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Enumerate the area of the regions, $a_1, \dots, a_n$, each region is bordered by two straight segments and a circular arc, enumerate the straight bordered regions $l_1, \dots, l_n$ so that the region with area $a_1$ is bordered by $l_1$ and $l_2$ (also $l_{n+1}=l_1$), enumerate the end points of $l_1, \dots, l_n$ as $e_1, \dots, e_n$. Define: $$L_n=\prod_{k=1}^na_k; L = \lim_{n \to \infty}L_n$$ We first prove the area of each section $a_k=\frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + o(|x^{-1}|)$. Decompose the area of each $a_k$ as follows:

We want to prove that the area of triangle $e_{k}e_{k+1}P$ is $\frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1}$ and area of the circular segment between $e_k$ and $e_{k+1}$ is $o(|n^{-1}|)$. The former is easy, just know that the area of a triangle is $\frac{ab}2\sin(\theta)$. The latter requires more work. Extend $Pe_{k+1}$ to intersect with the circle at $F$, it is apparent that the angle $2\pi/n=\angle e_kPe_{k+1}\geq \angle e_kFe_{k+1}$, and by inscribed angle theorem, we have $\angle 2e_kFe_{k+1}= \angle e_kAe_{k+1}$. Hence, $\angle e_kAe_{k+1} \leq 4\pi/n$. Now using the area for a cicular segment, we can derive: \begin{align} a_k & = \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{R^2}{2}(\angle e_kAe_{k+1}-\sin(\angle e_kAe_{k+1}))\\ & = \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{n}{2\pi(1-p^2)}(\angle e_kAe_{k+1}-\sin(\angle e_kAe_{k+1}))\\ & \leq \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{n(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)}\\ \end{align} All that remains to be proven is $A := \lim_{n \to \infty}\frac{n^2(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)} = 0$. Indeed, we may substitute $4\pi/n \to x$, remove constants, and evaluate instead at $0$ (because of the substitution) to get: \begin{align} A &= \lim_{n \to \infty}\frac{n^2(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)}\\ &=\text{Constant}\lim_{x \to 0} \frac{x-\sin(x)}{x^2}\\ &=\text{Constant} \lim_{x \to 0}\frac{1-\cos(x)}{x}\\ &=\text{Constant} \lim_{x \to 0} \frac{\sin(x)}{1}=0 \end{align} Now we may finally move onto the actual problem.

\begin{align} L_n & = \prod_{k=1}^na_k = \prod_{k=1}^n \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_k\prod_{k=2}^{n+1} l_k + o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\prod_{k=1}^n l_k\right)^2+ o(|1|)\\ \end{align} First we handle the even case, where $n = 2m$, in this case, as mentioned in the comment by @Dan, we simply observe that the $l_k$ and $l_{k+m}$ connect to form a chord for $k \leq m$, hence via power of a point, we have: $$\left(\prod\limits_{k=1}^{n}l_k\right)^2=(a^2(1-p^2))^n=\left(\dfrac{n}{\pi(1-p^2)}(1-p^2)\right)^n=\left(\dfrac{n}{\pi}\right)^n$$ Replacing this in the original expression: $$\lim_{n \to \infty}L_n = \lim_{n \to \infty} \left(\frac{n\sin(2\pi/n)}{2\pi}\right)^n = 1$$ Now onto the odd case, we denote $L_{n,\theta}$ for $0\leq \theta \leq 2\pi/n$ as $\prod_{k=1}^na_k$, where $a_1$ is $\theta$ degrees away from the line created by connecting $P$ and the center of the circle. Likewise denote $a_{k,\theta}$ and $l_{k,\theta}$ as the regions and segments respectively corresponding to $L_{n,\theta}$.

