Pi down low two

In the diagram, the distance from $P$ to any point on the blue arc is less than the length of the grey line; the distance from $P$ to any point on the greater; and the arcs have the same length when the grey line has length $\sqrt2$ . Since a randomly chosen point has equal chance of being on either the blue or yellow arc, it follows that the median is $\sqrt2$ .

To be a little more rigorous with that assertion, say the points we're joining are $P(1,0)$ and $Q(\cos t,\sin t)$ . Then $PQ=\sqrt{(1-\cos t)^2+\sin^2 t}=\sqrt{2-2\cos t}=2\sin\frac{t}{2}$

which proves that points on the yellow arc are indeed further from $P$ than those on the blue arc.

It does not matter where the first particle lies; what is important is the angle $-\pi < \theta < \pi$ between the first particle and the second. The angle $\theta$ is uniformly distributed on $(-\pi,\pi)$ . The distance between the two particles is $Z = 2\sin\tfrac12|\theta|$ . Now $\begin{aligned} P[Z \le z] & = \; P[\sin\tfrac12|\theta| \le \tfrac12z] \; =\; P[|\theta| \le 2\sin^{-1}\tfrac12z] \\ & = \; \frac{2 \times 2\sin^{-1}\frac12z}{2\pi} \; = \; \tfrac{2}{\pi}\sin^{-1}\tfrac12z \end{aligned}$ for any $0 < z < 2$ , and so the median distance $m$ is such that $\tfrac{2}{\pi}\sin^{-1}\tfrac12m = \tfrac12$ and hence the median distance is $m=\boxed{\sqrt{2}}$ .