4. Let us have $n$ points in a plane, $n \geqq 3$, such that no three of them lie on a straight line. Consider the interior angles of all triangles $\mathrm{s}$ with vertices $\mathrm{v}$ at the given points, and let the size of the smallest such angle be $\varphi$. For a given $n$, find the largest possible $\varphi$.

The regional round of category A takes place

## on Tuesday, January 13, 2015

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. Allowed aids are writing and drawing supplies and school MF tables. Calculators, laptops, and any other electronic aids are not allowed. Each problem can earn the contestant 6 points; the score threshold $\mathrm{k}$ for determining successful solvers will be set centrally after evaluating the statistical score results from all regions. This information will be communicated to the students before the competition begins.
4. We will show that the largest possible $\varphi$ is obtained when the points are placed at the vertices of a regular $n$-gon (Fig.3). In this case, each considered triangle has a circumcircle identical to the circumcircle of the chosen $n$-gon, so all considered internal angles are inscribed angles corresponding to some chord of this circle. The smallest angle will be the acute angle corresponding to the shortest chord, i.e., any of the $n$ congruent sides of the chosen $n$-gon (all other chords are diagonals, and are therefore longer). Since the central angle corresponding to such a side has a measure of $360^{\circ} / n$, the corresponding smallest angle in this case has a measure of $\varphi=180^{\circ} / n$.

![](https://cdn.mathpix.com/cropped/2024_04_17_7e8ab0ec2c7128d02808g-5.jpg?height=491&width=491&top_left_y=1942&top_left_x=791)

Fig. 3

It remains to show that for any arrangement of $n$ points, $\varphi \leqq 180^{\circ} / n$. We will prove this by contradiction. Assume that the $n$ given points are arranged such that all considered angles are greater than $180^{\circ} / n$. Construct the convex hull of this set of points, i.e., the smallest convex polygon $M$ that contains all the given points (which, according to the problem, do not lie on a straight line). Let $m$ be the number of points that lie on the boundary of the polygon $M$. Clearly, $m \leqq n$ (the vertices of the convex hull form a subset of the given set of $n$ points).

The sum of the interior angles of the polygon $M$ is $(m-2) \cdot 180^{\circ} .{ }^{1}$ Construct a segment between every pair of the given $n$ points. From each vertex of the polygon $M$, $n-1$ such segments emanate. Two of them are sides of $M$ and the remaining segments "divide" the interior angle at that vertex into $n-2$ smaller angles (Fig. 4). According to the assumption, each of these

![](https://cdn.mathpix.com/cropped/2024_04_17_7e8ab0ec2c7128d02808g-6.jpg?height=377&width=592&top_left_y=717&top_left_x=732)

Fig. 4

angles has a measure greater than $180^{\circ} / n$, so each interior angle of the polygon $M$ is greater than $(n-2) \cdot 180^{\circ} / n$. There are $m$ vertices, so the total sum of the interior angles in $M$ is greater than $m(n-2) \cdot 180^{\circ} / n$. Therefore, the following inequality holds, which we will further simplify:

$$
\begin{aligned}
(m-2) \cdot 180^{\circ} & >m(n-2) \cdot \frac{180^{\circ}}{n}, \\
n(m-2) & >m(n-2), \\
m n-2 n & >m n-2 m, \\
m & >n .
\end{aligned}
$$

This, however, contradicts the inequality $m \leqq n$.

For a complete solution, award 6 points, of which 2 points for proving that if the points are the vertices of a regular $n$-gon, then $\varphi=180^{\circ} / n$, and 4 points for proving that $\varphi \leqq 180^{\circ} / n$ always holds. If the student derives the relation $\varphi \leqq 180^{\circ} / n$ only for the case where $m=n$ (i.e., for configurations where all points lie on the boundary of the convex hull), award only 2 out of 4 points.

1 This well-known statement can be easily proven by dividing the $m$-gon into $m-2$ triangles having vertices at its vertices, or by noting that the sum of the exterior angles at all $m$ vertices is exactly $360^{\circ}$, so the sum of the interior angles of the $m$-gon is $m \cdot 180^{\circ}-2 \cdot 180^{\circ}$.