Solutions to the Shortlisted Problems of IMO 2009

4.50 Solutions to the Shortlisted Problems of IMO 2009

1. Notice that b 2009 ,r 2009 , and w 2009 always form a triangle. In order to prove this fact it suffices to show that the largest of these three numbers (say w 2009 ) is less than the sum of the other two. Since there are b i and r j such that w 2009 ,b i , and r j form a triangle, we get w 2009 <b i +r j ≤b 2009 +r 2009 . This proves that k ≥ 1. We will now provide an example in which only one triple (b i ,r i ,w i ) forms a triangle. Let us set w i = 2i, r i = i, for i = 1, 2, . . . , 2009; and let us define b i =i for i ∈ {1,2,...,2008}, and b i = 2i for i = 2009. We will now form triangles

T 1 , ...,T 2009 such that each has one blue, one red, and one white side. For j =

1 , 2, . . . , 2008 we can define the triangle T j to be the one with sides w j ,r j +1 , and

b j because 2 j < j + j + 1. The sides of T 2009 can be w 2009 = 4018, r 1 = 1, and

b 2009 = 4018. The conditions of the problem are clearly satisfied.

2. The given conditions on a, b, and c imply that abc (a + b + c) = ab + bc + ca. Applying the inequality (X + Y + Z) 2 ≥ 3(XY +Y Z + ZX) to X = ab, Y = bc, Z = ca gives us (ab + bc + ca) 2 ≥ 3abc(a + b + c) = 3(ab + bc + ca), which means that ab + bc + ca ≥ 3. p From 2a +b+c≥2 (a + b)(a + c) and two analogous inequalities we deduce

1 1 1 2 (a + b + c)

2 (2a + b + c) ≤ (2b + c + a) (2c + a + b) 4 (a + b)(b + c)(c + a)

9 (a + b + c)

16 (ab + bc + ca)(a + b + c) The last inequality follows from 9 (a + b)(b + c)(c + a) = 9(a 2 b + ab 2 +b 2 c +

bc 2 +c 2 a +ca 2 )+18abc = 8(ab+ bc+ ca)(a + b +c)+(a 2 b +ab 2 +b 2 c + bc 2 +

c 2 a + ca 2 − 6abc) ≥ 8(ab + bc + ca)(a + b + c). The required inequality now follows from ab + bc + ca ≥ 3. Equality holds if and only if a = b = c = 1.

3. Assume that f satisfies the given requirements. Let us prove that f (1) = 1. As- sume that K = f (1) − 1 > 0. Let m be the minium of f , and b any number for which f (b) = m. Since 1, m = f (b), and f (b + f (1) − 1) = f (b + K) form

a triangle we must have f (b + K) < 1 + f (b). The minimality of m implies

f (b + K) = m, and by induction f (b + nK) = m for all n ∈ N. There exists a triangle with sides b + nK, f (1), and f (m); hence b + nK < f (1) + f (m) for each n. This contradiction implies f (1) = 1. The numbers a, 1 = f (1), and f ( f (a)) form a triangle for every a. Therefore

a − 1 < f ( f (a)) < a + 1; hence f ( f (a)) = a, and f is a bijection. We now have that f (a), f (b), and f (b + a − 1) determine a triangle for all a,b ∈ N. Let z = f (2). Clearly, z > 1. Since f (z), f (z), and f (2z − 1) form a triangle, we get f (2z − 1) < f (z) + f (z) = 2 f ( f (2)) = 4. This implies that f (2z − 1) ∈ {1,2,3}. Since f is a bijection and f (1) = 1, f (z) = 2, we must have f (2z−1) =

3. Let us prove that f (k) = (k − 1)z − k + 2 for all k ∈ N. The statement is true for 1 and 2. Assume that it holds for all of 1, 2, . . . , k. Since

f ((k − 1)z − k + 2), f (z), and f (kz − k + 1) form a triangle, we have f (kz − k +

778 4 Solutions

1 ) ≤ k +1. The function f is injective; hence f (kz−k+1) 6= i unless kz−k+1 = (i − 1)z − i + 2, i.e., k + 1 = i. Therefore f (kz − k + 1) = k + 1, or f (k + 1) = kz − k + 1, and the induction is complete. Furthermore, f is increasing. If z > 2 then 2 = f (z) > f (2) = z, a contradiction. Thus z = 2 and f (k) = 2(k −

1 ) − k + 2 = k. It is easy to verify that f (k) = k satisfies the given condition.

