Tổng hợp đề thi và lời giải của kỳ thi HSG Châu Á - Thái Bình Dương APMO từ năm 1989 đến 2019

March, 2018 Time allowed: 4 hours Each problem is worth 7 points The contest problems are to be kept confidential until they are posted on the official APMO website http://apmo.ommenline[r]

(1)

1989 - 2019

COLLECTION OF MATHEMATICS OLYMPIAD A

P M

(2)

THE 1989 ASIAN PACIFIC MATHEMATICAL OLYMPIAD Time allowed: hours

NO calculators are to be used. Each question is worth seven points. Question

Letx1,x2, , xn be positive real numbers, and let S=x1+x2+· · ·+xn. Prove that

(1 +x1)(1 +x2)· · ·(1 +xn)≤1 +S+ S2

2! + S3

3! +· · ·+ Sn

n! . Question

Prove that the equation

6(6a2+ 3b2+c2) = 5n2 has no solutions in integers except a=b =c=n= Question

Let A1, A2, A3 be three points in the plane, and for convenience, let A4 = A1, A5 = A2 For n = 1, 2, and 3, suppose that Bn is the midpoint of AnAn+1, and suppose that Cn is the midpoint of AnBn Suppose that AnCn+1 and BnAn+2 meet at Dn, and that AnBn+1 and CnAn+2 meet at En Calculate the ratio of the area of triangle D1D2D3 to the area of triangleE1E2E3.

Question

Let S be a set consisting of m pairs (a, b) of positive integers with the property that 1≤a < b≤n Show that there are at least

4m·(m− n2

4 ) 3n

triples (a, b, c) such that (a, b), (a, c), and (b, c) belong to S Question

Determine all functions f from the reals to the reals for which (1) f(x) is strictly increasing,

(2) f(x) +g(x) = 2x for all real x,

(3)

Given triagnle ABC, let D, E, F be the midpoints of BC, AC, AB respectively and let G be the centroid of the triangle

For each value of ∠BAC, how many non-similar triangles are there in which AEGF is a cyclic quadrilateral?

Question

Leta1, a2, , an be positive real numbers, and letSk be the sum of the products of a1, a2, , an takenk at a time Show that

SkSn−k ≥

à n k

ả2

a1a2Ã Ã Ãan

for k = 1, 2, ,n−1 Question

Consider all the trianglesABC which have a fixed base AB and whose altitude from C is a constant h For which of these triangles is the product of its altitudes a maximum?

Question

A set of 1990 persons is divided into non-intersecting subsets in such a way that No one in a subset knows all the others in the subset,

2 Among any three persons in a subset, there are always at least two who not know each other, and

3 For any two persons in a subset who not know each other, there is exactly one person in the same subset knowing both of them

(a) Prove that within each subset, every person has the same number of acquaintances (b) Determine the maximum possible number of subsets

Note: It is understood that if a person A knows person B, then person B will know person A; an acquaintance is someone who is known Every person is assumed to know one’s self. Question

(4)

NO calculators are to be used. Each question is worth seven points.

Question

Let G be the centroid of triangle ABC and M be the midpoint of BC Let X be on AB

and Y onAC such that the pointsX, Y, and Gare collinear and XY and BC are parallel Suppose thatXCandGB intersect atQandY B andGC intersect atP Show that triangle

MP Qis similar to triangle ABC Question

Suppose there are 997 points given in a plane If every two points are joined by a line segment with its midpoint coloured in red, show that there are at least 1991 red points in the plane Can you find a special case with exactly 1991 red points?

Question

Let a1, a2, , an, b1, b2, , bn be positive real numbers such that a1 +a2 +· · ·+an =

b1 +b2+· · ·+bn Show that

a2

a1 +b1

+ a

2

a2+b2

+· · ·+ a n

an+bn

≥ a1+a2+· · ·+an

2 .

Question

During a break, n children at school sit in a circle around their teacher to play a game The teacher walks clockwise close to the children and hands out candies to some of them according to the following rule He selects one child and gives him a candy, then he skips the next child and gives a candy to the next one, then he skips and gives a candy to the next one, then he skips 3, and so on Determine the values of n for which eventually, perhaps after many rounds, all children will have at least one candy each

Question

(5)

Question

A triangle with sides a, b, and c is given Denote by s the semiperimeter, that is s = (a+b+c)/2 Construct a triangle with sidess−a,s−b, and s−c This process is repeated until a triangle can no longer be constructed with the side lengths given

For which original triangles can this process be repeated indefinitely? Question

In a circle C with centre O and radius r, let C1, C2 be two circles with centres O1, O2 and radii r1, r2 respectively, so that each circle Ci is internally tangent to C at Ai and so that

C1,C2 are externally tangent to each other at A

Prove that the three lines OA, O1A2, and O2A1 are concurrent Question

Let n be an integer such that n > Suppose that we choose three numbers from the set {1,2, , n} Using each of these three numbers only once and using addition, multiplication, and parenthesis, let us form all possible combinations

(a) Show that if we choose all three numbers greater than n/2, then the values of these combinations are all distinct

(b) Let pbe a prime number such that p≤√n Show that the number of ways of choosing three numbers so that the smallest one is p and the values of the combinations are not all distinct is precisely the number of positive divisors of p−1

Question

Determine all pairs (h, s) of positive integers with the following property: If one draws h horizontal lines and another s lines which satisfy

(i) they are not horizontal, (ii) no two of them are parallel,

(iii) no three of the h+s lines are concurrent,

then the number of regions formed by these h+s lines is 1992 Question

(6)

(7)

Question

LetABCD be a quadrilateral such that all sides have equal length and angleABC is 60 deg Let l be a line passing through D and not intersecting the quadrilateral (except at D) Let

E andF be the points of intersection ofl withAB and BC respectively LetM be the point of intersection of CE and AF

Prove that CA2 =CM ×CE. Question

Find the total number of different integer values the function

f(x) = [x] + [2x] + [5x

3 ] + [3x] + [4x] takes for real numbers x with 0≤x≤100

Question Let

f(x) = anxn+an−1xn−1+· · ·+a0 and

g(x) = cn+1xn+1+cnxn+· · ·+c0

be non-zero polynomials with real coefficients such that g(x) = (x+r)f(x) for some real number r If a= max(|an|, ,|a0|) and c= max(|cn+1|, ,|c0|), prove that ac ≤n+ Question

Determine all positive integers n for which the equation

xn+ (2 +x)n+ (2−x)n= 0 has an integer as a solution

Question

LetP1, P2, , P1993 =P0 be distinct points in the xy-plane with the following properties: (i) both coordinates of Pi are integers, for i= 1,2, ,1993;

(ii) there is no point other than Pi and Pi+1 on the line segment joining Pi with Pi+1 whose coordinates are both integers, for i= 0,1, ,1992

(8)

Question

Letf :R→R be a function such that (i) For all x, y ∈R,

f(x) +f(y) + 1≥f(x+y)≥f(x) +f(y),

(ii) For all x∈[0,1), f(0) ≥f(x), (iii) −f(−1) = f(1) =

Find all such functions f Question

Given a nondegenerate triangleABC, with circumcentreO, orthocentreH, and circumradius

R, prove that |OH|<3R Question

Letn be an integer of the form a2+b2, where a and b are relatively prime integers and such that if p is a prime, p≤√n, then pdivides ab Determine all such n

Question

Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational?

Question

You are given three lists A, B, and C List A contains the numbers of the form 10k in base 10, with k any integer greater than or equal to Lists B and C contain the same numbers translated into base and respectively:

A B C

10 1010 20

100 1100100 400 1000 1111101000 13000

(9)

Question

Determine all sequences of real numbers a1, a2, , a1995 which satisfy: 2pan−(n−1)≥an+1−(n−1), for n= 1,2, 1994, and

2√a1995−1994≥a1 + 1. Question

Leta1, a2, , an be a sequence of integers with values between and 1995 such that: (i) Any two of the ai’s are realtively prime,

(ii) Each ai is either a prime or a product of primes

Determine the smallest possible values of n to make sure that the sequence will contain a prime number

Question

Let P QRS be a cyclic quadrilateral such that the segments P Q and RS are not paral-lel Consider the set of circles through P and Q, and the set of circles through R and S Determine the set A of points of tangency of circles in these two sets

Question

Let C be a circle with radius R and centre O, and S a fixed point in the interior ofC Let

AA0 and BB0 be perpendicular chords throughS Consider the rectanglesSAMB,SBN0A0,

SA0M0B0, andSB0NA Find the set of all points M, N0, M0, andN whenA moves around the whole circle

Question

(10)

Let ABCD be a quadrilateral AB =BC =CD =DA Let MN and P Q be two segments perpendicular to the diagonal BD and such that the distance between them is d > BD/2, with M ∈ AD, N ∈ DC, P ∈ AB, and Q ∈ BC Show that the perimeter of hexagon AMNCQP does not depend on the position ofMN andP Qso long as the distance between them remains constant

Question

Letm and n be positive integers such that n≤m Prove that 2nn!≤ (m+n)!

(m−n)! ≤(m

2+m)n .

Question

LetP1,P2,P3,P4 be four points on a circle, and letI1 be the incentre of the triangleP2P3P4; I2 be the incentre of the triangle P1P3P4;I3 be the incentre of the triangle P1P2P4;I4 be the incentre of the triangleP1P2P3 Prove that I1, I2, I3, I4 are the vertices of a rectangle Question

The National Marriage Council wishes to inviten couples to form 17 discussion groups under the following conditions:

1 All members of a group must be of the same sex; i.e they are either all male or all female

2 The difference in the size of any two groups is or All groups have at least member

4 Each person must belong to one and only one group

Find all values of n, n≤1996, for which this is possible Justify your answer Question

Leta,b, cbe the lengths of the sides of a triangle Prove that √

(11)

Given

S = + 1 +

3

+

1 + +16

+· · ·+

1 +

3 + 16 +· · ·+19930061 ,

where the denominators contain partial sums of the sequence of reciprocals of triangular numbers (i.e k =n(n+ 1)/2 for n= 1, 2, , 1996) Prove that S >1001

Question

Find an integer n, where 100≤n ≤1997, such that 2n+ 2

n is also an integer

Question

LetABC be a triangle inscribed in a circle and let la =

ma

Ma

, lb =

mb

Mb

, lc =

mc

Mc

,

where ma, mb, mc are the lengths of the angle bisectors (internal to the triangle) and Ma,

Mb,Mcare the lengths of the angle bisectors extended until they meet the circle Prove that

la

sin2A+ lb

sin2B + lc

sin2C ≥3, and that equality holds iff ABC is an equilateral triangle Question

TriangleA1A2A3has a right angle atA3 A sequence of points is now defined by the following iterative process, where n is a positive integer From An (n ≥ 3), a perpendicular line is

drawn to meet An−2An−1 at An+1

(a) Prove that if this process is continued indefinitely, then one and only one point P is interior to every triangle An−2An−1An, n ≥3

(b) Let A1 and A3 be fixed points By considering all possible locations of A2 on the plane, find the locus of P

Question

(12)

objects such that

a1+a2+· · ·+an=nN,

whereN is a positive integer In order that each person has the same number of objects, each person Ai is to give or to receive a certain number of objects to or from its two neighbours

(13)

10th Asian Pacific Mathematics Olympiad

March 1998

Time allowed: hours No calculators to be used Each question is worth points

1 Let F be the set of all n-tuples (A1, A2, …, An) where each Ai, i = 1, 2, …, n is a subset of {1, 2, …, 1998} Let |A| denote the number of elements of the set A

Find the number ∑ ∪ ∪ ∪

) , , , ( 2 | | n A A A n A A A

2 Show that for any positive integers a and b, (36a+b)(a+36b) cannot be a power of

3 Let a, b, c be positive real numbers Prove that

    + + + ≥       +       +     

 +1 1 3

abc c b a a c c b b a

4 Let ABC be a triangle and D the foot of the altitude from A Let E and F be on a line through D such that AE is perpendicular to BE, AF is perpendicular to CF, and E and F are different from D Let M and N be the midpoints of the line segments BC and EF, respectively Prove that AN is perpendicular to NM

5 Determine the largest of all integers n with the property that n is divisible by all positive integers that are less than

n

(14)

11th Asian Pacific Mathematical Olympiad March, 1999

1 Find the smallest positive integer n with the following property: there does not exist an arithmetic progression of 1999 real numbers containing exactly n integers

2 Leta1, a2, be a sequence of real numbers satisfyingai+j ≤ai+aj for alli, j = 1,2, Prove that

a1+ a2

2 + a3

3 +· · ·+ an

n ≥an for each positive integer n.

3 Let Γ1 and Γ2 be two circles intersecting at P and Q The common tangent, closer to P, of Γ1 and Γ2 touches Γ1 atA and Γ2 at B The tangent of Γ1 at P meets Γ2 atC, which is different from P, and the extension of AP meets BC at R Prove that the circumcircle of triangle P QR is tangent to BP and BR.

4 Determine all pairs (a, b) of integers with the property that the numbers a2+ 4b and b2+ 4a are both perfect squares.

(15)

12th Asian Pacific Mathematics Olympiad

March 2000

1 Compute the sum

3 101

2 01 3

i

i i i

x S

x x

=

− +

∑ for

101

i

i x =

2 Given the following triangular arrangement of circles:

Each of the numbers 1, 2, …, is to be written into one of these circles, so that each circle contains exactly one of these numbers and

(i) the sums of the four numbers on each side of the triangle are equal;

(ii) the sums of the squares of the four numbers on each side of the triangle are equal Find all ways in which this can be done

3 Let ABC be a triangle Let M and N be the points in which the median and the angle bisector, respectively, at A meet the side BC Let Q and P be the points in which the perpendicular at N to NA meets MA and BA, respectively, and O the point in which the perpendicular at P to BA meets AN produced Prove that QO is perpendicular to BC

4 Let n, k be given positive integers with n > k Prove that

1 !

1 ( ) !( )! ( )

n n

k n k k n k

n n n

n+ ⋅k n k− − <k n k− <k n k− −

5 Given a permutation (a a0, 1,,an) of the sequence 0, 1, …, n A transposition of ai with j

a is called legal if ai =0 for i>0, and ai−1+ =1 aj The permutation (a a0, 1,,an) is

called regular if after a number of legal transpositions it becomes (1, 2,, , 0)n For which numbers n is the permutation (1, ,n n−1,, 3, 2, 0) regular?

(16)

XIII Asian Pacific Mathematics Olympiad March, 2001

Problem

For a positive integernletS(n) be the sum of digits in the decimal representation ofn Any positive integer obtained by removing several (at least one) digits from the right-hand end of the decimal representation of n is called a stump of n Let T(n) be the sum of all stumps of n Prove that n=S(n) + 9T(n)

Problem

Find the largest positive integerN so that the number of integers in the set {1,2, , N} which are divisible by is equal to the number of integers which are divisible by or (or both)

Problem

Let two equal regularn-gons S and T be located in the plane such that their intersection is a 2n-gon (n≥3) The sides of the polygonS are coloured in red and the sides ofT in blue

Prove that the sum of the lengths of the blue sides of the polygon S∩T is equal to the sum of the lengths of its red sides

Problem

A point in the plane with a cartesian coordinate system is called amixed pointif one of its coordinates is rational and the other one is irrational Find all polynomials with real coefficients such that their graphs not contain any mixed point

Problem

(17)

XIV Asian Pacific Mathematics Olympiad March 2002

Time allowed: hours No calculators are to be used Each question is worth points

Problem

Leta1, a2, a3, , an be a sequence of non-negative integers, wherenis a positive integer Let

An= a1+a2+· · ·+an

n .

Prove that

a1!a2! an!≥(bAnc!)n,

wherebAncis the greatest integer less than or equal to An, anda! = 1×2× · · · ×afor a≥1 (and 0! = 1)

When does equality hold? Problem

Find all positive integersaandb such that a2+b b2−a and

b2+a a2−b are both integers

Problem

LetABCbe an equilateral triangle LetP be a point on the sideAC andQbe a point on the sideABso that both trianglesABP andACQare acute LetRbe the orthocentre of triangleABP andSbe the orthocentre of triangleACQ Let T be the point common to the segmentsBP andCQ Find all possible values of6 CBP and6 BCQsuch that triangleT RS is equilateral.

Problem

Letx, y, z be positive numbers such that x+

1 y +

1 z =

Show that √

x+yz+√y+zx+√z+xy≥√xyz+√x+√y+√z. Problem

LetRdenote the set of all real numbers Find all functionsf fromRtoRsatisfying: (i) there are only finitely manysinRsuch thatf(s) = 0, and

(18)

XV Asian Pacific Mathematics Olympiad March 2003

Problem

Leta, b, c, d, e, f be real numbers such that the polynomial

p(x) =x8−4x7+ 7x6+ax5+bx4+cx3+dx2+ex+f

factorises into eight linear factorsx−xi, withxi >0 for i= 1,2, ,8 Determine all possible values off

Problem

SupposeABCD is a square piece of cardboard with side lengtha On a plane are two parallel lines`1 and`2, which are alsoaunits apart The squareABCDis placed on the plane so that sidesAB andADintersect`1 atE andF respectively Also, sidesCB andCD intersect`2 at GandH respectively Let the perimeters of

4AEF and 4CGH be m1 and m2 respectively Prove that no matter how the square was placed,m1+m2 remains constant

Problem

Letk≥14 be an integer, and letpk be the largest prime number which is strictly less thank You may assume thatpk ≥3k/4 Letnbe a composite integer Prove:

(a) ifn= 2pk, thenndoes not divide (n−k)! ; (b) ifn >2pk, then ndivides (n−k)!

Problem

Leta, b, cbe the sides of a triangle, with a+b+c= 1, and letn≥2 be an integer Show that n

√

an+bn+√n

bn+cn+ √n

cn+an<1 +

n

√

2 .

Problem

(19)

XVI Asian Pacific Mathematics Olympiad March 2004

Problem

Determine all finite nonempty setsS of positive integers satisfying

i+j

(i, j) is an element ofS for alli, j inS, where (i, j) is the greatest common divisor ofiandj

Problem

LetObe the circumcentre andH the orthocentre of an acute triangleABC Prove that the area of one of the trianglesAOH,BOH andCOH is equal to the sum of the areas of the other two

Problem

Let a setSof 2004 points in the plane be given, no three of which are collinear LetLdenote the set of all lines (extended indefinitely in both directions) determined by pairs of points from the set Show that it is possible to colour the points ofS with at most two colours, such that for any pointsp, qofS, the number of lines inL which separatepfromqis odd if and only ifpandqhave the same colour

Note: A line` separatestwo pointspandqifpandqlie on opposite sides of` with neither point on`

Problem

For a real numberx, letbxcstand for the largest integer that is less than or equal tox Prove that

¹

(n−1)!

n(n+ 1)

º

is even for every positive integern

Problem Prove that

(20)

XVII Asian Pacific Mathematics Olympiad Time allowed : hours

Each problem is worth points

∗The contest problems are to be kept confidential until they are posted on the official APMO website Please not disclose nor discuss the problems over the internet until that date. No calculators are to be used during the contest.

Problem Prove that for every irrational real numbera, there are irrational real numbers b and b0 so that a+b and ab0 are both rational while aband a+b0 are both irrational.

Problem Let a, band c be positive real numbers such thatabc= Prove that a2

p

(1 +a3)(1 +b3)+

b2

p

(1 +b3)(1 +c3) +

c2

p

(1 +c3)(1 +a3) ≥

4 3.

Problem Prove that there exists a triangle which can be cut into 2005 congruent triangles

(21)

XVIII Asian Pacific Mathematics Olympiad

Time allowed : hours

Problem Let n be a positive integer Find the largest nonnegative real number f(n) (depending on n) with the following property: whenever a1, a2, , an are real numbers

such that a1+a2+· · ·+an is an integer, there exists some i such that|ai−12 | ≥f(n) Problem Prove that every positive integer can be written as a finite sum of distinct integral powers of the golden mean τ = 1+√5

2 Here, an integral power of τ is of the form τi, wherei is an integer (not necessarily positive).

Problem Let p≥5 be a prime and let r be the number of ways of placing p checkers on a p×p checkerboard so that not all checkers are in the same row (but they may all be in the same column) Show thatr is divisible byp5 Here, we assume that all the checkers

are identical

Problem LetA, B be two distinct points on a given circleOand letP be the midpoint of the line segment AB Let O1 be the circle tangent to the line AB at P and tangent to

the circle O Let ` be the tangent line, different from the line AB, to O1 passing through

A Let C be the intersection point, different from A, of ` and O Let Q be the midpoint of the line segment BC and O2 be the circle tangent to the line BC atQ and tangent to

the line segment AC Prove that the circle O2 is tangent to the circle O

(22)

XIX Asian Pacific Mathematics Olympiad

G

Time allowed : hours

Problem LetS be a set of distinct integers all of whose prime factors are at most Prove that S contains distinct integers such that their product is a perfect cube

Problem LetABC be an acute angled triangle with∠BAC = 60◦ andAB > AC Let

I be the incenter, and H the orthocenter of the triangleABC Prove that 2∠AHI = 3∠ABC.

Problem Consider n disksC1, C2, , Cn in a plane such that for each 1≤i < n, the center of Ci is on the circumference of Ci+1, and the center ofCn is on the circumference of C1 Define the score of such an arrangement of n disks to be the number of pairs (i, j) for which Ci properly contains Cj Determine the maximum possible score

Problem Letx, y and z be positive real numbers such that√x+√y+√z = Prove that

x2+yz p

2x2(y+z)+

y2+zx p

2y2(z+x) +

z2 +xy p

2z2(x+y) ≥1.

Problem A regular (5×5)-array of lights is defective, so that toggling the switch for one light causes each adjacent light in the same row and in the same column as well as the light itself to change state, from on to off, or from off to on Initially all the lights are switched off After a certain number of toggles, exactly one light is switched on Find all the possible positions of this light

(23)

XX Asian Pacific Mathematics Olympiad

March, 2008 Time allowed : hours

∗The contest problems are to be kept confidential until they are posted on the official APMO website Please not disclose nor discuss the problems over the internet until that date No calculators are to be used during the contest.

Problem LetABC be a triangle with∠A <60◦ LetX andY be the points on the sides

ABand AC, respectively, such thatCA+AX =CB+BX and BA+AY =BC+CY Let

P be the point in the plane such that the lines P X and P Y are perpendicular to AB and

AC, respectively Prove that ∠BP C <120◦.

Problem Students in a class form groups each of which contains exactly three members such that any two distinct groups have at most one member in common Prove that, when the class size is 46, there is a set of 10 students in which no group is properly contained

Problem Let Γ be the circumcircle of a triangle ABC A circle passing through points

Aand C meets the sidesBC and BAatDand E, respectively The linesAD andCE meet Γ again atG andH, respectively The tangent lines of Γ atA and C meet the line DE atL

andM, respectively Prove that the linesLH and M Gmeet at Γ

Problem Consider the function f : N0 → N0, where N0 is the set of all non-negative

integers, defined by the following conditions :

(i) f(0) = 0, (ii) f(2n) = 2f(n) and (iii)f(2n+ 1) =n+ 2f(n) for alln≥0.

(a) Determine the three sets L:={n|f(n)< f(n+ 1)}, E :={n|f(n) =f(n+ 1)}, and

G:={n|f(n)> f(n+ 1)}.

(b) For eachk≥0, find a formula forak:= max{f(n) : 0≤n≤2k} in terms ofk

(24)

XXI Asian Pacific Mathematics Olympiad

March, 2009

Time allowed : hours Each problem is worth points

∗The contest problems are to be kept confidential until they are posted on the official APMO website (http://www.kms.or.kr/Competitions/APMO) Please not disclose nor discuss the problems over the internet until that date Calculators are not allowed to use.

Problem Consider the following operation on positive real numbers written on a black-board: Choose a number r written on the blackboard, erase that number, and then write a pair of positive real numbers aand b satisfying the condition2r2 =ab on the board.

Assume that you start out with just one positive real number r on the blackboard, and apply this operation k2−1 times to end up with k2 positive real numbers, not necessarily

distinct Show that there exists a number on the board which does not exceedkr. Problem Leta1, a2, a3, a4, a5 be real numbers satisfying the following equations:

a1 k2+ 1+

a2 k2+ 2+

a3 k2+ 3+

a4 k2+ 4+

a5 k2+ 5 =

1

k2 fork= 1,2,3,4,5

Find the value of a1 37+

a2

38+ a3

39+ a4

40+ a5

41 (Express the value in a single fraction.)

Problem Let three circles Γ1,Γ2,Γ3, which are non-overlapping and mutually external,

be given in the plane For each point P in the plane, outside the three circles, construct six pointsA1, B1, A2, B2, A3, B3 as follows: For each i= 1,2,3,Ai, Bi are distinct points on

the circle Γi such that the lines P Ai and P Bi are both tangents to Γi Call the point P

exceptional if, from the construction, three lines A1B1, A2B2, A3B3 are concurrent Show

that every exceptional point of the plane, if exists, lies on the same circle

Problem Prove that for any positive integerk, there exists an arithmetic sequence a1

b1 ,

a2

b2 , ,

ak bk

of rational numbers, whereai, bi are relatively prime positive integers for eachi= 1,2, , k,

such that the positive integersa1, b1, a2, b2, , ak, bk are all distinct

Problem Larry and Rob are two robots travelling in one car from Argovia to Zillis Both robots have control over the steering and steer according to the following algorithm: Larry makes a 90◦ left turn after every` kilometer driving from start; Rob makes a 90◦ right turn after everyr kilometer driving from start, where`andr are relatively prime positive integers In the event of both turns occurring simultaneously, the car will keep going without changing direction Assume that the ground is flat and the car can move in any direction

(25)

XXII Asian Pacific Mathematics Olympiad

G

March, 2010

Time allowed: hours Each problem is worth points

*The contest problems are to be kept confidential until they are posted on the official APMO website (http://www.mmjp.or.jp/competitions/APMO) Please not disclose nor discuss the problems over the internet until that date Calculators are not allowed to use.

Problem Let ABC be a triangle with 6 BAC 6= 90◦. LetO be the circumcenter of the

triangleABC and let Γ be the circumcircle of the triangle BOC Suppose that Γ intersects the line segmentAB atP different fromB, and the line segmentAC atQdifferent from C LetON be a diameter of the circle Γ.Prove that the quadrilateralAP N Qis a parallelogram

Problem For a positive integerk, call an integer apurek-th powerif it can be represented asmkfor some integerm Show that for every positive integernthere existndistinct positive

integers such that their sum is a pure 2009-th power, and their product is a pure 2010-th power

Problem Letnbe a positive integer npeople take part in a certain party For any pair of the participants, either the two are acquainted with each other or they are not What is the maximum possible number of the pairs for which the two are not acquainted but have a common acquaintance among the participants?

Problem Let ABC be an acute triangle satisfying the condition AB > BC and AC > BC Denote by O and H the circumcenter and the orthocenter, respectively, of the triangleABC. Suppose that the circumcircle of the triangleAHC intersects the line AB at M different fromA,and that the circumcircle of the triangleAHB intersects the lineAC at N different fromA.Prove that the circumcenter of the triangle M N H lies on the lineOH.

Problem Find all functions f from the set R of real numbers into R which satisfy for allx, y, z ∈R the identity

(26)

2011 APMO PROBLEMS

Time allowed: hours Each problem is worth points *The contest problems are to be kept confidential until they are posted on the offi-cial APMO website (http://www.mmjp.or.jp/competitions/APMO) Please not disclose nor discuss the problems over the internet until that date Calculators are not allowed to use

Problem Leta, b, c be positive integers Prove that it is impossible to have all of the three numbersa2+b+c, b2+c+a, c2+a+b to be perfect squares Problem Five points A1, A2, A3, A4, A5 lie on a plane in such a way that no three among them lie on a same straight line Determine the maximum possible value that the minimum value for the angles ∠AiAjAk can take where i, j, k are distinct integers between and

Problem LetABCbe an acute triangle with∠BAC= 30◦ The internal and external angle bisectors of∠ABC meet the lineACatB1andB2, respectively, and the internal and external angle bisectors of∠ACB meet the lineABatC1andC2, respectively Suppose that the circles with diametersB1B2 and C1C2 meet inside the triangleABC at pointP Prove that∠BP C= 90◦

Problem Letn be a fixed positive odd integer Takem+ 2distinct points

P0, P1,· · ·, Pm+1 (where m is a non-negative integer) on the coordinate plane in such a way that the following conditions are satisfied:

(1) P0= (0,1), Pm+1= (n+ 1, n), and for each integeri, 1≤i≤m, bothx- and

y- coordinates ofPi are integers lying in between andn(1 andninclusive) (2) For each integeri, 0≤i≤m,PiPi+1is parallel to thex-axis ifiis even, and is parallel to they-axis ifiis odd

(3) For each pair i, j with 0≤i < j ≤m, line segmentsPiPi+1 and PjPj+1 share at most point

Determine the maximum possible value thatmcan take

Problem Determine all functions f :R→R, where Ris the set of all real numbers, satisfying the following conditions:

(1) There exists a real numberM such that for every real numberx,f(x)< M is satisfied

(2) For every pair of real numbersxandy,

f(xf(y)) +yf(x) =xf(y) +f(xy) is satisfied

(27)

2012 APMO PROBLEMS

Time allowed: hours Each problem is worth points *The contest problems are to be kept confidential until they are posted on the offi-cial APMO website (http://www.mmjp.or.jp/competitions/APMO) Please not disclose nor discuss the problems over the internet until that date Calculators are not allowed to use

Problem LetP be a point in the interior of a triangleABC, and letD, E, F be the point of intersection of the lineAP and the side BC of the triangle, of the lineBP and the sideCA, and of the lineCP and the sideAB, respectively Prove that the area of the triangle ABC must be if the area of each of the triangles P F A,P DBandP EC is

Problem Into each box of a 2012×2012 square grid, a real number greater than or equal to and less than or equal to is inserted Consider splitting the grid into non-empty rectangles consisting of boxes of the grid by drawing a line parallel either to the horizontal or the vertical side of the grid Suppose that for at least one of the resulting rectangles the sum of the numbers in the boxes within the rectangle is less than or equal to 1, no matter how the grid is split into such rectangles Determine the maximum possible value for the sum of all the 2012×2012 numbers inserted into the boxes

Problem Determine all the pairs (p, n) of a prime number pand a positive integernfor which np+1

pn+1 is an integer

Problem Let ABC be an acute triangle Denote by D the foot of the perpendicular line drawn from the point Ato the sideBC, by M the midpoint of BC, and byH the orthocenter of ABC LetE be the point of intersection of the circumcircle Γ of the triangle ABC and the half line M H, andF be the point of intersection (other thanE) of the lineED and the circle Γ Prove that BFCF = ABAC must hold

Here we denote byXY the length of the line segmentXY

Problem Let n be an integer greater than or equal to Prove that if the real numbersa1, a2,· · ·, an satisfya21+a22+· · ·+a2n=n, then

X

1≤i<j≤n

1 n−aiaj

≤n must hold

(28)

XXV Asian Pacific Mathematics Olympiad

Time allowed: hours Each problem if worth points Problem LetABCbe an acute triangle with altitudesAD, BE andCF, and letO

be the center of its circumcircle Show that the segments OA, OF, OB, OD, OC, OE

dissect the triangle ABC into three pairs of triangles that have equal areas Problem Determine all positive integersn for which n

2+ 1

[√n]2+ 2 is an integer Here [r] denotes the greatest integer less than or equal tor

Problem For 2k real numbers a1, a2, , ak, b1, b2, , bk define the sequence of numbers Xn by

Xn = k X

i=1

[ain+bi] (n = 1,2, ) If the sequence Xn forms an arithmetic progression, show that

Pk

i=1ai must be an integer Here [r] denotes the greatest integer less than or equal to r

Problem Let a and b be positive integers, and let A and B be finite sets of integers satisfying:

(i)A and B are disjoint;

(ii) if an integer i belongs either to A or to B, then i+a belongs to A or i−b

belongs to B

Prove thata|A|=b|B| (Here |X| denotes the number of elements in the setX.)

Problem Let ABCD be a quadrilateral inscribed in a circle ω, and let P be a point on the extension ofAC such thatP B andP D are tangent toω The tangent at

(29)

XXVI Asian Pacific Mathematics Olympiad

Time allowed: hours Each problem if worth points Problem For a positive integer m denote by S(m) and P(m) the sum and product, respectively, of the digits of m Show that for each positive integern, there exist positive integers a1, a2, , an satisfying the following conditions:

S(a1)< S(a2)<· · ·< S(an) and S(ai) =P(ai+1) (i= 1,2, , n)

(We let an+1 =a1.)

(Proposed by the Problem Committee of the Japan Mathematical Olympiad Foundation) Problem Let S = {1,2, ,2014} For each non-empty subset T ⊆ S, one of its members is chosen as its representative Find the number of ways to assign representatives to all non-empty subsets of S so that if a subset D⊆ S is a disjoint union of non-empty subsets A, B, C ⊆ S, then the representative of D is also the representative of at least one of A, B, C

(Proposed by Warut Suksompong, Thailand) Problem Find all positive integers n such that for any integer k there exists an integer a for which a3+a−k is divisible by n

(Proposed by Warut Suksompong, Thailand) Problem Letn andb be positive integers We sayn isb-discerning if there exists a set consisting of n different positive integers less than b that has no two different subsetsU andV such that the sum of all elements inU equals the sum of all elements in V

(a) Prove that is a 100-discerning (b) Prove that is not 100–discerning

(Proposed by the Senior Problems Committee of the Australian Mathematical Olympiad Committee) Problem Circles ω and Ω meet at points A and B Let M be the midpoint of the arc AB of circle ω (M lies inside Ω) A chord M P of circle ω intersects Ω atQ

(Q lies inside ω) Let `P be the tangent line to ω at P, and let `Q be the tangent

line to Ω at Q Prove that the circumcircle of the triangle formed by the lines`P,`Q,

and AB is tangent to Ω

(30)

XXVII Asian Pacific Mathematics Olympiad

Time allowed: hours Each problem if worth points

Problem Let ABC be a triangle, and let D be a point on side BC A line through D intersects side AB at X and ray AC at Y The circumcircle of triangle

BXD intersects the circumcircleω of triangle ABC again at point Z 6=B The lines

ZD and ZY intersect ω again at V and W, respectively Prove that AB=V W Proposed by Warut Suksompong, Thailand

Problem Let S ={2,3,4, } denote the set of integers that are greater than or equal to Does there exist a function f :S →S such that

f(a)f(b) =f(a2b2) for alla, b∈S with a6=b?

Proposed by Angelo Di Pasquale, Australia

Problem A sequence of real numbers a0, a1, is said to be good if the following

three conditions hold

(i) The value of a0 is a positive integer

(ii) For each non-negative integer i we have ai+1 = 2ai+ or ai+1 =

ai

ai+

(iii) There exists a positive integer k such thatak= 2014

Find the smallest positive integer n such that there exists a good sequencea0, a1,

of real numbers with the property that an= 2014

Proposed by Wang Wei Hua, Hong Kong

Problem Letn be a positive integer Consider 2n distinct lines on the plane, no two of which are parallel Of the 2n lines,n are colored blue, the other n are colored red Let B be the set of all points on the plane that lie on at least one blue line, and R the set of all points on the plane that lie on at least one red line Prove that there exists a circle that intersects B in exactly 2n−1 points, and also intersects R

in exactly 2n−1 points

Proposed by Pakawut Jiradilok and Warut Suksompong, Thailand

Problem Determine all sequencesa0, a1, a2, of positive integers witha0 ≥2015

such that for all integers n ≥1: (i)an+2 is divisible by an;

(ii) |sn+1−(n+ 1)an|= 1, where sn+1 =an+1−an+an−1− · · ·+ (−1)n+1a0

(31)

XXVIII Asian Pacific Mathematics Olympiad

March, 2016

Time allowed: hours Each problem is worth points

The contest problems are to be kept confidential until they are posted on the official APMO website http://apmo.ommenlinea.org

Please not disclose nor discuss the problems over online until that date The use of calculators is not allowed

Problem We say that a triangle ABC is great if the following holds: for any point Don the side BC, ifP and Qare the feet of the perpendiculars fromD to the lines AB and AC, respectively, then the reflection of D in the line P Q lies on the circumcircle of the triangle ABC

Prove that triangle ABC is great if and only if∠A= 90◦ and AB =AC.

Problem A positive integer is calledfancy if it can be expressed in the form 2a1

+ 2a2

+· · ·+ 2a100,

where a1, a2, , a100 are non-negative integers that are not necessarily distinct

Find the smallest positive integern such that no multiple of n is a fancy number Problem LetAB andAC be two distinct rays not lying on the same line, and let ω be a circle with center O that is tangent to rayAC at E and ray AB atF Let R be a point on segment EF The line through O parallel toEF intersects line AB at P Let N be the intersection of lines P R and AC, and let M be the intersection of line AB and the line throughR parallel to AC Prove that lineM N is tangent to ω Problem The country Dreamland consists of 2016 cities The airline Starways wants to establish some one-way flights between pairs of cities in such a way that each city has exactly one flight out of it Find the smallest positive integer k such that no matter how Starways establishes its flights, the cities can always be partitioned into k groups so that from any city it is not possible to reach another city in the same group by using at most 28 flights

Problem Find all functionsf :R+→R+ such that

(32)

XXIX Asian Pacific Mathematics Olympiad

March, 2017

Problem We call a 5-tuple of integers arrangeable if its elements can be labeled

a, b, c, d, e in some order so that a−b+c−d+e= 29 Determine all 2017-tuples of integers n1, n2, , n2017 such that if we place them in a circle in clockwise order,

then any 5-tuple of numbers in consecutive positions on the circle is arrangeable

Proposed by Warut Suksompong, Thailand

Problem LetABC be a triangle withAB < AC LetDbe the intersection point of the internal bisector of angle BAC and the circumcircle of ABC Let Z be the intersection point of the perpendicular bisector of AC with the external bisector of angle ∠BAC Prove that the midpoint of the segment ABlies on the circumcircle of triangle ADZ

Proposed by Equipo Nicaragua, Nicaragua

Problem LetA(n) denote the number of sequencesa1 ≥a2 ≥ .≥ak of positive

integers for which a1+· · ·+ak =nand each ai+ is a power of two (i= 1,2, , k)

Let B(n) denote the number of sequences b1 ≥b2 ≥ .≥bm of positive integers for

which b1+· · ·+bm =n and each inequality bj ≥2bj+1 holds (j = 1,2, , m−1)

Prove thatA(n) =B(n) for every positive integer n

(33)

Problem Call a rational numberr powerful if r can be expressed in the form pqk for some relatively prime positive integers p, q and some integer k >1 Leta, b, c be positive rational numbers such that abc = Suppose there exist positive integers

x, y, z such that ax+by+cz is an integer Prove thata, b, c are all powerful.

Proposed by Jeck Lim, Singapur

Problem Let n be a positive integer A pair of n-tuples (a1, , an) and

(b1, , bn) with integer entries is called an exquisite pair if |a1b1+· · ·+anbn| ≤1

Determine the maximum number of distinct n-tuples with integer entries such that any two of them form an exquisite pair

(34)

XXX Asian Pacific Mathematics Olympiad

March, 2018 Time allowed: hours Each problem is worth points The contest problems are to be kept confidential until they are posted on the official APMO website http://apmo.ommenlinea.org

Problem Let H be the orthocenter of the triangle ABC Let M and N be the midpoints of the sides AB and AC, respectively Assume that H lies inside the quadrilateral BM N C and that the circumcircles of triangles BM H and CN H are tangent to each other The line through H parallel toBC intersects the circumcircles of the triangles BM H and CN H in the points K and L, respectively Let F be the intersection point ofM K andN Land letJ be the incenter of triangleM HN Prove that F J =F A

Proposed by Mahdi Etesamifard, Iran

Problem Letf(x) and g(x) be given by

f(x) =

x +

1

x−2 +

x−4 +· · ·+

x−2018 and

g(x) =

x−1 +

x−3 +

x−5 +· · ·+

x−2017 Prove that

|f(x)−g(x)|>2

for any non-integer real number xsatisfying 0< x < 2018

Proposed by Senior Problems Committee of the Australian Mathematical Olympiad Committee

Problem A collection of n squares on the plane is called tri-connected if the following criteria are satisfied:

(35)

(ii) If two squares have a point P in common, then P is a vertex of each of the squares

(iii) Each square touches exactly three other squares

How many positive integers n are there with 2018≤n ≤3018, such that there exists a collection of n squares that is tri-connected?

