Let $ABCD$ be a square and let $S$ be a point inside the square such that the triangle $ABS$ is equilateral. Determine the angle $\widehat{DSC}$.
Since triangle $ABS$ is equilateral, $\widehat{BAS}=60^{\circ}$. Since $\widehat{BAD}=90^{\circ}, \widehat{SAD}=90^{\circ}-$ $60^{\circ}=30^{\circ}$.

Moreover, we have $AS=AB$, because triangle $ABS$ is equilateral, and $AB=AD$, because quadrilateral $ABCD$ is a square. Therefore, $AS=AD$, which means that triangle $DAS$ is isosceles at $A$.

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

$$
180^{\circ}=\widehat{DAS}+\widehat{ADS}+\widehat{ASD}=30^{\circ}+2 \widehat{DSA}
$$

We deduce that $\widehat{DSA}=75^{\circ}$.

Similarly, we can prove that $\widehat{BSC}=75^{\circ}$. We deduce that

$$
\widehat{DSC}=360^{\circ}-\widehat{DSA}-\widehat{BSC}-\widehat{ASB}=360^{\circ}-75^{\circ}-75^{\circ}-60^{\circ}=150^{\circ}
$$

and the angle $\widehat{DSC}$ measures $150^{\circ}$.

## Alternative Solution

Once we have calculated the angle $\widehat{ADS}$, we can note that triangle $DSC$ is isosceles at $S$, given that the figure is symmetric with respect to the perpendicular bisector of segment $[AB]$. The angle $\widehat{SDC}$ is $90^{\circ}-\widehat{ASD}=15^{\circ}$. Similarly, $\widehat{DCS}=15^{\circ}$. Using the fact that the sum of the angles in triangle $DSC$ is $180^{\circ}$, we find

$$
\widehat{DSC}=180^{\circ}-\widehat{SDC}-\widehat{SCD}=180^{\circ}-15^{\circ}-15^{\circ}=150^{\circ}
$$

which is again the expected answer.

Comment from Examiners: The exercise was very well solved overall. Some students tried to use a very computational approach, based on trigonometric formulas. While these solutions sometimes work, 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.