Tài liệu Math Problem Book I pptx

1999 IMO Two circles Dy; and F2 are contained inside the circle LP, and are tangent to [ at the distinct points Af and N, respectively.. and N, respectively, so that AC CE — Determine 7

Math Problem Book I

compiled by

Kin Y Li

Department of Mathematics

Hong Kong University of Science and Technology

Copyright © 2001 Hong Kong Mathematical Society IMOCHK) Committee

Printed in Hong Kong

Preface

‘There are over fifty countries in the world nowadays that hold math-

ematical olympiads at the secondary school level annually in Hungary,

Russia and Romania, mathernatical competitions have a long history, dat-

ing back to the late 1800’s in Hungary’s case Many professional or ama-

teur mathematicians developed their interest in math by working on these

olympiad problems in their youths and some in their adulthoods as well

The problems in this book came from many sources For those involved

in international math competitions, they no doubt will recognize many of

these problems We tried to identify the sources whenever possible, but

there are still some that escape us at the moment Hopefully, in future

editions of the book we can fill in these missing sources with the help of the

knowledgeable readers

This book is for students who have creative minds and are interested in

mathematics Through problem solving, they will learn a great deal more

than school curricula can offer and will sharpen their analytical skills We

hope the problems collected in this book will stimulate them and seduce

them to deeper understanding of what mathematics is all about We hope

the international math communities support our efforts for using these bril-

liant problems and solutions to attract our young students to mathematics

Most of the problems have been used in practice sessions for students

participated in the Hong Kong IMO training program We are especially

pleased with the efforts of these students Im fact, the original motivation

for writing the book was to reward them in some ways, especially those wha k7 «

worked so hard to become reserve or team members It is only fitting to

list their names along with their solutions Again there are unsung Oo ¬ > heros

1H

who contributed solutions, but whose names we can only hope to identify

in future editions

As the title of the book suggest, this is a problem book So very little

introduction materials can be found We do promise to write another book

presenting the materials covered in the Hong Kong IMO training program This, for certain, will involve the dedication of more than one person Also, this is the first of a series of problem books we hope From the results of the Hong Kong IMO preliminary contests, we can see waves of new creative minds appear in the training program continuously and they are younger and younger Maybe the next problem book in the series will be written by our students

Finally, we would like to express deep gratitude to the Hong Kong Quality Education Fund, which provided the support that made this book

possible

Kin Y L¡ Hong Kong

April, 2001

iv

Advices to the Readers

The only way to learn mathematics is to do mathematics In this

book, you will find many math problems, ranging from simple to challenging

problerns You may not succeed in solving all the problems Very few

people can solve them all The purposes of the book are to expose you to

many interesting and useful mathematical ideas, to develop your skills in

analyzing problems and most important of all, to unleash your potential

of creativity While thinking about the problems, you may discover things

you never know before and putting in your ideas, you can create something

you can be proud of

To start thinking about a problem, very often it is helpful to look at

the initial cases, such as when 7 = 2,3,4,5 These cases are sirnple enough

to let you get a feeling of the situations Sometimes, the ideas in these

cases allow you to see a pattern, which can solve the whole problem For

geometry problems, always draw a picture as accurate as possible first

Have protractor, ruler and compass ready to measure angles and lengths

Other things you can try in tackling a problem include changing the

given conditions a little or experimenting with some special cases first

Sometimes may be you can even guess the answers from some cases, then

you can study the form of the answers and trace backward

Finally, when you figure out the solutions, don’t just stop there You

should try to generalize the problem, see how the given facts are necessary

for solving the problem This may help you to solve related problems later

on Always try to write out your solution in a clear and concise manner

Along the way, you will polish the argument and see the steps of the so-

lutions more clearly This helps you to develop strategies for dealing with

other problems

Vv

The sobitions presented in the book are by no means the only ways

to do the problems H you have a nice elegant solution to a problem and would like to share with others (in future editions of this book), please send

it to us by email at maky@ust.hk Also if you have something you cannot

understand, please feel free to contact us by email We hope this book will

increase your interest in math

Finally, we will offer one last advice Don’t start with problem 1, Read the statements of the problems and start with the ones that interest you the most We recornmend inspecting the list of miscellaneous problems first

Have a fun time

VI

Number Theory v0 Tô n /ẽằHa.nằB i( 18 Combinatorlcs ProblOTR vn nh ng nh nu kg kg nà ky kg va 24 Miscellaneous Problems 0.0.0.0 ec c ccc ce nn cnc e ener ees 28 Solutions to Algebra Problems 00.0 000 c cece eee eens dỗ 5olutions to Geomefry PrODÌOTHS cu kh nh nh nh eee ee 69 Solutions to Number Pheory Problefm8 cv ch nh khen 88 5olutions to Combinatorios Pro blGEHR ch nh vn kh kg ray 121

Solutions to Miscellaneous Problems 0 0.00.0 0 cc eee eee ec nees 135

Contributors

Chan Kin Hang, 1998, 1999, 2000, 2001 Hong Kong team member

Chan Ming Chin, 1997 Hong Kong team reserve member

Chao Khek Lun, 2001 Hong Kong team member

Cheng Kei T'si, 2001 Hong Kong team member

Cheung Pok Man, 1997, 1998 Hong Kong team member

Fan Wai Tong, 2000 Hong Kong team member

Fung Ho Yin, 1997 Hong Kong team reserve member

Ho Wing Yip, 1994, 1995, 1996 Hong Kong team member

Kee Wing Tao, 1997 Hong Kong team reserve member

Lam Po Leung, 1999 Hong Kong team reserve member

Lam Pei Fung, 1992 Hong Kong team member

Lau Lap Ming, 1997, 1998 Hong Kong team member

Law Ka Ho, 1998, 1999, 2000 Hong Kong team mernber

Law Siu Lung, 1996 Hong Kong team member

Lee Tak Wing, 1993 Hong Kong team reserve member

Leung Wai Ying, 2001 Hong Kong team mernber

Leung Wing Chung, 1997, 1998 Hong Kong team member

Mok Tze Tao, 1995, 1996, 1997 Hong Kong team member

Ng Ka Man, 1997 Hong Kong team reserve member

Ng Ka Wing, 1999, 2000 Hone Kong team member

Poon Wai Hoi, 1994, 1995, 1996 Hong Kong team member Poon Wing Chi, 1997 Hong Kong team reserve member

Tam Siu Lung, 1999 Hong Kong team reserve member

To Kar Keung, 1991, 1992 Hong Kong team member

Wong Chun Wai, 1999, 2000 Hong Kong team member

Wong Him Ting, 1994, 1995 Hong Kong team member

Yu Ka Chun, 1997 Hong Kong team member

Yung Fai, 1993 Hong Kong team member

4 1

Problems

or

Aigebra Problems

Polynomials

(Crux Mathematicorum, Problem 7) Find (without calculus) a fifth

degree polynomial p(x) such 1 that ple n) 4 - | is divisible by (2 — 1)° and

p(x) — 1 is divisible by (2 + 1)%

A polynomial P(x) of the n-th degree satisfies P(k) = 2” for k =

O,41,2, ,m Find the value of P(n + i)

(1999 Putnam Exam) Let P(z} be a polynomial with real coefficients

such that P(a) > 0 for e every real z Prove that

P(ø) = ñ()ˆ + 20)“ + wee + at)

for some polynomials fy (x), fo(x), , fn(x) with real coefficients

(1995 Russian Math Olympiad) Is it possible to find three quadratic

polynomials f(x}, g(a), h{x) such that the equation ƒ(a(h(z))) = 0 has

the eight roots 1,2,3,4,5,6,7, 8?

oom 1968 Putnam Hixam) Determine all polynomials whose coefficients are

all 1 that have only real roots

(1990 Putnam Exam) Ís there an infinite sequence a9,a1,d2, of

nonzero real mumbers such that for mn = 1,2,3 , the polynomial

P(x) = a +04@ +0927 +-+-+a,,2” has exactly n distinct real roots?

(1991 Austrian-Polish Math Competition} Let P(2} be a polynomial

with real coefficients such that P(x) > 0 for 0 < x < 1 Show that

there are polynomials A(x), B(x), C(x) with real coefficients such that

(a) A{a) > 0, B(x) > 0,C(a) > 0 for all real x and

(b) Pla) = Ala) + e2Bla)+ (1 — 2}C (x) for all real a

(For example, if P(x) = z(1—z), then (+) = 0+2(1—-2)*+(1—-2)2".)}

12

14,

(1993 IMO) ret fle) = 2" +5a"- + +3, where n > 1 is an integer Prove that f(a) cannot be expressed as a product of two polynomials, each has integer coefficients and degree at least 1

Prove that if the integer a is not divisible by 5, then f(x) = x? —aw@+a

cannot be factored as the product of two nonconstant poly nomials with

integer coefficients

(1991 Soviet Math Olympiad) Given 2n distinct mumbers a1, a2, ,@n,

by, bo, ,b,, an 7 x 7 table is filled as follows: inte the cell in the ¢-th row and 7 -th column is written the number a; + 6; Prove that if the product of each column is the same, then also the product of each row

is the same

Let a4,@9, ,@, and 6, b9, ,6, be two distinct collections of n pos-

itive mbogors where each collection may cortain repet itions If the two

collections of integers a;-+ asl < # < j <S n) and bị +b | A<i<j<n} are the same, then show that n is a power of 2

Recurrence Relations

The sequence z,, is defined by

Prove that 2, 3 4 or 0 for all n and the terms of the sequence are all distinct,

(1988 Nanchang City Math Competition) Define ay = 1, ag = 7 and

2

Gn4iQ = = for positive integer mn Prove that Ga,an4_+ 1 is a

an, perfect square for every positive integer

(Proposed by Bulgaria for 1988 IMO) Define ag = 0,a, == 1 and a, =

20n—-1+y—2 for nm > 1 Show that for positive integer k, a, is divisible

by 2" if and only if m is divisible by ạt

2

1 (American Mathematical Monthly, Problem E2998} Let x and y be

four consecutive positive integers m Show that is an integer

for all positive integers 7

Inequalities

For real numbers a;,@0,d, , if Qn_-7 + @n41 > 2a, for mn = 2,3, ,

then prove that

Any + Anis 2 2Ay, for m= 2,3, , where A,, is the average of a1,a2, ,@n

Let a,b,c > Q and abc < 1 Prove that

Can equality occur?