We wish to prove that $\prod_{k=1}^n l_{k,\theta} = \prod_{k=1}^n l_{k,\theta+\pi/n} + o(|n^{n/2}|)$. Indeed if we have this then we may break down the product of $l_k$s, via once again power of a point: \begin{align} L_{n,\theta} & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\prod_{k=1}^n l_{k,\theta}\right)^2+ o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_{k,\theta}\prod_{k=1}^n l_{k,\theta+\pi/n} + \prod_{k=1}^n l_{k,\theta}o(|n^{n/2}|)\left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n+ o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_{k,\theta}\prod_{k=1}^n l_{k,\theta+\pi/n} + o(|1|)\\ & = \left(\frac{n\sin(2\pi/n)}{2\pi}\right)^n +o(|1|) \end{align} Returning us to the even case. However this proves to be rather difficult, as we must now derive a formula for $l_{k,\theta}$. Let $Pe_{k}$ be $l_{k,\theta}$ and $Pe_{k}'$ is $l_{k,\theta+\pi/n}$. We know that $\angle e_k'Pe_k$ is $\pi/n$ and that $\angle e_kPA$ is some arbitrary $\gamma = \theta + 2\pi/n$ and so we may use law of sines to derive: $$\frac{\sin(\gamma)}{a} = \frac{\sin(\angle Ae_kP)}{ap} = \frac{\sin(PAe_k)}{l_{k,\theta}}$$ and solving: \begin{align} \angle Ae_kP & = \sin^{-1}(p\sin(\gamma))\\ \sin(PAe_k) & = \sin(180-(\sin^{-1}(p\sin(\gamma)) + \gamma)) \\ & = p\sin(\gamma)\cos(\gamma) + \cos(\sin^{-1}(p\sin(\gamma)))\sin(\gamma)\\ & = \sin(\gamma)\left(p\cos(\gamma)+\sqrt{1-p^2\sin^2(\gamma)}\right)\\ l_{k,\theta} & = a\left(p\cos(\gamma)+\sqrt{1-p^2\sin^2(\gamma)}\right) \end{align} Likewise, setting up the analagous relations for $l_{k,\theta+\pi/n}$: $$\frac{\sin(\gamma+\pi/n)}{a} = \frac{\sin(\angle Ae_k'P)}{ap} = \frac{\sin(PAe_k')}{l_{k,\theta+\pi/n}}$$ and solving to get: $$l_{k,\theta+\pi/n} = a\left(p\cos(\gamma+\pi/n)+\sqrt{1-p^2\sin^2(\gamma+\pi/n)}\right) $$ This shows that $l_{k,\theta}-l_{k,\theta+\pi/n} \leq a(1+p)=\sqrt{n}\sqrt{\frac{1+p}{(1-p)\pi}}$ and that $\prod_{k=1}^n l_{k,\theta} = \prod_{k=1}^n l_{k,\theta+\pi/n} + o(|n^{n/2}|)$. Proven as required.

Enumerate the area of the regions, $a_1, \dots, a_n$, each region is bordered by two straight segments and a circular arc, enumerate the straight bordered regions $l_1, \dots, l_n$ so that the region with area $a_1$ is bordered by $l_1$ and $l_2$ (also $l_{n+1}=l_1$), enumerate the end points of $l_1, \dots, l_n$ as $e_1, \dots, e_n$. Define: $$L_n=\prod_{k=1}^na_k; L = \lim_{n \to \infty}L_n$$ We first prove the area of each section $a_k=\frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + o(|x^{-1}|)$. Decompose the area of each $a_k$ as follows:

We want to prove that the area of triangle $e_{k}e_{k+1}P$ is $\frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1}$ and area of the circular segment between $e_k$ and $e_{k+1}$ is $o(|n^{-1}|)$. The former is easy, just know that the area of a triangle is $\frac{ab}2\sin(\theta)$. The latter requires more work. Extend $Pe_{k+1}$ to intersect with the circle at $F$, it is apparent that the angle $2\pi/n=\angle e_kPe_{k+1}\geq \angle e_kFe_{k+1}$, and by inscribed angle theorem, we have $\angle 2e_kFe_{k+1}= \angle e_kAe_{k+1}$. Hence, $\angle e_kAe_{k+1} \leq 4\pi/n$. Now using the area for a cicular segment, we can derive: \begin{align} a_k & = \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{R^2}{2}(\angle e_kAe_{k+1}-\sin(\angle e_kAe_{k+1}))\\ & = \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{n}{2\pi(1-p^2)}(\angle e_kAe_{k+1}-\sin(\angle e_kAe_{k+1}))\\ & \leq \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{n(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)}\\ \end{align} All that remains to be proven is $A := \lim_{n \to \infty}\frac{n^2(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)} = 0$. Indeed, we may substitute $4\pi/n \to x$, remove constants, and evaluate instead at $0$ (because of the substitution) to get: \begin{align} A &= \lim_{n \to \infty}\frac{n^2(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)}\\ &=\text{Constant}\lim_{x \to 0} \frac{x-\sin(x)}{x^2}\\ &=\text{Constant} \lim_{x \to 0}\frac{1-\cos(x)}{x}\\ &=\text{Constant} \lim_{x \to 0} \frac{\sin(x)}{1}=0 \end{align} Now we may finally move onto the actual problem.

