# Regular tessellations

## Main Question or Discussion Point

Show that the only regular figures that fills the plane are the triangle, the square and the hexagon.

This can be done with algebra after the problem has been restated.

Note that the angle of a corner of a regular n-gon equals (n-2).Pi/n.

To fit the plane, k copies of a regular n-gon must be able to touch with their corners and therefore k angles should make up for a total arc of 2.Pi.
So, k should be chosen such that

k(n-2)Pi/n = 2Pi eq.
k(n-2) = 2n (*) eq.
k = 2n/(n-2).

So, we must choose n such that n-2 | 2n.
This looks already as if there are only few possibities. First note that n must be larger than 2. (Otherwise we don't even have a polygon).

n=3 gives 1|6 which is true, and k = 6/1 = 6
n=4 gives 2|8 which is true, and k = 8/2 = 4
n=5 gives 3|10 which is NOT true
n=6 gives 4|12 which is true, and k=12/4 = 3

How does it go on?

Well, the next quotient will be smaller than 3, so it must be 2, but then this would mean that just two corners n-gon fill an arc of 2Pi and this corner should be Pi, but this does not happen for a finite n.
Therefore, there are no other regular n-gons that tesselate the plane.

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You can also look at solving the same equation (*) for n, we get:
n = 2k/(k-2)

So k should satisfy k-2 | 2k
So either
(i) k-2 = 1 or
(ii) k-2 = 2 or
(iii) k-2 is odd and k-2|k
(iv) k-2 is even and (k-2)/2 | k