# How to prove that a hexagon is the regular polygon with the most sides that can tile a plane [duplicate]

I need to know the answer to this question to find out why bees use hexagonal cells in hives. I know that a circle takes up the most area using the least perimeter, so bees would try to make shapes as close to a circle as possible to use the most space without wasting too much material on walls. However, bees don't use circles because using circles creates a lot of waste space between cells, so bees use a shape that can tile a plane without overlap. The shape that meets the conditions of having a lot of sides and being regular to look like a circle, and being able to tile a plane is the hexagon, so bees use this in their hives. However, I want to know whether or not this is the polygon with the most sides that fits this requirement. I try to prove there is no bigger polygon like this.

First, I note that an integer number of interior angles must meet at a site and add up to $$360$$ degrees to tile a plane. For example, squares can tiles a plane because they all have $$90$$ degree angles, and $$4$$ of these make $$360$$. All polygons have exterior angles adding up to $$360$$, so a polygon with $$n$$ sides has $$\frac{360}n$$ degrees per side. However, interior and exterior angles are supplementary, thus the interior angle of an $$n$$ side polygon measures $$180- \frac{360}n$$ degrees. Now, let's test whether this quantity divides $$360$$ degrees. We get $$360 \over 180- \frac{360}n$$, which simplifies down to $$\frac {2n}{n-2}$$. Now, I need to prove that $$n=6$$ is the biggest number such that this quanitity is an integer, but I am not sure how I would go about doing that.

• For $n>6$, you have $2<\frac{2n}{n-2}<3$, so it can't be an integer Feb 12 at 1:46
• Alternatively, $\frac{2n}{n-2}=\frac{(2n-4)+4}{n-2}=2+\frac4{n-2}$, so $n-2$ needs to be a factor of $4$ Feb 12 at 1:47
• Actually, bees do use circles. They initially make circular holes in the honeycomb. The walls between the holes later get reshaped so the hole is almost a hexagon, but still rounded at the corners. nature.com/articles/srep28341 Feb 12 at 2:00
• Interestinggggg Feb 12 at 2:01
• Well, you must have the angle of the polygon dividing evenly into $360$. That is to say if you have $k$ polygons meeting at of vertex you must have $k$ of those angles adding up to $360$. The angle of the hexagon are $120 = \frac {360}3$. So to have any larger angle we can only have $\frac {360}2 = 180$ or $\frac {360}1 = 360$. As no regular polygon can have those angles $120$ is the largest possible angle. And any regular polygon with more than $6$ sides will have angles that are too bing. Feb 12 at 2:16

The simplest way is to begin as you did, and consider how many polygons are meeting at a corner. It has to be at least 3 to be a corner; if it were 2 it would be an edge, not a vertex. The minimum is 3. If all the angles are the same size, then the size of each interior angle that joins at the vertex must be 360/3 = 120. Therefore a hexagon has the most sides.

• I am not following, how did you go from "The minimum is 3, and 360/3 = 120, so that is the largest interior angle possible" to "Therefore a hexagon has the most sides." Feb 12 at 1:53
• Edited hopefully for clarity. Feb 12 at 1:59
• oh i see thank you Feb 12 at 2:00

Well, $$\lim_{n \rightarrow \infty} \frac{2n}{n-2} = 2$$. Its derivative, $$\frac{-4}{(n-2)^2}$$ is negative everywhere, so this expression decreases asymptotically to $$2$$. So $$3$$ is the least integer it can take and you have shown that it does.

which simplifies down to $$\frac {2n}{n-2}$$. Now, I need to prove that n=6 is the biggest number such that this quanitity is an integer

Well, $$\frac {2n}{n-2} = \frac {2n -4 + 4}{n-2} = 2 + \frac 4{n-2}$$.

That can only be an integer if $$n-2|4$$. Or in other words if $$n-2 = 1,2$$ or $$4$$ or if $$n = 3, 4$$ or $$6$$.

In particular if $$n > 6$$ then $$4{n-2} < 1$$ and can't be an integer.

......

Alternatively $$\frac {2n}{n-2} > \frac {2n}{n} =2$$ so $$3$$ is the smallest possible integer option. And $$\frac {2n}{n-2} -3$$ occurs if $$n = 6$$.

If $$n > 6$$ then $$\frac {2n}{n-2} < 3$$