4. We first prove that for a , b > 0 the following inequality holds:

After dividing both sides by ab and substituting p a b = x it becomes

x 2 + 1 2 +1≤ x +

x Taking squares transforms this inequality into x + 1 x 2 −2 ≥ 0, which obviously

holds. Equality occurs if and only if x = 1, or a = b. q

a 2 +b 2 √ q ab √ Rewriting (1) in the form √

a +b + 2 a +b ≤ 2 a + b and summing it with the two analogous inequalities for the pairs (b, c) and (c, a), we obtain

It suffices to prove that

c +a ≥ √ 3 2 . Applying the mean in-

c +a gives us r

equality M 1 ≥M −2 to the sequence

2 This completes the proof of the required inequality. Equality holds if and only if

a = b = c.

5. Assume the contrary, that f (x − f (y)) ≤ y f (x) + x for all x,y ∈ R. Setting y = 0, x = z+ f (0) gives f (z) ≤ z+ f (0). For x = f (y) we get f (0) ≤ y f ( f (y))+ f (y) ≤ yf ( f (y)) + y + f (0); hence we can conclude that y( f ( f (y)) + 1) ≥ 0. This means that if y > 0, then f ( f (y)) ≥ −1, and if y < 0, then f ( f (y)) ≤ −1. If f (x) > 0 for some x, then each y < x − f (0) must satisfy f (y) − f (0) ≤ y < x − f (0), therefore f (y) < x. From this we conclude that −1 ≤ f ( f (x − f (y))) ≤

4.50 Shortlisted Problems 2009 779

f (x − f (y)) + f (0) ≤ y f (x) + x + f (0) and y ≥ −1−x− f (0) f (x) . We have concluded that each y

, which is impossible. Therefore f (x) ≤ 0 for all x. In particular, for all real x we have

< x − f (0) must be greater than or equal to −1−x− f (0)

f (x)

f (x) ≤ x + f (0) ≤ x. For any z > 0 we now have f (−1) = f [( f (z)−1)− f (z)] ≤ zf ( f (z) − 1) + f (z) − 1 ≤ z( f (z) − 1) + f (z) − 1 = (z + 1)( f (z) − 1) ≤ −z − 1. Thus each z > 0 satisfies z ≤ − f (−1) − 1, which is a contradiction.

6. Assume that s s n = an + b and s s n +1 = cn + d for some a, b, c, d ∈ Z. If m > n > 0 are two integers, then s m −s n = (s n +1 −s n ) + (s n +2 −s n +1 ) + ···+(s m −s m −1 )≥ m − n because the sequence s n is increasing. Hence s n +1 −s n ≤s s n +1 −s s n = a. Denote by m and M the minimal and maximal values of s n +1 −s n for n ∈ N. Our goal is to prove that M = m. Assume the contrary, m < M. If s k +1 −s k =m for some k ∈ N, we get a = s s k +1 −s s k = (s s k +1 −s s k ) + (s s k +2 −s s k +1 ) + ··· + (s s k +1 −s s k +1 −1 ) ≤ m · M. Similarly, if s l +1 −s l = M for some l ∈ N, we get that

a =s s l +1 −s s l ≥ m · M. In particular, these two inequalities imply that M · m = a,

s s k +1 −s s k = M, whenever s k +1 −s k = m, and

s s l +1 −s s l = m, whenver s l +1 −s l = M. Take any k ∈ N such that s k +1 −s k = m. Then M = s s k +1 −s s k = ck + d − (ak +

b ) = (c−a)k +d −b. Furthermore, we have m = s s sk +1 −s s sk = (c−a)s k + d −b. Repeating the same argument yields M = (c − a)s s k + d − b. Consequently the equation (c − a)x + d − b = M has two solutions, x = k and x = s s k , which yields

s k = k. Since s k =s 1 + (s 2 −s 1 ) + ··· + (s k −s k −1 ) ≥ k, we conclude that s i =i for i = 1, 2, . . . , k. Thus M = s s k +1 −s s k =s k +1 −s k = m, a contradiction.