(1997 IMO shortlisted problem) Let ay > + + > ay,

sequence of real numbers Prove that

¬ -Š

› đị; › VW(Vũy — Vk+1)- k== k=1

(1994 Chinese Team Selection Test} For 0 <a <b<c<d<e and

a+b+e+d+e= 1, show that

(1999 IMO) Let n be a fixed integer, with nm > 2

(a) Determine the least constant C such that the inequality

> 43 (gƒ + +7) < of Ñ x; |

⁄

holds for all nonnegative real numDer§ Z1,22, , an

(b) For this constant C, determine when equality holds

(1995 Bulgarian Math Competition) Let n > 2 and 0 < x; < 1 for i== 1,2, ,n Prove that

where [a] is the greatest integer less than or equal to x

For every triplet of functions f,g,h : [0,1] — R, prove that there are nurabers x,y, z in (0, 1] such that

f{@) + oy) + A(2) ~ eval 2 =

(Proposed by Great Britain for 1987 IMO) Hf x,y, z are real nurnbers

= + 2 2, 2 | a 4© LÔ KV 4” xay ể such that 2 + y* + 2° = 32, then show that ø + + z € #øyz+ 3

4

(Proposed by USA for 1993 IMO) Prove that for positive real numbers

a, b,¢, d,

Let @1,@9, ,@, and by,69, ,5, be 3n positive real numbers such

that

(a) ay > @g >-++ > an and

(b) 6)62°-+ by, > ayag-+- ax for allk,i <k <n

Show that 6; +69 + -+6, >a; +a9+ -+4y

(Proposed by Greece for 1987 IMO) Let a,b,c > 0 and m be a positive

integer, prove that

(1982 West German Math Olympiad) Hf ay,ae, ,a@, > Ganda =

ay ag + + +a, then show that

Prove that abcd > 3

(Due to Paul Erdés) Each of the positive integers a,, ,@,, is less than

1951 The least common multiple of any two of these is greater than

for every nonnegative integer n and all x in (0,1)

eal polynomial tune-

(1996 IMO shortlisted problem) Let P(2) be the

P <1 for all x such

4 in terms of positive integers; namely

EHI+141, 14142, 143, 242, and 4

6

44

Prove that p(n + 1) — 2p(m) + p(n — 1} > 0 for each n > 1

Functional Equations

Find all polynomials f satisfying f(27) + f(x) fla +1) = 0

(1997 Greek Math Olympiad) Let f : (0,00) — & be a function such

that

(a) f is strictly increasing,

(b) f(z) > —+ for all a :> O and

2

(c) ƒ@)ƒŒ(Œ) + $) =1 for all z >0,

Find (1)

(1979 Eétv6s-Kiirschak Math Competition) The function f is defined

for all real numbers and satisfies f(z) <a and fix +y} < fla) + f@)

for all real x,y Prove that f(x) = x for every real number x

(Proposed by Ireland for 1989 IMO) Suppose ƒ : RÑ — Ññ satisfies

y= =1,f(a+b) = ia (a) + f(b) for all a,b € Rand f(2)f(4) = 1 for

ve 0 Show that f(x) = ax for all x

(1992 Polish Math Olympiad) Let QT be the positive rational numbers

Determi ine all functions ƒ : Qt — QT such that ƒ(œ + 1) = ƒ(z)+ 1

and ƒ(”) = f(a)? for every 2 € QT, _——

(1996 IMO shortlisted problem) Let A denote the real numbers and

ƒ:R — [1,1] satisfy

plat te) 4 pay rí»+3 c#Íz+2

Ì ấ + —- {17 == j 4 +- Fo j ấ +

for every x € R Show that f is a periodic function, i.e there is a

nonzero real number 7 such that f(@ +7) = f(x) for every x € Re

Let N denote the positive integers Suppose s : N —» N is an increasing

function such that s(s()) = ẩn for all n € N Find all possible values

of s(1997)

4i

Let N be the positive integers Is there a function f : N -+ N such that

0996) m„) = In for all n € N, where fO(x) = fle) and ft) (a) =

fsa)?

1à (American Mathematical Monthly, Problem E984) Let A denote tfÌ

real numbers Find all functions f : R - R such that f(f(@)) = 2? -3

or show no such function can exist

Let & be the real numbers Find all functions f : R —- A such that for all real mumnbers w and y,

flatly) +2) = vy + fla)

(1999 IMO) Determine all functions f: R-—- A such that

f@— fy) = ƒŒ()) + z/@0) + ƒ0ø) — 1

for all z,y in R

(1995 Byelorussian Math Olympiad) Let A be the real numbers Find

all functions f : ?# —› A such that

ƒŒ@Œœ +ø)) = ƒŒœ +1) + ƒ()ƒ/G@) — #0 for all z,y € R

(1993 Czechoslovak Math Olympiad) Let Z be the integers Find all functions f : Z — Z such that

f(-ij= fG) and f(z) + fly) = fle + 2ry) + fly — 2xy) >

for all integers x, y

(1995 South Korean Math Olympiad) Let A be the set of non-negative integers Find all functions f : A —+ A satisfying the following two conditions:

(a) For any m,n € A, 2f(me tn

oT

(b) For any m,n € A with m >n, ƒ(mn2) > ƒ(n)

ot D

(American Mathematical Monthly, Pre yblem £2176) Let Q@ denote

rational numbers Find all functions f : Q — Q such that

‘ety flr) + fly) FQ) =2 and f cu) 7) = fin) + Fy) for a 4 y, fle)—fy)

(Mathematics Magazine, Problem 1552) Find all functions f: R— &

What is the minimum of f(n + 1)/f(n}?

(1996 Putnam Exam) Given that {74,%9, ,0@,$ = {1,2, ,n}, find

the largest possible value of wyxq+ ro%e+ e+ + y_ 1p, 8n in terms

(1988 Leningrad Math Olympiad) Squares ABDE and BCG are

drawn outside of triangle ABC’ Prove that triangle ABC is isosceles if

DG is parallel to AC

AB is a chord of a circle, which is not a diameter Chords A,B; and

Ag Bo intersect at the midpoint P of AB Let the tangents to the circle

at A, and & intersect at Cy) Similarly, let the tangents to the circle

at Ao and &o intersect at Co Prove that Cy)Co is parallel to AB

(1991 Hunan Province Math Competition) Two circles with centers Oy and Og intersect at points A and B A line through A intersects the

circles with centers OQ; and Og at points Y,4, respectively Let the tangents at Y and Z intersect at X and lines YO; and 7Opo intersect

at P Let the circumcircle of A©,OoB have center at C and intersect

line XB at B and Q Prove that PQ is a diameter of the circumcircle

of LO VOOR

(1981 Beijing City Math Competition) In a disk with center O, there are four points such that the distance between every pair of them is greater than the radius of the disk Prove that there is a pair of per- pendicular diameters such that exactly one of the four points lies inside each of the four quarter disks formed by the diameters

The lengths of the sides of a quadrilateral are positive integers The length of each side divides the sum of the other three lengths Prove that two of the sides have the same length

(1988 Sichuan Province Math Competition} Suppose the lengths of the three sides of AABC are integers and the inradius of the triangle is 1 Prove that the triangle is a right triangle

10

Geometric Equations

64 (1985 IMO) A circle has center on the side AB of the cyclic quadri-

lateral ABCD The other three sides are tangent to the circle Prove

that AD+ BC = AB

65 (1995 Russian Math Olympiad) Circles S$; and Sg with centers O1, Oz

espectively intersect each other at points A and B Ray O; B intersects

So at point & and ray CoB intersects 5; at point # The line parallel

to EF and passing through Ø intersects S$; and So at points M and

N, respectively Prove that (B is the incenter of AFAF and) MN =

AB + AF

66 Point C hes on the minor arc AB of the circle centered at O Suppose

the tangent line at C cuts the perpendiculars to chord AB through A

at & and through B at F Let D be the intersection of chord A& and

radius OC Prove that CE -CF = AD-BD and CD? = AE - BF

67 Gnadrilaterals ABCP and A’B'C'P’ are inscribed in two concentric

circles Tf triangles ABC and A’B’C" are equilateral, prove that

PA? + PIB? 4 P'C? = PA? + PB? + PC”

68 Let the inscribed circle of triangle ABC touchs side BC at D, side C'A

at & and side AB at F Let G' be the foot of perpendicular from D to

EF Show that PC _ BE

EF Show that EG =, CE

69 (1998 IMO shortlisted problem) Let ABCDEF be a convex hexagon

Let tP OF R be the points of intersection of AC and BD ( CE and DF

HA and FB respectively Prove that triangles PQR and BDF are similar

(1998 IMO shortlisted problem} Let ABC'D be a cyclic quadrilateral

Let & and & be variable points on the sides 4B and CD, respectively,

such that AB: BB = CF: FD Let P be the point on the segment

EF such that PE: PF = AB:CD Prove that the ratio between the areas of triangles APD and BPC does not depend on the choice of £ and #2

Tangent Lines

Two circles intersect at points A and 6 An arbitrary line through B intersects the first circle again at C and the second circle again at D The tangents to the first circle at C and to the second circle at D

intersect at M4 The parallel to CAf which passes through the point,

of intersection of AM and CD intersects AC at K Prove that BK is tangent to the second circle

(1999 IMO) Two circles Dy; and F2 are contained inside the circle LP, and are tangent to [ at the distinct points Af and N, respectively I; passes through the center of Pg The line passing through the two points of intersection of Fy and [2 meets (at A and B, respectively The lines MA and M16 meets [y at C and 2, respect ively Prove that

CD is tangent to Po

(Proposed by India for 1992 IMO) Circles Gy and Ge touch each other externally at a point W and are inscribed in a circle G A,B,C are

12

points on G such that A, G; and G's are on the same side of chord BC,

which is also tangent to G; and Go Suppose AW is also tangent to

Gy, and Ga Prove that W is the incenter of triangle ABC

LOCUS

Perpendiculars from a point P on the circumcircle of AABC are drawn

to lines AB, BC with feet at D,&, respectively Find the locus of the

circumcenter of APDE as P moves around the circle

Suppose A is a point inside a given circle and is different from the

center Consider all chords (excluding the diameter) passing through

A What is the locus of the intersection of the tangent lines at the

endpoints of these chords?

