Third Force

Since the system is symmetrical about the $y$ -axis, let us consider the right half. Let the radius the red circle be $r$ and its center be $C$ , the centers of the middle and the right blue circles be $O$ and $Dr$ respectively, $DE$ be perpendicular to $BF$ , and $\angle OFB = \theta$ . Then we have:

$\begin{aligned} \tan \theta & = \frac {OB}{OF} = \frac {\sqrt{BC^2-OC^2}}{OF} = \frac {\sqrt{r^2-(r-1)^2}}{2r-1} = \frac {\sqrt{2r-1}}{2r-1} = \frac 1{\sqrt{2r-1}} \\ \implies \sin \theta & = \frac 1{\sqrt{2r}} \end{aligned}$

We note that

$\begin{aligned} \sin \theta & = \frac {DE}{DF} \\ \frac 1{\sqrt{2r}} & = \frac 1{2r-3} \\ \sqrt{2r} & = 2r-3 & \small \blue{\text{Squaring both sides}} \\ 2r & = 4r^2 - 12r+9 \\ 4r^2 - 14r + 9 & = 0 \\ \implies r & = \frac {7+\sqrt{13}}4 = \frac 9{7-\sqrt{13}} \end{aligned}$

Therefore $a+b+c = 9+7+13 = \boxed{29}$ .

Label the diagram as follows:

Let the red radius be $r$ , so that $CF = 2r$ . Since the blue radius is $1$ , $CD = GE = 1$ , $DF = 2r - 1$ , and $EF = 2r - 3$ .

By the Pythagorean Theorem on $\triangle GEF$ , $GF = \sqrt{(2r - 3)^2 - 1^2} = 2\sqrt{r^2 - 3r + 2}$ .

By Thales's Theorem , $\angle CBF = 90°$ , so triangles $\triangle BCF$ , $\triangle GEF$ , $\triangle CDB$ , and $\triangle BDF$ are all similar to each other by AA similarity.

Since $\triangle CDB \sim \triangle BDF$ , $\cfrac{CD}{BD} = \cfrac{BD}{DF}$ , or $\cfrac{1}{BD} = \cfrac{BD}{2r - 1}$ , which solves to $BD = \sqrt{2r - 1}$ .

Since $\triangle BDF \sim \triangle GEF$ , $\cfrac{DF}{BD} = \cfrac{GF}{GE}$ , or $\cfrac{2r - 1}{\sqrt{2r - 1}} = \cfrac{2\sqrt{r^2 - 3r + 2}}{1}$ , which solves to $r = \cfrac{7 + \sqrt{13}}{4} = \cfrac{9}{7 - \sqrt{13}}$ for $r > 1$ .

Therefore, $a = 9$ , $b = 7$ , $c = 13$ , and $a + b + c = \boxed{29}$ .