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A rational triangle is one in which all side lengths are rational numbers.

Question: Can we tile the Euclidean plane with rational triangles that are pairwise non-congruent? No further requirements on the triangles. Their perimeters are not upper or lower bounded.

The answer seems "no" but I have no proof. It is known that there is no tiling of the plane by pairwise non-congruent triangles all of same area and same perimeter (https://arxiv.org/abs/1711.04504) - not sure if that is relevant here.

Further, we can ask if the plane can be tiled by quadrangles and so forth whose sides are all rational (and if the answer is "yes", add further constraints on diagonals etc.).

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    $\begingroup$ Worth noting: if you can tile arbitrarily large regions of the plane in consistent fashion, then you can tile the whole plane by the compactness theorem. I feel like there's a very good chance of constructing a triangle spiral that will accomplish this. $\endgroup$ Commented Apr 27, 2023 at 4:56
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    $\begingroup$ This answer to another question seems to answer your question as well: it gives a tiling by integer triangles that do not share an edge length. $\endgroup$ Commented Apr 27, 2023 at 7:02
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    $\begingroup$ Thanks for both comments! The spiral construction given by Prof O'Rourke at mathoverflow.net/questions/307130/… seems quite a promising candidate. But what I am not quite sure is whether rationals are dense enough for this spiral to be continued indefinitely - all edges need to be rational throughout. $\endgroup$ Commented Apr 27, 2023 at 9:41
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    $\begingroup$ Sorry, I misread the question as asking for triangles with rational vertices. $\endgroup$ Commented Apr 27, 2023 at 14:23
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    $\begingroup$ But I suspect the spiral construction on the page you pointed out too might lead to an answer to the present question - with the added benefit that every edge on every triangle has a unique length. The downside is that the triangles would be unbounded in size! $\endgroup$ Commented Apr 27, 2023 at 14:32

4 Answers 4

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First answer: The plane can be tiled as requested. First, we tile the plane with equilateral triangles with side lengths $1$. Now each such triangle can be tiled into two rational-sided triangles in infinitely many different ways: Choose a point on a side which dissects it into lengths $x$ and $1-x$. (Sorry, too lazy to produce a picture :-).) Then we get a tiling of the unit triangle into the two triangles of side lengths $1, x, y$ and $1, 1-x, y$, respectively, where $y^2=1-x+x^2$. Setting $x=(1-t^2)/(1+2t)$ for a rational $t$ yields $y=(1+t+t^2)/(1+2t)$. So any rational $0<t<1$ yields a tiling into two rational-sided triangles.

Second answer: There is even a solution with all sides integers! Extend the equilateral triangle with side lengths $m^2-1$ to the one with side lengths $(m+1)^2-1$. So the new piece is a trapezoid with side lengths $m^2-1$, $2m+1$, $(m+1)^2-1$, $2m+1$. Pythagoras tells us that the diagonal has integral length $m^2+m+1$. So we have an obvious inductive procedure to tile the plane. The following picture should illustrate the procedure, starting with the triangle with side lengths $3=2^2-1$:

enter image description here

Third answer: A minor variant to Rosie F's answer, where we arrive one step earlier at a triangle which is similar to the starting one:

enter image description here

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    $\begingroup$ Really nice. The bonus being all triangles being of bounded size. Thanks! $\endgroup$ Commented Apr 27, 2023 at 14:34
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    $\begingroup$ Indeed, the process of adding your 2nd to 5th triangles to your 3-3-3 produces a 15-15-15, which is the same enlarged by an integer factor. One may then repeat the process, enlarged by factors of successive powers of 5. $\endgroup$
    – Rosie F
    Commented Apr 27, 2023 at 18:13
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    $\begingroup$ @RosieF Nice observation! $\endgroup$ Commented Apr 27, 2023 at 19:15
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    $\begingroup$ Your second solution is more economical than my suggestions because the combined area of the first $n$ triangles is $O(n^4)$, whereas, in a solution involving scaling up, it is exponential in $n$. What's more, it can be easily adapted to make a solution where none of the component triangles are even similar. $3,5,7$ is similar to $9,15,21$, but we avoid the former by starting with the $3,7,8$ or the $8,8,8$. $\endgroup$
    – Rosie F
    Commented Apr 28, 2023 at 5:59
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Yes, it is possible; in fact, we can do it entirely with $5-12-13$ right triangles at different scales.

