25. The point of intersection of the medians of a right triangle lies on the circle inscribed in this triangle. Find the acute angles of the triangle.
25. Let in triangle $ABC$ angle $C$ be right, $M$ be the point of intersection of the medians, $O$ be the center of the inscribed circle, $r$ be its radius, and $\widehat{CBA} = \alpha$; then

$$
\begin{aligned}
& |AB| = r \left( \operatorname{ctg} \frac{\alpha}{2} + \operatorname{ctg} \left( \frac{\pi}{4} - \frac{\alpha}{2} \right) \right) = \\
& = \frac{r \sin \frac{\pi}{4}}{\sin \frac{\alpha}{2} \sin \left( \frac{\pi}{4} - \frac{\alpha}{2} \right)} = \frac{r \sqrt{2}}{\cos \left( \frac{\pi}{4} - \alpha \right) - \frac{\sqrt{2}}{2}}
\end{aligned}
$$

$$
\begin{aligned}
& |CM| = \frac{1}{3} |AB| = \frac{2 r \sqrt{2}}{3 \left[ 2 \cos \left( \frac{\pi}{4} - \alpha \right) - \sqrt{2} \right]} \\
& |CO| = r \sqrt{2}, \quad |OM| = r, \quad \widehat{OCM} = \alpha - \frac{\pi}{4}
\end{aligned}
$$

Writing the cosine theorem for $\triangle COM$, denoting $\cos \left( \frac{\pi}{4} - \alpha \right) = x$, we get

$$
1 = 2 + \frac{8}{9 (2 x - \sqrt{2})^2} - \frac{8 x}{3 (2 x - \sqrt{2})}
$$

from which $x = \frac{4 \sqrt{6} - 3 \sqrt{2}}{6}$.

Answer: the angles of the triangle are $\frac{\pi}{4} \pm \arccos \frac{4 \sqrt{6} - 3 \sqrt{2}}{6}$.