Pлотников M.

Let $T_{1}, T_{2}$ be the points of tangency of the excircles of triangle $ABC$ with sides $BC$ and $AC$ respectively. It turns out that the point symmetric to the center of the inscribed circle of the triangle with respect to the midpoint of $AB$ lies on the circumcircle of triangle $C T_{1} T_{2}$. Find the angle $BCA$.

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Let $D$ be the fourth vertex of the parallelogram $A C B D, J$ be the center of the inscribed circle of triangle $A B D, S_{1}$, $S_{2}$ be the points of tangency of this circle with $A D$ and $B D$. Then $D S_{1}=B T_{1}$ (see problem $\underline{55404}$), so $S_{1} T_{1} \| A C$.

Similarly, $S_{2} T_{2} \| B C$, which means $C T_{1} J T_{2}$ is a parallelogram. But by the condition, it is inscribed, hence it is a rectangle.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3ff818e52eab4ca192bg-25.jpg?height=844&width=1018&top_left_y=527&top_left_x=560)

## Answer

$90^{\circ}$.