What I know about {5,4}

[Marek14]

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}.

a5.png

{5,4}

(946.09 KiB) Not downloaded yet

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:

a3a4 small.png

(452.26 KiB) Not downloaded yet

(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...

a3a4a10a20 b.png

(1016.76 KiB) Not downloaded yet

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.

a3a4a5 b small.png

(463.7 KiB) Not downloaded yet

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.

a4a5b4 b.png

(429.08 KiB) Not downloaded yet

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.

a3a4boo b.png

(341.67 KiB) Not downloaded yet

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.

[Marek14]

Seriously, is there a way to make those attachments display?

I have showed how the triangles and squares lend themselves to double-edge polygons. Clearly, anything bigger would only form a pseudogon; which might also be an interesting thing to see, but not really what I need.
But: some other double-edge polygons arise anyway.

{5,4} is one of the handful of regular tilings where you can construct another regular tiling with double of its edge. In fact, this double is the only edge (as far as I know) that allows construction of three different regular tilings: {3,10}, {5,6}, and {10,5}.

A relation between {3,10} and {10,5} is easy to see: if you assemble 10 equilateral triangles with angle 36 degrees around a vertex, they will clump into a regular decagon with 72-degree angle, like in this tiling that is basically a double of decagonal antiprism.

b3b10 b.png

(380.97 KiB) Not downloaded yet

{5,6} having the same edge is actually a part of another infinite family of hybrids.

So, we can combine small and large polygons in various ways. I'll just show a simple one here:

a5b3 b.png

(481.74 KiB) Not downloaded yet

So, that's 10 distinct polygons we can use: small triangle, square, pentagon, decagon, and 20-gon, and large triangle, square, pentagon, decagon, and apeirogon.

There are two more I know of.

First, this one:

a3a4doo b.png

(425.12 KiB) Not downloaded yet

a4a5doo b.png

(412.56 KiB) Not downloaded yet

Basically, two small squares don't quite add to a straight angle. But the gap they create is just enough for a quadruple-edge apeirogon. I like the second solution here, with its infinite square strips and pentagons haphazardly constructed between the apeirogonal pits.

And of course, this:

image4.png

(835.82 KiB) Not downloaded yet

This is still the only known family that can incorporate a septuple-edge polygon. I am currently running a search that should find this solution again in a more inclusive group, but for now, it has only found this:

a3a4b4boog3 c.png

(414.98 KiB) Not downloaded yet

It might not look as nice, but it still lets you see the very tiny angle of this triangle. This angle is what is left over if you put one small triangle and four squares around a vertex. In this solution, two of those squares are replaced with double-edge apeirogons, which, as you already know, have the same inner angle.

One more picture, showcasing a center of 4-fold rotation here:

a3a4b4boog3 d.png

(419.23 KiB) Not downloaded yet

[mr_e_man]

Seriously, is there a way to make those attachments display?

I think the image dimensions are too big.

It seems that the acceptable limits are 800 pixels (width) and 600 pixels (height) (or 344 for no scrollbar). But this may depend on your browser.

[mr_e_man]

So, can you edit those posts, and upload smaller images?

[Marek14]

So, can you edit those posts, and upload smaller images?

Maybe, but not sure if it's necessary.

[quickfur]

I think your images are too big. I tried quoting your original post and replacing the attachment tag with a direct link to the image download URL, and here's what I got:

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:

a3a4 small.png

(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...

a3a4a10a20 b.png

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.

a3a4a5 b small.png

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.

a4a5b4 b.png

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.

a3a4boo b.png

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.

At least on my browser the image is just way too big and gets clipped off the right edge of the window. You should try to reduce the size, and maybe edit your post afterwards to replace the [ attachment ] tags with [ img ] tags instead.

[Marek14]

I might try, but first I want the search program to finally finish, which hasn't happened so far

[wendy]

There is a compound that the faces of (5,4) divide into six copies of (10,4).

The (5,4) and (4,5) are not in the main sequence, the only duals to do so.

[Marek14]

The search has finally finished.

I made a short reddit writeup:

https://www.reddit.com/r/Geometry/comme ... c_tilings/

[quickfur]

Nice, very interesting! I've been fooling around with Hyperrogue exploring hyperbolic geometry... it's very strange indeed!