Dynamic Geometry: P82

Let the centers of the unit, cyan, green, and pink semicircles be $O$ , $A$ , $B$ , and $C$ , their radii $1$ , $r_1$ , $r_2$ , and $r$ , Then $r_1 + r_2 = 1$ , $OA = 1-r_1 = r_2$ , and $OB = 1-r_2 = r_1$ .

Let diameters $DE$ and $FG$ through centers $C$ and $O$ respectively. By intersecting chords theorem ,

$\begin{aligned} DC\cdot CE & = FC \cdot CG \\ r^2 & = (1-k)(1+k) = 1 - k^2 \\ \implies k^2 & = 1 - r^2 \end{aligned}$

Let $\angle COB = \theta$ . By cosine rule ,

$\begin{cases} BC^2 = OB^2 + OC^2 - 2 \cdot OB \cdot OC \cdot \cos \theta \\ CA^2 = OA^2 + OC^2 + 2 \cdot OA \cdot OC \cdot \cos \theta \end{cases} \\ \begin{cases} (r_2+r)^2 = r_1^2 + (1-r^2) - 2r_1\sqrt{1-r^2} \cos \theta & ...(1) \\ (r_1+r)^2 = r_2^2 + (1-r^2) + 2r_2\sqrt{1-r^2} \cos \theta & ...(2) \end{cases} \\ \begin{aligned} r_2 \times (1) + r_1 \times (2): \quad r_1^3+r_2^3 + 2(r_1^2+r_2^2)r + \blue{(r_1+r_2)} r^2 & = r_1r_2 \blue{(r_1+r_2)} +\blue{(r_1+r_2)}(1-r^2) \\ \blue{(r_1+r_2)}(\blue{(r_1+r_2)}^2-3r_1r_2) + 2(\blue{(r_1+r_2)}^2 - 2r_1r_2)r + r^2 & = r_1r_2 + 1 - r^2 \quad \quad \small \blue{\text{Note that }r_1+r_2 = 1} \\ 1 - 3r_1r_2 + 2r - 4r_1r_2r + r^2 & = r_1r_2 + 1 - r^2 \\ r + r^2 & = 2r_1r_2 + 2r_1r_2r \\ \implies r & = 2r_1r_2 \end{aligned}$

When the product of the inradius $r_i$ and the circumradius $R$ of the $\triangle ABC$ :

$\begin{aligned} r_i R & = \frac {297}{1850} \\ \frac \blue A {r_1+r_2+r} \cdot \frac {(r_1+r)(r_2+r)(r_1+r_2)}{4 \blue A} & = \frac {297}{1850} & \small \blue{\text{where }A \text{ is the area of the }\triangle ABC.} \\ \frac {(r_1+r)(r_2+r)}{r_1+r_2 + r} & = \frac {594}{925} \\ 925(r_1r_2 + (r_1+r_2)r + r^2) & = 594 (1+r) \\ 925 \left(\frac r2 + r + r^2 \right) & = 594 + 594 r \\ 1850r^2 + 1587r - 1188 & = 0 \\ (25r-12)(74r+99) & = 0 & \small \blue{\text{Since }r >0} \\ \implies r & = \frac {12}{25} \end{aligned}$

Therefore $p+q=12+25 = \boxed{37}$ .