The diameter of a convex region is the greatest distance between any pair of points in the region. The least width of a 2D convex region can be defined as the least distance between any pair of parallel lines that touch the region.

  1. Given a positive integer $n$, can every 2D convex region $C$ be divided into $n$ convex pieces, all of the same diameter? The pieces ought to be non-degenerate and have finite area.

  2. If the answer to 1 is yes, how does one minimize the common diameter of the $n$ pieces?

  3. For any $n$, can any $C$ be divided into $n$ convex nondegenerate pieces, all of the same least width?

  4. If 4 has a "yes" answer, how does one maximize the common least width of the $n$ pieces?

These questions have obvious analogs in higher dimensions and other geometries.

Note added on 15th November 2020: As I have just come to know, both question 1 and 3 (existence of partitions into n pieces all of same diameter and into n pieces all of equal least width) have affirmative answers. They follow from the work of Avvakumov, Akopyan and Karasev: Convex fair partitions into an arbitrary number of pieces.

However, the existence proof for $n$ pieces all of same diameter (or same least width) does not directly yield an algorithm to determine a partition with that property.

  • 4
    $\begingroup$ This is too many questions for one post. I suggest moving the questions about widths to another post instead. $\endgroup$
    – Matt F.
    Nov 12, 2020 at 2:50
  • $\begingroup$ The questions are closely related and as I just learned, 2 of them have answers at the same source - am editing the question above accordingly. Hope that would be okay. $\endgroup$ Nov 14, 2020 at 19:02
  • $\begingroup$ ok. questions 3 and 6 which seem simpler and more intuitive can be shifted into another question (basically qns 1-3 are on diameter and 4-6 on least width. so one from each set goes to another post) $\endgroup$ Nov 17, 2020 at 11:03

1 Answer 1


This is not a complete answer to all 6 questions, but provides some progress:

1. I assume that you want $C$ to be bounded here? Your language throughout the post seems to assume it. If not, then either $C$ is either a ray or a line or the pieces can be chosen to all have infinite diameter.

If $C$ is bounded, then when $n=2$, this can always be done*; fix a direction of a line, and consider the family of dissections given by cutting $C$ along translates of this line. The diameters of the two resulting pieces are monotonic with the position of the line, and continuous except possibly where part of $C$'s boundary forms a line segment parallel to the line (which happens in at most countably many cases), so there is some line orientation for which the diameters of the pieces must be equal as it slides along $C$.

3. This is not the case; take $C$ to be a unit equilateral triangle, and $n=2$. Then some piece must contain at least two vertices of the triangle, so the diameter must be at least $1$. But there are dissections attaining this bound where one piece is arbitrarily small: just cut it in two via a line parallel to one side and very close to the opposite vertex.

*I am not sure what you take to be a "nondegenerate" convex set, so I can't guarantee that the resulting pieces will have this property. If you mean something like "its intersection with any open ball is either empty or has positive area", then this will only happen if the convex set is a line segment or a point, and the construction given above avoids such dissections unless $C$ itself is similarly degenerate.

  • $\begingroup$ I had not really thought about unbounded convex regions. Point 2 you make above I think is fine - that for n = 2 (2 pieces) one can cut any C into 2 pieces of same diameter. But point 3 I do not understand. Pls note I have added a bit to the question. Both existence questions (1 and 4) do have "yes" as answer. $\endgroup$ Nov 14, 2020 at 19:15

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