Why are triangles, squares and hexagons the only polygons with which it is possible to tile a plane?

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?

Regards

Solution 1:

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.

Solution 2:

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 $6$.

Solution 3:

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.

Solution 4:

To the arithmetic of roger/#347412 (Mar 31 '13 16:37) above: I didn't immediately see why $\{n \in \mathbb{N}_{>0} \mid (n-2)|2n \} = \{3,4,6\}$.

Here is my explanation:

The comprehension for all $n>6$ is equivalent to the following statement: $2n \not\equiv 0 \pmod{(n-2)}$. It is even true: $2n \equiv 4 \pmod{(n-2)}$, because

\begin{align} 2n \pmod{(n-2)} & = (\underbrace{2 \pmod{(n-2)}}_{2 \ \text{(too much)}})\cdot(\underbrace{n \pmod{(n-2)}}_{2 \ \text{(too less)}}) \pmod{(n-2)} \\ & = 2 \cdot 2 \pmod{(n-2)}) \\ & = 4 \pmod{(n-2)})\\ & \not= 0\pmod{(n-2)} \quad \text{because $n-2 > 4$ and $(n-2) \not|\ 4$} \end{align}

And for $n=3,4,6$ applies: $2 \cdot 3 \equiv 0 \pmod 1$, $2 \cdot 4 \equiv 0 \pmod 2$ and $2 \cdot 6 \equiv 2 \cdot 2 \equiv 0 \pmod 4$. And for $n=5$ applies: $2 \cdot 5 \equiv 1 \cdot 1 \equiv 1 \pmod 3 $.

Is the argumentation correct? Do you have to show this with induction evidence?

But yes, I confess that the geometric arguments above are more pictorial and much simpler. But the arithmetic "$(n-2)|2n$ only for $n=3,4,6$" rumbled in my head.