Given AABC Let line EF bisects ZBAC and AH AF = AB AC

Find the locus of the intersection P of ines BE and CF

(1996 Putnam Exam) Let Cy and Co be circles whose centers are 10

units apart, and whose radii are 1 and 3 Find the locus of all points

M for which there exists points X on Cy and Y on Co such that M is

the midpoint of the ne segment XY

Collinear or Concyclic Points

(1982 IMO) Diagonals AC’ and CF of the regular hexagon ABCDEF

are divided by the inner points A? and N, respectively, so that

AC CE —

Determine 7 if B, Mand N are collinear,

(1965 Putnam Exam} ff A, B,C,D are four distinct points such that

every circle through A and & intersects or coincides with every circle

through C and D, prove that the four points are either collinear or

coneyclic,

13

(1957 Putnam Exam) Given an infinite number of points in a plane,

prove that if all the distances between every pair are integers, then the points are collinear

(1995 IMO shortlisted problem) The incircle of triangle ABC touches BC,CA and AB at D,E and F respectively Ä is a point inside triangle ABC such that the incircle of triangle X BC touches BC at

D also, and touches CX and X 8 at Y and Z respectively Prove that LFZY is a cyclic quadrilateral

(1998 IMO) In the convex quadrilateral ABC'D, the diagonals AC and

GBD are perpendicular and the opposite sides AB and DC are not

parallel Suppose the point P, where the perpendicular bisectors of

AB and DC meet, is inside ABCD Prove that ABCD is a cyclic quadrilateral if and only if the triangles ABP and CDP have equal areas,

(1970 Putnam Exam) Show that if a convex quadrilateral with side-

lengths a,6,c,d and area V/ ‘abcd has an inscribed circle, then it is a

cyclic quadrilate toral

(1990 Chinese National Math Competition) Diagonals AC and BD

of a cyclic quadrilateral ABCD meets at P Let the circumecenters of

ABCD, ABP, BCP, CDP and DAP be O, 0), O02, O03 and O;, respec- tively Prove that OP, O,O03, OgQ4 are concurrent

(1995 IMO} Let A, B,C and D be four distinct points on a line, in that order The circles with diameters AC and BD intersect at the points

X and Y The line XY meets BC at the point Z Let P be a point on the line XY different from Z The line CP intersects the circle with

14

90

91

93

diameter AC’ at the points C and Ã/, and the line BP intersects the

cirele with diameter BD at the points B and N Prove that the hnes

AM,DN and XY are concurrent

AD, BE,CF are the altitudes of AABC Y P,@,R are the midpoints

ot DE, EF, FD, respectively, then show that the perpendicular from +

Đ,Q,Rto AB, BƠŒ,CA, respectively, are concUrrent

(1988 Chinese Math Olympiad Training Test) ABC DEF is a hexagon

inscribed in a circle Show that the diagonals AD, BE,CF are concur-

rent ifand only if AB- CD HF = BC.DE FA,

A circle intersects a triangle ABC at six points Ay, As, By, Bo, Cy, Co,

where the order of appearance along the triangle is A, Cy, Co, B, Ay, 4a,

C', By, Bo, A Suppose #,C,, BoCo meets at X, Cy Ai, CoAq mects at

Y and A;.8,, Ao Be meets at Z Show that AX, BY,CZ are concurrent

(1995 IMO shortlisted problem) A circle passing through vertices B

and C of triangle ABC intersects sides AB and AC at C’ and 8B’,

respectively Prove that BB’,CC’ and HH’ are concurrent, where H

and 7 are the orthocenters of triangles ABC and AB’C’, respectively

Perpendicular Lines

(1998 APMO) Let ABC be a triangle and D the foot of the altitude

from A Let # and F be on a line passing through D such that AF

is perpendicular to BE, AF is perpendicular to CF, and & and F are

different from 0 Let M and N be the midpoints of the line segments

BC and EF, respectively Prove that AN is perpendicular to NAZ,

(2000 APMO) Let ABC be a triangle Let Af and N be the points

in which the median and the angle bisector, respectively, at A meet

the side BC Let @ and P be the points in which the perpendicular at

N to NA meets AVA and BA, respectively, and © the point in which

the perpendicular at P to BA meets AN produced Prove that QO is

perpendicular to BC

15

100

101

Lẹt ĐH and CC? be altitudes of triangle AC Assume that 4Ö #

AC, Let M be the midpoint of BC, H the orthocenter of ABC and D the intersection of B’C’ and BC Prove that DH L AM,

(1996 Chinese Team Selection Test) The semicircle with side BC of AABC as diameter intersects sides AB, AC at points D,F, respec- tively Let #\G be the feet of the perpendiculars from D,E to side

BC respectively Let Af be the intersection of DG and EF Prave that,

AM 1 BC

(1985 IMO) A circle with center O passes through the vertices A and

C of triangle ABC’ and intersects the segments AB and AC again at distinct points WK and N, respectively The circumcircles of triangles ABC and K BN intersect at exactly two distinct points B and M Prove that OMS 1 MB

(1997 Chinese Senoir High Math Competition} A circle with center O

is internally tangent to two circles inside it at points S and 7 Suppose

the two circles inside intersect at Mf and N with N closer to ST Show that OM | MN if and only if S,N,7 are collinear

AD, BE,CF are the altitudes of AABC Lines BF, FD, DFE meot lines BO,CA, AB in points 0, M,N, respectively Show that 0, W,N are 1 ? ? ? ? a 1 ? collinear and the line through them is perpendicular to the line joining the orthocenter 4 and circumcenter O of AARC

Geometric Inequalities, Marimum/Minamum

(1973 EMO) Let Pi, Po, , Pons, be distinct points on some half of the unit circle centered at the origin QO Show that

Gi +Ö + - Ø7] >1

Let the angle bisectors of 2A,2B,ZC of triangle ABC intersect its

circumceircle at P,Q, R, respectively Prove that

AP+ BQ4+CR> BC+CA4 AB

16

(1997 APMO) Let ABC be a triangle inscribed in a circle and let l, =

Ma/Ma,ly = mp/My le = me/M., where m,,ms, 7m are the lengths

of the angle bisectors (internal to the triangle) and A4,, My, M, are

the lengths of the angle bisectors extended until they meet the circle

ae fe mee fe nee DB,

sin“ A sin’ B sin* C

and that equality holds iff ABC is equilateral

(Mathematics Magazine, Problem 1506) Let J and O be the incen-

ter and circumecenter of AABC, respectively Assume AABC is not

equilateral (so f 3 O) Prove that

LAIO <90° ifand only if 2BC< AB+CA.,

Squares ABDE and AC FG are drawn outside AAGC Let P,Q be

points on #G such that BP and C'@ are perpendicular to BC Prove

that BP+CQ> BƠ + £#G When does equality hold?

Point P is inside AABC Determine points D on side AB and F on

side AC such that BD = CE and PD-+ PE is minimum

Solid or Snace Geometry

(Proposed by Italy for 1967 IMO) Which regular polygons can be ob-

tained {and how) by cutting a cube with a plane?

(1995 Israeli Math Olympiad) Four points are given in space, in general

position (.e., they are not coplanar and any three are not collinear)

A plane 7ø is called an equalizing plane if all four points have the same

distance from 7 Find the number of equalizing planes

(1956 Putnam Exam) Prove that every positive integer has a multiple

whose decimal representation involves all ten digits

Does there exist a positive integer a such that the sum of the digits

(in base 10) of a is 1999 and the sum of the digits (in base 10) of a® is

199927 (Proposed by USSR for 1991 IMO) Let a, be the last nonzero digit

in the decimal representation of the number n! Does the sequence

&4,02, ,@,, become periodic after a finite number of terms?

Modulo Arithmetic

(1956 Putnam Exam) Prove that the number of odd binomial coeff-

cients in any row of the Pascal triangle is a power of 2

Let @1,@9,43, ,@ 4, and 5;, bo, b3, ,b1, be two permutations of the

natural numbers 1,2,3, ,11 Show that if each of the numbers a,6;,

agbo, a3h3, ,44404, is divided by 11, then at least two of them will have the same remainder

(1995 Czech-Slovak Match) Let a;,a2, be a sequence satisfying a, =

2, Gg = 5 and

Ong = (2-1 Jong + 2+ 07 )an

for all > 1 Do there exist indices p,g and r such that quaa = a;?

Prime Factorization

(American Mathematical Monthly, Problem E2684) Let A, be the set

of positive integers which are less than m and are relatively prime to n For which > 1, do the integers in A,, form an arithmetic progression?

18

(1971 IMO) P e that the set of integers of the form 2* — 3 (k =

2,3, ) con tains ns an infinite subset in which every two members are

relativ ely prime

(1988 Chinese Math Olympiad Training Test} Determine the smallest ¬

value of the natural number 7m > 3 with the property that whenever

the set S,, = {3,4, ,n} is partitioned into the union of two sub-

sets, at least one of the subsets contains three mumbers a, 6 and c (not

necessarily distinct) such that ab = c

Base n Representations

(1983 IMO) Can you choose 1983 pairwise distinct nonnegative integers

less than 10° such that no three are in arithmetic progression?

‘American Mathematical Monthly, y Problem 2486) Let p be an odd

prime number and r be a positive integer not divisible by p For any

positive integer k, show that there exists a positive integer m such that

the rightmost & digits of m”, when expressed in the base ø, are al 1s

(Proposed by Romania for 1985 IMO) Show that the sequence {a,,

defined by a,, = [n/2} for n = 1,2,3, (where the brackets denote

the greatest integer function) contains an infinite number of integral

powers of 2

Representations

Find all (even) natural numbers n which can be written as a sum of

two odd composite numbers

Find all positive integers which cannot be written as the sum of two

or more consecutive positive integers

(Proposed by Australia for 1990 IMO) Observe that 9 = 445 == 243+4

Is there an integer N which can be written as a sum of 1990 consecutive

positive integers and which can be written as a sum of (more than one)

consecutive integers in exactly 1990 ways? ey

ay, is called guadratic ÏŸ for cách ¡ € {1,2, ,n}, lai = ¡| = Ê

(a) Prove that for any two integers 6 and c, there exists a natural

number m and a quadratic sequence with ag = 6 and a, =6

(b) Find the least natural number n for which there exists a quadratic

sequence with ag = 0 and a,, = 1996

Prove that every integer greater than 17 can be represented as a sum of

three integers > | which are pairwise relatively prime, and show that

17 does not have this property

Chinese Remainder Theorem (1988 Chinese Team Selection Test) Define w,, == 32,1 + 2 for all positive integers n Prove that an integer value can be chosen for xg so that 2109 1s divisible by 1998

(Proposed by North Korea for 1992 IMO) Does there exist a set AL

with the following properties:

(a) The set Af consists of 1992 natural numbers

(b) Every element am M and the sum of any number of elernents in Af

have the form m”, where m,k are positive integers and k > 2?

Divisibilaty

1 Find all positive integers a,b such that 6 > 2 and 2° + 1 is divisible by 2° — 1

20)

Prove that there are infinitely many positive integers n such that 2”-+Í

is divisible by n Find all such n’s that are prime numbers

(1998 Romanian Math Olympiad) Find all positive integers (a,n) such

that 2” +2” + 1is a divisor of 2+! 4274) 4 1

(1995 Bulgarian Math Competition) Find all pairs of positive integers

%1 x,y) for which is an integer and divides 1995 =

(1995 Russian Math Olympiad) Is there a sequence of natural numbers

in which every natural number occurs just once and moreover, for any

& = 1,2,3, the sum of the first & terms is divisible by &?

