18. (MON 4) Given a convex polygon $A_{1} A_{2} \ldots A_{n}$ with area $S$, and a point $M$ in the same plane, determine the area of polygon $M_{1} M_{2} \ldots M_{n}$, where $M_{i}$ is the image of $M$ under rotation $\mathcal{R}_{A_{i}}^{\alpha}$ around $A_{i}$ by $\alpha, i=1,2, \ldots, n$.
18. Consider the triangle $M A_{i} M_{i}$. Obviously, the point $M_{i}$ is the image of $A_{i}$ under the composition $C$ of rotation $R_{M}^{\alpha / 2-90^{\circ}}$ and homothety $H_{M}^{2 \sin (\alpha / 2)}$. Therefore, the polygon $M_{1} M_{2} \ldots M_{n}$ is obtained as the image of $A_{1} A_{2} \ldots A_{n}$ under the rotational homothety $C$ with coefficient $2 \sin (\alpha / 2)$. Therefore $S_{M_{1} M_{2} \ldots M_{n}}=4 \sin ^{2}(\alpha / 2) \cdot S$.