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Inspired by Wikipedia's article on pentagonal tiling, I made my own attempt.

4-tile lattice

I believe this belongs to the 4-tile lattice category, because it's composed of pentagons pointing towards 4 different directions. According to Wikipedia, 3 out of the 15 currently discovered types of pentagonal tiling belongs to the 4-tile lattice category (type 2, 4 & 6).

https://en.wikipedia.org/wiki/Pentagonal_tiling

I don't think my attempt is an instance of type 4 or 6, since in both type 4 & 6, any side of all pentagons overlaps with only one side of another pentagon, while in my attempt, half of the long sides overlap with two shorter sides.

At the same time, I can't figure out how my attempt is an instance of type 2... Please kindly offer your insight.

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5 Answers 5

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There are two questions here:

Q1) Which convex pentagons tile the plane?

Q2) What are all tilings of the plane by copies of a single convex pentagon?

The Wikipedia page you cite concerns Question 1 (though it could make this more explicit); Q1 is contained in Q2, and likely more tractable: once we know that a pentagon tiles the plane, it might still be hard to describe all tilings.

That is the case for your pentagon, which has two parallel sides and is thus contained in Type 1. It is a special case of Type 1 that allows further tilings such as the one you found, but that's a Q2 distinction and doesn't affect Q1.

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  • $\begingroup$ Thanks for the answer. May I further enquire that is it a 2-tile lattice or a 4-tile lattice? If it's an example of a 4-tile lattice, can it belongs to type 1? $\endgroup$
    – Jacky
    Commented Aug 24, 2015 at 3:52
  • $\begingroup$ Your tiling is "4-tile" (the pentagon appears in 4 orientations). But your pentagon also has the Type-1 tiling with a "2-tile lattice" that's shared by all convex pentagons with two parallel sides. It looks like this pentagon also fits into Type 4, giving yet another 4-tiling. $\endgroup$ Commented Aug 24, 2015 at 4:00
  • $\begingroup$ Thanks again Noam. Do you think there's fundamental difference between the pattern I posted and type 4 on Wikipedia (that in type 4, each side of all pentagons overlaps with one side of another pentagon, while in my pattern, there are occasions when a side of a pentagon overlaps with 2 shorter sides of 2 pentagons)? $\endgroup$
    – Jacky
    Commented Aug 24, 2015 at 4:09
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    $\begingroup$ Yes, the argument you give proves that your tiling is not the same as the generic Type 4 tiling. But unless you can generalize your pentagon to one that's not already known to tile the plane in some other way, it's only a "Q2" distinction. $\endgroup$ Commented Aug 24, 2015 at 4:25
  • $\begingroup$ Sorry for the bad notations... Do you think any of the two generalizations (below) is instance of a currently discovered type of convex pentagonal tiling? $\endgroup$
    – Jacky
    Commented Aug 24, 2015 at 10:18
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The pentagon tiling problem is just settled by Michaël Rao: The list of $15$ types is complete. See the Natalie Wolchover article in Quanta.

"Exhaustive search of convex pentagons which tile the plane": link.

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2 generalizations of the tiling I posted in the original question:

enter image description here


enter image description here

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I helped work on the wikipedia article, and also can't identify this tiling as one of the 15. I agree its 2-isohedral, so not types 1-5, and its contraints don't appear to match the 2-isohedral types. I made an image copy and posted in the Wikipedia talk page: https://en.wikipedia.org/wiki/Talk:Pentagonal_tiling#Mystery_convex_monohedral_type

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    $\begingroup$ Thanks Tom. I agree that there is difference between "a tile and a tiling". Thanks again! $\endgroup$
    – Jacky
    Commented Sep 4, 2015 at 3:31
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Your tiling is 1-isohedral, according to wikipedia.

EDIT: Yoav points out that it is actually not 1-isohedral...

Reinhardt (1918) found the first five pentagonal tilings. These all share the property to be isohedral, or "tile transitive", meaning that the symmetries of the tiling can take any tile to any other tile (more formally, the automorphism group acts transitively on the tiles). By contrast, all subsequently found tilings are k-isohedral, with k>1.

Thus, your tiling, if not new, is one of these five. My guess is that it is p4, (wikipedia notation), and it looks very similar.

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    $\begingroup$ Thanks. I guess by p4 you were referring to type 4? If you take a look at the 2nd generalization in my answer to my own question, you would find out that only 1 right angle is needed, while type 4 pentagon needs 2 right angle. As answered by Noam, it actually belongs to type 1 I guess, since type 1 pentagons summarizes all pentagons with parallel sides, which include both of my generalizations. $\endgroup$
    – Jacky
    Commented Aug 25, 2015 at 3:18
  • $\begingroup$ Yeah, it seems so simple, that it is most likely in the original family of 5, since it has the 1-isohedral property, and these are probably easy to exhaust/verify is complete by computer search (but I might be completely wrong on this, maybe suitable as a follow-up question?). $\endgroup$ Commented Aug 25, 2015 at 3:32
  • $\begingroup$ Added: According to this source, jaapsch.net/tilings/Tilings.pdf all 1-isohedral tilings are fully understood.... $\endgroup$ Commented Aug 25, 2015 at 3:34
  • $\begingroup$ Again, Wikipedia does not answer quite the same question. The five Reinhardt classes of pentagons all have 1-isohedral tilings. Some of them may also have other 1-isohedral tilings. So the fact that Jacky's tiling is 1-isohedral doesn't force it to be either new or one of the five types of tilings shown, even though the pentagon does fall in at least one of Reinhardt's classes. $\endgroup$ Commented Aug 25, 2015 at 4:06
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    $\begingroup$ Jacky's tiling is not isohedral. Some pentagons have the long edge (the one between the two parallel edges) meeting a long edge, while some pentagons have the long edge meeting two short edges. Therefore, those two different pentagons cannot be equivalent. $\endgroup$ Commented Aug 25, 2015 at 5:29

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