The 53rd International Mathematical Olympiad: Problems and Solutions

(1)

Log In Register

IMO (Table of contents) PDF

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 , , and at , , and , respectively. The lines and meet at , and the lines and meet at . Let be the intersection of and , and let be the intersection of and . Prove that is the midpoint of .

Hide solution

We have hence is parallel to the bisector of . Therefore and

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

The points and are symmetric with respect to the line hence ,

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

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

(tangents from to the excircle) we get .

Problem 2 (Angelo di Pasquale, Australia)

Let , , , be positive real numbers that satisfy . Prove that

Hide solution

The inequality between arithmetic and geometric mean implies

The inequality is strict unless . Multiplying analogous inequalities for , , , yields

IMOmath Olympiads Book Training IMO Results Forum

ABC

J

A

AB AC

BC K L

M

BM

JF

F

KM

CJ

G

S

AF

BC

T

AG

BC

M

BC

KM ⊥ BJ

BM

∠ABC

∠BMK = ∠ABC

1

2

∠FMB = ∠ACB

1

2

X

KM

FJ

FXM

∠XFM =

90

∘

− ∠FMB − ∠BMK = ∠BAC.

1

2 K

M

FJ

∠KFJ = ∠JFM = ∠BAC = ∠KAJ

1

2

K J A

F

∠JFA = ∠JKA = 90

∘

KM∥AS

SKMA

∠SMK = ∠AKM

SM = AK

AL = TM

AK = AL

A

SM = TM

a

2

a

3

… a

n

a

2

⋅

a

3

⋯

a

n

= 1

⋅

⋯

>

.

( + 1)

a

2 2

( + 1)

a

3 3

( + 1)

a

n n

n

=

≥

⋅

=

⋅ .

( + 1)

a

k k

( +

a

k

_{k − 1}

1 +

_{k − 1}

1 + ⋯ +

_{k − 1}

1 )

k

_a

k

_{(k − 1)}

1

_k−1

k

(k − 1)

k−1

a

k

=

a

k _k−11

k = 2 3 … n

⋅

⋯

>

⋅

⋯

⋅

⋯

=

.

( + 1)

a

2 2

( + 1)

a

3 3

( + n)

a

n n

2

1

3

2

4

3

n

(2)

The inequality is strict because at least one of , , has to be greater than or equal than and thus holds

for at least one integer .

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 integers and which are known to both players.

At the start of the game the player chooses integers and with . Player keeps secret, and truthfully tells to the player . The player now tries to obtain information about by asking player questions as follows: each question consists of specifying an arbitrary set of positive integers (possibly one specified in some previous question), and asking whether belongs to . Player may ask as many questions as he wishes. After each question, player 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.

After has asked as many questions as he wants, he must specify a set of at most positive integers. If , then wins; otherwise, he loses. Prove that:

(a) If then has a winning strategy.

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

Hide solution

The game can be reformulated in an equivalent one: The player chooses an element from the set (with ) and the player asks the sequence of questions. The -th question consists of choosing a set and player selecting a set . The player has to make sure that for every the following relation holds:

The player wins if after a finite number of steps he can choose a set with such that .

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

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

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

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

such that and .

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

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

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

The player selects . Now the player repeats the previous procedure on the set to obtain the sequence of sets , , , . The following inequality holds:

since . However, now we have

and we may take .

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

a

2

… a

n

1 a

k

>

_k−11

k ∈ {2, … , n}

A

B

k

n

A

x

N

1 ≤ x ≤ N

A

x

N

B

x

A

B

S

A

x

S

B

A

k + 1

B

X

n

x ∈ X

B

n ≥ 2

k

_B

k

0

k ≥ k

0

n ≥ 1.99

k

B

A

x

S

|S| = N

B

j

B

D

j

⊆ S

A

∈ { ,

}

P

j

Q

j

Q

Cj

A

j ≥ 1

x ∈

P

j

∪

P

j+1

∪ ⋯ ∪

P

j+k

.

