$n$ particles hunting each other If $n$ particles $A_1,A_2,...,A_n$ are situated at the vertices (at time $t=0$ $s$ of an $n$ sided regular polygon $\overline{A_1A_2...A_n}$ such that the velocity of the $n^{th}$ particle is always along the side $\overline{A_{n-1}A_n}$ of the polygon. Each particle has a constant velocity $v$.
I want to model a general equation to find the time at which these particles would collide with each other. One example is that of an equilateral triangle's case. Any suggestions would be helpful. (One idea is to take the angle-component of one particle toward another side, and use relative velocity)
 A: Because the particles are situated on an n-sided regular polygon at $t=0$ they admit a radial symmetry at an angle of $\alpha = \frac{2\pi}{n}$. The velocity vectors preserve this symmetry, so the particles will stay on an n-sided regular polygon.
Therefore it suffices to consider one point $(x, y)$, or in the complex plane $z=x+yi$, if we take $0$ as the center of the polygon, the next point is at $e^{-\alpha i} z$. So the differential equation to figure out the path of $z$ is $$\frac{dz}{dt} = z - e^{-\alpha i} z = (1 - e^{-\alpha i}) z.$$
We won't care about the velocity of the point for now, because this differential equation is easy to solve. The solutions are $$z(t) = C e^{(1-e^{-\alpha i})t}$$ for some $C$ in $\mathbb{C}$ which is the equation for a spiral. We can rewrite this by noticing $e^{-\alpha i} = \cos(\alpha) - i\sin(\alpha)$ so that
$$z(t) = C e^{(1-\cos(\alpha))t}(e^{i\sin(\alpha)t}).$$
In polar coordinates $$r(\theta) = R e^{\frac{1-\cos(\alpha)}{\sin(\alpha)}\theta}.$$ Let's introduce the variable $\beta = \frac{1-\cos(\alpha)}{\sin(\alpha)} = \tan(\frac{\alpha}{2})$, then we see that the trajectory of the point is a classic logarithmic spiral $r(\theta) = R e^{\beta \theta}$.
The arc length parametrization of this function is known (check Wikipedia for a full explanation). This also explains the way to calculate the arc length to the origin, it requires the following calculation.
We know that $\beta = \tan(\frac{\alpha}{2}) = \tan(\frac{\pi}{n})$ so that the formula for the arc length between a point on the curve $(r, \theta)$ and the origin is equal to $\frac{r}{\sin(\frac{\pi}{n})}$. Meaning that the time it takes to reach the origin from a point at distance $r$ is equal to $\frac{r}{v\sin(\frac{\pi}{n})}$
Some values of $\frac{1}{\sin(\frac{\pi}{n})}$ for specific values of $n$.

*

*For $n=3$ the value is $\frac{2}{\sqrt 3} = 1.1547...$

*For $n=4$ the value is $\sqrt 2 = 1.4142...$

*For $n=5$ the value is around 1.7013,

*For $n=6$ the value is exactly 2, etc...

