Random Circle Problem

In this mathematical puzzle, we’ll exercise our minds with basic probability and geometry, making it an ideal exercise for learners of all levels. Brace yourself for a short yet thought-provoking solution that will challenge your understanding of probabilities within probabilities. Are you ready to take on this captivating challenge? Let’s dive right in!

Statement

Given a circle \(C_a\)​ of radius \(a\) centered at the origin, inscribed in a square \(S\) of side \(2a\).

A circle \(C_a\) inscribed in a square \(S\).

Consider a random variable \(B \sim \text{Uniform}(0, a)\), the random variables \(X, Y \sim \text{Uniform}(-a, a)\), the outcomes \(b \in B\), \(x \in X\) and \(y \in Y\), and a real number \(\rho \in [0, 1)\).

Find the probability of observing an outcome \(b\) such that the probability of a circle \(C_b\) with radius \(b\) centered at \(P = (x, y)\) being fully contained within \(C_a\) is greater than \(\rho\).

Example of sampling random \(C_b\) circles for a fixed value of \(b\).

That is, find the probability of observing an outcome \(b\) such that:

\[\mathbb{P}\{C_b \subseteq C_a\} > \rho\]

Note that \(C_b\) is not necessarily fully contained in \(S\).

Solution

Probability of \(C_b \subseteq C_a\)

Let’s consider a point \(P\) inside the square \(S\). The circle \(C_b\) is described by the circumference of radius \(b\) centered at \(P\). If \(C_b\) is fully contained in \(C_a\), then \(P\) must be inside \(C_a\), and the distance between \(P\) and the circumference of \(C_a\) must be greater than \(b\).

This means that \(C_b\) is fully contained in \(C_a\) if and only if its center \(P\) is inside the circle \(C_{ab}\) of radius \(a-b\) centered at the origin.

Then the probability that \(C_b\) is fully contained in \(C_a\) is the ratio between the area of \(C_{ab}\) and the area of \(S\):

\[\mathbb{P}\{C_b \subseteq C_a\} = \frac{\pi (a-b)^2}{(2a)^2} = \frac{\pi}{4} \frac{(a-b)^2}{a^2}\]

Probability of \(\mathbb{P}(C_b \subseteq C_a) > \rho\)

We are looking for the probability of observing an outcome \(b\) such that

\[\mathbb{P}\{C_b \subseteq C_a\} > \rho\]

that is

\[\frac{\pi}{4} \frac{(a-b)^2}{a^2} > \rho\]

By clearing \(b\) from the inequality, we get

\[b < a \left(1 - \sqrt{\frac{4}{\pi} \rho }\right)\]

As \(B\) is a uniformly distributed within the interval \([0, a]\), its probability density function is:

\[f(x) = \begin{cases} \frac{1}{a} &\quad\text{if } 0 \le x \le a\\ 0 &\quad\text{otherwise.} \\ \end{cases}\]

Then the probability of \(b < a \left(1 - \sqrt{\frac{4}{\pi} \rho }\right)\) is simply the area of the rectangle with height \(\frac{1}{a}\) and base \(a \left(1 - \sqrt{\frac{4}{\pi} \rho }\right)\)

\[P\Big\{b < a \left(1 - \sqrt{\frac{4}{\pi} \rho }\right)\Big\} = \frac{1}{a} \cdot a \left(1 - \sqrt{\frac{4}{\pi} \rho }\right) = 1 - \sqrt{\frac{4}{\pi} \rho }\]

Therefore, the probability of observing an outcome \(b\) such that \(\mathbb{P}\){\(C_b \subseteq C_a\)}\(> \rho\) is

\[\therefore \mathbb{P}\Big\{\mathbb{P}\{C_b \subseteq C_a\} > \rho\Big\} = 1 - \sqrt{\frac{4}{\pi} \rho }\]