TITLE: Planar sets where any line through the center of mass divides the set into two regions of equal area.
QUESTION [17 upvotes]: 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!

REPLY [7 votes]: This is to answer a natural and interesting question raised 
by Joseph O'Rourke's in a comment above. Indeed we have:

Any open subset $A$ of $\mathbb{R}^n$, star-shaped wrto a point $P$ and which is  partitioned into two pieces of equal measure by each hyperplane through $P$, is center-symmetric wrto $P$.

Here below I'm describing (with some freedom) the main argument of the proof, that I extracted by this short paper by K.J.Falconer, for those who have no access to Jstor. The key-point is a consequence of the Funk-Hecke theorem: the integral mean of a spherical harmonic $S$ over the hemi-sphere centered at $\theta\in \mathbb{S}^m$, as a function of $\theta$, is a non-zero scalar multiple of $S$ (the F-H theorem says much more; so I think hopefully there is also a  short proof of this fact).
Assume $P$ is the origin, and let $f:\mathbb{S}^{n-1}\to \mathbb{R}_+$ describe the boundary of $A$ in polar coordinates (that is, for $x=r\theta \in \mathbb{R}^n$ with $r\ge 0$ and $\theta\in \mathbb{S}^{n-1}:=\partial B _ {\mathbb{R}^n }(0,1 )$, then $x\in A$ if and only if $r< f(\theta)\, $ ).
By integrating in polar coordinates, the condition on $A$ writes:
$$\int_{(\psi\cdot \theta)\ge0} f(\theta)^n d\theta=\int_{(\psi\cdot \theta)\ge0} f(-\theta)^n d\theta\, , \quad\forall\psi\in\mathbb{S}^{n-1}   $$
and we are to show that this implies that $f^n$, thus $f$ itself, is an even function (the other implication is of course quite obvious, and reflects the fact that a center-symmetric $A$ is equi-partitioned by any hyperplane through the origin). To this end, consider the transformation 
$u\in L^2(\mathbb{S}^{n-1})\mapsto \tilde u\in L^2(\mathbb{S}^{n-1})$ defined by 
$$\tilde u (\psi):=\int_{(\psi\cdot \theta)\ge0} u(\theta)d\theta=\int_\mathbb{{S}^{n-1}} \chi_ { \mathbb{R}_+}(\psi\cdot \theta) u(\theta) d\theta\, , \quad\forall\psi\in\mathbb{S}^{n-1}  \, . $$
Due to the symmetry of the integral kernel $\chi_ { \mathbb{R}_+}(\psi\cdot \theta)$ we have 
$(\tilde u\cdot v) _ {L^2}=(u\cdot \tilde v) _ {L^2}$; moreover, as recalled, spherical harmonics are eigenfunctions of this transformation, with non-zero eigenvalues. Therefore, if $\tilde u$ is even then for any odd spherical harmonic $S$ we have $(u\cdot \tilde S) _ {L^2}=(\tilde u\cdot S) _ {L^2}=0$, thus also $(u\cdot S) _ {L^2}=0$, and $u$ is an even function too, ending the proof.
Moreover, just differentiating with respect to $\psi$ it is easy to see that for a star-shaped open subset $A$ the above property of equipartition by all hyperplanes is equivalent to: any section of $A$ by a hyperplane through $P$ has $n-1$ dimensional barycenter located in $P$ (note that in dimension 2 this immediately implies that $A$ is center-symmetric).