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AIME 1987 Problem 9

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Triangle ABCABC has right angle at BB, and contains a point PP for which PA=10PA = 10, PB=6PB = 6, and ∠APB=∠BPC=∠CPA\angle APB = \angle BPC = \angle CPA. Find PCPC. [asy] unitsize(0.2 cm); pair A, B, C, P; A = (0,14); B = (0,0); C = (21*sqrt(3),0); P = intersectionpoint(arc(B,6,0,180),arc(C,33,0,180)); draw(A--B--C--cycle); draw(A--P); draw(B--P); draw(C--P); label("AA", A, NW); label("BB", B, SW); label("CC", C, SE); label("PP", P, NE); [/asy]

Dataset text AIME 1983 to 2026. Quoted with source.

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