360 Problems for Mathematical Contests

(1)

TITD ANDREESCD

DORIN ANDRICA

360 Problems for

(2)

TITU ANDREESCU DORIN ANDRICA

360 Problems

for

Mathematical Contests

(3)

360 Problems for Mathematical Contests

Authors: Titu Andreescu, Dorin Andrica

National Library of Romania CIP Description

ANDREESCU, TITU

360 Problems for Mathematical Contests/ Titu Andreescu, Dorin Andrica. - Zalau: Gil, 2003

p. ;cm.

Bibliogr.

ISBN 973-9417-12-4

1. Andrica, Dorin 51(075.35)(076)

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 $erbanVodft _I33,S.4,Cod 70517,BUCURE$TI Tel.:3362911 Fax: 337 0735

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.

(5)

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

4

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.

(6)

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

6

Chapter

₁

(7)

PROBLEMS

1. Let C be a set of n characters

{

Cl '

C2,

. . •

, cn}.

We call word a string of at

most m characters, m ::; n, that does not start nor end with Cl .

How many words can be formed with the characters of the set C?

2. The numbers 1, 2, . . .

, 5n

are divided into two disjoint sets. Prove that these

sets contain at least

n

pairs

(x, y), x

>

y,

such that the number

x - y

is also an element of the set which contains the pair.

3. Let

al, a2, .. . ,an

be distinct numbers from the interval

[a, b]

and let 0' be a

permutation of {I, 2, ... , n} .

Define the function

f

:

[a, b]

-t

[a, b]

as follows:

f(x)

=

{

a

x

q(

i

) if _otherwIse

x

=

�i'

i = I, n

Prove that there is a positive integer

h

such that

flh](X)

fafa ... aJ.

�

htimes

x,

where

flh]

4. Prove that if

x, y, z

are nonzero real numbers with

x

+

y

+

z

= 0, then

x2

+

y2 y2

+

Z2 Z2

+

x2 x3 y3 Z3

-- +

x

--+ -- = - + - +-.

+

y y

+

Z Z

+

x yz zx xy

5. Let

a, b, c, d

be complex numbers with

a

+

b

+ C +

d

_{= o.}Prove that

a3

+

b3

+

c3

+

d3

=

3(abc

+

bcd

+

cda

+

dab).

6. Let

a, b, c

be nonzero real numbers such that

a

+

b

+

c

= 0 and

a3

+

b3

+

c3

=

a5

+

b5

+

c5•

Prove that

7.

Let

a, b, c, d

be integers. Prove that

a

+

b

+

c

+

d

divides

(8)

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

yX

+

VY

+ 2v'z=2 +

v'u

+

Vv

= x + y + z + u + v.

10.

Find the real solutions to the equation

(x + y) 2 = (X + 1) (y - 1).

11.

Solve the equation

x +

V

4X +

V

16X +

J. .

.

+ J4nx + 3 -

JX

= 1.

12.

Solve the equation

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

where

a, b, c

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

13.

Let

a

and

b

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

(x, y),

solutions to the system of equations

14.

Solve the equation

{

x4 - y4 = ax - by

x2 - y2 =

f/

a2 - b2.

[

25x -2]

₄

_{= 13x + 4}

_{3 '}

where

[a]

denotes the integer part of a real number

a.

. 1 +v1s

15.

Prove that If

a

�

--2-'

then

[

1 +

[�]l

= n, n = O, 1,2, .. .

1. 1 . PROBLEMS 11

16.

Prove that if

x, y, z

are real numbers such that

xS + yS + ZS

i=

0,

then the ratio

2xyz - (x + y + z)

xS + yS + ZS

equals

�

if and only if

x + y + z = O.

17.

Solve in real numbers the equation

.vx;:-=-r

+ 2-JX2 - 4 + ... + nJXn - n2 = �(Xl + X2 +

. . . +

xn).

2

18.

Find the real solutions to the system of equations

1

1 1

- +- = 9

x y

1 1

1

1 - + - 1 + - 1 + - - 18

(� ?'Y)( �)( ?'Y)

-19.

Solve in real numbers the system of equations

y2 + u2 + v2 + w2 = 4x - 1

x2 + u2 + v2 + w2 = 4y - 1

x2 + y2 + v2 + w2 = 4u - 1

x2 + y2 + u2 + w2 = 4v - 1

x2 + y2 + u2 + v2 = 4w - 1

20.

Let

aI, a2, as, a4, a5

be real numbers such that

al + a2 + as + a4 + a5 = 0

and max

lai - ajl

<

1.

Prove that

ai + a� + a5 + a� + a�

<

10. l�i<j�5

-

-21.

Let

a, b, c

be positive real numbers. Prove that

1 1 1 1

1 1

- + - + - > -- +-- + --

_{2a 2b 2c - a + b b + c c + a}

22.

Let

a, b, c

be real numbers such that the sum of any two of them is not equal to zero. Prove that

_

a5:-+_b

-::-5 _

+_

c

--:-5_

----.:.,.( a_

+_b_

+

---..:c )

_

5 > 10 (a + b + C)2

as + bS +

c3 -

(a + b + c)S - 9

23.

Let

a, b, c

be real numbers such that

abc = 1.

Prove that at most two of the

numbers

are greater than

1.

1

(9)

1 2 1. ALGEBRA

24.

Let

a,

b, c,

d

be real numbers. Prove that

min(a

- b2,

b - c2, C

-

d}, d -

a2)

�

�.

25.

Let

aI, a2,' .. ,an

be numbers in the interval

(0, 1)

and let

k ;::: 2

be an integer. Find the maximum value of the expression

n

L

_{ytai(I - ai+I),}

i=l

where

an+1

=

a

1

.

26.

Let m and

n

be positive integers. Prove that xmn - 1 x

n - I

---

_m

>

-

--

_x

for any positive real number x.

27.

Prove that m! ;::: (n!) [�] for all positive integers m and

n.

28.

Prove that

1 1

1

�

2 I +- +-+"' +- > n - -

y'2 v'3 \Iii

n +

1

for any integer

n ;::: 2.

29.

Prove that

n

(1 - 1/

vn) + 1 > 1 +

�

_{2 3}

+ � + .. . +

.!.

_n

> n (\In + 1 - 1)

for any positive integer

n.

( ) d

na

1

a2 .. . an

30.

Let

aI, a2, .. . , an

E

0, 1

an let

tn

=

al + a2 + .. . + an

.

Prove that

n

L

loga;

tn ;::: (n - I

)

n

.

i=l

31.

Prove that between n and

3n

there is at least a perfect cube for any integer

n ;::: 1.

0.

32.

Compute the sum

33.

Compute the sums:

n

� Io(k+l) Sn =

L-(-I)

2

k.

k=l

a) Sn =

�

(k + l)l(k + 2)

(�);

_b)

Tn =

�

(k + l)(k + 2)(k + 3)

(�).

1.1. PROBLEMS

34.

Show that for any positive integer

n

the number

C

n

;

1 )

22n +

C

n

:

1

)

22n-2 .3+ .. . +

C

n2: 1

)

3n

is the sum of two consecutive perfect squares.

