This question is influenced by the following riddle:



> 
> You are given a rectangular set in the plane with a rectangular hole cut out (in any orientation). How do you cut the region into two sets of equal area?
> 
> 
> 


*SPOILER ALERT!!* - The answer is that you can cut through the center of both rectangles, and because any line through the center of a rectangle divides it into two pieces of equal area, this cut works. 




---


I have been wondering about the following question - What sort of conditions on a set guarantee that it has this property, that any line through the center of mass divides it into two regions of equal area?


The only thing that I have been able to think of is $\pi$ rotational symmetry around the center of mass. For example the rectangle has this symmetry. This symmetry means that in fact the two regions cut by any line through the center of mass are congruent, and not just equal area. 


Thus, my question is:



> 
> Suppose that we have a planar (measurable) set $A \subset \mathbb{R}^2$ (with positive measure). If there is a point $a\in \mathbb{R}^2$ such that: for any line $\ell \subset \mathbb{R}^2$ through $a$, denoting the regions of $A$ on either side of the line $B$ and $C$ then we have $|B| = |C|$ (Lebesgue measure), then is it necessarily true that
>  (1) $a$ is the center of mass of $A$ and (2) that $A$ has $\pi$ rotational symmetry around $A$ in the a.e. sense, i.e. if $\tilde A$ is $A$ rotated by $\pi$ around $a$ then the symmetric difference between $A$ and $\tilde A$ has measure zero, i.e. $ |A \ \Delta \ \tilde A| = 0$.
> 
> 
> 


I feel like (1) should be true, but I'm not so sure about (2). If the answer to (2) is no, then what sort of sufficient conditions are there? I'm mostly just curious about the answer, so by all means feel free to strengthen the assumptions on $A$, like requiring it to be a region bounded by a smooth boundary, etc. Thanks!



Planar sets where any line through the center of mass divides the set into two regions of equal area.
Assume that $A$ is compact and convex. If there is a point $P$ such that any line through it is a bisector of $A$ then $A$ has to be centrally symmetric. In fact a stronger result is known (see the [paper](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-16/issue-1/A-theorem-on-partitions-of-mass-distribution/pjm/1102995091.full) by V. Menon):



> 
> **Theorem.** Let $K$ be a compact convex figure. The following
> four statement are equivalent:
> 
> 
> 



> 
> * the point $P$ through which three bisectors of $K$ pass is unique,
> * all bisectors of $K$ are concurrent in $P$;
> * there exists a point $P$ such that any line through it is a bisector of $K$;
> * $K$ is a centrally symmetric figure with $P$ as its centre.
> 
> 
> 


