So, my research about hybrid hyperbolic tilings is advancing nicely. I now have a large set of solutions from various "families". Each family can be defined by its "base edge", i.e. the greatest common divisor of all edge lengths present in the tiling.

I still know desperately little. Mostly the reasons. My numerical searches have found some amazing identities, but the reasons why those identities exist are still mostly unclear. I always say that it's like the "monkey paw" situation: I was able to discover something genuinely new in mathematics, but I don't know enough to truly understand my discoveries.

And one of the greatest mysteries is related to one of the very simplest hyperbolic tilings, one that is almost always mentioned: {5,4}.

As a regular tiling, there's a pretty simple formula to compute its edge length. Well... there is a formula, in any case.

edge = 2*arccosh(cos(pi/5)/sin(pi/4))

This is about 1.0612; I started to use four decimal places for my edges, and so far I didn't get into anything ambiguous. The hybrid families are pretty sparse.

A pentagon is the only regular polygon with this edge that gets a "nice" inner angle, a rational multiple of pi (at least, I think so). But, inner angles of several other regular polygons, while irrational, can nevertheless get rid of their irrationalities in proper combinations.

Exhibit 1:

(I apologize for the inconsistent borders.)

Turns out that four triangles and two squares around a vertex fit exactly at this edge length. But that's not all...

This tiling shows that a triangle, a square, a decagon, an a 20-gon fit as well! Admittedly, they are hard to incorporate. This particular solutions showcases how the large polygons form a "skelet" (this is not in all tilings, but it's common) that will then get a pseudogonal row of triangles and squares attached, followed by a straight row of triangles. This is a "funhouse mirror" tiling where the central line seems to be a mirror, but a closer look reveals that the halves are not actually the same.

These five polygons are all you can get. But you can combine them in various ways.

Here, you can see the simplest tiling that combines vertices with four pentagons and vertices with four triangles and two squares. To connect them, you need some "interim" vertices, of course, which have two triangles, one square, and two pentagons. Since both {5,4} and (3,3,3,3,4,4) are even combinations (they can be cleanly split in two equal halves), those halves can combine into a new vertex.

Note that this can't be done with (3,4,10,20). No clean split there.

Now, you might have noticed something about the squares in the previous tiling. If two triangles and a square add up to a straight angle -- and they do -- you can combine those five polygons together to get one larger square.

Now, this is, admittedly, a bit hacky. There's not a big difference, geometrically speaking, between this square and the structure made from four triangles and one smaller square. This is, in fact, an example of one of the infinite families: whenever you have four triangles and two n-gons at a vertex, the n-gon can be augmented by triangles to form an n-gon with double edge.

(There is also a similar trick that lets you construct triple-edge regular polygons.)

So the fact that the inner angle of the triangle in (3,3,3,3,4,4) is equal to the inner angle of a double-length square is fairly obvious. What is not obvious is that the inner angle of the small square is also equal to the inner angle of a double-edge polygon.

Here, every vertex that would normally have two squares has one of them replaced with a double-edge apeirogon. And unlike the square example, there apeirogons cannot be subdivided. You would have to put a square in every vertex, but then the middle of their edge would be a vertex with two triangles, three squares and a little bit of missing space.

We're still not done, but let's continue in next post.