Finding out the trajectory of the particle in polar coordinates The question goes as follows
Consider n identical particles resting on the n vertices of a n sided regular convex polygon such that at $t=0$ each particle starts following its immediate neighbour. Suppose that $a_1, a_2,..., a_n$ are the verticles of the polygon in order then according to question $a_1$ follows $a_2$,$a_2$ follows $a_3$ and so on. The question is to find out the trajectory of any one of the particle taking the origin as the point where all the particles finally meet. 
I could find out that the particles would meet at the centre of the polygon by symmetry of the situation. I guess that the path would be spiral. However I have no idea on how  to find out the equation that determines such path. Any help is appreciated. Thanks. 
 A: Let's say that the position of particle 1 at some time $t$ is $(x_1,y_1)$, which we can write in polar coordinates as $(r_1\cos{\theta_1},r_1\sin{\theta_1})$. By symmetry, particle 2 is going to be at the same distance from the origin ($r_2=r_1=r$), but at an angle $\theta_2=\theta_1+\frac{2\pi}{n}$. The coordinates of particle 2 are $(x_2,y_2)=(r\cos{\theta_2},r\sin{\theta_2})$. The tangent to the trajectory at particle 1 location points towards particle 2, so 
$$\frac{dy}{dx}=\frac{y_2-y_1}{x_2-x_1}=\frac{r(\sin\theta_2-\sin\theta_1)}{r(\cos\theta_2-\cos\theta_1)}=\frac{\sin(\theta+\frac{2\pi}{n})-\sin\theta}{\cos(\theta+\frac{2\pi}{n})-\cos\theta}$$
The next step is writing $\frac{dy}{dx}$ in terms of $r$ and $\theta$, and let's say that $r=f(\theta)$. It is relatively straightforward to get this value:
$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{r\cos\theta+r'\sin\theta}{-r\sin\theta+r'\cos\theta}$$
where $r'=\frac{dr}{d\theta}$. Let's call for now $\frac{dy}{dx}=\alpha$. Then $$\alpha(-r\sin\theta+r'\cos\theta)=r\cos\theta+r'\sin\theta\\r'(\sin\theta-\alpha\cos\theta)=-r(\alpha\sin\theta+\cos\theta)$$
and therefore:
$$\frac{dr}{r}=-\frac{\alpha\sin\theta+\cos\theta}{\sin\theta-\alpha\cos\theta}d\theta$$
Plug in the value of $\alpha$ in terms of $\theta$ and $n$, and integrate both sides. For n=4, the last equation reduces to $$\frac{dr}{r}=-d\theta$$ which is easy to integrate, yielding an exponential spiral