7. Substituting x = 0 in the given relation gives f (0) = f (y f (0)) for all y. Therefore

f (0) = 0, because otherwise for each z ∈ R we could take y = z/ f (0) to get

f (z) = f (0), meaning that f is constant (which is obviously impossible). We now have f (x f (x)) = f (x f (x +0)) = f (0 f (x))+ x 2 =x 2 and 0 = f (x f (x −x)) =

f (−x f (x)) + x 2 , implying f (−x f (x)) = −x 2 . Hence f is onto. If f

(z) = 0 for some z 6= 0 we would have 0 = f (z f (z)) = z 2 , a contradiction. Assume that f

(x) = f (y) for some x, y ∈ R. Then x 2 = f (x f (x)) = f (x f (y)) =

f ((y − x) f (x)) + x 2 , giving that f ((y − x) f (x)) = 0; hence f (x) = 0 or x − y = 0. Both cases now yield x = y; therefore f is one-to-one. Now we will prove that f (−x) = − f (x) for all x ∈ R. Assume that x 6= 0 (the

other case is trivial). If f (x) > 0, there exists z such that f (x) = z 2 . Since f is injective and f (z f (z)) = z 2 , we conclude that x = z f (z); hence f (−x) =

f (−z f (z)) = −z 2 = − f (x). The case f (x) < 0 is similar. On the other hand, we have

f (y f (x)) = −x 2 + f (x f (x + y)) = −x 2 + (x + y) 2 − [(x + y) 2 + f (−x f (x + y))] =y 2 + 2xy − f ((x + y) f (y)) = 2xy + [(−y) 2 + f ((x + y) f (−y))] = 2xy + f (−y f (x)),

780 4 Solutions which implies that f (x f (y)) = xy. Analogously, f (y f (x)) = xy; hence x f (y) =

yf (x). Hence f (x) = cx for some c ∈ R. The equation f (x f (x)) = x 2 implies that

c ∈ {−1,1}. Clearly, both f (x) = x and f (x) = −x satisfy the given conditions.

8. (a) If we denote gold cards by 1, and black by 0, the entire sequence of cards corresponds to a number in binary representation. After each of the moves, the number decreases; hence the game has to end.

(b) We will show that the second player wins game no matter how the players play. Consider the cards whose position (counted from the right) is divisible by 50. There is a total of 40 such cards, and in each move exactly one of these cards is turned over. In the beginning, all 40 of these cards are 1, and in the end all 40 are 0; hence the second player must win.

9. Assume that (a i ,b i ,c i ) N i =1 satisfy the conditions of the problem. Then

N ∑ (N − 1)

i =1

and similarly the two analogous inequalities hold for the sequences (b i ) and (c i ). Hence 3N

i =1 (a i +b i +c i ) = nN, which implies that N ≤ 3 + 1. To prove that there are sequences of length 2n 3 + 1 with the given properties, let us consider the following cases:

(N −1)/2 ≤ ∑ 2n N

1 ◦ n = 3k for some k ∈ N. We can take (a i ,b i ,c i ) = (i−1,k +i−1,2k −2i+2) for i = 1, 2, . . . , k + 1, and (a i ,b i ,c i ) = (3k − i + 2,2k − i + 1,2(i − k) − 3) for i = k + 2, . . . , 2k + 1.

2 ◦ n = 3k + 1 for some k ∈ N. Take (a i ,b i ,c i ) = (i − 1,k + i − 1,2k − 2i + 3) for i = 1, 2, . . . , k + 1, and (a i ,b i ,c i ) = (3k − i + 2,2k − i + 1,2(i − k) − 2) for i = k + 2, . . . , 2k + 1.

