Some Further Details on Fair Partitions...

A detail of the fair partition / spicy chicken problem. Thanks to Prof. Roman Karasev.

Question: In the paper by Karasev, Hubard and Aronov, (https://arxiv.org/pdf/1306.2741.pdf) theorem 1.3 says that "in 2D, a convex region can be divided into prime power number of convex pieces such that an absolutely continuous finite measure (eg. area) is equally divided and and one continuous functional ( eg: perimeter or diameter) has equal value for all the pieces". Qn: What will happen if we try to partition a 2D convex region into n convex pieces such that 2 continuous functionals have to have same value for all pieces - for example, diameter and perimeter should be equal for all convex pieces with no constraint on area? Eg: For n =2, this is possible from a simple continuity argument.

Note: The fair partition claim has been recently extended to all values of the number of convex pieces, not only prime powers.

Answer:

There is no guarantee that diameter and perimeter can be simultaneously equalized among pieces. For example, as the pair of the continuous functionals defined on the pieces you may take the $x$ coordinate of the mass center and the $y$ coordinate of the mass center. It seems unlikely possible to equalize the mass centers of the convex parts in a partition, right?

The crucial property used in the proof is that at least one of the functionals need to be "continuous on empty sets", that is tend to zero when a part is about to disappear, and at the same time be positive on parts with nonempty interior. Those are crucial properties we need from the area functional.

Further question:

One wonders if, from among continuous functionals defined on the pieces, there could be qualitative differences between those functionals that depend on coordinates (such as the x coordinate of center of mass of a piece - it depends on the coordinates and orientation of the input region) and those functionals that don't (such as perimeter or diameter of a piece).

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Another question: If one insists in a fair partition that the equal area and perimeter pieces ought to be also mutually non-congruent, what could be the implications? For N =2 and the input convex polygon a square, the 2 pieces are necessarily congruent so such a fair partition is obviously not always possible. But can one say more about situations and values of N where such partitions necessarily exist? Remark: Special cases of this question (eg: fair partition of square into 3 mutually non-congruent pieces, equilateral triangle into 5 pieces etc..) could be of interest.

I just learned from Prof. Karasev, that the solution space, of fair partitions (partitions into convex pieces of equal area and equal perimeter), at least when N is a prime power, is rather big, its dimension must be of order N. But it is not clear if the large number of solutions guarantee that for one of them the parts will be pairwise non-congruent.

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An unrelated question: Are there rep-tiles on the Hyperbolic plane?

Answer: Scaling makes no sense on Hyperbolic plane - among for example equilateral triangles, the bigger triangles have smaller angles than smaller ones. So, no way to build a scaled up copy of any tile with copies of the tile - the scaled up copy has smaller angles.

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Material added on 6th September 2018:

Questions:

1. If a convex 2D region is partitioned into N pieces such that the pieces all have the same perimeter (with no constraint on the area) such that we also minimize the total length of cuts, will the pieces necessarily be convex?

2. If the answer is "yes", does the same (convexity of pieces) hold if we equalize and minimize any other continuous functional over the pieces? And does such equalize+minimize operation on any other functional automatically yield pieces of equal area?

Note: Indeed, if one does an equal area partitioning into N pieces using least total length of cuts, the cuts could be arcs and so pieces are non-convex.

Here is an intuitive argument on question 1 above from Prof. Karasev:

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If we have a partition into equal perimeter parts (call this perimeter P) that are not necessarily convex then we may first straighten the cuts. After that the parts become polygons, but still may have concave vertices. It then seems possible to push out the concave vertices so that the perimeter of the current polygon only decreases and the perimeters of neighboring polygons also decrease. Of course, there are technicalities in this process, but eventually we seem to end up in a convex partition into polygons with all perimeters at most $P$, but possibly not equal.

Then we try to move the cuts to equalize the perimeters and it seems that the maximal perimeter does not increase during this process, so we are likely to end up with equal perimeters no greater than the original value P. Since the common value of the perimeter is expressed in the perimeter of the total convex body and the total length of the cuts, the total length of the cuts will also be no greater in this convex situation then in the original non-convex one. In other words, given a partition of a convex region into convex pieces all with perimeters of pieces less than or equal to P but possibly unequal to one another (achieved by the previous steps) , it is very conceivable the pieces can be deformed to equalize their perimeter to a value below P and also keeping them convex.

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Another question: For any convex 2D region, will the shortest perimeter bisector also be a fair bisector?

Answer: Not necessarily. A shortest perimeter bisector of a convex region will meet its two end edges at the same angle. This perimeter bisector need not be an area bisector too of the full region - indeed, assuming this shortest perimeter bisector to also be an area bisector of the full region, either piece resulting from this bisector can be altered to have an altered area without changing either its perimeter or the angle at which it meets the perimeter bisector; then, the perimeter bisector remains a shortest perimeter bisector but is no more an area bisector.