Dynamic Geometry: P80

Let the centers of the unit, green, and cyan semicircles, and the pink circle be $O$ , $A$ , $B$ , and $C$ respectively; the radii of the green and cyan semicircles, and the pink circle be $r_1$ , $r_2$ , and $r$ respectively; $\angle COB = \theta$ : and $CN$ be perpendicular to $AB$ . We note that

$\begin{aligned} r + r \csc \theta & = 1 \\ r & = \frac 1{1+\csc \theta} & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ & = \frac 1{1+\frac {1+t^2}{2t}} = \frac {2t}{(1+t)^2} \end{aligned}$

By Pythagorean theorem ,

$\begin{aligned} AC^2 - CN^2 & = AN^2 = (AO+ON)^2 \\ (r_1+r)^2 - r^2 & = (1 - r_1 + r \cot \theta)^2 \\ r_1^2 + 2r_1r + r^2 - r^2 & = \left(1 - r_1 + \frac {2t}{(1+t)^2} \cdot \frac {1-t^2}{2t} \right)^2 \\ r_1^2 + 2r_1r & = \left(\frac 2{1+t}- r_1 \right)^2 \\ & = \frac 4{(1+t)^2} - \frac {4r_1}{1+t} + r_1^2 \\ \frac {4tr_1}{(1+t)^2} & = \frac 4{(1+t)^2} - \frac {4r_1}{1+t} \\ \implies r_1 & = \frac 1{1+2t} \end{aligned}$

Similarly, from $BC^2 - CN^2 = BN^2$ , we get $r_2 = \dfrac t{2+t}$ . When the ratio of the area to the perimeter is:

$\begin{aligned} \frac Ap & = \frac 3{28} \\ 28A & = 3p \\ 28 \cdot \frac {AB \cdot CN}2 & = 3(AB + BC + CA) \\ 14(2-r_1-r_2)r & = 3(2+2r) \\ 14 \left(2 - \frac 1{1+2t} - \frac t{2+t} \right) \frac {2t}{(1+t)^2} & = 6\left(1 + \frac {2t}{(1+t)^2}\right) \\ \frac {14t(2t^2+8t+2)}{2t^2+5t+2} & = 3(t^2 + 4t+1) \\ 28 t & = 6t^2 + 15t + 6 \\ 6t^2 - 13t + 6 & = 0 \\ (3t-2)(2t-3) & = 0 \\ t & = \frac 23, \frac 32 \end{aligned}$

Since the system is symmetrical about the $y$ -axis, we can use either of the values of $t$ . Let us use $t=\frac 23$ ; then $r_1 = \dfrac 1{1+2t} = \dfrac 37$ , $r_2 = \dfrac t{2+t} = \dfrac 14$ , and $\dfrac {r_1}{r_2} = \dfrac {12}7$ . Therefore $p+q = 12 + 9 = \boxed{19}$ .