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I was playing with a geogebra applet that shows regular $n$-gons of radius $1$ with their diagonals. For example, here is the $12$-gon with its diagonals:

enter image description here

For any value of $n$, when I shrink the image, the image becomes darkened (due to the thickness of the lines), but sometimes there remains a faint single white ring, indicating a ring of exceptionally large cells. For example:

$n=20$ enter image description here

$n=45$ enter image description here

$n=50$ enter image description here

Now here's the interesting thing: The radius of the white ring always seems to be approximately $1/e$. (I used Perfect Screen Ruler.)

For some $n$-values, I cannot perceive a single white ring; I guess it still exists but is not perceivable due to limitations in pixelation and/or visual acuity.

I can formalize my conjecture as follows:

In a regular $n$-gon of radius $1$ with it diagonals, if $d_n=$ distance between the centre and the centroid of one of the cells with the greatest area (excluding the centre cell when $n$ is odd), then $$\lim\limits_{n\to\infty}d_n=\frac{1}{e}$$

Question: Is my conjecture true?

(This question was inspired by another question: Distribution of areas in regular $n$-gon with diagonals, as $n\to\infty$.)

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    $\begingroup$ Extremely interesting conjecture! I hope someone manages to provide some insight. $\endgroup$ Dec 3, 2022 at 2:45
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    $\begingroup$ Extrapolating from the $12$-gon, I'll guess that (for $2n$-gons) the largest quadrilateral cell is one determined by chords joining four consecutive vertices to their appropriate "almost-opposites" (ie, a vertex adjacent to the diametric opposite). If my hasty Mathematica analysis is correct, the limiting position of the centroid (for circumradius $1$) is at distance $13/16=0.3611\ldots$ from the center. Not quite the conjectured value of $1/e=0.3678\ldots$, but still interesting. (My guess and/or analysis could be wrong. Even so, I don't see how $e$ could creep into the result.) $\endgroup$
    – Blue
    Dec 3, 2022 at 3:41
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    $\begingroup$ Just as a preliminary analysis, we should expect dark rings to occur at radii of the distance between its center and the halfway-point between any two vertices. These occur at $r=|\cos\left(\frac{\pi k}{n}\right)|$ for $k\in\{0,\cdots,n-1\}$ where $n$ is the number of sides. We then should expect any light ring to occur between these dark-rings. Although the inner-most light-ring should be the largest, and hence the brightest, the difficulty now becomes what rings have the least-density of overlapping chords. Unfortunately, the radius of the least-dense ring appears to change, varying $n$. $\endgroup$
    – Graviton
    Dec 3, 2022 at 6:13
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    $\begingroup$ I've written a program to see how much in-effect the intersection point play into the picture. Here's an image of all the intersection points (in white, with black background) of the first $n=3$ to $n=40$ polygons overlaid on top of each other. Because the brightness is inverted, any light-rings you're looking for should be dark-rings in my image. Just from looking at my image, there doesn't appear to really be any rings converging to the same radius when overlaid (besides the large dark ring which is where the unit-circle is). $\endgroup$
    – Graviton
    Dec 3, 2022 at 8:33
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    $\begingroup$ Just for the sake of following up, I spent several hours waiting for the $n=80$ to $n=100$ calculation, and forgot to consider that there'd basically be no space from the millions of intersection points to see any structure. I re-did the computation from $n=50$ to $n=80$ with some transparency and a larger resolution, but the structure is still pretty washed out. Not sure if it's useful at this point, but here was the result. There does vaguely appear to be some circular structures for $r<0.3$ but my lack of foresight makes it quite subjective. I may try again. $\endgroup$
    – Graviton
    Dec 5, 2022 at 7:48

1 Answer 1

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I have two graphs. They refer to white dots, which are small discs in the unit circle not crossed by any chord.

The first graph shows the size of the largest white dots, for each polygon up to 200 sides. It seems to match $3/n^{1.5}$ well, where $n$ is the number of sides of the polygon.
enter image description here
I did the following calculation for each value of $n$ from $n=1$ to $n=200$. (I don't know what the polygon is for $n=1$ and $n=2$.)
I took an array of points in polar coordinates, where $r$ ranged from $0.1$ to $1$ in steps of $0.0001$, and $\theta$ ranged in $100$ steps from $0$ to $2\pi/n$. That covers one wedge of the unit circle except for the centre of the circle; the calculations are periodic in theta. My first graph, above, shows that the grid is fine enough to land near the middle of the largest white dot. $r$ steps by $0.0001$, while the largest white dot has size at least $0.001$. For each array point, I calculated the minimum distance to any of the chords. The largest white dot is the array point where that minimum distance is greatest.

The second graph shows the $r$ position of the largest white dots. These positions are often near $0.35r$, $0.5r$ and $0.6r$, but by no means precisely. Perhaps I need a finer grid when $n>160$.

enter image description here

The largest white dots stand out well when they are much larger than the second-largest white dots. OP found that was true for $n=20,45,50$; it is also true for $n=143$

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  • $\begingroup$ That is a very clever way to, in effect, measure the distance between the centre and the largest cell. In your second graph, for the higher values of $n$ there is no significant pattern, so I guess my conjecture is not true. But I do wonder why, for the lower values of $n$, there is that concentration around $r\approx 0.36$. $\endgroup$
    – Dan
    Dec 4, 2022 at 13:27
  • $\begingroup$ If I'm reading your second graph correctly, it shows $r=1$ when $n=4$. But shouldn't $r$ be approximately $0.414$ when $n=4$? I drew a square with "radius" $1$, with diagonals, and drew the largest disc that fits in one of the regions. The distance between the centre of the square, and the centre of the disc, is $\sqrt{2}-1\approx0.414$. $\endgroup$
    – Dan
    Dec 5, 2022 at 4:40
  • $\begingroup$ Hmm the point on the circumference is the same distance from the chord. I didn't check for that. $\endgroup$
    – Empy2
    Dec 5, 2022 at 5:19
  • $\begingroup$ Your first graph has inspired another question. $\endgroup$
    – Dan
    Dec 7, 2022 at 4:56

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