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I’m working through Grünbaum and Shepard’s Tilings & Patterns and I have been unable to make much progress on this problem.

Does there exist a tiling that is monohedral and monogonal, but is neither isohedral nor isogonal?

I suspect there is not a tiling that is both monohedral and monogonal but neither isohedral nor isogonal based on all the monohedral and monogonal tilings I’ve been able to find. However, I’m unable to formulate any sort of proof as to why.

The way I’ve been attempting to prove it is by starting with a vertex and its incident edges and seeing when it is possible to “tile” the plane with it while producing a monohedral tiling. For example, I’ve found that every possible configuration of three edges that meet together at a single vertex can “tile” the plane monohedrally (as well as isohedrally and isogonally) by forming hexagons. However, I’m not sure this is a feasible route especially since I’m unsure if there exist k-valent tilings in which k is not 3, 4, or 6.

Something I think may be useful but am not sure how to use is that due to the monogolality of the tiling all angles and edges of the prototile must meet at every vertex and therefore be present in every vertex-incident edge “tile”.

Edit Definitions:

Monohedral: every tile in the tiling is the same.

Monogonal: every vertex and its incident edges (the edges connected to the vertex) forms a figure that is congruent to every other vertex and its incident edges.

Isohedral: every tile can be mapped to any other tile using only symmetries of the tiling (imagine tracing out the tiling on a separate sheet of paper and then being able to line up any tile on your paper to any tile on the original and still have the tracing line up with the original everywhere).

Isogonal: every vertex can be mapped to any other vertex of the tiling using only symmetries of the tiling (like isohedral but with vertices).

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    $\begingroup$ Can you define the terms (monohedral, monogonal, etc.) here? Grünbaum and Shepard define a lot of specialized terminology and it is not easy to remember them. $\endgroup$ – Ted Dec 25 '20 at 6:51
  • $\begingroup$ @Ted Done. I’m a bit new so I’m not sure how much info I need to provide but I assume terms like vertex and edge aren’t necessary to define? $\endgroup$ – Zach Traul Dec 25 '20 at 7:54
  • $\begingroup$ Can you clarify if the mapping allows for non orientation preserving transformation (can flip the paper)? $\endgroup$ – Calvin Lin Dec 25 '20 at 14:45
  • $\begingroup$ @CalvinLin Yes, reflections (as well as glide reflections) are allowed $\endgroup$ – Zach Traul Dec 25 '20 at 15:28
  • $\begingroup$ As an aside, the answer appears to be no if "tiling" is replaced with "convex polyhedron"; of the known monohedral but not isohedral convex polyhedra, none seem to be monogonal (regardless of isogonality). $\endgroup$ – RavenclawPrefect Dec 26 '20 at 8:11
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Let $\alpha_4 = 0. a_1 a_2 a_3 \ldots_4 $ be an irrational number in base 4 subject to $|a_i - a_{i+1} | = 1$.
(There are many ways to construct this, like starting from any irrational number in base 4 to determine the "height of the peaks".)

Consider unit squares laid out in the plane, where the lower left corner is $(m + \frac{1}{4} a_{|n| + 1 } , n )$, for integer $m, n$.
This is essentially rows of unit squares stacked upon each other, and shifted left/right by $1/4$ from the row above/below it, though in a aperiodic way.

  • It is monohedral as all tiles are unit squares.
  • It is monogonal as each vertex is a T with incident edges of length $1$ (for the leg of the T), and lengths $3/4, 1/4$ (for the arms of the T). We allow for reflections of the T.
  • It is neither isohedral nor isogonal as the only symmetries of the tiling are left/right translations due to the aperiodic nature.
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