Dynamic Geometry: P129

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 $CO$ and $C'O$ make with the $x$ -axis. From the previous problem Dynamic Geometry: P128 , the inradius of $\triangle ABC$ is given by:

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

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 $\dfrac r{r'} = \dfrac {21}{40}$ ,

$\implies \frac {(78\sin \theta - 30)(\sqrt{32.5\sin \theta + 78\cos \theta +84.5} + \sqrt{32.5\sin \theta - 78\cos \theta +84.5} + 12)}{(78\sin \theta + 30)(\sqrt{-32.5\sin \theta + 78\cos \theta +84.5} + \sqrt{-32.5\sin \theta - 78\cos \theta +84.5} + 12)} = \frac {21}{40}$

Solving the equation above, we have $\theta = \sin^{-1} \dfrac {2035}{2197}$ . Since the upper and lower triangles have the same base $AB$ , their area is proportional to their height. Then we have:

$\frac A{A'} = \frac h{h'} = \frac {6.5\sin \theta -2.5}{6.5\sin \theta + 2.5} = \frac {2035-845}{2035+845} = \frac {1190}{2880} = \frac {119}{288}$

Therefore $\sqrt{q-p} = \sqrt{288-119} = \sqrt{169} = \boxed{13}$ .