Difference between revisions of "2010 AIME II Problems/Problem 14"
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Revision as of 16:13, 2 April 2010
Label the center of the circumcircle of as and the intersection of with the circumcircle as . It now follows that . Hence is isosceles and .
Denote the projection of onto . Now . By the pythagorean theorem, . Now note that . By the pythagorean theorem, . Hence it now follows that,
This gives that the answer is .