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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.

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    $\begingroup$ For $n>6$, you have $2<\frac{2n}{n-2}<3$, so it can't be an integer $\endgroup$ Feb 12 at 1:46
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    $\begingroup$ Alternatively, $\frac{2n}{n-2}=\frac{(2n-4)+4}{n-2}=2+\frac4{n-2}$, so $n-2$ needs to be a factor of $4$ $\endgroup$ Feb 12 at 1:47
  • $\begingroup$ 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 $\endgroup$
    – David K
    Feb 12 at 2:00
  • $\begingroup$ Interestinggggg $\endgroup$
    – Some Guy
    Feb 12 at 2:01
  • $\begingroup$ 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. $\endgroup$
    – fleablood
    Feb 12 at 2:16
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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.

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    $\begingroup$ 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." $\endgroup$
    – Some Guy
    Feb 12 at 1:53
  • $\begingroup$ Edited hopefully for clarity. $\endgroup$ Feb 12 at 1:59
  • $\begingroup$ oh i see thank you $\endgroup$
    – Some Guy
    Feb 12 at 2:00
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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.

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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 $\frac {2n}{n-2} < 3$

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