Toán Olympic quốc tế 2004 Tiếng Anh

Prove that ABCD is a cyclic quadrilateral if and only if AP = CP.. Every two of them intersect at two distinct points, and allpoints of intersection they determine are distinct.. For a f

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Problems

1.1 The Forty-Fifth IMO

Athens, Greece, July 7–19, 2004

1.1.1 Contest Problems

First Day (July 12)

1 Let ABC be an acute-angled triangle with AB 6= AC The circle withdiameter BC intersects the sides AB and AC at M and N , respectively.Denote by O the midpoint of BC The bisectors of the angles BAC and

M ON intersect at R Prove that the circumcircles of the triangles BM Rand CN R have a common point lying on the line segment BC

2 Find all polynomials P (x) with real coefficients that satisfy the equality

P (a− b) + P (b − c) + P (c − a) = 2P (a + b + c)for all triples a, b, c of real numbers such that ab + bc + ca = 0

3 Determine all m× n rectangles that can be covered with hooks made up

of 6 unit squares, as in the figure:

Rotations and reflections of hooks are allowed The rectangle must becovered without gaps and overlaps No part of a hook may cover areaoutside the rectangle

Second Day (July 13)

2 1 Problems

4 Let n≥ 3 be an integer and t1, t2, , tn positive real numbers such that

n2+ 1 > (t1+ t2+· · · + tn)

1

Show that ti, tj, tk are the side lengths of a triangle for all i, j, k with

1≤ i < j < k ≤ n

5 In a convex quadrilateral ABCD the diagonal BD does not bisect theangles ABC and CDA The point P lies inside ABCD and satisfies

∠P BC = ∠DBA and ∠P DC = ∠BDA

Prove that ABCD is a cyclic quadrilateral if and only if AP = CP

6 We call a positive integer alternate if its decimal digits are alternatelyodd and even Find all positive integers n such that n has an alternatemultiple

Show that ti, tj, tk are the side lengths of a triangle for all i, j, k with

1≤ i < j < k ≤ n

2 A2 (ROM) An infinite sequence a0, a1, a2, of real numbers satisfiesthe condition

an=|an+1− an+2| for every n ≥ 0with a0 and a1 positive and distinct Can this sequence be bounded?

3 A3 (CAN) Does there exist a function s : Q → {−1, 1} such that if xand y are distinct rational numbers satisfying xy = 1 or x + y∈ {0, 1},then s(x)s(y) =−1? Justify your answer

4 A4 (KOR)IMO2 Find all polynomials P (x) with real coefficients thatsatisfy the equality

P (a− b) + P (b − c) + P (c − a) = 2P (a + b + c)for all triples a, b, c of real numbers such that ab + bc + ca = 0

5 A5 (THA) Let a, b, c > 0 and ab + bc + ca = 1 Prove the inequality

3

r1

a+ 6b +

3

r1

b + 6c +

3

r1

c + 6a≤abc1

1.1 Copyright c 3

6 A6 (RUS) Find all functions f : R→ R satisfying the equation

f x2+ y2+ 2f (xy)

= (f (x + y))2 for all x, y∈ R

7 A7 (IRE) Let a1, a2, , an be positive real numbers, n > 1 Denote by

gntheir geometric mean, and by A1, A2, , Anthe sequence of arithmeticmeans defined by Ak = a1 +a 2 +···+a k

k , k = 1, 2, , n Let Gn be thegeometric mean of A1, A2, , An Prove the inequality

8 C1 (PUR) There are 10001 students at a university Some students jointogether to form several clubs (a student may belong to different clubs).Some clubs join together to form several societies (a club may belong

to different societies) There are a total of k societies Suppose that thefollowing conditions hold:

(i) Each pair of students are in exactly one club

(ii) For each student and each society, the student is in exactly one club

of the society

(iii) Each club has an odd number of students In addition, a club with2m + 1 students (m is a positive integer) is in exactly m societies.Find all possible values of k

9 C2 (GER) Let n and k be positive integers There are given n circles

in the plane Every two of them intersect at two distinct points, and allpoints of intersection they determine are distinct Each intersection pointmust be colored with one of n distinct colors so that each color is used

at least once, and exactly k distinct colors occur on each circle Find allvalues of n≥ 2 and k for which such a coloring is possible

10 C3 (AUS) The following operation is allowed on a finite graph: Choose

an arbitrary cycle of length 4 (if there is any), choose an arbitrary edge

in that cycle, and delete it from the graph For a fixed integer n≥ 4, findthe least number of edges of a graph that can be obtained by repeated ap-plications of this operation from the complete graph on n vertices (whereeach pair of vertices are joined by an edge)

11 C4 (POL) Consider a matrix of size n×n whose entries are real numbers

of absolute value not exceeding 1, and the sum of all entries is 0 Let n be

an even positive integer Determine the least number C such that everysuch matrix necessarily has a row or a column with the sum of its entriesnot exceeding C in absolute value

12 C5 (NZL) Let N be a positive integer Two players A and B, takingturns, write numbers from the set{1, , N} on a blackboard A beginsthe game by writing 1 on his first move Then, if a player has written n on

4 1 Problems

a certain move, his adversary is allowed to write n + 1 or 2n (provided thenumber he writes does not exceed N ) The player who writes N wins Wesay that N is of type A or of type B according as A or B has a winningstrategy

(a) Determine whether N = 2004 is of type A or of type B

(b) Find the least N > 2004 whose type is different from that of 2004

13 C6 (IRN) For an n× n matrix A, let Xi be the set of entries in row

i, and Yj the set of entries in column j, 1≤ i, j ≤ n We say that A isgolden if X1, , Xn, Y1, , Yn are distinct sets Find the least integer nsuch that there exists a 2004× 2004 golden matrix with entries in the set{1, 2, , n}

14 C7 (EST)IMO3 Determine all m× n rectangles that can be covered withhooks made up of 6 unit squares, as in the figure:

Rotations and reflections of hooks are allowed The rectangle must becovered without gaps and overlaps No part of a hook may cover areaoutside the rectangle

15 C8 (POL) For a finite graph G, let f (G) be the number of trianglesand g(G) the number of tetrahedra formed by edges of G Find the leastconstant c such that

g(G)3≤ c · f(G)4 for every graph G

16 G1 (ROM)IMO1 Let ABC be an acute-angled triangle with AB 6= AC.The circle with diameter BC intersects the sides AB and AC at M and

N , respectively Denote by O the midpoint of BC The bisectors of theangles BAC and M ON intersect at R Prove that the circumcircles of thetriangles BM R and CN R have a common point lying on the line segmentBC

17 G2 (KAZ) The circle Γ and the line ` do not intersect Let AB be thediameter of Γ perpendicular to `, with B closer to ` than A An arbitrarypoint C 6= A, B is chosen on Γ The line AC intersects ` at D The line

DE is tangent to Γ at E, with B and E on the same side of AC Let

BE intersect ` at F , and let AF intersect Γ at G6= A Prove that thereflection of G in AB lies on the line CF

18 G3 (KOR) Let O be the circumcenter of an acute-angled triangle ABCwith ∠B < ∠C The line AO meets the side BC at D The circumcenters

of the triangles ABD and ACD are E and F , respectively Extend thesides BA and CA beyond A, and choose on the respective extension points

G and H such that AG = AC and AH = AB Prove that the quadrilateral

EF GH is a rectangle if and only if ∠ACB− ∠ABC = 60◦

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19 G4 (POL)IMO5 In a convex quadrilateral ABCD the diagonal BD doesnot bisect the angles ABC and CDA The point P lies inside ABCD andsatisfies

∠P BC = ∠DBA and ∠P DC = ∠BDA

Prove that ABCD is a cyclic quadrilateral if and only if AP = CP

20 G5 (Duˇsan Djuki´c, SMN) Let A1A2 An be a regular n-gon Thepoints B1, , Bn−1are defined as follows:

(i) If i = 1 or i = n− 1, then Bi is the midpoint of the side AiAi+1.(ii) If i 6= 1, i 6= n − 1, and S is the intersection point of A1Ai+1 and

AnAi, then Bi is the intersection point of the bisector of the angle

AiSAi+1 with AiAi+1

Prove the equality

∠A1B1An+ ∠A1B2An+· · · + ∠A1Bn−1An= 180◦

21 G6 (GBR) Let P be a convex polygon Prove that there is a convexhexagon that is contained inP and that occupies at least 75 percent ofthe area ofP

22 G7 (RUS) For a given triangle ABC, let X be a variable point onthe line BC such that C lies between B and X and the incircles of thetriangles ABX and ACX intersect at two distinct points P and Q Provethat the line P Q passes through a point independent of X

23 G8 (Duˇsan Djuki´c, SMN) A cyclic quadrilateral ABCD is given Thelines AD and BC intersect at E, with C between B and E; the diagonals

AC and BD intersect at F Let M be the midpoint of the side CD, andlet N 6= M be a point on the circumcircle of the triangle ABM such thatAN/BN = AM/BM Prove that the points E, F , and N are collinear

24 N1 (BLR) Let τ (n) denote the number of positive divisors of the positiveinteger n Prove that there exist infinitely many positive integers a suchthat the equation

