Toán Olympic quốc tế 2007 Tiếng Anh

Call a group of competitors a clique if each two of themare friends.. The number of members of a clique is called its size.. Given that, in this competition, the largest size of a clique

Duˇsan Djuki´c Vladimir Jankovi´c

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Problems

1.1 The Forty-Eighth IMO

Hanoi, Vietnam, July 19–31, 2007

1.1.1 Contest Problems

First Day (July 25)

1 Real numbers a1, a2, , an are given For each i (1≤ i ≤ n) define

di= max{aj | 1 ≤ j ≤ i} − min{aj | i ≤ j ≤ n}

and let d = max{di| 1 ≤ i ≤ n}

(a) Prove that, for any real numbers x1≤ x2≤ · · · ≤ xn,

max{|xi− ai| | 1 ≤ i ≤ n} ≥ d2 (∗)(b) Show that there are real numbers x1 ≤ x2 ≤ · · · ≤ xn such thatequality holds in (∗)

2 Consider five points A, B, C, D and E such that ABCD is a parallelogramand BCED is a cyclic quadrilateral Let ℓ be a line passing through

A Suppose that ℓ intersects the interior of the segment DC at F andintersects line BC at G Suppose also that EF = EG = EC Prove that

ℓ is the bisector of angle DAB

3 In a mathematical competition some competitors are friends Friendship

is always mutual Call a group of competitors a clique if each two of themare friends (In particular, any group of fewer than two competitors is aclique.) The number of members of a clique is called its size

Given that, in this competition, the largest size of a clique is even, provethat the competitors can be arranged in two rooms such that the largestsize of a clique contained in one room is the same as the largest size of aclique contained in the other room

2 1 Problems

Second Day (July 26)

4 In triangle ABC the bisector of angle BCA intersects the circumcircleagain at R, the perpendicular bisector of BC at P , and the perpendicularbisector of AC at Q The midpoint of BC is K and the midpoint of AC

is L Prove that the triangles RP K and RQL have the same area

5 Let a and b be positive integers Show that if 4ab− 1 divides (4a2

− 1)2,then a = b

6 Let n be a positive integer Consider

S =(x, y, z)| x, y, z ∈ {0, 1, , n}, x + y + z > 0

as a set of (n + 1)3

− 1 points in three-dimensional space Determine thesmallest possible number of planes, the union of which contains S butdoes not include (0, 0, 0)

1.1.2 Shortlisted Problems

1 A1 (NZL)IMO1Given a sequence a1, a2, , an of real numbers, for each

i (1≤ i ≤ n) define

di= max{aj : 1≤ j ≤ i} − min{aj : i≤ j ≤ n}

and let d = max{di: 1≤ i ≤ n}

(a) Prove that for arbitrary real numbers x1≤ x2≤ · · · ≤ xn,

max{|xi− ai| : 1 ≤ i ≤ n} ≥ d2 (1)(b) Show that there exists a sequence x1≤ x2≤ · · · ≤ xn of real numberssuch that we have equality in (1)

2 A2 (BUL) Consider those functions f : N→ N which satisfy the tion

condi-f (m + n)≥ f(m) + f(f(n)) − 1, for all m, n ∈ N

Find all possible values of f (2007)

3 A3 (EST) Let n be a positive integer, and let x and y be positive realnumbers such that xn+ yn= 1 Prove that

4 A4 (THA) Find all functions f : R+→ R+ such that

f (x + f (y)) = f (x + y) + f (y)for all x, y∈ R+

Prove that the sequence a(n) is bounded.

