## Problem III - 3

In triangle $A B C$, the internal angle bisector $A L$ (where $L$ lies on side $B C$), the altitude $B H$ (where $H$ lies on side $A C$), and the median $C M$ (where $M$ lies on side $A B$) are drawn.

It is known that the angles $\angle C A L, \angle A B H$, and $\angle B C M$ are equal.

Determine, with reasoning, the measures of the angles of triangle $A B C$.
Solution

Let $\alpha=\angle C A L=\angle A B H=\angle B C M$. Then $\angle B A C=2 \alpha$.

If the angle $B A C$ were obtuse, the point $H$ would be on the extension of side $A C$ beyond $A$, and therefore $2 \alpha=\alpha+90^{\circ}$, from which $\alpha=90^{\circ}$. This is impossible, because all half-angles of a triangle are acute (they sum up to $90^{\circ}$ among the three). Also, $\angle B A C \neq 90^{\circ}$, since $\alpha \neq 0$. Therefore, $A$ is acute, and then, considering the triangle $A B H$, we obtain

$$
\alpha+2 \alpha=90^{\circ} \Leftrightarrow \alpha=30^{\circ}
$$

It is easy to see that the converse is also true, that is, if in a triangle the angle $A$ is $60^{\circ}$, then $\angle C A L=\angle A B M$.

Consequently, the triangles sought are those that have the angle $\angle B A C=60^{\circ}$, and whose median $C M$ forms an angle of $30^{\circ}$ with the side $B C$.

The set of points from which the segment $B M$ is seen under an angle of $30^{\circ}$, located on one side of the line $A B$, is an arc of a circle that can intersect $A C$ in at most two points.

This condition is satisfied by the points $C$ with $A B C$ equilateral, and $H$, with $H M$ the midline parallel to $A B C$ equilateral ( $\left.\angle B H M=\angle H B C=30^{\circ}\right)$.

Therefore, the only solutions to the problem are

$$
A=B=C=60^{\circ} \text { or } A=60^{\circ}, B=30^{\circ}, C=90^{\circ}
$$