When you draw a circle in a plane you can perfectly surround it with 6 other circles of the same radius. This works for any radius. What's the significance of 6? Why not some other numbers?

I'm looking for an answer deeper than "there are $6\times60^\circ=360^\circ$ in a circle, so you can picture it".


5 Answers 5


The short answer is "because they don't work," but that's kind of a copout. This is actually quite a deep question. What you're referring to is sphere packing in two dimensions, specifically the kissing number, and sphere packing is actually quite a sophisticated and active field of mathematical research (in arbitrary dimensions).

Here's one answer, which isn't complete but which tells you why $6$ is a meaningful number in two dimensions. The packing you refer to is a special type of packing called a lattice packing, which means it comes from an arrangement of regularly spaced points; in this case, the hexagonal lattice. The number $6$ appears here because the hexagonal lattice has $6$-fold symmetry. So a natural question might be whether one can find lattices in two dimensions with, say $7$-fold or $8$-fold symmetry, since these might correspond to circle packings with more circles around a given circle. (Intuitively, we expect more symmetric lattices to give rise to denser packings and to packings where each circle has more neighbors.)

The answer is no: $6$-fold symmetry is the best you can do! This is a consequence of the crystallographic restriction theorem. The generalization of the theorem to $n$ dimensions says this: it is possible for a lattice to have $d$-fold symmetry only if $\phi(d) \le n$, where $\phi$ is Euler's totient function.

The generalization implies that you still cannot do better than $6$-fold symmetry in $3$ dimensions. There are two natural lattice packings in $3$ dimensions, which both occur in molecules and crystals in nature and which both have $6$-fold symmetry, and it turns out that these are the densest sphere packings in $3$ dimensions. It also turns out that they give the correct kissing number in $3$ dimensions, which is $12$ (see the wiki article).

In $4$ dimensions, the kissing number is $24$, and I believe the corresponding packing is a lattice packing coming from a lattice with $8$-fold symmetry, which is possible in $4$ dimensions. In higher dimensions, only two other kissing numbers are known: $8$ dimensions, where the $E_8$ lattice gives kissing number $240$, and $24$ dimensions, where the Leech lattice gives kissing number $196560$! These lattices are really mysterious objects and are related to a host of other mysterious objects in mathematics.

A great reference for this stuff, although it is a little dense, is Conway and Sloane's Sphere Packing, Lattices, and Groups (Wayback Machine). Edit: And for a very accessible and engaging introduction to symmetry in the plane and in general, I highly recommend Conway, Burgiel, and Goodman-Strauss's The Symmetries of Things.

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    $\begingroup$ +1, but... this is more an answer to "why is a sphere surrounded by 12 other spheres", i.e., the kissing number problem without regard to rigidity. For the planar 6-circles problem I think the point of the question was that the inner circle is perfectly surrounded, which means an optimal rigid packing. Disordered sphere packings (or kissing arrangements) are likely to beat lattice packings except in special dimensions, but in the superspecial dimensions (such as 2,8,24) with rigid kissing configurations, rigidity probably implies that the configuration must come from a lattice packing. $\endgroup$
    – T..
    Commented Nov 17, 2010 at 18:54
  • $\begingroup$ @T..: sure. I was just trying to provide some context to make sense of the 2-dimensional case. $\endgroup$ Commented Nov 17, 2010 at 19:01
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    $\begingroup$ The answer is excellent (enjoy the stream of upvotes!), but since we don't have multidimensional ratings to separate quality-of-answer from was-question-addressed, I thought it worth adding the posting and comment emphasizing the role of rigidity and flatness. $\endgroup$
    – T..
    Commented Nov 17, 2010 at 19:27
  • $\begingroup$ +1 for the great explanation including the relevant references! $\endgroup$
    – Jens
    Commented Oct 26, 2014 at 20:06
  • $\begingroup$ How many spheres of equal radius can you pack around a sphere in 3D? $\endgroup$ Commented Feb 25, 2016 at 19:49

The question is equivalent to asking why 6 equilateral triangles fit together exactly around a point, with no additional room left over. The answer is "because the Euclidean plane is flat", a condition equivalent to triangles having angle sum of 180 degrees (half the angle around a point), so that each vertex of a symmetrical triangle has 1/3 of half of a full rotation = 1/6. That the six-circle arrangement exists for any radius is also a special feature of Euclidean geometry: scale invariance.

