101 Algebra

Published by
AMT PUBLISHiNG
Mathernafics Trust
University of Canberra ACT 2601
AUSTRALIA
Copyright C2001 AMT Publishing
Telephone: +61 2 6201 5137
AMTT Limited ACN 083 950 341
National Library of Australia Card Number and ISSN
Australian Mathematics Trust Enrichment Series 1SSN 1 326-0170
101 Problems in Algebra iSBN 1 876420 12 X
THE AUSTRALIAN MATHEMATICS TRUST
ENRiCHMENT SERIES
MiTT E EJ
•
•
Chairman
Editor
GRAHAM H POLLARD, Canberra AUSI'RALiA
PETER J TAYLOR, Canberra AUsTRALiA
WARREN J ATKiNs, Canberra AUSTRALIA
Eo J BARBEAU, Toronto CANADA
GEORGE BERZSENYI, Terra Haute USA
RON DUNKLEY, Waterloo CANADA
WALTER E MIENTKA, lincoln USA
N1KOLAY K0NsrANT1N0v, Moscow RussiA
ANDY Liii,
Edmonton
CANADA
JORDAN B TABOV, Sofia BULGARiA
JOHN WEBB, Cape Town SouFH AFRiCA
The books in this series are selected for their motivating, interesting
and stimulating sets of quality problems, with a lucid expository style
in their solutions. Typically, the problems have occurred in either
national or international contests at the secondary school level.
They are intended to be sufficiently detailed at an elementary level
for the mathematically inclined or interested to understand but, at
the same time, be interesting and sometimes challenging to the
undergraduate and the more advanced mathematician. lt is believed
that these mathematics competition problems are a positive
influence on the learning and enrichment of mathematics.
PREFACE
This book contains one hundred highly rated problems used in the training and testing of the USA international Mathematical Olympiad (IMO)
team. It is not a collection of one hundred very difficult, impenetrable
questions. Instead, the book gradually builds students' algebraic skills
and techniques. This work aims to broaden students' view of mathematics and better prepare them for possible participation in various mathematical competitions. It provides in-depth enrichment in important areas
of algebra by reorganizing and enhancing students' problem-solving tac-
tics and strategies. The book further stimulates students' interest for
future study of mathematics.
INTRODUCTION
In the United States of America, the selection process Leading to par-
ticipation in the International Mathematical Olympiad (IMO) consists
of a series of national contests called the American Mathematics Contest 10 (AMC 10), the American Mathematics Contest 12 (AMC 12),
the American Invitational Mathematics Examination(AIME), and the
United States of America Mathematical Olympiad (USAMO). Participation in the AIME and the USAMO is by invitation only, based on
performance in the preceding exams of the sequence. The Mathematical Olympiad Summer Program (MOSP) is a four-week. intense training of 24-30 very promising students who have risen to the top of the
American Mathematics Competitions. The six students representing the
United States of America in the IMO are selected on the basis of their
USAMO scores and further IMO-type testing that takes place during
MOSP. Throughout MOSP, full days of classes and extensive problem
sets give students thorough preparation in several important areas of
mathematics. These topics include combinatorial arguments and identities, generating functions, graph theory. recursive relations, telescoping
sums and products, probability, number theory, polynomials, theory of
equations. complex numbers in geometry, algorithmic proofs, combinatorial and advanced geometry, functional equations and classical inequalities.
Olympiad-style exams consist of several challenging essay problems. Cor-
rect solutions often require deep analysis and careful argument. Olympiad questions can seem impenetrable to the novice, yet most can he
solved with elementary high school mathematics techniques, cleverly applied.
Here is some advice for students who attempt the problems that follow.
• Take your time! Very few contestants can solve all the given problems.
• Try to make connections between problems. A very important
theme of this work is; all important techniques and ideas featured
in the book appear more than once!
• Olympiad problems don't "crack" immediately. Be patient. Try
different approaches. Experiment with simple cases. In some cases,
working backward from the desired result is helpful.
• Even if you can solve a problem, do read the solutions. They may
contain some ideas that did not occur in your solutions, and they
VIII
may discuss strategic and tactical approaches that can be used else-
where. The formal solutions are also models of elegant presentation that you should emulate, but they often obscure the torturous
process of investigation, false starts, inspiration and attention to
detail that led to them. When you read the solutions, try to reconstruct the thinking that went into them. Ask yourself. "What
were the key ideas?" 4How can I apply these ideas further?"
• Go back to the original problem later, and see if you can solve it
in a different way. Many of the problems have multiple solutions,
but not all are outlined here.
• All terms in boldface are defined in the Glossary. Use the glossary
and the reading list to further your mathematical education.
• Meaningful problem solving takes practice. Don't get discouraged
if you have trouble at first. For additional practice, use the books
on the reading list.
ACKNOWLEDGEM ENTS
Thanks to Tiankai Liu who helped in proof reading and preparing solu-
Many problems are either inspired by or fixed from mathematical contests
in different countries and from the following journals:
High-School Mathematics, China
Revista Maternaticã
Romania
Kvant, Russia
cite all the original sources of the problems in the solution part. We express our deepest appreciation to the original proposers
of the problems.
We did our best
ABBREVIATIONS AND NOTATIONS
Abbreviations
AHSME
AIME
AMC1O
AMC12
ARML
IMO
USAMO
MOSP
Putnam
St. Petersburg
American High School Mathematics
Examination
American Invitational Mathematics
Examination
American Mathematics Contest 10
American Mathematics Contest 12,
which replaces AIISME
American Regional Mathematics League
International J\Iathematical Olympiad
United States of America Mathematical Olympiad
Mathematical Olympiad Summer Program
The William Lowell Putnam Mathematical
Competition
St. Petersburg (Leningrad) Mathematical
Olympiad
Notations for Numerical Sets and Fields
Z
N
N0
Q
Q+
Q°
R
IR+
R°
R'2
C
the set of integers
the set of integers modulo n
the set of positive integers
the set of nonnegative integers
the set of rational numbers
the set of positive rational numbers
the set of nonnegative rational numbers
the set of n-tuples of rational numbers
the set of real numbers
the set of positive real numbers
the set of nonnegative real numbers
the set of n-tuples of real numbers
the set of complex numbers
C
1
• PREFACE
vii
• INTRODUCTION
ix
• ACKNOWLEDGEMENTS
xi
• ABBREViATiONS AND NOTATiONS
• 1. INTRODUCTORY PROBLEMS
• 2. ADVANCED PROBLEMS
xiii
1
13
• 3. SOLUTIONS TO INTRODUCTORY PROBLEMS 27
• 4. SOLUTIONS TO ADVANCED PROBLEMS
65
• GLOSSARY
131
• FURTHER READING
137
iNTRODUCTORY PROBLEMS
INTRODUCTORY PROBLEMS
1
problem
1
Let a. b. and c be real and positive parameters. Solve the equation
ax =
CX-r
Problem 2
Find the genera! terni of the sequence
=
for
bi +
ax +
by
=
3.
- cx.
x1 =
—
alln E N.
Problem 3
x,, be a sequence of integers such that
Let x1. x2
n:
(i)
(ii) x1 + x2 +
+
= 19:
=99.
(iii)
Determine the minimum aiid niaxiniuni possible values of
Problem 4
The function f. defined by
1(x)
where a.
b.
c. and
h
= c.r+ d
d are nonzero real nunibers. has the j)rOperties
f(19) =
for
ax +
19.
all values of x. eXCCl)t
Find
range of f,
f(97) =
--
97.
and
f(f(x)) =
and
1. Introductory Problems
2
Problem 5
Prove that
(a—b)2 <
8a
for all a
a+b
—
(a—b)2
2
8b
—
b> 0.
Problem 6
Several (at least two) nonzero numbers are written on a board. One may
erase any two numbers. say a and b, and then write the numbers a +
and b —
instead.
Prove that the set of numbers on the board, after any number of the
preceding operations, cannot coincide with the initial set.
Problem 7
The polynomial
may be written in the form
ao + a1y + a2y2
where y = x + 1 and
Find a2.
+ a15y16 + a17y17,
are constants.
Problem 8
Let a, b, and c be distinct nonzero real numbers such that
a+-=b+-c
1
1
b
Prove that
=
1
a
1.
Problem 9
Find polynomials 1(x), g(x), and h(x), if they exist, such that for all x,
(
—1
3x+2
—2x+2
ifx<—1
if—i
ifx>0.
i. Introductory Problems
Problem 10
Find all real numbers x for which
8x+27x
7
12x + 18x
Problem 11
Find the least positive integer m such that
(2n'\
I
I
for all positive integers n.
Problem 12
Let a, b, c, d, and e be positive integers such that
Find the maximum possible value of max{a, b, c, d, e}.
Problem 13
Evaluate
3
4
2001
1!+2!+3!
1999!+2000!+2001!
Problem 14
Letx=
1, a ER.
Find all possible values of x.
Problem 15
Find all real numbers x for which
+ lix + 12T =
13X
+ 14x.
1. Introductory Problems
4
Problem 16
Let f N x N
N
be a function such that f(1, 1) =
2,
f(rn + 1. n) = f(rn. n) + rn and f(rn, n + 1) = f(n2, n)
—
n
for all rn, n E N.
Find all pairs (pq) such that f(p,q) = 20th.
Problem 17
Let f be a function defined on [0, 1] such that
f(0) =
for all a
b
1(1)
and (f(a) — f(b)I < a— bI.
1
in the interval [0, 1].
Prove that
1
1(a) — f(b)( <
Problem 18
Find all pairs of integers (x, y) such that
x3 + y3 = (x + y)2.
Problem 19
2
Let f(x) =
for real numbers x.
4x + 2
Evaluate
1
/
\
1
2
/2000
•\
Problem 20
Prove that for ii
6 the equation
1
1
x1
x2
1
has integer solutions.
Problem 21
Find all pairs of integers (a, b)
is divisible by x2 — x
—
1.
such
that the polynomial ax'7 + bx16 + I
1. IntroductorY Problems
Problem 22
Given a positive integer n, let p(n) be the product of the non-zero digits
of n. (If n has only one digit, then p(n) is equal to that digit.) Let
+p(999).
What is the largest prime factor of S?
Problem 23
Let
be a sequence of nonzero real numbers such that
xn
=
—
n = 3,4
Establish necessary and sufficient conditions on x1 and x2 for
for
to be
an integer for infinitely many values of ri.
Problem 24
Solve the equation
—
3x
=
Problem 25
For any sequence of real numbers A = {ai,a2,as,.
the sequence {a2 — a1, a3 — a2, a4 — a3..
of the sequence
Find a1.
.
.
}, define
to be
}. Suppose that all of the terms
are 1, and that a1g =
a92
=
0.
Problem 26
Find all real numbers x satisfying the equation
2x+3x
4X+6X9X =1.
Problem 27
Prove that
16
<17.
Problem 28
Determine the number of ordered pairs of integers (rn n) for which mn
m3 + n3 + 99mn =
1. Introductory Problems
6
Problem 29
Let a, b, and c be positive real numbers such that a + b + c
4 and
ab 4-bc+ca 4.
Prove that at least two of the inequalities
a—bl
2,
lb—cl 2,
Ic—al
are true.
Problem 30
Evaluate
1
Problem 31
Let 0 < a < 1. Solve
=
ax
for positive numbers x.
Problem 32
What is the coefficient of x2 when
is expanded?
Problem 33
n be distinct positive integers.
Find the maximum value of lxm —
where x is a real number in the
interval (0,1).
Let
Problem 34
Prove that the polynomial
are distinct integers, cannot be written as the product of two non-constant polynomials with integer coefficients, i.e., it is
where ai. a2, ..
irreducible.
,
Problem 35
Find all ordered pairs of real numbers (x. y) for which:
(1+x)(1+x2)(1+x4)
and
(1+y)(1+y2)(1+y4) =
1+x7.
Problem 36
Solve the equation
2(2x
—
1)x2 + (2z2 — 2)x =
2
for real numbers x.
Problem 37
Let a be an irrational number and let n be
an integer greater than 1.
Prove that
is an irrational number.
Problem 38
Solve the system of equations
= x2(x4+x5—x2)
(x1 — x2 -I-- x3)2
= x3(x5+x1—x3)
=
(x4 — x5 + x1)2 = x5(x2+x3—x5)
(x5 — xi +
= x1(x3+x4—x1)
(x2 — x3 + x4)2
(x3 — x4 + x5)2
for real numbers Xl, x2,
x4, x5.
Problem 39
Let x, y, and z be complex numbers such that
x
y
+z=
2,
+ z2
=
+
and
STAATS.UL.NIV)
11
=
3
4.
Xy+z—1 + yz+x—1 + zx+y—1
—
1. Introductory Problems
8
Problem 40
Mr. Fat is going to pick three non-zero real numbers and Mr. Taf is going
to arrange the three numbers as the coefficients of a quadratic equation
_x2 +_x+_=O.
Fat wins the game if and
distinct rational solutions.
Who has a winning strategy?
Mr.
only if the resulting equation
has two
Problem 41
Given that the real numbers a, b, c, d, and e satisfy simultaneously the
relations
a+b+c+d+e=8anda2-j-b2+c2+d2+c2 =16,
determine the maximum and the minimum value of a.
Problem 42
Find the real zeros of the polynomial
Pa(S) = (x2 + 1)(x —
1)2 — ax2,
where a is a given real number.
Problem 43
Prove that
1
3
2n—1
2
4
2n
1
for all positive integers n.
Problem 44
Let
P(x) =
a nonzero polynomial with integer coefficients such that P(r) =
P(s) = 0 for some integers r and s. with 0 < r < s.
Prove that ak < —s for some k.
Problem 45
Let rn be a given real number.
Find all complex numbers x such that
(S
\
(X\
/
+(
2
=m +772.
Problem 46
The sequence given by xo =
a,
.r1 =
b,
and
11
1
is periodic.
Prove that ab =
1.
Problem 47
Let a, b, c, and d be real numbers such that
(a2 +b2 — i)(c2 +d2 —1)>
(ac+bd—
1)2.
Prove that
a2
+ b2> 1
and c2
+ d2> 1.
Problem 48
Find
all complex numbers z such that
(3z + 1)(4z + 1)(6z + 1)(12z + 1) =
2.
Problem 49
be the zeros different from 1 of the polynomial
Let
P(x) =
— 1,
n 2.
Prove that
1
i—x1
+
1
i—x2
+...+
n—i
1
2
Problem 50
Let a and b be given real numbers. Solve the system of equations
V1_x2+y2
for real numbers x and y.
=
a
—
b
ADVANCED PROBLEMS
Z ADVANCED PROBLEMS
Problem 51
Evaluate
(2000'\
(2000\
(2000\
(2000
Problem 52
Let x, y, z be positive real numbers such that x4 + y4 -i- z4 =
Determine with proof the minimum value of
x3
1—x 8+
l—y8
z3
±
1—z8
Problem 53
Find all real solutions to the equation
2X
+ 3X + 6X = x2.
Problem 54
Let {an}ni be a sequence such that
=
2
and
1
=
+—
for all n E N.
Find an explicit formula for
Problem 55
Let x, y, and z be positive real numbers. Prove that
2;
+
+
z
z+
y)
1.
1.
2. Advanced Problems
14
Problem 56
Find, with proof, all nonzero polynomials f(z) such that
f(z2) + f(z)f(z +1) =0.
Problem 57
Let f N
N
be a function such that f(n + 1) > f(n) and
f(f(n)) =
3n
for all n.
Evaluate 1(2001).
Problem 58
Let F be the set of all polynomials f(x) with integers coefficients sueh
that f(x) = 1 has at least one integer root.
For each integer k > 1. find
the least integer greater than 1 for
which there exists f E P such that the equation f(x) = rnk has exactly
k distinct integer roots.
Problem 59
Let x1 =
2
and
=
—
÷
1,
i.
Prove that
for ii
1
1—
Problem 60
Suppose that f
22
1
1
.r1
x2
1
1
22
is a decreasing function such that for all
:
x,y E
f(x + y) + f(f(x) + 1(y)) = 1(1(1 + f(y)) + f(y + 1(x))).
Prove
that f(f(x)) =
x.
15
Problem 61
Find all functions f : Q
Q such that
f(x + y) + f(x
for
—
y)
= 2f(x) + 2f(y)
all x,y EQ.
Problem 62
Let <a < 1.
Prove that the equation
x3(x + 1) = (x + a)(2x + a)
has
four distinct real solutions and find these solutions in explicit form.
Problem 63
Let a, b, and c be positive real numbers such that abc =
Prove that
1
1
I
<1.
+
±
1.
1
a+b+1
c+a+1
b+c-I-1
Problem 64
Find all functions f, defined on the set of ordered pairs of positive integers, satisfying the following properties:
f(x, x) = x, f(x, y) = f(y, x). (x + y)f(x. y)
yf(x. x + y).
Problem 65
Consider n complex numbers zk. such that IzkI
2
1, Ic =
Prove that there exist c1. a2....
E f—i. 1} such that, for any vi
e1z1
+
+
Problem 66
Find a triple of rational numbers (a.
+
<2.
c) such that
n.
fl,
2. Advanced
16
Problem 67
Find the minimum of
+ log12
(X2-_
log11
+
—
where
x1,x2,. ..
—
are real numbers in the interval
1).
Problem 68
Determine x2 +
+ z2 + w2 if
22_12+ 22_32+22_52
=1,
+4252
=1,
42_12 + 42_32
62_12
+6232+6252+6272 =L
82_12 + 82_32
-L
82_52 + 82_72 =1
Problem 69
Find all functions f R
:
R
such that
f(xf(x) + f(y)) = (f(x))2 +
for all x,y E
Problem 70
The numbers 1000, iOOl,
,2999 have been written on a board.
Each time, one is allowed to erase two numbers, say, a and b, and replace
them by the number
min(a. b).
After 1999 such operations, one obtains exactly one number c on the
board. Prove that c < 1.
Problem 71
Let a1,a2,. .
be real numbers, not all zero.
Prove that the equation
.
has at most one nonzero real root.
2.
Advanced Problems
Problem '72
real numbers defined by a1 = t and
Let {an) be the sequence of
=
for n
1.
For bow many distinct values
oft
—
do we have a1998
=
0?
Problem '73
R
(a) Do there exist functions f :
f(g(x)) = x2
and
and g R
R
such that
g(f(x)) =
for all x E R?
and 9:
(b) Do there exist functions f: R
f(g(x)) = x2
and
R
g(f(x))
such that
=
for all x E R?
Problem 74
LetO<a1 a2
<
O<b1 <b2
Suppose that there exists 1
b2
for
i >
k n such
berealnumberssuchthat
for 1
that
i
k and
k.
Prove that
a1a2
b1b2
Problem 75
Given
eight non-zero real numbers a1, a2,.
a8. prove that at least one
of the following six numbers: a1a3 + a2a4, a1a5 + a2a6, a1a7 + a2a8,
a3a5 + a4a6, a3a7 + a4a8.
