Curvature Charm

Figure 1 Let ${{C}_{p}}\left( E,{{r}_{p}} \right)$ , ${{C}_{g}}\left( G,{{r}_{g}} \right)$ , ${{C}_{v}}\left( F,{{r}_{v}} \right)$ and ${{C}_{b}}\left( K,{{r}_{b}} \right)$ be the left pink semicircle, the green, the violet and the blue circles respectively (figure 1). Then, ${{r}_{p}}=\dfrac{1}{2} \ \ \ \ and \ \ \ \ {{r}_{g}}=\dfrac{1}{4}$

By Pythagorean theorem on $\triangle EGF$ , $\begin{aligned} F{{G}^{2}}=E{{G}^{2}}+E{{F}^{2}} & \Rightarrow {{\left( {{r}_{g}}+{{r}_{v}} \right)}^{2}}={{r}_{g}}^{2}+{{\left( {{r}_{p}}-{{r}_{v}} \right)}^{2}} \\ & \Rightarrow {{\left( \dfrac{1}{4}+{{r}_{v}} \right)}^{2}}=\dfrac{1}{16}+{{\left( \dfrac{1}{2}-{{r}_{v}} \right)}^{2}} \\ & \Rightarrow {{r}_{v}}=\dfrac{1}{6} \\ \end{aligned}$ Using Descart’s circle theorem we calculate the radius of the blue circle: $\begin{aligned} {{k}_{b}} & ={{k}_{v}}+{{k}_{g}}+{{k}_{p}}-2\sqrt{{{k}_{v}}{{k}_{g}}+{{k}_{g}}{{k}_{p}}+{{k}_{p}}{{k}_{v}}} \\ & =6+4-2-2\sqrt{6\times 4+4\times \left( -2 \right)+\left( -2 \right)\times 6} \\ & =12 \\ & \\ \Rightarrow {{r}_{b}}=\dfrac{1}{12} \\ \end{aligned}$

Figure 2 We place the configuration on a coordinate system, so that the center of the ellipse is at the origin and its major axis is on the $x$ -axis (figure 2). Then, the equation of ${{C}_{p}}$ is
${{\left( x+\dfrac{1}{2} \right)}^{2}}+{{\left( y-\dfrac{5}{12} \right)}^{2}}=\dfrac{1}{4}$ The semi-minor axis of the ellipse is $b={{r}_{b}}=\dfrac{1}{12}$ .
If $a$ is the major semi-axis of the ellipse, then its equation is $\frac{{{x}^{2}}}{{{a}^{2}}}+144{{y}^{2}}=1$ Let $l$ be the common tangent of the left pink semicircle ${{C}_{p}}$ and the ellipse and $N\left( {{x}_{N}},{{y}_{N}} \right)$ their point of tangency. Then, the equation of $l$ as a tangent to the ellipse at $N$ is $\frac{{{x}_{N}}}{{{a}^{2}}}x+144{{y}_{N}}y=1$ The slope of $l$ is ${{m}_{l}}=-\dfrac{{{x}_{N}}}{144{{a}^{2}}{{y}_{N}}}$ and the slope of line $EN$ is ${{m}_{EN}}=\frac{{{y}_{N}}-{{y}_{E}}}{{{x}_{N}}-{{x}_{E}}}=\dfrac{{{y}_{N}}-\dfrac{5}{12}}{{{x}_{N}}+\dfrac{1}{2}}=\dfrac{12{{y}_{N}}-5}{12{{x}_{N}}+6}$ So, we have $EN\bot l\Rightarrow {{m}_{EN}}{{m}_{l}}=-1\Rightarrow \dfrac{12{{y}_{N}}-5}{12{{x}_{N}}+6}\cdot \left( -\dfrac{{{x}_{N}}}{144{{a}^{2}}{{y}_{N}}} \right)=-1\Leftrightarrow 144{{a}^{2}}{{y}_{N}}\left( 12{{x}_{N}}+6 \right)={{x}_{N}}\left( 12{{y}_{N}}-5 \right)$ $N$ belongs to both ${{C}_{p}}$ and the ellipse, thus we get the system $\left\{ \begin{matrix} {{\left( {{x}_{N}}+\dfrac{1}{2} \right)}^{2}}+{{\left( {{y}_{N}}-\dfrac{5}{12} \right)}^{2}}=\dfrac{1}{4} \\[1em] \dfrac{{{x}_{N}}^{2}}{{{a}^{2}}}+144{{y}_{N}}^{2}=1 \\[1em] 144{{a}^{2}}{{y}_{N}}\left( 12{{x}_{N}}+6 \right)={{x}_{N}}\left( 12{{y}_{N}}-5 \right) \\ \end{matrix} \right.$ Solving this system for the positive number $a$ we get $a=\frac{1}{8}\sqrt{\frac{1}{15}\left( 299-41\sqrt{41} \right)}$ For the answer, $abcdefg=1\times 8\times 1\times 15\times 299\times 41\times 41=\boxed{60314280}$ .