On Triangulations of the Plane

Back in 2014, I had asked here if the plane can be tiled by pairwise noncongruent triangles all of same area and perimeter and that question has been answered in the negative recently (https://arxiv.org/abs/1711.04504) and further aspects explored in for example, https://arxiv.org/abs/1905.08144. Now, let me record another line of thinking...

Basic question: Can the plane be divided into a vertex to vertex arrangement of non-overlapping triangles such that every edge has a unique rational length r that lies between 1 and some specific rational greater than 1?

If this is possible, one can apply further constraints such as "all triangles should have equal area (OR equal perimeter)". Alternatively, one can relax the vertex-to-vertex requirement to frame another question.

Note 1: Requiring the lengths of all edges to be integers rather than rationals would lead to the lengths of the triangles being unbounded even if a triangulation with all edges having unique lengths is possible (not sure if this is possible).

Here is a discussion page on this subject:
https://mathoverflow.net/questions/351047/triangulating-the-plane-using-edges-of-unique-rational-lengths

A Question on Dissecting Triangles

The Wallace-Bolyai-Gerwein theorem answers the question when one polygon can be formed from another by cutting it into a finite number of pieces and recomposing these by translations and rotations. The answer: this can be done if and only if two polygons have the same area.

It is easy to see this: An equilateral triangle can be cut into 2 identical 30-60-90 degrees right triangles which can be patched together to form a 30-30-120 degrees triangle. So, via 2 intermediate pieces, we can go from an equilateral triangle to the 30-30-120 triangle of the same area.

Question: Find two triangles equal area T1 and T2 such that T1 can be cut into 3 (and not less than 3) pieces which can be reassembled into T2. Not sure if such a T1-T2 pair exists.

One can readily generalize and ask for equal area {T1-T2} pairs which can be dissected into each other via n (and not less than n) intermediate pieces. Note that we are not asking about the general triangle to any other equal area triangle dissection but for specific triangle pairs.

Remark: n cannot be arbitrarily large as per the W-B-G theorem. The method based on Montucla's dissection of one rectangle to another equal area rectangle with specified length (as given in Greg Frederickson's 'Dissections - Plane and Fancy') might give a large (but bounded) number of intermediate pieces.

Further question: What can one say about equal area {T1-T2} pairs which are the worst for mutual dissection - ie for which the number of intermediate pieces is the highest possible? A guess is that a triangle pair with equal area and equal perimeter is bad for mutual dissection.

Further Note - Going to quadrilaterals from triangles: we first note the problem is trivial for rectangles; indeed, for any rectangle R and any n, there is a different rectangle with same area as R and n times as long which can be got by cutting R into n equal, length-parallel strips and attaching them end to end. For general quadrilaterals, things seem less trivial. For any cyclic quad Q (each pair of opposite vertices add to 180 degrees), it appears that we can cut Q into 2 pieces and patch these to get another quad Q' different from Q. But for non-cyclic Q's, one is not sure of finding such a Q' even for n =2. For general n, things appear harder.

Here is a discussion page on these questions:
https://mathoverflow.net/questions/350735/on-dissecting-a-triangle-into-another-triangle

Melzak, Aberth and Beyond

Melzak's problem: Among all polyhedrons with edge length sum = 1, which one has the max volume?

Aberth's problem: Among all polyhedrons with edge length sum = 1, which one has max total surface area?

Both these questions are not fully answered. For a detailed survey there is the PhD thesis of Scott Berger (2001). The conjectured answer to Melzak is a right triangular prisms. For Aberth, the guess is that there is no such well defined polyhedron (in the same sense that there is no unique smallest positive real number).

I couldn't find references on these variants of the above questions:

1. With fixed edge length sum AND surface area, which polyhedron maximizes the volume?

2. With fixed edge length sum AND volume, which polyhedron maximizes surface area?

Note: With given area and perimeter, there are in general, infinitely many triangles. This implies for a given edge length sum and surface area, there can be infinitely many right triangular prisms with the same volume. Does this mark a useful starting point in answering question 1?