14.9. (SFRY, 74). For arbitrary $n \geqslant 3$ points $A_{1}$, $A_{2}, \ldots, A_{n}$ on a plane, no three of which lie on the same line, let $\alpha$ denote the smallest of the angles $\angle A_{i} A_{j} A_{k}$ formed by triples $A_{i}, A_{j}, A_{k}$ of distinct points. For each value of $n$, find the greatest value of $\alpha$. Determine for which arrangements of points this value is achieved.
14.9. We will prove that the greatest value of $\alpha$ is $180^{\circ} / n$. Indeed, let some arrangement of points on a plane correspond to the value $\alpha$. Consider a line, say $A_{1}^{\prime} A_{2}^{\prime}$, such that all points are located in one half-plane relative to this line, and choose a point $A_{3}^{\prime}$ for which the angle $\angle A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime}$ is maximal (Fig. 98). Then all other points lie inside this angle and

$$
\angle A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime} \geqslant \alpha(n-2)
$$

since each of the $n-2$ angles between adjacent rays $A_{2}^{\prime} A_{i}$ ( $A_{i} \neq A_{2}^{\prime}$ ) is not less than $\alpha$. Next, choose a point $A_{4}^{\prime}$ for which the angle $\angle A_{2}^{\prime} A_{3}^{\prime} A_{4}^{\prime}$ is maximal. Then all points

$$
A_{i} \in\left\{A_{2}^{\prime} ; A_{3}^{\prime} ; A_{3}^{\prime}\right\}
$$

lie inside this angle and

$$
\angle A_{2}^{\prime} A_{3}^{\prime} A_{1}^{\prime} \geqslant \alpha(n-2)
$$

If $A_{4}^{\prime} \neq A_{1}^{\prime}$ and $A_{4}^{\prime} \neq A_{2}^{\prime}$, then we

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-250.jpg?height=513&width=511&top_left_y=249&top_left_x=545)

Fig. 98 take analogously the point $A_{5}^{\prime}$ and so on. Since the number of points is $n$, one of the points in the sequence $A_{1}^{\prime}, A_{2}^{\prime}, A_{3}^{\prime}, \ldots$ will necessarily repeat for the first time. Let this happen after choosing the point $A_{k}^{\prime}$, i.e., the angle $\angle A_{k-1}^{\prime} A_{k}^{\prime} A_{j}$ is maximal for some point

$$
A_{j}=A_{i}^{\prime} \in\left\{A_{1}^{\prime} ; \ldots ; A_{k-2}^{\prime}\right\}
$$

Then, if $i \neq 1$, the point $A_{1}^{\prime}$ lies inside the angle $\angle A_{k-1}^{\prime} A_{k}^{\prime} A_{i}^{\prime}$, and hence inside the convex polygon $A_{i}^{\prime} A_{i+1}^{\prime} \ldots A_{k-1}^{\prime} A_{k}^{\prime}$, which contradicts its choice. Therefore, $i=1$ and the convex $k$-gon $A_{1}^{\prime} A_{2}^{\prime} \ldots A_{k}^{\prime}$ has the sum of angles

$$
180^{\circ}(k-2) \geqslant k \cdot \alpha(n-2)
$$

from which

$$
\alpha \leqslant \frac{180^{\circ}(k-2)}{(n-2) k}=\frac{180^{\circ}}{n-2}\left(1-\frac{2}{k}\right) \leqslant \frac{180^{\circ}}{n-2}\left(1-\frac{2}{n}\right)=\frac{180^{\circ}}{n} .
$$

Equality $\alpha=180^{\circ} / n$ is possible only if $k=n$,

$$
\angle A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime}=\angle A_{2}^{\prime} A_{3}^{\prime} A_{4}^{\prime}=\ldots=\angle A_{k}^{\prime} A_{1}^{\prime} A_{2}^{\prime}=\alpha(n-2)
$$

and all diagonals emanating from any angle of the $n$-gon $A_{1}^{\prime} \ldots A_{n}^{\prime}$ divide it into equal angles $\alpha$. Such an $n$-gon can only be regular, since for any value $l=1, \ldots, n$ we have

$$
\angle A_{l}^{\prime} A_{l+2}^{\prime} A_{l+1}^{\prime}=\angle A_{l+2}^{\prime} A_{l}^{\prime} A_{l+1}^{\prime} \text {, i.e., } \quad A_{l}^{\prime} A_{l+1}^{\prime}=A_{l+1}^{\prime} A_{l+2}^{\prime}
$$

(we assume $A_{n+1}^{\prime}=A_{1}^{\prime}, A_{n+2}^{\prime}=A_{2}^{\prime}$ ) (Fig. 98), hence, the $n$-gon
must have not only equal angles but also equal sides. Finally, the vertices of a regular $n$-gon indeed satisfy the condition $\alpha=180^{\circ} / n$. This becomes clear if we circumscribe a circle around the $n$-gon and note that all angles between adjacent diagonals emanating from any vertex subtend an arc of $360^{\circ} / n$, and thus are equal to $180^{\circ} / n$.

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-251.jpg?height=453&width=423&top_left_y=342&top_left_x=139)

Fig. 99