Pi down low

This is a folklore.

First, set up the cartesian coordinate system such that the center of the unit circle, $O$ is centered at $(0,0)$ . Consider two points, $A$ and $B$ , then fix one of the points at $(1,0)$ (let's say $A$ ). We compute the average distance between that point and the other point, $B$ that varies along the unit circle. (The average is the same as the average distance between any two pairs of points on the unit circle.)

Also, since the distance of $AB$ is also the same as $AB'$ where $B'$ is the point reflected above the $x$ -axis, we can let $B$ varies along the upper semicircle and calculate the average distance based on it. In $\Delta AOB$ , let $\theta$ equals to $\angle AOB$ and it ranges from $0$ to $\pi$ , the distance $AB$ is given by the cosine rule, $AB=\sqrt{2-2\cos \theta}=2\sin \frac{\theta}{2}$ . So, the average value is $\frac{1}{\pi} \int_{0}^\pi 2\sin \frac{\theta}{2} \, d\theta = \frac{2}{\pi}\left. \left(-2 \cos \frac{\theta}{2}\right) \right|_{0}^\pi=\boxed{\frac{4}{\pi}}.$

Extra: What is the average distance of two points in a unit square?