D15. Eccentric Repulsion

Contributed by Paul Morton, PMP Participant

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).

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

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 = 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 will be accepted through 2024 Mar 1.