Partitioning a rectangle into different isosceles triangles

Question

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!

Answer 1

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.

Answer 2

Let $\ A\ B\ C\ D\ \in\ \mathbb R^2\ $ be the vertices of a rectangle, where $\ A+C=B+D=\mathbb 0\ $ is the origin. Let $\ E\ $ belong to the interval $\ BD,\ $ and be such that $\ AE\ $ and $\ BD\ $ are perpendicular one to another.

Then we get the following partition of the square into six isosceles triangles (the vertices on the symmetric line are listed as the middle of the three vertices):

$$ A\ \ \frac{A+B}2\ \ E $$ $$ B\ \ \frac{A+B}2\ \ E $$ $$ A\ \ \frac{A+D}2\ \ E $$ $$ D\ \ \frac{A+D}2\ \ E $$ $$ B\quad \mathbb 0\quad C $$ $$ C\quad \mathbb 0\quad D $$

Thus the problem is solved, with $\ \mathbf 6\ $ triangles, in the case of all rectangles but squares--only in the case of a square some of the given $\ \mathbf 6\ $ triangles are congruent. Otherwise, we get three pairs of triangles which have the same area within the pair but different for the different pairs. And within the pair, one triangle is acute (i.e. all its angles are acute), and one is obtuse. Thus no two of the six are congruent.

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In the case of square, @Wolfgang's construction provides $7$ triangles. However, $\ \mathbf 5\ $ is enough.

Indeed, let

$$ a\ :=\ 2-\sqrt{2}\ =\ \sqrt{2}\cdot(\sqrt{2}-1) $$

Then, consider the following isosceles triangle decomposition of square $[0;1]^2$:

$$ (0\ 0)\quad (0\ 1)\quad (1\ 1) $$ $$ (0\ 0)\quad (a\ 0)\quad (a\ a) $$ $$ (a\ 0)\quad (a\ a)\quad (1\ 1) $$ $$ (a\ 0)\quad (\frac{a+1}2\,\ \frac 12)\quad (1\ 0) $$ $$ (1\ 0)\quad (\frac{a+1}2\,\ \frac 12)\quad (1\ 1) $$

Not any two of them are congruent.

Answer 3

Thinking at it again: Even more simply, for the general case, it can be done with 7 non-congruent isosceles triangles as in the picture, using 3 right triangles $CDE, ABF, BEF$.
Segments of same color have same lengths.
$E$ is the midpoint of $BC$, and $BF$ is the height in $ABE$.