The 53rd International Mathematical Olympiad: Problems and Solutions

The 53rd International Mathematical Olympiad: Problems and Solutions Day 1 (July 10th, 2012) Problem 1 (Evangelos Psychas, Greece)

Given a triangle , let be the center of the excircle opposite to the vertex . This circle is tangent to lines , ,

ABC J A AB AC

and at , , and , respectively. The lines and meet at , and the lines and meet at . Let

BC K L M BM JF F KM CJ G S

be the intersection of and , and let be the intersection of and . Prove that is the midpoint of .

AF BC T AG BC M BC

1 We have hence is parallel to the bisector of . Therefore and KM ⊥ BJ BM ∠ABC ∠BMK = ∠ABC

. Denote by the intersection of and . From the triangle we derive:

∠FMB = ∠ACB

X KM FJ FXM

1 ∘

∠XFM = 90 − ∠FMB − ∠BMK = ∠BAC.

1 The points and are symmetric with respect to the line hence , K M FJ ∠KFJ = ∠JFM = ∠BAC = ∠KAJ

2 ∘

therefore , , , and belong to a circle which implies that and . The quadrilateral

K J A F ∠JFA = ∠JKA = 90 KM∥AS

is a trapezoid and from we obtain . Similarly, we get . Since

SKMA ∠SMK = ∠AKM SM = AK AL = TM (tangents from to the excircle) we get . AK = AL A SM = TM

Problem 2 (Angelo di Pasquale, Australia) Let , , , be positive real numbers that satisfy . Prove that

a a … a a ⋅ a ⋯ a = 1

2 3 n

2 3 n n ( + 1) a ⋅ ( + 1) a ⋯ ( + 1) a > n .

2 3 n

The inequality between arithmetic and geometric mean implies

k k

1 1 k k k

( + 1) a ( + a + ⋯ + ) ≥ + = k ⋅ a ⋅ = ⋅ . a k k k k k−1 k−1 k − 1 k − 1 k − 1 (k − 1) (k − 1)

1 The inequality is strict unless . Multiplying analogous inequalities for , , , yields a = k = 2 3 … n k k−1

3 4 n

2 3 n n ( + 1) a ⋅ ( + 1) a ⋯ ( + n) a > ⋅ ⋅ ⋯ ⋅ a a ⋯ a = n .

2 3 n

3 n−1 (n − 1)

1 a … a

1 a > 2 n k k−1

for at least one integer .

k ∈ {2, … , n}

Problem 3 (David Arthur, Canada)

The liar’s guessing game is a game played between two players and . The rules of the game depend on two positive

A B integers and which are known to both players. k n

At the start of the game the player chooses integers and with . Player keeps secret, and truthfully

A x N 1 ≤ x ≤ N A x

tells to the player . The player now tries to obtain information about by asking player questions as follows: each

N B B x A

question consists of specifying an arbitrary set of positive integers (possibly one specified in some previous question),

B S

and asking whether belongs to . Player may ask as many questions as he wishes. After each question, player

A x S B A

must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that,

among any consecutive answers, at least one answer must be truthful.

k + 1

After has asked as many questions as he wants, he must specify a set of at most positive integers. If , then

B X n x ∈ X B

wins; otherwise, he loses. Prove that:

k (a) If then has a winning strategy. n ≥ 2 B k

(b) There exists a positive integer such that for every there exists an integer for which cannot

k k ≥ k n ≥ 1.99 B guarantee a victory.

The game can be reformulated in an equivalent one: The player chooses an element from the set (with )

A x S |S| = N

and the player asks the sequence of questions. The -th question consists of choosing a set and player

B j B D j ⊆ S A C

selecting a set . The player has to make sure that for every the following relation holds:

P ∈ { , Q Q } A j ≥ 1 j j j x ∈ P ∪ P ∪ ⋯ ∪ P . j j+1 j+k The player wins if after a finite number of steps he can choose a set with such that .