For flat surfaces such as a cylinder (rolled up plane) or flat torus (as in the Asteroids video game) the perfect 6-circle configurations exist only for small enough radius of the circles. These geometries are, in small regions, the same as the Euclidean plane but differ "globally", e.g. there is a maximum distance between points of the torus. So the magic number of 6 is really about local flatness (absence of curvature) and having this configuration for circles of all size is a global question about the space in which the circles live.

In hyperbolic geometry there are tesselations of the plane by equilateral triangles with angle $180/n$ at each vertex, for any $n \geq 7$. In the picture of the $n=7$ triangular tesselation at http://www.plunk.org/~hatch/HyperbolicTesselations/3_7_trunc0_512x512.gif (triangles in white with the dual tesselation by heptagons shown in blue) if at each triangle vertex you draw a circle inscribed in its heptagon, there will be 7 circles exactly surrounding each circle, with all circles of the same radius. The same type of configuration exists for any $n$ and suitable radius of the circles. So the flatness condition is necessary; the theorem is false in negatively curved two-dimensional geometry. Six is also not the correct number for spherical geomety, and both spherical and hyperbolic geometry lack a radius-independent "number of circles that fit around one circle".

In the geometry of surfaces, having 5-or-fewer as the local number of circles that can be fit around a single circle characterizes positive curvature, as in spherical geometry. Having more-than-6 fit, or extra room when surrounding one circle by six, is a characterization of negative curvature, as in hyperbolic geometry. This is a statement about the local geometry of general 2-dimensional surfaces, and does not assume the surface has the same amount of curvature at all points, as would be the case for the spherical and hyperbolic analogues of Euclidean plane geometry, where there is a homogeneity assumption that "geometry is the same at all points". Having exactly 6 circles fit perfectly is a characterization of locally Euclidean (that is, zero curvature) geometry. If you don't know what curvature is in a technical sense, for purposes of this discussion it is (for surfaces) a number that can be associated to any point on the surface, and curvature being zero in a region of surface is equivalent to the ability to make a distance-preserving planar map of that region. Impossibility of doing this for surfaces of nonzero curvature, such as the sphere which is positively curved, was Gauss' Theorema Egregium which ruled out perfect flat cartography of the Earth.

Flatness plays the same role in the higher dimensional, sphere-packing interpretation of the question that Qiaochu suggested. The necessary packings exist only in a limited set of dimensions. The reason is that exactly surrounding a sphere of radius $r$ by $k$ spheres of that radius means more than a kissing number of $k$, the optimal kissing configuration should also be rigid (tight enough that the spheres cannot be moved). Rigidity is false in three dimensions and is believed to be false in most other dimensions but in those cases where it is known or suspected to be true (i.e., dimensions $d = 2, 8, 24$ and possibly a few others) the existence of a configuration with the same set of equalities between various interpoint distances as in the optimal kissing arrangement (in flat Euclidean space $\mathbb{R}^d$)should force a homogeneous $d$-dimensional geometry to be flat. This is because deforming the curvature of the space in the positive direction would reduce, and negative curvature would increase, the freedom to position such spheres around a central sphere.

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    $\begingroup$ I wish I knew the word for the difference between this question and the accepted one. So much easier to keep a model in my head with this one. Thanks. +1, obviously. $\endgroup$ Commented Jul 21, 2011 at 11:07
  • $\begingroup$ Do ideas like yours also capture the rigorous aspects of the arguments involved, or is it purely intuition while a rigorous argument would involve both different statements and proofs of those statements? I'm worried because you say things like angles, curvature, so I'm thinking you must mean something about Riemannian metrics and the curvature tensor. But intuitive geometry is one thing people talk about, and then manifold theory is another thing. Nobody ever seems to tell me how the latter formalizes the former. $\endgroup$
    – Jeff
    Commented Jun 1, 2013 at 23:31
  • $\begingroup$ It seems like the logic of the first statement here could be extended to say "The number of spheres touching a single sphere in a dense 3d sphere packing is 12 because 12 equilateral tetrahedrons fit together exactly around a point." Is that correct? $\endgroup$ Commented Oct 19, 2018 at 19:14

The centres of three circles of radius r just touching one another will form an equilateral triangle of side length 2r. Exactly 6 such non-overlapping equilateral triangles will be formed in this way as you keep adding more circles around the edge of the central circle. Why 6? These equilateral triangles will have a common vertex at the centre, each forming a 60 degree angle there. And as, you pointed out, "there are 6*60=360 degrees in a circle."

