Pentagonal Spin Cycle

The position of the particles can be described by the five complex numbers $z\;,\; z\zeta\;,\; z\zeta^2\;,\; z\zeta^3\;,\;z\zeta^4$ where $\zeta = e^{\frac{2\pi i}{5}}$ . We then have the differential equation $\dot{z} \; = \; v \frac{z \zeta - z}{|z \zeta - z|} \; = \; ive^{\frac{\pi i}{5}} \frac{z}{|z|}$ Putting $z = re^{i\theta}$ we deduce that $(\dot{r} + ir\dot{\theta})e^{i\theta} \; = \; ive^{\frac{\pi i}{5}}e^{i\theta}$ and so $\dot{r} + ir\dot{\theta} \; = \; ive^{\frac{\pi i}{5}}$ In particular, $\dot{r} = -v\sin\tfrac{\pi}{5}$ . Since the initial pentagon has side $a$ , the initial value of $r$ is $\tfrac12a \mathrm{cosec}\tfrac{\pi}{5}$ . Thus we have $r \; = \; \tfrac12a\mathrm{cosec}\,\tfrac{\pi}{5} - vt\sin\tfrac{\pi}{5}$ and hence the particles meet after time $\tfrac{a}{2v}\mathrm{cosec}^2\tfrac{\pi}{5}$