Proposed by Senior Problems Committee of the Australian Mathematical Olympiad Committee

Problem Let ABC be an equilateral triangle From the vertex A we draw a ray towards the interior of the triangle such that the ray reaches one of the sides of the triangle When the ray reaches a side, it then bounces off following the law of reflection, that is, if it arrives with a directed angleα, it leaves with a directed angle 180◦−α Aftern bounces, the ray returns to A without ever landing on any of the other two vertices Find all possible values of n

Proposed by Daniel Perales and Jorge Garza, Mexico

Problem Find all polynomialsP(x) with integer coefficients such that for all real numbers s and t, if P(s) andP(t) are both integers, then P(st) is also an integer

(36)

XXXI Asian Pacific Mathematics Olympiad

March, 2019

Problem LetZ+ be the set of positive integers Determine all functionsf :

Z+→

Z+ such thata2+f(a)f(b) is divisible byf(a) +b for all positive integers a and b

Problem Let m be a fixed positive integer The infinite sequence {an}n≥1 is defined in the following way: a1 is a positive integer, and for every integer n≥ we have

an+1 =

(

a2

n+ 2m if an<2m

an/2 if an≥2m

For each m, determine all possible values ofa1 such that every term in the sequence is an integer

Problem Let ABC be a scalene triangle with circumcircle Γ Let M be the midpoint of BC A variable point P is selected in the line segment AM The circumcircles of triangles BP M and CP M intersect Γ again at points D and E, respectively The lines DP and EP intersect (a second time) the circumcircles to triangles CP M and BP M at X and Y, respectively Prove that as P varies, the circumcircle of 4AXY passes through a fixed point T distinct from A

Problem Consider a 2018×2019 board with integers in each unit square Two unit squares are said to be neighbours if they share a common edge In each turn, you choose some unit squares Then for each chosen unit square the average of all its neighbours is calculated Finally, after these calculations are done, the number in each chosen unit square is replaced by the corresponding average Is it always possible to make the numbers in all squares become the same after finitely many turns? Problem Determine all the functionsf :R→R such that

(37)

1st APMO 1989

Problem

ai are positive reals s = a1 + + an Prove that for any integer n > we have (1 + a1) (1 + an) < + s + s2/2! + + sn/n!

Solution

We use induction on n For n = the rhs is + a1 + a2 + a1a2 + (a12 + a22)/2 > lhs Assume the result is true for n We note that, by the binomial theorem, for s and t positive we have sm+1 + (m+1) t sm < (s + t)m+1, and hence sm+1/(m+1)! + t sm/m! < (s + t)m+1/(m+1)! Summing from m = to n+1 we get (s + t) + (s2/2! + t s/1!) + (s3/3! + t s2/2!) + + (sn+1/(n+1)! + t sn/n!) < (s + t) + (s + t)2/2! + + (s + t)n+1/(n+1)! Adding to each side gives that (1 + t)(1 + s + s2/2! + + sn/n!) < (1 + (s+t) + + (s+t)n+1/(n+1)! Finally putting t = an+1 and using the the result for n gives the result for n+1

Problem

Prove that 5n2 = 36a2 + 18b2 + 6c2 has no integer solutions except a = b = c = n = Solution

The rhs is divisible by 3, so must divide n So 5n2 - 36a2 - 18b2 is divisible by 9, so must divide c We can now divide out the factor to get: 5m2 = 4a2 + 2b2 + 6d2 Now take m, a, b, d to be the solution with the smallest m, and consider residues mod 16 Squares = 0, 1, 4, or mod 16 Clearly m is even so 5m2 = or mod 16 Similarly, 4a2 = or mod 16 Hence 2b2 + 6d2 = 0, or 12 mod 16 But 2b2 = 0, or mod 16 and 6d2 = 0, or mod 16 Hence 2b2 + 6d2 = 0, 2, 6, 8, 10 or 14 mod 16 So it must be So b and d are both even So a cannot be even, otherwise m/2, a/2, b/2, d/2 would be a solution with smaller m/2 < m

(38)

Problem

ABC is a triangle X lies on the segment AB so that AX/AB = 1/4 CX intersects the median from A at A' and the median from B at B'' Points B', C', A'', C'' are defined similarly Find the area of the triangle A''B''C'' divided by the area of the triangle A'B'C' Solution

Answer: 25/49

Let M be the midpoint of AB We use vectors center G Take GA = A, GB = B, GC = C Then GM = A/2 + B/2 and GX = 3/4 A + 1/4 C Hence GA' = 2/5 A (showing it lies on GA) = 4/5 (3/4 A + 1/4 B) + 1/5 C, since A + B + C = (which shows it lies on CX) Similarly, GB'' = 4/7 (1/2 A + 1/2 C) (showing it lies on the median through B) = 2/7 A + 2/7 C = 5/7 (2/5 A) + 2/7 C (showing it lies on CA' and hence on CX) Hence GB'' = -2/7 B So we have shown that GB'' is parallel to GB' and 5/7 the length The same applies to the distances from the centroid to the other vertices Hence triangle A''B''C'' is similar to triangle A'B'C' and its area is 25/49 times the area of A'B'C'

Problem

Show that a graph with n vertices and k edges has at least k(4k - n2)/3n triangles Solution

Label the points 1, 2, , n and let point i have degree di (no of edges) Then if i and j are joined they have at least di + dj - other edges between them, and these edges join them to n - other points So there must be at least di + dj - n triangles which have i and j as two vertices Hence the total number of triangles must be at least ∑edges ij (di + dj - n)/3 But ∑edges ij (di + dj) = ∑ di2

(39)

Problem

f is a strictly increasing real-valued function on the reals It has inverse f-1 Find all possible f such that f(x) + f-1(x) = 2x for all x

Solution

Answer: f(x) = x + b for some fixed real b

Suppose for some a we have f(a) ≠ a Then for some b ≠ we have f(a) = a + b Hence f(a + b) = a + 2b (because f( f(a) ) + f-1( f(a) ) = f(a), so f(a + b) + a = 2a + 2b ) and by two easy inductions, f(a + nb) = a + (n+1)b for all integers n (positive or negative) Now take any x between a and a + b Suppose f(x) = x + c The same argument shows that f(x + nc) = x + (n+1)c Since f is strictly increasing x + c must lie between f(a) = a + b and f(a+b) = a + 2b So by a simple induction x + nc must lie between a + nb and a + (n+1)b So c lies between b + (x-a)/n and b + (a+b-x)/n or all n Hence c = b Hence f(x) = x + b for all x

(40)

2nd APMO 1990

Problem

Given θ in the range (0, π) how many (incongruent) triangles ABC have angle A = θ, BC = 1, and the following four points concyclic: A, the centroid, the midpoint of AB and the midpoint of AC?

Solution

Answer: for θ ≤ 60 deg Otherwise none

Let O be the circumcenter of ABC and R the circumradius, let M be the midpoint of BC, and let G be the centroid We may regard A as free to move on the circumcircle, whilst O, B and C remain fixed Let X be the point on MO such that MX/MO = 1/3 An expansion by a factor 3, center M, takes G to A and X to O, so G must lie on the circle center X radius R/3

The circle on diameter OA contains the midpoints of AB and AC (since if Z is one of the midpoints OZ is perpendicular to the corresponding side) So if G also lies on this circle then angle OGA = 90 deg and hence angle MGO = 90 deg, so G must also lie on the circle diameter OM Clearly the two circles for G either not intersect in which case no triangle is possible which satisfies the condition or they intersect in one or two points But if they intersect in two points, then corresponding triangles are obviously congruent (they just interchange B and C) So we have to find when the two circle intersect

(41)

Problem

x1, , xn are positive reals sk is the sum of all products of k of the xi (for example, if n = 3, s1 = x1 + x2 + x3, s2 = x1x2 + x2x3 + x3x1, s3 = x1x2x3) Show that sksn-k ≥ (nCk)2 sn for < k < n

Solution

Each of sk and sn-khave nCk terms So we may multiply out the product sksn-k to get a sum of (nCk)2 terms We now apply the arithmetic/geometric mean result The product of all the terms must be a power of sn by symmetry and hence must be sn to the power of (nCk)2 So the geometric mean of the terms is just sn Hence result

Problem

A triangle ABC has base AB = and the altitude from C length h What is the maximum possible product of the three altitudes? For which triangles is it achieved?

Solution

Answer: for h ≤ 1/2, maximum product is h2, achieved by a triangle with right-angle at C; for h > 1/2, the maximum product is h3/(h2 + 1/4), achieved by the isosceles triangle (AC = BC)

Solution by David Krumm

Let AC = b, BC = a, let the altitude from A have length x and the altitude from B have length y Then ax = by = h, so hxy = h3/ab But h = a sin B and b/sin B = 1/sin C, so h = ab sin C and the product hxy = h2 sin C

The locus of possible positions for C is the line parallel to AB and a distance h from it [Or strictly the pair of such lines.] If h ≤ 1/2, then there is a point on that line with angle ACB = 90 deg, so in this case we can obtain hxy = h2 by taking angle ACB = 90 deg and that is clearly the best possible

If h > 1/2, then there is no point on the line with angle ACB = 90 deg Let L be the perpendicular bisector of AB and let L meet the locus at C Then C is the point on the locus with the angle C a maximum For if D is any other point of the line then the

(42)

My original, less elegant, solution is as follows.

Take AP perpendicular to AB and length h Take Q to be on the line parallel to AB through P so that BQ is perpendicular to AB Then C must lie on the line PQ (or on the corresponding line on the other side of AB) Let a(A) be the length of the altitude from A to BC and a(B) the length of the altitude from B to AC If C maximises the product h a(A) a(B), then it must lie on the segment PQ, for if angle ABC is obtuse, then both a(A) and a(B) are shorter than for ABQ Similarly if BAC is obtuse So suppose PC = x with ≤ x ≤ Then AC = √(x2

+ h2), so a(B) = h/√(x2 + h2) Similarly, a(A) = h/√( (1-x)2 + h2) So we wish to minimise f(x) = (x2 + h2)( (1-x)2 + h2) = x4 - 2x3 + (2h2 + 1)x2 - 2h2x + h4 + h2 We have f '(x) = 2(2x-1)(x2 - x + h2), which has roots x = 1/2, 1/2 ± √(1/4 - h2) Thus for h >= 1/2, the minimum is at x = 1/2, in which case CA = CB For h < 1/2, the minimum is at x = 1/2 ± √(1/4 - h2

) But if M is the midpoint of AB and D is the point on AB with AD = 1/2 ± √(1/4 - h2), then DM = √(1/4 - h2

) But DC = h, and angle CDM = 90, so MC = 1/2 and hence angle ACB = 90

(DK's solution added Jun 2002)

Problem

A graph with n points satisfies the following conditions: (1) no point has edges to all the other points, (2) there are no triangles, (3) given any two points A, B such that there is no edge AB, there is exactly one point C such that there are edges AC and BC Prove that each point has the same number of edges Find the smallest possible n

Solution Answer:

We say A and B are joined if there is an edge AB For any point X we write deg X for the number of points joined to X Take any point A Suppose deg A = m So there are m points B1, B2, , Bm joined to A No Bi, Bj can be joined for i ≠ j, by (2), and a point C ≠ A cannot be joined to Bi and Bj for i ≠ j, by (3) Hence there are deg Bi - points Cij joined to Bi and all the Cij are distinct

(43)

But now if we started with Bi instead of A and repeated the whole argument we would establish that deg Bi is the same as the deg Chk, where Chk is one of the points joined to Ci1 Thus all the points have the same degree

Suppose the degree of each point is m Then with the notation above there is point A, m points Bi and m(m-1) points Cjk or m2 + in all So n = m2 + The smallest possible m is 1, but that does not yield a valid graph because if does not satisfy (1) The next smallest possibility is m = 2, giving points It is easy to check that the pentagon satisfies all the conditions

Problem

Show that for any n ≥ we can find a convex hexagon which can be divided into n congruent triangles

Solution

We use an isosceles trianglea as the unit The diagram shows n = and n = We can get any n ≥ by adding additional rhombi in the middle

(44)

3rd APMO 1991

Problem

ABC is a triangle G is the centroid The line parallel to BC through G meets AB at B' and AC at C' Let A'' be the midpoint of BC, C'' the intersection of B'C and BG, and B'' the intersection of C'B and CG Prove that A''B''C'' is similar to ABC

Solution

Let M be the midpoint of AB and N the midpoint of AC Let A''M meet BG at X Then X must be the midpoint of A''M (an expansion by a factor center B takes A''M to CA and X to N) Also BX/BN = 1/2 and BG/BN = 2/3, so XG = BX/3 Let the ray CX meet AB at Z Then ZX = CX/3 (There must be a neat geometric argument for this, but if we take vectors origin B, then BX = BN/2 = BA/4 + BC/4, so BZ= BA/3 and so XZ = 1/3 (BA/4 - 3BC/4) = CX/3.) So now triangles BXC and ZXG are similar, so ZG is parallel to BC, so Z is B' and X is C'' But A''X is parallel to AC and 1/4 its length, so A''C'' is parallel to AC and 1/4 its length Similarly A''B'' is parallel to AB and 1/4 its length Hence A''B''C'' is similar to ABC

Problem

There are 997 points in the plane Show that they have at least 1991 distinct midpoints Is it possible to have exactly 1991 midpoints?

Solution

Answer: yes Take the 997 points collinear at coordinates x = 1, 3, , 1993 The midpoints are 2, 3, 4, , 1992

Take two points A and B which are the maximum distance apart Now consider the following midpoints: M, the midpoint of AB, the midpoint of each AX for any other X in the set (not A or B), and the midpoint of each BX We claim that all these are distinct Suppose X and Y are two other points (apart from A and B) Clearly the midpoints of AX and AY must be distinct (otherwise X and Y would coincide) Similarly the

midpoints of BX and BY must be distinct Equally, the midpoint of AX cannot be M (or X would coincide with B), nor can the midpoint of BX be M Suppose, finally, that N is the midpoint of AX and BY Then AYXB is a parallelogram and either AX or BY must exceed AB, contradicting the maximality of AB So we have found 1991 distinct

(45)

Problem 3

xi and yi are positive reals with ∑1n xi = ∑1n yi Show that ∑1n xi2/(xi + yi) ≥ (∑1n xi)/2 Solution

We use Cauchy-Schwartz: ∑ (x/√(x+y) )2 ∑ (√(x+y) )2 ≥ (∑ x )2 So ∑ x2/(x+y) >= (∑ x)2/(∑(x+y) = 1/2 ∑ x

Problem

A sequence of values in the range 0, 1, 2, , k-1 is defined as follows: a1 = 1, an = an-1 + n (mod k) For which k does the sequence assume all k possible values?

Solution

Let f(n) = n(n+1)/2, so an = f(n) mod k If k is odd, then f(n+k) = f(n) mod k, so the sequence can only assume all possible values if a1, , ak are all distinct But f(k-n) = f(n) mod k, so there are at most (k+1)/2 distinct values Thus k odd does not work If k is even, then f(n+2k) = f(n) mod k, so we need only look at the first 2k values But f((2k-1-n) = f(n) mod k and f(2k-1) = mod k, so the sequence assumes all values iff a1, a2, , ak-1 assume all the values 1, 2, , k-1

Checking the first few, we find k = 2, 4, 8, 16 work and k = 6, 10, 12, 14 not So this suggests that k must be a power of Suppose k is a power of If f(r) = f(s) mod k for some < r, s < k, then (r - s)(r + s + 1) = mod k But each factor is < k, so neither can be divisible by k Hence both must be even But that is impossible (because their sum is 2r+1 which is odd), so each of f(1), f(2), , f(k-1) must be distinct residues mod k Obviously none can be mod k (since 2k cannot divide r(r+1) for < r < k and so k cannot divide f(r) ) Thus they must include all the residues 1, 2, k-1 So k a power of does work

(46)

Problem

Circles C and C' both touch the line AB at B Show how to construct all possible circles which touch C and C' and pass through A

Solution

Take a common tangent touching C' at Q' and C at Q Let the line from Q to A meet C again at P Let the line from Q' to A meet C' again at P' Let the C have center O and C' have center O' Let the lines OP and O'P' meet at X Take X as the center of the required circle There are two common tangents, so this gives two circles, one enclosing C and C' and one not

(47)

4th APMO 1992

Problem

A triangle has sides a, b, c Construct another triangle sides (-a + b + c)/2, (a - b + c)/2, (a + b - c)/2 For which triangles can this process be repeated arbitrarily many times? Solution

Answer: equilateral

We may ignore the factor 1/2, since clearly a triangle with sides x, y, z can be constructed iff a triangle with sides 2x, 2y, 2z can be constructed

(48)

Problem

Given a circle C centre O A circle C' has centre X inside C and touches C at A Another circle has centre Y inside C and touches C at B and touches C' at Z Prove that the lines XB, YA and OZ are concurrent

Solution

We need Ceva's theorem, which states that given points D, E, F on the lines BC, CA, AB, the lines AD, BE, CF are concurrent iff (BD/DC) (CE/EA) (AF/FB) = (where we pay attention to the signs of BD etc, so that BD is negative if D lies on the opposite side of B to C) Here we look at the triangle OXY, and the points A on OX, B on OY and Z on XY (it is obvious that Z does lie on XY) We need to consider (OA/AX) (XZ/ZY) (YB/BO) AX and BY are negative and the other distances positive, so the sign is plus Also OA = OB, AX = XZ, and ZY = YB (ignoring signs), so the expression is Hence AY, XB and OZ are concurrent as required

Problem

Given three positive integers a, b, c, we can derive numbers using one addition and one multiplication and using each number just once: a+b+c, a+bc, b+ac, c+ab, (a+b)c, (b+c)a, (c+a)b, abc Show that if a, b, c are distinct positive integers such that n/2 < a, b, c, <= n, then the derived numbers are all different Show that if p is prime and n ≥ p2, then there are just d(p-1) ways of choosing two distinct numbers b, c from {p+1, p+2, , n} so that the numbers derived from p, b, c are not all distinct, where d(p-1) is the number of positive divisors of p-1

Solution

If < a < b < c, we have a + b + c < ab + c < b + ac < a + bc and (b+c)a < (a+c)b < (a+b)c < abc We also have b + ac < (a+c)b So we just have to consider whether a + bc = (b+c)a But if a > c/2, which is certainly the case if n/2 < a, b, c ≤ n, then a(b + c - 1) > c/2 (b + b) = bc, so a + bc < a(b + c) and all numbers are different

(49)

Problem

Find all possible pairs of positive integers (m, n) so that if you draw n lines which intersect in n(n-1)/2 distinct points and m parallel lines which meet the n lines in a further mn points (distinct from each other and from the first n(n-1)/2) points, then you form exactly 1992 regions

Solution

Answer: (1, 995), (10, 176), (21, 80)

n lines in general position divide the plane into n(n+1)/2 + regions and each of the m parallel lines adds a further n+1 regions So we require n(n+1)/2 + + m(n+1) = 1992 or (n+1)(2m+n) = 3982 = 2·11·181 So n+1 must divide 3982, also (n+1)n < 3982, so n ≤ 62 We are also told that n is positive Thus n = is disallowed The remaining

possibilities are n+1 = 2, 11, 2·11 These give the three solutions shown above Problem

a1, a2, a3, an is a sequence of non-zero integers such that the sum of any consecutive terms is positive and the sum of any 11 consecutive terms is negative What is the largest possible value for n?

Solution Answer: 16

We cannot have 17 terms, because then:

a1 + a2 + + a11 <

a2 + a3 + + a12 <

a3 + a4 + + a13 <

a7 + a8 + + a17 <

So if we add the inequalities we get that an expression is negative But notice that each column is positive Contradiction

On the other hand, a valid sequence of 16 terms is: -5, -5, 13, -5, -5, -5, 13, -5, -5, 13, -5, -5, -5, 13, -5, -5 Any run of terms has two 13s and five -5s, so sums to Any run of 11 terms has three 13s and eight -5s, so sums to -1

(50)

5th APMO 1993

Problem

A, B, C is a triangle X, Y, Z lie on the sides BC, CA, AB respectively, so that AYZ and XYZ are equilateral BY and CZ meet at K Prove that YZ2 = YK·YB

Solution

Use vectors Take A as the origin Let AZ = b, AY = c We may take the equilateral triangles to have side 1, so b2 = c2 = and b.c = 1/2 Take AB to be k b AX is b + c, so AC must be k/(k-1) c (then AX = 1/k (k b) + (1 - 1/k) ( k/(k-1) c), which shows that X lies on BC)

Hence AK = k/(k2 - k + 1) (b + (k-1) c) Writing this as (k2-k)/(k2-k+1) c + 1/(k2-k+1) (k b) shows that it lies on BY and writing it as k/(k2-k+1) b + (k2-2k+1) ( k/(k-1) c) shows that it lies on CZ Hence YK.YB = YK.YB= ( k/(k2-k+1) b - 1/(k2-k+1) c) ( k b - c) = (k b - c)2/(k2-k+1) = = YZ2

Thank to Achilleas Porfyriadis for the following geometric proof

BZX and XYC are similar (sides parallel), so BZ/ZX = XY/YC But XYZ is equilateral, so BZ/ZY = ZY/YC Also ∠BZY = ∠ZYC = 120o, so BZY and ZYC are similar Hence

(51)

Problem

How many different values are taken by the expression [x] + [2x] + [5x/3] + [3x]+ [4x] for real x in the range ≤ x ≤ 100?

Solution Answer: 734

Let f(x) = [x] + [2x] + [3x] + [4x] and g(x) = f(x) + [5x/3] Since [y+n] = n + [y] for any integer n and real y, we have that f(x+1) = f(x) + 10 So for f it is sufficient to look at the half-open interval [0, 1) f is obviously monotonic increasing and its value jumps at x = 0, 1/4, 1/3, 1/2, 2/3, 3/4 Thus f(x) takes different values on [0, 1)

g(x+3) = g(x), so for g we need to look at the half-open interval [0, 3) g jumps at the points at which f jumps plus additional points: 3/5, 1/5, 4/5, 2/5 So on [0, 3), g(x) takes x + = 22 different values Hence on [0, 99), g(x) takes 33 x 22 = 726 different values Then on [99, 100] it takes a further + + (namely g(99), g(99 1/4), g(99 1/3), g(99 1/2), g(99 3/5), g(99 2/3), g(99 3/4), g(100) ) Thus in total g takes 726 + = 734 different values

Problem

p(x) = (x + a) q(x) is a real polynomial of degree n The largest absolute value of the coefficients of p(x) is h and the largest absolute value of the coefficients of q(x) is k Prove that k ≤ hn

Solution

Let p(x) = p0 + p1x + + pnxn, q(x) = q0 + q1x + + qn-1xn-1, so h = max |pi|, k = max |qi|

If a = 0, then the result is trivial So assume a is non-zero We have pn = qn-1, pn-1 = qn-2 + aqn-1, pn-2 = qn-3 + aqn-2, , p1 = q0 + aq1, p0 = aq0

We consider two cases Suppose first that |a| ≥ Then we show by induction that |qi| ≤ (i+1) h We have q0 = p0/a, so |q0| ≤ h, which establishes the result for i = Suppose it is true for i We have qi+1 = (pi+1 - qi)/a, so |qi+1| ≤ |pi+1| + |qi| ≤ h + (i+1)h = (i+2)h, so it is true for i+1 Hence it is true for all i < n So k ≤ max(h, 2h, , nh) = nh

(52)

true for i We have qn-i-1 = pn-i - aqn-i, so |qn-i-1n-i| + |qn-i| ≤ h + ih = (i+1)h, so it is true for i+1 Hence it is true for all ≤ i ≤ n Hence k ≤ max(h, 2h, , nh) = nh

Problem

Find all positive integers n for which xn + (x+2)n + (2-x)n = has an integral solution Solution

Answer: n =

There are obviously no solutions for even n, because then all terms are non-negative and at least one is positive x = -4 is a solution for n = So suppose n is odd n and > If x is positive, then xn + (x+2)n > (x+2)n > (x-2)n, so xn + (x+2)n + (2-x)n > Hence any solution x must be negative Put x = -y Clearly x = -1 is not a solution for any n, so if x = -y is a solution then (x+2) = -(y-2) ≤ we have (y+2)n = yn + (y-2)n Now = ( (y+2) - (y-2) ) divides (y+2)n - (y-2)n Hence divides y Put y = 2z, then we have (z+1)n = zn + (z-1)n Now divides (z+1)n - (z-1)n so divides z, so z+1 and z-1 are both odd But an - bn = (a - b)(an-1n-2b + an-3b2 + + bn-1) If a and b are both odd, then each term in (an-1 n-2b + an-3b2 + + bn-1) is odd and since n is odd there are an odd number of terms, so (a n-1

n-2b + an-3b2 + + bn-1) is odd Hence, putting a=z+1, b=z-1, we see that (z+1)n - (z-1)n = 2(an-1n-2b + an-3b2 + + bn-1) is not divisible by But it equals zn with z even Hence n must be

Problem

C is a 1993-gon of lattice points in the plane (not necessarily convex) Each side of C has no lattice points except the two vertices Prove that at least one side contains a point (x, y) with 2x and 2y both odd integers

Solution

We consider the midpoint of each side We say that a vertex (x, y) is pure if x and y have the same parity and impure if x and y have opposite parity Since the total number of vertices is odd, there must be two adjacent pure vertices P and Q or two adjacent impure vertices P and Q But in either case the midpoint of P and Q either has both coordinates integers, which we are told does not happen, or as both coordinates of the form an integer plus half, which therefore must occur

(53)

6th APMO 1994

Problem

Find all real-valued functions f on the reals such that (1) f(1) = 1, (2) f(-1) = -1, (3) f(x) ≤ f(0) for < x < 1, (4) f(x + y) ≥ f(x) + f(y) for all x, y, (5) f(x + y) ≤ f(x) + f(y) + for all x, y

Solution

Answer: f(x) = [x]

f(x+1) >= f(x) + f(1) = f(x) + by (4) and (1) But f(x) ≥ f(x+1) + f(-1) = f(x+1) - by (4) and (2) Hence f(x+1) = f(x) +

In particular, = f(1) = f(0+1) = f(0) + 1, so f(0) = Hence, by (3), f(x) ≤ for < x < But, by (5), = f(1) = f(x + 1-x) ≤ f(x) + f(1-x) + 1, so f(x) + f(1-x) ≥ But if < x < 1, then also < 1-x < 1, so f(x) = f(1-x) =

Thus we have established that f(x) = for ≤ x < 1, and f(x+1) = f(x) + It follows that f(x) = [x] for all x

Problem

ABC is a triangle and A, B, C are not collinear Prove that the distance between the orthocenter and the circumcenter is less than three times the circumradius

Solution

We use vectors It is well-known that the circumcenter O, the centroid G and the

orthocenter H lie on the Euler line and that OH = OG Hence taking vectors with origin O, OH = OG = OA + OB + OC Hence |OH| ≤ |OA| + |OB| + |OC| = x circumradius We could have equality only if ABC were collinear, but that is impossible, because ABC would not then be a triangle

(54)

Problem

Find all positive integers n such that n = a2 + b2, where a and b are relatively prime positive integers, and every prime not exceeding √n divides ab

Solution

Answer: = 12 + 12, = 12 + 22, 13 = 22 + 32

The key is to use the fact that a and b are relatively prime We show in fact that they must differ by (or 0) Suppose first that a = b Then since they are relatively prime they must both be That gives the first answer above So we may take a > b Then (a - b)2 < a2 + b2 = n, so if a - b is not 1, it must have a prime factor which divides ab But then it must divide a or b and hence both Contradiction So a = b +

Now (b - 1)2 < b2 < n, so any prime factor p of b - must divide ab = b(b + 1) It cannot divide b (or it would divide b and b - and hence 1), so it must divide b + and hence must be But if divides b - 1, then does not divide b(b - 1), so b - must be 0, or But it is now easy to check the cases a, b = (4, 3), (3, 2), (2, 1)

Problem

Can you find infinitely many points in the plane such that the distance between any two is rational and no three are collinear?

Solution Answer: yes Let θ = cos-1

3/5 Take a circle center O radius and a point X on the circle Take Pn on the circle such that angle XOPn = 2nθ We establish (A) that the Pn are all distinct and (B) that the distances PmPn are all rational

(55)

It follows that (1) if cos x is rational, then so is cos nx, and (2) if cos x is rational, then x/π is irrational To see (2), suppose that x/π = m/n, with m and n integers Then nx is a multiple of π and hence cos nx = 0, so pn(2 cos x) = Now we may write pn(y) = yn + an-1yn-1 + + a0 Now if also cos x = r/s, with r and s relatively prime integers, then we have, pn(2 cos x) = rn + an-1s rn-1 + + a0sn = But now s divides all terms except the first Contradiction

Thus we cannot have cos mθ = cos nθ for any distinct integers m, n, for then θ/π would be rational as well as cos θ So we have established (A)

We have also established that all cos nθ are rational But since sin(n+1)x = sin nx cos x + cos nx sin x and sin θ = 4/5, it is a trivial induction that all sin nθ are also rational Now PmPn = |sin(m - n)θ|, so all the distances PmPn are rational, thus establishing (B) Problem

Prove that for any n > there is either a power of 10 with n digits in base or a power of 10 with n digits in base 5, but not both

Solution

10k has n digits in base iff 5n-1 < 10k < 5n Similarly, 10h has n digits in base iff 2n-1 < 10h < 2n So if we can find both 10k with n digits in base and 10h with n digits in base 2, then, multiplying the two inequalities, we have 10n-1 < 10h+k < 10n, which is clearly impossible This establishes the "but not both" part

Let S be the set of all positive powers of or Order the members of S in the usual way and let an be the n-1th member of the set We claim that if an = 2k, then 10k has n digits in base 5, and if an = 5h, then 10h has n digits in base We use induction on n

a2 = 21, a3 = 22, a4 = 51, a5 = 23, Thus the claim is certainly true for n = Suppose it is true for n

(56)

Jacob Tsimerman pointed out that the second part can be done in a similar way to the first - which is neater than the above:

If no power of 10 has n digits in base or 5, then for some h, k: 10h < 2n-1 < 2n < 10h+1 and 10k < 5n-1 < 5n < 10k+1 Hence 10h+k < 10n-1 < 10n < 10h+k+2 But there is only one power of 10 between h+k and h+k+2

(57)

7th APMO 1995

Problem

Find all real sequences x1, x2, , x1995 which satisfy 2√(xn - n + 1) ≥ xn+1 - n + for n = 1, 2, , 1994, and 2√(x1995 - 1994) ≥ x1 +

Solution

Answer: the only such sequence is 1, 2, 3, , 1995

Put x1995 = 1995 + k We show by induction (moving downwards from 1995) that xn ≥ n + k For suppose xn+1 ≥ n + k + 1, then 4(xn - n + 1) ≥ (xn+1- n + 1)2 ≥ (k+2)2 ≥ 4k + 4, so xn ≥ n + k So the result is true for all n ≥ In particular, x1 ≥ + k Hence 4(x1995 - 1994) = 4(1 + k) ≥ (2 + k)2

= + 4k + k2, so k2 ≤ 0, so k =

Hence also xn ≥ n for n = 1, 2, , 1994 But now if xn = n + k, with k > 0, for some n < 1995, then the same argument shows that x1 ≥ + k and hence = 4(x1995 - 1994) ≥ (x1 + 1)2 ≥ (2 + k)2 = + 4k + k2> Contradiction Hence xn = n for all n ≤ 1995 Problem

Find the smallest n such that any sequence a1, a2, , an whose values are relatively prime square-free integers between and 1995 must contain a prime [An integer is square-free if it is not divisible by any square except 1.]

Solution

Answer: n = 14

We can exhibit a sequence with 13 terms which does not contain a prime: 2·101 = 202, 3·97 = 291, 5·89 = 445, 7·83 = 581, 11·79 = 869, 13·73 = 949, 17·71 = 1207, 19·67 = 1273, 23·61 = 1403, 29·59 = 1711, 31·53 = 1643, 37·47 = 1739, 41·43 = 1763 So certainly n ≥ 14

(58)

Problem

ABCD is a fixed cyclic quadrilateral with AB not parallel to CD Find the locus of points P for which we can find circles through AB and CD touching at P

Solution

Answer: Let the lines AB and CD meet at X Let R be the length of a tangent from X to the circle ABCD The locus is the circle center X radius R [Strictly you must exclude four points unless you allow the degenerate straight line circles.]

Let X be the intersection of the lines AB and CD Let R be the length of a tangent from X to the circle ABCD Let C0 be the circle center X radius R Take any point P on C0 Then considering the original circle ABCD, we have that R2 = XA·XB = XC·XD, and hence XP2 = XA·XB = XC·XD

If C1 is the circle through C, D and P, then XC.XD = XP2, so XP is tangent to the circle C1 Similarly, the circle C2 through A, B and P is tangent to XP Hence C1 and C2 are tangent to each other at P Note that if P is one of the points on AB or CD and C0, then this construction does not work unless we allow the degenerate straight line circles AB and CD

(59)

Problem

Take a fixed point P inside a fixed circle Take a pair of perpendicular chords AC, BD through P Take Q to be one of the four points such that ASCQ, ASDQ, BSCQ or BSDQ is a rectangle Find the locus of all possible Q for all possible such chords

Solution

Let O be the center of the fixed circle and let X be the center of the rectangle ASCQ By the cosine rule we have OQ2 = OX2 + XQ2 - 2·OX·XQ cos θ and OP2 = OX2 + XP2 - 2·OX·XP cos(θ+π), where θ is the angle OXQ But cos(θ+π) = -cos θ, so OQ2

+ OP2= 2OX2 + 2XQ2 But since X is the center of the rectangle XQ = XC and since X is the midpoint of AC, OX is perpendicular to AC and hence XO2 + XC2 = OC2 So OQ2 = 2OC2 - OP2 But this quantity is fixed, so Q must lie on the circle center O radius √(2R2 - OP2), where R is the radius of the circle

Conversely, it is easy to see that all points on this circle can be reached For given a point Q on the circle radius √(2R2

- OP2) let X be the midpoint of PQ Then take the chord AC to have X as its midpoint

Problem

f is a function from the integers to {1, 2, 3, , n} such that f(A) and f(B) are unequal whenever A and B differ by 5, or 12 What is the smallest possible n?

Solution Answer: n =

Each pair of 0, 5, 12 differ by 5, or 12, so f(0), f(5), f(12) must all be different, so n ≥

We can exhibit an f with n = Define f(m) = for m = 1, 3, 5, 7, 9, 11 (mod 24), f(m) = for m = 2, 4, 6, 8, 10, 12 (mod 24), f(m) = for m = 13, 15, 17, 19, 21, 23 (mod 24), f(m) = for m = 14, 16, 18, 20, 22, (mod 24)

(60)

8th APMO 1996

Problem 1

ABCD is a fixed rhombus P lies on AB and Q on BC, so that PQ is perpendicular to BD Similarly P' lies on AD and Q' on CD, so that P'Q' is perpendicular to BD The distance between PQ and P'Q' is more than BD/2 Show that the perimeter of the hexagon APQCQ'P' depends only on the distance between PQ and P'Q'

Solution

BPQ and DQ'P' are similar Let PQ meet BD at X and P'Q' meet BD at Y XY is fixed, so BX + DY is fixed Hence also, BP + DQ' and BQ + DP' and PQ + P'Q' are fixed So PQ + P'Q' - BP - BQ - DP' - DQ' is fixed, so PQ + P'Q' + (AB - BP) + (BC - BQ) + (CD - DP') + (DA - DQ') is fixed, and that is the perimeter of the hexagon

Problem

Prove that (n+1)mnm ≥ (n+m)!/(n-m)! ≥ 2mm! for all positive integers n, m with n ≥ m Solution

For any integer k ≥ 1, we have (n + k)(n - k + 1) = n2

+ n - k2 + k ≤ n(n + 1) Taking the product from k = to m we get (n + m)!/(n - m)! ≤ (n + 1)mnm

(61)

Problem

Given four concyclic points For each subset of three points take the incenter Show that the four incenters from a rectangle

Solution

Take the points as A, B, C, D in that order Let I be the incenter of ABC The ray CI bisects the angle ACB, so it passes through M, the midpoint of the arc AB Now ∠MBI = ∠MBA + ∠IBA = ∠MCA + ∠IBA = (∠ACB + ∠ABC)/2 = 90o - (∠CAB) /2 = 90o -

∠CMB/2 = 90o - ∠IMB/2 So the bisector of ∠IMB is perpendicular to IB Hence MB = MI Let J be the incenter of ABD Then similarly MA = MJ But MA = MB, so the four points A, B, I, J are concyclic (they lie on the circle center M) Hence ∠BIJ = 180o -

∠BAJ = 180o - ∠BAD/2

Similarly, if K is the incenter of ADC, then ∠BJK = 180o - ∠BDC/2 Hence ∠IJK = 360o - ∠BIJ - ∠BJK = (180o - ∠BIJ) + (180o - ∠BJK) = (∠BAD + ∠BDC)/2 = 90o Similarly, the other angles of the incenter quadrilateral are 90o, so it is a rectangle Problem

For which n in the range to 1996 is it possible to divide n married couples into exactly 17 single sex groups, so that the size of any two groups differs by at most one

Solution

Answer: 9, 10, 11, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22, 23, 24, 27, 28, 29, 30, 31, 32, 36, 37, 38, 39, 40, 45, 46, 47, 48, 54, 55, 56, 63, 64, 72

If n = 17k, then the group size must be 2k Hence no arrangement is possible, because one sex has at most groups and 8.2k < n

If 2n = 17k+h with < h < 17, then the group size must be k or k+1 One sex has at most groups, so 8(k+1) ≥ n Hence 16k + 16 ≥ 17k + h, so 16 - h ≥ k (*) We also require that 9k ≤ n Hence 18k < 2n = 17k + h, so k ≤ h (**) With (*) this implies that k ≤ So n ≤ 75

(62)

giving n = 27-32 For k = 4, we can have h = 4, 6, 12, giving n = 36-40 For k = we can have h = 5, 7, 9, 11, giving n = 45-48 For k = 6, we can have h = 6, 8, 10, giving n = 54, 55, 56 For k = 7, we can have h = 7, 9, giving n = 63, 64 For k = 8, we can have h = 8, giving n = 72

Problem 5

A triangle has side lengths a, b, c Prove that √(a + b - c) + √(b + c - a) + √(c + a - b) ≤ √a + √b + √c When you have equality?

Solution

Let A2 = b + c - a, B2 = c + a - b, C2 = a + b - c Then A2 + B2 = 2c Also A = B iff a = b We have (A - B)2 ≥ 0, with equality iff A = B Hence A2 + B2 ≥ 2AB and so 2(A2 + B2) ≥ (A + B)2

(63)

9th APMO 1997

Problem

Let Tn = + + + n = n(n+1)/2 Let Sn= 1/T1 + 1/T2 + + 1/Tn Prove that 1/S1 + 1/S2 + + 1/S1996 > 1001

Solution

1/Tm = 2(1/m - 1/(m+1) ) Hence Sn/2 = - 1/(n+1) So 1/Sn = (1 + 1/n)/2 Hence 1/S1 + 1/S2 + + 1/Sn = 1996/2 + (1+ 1/2 + 1/3 + + 1/1996)/2

Now + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + + 1/16) + (1/17 + + 1/32) + (1/33 + + 1/64) + (1/65 + + 1/128) + (1/129 + + 1/256) + (1/257 + + 1/512) + (1/513 + + 1/1024) > + 1/2 + 1/2 + + 1/2 = So 1/S1 + 1/S2 + + 1/Sn = 1996/2 + 6/2 = 998 + = 1001

Problem

Find an n in the range 100, 101, , 1997 such that n divides 2n + Solution

Answer: the only such number is 946

We have 2p-1 = mod p for any prime p, so if we can find h in {1, 2, , p-2} for which 2h = -2 mod p, then 2k= -2 mod p for any h = k mod p Thus we find that 2k = -2 mod for k = mod 4, and 2k = -2 mod 11 for k = mod 10 So we might then hope that 5·11 = mod and = mod 10 Unfortunately, it does not! But we try searching for more examples

(64)

3 mod 4), 5·61 (not = mod 4), 13·29 (not = mod 12), 13·37 (not = mod 12), 13.53 (not = mod 12), 13·61 (not = mod 12), 29·37 (not = 15 mod 28), 29·53 (not = 15 mod 28), 29·61 (not = 15 mod 28), 37·53 (not = 19 mod 36) So that does not advance matters much!

