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

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

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

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

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 $u = \tan(\dfrac{\pi}{10}) = \dfrac{1}{5}\sqrt{25 - 10\sqrt{5}} \implies R = (\dfrac{1}{2})(\dfrac{5 + \sqrt{5}}{\sqrt{5}})r_{1} = (\dfrac{1 + \sqrt{5}}{2})r_{1} \implies \dfrac{R}{r_{1}} = \boxed{\dfrac{1 + \sqrt{5}}{2}} = \dfrac{a + \sqrt{b}}{c}$ $\implies a + b + c = \boxed{8}$

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Note: To show $u = \tan(\dfrac{\pi}{10}) = \dfrac{1}{5}\sqrt{25 - 10\sqrt{5}}$ :

$(1):$ Obtain $\sin(5\theta) = 16\sin^5(\theta) - 20\sin^3(\theta) + 5\sin(\theta)$ using $\cos(5\theta) + i\sin(5\theta) = (\cos(\theta) + i\sin(\theta))^5$ and equate imaginary terms.

$(2):$ Let $\theta = \dfrac{\pi}{5}$ and $x = \sin(\theta)$ obtaining $x(16x^4 - 20x^2 + 5) = 0$

and $x \neq 0 \implies x = \dfrac{\sqrt{5 - \sqrt{5}}}{8}$ (dropping the root for which $x > 1$ )

$\implies \cos(\theta) = \sqrt{\dfrac{3 + \sqrt{5}}{8}} = \sqrt{\dfrac{(1 + \sqrt{5})^2}{16}} = \dfrac{1}{2}\phi$

$\implies \tan(\dfrac{\pi}{10}) = \sqrt{\dfrac{1 - \dfrac{\phi}{2}}{1 + \dfrac{\phi}{2}}} =$ $\sqrt{\dfrac{3 - \sqrt{5}}{5 + \sqrt{5}}} =$

$\sqrt{\dfrac{20 - 8\sqrt{5}}{20}} = \sqrt{\dfrac{5 - 2\sqrt{5}}{5}} = \dfrac{1}{5}\sqrt{25 - 10\sqrt{5}}$ .