The Prisoner's Dilemma

PMP participants Chris Bistryski of Monroe, WA, William Jones of FCI Loretto, William Keehn, and Jesse Waite submitted solutions to this problem. As Mr. Keehn pointed out, it’s smoothest to work with polar coordinates; see Figure 2 for an illustration of the coordinates $(r,\theta )$ of a general point $M$ on the mullipse. We have by the Law of Cosines that$$M{G}^2=1^2+r^2-2r\cos \theta .$$ Similarly, because $\cos ({180}^{\circ }-\theta )=-\cos \theta$, $$F{M}^2=1^2+r^2+2r\cos \theta .$$

The definition of the curve we want is that $FM\cdot MG=1$, so we also have that $F{M}^2\cdot M{G}^2=1.$ Substituting and using the difference of squares, these observations mean that $1=(1+r^2{)}^2-(2r\cos \theta {)}^2.$ Finally, multiplying out, collecting terms, and using the cosine double-angle formula gives us:

This last equation gives us a pleasantly simple (polar) equation for the mullipse. You can see it as the green curve in Figure 3. (An equation for it in Cartesian coordinates is$$4x^2+1=(x^2+y^2+1{)}^2,$$ which you can find by a similar process.)

As for what you get when the desired product 1 is replaced by some other constant $c$, you can re-do the same derivation starting from $F{P}^2\cdot P{G}^2=c^2$ to get the formula$$c^2-1=r^2(r^2-2\cos 2\theta ).$$ These curves for $c$ equal to 0.5, 0.75, 1, 1.25, and 1.5 are shown in Figure 3. In case you’d like to look up more information about them, the mullipse is more widely known as the “lemniscate of Bernoulli” and the whole family of curves created by varying the constant $c$ are called “Cassini ovals.”