57. Assume by way of contradiction that P(x)

has less than two nonreal zeros. As a polynomial with real coefficients

P(x)

cannot have only one nonreal zero, hence all of its are real. Let Xl , X2 , . . . , Xn be the zeros of

P(x).

Then P' (x)

_ n

_1_ P(x) - � X - x ' �=l �

and differentiating we obtain P"(

x)P(

x) -[P' (

X)] 2 n 1 P2(

X) = - tt (x - Xi

)2

Setting x = a we reach a contradiction, therefore P(x)

has at least two nonreal zeros, as claimed. ( Ti

tu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1-2(1979) , pp. 59,

Problem 3883)

58. Let P(x)

= aoxn + alxn-l + . . . + an be a polynomial with real coefficients. If all of its zeros are real, then the same is true for the polynomials

P', P"

, . . . , p(n-2) .

Because ( 2)

(n -

2) ! 2 P

n- (x)

= 2 [n (n -

l)aox + 2

(n -

l)alx + 2a2 ]

is a quadratic polynomial with real zeros, we have

or (n -

l)a�

2: 2naOa2 .

The reciprocal is not always true, as we can see from the following example: P(x)

= x3 + (a + 1

)x2 + (a +

l)x + a,

with a E (- 00 , -1] U [2, (0).

46 1 . ALGEBRA

Observe that 2(a + 1)2 ;::: 2 . 3(a + 1) , or

(a + l) (a - 2) ;::: 0, so the inequality

holds. On the other hand, P(x) = (x + a

) (x2

+ x + 1) does not have all zeros real. (Dorin Andrica)

59. For m = ° the equation becomes

X4 - x3 - x

2 + X = °

and has roots Xl = 0, X2 = -1 , X3 = X4 = 1. If m -:j:. 0, we will solve the equation in terms of m. We have

2xm2 + (x2 -2x3 + l)m + X

4 - x3 - x

2 + X = °

and

It follows that

The initial equation becomes 2 [ x

2 - 1 ] [m -(x - x

)] m - 2X = 0.

Hence

and x2 - x - m = 0, with solutions I ± Jl + 4m

Xl ,2 = 2

x2

- 2mx - 1 = 0, with solutions X3 ,4 = m ± VI

+ m2 . (Dorin Andri

ca, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 75, Problem 3121)

60. From the relations between the zeros and the coefficients we obtain L

XIX2 · · · X2n-1 = 2n and XIX2 · · · X2n = 1.

Hence 2n 1 L- = 2n,

k=l Xk

so we have the equality case in the AM-GM inequality. Therefore Xl = X2 = . . . = X2n ' Since Xl X2 . . . X2n = 1 and 2n 1 � Xk > 0,

we have Xl = X2 = . . . = X2n = 1. ( Ti

tu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 52,

Problem 2299)

Chapter 2 NUMBER THEORY

PROBLEMS

1. How many 7-digit numbers that do not start nor end with 1

are there?

2. How many integers are among the numbers l · m 2 · m p · m -n' -n-" " ' -n

where p, m, n are given positive integers?

3. Let p > 2 be a prime number and let n be a positive integer. Prove that p divides 1pn + 2pn + . . . + (p _ l)pn .

4. Prove that for any integer n the number 55n+1 + 55n

+ 1

is not prime.

5. Let n be an odd integer greater than or equal to 5. Prove that

is not a prime number.

6. Prove that 345 + 456

is a product of two integers, each of which is larger than 102002

.

7. Find all positive integers n such that [ \1'111]

divides 111

.

8. Prove that for any distinct positive integers a and b

the number 2a( a2 + 3b2) is not a perfect cube.

9. Let p be a prime greater than 5. Prove that p - 4 cannot be the fourth power

of an integer.

10. Find all pairs (x, y) of nonnegative integers such that x2 + 3y and y2 + 3x are simultaneously perfect squares.

49

50 2. NUMBER THEORY

11 . Prove that for any positive integer n the number

(17 + 120) n _ (17 _ 120) n

4v'2 is an integer but not a perfect square.

12. Let (Un)n�l be the Fibonacci sequence:

Prove that for all integers n 2:: 6 between Un and un+ 1 there is a perfect square.

13. Prove that for all positive integers n the number n! +5 is not a perfect square.

14. Prove that if n is a perfect cube then n2 + 3n + 3 cannot be a perfect cube.

15. Let p be a prime. Prove that a product of 2p+ 1 positive consecutive numbers cannot be the 2p + I-power of an integer.

1 16. Let p be a prime and let a be a positive real number such that pa2 < 4'

Prove that [n0i - �] = [n0i + �]

for all integers n 2:: [ a ] + 1 y'1 - 2avp

.

17. Let n be an odd positive integer. Prove that the set

{ (�) , (�) , • • • , ( n� 1 ) } contains an odd number of odd numbers.

18. Find all positive integers m and n such that (:) = 1984.

19. Solve in nonnegative integers the equation

x2 + 8y2 + 6xy - 3x - 6y = 3

20. Solve in integers the equation

(x2 + 1 ) (y2 + 1) + 2(x - y) (1 - xy) = 4(1 + xy)

21 . Let p and q be prime numbers. Find all positive integers x and y such that 1 1 1 - + - = -. x y pq

2. 1 . PROBLEMS

22. Prove that the equation

has infinitely many solutions in positive integers such that u and v are both p:

23. Find all triples (x, y, z

) of integers such that

x2 (y - z

) + y

2 (Z - x

) + Z

2 (X - y

) = 2.


x + y + z + xyz = xy + yz + zx + 2

25. Solve in integers the equation

xy(x2 + y2

) = 2z4.

26. Prove that for all positive integers n the equation

x2

+ y2

+ Z2

= 59n

has integral solutions.

27. Let n be a positive integer. Prove that the equations

and xn + yn + zn + un = vn+1

have infinitely many solutions in distinct positive integers .

28. Let n be a positive integer. Solve in rational numbers the equation

xn + yn = xn-1 + yn-l .

29. Find all nonnegative integers x and y such that

x(x +

2) (x + 8) = 3Y.


(1 + xl ) (l

+ yl)

= (x + y

) !.

31. Solve the equation I x

l + y

l + z

l = 2v . .

52 2, NUMBER THEORY

32 . Find all distinct positive integers Xl , X2 , . . . , Xn such that 1 + Xl +

2XIX2 + . . . +

(n -

1)XIX2 . . . Xn-l = XIX2 . . . Xn ·

33. Prove that for all positive integers n and all integers aI , a2 , • . . , an , bl , b2 , . . . , b

n the number n II (

a� -b�)

k=l can be written as a difference of two squares.

34. Find all integers x, y, z, v, t such that

X + y + z + v + t = xyvt + (x + y

) (v + t

)

xy + z + vt = xy(v + t

) + vt

(x + y

).

35 . Prove that for all nonnegative integers a, b, c,

d such that a and

b are relatively

prime, the system ax - yz - c = 0 bx - yt +

d = 0

has at least a solution in nonnegative integers.

36. Let p be a prime and let Xl , X2 , . . • , xp be nonnegative integers. Prove that if

Xl + X2 + . . . + xp == 0 (mod p)

x� + x

� + . . . + x

� == 0 (mod p

)

Xf-l + X

�-l + . . . + X�-l == 0 (mod p

)

then there are k, l

E {I, 2, . . . ,p

}, k

=I l, such that Xk - Xl == 0 (mod p).

37. Prove that for any odd integers n, aI , a2 , . . . , an , the greatest common divisor of numbers aI , a2 , . . . , an is equal to the greatest common divisor of al + a2 a2 + a3 an + al 2 2 ' " ' ' 2

3,8. Let <p(n)

be the number of numbers less than n and relatively prime with n. Prove that there are infinitely many positive integers n such that

n <p(n)

= 3 '

39. Let 7r (x)

the number of primes less than or equal to x . Prove that n 7r

(n)

< 3 + 2


2. 1 . PROBLEMS

40. Let Pk denote the k-th prime number. Prove that

pf

+ pr

+ . . . + pr;:

> n m+l

for all positive integers m and n.

53

41 . Let n be a positive integer. Find the sum of all positive integers less than 2n and relatively prime with n.

42. Prove that any number between 1

and n! can be written as a sum of at most n distinct divisors of n! .

43. Find the largest value of n such that the complementary set of any subset

with n elements of {I, 2, . . . ,

1984} contains at least two elements that are relatively

prime.

44. Find all positive integers n such that for all odd integers a, if a2 � n then

a ln .

45 . Consider the sequences (Un)n�l '

(Vn)n�l defined by Ul =

3, Vl =

2 and

Un+l = 3un +

4vn, Vn+l =

2un +

3vn, n �

1. Define Xn = Un + Vn , Yn = Un +

2vn ,

n � 1. Prove that Yn =

[xnV2J

for all n � 1.

46. Define Xn = 22,, -1 +

1 for all positive integers. Prove that

(i) Xn = XIX2 . . . Xn-l + 2, n E N

(ii) (Xk ' xz

) = 1, k,l

E N, k -:f. l

(iii) Xn ends in 7 for all n � 3.

47. Define the sequence (an)n�l by al =

1 and

an+l = 2an +

J3a� -2

for all integers n � 1. Prove that an is an integer for all n.

48. Define the sequences (an)n�o and

(bn)n�o, by ao =

1,

2an-l 1

an = 2 and bn = 1 _ 2 2 1

+ 2an_l an-l

for all positive integers n. Prove that all terms of the sequence (an)n�O are irreducible

fractions and all terms of the sequence (bn)n�o are squares.

49. Define the sequences (Xn)n�o and

(Yn)n�O by Xo =

3, Yo =

2,

Xn = 3Xn-l +

4Yn-l and Yn =

2Xn-l +

3Yn-l

for all positive integers n. Prove that the sequence (zn)n�o, where Zn =

1 + 4x;y;,

contains no prime numbers.

54 2. NUMBER THEORY

50. Let p be a positive integer and let Xl be a positive real number. Define the sequence

(Xn)n�l by

Xn+l = Vp2 + 1xn + p

vx;

+ 1 for all positive integers n. Prove that among the first m terms of the sequence there are at least [�] irrational numbers.

51 . Define the sequence (xn)n�o by

1) Xn = 0 if and only if n = 0 and 2) Xn+l = X

[�

] + (-l)nx

[�] for all n 2:: o.

Find Xn in closed form.

52. Define the sequence (an)n�O by ao = 0, al = 1, a2 =

2, a3 = 6 and

Prove that n divides an for all n > O.

53. Let Xl = X2 = X3 = 1 and Xn+3 = Xn + Xn+IXn+2 for all positive integers n. Prove that for any positive integer m there is an integer k > 0 such that m divides Xk ·

54. Let (an)n�O be the sequence defined by ao = 0, al = 1 and

an+l -3an + an-l =

(_l)n 2

for all integers n > O. Prove that an is a perfect square for all n 2:: O.

55. Let al = a2 = 97 and

Prove that

an+l = anan-l + V(

a; -l) (a;_1 - 1) , n > 1.

a) 2

+ 2an is a perfect square.

b) 2

+ J2

+ 2an is a perfect square.

56. Let k 2:: 2

be an integer. Find in closed form for the general term an of the sequence defined by ao = 0 and an - a[ �

] = 1 for all n > O.

57. Let ao = al = 3

and an+l = tan - an-l for n 2:: 1 . Prove that an -

2 is a

perfect square for all n 2:: 1.

58. Let a and (3 be nonnegative integers such that a2 + 4(3 is not a perfect square. Define the sequence

(xn)n�O by

2. 1 . PROBLEMS

for all integers n 2:: 0, where Xl and X2 are positive integers. Prove that there is no positive integer no such that

2 _ Xno - Xno -IXno+l .

55

59. Let n > 1 be an integer. Prove that there is no irrational number a such that Va + Ja2 - 1 + Va - Ja2 - 1 is rational.

60. Prove that for different choices of signs + and - the expression

± 1 ± 2 ± 3 ± . . . ± (4n + 1) ,

yields all odd positive integers less than or equal to (2n + 1) ( 4n + 1) .

SOLUTIONS

1 . The problem is equivalent to finding the number of functions 1 : {I, 2, 3,4,5,6, 7}

-+ {O, 1, 2, . . . ,

9}

such that 1(1) f. 0,

1(1) f. 1 and

1(7) f. l.

Because 1(1)

E {2, 3, . . . ,

9}, there are

8 possibilities to define

1(1). For

1 (7) there

are 9

possibilities and for 1(2)

,1 (3)

,1 (4)

,1 (5)

,1 (6)

there are 10.

To conclude, there are 8.9.105

= 72

.105

numbers with the desired property. (Dorin Andri

ca, Gazeta Matematica (GM-B) , No. 1 1 (1979)

, pp. 421

, Problem 17999)

2 . Let d

be the greatest common divisor of m and n. Hence m = ml d

and n = nl d

for some integers ml and nl . The numbers are 1

· ml 2· ml p · ml

-- , -- , . . . , nl nl nl and, since ffi" n, are relatively prime, there are

[:,]

integers among them. Because

n n . [gCd(m, n)p] . nl = d =

gcd(m, n)

It follows that there are n mtegers. (Dorin Andri

ca, Gazeta Matematica (GM-B) , No. 11 (1979)

, pp. 429

, Problem 0:89)

3 . Define k = pn and note that k is odd. Then dk + (p -

d)k = p[dk-1 _

dk-2(p _

d) + . . . + (p _

d)k-l ] Summing up the equalities from

d = 1

to d

= [�] implies that p divides 1 k +

2k + . . . + (p - l)k , as claimed. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No.

1-2(1979)

, pp. 49,

Problem 3813)

4. Define m = 55n

• Then 55n+1 +

55n + 1

= m5

+ m + 1

= (m2 + m +

1) (m3 - m

2 + 1)

57

58 2. NUMBER THEORY

and, since both factors are greater than 1, the conclusion follows. ( Ti

tu Andreescu, Korean Mathematics Competition,

2001)

5. Let N =

(�) -5(�) + 52 (�) - ' " +

5n-' (�) . Then

5N=1-1+5(�) _ 52(�) + 53(�) _ . . . + 5n(�) = 1

+(-1+W

.

Hence N

= �W+l)= � [C2n + 1)2

- (2�r] = � [2n - 2 �

+ 1] [2n + 2�

+ 1] =

= � [(2 ";-' - 1 r + 2n-' ] [( 2 ";-' + 1 r + 2n-' ] . Because n is greater than or equal to

5, both factors of the numerator are greater

than 5. One of them is divisible by

5, call it

5N1 , N1 >

1, the other being

N2 . Then N

= N1N2, where

N1 and

N2 are both integers greater than

1, and we are done. ( Ti


2001)

6. The given number is of the form m4 + �n4 , where m = 344

and

� � n =

4 4 =2 2 •

The conclusion follows from the identity

and the inequalities 2

1 2 ( 1 ) � ( 56_1 4

4) m - mn + in > n in - m =

2 2

2 2 -

3 >

>2

2 2

2 -2

>22 .

2 22 -

1 > � ( 56- 1 2

.44) � 512 ( 56 -1 -512 )

> 210. 5�t1 .

2512 > 210.5

4 . 210.50 >

103.54 . 103.50 >

102002 (Ti


2002)

7. The positive divisors of 111

are 1, 3, 37, 111

. So we have the following cases: 1) [yl111]

= 1

or 1 $ 111 <

2n, hence n � 7. 2) [yl111]

= 3, or

3n $ 111 <

4n, so n =

4. 3) [yl111]

= 37, or

37n $ 111 <

38n, impossible. 4) [yl111]

= 111

, or 111n $ 111 <

112n, and so n = 1.

Therefore n = 1, n =

4 or n �

7.

2.2. SOLUTIONS

( Titu Andreescu

)

8. Note that 2a(a2 +

3b2 ) = (a +

b)3 + (a -

b)3

The Fermat equation for n = 3

x3 + y3 = z3

59

has no solution in positive integers (see T. Andreescu, D. Andrica, "An Introduction to Diophantine Equations" , GIL Publishing House,

2002, pp.

87-93)

. Hence there is no integer c such that

if a > b.

On the other hand, if b > a then there is no integer c such that

This concludes the proof. ( Titu

Andreescu, Revista Matematidi Timi§oara (RMT) , No.

1 (1974), pp.

24,

Problem 191 1)

9 . Assume that p -

4 = q

4 for some positive integer q. Then p = q

4 + 4

and q > 1.

We obtain p =

(q2

_

2q +

2) (q2 +

2q +

2),

a product of two integers greater than 1, contradicting the fact that p is a prime. ( Ti

tu Andreescu, Math Path Qualifying Quiz,

2003)

10. The inequalities

cannot hold simultaneously because summing them up yields 0 � x + y +

8, which is

false. Hence at least one of x2 + 3y < (x +

2)2 or y2 + 3x < (y +

2)2 is true. Without loss of generality assume that x2 +

3y < (x +

2)2.

From x2 < x2 + 3y < (x +

2)2 we derive x2 + 3y = (x +

1)2, hence

3y =

2x +

1.

Then x = 3k

+ 1 and y = 2k

+ 1

for some integer k � 0

and so y2 + 3x =

4k2 + 13k

+ 4.

If k > 5, then

(2k + 3)2

< 4k2 +

13k + 4

< (2k

+ 4)2

so y2 + 3x cannot be a square. It is easy to check that for

k E {O,

1,2,3,4}, y2 +

3x

is not a square but for k

= 0, y2 +

3x =

4 =

22. Therefore the only solution is

(x, y) = (1,1). ( Ti

tu Andreescu

)

60 2. NUMBER THEORY

11 . Note that 17 +

12v'2 = (v'2

+ 1) 4

and 17

-12v'2

= (v'2

_1) 4

, so (17 + 1

2v'2)n -(17

-12v'2)n

_

(v'2 + 1

) 4n -(v'2

_1) 4n

Define

4v'2 - 4v'2

_

(v'2 + 1

) 2n + (v'2

-1) 2n

.

(v'2 + 1) 2n _

(v'2 _ 1) 2n

- 2 2v'2

( v'2 + 1) 2n +

(v'2 -1) 2n ( v'2

+ 1) 2n

-(v'2

_ 1) 2n A

= 2 and B = 2v'2

U sing the binomial expansion formula we obtain positive integers x and y such that ( v2

+ 1) 2n = X + yv2

, ( v2 -1) 2n = X - y

v2

Then ( v'2

+ 1) 2n +

(v'2 -1) 2n ( v'2

+ 1) 2n _

(v'2 _ 1) 2n

X = 2 and y = 2v'2

and so AB

is as integer, as claimed. Observe that

so A

and B

are relatively prime. It is sufficient to prove that at least one of them is not a perfect square.

and

We have A =

(v'2 + 1

) 2n + (v'2

_1) 2n

= [ (v'2 + l

) n + (v'2

_1) n ] 2

-1 (1)

2 v'2

Since only one of the numbers

(v2 + l) n + (v2 _ l) n

v'2 (v2 +

l) n - (v2 _ l) n v'2

is an integer - depending on the parity of n - from the relations (1)

and (2)

we derive that

A is not a square. This completes the proof. (Dorin Andri

ca, Revista Matematica Timi§oara (RMT) , No. 1 (1981)

, pp. 48,

Problem 4285)

12. The claim is true for n = 6

and n = 7, because U6 =

8 < 9

< U7 = 13

< 16

<

U8 = 21.

If n � 8, then

2.2. SOLUTIONS

. r,;-;--:-: . � _ Un+1 - Un > Un-1 > y Un+1 - yUn - ---'---= v'Un+1 + VU;; - 2v'Un+1 -

Un-1 1 � 2 . � = . Fiv'Un-1 > 1 YoJUn-1 2y3

and so between Un and Un+1 there is a perfect square. (Dorin Andrica)

61

13. If n = 1 , 2 , 3 or 4 then n! + 5 = 6, 7, 11 or 29, so it is not a square. If n � 5, then n! + 5 = 5(5k + 1) for some integer k and therefore is not a perfect square, as desired. (D

orin Andri

ca, Gazeta Matematidl. (GM-B ) , No. 8 (1977) , pp. 321, Problem 16781; Revista Matematica Timi§oara (RMT) , No. 1 (1978) , pp. 61, Problem 3254)

14. Suppose by way of contradiction that n2 +3n+3 is a cube. Hence n(n2 +3n+3) is a cube. Note that

n (n 2 + 3n + 3) = n 3 + 3n 2 + 3n = (n + 1)3 - 1

and since (n + 1)3 - 1 is not a cube, we obtain a contradiction. (Dorin Andri

ca, Gazeta Matematica (GM-B) , No. 8 (1977) , pp. 312, Problem E5965; Revista Matematica Timi§oara (RMT) , No. 1-2(1979) , pp. 28, Problem 3253)

15 . Consider the product of 2p + 1 consecutive numbers

P(n) = (n + 1) (n + 2) . . . (n + 2p + 1)

Observe that P(n) > (n + 1)2P+1 . On the other hand,

P( ) [ (n + l) + (n + 2) + . . . + (n + 2P + l) ] 2P+1 ( 1)2p+1 n < 2p + 1

= n + p +

from the AM-GM inequality. If P(n) = m2p+1 , then m E {n + 2, . . . , n + pl . Assume by way of contradiction

that there is k E {2, 3 , . . . , p} such that P(n) = (n + k)2p+1 . Then

(n + 1) (n + 2) . . . (n + k - 1) (n + k + 1) . . . (n + 2p + 1) = (n + k)2P . (1)

We have two cases: I. k = p 1) If n ==

0 (mod p) , then (n + k)2p is divisible by p2p . The left-hand side of the equality (1) is clearly not divisible by p2p, hence we reach a contradiction.

2) If n == r (mod p) , r f. 0, then the left-hand side of the equality (1) is divisible by p2 , because of the factors n + p - r and n + 2p - r, while the right-hand side is not, since (n + k)2p == r2

P. This is a contradiction.

II. k E {2, 3, . . . ,p - l}

62 2. NUMBER THEORY

1) If n == -

k (mod p), then the right-hand side of

(1) is divisible by p2p , but the

left-hand side is not. 2) If n == -q (mod p

), q

f. k and q E

{O, 1, . . . ,p -

I}, then the left-hand side of (1)

is divisible by p. On the other hand (n +

k)2p == (q -

k)2 (mod p)

� 0 (mod p

),

because 0 < Iq -

kl < p.

Both cases end up in contradictions, so the problem is solved. (Dorin Andrica)

16. It suffices to prove that there are no integers in the interval

(nvp -�, ny'P

+ �] for n � [ a ] + 1.

n n JI

-2avp

Assume by way of contradiction that there is integer k

such that a a

ny'P

- - <k< n

y'P+ - .

n - n Hence

so

a2

a2

n2p + 2" -

2ay'P

< k2

< n2p + 2" + 2a

y'P.

n - n a2 [ a ] a2

Observe that 2" -2a..,fP

> -1. If n � J +

1, then 2" +

2a..;p

< 1

n 1-2avp

n

n2p -

1 < k2

< n2p +

1

It follows that k2 = pn 2 or

vp = � , which is false, since p is prime. n (D

orin Andri

ca, Gazeta Matematica (GM-B) , No. 8(1977)

, pp. 324

, Problem 16804)

17. For n = 1

the claim is clear, so let n � 3.

Define Sn = (�) + (;) + . . . + (n�l) . Then

or S!,- = 2n-1 - 1

. Because Sn is odd it follows that the sum Sn contains an odd number of odd terms, as desired. ( Ti


2(1984), pp.

71,

Problem 5346)

18. Because (:) = (m': n

). we can assume that m :O; [iT

If n = 0, then

1 = 1984

, false. If n =

1, then m =

1984.

2.2. SOLUTIONS

m(m - 1) If n = 2, then 2 = 1984, with no integral solutions.

63

If n :::: 3, then (:) :::: (�) , so 1984 :::: m(m - � (m - 2)

. Hence 11904 ::::

m3 - 3m2 + 2m or (m - 30) (m2 + 27m + 812) � -12456 < 0, and so m < 30. This . h (m) -I- 4 m (m - 1) . . . (m - n + 1) . implIes t at r 198 , because , does not contam the

n n. factor 31 of 1984.

To conclude, the solutions are m = 1984, n = 1 and m = 1984, n = 1983. ( Titu


Problem 5)

or

or

19 . The equation is equivalent to

(x + 2y) (x + 4y) - 3 (x + 2y) = 3,

(x + 2y) ( x + 4y - 3) = 3. We have

(I') { X + 2y = 3 ( ) ( 1 ) , with solution x, y = 2, -2 x + 4y - 3 = 1

(" ) { x + 2y = 1 . h I ' ( ) ( 4 5) 11 , WIt so utlOn x, y = - , - . x + 4y - 3 = 3 2

Note that there are no solutions in integers, as claimed. ( Titu


Problem 312)


x2y2 - 2xy + 1 + x2 + y2 - 2xy + 2(x - y) (l - xy) = 4,

or (xy - 1) 2 + (x - y) 2 + 2 (x - y) (1 - xy) = 4.

Hence (1 - xy + x - y)2 = 4 and, consequently, 1 (1 + x) (l - y) 1 = 2. We have two cases: I. (1 + x) (l - y) = 2. Then a) 1 + x = 2, 1 - y = 1, so x = 1, y = 0. b) 1 + x = -2, 1 - y = -1 , so x = -3, y = 2. c) 1 + x = 1 , 1 - y = 2, so x = 0, y = -1 . d) 1 + x = -1 , 1 - y = -2, so x = -2, y = 3. II. (1 + x) (l - y) = -2. Then

64 2. NUMBER THEORY

a) 1 + x = 2, 1 - y = -1 , so x = 1, y = 2. b) 1 + x = -2, 1 - y = 1 , so x = -3, y = O. c) 1 + x = 1 , 1 - y = -2, so x = 0, y = 3. d) 1 + x = -1, 1 - y = 2, so x = -2, Y = -l. (Titu

Andreescu, Revista Matematica Timi§oara (RMT) , No. 4-5(1972) , pp. 43,

Problem 1383)

21 . The equation is equivalent to

We have the cases: 1) x - pq = 1, y - pq = p2q2 , so X = 1 + pq, y = pq(l + pq) . 2) x - pq = p, y - pq = pq2 , so X = p(l + q) , y = pq(l + q) . 3) x - pq = q, y - pq = p2 q, so x = q(l + p) , y = pq(l + p) . 4) x - pq = p2 , Y - pq = q2 , so X = p(p + q) , Y = q(p + q) . 5) x - pq = pq, y - pq = pq, so x = 2pq, Y = 2pq. The equation is symmetric, so we have also: 6) x = pq(l + pq) , y = 1 + pq. 7) x = pq(l + q) , y = p(l + q) . 8) x = pq(l + p) , Y = q(l + p) . 9) x = q(l + q) , Y = p(p + q) . (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp. 45,

Problem 3486)

for all n � l.

Xn+l = 2xn + Yn and Yn+l = Xn + 2Yn

By induction we obtain that 3n = x2 - y

2 n >_ 1 n n '

Denote by Pk the k-th prime number. Then x = xPlo ' Y = YPlo ' U = 3, v = Pk is a solution of equation x2 - y

2 = UV for any integer k > O. Al

ternative so

lution. Let p and q be two arbitrary primes, p � 3. Then pq = 2k + 1,

for some positive integer k. Because 2k+1 = (k+1)2 -k2 , it follows that all quadruples (pq + 1 pq - 1 ) . . (x, y, u, v) = -2-' -2- ' P, q satIsfy the equatIOn. (Dorin Andrica) 23. The equation is equivalent to

(x - y) (x - z) (y - z

) = 2.

2.2. SOLUTIONS 65

Observe that (x - y) + (y - z) = x - z. On the other hand, 2

can be written as a product of three distinct integers in the following ways

i) 2= (-1) · (-1) · 2,

ii) 2 = 1 . 1 . 2,

iii) 2 = (-1) · 1 · (-2) .

Since in the first case any two factors do not add up to the third, we only have three possibilities :

a) { x - y =

l

x - z = 2

so (x, y, z) = (k + 1, k, k -1) for some integer

k;

y - z =l

b) { X - y = -

2

x - z = -1

so (x, y, z) = (k - 1, k

+ 1, k) for some integer

k;

y - z =l

{ x - y =l

c) x - z = -1

so (x, y, z) = (k, k - 1, k

+ 1) for some integer

k.

y - z = -2. ( Ti

tu Andreescu, Revista Matematidl. Timi§oara (RMT) , No. 1-2(1989) , pp. 97,

Problem 2)

24. We have xyz - (xy + yz + zx) + x + y + z -

1 = 1,

and, consequently, (x - 1) (y - 1) (z - 1) = 1.

Because x, y, z are integers , we obtain

x -I = y -

1 = z -

1 = 1,

so x = y = z = 2. ( Ti

tu Andreescu, Revista Matematica Timi§oara (RMT) , No. 3 (1971) , pp.

26,

Problem 487)

25. Multiplying by 8 yields 8xy (x2 + y2 ) =

16z4

or (x + y)4 _ (x _ y)4 = (2z)4 ,

and so (x - y)4 + (2z)4 = (x + y)4 .

This is Fermat's equation for the case n = 4 and it is known that this equation has solutions only if x - y =

0 or

2z =

0 (see T. Andreescu, D. Andrica, "An Introduction

to Diophantine Equations" , GIL Publishing House, 2002

, pp. 85-87) .

66 2. NUMBER THEORY

Case I. If x - Y = 0, then x = Y and Z =

±x. The solutions are

x = y = m, z = ±m

for any integer m. Case II. If z =

0, then

(x - y

)4 = (x + y

)4 and so x =

0 or y =

O. The solutions

are x =

O, y = m, z =

o

and x = m, y =

O, z =

o

for any integer m. (Titu


1 (1978), Problem 2813

; Gazeta Matematica (GM-B) , No. 11 (1981)

, pp. 424

, Problem 0:264)

26. Consider the sequences (Xn)n�l '

(Yn)n�l '

(Zn)n�l , defined by

Xn+2 = 592

xn ' Xl = 1, X2 =

14

for all n � 1.

Yn+2 = 592

Yn' YI = 3, Y2 =

39

Zn+2 = 592

zn , Zl = 7, Z2 =

42

It is easy to check that x�

+ y�

+ z; = 59n, for all integers n �

1. (D

orin Andrica, Romanian, Mathematical Olympiad - second round,

1979, Revista

Matematica Timi§oara (RMT) , No. 1-2(1980)

, pp. 58, Problem

4075)

27. Observe that the equation

(1) has infinitely many solutions in distinct nonnegative integers, for example

for any integer k

� 0.

Let (Xkl , Ykl , Zkl

) and

(Xk2 ' Yk2 ' Zk2

) be two solutions of equation (1) with

kl f. k

2 . Then

and multiplying yields (Xk1 Xk2

)n + (Xkl Yk2

)n + (Ykl Xk2

)n + (Ykl Yk2

)n = (Zkl Zk2

)n-1

This means that

2.2. SOLUTIONS 67

is a solution to the equation

Since kl f. k2 are arbitrary positive integers, the conclusion follows.

For the second equation, the proof is similar, based on the fact that the equation

has infinitely many solutions in distinct nonnegative integers, for example

Xk = 1

+ kn, Yk =

k(l + kn) , Zk =

1 + kn,

for any integer k � O.

so

(Dorin Andrica)

28. It is clear that x = 0, Y =

0 is a solution to the equation

xn + yn = xn-1 + yn-l

Let a f. -1

be a rational number such that y = ax. Hence

1 + an-1

1 + an-1

x = 1 + an ' y = a 1

+ an U sing the symmetry of the equation, we also have the solution

1 + an-1

x = a , 1 + an

1 + an-1

y = 1 + an

with a f. -1

rational. If a = - 1 and n >

1, then again x = y = O. This concludes the solution.

(2)

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No.

2(1981), pp.

62,

Problem 4578)

29. Let x = 3u, x +

2 = 3v, x +

8 = 3t so U + v + t = y. Then

3v -3u =

2 and

3t -3u =

8.

It follows that 3U(3V

-U -1) = 2

and 3U(3t-u - 1)

= 8.

Hence u = 0

and 3v - 1 = 2, 3t - 1 = 8, therefore v =

1, t =

2.

The solution is x = 1, y =

3. ( Ti


2(1978), pp.

47,

Problem 2812


, pp. 496

, Problem 18541)

30. If x, y � 2

then 1 + x

l and

1 + y

l are both odd and (x + y) l is even. Hence

the equation has no solutions.

68 2. NUMBER THEORY

Consider the case x = 1. The equation becomes

2(1 + y

l) = (1 + y

) l

and it is not difficult to notice the solution y = 2. If y �

3, then

3 divides

(1 + y

) l but

not 2(1

+ yl )

and y = 1

does not satisfy the equality. Hence x =

1, y =

2 or x =

2, Y =

1 due to the symmetry of the equation. ( Ti

tu Andreescu, Revista Matematidl. Timi§oara (RMT) , No.

2(1977), pp.

60,

Problem 3028


, pp. 75, Problem 0:118)

31 . Without loss of generality we may assume that x � y � z. The equation is equivalent to

zl [x(x -

1) . . .

(z +

1) + y

(y -

1) . . .

(z +

1) + 1]

= 2vl .

If z � 3, then the right-hand side is divisible by

3 but the left-hand side is not,

so z � 2. We have two cases.

I. z = 1. Then we have

xl + y

! = 2vl -

1,

or y! [x(x -

1) . . .

(y +

1) + 1]

= 2vl -

1.

If y � 2, then the right-hand side is an even number but the left-hand side is odd,

so y = 1. Then

or

xl

= 2(2VI -1 -

1).

If x � 4, then

2(2VI -1 -1)

== 0

(mod 8), false.

It remains to examine the cases x = 1, x =

2, and x =

3.

If x = 1, then

1 = 2(2V I-1 -

1), impossible.

If x = 2, then

2 = 2(2VI-1 -

1) or v

! = 2, so v =

2.

If x = 3, then

2vl-1 -

1 = 3

or vI

= 3, false.

Hence the only solution in this case is

x = 2, Y =

1, z =

1, v =

2

II. z = 2. Now we have

xl + y

l = 2vl -

2,

yl [x(x -

1) . . .

(y +

1) + 1]

= 2(2VI -1 -

1).

If y � 4, the

2(2VI-1 -1)

== 0

(mod 8), false.

It follows that y = 1, y =

2, or y =

3.

If y = 1, then x

l = 2vl -

3. Since x �

2 implies

2vl -

3 == 0

(mod 2)

false, then x = 1 , v =

2.

If y = 2, then x

l = 2vl -

4. We must have v �

3 so x

l = 4(2VI -2 -

1).

If x � 3, then

4(2VI-2 -1)

== 0

(mod 8), false.

2.2. SOLUTIONS 69

Hence x = 1, x =

2 or x =

3 and all those cases lead to a contradiction.

If y = 3, then x

l =

2vl -

8. Then v �

3 and x

l =

23 (2VI-3 -1)

� 23

•

7. It

follows that x � 5

and because x = 5

does not yield a solution and x � 6

implies 23 (2v!-3 -1)

== 0

(mod 16)

, which is false, we do not obtain a solution here. In case II we have found only

x = 1, y =

1, z =

2, v =

2,

which does not satisfy the condition x � y � z.

To conclude, we have the solution - from case I -

x = 2, Y =

1, z =

1, v =

2

and, due to the symmetry of the equation, we also have

and

x = 1, y =

2, z =

1, v =

2

x = 1, y =

1, z =

2, v =

2.

( Titu

Andreescu, Revista Matematidl. Timi§oara (RMT) , No.

2(1981), pp. 62,

Problem 4576)


or XI[X2 . . . Xn - (n -

l)x2 . . . Xn-l - . . . -

2X2 -

1] = 1.

Hence Xl = 1

and

X2[X3 . " Xn - (n -

1)x3 • • , Xn-l - • . • -

3X3 -

2] = 2.

Since X2 f:. Xl , it follows that X2 =

2 and

X3[X4 . . • Xn - (n -

l)x4 . • . Xn-l - • . • -

4X4 -

3] = 3.

Because X3 f:. X2 and X3

f:. Xl , we obtain X3 =

3.

Continuing with the same procedure we deduce that Xk = k

for all k. R

emark. Turning back to the equation we find the identity

1 + 11 . 1 + 2! . 2 + . . . + (n -

1) ! (n -

1) = nIno

( Titu

Andreescu, Revista Matematidl. Timi§oara (RMT) , No.

3(1973), pp.

23,

Problem 1509)

33. We proceed by induction on n. For n = 1

the claim is true. Using the identity (X2 - y

2) (U2

_ v2 )

= (xu + YV)2 -(xv + YU

)2

70 2. NUMBER THEORY

and the fact that the claim holds for n we deduce that the property is valid for n + 1,

as desired. Alternat

ive so

lution. We have

n n n Pn =

II (a�

-

b�) = II (

ak -bk) II (

ak + bk)

= k=l k=l k=l

where An , Bn are integers. Hence

as claimed. (Dorin Andri

ca, Revista Matematica Timi§oara (RMT

) , No.

2(1975), pp.

45,

Problem 2239

; Gazeta Matematica (GM-B

), No.

7(1975), pp.

268, Problem

15212)

34. Subtracting the equalities yields (x + y - xy

) + (v + y - vt

) = (x + y - xy

) (v + t - vt

),

or [( x + y - xy

) -

1] [ (v + t - vt

) -1]

= 1,

so (1 - x

) (1 - y

) (1 - v

) (1 - t

) = 1.

It follows that 1 1 - x

l = 1 1

- yl = 1 1

- vi

= 1 1 - t

l = 1,

and using (1)

we obtain (x, y, z, t

) = (0,0,0,0), (0,0,2,2), (0,2,0,2),

(0,2,2,0), (2,0,0,2), (2,0,2,0), (2,2,0,0)

and (2,2,2,2).

Turning back to the system we obtain (x, y, z, v, t

) = (0,0,0,0,0), (0,0, -4, 2, 2), (0,2,0,0,2),

(0,2,0,2,0), (2,0,0,0,2), (2,0,0,2,0), (2,2, -4,0,0)

and (2,2,24,2,2)

(1)

(Titu Andreescu, Revista Matematica Timi§oara (RMT

) , No.

2(1978), pp.

46,

Problem 3431

; Gazeta Matematica (GM-B

), No.

5(1981), pp.

216, Problem

18740)

35. We start with a useful lemma. Lemma. If a an

d b are re

latively pr

ime pos

itive

integers, t

hen t

here are pos

itive i

ntegers u and v suc

h that

au -bv =

1.

2.2. SOLUTIONS 71

Proof. Consider the numbers 1 . 2, 2 · a, . . . ,

(b -1) . a (1)

When divided by b

the remainders of these numbers are distinct. Indeed, otherwise we have

kl f. k

2 E {I , 2, . . . , b - I} such that

kla = Pl

b+ r,

k2a = P2

b+ r

for some integers PI ,P2 ' Hence

Since a and b

are relatively prime it follows that I kl -

k21

== 0 (mod b), which is

false because 1 ::;

I kl -

k21

< b.

On the other hand, none of the numbers listed in (1) is divisible by b. Indeed, if

so, then there is k

E {I , 2, . . . , n - I} such that k . a = P .

b for some integer p.

Let d

be the greatest common divisor of k

and p. Hence k = kl d, P = PI

d, for

some integers PI , kl with ge

d(pI '

kl) = 1. Then

kl a = PI

b and since ge

d( a, b) = 1 , we

have kI =

b, PI = a. This is false, because

kl <

b.

It follows that one of the numbers from (1) has the remainder 1 when divided by b so there is U E {I , 2, . . . ,

b - I} such that au =

bv + 1 and the lemma is proved.

We prove now that the system { ax - yz - e =O

bx - yt +

d = 0

with a, b, e,

d nonnegative integers and ge

d(a, b) = 1 has at least a solution in nonneg

ative integers. Because ge

d(a, b) = 1

using the lemma, there are positive integers u and v such that au -

bv =

1. Hence

x = eu + dv, y = a

d + be, z = v, t = u,

is a solution to the system. ( Titu

Andreescu, Revista Matematidl. Timi§oara (RMT) , No. 2(1977) , pp. 60,

Problem 3029)

36. Consider the determinant

1 1 1 Xl X2 P �

= Xp =

II (Xi - Xj ) i ,i=l p-l Xl

p-l X2 p-l Xp i >i

72 2. NUMBER THEORY

Summing up all columns to the first one and applying the hypothesis yields �

== 0 p

(mod p), hence

II (Xi - Xj ) == 0 (mod p).

i,j=l i>j

Because p is a prime number, it follows that there are distinct positive integers k, 1 E

{I, 2, . . . ,p

} such that Xk - Xl == 0 (mod p

). (D

orin Andrica)

37. Let

d b -

d (al + a2 a2 + a3 an + al ) a = gcd(al ' a2 , . . · , an) an - gc 2 ' 2 , . . . , 2

Then ak = aka, for some integers ak, k = 1,2, . . . , n. It follows that

ak + ak+l ak + ak+l 2 = 2 a, (1)

where an+l = al and an+l = al ' Since ak are odd numbers, ak are also odd, so ak + ak+l . 2 are mtegers.

From relation (1) it follows that a divides ak +2ak+l for all so a divides

b.

ak + ak+l . On the other hand, 2 = f3kb, for some mtegers

f3k . Then

ak + ak+l == 0 (mod 2b)

for all k E {I, 2, . . . , n

}. Summing up from k = 1 to k = n yields 2(al + a2 + . . . + an

) == 0 (mod

2b)

Since n, al , a2 , . . . , an are all odd al + a2 + . . . + an � 0 (mod 2), hence

al + a2 + . . . + an == 0 (mod b).

Summing up for k = 1 , 3, . . . , n -2

implies

and furthermore

al + a2 + . . . + an-l == 0 (mod 2b)

al + a2 + . . . + an-l == 0 (mod b).

(3)

(4)

Subtracting (4) from (3) implies an == 0 (mod b), then using relation

(2) we obtain

ak == 0 (mod b)

for all k. Hence bla and the proof is complete. ( Ti


Problem 2814)

38. It is known that cp(kl) = cp(k)cp(l)

for any relatively prime positive integers k and l.

2.2. SOLUTIONS

On the other hand, it is easy to see that if P is a prime number, then

<p(pl)

= pI _ pl-I for all positive integers

l.

Let n = 2· 3m

, where m is a positive integer . Then

<p(n)

= <p(2· 3m) = <p

(2)<p(3m) =

3m -3m-1 =

2· 3m-1 = � 3

for infinitely many values of n, as desired.

73

(Dorin Andri

ca, Revista Matematica Timi§oara (RMT) , No. 1(1978) , pp. 61, Problem

3255)

39. For n = 1 to n = 6

it is easy to check the claim. For n � 7 note that the number of even positive integers less than n is [�] . Moreover, the number of positive

multiples of 3

less than n which are odd is [i] - [�] . Then

7r(n)

< n - [�] - ( [i] - [�] ) , n � 7.

Since x - I < [x] ::; x, it follows that

7r( n)

< n - (� - 1) - (� - 1) + � = � + 2

2 3 6 3 '


ca, Revista Matematica Timi§oara (RMT) , No. 1 (1978) , pp. 61, Problem

3256)

40. Summing the inequalities Pk+1 - Pk � 2

from k

= 1 to k

= n - 1 yields Pn -

2 � 2 (n - 1) and so Pn >

2n - 1, n � 1 .

Then n(n + 1) 2

PI + P2 + . . . + Pn > 2

2 - n = n .

The inequality al"

+ a;" : . . . + a;:' ;::

(a1 + "" : . . . + an ) m holds for any positive real numbers aI , a2 , . . . , an . Hence

Pl"

+ pr + . . . + P::' ;:: n (pJ + 1'2 : . . . + Pn r > n (:) m = nm+1 ,


ca, Revista Matematica Timi§oara (RMT) , No. 2 (1978) , pp. 45,

Pro blem 3483)

41. Let S

= L d

and let <p(n)

be the number of numbers less than n and d <2n

gcd(d, n)=l

relatively prime with n.

