24. (POL 5) For points $A_{1}, \ldots, A_{5}$ on the sphere of radius 1, what is the maximum value that $\min _{1 \leq i, j \leq 5} A_{i} A_{j}$ can take? Determine all configurations for which this maximum is attained. (Or: determine the diameter of any set $\left\{A_{1}, \ldots, A_{5}\right\}$ for which this maximum is attained.)
24. Instead of Euclidean distance, we will use the angles $\angle A_{i} O A_{j}$, with $O$ denoting the center of the sphere. Let $\left\{A_{1}, \ldots, A_{5}\right\}$ be any set for which $\min _{i \neq j} \angle A_{i} O A_{j} \geq \pi / 2$ (such a set exists: take for example five vertices of an octagon). We claim that two of the $A_{i}$'s must be antipodes, thus implying that $\min _{i \neq j} \angle A_{i} O A_{j}$ is exactly equal to $\pi / 2$, and consequently that $\min _{i \neq j} A_{i} A_{j}=\sqrt{2}$. Suppose no two of the five points are antipodes. Visualize $A_{5}$ as the south pole. Then $A_{1}, \ldots, A_{4}$ lie in the northern hemisphere, including the equator (but excluding the north pole). No two of $A_{1}, \ldots, A_{4}$ can lie in the interior of a quarter of this hemisphere, which means that any two of them differ in longitude by at least $\pi / 2$. Hence, they are situated on four meridians that partition the sphere into quarters. Finally, if one of them does not lie on the equator, its two neighbors must. Hence, in any case there will exist an antipodal pair, giving us a contradiction.