\begin{align} L_n & = \prod_{k=1}^na_k = \prod_{k=1}^n \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_k\prod_{k=2}^{n+1} l_k + o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\prod_{k=1}^n l_k\right)^2+ o(|1|)\\ \end{align} First we handle the even case, where $n = 2m$, in this case, as mentioned in the comment by @Dan, we simply observe that the $l_k$ and $l_{k+m}$ connect to form a chord for $k \leq m$, hence via power of a point, we have: $$\left(\prod\limits_{k=1}^{n}l_k\right)^2=(a^2(1-p^2))^n=\left(\dfrac{n}{\pi(1-p^2)}(1-p^2)\right)^n=\left(\dfrac{n}{\pi}\right)^n$$ Replacing this in the original expression: $$\lim_{n \to \infty}L_n = \lim_{n \to \infty} \left(\frac{n\sin(2\pi/n)}{2\pi}\right)^n = 1$$ Now onto the odd case, we denote $L_{n,\theta}$ for $0\leq \theta \leq 2\pi/n$ as $\prod_{k=1}^na_k$, where $a_1$ is $\theta$ degrees away from the line created by connecting $P$ and the center of the circle. Likewise denote $a_{k,\theta}$ and $l_{k,\theta}$ as the regions and segments respectively corresponding to $L_{n,\theta}$.

We wish to prove that $\prod_{k=1}^n l_{k,\theta} = \prod_{k=1}^n l_{k,\theta+\pi/n} + o(|n^{n/2}|)$. Indeed if we have this then we may break down the product of $l_k$s, via once again power of a point: \begin{align} L_{n,\theta} & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\prod_{k=1}^n l_{k,\theta}\right)^2+ o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_{k,\theta}\prod_{k=1}^n l_{k,\theta+\pi/n} + \prod_{k=1}^n l_{k,\theta}o(|n^{n/2}|)\left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n+ o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_{k,\theta}\prod_{k=1}^n l_{k,\theta+\pi/n} + o(|1|)\\ & = \left(\frac{n\sin(2\pi/n)}{2\pi}\right)^n +o(|1|) \end{align} Returning us to the even case. We can bound $De_k=l_{k,\theta}-l_{k,\theta+\pi/n}$ by the length of the cord between $e_k$ and $e_k'$. Which can be calculated by $$l_{k,\theta}-l_{k,\theta+\pi/n}\leq 2\sin(2\pi/n)$$ This means that: \begin{align} \prod_{k=1}^{n}l_{k,\theta+\pi/n} & \leq \prod_{k=1}^{n}(l_{k,\theta} + 2\sin(2\pi/n))\\ & = \prod_{k=1}^{n}(l_{k,\theta}) + \sum_{i=1}^n \prod_{j\neq i}(l_j)2\sin(2\pi/n))+\dots\\ & = \prod_{k=1}^{n}(l_{k,\theta}) + o(|n^{\frac{n}{2}}|) \end{align} Proven as required.

Enumerate the area of the regions, $a_1, \dots, a_n$, each region is bordered by two straight segments and a circular arc, enumerate the straight bordered regions $l_1, \dots, l_n$ so that the region with area $a_1$ is bordered by $l_1$ and $l_2$ (also $l_{n+1}=l_1$), enumerate the end points of $l_1, \dots, l_n$ as $e_1, \dots, e_n$. Define: $$L_n=\prod_{k=1}^na_k; L = \lim_{n \to \infty}L_n$$ We first prove the area of each section $a_k=\frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + o(|x^{-1}|)$. Decompose the area of each $a_k$ as follows:

We want to prove that the area of triangle $e_{k}e_{k+1}P$ is $\frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1}$ and area of the circular segment between $e_k$ and $e_{k+1}$ is $o(|n^{-1}|)$. The former is easy, just know that the area of a triangle is $\frac{ab}2\sin(\theta)$. The latter requires more work. Extend $Pe_{k+1}$ to intersect with the circle at $F$, it is apparent that the angle $2\pi/n=\angle e_kPe_{k+1}\geq \angle e_kFe_{k+1}$, and by inscribed angle theorem, we have $\angle 2e_kFe_{k+1}= \angle e_kAe_{k+1}$. Hence, $\angle e_kAe_{k+1} \leq 4\pi/n$. Now using the area for a cicular segment, we can derive: \begin{align} a_k & = \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{R^2}{2}(\angle e_kAe_{k+1}-\sin(\angle e_kAe_{k+1}))\\ & = \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{n}{2\pi(1-p^2)}(\angle e_kAe_{k+1}-\sin(\angle e_kAe_{k+1}))\\ & \leq \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{n(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)}\\ \end{align} All that remains to be proven is $A := \lim_{n \to \infty}\frac{n^2(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)} = 0$. Indeed, we may substitute $4\pi/n \to x$, remove constants, and evaluate instead at $0$ (because of the substitution) to get: \begin{align} A &= \lim_{n \to \infty}\frac{n^2(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)}\\ &=\text{Constant}\lim_{x \to 0} \frac{x-\sin(x)}{x^2}\\ &=\text{Constant} \lim_{x \to 0}\frac{1-\cos(x)}{x}\\ &=\text{Constant} \lim_{x \to 0} \frac{\sin(x)}{1}=0 \end{align} Now we may finally move onto the actual problem.

\begin{align} L_n & = \prod_{k=1}^na_k = \prod_{k=1}^n \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_k\prod_{k=2}^{n+1} l_k + o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\prod_{k=1}^n l_k\right)^2+ o(|1|)\\ \end{align} First we handle the even case, where $n = 2m$, in this case, as mentioned in the comment by @Dan, we simply observe that the $l_k$ and $l_{k+m}$ connect to form a chord for $k \leq m$, hence via power of a point, we have: $$\left(\prod\limits_{k=1}^{n}l_k\right)^2=(a^2(1-p^2))^n=\left(\dfrac{n}{\pi(1-p^2)}(1-p^2)\right)^n=\left(\dfrac{n}{\pi}\right)^n$$ Replacing this in the original expression: $$\lim_{n \to \infty}L_n = \lim_{n \to \infty} \left(\frac{n\sin(2\pi/n)}{2\pi}\right)^n = 1$$ Now onto the odd case, we denote $L_{n,\theta}$ for $0\leq \theta \leq 2\pi/n$ as $\prod_{k=1}^na_k$, where $a_1$ is $\theta$ degrees away from the line created by connecting $P$ and the center of the circle. Likewise denote $a_{k,\theta}$ and $l_{k,\theta}$ as the regions and segments respectively corresponding to $L_{n,\theta}$.

We wish to prove that $(\prod_{k=1}^n l_{k,\theta} = (\prod_{k=1}^n l_{k,\theta+\pi/n} + o(|n^{n/2}|)$. Indeed if we have this then we may break down the product of $l_k$s, via once again power of a point: \begin{align} L_{n,\theta} & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\prod_{k=1}^n l_{k,\theta}\right)^2+ o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_{k,\theta}\prod_{k=1}^n l_{k,\theta+\pi/n} + o(|n^{n/2}|)\left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n+ o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_{k,\theta}\prod_{k=1}^n l_{k,\theta+\pi/n} + o(|1|)\\ & = \left(\frac{n\sin(2\pi/n)}{2\pi}\right)^n +o(|1|) \end{align} Returning us to the even case. However this proves to be rather difficult, as we must now derive a formula for $l_{k,\theta}$. Let $Pe_{k}$ be $l_{k,\theta}$ and $Pe_{k}'$ is $l_{k,\theta+\pi/n}$. We know that $\angle e_k'Pe_k$ is $\pi/n$ and that $\angle e_kPA$ is some arbitrary $\gamma = \theta + 2\pi/n$ and so we may use law of sines to derive: $$\frac{\sin(\gamma)}{a} = \frac{\sin(\angle Ae_kP)}{ap} = \frac{\sin(PAe_k)}{l_{k,\theta}}$$ and solving: \begin{align} \angle Ae_kP & = \sin^{-1}(p\sin(\gamma))\\ \sin(PAe_k) & = \sin(180-(\sin^{-1}(p\sin(\gamma)) + \gamma)) \\ & = p\sin(\gamma)\cos(\gamma) + \cos(\sin^{-1}(p\sin(\gamma)))\sin(\gamma)\\ & = \sin(\gamma)\left(p\cos(\gamma)+\sqrt{1-p^2\sin^2(\gamma)}\right)\\ l_{k,\theta} & = a\left(p\cos(\gamma)+\sqrt{1-p^2\sin^2(\gamma)}\right) \end{align} Likewise, setting up the analagous relations for $l_{k,\theta+\pi/n}$: $$\frac{\sin(\gamma+\pi/n)}{a} = \frac{\sin(\angle Ae_k'P)}{ap} = \frac{\sin(PAe_k')}{l_{k,\theta+\pi/n}}$$ and solving to get: $$l_{k,\theta+\pi/n} = a\left(p\cos(\gamma+\pi/n)+\sqrt{1-p^2\sin^2(\gamma+\pi/n)}\right) $$ This shows that $l_{k,\theta}-l_{k,\theta+\pi/n} \leq a(1+p)=\sqrt{n}\sqrt{\frac{1+p}{(1-p)\pi}}$ and that $(\prod_{k=1}^n l_{k,\theta} = (\prod_{k=1}^n l_{k,\theta+\pi/n} + o(|n^{n/2}|)$. Proven as required.

