Fair Partitions into non-Congruent pieces
We continue the last post here and the old fair partitions story.
Basic question: Given a planar convex region C and integer n, when is it possible to cut the region into n mutually non-congruent convex pieces such that the pieces all have the same area and same perimeter?
Remarks:
- It is quite easy to cut C into n mutually non-congruent pieces of equal area (just slide a line initially tangent to C perpendicular to itself and chop off a piece when the line has passed over 1/n of the area of C. Then, with the remainder of C, start with a tangent inclined at some other odd angle and repeat). Moreover, even for closed surfaces such as a sphere, such an equal area partition into non-congruent pieces can be done (the last post here, linked above).
- However, it is not immediately clear how to achieve 'fair non-congruent partitions' even for a square. It might well be the case that for any C, there are at least some values of n for which such partitions are NOT possible - indeed, for a disk or for that matter, any centrally symmetric region, such a partition obviously does not exist for n =2; maybe for disk, it is not possible for any n>2.
- A weaker condition on the pieces would be: "not all pieces should be mutually congruent". And apart from the original or basic fair partition question, the condition of the non-congruence of pieces could be added to other variants such as the one where diameter and perimeter ought to be equal among n pieces.
- It is known that for example, a quadrilateral with all angles irrational fractions of pi cannot be cut into n mutually congruent pieces for any n (even if the pieces are allowed to be non-convex) (see this link). So, since fair convex partitions do exist for any C for any n (proved by Avvakumov, Akopyan and Karasev), fair convex partition of such a quad for any n will obviously not have all pieces congruent. Some pairs might be congruent though - at least that cannot be ruled out upfront. Or maybe there are indeed regions for which any fair convex partition into any number of pieces will contain only mutually non-congruent pieces (that would probably take some proving!).
Basic question: Given a planar convex region C and integer n, when is it possible to cut the region into n mutually non-congruent convex pieces such that the pieces all have the same area and same perimeter?
Remarks:
- It is quite easy to cut C into n mutually non-congruent pieces of equal area (just slide a line initially tangent to C perpendicular to itself and chop off a piece when the line has passed over 1/n of the area of C. Then, with the remainder of C, start with a tangent inclined at some other odd angle and repeat). Moreover, even for closed surfaces such as a sphere, such an equal area partition into non-congruent pieces can be done (the last post here, linked above).
- However, it is not immediately clear how to achieve 'fair non-congruent partitions' even for a square. It might well be the case that for any C, there are at least some values of n for which such partitions are NOT possible - indeed, for a disk or for that matter, any centrally symmetric region, such a partition obviously does not exist for n =2; maybe for disk, it is not possible for any n>2.
- A weaker condition on the pieces would be: "not all pieces should be mutually congruent". And apart from the original or basic fair partition question, the condition of the non-congruence of pieces could be added to other variants such as the one where diameter and perimeter ought to be equal among n pieces.
- It is known that for example, a quadrilateral with all angles irrational fractions of pi cannot be cut into n mutually congruent pieces for any n (even if the pieces are allowed to be non-convex) (see this link). So, since fair convex partitions do exist for any C for any n (proved by Avvakumov, Akopyan and Karasev), fair convex partition of such a quad for any n will obviously not have all pieces congruent. Some pairs might be congruent though - at least that cannot be ruled out upfront. Or maybe there are indeed regions for which any fair convex partition into any number of pieces will contain only mutually non-congruent pieces (that would probably take some proving!).
posted by R.Nandakumar @ 10:12 PM
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