Dynamic Geometry: P73

Let the centers of the large, yellow, and cyan semicircle, and the green circle be $O$ , $A$ , $B$ , and $C$ , and the radii of the yellow and cyan semicircles, and the green circle be $r_1$ , $r_2$ , and $r$ respectively. Then $r_1 + r_2 = 1$ . Since $AO = 1-r_1 = r_2$ and $OB = 1 - r_2 = r_1$ . By cosine rule ,

$\begin{aligned} BC^2 & = OB^2 + OC^2 - 2\cdot OB \cdot OC \cdot \cos \blue{\angle BOC} & \small \blue{\text{Let }\angle BOC = \theta.} \\ (r_2+r)^2 & = r_1^2 + (1-r)^2 - 2 r_1 (1-r) \cos \blue \theta \quad ...(1) & \small \blue{\text{Similarly for }CA^2} \\ (r_1+r)^2 & = r_2^2 + (1-r)^2 + 2 r_2 (1-r) \cos \theta \quad ...(2) \end{aligned}$

From $r_2 \cdot (1) + r_1 \cdot (2)$ :

$\begin{aligned}r_2(r_2+r)^2 + r_1(r_1+r)^2 & = r_2r_1^2 + r_1r_2^2 + (r_2+r_1)(1-r)^2 \\ r_1^3 + r_2^3 + 2r(r_1^2+r_2^2) + \blue{(r_1+r_2)}r^2 & = r_1r_2\blue{(r_1+r_2)} + \blue{(r_1+r_2)}(1 - 2r + r^2) & \small \blue{\text{Note that }r_1+r_2 = 1} \\ (r_1+r_2)((r_1+r_2)^2 - 3r_1r_2) + 2r((r_1+r_2)^2 - 2r_1r_2) + r^2 & = r_1r_2 + 1 - 2r + r^2 \\ 1 - 3r_1r_2 + 2r - 4r_1r_2r + r^2 & = r_1r_2 + 1 - 2r + r^2 \\ 4r - 4r_1r_2r & = 4r_1r_2 \\ \implies r_1r_2 & = \frac r{1+r} \end{aligned}$

By Heron's formula , the area of $\triangle ABC$ , the triangle of centers of three mutually tangent (kissing) circles,

$\begin{aligned} A_c & = \sqrt{(r_1+r_2+r)\blue{r_1r_2}r} = \sqrt{(1+r) \cdot \frac {r^2}{1+r}} = r & \small \blue{\text{Note that }r_1r_2 = \frac r{1+r}} \end{aligned}$

The area of triangle of tangent points of three kissing circles is given by:

$\begin{aligned} A_t & = \frac {2r_1r_2r}{(r_1+r_2)(r+r_1)(r+r_2)}A_c \\ & = \frac {\frac r{1+r} \cdot 2 r^2}{1(r_1r_2+r_1r+r_2r+ r^2)} \\ & = \frac {2r^3}{(1+r)\left(\frac r{1+r} + r + r^2\right)} \\ & = \frac {2r^3}{r+r+r^2+r^2 + r^3} \\ \implies \frac 4{65} & = \frac {2r^2}{2+2r + r^2} & \small \blue{\text{Putting }A_t = \frac 4{65}} \\ 63r^2 - 4r - 4 & = 0 & \small \blue{\text{and rearrange.}} \\ (7r-2)(9r+2) & = 0 & \small \blue{\text{Since }r > 0} \\ \implies r & = \frac 27 \end{aligned}$

Therefore $\sqrt{p+q} = \sqrt{2+7} = \boxed 3$ .