Extending Fair Partitions to Spiral Planar Regions
The original fair partition (spicy chicken) question, posed in 2006, asks if given any positive integer n, every planar convex region can be cut into n convex pieces all of same area and perimeter. After a long series of developments, the question was fully answered in the affirmative in 2018 by Avvakumov, Akopyan and Karasev. The results they obtained are far more general than for area-perimeter.
Definition: a spiral polygon is a simple polygon whose boundary chain contains exactly one concave subchain. We can classify the points on the boundary of a closed planar region as either convex or concave depending on where the center of curvature lies. By generalization, we can define a spiral region which has only one continuous stretch of concave points on its boundary.
Question: For any n, an every planar spiral region be cut into n spiral regions all of the same area and perimeter?
For n = 2, the answer seems "yes". We now try to prove this.
Proof: We use intermediate value theorem (as was done in the n=2 case of the basic fair partition question). We only need to show that there is a continuum of cuttings of the spiral planar region S into 2 spiral regions of same perimeter (not area) and thereafter, we can always say that due to continuity, at least one partition in the continuum also yields 2 equal area pieces.
For every point P on the boundary of S, there is another boundary point Q such that the perimeter of S is divided into 2 equal portions by P and Q. Case 1: If P is somewhere on the concave portion of S, Q is not on this concave portion (easy to see) whereas
Case 2: if P is not on the concave portion, Q may or may not be on the concave portion.
(1) If both P and Q are not on the concave portion, P can always be connected to Q with a curve such that S is divided into 2 sprial regions (easy to see).
(2) If one of P and Q are on the concave portion, we assume without loss of generality, that P is on the non-convex portion of the boundary. Now, it seems clear that the two points can be joined by either a straight line or with a suitable curve such that the two resulting pieces are both spiral. Indeed, If P and Q are mutually visible thru interior of S, we connect P and Q with a straight line; then, the two resulting pieces will obviously have one concave portion each and these portions join at Q in S (so both are spiral); else (if P and Q are not mutually visible), we connect P and Q with a dividing curve that has to meet the concave portion of S at Q at a tangent and should also be bent suitably.
I am not sure about this but our proof of the spicy chicken problem for values of n that are powers of 2 (https://www.ias.ac.in/article/fulltext/pmsc/122/03/0459-0467) *might* generalize for the above spiral version of the question as well.
A Weaker version: If we slightly relax the definition of spiral region to one with *at most one concave subchain* thereby making convex regions trivially spiral, can we say that for any n, we can cut any spiral region into n spiral (in this relaxed sense) regions of same area and perimeter?
Definition: a spiral polygon is a simple polygon whose boundary chain contains exactly one concave subchain. We can classify the points on the boundary of a closed planar region as either convex or concave depending on where the center of curvature lies. By generalization, we can define a spiral region which has only one continuous stretch of concave points on its boundary.
Question: For any n, an every planar spiral region be cut into n spiral regions all of the same area and perimeter?
For n = 2, the answer seems "yes". We now try to prove this.
Proof: We use intermediate value theorem (as was done in the n=2 case of the basic fair partition question). We only need to show that there is a continuum of cuttings of the spiral planar region S into 2 spiral regions of same perimeter (not area) and thereafter, we can always say that due to continuity, at least one partition in the continuum also yields 2 equal area pieces.
For every point P on the boundary of S, there is another boundary point Q such that the perimeter of S is divided into 2 equal portions by P and Q. Case 1: If P is somewhere on the concave portion of S, Q is not on this concave portion (easy to see) whereas
Case 2: if P is not on the concave portion, Q may or may not be on the concave portion.
(1) If both P and Q are not on the concave portion, P can always be connected to Q with a curve such that S is divided into 2 sprial regions (easy to see).
(2) If one of P and Q are on the concave portion, we assume without loss of generality, that P is on the non-convex portion of the boundary. Now, it seems clear that the two points can be joined by either a straight line or with a suitable curve such that the two resulting pieces are both spiral. Indeed, If P and Q are mutually visible thru interior of S, we connect P and Q with a straight line; then, the two resulting pieces will obviously have one concave portion each and these portions join at Q in S (so both are spiral); else (if P and Q are not mutually visible), we connect P and Q with a dividing curve that has to meet the concave portion of S at Q at a tangent and should also be bent suitably.
I am not sure about this but our proof of the spicy chicken problem for values of n that are powers of 2 (https://www.ias.ac.in/article/fulltext/pmsc/122/03/0459-0467) *might* generalize for the above spiral version of the question as well.
A Weaker version: If we slightly relax the definition of spiral region to one with *at most one concave subchain* thereby making convex regions trivially spiral, can we say that for any n, we can cut any spiral region into n spiral (in this relaxed sense) regions of same area and perimeter?
posted by R.Nandakumar @ 12:59 AM
0 comments
