olymmath

ISL 2015 G6

Let ABCA B C be an acute triangle with AB>ACA B>A C, and let Γ\Gamma be its circumcircle. Let HH, MM, and FF be the orthocenter of the triangle, the midpoint of BCB C, and the foot of the altitude from AA, respectively. Let QQ and KK be the two points on Γ\Gamma that satisfy ∠AQH=90∘\angle A Q H=90^{\circ} and ∠QKH=90∘\angle Q K H=90^{\circ}. Prove that the circumcircles of the triangles KQHK Q H and KFMK F M are tangent to each other. (Ukraine)

Dataset text NuminaMath olympiads. Quoted with source.

Official text: IMO 2015 Shortlist booklet, page 7.

Solutions

Official solution Solution 1 by the IMO Problem Selection Committee

The terms of the source let olymmath link to this solution but not copy it. Read the solution (IMO official site).

Locator isl/2015/g6 · Tag TX3X2JCSF6 · Edit this page on GitHub · Keys: j next, k previous