5. In a regular 100-gon $A_{1} A_{2} A_{3} \ldots A_{100}$, the diagonals $\overline{A_{2} A_{100}}$ and $\overline{A_{1} A_{3}}$ intersect at point $S$. Determine the measure of the angle $\angle A_{2} S A_{3}$.

## Problems worth 10 points:
First method:

Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_4e3bea37fa3879b468d2g-4.jpg?height=363&width=925&top_left_y=1897&top_left_x=591)

The measure of one interior angle of a regular 100-gon is

$\alpha_{100}=\frac{(100-2) \cdot 180^{\circ}}{100}=176.4^{\circ}$

1 POINT

Triangle $\Delta A_{100} A_{1} A_{2}$ is isosceles. The angle $\alpha_{100}$ is the angle opposite the base.

The base angles are equal in measure $\beta$ and it holds that $2 \beta=180^{\circ}-\alpha_{100}$.

It follows that $\beta=1.8^{\circ}$.

2 POINTS

Similarly, due to the isosceles nature of triangle $\triangle A_{3} A_{1} A_{2}$, the angle $\left|\angle A_{3} A_{1} A_{2}\right|=1.8^{\circ}$.

1 POINT

The angle $\angle A_{2} S A_{3}$ is an exterior angle of triangle $\triangle A_{1} A_{2} S$, so it is equal to the sum of the two non-adjacent interior angles, i.e.,

$\left|\angle A_{2} S A_{3}\right|=\left|\angle S A_{1} A_{2}\right|+\left|\angle A_{1} A_{2} S\right|=1.8^{\circ}+1.8^{\circ}=3.6^{\circ}$.

TOTAL 6 POINTS

##