# “Transcendental tilings”: Do they exist?

Let $T$ be a tiling of the plane. Fix an origin and shoot a ray $r$ from the origin. Mark off points $p_i$ along $r$ separated by unit distance. Compute from $r$ a binary number $0 < b(r) < 1$ that alternates $0$'s and $1$'s for each marked point $p_i$ as the ray enters a new tile of $T$. For example, the square tiling and illustrated ray below lead to $$.00011011100010011011100 \ldots$$ Square tiles side length $=7/3$. Ray slope $= 1/\sqrt{2}$.
To avoid thin tiles, assume every tile includes a disk of diameter $> 1$ so that more than one $p_i$ could land in a tile. One needs a rule when $p_i$ is on the boundary of a tile to make $b(r)$ well-defined, but I think that detail is not relevant to my question.

It is not difficult to find tilings and rays where $b(r)$ is rational, irrational, or transcendental, for example, by selecting the slope appropriately in the above example.

Q1. Is there a tiling $T$ such that every $b(r)$, for all origins and rays $r$, is transcendental?

If the answer to Q1 is No, the following two questions are superfluous:

Q2. What is the class of all such transcendental tilings (if I may coin a term)?

Q3. How does this class relate to the aperiodic tilings?

• As for Q1, I would imagine that a regular rectangular tiling, where the rectangles all have side lengths $e$ and $\pi$ (or something like that) should do the trick. (Proving this works could be a bit of a chore, though.) – Will Brian Apr 21 '17 at 16:03
• @WillBrian: If the rectangles are $\pi \times e$ and the slope of the ray is $e/\pi$, then ... – Joseph O'Rourke Apr 21 '17 at 16:28
• If the slope of the ray is $e / \pi$ and the ray happens to originate at a corner of one of the rectangles, then I think $b(r)$ should be transcendental if and only if $e \cdot \pi$ is. (If the ray originates elsewhere I'm less sure, but I think this should still be true.) It is an open question whether $e \cdot \pi$ is transcendental, so I guess we don't want to work with this tiling (in this case, "a bit of a chore" is a bit of an understatement). What we want instead is rectangles of side length $\alpha$ and $\beta$, where $\alpha$, $\beta$, and $\alpha \cdot \beta$ are all transcendental. – Will Brian Apr 21 '17 at 17:17
• @WillBrian: Oh, you're right; sorry. – Joseph O'Rourke Apr 21 '17 at 18:14

Answering Q1, I believe there is a transcendental tiling. Let us begin with this tiling with congruent convex pentagons:

Notice that one can rearrange any of the triples of pentagons that form the regular hexagon, by rotating any triple we want by 180 degrees. Thus we can have two kinds of triples: pointing up or down. Now, the entire tiling can be viewed as formed by triples, each filling a hexagon, and we can view the tiling by hexagons as the union of concentric "annuli" with disjoint interiors. Make it so that the annuli's thickness (not just diameter) is strictly increasing, if needed, increasing exponentially. Then arrange all triples in the same annulus to be pointing in the same direction, alternating the direction when passing from an annulus to the adjacent one. 