Last time, on Ask Doctor Math:
"... it's true that hexagons do make an appearance in a large variety of different contexts."
"Primarily, I'm referring to regular hexagons--hexagons with 6 equal sides..."
"... by putting a bunch of identical hexagons together as tiles, we can cover an entire plane surface."
"... this way of packing circles has the property of being optimal..."
"... hexagon madness..."
And now, the conclusion!
While regular hexagons have all those great qualities that make them perfect for the needs of bees and frat-boys alike, it happens that irregular (that is, not necessarily equal-sided) hexagons have a somewhat amazing property, too. That is, in some sense they're the average random polygon. What I mean here is that if we generate a random sectioning of a flat surface into a large number of polygonal shapes, the average number of sides per polygon will be about 6. (Side note: this, of course, doesn't mean that there actually are any hexagons; it could be that the shapes are composed of equal parts squares and octagons, say, but in practice, this is unlikely.)
First, we should be clear on some terminology: a "polygon," from the Greek "poly" = "many" and "gon" = "side," is a collection of points connected together by line segments that enclose an area of the plane. The points are called "vertices" (singular "vertex," not "vertice"!), and the segments are called "sides" of the polygon. Notice that it's always true that polygons have the same number of sides as vertices. When we put some polygons together in the plane, their sides are now called "edges," except if two sides overlap each other, we only count that as one edge. Also, when vertices overlap each other, we only count that as one vertex. For example:
Now, before I demonstrate that hexagons are the average, I'll need to lay out a couple of house rules for generating these random polygons:
Rule 1. All the vertices of the shapes line up with each other, so no brick-patterns, like this:
This is disallowed because it has a vertex of one brick in the middle of the side of another brick.
Rule 2. Every vertex is the junction of exactly 3 polygons (except for a few on the boundary, which we'll ignore).
Now, why are these reasonable? Well, the most common setting for this kind of random polygon-generation is the formation of cracks in some surface, like mud or peanut brittle:
(Note: we're approximating a crack here as a straight line, which takes some imagination.)
Rule 1 essentially states that no cracks spontaneously form in the middle of already existing sides; instead, they emanate from junctions between existing cracks. This is reasonable because a junction between cracks is likely to be a weaker point in the surface than any point along an existing crack. Another way to think about it is that cracks occasionally split into more than one crack, but when they do, both cracks generally change direction, instead of one continuing on like nothing had happened. In pictures, this:
is much more likely than this:
Similarly, rule 2 states that cracks tend to only split off into pairs. To see why that's reasonable, imagine if a crack were trying to split into 3 cracks. If any one of the three were just the slightest bit late to form, it would end up splitting off from one of the already existing 2 cracks, instead of the original one. In pictures, again, this:
is much more likely to occur naturally than this:
There is actually physics that could back me up here, but for now we'll just take these as givens.
It turns out that these same rules make sense in other settings, as well--for example(s), the formation of soap bubbles:
the shape of storm clouds, like this one on Saturn:
the sections on a tortoise shell:
and even France!
Can you spot the hexagons?
OK, before we get to crack apart soapy French turtles on Saturn, we need to take a little side trip to talk about a fundamental fact about polygons in the plane, called Euler's formula.
Euler's formula says that, subject to the rules above, no matter how we arrange a collection of polygons in the plane, the number of vertices minus the number of edges plus the number of polygons (also called faces) is a constant. For our purposes, the constant is 1.* In equation form,
V - E + F = 1
It actually makes quite a lot of sense if you think about it for a second. Imagine we only had one shape, a lonely little triangle all alone in the plane. So, V, the number of vertices, would be 3, as would the number of edges, E. And F, the number of faces, in this case a frowny face, would be 1:
V - E + F = 3 - 3 + 1 = 1
Now, if we tacked on a friend for the triangle, for example a square, the resultant shape would have 5 total vertices, 6 total edges, and 2 (now smiley) faces:
So again, V - E + F = 5 - 6 + 2 = 1. Essentially, the square gobbled up 2 of the triangle's vertices and 1 of its edges, while adding 4 of both edges and vertices and 1 extra face. As a result, the quantity V - E + F stayed constant. By similar reasoning, you could convince yourself that the same would happen no matter what shape we tacked on. And we can now repeat the process by adding a third polygon, and a fourth, and so on, until we have a whole polygon party on our hands.
OK now, at long last, we come back to random shapes. In our random polygonal mix, let's call S the total number of sides the polygons have altogether (different from E because 2 sides overlap to form 1 edge). Polygons individually always have the same number of vertices and sides, and since each vertex is shared by 3 polygons, the net total number of vertices is S/3. Also, each edge is shared by 2 sides (except for a small number of boundary edges), so E, the total number of edges, is the same as S/2. Putting these into Euler's formula gives us:
S/3 - S/2 + F = 1
Equivalently, F = 1 + S/2 - S/3.
We can combine the S/2 - S/3 to get S/6, so we have:
F = 1 + S/6
Multiplying by 6 and dividing by F gives us:
6 = 1/F + S/F
Now if we imagine the number of faces being very large, this tells us that 1/F is very small, so S/F is very close to 6. In other words, the ratio of sides to polygons is about 6, so the average polygon is a hexagon!
Next time you're out, keep your compound eyes peeled for hexagons, regular and otherwise, and someday you too can catch the hexagon madness!
*Those in the know, take note: the reason the constant is 1 and not 2 is that I'm not counting the unbounded component as a face. However, it doesn't matter in the end, because the constant gets divided by F, so this same property would be true in a topological space with any Euler characteristic.