Some Fair Partition Extensions

Some raw claims:

1. For a circular disk, for any n, the only convex fair partition is the one into n sectors.

2. For any convex region, there are at least some values of n for which there is only one fair partition.

3. For a quadrilateral with all angles irrational fractions of pi and mutually linearly independent, there is exactly 1 fair partition for any value of n.
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What is known (as I gathered from Prof. Roman Karasev):


The technique of https://arxiv.org/abs/1306.2741 essentially establishes that the solution set is $(n-2)$-dimensional for prime power $n$. In particular, when you divide the round disk into $n\ge 4$ parts, you must have other partitions than the obvious splitting into sectors.

This does not imply that there must be different values of the perimeter, but a further investigation may show something like this.
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Observation:

If one considers pairs of quantities like {diameter, perimeter} instead of {area, perimeter} there could even be partitions where both quantities are equal over pieces in a given partition but different between different partitions.

If one equalizes {perimeter, diameter} among 4 pieces from a disk, there seem to be one more 1-d space of solutions in addition to the partition into sectors.

See the attached figure (not to scale). It shows 4 different ways of the same type of partition of a disk into 4 pieces all of equal perimeter (each stage shown in diff color).

The partition consists of (1) 2 identical segments of the disk cut off by by two lines of same color that are parallel to a diameter and also equidistant from it and (2) the remaining central portion cut by a tilted green dashed line into 2 identical pieces with their diameter also equal to that of the two circle segments. In each 4-partition, all pieces thus have equal diameter.

Now, with the yellow pair of parallels, the (very thin) central pieces have much less perimeter than the circular segments which are very big. With the violet parallel lines, the central pieces have more perimeter than circular segments which are thin. So, in between there is a pair of equally spaced parallels for which the perimeter and diameter are both equal among all 4 pieces.

There appears to be a 1D space of these partitions. But I don't know if they plus the space of partitions into sectors will give a 2d space.
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What can be rigorously proved (as I learned from Prof. Karasev):

If a disk is to be partitioned into n >2 convex pieces of equal area and sharing the outer boundary of the disk equally, then, the pieces have to be necessarily sectors.

If a disk is cut into m >2 convex pieces of equal perimeter and sharing the outer boundary of the disk equally, then, the parts are necessarily sectors.