3
$\begingroup$

The general claim goes something like this

the best regular polygon that tiles the 2D (Euclidean?) plane with equal size units and leaves no wasted space is the hexagon

I have seen similar claims like this but I do not get what I have to do to have an idea of what "best" means here.

I would have thought that I have to make some kind of circle, tile it with hexagons and compare it with another circle of the same size tiled with squares. But should I use squares of the same area as the hexagons? The same radius/side lenght? What should I count? number of polygons? I could ask questions many more questions. Should I find the limit for large areas of the circle or should I verify it for other finite spaces. I have sometimes seen that it has to do with the perimeter of the tiling, in what way?

The closest I could find is the honeycomb conjecture

Any partition of the plane into regions of equal area has perimeter at least that of the regular hexagonal grid

but I would like to still understand what partition and perimeter mean here.

Could somebody provide an example on how hexagons are better than squares?

$\endgroup$
2
  • $\begingroup$ The claim is quite vague. It can tesselate the plane (you can cover it all with hexagons), but the same is true of squares and triangles. My best guess is that, in terms of "best", if we use a fixed side length for all shapes, the hexagons will occupy the most area, and you can tile the plane using fewer of them (in a limiting, loose sense), or rather tile a fixed area with fewer. But without prior definition, yeah, it's kind of meaningless to think about. $\endgroup$ Commented Aug 18, 2022 at 17:12
  • 1
    $\begingroup$ @PrincessEev I think the idea is called the honeycomb conjecture for more information. $\endgroup$
    – Mauricio
    Commented Aug 18, 2022 at 17:14

2 Answers 2

3
$\begingroup$

the best regular polygon that tiles the 2D (Euclidean?) plane with equal size units and leaves no wasted space is the hexagon

There are three regular polygons that tile the plane by themselves, the triangle, square, and hexagon. If you make them all have the same area, say $1$ square centimetre, then the hexagon has the smallest perimeter. For example, a square of area $1$ has a perimeter of $4$, whereas a regular hexagon of area $1$ has a perimeter of $2\cdot 12^{1/4} \approx 3.7224$.

This is because in a sense the hexagon is more circular, it looks more like a circle than the square or triangle. The circle is special because it has the smallest perimeter for its size than any other shape. This is a 2-dimensional analog for why soap bubbles are spherical. However, circles of the same size cannot tile the plane without gaps.

The honeycomb conjecture is more general than this, as it is not restricted to regular polygons, nor even just a single shape.

The honeycombs that bees make consist of lots of storage compartments of equal size. They use wax to make the walls between the compartments. They don't have to make them all the same shape, and indeed they don't when near the edge of the space that the honeycomb must fit into. However, towards the middle of the honeycomb the cells are all hexagonal. A reason for this is that the hexagonal tiling pattern uses less wax on average than any other pattern.

The honeycomb conjecture says exactly this a bit more formally. It says that if you divide the infinite plane into regions of the same area (without any gaps between regions) then the most efficient way to do this is with hexagons. The infinite plane is used so that there are no edge effects - there is no surrounding boundary that cells will have to comform to like in a real honeycomb. Efficiency is a little tricky to define rigorously on an infinite plane, but you could still see it as the average amount of wax needed per cell, or equivalently the average perimeter of the regions. To define that rigorously you will need to use limits - calculate the average perimeter for a finite region and see what happens to that as the region gets infinitely large.

$\endgroup$
7
  • $\begingroup$ When you say calculate the average perimeter, do you mean the external perimeter or does it also include all the internal lines of the hexagons? $\endgroup$
    – Mauricio
    Commented Aug 19, 2022 at 16:14
  • $\begingroup$ Also, what do you mean by "three regular polygons that tile the plane by themselves"? And why this is enough to rule out shapes with more than 6 sides? $\endgroup$ Commented Aug 19, 2022 at 16:24
  • 2
    $\begingroup$ @InanimateBeing the crystallographic theorem states that the only regular polygons that tile 2D space without leaving holes are the triangle, square and hexagon. $\endgroup$
    – Mauricio
    Commented Aug 19, 2022 at 16:40
  • $\begingroup$ The angles around any vertex of the tessellation have to add up to 360°, so as to not leave any holes. So any regular polygon that tessellates the plane has to have interior angles that are factors of 360°. This includes the triangle (60°), square (90°), and hexagon (120°). The regular pentagon with its 108° angles doesn't work (after you put three of them around a vertex, there's a 36° gap that can't be filled). And any shape with 7 or more sides could only fit two polygons with a gap left over. $\endgroup$
    – Dan
    Commented Aug 19, 2022 at 16:46
  • 1
    $\begingroup$ @Dan wow! that's even more lucid reasoning. Thank you! $\endgroup$ Commented Aug 19, 2022 at 16:49
0
$\begingroup$

There are only three regular polygons that will tesselate the plane:

  • The equilateral triangle, with 60° interior angles (6 of which will fit around a vertex)
  • The square, with 90° interior angles (4 of which will fit around a vertex)
  • The regular hexagon, with 120° interior angles (3 of which will fit around a vertex)

Let $s$ be the length of a side of the polygon. Then it can be shown that:

  • The area of the equilateral triangle is $A = \frac{\sqrt{3}}{4}s^2$
  • The area of the square is $A = s^2$.
  • The area of the hexagon is $A = \frac{3\sqrt{3}}{2}s^2$.

Or equivalently, if the area $A$ of each shape is fixed, then:

  • The equilateral triangle has side length $s = \sqrt{\frac{4A}{\sqrt{3}}} = \frac{2}{\sqrt[4]{3}} \sqrt{A}$
  • The square has side length $s = \sqrt{A}$.
  • The hexagon has side length $s = \sqrt{\frac{2A}{3\sqrt{3}}} = \frac{\sqrt{2}}{3^{3/4}} \sqrt{A}$.

To find the perimeter of each shape, simply multiply by its number of sides:

  • The equilateral triangle has perimeter $\frac{6}{\sqrt[4]{3}} \sqrt{A} \approx 4.559 \sqrt{A}$
  • The square has perimeter $4 \sqrt{A}$.
  • The hexagon has perimeter $\frac{6\sqrt{2}}{3^{3/4}} \sqrt{A} \approx 3.722$.

So, if you want to divide a plane into fixed-area cells by putting some kind of physical material (like concrete pavement, fencing, or beeswax) along the edges of those cells, then a hexagonal tiling will use 7% less material than the equivalent square tiling, and 17% less material than the equivalent triangular tiling. In that sense, hexagons are the bestagons.

$\endgroup$
3
  • $\begingroup$ Just to be clear, so the key component is the number of fences? Why is this sometimes called the perimeter? $\endgroup$
    – Mauricio
    Commented Aug 19, 2022 at 20:48
  • 1
    $\begingroup$ @Mauricio: Not the number of fences, but their combined length. (Technically, I'm double-counting because each fence is shared between two cells, but this applies equality to all shapes and so doesn't affect their relative amount of fencing.) $\endgroup$
    – Dan
    Commented Aug 19, 2022 at 20:54
  • $\begingroup$ so I count the number of hexagons and multiply by 6 times the side length (regardless of double counting)? $\endgroup$
    – Mauricio
    Commented Aug 20, 2022 at 11:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .