# Fair convex partitioning |26 September 2021|tags: math.OC

When learning about convex sets, the definitions seem so clean that perhaps you think all is known what could be known about finite-dimensional convex geometry. In this short note we will look at a problem which is still largely open beyond the planar case. This problem is called the fair partitioning problem.

In the -dimensional case, the question is the following, given any integer , can a convex set be divided into convex sets all of equal area and perimeter. Differently put, does there exist a fair convex partitioning, see Figure 1.

 Figure 1: A partition of into fair convex sets.

This problem was affirmatively solved in 2018, see this paper.

As you can see, this work was updated just a few months ago. The general proof is involved, lets see if we can do the case for a compact set and .

First of all, when splitting (into just two sets) you might think of many different methods to do so. What happens when the line is curved? The answer is that when the line is curved, one of the resulting sets must be non-convex, compare the options in Figure 2.

 Figure 2: A partitioning of the square in into two convex sets can only be done via a straight cut.

This observation is particularly useful as it implies we only need to look at two points on the boundary of (and the line between them). As is compact we can always select a cut such that the resulting perimeters of and are equal.

Let us assume that the points and in Figure 3.(i) are like that. If we start moving them around with equal speed, the resulting perimeters remain fixed. Better yet, as the cut is a straight-line, the volumes (area) of the resulting set and change continuously. Now the result follows from the Intermediate Value Theorem and seeing that we can flip the meaning of and , see Figure 3.(iii).

 Figure 3: By moving the points and we continuously change and .