Comparing Stars.

Let $OP = r$ and $OC = x \implies PC = r - x$ .

For convenience let $n = 2j + 3$ .

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

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

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

$\implies A(n) = 2n(\dfrac{2\tan(\dfrac{\pi}{2n})}{3 - \tan^2(\dfrac{\pi}{2n})}) r^2$

For $A_{O} = \lim_{n \rightarrow \infty} A(n)$

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 $\pi\lim_{n_\rightarrow \infty} \sec(\dfrac{\pi}{2n}) = \pi \implies A_{0} = \lim_{n_\rightarrow \infty} A(n) = \dfrac{1}{3}\pi r^2$ .

For convenience let $n = 2j + 4$

. $\dfrac{r_{1}}{2} = r_{2}\cos(\dfrac{\pi}{n})) \implies r_{2} = \dfrac{r_{1}}{2\cos(\dfrac{\pi}{n})} \implies h = \dfrac{r_{1}}{2}\tan(\dfrac{\pi}{n}) \implies A^{*}(n) = \dfrac{n}{2}\tan(\dfrac{\pi}{n})r_{1}^2$ .

Using the inequality $\cos(x) < \dfrac{\sin(x)}{x} < 1 \implies \pi < n\sin(\dfrac{\pi}{n}) < \pi \implies \pi < n\tan(\dfrac{\pi}{n}) < \pi\sec(\dfrac{\pi}{n})$ and $\pi\lim_{n \rightarrow \infty} \sec(\dfrac{\pi}{n}) = \pi \implies A_{E} = \lim_{n \rightarrow \infty} A^{*}(n) = \dfrac{1}{2}\pi r_{1}^2$

$\implies \dfrac{A_{O}}{A_{E}} = \dfrac{2}{3} = \dfrac{a}{b} \implies a + b = \boxed{5}$ .