B X |X| ≤ n x ∈ X k ′ ′

(a) It suffices to prove that if then the player can determine a set with such that

N ≥ 2 + 1 B S ⊆ S | | ≤ N − 1 S ′ . x ∈ S n k−1 C k−1

Assume that . In the first move selects any set such that and . After

N ≥ 2 + 1 B D ⊆ S | | ≥ D 2 | D | ≥

receiving the set from , makes the second move. The player selects a set such that

P A B B D ⊆ S

1 2 k−2 k−2

C C C

and . The player continues this way: in the move he/she chooses a set

| D 2 ∩ P | ≥ 2 | D ∩ P | ≥

2 B j D j

1 k−j k−j C C C such that and .

| D ∩ P | ≥ 2 | D ∩ P | ≥

2 j j j j

In this way the player has obtained the sets , , , such that . Then chooses the set

B P

1 P 2 … P k ( P 1 ∪ ⋯ ∪ P k ) ≥ 1 B

to be a singleton containing any element outside of . There are two cases now:

D P ∪ ⋯ ∪ P k+1

1 k ∘ C ′

The player selects . Then can take and the statement is proved.

1 A P = D B S = S ∖ D k+1 k+1 k+1

∘

The player selects . Now the player repeats the previous procedure on the set

2 A P k+1 = D k+1 B

to obtain the sequence of sets , , , . The following inequality holds:

S = S ∖ D P P … P 1 k+1 k+2 k+3 2k+1

| ∖ ( S P ⋯ P )| ≥ 1, 1 k+2 2k+1 k

since . However, now we have

| | ≥ S

1 C P ∪ P ∪ ⋯ ∪ P ) ≥ 1, k+1 k+2 2k+1

∣∣( ∣∣ ′ and we may take .

S = P ∪ ⋯ ∪ P k+1 2k+1

(b) Let and be two positive integers such that . Let us choose such that

p q 1.99 < p < q < 2 k k p q k k

( ) ≤ 2 ⋅ (1 − ) and p − 1.99 > 1. q

2 k k k ≥ k |S| ∈ (

1.99 , ) p A P P …

(based on sets , , provided by ) such that for each the following relation holds:

1 D 2 … B j P ∪ P ∪ ⋯ ∪ P = S. j j+1 j+k

Assuming that , the player will maintain the following sequence of -tuples:

S = {1, 2, … , N} A N j j j

∞ j+1

. Initially we set . After the set is selected then we define

(x ) = ( , , … , x x x ) x = x = ⋯ = x = 1 P x j j=0

2 N

2 N j

based on as follows:

x 1, if i ∈ P j j+1 x = { i j q ⋅ , x if i ∉ P . j i j

N k

The player can keep from winning if for each pair . For a sequence , let us define . It

A B x ≤ q (i, j) x T(x) = ∑ x i i i=1 j k

suffices for player to make sure that for each .

A T ( ) ≤ x q j k k Notice that .

T ( ) = N ≤ x p < q j j k

We will now prove that given such that , and a set the player can choose

x T ( ) ≤ x q D A P ∈ { D , D } j+1 j+1 j+1 j+1 j+1 k

such that . Let be the sequence that would be obtained if , and let be the sequence that

T ( x ) ≤ q y P = D z j+1 j+1

would be obtained if . Then we have

P = D j+1 j+1 j

T (y) = ∑ q + | x D | j+1 _C i i∈D _j+1 j

C T (z) = ∑ q + x . i ∣∣D j+1 ∣∣ i∈D _j+1

Summing up the previous two equalities gives:

j k+1 k T (y) + T (z) = q ⋅ T ( ) + N ≤ x q + , hence p k q p k k min {T (y) , T (z)} ≤ ⋅ q ≤ , q +

2 because of our choice of . k

Day 2 (July 11th, 2012) Problem 4 (Liam Baker, South Africa)

Find all functions such that, for all integers , , with the following equality holds:

f : Z → Z a b c a + b + c = 0

2 f(a + f(b + f(c = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a). ) ) )

2 Substituting yields which implies . Now we can place , to a = b = c = 0 3f(0 = 6f(0 ) ) f(0) = 0 b = −a c = 0

2 obtain , or, equivalently which implies . f(a + f(−a = 2f(a)f(−a) ) ) (f(a) − f(−a)) = 0 f(a) = f(−a)

Assume now that for some . Then for any we have hence

f(a) = 0 a ∈ Z b a + b + (−a − b) = 0

2 , which is equivalent to , or . f(a + f(b + f(a + b = 2f(b)f(a + b) ) ) ) (f(b) − f(a + b)) = 0 f(a + b) = f(b) Therefore if for some , then is a periodic function with period . f(a) = 0 a ≠ 0 f a

Placing and in the original equation yields . Choosing we get

b = a c = 2a f(2a) ⋅ (f(2a) − 4f(a)) = 0 a = 1 or . f(2) = 0 f(2) = 4f(1) If , then is periodic with period 2 and we must have for all odd . It is easy to verify that for each the function

c ∈ Z 0, 2 ∣ n, f(x) = { c, 2 ∤ n satisfies the conditions of the problem.

