Sets cut into two halves of equal size by any straight line through a particular point Is there an easy characterization of all sets $M \subseteq \mathbb{R}^2$ with the following property?
A point $(x_M,y_M)$ (which may depend on $M$) exists such that each straight line through $(x_M,y_M)$ cuts $M$ into two halves of equal Lebesgue measure.
 A: Let's use a polar coordinate system with $M$ as the  origin.  Then the property can be stated as: the area of $M$ within the angular range $\alpha\le \theta\le \alpha+\pi$ is independent of $\alpha$. This area is 
$$\frac12 \int_\alpha^{\alpha+\pi} \int_0^\infty \chi_M(r,\theta) r\,dr\,d\theta$$
where $\chi_M$ is the characteristic function of $M$. Differentiation with respect to $\alpha$ yields
$$I(\alpha) =I(\alpha+\pi) \quad \text{for a.e. } \alpha,  \text{ where }\ I(\alpha) = \int_0^\infty \chi_M(r,\alpha) r\,dr$$
This is a necessary and sufficient condition. It doesn't look particularly geometric, but that's what it is. 
In the special case when the region is star-shaped about $M$, i.e., is described by $r\le f(\theta)$, the condition becomes $f(\theta)=f(\theta+\pi)$ a.e., which is central symmetry (up to a set of zero measure).
A non-centrally symmetric example: annulus-type domain 
$$
\sqrt{1+\cos\theta}\le r\le \sqrt{2+\cos\theta}
$$
pictured below:

Of course,   $\sqrt{f(\theta)}\le r\le \sqrt{A+f(\theta)}$ works as well, for any function $f$ and any constant $A$.