3 ◦ n = 3k − 1 for some k ∈ N. Define (a i ,b i ,c i ) = (i − 1,k + i − 1,2k − 2i + 1) for i = 1, 2, . . . , k, and (a i ,b i ,c i ) = (3k − i + 1,2k − i,2i − 2k − 2) for i = k + 1, . . . , 2k.

10. For a binary sequence ( ε )

n −1 =( ε i ) −1 i =1 , let us define f (u, v, ( ε ) n −1 )=c n where

the sequence (c i ) n i =1 is defined as: c 0 = u, c 1 = v, and

c i i +1 = −1

2c i + 3c i , if ε = 0,

= 1, . . . , n − 1. The given sequences (a n ) and (b n ) can be now rewritten as a n = f (1, 7, ( ε ) n −1 ),

3c i −1 +c i , if ε i = 1,

for i

b n = f (1, 7, ( ε ) n −1 ) where ( ε ) n −1 i =1 is defined as ε i = ε n −i . Using induction on n we will prove that f (1, 7, ( ε ) n −1 ) = f (1, 7, ( ε ) n −1 ). This is straightforward to verify for n = 2 and n = 3, so assume that n > 3 and that the statement is

true for all binary sequences ( ε ) k of length smaller than n. Notice that f ( α u 1 + β u 2 , α v 1 + β v 2 ,( ε ) m )= α f (u 1 ,v 1 ,( ε ) m )+ β f (u 2 ,v 2 ,( ε ) m ) (this easily follows by induction on m). Assuming that ε n = 0, we obtain

4.50 Shortlisted Problems 2009 781

f (1, 7, ( ε ) n ) = 2 f (1, 7, ( ε ) n −2 ) + 3 f (1, 7, ( ε ) n −1 )

= 2 f (1, 7, ( ε ) n −2 ) + 3 f (1, 7, ( ε ) n −1 )

= 2 f (1, 7, ( ε ) n −2 ) + 3 f (7, f (1, 7, ( ε n −1 ) 1 ), ( ε ) n −2 ) = f (23, 14 + 3 f (1, 7, ( ε n −1 ) 1 ), ( ε ) n −2 ).

Using 14 + 3 f (1, 7, ( ε n −1 ) 1 ) = f (1, 7, (0, ε n −1 ) 2 ) and 23 = f (1, 7, (0) 1 ), we get

f (23, 14 + 3 f (1, 7, ( ε n −1 ) 1 ),( ε ) n −2 )

= f ( f (1, 7, (0) 1 ), f (1, 7, (0, ε n −1 ) 2 ), ( ε ) n −2 )

= f (1, 7, ( ε ) n ).

To finish the proof, it remains to see that for ε n = 1 we have

f (1, 7, ( ε ) n ) = 3 f (1, 7, ( ε ) n −2 ) + f (1, 7, ( ε ) n −1 ) = f ( f (1, 7, (1) 1 ), f (1, 7, (1, ε n −1 ) 2 ), ( ε ) n −2 ) = f (1, 7, ( ε ) n ).

11. Denote by (i, j) the cells of the table, and assume that the diagonal cells (i, i) form separate rectangles. We will prove by induction that it is possible to have

a perimeter of p = (m + 1)2 m +2 m . The case m = 0 is obvious, and assume that the statement holds for some m

m ≥ 0. Divide the 2 +1 ×2 board into four equal boards. Each of the two off-diagonal squares has perimeter 4 m ·2 , while

m +1

the other two can be partitioned into rectangles of total perimeter p m each. The total perimeter is therefore equal to 2

m ·4·2 +3 + 2p m = (m + 2)2 . Let us now prove the other direction, that the total perimeter P satisfies P ≥ (m+

1 )2 m +2 . Assume that the table is partitioned into n rectangles in the described way. Denote by R i the set of those rectangles that contain at least one square from the ith row. Similarly, let C i

be the set of rectangles that contain the squares from the ith column. Clearly the intersection C i ∩R i contains only the diagonal square (i, i). We certainly have

P =2 ∑ |R i |+ ∑ |C i | .

