$10 \cdot 88$ Let points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ all lie on the unit sphere, find $\min \left\{A_{i} A_{j} \mid 1 \leqslant i<j \leqslant 5\right\}$'s maximum value and determine all cases where the maximum value is achieved.
[Solution] Let the center of the sphere be $O$. Since the 5 points are all on the sphere, the size of $\angle A_{i} O A_{j}$ can be used to replace the distance $A_{i} A_{j}$. Clearly, when the 5 points are taken as 5 vertices of a regular octahedron inscribed in the sphere, we have
$$
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\min \left\{\angle A_{i} O A_{j} \mid 1 \leqslant i0$, that is, point $A_{1}$ is on the boundary plane dividing the upper half-space into two quadrants. By the pigeonhole principle, there must be 1 quadrant (including the boundary) that contains two given points. According to (3), one of these two points must be at the "corner" of the quadrant (i.e., the intersection of the equator and the quadrant plane on the sphere), and the other point must be on the opposite radius. In this way, each of these two points is a common point of two quadrants. Similarly, points $A_{1}, A_{2}, A_{3}$, $A_{4}$ are each common points of two quadrants, meaning they are located on the 4 radii that divide the northern hemisphere into 4 quadrants. If one of these points is not on the equator, then its two adjacent points are on the equator (each in the corner of a quadrant), forming a pair of antipodal points.

Since there are always two points that are antipodal, let them be $A_{1}, A_{2}$. Then, for any third point $A_{3}$, it cannot simultaneously satisfy $\angle A_{1} O A_{3}>\frac{\pi}{2}, \angle A_{2} O A_{3}>\frac{\pi}{2}$. Therefore, the maximum value of $\min \angle A_{i} O A_{j}$ is $\frac{\pi}{2}$. When the 5 points achieve this maximum value, there must be two points that are antipodal. When these two points are taken as the poles, the other 3 points are all on the equator and the angles between each pair of them with respect to the center $O$ are not less than $\frac{\pi}{2}$.