Problem 1089

a) The line on which a side of the partition polygon lies passes through two vertices of the original polygon, and no more than two such lines can pass through each vertex of the original polygon. Therefore, the number of sides of the partition polygon is not greater than the number of vertices of the original polygon.

b) The same reasoning as in part a) shows that the resulting polygon has no more than $n$ sides, and if the number of its sides is $n$, then exactly two diagonals emanate from each vertex of the original polygon, bounding the resulting polygon. Let the two diagonals emanating from vertex $A_{1}$ be $A_{1} A_{p}$ and $A_{1} A_{q}$, bounding the resulting polygon. Then $A_{p}$ and $A_{q}$ are adjacent vertices, since otherwise there would be a diagonal inside the angle $A_{\mathrm{p}} A_{1} A_{\mathrm{q}}$ that would cut the resulting polygon.

Indeed, a vertex lying between $A_{\mathrm{p}}$ and $A_{\mathrm{q}}$ would have to be connected to a vertex lying between $A_{1}$ and $A_{\mathrm{p}}$ or between $A_{1}$ and $A_{\mathrm{q}}$. By changing the direction of the vertex numbering if necessary, we can assume that $q=p+1$ and $p \leq n / 2$. If we exclude the diagonal $A_{1} A_{p}+1$, then any other diagonal bounding the resulting polygon connects one of the vertices numbered from 2 to $p$ with some vertex. Therefore, the resulting polygon can have no more than $1+\left(\frac{n}{2}-1\right) \cdot 2=n-1$ sides. To obtain an example of an $n$-gon, the cutting of which results in an $(n-1)$-gon, one can take a regular $(n-1)$-gon and cut off a small triangle, i.e., instead of vertex $A_{1}$, take two vertices $A_{1}'$ and $A_{\mathrm{n}}$, located on the sides $A_{1} A_{2}$ and $A_{1} A_{\mathrm{n}-1}$ near vertex $A_{1}$.