3. Let $S$ be an arbitrary set of $n$ points $P_{1}, P_{2}, \ldots, P_{n}$ in the plane such that no three are collinear, and let $\alpha$ be the smallest of the angles $P_{i} P_{j} P_{k}$ where
$i \neq j \neq k \neq i$ and $i, j, k \in\{1,2, \ldots, n\}$. Determine $\max _{S} \alpha$ and find the sets $S$ for which this maximum value of the angle $\alpha$ is achieved.
Solution. Let $T$ denote the convex hull of the given points $P_{1}, P_{2}, \ldots, P_{n}$, i.e., the smallest convex polygon such that all points are inside it or on its boundary. We can assume that the given points are numbered such that $\measuredangle P_{n} P_{1} P_{2}$ is an angle of the polygon $T$ and that the rays $P_{1} P_{2}, P_{1} P_{3}, \ldots, P_{1} P_{n}$ are arranged in the given order, as shown in the diagram on the right. Let $Q$ be an arbitrary point on the line $P_{1} P_{n}$ such that $P_{1}$ is between $P_{n}$ and $Q$. Then

![](https://cdn.mathpix.com/cropped/2024_06_05_d0eb44e15bc267091b4bg-12.jpg?height=502&width=509&top_left_y=412&top_left_x=959)

\[
\begin{aligned}
180^{\circ} & =\measuredangle Q P_{1} P_{2}+\measuredangle P_{2} P_{1} P_{n} \\
& =\measuredangle P_{1} P_{2} P_{n}+\measuredangle P_{2} P_{n} P_{1}+\measuredangle P_{2} P_{1} P_{3}+\measuredangle P_{3} P_{1} P_{4}+\ldots+\measuredangle P_{n-1} P_{1} P_{n}
\end{aligned}
\]

In the last sum, there are exactly $n$ angles, and since none of them is less than $\alpha$, we get that $n \alpha \leq 180^{\circ}$, i.e., $\alpha \leq \frac{180^{\circ}}{n}$. Therefore, $\max _{S} \alpha \leq \frac{180^{\circ}}{n}$.

We will prove that $\max _{S} \alpha=\frac{180^{\circ}}{n}$ and that this value for $\alpha$ is achieved if and only if the points $P_{1}, P_{2}, \ldots, P_{n}$ of the set $S$ are the vertices of a regular $n$-gon. Indeed, if the equality $n \alpha=180^{\circ}$ holds, then the following conditions are satisfied (and in every vertex of the polygon $T$):

1) $\measuredangle P_{1} P_{2} P_{n}=\measuredangle P_{2} P_{n} P_{1}=\alpha$, i.e., $P_{1} P_{2}=P_{n} P_{1}$, which means that all sides of the polygon $T$ are equal.
2) $\measuredangle P_{2} P_{1} P_{n}=\measuredangle P_{2} P_{1} P_{3}+\measuredangle P_{3} P_{1} P_{4}+\ldots+\measuredangle P_{n-1} P_{1} P_{n}=(n-2) \alpha$, which means that all angles of the polygon $T$ are equal.

Thus, the equality $n \alpha=180^{\circ}$ is possible only if $P_{1}, P_{2}, \ldots, P_{n}$ are the vertices of a regular polygon. The converse statement is obvious.