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It is known that the plane cannot be tiled by pair-wise non-congruent triangles all having same area and same perimeter (https://arxiv.org/abs/1711.04504).

Question: Can a square be partitioned into some finite number of pair-wise non-congruent triangles all with same area and perimeter? If "no", does any regular polygon allow such a triangulation?

Note: If one needs the non-congruent triangular pieces only to have same area, a square can be cut into at least some finite number n of such triangles (as just an example, a 20X20 square centered on the origin can be cut into 40 mutually non-congruent triangles all of area 10 and all sharing a vertex at the point (3,1) and the cut lines radiating from there). I don't know for what values n a square can be so partitioned - for odd n, the answer is certainly "impossible" by Monsky's theorem but beyond that, what?

An earlier and wider question: To partition planar convex regions into n mutually non-congruent convex pieces of equal area and perimeter

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    $\begingroup$ definitely not an odd number of triangles with the same area (to say nothing of perimeter). cf. Monsky's Theorem, which I describe a learning path for briefly here. $\endgroup$
    – Benjamin Dickman
    Commented Dec 26, 2023 at 16:17
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    $\begingroup$ I'm probably missing something, but doesn't this follow from theorem 6 in the paper? $\endgroup$
    – 1001
    Commented Dec 30, 2023 at 20:04
  • $\begingroup$ belated thanks! i was the one who missed that theorem and its implication. indeed, "Let k ≥ 4. In any tiling of a convex k-gon with finitely many triangles, there are two triangles that share an edge" - so the answer is clearly NO. $\endgroup$
    – Nandakumar R
    Commented Jan 9 at 6:51
  • $\begingroup$ to mark the question as answered, your observation may be given as an answer. $\endgroup$
    – Nandakumar R
    Commented Jan 9 at 6:53

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