4. Given a convex 2004-gon whose side lengths are greater than 2. Around each vertex, a circle with radius 1 is described. Calculate the sum of the areas of the parts of the circles enclosed by the interior angles of the polygon.
Solution. The sum of the exterior angles of a polygon is $2 \pi$. Let $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{2004}$ be the angles of the polygon and $\beta_{1}, \beta_{2}, \ldots, \beta_{2004}$ be their corresponding exterior angles. We have $\alpha_{i}+\beta_{i}=\pi$ and $\sum_{i=1}^{2004} \beta_{i}=2 \pi$. Furthermore,

$$
\sum_{i=1}^{2004}\left(\alpha_{i}+\beta_{i}\right)=2004 \pi
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_e2113f1d4cf0d5f382ffg-2.jpg?height=423&width=407&top_left_y=1252&top_left_x=1075)

therefore,

$$
\sum_{i=1}^{2004} \alpha_{i}=2004 \pi-\sum_{i=1}^{2004} \beta_{i}=2002 \pi
$$

The desired sum is

$$
P=\sum_{i=1}^{2004} \frac{r^{2} \alpha_{i}}{2}=\frac{1}{2} \sum_{i=1}^{2004} \alpha_{i}=1001 \pi
$$