It transpires that the world didn’t actually end yesterday. At the very least, Descartes’ famous deduction *‘cogito ergo sum’* seems to imply that. To summarise, the Mayan calendar has finished its 13th long count cycle; equivalently, 13×20×20×18×20 days have passed since the beginning of the aforementioned calendar. Suffice it to say, it was rather anticlimatic, with nothing particularly interesting happening.

Depicted below is a tiling of a triangular arrangement of hexagons of side length *s* = 9 with trihexes. An interesting problem, solved by John Conway, asks which side lengths can actually be tiled with trihexes.

Obviously, it can be done in the cases *s* = 2 and *s* = 9, and it’s vacuously true for *s* = 0. Can we do it for other sizes? Yes, we can attain 11 and 12 by augmenting the previous tiling:

You can also add multiples of 12 rows by the following scheme, thereby attaining any side lengths congruent to {0, 2, 9, 11} mod 12.

It’s trivially obvious that you can’t tile a triangle with side length congruent to {1, 4, 7, 10} mod 12, since the number of hexagons is not divisible by 3. This leaves the cases of {3, 5, 6, 8} mod 12, which are impossible for more subtle reasons. We can convert this problem into one on a square grid by applying an affine transformation:

Can we tile this with L-trominos in two orientations? If we consider the perimeter of the region to be a group word (where a = up, b = left, a’ = down, b’ = right), then the two types of tromino are equivalent to the relations bbaa = abab and aabb = baba. Under these relations, the perimeter (which reduces to bbbaaab’a’b’a’b’a’) is not equivalent to the identity, so we can’t tile the entire area with trominos. This can be proved by constructing a group where the relations bbaa = abab and aabb = baba hold, but where bbbaaa ≠ ababab. Some example groups are mentioned here.

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