Cutting the unit square into pieces with rational length sides

Question

The following questions seem related to the still open question whether there is a point(s) whose distances from the 4 corners of a unit square are all rational.

To cut a unit square into n (a finite number) triangles with all sides of rational length. For which values of n can it be done if at all?
To cut a unit square into n right triangles with all sides of rational length. For which values of n can it be done, if at all?

Remark: If one can find a finite set of 'Pythagorean rectangles' (rectangles whose sides and diagonal are all integers) that together tile some square (of integer side), that would answer this question.

3.To cut a unit square into n isosceles triangles with all sides of rational length. For which values of n can it be done, if at all?

Now, one can add the requirement of mutual non-congruence of all pieces to all these questions. Further, one can demand rationality of area of pieces or replace the unit square with other shapes (including asking for a triangulation of the entire plane into mutually non-congruent triangles all with finite length rational length sides)...

Note: From what has been shown by Yaakov Baruch in the discussion below, cutting the unit square into mutually non-congruent rational sided-right triangles can be done for all n>=4. Indeed, he has shown n=4 explicitly; for higher n, one can go from m non-congruent pieces to m+1 pieces by recursively cutting any of the m right triangular pieces n by joining its right angle to the hypotenuse to cut it into two smaller and mutually similar but non-congruent pieces. That basically settles questions 1 and 2 - the non-congruent pieces case. However, if we need all pieces to be non-congruent and non-similar, the n=4 answer has no obvious generalization to higher n.

References: 1. https://nandacumar.blogspot.com/2016/06/non-congruent-tiling-ongoing-story.html?m=1

On dissecting a triangle into another triangle

Why stop there, and not ask for all ways to cut arbitrary shapes into an arbitrary number of other arbitrary shapes with arbitrary conditions on arbitrary numerical invariants of the shapes? — Gerry Myerson, Feb 03 '20 at 21:06
There are questions on point sets where all or maximally many distances have to be rational. Here the attempt was to think of a family of questions that apply rationality of distances only locally - only sides of each single piece need to be rational; the rest of the distances are free. — Nandakumar R, Feb 04 '20 at 04:17
Thus for 1. it's easy to achieve $n=5$, and from there any $n\ge 5$. $n=3$ would solve the famous open problem, so the only remaining question for 1. is whether $n=4$ is possible. — Yaakov Baruch, Feb 04 '20 at 09:28
As for $n=4$, one can split a $360\times 360$ square into 3 right triangles along the perimeter (360-224-424, 360-105-375, 255-136-289) and a middle triangle (375-289-424). — Yaakov Baruch, Feb 04 '20 at 11:34

score 8 · Accepted Answer · edited Feb 05 '20 at 15:24

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@YaakovBaruch's beautiful construction for $n=4$:

Edit by Yaakov Baruch. A partition with 8 isoceles triangles (I'm sure not minimal):

edited Feb 05 '20 at 15:24

Yaakov Baruch

4,865

answered Feb 04 '20 at 13:00

Joseph O'Rourke

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So the triangles are (8-15-17) * 17, (7-24-25) * 15, and (28-45-53) * 8. That's the first time I remember seeing 28-45-53. – Feb 04 '20 at 13:17
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Drawing heights in right triangles splits them to produce solutions for $n\ge 5$ too. – Yaakov Baruch Feb 04 '20 at 13:47
Beautiful example! Thanks very much for the illustration too! And the nice thing is all right triangles are mutually-noncongruent. Perhaps such examples can be found for other values of n as well. And yes, the isosceles pieces case... – Nandakumar R Feb 04 '20 at 13:51
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(It's always amazing what mathematicians find beautiful...) – Yaakov Baruch Feb 04 '20 at 13:58
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Searching around the internet I see here that the rational distance problem is known to not have solutions on the square's boundary. So that rules out dissections for $n=3$. – Yaakov Baruch Feb 04 '20 at 14:25
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Notice that the middle triangle has rational area, since the other 3 do and the tota areal is 1; any height of the middle triangle therefore splits it into rational ones (as the sum of 2 square roots is only rational if both roots are). – Yaakov Baruch Feb 05 '20 at 08:11
Thanks... for right triangles, if sides are rational, area is automatically rational. So, requiring area too to be rational could be of interest only in the cases such as isosceles triangle pieces. – Nandakumar R Feb 05 '20 at 15:21
@NandakumarR. I don't follow your last comment. But what I meant earlier is that if any triangle has rational sides and area, then its heights are rational and they split the bases at rational point. I think though that Gerry Myerson early comment about the proliferation and arbitrariness of possible questions here has merit. – Yaakov Baruch Feb 05 '20 at 15:38
Thanks for the isosceles example. It would be a nice curiosity if there exists a cutting of the unit square into a finite number of mutually noncongruent isosceles triangles. And yes, my last comment only mentioned that if for a right triangle, all sides are of rational length, its area is automatically rational and this property does not hold in general for isosceles triangles. – Nandakumar R Feb 06 '20 at 09:08
@NandakumarR, after a while this continual adding of criteria stops being fun and starts to seem rude instead. I advise against asking additional questions if you don't care about the answers enough to spend some time trying to answer them yourself beforehand. – Feb 06 '20 at 17:55

Cutting the unit square into pieces with rational length sides

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