Dynamic Geometry: P94

Let the radii of the large, left, and right semicirles, and the left and right circles be $1$ , $r_1$ , $r_2$ , $r_3$ , and $r_4$ respectively. Note that $r_1+r_2=1$ .

From the previous problem Dynamic Geometry: P77 , we find that $r_3=r_1r_2=r_4$ and the heights of the centers of the left and right circles are $y_1 = 2r_1 \sqrt{r_2}$ and $y_2 = 2r_2 \sqrt{r_1}$ respectively. Therefore the quadrilateral in question is a parallelogram and its area is:

$\begin{aligned} A & = r_3(y_1-y_2) \\ & = r_1r_2 \left(2r_1\sqrt{r_2}-2r_2\sqrt{r_1}\right) \\ & = 2r_1^2r_2^\frac 32 - 2r_2^2r_1^\frac 32 & \small \blue{\text{To find maximum }A} \\ \frac {dA}{dr_1} & = 4r_1 r_2^\frac 32 - 3r_1^2 r_2^\frac 12 + 4r_2r_1^\frac 32 - 3r_2^2r_1^\frac 12 & \small \blue{\text{Putting }\frac {dA}{dr_1}=0} \\ 4r_1 r_2^\frac 32 + 4r_2r_1^\frac 32 & = 3r_1^2 r_2^\frac 12 + 3r_2^2r_1^\frac 12 & \small \blue{\text{Divide both sides by }r_1^\frac 12 r_2^\frac 12} \\ 4 r_2\sqrt{r_1} + 4 r_1\sqrt{r_2} & = 3r_1 \sqrt{r_1} + 3r_2\sqrt{r_2} \\ (4r_2 - 3r_1)\sqrt {r_ 1} & = (3r_2-4r_1)\sqrt{r_2} & \small \blue{\text{Note that }r_1+r_2 = 1} \\ (4-7r_1)\sqrt {r_ 1} & = (3-7r_1)\sqrt{1-r_1} & \small \blue{\text{Squaring both sides}} \\ (4-7r_1)^2r_ 1 & = (3-7r_1)^2(1-r_1) \\ 98r_1^3-147r_1^2+67 r_1-9 & = 0 \\ (2r_1-1)(49r_1^2 - 49r_1 + 9) & = 0 & \small \blue{\text{Since }r_1 \text{ and }r_2 \text{ are interchangeable,}} \\ \implies r_1, r_2 & = \frac 12, \frac 12 \pm \frac {\sqrt{13}}{14} \end{aligned}$

Since when $r_1 = r_2 = \dfrac 12$ , $A =0$ , the minimum, $A$ is maximum when $r_1, r_2 = \dfrac 12 \pm \dfrac {\sqrt{13}}{14}$ . And the ratio of the larger radius to the smaller one is

$\frac {\frac 12 + \frac {\sqrt{13}}{14}} {\frac 12 - \frac {\sqrt{13}}{14}} = \frac {7+\sqrt{13}}{7-\sqrt{13}} = \frac {31+7\sqrt{13}}{18}$

Therefore $p+q+m+n = 31+7+13+18 = \boxed{69}$ .