6 A6 (POL) Let a1, a2, , a100 be nonnegative real numbers such that

8 C1 (SER) Let n be an integer Find all sequences a1, a2, , an 2 +n

satisfying the following conditions:

(i) ai∈ {0, 1} for all 1 ≤ i ≤ n2+ n;

(ii) ai+1+ ai+2+· · · + ai+n < ai+n+1+ ai+n+2+· · · + ai+2n for all 0≤

i≤ n2− n

9 C2 (JAP) A unit square is dissected into n > 1 rectangles such thattheir sides are parallel to the sides of the square Any line, parallel to aside of the square and intersecting its interior, also intersects the interior

of some rectangle Prove that in this dissection, there exists a rectanglehaving no point on the boundary of the square

10 C3 (NET) Find all positve integers n, for which the numbers in the set

S ={1, 2, , n} can be colored red and blue, with the following conditionbeing satisfied: the set S× S × S contains exactly 2007 ordered triples(x, y, z) such that

(i) x, y, z are of the same color and

(ii) x + y + z is divisible by n

11 C4 (IRN) Let A0 ={a1, , an} be a finite sequence of real numbers.For each k≥ 0, from the sequence Ak = (x1, , xn) we construct a newsequence Ak+1in the following way:

(i) We choose a partition {1, , n} = I ∪ J, where I and J are twodiscjoint sets, such that the expression

X

4 1 Problems

attains the smallest possible value (We allow the sets I or J to beempty; in this case the corresponding sum is 0.) If there are severalsuch partitions, one is chosen arbitrarily

(ii) We set Ak+1= (y1, , yn), where yi= xi+ 1 if i∈ I, and yi= xi− 1

13 C6 (RUS)IMO3 In a mathematical competition some competitors arefriends; friendship is always mutual Call a group of competitors a clique

if each two of them are friends The number of members in a clique iscalled its size

It is known that the largest size of a clique is even Prove that the petitors can be arranged in two rooms such that the largest size of a clique

com-in one room is the same as the largest size of a clique com-in the other room

14 C7 (AUT) Let α < 3−2√5 be a positive real number Prove that thereexist positive integers n and p such that p > α· 2n and for which onecan select 2p pairwise distinct subsets S1, , Sp, T1, , Tp of the set{1, 2, , n} such that Si∩ Tj6= ∅ for all 1 ≤ i, j ≤ p

15 C8 (UKR) Given a convex n-gon P in the plane, for every three vertices

of P , consider the triangle determined by them Call such a triangle good

if all its sides are of unit length Prove that there are not more than 2n/3good triangles

16 G1 (CZE) IMO4In a triangle ABC the bisector of angle BCA intersectsthe circumcircle again at R, the perpendicular bisector of BC at P , andthe perpendicular bisector of AC at Q The midpoint of BC is K and themidpoint of AC is L Prove that the triangles RP K and RQL have thesame area

17 G2 (CAN) Given an isosceles triangle ABC, assume that AB = AC.The midpoint of the side BC is denoted by M Let X be a variable point

on the shorter arc M A of the circumcircle of triangle ABM Let T be thepoint in the angle domain BM A for which ∠T M X = 90◦and T X = BX.Prove that ∠M T B− ∠CT M does not depend on X

18 G3 (UKR) The diagonals of a trapezoid ABCD intersect at point P Point Q lies between the parallel lines BC and AD such that ∠AQD =

∠CQB, and the line CD separates the points P and Q Prove that

∠BQP = ∠DAQ

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19 G4 (LUX)IMO2Consider five points A, B, C, D and E such that ABCD

is a parallelogram and BCED is a cyclic quadrilateral Let ℓ be a linepassing through A Suppose that ℓ intersects the interior of the segment

DC at F and intersects line BC at G Suppose also that EF = EG = EC.Prove that ℓ is the bisector of angle DAB

20 G5 (GBR) Let ABC be a fixed triangle, and let A1, B1, C1 be themodpoints of sides BC, CA, AB respectively Let P be a variable point

on the circumcircle Let lines P A1, P B1, P C1meet the circumcircle again

at A′, B′, C′respectively Assume that the points A, B, C, A′, B′, C′aredistinct, and lines AA′, BB′, CC′ form a triangle Prove that the area ofthis triangle does not depend on P