35.

Evaluate the sums:

36.

Pr.ove that

12 (�)

+ 32

(�)

+ 52

(�)

+ .

.

=

n(n + 1)2n-3

for all integers

n

;:::

3.

37.

Prove that

2n

L

[log2

kJ = (n - 2)2n + n + 2

k=l

for all positive integers

n.

38.

Let

Xn

=

22n +

1,

n

=

1,2,3,

. . . Prove that for all positive integers

n.

1 2 22

2n-1 1

- + - + - + .. . +-- <

_{Xl X2 X3}

_{Xn 3}

39.

Let f : C -+ C be a function such that f(z)f(iz)

= Z2

for all z E C.

Prove that

f(z)

+

f( -z) =

0

for all z E C

13

40.

Consider a function f

: (0,00)

-+ 1R and a real number

a > 0

such that

f(

a

) = 1. Prove that if

f(x)f(y)

+

f

(;)

f

(�)

= 2f(xy) for all x, y E

(0, 00),

then f is a constant function.

41.

Find with proof if the function f·: 1R -+ [- 1, 1), f(x) = sin[x) is periodical.

n

42.

For all i,j =

l,n

define S(i,j)

=

L

ki

+i. Evaluate the determinant

� =

k=l

(10)

14

43.

Let

1.

ALGEBRA

{

a

i if i = j Xij = 0 if i _ij, i + j _i2n + 1

bi

if i + j ₌2n + 1

where

a

i ,

bi

are real numbers.

Evaluate the determinant

�2n

= IXij l .

44.

a) Compute the determinant

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

b) Prove that if the numbers

abed, bade, cdab, deba

are divisible by a prime

p,

then at least one of the numbers

a

+

b

+

e

+

d, a

+

b

-e -

d, a - b

+

e - d, a - b - e

+

d,

is divisible by

p.

45.

Consider the quadratic polynomials

t1

(x) = x2 + PI

X

+ q

r

and

t2

(x) = x2 + P2X + q�, where Pl , P2, ql , q2 are real numbers.

Prove that if polynomials

tl

and

t2

have zeros of the same nature, then the polynomial

has real zeros.

46.

Let

a, b, e

be real numbers with

a

> 0 such that the quadratic polynomial

T(x) =

ax2

+

bex

+

b3

+

e3 - 4abe

has nonreal zeros.

Prove that exactly one of the polynomials Tl (x) =

ax2

+ bx +

e

and T2 (x)

ax2 +

ex

+

b

has only positive values.

47.

Consider the polynomials with complex coefficients

P(x)

=

xn

+

alxn-l

+ . . . +

an

and Q(x) =

xn

+

blxn-l

+ . . . +

bn

having zeros

Xl, X2, .. . ,Xn

and x

?

, x

�

, . . . , x

;

respectively.

Prove that if

al

+

a3

+

a5

+ . . . and a2 +

a4

+

a6

+ . . . are real numbers, then bl +

b2

+ . .. +

bn

is also a real number.

1.1.

BROBLEMS

48.

Let

P(x)

be a polynomial of degree n. If

k

P(k)

=

-k-

for

k

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

evaluate

P(m),

where

m

> n.

49.

Find all polynomials

P(x)

with integral coefficients such that for all real numbers x.

15

50.

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

P(x)

= PI

(xn+l)

+

XP2(xn+l)

+ . . . +

xn-lpn(xn+1),

is divisible by

xn+xn-l

+ . . ·+ 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

+

alxn-1

+ ... +

an

be a polynomial with integral coefficients such that

an

=1= 0 (mod

p).

Prove that if there are n + 1 integers

0'1, 0'2,

• • • ,

an+ 1

such that

P ( ar)

== 0 (mod

p)

for all

r = 1, 2, . . . ,n + 1, then there exist i,j with i i j such that

a

i ==

a

j (mod

p).

52.

Determine all polynomials

P

with real coefficients such that

pn(x)

=

P(xn)

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

53.

Let

P(x)

=

aoxn

+

alxn-l

+ . . . +

an, an

i 0,

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

m

with Prove that the polynomial

P

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

54.

Find all polynomials

P

of degree n having only real zeros Xl , X2 , . . .

, Xn

such

that

_n

1 n2

tt

P(x)

- Xi =

XPI(X) ,

for all nonzero real numbers x.

55.

Consider the polynomial with real coefficients

(11)

16

1.

ALGEBRA

and

an

f:. O.

Prove that if the equation

P(x)

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

equation

x2 PII(x)

+

3xP'(x)

+

P(x)

= 0 has the same property.

56.

Let R

�f

and R

�

) 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

�

j

and

P(Q(x))

E R

�j

, then

Q'(x)

E R

�

]-l).

57.

Let

P(x)

be a polynomial with real coefficients of degree at least 2. Prove that if there is a real number

a

such that

P(a)plI(a)

>

(P'(a))2,

then

P

has at least two nonreal zeros.

58.

Consider the equation

aoxn

+

alxn-1

+ ... +

an

= 0

with real coefficients

ai.

Prove that if the equation has all of its roots real, then (n

- l)ar

2:: 2n

a

O

a2

. Is the reciprocal true?

59.

Solve the equation

X4 - (

2m +

l)x3

+ (m

- l)x2

+

(

2m

2

+

l)x

+ m = 0,

where m is a real parameter.

60.

Solve the equation

x2n

+

a1x2n-1

+ ... +

a2n_2X2 -

2n

x

+ 1 = 0,

if all of its roots are real.

SOLUTIONS

1.

Let

Nk

be the number of words having exactly k characters from the set C,

1 _{� k �}m . Clearly,

N

l = n

-

1. The number that we seek is

N

l +

N

2

+ ... +

Nm·

Let A

k

=

{I,

2, ... ,k}, 1 _{� k �}m. We need to find out the number of functions

f : A

k

-+ A, k = 2, n with the properties

f

(l)

f:.

a1

and f(k) f:.

a1

For f(l) and f(k) there are n

-

1 possibilities of choosing a character from

C2, ... ,en

and for f(i), 1

<

i

<

k there are n such possibilities. Therefore the number

of strings f

(l)

f

(

2

)

... f(k

- l)

f

(

k

)

is

Nk

= (n

-

1

)2

nk-

2

It follows that

N

l +

N

2

+ ... +

Nm

= (n - 1) + (n

-

1

)2

nO + (n

-

1

)2

n

1

+ ... + (n -1

)2

n

m

-

2

=

(Dorin Andrica)

2.

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

{I,

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

(x,

y),

_x

>

y,

with the desired property.

Let k be a given number, k = 1, n. If k and 2k are in the same set

-

A or B

-the same can be said about -the difference 2k - k = k. The same argument is applied

for 4k and 2k. Consider the case when k and 4k are elements of A and 2k is an

element of B. If 3k is an element of A, then 4k - 3k = k E A, so let 3k E B. Now if 5k E A, then 5k - 4k = k E A and if 5k E B, then 5k - 3k = 2k E B; so among the

numbers k, 2k, 3k, 4k, 5k there is at least a pair with the desired property. Because k = 1, 2, ... , n, it follows that there are at least n pairs with the desired property.

