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Geometry Difficulty 4.8 AIME Find the answer

Consider triangle ABCA B C where BC=7,CA=8B C=7, C A=8, and AB=9A B=9. DD and EE are the midpoints of BCB C and CAC A, respectively, and ADA D and BEB E meet at GG. The reflection of GG across DD is GG^{\prime}, and GEG^{\prime} E meets CGC G at PP. Find the length PGP G.

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

Solution

Observe that since GG^{\prime} is a reflection and GD=12AGG D=\frac{1}{2} A G, we have AG=GGA G=G G^{\prime} and therefore, PP is the centroid of triangle ACGA C G^{\prime}. Thus, extending CGC G to hit ABA B at F,PG=13CG=29CF=292(82+72)924=1459F, P G=\frac{1}{3} C G=\frac{2}{9} C F=\frac{2}{9} \sqrt{\frac{2\left(8^{2}+7^{2}\right)-9^{2}}{4}}=\frac{\sqrt{145}}{9} by the formula for the length of a median.

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