In a circle of radius $a$, 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 in unison; any of these ways is fine.)

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 polar equation of the circle, with $P$ at the origin, is $(r\cos{\theta}-pa)^2+(r\sin{\theta})^2=\dfrac{n}{\pi(1-p^2)}$

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 in unison.)

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 $L=1$, but I have not been able to prove this. (Wolfram evaluates the indefinite integral, but it seems hopelessly complicated.)

(Context: I am interested in questions like this about the product of areas in a circle. Here is another one.)

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.)

In a circle of radius $a$, 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 a constant angle between them. (There is more than one many way to do this, because all the line segments can be rotated in unison; any of these ways is fine.)

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).

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

My attempt:

The polar equation of the circle, with $P$ at the origin, is $(r\cos{\theta}-pa)^2+(r\sin{\theta})^2=\dfrac{n}{\pi(1-p^2)}$

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 in unison.)

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 $L=1$, but I have not been able to prove this. (Wolfram evaluates the indefinite integral, but it seems hopelessly complicated.)

(Context: I am interested in questions like this about the product of areas in a circle. Here is another one.)

In a circle of radius $a$, 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 in unison; any of these ways is fine.)

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 polar equation of the circle, with $P$ at the origin, is $(r\cos{\theta}-pa)^2+(r\sin{\theta})^2=\dfrac{n}{\pi(1-p^2)}$

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 in unison.)

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 $L=1$, but I have not been able to prove this. (Wolfram evaluates the indefinite integral, but it seems hopelessly complicated.)

(Context: I am interested in questions like this about the product of areas in a circle. Here is another one.)