i =1

i =1

Let F be the collection of all subsets of rectangles in the partition. Since there are n rectangles, we have |F | = 2 n . Let F i denote the collection of those subsets

S that satisfy (R i \ (i,i)) ⊆ S and C i ∩ S ⊆ {(i,i)}. Since F 1 ,F 2 , ...,F 2 m are disjoint sets and |F i |=2 n −|C i |−|R i |+2 , using Jensen’s inequality applied to f (x) =

2 −x , we obtain

n +2

2 −|C i |−|R i |

n +2

2m ∑( |C i |+|R i |)

· ∑ P =2 n +m+2− 1 2m 2 .

i =1

This yields P 2 ≥ (m + 1) · 2 m +1 , which is the relation we wanted to prove.

782 4 Solutions

12. We will show that Cinderella can always make sure that after each of her moves: (i) The total amount of water in all buckets is less than 3 /2; and (ii) The amount of water in each pair of nonadjacent buckets is smaller than 1. The condition (ii) ensures that the stepmother won’t be able to make a bucket overflow. Both (i) and (ii) hold in the beginning of the game. Assume that they are satisfied after the kth round, and let us prove that in round k + 1, Cin- derella can make them both hold again. Denote the buckets by 1, . . . , 5 (counter- clockwise), and let x i

be the amount of water in bucket i. After the (k + 1)st move of the stepmother we have x 1 + ··· + x 5 < 5 2 and x i +x i +2 < 2 for each i (summation of indices is modulo 5). It is impossible that x i +x i +2 ≥ 1 for each

i ; hence we may assume that x 2 +x 5 < 1. If x 1 +x 2 +x 5 < 3 2 , it would be safe for Cinderella to empty buckets 3 and 4. Assume therefore that x 1 +x 2 +x 5 ≥ 3 2 . Hence x 1 > 1 2 . If both x 2 +x 4 ≥ 1 and x 3 +x 5 ≥ 1, hold then we must have x

1 ≤ 2 , a contradiction. Assume therefore that x 2 +x 4 < 1. If x 2 +x 3 +x 4 < 2

4 ≥ 2 . This gives x 3 > 2 , x 2 + 5 2 > (x 1 +x 2 +x 5 ) + (x 2 +x 3 +x 4 ) ≥ 6, which yields x 2 > 1 2 . Therefore at least one of x 1 +x 4 < 1 and x 3 +x 5 < 1 holds, say the first one. Thus, if Cin- derella empties buckets 2 and 3, condition (ii) will be satisfied. Condition (i) will

she could empty 1 and 5, so assume that x 2 +x 3 +x

hold as well because x 2 +x 3 > 1 2 + 1 2 > 1.

13. Let k ≥ 3. We will prove that the longest cyclic route in a (4k − 1) × (4k − 1) board has length 4 · [(2k − 1) 2 − 1]. Let us label the cells 1, 2, 3, 4 using the pattern from Figure 1 , so that the top-left corner is labeled 1.

Figure 1 Figure 2 Any four consecutive moves land on squares with different labels. Therefore,

each cyclic path has equal numbers of squares of each label. The label 4 appears exactly

(2k − 1) 2 times, but we will prove now that the limp rook can’t visit all of them. Assume the contrary, that it is possible for a cyclic route to pass through

all the squares labeled 4. Let us paint all squares labeled 4 alternately black and white so that the top left square is black. We see that the number of black squares is bigger by 1 than the number of white ones. Therefore, a cyclic route has two consecutive black squares. Assume that these squares are those denoted by •.

4.50 Shortlisted Problems 2009 783 Without loss of generality we may assume that the part of route is as shown in

Figure 1 . Since the route always visits 4 , 3, 1, 2 in that order, immediately before visiting ⋆, the rook has to land on the cell labeled 2 that is exactly below ⋆. The rook has to leave ⋆ by visiting 3 exactly to the right of it. Each point of the two- dimensional plane must be all the time either to the left or to the right of the rook when it is passing next to it. However, this is not the case with the point marked

by ◦. A contradiction. Therefore the rook can visit at most 4 · ((2k − 1) 2 − 1) squares in a cyclic route. Figure 2 shows how it is possible to recursively make a route that omits only the central square and visits all the other squares labeled 4.

