olymmath

HMIC 2013 Problem 3

  • Answer

Triangle ABCA B C is inscribed in a circle ω\omega such that ∠A=60∘\angle A=60^{\circ} and ∠B=75∘\angle B=75^{\circ}. Let the bisector of angle AA meet BCB C and ω\omega at EE and DD, respectively. Let the reflections of AA across DD and CC be D′D^{\prime} and C′C^{\prime}, respectively. If the tangent to ω\omega at AA meets line BCB C at PP, and the circumcircle of APD′A P D^{\prime} meets line ACA C at F≠AF \neq A, prove that the circumcircle of C′FEC^{\prime} F E is tangent to BCB C at EE.

Dataset text NuminaMath olympiads. Quoted with source.

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A F=\frac{3-\sqrt{3}}{2}

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