olymmath

BALKAN-SL 2017 G7

    Let ABCA B C be an acute triangle with AB≠ACA B \neq A C and circumcirle Γ\Gamma. The angle bisector of BACB A C intersects BCB C and Γ\Gamma at DD and EE respectively. Circle with diameter DED E intersects Γ\Gamma again at F≠EF \neq E. Point PP is on AFA F such that PB=PCP B=P C and XX and YY are feet of perpendiculars from PP to ABA B and ACA C respectively. Let HH and H′H^{\prime} be the ortocenters of ABCA B C and AXYA X Y respectively. AHA H meets Γ\Gamma again at QQ. If AH′A H^{\prime} and HH′H H^{\prime} intersect the circle with diameter AHA H again at points SS and TT, respectively, prove that the lines AT,HSA T, H S and FQF Q are concurrent.

    Dataset text NuminaMath olympiads. Quoted with source.

    Solutions

    No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

    Add a solution

    Locator balkan-sl/2017/g7 · Tag THPRDE4P55 · Edit this page on GitHub · Keys: j next, k previous