Nested Finale

For Circles:

$m\angle{AOP} = \dfrac{\pi}{5}$ and by Inscribed Angle Theorem $m\angle{EPF} = \dfrac{1}{2}m\angle{EOF} \implies m\angle{APO} = \dfrac{\pi}{10}$

$\dfrac{AC}{OC} = \tan(\dfrac{\pi}{5}) \implies OC = \dfrac{AC}{\tan(\dfrac{\pi}{5})}$ and $PC = \dfrac{AC}{\tan(\dfrac{\pi}{10})} = r_{1} - OC = r - \dfrac{AC}{\tan(\dfrac{\pi}{5})} \implies$

$h = AC = \dfrac{\tan(\dfrac{\pi}{10})\tan(\dfrac{\pi}{5})}{\tan(\dfrac{\pi}{5}) + \tan(\dfrac{\pi}{10})}r_{1}$

$\tan(\dfrac{\pi}{5}) = \tan(\dfrac{2\pi}{10}) = \dfrac{2\tan(\dfrac{\pi}{10})}{1 - \tan^2(\dfrac{\pi}{10})} \implies$ $h = \dfrac{2\tan(\dfrac{\pi}{10})}{3 - \tan^2(\dfrac{\pi}{10})}r_{1}$

Let $\beta = \dfrac{\tan(\dfrac{\pi}{10})}{3 - \tan^2(\dfrac{\pi}{10})} \implies h_{1} = 2\beta r_{1} = r_{2}\sin(\dfrac{\pi}{5})$ $\implies \boxed{r_{2} = \dfrac{2\beta}{\sin(\dfrac{\pi}{5})}r_{1}}$

In General $\boxed{r_{n} = (\dfrac{2\beta}{\sin(\dfrac{\pi}{5})})^{n - 1}r_{1}}$ and $R = \displaystyle\sum_{n = 1}^{\infty} r_{n} = \dfrac{\sin(\dfrac{\pi}{5})}{\sin(\dfrac{\pi}{5}) - 2\beta}r_{1}$ .

Let $u = \tan(\dfrac{x}{2}) \implies u^2 = \dfrac{1 - \cos(x)}{1 + \cos(x)} = \dfrac{\sin^2(x)}{(1 + \cos(x))^2} \implies u = \dfrac{\sin(x)}{1 + \cos(x)}$ and $u^2 = \dfrac{1 - \cos(x)}{1 + \cos(x)} \implies \cos(x) = \dfrac{1 - u^2}{1 + u^2} \implies \sin(x) = \dfrac{2u}{1 + u^2}$ and $u = \tan(\dfrac{x}{2})$

Let $x = \dfrac{\pi}{5} \implies u = \tan(\dfrac{\pi}{10}) \implies \beta = \dfrac{u}{3 - u^2}$ and $\sin(\dfrac{\pi}{5}) = \dfrac{2u}{1 + u^2}$

$\implies R = \dfrac{3 - u^2}{2(1 - u^2)}r_{1}$ and $u = \tan(\dfrac{\pi}{10}) = \dfrac{1}{5}\sqrt{25 - 10\sqrt{5}} \implies R = (\dfrac{1}{2})(\dfrac{5 + \sqrt{5}}{\sqrt{5}})r_{1} = (\dfrac{1 + \sqrt{5}}{2})r_{1} \implies \dfrac{R}{r_{1}} = \boxed{\dfrac{1 + \sqrt{5}}{2}}$ .

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For Hexagons:

Let $x_{1}$ be the length of a side of the initial Hexagon.

$\dfrac{x_{1}}{2} = r_{1}\sin(\dfrac{\pi}{5}) \implies r_{1} = \dfrac{x_{1}}2\sin(\dfrac{\pi}{5})$

From above we have $h_{1} = 2\beta r_{1} = \dfrac{\beta}{\sin(\dfrac{\pi}{5})}x_{1}$ and $x_{2} = 2h_{1} = \dfrac{2\beta}{\sin(\dfrac{\pi}{5})}x_{1} \implies x_{3} = 2h_{2} =\dfrac{2\beta}{\sin(\dfrac{\pi}{5})}x_{2} = (\dfrac{2\beta}{\sin(\dfrac{\pi}{5})})^2 x_{1}$

In General $\boxed{x_{n} = (\dfrac{2\beta}{\sin(\dfrac{\pi}{5})})^{n - 1}x_{1}}$ and from above $\implies \boxed{\dfrac{X}{x_{1}} = \dfrac{R}{r_{1}} = \dfrac{1 + \sqrt{5}}{2}}$

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For ratio $\dfrac{PB}{PA}$ :

Let $PA = z$ and $AB = y$

From above $\beta = \dfrac{\tan(\dfrac{\pi}{10})}{3 - \tan^2(\dfrac{\pi}{10})}, \:\ h_{1} = 2\beta r_{1}$ and $r_{2} = \dfrac{2\beta}{\sin(\dfrac{\pi}{5})}r_{1}$

Using law of cosines on $\triangle{AOB} \implies y = 2\sin(\dfrac{\pi}{5})r_{2} = 4\beta r_{1}$ and $h = 2\beta r_{1} = z\sin(\dfrac{\pi}{10}) \implies z = \dfrac{2\beta}{\sin(\dfrac{\pi}{10})}r_{1}$

$\implies PB = y + z = 2\beta r_{1}(\dfrac{2\sin(\dfrac{\pi}{10}) + 1}{\sin(\dfrac{\pi}{10})}) \implies \dfrac{PB}{PA} = 2\sin(\dfrac{\pi}{10}) + 1$ and $\sin(\dfrac{\pi}{10}) = \dfrac{1}{4}(\sqrt{5} - 1) \implies \dfrac{PB}{PA} = \dfrac{1 + \sqrt{5}}{2} = \dfrac{R}{r_{1}} = \dfrac{X}{x_{1}} = \dfrac{a + \sqrt{b}}{c} \implies a + b + c = \boxed{8}$ .