# Why a tesselation of the plane by a convex polygon of 7 or more sides is not possible?

I read in several places, including Wikipedia, that a tessellation of the plane by a single, convex, $n-$sided polygon is not possible for $n\geq7$. I was not able to locate a proof, or a paper that discusses this. Maybe it is too trivial, but I am not able to figure it out myself.

So I would like to ask you for help. Where can I find a proof for this?

Also, I am interested in the mathematical background behind plane tiling (especially periodic), If you can suggest papers on the topic it would really help.

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Look at Tilings and Patterns by Branko Grunbaum and Geoffrey Shephard (Freeman, 1987). Tilings are "infinite graphs" but ideas related to extending Euler's formula: Vertices + Faces - Edges = 2 come into play. – Joseph Malkevitch Dec 15 '11 at 18:51
(Deleted my earlier comment because it was misinformed. Note that the convexity comes into play here in forcing there to be at least three polygons meeting at each vertex, as Jyrki notes, and I believe that's all that's needed for a 'topological' proof to go through.) – Steven Stadnicki Dec 15 '11 at 20:02

In a tessellation of the entire plane by convex tiles at least three tiles will meet in each corner of each tile. Therefore the average of inner angles of the tile cannot be higher than 120 degrees. The sum of the inner angles of an $n$-gon is $(n-2)180$. If $n>6$, then their average $180(n-2)/n=180(1-\frac2n)>180(1-\frac26)=120$.

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Note that this argument does not rely on regularity of the $n$-gon. – Jyrki Lahtonen Dec 15 '11 at 15:04
Can you explain why the average of inner angles of the tile cannot be higher than 120 degrees? – Artium Dec 15 '11 at 15:42
In fact, there are tilings for which these averages don't exist - they're known as 'unbalanced'. I think this argument could be made to work, but there are a lot more details to be filled in than you might think at first glance. As Joseph Malkevitch points out above, the standard reference on this is "Tilings And Patterns", and I highly recommend it; while it's a few years (decades!) old now, it's an absolutely beautiful book. – Steven Stadnicki Dec 15 '11 at 19:25
@StevenStadnicki: Wow. Surprised. Multiple tiles, surely, but can you find such a tesselation with all tiles congruent such that this average wouldn't actually exist as a limit of averages within a growing sequence of, say, squares or circles? Agree that more details are missing than I initially thought. – Jyrki Lahtonen Dec 15 '11 at 22:09
The gap can be repaired as follows. Enumerate the tiles, and assign tile $k$ the weight $1/2^k$. Divide each tile's weight evenly among all the vertices of the tiling that it touches; let each vertex's weight be the sum of its share of the weights of the tiles that it touches. Then the weighted averages exist and are equal; but still the vertex-by-vertex average is at most 120 degrees and the tile-by-tile average is at least $180(1-\frac2n)$ degrees. – user21467 Sep 16 '15 at 0:04

There is a way to cover the entire plane with convex heptagons. The one in the center is regular, then a ring of seven of them are almost regular. The next ring out are already pretty long and narrow, and so on.

From the point of view of your question, the key fact is that these are not all congruent. I'm not sure the word "tessellation" would be correct here.

This construction has been known for a very long time, it is in page 77 in Mathematical Snapshots by H. Steinhaus, probably plenty of other books and web pages. STEINHAUS

Evidently it is also discussed on pages 248-249 of GARDNER

I'm afraid I have been unable to find an image on line. The closest I have found in in $\mathbb H^2,$ see the tiling called (7,3) at HYPERBOLIC and imagine what would happen if the whole thing were stretched radially, in polar coordinates $(r,\theta) \rightarrow \left( \tan \frac{\pi r}{2},\; \theta \right).$ I should admit that I do not really know whether this gives a good approximation of the covering of $\mathbb R^2$ beyond the first three or four rings depicted. Probably not.

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Do you know if this is explicitly based around the tiling of the hyperbolic plane? I could easily see some scheme that would start with the (7,3) tiling as represented in the Klein disc model (where all edges are straight lines) and then blow it up non-conformally in some fashion that maintains both the linearity (maybe by simply keeping track of points and redrawing the edges?) and the convexity... – Steven Stadnicki Dec 15 '11 at 23:15
@StevenStadnicki , I really have no idea. There is a clear resemblance near the center, as there must be. I do not know how much work would be required either to confirm your idea or rule it out. If you know anyone with the software, maybe do a larger subset of of (7,3) and see if you can get some radial mapping to work out well. I'm uneasy about the behavior of geodesics that never get near the origin. Even the arrangement of polygons in obvious rings seems unclear in (7,3) as we go farther out, while that is the most important feature of the Steinhaus diagram. – Will Jagy Dec 15 '11 at 23:40
Will: do you know of a good diagram online? The easiest way to confirm would likely be via counting the number of heptagons at distance 'n' (measured in the dual graph, i.e. how many tiles need to be stepped through) from the center. I've looked at similar sequences for the (5, 4) tiling and I don't imagine they'd be too hard to tally here. – Steven Stadnicki Dec 16 '11 at 0:00
@StevenStadnicki no, I'm afraid I was unable to find this online with a diagrm, the only mention so far is at mathworld.wolfram.com/Tiling.html and quite brief. – Will Jagy Dec 16 '11 at 0:11

As a supplement to Will Jagy's answer, permit me to include two figures. The first image below is from Jay Kappraff's 1990 book, Connections: the geometric bridge between art and science. The second image is from a 1983 paper by Danzer, Grünbaum, Shephard, "Does Every Type of Polyhedron Tile Three-Space?" in Structural Topology 8 (perhaps the source for Kappraff?).

