¶ 2011 USAMO Day 2 Problems and Solutions
Individual Problems and Solutions
For problems and detailed solutions to each of the 2011 USAMO Day 2 problems, please refer below:
Problem 4: Consider the assertion that for each positive integer n≥2, the remainder upon dividing 22n by 2n−1 is a power of 4 . Either prove the assertion or find (with proof) a counterexample.
Solution:
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Problem 5: Let P be a given point inside quadrilateral ABCD. Points Q1 and Q2 are located within ABCD such that
∠Q1BC=∠ABP,∠Q1CB=∠DCP,∠Q2AD=∠BAP,∠Q2DA=∠CDP
Prove that Q1Q2∥AB if and only if Q1Q2∥CD.
Solution:
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Problem 6: Let A be a set with ∣A∣=225, meaning that A has 225 elements. Suppose further that there are eleven subsets A1,…,A11 of A such that ∣Ai∣=45 for 1≤i≤11 and ∣Ai∩Aj∣=9 for 1≤i<j≤11. Prove that ∣A1∪A2∪⋯∪A11∣≥165, and give an example for which equality holds.
Solution:
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The problems on this page are the property of the MAA's American Mathematics Competitions