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 $n=12$.

enter image description here

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


$$\therefore\prod\limits_{k=1}^n l_k=1$$

$$\therefore \lim\limits_{n\to\infty}\prod\limits_{k=1}^n l_k=1$$

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.)

  • $\begingroup$ When $n$ is odd you can extend the chords and this procedure gives $2n$-chords case. And the product of the added chords must be close to the product of $n$ chords we started. $\endgroup$
    – Bob Dobbs
    Sep 15 at 22:17
  • 1
    $\begingroup$ @Bob Dobbs The problem is that you an multiply an ever-growing number of multiples close to starting ones. E.g. $e^{1/n}$ tends to $1$, but a product of $n$ copies of $e^{1/n}$ does not. $\endgroup$ Sep 15 at 23:25
  • $\begingroup$ You should first try to prove that the answer is independent of $r$, then prove that it is independent of the ways of rotating the lines so that equiangularity is ensured. $\endgroup$ Sep 16 at 9:35
  • $\begingroup$ For odd $n$ product can be far from 1 for big $r$, for example at $r=100$, $n=9$ product can be from $1/a$ to $a$, where $a>16$. But I believe $a$ value is approaching 1 with $n$ increase. $\endgroup$ Sep 16 at 10:05

1 Answer 1


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.

  • 1
    $\begingroup$ I think there is no need to separate the two odd cases. Let $x_0$ be the length of the line segment that is closest to being perpendicular to the line through $P$ and the centre of the circle. So $\lim\limits_{n\to\infty}x_0=1$, and for any odd $n$, $\prod\limits_{k=1}^n x_k = x_0\prod\limits_{k=1}^{\frac{n-1}{2}} x_k x_{\left(k+\frac{n-1}{2}\right)}$ . If you simplify your answer like this, then in your expression for $x_k$, you would need to change cos to sin. (The big-O thing would still work out.) $\endgroup$
    – Dan
    Sep 18 at 3:17
  • $\begingroup$ @Dan I am still trying to accept my answer. I must learn more about O. Thanks for your comments. $\endgroup$
    – Bob Dobbs
    Sep 18 at 7:42

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