In a triangle $ABC$, we connected the incenter of the inscribed circle to the vertices of the triangle. Among the triangles thus formed, one is similar to triangle $ABC$. What are the angles of triangle $ABC$?
Solution. Let the angles of the triangle be $\alpha, \beta, \gamma$, and the center of the inscribed circle be $K$. We can assume that the triangle $BCK$ is similar to the triangle $ABC$. Since $K$ is the intersection of the angle bisectors, the angles of the triangle $BCK$ are $\frac{\beta}{2}, \frac{\gamma}{2}$, and $\pi - \left(\frac{\beta}{2} + \frac{\gamma}{2}\right) = \frac{\pi}{2} + \frac{\alpha}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c50f56640379dbd64a28g-1.jpg?height=325&width=593&top_left_y=382&top_left_x=755)

These angles must match the angles of the triangle $ABC$ in some order. It cannot be that $\frac{\pi}{2} + \frac{\alpha}{2} = \alpha$, because that would imply $\alpha = \pi$. By symmetry, we can assume that $\frac{\pi}{2} + \frac{\alpha}{2} = \gamma$. Since $\beta = \frac{\beta}{2}$ is not possible, it must be that $\beta = \frac{\gamma}{2}$ and $\alpha = \frac{\beta}{2}$, from which we get

$$
\frac{\pi}{2} + \frac{\alpha}{2} = \gamma = 2 \beta = 4 \alpha
$$

from which $\alpha = \frac{\pi}{7}$, and thus $\beta = \frac{2 \pi}{7}$ and $\gamma = \frac{4 \pi}{7}$. It is easy to verify that the angles of the triangle $BCK$ are also the same.

Therefore, the angles of the triangle $ABC$ are:

$$
\frac{\pi}{7}, \quad \frac{2 \pi}{7} \quad \text{and} \quad \frac{4 \pi}{7}
$$