(1998 Putnam Exam) Let Ay = 0 and Ay = 1 For n > 2, the number

A, is defined by concatenating the decimal expansions of 4,— 1 and

Ay —2 from left to right For example, Ag = AoA; = 10,Aq = AgAo =

101, As =: AqAg == 10110, and so forth Determine all nm such that A,,

is divisible by 11,

(1995 Bulgarian Math Competition) If k > 1, show that k does not

divide 2°~! + 1 Use this to find all prime numbers p and g such that

2? + 29 is divisible by pq

Show that for any positive integer », there is a mumber whose decimal

representation contains 7 digits, each of which is 1 or 2, and which is

divisible by 2”

For a positive integer n, let f(m) be the largest integer k such that 2*

divides 7 and g{n} be the sum of the digits in the binary representation

of 7 Prove that for any positive integer n,

(a) x and y are relatively prime;

(1972 Putnam Exam) Show that ifm is an integer greater than 1, then

n does not divide 2” — |,

(Proposed by Romania for 1985 IMO) For k > 2, let rị,ma, ,ay be positive integers such that

whose sum is divisible by 18

Perfect Squares, Perfect Cubes

Ne

(1998 Putnam Exam) Prove that, for any integers a,b,c, there exists a geen

positive integer n such that Vn? + an? +bn+c is not an integer

(1995 IMO shortlisted problem) Let & be a positive integer Prove that,

there are infinitely many perfect squares of the form n2” — 7, where n

iS & positive integer

Find all sets of positive integers x,y and 2 such that « <y < z and

(Due to W Sierpinski in 1955) Find all positive integral solutions of

BY 4 49 = 57%,

(Due to Euler, also 1985 Moscow Math Olympiad} in > 3, then prove

that 2° can be represented in the form 2” = 7x7 + y? with z,y odd

positive integers

(1995 IM 0 shortlisted problem) Pind all positive integers x and y such

that 2 + y° + 2° = xyz, where z is the greatest common divisor of x

(1995 Czech-Slovak Match) Find all pairs of nonnegative integers ø and

y which solve the equation p” — y? == 1, where p is a given odd prime

Find all integer solutions of the systerm of equations

‹ - 3, „3 rou+ez=ä3 and z? +?) +z/ =3

(1996 Italian Mathematical Olympiad) Given an alpha bet with three letters a,b,c, find the number of words of n letters which contain an even number of a’s

two neighbors can differ by a t most L (1995 Romanian Math Olympiad) Let Ay, A4o, ,A, be points on a

circle Find the number of possible colorings of these points with p

colors, p > 2, such that any two neighboring points have distinct colors

Pigeonhole Principle

(1987 Austrian-Polish Math Competition} Does the set {1,2, , 3600}

contain a subset A consisting of 2000 numbers such that € 4 implies 2x ¢ A?

(1989 Polish Math Olympiad} Suppose a triangle can be placed inside

a square of unit area in such a way that the center of the square is not

inside the triangle Show that one side of the triangle has length less than 1

The cells of a 7 x 7 square are colored with two colors Prove that,

there exist at least 21 rectangles with vertices of the same color and with sides parallel to the sides of the square

For n > 1, let 2n chess pieces be placed at the centers of 2n squares of

an nxn ch essboard Show that there are four pieces among them that,

formed the vertices of a parallelogram If 2n is replaced by 2n — 1, is

the statement still true in general?

The set {1,2, ,49} is partitioned into three subsets Show that at least one of the subsets contains three different murmmbers a,b,c such

that @a+tb=e

24

Tmeciusiom- Emciuston Principle

Let m > > 0 Find the number of surjective functions from B,, =

{1,2, ,m} to B, = {1,2, ,n}

Let A be a set with & elements Find the maximal mumber of 3-element

subsets of A, such that the intersection of any two of them is not a 2-

element set

(a) (1999 China Hong Kong Math Olympiad) Students have taken a

test paper in each of nm (2 > 3) subjects [t is known that for any

subject exactly three students get the best score in the subject, and

for any two subjects excatly one student gets the best score in every

one of these two subjects Determine the smallest n so that the above

conditions imply that exactly one student gets the best score in every

one of the m subjects

(b) (1978 Austrian-Polish Math Corpetition) There are 1978 clubs

Fach has 40 merbers If every two clubs have exactly one common

member, then prove that all 1978 clubs have a common member

Combinatorial Designs

(1995 Byelorussian Math Olympiad} In the begining, 65 beetles are

placed at different squares of a 9 x 9 square board In each move, every

beetle creeps to a horizontal or vertical adjacent square If no beetle

rnakes either two horizontal moves or two vertical moves in succession,

show that after some moves, there will be at least two beetles in the

same square

1995 Greek Math Olympiad) Lines /,,lo, ,d, are on a plane such } t2) ) +

that no two are paraHel and no three are concurrent Show that we

can label the Cy intersection points of these lines by the numbers

1,2, ,k — Í so that in each of the lines /,,/5, ,/, the numbers

1,2, ,4—1 appear exactly once if and only if & is even

om 1996 Tournaments of the Towns) In a lottery game, a person must,

six distinct numbers from 1,2,3, ,36 to put on a ticket The

9 tickets guaranteeing at least a winning ticket, but 8 tickets is not enough to guarantee a winning ticket in general

(1995 Byelorussian Math Olympiad) By dividing each side of an equi-

lateral triangle into 6 equal parts, the triangle can be divided into 36 smaller equilateral triangles A beetle is placed on each vertex of these triangles at the same time Then the beetles move along different edges

with the same speed When thev get to a vertex, they must make a 60° or 120° turn Prove that at some moment two beetles must meet

at some vertex Is the statement true if 6 is replaced by 5?

Covering, Conver Hull

(1991 Australian Math Olympiad) There are n points given on a plane such that the area of the triangle formed by every 3 of them is at most

i Show that the n points lie on or inside some triangle of area at most

three times the radius

(1995 IMO} Determine all integers n > 3 for which there exist n points

Ay, Ag, , An in the plane, and real numbers ry,7re, ,7,, satisfying the following two conditions:

) no three of the points Ay, Ae, ,Ap lie on a line;

(B) for each triple i,j,k (A <i<j<k <n) the triangle A;A;A;, has

area equal tor; +rj +7

26

177 (14999 IMO} Determine all finite sets S$ of at least three points in the

plane which satisfy the following condition: for any two distinct points

A and B in S, the perpendicular

(1995 Russian Math Olympiad) There are n seats at a merry-go-around

A boy takes 7 rides Between each ride, he moves clockwise a certain number (less than Ø0) of places to a new horse Each time he moves a different number of places Find all n for which the boy ends up riding each horse

(1995 Israeli Math Olympiad) Two players play a game on an infinite board that consists of 1 x 1 squares Player I chooses a square and

marks it with an O Then, player IT chooses another square and marks

it with X They play until one of the players marks a row or a column

of 5 consecutive squares, and this player wins the game If no player

can achieve this, the game is a tie Show that player Hf can prevent,

player I from winning

(1995 USAMO) A calculator is broken so that the only keys that still

work are the sin, cos, tan, sin™?,cos7!, and tan~! buttons The dis- play initially shows 0 Given any positive rational number g, show that

pressing some finite sequence of buttons will yield q Assume that the

calculator does real number calculations with infinite precision All

functions are in terms of radians

(1977 Eotvés-Kiirschak Math Competition) Each of three schools is attended by exactly n students Each student has exactly n+ 1 ac-

quaintances in the other two schools Prove that one can pick three students, one from each school, who know one another It is assumed that acquaintance is mutual

Is there a way to pack 250 1 x I « 4 bricks into a 10 x 10 x 10 box?

Is it possible to write a positive integer into each square of the first, quadrant such that each column and each row contains every positive

integer exactly once?

There are 1 identical cars on a circular track Among all of them, they have just enough gas for one car to complete a lap Show that there is >

28

a car which can complete a lap by collecting gas from the other cars

on its way around the track in the clockwise direction

(1996 Russian Math Olympiad) At the vertices of a cube are written

eight pairwise distinct natural numbers, and on each of its edges is

written the greatest common divisor of the numbers at the endpoints

of the edge Can the sum of the numbers written at the vertices be the

same as the sum of the numbers written at the edges?

Can the positive integers be partitioned into infinitely many subsets

such that each subset is obtained from any other subset by adding the

same integer to each element of the other subset?

(1995 Russian Math Olympiad) Is it possible to fill in the cells of a

9x 9 table with positive integers ranging g¢ from 1 to 81 in such a way

that the sum of the elernents of every 3 x 3 square is the same?

(1991 German Mathematical Olympiad) Show that for every positive

integer n > 2, there exists a permutation pi,po, ,pn of 1,2, ,?

such that prii divides py + po+ -+p, for k= 1,2, ,n—1

Each lattice point of the plane is ý lap eled by a positive integer Each

of these numbers is the arithmetic mean of its four neighbors (above,

below, left, right) Show that all ‘the numbers are equal

(1984 Tournament of the Towns) [n a party, m boys and n girls are

paired It is observed that in each pair, the difference in height is less

than 10 cm Show that the difference in height of the k-th tallest boy

and the k-th tallest girl is also less than LŨ em for k= 1,2, ,7

(1991 Leningrad Math Olympiad) One may perform the following two

operations on a positive integer

(a) raultiply it by any positive integer and

(b) delete zeros in its decimal representation

, one can perform a sequence of these operations that will transform X to a one-digit number

Prove that for every positive integer X,

(1996 IMO shortlisted problem) Four integers are marked on a circle

On each step we simuli taneousÌy replace each number by the difference between this number and next number on the circle in a given direction (that is, the numbers a,b, c,d are replaced by a —b,b—c,c—d,d—a)

Is it possible after 1996 such steps to have numbers a,b,c, d such that

the numbers | , MẠC — , lak are primes?

(1989 Nanchang City Math Competition) There are 1989 coins on a table Same are placed with the head sides up and some the tail sides

up A group of 1989 persons will perform the following operations:

the first person is alowed turn over any one coin, the second person is allowed turn over any two coins, ., the k-th person is allowed turn over anv & coins, ., the 1989th person is allowed to turn over every

coin Prove that (1) no matter which sides of the coins are up initially, the 1989 persons can come up with a procedure turning all coins the same sides up

at the end of the operations, (2) in the above procedure, whether the head or the tail sides turned

up at the end will depend on the initial placement of the coins (Proposed by India for 1992 IMO) Show that there exists a convex polygon of 1992 sides satisfying the following conditions:

a) its sides are 1,2,3, ,1992 in some order;

(b) the polygon is circumscribable about a circle

There are 13 white, 15 black, 17 red chips on a table In one step, you

may choose 2 chips of different colors and replace each one by a chip of

the third color Can all chips become the same color after some steps?

