360 Problems for Mathematical Olympiads 9739417124

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'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 e-mail: [email protected]

www.gil.ro

IMPRIMERIA

�

ARTA IV GRAFICA I ��LIBRIS

Calea $erbanVodti 133,S.4,Cod 70517,BUCURE$TI Tel.: 336 29 11 Fax: 337 07 35

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 math­ ematics 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 Math­ ematical 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 recog­ nized 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 Ro­ manians 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 under­ graduate 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

{Cl ' C2 , , cn } .

Let C be a set of characters We call word a string of at most m characters, m :::; that does not start nor end with How many words can be formed with the characters of the set C? 1.

n n,

. . •

Cl.

. . . , 5(n ) n x, y , x y,

The numbers 1, 2, are divided into two disjoint sets. Prove that these > sets contain at least pairs is also an such that the number element of the set which contains the pair. 2.

x-y [ b] 3. Let a l , a2 ,I. . . ,an be distinct numbers from the interval a, and let be a } { permutation of , 2, . . . , n [. b] [ b] Define the function f a, a, as follows: 1 ,n a { q i if x = �i' x = f( ) x otherwIse x, where f[h] Prove that there is a positive integer h such that f[h](X) a

:

-t

()

i

=

fafa ... af. htimes

'----v-----"

2x x,+y,yz2 areynonzero 2Z +numbers 2x x3withy3x + yZ3+ z = 0, then 2 + Z2 real x + y + y + Z + Z + x = -yz + -zx + xy-. b d b d Prove that Let a, , c, be complex numbers with a + + C + a3 + b3 + c3 + d3 = 3 (abc + bcd + cda + dab) . b3 3 b b 3 6.b Let a, , c be nonzero real numbers such that a + + c = 0 and a + + c = a5 + 5 + c5• Prove that 4.

Prove that if

---

5.

7. Let

=

O.

b d a, b, c,2d( be4 integers. Prove that a + + c + divides a + b4 + c4 + �) _ (a2 + b2 + c2 + 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 ) . =

1 1 . Solve the equation

� x + J4X + V16X + l·· + J4nx + 3

-..;x =

1.

1 2 . Solve the equation

b where a, ,

x + a + -Jx + b + -Jx +

-J c

c =

x+a+b-

-J

c,

are real parameters. Discuss the equation in terms of the values of the

parameters.

b be distinct positive real numbers. Find all pairs of positive real a and ( ) numbers x, y , solutions to the system of equations { xx24 - yy24 ax{/a-2 byb2 . 13. Let

=

_

14. Solve the equation

=

_

[ 25x4- 2 ] l3x3+ 4 ' =

[a] denotes the integer part of a real number a. 1 + V5 . 15. Prove that If a � -- -' then

where

2

11

1. 1. PROBLEMS

x, y, z are real numbers such that x S + y S + Z S f:. 0, then the 2xyz - (x + y + z) ratio xS+yS+ZS equals � if and only if x + y + z 16.

Prove that if

=

17.

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

18.

1 9.

O.

2 .JX2 - 4 + . . . + n JX n - n2 "21 (xl + X2 + . . . + xn) .

1 -+x -y = 9

=

Find the real solutions to the system of equations 1 1

(

1

-

?Ix

+

1

-

�

)(

1+

1

-

?Ix

)(

1+

1

-

�

Solve in real numbers the system of equations

) - 18 -

y22 + u22 + v22 + w22 44x - 1 x2 + u2 + v2 + w2 = 4y 1 x2 + y2 + v 2 + w 2 4u - 1 x2 + y2 + u2 + w2 4 v - 1 x +y +u +v w- 1 20. Let aI, a2 , a s, a4 , a5 be real numbers such that al� + a2 + a s + a4 + a5 0 and � � � i l l l�max i n � -n +2 -1 v'2 v'3 \Iii

n ;::: 2. 2 9. Prove that n ( 1 - 1 / vn) + 1 > 1 + �2 + �3 + . . . + .!.n > n (\In + 1 - 1 ) for any positive integer n . (0 1 ) na12a2 ' ." . .an Prove that 30. Let a I , a2 , . . . , an E , and let tn = a1 + a + + an Ln loga. tn ;::: (n - l) n. i= l 3n there is at least a perfect cube for any intege 31. Prove that between n and n ;::: 1 0. 32 . Compute the sum Sn = �n (- 1 ) k. k=l 33. Compute the sums: lk S T k ) ( ( l a) n = � b n = � (k l )(k 2)(k 3) + + 2) (�); ) + + + (�). .

"'"

1c(Ic+l) 2

13

1 1. . PROBLEMS

Show that for any positive integer n the number n n 1 1 .3+ . . . + 2 +

34.

: ) 22 n - 2

; ) 2n

C C is the sum of two consecutive perfect squares. 35.

Evaluate the sums:

36.

Prove that

12 (�) 32 (�) 52 (�) +

+

+...

C

n

2

: 1) 3n

( + 1)2

= n n

for all integers n � 3. Prove that

37.

for all positive integers n. Let

38.

Xn 22n + 1, =

for all positive integers n.

L2n [log k] k=l 2

=

(n - 2)2 n + n + 2

n =

1, 2, 3, . . . Prove that 1 2 2 - + - + - + . . . + -- < ­1 3

2n -l Xn

2 Xl X2 X3

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

0

40.

f(a)

=

n-3

Consider a function f 1. Prove that if f(x)f(y) + f

:

(0, 00)

(�) f (�)

-t

=

�

Z2 for all z E

and a real number a

2f(xy) for all x, y E

then f is a constant function. 41.

Find with proof if the function t: � -t [- 1, 1], f(x)

42 .

For all i, j

IS(i, j ) l ·

=

1, n define S (i, j)

=

=

>

C.

0 such that

(0 , 00) , sin[x] is periodical.

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

14

43.

Let

xii

=

{ abi 0

i where ai , bi are real numbers. � Evaluate the determinant 2 n 44. a)

1. ALGEBRA

if if if =

Compute the determinant

j i j, + j i 2n + 1 j 2n + 1 + IXii l · i

i

=

=

i

x y y x z

i

v

z

v

v z

x y y cdab x b b b d d b ) Prove that if the numbers a e , a e, , de a are divisible by a prime then at least one of the numbers a + b + e + d, a + b - e - d, a - b + e - d, a - b - e + d, v

z

p,

is divisible by p.

tl (x) x2 PI X qr t2 (x) x2

= = Consider the quadratic polynomials + + and + where + are real numbers. Prove that if polynomials and have zeros of the same nature, then the polynomial

45.

P2 X q� ,

PI , P2 , ql, q2

tl

t2

has real zeros.

a, b, e be realTnumbers withba 0bsuch that the quadratic polynomial (x) ax2 + ex + S + e S - 4abe has nonreal zeros. T ( ) ax2 + bx + e and T2 (x) Prove that exactly one of the polynomials I x b ax2 + ex + has only positive values. 46.

>

Let

=

=

47.

Consider the polynomials with complex coefficients

P (x) xn + alxn-l + . . . + an and Q(x) xn + blxn- l + . . . + bn i � ; having zeros Xl, X2 , . . . ,Xn and x , x , . . . ,x respectively. if al + a s + a5 + . . . and a2 + a4 + a6 + . . . are real numbers, then bl +Prove b2 + . that b . . + n is also a real number. =

=

15

1. 1 . BROBLEMS P (x) be a polynomial of degree n. If 48. Let k k P (k) -k0, 1, . . . , n for +1 P ) ( evaluate m , where m n. P( ) 49. Find all polynomials x with integral coefficients such that =

=

>

for all real numbers

x.

Consider the polynomials Pi, i that if the polynomial 50.

=

1, 2, . . . , n with degrees at least 1. Prove

P (x) P1 (Xn+1 ) + XP2 (xn+1 ) + . . . n- 1 ( n+1 ) + x pn x , 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 + a1 xn-1 + . . . + an ) . Prove that be a polynomial with integral coefficients such that anP =1= 0 (mod p ( ) 0 (mod p) for all if there are n + 1 integers 0'1 , 0'2 , , an+ 1 such that ar ) 1, 2, . . . , n + 1, then there exist i , j with i i- j such that a i aj (mod p . P ( ) P (xn ) 52. Determine all polynomials with real coefficients such that p n x =

==

• • •

r =

==

=

for all real numbers x, where n > 1 is a given integer. 53.

Let

P (x) aoxn + a1 xn-1 + . . . + an , an i- 0, =

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

m with

P has at least a zero with the absolute value less than 1. P 54. Find all polynomials of degree n having only real zeros Xl, X2 , . . . , X n such that n P ( )1 n2I( ) P � x - Xi X X ' Prove that the polynomial

=

for all nonzero real numbers x. 55.

Consider the polynomial with real coefficients = + +...+

P (x) aoxn a1 xn-1

an ,

16

1. ALGEBRA

f:. and an

O.

Prove that if the equation P(x) equation

=

0 has all of its roots real and distinct, then the

x2 PII(x) + 3xP' (x) + P(x)

=

0

has the same property. 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 ) . 56.

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 aox n + a 1 Xn - 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? 59.

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

where m is a real parameter. 60.

Solve the equation x2 n + a1 x2 n - 1 + . . . + a2 n _2X2 - 2nx + 1 = 0,

if all of its roots are real.

SOLUTIONS

Let Nk be the number of words having exactly k characters from the set C, = n - 1. The number that we seek is 1 :s; k :s; m. Clearly, + 2+...+ = Let Ak 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 1.

N {I , 1

N1 N

f(l)

f:. a1

and f(k)

Nm·

f:. a1

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

Nk

N1It+follows N2 + . .that N .+ m (Dorin Andrica)

=

=

(n - 1) 2 nk-2

(n - 1) + (n - 1) 2 nO + (n - 1) 2 n + . . . + (n - 1) 2 n - 2

1

m

=

2. Suppose, for the sake of contradiction, that there are two sets A and B such that Au B = 2, . . . , 5n }, An B = 0 and the sets contain together less than n pairs ( x, x > 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. Revista Matematica Timi§oara (RMT), No. 2(1978) , pp. 75, Problem 3698)

y) , y,

{I ,

(Dorin Andrica,

3.

Note that (1)

17

18

1. ALGEBRA

and furthermore

for all integers 2:: Suppose that for all integers k 2:: we have Because there are n ! positive integers > such that

m l , m2 1 .

1

n l n2

(2)

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

(3) [ ] I Let h n l - in2 0 and observe that for all k the functions k are injective, 1 , n are[ distinct. since numbers ai , From relation (3) we derive that h ] ( ) + n l 2 x l[n2 ](x) , x E [a, b] , or [ h ]) ( ) (10 l[n2 -1 ]) (x) , X E [a, b] . (10 l n2 + - l x I Because is injective, we obtain l[n2+h- l ] (x) l[n2 - 1 ] (x) , x E [a, b] and in the same manner l [h+l ](X) I(x) , x E [a, b] or [h]( ) [ b] l x, x E a, . X Alternative solution. Let Sn be the symmetric group of order n and Hn the cyclic subgroup generated by h] a. It is clear that Hn is a finite group and therefore there is [ integer h such that a is identical permutation. Notice that I[k ](x) { aq[kJ (i) if x �i' i 1 , n x otherwIse h]( ) [ I Then and the solution is complete. (Dorin Anxdrica,x Revista ( 978) , pp. 53, Matematica Timi§oara (RMT) , No. 2 1 =

=

>

=

=

=

=

=

=

=

=

=

Problem 3540) 4.

=

x y z 0, we obtain x + y -z, y + z -x, z + x -y, or, by squaring and rearranging, x2 + y2 Z2 2xy, y2 + Z2 x2 2yz, Z2 + x2 y2 - 2zx. The given equality is equivalent to Z3 Z2 - 2xy + x2 - 2yZ + y2 - 2zx = -x3 + -y3 + -, -z -x -y yz zx xy Because + +

=

=

_

---

=

=

=

_

=

19

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 + y2z2 + Z2 X2 ) x4 + y4 + Z4. 0 we obtain (x + y + Z)2 0 or On the other side, from x + y + z x2 + y2 + Z2 - 2 (xy + yz + zx) . Squaring yields 4X + y4 + Z4 + 2 (X2 y2 + y2 z2 + Z2 X2 ) 4(X2 y2 + y2z2 + Z2 X2 ) + 8xyz (x + y + z) =

=

=

=

=

=

or

as desired.

( Titu A ndreescu, Revista Matematica Timi§oara (RMT) , No. 3 ( 1971) , pp. 25, 483; Gazeta Matematica (GM-B), No. 12 (1977) , pp. 501 , Problem 6090) Problem b d 5. We assume that numbers a, , c, are different from zero. Consider the equation xb 4 d- (2: a) x3 + (2: ab) xb2 - (2:dabc) x + abed 0 b d f:. 0 with roots a, , c, . Substituting x with a, , e and and simplifying by a, , c, , =

after summing up we obtain

Because

2: a

=

0, it follows that

b +a,c+thend = 0 , b b d 3bcd. Now d or + c - . It is left to prove that 3 + c3 + 3 b3 + c3 + d3 b3 + c3 - (b + c) 3 -3bc(b + c) 3bcd as desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2 (1979) , pp. 47, 3803) Problem b 0 , we obtain 6. Because a + + c a3 + b3 + c3 3abc and a5 + b5 + c5 5abc(a2 + b2 + c2 + ab + bc + ca) If

one of the numbers is zero, say

=

=

=

=

=

=

=

=

20

1. ALGEBRA

The given relation becomes

3abc 5abc(a2 + b2 + c2 + ab + bc + ca) or 2a + b2 + c2 + ab + bc + ca 5'3 b since a, , c are nonzero numbers. follows that 1- [(a + b + C)2 + a2 + b2 + c2] 35 2 b 0 and, using again the relation a + + c = , we obtain a2 + b2 + c2 �5 ' as desired. ( Titu A ndreescu, Revista Matematica Timi§oara (RMT), No. 2 (1977) , pp. 59 , 30 16) =

=

It

= -

=

Problem 7.

Consider the equation

xb 4 -d (L: a) x3 + (L: ab) x2 b- (L: adbc) x + abed 0, with roots a, , e, . Substituting x with a, , e and , respectively, we obtain after summation that L: a4 + (L: ab) L: a2 + 4abed is divisible by L: a. Taking into account that =

we deduce that

L: a. 4Henceb4 4 d ) ( b 2 (ba + + e + 4 - a2 + 2 + e2 + d2 )2 + 8abed d, desired. + + e + is divisible by a (Dorin Andrica) is divisible by

as

8.

The equation is equivalent to

(x2 + 6x + 5)(x2 + 6x + 8)(x2 + 6x + 9) 360 . 2 6 Setting x + x y yields (y + 5)(y + 8)(y + 9) 360 , =

=

=

1.2. 21 or y3 +22y2 +157y = 0, with solutions Yl = 0, Y2 = - 11 +6i , = - 11 - 6i . 6 0 Turning back to the 6substitution, we obtain a first equation, X2 + x = , with 0 solutions = , X2 = -6 . 6i is equivalent to ( +3)2 = - 2 +6i. Setting 11 x = The equation X2 + + +3 = u+i v, u, v we obtain the system { U22uv-=v26= - 2 4 36 40 (2 (2 It follows that u +V2)2 = u -V2)2 + (2UV)2 = + = . Therefore 2 - v2 = -v'W { U2u2 +V2 2 = 2 v'W - 1, v2 = + 1, yielding the solutions and u = = - 3 ± V - 1 ± i V +1 where the signs + and -6 correspond.6i 1 The equation X2 + x = - 1 - can be solved in a similar way and it has the solutions 3 + J - 1 - i V +1 , = -3 - V.Jill - 1 +i V + = 3 ( 97 ) 6 ( Ti A d SOLUTIONS

Y3

Xl

x

E lR,

X

y'lO

V16

X 3 ,4

X5

y'lO

y'lO

V16

y'lO

X6

l.

tu1 55n )reescu, Revista Matematica Timi§oara (RMT) , No. 1 2 , pp. 2 , Problem 2 9. The equation is equivalent to ..;x+Y - + -2vz-=-2+u - +v - ..;v = 0, XVY

or

Vu

z

( - D2 + ( - D2 + (vz-=-2_1)2+ + (.ru-D\ (VV-D2 = 0 ..;xU, v are real numbers, ..;v it follows that vx

v'Y

Because x, y, Z,

= VY = Vu = = � and vz-=-2 = 1 . Hence 1 = 3. X = Y = U = v = 4' ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2 (1974) , pp. 47, 00 0 ( 974) , pp. 560, Problem 14536) Problem 2 2 ; Gazeta Matematica ( GM-B ) , No. 1 1 Z

22

1. ALGEBRA

10.

Setting

or

X = x + 1 and Y = y - 1 yields (X + y)2 = Xy

0 �[X2 + y2 + (X + y)2] = . 2 0 , so the solution is x - 1 and y = 1 . Hence ( 977) , pp. 40, ( TiXtu =AYnd=reescu, Revista Matematidl Timi§oara (RMT) , No. 1 1 8 ) Problem 2 11 =

1 1.

The equation is equivalent to

� x + V4x + V16x + V/

Squaring the equation yields

..

+ v'4nx + 3 = ..jii + 1

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' 4 2 and 2Ti. 2n Vx Hence x = . n 4-5 (1972) , pp. 43, ( tu Andreescu, Revista Matematidl Timi§oara ( RMT ) , No. 385 ) Problem 1 . .

=

12.

We distinguish two cases:

1) b = c. The equation becomes Jx + a + Jx + b = Jx + a, b so x;:=) b-f:.. 2 c. The equation Jx + b +is equivalent Jx + c = Jtox + a + b - c - Jx + a, b-c b-c or Jx + b - Jx + c - Jx + a + b - c+ Jx + a ' so Jx + b - Jx + c = Jx + a + b - c+ Jx + a.

(1 ) (2)

23 1 .2. SOLUTIONS () Summing up relations 1 and (2) we obtain v'x + b = v'x + a + b and then a = To conclude, we have found that b = then b (i ) the equation has the solution x = - . b ( ii ) If b f=. and a f=. there is no solution. ((iii and a = then x = -a is the only solution. Ti)tu Anf=.dreescu, ( 978) , pp. 26, Revista Matematica Timi§oara (RMT ) , No. 2 1 7 30 ) Problem 1 b 1 3. Because a and are distinct numbers, x and y are distinct as well. The second c,

c.

If

If

c,

c

c

c,

c,

equation could be written as

b. We have a a22 b22 = bb22y22 + 2b2y((x24 _ y24))3 + (X4 _ y4 ) 2 a x = x +X X _y .

and the system could be solved in terms of and

Subtracting the first equation from the second yields

which reduces to

3xy2 ) 3 + a = x b = y3 - x2 y a = x3 - xy2 ) . 3 + 3xy2 = x a = xb (x2 _3 y22 ) 3b = y (y2 _ X2 )b ( ) 3 a b ) ( = x y + y . a + = x + y a - = x - y 3. x + y = {!� a + bb x -b y =� a -b)/ (� b � b)/ � ( and (its solution is x = a + + a - 2 , y = 00 a +1 ) - a - 2. Andreescu, Tituunique Korean Mathematics Com petitions, 2 3x + 4 1 3 = y, Y E Z. It follows that 14. Let 3y - 4 13 x= b

b = y3 + 3x2 y

Solving the quadratic equation in yields ( and or ( and The second alternative is not possible because and cannot be both positive. It follows that and Hence The system now and becomes

24

1. ALGEBRA

[ � (3Y : 4) - 2 ] = Y,

and the equation is equivalent to

or

75Y - 126 [ [52] ] =[ Y] Using that for any real number a, a ::; a < a + 1 , we obtain 75y - 126 1 ::; 78 52 < y + , Y 6

--

>

>

--

--

.

2 2 . Using the identities 5a + b5 + e5 = (a + b + e) 5 - 5 (a + b)(b + e)(e + a)(a2 + b2 + e2 + ab + be + 00)

and

we obtain

a53 + bb53 + e5 - (a + bb + e))53 = 5 (a2 + b2 + e2 + ab + be + ca) a + + - (a+ + e It suffices now to prove that b )2 b b 5 ( 2 b2 2 10 3 a + + e + a + e + 00) � 9 ( a + + e 3 (a2 + b2 + e2 + ab + be + ca) � 2 (a2 + b2 + e2 + 2ab + 2be + 2ca). -3

c3

or

The last inequality is equivalent to

which is clearly true.

a2 + b2 + e2 � ab + be + 00,

( 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)

1. ALGEBRA

28

2a- �b, 2b- �,c 2c-�a are greater (1) (2a-D (2b-D (2C-D >1 and (2) (2a-D + (2b-D+ (2C-D >3. From the relation (1) and using abc 1 we obtain (3) 3>2(a+b+c)-G+�+D. On the other hand, relation (2) gives 2(a+b+c)-G+�+D >3

Assume by contradiction that all numbers than 1. Then 23.

=

which is a contradiction. The proof is complete. Revista Matematica Timi§oara (RMT) , No. 2(1986) , pp. 72, Problem 5982)

( Titu Andreescu,

24.

Assume by contradiction that all numbers are greater than 1/4. Then 1 1 1 1 �J2 4 4 4 4

a-b2 +b-c2 +c-a-+d-a2> - +- + - +0> G-ar + G-br + G-cr+ G-dr

hence

This is a contradiction so the claim holds. Revista Matematica Timi§oara (RMT) , No. 1(1985), pp. 59, Problem 5479)

( Titu Andreescu,

ai

Setting = sin2 expression becomes 25.

E

=

ai for i

=

1, n, where

aI, a2 , . . . , an are real numbers, the

n

\!sin2 cos2 ai a al· 2: i=1 ai +b n+l =

Using the AM-GM inequality yields

k bf ;::: b1 b2 -k1 2: i=1

bk, bi>0, i U . For b1 sinf ai, b2 cosf ai+l and b3 b4 bk {/2� we obtain k 1 . 2 (sm ai +cos2 ai+l + -2-2-) ;::: 2¥1 \I'sm2 ai cos2 ai+l· =

=

k

=

• • .

=

= ... =

=

k

1.2. SOLUTIONS

29

k 1 , 2, . . . , n yields (k ) l E �k (n + n - 2 ) 2 - 2_-/0 /0_ /ok2 1 � and so E < --2 2- n ·E2 n · 2 {14n Hence the maximum value of is ;; and it is reached if and only if 1 = a2 = . . . = an = 2"' al (Dorin A ndrica, Revista Matematidl Timi§oara (RMT) , No. 1 (1978) , pp. 63, Summing up these relations for =

>

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-I x + (xn ) m-2 x + . . . + x ] � 0. E( ) ( ) ( ) 2 Define E (x ) = x n m- I x + xn m - x + . . . + x and note that if x � 1 , then xnn �< 11 and E (x ) �< 0, so the inequality holds. In the other case, when x < 1 , we have x ( TiandA dx and again the inequality holds, as claimed. (1978) tu n reescu, Revista Matematica Timi§oara (RMT) , No. 2 , pp. 45, -

-m

-m

°

Problem 3480) 27.

For

m

:::;

can be written as We have

m >

n the inequality is clearly true,{O so1 considerI } n and define = + q with q E , , . . . ,n and the inequality ( !)P (pn + q)! � n .

p = [:]. This implies that

m

pn

(pn + q)! � (pn)! = ( !)P )( ) ( ) ( 1 1 l) ) = · 2 . . . n n + . . . 2 n ... ((P - n + 1 . . . (pn) � n , and (we TituareAdone. ndreescu, Revista Matematica Timi§oara (RMT) , No. 2 ( 1977) , pp. 61,

Problem 3034) 28.

