Dynamic Geometry: P128

From the compiled calculations in Dynamic Geometry: P96 Series , we know that the radius of the large circle is $6.5$ , the dividing chord $AB=12$ , $\angle ACB = \pi - \tan^{-1} \frac {12}5$ , and $\angle AC'B = \tan^{-1} \frac {12}5$ .

Let the center of the large circle $O$ be the origin of the $xy$ -plane. Then $C=(6.5\cos \theta, 6.5 \sin \theta)$ and $C'=(-6.5\cos \theta, -6.5 \sin \theta)$ , where $\theta$ is the angle $CC'$ makes with the $x$ -axis. The inradius of $\triangle ABC$ is $r = \dfrac As$ , where $A$ and $s$ are the area and semiperimeter or $\triangle ABC$ respectively, then:

$\begin{aligned} r & = \frac {AB \cdot \blue{CD}}{CA+BC+AB} \qquad \qquad \small \blue{\text{where }CD \text{ is the height of }\triangle ABC} \\ & = \frac {12(6.5\sin \theta - 2.5)}{\sqrt{(6.5\cos \theta +6)^2+(6.5\sin \theta -2.5)^2} + \sqrt{(6.5\cos \theta -6)^2+(6.5\sin \theta -2.5)^2} + 12} \\ & = \frac {78\sin \theta - 30}{\sqrt{-32.5\sin \theta + 78\cos \theta +84.5} + \sqrt{-32.5\sin \theta - 78\cos \theta +84.5} + 12} \end{aligned}$

Similarly for $\triangle ABC'$ :

$\begin{aligned} r' & = \frac {78\sin \theta + 30}{\sqrt{32.5\sin \theta + 78\cos \theta +84.5} + \sqrt{32.5\sin \theta - 78\cos \theta +84.5} + 12} \end{aligned}$

When $\dfrac r{r'} = \dfrac 9{16}$ ,

$\implies \frac {(78\sin \theta - 30)(\sqrt{32.5\sin \theta + 78\cos \theta +84.5} + \sqrt{32.5\sin \theta - 78\cos \theta +84.5} + 12)}{(78\sin \theta + 30)(\sqrt{-32.5\sin \theta + 78\cos \theta +84.5} + \sqrt{-32.5\sin \theta - 78\cos \theta +84.5} + 12)} = \frac 9{16}$

Solving the equation above, we have $\theta = \sin^{-1} \dfrac {323}{325}$ . Since the upper and lower triangles have the same base $AB$ , their area is proportional to their height. Then we have:

$\frac A{A'} = \frac h{h'} = \frac {6.5\sin \theta -2.5}{6.5\sin \theta + 2.5} = \frac {323-125}{323+125} = \frac {198}{448} = \frac {99}{224}$

Therefore $p+q = 99+224 = \boxed{323}$ .