2014IMOmath The 55th International Mathematical Olympiad Problems and Solutions

(1)

Log In Register

IMO (Table of contents)

The 55th International Mathematical Olympiad:

Problems and Solutions

Day 1 (July 8th, 2014)

Problem 1

Let be an infinite sequence of positive integers. Prove that there exists a unique integer such that

Hide solution

For we will say that the term is large if . Notice that the last inequality is equivalent to for . If is not large we say that it is

small. We need to prove that there exists a unique such that is small and is large. Since we have that is small. We will first prove that there is an integer such that is large. Assume the contrary, that is small for all . Then for we have

Continuing in the same way we obtain

which means that is a bounded sequence.

IMOmath Olympiads Book Training IMO Results Forum

<

< ⋯

1

+ ⋯ +

_n

a

n

a

n+1

k ≥ 1

a

k ≥ 1

k ≥ 2

a

k

<

=

( +

+ ⋯ +

)

1 k − 1

a

0

a

1

a

k−1

( +

+ ⋯ +

+

( +

+ ⋯ +

))

1 k − 1

a

0

a

1

a

k−2

1 k − 2

<

( +

+ ⋯ +

) < ⋯ < ( +

+

+ ) ,

a

k

_{k − 2}

1 a

0

a

1

a

k−3

1 ₃

a

0

a

1

a

2

a

3

(2)

inequalities are satisfied. Then we have

This completes the proof of the required statement.

Problem 2

Let be an integer. Consider an chessboard consisting of unit squares. A configuration of rooks on this board is peaceful if every row and every column contains exactly one rook. Find the greatest positive integer such that, for each peaceful configuration of rooks, there is a square which does not contain a rook on any of its unit squares.

Hide solution

We will prove that the maximal such is equal to . For a given board and given , denote by the square whose top-left corner is (we use the usual matrix notation here).

Assume that . Consider a peaceful configuration of rooks in the board , and assume that the rook in the first column belongs to the -th row. Let us denote by an

integer such that and . Consider now the

disjoint squares , , , , each of which is

completely contained in the square . There are of them hence each of the rows , , , must contain a rook in one of these squares. The cell already contains a rook, which contradicts the assumption that the configuration is peaceful.

We will now prove that for there is a configuration of rooks such that each square contains a rook. Assume first that for some . Consider the following configuration of rooks: for each pair we place a rook in the cell

. Assume that and are two pairs of integers from

such that . Since we

conclude that . This implies that and we have proved that no two rooks are in the same row. Similarly we prove that no two rooks are in the same column, and the configuration of rooks is peaceful.

Assume that . We will prove that contains at least one

rook from the configuration . First, in order for to be within the board we must have . There are pairs of integers

such that and

. We know that a rook is located on each of the squares in the set

(3)

We will prove that there exist such that . We will form a set of ordered pairs obtained by subtracting from each element of .

We need to prove that at least one element of belongs to . Assume that

or . We have that and

. It suffices to prove that and .

Assume that . Then we would have and which together with

would imply that . In this case we would have

and , which is not possible. In a similar way we prove that . This

completes the proof of the statement for the case when is a perfect square.

If is not a perfect square, denote by the smallest perfect square larger than . Let . There is a peaceful configuration of rooks on board in which every

square contains a rook. Consider a arbitrary rooks in this configuration. For each of these rooks we remove the entire column and row of the board that contains the rook. We obtain an

square with the desired properties.

Problem 3

Convex quadrilateral has . Point is the foot of the

perpendicular from to . Points and lie on sides and , respectively, such that lies inside triangle and

Prove that line is tangent to the circumcircle of triangle .

Hide solution

It suffices to prove that the circumcenter of lies on . Denote by and the intersection of the perpendicular bisectors of and with , respectively. We need to prove that and coincide. Let us denote by and the intersections of perpendicular bisectors of and with the lines and , respectively. From the angle-bisector

theorem we know that and .

Therefore it suffices to prove that .

Assume that the line intersects the circumcircle of at point . Then

Therefore and the center of the circumcirle must belong to . Since the center belongs to the bisector of we conclude that is the center of the circumcircle of . In an analogous way we prove that is circumcenter of .

(x, y) ∈ {0, 1, … , m − 1}

2

_{(p + x, q + y) ∈ S}

T

(p, q)

S

T = {(c + 1 − b, a + 1 − d), (m + c + 1 − b, a + 2 − d) ,

(c + 2 − b, m + a + 1 − d) , (m + c + 2 − b, m + a + 2 − d)} .

T

{0, 1, … , m − 1}

2

c + 1 − b < 0 a + 1 − d < 0

1 − m ≤ c + 1 − b ≤ m − 1

1 − m ≤ a + 1 − d ≤ m − 1

c + 2 − b < m

a + 2 − d < m

c + 2 − b = m

c = m − 1

b = 1

cm + d ≤

m

2

− m + 1

_{d = 1}

_{c + 1 − b ≥ 0}

a + 1 − b ≥ 0

a + 2 − d < m

n

′

_n

k = n

∠ABC = ∠CDA = 90

∘

_H

A BD

S

T

AB

AD

H

SCT

∠CHS − ∠CSB =

90

∘

,

∠THC − ∠DTC =

₉₀

∘

.

BD

TSH

△TSH

AH

O

D

OB

TH

TS

HA

O

D

O

B

′

D

′

HS

HT

AB

AD

A

O

D

:

O

D

H =

D

′

A :

D

′

H

A

O

B

:

O

B

H =

B

′

A :

B

′

H

A :

H =

A :

H

D

′

− ∠THC =

90

∘

− (∠THC −

90

∘

) =

90

∘

− ∠DTC.

