User:Double sharp/List of uniform tilings by Schwarz triangle

Yes, I'm fully aware that this is a misnomer: Schwarz triangles are spherical. I should really be referring to Coxeter polytopes (since we're referring to tilings of E2/H2, I could say Coxeter polygons.)

This is the Euclidean and hyperbolic version of List of uniform polyhedra by Schwarz triangle.

What exactly sabotages {5,10/3}, {5/2,10}, {7,14/5}, {7/2,14/3}, {7/3,14} etc. and makes them have infinite density though they'd fit in the plane? – ah, I see, from Regular Polytopes (3rd ed.), p.108: they would have rotational symmetries that are not 2-, 3-, 4-, or 6-fold, which we know to be impossible.

Basically what we need is:

Add and subtract interior angles of {3} (60°), {4} (90°), {6} (120°), {8} (135°), {8/3} (45°), {12} (150°), {12/5} (30°), {&infin;} (180°) to yield some multiple of 360°. This will determine a candidate vertex figure. We may reject everything containing consecutive positive and negative terms of the same angle as degenerate. Some candidates may also be excluded by symmetric concerns, such as {8/3, 8} above. (Though we need to prove that only the above tiles are possible.) However such a bald listing ends up creating lots of junk as well...

Summary table
There are seven generator points with each set of p,q,r (and a few special forms):

There are three special cases:
 * p q (r s) &#124; – This is a mixture of p q r &#124; and p q s &#124;.
 * &#124; p q r – Snub forms (alternated) are give this otherwise unused symbol.
 * &#124; p q r s – A unique snub form for U75 that isn't Wythoff-constructible.

Euclidean tilings
The only plane triangles that tile the plane once over are (3 3 3), (4 2 4), and (3 2 6): they are respectively the equilateral triangle, the 45-45-90 right isosceles triangle, and the 30-60-90 right triangle. It follows that any plane triangle tiling the plane multiple times must be built up from multiple copies of one of these. The only possibility is the 30-120-120 isosceles triangle (3/2 6 6) = (6 2 3) + (2 6 3) tiling the plane twice over. Each triangle counts twice with opposite orientations, with a branch point at the 120° vertices.

The tiling {∞,2} made from two apeirogons is not accepted, because its faces meet at more than one edge. Here ∞' denotes the retrograde counterpart to ∞.

The degenerate named forms are:
 * chatit: compound of 3 hexagonal tilings + triangular tiling
 * chata: compound of 3 hexagonal tilings + triangular tiling + double covers of apeirogons along all edge sequences
 * cha: compound of 3 hexagonal tilings + double covers of apeirogons along all edge sequences
 * cosa: square tiling + double covers of apeirogons along all edge sequences

The tiling 6 6/5 | ∞ is generated as a double cover by Wythoff's construction:

Also there are a few tilings with the mixed symbol p q $r s$ |:

There are also some non-Wythoffian tilings:

Hyperbolic
OK, apparently the hyperbolic fundamental domains are called Lannér triangles (compact) per Coxeter–Dynkin diagram, Koszul triangles (paracompact) and Vinberg triangles (noncompact). But these are only right for simplices, no? So in general I'd write "Coxeter polygons" again.

Symmetry mutations
(should really also add *333, but this is a start, from Mandara: The World of Uniform Tessellations

Families which contain only degenerate members (e.g. the quasitruncated {3,n}) are not shown; neither are those Wythoff symbols that already contain reducible fractions. Those that turn out to be degenerate anyway but do not satisfy either criterion are still shown. In some cases I have naughtily silently corrected the "doubled" constructions of the hemipolyhedra. Some of the Euclidean families involving {∞} correspond quite nicely to the hemipolyhedra, taking {∞} as an equator of r{4,4} or r{3,6}. However, some others do not have clear spherical analogues.