2p will not work (at least with h = (p+1)/2) because we cannot have 2p = (p+1)/2 mod p-1 So we try looking at 2pq This requires that p and q = mod So searching for suitable p we find mod 11, 10 mod 19, 22 mod 43, 30 mod 59, 34 mod 67, 42 mod 83 So we look at 2·11·43 = 946, which works

Proving that it is unique is harder The easiest way is to use a computer to search (approx to write a Maple program or similar and a few seconds to run it)

Problem

ABC is a triangle The bisector of A meets the segment BC at X and the circumcircle at Y Let rA = AX/AY Define rB and rC similarly Prove that rA/sin2A + rB/sin2B + rC/sin2C >= with equality iff the triangle is equilateral

Solution

AX/AB = sin B/sin AXB = sin B/sin(180 - B - A/2) =sin B/sin(B + A/2) Similarly, AB/AY = sin AYB/sin ABY = sin C/sin(B + CBY) = sin C/sin(B + A/2) So AX/AY = sin B sin C/sin2(B + A/2) Hence rA/sin2A = sA/sin2(B + A/2), where sA = sin B sin C/sin2A Similarly for rB and rC Now sAsBsC = 1, so the arithmetic/geometric mean result gives sA + sB + sC ≥ But 1/sin k ≥ for any k, so rA/sin2A + rB/sin2B + rC/sin2C ≥

(65)

Problem

P1 and P3 are fixed points P2 lies on the line perpendicular to P1P3 through P3 The sequence P4, P5, P6, is defined inductively as follows: Pn+1 is the foot of the

perpendicular from Pn to Pn-1Pn-2 Show that the sequence converges to a point P (whose position depends on P2) What is the locus of P as P2 varies?

Solution

PnPn+1Pn+2 lies inside Pn-1PnPn+1 So we have sequence of nested triangles whose size shrinks to zero Each triangle is a closed set, so there is just one point P in the

intersection of all the triangles and it is clear that the sequence Pn converges to it

Obviously all the triangles PnPn+1Pn+2 are similar (but not necessarily with the vertices in that order) So P must lie in the same position relative to each triangle and we must be able to obtain one triangle from another by rotation and expansion about P In particular, P5P4P6 is similar (with the vertices in that order) to P1P2P3, and P4P5 is parallel to P1P2, so the rotation to get one from the other must be through π and P must lie on P1P5 Similarly P3P4P5 must be obtained from P1P2P3 by rotation about P through π/2 and expansion But this takes P1P5 into a perpendicular line through P3 Hence P1P is perpendicular to P3P Hence P lies on the circle diameter P1P3

(66)

Problem

n people are seated in a circle A total of nk coins are distributed amongst the people, but not necessarily equally A move is the transfer of a single coin between two adjacent people Find an algorithm for making the minimum number of moves which result in everyone ending up with the same number of coins?

Solution

Label the people from to n, with person i next to person i+1, and person n next to person Let person i initially hold ci coins Let di = ci - k

It is not obvious how many moves are needed Clearly at least 1/2 ∑ |di| are needed But one may need more For example, suppose the starting values of di are 0, 1, 0, -1, Then one needs at least moves, not

Obviously ∑ di = 0, so not all di can be negative Relabel if necessary so that d1 ≥= Now consider X = |d1| + |d1 + d2| + |d1 + d2 + d3| + + |d1 + d2 + + dn-1| Note first that X is zero iff all di are zero Any move between i and i+1, except one between n and 1, changes X by 1, because only the term |d1 + d2 + + di| is affected Thus if we not make any moves between n and 1, then we need at least X moves to reach the desired final position (with all di zero)

Assume X > We show how to find a move which reduces X by This requires a little care to avoid specifying a move which might require a person with no coins to transfer one We are assuming that d1≥ Take the first i for which di+1 < There must be such an i, otherwise all di would be non-negative, but they sum to 0, so they would all have to be zero, contradicting X > If d1 + + di > 0, then we take the move to be a transfer from i to i+1 This will reduce |d1 + + di| by and leave the other terms in X

unchanged, so it will reduce X by If d1 + + di is not strictly positive, then by the minimality of i we must have d1 = d2 = = di = We know that di+1 < Now find the first j > i+1 such that dj ≥ There must be such a j, otherwise we would have ∑ dm < We have d1 + + dj-1 < 0, so a transfer from j to j-1 will reduce |d1 + + dj-1| and hence reduce X Finally note that the move we have chosen leaves d1 ≥ Thus we can repeat the process and reduce X to zero in X moves

(67)

10th APMO 1998

Problem

S is the set of all possible n-tuples (X1, X2, , Xn) where each Xi is a subset of {1, 2, , 1998} For each member k of S let f(k) be the number of elements in the union of its n elements Find the sum of f(k) over all k in S

Solution

Answer: 1998(21998n - 21997n)

Let s(n, m) be the sum where each Xi is a subset of {1, 2, , m} There are 2m possible Xi and hence 2mn possible n-tuples We have s(n, m) = 2ns(n, m-1) + (2n - 1)2n(m-1) (*) For given any n-tuple {X1, , Xn} of subsets of {1, 2, , m-1} we can choose to add m or not (2 choices) to each Xi So we derive 2n n-tuples of subsets of {1, 2, , m} All but of these have f(k) incremented by The first term in (*) gives the sum for m-1 over the increased number of terms and the second term gives the increments to the f(k) due to the additional element

Evidently s(n, 1) = 2n - It is now an easy induction to show that s(n, m) = m(2nm - n(m-1)

)

Putting m = 1998 we get that the required sum is 1998(21998n - 21997n)

Problem

Show that (36m + n)(m + 36n) is not a power of for any positive integers m, n Solution

(68)

Problem

Prove that (1 + x/y)(1 + y/z)(1 + z/x) ≥ + 2(x + y + z)/w for all positive reals x, y, z, where w is the cube root of xyz

Solution

(1 + x/y)(1 + y/z)(1 + z/x) = + x/y + y/x + y/z + z/y + z/x + x/z = (x + y + z)(1/x + 1/y + 1/z) - ≥ 3(x + y + z)/w - 1, by the arithmetic geometric mean inequality,

= 2(x + y + z)/w + (x + y + z)/w - ≥ 2(x + y + z) + - 1, by the arithmetic geometric mean inequality

Problem 4

ABC is a triangle AD is an altitude X lies on the circle ABD and Y lies on the circle ACD X, D and Y are collinear M is the midpoint of XY and M' is the midpoint of BC Prove that MM' is perpendicular to AM

Solution

Take P, Q so that PADB, AQCD are rectangles Let N be the midpoint of PQ Then PD is a diameter of the circumcircle of ABC, so PX is perpendicular to XY Similarly, QY is perpendicular to XY N is the midpoint of PQ and M' the midpoint of XY, so NM is parallel to PX and hence perpendicular to XY NADM' is a rectangle, so ND is a diameter of its circumcircle and M must lie on the circumcircle But AM' is also a diameter, so ∠AMM' = 90o

Thanks to Michael Lipnowski for the above My original solution is below.

Let P be the circumcenter of ABD and Q the circumcenter of ADC Let R be the

midpoint of AM' P and Q both lie on the perpendicular bisector of AD, which is parallel to BC and hence also passes through R We show first that R is the midpoint of PQ Let the feet of the perpendiculars from P, Q, R to BC to P', Q', R' respectively It is sufficient to show that BP' = BD/2 BR' = BM' + M'R' = (BD + DC)/2 + M'D/2 = (BD + DC)/2 + ( (BD + DC)/2 - DC)/2 = 3BD/4 + DC/4, so P'R' = (BD + DC)/4 Q'C = DC/2, so BQ' = BD + DC/2 and P'Q' = (BD + DC)/2 = 2P'R'

(69)

Let the feet of the perpendiculars from P, Q, R to XY be P'', Q'', R'' respectively So on XY we have, in order, X, P'', M'', R'', D, Q'', Y Since R is the midpoint of PQ, R'' is the midpoint of P''Q'' Now P'' is the midpoint of XD and Q'' is the midpoint of DY, so P''Q'' = XY/2 = 2(x+y), so R''Q'' = x+y But DQ'' = 2y, so R''D = x-y R'' is the midpoint of M''D, so M''D = 2(x-y) and hence M''Y = M''D + DY = 2(x-y) + 4y = 2(x+y) = XY/2 So M'' is just M the midpoint of XY Now AM' is a diameter of the circle center R, so AM is perpendicular to MM'

Problem

What is the largest integer divisible by all positive integers less than its cube root Solution

Answer: 420

Let N be a positive integer satisfying the condition and let n be the largest integer not exceeding its cube root If n = 7, then 3·4·5·7 = 420 must divide N But N cannot exceed 83 - = 511, so the largest such N is 420

If n ≥ 8, then 3·8·5·7 = 840 divides N, so N > 729 = 93

Hence divides N, and hence 3·840 = 2520 divides N But we show that no N > 2000 can satisfy the condition Note that 2(x - 1)3 > x3 for any x > Hence [x]3 > x3/2 for x > So certainly if N > 2000, we have n3 > N/2 Now let pk be the highest power of k which does not exceed n Then pk > n/k Hence p2p3p5 > n3/30 > N/60 But since N > 2000, we have < 11 < n and hence p2, p3, p5, 7, 11 are all ≤ n But 77 p2p3p5 > N, so N cannot satisfy the condition

(70)

11th APMO 1999

Problem

Find the smallest positive integer n such that no arithmetic progression of 1999 reals contains just n integers

Using a difference of 1/n , where n does not divide 1999, we can get a progression of 1999 terms with m = [1998/n] or m = [1998/n] - integers Thus {0, 1/n, 2/n, ,

1998/n} has m+1 integers, and {1/n, 2/n, , 1999/n} has m integers So we are ok until n gets so large that the gap between [1998/n] and [1998/(n+1)] is or more This could happen for 1998/n - 1998/(n+1) just over or n > 31 So checking, we find [1998/31] = 64, [1998/30] = 66, [1998/29] = 68, [1998/28] = 71

We can get 68 integers with {1/29, 2/29, , 1999/29} and 69 with {0, 1/29, 2/29, , 1998/29} We can get 71 with {1/28, 2/28, , 1999/28}, but we cannot get 70 Note that a progression with irrational difference gives at most integer A progression with

difference a/b, where a and b are coprime integers, gives the same number of integers as a progression with difference 1/b So it does not help to widen the class of progressions we are looking at

Problem

The real numbers x1, x2, x3, satisfy xi+j <= xi + xj for all i, j Prove that x1 + x2/2 + + xn/n >= xn

Solution

We use induction Suppose the result is true for n We have: x1 >= x1

x1 + x2/2 >= x2

x1 + x2/2 + + xn/n >= xn

Also: x1 + 2x2/2 + + nxn/n = x1 + + xn

(71)

Problem

Two circles touch the line AB at A and B and intersect each other at X and Y with X nearer to the line AB The tangent to the circle AXY at X meets the circle BXY at W The ray AX meets BW at Z Show that BW and BX are tangents to the circle XYZ Solution

Let angle ZXW =  and angle ZWX =  XW is tangent to circle AXY at X, so angle AYX =  AB is tangent to circle AXY at A, so angle BAX =  AB is tangent to circle BXY at B, so angle ABX =  Thus, considering triangle ABX, angle BXZ =  Considering triangle ZXW, angle BZX = 

BXYW is cyclic, so angle BYX = angle BWX =  Hence angle AYB = angle AYX + angle XYB =  = angle AZB So AYZB is cyclic Hence angle BYZ = angle BAZ =  So angle XYZ = angle XYB + angle BYZ =  Hence angle BZX = angle XYZ, so BZ is tangent to circle XYZ at Z Similarly angle BXY = angle XYZ, so BX is tangent to circle XYZ at X

Problem

Find all pairs of integers m, n such that m2 + 4n and n2 +4m are both squares Solution

Answer: (m, n) or (n, m) = (0, a2), (-5, -6), (-4, -4), (a+1, -a) where a is a non-negative integer

Clearly if one of m, n is zero, then the other must be a square and that is a solution If both are positive, then m2 + 4n must be (m + 2k)2 for some positive k, so n = km + k2 > m But similarly m > n Contradiction So there are no solutions with m and n positive

(72)

But N2 - 4M is also a square, so (M-1)2 - 4M = M2 - 6M + is a square But (M-3)2 > M2 - 6M + and (M-4)2 < M2 - 6M + for M >= 8, so the only possible solutions are M = 1, 2, , Checking, we find that only M = gives M2 - 6M + a square This gives the soluton (m, n) = (-6, -5) Obviously, there is also the solution (-5, -6)

Finally, consider the case of opposite signs Suppose m = M > 0, n = -N < Then N2 + 4M is a square, so by the argument above M > N But M2 - 4N is a square and so the argument above gives N = M-1 Now we can easily check that (m, n) = (M, -(M-1) ) is a solution for any positive M

Problem

A set of 2n+1 points in the plane has no three collinear and no four concyclic A circle is said to divide the set if it passes through of the points and has exactly n - points inside it Show that the number of circles which divide the set is even iff n is even Solution

Take two of the points, A and B, and consider the 2n-1 circles through A and B We will show that the number of these circles which divide the set is odd The result then follows almost immediately, because the number of pairs A, B is (2n+1)2n/2 = N, say The total number of circles which divide the set is a sum of N odd numbers divided by (because each such circle will be counted three times) If n is even, then N is even, so a sum of N odd numbers is even If n is odd, then N is odd, so a sum of N odd numbers is odd Dividing by does not change the parity

Their centers all lie on the perpendicular bisector of AB Label them C1, C2, , C2n-1, where the center of Ci lies to the left of Cj on the bisector iff i < j We call the two half-planes created by AB the left-hand half-plane L and the right-hand half-plane R

correspondingly Let the third point of the set on Ci be Xi Suppose i < j Then Ci contains all points of Cj that lie in L and Cj contains all points of Ci that lie R So Xi lies inside Cj iff Xi lies in R and Xj lies inside Ci iff Xj lies in L

(73)

n-1 from one side to the other We now want to say that f starts on one side of the line n-n-1 and ends on the other, so the final parity must be odd Suppose there are m points in L and hence 2n-1-m in R Without loss of generality we may take m <= n-1 The first circle C1 contains all the points in L except X1 if it is in L So f(1) = m or m-1 Similarly the last circle C2n-1 contains all the points in R except X2n-1 if it is in R So f(1) = 2n-1-m or 2n-2-m Hence if m < n-1, then f(1) = m or m-1, so f(1) < n-1 But 2n-2n-1-m >= n+1, so f(2n-1) > n-1 So in this case we are done

However, there are complications if m = n-1 We have to consider cases Case (1): m = n-1, X1 lies in R, X2n-1 lies in L Hence f(1) = n-1, f(2n-1) = n > n-1 So f starts on the line n-1 If it first leaves it downwards, then for the previous point i, Xi is in L and hence there were an even number of points up to i on the line So the parity is the same as if f(1) was below the line f(2n-1) is above the line, so we get an odd number of points on the line If f first leaves the line upwards, then for the previous point i, Xi is in R and hence there were an odd number of points up to i on the line So again the parity is the same as if f(1) was below the line

Case (2): m = n-1, X1 lies in R, X2n-1 lies in R Hence f(1) = f(2n-1) = n-1 As in case (1) the parity is the same as if f(1) was below the line If the last point j with f(j) not on the line has f(j) < n-1, then (since X2n-1 lies in R) there are an odd number of points above j, so the parity is the same as if f(2n-1) was above the line Similarly if f(j) > n-1, then there are an even number of points above j, so again the parity is the same as if f(2n-1) was above the line

Case (3): m = n-1, X1 lies in L, X2n-1 lies in L Hence f(1) = n-2, f(2n-1) = n So case has already been covered

(74)

12th APMO 2000

Problem

Find a13/(1 - 3a1 + 3a12) + a23/(1 - 3a2 + 3a22) + + a1013/(1 - 3a101 + 3a1012), where an = n/101

The nth term is an3/(1 - 3an + 3an2) = an3/( (1 - an)3 + an3) = n3/( (101 - n)3 + n3) Hence the sum of the nth and (101-n)th terms is Thus the sum from n = to 100 is 50 The last term is 1, so the total sum is 51

Problem

Find all permutations a1, a2, , a9 of 1, 2, , such that a1 + a2 + a3 + a4 = a4 + a5 + a6 + a7 = a7 + a8 + a9 + a1 and a12 + a22 + a32 + a42 = a42 + a52 + a62 + a72 = a72 + a82 + a92 + a12

Solution

We may start by assuming that a1 < a4 < a7 and that a2 < a3, a5 < a6, a8 < a9

Note that + + = 45 and 12 + + 92 = 285 Adding the three square equations together we get (a12 + + a92) + a12 + a42 + a72 = 285 + a12 + a42 + a72 The total must be a multiple of But 285 is a multiple of 3, so a12 + a42 + a72 must be a multiple of Now 32, 62 and 92 are all congruent to mod and the other squares are all congruent to mod Hence either a1, a4 and a7 are all multiples of 3, or none of them are Since 45 is also a multiple of three a similar argument with the three linear equations shows that a1 + a4 + a7 is a multiple of So if none of a1, a4, a7 are multiples of 3, then they are all

(75)

In case (1), we have that each sum of squares equals 137 Hence a82 + a92 = 47 But 47 is not a sum of two squares, so this case gives no solutions

In case (2), we have that each sum of squares is 117 Hence a52 + a62 = 52 But the only way of writing 52 as a sum of two squares is 42 + 62 and is already taken by a4, so this case gives no solutions

In case (3), we have that each sum of squares is 126 and each linear sum 20 We quickly find that the only solution is 2, 4, 9, 5, 1, 6, 8, 3,

Obviously, this generates a large number of equivalent solutions We can interchange a2 and a3, or a5 and a6, or a8 and a9 We can also permute a1, a4 and a7 So we get a total of x x x =48 solutions

Problem

ABC is a triangle The angle bisector at A meets the side BC at X The perpendicular to AX at X meets AB at Y The perpendicular to AB at Y meets the ray AX at R XY meets the median from A at S Prove that RS is perpendicular to BC

Solution

Let the line through C parallel to AX meet the ray BA at C' Let the perpendicular from B meet the ray C'C at T and the ray AX at U Let the line from C parallel to BT meet BA at V and let the perpendicular from V meet BT at W So CVWT is a rectangle

AU bisects ∠CAV and CV is perpendicular to AU, so U is the midpoint of WT Hence the intersection N of AU and CW is the center of the rectangle and, in particular, the midpoint of CW Let M be the midpoint of BC Then since M, N are the midpoints of the sides CB and CW of the triangle CBW, MN = BW/2

Since CC' is parallel to AX, ∠CC'A = ∠BAX = ∠CAX = ∠C'CA, so AC' = AC Let A' be the midpoint of CC' Then AU = C'T - C'A' But N is the center of the rectangle CTWV, so NU = CT/2 and AN = AU - NU = C'T - C'A' - CT/2 = C'T/2 Hence MN/AN = BW/C'T But MN is parallel to BW and XY, so SX/AX = MN/AN = BW/C'T

Now AX is parallel to VW and XY is parallel to BW, so AXY and VWB are similar and AX/XY = VW/BW = CT/BW Hence SX/XY = (SX/AX) (AX/XY) = CT/C'T

(76)

= (SX/XY) (XY/XR) = (CT/C'T) (C'T/BT) = CT/BT But angles CTB and SXR are both right angles, so SXR and CTB are similar But XR is perpendicular to BT, so SR is perpendicular to BC

Problem

If m < n are positive integers prove that nn/(mm (n-m)n-m) > n!/( m! (n-m)! ) > nn/( mm(n+1) (n-m)n-m)

Solution

The key is to consider the binomial expansion (m + n-m)n This is a sum of positive terms, one of which is nCm mm(n-m)n-m, where nCm is the binomial coefficient n!/( m! (n-m)! ) Hence nCm mm(n-m)n-m < nn, which is one of the required inequalities

We will show that nCm mm(n-m)n-m is the largest term in the binomial expansion It then follows that (n+1) nCm mm(n-m)n-m > nn, which is the other required inequality

Comparing the rth term nCr mr(n-m)n-r with the r+1th term nCr+1 mr+1(n-m)n-r-1 we see that the rth term is strictly larger for r ≥ m and smaller for r < m Hence the mth term is larger than the succeeding terms and also larger than the preceding terms

Problem

Given a permutation s0, s2, , sn of 0, 1, 2, , n, we may transform it if we can find i, j such that si = and sj = si-1 + The new permutation is obtained by transposing si and sj For which n can we obtain (1, 2, , n, 0) by repeated transformations starting with (1, n, n-1, , 3, 2, 0)?

Solution

Experimentation shows that we can it for n=1 (already there), n = (already there), 3, 7, 15, but not for n = 4, 5, 6, 8, 9, 10, 11, 12, 13, 14 So we conjecture that it is possible just for n = 2m - and for n =

(77)

1 n n-2 n-1 n-4 n-3

and we can go no further So n even fails for n > If n = 15 we get successively:

15 14 13 12 11 10 start

14 15 12 13 10 11 after moves 12 13 14 15 10 11 after more moves

10 11 12 13 14 15 after more moves

This pattern is general Suppose n = 2m - Let P0 be the starting position and Pr be the position:

1 R-1 0, n-R+1 n-R+2 n-R+3 n, n-2R+1 n-2R+2 n-R, , R R+1 2R-1

Here R denotes 2r and the commas highlight that, after the initial R-1 0, we have increasing runs of R terms If we start from Pr, then the is transposed successively with R, 3R, 5R, , n-R+1, then with R+1, 3R+1, , n-R+2, and so on up to 2R-1, 4R-1, , n But that gives Pr+1 It is also easy to check that P0 leads to P1 and that Pm is the

required finishing position Thus the case n = 2m - works

Now suppose n is odd but not of the form 2m - Then we can write n = (2a + 1)2b - (just take 2b as the highest power of dividing n + 1) We can now define P0, P1, , Pb as before As before we will reach Pb:

1 ¼ B-1 0, 2aB 2aB+1¼ (2a+1)B-1, (2a-1)B ¼ 2aB-1, ¼ , 3B, 3B+1, ¼ 4B-1, 2B, 2B+1, ¼ , 3B-1, B, B+1, ¼ , 2B-1

where B = 2b - But then the is transposed successively with B, 3B, 5B, , (2a-1)B, which puts it immediately to the right of (2a+1)B-1 = n, so no further transformations are possible and n = (2a+1)2b - fails

(78)

13th APMO 2001

Problem

If n is a positive integer, let d+1 be the number of digits in n (in base 10) and s be the sum of the digits Let n(k) be the number formed by deleting the last k digits of n Prove that n = s + n(1) + n(2) + + n(d)

Solution

Let the digits of n be ad, ad-1, , a0, so that n = ad 10d + + a0 Then n(k) = ad 10d-k + a d-1

10d-k-1 + + ak Obviously s = ad + + a0 Hence s + n(1) + + n(d) = ad(9.10d-1 + 9.10d-2 + + + 1) + ad-1(9.10d-2 + + + 1) + + ad-k(9.10d-k-1 + + + 1) + a1(9 + 1) + a0 = ad 10d + + a0 = n

Problem

Find the largest n so that the number of integers less than or equal to n and divisible by equals the number divisible by or (or both)

Let f(n) = [n/3] - [n/5] - [n/7] + [n/35] We are looking for the largest n with f(n) = Now [n/5] + [n/7} <= [n/5 + n/7] = [12n/35] = [n/3 + n/105] So for [n/5] + [n/7] to exceed [n/3] we certainly require n/105 ≥ 1/3 or n ≥ 35 Hence f(n) ≥ for n ≤ 35 But f(n+35) = [n/3 + 11 + 2/3] - [n/5 + 7] - [n/7 + 5] + [n/35 + 1] = [n/3 + 2/3] - [n/5] - [n/7] + [n/35] ≥ f(n) (*) Hence f(n) ≥ for all n But f(n+105) = [n/3 + 35] - [n/5 + 21] - [n/7 + 15] + [n/35 + 3] = f(n) + Hence f(n) ≥ for all n ≥ 105

Referring back to (*) we see that f(n+35) > f(n), and hence f(n+35) > 0, unless n is a multiple of But if n is a multiple of 3, then n + 35 is not and hence f(n+70) > f(n+35) > So f(n) > for all n ≥ 70

(79)

Problem

Two equal-sized regular n-gons intersect to form a 2n-gon C Prove that the sum of the sides of C which form part of one n-gon equals half the perimeter of C

Solution

Let one regular n-gon have vertices P1, P2, , Pn and the other have vertices Q1, Q2, , Qn Each side of the 2n-gon forms a triangle with one of the Pi or Qi Note that Pi and Qj must alternate as we go around the 2n-gon For convenience assume that the order is P1, Q1, P2, Q2, , Pn, Qn Let the length of the side which forms a triangle with Pi be pi, and the length of the side which forms a triangle with Qibe qi Each of these triangles has one angle (180o - 360o/n) Adjacent triangles have one of their other angles equal

(alternate angles), so all the triangles are similar If the sides of the triangle vertex Pi have lengths ai, bi, pi, then the side PiPi+1 is bi + qi + ai+1, and ai/ai+1 = bi/bi+1 = pi/pi+1 Similarly, if the sides of the triangle vertex Qi have lengths ci, di, qi, then the side QiQi+1 is di + pi+1 + ci+1 and ci/ci+1 = di/di+1 = qi/qi+1 But ai/bi = di/ci (not ci/di), because the triangles alternate in orientation

Put ai/pi = h, bi/pi = k Note that + bi > pi, so h + k - > We have also ci/qi = k, di/qi = h Adding the expressions for PiPi+1 we get perimeter Pi = ∑(bi + qi + ai+1) = k ∑ pi + ∑ qi + h ∑ pi Similarly, perimeter Qi = (h + k) ∑ qi + ∑ pi The two n-gons are equal, so (h + k - 1) ∑ pi = (h + k - 1) ∑ qi Hence ∑ pi = ∑ qi, which is the required result

Problem

Find all real polynomials p(x) such that x is rational iff p(x) is rational Solution

It is necessary for all the coefficients of x to be rational This is an easy induction on the number of coefficients For p(0) must be rational, hence ( p(x) - p(0) )/x must be rational for all rational x and so on

Clearly this condition is also sufficient for polynomials of degree or

There are obviously no quadratics, for if p(x) = ax2 + bx + c, with a, b, c rational, then p(√2 - b/2a) = 2a - b2

(80)

We prove that if there are also no higher degree polynomials The idea is to show that there is a rational value k which must be taken for some real x, but which cannot be taken by any rational x

Suppose p(x) has degree n > Multiplying through by the lcm of the denominators of the coefficients, we get p(x) = (a xn + b xn-1 + + u x + v)/w, where a, b, , w are all integers Put x = r/s, where r and s are coprime integers, then p(r/s) = (a rn + b rn-1s + + u r sn-1 + v sn)/( w sn) Let q be any prime which does not divide a or w Consider first a > p(x) must assume all sufficiently large positive values So it must in particular take the value k = m + 1/q, where m is a sufficiently large integer So k = (mq + 1)/q The denominator is divisible by q, but not q2 and the numerator is not divisible by q Suppose p(r/s) = k for some integers r, s The denominator of p(r/s) is w sn We know that w is not divisible by q, so q must divide s But n > 1, so q2 divides w sn The numerator of p(r/s) has the form a rn + h s Neither a nor r is divisible by q, so the numerator is not divisible by q Thus no cancellation is possible and we cannot have p(r/s) = k Thus there must be some irrational x such that p(x) = k

If a < 0, then the same argument works except that we take k = m + 1/q, where m is a sufficiently large negative integer

Problem 5

What is the largest n for which we can find n + points in the plane, A, B, C, D, X1, , Xn, so that AB is not equal to CD, but for each i the two triangles ABXi and CDXi are congruent?

Answer

Solution

Many thanks to Allen Zhang for completing the proof

Assume AB = a, CD = b with a > b If ABX and CDX are congruent, then either AX or BX = CD = b, so X lies either on the circle SA center A radius b, or on the circle

(81)

However, we show that if two points of intersection of (SA, SC) are included,

then no points of (SB, SD) can be included The same applies to each pair of circles, so at most points X are possible Finally, we will give an example of n = 4, showing that the maximum of is achieved

So suppose (SA, SC) intersect at X and Y We must have BX = DX and BY = DY, so X and Y both lie on the perpendicular bisector of BD In other words, XY is the

perpendicular bisector of BD, so D is the reflection of B in the line XY There is no loss of generality in taking B (and D) to be on the same side of AC as X Let A' be the

reflection of A in the line XY Since B lies on the circle center A radius a, D must lie on the circle center A' radius A Thus the triangles A'XC and CDA' are congruent

(Note that A and C can be on the same side of XY or opposite sides.) Hence D is the same height above AC as X, so DX is perpendicular to XY Hence X is the midpoint of BD Also ∠A'CD = ∠CA'X = 180o - ∠CAX, so AX and CD are parallel They are also equal, so ACDX is a parallelogram and hence AC = DX = BD/2 In the second

configuration above, both A and C are on the same side of XY as D, so the midpoint M of AC lies on the same side of XY as D In the first configuration, since AX = b < a = CX, M lies to the right of XY

Now suppose there is a solution for the configuration (SB, SD) Thus there is a point Z such that ABZ and ZDC are congruent Then AZ = CZ, so Z lies on the perpendicular bisector of AC and hence on the same side of XY as D But it is also a distance a from D and a distance b from B, and a > b, so it must lie on the same side of XY as B

Contradiction So there are no solutions for the configuration (SB, SD), as required That completes the proof that n ≤

For an example with n = 4, take a regular hexagon ACDBX3X2 Extend the side X2X3 to X1X4, with X1, X2, X3, X4 equally spaced in that order, so that X1AX2 and X3BX4 are equilateral Then ABX1 and CX1D are congruent, ABX2 and DX2C are congruent, ABX3 and X3CD are congruent, and ABX4 and X4DC are congruent

(82)

14th APMO 2002

Problem

xi are non-negative integers Prove that x1! x2! xn! ≥ ( [(x1 + + xn)/n] ! )n (where [y] denotes the largest integer not exceeding y) When you have equality?

Solution

Answer: Equality iff all xi equal

For given 2m the largest binomial coefficient is (2m)!/(m! m!) and for 2m+1 the largest binomial coefficient is (2m+1)!/( m! (m+1)!) Hence for fixed xi + xj the smallest value of xi! xj! is for xi and xj as nearly equal as possible

If x1 + x2 + + xn = qn + r, where < r < n, then we can reduce one or more xi to reduce the sum to qn This will not affect the rhs of the inequality in the question, but will

reduce the lhs Equalising the xiwill not increase the lhs (by the result just proved) So it is sufficient to prove the inequality for all xi equal But in this case it is trivial since k! = k!

Problem

Find all pairs m, n of positive integers such that m2 - n divides m + n2 and n2 - m divides m2 + n

Solution Assume n ≥ m

(83)

Suppose n = m Then we require that n2 - n divides n2 + n If n > 3, then n2 > 3n, so 2(n2 - n) > n2 + n Obviously n2 - n < n2 + n, so if n > 3, then n2 - n cannot divide n2 + n It is easy to check that the only solutions (with n = m) less than are n = and n = Finally suppose n = m+1 We require m2 - m - divides m2 + 3m + If m >= 6, then m(m - 5) > 3, so 2(m2 - m - 1) > m2 + 3m + Obviously m2 - m - < m2 + 3m + 1, so m2 - m - cannot divide m2 + 3m + for m >= Checking the smaller values, we find the solutions less than are m = and m =

Summarising, the only solutions are: (n, m) = (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)

Problem 3

ABC is an equilateral triangle M is the midpoint of AC and N is the midpoint of AB P lies on the segment MC, and Q lies on the segment NB R is the orthocenter of ABP and S is the orthocenter of ACQ The lines BP and CQ meet at T Find all possible values for angle BCQ such that RST is equilateral

Answer 15o Solution

(84)

Let L be the midpoint of BC Put ∠CBP = x and ∠RAM = y So RM = AM tan y, TL = BL tan x = AM tan x But ∠APB = 60o + x, so y = 30o - x So if x ≠ 15o, then TL ≠ RM However, RST equilateral implies TL = RM, so x and hence also ∠BCQ = 15o

Problem

The positive reals a, b, c satisfy 1/a + 1/b + 1/c = Prove that √(a + bc) + √(b + ca) + √(c + ab) ≥ √(abc) + √a + √b + √c

Solution

Thanks to Suat Namli for this.

Multiplying by √(abc), we have √(abc) = √(ab/c) + √(bc/a) + √(ca/b) So it is sufficient to prove that √(c + ab) ≥ √c + √(ab/c)

Squaring, this is equivalent to c + ab ≥ c + ab/c + 2√(ab) or c + ab ≥ c + ab(1 - 1/a - 1/b) + 2√(ab) or a + b >= 2√(ab) or (√a - √b)2

≥

(85)

Problem

Find all real-valued functions f on the reals which have at most finitely many zeros and satisfy f(x4 + y) = x3f(x) + f(f(y)) for all x, y

Solution

Putting x = 0, we get f(y) = f(f(y)) for all y (*)

Putting x = 1, y = 0, we get f(0) = Hence putting y = 0, f(x4) = x3f(x) (**)

If f(1) = 0, then f(2) = f(1 + 1) = f(1) + f(1) = 0, so f(3) = f(1 + 2) = f(1) + f(2) = and so on Contradiction (only finitely many zeros) So f(1) = k for some non-zero k By (*), f(k) = k

Suppose f(h) = for some h not or Then f(h4) = h3f(h) = 0, so f(x) = for any of the distinct values x = h, h4, h16, h64, Contradiction (only finitely many zeros) So f(h) is not for any non-zero h

(86)

15th APMO 2003

Problem 1

The polynomial a8x8 +a7x7 + + a0 has a8 = 1, a7 = -4, a6 = and all its roots positive and real Find the possible values for a0

Answer 1/28 Solution

Thanks to Jonathan Ramachandran

Let the roots be xi We have Sum xi2 = 42 - 2·7 = By Cauchy we have (x1·1 + + x8·1) ≤ (x12 + + x82)1/2(12 + + 12)1/2 with equality iff all xi are equal Hence all xi are equal So they must all be 1/2

Problem 2

A unit square lies across two parallel lines a unit distance apart, so that two triangular areas of the square lie outside the lines Show that the sum of the perimeters of these two triangles is independent of how the square is placed

(87)

Let the lines be L, L' Let the square be ABA'B', with A, A' the two vertices not between L and L' Let L meet AB at X and AB' at Y Let L' meet A'B' at X' and A'B at Y' So AXY and A'X'Y' are similar Suppose angle AXY = x If we move L towards A by a distance d perpendicular to itself, then AX is shortened by d cosec x If L' remains a distance from L, then A'X' is lengthened by d cosec x The new triangle AXY is similar to the old Suppose that perimeter AXY = k·AX, then perimeter AXY is increased by kd cosec x Since AXY and A'X'Y' are similar, perimeter A'X'Y' is

shortened by kd cosec x, so the sum of their perimeters is unchanged It remains to show that the sum of the perimeters does not depend on the angle x

Let us move L towards A until L' passes through A', at which point the perimeter of A'X'Y' is zero Now if h is the height of AXY (from the base XY), then + h = AA' sin(45o + x) = sin x + cos x The perimeter of AXY is h/sin x + h/cos x + h/(sin x cos x) = h(sin x + cos x + 1)/(sin x cos x) = (sin x + cos x - 1)(sin x + cos x + 1)/(sin x cos x) = 2, which is independent of x

Problem 3

k > 14 is an integer and p is the largest prime smaller than k k is chosen so that p ≥ 3k/4 Prove that 2p does not divide (2p - k)!, but that n does divide (n - k)! for composite n > 2p

Solution

Since k > p, we have p > 2p-k and hence p does not divide any of 1, 2, 3, , 2p-k But p is prime, so it does not divide (2p - k)! So 2p does not either

Now consider composite n > 2p Note that k > 14, so p ≥ 13, so n > 26 Take q to be the largest prime divisor of n and put n = qr We now have three cases

(1) q > r ≥ Then n > 2p ≥ 3k/2, so 2n/3 > k Hence n-k > n-2n/3 = n/3 ≥ n/r = q > r So q and r are distinct integers < n-k Hence n = qr divides (n-k)!

(2) r = Then n = 2q > 2p But p is the largest prime < k Hence q ≥ k, so n-k ≥ n-q = q Obviously q > (since n > 26), so q and are distinct integers < n-k Hence n = 2q divides (n-k)!

(3) The final case is n = q2 We show that 2q ≤ n-k Suppose not Then 2q ≥ n-k+1 ≥ (2p+1)-k+1 ≥ 3k/2 + - k + = k/2 + So q ≥ k/4 + Also since 2q ≥ n-k+1, we have k ≥ n-2q+1 = (q-1)2

≥ k2/16 If k > 16, this is a contradiction If k = 15 or 16, then p = 13 and k ≥ (q-1)2

(88)

established that 2q ≤ n-k But that means that q and 2q are distinct integers ≤ n-k and so their product 2n divides (n-k)!

Thanks to Gerhard Woeginger for this.

Problem 4

Show that (an + bn)1/n + (bn + cn)1/n + (cn + an)1/n < + (21/n)/2, where n > is an integer and a, b, c are the sides of a triangle with unit perimeter

Solution

Thanks to Olena Bormashenko

We may take a ≥ b ≥ c Since a + b + c = and a < b+c, we have b ≤ a < 1/2 Hence (an + bn)1/n < 21/n/2 (*)

We have (b + c/2)n = bn + n/2 c bn-1 + other positive terms > bn + cn Hence (bn + cn)1/n < b + c/2 Similarly, (cn + an)1/n < a + c/2 Adding we get (bn + cn)1/n + (cn + an)1/n < a+b+c = Adding to (*) gives the required result

Problem 5

Find the smallest positive integer k such that among any k people, either there are 2m who can be divided into m pairs of people who know each other, or there are 2n who can be divided into n pairs of people who not know each other

Answer

k = m + n + max(m, n) - Solution

We have to find the smallest k so that any graph with k points has either m disjoint edges or n disjoint pairs of points with each pair not joined by an edge Let the smallest k be f(m, n) For any graph G, there is another graph G' with the same points and the complementary set of edges (so that A, B are joined by an edge in G' iff they are not joined in G) This shows that f(m, n) = f(n, m), so there is no loss of generality in assuming that m ≤ n

(89)

(in A or B) and there are no other edges Now since all points in A are joined to every other point in G, if two points are not joined then they must both be in B So there can be at most n-1 pairs of unjoined points If two points are joined then at least one of them must be in A (since if both are in B they are unjoined) But A has less than m points, so if there are m pairs of joined points, then they must overlap Thus G shows that f(m, n) > m+2n-2

We now prove by induction on m that any graph with m-1+2n points (and m <= n) has either m joined pairs or n unjoined pairs It is trivial for m = (if the graph has an edge, then we have the pair of joined points, if not then we can divide the 2n points arbitrarily into n pairs of unjoined points) So suppose it is true for m-1 We now use induction on n We suppose that a graph G with m-1+2n points does not have m pairs of joined points or n pairs of unjoined points We seek a contradiction

Ignore for a moment the case n = m Suppose that n > m and we have already established the result for n-1 Since m-1+2(n-1) < m-1+2n, and G does not have m pairs of joined points it must have n-1 pairs of unjoined points

So let B be a set of n-1 pairs of unjoined points Let A be the remaining m+1 points Take any two points P and Q in A and any pair P', Q' in B If P is not joined to P' and Q is not joined to Q', then we have a contradiction, because we could remove the pair P', Q' from B and replace it by the two pairs P, P' and Q, Q', thus getting n pairs of unjoined points So we may assume that P is joined to P' Now remove P from A, leaving m points and mark the pair P', Q' in B as used We now repeat Take any two points in the reduced A and any unmarked pair in B and conclude that one of the pair in A is joined to one of the pair in B Remove the point from A and mark the pair in B Since n-1 >= m we can continue in this way until we obtain m pairs of joined points (one of each pair in A and the other in B) Contradiction So the result is true for n also

It remains to consider the case n = m

To get the n-1 pairs we use the induction on m G has m-1+2m = 3m-1 points and does not have either m pairs of joined points or m pairs of unjoined points We know by induction that f(m-1, m) = m-1+2m-1 = 3m-2, so f(m, m-1) = 3m-2 also Since m-1+2m > 3m-2, G has either m pairs of joined points or m-1 points of unjoined points It does not have the former (by assumption) so it must have the latter

(90)

with the pairs in B they would give n pairs of unjoined points So they are joined and hence give us an mth joined pair Contradiction

Comment This all seems complicated, and certainly much harder work than questions 1-3 So maybe I am missing something Does anyone have a simpler solution?