(Dorin Andrica,

Revista Matematica Timi§oara (RMT), No. 2(1978), pp. 75, Problem 3698)

3.

Note that

(1)

(12)

18 1. ALGEBRA

and furthermore

(2)

for all integers ml, m2 2:: 1.

Suppose that for all integers k 2:: 1 we have I[k](x) i-

x.

Because there are n! permutations, it follows that for k > n ! there are distinct

positive integers nl >

n2

such that

(3)

Let

h

= nl

- n2

> 0 and observe that for all k the functions I[k] are injective, since numbers

ai,

i =

1 , n are distinct. From relation

(3)

we derive that

or

l[n2+h](x) = l[n2] (x), x _E

[a,

b],

(10 l[n2+h-l])(x)

=

(10 l[n2-1])(x), X E

[a,

b].

Because I is injective, we obtain

l[n2+h-l]

(x)

= l[n2-1]

(x), x

E

[a,

b]

and in the same manner

or

l[h+l](X)

=

I(x), x E

[a,

b]

l[h](X)

= x, x

E

[a,

b].

Alternative solution.

Let Sn be the symmetric group of order n and Hn the cyclic subgroup generated by a. It is clear that Hn is a finite group and therefore there is integer

h

such that a[h) is identical permutation.

Notice that

Ilk](X) =

{ aq[k1(i)

if

x

=

�i'

i =

1, n

x

otherwIse

Then I[h](x)

= x

and the solution is complete.

(Dorin Andrica,

Revista Matematidi Timi§oara (RMT) , No.

2

(197

8

)

,

pp.

53,

Problem 3540)

4. Because

x + y + z =

0, we obtain

x + y

=

-z, y + z

=

-x, z + x

=

-y,

or, by squaring and rearranging,

x2 + y2 = Z2

_

2xy, y2 + Z2 = x2

_

2yz, Z2 + x2

=

y2 - 2zx.

The given equality is equivalent to

Z2 - 2xy x2 - 2yz y2 - 2zx x3 y3 Z3

--

--=-

+

= -

+ - + -,

-z

-x

-y yz zx xy

1.2. SOLUTIONS and consequently to

(

Xy yz zx

)

x3 y3 Z3

-(x +y + z) + 2 - + - +

_{z x}

-

_Y

=

-

_{yz zx xy}

+ - + -.

The last equality is equivalent to

2(X2y2 + y2z2 +

Z

2

X

2

)

= x4 + y4 + Z4.

On the other side, from

x + y + z =

0 we obtain

(x

+ y + Z)2 =

0 or

x2 + y2 + Z2 = -2( xy + y z + zx).

Squaring yields

X4 + y4 + Z4 + 2(X2y2 + y2z2 + Z2X2)

=

4(X2y2 + y2z2 + Z2X2) + 8xyz(x + y + z)

or

as desired.

19

(Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No. 3(1971), pp. 25, Problem 483; Gazeta Matematica (GM-B), No. 12(1977) , pp. 501, Problem 6090)

5.

We assume that numbers

a,

b,

c,

d

are different from zero. Consider the equation

x4

- (L

_a

)

_{x3 +}

(L

_ab

)

_x2

- (L

_abc

)

_{x + abed}

= 0

with roots

a,

b, e, d.

Substituting

x

with

a,

b, e

and

d

and simplifying by

a,

b, e, d

i- 0, after summing up we obtain

Because

L

a

=

0, it follows that

L

a

_{3 = 3}

L

_abe.

If one of the numbers is zero, say

a,

then

b+ e + d =

0,

or

b

+

e = -d.

It is left to prove that

b3

+

e3 + d3 = 3bed.

Now

b3 + e3 + d3 = b3 + e3 - (b + e)3

=

-3be(b +

c)

= 3bed

as desired.

(Dorin Andrica,

Revista Matematica Timi§oara (RMT), No. 1-2(1979) , pp. 47, Problem 3803)

6.

Because

a

+ b + e

= 0, we obtain

(13)

20

or

1.

ALGEBRA

The given relation becomes

3abc = 5abc(a2 + b2 + c2 + ab + bc + ca)

3 a2 + b2 + e2 + ab + be +

ca

=

"5'

since

a, b, e

are nonzero numbers. It follows that

1

3 "2 [(a + b + e)2 + a2 + b2 +e2] ="5

and, using again the relation

a + b + e = 0,

we obtain

a2 + b2 + e2 =

�

_5'

as desired.

(Titu Andreescu,

Revista Matematica Timi§oara (RMT), No.

2(1977),

pp.

59,

Problem

3016)

7.

Consider the equation

x4 - (L: a)

x3

_{+ (L: ab) X2 - (L: abc) x + abed = 0,}

with roots

a, b, e, d.

Substituting

x

with

a, b, e

and

d,

respectively, we obtain after summation that

L:a4 + (L:ab) L:a2 +4abed

is divisible by

L: a.

Taking into account that we deduce that

is divisible by

L: a.

Hence

2(a4 + b4 + e4 + d4) - (a2 + b2 + e2 + d2)2 + 8abed

is divisible by

a + b + e + d,

as

desired.

(Dorin Andrica)

8.

The equation is equivalent to

(x2 + 6x + 5)(x2 + 6x + 8)(x2 + 6x + 9) = 360.

Setting

x2 + 6x = y

yields

(y + 5)(y + 8)(y + 9) = 360,

1.2.

SOLUTIONS

21

or

y3 + 22y2 + 157y = 0,

with solutions

Yl

_{= 0, Y2 = -11 + 6i,}

Y3

_{= -11 - 6i.}

Turning back to the substitution, we obtain a first equation,

x2 + 6x = 0,

with solutions

Xl = 0, X2 = -6.

The equation

x2 + 6x

=

-11 + 6i

is equivalent to

(x + 3)2 = -2 + 6i.

Setting

x + 3 = u + iv, u, v

E lR, we obtain the system

{

U2 - v2 = -2

_{2uv = 6}

It follows that

(u2 + V2)2 = (u2 - V2)2 + (2UV)2 = 4 + 36 = 40.

Therefore

{

U2 - v2 = -2

u2 + v2 = 2y'lO

and

u2 = y'lO - 1, v2 = vTO + 1,

yielding the solutions

X3,4 = -3

±

V..flO - 1

±

iV..flO + 1

where the signs

+

and - correspond.

The equation

x2 + 6x = -11 - 6i

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

X5 = -3 + VvTO - 1 - iVvTO + 1, X6 = -3 -V.J[O - 1 +iVvTO + 1.

(Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No.

3(1972),

pp.

26,

Problem

1255)

or

9. X

- ..;x

+ y

-VY

+

z

- 2vz-=-2 + u

- ..jU

+ v -

..;v

= 0,

(

VX-

D

2 +

(

VV -

D

2 + (vz-=-2 - 1)' +

+

(

VU -

D

2 +

(

vv -

D

2 = 0

Because

x, y,

Z ,

U, v

are real numbers, it follows that

1 ..;x

=

VY

=

..jU

=

..;v

="2

and

vz-=-2 = 1.