The following operations are permitted with the quadratic polynomial

¬ an” + ba +:

(a) switch a and c, (b) replace x by « +4, where ¢ is a real mumber

197

198

199

200

Five numbers 1,2,3,4,5 are written on a biackboard A student may

erase any two of the numbers a and b on the board and write the

numbers a+6 and ab replacing them If this operation is performed re-

peatedly, can the numbers 21, 27, 64, 180, 540 ever appear on the board? Nine 1 « 1 cells of a 10 « 10 square are infected In one unit time, the cells with at least 2 infected neighbors (having a common side) become infected Can the infection spread to the whole square? What if nine

is replaced by ten’?

(1997 Colombian Math Olympiad) We piay the following game with

an equilateral triangle of n(n + 1)/2 dollar coins (with n coins on each side) Initially, all of the coins are turned heads up On each turn, we may turn over three coins which are mutually adjacent; the goal is to

make all of the coins turned tails up For which values of n can this be

done?

(1990 Chinese Team Selection Test) Every integer is colored with one

of 100 colors and all 100 colors are used For intervals ja, |, |c, d] having

integers endpoints and same lengths, if a,c have the same color and

b,d have the same color, then the intervals are colored the same way,

which means a+« and c+2z have the same color for x = 0,1, ,5-a

Prove that —1990 and 1990 have different colors

3]

Solutions

Solutions to Algebra Problems

(Crux Mathematicorum, Problem 7} F ind ( (without calculus) a fifth

degree polynomial p(x) such that p(x) + 1 is divisible by (~7 — 1)? and

p(x) — 1 is divisible by (a + 1)°

Solution (Due to Law Ka Ho, Ng Ka Wing, đạm Siu Lung} Note

(x — 1)° divides p(x} + 1 and p(—x) — 1, so (2 —1)8 divides their sum

p(t) +p(—x) Also (+ 1)8 divides p(x) — 1 and p(—a) +1, so (x +1)

divides p(x) +p(—a Then (z—1)?(z-+-1) divides p(x)-+p(—2), which is

of degree at roost 5 So p(x}+p(—x) = 0 for all x Then the even degree

term coefficients of p(x) are zero Now p(a)+1 = (v—1)° (Ar? +Ba-l)

_ the degree 2 and 4 coeficienis, we gọt # — 3A = 0 and

3+ 3B — A = 0, which bes A= —3/8 and B= 9/8 This yields

A polynomial P(x} of the n-th degree satisfies P(k) = 2° for k =

0,1,2, , Find the value of P(n + 1)

By the binomial theorem, Q(k) = +1)" = 2* for k = 0,1,2, 7

So P(x) = Q(x) for all x Then

Pint)) = G(n+1) = “ )+ (" ' ).+ A ) =2"! =1,

\

1999 Putnam Exam) Let P(x} be a polynomial with real coefficients } ey y

such that P(x) > 0 for every real x Prove that

Pla) = fila)’ + fole? + + fe

30

for some polynomials fi (a), fo(x), , fa{v) with real coefficients

+ is the coefficient of the highest degree term, R( (x) § is s the anoduet of all

real root factors (c—r) repeated according to multiplicities and C(x) is

the product of all conjugate pairs of nonreal root factors (x—- 2, )(a—ZE} Then a > 0 Since P(x) > 0 for every real z and a factor (2 — r}?"+1 would change sign near a real root r of odd multi sàn each real root

of P must have even multiplicity So R(x) = f(x)? for some polynomial f(x) with real coefficients

Next pick one factor from each conjugate pair of nonreal factors

and let the product of these factors (a — z,) be equal to U(x) + iV (2), where U(x), V(x} are polynomials with real coefficients We have

P(x} = af (2)? (U(x) + 2V (x) )(U (a) — ¡V())

= (Vøƒf(œ)U())° + (Wa†(œ)V(a))?

or (1995 Russian Math Olympiad) Is it possible to find three quadratic

polynomials f(x}, g(a), h(x) such that the equation ƒ(ø(h(z))) = 0 has

the eight roots 1, 2,3, 4, 5,6, 7,8? £

Solution Suppose there are such ƒ, ø, h Then h(1),b(2), ,h(8) wil

be the roots of the 4-th degree polynomial f(g(2)) Since hia) =

h(b),a # 6 if and only if a,b are symmetric with respect to the axis

of the parabola, it follows that ACL) = h(8),h(2) = h(Œ7),h(3) =

ACG), hf ty =: A{5) and the parabola y = Me is symmetric with re- spect to a = 9/2 Also, we have either h(1) < h{23) < h(3) < A(A) or ACL) > (2) > A(3) > A(4)

Now g(h(1)), g(h(2)), g(h5(8)), g(5(8)) are the roots of the quadratie polynomial fle), so 0(h(1)) —= g(h(4)) and ø g(h(2 = g(h(3)), which implies h(1)-+A(4) = A(2)-+h(3) For h(x) = Ax? + Br +C, this would

force A == 0, a contr radiction

(1968 Putnam Exam) Determine all polynomials whose coefficients are all +1 that have only real roots

36

Solution Ha polynomial agz” +a,2""'+ -+a,, is such a polynomial,

then so is its negative Hence we may assume ag = 1 Let ry, 7, be

the roots Then r? + -+r2 = af —2ae and rj -r? = a2 If the roots

are all real, then by the AM-GM inequality, we get (aj —2ag}/n >a in

Since @),@g == -£1, we must have ag = —i and n < 3 By simple

checking, we get the list

+ =1), t@+)), +(e? +e2-1), tla? -2x-1),

+(#° +ø2 m1), +e? - 2? -2 41)

(1990 Putna

nonzero real mumbers such that for n =

n Exam) Is there an infinite sequence @9,G1,@2, of

1,2,3, , the polynomial

P, (a) = ag +ay@ + a927 +-+-++a,20" has exactly n distinct real roots?

Solution Yes ‘Take ag == l,a, == ~—1 and proceed by induction

Suppose ag, ,@, have been chosen so that F,,{«) has n distinct real

roots and P, (a) > co or oo as @ -+ co depending upon wheth ern

is even or odd Suppose the roots of P,,(z)} is in the interval (—T,T)

Let đuaai = (-1)"t!/44, where M is chosen to be very large so that

T"*}/M is very small Then P,44(2} = inte) 4 + (=z)??)/ /M is very

close to Ø„,(z) on |~T,T} because |P,44(2) = Ð,(+)| šs 5 PPM for

every « on [-T,T] So, Đ„ua( x} has a sign change very close to every

root of „(z) and has the same sign as P, (x2) at T Since Ð,(z) and

Py4i(e) take on different sign when 2 — oo, there must be another

sign change beyond T So P,.,(a) must have n+ 1 real roots

(1991 Austrian-Polish Math Competition} Let P(2} be a polynomial

with real coefficients such that P(x) > 0 for 0 < x < L Show that

there are polynomials A(x), B(x), C(x) with real coefficients such that

(a) A(œ) > ' Bla) 2 0,C(x) > 0 for all real x and

(b) Pla) = 4Œ) +zB(ø) + (L— z)C(z) for all real x

(For example, if {z) = z(—z}), then PŒ) = 0+z(1—z)?r(—=z)22.)

Solution (Below all polynomials have real coefficients.) We induct

on the degree of P(x) If P(x) is a constant polynomial c, then c > 0

oF

nm case is true For the case Pir) is 5 of degree 1 n Lh If f Pla ) > for

all real x, then simply let A(z) = P(x), B(z) = C(x) = 0 Othe erwise, P(x) has a root % in (—20, 0) or 1, +00)

{= — tl (Then bts = TH ân (x) and Q(z) is of degree n

where the polynomials A(x), B(v)}, C(x) > 0 for all x in [0, 1] `

Case Zp in |1 +) Consider Q(z) = P1 — x) This reduces to the

Drevious case WVe have Q2) = Ai() +œBt() + (T— #)C (+), where

the polynomials Ai(z), 5¡(2),C:(z) >0 for all ø in 1.1], Then c H HỆ

‹

— 2) By ~ 2),

Snare, oman! C(x)

P(x) = QU —2) = A\4—2) 420,01 —2) +01

mm

A fap A(z)

where the polynomials A(ø), Ð(z),C(z) > 9 for aH z in |0, 1]

(1993 IMO) Let ƒ(œ) = ø” + 5z"”! +3, where ø > I is an integer

Prove that f(x) cannot be expressed as a product of two polynomials,

each has integer coefficients and degree at least 1

Solution Suppose f(a) == b{x)c(x) for nonconstant polynomials (x)

and c(z) with integer coefficients Since f(0) = 3, we may: BSSUINe

b(0) = +l and b(z) = 2” - + 1, Sinee ƒŒ1) z O,r > 1 Let

21, ,2, be the roots of b(z) Then |zy + 2.) = = 1 and

-—- |(—5 _—— Z1) wae (—5 — zy} — I Hiện (2; +- 5} oc 237 > Gg

t=)

38

ii

However, b{(—5} also divides f{—5} = 3, a contradiction

Prove that if the integer a is not divisible by 5, then f(@) = a —ax+a

cannot be factored as the product of two nonconstant polynomials with

integer coefficients

Solution Suppose f can be factored, then ffm) = (@ — b)g(x) or

ia) = = (z?—bz +c)g() In the fomer case, bŠ —b-+a = f(b) = 9 Now

b” = b (mod 5) by Fermat s little theorem or simply checking the cases

== 0,1, 2,3,4 (mod 5) Then 5 divides p — bP = a, a contradiction In

the latter case, thethes ƒ G =e -eta by œ2 — bạ + c, we get the

remainder (b† + 302e + đc? — 1) + (be + 2be? +a} Since x* — br +c is

a factor of f(x), both cvcfficionts equal 0 Finally,

c —1)

implies 3a = 6° — b — Sbc* is divisible by 5

by 5, a contradiction

0 = bí? + 30 e+ — 3(bÖe + 2be? + a) = bŠ — b— 5be? — 3a

Then a would be divisible

(1991 Soviet Math Olympiad) Given 2n distinct mumbers a1, a2, ,@n,

by, bo, ,b,, an 7 x 7 table is filled as follows: inte the cell in the ¢-th

row and j-th column is written the number a; + 6; Prove that if the

prowess of each column is the same, then also the product of each row

the same

Solution Let

P(x) = xT—— 8 ay )(a@ + @g})-+- (a + a,) — (a — by )(a — bg) + + (@ — By),

then deg P <n Now P63) —=Í (b; +

constant, for j =1, 2,

Therefore, Pl ) =

đ1)(b¿ + aa) - (bị TT a„) —= ¢, some

vn So P(œ)—c has distinct roots by, bo, -+, bn

c for all x and so _—1)m {a; + by (a; + bo) ++ (0; + by)

,n Then the product of each row is (—1)"*1e

.b, be two distinct collections of m pos- ere each cc sllection may contain repetitions H the two

