Consider the following problem:

How many regular tetrahedra of edge length 1 can be packed inside a unit sphere with each one has a vertex located at the origin?

The answer is at least 20, forming an icosahedron. On the other hand, it is at most 22, which can be shown by dividing the surface area of the unit sphere by that of a spherical triangle generated from the three vertices of a tetrahedron. Can further progress be made? Thanks in advance.

Edit: as suggested by Wlodek Kuperberg, an equivalent formulation is:

What is the maximum number of equilateral spherical triangles of edge length $π/3$ that can be packed on the unit sphere?

  • 5
    $\begingroup$ The phrase "forming an icosahedron" is not quite accurate, but it is clear what you mean: if a regular icosahedron inscribed in the sphere is projected to the surface of the sphere, then each of the resulting 20 spherical equilateral triangles has edge-lenth greater than 1, hence it contains in its interior the vertices of an equilateral triangle of edge-length 1. Then one clearly can see the 20 regular tetrahedra of edge-length 1 packed in the sphere, each with a vertex at the sphere's center. It is also clear that the 20 tetrahedra, while joined at the center, have some gaps between them. $\endgroup$ Jan 5, 2019 at 6:34
  • 6
    $\begingroup$ Here is a paper showing that the number is between 20 and 22; it seems the problem is open: math.uchicago.edu/~may/VIGRE/VIGRE2008/REUPapers/Wilhelm.pdf $\endgroup$ Jan 5, 2019 at 8:58
  • $\begingroup$ Arrange 7 equilateral triangles in the plane with four forming a large equilateral triangle, and for each side of this large triangle, externally place a smaller triangle so that midpoint of sides agree. On a sphere, these last three triangles get pushed away so that side midpoints separate. Is there a sphere packing which has two of these antipodally placed, and room for seven more triangles? Gerhard "Antipodally, Not Ant IPod Ally" Paseman, 2019.01.05. $\endgroup$ Jan 5, 2019 at 20:15
  • 1
    $\begingroup$ This question was asked at MSE before. $\endgroup$ Jan 14, 2019 at 6:48
  • $\begingroup$ yeah, but no satisfactory answers are given $\endgroup$ Jan 14, 2019 at 9:54

1 Answer 1


I'm not sure how to answer this question, but I'll suggest another approach to get an upper bound.

Considering the problem of packing equilateral $\pi/3$ triangles on a unit sphere, one may convert this to a packing problem in the unit tangent bundle $T^1(S^2)$. Take a unit vector $v$ on the equilateral triangle (say at the center, pointing toward a vertex). Then the unit vector associated to a disjoint equilateral triangle will lie outside of some region $R\subset T^1(S^2)$. The group of rotations $SO(3)$ acts transitively on $T^1(S^2)$, homeomorphic to $\mathbb{RP}^3$. There is a biinvariant metric on $SO(3)$ which we may therefore transfer to $T^1(S^2)$. Then we get a region $R'\subset R$ of half size with respect to this metric surrounding the vector $v$ at the center of a triangle which must be disjoint from the $R'$ regions surrounding the vectors of the other equilateral triangles in some packing. If one could estimate the volume of $R'$, then one would be able to get a packing estimate. I don't know if this gives an improvement, but it's possible that it could given that it takes into account the orientation of triangles, not just their area. The region $R'$ should be described piecewise by algebraic equations, and hence one ought to be able to compute it to arbitrary approximation to get an estimate of the volume of $R'$.

  • $\begingroup$ +1. Although I'm not familiar with differential geometry, I think your answer looks promising. $\endgroup$ Jan 16, 2019 at 7:32
  • $\begingroup$ @YuiToCheng: I'm now pessimistic about this approach. For example, this probably doesn't give the optimal answer when one takes a spherical triangle with angles $2\pi/5$, for which one can pack 20 as the faces of a spherical icosahedron. This estimate will miss some of the volume in $SO(3)$, so can't be sharp. On the other hand, viewing this as a packing or covering problem in $SO(3)$ could possibly be useful, so I'll leave this answer up. $\endgroup$
    – Ian Agol
    Jan 17, 2019 at 18:40
  • $\begingroup$ Yeah, I believe that's a hard problem. Anyway, I will be generous here. $\endgroup$ Jan 20, 2019 at 2:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.