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Geometry Difficulty 4.9 AIME Find the answer

In triangle ABCA B C, points MM and NN are the midpoints of ABA B and ACA C, respectively, and points PP and QQ trisect BCB C. Given that A,M,N,PA, M, N, P, and QQ lie on a circle and BC=1B C=1, compute the area of triangle ABCA B C.

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

Solution

Note that MPAQM P \parallel A Q, so AMPQA M P Q is an isosceles trapezoid. In particular, we have AM=MB=BP=PQ=13A M=M B=B P=P Q=\frac{1}{3}, so AB=23A B=\frac{2}{3}. Thus ABCA B C is isosceles with base 1 and legs 23\frac{2}{3}, and the height from AA to BCB C is 76\frac{\sqrt{7}}{6}, so the area is 712\frac{\sqrt{7}}{12}.

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