Enumerate the area of the regions, $a_1, \dots, a_n$, each region is bordered by two straight segments and a circular arc, enumerate the straight bordered regions $l_1, \dots, l_n$ so that the region with area $a_1$ is bordered by $l_1$ and $l_2$ (also $l_{n+1}=l_1$), enumerate the end points of $l_1, \dots, l_n$ as $e_1, \dots, e_n$. Define: $$L_n=\prod_{k=1}^na_k; L = \lim_{n \to \infty}L_n$$ We first prove the area of each section $a_k=\frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + o(|x^{-1}|)$. Decompose the area of each $a_k$ as follows:

We want to prove that the area of triangle $e_{k}e_{k+1}P$ is $\frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1}$ and area of the circular segment between $e_k$ and $e_{k+1}$ is $o(|n^{-1}|)$. The former is easy, just know that the area of a triangle is $\frac{ab}2\sin(\theta)$. The latter requires more work. Extend $Pe_{k+1}$ to intersect with the circle at $F$, it is apparent that the angle $2\pi/n=\angle e_kPe_{k+1}\geq \angle e_kFe_{k+1}$, and by inscribed angle theorem, we have $\angle 2e_kFe_{k+1}= \angle e_kAe_{k+1}$. Hence, $\angle e_kAe_{k+1} \leq 4\pi/n$. Now using the area for a cicular segment, we can derive: \begin{align} a_k & = \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{R^2}{2}(\angle e_kAe_{k+1}-\sin(\angle e_kAe_{k+1}))\\ & = \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{n}{2\pi(1-p^2)}(\angle e_kAe_{k+1}-\sin(\angle e_kAe_{k+1}))\\ & \leq \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{n(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)}\\ \end{align} All that remains to be proven is $A := \lim_{n \to \infty}\frac{n^2(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)} = 0$. Indeed, we may substitute $4\pi/n \to x$, remove constants, and evaluate instead at $0$ (because of the substitution) to get: \begin{align} A &= \lim_{n \to \infty}\frac{n^2(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)}\\ &=\text{Constant}\lim_{x \to 0} \frac{x-\sin(x)}{x^2}\\ &=\text{Constant} \lim_{x \to 0}\frac{1-\cos(x)}{x}\\ &=\text{Constant} \lim_{x \to 0} \frac{\sin(x)}{1}=0 \end{align} Now we may finally move onto the actual problem.

\begin{align} L_n & = \prod_{k=1}^na_k = \prod_{k=1}^n \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_k\prod_{k=2}^{n+1} l_k + o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\prod_{k=1}^n l_k\right)^2+ o(|1|)\\ \end{align} First we handle the even case, where $n = 2m$, in this case, as mentioned in the comment by @Dan, we simply observe that the $l_k$ and $l_{k+m}$ connect to form a chord for $k \leq m$, hence via power of a point, we have: $$\left(\prod\limits_{k=1}^{n}l_k\right)^2=(a^2(1-p^2))^n=\left(\dfrac{n}{\pi(1-p^2)}(1-p^2)\right)^n=\left(\dfrac{n}{\pi}\right)^n$$ Replacing this in the original expression: $$\lim_{n \to \infty}L_n = \lim_{n \to \infty} \left(\frac{n\sin(2\pi/n)}{2\pi}\right)^n = 1$$ Now onto the odd case, we denote $L_{n,\theta}$ for $0\leq \theta \leq 2\pi/n$ as $\prod_{k=1}^na_k$, where $a_1$ is $\theta$ degrees away from the line created by connecting $P$ and the center of the circle. Likewise denote $a_{k,\theta}$ and $l_{k,\theta}$ as the regions and segments respectively corresponding to $L_{n,\theta}$.

