2014IMOmath The 55th International Mathematical Olympiad Problems and Solutions

  

  

  

  

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

  a < a < a < ⋯

  1

  2 unique integer such that

  n ≥ 1 a a + ⋯ + a +

  1 n a < ≤ a .

n n+1

  n

   a + +⋯+ a ^{1 a k}

  For we will say that the term is large if . Notice that the last

  k ≥ 1 a a ≥

  k k k a + +⋯+ a ^{1 a k−1} inequality is equivalent to for . If is not large we say that it is

  a ≥ k ≥ 2 a

  k k k−1 such that is small and is small. We need to prove that there exists a unique n ≥ 1 a a n n+1

large. Since we have that is small. We will first prove that there is an integer such

  a > 0 a n

  1 that is large. Assume the contrary, that is small for all . Then for we have

  a a k ≥ 1 k ≥ 2

  n k

  1 < ( + + ⋯ + ) a a a a

  1 k−1 k

  k − 1

  1

  1

• < ( + + ⋯ + ( + + ⋯ + )) a a a a a a

  1 k−2 1 k−2

  k − 1 k − 2

  1 = ( + + ⋯ + ) . a a a

  1 k−2

  k − 2

  Continuing in the same way we obtain

  1

  1

• a < ( + a a + ⋯ + a ) < ⋯ < ( + a a a + ) , a

  k 1 k−3

  1

  2

  3

  k − 2

  3 which means that is a bounded sequence. a

  k This contradiction proves that there is at least one for which is large. Let us assume that

  n a

  n there are two positive integers and such that for which the given two

  n m 1 ≤ n < m inequalities are satisfied. Then we have

  ( + ⋯ + ) + ( a a a + ⋯ + a ) n a + ( a + ⋯ + a )

  n n+1 m n+1 n+1 m

  < ≤ a

  m

  m m n a + ( a + ⋯ + a )

  m m m < = .

  a

  m

  m This completes the proof of the required statement.

  Problem 2

  2 Let be an integer. Consider an chessboard consisting of unit squares. A

  n ≥ 2 n × n n

  

configuration of rooks on this board is peaceful if every row and every column contains exactly

  n

  one rook. Find the greatest positive integer such that, for each peaceful configuration of

  k n

  2 rooks, there is a square which does not contain a rook on any of its unit squares.

  k × k k

  

We will prove that the maximal such is equal to . For a given board and given ,

  k ⌊ √ n − 1⌋ A k

  denote by the square whose top-left corner is (we use the usual matrix

  A^ (k) k × k (i, j)

  i,j notation here).

  − − −−−

  

Assume that √ . Consider a peaceful configuration of rooks in the board

  k ≤ ⌊ n − 1 ⌋ n n × n A

  , and assume that the rook in the first column belongs to the -th row. Let us denote by an

  r s

  integer such that and . Consider now the

  s + k − 1 ≤ n r ∈ {s, s + 1, … , s + k − 1}

  disjoint squares , , , , each of which is

  A^ (k) A^ (k) A^ (k) … A^ (k)

  2,s k+2,s 2k+2,s (k−1)k+2,s

completely contained in the square . There are of them hence each of the rows , , ,

  A k s s + 1 …

  must contain a rook in one of these squares. The cell already contains a rook,

  s + k − 1 (1, r) which contradicts the assumption that the configuration is peaceful.

  − − −−−

  We will now prove that for √ there is a configuration of rooks such that each

  k = ⌊ n − 1 ⌋ k × k

  2 square contains a rook. Assume first that for some . Consider the following

  n = m m ∈ N

  2 configuration of rooks: for each pair we place a rook in the cell

  C (i, j) ∈ {1, 2, … , m}

  

′ ′

. Assume that and are two pairs of integers from

  A (i, j) ( , ) i j

  (i−1)m+j,(j−1)m+i 2 ′ ′ ′ such that . Since we

  {1, 2, … , m} (i − 1)m + j = ( − 1)m + i j j, ∈ {1, 2, … , m} j

  ′ ′ conclude that . This implies that and we have proved that no two rooks are in the

  j = j i = i

  

same row. Similarly we prove that no two rooks are in the same column, and the configuration of

rooks is peaceful.

2 Assume that . We will prove that contains at least one

  p, q ∈ {1, … , n − m + 1} A^ (m)

  p,q

rook from the configuration . First, in order for to be within the board we must have

  C (m) A^

  p,q

  2 . There are pairs of integers

  p, q ≤ m − m + 1

  such that and

  (a, b), (c, d) ∈ {0, 1, … , m − 1} × {1, 2, … , m} p = am + b

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

  q = cm + d S = {(am + c + 1, cm + a + 1), ((a + 1)m + c + 1, cm + a + 2) , (am + c + 2, (c + 1)m + a + 1) , ((a + 1)m + c + 2, (c + 1)m + a + 2)} .