+ a6a8 is non-negative.
Problem '76
Let a, b and
Prove that
c
be
positive real numbers such that abc
ab
bc
ca
=
1.
a5+b5+ab+b5+cS+bc+c5+a5+caL
2. Advanced Problems
18
Problem 77
Find all functions f: R —* R such that the equality
f(f(x) + y) = f(x2
y)
—
+ 4f(x)y
holds for all pairs of real numbers (x. y).
Problem 78
Solve the system of equations:
3x — y
x2
—
+
=3
y2
x+3y
11
=0
Problem 79
Mr. Fat and Mr. Taf play a game with a polynomial of degree at least 4:
+... +_x + 1.
They fill in real numbers to empty spaces in turn. If the resulting polynomial has no real root, Mr. Fat wins; otherwise, Mr. Taf wins.
If Mr. Fat goes first, who has a winning strategy?
Problem 80
Find all positive integers k for which the following statement is true:
F(x) is a polynomial with integer coefficients satisfying the condition
for
then F(0)= F(1)=
c=0.1
k+1,
=F(k+ 1).
Problem 81
The Fibonacci sequence
is given by
(nEN).
Prove that
2n+2 '
for all n 2.
2n—2
bf2n
if
19
ced ProblemS
Problem 82
R for which there exists a strictly monotonic
Find all functions u :
function f : R —p R such that
f(x + y) = f(x)u(y) + f(y)
for all x, y c R.
Problem 83
Zn be
Let
complex numbers such that
IzlI+Iz2H+IzflI=1.
Prove
that there exists a subset S of {z1, Z2,..
.
.
such that
Problem 84
A polynomial P(x) of degree n 5 with integer coefficients and n distinct
integer roots is given.
Find all integer roots of P(P(x)) given that 0 is a root of P(x).
Problem 85
Two real sequences x1, x2,
and Yi
.
Y2.
.
are defined in the following
way:
and
Yn+1
for all n 1. Prove that 2 <
yn
=
1
+ yi +
<3 for all a > 1.
Problem 86
For a polynomial P(x), define the difference of P(x) on the interval [a. b]
([a,b), (a,b). (a.b]) as P(b) P(a).
Prove that it is possible to dissect the interval [0. 1[ into a finite number
of intervals and color them red and blue alternately such that. for every
polynomial P(x). the total difference of P(.r) on red intervals
is equal to that of P(x) on blue intervals.
What about cubic polynomials?
-
2. Advanced Problems
20
Problem 87
Given a cubic equation
x3 + _x2 + _x + — = 0,
Mr. Fat and Mr. Taf are playing the following game. In one move, Mr.
Fat chooses a real number and Mr. Taf puts it in one of the empty spaces.
After three moves the game is over. Mr. Fat wins the game if the final
equation has three distinct integer roots.
Who has a winning strategy?
Problem 88
Let n > 2 be an integer and let f R2
any regular n-gon A1A2 .
R
be a function such that for
. .
f(A1) + f(A2) + .
+
f is the zero function.
Problem 89
Let p be a prime number and let f(x) be a polynomial of degree d with
integer coefficients such that:
(i) f(0) =
0,
f(l) =
1;
(ii) for every positive integer n, the remainder upon division of
by p is either 0 or 1.
Prove that d p —
1.
Problem 90
Let n be a given positive integer.
with a0 =
Consider the sequence a0,a1,•
ak =
4
TI
fork=l,2..n,
Prove that
1
_!
TI
<
1.
and
2.
21
Advanced problems
Problem 91
Let ai,a2,."
be nonnegative real numbers. not all zero.
—
(a) Prove that
positive real root R.
(b) Let A =
a2 and
Prove that AA
—
—
= 0 has precisely one
ja3.
B=
<RB.
Problem 92
Prove that there exists a polynomial P(x, y) with real coefficients such
that P(x, y) 0 for all real numbers .r and y. which cannot be written
as the sum of squares of polynomials with real coefficients.
Problem 93
For each positive integer Ti. show that there exists a positive integer k
such that
/c = f(x)(x + 1)2n +
+ 1)
for some polynomials f.g with integer coefficients, and find the smallest
such Ic as a function of n.
Problem 94
Let x be a positive real number.
(a) Prove that
(x
(ii— 1)!
1
1).. (x + n) — I
.
(b) Prove that
(n—i)!
1
2. Advanced
22
Problem 95
Let ii 3 be an integer, and let
Xcs={1.2
n3}
be a set of 3n2 elements.
Prove that one can find nine distinct numbers
such that the system
c7,
=
0
a2x+h2y+c2z =
0
a3x+b3y+c3z
0
=
(1 =
1.
2, 3) in X
has a solution (XO, yo, zO) in nonzero integers.
Problem 96
he positive real numbers.
Let n 3 be an integer and let Xi. i2.
Suppose that
=
1
Prove
that
Problem 97
Let x1,x2
be
distinct real numbers. Define the polynomials
P(x) =
(x — x1)(.r
(1
and
Q(x)=P(x)(
\.i—X1
+
•..
—
1
X—X2
-+.
Let y1,y2.. . . 'Yn—l be the roots of Q. Show
tJ
<
—
+
that
2.
Advanced Problems
Problem 98
Show that for any positive integer fl. the polynomial
f(x) =
(x2
+
+1
cannot be written as the product of two non-constant polynomials with
integer coefficients
Problem 99
be functions such that
R
Let 11,12.13 R
a1f1 + a2]'2 + a3f3
monotonic for all a1,a2,a3
Prove that there exist c1, c2. C3
is
R.
R. not all zero. such that
cifi(x) +c2f2(x)
for all x
+c3f3(x) = 0
Ift.
Problem 100
be the sums of
be variables, and let
Let x1.x2,
nonempty subsets of
elementary symmetric polynomial in
be the
Let pk(xl
the yj (the sum of every product of A- distinct yjs).
For which k and n is every coefficient of Pk (as a polynomial in x1,. ..
. .
,
even?
For example, if n =
2.
Pi =
P2 =
P3 =
+ s2 and
then yi' Y2. y3 are Xl.
=
+ Y2 +
Y1Y2
+ 2x2,
+ 112Y3 + Y3Y1 =
Y1Y2Y3
=
X1X2
+
+
+
Problem ioi.
Prove that there exist 10 distinct real numbers al,a2..
the equation
(x—ai)(x_a2)...(xaio)
has exactly 5 different real roots.
= (X+al)(X+a2)
.
aio
such that
(x+aio)
SOLUTiONS TO
iNTRODUCTORY PROBLEMS
3. SOLUTIONS TO
PROBLEMS
problem 1 [Romania 1974]
Let a, b, and c be real and positive parameters.
Solve the equation
\/c+ax= v'b—ax -f-s/c—
Solution 1
It is easy to see that x =
0 is a solution. Since the right hand side is a
decreasing function of x and the left hand side is an increasing function
of x, there is at most one solution.
Thus x = 0 is the only solution to the equation.
Problem 2
Find the general term of the sequence defined by x0 =
_2
—
3,
=
4
and
—
for all n C N.
Solution 2
We shall prove by induction that
=
n
+ 3. The claim is evident for
n=O,1.
Fork 1, if Xk_1 =k+2andxk=k+3, then
—kxk =(k+2)2 —k(k+3)=k+4,
as desired.
This completes the induction.
3. Solutions to Introductory Problems
28
Problem 3 [AHSME 1999]
Let x1. x2,
, x7, be a sequence of integers such that
.
.
n;
(i)
(ii)
19;
X1
(iii)
Determine the minimum and maximum possible values of
Solution 3
Let a, b, and c denote the number of —is, is, and 2s in the sequence,
respectively. We need not consider the zeros. Then a, b, c are nonnegative
integers satisfying
—a+b+2c= l9anda+b+4c=99.
It follows that a = 40—c and b
59—3c, where 0
c
19 (since b 0),
so
19+6c.
(a = 40,b = 59), the lower bound (19) is achieved.
When c = 19 (a = 21, b = 2), the upper bound (133) is achieved.
When c =
0
Problem 4 [AIME 1997]
The function f, defined by
f(x)
ax + b
cx+d'
where a, b, c, and d are nonzero real numbers, has the properties
1(19) =
19,
f(97) =
97,
and
for all values of x, except —
Find the range of f.
Solution 4, Alternative 1
For all x, f(f(x)) = .r, i.e.,
(ax + b'\
a(
\cx+d)
(ax+b\
cx + dj
—x
f(f(x)) =
to Introductory Problems
3
i.e.
29
(a2 + bc)x + b(a + d)
c(a+d)x+bc+d2
i.e.
—
X.
c(a + d)x2 + (d2 — a2)x — b(a + d) =
0,
implies that c(a + d) = 0. Since c 0, we must have a = —d.
The conditions f(19) = 19 and f(97) = 97 lead to the equations
which
and
Hence
(972
It follows that a =
58c,
192)c
—
972c=2•97a+b.
= 2(97
—
19)a.
which in turn leads to b =
— 1843
f(x)= 58x.x—58
—1843c.
Therefore
1521
which never has the value 58.
Thus the range of f is R — {58}.
Solution 4, Alternative 2
The statement implies that f is its own inverse. The inverse may be
found by solving the equation
ay + b
cy + d
fory. Thisyields
dx—b
—cx +
a
The nonzero numbers a. b, c, and d
must therefore be proportional to d,
—c, and a, respectively; it follows that a = —d, and the rest is the
same as in the first solution.
Problem 5
Prove that
forallab>
(a—b)2 <
8a
a+b
2
0.
Alternative
1
that
) -
(a—b)2
—
8b
3. Solutions to Introductory Problems
30
i.e.
4a
4b
—
i.e.
(a—b)2 <
< (a—b)2
from which the result follows.
Solution 5, Alternative 2
Note that
(a-t-b\2
2
-ab
+
(a-b)2
2(a + b) +
Thus the desired inequality is equivalent to
4a
a+b+
which is evident as a b> 0 (which implies a
4b.
b).
Problem 6 [St. Petersburg 1989]
Several (at least two) nonzero numbers are written on a board. One may
erase any two numbers, say a and b, and then write the numbers a +
and b —
instead.
Prove that the set of numbers on the board. after any number of the
preceding operations, cannot coincide with the initial set.
Solution 6
Let S be the sum of the squares of the numbers on the board. Note that
S increases in the first operation and does not decrease in any successive
operation, as
a)2 =
(a+
with equality only if a = b =
This completes the proof.
0.
to Introductory Problems
31
problem 7 [AIME 1986]
The
may be
written in the form
ao + aiy + a2y2 + .
+ ai6y'6 + a17y17,
-
are constants. Find a2.
= x + 1 and
SolutiOn 7, Alternative 1
where y
Let f(x) denote the given expression. Then
xf(x)=x—x2+x3—
.—x18
and
(1 + x)f(x) = 1
Hence
f(x)=f(y—1)=
—
1— (y— 1)18
—
1— (y— 1)18
l+(y—l)
y
Therefore a2 is equal to the coefficient of y3 in the expansion of
(y
1 —
—
1)18,
i.e.,
a2
= 816.
=
Solution 7, Alternative 2
Let 1(x) denote the given expression. Then
f(x) =f(y—1)
l—(y— l)-f(y—
Thus
/2\
/3\.
(17\
Here we used the
I
and the fact that
7i
'\
/n+1"
(18"
3. Solutions to Introductory Problems
32
Problem 8
Let a. b, and c be distinct nonzero real numbers such that
-1 = b+ -1 = c+ -.1
c
b
a
Prove that
=
1,
Solution 8
From the given conditions it follows that
b—c
c—a
a—b
a—b=—.b—c———,andc—a=—.
bc
ca
ab
Multiplying the above equations gives (abc)2 =
result follows.
1,
from which the desired
Problem 9 [Putnam 1999]
Find polynomials f(s), g(x), and h(s). if they exist. such that for all x
— g(')I + h(s) =
(
ifr<—1
—1
3x
+
2
—2x+2
if —1 <
.r
ifx>0.
<
0
Solution 9, Alternative 1
Since x = —1 and x = 0 are the two critical values of the absolute
functions, one can suppose that
F(s) =
=
I (c—a—b)x+d—a
ifs <—1
if—1<x<0
1(a+b+c)x+a+d ifx>0.
which implies that a 3/2, b = —5/2. c = —1. and d = 1/2.
Hence f(s) = (3s + 3)/2, g(x) = 5x/2, and h(s) = —x +
Solution 9, Alternative 2
Note that if r(x) and 8(x) are any two functions, then
max(r,s)
r-l-s+lr—sj
=
2
Therefore, if F(s) is the given function. we have
F(x) = max{—3x — 3. 0} — max{5x, 0} + 3x + 2
= (—3x—3-'-j3x+31)/2—(5x+15x1)/2-4-3x--i-2
=
colutions to Introductory Problems
33
Problem 10
Find all real numbers x for which
8X
— 7
12x + 18x —
6
Solution 10
=
= a and
By setting
the equation becomes
b.
a3+b3 _7
a2b+ b2a —
i.e.
6
a2—ab+b2 _7
6'
ab
i.e.
Ga2
—
l3ab+6b2 =
0.
i.e.
(2a — 3b)(3a — 2b) = 0.
Therefore
2x+i = 3x+i or
2x—1
x = 1.
It is easy to check that both .r =
= 3x—1, which
—1
and x =
1
implies that x =
such
(2n\
\n)
that
<ni
for all positive integers n.
ii
Note that
+
+
<
... +
and for n
(10)
Thus in
= 252 >
and
satisfy the given equation.
Problem 11 [Romania 1990]
Find the least positive integer in
—1
= (1+ i)2n =
3. Solutions to Introductory Problems
34
Problem 12
Let a. b, c, d, and e be positive integers such that
abcde=a+b+c+d+e.
Find the maximum possible value of max{a, b, c. d, e}.
Solution 12, Alternative 1
Suppose that a b c d
e. We need to find the maximum
e. Since
value of
c<a+b+c+d+c<5c.
5e, i.e. < abcd 5.
Hence
= (1.1,1,2), (1,1,1,3).
(1,1, 1.5), which leads to max{e} = 5.
then e < abcdc
1
(1. 1,1,4), (1, 1,2,2), or
Solution 12, Alternative 2
As
before,
suppose that a
c
b
d c. Note that
1
1
1
1
1
1
de
d
d = 1, then a = b = c = 1 and 4 + c = e, which is impossible.
e 5.
Thusd—1 land c—i
It is easy to see that (1, 1,1,2,5) is a solution.
Therefore
max{e} =
Comment:
5.
The second solution can be used to determine the maxiwhen x1, x2, .
are positive integers
mum value of {x1, x2,. .
such
. ,
.
.
.
that
x1x2
.
= xi+
.
+
+ xm.
Problem 13
Evaluate
3
1!+2!+3!
+
4
2!+3!+4!
++
2001
to Introductory Problems
35
13
Note that
k+2
= k![1+k+1+(k+1)(k+2)]
k+2
1
—
k!(lc+2)
—
(k+2)!
—
(k+2)—1
—
(k-i--2)!
k
—
1
—
(k+1)!
—
(k+2)!
By telescoping sum, the desired value is equal to
2
2001!
Problem 14
Find all possible values of x.
solution 14, Alternative 1
Since
\/a2 + al + 1>
al
and
2a
Va2
+a+ 1+
Va2
-a+ 1'
we have
lxi <
12a/ai
= 2,
Squaring both sides of
Va2+a+1
Yields
—
a + 1 = 2a — x2.
3. Solutions to Introductory Problems
36
Squaring both sides of the above equation gives
4(x
2
—1)a =x (x2'—4)ora =
2
2
2
x2(x2—4)
4(21)'
Since a2 0, we must have
x2(x2
—
4)(x2 — 1)
0,
Since xI < 2, x2 —4 < 0 which forces x2 —1 < 0. Therefore, —1 < x < 1.
Conversely, for every x C (—1, 1) there exists a real number a such that
x
Va2
+a+1 —
—a+1.
Solution 14, Alternative 2
Let A =
B=
and P = (a.O). Then P
is a point on the x-axis and we are looking for all possible values of
d=PA-PB.
By the Triangle Inequality, IPA — PBJ <
= 1. And it is clear
that all the values —1 <d < 1 are indeed obtainable. In fact, for such
a d, a half hyperbola of all points Q such that QA — QB = d is well
defined, (Points A and B are foci of the hyperbola.)
Since line AB is parallel to the x-axis, this half hyperbola intersects the
x- axis, i.e., P is well defined.
Problem 15
Find all real numbers x for which
lOx + lix + 12x =
13X
+ 14x.
Solution 15
It is easy to check that x = 2 is a solution. We claim that it is the only
one. In fact, dividing by 13X on both sides gives
112\x
The left hand side is a decreasing function of x and the right hand side
is an increasing function of x.
Therefore their graphs can have at most one point of intersection.
Solutions to Introductory Problems
37
Comment: More generally.
a2+(a+1)2+.+(a+k)2
=(a+k+1)2+(a+k+2)2+"+(a+2k)2
forak(2k+'). kcN.
problem 16 [Korean Mathematics Competition 2001]
N be a function such that f(1, 1) = 2.
Let f: N x N
f(m + 1. n) = f(ni, n) + ni and f(772,
f(n2. n) —
72 + 1)
for all rn, n c N.
Find all pairs (p, q) such that f(p. q) =
2001.
Solution 16
We have
f(pq) = f(p—l,q)+p—1
=
=
=
2
= f(1,l)
=
q(q—1)
p(p—l)
2
2
2001.
Therefore
p(p— 1) —q(q—
2
2
1) —
1999,
—
i.e.
(p—q)(p+q— 1) =2. 1999.
ROte
that
1999
is a prime number and thatp—q < p+q— 1 for p.q
We have the following two cases:
1
= 1999. Hence p= 1001 and q= 999.
c N
3. Solutions to Introductory Problems
38
Therefore (p. q) = (2000, 1999) or (1001,999).
Problem 17 [China 1983]
Let f be a function defined on [0, 1] such that
f(0) = f(1) = land f(a)—f(b)I < la—bi,
bin the interval [0, 1].
for all a
Prove that
f(a) — f(b)j <
Solution
1.7
We consider the following cases.
1. a —
bI
< 1/2. Then If(a) — f(b)f < a
- bI
as desired.
2. Ia — bi > 1/2. By symmetry, we may assume that a> b. Then
f(a) — f(b)I
= If(a) — 1(1) + f(0) — f(b)I
<
f(a) — f(')I + 1(0) — f(b)I
< Ia—lI+I0—bI
= 1—a+b—0
=
1—(a-b)
as desired.
Problem 18
Find all pairs of integers (x, y) such that
+
= (x + y)2.
Solution 18
Since x3+y3 = (x+y)(x2—xy+y2), all pairs of integers (n—ri), n C 7Z.
are solutions.
Suppose that x + y
0. Then the equation becomes
S2
i.e.
