SOLUTIONS FOR IMO 005 PROBLEMS AMIR JAFARI AND KASRA RAFI Problem. Six points are chosen on the sides of an equilateral triangle AB: A, A on B; B, B on A;, on AB. These points are the vertices of a convex hexagon A A B B with equal side lengths. Prove that the lines A B, B and A are concurrent. A B B A A B Solution: In triangles AB and B A, A = B. Therefore, if B A > AB, then BA < A B. But A = B. Therefore, the law of sines implies BA > A B < AB. On the other hand, we have B + A = AB + B. Therefore, By a similar argument, we have: B < AB A > B. BA > A A > B B > BA. The contradiction shows that BA = A = B. Thus, the three triangles AB, B A and A B are congruent. This implies that the triangle A B is equilateral and A B, B and A are its heights. Therefore they are concurrent.
AMIR JAFARI AND KASRA RAFI Problem. Let a, a,... be a sequence of integers with infinitely many positive terms and infinitely many negative terms. Suppose that for each positive integer n, the numbers a, a,..., a n leave n different remainders on division by n. Prove that each integer occurs exactly once in the sequence. Solution: Let A n = {a,..., a n }. Elements of A n are distinct, because they are distinct modulo n. Observe that, for a i, a j A n, k := a i a j < n, because, otherwise, a i, a j A k and a i a j mod k. Therefore, max A n min A n < n. But A n consists of n distinct integers. Therefore, for m n = min A n, A n = {m n, m n +,..., m n + n }. There are infinitely many negative and positive numbers in the sequence; therefore, all integers have to appear in our sequence. This finishes the proof.
SOLUTIONS FOR IMO 005 PROBLEMS 3 Problem 3. Let x, y and z be positive real numbers such that xyz. Prove that x 5 x x 5 + y + z + y5 y x + y 5 + z + z5 z x + y + z 5 0. Solution: The above inequality is equivalent to () x 5 + y + z + x + y 5 + z + x + y + z 5 3 x + y + z. We have (auchy-schwarz) (xyz ) Therefore, Similarly, (x 5 + y + z ) (y z + y + z ) ( x 5 y z + y + z ) (x + y + z ). x 5 + y + z y z + y + z y+z (x + y + z ) + y + z (x + y + z ). x+z x + y 5 + z + x + z (x + y + z ) and Adding the above three inequalities proves (). x+y x + y + z 5 + x + y (x + y + z ).
4 AMIR JAFARI AND KASRA RAFI Problem 4. onsider the sequence a, a,... defined by a n = n + 3 n + 6 n (n =,,...). Determine all positive integers that are relatively prime to every term of the sequence. Solution: If p > 3, then p +3 p +6 p mod p. To see this, multiply both sides by 6 to get : 3 p + 3 p + 6 p 6 mod p, which is a consequence of Fermat s little theorem. Therefore p divides a p. Also, divides a and 3 divides a. So, there is no number other than that is relatively prime to all the terms in the sequence.
SOLUTIONS FOR IMO 005 PROBLEMS 5 Problem 5. Let ABD be a given convex quadrilateral with sides B and AD equal in length and not parallel. Let E and F be interior points of the sides B and AD respectively such that BE = DF. The lines A and BD meet at P, the lines BD and EF meet at Q, the lines EF and A meet at R. onsider all the triangles P QR as E and F vary. Show that the circumcircles of these triangles have a common point other than P. A B P R E Q T F D Solution: The circumcircles of the triangles P AD and P B intersect in points P and T. We claim that T is the desired point, i.e., P, Q, R and T lie on a circle. To prove this we show that the angles T P R and T QR are equal. The angles ADT and AP T are complimentary, therefore ADT = T P. But T P is also equal to T B. Therefore, ADT = T B. Similarly, T AD = T B. This implies that the triangles AT D and BT are equal. In particular, T F D = T EB. This, in turn, implies that the isosceles triangles ET F and BT D are similar. Therefore, QF T = QDT. This means that D, Q, F and T lie on a circle, and thus F DT = RQT. But F DT was also equal to T P R. Hence, T P R = T QR which is as we claimed.
6 AMIR JAFARI AND KASRA RAFI Problem 6. In a mathematical competition 6 problems were posed to the contestants. Each pair of problems was solved by more than 5 of the contestants. Nobody solved all 6 problems. Show that there were at least contestants who each solved exactly 5 problems. Solution: Let n be the number of contestants, c be the number of contestants who solved exactly 5 problems and p ij be the number of contestants who solved problems i and j, for i, j 6. We know that: ( ) 6 n + p ij = 6 n + 3. 5 Also, i,j p ij i,j ( ) 5 c + ( ) 4 (n c) = 6 n + 4 c. Therefore, 4c 3. This shows that there is at least one contestant who solved exactly 5 problems. If n + is not divisible by 5, then we can replace n+ 5 in the above argument by n+ 5 and this will imply that 4c 6 and hence there are at least two contestants who have solved 5 problems. Now assume that n + is divisible by 5, i.e., n = 5 k +, for some positive integer k. Assuming that there is exactly one person who has solved 5 problems and the rest have solved exactly 4 problems will lead to a contradiction, as we now argue in two cases. We call the only person who has solved 5 problems the champion. ase. Assume that n is not divisible by 3. Let a i be the number of contestants besides the champion who have solved problem i. Then a i = 4 (n ) = 4 n 4. i Let problem be the problem that the champion missed. There are 5 pairs of problems containing problem, and they have been solved by at least 5 n+ 5 = n + contestants. Since each person who has solved problem has solved exactly 3 other problems, every such person has solved 3 of above 5 pairs of problems. Thus 3 a n +. For i >, the champion has solved 4 pairs that include i. The above argument implies, 3 a i n 3. But, n is not divisible by 3. Therefore 3 a i n. Adding the above inequalities we get: 3 ai ( n + ) + 5 ( n ) = n 9, which is a contradiction because the left hand side is n.
SOLUTIONS FOR IMO 005 PROBLEMS 7 ase. We are left with the case where n is divisible by 3 and is of the form 5 k +, i.e., n = 5 h 3, and each pair of problems is solved by at least 6 h contestants. As before, assume that the champion has not solved problem and that a be the number of people who have solved this problem. Each of them has solved 3 other problems. So they have each solved 3 pairs of problems containing problem. That is: But a is an integer; therefore, 3 a 5 (6 h ) = 30 h 5. () a 0 h. Restricting our attention to 0 pairs of problems that do not contain, we observe that there are at least 0 (6 h ) contestants who have solved at least one of these pairs. On the other hand, the champion has solved 0 pairs, the a contestants who have solved problem have solved 3 a pairs and the rest have solved ( 4 ) (5 h 4 a ) pairs. That is, ( ) 4 0 + 3 a + (5 h 4 a ) 0 (6 h ) 3 a 30 h 4. This contradicts the inequality (). Therefore, more than one contestant solved 5 problems.