21 G6 (USA) Let ABCD be a convex quadrilateral, and let points A1,

B1, C1, and D1 lie on sides AB, BC, CD, and DA respectively Considerthe areas of triangles AA1D1, BB1A1, CC1B1, and DD1C1; let S be thesum of the two smallest ones, and let S1be the area of the quadrilateral

23 G8 (POL) A point P lies on the side AB of a convex quadrilateralABCD Let ω be the incircle of the triangle CP D, and let I be its incenter.Suppose that ω is tangent to the incircles of triangles AP D and BP C atpoints K and L, respectively Let the lines AC and BD meet at E, andlet the lines AK and BL meet at F Prove that the points E, I, and Fare colinear

24 N1 (AUT) Find all pairs (k, n) of positive integers for which 7k

− 3n

divides k4+ n2

25 N2 (CAN) Let b, n > 1 be integers Suppose that for each k > 1 thereexists an integer ak such that b− an

k is divisible by k Prove that b = An

for some integer A

26 N3 (NET) Let X be a set of 10000 integers, none of which is divisible

by 47 Prove that there exists a 2007-element subset Y of X such that

a− b + c − d + e is not divisible by 47 for any a, b, c, d, e ∈ Y

27 N4 (POL) For every integer k≥ 2, prove that 23k divides the number

6 1 Problems

but 23k+1 does not

28 N5 (IRN) Find all surjective functions f : N→ N such that for every

m, n∈ N and every prime p, the number f(m + n) is divisible by p if andonly if f (m) + f (n) is divisible by p

29 N6 (GBR) IMO5 Let k be a positive integer Prove that the number(4k2

− 1)2 has a positive divisor of the form 8kn− 1 if and only if k iseven

30 N7 (IND) For a prime p and a positive integer n, denote by νp(n) theexponent of p in the prime factorization of n! Given a positive integer dand a finite set{p1, , pk} of primes, show that there are infinitely manypositive integers n such that d|νp i(n) for all 1≤ i ≤ k

Solutions

8 2 Solutions

2.1 Solutions to the Shortlisted Problems of IMO 2007

1 (a) Assume that d = dm for some index m, and let k and l (k≤ m ≤ l)

be the indeces such that dm= ak− al Then dm = ak− al ≤ (ak −

xk) + (xl− al) hence ak− xk≥ d/2 or xl− al≥ d/2 The claim followsimmediately

(b) Let Mi = max{aj : 1 ≤ j ≤ i} and mi = min{aj : i ≤ j ≤ n} Set

2 Placing n = 1 we get f (m + 1) ≥ f(m) + f(f(1)) − 1 ≥ f(m) hencethe function is non-decreasing Let n0 be the smallest integer such that

f (n0) > 1 If f (n) = n + k for some k, n ≥ 1 then placing m = 1 givesthat f (f (n)) = f (n + k)≥ f(k) + f(f(n)) − 1 which implies f(k) = 1

We immediately get k < n0 Choose maximal k0 such that there exists

n ∈ N for which f(n) = n + k0 Then we have 2n + k0 ≥ f(2n) ≥

f (n)+f (f (n))−1 = n+k0+f (n+k0)−1 ≥ n+k0+f (n)−1 = 2n+(2k0−1)hence 2k0−1 ≤ k0, or k0≤ 1 Therefore f(n) ≤ n+1 and f(2007) ≤ 2008.Now we will prove that f (2007) can be any of the numbers 1, 2, , 2008.Define the functions

3 The inequality 1+t1+t24 <1t holds for all t∈ (0, 1) because it is equivalent to

0 < t4− t3− t + 1 = (1 − t)(1 − t3) Applying it to t = xk and summingover k = 1, , n we get Pnk=11+x 2k

1+x 4k < Pnk=1 1

x k = x n

−1

x n (x−1) = xn (1−x)yn Writing the same relation for y and multiplying by this one gives thedesired inequality