We will use the inequality

Xml + xm2 + . . . + xmn - n_m1-_l (Xl + X2 + . . . + Xn ) m , which holds for all positive real numbers X I ,X2 , . . . ,Xn and all E ( >

m

-00 ,

] [1 ,00) .

0 U

1. ALGEBRA

30

.!. 2 = _ = Now set X l = 1 , X = 2, . . . , X n n and n . We obtain 1 1 1 1 [ n (n + 1) ]�2 = n - -' > -� l 1+v'2 + y'3 + · · · + \Iii -n+1 2 n as desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1974), pp. 52, m

1

n

Problem 2035) 29.

From the AM-GM inequality we deduce

i � IT � 1 = \in + 1, i=l or 1 n 1 \i 1 + - 2: -;- � n + 1, n i= l and so ( \i > n n + 1 - 1) ' 1 + ! + ! + . . . + .!. n 2 3 i+1 as desired. . · IS · stnct b ecause the numb ers -. · Ob serve that the mequa 1Ity nt i=l

.!.

i

�1 �

n

�

�

�

.

�

=

1 - are

,n,

distinct. In order to prove the first inequality we apply the AM-GM inequality in the form 1 2 1 + - + - + · · · + -1 2 3 > n \Iii

n

Therefore

or

n-1 n

if!n = _

_

.

n (1 - _\Iii1_ ) + 1 > 1 + �2 + . . . + .!..n

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 62,

Problem 3037)

al ,a2 , . . . ,an are positive, from the AM-GM inequality al a2 + . . . + an--- � -tlal a2 . . . an -n ) ( we deduce that tn ::; ala2 . . . an . Using that numbers ai are less than 1 we obtain ( ) n-1 logai tn � -- logai ala2 . . . an . n 30.

Because the numbers +

n

;;: l

1. 2. SOLUTIONS 31 Summing up these inequalities yields n n ) n- 1 loga; tn loga; (a1 a2 . . . an = 2: 2: n i=1 i= 1 ) ( ) ( n-1 -- [n + logal a2 + loga2 a1 + . . . + logal an + logan a1 + . . . + n )] ( + logan an - 1 + logan_l an · 1 Note that a + a > 2 for all a > 0, so 2] (n - l)n, n-1 � logai t n -[n + 2(n - 1) + 2(n - 2) + . . . + n i=1 as claimed. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 62, 2::

=

-

L.J

2::

=

Problem 3038)

31 . We begin with the following lemma. Lemma. >

Let a b be two positive integers such that 1. Then between numbers a and b �there- � is at>least perfect cube. Proof. Suppose, for the sake of contradiction, thata there is no perfect cube between that a and b. Then there is an integer c such b c3 � < 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 1 1 n > (2, 5) 3 = > �_ (1, 4 - 1) 3 ( 1) 3 ' or 1 aC y n > �_( Hence

{I3n - -rn > 1,

and using the above lemma the problem is solved. Revista Matematica Timi§oara RMT , No. 1-2(1990), pp . 59, Problem 4080)

( Titu Andreescu,

( )

32

1.

k(k2+

ALGEBRA

1) IS odd for 1 or Note that the number even for where is a positive integer. or We have the following cases: then i) if 32 .

k= 4p,

k=4p + 3 n= 4m,

k= 4p + k= 4p + 2

.

p

Sn= n k= m-lL..J (-4p - 1 - 4p - 2 + 4p+ 3 + 4p + 4) = 4m. n= 4m + 4m - (4m + = S = n n= 4m + 2, 4 (4 (4 (4m + 3) . m + m m + 2) S = = n n=4m + 3 4 (4 (4 (4 3) O Sn= m - m + - m + 2) + m + = . 4m n=4m 4m + n= 4 Sn= �( m + 3) n=4m+2 4m + 3. n = (D i A d i "".,. (-1)

L..

k=1

ii) if

and IS .

""

k(k+l) 2

p=O

1, then

1)

- 1.

then

iii) if

1)

iv) if

then

Hence

or 4303 n n) r ca,

{

1)

if if 1 if if Revista Matematidl Timi§oara (RMT) , No. 1(1981), pp. 50, -1

Problem 33.

for

a) Summing up the identities

(n +k 2) - (n +(n2 + 2k)(n)(n ++ k) (n)k k= 0 k=n 1 Sn= (n + 2)(n + (�(n;2) - (�:�) - (�:�)) = 3) n 2 + +2 n ( n (n + 2)(n + [2 +2 - (n + 2) - (n + 2)(n + to

1) 1-

yields

1)

1

1) b) Summing up the identities

-

1]

_

1) .

(n +k 3) - (n + 3(n +k)(3)(n +n 2+ -2)(nk)(n+ + k) (n)k k= 0 k= n n - (n + 3)(n + 2)(n + 1)

1-

-

for

to

yields

1'. _

1

1)

1 . 2.

SOLUTIONS

. (� ( ) ( )

(

33

) ( )) )

n+3 n+3 n+3 n+3 = 3 k n + 1 n + 2 n + k=O 1 2n 3 � (n2 + 3n + 2) = = 2 (n + 3) (n + 2) (n + 1) n+ 2 2 - (n + 3n + 2) - 2(n + 3) (n 2) (n + 1) . Revista Matematica Timi§oara (RMT) , No. 2(1975), pp. 43, _

_

(Dorin Andrica,

Problem 2116)

( 4

_

+

_

_

+

Let Sn be the number in the statement. It is not difficult to see that l Sn = 2 + v'3 + 2 - v'3 34.

� [(

t 0 such that Sn = (k - 1) 2 + k2 , or, equivalently, 2k2 - 2k + 1 - Sn = The discriminant of this equation is Ll = 4(2Sn - 1), and, after usual computations, we obtain n+ l nH 2 ( 1 - v3) 2 ( 1 + v3) 2 �=

(

O.

;

)

Solving the equation, we find n+ l n+ l 2n 1 + ( 1 + v'.3) 2 + ( 1 - v'3 ) 2 k= 2n+2 Therefore, it is sufficient to prove that k is an integer. Let us denote is an integer for ( 1 + .J3) + ( 1 - .J3) where m is a positive integer. Clearly, [ all m. We will prove that 2 divides m = 1, 2, 3, . . . Moreover, the numbers satisfy the relation = 2 -1 + The property now follows by induction. Romanian IMO Selection Test, 1999)

+

m

Em

(Dorin Andrica,

35.

where

Em

Em

Differentiating the identity

sin nx = sinn X yields

m, ] � Em , Em Em 2Em-2•

((�) cotn-1 X - (�) cotn-3 x + (�) cotn-5 X )

n cos nx = n sinn - 1 x cos xP(cot x) - sinn X

-

� PI (cot X), sm X

• • •

34 For

1. ALGEBRA

x = � we obtain

7f

n cos n "4 Because

n ( P (l) P' ( )) ( ) 0. n -2 1 . = 2

nP (l) = n (�) - n (;) + n (�) - . . . and ' - 2P (1) = - 2 (n - 1) (�) + 2 (n - 3) (;) - 2 (n - 5) (�) + . . . , we have ' nP( 1) - 2P (1) = - [(n - 2) (�) - (n - 6)(;) + (n - 10) (�) - . . . J = = -n ((�) - (;) + (�) - . . . ) + 28n· To conclude, use that

hence

Sn = n (0.) n (cos n + sinn) 2 4 4 7f

7f

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2 ( 1977) , pp. 3 00) Problem 2 36. Differentiating with respect to x the identities (x + l) n = (�) + (�) X + . . . + (�) xn and (x - l) n = (�) xn - ( � ) xn- 1 + . . . + (- 1) n (�) n 1 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.

Differentiating again we obtain

1.2. SOLUTIONS

35

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 -

and

()

Summing up the last two identities gives

as desired.

Sn = 12 (�) + 32 (�) + . . . = n (n + 1)2n- a,

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1 (1978) , pp. 90, 3438) Problem 37.

Note that

2n k] n_ 1 2i+1_1 k] ] 2: [log2 = 2: 2: [log2 + [log2 2 n , k=lk 2 i=O k=2i i k] 2 and [log2 = for i < i+1 . Hence 2n � k] n-1 i 2i 2n] 2)2n 2 k2:=l g2 = 2: i=O . + [log2 = (n - + n + as claimed. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2 ( 1981 ) , pp. 63, 4585; Gazeta Matematica (GM-B) , No. 2-3 (1982) , pp. 83, Problem 19113) Problem 22n - 1 for all positive integers n. Then 38. Let Yn = (22n )2 - 2 . 22n + 1 1 2 2 � _ __ 22n Yn Yn+1 ( 2n 1-) 1 22n +1 -2 1n (22n - 1)(22n+ 1 - 1 ) _- (22n _21 )(_22n +21 - 1 ) - (222n ) 2--1 1 _- 22n 1+ 1 -_ x1n and therefore 1 = -1 - -2 ::;

O

_

1. ALGEBRA

36

Xn Yn Yn+l Summing up these relations yields 2 2n-1 = 1 - -2n < ­1 -X1l + -X22 + -X32 + . . . + -Xn YI Yn+l YI for all integers n, desired. (Dpositive ori4n135)Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2 ( 1980) , pp. 67, Problem i 39. Substituting z with z in the relation J(z)J(iz) = Z2 yields J(iz )J( -z) _ Z2 . Summing up gives J(iz)(J(z) + J( -z)) = 0, J(i ) 0 J( ) J( ) o. so z = or z + J(-z)J(= J(z) = 0 if and only if z = From the relation that zJ(i i)z) =0 Z2 we deduce O. Hence 0 J( ) J( ) = 0 and, if z = 0, then + -:f. and so if z -z , then z z -:f. J(z) + J( -z) = 2J(0) = O. Clearly, J(z) + J( -z) = 0 for all numbers z E desired. Remark. A function J J( )J(i ) 2 J( ) satisfying the relation z z = Z is z = (- �( Ti + iA�d) z. tu 583n )reescu, Revista Matematica Timi§oara (RMT) , No. 2 ( 1976) , pp. 56 , Problem 2 40. Setting X = Y = 1 yields J2 ( 1 ) + J2 ( a ) = 2 J ( 1 ) (J(l) - 1) 2 = 0 so J(l) = 1. Substituting Y = 1 gives and . J(x)J(l) + J (�) J(a) = 2J(x) -

as

=

C,

:

C

-+

C

or

Y Xa and observe that J(x)J (�) + J (�) J(x) = 2J(a) .

Take now =

-

as

1. 2. SOLUTIONS f(x)f (;) = 1,

Consequently,

37

f2 ( ) = 1 , x > o. = y = 0, that gives 12 (,ji) + 12 (�) = 2/(t)

therefore x Now set x

f( ) = 1 ( ) 12 ( 1977) , 45, 10 ( 1980) , 439, 18455) f

and because the left-hand side is positive, it follows that is positive and x for all x. Then is a constant function, as claimed. Revista Matematica Timi§oara RMT , No. pp. Problem Gazeta Matematica (GM-B , No. pp. Problem

f ( Titu Andreescu, 2849;

)

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

T 0 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 all x E JR and therefore

7r

is irrational, we deduce that

[x] = [x + T] for

all

x E JR

which is false, since the greatest integer function is not periodical. Revista Matematica Timi§oara RMT , No. Pro blem

(Dorin Andrica, 3430)

42 .

k(x) = 0 for

( )

1 ( 1978) , pp. 89,

Considering the determinant

3 32 nn2 3n nn we have In 1 122 13 I S (i , j)1 = 8 · 21 2 23 2 n = 82 , �= 2 n3 n n n1 n 8 because the second determinant is obtained from by interchanging rows and columns. 12 8= 1 In

2 22 2n

38

1 . ALGEBRA

On the other hand, 1 1

8 = n!

1 2

1

1 3

n

1n- 1 2n - 1 3 n- 1

n

n- 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 � 2n =

0 0

0 0

0

a2 0

b2n0 l b2n 0-

0 0

a3 b2n 2 0 0

0 0

b3 a2n-2 0 0

b02 0

0

b1 0 0 0 0

a2 n - l 2 0 an

Expanding along the first and then the last row we obtain which gives

�2n = IIn (ak a2n k+l - bk b2n k+d k= l -

(Dorin Andrica, 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 divides the determinant. Analogously we can check that and divide the determinant, and taking into account that it has degree in each of the variables, the determinant equals

x+y- z- v

where

..\

x-y+z-v 4 x-y-z+v ..\(x + y + z + v)(x + y - z - v)(x - y + z - v)(x - y - z + v) ,

is a constant.

39 1 . 2. SOLUTIONS 4 Because the coefficient of X is equal to 1 , we have A = 1 and so x y y x v = (x + y + + v)(x + y - - v)(x - y + - v)(x - y - + v) v x y v y x b ) As shown above, we have b e d a 6.. = b ad d eb = (a + b + e + d) (a + b - e - d) (a - b + e - d) (a - b - e + d) de e ab a 000, the second by 100, 1 On the other hand, mUltiplying the first column by 0 alld these to the fourth, we obtain on the last column the third by 1 b dandb dadding d b b all those numbers are divisible by the the numbers a e , a e, e a , e ao Because 6.. prime numberb it follows d. at least one of the d, a +that d, a - b +ande -therefore d, a - b - divides b - e -divides + e + e + numbers a + ( Titu Andreescu) () () 45. Because the quadratic polynomials h x and t2 X have zeros of the same z

z

w z

z

z

z

z

z

p,

p

p

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

(P1P2 + 4q1q2 )2 - 4(P1q2 + P2Q 1 )2 2:: O .

Note now that the left-hand side of the inequality is the discriminant of the quadratic polynomial and the conclusion follows. Revista Matematica Timi§oara (RMT ) , No. pp. Problem Gazeta Matematica ( GM-B ) , No. Problem pp.

( Titu Andreescu, t 3 2 67;

( 1978) , 63 , 1 5 ( 1979) , 19 1, 1 7740) T 46. Because the quadratic polynomial has nonreal zeros, the discriminant 6.. = b2e2 - 4a(b3 + e3 - 4abe)

is negative. Observe that

6.. = (b2 - 4ae)(e2 - 4ab) < 0 , 6.. b2 4 6.. 2 4 b where 1 =T - Tae and 2 = e - a are the discriminants of the quadratic 6.. 6.. 2 polynomials 1 and • Hence exactly one of the numbers 1 and 2 is negative and 0 since( Ti a > A, thed conclusion follows. tu 8 1n0)reescu, Revista Matematica Timi§oara (RMT) , No. 1 ( 1977) , pp. 40 , Problem 2

1 . ALGEBRA (_ l) n- I an are real numbers, 2 2 . . . . +a + + 47.PObserve that a ·+a and a + . -a l l n (l ) and P( - 1) are real numbers. Hence that is P (l) = P( l) and P(- l) = P( - l) (1) P( ) ( ) ) () Because x = x - Xl . . . (X - x n , the relations 1 become ( 1 - xI ) . . . ( 1 - xn ) = (1 - xI ) . . . ( 1 - xn ) and ( 1 + xI ) . . . (1 + xn ) ( 1 + xI ) . . . ( 1 + xn ) 40

=

Multiplying these relations yields

(1 - x� ) . . . ( 1 - x� ) = ( 1 - x�) . . . ( 1 - x�) , b b b Q (l) Q (l) or ( Ti A d . Therefore l + 2 + . . . + n is a real number. tu 86n4)reescu, Revista Matematica Timi§oara (RMT) , No. 1 ( 1977) , pp. 47, Pro blem 2 P (O) 0 , there is a polynomial Q with P (x) xQ (x) . Then 48. Because Q (k) = k 1 l ' k = 1,n. + d H( ) H(k) 0 for ) H( ( ) ) ( Q l 1 It is clear that eg x = n and Define x = x + x . k all = 1 , n, hence ( 1) H(x) = (x + l ) Q (x) - 1 = ao (x - l)(x - 2) . . . (x - n) (1) yields Setting x = m > n in relation Q ( m) = ao ( m - 1 ) ( m - 2 ) . . . ( m - n ) + 1 . m+1 1 On the other hand, setting x = - in the same relation implies ( _ l ) n+ 1 ao = (n + I) ! Therefore Q (m) (_ l ) n+l (m( - ll)()!(m - 2)1.). . (m - n) + _1_ n+ m+ m+1 and then P(m) = (_ l) n+(l m (ml)-! ( 1) . . . )(m - n) + �. n+ m+1 m977+ )1 3 9 8 (Dorin Andrica, Gazeta Matematica ( 1 2 , Problem ( GM-B ) , No. 16833 ; Revista Matematica Timi§oara (RMT) , No. 1-2 (1980) , p. ,67pp., Problem 4133) 49. We are looking for a polynomial with integral coefficients P (x) = aoxn + alxn-l + . . . + an , ao -:j:. O. =

=

=

=

m,

=

41 1 .2. SOLUTIONS We have ( l) 2 P' (x) = naOxn -l + n - alXn - + . . . + 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

a0 - nn1- l and since ao is an integer, we deduce that n 1 and ao = 1. Then P(x) x + aI, 1 and P(P'(x)) = P' (P(x)) yields 1 + al = 1 or al = P' (x) Therefore P(x) = x is the only polynomial with the desired property. ( Titu An dreescu, Revista Matematica Timi§oara (RMT), No. 1-2(1979), Problem 3902) O2 , . . . , On be the roots of the equation 50. Let xn + Xn -l + . . + x + 1 = Of i 1, n. They are all distinct and +l = 1, P (Oi) = 0, l n n . . . Because P(x) is divisible by x + + + x + 1, it follows that x i 1, n, hence Pl ( l) + OlP2 (I) + . . . + Or- lpn ( l) = 0 PI (1) + 02P2 ( 1) + . . . + O�- lPn (l ) = 0 --

=

=

O.

=

(h,

.

O.

=

=

The above system of equations has the determinant

0 on- l O21 O�1 -l V= On on- l 1 n 0 O 0 Because all of the numbers 1 , 2 , . . . , n are distinct, it follows that V -:f. 0 and I ) ( ) ( l . . . 2 so the system has only the trivial solution I dividesPI Pi(1)(X=) forP all =i = 1,=n. Pn = This is just (another way of saying that x Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 75, 1 1

O.

Problem 3120; Gazeta Matematica (GM-B) , No. 8(1977), pp. 329, Problem 16834)

42

51.

Consider the determinant

1. ALGEBRA nII+1 ( ) ak - al = an

v=

"' , 1 = 1 "' > 1

an-I , the third by an-2 , ... , the last by ao and P ( a d P ( a2 ) P ( an+ d an2 +1 al� a2� nII+1 (a - a ) v= an + 1 = an k t a a a) �+1 a? P( )a� On the II other hand, ar 0 (mod) p , for all r = 1, n + 1 and an ;j. 0 (mod p) ) ( implies ak - al 0 (mod p . Therefore there are at least two numbers + k 1 1:Sl �n < ai, aj,(D i =Ii j Asuchd ithat ai - aj 0 (mod p) and so ai aj (mod p) , as desired. or n n r ca, Gazeta Matematica (GM-B), No. 8(1977) , pp. 329, Problem 16835) d P( ) 52 . Let m = eg x and let P (x) = aoxm + Q (x) , ao =I 0 d Q( ) If follows that( )eg Px( =) r � m - 1. From pn x = xn we obtain ( a�xmn + (�) a�- l xm n-l ) Q (x) + . . + Qn (x) = aoxmn + Q (xn ) , or S a�xmn + R (x) = aoxmn + (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, degR(x) = degS (x) or m(n - 1) + r = nr and so (n - l) (m - r) = Multiplying the second row by adding all to the first yields

"' , 1 =1 "' > 1

==

==

==

==

.

O.

43 1. 2. SOLUTIONS Q( ) Q( ) This is impossible if x -:f. 0, because n > 1 and m > therefore x 0 and the polynomials are P (x) = xm , for n even and P (x) ±xm, for n odd. Alternative solution. Let degP (x) = m and let P(x) aoxm + alxm- 1 + . . . + am . P( ) Q( ) k If x = xk x with a positive integer, then kn Qn (x) = xkn Q (xn ) or Qn (x) Q (xn ) x P. Assume that P (O) -:f. O . Q Note that satisfies the same condition � am ' Then am 1 if n is even Setting±x 0 in the initial condition yields a and am 1 if n is odd. Differentiating the relations implies pn-1 (x) P' (x) = nP' (xn )xn- l . n (1) ' P O O ( ) Setting now x 0 gives 0 and so am- l . Differentiating again in relation (1) yields analogously a m - 2 0 and then am-3 = am-4 = . . . = ao O. The polynomials are P (x) xm , if n is even and P(x) = ±xm , if n is odd. ( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1-2(1979) , pp. 59, r,

=

=

=

=

as

=

=

=

=

=

=

=

=

=

=

Problem 3884) 53.

From the relations between the zeros and the coefficients we obtain

am = ( _ l) m L XIX2 . . ' Xm ao

and It follows that

and by applying the triangle's inequality for complex numbers, we deduce that 1

L I ll 1 I l > ( n ) . Xl1 x2 I · · . xnl}-m m I { Consider Xo = min l x l l , X2 , · · . , xn- m . Then

so

Xo < 1,

as

claimed.

44

1. ALGEBRA

( Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 2(1978) , pp. 52,

Problem 3531) 54.

We have

Ln

PI (x) n2 = . -;X · P(x) � i=l Integrating the equation yields n ln IP(x) - xi i = n2 In C lxl ,

L

C>0

i=l

or

n 2 2 In IT IP(x) - xi i = In Cn Ixl n i=l

Hence or

I ll. (F(x) -Xi) I = klxl n' ,

k>0

2 IP(P(x)) 1 = k l xl n . Eliminating the modules gives 2 P(P(x)) = AXn , A E JR.

Therefore P (x) = axn with a E R Revista Matematica Timi§oara (RMT) , No. 1(1977), pp. 47, Problem 2863; Gazeta Matematica (GM-B), No . 1(1977) , pp . 22, Problem 17034)

( Titu Andreescu,

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) 55.

the conclusion follows. Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 52, Problem 3530)

(Dorin Andrica,

56.

Let m = degP(x) and let P(x) = ao (x - x I ) (x - X2 ) ' " (x - xm ).

Because P(x) E R[x], X l , X 2 , . . . , X m are distinct zeros. Now P(Q(x)) = (Q (X) - x I ) (Q(X) - X2 ) . . . (Q(X) - Xm )

45 1.2. SOLUTIONS P (Q (a)) 0, we have has a multiple zero a of order k. Since Q (a) Xi ) 0, ao iII( =l Q(a) - xp 0. Observe that and so there is an integer m, such that 1 � � ) Q (x) - Xj has Q (a - xp -:f. 0, for all j -:f. otherwise X'j xp , which is false. Hence ' ' ( ) ) ) ( ( ) ( Q Q Q x has a multiple x x - xp the mUltiple zero a of order k and so concludes the proof. zero (of Dororder in Ankdri- ca,1. This Romanian Mathematical Olympiad - final round, 1978; Revista Matematica. Timi§oara (RMT), No. 2(1978) , pp. 67, Problem 3614) P (x) has less than two nonreal zeros. 57. Assume by way of contradiction that P( ) As a polynomial with real coefficients x cannot have only one nonreal zero, hence P ) ( 2 all of its are real. Let Xl, X , . . . ,Xn be the zeros of x . Then P' ( x ) n 1 P (x ) - � _ _ X �=l x'� and differentiating we obtain P" (x) P (x) - [P' (X)] 2 n 1 P2 ( X ) tt (xP-( X) i )2 Setting x a we reach a contradiction, therefore x has at least two nonreal zeros,( Tias claimed. tu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1-2(1979) , pp. 59, =

m

=

-

p,

p

=

p,

=

=

=

_

=

=

Problem 3883)

P (x) aoxn + alxn- l + an P" , . . . , p(n-2) . Because P (n- 2) (x) (n - 2)! [n (n - l)aox2 + 2 (n - l )alx + 2a2] 2

. + be a polynomial with real coefficients. = Let If all of its zeros are real, then the same is true for the polynomials

58.