74 2. NUMBER THEORY

Note that

gcd(n, d) = 1 {:} gc

d(n, n -

d) = 1

{:} gcd(n, n +

d) = 1

(1)

Let d1 , d2 , • • • ,

dcp(n) be the numbers less than n and relatively prime with n. From (1)

we deduce that d1 +

dcp(n) = n

d2 +

dcp(n)-l = n

hence

On the other hand,

d<n gcd(d, n ) = l

d = ncp

(n)

. 2

L d=

L (n +

d) = ncp

(n) +

L d=

Therefore

n < d <2n gcd(d,n)=l

d<n gcd(d ,n)=l

_ ( ) ncp(n)

_

3ncp

(n)

- ncp n + -- - . 2 2

d<n gcd(d ,n)=l

s = ncp(n)

+ 3ncp

(n)

= 2ncp

(n). 2 2


2(1981), pp.

61,

Problem 4574)

42. We proceed by induction. For n = 3

the claim is true. Assume that the hypothesis holds for n -

1. Let

1 < k < n

! and let kl, q be the quotient and the

remainder when k is divided by n. Hence k = kin + q, 0

� q < n and 0

� kl < � < n

n!

- = (n -

I) !.

n From the inductive hypothesis, there are integers

di <

d� < . . . <

d�, s �

n -1, such that

di l (n -

I) !, i =

1,2, . . . , s and kl = di + d� + . . . +

d�. Hence

k = ndi + nd� + . . . + nd� + q. If q =

0, then k = d1 + d2 + . . . +

ds , where

di = n

di, i

= 1,2, . . . , s, are distinct divisors of n

!.

If q f. 0, then k = d1 + rh + . . . +

ds+1 , where

di = n

dL iI, 2, . . . , s, and

ds+1 = q.

It is clear that diln!, i =

1,2, . . . , s and

ds+1

In!, since q < n. On the other hand, d

S+1 < d1 <

d2 < . . . <

ds , because

dS+1 = q < n � n

di = d1 . Therefore k can be

written as a sum of at most n distinct divisors of n!, as claimed. ( Ti


2(1983), pp.

88,

Problem C4

:10)

2.2. SOLUTIONS 75

43. If n � 992, take the set with all 992 odd numbers from {I , 2, . . . , 1984} . Its complementary set has only even numbers, any two of them not being relatively prime. Hence n � 991 . Let c be the complementary set of a subset with 991 elements of the set {I , 2, . . . , 1984}. Define D = {c + 1 1 c E C} . If C n D = 0, then C U D has 2 · 993 = 1986 elements but C u D C {I , 2, . . . , 1985}, which is false.

Hence C n D f. 0, so there is an element a E C n D. It follows that a E C and a + 1 E C and since a and a + 1 are relatively prime we are done. ( Ti


Problem C4:7)

44. Let a be the greatest odd integer such that a2 < n, hence n < (a + 2)2 . If a � 7, then a - 4, a - 2, a are odd integers which divide n. Note that any two of these numbers are relatively prime, so (a - 4) (a - 2)a divides n. It follows that (a - 4) (a - 2)a < (a + 2)2 so a3 - 6a2 + 8a < a2 + 4a + 4. Then a3 - 7a2 + 4a - 4 � 0 or a2 (a - 7) + 4(a - 1) �

O. This is false, because a � 7, hence a = 1 , 3 or 5.

If a = 1, then 12 � n < 32 , so n E {I , 2, . . . , 8} . If a = 3, then 32 � n < 52 and 1 . 3 1n, so n E {9, 12, 15, 18, 21, 24} . If a = 5, then 52 � n < 72 and 1 . 3 . 51n so n E {30, 45} . Therefore n E

{2, 3, 4, 5 , 6, 7, 8, 9, 12, 15, 18, 21, 24, 30, 35}. (Dorin Andrica and

Adrian P.

Ghioca, Romanian Winter Camp 1984; Revista

Matematica Timi§oara (RMT) , No. 1 (1985) , pp. 78, Problem T.21)

45. We prove by induction that

u� - 2v� = 1 , n � 1 . (1)

For n = 1 the claim is true. Assuming that the equality is true for some n, we have

U�+l - 2V�+1 = (3un + 4Vn)2 - 2(2un + 3Vn )2 = u� - 2v� = 1 hence (1) is true for all n � 1 .

We prove now that

Indeed, 2x� - Y� = 1, n � 1

2x� - Y� = 2(un + Vn )2 - (un + 2vn)2 = u� - 2v� = 1 ,

as claimed. It follows that

( Xn V2 - Yn) (Xn V2 + Yn) = 1, n � 1.

Notice that xn..,fi + Yn > 1 so

0 < xnV2 - Yn < 1, n � 1.

Hence Yn = [Xn V2] , as claimed.

(2)

76 2. NUMBER THEORY


11 (1979), pp. 430, Problem 0

:97)

46. (i) We have 2k-1 2k-2 .2 2 2 Xk = 2

+ 1 = 2

+ 1

= (Xk-l -

1) + 1

= Xk-l -2Xk-1 +

2,

hence Xk -

2 = Xk-l

(Xk-l -

2).

Multiplying the relations (1)

from k

= 2

to k

= n yields

and since Xl = 3, we obtain

Xn = Xl X2 . . . Xn-l + 2

for all n 2:: 2.

For a different proof use the identity 2n- 1 n-l x -

I II ( 2k- 1 ) 1 = X +

1.

X - k=l

(1)

(2)

(ii) From relation (2) we obtain gcd(xn ' x

d gcd(xn' X2

) = . . . =

gcd(xn , Xn-l

) =

1 for all n, hence gc

d(Xk ' xz

) =

1 for all distinct positive integers k

and l.

(iii) Since X2 = 5 and XIX2 . . . Xn-l is odd, using relation (2)

it follows that Xn end with digit 7 for all integers n 2:: 3.

and

and

(Dorin Andrica)

47. We have

Then

a�+2 + a

�+l -

4an+2an+1 = -

2,

Subtracting these relations yields

a�+2 - a

� -4an+l

(an+2 - an

) = °

(an+2 - an

) (an+2 + an -

4an+

d = 0, n 2::

1.

Since the sequence (an)n2:1 is increasing, it follows that

2.2. SOLUTIONS 77

Taking into account that al = 1, a2 = 3 and inducting on n we reach the conclusion. ( Ti

tu Andreescu, Gazeta Matematica (GM-B) , No. 11(1977) , pp. 453, Problem

16947; Revista Matematica Timi§oara (RMT) , No. 1 (1978) , pp. 51, Problem 2840)

2 2

48. For n = 2

we have a2 = 1 + 2 = 3" '

Suppose that an = Pn , where Pn, qn are integers and gcd(Pn' qn

) = 1 .

qn Then 2

pnqn an+1 = 2 + 2 2 ' qn Pn

and it suffices to prove that gcd(2

Pnqn' q� + 2p�)

= 1 . Assume by way of contradiction that there are integers

d and

kl ' k2 such that

2Pnqn =

kId

and q� + 2p�

= k2d.

If d

= 2, then q

� =

2k2 -

2p�, so qn is even. This is a contradiction since

ql , q2 , · . . , qn are all odd. If d

> 2

then q�

= (k2qn -

klPn

)d so

dlqn '

Since q� + 2p�

= k2d

it follows that dlPn , which is false, because gc

d(Pn' qn

) = 1 .

This proves our claim. We prove that

q� -2p�

= 1, n 2:: 2.

For n = 2

we have q� -2p�

= 32 -2 . 22 = 1.

Suppose that q� -2p�

= 1. Because

is irreducible, we obtain

Hence

as claimed. Hence

2 2 2 ( 2 2 2 )2 8 2 2 ( 2 2 2 )2 1 qn+1 - Pn+1 = qn + Pn - pnqn = qn - Pn = ,

1 bn =

1 -2 2 an-I

1 q�-I --::-_.:.:-=-= _ _ q2 2 = 2 2

2 - n-I 1 _

2Pn-1 qn-I - Pn-I 2 q'rl-I

and so bn is a perfect square for all n 2::

2.

(1)

( Titu


Problem 3083; Gazeta Matematica (GM-B) , No. 1 (1981) , pp. 44, Problem C :88)

78 2. NUMBER THEORY

49. Inducting on n we obtain

Hence

2 2 2 - 1 xn - Yn - ,

Z = 1 + 4X2 2 = (X2 _ 2y2 )2 + 4X2 y2 = (X2 + 2y2 )2 _ 4X2 y2 = n nYn n n n n n n n n

= (x; + 2y� + 2xnYn) (x; + 2y� - 2xnYn) , for all n � 1 .

Since both factors are greater than 1, it follows that all numbers Zn are not prime. ( Titu


Problem 2571)

50. We have

or

Hence

and Xn =

Jp2 +

1Xn+1

±PJX;+1 + 1 .

From hypothesis we have Xn+l > Xn , hence

Xn = Jp2

+ 1Xn+1 - p

JX�+1 + 1. (1)

But (2)

and so (3)

by summing up the relations (1) and (2) . Because p is a positive integer, it follows that

Jp2 + 1 is irrational. From (3)

we deduce that among any three consecutive terms of the sequence there is at least an irrational term. Hence there are at least [;] irrational terms among the first m

terms of the sequence. ( Titu

Andreescu, Revista Matematica Timi§oara (RMT) , No. 1-2(1980) , pp. 68,

Problem 4139; Gazeta Matematica (GM-B) , No. 6 (1980), pp. 281, Problem C:48)

51 . Setting n = 0 and n = 1 yields Xl = x� and X2 = x� , hence Xl = X2 = 1 . From the given condition we obtain

2 2 d 2 2 X2n+ 1 = Xn+l + Xn an X2n = Xn+l - Xn-l .

2.2. SOLUTIONS

Subtracting these relations implies

hence X2n+1 = X2n + X2n-l , n �

1.

We induct on n to prove that

X2n = X2n-1 + X2n-2 , n � 1.

Indeed, X2 = Xl + Xo and assume that (2)

is true up to n. Then 2 2 2 2 (*) ( )2 2 2 X2n+2 - X2n = Xn+2 - Xn - Xn+l + Xn-l = Xn+l + Xn - Xn - Xn+l + +(Xn+l - Xn)2 = X

�+l + X� = X2n+l ,

as claimed (the equality (*) holds because of (1)

and the induction hypothesis) .

79

(1)

(2)

From relations (1)

and (2)

it follows that Xn+2 = Xn+l + xn for all n � O. Because

Xo = 0

and Xl = 1

the sequence (xn)n2:0 is the Fibonacci's sequence, hence

( Titu

Andreescu, Romanian Winter Camp

1984; Revista Matematidi Timi§oara

(RMT) , No. 1 (1985)

, pp. 73, Problem T.3)

52. From the hypothesis it follows that a4 = 12, a5 =

25, a6 =

48. We have �l ,

a2 a3 a4 a5 a6 an - = 1, - =

2, - =

3, - =

5, - =

8 so - =

Fn for all n =

1,2,3,4,5,6, where 2 3 4 5 6

n (Fn)n2:1 is the Fibonacci's sequence. We prove by induction that an = n

Fn for all n. Indeed assuming that ak =

kFk

for k � n +

3, we have

an+4 = 2(n + 3)Fn+3 +

(n + 2)Fn+2 -

2(n + l

)Fn+1 - n

Fn =

= 2(n +

3)Fn+3 +

(n +

2)Fn+2 -

2(n + l

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

= 2(n +

3)Fn+3 +

2Fn+2 -

(n + 2)

Fn+1 =

as desired.

= 2(n +

3)Fn+3 +

2Fn+2 -

(n +

2) (Fn+3 - Fn+2) =

= (n +

4)(Fn+3 +

Fn+2 ) =

(n +

4)Fn+4 '

(Dorin Andrica, Revista Matematidl. Timi§oara (RMT) , No.

1 (1986), pp.

106,

Problem C8

:2)

53. Observe that setting Xo = 0 the condition is safisfied for n = O.

We prove that there is integer k

� m3 such that Xk divides m. Let rt be the remainder of Xt when divided by m for t =

0,1, . . . , m3 +

2. Consider the triples

(ro , rl , r2 ) , (rl ' r2 , r3 ) , ' . . , (rm3 , rm3+1 , rm3+2 ) ' Since rt can take m values, it follows

80 2. NUMBER THEORY

by the Pigeonhole Principle that at least two triples are equal. Let p be the smallest number such that triple

(rp , rp+l , rp+2

) is equal to another triple

(rq , rq+l , rq+2

),

p < q :::; m3 . We claim that p = O.

and

Assume by way of contradiction that p 2:: 1. Using the hypothesis we have

rp+2 == Tp-l + rprp+l (mod m)

rq+2 == rq-l + rqrq+l (mod m) .

Since rp = rq , rp+l = rq+l and rp+2 = rq+2 , it follows that rp-l = rq-l so (rp-l , rp , rp+

d = (rq-l , rq , rq+l

), which is a contradiction with the minimality of p.

Hence p = 0, so rq = ro =

0, and therefore Xq ==

0 (mod m) . ( Ti

tu Andreescu and

Dore

l Mihet, Revista Matematica Timi§oara (RMT) , No.

1 (1986) , pp. 106

, Problem C8: 1)

54. Note that a2 = 1 , a3 = 4, a4 = 9, a5 = 25, so ao = F6

, al = F12 , a2 =

F?,

a3 = Fi

, a4 = F;

, a5 = Fi

, where (Fn)n2:o is the Fibonacci sequence.

We induct on n to prove that an = F�

for all n 2: O. Assume that ak =

Ff for all

k :::; n. Hence

and

-F2 an - n ' -

F2 an-l - n-l '

From the given relation we obtain an-2 =

F�_2 '

an -3an-l + an-2 = 2( -

l )n-r , n 2: 2 .

Summing up these equalities yields

Using the relations (1) and (2) we obtain

an+l = 2F�

+ 2F�_1 -

F�_2 =

(Fn +

Fn_1

)2 + (Fn -

Fn_1

)2 -F�_2 =

as desired.

(1)

(2)

( Titu


Problem C8:8)

55. The expressions a2 - 1 and 2 + 2a ask for the substitution a = � (b2 + b� ) . The equality � (b2 + b� ) = 97 implies (b

+ �) 2 = 196, hence b + � = 14. Setting

2.2. SOLUTIONS 81

b = c2 yields (c + �) 2 = 16, thus c + � = 4. Let c = 2 + va. We will prove by

induction that an = � (C4F. +

C4�. ) , n 2: 1 ,

where Fn is the nth Fibonacci number.

Indeed, this is true for n = 1, n = 2 and, assuming that

a = � (e4Fk + _1_) k < _ n k 2 e

4Fk '

implies

and 1 ( 1 ) 2 2

+ J2

+ 2an = 2 + e2Fn + -- = eFn + -e2Fn eFn

Let us mention that em + � is an integer for all positive integers m. em ( Ti

tu Andreescu, USA Mathematical Olympiad Shortlist, 1997)

56. We induct on m to prove that

if km � n <

km+1 , then an = 1 + m. (1)

For m = 0 the claim is true, since if 1 � n < k, then [�] = 0 hence an = l +ao =

1 + 0.

so Assume that (1) holds for m. Then if

km+1 � n < km+2 , we have

km � � < km+1 ,

km � [�] < km+1 .

Using the inductive hypothesis, we deduce that a[�] = 1 + m so

an = 1 + a[ �] = 1 + (m + 1) as claimed.

Therefore for any positive integer n with m � logk n < m + 1, we have an = 1 + m thus an = 1 + [log kn] . ( Ti

tu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1982) , pp.

66,

Problem 4997)

57. We prove that an -2 = b�, where bo = -1,

b1 = 1 and

bn+1 =

3bn -

bn-1 for

all n 2:: 1 . This is clear for n = 0 and n = 1 .

82 2. NUMBER THEORY

Assume that for any k

� n we have ak - 2 = b�. Subtracting the relation an = 7

an-l - an-2 from an+l = tan - an-l yields an+l =

8an -

8an-l + an-2 or

an+l - 2 = 8(an - 2) - 8(

an-l - 2) + (an-2 - 2) Hence

an+l - 2 = 8b�

-8b�

_1 + b�_2 =

8b� -8b�

_1 + (3b

n-1 -bn)2 =

= 9b�

-6bnbn-1 + b�-l =

(3bn -

bn_1

)2 = b�+l

Alternat

ive so

lution. The general term of the sequence is given by

Hence

_ ( 1 + v'5) 4n-2 ( 1 _

v'5) 4n-2 an - 2 + 2 ' n 2:: 0

and an inductive argument proves that ( 1 +2V5) 2n-l + (1 -2V5) 2n-l is an integer for all n. ( Ti

tu Andreescu

)

58. Note that all terms of the sequence are positive integers. Assume by way of contradiction that there is a positive integer no such that

Then

X�o = Xno -lXno+l '

Xno+l _ xno - t -- - -- - , xno Xno-l

where t is rational.

or

Because Xno+l = o:xno + (3Xno-l , it follows that Xno+l = 0: +

(3� xno Xno-l

t2 - o:t -

(3 = O.

The last equation has no rational roots since the discriminant �

= 0:2 + 4(3

is not a perfect square. This is a contradiction and hence the problem is solved. Al

ternative so

lution. From the condition of the problem we obtain (1)

2.2. SOLUTIONS

Assume by way of contradiction that there is a positive integer no such that 2 _ Xno - Xno -lXno+l '

From the relations (1)

and (2) we deduce that (a2 +

4,8)x�o =

(Xno+l +

,8xno-

d2

which is false, since a2 + 4,8

is not a square. This completes the proof. (Dorin Andrica)

83

(2)

59. Assume by way of contradiction that there is an irrational number a such that

is rational. Define a = \j a +

Ja2 - 1 and observe that \j a -

Ja2 - 1 = � .

a Hence A = a + � is rational.

a We prove that an + � is rational. an Indeed,

is rational and

a2 + _ = a + - - 2 1 ( 1 ) 2

a2 a

a3 + :3 = (a + � r - 3 ( a + �) is rational.

U sing the identity

ak + � = (ak-1 + _1 ) (a + �) _ (ak-2 + _1 ) ak ak-1 a ak-2

it follows by induction that ak + -\- is rational for all positive integers k, hence

a n 1 . . 1 a + - IS ratlOna .

an

Thus a + J

a2 - 1 + a -J

a2 - 1 = 2a is rational, which is false. The solution is complete. ( Ti

tu Andreescu, Romanian IMO Selection Test 1977; Revista Matematidi

Timi§oara (RMT

), No. 1

(1978

), pp. 78, Problem

3344)

60. We induct on n. For n = 1, from ±1 ± 2 ± 3 ± 4 ± 5 we obtain all odd positive

integers less than or equal to (2 + 1) (

4 + 1) = 15:

+1

-2+3+4

-5

=1

-1+2+3+4

-5

=3

-

1+2+3

-4+5

=5

84 2. NUMBER THEORY

-1+2-3+4+5=7

-1-2+3+4+5=9

+1 -2 + 3 + 4 + 5 = 11

-1 + 2 + 3 + 4 + 5 = 13

+1 + 2 + 3 + 4 + 5 = 15

Assume that from ± 1 ± 2 ±

. . . ± (4

n + 1)

with suitable choices of signs + and -we obtain all odd positive integers less than or equal to

(2n +

1) (4n +

1).

Observe that -(4n +

2) + (4n +

3) + (4n +

4) -(4n +

5) = O. Hence from

±1 ± 2 ±

. . . ± (4

n + 5)

for suitable choices of signs + and - we obtain all odd positive numbers less than or equal to

(2n +

1) (4n +

1).

It suffices to obtain all odd integers m such that

There are

(2n +

1) (4n +

1) < m :s;

(2n +

3) (4n +

5)

(2n +

3) ( 4n +

5) -(2n +

1) ( 4n +

1)

= 8n +

7 2

such odd integers m.

We have (2n +

3) (4n +

5) = +

1 + 2 + . . . +

(4n +

5)

(2n +

3) (4n +

5) -2k

= +1 + 2 + . . . +

(k -1) -k+

+ (k + 1) + . . . +

(4n +

4) + (4n +

5),

k = 1, 2, . . . ,

4n +

5

and (2n +

1) ( 4n +

5) -2l = +

1 + 2 + . . . +

(l -1) -l +

+ (l + 1) + . . . +

(4n +

4) + (4n +

5),

l = 1, 2, . . . ,

4n + l.

Hence all numbers m from (1)

are obtained, as desired.

(1)


2(1986), pp.

63, Problem

C570)

Chapter 3

GEOMETRY

PROBLEMS

1. Triangle ABC

has side lengths equal to a, b, c. Find a necessary and sufficient condition for the angles of the triangle such that a2 , b2 , e can be the side lengths of a triangle.

2. Prove that the triangle whose side lengths are equal to the length of the medians of a triangle

ABC has area equal to 3/4 of the area of triangle

[ABC].

3. In triangle ABC

the median AM meets the internal bisector BN

at P. Let Q

be the point of intersection of lines C

P and AB

. Prove that triangle B N

Q is isosceles.

4. In triangle ABC

the midline parallel to AB

meets the altitudes from A

and B

at points D

and E. The midline parallel to

AC intersects the altitudes from

A and

C

at points F

and G.

Prove that DCI IBFI IGE

.

5. Let M be a point in the interior of triangle ABC

. Lines AM

, B

M, C

M intersect sides

BC,CA

,AB

at points A',B',C', respectively. Denote by 81 , 82 , 83 , 84 , 85 , 86

the areas of triangles M A' B

, M A' C

, M B' C

, M B' A

, M C' A

and M C' B

, respectively. Prove that if

81 83 85 - 3 82 + 84 + 86 - ,

then M is the centroid of the triangle ABC

.

6. Let M be a point in the interior of triangle ABC

and let M, P, Q be three

colinear points on sides AB

, BC

and line CA

. Prove that if

SMAN + 5MBP = 2J SMAQ ,

8MBN 8MCP 8McQ

then MP is skew-parallel with AC

.

7. Let a, b, c and 8 be the sides lengths and the area of a triangle ABC

. Prove that if

P is a point interior to triangle

ABC such that

aPA + bPB + cPC = 48

87

88 3. GEOMETRY

then P is the orthocenter of the triangle.

8. Let 11 , 12 be the incenters of triangles AlBl Cl and A2B2C2 . Prove that if II and 12 divide the internal bisectors Al A� , A2A� in the same ratios they divide the internal bisectors BlBL B2B� , then triangles AlBl Cl and A2B2C2 are similar.

9. On side BC of a triangle ABC consider points M and N such that ifAjJ == CAN.

Prove that MB NB AB MC + NC

2: 2

AC

10 . Let M be a point on the hypotenuse BC of a right triangle ABC and let points N, P be the feet of the perpendiculars from M to AB and AC.

Find the position of M such that the length N P is minimal.

1 1 . Let ABC be an equilateral triangle and let P be a point in its interior. Let the lines AP, BP, CP meet the sides BC, CA, AB at the points AI , Bl , Cl respectively. Prove that

12 . Let S be the set of all triangles ABC for which ( 1 1 1 ) 3 6 5

AP + BQ + CR - min{P, BQ, CR} = r ' where r i s the imradius and P, Q, R are the points of tangency of the in circle with sides AB, BC, C A, respectively. Prove that all triangles in S are isosceles and similar to one another.

13. Let ABC be a triangle inscribed in a circle of radius R, and let P be a point in the interior of ABC. Prove that

PA PB PC 1

BC2 + CA2 + AB2 2:

R' 14 . Let I be the incenter of an acute triangle ABC. Prove that

AI · BC + BI · CA + CI · AB = SABe

if and only if triangle ABC is equilateral.

15 . Let ABC be a triangle such that

(8AB - 7BC - 3CA)2 = 6 (AB2 - BC2 - CA2) .

Prove that A = 60° .

3 . 1 . PROBLEMS 89

16. Let P

be a point in the plane of triangle ABC

such that the segments PA

, P B and

PC are the sides of an obtuse triangle. Assume that in this triangle the

obtuse angle opposes the side congruent to P A

. Prove that BAlJ is acute.

17. Triangle ABC

has the following property: there is an interior point P

such that PAJj = 10

°, PEA = 20

°, PcA = 30

°, and PAC = 40

°. Prove that the triangle ABC

is isosceles.

18. Let ABC

be a triangle such that max{A

,B} = C + 30

°. Prove that

ABC is

right angled if and only if !!. = J3 + 1 . r

19 . Prove that in the interior of any triangle ABC

there is a point P

such that the circum radii of triangle

PAB,PBC

,PCA

are equal.

20. The incircle of the triangle ABC

touches the sides AB

, BC

, CA

at the points

C', A', B', respectively. Prove that the perpendiculars from the midpoints of A'B'

, B'C'

, C'A'

to AB

, BC

, CA

, respectively, are concurrent.

21 . Let AIA2A3 be a non-isosceles triangle with the incentre

I. Let

Ci , i = 1 , 2, 3 ,

be the smaller circle through I tangent to AiAi+1 and

AiAi+2 (the addition of indices

being mod 3). Let

Bi , i = 1 , 2, 3, be the second point of intersection of

CHI and

Ci+2 .

Prove that the circumcenters of the triangles AIBII, A2B2I, A3B3I

are collinear.

22. On sides AB

and AC

of a triangle ABC

consider points B'

and C'

such that AB' AC'

B' B + C' C = k is constant.

Find the locus of the intersection point of lines BC'

and C B'

.

23. Let ABC

be an equilateral triangle of side length 1. Find the locus of points P

such that 2PA

·PB

· PC

max{PA

,PB

,PC} = PA .

PB+PB

. PC

+PC

· PA

-1

24. Prove that in any acute triangle Ja2b2 -482 +

Ja2c2 -

482 = a2 .

25. Prove that a triangle in which

is equilateral.

. r,;.- . r,;:: . � Jrarbrc y Ta + y rb + y rc = -'----r

90 3. GEOMETRY

26. Prove that a triangle is equilateral if and only if 1 1 1 1 1 1

- + - + - - - + - + -m2 m2 m2 - r2 r2 r2 . a b c a b c

27. A triangle with side lengths a, b, c has circum radius equal to

1. Prove that

a + b + c ;::: abc. 28. Prove that in any triangle

a2 6R

� (p -b) (p - c

) � -;:-.

29. Prove that in any triangle 1 1 1 1

h2 + h2 + h2 ;::: 3r2 a b c

30. Prove that in any triangle

Jrarb + Jrbrc + Jrcra ;::: gr.

31 . Prove that in any triangle

32. Prove that

for any triangle.

33. Let ABC be a triangle. Prove that A � i if and only if be (

p -b) (p - c

) � 4'

34. Three equal circles of radii r are given such that each one passes through the centers of the other two.

Find the area of the common region.

35. Let ABCD be a nonisosceles trapezoid with bases AB and CD. Prove that AC2 - BD2 AB + CD AD2 _ BC2 - AB - CD '

36. Let ABC D be a trapezoid with bases AB and CD. Prove that if

(AB + CD)2 = BD2 + AC2

3 .1 . PROBLEMS 91

then the diagonals of the trapezoid are perpendicular.

37. Prove that if in a trapezoid with perpendicular diagonals the altitude is equal to the midline, then the trapezoid is isosceles.

38. Let ABCD be a trapezoid with bases AB and CD. Prove that

AB2 - BC2 + AC2 AB2 - AD2 + BD2 AB CD2 - AD2 + AC2 = CD2 _ BC2 + BD2 = CD '

39. Prove that a trapezoid whose difference of the diagonal lengths is equal to the difference of nonparallel side lengths is isosceles.

40. Let ABCD be a cyclic quadrilateral. Prove that

lAB - CDI + lAD - BCI 2: 2 1AC - BDl ·

41 . Prove that a right trapezoid whose altitude length is equal to the geometrical mean of the lengths of its bases has perpendicular diagonals.

42. Let E

and F be the projections of the vertices A and B of a trapezoid ABCD on line CD. Let M and N be the projections of

E and F onto BD and AC, respectively

and let P and Q be projections of E

and F onto BC and AD, respectively. Prove that the quadrilateral M N PQ is cyclic.

43. Let ABC D be a convex quadrilateral that is not a parallelogram and let M and N be the midpoints of the diagonals AC and BD. Prove that numbers

AB + CD, BC + AD, AC + BD, 2MN

can be the side lengths of a cyclic quadrilateral.

44. Let ABCD be a cyclic quadrilateral and let lines AB and CD meet at point E.

Point F is the reflection of C across E. Prove that lines AF and B D are perpendicular

if and only if lines AB and CD are perpendicular.

45. Find all cyclic quadrilaterals having the side lengths odd integers and the area a prime number.

46. A cyclic quadrilateral has the side lengths a, b, c, d, the diagonal lengths e, f

and the semiperimeter p. Prove that

92 3. GEOMETRY

47. A cyclic quadrilateral has the side lengths a, b, c, d and the diagonal lengths

e and I.

Prove that if max{la - c l ,

I b - dl }

:::; 1 , then

Ie -I I :::; V2.

48. A cyclic quadrilateral has area S and semiperimeter p. Prove that if S = (�) 2 , then the quadrilateral is a square.

49. Let ABCD be a convex quadrilateral and let P be the intersection point of its diagonals.

Prove that SPAB + SPCD = SPBC + SPDA

if and only if P is the midpoint of AC or BD.

50. Let M be a point on the circumcircle of the cyclic quadrilateral ABCD and let points A' , B', C' , D' be the projections of M onto AB, BC, CD, DA, respectively. Prove that

(i) lines A' B' , C'D' and AC are concurrent (ii) lines B'C' , D' A' and BD are concurrent.

51 . A convex quadrilateral ABCD with area 2002 contains a point P

in its interior such that P A = 24, P B = 32, PC = 28, and P D = 45. Find the perimeter of ABC D.

52. Find the locus of points P in the plane of a square ABC such that 1

max(PA, PC) = V2

(PB + PD) .

53. Let P be the set of all quadrialterals with the same diagonal lengths and let >'1 (P)

and A2 (P)

be the lengths of the segments determined by the midpoints of two opposite sides of a quadrilateral p E P.

Prove that for all p E P the sum A�(P)

+ A�(P)

is constant and find the value of this constant.

54. Let ABCDEFGHIJKL be a regular dodecagon and let R be the circumradius.

Prove that AB AF AF

+ AB

= 4

and AB2 + AC2 + AD2 + AE2 + AF2 = 10R2 .

3. 1 . PROBLEMS 93

55. Prove that if inside a convex poligon there is a point such that the sum of the squares of its distances to the vertices of the poligon is twice the area of the poligon, then the poligon is a square.

56. Let AIA2 . . . An be a cyclic polygon and let P be a point on its circumcircle. Let PI , P2 , . . . , Pn be the proj ections of P onto the sides of the polygon. Prove that the product n PA� II

P�� i=l �

is constant.

57. Let AI A2 . . . A2n be a cyclic polygon and let M be a point on its circumcircle. Points KI , K2 ' . . . , K2n are the projections of M onto sides AIA2 , A2A3 , . . . , A2nAI and points HI , H2 , . . . , Hn are the projections of M onto diagonals AI An+I , A2An+2 , . . . , AnA2n · Prove that

MKI · MK3 . . . MK2n-l = MK2 · MK4 . · . MK2n = MHI · MH2 . . . MHn·

58. Find the circumradius of a cyclic polygon with 2n sides if n sides have the length a and n sides have the length

b.

59. Let P be a point in the interior of a tetrahedron ABCD such that its projections AI , BI , CI , DI onto the planes (BCD) , (CDA) , (DAB) , (ABC), respectively, are all situated in the interior of the faces. If S is the total area and r the inradius of the tetrahedron, prove that

SBeD + SeDA + SDAB + SABe > � PAl PBI PCI PDI - r

When does the equality hold?

60. Let AIA2A3A4 be a tetrahedron, G its centroid, and A� , A� , A� , A� the points where the circumsphere of AIA2A3A4 intersects GAl , GA2 , GA3 , GA4 respectively. Prove that

and GAl · GA2 · GA3 · GA4 � GA� . GA� · GA� . GA� 1 1 1 1 1 1 1 1

GA� + GA� + GA� + GA4 � GAl + GA2 + GA3 + GA4 ·

SOLUTIONS

1 . The positive real numbers a2, b2, c2 can be the side lengths of a triangle if and

only if

Because

c2

+ a2 -b2

cos B = 2 ' ca a2 + b2 - c2 cos C = 2ab '

(1)

relation (1) is equivalent to cos A > 0, cos B >

0, cos C >

O. Hence the necessary and

sufficient condition is that triangle ABC is acute. (Dorin Andrica, Revista Matematidl Timi§oara (RMT) , No. 2

(1978), pp.

48,

Problem 3507

)

2 . Let A' , B' , C' be the midpoints of sides BC, CA, AB respectively. Construct point M such that BC M C' is a parallelogram. Note that AC' C M and B B' M A' are also parallelograms hence AM = CC' and A'M = BB' . Hence the triangle determined by the medians is AA' M.

A M

B A' C Let N be the intersection point of lines B' C' and AA' . Because AC' A' B' is a

parallelogram, we have

C'N = B'N = !B'M 2 '

95

96 3. GEOMETRY

SO BI

is the centroid of triangle AAI M

. Hence

Alternat

ive so

lution: Consider the vectors a = BC, b = AC, c = AB, ma = AA

I,

mb = BBI

and observe that

We obtain 1 - -

3-

ma x mb = 4 (b + c) x (b - 2c) = 4 (c x b) ,

3 -hence

Ima x mb

l = 4 1c x bl and the conclusion follows. ( Ti

tu Andreescu

)

3. If B

P meets AC

at P, then by Ceva's theorem we obtain QA MB NC

QB ' MC ' NA = 1 .

Because M B = M C

, we have QA NA

QB NC '

Therefore QNI IBC

and then NBC = CiiiB. On the other hand, because

BN bisects angle --xiiC, it follows that NBC = QiiN.

Hence CiiiB = QiiN and triangle BNQ

is isosceles, as desired. ( Titu


1 (1978), pp.

66,

Problem 3286)

4. Let H

be the orthocenter of triangle ABC

and let AI

be the midpoint of side BC. Because

FAI I IAC, it follows that

FAI 1-BH

. Moreover BAI

1-HF

, so AI

is the orthocenter of triangle

B H F. Hence

HAI1-BF

. (1)

On the other hand, AI D I IAB

, so AID

1-H C

. Since H D

1-AI D

it follows that D

is the orthocenter of triangle HAl C

thus DC

1-HAI

. (2)

3.2. SOLUTIONS 97

A

B �------�-*�----------�

F

Note that GH 1- AB and EH 1- AG. Since ABI IA'E and AGI IGA' , we have GH 1- A' E and EH 1- GA' , so H is the orthocenter of triangle A' EG. Therefore

A'H 1- GE.

From relations (1) , (2) and (3) we obtain

DGI IBFI IGE,

as desired. ( Titu Andreescu

)

5 . Using Ceva's theorem, we obtain G'A A'B B'G - · _ · - = 1 G'B A'G B'A '

hence

so

S1 . S3 . S5 =

l . S2 S4 S6

The given condition now reads

This implies

S1 + S3 + S5 = 3 .3/ S1 . S3 . S5

S2 S4 S6 V S2 S4 S6

G'A A'B B'G G'B = A'G

= B'A = l .

(3)

It follows that A' , B' , G' are the midpoints of the triangle's sides and M is the centroid, as claimed.

( Titu

Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 (1974) , pp.

23,

Problem 1904; Gazeta Matematica (GM-B ) , No. 2 (1979) , pp. 63, Problem 0:16)

98 3. GEOMETRY

6. Using Menelaus' theorem yields NA AB QC NB

. BC

. QA

= 1.

Hence

or

B

SMAN 5MBA SMCQ -1

5MBN .

SMCA .

SMAQ - ,

SMAN 5MBP SMQA 5MBN

. SMCP

= SMCQ

'

A

From the condition in the hypothesis it follows that

SMAN + 5MBP = 2 . jSMAN .

5MBP , SBMN SMCP V 5MBN SMCP

and so SMAN 5MBP 5MBN - SMCP

'

Thus NA PB NB PC

hence N P and AC are skew-parallel. ( Titu


1 (1978), pp.

66,

Problem 3287; Gazeta Matematica (GM-B) , No. 2

(1979), pp.

56, Problem

17607)

7. Let AI , Bl , Cl be the feet of triangle's altitudes and let A', B' , C' be the projections of P onto its sides.

If P is the orthocenter of the triangle, then the equality holds. Assume by way of contradiction that P is not the orthocenter of the triangle and

aPA + bPN + cPC = 4S.

Then at least two of the inequalities

PA + PA � AAl , PB + PB' � BBl , PC + PC' � CCI

3.2. SOLUTIONS 99

are strict. A

It follows that

aPA

+ bPB

+ cPC

> a(AA

l -PAl)

+ b(BB

l -PB

I) + c

(CCl -

PC/)

or 4S > a

AAl +

bBBl + c

CCl -

(aP A

I + bP B

I + cPC/)

.

Hence 4S > 6S - 2S = 4S, which is false. ( Titu


(1978), pp.

74,

Problem 3689


, pp. 64, Problem 181

22)

and

Let

8. Using the angle bisector theorem and Van Aubel's theorem, it follows that IA BIA C

IA b

+ c IAI = B

IC + C

IB = -a-

= kl

IB AlB CIB

c + a IBI = A

IC + C

IA = -

b- =k2

IC _ a +

b _ k

ICI - C - 3

and note that 2p 2p k

l + 1 = -;;:

' k2 +

1 = b' 2p k

3 + 1 = -,

c where p is the semiperimeter. Hence

so

Furthermore, since

1 1 1

-- + -- + -- =1 k

l + 1 k

2 + 1 k

3 + 1

cos A = !

[k2 +

1

+

k3 +

1 _ (k2 +

1) (k3 +

1) 1 2

k3 +

1 k2 +

1 (kl +

1)2 '

100 3. GEOMETRY

we observe that cos A depends only on kl and k3 • Hence cos A = cos Al and, analogously, cos B = cos BI , thus triangles ABC and AIBICI are similar. (D

orin Andrica, Romanian Regional Mathematical Contest "Grigore Moisil" ,

1999; Revista Matematica Timi§oara (RMT) , 1999, pp. 87)

9. Using Steiner's theorem, we obtain

MB · NB AB2 = MC · NC AC2

Applying the AM-GM inequality yields

MB NB . /MB · NB AB MC + NC � 2V MC . NC = 2AC '

as desired. Note that the equality case occurs only if AM and AN coincide with the internal

bisector of angle A. ( Titu Andreescu

)

10. The quadrilateral APMN is a rectangle, therefore NP = AM. Hence NP is minimal if AM ..1 BC, so M is the foot of the altitude from A. ( Ti

tu Andreescu

)

11 . Applying the law of cosines to triangle Al BI C we obtain

and using the inequality x2 + y2 - xy � xy, which holds for all real numbers x, y, we get

AIB� � AI C , BI C.

A

3.2. SOLUTIONS 101

Similarly, we obtain Bl Cr � Bl A · CIA and CI A� � ClB . AlB. MUltiplying the three inequalities together, we obtain

Now the lines AAI , BBI , CCI concur, so

and after substituting and taking square roots, we have

AlBl . BlCI . CIAI � AlB · BIC · CIA,

the desired inequality. Equality holds if and only if CAl = CBl , ABI = ACI and BCI = BAI , which in turn holds if and only if P is the center of triangle ABC. ( Ti

tu Andreescu, IMO 1996 Shortlist)

12 . We start with the following lemma. Lemma.

Let A, B, C

be the ang

le of a tr

iang

le ABC.

Then

A B B C C A tan - + tan - + tan - tan - + tan - tan - = 1 . 2 2 2 2 2 2

Proo

f. We present two arguments. Fi

rst approach. Since

tan(a + ,8) [1 - tan atan,8] = tan a + tan,8,

tan(90° - a) = cot a = l/ tana, and A/2 + B/2 + C/2 = 90° , the desired identity follows from

tan - tan - + tan - tan - = tan - tan - + tan - = A B B C B ( A C) 2 2 2 2 2 2 2

= tan B tan (� + C) [l - tan � tan C J = 2 2 2 2 2

= tan � tan (90° - �) [1 - tan � tan �J = A C

= 1 - tan '2 tan '2 ' Secon

d approac

h. Let a,

b, c, r, s denote the side lengths, inradius and semiperime

ter of triangle ABC, respectively. Then SABe = rs, AP = s - a, and tan(A/2) = r/(s - a) . Hence

Likewise,

A SABe tan - = .

B SABe tan - = -:---2 s(s -

b)

2 s(s - a)

and tan C = SABe . 2 s

(s - c)

102

Hence

3. GEOMETRY

A B B C C A tan - tan - + tan - tan - + tan - tan - = 2 2 2 2 2 2

= S�BC ( (8 - c) + (8 - a) + (8 - b) ) = 82 (8 - a) (8 - b)

(8 - c) S�BC = 1 ,

8 (8 - a) (8 - b)(8 - c)

by Heron's formula. 0 A

B Q C

Without loss of generality assume that AP = min{AP, BQ, CR} . Let x = tan(A;2) , y = tan(ii/2) , and z = tan(C;2) . Then AP = r/x, BQ = r/y, and CR = r/z . The condition given in the problem statement becomes

2x +

5y +

5z = 6,

and the equation in the lemma is

xy + yz + zx = 1 . Eliminating x from

(1) and

(2) yields

5y2 + 5z2 + 8yz -

6y -

6z +

2 = O. Completing squares, we obtain

(3y _ 1

)2 + (3z _ 1

)2 = 4(y - Z)2 .

Setting 3y - 1 = u, 3z - 1 = v (Le. , y = (u + 1)/3, z = (v + 1)/3) gives

5u2 + 8uv +

5v2 = O.

(1)

(2)

Because the discriminant of this quadratic equation is 82

-4

x 25

< 0, the only real solution to the equation is u = v = O. Hence there is only one possible set of values for the tangents of half-angles of ABC (namely x = 4/3, y = z = 1/3) . Thus all triangles in S are isosceles and similar to one another.

Indeed, we have x = r/AP = 4/3 and y = z = r/BQ = r/CQ = 1/3 = 4/12, so we can set r = 4, AP = AR = 3, and BP = BQ = CQ = CR = 12. This leads to AB = AC = 15 and BC = 24 . By scaling, all triangles in S are similar to the triangle with side lengths

5, 5, 8.

3.2. SOLUTIONS

We can also use half-angle formulas to calculate

C 2 tan -. B . C

2 sm = sm = C 1

+ tan2 "2

103

From this it follows that AQ : QB : BA = 3 : 4: 5 and BC : AC : BC = 5 : 5 :

8. Al

ternative so

lution. By introducing the variables p = y + z and q = yz -

1,

relations (1)

and (2)

become 2x + 5p = 6 and xp + q =

0, respectively. Eliminating x

yields

p(6 - 5p) + 2q =

O.

(3)

Note that y and z are the roots of the equation

t2

- pt + (q +

1) = O.

(4)

Expressing q in terms of p in (3)

, and substituting in (4)

, we obtain the following quadratic equation in t:

t2

_ t 5p2 - 6p +

2 -0

p + - . 2

This equation has discriminant - (3p - 2)2 $ O. Hence the equation has real solutions only if p = 2/3, and y = z =

1/3. N

ote. We can also let x = AP, Y = BQ, z = CR and use the fact that

r(x + y + z

) = SABC =

Jxyz

(x + y + z

)

to obtain a quadratic equation in three variables. Without loss of generality, we may set x =

1. Then the solution proceeds as above.

(Titu Andreescu, USA Mathematical Olympiad, 2000)

13. Let a, b, c, A, B, C be the side lengths and angles of triangle ABC. Let X, Y,

Z

be the feet of the perpendiculars from P to lines BC, CA, AB, respectively. Recall the inequality (the key ingredient in the proof of the Erdos-Mordell inequality) :

P A sin A � PY sin C + P Z

sin B. (1)

104 3. GEOMETRY

This says that the length of Y Z

is greater than equal to its projection onto BC

, the latter being equal to the sum of the lengths of the projections of

PY and

P Z

onto BC

. In fact, since AYP = Azp = 90°, AZPY

is cyclic with AP

as a diameter of its circumcircle. By the Extended Law of Sines,

Y Z = P A

sin A. Let

M and N be

the feet of perpendiculars from Z

and Y

to the line P X

. Since 1fZp = BxP = 90° , . P Z B X

is cyclic. Hence x:iPZ = 13 and Z M

= P Z

sin B. Similarly,

Y N =

PY sin

C.