4 Notice that f (x) > x for all x Indeed, f (x + f (y)) 6= f(x + y) and if

f (y) < y for some y, setting x = y− f(y) yields to a contradiction.Now we will prove that f (x)− x is injective If we assume that f(x) − x =

f (y)− y for some x 6= y we would have x + f(y) = y + f(x) hence

f (x+y)+f (y) = f (x+y)+f (x) implying f (x) = f (y), which is impossible.From the functional equation we conclude that f (f (x) + f (y))− (f(x) +

f (y)) = f (x+y), hence f (x)+f (y) = f (x′)+f (y′) whenever x+y = x′+y′

In particular, we have f (x) + f (y) = 2f (x+y2 )

Now we will prove that f is injective If f (x) = f (x + h) for some h > 0then f (x) + f (x + 2h) = 2f (x + h) = 2f (x) hence f (x) = f (x + 2h), and

2.1 Copyright c 9

by induction f (x + nh) = f (x) Therefore, 0 < f (x + nh)− (x + nh) =

f (x)− x − nh for every n, which is impossible

We now have f (f (x) + f (y)) = f (f (x) + y) + f (y) = 2f (f (x)2 + y) and bysymmetry f (f (x) + f (y)) = 2f (f (y)2 + x) Hence f (x)2 + y = f (y)2 + x, thus

f (x)− 2x = c for some c ∈ R The functional equation forces c = 0 It iseasy to verify that f (x) = 2x satisfies the given relation

5 Defining a(0) to be 0 the relations in the problem remain to hold It follows

by induction that a(n1+ n2+· · ·+ nk)≤ 2a(n1) + 22a(n2) +· · ·+ 2ka(nk)

We also have a(n1+ n2+· · · + n2 i)≤ 2a(n1+· · · + n2 i−1) + 2a(n2 i−1 +1+

· · · + n2 i)≤ · · · ≤ 2i(a(n1) +· · · + a(n2 i)) For integer k ∈ [2i, 2i+1) wehave

− 1 we get a(N) ≤ 22α

· 22P

n2n−1−α(c−2)(n−1) Thus,choosing any α > 1

c−2 would give us the sequence τn for which the seriesP

n2n−1−α(c−2)(n−1)is bounded, which proves the required statement

6 Using the Cauchy-Schwarz inequality we can bound the left-hand side inthe following way: 13[a1(a2

100+ 2a1a2) + a2(a2+ 2a2a3) +· · · + a100(a2

99+2a100a1)] ≤ 1

3 a2+· · · + a2

100

1/2

·P100k=1(a2+ 2ak+1ak+2)1/2 (the deces are modulo 100) It suffices to show

ak+2)≤ (a4+ 2a2a2 + 2a2a2 ) + 4a2 a2 The required inequality

Let S =Pni=0Pnj=0Pnk=0δiδjδkP (i, j, k) By the construction of P weknow that P (0, 0, 0)6= 0 and P (i, j, k) = 0 for all other choices of i, j, k ∈{0, 1, , n} Therefore S = δ3P (0, 0, 0) On the other hand expanding P

8 Let Skm = ak+ ak+1+· · · + am Since Sn1 < Sn+12n < · · · < Sn 2

+n

n 2 +1 andsince each of these n + 1 numbers belongs to {0, 1, , n} we have that

Sin+1(i+1)n = i We immediately get a1 = a2 = · · · = an = 0 and an 2 +1 =

an 2 +2 =· · · = an 2 +n = 1 For every 0 ≤ k ≤ n, consider the sequence

lk = (Sk+nk+1, Sk+n+1k+2n , , Sk+nk+n22−n+1) The sequence is strictly increasing,and its elements are from the set{0, 1, 2, , n} Let m be the numberthat doesn’t appear in lk, and Ukthe total sum of the elements of lk Since