(91)

16th APMO 2004

Problem 1

Find all non-empty finite sets S of positive integers such that if m,n ∈ S, then (m+n)/gcd(m,n) ∈ S

Answer {2} Solution

Let k ∈ S Then (k+k)/gcd(k,k) = ∈ S Let M be the largest odd element of S Then (M+2)/gcd(M,2) = M+2 ∈ S Contradiction So all elements of S are even

Let m = 2n be the smallest element of S greater than Then (m+2)/2 = n+1 ∈ S But n must be > (or m = 2), so 2n > n+1 Hence 2n = (by minimality of m), so n = Contradiction So S has no elements apart from

Last corrected/updated 22 Mar 04 Problem 2

ABC is an acute-angled triangle with circumcenter O and orthocenter H (and O ≠ H) Show that one of area AOH, area BOH, area COH is the sum of the other two

Solution

(92)

Problem 3

2004 points are in the plane, no three collinear S is the set of lines through any two of the points Show that the points can be colored with two colors so that any two of the points have the same color iff there are an odd number of lines in S which separate them (a line separates them if they are on opposite sides of it)

Solution

Let us denote by dXY the number of points separating the points X and Y If the result is true, then the coloring is effectively determined: take a point X and color it blue Then for every other point Y, color it blue iff dXY is odd This will work provided that given any three points A, B, C, we have dAB + dBC + dCA is odd (For then if Y and Z are the same color, dXY and dXZ have th same parity, so dYZ is odd, which is correct Similarly, if Y and Z are opposite colors, then dYZ is even, which is correct.)

We are interested in lines which pass through an interior point of one or more of AB, BC, CA Lines cannot pass through all three If they pass through two, then they not affect the parity of dAB + dBC + dCA So we are interested in lines which pass through A and the (interior of) BC, and similarly for B, C Let n1, n2, , n7 be the number of points (excluding A, B, C) in the various regions (as shown) The number of lines through A and BC is n1 + n2 + n3 So dAB + dBC + dCA = (n1 + n2 + n3) + (n1 + n4 + n5) + (n1 + n6 + n7) = n1 + n2 + n3 + n4 + n5 + n6 + n7 = (2004 - 3) = mod

Thanks to Dinu Razvan

(93)

Problem 4

Show that [(n-1)!/(n2+n)] is even for any positive integer n Solution

Thanks to Juan Ignacio Restrepo

For n = 1, 2, 3, 4, we have [(n-1)!/(n2+n)] = 0, which is even So assume n ≥ If n and n+1 are composite, then they must divide (n-1)! They are coprime, so their product must divide (n-1)! Note that just one of n, n+1 is even For m ≥ 6, (m-2)! is divisible by more powers of than m, so (n-1)!/(n2+n) is even It remains to consider the two cases n+1 = p, a prime, and n = p a prime

If n+1 = p, then p-1 is composite, so p-1 divides (p-2)! Let k = (p-2)!/(p-1) By Wilson's theorem we have (p-2)! = mod p, so k(p-1) = mod p and hence k = -1 mod p So (k+1)/p is an integer But k is even, so k+1 is odd and hence (k+1)/p is odd Now [k/p] = (k+1)/p - 1, so [k/p] = [(n-1)!/(n2+n)] is even

If n = p, then k = (p-1)!/(p+1) is an even integer, so k+1 is an odd integer By Wilson's theorem, k(p+1) = -1 mod p, so (k+1)/p is an integer and hence an odd integer Hence [(n-1)!/(n2+n)] = [k/p] = (k+1)/p - is even

Wilson's theorem states that p is prime iff (p-1)! = -1 mod p If p is composite, then it has a factor ≤ p-1, which divides (p-1)! and so does not divide (p-1)! + Hence (p-1)! ≠ -1 mod p Now suppose p is prime If p = 2, then (p-1)! = = -1 mod So assume p is odd Note first that if a2 = mod p and < a < p, then a = or p-1 For p divides (a+1)(a-1) and so it divides either a+1, giving a = p-1, or it divides a-1, giving a = Now for each 0 < a < p, there is a unique a' such that aa' = mod p We have just shown that a = a' iff a = or p-1 So we can divide 2, 3, , p-2 into pairs with the product of each pair being mod p Hence (p-2)! = mod p, as required.

On the powers of 2, note that 2, 22, 23, , 2k-1 < 2k and their product is 2k(k-1)/2 If 2k < n < 2k+1, then n is divisible by at most 2k-1 So for n ≥ 16, for example, (n-8)!/n is even For n = 8, 9, , 15, (n-2)! is divisible by 2·4·6 and n is divisible by at most 8, so (n-2)!/n is even Finally, we can easily check n = 6, 7.

(94)

Problem 5

Show that (x2 + 2)(y2 + 2)(z2 + 2) ≥ 9(xy + yz + zx) for any positive reals x, y, z Solution

Expanding lhs we get x2y2z2 + 2(x2y2 + y2z2 + z2x2) + 4(x2 + y2 + z2) +

By AM/GM we have 3(x2 + y2 + z2) ≥ 3(xy + yz + zx) and (2x2y2 + 2) + (2y2z2 + 2) + (2z2x2 + 2) ≥ 4(xy + yz + zx) That leaves us needing x2y2z2 + x2 + y2 + z2 + ≥ 2(xy + yz + zx) (*) That is hard, because it is not clear how to deal with the xyz on the lhs By AM/GM we have (a+b)(b+c)(c+a) ≥ 8abc So if (u+v-w), (u-v+w), (-u+v+w) are all non-negative, we can put 2a = -u+v+w, 2b = u-v+w, 2c = u+v-w and get uvw ≥

u+v+w)(u-v+w)(u+v-w) If u, v, w are positive, then at most one of (u+v-w), (u-v+w), (-u+v+w) is negative In that case uvw ≥ (-(-u+v+w)(u-v+w)(u+v-w) is trivially true So it holds for all positive u, v, w Expanding, we get u3 + v3 + w3 + 3uvw ≥ uv(u+v) +

vw(v+w) + wu(w+u), which is a fairly well-known inequality Applying AM/GM to u+v etc we get, u3 + v3 + w3 + 3uvw ≥ 2(uv)3/2 + 2(vw)3/2 + 2(wu)3/2 Finally, putting u = x2/3, v = y2/3, w = z2/3, we get x2 + y2 + z2 + 3(xyz)2/3 ≥ 2(xy + yz + zx) (**)

Thus we have x2y2z2 + x2 + y2 + z2 + ≥ ≥ (x2 + y2 + z2) + 3(xyz)2/3, by applying AM/GM to x2y2z2, 1, 1, and now (**) gives the required (*)

(95)

XVII APMO - March, 2005 Problems and Solutions

Problem Prove that for every irrational real numbera, there are irrational real numbers

b and b0 so that a+b and ab0 are both rational while aband a+b0 are both irrational.

(Solution) Let a be an irrational number If a2 is irrational, we let b = −a Then,

a+b= is rational and ab=−a2 is irrational If a2 is rational, we let b=a2 −a Then,

a+b=a2 is rational and ab=a2(a−1) Since

a = ab a2 +

is irrational, so is ab Now, we let b0 =

a or b

0 =

a Thenab

0 = or 2, which is rational Note that

a+b0 = a

2+ 1

a or a+b

0 = a2+

a .

Since,

a2+ 2

a −

a2+ 1

a =

1 a, at least one of them is irrational

(96)

Problem Let a, band c be positive real numbers such thatabc= Prove that a2

p

(1 +a3)(1 +b3)+

b2

p

(1 +b3)(1 +c3) +

c2

p

(1 +c3)(1 +a3) ≥

4 3.

(Solution)Observe that

1 √

1 +x3 ≥

2

2 +x2 . (1)

In fact, this is equivalent to (2 +x2)2 ≥4(1 +x3), orx2(x−2)2 ≥0 Notice that equality

holds in (1) if and only if x=

We substitute x bya, b, c in (1), respectively, to find a2

p

(1 +a3)(1 +b3) +

b2

p

(1 +b3)(1 +c3) +

c2

p

(1 +c3)(1 +a3)

≥ 4a2

(2 +a2)(2 +b2) +

4b2

(2 +b2)(2 +c2)+

4c2

(2 +c2)(2 +a2). (2)

We combine the terms on the right hand side of (2) to obtain Left hand side of (2) ≥ 2S(a, b, c)

36 +S(a, b, c) =

2

1 + 36/S(a, b, c), (3) where S(a, b, c) := 2(a2+b2+c2) + (ab)2+ (bc)2+ (ca)2 By AM-GM inequality, we have

a2+b2+c2 ≥ 3p3 (abc)2 = 12,

(ab)2+ (bc)2 + (ca)2 ≥ 3p3 (abc)4 = 48.

Note that the equalities holds if and only ifa =b =c= The above inequalities yield S(a, b, c) = 2(a2+b2+c2) + (ab)2+ (bc)2+ (ca)2 ≥72. (4)

Therefore

2

1 + 36/S(a, b, c) ≥

1 + 36/72 =

3, (5)

which is the required inequality

(97)

Problem Prove that there exists a triangle which can be cut into 2005 congruent triangles

(Solution) Suppose that one side of a triangle has length n Then it can be cut into n2

congruent triangles which are similar to the original one and whose corresponding sides to the side of length n have lengths

Since 2005 = 5×401 where and 401 are primes and both primes are of the type 4k+ 1, it is representable as a sum of two integer squares Indeed, it is easy to see that

2005 = 5×401 = (22+ 1)(202+ 1)

= 402+ 202+ 22+ 1

= (40−1)2+ 2×40 + 202+ 22

= 392+ 222.

Let ABC be a right-angled triangle with the legs AB and BC having lengths 39 and 22, respectively We draw the altitude BK, which dividesABC into two similar triangles Now we divide ABK into 392 congruent triangles as described above and BCK into 222

congruent triangles SinceABK is similar to BKC, all 2005 triangles will be congruent

(98)

Problem In a small town, there aren×n houses indexed by (i, j) for 1≤i, j ≤nwith (1,1) being the house at the top left corner, where iand j are the row and column indices, respectively At time 0, a fire breaks out at the house indexed by (1, c), where c ≤ n2 During each subsequent time interval [t, t+ 1], the fire fighters defend a house which is not yet on fire while the fire spreads to all undefended neighbors of each house which was on fire at timet Once a house is defended, it remains so all the time The process ends when the fire can no longer spread At most how many houses can be saved by the fire fighters? A house indexed by (i, j) is a neighbor of a house indexed by (k, `) if |i−k|+|j −`|=

(Solution)At mostn2+c2−nc−chouses can be saved This can be achieved under the

following order of defending:

(2, c),(2, c+ 1); (3, c−1),(3, c+ 2); (4, c−2),(4, c+ 3); .

(c+ 1,1),(c+ 1,2c); (c+ 1,2c+ 1), ,(c+ 1, n). (6) Under this strategy, there are

2 columns (column numbers c, c+ 1) at which n−1 houses are saved columns (column numbers c−1, c+ 2) at which n−2 houses are saved · · ·

2 columns (column numbers 1,2c) at which n−c houses are saved

n−2ccolumns (column numbersn−2c+ 1, , n) at which n−c houses are saved Adding all these we obtain :

2[(n−1) + (n−2) +· · ·+ (n−c)] + (n−2c)(n−c) = n2+c2−cn−c. (7)

We say that a house indexed by (i, j) is at levelt if|i−1|+|j−c|=t Let d(t) be the number of houses at level t defended by timet, andp(t) be the number of houses at levels greater thant defended by time t It is clear that

p(t) +

t

X

i=1

d(i)≤t and p(t+ 1) +d(t+ 1) ≤p(t) + 1.

Lets(t) be the number of houses at level twhich are not burning at timet We prove that s(t)≤t−p(t)≤t

for ≤ t ≤ n−1 by induction It is obvious when t = Assume that it is true for t = k The union of the neighbors of any k−p(k) + houses at level k+ contains at least k−p(k) + vertices at level k Since s(k) ≤ k−p(k), one of these houses at level k is burning Therefore, at mostk−p(k) houses at level k+ have no neighbor burning Hence we have

s(k+ 1) ≤k−p(k) +d(k+ 1)

= (k+ 1)−(p(k) + 1−d(k+ 1)) ≤(k+ 1)−p(k+ 1).

(99)

We now prove that the strategy given above is optimal Since

n−1

X

t=1

s(t)≤

à

n

ả

,

the maximum number of houses at levels less than or equal to n−1, that can be saved under any strategy is at most ¡n2¢, which is realized by the strategy above Moreover, at levels bigger than n−1, every house is saved under the strategy above

The following is an example when n = 11 and c = The houses with ° mark are burned The houses with Nmark are blocked ones and hence those and the houses below them are saved

•

J J J J J J J J J J J

J J J N N J J J J J J

J J N N J J J J J

J N N J J J J

N N N N N

(100)

Problem In a triangle ABC, points M and N are on sides AB and AC, respectively, such that MB =BC =CN LetR and r denote the circumradius and the inradius of the triangleABC, respectively Express the ratio MN/BC in terms of R and r

(Solution)Letω, OandI be the circumcircle, the circumcenter and the incenter ofABC,

respectively Let D be the point of intersection of the line BI and the circle ω such that D6=B ThenD is the midpoint of the arc AC Hence OD ⊥CN and OD =R

We first show that triangles MNC and IOD are similar BecauseBC =BM, the line BI (the bisector of ∠MBC) is perpendicular to the line CM Because OD ⊥ CN and ID⊥MC, it follows that

∠ODI =∠NCM (8)

Let∠ABC = 2β In the triangle BCM, we have CM

NC =

CM

BC = sinβ (9)

Since ∠DIC = ∠DCI, we have ID = CD = AD Let E be the point of intersection of the line DO and the circle ω such that E 6= D Then DE is a diameter of ω and

∠DEC =∠DBC =β Thus we have DI

OD =

CD

OD =

2Rsinβ

R = sinβ. (10)

Combining equations (8), (9), and (10) shows that triangles MNC and IOD are similar It follows that

MN

BC =

MN

NC =

IO

OD =

IO

R . (11)

The well-known Euler’s formula states that

OI2 =R2−2Rr. (12)

Therefore,

MN

BC =

r

1− 2r

R . (13)

(Alternative Solution) Leta (resp., b, c) be the length of BC (resp., AC, AB) Let α

(resp.,β, γ) denote the angle∠BAC (resp.,∠ABC, ∠ACB) By introducing coordinates B = (0,0), C = (a,0), it is immediate that the coordinates of M and N are

M = (acosβ, asinβ), N = (a−acosγ, asinγ), (14)

(101)

respectively Therefore,

(MN/BC)2 = [(a−acosγ−acosβ)2 + (asinγ−asinβ)2]/a2

= (1−cosγ −cosβ)2+ (sinγ−sinβ)2

= 3−2 cosγ−2 cosβ+ 2(cosγcosβ−sinγsinβ) = 3−2 cosγ−2 cosβ+ cos(γ+β)

= 3−2 cosγ−2 cosβ−2 cosα = 3−2(cosγ+ cosβ+ cosα).

(15)

Now we claim

cosγ+ cosβ+ cosα = r

R + 1. (16)

From

a =bcosγ+ccosβ b =ccosα+acosγ c =acosβ+bcosα

(17) we get

a(1 + cosα) +b(1 + cosβ) +c(1 + cosγ) = (a+b+c)(cosα+ cosβ+ cosγ). (18) Thus

cosα+ cosβ+ cosγ

=

a+b+c(a(1 + cosα) +b(1 + cosβ) +c(1 + cosγ))

=

a+b+c

µ

a

µ

1 + b

2+c2−a2

2bc

¶

+b

µ

1 + a

2+c2−b2

2ac

ả

+c

à

1 + a

2+b2−c2

2ab

¶¶

=

a+b+c

µ

a+b+c+ a

2(b2+c2 −a2) +b2(a2+c2−b2) +c2(a2+b2−c2)

2abc

¶

= + 2a

2b2+ 2b2c2+ 2c2a2−a4−b4−c4

2abc(a+b+c) .

(19) On the other hand, from R= a

2 sinα it follows that

R2 = a2

4(1−cos2α) =

a2

4

Ã

1−

µ

b2+c2−a2

2bc

¶2!

= a2b2c2

2a2b2+ 2b2c2+ 2c2a2−a4−b4−c4 .

(20)

(102)

Also from

2(a+b+c)r=

2bcsinα, it follows that

r2 = b2c2(1−cos2α)

(a+b+c)2 =

b2c2

Ã

1−

µ

b2 +c2−a2

2bc

¶2!

(a+b+c)2

= 2a2b2+ 2b2c2+ 2c2a2−a4−b4−c4 4(a+b+c)2 .

(21)

Combining (19), (20) and (21), we get (16) as desired Finally, by (15) and (16) we have

MN

BC =

r

1− 2r

R . (22)

Another proof of (16) from R.A Johnson’s “Advanced Euclidean Geometry”1:

Construct the perpendicular bisectors OD, OE, OF, where D, E, F are the midpoints of BC, CA, AB, respectively By Ptolemy’s Theorem applied to the cyclic quadrilateral OEAF, we get

a ·R =

b

2 ·OF + c 2·OE. Similarly

b 2·R =

c

2 ·OD+ a ·OF,

c ·R=

a

2 ·OE+ b ·OD. Adding, we get

sR=OD· b+c

2 +OE· c+a

2 +OF · a+b

2 , (23)

where s is the semiperimeter But also, the area of triangle OBC is OD· a

2, and adding similar formulas for the areas of triangles OCA and OAB gives

rs=4ABC =OD·a

2+OE· b

2+OF · c

2 (24)

Adding (23) and (24) gives s(R+r) =s(OD+OE+OF), or OD+OE+OF =R+r. Since OD =RcosA etc., (16) follows

1This proof was introduced to the coordinating country by Professor Bill Sands of Canada.

(103)

Problem Let n be a positive integer Find the largest nonnegative real number f(n) (depending on n) with the following property: whenever a1, a2, , an are real numbers

such that a1+a2+· · ·+an is an integer, there exists some i such that|ai−12 | ≥f(n)

(Solution)The answer is

f(n) =

½

0 if n is even,

1

2n if n is odd

First, assume that n is even If ai = 12 for all i, then the sum a1 +a2+· · ·+an is an

integer Since |ai− 12 |= for alli, we may conclude f(n) = for any even n

Now assume that n is odd Suppose that |ai− 12 |< 21n for all 1≤ i≤n Then, since Pn

i=1ai is an integer,

1 ≤ ¯ ¯ ¯ ¯ ¯ n X i=1

ai−

n ¯ ¯ ¯ ¯ ¯≤ n X i=1 ¯ ¯ ¯ ¯ai−

1

¯ ¯ ¯

¯< 21n ·n = 12,

a contradiction Thus |ai− 12| ≥ 21n for some i, as required On the other hand, putting

n= 2m+ and ai = 2mm+1 for all i gives P

ai =m, while ¯

¯ ¯ ¯ai−

1

¯ ¯ ¯

¯= 12− 2mm+ 1 = 2(2m1+ 1) = 21n

for all i Therefore, f(n) =

2n is the best possible for any odd n

Problem Prove that every positive integer can be written as a finite sum of distinct integral powers of the golden mean τ = 1+√5

2 Here, an integral power of τ is of the form

τi, wherei is an integer (not necessarily positive).

(Solution)We will prove this statement by induction using the equality

τ2 =τ + 1.

Ifn= 1, then = τ0 Suppose thatn−1 can be written as a finite sum of integral powers

of τ, say

n−1 =

k X

i=−k

aiτi, (1)

where ai ∈ {0,1} and n ≥2 We will write (1) as

n−1 = ak· · ·a1a0.a−1a−2· · ·a−k. (2)

For example,

(104)

Firstly, we will prove that we may assume that in (2) we have aiai+1 = for all i with −k ≤ i ≤ k −1 Indeed, if we have several occurrences of 11, then we take the leftmost such occurrence Since we may assume that it is preceded by a 0, we can replace 011 with 100 using the identity τi+1+τi =τi+2 By doing so repeatedly, if necessary, we will

eliminate all occurrences of two 1’s standing together Now we have the representation

n−1 =

K X

i=−K

biτi, (3)

where bi ∈ {0,1}and bibi+1 =

If b0 = in (3), then we just add =τ0 to both sides of (3) and we are done

Suppose now that there is in the unit position of (3), that is b0 = If there are two

0’s to the right of it, i.e

n−1 = · · ·1.00· · · ,

then we can replace 1.00 with 0.11 because = τ−1 +τ−2, and we are done because we

obtain in the unit position Thus we may assume that

n−1 =· · ·1.010· · · .

Again, if we have n−1 =· · ·1.0100· · ·, we may rewrite it as

n−1 =· · ·1.0100· · ·=· · ·1.0011· · ·=· · ·0.1111· · · and obtain in the unit position Therefore, we may assume that

n−1 = · · ·1.01010· · · .

Since the number of 1’s is finite, eventually we will obtain an occurrence of 100 at the end, i.e

n−1 = · · ·1.01010· · ·100.

Then we can shift all 1’s to the right to obtain in the unit position, i.e

n−1 =· · ·0.11· · ·11,

and we are done

Problem Let p≥5 be a prime and let r be the number of ways of placing p checkers on a p×p checkerboard so that not all checkers are in the same row (but they may all be in the same column) Show thatr is divisible byp5 Here, we assume that all the checkers

are identical

(Solution)Note that r=

µ

p2

p

¶

−p Hence, it suffices to show that

(105)

Now, let

f(x) := (x−1)(x−2)· · ·(x−(p−1)) =xp−1+s

p−2xp−2+· · ·+s1x+s0. (2)

Then the congruence equation (1) is same asf(p2)−s

0 ≡0 (modp4) Therefore, it suffices

to show that s1p2 ≡0 (mod p4) or s1 ≡0 (mod p2)

Since ap−1 ≡1 (mod p) for all 1≤a≤p−1, we can factor

xp−1−1≡(x−1)(x−2)· · ·(x−(p−1)) (mod p). (3)

Comparing the coefficients of the left hand side of (3) with those of the right hand side of (2), we obtain p|si for all ≤ i ≤ p−2 and s0 ≡ −1 (mod p) On the other hand,

plugging p for x in (2), we get

f(p) = (p−1)! =s0 =pp−1+sp−2pp−2+· · ·+s1p+s0,

which implies

pp−1+s

p−2pp−2+· · ·+s2p2 =−s1p.

Since p≥5,p|s2 and hence s1 ≡0 (mod p2) as desired

Problem LetA, B be two distinct points on a given circleOand letP be the midpoint of the line segment AB Let O1 be the circle tangent to the line AB at P and tangent to

the circle O Let ` be the tangent line, different from the line AB, to O1 passing through

A Let C be the intersection point, different from A, of ` and O Let Q be the midpoint of the line segment BC and O2 be the circle tangent to the line BC atQ and tangent to

the line segment AC Prove that the circle O2 is tangent to the circle O

(Solution)LetS be the tangent point of the circlesOandO1and letT be the intersection

point, different from S, of the circle O and the line SP Let X be the tangent point of

` to O1 and let M be the midpoint of the line segment XP Since ∠T BP =∠ASP, the

triangleT BP is similar to the triangleASP Therefore,

P T P B =

P A P S .

Since the line ` is tangent to the circle O1 atX, we have

∠SP X = 90◦−∠XSP = 90◦ −∠AP M =∠P AM

which implies that the triangle P AM is similar to the triangle SP X Consequently,

XS XP =

MP MA =

XP

2MA and

XP P S =

(106)

From this and the above observation follows

XS XP ·

P T P B =

XP

2MA · P A P S =

XP

2MA · MA

XP =

1

2. (1)

LetA0 be the intersection point of the circleO and the perpendicular bisector of the chord

BC such that A, A0 are on the same side of the line BC, and N be the intersection point

of the lines A0Qand CT Since

∠NCQ=∠T CB =∠T CA=∠T BA=∠T BP

and

∠CA0Q= ∠CAB

2 =

∠XAP

2 =∠P AM =∠SP X,

the triangle NCQ is similar to the triangle T BP and the triangle CA0Q is similar to the

triangleSP X Therefore

QN QC =

P T

P B and QC QA0 =

XS XP .

and hence QA0 = 2QN by (1) This implies that N is the midpoint of the line segment

QA0 Let the circle O

2 touch the line segmentAC at Y Since

∠ACN =∠ACT =∠BCT =∠QCN

and |CY|=|CQ|, the trianglesY CN and QCN are congruent and hence NY ⊥AC and

NY =NQ=NA0 Therefore, N is the center of the circle O

2, which completes the proof

Remark: Analytic solutions are possible : For example, one can prove for a triangleABC

inscribed in a circle O that AB = k(2 + 2t), AC = k(1 + 2t), BC = k(1 + 4t) for some positive numbers k, t if and only if there exists a circle O1 such that O1 is tangent to the

side AB at its midpoint, the side AC and the circle O

One obtains AB = k0(1 + 4t0), AC = k0(1 + 2t0), BC = k0(2 + 2t0) by substituting

t= 1/4t0 and k = 2k0t0 So, there exists a circle O

2 such thatO2 is tangent to the sideBC

at its midpoint, the sideAC and the circle O In the above, t = tan2α and k = 4Rtanα

(1+tan2α)(1+4 tan2α), where R is the radius of O and

∠A= 2α Furthermore, t0 = tan2γ and k0 = 4Rtanγ

(1+tan2γ)(1+4 tan2γ), where∠C = 2γ Observe

that √tt0 = tanα·tanγ = XS XP ·

P T P B =

1

2, which implies tt0 =

4 It is now routine easy to

check that k = 2k0t0.

(107)

of colours and no more than 20 clowns may use any one particular colour Find the largest number n of clowns so as to make the ringmaster’s order possible

(Solution) Let C be the set of n clowns Label the colours 1,2,3, ,12 For each

i = 1,2, ,12, let Ei denote the set of clowns who use colour i For each subset S of

{1,2, ,12}, letES be the set of clowns who use exactly those colours inS Since S 6=S0

implies ES∩ES0 =∅, we have X

S

|ES|=|C|=n,

where S runs over all subsets of {1,2, ,12} Now for each i,

ES ⊆Ei if and only if i∈S,

and hence

|Ei|= X

i∈S

|ES|.

By assumption, we know that |Ei| ≤ 20 and that if ES 6= ∅, then |S| ≥ From this we

obtain

20×12≥ 12

X

i=1

|Ei|=

12

X

i=1

Ã X

i∈S

|ES| !

≥5X

S

|ES|= 5n.

Therefore n≤48

Now, define a sequence {ci}52i=1 of colours in the following way:

1 | 8| 10 11 12 | | 7| 12 10 11 | | 6| 11 12 10 |

2 | 5| 10 11 12 |

The first row lists c1, , c12 in order, the second row lists c13, , c24 in order, the third

row lists c25, , c36 in order, and finally the last row lists c37, , c52 in order For each

j, 1≤j ≤48, assign colours cj, cj+1, cj+2, cj+3, cj+4 to the j-th clown It is easy to check

that this assignment satisfies all conditions given above So, 48 is the largest for n

Remark: The fact thatn ≤48 can be obtained in a much simpler observation that 5n≤12×20 = 240.

There are many other ways of constructing 48 distinct sets consisting of colours For example, consider the sets

{1,2,3,4,5,6}, {3,4,5,6,7,8}, {5,6,7,8,9,10}, {7,8,9,10,11,12},

{9,10,11,12,1,2}, {11,12,1,2,3,4}, {1,2,5,6,9,10}, {3,4,7,8,11,12}.

(108)

XIX Asian Pacific Mathematics Olympiad

Problem LetS be a set of distinct integers all of whose prime factors are at most Prove that S contains distinct integers such that their product is a perfect cube

Solution Without loss of generality, we may assume thatScontains only positive integers Let

S ={2ai3bi |a

i, bi ∈Z, ai, bi ≥0, 1≤i≤9}.

It suffices to show that there are 1≤i1, i2, i3 ≤9 such that

ai1 +ai2 +ai3 ≡bi1 +bi2 +bi3 ≡0 (mod 3). (†) Forn= 2a3b ∈S, let’s call (a (mod 3), b (mod 3)) the typeof n Then there are possible

types :

(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2).

Let N(i, j) be the number of integers in S of type (i, j) We obtain distinct integers whose product is a perfect cube when

(1) N(i, j)≥3 for somei, j, or

(2) N(i,0)N(i,1)N(i,2)6= for some i= 0,1,2, or

(3) N(0, j)N(1, j)N(2, j)6= for some j = 0,1,2, or

(4) N(i1, j1)N(i2, j2)N(i3, j3)6= 0, where{i1, i2, i3}={j1, j2, j3}={0,1,2}

(109)

Second solution Up to (†), we the same as above and get possible types : (a (mod 3), b (mod 3)) = (0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2) for n= 2a3b ∈S.

Note that (i) among any integers, there exist whose sum is (mod 3), and that (ii) ifi, j, k ∈ {0,1,2}, theni+j+k ≡0 (mod 3) if and only ifi=j =k or{i, j, k}={0,1,2}

Let’s define

T : the set of types of the integers inS;

N(i) : the number of integers in S of the type (i,·) ;

M(i) : the number of integers j ∈ {0,1,2} such that (i, j)∈T

IfN(i)≥5 for somei, the result follows from (i) Otherwise, for some permutation (i, j, k) of (0,1,2),

N(i)≥3, N(j)≥3, N(k)≥1.

If M(i) or M(j) is or 3, the result follows from (ii) Otherwise M(i) = M(j) = Then either

(i, x),(i, y),(j, x),(j, y)∈T or (i, x),(i, y),(j, x),(j, z)∈T

(110)

Problem LetABC be an acute angled triangle with∠BAC = 60◦ andAB > AC Let

I be the incenter, and H the orthocenter of the triangleABC Prove that 2∠AHI = 3∠ABC.

Solution Let D be the intersection point of the lines AH and BC Let K be the intersection point of the circumcircleO of the triangleABC and the line AH Let the line through I perpendicular to BC meet BC and the minor arc BC of the circumcircle O at

E and N, respectively We have

∠BIC = 180◦−(∠IBC+∠ICB) = 180◦−

2(∠ABC+∠ACB) = 90

◦+1

2∠BAC = 120

◦

and also ∠BNC = 180◦ −∠BAC = 120◦ = ∠BIC Since IN ⊥ BC, the quadrilateral

BICN is a kite and thus IE =EN

Now, since H is the orthocenter of the triangle ABC, HD = DK Also because

ED ⊥ IN and ED ⊥ HK, we conclude that IHKN is an isosceles trapezoid with

IH =NK Hence

∠AHI = 180◦−∠IHK = 180◦−∠AKN =∠ABN.

Since IE =EN and BE ⊥IN, the triangles IBE and NBE are congruent Therefore ∠NBE =∠IBE =∠IBC =∠IBA=

2∠ABC and thus

∠AHI =∠ABN =

2∠ABC.

Second solution Let P, Q and R be the intersection points of BH, CH and AH with

AC, AB and BC, respectively Then we have ∠IBH =∠ICH Indeed, ∠IBH =∠ABP −∠ABI = 30◦−

2∠ABC and

∠ICH =∠ACI−∠ACH =

2∠ACB−30

◦ = 30◦−

2∠ABC,

because∠ABH =∠ACH = 30◦ and∠ACB+∠ABC = 120◦ (Note that∠ABP >∠ABI

and ∠ACI >∠ACH because AB is the longest side of the triangle ABC under the given conditions.) Therefore BIHC is a cyclic quadrilateral and thus

∠BHI =∠BCI =

(111)

On the other hand,

∠BHR= 90◦−∠HBR= 90◦ −(∠ABC −∠ABH) = 120◦−∠ABC.

Therefore,

∠AHI = 180◦−∠BHI−∠BHR= 60◦−1

2∠ACB+∠ABC = 60◦−

2(120

◦−∠ABC) +∠ABC =

(112)

Problem Consider n disksC1, C2, , Cn in a plane such that for each 1≤i < n, the

center of Ci is on the circumference of Ci+1, and the center ofCn is on the circumference

of C1 Define the score of such an arrangement of n disks to be the number of pairs (i, j)

for which Ci properly contains Cj Determine the maximum possible score

Solution The answer is (n−1)(n−2)/2

Let’s call a set of n disks satisfying the given conditions an n-configuration For an n -configuration C ={C1, , Cn}, let SC ={(i, j)| Ci properly contains Cj } So, the score

of ann-configuration C is |SC|

We’ll show that (i) there is ann-configuration C for which|SC|= (n−1)(n−2)/2, and

that (ii) |SC| ≤(n−1)(n−2)/2 for anyn-configuration C

Let C1 be any disk Then for i = 2, , n−1, take Ci inside Ci−1 so that the

cir-cumference of Ci contains the center of Ci−1 Finally, let Cn be a disk whose center is on

the circumference of C1 and whose circumference contains the center of Cn−1 This gives

SC ={(i, j)|1≤i < j ≤n−1} of size (n−1)(n−2)/2, which proves (i)

For any n-configuration C, SC must satisfy the following properties: (1) (i, i)6∈SC,

(2) (i+ 1, i)6∈SC, (1, n)6∈SC,

(3) if (i, j),(j, k)∈SC, then (i, k)∈SC, (4) if (i, j)∈SC, then (j, i)6∈SC

Now we show that a set G of ordered pairs of integers between and n, satisfying the conditions (1)∼(4), can have no more than (n−1)(n−2)/2 elements Suppose that there exists a set G that satisfies the conditions (1)∼(4), and has more than (n−1)(n−2)/2 elements Let n be the least positive integer with which there exists such a set G Note that G must have (i, i+ 1) for some ≤ i ≤ n or (n,1), since otherwise G can have at

most µ

n

2 ¶

−n= n(n−3) <

(n−1)(n−2)

elements Without loss of generality we may assume that (n,1)∈G Then (1, n−1)6∈G, since otherwise the condition (3) yields (n, n−1)∈Gcontradicting the condition (2) Now letG0 ={(i, j)∈G|1≤i, j ≤n−1}, then G0 satisfies the conditions (1)∼(4), withn−1.

We now claim that |G−G0| ≤n−2 :

Suppose that |G−G0|> n−2, then |G−G0|=n−1 and hence for each 1≤i≤n−1,

either (i, n) or (n, i) must be in G We already know that (n,1) ∈ G and (n−1, n) ∈ G

(113)

Since |G−G0| ≤n−2, we obtain

|G0| ≥ (n−1)(n−2)

2 −(n−2) =

(n−2)(n−3)

2 .

(114)

Problem Letx, y and z be positive real numbers such that√x+√y+√z = Prove that

x2+yz

p

2x2(y+z)+

y2+zx

p

2y2(z+x) +

z2 +xy

p

2z2(x+y) ≥1.

Solution We first note that

x2+yz

p

2x2(y+z) =

x2−x(y+z) +yz

p

2x2(y+z) +

x(y+z) p

2x2(y+z)

= (px−y)(x−z) 2x2(y+z) +

r

y+z

2 ≥ (px−y)(x−z)

2x2(y+z) + √

y+√z

2 . (1)

Similarly, we have

y2+zx

p

2y2(z+x) ≥

(y−z)(y−x) p

2y2(z+x) + √

z+√x

2 , (2)

z2 +xy

p

2z2(x+y) ≥

(z−x)(z−y) p

2z2(x+y) + √

x+√y

2 . (3)

We now add (1)∼(3) to get

x2+yz

p

2x2(y+z) +

y2+zx

p

2y2(z+x)+

z2+xy

p

2z2(x+y) ≥ (px−y)(x−z)

2x2(y+z) +

(y−z)(y−x) p

2y2(z+x) +

(z−x)(z−y) p

2z2(x+y) + √

x+√y+√z

= (px−y)(x−z) 2x2(y+z) +

(y−z)(y−x) p

2y2(z+x) +

(z−x)(z−y) p

2z2(x+y) + 1.

Thus, it suffices to show that (x−y)(x−z)

p

2x2(y+z) +

(y−z)(y−x) p

2y2(z+x) +

(z−x)(z−y) p

2z2(x+y) ≥0. (4)

Now, assume without loss of generality, that x≥y≥z Then we have (x−y)(x−z)

p

(115)

and

(z−x)(z−y) p

2z2(x+y) +

(y−z)(y−x) p

2y2(z+x) =

(y−z)(x−z) p

2z2(x+y) −

(y−z)(x−y) p

2y2(z+x) ≥ (py−z)(x−y)

2z2(x+y) −

(y−z)(x−y) p

2y2(z+x) = (y−z)(x−y)

Ã

1 p

2z2(x+y) −

1 p

2y2(z+x)

!

.

The last quantity is non-negative due to the fact that

y2(z+x) = y2z+y2x≥yz2+z2x=z2(x+y).

This completes the proof

Second solution By Cauchy-Schwarz inequality, Ã

x2

p

2x2(y+z) +

y2

p

2y2(z+x)+

z2

p

2z2(x+y)

!

(5)

×(p2(y+z) +p2(z+x) +p2(x+y))≥(√x+√y+√z)2 = 1,

and

Ã

yz

p

2x2(y+z)+

zx

p

2y2(z+x) +

xy

p

2z2(x+y)

!

(6)

ì(p2(y+z) +p2(z+x) +p2(x+y))

àr yz x + r zx y + r xy z ¶2 .

We now combine (5) and (6) to find Ã

x2+yz

p

2x2(y+z) +

y2+zx

p

2y2(z+x) +

z2+xy

p

2z2(x+y)

!

×(p2(x+y) +p2(y+z) +p2(z+x))

≥1 +

µr yz x + r zx y + r xy z ả2 2 àr yz x + r zx y + r xy z ¶ .

Thus, it suffices to show that µr yz x + r zx y + r xy z ¶

≥p2(y+z) +p2(z+x) +p2(x+y). (7)

Consider the following inequality using AM-GM inequality ·r yz x + µ r zx y + r xy z ảá2 4 r yz x µ r zx y + r xy z ¶

(116)

or equivalently r

yz x +

µ

r

zx y +

1

r

xy z

¶

≥p2(y+z).

Similarly, we have

r

zx y +

µ

r

xy z +

1

r

yz x

¶

≥p2(z+x),

r

xy z +

µ

r

yz x +

1

r

zx y

¶

≥p2(x+y).

Adding the last three inequalities, we get

µr

yz x +

r

zx y +

r

xy z

¶

≥p2(y+z) +p2(z+x) +p2(x+y).

