It's All Circles!

For vertical stack:

Let $H_{1} = \dfrac{\sqrt{3}}{2}a$ be the height of equilateral $\triangle{ABC}$ .

$\dfrac{2h_{1}}{a} = \tan(30^\circ) = \dfrac{1}{\sqrt{3}} \implies h_{1} = \dfrac{a}{2\sqrt{3}}$ and $H_{2} = H_{1} - 2h_{1} = \dfrac{a}{2\sqrt{3}}$ and

$\dfrac{H_{1}}{H_{2}} = 3 \implies H_{2} = \dfrac{H_{1}}{3} \implies h_{2} = \dfrac{1}{3}h_{1},$ $h_{3} = \dfrac{1}{3}h_{2} = \dfrac{1}{3^2}h_{1}$ and in general

$h_{n} = (\dfrac{1}{3})^{n - 1}h_{1} \implies A_{v}(n) = \pi h_{1}^2(\dfrac{1}{9})^{n - 1} \implies$

$A_{v} = \pi h_{1}^2\sum_{n = 1}^{\infty} (\dfrac{1}{9})^{n - 1} = \pi h_{1}^2(\dfrac{9}{8}) =$ $\pi(\dfrac{a}{2\sqrt{3}})^2(\dfrac{9}{8}) = \dfrac{3a^2\pi}{32}$ .

For other two stacks let $A_{d} = A_{v} - A(1) = \dfrac{3a^2\pi}{32} - \dfrac{a^2}{12}\pi =$ $\dfrac{a^2\pi}{96} \implies 2A_{d} = \dfrac{a^2\pi}{48}$

$\implies A_{T} = A_{v} + 2A_{d} = \dfrac{11}{96}\pi a^2 = \dfrac{96}{11}\pi \implies a = \dfrac{96}{11} = \dfrac{\alpha}{\beta} \implies \alpha + \beta =\boxed{107}$ .