On a circular running track, two people are running in the same direction at constant speeds. At a certain moment, runner $A$ is 10 meters ahead of runner $B$, but after $A$ runs 22 meters, $B$ catches up.

How many points on the track can $B$ later overtake $A$?
Let the speeds of the two runners be denoted by $v_{A}$ and $v_{B}$, respectively. According to the problem,

$$
\frac{v_{A}}{v_{B}}=\frac{22}{32}=\frac{11}{16}
$$

This means that while runner $A$ completes 11 laps, runner $B$ completes 16. If the time required for this is denoted by $T$, and we start observing the events when $B$ overtakes $A$ as specified in the problem, then after time $T$ has elapsed, the runners will be at the same point $S$ on the track. Therefore, their combined movement is periodic with a period of $T$, and a meeting can only occur at the points where this happens within the time $T$.

During this given time, $B$ runs 5 more laps than $A$, so $B$ overtakes $A$ five times, the last of which is exactly at point $S$. Since 11 and 16 have a greatest common divisor of 1, the movement does not have a period shorter than $T$. Otherwise, the runners would also reach point $S$ after a time $T^{\prime}(<T)$, and the number of laps they would complete in this time would be a positive integer smaller than 11 and 16, respectively, with a ratio of $\frac{v_{A}}{v_{B}}=\frac{11}{16}$, which is not possible. If, however, $T$ is the shortest period of the movement, then the five meetings occur at different points on the track.

Therefore, there are five points on the track where $B$ can overtake $A$.

Remarks. 1. It is obvious that the five meeting points form the vertices of a regular pentagon.

2. It is clear from the solution that the shape of the track is irrelevant, and the result remains valid for any track whose shape is a closed curve that does not intersect itself.