Hence

1 X =

Y

_{= u = v =}

4' Z

= 3.

(Titu Andreescu,

2(1974),

pp.

47,

Problem

2002;

Gazeta Matematica (GM-B), No.

10(1974),

pp.

560,

Problem

14536)

(14)

22

or

1. ALGEBRA

10.

Setting

X = x + 1

and

Y = y - 1

yields

(X + y)2 = XY

�

[X2 + y2 + (X + y)2] =

o.

Hence

X = Y = 0,

so the solution is

x

=

-1

and

y = 1.

(Titu Andreescu,

Revista Matematica Timi§oara (RMT

)

, No.

1(1977),

pp.

40,

Problem

2811)

11. �

x +

V

4X+

V

16X +

l··

+ v'4nx + 3 = v'Z + 1

Squaring the equation yields

V

4x +

V

16x +

l··

+ v'4nx + 3 =

2 _{v'Z +}

1 Squaring again implies

V

16X+

l··

+ v'4nx + 3 = 4v'Z + 1

Continuing this procedure yields

4nx + 3 = 4nx + 2· 2nVX + 1

1

and

2 .

2n Vx =

2.

Hence

x = 4n'

(Titu Andreescu,

Revista Matematica Timi§Oara (RMT), No.

4-5(1972),

pp.

43,

Problem

1385)

12.

We distinguish two cases:

1) b = c.

The equation becomes

Jx + a+ Jx + b = Jx + a,

so

x:= -b.

or

so

2) b

f:.

c.

Jx + b + Jx + c = Jx + a + b - c - Jx + a,

b - c

Jx + b - Jx + c Jx + a + b - c + Jx + a'

Jx + b - Jx + c = Jx + a + b - c+ Jx + a.

(1)

(2)

1.2.

SOLUTIONS Summing up relations

(1)

and

(2)

we obtain

J x + b

=

J x + a + b - c,

and then

a = c.

To conclude, we have found that

(i) If

b = c,

then the equation has the solution

x = -b.

(ii) If

b

f:. c and

a

f:.

c,

there is no solution.

(iii) If

b

f:.

c

and

a = c,

then

x = -a

is the only solution.

23 (Titu Andreescu,

2(1978),

pp.

26,

Problem

3017)

13.

Because

a

and

b

are distinct numbers,

x

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

and the system could be solved in terms of

a

and

b.

We have

a2b2 = b2y2 + 2bY(X4 _ y4) + (X4 _ y4)2

a2x2 = b2x2 + X2(X2 _ y2)3.

Subtracting the first equation from the second yields

which reduces to

Solving the quadratic equation in

b

yields

b = y3 + 3x2y

(and

a

=

x3 + 3xy2)

or

b

=

y3 - x2y

(and

a = x3 - xy2).

The second alternative is not possible because

a

=

x(x2 _y2)

and

b

=

y(y2_X2)

cannot be both positive. It follows that

a = x3+3xy2

and

b

=

3x2y + y3.

Hence

a + b = (x + y)3

and

a - b = (x - y)3.

The system now becomes

x+y = -Va + b

x - y = -Va - b

and its unique solution is

x = (-Va + b + -Va - b)/2, y = (-Va + b - -Va - b)/2.

(Titu Andreescu,

Korean Mathematics Competitions,

2001)

13x + 4

14.

Let

--3- = y,

Y

E Z. It follows that

3y - 4

x

=

(15)

---u-24

1. ALGEBRA and the equation is equivalent to

or

[

�(3Y : 4) - 2

]

_{= y,}

[75Y - 126] =

₅₂

Y

Using that for any real number

a, [a]

:s;

a < [a] + 1,

we obtain

75y - 126

₁

y:S; 52 < y + ,

126

178

or

126

:s;

23y < 178,

so 23 :s;

y <

23·

Note that

y

E Z, therefore

y = 6

or

y = 7,

thus

14

17 Xl = 13

and

X2 = 13'

are the desired solutions.

(Titu Andreescu,

3(1972),

pp.

25,

Problem

1552)

1 +V5

.

1

15.

From

a >

- 2

--- we obtam

a2 - a - I > 0,

-

or

a > - + 1.

- a

We have

Hence

That is because and

[1 + na2] 1

_{--a- =}

a

+ na - a,

O:S;a < 1.

[

1 +[

�

]

l

=

[

l + �:na - a

]

= [(1 +

�

- a

H

+ n] = n, n

�

O. ( 1 ) 1

1 + - - a

_{a a}

-:s;

(a - a)- = 1 - - < 1

1 _a

a

_a

1 +

-- Q

- > - > 0.

( 1 ) 1 1

_{a a a2}

(Titu Andreescu,

2(1978),

pp.

45,

Problem

3479)

16.

First we consider the case when

X + y + z = 0.

Then

x3 + y3 + Z3 = 3xyz

and the ratio equals

�,

as desired.

1.2.

SOLUTIONS

25

Conversely, if

2xy z - (x + y + z) 2

-�---'--::---�-

x3 + y3 + Z3 3

= -,

then and so

2(x3 + y3 + Z3 - 3xyz) + 3(x + y + z) = 0.

U sing the formula

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

we obtain by factorization that

(x + y + z)[2(x2 + y2 + Z2 - xy - yz - zx) + 3] =

°

and so

(x + y + z)[(x - y)2 + (y - Z)2 + (z - X)2 + 3] = 0.

Because the second factor is positive, it follows that

x + y + z = 0,

as desired.

(Titu Andreescu,

Revista Matematica Timi§oara (RMT)

,

No.

1(1973),

pp.

30,

Problem

1513)

or

17.

We write the equation as

Xl - 2VXI - 1 + X2 - 2 · 2JX2 - 22 + .. . + Xn - 2nJxn - n2 = 0,

(VXI - 1 - 1)2 + (.JX2 - 22 - 2)2 + .. . + (Jxn - n2 - n)2 = 0.

Because the numbers

Xi,

i

= 1, n

are real, it follows that

Xl = 2, X2 = 2 . 22, .. . ,xn = 2n2

(Titu Andreescu,

1(1977),

pp.

14,

Problem

2243)

18.

Using the identity

(a + b + C)3 = a3 + b3 + c3 + 3(a + b)(b + c)(c + a)

we obtain

(�+ �+ l) " = �+�+ 1 + 3 (�+ �) (1 + �) (1 + �) =

= 9 + 1 + 54 = 64

(16)

26

1.

ALGEBRA and so

1 1

- +- = 3.

_fIX

_*

The system is now reduced to

{

.!.+.!. = 9

_{x y}

1 1

fIX

+ * = 3,

which is a symmetric system, having the solution

1

1 x = -, y = 1

₈

and

x = 1,

Y

= -.

₈

(Titu Andreescu,

4-5(1972),

pp.

43,

Problem

1386)

19.