τ (an) = ndoes not have a positive integer solution n

25 N2 (RUS) The function ψ from the set N of positive integers into itself

is defined by the equality

where (k, n) denotes the greatest common divisor of k and n

(a) Prove that ψ(mn) = ψ(m)ψ(n) for every two relatively prime m, n∈N

(b) Prove that for each a∈ N the equation ψ(x) = ax has a solution

6 1 Problems

(c) Find all a∈ N such that the equation ψ(x) = ax has a unique solution

26 N3 (IRN) A function f from the set of positive integers N into itself issuch that for all m, n∈ N the number (m2+ n)2 is divisible by f2(m) +

f (n) Prove that f (n) = n for each n∈ N

27 N4 (POL) Let k be a fixed integer greater than 1, and let m = 4k2

− 5.Show that there exist positive integers a and b such that the sequence(xn) defined by

x0= a, x1= b, xn+2= xn+1+ xn for n = 0, 1, 2, has all of its terms relatively prime to m

28 N5 (IRN)IMO6 We call a positive integer alternate if its decimal digitsare alternately odd and even Find all positive integers n such that n has

an alternate multiple

29 N6 (IRE) Given an integer n > 1, denote by Pn the product of allpositive integers x less than n and such that n divides x2− 1 For each

n > 1, find the remainder of Pn on division by n

30 N7 (BUL) Let p be an odd prime and n a positive integer In thecoordinate plane, eight distinct points with integer coordinates lie on acircle with diameter of length pn Prove that there exists a triangle withvertices at three of the given points such that the squares of its side lengthsare integers divisible by pn+1

Solutions

2.1 Solutions to the Shortlisted Problems of IMO 2004

1 By symmetry, it is enough to prove that t1+ t2> t3 We have

All the summands on the RHS are positive, and therefore the RHS is notsmaller than n2+ T , where T = (t1/t3+ t3/t1− 2) + (t2/t3+ t3/t2− 2)

We note that T is increasing as a function in t3 for t3 ≥ max{t1, t2} If

t1+t2= t3, then T = (t1+t2)(1/t1+1/t2)−1 ≥ 3 by the Cauchy–Schwarzinequality Hence, if t1+ t2 ≤ t3, we have T ≥ 1, and consequently theRHS in (1) is greater than or equal to n2+ 1, a contradiction

Remark In can be proved, for example using Lagrange multipliers, that

if n2+ 1 in the problem is replaced by (n +√

10− 3)2, then the statementremains true This estimate is the best possible

2 We claim that the sequence{an} must be unbounded

The condition of the sequence is equivalent to an> 0 and an+1= an+an−1

or an− an−1 In particular, if an< an−1, then an+1 > max{an, an−1}.Let us remove all ansuch that an< an−1 The obtained sequence (bm)m∈N

is strictly increasing Thus the statement of the problem will follow if weprove that bm+1− bm≥ bm− bm−1for all m≥ 2

Let bm+1 = an+2 for some n Then an+2 > an+1 We distinguish twocases:

(i) If an+1 > an, we have bm = an+1 and bm−1 ≥ an−1 (since bm−1 iseither an−1 or an) Then bm+1− bm = an+2 − an+1 = an = an+1−

an−1= bm− an−1≥ bm− bm−1

(ii) If an+1 < an, we have bm = an and bm−1 ≥ an−1 Consequently,

bm+1−bm= an+2−an= an+1 = an−an−1= bm−an−1≥ bm−bm−1

8 2 Solutions

3 The answer is yes Every rational number x > 0 can be uniquely expressed

as a continued fraction of the form a0+ 1/(a1+ 1/(a2+ 1/(· · · + 1/an)))(where a0 ∈ N0, a1, , an ∈ N) Then we write x = [a0; a1, a2, , an].Since n depends only on x, the function s(x) = (−1)n is well-defined For

x < 0 we define s(x) =−s(−x), and set s(0) = 1 We claim that this s(x)satisfies the requirements of the problem

The equality s(x)s(y) =−1 trivially holds if x + y = 0

Suppose that xy = 1 We may assume w.l.o.g that x > y > 0 Then

x > 1, so if x = [a0; a1, a2, , an], then a0 ≥ 1 and y = 0 + 1/x =[0; a0, a1, a2, , an] It follows that s(x) = (−1)n, s(y) = (−1)n+1, andhence s(x)s(y) =−1

Finally, suppose that x + y = 1 We consider two cases:

