Recently I stumbled upon this question Similar Triangles in Tiling a Plane, and I considered the same problem after imposing some restrictions. In particular I considered the following cases.
- Find a triangular tiling of the plane where no triangle is similar to another one, and each triangle is rational-sided.
Sketch of the proof: we start by tiling the plane with $3-4-5$ right-triangles and then it is possible to split each right-triangle into two rational-sided triangles in infinitely many different ways.
- Find a triangular tiling of the plane where no triangle is similar to another one, and each triangle is integer-sided.
Here I feel quite lost. Any idea? Does this kind of triangular tiling exist?
Best Answer
This is not an answer, but a quarter of it.
Upd. OK, the initial description was probably too terse, so I'll expand it a bit. Define a sequence as $a_1=6$ and $a_{n+1}$ is the greatest possible leg that would form a Pythagorean triangle with $a_n$ for another leg, that is: $$a_{n+1}=\begin{cases} \dfrac{a_n^2}4-1, &\text{if $a_n$ is even}\\ \dfrac{a_n^2-1}2, &\text{if $a_n$ is odd} \end{cases}$$
Mark the points $a_n$ with odd $n$ on one axis and those with even $n$ on another. Then connect them in a zigzag fashion as follows. (The drawing is not to scale!)
I can claim with a good deal of confidence that the sequence never stops and never repeats. The problem is with the other three quarters. Well, maybe we can do them in a similar way, but starting with other Pythagorean triangles.
So it goes.