Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $50^{\circ}$. Find the other two angles of triangle $A B C$.
Answer: $60^{\circ}$ and $70^{\circ}$.

Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}$. We aim to prove that triangle $H B C_{1}$ is equilateral. Then $\angle B A C = \angle B C_{1} C = 60^{\circ}$, from which the answer follows given the condition.

First, note that $H$ is the intersection of the angle bisectors of triangle $A_{1} B_{1} C_{1}$. Indeed, for example, $\angle B_{1} C_{1} C = \angle B_{1} B C = 90^{\circ} - \angle C = \angle C A A_{1} = \angle C C_{1} A_{1}$, so $C_{1} H$ is the angle bisector of $\angle A_{1} C_{1} B_{1}$; similarly, it can be shown that $A_{1} H$ and $B_{1} H$ are also angle bisectors of the corresponding angles. Therefore, $H$ is the incenter of triangle $A_{1} B_{1} C_{1}$ and $H K = H L$. Moreover, it has been shown that $\angle H C_{1} L = \angle H B K$, so the right triangles $H C_{1} L$ and $H B K$ are congruent by a leg and an acute angle. Therefore, $H C_{1} = H B$.

It remains to note that $H B = B C_{1}$. This fact is well-known and can be proven in various ways. Here, we will present just one of them. Note that triangles $A H B$ and $A C_{1} B$ are congruent by a side (common side $A B$) and two angles ($\angle H A B = \angle A_{1} B_{1} B = \angle C_{1} B_{1} B = \angle C_{1} A B$; similarly, it can be shown that $\angle H B A = \angle C_{1} B A$.

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