ΒΆ 2016 USAMO Day 1 Problems and SolutionsIndividual Problems and Solutions
For problems and detailed solutions to each of the 2016 USAMO Day 1 problems, please refer below:
Problem 1: Let X 1 , X 2 , β¦ , X 100 X 1 β , X 2 β , β¦ , X 1 0 0 β S S X i X i β X i + 1 X i + 1 β S S X i β© X i + 1 = β
X i β β© X i + 1 β = β
X i βͺ X i + 1 β S X i β βͺ X i + 1 β ξ = S i β { 1 , β¦ , 99 } i β { 1 , β¦ , 9 9 } S S
Solution:
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Problem 2: Prove that for any positive integer k k
( k 2 ) ! β
β j = 0 k β 1 j ! ( j + k ) ! ( k 2 ) ! β
j = 0 β k β 1 β ( j + k ) ! j ! β
is an integer.
Solution:
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Problem 3: Let β³ A B C β³ A B C I B , I C I B β , I C β O O B B C C E E Y Y A C βΎ A C β A B Y = β C B Y β A B Y = β C B Y B E βΎ β₯ A C βΎ B E β₯ A C F F Z Z A B βΎ A B β A C Z = β B C Z β A C Z = β B C Z C F βΎ β₯ A B βΎ C F β₯ A B
Lines I B F β I B β F β I C E β I C β E β P P P O βΎ P O Y Z βΎ Y Z
Solution:
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The problems on this page are the property of the MAA's American Mathematics Competitions