Shvecov D.V.

On the hypotenuse $AC$ of the right triangle $ABC$, a point $C_{1}$ is marked such that $BC = CC_{1}$. Then, on the leg $AB$, a point $C_{2}$ is marked such that

$AC_{2} = AC_{1}$; the point $A_{2}$ is defined similarly. Find the angle $AMC$, where $M$ is the midpoint of the segment $A_{2}C_{2}$.
Let $I$ be the center of the inscribed circle of triangle $ABC$. Since point $C_{1}$ is symmetric to $B$ with respect to $CI$, and $C_{2}$ is symmetric to $C_{1}$ with respect to $AI$, then

$BI = IC_{2}$ and $\angle BIC_{2} = 90^{\circ}$. Similarly, $BI = IA_{2}$ and $\angle BIA_{2} = 90^{\circ}$ (see figure). Therefore, $I$ coincides with $M$, and $\angle AIC = 135^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_65daf5394c369f17193bg-47.jpg?height=569&width=1380&top_left_y=793&top_left_x=339)

## Answer

$135^{\circ}$.