4 Circles in a Triangle

Nice problem!
Fun fact: it is a $30{}^\circ$ - $60{}^\circ$ - $90{}^\circ$ triangle.

Referring to the figure above, $\triangle ABD$ and $\triangle CMF$ are similar. Their similarity ratio is equal to the ratio of their inradii, which is $1$ , hence the triangles are congruent.

Let $AB=c$ , $AC=b$ , $\angle ABC=B$ , $\angle ACB=C$ , $\cot \dfrac{B}{2}=x$ . Denote the radius of each one of the four circles by $r$ .
$\begin{aligned} \dfrac{B}{2}+\dfrac{C}{2}=\dfrac{\pi }{4}& \Rightarrow \cot \left( \dfrac{B}{2}+\dfrac{C}{2} \right)=1 \\ & \Rightarrow \dfrac{x\cot \dfrac{C}{2}-1}{x+\cot \dfrac{C}{2}}=1 \\ & \Rightarrow \cot \dfrac{C}{2}=\dfrac{x+1}{x-1} \\ \end{aligned}$ Since $\triangle ABD$ and $\triangle CMF$ are congruent, $\begin{aligned} AD=FC& \Rightarrow c\cdot \sin B=r\cot \dfrac{C}{2}+r \\ & \Rightarrow c\cdot \dfrac{2\cot \dfrac{B}{2}}{{{\cot }^{2}}\dfrac{B}{2}+1}=r\cot \dfrac{C}{2}+r \\ & \Rightarrow c\cdot \dfrac{2x}{{{x}^{2}}+1}=r\left(\frac{x+1}{x-1}+1 \right) \\ & \Rightarrow c=r\cdot \dfrac{{{x}^{2}}+1}{x-1} \ \ \ \ (1)\\ \end{aligned}$ Moreover, $b=AE+EG+GC=r\cot \dfrac{B}{2}+4r+r\cot \dfrac{C}{2}=r\left( x+4+\dfrac{x+1}{x-1} \right)=r\cdot \dfrac{{{x}^{2}}+4x-3}{x-1} \ \ \ \ (2)$ These results yield a restriction for $x$ : $\left. \begin{matrix} b>0 \\ c>0 \\ \end{matrix} \right\}\Rightarrow \left. \begin{matrix} \dfrac{{{x}^{2}}+4x-3}{x-1}>0 \\ \ \ \\ x>1 \\ \end{matrix} \right\}\Rightarrow x>1 \ \ \ \ (3)$ Now we proceed to an equation for finding $x$ $\cot B=\dfrac{c}{b}\Leftrightarrow \dfrac{{{\cot }^{2}}\dfrac{B}{2}-1}{2\cot \dfrac{B}{2}}=\dfrac{c}{b}\Leftrightarrow \dfrac{{{x}^{2}}-1}{2x}=\dfrac{r\cdot \dfrac{{{x}^{2}}+1}{x-1}}{r\cdot \dfrac{{{x}^{2}}+4x-3}{x-1}}$ which simplifies to $\begin{aligned} & {{x}^{4}}+2{{x}^{3}}-4{{x}^{2}}-6x+3=0 \\ & \Leftrightarrow \left( {{x}^{2}}-3 \right)\left( {{x}^{2}}+2x-1 \right)=0 \\ & \Leftrightarrow x=\pm \sqrt{3} \ \ \ \ or \ \ \ \ x=-1+\sqrt{2} \ \ \ \ or \ \ \ \ x=-1-\sqrt{2} \\ \end{aligned}$ Due to the restriction $(3)$ , the only accepted solution is $x=\sqrt{3}$ .

Finally, we go for the required ratio:

$\begin{aligned} \left( 1 \right),\left( 2 \right) & \Rightarrow bc=\left( r\dfrac{{{x}^{2}}+4x-3}{x-1} \right)\cdot \left( r\dfrac{{{x}^{2}}+1}{x-1} \right) \\ & \overset{x=\sqrt{3}}{\mathop{\Rightarrow }}\,bc=8{{r}^{2}}\left( 2\sqrt{3}+3 \right) \\ & \Rightarrow \dfrac{{{r}^{2}}}{bc}=\dfrac{2\sqrt{3}-3}{24} \\ \end{aligned}$ Hence, $ratio=\frac{4\pi {{r}^{2}}}{\frac{1}{2}bc}=8\pi \cdot \frac{2\sqrt{3}-3}{24}=\pi \left( \frac{2\sqrt{3}}{3}-1 \right).$ For the answer, $a=2$ , $b=3$ , $c=1$ , thus $a+b+c=\boxed{6}$ .

$\ \ \$

PS: $x=\sqrt{3}\Rightarrow \cot \dfrac{B}{2}=\sqrt{3}\Rightarrow \dfrac{B}{2}=30{}^\circ \Rightarrow B=60{}^\circ$ . $\triangle ABC$ is indeed a $30{}^\circ$ - $60{}^\circ$ - $90{}^\circ$ triangle.