Difference between revisions of "2013 USAMO"
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===Problem 3=== | ===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. | + | Let <math>n</math> be a positive integer. There are <math>\tfrac{n(n+1)}{2}</math> marks, each with a black side and a white side, arranged into an equilateral triangle, with the biggest row containing <math>n</math> marks. Initially, each mark has the black side up. An [i]operation[/i] is to choose a line parallel to the sides of the triangle, and flipping all the marks on that line. A configuration is called [i]admissible [/i] if it can be obtained from the initial configuration by performing a finite number of operations. For each admissible configuration <math>C</math>, let <math>f(C)</math> denote the smallest number of operations required to obtain <math>C</math> from the initial configuration. Find the maximum value of <math>f(C)</math>, where <math>C</math> varies over all admissible configurations. |
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[[2013 USAMO Problems/Problem 3|Solution]] | [[2013 USAMO Problems/Problem 3|Solution]] | ||
Revision as of 18:22, 11 May 2013
Contents
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 [i]operation[/i] is to choose a line parallel to the sides of the triangle, and flipping all the marks on that line. A configuration is called [i]admissible [/i] 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 | ||
