Krekov D:

In an acute scalene triangle $ABC$, the altitudes $AA'$ and $BB'$ intersect at point $H$, and the medians of triangle $AHB$ intersect at point $M$. The line $CM$ bisects the segment $A'B'$. Find the angle $C$.
Let $C_{0}$ be the midpoint of $AB$, and $H^{\prime}$ be the point symmetric to $H$ with respect to $C_{0}$ (it is known that $H^{\prime}$ is a point on the circumcircle of triangle $ABC$, diametrically opposite to $C$). The medians $CC_{0}$ and $CM$ of similar triangles $ABC$ and $A^{\prime}B^{\prime}C$ are symmetric with respect to the angle bisector of $\angle C$. Also, the altitude $CH$ and the diameter of the circumcircle $CH^{\prime}$ are symmetric with respect to this angle bisector. Therefore, $\angle H^{\prime}CC_{0} = \angle MCH$, which means that $CM$ is the symmedian of triangle $CHH^{\prime}$ (see the figure). From this, $\left(\frac{CH^{\prime}}{CH}\right)^2 = \frac{H^{\prime}M}{MH} = 2$ (see problem $\underline{56978}$), and since $CH = CH^{\prime} \cos \angle C$, it follows that $\angle C = 45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4bea0374a35e8bb7d318g-57.jpg?height=723&width=723&top_left_y=-1&top_left_x=673)

## Answer

$45^{\circ}$.

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