(117)

Problem A regular (5×5)-array of lights is defective, so that toggling the switch for one light causes each adjacent light in the same row and in the same column as well as the light itself to change state, from on to off, or from off to on Initially all the lights are switched off After a certain number of toggles, exactly one light is switched on Find all the possible positions of this light

Solution We assign the following first labels to the 25 positions of the lights: 1 1

0 0 0 1 1 0 0 1 1

For each on-off combination of lights in the array, define its first value to be the sum of the first labels of those positions at which the lights are switched on It is easy to check that toggling any switch always leads to an on-off combination of lights whose first value has the same parity(the remainder when divided by 2) as that of the previous on-off combination

The 90◦ rotation of the first labels gives us another labels (let us call it the second labels) which also makes the parity of the second value(the sum of the second labels of those positions at which the lights are switched on) invariant under toggling

1 1 1 0 0 1 1 1

Since the parity of the first and the second values of the initial status is 0, after certain number of toggles the parity must remain unchanged with respect to the first labels and the second labels as well Therefore, if exactly one light is on after some number of toggles, the label of that position must be with respect to both labels Hence according to the above pictures, the possible positions are the ones marked with∗i’s in the following picture:

∗2 ∗1

∗0

(118)

Now we demonstrate that all five positions are possible :

Toggling the positions checked by t (the order of toggling is irrelevant) in the first picture makes the center(∗0) the only position with light on and the second picture makes

the position∗1 the only position with light on The other ∗i’s can be obtained by rotating

the second picture appropriately

t t t

t t

t t t

t t

t t t t

t

(119)

XX Asian Pacific Mathematics Olympiad

March, 2008

Problem 1. LetABC be a triangle with∠A <60◦ LetX andY be the points on the sides

ABand AC, respectively, such thatCA+AX =CB+BX and BA+AY =BC+CY Let

P be the point in the plane such that the lines P X and P Y are perpendicular to AB and

AC, respectively Prove that ∠BP C <120◦

(Solution) Let I be the incenter of △ABC, and let the feet of the perpendiculars from I

to AB and to AC be D and E, respectively (Without loss of generality, we may assume that AC is the longest side Then X lies on the line segment AD Although P may or may not lie inside△ABC, the proof below works for both cases Note that P is on the line perpendicular toAB passing through X.) Let O be the midpoint ofIP, and let the feet of the perpendiculars fromO toAB and toAC beM and N, respectively ThenM and N are the midpoints ofDX and EY, respectively

(120)

The conditions on the pointsX and Y yield the equations

AX= AB+BC−CA

2 and AY =

BC+CA−AB

2 .

FromAD=AE= CA+AB−BC

2 , we obtain

BD=AB−AD=AB− CA+AB−BC

2 =

AB+BC−CA

2 =AX.

SinceM is the midpoint of DX, it follows thatM is the midpoint of AB Similarly,N is the midpoint ofAC Therefore, the perpendicular bisectors of AB and AC meet at O, that is,

O is the circumcenter of △ABC Since ∠BAC <60◦,O lies on the same side of BC as the pointA and

∠BOC= 2∠BAC.

We can compute∠BIC as follows :

∠BIC = 180◦−∠IBC−∠ICB = 180◦−1

2∠ABC−

2∠ACB = 180◦−1

2(∠ABC+∠ACB) = 180

◦−1

2(180

◦−∠BAC) = 90◦+1

2∠BAC It follows from∠BAC <60◦ that

2∠BAC <90◦+1

2∠BAC, i.e., ∠BOC <∠BIC.

From this it follows thatI lies inside the circumcircle of the isosceles triangle BOC because

Oand I lie on the same side ofBC However, asO is the midpoint ofIP,P must lie outside the circumcircle of triangleBOC and on the same side ofBC asO Therefore

∠BP C <∠BOC= 2∠BAC <120◦.

Remark. If one assumes that ∠A is smaller than the other two, then it is clear that the line P X (or the line perpendicular to AB at X if P = X) runs through the excenter IC of the excircle tangent to the side AB Since 2∠ACIC = ∠ACB and BC < AC, we have 2∠P CB >∠C Similarly, 2∠P BC >∠B Therefore,

∠BP C= 180◦−(∠P BC+∠P CB)<180◦− (

∠B+∠C

2 )

= 90 + ∠A <120

◦.

In this way, a special case of the problem can be easily proved

(121)

Problem 2. Students in a class form groups each of which contains exactly three members such that any two distinct groups have at most one member in common Prove that, when the class size is 46, there is a set of 10 students in which no group is properly contained

(Solution)We letC be the set of all 46 students in the class and let

s:= max{ |S| : S ⊆C such that S contains no group properly}.

Then it suﬃces to prove thats≥10 (If|S|=s >10, we may choose a subset ofS consisting of 10 students.)

Suppose thats≤9 and let S be a set of sizes in which no group is properly contained Take any student, sayv, from outsideS Because of the maximality of s, there should be a group containing the studentv and two other students in S The number of ways to choose two students fromS is (

s

2 )

≤ (

9 )

= 36.

On the other hand, there are at least 37 = 46−9 students outside ofS Thus, among those 37 students outside, there is at least one student, sayu, who does not belong to any group containing two students inSand one outside This is because no two distinct groups have two members in common But then,S can be enlarged by including u, which is a contradiction

Remark. One may choose a subsetS ofC that contains no group properly Then, assuming |S|<10, prove that there is a student outside S, sayu, who does not belong to any group containing two students in S After enlarging S by including u, prove that the enlarged S

still contains no group properly

(122)

Problem 3. Let Γ be the circumcircle of a triangle ABC A circle passing through points

Aand C meets the sidesBC and BAatDand E, respectively The linesAD andCE meet Γ again atG andH, respectively The tangent lines of Γ atA and C meet the line DE atL

andM, respectively Prove that the linesLH and M Gmeet at Γ

(Solution)LetM G meet Γ atP Since∠M CD =∠CAE and ∠M DC =∠CAE, we have

M C =M D Thus

M D2=M C2 =M G·M P

and henceM D is tangent to the circumcircle of △DGP Therefore∠DGP =∠EDP Let Γ′ be the circumcircle of△BDE IfB =P, then, since∠BGD=∠BDE, the tangent lines of Γ′ and Γ at B should coincide, that is Γ′ is tangent to Γ from inside LetB ̸=P IfP lies in the same side of the lineBC asG, then we have

∠EDP +∠ABP = 180◦

because∠DGP+∠ABP = 180◦ That is, the quadrilateralBP DE is cyclic, and henceP is on the intersection of Γ′ with Γ

(123)

Otherwise,

∠EDP =∠DGP =∠AGP =∠ABP =∠EBP.

Therefore the quadrilateral P BDE is cyclic, and hence P again is on the intersection of Γ′ with Γ

Similarly, if LH meets Γ at Q, we either haveQ = B, in which case Γ′ is tangent to Γ from inside, or Q ̸= B In the latter case, Q is on the intersection of Γ′ with Γ In either case, we haveP =Q

(124)

Problem 4. Consider the function f : N0 → N0, where N0 is the set of all non-negative integers, deﬁned by the following conditions :

(i) f(0) = 0, (ii) f(2n) = 2f(n) and (iii)f(2n+ 1) =n+ 2f(n) for alln≥0.

(a) Determine the three sets L:={n|f(n)< f(n+ 1)}, E :={n|f(n) =f(n+ 1)}, and

G:={n|f(n)> f(n+ 1)}.

(b) For eachk≥0, ﬁnd a formula forak:= max{f(n) : 0≤n≤2k} in terms ofk

(Solution)(a) Let

L1 :={2k:k >0}, E1 :={0} ∪ {4k+ :k≥0}, and G1 :={4k+ :k≥0}.

We will show thatL1 =L, E1 =E, and G1 =G It suﬃces to verify that L1 ⊆E,E1 ⊆E,

andG1 ⊆Gbecause L1,E1, and G1 are mutually disjoint andL1∪E1∪G1=N0

Firstly, ifk >0, thenf(2k)−f(2k+ 1) =−k <0 and therefore L1⊆L

Secondly, f(0) = and

f(4k+ 1) = 2k+ 2f(2k) = 2k+ 4f(k)

f(4k+ 2) = 2f(2k+ 1) = 2(k+ 2f(k)) = 2k+ 4f(k) for allk≥0 Thus,E1 ⊆E

Lastly, in order to proveG1 ⊂G, we claim that f(n+ 1)−f(n)≤n for all n (In fact,

one can prove a stronger inequality : f(n+ 1)−f(n)≤n/2.) This is clearly true for evenn

from the deﬁnition since forn= 2t,

f(2t+ 1)−f(2t) =t≤n.

If n = 2t+ is odd, then (assuming inductively that the result holds for all nonnegative

m < n), we have

f(n+ 1)−f(n) =f(2t+ 2)−f(2t+ 1) = 2f(t+ 1)−t−2f(t) = 2(f(t+ 1)−f(t))−t≤2t−t=t < n.

For allk≥0,

f(4k+ 4)−f(4k+ 3) =f(2(2k+ 2))−f(2(2k+ 1) + 1)

= 4f(k+ 1)−(2k+ + 2f(2k+ 1)) = 4f(k+ 1)−(2k+ + 2k+ 4f(k)) = 4(f(k+ 1)−f(k))−(4k+ 1)≤4k−(4k+ 1)<0.

This provesG1 ⊆G

(b) Note thata0 =a1 =f(1) = Let k ≥2 and let Nk ={0,1,2, ,2k} First we claim that the maximum ak occurs at the largest number in G∩Nk, that is, ak = f(2k−1) We use mathematical induction on kto prove the claim Note that a2 =f(3) =f(22−1)

Now let k≥3 For every even number 2t with 2k−1+ 1<2t≤2k,

f(2t) = 2f(t)≤2ak−1= 2f(2k−1−1) (†)

by induction hypothesis For every odd number 2t+ with 2k−1+ 1≤2t+ 1<2k,

f(2t+ 1) =t+ 2f(t)≤2k−1−1 + 2f(t) ≤2k−1−1 + 2a

k−1= 2k−1−1 + 2f(2k−1−1)

(‡)

(125)

again by induction hypothesis Combining (†), (‡) and

f(2k−1) =f(2(2k−1−1) + 1) = 2k−1−1 + 2f(2k−1−1),

we may conclude thatak =f(2k−1) as desired Furthermore, we obtain

ak = 2ak−1+ 2k−1−1

for allk≥3 Note that this recursive formula forak also holds fork≥0,1 and Unwinding this recursive formula, we ﬁnally get

ak = 2ak−1+ 2k−1−1 = 2(2ak−2+ 2k−2−1) + 2k−1−1

= 22ak−2+ 2·2k−1−2−1 = 22(2ak−3+ 2k−3−1) + 2·2k−1−2−1

= 23ak−3+ 3·2k−1−22−2−1

= 2ka0+k2k−1−2k−1−2k−2− −2−1

=k2k−1−2k+ for all k≥0.

(126)

Problem 5. Let a, b, c be integers satisfying < a < c−1 and < b < c For each k, 0≤k≤a, let rk, 0≤rk < c, be the remainder of kbwhen divided by c Prove that the two sets{r0, r1, r2, , ra} and{0,1,2, , a}are diﬀerent

(Solution)Suppose that two sets are equal Then gcd(b, c) = and the polynomial

f(x) := (1 +xb+x2b+· · ·+xab)−(1 +x+x2+· · ·+xa−1+xa)

is divisible byxc−1 (This is because : m=n+cq =⇒ xm−xn=xn+cq−xn=xn(xcq−1) and (xcq−1) = (xc−1)((xc)q−1+ (xc)q−2+· · ·+ 1).) From

f(x) = x

(a+1)b−1

xb−1 −

xa+1−1

x−1 =

F(x) (x−1)(xb−1) , whereF(x) =xab+b+1+xb+xa+1−xab+b−xa+b+1−x, we have

F(x)≡0 (modxc−1).

Sincexc≡1 (modxc−1), we may conclude that

{ab+b+ 1, b, a+ 1} ≡ {ab+b, a+b+ 1,1} (mod c). (†) Thus,

b≡ ab+b, a+b+ or (modc).

But neither b ≡ (mod c) nor b ≡ a+b+ (mod c) are possible by the given conditions Therefore,b≡ab+b(modc) But this is also impossible because gcd(b, c) =

(127)

XXI Asian Pacific Mathematics Olympiad

March, 2009 Problem 1. Consider the following operation on positive real numbers written on a black-board:

Choose a numberr written on the blackboard, erase that number, and then write a pair of positive real numbers aand bsatisfying the condition 2r2 =abon the board Assume that you start out with just one positive real numberron the blackboard, and apply this operationk2−1 times to end up withk2 positive real numbers, not necessarily distinct Show that there exists a number on the board which does not exceedkr

(Solution)Using AM-GM inequality, we obtain

r2 =

2

ab =

2ab a2b2 ≤

a2+b2 a2b2 ≤

1

a2 +

1

b2 . (∗)

Consequently, if we letSℓbe the sum of the squares of the reciprocals of the numbers written on the board after ℓoperations, then Sℓ increases as ℓincreases, that is,

S0≤S1 ≤ · · · ≤Sk2−1. (∗∗)

Therefore if we letsbe the smallest real number written on the board afterk2−1 operations,

then

s2 ≥

1

t2 for any number tamong k

2 numbers on the board and hence

k2×

s2 ≥ Sk2−1 ≥ S0=

1

r2 ,

which implies thats≤kr as desired

Remark. The nature of the problem does not change at all if the numbers on the board are restricted to be positive integers But that may mislead some contestants to think the problem is a number theoretic problem rather than a combinatorial problem

(128)

Problem 2. Leta1, a2, a3, a4, a5 be real numbers satisfying the following equations:

a1

k2+ 1+

a2

k2+ 2+

a3

k2+ 3+

a4

k2+ 4+

a5

k2+ 5 =

1

k2 fork= 1,2,3,4,5.

Find the value of a1 37+ a2 38+ a3 39+ a4 40+ a5

41 (Express the value in a single fraction.)

(Solution) Let R(x) := a1

x2+ 1 +

a2

x2+ 2 +

a3

x2+ 3 +

a4

x2+ 4 +

a5

x2+ 5 Then R(±1) = 1,

R(±2) =

4, R(±3) =

9, R(±4) =

16, R(±5) =

25 and R(6) is the value to be found Let’s putP(x) := (x2+ 1)(x2+ 2)(x2+ 3)(x2+ 4)(x2+ 5) and Q(x) :=R(x)P(x).Then for

k=±1,±2,±3,±4,±5, we getQ(k) =R(k)P(k) = P(k)

k2 , that is, P(k)−k

2Q(k) = Since

P(x)−x2Q(x) is a polynomial of degree 10 with roots ±1,±2,±3,±4,±5, we get

P(x)−x2Q(x) =A(x2−1)(x2−4)(x2−9)(x2−16)(x2−25). (∗)

Puttingx = 0, we get A= P(0)

(−1)(−4)(−9)(−16)(−25) =−

120.Finally, dividing both sides of (∗) byP(x) yields

1−x2R(x) = 1−x2Q(x) P(x) =−

1 120·

(x2−1)(x2−4)(x2−9)(x2−16)(x2−25) (x2+ 1)(x2+ 2)(x2+ 3)(x2+ 4)(x2+ 5)

and hence that

1−36R(6) =− 35×32×27×20×11

120×37×38×39×40×41 =−

3×7×11

13×19×37×41 =− 231 374699, which implies R(6) = 187465

6744582

Remark. We can geta1 =

1105

72 , a2=− 2673

40 , a3 = 1862

15 , a4=− 1885

18 , a5 = 1323

40 by solving the given system of linear equations, which is extremely messy and takes a lot of time

(129)

Problem 3. Let three circles 1, 2, 3, which are non-overlapping and mutually external,

be given in the plane For each point P in the plane, outside the three circles, construct six points A1, B1, A2, B2, A3, B3 as follows: For each i = 1,2,3, Ai, Bi are distinct points on the circle i such that the lines P Ai and P Bi are both tangents to i Call the point

P exceptional if, from the construction, three linesA1B1, A2B2, A3B3 are concurrent Show

that every exceptional point of the plane, if exists, lies on the same circle

(Solution)Let Oi be the center andri the radius of circle i for eachi= 1,2,3 Let P be an exceptional point, and let the three corresponding lines A1B1, A2B2,A3B3 concur at Q

Construct the circle with diameter P Q Call the circle , its center O and its radius r We now claim that all exceptional points lie on

(130)

Let P O1 intersect A1B1 in X1 As P O1 ⊥ A1B1, we see that X1 lies on As P A1 is a

tangent to 1, triangleP A1O1 is right-angled and similar to triangleA1X1O1 It follows that

O1X1

O1A1

= O1A1

O1P

, i.e., O1X1·O1P =O1A12 =r12.

On the other hand,O1X1·O1P is also the power of O1 with respect to , so that

r21 =O1X1·O1P = (O1O−r)(O1O+r) =O1O2−r2, (∗)

and hence

r2=OO12−r12= (OO1−r1)(OO1+r1).

Thus, r2 is the power of O with respect to

1 By the same token, r2 is also the power of

O with respect to and Hence O must be the radical center of the three given circles

Since r, as the square root of the power ofO with respect to the three given circles, does not depend on P, it follows that all exceptional points lie on

Remark. In the event of the radical point being at in nity (and hence the three radical axes being parallel), there are no exceptional points in the plane, which is consistent with the statement of the problem

(131)

Problem 4. Prove that for any positive integerk, there exists an arithmetic sequence

a1

b1

, a2 b2

, , ak bk

of rational numbers, whereai, bi are relatively prime positive integers for eachi= 1,2, , k, such that the positive integersa1, b1, a2, b2, , ak, bk are all distinct

(Solution)Fork= 1, there is nothing to prove Henceforth assume k≥2 Letp1, p2, , pk be kdistinct primes such that

k < pk <· · ·< p2 < p1

and let N = p1p2· · ·pk By Chinese Remainder Theorem, there exists a positive integer x satisfying

x≡ −i (mod pi)

for alli= 1,2, , k and x > N2 Consider the following sequence :

x+

N ,

x+

N , , ,

x+k N .

This sequence is obviously an arithmetic sequence of positive rational numbers of lengthk For each i = 1,2, , k, the numerator x+i is divisible by pi but not by pj for j ̸= i, for otherwisepj divides|i−j|, which is not possible because pj > k >|i−j| Let

ai :=

x+i pi

, bi:=

N pi

for all i= 1,2, , k.

Then

x+i N =

ai

bi

, gcd(ai, bi) = 1 for all i= 1,2, , k, and allbi’s are distinct from each other Moreover,x > N2 implies

ai =

x+i pi

> N

2

pi

> N > N pj

=bj for all i, j= 1,2, , k

and hence allai’s are distinct frombi’s It only remains to show that allai’s are distinct from each other This follows from

aj =

x+j pj

> x+i pj

> x+i pi

=ai for all i < j by our choice ofp1, p2, , pk Thus, the arithmetic sequence

a1

b1

, a2 b2

, , ak bk

of positive rational numbers satis es the conditions of the problem

(132)

Remark. Here is a much easier solution :

For any positive integerk≥2, consider the sequence (k!)2+

k! ,

(k!)2+

k! , ,

(k!)2+k k! . Note that gcd(k!,(k!)2+i) =i for alli= 1,2, , k So, taking

ai :=

(k!)2+i

i , bi := k!

i for all i= 1,2, , k,

we have gcd(ai, bi) = and

ai =

(k!)2+i

i > aj =

(k!)2+j

j > bi = k!

i > bj = k!

j

for any 1≤i < j≤k Therefore this sequence satis es every condition given in the problem

(133)

Problem 5. Larry and Rob are two robots travelling in one car from Argovia to Zillis Both robots have control over the steering and steer according to the following algorithm: Larry makes a 90◦ left turn after everyℓ kilometer driving from start; Rob makes a 90◦ right turn after everyr kilometer driving from start, whereℓandr are relatively prime positive integers In the event of both turns occurring simultaneously, the car will keep going without changing direction Assume that the ground is flat and the car can move in any direction

Let the car start from Argovia facing towards Zillis For which choices of the pair (ℓ, r) is the car guaranteed to reach Zillis, regardless of how far it is from Argovia?

(Solution)Let Zillis bed kilometers away from Argovia, where dis a positive real number For simplicity, we will position Argovia at (0,0) and Zillis at (d,0), so that the car starts out facing east We will investigate how the car moves around in the period of travelling the rstℓr kilometers, the secondℓr kilometers, ., and so on We call each period of travelling

ℓr kilometers a section It is clear that the car will have identical behavior in every section except the direction of the car at the beginning

Case 1: ℓ−r≡2 (mod 4) After the rst section, the car has made ℓ−1 right turns and

r−1 left turns, which is a net of 2(≡ℓ−r (mod 4)) right turns Let the displacement vector for the rst section be (x, y) Since the car has rotated 180◦, the displacement vector for the second section will be (−x,−y), which will take the car back to (0,0) facing east again We now have our original situation, and the car has certainly never travelled further thanℓr

kilometers from Argovia So, the car cannot reach Zillis if it is further apart from Argovia

Case 2: ℓ−r≡1 (mod 4) After the rst section, the car has made a net of right turn Let the displacement vector for the rst section again be (x, y) This time the car has rotated 90◦ clockwise We can see that the displacements for the second, third and fourth section will be (y,−x), (−x,−y) and (−y, x), respectively, so after four sections the car is back at (0,0) facing east Since the car has certainly never travelled further than 2ℓr kilometers from Argovia, the car cannot reach Zillis if it is further apart from Argovia

Case 3: ℓ−r ≡3 (mod 4) An argument similar to that in Case 2(switching the roles of left and right) shows that the car cannot reach Zillis if it is further apart from Argovia

Case 4: ℓ≡r (mod 4) The car makes a net turn of 0◦ after each section, so it must be facing east We are going to show that, after traversing the rst section, the car will be at (1,0) It will be useful to interpret the Cartesian plane as the complex plane, i.e writing

x+iy for (x, y), wherei=√−1 We will denote the k-th kilometer of movement by mk−1,

(134)

which takes values from the set{1, i,−1,−i}, depending on the direction We then just have to show that

ℓ r∑−1

k=0

mk= 1,

which implies that the car will get to Zillis no matter how far it is apart from Argovia

Case 4a: ℓ≡r ≡1 (mod 4) First note that fork= 0,1, , ℓr−1,

mk =i⌊k/ℓ⌋(−i)⌊k/r⌋

since ⌊k/ℓ⌋ and ⌊k/r⌋ are the exact numbers of left and right turns before the (k+ 1)st kilometer, respectively Let ak(≡ k (mod ℓ)) and bk(≡ k (mod r)) be the remainders of k when divided byℓand r, respectively Then, since

ak=k− ⌊

k ℓ

⌋

ℓ≡k−

⌊

k ℓ

⌋

(mod 4) and bk=k− ⌊

k r

⌋

r≡k−

⌊

k r

⌋

(mod 4),

we have⌊k/ℓ⌋ ≡k−ak (mod 4) and⌊k/r⌋ ≡k−bk (mod 4) We therefore have

mk=ik−ak(−i)k−bk = (−i2)ki−ak(−i)−bk = (−i)akibk.

Asℓandr are relatively prime, by Chinese Remainder Theorem, there is a bijection between pairs (ak, bk) = (k(modℓ), k(modr)) and the numbersk= 0,1,2, , ℓr−1 Hence

ℓ r∑−1

k=0

mk = ℓ r∑−1

k=0

(−i)akibk =

(ℓ−1 ∑ k=0

(−i)ak

) (r−1 ∑ k=0

ibk

)

= 1×1 =

as required becauseℓ≡r≡1 (mod 4)

Case 4b: ℓ≡r ≡3 (mod 4) In this case, we get

mk =iak(−i)bk,

where ak(≡ k (mod ℓ)) and bk(≡k (modr)) for k = 0,1, , ℓr−1 Then we can proceed analogously toCase 4a to obtain

ℓ r∑−1

k=0

mk= ℓ r∑−1

k=0

(−i)akibk =

(ℓ−1 ∑ k=0

(−i)ak

) (r−1 ∑ k=0

ibk

)

=i×(−i) = as required becauseℓ≡r≡3 (mod 4)

Now clearly the car traverses through all points between (0,0) and (1,0) during the rst section and, in fact, covers all points between (n−1,0) and (n,0) during the n-th section Hence it will eventually reach (d,0) for any positived

(135)

To summarize: (ℓ, r) satis es the required conditions if and only if

ℓ≡r ≡1 or ℓ≡r ≡3 (mod 4).

Remark. In case gcd(ℓ, r) =d̸= 1, the answer is :

ℓ d ≡

r

d ≡1 or ℓ d ≡

r

d ≡3 (mod 4).

(136)

SOLUTIONS FOR 2010 APMO PROBLEMS

Problem LetABC be a triangle with ∠BAC 6= 90◦.Let O be the circumcenter of the triangle ABC and let Γ be the circumcircle of the triangle BOC Suppose that Γ intersects the line segment AB atP different from B,and the line segmentAC atQdifferent fromC

LetON be a diameter of the circle Γ.Prove that the quadrilateralAP N Qis a parallelogram

Solution: From the assumption that the circle Γ intersects both of the line segmentsAB

and AC, it follows that the points N, C, Q, O are located on Γ in the order of N, C, Q, O

or in the order of N, C, O, Q The following argument for the proof of the assertion of the problem is valid in either case Since ∠N QC and ∠N OC are subtended by the same arc

_

N C of Γ at the points Q and O, respectively, on Γ, we have ∠N QC = ∠N OC We also have∠BOC = 2∠BAC, since∠BOC and ∠BAC are subtended by the same arcBC_ of the circum-circle of the triangle ABC at the center O of the circle and at the point A on the circle, respectively FromOB =OC and the fact that ON is a diameter of Γ, it follows that the triangles OBN and OCN are congruent, and therefore we obtain 2∠N OC = ∠BOC

Consequently, we have∠N QC = 12∠BOC = ∠BAC, which shows that the lines AP, QN

are parallel

In the same manner, we can show that the lines AQ, P N are also parallel Thus, the quadrilateralAP N Q is a parallelogram

Problem For a positive integerk,call an integer apurek-th powerif it can be represented asmkfor some integerm.Show that for every positive integernthere existndistinct positive integers such that their sum is a pure 2009-th power, and their product is a pure 2010-th power

Solution: For the sake of simplicity, let us set k= 2009

First of all, choosendistinct positive integersb1,· · · , bn suitably so that their product is a purek+1-th power (for example, letbi =ik+1fori= 1,· · ·, n) Then we haveb1· · ·bn=tk+1

for some positive integert Setb1+· · ·+bn=s Now we set = bisk

2−1

for i = 1,· · · , n, and show that a1,· · · , an satisfy the required

conditions Since b1,· · ·, bn are distinct positive integers, it is clear that so are a1,· · ·, an From

a1+· · ·+an=sk 2−1

(b1+· · ·+bn) =sk

= (sk)2009, a1· · ·an= (sk

2−1

)nb1· · ·bn= (sk

2−1

)ntk+1= (s(k−1)nt)2010

we can see thata1,· · · , an satisfy the conditions on the sum and the product as well This

ends the proof of the assertion

Remark: We can find the appropriate exponent k2−1 needed for the construction of the

ai’s by solving the simultaneous congruence relations: x≡0 (mod k+ 1), x≡ −1 (mod k) Problem Let n be a positive integer n people take part in a certain party For any pair of the participants, either the two are acquainted with each other or they are not What is the maximum possible number of the pairs for which the two are not acquainted but have a common acquaintance among the participants?

(137)

Solution: When participant, say the personA, is mutually acquainted with each of the remaining n−1 participants, and if there are no other acquaintance relationships among the participants, then for any pair of participants not involving A, the two are not mutual acquaintances, but they have a common acquaintance, namelyA, so any such pair satisfies the requirement Thus, the number desired in this case is (n−1)(2n−2) = n2−32n+2

Let us show that n2−32n+2 is the maximum possible number of the pairs satisfying the requirement of the problem First, let us observe that in the process of trying to find the maximum possible number of such pairs, if we split the participants into two non-empty subsets T and S which are disjoint, we may assume that there is a pair consisting of one person chosen from T and the other chosen from S who are mutual acquaintances This is so, since if there are no such pair for some splittingT andS, then among the pairs consisting of one person chosen fromT and the other chosen fromS, there is no pair for which the two have a common acquaintance among participants, and therefore, if we arbitrarily choose a personA∈T andB ∈S and declare that AandB are mutual acquaintances, the number of the pairs satisfying the requirement of the problem does not decrease

Let us now call a set of participants a groupif it satisfies the following conditions:

• One can connect any person in the set with any other person in the set by tracing a chain of mutually acquainted pairs More precisely, for any pair of people A, B in the set there exists a sequence of people A0, A1,· · ·, An for which A0 = A, An = B and, for each

i: 0≤i≤n−1,Ai and Ai+1 are mutual acquaintances

•No person in this set can be connected with a person not belonging to this set by tracing a chain of mutually acquainted pairs

In view of the discussions made above, we may assume that the set of all the participants to the party forms a group ofnpeople Let us next consider the following lemma

Lemma In a group of npeople, there are at leastn−1 pairs of mutual acquaintances

Proof: If you choose a mutually acquainted pair in a group and declare the two in the pair are not mutually acquainted, then either the group stays the same or splits into groups This means that by changing the status of a mutually acquainted pair in a group to that of a non-acquainted pair, one can increase the number of groups at most by Now if in a group of n people you change the status of all of the mutually acquainted pairs to that of non-acquainted pairs, then obviously, the number of groups increases from ton Therefore, there must be at least n−1 pairs of mutually acquainted pairs in a group consisting of n

people

The lemma implies that there are at most n(n2−1) −(n−1) = n2−32n+2 pairs satisfying the condition of the problem Thus the desired maximum number of pairs satisfying the requirement of the problem is n2−32n+2

(138)

Alternate Solution 1: The construction of an example for the case for which the number

n2−3n+2

2 appears, and the argument for the case where there is only group would be the same as in the preceding proof

Suppose, then,nparticipants are separated intok(k≥2) groups, and the number of people in each group is given byai, i= 1,· · ·, k.In such a case, the number of pairs for which paired people are not mutually acquainted but have a common acquaintance is at mostPk

i=1 aiC2,

where we set 1C2 = for convenience Since aC2 + bC2 ≤ a+bC2 holds for any pair of positive integersa, b, we havePk

i=1 aiC2 ≤ a1C2+ n−a1C2.From

a1C2+ n−a1C2 =a

1−na1+

n2−n

2 = a1−

n

2

+n 2−2n

4

it follows that a1C2+ n−a1C2 takes its maximum value when a1 = 1, n−1 Therefore, we have Pk

i=1 aiC2 ≤ n−1C2, which shows that in the case where the number of groups are

or more, the number of the pairs for which paired people are not mutually acquainted but have a common acquaintance is at most n−1C2 = n

2−3n+2

2 , and hence the desired maximum number of the pairs satisfying the requirement is n2−32n+2

Alternate Solution 2: Construction of an example would be the same as the preceding proof

For a participant, say A, call another participant, say B, a familiar face if A and B are not mutually acquainted but they have a common acquaintance among the participants, and in this case call the pairA, B a familiar pair

Suppose there is a participant P who is mutually acquainted withd participants Denote by S the set of these d participants, and by T the set of participants different from P and not belonging to the setS.Suppose there are epairs formed by a person in S and a person inT who are mutually acquainted

Then the number of participants who are familiar faces to P is at most e The number of pairs formed by two people belonging to the set S and are mutually acquainted is at most

dC2 The number of familiar pairs formed by two people belonging to the set T is at most

n−d−1C2 Since there are e pairs formed by a person in the set S and a person in the set

T who are mutually acquainted (and so the pairs are not familiar pairs), we have at most

d(n−1−d)−efamiliar pairs formed by a person chosen fromS and a person chosen fromT

Putting these together we conclude that there are at moste+dC2+n−1−dC2+d(n−1−d)−e familiar pairs Since

e+ dC2+ n−1−dC2+d(n−1−d)−e=

n2−3n+

2 ,

the number we seek is at most n2−32n+2, and hence this is the desired solution to the problem

Problem Let ABC be an acute triangle satisfying the condition AB > BC and

AC > BC Denote by O and H the circumcenter and the orthocenter, respectively, of the triangleABC Suppose that the circumcircle of the triangleAHC intersects the line AB at

M different fromA,and that the circumcircle of the triangleAHB intersects the lineAC at

N different fromA.Prove that the circumcenter of the triangle M N H lies on the lineOH Solution: In the sequel, we denote ∠BAC = α,∠CBA = β,∠ACB = γ Let O0 be the circumcenter of the triangleM N H The lengths of line segments starting from the point H

(139)

Let us denote by M0, N0 the point of intersection of CH, BH, respectively, with the cir-cumcircle of the triangle ABC (distinct from C, B, respectively.) From the fact that points A, M, H, C lie on the same circle, we see that ∠M HM0 = α holds Furthermore, ∠BM0C,∠BN0C and α are all subtended by the same arc BC_ of the circumcircle of the triangle ABC at points on the circle, and therefore, we have ∠BM0C=α, and ∠BN0C=α

as well We also have∠ABH =∠ACN0 as they are subtended by the same arc

_

AN0 of the circumcircle of the triangleABC at points on the circle SinceHM0 ⊥BM, HN0 ⊥AC,we conclude that

∠M0HB = 90◦−∠ABH = 90◦−∠ACN0 =α

is valid as well Putting these facts together, we obtain the fact that the quadrilateral

HBM0M is a rhombus In a similar manner, we can conclude that the quadrilateralHCN0N

is also a rhombus Since both of these rhombuses are made up of right triangles with an angle of magnitude α,we also see that these rhombuses are similar

Let us denote by P, Q the feet of the perpendicular lines on HM and HN, respectively, drawn from the point O0 Since O0 is the circumcenter of the triangle M N H, P, Q are re-spectively, the midpoints of the line segments HM, HN Furthermore, if we denote by R, S

the feet of the perpendicular lines onHM and HN, respectively, drawn from the point O,

then since O is the circumcenter of both the triangle M0BC and the triangle N0BC, we see that R is the intersection point of HM and the perpendicular bisector ofBM0, andS is the intersection point ofHN and the perpendicular bisector ofCN0

We note that the similarity map φbetween the rhombusesHBM0M and HCN0N carries the perpendicular bisector ofBM0 onto the perpendicular bisector of CN0, and straight line

HM onto the straight lineHN,and henceφmapsRontoS,andP ontoQ Therefore, we get

HP :HR=HQ:HS.If we now denote byX, Y the intersection points of the lineHO0 with the line throughRand perpendicular to HP, and with the line throughS and perpendicular toHQ, respectively, then we get

HO0:HX =HP :HR=HQ:HS =HO0 :HY

so that we must have HX = HY, and therefore, X = Y But it is obvious that the point of intersection of the line throughR and perpendicular to HP with the line through S and perpendicular toHQ must beO, and therefore, we conclude that X =Y =O and that the pointsH, O0, O are collinear

Alternate Solution: Deduction of the fact that both of the quadrilaterals HBM0M and

HCN0N are rhombuses is carried out in the same way as in the preceding proof

We then see that the point M is located in a symmetric position with the point B with respect to the lineCH,we conclude that we have∠CM B=β.Similarly, we have∠CN B =γ

If we now putx=∠AHO0,then we get

∠O0 =β−α−x, ∠M N H = 90◦−β−α+x,

from which it follows that

∠AN M = 180◦−∠M N H−(90◦−α) =β−x

Similarly, we get

(140)

Using the laws of sines, we then get sin(γ+x)

sin(β−x) =

AN AM =

AC AM ·

AB AC ·

AN AB

= sinβ sin(β−α) ·

sinγ

sinβ ·

sin(γ−α) sinγ =

sin(γ−α) sin(β−α) On the other hand, if we lety=∠AHO,we then get

∠OHB = 180◦−γ−y, ∠CHO = 180◦−β+y,

and since

∠HBO=γ−α, ∠OCH =β−α,

using the laws of sines and observing that OB =OC, we get sin(γ−α)

sin(β−α) =

sin∠HBO

sin∠OCH =

sin(180◦−γ−y)·OH OB

sin(180◦−β+y)·OH OC

= sin(180

◦−γ−y) sin(180◦−β+y) =

sin(γ+y) sin(β−y)

We then get sin(γ+x) sin(β−y) = sin(β−x) sin(γ+y) Expanding both sides of the last identity by using the addition formula for the sine function and after factoring and using again the addition formula we obtain that sin(x−y) sin(β+γ) = 0.This implies that x−y

must be an integral multiple of 180◦, and hence we conclude thatH, O, O0 are collinear

Problem Find all functions f from the set R of real numbers into R which satisfy for allx, y, z ∈R the identity

f(f(x) +f(y) +f(z)) =f(f(x)−f(y)) +f(2xy+f(z)) + 2f(xz−yz)

Solution: It is clear that if f is a constant function which satisfies the given equation, then the constant must be Conversely, f(x) = clearly satisfies the given equation, so, the identically function is a solution In the sequel, we consider the case where f is not a constant function

Lett∈Rand substitute (x, y, z) = (t,0,0) and (x, y, z) = (0, t,0) into the given functional equation Then, we obtain, respectively,

f(f(t) + 2f(0)) =f(f(t)−f(0)) +f(f(0)) + 2f(0), f(f(t) + 2f(0)) =f(f(0)−f(t)) +f(f(0)) + 2f(0),

from which we conclude thatf(f(t)−f(0)) =f(f(0)−f(t)) holds for allt∈R.Now, suppose for some pairu1, u2,f(u1) =f(u2) is satisfied Then by substituting (x, y, z) = (s,0, u1) and (x, y, z) = (s,0, u2) into the functional equation and comparing the resulting identities, we can easily conclude that

f(su1) =f(su2) (∗)

holds for alls∈R.Sincef is not a constant function there exists ans0such thatf(s0)−f(0)6= 0.If we putu1=f(s0)−f(0), u2 =−u1,thenf(u1) =f(u2),so we have by (∗)

f(su1) =f(su2) =f(−su1) for all s∈R.Since u1 6= 0,we conclude that

(141)

holds for allx∈R

Next, if f(u) = f(0) for some u 6= 0, then by (∗), we have f(su) = f(s0) = f(0) for all

s, which implies that f is a constant function, contradicting our assumption Therefore, we must havef(s)6=f(0) whenevers6=

We will now show that if f(x) = f(y) holds, then either x = y or x = −y must hold Suppose on the contrary thatf(x0) =f(y0) holds for some pair of non-zero numbers x0, y0 for whichx0 6=y0, x0 6=−y0 Since f(−y0) =f(y0), we may assume, by replacingy0 by −y0 if necessary, that x0 and y0 have the same sign In view of (∗), we see thatf(sx0) =f(sy0) holds for alls, and therefore, there exists somer >0, r6= such that

f(x) =f(rx)

holds for allx.Replacing x by rxand y by ry in the given functional equation, we obtain

f(f(rx) +f(ry) +f(z)) =f(f(rx)−f(ry)) +f(2r2xy+f(z)) + 2f(r(x−y)z) (i),

and replacingx by r2x in the functional equation, we get

f(f(r2x) +f(y) +f(z)) =f(f(r2x)−f(y)) +f(2r2xy+f(z)) + 2f((r2x−y)z) (ii)

Sincef(rx) =f(x) holds for allx∈R, we see that except for the last term on the right-hand side, all the corresponding terms appearing in the identities (i) and (ii) above are equal, and hence we conclude that

f(r(x−y)z) =f((r2x−y)z)) (iii)

must hold for arbitrary choice ofx, y, z ∈ R For arbitrarily fixed pair u, v ∈ R, substitute (x, y, z) = (rv2−−1u,v−r

2u

r2−1 ,1) into the identity (iii) Then we obtain f(v) =f(ru) =f(u),since

x−y =u, r2x−y=v, z= 1.But this implies that the functionf is a constant, contradicting our assumption Thus we conclude that iff(x) = f(y) then either x = y or x = −y must hold

By substituting z= in the functional equation, we get

f(f(x) +f(y) +f(0)) =f(f(x)−f(y) +f(0)) =f((f(x)−f(y)) +f(2xy+f(0)) + 2f(0)

Changing y to −y in the identity above and using the fact that f(y) = f(−y), we see that all the terms except the second term on the right-hand side in the identity above remain the same Thus we conclude that f(2xy +f(0)) = f(−2xy +f(0)), from which we get either 2xy+f(0) = −2xy +f(0) or 2xy +f(0) = 2xy −f(0) for all x, y ∈ R The first of these alternatives says that 4xy= 0,which is impossible ifxy 6= 0.Therefore the second alternative must be valid and we get thatf(0) =

Finally, let us show that if f satisfies the given functional equation and is not a constant function, thenf(x) =x2.Letx=y in the functional equation, then since f(0) = 0,we get

f(2f(x) +f(z)) =f(2x2+f(z)),

from which we conclude that either 2f(x) +f(z) = 2x2+f(z) or 2f(x) +f(z) =−2x2−f(z) must hold Suppose there existsx0 for which f(x0)6=x20,then from the second alternative, we see thatf(z) =−f(x0)−x20 must hold for all z, which means thatf must be a constant function, contrary to our assumption Therefore, the first alternative above must hold, and we have f(x) =x2 for all x,establishing our claim

(142)

SOLUTIONS FOR 2011 APMO PROBLEMS

Problem

Solution: Suppose all of the numbersa2+b+c, b2+c+aandc2+a+bare perfect squares Then from the fact thata2+b+cis a perfect square bigger than a2 it follows thata2+b+c≥(a+ 1)2, and therefore,b+c≥2a+ 1.Similarly we obtainc+a≥2b+ and a+b≥2c+

Adding the corresponding sides of the preceding inequalities, we obtain 2(a+b+c) ≥ 2(a+b+c) + 3, a contradiction This proves that it is impos-sible to have all the given numbers to be perfect squares

Alternate Solution: Since the given conditions of the problem are symmetric in a, b, c, we may assume thata≥b≥cholds From the assumption thata2+b+cis a perfect square, we can deduce as in the solution above the inequalityb+c≥2a+ But then we have

2a≥b+c≥2a+ 1, a contradiction, which proves the assertion of the problem

Problem

Solution: We will show that 36◦ is the desired answer for the problem

First, we observe that if the given points form a regular pentagon, then the minimum of the angles formed by any triple among the five vertices is 36◦, and therefore, the answer we seek must be bigger than or equal to 36◦

Next, we show that for any configuration of points satisfying the condition of the problem, there must exist an angle smaller than or equal to 36◦ formed by a triple chosen from the given points For this purpose, let us start with any points, sayA1, A2, A3, A4, A5, on the plane satisfying the condition of the problem, and consider the smallest convex subset, call it Γ, in the plane containing all of the points Since this convex subset Γ must be either a triangle or a quadrilateral or a pentagon, it must have an interior angle with 108◦ or less We may assume without loss of generality that this angle is∠A1A2A3 By the definition of Γ it is clear that the remaining pointsA4andA5lie in the interior of the angular region determined by ∠A1A2A3, and therefore, there must be an angle smaller than or equal to

3·108

◦= 36◦,which is formed by a triple chosen from the given points, and this proves that 36◦ is the desired maximum.