By summing up the equations of the system we obtain

It follows that

(4x2 - 4x + 1) + (4y2 - 4y + 1) + (4u2 - 4u + 1) +

+(4v2 - 4v + 1) + (4w2 - 4w + 1) =

O.

Due to the fact that

x,y,u,v,w

are real numbers, we obtain

1 x = y = u = v = w = 2

(Dorin Andrica,

8(1977),

pp.

321,

Problem

16782)

20.

From the triangle inequality we deduce

Hence So

l

ai - aj

l

::;

l

ai - ai+l

l

+ .. . +

l

aj-l -

a

j

l

::;

I:

_{(ai - aj}

)

2::;

I:

(i -j)2 =

1�i<j�5

= (12 + 22 + 32 + 42) + (12 + 22 + 32) + (12 + 22) + 12 = 50

5

4

� _�

a� - 2

_l � _{� ' 3 -}

a·a·

<

50 i=l

i,i=l _i'i'i

1.2.

SOLUTIONS and consequently

5 t

a� -

(t

al

)

"

�

50

5

Note that

I:

ai = 0

and so

I:

ar ::; 10,

as claimed.

i=l

27 (Titu Andreescu,

Romanian Mathematica Olympiad - second round

1979;

1-2(1980),

pp.

61,

Problem

4094)

21.

The inequality

(a + b)2 ;::: 4ab

yields

1 1 4

- + - >

--a b - --a+b

. . 1 1 4 1 1 4

and sImIlarly

,

' b e - b + e' e a - e + a

- +

- >

--

- +

- > --.

Summing up these inequalities yields

1 1 1 1

1

1 2a + 2b + 2e ;::: a + b + b + e + e +

a' as desired.

(Dorin Andrica,

8(1977),

Problem

5966)

22.

Using the identities

a5 + b5 + e5 = (a + b + e)5 - 5(a + b)(b + e)(e +

a

)

(a2

+ b2 + e2 + ab + be +

00

)

and we obtain or

a5 + b5 + e5 - (a + b + e

)

5 5

2 2 2

3 b3

c3

( b )3 = -3 (a + b + e + ab + be +

00

)

a + + - a+ + e

It suffices now to prove that

5

10 3(a2 + b2 + e2 + ab + be +

00

) ;::: g

₍a

+ b + e)2

3(a2 + b2 + e2 + ab + be +

00

)

;::: 2(a2 + b2 + e2 + 2ab + 2be +

200

)

.

The last inequality is equivalent to

a2 + b2 + e2 ;::: ab + be +

00,

which is clearly true.

(Titu Andreescu,

1(1981),

pp.

49,

Problem

4295;

6(1980),

pp.

280,

Problem

0-148;

No.

11(1982),

pp.

422,

Problem

19450)

(17)

28

1.

ALGEBRA

23.

Assume by contradiction that all numbers

2a -

�,

2b -

�,

2c -

�

are greater

b c a

than

1.

Then

and

(

2a -

D (

2b -

D (

20 -

D

>

1 (

2a -

D

+

(

2b -

D

+

(

20-

D

> 3.

From the relation

(1)

and using

abc

=

1

we obtain

3 >

2

(

a

+

b

+

c

)

-

(

�

+

�

+

�

)

.

a b c

On the other hand, relation (

2

) gives

2

(

a

+

b

+

c

) -

(

- +

1 1 1

-

+

-

)

> 3

a b c

which is a contradiction. The proof is complete.

(

1

)

(

2

)

(3)

(Titu Andreescu,

2

(

1986

)

,

pp.

72,

Problem

5982

)

24.

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

2

�J2

2

1 1 1 1

a - b

+

b - c

+

c

- a- +

d

- a

> + + +

-4 -4 -4 -4 0 >

G-ar

+

G-br

+

G-or

+

G-dr

This is a contradiction so the claim holds.

(Titu Andreescu,

1

(

1985

)

,

pp.

59,

Problem

5479

)

25.

Setting

a

i

=

sin2

ai

for i

=

1,

n, where

aI, a2, .. . , an

are real numbers, the expression becomes

n

E

=

2::

t/

sin2

ai

cos2

ai+b an+1

=

a1·

i=l

Using the AM-GM'inequality yields

1 _k

- 2::

_k

i=l

_b

f

;::: b1 b2 .. . bk,

bi > 0, i =

U.

For

b1

= sint

ai,

b

2 =

cost

ai+1

and

b3

=

b4

= ... =

bk

=

�

we obtain

ij2

1.2.

SOLUTIONS Summing up these relations for k =

1 ,2

, ... , n yields

and so

�

(

n(k

- 2

)

_>

_{_l}

_{_E}

k n +

2

-

2-;;-

1e-2 , n.21e;2 n

·

2

1

-t

E <

---

=

---- 2

2

{14 n

Hence the maximum value of

E

is

;;

and it is reached if and only if

1 a1

=

a

2 =

. .

.

=

an

= 2'

29 (Dorin Andrica,

1

(

1978

)

,

pp. 63, Problem

3266

)

26.

Because

x

and m are positive, we have to prove that

x(xmn - 1) - m(xn - 1)

;::: 0,

or

(xn - 1)[(xn)m-1x

+

(xn)m-2x

+ ... +

x

-

m

]

�

0.

Define

E(x)

=

(xn)m-1x

+

(xn)m-2x

+ . . . +

x

-

m and note that if

x

�

1,

then

xn ;::: 1

and

E(x)

�

0, so the inequality holds. In the other case, when

x

<

1,

we have

xn

<

1

and

E(x)

<

° and again the inequality holds, as claimed.

(Titu Andreescu,

2

(

1978

)

,

pp.

45,

Problem 3480)

27.

For m ::; n the inequality is clearly true, so consider m > n and define

p =

[:].

This implies that m = pn +

q

with

q

E {O,

1,

... ,n - I} and the inequality

can be written as

(pn +

q

)

!

�

(n!)P.

We have

(pn +

q

)

!

�

(pn) ! =

=

(

1

·

2 ... n)(n +

1

) ... (2n)

... ((P

- l)n +

1) .

.

(pn) ;::: (n!)P,

and we are done.

(Titu Andreescu,

2

(

1977

)

,

pp.

61,

Problem 3034)

28.

We will use the inequality

Xm1

+

x2m

+ .. . +

xm

n - nm-1

>

_1_ (Xl

+

X2

+ .. . +

X )m

n ,

(18)

30

1. ALGEBRA

Now set

Xl

=

1, X2

=

2, .

.

, Xn

= n and m =

_.!..

We obtain

n

1 1

1 1 [n(n+l)l-�

{;f;

1 +-+-+

..

.

+->--

=

nn

_ _

y'2 V'3 vn

n-�

-

l 2

n+l'

as desired.

(Titu Andreescu,

Revista Matematica Timi§oara

(

RMT

)

, No.

2(1974),

pp.

52,

Problem

2035)

or

29.