— Xy + y2 = x + y,
3
solutions to Introductory Problems
39
a quadratic equation in x, we calculate the discriminant
A = y2 + 2y + 1 — 4y2 + 4y =
—3y2 + 6y + 1.
solving for A 0 yields
3
3
Thus the possible values for y are 0, 1, and 2. which lead to the solutions
(1,0), (0,1), (1,2), (2.1), and (2,2).
Therefore, the integer solutions of the equation are (x. y) =
(1,2), (2,1), (2,2), and (n, —n). for all nEZ.
(1,
0), (0, 1),
Problem 19 [Korean Mathematics Competition 2001]
Let
2
f(x)= 4X+2
for real numbers x. Evaluate
/
1
\
/
2
/2000
'\
f has a half-turn symmetry about point (1/2, 1/2). Indeed,
2.4x
2
from which it follows that 1(x)
f(1
-t-
—
x)
4X
1.
Thus the desired sum is equal to 1000.
Problem 20
Prove that for n 6 the equation
1
1
1
x1
has
integer solutions.
Solution 20
Note that
1
1
1
1
1
3. Solutions to Introductory Problems
40
from which it follows that if (x1, x2,
=
,
(a1
.a2,
is an inte.
ger solution to
1
1
x1
x2
1
then
(x1,x2,. .
is an integer solution to
1
1
1
xl
x2
Therefore we can construct the solutions inductively if there are solutions
for n = 6, 7, and 8.
Since xi = 1 is a solution for n = 1, (2,2,2,2) is a solution for n =
and (2, 2, 2,4,4.4,4) is a solution for n = 7.
4,
It is easy to check that (2, 2, 2,3, 3, 6) and (2, 2. 2, 3,4,4, 12, 12) are solutions for n = 6 and n = 8, respectively. This completes the proof.
Problem 21 [AIME 1988]
Find all pairs of integers (a, b) such that the polynomial
ax17 + bx'6 -i- 1
is divisible by x2 — x
— 1.
Solution 21, Alternative 1
Let p and q be the roots of x2 — x — I = 0. By Vieta's theorem,
p + q = 1 and pq —1. Note that p and q must also be the roots of
ax17 + bx'6 + 1 = 0. Thus
ap17 + bp'6 =
—1
and aq17 + bq'6
—1.
Multiplying the first of these equations by q'6, the second one by p16,
and using the fact that pq = —1, we find
ap+b=—q16andaq+b=—p'6.
Thus
a=
I
q'6
=
(p8 + q8)(p4 + q4)(p2 + q2)(p ± q).
(1)
solutions to Introductory Problems
41
since
p+q =
1,
p2+q2 = (p+q)2—2pq=1+2=3,
p4+q4 = (p2+q2)2—2p2q2=9—2=7,
(p4+q4)2—2p4q4=49—247,
=
p8+q8
follows that a 1 3 7 . 47 = 987.
Likewise, eliminating a in (1) gives
—b = p'7—q'7
p—q
= p'6+p'5q+p'4q2++q16
=
(p'6 + q16) + pq(p'4 + q14) + p2q2 (p'2 + q'2)
=
+ p7q7(p2 + q2 ) + p8q8
(p16 +q'6) —(p'4 +q'4)+
2n+4
—
=
+q2)+ 1.
2n+4
2n+4
P
(p2fl+2
=
for
(p2
k2 = 3 and k4 = 7, and
For n 1, let
1.
—
+
+ q2Th)
+ q2) —
—
n 3. Then k6
18, k8 = 47, k1,, = 123, k,2 = 322, k,4
843,
k16 = 2207.
Hence
—b=2207—843+322—123+47— 18+7—3+1=1597
or
(a, b) = (987, —1597).
solution 21, Alternative 2
The other factor is of degree 15 and we write
(C15x15
—
C14X14 +
.. + cix —
—
x—
1)
= ax17 + bx'6 + 1.
Comparing coefficients:
x0:
c0=1.
c0—c,=0,c,=1
x2:
andfor3<k<15, xk:
—co—c,+c2=0.c2=2,
Ck_2Ck_,+CkO.
3. Solutions to Introductory
42
It follows that for k 15,
= Fk+1 (the Fibonacci number).
Thus a =
—c14 — c15 = —F17 = —1597
= F16 = 987 and b
(a, b) = (987, —1597).
Comment:
Combining the two methods, we obtain some interesting
and
Since
facts about sequences
=
—
=
—
=
—
it follows that
and
satisfy the same recursive relation.
easy to check that k2 = F1 + F3 and k4 = F3 + F5.
=
and
Therefore
+
—I.2m
2n+1 —
2n—2 -rLb2n—4
—
Lf
2
It ,i5
L1
Problem 22 [AIME 1994]
Given a positive integer n, let p(n) be the product of the non-zero digits
of n. (If n has only one digit, then p(n) is equal to that digit.) Let
S=p(l)+p(2)+
+p(999).
What is the largest prime factor of 8?
Solution 22
positive integer less than 1000 to be a three-digit number
by prefixing Os to numbers with fewer than three digits. The sum of the
products of the digits of all such positive numbers is
Consider
each
(0O.O+OO.1+...+9.99)—O.O.O
However, p(n) is the product of non-zero digits of n.
products
The sum of these
can be found by replacing 0 by 1 in the above expression, since
ignoring 0's is equivalent to thinking of them as l's in the products. (Note
that the final 0 in the above expression becomes a 1 and compensates
for the contribution of 000 after it is changed to 111.)
Hence
S
—1 = (46— 1)(462+46+ 1) = 33 5• 7.103,
and the largest prime factor is 103.
Solutions to Introductory Problems
43
problem 23 [Putnam 1979]
of nonzero real numbers such that
Zn be a sequence
—
xn —'
Xn_2Xm_1
—
for n = 3,4
establish necessary and sufficient conditions on x1 and x2 for
an integer for infinitely many values of n.
to be
Solution 23, Alternative 1
We have
—
=
Let
2
—
1
Xn_2Xn_1
—
—
= Yn—1 — Yn—2. i.e.. y,-, is an arithmetic
is a nonzero integer when n is in an infinite set 8, the
Then
sequence. If
1.
yb's for n C S satisfy —1
Since an arithmetic sequence is unbounded unless the common difference
is 0, 7/n — Yn—i
= 0 for all n, which in turn implies that x1 =
= m, a
nonzero integer.
Clearly, this condition is also sufficient.
Solution 23, Alternative 2
An easy induction shows that
—
xn_
(n
—
x1x2
— (n
1)xi
—
2)x2
—
—
x1x2
(x1 — x2)n + (2x2 — x1)
for n
3,4
In this form we see that
of n if and only if x1 =
will be an integer for infinitely many values
= in for some nonzero integer m.
Problem 24
Solve the equation
—
3x
=
Solution 24, Alternative 1
It is clear that
1. —2
—2. We consider the following cases.
x < 2. Setting x = 2cosa, 0 < a < ir, the equation becomes
8cos3 a
or
—
6cosa
=
+ 1).
3. Solutions to Introductory Problems
44
from which it follows that cos 3a =
Then 3a —
or 3a + =
= 2m7r, rn
Since 0 a
cos
2n7r,
n E Z.
the solution in this case is
4ir
x=2cos0=2,
2. x> 2. Then x3 — 4x = x(x2
—
4)
>
0
4ir
andx=2cosT.
and
x2—x—2=(x—2)(x4-1) >0
or
x>
It follows that
—
3x >
x>
Hence there are no solutions in this case.
Therefore, x = 2, x = 2cos4ir/5, and x = 2cos4ir/7.
Solution 24, Alternative 2
For x> 2. there is a real number t > 1 such that
x=t 21
The equation becomes
(t7 — 1)(t5
—
1)
=
which has no solutions for t> 1.
Hence there are no solutions for x > 2.
For —2 x 2, please see the first solution.
0,
Solutions to Introductory Problems
45
problem 25 [AIME 1992]
sequence of real numbers A = {ai, a2, a3,
any
For
}, define
AA
to
be
are 1, and that
Suppose that all of the terms of the sequence
=
Find al
a92 — 0.
SolutiOn 25
Suppose that the first term of the sequence
Then
is d.
AA={d,d+1,d+2,..}
term given by d + (n —
with the
1).
Hence
with
term given by
the
=a1 +
(n — 1)d+
— 1)(n —2).
a quadratic polynomial in ii with leading coefficient
This shows that
1/2,
Since a19 =
a92
=
0.
we must have
=
— 19)(n —92),
soa1 =(1—19)(1 —92)/2=819.
Problem 26 [Korean Mathematics Competition 2000]
Find all real numbers x satisfying the equation
2X + 3X — 4X + 6X — 9X
solution 26
Setting
= a and
=
b,
=
the equation becomes
1-i-a2 -i-b2 —a —b—ab= 0.
Multiplying
Square5 gives
both sides of the last equation by 2 and completing the
(1—a)2+(a—b)2+(b—- 1)2
3. Solutions to Introductory Problem
46
= 3X and x =
Therefore 1 =
0
is the only solution.
Problem 27 [China 1992]
Prove that
16<
<17.
Solution 27
Note that
Therefore
which proves the lower bound.
On the other hand.
Therefore
17.
which proves the upper bound. Our proof is complete.
Problem 28 [AHSME 1999]
Determine the number of ordered pairs of integers (m. m) for which mm
O and
in3 + n3 + 99mn =
Solution 28
Note that (m + n)3 =
333
m3
= (in + n)3 =
Hence m + n —
33
+
=
m3
+ n3 + 3mn(m + n). If m + n = 33. then
n3
+ 3nin(m + n) = in3 +
is a factor of in3 + n3 + 99inn —
n3
+
+ 99mm.
We have
99mm — 333
n — 33)[(m
—
m)2 + (m + 33)2
+ (n + 33)2]
5olution5 to Introductory Problems
Uence
47
there are 35 solutions altogether: (0,33), (1.32),
(33,0), and
(33, -33).
corn ient:
More generally, we have
a3 +
+ c3
—
3abc
problem 29 [Korean Mathematics Competition 2001]
Let a, b, and c be positive real numbers such that a + b c 4 and
ab+bc+ca 4.
Prove that at least two of the inequalities
a—bI2,
are true.
Solution 29
We have
(a+b+c)2 <16.
i.e.
a2+b2+c2+2(ab+bc+ca)<16,
i.e.
a2 + b2
-f
c2
8,
i.e.
a2+b2+c2—(ab+bc+ca) 4,
i.e.
(a—b)24-(b—c)2+(c—a)2 8,
and the desired result follows.
Problem 30
Evaluate
1
3. Solutions to Introductory
48
Solution 30
denote the desired sum. Then
Let
s
—
—
(2n)!
1
(2n)!k(n_k)!(n+k)!
n
2n
1
,'
1
(272
—
/
—
(2n)!
2
(2n)!
2(n!)
Problem 31 [Romania 1983]
Let
0 <
a < 1.
Solve
=
aX
for positive numbers x.
Solution 31
Taking loge yields
aX loge x =
Consider
Then
functions from
both f
f(x)g(x) =
f(x)=aX. g(x)=Iogax,
h(x)=xa.
and are decreasing and h is
has unique solution x = a.
increasing.
h(x)
Problem 32
What is the coefficient of x2 when
is expanded?
It follows that
to Introductory Problems
49
3
SolUtbohl 32
Let
It
= an.o + afl1X+
=
is eaSY to see that
and
1
Since
=
=
=
we
(1
+
—
+...) (1 + 2Thx)
1)x +
have
+
+
=
a1,2
—
—
—
+ 26 +
+
+ 22n) — (22 +
=2+ 24(2272_2 — 1) -
—
3
—
22Th+2_3.2n+1 +2
—
—
3
—
+
+
1)
(2n+1 —1) (2m+1
2)
3
Problem 33
Let in and n be distinct positive integers.
Find the maximum value of
x is a real number in the
interval (0,
Solution 33
By symmetry, we can assume that rn > n. Let y =
0 < x < 1,
< xTh and 0 < y <1. Thus
Since
— Xmn) =
=
APplying the AM-GM inequality yields
I
—
=
=
—
(
ri
—
_y)m_fl
72
(n.
/
+(m
—
—
n)(1
n+m—n
=
—
mm
3. Solutions to Introductory
50
Therefore
Is
—x
(
=(m—n)(—
—
It \
I
mm
J
Equality holds if and only if
(rn—n)y
1
or
In \
\m
Comment: For m = n + 1, we have
—
5n+1
<
—
for real numbers 0 <x <1. Equality holds if and only if x = n/(n + 1).
Problem 34
Prove that the polynomial
,
are distinct integers, cannot be written as the product of two non-constant polynomials with integer coefficients, i.e., it is
where a1, a2,•
irreducible.
Solution 34
For the sake of contradiction, suppose that
is not irreducible. Let f(s) = p(x)q(x) such that p(x) and q(x) are two
polynomials with integral coefficients having degree less than n. Then
g(x) = p(x) + q(x)
is a polynomial with integral coefficients having degree less than n.
Since
p(aj)q(aj) = f(aj) = —1
and both p(aj) and q(aj) are integers,
=
=
1
to Introductory Problems
and
Thus
51
p(aj) + q(aj) =
0.
g(x) has at least ri roots. But degg <n, so g(x) = 0.
p(x)
= —q(x)
and
f(x)
=
Then
—p(x)2,
which implies that the leading coefficient of 1(x) must be a negative
integer, which is impossible, since the leading coefficient of f(x) is 1.
problem 35
Find
all ordered pairs of real numbers (x, y) for which:
and
(1+x)(1+x2)(1+x4) =
1+y7
(1+y)(1+y2)(1+y4)
1-f-x7.
35
We consider the following cases.
Solution
1. xy =
Then it
0.
clear that x =
is
y
=
0 and
(x, y) = (0,0)
is
a
solution.
2. xy <0. By the symmetry, we can assume that x >
0 > y. Then
(1+x)(1+x2)(1+x4) >1 and 1+y7 <1. Therearenosolutions
in this case.
3. x, y >
0 and x
y. By the symmetry, we can assume that x >
y > 0. Then
(1 +x)(1 +x2)(1 +x4) >1 +x7>1 +y7,
showing that there
4. x,y
<0
and x
are no solutions in this case.
y. By the symmetry, we can assume that x <
0. Multiplying by 1
— x and 1 — y
y <
the first and the second equation,
respectively, the system now reads
1—x8 = (1+y7)(1—x)=1—x+y7—xy7
1—y8 = (1+x7)(1—y)=1—y+x7—x7y.
Subtracting the first equation from the second yields
xS_yS=(x_y)-f-(x?_y7)_xy(x6_y6).
Since x
<y <0, xs —
X
> 0.
> 0,
x— y
<
0, x7
—
y7
<0,
—xy
(1)
<0, and
Therefore, the left-hand side of (1) is positive while
the right-hand side of (1) is negative.
—
Thus
there are no solutions in this case.
3. Solutions to Introductory Problems
52
5.
x=
y.
Then solving
1—
leads
to x =
0,
1
x
—
xy7
1— x + X7
X8
1, —1, which implies that (x, y) = (0.0) or (—1, —1).
Therefore, (x, y) = (0,0) and (—1, —1) are the only solutions to the
system.
Problem 36
Solve the equation
—
1)x2
+
(2x2 — 2)x
=
2x+1 —2
for real numbers x.
Solution 36
Rearranging terms by powers of 2 yields
1)—2(x2+x—1) =0.
Setting y =
—1
(1)
and dividing by 2 on the both sides, (1) becomes
2Yx+2xy_(x+y) =0
or
(2)
Since f(x) =
—1
and x always have the same sign,
0.
Hence if the terms on the left-hand side of (2) are nonzero, they must
have the same sign, which in turn implies that their sum is not equal to
0.
Therefore (2) is true if and only if x =
x = —1,0, and
0
or y =
0.
which leads to solutions
1.
Problem 37
Let a be an irrational number and let n be an integer greater than 1.
Prove that
(a+ Va2_
+ (ais
an irrational number.
to Introductory Problems
3
53
37
SolUtbol1
Let
.1.
N= (a+
+ (a—
and let
—
1)*,
For the sake of contradiction, assume that N
b + 1/b.
Then by using the identity
Then N =
rational.
=
+
repeatedly for m
m E N.
In particular,
+
-
+
+
1)
we obtain that btm + 1/btm is rational for all
1, 2
=a+
1+a- Va2 - 1=2a
is rational, in contradiction with the hypothesis.
Therefore our assumption is wrong and N is irrational.
Problem 38
Solve the system of equations
(x1 —
X2
=
x3)2
x2(x4 + X5 — x2)
(x2—x3+x4)2 = x3(x5+x1—x3)
(x3—x4+x5)2
=
=
=
(x5—x1-f-x2)2
for real numbers x1, x2,
x4(x1+x2—x4)
x5(x2+x3—x5)
xi(x3+x4—x1)
x3, x4, S5.
38
Let 5k+5 =
Adding the five equations gives
5
5
—
is
+ 2XkXk+2) =
+ 2XkXk+2).
k=1
It follow5 that
—
= 0.
3. Solutions to Introductory Problems
54
Multiplying both sides by 2 and completing the squares yields
—
from which x1 =
system are
x2
=
X3
=
x4
= 0,
Xk+1)2
=
Therefore the solutions to the
(xi,x2.x3,x4,xs) =
(a, a, a. a, a)
for a C R.
Problem 39
Let x. y, and z be complex numbers such that x+y+z =
2,
x2+y2+z2
3, and xyz = 4.
Evaluate
1
xy+z—1
+
1
yz+x—l
+
1
zx+y—l
Solution 39
Let S be the desired value. Note that
xy+z—l =xy+1—x--y=(x—l)(y—1).
Likewise,
yz+x—l
(y— 1)(x— 1)
zx+y— 1 =
(z— 1)(x —1).
and
Hence
1
S
=
—
-
1
1
(x_1)(y_1)+(y_1)(z_1)+(z_1)(x_1)
—
—1
x+y+z—3
(x-l)(y-l)(z--l)
—1
—
xyz—(xy+yz+zx)+x+y+z--
1
—1
—
5—(xy+yz+zx)
But
2(xy + yz + zx) =
Therefore S = —2/9.
(x
+ y
+
z)2 —
(x2
+
+ z2) =
1.
5oiutions to Introductory Problems
55
problem 40 [USSR 1990]
Mr Fat is going to pick three non-zero real numbers and Mr. Taf is going
8rrange the three numbers as the coefficients of a quadratic equation
x2+ x+ =0.
Mr. Fat wins the game if and only if the resulting equation has two
distinct rational solutions.
Who has a winning strategy?
Solution 40
Mr. Fat has the winning strategy. A set of three distinct rational nonzero
numbers a, b, and c, such that a + b + c = 0, will do the trick. Let A, B,
and C be any arrangement of a, b, and c, and let f(x) = Ax2 + Bx + C.
Then
f(1) = A
+ B + C = a+b + c
=0.
which implies that 1 is a solution.
Since the product of the two solutions is C/A, the other solution is C/A.
and it is different from 1.
