Tiling the plane with incongruent isosceles triangles

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.

Q1: Yes. Any acute non-isosceles triangle can be tiled by three pairwise incongruent isosceles triangles, by connecting each vertex to the circumcenter. Start from some isosceles $T_0$ with repeated side $s$; inscribe $T_0$ into a larger triangle $T_1$ such that $T_1 - T_0$ is the union of three acute, non-isosceles triangles with circumradii distinct from each other and from $s$; likewise inscribe $T_1$ into $T_2$, and $T_2$ into $T_3$, etc., tiling the complement of $T_0$ with ever-larger acute, non-isosceles triangles with all circumradii pairwise distinct and different from $s$. Now connect each of these triangles' vertices to its circumcenter to obtain a tiling of the plane by isosceles triangles any two of which have distinct repeated sides, and thus a fortiori are not congruent, QEF.

[Joseph O'Rourke asks how to find $T_k$ so that the three components of $T_k-T_{k-1}$ are acute and avoid circmuradius coincidences. One way is to deform the triangle, call it $T'_k$, each of whose sides contains a vertex of $T_{k-1}$ and is parallel to the opposite side of $T_{k-1}$ (so $T_{k-1}$ is the median triangle of $T'_k$). Then each component of $T'_k - T_{k-1}$ is congruent to $T_{k-1}$, and thus acute. Now form $T_k$ by slightly moving each vertex of $T'_k$, keeping all angles acute but removing any coincidences among the circumradii and $s$. While you're at it, you can make sure that none of the angles is $30^\circ$ if you don't accept an equilateral triangle as isosceles.]

• Nice construction! Oct 20 '15 at 23:43
• Perhaps you meant $T_1-T_0$? Oct 21 '15 at 19:37
• There is a little piece of the proof implicit: Why does there always exist a $T_k$ such that $T_k-T_{k-1}$ is three acute triangles? Oct 21 '15 at 19:39
• Thanks (also for the illustration), and good questions. Yes, $T_1-T_0$ not $T_0-T_1$; and I'll explain the existence of $T_k$ after removing this typo. Oct 21 '15 at 23:43
• In the last sentence, did you mean $60^\circ$ instead of $30^\circ$? Oct 23 '15 at 13:01

Google soon finds that Q2 is problem C11 in Unsolved Problems in Geometry by Croft, Falconer, and Guy. But perhaps it's been solved during the intervening decades.

URL

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)

Illustration of Noam's construction: • Wouldn't it make more sense to ask Noam if this figure could be included in his answer? After reading his answer and understanding it, it's little use to find this figure after scrolling down to the bottom... Nov 24 '17 at 4:23
• Both constructions shown above appear to use arbitrarily large isosceles triangles. Can one have tilings with incongruent isosceles triangles with bounds on the size of tiles? One can also further constrain the question with requirements such as "all tiles should have same area(same perimeter)" . Jun 9 '20 at 18:21
• Indeed, we can tile plane with non-congruent isosceles triangles with bounds only on edge lengths: perturb each vertex of an equilateral triangle tiling infinitesimally in a unique direction, to get a tiling with non-congruent acute triangles.with every side unique and with both upper and lower bounds on max edge length.. Then, divide each triangle into isosceles triangles by joining its circumcenter with the vertices - as in one of the constructions given above. That should be it. However, it looks like further constraints such as equality of area/perimeter would make the problem harder. Jun 10 '20 at 12:24
• A further variant one can think of is: to tile the plane with mutually non-congruent ACUTE isosceles triangles ( without or with area/perimeter constraints), Jun 11 '20 at 16:21

Q2: The answer is yes (unless we put some restrictions on tilings, see below), this has been demonstrated in this paper.

In the above construction, there are three points such that any neighbourhood of these contains infinitely many triangles, which might make it not count as a "valid tiling".

One way to exclude such situations is to enforce the triangles' areas to be uniformly bounded away from zero. I don't know of the answer in this case, but if we moreover assume that infimum of triangles' areas is achieved, i.e. there is a minimal triangle, then the answer is known to be no. This is shown in an article of Karl Scherer of very descriptive title "The impossibility of a tesselation of the plane into equilateral triangles whose sidelengths are mutually different, one of them being minimal".

• Scherer, Karl. "The impossibility of a tesselation of the plane into equilateral triangles whose sidelengths are mutually different, one of them being minimal." Elemente der Mathematik 38 (1983): 1-3. Oct 21 '15 at 20:19
• Richter, Christian. "Tiling by incongruent equilateral triangles without requiring local finiteness." Elemente der Mathematik 67.4 (2012): 157-163. Oct 21 '15 at 20:26
• The link given in the post does not work for me: caj.informatik.uni-jena.de/caj/file/… (In fact, considering that the website caj.informatik.uni-jena.de requires login and password, I wonder whether it works for users outside this specific university at all.) Nov 24 '17 at 5:06
• I think the Scherer paper can be accessed freely by way of eudml.org/doc/141300 Nov 24 '17 at 5:38