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Geometry Difficulty 4.8 AIME Find the answer

In acute ABC\triangle A B C with centroid G,AB=22G, A B=22 and AC=19A C=19. Let EE and FF be the feet of the altitudes from BB and CC to ACA C and ABA B respectively. Let GG^{\prime} be the reflection of GG over BCB C. If E,F,GE, F, G, and GG^{\prime} lie on a circle, compute BCB C.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Note that B,C,E,FB, C, E, F lie on a circle. Moreover, since BCB C bisects GGG G^{\prime}, the center of the circle that goes through E,F,G,GE, F, G, G^{\prime} must lie on BCB C. Therefore, B,C,E,F,G,GB, C, E, F, G, G^{\prime} lie on a circle. Specifically, the center of this circle is MM, the midpoint of BCB C, as ME=MFM E=M F because MM is the center of the circumcircle of BCEFB C E F. So we have GM=BC2G M=\frac{B C}{2}, which gives AM=3BC2A M=\frac{3 B C}{2}. Then, by Apollonius's theorem, we have AB2+AC2=2(AM2+BM2)A B^{2}+A C^{2}=2\left(A M^{2}+B M^{2}\right). Thus 845=5BC2845=5 B C^{2} and BC=13B C=13.

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