Expected perimeter of a random triangle in a disc, Geometric Probability

Problem 47.1

Let $P_1, P_2, P_3$ be three points chosen independently and uniformly at random in the closed unit disc. What is the expected value of the perimeter of the triangle they form?

Solution. The perimeter of $\triangle P_1 P_2 P_3$ is $|P_1 P_2| + |P_2 P_3| + |P_3 P_1|$ . By linearity of expectation, $\mathbb{E}[\text{perimeter}] = 3 \cdot \mathbb{E}[|P_1 P_2|],$ where $P_1, P_2$ are iid uniform in the unit disc.

The constant $\mathbb{E}[|P_1 P_2|]$ is a classical “line-picking” constant, first computed by Czuber in 1884: $\mathbb{E}[|P_1 P_2|] = \frac{128}{45 \pi} \approx 0.9054.$ (The derivation is a polar-coordinate integration that evaluates, after some effort, to this closed form.) Accepting this result, $\boxed{\mathbb{E}[\text{perimeter}] = 3 \cdot \frac{128}{45 \pi} = \frac{128}{15 \pi} \approx 2.716.} \qedhere$

Two remarks. First, the perimeter does not depend on the correlations between sides — linearity of expectation works even for dependent random variables — so we effectively get the three sides “for the price of one.” Second, the analogous mean distance between two random points of the unit square is $\frac{2 + \sqrt{2} + 5 \ln(1 + \sqrt 2)}{15} \approx 0.5214,$ which makes the disc-perimeter $128 / (15 \pi)$ look positively elegant. (The three-dimensional cousin, the mean distance in a unit cube, is the genuinely notorious Robbins constant $\approx 0.6617$ ; see Chapter 60.) The disc’s rotational symmetry is again the key.

import numpy as np
# Three uniform points in unit disc; compute triangle perimeter, average
r = np.sqrt(np.random.rand(3, 10**6))
t = np.random.rand(3, 10**6) * 2*np.pi
x, y = r*np.cos(t), r*np.sin(t)
d = lambda i, j: np.sqrt((x[i]-x[j])**2 + (y[i]-y[j])**2)
perim = d(0,1) + d(1,2) + d(2,0)
print(f"sim: {perim.mean():.5f}   exact: {128/(15*np.pi):.5f}")
# sim: 2.71704   exact: 2.71624