Dynamic Geometry: P52

Label the centers of the large and cyan circles $O$ and $P$ respectively, the triangle $ABC$ , its median $AM$ , and $\angle PLC = \angle PNO = 90^\circ$ . Let the radius of the green circle be $r$ , then the radius of the cyan circle is $\frac 67r$ .

Since the center of the green circle is centroid of the equilateral $\triangle ABC$ , the median $AM = 3r$ . Let $OM = a$ . Then $3r+a = 1 \implies a = 1-3r$ .

Consider the right $\triangle OPN$ , by Pythagorean theorem ,

$\begin{aligned} ON^2+PN^2 & = OP^2 \\ ML^2 + (PL+LN)^2 & = OP^2 \\ (MC+CL)^2 + (PL+LN)^2 & = OP^2 \\ \left(\sqrt 3r + \frac 6{7\sqrt 3}r\right)^2 + \left(\frac 67r + \blue a \right)^2 & = \left(1 - \frac 67 r\right)^2 & \small \blue{\text{Note that }a = 1-3r} \\ \left(\frac {21+6}{7\sqrt 3}r\right) + \left(1 - \frac {15}7r \right)^2 & = \left(1 - \frac 67 r\right)^2 \\ \frac {243}{49} r^2 + 1 - \frac {30}7r + \frac {225}{49}r^2 & = 1 - \frac {12}7 r + \frac {36}{49}r^2 \\ \frac{432}{49} & = \frac {18}7 r & \small \blue{\text{Since }r > 0} \\ \implies r & = \frac 7{24} \end{aligned}$

Therefore the sum of areas of the two circles is $\pi r^2 + \pi \left(\dfrac 67r\right)^2 = \dfrac {49+36}{49} \times \dfrac {49}{576} \pi = \dfrac {85}{576}\pi$ . The required answer $p+q = 85+576 = \boxed{661}$ .