Thus (1) is equivalent to Y Z

� Y

N +M Z

. Multiplying by 2R and using the Extended

Law of Sines, (1) becomes

aPA

� cPY

+ bPZ

.

Likewise, we have bPB

� aPZ

+ cPX

and cPC

� bPX

+ aPY

. Using these inequalities, we obtain

PA P B PC ( b c ) ( c a ) ( a b ) a2 + b2 + 7 �

P X

c3 + b3 +

PY

a3 + c3 + P Z

b3 + a3 �

2PX 2PY 2PZ

� be + co: + � (AM-GM inequality) 4SABC

1

abc

= R '

Equality in the first step requires that Y Z

be parallel to BC

and so on. This occurs if and only if

P is the circumcenter of

ABC. Equality in the second step

requires that a = b = c. Thus equality holds if and only if

ABC is equilateral and

P

is its center. ( Titu Andreescu, USA IMO Team Selection Test,

2000)

14. It is known that if H

is the orthocenter of a triangle ABC

then AH

· BC

+ BH

· CA

+ CH

·AB

=4SABC .

3.2. SOLUTIONS 105

A

We prove that I

== H.

Assume by way of contradiction that points I

and H

are distinct. Let Al , Bl , C1

be the feet of the altitudes from A, B, C

and let A', B' ,C'

be the projection of I

onto the sides

BC, CA

, AB

, respectively. Hence at least two of the inequalities IA

+ lA'

� AA

1 , IB

+ IB'

� BB

1 , IC

+ IC'

� CC

1

are strict. It follows that

a · IA

+ b· IB

+ c · IC

> a(AA

1 -IA')

+ b(BB

l -IB')

+ c(CC

l -IC')

,

or

which is false. Therefore 1 = H and

ABC is an equilateral triangle, as desired. ( Ti


2(1981)

, pp. 67,

Problem 4616)

15 . Using standard notations, we have 64c2 + 49a2 + 9b2

-112

ac -48b

c + 42ab = 6c2 - 6

a2 -6b2

.

This is equivalent to 15b2

+ 2b(21

a -24c) + 55a2 - 11

2ac +

58c2 = O.

Viewing this as a quadratic equation in b, the condition

� � 0

is satisfied. That is 441

a2 -1008

ac + 576

c2 -825

a2 + 1680

ac -870

c2

� O. The last relation is equivalent to

-6(64

a2 -112

ac + 49c2)

� 0,

or (8a -

7C)2

� O. It follows that 8a =

7c. Substituting back into the given condition

yields 3a = 7b. We obtain

106 3. GEOMETRY

hence triangle ABC is similar to triangle A' B' C' having sides 7, 3, 8. In this triangle

I 32 + 82 - 72 1

cos A = 2 . 3 . 8 = 2 '

It follows that A = A7 = 60°

. ( Titu Andreescu, Korean Mathematics Competition,

2002)

16. First so

lution. By the Cauchy-Schwarz Inequality,

..jPB2 + PC2..j AC2 + AB2 2:: PB · AC + PC · AB.

Applying the (Generalized) Ptolemy's Inequality to quadrilateral ABPC yields

P B . AC + PC . AB 2:: P A . BC.

Because PA is the longest side of an obtuse triangle with side lengths PA, PB, PC, we have PA > y'PB2 + PC2 , and hence

P A . BC 2:: ..j P B2 + PC2 . BC.

Combining these three inequalities yields

..j AB2 + AC2 2:: BC,

implying that angle BAC is acute. Note. With some careful argument, it can be proved that quadrilateral ABPC is

indeed convex. We leave it as an exercise for the reader. Secon

d solution. Let D and Q be the feet of the perpendiculars from B and P to

line AC, respectively. Then DQ � BP. Furthermore, the given conditions imply that Ap2 > Bp2 + PC2 , which can be written as Ap2 - PC2 > BP2 . Hence,

AQ2 2:: AQ2 _ QC2 = (Ap2 _ PQ2 ) _ (Cp2 _ PQ2) =

= AP2 - PC2 > BP2 2:: DQ2. Let I be the ray AC minus the point A. Note that, since PA > PC, Q lies on ray l

. If D did not lie on l, then AQ would be less than or equal to DQ, a contradiction.

Thus, D lies on l, and angle BAC is acute.

B o

P o

o�----�o�----------�o�------�o A D Q C

3. 2. SOLUTIONS 107 Third

solution. Set up a coordinate system on the plane with A =

(0,0), B = (

a,O), C =

(b, c) , and P =

(x, y

). Without loss of generality, we may assume that

a > 0

and that c > O. Proving that angle BAC is acute is equivalent to proving that b

> O. Since PA2 > PB2 + PC2,

x2 + y2 > (x _ a

)2 + y2 + (x _

b)2 +

(y _ C)2 .

Hence o >

(x - a

)2 -2bx +

b2 +

(y - C)2 � -

2bx.

Since P A > P B, we have x > � > O. It follows that b > 0, as desired. F

ourth solution. We first prove the following Lemma.

Lemma. For any

four po

ints

W, X, Y, Z i

n the p

lane,

Proo

f. Pick an arbitrary origin

0 and let w, x, y, z denote the vectors from

0 to W

,X,Y,Z, respectively. Then

= Iw - x l2 +

Ix - Yl2 +

Iy - Z l2 +

Iz - w l2 -

Iw - Yl2 -

Ix - Z l2 =

= w · w + x · x + y . y + z · z -2(w , x + x · y + y . z + z · w - w · y - x · z

) =

= Iw + y - x - Z 12 ,

which is always nonnegative. Equality holds if and only if w + y = x + z, which is true if and only if

W XY Z is a (possibly degenerate) parallelogram. 0

Applying the Lemma to points A, B, C, P gives 0 :::; AB2 + Bp2 + PC2 + CA2 _ AP2 - BC2 =

= (PB2 + PC2 _ PA2) + (AB2 + AC2 - BC2 ) < < 0 + (AB2 + AC2 - BC2) = AB2 + AC2 - BC2.

Therefore angle B AC is acute. Fifth solution. In this solution, sin-1 takes on values between

0° and

90°. Note

that -p;fjj < 90°

, since P B < P A. Applying the Law of Sines to triangle P AB yields

It follows that

- PB - PB sin PAB = PA sin ABP :::; PA '

- • _1 PB PAB :::; sm PA ' Since P A 2 > P B2 + PC2 , we have similarly

P-AC < ' -1 PC . _1 VPA2 _ PB2 _ sm PA < sm PA .

108

Thus

3. GEOMETRY

- - -- 1 PB 1 J

PA2 - PB2 BAC � BAP + PAC < sin-

PA + sin-PA

If ()

= sin-1 � � , then

Hence

JPA2 - PB2

sin (90° -())

= cos ()

= V'l - sin2 ()

= P A .

- . -1 PB . -1 J

PA2 _ PB2 ° BAC < sm P A + sm P A = 90 ,

and angle B AC is acute. C

P

A

B B

P

As we mentioned at the end of the first solution, the conditions in the problem imply that quadrilateral ABPC is indeed convex. Thus, the diagram on the right-hand side is not possible, but this solution does not depend on this fact. Si

xth solution.

--...-- "" /'

I _ _ B

/ ,/ I ;'

I /

( I I P \

\ \ '\. C ,/

\ -.... I "- ./ '-... -- ...--

Note that PA2 > PB2 + PC2 . Regard P as fixed and A, B, C as free to rotate on circles of radii PA, PB, PC about P, respectively. As A, B, C vary,

IfA1J will be

maximized when B and C are on opposite sides of line P A and Jfijp

and ;[(j'p

are right angles, i .e . , lines AB and AC are tangent to the circles passing through B and C.

Without loss of generality, we assume that PA > PB � PC. In this case, ABPC is cyclic and AB2 = P A 2 - P B2 > PC2 , and similarly AC2 > P B2 . Hence on

3.2. SOLUTIONS 109

the circumcircle of ABPC, arcs AB and AC are bigger than arcs PC and PB, respectively. Thus, IiPC > BAiJ. Because these two angles are supplementary, angle BAC is acute.

P

B -,., A

/'

( Titu Andreescu, USA Mathematical Olympiad, 2001)

17. All angles will be in degrees . Let x = F0B. Then PiiC = 80 - x. By the Law of Sines (or the trigonometric form of Ceva's Theorem) ,

1 = PA PB PC =

sin� sin!!.!!. sin� = PB PC PA sin PAB sin PBC sin PCA

sin 20 sin x sin 40 = ------�---------

sin 10 sin(80 - x) sin 30

B

C

4 sin x sin 40 cos 10 sin(80 - x)

A

The identity 2 sin a cos b

= sin( a -b) + sin( a +

b) (a consequence of the addition

formula) now yields

so

1 = 2 sin x(sin 30 + sin 50)

= sin x(1 + 2 cos 40)

sin(80 - x) sin(80 - x) ,

2 sinx cos 40 = sin(80 - x) - sin x = 2 sin(40 - x) cos 40.

This gives x = 40 - x and thus x = 20. It follows that AcE = 50 = BAiJ, so triangle ABC is isosceles.

1 1 0 3. GEOMETRY

Alternat

ive so

lution. Let D be the reflection of A across the line BP. Then triangle

AP D is isosceles with vertex angle

APi5 = 2(180 - EPA) = 2(PAii

+ AiiP) = 2(10 + 20) = 60, and so is equilateral. Also, ISiiA = 2

JiBA = 40. Since 1fAC = 50, we have DB .l AC.

B

C

D

Let E be the intersection of DB with CPo Then

PEi5 = 180 - CEi5 = 180 - (90 - AcE) = 90 + 30 = 120

and so PEi5 + I5A.J5 = 180. We deduce that the quadrilateral APED is cyclic, and - -

therefore DEA = DPA = 60.

Finally, we note that I5EA = 60 = 15EC. Since AC .l DE, we deduce that A and C are symmetric across the line DE, which implies that BA = BC, as desired. ( Ti

tu Andreescu, USA Mathematical Olympiad,

1996)

18. Let max{ A, B} = A. If triangle ABC is right-angled, then A = 90°

, B = 30°

and C = 60°

. In order to find !!:., we may assume that ABC is the triangle with sides r

a = 2, b = 1, c = y'3. We have R =

1 and

y'3 _

S ABC _

2 _ y'3 r - -- - - ---

8 2+1+J3 3

+J3'

2

R 3+J3

so - = -- =v'3

+1. r y'3

R Conversely, assume that - = J3

+ 1. From the identity

r

4R · A . B . C r = sm 2 sm "2 sm "2

it follows that

or

Then

3.2. SOLUTIONS

r = 4(V3 + l)r sin � sin � sin �

0-1 ( A

-C A

+C) . B --4- = cos -2- - cos -2- sm 2"

and, since A -C

= 30°

, we obtain

o - 1 = (J6 + V2 _ sin

B) sin B

4 4 2 2 '

L · . B . ld ettmg sm 2" = x Yle s

2 _ J6 + V2

0-1 _ 0

x 4 x + 4 - ,

1 1 1

. J6 - V2 V2 B B

whose solutlOns are x = 4 and x = 2 ' It follows that 2" = 15°

or 2" = 45°

. The second solution is not acceptable, because

A � B. Hence

B = 30°

, A

= 90°

and C = 60°. Thus triangle

ABC is right angled. ( Ti


2002)

1 9. Construct in the exterior of triangle ABC

three circles equal to the circumcircle

ABC that pass through two vertices of the triangle. By the five-coin theorem

the circles will have a common point P, as desired

(see Dorin Andrica, Csaba Varga,

Daniel Vacare�u, "Selected Topics and Problems in Geometry" , PLUS, Bucharest, 2002, pp.

51-56)

. Alternat

ive so

lution. Let

H be the orthocenter of triangle

ABC. The reflections

of H

across the sides of the triangle are points of the circumcircle of triangle ABC

. Therefore the circum circles of

HAB,HBC

,HCA

are equal to the circumcircle of ABC and for

P = H

the claim holds. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No.

2(1978)

, pp. 74)

20. Denote the midpoints of A'B'

,B'C'

,C'A' b

y Co ,Ao ,Bo , respectively, and

the three perpendiculars in question by le , lA , lB . Consider the centroid of triangle A'B'C'

.

1 12 3. GEOMETRY

C

Since AoG : GA' = BoG : GB' = CoG : GC' = 1 : 2, the dilatation h with center G and coefficient -

2 takes Ao, Bo , Co to A' , B' , C', respectively. Since dilatations carry

straight lines into parallel lines, h transforms le into the line through C' perpendicular to AB. But C' is the point of tangency of the incircle and AB, so this line passes through the incenter of triangle ABC. The same applies to the images of lA and lB under h. Since the images of lA , lB , le under h are concurrent, so are lA , IB , le themselves.

( Titu Andreescu, Romanian IMO Selection Test, 1986)

21 . Because triangle AI A2A3 is not isosceles, it is not difficult to see that the circumcenter of the triangles AI BI I, A2B2I, A3B31 are defined. We start with a sim pIe lemma.

Lemma. Let ABC be a triangle with the incenter I. Let T be the circumcenter of the triangle BIC. Then T lies on the internal bisector of the angle A.

Proof. Let us draw the external bisectors of the angles B and C as shown in the figure below.

B

A C

They intersect at the excenter E, which lies on the internal bisector of the angle A. Since BE ..1 BI and CE ..1 CI, the quadrilateral BECI is cyclic with the center of the circumscribed circle on IE. This center will be also the circumcenter of BIC. The lemma is proved.

3.2. SOLUTIONS 1 13

Let us prove the main statement. For i = 1, 2, 3 we denote by

Qi the center of the

circle Ci and by

Ti the circumcenter of the triangle

Ai+1

IAi+2 . Clearly,

Oi lies on the

internal bisector of the angle Ai ' By the lemma,

Ti also lies on the same bisector. Thus

the triangles 010203 and

T1T2T3 are perspective from the point

I. By Desargues'

theorem these triangles are perspective from a line. This is to say that if we denote Qi , i = 1 , 2 , 3 , to be the point of intersection of the lines

0i+1

0i+2 and

Ti+1

Ti+2 ' then

the points Q1 , Q2 , Q3 are collinear . But since

Ti+1

Ti+2 is the perpendicular bisector

of Ail

and 0i+1

0i+2 is the perpendicular bisector of

Bili these points are exactly

the circumcenter of the triangles A1B1I, A2B2I, A3B3I, respectively.

Remar

k. A student not familiar with Desargues' theorem may proceed from the

point as follows. Applying Menelaus' theorem to the triangles 10

102 ,10

203 , 10

301

and to the triples of points (T1 ' T2 , Q3), (T2 ' T3 , Q1), (T3 ' Tt , Q2), respectively, one

can, observing usual agreement about the signs, write: 01 T1

IT2

02Q3 _ 1 IT

1 . 0

2T2 . 0

1Q3 - ,

IT3

02T2

Q3Q1 _ 1

-- . -- . -- - , 03T3

IT2

02Q1 IT

1 03T3

01 Q2

_ 1 01T1 . IT

3 . 0

3Q2 - .

MUltiplying them all one gets 02Q3 .

03Q1 .

01 Q2 = 1 , 0

1 Q3

02Q1

03Q2

which means that the points Q1 , Q2 , Q3 are collinear. Al

ternative so

lution. This proof will be based on inversion. We take the incenter I

to be the center of the inversion and the power of the inversion is arbitrary. Using primes to denote images of points under the inversion we have the following " dual" figure shown below.

1 14

B' I

3. GEOMETRY

B' 2

Indeed, the image of the circle Gi is a straight line

B�+I B�+2 ' with these lines

forming the triangle B� B�B�

. The line AiAi+1 will be transformed into the circle r

i+2 , with the side AiAi+1 becoming the arc

AiAi+1 which does not contain

I. Note

that all these circles have equal radii since the distances from I

to the sides of AIA2A3

were equal. Let us note that if

EI , E2 , E3 are three circles passing through the common point I

and no two of them touch, then their centres are collinear if and only if there is another common point J

i- I through which all these three circles pass.

We will use this observation for Ei being the circumcircle of

AiBiI. Since the

inversion takes Ei to the line

AiB�, the desired result is to show that the lines

A� BL A�B�

, A�B�

are concurrent. For this, it suffices to show that the triangles A�A�A�

and B� B�B�

are homothetic, which is the same to say that their corresponding sides are parallel. Since the radii of the circles

r I , r 2 , r 3 are equal, the triangle

PI P2 P3

formed by their centre has its sides parallel to the corresponding sides of the triangle B� B�B�. The homothety of ratio

1/2 centred at

I takes the triangle

A� A�A� into

the triangle whose vertices are the midpoints of the triangle PIP2P3 . Therefore the

corresponding sides of the triangles A� A�A�

and PIP2P3 are also parallel and the

result follows. ( Titu Andreescu, IMO

1997 Shortlist

)

22. Let I

be the intersection point of lines GB'

and G' B

and let A'

be the intersection point of lines

AI and

BG.

We have AB' AG' AI B'B + G'G = lA"

from Van Aubel's theorem, therefore ::, is constant. Hence the locus of point I

is a line segment parallel to

BG. ( Ti

tu Andreescu

)

or

3.2. SOLUTIONS

23. Without loss of generality assume that PC

= max{PA

,PB

,PC}

.

The condition in the hypothesis is P B

. PC

+ P A

. PC

= P A

. P B

+ 1

PC PA·PB

+ 1 · 1 1 - P A . 1 + PC

. 1 .

1 1 5

From the converse of the second theorem of Ptolemy it follows that P AC B

is a cyclic quadrilateral. Note that

P cannot be

A, B

or C

otherwise the denominator of the right-hand side equals O. Hence the locus of point

P is the circumcircle of triangle ABC

without the vertices A, B, C. ( Ti


(1985), Problem C7

:3)

24. We have J

a2 b2 -4S2

+ J

a2c2 -4S2

= J

a2 b2 - a2 b2 sin2 C+

+J

a2c2 - a2c2 sin2 B = ab cos

C + ac cos B =

a2 + b2 _ c2 a2 + c2 - b2 = ab + ac = a2

2ab 2ac as desired. ( Ti


8(1971)

, pp. 25,

Problem 1006)

or

so

25. The relation is equivalent to Fa, + y'rb + Fc 1

= - ,

On the other hand,

..jrarbre r

1 1 1 1

--- + + = - . Fa y'rb FcFa Fa y'rb r

1 1 1 1 - + - + - = - , ra rb re r

1 1 1 1 1 1 --- + + = - + - + -. Fa,y'rb y'rbFc FcFa, ra rb re

Then ( 1 1 ) 2 ( 1 1 ) 2 ( 1 1 ) 2

Fa, - ..fib + y'rb - Fc + Fc - Fa =

0,

so ra = rb = re' It follows that the triangle is equilateral, as desired. ( Titu


(1974)

, pp. 21,

Problem 1903)

1 1 6 3 . GEOMETRY

26. If the triangle is equilateral the conclusion is true, To prove the converse, we assume by way of contradiction that the triangle is not

equilateral and say that b f:.

c, Then

b2 + c2 a2 (b

+ C)2 - a2

m� = -2- - 4 > 4

= p(p - a) and likewise m

� 2: p(p -

b), m

� 2: p(p - c) ,

It follows that 1 1 1 1 ( 1 1 1 ) - + - + - < - -- + -- + -- = m�

m�

m�

p p - a p -b

p - c

On the other hand

-p2 + ab + bc + ca

S2

1 1 1 1 _p2 + a2 + b2 + c2 - + - + - = -[(p _ a)2 + (p _ b)2 + (p _ C)2] = ---=-____ _ r�

r�

r�

S2 S2

Since b f:.

c then ab + bc + ca < a

2 + b2 + c2 hence

1 1 1 1 1 1

-2 + -2 + -2 < 2 + 2" + 2' ma mb me r a rb re

which is a contradiction, ( Titu

Andreescu, Revista Matematidi Timi§oara (RMT) , No,

2(1977), pp,

66,

Problem 3063)

so

27. We know that

On the other hand,

, A

, B

,C 1

sm - sm - sm - < -2 2 2 -8 '

A B C ' A ' B ' C

cos - cos - cos - > sm sm sm , 2 2 2 -

' A ' B ' C 4

A B C

sm + sm + sm = cos "2 cos 2" cos 2" ' so inequality

(1) gives

sin A + sin

B + sin

C 2: 4 sin

A sin

B sin

C,

Since the circumradius is 1, we have

and relation (2)

yields

as claimed,

a = 2 sin A, b = 2 sin B, c =

2sin

C,

a + b + c 2: a

bc,

(1)

(2)

3.2 . SOLUTIONS 1 17 Alternat

ive so

lution. Let Zl , Z2 , Z3 be the afixes of points

A, B, C

such that Izd = IZ2 1 = I Z3 1 = 1 . We have

a = BC = I Z2 - z3 1 , b = AC = IZ3 - zI i , c = AB = IZI - z2 1 ·

U sing the identity

Z; (Z2 - Z3 ) + Z� (Z3 - zd + Z� (ZI - Z2 ) = (Zl - Z2 ) (Z2 - Z3 ) (Z3 - Zl )

and triangle inequality it follows

abc = I ZI - z2 1 1 z2 - z3 1 1z3 - zd ::; IZl l2 1z2 - z3 1 + I Z2 12 1 1 z3 - zl l + I Z3 1 2 1 z1 - z2 1 = = I Z2 - z3 1 + IZ3 - zl l + I ZI - z2 1 = a + b + c.

( Titu


2(1978), pp.

49,

Problem 3513

; Gazeta Matematidi (GM-B) , No. 11 (1981)

, Problem 0258

; No. 2(1988), pp.

78, Problem

21353)

28. For any positive real numbers x, y, Z we have

Setting

gives

so

8xyz ::; (x + y

) (y + z) (z + x)

x = -a + b + c, Y = a - b + c, Z = a + b - c,

( -a + b + c) (a - b + c) (a + b - c) ::; abc,

It follows that

then

Hence

as desired.

" a2 < 3

abc = � . pabc = L...J (p - b) (p - c) -2 (p

- a) (p - b) (p

- c) 2 S2

= 6abc . E = 6R 4S S r'

( Titu


2(1974), pp.

51,

Problem 2028)

29. Using the inequality 3(a2 + b2 + c2 ) � (a + b + C)2 ,

1 1 8

we obtain

Hence

as desired.

3. GEOMETRY

a2 + b2 + C

2 p2

4S2 � 3S2 '

1 1 1 1 h� + h� + h� � 3

r2 '

( Titu


2(1977), pp.

66,

Problem 3062)

30. From the inequality

we obtain

Then

hence

as claimed.

1 1 1 1 1 1 - + - + - > -- + -- + --ra rb rc -

Jrbrc

Jrcra

Jrarb

'

1 1 1 1 - > -- + -- + --. r -

Jrbrc

Jrcra

Jrarb

Jrarb +

Jrbrc +

Jrcra �

9r,

( Titu

A ndreescu, Revista Matematica Timi§oara (RMT) , No.

2(1978), pp.

64,

Pro blem 3277)

31. We have

and likewise

Jb2+ C

2 a2 J(b

+ C)2 _ a

2 ma = -2- - 4 � 4 = "';p(p - a

)

It follows that

mambmc � p"';p(p - a

) (p -

b)(P - c

) = pS = rarbrc,

as desired. ( Titu


1 (1978)

, pp. 64,

Problem 3276)

32 . By the AM-GM inequality,

so 8p3 �

27 abc. Hence

(a +

b + C

)3 � 27abc,

2 abc

S 2p >

27- · - =

27Rr -

4S p

,

3.2. SOLUTIONS 119

as desired. ( Titu


1 (1973), pp.

43,

Problem 1585)

33. We have

It follows that

hence

as desired.

_

A< _7r .

A 1 2 -6

{:} sm 2 S; "2 ' V(

P -b)(P

- C) < � b

e -2 '

be (

p -b) (P - e

) S; 4'

( Titu

A ndreescu, Revista Matematica Timi§oara (RMT) , No . 1

(1984)

, pp. 67,

Problem 5221)

34. Let 01 , O2 , 03 be the centers of the three circles and S the area of the common

region.

The three sectors with centers 01 , O2 , 03 which subtend the arcs 0

203 ,0103 ,0201 , respectively, cover the surface of area S and twice more

the surface of triangle 010203 (which is r

2 � ) . On the other hand, the area of

these three circular sectors equals the area of a semicircle, which is �7rr2 . Hence

therefore

1 2 r

2J3 -7rr = S +

2 . --

2 4 '

S = �r2 (7r - V3). 2


2(1978)

, pp. 50,

Problem 3522)

120 3. GEOMETRY

35. The parallel BD through C meets AB at point E. By Stewart's formula, we obtain

AC2 · BE + CE2 . AB - CB2 · AE = AB · BE · AE Because CE = BD and BE = CD, we deduce

AC2 . CD + BD2 . AB - BC2 . (AB + CD) = AB . CD . (AB + CD) (1)

A

D C

B E

Drawing the parallel to AC through D and using similar computations yields

BD2 . CD + AC2 · AB - AD2 . (AB + CD) = AB · CD · (AB + CD) (2)

Subtracting the relation (2)

from (1)

gives

(AC2 - BD2) (AB - CD) = (AD2 - BC2 ) (AB + CD) ,

as desired. (Dorin Andrica, Gazeta Matematidi (GM-B) , No.

9(1977), Problem

6852; Revista

Matematidi Timi§oara (RMT) , No. 1-2(1980)

, pp. 64, Problem

4119)

36. Let I be the intersection point of the diagonals AC and BD. Since AB · AC AB · BD IA = AB + CD

and IB = AB + CD the condition in the statement becomes

Hence JfilJ = 90°

, as desired. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No.

2(1978), pp.

59,

Problem 3524)

37. Let ABCD be the trapezoid. Point I is the intersection of diagonals and M, N are the midpoints of AB and DC. In a right triangle the length of the median corresponding to the hypothenuse is half of lenght of the the hypothenuse. Hence

IM = AB and IN = CD . 2 2

A

Then

3.2. SOLUTIONS

D F

N

M E

IM+IN

=

AB+CD

=E

F, 2

121

C

B

which is the length of the midline and hence the length of the altitude. It follows that IM and

IN are also altitudes in triangles

lAB and

ICD therefore

lAB and

ICD

are isosceles. Thus ABCD

is isosceles, as claimed. ( Titu

Andreescu, Revista Matematidl Timi§oara (RMT) , No.

1 (1978), pp. 48,

Problem 2817)

38. From the Law of Cosines we deduce that

Note that

2AB . AC

cos 1fAC = AB2

-BC2

+ AC2

2DC· AC

cos DcA = CD2

-AD2

+ AC2

2AB . DB

cos DiiA = AB2

-AD2

+ DB2

2DC· DB

cos Cfiiij = CD2

-BC2

+ DB2

1fAC = DcA, DiiA = Cfiiij, so dividing relations

(1) and

(2), (3)

and (4)

yields

as desired.

AB2 -BC2

+ AC2 AB2

-AD2

+ DB2 AB

CD2 -AD2

+ AC2 = CD2

-BC2

+ DB2 = DC

'

(Dorin Andrica)

(1)

(2)

(3)

(4)

39. Let a, b

the lengths of the bases, c, d

be the lengths of the nonparallel sides and

d1 , d2 be the lengths of the diagonals. From Euler's theorem for quadrilaterals, it

follows that

Hence (d1 -

d2)2 + 2

d1d2 =

(c -

d)2 + 2(ab + c

d),

122 3. GEOMETRY

and d1 - d2 = C - d implies d1 d2 = ab + cd.

From Ptolemy's Theorem, we deduce that the trapezoid is cyclic and so isosceles, as claimed. ( Ti


1 (1978)

, pp. 48, Problem

2817)

40. First so

lution. Assume the opposite. Then lAC - BDI > lAB - CDI or

lAC - BDI > lAD - BCI . Without loss of generality, lAC - BDI > lAB - CDI , otherwise switch B and D. We have

AC2 - 2AC . BD + BD2 > AB2 - 2AB . CD + CD2 (1)

and, from Euler's relation,

(2)

where M and N are the midpoints of AC and BD, respectively. From

(1) and (2) ,

AD2 + BC2 - 2AC · BD > 4MN2 - 2AB · CD. (3)

Let P be the midpoint of AB. Then N P = AD /2, M P = BC /2 and since MN � INP - MPI , it follows that

4MN2 � (AD - BC)2 . (4)

From (3)

and (4) , -2AC · BD > -2AB · CD - 2AD · BC, in contradiction with Ptolemy's Theorem. We are done. N

ote. The cyclicity is essential. The inequality fails if ABCD is a parallelogram. Secon

d solution. Let E be the intersection of AC and BD. Then the triangles

ABE and DCE are similar, so if we let x = AE, y = BE, z = AB, then there exists k such that kx = l)E,

ky = CE,

kz = CD. Now

lAB - CDI = I k -l iz

and lAC - BDI = I (kx + y

) -(ky + z

) 1 = Ik -1 1 . Ix - y

l·

Since I x - y l ::; z by the triangle inequality, we conclude that lAB - CDI �

lAC - BDI , and similarly IBC - DAI � lAC - BDl . These two inequalities imply the desired result. Thi

rd solution. Let 20', 2,8, 2,, 28 be the measures of the arcs subtended by

AB, BC, CD, DA, respectively, and take the radius of the circumcircle of ABCD to be

1. Assume without loss of generality that ,8 � 8. Then 0' + ,8 + , + 8 = 7r, and

(by the Extended Law of Sines)

lAB - CDI = 21 sin a - sin 1'1 = ISin a ; l' cos a � l' I

3.2. SOLUTIONS 123

and lAC

-BDI

= 21

sin(o: + t3l

- sin(t3 + "Il l

= I Sin 0: ; "I cos (0:; "I + t3) I . Since

0 ::; (a + ,

) /2 ::; (a + ,

) /2 + (3 ::; 7r

/2 (by the assumption (3 ::; 6) and the

cosine function is nonnegative and decreasing on [0, 7r /2]

, we conclude that lAB

-CDI � lAC

-BDI

, and similarly lAD

-BCI

� lAC

-BDl

. ( Titu

Andreescu, USA Mathematical Olympiad,

1999)

41 . Let E

be the intersection point of the diagonals. Consider AD

< BC

the basis of the trapezoid and

AB the altitude. Since

then

AB2 =AD

.BC

,

AB BC AD = AB '

so the right triangles ABC

and ABD

are similar. On the other hand we have

- -BCA + CAB

= 90°

hence

-It follows that

AEB = 90°

so the diagonals are perpendicular, as claimed. ( Titu


2(1972)

, pp. 28,

Problem 1 164)

42. Let I

be the intersection point of the diagonals AF

and BE

of the rectangle ABF E. Notice that

N I is the median of the right triangle

AN F

with hypothenuse AF, so

Likewise, IM

=

BE

=IE

2 '

IN=

AF

=IA

2

IQ =

AF

= IF

, 2 BE

fP= - =

IB. 2

Since fA

= IE

= IF

= IB

, it follows that 1M

= IN

= IP =

IQ. Hence MNPQ

is cyclic, as desired.

124 3. GEOMETRY

A B

Q

D E F C

(Titu Andreescu

)

43. By Sturm's theorem, we know that if 0 < al :::; a2 :::; a3 :::; a4 < al + a2 + a3 , then there is a cyclic quadrilateral having side lengths aI , a2 , a3 , a4 .

Denote by a, b, c,

d, e,

j, m the lengths of the segments

AB, BC

, CD

, DA

, AC

, BD, M N

, respectively. Without loss of generality assume that b + d � a + c.

Let P

be the midpoint of the side BC

. The segments M P

and N P

are midlines in triangles

CAB and

BDC, so

so

Then

1 1 MP = "2a and NP = "2c.

2m = 2MN

< 2MP

+2NP = a + c

2m < a + c <

b + d.

On the other hand, if 0

is the intersection point of the diagonals, we have b+ d = BC + DA

< BO

+ OC

+DO

+ OA = AC + BD

= e + j,

hence 2m < a + c <

b + d < e +

j.

It suffices to prove that e + j < 2m + a + c +

b + d. Note that

e < c + d,

j < b + c, e < a +

b,

j < a +

d.

Summing up these inequalities yields

e +j< a +

b+ c +

d< a +

b+ c +

d+2m

and the proof is complete. ( Titu

Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1 (1978) , pp.

66,

Problem 3288; Gazeta Matematidi (GM-B) , No. 10 (1981) , pp. 402, Problem C148)

3.2. SOLUTIONS 125

44. Let I be the intersection point of lines BD and AF. The parallel to BD through C meets line AF at point T. First we consider AF .l BD and prove that AB .l CD.

F

F

A �----------� D

C

i) Assume that D lies on the segment CEo Then .ArC = 90°. Since 1fAC : Ifi5C,

we obtain

On the other hand CTI IBD so

Relations (1)

and (2)

imply ifAE : EaT, therefore EATC is cyclic. It follows that BEG = 90

°, hence AB .l CD, as desired.

(1)

(2)

ii) Assume that C lies on the segment DE. In the right triangle CT F, T E is the median, so

Ere : EaT. (3)

Because CTI IBD, we have

EaT : CnB . (4)

Also, (5)

so from (3)

, (4)

, (5)

, we obtain Ere : 1fAC. Hence ATEC is cyclic, then AEC = JfiiC = 90

°, and AB .l CD, as desired.

Conversely, consider that AB .l CD. i) If D is on the segment CE, then ME == AcE . On the other hand AcE ==

Aci5 == AiiJ5 , so ME == AiiJ5 , and FBI E is cyclic. It follows that BiF = 1iEF = 90°, hence B D .l AF, as desired. ii) If C is on the segment DE, then

(6)

126 3. GEOMETRY

AcE := AiiJ. (7)

From (6) and (7)

we obtain ME := AiiJ, hence FEB! is cyclic. Note that lfEF = 90° , so ifiF = 90° and BD ..L AF, as desired. ( Ti


Problem C6:4)

45. Let a, b, c,

d be the side lengths of the quadrilateral and let

S be its area.

Because the quadrilateral is cyclic, we have

S2 = (p - a

)(p -b)(P

- c)(P

-d).

N umbers a, b, c,

d are odd, hence

is an integer.

a +b+ c +

d

p = -----2

(1)

If p is odd, then p - a, p -b, p - c, p -

d are even and so 82 is divisible by 16,

which is false. Hence p is even.

3.2. SOLUTIONS 127

1 1

Without loss of generality assume that a � b

� c � d. Since S is a prime number,

from relation (1)

we obtain

p -d= p - c =

lmdp - a = p -

b=£

Summing up these equalities yields 4p -

2p =

2 + 2S so p = S + 1 . Hence a =

b = 1

and c = d= S.

The required quadrilaterals are either rectangles or kites. ( Titu


2(1977), pp.

66,

Problem 3067)

46. From the AM-GM inequality it follows that b d (a + b

+ c +d) 4 _

1 4 a c < - -p . -

4 4

Hence 16abcd

� p4 ,

or 8 (ac + bd)2 - p4 � 8 (a2c2 + b2d2 ) .

The desired inequality is now obtained from Ptolemy's Theorem:

ac + bd

= eJ.

( Titu


3(1973)

, pp. 36,

Problem 1811

; Gazeta Matematidi (GM-B) , No. 8(1980)

, pp. 364

, Problem 18370)

47. Let m be the length of the segment determined by the midpoints of the diagonals.

From Euler's Theorem for quadrilaterals we have (1)

128 3. GEOMETRY

and from Ptolemy's Theorem,

ac + bd

= ef.

(2)

From relations (1)

and (2)

we obtain

Since max{/a - c/, / b -d/}

� 1, then

2 = 1 + 1

� (a - C)2 + (b -d)2

= (e -

f)2 + 4m2

� (e -

f)2.

Hence le -

f l � V2,

as desired. ( Titu


2(1978), pp.

51,

Problem 3527)

48. The quadrilateral is cyclic so

S = J(

p - a) (p -

b)(P - c

)(P -d)

Since S = (�) 2 , we have

{j(p - a

) (p -

b)(P - c

) (P -

d) = P. = 2

(p - a) +

(p -

b) + (p - c

) + (p -d)

4

Note that this is the equality case in the AM-G M inequality, hence p -a = p -b

=

p - c = p -d. It follows that a =

b = c =

d, so the quadrilateral is a square. ( Ti


(1977), pp.

24,

Problem 2136)

4�. Observe that SPAB ' SPCD = SPBC ' SPDA , since both are equal to �PA . P B · PC · PD · sinP. The numbers SPAB , SPCD and SPBC , SPDA have the same sum and the same product, thus SPAB = SPBC and SPCD = SPDA or SPAB = SPDA and SPBC = SPCD , i .e. P is the midpoint of AC or BD, as desired. ( Ti

tu Andreescu, Korean Mathematics Competitions,

2001)

50. (i) Let MI and Mil be the projections of point M onto diagonals AC and BD.

3.2. SOLUTIONS 129 D

We recall the Simpson's theorem: the projections of a point of the circumcircle of a triangle onto the sides of the triangle are collinear. Applying this result to triangles ABC

and DAC

yields that A', M',

B' are collinear and

C', M'

, D'

are collinear. Hence the lines

A' B', C'D'

and AC

meet at M', as claimed. (ii) From Simpson's Theorem for triangles

ABD and

BDC we deduce that Mil

is on the lines B' C'

and D' A'

. Since Mil is a point of AC

, the conclusion follows. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 1

(1979), pp.

54,

Problem 3855

; Romanian Regional Mathematical Contest " Grigore MoisH" , 1995)

51 . We have 1 SABCD S; 2

AC. BD

,

with equality if and only if AC

1-BD

. Since 1 2002

= SABCD S; 2AC

. BD

S; 1 52

·77

S; 2(AP

+PC)

. (BP

+PD)

= -2- =

2002,

it follows that the diagonals AC

and BD

are perpendicular and intersect at P. Thus, AB

= V242 +

322 = 40, BC

= V282 +

322 = 4Vl13

, CD

= V282

+ 452

= 53, and DA

= V452 + 2

42 = 51. The perimeter of

ABCD is therefore

144 + 4V113 = 4(36 + V113) .

( Titu Andreescu, American Mathematics Contest 12 (AMC 12 - Contest

B), 2002

, Problem

24)

52. Let a be the side length of the square ABCD

. Assume without loss of generality that max (

P A, PC)

= P A

. We have y'2PA

=PB

+PD

.

130 3. GEOMETRY

Then a.j2P A

= aP B

+ aP D

, hence BD

· PA

= AD

· PB

+AB

· PD

.

From the converse of the Ptolemy's Theorem it follows that PDAB

is a cyclic quadrilateral, therefore

P lies on the circumcircle of square

ABCD.

Conversely, using the Ptolemy's Theorem we deduce that any point of the circumcircle of square

ABC D has the given property. P

It follows that the locus of point P

is the circumcircle of square ABCD

. Alternat

ive so

lution. Let

P(x, y

) be a point with the given property and assume

y > O. Point

A, B, C, D

are considered like in the diagram. Note that

PC > PA

, so V2PC

=PB

+PD

Squaring both sides yields 2X2 + 2(y +

b)2 = (x -

b)2 + (x +

b)2 + 2y2

+ 2y'(

x -b)2

+ y2 + 2y'(

x + b)2 + y2 ,

then 2by =

y'(x -

b)2 + y

2 + y'(

x + b)2 + y2 ,

hence (x2 + y

2 + b2 _ 2bx

) (x2 + y

2 + b2

+ 2bx) = 4b2

y2.

It follows that

so

Thus x2 + y

2 = b2

and so point P

lies on the semicircle of diameter BD

that contains A. Likewise, for y :::;

0 we deduce that

P lies on the semicircle of diameter

BD that contains

C and

finally we obtain the circumcircle of square ABCD

.

3.2. SOLUTIONS 131

y

A(O, a) P(x, y

)

D(a, 0)

x

C(O, -a)

( Titu

Andreescu, Romanian IMO Selection Test, 1981; Revista Matematica

Timi§oara (RMT) , No. 2 (1981) , pp. 87, Problem 4751)

53. Let ABCD be a quadrilateral from the set P and let M,N,P,Q

be the midpoints of sides AB, CD, BC, AD, respectively.

or

so

The Euler's relation for the parallelogram M N PQ

is MP2

+PN2

+NQ2

+QM2

= MN2

+PQ2

,

A

On the other hand, we have

Hence

MQ=

DB , 2

M

MP=

AC 2

C

B

132 3. GEOMETRY

which is, clearly, a constant. (Dorin Andrica)

or

54. From the law of cosines we derive AB = 2R sin ;2 and

AF = 2R sin �;.

The first equality is equivalent to

Furthermore,

. rr . 5rr sm 12 sm 12 --5- + --rr- = 4, rr .

sin 12 sm 12

1 4 ' 5rr . rr = sm 12 sm 12 '

1 4 rr . rr 1 2 ' rr = cos 12 sm 12 or = sm '6 which is clear.

For the second equality, we have

It reduces to

AC2 - 4R2 . 2

2rr

- sm 12 ' AE2

- 4R2 . 2 4rr

- sm 12 '

AD2 - 4R2 . 2

3rr

- sm 12 ' AF2 -

4R2 . 2 5rr

- sm 12 '

• 2 rr . 2 2rr . 2

3rr . 2 4rr . 2

5rr

5

sm - + sm - + sm - + sm - + sm - = -

12 12 12 12 12 2 ' which is also clear. ( Ti

tu Andreescu, Revista Matematica Timi§oara

(RMT) , No. 6(1971) , pp. 27,

Problem 821)

55. Let A1A2 . . .

An be the given polygon and let S be the area of

AIA2 . . .

An .

There is a point M

inside the polygon such that

We write

Hence

n LMA% = 2S. k=1

n S = L

SAkMAk+l ' An+l = AI .

k=1

3 .2 . SOLUTIONS 133

In order to have equality everywhere we must have

sin Ak 7i'i.A'k+1 =

1 and

M Ak =

M Ak+l ,

k = 1,2, . . . , n.

It follows that M

is the circumcenter of a cyclic polygon and all sides subtent arcs of

90°. That is the polygon is a square, as desired. ( Ti

tu Andreescu, Revista Matematidi Timi§oara (RMT) , No.

2(1978)

, pp. 50,

Problem 3528)

56. In a triangle ABC

with the altitude AA'

and the circumradius R the following equality holds: AB

·AC

= 2R

· AA'

. (1)

Let R

be the circum radius of the polygon AIA2 . . .

An and let

PI , P2 , . . . ,

Pn be

the projections of a point P

on the circumcircle onto the sides Al A2 , A2 A3 , . . . ,

AnAl ,

respecti vely. Applying

(1) for triangles

PAiAi+I , where

An+l =

AI , yields

Hence

PAi ·PA

i+1 =2R

·PP

i , i=1,2, . . . , n.

which is a constant, as claimed. (Dorin Andri

ca, Revista Matematidi Timi§oara (RMT) , No. 1-2(1980)

, pp. 65,

Problem 4123)

57. Applying relation (1)

from the previous problem to triangles M A

IA2 , M A

3A4 , M A

5A6 , • • • ,

M A2n-I

A2n , we obtain

MA2n-1 .

MA2n =

2R· MK

2n-l .

Multiplying these equalities yields MK

.MK MK

_MA

I ·MA

2 . . . MA

2n I 3 · · · 2n-1 - 2nRn

(2)

For the triangles

relation (1)

yields MA

2 · MA

3 = 2R

· MK

2 , MA

4 • MA

5 = 2R

· MK

4, . . . ,

134 3. GEOMETRY

Multiplying these equalities gives M

. MK M K

=

M Al . M A

2 . . . M A

2n K2 4 · · · 2n 2

nRn

Similarly, we obtain MH

.MH MH

_MA

I ·MA

2 ·MA

2n 1 2 · · · n - 2

nRn

by applying relation (1) to triangles M A

IAn+I '

M A2An+2 , . . . ,

M AnA2n .

