Exercise 15. Let $A B C D$ be a square and let $S$ be a point outside the square $A B C D$ such that the triangle $B C S$ is equilateral. Let $N$ be the midpoint of the segment $[A S]$ and $H$ the midpoint of the segment $[C D]$. Let $P$ be the midpoint of the segment $[B S]$.

1) Calculate the angle $\widehat{B P N}$.
2) Calculate the angle $\widehat{N H C}$.
## Solution to Exercise 15

![](https://cdn.mathpix.com/cropped/2024_06_04_8ce8798f2c48d3d66a92g-24.jpg?height=711&width=1350&top_left_y=684&top_left_x=250)

1) 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 that are 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 from the graders The problem consisted of two questions. The first question was solved by a large 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 approach 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 points. Finally, a few students misread the statement and incorrectly placed point $S$. While the rest of their reasoning was often correct relative to their figure, they could not earn all the points for the exercise.