Let a1, @2, ,@, and by, be,

itive integers, whie

39

`

S2<J<nm) and b+b¿ (1 S¿< ¿

then show that 7m is a power of 2

collections of integers a;+a,{1 : are the same,

Solution (Due to Law Sin Lung) Consider the functions f(z) =

2 + Lin

L— 2e,,’ 12,3)

⁄

Prove that 2, 4 4 or distinct

0 for all n and the terms of the sequence are all

Solution (Due to Wong Chun Wai) The terms z,,’s are clearly rational

by induction Write @,, = py/dm, where p,,, Gy, are relatively prime inte-

gers and gq, > 0 Then q; = 1 and Pr4i/Gnta = 2dr + Pal/ dn — 2Pn)

SO dn41 divides g, — 2p,, which implies every q, is odd by induction Hence, every x, 34 4,

Next, to show every zx, # 0, let a = arctan 2, then v7, = tanna

by induction Suppose x, = 0 and mis the least such index If n =

40

2m is even, then 0 = Ga, = tan 2œ = 2Z„,/(1 — re „) would imply

Um = 0, a contradiction to nm being least If n = 2m +1 is odd, then

0 = Lomi, = tanfa + 2ma) = vi + stom) ‘(1 — 205m) would imply

Lam = —2 Then —2 = 22,,/(1 —x7,) would imply z,, = (1+ V5)/2 is

irrational, a contradiction F inally” if 2, == ©, for some m > n, then

Lm —n = tanứma@ —- na) = (Lm = ø„)/(1 + ¬- = Ú, a contradiction

Therefore the terms are nonzero and distinct

(1988 Nanchang City Math Competition) Define aj = 1, ag = 7 and

2 1

Qn 1g = ———— for positive integer n Prove that 9a,a,44 +1 is a

Th

perfect square for every positive integer n

Solution (Due to Chan Kin Hang) (Since an depends on a,,14 and

an, it is plausible that the sequence satisfies a linear recurrence relation

đụ L2 — Cin +ca, If this is so, then using the first 4 terms, we find

ese 7,co = —1.) Define dy = ay, bạ = đa, Da mu kt by ae n> 4k

Then bg == 48 == ag Suppose ay == by for kK < n+ 1, then

Next, writing out the first few terms of 9an0ni4 +1 will muss rest

that el a (an + On44} \?_ The case n =: 1 is true as 9-7-+ =

(1+7)* Suppose this is true for n = k Using the recurrence r stations

and £ 9 302 +—3 —= 2q_0kL2 — l4apd,i—242, we get the case 0£ = k+1

as follow:

96n41Gn42 +1 = Dane (Tapey - an} +4

r2 € 2 1 \2 ‘

= 6361 ì —_ (ty TC G1) + 2 soi 2 £ : :

(1 — V2)" = A, - Bav ‘9 Multiplying these 2 equations, we get 42 —

282 = CD, This implies A, is always odd Using characteristic

equation method to solve the given recurrence relations on @,, we find

that Gn = = §, Now write 7 = 2%m, wee m is odd We have k = 0

(i.e m is odd) if and only if 2B2 = A* +1 = 2 (mod 4), (Le Bn is

odd} Next suppose case k is true Since (1- + V⁄2)2" = (A,d +-B,v⁄2 = Aon + BanV2, 80 Bo, = 2A,B, Then it follows case k implies case

for all positive integers n

Polution For nonnegative integer n, let t, = (2" — y")/(@ — y) So

=: G,t, =: 1 and we have a recurrence relation

trig tbtnsy tect, =O, where b= —(e@+y), c= xy

1? Suppose t,, is an integer for m,m-+1,m+2,m+3 Since c” = (xy)

2 49 — Eu„+a Is an integer for n = 7n,fm + 1, so 6 1s rational Since ct? is integer, c must, in fact, be an integer Next:

be imbmsa = bmritm $2

So 6 is rational From the recurrence relation, it follows by induction

that t,, = f,—1(b) for some polynomial f, 1 of degree n—1 with integer coefficients Note the coefficient of 2"! in f,1 is 1, Le fy—1 is monic

Since 6 is a root of the integer coefficient polynomial f,,(2) ~tm+oi = 9,

6 must be an integer So the recurrence relation implies all #„'s are integers

42

17

Inequalities

For real numbers 41, @2,d3,

then prove that

Any + Anty 2 2Ay, for m= 2,3,

where A,, is the average of a1,d9, ,4n

Solution Expressing in a,, the required inequality is equivalent to

Solution (Due to Leung Wai Ying) Since abc < 1, we get 1/(be) > a

1/(ac) > b and 1/(ab} > c By the AM-GM inequality,

Similarly, 2b/a + a/c > 36 and 2c/b + 6/a > 3c Adding these and

dividing by 3, we get the desired inequality

AS

Alternat ively, let a = §/ath/e?, y = 8/cta/b? and 2 = °/blc/a?

We have a = 27y,b = 22a,¢ = y’z and ays = Wabe and the rearrangement inequality, we get

Solution For n = 1, af + af — 2(a3)? = ala, — 1)

a= 1 is true Suppose the case n = k is true For the case n =k +1,

without loss of generality, we may assume a, < ag < < ap41 Now

Equality occurs if and only if a,,@2, ,@, are 1,2, ,n

(1997 IMO shortlisted problem) Let ay > - > a) > Gnii =O bea sequence of real numbers Prove that

\ ` ay < <S vk(V% — V/Gk+1)- k;-== k=1

44

Solution (Due to Lee Tak Wing) Let a, = v/@— V8

(te + Bega too + on)* So,

Then ap ==

` Qk = ST (en + tet (#+ + #k+t tr: Đa)“ „À2 = SO = kay, +2 be? 5 S im ee

Th

< > kat +2 »_ Vij#¡# = S V#y)

(1994 Chinese Team Selection Test} For 0 <a <b<e<d<e and

a+t-b+ceo+d+e= 1, show that

ad-+de+cb+he+ea< =

5 Solution [pee to Lau Lap Ming) Since a < 6 < ce < d < 6e, so

d+e>c+e>b4+d>a+e>at b By Chebysev’s inequality,

(1985 Wuhu City Math Competition) Let «a, y, 2 be real numbers such

that œ + +z =0 Show that

3\2 <

6(z3 + 0° + z3)° < (z?° +? + 22)

Solution (Due to Ng Ka Wing) We have z = —(2 + y) and so

(z? + y? 4 zy — (a* + yf" Pt (x +y)*)

> (Sứ +) aye = Ln eye $y)?

= 6z "+ ựẺ a (x -+ 1) ) “ 6(2° 4 y? 4 z7)2,

Comments Let f(w) = (w—av\lw — )(40 — 2) = we + bw +c Thei

#2 u2+z2 = cán + ut 2) —2ayt+yetezr) = —2band0= f0): /0)+ f(s) = (a8 + yP 429) 4 ble + yt 2) + 3c implies 2° + y? + 24 = —äc,

So the inequality is the same as 2(—4b? — 27c?) > 0 For the cubic

polynomial f(w) = w® + bw +c, it is well-known that the discriminant

x)" equals —4b? —27¢? The inequality follows /

ad da VBA

A = (œ0) =3)

easily fromm this

(1999 IMO) Let » be a fixed integer, with m > 2

(a) Determine the least constant C such that the inequality

>

1<¿<j<n

ajar; (x? +

holds for all nonnegative real numbers 71, £9, ,0n.-

(b) For this constant C, determine when equality holds

Solution (Due to Law Ka Ho and Ne Ka Wing} We will show the least C is 1/8 By the AM-GM inequality,

(Equality holds in the second inequality if and only if at least n — 2

of the w,’s are zeros Then equality holds in the first inequality if and only if the remaining pair of w;’s are equal.) Overall, equality holds if and only if two of the x,;’s are equal and the others are zeros

46

23 (1995 Bulgarian Math Competition) Let m > 2 and 0 < a; < 1 for

i== 1,2, ,n Prove that

(ay đa th E8) — (0103 cE 820 cE ch ha -1a a1) SÌa

where |Z| is the greatest integer less than or equal to x

Solution When xo, ,%, are fixed, the left side is a degree one poly-

nomial in «,, so the maximum value is attained when 2, = 0 or i

The situation is similar for the other 2,’s So when the left side is

maximum, every x;’s is 0 or | and the value is an integer Now

2((ay +++ 4+ hy) — (wimg + mgag + + in 18a a3) )

24, For every triplet of functions f,g,h : [0,1] - &, prove that there are

numbers x,y,z in (6,1) such that

(Proposed by Great Britain for 1987 IMO) Ef 2,y, 2 are real numbers

such that o° + y* + 27 = 2, then show that 2 + y+ 2 < xyz +2 Solution (Due to Chan Ming Chiu) [If one of x, y,2 is nonpositive, say z, then

- ¢ ~ oO, 2) a“ 4 toy

because + y < v⁄202 +2) < 2 and zw < (#7 + y*}/2 < 1 So we

may assume @,y, 2 are positive, say O< es y<z the <1, then

2-+Zz— #—— z=(1—#)(1— 0) + (1— z)(1T— zụ) 3 9

If z > 1, then (œ +) +z<: = 34/0 -E 1 S #ụ + 2 <S muz + 2

The inequality to be proved | is ty +23+a3+a7 > 2/3 By the Cauchy-

Schwarz inequality, (a7 + »+ 27 Deut feeb yt) > AG + be e+d)* To

finish, it suffices to show (a + 6+ e+ d)*/(y} + y§ tuệ + 0ì) > 2/3

This follows from

(B) ñịba -by > ayag: ay forall k, i <k <n

Show that by + by ++ + by > ay +g) đạc,

Solution Let c, = 6; /a, and dy, = (¢,-1)+ (cg —1)~+: -+(øy— 1) for

1<k<n SS the AM-GM inequality and (b), (¢, +9 +++ +en)/k >

&/C4C2 > 1, which implies d, > 0 Finally,

c+a > b+ ec, which implies > > By the AM-HM

This yields + + >3 tường - Hy the Chebysev In-

b+e e+a atb™~ 2(a-~+b+c) ( )

equality and the power mean inequality respectively, we have

30 (1982 West German Math Olympiad) H ax,ds, ,„ > Ô and a

ay + ag + + + a, then show that

Solution By symmetry, we may assume a, > @g > > ay Then

For m = 0,1, ,n—1, by the rearrangement inequality, we get

31 Prove that if a,b,c > 0, then —— +—— +—— >

Solution (Due to Ho Wing Yip) By symmetry, we may assume a <

cta ath b+e7 b+e cra a+b

<< For convenience, let a; = a, ifi = 7 (mod n)