Problem 41 [USAMO 1978]
Given that the real numbers a, b, c, d,
and e
satisfy
simultaneously the
relations
a+b+c+d+e=Sanda2+b2+c2+d2+c2= 16,
determine the maximum and the minimum value of a.
Solution 41, Alternative 1
Since the total of b.c. d.
and
c is S — a, their average is x =
a)/4.
(S —
Let
b=x+b1, c=x+c1, d=x+d1, e=x+c1.
Then
b1
+ d1 + c1 =
0
and
(8—a)2
(1)
or
0 5a2 —
16a
= a(5a
—
16).
Therefore 0 a 16/5, where a = 0 if and only if b =
nd a 16/5 if and only if
c—d
6/5
c
=
d
=
c
=
2
3. Solutions to Introductory Problem5
56
Solution 41, Alternative 2
By the EtMS-AM inequality, (1) follows from
and the rest of the solution is the same.
Problem 42
Find the real zeros of the polynomial
Pa(X) = (x2 + 1)(x
where
1)2
—
ax2.
2x + 1) — ax2
0.
1'\
x + — I (x —2 + — I — a
0.
—
a is a given real number.
Solution 42
We have
(x2 + 1)(x2
—
Dividing by x2 yields
I
I
1\I
xj\
xj
By setting y = x + 1/x, the last equation becomes
y2
—a
—
=
0.
It follows that
x+- =
x
1±
which in turn implies that, if a 0, then the polynomial Pa(X') has the
real zeros
±
1+
2
In addition, if a 8, then Pa(X) also has the real zeros
1- fiPi ±
2
Problem 43
Prove that
1
3
2
4
for all positive integers n.
2n—1
2n
1
Solutions to Introductory Problems
57
1
4
We use induction
For n = 1, the result is evident.
SuPPOSe the statement is true for some positive integer k, i.e.,
2k—i
3
1
1
2k
Then
3
2k—1 2k+1
2k+1
1
2k
In order for the induction step to pass it suffices to prove that
2k+1
1
I
2k+2 < y3k+4
This reduces to
(2k+1\2
3k+1
<3k+4'
i.e.
(4k2+4k+1)(3k+4) <(4k2+8k+4)(3k+1),
i.e.
0< k,
which is evident. Our proof is complete.
Comment:
proved to
By using Stirling numbers, the upper bound can be imfor sufficiently large n.
Problem 44 [USAMO Proposal, Gerald Heuer]
Let
P(x) =
a nonzero
polynomial with integer coefficients such that
P(r) =
for Some integers
P(.s)
=
r and s, with 0 <r <s.
that ak —s for some k.
0
3. Solutions to Introductory
58
Solution 44
Write P(x) = (x — S)XCQ(X) and
+ ... +
Q(x) = b0xm +
0. Since Q has a positive root, by Descartes' rule of signs,
either there must exist some k for which bk > 0 bk4l, or bm > 0.
If there exists a k for which bk > 0 bk+1, then
where bm
ak+1 = —sbk
+ bk+1 —S.
If bm > 0, then
—S.
= —sbm
In either case, there is a k such that ak —s, as desired.
Problem 45
Let rn be a given real number, Find all complex numbers x such that
'.2
'.2
/
/
IX
IX
\
\
+I—1 =772
I—1
\x+1J \X—1J +712.
2
Solution 45
Completing the square gives
\2
2x2
x—lj = x 2—1 +m2+rn,
/ x
\,x+l
i.e.
/
x
2x2
\2
I
=
2x2
x2—1
2
+712 +772.
Setting y = 2x2/(x2 — 1), the above equation becomes
—
y
—
(rn2
+ m) = 0,
i.e.
(y—m—1)(y+m)=0.
Thus
2x2
x2—1
=—772or
2x2
x2—1
=712+1,
which leads to solutions
-2andx
1.
solutions to Introductory Problems
59
problem 46
=
The sequence given by xo = a,
=
b,
and
+—
2\
—
js periodic.
prove that ab =
1.
solution 46
on both sides of the given recursive relation yields
by
=
+1
or
—
=
1)
— 1.
is a geometric
— 1 for n C N. Since Yn+i =
xn is periodic, then so is
which implies that
= 0 for
all n E N. Therefore
=
Let
sequence.
If
ab =
XçjXl
=
Yi
+ 1=
1.
Problem 47
Let a, b, c, and d be real numbers such that
(a2 + b2
1)(c2
—
+
d2
—1) >
(ac+ bd
—
1)2.
Prove that
a2
+ b2> I and c2 + d2 >
1.
Solution 47
For the sake of the contradiction, suppose that one of a2 + b2 or c2 + d2
'S less than or equal to 1. Since (ac+bd— 1)2
0, a2 +b2 —land
C2 + d2 — 1
must have the same sign. Thus both a2 ± b2 and c2 + d2 are
less than 1. Let
x=1—a2—b2andy=1—c2---d2.
Thefl 0 < x, y
4xy
1. Multiplying by 4 on both sides of the given inequality
+ 2bd — 2)2 = (2 — 2ac — 2bd)2
= (a2+b2+x-i-c2+d2+y—2ac—2bd)2
= [(a—c)2
>
(2ac
(x+y)2=x2+2xy+y2,
3. Solutions to Introductory Problems
60
or 0> x2 — 2xy f y2 = (x
—
y)2, which is irhpossible.
Thus our assumption is wrong and both
a2 + b2
and
c2
+ d2 are
than 1.
Problem 48
Find all complex numbers 2 such that
(3z+1)(4z+
1)(6z+ 1)(12z+ 1) =2.
Solution 48
Note that
8(3z
=
+ 1)6(4z + 1)4(6z 4- l)2(12z -1- 1)
768,
i.e.
+4)(24z+2) = 768.
(24z+8)(24z +6)(24z
Setting u
=
24z
+ 5 and w =
u2
yields
1)(u— 1)(u—3).= 768.
(u+3)(u+
i.e.
(u2 —
1)(u2 — 9)
=
i.e.
— lOw
—
759
=
0,
i.e.
(w—33)(w+23)
Therefore
the
solutions
to the
Z
24
=0.
given equation are
andz=
24
Problem 49
Let x1,x2,
P(x) = —
Prove that
be the zeros different from
1,
n
1
of the polynomial
2.
1
1—x1
+
1
1—x2
1
n—i
1-
2
to Introductory Problems
61
Alternative 1
5olutiofl
Q(x) = P(1— x) =
(1
—
—
Then
Q(x) = (_1)nxnl +
and
+•• +
-
are the nonzero roots of the polynomial Q(x). as
(1—a—i
=0.
Thus the desired sum is the sum of .the reciprocals of the roots of polynomial Q(x), that is,
1
1
— X1
+
1
I
1
I
—
—
=—+—+4-—
a2
1
1
1
a1
= a2a3
By
+ ala3
--
+ a1a2
the Vieta's Theorem, the ratio between
and
is equal to the additive inverse of the ratio between the coefficient of x
and the constant term in Q(x), i.e., the desired value is equal to
(n)
as desired.
Solution 49, Alternative 2
For any polynomial R(x) of degree n—i, whose zeros are Xi.
the following identity holds:
1
X—X1
+
1
+...+
1
=
R'(x)
R(x)
Xn_ 1'
3. So'utions to Introductory Problems
62
For
X—l
—2
1
x—1
+...+x+1,
R(1) = n and
R'(i)=(n— 1)+(n—2)+'+ =
n(n—
1)
It follows that
1
1—x1
1
1
+...+
1—x2
+
R'(l)
n—i
R(1)
2
Problem 50
Let a and b be given real numbers.
Solve the system of equations
—
—
y
—
a
—
s/f— x2 + y2
for real numbers x and y.
Solution 50
Let u = x,+ y and v = x —
y.
Then
U+V
2
2
U—V
Adding the two equations and subtracting the two equations in the original system yields the new system
(a+b)s/1—uv
=
=
Multiplying the above two equations yields
uv(1 — uv)
hence uv =
a2
—
b2.
=
(a2
—
b2)(1
—
uv).
It follows that
u= (a+b)i/1—a2+b2 andv= (a—b)V1—a2+b2
which in turn implies that
(a+bV'a2_b2 b+a'1a2—b2
-i-b2
whenever 0 < a2 —
b2
< 1.
SOLUTiONS TO
ADVANCED PROBLEMS
4. SOLUTIONS TO
ADVANCED PROBLEMS
problem 51
f2000\
f2000\
1 2000\
(2000
Solution 51
Let
2000
1(x) = (1 + x)200° =
Let w = (—1
Then w3 = 1 and
+
((2000\
w
(2000\
+
1
= 0. Hence
(2000
= f(1) + wf(w) + w2f(w2)
= 22000 +
+
+
+
w2)200°
= 22000 + w(—w2)200° + w2(—w)200°
=
Thus
22000
+
= 22000 — 1.
+
the desired value is
22000 —
1
3
Problem 52
Let x, y, z be positive real numbers such that x4 + y4 + z4 =
Determine with proof the minimum value of
1— x8
+
52
For 0 < u < 1, let 1(u) = u(1
RY the AM-GM inequality,
—
1— y8
u8).
+
1
1— z8
Let A be a positive real number.
A(f(u))8 =Au8(1 _u8)..(1 _u8) <
[Au8 +8(1 _u8)]9
4. Solutions to Advanced
66
Setting A =
in the above inequality yields
8
8(f(u))8 <
(8)
or
It follows that
1 — x8
+
1 —
y8
+
=
1—
x(1
—
x8)
+
y(1
—
y8)
+
z(1 — z8)
>
8
with equality if and only if
x=y
=
z
1
=
Comment: This is a simple application of the result of problem 33 in
the previous chapter.
Problem 53 [Romania 1990]
Find all real solutions to the equation
2X + 3X + 6X
=
x2.
Solution 53
x < 0, the function f(x) = + + — x2 is increasing, so the
equation 1(x) = 0 has the unique solution x = —1.
Assume that there is a solution s 0. Then
For
—
so s
But then s
and hence
LsJ
+
3S + 6S
3,
1.
yields
2S 2LsJ = (1 +
which in turn implies that
GS > 4S = (23)2
I+
s,
_____________
Solutions to Advanced Problems
28
+
x
67
> s2, a contradiction.
—1 is the only solution to the equation.
problem 54
j4et {an}n1 be a sequence such that al =
and
2
1
+—
=
for all n E N.
Find an explicit formula for
Solution 54
Solving the equation
Note
leads to x
x
1
2
x
that
Therefore,
2+1)
—
and
+i]
1
Problem 55
Let x, y, and z be positive real numbers. Prove that
x
+
y
y+f(y+z)(y+x)
1.
55
Note
that
.J(x+y)(x+z)
4. Solutions to Advanced Problems
68
In fact, squaring both sides of the above inequality yields
x2+yz
which
is evident by the AM-GM inequality. Thus
x
x
-
—
Likewise,
v'V
Y
and
z
<
Adding the last three inequalities leads to the desired result.
Problem 56
Find, with proof, all nonzero polynomials f(z) such that
f(z2) + f(z)f(z +
1)
=
0.
Solution 56
Let
1(z) =
azm(z
where m and n are non-negative integers
—
and
g(z) = (z— zi)(z—
0 and ;
1, for i = 1,2,..
az2m(z
. ,
+
—
=
k. The given condition becomes
—
1)m(z
Zk),
zi)(z2 —
(z2 — Zk)
—z2)...(z —zk)
—
.(z+1z1)(z+ 1—z2)(z+1—zk).
a=
and f is nonzero, so a =
that is, m = n.
Thus f is of the form
Thus
—a2,
—1.
Since z2
1, 1— z2
Then 22m =
—zm(z — 1)mg(z).
Dividing by Z2m(2
—
+
the last equation becomes
g(z2) = g(z)g(z + 1).
0.
,lutions to Advanced Problems
claim that g(z)
one complex root r
We
69
1. Suppose not; then clearly g must have at least
0. Now
= g(r)g(r +
= 0,
= 0.
g(r2)
g(r4)
g(r8)
1)
=
0,
and so on.
Since g cannot have infinitely many roots, all its roots must have absolute
value 1.
Now,
g((r — 1)2) = g(r
soj(r—1)21 =
—
1)g(r) =
0,
1.
Clearly, if
then
2
'
2
But r2 is also a root of g, so the same should be true of r2:
7'
2
fi+vi
2
2
This is absurd. Hence, g cannot have any roots, and g(z)
Therefore, the f(z) are all the polynomials of the form
—
for
mEN,
Problem 57
Letf: N —* N be a function such that f(n+ 1) > f(n') and f(f(n)) = 3n
for all ri.
Evaluate f(2001).
57, Alternative 1
We prove the following lemma.
Lemma For ri=0,1,2,...,
:1.
2.
and
4. Solutions to Advanced Problems
70
We use induction.
Proof
For n = 0, note that f(1) 1, otherwise 3 = f(f(1)) = 1(1) = 1, which
N, f(1) > 1. Since f(n + 1) > 1(n),
is impossible. Since f N
f is increasing. Thus 1 < 1(1) < f(f(1)) = 3 or 1(1) = 2. Hence
I (2) = 1 (1 (1)) = 3.
Suppose that for some positive integer ri 1,
f
and
f (f(3n+1))
desired. This completes the induction.
There are
— 1 integers m such that
3fl — 1 integers m' such that
= 3n+2,
0
as
< in <
2
and there are
.
f
2
for 0 m 3n•
3fl
+in,
Therefore
+m) =f(f (3fl +m)) =
Hence
1(2001) = 1(2
.
36
+ 543) =
3
(36
+ 543) =
3816.
Solution 57, Alternative 2
For integer n, let fl(3) =
denote the base 3 representation of
a1a2•
Using similar inductions as in the first solution, we can prove that
ifa1 =1.
I
=
1a2
if al =
.
2.
Since 2001(3) = 2202010, f(2001)(3) = 12020100 or
1 (2001) = 1• 32
+2
34
+2
36
+
37
=
3816.
Advanced Problems
71
problem 58 [China 1999]
Let F be the set of all polynomials f(x) with integers coefficients such
f(x) = 1 has at least one integer root.
the least integer greater than 1 for which
For each integer k> 1, find
F
such
that
1(x)
= mk has exactly k distinct integer
there exists an f E
roots.
Solution 58
has exactly
Suppose that fk E F satisfies the condition that 1k (x) =
k distinct integer roots, and let a be an integer such that fk(a) = 1. Let
be the polynomial in F such that
gk(X)
for
fk(x+a)
all x.
= fk(a) = 1, so the constant term of is 1. Now gk(x) = mk
exactly
k distinct integer roots r1, r2
rk, so we can write
has
Now
gk(x)
where qk(x) is
—
an integer polynomial.
Note that r1r2•
equals 1- mk.
Since rnk > 1,
mk = (x — r1)(x — r2) ... (x
—.
.
. Tk
divides the constant term of gk(x)
1 — 172k cannot
1—
Now
are
— mk, which
be 0,
mk! rir2
-
-
distinct integers, and none of them isO. so
rlr2..rkl
hence
mk
Thi5 value of mk
gk(x)
is
=
Ik/21! + 1.
attained by
1)(x+1)(x-2)(x+2)
(x+ (_1)klk/2]) + Lk/2i! [k/2]! + 1.
Thus,
mk = Lk/2Th [k/2]! + 1.
4. Solutions to Advanced
72
Problem 59
Let x1 = 2 and
=
1
1
Xi
X2
1
1—
22
— xn
+
1
2
Solution 59
Since x1 =
2
and
—
1
=
— 1),
is increasing.
Then
— 1
0.
_1
Hence
—
1
1
— 1) —
— 1 —
1
—1 —
or
—+—±...+—=1x2
x1
1
Thus it suffices to prove that, for m
1—
1
22
1
1
1
xn+1—1
N,
1
1
<1_
22
or
—
(1)
1
We use induction to prove (1).
For ri = 1, X2 =
— x1 + 1 = 3 and (1) becomes 2 < 3 <4, which is
true.
Now suppose that (1) is true for some positive integer ii = k, i.e.,
22k1
22k
(2)
—
Then for ri =
k + 1,
Xk+2 —
1
the lower bound of (1) follows from
=
— 1)
22k—1
22k
Solutions to Advanced Problems
Since xk4-1 is
73
an integer, the lower bound of (2) implies that
xk+1
22k
and
— 1,
— 1
from which it follows that
xk+2 —
1
=
22k
Xk+1(Xk+1
—
(22k
1)
—
i)
22k+1
desired.
This finishes the induction and we are done.
problem 60 [Iran 1997]
R+ is a decreasing function such that for all
Suppose that f
:
z,y E R+,
f(x +
y)
+ f(f(x) + 1(y)) =
f(f(x + f(y)) + f(y + 1(x))).
Prove that f(f(x)) = x.
Solution 60
Setting y = x gives
f(2x) + f(2f(x)) = f(2f(x + f(x))).
Replacing x with f(x) yields
f(2f(x)) + f(2f(f(x))) = 1(21(1(x) + f(f(x)))).
Subtracting these two equations gives
1(21(1(x))) - f(2x) = f(2f(f(x) + f(f(x)))) - f(2f(x + 1(x))).
If 1(1(x)) > x, the left hand side of this equation is negative, so
f(f(x) +1(1(x))> f(x+ f(x))
and
f(x)+f(f(x)) <x 1-f(x),
contradiction A similar contradiction occurs if f(f(x)) <x.
Thus f(f(x)) = x as desired.
a
Comment: In the original formulation I was meant to be a continous
function. The solution above shows that this condition is not necessary.
Solutions to Advanced
74
Problem 61 [Nordic Contest 1998]
Find all functions f : Q —i Q such that
f(x+y)+f(x -y)——2f(x)+2f(y)
for all x,y EQ.
Solution 61
The only such functions are 1(x) = kx2 for rational k. Any such function
works, since
f(x+y)+f(x—y)=k(x+y)2+k(x--y)2
=kx2+2kxy+ky2+kx2 —2kxy+ky2
= 2kx2 + 2ky2
= 21(x) + 21(y).
Now suppose f is any function satisfying
f(x+ y)+ f(x—y) =2f(x)+2f(y).
Then letting x = y = 0 gives 21(0) = 41(0), so 1(0) = 0.
We will prove by induction that f(nz) = n2f(z) for any positive integer
ri and any rational number z.
The claim holds for ri = 0 and ri = 1; let ri 2 and suppose the claim
holds for n —
and n — 2.
Then letting x = (n — 1)z, y
1
z in the given equation we obtain
f(riz) + f((n — 2)z) = f((n
—
1)z
+ z) + f((n — 1)z — z)
=2f((n—1)z)+2f(z)
so
f(nz) = 2f((ri — 1)z) + 2f(z) — f((ri — 2)z)
= 2(n — 1)21(z) + 2f(z) — (n — 2)21(z)
—(2n2—4n+2+2—n2+4n—4)f(z)
= ri2f(z)
and the claim holds by induction.
