# 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 tesselation 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
The angles (belonging to different tiles) meeting at a given point must sum up to 360. There are at least three tiles meeting at each corner of each tile (If there were only two, then one of the tiles would not be convex). Thefore the average of those angles is at most 360/3=120 degrees. So basically we are counting the average in two different ways: one way is per a tile, the other is per a meeting point. The two averages must be equal, when we tile the entire plane. –  Jyrki Lahtonen Dec 15 '11 at 15:54
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

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

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