From equalities (2) , (3)

and (4) we draw the conclusion.

(3)

(4)

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1981 ) , pp. 68,

Problem 4622)

58. Let x and y be the measures of the arcs subtended by the sides a and b, respectively. We have

or

nx + ny = 27r,

27r x + y = - .

n Let R be the circumradius of the polygon. Then

27r Now y = - - x so n

. x a d ' y b sm 2 = 2R an sm 2 = 2R '

. y . (7r X ) . 7r X 7r . x b SIn 2 = sm ;; - 2 = sm ;; cos 2 - cos ;; sm 2 = 2R '

hence . 7r X b 7r . X sm - cos - = - + cos - SIn -. n 2 2R n 2

Squaring boths sides yields

or

. 2 7r 2 X b2 b 7r . X 2 7r . 2 X sm - cos - = -- + 2- cos - sm - + cos - sm -

n 2 4R2 2R n 2 n 2

. 2 7r ( . 2 X ) b2 b 7r . X 2 7r . 2 X sm - 1 - sm - = -- + 2- cos - sm - + cos - sm - . n 2 4R2 2R n 2 n 2

S · . x a b ' mce sm 2 = 2R ' we 0 tam

sin2 � (4R2 - a2 ) = b2 + 2ab cos � + a2 cos2 � . n n n

Therefore

(Dorin Andrica)

R = �Ja2 + 2ab cos :': + b2 • 2 sin - n

n

3.2. SOLUTIONS

59. We have from the Cauchy-Schwarz Inequality that

for any positive numbers ai , bi , i = 1 , 2, . . . , n.

Setting n = 4,

yields 3V"

SBCD > S2

L....J PA - ,

where V

is the volume of the tetrahedron. Then

as desired.

" S BCD >

S2 =

S2 = �

L....J P A - 3V

rS

r '

The equality occurs if and only if

� = aJaibi V b; for

i = 1, 2, 3, 4 hence bi = a-I , i = 1, 2, 3, 4. Then

PAl = PBI = PCI = PDI ,

so P

is the incenter of tetrahedron AB CD

.

135

Remar

k. The inequality holds for convex polyhedra circumscribed about a sphere. ( Ti

tu Andreescu, Romanian IMO Selection Test,

1982; Revista Matematidi

Timi§oara (RMT) , No.

1 (1982), pp.

82, Problem

4910)

60. All summations here range from i = 1 to

i = 4. Let 0

be the circumcenter and R

be the circum radius of AIA2A3A4 . By the Power-of-a-point Theorem,

GAi ·GA� =

R2 - OG2, for

1 � i

� 4. Hence the desired inequalities are equivalent to

and (R2 - OG2 ) L G�

i � L GA

i •

Now (1)

follows immediately from

(1)

(3)

136 3. GEOMETRY

by the Arithmetic-Geometric-Mean Inequality. To prove (3) , let P denote the vector from

0 to the point

P. Then

(4)

This is equivalent to (3) , since the last term of (4)

vanishes. By Cauchy-Schwarz Inequality,

so

1 and LGA

i L GA

i 2:: 16,

1 2 1 1 ( 2) 2 1 -"GA

."

- > -"(GA

·) "

- >"GA

· . 4 6 � 6

GAi - 16 6 Z 6

GAi - 6 �

Hence (2) also follows from (3) . ( Titu Andreescu, IMO 1995 Shortlist

)

Chapter 4 TRIGONOMETRY

1 . Prove that

2. Prove that

for all x E JR.

PROBLEMS

2 7r 27r 37r ctg 7 + ctg27 + ctg2 7 = 5.

3 X 3 X + 27r 3 X + 47r 3 cos - + cos -- + cos -- = - cos x 3 3 3 4

3. Evaluate the sum n-1

Sn = L

sin kx cos(n - k)x. k=1

4. Evaluate the sums

S1 = sin x cos 2y + sin 2x cos 3y + . . . + sin(n - l)x cos ny,

S2 = cos x sin 2y + cos 2x sin 3y + . . . + cos(n - l)x sin ny.

5. Evaluate the products 1) PI = (1 - tg1°) (1 - tg2°) . . . (1 - tg8g0) ; 2 ) P2 = ( 1 + tg1 °) (1 + tg2°) . . . ( 1 + tg44°) .

6. Prove that (4 cos2 go - 3) (4 cos2 27° - 3) = tan go.

7. Let x be a real number such that sec x - tan x = 2. Evaluate sec x + tan x.

8. Evaluate the product

7r where Ixl < 2n+2 •

139

140

and

4. TRIGONOMETRY

9. Let a, b, c, d, x be real numbers such that x f:. k7r, k E Z and sin x sin 2x sin 3x = -- =

a b c sin 4x -

d- '

10. Let a, b, c, d E [0, 7r] such that

2 cos a + 6 cos b + 7 cos c + 9 cos d = °

2 sin a - 6 sin b + 7 sin c - 9 sin d = 0.

Prove that 3 cos(a + d) = 7 cos(b + c) ,

1 1 . Prove that if arccos a + arccos b + arccos c = 7r,

then

12 . Let a, b, c be positive real numbers such that

Prove that

ab + bc + ca = 1 . 1 1 1

arctg - + arctg -b + arctg - = 7r, a c

13. Let x and y be real numbers from the interval (o , �) such that

cos2 (x - y) = sin 2x sin 2y

7r Prove that x + y = 2 '

14. Consider the numbers a , (3" E (0, i) such that 1 2 (1 - tga) (l - tg(3)

(1 - tg,) = 1 - (tga + tg(3 + tg,) ,

7r Prove that a + (3 + , = 4"

15. Let a, b E (0, �) , Prove that

if and only if a = b.

( Si�2 a) 2 + (cos2 a) 2 = 1 sm b cos b

16. Prove that

7r for all 0 < a,

b < 2" '

4. 1 . PROBLEMS

sin3 a cos3 a -- + -- > sec(a -

b)

sin b

cos b -

17. Let a, (3 be real numbers with (3 2:: 1. Prove that

(1 + 2 sin2 a)(3 + (1 + 2 cos2 a)(3 2:: 2(3+1

for all a E JR.

18. Let x be a real number, x E [- 1, 1] . Prove that

_1_ < x2n + (1 _ x2 )n < 1 2n-1 - -for all positive integers n.

19. Prove that sec2n x + cosec2nx 2:: 2n+1

for all integers n 2:: 0 and for all x E (0, %) .

20. Prove that (1 + sin x) (1 + cos x) :s � +

V2

for all real numbers x.

21 . Find the maximal value of the expression

E = sin Xl COS X2 + sinx2 cos x3 + . . . + sin xn cos Xl ,

when Xl , X2 , . . . , Xn are real numbers.

22. Find the extreme values of the function f : JR -t 1R,

f (x) = a cos 2x + b cos x + c,

where a, b, c are real numbers and a,

b > O.

23. Let ao, aI , . . . , an be numbers from the interval (0, 7r /2) such that

tan ( ao - �) + tan ( a1 - �) + . . . + tan ( an - �) 2:: n - 1 .

Prove that tan ao tan a1 . . . tan an 2:: n n+1 .

24. Find the period of the function

if p, q are positive integers.

f(x) = cospx + cos qx, x E JR

141

142 4. TRIGONOMETRY

a2 - 5 25. Let aD = � + V3 + v'6 and let an+1 = ( n ) for n 2::

O. Prove that 2 an+2

(2n-37r) an = cot -3- -2 for all n.

26. Let n be an odd positive integer. Solve the equation

cos nx = 2n-1 cos x.


A sin2 x + B sin 2x + C = 0,

where A, B, C are real parameters.


. . . 3 Slll X cos y + sm y cos z + sm z cos x = '2

29. Prove that the equation

. . 2 . 3 . 4 3 sm x sm x sm x sm x = 4 has no real solutions.

30. Solve the system of equations { 2 sin x + 3 cos y = 3 3 sin y + 2 cos x = 4.

31. Solve the system of equations

{ x sin y + :1- x2 cos y = V;

x+y = 4'

32. Solve the system of equations

1

37r x + y + Z = 4 tgx + tgy + tgz = 5 tgx . tgy . tgz = 1 .

33. Prove that in any triangle abc a cosA + b cosB + ccos C = 2R2

and

and

and

4. 1 . PROBLEMS


� 3A B C A B C

� . 2 A � cos "2 sin "2 sin "2 = cos "2 cos "2 cos "2 � sm "2 '

35. Let n be a positive integer. Prove that in any triangle L

sin nA sin nB sin nC = (_l)n+l + cos nAcos nB cos nC

L cos nA cos n

B sin n

C = sin nA sin nB sin nCo

36. Consider a triangle ABC

such that

sin A

sin B

+ sin B

sin C

+ sin C

sin A

= A

(1 + sin

A) (l + sin B) (l + sin

C) = 2 (A +

1)

Prove that triangle AB C

has a right angle.

37. Let A > 1

be a real number and let ABC

be a triangle such that

a>' cos B

+ b>' cos A = c>'

a2>.-1 cos

B + b2>.-1 cos A = C

2>'-1 .

Prove that the triangle is isosceles.

38. Prove that the triangle ABC

is equilateral if and only if A B C 1

2 2 2 ) tg"2 + tg"2 + tg"2 = 4S

(a +

b + c .

39. Let ABC


sin2 B

+ sin2 C

= 1

+ 2 sin B sin C cos A.

Prove that triangle ABC

has a right angle.

40. Let ABC


(cot � r + (2 cot �r + (3 cot �r = (�;) " ,

143

where s and r denote its semiperimeter and its inradius, respectively. Prove that triangle

ABC is similar to a triangle

T whose side lengths are all positive integers

with no common divisor and determine these integers.

144 4. TRIGONOMETRY

41. Prove that in any triangle . A B C

9 � sm - cos - cos - < - . L..J 2 2 2 - 8

42. Prove that in any triangle a2 b2

e2 4 ( . 2 A . 2

B . 2

C) - + - + - > sm - + sm - + sm - . be ea a

b - 2 2 2

43. Prove that in any triangle cos A

+ cos B +

cos C > �. a3

b3 e3 - 16p3

44. Prove that in any triangle 2A

2B

2C

sec - sec - sec - 9 __ 2 + __ 2_ + __ 2_ > _ . b

e ea ab

-p2


!!. > 3V3. r -

46. Let ABC

be a triangle. Prove that . 3A

. 3B

. 3C A

-B B

-C C

-A

sm - + sm - + sm - < cos -- + cos -- + cos -- . 2 2 2 - 2 2 2

47. Find the number of ordered pairs (a, b)

such that (a+

bi)2002 = a-

bi, a,

bE R

48. Find Imz5

min -zEC\1R Im5 z and the values of z for which the minimum is reached.

49. Let ZI , Z2 , . . . , Z2n be complex numbers such that IZll

= IZ21

= . . . = IZ2n

l and

arg ZI :::; arg Z2 :::; . . . :::; arg Z2n :::; 7r. Prove that

50. For all positive integers k define

Prove that for any integers m and n with 0 < m < n we have Al U

A2 U · · · U

Am C

An-m+1

U An-m+2

U· · ·

U An.

4. 1 . PROBLEMS 145

51 . Let Z1 , Z2 , Z3 be complex numbers, not all real, such that IZ11 = IZ2 1 = I Z3 1 = 1

and 2 (Z1 + Z2 + Z3) -3Z1Z2Z3 E JR.

Prove that

52. Let n be an even positive integer such that � is odd and co, C1 , . . . , cn-1 the complex roots of unity of order n. Prove that

n-l II (a +

bc%) = (a

� + b� )2

k=O

for any complex numbers a and b.

53. Let n be an odd positive integer and co, C1 , . . . , cn-l the complex roots of unity of order n. Prove that

n-1 II (a +

bc%) = an +

bn

k=O

for all complex numbers a and b.

54. Let Zl , Z2 , Z3 be distinct complex numbers such that IZ11 = IZ2 1 = IZ3 1 = r.

Prove that 1 1 1 1 ------ + + > -. IZ1 - z2

1 1z1 - z3

1 IZ2 - zd lz2 - z3

1 IZ3 - zl

l lz3 - z2

1 - r

2

55. Let Zl , Z2 , Z3 be distinct complex numbers such that IZ11 = IZ2 1 = I Z3 1 = r

and Z2 f:. Z3 ·

Prove that

56. If Z is a complex number satisfying Iz3 + z-3

1 :::; 2, the inequality show that I

z + z-11

:::; 2.

57. The pair (ZI , Z2

) of nonzero complex numbers has the following property:

there is a real number a E [-2, 2] such that z? - aZl Z2 + z� = O. Prove that all pairs (zf , z�) , n = 2, 3, . . . , have the same property.

58. Let A1A2 . . .

An be a regular polygon with the circumradius equal to

1. Find

n the maximum value of max

II P Aj when

P describes the circumcircle.

j=l

146 4. TRIGONOMETRY

59. Let n be an odd positive integer and let aI , a2 , . . . , an be numbers from the interval [0, 71'] .

Prove that 1 - n L cos(ai - aj ) 2:: -2- ·

l �i<j�n

60. Let n be a positive integer. Find the real numbers ao and akl , k, l = 1, n, k >

l, such that

sin2 nx -. -2- = ao + L

akl cos 2 (k -l)x

sm x 19<k�n for all real numbers x f:. m7l', m E Z.

SOLUTIONS

1 . We prove that � 2

k7r _ n

(2n - 1

)

� cot 21 - 3 ' k=l n +

for all integers n > O. Consider the equation

with roots

sin(2n + l)x = 0,

7r 27r n7r 2

n + 1 ' 2n + 1

' . . . , 2n + 1

.

Expressing sin(2n + l)x in terms of sin x and cos x, we obtain

sin(2n + l}x =

(2n: 1) cos2n x sin x _ en: 1) COs2n-2 sin3 X + . . . =

= Sin2n+1 X (en: 1) cot2n X - en: 1) cot2n-2 x + . . . ) S -

k7r k - 1

2 S· · 2n+l -I- 0 h et x - -- , - , , . . . , n. lnce SIn x -r , we ave 2

n + 1 (2n: 1) cat2n X _ en: 1) cot2n-2 x + . . . = O.

Substituting y = cot2 x yields en: l) yn - en: l) yn-l + . . . = 0,

(1)

with roots cot2 _7r_, cot2 �, . . . , cot2 2 n7r 1 . Using the relation between coef-2

n + 1 2n + 1 n +

ficients and roots, we obtain

n (2n + 1)

� 2 k7r 3 � cot 2

n + 1 = en: 1) Setting n = 3, the desired conclusion follows.

n(2n - 1

)

3

(Dorin Andrica, Revista Matematidi Timi§oara (RMT), No. 1-

2(1979)

, pp. 51,

Problem 3831)

147

148 4. TRIGONOMETRY

2 . Applying the identity t t cos t = 4 cos3 3" - 3 cos 3" ' t E

1R

for t = x, t = x + 271", t = x + 471" and summing up the three relations, we obtain ( X X + 271" X + 471") 3 cos x = 4 cos3 3' + cos3 -3- + cos3 -3

- -

-3 cos - + cos -- + cos -- . ( X X + 271" X + 471") 3 3 3

On the other hand, x x + 271" X + 471" 471" 2x + 471" X + 271" cos 3" + cos -3- + cos -3- = 2 cos 6 cos 6 + cos -3- =

= (2 cos 2; + 1) cos x � 2" = 0 and the desired identity follows. (D

orin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1975) , pp. 44,

Problem 2124)

3. We have 28n = 8n + 8n =

= sin x cos(n - l)x + sin 2x cos(n - 2)x + . . . + sin(n - l)x cos x+

+ sin(n - l)x cos x + sin(n - 2)x cos 2x + . . . + cos(n - l)x sin x = = sin nx + sin nx + . . . + sin nx = (n - 1) sin nx,

so n - 1 8n = -2- sin nx.

(Dorin Andrica, Gazeta Matematica (GM-B) , No. 8(1977) , pp. 324, Problem

16803; Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp. 30, Problem 3055)

and

4. Note that

81 + 82 = sin(x + 2y) + sin(2x + 3y) + . . . + sin[(n - l)x + ny]

82 - 81 = sin(2y - x) + sin(3y - 2x) + . . . + sin[ny - (n - l)x] . Setting

x + y = hI and y - x = h2 yields

81 + 82 = sin(y + hI ) + sin(y + 2hI ) + . . . + sin(y + (n - l)hI ) =

Hence

and

4.2. SOLUTIONS

. nhl . [

hI ] sm T sm y +

(n - I

)2

. hI sm -2 S

2 -SI = sin(y + h2) + sin(y + 2h2) + . . . + sin(y + (n -

I)h2) =

. nhl . [ (

1) hI ]

1 sm T sm y + n - 2 SI = h

2 . 1 sm 2

. nh2 . [ (

1) h2 ]

I sm T sm y + n - 2 2 . h

2 sm 2

sin nhl sin

[Y +

(n - 1)

hI ]

1 sin n�2 sin [Y + (n - 1)

h;]

S _ 1 2 2

2 - 2 . hI

+ 2 . h2

. sm2 sm2

149

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 65,

Problem 3056)

5 . We have PI = 0 because of the factor 1 - tan 45° = O. On the other hand we have

P2 = (cos 1 ° + sin 1 0) . . . (cos 44° + sin 44°) = cos 1 ° . . . cos 44 ° ( � cos 10 + � sin 10) • • •

( � cos 440 + � sm 440 ) = �--------------�--�----------------� (.;2) 44

2 cos 1 ° . . . cos 44°

= sin 46° . . . sin 89° (v'2) 44 = 222 . cos 1 ° . . . cos 440 ( Titu Andreescu

)

cos 3x 6. We have cos 3x = 4 cos3 x - 3 cos x, so 4 cos2 X - 3 = -- for all x f. cos x (2k + 1) . 90° , k E Z. Thus

(4 cos2 9° _ 3) (4 cos2 270 _ 3) = cos 27° . cos 81° = cos 81° = sin 9° = tan 9° cos go cos 270 cos 9° cos 9° ,

as desired. ( Titu Andreescu

)

150 4. TRIGONOMETRY

7. From the identity 1 + tan2 x = sec2 x it follows that 1 = sec2 x - tan2 x = (sec x - tan x) (sec x + tan x) = 2(secx + tanx) ,

so sec x + tan x = 0.5. ( Titu

Andreescu, American High School Mathematics Examination, 1999, Prob

lem 15) 8. Since 1

+ tan2 2kx cos2 2kx (1 - tan2 2kx)2 - cos2 2k+lX

for all Ix l < 2:+2 ' it follows that

(Dorin Andrica)

9. Let

Then

n cos2 2kx cos2 2x Pn = II

cos2 2k+lX = cos2 2n+1x ' k=l

sin x _ sin 2x _ sin 3x _ sin 4x _ A -a- - -b

- - -c- - -d

- - .

sin2 4x = 2 sin2 2x(l - sin2 2x) . Because sin2 4x = 2 sin2 2x(1 - sin2 2x) , we obtain

d2 = 2b2 (1 _ A2b2 ) .

On the other hand, sin 3x = AC, sin x =

Aa, and since

sin 3x = sin x (3 - 4 sin 2 x) ,

we have c = a(3 - 4A2a2) .

Eliminating A from the relations

(1) and (2) yields

2a3 (2b3 - �) = b4 (3a - c) ,

as desired. (Dorin Andrica)

10. Rewrite the two equalities as

2 sin a - 9 sin d = 6 sin b - 7 sin c

2 cos a + 9 cos d = -6 cos b - 7 cos c. By squaring the two relations and adding them up we obtain

85 + 18 cos(a + d) = 85 + 42 cos(b + c) ,

and the conclusion follows.

(1)

(2)

4.2. SOLUTIONS

( Titu Andreescu, Korean Mathematics Competition, 2002)

1 1 . From the hypothesis it follows that

sine arccos a + arccos b + arccos c) = o.

Using the identity L

cos a cos ,B sin'Y = sin(a + ,B + 'Y) + sin a sin ,B sin'Y and the formulas

we obtains

sin(arccos x) = � and sin(arcsinx) = x, x E [-1 , 1] ,

ab� + be� + eaJ1=b2 = )(1 - a

2 ) (I-b2) (I

- c2) , as desired.

151

( Titu


Problem 3054)

12 . The identity x + y + z - xyz arctgx + arctgy + arctgz = arctg ( ) + k7r 1 - xy + yz + zx

implies 1 1 1 a

b + be + ea - 1 arctg- + arctg-b + arctg- = arctg b (

b ) + k7r.

a e a e - a + + e Because a

b + be + ea = 1, we obtain

1 1 1 arctg- + arctg-b + arctg- = k7r, a e

where k is integer. 7r Note that 0 < arctgx < "2 for all real x > 0, hence

1 1 1 37r o < arctg� + arctgz; + arctgc < 2·

Therefore k = 1 and

as claimed.

1 1 1 arctg- + arctg-b + arctg- = 7r, a e

(Titu

Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1 (1977) , pp. 42,

Problem 2827)

13. The given relation is equivalent to

( cos x cos y + sin x sin y) 2 = 4 sin x sin y cos x cos y, or

( cos x cos y - sin x sin y) 2 = 0

152 4. TRIGONOMETRY

Hence cos2 (x + y) = 0,

and since x, y E (0, %) , we obtain x + y = %, as desired. ( Ti

tu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1(1977) , pp. 42,

Problem 2826)

or

14. Expanding yields 1 "2 (1 - tana - tan,B - tan')' + tan a tan,B + tan ,B tan,),+

+ tan,), tan a - tan a tan ,B tan,),) = = 1 - (tan a + tan,B + tan,),) ,

tan a + tan,B + tan')' - tan a tan ,B tan,), = = 1 - tana tan ,B - tan ,B tan,), - tan')' tan a.

Since a, ,B, ')' E (0, i) , we have ° < a + ,B + ')' < 7r , hence

tan a + tan ,B + tan')' f= tan a tan ,B tan,),. From relation (1) we derive

1 - tan a tan ,B - tan,B tan')' - tan,), tan a = 1 , tan a + tan,B + tan,), - tan a tan ,B tan ')'

therefore cot(a + ,B + ')') = 1 .

Hence a + ,B + ')' = i, as desired.

(1)

( Titu


Problem 1582)

15 . The relation in the statement is equivalent to

or

It follows that

. 2 2 (sin4 a cos4 a) (sm b + cos b) -.-2- + �b = 1 , sm b cos

• 4 cos2 b . 4 sin 2 b sm a + cos4 a + -.-2- sm a + �b cos4 a = 1 . sm b cos

. cos2 b . sin 2 b 1 - 2 sm2 a cos2 a + -.-2- sm4 a + �b cos4 a = 1 , sm b cos hence

Furthermore,

( COS b . 2 sin b 2 ) 2 --:--b sm a - --b cos a = 0. sm cos

cos b . 2 sin b 2 --:--b sm a = --b cos a, sm cos

4.2. SOLUTIONS

or tan2 a = tan2 b. Because a, b E (0, -i) , we obtain a = b. The converse is clear and we are done.

153

Alternat

ive so

lution. From the given relation we deduce that there is a number

e E (0, -i) such that sin2 a and cos2 a -. -b- = sin e --b- = cos e sm cos

Hence sin 2 a = sin b sin e and cos2 a = cos b cos e.

It follows that 1 = cos(b - e) and cos

2a = cos(b + e)

Since a, b, e E (0, i) , we have b - e = ° and 2a = b + e, hence a = b, as desired.

( Titu


1 (1977), pp.

41,

Problem 2825


, pp. 452

, Problem 16934)

16. Multiplying the inequality by sin a sin b + cos a cos b = cos(a - b) , we obtain the equivalent form

( Sin3 a cos3 a) -. -b- + --b- (sin a sin b + cos a cos b) 2::

1. sm cos

But this follows from Cauchy-Schwarz Inequality, because, according to this inequality, the left-hand side is greater than or equal to (sin2 a + cos2 a)2 =

1. ( Ti

tu Andreescu

)

17. Using the inequality

for m 2:: 1 we obtain

(1 + 2 sin2 a)p +

(1 + 2 cos2 a)p :::: 2 ( 2 + 2 sin2 � + 2 cos2 a) P = 2,8+1 ,

as desired. ( Titu


1 (1974), pp.

30,

Problem 1942)

18. Because x E [-1,1], there is a real number y such that x = sin y. It suffices

to prove that 1> sin2n y + cos2n y > �1 - -

2n-

154 4. TRIGONOMETRY

For the left-hand side note that I sin yl �

1 and

I cos yl � 1, hence

sin2n y + cos2n y � sin2n-2 y + COS2n-2 y � . . . � sin2 y + cos2 y = 1,

as desired. For the right-hand side we use the inequality

Hence

as claimed.

x? + x� > (Xl + X2 ) n 2 - 2

sin2n y + cos2n y > ( sin2 y + cos2 y) n 2 - 2

Alternat

ive

Solution. By setting u = x2 and v = 1 - x2 the inequality becomes

1 2n-l � un + vn � 1. Because u, v E

[0 ,1] and u + v = 1, we have un � U, vn � v,

implying un + vn � U + v = 1 . Also, by the power mean inequality,

un + vn > 2 (u + V ) n = 2 (!) n = _1_. - 2 2 2n-l

(Dorin Andrica)

19. We have

sec2 x = tan2 x + 1 2:: 2 tan x and cosec2 x = cot2 x + 1 2:: 2 cot x

by the AM-GM inequality. It follows that

sec2n x + cosec2nx 2:: 2n (tann x + cotn x) . Since tann x + cotn x 2:: 2, we obtain

sec2n x + cosec2nx 2:: 2n+l ,

as desired. Alternat

ive

Solution. Using the AM-GM inequality we obtain

sec2n x + cosec2nx > 2Jsec2n xcosec2nx = 2 1 - sinn x cosn x

= 2n+l __ 1_ > 2n+l sinn 2x -(D

orin Andrica, Gazeta Matematidi (GM-B) , No. 3 (1975) , pp. 104, Problem

14900)

20. We have

(1 . ) (1 ) < (1

+ sin x)2 + (1 + COS X)2 + mn x + coo x = - 2 _ 2 + 2(sin x + cos x) + (sin2 x + cos2 x) _ - 2 -

as desired.

4.2. SOLUTIONS

3 . 3 In . ( 7r ) 3 In = 2 + (sm x + cos x) = 2 + v 2 sm x + '4 � 2 + v 2,

Note that equality holds for x = � + 2k7r, where k is integer. Alternat

ive

Solution. Expanding the left-hand side, we see that

1 + (sin x + cos x) + � sin 2x =

1 + ../2 sin (x + �) + � sin 2x �

< 1 + ../2 + � = � + ../2. - 2 2

155

(Titu Andreescu, Revista Matematidl Timi§oara (RMT) , No. 2(1978)

, pp. 47, Problem 3500)

21 . We have E = sin Xl COS X2 + sin x2 COS X3 + . . . + sinxn cos Xl �

sin2 Xl + cos2 X2 sin2 X2 + cos2 X3 sin2 Xn + cos2 Xl n < + + . . . + = - 2 2

2 2

n Therefore the maximal value of E is and it is reached, for example, when 2 7r Xl = X2 = . . . = Xn = '4 '

(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1977) , pp. 65,

Problem 3058)

22. Because a, b

> 0, it follows that the maximal value of f is a +

b + c.

Setting y = cos X yields

f(x) = a(2y2 - 1) + by + c = 2ay2 +

by + c - a.

If - :a E [-1, 0) , then the minimal value of f is

� -b2 + 8a(c - a)

8a 8a b

If - 4a E (-00 , -1), then the minimal value of

f is

f ( -1)

= 2a -b + c - a = a -

b + c.

(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No.

1 (1981), pp. 52,

Problem 4315)

23. First

Solution. Let

bk = tan( ak - 7r / 4) , k = 0,

1, . . . , n . It follows from the

hypothesis that for each k, -1

< bk <

1 and

1 + bk �

L (1 -bl).

09¥k5;n (1)

156 4. TRIGONOMETRY

Applying the Arithmetic-Geometric-Mean Inequality to the positive numbers 1-b

z , I = 0, 1, . . . ,

k -1, k + 1, . . . , n, we obtain

L (1 -bl)

� n ( II (1 _

bl)) l in

O�l::fk�n 09::fk�n

From (1)

and (2)

it follows that

and hence that

Because

n ( n ) l in !! (1 + bk ) � nn+1 Do (1 - bz)n ,

1 + bk

1 + tan ( ak - �)

-- - = tan ( ( ak - �) + �) = tan ak , 1 -bk - 1

- tan (ak - �) the conclusion follows. S

econd S

olution. We first prove a short lemma:

(2)

Let W, X, Y, z

be rea

l num

bers w

ith x + y = w + z an

d Ix - y

l < Iw - z

l. Then

wz < xy . Proo

f. Let x + y = w + z =

2L. Then there are non-negative numbers r, s with

r < s and wz =

(L - s

) (L + s) <

(L - r

) ( L + r

) = xy.

We now use this lemma to solve the problem. For 0 ::; k

::; n, let bk = tan

(ak-7r /4)

and let

Then -1 <

bk <

1 and

tit. = G � :�) G � ::) = 1 + 1 +

bi�' _ 1 b

j + b

+ k

(1)

First note that because -1 <

bk <

1 and

bo +

b1 + . . . +

bn � n -

1, it follows

that bj +

bk > 0 for all 0 ::;

j, k

::; n with j f:. k

. Next note that if bj +

bk > 0 and b

j f:. b

k , then it follows from the lemma applied to (1)

that the value of tj tk can be made smaller by replacing

bj and

bk by two numbers closer together and with the

same sum. In particular, if bj < 0, then replacing

bj and

bk by their average reduces

the problem to the case where bi > 0 for all

i.

We . may now successively replace the bi 's by their arithmetic mean. As long as

the bi are not equal, one is greater than the mean and another one is less than the

mean. We can replace one of this pair by the arithmetic mean of all the bi 'S, and the

4.2. SOLUTIONS 157

other by a positive number chosen so that the sum of the pair does not change. Each such change decreases the product of the ti 'S. It follows that for a given sum of the bi 'S, the minimum product is attained when all of the bi 'S are equal. In this case we

n -1

have bi > --1 ' for each i, so - n +

tOt1 . . . tn;::: ( :: �)

n+1

= C;) n+l = nn+1 .

n +1

This completes the proof. Third S

olution. We present a solution based on calculus. We set

a = bo + b1 + . . . + bn , where -

1 < bi <

1, and assume that a ;::: n -

1. We then show that the product

lIn 1 + bk 1 - bk k=O

attains its minimum when all of the bk 's are equal, that is, their common value is al (n + 1) . The desired inequality will follow immediately. We proceed by induction. The case n = 1 was established in the discussion of

(1)

in the previous solution. For n ;::: 2, set

n-1 L bk = a

' = a - b

n > n -

2.

k=O The last inequality follows from a ;::: n - 1 and b

n < 1. Set b = b

n and c = a'

ln ,

so b + nc = a. By the induction hypothesis, (IT 1 + bk ) 1 + bn > ( 1 + c) n 1 + b . k=O 1 - bk 1 - bn - 1 - c

1- b

Thus we need to prove that ( 1 + c) n (�) >_

(n + 1 + a) n+1 , 1 - c 1 - b n + 1 - a

(1)

where the right-hand side is obtained by substituting al (n +

1) for each bk , k =

0, 1 , . . . , n , in the product. Next, recall that a is fixed, and that b + nc = a. Thus we can eliminate b from (1) to obtain the equivalent inequality

G � � r G � : � :�) � (:: � � :) n

+1 (2)

Now bring all terms in (2)

to the left-hand side of the inequality, clear denominators, and replace c by x. Let the expression on the left define a function f with

158 4. TRIGONOMETRY

To establish (2)

it is sufficient to show that for 0

� x < 1, I(x) attains its

minimum value at x = aJ (n +

1). Towards this end we differentiate to obtain

t(x) = n

(a -

(n +

l)x) ( (l

+ x)n-1 (n +

1 - a

)n+1 -(1 - x

)n-1 (n + 1 + a

)n+1 ) =

= n(a -

(n +

l)x)g(x),

where g(x) = (1 + x)n-1 (n + 1 - a

)n+1 -

(1 - x

)n-1 (n +

1 + a

)n+1 . It is clear that f' (_a

_) = 0, so we check the second derivative. We find

n +1

!" (n : l ) = -n(n +

llg (

n : l ) > 0,

so I has a local minimum at x = a

J (n +

1). But

I' (x) could have another zero, t,

obtained by solving the equation g(x) = O. Because

g' (x) = (n -

1) (1 + x

)n-2(n +

1- a

)n+1 + (n -

1) (1- x

)n-2(n +

1 + a

)n+1

is obviously positive for all x E [0,1), there is at most one solution to the equation

g(x) = 0 in this interval. It is easy to check that g(aJ (n + 1) ) < 0 and g

(l) > O. Thus

there is a real number t, aJ (n +

1) < t <

1, with g

(t) = O. For this t we have

I" (t) = n

(a -

(n +

l)t)g' (t)

< O. Thus, t is a local maximum for

I, and no other extrema exist on the interval (0

,1). The only thing left is to check that

1 (1) � l

(aJ (n +

1)). Note that the case x =

1

is also an extreme case with bo =

b1 = . . . = bn-1 =

1. This case does not arise in

our problem, but we must check to be sure that on the interval 0 � x < 1, I(x) has

a minimum at x = aJ (n +

1). We have

1(1) = 2n

(1 + a - n

) (n +

1 - a

)n+1 � 0,

since n -1

� a � n + 1, and l

(aJ (n +

1)) = 0 (by design) . Thus

I(x) indeed attains

a unique minimum at x = aJ(n +

1). ( Ti

tu Andreescu, USA Mathematical Olympiad,

1998, Problem 3)

24. Let d be the greatest common divisor of p and q. We prove that

T = � is the lowest positive period of the function f.

It is clear that I (x +

T1) = I(x)

for all real x, therefore T is a period of function

I.

Suppose there are T1 >

0 and an integer

A > 0 such that

T = AT1 and

l(x+

T1) = I (

x) for all real x. Then

I(T1) = 1(0) = 2, so

cos pT1 + cos q

T1 =

2,

4.2. SOLUTIONS

therefore COSp

T1 = COS q

T1 = 1 .

It follows that

for some integers k1 ' k2 >

O.

• 27r Smce T

= AT1 and

T1 = Ad '

T1 = 2

kl7r

= 2k27r

P q

k1

k2

1

P = q = Ad and so p = k1 (Ad)

, q = k2 (Ad) . On the other hand,

d = gc

d(p, q

), so

A = 1, hence

T = T1 as desired.

159

(Dorin Andrica, Revista Matematidt Timi§oara (RMT) , No.

2(1978), pp.

75,

Problem 3695)

25. We have 7r 2 7r

7r cos 24 2 cos 24 cot 24 = -.-7r- = . 7r 7r sm - 2 sm- cos -24 24 24 1 + cos (i - �) 1 + V; + v:

---�� = --�� sin (i - �) VB

_ v'2

4 4

7r 1 + cos 12

--=-==- = . 7r sm -12

4 + VB + v'2 VB -v'2

=

_ 4( VB + v'2)

+ (VB + v'2)2

_ 4( VB + v'2)

+ 8 + 4V3 _

- 6 - 2 - 4 -

= 2 + v'2 + V3 + VB = ao + 2. ( 2n-3 ) Hence an = cot T - 2 is true for n = O. ( 2n-3 ) It suffices to prove that bn = cot T , where bn

recursive relation becomes

b _ 2 _ (bn - 2)2 -5

n+1 - 2bn '

= an + 2, n � 1. The

b2 -1

2k-37r or bn+1 = ;bn

. Assuming, inductively, that bk = cot ck , where Ck = -3- ' yields

and we are done.

cot2 Ck -1

bk+1 = 2 t = cot (2Ck ) = cot Ck+1 , co Ck

( Titu Andreescu, Korean Mathematics Competition, 2002)

26. If n = 1, then all real numbers x are solutions to the equation.

160 4. TRIGONOMETRY

Let n > 1 and note that

cosnx = (�) cosn X - (�) cosn-2 x sin2 x + . . . + We have two cases:

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

-2- n _ 1 cos x sin n- x.

a) x f:. (2k + 1) �

for any integer k. Then 2

I cos nxl = I cos xl l (�) cosn-1 X - (�) cosn-3 sin2 X + . . . +

+( -1) n;

' (n � 1) sinn-1 x l ::; ::; I cos x l ((�) I cosn-1 x l + (�) i cosn-3 X sin2 xl + . . . + (n � 1) I sinn-1 x l) <

hence there are no solutions in this case. b) x = (2k + l)i for some integer k. Then

cos x = 0 and cosnx = 0, since n is odd, so { (2k +

1) i lk integer} is the set of solutions. Alternat

ive

Solution. With the substitution x = i - y the equation becomes

cos (n� - ny) = 2n-1 sin y.

Because n is odd, (1)

is equivalent to

± sinny = 2n-1 sin y. Taking modules gives

I sinnyl = 2n-1 1 sin y l · But I sin nYI � nl sin y l for all y in lR, hence

nl sin y l � 2n-1 I sin y l ·

(1)

(2)

If y f:. k7r, k E Il , then n � 2n-1 , which implies n E {1,3}. The case n = 1

is clear and for n = 3 the original equation reduces to cos 3x = 4 cosx, that is 4 cos3 X - 3 cos x = 4 cos x. Taking into account that cos x f:. 0, this yields cos2 x = � ,

which is not possible. It follows that y = k7r, which gives the solutions x = (2k + l)i , k E Il. ( Ti

tu Andreescu, Gazeta Matematidi (GM-B) , No.

7(1978), pp.

304, Problem

17297; Revista Matematidl Timi§oara (RMT) , No.

1-2(1980)

, pp. 63, Problem 4107)

4.2. SOLUTIONS

27. The equation is equivalent to (A

+ G)

sin2 x + 2B sin x cos x + G cos2 X = 0

We have the following cases: i) A

+ G = 0 and G 1= O. Then

2B cos x = 0 or cot x = -0'

hence x E { (2k ; 1

)1r I k

E Z } U { arcctg ( - 2g) + k1r1 k E Z } .

ii) A

+ G 1= 0

and G = O. Then

. 2B sm x = 0 or tan x = -A + G '

hence x E

{k1r1 k

E Z}

U { arctg ( -}: c) + k1r1 k

E Z } .

iii) A = B = G = O. Then any real number is a solution.

iv) A = G = 0 and B 1= O. Then sin 2x = 0 and so

x E {k� 1 k E Z } . v) A

+ G 1= 0

and G 1= O

. The equation is equivalent to (A

+ G)

tan2 x + 2B tan x + G = 0,

hence -B ± JB2 _

AG+G2

tan x = A+G

for B2 + G2 �

AG. It follows that

x E {arctgY1 +

k7rl k

E Z}

U {arctgY1 +

k7r1 k

E Z}

if B2 + G2 �

AG. Otherwise there are no solution.

161

(Dorin Andrica, Revista Matematica Timi§oara

(RMT

), No.

1 (1978), pp.

89,

Problem 3429)


or

It follows that

Hence

2 sin x cos Y + 2 sin Y cos z + 2 sin z cos x = 3, (sin x - cos y

)2 +

(sin y - COS Z

)2 + (sin z - COS X

)2 = O.

sin x = cos y, sin y = cosz , sin z = cos x.

162 4. TRIGONOMETRY

7r Z + X =

(4k3 + 1) '2

for some integers k1 ' k2 ' k3 and therefore

and

7r 7r X =

[4(k1 -

k2 +

k3)

+ 1] 4" ' y = [4(k

1 + k2 -

k3)

+ 1] 4" '

7r Z =

[4(-k1 +

k2 +

k3)

+ 1] 4" ' k1 ,k2 ,k3 E Z.

( Titu

Andreescu, Gazeta Matematica (GM-B ) , No. 11 (1977) , pp. 451, Problem

16931 ; Revista Matematica Timi§oara (RMT) , No. 1-2 (1979) , pp. 52, Problem 3835)

29. Note that

sin x sin 2x sin 3x sin 4x = � (cos 3x - cos 5x)

( cos x - cos 5x) =

1 = 4 (cos2 5x - cos 3x cos 5x - cos 5x cos x + cos x cos 3x) = 1 6 3 = 8" (2 cos2 5x - cos 2x + cos 8x - cos 4x + cos 6x) < 8" = 4'

hence the equation has no solution. ( Titu

Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 (1977) , pp.

41,

Problem 2923)

30. Squaring both equations and summing up yields 4(sin2 x + cos2 x) + 9 (cos2 Y + sin2 y) + 12 (sin x cos y + sin y cosx) = 25,

or 13 + 12 sin(x + y) = 25.

Hence sin(x + y) = 1 ,

and so 7r X + y =

(4k + 1 ) '2

for some integer k. It follows that

sin x = cos y and sin y = cos x.

Turning back to the system we obtain

hence

. 3 d . 4

sm x = cos y = 5 an smy = cos x = 5'

3 4

tan x = 4' tan y = 3 · Note that sin x, cos x, sin y, cos y are all positive, therefore

3 x = arctan 4 + 2k7r

4.2. SOLUTIONS 163

and 4

y = arctan 3' + 217r

,

for some integers k and

l. ( Ti

tu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp.

74,

Problem 3694)

31 . Observe that x E [-1,1]

and

x = sin( arcsin x) , � = cos( arcsin x) . From the first equation we obtain

then

. .j2 cos(y - arcsmx) = 2'

y - arcsin x = ±� +

2k1l".

U . 11" smg x + y = "4' we get x + arcsin x = � ± � +

2k1l",

for some integer k. C

ase 1 . x + arcsin x = i - 2k1l". Because x E

[-1,1] and arcsin x E [- i, iJ , we

have k = 0, hence

Therefore

• 11" x + arcsmx = 2'

• 11" arcsmx = 2' - x, or x = cos x.

For this equation there is only one solution Xo E (0, �) . The system has the solution

x = Xo , Case 2. x + arcsin x = 2k1l".

Using similar arguments, k

= 0, so

11" Y = - - Xo 4

arcsin x = -x.

This equation has the unique solution x = ° so the system has the solution

x = o, ( Ti

tu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 (1977) , pp.

41,

Problem 2824)

32. Using the formula

t ( ) tan x + tany + tan z - tan x tan y tan z an x + y + z = -------=�---------=�--1

- tan x tan y - tan y tan z - tan z tan x

164 4. TRIGONOMETRY

we have

Hence

37r 5 - 1 -1

= tan - = ---------------4 1- tan x tan y - tan y tan z - tan z tanx

tan x tan y + tan y tan z + tan z tan x = 5.

The equation t3 -

5t2

+ 5t - 1 = 0

has roots tan x, tan y, tan z, from the relations between the roots and the coefficients. On the other hand, the equation has the roots 1, 2 + V3, 2 - V3, hence {x, y, z} == { 7r 57r 7r } . 4' + k7r, 12 + h7r, 12 + p7r , for some mtegers

k,l, p.

