## Task A-1.5.

Given is a regular 2013-gon $A_{1} A_{2} \ldots A_{2013}$. Let $S$ be the intersection of the segments $\overline{A_{1} A_{4}}$ and $\overline{A_{3} A_{5}}$. Determine the measure of the angle $\varangle A_{3} S A_{4}$.
## Solution.

Let $\varphi$ be the exterior angle of a regular 2013-gon. We have $\varphi=\frac{360^{\circ}}{2013}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_abcf4bcd4243d7db1a46g-04.jpg?height=285&width=1219&top_left_y=2239&top_left_x=384)

Considering the isosceles triangle $A_{3} A_{4} A_{5}$, we get $\varangle A_{5} A_{3} A_{4}=\frac{1}{2} \varphi$.

Notice that $A_{1} A_{2} A_{3} A_{4}$ is an isosceles trapezoid.

Therefore, $\varangle A_{1} A_{4} A_{3}=180^{\circ}-\varangle A_{2} A_{3} A_{4}=\varphi$.

Now, considering the triangle $A_{3} A_{4} S$, we see that

$$
\begin{aligned}
\varangle A_{3} S A_{4} & =180^{\circ}-\varangle S A_{3} A_{4}-\varangle A_{3} A_{4} S & & 1 \text { point } \\
& =180^{\circ}-\frac{1}{2} \varphi-\varphi=180^{\circ}-\frac{3}{2} \varphi . & & 1 \text { point }
\end{aligned}
$$

Finally, the desired angle is

$$
\varangle A_{3} S A_{4}=180^{\circ}-\frac{3}{2} \cdot \frac{360^{\circ}}{2013}=180^{\circ}-180^{\circ} \cdot \frac{3}{2013}=\frac{2010}{2013} \cdot 180^{\circ}=\frac{670}{671} \cdot 180^{\circ} .
$$

Note: Here, for clarity, all angles are expressed in terms of $\varphi$ (which is also the central angle). Analogous results expressed in terms of another angle or explicitly written out should be credited. The final result should be recognized even if the last fraction is not simplified.