  2 We will prove that there exist such that . We

  (x, y) ∈ {0, 1, … , m − 1} (p + x, q + y) ∈ S

  will form a set of ordered pairs obtained by subtracting from each element of .

  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)} .

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

  T {0, 1, … , m − 1}

  or . We have that and

  c + 1 − b < 0 a + 1 − d < 0 1 − m ≤ c + 1 − b ≤ m − 1 . It suffices to prove that and . 1 − m ≤ a + 1 − d ≤ m − 1 c + 2 − b < m a + 2 − d < m

  

Assume that . Then we would have and which together with

  c + 2 − b = m c = m − 1 b = 1

  2 would imply that . In this case we would have

  cm + d ≤ m − m + 1 d = 1 c + 1 − b ≥ 0

  

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

  a + 1 − b ≥ 0 a + 2 − d < m completes the proof of the statement for the case when is a perfect square. n

  ′ If is not a perfect square, denote by the smallest perfect square larger than . Let

  n n n −−

  ′ ′ ′

  √ . There is a peaceful configuration of rooks on board in which every k = n n × n A k × k

  ′

square contains a rook. Consider a arbitrary rooks in this configuration. For each of these

  n − n

  rooks we remove the entire column and row of the board that contains the rook. We obtain an square with the desired properties.

  n × n

  Problem 3 ∘

  Convex quadrilateral has . Point is the foot of the

  ABCD ∠ABC = ∠CDA = 90 H

  

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

  A BD S T AB AD

  lies inside triangle and

  H SCT

  ∘ ∘

  ∠CHS − ∠CSB = 90 , ∠THC − ∠DTC = 90 .

  Prove that line is tangent to the circumcircle of triangle .

  BD TSH

   It suffices to prove that the circumcenter of lies on . Denote by and the

  △TSH AH O OB

  D intersection of the perpendicular bisectors of and with , respectively. We need to

  TH TS HA

  ′ ′

prove that and coincide. Let us denote by and the intersections of perpendicular

  O O B D

  D B bisectors of and with the lines and , respectively. From the angle-bisector

  HS HT AB AD

  ′ ′ ′ ′ theorem we know that and .

  A O : O H = D A : D H A O : O H = B A : B H

  

D D B B

′ ′ ′ ′

Therefore it suffices to prove that .

  D A : D H = B A : B H

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

  TD △CHT Q

  ∘ ∘ ∘ ∘

  ∠TQC = 180 − ∠THC = 90 − (∠THC − 90 ) = 90 − ∠DTC.

  ∘

Therefore and the center of the circumcirle must belong to . Since the center

  ∠QCT = 90 TD

  ′

belongs to the bisector of we conclude that is the center of the circumcircle of .

  TH D △CHT

  ′ In an analogous way we prove that is circumcenter of .

  △CHS B

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

  △THC △CHS C H D B

  ′ ′ ′ perpendicular bisector of . From and we conclude that

  CH CH ⊥ B D CB ⊥ BB

  ′ . Denote by and the intersections of the perpendicular bisector of

  ∠HCB = ∠X B

  X Y HA B

  ′ ′

with the lines and . Then is the center of the circle circumscribed around .

  B D HA

  X △CHA

1 Hence and

  ∠AXY = AXH = ∠ATH

  2 ′ ∘ ′

  ∠XAD = 90 − ∠AXY − ∠ AH = ∠ADH − ∠AXY D

  ′

  = ∠ACB − ∠ACH = ∠HCB = ∠X B. B

  ′ ′ Therefore is the tangent to the circumcircle of which means that is

  XA △ A D B k(X, XA)

  ′ ′ ′ ′ orthogonal to the circle . Since , the circle is an Apollonius circle of the

  l( A ) D B X ∈ D B k

  ′ ′ ′ ′ ′ ′ points and hence which implies the required statement.

  D B A D : A B = H D : H B

  Day 2 (July 9th, 2014) Problem 4 Points and lie on side of acute-angled triangle such that

  P Q BC ABC ∠PAB = ∠BCA

  

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

  ∠CAQ = ∠ABC M N AP AQ P

  

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

  AM Q AN BM CN the circumcircle of triangle .

