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Ok so I have heard that the only regular polygons which can completely fill the plane without overlapping are the 3,4 and 6 sided ones. I have also heard about Penrose tilings but this question ignores them. How can one prove that there isn't another polygon which can completely tile the plane only by itself?


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Interestingly, the only rectangles with whole-number side lengths for which the area equals the perimeter are $3\times6$ and $4\times4$. (The first has area and perimeter of $18$, the second has area and perimeter of $16$.) These are those numbers $3$, $4$, and $6$ again. I wonder if there's a connection… – Akiva Weinberger Nov 11 '15 at 16:02
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Well, if you tile the plane by congruent regular polygons, there must be $n$ polygons meeting at each vertex. Thus the interior angles of each polygon must be $2\pi/n$, for some positive integer $n$.

For $n=3$, we get polygons with angles of $2\pi/3$, which are regular hexagons. This tiling has three regular hexagons meeting at each vertex.

For $n=4$, we get polygons with angles of $2\pi/4 = \pi/2$, which are squares. This tiling has four squares meeting at each vertex.

For $n=5$, the polygons would need to have angles of $2\pi/5$. This is not possible for a regular polygon.

For $n=6$, the polygons would need to have angles of $2\pi/6 = \pi/3$, which are equilateral triangles. This tiling has six triangles meeting at each vertex.

For $n>6$, the polygons would need to have angles less than $\pi/3$, which is impossible.

Edit: As Blue points out below, this argument neglects tilings such as the brick wall tiling, where vertices of one polygon meet edges of another. See Steven Stadnicki's comment for the resolution of this case.

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The polygons don't have to meet exclusively at vertices. Consider the "brick wall" tiling of the plane by staggered rows of squares, and the counterpart triangular tiling. (Hexagons can't be staggered.) The vertices of some tiles meet other tiles along their edges. – Blue Mar 31 '13 at 16:42
@Blue that depends a lot on what you consider a tiling; often the brick wall tiling is treated as a tiling by degenerate hexagons, rather than regular squares! Fortunately, even with the expanded definition (and treating them as squares) the argument goes through, since any non-vertex point must be some number of vertices against a straight edge and the angles at the vertices must then sum to $\pi$, so the same proof goes through - for $n=2$ verts meeting along an edge the polygons have angles of $\pi/2$ and so are squares; for $n=3$ they're triangles; and $n\gt3$ is impossible again. – Steven Stadnicki Mar 31 '13 at 16:46
@StevenStadnicki: I agree that "that depends". That said, when the OP says that he "ha[s] heard" about this tiling result, it would seem helpful to the novice for us non-novices to be clear on our terminology. Besides, since it's easy enough to address the meet-along-edges case (as you've done), why not do so? In any case, the purpose of my comment has been achieved by your response, so: Thanks! – Blue Mar 31 '13 at 16:57
@Blue It's true that I neglected the case where the polygons are meeting along the edges. I'll edit the answer to point this out. – Jim Belk Mar 31 '13 at 18:22

Let $P$ be a regular polygon and $\alpha$ the value of the angle between two edges. To be able to tile the plane, you need that there exists a $k \in \mathbb N$ such that $k \alpha = 2 \pi$. The angle $\alpha(n)$ of a regular polygon of $n$ edges satisfies $n(\pi - \alpha(n)) = 2 \pi$, ie, $\alpha(n) = \pi - \frac {2 \pi} n$. The only possible $n$ for which $2 \pi/\alpha(n) = 2n/(n -2) \in \mathbb N$ are $n=3,4$ and $6$.

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Think of a tiles floor. The vertices always come together at some point. That is, with rectangular tiles there are always groups of 4 tiles, and the 4 corners are 90 degrees each so 4 of them add up to 360. A hexagon has an interior angle measure of 120, and 120 is the greatest factor of 360 (other than 180, which is a straight line, or 360). If you have polygons with 7 or more sides, their corners will not fit nicely to add up to 360 degrees.

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