Letting x =
0
in the given equation gives
f(y)
so
n.
f(—y) = 1(y)
for
+ f(—y) =
2f(0) + 2f(y)
all rational y; thus f(nz) = n2f(z)
for
all integers
Solutions to Advanced Problems
Now
75
let Ic = f(1): then for any rational number x = p/q,
q2f(x) = f(qx) = 1(p) = p2f(1) =
5°
Thus
kp2
1(x) = kp2/q2 = kx2.
the functions f(x) = kx2.
k E
Q. are the only solutions.
Problem 62 [Korean Mathematics Competition 2000]
Let
<a < 1.
Prove
that the equation
x3(x+1)=(x-i-a)(2x--a)
has four distinct real solutions and find these solutions in explicit form.
Solution 62
Look at the given equation as a quadratic equation in a:
a2+ 3xa+2x2 —x3 —x4 =0.
The discriminant of this equation is
9x2 — 8x2
+ 4x3 + 4x4 = (x + 2x2)2.
Thus
a—
The first choice a =
whose solutions are
—x
—3x±(x+2x2)
2
+ x2 yields the quadratic equation x2 — x — a =
—
(1±'/1+4a)
2
The second choice a =
2x
—
x2
yields the quadratic equation
x2 + 2x + a =
whose solutions are
The inequalities
Show that the four solutions are distinct.
0,
0.
4. Solutions to Advanced
76
Indeed
reduces to
< 3— v'l + 4a
which is equivalent to
6V1 + 4a < 6 + 8a,
or 3a <
4a2,
which is evident.
Problem 63 [Thurnament of Towns 1997]
Let a, b, and c be positive real numbers such that abc =
Prove that
1
1
1
<1.
+
+
a+b+1
1.
c+a+1
b+c+1
Solution 63, Alternative 1
Setting x =
a
+ b, y = b + c and z = c + a, the inequality becomes
x+1 + —
y+l+ —<1,
z+1
1
1
1
i.e.
i.e.
1
1
y+l
z+1
y+z+2
(y+1)(z+1)
x
x+1'
<
x
x+1'
i.e.
xy+xz+2x+y+z+2 xyz+xy+xz+x,
i.e.
x+y+z+2 xyz,
i.e.
2(a+b+c)+2
(a+b)(b+c)(c-4-a),
i.e.
2(a + b + c) a2b + ab2 + b2c + be2
c2a + ca2.
By the AM-GM inequality,
(a2b-t-a2c+1)
3a.
Advanced Problems
Likewise,
77
(b2c+b2a+ 1) 3b
(c2a+c2b+ 1) 3c.
Therefore we only need to prove that
2(a+b+c)+3 3(a+b+c),
i.e.
3 a+b
+ c,
which is evident from AM-GM inequality and abc =
1.
Solution 63, Alternative 2
b=
Let a =
C
=
Then a1b1c1 =
1.
Note that
0,
(ai
which implies that
a1b1(a1+bi).
Therefore,
1
1
a+b+1 =
1
aibi(ai + b1) +a1b1c1
a1b1c1
aibi(ai
+ b1 + c1)
Cl
—
a1
b1
a1+b1+c1
Adding
the three inequalities yields the desired result.
4. Solutions to Advanced Problems
78
Problem 64 [AIME 1988]
Find all functions f, defined on
gers,
the set of ordered pairs of positive inte..
satisfying the following properties:
f(x,x)—x, f(x,y)=f(y.x). (x+y)f(x.y)=yf(x,x-4-y).
Solution 64
We claim that f(x. y) = lcm(x. y). the least common multiple of x and
y. It is clear that
lcm(x, x) = x
and
lcm(x, y) = lcm(y, x).
Note that
lcm(x.y)=
y
gcd(x.y)
and
gcd(x.y) =gcd(x,x+y),
where gcd (u.
v) denotes
the greatest common divisor of u
(x+y)lcm(x,y)
and
v.
Then
=
—
x(x+y)
gcd(x,x+y)
= ylcm(x,x+y).
Now we prove that there is only one function satisfying the conditions of
the problem.
For the sake of contradiction, assume that there is another function
g(x, y) also satisfying the given conditions.
Let S be the set of all pairs of positive integers (x, y) such that f(x, y)
g(x, y), and let (in, ii) be such a pair with minimal sum m+ri. It is clear
that in n, otherwise
I (in, ii) = f(m. m) = in = g(in. in)
= g(ni. n).
By symmetry (f(x, y) = f(y, x)), we can assume that n — in >
Note that
nf(rn, n — in)
= [in + (n — m)]f(rn. ii — in)
= (m — ni)f (in, rn + (n — in))
=
(71. — in)f(m., n)
0.
Solutions to Advanced Problems
or
f(rn.ri—rn)=
[,ikewiSe,
g(rn, n — in)
79
n—rn
•f(rnn).
= ii
.
g(rn. n).
Since f(rn. ii) g(m. n), f(rn. n — in) g(rn. n — in).
Thus (in. n — rn) E S.
But (in. n — in) has a smaller sum in + (n — in) = n. a contradiction.
Therefore our assumption is wrong and f(x, y)
lcm(x. y) is the only
solution.
problem 65 [Romania 1990]
ii complex numbers zk. such that IzkI 1. k = 1.2
Prove that there exist c1.e2, . ..
E (—1. 1} such that, for any in
Consider
.
I
+ e2z2 +
ii,
+ emzrnl 2.
Solution 65
Call a finite sequence of complex numbers each with absolute value not
exceeding 1 a green sequence.
Call a green sequence
happy if it has a friend sequence
of is and — is, satisfying the condition of the problem.
We will prove by induction on ii. that all green sequences are happy.
For n = 2. this claim is obviously true.
Suppose this claim is true when n equals some number in. For the case
rn + 1. think of the zk as points in the complex plane.
For each k. let ek be the line through the origin and the point corresponding to zk. Among the lines
some two are within 600 of
each other; suppose they are
and
with the leftover one being 4.
The fact that
and
are within 600 of each other implies that there
exists some number c9 {—1, 1} such that z' =
+ e9z9 has absolute
value at most 1.
the sequence z'.
is a k-term green sequence. so,
z4. z5
by the induction hypothesis, it must be happy: let c'.
C4. C5
Ck+1
of ii =
be
its friend.
Let Ca
1.
Then the sequence
comp'ete
is the friend of
Induction is now
4. Solutions to Advanced Problems
80
Problem 66 [ARML 1997]
Find a triple of rational
(a, b, c) such that
Solution 66
Let x =
—1
and y =
Then y3 =
and x =
2
Note that
1 =y3 —1 =(y— 1)(y2+y+ 1).
and
2
—
3
3
x3 =y—l=
=
3
(y+l)3
or
(1)
y+1
On the other hand,
3=y3+1=(y+1)(y2—y+1)
from which it follows that
1
(2)
y+l
Combining
L1
2
3
(1) and (2), we obtain
x=
Consequently,
is
a desired triple.
(4
2
1
4
Solutions to Advanced Problems
81
problem 67 [Romania 1984)
Find the minimum of
+
(X2
+
+
—
are real numbers in the interval
where Xl.x2
1).
Solution 67
Since logs x is a decreasing function of x when 0 < a <
(x — 1/2)2 0 implies x2 x — 1/4, we have
=
xk+1
and. since
=
—
It follows that
+
2
(logx2
+
+
(X3
+
—
—
logx3
+
logx2
+
—
+
logxi
2n
by the AM-GM inequality.
Equalities hold if and only if
Problem 68 [AIME 1984]
Determine x2 + y2
22_12
z2
+ w2 if
+2232+2252 +22_72 =1,
42_ 12 + 42_32 + 42_52 + 42_72 =
62_12+62_32+
82_12
z2
62
52+62
w2
72
1,
1,
+8232+8252+8272=1.
4. Solutions to Advanced
82
Solution 68
The claim that the given system of equations is satisfied by
and w2 is equivalent to claiming that
z2
w2
t_12+t_32+i_52+t_72_l
(1)
satisfied by t = 4, 16. 36, and 64.
Multiplying to clear fractions, we find that for all values of t for which it
is defined (i.e., t 1, 9,25, and 49), (1) is equivalent to the polynomial
equation
is
P(t) =
0,
where
P(t) =
(t — 1)(t
9)(t — 25)(t — 49)
25)(t — 49) — y2(t — 1)(t — 25)(t
—
—x2(t — 9)(t
—z2(i
Since deg P(t) =
4,
—
—
l)(t —
9)(t
—
49)
—
w2(t
—
1)(t
—
P(t) = 0 has exactly four zeros t =
9)(t
4,
—
—
49)
25).
16, 36, and 64,
i.e.,
P(t) =
(t
4)(t
—
—
16)(t
—
36)(t
—
64).
Comparing the coefficients of t3 in the two expressions of P(t) yields
I +9+25+49+x2+y2+z2+w2 =4+16+36+64,
from which it follows that
+ y2 + z2 + w2 =
36.
Problem 69 [Balkan 1997]
Find all functions f : R
IR
such that
f(xf(x) + 1(y)) = (f(x))2 + y
for all x,y ER.
Solution 69
Let f(0) = a. Setting x =
0
in the given condition yields
I (f (y)) = a2 +
for all y E R.
Since the range of a2 +y consists of all real numbers, f must be surjective.
4
Solutions to Advanced Problems
83
there exists b C R such that f(b) = 0.
getting x = b in the given condition yields
Thus
f(f(y)) = f(bf(b) + 1(y)) =
for
(1(b))2
+y=
all y C IL It follows that, for all x, y
(f(x))2 + y = f(xf(x) +
f [1 (1 (x))f(x) + f(y)] = I [f(x)f(f (x)) +
= f(f(x))2 + y = x2 + y,
that is,
yJ
(f(x))2=x2.
(1)
It is clear that f(x) = x is a function satisfying the given condition.
Suppose that f(x) x. Then there exists some nonzero real number c
such that f(c) = —c. Setting x = cf(c) + f(y) in (1) yields
[f(cf(c) + 1(y))]2 = [cf(c) + 1(y)]2 =
for all y C R, and, setting x =
c
f—c2
+ 1(y)]2,
in the given condition yields
f(cf(c)+f(y)) = (f(c))2+y =c2+y,
for all y
C IL
that (f(y))2 =
y2.
It follows that
f—c2 +1(y)]2 =
(c2
+y)2,
or
f(y)=—y,
for all y C R, a function which satisfies the given condition.
Therefore the only functions to satisfy the given condition are 1(x) = x
or 1(x) = —x, for x C R.
Problem 70
The numbers 1000,
,2999 have been written on a board.
Each time, one is allowed to erase two numbers, say, a and b, and replace
them by the number
min(a, b)
After 1999 such operations, one obtains exactly one number c on the
board.
Prove that c <
1.
4. Solutions to Advanced
84
Solution 70
By symmetry, we may assume a b. Then
1.
=
11
We have
a
1
from which it follows that the sum of the reciprocals of all the numbers
on the board is nondecreasing (i.e.. the sum is a monovariant).
At the beginning this sum is
s———-+—+...+—<-.
1000
1001
2999 — c
1
where
1
1
1/c is the sum at the end. Note that, for I
1
2000—k
+
1
—
11
->
c — 1000
1
1
4000
4000
i\
2999)
as desired.
Problem 71 [Bulgaria 1998]
be real numbers, not all zero.
Prove that the equation
Let a1, a2
has at most one nonzero real root.
Solution 71
Notice that
=
jij1jfJ
/1
i,
>
1,
999,
k
2000+k — 20002—k2 > 20002 —
Rearranging terms in S yields
or c <
1
+ a2x is concave. Hence
2998)
45olutions to Advanced Problems
is
85
concave.
Since f'(x) exists. there can be at most one point on the curve y = f(x)
with derivative 0.
Suppose there is more than one nonzero root.
Since x = 0 is also a root, we have three real roots x1 < x2 < x3. Applying the Mean-Value theorem to 1(x) on intervals [xl.X2] and [x2,x3],
we can find two distinct points on the curve with derivative 0, a contradiction.
Therefore, our assumption is wrong and there can be at most one nonzero
real root for the equation f(x) =
n.
Problem 72 [Turkey 1998]
be the sequence of real numbers defined by a1 =
Let
=
—
1.
for n
For
t and
how many distinct values of t do we have a1998 =
0?
Solution 72, Alternative 1
Let 1(x) = 4x(1
—
x). Observe that
= {0. 1},
f1(1) = {1/2}.
and
1])
= [0,1],
=
2n_1
[0,1).
= {x E R:
Let
= 0}; then
=
=
We claim that for alln 1,
For n
c
[0. 1]. 1 E
=
+ 1.
and
1. we have
A1 = {x E R f(x) = 0} =
I
fO. 1},
and the claims hold.
NOW
8o
suppose n 1 and
c
[0. 1].
C [0. 1]. 1 e
and
+ 1. Then
4. Solutions to Advanced
86
Since
= 0 for all n 1, so I e
f(O) = f(1) = 0, we have
Now we have
=
=
aE
=
{x:f(x)=l}I+
=
1+
I{x:f(x)=a}I
>
2
aE
aE[O,1)
=
=
=
1+2(2n_1+1_1)
Thus the claim holds by induction.
Finally, a1998 = 0 if and only if I 1997(t) =
values of t.
0,
so there are 21996 + 1 such
Solution 72, Alternative 2
As in the previous solution, observe that if f(x) E
so if a1998 = 0 we must have t E [0. 1].
Observe that for any
e
f(sin2
=
(1 —
sin2 0.
a2 =
=
= 4sin2
sin2 20.
a1998 =
a3 =
a1ggg = sin2 219970.
sin2 40
0
sin 219970 = 0
7Z.
Thus the values of t which give a1998 =
0
are
sin2 (kir/21997),
k
sin2
it follows that
Therefore
for some k e
1] then x e
ir/2[ such that sin U =
Now choose U e
since
[0,
Z, giving 21996 + 1 such values of t.
0
=
[0, 1],
to Advanced Problems
problem 73 [IMO 1997 short list]
(a) Do there exist functions 1 : R —p R and g :
f(g(x))=x2
R
such that
R
such that
and
for all x e
(b)
Do there exist functions f R —i R and g R
f(g(x)) = x2
g(f(x)) =
and
for all x E R?
Solution 73
(a) The conditions imply that f(x3) = f(g(f(x))) = [1(x)]2, whence
XE f—i, 0, 1}
x3 = x
;
:
f(x) =
;
f(x) E
{O, 1}.
Thus, there exist different a, b E f—I, 0, 1} such that 1(a) = 1(b).
But then a3 = g(f(a)) = g(f(b)) = b3, a contradiction.
Therefore, the desired functions f and p do not exist.
(b) Let
p(x) =
if lxi 1
I
1
0
- In IxI if 0 < xi < I
ifx=0.
Note that g is even and al = bi whenever g(a) =
are
g(b); thus, we
allowed to define f as an even function such that
where y is such that p(+y)
= x.
1(x) =
We claim that the functions f, p described above satisfy the conditions of the problem.
It is clear from the definition of f that f(p(x)) = x2.
Now let y =
Then g(y) = x and
p(f(x)) = 9(y2)
(y2)Ifl(Y2)
=
y4Ifl Y =
= (yInY)4
(
=
=
0
[9(y)]4
x4.
if
i
if 0< y < 1
ify=0
4. Solutions to Advanced Problems
88
Problem 74 [Weichao Wu]
LetO< a1 a2•
Suppose
<b1
that there exists 1 k
n such that
a, for 1
i
k and
for 1> k.
b2
Prove that
a1a2
Solution 74, Alternative 1
We define two new sequences. For i = 1.2
=
/
and
n, Let
—.
=
Then
/
,
—
ak
= ak — — = —(a1 —
or
—
—
—
=
(ak
—
—
Therefore
nak =
a2 +
+
+
+ ... +
Applying the AM-GM inequality yields
(bib2.bnafl*
=
from which the desired result follows.
Solution 74, Alternative 2
We define two new sequences. For =
1.2
n, Let
and
Then
(1)
Note that, for cy(x
—
y)(y + c) 0
x+c x>yandc>0:
y—y+c, —
—
x
—>-———--.
45olutions to Advanced Problems
Setting
x=
aj/bi
Using
y=
and c = ak —
the above inequality implies that
for i = 1. 2, , n. Thus,
.
(1) and the AM-GM inequality yields
)
F
=
+... +
.
)
or
aia2
It is clear that the desired result follows from (2) and (3).
Problem 75
Given eight non-zero real numbers a1, a2
a8, prove that at least one
of the following six numbers: a1a3 + a2a4, aja5 + a2as, a1a7 + a2a8,
a3a5 + a4a5, a3a7 + a4a8,
+ a6a8 is non-negative.
Solution 75 [Moscow Olympiad 1978]
First, we introduce some basic knowledge of vector operations.
Let u = [a. b] and v = Im, nJ be two vectors.
Define their dot product u v = am + bn.
It is easy to check that
(i) vv = m2+n2 = 1v12,
that is, the dot product of vector with itself
is the square of the magnitude of v and v• v 0 with equality if
and only if v = [0.0];
(ii) u• v = v u;
(iii) u (v + w) = u v +
w. where w is a vector;
(iv) (cu) v = c(u . v), where c is a scalar.
.
When vectors u and v are placed tail-by-tail at the origin 0. let A and
.8 be the tips of u and v. respectively. Then
= v — u.
Let LAOB =0.
Applying the law of cosines to triangle AOB yields
= AB2
= 0A2-I-0B2-20A.OBcosO
=
juj2 + v]2
— 21u!JvJ cosO.
4. Solutions to Advanced Problem
90
It follows that
(v — u) (v
.
—
u) =
uu+v
v — 2IuIlvIcosO,
or
cosO=
Consequently, if 0
0
IuHvI
900, u v 0.
Consider vectors v1 = [ai, a2], v2 =
[a3,
a4], v3 =
[a5,
a6], and v4
[a7, as].
Note that the numbers a1a3+a2a4. a1a5+a2a6, a1a7+a2a8, a3a5+a4a6,
a3a7 + a4a8, a5a7 + a6a8 are all the dot products of distinct vectors Vj
and
Since there are four vectors, when placed tail-by-tail at the origin, at
least two of them form a non-obtuse angle, which in turn implies the
desired result.
Problem 76 [IMO 1996 short list]
Let a, b and c be positive real numbers such that abc =
Prove that
ab
ca
bc
1.
a5+b5+ab+b5+c5+bc+c5+a5+ca 1.
Solution 76
We have
a5 + b5
a2b2(a + b),
because
(a3 — b3)(a2
—
b2)
0,
with equality if and only if a = b. Hence
ab
ab
a2b2(a+b)+ab
1
—
ab(a+b)+1
—
ab(a+b+c)
abc
C
—
a+b+c
Likewise,
bc
b5 + c5 + bc
a
a+b+c
45olutions to Advanced Problems
91
and
b
+ a5 + ca — a + b + c
the last three inequalities leads to the desired result.
