Maximum of minimum angle

Let $S$ be the set of these points. Let $C$ be its convex hull . Let $p \in S$ be a vertex of $C$ that has the smallest interior angle, and call this interior angle $\theta$ . Essentially, $p$ is a point of $S$ such that the remaining vertices are in the smallest possible angle.

Suppose the convex hull contains $n$ vertices, then the sum of its interior angles is $180^\circ \cdot (n-2)$ . This sum is divided into $n$ interior angles (the $n$ vertices of the convex hull), thus the minimum among them must be not greater than $180^\circ \cdot \frac{n-2}{n}$ . But this minimum is precisely $\theta$ by definition, so $\theta \le 180^\circ \cdot \frac{n-2}{n}$ .

Now, $\theta$ encompasses the remaining $6$ points of $S$ . Drawing a ray from $p$ to each of the other $6$ points divides $\theta$ into $5$ angles (and one exterior angle). Again, applying this real version of Pigeonhole Principle, the sum of five angles is $\theta$ and thus there is at least one of them that is not greater than $\frac{\theta}{5}$ .

But this is one of the angles. Thus $\alpha$ cannot be greater than this. Thus, we have:

$\begin{aligned} \alpha^\circ &\le \frac{\theta}{5} \\ &\le \frac{1}{5} \cdot \left( 180^\circ \cdot \frac{n-2}{n} \right) \end{aligned}$

Since $n \le 7$ and $\frac{n-2}{n}$ is increasing for positive $n$ , we can bound this further:

$\begin{aligned} \alpha^\circ &\le \frac{1}{5} \cdot \left( 180^\circ \cdot \frac{n-2}{n} \right) \\ &\le \frac{1}{5} \cdot \left( 180^\circ \cdot \frac{7-2}{7} \right) \\ &= \frac{1}{5} \cdot 180^\circ \cdot \frac{5}{7} \\ &= \frac{180}{7}^\circ \end{aligned}$

So now we have an upper bound, $\alpha \le \frac{180}{7}$ . Can it be reached?

Indeed, it can. Just consider a regular heptagon.

Note that a regular polygon is circumscribable:

Thus, an angle made of three vertices of such polygon is an inscribed angle of the circle, whose measure we can simply compute as $180^\circ \cdot \frac{\text{arc length}}{\text{circle circumference}}$ .

The smallest angle made from the vertices of a circumscribable polygon must then be the angle that encloses the shortest arc (since the circle circumference is constant). For a regular heptagon, all the seven small arcs are equal in length, being 1/7 of a circle's circumference, so the smallest angle is $\frac{180}{7}^\circ$ . This matches with our upper bound, thus proving that the maximum value of $\alpha$ is $\boxed{\frac{180}{7} \approx 25.714}$ .