a1+· · ·+an 2 +n= Sn

1+S2n n+1+· · ·+Snn22+1+n= n(n+1)2 = Sk+Uk+Sk+nn2+n2 +1=

ak+(s+1)n = ak+snif either s + k < n or s + k≥ n + 1 If s + k = n then

2.1 Copyright c 11

Sk+snk−1+(s+1)n= s while Sk+sn+1k + (s + 1)n = s + 1 hence ak+(s+1)n= 1and ak+sn = 0 Now, by induction we can easily get that for 1≤ u ≤ nand 0≤ v ≤ n:

au+vn=



0, if u + v≤ n,

1, if u + v > n

It is easy to verify that the above sequence satisfies the required properties

9 Assume the contrary Consider the minimal such dissection of the squareABCD (i.e the dissection with the smallest number of rectangles) Notwo rectangles in this minimal dissection can share an edge Let AM N P

be the rectangle contianing the vertex A, and let U BV W be the rectanglecontaining B Assume that M N≤ BV Let MXY Z be another rectanglecontaining the point M (this one could be the same as U BV W ) We caneither have M N > M Z or M Z > M N In the first case the rectanglecontaining the point Z would have to touch the side CD (it can’t touch

BC because W U ≥ NM > MZ) The line MN doesn’t intersect any ofthe interiors of the rectangles Contradiction

If M Z > M N consider the rectangle containing the point N It can’ttouch AD because it can’t share the entire side with AM N P Hence ithas to touch CD and, again, M N would be a line that doesn’t intersectany of the interiors Contradiction

10 Let T ={(x, y, z) ∈ S × S × S : x + y + z is divisible by n} For any pair(x, y)∈ S × S there exists unique z ∈ S such that (x, y, z) ∈ T , hence

|T | = n2 Let M⊆ T be the set of those triples that have all elements ofthe same color Denote by R and B the sets of red and blue numbers andassume that the number r of red numbers is not less than n/2 Considerthe following function F : T \ M → R × B: If (x, y, z) ∈ T \ M, then

F (x, y, z) is defined to be one of the pairs (x, y), (y, z), (z, x) that belongs

to R× B (there exists exactly one such pair) For each element (p, q) ∈

R× B there is unique s ∈ S for which n|p + q + s Then F (p, q, s) =

F (s, p, q) = F (q, s, p) = (p, q) Hence|T \ M| = 3|R × B| = 3r(n − r) and

|T | = n2

− 3r(n − r) = n2

− 3rn + 3r2

It remains to solve n2− 3nr + 3r2 = 2007 in the set N× N First of all,

n = 3k for some k∈ N Therefore 9k2− 9kr + 3r2= 2007 and we see that

3|r Let r = 3s The equation becomes k2

− 3kr + 3r2 = 223 From ourassumption r≥ n/2 we get 223 = k2

−3kr+3r2= (k−r)(k−2r)+r2

≤ r2.Furthermore 4· 223 = (2k − 3r)2 + 3r2 ≥ 3r2

≥ 3 · 223 Hence r ∈{15, 16, 17} For r = 15 and r = 16, 4 · 223 − 3r2 is not a perfect square,and for r = 17 we get (2k− 3r)2= 25 hence 2k− 3 · 17 = ±5 Both k = 28and k = 23 lead to solutions (n, r) = (84, 51) and (n, r) = (69, 51)

11 Denote by ak,1, ak,2, , ak,n the elements of Ak, and let Qk =Pni=1a2

k,i.Assume the contrary, that |ak,i| < n/2 for all k, i This means that thenumber of elements inSk∈NAk is finite Hence there are different p, q∈ Nsuch that Ap= Aq For any I⊆ {1, 2, , n}, denote Sk(I) =Pi∈Iak,i

3... n2− 3nr + 3r2 = 2007 in the set N× N First of all,

n = 3k for some k∈ N Therefore 9k2− 9kr + 3r2= 2007 and we see that

3|r Let... X be a set of 10000 integers, none of which is divisible

by 47 Prove that there exists a 2007- element subset Y of X such that

a− b + c − d + e is not divisible by 47 for any a, b,