(i) Let x, y > 0 We may assume w.l.o.g that x > 1/2 Then thereexist natural numbers a2, , an such that x = [0; 1, a2, , an] =1/(1 + 1/t), where t = [a2, , an] Since y = 1− x = 1/(1 + t) =[0; 1 + a2, a3, , an], we have s(x) = (−1)n and s(y) = (−1)n−1,giving us s(x)s(y) =−1

(ii) Let x > 0 > y If a0, , an ∈ N are such that −y = [a0; a1, , an],then x = [1 + a0; a1, , an] Thus s(y) = −s(−y) = −(−1)n ands(x) = (−1)n, so again s(x)s(y) =−1

4 Let P (x) = a0+ a1x +· · · + anxn For every x ∈ R the triple (a, b, c) =(6x, 3x,−2x) satisfies the condition ab + bc + ca = 0 Then the condition

on P gives us P (3x) + P (5x) + P (−8x) = 2P (7x) for all x, implying thatfor all i = 0, 1, 2, , n the following equality holds:

i = 4 do we have K(i) = 0 It follows that P (x) = a2x2+ a4x4 for somereal numbers a2, a4

It is easily verified that all such P (x) satisfy the required condition

5 By the general mean inequality (M1≤ M3), the LHS of the inequality to

be proved does not exceed

E = √333

3

r1

From ab + bc + ca = 1 we obtain that 3abc(a + b + c) = 3(ab· ac +

ab· bc + ac · bc) ≤ (ab + ac + bc)2 = 1; hence 6(a + b + c) ≤ abc2 Since

f z2+ g(t)

= (f (z))2 for all t, z∈ R with z2≥ 4t (1)Let us set c = g(0) = 2f (0) Substituting t = 0 into (1) gives us

f (z2+ c) = (f (z))2 for all z∈ R (2)

If c < 0, then taking z such that z2+ c = 0, we obtain from (2) that

f (z)2= c/2, which is impossible; hence c≥ 0 We also observe that

u+ √

u 2 +d Therefore every valuefrom some suitably chosen segment [δ, 2δ] can be expressed as g(u)− g(v),with u and v bounded from above by some M

Consider any x, y with y > x≥ 2√M and δ < y2

− x2< 2δ By the aboveconsiderations, there exist u, v ≤ M such that g(u) − g(v) = y2

− x2,i.e., x2 + g(u) = y2+ g(v) Since x2

If k = 0, then f (x) ≡ 0, which is clearly a solution Assume k = 1.Then c = 2f (0) = 2 (because c ≥ 0), which together with (3) implies

f (x) = 1 for all x ≥ 2 Suppose that f(t) = −1 for some t < 2 Then

t− g(t) = 3t + 2 > 4t If also t − g(t) ≥ 0, then for some z ∈ R we have

z2= t−g(t) > 4t, which by (1) leads to f(z)2= f (z2+g(t)) = f (t) =−1,which is impossible Hence t− g(t) < 0, giving us t < −2/3 On theother hand, if X is any subset of (−∞, −2/3), the function f defined by

7 Let us set ck= Ak−1/Ak for k = 1, 2, , n, where we define A0= 0 Weobserve that ak/Ak = (kAk− (k − 1)Ak−1)/Ak = k− (k − 1)ck Now wecan write the LHS of the inequality to be proved in terms of ck, as follows:

C∈ S We shall count |T | in two different ways

Fix a student a and a societyS By (ii), there is a unique club C suchthat (a, C,S) ∈ T Since the ordered pair (a, S) can be chosen in nk ways,

2.1 Copyright c 11

On the other hand, for k = (n− 1)/2 there is a desired configuration withonly one club C that contains all students and k identical societies withonly one element (the club C) It is easy to verify that (i)–(iii) hold

9 Obviously we must have 2 ≤ k ≤ n We shall prove that the possiblevalues for k and n are 2≤ k ≤ n ≤ 3 and 3 ≤ k ≤ n Denote all colorsand circles by 1, , n Let F (i, j) be the set of colors of the commonpoints of circles i and j

Suppose that k = 2 < n Consider the ordered pairs (i, j) such that color jappears on the circle i Since k = 2, clearly there are exactly 2n such pairs

On the other hand, each of the n colors appears on at least two circles,

so there are at least 2n pairs (i, j), and equality holds only if each colorappears on exactly 2 circles But then at most two points receive each ofthe n colors and there are n(n− 1) points, implying that n(n − 1) = 2n,i.e., n = 3 It is easy to find examples for k = 2 and n = 2 or 3

Next, let k = 3 An example for n = 3 is given by F (i, j) ={i, j} for each

1≤ i < j ≤ 3 Assume n ≥ 4 Then an example is given by F (1, 2) ={1, 2}, F (i, i + 1) = {i} for i = 2, , n − 2, F (n − 1, n) = {n − 2, n − 1}and F (i, j) = n for all other i, j > i