( Titu

Andreescu, Revista Matematica Timi§oara (RMT), No. 8(1971) , pp. 27,

Problem 1018)

33. Using the Extended Law of Sines we obtain

a cos A + b cos B + c cos C = 2R( sin A cos

A + sin B cos B + sin C cos C) ==

= R(sin 2A + sin 2B + sin 2C) = R(2 sin(A + B) cos(A - B) + sin 2C) = A-B + C A-B-C

= 2R sin C(cos(A - B ) + cos C ) = 4R sin C cos 2 cos 2 =

= -4R sin C cos (i - B) cos (i - A) == 4R sin A sin B sin C == ;!�, as desired. ( Ti

tu Andreescu, Revista Matematica Timi§oara (RMT), No. 2(1977) , pp. 65,

Problem 3060)

34. Because A + B + C = 7r, we have A . B C . C B cos "2 = sm "2 cos "2 + sm "2 cos "2'

Hence

B . C A . A C cos - = sm - cos - + sm - cos -2 2 2 2 2 ' C . A B . B A cos - = sm - cos - + sm - cos -2 2 2 2 2 '

A cos -2

B cos -2

C cos -2

. B C sm - cos -2 2

. C A sm - cos -2 2

. A

B sm - cos -2 2

. C B sm - cos -2 2

. A C sm - cos -2 2

. B A sm - cos -2 2

= 0,

4.2. SOLUTIONS 165

because the first column is the sum of the other two. Computing the determinant, we obtain

L 3 A . B . C A B C L ' 2 A cos - sm - sm - = cos - cos - cos - sm -2 2 2 2 2 2 2 ' as desired. (D

orin Andrica)

35. Denote E1 = L

sin nA sin nB cos nC and & = L cos nA cos nB sin nCo

Observe that

(cos A + i sinA) (cos B + i sin B) (cos C + i sin C) = = cos(A + B + C) + i sin(A + B + C) =

= cos 7f + i sin 7f = -1. By de Moivre's formula,

(cos nA + i sin nA) (cos nB + i sin nB) (cos nC + i sin nC) = (_l)n .

Expanding the brackets yields

-E1 +

iE2 + cos nA cosnB cosnC - i sinnA sin nB sin nC = (_ l)n .

Hence E1 = (_ 1)n+1 + cos nA cos nB cosnC and & = sin nA sin nB sin nC.

(Dorin Andrica, Revista Matematidi Timi§oara (RMT), No. 1 (1978) , pp. 65,

Problem 3278)

36. Subtracting from the second equality the first multiplied by 2

yields

(sin A -1)

(sin B - l) (sin C - 1) = O. Hence sin A, sin B or sin C is

1, so AB C is a right triangle. ( Ti

tu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2(1978) , pp. 49,

Problem 3514)

37. Note that a cos B + b cosA - c = O.

The system of linear equations

{ a cos B + b cos A - c = 0

aA cos B + bA cos A - cA = 0

a2A-1 cos B +

b2A-1 cosA - C2A-1 = 0

166 4. TRIGONOMETRY

has the solution (cos B, cos A, - 1) and is homogeneous. Therefore the determinant

a b

c Ll = aA

bA cA

a2A-1

b2A-1 C2A-1

is zero. On the other hand,

1 1 1 Ll = a

bc aA-1

bA-1 cA-1 = (

aA-1 )2 (bA-1 )2 (

cA-1 )2

= abc(aA-1 _

bA-1 ) (aA-1 _ cA-1 ) (bA-1 _ cA-1

).

Therefore a = b, b

= c or c = a, hence the triangle is isosceles. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1977) , pp. 89,

Problem 3199)

or

38. The relation is equivalent to J(

P -b) (p - c

) J(P - c

) (p - a

) J(P - a

) (p -

b) 1 ( 2 b2 2 ) .....:..-..:--;=7='====;� + + = - a + + c J

p(P - a

) Jp(p -

b) Jp(p - c

) 4S

1 1 S L(

P - a) (p -

b) = 4S

(a2

+ b2

+ c2 ).

Expanding the brackets yields

then

It follows that

hence a = b

= c.

1 _p

2 + a

b + bc + ca = -

(a2

+ b2

+ c2 ), 4

4(ab

+ bc + ca

) = a

2 + b2

+ c2

+ (a +

b + C

)2.

(a -

b)2 + (b - C

)2 + (c - a

)2 = 0,

( Titu


(1972) , pp. 28,

Problem 1160)

39. By the Extended Law of Sines,

On the other hand,

so

a = 2Rsin A, b

= 2R sin B, c = 2Rsin C.

sin2 B + sin2 C = sin2 A + 2 sin B sin C cos A.

4.2. SOLUTIONS 167

From the hypothesis we have

sin2 B + sin2 C = 1 + 2 sin B sin C cosA,

therefore sin2 A = 1 . It follows that A = �, hence the triangle ABC is right, as desired. (D

orin Andrica, Revista Matematica Timi§oara (RMT) , No.

1-2(1977)

, pp. 52, Problem 3838)

40. Because 62 + 32 + 22 = 72 and

s A B C A B C - = cot - + cot - + cot - = cot - cot - cot -r 2 2 2 2 2

2 '

the given relation is equivalent to

(62 + 32 + 22) [ ( cot � ) 2 + (2 cot !) 2 + (3 cot �) 2] =

( A B C) 2 = 6 cot "2 + 6 cot 2" + 6 cot 2

(1)

This means that we have equality in the Cauchy-Schwarz inequality. It follows that

A cot __ 2

6

B 2 cot 2 3

C 3 cot 2

2

Plugging back into (1)

gives cot � = 7, cot � = � , and cot � = � . Hence by

the Double angle formulas, sin A = !.-, sin B = 28 , and sin C = �. Thus the side 25 65

130

lengths of T are 26, 40, and 45. ( Ti

tu Andreescu, USA Mathematical Olympiad, 2002, Problem 2)

41 . Summing up the formulas

yields

. A B C ra = 4R sm 2 cos 2 cos 2'

4R . B C A rb = sm 2 cos 2 cos 2'

4R . C B A r c = sm 2 cos 2 cos 2

'" . A B C r a + rb + r c 6 Slll 2" cos 2 cos 2 = 4R "

On the other hand ra + rb + rc = 4R + r, hence

'" . A B C 4R

+ r r 6 sm 2" cos 2 cos 2 = ----;w:- = 1 + 4R "

168 4. TRIGONOMETRY

Because � � r, it follows that

. A B C 1 9 '" sm - cos - cos - < 1 + - = -L...J 2 2 2 - 8 8 ' as desired. (D

orin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp. 49,

Problem 3510)

or

42. We have

a2 b e 2 cos A + - = - + - . be e b

Because � + � � 2, we obtain

a2 A be � 2(1 - cos A) = 4 sin2 "2

and likewise b2 4 ' 2 B - > sm -ae - 2 '


- + - + - > 4 sm - + sm - + sm -� � 2 ( . 2 A . 2 B . 2 C ) be ea ab - 2 2 2 '


43. We have

so 2be cosA b2 e2

a2 + 1 = 2 + 2 ' a a Likewise,

and

2ae cos B 1 _ a2 e2

b2 + - b2 + b2

2ab cos C 1 _ a2 b2 e2 + - 2 + 2 '

Summing up these equalities implies e e

3 2be cosA 2ae cosB 2ab cos C _ ( b2 a2 ) + 2 + b2 + 2 - 2 + b2 +

a e a

hence

and moreover

4.2. SOLUTIONS

be cos A ea cosB ab cos C > � a2 + b2 +

e2 - 2 ' '

cosA cos B cos C 3 -- + -- + -- > -- . a3 b3 e3 - 2abe

By the AM-G M inequality,

therefore

3 3 ( 3 ) 3 81 2abe � 2 a + b + e

= 16p3 '

cos A cosB cos C > 81 -as +

---,;s + � - 16pS '

169

(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1975), pp. 46,

Problem 2134)

44. We have

2 A 1 be sec 2" = --A- = (p - a)

, cos2 - P

2

2 B ea sec 2 = p(p - b) ,

so it suffices to prove that 1 1 1 9

-- + -- + -- > - . p - a p - b p - e - p Setting x = p - a, Y = P - b, Z = P - e in the inequality

yields

(x + Y + Z

) (.!. +�+�) � 9

x Y Z

( 1 1 1 ) P -- + -- + -- > 9 p - a p - b p - e - ,

and the solution is complete. (Dorin Andrica)

45. By the AM-GM inequality,

2 C ab sec 2 = p(p - e) ,

p = (p - a) + (p

- b) + (p

- e) � 3 \!(P - a)(p - b) (P - e) ,

Then p3 � 27(p - a) (p - b) (p - e)

so p4 � 27p(p - a) (p - b) (p - e) = 2782 •

It follows that p2 � 3V38, and since 8 = pr, !!. � 3va, as desired. r ( Ti


Problem 4993)

170 4. TRIGONOMETRY

A B C 46. Let a = 2' (3 = 2 ' , = 2 ' Then

0° < a, (3" <

90° and a + (3 + , =

90°.

We have

and

. 3A B - C . 3 ((3 ) ' 3 . ( 2 )

sm: "'2 - cos -2- = sm a - cos - , = sm a - sm a + , =

= 2 cos(2a + ,) sin(a - ,) = -2 sin(a - (3) sin(a - ,) . In exactly the same way, we can show that

. 3B C - A 2 ' ((3 ) . ((3 ) sm "'2 - cos -2- = - sm - a sm - ,

. 3C A - B 2 . ( ) . ( (3) sm T - cos -2- = - sm , - a Slll , - .

Hence it suffices to prove that

sin(a - (3) sin(a - ,) + sin((3 - a) sin((3 - ,) + sin(, - a) sin(, - (3) � O. Note that this inequality is symmetric with respect to a, (3" , we can assume

without loss of generality that 0°

< a < (3 < , < 90°

. Then regrouping the terms on the left-hand-side gives

sin(a - (3) sin(a - ,) + sin(, - (3) [sin(, - a) - sin((3 - a)] , which is positive as function y = sin x is increasing for

0° < x <

90°. Al

ternative

Solution. We keep the notation of the first solution. We have

sin 3a = sin a sin 2a + sin 2a cos a;

cos((3 - a) = sin( 2a + ,) = sin 2a cos, + sin , cos 2a; cos((3 - ,) = sin(2, + a) = sin 2, cosa + sin a cos 2,;

sin 3, = sin , cos 2, + sin 2, cos, . It follows that

sin 3a + sin 3, - cos((3 - a) - cos((3 - ,) =

= (sin a - sin ,) (cos 2a - cos 2,) + (cos a - cos,) (sin 2a - sin 2,) = = (sin a - sin ,) (cos 2a - cos 2,) + 2 (cos a - cos ,) cos(a + ,) sin(a - ,) .

Note that sin x is increasing and cos x is decreasing for 0

< x < 90°

. Since o < a" , a + , <

90°, each of the two products in the last addition is less than or

equal to O. Hence

sin 3a + sin 3, - cos((3 - a) - cos((3 - ,) � O.

In exactly the same way, we can show that

sin 3(3 + sin 3a - cos(, - (3) - cos(, - a) � 0

and sin 3, + sin 3(3 - cos(a - ,) - cos(a - (3) � O.

4.2. SOLUTIONS 171

Adding the last three inequalities gives the desired result. ( Titu

Andreescu, USA IMO Team Selection Test, 2002, Problem 1)

47. Let z = a + bi, Z = a - bi

, and Izl = Ja2 +

b2• The given relation becomes

Z2002 = z. Note that

from which it follows that Izl ( lzl 200l - 1) = O.

Hence Izl = 0, and

(a, b) = (0, 0) , or

Izi = 1 . In the case

Izl = 1, we have Z2

002 = Z, which is equivalent to Z2

003 = z · z = I z l2 = 1 . Since the equation Z2003 = 1 has 2003

distinct solutions, there are altogether 1

+ 2003 = 2004 ordered pairs that meet the required conditions. ( Ti

tu Andreescu, American Mathematics Contest 12A, 2002, Problem 24)

48. Let a, b

be real numbers such that z = a + bi, b i- O

. Then 1m z5 = 5a4b -

10a2b2

+ b5

and Imz5 = 5 (�) 4 - 10 (�) 2 + 1 . Im5 z

b b

Setting x = (�) 2 yields

1m Z5

-5- = 5x2

- lOx + 1 = 5(x -1)2

- 4. 1m z

The minimum value is -4 and is obtained for x = 1 Le. for z = a(1 ± i) , a i- O. ( Ti


Problem 5221)

49. Let Ml , M2 , . . . ,

M2n be the points with the complex coordinates

Zl , Z2 , . • . , Z2n and let AI , A2 , . . . ,

An be the midpoints of segments

MlM2n , M

2M2n-l , . . • ,

MnMn+l .

172 4. TRIGONOMETRY

The points Mi , i = 1, 2n lie on the upper semicircle centered in origin and with

radius 1. Moreover, the lengths of the chords MlM2n,

M2M2n-l , . . . ,

MnMn+l are

in a decreasing order, hence OA

l , OA

2 , . . . , OA

n are increasing. Thus

1 Z, � Z2n 1 :::; 1 Z2 + ;2n-' 1 :::; . . . :::; 1 Zn +2Zn+1 1 and the conclusion follows. Al

ternative Solution. Consider Zk = r (cos tk+i sin tk

), k = 1 , 2, . . . ,

2n and observe

that for any j = 1 , 2, . . . , n, we have

= r2 [(cos tj + Cos t2n_j+d2

+ (sin tj + sin t2n_j+l

)2] = = r2 [2 + 2(cos tj cos t2n-j+l + sin tj sin t2n-j+l

)] = 2 2 2 t2n-j+l - tj = 2r [1 + COS(t2n-j+l - tj

)] = 4r cos 2 .

Therefore IZj + Z2n-j+l

l = 2r cos t2n-j�1 - tj and the inequalities

IZI + z2n

l :::; IZ2 + Z2n-l

l :::; . . . :::; IZn + zn+l

l

are equivalent to t2n - tl � t2n- l - t2 � . . . � tn+l - tn · Because 0 :::; tl :::; t2 :::; . . . :::; t2n :::; 7r, the last inequalities are obviously satisfied. (D

orin Andrica, Revista Matematica Timi§oara (RMT), No. 1(1984) , pp. 67,

Problem 5222)

50. Let p = 1 , 2, . . . , m and let Z E Ap e Then zP = 1 .

Note that n-m+l, n -m+

2, . . . , n are m consecutive integers, and, since p :::; m,

there is an integer k

E {n - m + 1, n - m +

2, . . . , n

} such that p divides

k.

Let k = kip . It follows that zk = (zp)k

'

= 1 , so Z E Ak C

An-m+l

U An-m+2

U

. . . U An, as claimed. Remar

k. An alternative solution can be obtained by using the fact that

(an -

1) (an-l -

1) . . .

(an-k+l - 1) (

ak -1) (

ak-1 - 1) . . . (a -

1)

is an integer for all positive integers a > 1

and n > k.

(Dorin Andrica, Romanian Mathematical Regional Contest " Grigore Moisil" ,

1997

)

51 . Let Zk = cos tk + i sin tk ,

k E {I, 2, 3}.

The condition 2(Zl + Z2 + Z3

) -3ZlZ2Z3 E

1R implies

2(sin tl + sin t2 + sin t3) = 3 sin(tl + t2 + t3 ) . (1)

4.2. SOLUTIONS 173

Assume by way of contradiction that max(tl ' t2 , t3

) < i, hence tl , t2 , t3 < i' Let

tl + t2 + t3 (0 7r ) Th . f . . [0 7r ) t = 3 E , '6 ' e sme unctlOn IS concave on ' '6 ' so

1 ( . . . ) . tl + t2 + t3 '3 sm tl + sm t2 + sm t3 � sm 3 .

From the relations (1) and (2)

we obtain sin

(tl + t2 + t3

)

< . tl + t2 + t3 2 - sm 3 '

then sin

3t �

2 sin t.

It follows that 4 sin3 t - sin t �

0,

i .e. sin2 t � � . Hence sin t � � , then t � i, which contradicts that t E (0, i) , Therefore max

( tl , t2 , t3

) ;::: i , as desired.

(2)

( Titu

Andreescu, Revista Matematidi Timi§oara (RMT), No. 1(1986) , pp. 91,

Problem 5862)

52. Let n = 2(28 + 1) and b f:. 0

, otherwise the claim is obvious. Consider a complex number a such that a2 = � and the polynomial

f = xn - 1 = (X - eo

) (X - el

) . . .

(X - en-I

)'

We have

and f (-'n = (�1 r (a + i

oo)

• • • (a + ion-

d,

hence

Therefore n-l n-l n-l II (

a +be�)=bn II (� + e

�) = bn II (a2+ e

� )=

k=O k=O k=O

= bnf (T) f (- T) =

bn [(a2 )2s+1 + 1

]2 = bn [ (�t+l + If

= b2(2s+1) a

= ( � b� )2 ( 2s+1 +

b2S+1 ) 2 b2s+1 a + .

(Dorin Andrica, Romanian Mathematical Olympiad - second round,

2000)

53. If ab = 0, then claim is obvious, so consider the case when a f:.

0 and

b f:. O.

We start with a useful lemma.

174 4. TRIGONOMETRY

Lemma. If co , C1 , • • • , cn-l are t

he comp

lex roots o

f unity o

f order n, w

here n

is

an odd i

nteger, then

n-l II (A+ BCk) = An +

Bn, k=O f

or all comp

lex num

bers

A and B

. Proo

f. Using the identity

n-l xn - 1 = II

(x - ck) k=O A

. for x = -B Ylelds (An ) n-l (A ) - B

n + 1

= -!! B + Ck ,

and the conclusion follows. 0 Consider the equation

bX2 + a = 0 with roots Xl and X2 . Since

we have n-l n-l n-l n-l II (

a + bc�) = bn II (

ck - Xt} (ck - X2

) = bn II (

ck - Xl) II

(€k - X2)

k=O k=O k=O k=O Using the lemma for

A = -Xl ,

B = 1 and then for

A = -X2 ,

B = 1 gives

n-l II (ck - xt

} = (-Xl

)n + 1 = 1 - X

�,

k=O n-l II (

ck - X2) = (-X2

)n + 1 = 1 - x� .

k=O Hence

n-l II (a +

bcT:) = bn(l - x�) (l - x�) =

k=O

= bn [l + (XlX2

)n - (x� + x�)] = bn [1 + (�) n] = an + bn ,

since XlX2 = � and x? + x� = x? + (_xt}n = o.

(Dorin Andrica, Romanian Mathematical Olympiad - second round,

2000)

54. Consider the triangle with vertices of complex coordinates Zl , Z2 , Z3 and the circumcenter in the origin of the complex plane. Then the circumradius

R equals I

ZII = IZ21 = IZ31 = r and the side lengths are

a = IZ2 - z3

1,

b= IZl - Z3

1, c = IZI - Z2

1.

i .e .

4.2. SOLUTIONS

The desired inequality is equivalent to 1 1

1 1 - + - + - > ab b

e ea -R2

abc 48 4pr a +

b+ e � R2 = }f = ]f

or R � 2r, which is Euler's inequality for a triangle.

175

(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1985), pp. 82,

Problem 5720)

55. Let AI , A2 , A3 and

A be the points of complex coordinates ZI , Z2 , Z3 and let

Z = aZ2 + (1 - a)z3 , a E R Hence point

A lies on the line

A2A3 and the triangle A

lA2A3 has its circumcenter in the origin of the complex plane .

Point B

is the foot of the altitude from Al in the triangle

AlA2A3 . It follows that A

lA � AlB, so

We have IZ2 - z3

1

_

IZ2 - z3

1 h _

IZI - z2

1 1zl - z3

1 sin Al _

IZI - z2

1 1zl - z3

1·

2r 8A1A2A3 - 2 - 2 - 2 therefore

as desired.

h = IZI - z21 1zl - z3

1

2r '

y

(Dorin Andrica, Romanian Mathematical Olympiad - final round, 1984)

56. Denote Iz +

1/z1 by r. From the hypothesis,

176 4. TRIGONOMETRY

Hence r3 :::; 2 + 3r, which by factorization gives

(r - 2

) (r +

1)2 :::; O. This implies

r :::; 2, as desired. ( Titu Andreescu, Romanian Mathematical Olympiad - first round,

1987; Revista

Matematidl Timi§oara (RMT) , No. 1 (1987)

, pp. 75, Problem

6191)

57. Denote t = Zl , t E <C* . The relation z? - aZ1Z2 + z

� = 0 is equivalent Z2

a ± iv'4

- a2

to t2 - at + 1

= O. We have � = a2 - 4 :::; 0, hence t = 2 and

It I

==

. /a2 4 - a

2 zn

V 4 + -4- = 1. If t = cos a + i sin a, then

z� = tn = cos na +

i sin na and we can

write z?n - anz

fzg

+ z�n = 0, where an = 2 cos na E [-2, 2

]. Al

ternative

Solution. Because a E

[-2, 2

]' we can write a = 2 cos a. The relation

z?

- aZ1Z2 + z� = 0 is equivalent to

Zl Z2 - + - = 2 cosa Z2 Zl

and, by a simple inductive argument, from (1)

it follows that

zn zn � + � = 2 cos na, n = 1 , 2, . . . zg

zf

(1)

(Dorin Andrica, Romanian Mathematical Olympiad - second round, 2001 ; Gazeta

Matematidl (GM-B) , No. 4(2001) , pp. 166)

58. Rotate the polygon Al A2 . . .

An such that the complex coordinates of its

vertices are the complex roots of unity of order n, C1 , C2 , . . . , Cn ' Let z be the complex coordinate of point

P located on the circumcircle of the polygon and note that

Izl = 1.

The equality

yields

n zn -

1 = II (

z - Cj)

j=l

n n Izn -

1 1 = II Iz - cj

l = II PA

j . j=l j=l

n

Since Izn -

1 1 :::; Izln +

1 = 2, it follows that the maximal value of

II PA] is 2

j=l and is attained for zn = -

1, i .e. for middle points of arcs

AjAj+1 ' j = 1, . . . , n, where A

n+1 = A1 . (Dorin Andrica, Romanian Mathematical Regional Contest " Grigore MoisH" ,

1992)

59. We will use the following auxiliary result:

4.2. SOLUTIONS 177

Lemma. (IMO 1973, Problem 1)

Let C be a semicircle of unit radius and P1 , P2 , . . . ,

P n points on C, where n �

1 is an odd integer. Then

� � -::::-=+ IOP1 +

OP2 + . . . +

OPnl � 1,

where 0

is the center of C. Proof. The key idea is to show that the orthogonal projection of the vector sum

--=t � --=-=-t • OP

1 + OP

2 + . . . + OP

n onto some lme has length not less than 1

(see S. Savchev, T. Andreescu, "Mathematical Miniatures" , The Mathematical Association of America, 2003, pp. 75) . Let n = 2k -

1. From the considerations of symmetry, the line

l

containing the middle vector 0 P�

is a natural candidate for such a line (here we use the fact that n is odd!) .

A o---------��------� B

It is technically convenient to consider l as an axis with positive direction determined by

OP�. As is well known, the projection of the sum of several vectors is equal

to the sum of their projections. Hence it suffices to prove that the sum of the signed lengths

OP1 , OP

2 , . . . , OP

2k-1 of the projections of OP�

, OP�

, . . . ,OP

2k-!

onto l

is greater than or equal to

1. Denote the diameter of C by

AB and the orthogonal

projections of A

and B

onto l

by A1 and

B1 . We have

OPk = 1 and also

OP1 +

OP2 + . . . +

OPk-1 � (k -

1)OA1 '

OPk+1 +

OPk+2 + . . . +

OP2k-1 � (k -

1)OB1 .

This is because OP

j � OA

1 for j = 1 , . . . , k -1

and OP

j � OB

1 for j k +

1, . . . , 2k -

1. Since

OA1 +

OB1 = 0, the proof is complete. 0

Consider the complex numbers

Zk = cos O::k + i sin O::k , k = 1,2, . . . , n

and the points P1 , P2 , . . . ,

Pn with complex coordinates Zl , Z2 , . . . , Zn '

178 4. TRIGONOMETRY

Using the above Lemma we have IOP�

+ OP�

+ . . . + OP�I

� 1 , hence IZI + Z2 +

. . . + zn l � 1 , or

It follows that

as desired.

It COS O:k + it sin O:k I � 1 . k=1 k=l

(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1983) , pp. 90, Problem C:58)

and

60. Using the identities

S � 2 · sin nx cos(n + l )X 1 = L..J cos JX = ---.--'------'--

j=l sm x

S � . 2 · sin nx sin(n + l)x 2 = L..J sm JX = ---.----

j=l sm x

we obtain S2 + S2 = ( sin nx) 2 1 2 sin x

On the other hand,

hence

S� + si = (cos 2x + cos 4x + . . . + cos 2nx)2+

+ (sin 2x + sin 4x + . . . + sin 2nx)2 = = n + L: (cos 2kx cos 2lx + sin 2kx sin 2lx) =

= x + 2 L: cos 2 (k -l)x,

19<k�n

( sin nx ) 2 -- = n + sin x cos 2(k -

l)x .

l �l<k�n Set ao = n and akl = 2, 1 :::;

l < k :::; n, and the problem is solved.

(Dorin Andrica, Romanian Mathematical Regional Contest " Grigore Moisil" , 1995)

Chapter 5

MATHEMATICAL ANALYSIS

PROBLEMS

1 . Let 1 :::; a < f3 be real numbers. Prove that there are integers m, n > 1 such

that a < ifiii < f3.

2. Let (an)n�O and

(bn)n�o be the sequences of integers defined by ( . 1o

) 2n+1 . 10 1 + v 3 = an +

bn V 3, n E N.

Find a recursive relation for each of the sequences (an)n�O and

(bn)n�o ,

3. Study the convergence of the sequence (xn)n�O satisfying the following prop

erties: 1) Xn >

1, n = 0, 1, 2, . . .

2) - Xn+1 - -- = -- , n = 0, 1 , 2, . . . 1 ( 1 ) Xn + 1

2 Xn+1 Xn -1

4. Study the convergence of the sequence (Xn)n�l defined by Xl E (0, 2) and

Xn+1 = 1

+ V2xn - x�

for n � 1.

5 . Consider the sequence of real numbers (Xn)n�l such that

xi

+ x�

+ . . . + X2

lim n = O.

Prove that

Is the converse true?

n�oo n

1· Xl + X2 + . . . + Xn 0 1m = . n�oo n

6 . Let (an)n�l and

(bn)n�l be sequences of positive numbers such that an > n

bn

for all n > 1.

Prove that if (an)n�l is increasing and

(bn)n�l is unbounded, then the sequence (

Cn)n�l ' given by Cn = an+1 - an, is also unbounded.

7. Let 0 < a < a be real numbers and let (Xn)n�l be defined by Xl = a and

(a + 1)Xn-1 + 0'2

Xn = , n � 2. Xn-1 + (a +

1)

181

182

and

5. MATHEMATICAL ANALYSIS

Prove that the sequence is convergent and find its limit.

8. Find a sequence (an)n2:l of positive real numbers such that

lim (an+l - an

) = 00

n-+oo

lim (Jan+l -

�) = O. n-+oo

9. Let (Xn)n2:l be an increasing sequence of positive real numbers such that

1· Xn 0 1m 2 = . n-+oo n

Prove that there is a sequence (nkh2:l of positive integers such that

10. Let a, (3

be real numbers and let (Xn)n2:l ,

(Yn)n2:l ,

(Zn)n2:l be real sequences

such that max{x

� + O'Yn, y

� + (3xn} :::;

Zn for all n ;::: l. 1 a) Prove that Zn ;::: -8(0'2 +

(32 ) for all n ;::: 1 .

b) If lim Zn = --81 (0'2 +

(32 ), prove that the sequences

(Xn)n2:b

(Yn)n2:l are n-+oo

convergent and find their limits.

1 1 . The sequences (xn)n2:l and

(Yn)n2:l are defined by Xl 2, Yl = 1 and

Xn+l = x�

+ 1 , Yn+l = XnYn for all n ;::: 1 . a) Prove that Xn

/ Yn < V7 for all n ;::: 1 .

b) Prove that the sequence (Zn)n>l , Zn = xn

/Yn, is convergent and lim Zn < V7. - n-+oo

12. Let a ;::: 0 and a f= 0 be real numbers and let (Xn)n2:l be an increasing

sequence of real numbers such that

lim nC¥ (xn+l - xn

) = a. n-+oo

Prove that the sequence is bounded if and only if a > 1 .

13. Evaluate

14. Evaluate

1. l:n k( n - k) ! + (k + 1) 1m n-+oo k=O (k + l) ! (

n - k) ! n ( k ) �+l

lim -n-+oo

L

n2 k=l

15. Evaluate 00 1 (i) L -' q > 1 ; n=l nq

n

5 . 1 . PROBLEMS

00 1 (ii) � (4n +

l)qn+l '

q > 1 .

183

16. Let (Xn)n�l be an increasing sequence of positive integers such that Xn+2

+

Xn > 2Xn+l for all n � 1 .

Prove that the number

is irrational.

17. Prove that

is irrational for all n � 1 .

00 1 (J - "' -L... 10Xn n=l

00 1 An =

L (k!)n

k=l

18. Let k, s be positive integers and let aI , a2 , . . . , ak , bl , b2 , . . . , bs be positive real

numbers such that

for infinitely many integers n � 2.

Prove that 1) k = s; 2) ala2 . . . ak = bl

b2 . • .

bs .

19. Let (Xn)n� l be a sequence with Xl = 1 and let X be a real number such that

Prove that

IT (1 - �) = e-x • n=l Xn+l

20. Let A f:. ±1 be a real number. Find all functions f : R � R and 9 : (0, 00) � R

such that f(ln x + A lny) = 9 (JX) + 9 (..fij) for all x, y E (0, 00).

21. Let f be a continuous real-valued function on the interval [a, b] and let ml , m2 be real numbers such that ml m2 >

O. Prove that the equation

f(x) = � + m2 a - x b - x

has at least a solution in the interval (a, b).

184 5. MATHEMATICAL ANALYSIS

22. Let a and b

be real numbers in the interval (0,1/2), and let 9 be a continuous

real-valued function such that g(g(x)) = ag(x) + bx for all real x. Prove that g

(x)

= ex for some constant e.

23. Find all continuous functions f

: � � [0, 00

) such that

f2 (X + y) -f2 (X - y

) = 4f(x)f(y)

for all real numbers x, y.

24. ( i) Prove that if the continuous functions f

: � � (-00,

0] and g : � � [0

, 00)

have a fixed point, then f + 9 has a fixed point.

(ii) Prove that if the continuous functions rp : � � [0,1]

and 'ljJ : � � [1, 00

) have

a fixed point, then rp'ljJ

has a fixed point.

25. Let rp : � � � be a differentiable function at the origin and satisfying rp(O) = 0. Evaluate

;� � [rp(x) + rp (�) + ' " + rp (;)] , where n is a positive integer.

26. Let a be a positive real number. Prove that there is a unique positive real number f-£ such that

p? > aIL-x

for all x > 0.

XIL -

27. Let f

: [a, b)

� � be a twice differentiable function on [a, b)

such that f(a)

= f(b) and

f' (a) = j

' (b).

Prove that for any real number ,\

the equation f" (

x) -,\(f' (

X))2 = °


28. Find all functions f

: [0, 2]

� (0,1]

that are differentiable at the origin and satisfies f (

2x) = 2f2 (X) - 1, x E

[0,1]

29. Let ,\ be a positive integer. Prove that there is a unique positive real number f) such that

for all real number x > 0.

30. Let f

: � � � be a function continuous at the origin and let '\, f-£ be two distinct positive real numbers.

Prove that the limit

5. 1 . PROBLEMS

1. f(>..

x) -f(J-t

x)

1m -"----'-----'--"-a:�O X

exists and is finite if and only if f

is differentiable at the origin.

31 . The sequence (Xn)n�1 is defined by

Prove that lim nXn = -2. n�oo

185

32. Let Xo E (0,1]

and Xn+l = Xn - arcsin(sin3 xn), n �

O. Evaluate lim

Vnxn. n�oo

33. Let f : JR � JR be a twice differentiable function with the second derivative

nonnegati ve. Prove that f(

x + I'(x))

� f(x), x E R

34. Let a < b

be positive real numbers . Prove that the equation

(a; T+Y

= aXbY


35. Find with proof if there are differentiable functions cp : JR � JR such that cp( x )

and cp'(x)

are integers only if x is integer.

36. Let f : [a, b]

� JR be a differentiable function. Prove that for any positive integer n there are numbers fh <

O2 < . . . <

On in the

interval (a, b)

such that f(b)

-f(a) f' (O

I)

+ f' (0

2)

+ . . . + f' (On ) b- a n

37. Let f, 9 : JR � JR be differentiable functions with continuous derivatives such

that f (x)

+ 9 (x)

= l' (x) - g'

(x)

for all x E JR.

Prove that if Xl , X2 are two consecutive real solutions of the equation f(x) -g

(x)

= 0, then the equation

f(x)

+ g(x) = 0 has at least a solution in the interval

(Xl , X2

).

38. Let f: [- � , �] � (-1 , 1) be a differentiable function whose derivative

f' is

continuous and nonnegative. Prove that there exists Xo in [-� , �] such that (f (

XO) )2 + (1' (XO) )2 � 1.


39. Prove that there are no positive real numbers x and y such that

x2Y

+ y2-X = x + y.

( ) n(n+1) 40. a) Prove that if x � y � n : 1 for some integer n � 2, then

yX + n+¢Y � ifY + n+V'x.

b) Prove that

n n'n + � > 2n +

1, n _> 3. V 'H n+l n +

1 -

41 . Let Xl , X2 , . . . , Xn be positive real numbers such that Xl + X2 + . . . + Xn = 1. Prove that

42. Let f

: JR � JR be a function with a noninjective antiderivative. Prove that f(c) = 0 for some c E JR.

43. Let h , h, . . . ,

f n : JR � JR be continuous functions. Prove that

max(h (x), h (

x), . . . ,

fn(x))d

x

is a deri vati ve and evaluate

44. Evaluate

45 . Let p be a polynomial of odd degree such that p'

has no multiple zero and let

f : JR � JR be a function such that

fo p is a derivative.

Prove that f

is a derivative.

46. Let 1= (0, 00) and let

f : I

� I

be a function with an antiderivative F

that satisfies the condition

F(x)f G) = x,

for all x in I. Prove that g :

1-+

1R, g(x) = F(x)F (�) is a constant function and

then find f.

5. 1 . PROBLEMS 187

47. Let n > 1 be an integer and let f

: [0 , 1] � JR be a continuous function such that

11 1 1 f(

x)dx = 1 + - + . . . + - .

a 2 n

Prove that there is a real number Xo E (0, 1) , such that f(xo)

= 1 - xg

1 - Xo

48. Consider the continuous functions f, 9 :

[a, b]

� R Prove that the equation

!(x) /."

g(t)dt = g(x) f !(

t)dt


49. Let f : [a, b]

� JR be a continuous function such that {!(

X)dx

of O.

Prove that there are numbers a < a < f3 < b

such that

{ !(x)dx = (b - a

)! (p).

50. Let a, c be nonnegative real numbers and let f : [a, b]

� [c, dJ be a bijective

increasing function. Prove that there is a unique real number f-£ E

(a, b)

such that

{ !(t)dt = (IJ - a)e +

(b - lJ

)d.

51. Let <p : JR � JR be a continuous function such that

for all x, y E JR.

{x+Y {X Jx <p(t)dt = Jx-y <p

(t)dt,

Prove that <p is a constant function.

52. Let f : JR � JR be a differentiable function such that

for all real number x < y.

!:.±lL y i 2 f(t)dt 5 !!:.±lL f(t)dt

2

Prove that f

is a nondecreasing function.

53. Let f : JR � JR be an injective and differentiable function.


Prove that the function F

: (0, 00) � lR,

l

1a: F(

x) = -

f(t)dt

x a is monotone.

54. Prove that

55. Prove that there are no Riemann integrable functions f

: lR � lR \ {O} such that (Y f(

x)

J

a:

f(t)dt = f(

y) ,

for all real numbers x =J y. 56. Let

f : [0, 1]

� lR be a differentiable function with continuous derivative such that l' [i' (x)]2dx = 1.

Prove that I f(l)

-f(O) 1

< 1.

57. Find all continuous functions f

: [0, 1]

� lR such that {I 1 Ja

f(x) (x -

f(x) )d

x = 12 ·

58. Let fa :

[0, 1]

� lR be a continuous function and let the sequence (f n)n�l be

defined by fn(x) = { fn-l (t)dt, x E

[0,1].

Prove that if there is an integer m 2:: ° such that l' fm(t)dt = (m : 1) ! '

then the function fa has a fixed point.

5.9. Let f : [-1, 1]

� lR be a differentiable function with nondecreasing derivative. Prove that 1 ( I 2 J

- 1

f(x)dx �

f(-I) + f' (l)

.

60. Let f, g :

[a, b]

� lR be continuous functions. Prove that there is a real number C E

(a, b)

such that

[ f(x)dx + (e - a)g(e) = l g(x)dx +

(b - e

)f(e).

SOLUTIONS

1 . We prove that there is an integer n > 1 such that fJn - an >

1 . Let e = fJ - a. Then

fin - an = (a + e)n - an = (�) an-I e + . . . + en > nan-I e > ne,

because a > 1 . Take an integer n >

�

. Then fJn - an > 1 .

e The interval (an ,

fJn ) has length greater than 1, hence there is an integer m > 1 such that an < m <

fJn , or a < rm < fJ, as desired.

(Dorin Andriea, Revista Matematidi Timi§oara (RMT) , No. 1 (1982) , pp. 90, Problem 4955)

2. Note that (1 + V3) 2 (n+1)+1 = (1 + V3) 2n+3 =

(1 + V3) 2n+1 (1 + va) 2 = = (an + bn V3) (4 + 2V3) =

4an +

6bn + (2an +

4bn)V3

.

On the other hand,

( ) 2 (n+1)+1 1 + va = an+1 +

bn+1

V3,

and since an , bn are integers we derive that

or

(i) an+1 =

4an +

6bni (ii)

bn+1 =

2an +

4bn.

Fr 1 t" (. ) b · b an+1 -

4an S b t·t · . l · ( . . ) . r om re a lOn I we 0 tam n = 6 . u S I utmg m re atlOn 11 Imp Ies

an+2 -4an+1 2

4 an+1 -4an 6 = an + 6 '

an+2 =

8an+1 -

4an .

On the other hand, an = bn+l;

4bn , and the first relation gives

bn+2 -

4bn+1

=

4 bn+1 -

4bn 6b

2 2 + n ·

189


Thus bn+2 =

8bn+l -

4bn ·

It follows that the sequences (an)n2:l ,

(bn)n2:l are given by

for all n 2:: 1 . (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1981) , pp. 71,

Problem 4648)

3. Solving the quadratic equation in Xn+l and taking into account condition 1) yields

Xn + 1 + J2(

x;

+ 1) Xn+l = l ' n = 0, 1 , 2, . . .

Xn -(1)

That is Xn+l = !(xn), n = 0, 1 , 2, . . . , where ! : (1 , 00) �

1R is the function given

by

It is not difficult to check that ! is decreasing and /(2 + V3) distinguish three cases:

Case 1 . If Xo = 2 + V3, then Xn = 2 + V3 for all n.

2 + V3. We

Case 2 . If Xo E (1 , 2 + V3), then from the monotonicity of function ! it follows that

Xo < X2 < X4 < . . . < 2 + V3 < . . . < X5 < X3 < Xl · Case 3. If Xo E (2 + V3, 00), then

Xl < X3 < X5 < . . . < 2 + V3

< . . . < X4 < X2 < Xo .

In all cases the sequence (xn)n>O is convergent and lim Xn = 2 + V3. - n�oo

( Titu Andreescu and Dorin Andrica, Romanian Mathematical Regional Contest " Grigore Moisil" , 2003)

4. Note that Xn+2 = 1 +

JXn+l (2 - Xn+l

) =

= 1 + Jl

-2xn + x

; = 1 +

IXn - 1 1 ,

hence Xn 2:: 1 for all n 2:: 2. We study three cases.

5.2. SOLUTIONS

i) If Xl < 1, then

Xn = X2 if n is even { Xl if n = 1

X3 if n is odd and n > 1

So the sequence converges if and only if X2 = X3 . The equation

X2 = 1 + .j 2X2 - x�

has only the solution X2 = 2 +2 V2. In all other cases the sequence is divergent. ii) If Xl = 1, then

and the sequence is divergent. iii) If Xl > 1, then

X n = { 21 if n is odd

if n is even

if n is odd if n is even.

191

It follows that the sequence is convergent if and only if Xl = X2 Le. Xl = 2 +

2 V2 . ( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1-2(1979) , pp. 56,

Problem 3865)

5. We prove a more general statement: If p is a positive integer and (Xn)n�l is a

sequence such that

then

. x�P + x�P + . . . + X2p 11m n = 0, n-+oo n

. Xl + X2 + . . . + Xn 11m = 0 . n-+oo n For this, recall the inequality

It follows that (

) 2p 2p 2p 2p Xl + X2 + . . . + Xn < Xl + X2 + . . . + Xn •

n - n

I Xl + X2 : . • . + Xn I S " x�P + x�P : . . . + X;':'

(1)

(2)

Using the squeeze theorem and the hypothesis (1 ) , the conclusion (2) follows. For p = 1 we obtain the initial problem.

The converse is not true. Take Xn = (-l )n and observe that

if n is even

if n is odd

192 5. MATHEMATICAL ANALYSIS Hence

but

1· Xl + X2 + . . . + Xn 0 1m = n�oo n

XI + X

� + . . . + X

2 lim n = 1 . n�oo n

(Dorin Andrica, Revista Matematidl Timi§oara (RMT) , No. 2 (1977), pp. 47,

Problem 2570)

6. Assume by way of contradiction that there is M

> 0 such that an+l - an < M

for all n. Summing up these inequalities from 1 to n yields

or (1)

From an 2:: nbn it follows that

an+l n + 1 b --;- 2:: -n- n+ 1

Since the sequence (bn)n>l is not bounded from above we obtain that the sequence

( an + 1 ) is not bou:ded from above, which contradicts (1) . n n>l (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1(1978) , pp. 91,

Problem 3441)

7 N h (a + 1 )Xl + a2 Th 1 ft . 1" b ' d . ote t at 0 < X2 = (

) < a. e e mequa 1ty IS 0 V10US an Xl + a + 1

the right inequality is equivalent to Xl < a. Since 0 < X2 < a we obtain - likewise -o < X3 < a and then Xn E (0 , a) by inducting on n.

On the other hand,

_ (a + l)Xn-l + a2 _ a2 - X�_l 0 Xn - Xn-l - - Xn-l - > , Xn-l + (a + 1) Xn-l + (a + 1)

therefore the sequence is increasing and bounded. It follows that the sequence is convergent and let

1 = lim Xn . Then n�oo

so 1 = a.

1 = (a + l)l + a2 1 + (a + 1)

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1976) , pp. 53, Problem 2567)

8. We prove that an = n In n, n � 1 , satisfies the conditions.

so

5.2. SOLUTIONS

First,

an+! - an = (n + 1) In(n + 1) - n ln n = In (n + 1) + In (1 + �r Second,

v'an+l -

va;; = J(n + 1) In(n + 1) - yIn In n =

(n + 1) In(n + 1) - n ln n -r��7===��== = J (n + 1) In (n + 1) + v' n In n

In 1 + -( l ) n In(n + 1) n - + -- J(n + 1) In(n + 1) + v'n lnn J(n + 1) In(n + 1) + v'n ln n -

In(n + 1) 1 In (1 + �r n + 1

1 + , / n ln n +

J(n + 1) ln(n + 1) + v'n ln n '

Because V (n + 1) In(n + 1)

lim In (1 + .!.) n = 1, n-too n we have

. In (l + �r

11m JnTil'n = O. n�oo v(n + 1) ln(n + 1) + n ln n . In(n + 1) . n ln n . Because 11m

1 = 0 and 11m

( 1) I ( 1) = 1, It follows that n�oo n + n�oo n + n n +

lim (v'an+l -

va;;) = 0, n�oo

as desired. Remark. Another such sequence is given by an = n..jii, n � 1 .

193

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 70, Problem 3087)

9. Assume by way of contradiction that there is no sequence (nkh�l with the desired property. Then there is a > 0 such that

hence

It follows that

xn+ 1 - xn > a > 0 n -

Xn+l - xn � an, for all n � 1 .

(n - l)n Xn - Xl � a(l + 2 + . . . + (n - 1)) = a 2 '

194 5 . MATHEMATICAL ANALYSIS

then Xn n - 1 Xl a - > 0'-- + - > n2 - 2n n2 3

for all n � 1. This is in contradiction with

and we are done.