B

X

|X| ≤ n

x ∈ X

N ≥

2

k

+ 1

_B

_S

′

_{⊆ S}

_{| | ≤ N − 1}

_S

′

x ∈ S

′

N ≥

2

n

+ 1

_B

_D

_{⊆ S}

1

| | ≥

D

1

2

k−1

|

D

C1

| ≥

2

k−1

P

1

A B

B

D

2

⊆ S

|

D

2

∩

P

1C

| ≥

2

k−2

|

D

C2

∩

P

1C

| ≥

2

k−2

B

j

D

j

|

D

j

∩

P

jC

| ≥

2

k−j

|

D

Cj

∩

P

jC

| ≥

2

k−j

B

P

1

P

2

… P

k

(

P

1

∪ ⋯ ∪

P

k

)

C

≥ 1

B

D

k+1

P

1

∪ ⋯ ∪

P

k

1

∘

_A

_P

₌

k+1

D

Ck+1

B

S

′

= S ∖

D

k+1

2

∘

_A

_P

₌

k+1

D

k+1

B

= S ∖

S

1

D

k+1

P

k+2

P

k+3

… P

2k+1

| ∖ (

S

1

P

k+2

⋯

P

2k+1

)| ≥ 1,

| | ≥

S

1

2

k

≥ 1,

∣∣(

P

k+1

∪

P

k+2

∪ ⋯ ∪

P

2k+1

)

C

∣∣

=

∪ ⋯ ∪

S

′

_P

_k+1

_P

_2k+1

p

q

1.99 < p < q < 2

k

0

≤ 2 ⋅ (1 − )

and

−

> 1.

(3)

We will prove that for every if then there is a strategy for the player to select sets , , (based on sets , , provided by ) such that for each the following relation holds:

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

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

based on as follows:

The player can keep from winning if for each pair . For a sequence , let us define . It suffices for player to make sure that for each .

Notice that .

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

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

would be obtained if . Then we have

Summing up the previous two equalities gives:

because of our choice of .

Day 2 (July 11th, 2012)

Problem 4 (Liam Baker, South Africa)

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

Hide solution

Substituting yields which implies . Now we can place , to

obtain , or, equivalently which implies .

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

, which is equivalent to , or .

Therefore if for some , then is a periodic function with period .

Placing and in the original equation yields . Choosing we get

or .

k ≥ k

0

|S| ∈ (

1.99

k

, )

p

k

A

P

1

P

2

…

D

1

D

2

…

B

j

∪

∪ ⋯ ∪

= S.

P

j

P

j+1

P

j+k

S = {1, 2, … , N}

A

N

(x

)

∞

= ( , , … ,

)

j=0

x

j1

x

j2

x

jN

x

01

=

x

02

= ⋯ =

x

0N

= 1

P

j

x

j+1

x

j

= {

x

j+1_i

1,

q ⋅ ,

x

j_i

if i ∈ P

j

if i ∉

P

j

.

A

B

x

j_i

≤

q

k

_{(i, j)}

_x

_{T(x) = ∑}

N

i=1

x

i

A

T ( ) ≤

x

j

_q

k

_j

T ( ) = N ≤

x

0

_p

k

<

_q

k

x

j

_{T ( ) ≤}

_x

j

_q

k

_D

j+1

A

P

j+1

∈ {

D

j+1

,

D

Cj+1

}

T (

x

j+1

) ≤

_q

k

_y

_P

_j+1

₌

_D

_j+1

_z

=

P

j+1

D

Cj+1

T (y) =

∑

q + |

|

i∈DC j+1

x

j_i

D

j+1

T (z) =

∑

q +

.

i∈Dj+1

x

j_i

∣∣D

C j+1

∣∣

T (y) + T (z) = q ⋅ T ( ) + N ≤

x

j

q

k+1

+ , hence

p

k

min {T (y) , T (z)} ≤

q

₂

⋅

q

k

+

p

k

≤ ,

2 q

k

0

f : Z → Z

a b c

a + b + c = 0

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

)

2

₎

2

₎

2

a = b = c = 0

3f(0 = 6f(0

)

2

₎

2

_{f(0) = 0}

_{b = −a c = 0}

f(a + f(−a = 2f(a)f(−a)

)

2

₎

2

_{(f(a) − f(−a))}

2

_{= 0}

_{f(a) = f(−a)}

f(a) = 0

a ∈ Z

b

a + b + (−a − b) = 0

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

)

2

₎

2

₎

2

_{(f(b) − f(a + b))}

2

_{= 0}

_{f(a + b) = f(b)}

f(a) = 0

a ≠ 0

f

a

b = a

c = 2a

f(2a) ⋅ (f(2a) − 4f(a)) = 0

a = 1

(4)

If , then is periodic with period 2 and we must have for all odd . It is easy to verify that for each the function

satisfies the conditions of the problem.

