More Circles!

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

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

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

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

Note: I'm using the same notation $R_{n}$ in Diagram 2 to denote different radii.

$\overline{Ow_{0}} = \sqrt{4 + 2\sqrt{2}}R_{1}$

$\triangle{OA_{1}w_{0}} \sim \triangle{w_{1}A_{2}w_{0}} \implies \dfrac{\sqrt{4 + 2\sqrt{2}}R_{1}}{R_{1}} = \dfrac{R_{1} + R_{2}}{R_{1} - R_{2}} \implies$

$(\sqrt{4 + 2\sqrt{2}} - 1)R_{1} = (\sqrt{4 + 2\sqrt{2}} + 1)R_{2} \implies$ $R_{2} = \dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1}R_{1}$

$R_{3} = \dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1}R_{2} = (\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^2R_{1}$

In General: $R_{n} = (\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^{n - 1}R_{1}$

$\implies A_{n} = \pi R_{n}^2 = \pi R_{1}^2((\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^2)^{n - 1}$

$\implies A = \sum_{n = 1}^{\infty} A_{n} = \pi R_{1}^2\sum_{n = 1}^{\infty} ((\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^2)^{n - 1}$

$= \dfrac{(\sqrt{4 + 2\sqrt{2}} + 1)^2}{4\sqrt{4 + 2\sqrt{2}}}\pi R_{1}^2$

and

$C_{n} = 2\pi R_{n} = 2\pi R_{1}(\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^{n - 1}$

$\implies C = \sum_{n = 1}^{\infty} C_{n} = 2\pi R_{1}\sum_{n = 1}^{\infty} (\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^{n - 1} =$

$(\sqrt{4 + 2\sqrt{2}}) \pi R_{1} \implies C^2 = (\sqrt{4 + 2\sqrt{2}})^2 \pi^2 R_{1}^2$

$\implies \boxed{\dfrac{A}{C^2} = \dfrac{1}{4\pi\sqrt{4 + 2\sqrt{2}}}}$

$\implies \dfrac{S}{A_{\triangle{ABC}}} * \dfrac{A}{C^2} = \dfrac{\sqrt{3}}{2^5\sqrt{2^2 + 2\sqrt{2}}} =$ $\dfrac{\sqrt{\alpha}}{\beta^{\lambda}\sqrt{\beta^{\beta} + \beta\sqrt{\beta}}} \implies$

$\alpha + \beta + \lambda = \boxed{10}$ .