From the AM-GM inequality we deduce

1 _�

i + 1 >

�

_{TIn i + 1}

,.� -

�

-.-_ -.-=

v

n+l,

n i=l

�

i=l

_�

1 n 1

1 +

-

n� �

� -;-

> \In + 1,

-i=l

and so

1 l

I

n

1 + -+ - + ... + -> n (\In + 1 - 1)

_{2 3}

_n

_'

as desired.

Observe that the inequality is strict because the numbers

i

�

1, i

=

1, n,

are

distinct. �

or

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

Therefore

1 2

n-l

1 +-+ - + · · · +--

_{2 3}

_{n > n}

�

_ =

_1_.

n

vn

n (1 - _1_) + 1 > 1 +

_vn

!

₂

+ ... +

.!..

_n

(Dorin Andrica,

(

RMT

)

, No.

2(1977),

pp.

62,

Problem

3037)

30.

Because the numbers

aI, a2,. " ,an

are positive, from the AM-GM inequality

al + a2 + .. . + an

n

2: '\I'ala2 .. . an

we deduce that

tn

�

(ala2 '" an) n�l

. Using that numbers

ai

are less than

1

we obtain

n-l

logai tn 2: --loga. (ala2

. .

.

an

)

.

n

•

1.2. SOLUTIONS

Summing up these inequalities yields

n

_{1 n}

Llo

g

a

i

tn 2: n - Llogai(ala2 .. . an)

=

i=l

n i=l

n-l

=

-n-[n + (logal a2 + loga2 al) + ... +

(

logal

an +

log

an al) + ... +

+(logan an-l + logan_l an)].

1

Note that

a +

-

> 2

for all

a

>

0,

so

a

-�

n-l

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

=

(n -l)n,

i=l

n

as claimed.

31 (Dorin Andrica,

(

RMT

)

, No.

2(1977),

pp.

62,

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 a perfect cube.

Proof.

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

a

and

b.

Then there is an integer

c

such that

c3

�

b

<

a

�

(c

+

1)

3 .

This means

so

which is false. 0

Now we can easily check that for

n

=

10,11,12,13,14, 15

the statement holds. If

n

2:

16,

then or Hence

3

1

1 n > (2,5)

=

(1,4

- 1)3

_>

({13

_

1)

3 '

�

1 n>

{I3-r'

�

-

�

>

1,

and using the above lemma the problem is solved.

(Titu Andreescu,

(

RMT

)

, No.

1-2(1990),

pp.

59,

Problem

4080)

(19)

32 1. ALGEBRA k(k

+ 1) .

32.

Note that the number

₂

IS odd for k

= 4p + 1

or k

= 4p + 2

and is even for k

= 4p + 3

or k

= 4p,

where

p

is a positive integer.

We have the following cases: i) if n

= 4m,

then n m-l "'"' k(k+l) "'"'

Sn = �(-1)

2

k = �(-4p- 1 - 4p- 2 +4p+3+ 4p+4) = 4m.

k=1 p=O ii) if n

= 4m + 1,

then

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

iii) if n

= 4m + 2,

then

Sn = 4m - (4m + 1) - (4m + 2) = -(4m + 3).

iv) if n

= 4m + 3

then Hence

Sn = 4m - (4m + 1) - (4m + 2) + (4m + 3) =

o.

-1

{

4m

Sn =

_�

_{(4m + 3l}

if

n = 4m

if

n = 4m + 1

if

n = 4m + 2

if n

= 4m+3.

(Dorin Andrica,

1(1981),

pp.

50,

Problem

4303)

33.

a) Summing up the identities

(

n

+ 2

)

(n

+ 2)(

n

+ 1)

(

n

)

k

=

(n

+ 2 -

k)(n

+ 1 -

k) k for k

= 0

to k

=

n yields

S

n - (n _

+ 2)(n + 1)

1 (�

(

n

+ 2

)

_

(

n

+ 2

)

_

(

n

+ 2

)

=

k=O k n

+ 1

n

+ 2

1

[2n+2 _ (n

+ 2) _ 1]

_ 2n+2 - (n

+ 3)

(n

+ 2)(n + 1)

- (n

+ 2)(n + 1) .

b) Summing up the identities

(

n

+ 3

)

_ (n

+ 3)(n + 2)(n + 1)

(

n

)

k

-

(n

+ 3 -

k) (n

+ 2

-

k) (n

+ 1

- k) k for k

= 0

to k

=

n yields T. n - (n _

+ 3)(n + 2)(n + 1)

1

1.2. SOLUTIONS

.

(�(

n

;

3 ) -(�:D -(�:!) -(�:�))

=

_

1 (

n+3

1

( 2

)

_ - (n

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

2 n

+

3n

+ 2)

-_

2

n+

4

- (n2

+ 3n + 2)

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

33

(Dorin Andrica,

2(1975),

pp.

43,

Problem

2116)

34.

Let

Sn

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

1 [(

)

2n+l

(

)

2n+l

]

Sn =

4

2 +

V3

+ 2 -

V3 .

The required property says: there exists k >

0

such that

Sn

= (k

- 1)2 +

k2 , or, equi val entl y,

2k2 - 2k + 1 - Sn =

o.

The discriminant of this equation is

� = 4(2Sn - 1),

and, after usual

computa-tions, we obtain

A

=

e

l +

.J3)2n+l

;

(1-

.J3)2nH

r

Solving the equation, we find

k = 2n+1

+ (1 +

v'.3)2n+l

+ (1 -

V3)2n+l 2n+2

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

Em

(1 +

v'3)

m

+ (1 -

v'3)

m,

where

m

is a positive integer. Clearly,

Em

is an integer for all

m.

We will prove that

2[ �]

divides

Em,

m = 1,2,3, . . .

Moreover, the numbers

Em

satisfy the relation

Em

_{= 2Em-1 + 2}

Em

-2•

The property now follows by induction.

(Dorin Andrica,

Romanian IMO Selection Test,

1999)

35.

Differentiating the identity

yields

where

sin nx

=

sinn X

((�)

cotn-1 X -

(

;

)

cotn-3 X

+

(�)

cotn-5 X

-

• • •

)

n cos nx

=

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

�

PI (cot X), sm X

(20)

34

and

1.

ALGEBRA 7r • For

x =

"4 we obtam Because

7r

(v'2)

n

ncosn"4 =

2

(nP(l) - 2P'(1)).

nP(l) = n

(

�

)

- n

(

;

)

+ n

(

;

)

-

. . .

-2P'(1) = -2(n - 1)

(

�

)

+ 2(n - 3)

(

;

)

- 2(n - 5)

(

;

)

+ .. . ,

we have

nP(1) - 2P'(1) = -

[

(n - 2)

(

�

)

- (n - 6)

(

;

)

+ (n - lO)

(

;

)

-

... J

=

= -n

(G) - (

;

)

+

(

;

)

- .

. .

)

+ 28n•

To conclude, use that

(�)

_

(

;

)

+

(

;

)

_ '" =

(V2)

n

sm

n

;

,

hence

_

n

(

v'2

)

n

(

n7r . n7r

)

Sn

_{- 2}

cos ""4

+

sm ""4

(Dorin Andrica,

)

, No.

