The Prisoner's Dilemma

For convenience, since segment $AC$ is divided into four equal parts, we introduce coordinates scaled so that $AC=4$. We place the origin at the center of the (semi)ellipse, which is to say the midpoint between the foci $F$ and $G$. Thus the equation of the ellipse will have the standard form $x^2/a^2+y^2/b^2=1$ for some constants $a$ and $b$ giving the positive $x$– and $y$-intercepts of the ellipse. We’ve included this coordinate system in Figure 2.

The coordinates of some of the points are immediate:$$A(-\frac{5}{2},0),I(-\frac{3}{2},0),F(-\frac{1}{2},0),G(\frac{1}{2},0),\text{and}C(\frac{3}{2},0).$$Since $C$ is the $x$-intercept of the ellipse, we have $a=3/2$. We also added the $y$-intercept $D(0,b)$ in Figure 2. Since $FC+CG=3$, we have by the definition of an ellipse that also $3=FD+DG=2\sqrt{b^2+1/4}$, from which we solve for $b=\sqrt{2}$.

Now suppose the coordinates of point $B$ are $(3\text{/}2,h)$. Then the equation of line $AB$ is$$y=\frac{h}{4}(x+\frac{5}{2}).$$ Substituting this value for $y$ into the equation of the ellipse to solve for the $x$-coordinate of point $T$ (since the equations of both the line and the ellipse must be satisfied at point $T$), we get

We can write out the rightmost square and multiply both sides by $1152=9\cdot 32\cdot 4$ to obtain the quadraticHowever, it is given in the problem that the line $AB$ is tangent to the semiellipse, so they only intersect in one point. Therefore this quadratic must have only one real root. But by the quadratic formula for $A{x}^2+Bx+C=0$, namely $x=(-B\pm \sqrt{B^2-4AC})/2A$, there are two roots unless $B^2-4AC=0$. In other words, it must be the case thatAgain multiplying out, this equation simplifies to $h^2=8$, amazingly enough. Armed with this information, it is easy to compute that $AB=\sqrt{16+h^2}=2\sqrt{6},$ so $AB/AC=\sqrt{6}/2$.