Exercise 10. Let $A B C D$ be a square and let $S$ be a point inside the square such that the triangle $A B S$ is equilateral. Determine the angle $\widehat{D S C}$.
Solution to Exercise 10

![](https://cdn.mathpix.com/cropped/2024_06_04_472471a3fdefe9b0e85fg-17.jpg?height=605&width=606&top_left_y=360&top_left_x=728)

Since triangle $A B S$ is equilateral, $\widehat{B A S}=60^{\circ}$. Since $\widehat{B A D}=90^{\circ}, \widehat{S A D}=90^{\circ}-60^{\circ}=30^{\circ}$. Furthermore, we have $A S=A B$, because triangle $A B S$ is equilateral, and $A B=A D$, because quadrilateral $A B C D$ is a square. Therefore, $A S=A D$, which means that triangle $D A S$ is isosceles at $A$.

The sum of the angles in triangle $D A S$ is $180^{\circ}$, so

$$
180^{\circ}=\widehat{D A S}+\widehat{A D S}+\widehat{A S D}=30^{\circ}+2 \widehat{D S A}
$$

We deduce that $\widehat{D S A}=75^{\circ}$.

Similarly, we can prove that $\widehat{B S C}=75^{\circ}$. Therefore,

$$
\widehat{D S C}=360^{\circ}-\widehat{D S A}-\widehat{B S C}-\widehat{A S B}=360^{\circ}-75^{\circ}-75^{\circ}-60^{\circ}=150^{\circ}
$$

and the angle $\widehat{D S C}$ measures $150^{\circ}$.

Alternative Solution $n^{\circ} 1$ Once we have calculated the angle $\widehat{A D S}$, we can note that triangle $D S C$ is isosceles at $S$, given that the figure is symmetric with respect to the perpendicular bisector of segment $[A B]$. The angle $\widehat{S D C}$ is $90^{\circ}-\widehat{A S D}=15^{\circ}$. Similarly, $\widehat{D C S}=15^{\circ}$. Using the fact that the sum of the angles in triangle $D S C$ is $180^{\circ}$, we find

$$
\widehat{D S C}=180^{\circ}-\widehat{S D C}-\widehat{S C D}=180^{\circ}-15^{\circ}-15^{\circ}=150^{\circ}
$$

which is again the expected answer.

Graders' Comments: The exercise was generally very well done by a large majority of students. A few general remarks:

$\triangleright$ Several papers confuse the words "equilateral" and "isosceles." We recall that a triangle $X Y Z$ is said to be isosceles at $X$ if the sides $[X Y]$ and $[X Z]$ are of the same length, whereas a triangle $X Y Z$ is said to be equilateral if all three sides are of the same length. Thus, an equilateral triangle is always isosceles, but the converse is not true.

$\triangleright$ We emphasize the importance of reading the problem statement carefully, as several students placed point $S$ outside the square instead of inside, thus penalizing themselves despite often correct reasoning.
$\triangleright$ Regarding the writing, it is too often neglected, even by good students: we remind that a figure does not constitute a proof; a drawing is there to illustrate and accompany a reasoning, not to replace it. Be careful also not to over-write: it is not necessary to detail too much, at the risk of losing time that could have been used to solve other exercises; here, arguments such as the symmetry of the figure could have reduced the size of the proof.

$\triangleright$ Furthermore, some students ventured into analytic geometry calculations: this can sometimes work, but it is not recommended, as an analytic proof that does not work is not valued, and there are often simpler methods.

$\triangleright$ Finally, it is important to make a large and accurate figure, as several students made mistakes and did not realize their errors due to an absent or poorly made figure, which can be misleading.