First, note that we can three triangles at scales in the ratio $5:12:13$ to form a $5\times 12$ rectangle:

enter image description here

Then we place scaled copies of these rectangles in (an affine transformation of) the minimal perfect squared square:

                                         order 21 squared square

Thereafter, we can add rectangles at scales of $112, 2\cdot 112, 3\cdot 112, 5\cdot 112, 8\cdot 112, \ldots$ in a Fibonacci spiral:

                                         enter image description here

All that remains is to check that no pair of the rectangle scaling factors here have a ratio of $5/12$, $5/13$, or $12/13$; this is pretty straightforward to verify.

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    $\begingroup$ This doesn't quite work, because 50 is twice 25, 18 is twice 9, 33 is thrice 11, and so on. Also, your Fibonacci spiral starts off with two squares of the same size. $\endgroup$ Commented Apr 27, 2023 at 11:46
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    $\begingroup$ @DaveBenson: Why does the fact that $50$ is twice $25$ pose a problem? Which two triangles does it render congruent? And the smallest square in the spiral consists of the squared square described above. $\endgroup$ Commented Apr 27, 2023 at 15:14
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    $\begingroup$ I think this works, but I recommend that you flesh this out a little more. For example, in the squared square, it's not immediately obvious that one of the larger constituent triangles of a small square is not congruent to one of the smaller constituent triangles of a larger square. The number of details that are left to the reader to verify here is rather large. $\endgroup$ Commented Apr 27, 2023 at 16:08
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    $\begingroup$ @RavenclawPrefect Sorry, I misunderstood your construction. I now see that you're only subdividing one of the $1\times 1$ squares in the Fibonacci spiral, not the rest, and then stretching. $\endgroup$ Commented Apr 27, 2023 at 17:52
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    $\begingroup$ very nice. thanks. $\endgroup$ Commented May 1, 2023 at 9:21
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The diagram below shows an alternative approach, using nested Pythagorean triangles whose right angles are all at the origin:

nested Pythagorean triangles

The nested Pythagorean triangles shown here are 5-12-13, 12-16-20 (3-4-5 enlarged by a factor of 4), 16-30-34 (8-15-17 enlarged by a factor of 2) and 30-72-78, which is similar to the starting triangle. The same process of adding three triangles to the starting 5-12-13 may be repeated (enlarged by a factor of 6 and reflected in $y=x$) on the 30-72-78, and so on indefinitely.

This fills a quadrant. The other quadrants may be filled using the same construct, rotated about the origin as needed, and enlarged by factors of (say) 7, 11 and 13. These prime enlargement factors will prevent any triangles in one quadrant being congruent to any in another.

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    $\begingroup$ nice and intuitive. Thanks $\endgroup$ Commented May 1, 2023 at 9:21
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This is overkill for your question, but in Carl Pomerance's paper, On a tiling problem of R. B. Eggleton (Discrete Mathematics 18 (1977), 63–70), he shows that the plane can be tiled using precisely one triangle from each congruence class of rational triangles.

You also asked about quadrangles. For a long time, it was an open question whether the plane can be tiled using exactly one $1\times 1$ square, exactly one $2\times 2$ square, exactly one $3\times 3$ square, etc. This problem was solved affirmatively by Henle and Henle, Squaring the plane, American Mathematical Monthly 115 (2008), 3–12.

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    $\begingroup$ Thanks for pointing these out. $\endgroup$ Commented May 1, 2023 at 9:20
  • $\begingroup$ And hope someone could clarify if the next question here ( mathoverflow.net/questions/445762/…) gets answered by the results in these papers. $\endgroup$ Commented May 5, 2023 at 6:46

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