A sixth of a pi?

August 14, 2006 3:55 PM

An out-of-left-field question for the mathematicians regarding the possible link between pi and the number 6 in geometry.

If you take a group of seven identical (for all intents) objects with a circular base (try it with coins or bottles, for instance) you can arrange them most tightly by having six of the bottles or coins surround the central one, with each outer bottle also touching both adjacent outer bottles. Mathematically I can't understand why this would be the case with the number 6 specifically, and as I'm more of the right-brained-twinkie type, my background in math doesn't come anywhere near figuring it out. Can y'all help?

posted by Navelgazer to science & nature (22 comments total)

Because a hexagon subdivided into six triangles is composed of equilateral triangles, whereas with a pentagon or septagon you'd end up with isoceles triangles.

(Assuming that you're talking about identically-sized circular bases.)
posted by XMLicious at 4:01 PM on August 14, 2006

It has nothing to do with Pi.

It has to do with the fact that any two circles whose edges touch can always be joined by a third circle whose edge touches them both, and that the centers of the three circles will be the corners of an equilateral triangle. Since it's possible to fill a plane with tesselated equilateral triangles, it's also possible to densely pack circles on a plane, with their centers at the corners of those triangles.

The corner angle of an equilateral triangle is 60 degrees. Six triangles meet at each intersection, so six circles will surround the center circle. Pi has nothing to do with it.

Hexagons will do exactly the same thing, and pi doesn't relate to hexagons in any important way.
posted by Steven C. Den Beste at 4:03 PM on August 14, 2006

Put two coins next to each other, and you have a gap between them into which a third fits.

From there, you just have a tiling made up of that fundamental arrangement, of which the hexagon is just one subset.
posted by edd at 4:04 PM on August 14, 2006

Maybe this is what you mean:

The circumference of a circle it 2 * pi * r.

In a hexagonal grid the number of elements 1 away is 6 which is close to 2 * pi.

If you drew a circle of 100 units on the grid, and counted the number of elements just inside the circle and divided by 100 then that number would be even closer to 2 * pi.
posted by MonkeySaltedNuts at 4:08 PM on August 14, 2006

Thanks guys. I know this is of approximately zero importance, but I find I do most of my mathematical pondering while staring at bathroom tiles.
posted by Navelgazer at 4:12 PM on August 14, 2006

BTW, this kind of thing is referred to as a sphere packing problem in mathematics.
posted by XMLicious at 4:15 PM on August 14, 2006

In three space the same thing happens, because spheres can be packed densely. For any four adjacent spheres, their centers will be at the corners of regular tetrahedrons. Since it's possible to fill space with a three dimensional lattice of tetrahedrons, it's also possible to densely pack spheres. (But I don't remember if a given sphere in such a lattice touches 12 or 20 other spheres. Is it an icosahedron, or a dodecahedron? I think it's 12, an icosahedron...)
posted by Steven C. Den Beste at 4:28 PM on August 14, 2006

Thanks guys. I know this is of approximately zero importance

Not at all! Your question touches on a lot of fundamental stuff! Ignore these guys above who are giving you textbook "oh its so obvious" answers. :)
You might want to read up a bit on Kepler's conjecture and Thue's theorem:


Thue's theorem. No packing of non-overlapping discs of equal size in the plane has density higher than that of the hexagonal packing.

It is really impossible to imagine how it could be otherwise. We can also build the hexagonal packing in this way: we start with a single disc in the plane, and then place around it six others. In contrast to the similar construction in 3D, where spheres are placed around a sphere, it is clear that no more than six can be so placed. Furthermore this continues on for each of the new discs etc. to give a global packing, which has to be optimal - doesn't it? But no straightforward proof of the Theorem has yet been found.
posted by vacapinta at 4:36 PM on August 14, 2006

It's worth noting that the exact same thing doesn't happen in 3-space (that is, the spheres aren't tightly packed around the central sphere -- there was a debate between Newton and Some Other Guy about whether a 13th would fit). Reference "kissing number".

I think it's 12, an icosahedron...

Slip of the keys? It's 12, a dodecahedron.
posted by j.edwards at 4:39 PM on August 14, 2006

Ah, I've been thinking about it more, and I'm sure that it's 20 spheres, at the corners of a dodecahedron... (which has 12 faces and 20 vertices).
posted by Steven C. Den Beste at 4:51 PM on August 14, 2006

Rats. That can't be right, either. I don't know what the answer is for sphere packing.
posted by Steven C. Den Beste at 4:53 PM on August 14, 2006

But no straightforward proof of the Theorem has yet been found.

