Is it possible to determine particles' velocities from their crossings on a ring? 
2 particles are constrained to move in a ring. Both particles begin moving at $t=0$ from $p=0$, each particle moving in the opposite direction of the other. It is known that they are moving at different constant velocities. Both particles pass each other (without interaction) 4 times before simultaneously returning to $p=0$.

With this information, how can I find out the ratio between the particles' respective velocities?
It seems to me that this would hold true in any case where $x:y$ is the ratio of the particles' velocities and $|x - y| = 5$. In that case, all the following would qualify: $${-2:3},\ {-3:2},\ {1:6},\ {2:7}$$
If furthermore, I know that the order of the positions where they pass is (in radians):
$$\frac{4π}{5},\frac{8π}{5},\frac{2π}{5},\frac{6π}{5}$$
then can I definitively extrapolate the ratio?

Bonus points to anyone who can figure out what gave me the idea for this question in the first place.
 A: I am thinking about a simpler approach:
Let $\vec r_1 $ and $\vec r_2$ be the position vectors of the two particles.
The particles move on a circular ring so that the vectors will look like $$\vec r_i=\left ( \begin{matrix}
\cos\left ( \omega_i t \right )\\ 
\sin\left ( \omega_i t \right )
\end{matrix} \right )$$
where, WOLOG, $\omega_1>0$ and $\omega_2<0$ (since they move in opposite directions).
Equate the position vectors and solve for $t$ in the region $t\in\left [ 0,2\pi \right ]$, to get 2 equations to find crossings of the two:
$$\begin{array}{lcl}
\cos\left (\omega_1 t  \right )& = & \cos\left (\omega_2 t  \right ) \\
\sin\left (\omega_1 t  \right ) & = & \sin\left (\omega_2 t  \right )
\end{array}.$$
These are actually 2 difficult equations to solve, but you have a few constraints:


*

*$\vec r_1\left(0\right)=\vec r_2\left(0\right)=0$.

*$\vec r_1\left(T\right)=\vec r_2\left(T\right)=0$ where $T=2\pi$.


This limits you to integer $\omega$'s and if you "guess" that $\omega_1=-4 \omega_2$, the set of 2 equations has a solution at 
$$t=\frac{2\pi}{5},\ \frac{4\pi}{5},\ \frac{6\pi}{5},\ \frac{8\pi}{5},$$which is actually at multiples of $2\pi$ over $\omega_1+\omega_2$.
But beware, choosing different frequencies may lead to solutions of the form $\frac{n\pi}{\omega_1-\omega_2}$. This is strongly related to the aliasing effect.
I hope this helps a little.
A: The ratio should be 1:4, as the particles are moving in opposite directions (given in question), the way I perceive it is:
Suppose particles A and B with x and 4x velocity in opposite directions.
They start at point p. At the time A travels to 1/4th circumference B is at P again. They must have crossed somewhere between p and the point where A is now.
Similarly A travelling every quarter of circumference, B completes the round. After 4 rounds of, A and B both are on p.
If velocities are allowed to be negative, then there can be another answer (that also only one.)
I am really sorry if I have misinterpreted the question, but it seems really simple and your thesis was not understandable.
