Difference between revisions of "2013 USAMO Problems/Problem 3"
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| − | 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 | + | 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 ''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 <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 |
Revision as of 18:41, 11 May 2013
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