Equality holds if and only if a = b = c = 1.
c5
Comment:
Please compare the solution to this problem with the
second solution of problem 13 in this chapter.
Problem 77 [Czech-Slovak match 1997]
Find all functions f R —* R such that the equality
f(f(x)+y) =f(x2 —y) +4f(x)y
holds
for all pairs of real numbers (x, y).
Solution
77
Clearly. f(x) = x2 satisfies the functional equation.
Now assume that there is a nonzero value a such that f(a)
x2 —1(x)
Let y =
a2.
in the functional equation to find that
(f(x)+x2)
(f(x)+x2)
=
2f(x)(x2 —
In
1(x)
or 1(x) = x2.
both cases. f(0) = 0.
Setting x = a, it
follows from above that either f(a)
=
0 or 1(a) = 0
or
f(a)=a2.
=
The latter is false, so f(a)
Now,
let x = 0 and then x = a
1(u)
and
Let
in the
= f(—y),
functional equation to find that
1(y) = f(a2
—
so
1(n)
that
0.
= f(—y) = f(a2 + y);
is, the function is periodic with nonzero period a2.
y=
a2
in the original functional equation to obtain
f(f(x)) = 1(1(x) + a2) =
However, putting y =
for all x.
0
in
f(x2 — a2)
+ 4a2f(x) = f(x2) + 4a2f(x).
the functional equation gives 1(1(x))
= 1(x2)
4. Solutions to Advanced Problems
92
Thus, 4a2f(x) = 0 for all x. Since a is nonzero, 1(x) =
Therefore, either 1(x) = x2 or f(x) = 0
0
for all x.
Problem 78 [Kvantj
Solve the system of equations:
3x — y
x2 + y2
x+
=3
x+3y =0.
x2 + y2
Solution 78, Alternative 1
Multiplying the second equation by i and adding it to the first equation
yields
(3x—y)—(x+3y)i
or
i(x—yi)
x2+y2
—
x2+y2
Let z = x + yi. Then
=3,
!_ x—yi
z — x2 + y2
Thus the last equation becomes
z+
3—i
z
=
or
z2—3z+(3—i) =0.
Hence
3±v'—34-4i
—
3±(1+2i)
2
—
2
—
that is, (x,y) = (2,1) or (x,y) = (1,—i).
Solution 78, Alternative 2
Multiplying the first equation by y, the second by x, and adding up yields
2xy +
or 2xy —
1
=
3y.
(3x
—
y)y
(x + 3y)x
x2 + y2
It follows that y
—
0 and
3y + I
=
45olutions to Advanced Problems
this into the second equation of the given system gives
[(3y+ 1)2
+
(3Y+ 1)
—
= o.
or
4y4 — 3y2
it follows that y2
— 1
=
0.
1 and that the solutions to the system are (2, 1) and
(1,—i).
problem 79 [China 1995]
Mr. Fat and Mr. Taf play a game with a polynomial of degree at least 4:
+ ... +_x + i.
They fill in real numbers to empty spaces in turn.
If the resulting polynomial has no real root. Mr. Fat wins; otherwise, Mr.
Taf wins.
If Mr. Fat goes first, who has a winning strategy?
Solution 79
Mr. Taf has a winning strategy.
We say a blank space is odd (even) if it is the coefficient of an odd (even)
power of x.
First Mr. Taf will fill in arbitrary real numbers into one of the remaining
even spaces, if there are any.
Since there are only n
Space left after 2n —
3
— 1 even spaces, there will be at least one odd
plays, that is, the given polynomial becomes
p(x) = q(x)
where
+_x2t_1,
s and 2t — I are distinct positive integers and q(x) is a fixed
Polynomial.
We claim that there is a real number a such that
p(x) = q(x) +
Will
+_x2t1
always have a real root regardless of the coefficient of
Then
Mr. Taf can simply fill in a in front of x8 and win the game.
4. Solutions to Advanced
Now we prove our claim. Let b be the coefficient of x2t_1 in p(x). Note
that
+
p(l)
+ 2s_2t+la + b) + [q(—1) + (—1)8a —
=
Since s
=
+ q(_1)) +
2t — 1 2s—2t+1
+ (_1)8
+
o.
Thus
1
a=
is
+ q(—1)
— 2s2t+1 + (_1)8
well defined such that a is independent of b and
+p(—l) = 0.
It follows that either p(—l) = p(2) = 0 or p(—l.) and p(2) have different
signs, which implies that there is a real root of p(x) in between —1 and
2.
In either case, p(x) has a real root regardless of the coefficient of
as claimed.
Our proof is thus complete.
Problem 80 [IMO 1997 short list}
Find all positive integers k for which the following statement is true:
F(x) is a polynomial with integer coefficients satisfying the condition
for
if
c=0,1
then F(O) = F(1) = ... = F(k + 1).
Solution 80
The statement is true if and only if /c 4.
We start by proving that it does hold for each Ic 4.
Consider any polynomial F(x) with integer coefficients satisfying the
inequality 0 F(c) Ic for each c e {0. 1
k + 1}.
Note first that F(k + 1) = F(0), since F(k + 1) —F(0) is a multiple of
Ic + 1 not exceeding Ic in absolute value.
Hence
F(x) — F(0) = x(x —
Ic —
1)G(x).
45olutions to Advanced Problems
G(x) is a polynomial with integer coefficients.
consequently,
where
k jF(c) — F(O)j = c(k + I — c)IC(c)I
for each c E
2
k}.
The equality c(k + 1 — c) > k holds for each cC {2, 3
equivalent to (c — 1)(k — c) > 0.
k — 1},
as it is
that the set f2. 3
k — 1} is not empty if k 3, and for any c in
this set. (1) implies that
< 1.
Since C(c) is an integer, C(c) 0.
Note
Thus
F(x)—F(0)=x(x—2)(x—3)''(x—k+1)(x—k—1)H(x).
(2)
where H(x) is a polynomial with integer coefficients.
To complete the proof of our claim, it remains to show that 11(1) =
H(k) = 0.
Note that for c = 1 and c = k, (2) implies that
Ic
IF(c) — F(0)I = (Ic — 2)!
.
Ic
IH(c)I.
Fork4,(k—2)!>1.
Hence H(c) = 0.
We established that the statement in the question holds for any k 4.
But the proof also provides information for the smaller values of Ic as
well.
More exactly, if F(x) satisfies the given condition then 0 and k + 1 are
roots of F(x) and F(0) for any Ic
and if Ic 3 then 2 must also be
a root of F(x) — F(0).
Taking this into account, it is not hard to find the following counterexamples:
F(x)=x(2—x)
fork=1.
for Ic = 2.
F(x) = x(3 — x)
F(x)=x(4—x)(x—2)2 for k=3.
4. Solutions to Advanced Problems
96
Problem 81 [Korean Mathematics Competition 2001J
The Fibonacci sequence
is given by
(n.EN).
Prove
that
r'3
i'3
L 2n+2
£ 2n—2
9
for all n 2.
Solution 81
Note
that
=
—
=
—
=
—
whence
for
=0
—
—
(1)
all n 2.
Setting a =
and c =
b=
in the algebraic identity
—ab—bc—ca)
gives
rv,n3
htf2n
—
r,3
n3
—
n
—
—
—
Applying (1) twice gives
=
—
—
—
—
—
—
2
—
—
—
—
The desired result follows from
tlr2n+2r2nr2n_2
r'3 — r'
—
—
i
r'
r'
r'2
—
—
r2n)\ —
2
4. Solutions to Advanced Problems
97
problem 82 [Romania 1998]
Find all functions u : R —* R for which there exists a strictly monotonic
function f : R —* R such that
f(x +
= f(x)u(y) + f(y)
y)
for all x,y ER.
Solution 82
The solutions are u(x) = ax. a E R±.
To see that these work, take f(x) = x for a =
1.
1, take 1(x) =a —1; then
f(x + y) =
—1
=
(ax
—
=
1
+ f(y)
for all x, y E R.
Now suppose u R —*
R are functions for which f is strictly
R, f R
monotonic and f(x + y) = f(x)u(y) + f(y) for all x, y E R.
We must show that u is of the form u(x) = ax for some a E R+. First,
letting y = 0, we obtain f(x) = f(x)u(0) + f(0) for all x E R.
1 would imply f(x) = f(O)/(1 — u(O)) for all x, which
Thus, n(O)
would contradict the fact that f is strictly monotonic, so we must have
u(0) = 1 and f(0) = 0.
Then f(x) 0 for all x 0.
Next, we have
f(x)u(y) + f(y) = f(x + y) = f(x) + f(y)u(x),
or
f(x)(u(y) — 1) = f(y)(u(x)
—
1)
for all x,y ER. That is,
for all xy
u(x) — I
—
f(x)
—
u(y)
— 1
1(y)
0.
It follows that there exists C E R such that
u(x) — I
f(x)
for all x
=
0.
Thus, u(x) = 1+Cf(x) for x
also holds for x = 0.
0: sinceu(0) =
1,
f(0) = 0, this equation
4. Solutions to Advanced Problem
98
If C = 0, then u(x) = 1 for all x, and we are done.
Otherwise, observe
u(x+y) = 1+Cf(x+y)
+ Cf(x)u(y) + Cf(y)
= u(y)+Cf(th)u(y)
=
1
=
u(x)u(y)
for all x,y cR.
Thus u(nx) = u(x)°° for all n Z, x R.
Since u(x) = 1 + Cf(x) for all x, u is strictly monotonic, and u(—x) =
1/u(x) for all x, so u(x) >
Let a = u(1) >
0;
for all x as u(O) = 1.
then u(n) = a°° for all n N, and
0
u(p/q) = (u(p))lk =
for
allpe Z, q E N, so u(x) =
ax
for allxE Q.
monotonic and the rationals are dense in
for all x E R.
Thus all solutions are of the form u(x) = ax, a E
Since u
is
we have u(x) =
ax
Problem 83 [China 1986]
be complex numbers such that
Let z1 Z2,
,
,
Prove that there exists a subset S of {zi,
Z2,... ,
such
that
z€S
Solution 83, Alternative 1
Let
£2, and £3 be three rays from origin that form angles of 60°, 180°,
and 300°, respectively, with the positive
(here £4 =
For i = 1,2,3, let 7Z, denote the region between and
Then
including the ray
1=
jZkj+ >
ZkH
!ZkI
ZkEl?..3
By the Pigeonhole Principle, at least one of the above sums is not less
than 1/3.
4. Solutions to Advanced Problems
(otherwise, we apply a rotation, which does not effect the
it's
magnitude of a complex number), Let zk = xk + 1Yk. Then for Zk e 7Z3,
Say
XkI
Xk
ZkI/2.
>
as
desired.
Solution 83; Alternative 2
We prove a stronger statement: there is subset S of {zi, z2, .. .
that
,
z,.,}
such
zES
For
I k
n, let Zk
1
Xk +
iyk. Then
=
=
XkO
Xk<O
>
yk<O
By the Pigeonhole Principle, at least one of the above sums is not less
than 1/4. By symmetry, we may assume that
Xk>O
XkO
Consequently,
XkO
Comment:
Using advanced mathematics, the lower bound can be
further improved to 1/ir.
Problem 84 [Czech-Slovak Match 1998]
A polynomial P(x) of degree n 5 with integer coefficients and n distinct
integer roots is given.
Find all integer roots of P(P(x)) given that 0 is a root of P(x).
4. Solutions to Advanced Problems
100
Solution 84
The roots of P(x) are clearly integer roots of P(P(x)); we claim there
are no other integer roots.
We prove our claim by contradiction. Suppose, on the contrary, that
P(P(k)) = 0 for some integer k such that P(k) 0.
Let
P(x) =a(x—ri)(x— r2)(x—r3).. .(x— rn),
where a.rl,r2,...,rfl are integers,
Since P(k)
Since the
we must have 1k —
1 for all i.
are all distinct, at most two of 1k — r2j. k —
0,
equal
r31, 1k —
1, so
Ia(k—r2).(k—rn_i)I
IP(k)l
—
Also note that P(k) =
for some i0, so P(k)!
Now we consider the following two cases:
and
1.
Then P(k)I
rn!.
1k!
—
> rn!, a contradic-
rn)!
tion.
2.
1k!
a
< rn!, that is,
b. For
xe
[a,
1
1k!
b}, the
— 1.
Let a. b, c be real numbers,
function
f(x) =x(c—x)
reaches its minimum value at an endpoint x =
a
or x =
b,
or at
both endpoints.
Thus
k(k
—
rn)I = k!!rn — kI !k!(!Tn!
—
— 1.
1k!)
It follows that
rn!
P(k)!
— 1),
—
which implies that Tn! K 2. Since n 5, this is only possible if
P(x) = (x + 2)(x + 1)x(x — 1)(x
But then it is impossible to have k
diction.
—
2).
and 1k! rn!, a contra-
4
Solutions to Advanced Problems
101
our assumption was incorrect, and the integer roots of P(P(x)) are
exactlY all the integer roots of P(x).
Thus
problem 85 [Belarus 1999}
and
Two real sequences x1, x2
are defined in the following
Y2
way.
and
yn
— 1
+
+
for all n 1. Prove that 2 < XnYn <3 for all n> 1.
Solution 85, Alternative 1
Let
=
1
and note that the recursion for
is equivalent to
+
Zn+1 = Zn +
= x1; since the
and
Also note that Z2 =
recursion, this means that Zn = Xn_1 for all n> 1.
satisfy the same
Thus,
XnYn =
Xn
—=
2n
Xn
—.
Xn_1
Note that
>
Thus
2Xn_i and XnYn > 2, which is the lower bound of the desired
inequality.
Since Xn5 are increasing for n> 1, we have
= 3>
which implies that
2Xn_i >
Thus 3Xn_i > Xn. which leads to the upper bound of the desired inequality.
Solution 85, Alternative 2
Setting
= cot 0,, for 0 < 0,, <900 yields
= cot
+
+ cot2
= cot
+
= cot
4. Solutions to Advanced Problen
102
=
Since
300
30°, we have in general
Similar calculation shows
that
tan 2
2
tan
1
tan
0. tan2
And since for n>
1
is positive and XnYn > 2.
< 30°.
we have
<
tan2
so that XnYn
we also hare
<3.
Comment: From the closed
forms
for
and
in the second solution,
we can see the relationship
yn =
used
1
xfl—l
in the first solution.
Problem 86 [China 1995]
For a polynomial P(x), define the difference of P(x) on the interval 4 b)
({a,b), (a,b), (a,bJ) as P(b) — P(a).
Prove that it is possible to dissect the interval 9. 1] into a finite number
of
intervals and color them red and blue alternately such that, for every
the total difference
equal to that of P(x) on blue intervals.
What about cubic polynomials?
quadratic polynomial P(x),
of P(x)
on
red intervals
is
Solution 86
For
an interval i,
let
denote
the difference of polynomial P
on
i.
For a positive real number c and a set S ç R, let S + c denote the set
obtained by shifting S in the positive direction by c.
We
prove a more general result.
Lemma
Let £ be a positive real number, and let k be a positive integer. It is
always possible to dissect interval Ik
[0.
into
a finite number of
such that, for every
polynomial P(x) with degP k, the total difference of P(x) on the red
intervals is equal to that on the blue intervals.
intervals and color them red and blue alternatively
45olutions to Advanced Problems
103
proof
We induct on k.
For k = 1, we can just use intervals
and (t.2t]. It is easy to see
that a linear or constant polynomial has the same difference on the two
intervals.
Suppose that the statement is true for /c = n, where n is a positive
integer; that is. there exists a set
of red disjoint intervals and a set
fl
of blue disjoint intervals such that
U
=
and,
= @.
for any polynomials P(x) with deg P n. the total differences of P on
is equal to that of P on
consider polynomial f(x) with deg f n + 1. Define
g(x) = f(x +
Then deg h
and h(x) = f(x) — g(x).
n. By the induction hypothesis,
=
or
Arf+
It follows that
Arf,
where
11-n+1
and
—
—
D L)ii(D
T
=
U
+
)flD\
and
both contain the number
that number may be
removed from one of them.)
It is clear that
and
form a dissection of 'n+l and, for any
(If
Polynomial f with degf
equal to that of f on
n + 1, the total difference of f on
The only possible trouble left is that the colors in
be alternating (which can happen at the end of the
is
might not
and the beginning
But note that if intervals i1 = [a1, b1 J and i2 = [b1, c1 } are in the same
Color, then
+
Wherej3 = [ai,ci].
=
4. Solutions to Advanced Problems
104
we can simply put consecutive same color intervals
into one bigger interval of the same color.
Thus, there exists a dissection
Thus. in
=
such
that, for every polynomial f(x) with deg f n + 1.
This completes the induction and the proof of the lemma.
Setting first £ =
and then £ =
0
in the lemma, it is clear that the
answer to each of the given questions is
Problem 87 [USSR 1990]
Given a cubic equation
S3
+ _X2 + _X + — = 0,
Mr. Fat and Mr. Taf are playing the following game.
In one move, Mr. Fat chooses a real number ana Mr. Taf puts it in one
of the empty spaces.
After three moves the game is over.
Mr. Fat wins the game if the final equation has three distinct integer
roots.
Who has a winning strategy?
Solution 87
Mr. Fat has a winning strategy.
Let the polynomial be x3 + ax2 + bx + c. Mr. Fat can pick 0 first. We
consider the following cases:
(a) Mr. Taf chooses a = 0, yielding the polynomial equation
x3 + bx + c = 0.
Mr. Fat then picks the number —(inrip)2, where in. n, and p are
three positive integers such that
m2 + n2 =
If Mr. Taf chooses b = —(n2rip)2, then Mr. Fat will choose c = 0.
The given polynomial becomes
x(x — n2np)(x + rnnp).
4golutions to Advanced Problems
If Mr. Taf chooses c =
—(rnnp)2. then Mr. Fat will choose
fl2p2
b = ?TL2fl2 —
p27772.
—
The given polynomial becomes
(x + rn2)(x + n2)(x
—
p2).
(b) Mr. Taf chooses b = 0. yielding the equation
x3 + ax2
+
c
=
0.
Mr. Fat then picks the number
m2(m + 1)2(rn2 -+ m +
where rn is an integer greater than 1.
If Mr. Taf chooses
a = rn2(m + 1)2(m2 + m +
then Mr. Fat can choose
c = —m8(in + 1)8(in2 + rn + 1)6.
The polynomial becomes
(x
rnp)[x
(rn
1)pJ[x
+ m(rn + 1)pJ,
where
p = m2(n2 + 1)2(rn2 + rn + 1)2.
If Mr. Taf chooses
c=m2(m+ 1)2(m2 +m+ j)3,
then Mr. Fat can choose
a=
—(m2
+m+
1)2.
The polynomial becomes
(x+mq)[x—(rn+1)qffx—m(m+
where
q = in2 + in +
1.
4. Solutions to Advanced Problems
106
(c)
Mr. Taf chooses c =
0.
Then the problem reduces to problem 40 of the previous chapter.
