It's All Stars.

Let $n \geq 5$ be an odd positive integer.

Note: This is simpler to type then using $2n + 3$ for each positive integer $n$ .

Let $OP = r_{1}$ and $OC = x \implies PC = r_{1} - x$ .

$m\angle{AOP} = \dfrac{\pi}{n}$ and by Inscribed Angle Theorem $m\angle{EPF} = \dfrac{1}{2}m\angle{EOF} = \dfrac{1}{2}(\dfrac{2\pi}{n}) = \dfrac{\pi}{n} \implies m\angle{APO} = \dfrac{\pi}{2n}$ .

$\dfrac{h}{r_{1} - x} = \tan(\dfrac{\pi}{2n}) \implies h = (r_{1} - x)\tan(\dfrac{\pi}{2n})$ and $h = x \tan(\dfrac{\pi}{n}) \implies (r_{1} - x)\tan(\dfrac{\pi}{2n}) = x\tan(\dfrac{\pi}{n}) \implies r_{1}\tan(\dfrac{\pi}{2n}) = (\tan(\dfrac{\pi}{n}) + \tan(\dfrac{\pi}{2n}))x$ $\implies x = \dfrac{\tan(\dfrac{\pi}{2n})}{\tan(\dfrac{\pi}{n}) + \tan(\dfrac{\pi}{2n})} r_{1} \implies h = \dfrac{\tan(\dfrac{\pi}{n})\tan(\dfrac{\pi}{2n})}{\tan(\dfrac{\pi}{n}) + \tan(\dfrac{\pi}{2n})} r_{1}$

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

Let $B_{n} = \dfrac{\tan(\dfrac{\pi}{2n})}{3 - \tan^2(\dfrac{\pi}{2n})} \implies h = 2B_{n}r_{1} = r_{2}\sin(\dfrac{\pi}{n}) \implies r_{2} = \dfrac{2B_{n}}{\sin(\dfrac{\pi}{n})}r_{1}$

In General:

$r_{m} = (\dfrac{2B_{n}}{\sin(\dfrac{\pi}{n})})^{m - 1}r_{1} \implies A_{c}(m) = \pi r_{m}^2 = \pi(\dfrac{4B_{n}^2}{\sin^2(\dfrac{\pi}{n})})^{m - 1}r_{1}^2 \implies$ $A_{c}(n) = \sum_{m = 1}^{\infty} A_{c}(m) = \dfrac{\sin^2(\dfrac{\pi}{n})}{\sin^2(\dfrac{\pi}{n}) - 4B_{n}^2}\pi r_{1}^2$ = $\dfrac{\sin^2(\dfrac{\pi}{n})}{\sin^2(\dfrac{\pi}{n}) - 4B_{n}^2} A_{c}(1)$

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

Let $y = \dfrac{\pi}{n} \implies u_{n} = \tan(\dfrac{\pi}{2n}) \implies B_{n} = \dfrac{u_{n}}{3 - u_{n}^2}$ and $\sin(\dfrac{\pi}{n}) = \dfrac{2u_{n}}{1 + u_{n}^2} \implies A_{c}(n) = \dfrac{(3 - u_{n}^2)^2}{(3 - u_{n}^2)^2 - (1 + u_{n}^2)^2} A_{c}(1)$ and $\lim_{n \rightarrow \infty} u_{n} = 0 \implies A_{c} = \lim_{n \rightarrow \infty} A_{c}(n) = \dfrac{9}{8}A_{c}(1)$

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From above $h_{1} = 2B_{n}r_{1} \implies h_{2} = 2B_{n}r_{2} = 2B_{n}(\dfrac{2B_{n}}{\sin(\dfrac{\pi}{n})})r_{1} \implies h_{3} = 2B_{n}(\dfrac{2B_{n}}{\sin(\dfrac{\pi}{n})})^2 r_{1}$

In General:

$h_{m} = 2B_{n}(\dfrac{2B_{n}}{\sin(\dfrac{\pi}{n})})^{m - 1}r_{1}$ and from above $r_{m} = (\dfrac{2B_{n}}{\sin(\dfrac{\pi}{n})})^{m - 1}r_{1} \implies A_{p}(m) = 2n(\dfrac{1}{2})r_{m}h_{m} =$ $n(2B_{n})(\dfrac{4B_{n}^2}{\sin^2(\dfrac{\pi}{n})})^{m - 1}r_{1}^2$

$\implies A_{p}(n) = \sum_{m = 1}^{\infty} A_{p}(m) = 2nB_{n}(\dfrac{\sin^2(\dfrac{\pi}{n})}{\sin^2(\dfrac{\pi}{n}) - 4B_{n}^2})r_{1}^2$ .

Let $t_{n} = 2nB_{n} = 2n\dfrac{\tan(\dfrac{\pi}{2n})}{3 - \tan^2(\dfrac{\pi}{2n})}$ and $S_{n} = \dfrac{\sin^2(\dfrac{\pi}{n})}{\sin^2(\dfrac{\pi}{n}) - 4B_{n}^2} r_{1}^2$

From above it was shown that $\lim_{n_\rightarrow \infty} S_{n} = \dfrac{9}{8}$

For $t_{n}$ :

Using the inequality $\cos(x) < \dfrac{\sin(x)}{x} < 1$ we have:

$\pi\cos(\dfrac{\pi}{2n}) < 2n\sin(\dfrac{\pi}{2n}) < \pi \implies \pi < 2n\tan(\dfrac{\pi}{2n}) < \pi\sec(\dfrac{\pi}{2n})$ and $\pi\lim_{n_\rightarrow \infty} \sec(\dfrac{\pi}{2n}) = \pi \implies \lim_{n_\rightarrow \infty} t_{n} = \dfrac{\pi}{3}$

$\implies \lim_{n_\rightarrow \infty} A_{p}(n) = \lim_{n_\rightarrow \infty} t_{n} * S_{n} = \dfrac{\pi}{3} * \dfrac{9}{8} = \dfrac{3}{8}\pi r_{1}^2 = \dfrac{3}{8}A_{c}(1)$

$\implies \dfrac{A_{c} * A_{p}}{A_{c}(1)^2} = \dfrac{27}{64} = (\dfrac{3}{4})^3 = (\dfrac{a}{b})^{a} \implies a + b = \boxed{7}$

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