1· Xn 0 1m 2 = , n-too n

(Dorin Andrica, Romanian Mathematical Olympiad - final round, 1984)

10. a) Summing the inequalities x� + O'Yn ::; Zn and y� + fJxn ::; Zn we obtain

0 ::; (xn + �) 2 + (Yn + % r ::; 2 (zn + 0<2 � ,12 ) , for all n ;::: 1 .

The conclusion follows immediately. b) Notice that

I fJ I ( 0'2 + fJ2 ) � Xn + "2 ::; ...j2 Zn + 8 and I a I ( 0'2 + fJ2 ) �

Yn + "2 ::; ...j2 Zn + 8

From the squeeze theorem follows that

lim Xn = - � n-too 2 and 1. a 1m Yn = - -. n-too 2

(Dorin Andrica, Romanian Mathematical Regional Contest " Grigore Moisil" , 1997)

1 1 . By induction on n we obtain Yn = XlX2 • • • Xn-l and 1 1 1

for all n � 1 .

Zn = Xl + - + -- + . . . + -----Xl Xl X2 XlX2 " , Xn-l

Since Xn+l - Xn = x� - Xn + 1 > 0, the sequence (Xn)n�l is increasing. From

X2 = 5 we obtain

hence 1 < 1 --

k-l ' k E N* .

XlX2 . . . Xk - 2 · 5 -It follows that

as desired.

1 1 1 1 1 - Gr-I Zn < 2 + - + - + . . . + 2 . 5n-2 = 2 + -2 . 1 2 2 · 5 1 - -5

5 [ ( 1) n-l ] = 2 + 8 1 - 5 5 21 < 2 + - = - < V7 8 8 '

5 .2 . SOLUTIONS

On the other hand, Xn+1

Zn+l Yn+l Xn+lYn Xn+1Yn -- = -X- = = Zn � XnYn+l XnXnYn

Yn

Xn+l = X� + 1

= 1 + � > 1 x�

x� x� '

195

so (zn)n2:1 is an increasing and bounded sequence. Then (Zn)n2:1 is convergent and

as claimed.

lim Zn :::; 21

< ..[7,

n-too 8

(Dorin Andrica and $erban Buzeteanu, Romanian Mathematical Regional Contest " Grigore MoisH" , 1992)

12. Let c > 0 such that a - c > o. From

lim nC¥

(xn+l - xn) = a n-too

it follows that there is an integer nl such that 1 1 �

(a - c

) < Xn+l - Xn <

(a + c

)� (1)

n n for all n � nl .

Summing up inequality (1) from n = nl to n = nl + p - 1, p > 0 implies p-l 1 p-l

1 (a - c

) L(nl +

k)C¥ < xnl+p - xnl «

a + c) L

(nl +

k)a (2) k=O k=O

00 1 The series L

(nl + k)c¥ converges if and only if a > 1, therefore applying the k=O

squeeze theorem to the inequality (2) leads to the conclusion. (Dorin Andrica, Gazeta Matematidi (GM-B ) , No. 1 1 (1979) , pp. 422, Problem

18011)

13. Note that n k(

n _ k) !

+ (k + 1) n k n 1 L

(k + l) ! (

n -k) ! =

L (k + I) ! +

L k! (n -

k) ! = k=O k=O k=O

n k I n n! n ( 1 1 ) 1 n (n)

= L

(k + I) ! +

n! L

k! (n -

k) ! = L

k! - (k + I) ! +

n! L

k = k=O k=O k=O k=O

Since

1 2n

= 1 - + -. (n +

I) ! n!

1. 1 1m n-too

(n +

I) ! 2n lim -, = 0, n-too n.


it follows that

lim � k (n - k) ! + (k + 1)

= 1 . n -+ oo 6

(k + 1) ! (

n -k) !

k=O (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2 (1975) , pp. 52,

Problem 2281 ) 14. Note that

lim XX = 1 . ", -+ 0 ", > 0

Indeed, !�O ¥ l� (-x)

lim XX = lim ex 1n x = e "' >O '" = e '">O = 1, ", -+ 0 ", -+0 ," > 0 ", > 0

after applying L'Hospital rule. Then xx+1

lim -- = 1 . ",-+0 X ", > 0

Let c > O. There is an integer n

(c) > 0 such that for any integer n � n

(c), we

have

or

(:2 ) �+1 1 - c < k < 1 + c, k = 1, 2, . . . , n.

n2

Summing up from k = 1 to k = n and using algebraic manipulations yields

n (�) �+1 L

n2

1 - c < k= 1 n k < 1 + c, n � n

(c),

L

n2 k=1

1 1 ( 1 c ) n ( k ) �+1 1 1 ( 1 c ) - - - c - - + - < L - < - + - c + - + - , 2 2 n n n

2 2 2 n n k=l for any integer n � n

(c).

Therefore n ( k ) �+1 1 lim - = - . n-+oo � n2 2

(Dorin Andrica)

15. (i) We have

1 2 n 1 + x + x + . . . + x + . . . = -- , I

- x

Ixl < 1.

Hence

so

5.2. SOLUTIONS

11/q 11

/q dx (1

+ x + x2 + . . . )dx =

-1 -,

o 0 - x

00 1

q � nqn = In q _

1 ' q > l.

n=l (ii) For any

Ixl < 1 we have

Hence

so

1 4 8 4n 1

+ x + x + . . . + x + . . . = 1 _ x4 . 11

/q 11/q

dx (1

+ x4

+ x8 + . . . )dx = --4 '

o 0 1- x

00 1 1 1 ( 1 ) 1 1

� (4n + 1)q4n+l = - q + 4 ln

1-q2

+ "2arctgq .

197

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977)

, pp. 70,

Problem 3091


, pp. 40, Problem

18608)

16. The number () has the decimal representation () = 0.0 . . .

01 0 . . .

0 1 0 . . .

0 1 . . .

1 0 . . .

0 1 '-.;--' '-.;--' '-.;--' kl k2 kn

where kl ' k2 , . . . , kn are the number of zeros between two consecutive ones. Because

we have kl <

k2 <

k3 < . . . < kn < . . . ,

hence () does not have a periodical decimal representation. It follows that () is irrational, as claimed. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No.

2(1981), pp.

73,

Problem 4661)

17. Note that

and that 00

1 00

1 An = -

1 + � (k! )n < -

1 + � k! = -1 + e <

2.

k=O k=O Hence

1 < An <

2 for all n �

1. So

An is not an integer.


Assume by way of contradiction that there are positive integers n,p, q with q "# 1 such that P. = An. Then

q

pqn-1(q -

1) 1 = A + (

q: l)n + (q + 1)n1(q +

2)n + . . . <

1 1 <A+ + + . . . = (

q + l)n

(q +

l)n(q +

l)n

=A+

1+ + + . . . =

A+ ,

1 ( 1 1 ) 1 (q +

l)n (q +

l)n (q +

1)2n (q +

l)n -1

for some integer A � O. It follows that

1 1 B = (

q + l)n +

(q +

l)n(q + 2)n + . . . <

1,

so A + B

is not integer, which is false. Therefore

An is irrational for all n �

1.

(Dorin Andrica)

18. 1)

Using the limit lim z:ra = 1 n-+oo

and taking the limits in both sides of the equality, we obtain

so k = 8 . 2) Using the limit

and the relation

1+1+ · · · +

1=1+1+ · · · +

1

� �' k times s times

lim n ( z:ra - 1) = In a n-+oo

n ( \Ial - 1)

+ n ( y'(i2 - 1)

+ . . . + n ( ifiik - 1)

=

= n (\Ilh -1) + n (\Ib;-

l) + . . . + n ( v'b,; - l) we obt�in after taking limits:

This implies

as desired. (Dorin Andrica, Romanian Mathematical Regional Contest "Grigore Moisil" ,

1999)

5.2. SOLUTIONS

19. Let Pn = IT (1 - �) . Then

k=1 Xk+l

hence 1 1 Pn+1

Pn

It follows that

Because

Xn+2 Xn+l _ Xn+2 - (n + I)Xn+l _ xn+l

(n + 1) 1 - --;;,y- - (n + 1) 1 - (n + 1) 1 '

1 1 x2 x3 xn+l -- = - + - + - + . . . + = Pn+1

PI 2! 3 ! (n +

I) !

x x2 xn+l = 1 + I!

+ 2f + . . . + (n + 1) 1 '

( X x2 xn+l ) lim 1 + -I ' + -2'

+ . . . + ( 1) 1 = eX , n-+oo . . n + . it follows that lim

Pn+1 = e-x , as desired. n-+oo

199

( Titu Andreescu and

Dorin Andrica, Revista Matematica Timi§oara (RMT) , No.

1 (1977) , pp. 49, Problem 2843)

so

20. Interchanging x and y we obtain

Let

Then

f(ln x + >. ln y) = 9 (y'X) + g (..;y) , x, y E (0, 00)

f (In y + >' In x

) = 9 (y'X) + 9 ( ..;y) , x, y > O.

a = ln x + >' ln y and b

= ln y + >' ln x.

:>.b- a :>' a - b X = e :>.

2-1 and y = e :>.

2-1 ,

hence f(a) = f(b)

= 9 (e 2(:t_

a1 ) ) + 9 (e 2(:'2-:,b

1) ) , a, b E R It follows that

f is a constant and let

f (x) = C. Then for X = Y we have g(V'X) = � , so 9 is a constant and g

(x) = � . (D

orin Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2(1979) , pp. 51,

Problem 3827)

2 1 . Consider the function F

: I

-t lR, F

(x) = (x - a) (x -b)f (x) + ml (x -

b) + m2 (x - a) .


Note that F is continuous and

hence there is c E (a, b) such that F(c) = O. It follows that

and the solution is complete. (Dorin Andrica)

f ( ) - ml m2 c - -- + -a - c b - c

22. Note that g(x) = g(y) implies that g (g(x)) = g(g(y)) and hence x = y from

the given equation. That is, 9 is injective. Since 9 is also continuous, 9 is either strictly increasing or strictly decreasing. Moreover, 9 cannot tend to a finite limit L as x ----+ 00, or else we'd have g (g(x) ) - ag

(x) = bx, with the left side bounded and the right side

unbounded. Similarly, 9 cannot tend to a finite limit as x ----+ -00. Together with monotonicity, this yields that 9 is also surjective.

Pick Xo arbitrary, and define Xn for all n E Z

recursively by Xn+l = g (xn ) for n > 0, and Xn-l = g-l (xn) for n < O. Let rl = (1+Ja2 + 4b)j2 and r2 = (a-Ja2 + 4b)j2 be the roots of x2 - ax - b = 0, so that rl > 0 > r2 and 1 > Irl l > Ir2 1 . Then there exist Cl , C2 E lR such that Xn = cl r� + c2r� for all n E

Z.

Suppose 9 is strictly increasing. If C2 f:. 0 for some choice of xo , then Xn is dominated by r� for n sufficiently negative. But taking Xn and Xn+2 for n sufficiently negative of the right parity, we get 0 < Xn < Xn+2 but g (xn) > g(Xn+2 ) , contradiction. Thus C2 = O. Since Xo = Cl and Xl = Cl rl , we have g

(x) = rlX for all x. Analogously,

if 9 is strictly decreasing, then C2 = 0 or else Xn is dominated by r�

for n sufficiently positive. But taking Xn and Xn+2 for n sufficiently positive of the right parity, we get o < Xn+2 < Xn but g

(Xn+2

) < g

(xn), contradiction. Thus in that case, g(x) = r2X for

all x. ( Titu Andreescu, The "William Lowell Putnam" Mathematical Competition,

2001, Problem B-5)

23. Setting x = y = 0 yields f(O) = O. For x = y we obtain f2 (2x) = 4f2 (X) and

then f(2;C) = 2f(x) , since f(x) � O. We prove that

f(nx) = nf(x) , n � 1 .

Assume that f(kx) = kf(x) for all k = 1 , 2, . . . , n. We have

f2 ((n + l)x) -

f2 ((n - l)x) = 4f(nx)f(x) ,

then

5.2. SOLUTIONS 201

hence f( (n + l

)x) = (n +

l)f(x),

as desired. It follows that if p, q are positive integers then

qf (�) =

f(P)

= pf(l )

,

so f (�) = �f(l)

and f(r) = r f(l) for any positive rational r.

Setting x = 0 in the initial condition gives

then f(y) = f( -y

),

for all real y, hence f(r) = Ir lf (l) ,

for all rational numbers r. We prove that

f(x)

= Ixl f (l)

for all real numbers x. Let x be an arbitrary real number and let

(rn)n>l be a sequence of rational numbers with lim rn = x. Because - n--+oo

and f

is a continuous function, it follows that

hence

lim f(rn) = lim

Irnlf (l)

= f ( lim rn) , n--+oo n--+oo n--+oo

f(x)

= f(l) l

xl·

Note that a = f(l) � 0, therefore the desired functions are

f(x)

= alxl for some

a � O. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2 (1977) , pp. 90,

Problem 3203)

24. (i) Let a and b

be fixed points for f

and g, respectively. We have

a -f(a)

� 0, b

= g(b)

� 0

so a � b.

Consider the function <p : lR -t lR, <p(x)

= f(x)

+ g(x) - x.

The function <p is continuous because functions f

and 9 are continuous. Moreover,

<p(a) = f(a) + g

(a) - a = g

(a) � 0


and cp(b)

= I(b)

+ g(b)

-b = I(b)

::; O. By the Intermediate Value Theorem, there is an Xo E

(a, b)

such that cp(xo) = 0,

hence (I + g

) (xo) = Xo ,

as desired. (ii) Let a and

(3 be fixed points of the functions cp and

'l/J, respectively. We have

o ::; a = cp(a)

::; 1 , (3 = 'l/J((3)

� 1

so a ::; (3.

Consider the function w : lR -+ lR, w (x) = cp(x)'l/J(x) - x. The function w is continuous because functions cp and

'l/J are continuous. Moreover,

w(a) = cp

(a)'l/J(

a) - a = a

('l/J (a)

- 1) � 0,

w ((3) = cp((3)'l/J ((3) - (3

= (3( cp((3)

-1) ::; O.

Likewise, there is an ,0 E [a, (3]

such that w (,o ) = 0, hence (cp'l/J) (

,o) = ,0 , as

desired. ( Titu Andreescu, "Asupra unor functii cu punct fix" , Revista Matematica

Timi§oara (RMT), No. 1(1977)

, pp . 5-10)

25. We have �� (cp(x) + cp (�) + . . . + cp (�)) =

= lim (<p(X) - <p(0)

+ 1 <p (�) - <p

(0)

+ . . . + 1 <p (�) - <P(O))

= z-+o x -0

2 � _ 0

n � - 0 2 n

= <p' (0) (1 + � + . . . + �) , since cp

(O) = 0

and cp is differentiable at the origin. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 71,

Problem 3095

; Gazeta Matematica (GM-B) , No. 3(1979) , pp. 111, Problem 17671)

so

In ax 26. Consider the function 1 :

(0, 00

) -+ lR,

I(x) = -- .

x We have !' (

x)

=

1 - In ax x2

!, (x)

= 0

if and only if x = � . a

It follows that I-" = � is the maximum point of the function f so I-" is the only a

point such that f(x) ::; 1(1-") for all positive real numbers x.

Hence

5.2. SOLUTIONS 203

for all x > 0, as desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp. 54,

Problem 3547)

27. Consider the function F

: [a, b]

-t lR,

F(x) = I (x)e->'f(:r;) , ). E R Function

F is differentiable, since

I and

I' are differentiable, and

F(a)

= F(b)

. By Rolle's theorem it follows that there is c E (a,

b) such that

F' (c)

= O. On the other hand,

F' (x) = e->.f(:r;)

(I" (x) - )..(1' (X))2 ) ,

hence I

" (c) -)..(I' (

C))2 = 0, as desired.

(Dorin Andrica, Revista Matematica Timi§oara (RMT), No. 2(1981) , pp . 76, Problem 4677)

28. Let g : [0, 2] -t [0, �) , g(x) = arccos l(x) . Then

I(x) = cos g (x)

for all x E [0, 2] and the condition is equivalent to

cos g(2x) = cos 2g(x) .

It follows that g(2x) = 2g(x) + k(x)7r

for all x E [0, 1] , where k : [0, 1] --t Z.

7r On the other hand, 0 � g(x) < "2 for all x E [0, 2] , hence k(x) = 0 and

g(2x) = 2g(x), x E [0, 1]

By induction on n we obtain

Because 1 (0) = 1, g(O) = arccos 1 = 0, so

g (x) = 9 (�) - g(O)

x x 2n

for all x E (0, 2] . Since

I is differentiable at the origin 9 is differentiable at the origin and

for all x E (0, 2] .

g(x) . g (;) - g(O) , - = 11m x = g (0) X n-+oo _ 2

n


It follows that g(x) = fJ;x, where fJ; = g' (O) E [O, �) , because 0 � g(x) < i for all x E [0, 2] .

Therefore the desired functions are IJ1.(x)

= cOS fJ;X, I-" E [O , �) . ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 (1977) , pp. 45,

Problem 2852)

ln x 29. Consider the function

I : (0, 00) -t lR, I(x) = � . x

Then II ( ) = 1 - A In x x

X>.+l'

and I' (X)

= 0 for x = et.

It follows that () = ell>'

is the only maximum point of the function, so I(x) :::; I(et ), X > O.

Hence

only for () = et.

(Dorin Andrica, Gazeta Matematica (GM-B) , No. 3(1976) , pp. 104, Problem 15768; Revista Matematica Timi§oara (RMT), No. 1-2(1979), pp. 57, Problem 3870)

so

30. a) Let A > fJ; and let y = XfJ;. Then

. f lAx) -f(!-'x)

. f GY) - fl

y)

hm = hm y = A, z-tO X y-tO _

I-"

lim I(ay) -I(y)

= � = a, y-tO y fJ;

A where a = - > 1 . I-"

Let c > O. Then there is a 8 > 0 such that for any Iyl

< 8, we have

a - c < .

I (ay)

- I (y)

< a + c. (1) Y

Substituting y with Yk ' k = 1, 2, . . . , n, in the relation (1) we obtain a

1 I(y)

- I ( J!..) 1 -(a - c

)< a < - (a + c) ,

a y a

1 I (J!..) - I (JL) 1

_ (a - c ) < a a2

< (a + ) 2 '-' -2 c ,

a y a

5.2. SOLUTIONS

1 1 ( !-1 ) - 1 ( Yn ) 1

-(a - c) < a a < -

(a + c) .

an Y an


1 1 - � 1 (y)

- 1 ( Yn )

- a (a - c) < a a 1 -

� Y

a

1 I I - -an )

< a 1 (a + c . 1

- -a

Because 1 is continuous at the origin

and so

lim 1 (JL) = 1(0) , n-+oo an

_1_ (

a _ c) � I(

Y) - 1 (0) � _1_(a + c) . a - I Y a - I

It follows that I'

(0) = lim I(Y)

- 1 (0) = _a_ = �, y-+O Y a -1 A

- 1-£ so the function 1 is differentiable at the origin.

Conversely, if 1 is differentiable at the origin, then

Hence

as desired.

lim I(A

X) - 1(0) = 1' (0) and lim

I(I-£x) - 1(0) = 1' (0) . z-+o

AX z-+o I-£X

lim I(A

X) -I(I-£

x)

= (A

- 1-£)1' (0) , z-+o X

205

(Dorin Andrica, Revista Matematid1 Timi§oara (RMT) , No. 2(1978), pp . 76, Problem 3708)

31 . Since eZ > 1 + x for all real x 1=

0, we have Xn+1 - Xn = eZn - 1 - Xn > 0 so the sequence

(Xn)n�l is increasing. By induction on n we obtain Xn < 0 for all n � 1,

so the sequence (Xn)n�l is bounded. Therefore the sequence converges and let l be

its limit. The equation el = l + 1

has the unique solution l

= 0, hence lim Xn = O. n-+oo

Using lim (.! - �1) = -2

1 , it follows that z-+o x eZ -

1 1 1 1

1· Xn+1 Xn l' eZn - 1 1m = 1m n-+oo (n +

1) - n n-+oo

1

Cesaro-Stolz's Theorem implies

hence lim nXn = -2, as desired. n-+oo

1

lim n-+oo nXn


(Dorin Andrica, Revista Matematid1 Timi§oara (RMT) , No. 2(1982) , pp. 68, Problem 5004)

32. We first prove by induction that Xn > O. This is true for n = 0 and assuming Xk > 0 for some positive integer k yields 0 < sin Xk < 1, hence sin3 Xk < sin Xk, which implies Xk+l > Xk - arcsin(sin xk ) = O.

and

It is not difficult to see that (Xn)n>O is convergent and lim Xn = O. We have - n�oo n (vnsin xn)2 = --=-1-

sin2 Xn

n + 1 - n , 2 · 2 11'm l' sm Xn+l sm Xn -----::1�----:<1;--- = 1m . 2 • 2 n�oo n�oo SIn Xn - SIn Xn+l

sin2 Xn+l -

sin2 Xn . 2 . 2 . 2 . 2 1. sm Xn+l sm Xn l' sm Xn+l sm Xn = 1m = 1m n�oo sin(xn - Xn+1 ) sin (xn + xn+d n�oo X� sin(xn + xn+d

1· sm Xn+1 l' sm Xn l' Xn + Xn+l ( . ) 2 ( . ) 2 = 1m 1m -- 1m . n�oo Xn+1 n-+oo Xn n-+oo sin(xn + xn+d

1. Xn l' 1 1 1m -- 1m x = n-+oo Xn+1 n�oo 1 + n+l 2 Xn

From Cesaro-Stolz Theorem we obtain

lim n-+oo n 1

hence

lim (vnxn)2 = lim n lim (�) 2 = ! . n-+oo n-+oo 1 n-+oo sin Xn 2

Thus lim foxn = J2 . n-+oo 2

sin2 Xn

Alt�rnative Solution. We have seen above that (xn)n>O is convergent, lim Xn = 0 - n�oo and Xn > 0 for all n = 0, 1 , 2, . . . We will calculate

lim n-+oo n + 1 - n 1 1 = nl�� 1

1

-2- - X2 xn+1 n ------�--

(xn - arcsin(sin3 xn) )2

It is clear that the last limit is obtained from

1 .

x�

lim [ 1 _

1 1 = lim (arcsin t3 ) (2 arcsin t- arcsin t3 ) = t-+o (arcsin t - arcsin t3 ) 2 arcsin 2 t t-+O ( arcsin t - arcsin t3) 2 arcsin 2 t

(1)

5.2. SOLUTIONS (2 arcsin t 2 arcsin t3 )

_ l' t(2 arcsin t - arcsin t3 ) _ l ' t - t t3 - 1m ( . . 3 )2 - 1m 2 = 2. t-+o arCSIn t - arCSIn t t-+o (arcsin t

_ t2 arcsin t3 )

t t3

207

It follows that the limit (1) is � and the conclusion is obtained via Cesaro-Stoltz Theorem.

( Titu Andreescu, Gazeta Matematid1 (GM-B) , No. 10(2002) , pp. 409, Problem C :2557

)

33. Let x be a zero of 1'. Then J(x + f' (x) ) = J(x) and the conclusion follows. Let x be a real number such that J' (x) < O. Applying the Mean Value Theorem

on the interval [x + J' (x) , x] we obtain

J(x) - J(x + J' (x)) = -!, (x)J' (c) ,

for some c E (x + J' (x) , x) . Because the second derivative is nonnegative, J' is nondecreasing, hence

f'(c)

< !, (x) < 0,

and

J(x) - J(x + !' (x)) < 0,

as desired. Let x be a real number such that f' (x) > O. Likewise,

J(x + !' (x) ) - J(x) = !' (x)!' (c) ,

for some c E (x, x + f' (x

) ) and

f' (c) > J' (x) >

O. Hence J(x + J' (x)) � J(x) for all

real numbers x, as claimed. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2(1989), pp. 67,

Problem 6143)

34. Applying the Mean Value Theorem for the function J(t) = In t on the interval [a, a � b 1 yields

Hence

a + b In -- - In a 2 b - a

2

1 x '

( a + b) x E a' -2- .

In (a + b) :Z; = b - a

2a 2 . (1)


Using the same argnment for the interval [ a ; b , b

] gives

hence

a + b ln b - In --2 b - a

2

1 y '

(a + b ) y E -2-, b ,

I (�) Y =

b- a n a + b 2 .

From the equalities (1) and (2) we obtain

Then

as desired.

(a + b) Z = (�)Y

2a a + b

(2)

(Dorin Andrica, Revista Matematica Timi§oara (RMT

) , No. 2(1978), pp. 54,

Problem 3548)

35. We prove that no such function exists. Assume the contrary and let k be an integer . From the Mean Value Theorem we obtain

<p(k + 1) - <p(k) = <p' (e) , e E (k, k + 1) .

Since <p(k) and <p(k + 1) are integers, <p(k + 1) - <p(k) is also an integer and so is <pI (e) .

On the other hand, e is not an integer, hence <p' (e) is not integer, a contradiction. (Dorin Andrica, Revista Matematica Timi§oara

(RMT

), No. 2(1978), pp. 67,

Problem 3618)

k 36. Define Xk = a + -

(b - a

), k = 0, 1, . . . , n and note that

n b- a

Xk+l - Xk = -:;:;:-' k = 0, 1 , . . . , n.

The Mean Value Theorem yields b - a f(

Xl) -f(xo)

= --

f' ((h),

(h E

(xo, x

d

n

5.2. SOLUTIONS

Summing up these equalities implies

or

as desired.

f(b) -f(a)

= � t t((}i),

b - a n i=l

209

(Dorin Andrica, Revista Matematidl Timi§oara (RMT), No. 1-2(1979) , pp. 58, Problem 3878)

37. Let F : lR

--t lR, F(x) = f(x) - g

(x)

and note that F

is differentiable. Since F(xI) = F(X2) = 0 from Rolle 's Theorem there is c E

(Xl , X2

) such that

F' (c) = O.

On the other hand, F' (

x) = f' (

x) - g

' (x) = f

(x)

+ g(x), and therefore

f(c)

+ g(c) = 0,

as desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 74,

Problem 3113)

38.- Assume that (f (

X))2

+ (f' (

X))2

> 1 for all X in [-i , i] . Then f' (

x)

> 1 )1 -

(f(X))2

for all X E [- i , i] . Integrating from -� to i yields arcsin f (i) - arcsin f (- i) > 7r . On the other hand, it is clear that arcsin f (�) - arcsin f (-i) ::; i + i = 7r .

contradicting the previous inequality.

( Titu Andreescu, Mathematical Horizons, 2000)

39. If x = y, then x = y = 0, which is impossible, because x and y are positive. Assume that there are x

f:. y > 0 such that

x . 2Y + y . 2-3: = X + y,

and let y = Xl - X2 , X = X2 - X3 for some Xl > X2 > X3 > O.

or

Then 23:1 -3:2 - 1 Xl - X2 1 - 23:3-3:2 X2 - X3

,

23:1 _ 23:2 23:2 _ 23:3 Xl - X2 X2 - X3

By the Mean Value Theorem there are (}l E

(X2 ' xI

) and

(}2 E (X3 7 X2 ) such that

23:1 _ 23:2 () --- = 2 1 In 2,

(}l E

(X2 , XI

) and

Xl - X2 23:2 _ 23:3 --- = 2()2 In 2, (}2 E (X3 , X2 ) X2 - X3


Hence 291 In 2 = 292 In 2, that implies (}1 =

(}2 , a contradiction.

(Dorin Andrica, Revista Matematid1 Timi§oara (RMT) , No. 1-2 (1980) , pp. 70, Problem 4152)

40. a) Consider the differentiable function f

: (0, 00) -+ lR, f (t) = \Ii - n+\It.

We have

!, (t) = .!.t n�l -1 (t n(n\l) _ _ n_) , n n + 1

( ) n(n+l) so, if t � n : 1 ' then f'

(t) > O. Hence f

is increasing on

[(_n_) n(n+l) , 00) and for x > y > (�1 ) n(n+l) , we have f(x) >

f(y). There-n + 1 - - n + -

fore

as desired. b) We prove that

for all n � 3. The inequality is equivalent to

(1 + �r � n, which is clearly true, because

for all n � 1. Setting x = nn+l and y = (n + l)n , x � y in the inequality from a) yields

as desired.

n + 1 n\fii + n+\Yn+l � 2n + 1 , n � 3 (Dorin Andrica, Revista Matematid1 Timi§oara (RMT) , No. 2(1981 ) , pp. 74,

Problem 4668)

41 . Recall Jensen's inequality for a concave function f:

5.2. SOLUTIONS 21 1

Consider f(x) = ln x, Yi = �, P = YIY2 . · . Yn and

Ai = !!.. for i = 1 , 2, . . . , n. We

Xi Yi have

so

Hence

. 1 and smce Yi = - , i = 1, 2, . . . n, it follows that

Xi

thus

as desired.

0 = 1

1 n ( l tz;

--=----=---::-- < n = n,

X�l X�2 . . . X�n :2: Xi

i=l

(Dorin Andrica, Revista Matematid1 Timi§oara (RMT) , No. 2(1977) , pp. 74, Problem 3111)

42. Let F

be an antiderivative of the function f. Since

F is noninjective, there

are real numbers Xl < X2 such that F(xI)

= F(X2). Applying Rolle's Theorem on the

interval [XI , X2

] we obtain

f(c)

= F' (

c)

= 0 for some c E (XI , X2

), as desired.

( Titu Andreescu)

43. Because

max(h (x), h(

x) )

= h (x) + h(

x) + 1/1 (x)

- h (x) 1 2


and h , h are continuous, ma:x(h , h) is also continuous. Assume that if h , 12, · · . ,

fk-l are continuous, then ma:x(h , 12, . . . ,

fk-I

) is continuous. It follows

that

ma:x(h , 12 , · · .

fk) = ma:x(ma:x(h , · . . ,

fk-I

), fk)

and according to the 'first step, the function ma:x(h , . . . , fk)

is continuous. Hence ma:x(h , . . . , fn) is continuous and furthermore a derivative function.

and

and

Note that if n is even, then

{ Xn , X E (-oo, -I) ma:x(I, x, . . . , xn) = 1, x E [-1 , 1]

xn , X E (I , oo) , 1 xn+l - n 1

+C,

J max(l , x, . . . , xn)dx = x : �, xn+l + n

1 +C,

n + On the other hand, if n is odd, then

X E (-oo, -I)

X E [-I , I]

x E (1 , 00) .

{ xn-l , ma:x(I, x, . . . , xn) = 1,

xn ,

x E (-00, -1) X E [-I, I]

J max(l, x, . . . , xn)dx = 1 x E (1 , 00) ,

xn - n + 1 C

----

+ , n x +

C,

xn+l + n 1

+C,

n +

X E (-oo, - I)

x E [-1 , 1]

X E (I, oo).

(Dorin Andrica, Revista Matematidl Timi§oara (RMT), No. 2(1983) , pp. 62, Problem 5185)

44. Denote

We have

hence

so

5.2. SOLUTIONS

On the other hand, ! xearctg:c ! xl + x2earctg:c ! � 12 = dx = dx = xV 1 + x2d(earctgz ) =

VI + x2 1 + x2

= xVI + x2earctg:c - I - 2 dx ! x2earctg:c

1 VI + X2 '

! x2earctg:c dx = xJ1+X2earctg:c - (II + 12 ) =

VI + x2 2

(x - 1)V1 + x2earctg:c = 2 + C.

213

(Dorin Andrica, Romanian Mathematical Olympiad - final round, 1975; Revista Matematid1 Timi§oara (RMT) , No. 2(1978) , pp. 35, Problem 2125)

45. We start with the following lemma. Lemma. Let g, h :

lR --t

lR be functions such that:

1) h is a derivative; 2) 9 is differentiable with a continuous derivative.

Then 9 . h is a derivative function. Proof. Let H be an antiderivative of h and define u :

lR --t lR, u (x) = g(x)H(x) .

Then

ul (x) = g (x)h(x) + 9' (x)H

(x) , or g(x)h(x) = ul (x) - g' (x)H(x) .

The function ul is a derivative and gl . H

is continuous, hence the function 9 . h is a deri vati ve, as claimed. 0

Applying the lemma for h = f o p and g (x) = xk for a nonnegative integer k, it follows that xk (f 0 p) (x) is a derivative, hence p' (f 0 p

) is a deriva

tive. Since the degree of p is odd, p(lR) = lR. Assuming that lim p(x) = -00, z--+-oo

there are real numbers Xl , xi , x� , . . . , xm, x� such that p is increasing on each of

the interval (-00, xd, [xi , x2] , . . . , [X�_1 , Xm] ' [x� , 00) and p(xd

= p(xD = M1 , P(X2)

= p(x� )

= M2, . . . , p(xm) = p(x� ) = M

m . Let F1 , H1 , . . . , Hm be an antiderivative of p(f 0 p

) on the interval (-00, x

d, [x�, X2

], " . , [x� , 00) , respectively. It

follows that F1 0 p-I , HI 0 p-I , . . . , Hm 0 p-I are antiderivative of f on the intervals

(-00, MI], [M

I , M2], • • • , [Mm' 00) , respectively, hence f is a derivative function on all

R (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1985), pp. 76,

Problem 2)

1 46. Substituting x --t - in the given condition yields

x

F G) f (x) = � , (1)


for all x in I. We have

g(x) = F' (x)F G) + F(x)F' G) ( -:2 ) = f(x)F (D - :,1 G) F(x) � 1 1 = - - - · x = O X x2

for all x in I, so g is constant. Then there is a constant c > 0 such that F( x)F (�) = c

for all x in I, and from (1) we obtain

f(x) F(x)

1 ex ' x E I.

Integrating gives In F(x) = � ln x+ ln d, where d > O. It follows that F(x) = dx� , e for all x in I. The relation F(x)F (;�) = c becomes cP = c, so d = ,;c.

Finally, 1 1 1. -1 f(x) = -

XF---;(-�-;--) = -.;cXc , x E I,

where e is any positive real constant. ( Titu Andreeseu, Romanian Mathematical Olympiad - final round, 1987; Revista

Matematid1 Timi§oara (RMT) , No. 2(1987) , pp. 86, Problem 6307)

47. Consider the function 9 : [0, 1] -+ lR,

g(x) = f (x) - (1 + x + . . . + xn-1 ) and note that 9 is continuous. We have

r1 r1

r1 Jo g(x)dx = Jo g(x)dx - Jo (1 + x + . . . + xn-1 )dx =

= l' f (x)dx - (l + � + " ' + D = 0. From the Mean Value Theorem there is Xo E (0, 1) such that

hence

as desired.

g(xo ) = l' g(x)dx = 0, ) n-l 1 - x� f(xo = l + xo + . . . + xo = -- , 1 - Xo

( Titu Andreeseu, Revista Matematidl Timi§oara (RMT) , No. 1-2(1979) , pp. 33, Problem 3444)

5.2. SOLUTIONS 215

48. Consider the function F : [a, b]

--t lR,

F(x) = [ g(t)dt t f(t)dt.

Note that F

is differentiable and F(

a)

= F(b)

= O. Applying Rolle's Theorem,

we obtain c E (a, b)

such that F' (c) = 0 hence

ftc) [ g(t)dt = g

(c) t f(t)dt,


49. Consider the function F : [a, b]

--t lR,

F(t) = l f(x)dx - t f(x)dx

and observe that F

is continuous and F(a)F(b)

< O. Then there is a E (a,

b) such

that F(a) = 0 so

lQ f(x)dx = l f(x)dx From the Mean Value Theorem there is f3 E

(a, b)

such that

l f(x)dx = (b - a

)f({3),

therefore there are numbers a, f3 E (a, b), a < f3, with

{ f(x)dx = (b - a

)f({3),

as claimed. (Dorin Andrica, Revista Matematica Timi§oara (RMT), No. 1-2(1979) , pp. 61,

Problem 3897)

50. Since f

is an increasing bijective function, f

is continuous.

Denote

8, = I.' f(x)dx, 82 = f.d r' (y)dy,

and note from the diagram below that

(1)


y

d . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-----------------------7J .

.1! 1 ,/

C 1-.-.-.-_. "='-;....,. -T. �� • • • • • • �1 . . . . . . .

o a b

From the Mean Value Theorem there is e E (c, d) s.llch that

{ f-l (y)dy = (d - c)rl (e) .

x

Observe that e is unique and let fJ; = 1-1 (e) . The relation (1) gives

t f (t)dt = (" - a)e + (b

- ,,)d, " E (a, b),

as desired. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1978) , pp. 56,

Problem 3556)

51 . Consider the function F : � --+ �,

F(s) = [ cp(t)dt,

and note that F is differentiable. From the hypothesis we obtain F(x + y) - F(x) = F(x) - F(x - y

), so

F(x + y) + F(x - y) = 2F(x) , x, y E �. Differentiating with respect to y it follows that

F' (x + y) = F' (x - y) , x, y E lit Setting x = y = � , z E �, we obtain F' (z) = F' (O) , so c.p(z) = F' (O) for all z E lit Hence c.p is a constant function, as desired.

( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1 (1977) , pp. 47, Problem 2865)

52. Consider the function cp : JR -4 JR, cp(s) = /.8 f(t)dt. The condition is equiva

lent to x y .=.±.ll /. f(t)dt + /. f(t)dt :::: 2/. 2 f(t)dt,

hence

for all x, y E llt

5 .2 . SOLUTIONS

cp(x) + cp

(y)

> (x + y ) 2 - cp 2

217

(1)

Since I is differentiable, cp is twice differentiable and moreover is concave up, from

relation (1) . Hence cp" (x) � 0 or

f' (x) � 0 for all x, so I is a nondecreasing function.

( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1 (1976), pp. 56, Problem 2349; Gazeta Matematidi (GM-B) , No. 2(1980) , pp. 68, Problem 18154)

53. Consider the function h

: [0, 00

) --+

lR,

h(x)

= xf(x) - 1" f(t)dt.

Because I is differentiable, h

is differentiable and h(x)

= xf' (

x), x � O. (1)

Since I is injective and continuous, I is either increasing or decreasing, so f' (x)

� o for all x or I'

(x) � 0 for all real numbers x.

Case 1. If I' (x) � 0 for all x, then from (1) we deduce that h' (x) � 0, x � 0, hence

h is nondecreasing. It follows that

h(x)

� h(O)

= 0, x � O. Case 2 . If f

' (x) � 0 for all x, then

h' (x)

� 0, x � 0, and h is nonincreasing. It follows that

h(x)

� h(O)

= 0, x � O. Since F is differentiable and

xf(x) - (X f (t)dt

_

h(x) F' (

x)

= 10 x2 - --;2 '

we derive that F' (x) � 0 for all x > 0 or F' (x) � 0 for all x > 0 , hence F i s a monotonic function, as desired.

( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2(1978) , pp. 76, Problem 3709; Gazeta Matematidi (GM-B) , No. 1 (1980) , pp. 38, Problem 18115)

54. Recall from Problem 14 that

and let c > O. There is 0 > 0

such that for all x < 0, Ixx - 1 1 < c.

Then for n > � we obtain

I n° l'!. (x'+ ! - X)dX I � no l';; Ix·+1 - x

ldx =

l� l� c = n2 xlxx -l ld

x < cn2 xdx = - .

0 0 2


It follows that

hence

1

lim n2 {n (XX+1 - x)dx = 0, n-+oo 10 � 1

lim n2 (

xx+1dx = - , n-+oo 10 2

as desired.

Alternative Solution. Consider the function F(t) = l' x·+1dx. Then F(O) = 0

and we can write

lim n2 xx+ldx = lim n2F - = lim -- = lim -- = l� ( 1 ) F(u) F' (u) n-+oo 0 n-+oo n u-+O u2 u-+O 2u

. uu+l 1 . u 1 = hm -- = - hm u = - . u-+O 2u 2 u-+O 2

(Dorin Andrica, Gazeta Matematidi (GM-B) , No. 11 (1979) , pp. 424, Problem 18025; Revista Matematidi Timi§oara (RMT) , No. 1-2 (1980) , pp. 71, Problem 4160)

55. Assume the contrary and let x =f. y be real numbers. Then

hence

(Y

f(t)dt = f(x) and

Jx f (y) /.x f (t)dt =

f(y)

, Y f(x)

f(x) f(y) f (y) - f(x)

It follows that f2 (X) + f2 (y) = 0 , so f(x) = f(y) = 0, which is absurd since f(x) =f. 0

for all x. ( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2 (1978) , pp. 35,

Problem 3126)

56. It is clear that [ ( 1 /, (x) l - I)'dx ;::: O.

Hence [ (f' (x) )'dx - 2 [ 1 /, (x) ldx + [ dx ;::: 0,

and

1 ;::: [ 1/, (x) ldx ;::: I[ /' (X)dx l , as desired.

( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2 (1977) , pp. 77,

Problem 3130)

5.2. SOLUTIONS

57. The relation is equivalent to

Hence

or

r1 r1 X2 io (xf (x) - f2 (X) )dx = io 4dx.

l (f2 (X) - xf(x) + x:) dx = 0,

l (f(x) - �r dx = O.

219

Because f is continuous, f(x) - � = 0, for all x E [0, 1] , so f(x) = � , x E [0, 1] . ( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1 (1978) , pp. 72,

Problem 3319)

58. The relation l fm(t)dt = (m : 1) 1 ' is equivalent to l (fm (t) - :) dt = O.

Since fm is continuous by the Mean Value Theorem there is Xo E (0, 1) such that

or rco ( tm-1 ) io fm-1 (t) - (m _ I) ! dt = 0.

Using the same argument, we obtain Xl E (O, xo) such that m-1

fm-I (xd = (�1_ 1) 1 "

Continuing this procedure, we obtain Xm E (0, Xm-1 ) such that

fo (xm) = xm,

as desired.


59. From the Mean Value Theorem we deduce that for any X E [-1 , 1] there is Cx E (-1 , x) such that

f (x) - f ( -1) = (x + 1) f' ( cx ) . Since f' (cx ) :::; f' (I) , f(x) - f ( -1) :::; (x + 1)1' (1) , hence

r1 f(x)dx _ 2f( -1) :::; (x + 1)211

1'(1) i-I 2 -1



( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2(1981) , pp. 77, Problem 4686)

60. Consider the function h : [a, b]

--+ lR,

h(t) = (b - t) [ f (x)dx + (t - a) [ g(x)dx

and note that h

is differentiable and h'

lt) = -[ f(x)dx + (b - t)f(t) + [ g(x)dx - (t - a)g (t) .

Since h(a) = h(b) = 0, from Rolle's Theorem it follows that there is a real number

c E (a, b)

such that h' (c) = O. Then

- /." f(x)dx - (e - a)g (e) +

(b - e)f (e) + t g(x)dx = 0,

and the conclusion follows. ( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2 (1985) , pp. 56,

Problem 5628)

Chapter 6

COMPREHENSIVE P ROBLEMS

PROBLEMS

1 . Let A be a set with n elements and let B be a subset of A with m � 1 elements. Find the number of functions f : A --+ A such that f(B) � B.

2. Let A be a set with n elements and let X, Y

be subsets of A with p � 1

and q � 1 elements respectively. Find the number of functions f : A --+ A such that Y C f(X) .

3. Let A be a set with n elements and let X be a subset of A with k � 1 elements. Find the number of functions f : A --+ A such that f(X) = X.

4. Consider the sets A = {I, 2, . . . , n} , B = {I, 2 , . . . , m} and let k ::::; min{ n, m} .

Find the number of functions f : A --+ B having exactly k fixed points.

5. Let aI , a2 , . . . , an be positive real numbers and let m � 1. Prove that

al a2 an m ( ) m ( ) m ( )m 1 + a2 + 1 + a3 + . . . + 1 + al � n . 2 .

6. Let aI , a2 , . . . , an be positive real numbers and let k � O. Prove that

7. Let a, b, c be positive real numbers such that a

bc = 1 . Prove that

8. Let a, (3" be positive real numbers and let [a, b]

be an interval. Find the

numbers x, y, z E [a, b]

such that

E(x, y, z) = a(x - y)2 + {3(y - Z)2 + ,(z - X)2

is maximum.

