Four More Circles in a Box

In the figure, \MN\parallel BD) and \KL\parallel AB). Denote the lengths of $BD$ and $AB$ by $p$ and $q$ respectively.
Triangles $\triangle FBD$ and $\triangle FCH$ are similar, but since their ratio of similarity equals the ratio of their inradii, which is 1, in fact they are congruent and so are their altitudes to the hypotenuse, i.e. $FK=FL=\frac{q}{2}$ Likewise, $\triangle FBG\sim \vartriangle FJD$ , thus

$\begin{aligned} & \frac{FM}{FN}=\frac{\text{radius of green circle}}{\text{radius of red circle}}=\frac{1}{2} \\ & \Rightarrow FM=\frac{p}{3}\ \ \ \text{ and }\ \ \ FN=\frac{2p}{3} \\ \end{aligned}$ On right triangle $\triangle FBD$ ,

$\begin{aligned} F{{K}^{2}}=KB\cdot KD & \Rightarrow F{{K}^{2}}=FM\cdot FN \\ & \Rightarrow {{\left( \dfrac{q}{2} \right)}^{2}}=\frac{p}{3}\cdot \dfrac{2p}{3} \\ & \Rightarrow \dfrac{{{p}^{2}}}{{{q}^{2}}}=\dfrac{9}{8} \\ & \Rightarrow \frac{p}{q}=\dfrac{3}{2\sqrt{2}}=\dfrac{3\sqrt{2}}{4} \\ \end{aligned}$ Hence, $a+b+c=3+2+4=\boxed{9}$ .

Extend $GD$ and $AE$ to meet at $H$ , and extend $BC$ and $DE$ to meet at $I$ , and label the diagram as follows:

By alternate interior angles and vertical angles, $\triangle FHC \sim \triangle FDB$ by AA similarity, and since their incircles are congruent, $\triangle FHC \cong \triangle FDB$ , and $LF = FM$ .

Likewise, $\triangle FBG \sim \triangle FID$ by AA similarity, and since the radius of the red circle is twice the radius of the green circle, $FK = 2 \cdot JF$ , which means $LD = 2 \cdot BL$ .

$\triangle LFB \sim \triangle LDF$ by AA similarity, so $\cfrac{BL}{LF} = \cfrac{LF}{LD}$ , or $\cfrac{BL}{LF} = \cfrac{LF}{2 \cdot BL}$ , which solves to $LF = \sqrt{2} \cdot BL$ .

Therefore, the ratio $\cfrac{BD}{AB} = \cfrac{BL + LD}{2 \cdot LF} = \cfrac{BL + 2 \cdot BL}{2 \cdot \sqrt{2} \cdot BL} = \cfrac{3}{2\sqrt{2}} = \cfrac{3\sqrt{2}}{4}$ , so that $a = 3$ , $b = 2$ , $c = 4$ , and $a + b + c = \boxed{9}$ .