6
$\begingroup$

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 $k\ge5$.
enter image description here

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!

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Indeed. I think the following works. Create a linkage of three identical segments of length sufficient to fit inside the given rectangle when the linkage ends are attached to the diagonal vertices of the rectangle. Adjust the linkage so that one pivot point is closer to a rectangle corner than the other pivot point is to the opposite corner. You now get a decomposition into two different isosceles triangles and four acute triangles. I believe (thanks to Noam) this will give a 14 isosceles triangle decomposition as requested,. Gerhard "Even Fewer For Some Rectangles" Paseman, 2017.06.26. $\endgroup$ – Gerhard Paseman Jun 26 '17 at 21:44
  • $\begingroup$ In the above, there may be restrictions on the aspect ratio. In which case, divide a long rectangle into short enough incommensurable pieces. (Maybe three links is not enough and more links are needed. I should think more.) Gerhard "Construction Still In Alpha Phase" Paseman, 2017.06.26. $\endgroup$ – Gerhard Paseman Jun 26 '17 at 21:49
  • 2
    $\begingroup$ If you start with a triangulation of your rectangle into acute triangles such that no edge of the rectangle is an edge of a single triangle, and randomly perturb each vertex that is not a vertex of the rectangle (the vertices on the edges of the rectangle being required to move along that edge), for small enough perturbations the resulting triangles will still be acute, and with probability 1 no two triangles will have an edge of the same length with the exception of shared edges between neighbouring triangles, and all the circumradii will be distinct. $\endgroup$ – Robert Israel Jun 27 '17 at 1:18
  • $\begingroup$ Then when you do Noam Elkies's construction on these triangles, all the resulting triangles will be non-congruent. $\endgroup$ – Robert Israel Jun 27 '17 at 1:18
2
$\begingroup$

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.


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.

$\endgroup$
1
$\begingroup$

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 $ABE$.
enter image description here

$\endgroup$
  • $\begingroup$ This can be done with 6 isosceles triangles. When 6-construction encounters an exceptional situation then even fewer triangles will do. $\endgroup$ – Wlod AA Jun 27 '17 at 12:01
  • 1
    $\begingroup$ The diagram also has an exception: If the rectangle is twice as wide as it is high then the two "green triangles" are congruent. $\endgroup$ – Timothy Chow Jun 27 '17 at 17:00
  • $\begingroup$ It's because of that that I wrote "for the general case". :-) $\endgroup$ – Wolfgang Jun 27 '17 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.