223

224 6. COMPREHENSIVE PROBLEMS

9. Find the maximum number of nonzero terms of the sum n

L I f (i) - f (j) 1 i,j=l

where f : {I , 2, . . . , n} --+ {a, b, c} is one of the 3n possible functions.

10. Let aI , a2 , . . . , an, bl ::;

b2 ::; . . . ::; bn be positive numbers such that

al � bl , al + a2 ::; b1 + b2 , . . . , al + a2 + . . . + an � bl + b2 + . . . + bn ·

Prove that

0il + Viii. + . . . + ..;a;; ::; ..jb; + .jb; + . . . + V'bn. n

1 1 . Define S(n,p) = L(n + 1 - 2i)2p for all positive integers n and p. Prove that

i=l for all positive real numbers ai , i = 1, n the following inequality holds

12 . Prove that

4P n

rp.i� (ai - aj

)2p ::; -S( ) L a�P .

1 ��<J�n n,p i=l

for all integers n � k � 2.

13. a) Consider the real numbers aij , i = 1 , 2 , . . . , n - 2, j = 1, 2, . . . , n, n � 3,

and the determinants Ak , k = 1 , 2, . . . , n 1 1 1 au

an-2 ,k-1 an-2,k+1

Prove that Al +

A3 +

A5 + . . . =

A2 +

A4 +

A6 + . . .

b)' Let Xl , X2 , • • • , Xn be distinct real numbers and let

for k = 1 , 2, . . . , n. Prove that

n-(k+l )

Pk = II (Xn-i - Xk ) , i=O

k-l qk = II (Xk - Xi ) ,

i=l

1

c) Prove that

for all integers n � 3.

6.1. PROBLEMS

n (-l)k k2

'" - 0 � (n - k) ! (n + k) ! - , k=1

225

14. Let P(x) = xn + al xn-I + . . . + an be a polynomial with all zeros positive

real numbers. Prove that if there are m f:- p E

{I, 2, . . . , n} such that

am = (_l)m (�) and lip = (-1)" (;) , then

P(x) = (x -

l)n .

15 . Define the polynomials Po , PI ' " ' '

Pn by

Po(x) = 1 and

Pk+1

(x) = (n - k + x

)Pk(x)

+ xPk (

x)

for all k = 0, 1 , . . . , n - 1. Prove that deg Pk = k for all k and find the polynomial

Pn.

16. Let (Pn)n�o be a sequence of polynomials defined by

Pn+dx) = -

2xP� (

x) +

Pn(x)

and Po(x) = 1.

. Find Pn(O)

.

17. Consider the polynomial n 1 P(

x) = L

n + k + 1 xk .

k=O

Prove that the equation P(x2 ) = p

2 (x)

has no real roots.

18. Let P

be a polynomial with real coefficients. Find all functions f : 1R -+ 1R such that there is a real number

t such that

f(x + t) -

f(x)

= P(x)

for all x E R

and

19. Find the real numbers a, b, c, d, e E

[-2,2]

such that

a + b + c + d + e = O


20. Let p be an odd prime. The sequence (an)n?O is defined as follows: ao == 0, a1 = 1, . . . , ap-2 = p- 2 and, for all n � p- 1, an is the least integer greater than an-l that does not form an arithmetic progression of length p with any of the preceding terms. Prove that, for all n, an is the number obtained by writing n in base p - 1 and reading it in base p.

21 . i) Let a, c be nonnegative real numbers and let I : [a, b) -+ [c, d] be a bijective increasing function.

Prove that

L [! (k)] + L [1-1 (k)] - n(GJ) = [bJ [d] - a(a)a(c)

where k is integer, n (G J)

is the number of points with nonnegative integer coordinates on the graph of I and a : 1R -+

Z is defined by

ii) Evaluate

if x E 1R \ Z

{ [x] a (x) = 0 if x = 0

x - I if x E Z

\ {O}

22. i) Let a, c be nonnegative real numbers and let I : [a, b] -+ [c, d] be a bijective decreasing function.

Prove that

L [f (k)] - L [f-1 (k)] == [b]a(c) - [dJa(a) ,

where k is integer and a is the function defined in previous problem. ii) Prove that

t [n2 ] = f: [�]

k=l k2 k=1 VIi for all integers n � 1 .

23. Let 1 < nl < n2 < . . . < nk < . . . be a sequence of integers such that no two

are consecutive.

Prove that for all positive integers m between nl + n2 + . . . + nm and nl + n2 + . . . + nm+1 there is a perfect square.

24. Let (Xn)n?l be a sequence defined by Xl = 3 and Xn+l = x� - 2 for all positive integers n.

6. 1 . PROBLEMS 227

Taking into account that (xn ' xn+

d = 1 for all n 2:: 1, there are sequences (Un)n2:1 , (

vn)n2:1 of positive integers such that

Prove that

is a perfect square for all n 2:: 1 .

25. Find the sum of the series

� <pen) . � 2n - I n=l

26. In a coordinate system xOy consider the points Ak (k, n-I) , k = 0, 1 , . . . , n- I for a given positive integer n.

Find the number of open segments OAk that do not contain a point with integral coordinates.

27. Let (an)n2:1 ,

(bn)n2:1 '

(Cn)n2:1 be sequences of positive integers defined by

(1 + �

+ .ij4)n = an +

bn�

+ cn�

' n 2:: 1 .

Prove that

if if if

n =: O n =: 2 n =: 1

(mod 3)

(mod 3)

(mod 3)

28. Consider the sequences (an)n2:1 ,

(bn)n2:1 '

(Cn)n2:1 ,

(dn)n2:1 defined by al = 0, b

1 = 1 , Cl = 1 , d1 = 0 and an+l = 2bn + 3cn,

bn+1 = an +

3dn, Cn+l = an +

2dn, d

n+1 = bn + Cn, n 2:: 1 .

Find a closed formula for the general term of these sequences.

29. Let f : N* x N* --+ N* be a function such that f(I, 1) = 2,

f(m + I , n) = f(m, n) + m and f(m, n + 1) = f(m, n) - n

for all m, n E N* . Find all pairs (p, q) such that f(p, q) = 2001.

30. Determine all functions f : Z

--+ Z

satisfying

for all integers x, y, z.


31 . 1981 points lie inside a cube of side 9. Prove that there are two points within a distance less than 1 .

32. The squares of a chessboard are randomly labeled from 1 to 64. On the first 63 there is a knight. After some moves, the 64's square, initially unoccupied, is also

unoccupied. Let nk be the square number of the knight who was initially on the k's square.

Prove that 63 L ink - k l � 1984. k=l

33. The Fibonacci sequence (Fn)n2:1 is given by

Prove that

for all n 2:: 2.

FI =

F2 = 1,

Fn+2 =

Fn+l +

Fn, n 2:: 1.

34. Let n be a positive integer and let Nk be the number of increasing arithmetic progressions with k terms from the set {I , 2, . . . , n} .

Prove that

[n - 1 ] where q =

k _ 1 .

1 2 ( 1 ) Nk � - "2q + n + "2 q + 1 - k,

35. Let Fn be the nth Fibonacci number (that is,

FI =

F2 = 1 ,

Fn+l =

Fn +

Fn-l

for n 2:: 2) , and let P(x) be the polynomial of degree 990 such that P(k)

= Fk for

k = 992, 993, . . . , 1982. Prove that P(1983) = Fl983 - 1 .

36 . Let Xl , X2 , A. , f3 be real numbers and let the sequence (xn)n2:1 be given by

If x� - Xm+IXm-1 1:-

0 for all m > 1 , prove that there are real numbers AI , A2 such that

and

for all n > 2.

6. 1 . PROBLEMS 229

37. Consider b E [0 , 1) and the sequence (an)n�l defined by al = a2 = . . . =

ak-l = 0, k � 3, and

1 2 k-l an+l = k (b + an + an-l + . . . + an-k+2 ) for all k.

Prove that the sequence is convergent.

38. Let (an )n�1 and (bn )n2:1 be sequences such that

i) (bn}n>1 is strictly monotonic and unbounded;

ii) there-

exists lim abn ; n�oo n

. . . ) an+l + bn+1 - 2 > 1 111 b

- , n - . an n

Prove that lim abn = O. n�oo n

39. Let (Un)n�l ' be a sequence defined by Ul E 1R \ {O, I} and

Un+l = Ul U2 U3 °

If the sequence converges, evaluate

1 n lim - II (1 + Ul . . . Uk ) . n-+oo n k=2

40. Let (an)n2:1 and (bn )n2:1 be sequences such that i) 0 < b1 < b2 < . . . < bn < . . . ;

ii) b�: 1 � k > 1, n � 1;

iii) there exists lim abn . n�oo n

h . an+ 1 - an . d 1 l' an

Prove t at hm b b

eXIsts an is equa to 1m -b

. n�oo n+l - n n�oo n

41. Let k be a positive integer and let

Prove that

42. Let f : 1R � 1R n f(x) = L sin akx,

k=l


where ak are real numbers. Prove that if lail

-I lajl

for i

-I j, then there is a real number Xo such that f(xo) -I O .

43. Let f : lR

-t lR

be a differentiable function with continuous derivative such that

lim f(x) = lim t (x) = 00. x-too x-too

Prove that the function 9 : lR

-t lR,

g (x) = sin f(x) ,

is not periodical.

44. Let a be a real number and let f : N -t [0, 1 ) , f (n) = {an} i .e. the fractional part of the number an.

i) Prove that f is injective if and only if a is irrational. ii) If a is rational, find the number of elements of the set

M = {f(n) 1 n E N} .

45. Let f : lR

-t lR

be a function such that i) f has a period T > 0;

i i) f (x) � M

for all x;

iii) f (x) = M

if and only if x = kT, for some integer k. Prove that, for any irrational f) the function 9 :

lR -+ lR,

g (x) = f(x) + f(f)x) ,

i s not periodical.

46. Let f : lR

-t lR

be a continuous function with a period T

> O. a) Prove that if T is irrational, then for any

A E [min f(x) , max f(X)] there is a xER xER

sequence (Xn)n� l of integers such that

lim f(xn ) = A. n-too

b) ,Prove that if T is rational, then for any A

E [min f(X) , max f(x)] and for any xER xER irrational number f), there is a sequence (Xn )n�l of integers such that

47. i) Let x, y, z, v be distinct positive integers such that x + y = z + v. Prove that there is no

A > 1 such that

6. 1 . PROBLEMS 231

ii) Let p be a prime number and let a, b, c, d be distinct positive integers such that

aP + bP = cP + at. Prove that

48. i) Prove that

for any x � y > O. ii) Prove that

la - c l + I b - dl � p

(n + l)n 1 (n +

1)n+1 --- < n. < , n �

1.

en en

49. Let a, c be nonnegative real numbers and let f : [a, b) � [c, d), be a bijective function.

i) If f is increasing, prove that

l f (t)dt + ld r' (t)dt = bd - ac.

ii) If f is decreasing, prove that

f.' f (t)dt - ld r' (t)dt = bc - ad.

50. i) Let J-L : (0, 00) � lR

be a continuous function such that lim J-L(x) = o. z�oo

Prove that

lim e-z (Z et J-L(t)dt = 0 z-too 10

ii) Let f : [0, 00) � lR

be an n-time differentiable function with the n-derivative continuous such that there exists

n lim � C�f (k) (x) = A.

x-t= k=O Prove that lim f (x) exists and

z�=

lim f(x) = A. x-t=

51. Let f : [a, b] � [c, d) be a continuous function such that

a � b f.' f' (x)dx = cd.

Prove that if c + d =f. 0, then

0 ::; c� d l f(x)dx ::; b�

a G��r


52. Let I : [a , b] � lR be a continuous monotonic function and let

F: [a, b] � lR,

F(x) = (x - a) l' f(t)dt +

(x - b) [ f(t)dt.

Prove that all values of F have the same sign.

53. a) Consider the functions I : (0, 00) � lR and 9 : [1, 00) � lR such that

1) 9 is differentiable with continuous derivative; 2)

I is continuous and the function h : [1 , 00) � lR, h(x) = gl (X) - I(g(x)) is

nonincreasing. Denote

Prove that

b) Prove that

n an =

L h(x).

k=l

!.g(n+1) an+1 - h(l) ::;

I(x)dx ::; an,

g(l )

n I l

lim L

k2 cot -k = 00. n-+oo k=l

54. Let a be a positive real number and let I

: [0, 1] � lR+ be an integrable function.

Evaluate

55. Consider the functions In : lR � lR,

for all integers n � O.

( s�n r ) 2 ,

sm X

1 -1- 0 -I- k E '71 * X I , x I k7r ' /U

1 X = - k

E Z* k7r ' x =

o

1) Find the numbers an,k such that In is continuous on lR* .

2) Find the number an such that In is a derivative function.

56. Let p, q � 0 be integers. Find the numbers Cp,q such that Ip ,q : lR � lR,

f • . • (x) = {

is a deri vati ve function.

• p 1 q 1 sm _ . cos -X x

' if x =f- 0 if x = 0

6. 1 . PROBLEMS

57. Let q be a positive integer. Find the number an(q) such that fn : lR -+ lR,

f ( ) - { cos � cos !I . . . cos qn-l

, X =f. 0 n X - X X x '

an (q) , x = 0

is a derivative function for any integer n > O.

58. 1) Let f : lR -+ lR be a continuous function such that

Prove that the function

is a derivative function.

l /.Y lim - f(x)dx = M(f) .

I Y I-too Y 0

g(x) = { f G) , x ;l a M(f) , x = 0

n E N* ,

2) Let f : lR -+ lR be a continuous function with a period T > O. Prove that

l /.t l /.T M(f) = lim - f(x)dx = -T f(x)dx I t l-too

to o

233

59. If f : lR -+ lR is a derivative function, then is 9 : lR -+ 1R, g (x) = I f(x) 1 also a derivative function?

60. If h , h : lR -+ lR are derivative functions then f = max{h , h} is also a derivative function?

SOLUTIONS

1 . There are m m functions h

: B ---+ B. Each of them can be extended in n n-m

ways to a function f : A ---+ A which satisfies f (B ) � B. Hence the required number is m

m . nn-m. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2 (1981) , pp . 81,

Problem Cl

:l)

2. Let Xq be a subset with q elements of X. Since Y has q elements, it follows

that there are q! bijective functions 9 : Xq ---+ Y. Each of them can be extended in

nn-q ways to a function f : A ---+ A which satisfies Y � f(X) .

The number of subsets Xq of

X is (:) , hence the requested number is (:) nn-p.

Remark. If q > p, consider (:) = O.

(Dorin Andrica, Romanian Mathematical Regional Contest " Grigore MoisH" , 2000)

3. Since f(X) = X, it follows that f is bijective on X. There are k! such bijections

and each of them can be extended in n n-k ways to a function f : A ---+ A. Hence the

desired number is k!nn-k .

(Dorin Andrica)

4. We consider two cases. i) n ::; m. Let Ak be a subset with k elements of the set A. There is only one

function h

: Ak ---+ Ak that has the property h(i)

= i

for all i E Ak . This function can

be extended in (m - l)n-k ways to a function f : A ---+ B such that f(i) f:. i

for all i E A \ Ak •

The number of the subsets Ak of A is (�) , hence the desired number is (�) (ml)n-k .

ii) m ::; n. Let Bk be a subset with k

elements of the set B. There is only one

function h : Bk ---+ Bk such that

h(i) = i

for all i E Bk• This function can be extended

in (m - l)m-k ways to a function 9 : B ---+ B such that g(i) f:. i

for all i E B \ Bk•

235


Moreover, each function 9 can be extended in mn-m ways to a function f : A -4 B, that clearly has exactly

k fixed points .

The number of the subsets Bk of B is (7) hence the desired number is

(7)mn-

m(m -

W-k .

Therefore the number of functions f : A -t B with k

fixed points is (�) (m

l)n-k if n :O; m and (7)

mn-m (m _

l)m-k if m :O; n.

(Dorin Andrica, Romanian Mathematical Regional Contest "Marian 'farina, 2001)

5 . Using the inequality

xr + x� + . . . + x��

n

�

-l (Xl + X2 + . . . + Xn

)m

for

we obtain

On the other hand,

al a2 an al a2 an - + - + · · · + - > n n - ' - . . . - = n a2 a3 al - a2 a3 al

from the AM-GM inequality. Therefore (1 + al ) m

+ (1 + a2 ) m

+ . . . + (1 + an ) m

� � (2n)m = n . 2m

, a2 a3 al nm I

as desired.

( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1 (1974) , pp. 7,

Problem 1564; Gazeta Matematidi (GM-B) , No. 2 (1976), pp. 65)

6. We have n II(k + ai

) = kn + Lalkn-l + Lala2

kn-2 + . . . + ala2 · · . an· i=l

Using the AM-GM inequality gives

6 .2 . SOLUTIONS

From the previous relation we deduce 1 2

n(k + a;) � kn + (�) (n air kn-1 + (;) (n a;r kn-2+

+ . .. + (�) (nf (k + w.:) n

Thus

k + � n a; ::; n n (k + a;) . Using again the AM-GM inequality, we obtain

so

237

(1)

n 1 n n II (k

+ ai ) ::; k + - L ai . (2) i=l n i=l

( Titu Andreescu, Revista Matematidl Timi§oara (RMT) , No. 2 (1977) , pp. 63, Problem 3045)

7. First Solution. Since abc = 1 , this non-homogeneous inequality can be transformed into a homogeneous one by a suitable change of variables. In fact, there exist positive real numbers p, q, r such that

a = !!. , q q

b = - , r r c = - . p

Rewriting the inequality in terms of p, q, r, we obtain

(p - q + r) (q - r + p) (r - p + q) ::; pqr, where p, q, r > o.

(1)

At most one of the numbers u = p - q +r, v = q - r+p, W = r -p+q is negative,

because any two of them have a positive sum. If exactly one of the numbers u, v, w is negative, then uvw ::; 0 < pqr. If they are all nonnegative, then by the AM-GM

inequality, 1 ..;uv ::; 2 (u + v ) = p.

Likewise, ..;vw ::; q and ..jWu ::; r. Hence uvw ::; pqr, as desired.

Second Solution. Expanding out the left-hand side of (1) gives

(p- q+ r) (q - r +p) (r -p+ q) = [P(p- r) + (r - q)(P- q) + q(r - q) +pq] [r+ (q -p)] =

= pr(p - r) + r(r - q) (p - r) + rq(r - q) + pqr + p(p - r) (q - p)+ +(r - q) (p - r) (q - p) + q(r - q) (q - p) + pq(q - p) .


Note that

pr(p - r) + rq(r - q) + pq(q - p) + (r - q)(P - r) (q - p) = O.

Thus (1 ) is equivalent to

o ::; p(p - q) (p - r) + q(q - r) (q - p) + r(r - p) (r - q) ,

which is a special case of Schur's inequality.

Third Solution. Denoting the left-hand side of the desired inequality by L, we

have

L = abcL = b (a - 1 + D c (b - 1 + D a (c - 1 + D =

= (ab - b + l) (be - e + l) (ea - a + 1) = L1 • Also, since lib = ae, lie = ab, 11a = be,

L = (a - 1 + D (b - 1 + D (C - 1 + D =

= (a - 1 + ae) ( b - 1 + ab) ( e - 1 + be) = L2 . If u = a - I + lib ::; 0, then a < 1 and b > 1, implying that

v = b - 1 + lie > 0 and W = e - 1 + 11a > o.

Then L = uvw ::; 0, as desired. Similarly, either u ::; 0 or v ::; 0 yields the same

result. If u, v, W > 0, then all factors of Ll and L2 are positive. The AM-GM inequality gives

1 J(ab - b + l) (b - 1 + ab) ::; 2 [(ab - b + 1) + (b - 1 + ab)] = abo Likewise,

J(be - e + l) (e - 1 + be) ::; be,

J(ea - a + l) (a - 1 + ae) ::; ca. Hence L = VL1L2 ::; (ab) (be) (ea) = (abe)2 = 1 . Fourth Solution. Using the notations established in the third solution, i t is easy

to verify the equalities

beu + ve = 2, eav + aw = 2, abw + bu = 2.

As in the third solution, we only need to consider the case when u, v , w > O. The AM-GM inequality gives

2 2:: 2eVbuv, 2 2:: 2aVcvw, and 2 2:: 2vawu,

from which uvw ::; 1 . Fifth Solution. Let Ul = ab - b + 1, VI = be- e+ 1 , WI = ca - a+ 1 ; U2 = 1 - be+c,

V2 = 1 - ea + a, and W2 = 1 - ab + b. As in the third solution, we only need to consider the case in which Ui , Vi , Wi > 0 for i = 1 , 2 . Again, we have

6.2 . SOLUTIONS

Let X = a + b + e and Y = ab + be + ca. Then

Ul + VI + WI = Y

- X + 3 and U2 + V2 + W2 = X - Y + 3.

239

Hence either Ul + VI + WI :::; 3 or U2 + V2 + W2 :::; 3. In either case L :::; 1 follows from the AM -G M inequality.

( Titu Andreescu, IMO 2000, Problem 2)

8. Without loss of generality we can assume that a � (3

:::; 'Y. Let x, y, z be three

arbitrary numbers from the interval [a, b] such that x :::; y :::; z. Then

and

and

E(x, y, z)

- E(x, z, y) = (1' - a

) ( (z - X)2 -

(y - x2) ) � 0,

E(a, y, b) = a(y - a)2 +

(3(y - b)2 + 'Y(b - a)2

E(b, y, a) = a(y - b)2 + f3(y - a? + 'Y(b - a)2 .

We need to find the maximal values of the functions

fl (y)

= a(y - a)2 + f3(y - b)2

h (y) = a(y - b)2 +

(3(y - a)2

on the interval [a, b] . Since h (a) = f3(b - a)2 � a(b - a)2 = II (b) and the coefficient

of y2

in h is a + f3 � 0, it follows that the maximum value of II is obtained for y

= a. Likewise, h (b) � h (a) and the maximum value of h is obtained for

y = b.

Therefore

and

max E(a, y, b) = E(a, a, b) = (f3 + 'Y) (b - a)2 , yE[a,b]

max E(b, y, a) = E(b, b, a) = ({3

+ 'Y) (b - a)2 . yE[a,b] It follows that the maximum value of E is (f3 + 'Y) (b - a)2 and is obtained for

x = a, y = a, z = b or x = b, y = b, z = a. (Dorin Andrica and loan Ra§a, Revista Matematica Timi§oara (RMT) , No.

1 (1983) , pp. 66, Problem C5:2)

9. For a function f : { 1, 2, . . . , n} --+ {a, b, e} let Ma = f-l ({a}) , Mb = f-1 ({b}) , Me = f-1 ({ e}) and let p, q, r be the number of elements of sets Ma, Mb , Me, respectively. Obviously p + q + r = n and without loss of generality we may assume that p � q � r.

A term I f (i) - f(j ) 1 is different from ° if the pair (i, j ) is in one of the sets Ma x Mb , Mb X Ma , Ma x Me, Me X Ma , Mb X Me or Me X Mb. Hence the number

n

of nonzero terms in the sum L

I f (i) - f(j) 1 is 2(pq + qr + rp) . i ,j=l


The problem reduces to finding the maximal value of 2 (pq+qT+TP ) when p+q+T ==

n and p, q, T 2:: 0 are integers. Note that if (Po , qo , TO ) is a triplet that maximizes 2(pq+qT+Tp) then the absolute

value of any difference of two numbers from this triplet is at most 1. Indeed, assume

that Po - TO 2:: 2 and define

PI = Po - 1 , ql = qo , TI = TO + 1.

Then PI + ql + TI = n and

= Poqo + POTO + qoro + Po - ro - 1 > Poqo + POTO + qoTo , which contradicts the maximality of 2(Poqo + qoro + ropo) .

We have the following cases. 1) n = 3k. Then Po + qo + TO = 3k, Po 2:: qo 2:: ro 2:: Po - 1 , hence Po = k and then

qo = ro = k. In this case the maximal value is

2 (k2 + k2 + k2 )

= 6k2

= 2n2.

3 2) n = 3k + 1. Then Po + qo + ro = 3k + 1 , Po 2:: qo 2:: TO 2:: Po - 1, so 3po 2::

3k + 1 2:: 3po - 2. Hence Po = k + 1 and then qo = TO = k. In this case the maximal value is

2 2( (k + 1)k + (k + 1)k + k2 )

= 2(3k2 + 2k) = 3 (n2 - 1) .

3) n = 3k+ 2. Then Po +qo + ro = 3k+ 2, Po 2:: qo 2:: TO 2:= Po - 1, so 3Po 2:= 3k+ 1 2:: 3po - 2. It follows that Po = k + 1 and qo + ro = 2k + 1. Because k + 1 2:: qo 2:= TO 2:= k, qo = k + 1 and ro = k. The maximal value is in this case

n2 - 1 2[(k + l) (k + 1) + (k + l)k + (k + l)k] = 2(k + 1) (3k + 1) = 2 · -3

-

n2 n2 - 1

Therefore the requested number is 23 if 3 divides n and 2-3- otherwise.

Remark. The problem can be reformulated as follows: Suppose that n points in space are colored by three different colors. Find the maximum number of segments AB such that A and B are different colors.

(DoTin Andrica and Pal Dalyai, Romanian IMO Selection Test, 1982; Revista Matematica Timi§oara (RMT) , No. 1 (1982) , pp. 83, Problem 4917)

10. From the inequality

(.fiil

_ 4fb

l) 2 + (f02 _

4fb2) 2 + . . . + (va; _ 4 'b) 2 > 0 � Y UI � Y U2 � Y Un - ,

it follows that

6. 2. SOLUTIONS 241

We have

hence

va; + .Ja2 + . . . + va;;

� .jb; + Jb; + . . . + A, as claimed. ( Ti

tu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2

(1977), pp. 6

3,

Problem 3046)

1 1 1. By mUltiplying the numbers aI , a2 , . . . , an with a suitable factor M = -n--

n we may reduce the problem to the case when

L a;P =

1.

i=l Assume without loss of generality that

a = al � a2 � . . . � an · Let a = � and suppose by way of contradiction that

2 S (n ,p)

Then

hence n n L a;P >

L[a + a(i -

1)]2P . i=l i=l

Consider the function <p : JR -+ (0, 00) , n

<p(x) = L[

x + a(i _1)]2p .

i=l

'" 2p L-.t ai i=l


Then n <p' (x) = 2p I)x + a(i - 1)]2p-l

i=l and n

<p" (X) = 2p(2p - 1) I)x + a(i _1)]2p-2 > O.

i=l Because <p' (x) i s a polynomial of odd degree and <p" (x) > 0 for all real x, it follows

that <p' has a unique real zero: (1 - n) Xo = 2 a

The number Xo is also a minimum point of the function <p, so n n [ 1 ] 2P 2p 8 a;P > <p(xo ) = 8 ; n

a + a(i - 1) = � S(n,p) = 1 ,

a contradiction. Hence

as desired.

4P n

• ( . _ .)2p < __ '" 2p rp.l� a� aJ - S( ) L-.t ai , l ��<J�n n,p i=l

Remark. For p = 1 we obtain the MitrinoviC's inequality

• 2 12 � 2

l��f

�n

(ai - aj

) � n (n2 - 1)

�ai '

(Dorin Andrica)

12. Note that

for all positive real numbers x. Setting x = yfii - 1 > 0, implies

then

It follows that

and then

as desired.

(vn)n > 1 + (�) (vn _1)k

n - 1 > (�) (vn _1)k

,

n - 1 n k

(�) > (vn - l) ,

J� � (�)

> ifrl - l,

6.2. SOLUTIONS 243 (Dorin Andrica, Romanian Winter Camp, 1984; Revista Matematidi Timi§oara

(RMT) , No. 1 (1985) , pp. 72, Problem 1)

13. a) Clearly,

1 1 1 1 1 1 1 1

au al2 al ,n-l al ,n a2 l a22 a2,n-l a2 ,n

an-2, l an-2,2 an-2,n-l an-2,n Expanding the determinant after the first row yields

hence

as desired. b) Consider the determinants

1

Ak = Xl

n-2 Xl

From equality a) we obtain

1 1 Xk-l Xk+l

n-2 Xk-l n-2 Xk+l

n (-l)k L - = O. k=l Pkqk

1 Xn

X�-2

= 0

c) Set Xk = k2 , k = 1 , 2, . . . , n, in the previous equality. After some algebraic manipulations we obtain

as desired.

n (- l )k k2 "'"' = 0, n � 3, � (n - k) ! (n + k) !

(Dorin Andrica, Romanian Mathematical Regional Contest " Grigore MoisH" ,

1995)

244 6 . COMPREHENSIVE PROBLEMS

14. Let Xl , X2 , . . . , Xn > 0 be the zeros of polynomial P. The relations between zeros and coefficients yield

2: XlX2 • . ' Xm = (:) and 2: XlX2 • • . xp = �) . Then we have the equality case in the Generalized Mac Laurin's Inequality:

rn L XlX2 . . . Xm p L XlX2 . . . xp ------:-:---- > m � p C� - Ch '

hence X, = X2 = . . . = Xn - From 2: XlX2 . ' ' Xm = (:} it follows that Xi = 1 ,

i = 1 , 2 , . . . , n , hence P(x) = (x - l)n ,

as claimed. ( Titu Andreescu, Revista Matematidl Timi§oara (RMT) , No. 1 (1977) , pp. 24,

Problem 2300)

1 5 . It is obvious that degpo = 0 and degpl = 1 . Assuming that degpk = k, from the given relation we obtain degpk+l = k + 1 , hence by induction degpi = i for all i = 0 , 1 , . . . , n .

Consider the function f : 1R -+ lR, f(x) = xnex . It is easy to prove that

fCk) (x) = Xn-kQk (x)eX , where Qk (X) is a polynomial with real coefficients of degree k.

We prove that Qk (X) = Pk (X) for all k.

Note that Qo (x) = 1 - Po (x) and

xn-Ck+l ) Qk+l (x)eX = fCk+l ) (x) = (fCk) (x))' = (Xn-kQk (x)eX )' = = Xn-Ck+l) [(n - k + X)Qk (x) + xQ� (x)] ,

hence Qk+l (x) = (n - k + X)Qk (X) + xQ� (x) .

Since (Pk )k=O,n and (Qkh=o,n satisfy the same recursive relation and Po = Qo it follows that Pk = Qk for all k.

So

and, on the other hand, n

fCn) (x) = (xneX ) Cn) = L C�(xn ) Ck) ex . k=O

It follows that

6. 2. SOLUTIONS 245

(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 1 ( 1978) , pp. 67, Problem 3293)

16. We prove that

Note that

P� (x) = -2nPn-dx) , n � O.

P; (x) = -2 = -2 · 1 · Po (x) and assume that

Then P�-l (x) = -2(n - 1)Pn-2 (x) , n � 2.

Differentiating we obtain

P� (x) = -2Pn-1 (x) - 2XP�_1 (x) - 2(n - 1)P�_2 (x) = = -2Pn-1 (x) + 4(n - 1)xPn-2 (x) - 2(n - 1)P�_2 (x) =

= -2Pn-1 (x) - 2(n - 1 ) [-2xPn-2 (x) + P�_2 (X)] = = -2Pn- 1 (x) - 2(n - 1)Pn-1 (x) = -2nPn-1 (x) ,

as needed.

so

The initial relation becomes

Hence Pn (O) = -2(n - 1)Pn-2 (0) , n � 2.

Pn (O) = { �-l) � (�) ! if n is odd

if n is even

Alternative solution. Note that

(e-X2 ) (n) = Qn (x)e-X2

for a polynomial Qn . From

we obtain

(e-X2 ) (n+l) = [ (e-X2 )(n) ] , = (Qn (x)e-X2 ), = = [Q� (x) - 2xQn (x)]e-X2 = Qn+l (x)e-X2 ,

Qn+l (x) = -2xQn (x) - Q� (x) , n � O. Since Qo (x) = 1 = Po (x) , we note that

Qn (x) = Pn (X) for all n � 0,


hence

On the other hand,

_x2 X2 X4 X6 n X2n e = 1 - - + - - - + . . . + (-1) - + . . . 1 1 2 1 3! nl (1)

If n is odd, then by differentiating the series (1) for an odd number of times, we deduce that Pn (0) = o.

If n is even, set n = 2m. Differentiating (1 ) n times yields

(e-X' ) (2m) = (_l)m (:) + x [ (-W+1 (�:+;) ! + . . . J '

hence

Therefore

Pn (O) = { �- l) " (%) ! if n is odd

if n is even

(Dorin Andrica, Revista Matematidl Timi§oara (RMT), No. 2(1978) , pp. 76, Problem 3706)

n 17. Let P(x) = L akxk be a polynomial with nonnegative real coefficients. By

k=O the Cauchy-Schwarz Inequality we derive that

P(x) = (�va.. va.xk r � (�ak) (�akx2k) = P(1)P(x2 ) .

It suffices to prove that P(I) < 1 . Indeed n 1 L < 1

as needed. k=O n + l + 1

Therefore p2 (x) < P(x2 ) for all x, so the equation has no real roots. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2(1989) , pp. 107,

Problem C9:8)

18. If P = 0 , then for t = 0 and for any function f : 1R --+ 1R the claim holds. Let n

P(x) = Lakxk k=O

6.2 . SOLUTIONS

with an f:.

0, and let t E JR* . We search for a polynomial n+l

Qt (x) = I: bkXk k=l

such that Qt (x + t) - Qt (x) = P(x), x E JR.

Identifying the coefficients from both sides yields (n ; 1) tbn+1 = an

(n + 1) 2 (n) 2 t bn+1 + 1 tbn = an-l

(� : �) tn+1 bn+l + . . . + G) tb, = 1Io

247

Note that the system has unique solution, hence there is a unique polynomial Qt such that

and

Qt (x + t) - Qt (x) = P(x) , x E JR

Qt (O) = 0,

deg Qt = 1 + deg P. Set g(x) = f(x) -Qt (x) . Then f satisfies the claim if and only if g(x+t) -g (x) = 0

for all real x, i .e . 9 has period t. Therefore

f(x) = Qt (x) + g(x) , where 9 is a function of period t.

(Dorin Andrica, Revista Matematidl Timi§oara (RMT), No. 2(1984) , pp. 103, Problem C6:10)

that

1 1 1 1 1 19. Because 2a, 2b, 2c, 2d, 2e E [-1 , 1] , there are real numbers x, y, Z, t, u such

a = 2 cos x, b = 2 cos y, c = 2 cos z, d = 2 cos t, e = 2 cos u.

U sing the identity

2 cos 5a = (2 cos a) 5 - 5 (2 cos a) 3 + 5 (2 cos a) ,

we obtain 2 cos 5x = a5 - 5a3 + 5a

and the analogous relations. Summing them up yields

I: 2 cos 5x = I: a5 - 5 I: a3 + 5 I: a = 10


so

L cos 5x = 5.

Hence cos 5x = cos 5y = cos 5z = cos 5t = cos 5u = 1 , { V5 - 1 V5 + 1 }

and therefore a, b, c, d, e E 2, --2- ' - --2- .

From the relation a + b + c + d + e = 0, it follows that one of the numbers is 2, V5 - 1 V5 + 1 . two of them --- and the other two - --- . It 1S easy to check that for these 2 2

numbers L a3 = O. ( Titu Andreescu, Romanian Mathematical Olympiad - final round, 2002)

20. We will say that a subset of N

is p-progression-free if it does not contain an arithmetic progression of length p. Denote by bn the number obtained by writing n in base p - 1 and reading it in base p. One can easily prove that an = bn for all n = 0, 1 , 2 , . . . by induction, using the following properties of the set B = {bo , b1 , . . . , bn , . . . } (whose proofs we postpone) :

1 ° B is p-progression-free; 2° If bn-1 < a < bn for some n � 1, then the set {bo , b1 , . . . , bn-1 , a} is not

p-progression-free . Indeed, assume 1° and 2° hold. By the definition of ak and bk , we have ak = bk

for k = 0 , 1 , . . . , p - 2. Let al = bk for all k � n - 1 , where n � p - 1. By 1° , the set

is p-progression-free, so an � bn . Also, the inequality an < bn is impossible, in view of 2° . Hence an = bn and we are done.

So it suffices to prove 1° and 2° . Let us note first that B consists of all numbers whose base p representation does not contain the digit p - 1. Hence 1° follows from the fact that if a, a + d, . . . , a + (p - l)d is any arithmetic progression of length p, then all base p digits occur in the base p representation of its terms. To see this, represent d in the form d = pmk, where gcd(k, p) = 1 . Then d ends in m zeros, and the digit o preceding them is nonzero. It is easy to see that if a is the (m + l)st digit of a (from right to left) , then the corresponding digits of a, a + d, . . . , a + (p - l)d are the remainders of a, a + 0, . . . , a + (p - 1)0 modulo p, respectively. It remains to note that a, a + 0, . . . , a + (p - 1)0 is a complete set of residues modulo p, because 0 is relatively prime to p. This finishes the proof of 1 ° .

We start proving 2° by the remark that bn-1 < a < bn implies a � B. Since B consists precisely of the numbers whose base p representations do not contain the digit p - 1, this very digit must occur in the base p representation of a. Let d be the

6. 2 . SOLUTIONS 249

number obtained from a be replacing each of its digits by 0 if the digit is not p - 1 , and by 1 i f it i s p - 1 . Consider the progression

a - (p - l)d, a - (p - 2)d, . . . , a - d, a.

As the definition of d implies, the first p - 1 terms do not contain p - 1 in their base p representation. Hence, being less than a, they must belong to {bo , b1 , • • • , bn-I } . Therefore the set {bo , b1 , . . . , bn-1 , a} is not p-progression-free, and the proof i s finished. 0

( Titu Andreescu, USA Mathematical Olympiad, 1995)

21 . i) For a bounded region M of the plane we denote by n(M) the number of points with nonnegative integral coordinates in M.

Function f is increasing and bijective, hence continuous. Consider the sets

Then

We have

hence

M1 = { (x, y) E 1I�? I a :S x :S b, 0 :S y :S f (x) } ,

M2 = { (x, y) E 1I�? 1 c :S y :S d , O :S x :S f-1 (y) } ,

M3 = { (x, y) E ]R2 1 0 :S x :S b , O :S y :S d},

M4 = { (x, y) E ]R2 1 O :S x :S a, O :S y :S c} .

o a b

n (M1 ) = L [f(k)] , n (M2 ) = L [1-1 (k)] , a�k�b c�k�d

n (M3 ) = [b] [dJ , n(M4) = a(a)a(c) .



il) Consider the function f : [1 , nJ --t [1, n (n 2+ 1) ] ,

f(x) = x(x + 1) . 2 Function f is increasing and bijective. Note that n(Gf ) n and f-l (X)

-1 + VI + 8x A I ' £ 1 ' ) b ' 2 . pp ymg ormu a 1 we 0 tam

hence

n(n+l) t, [k(k; I) ] + t. [ -1 + �] _ n = n2 (n

2+ 1) ,

n (n+l) t. [ -1 +�] = n2 (n

2+ 1) + n - � t, k(k + l) =

n2 (n + 1) n(n + 1) n(n + 1) (2n + 1) n(n2 + 2) = 2 + n - 4 - 12 = 3

( Titu Andreescu and Dorin Andrica, "Asupra unor clase de identitati" , Gazeta Matematica (GM-B) , No. 11 (1978) , pp. 472-475)

22 . i) Function f is decreasing and bijective, hence continuous. Consider the sets

Then

Nl = { (x, y) E ]R2 1 a ::; x ::; b, c ::; y ::; f(x) } ,

N2 = { (x, y) E ]R2 1 c ::; y ::; d , a ::; x ::; f-1 (y) } , N 3 = { (x, y) E ]R2

I a ::; x ::; b, 0 ::; y ::; c} ,

N4 = { (x, y) E ]R2 1 0 ::; x ::; a, c ::; y ::; d}.

d

c

o a

L [f(k)] = n(N1 ) + n(N3 ) ,

L (f- l (k)] = n (N2) + n(N4) , c�k�d

b

6.2. SOLUTIONS

n(N1 ) = n(N2 ) , and

n(N3 ) = ( [b] - a(a) )a(c) , n (N4) = ( [d] - a(c) )a(a)

It follows that

as desired.

L [f (k)] - L [j-1 (k)] = n (N3) - n(N4) = a�k�b c�k�d

= [b]a(c) - [d]a(a) ,

ii) Consider the function f : [1 , n] --+ [1 , n2] , n2

f(x) = x2

Note that f is decreasing and bijective and

Using formula i) , we obtain

-1 n f (x) = Vx·

n n2 {; [�:] -(; [�] = na(l) - n2a(1) = 0 ,

hence

as desired.

251

(Dorin Andrica and Titu Andreescu, Gazeta Matematidl (GM-B), No. 6(1979) , pp. 254, Problem 0.48)

23. It is easy to prove that between numbers a > b � 0 such that Va - Vb > 1 there is a perfect square - take for example ( [Vb] + 1)2 .

It suffices to prove that

Jn1 + . . . + nm+1 - Jn1 + . . . + nm > 1 , m � 1 .

This i s equivalent to

n1 + . . . + nm + nm+1 > (1 + Jn1 + n2 + .. . + nm)2

and then nm+1 > 1 + 2Jn1 + n2 + . . . + nm, m � 1 .

We induct on m. For m = 1 we have to prove that n2 > 1 + 2J1i1. Indeed, n2 � n1 + 2 = 1 + (1 + n1 ) > 1 + 2.Jril. Assume that the claim holds for some m � 1 . Then


so (nm+l - 1 )2 > 4(nl + . . . + nm) hence

(nm+l + 1)2 > 4(nl + . . . + nm+d.

This implies nm+l + 1 > 2y'nl + . . . + nm+l , and since nm+2 - nm+l � 2, it follows that

as desired. ( Titu Andreescu, Gazeta Matematidi (GM-B) , No. 1 (1980) , pp. 41, Problem

0. 113)

24. Substituting Xn+l = x� - 2 in the relation

yields UnX� - VnXn - (2un + 1) = 0, n � 1 . (1)

For a given n � 1 the relation (1) is a quadratic equation with integral coefficient and with an integer root Xn - Hence the discriminant

is a square, as desired. (Dorin Andrica)

� = v� + 8u� + 4un = t� ,

25 . Let (an )n2:1 be a sequence of real numbers. From the equality xn --- = xn + x2n + . . . + xkn + . . . , Ix i < 1, n � 1 1 - xn

we derive

where

00 n 00

L � - LA n - nX , 1 - xn n=l n=l

An = Lad. din

Using Gauss' formula L <p(d) = n, yields d in

f <p(n)Xn = f nXn = X n=l 1 - xn n=l (1 - X)2 .

Setting X = -21

implies � <p(n) = 2 . L...J 2n - 1 n=l (Dorin Andrica)

26. We start with a useful lemma.

6.2 . SOLUTIONS 253

Lemma. There are gcd(k, n) - 1 integers among 1 . n 2 · n (k - l)n -k- ' -k-' " ' ' k

Proof. Let gcd(k, n) = d, k = kId, n = ni d and note that gcd(kl , nl ) = 1 . The numbers are

1 · ni 2 · ni (k - l)nl T ' T " ' " ki

The number of multiples of ki in the set 1 , 2, . . . , k - 1 is d - 1 , hence among the above numbers there are d - 1 = gcd(k, n) - 1 integers, as desired.

The line OAk has the equation n Y = k · x .

C == An- I B == Ao '-��--------'-----�

o D

From the lemma it follows that among numbers 1 . n 2 · n (k - l)n -k- ' -k- ' " ' ' k

there are gcd(k, n) - 1 integers. Hence the open segment OAk does not contain points with integral coordinates if and only if gcd(k, n) = 1 . There are <p(n) such numbers and we are done.

Remark. An alternative version of this problem can be: "A hunter stays at the point 0 in a forest where the trees are placed at points with integral coordinates. Deers stay at points Ao , AI , . . . , An-I . How many chances of success has the hunter?"