(143)

2

Problem

Solution: Since ∠B1BB2 = 90◦, the circle having B1B2 as its diameter goes through the points B, B1, B2 From B1A : B1C = B2A : B2C = BA : BC, it follows that this circle is the Apolonius circle with the ratio of the distances from the pointsAandC beingBA:BC Since the pointP lies on this circle, we have

P A:P C =BA:BC= sinC: sinA,

from which it follows that P AsinA = P CsinC Similarly, we have P AsinA = P BsinB, and therefore,P AsinA=P BsinB=P CsinC

Let us denote byD, E, F the foot of the perpendicular line drawn fromP to the line segmentBC, CAand AB, respectively Since the pointsE, F lie on a circle havingP Aas its diameter, we have by the law of sinesEF =P AsinA Similarly, we have F D = P BsinB and DE = P CsinC Consequently, we conclude that DEF is an equilateral triangle Furthermore, we have∠CP E =∠CDE, since the quadrilateralCDP E is cyclic Similarly, we have ∠F P B=∠F DB Putting these together, we get

∠BP C= 360◦−(∠CP E+∠F P B+∠EP F)

= 360◦− {(∠CDE+∠F DB) + (180◦−∠F AE)}

= 360◦−(120◦+ 150◦) = 90◦, which proves the assertion of the problem

Alternate Solution: LetO be the midpoint of the line segment B1B2 Then the pointsB andP lie on the circle with center atO and going through the point B1 From

∠OBC=∠OBB1−∠CBB1=∠OB1B−∠B1BA=∠BAC

it follows that the trianglesOCDandOBAare similar, and therefore we have that OC·OA=OB2=OP2.Thus we conclude that the trianglesOCP andOP Aare similar, and therefore, we have∠OP C =∠P AC.Using this fact, we obtain

∠P BC−∠P BA= (∠B1BC+∠P BB1)−(∠ABB1−∠P BB1) = 2∠P BB1=∠P OB1=∠P CA−∠OP C

=∠P CA−∠P AC,

from which we conclude that∠P AC+∠P BC = ∠P BA+∠P CA.Similarly, we get∠P AB+∠P CB =∠P BA+∠P CA Putting these facts together and taking into account the fact that

(∠P AC+∠P BC) + (∠P AB+∠P CB) + (∠P BA+∠P CA) = 180◦, we conclude that∠P BA+∠P CA= 60◦,and finally that

∠BP C= (∠P BA+∠P AB)+(∠P CA+∠P AC) =∠BAC+(∠P BA+∠P CA) = 90◦, proving the assertion of the problem

Problem

(144)

3

Call a point aturning pointif it coincides withPi for someiwith 1≤i≤m

Let us say also that points{P, Q} areadjacentif{P, Q}={Pi−1, Pi} for some i with 1≤i≤m, and vertically adjacentif, in addition, P Q is parallel to the y-axis

Any turning point is vertically adjacent to exactly one other turning point Therefore, the set of all turning points is partitioned into a set of pairs of points using the relation of ”vertical adjacency” Thus we can conclude that if we fix k∈ {1,2,· · · , n}, the number of turning points having thex-coordinatek must be even, and hence it is less than or equal to n−1 Therefore, altogether there are less than or equal to n(n−1) turning points, and this shows that m≤ n(n−1) must be satisfied

It remains now to show that for any positive odd number n one can choose a sequence for which m = n(n−1) We will show this by using the mathematical induction onn Forn= 1, this is clear Forn= 3, choose

P0= (0,1), P1= (1,1), P2= (1,2), P3= (2,2), P4= (2,1), P5= (3,1), P6= (3,3), P7= (4,3) It is easy to see that these points satisfy the requirements (See fig below)

figure

Let n be an odd integer ≥5, and suppose there exists a sequence satisfying the desired conditions forn−4 Then, it is possible to construct a sequence which gives a configuration indicated in the following diagram (fig 2), where the configuration inside of the dotted square is given by the induction hypothesis:

figure

(145)

4

number of turning points in this configuration is

(n−4)(n−5) + (n2−(n−4)2−4) =n(n−1),

showing that for this n there exists a sequence satisfying the desired properties, and thus completing the induction process

Problem

Solution: By substituting x= and y = into the given identity we obtain f(f(1)) =f(1) Next, by substituting x= and y =f(1) into the given identity and usingf(f(1)) =f(1), we getf(1)2=f(1), from which we conclude that either f(1) = or f(1) = But if f(1) = 1, then substituting y = into the given identity, we getf(x) =xfor allx, which contradicts the condition (1) Therefore, we must havef(1) =

By substitutingx = into the given identity and using the factf(1) = 0, we then obtainf(f(y)) = 2f(y) for ally This means that if a numbertbelongs to the range of the functionf, then so does 2t, and by induction we can conclude that for any non-negative integern, 2nt belongs to the range off ift does Now suppose

that there exists a real number a for whichf(a) >0, then for any non-negative integer n2nf(a) must belong to the range off, which leads to a contradiction to

the condition (1) Thus we conclude thatf(x)≤0 for any real numberx

By substituting x2 for xandf(y) for y in the given identity and using the fact thatf(f(y)) = 2f(y), we obtain

f(xf(y)) +f(y)fx

=xf(y) +fx 2f(y)

, from which it follows thatxf(y)−f(xf(y)) =f(y)f x

2

−f x

2f(y)

≥0,since the values of f are non-positive Combining this with the given identity, we conclude that yf(x)≥ f(xy) When x > 0, by letting y to be

x and using the fact that f(1) = 0, we getf(x)≥0.Sincef(x)≤0 for any real numberx, we conclude that f(x) = for any positive real numberx We also havef(0) =f(f(1)) = 2f(1) = Iff is identically 0, i.e.,f(x) = for allx, then clearly, thisf satisfies the given identity Iff satisfies the given identity but not identically 0, then there exists a b <0 for whichf(b)<0.If we setc=f(b), then we havef(c) =f(f(b)) = 2f(b) = 2c For any negative real number x, we have cx >0 so that f(cx) =f(2cx) = 0, and by substitutingy=cinto the given identity, we get

f(2cx) +cf(x) = 2cx+f(cx), from which it follows thatf(x) = 2xfor any negative real x

We therefore conclude that iff satisfies the given identity and is not identically 0, then f is of the form f(x) =

(

0 if x≥0

2x if x <0 Finally, let us show that the function f of the form shown above does satisfy the conditions of the problem Clearly, it satisfies the condition (1) We can check that f satisfies the condition (2) as well by separating into the following cases depending on whether x, yare non-negative or negative

• when bothxandy are non-negative, both sides of the given identity are

(146)

5

• when xis negative and y is non-negative, we havexy ≤0 and both sides of the given identity are 2xy

• when bothxandyare negative, we havexy >0 and both sides of the given identity are 2xy

Summarizing the arguments above, we conclude that the functionsf satisfying the conditions of the problem are

f(x) = and f(x) =

(

(147)

SOLUTIONS FOR 2012 APMO PROBLEMS Problem

Solution: Let us denote by 4XY Z the area of the triangle XY Z Let

x=4P AB,y=4P BC andz=4P CA From

y:z=4BCP :4ACP =BF:AF =4BP F :4AP F = (x−1) : follows that z(x−1) = y, which yields (z+ 1)x= x+y+z Similarly, we get (x+ 1)y=x+y+zand (y+ 1)z=x+y+z Thus, we obtain (x+ 1)y= (y+ 1)z= (z+ 1)x

We may assume without loss of generality thatx≤y, z If we assume thaty > z

holds, then we get (y+ 1)z >(z+ 1)x, which is a contradiction Similarly, we see that y < z leads to a contradiction (x+ 1)y <(y+ 1)z Therefore, we must have

y=z Then, we also get from (y+ 1)z= (z+ 1)xthat x=zmust hold We now obtain from (x−1) : =y : z = : that x= y =z = holds Therefore, we conclude that the area of the triangleABC equalsx+y+z=

Problem

Solution: If we insert numbers as in the figure below (00s are to be inserted in the remaining blank boxes), then we see that the condition of the problem is satisfied and the total number of all the numbers inserted is

0

1 1

0

We will show that the sum of all the numbers to be inserted in the boxes of the given grid cannot be more than if the distribution of the numbers has to satisfy the requirement of the problem Letn = 2012 Let us say that the row number (the column number) of a box in the given grid isi(j, respectively) if the box lies on thei-th row and thej-th column For a pair of positive integersxandy, denote byR(x, y) the sum of the numbers inserted in all of the boxes whose row number is greater than or equal toxand less than or equal toy(assign the value ifx > y) First let abe the largest integer satisfying 1≤a≤n andR(1, a−1)≤1, and then choose the smallest integer c satisfyinga≤c≤nand R(c+ 1, n)≤1 It is possible to choose such a paira, csinceR(1,0) = andR(n+ 1, n) = If a < c, then we have a < n and so, by the maximality of a, we must have R(1, a) >1, while from the minimality ofc, we must haveR(a+ 1, n)>1 Then by splitting the grid into rectangles by means of the horizontal line bordering the a-th row and thea+ 1-th row, we get the splitting contradicting the requirement of the problem Thus, we must havea=c

Similarly, if for any pair of integers x, ywe defineC(x, y) to be the sum of the numbers inserted in all of the boxes whose column number is greater than or equal toxand less than or equal toy (C(x, y) = ifx > y), then we get a numberbfor which

C(1, b−1)≤1, C(b+ 1, n)≤1, 1≤b≤n

(148)

2

If we letrbe the number inserted in the box whose row number isaand the column number is b, then since r≤1, we conclude that the sum of the numbers inserted into all of the boxes is

≤R(1, a−1) +R(a+ 1, n) +C(1, b−1) +C(b+ 1, n) +r≤5

Problem Solution

For integersa, b and a positive integerm, let us write a≡b (mod m) ifa−b

is divisible bym Since npnp+1+1 must be a positive integer, we see thatpn≤np must hold This means that if p= 2, then 2n≤n2must hold As it is easy to show by induction that 2n> n2 holds ifn≥5, we conclude that ifp= 2, thenn≤4 must be satisfied And we can check that (p, n) = (2,2),(2,4) satisfy the condition of the problem, while (2,3) does not

Next, we consider the case wherep≥3

Supposesis an integer satisfyings≥p Ifsp≤psfor such an s, then we have (s+ 1)p=sp

1 +1

s

p

≤ps

1 +

p

=ps

p

X

r=0 pCr

1

pr < p s p X r=0 r!

≤ps

+ p X r=1 2r−1

< ps(1 + 2)≤ps+1

Thus we have (s+ 1)p < ps+1, and by induction on n, we can conclude that if

n > p, thennp< pn This implies that we must haven≤pin order to satisfy our requirementpn≤np.

We note that sincepn+ is even, so isnp+ 1, which, in turn implies thatnmust be odd and therefore, pn+ is divisible by p+ 1, and np+ is also divisible by

p+ Thus we havenp≡ −1 (mod (p+ 1)), and therefore,n2p≡1 (mod (p+ 1)). Now, letebe the smallest positive integer for whichne≡1 (mod (p+ 1)) Then, we can write 2p=ex+y, where x, yare non-negative integers and 0≤y < e, and we have

1≡n2p= (ne)x·ny ≡ny (mod (p+ 1)),

which implies, because of the minimality ofe, that y = must hold This means that 2pis an integral multiple ofe, and therefore,emust equal one of the numbers 1,2, p,2p

Now, if e = 1, p, then we get np ≡ (mod (p+ 1)), which contradicts the fact that p is an odd prime Since n and p+ are relatively prime, we have by Euler’s Theorem thatnϕ(p+1)≡1 (mod (p+1)), whereϕ(m) denotes the number of integersj(1≤j≤m) which are relatively prime withm Fromϕ(p+1)< p+1<2p

and the minimality ofe, we can then conclude that e= must hold Fromn2≡1 (mod (p+ 1)), we get

(149)

3

which implies that p+ divides n+ Therefore, we must have p≤ n, which, together with the factn≤p, show thatp=nmust hold

It is clear that the pair (p, p) for any primep≥3 satisfies the condition of the problem, and thus, we conclude that the pairs (p, n) which satisfy the condition of the problem must be of the form (2,4) and (p, p) with any prime p

Alternate Solution Let us consider the case where p ≥ As we saw in the preceding solution, n must be odd if the pair (p, n) satisfy the condition of the problem Now, letq be a prime factor ofp+ Then, sincep+ dividespn+ 1,q

must be a prime factor ofpn+ and ofnp+ as well Supposeq≥3 Then, from

np≡ −1 (modq), it follows thatn2p ≡1 (modq) holds If we letebe the smallest positive integer satisfying ne ≡ 1 (modq), then by using the same argument as we used in the preceding solution, we can conclude that e must equal one of the numbers 1,2, p,2p Ife= 1, p, then we getnp≡1 (modq), which contradicts the assumption q≥3 Sincen is not a multiple ofq, by Fermat’s Little Theorem we getnq−1≡1 (modq), and therefore, we get by the minimality ofethate= must hold Fromn2≡1 (modq), we also get

np=n(2·p−21+1)≡n (modq),

and sincenp≡ −1 (modq), we haven≡ −1 (modq) as well.

Now, ifq= then since nis odd, we have n≡ −1 (modq) as well Thus, we conclude that for an arbitrary prime factorqofp+ 1, n≡ −1 (modq) must hold Suppose, for a primeq,qk for some positive integerk is a factor ofp+ Then

qk must be a factor ofnp+ as well But since

np+ = (n+ 1)(np−1−np−2+· · · −n+ 1) and

np−1−np−2+· · · −n+ 1≡(−1)p−1−(−1)p−2+· · · −(−1) + 16≡0 (modq),

we see thatqk must dividen+ By applying the argument above for each prime factorqof p+ 1, we can then conclude thatn+ must be divisible byp+ 1, and as we did in the preceding proof, we can conclude thatn=pmust hold

Problem

Solution: IfAB=AC, then we getBF =CF and the conclusion of the problem is clearly satisfied So, we assume thatAB6=AC in the sequel

Due to symmetry, we may suppose without loss of generality that AB > AC LetK be the point on the circle Γ such thatAKis a diameter of this circle Then, we get

∠BCK=∠ACK−∠ACB= 90◦−∠ACB=∠CBH

and

∠CBK=∠ABK−∠ABC = 90◦−∠ABC=∠BCH,

from which we conclude that the trianglesBCK andCBH are congruent There-fore, the quadrilateral BKCH is a parallelogram, and its diagonal HK passes through the centerM of the other diagonalBC Therefore, the points H, M, K

lie on the same straight line, and we have∠AEM=∠AEK= 90◦

From ∠AED = 90◦ = ∠ADM, we see that the points A, E, D, M lie on the circumference of the same circle, from which we obtain ∠AM B = ∠AED =

∠AEF =∠ACF Putting this fact together with the fact that ∠ABM=∠AF C, we conclude that the trianglesABM andAF Care similar, and we get AM

(150)

4

therefore, that AMCM = ABF B holds Noting thatBM =CM, we also get ACF C = ABF B, from which we can conclude that BF

CF = AB

AC, proving the assertion of the problem

Problem

Solution: Let us note first that ifi6=j, then sinceaiaj≤ a2

i+a2j

2 , we have

n−aiaj≥n−

a2i +a2j

2 ≥n−

n

2 =

n

2 >0

If we setbi=|ai|(i= 1,2, , n), then we getb21+b22+· · ·+b2n=nand n−aiaj ≤

n−bibj, which shows that it is enough to prove the assertion of the problem in the case where all ofa1, a2,· · ·, an are non-negative Hence, we assume from now on thata1, a2,· · ·, an are all non-negative

By multiplying bynthe both sides of the desired inequality we get the inequality:

X

1≤i<j≤n

n n−aiaj

≤ n

2

and since n−na

iaj = + aiaj

n−aiaj, we obtain from the inequality above by subtracting n(n−1)

2 from both sides the following inequality:

(i) X

1≤i<j≤n

aiaj

n−aiaj

≤n

2

We will show that this inequality (i) holds

If for someithe equalitya2i =nis valid, thenaj = must hold for allj6=iand the inequality (i) is trivially satisfied So, we assume from now on thata2i < nis valid for eachi

Let us assume that i 6= j from now on Since 0≤ aiaj ≤

a

i+aj

2

≤ a2i+a2j holds, we have

(ii) aiaj

n−aiaj

≤ aiaj n−a2i+a2j

2

≤

a

i+aj

2

n−a2i+a2j

=1 ·

(ai+aj)2 (n−a2

i) + (n−a2j)

Sincen−a2

i >0,n−a2j >0, we also get from the Cauchy-Schwarz inequality that

a2 j

n−a2 i

+ a

2 i

n−a2 j

!

((n−a2i) + (n−aj2))≥(ai+aj)2,

from which it follows that

(iii) (ai+aj)

2 (n−a2

i) + (n−a2j)

≤ a

2 j

n−a2 i

+ a

2 i

n−a2 j

(151)

5

holds Combining the inequalities (ii) and (iii), we get

X

1≤i<j≤n

aiaj

n−aiaj

≤

2

X

1≤i<j≤n

a2j n−a2

i

+ a

2 i

n−a2 j

!

=

X

i6=j

a2 j

n−a2 i

=

n

X

i=1

n−a2i n−a2 i = n

(152)

Solutions of APMO 2013

Problem Let ABC be an acute triangle with altitudes AD, BE and CF, and let O

be the center of its circumcircle Show that the segments OA, OF, OB, OD, OC, OE dissect the triangle ABC into three pairs of triangles that have equal areas

Solution Let M and N be midpoints of sides BC and AC, respectively Notice that

∠M OC = 12∠BOC = ∠EAB, ∠OM C = 90◦ = ∠AEB, so triangles OM C and AEB are similar and we get OMAE = OCAB For triangles ON A and BDA we also have ONBD = OABA Then

OM AE =

ON

BD or BD·OM =AE·ON

Denote by S(Φ) the area of the figure Φ So, we see that S(OBD) = 12BD ·OM =

1

2AE·ON =S(OAE) Analogously, S(OCD) = S(OAF) and S(OCE) =S(OBF)

Alternative solution LetR be the circumradius of triangleABC, and as usual write

A, B, C for angles ∠CAB, ∠ABC, ∠BCA respectively, and a, b, c for sides BC, CA, AB

respectively Then the area of triangle OCD is

S(OCD) = 12 ·OC·CD·sin(∠OCD) = 12R·CD·sin(∠OCD)

Now CD =bcosC, and

∠OCD = 180

◦−2A

2 = 90

◦−

A

(since triangle OBC is isosceles, and ∠BOC = 2A) So

S(OCD) = 12RbcosCsin(90◦−A) = 12RbcosCcosA

A similar calculation gives

S(OAF) = 12OA·AF ·sin(∠OAF) = 12R·(bcosA) sin(90◦−C) = 12RbcosAcosC,

soOCD and OAF have the same area In the same way we find that OBD and OAE have the same area, as OCE and OBF

Problem Determine all positive integers n for which [√n2+1

n]2+2 is an integer Here [r] denotes the greatest integer less than or equal to r

Solution We will show that there are no positive integersn satisfying the condition of the problem

Letm = [√n] anda=n−m2 We havem≥1 sincen≥1 Fromn2+1 = (m2+a)2+1≡ (a−2)2+ (mod (m2+ 2)),it follows that the condition of the problem is equivalent to the

fact that (a−2)2+ is divisible by m2+ Since we have

0<(a−2)2 + 1≤max{22,(2m−2)2}+ 1≤4m2+ <4(m2+ 2),

(153)

we see that (a−2)2+ = k(m2+ 2) must hold with k= 1,2 or We will show that none of these can occur

Case Whenk = We get (a−2)2−m2 = 1,and this implies thata−2 =±1, m = 0

must hold, but this contradicts with fact m≥1

Case When k = We have (a−2)2 + = 2(m2+ 2) in this case, but any perfect

square is congruent to 0,1,4 mod 8, and therefore, we have (a−2)2 + 1≡ 1,2,5 (mod 8),

while 2(m2+ 2)≡4,6 (mod 8).Thus, this case cannot occur either

Case When k = We have (a−2)2+ = 3(m2+ 2) in this case Since any perfect

square is congruent to or mod 3, we have (a−2)2+ 1≡1,2 (mod 3),while 3(m2+ 2)≡0

(mod 3), which shows that this case cannot occur either

Problem For 2k real numbers a1, a2, , ak, b1, b2, , bk define the sequence of numbers Xn by

Xn= k

X

i=1

[ain+bi] (n= 1,2, )

If the sequence Xn forms an arithmetic progression, show that Pki=1ai must be an integer Here [r] denotes the greatest integer less than or equal tor

Solution Let us writeA=Pk

i=1ai andB = Pk

i=1bi.Summing the corresponding terms

of the following inequalities over i,

ain+bi−1<[ain+bi]≤ain+bi,

we obtainAn+B−k < Xn < An+B Now suppose that{Xn}is an arithmetic progression with the common difference d,then we have nd=Xn+1−X1 and A+B−k < X1 ≤A+B

Combining with the inequalities obtained above, we get

A(n+ 1) +B −k < nd+X1 < A(n+ 1) +B,

or

An−k ≤An+ (A+B−X1)−k < nd < An+ (A+B−X1)< An+k,

from which we conclude that |A−d| < nk must hold Since this inequality holds for any positive integer n, we must have A=d Since {Xn} is a sequence of integers, d must be an integer also, and thus we conclude that A is also an integer

Problem Let a and b be positive integers, and let A and B be finite sets of integers satisfying:

(i) A and B are disjoint;

(ii) if an integer i belongs either to A or to B, then i+a belongs to A or i−b belongs toB

Prove that a|A|=b|B|.(Here |X| denotes the number of elements in the set X.)

Solution Let A∗ = {n − a : n ∈ A} and B∗ = {n +b : n ∈ B} Then, by (ii),

A∪B ⊆A∗∪B∗ and by (i),

(154)

Thus, A∪B =A∗∪B∗ and A∗ and B∗ have no element in common For each finite set X

of integers, let P

(X) =P

x∈Xx Then

X

(A) +X(B) =X(A∪B)

=X(A∗∪B∗) =X(A∗) +X(B∗)

=X(A)−a|A|+X(B) +b|B|, (2) which implies a|A|=b|B|

Alternative solution Let us construct a directed graph whose vertices are labelled by the members of A∪B and such that there is an edge from i toj iff j ∈A and j =i+a or

j ∈B andj =i−b From (ii), each vertex has out-degree≥1 and, from (i), each vertex has in-degree≤1 Since the sum of the out-degrees equals the sum of the in-degrees, each vertex has in-degree and out-degree equal to This is only possible if the graph is the union of disjoint cycles, say G1, G2, , Gn Let |Ak| be the number of elements of A in Gk and |Bk| be the number of elements of B inGk The cycleGk will involve increasing vertex labels by

a a total of|Ak| times and decreasing them by b a total of|Bk| times Since it is a cycle, we havea|Ak|=b|Bk| Summing over all cycles gives the result

Problem Let ABCD be a quadrilateral inscribed in a circleω, and let P be a point on the extension ofAC such thatP B andP D are tangent toω The tangent atC intersects

P D at Qand the line AD atR LetE be the second point of intersection between AQ and

ω Prove that B, E, R are collinear

Solution To show B, E, R are collinear, it is equivalent to show the lines AD, BE, CQ

are concurrent Let CQ intersect AD at R and BE intersect AD at R0 We shall show

RD/RA=R0D/R0A so that R=R0

Since 4P AD is similar to4P DC and4P AB is similar to4P BC, we haveAD/DC =

P A/P D = P A/P B = AB/BC Hence, AB ·DC = BC · AD By Ptolemy’s theorem,

AB·DC =BC·AD= 12CA·DB Similarly CA·ED=CE·AD= 12AE·DC Thus

DB

AB =

2DC

CA , (3)

and

DC

CA =

2ED

AE (4)

(155)

. ω A B C E D P Q

R(R0)

Since the triangles RDC and RCAare similar, we have RDRC = DCCA = RCRA Thus using (4)

RD

RA =

RD·RA

RA2 =

RC RA = DC CA = 2ED AE (5)

Using the similar triangles ABR0 and EDR0, we have R0D/R0B = ED/AB Using the similar triangles DBR0 and EAR0 we have R0A/R0B =EA/DB Thus using (3) and (4),

R0D R0A =

ED·DB

EA·AB =

2ED AE

2

(6)

It follows from (5) and (6) that R=R0

(156)

Problem For a positive integer m denote by S(m) and P(m) the sum and product, respectively, of the digits of m Show that for each positive integer n, there exist positive integers a1, a2, , an satisfying the following conditions:

S(a1)< S(a2)<· · ·< S(an) and S(ai) = P(ai+1) (i= 1,2, , n)

(We letan+1 =a1.) (Problem Committee of the Japan Mathematical Olympiad Foundation)

Solution Letk be a sufficiently large positive integer Choose for each i = 2,3, , n, to be a positive integer among whose digits the number appears exactly k+i−2 times

and the number appears exactly 2k+i−1−2(k+i−2) times, and nothing else Then, we

have S(ai) = 2k+i−1 and P(ai) = 2k+i−2 for eachi, 2≤i ≤n Then, we let a1 be a positive

integer among whose digits the number appears exactly k+n−1 times and the number appears exactly 2k−2(k+n−1) times, and nothing else Then, we see that a

1 satisfies

S(a1) = 2k and P(a1) = 2k+n−1 Such a choice of a1 is possible if we take k to be large

enough to satisfy 2k >2(k+n−1) and we see that the numbers a1, , an chosen this way

satisfy the given requirements

Problem Let S = {1,2, ,2014} For each non-empty subset T ⊆ S, one of its members is chosen as its representative Find the number of ways to assign representatives to all non-empty subsets of S so that if a subset D ⊆ S is a disjoint union of non-empty subsets A, B, C ⊆ S, then the representative of D is also the representative of at least one of A, B, C (Warut Suksompong, Thailand)

Solution Answer: 108·2014!

For any subset X let r(X) denotes the representative of X Suppose that x1 = r(S)

First, we prove the following fact:

Ifx1 ∈X and X ⊆S, then x1 =r(X)

If |X| ≤2012, then we can write S as a disjoint union of X and two other subsets of S, which gives that x1 =r(X) If |X| = 2013, then lety ∈X and y6=x1 We can write X as

a disjoint union of {x1, y} and two other subsets We already proved that r({x1, y}) = x1

(since |{x1, y}| = < 2012) and it follows that y 6= r(X) for every y ∈ X except x1 We

have proved the fact

Note that this fact is true and can be proved similarly, if the ground set S would contain at least elements

There are 2014 ways to choose x1 =r(S) and for x1 ∈ X ⊆S we have r(X) = x1 Let

S1 = S\ {x1} Analogously, we can state that there are 2013 ways to choose x2 = r(S1)

and for x2 ∈ X ⊆ S1 we have r(X) = x2 Proceeding similarly (or by induction), there

are 2014·2013· · ·5 ways to choose x1, x2, , x2010 ∈ S so that for all i = 1,2 ,2010,

xi =r(X) for each X ⊆S\ {x1, , xi−1} and xi ∈X

We are now left with four elements Y ={y1, y2, y3, y4} There are ways to choose r(Y)

Suppose that y1 = r(Y) Then we clearly have y1 =r({y1, y2}) = r({y1, y3}) = r({y1, y4})

The only subsets whose representative has not been assigned yet are {y1, y2, y3},{y1, y2, y4},

{y1, y3, y4},{y2, y3, y4},{y2, y3}, {y2, y4},{y3, y4}.These subsets can be assigned in any way,

hence giving 34·23 more choices.

(157)

In conclusion, the total number of assignments is 2014·2013· · ·4·34·23 = 108·2014!

Problem Find all positive integers n such that for any integer k there exists an integer a for which a3+a−k is divisible by n (Warut Suksompong, Thailand)

Solution Answer: All integersn = 3b, where b is a nonnegative integer.

We are looking for integersnsuch that the setA ={a3+a|a∈Z}is a complete residue

system by modulo n Let us call this property by (*) It is not hard to see that n = satisfies (*) and n = does not

If a ≡ b (mod n), then a3 +a ≡ b3 +b (mod n) So n satisfies (*) iff there are no

a, b∈ {0, , n−1} with a6=b and a3+a≡b3+b (mod n)

First, let us prove that 3j satisfies (*) for allj ≥1.Suppose thata3+a≡b3+b (mod 3j)

for a 6= b Then (a−b)(a2 +ab+b2 + 1) ≡ 0 (mod 3j). We can easily check mod that a2+ab+b2+ is not divisible by

Next note that if Ais not a complete residue system modulo integer r, then it is also not a complete residue system modulo any multiple of r Hence it remains to prove that any prime p >3 does not satisfy (*)

If p ≡ (mod 4), there exists b such that b2 ≡ −1 (mod p). We then take a = to

obtain the congruence a3+a≡b3+b (mod p).

Suppose now thatp≡3 (mod 4).We will prove that there are integers a, b6≡0 (mod p) such thata2+ab+b2 ≡ −1 (mod p) Note that we may suppose that a6≡b (mod p), since

otherwise if a≡ b (mod p) satisfies a2+ab+b2+ 1≡ 0 (mod p), then (2a)2+ (2a)(−a) +

a2 + ≡ (mod p) and 2a 6≡ −a (mod p) Letting c be the inverse of b modulo p (i.e

bc≡1 (mod p)), the relation is equivalent to (ac)2+ac+ 1≡ −c2 (mod p). Note that−c2

can take on the values of all non-quadratic residues modulo p If we can find an integer x

such that x2+x+ is a non-quadratic residue modulo p, the values of a and cwill follow immediately Hence we focus on this latter task

Note that if x, y ∈ {0, , p−1}=B,then x2+x+ 1≡y2+y+ (mod p) iff p divides

x+y+ We can deduce that x2+x+ takes on (p+ 1)/2 values as x varies in B Since there are (p−1)/2 non-quadratic residues modulo p, the (p+ 1)/2 values that x2+x+ take on must be and all the quadratic residues

Let C be the set of quadratic residues modulo p and 0, and let y ∈ C Suppose that

y≡ z2 (mod p) and letz ≡2w+ (mod p) (we can always choose such w) Then y+ 3≡ 4(w2 +w+ 1) (modp). From the previous paragraph, we know that 4(w2 +w+ 1) ∈ C.

This means that y∈C =⇒ y+ ∈C Unless p= 3, the relation implies that all elements of B are in C, a contradiction This concludes the proof

Problem Letn and b be positive integers We say n is b-discerning if there exists a set consisting of n different positive integers less than b that has no two different subsets U

and V such that the sum of all elements in U equals the sum of all elements in V (a) Prove that is a 100-discerning

(b) Prove that is not 100–discerning

(Senior Problems Committee of the Australian Mathematical Olympiad Committee) Solution

(a) Take S ={3,6,12,24,48,95,96,97}, i.e

S ={3·2k: 0≤k ≤5} ∪ {3·25−1,3·25+ 1}

(158)

As k ranges between to 5, the sums obtained from the numbers ·2k are 3t, where 1≤t≤63 These are 63 numbers that are divisible by and are at most 3·63 = 189

Sums of elements of S are also the numbers 95 + 97 = 192 and all the numbers that are sums of 192 and sums obtained from the numbers 3·2k with ≤ k ≤ These are 64 numbers that are all divisible by and at least equal to 192 In addition, sums of elements of S are the numbers 95 and all the numbers that are sums of 95 and sums obtained from the numbers 3·2k with 0≤k≤5 These are 64 numbers that are all congruent to −1 mod

Finally, sums of elements of S are the numbers 97 and all the numbers that are sums of 97 and sums obtained from the numbers 3·2k with ≤ k ≤ These are 64 numbers that are all congruent to mod

Hence there are at least 63 + 64 + 64 + 64 = 255 different sums from elements of S On the other hand, S has 28 −1 = 255 non-empty subsets Therefore S has no two different subsets with equal sums of elements Therefore, is 100-discerning

(b) Suppose that is 100-discerning Then there is a set S={s1, , s9}, si <100 that

has no two different subsets with equal sums of elements Assume that 0< s1 <· · ·< s9 <

100

Let X be the set of all subsets of S having at least and at most elements and let Y

be the set of all subsets of S having exactly or or elements greater than s3

The set X consists of + + +

= 84 + 126 + 126 + 84 = 420

subsets of S The set in X with the largest sums of elements is {s4, , s9}and the smallest

sums is in {s1, s2, s3} Thus the sum of the elements of each of the 420 sets in X is at least

s1+s2+s3 and at mosts4+· · ·+s9, which is one of (s4+· · ·+s9)−(s1+s2+s3) + integers

From the pigeonhole principle it follows that (s4+· · ·+s9)−(s1+s2+s3) + 1≥420, i.e.,

(s4+· · ·+s9)−(s1+s2+s3)≥419 (1)

Now let us calculate the number of subsets in Y Observe that {s4, , s9} has 62

2-element subsets, 63 3-element subsets and 64 4-element subsets, while {s1, s2, s3} has

exactly subsets Hence the number of subsets of S in Y equals + +

= 8(15 + 20 + 15) = 400

The set in Y with the largest sum of elements is {s1, s2, s3, s6, s7, s8, s9} and the smallest

sum is in{s4, s5} Again, by the pigeonhole principle it follows that (s1+s2+s3+s6+s7+

s8+s9)−(s4+s5) + ≥400, i.e.,

(s1+s2+s3+s6+s7+s8+s9)−(s4+s5)≥399 (2)

Adding (1) and (2) yields 2(s6 + s7 +s8 +s9) ≥ 818, so that s9 + 98 + 97 + 96 ≥

s9 +s8+s7+s6 ≥ 409, i.e s9 ≥ 118, a contradiction with s9 < 100 Therefore, is not

100-discerning

(159)

Problem Circles ω and Ω meet at points A and B Let M be the midpoint of the arc ABof circleω (M lies inside Ω) A chordM P of circleω intersects Ω at Q(Qlies inside

ω) Let `P be the tangent line to ω at P, and let `Q be the tangent line to Ω at Q Prove

that the circumcircle of the triangle formed by the lines `P, `Q, and AB is tangent to Ω

(Ilya Bogdanov, Russia and Medeubek Kungozhin, Kazakhstan)

Solution Denote X = AB∩`P, Y = AB ∩`Q, and Z = `P ∩`Q Without loss of

generality we have AX < BX Let F =M P ∩AB

A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A AAAAAAAAAAAAAAAAAAAAAAA

B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B BBBBBBBBBBBBBBBBBBBBBBBB

D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D DDDDDDDDDDDDDDDDDDDDDDDD

F M P Q R S T X Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y YYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Z

ω

Ω

`Q

`P

Denote byR the second point of intersection ofP Qand Ω; byS the point of Ω such that

SRkAB; and by T the point of Ω such that RT k`P SinceM is the midpoint of arc AB,

the tangent `M at M to ω is parallel to AB, so ∠(AB, P M) = ∠(P M, `P) Therefore we

have ∠P RT = ∠M P X = ∠P F X = ∠P RS Thus the point Q is the midpoint of the arc T QS of Ω, hence ST k `Q So the corresponding sides of the triangles RST and XY Z

are parallel, and there exist a homothety h mapping RST toXY Z

LetDbe the second point of intersection ofXR and Ω We claim that Dis the center of the homothetyh; sinceD∈Ω, this implies that the circumcircles of trianglesRST andXY Z

are tangent, as required So, it remains to prove this claim In order to this, it suffices to show that D∈SY

By ∠P F X = ∠XP F we have XF2 = XP2 = XA ·XB = XD· XR Therefore,

XF

XD =

XR

XF, so the trianglesXDF andXF Rare similar, hence∠DF X =∠XRF =∠DRQ= ∠DQY; thus the points D, Y, Q, and F are concyclic It follows that ∠Y DQ=∠Y F Q=

∠SRQ= 180◦−∠SDQwhich means exactly that the pointsY,D, andS are collinear, with

D between S and Y

(160)

Problem Let ABC be a triangle, and let D be a point on side BC A line through

D intersects sideAB atX and ray AC at Y The circumcircle of triangleBXD intersects the

circumcircleω of triangleABC again at pointZ 6=B The lines ZD andZY intersectω again

at V and W, respectively Prove that AB=V W

Solution SupposeXY intersectsωat pointsP andQ, whereQlies betweenX andY We

will show thatV andW are the reflections ofAandBwith respect to the perpendicular bisector

of P Q From this, it follows that AV W B is an isosceles trapezoid and hence AB =V W

First, note that

∠BZD =∠AXY =∠AP Q+∠BAP =∠AP Q+∠BZP,

so ∠AP Q=∠P ZV =∠P QV, and hence V is the reflection ofA with respect to the

perpen-dicular bisector of P Q

Now, suppose W0 is the reflection of B with respect to the perpendicular bisector of P Q,

and letZ0 be the intersection of Y W0 andω It suffices to show thatB,X,D,Z0 are concyclic

Note that

∠Y DC =∠P DB =∠P CB+∠QP C =∠W0P Q+∠QP C =∠W0P C =∠Y Z0C

SoD,C, Y,Z0 are concyclic Next, ∠BZ0D=∠CZ0B−∠CZ0D= 180◦−∠BXD and due to

the previous concyclicity we are done

Alternative solution Using cyclic quadrilaterals BXDZ and ABZV in turn, we have

∠ZDY =∠ZBA=∠ZCY So ZDCY is cyclic

Using cyclic quadrilateralsABZC andZDCY in turn, we have∠AZB =∠ACB =∠W ZV

(or 180◦−∠W ZV if Z lies between W and C)

So AB=V W because they subtend equal (or supplementary) angles in ω

Alternative solution Using cyclic quadrilaterals BXDZ and ABZV in turn, we have

∠ZDY =∠ZBA=∠ZCY So ZDCY is cyclic

Using cyclic quadrilaterals BXDZ and ABZV in turn, we have ∠DXA = ∠V ZB =

180◦−BAV SoXDkAV

Using cyclic quadrilaterals ZDCY and BCW Z in turn, we have ∠Y DC = ∠Y ZC =

∠W BC SoXDkBW

Hence BW kAV which implies that AV W B is an isosceles trapezium with AB=V W

Problem LetS ={2,3,4, }denote the set of integers that are greater than or equal

to Does there exist a function f :S →S such that

f(a)f(b) =f(a2b2) for all a, b∈S with a6=b?

