Problem

Source: 2014 XVII All-Ukrainian Tournament of Young Mathematicians named after M. Y. Yadrenko, Qualifying p22

Tags: geometry, incenter, orthcenter, Symmetric, Ukrainian TYM



In $\vartriangle ABC$ on the sides $BC, CA, AB$ mark feet of altitudes $H_1, H_2, H_3$ and the midpoint of sides $M_1, M_3, M_3$. Let $H$ be orthocenter $\vartriangle ABC$. Suppose that $X_2, X_3$ are points symmetric to $H_1$ wrt $BH_2$ and $CH_3$. Lines $M_3X_2$ and $M_2X_3$ intersect at point $X$. Similarly, $Y_3,Y_1$ are points symmetric to $H_2$ wrt $C_3H$ and $AH_1$.Lines $M_1Y_3$ and $M_3Y_1$ intersect at point $Y.$ Finally, $Z_1,Z_2$ are points symmetric to $H_3$ wrt $AH_1$ and $BH_2$. Lines $M_1Z_2$ and $M_2Z_1$ intersect at the point $Z$ Prove that $H$ is the incenter $\vartriangle XYZ$ .