Twice as Big

Referring to the figure, let $\angle CBK=\angle CBK=\theta$ , where $K$ is the center of the blue circle. Let $r$ and $2r$ be the radii the two circles. We have the relation $BC=BE+EC\Rightarrow 1=r\cot \theta +r\Rightarrow \cot \theta =\frac{1}{r}-1$ Hence, $\begin{aligned} BF & =r\cot \theta +r\cot \left( 45{}^\circ -\theta \right) \\ & =r\left( \cot \theta +\dfrac{\cot \theta -1}{\cot \theta +1} \right) \\ & =r\left( \left( \dfrac{1}{r}-1 \right)+\dfrac{\left( \dfrac{1}{r}-1 \right)+1}{\left( \dfrac{1}{r}-1 \right)-1} \right) \\ \Rightarrow BF=\dfrac{1-2r+2{{r}^{2}}}{1-2r} \ \ \ \ \ (1)\\ \end{aligned}$ Moreover, $\begin{aligned} FC=FG+GC & \Rightarrow FC=r\cot \left( 45{}^\circ -\theta \right)+r \\ & \Rightarrow FC=r\dfrac{\cot \theta +1}{\cot \theta -1}+r \\ & \Rightarrow FC=r\dfrac{\left( \dfrac{1}{r}-1 \right)+1}{\left( \dfrac{1}{r}-1 \right)-1}+r \\ & \Rightarrow FC=\dfrac{2r-2{{r}^{2}}}{1-2r} \\ \end{aligned}$ Hence, $DF=1-FC=1-\dfrac{2r-2{{r}^{2}}}{1-2r}=\dfrac{1-4r+2{{r}^{2}}}{1-2r}$ By Pythagorean theorem on $\triangle ADF$ , $A{{F}^{2}}=A{{D}^{2}}+D{{F}^{2}}=1+{{\left( \dfrac{1-4r+2{{r}^{2}}}{1-2r} \right)}^{2}}$ Thus, $AF=\sqrt{1+{{\left( \dfrac{1-4r+2{{r}^{2}}}{1-2r} \right)}^{2}}} \ \ \ \ \ (2)$ Using $(1)$ and $(2)$ we have an expression for the perimeter of $\triangle AFB$ : ${{P}_{AFB}}=AB+BF+AF=1+\dfrac{1-2r+2{{r}^{2}}}{1-2r}+\sqrt{1+{{\left( \dfrac{1-4r+2{{r}^{2}}}{1-2r} \right)}^{2}}}$ Finally, we use the area formula for $\triangle AFB$ to set and solve an equation for $r$ : $\begin{aligned} & \left[ AFB \right]=\dfrac{1}{2}{{P}_{AFB}}\cdot \left( 2r \right)\Rightarrow \dfrac{1}{2}=\dfrac{1}{2}{{P}_{AFB}}\cdot \left( 2r \right) \\ & 1=2r\left( 1+\dfrac{1-2r+2{{r}^{2}}}{1-2r}+\sqrt{1+{{\left( \dfrac{1-4r+2{{r}^{2}}}{1-2r} \right)}^{2}}} \right) \\ & 1=2r\left( \dfrac{\left( 1-2r \right)+\left( 1-2r+2{{r}^{2}} \right)}{1-2r}+\dfrac{\sqrt{{{\left( 1-2r \right)}^{2}}+{{\left( 1-4r+2{{r}^{2}} \right)}^{2}}}}{1-2r} \right) \\ & 1-2r=2r\left( 2-4r+2{{r}^{2}}+\sqrt{4{{r}^{4}}-16{{r}^{3}}+24{{r}^{2}}-12r+2} \right) \\ & 1-2r=4r-8{{r}^{2}}+4{{r}^{3}}+2r\sqrt{4{{r}^{4}}-16{{r}^{3}}+24{{r}^{2}}-12r+2} \\ & 1-6r+8{{r}^{2}}-4{{r}^{3}}=2r\sqrt{4{{r}^{4}}-16{{r}^{3}}+24{{r}^{2}}-12r+2} \\ & {{\left( 1-6r+8{{r}^{2}}-4{{r}^{3}} \right)}^{2}}=4{{r}^{2}}\left( 4{{r}^{4}}-16{{r}^{3}}+24{{r}^{2}}-12r+2 \right) \\ & \cancel{16{{r}^{6}}}-\bcancel{64{{r}^{5}}}+112{{r}^{4}}-104{{r}^{3}}+52{{r}^{2}}-12r+1=\cancel{16{{r}^{6}}}-\bcancel{64{{r}^{5}}}+96{{r}^{4}}-48{{r}^{3}}+8{{r}^{2}} \\ & 16{{r}^{4}}-56{{r}^{3}}+44{{r}^{2}}-12r+1=0 \\ & \left( 2r-1 \right)\left( 8{{r}^{3}}-24{{r}^{2}}+10r-1 \right)=0 \\ \end{aligned}$

Since $2r<1$ , we get $8{{r}^{3}}-24{{r}^{2}}+10r-1=0\Leftrightarrow {{\left( 2r \right)}^{3}}-6{{\left( 2r \right)}^{2}}+5\left( 2r \right)-1=0$ From this we see that the radius $2r$ of the yellow circle is a root of $f\left( x \right)={{x}^{3}}-6{{x}^{2}}+5x-1$ .

For the answer, $f\left( 6 \right)=\boxed{29}$ .