Dynamic Geometry: P74

Let the angle measuring $\tan^{-1} \dfrac {336}{527}$ be $\angle POQ = \theta$ , the centers of the green, red, cyan, and yellow circles be $A$ , $B$ , $C$ , and $D$ ; and their radii $r_0$ , $r_1$ , $r_2$ , and $r_3$ 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:

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

We can find $r_2$ using the equation, $\dfrac 1{\sqrt{r_2}} = \dfrac 1{\sqrt{r_0}} + \dfrac 1{\sqrt{r_1}} = \dfrac 1{\sqrt{r_0}} + \dfrac 3{4\sqrt{r_0}} \implies r_2 = \dfrac {16}{49} r_0$ . And $r_3$ by Descartes' theorem ,

$\begin{aligned} \frac 1{r_3} & = \frac 1{r_0} + \frac 1{r_1} + \frac 1{r_2} + 2 \sqrt{\frac 1{r_0r_1}+\frac 1{r_2r_3}+ \frac 1{r_2r_0}} \\ \implies r_3 & = \frac 4{37} \end{aligned}$

Let the triangle the yellow circle circumscribed be $EFG$ . We can first find the areas of $\triangle DEF$ , $\triangle DFG$ , and $\triangle DGE$ and then sum them up. We find that: the areas of $\triangle DAB$ and $\triangle DEF$ are given by:

$\begin{cases} [DAB] & = \dfrac {(r_0+r_3)(r_1+r_3)\sin \angle ADB}2 \\ [DEF] & = \dfrac {r_3^2\sin \angle ADB}2 \end{cases} \\ \begin{aligned} \implies [DEF] & = \frac {r_3^2}{(r_0+r_3)(r_1+r_3)}\blue{[DAB]} & \small \blue{\text{By Heron's formula}} \\ & = \frac {r_3^2\blue{\sqrt{(r_0+r_1+r_3)r_0r_1r_3}}}{(r_0+r_3)(r_1+r_3)} \end{aligned}$

Similarly for $[DFG]$ and $[DGE]$ , then

$\begin{aligned} [EFG] & = [DEF] + [DFG] + [DGE] \\ & = \frac {r_3^2\sqrt{(r_0+r_1+r_3)r_0r_1r_3}}{(r_0+r_3)(r_1+r_3)} + \frac {r_3^2\sqrt{(r_1+r_2+r_3)r_1r_2r_3}}{(r_1+r_3)(r_2+r_3)} + \frac {r_3^2\sqrt{(r_0+r_2+r_3)r_0r_2r_3}}{(r_0+r_3)(r_2+r_3)} \\ & = \frac {992}{238169}r_0^2 + \frac {6080}{1144373}r_0^2 + \frac {1632}{298849}r_0^2 = \frac {18944}{1268089}r_0^2 \quad \quad \small \blue{\text{When }[EFG] = \frac 8{8077}} \\ \implies r_0^2 & = \frac 8{8077} \times \frac {1268089}{18944} = \frac {157}{2368} \end{aligned}$

And the area of the cyan circle $A_{\rm cyan} = \pi r_2^2 = \left(\dfrac {16}{49} \right)^2 \times \dfrac {157}{2368} \pi = \dfrac {628}{88837} \pi$ . Therefore $p+q = \boxed{89465}$ .

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Reference: Heron's formula