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It is known that same-sized spheres can be packed at a density of $\pi/3\surd2$. If we uniformly stretch or compress the packed configuration as a whole, in any direction, the packing density does not change, because the gaps are rescaled in the same proportion as the (originally spherical) ellipsoids. However, if we start with ellipsoids, they do not have to be packed all with the same orientation. Thus, allowing freedom of orientation, the question arises:

Is there a suitably proportioned ellipsoid such that many copies of it can be packed at a density exceeding $\pi/3\surd2\,$?

For heuristic comparison, if instead of ellipsoids we take spherically capped cylinders, the answer is clearly positive; indeed, we do not even need to vary their orientation. There are also smoothly defined solids, for example one bounded by a surface of the type $$\frac{x^4}{a^4}+\frac{y^4}{b^4}+\frac{z^4}{c^4}=1,$$ which is on the way between an ellipsoid and a cuboid, that will pack more like cuboids, and so more densely than spheres. However, ellipsoids are a less obvious case.

If this question is too hard, perhaps one could make a start in the two-dimensional case:

Can many copies of an ellipse be packed at a density exceeding $\pi/2\surd3\,$?

John Bentin
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    The answer to the ellipsoid question is apparently yes: see https://en.wikipedia.org/wiki/Ellipsoid_packing – Chris Lewis Apr 03 '24 at 16:12
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    Interestingly, the ellipsoids do not need to have an especially high aspect ratio to get a denser packing than spheres (I had guessed that more needle-like ellipsoids might do best). The linked 2017 article on the Wikipedia page is very informative and has lots of references. Another point of interest the authors mention is that Ulam's packing conjecture (essentially, every 3D convex body packs better than spheres) is still an open problem: https://en.wikipedia.org/wiki/Ulam%27s_packing_conjecture – Chris Lewis Apr 03 '24 at 16:21
  • Mathworld has a page about ellipsoid packing. – Dan Apr 03 '24 at 22:57

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