Difference between revisions of "2013 AMC 12A Problems/Problem 11"
m (moved 2013 AMC 12A Problems/Problems 11 to 2013 AMC 12A Problems/Problem 11: Wrong title ("Problems 11"))
Revision as of 02:53, 7 February 2013
Let , and . We want to find , which is nothing but .
Based on the fact that , , and have the same perimeters, we can say the following:
Simplifying, we can find that
Since , .
After substitution, we find that , and = .
Again substituting, we find = .
Therefore, = , which is