olymmath

JBMO-SL 2008 G2

For a fixed triangle ABCA B C we choose a point MM on the ray CAC A (after AA ), a point NN on the ray ABA B (after BB ) and a point PP on the ray BCB C (after CC ) in a way such that AM−BC=BN−AC=CP−ABA M-B C=B N-A C=C P-A B. Prove that the angles of triangle MNPM N P do not depend on the choice of M,N,PM, N, P.

Dataset text NuminaMath olympiads. Quoted with source.

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