Task B-4.5. In a regular $n$-gon, the radius of the circumscribed circle is $2, S$ is its center, $A, B$ are consecutive vertices of the $n$-gon. Determine $n$ and the interior angle of the regular $n$-gon if the dot product $\overrightarrow{S A} \cdot \overrightarrow{S B}=2 \sqrt{2}$.
Solution. Let $\alpha$ be the central angle of a regular $n$-gon.

From $\overrightarrow{S A} \cdot \overrightarrow{S B}=2 \sqrt{2}$ it follows that

$$
|S A| \cdot|S B| \cos \alpha=2 \sqrt{2}
$$

which means $2 \cdot 2 \cdot \cos \alpha=2 \sqrt{2}$, so

$$
\cos \alpha=\frac{\sqrt{2}}{2}
$$

thus $\alpha=45^{\circ}$.

Since $\alpha=\frac{360^{\circ}}{n}=45^{\circ}$, we have $n=8$.

The interior angle is $2 \cdot\left(90^{\circ}-\frac{\alpha}{2}\right)=135^{\circ}$.