It is well-known that it is impossible to arrange 13 spheres of unit radius all tangent to another unit sphere without their interiors intersecting. This was apparently the subject of disagreement between Isaac Newton ("impossible") and David Gregory ("possible"). The cause of the dispute was, in part, that there seems to be a lot of room left over after 12 spheres.

How close of a call is it, so to speak? What is the largest possible $r$ so that it is possible to arrange 12 spheres of unit radius and a 13th sphere of radius $r$ all tangent to another unit sphere, without intersections? What does the optimal configuration look like?

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    $\begingroup$ or, also, one may ask for the minimum radius r so that there are 13 disjoint unit tangent balls to a ball of radius r -somehow more symmetric $\endgroup$ – Pietro Majer Apr 21 '12 at 15:08
  • $\begingroup$ @Pietro Majer: yes, that is in fact a good variant! I would be interested in an answer to either version, but hopefully I can get an answer to both! $\endgroup$ – Jamie J. Taylor Apr 21 '12 at 15:46

Pietro's version of this question is answered in a paper by Oleg Musin and Alexey Tarasov (to appear in Discrete & Computational Geometry, http://dx.doi.org/10.1007/s00454-011-9392-2, http://arxiv.org/abs/1002.1439). The configuration found by Schütte and van der Waerden (see Joseph O'Rourke's answer) is optimal and unique up to isometries.

The other version of the problem amounts to asking for the largest possible hole in a packing of 12 identical disks of radius $30^\circ$ on the 2-sphere. I don't know the answer offhand. One could certainly figure out what it must be by numerical optimization, but finding a rigorous proof would be difficult. (It might be possible using variants of the Musin-Tarasov approach, which is an enormous brute force search over small planar graphs.) I am sure someone must have looked at this problem, but I don't know of a place where the answer might be recorded.

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    $\begingroup$ Has anybody conjectured the minimum radius of a sphere which can kiss 14 unit spheres? and n spheres? just curious. $\endgroup$ – Piero D'Ancona Apr 22 '12 at 0:18
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    $\begingroup$ This is called the spherical code problem (since it's error-correcting codes on the surface of a sphere) or Tammes problem (named after a biologist who studied the distribution of pores on pollen). For $n$ points on $S^2$, the answer is known rigorously for $n \le 13$ and $n=24$, and you can find results of numerical optimization for more cases at neilsloane.com/packings/index.html. There's also a lot known in higher dimensions. $\endgroup$ – Henry Cohn Apr 22 '12 at 1:37

The recent paper by Hopkins, Stillinger, Turquato, "Densest local sphere-packing diversity. II. Application to three dimensions," Physical Review E 83, 011304 (2011) (PDF link), addresses the variant suggested by Pietro:

"The smallest radius spherical surface onto which the centers of 13 spheres of unit diameter can be placed is strongly conjectured to be $R = R_\min(13) = 1.045572\ldots$, with the centers arranged in a structure first documented in Ref. [5]. It appears that, although Gregory was incorrect in conjecturing $K_3$ to be 13, his guess was not particularly far off."

[5] K. Schütte and B. L. van der Waerden, Math Ann. 123, 96 (1951).

Here's a nice image (from MathWorld) that shows the gaps when 12 unit spheres, tangent at icosahedron vertices, surround one unit sphere:

          alt text (source)

Update. See Henry Cohn's answer, which cites a more recent paper that settles (positively) the conjecture noted above.

  • $\begingroup$ how not upvoting! $\endgroup$ – Pietro Majer Apr 21 '12 at 16:55
  • $\begingroup$ Sorry; I meant, I'm glad to learn that the variation I suggested has already been studied -therefore I couldn't help upvoting :) $\endgroup$ – Pietro Majer Apr 21 '12 at 17:36

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