∠QCT = 90

∘

TD

TH

D

′

_△CHT

(4)

The circumcircles of and intersect at and , hence the line is the

perpendicular bisector of . From and we conclude that

. Denote by and the intersections of the perpendicular bisector of with the lines and . Then is the center of the circle circumscribed around .

Hence and

Therefore is the tangent to the circumcircle of which means that is orthogonal to the circle . Since , the circle is an Apollonius circle of the

points and hence which implies the required statement.

Day 2 (July 9th, 2014)

Problem 4

Points and lie on side of acute-angled triangle such that

and . Points and lie on lines and , respectively, such that is

the midpoint of , and is the midpoint of . Prove that lines and intersect on the circumcircle of triangle .

Hide solution

Let us denote by the intersection of and . Let be the point symmetric to with

respect to . Then hence . Since and

we conclude that . Therefore

hence . This means that , , , and

belong to a circle which implies that . Thus belongs to the

circumcircle of .

△THC

△CHS

C

H

D

_HA

B

′

_D

′

_HA

_X

_△CHA

∠AXY = AXH = ∠ATH

1₂

∠XAD

′

₌

=

90 − ∠AXY − ∠ AH = ∠ADH − ∠AXY

∘

_D

′

∠ACB − ∠ACH = ∠HCB = ∠X B.

B

′

XA

△ A

′

∠PAB = ∠BCA

∠CAQ = ∠ABC

M

N

AP

AQ

P

AM

Q

AN

BM

CN

ABC

R

BM

CN

X

B

P

AX∥BM

∠BMP = ∠XAP

∠BAP = ∠ACQ

∠APB = ∠AQC = ∠BAC

△BAP ∼ △ACQ

△PAX ∼ △QCN

∠QCN = ∠PAX = ∠BMP

R M C

P

∠MRC = ∠MPC = ∠BAC

R

(5)

Problem 5

For each positive integer , the Bank of Cape Town issues coins of denominations . Given a finite collection of such coins (of not necessarily different denominations) with total value at most

, prove that it is possible to split this collection into or fewer groups, such that each group has total value at most .

Hide solution

We will use the induction on to prove the following statement: The set of coins whose total value does not exceed can be partitioned into at most subsets each of which has a total value of at most . The statement is easy to verify for , and assume it is true for . Assuming that the statement is true for all numbers smaller than , assume that it is false for , and consider the configuration with the smallest number of coins that can not be partitioned in the desired way. Denote by the set of the coins in this configuration. We can be certain that for each there are at most coins of value , as otherwise we could replace coins with a

is at most , therefore which implies that

. For each let us denote by the number of coins in that have value . We now have

This way we have obtained a configuration of coins of total value smaller than and by induction hypothesis it can be partitioned into subsets each of which has value smaller than . This is a contradiction.

Problem 6

A set of lines in the plane is in general position if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite area; we call these its finite regions. Prove that for all sufficiently large , in any set of

(6)

Note: Results with replaced by will be awarded points depending on the value of the constant .

Hide solution

Given a configuration of lines, denote by the maximal number of lines that can be colored in blue so that the conditions of the problem are satisfied, and denote by one such coloring. Let be the set of lines that are non-blue in the coloring . Each intersection point of two blue lines will be called blue point. Each region in this coloring with exactly one non-blue edge will be called

blue region. Let be the total number of blue regions.

For every blue region consider the internal angles corresponding to its blue

vertices and assign the value of to each of those angles. All other angles are assigned value .

The sum of the values of angles in each blue region is implying that the sum of all labels of all angles in the coloring is equal to .

For each blue point we calculate the sum of all labeled angles whose vertex is . If the sum of the labels is smaller than , we call such a point a good point. Otherwise, the point will be called

bad.

Lemma Assume that is a bad point. Then all four angles around belong to blue regions. Exactly one of these blue regions is a triangle (we will denote it by ) and at most two are quadrilaterals. The non-blue line that contains the edge of also contains the edge of two other blue regions whose vertex is .

Proof. If none of the regions is a triangle, then all four of the angles have value of at most hence their sum cannot be bigger than .

Assume that is a blue triangle with vertex . Let be the non-blue line that contains and . Denote by and the regions different from whose edges are and . Let us denote by the fourth region with vertex . The regions and cannot be triangles because no other line passes through . Assume that is a blue triangle. Assume and are vertices of this triangle that belong to lines and . Since the segments and cannot be intersected by any of the other lines, we conclude that is not blue. Similarly, is not blue, and the sum of labels of angles with vertex is exactly , contradicting the assumption that is bad. Therefore, is the only blue triangle with vertex . Consequently, each region that has vertex must be a blue region (otherwise would not be bad).

Let be the non-blue line that contains the edge of and let us denote by and the intersections of with the lines and . Since is blue, the segment is intersected by a blue line. Similarly, the segment is intersected by another blue line, and cannot be a quadrilateral.

A consequence of the previous lemma is that the sum of all labels of angles with vertex at any blue point is at most .

Let be the number of bad points and the number of good points. Then we have .

For denote by the number of blue regions with an edge on and by the set of bad points that are vertices of blue triangles with edges on . Fix a line for which

. Assume that , , , , , , are points on in that

order such that for each . According to the lemma we know

(7)

that for each there exist two non-triangular blue regions and whose one edge is on such that is an edge of and is an edge of . Since all are distinct we

conclude that . Let us denote . Therefore

We now obtain which implies the inequality

The consequence of the last relation is .

IMO (Table of contents)