Shveiov D.V.

The circle inscribed in the right triangle $ABC\left(\angle ABC=90^{\circ}\right)$ touches the sides $AB, BC, AC$ at points $C_{1}, A_{1}, B_{1}$ respectively. The excircle opposite to $A$ touches the side $BC$ at point $A_{2}$. $A_{0}$ is the center of the circumcircle of triangle $A_{1} A_{2} B_{1}$; the point $C_{0}$ is defined similarly. Find the angle $A_{0} B C_{0}$.
Since points $A_{1}$ and $A_{2}$ are symmetric with respect to the midpoint of segment $B C$ (see problem $\underline{55404}$), then $A_{0} B=A_{0} C$. On the other hand, $A_{0}$ lies on the perpendicular bisector of segment $A_{1} B_{1}$, which coincides with the bisector of angle $C$. Therefore, $\angle C B A_{0}=\angle A_{0} C B=1 / 2 \angle C$. Similarly,

$\angle A B C_{0}=1 / 2 \angle A$, and thus, $\angle A_{0} B C_{0}=\angle B-1 / 2(\angle C+\angle A)=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ccb5665c9ecd710ea159g-10.jpg?height=763&width=1397&top_left_y=1442&top_left_x=319)

## Answer

$45^{\circ}$.