Mr. Fat needs only to pick two integers a and b such that
ab(a— 1)(b—1)
and a+b= —1.
The polynomial becomes either x(x — 1)(x — a) or x(x — 1)(x
—
b).
Our proof is complete.
Below is an example of what Mr. Fat and Mr. Taf could do:
F
T
F
0
a
—3600
b
4.9.73
T
F
b
0
—481
c
Roots
—60,0.60
.38.76
a
—16, —9,25
—8.27.49,
4. 27 . 49,
89.
c
"
"
2
"
—49
a
—3
—14, 21,42
—3,0,1
b
—3
0,1.2
Problem 88 [Romania 1996]
Let n> 2 be an integer and let f : R2
any regular n-gon A1A2. .
49
c
R
be a function such that for
.
f is the zero function.
Solution 88
We identify R2 with the complex plane and let c =
Then the condition is that for any z C C and any positive real t,
In particular, for each of k
1,. ..
, n,
we obtain
4. Solutions to Advanced Problems
107
we have
gumming over
,n=1 k=1
Form = n the inner sum is nf(z); for other m, the inner sum again runs
over a regular polygon, hence is 0.
Thus f(z) = 0 for all z E C.
Problem 89 [IMO 1997 short list]
Let p be a prime number and let f(s) be a polynomial of degree d with
integer coefficients such that:
(i) 1(0) = 0, f(1) =
1;
(ii) for every positive integer n, the remainder upon division of 1(n)
by p is either 0 or 1.
Prove that dp— 1.
Solution 89, Alternative 1
For the sake of the contradiction, assume that d
p — 2.
Then by Lagrange's interpolation formula the polynomial f(s) is
determined by its values at 0, 1,
1 (x)
.
.
., p
—
2;
that is,
—
—
k=O
—
p-2
—
Setting x = p
k)5
k=O
— 1
f(p —
(—1) .. (k
1
—
xk
(x
k!(_1)P-k
p +2)
(x
P+2)
(p — k —2)!
gives
1)
=
=
>f(k)
—
—
—
k)
(_1)Pkk!
p—2
(modp).
It follows that
S(f) := f(o) + f(1) + ... + f(p -1)
0
(mod p)
(1)
4. Solutions to Advanced Problems
108
other hand, (ii) implies that S(f) j (mod p), where j
the number of those Ic E
1
1
p — 1} for which 1(k)
(mod p)
But (i) implies that 1 j < p — 1.
So S(f) 0 (mod p), which contradicts (1).
Thus our original assumption was wrong. and our proof is complete.
On the
Solution 89, Alternative 2
Again, we approach the problem indirectly.
Assume that d p — 2, and let
1(x) =
+
+ a1x + a0.
Then
p—i
S(f) =
p—i p—2
p—2
1(k) =
p—2
=
i=0
k=O i=0
k=0
p—i
=
k=O
i=0
p—i
where
=
(modp)foralli=0,1
p—2.
We use strong induction on I to prove our claim.
The statement is true for i 0 as So = p.
Now suppose that So S1
0 (mod p) for some 1 i
•..
p—2. Note that
p—i
p
p—I
k=Oj=0
(I
+
j—0
(mod p)
Since 0 < i + 1 <p, it follows that
0 (mod p). This completes the
induction and the proof of the claim. Therefore,
S(f)
0
(mod p).
=
The rest is the same as in the first solution.
4Solutions to Advanced Problems
109
problem 90
Let n be a given positive integer.
with a0 =
Consider the sequence a0, a1.
ak = ak_I
for k = 1, 2,
Prove that
,
and
+—
n
1
n.
1
< 1.
1— —
Ti
Solution 90, Alternative 1
We prove a stronger statement: For k = 1.
2
71
(1)
We use induction to prove both inequalities.
We first prove the upper bound, For k = 1, it is easy to check that
1
1
2n+1
n
2
4n
4n
2n—i
a1 = — + — =
Suppose that
<
2n— k'
for some positive integer k <n. Then
ak
= —(n-f-ak)
<
271_k(Ti+271_k)
—
n(2n—k+1)
—
(2n—k)2
<
2n—k—1'
(2n —k+ 1)(2n— k—i) = (2n—k)2 —1< (2n— k)2
Thus our induction step is complete. In particular, for k = n
afl=ak+1<
n
2n — (n
—
1)
—
I
=1,
—
1,
4. Solutions to Advanced Problems
110
as
desired.
Now we prove the upper bound. For k =
=
2n+1
1,
it is easy to check that
n+1
4n
Suppose that
n+1
2ri — k + 2'
ak>
for some positive integer k < n. Then
ak+l — ak +
(n -I- 1)2
ri + 1
n
> 2n_k+2+n(2n_k+2)2
It follows that
n+1
n+1
ak+i_2k+l
(ri.l-1)2
(2n_k+l)(2n_k+2)+n(2n_k+2)2
n+1
"ri+l
—
n+1
/1
—
2n—k+2)2
2n—k+2\
=
This
complete the induction step. In particular, for n
n+1
2n—(n—1)+1
as
1
2n_k+1) >0.
=
k
— 1,
72+1
n+2
n+2
ri
desired.
Solution
90, Alternative 2
Rewriting the given condition as
Ti
1
ak
1
a
ak
we obtain
rt+ak_1
It is clear that aks are increasing.
Thus
I
2
1
4. Solutions to Advanced Problems
111
Thus (2) implies that
1
———<—
n
ak_i
1
1
ak
for k = 1, 2, . .. , fl
Telescoping summation gives
1
1
———<1
a0
or
1
—>
that is,
1
——1 = 2—1 =
a0
<1, which gives the desired upper bound.
=
< 1, and, since aks are increasing,
Since
1,
< 1 for
<
k=1,2,...,n.
Then (2) implies
1
1
1
ak
n+ak_1
>
n+1
fork=1,2,...,ri.
Telescoping sum gives
1
1
—--—>
a0
n+1
or
1<1
n+ln+1
ao
that is,
n+2
72
n+1 =1— 1 >1——,
n
n+2
n-f-2
which is the desired lower bound.
Problem 91 [IMO 1996 short list]
Let a1, a2,..
.
,
be nonnegative real numbers. not all zero.
(a) Prove that
—
positive real root R.
(b) Let A =
—
and B =
Prove that AA RB.
..
—
—
jaj.
=
0
has precisely one
4. Solutions to Advanced Problem
112
Solution 91
(a) Consider the function
a1
a2
Note that f decreases from oo to 0 as x increases from 0 to oc.
Hence there is'a unique real number R such that f(R) = 1, that is,
there exists a unique positive real root R of the given polynomial.
(b) Let c3 = ad/A.
Then c3s are non-negative and
c2
=
1.
Since — ln x is a convex function on the interval (0, oo), by Jensen's
inequality,
Ecj
=-In(f(R))=0.
—ln
ft follows that
c3(—lnA+jlnR) 0
j=1
or
c3
= a3 /A, we obtain the desired inequality.
Comment: Please compare the solution of (a) with that of the problem
15 in the last chapter.
Problem 92
Prove that there exists a polynomial P(x, y) with real coefficients such
that P(x, y) 0 for all real numbers x and y, which cannot be written
as the sum of squares of polynomials with real coefficients.
Solution 92
We claim that
P(x,y) =
(x2 +y2 — 1)x2y2
+
is a polynomial satisfying the given conditions.
First we prove that P(x, y) 0 for all real numbers x and y.
4. Solutions to Advanced Problems
113
If x2 + y2 — I 0, then it is clear that P(x,y) > 0; if x2 + y2 —1 <0,
then applying the AM-GM inequality gives
or
(x2 + y2 — l)x2y2
It follows that P(x,y) 0.
We are left to prove that P(x, y) cannot be written as the sum of squares
of polynomials with real coefficients.
For the sake of contradiction, assume that
P(x,y) =
Since
degP =
3.
6,
Thus
=
+
+ C2y2 +
y)2, of terms x6
Comparing the coefficients, in P(x, y) and
and y6 gives
=0,
orA' =D2=Oforalli.
Then, comparing those of x4 and y4 gives
=
=
Next, comparing those of x2 and y2 gives
=
or H, =
0,
for all
Thus,
=
+ I2y +
+ F,.xy +
4. Solutions to Advanced Problems
114
But, finally, comparing the coefficients of the term x2y2, we have
= -1,
which is impossible for real numbers
Thus our assumption is wrong, and our proof is complete.
Problem 93 [IMO 1996 short list]
For each positive integer n, show that there exists a positive integer k
such that
k = f(x)(x + 1)2n +
g
with integer coefficients, and find the smallest
such /c as a function of n.
Solution 93
First we show that such a k exists.
Note that x + 1 divides 1 —
Then for some polynomial a(s) with
integer coefficients, we have
(1 + x)a(x) = 1 —
= 2— (1 +
or
2=(1+x)a(x)+(1 +x272).
Raising both sides to the (2n)th power, we obtain
= (1 +
+ (1 +
where b(s) is a polynomial with integer coefficients.
This shows that a k satisfying the condition of the problem exists. Let
/c0 be the minimum such k.
Let 2n = 2r . q, where r is a positive integer and q is an odd integer.
=
We claim that
First we prove that 2q divides k0. Let t =
(5t + 1)Q(x), where
Q(x) = 5t(q_l)
The roots of
—
Note that
+ ... —
+
+ 1.
+ 1 are
((2m—1)ir\
.
.
f(2m—1)ir\
),
m=1,2,...,t;
1
Solutions to Advanced Problems
that is,
Let
f(s)
115
R(x)=xt+l=(x_wi)(x_w2)..(x_wt).
and
g(x) be polynomials with integer
k0
coefficients
—
f(x)(x+1)272+g(x)(x272+1)
=
f(x)(x + 1)2n + g(x)Q(x)(xt + 1).
such that
It follows that
f(Wm)(Wm+1)Th=ko,
(1)
Since r is positive, t is even. So
2=R(—1) =
+w2)•(1+wt).
is a symmetric polynomial in
Since
..
,
with integer coefficients, it can be expressed as a polynomial with integer
coefficients in the elementary symmetric functions in w1, w2,.
and
..
,
therefore
F = f(wl)f(w2)•
.
is an integer.
Taking the product over m = 1,2,... , t, (1) gives 2272F = ku or
=
It follows that 2q divides k0.
It now suffices to prove that k0
Note that Q(—1) = 1.
It follows that
Q(x) =(x+1)c(x)+1,
where
is a polynomial with integer coefficients.
c(s)
Hence
(x +
1)2n
= Q(x)d(x) + 1,
for some polynomial d(x) with integer coefficients.
Also observe that, for any fixed m,
Thus
(1 +(iJm)(1
and
writing
(1 +w2t_1)
= R(—1) = 2,
(2)
4. Solutions to Advanced Problems
116
we
find that for some polynomial h(x), independent of m. with integer
such that
coefficients
(1 + w)th(w) = 2.
But then (x + 1)h(x) —
2
is divisible by xt + 1 and hence we can write
(x + 1)h(x) = 2+ (xt + 1)u(x),
for some polynomial u(x) with integer coefficients.
Raising both sides to the power q we obtain
= 2q + (xt + 1)v(x),
(x +
(3)
where v(x) is a polynomial with integer coefficients.
Using (2) and (3) we obtain
(x +
+ 1)v(x)
= Q(x)d(x)(xt + 1)v(x) + (xt + 1)v(x)
= Q(x)d(x)(xt -I- 1)v(x) + (x +
—
2q,
that is,
2q
where
(x)
= fi(x)(x
+ g1(x)
+
+ 1),
and 91(x) are polynomials with integer coefficients.
Hence k0 < 2q, as desired.
Our proof is thus complete.
Problem 94 [USAMO 1998 proposal, Kiran Kedlaya]
Let x be a positive real number.
(a) Prove that
(n—i)!
_!
(b) Prove that
(n—i)!
1
4. Solutions to Advanced Problems
117
Solution 94
We use infinite telescoping sums to solve the problem.
(a) Equivalently, we have to show that
1
1
that
n!x
(n—i)!
—
n!
and this telescoping summation yields the desired result.
(b) Let
(n-i)!
f( )X
Then,
by (a), 1(x)
In particular, 1(x)
write
f as
converges
to 0 as x approaches oo, so we can
an infinite telescoping series
=
+k
-1) - f(x + k)].
On the other hand, the result in (a) gives
f(x-1)-f(x)
=
—
—
(n—i)!
x1(x+1)•(x+n)
(1)
4. Solutions to Advanced Problems
118
Substituting the la.st equation to (1) gives
as desired.
Problem 95 [Romania 1996]
Let n 3 be an integer, and let
XcS={1,2,,,.,n3}
be a set of 3n2 elements.
Prove that one can find nine distinct numbers
such that the system
aix+biy+ciz =
a2x+b2y+c2z =
a3x+b3y+c3z =
(i = 1,2,3) in X
0
0
0
has a solution (xo, Yo, zo) in nonzero integers.
Solution 95
Label the elements of X in increasing order Xi
<<
and put
X1 = {x1,...
X2
X3
Define
=
=
the function f : X1 x X2 x X3
f(a, b, c) =
(b
S
—
x S as follows:
a, c — b).
The domain of f contains n6 elements,
The range of f, on the other hand, is contained in the subset of S x S
of pairs whose sum is at most n3, a set of cardinality
1
6
By the Pigeonhole Principle, some three triples (a2,
map to the same pair, in which case x = —
= ci
is a solution in nonzero integers.
—
cj) (i = 1,2,3)
a1,z = a1 — bi
4. Solutions to Advanced Problems
Note
ai
119
that
cannot equal
since X1 and X2 are disjoint, and that
a2 implies that the triples (ai,bi,c1) and (a2,b2,c2) are identical,
a contradiction.
Hence the nine numbers chosen are indeed distinct.
problem 96 [Xuanguo Huang]
Let n 3 be an integer and let x1, x2,
,
be positive real numbers.
Suppose that
= 1.
Prove that
(1
1
Solution 96
By symmetry, we may assume that x1 x2
following lemma.
Lemma
1
Prooft
+
—
+ x3
1
Since n 3, and, since
=
1,
1
we have
—
1
1
or
>
It follows that x2x3> 1. Thus
—
—
1+x2
1+x3
(1+x1)(1+x3)
—
-
(1+x2)(1+x3)
—
0,
We have the
4. Solutions to Advanced Problems
120
as
o
desired.
By the lemma, we have
>
1+xi
—
1+x2
—
—
and, since
it follows by the Chebyshev Inequality
1
—1
(1
—
By
the Cauchy-Schwartz Inequality, we have
2
or
(2)
Multiplying by
on both sides of (2) and applying (1) gives
which in turn implies the desired inequality.
4. Solutions to Advanced Problems
121
problem 97
be distinct real numbers.
Define the polynomials
Let X1,X2,. .
.
P(x) =
/1
and
Q(x)=P(x)(
\X—Xj
+
1
X—X2
+..,+
Yn—1 be the roots of Q.
Let Y1,Y2
Show that
minlx,,—x31 <minly2—y71.
Solution 97
By symmetry, we may assume that
d= minly, —Yjl=
Let 8k =
—
Xk, for
k = 1.2,..., n.
By symmetry, we may also assume that .s1 <
<
i.e., x1 >
For the sake of contradiction, assume that
= minx, —
d minlx2 —
= mins3 —
(1)
i(j
Since P has no double roots, it shares none with Q.
Then
1
(\Y,—X1
+
1
yj—X2
+
1
= Q(Yi)
0,
or
1
YiX1
+
1
1
Y%—X2
=0.
In particular, setting i = 1 and i = 2 gives
(2)
We
claim that there is a k such that sk(sk + d) <0, otherwise, we have
1
1
Sk+d
8k
4. Solutions to Advanced Problems
122
for. all
k, which in turn implies that
which contradicts (2).
< 0 < Sk+d.
Let j be the number of ks such that sk(sk+d) < 0, that is, Sk
the
same
sign. In
have
d
and
+
A simple but critical fact is that 8k
fact, suppose that
then
> 0 if and only if /c > i + 1, that is Sk + d> 0.
Then
From (1), we obtain Sk + d Sk+3, and, since sk + d and 3k+3 have the
same sign, we obtain
1
Sk
for all k =
.
,n
>1
+ d — 8k+j
— j. Therefore,
or
(3)
Also note that
E!<o<E 1•
+
k=1
d
Sk
Adding (3) and (4) yields
Ti
fl
which contradicts (2).
Thus our assumption is wrong and our proof is
4. Solutions to Advanced Problems
123
Problem 98 [Romania 1998]
Show that for any positive integer n, the polynomial
f(x)=(x2+x)2"+l
cannot be written as the product of two non-constant polynomials with
integer coefficients.
Solution 98
Note that 1(x) = g(h(x)), where h(x) =
x2
+ x and 9(y) = y2" + 1.
Since
and
is even for
criterion.
1 k
— 1,
g is irreducible, by Eisenstein's
Now let p be a non-constant factor of f, and let r be a root of p.
Then g(h(r)) = f(r) = 0, so .s h(r) is a root of g.
Since s = r2 -f- r C Q(r), we have Q(s) C Q(r), so
deg p deg(Q(r)/Q) deg(Q(s)/Q) =
deg
g=
Thus every factor of f has degree at least T'.
Therefore, if f is reducible, we can write f(x) = A(x)B(x) where A and
B have degree
5
Since
have
+5
2
+x+1) 2"
(mod2).
+ x + 1 is irreducible in Z2[x], by unique factorization we must
A(s)
B(s)
+ x + 1)2"'
+
+
1
(mod 2).
Thus, if we write
A(s)
=
B(s) =
+ ao,
+
+ .. + b0,
b0 are odd and all the other coefficients
then
ao,
are even. Since f is monic, we may assume without loss of generality
4. Solutions to Advanced Problems
124
= 1; also, a0bo = f(O) =
=
that
no real roots, so a0 = b0 = 1.
1.
but a0 > 0,
> 0 as f has
Therefore,
([x2n+2"1] + [x2"1])(g(x)h(x))
+
( i=2"1
+
+
EQ
as
But
+
+
(mod 4)
is even.
+ [x2"'])(f(x))
=
is odd by Lucas's theorem, so
and
([x2"+2"'] + [x2"')) (1(x))
2
(mod 4),
a contradiction.
Hence f is irreducible.
Problem 99 [Iran 1998]
Let Ii, 12,13 R R be functions such that
aifi + a212 + a3f3
is monotonic for all a1, a2, a3 E
Prove that there exist Ci, c2, c3 c R, not all zero, such that
cifi(x) + c2f2(x) + c3f3(x) =
0
for all x E R.
Solution 99, Alternative 1
We establish the following lemma.
Lemma: Let f, g R —p IR be functions such that f is nonconstant and
af + bg is monotonic for all a, b c R. Then there exists c R such that
g — cf is a constant function.
Proofi
Let s, t be two real numbers such that f(s)
1(t).
