Difference between revisions of "2013 USAMO"
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===Problem 2=== | ===Problem 2=== | ||
− | For a positive integer plot equally spaced points around a circle. Label one of them , and place a marker at . One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of distinct moves available; two from each point. Let count the number of ways to advance around the circle exactly twice, beginning and ending at , without repeating a move. Prove that for all . | + | For a positive integer <math>n\geq 3</math> plot <math>n</math> equally spaced points around a circle. Label one of them <math>A</math>, and place a marker at <math>A</math>. One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of <math>2n</math> distinct moves available; two from each point. Let <math>a_n</math> count the number of ways to advance around the circle exactly twice, beginning and ending at <math>A</math>, without repeating a move. Prove that <math>a_{n-1}+a_n=2^n</math> for all <math>n\geq 4</math>. |
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[[2013 USAMO Problems/Problem 2|Solution]] | [[2013 USAMO Problems/Problem 2|Solution]] | ||
Revision as of 18:21, 11 May 2013
Contents
[hide]Day 1
Problem 1
In triangle , points lie on sides respectively. Let , , denote the circumcircles of triangles , , , respectively. Given the fact that segment intersects , , again at respectively, prove that .
Problem 2
For a positive integer plot equally spaced points around a circle. Label one of them , and place a marker at . One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of distinct moves available; two from each point. Let count the number of ways to advance around the circle exactly twice, beginning and ending at , without repeating a move. Prove that for all . Solution
Problem 3
Let be a positive integer. There are marks, each with a black side and a white side, arranged into an equilateral triangle, with the biggest row containing marks. Initially, each mark has the black side up. An operation is to choose a line parallel to the sides of the triangle, and flipping all the marks on that line. A configuration is called admissible if it can be obtained from the initial configuration by performing a finite number of operations. For each admissible configuration , let denote the smallest number of operations required to obtain from the initial configuration. Find the maximum value of , where varies over all admissible configurations.
Day 2
Problem 4
Find all real numbers satisfying
Problem 5
Given postive integers and , prove that there is a positive integer such that the numbers and have the same number of occurrences of each non-zero digit when written in base ten.
Problem 6
Let be a triangle. Find all points on segment satisfying the following property: If and are the intersections of line with the common external tangent lines of the circumcircles of triangles and , then
See Also
2013 USAMO (Problems • Resources) | ||
Preceded by 2012 USAMO |
Followed by 2014 USAMO | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |