(given to Camille Dessirier and Maxime Martelli). In a triangle, the height, the median, and the angle bisector relative to vertex $A$ divide the angle $\widehat{B A C}$ into four angles of the same measure $\alpha$. Determine the angles of the figure in terms of $\alpha$, and calculate $\alpha$.

In a triangle, the height, the median, and the angle bisector relative to vertex $A$ share the angle $\widehat{B A C}$ into four angles of the same measure $\alpha$. Determine the angles of the figure in terms of $\alpha$, and calculate $\alpha$.
. If we call $D$ the foot of the bisector, $A^{\prime}$ the midpoint of $[B C]$, it is easy to see that $\widehat{A B C}=\frac{\pi}{2}-\alpha, \widehat{A D C}=\frac{\pi}{2}+\alpha$ and $\widehat{A A^{\prime} C}=\frac{\pi}{2}+2 \alpha$. The other angles can be deduced elementarily.

![](https://cdn.mathpix.com/cropped/2024_05_10_8d7011708e480f09f0d6g-064.jpg?height=480&width=1221&top_left_y=729&top_left_x=424)

If we use the fact that $A^{\prime}$ is the midpoint of $[B C]$, the law of sines in triangle $A A^{\prime} C$ gives $\frac{A^{\prime} C}{A C}=$ $\frac{\sin \alpha}{\sin \left(\frac{\pi}{2}+2 \alpha\right)}$, while in triangle $A B C$, the same law of sines gives $\frac{B C}{A C}=\frac{\sin (4 \alpha)}{\sin \left(\frac{\pi}{2}-\alpha\right)}$. Therefore, $\frac{\sin (\alpha) \cos (\alpha)}{\cos (2 \alpha) \sin (4 \alpha)}=$ $\frac{1}{2}$. Since $\sin (2 \alpha)=2 \sin \alpha \cos \alpha$ and $\sin (4 \alpha)=2 \sin (2 \alpha) \cos (2 \alpha)$, we finally find $2 \cos ^{2}(2 \alpha)=1$ from which $\alpha=\frac{\pi}{8}$. Therefore, triangle $A B C$ is a right triangle at $A$. A faster way to reach this result is to note that in any triangle, if $O$ is the center of the circumcircle and $H$ the orthocenter, $O$ and $H$ are isogonal, which means that the angles $\widehat{H A B}$ and $\widehat{C A O}$ are equal. It follows from the figure that $O$ lies on the median $\left(A A^{\prime}\right)$. But $O$ also lies on the perpendicular bisector of $[B C]$, which intersects the median only at point $A^{\prime}$. Therefore, the center of the circumcircle is the midpoint of $[B C]$, which implies that the triangle is a right triangle at $A$.