2010 AIME II Problems/Problem 14
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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 .