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Angle Bisector in Circle with Diameter Extensions

Source: 2012 AIME II Problem 15

March 29, 2012
trigonometryanalytic geometrygeometrycircumcirclegeometric transformationreflectionratio

Problem Statement

Triangle ABCABC is inscribed in circle ω\omega with AB=5AB = 5, BC=7BC = 7, and AC=3AC = 3. The bisector of angle AA meets side BCBC at DD and circle ω\omega at a second point EE. Let γ\gamma be the circle with diameter DEDE. Circles ω\omega and γ\gamma meet at EE and a second point FF. Then AF2=mnAF^2 = \frac mn, where m and n are relatively prime positive integers. Find m+nm + n.
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