A Pi Day Puzzle

The island of $\pi$-land is a half-disk: a semicircular arc (Semicircular Beach) closing off a straight diameter (Diametric Beach). You pick a spot for your picnic uniformly at random on the island. What is the probability that your spot is closer to Diametric Beach than to Semicircular Beach?

The Fiddler, Zach Wissner-Gross, March 14, 2025 (a guest puzzle from Jason Zimba)(original post)

Solution

Take radius $1$ with the diameter on the $x$-axis and the island in $y\ge0$. For an interior point $(x,y)$ the nearest point of the diameter is its foot $(x,0)$, so the distance to Diametric Beach is $y$. The nearest point of the arc lies radially outward, so the distance to Semicircular Beach is $1-r$ with $r=\sqrt{x^2+y^2}$. The picnic is closer to the diameter when $$y<1-r \iff \sqrt{x^2+y^2}<1-y \iff y<\frac{1-x^2}{2},$$ the last step by squaring $\sqrt{x^2+y^2}<1-y$. This is a downward parabola from $(-1,0)$ to $(1,0)$ peaking at $y=\tfrac12$. The favourable area is $$\int_{-1}^{1}\frac{1-x^2}{2}\,dx=\frac{2}{3},$$ against the half-disk area $\tfrac{\pi}{2}$, giving $$P=\frac{2/3}{\pi/2}=\frac{4}{3\pi}\approx\boxed{42.44\%}.$$

The polar route is less inviting: with $r\sim$ density $2r$ and $\theta\sim$ uniform on $[0,\pi]$, the same event $r\sin\theta<1-r$ becomes $r<\tfrac{1}{1+\sin\theta}$, so $P=\tfrac1\pi\int_0^\pi\!\tfrac{d\theta}{(1+\sin\theta)^2}$. Equating the two answers evaluates that integral as a by-product, $$\int_0^\pi\frac{d\theta}{(1+\sin\theta)^2}=\frac43,$$ a fact the Extra Credit will reuse.

The computation

Scatter random picnic spots over the island and compare the two distances directly: the spot is closer to the diameter when $y<1-r$. The hit rate lands on $\tfrac{4}{3\pi}$.

import numpy as np
rng = np.random.default_rng(0)
x = rng.uniform(-1, 1, 8_000_000); y = rng.uniform(0, 1, 8_000_000)
m = x*x + y*y <= 1; x, y = x[m], y[m]
r = np.hypot(x, y)
print((y < 1 - r).mean(), 4/(3*np.pi))   # 0.4243  0.42441 = 4/(3 pi)

Extra Credit

With radius $1$ mile, what is the expected distance from a random picnic spot to the nearest shore (either beach)?

Solution

The distance to shore is $X=\min(R\sin\Theta,\,1-R)$, with the two pieces splitting along the same equidistance locus $r=\tfrac{1}{1+\sin\theta}$. Conditioning on $\Theta=\theta$ and using the density $f_R(r)=2r$, $$E[X]=\frac1\pi\int_0^\pi\!\sin\theta\!\!\int_0^{\frac{1}{1+\sin\theta}}\!\!\!2r^2\,dr\,d\theta +\frac1\pi\int_0^\pi\!\int_{\frac{1}{1+\sin\theta}}^{1}\!\!(1-r)\,2r\,dr\,d\theta.$$ The two integrals collapse, and the leftover $\int_0^\pi\frac{d\theta}{(1+\sin\theta)^2}=\frac43$ from the main solution finishes it: $$E[X]=\frac13-\frac{4}{9\pi}\approx\boxed{0.1919\ \text{miles}},$$ which equals $\tfrac{3\pi-4}{9\pi}$. (The official extra-credit value is paywalled; this closed form is my own.)

The computation

Encode the distance itself: for each random spot take $\min(y,\,1-r)$, the gap to whichever beach is nearer, and average. No appeal to the collapsed integral.

x = rng.uniform(-1, 1, 20_000_000); y = rng.uniform(0, 1, 20_000_000)
m = x*x + y*y <= 1; x, y = x[m], y[m]
nearest = np.minimum(y, 1 - np.hypot(x, y))      # distance to the closer beach
print(nearest.mean(), 1/3 - 4/(9*np.pi))         # 0.1919