A disk of area $\dfrac{\pi}{a}$ is divided into $n$ regions of equal area by line segments from a point on the edge.

Here is an example with $n=8$.

enter image description here

Let $P(a,n)=\text{product of lengths of the line segments}$.

What is $\lim\limits_{n\to\infty}P(e,n)$ ? ($e$ is Euler's constant)

Why I chose $a=e$

Here are plots of $P(a,n)$ against $n$ with $a=0.99e$, $\space a=e$ and $\space a=1.01e$.

enter image description here

It seems that if $a<e$ then $\lim\limits_{n\to\infty}P(a,n)=\infty$, and if $a>e$ then $\lim\limits_{n\to\infty}P(a,n)=0$.

So $a=e$ seems to be a critical value.

Here are the $P(e,n)$ values that I got; some of these appear in the graph above.

$P(e,192)\approx 5.2476$

(I calculated these values "manually": that is, I used desmos to get an approximate solution to $x−\sin x=\frac{2k\pi}{n}$ with individual $k$ values, then I used those $x$ values and Excel to approximate $P(e,n)$.)

What makes this difficult

What makes my question difficult for me, is that I cannot find exact expressions for the lengths. For example, with $a=e$ and $n=8$, the length of the shortest line segments is $\frac{2}{\sqrt{e}}\sin \left(\frac{x}{2}\right)$ where $x-\sin x=\frac{\pi}{4}$.

I am aware of Kepler's equation, but that doesn't seem to help.


This question was inspired by the following remarkable fact: If $n$ evenly spaced points are drawn on a unit circle, and line segments are drawn from one point to each of the other points, then the product of lengths of the line segments equals $n$ (proof).

Related question: Conjectured connection between $e$ and $\pi$ in a semidisk


@Carl Schildkraut's answer shows that $\lim\limits_{n\to\infty}\frac{P(e,n)}{n^{1/3}}=\frac{e^{1/2}}{6^{1/3}}\approx0.9073$.

Here is a plot of $\frac{P(e,n)}{n^{1/3}}$ against $n$.

enter image description here

  • 2
    $\begingroup$ Your statement seems strange, if you increase the area, by a factor a, all lengths should in crease by a factor $\sqrt{a}$ so $p_a(n)=(\sqrt{a})^np(n)$ so still zero . $\endgroup$
    – trula
    Commented May 14 at 14:27
  • 1
    $\begingroup$ @trula Why still zero? For example, if $a=10000$ then $P_a(n)=100^n P(n)$. That $100^n$ factor could cause $P_a(n)$ to approach $\infty$ as $n\to\infty$. $\endgroup$
    – Dan
    Commented May 14 at 14:37
  • $\begingroup$ That's interesting. Seems like finding formulas for each length (then adding them) is hard, might be better to try to find just the formula for the sum - like in the product case. If it is indeed true, that assuming area $\frac{\pi}{k}$, the formula for $P_k(n)$ converges to $0$ for $k<e$ and diverges for higher values, maybe you should try to find inspiration from formulas with this property. I've definitely seen such interesting function before (maybe when playing with $(1+\frac{1}{x})^x$) but can't recall it, and on the top of my head can only think of something like $(x/e)^n$. $\endgroup$
    – me9hanics
    Commented May 14 at 14:47
  • $\begingroup$ I suspect that a similar problem with parallel lines will have a similar behavior, but would be more amenable $\endgroup$
    – user619894
    Commented May 14 at 16:19
  • 1
    $\begingroup$ @user619894 If the lines are parallel, it's just the same question in disguise. In both versions of the question, the line segments cut off segments of area $\frac{A}{n},\frac{2A}{n},\frac{3A}{n},\dots$, where $A$ is the area of the disk. So the sequence of lengths is exactly the same in both versions. $\endgroup$
    – Dan
    Commented May 14 at 21:11

2 Answers 2


Comment that's too long for a comment:

The proportion of the circle's area cut off by an arc with angle $\theta$ is $A(\theta) = \frac1{2\pi}(\theta-\sin\theta)$, and the length of the corresponding segment is $\frac2{\sqrt a}\sin\frac\theta2$. The geometric mean of the lengths of the segments should then be approximated by exponentiating the integral of the logarithm of the length function when the endpoint is parametrized by $A^{-1}(y)$, namely $$ \exp\biggl( \int_0^1 \log\Bigl( \frac2{\sqrt a} \sin\frac{A^{-1}(y)}2 \Bigr)\,dy \biggr) = \exp\biggl( \int_0^{2\pi} \log\Bigl( \frac2{\sqrt a} \sin\frac\theta2 \Bigr) \frac1{2\pi}(1-\cos\theta)\,d\theta \biggr). $$ It turns out that this integral equals $0$ precisely when $a=e$, which explains the critical point phenomenon. Since we're asking about the product itself and not the geometric mean, we would need to look at the rate of convergence of the Riemann sums of this integral to the integral itself.


