Let's first prove that if $A$ and $B$ are adjacent vertices of an $n$-gon, then a diagonal can be drawn from $A$ or from $B$. The case where the internal angle of the polygon at vertex $A$ is greater than $180^{\circ}$ is discussed in the solution to problem 22.20, part a). Now, let's assume that the angle at vertex $A$ is less than $180^{\circ}$. Let $B$ and $C$ be the vertices adjacent to $A$. If there are no other vertices of the polygon inside triangle $ABC$, then $BC$ is a diagonal. If $P$ is the vertex of the polygon closest to $A$ and lying inside triangle $ABC$, then $AP$ is a diagonal.

Therefore, the number of vertices from which a diagonal cannot be drawn does not exceed $[n/2]$ (i.e., the integer part of $n / 2$). On the other hand, there exist $n$-gons for which this estimate is achieved (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_57096a2c4cfa59fdac6dg-09.jpg?height=712&width=1240&top_left_y=1622&top_left_x=409)
[ Non-Convex Polygons ]

What is the maximum number of vertices of a non-convex $n$-gon from which no diagonal can be drawn?

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