That's sort of what I was getting at, actually. I know that it's true, but I can't think of any reason why it must have been that way. This one has been bugging me for years.

Still, Steven C. Den Best's answer above satisfies me, for the most part. But if anyone has any more to add, I'd love to hear it.

(And has anyone come up with a proof for the four-color map rule yet?)
posted by Navelgazer at 4:56 PM on August 14, 2006

The four color theorm was proved by computer. No short, elegant proof exists.
posted by Humanzee at 5:23 PM on August 14, 2006

Oh, bother; it isn't possible to pack space with tetrahedrons. I'm going to bed.
posted by Steven C. Den Beste at 5:41 PM on August 14, 2006

Steven C. Den Beste: It is absolutely possible to tile space with regular tetrahedra!

You put one in the middle, and glue one to each of it's four faces. Then you have a new bigger tetrahedra, and repeat with 4 more bigger tetrahedra. Keep going until your infinite space is full.
posted by aubilenon at 5:56 PM on August 14, 2006

Four tetrahedra packed together doesn't create a larger tetrahedron. The angle at the edge where two faces meet isn't 60 degrees. (But I can't tell you what it is precisely; high school geometry, which I never took anyway, was 35 years ago.)
posted by Steven C. Den Beste at 6:13 PM on August 14, 2006

It's worth noting that the exact same thing doesn't happen in 3-space (that is, the spheres aren't tightly packed around the central sphere

yes they are. it's called hexagonal close packing (hcp), where each sphere sits in the triangular depression created by 3 spheres in the layer below it. each sphere in the lattice touches 12 neighbors: 3 below, 6 on the same layer, and 3 above.

there are actually 2 variations on this: one called ABABAB, where the spheres in the third layer are directly above the spheres in the first layer, and another called ABCABC where the first, second, and third layers each occupy different sites. the ABC case of the hexagonal lattice is known as face-centered cubic (fcc) packing. (there is also a third type called random hcp, where you get maybe ABACBAC etc). more about the difference between ABA and ABC here.
posted by sergeant sandwich at 6:14 PM on August 14, 2006

Navelgazer, I think pi is related in a strange way to the six nearest neighbor packing of discs in two dimensions. If you imagine warping the plane into the shape of a sphere, for example, six circles touching another circle is only possible as the diameter of the circles approaches zero. As the diameter of the circles on a sphere increases, you can see that a stage is reached where each of three circles touch the other two and are 'locked' into place the way the seven are in the plane (aka flat two-space).

The connection to pi is that 'pi' has no fixed value for circles on a sphere; it goes from pi for circles of zero diameter to two for circles of maximum possible diameter (great circles). So the fact that you can arrange seven identical circles of arbitrary diameter so that six touch the central seventh implies that pi has the fixed familiar value, and that is the only situation in which that will be true.

Vacapinta, thank you for that link. It's very interesting that no straightforward proof of Thue's theorem has been found. As I said above in different words, it looks to me as if Thue's Theorem is equivalent to Euclid's fifth postulate, the parallel postulate.
posted by jamjam at 6:17 PM on August 14, 2006

also the solid polyhedron defined by the nearest-neighbors of a close-packed lattice of spheres is neither an icosahedron nor a dodecahedron but a cuboctahedron! (scroll to the bottom.)
posted by sergeant sandwich at 6:45 PM on August 14, 2006

uh, well, sorry. apparently i can't read - it's actually a triangular orthobicupola!!
posted by sergeant sandwich at 6:47 PM on August 14, 2006 [1 favorite]

Tetrahedrons can't fill space, but a mixture of tetrahedrons and octahedrons can. Also, cubes can fill space (trivially), but so can my favorite polyhedron, the rhombic dodecahedron (which I'm having trouble finding a picture of a space-filling of, but trust me, it's cool).
posted by wanderingmind at 1:20 AM on August 15, 2006

People who like this stuff would enjoy Buckminster Fuller's ideas: A Fuller Explanation, which talks a lot about sphere-packing.
posted by sonofsamiam at 7:06 AM on August 15, 2006