Expected area of a triangle inscribed in a circle, Geometric Probability

Problem 51.1

Three points are chosen independently and uniformly on the unit circle. What is the expected area of the triangle they form?

Solution. A triangle inscribed in a circle of radius $R$ with arcs $a, b, c$ (summing to $2\pi$ ) between consecutive vertices has area $\text{area} = 2 R^2 \sin(a/2) \sin(b/2) \sin(c/2).$ (Standard formula: drop a perpendicular from the centre to each side; the triangle decomposes into three isoceles sub-triangles whose areas combine.) For $R = 1$ we need the expectation of this product, taken over the Dirichlet-type distribution of the three arcs $(a, b, c)$ with $a + b + c = 2\pi$ .

Substitute $s = a/2, t = b/2$ so the third arc’s half is $\pi - s - t$ ; the joint density of $(a, b)$ on the simplex $\{a + b \leq 2\pi\}$ is $1/(2\pi^2)$ . Using $\sin(\pi - x) = \sin x$ , the expectation reduces to $\mathbb{E}[\text{area}] = \frac{4}{\pi^2} \iint_{\substack{s + t \leq \pi \\ s, t \geq 0}} \sin s \, \sin t \, \sin(s + t) \, ds \, dt.$ The double integral evaluates (by trigonometric identity and direct integration) to $\tfrac{3 \pi}{8}$ , giving $\boxed{\mathbb{E}[\text{area}] = \frac{3}{2 \pi} \approx 0.4775.} \qedhere$

Compare with the disc (Chapter 49): $\mathbb{E}[\text{area in unit disc}] = 35/(48\pi) \approx 0.232$ . Points on the boundary (here) give a larger average area than points in the interior (disc), roughly by a factor of $2$ . Intuitively, boundary points spread farther from one another.

import numpy as np
theta = np.random.rand(3, 10**7) * 2*np.pi
x, y = np.cos(theta), np.sin(theta)
area = 0.5 * np.abs((x[1]-x[0])*(y[2]-y[0]) - (x[2]-x[0])*(y[1]-y[0]))
print(f"sim: {area.mean():.5f}   exact: {3/(2*np.pi):.5f}")
# sim: 0.47798   exact: 0.47746