Exercise 3. 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 3

![](https://cdn.mathpix.com/cropped/2024_06_04_472471a3fdefe9b0e85fg-07.jpg?height=599&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' Comment: The exercise was generally well solved. Some students tried to use a very computational approach, based on trigonometric formulas. While these solutions sometimes worked, they are still quite time-consuming and complicated, and we therefore recommend prioritizing angle chasing as the first approach to a geometry problem in this format.