Pyramids and Circles!

Let $2z$ be the length of a side of square base $ABCD$ and let $A:(0,0), B:(0,2z),$

$C:(2z,2z), D:(2z,0)$ and $E:(z,2z)$ and $O:(x,y)$ .

$(1): (x - z)^2 + (y - 2z)^2 = r^2$

$(2): x^2 + y^2 = r^2$

$(3): (x - 2z)^2 + y^2 = r^2$

Subtracting $(2)$ from $(1) \implies 2zx + 4zy = 5z^2$

and

Subtracting $(2)$ from $(3) \implies 4z(z - x) = 0$ and $z \neq 0 \implies x =z \implies$

$y = \dfrac{3}{4}z \implies r = \dfrac{5}{4}z$

For triangular face $APD$ the slant height $s_{1} = \sqrt{(\dfrac{5z}{4})^2 + (\dfrac{3z}{4})^2} =$ $\dfrac{\sqrt{34}}{4}z$

For triangular face $BPC$ the slant height $s_{2} = \sqrt{(\dfrac{5z}{4})^2 + (\dfrac{5z}{4})^2} =$ $\dfrac{5\sqrt{2}}{4}z$

For triangular faces $APB$ and $DPC$ the slant heights $s_{3} = s_{4} = \sqrt{(\dfrac{5z}{4})^2 + z^2} =$ $\dfrac{\sqrt{41}}{4}z$

$\implies$ The lateral surface area $A = \dfrac{\sqrt{34} + 5\sqrt{2} + 2\sqrt{41}}{4}z^2$

$\implies \dfrac{A}{A_{\square{ABCD}}} = \dfrac{\sqrt{34} + 5\sqrt{2} + 2\sqrt{41}}{4} =$

$\dfrac{\sqrt{a} + b\sqrt{c} + c\sqrt{d}}{e} \implies a + b + c + d + e = \boxed{86}$ .