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We continue from Cutting convex regions into equal diameter and equal least width pieces - 2

Question: If a planar convex region C is to be cut into n convex pieces such that the average of the perimeter of the pieces is to be maximum, then will all pieces necessarily have the same perimeter? Or if there are many partitions which maximize the average perimeter, will at least one among them have all pieces having equal perimeter?

Note: If we consider geometric mean instead of arithmetic mean of the perimeter, we have a variant to the above question.

Remark: If we try to minimize the average perimeter, n-1 of the pieces will shrink to points and one piece will be C itself - not too interesting.

Note added on 5th April 2023: Some further thoughts on these lines are recorded at http://nandacumar.blogspot.com/2023/03/convex-partitions-averages-of-quantities.html

Nandakumar R
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    Can't all the pieces have perimeter arbitrarily close to the diameter $d$ of the original region $X$? For example for each $\varepsilon>0$ you can take points $x,y\in X$ at distance $\geq d-\varepsilon$, and then partition $X$ into $n$ convex pieces by $n-1$ lines very close to $x,y$ and parallel to $\overline{xy}$ – Saúl RM Mar 26 '23 at 19:05
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    I think I understand what you say. For example, if say, a circular disk is to be cut into 4 convex pieces such that the average perimeter (sum of perimeters) of pieces is to be maximum, it appears that the best way is to cut into 2 degenerate (zero thickness) pieces with perimeter 2*diameter each and 2 half-disks. So, the 4 pieces do not have the same perimeter. And the answer to the question is "no". Thanks. – Nandakumar R Mar 27 '23 at 07:17

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