.

.

=

is a quadratic polynomial with real zeros, we have

or

- l)a� a a2 P (x) x3 + (a + 1) x2 + (a + l) x + a, [

(n 2: 2n O . The reciprocal is not always true, as we can see from the following example: with

aE(

-

00 ,

-1] U 2, (0).

=

P' ,

46

1 . ALGEBRA ( 2 ( + 1),)( or (a + l) (a - 2) ;::: 0, so the inequality Observe that 2 a + 1) ;::: 2 . 3(a holds. drica) hand, P(x) = x + a x2 + x + 1) does not have all zeros real. D( OnorintheAnother 59.

For m = ° the equation becomes

X 4 - x3 - x2 + 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 (2 2 4 2 2xm2 + x - x3 + l)m + X - x3 - x + X = °

and

°

It follows that The initial equation becomes [m Hence

(x2 - x)] [m - 2X x2 - 1 ] = 0.

x2 - x - m = 0, with solutions Xl ,2 = I ± Jl2 + 4m and 2 2mx - 1 = 0, with solutions X3,4 m ± VI + m2 . x (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977), =

-

Problem 3121) 60.

pp .

From the relations between the zeros and the coefficients we obtain

75,

L XIX2 · · · X2n- 1 = 2n and XIX2 · · · X2n = 1 . Hence L2n -1 = 2n, Xk k=l 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, 2 = . . . X2n = 1. we have ( TituXlA=ndXreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977) , 52, =

Problem 2299)

pp.

Chapter 2 NUMBER THEO RY

PROBLEMS 1.

1

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

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

_

4.

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

+

55 n

is not prime. 5.

+

1

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

is not a prime number. 6.

Prove that

5 + 456

34

102002 . [ \1'111] divides 111 . 7. Find all positive integers n such that b 2 3 b2 ) 8. Prove that for any distinct positive integers a and the number 2a ( a

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

+

is not a perfect cube.

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

-

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

49

2. NUMBER THEORY

50

1 1 . Prove that for any positive integer

n (17 + 120)

n the number

(17 4v'2 _

_

120)

n

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.

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. 13. Prove that for all positive integers

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

for all integers n 2:: 1 7. Let

[

[n0i - �] = [n0i + �]

]

a + 1. y'1 - 2avp

n be an odd positive integer. Prove that the set

{ (�) , (�),

contains an odd number of odd numbers. 18. Find all positive integers m and

• • •

,

( n� ) } 1

n such that

19. Solve in nonnegative integers the equation

(:) = 1984.

x2 + 8y2 + 6xy - 3x - 6y = 3 20. Solve in integers the equation

(x2 + 1)(y2 + 1) + 2(x - y) (1 - xy) = 4(1 + xy) 2 1 . 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 23. Find all triples

u and v are both p:

(x, y, z) of integers such that x2 ( y - z ) + y 2 ( Z - x ) + Z 2 ( X - y ) = 2 .

24. Solve in nonnegative integers the equation

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

has integral solutions.

x2 + y2 + Z2 = 59n

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

and

x n + yn + z n + u n = vn + 1

have infinitely many solutions in distinct positive integers. 28. Let n be a positive integer. Solve in rational numbers the equation

x n + y n = x n - 1 + yn - l . 29. Find all nonnegative integers x and y such that x(x + 2)(x + 8) = 3Y. 30. Solve in nonnegative integers the equation

(1 + 31. Solve the equation

xl )( l + yl) = (x + y) ! . x l + y l + z l = 2v . . I

52

2, NUMBER THEORY 32 . Find all distinct positive integers Xl, X 2 , . . . ,Xn such that 1 + Xl + 2XIX2 + . . . + (n - 1 )XIX2 . . . Xn-l = XIX2 . . . Xn· Prove that for all positive integers n and all integers aI, a2 , , an , bl, b2 , . . . , bn the33.number IIn (a� - b� ) k=l • . .

can be written as a difference of two squares. z,

34. Find all integers

x, y, v, t such that ( )( ) X + y + + v + t(= xyv) t + x( + y )v + t xy + + vt = xy v + t + vt x + y . b d such that a and b are relatively 35 . Prove that for all nonnegative integers a, , prime, the system axb - y d 0 x - yt + = 0 z

z

c,

z - c=

has at least a solution in nonnegative integers.

Xl, X2 , ,xp be nonnegative integers. Xl� + X�2 + . . . + x�p 0 (mod p)) x + x + . . . + x 0 (mod p f- l + X�-l + . . . + X�- l 0 (mod p) X k l {I 2 } k ) then there are , E , , . . . ,p , =I l, such that Xk - Xl 0 (mod p . 37. Prove that for any odd integers n, aI , a2 , . . . ,an , the greatest com­ mon divisor of numbers a I, a2 , . . . ,an is equal to the greatest common divisor of al +2 a2 a2 +2 a3 ' " ' ' an +2 al () 3,8. Let

�

�

•

The conclusion follows from the identity

and the inequalities

4) 1 2 i1 3 2 2 2 _ 1 ( 56 4 i 2 - > m - mn + n > n ( n - m) = 2 4 > 2 2 (2 562- 1 - 22 4 )4 > 2 2 . 251 2 (42 562-1 -5 12 - 1) > 21 0. 5�t1 25 1 2 > 21 0.5 . 210.50 > 103 .5 . 103.50 > 102002 > (Titu Andreescu, Korean Mathematics Competition, 2002) 111 1 3 37 111 7.) The positive divisors of 2 are , , , 7 . So we have the following cases: yl [ ] 1 111 1 1 111 n �4 . 2) [ yl111] == 3, oror 3n 111< b.

On the other hand, if

b > a then there is no integer

c

such that

This concludes the proof. Revista Matematidi Timi§oara (RMT) , No. Problem

( Titu Andreescu, 1911)

1 (1974) , pp. 24,

4 = q4 for some positive integer q. Then p = q4 + 4 and q > 1 . We obtain ( 2 2q + 2)(q2 + 2q + 2) , p= q 1 a product integers greater than , contradicting dreescu, ( Titu Aofntwo ) fact that p is a prime. 2003the Math Path Qualifying Quiz, 9 . Assume that p

-

_

10. The inequalities

0 � x + y + 8, which is 3 3 2) x2 +2 y 3 (x + 2 2) 2 y2 + x (y + 2) 2 is true. Without 3y = 2 x + 1 . 2x3k 1x2 + 3y 2k(x +1x2)+2 y (x + x2 k+. 3y0 (x + 1 ) 2 , hence 3x = 4k2 + 13k + 4. 2 x y + � y + k> (2k + 3)2 < 4k2 + 13k + 4 < (2k + 4)2 3 k E {O, 1 , 2, 3, 4} , y2 + 3x so y2 + x cannot be a square. It is easy to check that for 0 4 3 k 22 is not a (square but for = , y2 + x = = . Therefore the only solution is ) 1 1 (x, y)( Ti= A, d. ) tu n reescu

cannot hold simultaneously because summing them up yields false. Hence at least one of or < < loss of generality assume that < From < < we derive = Then = + and = for some integer and so If 5, then

2. NUMBER THEORY 1 7 + 1 2v'2 = (v'2 + 1 ) 4 and 1 7 - 1 2v'2 = (v'2 _ 1 ) 4 , so 1 1 . Note that (17 1 2v'2) n - ( 1 7 - 1 2v'2) n ( v'2 1) 4n - (v'2 _ 1 ) 4n 4v'2 4v'2 (v'2 1) 2n (v'2 - 1 ) 2n (v'2 + 1 ) 2n _ (v'2 _ 1 ) 2n 2 . 2v'2 -

60

+

_

_

+

+

+

Define

A = (v'2 + 1) 2n +2 (v'2 - 1 ) 2n

and

B = (v'2 1) 2n2v'2 - (v'2 _ 1) 2n +

U sing the binomial expansion formula we obtain positive integers that

x

and

2n v2 1 + ( ) = X + y v2, (v2 - 1) 2n = X - y v2 Then (v'2 + 1 ) 2n + (v'2 - 1) 2n (v'2 1) 2n _ (v'2 _ 1) 2n 2 2v'2 and y = X= AB is as integer, as claimed. and so

y such

+

Observe that

A B

so and are relatively prime. not a perfect square. We have

It

is sufficient to prove that at least one of them is

A = (v'2 + 1) 2n +2 (v'2 _ 1) 2n = [ (v'2 l ) nv'2(v'2 _ 1) n ] 2 - 1 +

and

Since only one of the numbers

(1)

l ) n - (v2 _ l) n v'2 v'2 ( 1) and (2) we derive is an Ainteger - depending on the parity of n - from the relations that ( D isinotA a dri square. This completes the proof. or4n285n) ca, Revista Matematica Timi§oara (RMT) , No. 1 ( 1981 ) , pp. 48, Problem 7 8 9 6 13 < 16 < 12. The claim is true for n = and n = , because U 6 = < < U7 = 21 U8 = . (v2 +

l) n

+

(v2

_ l) n

+

(v2 +

If n

� 8, then

2. 2.

61

SOLUTIONS

Un - 1 > r,;-;--:-: . � -_ v'UnUn+1 --'-+1 +- UnVU;; >- 2v'Un +1 - 1 = .1Fi v'Un 1 > 1 � 2Y. Un� oJUn- 1 2y 3

y. Un+1 - y Un

-

--=

and so between Un and Un+ 1 there is a perfect square.

(Dorin Andrica)

13. If n = 1, 2, 3 or 4 then n! + 5 = 6, 7, 11 or 29, so it is not a square. If n

then n! + 5 desired.

� 5,

= 5(5k + 1) for some integer k and therefore is not a perfect square, as (Dorin Andrica, 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(n 2 +3n+3)

is a cube. Note that n (n2 + 3n + 3) = n3 + 3n2 + 3n = (n + 1) 3 - 1 and since (n + 1) 3 - 1 is not a cube, we obtain a contradiction. Gazeta Matematica ( GM-B ) , No. 8(1977) , pp. 312, Problem E5965; Revista Matematica Timi§oara (RMT) , No. 1-2(1979) , pp. 28, Problem 3253)

(Dorin Andrica,

1 5 . 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, (n + l) + (n + 2) + . . . + (n + 2P + l) 2P+ 1 = (n + p + 1) 2p+1 P(n ) < 2p + 1 from the AM-GM inequality. then E {n + 2, . . . , n + pl . Assume by way of contradiction If P(n) 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)

]

[

= m2p+1 ,

m

=

We have two cases: I. k p 1) If n == (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) , 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 == This is a contradiction. II. k E {2, 3, . . . , p - l }

= 0

r f. 0 , r2 P .

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 2) nside -qis not.(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) , 0 I kl because < q - < p. Both Andend ir ca)up in contradictions, so the problem is solved. (Dorincases 16. It suffices to prove that there are no integers in the interval � y'P � vp (n - n ,n + n] for n � [ JI -a2avp] + 1 . k Assume by way of contradiction that there is integer such that ny'P - -an < k Xn , hence Jp2 + 1Xn+ 1 -pJX�+ + 1. Xn = 1 =

(1)

But

(2) and so (3) by summing up the relations (1) and (2) . 2 + 1 is irrational. From (3) Because is a positive integer, it follows that 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. 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)

p

[;]

( Titu A ndreescu,

5 1 . Setting n = 0 and n = 1 yields From the given condition we obtain

Jp

Xl = x� and X2 = x� , hence Xl X2 X2n+ 1 = 2n+l + Xn2 and X2n Xn2 +l - X2n-l . X

=

=

=

1.

2.2. SOLUTIONS

79

Subtracting these relations implies hence We induct on

(1) (2)

X2n+1 = X2n + X2n- l , n � 1 .

n to prove that � 1. n , X2n = X2n-1 + X2n-2 (2)

X2 = Xl + Xo and assume that is true up to n . Then X2n+2 - X2n = Xn2 +2 - Xn2 - Xn2 + l + X�n2 - l (*)= (Xn+ l + Xn ) 2 - Xn2 - Xn2 +l + + (Xn+l - Xn ) 2 = X + l( 1+) X� = X2n+ l , and the induction hypothesis) . as claimed (the equality because of O. Because ( 1 ) and( )(2holds ) it follows � From relations that = for all n x X X + +2 + n n n l 0 Xo = and Xl = 1 the sequence (xn ) n 2:0 is the Fibonacci's sequence, hence Indeed,

*

( Titu Andreescu, Romanian Winter Camp 1984; Revista Matematidi Timi§oara 1 ( 1985) , pp. 73, Problem T.3) (RMT), No. 1 2 25 48 52. From the hypothesis it follows that a4 = , a5 = , a6 = . We have �l , a22 = 1 , a33 = 2 , a44 = 3 , a55 = 5 , a66 = 8 so an = Fn for all n = 1 , 2 , 3 , 4, 5 , 6 , where (Fn ) n 2: 1 is the Fibonacci's sequence. F n kFk We prove by induction that a for all n. Indeed assuming that a = = n k n n k 3 for � n + , we have 2 (n + l) Fn+1 - nFn = a2 n+4 = F2 (n + 3) Fn+32 +F(n + 2)F +2 n = (n + 3) n+32+( (n +3)F) n+2 -2F2 (n + l()Fn+1 -Fn(Fn+2 - Fn+l ) = = 3nF+ n+2F3 + n+( 2 - 2n +F 2) n+1 = 2 ( = n + ( ) n+4)3 +F n+2F- n + () ( n+4 3F- Fn+2 ) = = n + ( n+3 + n+2 ) = n + ) n+4 as desired. (Dorin Andrica, Revista Matematidl. Timi§oara (RMT) , No. 1 ( 1986) , pp. 106, C8 2) Problem : O 53. Observe that setting Xo = 0 the condition is safisfied for n = . k that Xk divides m. Let rt be the We prove that there is integer � m3 such 2 0 1 3 remainder of Xt when divided by m for t = , , . . . , m + . Consider the triples (ro , rl , r2 ) , (rl ' r2 , r3), ' . . , (rm3, rm3+ 1 , rm3 +2 ) ' Since rt can take m values, it follows -

-

-

-

-

-

'

2. NUMBER THEORY by the Pigeonhole Principle ( that at least) two triples are equal. Let p (be the smallest) number such that triple rp , rpO+l, rp+ 2 is equal to another triple rq, rq+l, rq+ 2 , p < q :::; m3 . We claim that p = . Assume by way of contradiction that p 2:: 1. Using the hypothesis we have rp+2 Tp- l + rprp+l (mod m) and rq+2 rq -l + rqrq+l (mod m) . rp d= rq,( rp+l = rq+l) and rp+2 = rq+2 , it follows that rp-l = rq - l so (rp- Since with the minimality of p. , which is a contradiction l ,rp ,rp+0 rq- l ,rq,rq+l 0 0 Hence = = ) m (mod Xq therefore and , r = rq so , ( Tiptu Andreescu ando Dorel Mihet, Revista Matematica. Timi§oara (RMT) , No. 106 , Problem C8: 1) 1(1986), pp. F6 , al = F12 , a2 = F? , 4 5 = so 25, = a 9, = 54. a 4, = a3 1, = a that Note ao 2 F F F F sequence. a3 = i , a4 = ; , a5 = i , where ( n )Fn2:o� is the Fibonacci Ff O We induct on n to prove that an = for all n 2: . Assume that ak = for all k :::; n . Hence (1) an - Fn2 ' an -l - Fn2- l ' an-2 = F�_2 ' 80

==

==

=

==

From the given relation we obtain

and

an - 3an- l + an-2 = 2( - l )n- r ,

n 2: 2.

Summing up these equalities yields

(2) 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.

( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1986), pp. 108,

Problem C8:8)

a2 - 1 and 2 + 2a ask for the substitution a � (b2 + b� ) . 2 The equality � ( 2 + � ) = 97 implies ( + �) = 196, hence + � = 14. Setting 55. The expressions b

b

=

b

b

81 2.2. SOLUTIONS 2 2 = c yields (c + �) = 16, thus c + � = 4. Let c = 2 + va. We will prove by induction that F a = � ( C4 . + 4�. ) , n 2: 1, n C F where n is the nth Fibonacci number. Indeed, this is true for n = 1, n = 2 and, assuming that 4 41_ ) ' k a l - + l - +c l or AA bBBl + cCCl - (aPAI + bPBI + cPC/ ) . 4S > a l + 6S - 2S = 4S , which is false. Hence ( 1978) , pp. 74, ( Titu A4Snd>reescu, Revista Matematica Timi§oara (RMT ) , No. 2 3 89 980 6 ) ( Problem ; Gazeta Matematica (GM-B) , No. 2 1 , pp. 64, Problem 18 122) 8. Using the angle bisector theorem and Van Aubel's theorem, it follows that IA BIA CIA b + c k IAI = BI C + CIB = -a- = l IB AlB CI B c + a k and IBI = AI C + CI A = -b- = 2 IC _ a + b k Let ICI - C - 3 and note that kl + 1 = -;;:2p k2 + 1 = b' 2p k3 + 1 = -2p , ' c where p is the semiperimeter. Hence kl 1+ 1 + -k2 1+ 1 + -k3 1+ 1 = 1 -_

so

Furthermore, since

k2 + 1 k3 + 1 (k2 + 1) (k3 + 1 ) 1 A [ ( kl + 1 ) 2 ' cos = !2 k 1 + k 1 _ 3+ 2+

1 00

3. GEOMETRY

we observe that cos A depends only on kl and k3 • Hence cos A = cos Al and, analo­ gously, cos B = cos BI , thus triangles ABC and AIB I CI are similar. Romanian Regional Mathematical Contest " Grigore Moisil" , 1999; Revista Matematica Timi§oara ( RMT ) , 1999, pp.

(Dorin Andrica,

87)

9. Using Steiner's theorem, we obtain

MB · NB AB2 = MC · NC AC2 Applying the AM-GM inequality yields MB MC

NB . / MB · NB AB + NC =2 � 2V MC . NC AC '

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.

( Titu Andreescu)

1 1 . Applying the law of cosines to triangle A l BI C we obtain

and using the inequality x2 get

+

y 2 - xy � xy, which holds for all real numbers x, y, we

AI B� � AI C , BI C. A

101

3.2. SOLUTIONS

�

Similarly, we obtain Bl Cr Bl A · CIA and CI A� three inequalities together, we obtain

� Cl B . Al B . MUltiplying the

Now the lines AAI , BBI , CCI concur, so and after substituting and taking square roots, we have Al Bl . Bl CI . CI AI

� Al B · BI C · 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 . IMO 1996 Shortlist)

=

( =Titu Andreescu,

=

1 2 . We start with the following lemma. Lemma. A, B, C

Let

tan

A

-

2

be the angle of a triangle ABC. Then

+ tan -2 + tan -2 tan -2 + tan -2 tan -2 B

B

C

A

C

= 1.

P f. We present two arguments. h. Since Firoorst approac

+ ,8) [1 - tan a tan ,8] = tan a + tan ,8, tan(90 ° - a) = cot a = l/ tan a, and A/2 + B/2 + C/2 = 90° , the desired identity follows from ( tan - tan - + tan - tan - = tan - tan - tan - ) = 2 2 2 2 2 2 2 = tan 2 tan (�2 + 2 ) [l - tan �2 tan C2 J = = tan � tan (90° - � ) [1 - tan � tan �J = tan(a

A

B

B

B

C

B

A

+

C

C

A C = 1 - tan '2 tan '2 ' Second approach. 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 A ( e tan - = SAB 2 s s - a) -

.

Likewise,

B

tan 2

(s -eb) = -:sSAB

--

and tan

C

2

( s ec) . = sSAB -

1 02

3. GEOMETRY

Hence

A B tan - tan -

by Heron's formula.

2 2 + tan -B2 tan C2 + tan -C2 tan A2 = b) ) + + ( a c ( 8 ( 8 ( 8 ) ) S �BC b) = = 82 ( 8 - a ) ( 8 - ( 8 - c) 1 S�BCb ) = , 8 ( 8 - a )( 8 - ( 8 - c) -

0

-

A

Q

B

C

Without loss of generality assume that AP = min{AP, BQ, CR}. Let = tan( A;2 ), = tan( ii/2) , and = tan( C;2 ) . Then AP = BQ = and CR = The condition given in the problem statement becomes

/r z. y

z

and the equation in the lemma is

r/x,

2x + 5y + 5z = 6 ,

1 xy + yz + zx = . (1) and (2) yields Eliminating x from 5y2 + 5z2 + 8yz - 6y - 6z + 2 =

Completing squares, we obtain

r/y,

x

( 1) ( 2)

O.

(3 y _ 1 ) 2 + (3 z _ 1 ) 2 = 4 ( y - Z ) 2 . 3 1 3 ( 1)/3, z = (v + 1 )/3) gives Setting y - 1 = u, z - = v ( Le. , y = u + 5u2 + 8uv + 5v2 = 82 4 25 < 0, the only Because the discriminant of this quadratic equation is real solution to the equation is u = v = Hence there is 4only / 3) set of /3, yone= zpossible = 1 . Thus values for the tangents of half-angles of ABC (namely x = all triangles in S are isosceles and 4similar / 3 4/ 2 /3 andtoy =onez another. Indeed, we have x = r/AP = = r/BQ = r/CQ =2 1 = 1 , so 3 4 1 we can set r =5 , AP = AR = , and BP = BQ = CQ = CR = . This leads to 24 1 BC = . By scaling, all triangles in S are similar to the triangle AB = AC = and 558 with side lengths , , . O.

O.

x

103

3.2. SOLUTIONS

We can also use half-angle formulas to calculate C

. C= . B = sm sm

2 tan -

2

1 + tan2 C "2 3 : 4 : 5 and BC : AC : BC = 5 : 5 : 8. From this it follows that AQ : QB : BA = Alternative solution. By introducing the variables p = y + z and q = yz - 1 , ( 1) and (2) become 2x + 5p = 6 and xp + q = 0, respectively. Eliminating x relations

yields

p(6 - 5p) Note that

(3 )

+ 2q = O .

y and z are the roots of the equation (4 ) t2 - pt + (q + 1) = O . (3 ) ( 4) Expressing q in terms of p in , and substituting in , we obtain the following quadratic equation in t: 2t pt + 5p2 - 26p + 2 - 0 . O 2 $ . Hence the equation has real This equation has discriminant - (3p 2) 1/3. solutions if p = 2/3, and y = z = Note.only We can also let x = AP, = BQ, z = CR and use the fact that r (x + y + z) = SABC = Jxyz (x + y + z) _

Y

to obtain a quadratic equation in three variables. Without loss of generality, we may set = Then the solution proceeds as above. ( USA Mathematical Olympiad,

x Ti 1 . A d tu n reescu, 13. Let a,

2000)

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): PA sin A � PY sin C

+ PZ sin B.

