Convex partitions - averages of quantities

We continue from the following posts:
- https://nandacumar.blogspot.com/2022/03/max-of-min-and-min-of-max-3.html
- https://nandacumar.blogspot.com/2021/04/convex-partitions-max-of-min-and-min-of.html
and this post at mathoverflow:
- https://mathoverflow.net/questions/376672/cutting-convex-regions-into-equal-diameter-and-equal-least-width-pieces-2
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Some background:

If a convex planar region C is cut into n convex pieces such that the average perimeter of pieces is maximized, will all pieces necessarily have same perimeter?

The answer appears to be "not necessarily". Consider a disk being cut into 3 pieces such that the sum of perimeters is maximized; the best seems to be two half disks and one infinitely thin piece along the diameter.

What if we try to maximize the geometric mean rather than arithmetic mean of all perimeters?

Remark 1: If we minimize the average perimeter, n-1 of the pieces will shrink to points and one piece will be C itself - not too interesting. If instead of perimeter, we try to maximize the average diameter, all pieces will try to contain a diameter of C and most will become degenerate.

Remark 2: Any partition whatever of C into n pieces will give the same arithmetic mean for *area* of pieces. Howeever if the geometric mean of area is to be maximized, then areas ought to be equal.
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Some possibly harder and more interesting questions:
Definition: 'width' of a convex region is the distance between the pair of parallel lines that touch it and are least apart.

1. If we try to maximize the average over pieces of the least width, will all pieces necessarily have equal least width? Again, average could be taken to mean arithmetic or geometric mean.
2. What if we consider the average width of each piece?

3. After perimeter and width, one can think of 'centralness' - maximizing the average of centralness. For definition, see https://mathoverflow.net/questions/396579/a-center-of-convex-planar-regions-based-on-chords

Inside out dissections - contd.

This post continues not an earlier post here but a query posted at mathoverflow in June 2021: https://mathoverflow.net/questions/394823/further-queries-on-inside-out-polygonal-dissections
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Earlier Definitions (https://mathoverflow.net/questions/186332/inside-out-polygonal-dissections): a polygon P has an *inside out dissection* into P' if P′ is congruent to P and the perimeter of P becomes interior to P' and so the perimeter of P'is composed of internal cuts in the dissection of P. The dissection to P' is *totally inside out* (https://mathoverflow.net/questions/394823/further-queries-on-inside-out-polygonal-dissections) if we further insist that no point on the boundary of P should be on the boundary of P'. These definitions have natural analogs in 3D.
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Let us drop the requirement that the final polygon (or solid) P' be congruent to P but only insist on the inside-out (or totally inside out) nature of the dissection of P into P'.

Question 1: Given a convex polygonal region, will it always have an inside out dissection into *some* convex polygonal region? If the answer is yes (as seems to be the case), how does one achieve such a dissection with least number of intermediate pieces (thse pieces could be nonconvex but ought to be simple polygons)? If the answer is no, how does one decide whether a given polygonal region has such a dissection?
Question 1A: same question as 1 with the'fully inside out' requirement.

For an inside out (totally inside out) dissection from P to P' (both convex), is the intermediate pieces being allowed to be non-convex of any consequence? Guess: for fully inside out dissection it might be.

Question 2: Given a cube, what is the convex solid into which it can be dissected inside out with the least number of intermediate pieces?

Note: A cube can be dissected inside out via 8 smaller cubes. What we ask is if the cube can be dissected inside out into some other convex solid via less number of intermediate pieces.

Question 2A: same question as 2 with the'fully inside out' requirement.

Question 3:How does one dissect a circular disk (spherical ball) inside out into a square (cube) of same area(volume)? this question has a negative answer in the Tarski circle squaring, as given in Wiki - even a general dissection of circle to square via a finite number of pieces is impossible.

Question 4: Can one think of a analogs of the Dehn invariant to determine if two polyhedra can be inside out/ fully inside out dissected into each other?

A deleted Mathoverflow post - stored

A post at mathoverflow can sometimes get deleted by a bot. The following was: https://mathoverflow.net/questions/416530/thinnest-3-fold-and-n-fold-coverings-of-the-plane-by-congruent-convex-shapes. So, saving a copy here.
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This post is totally based on https://mathoverflow.net/questions/183069/thinnest-2-fold-coverings-of-the-plane-by-congruent-convex-shapes (Thinnest 2-fold coverings of the plane by congruent convex shapes) - indeed, am only recording a couple of very natural further questions here.

Are there convex planar regions which do not tile the plane and do not form a perfect 2-cover of the plane but do form a perfect 3-cover (generalization from 3 to n is natural)?

In 3D, are there convex regions that cannot tile but can form a perfect 2-cover of 3-space?
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Note:This post linked above features a superb demonstration by Noam Elkies.

Cutting n-gons into triangles and quadrilaterals

Basically, we are trying to push the envelope beyond Monsky's theorem which states: a square cannot be cut into any odd number of equal area triangles.

Question: Given a convex n-gon. It is needed to cut it into equal area pieces that could be triangles or convex quadrilaterals but should not have more than 4 edges.

Claim: If for some integer m_0, a convex n-gon can be cut into m_0 equal area pieces that are all triangles or convex quads, then, it can be so partitioned into m pieces for all integer m > m_0.

The same question can be asked with perimeter replacing area.

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.

Trapeziums - Non Congruent Tiling (18) and Oriented Containers

This post marks a meeting of two tracks we have been pursuing - Oriented containers and Non-congruent tiling.

The previous posts in those series are here and here. We also build on this earlier post.

General remark: I had thought that the analog of 'isosceles triangles among triangles' is 'kites among quadrilaterals'. Had a series on them here But then trapeziums (being oriented) too could be considered among quads.

Questions:

1. How does one find the smallest (in terms of area OR perimeter) trapezium that contains a given set of points? It is of course enough to solve the problem of finding the smallest trapezium containing the convex hull of the points (A trapezium has an orientedness due to its pair of parallel edges - a property that I had not noticed earlier!).

2. The same question as above with the trapezium restricted to being isosceles (the two non-parallel edges are of equal length)?

3. For which convex region are the orientations of the smallest area trapezium container and the smallest perimeter trapezium container most different?

4. Can the plane be tiled with mutually non-congruent trapeziums (equally, isosceles trapeziums) having (1)same area and same perimeter OR (2) having same area and same diameter OR (3) same perimeter and diameter?
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And just as with containing trapeziums, we can ask a bunch of questions about largest trapeziums contained within a given convex region.

Some questions on partitioning into Triangles

1. For any n, can any triangle be cut into n non-degenerate triangles all of same diameter?

2. If the answer to (1) is yes, can any convex m-gon be cut into some finite number of triangles all of same diameter and if so, can the least number of such triangular pieces be given a lower bound?

Note: Any triangle can be trivially cut into n non degenerate triangles all of same area; it doesn't appear too difficult to cut any triangle into n triangles all of same perimeter either.

3. Can any triangle be cut into some finite number of triangles all of same perimeter?

Ref: https://math.stackexchange.com/questions/2822589/dissect-square-into-triangles-of-same-perimeter
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We can also consider partitioning triangles (and in general, n-gons) into triangles with:
1. equal circumradius
2.rź equal inradius
3. both radii equal
and so on...