We now prove by induction on k that a desired coloring exists for each

n≥ k ≥ 3 Let there be given n circles By the inductive hypothesis, circles

1, 2, , n− 1 can be colored in n − 1 colors, k of which appear on eachcircle, such that color i appears on circle i Then we set F (i, n) ={i, n}for i = 1, , k and F (i, n) ={n} for i > n We thus obtain a coloring ofthe n circles in n colors, such that k + 1 colors (including color i) appear

on each circle i

10 The least number of edges of such a graph is n

We note that deleting edge AB of a 4-cycle ABCD from a connectedand nonbipartite graph G yields a connected and nonbipartite graph, say

H Indeed, the connectedness is obvious; also, if H were bipartite withpartition of the set of vertices into P1 and P2, then w.l.o.g A, C ∈ P1

and B, D∈ P2, so G = H∪ {AB} would also be bipartite with the samepartition, a contradiction

Any graph that can be obtained from the complete n-graph in the scribed way is connected and has at least one cycle (otherwise it would

de-be bipartite); hence it must have at least n edges

Now consider a complete graph with vertices V1, V2, , Vn Let us removeevery edge ViVj with 3 ≤ i < j < n from the cycle V2ViVjVn Then for

i = 3, , n− 1 we remove edges V2Viand ViVn from the cycles V1ViV2Vn

and V1ViVnV2 respectively, thus obtaining a graph with exactly n edges:

V1Vi (i = 2, , n) and V2Vn

11 Consider the matrix A = (aij)ni,j=1such that aij is equal to 1 if i, j≤ n/2,

−1 if i, j > n/2, and 0 otherwise This matrix satisfies the conditions from

12 2 Solutions

the problem and all row sums and column sums are equal to±n/2 Hence

C≥ n/2

Let us show that C = n/2 Assume to the contrary that there is a matrix

B = (bij)ni,j=1 all of whose row sums and column sums are either greaterthan n/2 or smaller than−n/2 We may assume w.l.o.g that at least n/2row sums are positive and, permuting rows if necessary, that the first n/2rows have positive sums The sum of entries in the n/2× n submatrix B0

consisting of first n/2 rows is greater than n2/4, and since each column

of B0 has sum at most n/2, it follows that more than n/2 column sums of

B0, and therefore also of B, are positive Again, suppose w.l.o.g that thefirst n/2 column sums are positive Thus the sums R+ and C+ of entries

in the first n/2 rows and in the first n/2 columns respectively are greaterthan n2/4 Now the sum of all entries of B can be written as

X

aij = R++ C++ X

i>n/2 j>n/2

aij− X

i≤n/2 j≤n/2

a contradiction Hence C = n/2, as claimed

12 We say that a number n∈ {1, 2, , N} is winning if the player who is onturn has a winning strategy, and losing otherwise The game is of type A

if and only if 1 is a losing number

Let us define n0 = N , ni+1 = [ni/2] for i = 0, 1, and let k be suchthat nk = 1 Consider the sets Ai = {ni+1 + 1, , ni} We call a set

Ai all-winning if all numbers from Ai are winning, even-winning if evennumbers are winning and odd are losing, and odd-winning if odd numbersare winning and even are losing

(i) Suppose Ai is even-winning and consider Ai+1 Multiplying any ber from Ai+1 by 2 yields an even number from Ai, which is a losingnumber Thus x∈ Ai+1 is winning if and only if x + 1 is losing, i.e.,

num-if and only num-if it is even Hence Ai+1 is also even-winning

(ii) Suppose Ai is odd-winning Then each k ∈ Ai+1 is winning, since 2k

is losing Hence Ai+1 is all-winning

(iii) Suppose Ai is all-winning Multiplying x ∈ Ai+1 by two is then alosing move, so x is winning if and only if x + 1 is losing Since ni+1islosing, Ai+1is odd-winning if ni+1is even and even-winning otherwise

We observe that A0 is even-winning if N is odd and odd-winning wise Also, if some Ai is even-winning, then all Ai+1, Ai+2, are even-winning and thus 1 is losing; i.e., the game is of type A The game is of type

other-B if and only if the sets A0, A1, are alternately odd-winning and winning with A0 odd-winning, which is equivalent to N = n0, n2, n4, all being even Thus N is of type B if and only if all digits at the oddpositions in the binary representation of N are zeros

all-Since 2004 = 11111010100 in the binary system, 2004 is of type A Theleast N > 2004 that is of type B is 100000000000 = 211= 2048 Thus theanswer to part (b) is 2048