Bariz - Problem

Source: 2016 XIX All-Ukrainian Tournament of Young Mathematicians named after M. Y. Yadrenko, Qualifying p15

Tags: geometry, angles, Ukrainian TYM

parmenides51

06.05.2021 13:54

A non isosceles triangle $ABC$ is given, in which $\angle A = 120^o$. Let $AL$ be its angle bisector, $AK$ be it's median, drawn from vertex $A$, point $O$ be the center of the circumcircle of this triangle, $F$ be the point of intersection of the lines $OL$ and $AK$. Prove that $\angle BFC = 60^o$.

It suffices to show that the reflection of F over BC lies on (ABC). Note that if O’ is the reflection of O over BC, then the reflection of F is the intersection of O’L and the reflection of AK over BC, so it suffices to show that if J is the intersection of O’L and (ABC), then J, A, and the intersections of JA with BC, OK are harmonic. However, this is obvious by Menelaus and Ceva since J and A are the feet of altitudes from O’ to MJ and M to AO’, and MJ, BC, AO’ concur at the orthocenter of MLO’, so we are done.