( 1)

104

3. GEOMETRY

BC PY PZ, BC. = = 90° , AZPYY Z PA A AP M . Z Y PX=. 90° . = = PZBX ZM = P Z B . Y = PY C . = Y Z � Y + M Z. 2R PA � cPY + bPZ. a bPB � aPZ + cPX and cPC � bPX + aPY. Using these Likewise, we have inequalities, we obtain PA PB PC PX ( b c ) PY ( c a ) PZ ( a b ) a2 + b2 + 7 � c3 + b3 + a3 + + b3 + a3 � 2PX 2PY 2PZ � be + co: + � (AM-GM inequality) 4SABC 1 abc Y=ZR ' BC and so on. This Equality in the first step requires that be parallel to ABC. Equality in the second step P occurs if and onlyb if is the circumcenter of ABC is equilateral and P requires that a = = c. Thus equality holds if and only if is its( Ti center. tu Andreescu, USA IMO Team Selection Test, 2000) H is the orthocenter of a triangle ABC then 14. It is known that if AH · BC + BH · CA + CH · AB = 4SABC . YZ

This says that the length of is greater than equal to its projection onto the latter being equal to the sum of the lengths of the projections of and onto as a diameter In fact, since AYP Azp is cyclic with of its circumcircle. By the Extended Law of Sines, sin Let and N be the feet of perpendiculars from and to the line Since 1fZp BxP , sin Similarly, N is cyclic. Hence x:iPZ 13 and sin Thus (1) is equivalent to N Multiplying by and using the Extended Law of Sines, (1) becomes

c3

A

3. 2. SOLUTIONS

1 05

I H. Al, Bl, C1 I H ' ' ' A, B , C A , B , C I BC, CA, AB , IA + lA' � AA1 , IB + IB' � BB1, IC + IC' � CC1 are strict. It follows that ' ' ' ·a IA + b · IB + c · IC > a(AA1 - IA ) + b(BBl - IB ) + c(CCl - IC ) ,

We prove that == Assume by way of contradiction that points and are distinct. Let be the feet of the altitudes from and let be the projection of onto the sides respectively. Hence at least two of the inequalities

or

which is false. Therefore 1 =

H and ABC is an equilateral triangle, as desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2 ( 1981) , pp. 67, 46 16) Problem 1 5 . Using standard notations, we have

64c2 + 49a2 + 9b2 - 112ac - 48bc + 42ab = 6c2 - 6a2 - 6b2 .

This is equivalent to

15b2 + 2b(21a - 24c) + 55ab2 - 1 12ac + 58c2�= O.0 Viewing this as a quadratic equation in , the condition � is satisfied. That is 441a2 - 1008ac + 576c2 - 825a2 + 1680ac - 870c2 � O. The last relation is equivalent to 6(64a2 - 112ac + 49c2 ) � 0, 8 7c. Substituting back into the given condition (8 7 ) 2 or a 3- C 7b� O. follows that a yields a = . We obtain It

=

106

3. GEOMETRY

hence triangle ABC is similar to triangle A' B' C' having sides 7, 3, 8. In this triangle 3 2 82 - 72 cos AI = . 3 . 8 = 2' It follows that A = A7 = Korean Mathematics Competition,

1

2+

60 ° .

2002) ( Titu Andreescu, Fi l i 16. rst so ut on. By the Cauchy-Schwarz Inequality, ..jPB2 + PC2 ..jAC2 + AB 2 2:: PB · AC + PC · AB.

Applying the (Generalized) Ptolemy's Inequality to quadrilateral ABPC yields

+ PC . AB 2:: PA . 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:: ..jPB2 + PC2 . BC. P B . AC >

Combining these three inequalities yields

..jAB2 AC2

+

2:: BC,

implying that angle BAC is acute. With some careful argument, it can be proved that quadrilateral ABPC is indeed convex. We leave it as an exercise for the reader. 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, AQ 2 2:: AQ 2 QC2 = (Ap2 PQ 2 ) (Cp2 PQ 2 ) =

Note. Second solution. +

_

_

_

_

AP2 - PC2 > BP2 2:: DQ 2 . Let I be the ray AC minus the point A. Note that, since PA > PC, Q lies on ray If D did not lie on then AQ would be less than or equal to DQ, a contradiction. Thus, D lies on and angle BAC is acute. B =

l.

l,

l,

o

P o

o�----�o�----�o

A

D

Q

C

1 07

3 . 2. SOLUTIONS

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 ab > O0 and that c > O . Proving that angle BAC is acute is equivalent to proving that > . Since PA2 > PB2 + PC( 2 , ) x2 + y2 > x a 2 + y2 + (x _ b)2 + (y _ C)2 . Hence 2b ( ) 2b b ( o > x - a 2 - x + 2 + y - C) 2 � - x. O. follows that b > 0, as desired. Since P A > P B, we have x > � > Fourth solution. We first prove the following Lemma. F f i W X Y Z in the plane, Lemma. or any our po nts , , , =

=

=

=

_

It

Proof. Pick an arbitrary origin 0 and let w, x, y, z denote the vectors from 0 to W, X, Y, Z, respectively. Then I w - x l2 + I x - Yl2 + I y - Zl 2 + Iz - w l2 - Iw - Y l 2 - I x - Zl 2 w · w + x · x + y . y + z · z - 2I (w , x + x · y + y . z + z · w - w · y - x · z) w + y - x - Z 12 , which is always Equality holds if and only if w + y x + z, which is true Wnonnegative. XYZ is a (possibly if and only if degenerate) parallelogram. 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 AC is acute. Fifth solutangle ion.° InBthis 0° 90° . Note - 1 takes on values between and solution, sin 90 , since PB < PA. Applying the Law of Sines to triangle PAB yields that -p;fjj < =

=

=

=

=

=

0

=

=

_

=

-

sin PAB =

It

follows that

Since PA2

PB - PB sin ABP :::; PA PA '

-

> PB 2 +

_ PB PAB :::; sm 1 PA ' PC2 , we have similarly •

- < ' - 1 PC . _ 1 VPA2 _ PB 2 . PAC _ sm < sm PA PA

1 08

3. GEOMETRY

Thus

-

-

--

1 PB 1 BAC � BAP + PAC < sin- PA + sin then If = sin - 1

()

Hence

��,

()) sin (90° -

=

() cos

=

-

JPA2 - PB2 PA

JPA2 - PB2 () . V'l - sin2 PA J . 1 PA2 _ PB2 =

. - 1 PB + smBAC < sm PA and angle B AC is acute. C

=

PA

90° ,

P

P A B

B

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.

Sixth solution.

/

I

/'

I

(

\

I

\

\

"-

--

...--

I

\

'-...

_

,/

'\.

I

/

_

""

B

;'

P

-....

--

C

,/

...--

./

I

Note that PA2 > PB 2 + 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, will be maximized when B and C are on opposite sides of line PA and and 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 AB 2 = PA2 - PB 2 > PC2 , and similarly AC2 > PB 2 . Hence on

IfA1J Jfijp ;[(j'p

109

3.2. SOLUTIONS

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. B -

,.,

A

P /'

( 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) ,

PB PC sin � sin !!.!!. sin � = PA PB PC PA sin PAB sin PBC sin PCA x sin 40 cos 10 20 sin x sin 40 = sin sin10 sin(80 � - x) sin 30 4 sinsin(80 - x)

1

=

=

------

---------

B

A

C

b The identity 2 sin a cos

formula) now yields

1= so

=

b) b) sin(a - + sin(a + (a consequence of the addition

2 sin x(sin 30 sin 50) sin(80 - x)

+

=

sin x(1 2 cos 40) , sin(80 - x)

+

2 sin x cos 40 = sin(80 - x) - sin x = 2 sin(40 - x) cos 40. This gives x = 40 - x and thus x triangle ABC is isosceles.

=

20. It follows that AcE = 50

=

BAiJ, so

3. GEOMETRY

110

Alternative solution. Let D be the reflection of A across the line BP. Then triangle APD is isosceles with vertex angle 2 (180 - EPA) = 2 (PAii + AiiP) = 2 (10 + 20) = 60, APi5 = 2JiBA = 40. Since 1fAC = 50, we have DB .l AC. and so is equilateral. Also, ISiiA = B

C D Let E be the intersection of DB with CPo Then

= 180180CEi5 = 180 - (90 - AcE) = 90 + 30 = 120 and so PEi5 + I5A.J5 = 60. We deduce that the quadrilateral APED is cyclic, and therefore DEA = DPA = . 60 Finally, we note that I5EA = = 15EC. Since AC DE, we deduce that A across the line DE, which implies99that BA = BC, as desired. and (CTiare Asymmetric d 6) 1 tu n reescu, USA Mathematical Olympiad, 90° , B = 30° 18. Let max{ A, B} = A. If triangle ABC is right-angled, then A = 60° . In order to find !!:. , we may assume that ABC is the triangle with sides and C = r 1 2b 1 a = , = , c = y'3. We have R = and S y'3 2 2 r - - + 12+ J3 - 3 +y'3J3 ' 3 + J3 v'3 1 R so = -- = +. r y'3 R J3 1 Conversely, assume that = r + . From the identity r = 4R sm· 2A sm. "2B sm. "C2 PEi5

-

-

-

.l

_

ABC

--

8

-

-

_

_

---

3.2. SOLUTIONS

it follows that or

111

r = 4(V3 + l)r sin � sin � sin �

Then

0 - 1 ( A - C A + C) . B = cos -2- - cos -2- sm 2" ° , we obtain A C 30 = and, since B B 1 V2 + J6 o4- = ( sin 2 ) sin 2 ' B · Lettmg sm . 2" = x Yl.elds 0 1 + V2 + 4- _ 0 J6 x2 -, x B 5° B 5° V2 V2 J6 . whose solutlOns are x = and x = 2 ' AIt follows that B2" =301 ° Aor 2"90=° . B The is not because � . Hence = , = and ABCacceptable, C = second 60° . Thussolution is right angled. triangle ( Titu Andreescu, Korean Mathematics Competition, 2002) ABC three circles equal to the circum­ 1 9. Construct in the exterior of triangle ABC circle that pass through two vertices P of the triangle. ( By the five-coin theorem the circles will have a common point , as desired see Dorin Andrica, Csaba Varga, Daniel 2002 , pp.Vacare�u, 5 1-56) . " Selected Topics and Problems in Geometry" , PLUS, Bucharest, Alternative solution. Let H be the orthocenter of triangle ABC. The reflections ABC. H of across the sides of the triangle are points of the circumcircle of triangle HAB, HBC, HCA are equal to the circumcircle of Therefore circles of ABC and forthe Pcircum H the claim holds. 2 ( 978) , pp. 74) (Dorin Andr=ica, Revista Matematica Timi§oara (RMT), No. 1 A'B' , B' C' , C' A' by Co, Ao, Bo, respectively, and 20. Denote the midpoints of the ' C' perpendiculars in question by le, lA , lB. Consider the centroid of triangle A'Bthree . -4-

4

_

4

_

4

4

112

3 . GEOMETRY C

1 2,

Since AoG : GA' = BoG : GB' = CoG : GC' = : the dilatation h with center G and coefficient 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 l A and lB under h. Since the images of l A , lB, le under h are concurrent, so are lA , IB , le themselves. ( Titu Andreescu, Romanian IMO Selection Test,

-2

1986)

2 1 . Because triangle AI A2 A3 is not isosceles, it is not difficult to see that the

circumcenter of the triangles AI BI I, A2 B2 I, A3 B3 1 are defined. We start with a simpIe 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

113

i

Let us prove the main statement. For = 1, 2, 3 we denote by the center of the Clearly, lies on the circle and by the circumcenter of the triangle By the lemma, also lies on the same bisector. Thus internal bisector of the angle are perspective from the point By Desargues ' and the triangles theorem these triangles are perspective from a line. This is to say that if we denote = 1, 2, 3, to be the point of intersection of the lines and then are collinear. But since is the perpendicular bisector the points these points are exactly is the perpendicular bisector of and of respectively. the circumcenter of the triangles A student not familiar with Desargues' theorem may proceed from the point as follows. Applying Menelaus ' theorem to the triangles respectively, one and to the triples of points can, observing usual agreement about the signs, write:

Q Ai+1 IAi+2 . i Oi Ai' Ti I. T1 T2T3 0i+1 0i+2 Ti+1 Ti+2 ' Qi, i Ti+1 Ti+2 QQ Q Bi li Ail 01 i,+120, i+32 A1 B1 I, A2B2I, A3 B3 I, Remark. 101 02 , 102 03, 103 0 1 (T1' T2 , Q3 ) , (T2 ' T3, Q 1 ) , (T3 ' Tt, Q2 ) , 01 T1 IT2 02 Q3 IT1 . 02 T2 . 01 Q3 - 1 , IT3 02 T2 Q3 Q 1 03 T3 . IT2 . 02 Q 1 - , IT1 03T3 0 1 Q2 01 T1 . IT3 . 03 Q2 - . MUltiplying them all one gets 0 2 Q 3 03 Q 1 01 Q2 0 1 Q 3 . 0 2 Q 1 . 03 Q 2 = QQ Q meansi that lthei points 1 , 2 , 3 are collinear. which Alternat ve so ut on. This proof will be based on inversion. We take the incenter I Ci

Ti 0 1 02 03

_

--

--

--

_

_

1

1

1,

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.

11 4

B'I

3. GEOMETRY

B2' Gi is a straight line B�+I B�+2 ' with these lines Indeed, the imageBof the circle � B�B� . The line AiAi+1 will be transformed into the circle forming the triangle ri+2 , with the side AiAi+1 becoming the arc Ai Ai+1 which does not contain I. Note AAA I that all these circles have equal radii since the distances from to the sides of I 2 3 were equal. E I, E2 , E3 are three circles passing through the common point Let us note that if I and no two of them touch, then their centres are collinear if and only if there is i- I through which all these three circles pass. another common point Ei being the circumcircle of Ai Bi I. Since the We will useEthis observation for AiB� , the desired result is to show that the lines A� BL to the line Ainversion B� are iconcurrent. A� A� A� � B� , A�takes For this, it suffices to show that the triangles B� B� B� are homothetic, which is the same to say that their corresponding sides and r r r PPP are parallel. Since the radii of the circles I , 2 , 3 are equal, the triangle I 2 3 formed their centre has its sides parallel to the corresponding sides ofAthe triangle B� B�B�by. The A A� into I 2 � � / 1 centred at takes thePtriangle homothety of ratio P2 P3. Therefore the the triangle whose vertices are the midpoints of the triangle I A A A� � � and PI P2 P3 are also parallel and the corresponding sides of the triangles result( Tifollows. tu Andreescu, IMO 1997 Shortlist) I be the intersection point of lines GB' and G'B and let A' be the 22. Let AI and BG . intersection point of lines We have AB' AG' AI B'B + G' G = lA" I from Van Aubel 's theorem, therefore ::, is constant. Hence the locus of point is a BG line (segment dreescu) to . Titu Anparallel J

3 . 2 . SOLUTIONS

115

23. Without loss of generality assume that

PC = max{PA, PB, PC } .

The condition in the hypothesis is

PB . PC + PA . PC = PA . PB + 1 PC PA · PB + 1 · 1 or 1 - PA . 1 + PC . 1 . PACB is a From the converse of the second theorem of Ptolemy it follows that AB C P cyclic quadrilateral. Note that cannot be , or P otherwise the denominator of side equals OA. Hence the ABCright-hand B, C. the locus of point is the circumcircle of triangle without the vertices , ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 ( 1985) , Problem C7: 3) 24. We have Ja2 b2 4S2 + Ja2 c2 - 4S2 = Ja2 b2 - a2 b2 sin2 C+ + J-a2 c2 - a2 c2 sin2 B = ab cos C + ac cos B = = ab a2 +2ab2b c2 + ac a2 +2acc2 - b2 = a2 as desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 8 ( 1971) , pp. 25, 1006) Problem _

25. The relation is equivalent to

+..j y'rb + Fc = -,1 r rarbre 1 1 1 1 or + + = -. Fay'rb FcFa Fay'rb r On the other hand, 1 1 1 1 -ra + -rb + -re = -r , so 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 = 0, - ..fib y'rb - Fc Fc - Fa ' It follows so ra( Ti that the triangle is equilateral, as desired. ( 974) = rb A= rdeFa, tu1903n )reescu, Revista Matematica Timi§oara (RMT) , No. 1 1 , pp. 2 1 , Problem Fa,

---

116

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 Then

b f:. c, 2 a2 (b + C) 2 - a2 2 +� c b -2 =p(p - a) m= -4> 4 b � ) � and likewise m 2: p (p - , m 2: p(p - c) , It follows that 1 1 1 27 a 4S - · -p = r,

3.2. SOLUTIONS

11 9 desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 ( 1973) , pp. 43, 1 585) Problem as

33. We have

It follows that hence as

desired.

A 7r . A 1 _2

>

as desired. The equality occurs if and only if

a Jai bi i 1234 b i 1234 for = , , , hence i = a - I , = , , , . Then PAl = PBI = PCI = PDI , ABCD . P so is the kincenter of tetrahedron RTiemarA. The inequality holds for convex polyhedra circumscribed about a sphere. 982 ( tu ndreescu, 1 Romanian Selection49Test, ( 1 (1982) , pp.IMO82, Problem 10) ; Revista Matematidi Timi§oara RMT ) , No. i 1 i 4 0 and 60. All summations here range from = to = . Let be the circumcenter A A A A GA GA �= R be the 4 2 I · 3 circum radius of By the Power-of-a-point Theorem, . i 4 OG i 1 R2 - 2 , for � � . Hence the desired inequalities are equivalent to ( 1) and (R2 - OG2 ) L G� � L GAi • ( 1 ) follows immediately from i Now (3) �= V b;

3. GEOMETRY

136

) P denote the vector (4)

by the Arithmetic-Geometric-Mean Inequality. To prove (3 , let from to the point Then

0

P.

(4) vanishes. By Cauchy-Schwarz L GAi L GA1 2:: 16, and i GA1 i > "GA ·. GA1 i > -161 (" (GA · ) 2 ) 2 " 4-1 "GA2. " ). 3 ( from follows also (2) Hence ( Titu Andreescu, IMO 1995 Shortlist) )

This is equivalent to (3 , since the last term of Inequality, so

6

� 6

-

6

Z

6

-

6

�

C hapter 4 TRIGON OMETRY

PROBLEMS 1 . Prove that

27r 37r 7r ctg2 7 + ctg2 7 + ctg2 7

=

5.

2. Prove that

3 X + 47r X X + 2 7r cos3 - + cos3 -- + cos3 = - cos x 4 3 3 3 --

for all x E JR. 3. Evaluate the sum

Sn =

n- 1

L sin kx cos(n - k)x. k=1

4. Evaluate the sums

S1

=

sin x cos 2y + sin 2x cos 3y + . . . + sin(n

S2

=

cos x sin 2y + cos 2x sin 3y + . . . + cos(n - l )x sin ny.

-

l)x cos ny,

5. Evaluate the products

1) PI = (1 - tg1°) (1 - tg2 ° ) . . . (1 - tg8g 0 ) ; 2 ) P2 = ( 1 + tg1 ° ) (1 + tg2° ) . . . ( 1 + tg44°). 6. Prove that

(4 cos2 go - 3) (4 cos2 27° - 3)

=

7. Let x be a real number such that sec x - tan x 8. Evaluate the product

where Ixl

< 2n7r+2 • 1 39

tan go.

=

2. Evaluate sec x + tan x .

4. TRIGONOMETRY

140

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 sin 4x

=

a

--b

=

-

c

d

-'

10. Let a, b, c, d E [0, 7r] such that

2 cos a + 6 cos b + 7 cos c + 9 cos d = °

and

2 sin a - 6 sin b + 7 sin c - 9 sin d = 0 .

Prove that 3 cos(a + d) = 7 cos(b 1 1 . Prove that if

+ c) ,

arccos a + arccos b + arccos c = 7r,

then

1 2 . Let a, b, c be positive real numbers such that

ab + bc + ca = Prove that

-1

1

1.

-1

arctg + arctg - + arctg = 7r, b a c

13. Let x and y be real numbers from the interval

Prove that x + y = 2 ' 7r

cos2 (x - y) = sin 2x sin 2y

(o , �) such that

( i) such that 21 ( 1 - tga) (l - tg(3) ( 1 - tg,) = 1 - (tga + tg(3 + tg,) ,

14. Consider the numbers a , (3 "

E 0,

7r

Prove that a + (3 + , = 4"

( �), Prove that ( Si�2 a ) 2 + ( cos2 a ) 2 = 1

15. Let a, b E 0,

if and only if a = b.

sm b

cos b

141

4. 1. PROBLEMS 16. Prove that

sin3 a cos3 a + sin cos

b -> sec(a - b)

b

--

7r

b

for all < a, < 2" '

0

--

1 7. 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. 1 8 . Let x be a real number, x E [- 1, 1] . Prove that

for all positive integers n .

_1_ < x2n + (1 _ x2 ) n < 1 2n- 1 -

19. Prove that

sec2 n x + cosec2 n x 2:: 2 n+ 1 for all integers n 2:: and for all x E

0

(0, %) .

20. Prove that

(1 + sin x) (1 + cos x) for all real numbers x.

:s

� + V2

2 1 . Find the maximal value of the expression

E = sin Xl COS X2 + sin x2 cos x3 + . . + sin x n cos X l , .

when X l , X 2 , . . . , Xn are real numbers. 22. Find the extreme values of the function f : JR

b f (x) = a cos 2x + cos x + b O b where a, , are real numbers and a, > .

-t

1R,

c,

c

23. Let ao, a I , . . . , an be numbers from the interval

tan ao -

(

Prove that

(0,

7r

/2) such that

�) + tan (a1 - � ) + . . . + tan (an - �) tan ao tan a1 . . . tan an

2::

n

n+ 1 .

24. Find the period of the function

f(x) = cospx + cos qx, if p, q are positive integers.

x E JR

2::

n

- 1.

12

4. TRIGONOMETRY

4

25. Let

aD = + + �

V3

for all n .

an+1 = 2(aa2nn-+25 for 2:: O. Prove that an = cot ( -2n3--37r) -2

v'6 and let

)

n

26. Let n be an odd positive integer. Solve the equation

cosnx = 2n-1 cos x.

27. Solve the equation

A sin2

x + B sin 2x + C =

0,

where A, B, C are real parameters. 28. Solve the equation

. cosy + s.m y cos z+sm. z cos x = 3 Prove that the equation . . 2xsm. 3xsm. 4x = 3 smxsm

'2

Slll X

29.

4

has no real solutions. 30. Solve the system of equations

=3 { 32sisinn xy ++ 23cosy cos x = 4.

{ xsix+yn y=+:1 - x cosy = V;

31. Solve the system of equations

4'

2

32 . Solve the system of equations

1

x+y+ Z = 437r tgx + tgy +tgz = 5 1 tgx . tgy . tgz = .

33. Prove that in any triangle

acosA + b cos B + c cos C = 2R2abc

4. 1 . PROBLEMS 34. Prove that in any triangle A B C A B C � cos3 "2 sin "2 sin "2 cos "2 cos "2 cos "2 � sm. 2 "2A ' �

1 43

�

=

35. Let n be a positive integer. Prove that in any triangle

L sin nA sin nB sin nC (_l) n+l + cos nA cos nB cos nC and L cos nA cos nB sin nC sin nA sin nB sin nCo ABC such that 36. Consider a triangle A B B C C A A sin sin + sin sin + sin sin =

=

=

and

( 1 + sin A) (l + sin B) (l + sin C) 2 (A + 1) AB C has a right angle. Prove that triangle ABC be a triangle such that 1 37. Let A > be a real number and let a>' cos B + b>' cos A c>' and a2>.- 1 cos B + b2>.- 1 cos A C2>'- 1 . Prove that the triangle is isosceles. ABC is equilateral if and only if 38. Prove that the triangle A B C 1 ( 2 b2 2 ) tg "2 + tg "2 + tg "2 a + +c . ABC be a triangle such that 39. Let in2 B + sin2 C 1 + 2 sin B sin C cos A. s ABC =

=

=

= 4S

=

Prove that triangle 40. Let

has a right angle.