2(1977),

pp.

89,

Problem

3200)

36.

Differentiating with respect to

x

the identities

(x + l)n =

(�)

+

(

�

)

x + ... +

(

:

)

x

n

and

(x _l)n =

(

:

)

xn

_-

(

_n

�

_l

)

_x

n

_{-. + .. . + (-1)"}

(�)

yields and

n(x - 1}"-' = n

(

:

)

x

n

_{-. - (n - 1)}

(

_n

:

1 )

xn-2

_{+ .. . + (_1)"-'}

(�).

Multiplying by

x

gives

nx(x+ 1)"-' =

(

�

)

x + 2

(

;

)

x2 + .. . + n

(

:

)

x

n

and and

1.2.

SOLUTIONS

Differentiating again we obtain

n(x + l)n-. + n(n - l)x(x + 1)"-2 =

(�)

_{+ 22}

(

;

)

_{X + .. . + n2}

(

:

)

_x"-'

n(x - 1)"-' + n(n - l)x(x - 1)n-2 = n2

(

:

)

x"-' - (n - 1)2

(

n

�

1 )

X"-2+

+ .. . + (-1)"-'

(�)

Setting

x = 1

yields

12 (�) + 22

(

;

)

+ .. . + n2

(

:

)

= n(n + 1)2"-2

Summing up the last two identities gives

S

n

= 1 1 + 3 3 + .. . = n(n + 1)2 ,

2

(

n

)

2

(

n

)

n-3

as desired.

35 (Dorin Andrica,

(

RMT

)

, No.

1(1978),

pp.

90,

Problem

3438)

37.

Note that

2>1

n_1 2i+1_1

2:

[

log2

k] =

2: 2:

[

log2

k] +

[

log2

2 n

],

k=l

i=O

k=2i

and

[

log2

k]

=

i

for

2i

�

k < 2i+l.

Hence

as claimed.

2"

n-l

2:

�

og2

k] =

2:

i .

2i +

[

log2

2 n

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

k=l

i=O

(Dorin Andrica,

(

RMT

)

, No.

2(1981),

pp.

63,

Problem

4585;

Gazeta Matematica

(

GM-B

)

, No.

2-3(1982),

pp.

83,

Problem

19113)

38.

Let

Yn

=

22>1

- 1

n.

Then

1 2

1

2 (22)1)2 - 2 . 22" + 1

Yn - Yn+l =

22>1

_{- 1 -}

22>1+1

_{- 1 = (22)1 - 1)(22)1+1 - 1) =}

_

(22)1 _

1

)

2 _

2

2>1

- 1

1

1 - (22)1 _ 1)(22)1+1 _ 1) - (22)1)2 - 1 =

22>1

_{+ 1 = xn}

and therefore

1 1 2

(21)

36

1.

ALGEBRA

xn Yn Yn+l

Summing up these relations yields

1 2 2

2 2n-l

_{1 2n 1}

-+ -+ -+

. . .

+

--= -

-

--

<

Xl X2 X3

xn Yl Yn+l Yl

n, as

desired.

(Dorin Andrica,

)

, No.

1-2(1980),

pp.

67,

Problem

4135)

39.

Substituting

z

with

iz

in the relation

f(z)f(iz)

=

Z2

yields

f(iz)f( -z)

=

_Z2.

Summing up gives

f(iz)(f(z) + f( -z))

=

0,

so

f(iz)

=

0

or

fez) + f( -z)

= o.

From the relation

f(z)f(iz)

=

Z2

we deduce that

fez)

=

0

if and only if

z

= o. Hence if

z

f:.

0,

then

f(iz)

f:. 0

and so

fez) + f( -z)

=

0

and, if

z

=

0,

then

fez) + f( -z)

=

2f(0)

= O. Clearly,

fez) + f( -z)

=

0

for all numbers

z

E C,

as

desired.

Remark.

A function

f

: C -+ C satisfying the relation

f(z)f(iz)

=

Z2

is

fez)

=

(_

V2 +iV2

_{2 2 .}

)

z

(Titu Andreescu,

(

RMT

)

, No.

2(1976),

pp.

56,

Problem

2583)

40.

Setting

X

=

Y

=

1

yields

f

2 ₍₁₎

+ f

2 _{( a)}

=

2 f

(1 )

and

{f(l) -

1 )

2

=

0

so

f(l)

=

1.

Substituting

Y

=

1

gives

f(x)f(l) + f (�)

f(a)

=

2 f(

x

)

or

a

Take now

Y

= - and observe that

X

f(x)f (�) + f (�) f(x)

=

2f(a).

1.2.

SOLUTIONS

37

Consequently,

f(x)f (�)

=

1,

therefore

f

2 (

X

)

=

1,

x

> o.

Now set

x

=

Y

=

0,

that gives

12(0)

+

12 (�)

=

2/(t)

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

f

is positive and

f(x)

=

1

for all

x.

Then

f

is a constant function,

as

claimed.

(Titu Andreescu,

(

RMT

)

, No.

12(1977),

pp.

45,

Problem

2849;

Gazeta Matematica

(

GM-B

)

, No.

10(1980),

pp.

439,

Problem

18455)

41.

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

T

>

0

such that

f(

x

+ T) =

f(x)

or sin

[

x

+

T] = sin

[

x

]

, for all

x

E R

It follows that

[x

+

T] - [x]

=

2k(x)7r,

X

E lR,

where

k

: 1R -+ Z is a function. Because 7r is irrational, we deduce that

k(x)

=

0

for

all

x

E 1R and therefore

[x]

=

[x

+

T]

for all

x

E 1R

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

(Dorin Andrica,

(

RMT

)

, No.

1(1978),

pp.

89,

Problem

3430)

42.

Considering the determinant

1 2 3

n

8 =

1

2

3

2 n

2 In

_2n

₃

n

we have

1

2 ₁

3 In

�

=

IS(i,j)1

= 8·

2

1

2

3

2 n

=

82,

n

l

n

2 n

3 n

n

(22)

38

1.

ALGEBRA On the other hand,

1 1 1

1 8 = n! 1 2 3

n

1n-1

2 n-1

3 n

-

1

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

(Dorin Andrica,

1(1982),

pp.

52,

Problem

3862)

43.

The determinant is

al

0 0

0 a2

0

0 b2

0 0

a3

b3

0 �2n

=

0 0

b2n-2

a2n-2

0

0 b2n-l

0

0 a2n-l

b2n

0 0

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

�2n

=

(ala2n - bl b2n)�2n-2 '

n

�2n

=

IT

(aka2n-k+l - bkb2n-k+l)

k=l

b1

0

0 a2n

(Dorin Andrica,

2(1977),

pp.

90,

Problem

3201;

8(1977),

pp.

325,

Problem

16808)

44.

a) Adding the last three columns to the first one yields that

x + y + z + v

divides the determinant.

Adding the first and second columns and subtracting the last two columns implies that

x + y - z - v

divides the determinant.