Here's a sketch of an argument, following the paradigm given in Greg Martin's answer to compute the behavior of $P(e,n)$. Our conclusion is that $$\lim_{n\to\infty}\frac{P(e,n)}{n^{1/3}}=\frac{e^{1/2}}{6^{1/3}}\approx 0.9073.$$

Let $A(\theta)=\frac1{2\pi}(\theta-\sin\theta)$, so that $$\log P(e,n)=\sum_{k=1}^{n-1}\log\left(\frac 2{\sqrt e}\sin\frac{A^{-1}(k/n)}2\right).$$ We leave as an exercise the fact that $$\int_0^1\log\left(\frac 2{\sqrt e}\sin\frac{A^{-1}(y)}2\right)dy=0,$$ so that, if $\delta=1/(2n)$, $$\log P(e,n)=\sum_{k=1}^{n-1}\left[\log\left(\sin\frac{A^{-1}(k/n)}2\right)-\frac1{2\delta}\int_{k/n-\delta}^{k/n+\delta}\log\left(\sin\frac{A^{-1}(y)}2\right)dy\right]-E_0,$$ with $$E_0=\frac1\delta\int_0^\delta\log\left(\frac 2{\sqrt e}\sin\frac{A^{-1}(y)}2\right)dy.$$ Note that the summands do not change if $k$ is replaced by $n-k$. As in Part 2 of my previous answer to a similar question, one can show that the terms with $k$ far from $0$ or $n$ are small, so that $$\log P(e,n)=-E_0+o(1)+2\sum_{k=1}^\infty\lim_{n\to\infty}\left[\log\left(\sin\frac{A^{-1}(k/n)}2\right)-\frac1{2\delta}\int_{k/n-\delta}^{k/n+\delta}\log\left(\sin\frac{A^{-1}(y)}2\right)dy\right].$$ We now evaluate the behavior of these terms. Since $A(\theta)\approx \frac{\theta^3}{12\pi}$ near $\theta=0$, we have $$\sin\frac{A^{-1}(y)}2=\frac{(12\pi y)^{1/3}}2(1+o(1))=\left(\frac{3\pi y}2\right)^{1/3}(1+o(1));$$ since, for each fixed $k$ (or for $E_0$), the $y$-values we consider are tending to $0$ like some constant times $1/n$, this $o(1)$ term can be shown to be $O(n^{-2/3})$. Letting $c=(3\pi/2)^{1/3}$, $x=k/n$ and trusting that the errors turn out to be small, we have \begin{align*} E_k&:=\log\left(\sin\frac{A^{-1}(k/n)}2\right)-\frac1{2\delta}\int_{k/n-\delta}^{k/n+\delta}\log\left(\sin\frac{A^{-1}(y)}2\right)dy\\ &\approx \log(cx^{1/3})-\frac1{2\delta}\int_{k/n-\delta}^{k/n+\delta}\log(cy^{1/3})dy\\ &=\frac13\left(\log x-\frac1{2\delta}\int_{x-\delta}^{x+\delta}\log y dy\right)\\ &=\frac1{6\delta}\left[(x-\delta)\left(\log\left(1-\frac\delta x\right)-1\right)-(x+\delta)\left(\log\left(1+\frac\delta x\right)-1\right)\right]\\ &=\frac{2k+1}6\left(1-\log\left(1+\frac1{2k}\right)\right)+\frac{2k-1}6\left(\log\left(1-\frac1{2k}\right)-1\right). \end{align*} Applying a similar argument to $E_0$ gives that \begin{align*} E_0 &=\frac{\log 2-1}3+\log\left(\frac 2{\sqrt e}\right)+\log\left(\sin\frac{A^{-1}(\delta/2)}2\right)+o(1)\\ &=\frac13\log\left(\frac{6\pi}{e^{5/2}n}\right)+o(1). \end{align*} So, $$\log P(e,n)=o(1)+\frac13\log\left(\frac{e^{5/2}n}{6\pi}\right)+2\sum_{k=1}^\infty \frac{2k+1}6\left(1-\log\left(1+\frac1{2k}\right)\right)+\frac{2k-1}6\left(\log\left(1-\frac1{2k}\right)-1\right).$$ This sum can be found to be $\frac{\log\pi-1}6$, so that $$\log P(e,n)=\frac13\log\left(\frac{e^{3/2}n}{6}\right)+o(1).$$ This implies $$P(e,n)=\left(\frac{e^{3/2}n}{6}\right)^{1/3}(1+o(1)),$$ which gives the result.

  • $\begingroup$ I updated the OP with a plot of $\frac{P(e,n)}{n^{1/3}}$ against $n$. I also added the $P(e,n)$ values that I found. $\endgroup$
    – Dan
    Commented May 16 at 3:39
  • 1
    $\begingroup$ The second line above "Applying a similar argument..." is missing a negative sign in front of the whole line. The next line also should be entirely negated. In the line directly below "Applying a similar argument...", the $1-\log 2$ should be $\log 2-1$. This will give $\lim\limits_{n\to\infty}\frac{P(e,n)}{n^{1/3}}=\frac{e^{1/2}}{6^{1/3}}\approx0.9073$, which agrees with numerical results. $\endgroup$
    – Dan
    Commented May 16 at 8:35
  • 1
    $\begingroup$ I went ahead and corrected the mistakes. Thanks for your answer! $\endgroup$
    – Dan
    Commented May 17 at 23:25
  • 1
    $\begingroup$ @Dan Thanks for fixing it up! $\endgroup$ Commented May 18 at 5:20

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