We wish to prove that $\prod_{k=1}^n l_{k,\theta} = \prod_{k=1}^n l_{k,\theta+\pi/n} + o(|n^{n/2}|)$. Indeed if we have this then we may break down the product of $l_k$s, via once again power of a point: \begin{align} L_{n,\theta} & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\prod_{k=1}^n l_{k,\theta}\right)^2+ o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_{k,\theta}\prod_{k=1}^n l_{k,\theta+\pi/n} + \prod_{k=1}^n l_{k,\theta}o(|n^{n/2}|)\left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n+ o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_{k,\theta}\prod_{k=1}^n l_{k,\theta+\pi/n} + o(|1|)\\ & = \left(\frac{n\sin(2\pi/n)}{2\pi}\right)^n +o(|1|) \end{align} Returning us to the even case. However this proves to be rather difficult, as we must now derive a formula for $l_{k,\theta}$. Let $Pe_{k}$ be $l_{k,\theta}$ and $Pe_{k}'$ is $l_{k,\theta+\pi/n}$. We know that $\angle e_k'Pe_k$ is $\pi/n$ and that $\angle e_kPA$ is some arbitrary $\gamma = \theta + 2\pi/n$ and so we may use law of sines to derive: $$\frac{\sin(\gamma)}{a} = \frac{\sin(\angle Ae_kP)}{ap} = \frac{\sin(PAe_k)}{l_{k,\theta}}$$ and solving: \begin{align} \angle Ae_kP & = \sin^{-1}(p\sin(\gamma))\\ \sin(PAe_k) & = \sin(180-(\sin^{-1}(p\sin(\gamma)) + \gamma)) \\ & = p\sin(\gamma)\cos(\gamma) + \cos(\sin^{-1}(p\sin(\gamma)))\sin(\gamma)\\ & = \sin(\gamma)\left(p\cos(\gamma)+\sqrt{1-p^2\sin^2(\gamma)}\right)\\ l_{k,\theta} & = a\left(p\cos(\gamma)+\sqrt{1-p^2\sin^2(\gamma)}\right) \end{align} Likewise, setting up the analagous relations for $l_{k,\theta+\pi/n}$: $$\frac{\sin(\gamma+\pi/n)}{a} = \frac{\sin(\angle Ae_k'P)}{ap} = \frac{\sin(PAe_k')}{l_{k,\theta+\pi/n}}$$ and solving to get: $$l_{k,\theta+\pi/n} = a\left(p\cos(\gamma+\pi/n)+\sqrt{1-p^2\sin^2(\gamma+\pi/n)}\right) $$ This shows that $l_{k,\theta}-l_{k,\theta+\pi/n} \leq a(1+p)=\sqrt{n}\sqrt{\frac{1+p}{(1-p)\pi}}$ and that $\prod_{k=1}^n l_{k,\theta} = \prod_{k=1}^n l_{k,\theta+\pi/n} + o(|n^{n/2}|)$. Proven as required.

Work in progress

Enumerate the area of the regions, $a_1, \dots, a_n$, each region is bordered by two straight segments and a circular arc, enumerate the straight bordered regions $l_1, \dots, l_n$ so that the region with area $a_1$ is bordered by $l_1$ and $l_2$ (also $l_{n+1}=l_1$), enumerate the end points of $l_1, \dots, l_n$ as $e_1, \dots, e_n$. Define: $$L_n=\prod_{k=1}^na_k; L = \lim_{n \to \infty}L_n$$ We first prove the area of each section $a_k=\frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + o(|x^{-1}|)$. Decompose the area of each $a_k$ as follows:

We want to prove that the area of triangle $e_{k}e_{k+1}P$ is $\frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1}$ and area of the circular segment between $e_k$ and $e_{k+1}$ is $o(|n^{-1}|)$. The former is easy, just know that the area of a triangle is $\frac{ab}2\sin(\theta)$. The latter requires more work. Extend $Pe_{k+1}$ to intersect with the circle at $F$, it is apparent that the angle $2\pi/n=\angle e_kPe_{k+1}\geq \angle e_kFe_{k+1}$, and by inscribed angle theorem, we have $\angle 2e_kFe_{k+1}= \angle e_kAe_{k+1}$. Hence, $\angle e_kAe_{k+1} \leq 4\pi/n$. Now using the area for a cicular segment, we can derive: \begin{align} a_k & = \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{R^2}{2}(\angle e_kAe_{k+1}-\sin(\angle e_kAe_{k+1}))\\ & = \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{n}{2\pi(1-p^2)}(\angle e_kAe_{k+1}-\sin(\angle e_kAe_{k+1}))\\ & \leq \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{n(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)}\\ \end{align} All that remains to be proven is $A := \lim_{n \to \infty}\frac{n^2(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)} = 0$. Indeed, we may substitute $4\pi/n \to x$, remove constants, and evaluate instead at $0$ (because of the substitution) to get: \begin{align} A &= \lim_{n \to \infty}\frac{n^2(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)}\\ &=\text{Constant}\lim_{x \to 0} \frac{x-\sin(x)}{x^2}\\ &=\text{Constant} \lim_{x \to 0}\frac{1-\cos(x)}{x}\\ &=\text{Constant} \lim_{x \to 0} \frac{\sin(x)}{1}=0 \end{align} Now we may finally move onto the actual problem.

\begin{align} L_n & = \prod_{k=1}^na_k = \prod_{k=1}^n \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_k\prod_{k=2}^{n+1} l_k + o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\prod_{k=1}^n l_k\right)^2+ o(|1|)\\ \end{align} Now we bound the product by AM-GM, specifically: $$\prod_{k=1}^nl_k \leq \left(\frac{\sum_{k=1}^nl_k}{n}\right)^n$$ And so: \begin{align} L_n & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\prod_{k=1}^n l_k\right)^2+ o(|1|) \\ & \leq \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\frac{\sum_{k=1}^nl_k}{n}\right)^{2n}+ o(|1|)\\ \end{align} This is sort of where I am stuck, I suspect that the sum of all the $l_k$s might be bounded by $na$ but I haven't thought of a way to prove that. Indeed, the rest of the problem is just computation if that holds true.

Enumerate the area of the regions, $a_1, \dots, a_n$, each region is bordered by two straight segments and a circular arc, enumerate the straight bordered regions $l_1, \dots, l_n$ so that the region with area $a_1$ is bordered by $l_1$ and $l_2$ (also $l_{n+1}=l_1$), enumerate the end points of $l_1, \dots, l_n$ as $e_1, \dots, e_n$. Define: $$L_n=\prod_{k=1}^na_k; L = \lim_{n \to \infty}L_n$$ We first prove the area of each section $a_k=\frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + o(|x^{-1}|)$. Decompose the area of each $a_k$ as follows:

We want to prove that the area of triangle $e_{k}e_{k+1}P$ is $\frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1}$ and area of the circular segment between $e_k$ and $e_{k+1}$ is $o(|n^{-1}|)$. The former is easy, just know that the area of a triangle is $\frac{ab}2\sin(\theta)$. The latter requires more work. Extend $Pe_{k+1}$ to intersect with the circle at $F$, it is apparent that the angle $2\pi/n=\angle e_kPe_{k+1}\geq \angle e_kFe_{k+1}$, and by inscribed angle theorem, we have $\angle 2e_kFe_{k+1}= \angle e_kAe_{k+1}$. Hence, $\angle e_kAe_{k+1} \leq 4\pi/n$. Now using the area for a cicular segment, we can derive: \begin{align} a_k & = \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{R^2}{2}(\angle e_kAe_{k+1}-\sin(\angle e_kAe_{k+1}))\\ & = \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{n}{2\pi(1-p^2)}(\angle e_kAe_{k+1}-\sin(\angle e_kAe_{k+1}))\\ & \leq \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + \frac{n(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)}\\ \end{align} All that remains to be proven is $A := \lim_{n \to \infty}\frac{n^2(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)} = 0$. Indeed, we may substitute $4\pi/n \to x$, remove constants, and evaluate instead at $0$ (because of the substitution) to get: \begin{align} A &= \lim_{n \to \infty}\frac{n^2(4\pi/n+\sin(4\pi/n))}{2\pi(1-p^2)}\\ &=\text{Constant}\lim_{x \to 0} \frac{x-\sin(x)}{x^2}\\ &=\text{Constant} \lim_{x \to 0}\frac{1-\cos(x)}{x}\\ &=\text{Constant} \lim_{x \to 0} \frac{\sin(x)}{1}=0 \end{align} Now we may finally move onto the actual problem.

