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It is well known that, given a sphere, the maximum number of identical spheres that we can pack around it is exactly 12, corresponding to a face centered cubic or hexagonal close packed lattice.

My question is: given a sphere of radius $R$, how many spheres of radius $r<R$ can we closely pack around it?

With disks, the problem is rather easy to solve. Indeed, with reference to the picture at the bottom, we can see that we must have

$$\theta = \frac{2 \pi} n = 2 \arctan \left( \frac r {\sqrt{R^2+2 R r}} \right)$$

from which

$$n = \left \lfloor \frac \pi {\arctan \left( \frac r {\sqrt{R^2+2 R r}} \right)}\right \rfloor$$

The last expression gives the correct result for $R=r$, namely $n=6$ (hexagonal lattice). Moreover, when $R \gg r$, we get

$$n \simeq \left \lfloor \frac {\pi R} {r}\right \rfloor$$

which is completely reasonable.

How can I tackle the same problem in the 3D case (spheres)?

It is clear that for $R \gg r$ we must get

$$n \simeq \left \lfloor \frac {4 \pi R^2} {\pi r^2}\right \rfloor$$

and also that we must have $n(R=r)=12$.

Any hint/suggestion is appreciated.

enter image description here

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    $\begingroup$ Long story short, you can't. Even the kissing number of 12, though tentatively known for centuries, was notoriously difficult to prove. $\endgroup$ Jun 29, 2017 at 10:14
  • $\begingroup$ @valerio: While the two packings you list are the only lattice packings of space, twelve balls of radius $R$ touching a central ball of radius $R$ have "continuous flexibility", in the lax sense that individual balls can be moved while keeping others fixed. A natural approach for finding a lower bound is to surround a ball of radius $r < R$ with "as many $r$-balls as possible, leaving one gap", then to repeat the process (with as many pairwise-tangent triples as possible) until no more balls can be added. There is certainly no simple formula, and rigorous bounds are difficult. $\endgroup$ Jun 29, 2017 at 11:02
  • $\begingroup$ Actually, in the limit $R \gg r$ you shouldn't expect to get $n \simeq \lfloor \frac{4 \pi R^2}{\pi r^2} \rfloor$, but rather $n \simeq \lfloor \frac{\pi \sqrt{3}}{6} \frac{4 \pi R^2}{\pi r^2} \rfloor$, with $\frac{\pi \sqrt{3}}{6} \approx 0.9069$. This is because the problem becomes a circle-packing problem. $\endgroup$
    – m3tro
    Oct 11, 2018 at 16:41

1 Answer 1

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Two small spheres touching the large sphere fail to intersect if and only if their projections onto the surface of the large sphere do not overlap. Since there is a bijective map between small sphere radii and the radius of the corresponding circular discs on the sphere they get projected to, this problem is equivalent to finding the maximum number of points on a sphere that are mutually at a distance at least $d>0$ from each other.

Equivalently, for a fixed number of points $k$, we can ask for the maximum $d$ such that $k$ points can be so arranged; under this framing, the problem is known as the Tammes problem.

In general packing problems like this are incredibly difficult and have rather ugly solutions; you will find no elegant formulas, regular constructions, or easy proofs for the optimal solutions here, outside of isolated unusually nice cases like the case where the points form an inscribed Platonic solid with triangular faces.

That said, there exist pretty good numerical solutions to these problems for small values of $k$, and of course in the limit things will approach the packing density of circles in the plane at $\frac{\pi\sqrt{3}}{6}\approx 0.906$.

You can find a writeup on the problem here, but many others are available by searching online.

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