How many ellipsoids can be maximally inside a circle?

Question

This discussion is related to this discussion here where I want to deduce the area difference between such two circles filled with ellipsoids. Actually, to understand this difference is the main target.

The shape of the ellipsoid can vary if you want or not. I am thinking how much empty space can be left next to the borders of the circle. I know how to prove and/or derive similar cases for hexagons etc, but ellipsoid is more challenging.

Example of ellipsoid

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You can use an arbitrary given closed area, with freedom to copy, scale, rotate and translate to cover the circle and converge. In other words, an arbitrary region with piecewise smooth boundary.

How much of the circle area can non-overlapping ellipsoids fill maximally?

score 2 · Answer 1 · answered Jun 02 '15 at 20:23

2

I assume you're trying to fill a circle with non-overlapping ellipses.

If you allow an arbitrary number of ellipses of arbitrary sizes and any fixed shape and orientation, you can get arbitrarily close to the area of the circle.

For simplicity, first do a linear transformation so the ellipses become circles, and the original circle becomes an ellipse. Now, given any region $R$ (with boundary composed of elliptical arcs), start by overlaying a square grid of spacing $\delta$. For sufficiently small $\delta > 0$, more than $2/\pi$ of the area of $R$ consists of complete squares that are subsets of $R$. In each such square, you take a disk of radius $\delta/2$ centred at the centre of the square. This has area $\pi/4$ times the area of the square. Thus our disks total more than half the area of $R$. Replace $R$ by the region not covered by those disks, and repeat...

answered Jun 02 '15 at 20:23

Robert Israel

448,999

1

I wonder I we can use an arbitrary given closed area, with freedom to copy, scale, rotate and translate to cover the circle and converge., e.g. the Batman logo. – mvw Jun 02 '15 at 20:28
2

An arbitrary region with piecewise smooth boundary, yes. – Robert Israel Jun 02 '15 at 20:34
And thus, any open set (which is a disjoint countable union of such regions), and therefore any set with finite nonzero Lebesgue measure – hmakholm left over Monica Jun 02 '15 at 22:15
@HenningMakholm Not quite. For example, if you remove a countable dense set of points you can't fit any disks in at all. – Robert Israel Jun 02 '15 at 22:50
@RobertIsrael: Of course -- I take the "Lebesgue measure" criterion back; I had confused myself about which of the squeezing sets in the regularity theorem is open. Any open set should still work. – hmakholm left over Monica Jun 03 '15 at 10:38

How many ellipsoids can be maximally inside a circle?

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