Problem:
Triangle ABC is inscribed in circle Ο with AB=5,BC=7, and AC=3. The bisector of angle A meets side BC at D and circle Ο at a second point E. Let Ξ³ be the circle with diameter DE. Circles Ο and Ξ³ meet at E and a second point F. Then AF2=nmβ, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let M be the midpoint of BC. Because AE bisects β BAC, point E is the midpoint of BC and thus line ME is the perpendicular bisector of BC. Then line ME passes through the center W of circle Ο. Therefore β EMD=90β and M lies on Ξ³. Let rays AM and FM intersect Ο at F1β and A1β, respectively. Because EFDM is cyclic,
implying that BA1ββ=AC. Therefore A1β and A are symmetric across line ME, from which it follows that F1β and F are also symmetric across line ME and BF1β=CF. Thus β BAF1β=β FAC and
