In a circle of radius $r>1$, point $P$ is a distance $\sqrt{r^2-1}$ from the centre of the circle. From $P$, draw $n$ line segments to the circle, such that the angles between neighboring line segments are equal. (There is more than one way to do this, because all the line segments can be rotated together; any of these ways will do.)
Here is an example with $r=\sqrt2$ and $n=12$.
Call the lengths of the line segments $l_1, l_2, l_3, …, l_n$.
I am trying to prove (or disprove) the following conjecture:
$\lim\limits_{n\to\infty}\prod\limits_{k=1}^n l_k=1$
My attempt
If $n$ is even, then consider an arbitrary chord $AB$ through $P$, and diameter $CD$ through $P$. Using the intersecting chords theorem, we have
$$(AP)(PB)=(CP)(PD)=(r-\sqrt{r^2-1})(r+\sqrt{r^2-1})=1$$
$$\therefore\prod\limits_{k=1}^n l_k=1$$
$$\therefore \lim\limits_{n\to\infty}\prod\limits_{k=1}^n l_k=1$$
If $n$ is odd, then we can easily verify that $\prod\limits_{k=1}^n l_k$ does not always equal $1$. For example, let $r=\sqrt2$ and $n=3$, and let one of the line segments go through the centre of the circle. We can easily calculate that $\prod\limits_{k=1}^n l_k=\frac{1}{2}(3+3\sqrt2-\sqrt5-\sqrt{10})\approx 0.922$.
But since the product with even $n$ always equals $1$ (with or without taking the limit), it seems plausible that the product with odd $n$ approaches $1$ as $n\to\infty$. But I have not found a convincing way to prove this.
(This question was inspired by this related question and @person's answer.)
Best Answer
We can calculate the line segment lengths as $x_k=l_{k+1}=\sqrt{1+(r^2-1)\cos^2(\frac{2\pi k}{n})}-\sqrt{r^2-1}\cos(\frac{2\pi k}{n})$ where $k=0,1,2,...,n-1$.
For $n=4m+1$, I wrote the product in the form: $$\prod_{k=0}^{n-1}x_k=x_{\frac{n-1}{4}}\prod_{k=0}^{\frac{n-5}{4}}x_{k}x_{k+\frac{n+1}{2}}\prod_{k=\frac{n+3}{4}}^{\frac{n-1}{2}}x_{k}x_{k+\frac{n-1}{2}}.$$ Notice that $x_{\frac{n-1}{4}}\rightarrow 1$ as $n\rightarrow\infty$. Now, it is enough to show that $$x_{k+\frac{n\pm 1}{2}}=x_{k+\frac{n}{2}}+O(\frac{1}{n^2}),\tag{1}$$ since then $\prod_{k=0}^{n-1}x_k=(1+O(\frac{1}{n^2}))^{\frac{n-1}{2}}$ and when $n\rightarrow\infty$, it tends to $1$.
We can show that $$\cos(\frac{2\pi}{n}(k+\frac{n\pm 1}{2}))=-\cos(\frac{2\pi k}{n})+O(\frac{1}{n^2})$$ and since also $\sqrt{1+\frac{1}{n^2}}=1+O(\frac{1}{n^2})$, $(1)$ holds.
$n=4m+3$ case is similar.