14. We will prove the statement by induction. The case n = 1 is trivial, so let us assume that n > 1 and that the statement holds for 1, 2, . . . , n − 1. Assume that

a 1 < ··· < a n . Let m ∈ M be the smallest element. Consider the following cases:

1 ◦ m <a n : If a n 6∈ M, then if the grasshopper makes the first jump of size

a n the problem gets reduced to the sequence a 1 , ...,a n −1 and the set M \ {m}, which immediately follows by induction. Let us assume that a n ∈ M.

Consider the following n − 1 pairs: (a 1 ,a 1 +a n ), . . . , (a n −1 ,a n −1 +a n ). All numbers from these pairs belong to the (n − 2)-element set M \{a n }; hence one of these pairs, say (a k ,a k +a n ), has both of its members outside of M. If the first two jumps of the grasshopper are a k , and a k +a n , it has jumped over at least two members of M: m and a n . There are at most n − 3 more elements of M to jump over, and n − 2 more jumps, so the claim follows by induction.

2 ◦ m ≥a n : By the induction hypothesis the grasshopper can start from the point s =a 1 + ··· + a n , make n − 1 jumps of sizes a 1 , ...,a n −1 to the left, and avoid all the points of M \ {m}. If it misses the point m as well, then we are done (first make a jump of size a n and reverse the previously made jumps). Suppose that after making the jump a k the grasshopper landed at site m. If it changes the jump a k to the jump a n , it will miss the site m and all subsequent jumps will lend outside of M because m is the leftmost point.

15. For each i = 0, 1, . . . , 9 denote by N i the set of all finite strings whose terms are from {i,i+ 1,... ,9} (the empty string φ belongs to each of N i ). Define functions

m i :N i → N recursively: For each x ∈ N 9 we set m 9 (x) = 1+ the number of digits of x (set m 9 ( φ ) = 1). Once m 9 , ...,m i +1 are defined we construct m i as follows: Write each x ∈N i in the form x =x 0 ix 1 i ...x t −1 ix t where x 0 ,...,x t ∈N i +1 . We can now choose

m i (x) = ∑ 4 m i +1 (x s ) .

s =0

Let us prove that m =m 0 satisfies m (h(n)) < m(n) whenever n 6= φ . If the last digit of n is 0, then n =l 0 0l 1 . . . 0l t 0 for some l 0 ,...,l t ∈N 1 and m (n)−m(h(n)) =

4 m 1 (φ ) > 0. Assume that n = Ler(d + 1) for 9 ≥ d ≥ e ≥ 0 and r ∈ N d +1 . If L ∈N i for some i < e then L = l 0 il 1 i . . . il t with l 0 ,...,l t ∈N i +1 . Set n ′ =l t er (d + 1). Then h (n ′ )=l t erdrd and m i (n) − m i (h(n)) = 4 m i +1 (n ′ ) −4 m i +1 (h(n ′ )) . To prove that m (n) > m(h(n)) it suffices to prove that m i +1 (n ′ )>m i +1 (h(n ′ )). Repeating

784 4 Solutions this argument, we reduce our problem to the case i = e. There are l 0 , ...,l t ,

r ∈N d +1 such that n =l 0 e . . . el t er (d + 1). For e = d we need to prove that 0 < m d (n) − m d (h(n)) = 4 m d +1 (r(d+1)) −2·4 m d +1 (r) . This inequality suffices even in

the case e < d because r ∈ N d +1 implies

·· 4 md (r(d+1))

·· 4 md(rdrd)

m e +1 (r(d + 1)) = 4 · , and m e +1 (rdrd) = 4 · ; where 4 appears d − e − 1 times in the exponents. If d + 1 = 9, then m d (n) −

m d (h(n)) = 4 k +1 −2·4 k > 0, where k is the number of digits of r. If d < 8, then we can write r =r 0 (d + 1) ···(d + 1)r s for some r 0 ,...,r s ∈N d +2 . We get

