Let $ABCD$ be a square and let $S$ be a point outside the square $ABCD$ such that the triangle $BCS$ is equilateral. We denote $N$ as the midpoint of the segment $[AS]$ and $H$ as the midpoint of the segment $[CD]$. Let $P$ be the midpoint of the segment $[BS]$.

1) Calculate the angle $\widehat{BPN}$.
2) Calculate the angle $\widehat{NHC}$.
Since point $N$ is the midpoint of segment $[A S]$ and point $P$ is the midpoint of segment $[B C]$, according to Thales' theorem, the lines $(N P)$ and $(A B)$ are parallel. We deduce that $\widehat{N P S}=\widehat{A B S}$. Since the quadrilateral $A B C D$ is a square, $\widehat{A B C}=90^{\circ}$. Since the triangle $S B C$ is equilateral, $\widehat{S B C}=60^{\circ}$. Thus, $\widehat{A B S}=90^{\circ}+60^{\circ}=150^{\circ}$.

We deduce that

$$
\widehat{B P N}=180^{\circ}-\widehat{N P S}=180^{\circ}-\widehat{A B S}=180^{\circ}-150^{\circ}=30^{\circ}
$$

2) We use again the fact that the lines $(N P)$ and $(A B)$ are parallel. Since the sides $[A B]$ and $[C D]$ are parallel, the lines $(N P)$ and $(C D)$ are also parallel. Furthermore, according to Thales' theorem, $N P=\frac{1}{2} A B$ and since the quadrilateral $A B C D$ is a square, $A B=C D$. Thus, $N P=\frac{1}{2} C D=H C$ since point $H$ is the midpoint of segment $[C D]$.

The quadrilateral $H C P N$ has two opposite sides equal and parallel, so it is a parallelogram. Its opposite angles are equal, so we have that $\widehat{N H C}=\widehat{N P C}$.

Since point $P$ is the midpoint of segment $[B S]$ and the triangle $B C S$ is equilateral, the line $(C P)$ is the height from vertex $C$ in the triangle $B C S$. Therefore, $\widehat{C P S}=90^{\circ}$. We established in the previous question that $\widehat{N P S}=150^{\circ}$. Thus, $\widehat{N P C}=\widehat{N P S}-\widehat{C P S}=150^{\circ}-90^{\circ}=60^{\circ}$

We have therefore $\widehat{N H C}=60^{\circ}$.

Comment: The problem consisted of two questions. The first question was solved by the majority of students. The second question, more difficult, was still fully solved by a significant number of students. Several approaches were possible. The most elegant one involved using elementary geometry results such as Thales' theorem and identifying the quadrilateral $C P N H$ as a parallelogram. Several students attempted, with varying success, an analytical approach. This sometimes led to answers involving cosines or tangents. Such results, often easily simplifiable, could not earn full marks. Finally, a few students misread the statement and misplaced point $S$. Although the rest of their reasoning was often correct relative to their figure, they could not earn all the points for the exercise.