  • 1
    $\begingroup$ I think this is covered in the first paragraph of the other answer and to some extent in the question. The angle sum for one triangle is half a full rotation. For a symmetrical triangle the angles must be equal, so each angle is a third of the half-rotation, or one sixth of the angle measure around a point (360/6). The harder part is to explain the significance of six and maybe also the rigidity of the six-circle configuration. The OP was insightful to mention the radius-independence as a notable aspect of this. It is an excellent and deceptively simple question. $\endgroup$
    – T..
    Commented Nov 17, 2010 at 20:06
  • $\begingroup$ @Dan, you are simply rephrasing the question. $\endgroup$
    – John Smith
    Commented Nov 17, 2010 at 21:11
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    $\begingroup$ @John: The 6*60=360 argument combines two steps, 360=2*180 and 180=3*60. The first amounts to a rephrasing of "triangles have angle sum of 180 degrees (half of a circle or full rotation)", ie., the locally flat/Euclidean nature of the geometry. The second step, 3*60=180 is trivial and carries almost no information. It is a restatement of the symmetry of an equilateral triangle, which is a very general argument that holds in any geometry I can imagine, and thus says nothing in particular about the geometrical meaning of any special configuration such as the six surrounding circles. $\endgroup$
    – T..
    Commented Nov 17, 2010 at 21:30
  • $\begingroup$ FWIW, I gave this as a comment (now removed), which prompted his edit of "something deeper than 6*60=360". It's fundamentally related to the fact that we assume the parallel postulate (which turns up in the proof that the angles of a triangle sum up to 180 degrees, which means equilateral triangles can tile the plane, which means... you get the idea.) $\endgroup$ Commented Nov 17, 2010 at 22:24
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    $\begingroup$ An answer at any level is a contribution. Users with a broad range of backgrounds will read most threads. I think it is more informative if posters feel free to say whatever comes to mind at their actual knowledge level (rather than PhD's "dumbing down" or neophytes uploading from Wikipedia) as this allows everyone to spend the most time at their highest level of competence. This model would become more realistic when the user population becomes larger and there are enough "eyeballs" on each question to sample the whole spectrum of replies. $\endgroup$
    – T..
    Commented Nov 18, 2010 at 4:11

This question can be answered using right triangles. Draw $2$ identical circles tangent to each other. Draw $2$ tangent lines from the center of one circle to the edges the other circle. Now make two right triangles connecting the $2$ centers, $2$ radii, and the tangent lines. These $2$ triangles must be a $30-60-90$s because the line connecting the centers is $2r$, and the radius line is $r$, and the only right triangle where the hypotenuse is $2$ times the shorter leg, is a $30-60-90$. So the central angle formed by the center and the $2 $ tangent lines must be $30 + 30 =60$. If it is $60$, then there will be $6$ and only $6$ such setups to completely surround the circle.drawing of $2$ tangent identical circles


Im sorry guys to inform u all that the question itself is wrong... the number of circles surrounding a circle of same radius in a plane tangent to the central circle and tangent to each other is 2pi... ie, 2*3.14 = 6.28 circles....

  • $\begingroup$ that means 6 circles wont exactly fit around... there will always be a gap corresponding to the 0.28 circle $\endgroup$
    – user74165
    Commented Apr 24, 2013 at 15:16
  • $\begingroup$ I rather believe your interpretation of the question is wrong. The number of circles should be a natural number.. $\endgroup$
    – Mårten W
    Commented Apr 24, 2013 at 15:38
  • $\begingroup$ I realize you don't yet have enough reputation, but this is best as a comment. Regards $\endgroup$
    – Amzoti
    Commented Apr 24, 2013 at 15:38
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    $\begingroup$ This is not correct. 6 circles do fit exactly around a circle of the same size, with no gap. Get 7 coins and try it! $\endgroup$ Commented Sep 29, 2013 at 19:22

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