ABC be a triangle such that

(cot � r (2 cot �r + (3 cot �r (�;)" , +

=

where s and r denote its semiperimeter and its inradius, respectively. Prove that triangle is similar to a triangle whose side lengths are all positive integers with no common divisor and determine these integers.

ABC

T

4. TRIGONOMETRY

144

41. Prove that in any triangle

A B C 9 . � sm - cos - cos - < - . 2 2 2 -8 L..J

42 . Prove that in any triangle

a-b2 + -b2 + -e2b - (sm. 2 -A + sm. 2 -B + sm. 2 -C ) . e ea a 2 2 2 >

4

43. Prove that in any triangle

� e3 - 16p3 . 44. Prove that in any triangle A 2B 2C 2 sec - sec - sec be 2 + ea _2 + ab 2_ p29 45. Prove that in any triangle 3V3. !!. ABC be a triangle. Prove that 46. Let 3A . 3B . 3C A - B B - C C A .sm < cos -- + cos -- + cos -- . + sm - + sm - 2 2 2 2 2 2 ( b) ( bi) 2002 = a- bi , a, b E 47. Find the number of ordered pairs a, such that a + cos A

a3

+

cos B

b3

+

__

__

cos C

>

__

> _. -

> r -

-

R

48. Find

Imz 5 z EC\1R Im5 z and the values of z for which the minimum is reached . min

-­

ZI, Z2 , . . . , Z2n be complex numbers such that IZl l = IZ2 1 = . . . = IZ2n l and arg ZI arg Z2 . . . :::; arg Z2n Prove that 49. Let :::;

:::;

:::;

7r.

50. For all positive integers k define

Prove that for any integers m and n with 0 < m < n we have

Al A2 U

U···U

Am An m+1 U An m+2 U · · · U An . C

4. 1. PROBLEMS

145

Z1 ,Z2 ,Z3) be3 complex numbers, not all real, such that I Z1 1 = IZ2 1 = I Z3 1 = 1 Z2 + Z3 - Z1 Z2Z3 E JR.

5 1 . Let and 2 ( Z1 +

Prove that

� is odd and co, C1 , . . . , cn- 1 the complex roots of unity of order n. Prove that nII- l ( b % ) ( � b � ) 2 k=O ba + c = a + for any complex numbers a and . 53. Let n be an odd positive integer and co, C 1 , . . . , cn - l the complex roots of unity of order n. Prove that nII-1 ( b % ) n bn kb=O a + c = a + for all complex numbers a and . I 1 I 1 I 1 54. Let Zl,Z2 ,Z3 be distinct complex numbers such that Z1 = Z2 = Z3 = Prove that 1 1 1 1 I Z1 - z2 11z1 - z3 1 -- + I Z2 - zd lz2 - z3 1 + I Z3 - zl llz3 - z2 1 >- -. ---r2 I 1 I 1 I 1 55. Let Zl ,Z2 , Z3 be distinct complex numbers such that Z1 = Z2 = Z3 = r and Z2 f:. Z3· 52. Let n be an even positive integer such that

r.

Prove that

I 1 If Z is a complex number satisfying z3 + z - 3 :::; 2, the inequality show that I z + 56. 1 z -1 2. ( ) following property: 57. The pair ZI, Z2 of nonzero complex numbers has the O 2 there is a real2number 3 a E [-2, 2] such that z? - aZl Z + z� = . Prove that all pairs (zf, z�), n = , , . . , have the same property. AA A 1 58. Let 1 2 . . . n be a regular polygon with the circumradius equal to . Find IIn PAj when P describes the circumcircle. the maximum value of max j=l :::;

.

4.

146

TRIGONOMETRY

59. Let n be an odd positive integer and let interval [0, 71'] . Prove that

aI, a2, ... , an be numbers from the

L cos(ai - aj ) 2:: 1 --n · 2 ijn

l� < �

n be a positive integer. Find the real numbers ao and akl , k, 1 , n, l k > , such that L akl cos 2 (k - l) x sin2 -sm. -2nxx ao + 9 O. Consider the equation

n

(2n + l)x 0, 27r with roots 7r 2n + 1 ' 2n + 1 ' . . . , 2nn7r+ 1 . (2 Expressing sin n + l)x in terms of sin x and cos x, we obtain 2n : 1 2 ( ( ) cos2n x sin x en : 1) COs2n-2 sin3 X + . . . sin n + l}x n 1 n 1 Sin2 n+ 1 X ( e : ) cot2n - e : ) cot2 n -2 x + . . . ) k7r 2 2 · SIn · 2n+l x 0 , we have , k - 1 , , . . . , n. Slnce Set x n + 12 ( n : 1) cat2n en: 1) cot2n-2 x + . . sin

=

_

=

X

=

-

=

--

-I-r

-

X

Substituting y = cot2 x yields l n y

_

.

= O.

en : ) en : l) yn- l + .

. . = 0,

-

with roots cot2 cot2 �, ficients and roots, we obtain

2n7r+_,1 2n + 1 . . . , cot2 2nn7r+ 1 . Using the relation between coef_ 2n + 1 ( ) n ( 2n - 1 ) k7r n 3 2 � cot2 n + 1 en : 1 ) 3 desired conclusion follows. 2 ( 979) , 51 , Dorin An ndr3,ica,theRevista (Setting Matematidi Timi§oara (RMT), No. 1- 1 �

=

=

pp.

Problem 3831)

147

148

4. TRIGONOMETRY

Applying the identity cos t = 4 cos3 3"t - 3 cos 3"t ' t E 1R for t = x, t = x + 271", t = x + 471" and summing up the three relations, we obtain X +-271" + cos3 X +-471" ) 3 cos x = 4 (cos3 3'X + cos3 3 3 X + 471" ) . -3 (cos -X3 + cos X +3 271" + cos -3 On the other hand, X +-471" = 2 cos 6471" cos 2x + 471" + cos x +271" + cos X + 271" cos 3"x + cos 6 3 3- = 3 2 x � 2" = 0 = (2 cos ; + 1) cos and (the follows. ir ca, Revista Dordesired in Andidentity Matematica Timi§oara (RMT) , No. 2(1975), pp. 44, Problem 2124) 3. We have 2.

--

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, n-1 = -- sinnx. 8 n (Dorin Andrica, Gazeta Matematica2 (GM-B), No. 8(1977), pp. 324, Problem .

+

+

so

16803; Revista Matematica Timi§oara (RMT), No. 2(1978), pp. 30, Problem 3055) 4. Note that 81 + 82 = sin(x + 2y) + sin(2x + 3y) + . . . + sin[(n - l)x + ny] and 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 ) = .

.

149

4.2. SOLUTIONS h n l sm. [y + (n - I) 2hI ] .sm T .sm -h I S2 - SI = sin(y + h2 ) + sin(y + 2h22 ) + . . . + sin(y + (n - I ) h ) = 2

h l . [ ( h I ] . n h2 . [ ( h 2 ] n . SI = 1 sm T sm y +h n - 1) 2 I sm T sm y +h n - 1) 2 2 2 s. 22 sm. 21 and h; ] hl [ ( hI ] [ n n 2 � ( S 1 sin 2 sin Y +h n - 1) 2 + 1 sin sin Y +h n - 1) . 2-2 2 sm. 22 sm. 2I (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977), 65,

Hence

m

_

Problem 3056 ) 5 . We have PI = 0 because of the factor 1 - tan 45 ° = O. On the other hand we have 44° + sin 44°) = P2 = (cos 1 ° + sincos1 0)1 °. .. .. .(cos cos 44° � cos 10 + � sin 10 � cos 440 + � sm 440

(

(.;2)

) ( • • •

pp.

)

= �--------------� 44 --�----------------�

cos 1 ° . . . cos 44° sin 46° . . . sin 89° (v'2) 44 = 222 . = ( Titu Andreescu) cos 1 ° . . . cos 440 2

= 4cos3 x - 3cosx, so 4cos2 X - 3 = -- for all x f. (2k +6. 1)We. 90°,havek Ecos3x cos x Z. Thus 27° . cos 8 1° cos 81° sin 9 ° (4cos2 9 ° 3)(4 cos2 270 3) = cos cos go cos 270 = cos 9° = cos 9° = tan 9° , desired. ( Titu Andreescu) _

as

_

cos3x

1 50

4. TRIGONOMETRY

From the identity 1 + tan2 x = sec2 x it follows that 1 = sec2 x - tan2 x = (sec x - tanx)(secx + tan x) = 2(secx + tanx), = 0.5. so secx Andxreescu, ( Titu+ tan American High School Mathematics Examination, 1999, Prob­ lem 15) 8 . Since 1 + tan2 2kx cos2 2kx (1 - tan2 2k x)2 - cos2 2k+l X for all Ixl < 2:+2 ' it follows that cos2 2k xl = cos2 2x Pn = IIn cos2 k= l 2k+ X cos2 2n+1 x ' 7.

(Dorin Andrica) 9.

Let

4x - A sin3x - sin2x - sinsin-x - d b a Then sin2 4x = 2sin2 2x(l - sin2 2x). Because sin2 4x = 2 sin2 2x(1 - sin2 2x), we obtain dA2 = 2b2 (1 A A2 b2 ). On the other hand, sin 3x = C, sin x = a, and since sin 3x = sin x (3 - 4 sin2 x), we have = (a(31 ) - 4A2 a2 ). A Eliminating from the relations and (2) yields 2a3 (2b3 - � ) = b4 (3a ) desired. (D i A d i ) _

_

_

_

c

.

_

c

-c ,

as

or n n r ca

as

Rewrite the two equalities 2 sin a - 9 sin d = 6 sin b - 7 sin 2 cos a + 9 cos d = - 6 cos b - 7 cos By squaring the two relations and adding them up we obtain 1 85 + 8 cos(a + d) = 85 + 42 cos(b + ) and the conclusion follows. 10.

c

c.

c ,

(1)

(2)

4.2. SOLUTIONS

( Titu Andreescu, Korean Mathematics Competition, 2002)

15 1

From the hypothesis it follows that b sinearccos a + arccos + arccos c) = o. Using the identity L cosacos,Bsin'Y = sin(a + ,B + 'Y) + sin a sin,Bsin'Y and the formulas sin(arccosx) = � and sin(arcsinx) = x, x E [-1, 1], we obtainsb a � + be� + eaJ1=b2 = )(1 - a2 )(I - b2 )(I - c2 ) , as desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 64, Problem 3054) 1 2 . The identity arctgx + arctgy + arctgz = arctg 1 x-+(xyy ++zyz- +xyzzx) + k7r implies b + be + ea - 1 1 1 1 a b + e) + k7r. b = arctg arctg + arctg + arctg e ( a ea + b a Because ab + be + ea = 1, we obtain arctg -a1 + arctg -1b + arctg -1e = k7r, where k is integer. Note that 0 < arctgx < "27r for all real x > 0, hence 1 1 37r 1 o < arctg � + arctg z; + arctg c < 2· Therefore k = 1 and arctg -a1 + arctg -1b + arctg -e1 = 7r, as claimed. ( 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 11.

4. TRIGONOMETRY

152 Hence

cos2 (x + y) = 0, and (since (0, %), we obtain x + y = %, as desired. Titu x,AnydEreescu, Revista Matematidi Timi§oara (RMT) , No. 1(1977) , pp. 42, Problem 2826) 14. Expanding yields 1 (1 - tana - tan,B - tan')' + tan a tan,B + tan,Btan,),+ "2 + tan,), tan a - tan a tan ,B tan,),) = = 1 - (tan a + tan,B + tan,),), or tan a + tan,B + tan')' - tan atan,B tan,), = = 1 - tanatan,B - tan,Btan,), - tan')'tana. (1) Since a,,B, ')' E (0, i), we have ° < a + ,B + ')' < 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 tan a + tan,B + tan,), - tan a tan ,B tan ')' = 1, therefore cot(a + ,B + ')') = 1. Hence( Tia +A,B +d ')' = i, desired. tu n reescu, Revista Matematica Timi§oara (RMT) , No. 1(1973) , pp. 42, Problem 158 2) 1 5 . The relation in the statement is equivalent to 4 a cos4 a ) = 1, (sm. 2 b + cos2 b) ( -sin cos b sm. -2-b + � or sin2 b sm4 a + cos4 a + -cos .sm-22-bb sm. 4 a + � cos b cos4 a = 1. It follows that sin2 bb cos4 a = 1 , . 2 acos2 a + -cos.-2-b sm. 4 a + � 1 - 2sm cos sm2 b hence COS b . sin b cos2 a) 2 = 0. ( --:-2 a sm sm b cos b Furthermore, cos b sm. 2 a = -sin b cos b cos2 a, sm b 7r ,

as

•

--

--:--

153

4.2. SOLUTIONS

or tan2 a = tan2 b. Because a, b E (0, -i), we obtain a = b. The converse and we are done. Alterna it ve soislutclear ion. From the given relation we deduce that there is a number e E (0, -i) such that cos2 a sin2 a -sm. -b- = sine and -cos--b = cose Hence sin2 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) 2 Since a, b, e E (0, i), we have b e = ° and a = b + e, hence a = b, ( Titu A ndreescu, Revista Matematica Timi§oara (RMT) , No. 1 (1977) , desired.41 , Problem 2825 ; Gazeta Matematica (GM-B), No. 11 ( 1977) , 452, Problem 16934) 16. Multiplying the inequality by sinasinb + cosacosb = cos(a - b), we obtain the equivalent form 3 a cos3 -a ) (sin a sin b + cos a cos b) 2:: 1 . ( Sin -sm. -b- + -cos b But this follows from Cauchy-Schwarz Inequality, because, according 1to this in­ equality, d ) side is greater than or equal to (sin2 a + cos2 a)2 = . ( Ti theA left-hand as

pp .

pp.

tu n reescu

17.

Using the inequality

for m 2:: 1 we obtain (1 + 2 sin2 a)p + (1 + 2 cos2 a)p :::: 2 ( 2 + 2 sin2 � + 2cos2 a ) = 2,8+1 , desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 (1974) , 30, Problem 1942) [ 1 1] 18. Because x E - , , there is a real number y such that x = siny. It suffices to prove that 1 >- sin2n y + cos2n y > � - 2n 1 P

as

pp.

-

4. TRIGONOMETRY

154

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 + cos2 = 1, desired. For the right-hand side we use the inequality x? + x� - ( Xl + X2 ) n 2 2 Hence sin2n y + cos2n y - ( sin2 + cos2 y ) n 2 2 claimed. Alternative Solution. By setting u = x2 and v = 1 - x2 the inequality becomes [0, 1] and u + v = 1, we have un � U, vn � v, 1 n n E u,v Because 1. u v + � � 2n- l un + vn � U + v = 1. Also, by the power mean inequality, implying 1_ un + vn - 2 ( u +2 V ) n = 2 ( !2 ) n = _ -l . 2n (D i A d i ) y

y

as

>

y

>

as

>

or n n r ca

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 + cosec2n x 2:: 2n (tann x + cotn x) . Since tann x + cotn x 2:: 2, we obtain sec2n x + cosec2nx 2:: 2n+l , desired. Alternative Solution. Using the AM-GM inequality we obtain sec2n x + cosec2n x - 2Jsec2n xcosec2nx = 2 sinn x1cosn x 12 = 2 n+ l n _ x - 2n+lNo. 3(1975), 104, Problem sin (Dorin Andrica, Gazeta Matematidi (GM-B), 19.

as

>

_ _

>

pp.

14900)

20.

We have (1

.


. 27rA T AT Smce = 1 and 1 = d ' k1 k2 1 k (Ad) , q = k2 ( Ad) . = q = Ad and so p = 1 P d d(p, q) , so A = 1 , hence T = T1 desired. On the other hand, = gc (Dorin Andrica, Revista Matematidt Timi§oara (RMT) , No. 2 ( 1978) , 75, Problem 3695) l

•

as

25.

pp.

We have

7r 2 cos2 7r 1 + cos 17r2 cos 7r cot 24 = -sm.--247r- = 2sm. -7r cos24 -7r --=-==. -17r2 = sm 24 24 24 1 + cos (i - �) 1 + V; + 4 v: + VB + v'2 VB4 v'24 VB - v'2 sin (i - �) v'2) + ( VB + v'2)2 4(VB + v'2) + 8 + 4V3 4 ( VB + 6-2 4 = 2 + v'2 + V3 + VB = ao + 2. 2n-3 ) - 2 is true for n = O. Hence an = cot ( T 2n-3 ) , where b = an + 2, n � 1 . The It suffices to prove that bn = cot ( T n recursive relation becomes )2 5 ( bn - 2 bn+1 2 - 2b ' n b 2 -1 b 3 7r ' yields 2k -3or n+1 = ;b . Assuming, inductively, that bk = cotck, where Ck = n 1 bk+1 = cot2 2coCkt Ck- = cot(2Ck ) = cot Ck+1 , and (weTi areAdone. tu ndreescu, Korean Mathematics Competition, 2002) 1 26. If n = , then all real numbers x are solutions to the equation . ---���� = --����

=

_

_

_

_

_

_

1 60

4. TRIGONOMETRY

Let n > 1 and note that cosnx = (�) cosn X - (�) cosn-2 xsin2 x + . . . + 1 ) ( n 1) cos x sinn- 1 x. +(n We have two 1cases: � a) x f:. (2k + ) 2 for any integer k . Then I cosnxl = I cosxl l (�) cosn - 1 X - (�) cosn -3 sin2 X + . . . + ' ; ) n 1 (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 ) < n-l

-2-

_

hence there are no solutions in this case. b) x = (2k + l) i for some integer k. Then cos x = 0 and cosnx = 0, sinceAln is odd,i soS {l (2ki + 1 ) i l k integer } is the set of solutions. ternat ve o ut on. With the substitution x = i - y the equation becomes ( 1) cos (n � - ny) = 2n - 1 siny. Because n is odd, ( 1) is equivalent to ± sinny = 2n - 1 siny. Taking modules gives ( 2) I sinnyl = 2n -1 1 sinyl · But I sin nYI � n l sinyl for all y in lR, hence n l sinyl � 2n - 1 I sinyl · If y f:. k7r , k E Il , then n � 2n - 1, which implies n E { 1 , 3 } . The case n = 1 original equation reduces to cos3x = 4cosx, that is 4iscosclear3 X -and3 cosforx =n 4=cos3 x.theTaking 0 into account that cos x f:. , this yields cos2 x = �, which is not possible. It( Tifollows Il . that y = k7r , which gives the solutions x = (2k + l) i , k E304 A d 7 978 ( ) 1 tu n reescu, Gazeta Matematidi (GM-B),1 No2 ( 1. 980) , pp. , Problem 1 7297; Revista Matematidl Timi§oara (RMT), No. , pp. 63, Problem 4107)

4.2. SOLUTIONS

The equation(Ais equivalent to G ) + sin2 x + 2B sinxcosx + 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 1 )1r I k Z (2k 2 k 1 k E Z} . ; x E E { U { arcctg ( - g ) + 1r } ii) A + G 1= 0 and G = O . Then sm. x = 0 or tan x = - A2B+ G ' hence {k1r 1 k E Z } U { arctg (-}:c ) + k1r1 k E Z } . x E iii)) AA = GB = 0G = O . Then any real number is a solution. iv = = and B 1= O . Then sin 2x = 0 Zand so k� 1 k E } . x E { equation is equivalent to0 v) A + G 1= 0 and G (1=A O . The G G ) + tan2 x + 2Btanx + = , hence B ± JB2 AG + G2 A+G tan x = for B2 + G2 � AG . It{ follows thatk l k Z } { k 1 k E Z} arctgY + 7r E x E U arctgY1 + 7r 1 if B2(D+ Gi2 �AAGd .i Otherwise there are no solution. ( ) 1 ( 1978) , or n n r ca, Revista Matematica Timi§oara RMT , No. Problem 3429) 28. The equation is equivalent to 2 sin x cos Y + 2 sin Y cos z + 2 sin z cos x = 3, or (sin x - cosy)2 + (siny - COS Z)2 + (sinz - COS X)2 = O. It follows that sin x = cosy, siny = cosz, sinz = cosx. Hence

16 1

27.

-

_

pp.

89,

162

4. TRIGONOMETRY Z+X=

(4k3 + 1) '27r

for some integers k1 ' k2 ' k3 and therefore [4(k1 - k2 + k3 ) + 1] 7r ' y = [4(k1 + k2 - k3 ) + 1] 4"7r ' X= 4" and [4( k k k ) 7r k k k Z = - 1 + 2 + 3 + 1] 4" ' 1 , 2 , 3 E Z. ( Titu Andreescu, Gazeta Matematica (GM-B ) , No. 11 ( 1977) , 451, Problem 16931; Revista Matematica Timi§oara (RMT) , No. 1-2 ( 1979) , 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" ( 2cos2 5x - cos2x + cos8x - cos4x + cos6x ) < 8" = 4' hence( Tithe equation has no solution. A d tu n reescu, 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 + sin2 y) + 12 (sinxcosy + sinycosx) = 25, or 13 + 12 sin (x + y) = 25. Hence sin (x + y) = 1, and so (4k + 1 ) '27r X+y= for some integer k. It follows that sin x = cosy and siny = cosx. Turning back to the system we obtain 4 .smx = cosy = 53 and smy . = cos x = 5 ' 4 hence 3 tanx = 4' tany = 3 · Note that sinx, cosx, siny, cosy are all positive, therefore x = arctan 43 + 2k7r pp.

Y

pp.

4.2. SOLUTIONS

163

4 217r and y = arctan 3' + , l k for some . Andreescu,andRevista 2 ( 1978) , 74, ( Tituintegers Matematica Timi§oara ( RMT ) , No. Problem 3694) [ 1 1] 31 . Observe that x E - , and x = sin (arcsin x) , � = cos(arcsin x) . From the first equation we obtain . .j2 cos(y - arcsmx ) = 2' then y - arcsin x = ± � + 2k1l". . x + y = "4 ' we get Usmg x + arcsin x = � ± � + 2k1l", k. for some Caseinteger 2k1l". Because x E [- 1 , 1] and arcsin x E [-i, iJ , we 1 . x + arcsin x = i have k = 0, hence x + arcsmx = 2' Therefore arcsmx = 2' - x, or x = cosx. For this equation there is only one solution Xo E (0, � ). The system has the solution Case 2 . x + arcsin x = 2k1l".x = Xo, = 4 - Xo Using similar arguments, k = 0, so arcsin x = -x. This equation has the unique solution x = ° so the system has the solution pp.

11"

11"

•

•

11"

Y

11"

-

x = o, ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 ( 1 977) , 41 , 2824 ) Problem pp.

32 .

