Library / /1230 of 1394

, 2016

Combinatorics Difficulty 5.8 AIME, harder Prove it United States

Problem:

For $i = 0, 1, \ldots, 5$ let $l_{i}$ be the ray on the Cartesian plane starting at the origin, at angle $\theta = i \frac{\pi}{3}$ counterclockwise from the positive $x$ -axis. For each $i$ , point $P_{i}$ is chosen uniformly at random from the intersection of $l_{i}$ with the unit disk. Consider the convex hull of the points $P_{i}$ , which will (with probability 1) be a convex polygon with $n$ vertices for some $n$ . What is the expected value of $n$ ?

Solution

Solution:

A vertex $P_{i}$ is part of the convex hull if and only if it is not contained in the triangle formed by the origin and the two adjacent vertices. Let the probability that a given vertex is contained in the aforementioned triangle be $p$ . By linearity of expectation, our answer is simply $6(1-p)$ .

Say $|P_{0}| = a$ , $|P_{2}| = b$ . Stewart's Theorem and the Law of Cosines give that $p$ is equal to the probability that $|P_{1}| < \sqrt{a b - a b \frac{a^{2} + b^{2} + a b}{(a + b)^{2}}} = \frac{a b}{a + b}$ ; alternatively this is easy to derive using coordinate methods. The corresponding double integral evaluates to $p = \frac{2}{3}(1 - \ln(2))$ , thus telling us our answer.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.