4. Solutions to Advanced Problems
Let
-
125
g(s)—g(t)
f(s) - f(t)
Let h1 = g — d1f for some d1 E R.
Then h1 is monotonic. But
hi(s) — h1(t)
Since f(s) —
= g(s) — g(t)
f(t)
—
d1(f(s)
—
f(t)) = (f(s)
—
f(t))(u — d1).
0 is fixed, the monotonicity of h1 depends only on
thesignofu—d1.
Since f is nonconstant, there exist xl,x2 c
R
such that 1(x1)
f(x2).
Let
c= g(xl)—g(x2)
f(xi)
— f(x2)
= g — cf.
Then r = h(xi) = h(x2) and the monotonicity of h1 =
and h
— d1f, for each
d1, depends only on the sign of c — d1.
We claim that h = g — cf is a constant function.
We prove our claim by contradiction.
Suppose, on the contrary, that there exists x3 c R such that h(xs) r.
and f(x2) f(xi) is
Since f(xi) f(x2), at least one of f(xi)
true.
Without loss of generality, suppose that 1(x1)
Let
c' = g(x1)
—
f(x3).
g(x3)
f(xi) — f(x3)
Then the monotonicity of h1 also depends only on the sign of c' — d1.
Since h(x3)
r = h(xi),
f(xi) —
hence
c—d1
f(x3)
=c';
c' —d1.
So there exists some d1 such that h1 is both strictly increasing and decreasing, which is impossible.
Therefore our assumption is false and h is a constant function.
Now we prove our main result.
If fi f2, 13 are all constant functions, the result is trivial.
Without loss of generality, suppose that Ii is nonconstant.
4. Solutions to Advanced Problems
126
For a3 = 0, we apply the lemma to Ii and 12, so 12 = cf1 +d; for a2
we apply the lemma to fi and 13, so
Here c, c', d, d' are constant.
0,
= c'f1 + d'.
We have
(c'd — cd')f1 + d'f2 — df3 = (c'd — cd')fi + d'(cfi + d) — d(c'fi + d') =
If (c'd—cd',d',—d)
0.
(0,0,0), then let
(ci, c2, c3) = (c'd — cd', d',
—d)
we are done.
Otherwise, d = d' = 0 and 12,13 are constant multiples of Ii
Then the problem is again trivial.
and
Solution 99, Alternative 2
Define the vector
v(x) = (1i(x),12(x),13(x))
for x C R.
If the v(x) span a proper subspace of R3, we cart find a vector (ci, c2, c3)
orthogonal to that subspace, and then cifi(x) + c2f2(x) + c3f3(x) =
0
for all x R.
So suppose the v(x) span all of R3.
Then
exist x1 < X2 < X3 c R such that v(xi), v(x2), v(x3) are
has
linearly independent, and so the 3 x 3 matrix A with
=
linearly independent rows.
But then A is invertible, and its columns also span R3.
This means we can find
c2, c3 such that
= (0,1,0),
and the function cif1 +c2f2 +c3f3 is then not monotonic, a contradiction.
4. Solutions to Advanced Problems
127
Problem 100 [USAMO 1999 proposal, Richard Stong]
Let x1,x2,. .
be variables, and let Y1,Y2,.
be the sums of
nonempty subsets of
be the kth elementary symmetric polynomial in
Let pk(xl,.
the (the sum of every product of k distinct yj'S).
.
.
.
.
.
,
For which k and n is every coefficient of Pk (as a polynomial in x1,..
even?
For example, ifn = 2, then Y1,Y2,Y3 are
+x2 and
.
,
Pi =Y1+Y2+Y32X1+2X2,
Y1Y2+Y2Y3+Y3Y1
P2
P3
=
=
Y1Y2Y3
+
Solution 100
We say a polynomial pk is even if every coefficient of Pk is even.
Otherwise, we say Pk is not even.
For any fixed positive integer n, we say a nonnegative integer k is bad
for n if k = — for some nonnegative integer j.
We will show by induction on n that pk(xl, x2,
is not even if and
only if k is bad for ii.
,
.
For n =
=
1,
x1
k=1=21—2°=r—2°.
is not even and k =
1
is bad for n =
Suppose that the claim is true for a certain n.
We now consider pk(x1,x2,
Let
0k(yl, Y2,.
.. Ys) be
.
the
.
.
elementary symmetric polynomial.
We have the following useful, but easy to prove, facts:
1.
2.
For all 1 r
,y8)
3.
= i+j=k
>
,Yr)Uj(Yr+1,
.
0k(X+Y1,X+y2,...,X+Y&)
=
iI<22<"<ik
=
.
{si
Sl<82<<S,.
,ik}
1
as
4. Solutions to Advanced Problems
128
=
Hence
=
i+j=k
+
+X2 +
=
,
. ,
r.—O
By the induction hypothesis, every term of Pr(Xi, x2
,
xi,) is even un-
For such r, note that
2t
j - r) -
-
isevenunlessj—r=Oorj--r=2t.
Therefore, taking coefficients modulo 2,
,Xn)Pj(X1,X2,'
i+j=k
,Xn)p2n_2t(x1,X2,
By the induction hypothesis, the terms in the first sum are even unless
In the second sum, every term appears twice except the term
Pk/2(X1,
,
for k even.
By the induction hypothesis, this term is even unless k/2 =
2v+1
some 0 <v n, that is k =
It follows that Pk(X1, x2,••
is even unless k =
— 2w
—
for
for
some
4. Solutions to Advanced Problems
129
Furthermore, note that the odd coefficients in
Pk(X1,X2,
occur for different powers of
Therefore, the condition that k is bad for n + 1 is also sufficient for
to be odd.
Our induction is complete.
Problem 101 [Russia 20001
Prove that there exist 10 distinct real numbers a1, a2,..., a10 such that
the equation
(x —ai)(x—a2)...(x —a10) =(x-i--ai)(x-i-a2).(x+aio)
has exactly 5 different real roots.
Solution 101
We show that
6,. . —2} is a group of numbers
a2,. .. a10} =
satisfying the conditions given in the problem.
The given equality becomes
,
(x — 2)(x
—
.
,
1)x(x + 1)(x + 2)g(x2)
0,
where
g(u)
=
2[((7+6++3)u2+
(7.6•5+7.6.4+...--5.4.3)u+76•3].
no real solutions, then g(x2) = 0 has no real solutions.
If u1 and u2 are real solutions of g(u) = 0, then u1 -F-u2 <0 and U1U2 > 0,
that is, both u1 and u2 are negative.
It follows again that g(x2) = 0 has no real solutions.
Our proof is complete.
If g(u) =
has
GLOSSARY
Arithmetic-Geometric Mean Inequality (AM—GM Inequality)
If a1, a2,..
.
n nonnegative
are
,
numbers, then
1
with
equality if
and
only if
=
al = a2 =
Binomial Coefficient
in the expansion of (x +
The coefficient of
is
(n'\
= k!(n—k)!
Cauchy-Schwarz Inequality
For any
with
real
numbers
equality if
a1, a2,.. .
and only if
and b1, b2,..
,
and
,
are proportional,
i
= 1,2,.
..
, n.
Chebyshev Inequality
1. Let Xi, X2
.
.
,
and yi,
bers, such that x1 X2 <
.
.
,
be two sequences of real num-
and Yi y2
...
yn.
Then
+X2 +
2.
+Xn)(Yi +Y2+" +Yn)
+X2y2
.
Let Xl,X2...,Xn and Y1,Y2,.•.,Yn be two sequences of real numbers, such that x1 x2
..
xi-, and Yt y2
Then
+yn) Xlyl+X2y2+
+XnYn.
Glossary
132
De Moivre's Formula
For any angle a and for any integer n,
(cosa+
Elementary
= cosna+isinna.
Symmetric Polynomials (Functions)
Given indeterminates x1, ..., Xn, the elementary symmetric functions
are defined by the relation (in another indeterminate t)
Si,. ..
(t + x1) .. (t +
+
= tTh +
+
+ Sn.
taken k at a time. It
That is, Sk is the sum of the products of the
can be
is a basic result that every symmetric polynomial in x1
(uniquely) expressed as a polynomial in the
and vice versa.
Fibonacci Numbers
Sequence defined recursively by F1 = F2 = 1,
=
+
for
all
n E N.
Jensen's Inequality
1ff is concave up on an interval [a, b] and A1, A2, ...,
numbers with sum equal to 1, then
Aif(xi)+A2f(x2)+-..
are nonnegative
+).2X2+
in the interval [a. b]. If the function is concave
X2
down, the inequality is reversed.
for any x1 ,
Lagrange's Interpolation Formula
be ar. ,x7, be distinct real numbers, and let YO.Y1,. ..
bitrary real numbers. Then there exists a unique polynomial P(x) of
degree at most n such that
i = 0. 1,.. . , n. This is the
=
Let Xo,Xi,.
.
polynomial given by
—
i=O
x0)
—
—
Law of Cosines
Let ABC be a triangle. Then
BC2 = AB2 + AC2
- 2AB
.
ACcosA.
—
Glossary
133
Lucas' Theorem
Let p be a prime; let a and & be two positive integers such that
+ ak_lpkI +
a =
+ a0, b = bkpk +
b2 <p are integers for i =
where 0
(a'\
(ak\ (ak_1\
0,
1
+
bip + b0,
k. Then
(aj\ (ao\ (modp).
Pigeonhole Principle
If n objects are distributed among k < n boxes, some box contains at
least two objects.
Root Mean Square-Arithmetic Mean Inequality (RMS—AM Inequality)
For positive numbers x1, X2,.
.
.
,
\/X?+X2++XkXl+X2++Xk
be any positive numbers for which ai +
we define
For positive numbers x1, x2,...
More generally, let a1, a2,.. .
a2 +
+
=
1.
,
,
= min{x1,x2,. .. ,xk},
Moc, =max{xl,x2,...,xk},
—
aj a2
0—xl
x2
=
+
a non-zero real number. Then
<M8
for s
t.
M00
Glossary
134
Triangle Inequality
Let z =
a
+ bi be a complex number. Define the absolute value of z to
be
zI=
Let
is
and /3 be two complex numbers. The inequality
called the triangle inequality.
Let =
Then
and i9 = /9i
+/32i,
/92 are real numbers.
where
v = [/91,/92], and w =
Vectors u =
+192] form a
triangle with sides lengths
and + /31
The triangle inequality restates the fact that the length of any side of a
triangle is less than the sum of the lengths of the other two sides.
Vieta's Theorem
Let x1, X2,.
.
. ,
be the roots of polynomial
P(x) =
where
of
the
+... +
+
E C. Let 8k
#0 and a0,a1,
taken k at a time. Then
.
.
.
be
+ a0.
the sum of the products
=
that is,
x1x2
x1x2
+
+
=
=
a
sin a
cos a
1
a
—;
Glossary
135
addition and subtraction formulas:
sin(a±b) =sinacosb±cosasinb,
cos(a±b)
a ± tan b
tan(a±b)= tan tan
a tan b'
1
double-angle formulas:
sin2a = 2sinacosa,
cos 2a =
tan2a=
cos2 a — sin2
a = 2 cos2 a
— 1
=
1 —
2 tan a
1—tan2a'
triple-angle formulas:
sin3a = 3sina — 4s1n3 a,
cos3a = 4cos3a — 3cosa,
tan3
tan3a= 3tana — 2 a
1—3tan a
half-angle formulas:
2 tan
sina=
1
a
2
1
tan2
1
tan2
2
2
2
a+b
cosa+cosb=
tana +tanb =
a—b
sin(a+b)
cosacosb'
difference-to-product formulas:
sin a — sin b = 2 sin
a—b
cos
a+b
2 sin2
a,
Glossary
136
cos a — cos b — —2 sin
tana—tanb=
a—b
sin
a+b
sin(a — b)
cos a cos b
product-to-sum formulas:
2sinacosb = sin(a + b) +sin(a — b),
2cosacosb = cos(a + b) cos(a —
2sinusinb = —cos(a + b) + cos(a — b).
FURTHER READING
1.
Andreescu, T. Kedlaya, K.; Zeitz, P., Mathematical Contests
1995-1996: Olympiad Problems from around the World, with
Solutions, American Mathematics Competitions, 1997.
2. Andreescu, T. Kedlaya, K., Mathematical Contests 1996-1997:
Olympiad Problems from around the World, with Solutions,
American Mathematics Competitions, 1998.
3. Andreescu, T. Kedlaya, K., Mathematical Contests 1997-1998:
Olympiad Problems from around the World, with Solutions,
American Mathematics Competitions, 1999.
4. Andreescu, T. Feng, Z., Mathematical Olympiads: Problems and
Solutions from around the World, 1998-1999, Mathematical
Association of America, 2000.
5. Andreescu, T. Gelca, R., Mathematical Olympiad Challenges,
Birkhäuser, 2000.
6. Barbeau, E., Polynomials, Springer-Verlag, 1989.
7. Beckenbach, E. F., Bellman, R., An Introduction to Inequalities,
New Mathematical Library, Vol. 3, Mathematical Association of
America, 1961.
8. Chinn, W. C., Steenrod, N. E., First Concepts of Topology, New
Mathematical Library, Vol. 27, Random House, 1966.
9. Cofman, J., What to Solve?, Oxford Science Publications, 1990.
10. Coxeter, H. S. M., Greitzer, S. L., Geometry Revisited, New
Mathematical Library, Vol. 19, Mathematical Association of
America, 1967.
11. Doob, M., The Canadian Mathematical Olympiad 1969-1993,
University of Toronto Press, 1993.
12. Engel, A., Problem-Solving Strategies, Problem Books in
Mathematics, Springer, 1998.
13. Fomin, D., Kirichenko. A., Leningrad Mathematical Olympiads
1987-1991, MathPro Press, 1994.
Further Reading
138
14. Fomin, D., Genkin, S., Itenberg, 1., Mathematical Circles,
American Mathematical Society, 1996.
15. Graham, R. L., Knuth, D. E., Patashnik, 0., Concrete
Mathematics, Addison-Wesley, 1989.
16. Greitzer, S. L., International Mathematical Olympiads, 1959-1977,
New Mathematical Library, Vol. 27, Mathematical Association of
America, 1978.
17. Grossman, I., Magnus, W., Groups and Their Graphs, New
Mathematical Library, Vol. 14, Mathematical Association of
America, 1964.
18. Kazarinoff, N. D., Geometric Inequalities, New Mathematical
Library, Vol. 4, Random House, 1961.
19. Klamkin, M., International Mathematical Olympiads, 1978-1985,
New Mathematical Library, Vol. 31, Mathematical Association of
America, 1986.
20. Klamkin, M., USA Mathematical Olympiads, 1972-1986, New
Mathematical Library, Vol. 33, Mathematical Association of
America, 1988.
21. Kiirschák, J., Hungarian Problem Book, Volumes I & II, New
Mathematical Library, Vols. 11 & 12, Mathematical Association of
America, 1967.
22. Kuczma, M., 144 Problems of the Austrian-Polish Mathematics
Competition 1978-1998, The Academic Distribution Center, 1994.
23. Larson, L. C., Problem-Solving Through Problems, Springer-Verlag,
1983.
24. Lausch, H., Bosch Giral, C., The Asian Pacific Mathematics Olympiad 1989-2000, AMT Publishing, Canberra, 2001.
25. Liu, A., Chinese Mathematics Competitions and Olympiads
1981-1993, AMT Publishing, Canberra, 1998.
26. Lozansky, E., Rousseau, C. Winning Solutions, Springer, 1996.
27. Ore, 0., Graphs and their uses, Random House, 1963.
28. Ore, 0., Invitation to number theory, Random House, 1967.
29. Sharygin, I. F., Problems in Plane Geometry, Mir, Moscow, 1988.
Further Reading
139
30. Sharygin, I. F., Problems in Solid Geometry, Mir, Moscow, 1986.
31. Shklarsky, D. 0, Chentzov, N. N; Yaglom, I. M., The USSR
Olympiad Problem Book, Freeman, 1962.
32. Slinko, A., USSR Mathematical Olympiads 1989-1992, AMT
Publishing, Canberra, 1997.
33. Soifer, A., Colorado Mathematical Olympiad: The first ten years,
Center for excellence in mathematics education, 1994.
34. Szekely, C. J., Contests in Higher Mathematics, Springer- Verlag,
1996.
35. Stanley, R. P., Enumerative Combinatorics, Cambridge University
Press, 1997.
36. Taylor, P. J., Tournament of Towns 1980-1984, AMT Publishing,
Canberra, 1993.
37. Taylor, P. J., Tournament of Towns 1984-1 989, AMT Publishing,
Canberra, 1992.
38. Taylor, P. J., Tournament of Towns 1989-1993, AMT Publishing,
Canberra, 1994.
39. Taylor, P. .1., Storozhev, A., Tournament of Towns 1993-1997,
AMT Publishing, Canberra, 1998.
40. Tomescu, I., Problems in Combinatorics and Graph Theory, Wiley
1985.
41. Vanden Eynden, C., Elementary Number Theory, McGraw-Hill,
1987.
42. Wilf, H. S., Generatingfunctionology, Academic Press, 1994.
43. Wilson, R., Introduction to graph theory, Academic Press, 1972.
44. Yaglom, 1. M., Geometric Transformations, New Mathematical
Library, Vol. 8, Random House, 1962.
45. Yaglom, I. M., Geometric Transformations II, New Mathematic
Library, Vol. 21, Random House, 1968.
46. Yaglom , I. M., Geometric Transformations III, New Mathemati
Library, Vol. 24, Random House, 1973.
47. Zeitz, P., The Art and Craft of Problem Solving, John Wiley &
Sons, 1999.
Titu Andreescu is
Director of the
A met jean
tics
Competitions,
serves
as
Head
Coach of the USA
international
Mathem atical
Olympiad
(lMO)
Team, is Chair of the USA Mathematical
Olympiad Committee, and is Director of the
USA Mathematical Olympiad Summer
Program. Originally from Romania, Prof.
Andreescu received the Distinguished
Professor Award from the Romanian
Ministty of Education in 1983, then after
moving to the USA was awarded the Edyth
May Sliffe Award for Distinguished High
School Mathematics Teaching from the
MM in 1994. in addition, he received a
Certificate of Appreciation presented by the
President of the MM for This outstanding
service as coach of the USA Mathematical
Olympiad Program in prepanng the USA
team for its perfect performance in Hong
Xong at the 1994 1MO".
Zuming
Feng
graduated with a
PhD from Johns
Hopkins University
with an
on
Algebraic
Nuniber Theory
and
Elliptic
Curves. He teaches
at Phillips Exeter
Academy. He also serves as a coach of the
USA international Mathematical Olympiad
(IMO) Team, is a member of the USA
Mathematical Olympiad Committee, and is
an
assistant director of the USA
Mathematical Olympiad Summer Program.
He received the Edyth May Sliffe Award for
THF AUSTRALIAN MATHEMATiCS TRUST
Distinguished High School Mathematics
Teaching from the MM in 1996.
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