Using the formula + tany + tanz - tanxtanytanz tan (x + y + z) = 1tanx - tanxtany - tanytanz - tanztanx

---=�---------=�--

----

4. TRIGONOMETRY

164

we have

- 1 = tan -37r4 = 1 - tan x tan y - tan5 -y1tan z - tan z tan x

-------

Hence

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, + h7r, + p7r , for some mtegers 12 12 Revista Matematica Timi§oara (RMT), No. 8(1971), pp. 27, Problem 1018)

{ Ti A d ( tu n reescu,

}

k, l ,p.

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 = cos = 2R sin C(cos(A - B ) + cos C ) = 4R sin C cos 2 2 = -4R sin C cos as desired.

( i - B) cos ( i - A)

==

4R sin A sin B sin C ==

;!�,

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

Problem 3060)

34. Because A + B + C = 7r, we have

B A B C . C . "2 cos "2 = sm cos "2 + sm "2 cos "2 '

Hence

B C A . A C . cos - + sm cos - = sm - cos - ' 2 2 2 2 2 A B . B C A . cos - = sm cos - + sm - cos - ' 2 2 2 2 2 A . B B C . C - cos - sm cos - sm - cos 2 2 2 2 2 B . C A . A C cos - sm - cos - sm - cos 2 2 2 2 2

.A

C B . B A cos - sm - cos - sm - cos 2 2 2 2 2

= 0,

pp.

65,

4.2. SOLUTIONS

165

because the first column is the sum of the other two. Computing the determinant, we obtain A . B . C A A B C ' 2- sm - = cos - cos - cos sm cos3 - sm 2 2 2 2 2 2 2' as desired.

L (Dorin Andrica)

L

35. Denote

E1 = L sin nA sin nB cos nC

Observe that

and

&

=

L cos nA cos nB sin nCo

(cos A + i sin A) (cos B + i sin B) (cos C + i sin C) = = cos(A + B + C) + i sin(A + B + C) = = cos 7f + i sin 7f =

By de Moivre's formula,

- 1.

(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 cos nB cos nC - i sin nA sin nB sin nC = ( _ l)n . Hence E1 = ( _ 1) n+1 + cos nA cos nB cos nC and = sin nA sin nB sin nC. (Dorin Andrica, Revista Matematidi Timi§oara (RMT), No. 1(1978) , pp. 65, Problem 3278) 2 36. Subtracting from the second equality the first multiplied by yields O 1) (sin A - (sin B - l) (sin C - 1) = . 1 , so AB C is a right triangle. Hence sin A, sin B or sin C is A d Ti ( tu n reescu, Revista Matematidi Timi§oara (RMT), No. 2(1978), pp. 49, 1 &

Problem 35 4)

37. Note that

a cos B + b cos A The system of linear equations

{ aa2A

b

- c = O.

cos B + cos A - = 0 =0 cos B + cos A cos B + cos A

a A- 1

bA

c A b2 A- 1- c - C2 A- 1 = 0

4. TRIGONOMETRY

166

has the solution (cos B, cos A, - 1) and is homogeneous. Therefore the determinant Ll = is zero. On the other hand,

a a2A-A 1 a

b b b2AA- 1

c c2AA- 1 C

1 1 1 b b 1 A-1 A A Ll = a c ( a ) ( b ) ( c - 1) = A-1 2 A- 1 2 cA-1 2 a b ( b )( ) )(b = a c aA- 1 _ A- 1 aA - 1 _ cA - 1 A - 1 _ cA - 1 . b b or c = a, hence the triangle is isosceles. Therefore (Dorina A=nd,rica,= cRevista Matematidi Timi§oara (RMT) , No. 2(1977), pp. 89 , Problem 3199)

38. The relation is equivalent to

J(P - b)(p - c) J(P - c)(p - a) J(P - a)(p - b) 1 2 2 2 Jp(P - a) + Jp(p - b) + Jp(p - c) = 4S (a + b + c ) or 1 L( P - a)(p - b) = 4S1 (a2 + b2 + c2 ) . S Expanding the brackets yields _p2 + ab + bc + ca = 41 (a2 + b2 + c2 ) , then 4(ab + bc + ca) = a2 + b2 + c2 + (a + b + C)2 . It follows that (a - b)2 + (b - C)2 + (c - a)2 = 0, b hence( Tia = A =dc. tu n reescu, Revista Matematica Timi§oara (RMT) , No. 2 (1972) , pp. 28, ....--;= .:..-..: 7='====;�

-

-

Problem 1160)

39. By the Extended Law of Sines,

On the other hand, so

a = 2R sin A, b = 2R sin B, c = 2R sin 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 = + 2 sin B sin C cos A,

1

1 . It follows that A = �, hence the triangle ABC is right, as (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2( 1977) , 52,

therefore sin2 A = desired.

pp.

Problem 3838)

2 7

40. Because 6 2 + 32 + 2 = 2 and

s

A B C A B C + cot + cot = cot - cot - cot 2 2 2 2 2 the given relation is equivalent to 2 2 2 = (62 + 32 + 22 ) cot + 2 cot + 3 cot - = cot

r

-

-

2'

-

(1)

[ ( �) ( !) ( � ) ]

(

A

B

= 6 cot "2 + 6 cot 2"

+ 6 cot 2C ) 2

This means that we have equality in the Cauchy-Schwarz inequality. It follows that

A cot ­ 2 6

__

B 2 cot 2

C

3 cot 2

2 3 7 ) ( 1 Plugging back into gives cot � = , cot � = �, and cot � = �. Hence by 28 130 Thus the side the Double angle formulas, sin A = !.- , sin B = , and sin C = �. 45 25 65 lengths 26, 40, and . ( Tituof ATnaredreescu, USA Mathematical Olympiad, 2002, Problem 2) 41 . Summing up the formulas

ra = 4R sm. 2A cos 2B cos 2C ' rb = 4R sm. 2B cos 2C cos 2'A rc = 4R sm. 2C cos 2B cos 2A yields B C ra +4rb + rc '" Slll . A R " 6 2" cos 2 cos 2 = 4 R On the other hand ra + rb + rc = + r, hence 4R + r 1 r C '" sm . A cos B cos 2 = ----;w:- = + 4R " 2 6 2"

4. TRIGONOMETRY

168

Because � �

r,

it follows that A B C 1 9 . < 1 + '" sm cos - cos - = L...J 2 8 8' 2 2 -

as

desired.

-

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 49,

Problem 3510)

42. We have

or

Because � + �

a2 b e 2 cos A + - = - + - . be e b � 2, we obtain A a2 � 2(1 - cos A) = 4 sin2 "2 be

and likewise

b2 ' B -> 4 sm2 - ' 2 ae Summing up these inequalities yields

(

)

A . 2B . 2C � � 2 . 2> 4 sm + sm - + sm - ' -+-+-be ea ab 2 2 2 as

desired.

( Dorin Andrica)

43. We have

so Likewise, and

2be cos A b2 +1=2 2 a a 2ae cos B b2

+ ae2

2'

+ 1 - ab22 + eb22 _

2ab cos C 1 a2 b2 + - 2 + 2' e2 e e Summing up these equalities implies 2be cos A 2ae cos B 2ab cos C b2 3+ + + b2 a2 a2 e2 _

_

( + b2 ) + a2

4. 2. SOLUTIONS hence

169

be cos A

ea cos B ab cos C a2 + b2 + e2 -> �2 ' 3 cos A cos B cos C > -+ + -. 3 3 3 a b 2abe e

and moreover

--

By the AM-G M inequality,

'

--

( + ) = 16p813 '

3 3 3 3 � 2abe 2 a b + e

therefore

cos A

-as

cos B cos C > 81 + ---,;s + � - 16pS '

(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1975), pp. 46,

Problem 2134)

44. We have

= (p

2=

B 1 be ea A , sec2 , sec2 2" = --A p(p - b) cos2 - P - a) 2 so it suffices to prove that 1 1 1 9 > -. p-a p- b p-e - p Setting = p - a, P - b, = P - e in the inequality

Y=

x

yields

sec2

2 = p(p C

ab

- e)

,

-- + -- + -Z (x + Y + Z) ( .!. + � + � ) � 9 x Y Z 1 1 1 ) > 9, P ( -- + -- + -- p-a p-b p-e

and the solution is complete.

(Dorin Andrica)

45. By the AM-GM inequality,

p Then

= (p - a) + (p - b) + (p - e) � 3 \!(P -

a ) (p - b) (P - e),

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. Revista Matematica Timi§oara (RMT), No. 2(1982), pp. 66, Problem 4993)

( Titu Andreescu,

= r

4. TRIGONOMETRY

170

0° < a, (3" < 90° and a + (3 + , = 90° . We have ' B-C . ( a + 2, ) = . . 3A - cos sm: - = sm 3 a - cos ((3 - , ) = sm 3a - sm "'2 2 A

B

C

46. Let a = 2 ' (3 = 2 ' , = 2 ' Then

and

= 2 cos(2a + ,) sin(a - ,) = -2 sin(a - (3 ) sin(a - ,) . In exactly the same way, we can show that C-A . ( (3 - , ) . 3B - cos - = - 2 sm ' ((3 - a ) sm sm "'2 2

A-B . 3C - cos . (, - a) Slll . (, - (3 ) . - = - 2 sm sm T 2 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 < a < (3 < , < Then regrouping the terms on the left-hand-side gives

0°

90° .

a)] , ° ° 0 0 9 which is positive = sin x is increasing for < x < . Alternat ive Sasolutfunction ion. We ykeep the notation of the first solution. We have sin(a - (3 ) sin(a - ,) + sin(, - (3 ) [sin(, - a) - sin((3 -

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, cos a + 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 < x < Since o < a" , a + , < each of the two products in the last addition is less than or equal to O. Hence

90° ,

0

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) � and

0

sin 3, + sin 3(3 - cos(a - ,) - cos(a - (3) � O.

90° .

4.2. SOLUTIONS

Adding the last three inequalities gives the desired result. USA IMO Team Selection Test, 2002, Problem 1)

Z

171

( Titu Andreescu, bi bi I l J 2 b2 47. Let z = a + , = a - , and z = a + • The given relation becomes 2002 =

z.

Note that

Z

from which it follows that

I z l (l z l 200l - 1) = O. Il ( b) I z i = 1 . In the case I z l = 1, we have Z2002 = Hence z = 0, and a, = (0, 0) , or I l 200 2 1 200 1 which is equivalent to Z 3 = z = z = . Since the equation Z 3 = has 2003 1 distinct solutions, there are altogether + 2003 = 2004 ordered pairs that meet the required Andreescu, American Mathematics Contest 12A, 2002, Problem 24) ( Tituconditions. a, b be real numbers such that z = a + bi, b i- O. Then 1m z5 = 5a4b 01 a248.b2 +Letb5 and � ) 4 10 � ) 2 Imz5 = ( b - ( b + 1. Im5 z 2 Setting x = (�) yields 5 2 ( 1)2 - 4. 1m Z 5 = 5x - lOx + 1 = 5 x 1m z ( 1 ± i ) , a i- O. 1 The minimum value is -4 and is obtained for x = Le. for z = a ( Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 1(1984) , pp. 67, Problem 5221) Ml , M2 , . . . , M2n be the points with the complex coordinates 49. Let AA A MM Mn Mletn+l. I , 2 , . . . , n be the midpoints of segments l 2n , MZl 2, ZM2 ,2n l,, Z2n , and , Z

z·

5

. • .

.

. •

172

4. TRIGONOMETRY

M i 12 The points i , = , n lie on the upper semicircle Mn Mandn+lwitarhe M2n l ,in. . .origin Ml M2n , Mcentered 2 , radius 1. Moreover, the lengths of the chords OAl , OA2 , . . . , OAn are increasing.- Thus in a decreasing order, hence Z1 , � Z2n 1 1 Z2 + ;2n-' 1 :::; ... :::; 1 Zn +2Zn+ 1 1 and Al the conclusion follows. ternative S1olut2 ion. Consider Zk = r(cos tk +i sin tk ) , k = 1, 2, . . . , 2n and observe that for any j = , , . . . , n, we have i ) 2] d2 2 [( = r costj +Cost2 n _j+ + (sin tj + s nt2 n _j+l = )] 22 = r [ + 2(cos tj cos t2 n-j+l + sin tj sin t2 n - j+l = 2 2 [1 ( )] 4 2 2 t2n-j+l2 - tj . = r + COS t2 n - j+l - tj = r cos l 2 t2 j - tj and the inequalities I Therefore Zj + Z2 n-j+l = r cos n - �1 I ZI + z2n l :::; I Z2 + Z2n-l l :::; . . . :::; IZn + zn+l l :::; :::; :::; are equivalent to t2n - t l � t2 n- l - t2 � . . . � tn+l - tn · Because 0 tl t2 . . . t2n :::;(D7r, the last inequalities are obviously satisfied. n Andrica, Revista Matematica Timi§oara (RMT), No. 1(1984), pp. 67, ori5222 ) Problem A 12 1 50. Let p = , , . . . ,m and 2let Z E p e Then zP = . l ,n-m+ , . . . , n are m consecutive integers, and, since p :::; m, Note that n-m+ 2 } k. k 1 there is an integer p divides E { n - m + , n - m)+' , . . . ,n suchA that A U k = kip. It follows that zk = (zp k = 1 , so Z E k n-m+l An-m+2 U Let A . . . R n , ask claimed. can be obtained by using the fact that emar . An alternative( solution )( 1 an(-k a1n)(- lk -1 1) . . . (an(- k+l1-) 1) a - a - 1 - 1) . . . akis an( Dinteger integers a > and n > . in Aforndrallica,positive Romanian Mathematical Regional Contest " Grigore Moisil" , or 1997) tk + i sin tk,)k E3{ I , 2, 3 } . 1R 5 1 . Let Zk = cos 2 ( The condition Zl + Z2 + Z3 - ZlZ2 Z3 E implies ) 3 ( ) (1) 2(sin t l + sin t2 + sin t3 = sin tl + t2 + t3 . :::;

:::;

C

U

4.2. SOLUTIONS 173 ( ) Assume by way of contradiction that max tl' t2 , t3 < i, hence tl , t2 , t3 < i' Let . functl. On IS . concave on [0 ' '67r ) ' so t = tl + t32 + t3 E (0 , '67r ) ' The sme ( 2) 1( . '3 sm tl + sm(.2) t2 + sm. t3 ) � sm. tl + t32 + t3 . From the relations (1) and we obtain ) ( 2 sin t l + t + t3 < . t l + t32 + t3 ' - sm then 3 2 sin t � sin t. It follows that 0 4 sin3 t - sin t � , 0 i.e. sin2 t � �. Hence sin t � �, then t � i, which contradicts that t E ( , i), (tl , t2 , t3 ) i, as desired. Therefore ( Titu Andmax 2)reescu, Revista Matematidi Timi§oara (RMT), No. 1(1986), pp. 91, Problem 586 2 (28 + 1) and b f:. 0, otherwise the claim is obvious. Consider a 52. Let n = 2 � and the polynomial complex number a such that a f = xn - 1 = (X - eo )(X - el ) . . . (X - en- I ) ' 2

;:::

=

We have

and hence

f

ioo ) a ion d , 1 + a (-' n (� r =

Therefore

(

•

•

•

( +

nII( - l + b � ) bn nII-l � bn nII- l ( 2 � ) a e = (� + e ) = a + e = k=O k=O k=O b b b ] [( ) = n f (T) f (-T) n a2 2s+1 + 1 2 = n [(� t+l + If 2s+1b + b2S+1 ) 2 ( � b �)2 ( b 2 2 a ( ) s+1 2s+1 = a + . 2000) (Dorin Andrica, =Romanian Mathematical Olympiad - second round, 0 b b 53. If a = 0, then claim is obvious, so consider the case when a f:. and f:. O. =

We start with a useful lemma.

174

4. TRIGONOMETRY h l

If co, C , , cn-l are t e comp ex roots of unity of order n, where n is an odd integer, then 1 n-l A B A B II ( + Ck ) = n + n , for all complex numbers A ank=Od B . Proof. Using the identity n- l nx - 1 = II (x - ck ) k=O A . for x = - B Ylelds An 1) n- l ( A ) ( B - n + = - !! B + Ck , and the conclusion follows. b 2 Consider the equation X + a = 0 with roots Xl and X2 . Since Lemma.

• • •

0

we have

nII-l ( b � ) bn nII- l ( }( 2 ) bn nII- l ( ) nII- l 2 ) a + c = k=O ck - Xt ck - X = k=O ck - Xl k=O (€k - X k=O A B B A Using the lemma for = -Xl, = 1 and then for = -X2 , = 1 gives nII- l ( } ( ) n 1 c k - xt = -X l + = 1 X� , k=O nII- l ( 2 ) ( 2 )n 1 1 ck - X = -X + = - x�. -

Hence

k=O

nII- l ( b T:) n a + c = b ( l - x�)(l - x�) = k=O = bn [l + (XlX2 ) n - (x� + x�)] = bn [1 + (�) n] = an + bn , }n = o. since( DXlXi2 =A �dand x? + x� = x? + (_xt or n n rica, Romanian Mathematical Olympiad - second round, 2000) 54. Consider the triangle with vertices of complex coordinates Zl, Z2 , Z3 and the R circumcenter origin of the complex plane. Then the circumradius equals IZI I = I Z2 1 = I Zin3 1 the = r and the side lengths are b I a = IZ2 - z3 1 , = Zl - Z3 1 , c = IZI - Z2 1 .

4.2. SOLUTIONS 175 The desired inequality is equivalent to 1 1 R12 -ab + -be + e-1a >- ­ bc 48 4pr i. e. b a � a+ + e R2 = }f = ]f R � 2r,i which or ( D is Euler's inequality for a triangle. A d i or n n r ca, Revista Matematidi Timi§oara (RMT) , No. 2(1985), pp. 82, Problem 5720) AAA A 55. Let I, 2 , 3 and be the pointsAof complex coordinates ZI, Z2 , Z3 and let A A ) 2 2 3 ZA =A aZA + (1 - a z3, a E Hence point lies on the line and the triangle l 2 3 hasB its circumcenter in the origin ofA the complex planeA .A2 A 3 Al APoint � Al B ,issothe foot of the altitude from l in the triangle l . It follows that R

We have

I Z2 - z3 1 h I ZI - z2 11zl - z3 1 i A I ZI - z2 11zl - z3 1 · IZ2 2r- z3 1 8A 1 A 2 A 3 2 2 2 therefore h = IZI - z2 11zl - z3 1 ' 2r as desired. y _

_

s

n

l

_

(Dorin Andrica, Romanian Mathematical Olympiad - final round, 1984) I 1/ 1 56. Denote z + z by r. From the hypothesis,

4.

176

TRIGONOMETRY

3 ( )( 1 )2 :::; O . This implies Hence r3 :::; 2 + r, which by factorization gives r - 2 r + r 2,( Ti desired. 1987; Revista tu Andreescu, Romanian Mathematical Olympiad first round, 1 ( 1987) , pp. 75, Problem 6 191 ) Matematidl Timi§oara ( RMT) , No. � ? Zl 57. Denote t = 2 , t E 1 , n = 0,1 1, 2, . . . 1 ) Xn + 1 ( 2) - Xn+1 - -- = -2 Xn+1 Xn - 1 , n = 0, 1 , 2, . . . ( ) 4. Study the convergence of the sequence Xn n � l defined by Xl E (0, 2) and � 1 = + V2xn - x X n+1 1 for n � . ( ) 5 . Consider the sequence of real numbers Xn n� l such that i + x� + . . . + X2n x = O. lim n�oo n Prove that Xl + X2 + . . + Xn = 0. 1m n�oo n Is the converse true? b ( ) (b ) 6 . Let an n � l and n n � l be sequences of positive numbers such that an > n n 1 for all n > . (an) n� l is increasing and (bn)n� l is unbounded, then the sequence Prove that if (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 X l = a and 1) (a + Xn- 1 +10') 2 Xn = Xn- 1 + (a + , n � 2. 18 1 2. Let

. 10

v

1·

V

.

N.

182

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 n-+oo (Jan+l - �) O . lim n-+oo = 00

and

=

(Xn )n2: l be an increasing sequence of positive real numbers such that X 1·1m n 0 n-+oo n h ( Prove that there is a sequence nk 2: l of positive integers such that 9. Let

2 =

.

(3 be real numbers and let (Xn ) n2: l , (Yn )n2: l , (Zn ) n2: l be real sequences such that � (3 } � max{ x + O'Yn , y + xn :::; Zn for all n ;::: l . 1 ( 2 (32 ) for all n ;::: 1 a) Prove that Zn ;::: - 8 0' + . (3 ) ( ) ( ) ( 2 2 1 b ) If n-+oo lim Zn - -8 0' + , prove that the sequences Xn n 2:b Yn n 2: l are convergent and find their limits. ( ) ( ) 1 1 . The sequences x n n 2:l and Yn n 2: 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 ) ( ;::: 0 and a f= 0 be real numbers and let Xn n2:l be an increasing 1 2 . Let sequence of real numbers such that C¥ ( ) a. lim n x + x l n n n-+oo 10. Let

a,

=

=

=

a

=

Prove that the sequence is bounded if and only if a > 1 . 1 3. Evaluate

14. Evaluate

1n-+oom kl:n k((nk -+ k)!l) ! (+n (-k k)!+ 1) =O 1.

Ln ( -k2 ) � +l n-+oo k= l n lim

183

5.1. PROBLEMS

L -1 ' q > 1; ii (4 1l) +l ' q > 1 ( ) � n + qn . n=l nq( n ) 16. Let Xn n � l be an increasing sequence of positive integers such that X n+2 2 Xn > Xn+l for all n � 1. Prove that the number L... ­ 101Xn n=l 15. Evaluate

(i)

00

00

+

00

(J - "'

is irrational.

An L (k!1) k=l n

17. Prove that

is irrational for all 18. Let

k,

s

=

n � 1.

00

be positive integers and let numbers such that

aI, a2 , . . . , ak , bl , b2 , . . . , bs be positive real

n � 2. k 21)) al a . . . ak bl b bs. 2 2 ) ( 19. Let X n n � l be a sequence with X l 1 and let X be a real number such that

for infinitely many integers Prove that =

s;

=

. • .

=

Prove that

20. Let

such that

IT (1 - �) Xn+l n=l

= e -x •

A f:. ±1 be a real number. Find all functions f : R R and ( f ln x + A ln y ) 9 (JX) + 9 (..fij) for all x, y E (0, 00). �

9

:

(0, 00)

�

R

=

21. Let f be a continuous real-valued function on the interval

[a, b] and let ml , m2 O be real numbers such that m l m 2 > . Prove that the equation m f (x) � a b-) x b -2x has at least a solution in the interval (a, . =

+

184

5. MATHEMATICAL ANALYSIS

b (0 , 1 /2) , and let 9 be a continuous 22. Let a and be real numbers in the interval ( ( )) ( ) b () real-valued function such that g g x = ag x + x for all real x. Prove that g x ex for some constant e. f [0,00) such that 23. Find all continuous functions f2 (X + y) - f2 (X - y) = 4f(x) f (y) for all real numbers x, y. ( -00, 0] and g : f Prove that if the continuous functions 24. (i) [0 , 00) have a fixed point, then f + 9 has a fixed point. [0 , 1] and 'ljJ [1 ,00) have (ii) Prove that if'ljJthe continuous functions rp a fixed point, then rp 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? > for all >

f [a, b') 27. Let' b f( ) and f (a) = j (b) . :

�

�

XIL - aIL-

x

x 0.

be a twice differentiable function on

[a, b) such that f(a)

=

,\ the equation f" (x) - ,\(f' (X)) 2 = b) has at least a solution in the interval (a, . f [0, 2] (0 , 1] that are differentiable at the origin and 28. Find all functions satisfies f(2x) = 2f2 (X) - 1 , x E [0, 1] Prove that for any real number

°

:

�

29. Let ,\ be a positive integer. Prove that there is a unique positive real number

f) such that

x 0.

for all real number >

f

: � � � be a function continuous at the origin and let distinct positive real numbers.