(Dorin Andrica)

27. We have

a + b {12 + c {14 = (1 + 312 + {I4)n = ({I2(l + {12 + {/4))n = n n n V L, ({I2)n

= 2- i ( ij2 + {14 + 2)n = 2-i (1 + (1 + ij2 + {I4))n =


hence

an + bn� + cn ift! = 2-t f (�) ak+ k=l

(1)

We study three cases. i) If n == 0 (mod 3) , then 2-t E Q, hence

2- t f (�) ak = an , 2- t f (�) bk = bn , k=o k=O

(I)

ii) If n == 2 (mod 3) , then 2 -�± 2 E Q. Multiplying the relation (1 ) by 2� = �,

we obtain

then

Hence

- n ±2 n n ( ) 2 3 L k Ck = 2an ·

k=O

(2)

2-� � (�) ak = bn�' Ti � (�) bk = cn�' 2-i � (�) Ck = �. (II)

iii) if n == 1 (mod 3) , then 2 -;±1 E Q. By mUltiplying the relation (1 ) by 2� = {!2, we obtain

then

.=.!!±l n n ( ) 2 3 L k ak = 2cn ,

k=O Hence

(3)

6. 2 . SOLUTIONS 255

Relations (I) , (II) , (III) imply

TF tu (�) a. = { an , n = O (mod 3) bn �' n = 2 (mod 3) Cn .ij4, n = l (mod 3)

2-F tu (�) b. = 1

bn , n = O (mod 3) Cn {12, n = O (mod 3) an n = l (mod 3) {12 '

and { Cn , n = O (mod 3)

2-� � (�) c. = f' n = O (mod 3)

n = l (mod 3) {12' ( Titu Andreescu and Dorin Andrica, Revista Matematica Timi§oara (RMT), No.

1 (1984) , pp . 83, Problem C6:3)

28. Note that (\.1"2 + V3y� = an + bn\.1"2 + cnv'3 + dnV6, n � 1 . Let n = 2k and

Xk = � [(5 + 2v6)k + (5 - 2v6)k] ,

Then

hence

Yk = �v6 [(5 + 2v6)k - (5 - 2v6)k) .

1 an = -x.!!. , bn = Cn = 0 and dn = Y.!!. . 2 2 2 Let n = 2k + 1 . Then

(v0 + -/3)2k+l = (v0 + v'3) (v0 + v'3)2k = (Xk + 3Yk )v0 + (Xk + 2Yk )v'3, for all k � 1, hence

(Dorin Andrica)

29. We have

an = 0 , bn = X n-l + 3Y n- l , 2 2

Cn = X n-l + 2Y n-l , dn = 0 2 2

f(P, q) = f(p - 1, q) + p - 1 =

= f(P - 2, q) + (P - 2) + (p - 1) = . . . = f(l , q) + p(p; 1) = p(p - 1) = f (l , q - 1) + (q - 1) + = . . . = 2

= f(l , 1) _ q(q; 1) + p(p; 1) = 2001 .

256

Therefore

6 . COMPREHENSIVE PROBLEMS

p(p - 1) _ q(q - 1) = 1999 2 2

(p - q) (p + q - 1) = 2 · 1999. Note that 1999 is a prime number and that p - q < p + q - 1 for p, q E N* . We

have the following two cases: (a) p - q = 1 and p + q - 1 = 3998. Hence p = 2000 and q = 1999. (b) p - q = 2 and p + q - 1 = 1999. Hence p = 1001 and q = 999. Therefore (p, q) = (2000, 1999) or (1001, 999) . ( Titu Andreescu, Korean Mathematics Competition, 2001)

30. The only solutions are I(x) = 0, I (x) = x, and I(x) = -x. First, it is clear that these three are solutions . Next, setting x = y = z = 0, we find 1(0) = 3(/(0))3 , the only integer solution of which is 1(0) = O. Next, with y = -x and z = 0, we obtain 1 (0) = (/ (X))3 + (/( -X))3 + (/(0) )3 . This yields I(-x) = -I(x) , so 1 is an odd function. With (x, y, z) = (1 , 0, 0) , we obtain 1(1) = (/(1))3 + 2(/(0))3 = 1(1)3 ; thus 1 (1) E {-1, 0, 1} . Continuing with (x, y, z) = (1 , 1 , 0) and (x, y , z) = (1 , 1 , 1) yields 1 (2) = 2(/(1))3 = 21(1) and 1(3) = 3 (/(1))3 = 31(1) . To continue, we need a lemma.

Lemma. If x is an integer greater than 3, then x3 can be written as the sum 01 five cubes that are smaller in magnitude than x3 .

Proof. We have 43 = 33 + 33 + 23 + 13 + 13 , 53 = 43 + 43 + (_1)3 + (_1)3 + (_1)3 , 63 = 53 + 43 + 33 + 03 + 03 , and 73 = 63 + 53 + 13 + 13 + 03 . If x = 2k + 1 with k > 3, then

x3 = (2k + 1)3 = (2k _ 1)3 + (k + 4)3 + (4 - k)3 + (_5)3 + (_ 1)3 , and all of {2k - 1, k + 3 , 14 - k l , 5, I} are less than 2k + 1 . If x > 3 is an arbitrary integer , then write x = my, where y is 4 or 6 or an odd number greater than 3 and m is a natural number. Express y3 as yr + y� + y� + y� + y� . The number x3 can then be expressed as (mYl )3 + (mY2)3 + (mY3)3 + (mY4)3 + (mY5)3 . 0

Since 1 is an odd function and 1 (1) E {-I , 0, 1} , it suffices to prove that I(x) = xl(l) for every integer x. We have proved this for I x l ::; 3. For x � 4, suppose that the claim is true for all values with magnitude smaller than x. By the lemma, x3 = xr +x�+x�+x�+x� , where I Xi l < x for all i. After writing x3+ ( -X4)3+ ( -X5 )3 = xr + x� + x� , we apply 1 to both sides. By the stated condition of 1 and the oddness of I, we have

Therefore, the inductive hypothesis yields 5 5

(/(X))3 = I:(xi/(1))3 = (/(1))3 I: xr = (/ (1))3x3 . i=l i=l

6.2 . SOLUTIONS 257

Thus f (x) = xf(l) , and the result follows by induction. ( Titu Andreescu, The American Mathematical Monthly, Volume 108, No. 4(2001) ,

pp. 372, Problem 10728)

31 . Assume by way of contradiction that the distance between any two points is greater than or equal to 1. Then the spheres of radius 1/2 with centers at these 1981 points have disjoint interiors and are included in the cube of side 10 determined by the six parallel planes to the given cube's faces and situated in the exterior at a distance of 1/2. It follows that the sum of the volumes of the 1981 spheres is less than the volume of the cube of side 10, hence

4" . 1 � 1 3 1981 · 3

2 = 1981 . � > 1000,

a contradiction. The proof is complete. Remark. The pigeonhole principle does not help us here. Indeed, dividing each side

of the cube in [{!1981] = 12 congruent segments we obtain 123 = 1728 small cubes of side � = � . In such a cube there will be two points from the initial 1981 points. 12 4 The distance between them is less than � J3 which is not enough, since � J3 > 1 .

( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1981) , pp. 68, Problem 4627)

32. Note that

hence

63 63 2: Ink - k l = 2:ck

(nk - k) , where Ck E

{-I, I},

k=l k=l

63 S = 2: I

nk - k l = ±63 ± 63 ± 62 ± 62 ± . . . ± 2 ± 2 ± 1 ± 1 , k=l

with 63 signs of + and 63 signs of -. Then

S ::; (63 + 63 + 62 + 62 + . . . + 33 + 33 + 32) - (32 + 31 + 31 + . . . + 1 + 1) = 1984,

as desired. Remark. We prove that, for some labeling, S = 1984. It is known that a knight

can pass through all the 64 squares of the board only once and then come back to the initial square. Now label the squares from 1 to 64 in the order given by these knight moves. The free position can be made successively 64, 1 , 2, . . . , 63, 64, . . . so we can reach the situation nl = 32, n2 = 33, . . . , n32 = 63, n33 = 1 , n34 = 2, . . . , n63 = 31 . For this diagram we have S = 1984.

( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1984) , pp. 103, Problem C6:6)


33. Note that

hence (1)

for all n � 2. Setting a = 3F2n , b = -F2n+2 ' and e = -F2n-2 in the algebraic identity

a3 + b3 + e3 - 3abe = (a + b + e) (a2 + b2 + e2 - ab - be - ea) gives

27 F:tn - F:tn+2 - F:tn-2 - 9F2n+2F2nF2n-2 = O. Applying (1) twice gives

= F2n (3F2n-2 - F2n ) - F:jn-2 = F2nF2n-4 - D�n-2 = = . . . = F6F2 - Fi = -l.

The desired result follows from

9F2n+2F2nF2n-2 - 9F:tn = 9F2n (F2n+2F2n-2 - Fin) = -9F2n · ( Titu Andreescu, Korean Mathematics Competition, 2000)

34. Let r be the ratio of an arithmetic progression. We have 1 + r(k - 1) :S n, so

r ::; � = � _ It follows that the maximum ratio is q = 1 � = � I -Let r E

{I, 2, . . . , q } and let ar E

{I, 2 , . . . , n } b e the greatest first term of a

progression with ratio r. Then

ar + (r - 1) k :S n, (1) so ar is equal to the number of arithmetic progressions of ratio r from the set {I, 2, . . . , n} . Hence

N(n, k) = al + a2 + . . . + aq • Because al = n - k + 1 , using (1) gives

N :S al + n - k + n - 2k + . . . + n - (q - l)k = (q - l)qk = nq - k + 1 - k (1 + 2 + . . . + (q - 1) ) = nq - k - + 1 = 2

q2 ( k) = -2k + n + '2 q + 1 - k.

It suffices to prove that

� k + (n + �) q + 1 - k ::; -�q2 + ( n + �) q

6. 2. SOLUTIONS 259

which is equivalent to k � 1 q ::; k � 1 q2 . This inequality is clearly true, so we are done.

Alternative proof. Note that n � k � 2. The arithmetic progressions with k terms and ratio r = 1 are

1 , 2, . . . , k 2, 3 , . . . , k + 1

n - (k - 1 ) , . . . , n so there are n - k + 1 such progressions.

The arithmetic progressions with k terms and ratio r = 2 are

1 , 3 , . . . , 2k - 1

2, 4, . . . , 2k

n - 2(k - 1) , . . . , n so a total of n - 2(k - 1) progressions.

It follows that q Nk = I: [n - d(k - 1)] ,

d=l where q is the greatest ratio of an arithmetic progression with k terms from the set {I , 2, . . . , n} . We have proved that q = [� = �] , hence

Nk = t (n - d(k - 1) ) = nq _ q(q + l�(k - 1) .

d= l It suffices to show that

(k - 2)q2 + kq + 2 - 2k � O . (1)

The roots of the quadratic polynomial from the left-hand side of the inequality are

-2(k - 1) ql = k _ 2 and q2 = l.

Note that ql < 0 for k > 2. Since q = [� = �] :2: 1, the inequality (1) is true for

all k > 2. If k = 2, then is easy to check that the claim holds as well. ( Titu Andreescu and Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No.

1 ( 1982) , pp. 104, Problem C4:2)

35. Denote by Pn (x) the (unique) polynomial of degree n such that

Pn (k) = Fk for k = n + 2, n + 3 , . . . , 2n + 2. (1)


We are going to show that Pn (2n + 3) = F2n+3 - 1 for all n � O . Clearly, Po (x) = 1 and the claim is true for n = O. Suppose it holds for Pn-I (x) ,

and consider P n (x) . The polynomial

has degree at most n - 1 . In view of (1 ) ,

for each k = n+1, n+2, . . . , 2n. Therefore Q (x) and Pn-I (x) agree at n distinct points , and hence Q (x) = Pn-1 (x) for all x. In other words, Pn (x + 2) = Pn (x + 1) + Pn-I (x) for all x. Combined with the inductive hypothesis P2n-1 (2n + 1) = F2n+1 - 1, this implies

Pn (2n + 3) = F2n+2 + F2n+1 - 1 = F2n+3 - l . ( Titu Andreescu, IMO 1983 Shortlist)

36. For an arbitrary integer n we have

Xn+1 = aXn + (3Xn-l , Xn = aXn-1 + (3Xn-2 , n � 2 .

(1)

This is a system of linear equations with unknowns a and (3. The solutions are

and

and

�a XnXn-1 - Xn+IXn-2 a = - = 2 � Xn-1 - XnXn-2

-(3 - _ �(3 _ X� - Xn-IXn+1 - � - X�_I - XnXn-2 Since a and -(3 are constant, the conclusion follows. (Dorin Andrica)

37. Let �n = an - an-l for n � 2. Because 1 2 k-I an+1 = k (b

+ an + an-I + . . . + an-k+2 )

we have

where Al = an-I + an-2 A2 = a�_2 + an-2an-3 + a�_3

\ k-2 k-3 k-2 Ak-I = an-k+2 + . . . + an-k+2an_k+1 + an-k+l ·

(1)

6. 2. SOLUTIONS 261

Note that a1 = a2 = . . . = ak-1 = 0 implies an > 0 for n � k , so A1 , A2 , . . . , Ak-1 � O .

On the other hand, 6.1 = 6.2 = . . . , 6.k-1 = 0 and from relation (1) it follows that 6.n > 0 for all n � k. Hence the sequence (an)n�k is increasing.

We prove that an < 1 for all n � 1 . Assume that

so

since b < l . Therefore the sequence is upper bounded, so is convergent. Let x = lim an . Then n-+oo

Xk-1 + Xk-2 + . . . + x2 - (k - l)x + b = 0 If b = 0, then an = 0 for all n � 1 , hence x = O.

(2)

If b E (0 , 1) , we prove that the equation (2) has a unique solution in the interval (0, 1) .

Let I : [0 , 1] -+ JR,

I(x) = Xk-1 + Xk-2 + . . . + x2 - (k - l)x + b. Then 1(0) = b, 1 (1) = b - 1 ,

1 (0)/ (1 ) = b(b - 1) < 0

hence the equation (2) has an odd number of solutions in the interval (0 , 1 ) . The function I is twice differentiable and since

I" (x) - (k - l) (k - 2)Xk-3 + 2 > 0, x E (0, 1) ,

I is concave up on (0, 1) . It follows that equation (2) has at most two solution in (0, 1) therefore the conclusion follows.

Remark. The claim that equation (2) has a unique positive solution it follows from

1(0)/(1 ) = b(b - 1) < 0 and from the fact that I(x) = xk-1 + Xk-2 + . . . + x2 - (k - l)x + b has a unique variation of sign (Descartes) .

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2 (1979) , pp . 56, Problem 3866)

38. From the relation iii) we obtain an+1 - an bn+1 - bn - 0 an + bn - , n � 1 ,

262

then

Using ii) yields

6. COMPREHENSIVE PROBLEMS

1· ( an ) l' an+l - an 1m -- = 1m n-+oo bn n-+oo bn+l - bn On the other hand, by the Stolz-Cesaro theorem we have

1· an+l - an l' an 1m = lm -n-+oo bn+l - bn n-+oo bn It is easy to see that relations (1 ) and (2) imply

1· an 0 lm -b = , n-+oo n as desired.

(1)

(2)


39. Consider the determinant 0 Xl X2 Xn Xl 0 X2 Xn

� (Xl , X2 , . . . , X n) = Xl X2 0 Xn

Xl X2 X3 0 Note that

� (0, X2 , . . . , Xn) =�(XI , 0 , X3 , . . . , Xn) = . . . =�(Xl , X2 , . . . , Xn-I ' 0) = O. Moreover, if Xl + X2 + . . . + Xn = 0, then �(XI ' X2 , . . . , xn) = 0, therefore

�(XI ' X2 , · · . , xn) = aXIX2 . . . Xn (XI + X2 + . . . + xn) , for some real number a . By identifying the coefficients of XIX2 . • . Xn from both sides we obtain a = l .

Hence

We have

then

Uk+l 1 + UI . . . Uk = 1 + -----'--

UI + . . . + Uk UI + U2 + . . . + Uk+l UI + U2 + . . . + Uk

n II UI + U2 + . . . + Un+l (1 + UI . . . Uk ) = ----------'--k=2 UI + U2

Since U2 = -uI , we obtain

1 IIn UI + . . . + Un+l - (1 + UI . . . Uk) = ( ) n nUl 1 - UI k=2

6.2 . SOLUTIONS

Let U = lim Un and note that n-+oo

It follows that

1. Ul + . . . + Un+l 1m = U. n-+oo n

1 n lim - II (1 + Ul . . . Uk) = (U

) . n-+oo n Ul 1 - Ul k=2

263

( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2 (1978) , pp. 52, Problem 3533)

40. Let a = lim an and let c > O. There is an integer n(c) > 0 such that n-+oo bn k + 1 an k + 1 ( a - c -- < - < a + c -- for n > n c) k - 1 bn k - 1 -

Since bn > 0 it follows that k + 1 k + 1 abn - c k _ 1 bn < an < abn + c k _ 1 bn for all n ;::: n (c) .

Then k + 1 a(bn+1 - bn ) - c -k -(bn+1 + bn) < an+l - an < - 1

k + 1 < a(bn+1 - bn) + c k _ 1 (bn+1 + bn ) ,

and, dividing by bn+1 - bn > 0, we obtain

k + 1 bn+1 + bn an+l - an k + 1 bn+1 + bn a - c -- · < < a + c -- - , k - 1 bn+ 1 - bn bn+ 1 - bn k + 1 bn+ 1 - bn

for all n ;::: n(c) .

From relation ii) we deduce

hence

Therefore

as desired.

bn+1 + bn k - 1 -'-------::-- < -- , n ;::: 1 , bn+1 - bn - k + 1

an+l - an ( a - c < b < a + c for all n ;::: n c) n+l - bn

(Dorin Andrica, " 0 reciproca a teoremei Stolz-Cesaro §i aplicatii ale acesteia" , Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp. 6-12)

41. Let Xl = k + Jk2 + 1 and X2 = k - Jk2 + 1. We have 1 1 1

IX2 1 = - < - < -Xl 2k - 2 '

264

so


- 0) n � x� � 0) n Hence

n n n ( l) n n ( l ) n

. n n Xl + X2 - 1 < Xl + 2" - 1 < an < Xl - 2" + 1 < Xl + X2 + 1

for all n � 1 . The identities

X�+l + X�+l = (Xl + X2 ) (Xr + x�) - XIX2 (X�-1 + X�-l ) =

= 2k(xr + x�) + (X�-l + X�-l ) , n � 1

show that xr + x� is an integer for all n, and since an is an integer, it follows that an = xr + x� for all n � 0, and that an+l = 2kan + an-l for all n � 1 .

Then

and

Therefore

( Titu Andreescu)

42 . Assume by way of contradiction that f(xo) =I- 0 for all X E R Clearly, f(p) (x) = 0 for all integers p > O . For p = 0, 2 , 4, . . . , 2 (n - 1) , we obtain

for all X e R

sin alX + sin a2x + . . . + sin anx = 0 ai sin alx + a§ sin a2x + . . . + a� sin anx = 0

2(n-l) . 2(n-l) . 2(n-l) . 0 al sm alX + a2 sma2X + . . . + an sm anx = ,

(1)

Consider a number X such that at least one of sin al x, sin a2X, . . . , sin anx is not zero. Then the homogeneous system of linear equation (1) has a nontrivial solution, hence the determinant is zero:

�8 =

1 a§

2 (n-l) 2 (n-l ) al a2 2(n-l) an

= II (a; - a; ) = O. i , i=l i >i

6.2 . SOLUTIONS 265

It follows that a� = aI for some k =I- l i.e. lak I = l ad , which is a contradiction. The solution is complete.

(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 1-2 (1980) , pp. 69, Problem 4148)

43. Assume by way of contradiction that g : JR -+ JR, g(x) = sin f(x) , is periodical. We have g' (x) = f' (x) cos f(x) , and since f and f' are continuous then g' is also continuous.

Note that if g is periodical, then g' is periodical. Moreover, g' is continuous, so it is bounded.

Consider the sequence Yn = (4n + 1) �, n � 1. Function f is continuous and lim Xn = 00 , hence f(xn) = Yn for n sufficiently large. n-+oo

Then g' (xn) = f' (xn) so

lim g' (xn) = lim f' (xn) = 00 n-+oo n-+oo which is a contradiction, since g' is bounded. This concludes the proof.

(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1978) , pp. 54, Problem 3544)

44. i) Let f be injective and assume by way of contradiction that a is rational. Hence there are integers p, q with a = P. and gcd(p, q) = 1 . q

Then f(q) = f (2q) = 0 a contradiction. Therefore a is irrational. Conversely, let a be irrational and assume by way of contradiction that f (m) =

f(n) for some integers m =I- n � O. Then {am} = {an} so am - [am] = an - [an] . We obtain

a = [am] - [an] E Q, m - n

which is a contradiction, and the conclusion follows. ii) Let p, q be relatively prime integers such that a = P. . We have q

f (n) = {an} = { r; } . By the division algorithm, there are integers q and r such that

Then

n = tq + r, r E {O, 1, 2, . . . , q - I} .

f(n) = { P(tqq+ r) } = {pt + r: } = r :} = f(r) .

We prove that f(O) , f(l) , . . . , f (q - 1) are all distinct. Indeed, if f (i) = f (j ) , then

{ i: } = { j;}


It follows that (i - j )p is an integer. Note that (p, q) = 1 and Ii - j l < q, hence q i - j = 0, so i = j .

Therefore { I 2 q - l } M = {f(O) , f (I) , . . . , f (q - I)} = 0, - , - , . . . , -q q q since M has q elements.

(Dorin Andrica, Romanian Winter Camp, 1984; Revista Matematica Timi§oara (RMT) , No. 1 (1985) , pp. 67, Problem 3)

45. Assume by way of contradiction that there is a number t > 0 such that

Then

g(x + t) = g(x) , x E R

f(x + t) + f(xO + to) = f(x) + f(xO) , x E JR, hence

f(t) + f(tO) = 2f(0) = 2M. From the relation iii) it follows that

f(t) = f(tO) = M

and then t = kI T and to = k2T,

for some integers kl ' k2 -:f. O. This gives

a contradiction. (Dorin Andrica)

k2 0 = ki E Q,

46. We start with an useful lemma. Lemma. If 0 is an irrational number, then the set

M = {mO + nl m, n integers}

is dense in JR. Proof. We prove that in any open bounded interval J � JR\ {O} there is an element

of M, i .e . I n M -:f. 0 . Let J be such an interval and without loss of generality consider J C (0, 00) .

There is an integer n (J) such that 1

n (J) J C (0, 1 ) . We consider two cases:

6 . 2. SOLUTIONS 1 . 1 . J1 = n(J) J = (O , e) , wIth 0 < e < 1 .

Let N be an integer such that � < e and consider the numbers

{B} , {2B}, . . . , {NB} . There are p, q E {I , 2, . . . , N, N + I} such that

On the other hand,

1 0 < {pB} - {qB} :::; N

{pB} - {qB} = [qB] - [PB] + (p - q)B E M,

hence JI n M -:f.

0. It follows that n(J) ( {pB} - {qB}) E J n M, as desired.

1 . 2. J1 = n(J) J = (a, b) wIth 0 < a < b < 1 .

Then 0 < b - a < 1 and by case 1) , there i s c E M

such that 0 < c < b - a. Let

no = [�] + 1 .

Then a < noc < b and noc E M

n J1 . Likewise, J n M -:f.

0, as desired. The lemma is now proved.

267

a) Let A E [min f(x) , max f(X)] . Hence there is Xo E 1R such that f(xo ) = A. xEIR xEIR From the lemma we deduce that there are sequences (Xn )n2:1 and (Yn)n2:1 such that

lim (xn + YnT) = Xo . n-+oo The function f is continuous, so

lim f(xn + YnT) = f(xo) = A. n-+oo Note that f (xn + YnT) = f(xn) , therefore

lim f(xn) = A, n-+oo as desired.

b) Let B be an irrational number and consider the function g(x) = f (xB) , x E R The number � is irrational and a period for the function g. Using the result from a) , there is a sequence (xn)n2:1 of integers such that

as desired.

A = lim g(xn) = lim f (Bxn) , n-+oo n-+oo

(Dorin Andrica, "Asupra unor §iruri care au multimea punctelor limita intervale" , Gazeta Matematica (GM-B) , No. 11 (1979) , pp. 404-406)

47. i) Let u = x + Y = z + v and assume that x < Y and z < v. Then we have u u x < - z < - Y = u - x and v = u - z . 2 ' 2 '


Suppose, by way of contradiction, that there is a number A > 1 such that

xA + yA = ZA + VA .

Consider the function f : (0, u) -+ (0, (0) ,

f (t) = tA + (u - t) >.., and note that f is differentiable. We have

l' (t) = A[tA-1 - (u - t) A-IJ , t E (0, u) ,

and since A > 1 , it follows that f is increasing on (0, �) . Both x, z are in (0, �) , so f(x) -:f. f(z) , because x -:f. z. This implies XA + yA -:f. ZA + v\ which is a contradiction ..

ii) Because p is a prime, by Little Fermat's Theorem we have

hence

aP - a == bP - b == cP - c == rF - d == ° (mod p) ,

- (aP - a) + (bP - b) - (cP - c) + (dP - d) == ° (mod p) . From aP + bP = cP + dP , we deduce that

a - c + b - d == ° (mod p)

By i) we note that a + b -:f. c + d, therefore

la - c + b - dl � p,

and then la - cl + I b - dl � p,

as desired.

(1)

( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp. 55, Problem 3550)

48. i) Consider the function <.p : (0, 00) -+ JR,

<.p (t) = t ln t - t.

The Mean Value Theorem yields

for some (J E (y , x) , then

It follows that

<.p(x) - <.p (y) = <.pI ((J) x - y

x In x - y In y - x + Y = In (J. x - y

and so

6. 2. SOLUTIONS

XX = (eB)X-Y . yY

The inequalities y < B < x, imply

as desired.

XX (ey)X-Y :::; y :::; (ex)X-Y , y

ii) Setting x = k + 1 and y = k yields ek (k + l)k

k + 1 < kk < e, k > O .

Multiplying these inequalities from k = 1 to k = n gives

and then

Hence

as claimed.

en 2 . 32 . 43 • • • (n + l)n n -- < < e n + 1 1 . 22 . 33 . . . nn '

en 1 en ---- < - < . (n + 1)n+1 n! (n + l)n

(n + l)n , (n + 1)n+1 -'-----'-- < n. < , en en

269

(Dorin Andrica, Gazeta Matematidi (GM-B), No. 8 (1977) , pp. 327, Problem 16820; Revista Matematica Timi§oara (RMT) , No. 1-2 (1980) , pp. 70, Problem 4153)

49. i) Note that f is bijective and increasing, ' therefore f is continuous. In this diagram, we have

S, = I.' f (t)dt, S2 = ld r' (t)dt,

hence Sl + S2 = bd - ac,

as desired. y

c

o a b x


ii) Again, f is continuous and the diagram shows that

S1 - S2 = (b - a)e - (d - e)a = be - ad,

as desired. y

d

e

o a b x

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1 (1981) , pp. 62, Problem 4363)

50. i) Let c > O . There is (>r > 0 such that I I1(t) 1 < c for all t > (>r and there is <52 > 0 such that

-x 1 � e < c ° ' x > U2 . 1 1 et ll1(t) ldt

For x > <5 = max (<51 , <52 ) , we have

I e-x 1" e'l' (t)dt l :'0 e-X 1" e' ll' (t) l dt + e-X 1� e' ll' (t) l dt <

< c + ce-X (eX - e01 ) < 2c,

and the conclusion follows. ii) We start with the following lemma. Lemma. If <.p : [0, 00) is a differentiable function with continuous derivative such

that lim (<.p(x) + <.pI (x)) = a, x-+oo

then the limit lim <.p(x) exists and equals a. x -+oo Proof. Without loss of generality we may assume that <.p(0) = O . Define w : [0 , 00) --+ JR, w (t) = <.p(t) + <.pl (t) - a. Function w is continuous and

lim w (t) = O. t-+oo Note that

6. 2. SOLUTIONS

then e"<p(x) = e"a - a + 1" etw (t)dt,

by integrating on [0, x] . It follows that

t.p(x) = a - - + e-X etw(t)dt, a lox eX 0

and from i) we obtain lim t.p(x) = a. x-+oo

Denote

and note that fm(x) = fm-l (X) + f:n-l (X) , m > O .

Using the lemma, lim fn (x) = A implies x-+oo lim fn-l (X) = A. X-+OO

Applying the same argument, we finally obtain

lim fo (x) = lim f(x) = A, x-+oo x-+oo as desired.

(Dorin Andrica)

5 1 . Note that

so

Then

(f(x) - c) (f (x) - d) :::; 0

f2 (X) + cd :::; (c + d)f(x) , x E [a, b] .

l' f2 (X)dx + (b - a)cd ::; (c + d) l' f(x)dx, and, since the left-hand side is 0, by hypothesis we obtain

and then

o ::; (c + d) t f(x)dx,

1 lb 0 :::; --d f(x)dx. c + a

On the other hand,

Then

( C + d) 2 (C + d)2 f (x) - -2- = f2 (X) - (c + d)f(x) + 4

� 0, x E [a, b]

lb (C + d)2 lb f2 (X)dx + (b - a) � (c + d) f (x)dx,

a 4 a

271

(1)

272

so

Hence


1 [ (c + d) 2 ] 1 r b ( C + d) 2 - (b - a) cd + (b - a)

4 � c + d } a

f (x )dx.

b - a (c - d) 2 1 lb - - � - f(x)dx. 4 c + d c + d a

From (1) and (2) the conclusion follows.

(2)

( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2(1986) , pp. 76, Problem 6004)

52. By the Mean Value Theorem there are numbers Cx E (x, b) and c� E (a, x) such that

F(x) = (x - a) (b - x)f(cx ) + (x - b) (x - a)f(c� ) = = (x - a) (b - x) [J(cx ) - f(c� )] .

We have c� < Cx for all x E [a, b] . Since f is monotonic f(cx ) - f(c� ) has constant sign on [a, b] . Moreover, (x - a) (b - x) � 0 for all x E [a, b] and the conclusion follows.

( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 (1985) , pp. 63, Problem 5505)

53. a) The function h

is continuous, hence it is a derivative. Let F be an antiderivative of

h. By the Mean Value Theorem we have

F(k + 1) - F(k) = h(Ck ) for some Ck E (k, k + 1) .

Since h

i s nonincreasing, h(k + 1) ::; h(Ck ) ::; h(k)

so h(k + 1) � F(k + 1) - F(k) �

h(k) .

Summing these inequalities from k = 1 to k = n yields

an+1 -h(l)

� F(n + 1) - F(l) � an , hence

Because

;:n+l an+1 -

h(l) � 1 g' (x)f (g(x))dx � an , n �

l.

;:n+1 19(n+1) g' (x)f (g(x))dx = f(x)dx,

1 g (l) the conclusion follows.

b) Setting f : (O, �) --+ JR, f(x) = - cot x and g : [1 , 00) --+ JR, g (x) = � , we

obtain h(x) = -; cot � , which is decreasing on the interval [1 , 00) . x x

We have

Hence

6 .2 . SOLUTIONS

!.g(n+1) ;: n�l I(x)dx = - cot xdx =

g (l) 1

. I n�l ( • 1 ) . = - In(sm x) = - In sm -- + In(sm 1) . 1 n + 1

lim !.g(n+1)

I(x)dx = lim [- In (Sin _1_) + In (sin 1)] = +00 n-+oo g(l ) n-+oo n + 1

Using the left-hand side inequality from a) it follows that

n I l lim an = lim � k2 cot - = 00.

(Dorin Andrica)

54. Recall that

n-+oo n-+oo L....t k k=l

at - 1 lim -- = ln a. t-+o t For c > 0 there is 8 > 0 such that

at - 1 In a - c < -- < In a + c, I t I < 8 t

273

(1)

Function 1 is integrable on [0 , 1] hence is bounded. Let M

> 0 such that I(x) ::; M

for all x E [0, 1] . There is an integer no such that

� f (�) < J for all n � no and k = 0, 1 , . . . , n .

The inequality (1) gives

a*f (�) - 1 In a - c < 1 ( k ) < In a + c, -I -n n

k = 0, 1 , . . . , n

then

It follows that

n L (a*f (� ) - l) In a - c <

k=l < In a + c. t � (�) n

k=l n n

L (a*f ( �) - 1) lim _k=_l ______ = In a. n-+oo t � (�)

k=l n n

274

On the other hand


n 1 ( k ) 11 lim

L -f - = f(x)dx, n-+oo k=l n n °

therefore

nl�moo (t, a�f (�) - n) = ([ f(X)dx) In a. (Dorin Andrica, Romanian Mathematical Regional Contest " Grigore Moisil" ,

1997)

55 . 1) We have

1. f ( ) - l' ( sin nk7rt) 2 _ ( 1' sin nk7rt) 2 1m n X - lm . - 1m ---x-+ k� t-+l sm k7rt t-+l sin k7rt

= ( lim nk7r cos nk7rt) 2 = n2 [( _1)nk-k] 2 = n2 . t-+l k7r cos k7rt

Hence fn is continuous on JR* if and only if an,k = n2 for all k E Z*. 2) It is known that ha : JR -+ JR, { a cos - x -:f. 0 ha (x) = a '

0, x = 0

is a derivative. Recall the identity

( Si�nx) 2 = n + 2

L cos 2 (k - l)x, x E JR \ {m7r, m E Z } . sm x 19<k:Sn

Then

fn (x) = { n, an , x -:f. 0 + 2 x = o 19<k:Sn

h2(k-l) (x) ,

so fn is a derivative function if and only if an = n and an,k = n2 for all k E Z*. (Dorin Andrica, Romanian Mathematical Regional Contest "Grigore Moisil" ,

1995)

56. We have

fo ,o (x) = { 1, cO ,o ,

if x -:f. 0 if x = 0

so fo ,o is a derivative if cO,o = l . Functions

( ) sm - , x -:f. 0 { . 1

u x = x 0 , x = 0

are derivatives , so Cl ,O = CO,l = o.

and v (x) = { 1 cos - , x

0,

x -:f. o x = o

6 . 2. SOLUTIONS

The function

",, (xl = { . 1 1 SIn - cos - , X X

Cl , l , x # 0

= � { sin � , x # 0

x = 0 2Cl ,l , X = 0

= - u 2x + -1 ( ) 1 { 0, 2 2 2Cl ,l ,

is a derivative if Cl , l = O . For p, q > 1 consider the differentiable function G : JR --+ JR,

with

G' (x) =

x2 smP - cosq - x # 0 G(x) = x x '

{ . 1 1 0, x = 0

1 1 2x sinP - cosq -x x . 1 1 -p smP-1 - cosq+l -+ x x

+q sinP+1 .!. cosq-l .!. , x # 0 0,

x x x = o

Function { 2x sinP .!. cosq .!. x 1--+ 0, x x ' x # O

x = O is continuous, hence a derivative. Therefore { -p sinp-l .!. cosq+l .!. + q sinp+l .!. cosq-l .!. , g(x) = x x x x 0,

is a derivative. Using the fact that sin2 t = 1 - cos2 t and cos2 t = 1 - sin2 t, we

g(x) = -p smP-l X cosq-I

X + (p + q) smP+l X cosq-I x ' x # 0 { . 1 1 . 1 1

Hence

Therefore

0, x = 0

g(x) = -P!p-l , q-l (x) + (p + q)!p+l ,q-l (x)+ { 0, x # 0 +

PCp-l ,q-l - (p + q)Cp+l , q-l , X # 0

= { 0, x # 0

(p + q)Cp-l ,q+l - qCp-l ,q-l , X = 0

PCp-l ,q-l = (p + q)Cp+l ,q-l qCp-l , q-l = (p + q)Cp-l ,q+l ,


and so P Cp+l q-l = -- Cp-l q- l , p + q , q

Cp-l q+l = -- Cp-l q-l · , p + q ,

For k , l 2: 1 we obtain 2k - 1 2k - 1 2k - 3

C2k ,2l = 2k + 2l C2k-2,2l =

2k + 2l .

2k + 2l _ 2 C2k-4,2l = . . . =

2k - 1 2k - 3 1 = 2k + 2l

. 2k + 2l - 2 . . . 2l + 2 CO,2l =

2k - 1 2k - 3 1 2l - 1 = 2k + 2l

. 2k + 2l - 2 . . . 2l + 2 . �CO,2l-2 = . . . =

2k - 1 2k - 3 1 2l - 1 2l - 3 1 = 2k + 2l . 2k + 2l - 2 ' " 2l + 2 . �

. 2l - 2 ' " "2co,o =

1 · 3 . . . (2l - 1) . 1 · 3 . . . (2k - 1) 2k+1 (k + l) !

Note that C2k,2l+1 = A· CO , I , C2k+I ,2l = B . CI ,O and C2k+I ,2l+1 = C · CI , I , where A

, B, C are rational numbers and CO, 1 = CI ,O = CI , 1 = 0, therefore

C2k,2l+1 = C2k+I ,2l = C2k+I ,2l+1 = 0, for all k, l 2: o. To conclude,

Cp,q = { (2k - 1) ! ! (2l - I) ! ! -'------'-----, if p = 2k and q = 2l

2k+1 (k + l ) ! 0, otherwise

(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 1 (1986) , pp. 78, Problem 5773)

57. We start with a useful lemma. Lemma. For any real numbers Xl , X2 , • • • , Xn , we have

1 cos Xl . COS X2 . · · COS Xn = - � COS(±XI ± X2 ± . . . ± Xn) 2n L.t where the sum is taken over all 2n possible choices of signs.

The proof can be made by induction on n. On the other hand recall that function ga : JR -+ JR,

is a derivative.

{ a cos - X =I- 0 ga (x) = x '

0, X = 0

We have two cases. 1 q qn-l

Case 1 . If q 2: 2, set Xl = - , X2 = - , . . . , Xn = -- , X =I- O. From the lemma we obtain

X X X 1 q qn-l 1 1 cos - . cos - . . . cos -- = - L cos(±l ± q ± . . . ± qn-l ) _ X X X 2n x

hence

6. 2. SOLUTIONS

1 � { �O!n (q) , X =1= 0 fn (x) = 2n L.t g±l±q±" ' ±qn - l (x) + 2n O!n (q) , X = 0

277

where O!n (q) is the number of choices of signs +, - such that ±1±q±q2 ± . . . ±qn-l = 0, because in the sum are considered only the choices of signs such that ± 1 ± q ± q2 ± . . . ± qn-l =1= O.

Note that if ±1 ± q ± q2 ± . . . ± qn-l = 0, then q < 1 , which is false. Hence O!n (q) = 0 and therefore fn is derivative if and only if an (q) = O.

Case 2. If q = 1 then for n odd we have O!n (l) = 0 because ±1 ± 1 ± . . . ± 1 cannot

be 0, having an odd number of terms. If n is even, let n = 2m. There are choices of m signs - and m signs + so

To conclude, we have

1 0,

O!n (q) = °i (n) 2n !! ' 2

if q � 2 if q = 1 and n odd

if q = 1 and n even

(2mm)

(Dorin A ndrica, Romanian Mathematical Regional Contest " Grigore MoisH" , 1999)

58. 1) Function f is continuous, so it is a derivative. Let F be an antiderivative of f. For x =1= 0 we have

( x2 F G) ) ' = 2xF G) - f G) therefore

f G) = 2xF G) - (X2 F G) ) ' for all x # O.

Consider the function h : JR -+ JR,

and note that

h(x) = { 2XF(1/X) , x =1= 0 2M(f) , x = O.

lim h(x) = lim 2xF (�) = lim � rY f (s)ds = 2M(f) . �;:g �;:g x y-HX) Y io Hence h is continuous so it is a derivative function. Let H be an antiderivative of h.

We have

g(x) = { 2xF(1/x) - (x2 F(l/x))' , x =1= 0 M(f) , x = 0


h(x) = { (X2 F(l/x))" X =1= 0 M(f) , X = o.

The function u : JR -+ JR,

u(x) = { (x2 F(l/x))" X =1= 0 M(f) , x = 0

is a deri vati ve since U : JR -+ JR,

U(x) = { �� F(l/x) , : � �. is differentiable and U' = u.

It follows that function G = H - U is an antiderivative of g , as desired. 2) We prove that

lim � rt f(x)dx = M(f) . t-*oo t io

For t > 0 there is an integer n = n(t) and a number a = a(t) E [0, T] such that t = nT + a.

Then j' t inT it f(x)dx = f (x)dx + f(x)dx

o 0 nT On the other hand

inT n-l /. (k+l)T f(x)dx = � f(x)dx.

o k=O kT Setting x = B + kT yields /. (k+l)T

f(x)dx = rT f (B + kT)dB = rT

f(B)dB kT io io

so relation (2) gives rnT rT

io f (x)dx = n

io f (x)dx

Setting B = x - nT yields

it it-nT ia(t) f(x)dx = f(B + nT)dB = f(B)dB

nT 0 0

The relation (1) becomes

and hence

rt rT ra (t) io

f(x)dx = n io

f (x)dx + io

f(x)dx

l it n(t) iT 1 ia(t) -t

f (x)dx = - f (x)dx + - f(x)dx for all t > O. o t o t 0

(1)

(2)

(3)

6 . 2. SOLUTIONS

We have

1 1 (aCt) I 1 (aCt) 1 (T o � t io f (x)dx � t io

I f (x) ldx � t io I f (x) l dx t4°o 0

hence

Moreover

therefore

as claimed.

1 ia(t) lim - f(x)dx = O. t-+-oo t 0

lim n(t) = lim n(t) t-+-oo t t-+-oo n(t)T + a(t)

= lim 1 1 t-+-oo a( t) T + n(t)

T '

l it l iT M(f) = lim - f(x)dx = -T f(x)dx, t-+-oo t o o

The proof is similar for the case t -+ - 00 .

(Dorin Andrica)

279

59. The answer is negative. Indeed, consider the derivative function f : JR -+ JR,

f (x) = { cos � , x =/- 0 0, x = 0

U sing the previous problem, the function

{ I cos � I x =/- O g(x) = :;;2 , x '

/ I x = O

is also a derivative. Therefore the function

x =/- 0 x = O

is not a deri vati ve and we are done. (Dorin Andrica, Romanian Mathematical Regional Contest " Grigore Moisil" ,

1992)

60. The answer is negative. For example, consider the derivative functions iI , h : JR -+ JR,

iI (x) = { 1 - cos -, x 0,

x =/- 0 x = O

h (x) = { 1 cos -x ' 0 ,

x =/- O x = O

280

Then


f(x) = ma:x(h (x) , 12 (x) ) = { Icos � I ' x # 0

0, x = 0 which is not a derivative function (see problem 59) .

Alternative proof. Consider the derivative functions

h (x) = { cos �, x =I- 0 0, x = 0

12 (x) = { - cos � , x =I- 0 0, x = 0

{ COS2 � x =I- 0 h (x) =

_1 x '

x = O 2 '

Assume for the sake of contradiction that the statement is valid. Then

9 = ma:x(h , 12 , h) - ma:x(h , h) is a derivative function, which is a contradiction, since { 0,

g(x) = � 2 '

Therefore the answer is negative.

x =l- O

x = O

(Dorin A ndrica, Romanian Mathematical Regional Contest " Grigore Moh 1997)

"I take great pleasure in recommending to all readers -Romanians or from abroad - the book of professors Titu Andreescu and Dorin Andrica. "

"All featured problems are interesting, with an increased level of difficulty; some of them are real gems that will give great satisfaction to any math lover ... "

"One more time, I strongly express my belief that the 360

mathematics problems featured in this book will reveal the beauty of mathematics to all students and it will be a guide to their teachers and professors. "

Professor loan Tomescu Department of Mathematics and Computer Science University of Bucharest Associate member of the Romanian Academy

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360 problems for mathematical contests [andreescu] 9739417124 (1)

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