Three Right Circles

Let $AB = x$ , $AC = 1$ , and $BC = y$ . Then the ratio of $\cfrac{AB}{AC} = \cfrac{x}{1} = x$ .

As an incircle of right $\triangle ABC$ , the radius of the red circle is $\frac{1}{2}(x + 1 - y)$ , which means $AI = AG = x + 1 - y$ .

That means $IB = AB - AI = x - (x + 1 - y) = y - 1$ and $GC = AC - AG = 1 - (x + 1 - y) = y - x$ .

Since $\triangle GHC \sim \triangle ABC$ by AA similarity, $GH = GC \cdot \cfrac{AB}{AC} = x(y - x)$ .

Since the diameter of the cyan circle is twice the diameter of the green circle and $\triangle GHC \sim \triangle IBJ$ by AA similarity, $GH = 2IB$ , so:

$x(y - x) = 2(y - 1)$

By the Pythagorean Theorem on $\triangle ABC$ , $y = \sqrt{x^2 + 1}$ , and substituting this into the above equation gives:

$x(\sqrt{x^2 + 1} - x) = 2(\sqrt{x^2 + 1} - 1)$

which solves to $x = \cfrac{9 - \sqrt{17}}{8}$ .

Therefore, $a = 9$ , $b = 17$ , $c = 8$ , and $a + b + c = \boxed{34}$ .

Let $\angle B = \theta$ . Then $\dfrac {AB}{AC} = \cot \theta$ . Let the radius of the red circle be $1$ . Then $AB = 1 + \cot \dfrac \theta 2$ . Let $t = \tan \dfrac \theta 2$ . Then $AB = \dfrac {1+t}t$ . We note that the $\triangle GHC$ and $\triangle IBJ$ are similar and that the diameter of the cyan circle is twice the diameter of the green circle, then $GH = 2 \cdot IB$ . Also $\triangle GHC$ is similar to $\triangle ABC$ then

$\begin{aligned} \frac {GC}{GH} & = \tan \theta \\ \frac {AC-2}{2\cdot IB)} & = \tan \theta \\ \frac {AB \cdot \tan \theta -\blue{AG}}{2(AB- \blue{AI})} & = \tan \theta & \small \blue{\text{Note that }AG=AI = 2} \\ AB \cdot \tan \theta - \blue 2 & = 2 (AB - \blue 2) \tan \theta \\ (4-AB) \blue{\tan \theta} & = 2 & \small \blue{\text{and }\tan \theta = \frac {2t}{1-t^2}} \\ \left(4 - \frac {1+t}t \right) \frac {2t}{1-t^2} & = 2 \\ \frac {3t-1}t \cdot \frac t{1-t^2} & = 1 \\ 3t - 1 & = 1 - t^2 \\ t^2 + 3t - 2 & = 0 \\ \implies t & = \frac {\sqrt{17}-3}2 & \small \blue{\text{Since }t > 0} \\ \frac {AB}{AC} & = \cot \theta = \frac {1-t^2}{2t} \\ & = \frac {9-\sqrt{17}}8 \end{aligned}$

Therefore $a+b+c = 9+17+8 = \boxed{34}$ .