360. Polygon on a chessboard. Let the chessboard consist of squares with a side length of 4. Furthermore, let a regular $4 n$-gon of "radius" $1^{*}$ be thrown onto this board. Determine the probability that this polygon will intersect the side of any square.
360. We can assume without loss of generality that the center of the polygon falls inside a certain square at a point which is at a distance $l$ from the nearest side of this square. Let us drop a perpendicular from the center of our polygon to this side and consider the angles formed by this perpendicular and the "radii" (i.e., the lines connecting the center with the vertices) of the polygon. One of these angles, the smallest in absolute value, we denote by $\theta$. The vertex lying on the corresponding "radius" is the closest to the given side, since the distance from it to this side is $l-\cos \theta$, and this value takes the minimum when $\cos \theta$ is maximal.

If the polygon intersects the given side of the square, then $l$ must be less than $\cos \theta$. Since the number of sides of the polygon is a multiple of 4, the situation is the same with respect to all sides of the square. Therefore, the center of the polygon must be no further than $\cos \theta$ from one of the sides of the square. The probability of such an event is

$$
\frac{16 \cos \theta-4 \cos ^{2} \theta}{16}
$$

This value represents the ratio of the area of the region where the center of the polygon, intersecting the square, can be located, to the area of the entire square.

It is clear that $0 \leqslant \theta$ and $\theta \leqslant \frac{\pi}{4 n}$. Therefore, the probability that our angle will fall into an infinitesimally small interval between some $\theta$ and $\theta+d \theta$ is

$$
\frac{4 n d \theta}{\pi}
$$

The probability that our polygon will intersect the square and at the same time the angle $\theta$ will fall into the required interval is the product of (1) and (2). The desired probability will now be obtained by integrating over all values of $\theta$, that is, it will be equal to

$$
\frac{4 n}{\pi} \int_{0}^{\frac{\pi}{4 n}}\left(\cos \theta-\frac{1}{4} \cos ^{2} \theta\right) d \theta=\frac{4 n}{\pi} \sin \frac{\pi}{4 n}-\frac{n}{4 \pi} \sin \frac{\pi}{2 n}-\frac{1}{8}
$$

[K. Kessel, A. M. M., 71, 81 (January 1970).]