\begin{align} L_n & = \prod_{k=1}^na_k = \prod_{k=1}^n \frac{\sin(\frac{2\pi}{n})}{2}l_kl_{k+1} + o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_k\prod_{k=2}^{n+1} l_k + o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\prod_{k=1}^n l_k\right)^2+ o(|1|)\\ \end{align} First we handle the even case, where $n = 2m$, in this case, as mentioned in the comment by @Dan, we simply observe that the $l_k$ and $l_{k+m}$ connect to form a chord for $k \leq m$, hence via power of a point, we have: $$\left(\prod\limits_{k=1}^{n}l_k\right)^2=(a^2(1-p^2))^n=\left(\dfrac{n}{\pi(1-p^2)}(1-p^2)\right)^n=\left(\dfrac{n}{\pi}\right)^n$$ Replacing this in the original expression: $$\lim_{n \to \infty}L_n = \lim_{n \to \infty} \left(\frac{n\sin(2\pi/n)}{2\pi}\right)^n = 1$$ Now onto the odd case, we denote $L_{n,\theta}$ for $0\leq \theta \leq 2\pi/n$ as $\prod_{k=1}^na_k$, where $a_1$ is $\theta$ degrees away from the line created by connecting $P$ and the center of the circle. Likewise denote $a_{k,\theta}$ and $l_{k,\theta}$ as the regions and segments respectively corresponding to $L_{n,\theta}$.

We wish to prove that $(\prod_{k=1}^n l_{k,\theta} = (\prod_{k=1}^n l_{k,\theta+\pi/n} + o(|n^{n/2}|)$. Indeed if we have this then we may break down the product of $l_k$s, via once again power of a point: \begin{align} L_{n,\theta} & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\left(\prod_{k=1}^n l_{k,\theta}\right)^2+ o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_{k,\theta}\prod_{k=1}^n l_{k,\theta+\pi/n} + o(|n^{n/2}|)\left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n+ o(|1|)\\ & = \left(\frac{\sin(\frac{2\pi}{n})}{2}\right)^n\prod_{k=1}^n l_{k,\theta}\prod_{k=1}^n l_{k,\theta+\pi/n} + o(|1|)\\ & = \left(\frac{n\sin(2\pi/n)}{2\pi}\right)^n +o(|1|) \end{align} Returning us to the even case. However this proves to be rather difficult, as we must now derive a formula for $l_{k,\theta}$. Let $Pe_{k}$ be $l_{k,\theta}$ and $Pe_{k}'$ is $l_{k,\theta+\pi/n}$. We know that $\angle e_k'Pe_k$ is $\pi/n$ and that $\angle e_kPA$ is some arbitrary $\gamma = \theta + 2\pi/n$ and so we may use law of sines to derive: $$\frac{\sin(\gamma)}{a} = \frac{\sin(\angle Ae_kP)}{ap} = \frac{\sin(PAe_k)}{l_{k,\theta}}$$ and solving: \begin{align} \angle Ae_kP & = \sin^{-1}(p\sin(\gamma))\\ \sin(PAe_k) & = \sin(180-(\sin^{-1}(p\sin(\gamma)) + \gamma)) \\ & = p\sin(\gamma)\cos(\gamma) + \cos(\sin^{-1}(p\sin(\gamma)))\sin(\gamma)\\ & = \sin(\gamma)\left(p\cos(\gamma)+\sqrt{1-p^2\sin^2(\gamma)}\right)\\ l_{k,\theta} & = a\left(p\cos(\gamma)+\sqrt{1-p^2\sin^2(\gamma)}\right) \end{align} Likewise, setting up the analagous relations for $l_{k,\theta+\pi/n}$: $$\frac{\sin(\gamma+\pi/n)}{a} = \frac{\sin(\angle Ae_k'P)}{ap} = \frac{\sin(PAe_k')}{l_{k,\theta+\pi/n}}$$ and solving to get: $$l_{k,\theta+\pi/n} = a\left(p\cos(\gamma+\pi/n)+\sqrt{1-p^2\sin^2(\gamma+\pi/n)}\right) $$ This shows that $l_{k,\theta}-l_{k,\theta+\pi/n} \leq a(1+p)=\sqrt{n}\sqrt{\frac{1+p}{(1-p)\pi}}$ and that $(\prod_{k=1}^n l_{k,\theta} = (\prod_{k=1}^n l_{k,\theta+\pi/n} + o(|n^{n/2}|)$. Proven as required.