Three particles $A,$ $B$ and $C$ are situated at the vertices of an equilateral $ΔABC$ with sides each of length $l.$ Particle $A$ moves towards particle $B$ with a speed of $s.$ Simultaneously, particle $B$ moves with the same speed towards particle $C$ and $C$ moves towards particle $A.$ Tracing out a pursuit curve, they meet each other.
What is the distance covered by particle $A$ when it revolves around by $2π$ radians?
I could find out the total time taken by the particle $A$ to collide with $B$ (which is $\frac{2l}{3s}$) and total distance $l$ covered by $A$ but I could not figure out the distance traced to revolove 360°.
The point is to recognize that the spiral is equiangular. The angle between the spiral and the radius is constant at $\frac {\pi} 3$. This means it is a logarithmic spiral of the form $r=ae^{b\theta}$ with $\arctan \frac 1b=\frac \pi 3, b=\frac 1{\sqrt 3}$. If we choose the origin of $\theta$ to go through one of the corners of the triangle we have $a=\frac {\sqrt 3}3l$