AIME 1992 Problem 14 PaperAnswer In triangle ABCABC^{}_{}ABC, A′A'A′, B′B'B′, and C′C'C′ are on the sides BCBCBC, ACAC^{}_{}AC, and ABAB^{}_{}AB, respectively. Given that AA′AA'AA′, BB′BB'BB′, and CC′CC'CC′ are concurrent at the point OO^{}_{}O, and that AOOA′+BOOB′+COOC′=92\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92OA′AO+OB′BO+OC′CO=92, find AOOA′⋅BOOB′⋅COOC′\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}OA′AO⋅OB′BO⋅OC′CO. Dataset text AIME 1983 to 2026. Quoted with source. Show the answer 94 Solutions No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label. Add a solution PreviousProblem 13 NextProblem 15 Locator aime/1992/p14 · Tag XA17K1MZ6P · Edit this page on GitHub · Keys: j next, k previous