olymmath

RMO 2013 D1

  • CRMO Paper 4

Let ω\omega be a circle with centre OO . Let γ\gamma be another circle passing through OO and intersecting ω\omega at points AA and BB . AA diameter CDCD of ω\omega intersects γ\gamma at a point PP different from OO . Prove that ∠APC=∠BPD\angle APC= \angle BPD

Dataset text AI-MO AoPS. Quoted with source.

Official text: Indian Regional Mathematical Olympiad 2013 official paper, page 1.

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