How to prove that a hexagon is the regular polygon with the most sides that can tile a plane 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.
 A: 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.
A: 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.
A: 
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 $\frac {2n}{n-2} < 3$