Adding these, then dividing by 2, we get

Prove that abed > 3

Solution Let a* = tana, bể = tan /đ, €? = tany, d? = tand Then

cos” a + cos? 3 + cos? -y + cos? ð — 1 By the AM-GM inequality,

sin? a = cos’ 6 + cos’ vy + cos? § > 3¢cos 2 cos -y cos 32/3 Multiplying this and three other similar inequalities, we have

sin? asin? @sin? y sin? § > Bi cos? acos” 6 cos” y cos? ổ

Then abed = /tanatan Gtanytand > 3

(Due to Paul Erdés) Each of the positive integers a,, ,@,, is less than

1951 The least common multiple of any two of these is greater than

34

Solution Observe that none of the numbers 1,2,

multiple of more than one a,’s The number of multiples of a; among

1,2, , 1951 is |1951/a;] So we have (1951 /a,)+ -+[1951/a,,| < 1954

the desired inequality

noving the negative terms to the right, we get

A sequence (P,,) of polynomials is defined recursively as follows:

Teạœ) =0 and forn >0, Pisile) = P,(£) + 5 ƒ———

, 1951 is a common 30 (1996 IMO shortlisted problem) ) Let P(a:) be the real polynomial func-

tion, P(a) = ax? + br? +ex+d Prove that if |P(x)| < 1 for

that |z| < 1, then

all x such

4

“

Comments Tracing the equality cases, we see that the maximum 7 is

obtained by P(a) = +(4x° — 32) only

(American Mati rematical Monthly, Problem 4426) Let P(z) = az? +

hz? -+ez-+d, where a,},c,d are complex numbers with |a| = |b] = el =

|đị = 1 Show that P(e) | > V6 for at least one complex number z satisfying |z| = t

Ww

b,c, d Using w + HH = 2Re w, we get

[P(2)P = (az? + be? + e2 + d\(ae? + bE? + 2 +d)

= 4+ 2Re (adz? + (a@ + bd)2” +a + bé+ cd)z)

Let Q(z) = adz? + (a@ + bả)z? + (ab + bẽ + cd)z, then LP{(z2)|° = 4+

2Re Q(z) Now we use the roots of unity trick! Let w be a cube root

of unity not equal to 1 Since l+w+w? =O and 1+w?+wt = 0, so

(1997 Hungarian-Israeli Math Competition) Find all real numbers a

with the following property: for anv positive integer n, there exists an

Solution The condition hoids if and only if « is an integer If x

is an integer, then for any n, take m = na Conversely, suppose the

condition holds for « Let m, be the integer corresponding to m =

Since the lelmost expression is |2m — ?m+il/2“T!, the inequalitie

imply it is 0, that : mn [2° =rnyi/2*#TÈ for every k Then |x — mo

la — (my, /2")| < 1/(3-2*) for every k Therefore, x = mo is an integer

(1979 British Math Olympiad) Hn is a positive integer, denote by p(n)

the number of ways of expressing 7 as the sum of one or more positive

40

integers Thus p(4) = s there are five different ways of expressing

4 in terms of positive i integers; namely

p(n) — p(n — 1) is the number of sums of m whose least summands are

at least 2 For every one of these p(n) — p(m — 1} sums of n, increasing

the largest summand by 1 will give a sum of 7+ 1 with least summand

at least 2 So p07 + 1) — p(n) > p(n} — p(n — 1)

Functional Lquations

tislying ƒ(z?) + ƒ(z)ƒ(œ + 1) =0 Find all polynomials f sa

Solution ff f is constant, then f is OQ or —1 H f is not constant, then let z be a root of f Setting « = z and x = z— 1, respectively,

we see that 2% and (2 — 1)’ are also roots, respectively Since f has

finitely many roots and z?” are all roots, so we must have |z| = 0 or

i Since z is a root implies (z — 1)% is a root, |z — 1] also equals 0

or 1 It follows that z = 0 or 1 Then ƒ( (2) = cr’ (2 — 1)” for some real ¢ and non negative integers m,n If ¢ 4 0, then after simplifving the functional equation, we will see that 7 =m and c = 1 Therefore,

f(x) = 0 or ~2" (1 — x)” for nonnegative integer n

(1997 Greek Math Olympiad) Let f : (0,co) - FR be a function such that

(a) f is strictly 1 increasing,

(b) f(x) > —+ for all x > 0 and

(c) T000) + 2} =m 1 for aÌl + > 0, Eind ƒ(1)

41

42

Solution a t = PCL) cons es 1 in ( (ce), we get tie +- 1) = ed So

eh) = = in Then Ae nt )ì== ƒ q) Since ƒ 18 ee increasing,

+ + H7 = , solving, we get be = (1+ V5 5)/2 ft = G+ V5) /2> 8,

then P< t= fi) < fG4tH = = 4 <1, a vontnadietion, Therefore,

fO) =t= a — By (Note (a) = (1= V)/(3#) is such a funetion.)

(1979 Eotvos-Kirsechák Math Comipetition) The hưection ƒ is defined

for all real numbers and satisles ƒ(z) < z and ƒ{(œ +) < ƒ(z) + f0)

for all real x,y Prove that f(a) = ø for every real number ø

Solution (Due to Ne Ka Wing) Since f(0 +0) < fF) + ƒ(0), so

< f(0} Since f(0) <0 also, we get f(0) = 9 For all real 2,

0 = ƒ# + (-2)) < fle) + ƒ(Cz) <z+(—z) =

So f(a} + f(-x) = Ô, henee —ƒ{(—=z) = ƒ(z) for alH real z Since

ƒ(—#) < —#, so ø S —ff{-x) = f(x) < zø Therefore, f(x) = «& for

all real a

(Proposed by Ireland for 1989 IMO) Suppose f : R — # satisfies

fQ) =1, fla+ 6) = f(a) + fd) for all a,b © Rand f(s) f(4) =1 for

x #0 Show that f(a) = a for all x

Solution (Due to Yung fy From f(0+0) = f(0) + f(0), we get

ƒ(0) = 0 From 0 = ƒ(z + (—=z)) = flv) + f(—x), we get f(—x) =

— f(x), By induction, fine) =nf lx ) for p0SiLl ive integer n For z=

my (+) = = So f(x) == ø for ratlonal z, “(The argument up to this point

is well known, The so- sealed Cauchy’s equation fla +b) = fla} + fb)

implies f(z) = ƒ()z for rational x.)

Next we will show f is continuous at 0 For O< jap < se we have

| + > 2 So there is w such that w+ + = - We have J (+) =

Now for every real x, let r, be a rational mumber agreeing with

x to nm places after the decimal point Then lim (@ —r,) = 0 By

n ie

continuity at 0, f(a) = tim { ƒ(œz—r,„) + Œr,)) = lũm r„ = z There-

Tb OO fore, f(a) = x for all z x “(This first and shird paragraphs show the

Cauchy equation with continuity at a point has the unique solution

fle) = faje.) (1992 Polish Math Olympiad) Let Q@* be the positive rational nang Determine all finetions f : QT => QT such that ƒ(z + l) = ƒ() +

and f(2°) = f(x) for every x € ỢT, Solution From f(a@ +1) = ƒ(z) + 1, we get ƒ(œ + n) = f(x) +n for

all positive integer 7 mn For : EQ, let t= f (2), On one hand,

ye

E + 3ø? + 3pq” + - g0) = = t+ 3p" oh 3pg” + q

and on the other hand,

Equating 5 the right sides and simoplifvine the equation to a quadratic in = ef cS

¢, we get the only positive root t= © So f(x) = ø for allw Ee Q™ (1996 IMO shortlisted problem) Let A denote the real numbers and

ƒ:R =— [=1,1] satisfy

for every « € A Show that f is a periodic function, i.e there is a nonzero real number 7 such that f(a +7) = f(x) for every x € fi Solution Setting 7 = w+ tor k = 0,1, ,4, we get 6 equations Adding these and ca ancelling ‘tert ms, we will get f(w + Ÿ) + f(w) =

' > cá ` É / k - 0

f(w+4)4+ flw + 4) for all w Setting w = 2+ 4 for k = 0,1,

58

AG,

in this new equation, we get 7 equations Adding these and cancelling

terms, we will get f(¢+2)+ f(z) = 2f(c+1) for all < Rewriting this as

f(e2+2)—fle4+1) = fle+l)— f(s), we see that f(e+n)—fl2+(n—1))

is a constant, say c Hc 4 0, then

flz + kỳ ron LAF +- n) TT fle oe in — 1) ))) + Fe z)

Let N denote the positive integers Suppose s: N —> N is an increasing

function such that s(s(m)) = ẩn for all n € N Find all possible values

of s(1997)

Solution (Due to Chan Kin Hang) Note that if s(m) = s(n), then

3m = s(s(m)) = s{s(n)) = 3n implies m = n From this, we see

that s is strictly increasing Next we have n < s(m) for all n (otherwise

s(n) <n for some n, which yields the contradiction that 8n = s(s(m)) <

s(n) <n.) Phen s(n) < s{s(n}} = 3n In particular, 1 < s(1) <3

implies s(1) == 2 and s(2) = s(s(1)) = 3 With the help of s

s(s(s(n))) = 3s(n), we get s(3") = 2-3“ and s(2-3#) = s(s(3Ÿ)) = 311,

Now there are 3° —1 integers in each of the open intervals (3*, 2-3)

an id (2: 3", 341) Since f is s me inereé asin: we must have s(3” +7) =

2.3% 44 for 7 = 1,2, ,3° —1 Then s(2-3" + 7) = s(s(3* + Tàn

3(3* +3) Since 1997 = 3.302 s39 < 3”, so (1997) = 3(35+539) = 380

Let N be the positive integers Is there a function ƒ: —+ Ñ such that

fO9%F (mn) = Qn for all n € N, where f(x) = f(a) and ft) (x) =

PFO)?

