Cutting a square into an infinite number of triangles constrained by two rules

Question

Can a square be cut into an infinite number of triangles so that

a) all of them are non-similar

and

b) only a finite number of them can have a common vertex?

Answer 1

I assume the good old classification of triangles from the elementary school. First divide the square $[0,1]\times [0,1]$ in countably many rectangles $R_n:=[0,1]\times [1/(n+1), 1/n]$, for all $n\ge1$. Draw three semi-diagonals of each $R_n$, thus dividing it into one rectangular, non-isosceles triangle $A_n$, one obtuse-angled isosceles triangle $B_n$ and one acute-angled isosceles triangle $C_n$. Of course, no two of them are similar.

Answer 2

There is also a pattern based on dividing an equilateral triangle into four similar triangles, and then repeating the subdivision on (say) the central triangle. Except, one can do this for an arbitrary triangle, and there is no reason to make the subdivision into similar triangles. Indeed, one can pick the subdivision so as to make an angle of each triangle unique and acute.

Gerhard "Triangles All The Way Down" Paseman, 2018.03.30.

Answer 3

I'm not sure if it's rigorous enough, but I think you can start with a regular triangulation with roughly $N \times M$ vertices, then shift the vertix at point $(x,y)$ by $(\epsilon x/N, \epsilon y/M)$ where $|\epsilon| < 1$, assuming the rectangle spans $[-1,1] \times [-1,1]$. You probably also have to add a term quadratic in $x$ and $y$ to remove the mirror symmetry. Then you take the $N,M \to \infty$ limit.