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A large disk of radius $R$ is divided into $n$ equal-width vertical strips. Here is an example with $n=6$.

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What value of $n$ maximizes the product of areas of the strips, in terms of $R$?

Numerical investigation suggests that the product of areas is maximized when $n\approx\dfrac{\pi R^2}{2.90}$.

I am looking for a closed form for the $2.90$.

Related results

  • If the disk is divided into $n$ regions of equal area, then the product of areas is maximized when $n\approx\dfrac{\pi R^2}{e}$. Proof.

  • If the disk is divided into $n$ regions like a seashell, then the product of areas is maximized when $n\approx\dfrac{\pi R^2}{2e}$. Proof.

My attempt

Let $f(k)$ be the total area of the first $k$ strips from the left. Using the formula for the area of a segment, we have

$$f(k)=\frac12R^2\left(2\arccos\left(1-\frac{2k}{n}\right)-\sin\left(2\arccos\left(1-\frac{2k}{n}\right)\right)\right)$$

$$=R^2\left(\arccos\left(1-\frac{2k}{n}\right)-\left(1-\frac{2k}{n}\right)\sqrt{1-\left(1-\frac{2k}{n}\right)^2}\right)$$

The product of areas is

$$P(n)=\prod_{k=1}^n(f(k)-f(k-1))$$

Alternatively,

$$\ln P(n)=\sum_{k=1}^n\ln(f(k)-f(k-1))$$

I don't know what to do with this.

Dan
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2 Answers2

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Let us estimate the leading-order answer under the assumption that $n$ is large. To do this, we define $x=\frac{2k}{n}-1$ that runs from $-1$ to $+1$ (if the index is is "off by one" it will be a sub-leading effect). The area of one of the slices is $$ f(x) \approx \frac{4R^2}{n}\sqrt{1-x^2}\, . $$ The log-product of the areas is $$ \begin{align} \log P(n)&=\log\prod_k f(k) \approx n\log \frac{4R^2}{n}+\frac 12\sum_k\log(1-x^2) \\&\approx n\log \frac{4R^2}{n}+\frac n4\int_{-1}^1 dx \,\log(1-x^2) =-n+n\log\frac{8R^2}{n}\,. \end{align} $$ Finally, extremizing this yields $$ \frac {d}{dn}\log p = 0\qquad \Rightarrow\qquad n \approx \frac{\pi R^2}{\alpha}\, , \qquad \alpha =\frac{e^2\pi}{8}\approx 2.90\,. $$

Ben
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  • You're using rectangles to approximate the slices, right? For the first slice on the left in OP's diagram (with large $n$), your approximation is off by a factor of about $\frac23$; for the second slice about $0.862$. We're taking the product of areas, so these discrepancies are compounded. So how to justify that your approximation works? I think we need to prove an asymptotic expression for $\prod_{k=0}^{n-1}\frac{\int_{k/n}^{(k+1)/n}\sqrt{1-x^2}dx}{\frac1n\sqrt{1-\left(\frac{k}{n}\right)^2}}$. – Dan May 01 '25 at 01:59
  • @Dan, similar to Ben, I was thinking of approximating the slice with a trapezium. For trapezium area, the average of the two parallel sides will touch the circle at some points and the other component is diameter divided by n. But I guess I will run into the same approximation problem you have highlighted – Srini May 01 '25 at 05:05
  • In the limit of large $n$ (and thus for self-consistency large $R^2$), these discrepancies will show up as $\mathcal O(n^0)$ in $\log P$ and can safely be neglected in the leading-order estimate. – Ben May 01 '25 at 12:02
  • @Ben Numerical investigation suggests that $\lim\limits_{n\to\infty}n^{1/4}\prod_{k=0}^{n-1}\dfrac{\int_{k/n}^{(k+1)/n}\sqrt{1-x^2}dx}{\frac1n\sqrt{1-\left(\frac{k}{n}\right)^2}}\approx 0.833887$, which would imply that the discrepancies show up as $\mathcal O(\log n)$ in $\log P$, which would imply that the answer to the OP is indeed $n\approx \dfrac{8R^2}{e^2}$. But we need to prove this limit (and I think that is the hard part). – Dan May 01 '25 at 23:16
  • I've asked about the approximation used in this answer. – Dan May 09 '25 at 14:59
4

Suggestions, not a complete answer

It might help to give $f$ another argument, so that $$ f(k, n) = R^2\left(\arccos\left(1-\frac{2k}{n}\right)-\left(1-\frac{2k}{n}\right)\sqrt{1-\left(1-\frac{2k}{n}\right)^2}\right). $$ Then you have $$ \ln P(n)=\sum_{k=1}^n\ln(f(k, n)-f(k-1, n)) $$ Now you can look at $\frac{\ln P(n+1) - \ln P(n)}{1}$, which is a discrete approximation of the derivative. When it's zero (or nearly zero), you've probably got the optimal $n$ (plus or minus one).

\begin{align} \ln P(n+1) - \ln P(n) & = \left( \sum_{k=1}^{n+1}\ln(f(k, n+1)-f(k-1, n+1))\right) - \left(\sum_{k=1}^n\ln(f(k, n)-f(k-1, n))\right) \\ & = \ln \left( f(n+1, n+1) - f(n, n+1) \right) + \left( \sum_{k=1}^{n}\ln(f(k, n+1)-f(k-1, n+1))\right) - \left(\sum_{k=1}^n\ln(f(k, n)-f(k-1, n))\right) \\ & = \ln \left( f(n+1, n+1) - f(n, n+1) \right)+ \left( \sum_{k=1}^{n}\ln(f(k, n+1)-f(k-1, n+1))- \ln(f(k, n)-f(k-1, n))\right) \\ & = \ln \left( f(n+1, n+1) - f(n, n+1)\right) + \sum_{k=1}^{n}\ln\left(\frac{f(k, n+1)-f(k-1, n+1)}{f(k, n)-f(k-1, n)}\right) \\ \end{align} Perhaps setting this to zero might help you get an estimate of the value of $n$ that's optimal.

Frankly, it all looks like a mess to me, but I don't have a lot of insight into the function $f$, and maybe there's some magical simplification that takes place.

Actually, there's a tiny insight to be had from that last formula: both the numerator and denominator of the fraction inside the log have a factor of $R^2$; cancelling those makes the entire sum independent of $R$, so that $R$ appears only in the first two terms (the $f$ terms) of the expression. I'm not sure whether this is valuable or not, of course.

John Hughes
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