Let $A M C$ be an isosceles triangle with $M$ as the vertex and such that the angle $\widehat{A M C}$ is acute. Let $B$ be the symmetric point of $A$ with respect to $M$ and let $H$ be the foot of the altitude from vertex $C$ in the triangle $A B C$. Suppose that $A H = H M$. Calculate the values of the three angles of the triangle $A B C$.
In the triangle $A M C$, the line $(C H)$ is both the altitude and the median from vertex $C$. Therefore, triangle $A C M$ is isosceles at $C$. Thus, $A C = C M$. Since $C M = M A$, triangle $A C M$ is equilateral and $\widehat{B A C} = 60^{\circ}$.

Since points $A, M$, and $B$ are collinear, we have $\widehat{C M B} = 180^{\circ} - \widehat{A M C} = 120^{\circ}$.

Since $A$ and $B$ are symmetric with respect to $M$, we have $M A = M B = M C$, so triangle $M B C$ is isosceles at $M$. Thus,

$$
180^{\circ} = \widehat{B M C} + \widehat{M B C} + \widehat{B C M} = \widehat{B M C} + 2 \widehat{M B C}
$$

Therefore, $2 \widehat{M B C} = 180^{\circ} - \widehat{B M C} = 60^{\circ}$ and $\widehat{A B C} = 30^{\circ}$.

We finally deduce that $\widehat{A C B} = \widehat{A C M} + \widehat{M C B} = 60^{\circ} + 30^{\circ} = 90^{\circ}$.

The angles of the triangle are therefore $(\widehat{A B C}, \widehat{B C A}, \widehat{C A B}) = \left(30^{\circ}, 90^{\circ}, 60^{\circ}\right)$.

Grader's Comment: The exercise is generally well done. The intuitions are good, and middle school students manage to show by several methods that $A M C$ is equilateral and then use this result effectively. Most of the time, students who did not solve the exercise did not fully exploit the equality $H M = H A$ in the problem statement, which is the key point to show that $A M C$ is isosceles at $C$.