Levitating Pentagon

Note that the curve $y=x^2$ is red and $y=2x^2$ is blue. Let $A=(u, 2u^2)$ and $B=(v, v^2)$ . Then the side length of the regular pentagon is $2u$ . Then we have:

$v =u+2u \sin 18^\circ = u + 2u \left(\frac {\sqrt 5-1}4\right) = \blue{\left(\frac {1+\sqrt 5}2\right)} u = \blue \varphi u$

where $\varphi$ denotes the golden ratio . And:

$v^2 = 2u^2 + 2u \cos 18^\circ = 2u^2 + 2u \sqrt{\frac {5+\sqrt 5}8} = 2u^2 + u \sqrt{\sqrt 5 \varphi}$

Therefore,

$\begin{aligned} \varphi^2 u^2 & = 2u^2 + u \sqrt{\sqrt 5 \varphi} \\ (\varphi^2 - 2)u & = \sqrt{\sqrt 5 \varphi} \\ (\varphi + 1 -2)^2u^2 & = \sqrt 5 \varphi & \small \blue{\text{Note that }\varphi - 1 = \frac 1\varphi} \\ \frac {u^2}{\varphi^2} & = \sqrt 5 \varphi \\ \implies u^2 & = \sqrt 5 \varphi^3 \end{aligned}$

The area of regular pentagon:

$\begin{aligned} A_{\rm 5gon} & = 5u^2\tan 54^\circ = 5 \sqrt 5 \varphi^3 \sqrt{1+\frac 2{\sqrt 5}} \\ & = 5 \sqrt 5 \varphi^3 \sqrt{\frac {\sqrt 5+1+1}{\sqrt 5}} = 5 \sqrt 5 \varphi^3 \sqrt{\frac \blue{2\varphi +1}{\sqrt 5}} & \small \blue{\text{Note that }\varphi^n = F_n \varphi + F_{n-1} \text{, where }F_n} \\ & = 5 \sqrt 5 \varphi^3 \sqrt{\frac {\varphi^3}{\sqrt 5}} = 5\sqrt{\sqrt 5 \varphi^9} & \small \blue{\text{denotes the }n\text{th Fibonacci number.}} \\ & = 5\sqrt{\sqrt 5 (34\varphi + 21)} = 5 \sqrt{5\cdot 17 + 17 \sqrt 5 + 21 \sqrt 5} \\ & = 5\sqrt{85+38\sqrt 5} \end{aligned}$

Therefore $A+B+C+D = 5+85+38+5 = \boxed{133}$ .

In the figure, the small yellow regular pentagon has circumradius 1, its center is at the origin and the y-axis is an axis of symmetry. $A$ and $B$ are two of its vertices. As we can see here , the coordinate of these vertices are ${{x}_{A}}=\dfrac{\sqrt{10+2\sqrt{5}}}{4}$ ${{y}_{A}}=\dfrac{\sqrt{5}-1}{4}$ ${{x}_{B}}=\dfrac{\sqrt{10-2\sqrt{5}}}{4}$ ${{y}_{B}}=\dfrac{\sqrt{5}+1}{4}$ If we enlarge the pentagon to get the blue one with circumradius $R$ , the corresponding vertices are ${A}'\left( R{{x}_{A}},\ R{{y}_{A}} \right) \ \ \ \ \ \text{and} \ \ \ \ \ {B}'\left( R{{x}_{B}},\ R{{y}_{B}} \right)$ “Levitating” this pentagon to the desired place by a vector $\left( \begin{matrix} 0 \\ c \\ \end{matrix} \right)$ , the vertices become ${A}''\left( R{{x}_{A}},\ R{{y}_{A}}+c \right) \ \ \ \ \ \text{and} \ \ \ \ \ {B}''\left( R{{x}_{B}},\ R{{y}_{B}}+c \right)$ for some positive real number $c$ .
Since ${A}''$ and ${B}''$ belong to the parabolas $y={{x}^{2}}$ and $y=2{{x}^{2}}$ respectively, we have the following system of equations $\begin{aligned} & R{{y}_{A}}+c={{\left( R{{x}_{A}} \right)}^{2}} \\ & R{{y}_{B}}+c=2{{\left( R{{x}_{B}} \right)}^{2}} \\ \end{aligned}$ Combining we get $R\left( {{y}_{A}}-{{y}_{B}} \right)={{R}^{2}}\left[ {{\left( {{x}_{A}} \right)}^{2}}-2{{\left( {{x}_{B}} \right)}^{2}} \right]\Rightarrow R=\dfrac{{{y}_{A}}-{{y}_{B}}}{{{\left( {{x}_{A}} \right)}^{2}}-2{{\left( {{x}_{B}} \right)}^{2}}}$ Plugging the corresponding values we find $R=3+\sqrt{5}$ Using the formula for the area of a regular pentagon with circumradius $R$ , we have $A=\dfrac{5{{R}^{2}}}{4}\sqrt{\dfrac{5+\sqrt{5}}{2}}=\dfrac{5{{\left( 3+\sqrt{5} \right)}^{2}}}{4}\sqrt{\frac{5+\sqrt{5}}{2}}=\ldots =5\sqrt{85+38\sqrt{5}}$ For the answer, $a=5$ , $b=85$ , $c=38$ , $d=5$ , thus, $a+b+c+d=\boxed{133}$