Isosceles Ratios

Let $DB = a$ . Then $CD=2a$ and $CA=3a$ . Let $DN=h$ be the altitude from $D$ to $AB$ . By similar triangle, we have the altitude $CM=3h$ . By Pythagorean theorem ,

$\begin{cases} a = \sqrt{DN^2+NB^2} = \sqrt{h^2 + \frac 19} & \implies h = \sqrt{a^2 - \frac 19} \\ b = \sqrt{DN^2+AN^2} = \sqrt{h^2 + \frac {25}9} = \sqrt{a^2+\frac 83} \end{cases}$

And the radii of the two circles are:

$\begin{cases} r_1 = \dfrac {\frac 12(2\cdot 3h-2 \cdot h)}{\frac 12 (3a + 2a + b)} = \dfrac {4h}{5a+b} = \dfrac {4\sqrt{9a^2-1}}{15a+ \sqrt{9a^2+24}} \\ r_2 = \dfrac {\frac 12 \cdot 2 \cdot h}{\frac 12(a+b+2)} = \dfrac {2h}{a+b+2} = \dfrac {2\sqrt{9a^2-1}}{3a+ \sqrt{9a^2+24}+6} \end{cases}$

Let the centers of the upper and lower circles be $O$ and $P$ respectively, the point where the two circles are tangent to each other be $L$ , $\angle CAD = \phi$ and $\angle DAB = \theta$ . Consider the length of $AL$ :

$\begin{aligned} OL \cdot \cot \frac {\angle CAD}2 & = LP \cdot \cot \frac {\angle DAB}2 \\ r_1 \cot \frac \phi 2 & = r_2 \cot \frac \theta 2 & \small \blue{\text{Using }\tan \frac \alpha 2 = \frac {1-\cos \alpha}{\sin \alpha}} \\ \frac 4{\sqrt{9a^2+24}-3a} & = \frac {2(9a^2-1)}{(3a+\sqrt{9a^2+24}+6)(\sqrt{9a^2+24}-5)} \\ 2(3a+\sqrt{9a^2+24}+6)(\sqrt{9a^2+24}-5) & = (9a^2-1)(\sqrt{9a^2+24}-3a) \\ (a-1)(3a+1)(\sqrt{9a^2+24}-3a-4) & = 0 \\ \implies a & = 1 , \pm \frac 13 \end{aligned}$

The acceptable solution is $a=1$ . Then

$\frac {r_1}{r_2} = \frac {2(3a+\sqrt{9a^2+24}+6)}{15+\sqrt{9a^2+24}} = \frac {2(9+\sqrt{33})}{15+\sqrt{33}}= \frac {17-\sqrt{33}}{16}$

The required answer is $17+33+16 = \boxed{66}$ .