2 Assume now that and that . Assume that holds for all (it f(2) = 4f(1) f(1) ≠ 0 f(i) = i ⋅ f(1) i ∈ {1, 2, … , n}

certainly does for ). We place , , in the original equation to obtain:

i ∈ {0, 1, 2} a = 1 b = n c = −n − 1

2 f(1 + f(1 + f(n + 1 = 2 f(1 + 2( + 1)f(n + 1)f(1) ) n ) ) n ) n

2 ⇔ (f(n + 1) − (n + 1 f(1)) ⋅ (f(n + 1) − (n − 1 f(1)) = 0. ) )

2 If then setting , , and in the original equation yields f(n + 1) = (n − 1 f(1) a = n + 1 b = 1 − n c = −2

)

4 2(n − 1 f(1 + 16f(1 = 2 ⋅ 4 ⋅ 2(n − 1 f(1 + 2 ⋅ (n − 1 f(1)

) ) ) ) ) )

which implies hence . Therefore . Placing , , and into the original

(n − 1 = 1 ) n = 2 f(3) = f(1) a = 1 b = 3 c = 4

equation implies that or . If we get

f(4) = 0 f(4) = 4f(1) = f(2) f(4) ≠ 0

2 f(2 + f(2 + f(4 = 2f(2 + 4f(2)f(4) ) ) ) )

hence . We already have that and this implies that , which is impossible according to

f(4) = 4f(2) f(4) = f(2) f(2) = 0

our assumption. Therefore and the function has period . Then , ,

f(4) = 0 f 4 f(4k) = 0 f(4k + 1) = f(4k + 3) = c and . It is easy to verify that this function satisfies the requirements of the problem. f(4k + 2) = 4c

2 The remaining case is for all , or for some . This function satisfies the given f(n) = n f(1) n ∈ N f(n) = cn c ∈ Z condition.

0, 2 ∣ n,

2 Thus the solutions are: for some ; for some ; and f(x) = cx c ∈ Z f(x) = { c ∈ Z

c, 2 ∤ n 0, 4 ∣ n,

⎧ ⎪ for some . f(x) = ⎨

c, 2 ∤ n, c ∈ Z ⎩ ⎪

4c, n ≡ 2 (mod 4)

Problem 5 (Josef Tkadlec, Czech Republic)

∘

Given a triangle , assume that . Let be the foot of the perpendicular from to , and let be any

ABC ∠C = 90 D C AB

point of the segment . Let and be the points on the segments and such that and

CD K L AX BX BK = BC , respectively. Let be the intersection of and . AL = AC M AL BK Prove that .

MK = ML

2 Since the triangles and are similar hence . Let be the point A L = A C = AD ⋅ AB ALD ABL ∠ALD = ∠XBA R

∘

on the extension of over point such that . Since we conclude

DC C DX ⋅ DR = BD ⋅ AD ∠BDX = ∠RDA = 90

hence , therefore and the points , , , and belong to

△RAD ∼ △BXD ∠XBD = ∠ARD ∠ALD = ∠ARD R A D L ∘

a circle. This implies that hence . Analogously we prove that

∠RLA = 90 R L = A R − A L = A R − A C

2 2 ∘

and . Since we have , therefore

R K = B R − B C ∠RKB = 90 RC ⊥ AB A R − A C = B R − B C ∘

hence . Together with we conclude hence

R L = R K RL = RK ∠RLM = ∠RKM = 90 △RLM ≅△LKM . MK = ML

Problem 6 (Dušan Djukić, Serbia) Find all positive integers for which there exist non-negative integers , , , such that

n a a … a

1 2 n

⋯ + = + + ⋯ + = 1.
1, a

1 a m+1
1 a m+1
1 a m+1
2, a
1, a
1, a ^j

_{2<3, a
2 a m+2
1 a 2m+j
1 a 2m+j
3 a m+2
(R ( ) − R ( )) , ∑}

∑ j=2

2 a 2m+j

∑ j=2

2 a m+2

L ( ) = L(a) − − + + + + 6 ⋅ = 1. a ′

if j = m + 2 if j ∈ {2m + 2, 2m + 3, 2m + 4, 2m + 5, 2m + 6} if j ∈ {4m + 4, 4m + 6, 4m + 8, 4m + 10, 4m + 12} if j ∈ {4m + 3, 4m + 5, 4m + 7, 4m + 9, 4m + 11, 4m + 13} otherwise.