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Joseph, thanks. That is Steven Stadnicki's question, is there a (radial) mapping of the hyperbolic (7,3) tiling that results in, more or less, the Steinhaus-Danzer-Grunbaum-Shephard diagram, which is arranged in obvious concentric rings. Steinhaus forces two edges of each heptagon to be radial, I see your diagram is not doing that. – Will Jagy Dec 18 '11 at 3:35

Here's an almost complete answer to the claim: No convex 7+ sided polygon can tessellate the plane. We don't need to compute interior angles or averages. We only need to count the number of interior angles in two ways.

We'll use contradiction of course. Let our polygon $P$ have $n$ sides and let it have area 1. Take a circle of area $A$. Keep every copy of $P$ that lies at least partially in the circle.

I will assume, without careful proof, that the number of polygons in our circle, which can be considered faces in the context of the Euler characteristic of the resulting planar graph, is $F=A+O(A^{1/2})$.

Now I will also assume, without careful proof, that the number of edges in our resultant graph is $E=nF/2+O(A^{1/2})$, as every edge is shared by two polygons except for the $O(A^{1/2})$ on the boundary. Thus $E=nA/2 + O(A^{1/2})$.

Using the Euler characteristic equation for a planar graph, we can compute $$V = E-F+1 = \left(\frac{n}{2}-1\right)A + O(A^{1/2}).$$

Now we can count the number of interior angles in two ways. First, it is easily seen that the number must be $nF$. But also,

$$nF = \sum_{\mbox{v in all vertices}} \mbox{(# of angles meeting at v)} \\ = \sum_{\mbox{v in interior vertices}} \mbox{(# of angles meeting at v)} + \sum_{\mbox{v in edge vertices}} \mbox{(# of angles meeting at v)}.$$

The number of edge vertices $V_e$ is assumed to be $O(A^{1/2})$ so that the number of interior vertices $V_i$ is assumed to be the same as $V=(n/2-1)A+O(A^{1/2})$

Because the number of angles meeting at a vertex must be bounded above given any polygon $P$, the second term in our split sum must be $O(A^{1/2})$.

Now using convexity, we note that for interior angles, the number of interior angles meeting at any interior vertex must be greater than or equal to 3. So,

$$nF \geq 3V_i + O(A^{1/2}) = 3V + O(A^{1/2}).$$

Substituting gives

$$nA \geq 3(n/2-1) A + O(A^{1/2}),$$

which cannot hold in the limit as $A \to \infty$ when $n > 6$. This gives us our desired contradiction.

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This is better than what I regurgitated back in the day :-) About those without careful proof. Can you argue as follows? Let $d$ be the maximum diameter of our polygon. Then all the polygons touching the circle of radius $R$ and center $O$ are completely contained in a circle of radius $R+d$ and center $O$. The difference of the areas of the two circles is, indeed, $\mathcal{O}(A^{1/2})$. Similarly, if any edge of any polygon intersects the perimeter of that circle, then the said polygon is completely inside the circle of radius $R+d$ and completely outside the circle of radius $R-d$... – Jyrki Lahtonen Jul 13 at 20:37
Thank you @JyrkiLahtonen for adding those details!! I've made my post community wiki if you would like to add them in to the actual post. – nayrb Jul 13 at 21:26
Also, I spent a good several hours last night wracking my brains to understand why this is true! Wonderfully, it paid off. – nayrb Jul 13 at 21:27

If you mean a regular polygon, then consider what happens at a vertex. If $k$ polygons meet at a vertex, then $2 \pi = k \alpha$, where $\alpha$ is the internal angle of the polygon, $\alpha = (n-2)\pi/n$. Then $2n = k(n-2)$. This equation has positive integer solutions only when $n\le6$. Indeed $$k = \frac{2n}{n-2} = 2 - \frac{4}{n-2}$$ and so $n-2$ is a divisor of $4$, i.e., $n-2=1,2,4$, which gives $n=3,4,6$.

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There might be a case where a vertex is a meeting point not only of polygon vertices, but also touches one of the edges. For example imagine a brick-wall with square bricks. – Artium Dec 15 '11 at 15:46
@Artium: That last point is something I overlooked. It amounts to adding another interior angle of size $\pi$ to one of the tiles. So for the purposes of my averaging argument it increases the average of the interior angles even further. Thus the argument is saved. If I think of a way of writing this more nicely, I will edit, but hopefully somebody else has seen this before, and will show up! – Jyrki Lahtonen Dec 15 '11 at 16:19