Solution We prove that there is no such function For arbitrary elements a and b of S,

choose an integer c that is greater than both of them Since bc > aand c > b, we have

f(a4b4c4) = f(a2)f(b2c2) = f(a2)f(b)f(c)

Furthermore, since ac > band c > a, we have

f(a4b4c4) = f(b2)f(a2c2) = f(b2)f(a)f(c)

Comparing these two equations, we find that for all elementsa and b of S,

f(a2)f(b) =f(b2)f(a) =⇒ f(a

2)

f(a) = f(b2)

(161)

It follows that there exists a positive rational number k such that

f(a2) =kf(a), for all a∈S (1)

Substituting this into the functional equation yields f(ab) = f(a)f(b)

k , for all a, b∈S with a 6=b (2)

Now combine the functional equation with equations (1) and (2) to obtain f(a)f(a2) = f(a6) = f(a)f(a

5)

k =

f(a)f(a)f(a4)

k2 =

f(a)f(a)f(a2)

k , for all a∈S

It follows thatf(a) = kfor all a∈S Substitutinga = andb = into the functional equation

yieldsk = 1, however 16∈S and hence we have no solutions

Problem A sequence of real numbersa0, a1, is said to be goodif the following three

conditions hold

(i) The value ofa0 is a positive integer

(ii) For each non-negative integeri we have ai+1 = 2ai+ orai+1 =

ai

ai+

(iii) There exists a positive integer k such that ak = 2014

Find the smallest positive integer n such that there exists a good sequence a0, a1, of real

numbers with the property that an= 2014

Answer: 60

Solution Note that

ai+1+ = 2(ai+ 1) orai+1+ =

ai+ai+

ai+

= 2(ai + 1)

ai+

Hence

1 ai+1+

=

2·

1 ai+

or

ai+1+

= ai+

2(ai+ 1)

=

2 ·

1 ai+

+1

2

Therefore,

1 ak+

=

2k ·

1 a0+

+

k

X

i=1

εi

2k−i+1, (1)

where εi = or Multiplying both sides by 2k(ak+ 1) and putting ak = 2014, we get

2k = 2015

a0+

+ 2015·

k

X

i=1

εi·2i−1

! ,

where εi = or Since gcd(2,2015) = 1, we havea0+ = 2015 and a0 = 2014 Therefore,

2k−1 = 2015·

k

X

i=1

εi ·2i−1

! ,

where εi = or We now need to find the smallest k such that 2015|2k−1 Since 2015 =

5·13·31, from the Fermat little theorem we obtain 5|24−1, 13|212−1 and 31|230−1 We also

have lcm[4,12,30] = 60, hence 5|260−1, 13|260−1 and 31|260−1, which gives 2015|260−1.

(162)

But 5-230−1 and sok = 60 is the smallest positive integer such that 2015|2k−1 To conclude,

the smallest positive integer k such that ak = 2014 is when k = 60

Alternative solution Clearly all members of the sequence are positive rational numbers

For each positive integer i, we have =

ai+1−1

2 or =

2ai+1

1−ai+1

Since > we deduce

that

ai =

      

ai+1−1

2 if ai+1 >1

2ai+1

1−ai+1

if ai+1 <1

Thus is uniquely determined fromai+1 Hence starting from ak = 2014, we simply run the

sequence backwards until we reach a positive integer We compute as follows

2014 , 2013 , 2011 , 2007 , 1999 16 , 1983 32 , 1951 64 , 1887 128, 1759 256, 1503 512, 991 1024, 1982 33 , 1949 66 , 1883 132, 1751 264, 1487 528, 959 1056, 1918 97 , 1821 194, 1627 388, 1239 776, 463 1552, 926 1089, 1852 163, 1689 326, 1363 652, 711 1304, 1422 593, 829 1186, 1658 357, 1301 714, 587 1428, 1174 841, 333 1682, 666 1349, 1332 683, 649 1366, 1298 717, 581 1434, 1162 853, 309 1706, 618 1397, 1236 779, 457 1558, 914 1101, 1828 187, 1641 374, 1267 748, 519 1496, 1038 977, 61 1954, 122 1893, 244 1771, 488 1527, 976 1039, 1952 63 , 1889 126, 1763 252, 1511 504, 1007 1008, 2014

There are 61 terms in the above list Thus k = 60

Alternative solution is quite computationally intensive Calculating the first few terms indicates some patterns that are easy to prove This is shown in the next solution

Alternative solution Start withak =

m0

n0

wherem0 = 2014 andn0 = as in alternative

solution By inverting the sequence as in alternative solution 1, we have ak−i =

mi

ni

for i≥0

where

(mi+1, ni+1) =

(m

i−ni,2ni) if mi > ni

(2mi, ni−mi) if mi < ni

Easy inductions show that mi +ni = 2015, ≤ mi, ni ≤ 2014 and gcd(mi, ni) = for

i ≥ Since a0 ∈ N+ and gcd(mk, nk) = 1, we require nk = An easy induction shows that

(mi, ni)≡(−2i,2i) (mod 2015) fori= 0,1, , k

Thus 2k ≡ 1 (mod 2015) As in the official solution, the smallest such k is k = 60 This

yieldsnk ≡1 (mod 2015) But since ≤nk, mk ≤2014, it follows that a0 is an integer

Problem Let n be a positive integer Consider 2n distinct lines on the plane, no two

of which are parallel Of the 2n lines,n are colored blue, the other n are colored red LetB be

the set of all points on the plane that lie on at least one blue line, and R the set of all points

on the plane that lie on at least one red line Prove that there exists a circle that intersects B

in exactly 2n−1 points, and also intersects R in exactly 2n−1 points

Solution Consider a line ` on the plane and a point P on it such that` is not parallel to

any of the 2nlines Rotate` aboutP counterclockwise until it is parallel to one of the 2n lines

Take note of that line and keep rotating until all the 2n lines are met The 2n lines are now

ordered according to which line is met before or after Say the lines are in order `1, , `2n

Clearly there must be k ∈ {1, ,2n−1} such that`k and `k+1 are of different colors

Now we set up a system ofX– andY– axes on the plane Consider the two angular bisectors

of`k and`k+1 If we rotate`k+1 counterclockwise, the line will be parallel to one of the bisectors

before the other Let the bisector that is parallel to the rotation of `k+1 first be the X–axis,

and the other theY–axis From now on, we will be using the directed angle notation: for lines

s and s0, we define ∠(s, s0) to be a real number in [0, π) denoting the angle in radians such

that whens is rotated counterclockwise by∠(s, s0) radian, it becomes parallel to s0 Using this

(163)

notation, we notice that there is no i = 1, ,2n such that ∠(X, li) is between ∠(X, `k) and

∠(X, `k+1)

Because the 2n lines are distinct, the setS of all the intersections between`i and `j (i6=j)

is a finite set of points Consider a rectangle with two opposite vertices lying on `k and the

other two lying on `k+1 With respect to the origin (the intersection of `k and `k+1), we can

enlarge the rectangle as much as we want, while all the vertices remain on the lines Thus,

there is one of these rectangles R which contains all the points in S in its interior Since each

side of R is parallel to either X– or Y– axis, R is a part of the four lines x = ±a, y = ±b

where a, b >0

M N

b

−b a

−a X

Y

lk

lk+1

Consider the circle C tangent to the right of the x=a side of the rectangle, and to both `k

and `k+1 We claim that this circle intersects B in exactly 2n−1 points, and also intersects R

in exactly 2n−1 points Since C is tangent to both`k and`k+1 and the two lines have different

colors, it is enough to show that C intersects with each of the other 2n−2 lines in exactly

points Note that no two lines intersect on the circle because all the intersections between lines

are in S which is in the interior of R

Consider any line Lamong these 2n−2 lines LetLintersect with`k and`k+1 at the points

M and N, respectively (M and N are not necessarily distinct) Notice that both M and N

must be inside R There are two cases:

(i) L intersectsR on thex=−a side once and another time on x=a side;

(ii) Lintersects y=−b and y =b sides

However, if (ii) happens, ∠(`k, L) and∠(L, `k+1) would be both positive, and then∠(X, L)

would be between ∠(X, `k) and ∠(X, `k+1), a contradiction Thus, only (i) can happen Then

L intersects C in exactly two points, and we are done

Alternative solution By rotating the diagram we can ensure that no line is vertical Let `1, `2, , `2nbe the lines listed in order of increasing gradient Then there is aksuch that lines

`k and `k+1 are oppositely coloured By rotating our coordinate system and cyclicly relabelling

our lines we can ensure that `1, `2, , `2n are listed in order of increasing gradient, `1 and `2n

are oppositely coloured, and no line is vertical

Let D be a circle centred at the origin and of sufficiently large radius so that

• All intersection points of all pairs of lines lie strictly inside D; and

• Each line`i intersectsDin two pointsAiandBi say, such thatAi is on the right semicircle

(the part of the circle in the positive x half plane) andBi is on the left semicircle

Note that the anticlockwise order of the pointsAi, BiaroundDisA1, A2, , An, B1, B2, , Bn

(164)

(IfAi+1 occurred beforeAi then raysri and ri+1 (as defined below) would intersect outside D.)

B1

A1

A2n

B2n

B2n−1

A2n−1

B2

A2

D

C

r1

r2n

r2n−1

r2

For each i, let ri be the ray that is the part of the line `i starting from point Ai and that

extends to the right Let C be any circle tangent to r1 and r2n, that lies entirely to the right

of D ThenC intersects each of r2, r3, , r2n−1 twice and is tangent tor1 and r2n ThusC has

the required properties

Problem Determine all sequences a0, a1, a2, of positive integers with a0 ≥ 2015

such that for all integers n≥1:

(i) an+2 is divisible byan;

(ii) |sn+1−(n+ 1)an|= 1, where sn+1=an+1−an+an−1− · · ·+ (−1)n+1a0

Answer: There are two families of answers:

(a) an=c(n+ 2)n! for alln ≥1 and a0 =c+ for some integer c≥2014, and

(b) an=c(n+ 2)n! for alln ≥1 and a0 =c−1 for some integer c≥2016

Solution Let {an}

∞

n=0 be a sequence of positive integers satisfying the given conditions

We can rewrite (ii) as sn+1 = (n+ 1)an+hn, where hn ∈ {−1,1} Substituting n with n−1

yields sn = nan−1 +hn−1, where hn−1 ∈ {−1,1} Note that an+1 =sn+1 +sn, therefore there

exists δn ∈ {−2,0,2}such that

an+1 = (n+ 1)an+nan−1+δn (1)

We also have|s2−2a1|= 1, which yieldsa0 = 3a1−a2±1≤3a1, and thereforea1 ≥ a30 ≥671

Substituting n = in (1), we find that a3 = 3a2 + 2a1+δ2 Since a1|a3, we have a1|3a2+δ2,

and therefore a2 ≥223 Using (1), we obtain that an ≥223 for alln ≥0

Lemma 1: For n≥4, we have an+2 = (n+ 1)(n+ 4)an

Proof For n≥3 we have

an=nan−1+ (n−1)an−2+δn−1 > nan−1+ (2)

By applying (2) with n substituted by n−1 we have forn ≥4,

an =nan−1+ (n−1)an−2+δn−1 < nan−1+ (an−1−3) +δn−1 <(n+ 1)an−1 (3)

Using (1) to write an+2 in terms ofan and an−1 along with (2), we obtain that for n≥3,

an+2 = (n+ 3)(n+ 1)an+ (n+ 2)nan−1+ (n+ 2)δn+δn+1

<(n+ 3)(n+ 1)an+ (n+ 2)nan−1+ 3(n+ 2)

<(n2+ 5n+ 5)an

(165)

Also forn ≥4,

an+2 = (n+ 3)(n+ 1)an+ (n+ 2)nan−1+ (n+ 2)δn+δn+1

>(n+ 3)(n+ 1)an+nan

= (n2+ 5n+ 3)an

Since an|an+2, we obtain that an+2 = (n2+ 5n+ 4)an= (n+ 1)(n+ 4)an, as desired

Lemma 2: For n≥4, we have an+1 =

(n+ 1)(n+ 3)

n+ an

Proof Using the recurrence an+3 = (n + 3)an+2 + (n+ 2)an+1 +δn+2 and writing an+3,

an+2 in terms ofan+1, an according to Lemma we obtain

(n+ 2)(n+ 4)an+1 = (n+ 3)(n+ 1)(n+ 4)an+δn+2

Hence n+ 4|δn+2, which yields δn+2 = and an+1 = (n+ 1)(nn+ 2+ 3)an, as desired

Suppose there exists n ≥1 such that an+1 =6 (n+ 1)(nn+ 2+ 3)an By Lemma 2, there exist a

greatest integer 1≤m≤3 with this property Then am+2 = (m+ 2)(mm+ 3+ 4)am+1 If δm+1 = 0,

we have am+1 =

(m+ 1)(m+ 3)

m+ am, which contradicts our choice ofm Thus δm+1 6=

Clearly m+ 3|am+1 Write am+1 = (m+ 3)k and am+2 = (m+ 2)(m+ 4)k Then (m+

1)am+δm+1 =am+2−(m+ 2)am+1 = (m+ 2)k So, am|(m+ 2)k−δm+1 But am also divides

am+2 = (m+ 2)(m+ 4)k Combining the two divisibility conditions, we obtainam|(m+ 4)δm+1

Since δm+1 6= 0, we have am|2m+ 8≤ 14, which contradicts the previous result thatan≥ 223

for all nonnegative integers n

So, an+1 = (n+ 1)(nn+ 2+ 3)an for n ≥ Substituting n = yields 3|a1 Letting a1 = 3c,

we have by induction that an = n!(n + 2)c for n ≥ Since |s2 −2a1| = 1, we then get

a0 = c±1, yielding the two families of solutions By noting that (n+ 2)n! = n! + (n+ 1)!,

we have sn+1 = c(n + 2)! + (−1)

n

(c−a0) Hence both families of solutions satisfy the given

conditions

(166)

Problem We say that a triangle ABC is great if the following holds: for any point

D on the side BC, if P and Q are the feet of the perpendiculars from D to the lines AB and

AC, respectively, then the reflection ofDin the lineP Qlies on the circumcircle of the triangle

ABC

Prove that triangle ABC is great if and only if ∠A= 90◦ and AB =AC

Solution For every pointD on the sideBC, letD0 be the reflection ofD in the lineP Q We will first prove that if the triangle satisfies the condition then it is isosceles and right-angled at A

Choose D to be the point where the angle bisector fromA meets BC Note thatP and Q

lie on the rays AB and AC respectively Furthermore, P and Q are reflections of each other in the line AD, from which it follows that P Q⊥ AD Therefore, D0 lies on the line AD and we may deduce that either D0 = A or D0 is the second point of the angle bisector at A and the circumcircle of ABC However, since AP DQ is a cyclic quadrilateral, the segment P Q

intersects the segment AD Therefore,D0 lies on the ray DA and therefore D0 =A By angle chasing we obtain

∠P D0Q=∠P DQ= 180◦ −∠BAC,

and since D0 =A we also know ∠P D0Q=∠BAC This implies that∠BAC = 90◦

Now we choose D to be the midpoint of BC Since ∠BAC = 90◦, we can deduce that

DQP is the medial triangle of triangle ABC Therefore, P Q||BC from which it follows that

DD0 ⊥ BC But the distance from D0 to BC is equal to both the circumradius of triangle

ABC and to the distance from A toBC This can only happen if A =D0 This implies that

ABC is isosceles and right-angled at A

We will now prove that ifABC is isosceles and right-angled atAthen the required property in the problem holds Let D be any point on side BC Then D0P = DP and we also have

DP = BP Hence, D0P = BP and similarly D0Q = CQ Note that AP DQD0 is cyclic with diameter P Q Therefore, ∠AP D0 = ∠AQD0, from which we obtain ∠BP D0 = ∠CQD0 So triangles D0P B and D0QC are similar It follows that ∠P D0Q = ∠P D0C +∠CD0Q =

∠P D0C +∠BD0P = ∠BD0C and D0P

D0Q =

D0B

D0C So we also obtain that triangles D

0P Q and

D0BC are similar But since DP Q and D0P Q are congruent, we may deduce that ∠BD0C =

∠P D0Q = ∠P DQ = 90◦ Therefore, D0 lies on the circle with diameter BC, which is the circumcircle of triangle ABC

Problem A positive integer is called fancy if it can be expressed in the form

(167)

2a1 + 2a2+· · ·+ 2a100,

where a1, a2, , a100 are non-negative integers that are not necessarily distinct Find the smallest positive integer n such that no multiple of n is a fancy number

Answer: The answer isn = 2101−1.

Solution Let k be any positive integer less than 2101 −1 Then k can be expressed in binary notation using at most 100 ones, and therefore there exists a positive integer r and non-negative integers a1,a2, ., ar such that r≤100 and k = 2a1 +· · ·+ 2ar Notice that for

a positive integer s we have:

2sk= 2a1+s+ 2a2+s+· · ·+ 2ar−1+s+ (1 + + +· · ·+ 2s−1)2ar = 2a1+s+ 2a2+s+· · ·+ 2ar−1+s+ 2ar + 2ar +· · ·+ 2ar+s−1.

This shows that k has a multiple that is a sum of r+s powers of two In particular, we may take s= 100−r ≥0, which shows that k has a multiple that is a fancy number

We will now prove that no multiple of n= 2101−1 is a fancy number In fact we will prove a stronger statement, namely, that no multiple of n can be expressed as the sum of at most 100 powers of

For the sake of contradiction, suppose that there exists a positive integer c such that cn is the sum of at most 100 powers of We may assume that c is the smallest such integer By repeatedly merging equal powers of two in the representation of cn we may assume that

cn= 2a1 + 2a2 +· · ·+ 2ar

wherer ≤100 anda1 < a2 < < arare distinct non-negative integers Consider the following

two cases:

• Ifar≥101, then 2ar−2ar−101 = 2ar−101n It follows that 2a1 + 2a2 +· · ·+ 2ar−1+ 2ar−101

would be a multiple ofn that is smaller than cn This contradicts the minimality of c

• Ifar ≤100, then {a1, , ar} is a proper subset of {0,1, ,100} Then

n≤cn <20+ 21+· · ·+ 2100 =n

This is also a contradiction

From these contradictions we conclude that it is impossible for cnto be the sum of at most 100 powers of In particular, no multiple of n is a fancy number

Problem Let AB and AC be two distinct rays not lying on the same line, and let ω

be a circle with center O that is tangent to ray AC at E and ray AB atF Let R be a point on segment EF The line through O parallel to EF intersects line AB at P Let N be the intersection of linesP Rand AC, and letM be the intersection of lineABand the line through

R parallel to AC Prove that line M N is tangent to ω

Solution We present two approaches The first one introduces an auxiliary point and studies similarities in the figure The second one reduces the problem to computations involving a particular exradius of a triangle The second approach has two variants

Solution

(168)

Let the line through N tangent to ω at pointX 6=E intersect AB at point M0 It suffices to show that M0R kAC, since this would yieldM0 =M

Suppose that the line P O intersects AC at Q and the circumcircle of AM0O at Y, respec-tively Then

∠AY M0 =∠AOM0 = 90◦−∠M0OP

By angle chasing we have ∠EOQ =∠F OP = 90◦−∠AOF = ∠M0AO=∠M0Y P and by symmetry ∠EQO =∠M0P Y Therefore 4M0Y P ∼ 4EOQ

On the other hand, we have

∠M0OP =∠M0OF +∠F OP =

2(∠F OX+∠F OP +∠EOQ) = =

2

180◦−∠XOE

2

= 90◦− ∠XOE

2

Since we know that ∠AY M0 and ∠M0OP are complementary this implies

∠AY M0 = ∠XOE

2 =∠N OE

Therefore, ∠AY M0 and ∠N OE are congruent angles, and this means that A and N are corresponding points in the similarity of triangles 4M0Y P and 4EOQ It follows that

AM0 M0P =

N E EQ =

N R RP

We conclude that M0RkAC, as desired

Solution 2a

(169)

As in Solution 1, we introduce point M0 and reduce the problem to proving RNP R = P M0

M0A

Menelaus theorem in triangle AN P with transversal line F RE yields

P R RN ·

N E EA ·

AF F P =

Since AF =EA, we have N EF P = RNP R, so that it suffices to prove

F P N E =

P M0

M0A (1)

This is a computation regarding the triangle AM0N and its excircle opposite A Indeed, setting a =M0N, b =N A, c= M0A, s = a+2b+c, x =s−a, y =s−b and z =s−c, then

AE = AF = s, M0F = z and N E = y From 4OF P ∼ 4AF O we have F P = r2a

s, where

ra = OF is the exradius opposite A Combining the following two standard formulas for the

area of a triangle

|AM0N|2 =xyzs (Heron’s formula) and |AM0N|=ra(s−a),

we haver2

a= yzs

x Therefore, F P = yzx We can now write everything in (1) in terms ofx, y, z

We conclude that we have to verify

yz x

y =

z+yzx

x+y ,

which is easily seen to be true

Note: Antoher approach using Menalaus theorem is to construct the tangent from M to create a point N0 in AC and then prove, using the theorem, that P, R and N0 are collinear This also reduces to an algebraic identity

Solution 2b

As in Solution 1, we introduce point M0 Let the line through M0 and parallel to AN

intersect EF at R0 Let P0 be the intersection of lines N R0 and AM It suffices to show that

P0O k F E, since this would yieldP =P0, and thenR =R0 and M =M0 Hence it is enough to prove that

AF F P0 =

AD

DO, (2)

whereDis the intersection ofAOandEF Once again, this reduces to a computation regarding the triangle AM0N and its excircle oppositeA

Letu=P0F and x, y, z, sas in Solution 2a Note that since AE =AF and M0R0 kAE, we have M0R0 =M0F =z Since M0R0 kAN, we have P0M0

P0A =

M0R0

N A , that is,

u+z

u+x+y+z = z x+z

From this last equation we obtain u= yzx Hence AF

F P0 =

xs

yz Also, as in Solution 2a, we have

r2

a = yzs

x

Finally, using similar triangles ODF, F DA and OF A, and the above equalities, we have

AD DO = AD DF · DF DO = AF OF · AF OF = s2 r2 a = s yzs x = xs yz = AF F P0 ,

as required

(170)

Problem The country Dreamland consists of 2016 cities The airline Starways wants to establish some one-way flights between pairs of cities in such a way that each city has exactly one flight out of it Find the smallest positive integer k such that no matter how Starways establishes its flights, the cities can always be partitioned into k groups so that from any city it is not possible to reach another city in the same group by using at most 28 flights

Answer: 57

Solution The flights established by Starways yield a directed graph G on 2016 vertices in which each vertex has out-degree equal to

We first show that we need at least 57 groups For this, suppose thatGhas a directed cycle of length 57 Then, for any two cities in the cycle, one is reachable from the other using at most 28 flights So no two cities in the cycle can belong to the same group Hence, we need at least 57 groups

We will now show that 57 groups are enough Consider another auxiliary directed graph H

in which the vertices are the cities of Dreamland and there is an arrow from city u to city v

if u can be reached fromv using at most 28 flights Each city has out-degree at most 28 We will be done if we can split the cities of H in at most 57 groups such that there are no arrows between vertices of the same group We prove the following stronger statement

Lemma: Suppose we have a directed graph on n ≥ vertices such that each vertex has out-degree at most 28 Then the vertices can be partitioned into 57 groups in such a way that no vertices in the same group are connected by an arrow

Proof: We apply induction The result is clear for vertex Now suppose we have more than one vertex Since the out-degree of each vertex is at most 28, there is a vertex, say v, with in-degree at most 28 If we remove the vertex v we obtain a graph with fewer vertices which still satifies the conditions, so by inductive hypothesis we may split it into at most 57 groups with no adjacent vertices in the same group Since v has in-degree and out-degree at most 28, it has at most 56 neighboors in the original directed graph Therefore, we may addv back and place it in a group in which it has no neighbors This completes the inductive step

Problem Find all functions f :R+→

R+ such that

(z+ 1)f(x+y) = f(xf(z) +y) +f(yf(z) +x), (3) for all positive real numbers x, y, z

Answer: The only solution isf(x) =x for all positive real numbers x

Solution The identity function f(x) = x clearly satisfies the functional equation Now, let f be a function satisfying the functional equation Plugging x = y = into (3) we get 2f(f(z) + 1) = (z+ 1)(f(2)) for all z ∈R+ Hence, f is not bounded above.

Lemma Leta, b, c be positive real numbers If cis greater than 1, a/b and b/a, then the system of linear equations

cu+v =a u+cv=b

has a positive real solution u, v

Proof The solution is

u= ca−b

c2−1 v =

cb−a c2−1

The numbers u and v are positive if the conditions on c above are satisfied

(171)

We will now prove that

f(a) +f(b) =f(c) +f(d) for all a, b, c, d∈R+ with a+b=c+d (4) Consider a, b, c, d ∈ R+ such that a +b = c+d Since f is not bounded above, we can choose a positive number e such that f(e) is greater than 1, a/b, b/a, c/d and d/c Using the above lemma, we can findu, v, w, t∈R+ satisfying

f(e)u+v =a, u+f(e)v =b f(e)w+t=c, w+f(e)t=d

Note that u+v = w+t since (u+v)(f(e) + 1) = a+b and (w+t)(f(e) + 1) = c+d Plugging x =u, y =v and z = e into (3) yields f(a) +f(b) = (e+ 1)f(u+v) Similarly, we have f(c) +f(d) = (e+ 1)f(w+t) The claim follows immediately

We then have

yf(x) =f(xf(y)) for all x, y ∈R+ (5) since by (3) and (4),

(y+ 1)f(x) =f

x

2f(y) +

x

2

+f

x

2 f(y) +

x

2

=f(xf(y)) +f(x)

Now, let a = f(1/f(1)) Plugging x = and y = 1/f(1) into (5) yields f(a) = Hence

a =af(a) and f(af(a)) =f(a) = Since af(a) =f(af(a)) by (5), we have f(1) =a = It follows from (5) that

f(f(y)) =y for all y∈R+. (6) Using (4) we have for all x, y ∈R+ that

f(x+y) +f(1) =f(x) +f(y+ 1), and

f(y+ 1) +f(1) =f(y) +f(2)

Therefore

f(x+y) = f(x) +f(y) +b for all x, y ∈R+, (7) where b=f(2)−2f(1) =f(2)−2 Using (5), (7) and (6), we get

4 + 2b = 2f(2) =f(2f(2)) =f(f(2) +f(2)) =f(f(2)) +f(f(2)) +b= +b

This shows that b= and thus

f(x+y) =f(x) +f(y) for all x, y ∈R+

In particular, f is strictly increasing

We conclude as follows Take any positive real number x If f(x) > x, then f(f(x)) > f(x) > x = f(f(x)), a contradiction Similarly, it is not possible that f(x) < x This shows that f(x) = x for all positive real numbers x

Marking Scheme:

(172)

• (2pt) Showing thatf(a) +f(b) = f(c) +f(d) whena+b =c+d

• (1pt) Showing thatyf(x) =f(xf(y))

• (1pt) Showing thatf(f(y)) = y

• (2pt) Showing thatf(x+y) =f(x) +f(y)

• (1pt) Conclusion

Note: Since f(x) = x clearly satisfies the functional equation, no points will be awarded or deducted for statements , or lack thereof, related to this fact

(173)

Problem We call a 5-tuple of integers arrangeable if its elements can be labeled a,

b, c, d, e in some order so that a−b+c−d+e = 29 Determine all 2017-tuples of integers

n1, n2, , n2017 such that if we place them in a circle in clockwise order, then any 5-tuple of numbers in consecutive positions on the circle is arrangeable

Answer: n1 =· · ·=n2017 = 29

Solution A valid 2017-tuple is n1 = · · · = n2017 = 29 We will show that it is the only solution

We first replace each number ni in the circle by mi :=ni−29 Since the condition a−b+

c−d+e= 29 can be rewritten as (a−29)−(b−29) + (c−29)−(d−29) + (e−29) = 0, we have that any five consecutive replaced integers in the circle can be labeled a, b, c, d, e in such a way that a−b+c−d+e = We claim that this is possible only when all of themi’s are

(and thus all of the original ni’s are 29)

We work with indexes modulo 2017 Notice that for every i, mi and mi+5 have the same parity Indeed, this follows from mi ≡ mi+1 +mi+2+mi+3 +mi+4 ≡ mi+5 (mod 2) Since gcd(5,2017) = 1, this implies that allmi’s are of the same parity Sincem1+m2+m3+m4+m5 is even, all mi’s must be even as well

Suppose for the sake of contradiction that not all mi’s are zero Then our condition still

holds when we divide each number in the circle by However, by performing repeated divisions, we eventually reach a point where some mi is odd This is a contradiction

Problem Let ABC be a triangle with AB < AC Let D be the intersection point of the internal bisector of angle BAC and the circumcircle of ABC Let Z be the intersection point of the perpendicular bisector of AC with the external bisector of angle ∠BAC Prove that the midpoint of the segment AB lies on the circumcircle of triangle ADZ

Solution Let N be the midpoint of AC Let M be the intersection point of the circumcircle of triangle ADZ and the segment AB We will show that M is the midpoint of

AB To this, let D0 the reflection of D with respect toM It suffices to show that ADBD0

is a parallelogram

The internal and external bisectors of an angle in a triangle are perpendicular This implies that ZD is a diameter of the circumcircle of AZD and thus ∠ZM D = 90◦ This means that

ZM is the perpendicular bisector of D0D and thus ZD0 = ZD By construction, Z is in the perpendicular bisector of AC and thusZA =ZC

Now, let α be the angle ∠BAD = ∠DAC In the cyclic quadrilateral AZDM we get

∠M ZD=∠M AD=α, and thus∠D0ZD= 2α By angle chasing we get

∠AZN = 90◦−∠ZAN =∠DAC =α,

which implies that ∠AZC = 2α Therefore,

∠D0ZA=∠D0ZD−∠AZD= 2α−∠AZD=∠AZC−∠AZD=∠DZC

Combining ∠D0ZA =∠DZC, ZD0 =ZD and ZA =ZC, we obtain by the SAS criterion that the triangles D0ZA and DZC are congruent In particular, D0A = DC and ∠D0AZ =

∠DCZ From here DB =DC =D0A

Finally, let β =∠ABC =∠ADC We get the first of the following equalities by the sum of angles around pointAand the second one by the sum of internal angles of quadrilateralAZCD

(174)

360◦ =∠D0AZ +∠ZAD+∠DAB+∠BAD0 =∠D0AZ+ 90◦+α+∠BAD0

360◦ =∠DCZ+∠ZAD+∠CZA+∠ADC =∠DCZ + 90◦+ 2α+β

By canceling equal terms we conclude that ∠BAD0 = α + β Also, ∠ABD = α +β Therefore, the segments D0Aand DB are parallel and have the same length We conclude that

ADBD0 is a parallelogram As the diagonals of a parallelogram intersect at their midpoints, we obtain that M is the midpoint of AB as desired

Variant of solution The solution above is indirect in the sense that it assumes thatM is in the circumcircle of AZD and then shows that M is the midpoint of AB We point out that the same ideas in the solution can be used to give a direct solution Here we present a sketch on how to proceed in this manner

Now we know that M is the midpoint of the side AB We construct the point D0 in the same way Now we have directly that ADBD0 is a parallelogram and thus D0A =DB =DC By constructionZA=ZC Also, the two sums of angles equal to 360◦ in the previous solution let us conclude that∠D0AZ =∠DCZ Once again, we use (differently) the SAS criterion and obtain that the triangles D0AZ and DCZ are congruent Thus, D0Z =DZ

We finish the problem by noting that ZM is a median of the isosceles triangle D0ZD, so it is also a perpendicular bisector This shows that ∠DM Z = 90◦ = ∠DAZ, and therefore M

lies in the circumcircle of DAZ

Solution We proceed directly As above, we name

α=∠DAC =∠AZN =∠CZN =∠DCB

LetLbe the midpoint of the segmentBC SinceM andN are midpoints ofABandAC, we have that M N =CL and that the segment M N is parallel to BC Thus, ∠AN M =∠ACB

(175)

Therefore,∠ZN M =∠ACB+ 90◦ Also,∠DCZ =∠DCB+∠ACB+∠ZCA=∠ACB+ 90◦ We conclude that ∠ZN M =∠DCZ

Now, the triangles ZN C and CLD are similar since∠DCL= 90◦ =∠CN Z and∠DCL=

α=∠CZN We use this fact to obtain the following chain of equalities

CD M N =

CD CL =

CZ ZN

We combine the equality above with ∠ZN M = ∠DCZ to conclude that the triangles

ZN M and ZCD are similar In particular, ∠M ZN = ∠DZC and ZMZN = ZDZC Since we also have ∠M ZD = ∠M ZN +∠N ZD = ∠DZC +∠N ZD = ∠N ZC, we conclude that the triangles M ZD and N ZC are similar This yields that ∠ZM D = 90◦ and therefore M is in the circumcircle of triangle DAZ

Solution Letm be the perpendicular bisector ofAD; thus m passes through the center

O of the circumcircle of triangle ABC Since AD is the internal angle bisector of A and OM

andON are perpendicular toAB andACrespectively, we obtain thatOM andON form equal angles with AD This implies that they are symmetric with respect to M

Therefore, the line M O passes through the point Z0 symmetrical to Z with respect to m Since ∠DAZ = 90◦, then also ∠ADZ0 = 90◦ Moreover, since AZ = DZ0, we have that

∠AZZ0 = ∠DZ0Z = 90◦, so AZZ0D is a rectangle Since ∠AM Z0 = ∠AM O = 90◦, we conclude that M is in the circle with diameter AZ0, which is the circumcircle of the rectangle Thus M lies on the circumcircle of the triangle ADZ

Problem Let A(n) denote the number of sequences a1 ≥ a2 ≥ ≥ ak of positive

integers for which a1 +· · · + ak = n and each + is a power of two (i = 1,2, , k)

Let B(n) denote the number of sequences b1 ≥ b2 ≥ ≥ bm of positive integers for which

b1+· · ·+bm =n and each inequality bj ≥2bj+1 holds (j = 1,2, , m−1) Prove that A(n) = B(n) for every positive integer n

Solution We say that a sequence a1 ≥ a2 ≥ ≥ ak of positive integers has type A if

ai + is a power of two for i = 1,2, , k We say that a sequence b1 ≥ b2 ≥ ≥ bm of

positive integer has type B if bj ≥2bj+1 for j = 1,2, , m−1

Recall that the binary representation of a positive integer expresses it as a sum of distinct powers of two in a unique way Furthermore, we have the following formula for every positive integer N

2N −1 = 2N−1+ 2N−2+· · ·+ 21+ 20

Given a sequence a1 ≥ a2 ≥ .≥ak of type A, use the preceding formula to express each

term as a sum of powers of two Write these powers of two in left-aligned rows, in decreasing order of size By construction, is the sum of the numbers in the ith row For example, we

obtain the following array when we start with the type A sequence 15,15,7,3,3,3,1

8 4 2 2 1

27 13

(176)

Define the sequence b1, b2, , bn by setting bj to be the sum of the numbers in the jth

column of the array For example, we obtain the sequence 27,13,5,2 from the array above We now show that this new sequence has type B This is clear from the fact that each column in the array contains at least as many entries as the column to the right of it and that each number larger than in the array is twice the number to the right of it Furthermore, it is clear that a1 +a2 +· · ·+ak = b1 +b2 +· · ·+bm, since both are equal to the sum of all the

entries in the array

We now show that we can this operation backwards Suppose that we are given a type B sequence b1 ≥b2 ≥ .≥bm We construct an array inductively as follows:

• We fill bm left-aligned rows with the numbers 2m−1,2m−2, ,21,20

• Then we fill bm−1−2bm left aligned rows with the numbers 2m−2,2m−3, ,21,20

• Then we fill bm−2−2bm−1 left aligned rows with the numbers 2m−3,2m−4, ,21,20, and so on

• In the last step we fill b1−2b2 left-aligned rows with the number

For example, if we start with the type B sequence 27,13,5,2, we obtain once again the array above We define the sequence a1, a2, , ak by setting to be the sum of the numbers in the

ith row of the array By construction, this sequence has type A Furthermore, it is clear that

a1+· · ·+ak=b1+· · ·+bm, since once again both sums are equal to the sum of all the entries

in the array

We have defined an operation that starts with a sequence of type A, produces an array whose row sums are given by the sequence, and outputs a sequence of type B corresponding to the column sums We have also defined an operation that starts with a sequence of type B, produces an array whose column sums are given by the sequence, and outputs a sequence of type A corresponding to the row sums The arrays produced in both cases comprise left-aligned rows of the form 2N−1,2N−2, ,21,20, with non-increasing lengths Let us refer to arrays obeying these properties as marvelous

To show that these two operations are inverses of each other, it then suffices to prove that marvelous arrays are uniquely defined by either their row sums or their column sums The former is obviously true and the latter arises from the observation that each step in the above inductive algorithm was forced in order to create a marvelous array with the prescribed column sums

Thus, we have produced a bijection between the sequences of type A with sum n and the sequences of type B with sum n So we can conclude that A(n) = B(n) for every positive integer n

RemarkThe solution above provides a bijection between type A and type B sequences via an algorithm There are alternative ways to provide such a bijection For example, given the numbers a1 ≥ .≥ak we may define the bi’s as

bj =

X

i

ai+

2j

Conversely, given the numbers b1 ≥ ≥ bm, one may define the ai’s by taking, as in the

solution, bm numbers equal to 2m−1,bm−1−2bm numbers equal to 2m−1−1, , and b1−2b2 numbers equal to 21 −1 One now needs to verify that these maps are mutually inverse

Problem Call a rational number r powerful if r can be expressed in the form pqk for some relatively prime positive integers p, q and some integer k > Let a, b, c be positive

(177)

rational numbers such that abc = Suppose there exist positive integers x, y, z such that

ax+by +cz is an integer Prove thata, b, c are all powerful.

