A simple polygon's plane is divided into a positive and a negative half-plane by each side line of the polygon. When traversing the perimeter of the polygon in a counterclockwise direction, we call the half-plane on the left side of our hand the positive one.

For which $n$ is the following statement true: the intersection of the positive half-planes lying on the side lines is a single $n$-sided, non-self-intersecting polygon and is not empty?
The statement is true for the case $n=3$. For convex polygons, indeed, the intersection of the positive half-planes is the polygon itself, and every triangle is convex.

$1987-10-295-1$. eps

Figure 1

We will show that the statement is true for the cases $n=4$ and $n=5$ as well. Let $n=5$ first. According to our previous observation, the statement is true for a convex pentagon. If the pentagon is not convex, then it can have at most two concave angles, which are either adjacent or not. In the first case - using the notation of Figure 1 - points $B$ and $C$ are interior points of the convex hull of the pentagon $A D E$. Therefore, $A B$ and $D C$ intersect at an interior point $F$ of the pentagon, and the segments $E D$ and $A E$ intersect at interior points $H$ and $G$, respectively. Thus, the intersection of the positive half-planes is the convex quadrilateral $F H E G$, which is certainly not empty.

$$
\text { 1987-10-295-2.eps }
$$

Figure 2

In the second case, let the angles at vertices $E$ and $C$ be concave (Figure 2). Then the intersection of the positive half-planes defined by the sides contains the segment $E C$, so it is not empty.

The case $n=4$ can be examined similarly to the previously discussed first case.

$1987-10-296-1$. eps

Figure 3

If $n \geqq 6$, then the statement is not true, as shown by the positive half-planes defined by the sides $B C$ and $E D$ of the polygon in Figure 3.

Thus, the statement of the problem is true for the cases $5 \geqq n(\geqq 3)$.