m+2 , a j

m+2

⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪

= a ′ j

′ 4m + 13

L ( ) = R ( ) = 1 a ′ a

′ 4m+13

2 a

4m + 11, a ′ 4m+13

3 c ¹ d k

1 … , … , c k d k d

1 , d

2m + j R ( ) = + ⋯ + . , c

4m + j a 2m+j

2m + j, a ′ 4m+2j

′ 2m+j

6 , a

j=2

4m + 13 a m+2 m + 2

′ 4m+11

4m + 9, , a

′ 4m+9

4m + 7, , a

′ 4m+7

4m + 5, , a

′ 4m+5

4m + 3, , a

, a ′ 4m+3

′ m+2 m + 2,

′ , a

R ( ) − R(a) = R ( ) − R ( ) a

R ( ) = R(a) = 1 a ′

′

1 a

3 M

2 a ¹

2 a n L(a) = + + ⋯ + and R(a) = + + ⋯ + .

1 a

1 … a n a = ( , , … , ) a

2 n ≡ 1 n ≡ 2 n ∈ N 4k + 1 4k + 2 k ∈ N a

2 n n(n+1)

3 M−a n n(n+1)

3 M−a ²

3 M−a ¹

M = max{ , … , } a 1 a n = 1 ⋅ + 2 ⋅ + ⋯ + n ⋅ ≡ 1 + 2 + ⋯ + n =

3 a n

3 a ² n

3 a ¹

2 a n

2 a ²

2 a ¹

Let . Then we have (mod ). Therefore, the number must be odd and hence (mod 4) or (mod 4).

We will now prove that each of the form or (for some ) there exist integers , , with

the described property. For a sequence let us introduce the following notation: Assume that for there exists a sequence of non-negative integers with . Consider the sequence defined in the following way: Then we have This implies that if the statement holds for , then it holds for .

Assume now that the statement holds for for some , and assume that is the

corresponding sequence of non-negative integers. We will construct a following sequence that satisfies thus proving that the statement holds for . Define: We now have It remains to verify that . We write where For each we have

2 a ²

2 a n

= ( , , … , ) a ′ a

4m+2 n

1 a

2m + 1 2m + 2 n = 4m + 2 m ≥ 2 a = ( , … , ) a

2m + 2

3 a _m+1 m + 1

a ′ m + 1

2 a _m+1

L ( ) = L(a) − + 2 ⋅ = 1 and R ( ) = R(a) − + + = 1. a ′

m+1 if j ∉ {m + 1, 2m + 2} if j ∈ {m + 1, 2m + 2}.

′ j , a j

′ n+1 = { a

1 a

′

′ a

L(a) = R(a) = 1 = ( , … , ) a

1 a n

3 a n n = 2m + 1 a = ( , … , ) a

3 a ² n

3 a ¹

3 c k j ∈ {2, 3, 4, 5, 6}

′ ′ a a 2m + j 4m + 2j 2m + j a 2m+j

2m+j 4m+2j R ( ) − R ( ) = + − = 0. a

1 +1 2m+j a 2m+j a 2m+j

3 2m + j

3 2m + j, 4m + j

′

The first term in the expression for is also equal to because

R ( ) − R(a) a ′ ′ ′ ′ ′ ′ ′ a , a , a , a , a , a , a a m+2 4m+3 4m+5 4m+7 4m+9 4m+11 4m+13 m+2

R ( ) − R ( ) = m + 2 m + 2, 4m + 3, 4m + 5, 4m + 7, 4m + 9, 4m + 11, 4m + 13

6 m + 2 4m + 2j + 1 m + 2 − + = ∑ = 0. a m+2 a m+2 +2 a m+2 +3

3 3 j=1

3 ′

Thus and the statement holds for . It remains to verify that there are sequences of lengths , , , ,

R ( ) = 0 a 4m + 13 1 5 9 13

and . One way to choose these sequences is:

17 (1), (2, 1, 3, 4, 4), (2, 3, 3, 3, 3, 4, 4, 4, 4), (2, 3, 3, 4, 4, 4, 5, 4, 4, 5, 4, 5, 5), (3, 2, 2, 4, 4, 5, 5, 6, 5, 6, 6, 6, 6, 6, 6, 6, 5).

The 53rd International Mathematical Olympiad: Problems and Solutions