In the light of Cubing the cube - as 'perfectly' as possible, We try to slightly 'relax' the main theorem proved by Kupaavski, Pach and Tardos here: https://arxiv.org/pdf/1711.04504.pdf
Theorem: There is no tiling of the plane with pairwise noncongruent triangles all of the same area and the same perimeter
Question: Can the plane be tiled by triangles all of same area and perimeter that are pair-wise non-congruent except that exactly 2 of them are mutually congruent? Of course, '2' can be generalized to 'finitely many mutually congruent triangles' or 'finitely many pairs of congruent triangles' etc.
