Random chord through an inner concentric disc, Geometric Probability

Problem 23.1

A chord of the unit circle is determined by two points chosen independently and uniformly on the circle. What is the probability that the chord passes through a concentric disc of radius $r \in (0, 1)$ ?

Solution. Let the endpoints be at angles $\alpha, \beta \in [0, 2\pi)$ . Define $\phi = |\alpha - \beta| \bmod 2\pi$ , reduced to $[0, \pi]$ (the smaller of the two arcs). The midpoint of the chord lies at distance $d = \cos(\phi / 2)$ from the centre, with $d \in [0, 1]$ since $\phi / 2 \in [0, \pi/2]$ .

The chord intersects the concentric disc of radius $r$ iff $d < r$ , i.e., $\cos(\phi/2) < r$ , i.e., $\phi/2 > \arccos r$ , i.e., $\phi > 2 \arccos r$ .

For uniform $\alpha, \beta$ on $[0, 2\pi)$ , the reduced angular separation $\phi = |\alpha - \beta| \wedge (2\pi - |\alpha - \beta|)$ is uniform on $[0, \pi]$ . Hence $\mathbb{P}(\phi > 2 \arccos r) = \frac{\pi - 2 \arccos r}{\pi},$ and therefore $\boxed{\mathbb{P}(\text{chord through inner disc}) = 1 - \frac{2 \arccos r}{\pi}.} \qedhere$

The unit disc with the inner concentric disc of radius r = \tfrac12 shaded (sage). Three sample chords: the copper chord’s midpoint is outside the inner disc (chord misses it); the rose chord passes through (midpoint distance small); the plum chord grazes near the inner-disc boundary. — The unit disc with the inner concentric disc of radius $r = \tfrac12$ shaded (sage). Three sample chords: the copper chord’s midpoint is outside the inner disc (chord misses it); the rose chord passes through (midpoint distance small); the plum chord grazes near the inner-disc boundary.

For $r = \tfrac12$ (the inscribed-equilateral-triangle-side case from Bertrand: a chord is longer than the side $\sqrt3$ of the inscribed equilateral triangle iff its midpoint distance $d < \tfrac12$ ): $\mathbb{P} = 1 - \tfrac{2 \cdot \pi/3}{\pi} = \tfrac13$ . For $r = \tfrac{\sqrt 3}{2}$ (the chord-longer-than-the-radius case, equivalently longer than the side $1$ of the inscribed regular hexagon): $\mathbb{P} = 1 - \tfrac{2 \cdot \pi/6}{\pi} = \tfrac23$ . For $r = \tfrac{\sqrt 2}{2}$ : $\mathbb{P} = 1 - \tfrac{2 \cdot \pi/4}{\pi} = \tfrac12$ . For $r = 1$ : $\mathbb{P} = 1 - 0 = 1$ (the chord is always within the unit disc). For $r \to 0^+$ : $\mathbb{P} = 1 - \tfrac{2 \cdot \pi/2}{\pi} = 0$ (a chord almost never passes exactly through a single point).

import numpy as np
r = 0.4
theta = np.random.rand(2, 10**7) * 2*np.pi
# Midpoint distance from centre for endpoints at angles theta[0], theta[1]:
mid = np.abs(np.cos((theta[0] - theta[1]) / 2))
print(f"sim: {(mid < r).mean():.5f}   exact: {1 - 2*np.arccos(r)/np.pi:.5f}")
# sim: 0.26215   exact: 0.26198