Semi Inellipse – The Prisoner's Dilemma

D9. Semi Inellipse

Contributed by Paul Morton, PMP Participant

Figure 1: A semiellipse inscribed in right triangle $C B A$ so that its foci and one vertex cut a leg in four equal parts.

PMP participant Paul Morton has also investigated the properties of ellipses. As shown in Figure 1, semiellipse $C T I$ is inscribed in right triangle $C B A$ so that it is tangent to hypotenuse $A B$ at $T$ , and so that points $I$ and the two foci $F$ and $G$ of the ellipse cut leg $A C$ of the triangle into four equal segments. What is the ratio of the length of hypotenuse $A B$ to the length of leg $A C$ , or in symbols, $A B / A C$ ?

This problem originally appeared in the Prisoner’s Dilemma in the 2022 Fall issue of the PMP Newsletter. Solutions are no longer being accepted for this Dilemma.

Show solution?

Solution.

For convenience, since segment $A C$ is divided into four equal parts, we introduce coordinates scaled so that $A C = 4$ . 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.

Figure 2: Diagram for Semi Inellipse with coordinate system imposed and auxiliary point $D$ .

The coordinates of some of the points are immediate: $A (- \frac{5}{2}, 0), I (- \frac{3}{2}, 0), F (- \frac{1}{2}, 0), G (\frac{1}{2}, 0), and C (\frac{3}{2}, 0) .$ 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 $F C + C G = 3$ , we have by the definition of an ellipse that also $3 = F D + D G = 2 \sqrt{b^{2} + 1 / 4}$ , from which we solve for $b = \sqrt{2}$ .

Now suppose the coordinates of point $B$ are $(3 / 2, h)$ . Then the equation of line $A B$ 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 $29 + \frac{h^{2}}{32} {(x + \frac{5}{2})}^{2} = 1.$

We can write out the rightmost square and multiply both sides by $1152 = 9 \cdot 32 \cdot 4$ to obtain the quadratic $(36 h^{2} + 512) x^{2} + 180 h^{2} x + (225 h^{2} - 1152) = 0.$ However, it is given in the problem that the line $A B$ 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 $A x^{2} + B x + C = 0$ , namely $x = (- B \pm \sqrt{B^{2} - 4 A C}) / 2 A$ , there are two roots unless $B^{2} - 4 A C = 0$ . In other words, it must be the case that $(180 h^{2})^{2} = 4 (36 h^{2} + 512) (225 h^{2} - 1152) .$ Again multiplying out, this equation simplifies to $h^{2} = 8$ , amazingly enough. Armed with this information, it is easy to compute that $A B = \sqrt{16 + h^{2}} = 2 \sqrt{6},$ so $A B / A C = \sqrt{6} / 2$ .

September 15, 2022

Main Maker