National Math Olympiad – First Round, 2015

Let $D$ be the midpoint of the side $BC$, $E$ be the foot of the altitude from vertex $A$, and $S$ be the intersection of the angle bisector at vertex $A$ with the side $BC$. Since the angle bisector divides the angle into two equal parts it must lie between the altitude and the median. Thus the point $S$ lies between the points $E$ and $D$. We have two possibilities. Either $D$ lies between $B$ and $S$, and $E$ lies between $S$ and $C$, or $E$ lies between $B$ and $S$, and $D$ lies between $S$ and $C$. Due to symmetry we may consider only the first case.

Denote the angle $\angle BAC$ by $4\alpha$ where $0 < \alpha < \frac{\pi}{4}$. Then
$$|BE| = |AE| \tan 3\alpha, \quad |DE| = |AE| \tan 2\alpha \quad \text{in} \quad |CE| = |AE| \tan \alpha.$$
From this we deduce
$$|BD| = |BE| - |DE| = |AE|(\tan 3\alpha - \tan 2\alpha)$$
and
$$|CD| = |CE| + |DE| = |AE|(\tan \alpha + \tan 2\alpha).$$
Since $D$ is a midpoint of the side $BC$ it follows $\tan 3\alpha - \tan 2\alpha = \tan \alpha + \tan 2\alpha$ and thus
$$\tan 3\alpha - \tan \alpha - 2 \tan 2\alpha = 0.$$
We can rearrange the left side
$$\begin{align*} \tan 3\alpha - \tan \alpha - 2 \tan 2\alpha &= \frac{\sin 3\alpha}{\cos 3\alpha} - \frac{\sin \alpha}{\cos \alpha} - 2 \frac{\sin 2\alpha}{\cos 2\alpha} = \\ &= \frac{\sin 3\alpha \cos \alpha - \sin \alpha \cos 3\alpha}{\cos \alpha \cos 3\alpha} - 2 \frac{\sin 2\alpha}{\cos 2\alpha} = \frac{\sin 2\alpha}{\cos \alpha \cos 3\alpha} - 2 \frac{\sin 2\alpha}{\cos 2\alpha} = \\ &= \frac{\sin 2\alpha(\cos 2\alpha - 2 \cos \alpha \cos 3\alpha)}{\cos \alpha \cos 2\alpha \cos 3\alpha} = - \frac{\sin 2\alpha \cos 4\alpha}{\cos \alpha \cos 2\alpha \cos 3\alpha}. \end{align*}$$
To get the fourth equality we used the addition formula for sine
$$\sin 2\alpha = \sin(3\alpha - \alpha) = \sin 3\alpha \cos \alpha - \sin \alpha \cos 3\alpha$$
and to get the last equality we used the product formula for cosine
$$2 \cos \alpha \cos 3\alpha = \cos(3\alpha + \alpha) + \cos(3\alpha - \alpha) = \cos 4\alpha + \cos 2\alpha.$$
It follows that $\sin 2\alpha = 0$ or $\cos 4\alpha = 0$, thus $2\alpha = k\pi$ or $4\alpha = \frac{\pi}{2} + k\pi$, where $k \in \mathbb{Z}$. Due to $0 < \alpha < \frac{\pi}{4}$ the only possibility is $\alpha = \frac{\pi}{8}$. Thus $\angle BAC = 4\alpha = \frac{\pi}{2}$, $\angle ACB = \frac{\pi}{2} - \alpha = \frac{3\pi}{8}$, and $\angle CBA = \frac{\pi}{8}$.

Taking into account both symmetrical cases we conclude that the angles of the triangle $ABC$ are either $\angle BAC = \frac{\pi}{2}$, $\angle ACB = \frac{3\pi}{8}$, and $\angle CBA = \frac{\pi}{8}$ or $\angle BAC = \frac{\pi}{2}$, $\angle ACB = \frac{\pi}{8}$, and $\angle CBA = \frac{3\pi}{8}$.