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AIME 2010b Problem 15

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In triangle ABCABC, AC=13AC = 13, BC=14BC = 14, and AB=15AB=15. Points MM and DD lie on ACAC with AM=MCAM=MC and ∠ABD=∠DBC\angle ABD = \angle DBC. Points NN and EE lie on ABAB with AN=NBAN=NB and ∠ACE=∠ECB\angle ACE = \angle ECB. Let PP be the point, other than AA, of intersection of the circumcircles of △AMN\triangle AMN and △ADE\triangle ADE. Ray APAP meets BCBC at QQ. The ratio BQCQ\frac{BQ}{CQ} can be written in the form mn\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m−nm-n.

Dataset text AIME 1983 to 2026. Quoted with source.

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