Exercise 2. 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 $A B C$.
Solution to Exercise 2

![](https://cdn.mathpix.com/cropped/2024_06_04_01f3cce223794094ee24g-04.jpg?height=509&width=780&top_left_y=405&top_left_x=638)

In 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 \(C\). Hence, \(A C = C M\). 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 \((\widehat{A B C}, \widehat{B C A}, \widehat{C A B}) = (30^{\circ}, 90^{\circ}, 60^{\circ})\).

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 \(H M = H A\) from the problem statement, which is the key point to show that \(A M C\) is isosceles at \(C\).