Good Things Come in Threes

Extend $DC$ and $AE$ to meet at $G$ . Then we note that $\triangle DFG$ and $\triangle ABE$ both have congruent incircles and same three vertex angles. The two triangles are congruent. Therefore $DF=EB=x$ and let $ABCD$ be a unit square. We also note that $\triangle ADF$ and $\triangle ABE$ are similar. Therefore,

$\begin{aligned} \frac {DF}{DA} & = \frac {AB}{AE} \\ \frac x1 & = \frac 1{\sqrt{1+x^2}} \\ x^2 & = \frac 1{1+x^2} \\ x^4 + x^2 - 1 & = 0 \\ \implies x & = \sqrt{\frac {\sqrt 5-1}2} \end{aligned}$

The ratio of the radius of the blue circle to that of the red circle, $\dfrac {r_\blue{\rm blue}}{r_\red{\rm red}} = \dfrac {AB}{DF} = \dfrac 1x = \sqrt{\dfrac 2{\sqrt 5-1}} = \sqrt{\dfrac {1+\sqrt 5}2} = \sqrt \varphi$ , where $\varphi$ denotes the golden ratio . Therefore $a+b+c = 1 + 5 + 2 = \boxed 8$ .

Let the square be in a Coordinate plane with origin at point $A$ and $\overline{AD}, \overline{AB}$ lie on y axis and x axis respectively and $|\overline{AB}|=1$ and let $|\overline{BE}|=h$

Let the coordinates of point $F$ be $x_1, y_1$

Now, we know that line $\overline{AB}$ coincides with line $y=hx$ and $\overline{DF}$ coincide with line $y=1-\dfrac{x}{h}$ $\therefore hx_1=1-\dfrac{x_1}{h}$ $\Rightarrow x_1=\dfrac{h}{h^2+1}$ $\Rightarrow y_1=\dfrac{h^2}{h^2+1}$ Now, Applying distance formula $|\overline{DF}|=\sqrt{\dfrac{h^2}{(h^2+1)^2}+\dfrac{1}{(h^2+1)^2}}$ $=\sqrt{\dfrac{1}{h^2+1}}$ $|\overline{DF}|=\dfrac{1}{\sqrt{h^2+1}}$ Similarly, $|\overline{FG}|=\dfrac{1}{h\sqrt{1+h^2}}$ $|\overline{DG}|=\dfrac{1}{h}$

Now, let extended $\overline{DC}$ and extended $\overline{AE}$ intersect each other at point $G$ with coordinates $x_2,y_2$

As line $\overline{AB}$ coincides with line $y=hx$ and $\overline{DC}$ coincies with line $y=1$ $\therefore x_2=\dfrac{1}{h}, y_2=1$

Now, as we know that inradius of right triangle with side $a,b,c$ where hypotenuse is $c=\dfrac{a+b-c}{2}$

$\therefore$ inradius of $\triangle DFG = \dfrac{1+h-\sqrt{1+h^2}}{2h\sqrt{1+h^2}}$

$\therefore$ inradius of $\triangle AFE=\dfrac{1+h-\sqrt{1+h^2}}{2}$

As there inradius are equal according to the question $\therefore\dfrac{1+h-\sqrt{1+h^2}}{\cancel{2}h\sqrt{1+h^2}}=\dfrac{1+h-\sqrt{1+h^2}}{\cancel{2}}$ $\Rightarrow\dfrac{1+h-\sqrt{1+h^2}}{h\sqrt{1+h^2}}=1+h-\sqrt{1+h^2}$ $\Rightarrow 1+h-\sqrt{1+h^2}=(h^2+h)\sqrt{1+h^2}-(h+h^3)$ $\Rightarrow (h^2+h+1)\sqrt{1+h^2}=(h^3+2h+1)$ Squaring both the sides $\Rightarrow (h^2+h+1)^2(h^2+1)=(h^3+2h+1)^2$ $\Rightarrow (h^4+h^2+1+2h^3+2h^2+2h)(h^2+1)=(4h^2+h^6+1+4h+4h^4+2h^3)$ $h^4+\red{\cancel{h^2}}+1+2h^3+\red{\cancel{2h^2}}+2h+h^6+h^4+\red{\cancel{h^2}}+2h^5+2h^4+2h^3=\red{\cancel{4h^2}}+h^6+1+4h+4h^4+2h^3$ $\green{\cancel{h^4}}+1+2h^3+2h+h^6+\green{\cancel{h^4}}+2h^5+\green{\cancel{2h^4}}+2h^3=h^6+1+4h+\green{\cancel{4h^4}}+2h^3$ $\cancel{1}+\blue{\cancel{2h^3}}+2h+\green{\cancel{h^6}}+2h^5+2h^3=\green{\cancel{h^6}}+\cancel{1}+4h+\blue{\cancel{2h^3}}$ $\Rightarrow 2h+2h^5+2h^3=4h$ Dividing both sides by $2h$ $1+h^4+h^2=2$ $(h^2)^2+h^2-1=0$ From quadratic formula $h^2=\dfrac{-1{^+_-}\sqrt{5}}{2}$ as $h\in\mathbb{R}$ $\therefore h^2=\dfrac{\sqrt{5}-1}{2}$ $h=\sqrt{\dfrac{\sqrt{5}-1}{2}}$ Now, inradius of $\triangle ADF = \dfrac{h+1-\sqrt{1+h^2}}{2\sqrt{1+h^2}}$ $Ratio=\dfrac{inradius of\triangle DFG}{inradius of \triangle ADF}=\dfrac{\dfrac{\cancel{1+h-\sqrt{1+h^2}}}{\cancel{2}\red{h}\cancel{\sqrt{1+h^2}}}}{\dfrac{\cancel{h+1-\sqrt{1+h^2}}}{\cancel{2\sqrt{1+h^2}}}}$ $=\dfrac{1}{h}=\sqrt{\dfrac{1}{\dfrac{\sqrt{5}-1}{2}}}$ $=\sqrt{\dfrac{2}{\sqrt{5}-1}}$ $=\sqrt{\dfrac{\sqrt{5}+1}{2}}$ $Ans=5+2+1=\boxed{8}$

Extend $AE$ and $DC$ , intersecting at $G$ : $\\ \triangle_{EBA} \sim \triangle_{DFG}$

As the blue circles are congruent，so $\\ \triangle_{EBA} \cong \triangle_{DFG} \implies DF = BE \\ \triangle_{EBA} \sim \triangle_{AFD} \\ \therefore \dfrac {AF}{FD} = \dfrac {EB}{BA} = \dfrac {FD}{AD} \\ \therefore FD^2 = AF \times AD \\ AF^2 + FD^2 = AD^2 \\ AF^2 + AF \times AD = AD^2 \\ \implies AF = \dfrac {\sqrt 5 - 1}{2} AD \\ FD = \sqrt {AF \times AD} = \sqrt {\dfrac {\sqrt 5 - 1}{2}} AD \\ \dfrac {r_{blue}}{r_{red}} = \dfrac {BE}{AF} = \dfrac {FD}{AF} \\ = \dfrac {\sqrt {\dfrac {\sqrt 5 - 1}{2}}}{\dfrac {\sqrt 5 - 1}{2}} = \sqrt {\dfrac {\sqrt 5 + 1}{2}} \\ \implies a+ b + c = 1 + 5 + 2 = \boxed 8$