Solution For such a function f(2r) = fC (n) = fO°9(F(n)) =

27f{n} So ifn = 2°q, where e,g are nonnegative integers and g odd,

then f(n) = 2°ƒQq) To define such a function, we need to define it at

odd integer g Now define

(American Mathematical Monthly, Problem E984) Let R denote the

real numbers Find all functions f: R—-> R such that ƒ(ƒ(+)) = #2 =2

r show no such function can exist

Solution Let ae = 2 — 2 and suppose f(f(x)) == g(x) Put A(x) = gigla)) = v* ~ 4a? +2 The fixed points of g (i.e the solutions of the equation (2) = _ are 2-1 1 and 2 The set of fixed points of i contains

the fixed points of l An is S = {—-1, 2, CT + _v5)/3) Now observe that

€ øimplies h(ƒ(2)) = ƒ/(2)) = F(a), \eS Also, x, yes

and fix) = fly rae = h(ø) = Af (U) = y ` fisa bijeclion S§ — S

{fF ¢ =: —1 or 2, then g(ƒ(c)) = ƒŒ(©))) = flgle)) = fle) and

consequently 4 f{ mì ƒ2)} = {—1,2} Eor ø = (1+ VB 5)/2, since f

induces a bijection S and gla) = = ẹ —2#aimnplies ƒ(a) # a, we

rust have f(a) = b = (-1—- V5 vỗ) ‘2 It follows that f(b) = a and we have a contradiction a = f(b) Nai = g(a)

Let R be the real numbers Find all functions f : R -» & such that for

all real numbers # and y,

Putting = ø and letting b = ƒ(0), we get

b= f(x f(a) +2) =a0 + f(a),

so f{x) = —~ax + Putting this into the equation, we have

ara 1 — gồø — an + b= ey - ax + b

Equating coefficients, we get @ = +1 and 6 = 0,80 f(x) = x or fle) =

—x We can easily check both are solutions,

Solution 2 Setting x = 1, we get

Soc dand f is injective Taking x = 1,y = 0, we get f (7 (0) + 1) =

fl) Since f is injective, we get f(0) = 0

For 2 #0, let y = —f(r)/2, then

ƒ(ƒ0) +z) = 0= (0)

By injectivity, we øget øƒ() + =0 Then

ƒ(-f()/+) = #() =—1= /(Ð

and so —f(r)/a = b for cvery œø z0 That is, ƒ(z) = —bx Putting

this into the given equation, we find f(x) = x or f(x) = —z, which are

easily checked to be solutions

49 (1999 IMO) Determine all functions f: R = R such that

Fle — FQ) = AFQ)) + ef + f6) — Ì

61

for all x,y in R

Solution Let A be the range of f andc= = LO) Setting « = y = 0, we

get f(—e) = fle} +e-1 Soc #0 For x = fly) € A, f(z) = St - oe

Next, if we set y = 0, we get

{ƒ( = c) = ƒ) ra G TH} m {em + ƒ(@) =1 :ac R} m1

because c #0 This means A- A = {01 — 1a LỰA Uạ CA} = Ìà

Now for an arbitrary « © #Ÿ, let \,a € A be such tha£ ø = Uìị 1a Then

(L+ fO)fe+y = /ƯŒ + 0)) = ƒŒ + 0) + ƒ)ƒ(W) = zy,

which simplifies to (*) ƒ(0)fŒ + 0) = fla) f(y) — cy Putting y = 1 in

(5), we get ƒ(0)ƒ(œ + 1) = fla) fC) —-2 Putting y = —1 and replacing

62

z by ø-+1in (®), we get ƒ(0)ƒ(z) = ƒ(ø+.L)ƒ(—~1)+z+1 Eliminating

f(x +1) in the last two oquations, we get,

(F°(0) — FFD) F@) =

tf f2(0) — fC) fF (-1) = 0, then putting « = 0 in the last equation,

we get f(0) = 0 By (*), fl@)fy) = ay ‘Then F(a\fQ) = 2 for

(FO) — ƒ(—=1))z + /00)

all x € R So f*(0) - ƒ)ƒ(U =-=L resulting in a contradiction

Therefore , £200) ~ fO)F(-1) 40 and £ () is a degree Ì polynomial,

Einally, substituting ƒ(œ) = ex + 6 into the original equation, we

lnd a = 1 and b =0, lo ƒ(z) = ø for alla € R

(1993 Czechosiovak Math Olympiad) Let Z be the integers Find all

fumetions f: Z—- # such that

fl-R) =f) and f(a) + fly) = fe + 20y) + f(y ~ 20y)

for all integers x, y

Solution We have () FO) + fim) = #Œ +23n) + f(—n) and ƒ(n) +

ƒ(<1) = ƒ(n)+f(~1+2n) Since ƒ(—1) = (1), thịa gives ƒ(1+2n) =

f(-1+ 2n) for every integer n So f(k} has the same value for every

odd k Then equation (*) implies (2) = f{—n) for every integer n So

we need to find f(n) for nonnegative integers n only

TH we let œ = —(2k -+- 1), = n, then ø and z + 2# are odd 'Phe

functional equation gives F(n) = = ƒ() = ƒŒ/T— 2z) = ƒ(n(4k + 3))

It we let 2 = 2, y = —(2k + 1), then similarly, v we get f(n) = f(x) =

fle+2ay) = f(n(—4k—1)) = f(m(4k+1)} So f(n} = f(mm) for every

odd m

2 € For a positive integer n, we can factor 7 = 2N, where €,m are

tion fi is determined by the values ?(0) ở), fla), flay ƒ(8), FQ),

(which may be arbitrary} All other values are given by f(m} = 2°) a as

above Finally, we check such functions satisfv the equations Clearly,

f(-1) = f(t) If & or y = 0, then the functional equation is clearly

satisfied IE a x= 2°m,y = 2¢n, where m,n are odd, then

“

Fle) + fy) = F029) + FQ") = Fla + 2y)) + Fy -+ — 2z)

52 (1995 South Korean Math Olympiad) Let A be the set of non-negative integers Find all functions f : A - A satisfying the following two conditions:

(a) For any m,n € A, 2ƒ(m2 +mn”) = (ƒ(mn)) (b) For any m,n € A with m >n, f(m7) > f(n*)

Solution For m = 0, we me 2/0?) = ƒ(0) 2+ ƒf(n)° Let m > n, then

f0n)2 — ƒ(n)? = 2(f(m?) — f(n?)) > 0 So itr n) > f(m) This means

f is m— Sung m = Ô = n, we get 2ƒ(9) = ƒ(0)2 + ƒ(0)2,

which ae ƒ(0) =9 or 1

Case ƒ(0) = 1 Then 2f(n*) = P+ Fay" For 7 = 1, we f(

Por m = 1 = n, we get f(2) = 1 Assume ƒ( 322) = ĩ Then for

aki, n= 22, we got 2ƒ( 227 )=l+ pe \f = 2 So f(2"") = 1 Since

lim 2° = oo and f is nondecreasing, so f(n) = 1 for all n

Case_f(0) = 6 Then 2 2f{n*) = =: f(n}* So f(n) is even for all n For m =

1 =n, we get 2f(2) = f(1)? +f)", which implies f(2) = f(1}* Using

then lim 22° = oo and f nondecreasing imply f(m) = 0 for all n If

FQ) = 2, then fa") =: a= 22°" Now

fine 2f (On + 1°) = 2flm? + 2m + 1)

>3ƒ0m? + 1) = đụ ny + fy > fin)?

As f (7) is always even, we get f(m+1) > fm) +2 By induction, we

get, fin) > 2n Since fe y= 92 Tì =9 x- for all &, im 92" = 90

\ and f is nondecreasing, so f{m) = 2n for all n

It is easy to check that f(m) = 1, fin) = 6 and f(n) = 2n are

`

solutions Therefore, they are the only solutions

(American Mathematical Monthly, Problem 2176) Let Q denote the rational numbers Find all functions f : Q—- Q such that

(Q)=2 and f ~Ì =————- fore xy

64

Solution We will show f(z) = ø 1s the only solution by a series of

observations

(1) Setting y = 0, we get f f1) = (() + FOIE f(x) — f(0)), which

yields (f(1) — I f(z) = f()(1+ fG)) (Now f is not constant

TH n the non nao in the equation cannot equal 0.) So,

F(1) = 1 and then f(0} =

(2) Setting y= —z, we get O= ec }+ f{—x}, so f(—x) = —f(z)

(3) Setting y = cn, cz21,xz z0, we get

rat #) + flee), (; + 2 1+ flo

which implies flca) = = ƒ(c)f() Taking © = ga = p/q, we get

f0/4) = ƒ0)/744):

(4) Setting y = 2 — tơ we G et fla — 1) = (Fle) + fl@ — 2))/Fla) —

f

i œ — 2)) H ƒ{n — 3) =n — 2 # 0 and ƒ{n — 1) = n— b then

us equation plies / (n) = 1, Sinee F(t) = 1 and f (2) sx 2, then

fn = n for all positive integers by induction and (2), (3) will

imply f(e) = x for ail a ved

Comments The condition f(2) = 2 can also be deduced from the

functional equation as shown below in (5) Ef rational numbers are

replaced by real numbers, then again the only solution is still f(a) = x

as shown below in (6) ) and (7)

Also, f(2)° = f(4) = (f(5)4+ f (ƒ(5) — ƒ(3)) Substituting the

equations for ; ƒ(3) and f(5) in sae of 7 (2) and Psimpriying, we

get f(2)* = ai) (Now f(2) 46, otherwise ie + tke ~2))=

(f(s) +0)/(f(a) — 0) = 1 will farce f to be constant.) Therefore,

f(2) = 2

(6) Note f(a) #0 for x > 0, otherwise f{ca) = 0 for c 4 1 will force f

to be constant So, if 2 > 0, then f(x) = f(./x)* > 0 Ife > y > 0,

then f(x) — f(y) = (fla + fai / fle + y) fla - y) > © This

implies f is strictly increasing for positive real numbers

6D

(7) For a > 0, ifa < f(x), then picking r € Q such thata<r< f(x)

will give the contradiction that ƒ() < #ứ) =r < f(x) Similarly, F(a) < x will also lead to a contradiction So, 7 gì = xr for all x (Mathematics Magazine, Problem 1552) Find all functions f: R= R such that

Fla+yf(e)) = flaj+eaefly) forall az,y in R

Solution [t is easy to check that f(z) =Qand f(a) = x are solutions

Suppose f is a solution that is not the zero functio: (We will show f(x) = x for all x.)

Step 1 Setting y = 0,2 = 1, we get f(0) = 0 If f(x) = 0, then

Ox & MM {y} for all y, which implies x = 6 as f is not the zero function

So f(x) = 0 if and only if a = 6

Step 2 Setting x = 1, we get the equation (*) fG+yfG)) = FO)+fy) for all y If fl) # 1, then setting y = 1/(1 ~ f(41)) in (), we get

fly) = fG) + fy), resulting in f(1) = 0, contradicting step 1 Soe

Ff) = 1 and (*) becomes f(1 + y) = f(4} + f(@), which implies f(m) =n for every integer n

Step 3 For integer n, real z, setting © = 7,y == z- 1 in the functional equation, we get,

f{nz) = fint(e—-lAftn) =n + nƒƑ(z — 1) = nƒŒ)

sở

sách 4 lÝÍa = —b, then f(a) = f(—b) = —f{b) implies Fla) + f(b) =

= f(a+b) Ifa #4 —b, then a+b #0 and f(a+ 6) # 0 by step 1

Setting v= (a +b)/2,y = (a — b)/(2f( 48"), we get

Ft ) It 2 2ƒ(1%) ` 9 )) Fe 2 ) 2

Ằ b-a ,,at+b,, atb ath