TITLE: Tiling the plane with incongruent isosceles triangles
QUESTION [34 upvotes]: It is not difficult to tile the plane with incongruent triangles.
One could tile with equilateral triangles, and then partition
each equilateral into three triangles, displacing their common
centerpoint so that no two triangles are congruent (left below).

          



Q1.
  Is it possible to tile the plane with isosceles triangles,
  no two of which are congruent?

It is easy to tile the plane with congruent isosceles triangles,
as illustrated right above.
But I don't see how to achieve a tiling with incongruent isosceles triangles.
Perhaps it is easier to answer this question:

Q2.
  Is it possible to tile the plane with equilateral triangles,
  no two of which are congruent?


Added 14Feb2020:
Q2 has been answered (negatively) in two papers
(independently). These results were presaged 
(by @Wojowu)
below.

(1) Pach, János, and Gábor Tardos. "Tiling the plane with equilateral triangles." arXiv:1805.08840 abstract (2018).

Corollary 4. There is no tiling of the plane with pairwise noncongruent equilateral triangles
whose side lengths are bounded from below by a positive constant.

(2) Richter, Christian, and Melchior Wirth. "Tilings of convex sets by mutually incongruent equilateral triangles contain arbitrarily small tiles." Discrete & Computational Geometry 63, no. 1 (2020): 169-181.
  Springer link.



Question Q1 was inspired by (but not addressed in) this paper:

Malkevitch, J. "Convex isosceles triangle polyhedra." Geombinatorics 10 (2001): 122-132.

REPLY [17 votes]: Simpler construction: a non-isosceles right triangle $T_0$
can be divided into two isosceles triangles not congruent to each other,
or into two right triangles similar to $T_0$.  Reversing the latter
construction, let $T'_n$, $T_n$ ($n = 1, 2, 3, \ldots$) be right triangles
similar to $T_0$ such that $T_n = T_{n-1} + T'_n$, and divide $T_0$ and
each $T'_n$ into a pair of isosceles triangles.  There are two choices
at each stage, most of which yield a tiling of the plane; alternatively,
choose the $T'_n$ and $T_n$ to all share a vertex with $T_0$, getting
a geometric progression [sic] of triangles that tiles a wedge,
and then arrange several wedges around a common vertex to fill the plane.
Here's a picture of such a wedge using $(3:4:5)$ triangles:

    (source)