Dynamic Geometry: P69

Let the angle measuring $\tan^{-1} \dfrac {336}{527}$ be $\angle POQ = \theta$ , the centers of the green, red, and cyan circles be $A$ , $B$ , and $C$ and their radii $r_0$ , $r_1$ , and $r_2$ respectively, and $AM$ and $BN$ be perpendicular to $OQ$ .

We note that $\angle BON = \dfrac \theta 2$ and $\tan \dfrac \theta 2 = \dfrac {1-\cos \theta}{\sin \theta} = \dfrac 7{24}$ . This means that similar $\triangle AOM$ and $\triangle BON$ are $7$ - $24$ - $25$ triangles. By similar triangles:

$\begin{aligned} \frac {BN}{AM} & = \frac {OB}{OA} = \frac {OA+AB}{OA} \\ \frac {r_1}{r_0} & = \frac {\frac {25}7r_0+r_0+r_1}{\frac {25}7r_0} \\ 25r_1 & = 32 r_0 + 7 r_1 \\ \implies r_1 & = \frac {16}9r_0 \end{aligned}$

Using the equation, $\dfrac 1{\sqrt{r_2}} = \dfrac 1{\sqrt{r_0}} + \dfrac 1{\sqrt{r_1}} = \dfrac 1{\sqrt{r_0}} + \dfrac 3{4\sqrt{r_0}} \implies r_2 = \dfrac {16}{49} r_0$ . By Heron's formula , the area of $\triangle ABC$ , $\Delta = \sqrt{(r_0+r_1+r_2)r_0r_1r_2} = \dfrac {592}{441}r_0^2$ . Let the area of the "black" triangle be $\Delta_b$ . Then

$\Delta_b = \frac {2r_0r_1r_2}{(r_0+r_1)(r_1+r_2)(r_2+r_0)} \Delta = \dfrac {7056}{47125} \Delta$

Putting $\delta = \dfrac {1813}{16965}$ , we have

$\begin{aligned} \frac {7056}{47125} \Delta & = \frac {1813}{16965} \\ \frac {7056}{47125} \cdot \frac {592}{441}r_0^2 & = \frac {1813}{16965} \\ \implies r_0 & = \frac {35}{48} \\ r_0 + r_1 + r_2 & = \left(1 + \frac {16}9 + \frac {16}{49} \right)r_0 = \frac {6845}{3024} \\ \implies p+q & = \boxed{9869} \end{aligned}$