Solution Let a = a1

b1 , b =

a2

b2, where gcd(a1, b1) = gcd(a2, b2) = Then c=

b1b2

a1a2 The condition that ax+by +cz is an integer becomes

ax1+zaz

2b

y

2+az1a

y+z

2 bx1 +b

x+z

1 b

y+z

2

az

1az2bx1b

y

2

∈Z,

which can be restated as

az1az2bx1by2 |ax1+zaz2by2 +az1ay2+zbx1 +bx1+zby2+z (1) In particular, az

1 divides the right-hand side Since it divides the first and second terms in the sum, we conclude that az1 |bx1+zb2y+z Since gcd(a1, b1) = 1, we have az1 |b

y+z

2

Let p be a prime that divides a1 Let m, n ≥ be integers such that pnka1 (i.e pn|a1 but pn+1

- a1) and pmkb2 The fact that az1 | b

y+z

2 implies nz ≤ m(y+z) Since gcd(a1, b1) = gcd(a2, b2) = 1, we have p does not divide b1 and does not divide a2 Thus

pnzkaz

1a

y+z

2 b

x

1 and p

m(y+z)kbx+z

1 b

y+z

2 (2)

On the other hand, (1) implies that

pnz+my |az1ay2+zbx1 +bx1+zby2+z (3)

If nz < m(y+z), then (2) gives pnzkaz

1a

y+z

2 bx1 +b

x+z

1 b

y+z

2 , which contradicts (3) Thus

nz = m(y+z) so n is divisible by k := y+z

gcd(z, y+z) > Thus each exponent in the prime

decomposition of a1 must be divisible by k Hence a1 is a perfect k-power which means a is powerful Similarly, b and care also powerful

Problem Let n be a positive integer A pair of n-tuples (a1, , an) and (b1, , bn)

with integer entries is called an exquisite pair if

|a1b1+· · ·+anbn| ≤1

Determine the maximum number of distinctn-tuples with integer entries such that any two of them form an exquisite pair

Answer: The maximum isn2+n+

Solution First, we construct an example with n2 +n+ 1 n-tuples, each two of them forming an exquisite pair In the following list, ∗represents any number of zeros as long as the total number of entries is n

• (∗)

• (∗,1,∗)

• (∗,−1,∗)

• (∗,1,∗,1,∗)

• (∗,1,∗,−1,∗)

(178)

For example, forn = we have the tuples (0,0), (0,1), (1,0), (0,−1), (−1,0), (1,1), (1,−1) The total number of such tuples is +n+n+ n2+ n2=n2+n+ For any two of them, at most two of the productsaibi are non-zero The only case in which two of them are non-zero

is when we take a sequence (∗,1,∗,1,∗) and a sequence (∗,1,∗,−1,∗) with zero entries in the same places But in this case oneaibi is and the other −1 This shows that any two of these

sequences form an exquisite pair

Next, we claim that among anyn2+n+ tuples, some two of them not form an exquisite pair We begin with lemma

Lemma Given 2n + distinct non-zero n-tuples of real numbers, some two of them (a1, , an) and (b1, , bn) satisfya1b1+· · ·+anbn >0

Proof of Lemma We proceed by induction The statement is easy for n= since for every three non-zero numbers there are two of them with the same sign Assume that the statement is true for n−1 and consider 2n+ tuples with n entries Since we are working with tuples of real numbers, we claim that we may assume that one of the tuples is a= (0,0, ,0,−1) Let us postpone the proof of this claim for the moment

If one of the remaining tuples b has a negative last entry, then a and b satisfy the desired condition So we may assume all the remaining tuples has a non-negative last entry Now, from each tuple remove the last number If two n-tuplesb and cyield the same (n−1)-tuple, then

b1c1+· · ·+bn−1cn−1+bncn=b21+· · ·+b

n−1+bncn >0,

and we are done The remaining case is that all then-tuples yield distinct (n−1)-tuples Then at most one of them is the zero (n−1)-tuple, and thus we can use the inductive hypothesis on 2n−1 of them So we find b and c for which

(b1c1+· · ·+bn−1cn−1) +bncn>0 +bncn>0

The only thing that we are left to prove is that in the inductive step we may assume that one of the tuples is a = (0,0, ,0,−1) Fix one of the tuples x = (x1, , xn) Set a real

number ϕ for which tanϕ = x1

x2 Change each tuple a = (a1, a2, , an) (including x), to the tuple

(a1cosϕ−a2sinϕ, a1sinϕ+a2cosϕ, a3, a4, , an)

A straightforward calculation shows that the first coordinate of the tuple x becomes 0, and that all the expressions of the forma1b1+· · ·+anbn are preserved We may iterate this process

until all the entries of xexcept for the last one are equal to We finish by multiplying all the entries in all the tuples by a suitable constant that makes the last entry ofx equal to−1 This preserves the sign of all the expressions of the form a1b1+· · ·+anbn

We proceed to the proof of our claim Let A be a set of non-zero tuples among which any two form an exquisite pair It suffices to prove that |A| ≤n2+n We can write A as a disjoint union of subsets A1 ∪A2 ∪ .∪An, where Ai is the set of tuples in A whose last non-zero

entry appears in theith position We will show that |Ai| ≤2i, which will finish our proof since

2 + +· · ·+ 2n =n2+n.

Proceeding by contradiction, suppose that |Ai| ≥ 2i+ If Ai has three or more tuples

whose only non-zero entry is in the ith position, then for two of them this entry has the same sign Since the tuples are different and their entries are integers, this yields two tuples for which

|P

aibi| ≥2, a contradiction So there are at most two such tuples We remove them fromAi

Now, for each of the remaining tuples a, if it has a positive ith coordinate, we keep a as it is If it has a negative ith coordinate, we replace it with the opposite tuple −a with entries with opposite signs This does not changes the exquisite pairs condition

(179)

After making the necessary changes, we have two cases The first case is that there are two tuples a and b that have the same first i−1 coordinates and thus

a1b1+· · ·+ai−1bi−1 =a21+· · ·+a2i−1 >0,

and thus is at least (the entries are integers) The second case is that no two tuples have the same first i−1 coordinates, but then by the Lemma we find two tuples a and b for which

a1b1+· · ·+ai−1bi−1 ≥1 In any case, we obtain

a1b1+· · ·+ai−1bi−1+aibi ≥2

This yields a final contradiction to the exquisite pair hypothesis

(180)

Problem Let H be the orthocenter of the triangle ABC Let M and N be the

midpoints of the sides AB and AC, respectively Assume that H lies inside the quadrilateral

BM N C and that the circumcircles of triangles BM H and CN H are tangent to each other

The line throughH parallel toBCintersects the circumcircles of the trianglesBM HandCN H

in the points K and L, respectively Let F be the intersection point of M K and N L and let J

be the incenter of triangle M HN Prove that F J =F A

Solution

Lemma In a triangle ABC, let D be the intersection of the interior angle bisector at A

with the circumcircle of ABC, and let I be the incenter of 4ABC Then

DI =DB =DC

Proof

∠DBI = ∠BAC

2 +

b B

2 =∠DIB ⇒ DI =DB

Analogously DI =DC

We start solving the problem First we state some position considerations Since there is

an arc of the circumcircle of BHM outside the triangle ABC, it must happen that K and N

lie on opposite sides of AM Similarly, L and M lie on opposite sides of AN Also, K and L

lie on the same side of M N, and opposite to A Therefore, F lies inside the triangle AM N

Now, since H is the orthocenter of 4ABC and the circumcircles of BM H and CN H are

tangent we have

∠ABH = 90◦−∠BAC =∠ACH ⇒ ∠M HN =∠M BH+∠N CH = 180◦−2∠BAC (1)

So∠M BH =∠M KH =∠N CH =∠N LH = 90◦−∠BAC and, since M NkKL, we have

∠F M N =∠F N M = 90◦−∠BAC ⇒ ∠M F N = 2∠BAC (2) The relations (1) and (2) yield that the quadrilateral M F N H is cyclic, with the vertices

in this order around the circumference Since F M = F N, ∠M F N = 2∠BAC and F is the

correct side of M N we have that the point F is the circumcenter of triangle AM N, and thus

F A=F M =F N

(181)

Since the quadrilateral M F N H is cyclic, F M = F N and H lies on the correct side of

M N, we have that H, J and F are collinear According to Lemma 1, F J = F M = F N So

F J =F A

Solution 2: According to Solution 1, we have ∠M HN = 180◦ −2∠BAC and since the

point J is the incenter of 4M HN, we have ∠M J N = 90◦ +12∠M HN = 180◦−∠BAC So

the quadrilateral AM J N is cyclic

According to Solution 1, the point F is the circumcenter of 4AM N So F J =F A

Problem Let f(x) and g(x) be given by

f(x) =

x +

1

x−2 +

x−4 +· · ·+

x−2018 and

g(x) =

x−1 +

x−3+

x−5+· · ·+

x−2017 Prove that

|f(x)−g(x)|>2 for any non-integer real number x satisfying 0< x <2018

Solution There are two cases: 2n− < x < 2n and 2n < x < 2n + Note that

f(2018−x) = −f(x) and g(2018−x) = −g(x), that is, a half turn about the point (1009,0) preserves the graphs of f and g So it suffices to consider only the case 2n−1< x <2n

Let d(x) = g(x)−f(x) We will show that d(x) > whenever 2n −1 < x < 2n and

n ∈ {1,2, ,1009}

For any non-integer x with 0< x <2018, we have

d(x+ 2)−d(x) =

1

x+ −

x+

+

1

x−2018 −

x−2017

>0 + =

Hence it suffices to prove d(x)>2 for 1< x <2 Since x <2, it follows that x−21i−1 >

x−2i fori= 2,3, ,1008 We also have x−12018 <0 Hence it suffices to prove the following

(182)

for 1< x <2

1

x−1 +

x−3 −

x −

1

x−2 >2

⇔

1

x−1 +

2−x

+

1

x−3−

x

>2

⇔

(x−1)(2−x) +

x(x−3) >2

By the GM −HM inequality (alternatively, by considering the maximum of the quadratic

(x−1)(2−x)) we have

x−1 · 2−x >

2

(x−1) + (2−x)

2

=

To find a lower bound for

x(x−3), note that x(x−3)<0 for 1< x < So we seek an upper

bound for x(x−3) From the shape of the quadratic, this occurs at x = or x = 2, both of

which yield

x(x−3) >−

It follows that d(x)>4− 32 >2, as desired

Solution

As in Solution 1, we may assume 2n −1 < x < 2n for some ≤ n ≤ 1009 Let d(x) =

f(x)−g(x), and note that

d(x) =

x +

1009

X

m=1

1

(x−2m)(x−2m+ 1)

We split the sum into three parts: the terms before m=n, after m=n, and the term m=n The first two are

0≤

n−1

X

m=1

1

(x−2m)(x−2m+ 1)

≤

n−1

X

m=1

1

(2n−1−2m)(2n−2m) =

n−1

X

i=1

1

(2i)(2i−1) ≤

1008

X

i=1

1 2i−1−

1 2i,

0≤

1009

X

m=n+1

1

(2m−x)(2m−1−x)

≤

1009

X

m=n+1

1

(2m−2n+ 1)(2m−2n) =

1009−n

X

i=1

1

(2i+ 1)(2i) ≤

1008

X

i=1

1 2i −

1 2i+ When we add the two sums the terms telescope and we are left with

0≤ X

1≤m≤1009,m6=n

1

(x−2m)(x−2m+ 1) ≤1−

2017 <1,

For the term m=n, we write

0<−(x−2n)(x−2n+ 1) = 0.25−(x−2n+ 0.5)2 ≤0.25,

(183)

whence

−4≥

(x−2n)(x−2n+ 1) Finally, x1 <1 sincex >2n−1≥1 Combining these we get

d(x) =

x+

1009

X

m=1

1

(x−2m)(x−2m+ 1) <1 + 1−4<−2

Solution

First notice that

f(x)−g(x) =

x −

1

x−1 +

x−2 − · · · −

x−2017 +

x−2018

As in Solution 1, we may deal only with the case 2n < x < 2n+ Then x−2k+ and

x−2k never differ in sign for any integer k Then

−

x−2k+ +

x−2k =

1

(x−2k+ 1)(x−2k) >0 for k= 1,2, , n−1, n+ 2, ,1009

x−2n −

1

x−2n−1 =

1

(x−2n)(2n+ 1−x) ≥

2

x−2n+ 2n+ 1−x

= 4,

Therefore, summing all inequalities and collecting the remaining terms we findf(x)−g(x)>

4 + x−1 2 >4−1 = for 0< x <1 and, forn >0,

f(x)−g(x)> x −

1

x−2n+ + +

1

x−2n−2

=

x −

1

x−2n+ + 4−

1 2n+ 2−x >

x −

1

2n−2n+ + 4−

1

2n+ 2−2n−1

= +

x >2

Problem A collection of n squares on the plane is called tri-connected if the following criteria are satisfied:

(i) All the squares are congruent

(ii) If two squares have a point P in common, then P is a vertex of each of the squares (iii) Each square touches exactly three other squares

How many positive integersnare there with 2018≤n ≤3018, such that there exists a collection of n squares that is tri-connected?

Answer: 501

Solution We will prove that there is no tri-connected collection if n is odd, and that

tri-connected collections exist for all even n ≥ 38 Since there are 501 even numbers in the

(184)

For any two different squares A and B, let us write A ∼B to mean that square A touches

square B Since each square touches exactly three other squares, and there are n squares in

total, the total number of instances of A∼ B is 3n But A∼ B if and only if B ∼ A Hence

the total number of instances ofA ∼B is even Thus 3n and hence also n is even

We now construct tri-connected collections for each even n in the range We show two

Construction The idea is to use the following two configurations Observe that in each configuration every square is related to three squares except for the leftmost and rightmost squares which are related to two squares Note that the configuration on the left is of variable length Also observe that multiple copies of the configuration on the right can be chained together to end around corners

Putting the above two types of configurations together as in the following figure yields a tri-connected collection for every even n≥38

Construction Consider a regular 4n−gon A1A2· · ·A4n, and make 4n squares on the

outside of the 4n−gon with one side being on the 4n−gon Reflect squares sharing sides

A4m+2A4m+3, A4m+3A4m+4 across line A4m+2A4m+4, for ≤ m ≤ n−1 This will produce a

tri-connected set of 6n squares, as long as the squares inside the 4n−gon not intersect When n ≥4, this will be true The picture for n= 24 is as follows:

To treat the other cases, consider the following gadget

(185)

Y X

Two squares touch other squares, and the squares containing X, Y touch other squares

Take the 4n−gon from above, and break it into two along the line A1A2n, moving the two

parts away from that line Do so until the gaps can be exactly filled by inserting two copies of the above figure, so that the vertices X, Y touch the two vertices which used to be A1 in one

instance, and the two vertices which used to be A2n in the other

This gives us a valid configuration for 6n+ squares, n≥4 Finally, if we had instead spread the two parts out more and inserted two copies of the above figure into each gap, we would get 6n+ 16 for n ≥4, which finishes the proof for all even numbers at least 36

Problem Let ABC be an equilateral triangle From the vertex A we draw a ray towards the interior of the triangle such that the ray reaches one of the sides of the triangle When the ray reaches a side, it then bounces off following the law of reflection, that is, if it arrives with a directed angle α, it leaves with a directed angle 180◦ −α After n bounces, the ray returns to Awithout ever landing on any of the other two vertices Find all possible values of n

Answer: Alln ≡1,5 mod with the exception of and 17

Solution Consider an equilateral triangleAA1A2 of side lengthm and triangulate it with

unitary triangles See the figure To each of the vertices that remain after the triangulation we can assign a pair of coordinates (a, b) where a, b are non-negative integers, a is the number of edges we travel in theAA1direction and bis the number of edges we travel in theAA2 direction

to arrive to the vertex, (we haveA= (0,0),A1 = (m,0) andA2 = (0, m)) The unitary triangle

with vertex Awill be our triangle ABC, (B = (1,0), C = (0,1)) We can obtain every unitary

triangle by starting with ABC and performing reflections with respect to a side (the vertex

(1,1) is the reflection ofA with respect to BC, the vertex (0,2) is the reflection of B = (1,0) with respect to the side formed byC = (1,0) and (1,1), and so on)

When we reflect a vertex (a, b) with respect to a side of one of the triangles, the congruence

ofa−bis preserved modulo Furthermore, an induction argument shows that any two vertices

(a, b) and (a0, b0) with a−b ≡a0−b0 mod can be obtained from each other by a series of such reflections Therefore, the set of vertices V that result from the reflections of A will be those of the form (a, b) satisfying a≡b mod See the green vertices in the figure

Now, let U be the set of vertices u that satisfy that the line segment between u and A

does not pass through any other vertex A pair (a, b) is in U if and only if gcd(a, b) = 1, since otherwise for d = gcd(a, b) we have that the vertex (a/d, b/d) also lies on the line segment

between u and A

Observe that the rays that come out from Aand eventually return toAare those that come

out towards a vertex in V ∩U (they would be in V to be able to come back to A and in U

so that they not reach a vertex beforehand) In the diagram, a ray toward one such vertex (a, b) will intersect exactly (a−1) + (b−1) + (a+b−1) = 2(a+b)−3 lines: a−1 of them parallel to AB, b−1 parallel to AC and a+b−1 parallel to BC Therefore, in the triangle

ABC the ray will bounce 2(a +b)−3 times before returning to A So we want to find all

(186)

n = 2(a+b)−3 wherea≡b mod and gcd(a, b) =

Ifa+bis a multiple of then we cannot satisfy both conditions simultaneously, thereforen

is not a multiple of We also know thatn is odd Therefore n ≡1,5,7,11 mod 12 Note that the pair (1,3k+ 1) satisfies the conditions and we can create n = 2(3k+ 2)−3 = 6k+ for all k ≥ (this settles the question for n ≡ 1,7 mod 12) For n ≡ mod 12 consider the pair (3k−1,3k+ 5) when k is even or (3k−4,3k+ 8) when k is odd This gives us all the integers of the form 12k + for k ≥ For 11 mod 12, take the pairs (3k−1,3k + 2) (with k ≥ 1), which yield all positive integers of the form 12k−1

Finally, to discard and 17 note that the only pairs (a, b) that are solutions to 2(a+b)−3 =

or 2(a+b)−3 = 17 with the same residue mod in this range are the non-relatively prime

pairs (2,2), (2,8) and (5,5)

Problem Find all polynomials P(x) with integer coefficients such that for all real numbers s and t, if P(s) and P(t) are both integers, then P(st) is also an integer

Answer: P(x) = xn+k,−xn+k for n a non-negative integer and k an integer.

Solution 1: P(x) = xn+k,−xn+k for n a non-negative integer and k an integer.

Notice that if P(x) is a solution, then so is P(x) +k and −P(x) +k for any integer k, so we may assume that the leading coefficient of P(x) is positive and that P(0) = 0, i.e., we can assume that P(x) = Pn

i=1aix

i witha

n >0 We are going to prove thatP(x) =xn in this case

Letp be a large prime such thatp > Pn

i=1|ai| Because P has a positive leading coefficient

and p is large enough, we can find t ∈ R such that P(t) = p Denote the greatest common

divisor of the polynomial P(x)−p and P(2x)−P(2t) as f(x), and t is a root of it, so f is a non-constant polynomial Notice thatP(2t) is an integer by using the hypothesis fors= and

t SinceP(x)−pand P(2x)−P(2t) are polynomials with integer coefficients,f can be chosen as a polynomial with rational coefficients

In the following, we will prove that f is the same as P(x)−p up to a constant multiplier Say P(x)−p=f(x)g(x), where f andg are non-constant polynomials By Gauss’s lemma, we can get f1, g1 with P(x)−p= f1(x)g1(x) where f1 is a scalar multiple of f and g1 is a scalar

multiple of g and one of f1, g1 has constant term ±1 (this is because−p=P(0)−p=f(0)g(0)

withpprime) SoP(x)−phas at least one rootrwith absolute value not greater than (using

(187)

Vieta, the product of the roots of the polynomial with constant term ±1 is ±1), but

|P(r)−p|=

n X i=1

airi−p

> p−

n

X

i=1

|ai|>0,

hence we get a contradiction!

Therefore f is a constant multiple of P(x)−p, so P(2x)−P(2t) is a constant multiple

of P(x)−p because they both have the same degree By comparing leading coefficients we

get that P(2x)−P(2t) = 2n(P(x)−p) Comparing the rest of the coefficients we get that

P(x) = anxn If we let a = b = (1/an)1/n, then P(a) = P(b) = 1, so P(ab) must also be an

integer But P(ab) = an1 Therefore an= and the proof is complete

Solution 2: Assume P(x) = Pn

i=0aixi Consider the following system of equations

a0 =P(0)

antn+an−1tn−1+· · ·+a0 =P(t)

2nantn+ 2n−1an−1tn−1+· · ·+a0 =P(2t)

nnantn+nn−1an−1tn−1+· · ·+a0 =P(nt)

viewing aktk as variables Note that if P(t) is an integer, then by the hypothesis all the terms

on the right hand side of the equations are integers as well By using Cramer’s rule, we can get that aktk =D/M, where D is an integer and M is the following determinant

1 0 · · · 1 · · · 1 · · · 2n

1 n n2 · · · nn

=

Thus, if we let r be the smallest positive index such thatar 6= 0, we can express each t∈R

with P(t)∈Z in the form (m

M0)

1/r for some integer m, and where M0 =M ×a

r is a constant

We can chooseLlarge enough such thatP|R≥L is injective, and for any largerN, the growth order of the number of values in the form ( m

M0)

1/r is Nr, while the growth order of the number

of integers in [P(L), P(N)] is Nn, so r = n Therefore P(x) is of the form anxn +k The

problem can be finished as in Solution

(188)

Problem Let Z+ be the set of positive integers Determine all functions f :

Z+→Z+

such that a2+f(a)f(b) is divisible by f(a) +b for all positive integers a and b

Answer: The answer isf(n) = n for all positive integers n

Clearly, f(n) = n for all n ∈ Z+ satisfies the original relation We show some possible

approaches to prove that this is the only possible function

Solution First we perform the following substitutions on the original relation: Witha=b = 1, we find that f(1) + 1|f(1)2+ 1, which implies f(1) = 1.

2 Witha= 1, we find that b+ |f(b) + In particular, b≤f(b) for all b ∈Z+.

3 With b = 1, we find that f(a) + 1|a2+f(a), and thus f(a) + 1|a2−1 In particular,

f(a)≤a2−2 for alla ≥2

Now, let p be any odd prime Substituting a =p and b =f(p) in the original relation, we find that 2f(p)|p2+f(p)f(f(p)) Therefore,f(p)|p2 Hence the possible values of f(p) are 1, p

and p2 By (2) above, f(p)≥p and by (3) abovef(p)≤p2−2 So f(p) = pfor all primes p.

Substituting a=p into the original relation, we find thatb+p|p2+pf(b) However, since (b+p)(f(b) +p−b) =p2−b2+bf(b) +pf(b), we have b+p|bf(b)−b2 Thus, for any fixedb

this holds for arbitrarily large primespand therefore we must have bf(b)−b2 = 0, or f(b) = b,

as desired

Solution 2: As above, we have relations (1)-(3) In (2) and (3), forb= we have 3|f(2) + and f(2) + 1|3 These implyf(2) =

Now, using a= we get +b|4 + 2f(b) Let f(b) =x We have +x≡0 (mod b+ 1)

4 + 2x≡0 (mod b+ 2)

From the first equation x≡b (mod b+ 1) so x=b+ (b+ 1)t for some integert ≥0 Then 0≡4 + 2x≡4 + 2(b+ (b+ 1)t)≡4 + 2(−2−t)≡ −2t (mod b+ 2)

Also t≤b−2 because +x|b2−1 by (3)

If b+ is odd, then t≡0 (mod b+ 2) Then t= 0, which implies f(b) =b

If b+ is even, then t ≡ (mod (b+ 2)/2) Then t = or t = (b+ 2)/2 But if t 6= 0, then by definition (b+ 4)/2 = (1 +t) = (x+ 1)/(b+ 1) and since x+ 1|b2−1, then (b+ 4)/2 divides b−1 Therefore b+ 4|10 and the only possibility isb = So for evenb,b 6= we have

f(b) =b

Finally, by (2) and (3), for b = we have 7|f(6) + and f(6) + 1|35 This means f(6) = orf(6) = 34 The later is discarded as, fora= 5, b = 6, we have by the original equation that 11|5(5 +f(6)) Therefore f(n) = n for every positive integer n

Solution 3: We proceed by induction As in Solution 1, we have f(1) = Suppose that

f(n−1) =n−1 for some integer n≥2

With the substitution a =n and b = n−1 in the original relation we obtain that f(n) +

n−1|n2+f(n)(n−1) Since f(n) +n−1|(n−1)(f(n) +n−1), then f(n) +n−1|2n−1.

With the substitution a = n−1 and b = n in the original relation we obtain that 2n− 1|(n−1)2 + (n−1)f(n) = (n−1)(n−1 +f(n)) Since (2n−1, n−1) = 1, we deduce that 2n−1|f(n) +n−1

(189)

Therefore, f(n) +n−1 = 2n−1, which implies the desired f(n) = n

Problem Letm be a fixed positive integer The infinite sequence{an}n≥1 is defined in

the following way: a1 is a positive integer, and for every integer n≥1 we have

an+1 =

(

a2n+ 2m if an<2m

an/2 if an≥2m

For each m, determine all possible values of a1 such that every term in the sequence is an

integer

Answer: The only value of m for which valid values of a1 exist ism = In that case, the

only solutions are a1 = 2` for `≥1

Solution Suppose that for integers m and a1 all the terms of the sequence are integers

For each i ≥ 1, write the ith term of the sequence as = bi2ci where bi is the largest odd

divisor of (the “odd part” of ai) and ci is a nonnegative integer Lemma The sequence b1, b2, is bounded above by 2m

Proof Suppose this is not the case and take an index i for which bi >2m and for which ci is

minimal Sinceai ≥bi >2m, we are in the second case of the recursion Therefore, ai+1 =ai/2

and thus bi+1 =bi >2m and ci+1 =ci−1< ci This contradicts the minimality of ci Lemma The sequence b1, b2, is nondecreasing

Proof If ≥2m, thenai+1 =ai/2 and thus bi+1 =bi On the other hand, if <2m, then

ai+1 =a2i + m =b2

i2

2ci+ 2m,

and we have the following cases:

• If 2ci > m, then ai+1 = 2m(b2i22ci−m+ 1), so bi+1 =b2i22ci−m+ 1> bi

• If 2ci < m, then ai+1 = 22ci(b2i +

m−2ci), so b

i+1 =b2i +

m−2ci > b i

• If 2ci =m, then ai+1 = 2m+1·

b2 i+1

2 , so bi+1 = (b

i + 1)/2≥bi since b2i + 1≡2 (mod 4)

By combining these two lemmas we obtain that the sequenceb1, b2, is eventually constant

Fix an index j such that bk = bj for all k ≥ j Since an descends to an/2 whenever an ≥ 2m,

there are infinitely many terms which are smaller than 2m Thus, we can choose ani > j such

that <2m From the proof of Lemma 2, < 2m and bi+1 = bi can happen simultaneously

only when 2ci =m and bi+1 =bi = By Lemma 2, the sequenceb1, b2, is constantly and

thus a1, a2, are all powers of two Tracing the sequence starting from = 2ci = 2m/2 <2m,

2m/2 →2m+1 →2m →2m−1 →22m−2+ 2m

Note that this last term is a power of two if and only if 2m−2 =m This implies that m

must be equal to When m= and a1 = 2` for `≥1 the sequence eventually cycles through

2,8,4,2, When m= and a1 = the sequence fails as the first terms are 1,5,5/2

Solution 2: Let m be a positive integer and suppose that {an} consists only of positive

(190)

after a small number we have a large one and after a large one we successively divide by until we get a small one

First, we note that {an} is bounded Indeed, a1 turns into a small number after a finite

number of steps After this point, each small number is smaller than 2m, so each large number is smaller than 22m+ 2m Now, since {a

n} is bounded and consists only of positive integers, it

is eventually periodic We focus only on the cycle

Any small number an in the cycle can be writen as a/2 for a large, so an ≥ 2m−1, then

an+1 ≥ 22m−2+ 2m = 2m−2(4 + 2m), so we have to dividean+1 at least m−1 times by until

we get a small number This means that an+m = (a2n+ 2m)/2m−1, so 2m−1|a2n, and therefore

2d(m−1)/2e | an for any small number an in the cycle On the other hand, an ≤ 2m −1, so

an+1 ≤ 22m −2m+1 + + 2m ≤ 2m(2m −1), so we have to divide an+1 at most m times by

two until we get a small number This means that after an, the next small number is either

N =am+n= (a2n/2m−1) + or am+n+1 =N/2 In any case, 2d(m−1)/2e divides N

If m is odd, then x2 ≡ −2 (mod 2d(m−1)/2e) has a solution x=a

n/2(m−1)/2 If (m−1)/2≥

2 ⇐⇒ m≥5 then x2 ≡ −2 (mod 4), which has no solution So if m is odd, then m≤3.

If m is even, then 2m−1 | a2n =⇒ 2d(m−1)/2e | an ⇐⇒ 2m/2 | an Then if an = 2m/2x,

2x2 ≡ −2 (mod 2m/2) ⇐⇒ x2 ≡ −1 (mod 2(m/2)−1), which is not possible for m≥6 So if m

is even, then m≤4

The cases m = 1,2,3,4 are handed manually, checking the possible small numbers in the cycle, which have to be in the interval [2m−1,2m) and be divisible by 2d(m−1)/2e:

• Form = 1, the only small number is 1, which leads to 5, then 5/2

• For m = 2, the only eligible small number is 2, which gives the cycle (2,8,4) The only way to get to is by dividing by 2, so the starting numbers greater than are all numbers that lead to 4, which are the powers of

• Form = 3, the eligible small numbers are and 6; we then obtain 4,24,12,6,44,22,11,11/2 • Form= 4, the eligible small numbers are and 12; we then obtain 8,80,40,20,10, or

12,160,80,40,20,10, , but in either case 10 is not an elegible small number

Problem Let ABC

be a scalene triangle with circumcircle Γ Let M be the midpoint of BC A variable point P

is selected in the line segment AM The circumcircles of triangles BP M and CP M intersect Γ again at points D and E, respectively The linesDP and EP intersect (a second time) the circumcircles to triangles CP M and BP M atX and Y, respectively Prove that as P varies, the circumcircle of 4AXY passes through a fixed point T distinct from A

Solution Let N be the radical center of the circumcircles of triangles ABC, BM P and

CM P The pairwise radical axes of these circles areBD, CE and P M, and hence they concur atN Now, note that in directed angles:

∠M CE =∠M P E=∠M P Y =∠M BY

(191)

It follows thatBY is parallel toCE, and analogously thatCX is parallel toBD Then, ifL

is the intersection ofBY andCX, it follows thatBN CLis a parallelogram SinceBM =M C

we deduce thatLis the reflection of N with respect toM, and thereforeL∈AM Using power of a point from L to the circumcircles of trianglesBP M and CP M, we have

LY ·LB =LP ·LM =LX·LC

Hence, BY XC is cyclic Using the cyclic quadrilateral we find in directed angles: ∠LXY =∠LBC =∠BCN =∠N DE

Since CX kBN, it follows that XY kDE

Let Q and R be two points in Γ such that CQ, BR, and AM are all parallel Then in directed angles:

∠QDB =∠QCB =∠AM B =∠P M B =∠P DB

Then D, P, Q are collinear Analogously E, P, R are collinear From here we get ∠P RQ= ∠P DE = ∠P XY, since XY and DE are parallel Therefore QRY X is cyclic Let S be the radical center of the circumcircle of triangle ABC and the circles BCY X and QRY X This point lies in the lines BC, QR and XY because these are the radical axes of the circles Let

T be the second intersection ofAS with Γ By power of a point from S to the circumcircle of

ABC and the circle BCXY we have

SX ·SY =SB·SC =ST ·SA

(192)

Therefore T is in the circumcircle of triangle AXY Since Q and R are fixed regardless of the choice of P, then S is also fixed, since it is the intersection of QR and BC This implies

T is also fixed, and therefore, the circumcircle of triangle AXY goes through T 6= A for any choice of P

Now we show an alternative way to prove that BCXY and QRXT are cyclic

Solution Let the lines DP and EP meet the circumcircle of ABC again at Q and R, respectively Then ∠DQC∠DBC =∠DP M, so QC kP M Similarly,RB kP M

Now, ∠QCB = ∠P M B = ∠P XC = ∠(QX, CX), which is half of the arc QC in the circumcircle ωC of QXC So ωC is tangent to BS; analogously, ωB, the circumcicle of RY B,

is also tangent to BC Since BR k CQ, the inscribed trapezoid BRQC is isosceles, and by symmetry QR is also tangent to both circles, and the common perpendicular bisector of BR

and CQpasses through the centers ofωB and ωC SinceM B =M C and P M kBRkCQ, the

line P M is the radical axis of ωB and ωC

However, P M is also the radical axis of the circumcircles γB of P M B and γC of P M C

Let CX and P M meet at Z Let p(K, ω) denote the power of a point K with respect to a circumference ω We have

p(Z, γB) =p(Z, γC) =ZX ·ZC =p(Z, ωB) = p(Z, ωC)

Point Z is thus the radical center of γB, γC, ωB, ωC Thus, the radical axesBY, CX, P M meet

atZ From here,

ZY ·ZB =ZC ·ZX ⇒BCXY cyclic

P Y ·P R=P X·P Q⇒QRXT cyclic

We may now finish as in Solution

Problem Consider a 2018×2019 board with integers in each unit square Two unit squares are said to be neighbours if they share a common edge In each turn, you choose some unit squares Then for each chosen unit square the average of all its neighbours is calculated Finally, after these calculations are done, the number in each chosen unit square is replaced by the corresponding average Is it always possible to make the numbers in all squares become the same after finitely many turns?

Answer: No

(193)

Solution Letnbe a positive integer relatively prime to and We may study the whole process modulo n by replacing divisions by 2,3,4 with multiplications by the corresponding inverses modulon If at some point the original process makes all the numbers equal, then the process modulo n will also have all the numbers equal Our aim is to choose n and an initial configuration modulon for which no process modulo n reaches a board with all numbers equal modulo n We split this goal into two lemmas

Lemma There is a 2×3 board that stays constant modulo and whose entries are not

all equal

Proof Here is one such a board:

The fact that the board remains constant regardless of the choice of squares can be checked square by square

Lemma If there is anr×s board with r≥2, s≥2, that stays constant modulo 5, then

there is also akr×ls board with the same property

Proof We prove by a case by case analysis that repeateadly reflecting the r×s with respect

to an edge preserves the property:

• If a cell had neighbors, after reflections it still has the same neighbors

• If a cell with a had neighbors b, c, d, we have by hypothesis that a ≡ 3−1(b+c+d)≡

2(b+c+d) (mod 5) A reflection may add a as a neighbor of the cell and now 4−1(a+b+c+d)≡4(a+b+c+d)≡4a+ 2a≡a (mod 5)

• If a cell with ahad neighbors b, c, we have by hypothesis that a≡2−1(b+c)≡3(b+c) (mod 5) If the reflections add one a as neighbor, now

3−1(a+b+c)≡2(3(b+c) +b+c)≡8(b+c)≡3(b+c)≡a (mod 5)

• If a cell with a had neighbors b, c, we have by hypothesis that a≡2−1(b+c) (mod 5) If the reflections add two a’s as neighbors, now

4−1(2a+b+c)≡(2−1a+ 2−1a)≡a (mod 5)

In the three cases, any cell is still preserved modulo after an operation Hence we can fill in the kr×ls board byk×l copies by reflection

Since 2|2018 and 3|2019, we can get through reflections the following board:

(194)

By the lemmas above, the board is invariant modulo 5, so the answer is no

Problem Determine all the functions f :R→R such that

f(x2+f(y)) =f(f(x)) +f(y2) + 2f(xy) for all real number x and y

Answer: The possible functions are f(x) = for all x and f(x) =x2 for all x.

Solution By substitutingx=y= in the given equation of the problem, we obtain that

f(0) = Also, by substituting y= 0, we get f(x2) =f(f(x)) for anyx.

Furthermore, by letting y= and simplifying, we get

2f(x) =f(x2+f(1))−f(x2)−f(1),

from which it follows that f(−x) = f(x) must hold for every x

Suppose now that f(a) =f(b) holds for some pair of numbers a, b Then, by letting y =a

and y=b in the given equation, comparing the two resulting identities and using the fact that

f(a2) =f(f(a)) =f(f(b)) =f(b2) also holds under the assumption, we get the fact that

f(a) = f(b)⇒f(ax) =f(bx) for any real number x (1) Consequently, if for some a 6= 0, f(a) = 0, then we see that, for any x, f(x) = f(a· x

a) =

f(0· x

a) =f(0) = 0, which gives a trivial solution to the problem

In the sequel, we shall try to find a non-trivial solution for the problem So, let us assume from now on that if a 6= then f(a)6= must hold We first note that since f(f(x)) = f(x2)

for all x, the right-hand side of the given equation equals f(x2) +f(y2) + 2f(xy), which is invariant if we interchange x and y Therefore, we have

f(x2) +f(y2) + 2f(xy) =f(x2+f(y)) =f(y2+f(x)) for every pair x, y (2) Next, let us show that for any x,f(x)≥0 must hold Suppose, on the contrary,f(s) =−t2

holds for some pair s, t of non-zero real numbers By setting x = s, y = t in the right hand side of (2), we get f(s2 +f(t)) = f(t2 +f(s)) = f(0) = 0, so f(t) = −s2 We also have

f(t2) =f(−t2) = f(f(s)) =f(s2) By applying (2) with x=√s2+t2 and y=s, we obtain

f(s2+t2) + 2f(s·√s2+t2) = 0,

and similarly, by applying (2) with x=√s2+t2 and y=t, we obtain

f(s2+t2) + 2f(t·√s2+t2) = 0.

Consequently, we obtain

f(s·√s2+t2) = f(t·√s2+t2).

By applying (1) with a = s√s2+t2, b = t√s2+t2 and x = 1/√s2+t2, we obtain f(s) =

f(t) =−s2, from which it follows that

0 =f(s2+f(s)) =f(s2) +f(s2) + 2f(s2) = 4f(s2),

a contradiction to the fact s2 > 0 Thus we conclude that for all x 6= 0, f(x) > 0 must be

satisfied

Now, we show the following fact

(195)

Letk >0 for which f(k) = We havef(k2) = f(f(k)) = f(1), so by (1),f(1/k) =f(k) = 1, so we may assumek ≥1 By applying (2) withx=√k2−1 and y=k, and usingf(x)≥0,

we get

f(k2−1 +f(k)) = f(k2−1) +f(k2) + 2f(k√k2−1)≥f(k2−1) +f(k2).

This simplifies to 0≥f(k2−1)≥0, so k2−1 = and thus k = 1.

Next we focus on showing f(1) = If f(1) = m ≤ 1, then we may proceed as above by setting x = √1−m and y = to get m = If f(1) = m ≥ 1, now we note that

f(m) =f(f(1)) =f(12) = f(1) =m ≤m2 We may then proceed as above withx=√m2 −m

and y= to show m2 =m and thus m=

We are now ready to finish Let x >0 andm=f(x) Since f(f(x)) =f(x2), thenf(x2) =

f(m) But by (1),f(m/x2) = Therefore m=x2 Forx <0, we have f(x) = f(−x) = f(x2)

as well Therefore, for all x,f(x) =x2

Solution After proving that f(x)>0 forx6= as in the previous solution, we may also proceed as follows We claim that f is injective on the positive real numbers Suppose that

a > b > satisfy f(a) = f(b) Then by setting x = 1/b in (1) we have f(a/b) = f(1) Now, by induction on n and iteratively setting x =a/b in (1) we get f((a/b)n) = for any positive

integer n

Now, let m = f(1) and n be a positive integer such that (a/b)n > m By setting x = p

(a/b)n−m and y= in (2) we obtain that

f((a/b)n−m+f(1)) =f((a/b)n−m) +f(12) + 2f(p(a/b)n−m))≥f((a/b)n−m) +f(1).

Since f((a/b)n) = f(1), this last equation simplifies to f((a/b)n−m) ≤ 0 and thus m =

(a/b)n But this is impossible since m is constant and a/b > 1 Thus, f is injective on the

positive real numbers Since f(f(x)) = f(x2), we obtain that f(x) =x2 for any real value x

Định dạng
Số trang	195
Dung lượng	4,98 MB