What is the maximum number of spheres that can be placed in 3D such that all inter-touch?

One can of course place four unit spheres tetrahedrally and then add a smaller sphere in the middle, so this number must be at least 5.

[By the way, I was trying to extend the "five points in 2D cannot be inter-connected without a crossing" limitation to 3D with a simple statement, but this was sadly the best I could do. If anyone knows a better simple extension, please comment.]

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    $\begingroup$ See Ian Agol's answer to the earlier MO question "Extensions of the Koebe–Andreev–Thurston theorem to sphere packing?" He shows that $K_6$ is not realizable in $\mathbb{R}^3$, using the same argument that Anton provides for $\mathbb{R}^n$: mathoverflow.net/questions/85547/… $\endgroup$ Sep 1, 2012 at 18:35

1 Answer 1


In $\mathbb R^n$, the answer is $n+2$.

You can apply an inversion which sends two of the spheres in to two parallel hyperplanes. The rest of the spheres will have the same radii and their centers lie in a hyperplane. Hence everything follows.

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    $\begingroup$ By "inverse" I imagine you mean "inversion"? $\endgroup$
    – Lee Mosher
    Sep 1, 2012 at 18:28
  • $\begingroup$ Since the two spheres touch each other at only one point (before the inversion), the two parallel hyperplanes would similarly touch for a contradiction, right? I'm probably mistaken, but it seems you can only send two of the spheres in to "a hyperplane plus a touching sphere". $\endgroup$
    – bobuhito
    Sep 1, 2012 at 23:05
  • $\begingroup$ @bobuhito, the touch-point goes to infinity. $\endgroup$ Sep 2, 2012 at 1:35
  • $\begingroup$ Thanks, very slick! I now wonder about other shapes, so I started another question: mathoverflow.net/questions/106163 $\endgroup$
    – bobuhito
    Sep 2, 2012 at 7:35

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