Calisson

Let the radius of the small circle be $r$ and the radius of the large circle be $R$ . For any triangle, the line joining a vertex and the incenter bisects the angle of the vertex. Consider $\triangle ABE$ and its side $AB$ ,

$\begin{aligned} r \cot \frac {\angle EAB}2 + r \cot \frac {\angle EBA}2 & = AB & \small \blue{\text{Let }\angle EAB = \theta} \\ r \cot \frac \theta 2 + r \cot 60^\circ & = 4 & \small \blue{\text{and }t = \tan \frac \theta 2} \\ \frac rt + \frac r{\sqrt 3} & = 4 \\ \implies t & = \frac r{4-\frac r{\sqrt 3}} \end{aligned}$

Let $\angle CDE = \phi$ and $u = \tan \dfrac \phi 2$ . Similarly for $\triangle CDE$ and side $CD$ , $\dfrac ru + \sqrt 3 r = 4 \implies u = \dfrac r{4-\sqrt 3r}$ .

Now consider $BE+EC=BC$ :

$\begin{aligned} r\cot 60^\circ + r \cot \frac {60^\circ - \theta}2 + r \cot \frac {120^\circ - \phi}2 + r\cot 30^\circ & = 4 \\ \frac r{\sqrt 3} + \blue{\frac {\sqrt 3 + t}{1-\sqrt 3t}r + \frac {1+\sqrt 3u}{\sqrt 3-u} r} + \sqrt 3 r & = 4 \\ \frac r{\sqrt 3} + \frac {3r}{\sqrt 3-r} + \frac r{\sqrt 3-r} + \sqrt 3 r & = 4 \\ \frac {4r}{\sqrt 3} + \blue{\frac {4r}{\sqrt 3-r}} & = 4 \\ \frac r{\sqrt 3} + \frac r{\sqrt 3-r} & = 1 \\ r^2 - 3\sqrt 3 r + 3 & = 0 \\ (\sqrt 3 - r)^2 - \sqrt 3r & = 0 & \small \blue{\text{Divide both sides by }r^2} \\ \left(\frac {\sqrt 3}r - 1\right)^2 - \left(\frac {\sqrt 3}r - 1\right) - 1 & = 0 \\ \implies \frac {\sqrt 3}r - 1 & = \varphi & \small \blue{\text{where }\varphi = \frac {1+\sqrt 5}2 \text{ denotes the golden ratio.}} \end{aligned}$

Now consider $\triangle ADE$ and side $DA$ .

$\begin{aligned} R \cot \frac {60^\circ - \theta}2 + R \cot \frac {120^\circ - \phi}2 & = 4 \\ R \left(\frac {\sqrt 3+t}{1-\sqrt 3t} + \frac {1+\sqrt 3u}{\sqrt 3-u} \right) & = 4 \\ \frac Rr \left(\blue{\frac {\sqrt 3+t}{1-\sqrt 3t} r + \frac {1+\sqrt 3u}{\sqrt 3-u} r} \right) & = 4 & \small \blue{\text{See above}} \\ \frac Rr \left(\blue{\frac {4r}{\sqrt 3-r}} \right) & = 4 \\ \implies \frac Rr & = \frac {\sqrt 3}r - 1 = \varphi = \frac {1+\sqrt 5}2 \end{aligned}$

Therefore $a+b+c = 1+5+3 = \boxed 8$ .