30. Let

'\, f-£

be two

185

5. 1. PROBLEMS

Prove that the limit

f( >..x) - f(J-tx) 1�O -"----'-- X --'--"a: f exists and is finite if and only if is differentiable at the origin. ( ) 31 . The sequence Xn n � 1 is defined by 1. m

n�oo(0 nX] n = - 2. Vnxn . ) O 1 32. Let Xo E , and Xn+ l Xn - arcsin(sin 3 x n , n � . Evaluate lim n�oo f JR be a twice differentiable function with the second derivative 33. Let : JR nonnegati ve. Prove that f(x + I' (x)) � f(x) , x E b 34. Let a < be positive real numbers. Prove that the equation Y a( ; T+ aX bY ( b) has at least a solution in the interval a, . ) 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. f [ b] JR be a differentiable function. 36. Let : a, O O Prove that for any positive integer n there are numbers < 2 < . . . < n in the b ) ( interval a, such that f(b) - f(a) f' (OI ) + f' (02 ) + . . . + f' (On ) b-a n f 37. Let , 9 : JR JR be differentiable functions with continuous derivatives such that f ( x) + 9 ( x ) l ' ( x ) - g' ( x ) for all x E JR. f(x) - g (x) Prove that if Xl, Xf( real solutions of the equation 2 are) two( consecutive ) ( ) 0, then the equation x + g x = 0 has at least a solution in the interval X l , X2 . f' f 38. Let : [-�, �] (-1, 1) be a differentiable function whose derivative is continuous and nonnegative. Prove that there exists Xo in [-�, �] such that (f(XO ))2 + (1' (XO ))2 � 1. Prove that lim

=

�

R

=

�

�

fh

�

=

=

�

186

5.

MATHEMATICAL ANALYSIS

39. Prove that there are no positive real numbers

x and y such that Y 2 X x2 y - = x + y . (n+1) ( n 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 2n + 1 , n > 1 > n n +l� _ n+ 4 1 . Let Xl , X2 , . . . , Xn be positive real numbers such that Xl + X2 + . . . + Xn +

+

+ n'n V 'H

3.

Prove that

42. Let

f

:

=

1.

�

f(c) = 0 for someJR c E JRJR. be a function with a noninjective antiderivative. Prove that h f JR be continuous functions. Prove that 43. Let h , , . . . , n JR ( ) h( ) f ( ))d max( h x , x , . . . , n x x �

:

is a derivative and evaluate

44. Evaluate

p'

45 . Let be a polynomial of odd degree such that has no multiple zero and let : JR � JR be a function such that is a derivative.

f

p

f op f Prove that is a derivative. 1 f I I be a function with an antiderivative F that 46. Let = (0, 00) and let satisfies the condition F (x)f G) = x, 1 -+ 1R, g (x) = F (x)F (�) is a constant function and I for all x in . Prove that f then find . :

g

:

�

187

5. 1. PROBLEMS 47. Let n >

1 be an integer and let

that

f

:

[0, 1]

�

JR be a continuous function such

f(x) dx = 1 + -1 + . . . + -1 . 1 n 2 1 a

Prove that there is a real number

Xo E (0, 1), such that f(xo ) 1 - xg 1 - Xo f [a, b] 48. Consider the continuous functions , =

:

9

Prove that the equation

�

R

!(x) /." g(t)dt = g (x) f !(t) dt ( b) has at least a solution in the interval a, . [ b] JR be a continuous function such that 49. Let f a, { !(X) dx of O. b Prove that there are numbers a < a < f3 < such that { !(x) dx (b - a)!(p) . [ b] [c, dJ be a bijective 50. Let a, c be nonnegative real numbers and let f a, increasing function. ( b) Prove that there is a unique real number E a, such that { !(t)dt = (IJ - a)e + (b - lJ) d. 5 1 . Let

(- x )

xx + 1

= 1. X O () () Let c > . There is an integer n c > 0 such that for any integer n � n c , we

lim ", -+ 0

--

", > 0

have

� +1

- c < (:2)k < 1 + c, k = 1, 2, . . . , n. 2 n k Summing up from k = 1 to = n and using algebraic manipulations yields Ln (�)2 �+1 n k < 1 + c, n � n (c) , 1 - c < k= 1 L 2 k=1 n or k ) �+1 1 1 ( 1 c ) 1 1 1 c) < ( -2 - -2 (c - n + -n L -n2 < -2 + -2 c + -n + -n , k=l () for any integer n � n c . Therefore n ( k ) �+ 1 1 lim � -n2 = -.2 1

n

n

n-+oo

(Dorin Andrica)

()

15. i We have

1

I -1 x , Ix l < 1. + x + x2 + . . . + x + . . = -n

.

5. 2.

SOLUTIONS

197

Hence

11/q (1 + x + x2 + . . . )dx = 10 1 /q -1dx- x, o so 1 q 1 ' q > l. In = � n q l nq n= I l 1 (ii) For any x < we have 1 + x4 + x 8 + . . . + x4 n + . . . = 1 1 4 . _x Hence dx 11 /q (1 + x4 + x8 + . . . ) dx = 11/q -1 - x4 ' 0 o so 1 1 1 (1 1 ) 1 1 4 ) ( 1 � n + q4n+ l = - q + 4 ln - q2 + "2 arctg q2(. 1977) 70 (Dorin Andrica, Revista Matematica Timi§oara (RMT) ,40No. ,86pp.08) , 309 98 ) ( 1 1 1 1 1 Problem ; Gazeta Matematica (GM-B), No. , pp. , Problem 00

_

00

16. The number () has the decimal representation

() = 0. 0 . . . 0 1 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. Problem

4661 )

17. Note that

and that

An = - 1 + (k!)1 < - 1 + 1 = - 1 + < 2. � � n k! k O k = O = A 2 A 1 1 Hence < n < for all n � . So n is not an integer. 00

00

e

2(1981 ) , pp. 73,

5. MATHEMATICAL ANALYSIS

198

Assume by way of contradiction that there are positive integers n,p, with "# such that P. Then

q q 1 A q = n. 1( 2) 1 ( ) 1 n 1 pq q - = A + (q : l )n + (q + 1 )n q + n + . . . < 1 < A + (q +1 l) n + (q + l) n (q + l) n + . . . = 1 1 1 = A + (q + l )n (1 + (q + l) n + (q + 1 )2n + . . . ) = A + (q + l1)n - 1 , O for some integer A � . It follows that B = ( 1 l ) + ( l ) 1( 2) + . < 1 , q+ n q+ n q+ n . . B A so + is notA integer, which is false. 1 Therefore n is irrational for all n � . (Dorin Andrica) 1 ) Using the limit 18. 1 lim z:ra = n-+oo and taking the limits in both sides of the equality, we obtain 1+ 1+··· +1 = 1+1+···+1 so k 2)= Using the limit 1) = In a lim n ( z:ra n-+oo and the relation n ( \Ial - 1 ) + n ( y'(i2 - 1 ) + . . . + n ( ifiik - 1 ) = = n ( \Ilh - 1) +n ( \Ib; - l) + . . . + n ( v'b,; - l) 8.

� k times

�' s tim es

we obt�in after taking limits:

This implies as

desired .

) Andrica, Romanian Mathematical Regional Contest " Grigore Moisil" , 1999(Dorin

5. 2. SOLUTIONS

19.

Let

199

Pn = IT (1 - �) . Then k=1 Xk+l

hence

I) X Xn+2 xn+l ' - (n + Xn+l n+l Pn+11 P1n (nXn+2 + 1)1 --;;,y(n + 1)1 ( n + 1)1 It follows that Pn+11 = -P1I + -x2!2 + -x33! + . . + (nxn++lI ) ! x2 + . . . + xn+l ' = 1 + I!x + 2f (n + 1)1 Because 2 + . . . + ( xn+l ) eX , X x lim ( 1 + + n-+oo x . 2 .' n + 1) . P it follows lim n+1 D = e-i ,A ddesired. Andreescu n-+oo i ( Titu that and or n n r ca, Revista Matematica Timi§oara (RMT), No. _

_

-

.

--

=

I'

1

=

as

1(1977) , p p . 49, Problem 2843) 20.

Interchanging and y we obtain

x f (ln x + >. ln y) = 9 (y'X) + g (..;y) , x, y E (0, 00) so f (In y + >' In x) = 9 (y'X) + 9 (..;y) , x, y > O. Let b a = ln x + >' ln y and ln y + >' ln x. Then X = e:>.:>.2b--1a and y = e :>.2 - 1 , hence f(a) = f(b) = 9 (e2(:t_a1 ) ) + 9 (e 2(:'2-:,b1 ) ) , a, b E f f ) C. Then for X we have It follows that is a constant and let (x ( constant and g x ) �. g (V'X) D( =ori�,n Asond9riisca,aRevista Matematica Timi§oara (RMT) , No. 1-2(1979), 51, Problem 3827) F I lR, 2 1 . Consider the function F ( x ) ( x - a ) ( x - b ) f ( x ) + m l ( x b ) + m2 ( x a ) . =

:>' a - b

R

=

= Y

=

pp.

:

=

-t

-

-

5 . MATHEMATICAL ANALYSIS

20 0

Note that F is continuous and

hence there is c E (a, b) such that F c) = O. It follows that

( - aml- c + bm-2c f(c) --

and the solution is complete.

-­

(Dorin Andrica)

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 or else we'd have g (g (x)) - ag x) = bx, with the left side bounded and the right side 22. Note that

----+ 00 ,

unbounded. Similarly, 9 cannot tend to a finite limit as x ----+ - 00 . Together with monotonicity, this yields that 9 is also surjective. Pick arbitrary, and define for all n E recursively by + l = g (x ) for n 0, and for n < O. Let rl = 1 + Ja + 4b)j2 and = (a- Ja2 + 4b)j2 = be the roots of - ax - b = 0, so that rl 0 and 1 I rl l I r2 1 . Then there = clr� + c2 r� for all n E exist E lR such that Suppose 9 is strictly increasing. If f:. 0 for some choice of xo, then is dominated by r� for n sufficiently negative. But taking and Xn+2 for n sufficiently negative of the right parity, we get 0 < < but g (x ) g(Xn+2 ), contradiction. = Thus = O . Since = l and l Clrl , we have = l for all x. Analogously, if 9 is strictly decreasing, then = 0 or else is dominated by for n sufficiently positive. But taking and for n sufficiently positive of the right parity, we get o< contradiction. Thus in that case, g (x) = < for but < all x. ( Titu Andreescu, The " William Lowell Putnam" Mathematical Competition, 2001, Problem B-5)

Xo - l ( ) Xn Xn- l g 2 xn x Cl ,C2 Xn

Z

Xn 2 r > > r2 Z > 2 > . C2 Xn Xn Xn+2 ( ) n > C2 Xo C X gx rX � Xn r C2 Xn Xn+2 X n+ 2 Xn g ( Xn + 2 ) g ( x n ) , (

>

n

Xn

r2 X

x = = 0 yields f O) = O. For x = we obtain f2 (2x) = 4f2 (X) and then f(2;C) = 2f(x), since f(x) � O. 23. Setting

We prove that

(

y

y

(

f nx) = nf(x) , n � 1. Assume that f(kx) = k f (x) for all k = 1, 2 , . . . , n. We have

2(

f ( n + l )x) then

- f2 ((n - l )x) = 4f(nx)f(x),

201

5.2. SOLUTIONS hence as

f((n + l) x) = (n + l )f(x) ,

desired. It follows that if p, are positive integers then

q

qf (�) = f (P) = pf(l ) , so f (� ) = � f(l ) f( ) f(l ) for any positive rational r. and r = r Setting x = 0 in the initial condition gives then for all real

y, hence

f(y) = f( -y) , f(r) = Ir l f ( l ) ,

for all rational numbers We prove that = for all real numbers Let be an arbitrary real number and let be a sequence of rational numbers with lim = Because

f(x) r. I xlf( l ) (rn ) n>l -

x. x n--+oo rn x.

f is a continuous function, it follows that f(rn ) = lim I rn l f ( l ) = f ( lim rn) , lim n--+oo n--+oo n--+oo hence f(x) = f( l )lx l · f( l) � 0, therefore the desired functions are f(x) = al x l for some Note that a = a � O.

and

( Titu Andreescu, Revista Matematica Timi§oara ( RMT) , No. 2 ( 1977) , pp. 90,

Problem 3203)

f for and g, respectively. We have a and b be fixed points f(a) � 0, b = g (b) � 0 a b so a � . ( ) f( ) g(x) - x. Consider the function

0 ( Xn ) n � l n 0 n X n (X n ) n � l l l 1 n-+oo Xn � 1) z-+o (.!x 1 1 1 n-+oo (Xnn+1+ 1 ) -Xnn n-+oo 1 Cesaro-Stolz ' s Theorem implies 1 nlim-+oo nXn hence lim nXn = 2, as desired. n-+oo

eZ n - 1 Since eZ + for all real 0, we have so is increasing. By induction on we obtain < for all � 1, the sequence so the sequence is bounded. Therefore the sequence converges and let l be its limit. The equation el = + has the unique solution 0, hence lim = O. 1 = - , it follows that Using lim 2 eZ 1 1· 1 eZn ' = l1m 1m =

31 .

=

-

5. MATHEMATICAL ANALYSIS

206

(Dorin Andrica, Revista Matematid1 Timi§oara (RMT) , No. 2(1982), pp. 68, Problem 5004) 32. We first prove by induction that X n 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 X k - arcsin(sin xk ) = O. It is not difficult to see that (X n ) n>O is convergent and lim X n = O. We have n�oo

>




>

(vn sin xn ) 2 = and


O. F(x) = dx�, x I. = d = ,;c. F(x)F (;�) = 1 1. - 1 x I, Xc , f(x) = -XF---;1 -�-;--) = .;c (

for all

in

for all

in

for all

in

We have

so is constant. Then there is a constant c

such that

c

we obtain

and from

E

Integrating gives

for all

in

ln

ln

where

It follows that c, so

c becomes cP

The relation

Finally,

E

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)

lR, g(x) = f (x) + x + . . . + xn - 1 )

47. Consider the function

and note that

9

9

:

[0, 1] -+ - (1

is continuous. We have

1

Jro 1 g (x)dx = Jor 1 g(x)dx - Jro (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 g(xo) = l' g(x)dx = 0, hence 1 - x� , f (xo ) = l + xo + . . . + xno - l = -1 Xo

as

-

desired.

( Titu Andreeseu, Revista Matematidl Timi§oara (RMT) , 3444)

Problem

No.

1-2 ( 1979),

pp.

33,

5. 2 .

SOLUTIONS

215

F : [a, b] lR, F (x) = [ g (t)dt t f(t)dt. F is differentiable' and F(a) = F (b) = O. Applying Rolle's Theorem, Note that b) F ) we obtain c E (a, such that ( c = 0 hence ftc) [ g (t)dt = g (c) t f(t)dt, 48. Consider the function

--t

as desired.

(Dorin Andrica)

F : [a, b] lR, F (t) = l f(x)dx - t f(x)dx F is continuous and F (a)F (b) < O . Then there is a E (a, b) such and observe that F( ) that a = 0 so lQ f(x)dx = l f(x)dx ( b) From the Mean Value Theorem there is f3 E a, such that l f(x)dx = (b - a)f({3) , ( b) therefore there are numbers a, f3 E a, , a < f3, with { f(x)dx = (b - a)f({3) , 49. Consider the function

--t

as claimed.

(Dorin Andrica,

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

Problem 3897) 50. Since Denote

pp.

61,

f is an increasing bijective function, f is continuous. ' ( )d r )d f( y y, 8, = I.' x x, 82 = f.d

and note from the diagram below that

(1)

5. MATHEMATICAL ANALYSIS

216 y

d . . . .. . . . . . . . . . . . . . . . . . . . . . . ---7J .

1-.1! 1

C

��

1-. .-.._"=' -;...., . -T .

•

•

•

•

,/

•

•

� 1. . . . . . .

b

a

o

x

( From the Mean Value Theorem there is e E c, d) s.llch that

{ f - l (y)dy

Observe that e is unique and let

as

desired.

=

(d - c)rl (e).

fJ; = 1- 1 (e) . The relation ( 1)

t f (t) dt = (" - a)e + (b - ,,)d, " E (a, b) ,

(Dorin Andrica, 3556)

gives

Revista Matematidi Timi§oara (RMT) , No.

2(1978),

pp.

56,

Problem

5 1 . Consider the function F : �

--+ �,

[ 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 F( s ) =

�.

Differentiating with respect to y it follows that

F'(x y) = F'( x - y), x, y lit z �, we obtain F'(z) = F'( O), so c.p(z) = F' (O) for all z E lit

�, E

+

E

Setting x = y = Hence c.p is a constant function, as desired. ( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. Problem 2865)

pp.

47,

/.8 f(t)dt. The condition is equiva­ /. f(t)dt + /. y f(t)dt :::: 2 /. f(t)dt,

52. Consider the function

lent to

1 ( 1977),

x

cp

:

JR

-4

JR,

cp(s) =

.=.±.ll 2

5.2.

SOLUTIONS

217

cp(x) + cp(y) -> cp ( x + y )

hence

(1)

2

2

for all E llt Since I is differentiable, is twice differentiable and moreover is concave up, from � 0 or � 0 for all so I is a nondecreasing function. relation (1) . Hence ( 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)

x, y

"cp (x) cp f' (x)

x,

h [0 ,00 ) --+ lR, h(x) xf(x) - 1" f(t)dt. h Because I is differentiable, 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. 0, x � 0, Case 1. If I ' (x) � 0 for all x, then hfrom (1)h we deduce that h (x ) O h ) ) ( ( 0, x � O. hence is nondecreasing. It follows that x '( h' (x) � 0, x � 0, and h is nonincreasing. It Case 2 . hIf f x )h �O 0 for all x, then ( ) ( ) 0, x � O. follows that x � Since F is differentiable and F' (x) xf(x) - 10(X f (t)dt h(x) 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 :

53. Consider the function

=

=

�

'

=

�

�

=

=

_

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

0 such that for all x < I xx - 11 < c. Then for n > � we obtain I n° l '!. (x'+ ! - X)dX I � nol';; Ix·+1 - xldx n2 l� xlxx - l ldx < cn2 l� xdx c2

and let c > O. There is 0 >

0,

=

=

0

0

= -.

21

5. MATHEMATICAL ANALYSIS

8

It follows that

1

lim n-+oo

hence

n2 {n (XX+ 1 - x)dx 10

lim

as

n-+oo

desired.

Alternative Solution .

n2

(�

10

xx+ 1 dx

Consider the function

= 0,

= -21 ,

F (t )

and we can write

= l' x·+1 dx. Then F(O)

=

0

� xx+ l dx = lim n2 F ( -1 ) = lim F(u) = lim F' (u) = l n-+oo n u-+O u2 u-+O 2u n-+oo 1 . uu+ l = -1 hm = hm -+u O 2u 2 u-+O. uu = -.2 lim

n2

--

0

--

--

( Dorin Andrica, Gazeta Matematidi (GM-B) , No. 1 1 (1979) , pp. 424, Problem 18025; Revista Matematidi Timi§oara (RMT) , No. 1-2 (1980) , pp. 71, Problem 4160)

x =f. y be real numbers. Then x f(x) and f (t) dt f(y)

55. Assume the contrary and let

(Y

Jx

f(t)dt

hence

=

0,

f(y) f( = x) ,

/.Y

f(y) f(x) f(y) - f(x) so f(x) = f(y)

It follows that f 2 (X) + f 2 (y) = 0, which is absurd since f(x) =f. 0 for all x. ( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2 (1978) , p p . 35, Problem 3126)

=

56. It is clear that

[ (1 /,(x) l - I)'dx O. [ (f'(x))'dx [ 1/, (x) ldx + [ dx [ 1 /,(x) l dx I [ /' X)dx l ;:::

Hence

-2

and

as

1 ;:::

desired.

( Titu Andreescu,

Problem 3130)

;:::

(

;::: 0,

,

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

pp.

77,

5. 2. SOLUTIONS

2 19

57. The relation is equivalent to

Hence or

r 1 (xf(x) f2 (X) )dx = r 1 X2 dx . io io 4 (f2 (X) - xf(x) + x dx = 0,

l

l (f(x) - �r

:)

= O. Because f is continuous, f ( x) - � = 0, for all x E [0, 1], so f (x) = � , x E [0, 1].

( Titu Andreescu, Problem 3319 )

dx

Revista Matematidi Timi§oara (RMT) , No.

58. The relation

1(1978),

pp.

72,

l fm(t)dt = m : 1) 1 ' l (fm(t) - : ) dt = O. (

is equivalent to

Since

fm is continuous by the Mean Value Theorem there is Xo E (0, 1) such that

or

rco

io

(fm-1 (t) - mtm-1_ I)! ) dt = 0. (

Xl E (O, xo) such that m-1 fm-I (xd = (�1_ 1) 1 " Continuing this procedure, we obtain Xm E (0, X m-1 ) such that fo (xm ) = xm ,

Using the same argument, we obtain

as

desired.

( Titu Andreescu, Problem 33 2 0)

Revista Matematidi Timi§oara (RMT) , No.

1 (1978),

pp.

72,

the Mean Value Theorem we deduce that for any X E [-1, 1] there is xC E59.(-1,From 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

r 1 f (x) dx _ 2f ( - 1) :::; (x + 1 )2 1 1' (1) i-I 2 1 -1

220

5 . MATHEMATICAL ANALYSIS

and the conclusion follows. ( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2(1981) , Problem 4686)

pp.

77,

h : [a, b] --+ lR, h (t) = (b - t) [ f (x)dx + (t - a) [ g(x)dx h and note that is differentiable and h' lt) = - [ f (x)dx + (b - t)f(t) + [ g (x)dx - (t - a)g (t) . h (a) = h (b)' = 0, from Rolle's Theorem it follows that there is a real number Since b c E (a, ) such that h (c) = O. Then - /." f(x)dx - (e - a)g (e) + (b - e)f (e) + t g(x)dx = 0, 60. Consider the function

and the conclusion follows. ( Titu Andreescu, Revista Matematidi Timi§oara (RMT ) , No. 2 (1985) , Problem 5628)

pp .

56,

C hapter 6 C O MP REHEN SIVE P RO B LEMS

P ROBLEMS 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.

Y

2. Let A be a set with n elements and let X, be subsets of A with p � 1 and q � 1 elements respectively. Find the number of functions f : A --+ A such that C f(X ) .

Y

3. Let A b e a set with n elements and let X b e a subset of A with k � 1 elements. Find the number of functions f : A --+ A such that f(X) X.

=

= {I, .

4. Consider the sets A 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.

=

aI, a2 , . . . , an be positive real numbers and let m � 1. Prove that (1 al )m ( a2 )m . . + (1 an ) � n . 2 . + a2 + 1 + a3 + . + al 6. Let a I , a2 , . . . ,a n be positive real numbers and let k � O. Prove that

5. Let

7. Let

m

m

a, b, c be positive real numbers such that abc = 1 . Prove that

8. Let numbers x ,

[a, b] be an interval. Find the y, z [a, b] {3(y - Z)2 + ,(z - X)2 E( x , y, z) = a( x - y ) 2 + a, (3"

be positive real numbers and let E such that

is maximum.

