Dynamic Geometry: P71

Let the origin of the $xy$ -plane be $O$ , the centers of the yellow, green, and cyan circles be $A$ , $B$ , and $C$ respectively, $CF$ be perpendicular to the $x$ -axis, $CD$ and $AE$ be perpendicular to $CF$ and the $y$ -axis, and the radius of the cyan circle be $R$ .

By Pythagorean theorem ,

$\begin{cases} BC^2-BD^2 = CD^2 \\ AC^2 - CE^2 = AE^2 = CD^2 \end{cases} \\ \begin{aligned} \implies BC^2-BD^2 & = AC^2 - CE^2 \\ BC^2-(OB-OD)^2 & = AC^2 - (CF-EF)^2 \\ (r+R)^2 - (r+2-R)^2 & = (R+1)^2 - (R-1)^2 \\ 4rR - 4r - 4 + 4R & = 4R \\ \implies R & = \frac {r+1}r \end{aligned}$

The ratio of the area of triangle of tangential points (purple) $A_t$ to the area of triangle of centers (red) $A_c$ of three mutually externally tangent circles is given by:

$\frac {A_t}{A_c} = \frac {2abc}{(a+b)(b+c)(c+a)}$

where $a$ , $b$ , and $c$ are the radii of the three circles. Therefore in this case, we have:

$\begin{aligned} \frac {2rR}{(1+r)(r+R)(R+1)} & = \frac 9{38} & \small \blue{\text{Putting }R = \frac {r+1}r} \\ \frac {2r \cdot \frac {r+1}r}{(1+r)(r+\frac {r+1}r)(\frac {r+1}r+1)} & = \frac 9{38} \\ \frac {2r^2}{(r^2+r+1)(2r+1)} & = \frac 9{38} \\ 18r^3-49r^2+27r+9 &=0 \\ (2r-3)(9r^2 - 11r-3) & = 0 & \small \blue{\text{For rational value of }r,} \\ \implies r & = \frac 32 \\ 1 + r + R & = 1 + \frac 32 + \frac 53 = \frac {25}6 \end{aligned}$

Therefore $p+q = 25+6 = \boxed{31}$ .