  ABC

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

  R BM CN

  X B

  respect to . Then hence . Since and

  P AX∥BM ∠BMP = ∠XAP ∠BAP = ∠ACQ

  we conclude that . Therefore

  ∠APB = ∠AQC = ∠BAC △BAP ∼ △ACQ

  hence . This means that , , , and

  △PAX ∼ △QCN ∠QCN = ∠PAX = ∠BMP R M C P

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

  ∠MRC = ∠MPC = ∠BAC R circumcircle of . △ABC

  Problem 5

  1 For each positive integer , the Bank of Cape Town issues coins of denominations . Given a

  n

  n

finite collection of such coins (of not necessarily different denominations) with total value at most

  1 , prove that it is possible to split this collection into or fewer groups, such that each

  99 + 100

  2 group has total value at most .

  1

  We will use the induction on to prove the following statement: The set of coins whose total

  n

  1

value does not exceed can be partitioned into at most subsets each of which has a total

  n − n

  2 value of at most . The statement is easy to verify for , and assume it is true for .

  1 n ≤ 2 n − 1

  

Assuming that the statement is true for all numbers smaller than , assume that it is false for ,

  n n

  

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

  S

  

1

each there are at most coins of value , as otherwise we could replace coins with a

  k k − 1 k

  

k

  1 coin of value . Moreover, if is even number, there is at most one coin of value since two

  1 k

  k

  1

  1 coins of value could be replaced by one coin of value . k k/2

  1

  1 Let be the largest integer for which . Then the set can be partitioned into at

  m ∈ S S ∖ { }

  m m

  1 most subsets such that the total value of the -th subset is at most . Since the coin

  n γ i

  1

  i m

  1 cannot be added to the th subset we must have . The total sum of coins in

  i 1 − < γ ≤ 1 S

  i m

  1

  1

  1

  1

  1

  which implies that

  n − n − ≥ ∑ γ > + (1 − ) ⋅ n

• is at most , therefore

  i

i

m m m m

  2 . For each let us denote by the number of coins in that have

  m < 2(n − 1) 1 ≤ i ≤ m f(i) S

  1 value . We now have i n−1 n−2

  f(2i) f(2i + 1) ∑ c ≤ + ∑ ∑

  2i 2i + 1

  i=1 i=1 c∈S n−1 n−1 n−1

  1 1 2i

  1 1 2i

•   ≤
• ∑ ∑ ≤ ∑ ( ) 2 2i 2i + 1

  2 2i 2i + 1

  i=2 i=2 i=2

  1

  1 < + n − 2 = (n − 1) − .

  2

  2

  1 This way we have obtained a configuration of coins of total value smaller than and

  (n − 1) −

  2

by induction hypothesis it can be partitioned into subsets each of which has value smaller

  n − 1 than . This is a contradiction.

  1 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

  n n

  lines in general position it is possible to color at least of the lines blue in such a way that

  n √ none of its finite regions has a completely blue boundary.

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

1 B

  3

  1 X m

  X

  l

  l l = { , … , } ≠ ∅ P

  l

  l ∈ N β(l) l P

  β ≤ 2G + B

  7

  1 Q

  7

  XV Y

  XU

  XQ Q

  XP

  ′

  P

  

1

P

  Y U

  X

  1 ≤ i ≤ m

  i

  Q

  i

  P

  i

  i

  2 Q

  X

  l T( ) = △

  m

  Q

  m

  … P

  2

  V l

  ′

  

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.

  √ c n x

  X T(X) T(X)

  2 X

  B

  1 C β B

  2 B k 1 k−2

  … B

  C N C β

  n √ c n

  1

  . Assume that , , , , , , are points on in that order such that for each . According to the lemma we know

  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 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.

  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 .

  

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 .

  

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.

  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 .

  X

  2

  X l

  XQ

  X X

  2 X △XPQ

  X

  P Q

  ′

  XQ

  XP

  XP

  2 PQX X l P Q P Q △PQX

  ′

  P

  ′

  X Y Q

  X P Q

  XQ Y

  XP

3 B G

  that for each there exist two non-triangular blue regions and whose one edge is on

  i P Q l

  i i such that is an edge of and is an edge of . Since all are distinct we

  X P P

  X Q Q P

  i i i i i i i conclude that . Let us denote . Therefore

  β(l) ≥ 2m B = {l ∈ N : P ≠ ∅}

  l

  β = ∑ β(l) = ∑ β(l) + ∑ β(l) ≥ ∑ 2 | | + |N ∖ B| P

  l l∈N l∈B l∈N∖B l∈B

  ≥ ∑ (| | + 1) + |N ∖ B| = P ∑ | | + |B| + |N ∖ B| P

  l l l∈B l∈B

  = B + n − x.

7 We now obtain which implies the inequality

  B + n − x ≤ 2G + B

  3

  4 x n − x ≤ 2G + B ≤ 2(G + B) = 2( ).

  3

  2 The consequence of the last relation is . x ≥ n √

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