This post continues Cutting a spherical surface into mutually non-congruent pieces of equal area.
Question: Given a planar convex region C and an integer n, how does one decide if C can be divided into n pair-wise non-congruent convex pieces all with same area and perimeter (such a partition could be called a 'totally non-congruent convex fair partition')?
Guesses: a circular disk might not allow such a partition for any n. A square (or maybe some rectangles) might for some n.
Note: Every convex region allows partition into n convex pieces of same area and perimeter (https://arxiv.org/abs/1804.03057). It is also known that there are convex regions (for example, a quadrilateral with angles that are irrational fractions of pi) that do not allow partition into n pieces that are all congruent for any value of n (https://research.ibm.com/haifa/ponderthis/challenges/December2003.html). Some further thoughts on the present question have been recorded at http://nandacumar.blogspot.com/2022/02/fair-partitions-into-non-congruent.html
