A circle is inscribed in a triangle, and the points of tangency with the sides of the triangle are connected to each other. A new circle is inscribed in the resulting triangle, and the points of tangency with the sides form the vertices of a third triangle, which has the same angles as the original triangle. Find these angles.
Let the inscribed circle touch the sides $A B, B C, C A$ at points $C_{1}, A_{1}, B_{1}$. Triangles $A_{1} B C_{1}$ and $A_{1} C B_{1}$ are isosceles; their angles at the bases are $1 / 2\left(180^{\circ}-\angle B\right)$ and $1 / 2\left(180^{\circ}-\angle C\right)$. Therefore, $\angle A_{1}=1 / 2$ $(\angle B+\angle C)$. Similarly, $\angle B_{1}=1 / 2(\angle A+\angle C)$ and $\angle C_{1}=1 / 2(\angle A+\angle B)$. Similar calculations for the second triangle show that $\angle A_{2}=1 / 2\left(\angle B_{1}+\angle C_{1}\right)=1 / 2(2 \angle A+\angle B+\angle C), \angle B_{2}=1 / 2(2 \angle B+\angle A+\angle C)$ and $\angle C_{2}=1 / 2(2 \angle C+\angle A+\angle B)$. Let for definiteness $\angle A \leq \angle B \leq \angle C$. Then $\angle A_{2} \leq \angle B_{2} \leq \angle C_{2}$. Thus, from the given condition it follows that

$\angle A=\angle A_{2}$, that is, $2 \angle A=\angle B+\angle C$. Considering the inequality $\angle A \leq \angle B \leq \angle C$, we get $\angle A=\angle B=\angle C$.

## Answer

All angles are $60^{\circ}$.