Olympiad problems

2017 Sharygin Geometry Olympiad, P18

Let $L$ be the common point of the symmedians of triangle $ABC$, and $BH$ be its altitude. It is known that $\angle ALH = 180^o -2\angle A$. Prove that $\angle CLH = 180^o - 2\angle C$.

2022 Yasinsky Geometry Olympiad, 3

Tags: geometry , lemoine point , perpendicular

Given a triangle $ABC$, in which the medians $BE$ and $CF$ are perpendicular. Let $M$ is the intersection point of the medians of this triangle, and $L$ is its Lemoine point (the intersection point of lines symmetrical to the medians with respect to the bisectors of the corresponding angles). Prove that $ML \perp BC$. (Mykhailo Sydorenko)

2021 239 Open Mathematical Olympiad, 4

Tags: geometry , perpendicular , concurrency , lemoine point , 3d geometry

Symedians of an acute-angled non-isosceles triangle $ABC$ intersect at a point at point $L$, and $AA_1$, $BB_1$ and $CC_1$ are its altitudes. Prove that you can construct equilateral triangles $A_1B_1C'$, $B_1C_1A'$ and $C_1A_1B'$ not lying in the plane $ABC$, so that lines $AA' , BB'$ and $CC'$ and also perpendicular to the plane $ABC$ at point $L$ intersected at one point.