After all the discussion raised by this old question, I am wondering about a somewhat complementary one:
For any given rectangle, does there exist a finite set of pairwise different isosceles triangles which tile it?
It is easy to tile e.g. a $1\times a$ rectangle for $1<a<2$ by four isosceles triangles, but with two of them being equal. In the case that $a=\sqrt{\frac{5-\sqrt{5}}2}$, we are lucky and can split one of those into two smaller ones, obtaining a tiling into 5 different isosceles triangles (with all occurring angles being multiples of $\frac\pi{10}$). BTW, we can iterate that by splitting the blue triangle again etc., getting tilings of the same rectangle into $k$ different isosceles triangles for all $k\ge5$.
I am quite sure the answer to the initial question is no, and it may even be interesting to restrict it to the following:
For which other rectangles is such a tiling known to exist?
And possibly, it doesn't even make a difference if we allow an infinite set of pairwise different isosceles triangles!
Best Answer
As Noam Elkies has observed, any acute non-isosceles triangle can be tiled by three pairwise non-congruent isosceles triangles, by connecting each vertex to the circumcenter. There are lots of ways to partition any rectangle into non-congruent non-isosceles triangles, each of which can be replaced by three isosceles triangles, and I think it should be easy to find a partition for which this construction produces non-congruent isosceles triangles.