Analogously we can check that

x-y+z-v

and

x-y-z+v

divide the determinant, and taking into account that it has degree

4

in each of the variables, the determinant equals

..\

(

x + y + z + v

)(

x + y - z - v

)

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

where ..\ is a constant.

1.2.

SOLUTIONS

Because the coefficient of X4 is equal to

1,

we have ..\

= 1

and so

x y z

w

y x v z

z v x y

v z y x

= (

x + y + z + v

)(

x + y - z - v

)(

x - y + z - v

)

(x - y - z + v

)

b) As shown above, we have

a b e d

39 �

=

b a d e

e d a b

d e b a

=

(a

+

b

+

e

+

d) (a

+

b - e - d) (a - b

+

e - d) (a - b - e

+

d) On the other hand, multiplying the first column by

1000,

the second by

100,

the third by

10

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

abed, bade, edab, debao

Because all those numbers are divisible by the prime number p, it follows that p divides

�

and therefore p divides at least one of the numbers

a

+

b

+

e

+

d, a

+

b - e - d, a - b

+

e - d, a - b - e

+

d.

(Titu Andreescu)

45.

Because the quadratic polynomials h (

x

) and

t2

(

x

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

Consequently,

(pIP2

+

4ql q2 )2 - 4(plq2

+

P2Ql )2

2:: O.

Note now that the left-hand side of the inequality is the discriminant of the quadratic polynomial

t

and the conclusion follows.

(Titu Andreescu,

1(1978),

pp.

63,

Problem

3267;

5(1979),

pp.

191,

Problem

17740)

46.

Because the quadratic polynomial

T

has nonreal zeros, the discriminant

�

=

b2e2 - 4a(b3

+

e3 - 4abe)

is negative.

a bserve that

�

=

(b2 - 4ae) (e2 - 4ab)

<

0,

where

�l

=

b2 - 4ae

and

�2

=

e2 - 4ab

are the discriminants of the quadratic polynomials

Tl

and

T2 •

Hence exactly one of the numbers

�l

and

�2

is negative and since

a

>

0,

the conclusion follows .

(Titu Andreeseu,

1(1977),

pp.

40,

Problem

2810)

(23)

40

1.

ALGEBRA

47.

Observe that

al +a2 + .. ·+an

and

al -a2

+

.. . + (_l)n-lan

are real numbers,

that is

P(l)

and

P( -1)

are real numbers. Hence

and

P(l) = P(l)

and

P( -1) = P( -1)

Because

P(x) = (x - xd .. . (x - xn),

the relations

(1)

become

(1 - xd

. . .

(1 - xn) = (1 - xd .. . (1 -xn)

(1 + xd ..

.

(1

+

xn) = (1

+

xd .. . (1 + xn)

Multiplying these relations yields

(1 - xi)

. ..

(1 - x

�

) = (1 -xi)

. . .

(1

- x�),

or

Q(l) = Q(l).

Therefore

b1

+

b2 + .. . + bn

is a real number.

(1)

(Titu Andreescu,

1(1977),

pp.

47,

Pro blem

2864)

48.

Because

P(O)

=

0,

there is a polynomial

Q

with

P(x)

=

xQ(x).

Then

1 Q(k) =

k

+ l'

k

= 1,n.

Define

H(x) = (x + l)Q(x) - 1.

It is clear that degH(x)

= n

and H(k)

= 0

for all k

=

!,n,

hence

H(x) = (x + l)Q(x) - 1 = ao(x - l)(x - 2)

.

(x - n)

Setting

x =

m,

m

>

n

in relation

(1)

yields

Q (m) = ao (m - 1) (m - 2) .

_{m + 1}

.

. (m - n) + 1 .

On the other hand, setting

x = -1

in the same relation implies

( _l)n+l

ao = (n + I)!

Therefore and then

Q( ) (_l)n+l(m - l)(m - 2)

m =

... (m

- n) 1

+

--(n

+

l)!(m + 1)

m

+

1 P( ) (_l)n+lm(m - 1)

m -

-

_{(n + l)!(m + 1)}

. .

. (m - n) m

+

--

_{m + l'}

(1)

(Dorin Andrica,

8(1977),

pp.

329,

Problem

16833;

1-2(1980),

p.

67,

Problem

4133)

49.

We are looking for a polynomial with integral coefficients

P(x) = aoxn + alxn-1 + .. . + an, ao i= O.

1.2.

SOLUTIONS 4

1

We have

P'(x) = naoxn-1 + (n - 1)alxn-2 + .. . + an-l

and by identifying the coefficient of

x(n-l)n

in the relation

P(P'(x)) = P'(P(x)),

we obtain

or

aonn-1 = 1.

Hence

1 ao = nn-l

and since

ao

is an integer, we deduce that

n = 1

and

ao = 1.

Then

P(x) = x

+

aI,

P'(x) =

1

and

P(P'(x)) = P'(P(x))

yields 1

+ al = 1

or

al = O.

Therefore

P(x) = x

is the only polynomial with the desired property.

(Titu Andreescu,

1-2(1979),

Problem

3902)

50.

Let

(}l, (}2, .. . , (}n

be the roots of the equation

xn + xn-1

+ . ..

+ x + 1 = O.

They are all distinct and

Or+1 = 1,

i

=

!,n.

Because

P(x)

is divisible by

xn + xn-1 + .. . + x + 1,

it follows that

P(Oi) = 0,

i

₌

!,n,

hence

PI

(1)

+

(}1P2(1) + .. . + O�-lPn(l) = 0

PI (1) + 02P2(1) + .. . + O�-lPn(l) = 0

The above system of equations has the determinant

1

01 on-l

₁

V = 1

02 0'2-1

1 On

(}n-l

n

Because all of the numbers

01, 02, .. . ,On

are distinct, it follows that

V i= 0

and so the system has only the trivial solution

PI (1) = P2(1) =

. .

. = Pn(l) = O.

This is just another way of saying that

x - I

divides

pi(X)

for all

i

= 1, n.

(Dorin Andrica,

2(1977),

pp.

75,

Problem

3120;

8(1977),

pp.

329,

Problem

16834)

360 Problems for Mathematical Contests

TITD ANDREESCD

DORIN ANDRICA

360

Problems for

TITU ANDREESCU DORIN ANDRICA

360

Problems

for

Mathematical Contests

�

ARTA

IV GRAFICA

I ��LIBRIS

Contents

FOREWORD

FROM THE AUTHORS

Chapter

1

PROBLEMS

{

C2,

, cn}.

, 5n

n

(x, y), x

>

y,

x - y

al, a2, .. . ,an

[a, b]

f

[a, b]

[a, b]

f(x)

{

a

x

i

x

�i'

h

flh](X)

fafa ... aJ.

htimes

x,

flh]

x, y, z

x

y

z

x2

y2 y2

Z2 Z2

x2 x3 y3 Z3

x

y y

Z Z

x yz zx xy

a, b, c, d

a

b

d

a3

b3

c3

d3

3(abc

bcd

cda

dab).

a, b, c

a

b

c

a3

b3

c3

a5

b5

₁

₄

_{= 13x + 4}

_{3 '}