Problem 1003

Solution. Let $A_{1}, A_{2}, \ldots, A_{n}$ be the vertices of the polygon $M$, and $B_{1}, B_{2}, \ldots, B_{n}$ be the midpoints of the segments $A_{1} A_{2}, A_{2} A_{3}, \ldots, A_{n} A_{1}$, as shown in the diagram below. Note that, due to the convexity of the polygon $M$, no three of the triangles

$$A_{1} A_{2} A_{3}, A_{2} A_{3} A_{4}, \ldots, A_{n-1} A_{n} A_{1}$$

have an interior common point. Therefore, the sum of the areas of these triangles is at most twice the area $P$ of the polygon $M$. If we denote the area of the polygon $B_{1} B_{2} \ldots B_{n}$ by $P_{1}$, then

$$\begin{aligned} P-P_{1} & =P_{B_{n} A_{1} B_{1}}+P_{B_{1} A_{2} B_{2}}+\ldots+P_{B_{n-1} A_{n} B_{n}} \\ & =\frac{1}{4}\left(P_{A_{n} A_{1} A_{2}}+P_{A_{1} A_{2} A_{3}}+\ldots+P_{A_{n-1} A_{n} A_{1}}\right) \\ & \leq \frac{2 P}{4}=\frac{P}{2} \end{aligned}$$

from which it follows that $P_{1} \geq \frac{P}{2}$. Equality holds if and only if every point of the polygon $M$ belongs to at least two of the triangles (1). If $n \geq 5$, let $C$ and $D$ be the intersections of the segment $A_{2} A_{4}$ with the segments $A_{1} A_{3}$ and $A_{3} A_{5}$, respectively. Then every interior point of the triangle $A_{3} C D$ is contained in exactly one of the triangles (1). It is easily verified that for $n=4$, $P_{1}=\frac{P}{2}$. Therefore, the equality $P_{1}=\frac{P}{2}$ holds if and only if $M$ is a quadrilateral.