With what probability does an inscribed triangle contain a given point?

TL;DR

This work studies the probability that three random points on a unit circle form a triangle containing a fixed interior point X, depending only on $r=\frac{OX}{R}$ with $0\le r\le 1$. It provides an alternative proof of the known result $P(r)=\frac{1}{4}-\frac{3}{2\pi^2}\mathrm{Li}_2(r^2)$, using a calculation of conditions and a contour-integral evaluation that leads to a dilogarithm expression. The approach derives a linear relation for $P(r)$, rewrites it as a geometric integral involving $\mathcal{R}$ and arc differences, and reduces the problem to evaluating an integral of $\arg(1-r e^{i\theta})^2$, ultimately yielding the Li_2 form. The paper highlights the appearance of the dilogarithm in geometric probability and connects the result to related problems involving random chords and circle configurations.

Abstract

Three points uniformly selected on the unit circle form a triangle containing a point $X$ at distance $r \in [0; 1]$ from its center with probability $P(r) = \frac{1}{4} - \frac{3}{2 π^2}\textrm{Li}_2(r^2)$, where $\textrm{Li}_2$ is the dilogarithm function (Jeremy Tan Jie Rui, 2018). In this paper we present an alternative proof of this fact. We also discuss a couple of other geometric probability problems where the dilogarithm function arises.

Figures (7)

• Figure 1:
• Figure 2: Think of $(\alpha, \beta, \gamma)$ as of barycentric coordinates
• Figure 3: Plot of $P(r)$
• Figure 4: Three tangent circles
• Figure 5:

Theorems & Definitions (1)

• Theorem 1.1