I originally thought that this problem would be trivial to solve, but it has proven more difficult than I expected.
Suppose there are two runners on a circular track with radius r. Runner a runs at a speed $V_a$, which corresponds to an angular velocity $\omega_a=V_a/2\pi*r$. Runner b has velocity $V_b$ which normally would correspond to an angular velocity $\omega_b=V_b/2\pi*r$.
However, runner b has a peculiar habit. He is slower than runner a, and wants to minimize the distance between himself and his friend. Therefore, when runner a reaches 180 degrees around the track from him, he turns around and starts running in the opposite direction around the track. When he reaches runner a, he turns back around and starts running in the same direction until runner a is 180 degrees ahead, after which he does the same. And the cycle repeats.
Here is the question: What is the average angular velocity ${\omega_b}_{avg}$ of runner b? It's tempting to try and think in terms of how much distance runner b will lose with each revolution, but this is complicated because it depends on runner a's position. The result would seem to depend on the least common multiple of ${\omega_b}_{avg}$ and $\omega_a$ but since we don't know ${\omega_b}_{avg}$ this is not trivial to compute.
Is there a simple solution that I am missing? How would one go about solving this problem?
We can find the answer to be:
$$\vec{\omega}_b = \frac{\omega_b^2}{\omega_a}$$
where $\vec\omega_b$ is the average angular velocity (a vector), and $\omega_a$ and $\omega_b$ are the runner's speeds (scalars).
Imagine the runners on a circular track that, for simplicity's sake, is the unit circle. As a starting point, place the runners opposite from one another as if they've just had their second alignment, so they're both moving CCW.
It's clear that to realign, runner A will have to overtake B, then keep going to a point opposite from B. Specifically, A must move by an angle of $\tau + x$ while B moves by an angle of $x$--A moves exactly a full circle farther. A will pass B at angle $x/2$ CCW from B's starting position.
To simplify a bit, let $\omega_a = k\omega_b$. (Hence $k>1$.) Then taking these together, it's clear that when they next align, we will have:
$$\tau + x = kx \implies x=\tau/(k-1)$$
Note that the angle $x$ traveled by B will be the same no matter where on the track the runners are when they're aligned. We didn't specify a starting point because we didn't need to. And this solution passes a common sense test: if $k=2$, so A is twice as fast as B, we get $x=\tau$; B makes a full circle, while A runs two laps. Visualize it or draw it out to convince yourself of how it works.
Once they've aligned, B instantly turns around. This time B must move by an angle of $-y$ while A moves by $\tau -y$; that means this angle (and the time taken) will be shorter. Hence by the prior algebra we have
$$\tau - y = -ky \implies y = -\tau/(k+1)$$
And again, common sense test: for $k=2$, B will move by $-\tau/3$, one-third of the circle, while A moves $2\tau/3$ in the opposite direction. If B started at angle $0$, then B has moved to $2\tau/3$, and A has moved to $\tau/6$ (starting from $\tau/2$ and passing
go$0$).This means that every two alignments, B will move an angle (angular displacement) of
$$\vec{\theta} = x+y = \frac{\tau}{k-1} - \frac{\tau}{k+1} = \frac{2\tau}{k^2 -1}$$
But note their total angular distance traveled is the difference rather than the sum (as we made $y$ negative to begin with). So the total angular distance (scalar) is
$$\theta = x-y = \frac{2k\tau}{k^2-1}$$
How far does A move? We weren't asked, so we won't bother! However, we were asked for an average velocity, and we've found... an angle. Two angles, actually. But here's the cool thing: our angle is independent of the velocities; it depends only on k, their ratio. And runner B will traverse this angle (which, remember, includes both the forward and backward part) in a certain time related to their speed. That time is $t=\theta / \omega_b$. But $\vec{\omega}_b = \vec\theta /t$, hence
$$\vec\omega_b = \omega_b \cdot \frac{\vec\theta}{\theta} = \omega_b \frac{2\tau}{2k\tau} = \frac{\omega_b}{k} = \frac{\omega_b^2}{\omega_a}$$
Hopefully this helps you!
That said, it really shouldn't be tagged number-theory or modular-arithmetic, so I'll edit your tags a bit.