The Prisoner's Dilemma

The Prisoner’s Dilemma received submissions from three PMP participants: a full one from Christopher West, a nearly complete solution from Blazej Kot, and one from Andrew Trentacoste II. The Eagle Problem Solvers of Georgia Southern University also submitted a very thorough solution.

Every complete solution received relied on methods of calculus, so we follow such an approach here. Raising both sides of the equation defining the experbola to the $1/xy$ power, we see that it is also the curve satisfying $x^{1/x}=y^{1/y}$. So determining points on the experbola (and properties of the curve as asked about here) becomes entirely about finding distinct values of $t$ for which the function $E(t)=t^{1/t}$ (shown in Figure 6) takes on the same value. In other words, we want to find horizontal lines which intersect the graph of $E$ twice. Now, if , then , and so . Therefore, any such horizontal line must satisfy . Relating this observation back to the experbola, all points on it must satisfy and . Moreover, it appears from the graph in Figure 6 that $E$ is strictly increasing up to some value of $x$, and strictly decreasing thereafter. Presuming so, it means that any horizontal line can intersect the graph of $E$ at most twice, once to the left of that maximum value, and once to its right.

We verify this property by considering the first derivative

$$E^{\prime }(t)=t^{-2+1/t}(1-\ln t).$$

Since $t^{-2+1/t}$ is always positive, this derivative is positive (i.e., $E$ is increasing) when , or in other words, for . Similarly, $E$ is strictly decreasing for , and has a unique maximum at $t=e$. Translating these facts back to the experbola, if $(x,y)$ is any point on the experbola, one of $x$ or $y$ is less than $e$ and the other is larger.

So for question (a), we already know that in any integer solution, both $x$ and $y$ must be larger than one, and one or the other must be less than $e$, or in other words, equal to 2. Thus the only integer solutions are $x=2$, $y=4$ and $x=4$, $y=2$.

As the original proposer William Keehn points out, it is easiest to next consider part (d): one can verify by direct computation that for all ,

$(t^{1/(t-1)},t^{t/(t-1)})$

lies on the experbola. So the desired functions are $f(t)=t^{1/(t-1)}$ and $g(t)=t^{t/(t-1)}$.

Turning back to question (b), we want to find asymptotes of the experbola, so we let $t$ tend to infinity in $f(t)$ and $g(t).$ It is easier to work with $F(t)=\ln f(t)=(\ln t)/(t-1)$ and $G(t)=\ln g(t)=t(\ln t)/(t-1).$ Using L’Hôpital’s rule, the limits of $F$ and $G$ as $t$ goes to infinity are the limits of $1/t$ and $\ln t+1$, respectively. So $F$ tends to 0 and $G$ to infinity, or in other words, $x=f(x)=e^{F(t)}$ tends to 1 and $y=g(x)=e^{G(t)}$ tends to infinity. Thus, the experbola has a vertical asymptote at $x=1$, and by symmetry, a horizontal asymptote at $y=1.$

Finally, we can find the point that should be added to restore the continuity of the experbola by taking the limit in the parametrization of the experbola as $t$ goes to 1. Using L’Hôpital’s rule again in exactly the same way, we see that both and $F$ and $G$ approach 1 as $t$ tends to 1, which in turn means that $f$ and $g$ approach $e$, so the “missing” point is $(e,e)$.