2013 AMC 12A Problems/Problem 11
Revision as of 02:53, 7 February 2013 by Epicwisdom (talk | contribs) (moved 2013 AMC 12A Problems/Problems 11 to 2013 AMC 12A Problems/Problem 11: Wrong title ("Problems 11"))
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