22 3

224

6. COMPREHENSIVE PROBLEMS 9. Find the maximum number of nonzero terms of the sum n i L If ( ) - f (j ) 1 i ,j = l where f : {I, 2, . . . , n } --+ { a , b , c } is one of the 3 n possible functions. b b 10. Let a I , a2 , . . . , an , l ::; 2 ::; . . . ::; bn be positive numbers such that al � bl , al + a2 ::; b1 + b2 , . . . ,al + a2 + . . . + an � bl + b2 + . . . + bn · Prove that

+ Viii. + . . . + ..;a;; ::; ..jb; + .jb; + . . . + V'bn. n i2 1 1 . Define S (n,p) = L (n + 1 - 2 ) p for all positive integers n and p. Prove that i= l for all positive real numbers a i , i = 1, n the following inequality holds ) ( 4P n 2 p rp.i� ai - aj ::; S ( n,p) L 1�� 1, n � 1; ii) an iii) there exists lim . n �oo b n an an+ 1 - an . . Prove t hat hm eXIsts and is equa1 to nl'1m - . n�oo bn+l - bn �oo bn

< < < ... < < ...; �:

41. Let k be a positive integer and let

Prove that

42. Let f : 1R

�

1R f (x) =

n

L sin akx,

k =l

=

a2

= ... =

23 0

6. COMPREHENSIVE PROBLEMS

where are real numbers. Prove that if number Xo such that f (xo) -I O.

ak

43. Let that

f

:

l ai l -I l aj l for i -I j , then there is a real

lR lR be a differentiable function with continuous derivative such -t

lRf(x)lR=, lim t (x) = g (x) = sin f (x) ,

lim

x-too

Prove that the function

is not periodical.

9

:

x -too

-t

00 .

44. Let a be a real number and let f : N -t [0, 1 ) , f (n) part of the number an. i) Prove that f is injective if and only if is irrational. ii) If a is rational, find the number of elements of the set

= {an} i.e. the fractional

a

M = {f(n) 1 n E N}.

lR lR M =M

45. Let f : -t be a function such that i) f has a period T > 0; ii) f (x) � for all x; iii) f (x) if and only if x = kT, for some integer k. Prove that, for any irrational f) the function 9 : -+ lR,

lR

g (x) i s not periodical.

= f(x) + f (f)x) ,

lR lR be a continuous function with a period T > O. A a) Prove that if T is irrational, then for any E [min f(x) , max f (X)] there is a 46. Let f

-t

:

sequence (X n ) n � l of integers such that

xER

xER

= A. A b) , Prove that if T is rational, then for any E [min f(X) , max f(x) ] and for any lim f (x n ) n -too

x ER

xER

irrational number f), there is a sequence (X n ) n �l of integers such that

47. i) Let x, y, z, v be distinct positive integers such that that there is no > 1 such that

A

x

+y

=

z + v. Prove

6. 1 . PROBLEMS

23 1

a, b, c, d be distinct positive integers such that aP + bP = cP + at .

ii) Let p be a prime number and let

Prove that

l a - c l + I b - dl �

p

48. i) Prove that

for any x � y O. ii) Prove that

>

1) l) n (n + n+ 1 --en < n. < en , (n +

1

n�

49. Let c be nonnegative real numbers and let f function. i) If f is increasing, prove that

a,

1.

: [a, b) � [c, d), be a bijective

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. lR be a continuous function such that lim J-L(x) = o. 50. i) Let (0, 00) Prove that t lim e - (Z e J-L(t)dt = 0 lR be an n-time10differentiable function with the n-derivative ii) Let f : [0, 00) continuous such that there exists n � k J-L

�

:

z�oo

z

z-too

�

lim

x-t=

Prove that lim

z�=

=Ok C f ( ) (x) = �

f (x) exists and

lim

x-t=

5 1 . Let

f

:

A.

f(x) = A.

[a, b] [c, d) be a continuous function such that �

Prove that if c + d =f. 0 ::;

' f' (x)dx = cd. f. � a b

0, then

�

c d

l f(x)dx ::; b � a G� �r

232

6. 52. Let

COMPREHENSIVE PROBLEMS

I : [a, b] lR be a continuous monotonic function and let F : [a, b] F (x) = (x - ) l' f(t)dt + (x - b) [ f(t)dt. �

�

lR,

a

Prove that all values of F have the same sign.

53. a) Consider the functions : (0, 00) � lR and 9 : [1, 00 ) 1) 9 is differentiable with continuous derivative; 2) is continuous and the function h [1, 00) � lR, h(x) nonincreasing. Denote

I

I

:

n = kL= l h(x) . Prove that ( 1) an+ 1 - h( l ) ::; !.gg(l )n+ I(x)dx ::; an , b) Prove that n kI kl L lim n-+oo k=l 2 cot - = 00. I [0, 1] 54. Let be a positive real number and let

�

lR such that

= gl(X) - I(g(x)) is

an

:

a

function. Evaluate

( r)

�

lR+ be an integrable

In : lR lR, 2 , X -1- 0, x 1 k E '71 * s�n k7r ' sm ­ X X = o-k7r1 ' k E Z* x= for all integers n � O. I 1) Find the numbers an , k such thatI n is continuous on lR* . 2) Find the number an such that n is a derivative function. 55. Consider the functions

56. Let p, q

�

-II

I

/U

� 0 be integers. Find the numbers Cp, q such that Ip , q : lR � lR, 1 cosq 1 sm - if x =f- 0 f if x 0 •.•

is a deri vative function.

(x) = {

•

p

_

X

.

x'

=

6. 1.

n X) -- { 'an ( X) , X

PROBLEMS

57. Let q be a positive integer. Find the number an (q) such that fn : lR n- l , =f. 0 cos q cos � cos !I

f(

.

. .

q is a derivative function for any integer n > O. 58. 1) Let

x

X

x

-+

lR,

233

n E N* ,

=0

f : lR -+ lR be a continuous function such that l

lim

-

I Y I -too Y

Prove that the function

g(x)

/. f(x)dx = M (f). 0 Y

= { f G) , M(f),

x ;l a x 0

=

is a derivative function. T 2) Let f : lR -+ lR be a continuous function with a period > O. Prove that

M(f)

=

lim

Itl -too

lt /.t f(x)dx = l /.T f(x)dx o o

-

T

59. If f : lR -+ lR is a derivative function, then is derivative function? 60. If h , h : lR derivative function?

-+

9

: lR -+ 1R,

lR are derivative functions then f

g(x)

= If(x) 1 also a

= max{h , h } is also a

SOL UTIONS

h

functions : B ---+ B. Each of them can be extended in n n - m 1 . There are ways to a function f : A ---+ A which satisfies f (B ) � B . Hence the required number is (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2 (1981), pp . 81, Problem C : )

mm

mm . n n - m .

ll

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 : ---+ Y . Each of them can be extended in n n- q ways to a function f : A ---+ A which satisfies Y � f(X) .

Xq X X The number of subsets q of 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) ::; m. Let A k be a subset with k elements of the set A. There is only one function : A k ---+ A k that has the property for all E A k . This function can be extended in (m ways to a function f : A ---+ B such that f(i) f:. for all

nh

i E A \ Ak •

h (i ) = i

- l )n- k

The number of the subsets Ak of A is

l ) n- k .

(�),

k

i

i

hence the desired number is

(�) m­ (

ii) m ::; n. Let Bk be a subset with elements of the set B. There is only one function B k ---+ Bk such that = for all E Bk• This function can be extended in ( m - l ) m - k ways to a function 9 : B ---+ B such that g (i) f:. for all i E B \ Bk •

h:

h (i ) i

235

i

i

6. COMPREHENSIVE PROBLEMS

23 6

mn-m ways to a function f : A

Moreover, each function 9 can be extended in that clearly has exactly fixed points .

k

The number of the subsets Bk of B is

(7) mn -m (m - W -k .

(7)

-4

B,

hence the desired number is

: A 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. Therefore the number of functions f

(Dorin Andrica,

-t

Romanian Mathematical Regional Contest "Marian 'farina,

2001) 5 . Using the inequality

x r + x� +

.

.

.

+ x��

�

for

� (Xl X2 n -l +

+ ... +

Xn ) m

we obtain

On the other hand,

an al a2 an al a2 a2 + a3 + · · · + -al >- n n a2 a3 . . . -al = n from the AM-GM inequality. Therefore (2n) m (1 aa2l ) m (1 aa32 ) m (1 aaln ) m � � m I n -

+

+

-

-

+ . . . +

+

'

-

+

as desired.

= n.

2m ,

( 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 k ) kn L kn-l L 2 kn-2 II( al al a i = l ai = +

+

Using the AM-GM inequality gives

+

+ ... +

ala2 · · .an·

237

6.2. SOLUTIONS From the previous relation we deduce

(n a;r kn-2 + n ( k + w.:) + . . + (�) . (nf k + � n a; ::; n (k + a;). 1

2

n ( k + a;) � kn + (�) (n ai r kn- 1 + (;) Thus

n

(1)

Using again the AM-GM inequality, we obtain

so

n

n

II

i=l

(k + ai) ::; k + -1 Ln ai. n i=l

(2)

(Titu Andreescu, Revista Matematidl Timi§oara (RMT) , No. 2 (1977), pp. 63,

Problem 3045)

First Solution.

7. Since abc = 1, this non-homogeneous inequality can be trans­ formed into a homogeneous one by a suitable change of variables. In fact, there exist positive real numbers such that

p, q, r

q c = -r . q p r Rewriting the inequality in terms of p, q, r, we obtain (1) (p - q + r) (q - r + p)(r -p + q) ::; pqr, where p, q, r > o. At most one of the numbers = p -q+r, = q -r+p, W = r-p+q is negative, because any two of them have a positive sum. If exactly one of the numbers is negative, then ::; 0 < pqr. If they are all nonnegative, then by the AM-GM inequality, 1 ..;uv ::; 2 ( + ) = p. Likewise, ..;vw ::; q and ..jWu r. Hence pqr, desired. Second Solution. Expanding [P(out the left-hand side of (1) gives (p-q+r)(q -r +p)(r -p+q) = 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). a = !!. ,

b = -,

v

u

u, v, w

uvw

u

::;

v

uvw

::;

as

=

6. COMPREHENSIVE PROBLEMS

23 8 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 + c b-1+ = a c- 1 +

(

D(

D (

D

= (ab - b + l ) (be - e + l) (ea - a + 1) = L 1 • Also, since lib = ae, lie L=

(

= ab, 11a = be,

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) ::; [ (ab - b + 1) + (b - 1 + ab)] = abo 2 Likewise, J(be - e + l) (e - 1 + be) ::; be,

J(ea - a + l) ( a - 1 + ae) ::; ca. Hence L = VL 1 L2 ::; (ab) (be) (ea) (abe) 2 = 1. Fourth Solution. Using the notations established in the third solution, it 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 AM-GM inequality gives 2 2:: 2eVbuv,

2 2:: 2aVcvw,

> O. The

and 2 2:: 2vawu,

from which uvw ::; 1 . Fifth Solution. Let ab - b + 1, = be - e + 1 , = ca - a + 1; U2 = 1 - be + c, V2 = 1 - ea + a, and 2 = 1 - ab + b. As in the third solution, we only need to consider the case in which > 0 for i = 1, 2. Again, we have

Ul = W Ui , Vi , Wi

VI

WI

6.2. SOLUTIONS

239

Y Let X = a + b + e and = ab + be + ca. Then Y U l + VI + WI = - X + 3 and U + V + W 2

2

2

=

X - Y + 3.

Hence either U l + VI + WI :::; 3 or U2 + V2 + W2 :::; 3. In either case L :::; from the AM-G M inequality. ( Titu Andreescu, IMO 2000, Problem 2 )

1 follows

a �y (3 'Y. Let x, y, be three Then ( ) )(( ) y E (x, y, z - E ( , z, y = (1' - a z - X ) 2 - - x2 ) ) � 0, and (3( ( E( a, y, b) = a y - a) 2 + 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 ( ) a(y - a)2 + f3(y - b)2 fl y and (3( ( h (y) = a y - b) 2 + y - a) 2 on the interval [a , b] . Since h (a) = f3(b - a) 2 � a ( b - a) 2 II (b) and the coefficient y a. 2 of y in h is a + f3 � 0, it follows that the maximum value of II is obtained for y b. Likewise, h (b) � h (a) and the maximum value of h is obtained for 8. Without loss of generality we can assume that arbitrary numbers from the interval [a, b] such that x :::; x

:::;

z

:::;

z.

=

=

=

Therefore

=

y, b) = E(a, a, b) = (f3 + 'Y) (b - a)2 , and ({3 + 'Y) (b - a)2 . max E (b , y, a) = E (b , b , a) It follows that the maximum value of E is ( f3 + 'Y) (b - a) 2 and is obtained for a , y = a , z = b or = b, y = b, z a. max E (a,

yE[ a , b]

=

yE[ a , b]

x =

(Dorin Andrica

1 (1983) ,

pp.

x

and loan 66, Problem C5:2)

Ra§a,

=

Revista Matematica Timi§oara (RMT) , No.

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, respec­ tively. Obviously p + q + r n and without loss of generality we may assume that p � q � r. A term If(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

6. COMPREHENSIVE PROBLEMS

24 0

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 Then PI

+ ql + TI

=

= n and

= qo , T = TO + 1. I

= Poqo + POTO + qo ro + 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

= qo = ro = k.

=

= k and then

In this case the maximal value is

2 k k ) 6 2 2 2 n + = = 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 k) 2((k + 1)k + (k + 1) k + 2 = 2(3k2 + 2k) = 32 (n2 - 1). 3) = 3k + 2. Then Po + qo + ro = 3k + 2, Po 2:: qo 2:: TO Po - 1, so 3Po 3k + 1 2:: 3po - 2. It follows that Po = k + 1 and qo + ro = 2k + 1. Because k + 1 qo TO 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 2 2 n n -1 Therefore the requested number is 2 3 if 3 divides n and 2-- otherwise. 2 ( k2

+ k2

n

2:=

2::

2:=

3

Remark.

2:=

2:=

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

4 fb 2 4 fb2 2 . . . ( .fiil ) + + (va; l ) + ( f02 �_ � � Y UI

it follows that

_

Y

U

_

'b)

4 Un Y

2

> - 0,

6. 2. SOLUTIONS

241

We have

hence

va;; � .jb; + Jb; + . . . + A,

va; + .Ja2 + . . . +

as claimed.

( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2 (1977) , 63, 3046) Problem 1 11 . By mUltiplying the numbers aI, a2 , . . . , an with a suitable factor '" ai2p i=l n ;P 1 L we may reduce the problem to the case when a . i l = Assume without loss of generality that a al � a2 � . an· Let a = � and suppose by way of contradiction that S ( n , p) pp.

M =

-n

L-.t

=

=

.

.

�

2

Then

hence

n

n L a;P L[ 1 )] 2P . + a(i a i= l (0 i= l Consider the function

0 for all real x, it follows that

m

1,

�

>

1

1.

This is equivalent to

n1 + . . . + nm + nm+ 1 and then

>

(1 + Jn1 + n2 + . . + nm )2 .

1 + 2Jn1 + n2 + . . . + n m , m � 1 . We induct on m. For m = 1 we have to prove that n 2 > 1 + 2J1i1. Indeed, n2 � n1 + 2 = 1 + (1 + n1) > 1 + 2.Jril. Assume that the claim holds for some m � 1 .

nm+ 1

Then

>

252

6. COMPREHENSIVE PROBLEMS so (n m+l - 1 ) 2 4(nl nm ) hence (nm+l 1)2 4(nl + nm+d. This implies n m+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 � v� 8u� 4u n t� +

=

+

=

,

is a square, as desired.

(Dorin Andrica)

(an ) n2:n1 be a sequence of real numbers. From the equality x n xn x2n . . . xkn . . , I x i < 1, n � 1 1-x we derive n - L An Xn , L � 1 - xn n= l l n= where An La ndi d . Using Gauss' formula L

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 + X k-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 3 8 66) 38.

From the relation iii) we obtain

an + 1 - a n an

+

b n+ 1 - b n - 0 bn

, n � 1,

6. COMPREHENSIVE PROBLEMS

262 then

' an+l - an 1·1m ( - -ab n ) = nl-+oo 1m - bn b n-+oo n+l n On the other hand, by the Stolz-Cesaro theorem we have ·1m abn+l -- anbn = nl-+oo ' m an l -bn n1-+oo n+l It is easy to see that relations (1) and (2) imply l· m -anbn = 0, n1-+oo

Using ii) yields

(1) (2)

as desired.

( Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1 (1978 ), pp. 69 ,

Problem 39.

3304)

Consider the determinant

0 X l X2 Xn X l 0 X2 Xn � (Xl, X2 , . . . , Xn ) = Xl X2 0 Xn X l X 2 X3 0 Note that �(0,X2, . . . ,Xn) = �(XI,0,X3 , . .. , Xn ) = . . . =�(Xl,X2 , . . . ,Xn-I ' 0) = O. Moreover, if Xl + X 2 + . . . + X n = 0 , then �(XI ' X 2 , . . . , x n ) = 0 , therefore �(XI ' X2 ,·· . , xn ) = aXI X2 . . . 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

---'-. . . + Uk UIUI++U2U2++. .. .. +. +Uk+l UI . . . Uk = 1 + UI +Uk+l Uk then n + U 2 . . . + Un+l UI ----II (1 + UI . . . Uk) = UI + U2 ----'-k=2 Since U 2 = -uI , we obtain UI + . . . + Un+l 1 n -n II (1 + UI . . . Uk) = nU ( 1 - UI l ) k=2 1+

--

-­

+

263

6.2. SOLUTIONS U = nlim-+oo Un and note that + . . . + Un+l = U. U l 1m n-+oo It follows that 1 IIn (1 + Ul . . . Uk ) = U lim n-+oo k=2 Ul ( 1 - Ul ) Let

1.

n

-

n

.

( Titu Andreescu, Revista Matematica Timi§oara (RMT), No. Problem 3533)

40. Let

a = nlim-+oo anbn and let

bn

>

O . There is an integer n (c)

>

0 such that

k+1 < an k + 1 k - 1 bn < a + c k - 1 for n - n (c) 0 it follows that abn - c kk +_ 11 bn < an < abn + c kk +_ 11 bn for all n ;::: n (c) . + 1 n+1 + bn ) < an+l - an < a(bn+1 - bn ) - c kk -(b -1 k + 1 (bn+1 + bn ) , < a(bn+ 1 - b n ) + c k 1 a-c

Since

c

2(1978), pp. 52,

>

--

--

>

Then

-

_

and, dividing by

bn+1 - bn 0, we obtain k + 1 bn+1 + bn an+l - an k - 1 bn+ bn bn+1 - bn >

a - c -- ·

1

-


i

6.2. SOLUTIONS

265

= = It follows that for some k =I- l i.e. which is a contradiction. The solution is complete. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. pp. Problem

a� aI

l ak I lad,

1-2(1980),

69,

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 = (4 � Function f is continuous and = nlim -+oo Xn 00 , hence f(xn) = Yn for sufficiently large. Then g' (xn ) = f' (xn) so

n + 1) �, n 1. n

nlim -+oo g'(x n) = nlim -+oo f'(x n ) = 00 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 is rational. Hence there are integers with = P. and gcd(p, = = 0 a contradiction. Therefore is irrational. Then f(q) = = Conversely, let be irrational and assume by way of contradiction that f = = f(n) for some integers =I- � O. Then so We obtain

f (2q) a

p, q

q) 1.

a q

a

a

(m) m n {am} {an} am - [am] an - [an]. a = [am]m -- n[an] Q, 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; } . E

q

By the division algorithm, there are integers and r such that

n = tq + r,

Then

r E {O, 1, 2,

. . . , q - I}.

{ 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;} f(n)

=

6.

266 It follows that i - j 0, so i j . Therefore

=

=

COMPREHENSIVE PROBLEMS

(i - j )p is an integer. Note that (p, q) q

M = {f(O), f(I), . . . , f(q - I)}

= 1 and Ii - j l < q, hence

= { 0 , qI , q2 , . . . -

-

q-l 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 g(x + t) g(x) , x E R

=

Then

f( x + t) + f(xO + to) = f(x ) + f(xO ), x E JR,

hence

f(t) + f(tO)

From the relation iii) it follows that

f(t) and then

t

= kI T

= 2f(0) = 2M.

= f(tO) = M

and to = k2 T,

for some integers k l ' k2 -:f. O. This gives

0 = kik2 E Q,

a contradiction.

(Dorin Andrica)

46. We start with an useful lemma. Lemma. If 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

0

=

1

n (J) J C (0, 1 ) . We consider two cases:

267

6. 2. SOLUTIONS 1 J = (O, e) , wI. th 0 < e < 1. n(J) Let N be an integer such that < e and consider the numbers 1 . J1

=

�

{B}, {2B}, . . . , {NB}. There are p, q E {I, 2, . . . , N, N + I} such that 0 < {pB} - { qB} :::; 1 N On the other hand,

M, M,

{pB} - { qB} = [qB] - [PB] + (p - q)B E hence JI n 0. It follows that n(J) ( {pB} - { qB}) E J n 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 is c E such that 0 < c < b - a.

M -:f.

Let

M

= [�]

no + 1. n J1 . Likewise, J n

M

M -:f.

Then a < noc < b and no c E 0, as desired. The lemma is now proved. a) Let A E min f(x), max f(X) . Hence there is Xo E 1R such that f(xo) = A. x E IR x EIR From the lemma we deduce that there are sequences (Xn ) n 2: 1 and (Yn) n 2: 1 such that

[

]

nlim -+oo (xn + YnT) = Xo .

The function f is continuous, so

nlim -+oo f(xn + YnT) = f(xo) = A. Note that f(xn + YnT) = f(xn ), therefore lim f(x n ) = 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 ) n 2: 1 of integers such that

�

A = nlim -+oo f(Bxn), -+oo g(xn ) = nlim as desired.

(Dorin Andrica, " Asupra unor §iruri care au multimea punctelor limita intervale" , Gazeta Matematica (GM-B) , No. 11 (1979) , pp. 404-406)

47. i) Let = x + Y = z + v and assume that x < Y and z < v . Then we have x < -2 ' z < -2 ' Y = - x and v = - z . u

u

u

u

u

268

6. COMPREHENSIVE PROBLEMS Suppose, by way of contradiction, that there is a number A > 1 such that

xA + y A Consider the function f : (0, u )

-+

f (t)

=

=

Z A + VA .

(0, (0) ,

tA + ( - t) >.., u

and note that f is differentiable. We have l' (t) = A [t A -1 - ( u - t) A - I J ,

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. Z A + v\ which is a contradiction. ii) Because p is a prime, by Little Fermat ' s Theorem we have

aP - a == bP - b == cP - c ==

hence

rF

- d == ° (mod p),

- ( aP - ) + (bP - b) - (cP - c) + (dP - d) == ° (mod p).

From P + bP

a

=

a

cP + dP , we deduce that

a - c + b - d ==

°

(mod p)

(1)

By i) we note that + b -:f. c + d, therefore

a

and then as desired.

l - c + b - d l � p,

a l a - c l + I b - d l � p,

( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978 ) , pp. 55,

Problem 3550)

48. i) Consider the function

at - 1 = ln a. t-+limo -t 0 there is 8 0 such that at - 1 In a - c -- In a + c , >