Eccentric Repulsion – The Prisoner's Dilemma

D15. Eccentric Repulsion

Contributed by Paul Morton, PMP Participant

Figure 1: Avoiding the smell at $S$ by hiding at $H$ .

Paul is back with another elliptical quandary. Suppose you are constrained to a circular track, which we will model as the unit circle in the plane, and there is something smelly at one point on the track, say where it intersects the $x$ -axis at the point $(1, 0)$ . If you want to get as far away as possible from the smell, clearly you should run to the diametrically opposite point on the track, namely the point $(- 1, 0)$ . Now suppose you change the shape of the track by stretching it in the $y$ -direction into an ellipse. If you stretch it far enough, then the diametrically opposite point will no longer be the farthest away: as shown in Figure 1, when you have stretched it by a factor of four, most of the ellipse lies more than two units away (the portion shown in green).

What is the most you can stretch the track so that the opposite point $(- 1, 0)$ remains the farthest place to hide from the smell? (Paul originally asked this question in terms of the eccentricity of the elliptical track: if the circle is stretched by a factor of $a > 1$ in the $y$ -direction, then the eccentricity of the resulting ellipse is $e = \sqrt{1 - 1 / a^{2}}$ . You may phrase your answer in terms of either the eccentricity $e$ or the stretch factor $a$ .)

This problem originally appeared in the Prisoner’s Dilemma in the 2023 Fall issue of the PMP Newsletter. Solutions are no longer being accepted for this Dilemma.

Show solution?

Solution.

Robert Noll (PMP), Randy Schwartz (Schoolcraft College), and the Eagle Problem Solvers of Georgia Southern University submitted solutions.

The key observation is that if the elliptical track ever lies outside of the circle of radius two centered at (1, 0), then it will have to intersect that circle again for some $x$ value greater than -1. So we solve for the point of intersection of that circle and the ellipse that is stretched by a factor of $a$ in the $y$ -direction, and determine for which $a$ there is an intersection with $x$ -coordinate greater than -1.

The equation of the circle is $(x - 1)^{2} + y^{2} = 2^{2}$ , which simplifies to $y^{2} = 3 + 2 x - x^{2}$ . The equation of the ellipse is $x^{2} + y^{2} / a^{2} = 1$ , which simplifies to $y^{2} = a^{2} - a^{2} x$ . Equating the two expressions for $y^{2}$ to find the points (or possibly point) of intersection yields $(a^{2} - 1) x^{2} + 2 x + (3 - a^{2}) = 0$ , with right-hand root

$\begin{matrix} \frac{- 2 + \sqrt{4 - 4 (a^{2} - 1) (3 - a^{2})}}{2 (a^{2} - 1)} = \frac{- 2 + \sqrt{4 a^{4} - 16 a^{2} + 16}}{2 (a^{2} - 1)} \\ = \frac{- 2 + 2 (a^{2} - 2)}{2 (a^{2} - 1)} = \frac{a^{2} - 3}{a^{2} - 1} . \end{matrix}$

Solving for the condition that this root is greater than -1 yields the condition that $a^{2} - 3 > 1 - a^{2}$ , which simplifies to $a^{2} > 2$ . Thus, we conclude that the largest vertical stretch for which (-1,0) is the farthest point on the ellipse from (0, 1) occurs for $a = \sqrt{2}$ . This stretch corresponds to an eccentricity of $e = \sqrt{1 / 2}$ .

November 15, 2023

Main Maker