Olympiad problems

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Found problems: 2

2020 Yasinsky Geometry Olympiad, 4

The altitudes of the acute-angled triangle $ABC$ intersect at the point $H$. On the segments $BH$ and $CH$, the points $B_1$ and $C_1$ are marked, respectively, so that $B_1C_1 \parallel BC$. It turned out that the center of the circle $\omega$ circumscribed around the triangle $B_1HC_1$ lies on the line $BC$. Prove that the circle $\Gamma$, which is circumscribed around the triangle $ABC$, is tangent to the circle $\omega$ .

Ukrainian TYM Qualifying - geometry, 2014.22

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 $\vartriangle XYZ$ .