# Planar sets where any line through the center of mass divides the set into two regions of equal area.

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!

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You want to assume that A has positive Lebesgue measure, of course. Nice question. –  Jeff Schenker Jul 20 '10 at 20:48
Thanks! I fixed that. –  Otis Chodosh Jul 20 '10 at 21:15

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 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.
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Thanks! This is exactly what I was looking for –  Otis Chodosh Jul 20 '10 at 21:16
You're welcome. –  Andrey Rekalo Jul 20 '10 at 21:30
@Andrey (or whomever): Is there a higher-dimensional version of this theorem? –  Joseph O'Rourke Jul 20 '10 at 23:25
This may be a naive question, but is there any reason to believe that central symmetry continues to hold for non-convex shapes? –  Victor Protsak Jul 21 '10 at 1:47
I just read the paper, which was pretty neat. I wasn't quite sure what it meant by "continuous mass distribution." Is that just a $\rho:\mathbb{R}^2 \to \mathbb{R}^+$ continuous, and supported in a compact set? Then he is approximating the indicator function of the convex set by continuous functions? I can't quite see how this approximation would work, i.e. how do you prove that the continuous approximate indicator functions have the first property in your list... –  Otis Chodosh Jul 21 '10 at 16:16

For those who found this problem of interest there is a large literature dealing with area bisectors (and a separate literature for perimeter bisectors) of polygons, convex or otherwise (including the allowing of holes in the area case). For example for a triangle it makes a nice exercise to think through where the points in the interior of any triangle are which allow the drawing of exactly one, exactly two, and exactly three area bisecting lines. (Hint: Remember some of the basic properties of the conic sections.) There are also generalizations related to finding points which admit lines which divide the area of the regions cut off by the lines into parts of equal area.

Here are some references that take off from this circle of ideas but there are many more:

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This argument does not require any additional hypothesis, and the proof is cute, so I thought I'd include it. I claim: There is at most one point satisfying the condition you require, that is, that any line through it divides $A$ into areas of equal measure. (We assume that $A$ has positive measure.)

Assume the contrary, and let $a, b$ be two such points. Let $l_a$ be a line passing through $a$ and let $l_b$ be a line passing through $b$, with $l_a, l_b$ parallel, such that the region between $l_a$ and $l_b$ contains a subset $S$ of $A$ with non-zero measure. (Such a pair of lines exists as otherwise $A$ is concentrated on the line connecting $a, b$ and thus has zero measure.) But then in your notation we have $|A|=|A|/2+|A|/2+|S|$ which is impossible. So $a=b$.

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Thats pretty slick, thanks! –  Otis Chodosh Jul 20 '10 at 21:18

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).

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