# Case of the 'mice problem' for $n=3$ [duplicate]

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°.

• What is their angular velocity? This is constant, so once you figure that out you can find out how long it takes, and hence how far it has gone. Commented Jun 1, 2018 at 14:12
• @JaapScherphuis Logic dictates that the mice are moving with a constant speed, only their direction changes. Commented Jun 1, 2018 at 14:22
• @peterh and their direction changes at a constant rate, which is what you can use to solve this particular question.. Commented Jun 1, 2018 at 14:32
• @JaapScherphuis I think it is not, their direction change is accelerating. They make infinite many loops in finite time. Commented Jun 1, 2018 at 14:34
• Oh wait, I think you are right. At any moment the relative locations and angles all stay the same, everything is just scaled down and rotated, but the linear speed is not scaled accordingly so everything is relatively speeded up, including the angular velocity. So ignore my previous comments, sorry. Thanks @peterh. Commented Jun 1, 2018 at 14:40

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 $a=\frac {\sqrt 3}3l$
• So, is the answer $l(1-e^{\frac{-2π}{\sqrt3}})$ ? Commented Jun 1, 2018 at 14:38
• No, you need to integrate the arc length along the curve from $0$ to $-2\pi$. I couldn't convince Alpha to do it for me. Commented Jun 1, 2018 at 14:39