Circles inscribed in equilateral triangle!

Using the above diagram we obtain:

$\dfrac{2R_{1}}{a_{1}} = \tan(30^{\circ}) = \dfrac{1}{\sqrt{3}} \implies \boxed{R_{1} = \dfrac{a_{1}}{2\sqrt{3}}}$

and $H_{2} = H_{1} - 2R_{1} = \dfrac{\sqrt{3}}{2}a_{1} - \dfrac{1}{\sqrt{3}}a_{1} = \dfrac{a_{1}}{2\sqrt{3}}$

$\dfrac{H_{1}}{H_{2}} = 3 = \dfrac{a_{1}}{a_{2}} \implies a_{2} = \dfrac{a_{3}}{3}$

$\dfrac{2R_{2}}{a_{2}} = \dfrac{1}{\sqrt{3}} \implies \boxed{R_{2} = \dfrac{a_{2}}{2\sqrt{3}} = \dfrac{a_{1}}{6\sqrt{3}}}$

$H_{3} = H_{2} - 2R_{2} = \dfrac{a_{1}}{2\sqrt{3}} - \dfrac{a_{1}}{3\sqrt{3}} = \dfrac{\sqrt{3}}{18}a_{1}$ $= \dfrac{a_{1}}{6\sqrt{3}}$

$\dfrac{H_{2}}{H_{3}} = 3 = \dfrac{a_{2}}{a_{3}} \implies a_{3} = \dfrac{a_{2}}{3} = \dfrac{a_{1}}{9}$ $\implies$

$\dfrac{2R_{3}}{a_{3}} = \dfrac{1}{\sqrt{3}} \implies \boxed{R_{3} = \dfrac{a_{3}}{2\sqrt{3}} = \dfrac{a_{1}}{18\sqrt{3}}}$

In General:

$R_{n} = \dfrac{a_{1}}{2\sqrt{3}}(\dfrac{1}{3})^{n - 1}$

Note: $H_{n} = \dfrac{a_{1}}{2\sqrt{3}}(\dfrac{1}{3})^{n - 2}$

$\implies A_{n} = \pi R_{n}^2 = \dfrac{\pi}{12}(\dfrac{1}{9})^{n - 1}a_{1}^2$

$\implies S = \sum_{n = 1}^{\infty} A_{n} = \dfrac{\pi}{12}a_{1}^2(\dfrac{9}{8}) = \dfrac{3}{32}\pi a_{1}^2$ $= \dfrac{\sqrt{3}}{4}(\dfrac{3\pi}{8}(\dfrac{1}{\sqrt{3}})a_{1}^2 = \dfrac{3\pi}{8\sqrt{3}}A_{\triangle{ABC}} =$ $\dfrac{\sqrt{3}}{8}\pi A_{\triangle{ABC}}$

$\implies \dfrac{S}{A_{\triangle{ABC}}} = \dfrac{\sqrt{3}}{8}\pi = \dfrac{\sqrt{\alpha}}{\beta}\pi$ $\implies \alpha + \beta = \boxed{11}$ .