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SHARYGIN 2013 Problem 2

    Let ABCABC be an isosceles triangle (AC=BC)(AC = BC) with ∠C=20∘\angle C = 20^\circ. The bisectors of angles AA and BB meet the opposite sides at points A1A_1 and B1B_1 respectively. Prove that the triangle A1OB1A_1OB_1 (where OO is the circumcenter of ABCABC) is regular.

    Official text Sharygin Geometry Olympiad 2013 official paper, page 2. Quoted with source.

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