Assume now that and that . Assume that holds for all (it

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

If then setting , , and in the original equation yields

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

equation implies that or . If we get

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

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

and . It is easy to verify that this function satisfies the requirements of the problem.

The remaining case is for all , or for some . This function satisfies the given

condition.

Thus the solutions are: for some ; for some ; and

for some .

Problem 5 (Josef Tkadlec, Czech Republic)

Given a triangle , assume that . Let be the foot of the perpendicular from to , and let be any point of the segment . Let and be the points on the segments and such that and

, respectively. Let be the intersection of and .

Prove that .

Hide solution

Since the triangles and are similar hence . Let be the point

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

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

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

and . Since we have , therefore

hence . Together with we conclude hence

.

Problem 6 (Dušan Djukić, Serbia)

Find all positive integers for which there exist non-negative integers , , , such that

f(2) = 0

f

f(n) = f(1)

n

c ∈ Z

f(x) = {

0,

c,

2 ∣ n,

2 ∤ n

f(2) = 4f(1)

f(1) ≠ 0

f(i) =

i

2

⋅ f(1)

_{i ∈ {1, 2, … , n}}

i ∈ {0, 1, 2}

a = 1 b = n c = −n − 1

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

)

2

_n

4

₎

2

₎

2

_n

2

₎

2

_n

2

⇔

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

)

2

₎

2

f(n + 1) = (n − 1 f(1)

)

2

_{a = n + 1 b = 1 − n}

_{c = −2}

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

)

4

₎

2

₎

2

₎

2

₎

2

₎

4

(n − 1 = 1

)

2

_{n = 2}

_{f(3) = f(1)}

_{a = 1 b = 3}

_{c = 4}

f(4) = 0 f(4) = 4f(1) = f(2) f(4) ≠ 0

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

)

2

₎

2

₎

2

₎

2

f(4) = 4f(2)

f(4) = f(2)

f(2) = 0

f(4) = 0

f

4 f(4k) = 0 f(4k + 1) = f(4k + 3) = c

f(4k + 2) = 4c

f(n) =

n

2

f(1)

_{n ∈ N}

_{f(n) = cn}

2

_{c ∈ Z}

f(x) = cx

2

_{c ∈ Z f(x) = {}

0,

c,

2 ∣ n,

2 ∤ n

c ∈ Z

f(x) =

⎧

⎩

⎨

⎪

0,

c,

4c,

4 ∣ n,

2 ∤ n,

n ≡ 2 (mod 4)

c ∈ Z

ABC

∠C = 90

∘

D

C AB

X

CD

K

L

AX

BX

BK = BC

AL = AC

M

AL

BK

MK = ML

A

L

2

= A

_C

2

= AD ⋅ AB

_ALD

_ABL

_{∠ALD = ∠XBA}

_R

DC

C

DX ⋅ DR = BD ⋅ AD

∠BDX = ∠RDA = 90

∘

△RAD ∼ △BXD

∠XBD = ∠ARD

∠ALD = ∠ARD

R A D

L

∠RLA = 90

∘

R

L

2

= A

_R

2

− A

_L

2

= A

_R

2

− A

_C

2

R

K

2

= B

_R

2

− B

_C

2

_{∠RKB = 90}

∘

_{RC ⊥ AB}

_A

_R

2

_{− A}

_C

2

_{= B}

_R

2

_{− B}

_C

2

R

L

2

= R

_K

2

_{RL = RK}

_{∠RLM = ∠RKM = 90}

∘

_{△RLM ≅△LKM}

MK = ML

(5)

Hide solution

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

(6)

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

Thus and the statement holds for . It remains to verify that there are sequences of lengths , , , , and . One way to choose these sequences is:

IMO (Table of contents) PDF