Product of areas in a circle: Why does it approach $1$? In a circle of radius $a$, point $P$ is located $pa$ from the centre, where $0<p<1$. Draw $n$ line segments from $P$ to the circle, such that neighboring line segments have equal angles between them. (There is more than one many way to do this, because all the line segments can be rotated together; any of these ways will do.)
Here is an example with $p=\frac{1}{3}$ and $n=12$.

I am trying to prove (or disprove) the following conjecture (which I worked out by playing with desmos).

For constant $p$, if the mean area of the regions is always $\dfrac{1}{1-p^2}$, then $L=\lim\limits_{n\to\infty}{(\text{product of areas})}=1$.

My attempt:
The cartesian equation of the circle, with $P$ at the origin, is
$$(x-pa)^2+y^2=a^2$$
Changing this to a polar equation, and using $\dfrac{\pi a^2}{n}=\dfrac{1}{1-p^2}\implies a=\sqrt{\dfrac{n}{\pi(1-p^2)}}$, we get
$$\left(r\cos{\theta}-p\sqrt{\dfrac{n}{\pi(1-p^2)}}\right)^2+(r\sin{\theta})^2=\dfrac{n}{\pi(1-p^2)}$$
Using the formula for area in polar coordinates, we have
$$L=\lim\limits_{n\to\infty}{\prod_{k=1}^n{\dfrac{1}{2}\int_{2(k-1)\pi/n+t}^{2k\pi/n+t}r^2{\text{d}\theta}}}$$
($t$ is there because all the line segments can be rotated together.)
Extracting $r$ from the polar equation of the circle and putting it into the integral, and simplifying, we get
$$L=\lim_{n\to\infty}{\prod_{k=1}^n{\dfrac{n}{2\pi(1-p^2)}\int_{2(k-1)\pi/n+t}^{2k\pi/n+t}\left(p\cos{\theta}+\sqrt{1-(p\sin{\theta})^2}\right)^2{\text{d}\theta}}}$$
Desmos suggests that $L=1$ (for $0<p<1$ and $t\in \mathbb{R}$), but I have not been able to prove this. Wolfram evaluates the indefinite integral, but it seems hopelessly intractable.
(Context: I am interested in questions like this about the product of areas in a circle. Here is another one.)
 A: 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.
A: As @person has shown, the circular segment portion of each region is negligible in the infinite product, so we can regard each region as the remaining triangle.
Call the lengths of the line segments $m_1, m_2, m_3, ..., m_n$.
$L=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n\left(\prod\limits_{k=1}^n m_k \right)^2$
It has been shown in a related question, that if $p=\dfrac{\sqrt{a^2-1}}{a}$, which means $\color{blue}{a=\sqrt{\dfrac{1}{1-p^2}}}$, then
$$\color{blue}{\lim\limits_{n\to\infty}\prod\limits_{k=1}^n m_k=1}$$
In this question, the mean area of the regions is $\dfrac{1}{1-p^2}$, which means $\color{red}{a=\sqrt{\dfrac{n}{\pi}}\sqrt{\dfrac{1}{1-p^2}}}$, so
$$\color{red}{\lim\limits_{n\to\infty}\dfrac{\prod\limits_{k=1}^n m_k}{\left(\sqrt{\dfrac{n}{\pi}}\right)^n}=1}$$
$L=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n \left(\sqrt{\dfrac{n}{\pi}}\right)^{2n}\left(\dfrac{\prod\limits_{k=1}^n m_k}{\left(\sqrt{\dfrac{n}{\pi}}\right)^n}\right)^2$
$=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n \left(\sqrt{\dfrac{n}{\pi}}\right)^{2n}$
$=\lim\limits_{n\to\infty}\dfrac{\left(\sin{\dfrac{2\pi}{n}}\right)^n}{\left(\dfrac{2\pi}{n}\right)^n}$
By L'Hopital's rule, the last expression equals $1$.
