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AIME 2012b Problem 15

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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 BC‾\overline{BC} at DD and circle ω\omega at a second point EE. Let γ\gamma be the circle with diameter DE‾\overline{DE}. Circles ω\omega and γ\gamma meet at EE and a second point FF. Then AF2=mnAF^2 = \frac mn, where mm and nn are relatively prime positive integers. Find m+nm+n.

Dataset text AIME 1983 to 2026. Quoted with source.

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