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Problem 2. Let P be a regular 2006-gon. A diagonal of P is called good if its endpoints divide the boundary of P into two parts, each composed of an odd number of sides of P. The sides of P are also called good. Suppose P has been dissected into triangles by 2003 diagonals, no two of which have a common point in the
IMO 2015 Thailand. Solutions. Algebra. A1. Suppose that a sequence a1,a2, of positive real numbers satisfies ak`1 e kak a2 k ` pk ? 1q. (1) for every positive integer k. Prove that a1 ` a2 `???` an e n for every n e 2. (Serbia). Solution. From the constraint (1), it can be seen that k ak`1 d a2 k ` pk ? 1q ak. “ ak ` k ? 1 ak. , and so.
IMO 2013 Colombia. Solutions. Algebra. A1. Let n be a positive integer and let a1,,an?1 be arbitrary real numbers. Define the sequences u0,,un and v0,,vn inductively by u0 “ u1 “ v0 “ v1 “ 1, and uk`1 “ uk ` akuk?1, vk`1 “ vk ` an?kvk?1 for k “ 1,,n ? 1. Prove that un “ vn. (France). Solution 1. We prove by induction on k that.
strictly confidential until IMO 2011. Contributing Countries. The Organizing Committee and the Problem Selection Committee of IMO 2010 thank the following 42 countries for contributing 158 problem proposals. Armenia, Australia, Austria, Bulgaria, Canada, Columbia, Croatia,. Cyprus, Estonia, Finland, France, Georgia,
SOLUTIONS FOR IMO 2005 PROBLEMS. AMIR JAFARI AND KASRA RAFI. Problem 1. Six points are chosen on the sides of an equilateral triangle. ABC: A1, A2 on BC; B1, B2 on CA; C1, C2 on AB. These points are the vertices of a convex hexagon A1A2B1B2C1C2 with equal side lengths. Prove that the lines A1B2, B1C2
A SOLUTION TO 1988 IMO QUESTION 6. (The Most Difficult Question Ever Set at an IMO). Theorem. If a, b are integers > 0 such that. (a2 + b2) q = (ab + 1) is integral, then q = (GCD(a,b))2. Proof. If ab = 0 (i.e. if a = 0 or b = 0) the result is plain. This suggests using induction on ab. If ab > 0, we may suppose (from the
Solution. First we shall prove that a a2. 8b c a. 4. 3 a. 4. 3 b. 4. 3 c. 4. 3. , or equivalently, that a. 4. 3 b. 4. 3 c. 4. 3. 2 a. 2. 3 a. 2. 8b c . The AM-GM inequality yields a. 4. 3 b. 4. 3 c. 4. 3. 2 a. 4. 3. 2 b. 4. 3 c. 4. 3 a. 4. 3 a. 4. 3 b. 4. 3 c. 4. 3. 2b. 2. 3 c. 2. 3. 4a. 2. 3 b. 1. 3 c. 1. 3. 8a. 2. 3 b c. Thus. 4. IMO 2001 Competition Problems.
ous problems and novel ideas presented in the solutions and emerge ready to tackle even the most difficult problems on an IMO. In addition, the skill ac- quired in the process of successfully attacking difficult mathematics problems will prove to be invaluable in a serious and prosperous career in mathematics. However, we
International Mathematical Olympiad. Problems and Solutions. 1959 - 2009. IMO. The most important and prestigious mathematical competition for high-school students
The Problem Selection Committee and the Organising Committee of IMO 2003 thank the following thirty-eight countries for contributing problem proposals. Armenia. Greece. New Zealand. Australia. Hong Kong. Poland. Austria. India. Puerto Rico. Brazil. Iran. Romania. Bulgaria. Ireland. Russia. Canada. Israel. South Africa.
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