4 md+2( φ m ) d +1 (r) = 4 + ··· + 4 and m d +1 (r(d + 1)) = m d +1 (r) · 4 . Hence m

m d +2 (r 0 )

m d +2 (r s )

d (n) − m d (h(n)) = 4 m d +1 (4 4 md+2( φ ) − 2) > 0. Therefore the function m is positive and decreasing on the sequence n, h (n),

(r)

h (h(n)), . . . , forcing this sequence to eventually become equal to φ . The only way this can occur is if the last terms of the sequence are 1, 00, 0, φ . Remark. The number of operations required can be huge. For example, n =5

will reduce to 1 after approximately 0 .828753 · 2 2 127 iterations.

16. Denote by I and L the incenters of △ABC and △BDA. From ∠ALI = ∠LBA + ∠LAB = 45 ◦ we see that AL kEK. Let L ′

be the intersection of DK and BI. From ∠DL ′ I = ∠BID − ∠IDK = ∠A/4 = ∠LAI we conclude that A, L, D, and L ′ belong to a circle. Hence ∠LAL ′ = 180 ◦ − ∠LDL ′ = 90 ◦ . Now consider △AKL ′ . The segment KE is the altitude from K and ∠KAE = ∠KL ′ E . We will now prove that this is equivalent to E being the orthocenter or △AKL ′ being isosceles (with KA = KL ′ ). Denote by P and Q the intersections of L ′ E and AE with AK and KL ′ respectively. If KA = KL ′ , then the statement is obvious. If this is not the case, then PQ intersects AL ′ at some point M. Then KE is the polar line of the point M

with respect to the circumcircle k of PQL ′ A , and since MA ⊥ KE, we conclude

that MA contains the center of k. Then we must have ∠APL ′ = ∠AQL ′ = 90 ◦ .

If E is the orthocenter of △KL ′ A then from △ABP we conclude that 3∠A/4 +

45 ◦ − ∠A/4 = 90 ◦ which yields ∠A = 90 ◦ . If KA = KL ′ , this together with KA = AL = AL ′ implies that △AKL ′ is equilateral and ∠A /4 = ∠KL ′ E =

60 ◦ − ∠LL ′ A = 15 ◦ . This means that ∠A = 60 ◦ . It is easy to verify that ∠A ∈ {60 ◦ , 90 ◦ } implies ∠BEK = 45 ◦ .

17. From MK kAB and MLkAC we get ∠KML = ∠BAC. Also, ∠AQP = ∠KMQ = ∠MLK, because of the assumption that PQ is a tangent to Γ . Therefore △AQP ∼ △MLK; hence

and AQ · QB = AP · PC. The quantity on the left-hand side of the last equality represents the power of the point Q with respect to the circumcircle of △ABC and

it is equal to OA 2 2 2 2 2 − OQ 2 . Similarly, AP · PC = OA − OP . Thus OA − OP = OA 2 − OQ 2 , implying OP = OQ.

4.50 Shortlisted Problems 2009 785

18. Consider the excircle k a corresponding to the vertex A. Let X be the point of tangency of the incircle k and BC, and X a the point of tangency of k a and BC. Similarly, let us denote by Z a and Y a the points of tangency of k a with AB and AC. Then we have ZZ a = ZB + BZ a = BX + BX a = BX + CX = BC = ZS. We used the fact that BX a = CX. On the other hand, CY a = CX a = BX = BZ = CS. Denote by s the degenerate circle with center S and radius 0. Points Z and C belong to the radical axis of the circles s and k a . Similarly, we prove that B and Y belong to the radical axis of the circles k a and r, where r is the circle with center R and radius 0. Thus G is the radical center of r, s, and k a ; hence GS = GR.

19. Denote by M and N respectively the points symmetric to E with respect to G and H. Notice that ∠FAM = ∠FAC + ∠BEC = ∠EBC + ∠CEB = ∠FCE. From

FA AB △FDC ∼ △FBA we have AB FC = CD , and from △AEB ∼ △DEC we have DC =

BE

CE . Therefore