Under what conditions does every line through the centroid divide in half?

Question

Let $G$ be an enclosed region of the complex plane, such that the property $\iint_G z dA=0$ holds; in other words, such that the center of mass is the origin.

Draw an arbitrarily line $C$ passing through the origin. Now, let the regions $G_1$ and $G_2$ be the parts of $G$ that fall on each side of $C$.

The question I'm trying to figure out is under what conditions defining the shape of $G$ result in $G_1$ and $G_2$ having the same area, regardless of the choice of $C$.

I tried looking at some different possibilities for $G$. Trivially, for a circle, any choice of $C$ is a diameter, thus dividing the circle in half. On the other hand, if $G$ is an equilateral triangle and $C$ runs parallel to one of its sides, then the ratio of areas of $G_1$ and $G_2$ is $5:4$ (or $4:5$).

If $A$ is any shape such that the mapping $\phi: z\rightarrow -z$ maps $A$ to itself, then $A$ is a solution as for any $C$ this mapping will map $G_1$ to $G_2$ and vice versa.

I tried unsuccessfully to find solutions that lacked this rotational symmetry of $\pi$ radians, and came to the hypothesis that none exist. Suppose one such solution $H'$ exists. Because for any two solutions $A$ and $B$, we have that $A\setminus B$ is a solution, then the set of points $P=\{z: z\in H'\land -z\in H'\}$ gives a solution subset of $H'$, and we can consider only the supposed solution $H\equiv H'\setminus P$. However, I'm not sure how to prove that $H$ is an empty set.

Answer 1

Here is an explicit counterexample:

The upper ring has inner and outer radii equal to $1$ and $2$.
The smaller lower ring has inner and outer radii equal to $\frac12$ and $1$.
The larger lower ring has inner and outer radii equal to $a$ and $b$, where $a\approx 1.483542$ and $b\approx 2.109715$ are chosen so that

• $b^2-a^2+1^2-(\frac12)^2=2^2-1^2$, so that the areas match;
• $b^3-a^3+1^3-(\frac12)^3=2^3-1^3$, so that the centroids balance out.

Answer 2

In case your region is a star-shaped area bounded by a curve, the only solutions are those where that curve (and equivalently, the shape) is point-symmetric with respect to the origin. Indeed, if you pick two lines through the origin that differ only by a tiny angle (as in the image below), then we want the two triangli-ish shapes to have equal area. But that does not work out if the two radii differ.

If we drop "star-shaped", additional solutions appear and do not seem to allow a simple description.