On a "Car-Talk" problem

While driving back home I heard an interesting math question from all places, the Car Talk show on NPR.  This   made me think of a generalization of the trick they used and in particular, formulate the following problem.  $\newcommand{\bR}{\mathbb{R}}$

Problem  Determine all, reasonably well  behaved compact domains  $D\subset \bR^2$  with the following  property: any line through the origin divides  $D$ into two regions of equal areas.  We will refer to this as property $\boldsymbol{C}$ (for cut).


I know that "reasonably well behaved" is a rather fuzzy   requirement.  At this moment I don't want to think of Cantor like weirdos.  So let's assume that $D$ is semialgebraic.


We say that a domain $D$ satisfies property $\boldsymbol{S}$ (for symmetry) if it is invariant with respect to the  involution

$$\bR^2\ni (x,y)\mapsto (-x,-y) \in \bR^2. $$


It is not hard to see that

$$\boldsymbol{S}\Rightarrow \boldsymbol{C}. $$

Is the converse true?

 I describe below one situation when this happens.


A special case.    I'll assume that $D$ is semialgebraic,  star-shaped with respect to the origin and satisfies $\boldsymbol{C}$.  We can  then describe $D$ in polar coordinates by an inequality of the form

$$(r,\theta)\in D \Longleftrightarrow  0\leq f(\theta),\;\;\theta\in[0,2\pi], $$

where $f:[0,2\pi]\to (0,\infty)$ is a semialgebraic function  such that $R(0)=R(2\pi)$. We can extend $f$ by $2\pi$-periodicity to a function  $f:\bR\to [0,\infty)$ whose restriction to any finite interval is semialgebraic.

For any $\phi\in[0,2\pi]$ denote by $\ell_\phi:\bR^2\to \bR$ the linear functiondefined by

$$\ell_\phi (x,y)= x\cos\phi +y\sin\phi. $$

Denote by  $A(\phi)$ the area of the region $D\cap \bigl\{ \ell\phi\geq 0\bigr\}$.    Since $D$ satisfies $\boldsymbol{C}$ we deduce

$$A(\phi)=\frac{1}{2} {\rm area}\;(D), $$

so that

$$A'(\phi)=0,\;\;\forall \phi. $$

Observe that

$$ A(\phi+\Delta \phi)-A(\phi) =\int_{\phi+\frac{\pi}{2}}^{\phi+\frac{\pi}{2}+\Delta\phi} \left(\int_0^{f(t)} r dr\right) dt -\int_{\phi+\frac{3\pi}{2}}^{\phi+\frac{3\pi}{2}+\Delta\phi} \left(\int_0^{f(t)} r dr\right) dt. $$

For simplicity we set $\theta=\theta(\phi)=\theta+\frac{\pi}{2}$. We can then rewrite the  above equality  as


$$  A(\phi+\Delta \phi)-A(\phi)=\int_\theta^{\theta+\Delta\theta}\left(\int_0^{f(t)} r dr\right) dt -\int_{\theta+\pi}^{\theta+\pi+\Delta\theta} \left(\int_0^{f(t)} r dr\right) dt.  $$

Hence

$$0=A'(\phi)= \frac{}{2}\Bigl( f(\theta)^2-f(\theta+\pi)^2\Bigr). $$

Hence $f(\theta)= f(\theta+\pi)$, $\forall \theta$. This shows that   $D$ satisfies the symmetry condition $\boldsymbol{S}$.

$$ \ast\ast\ast $$

Here is a simple instance when $\boldsymbol{C}$   does not imply $\boldsymbol{S}$.

Suppose that $D$ is semialgebraic and has the annular description

$$ f(\theta)\leq r\leq  F(\theta). \tag{1}\label{1}$$.


Using the same notations as above we deduce that

$$ 0=A'(\phi)= \frac{1}{2} \Bigl(F^2(\theta)-f^2(\theta)\Bigr)-  \frac{1}{2} \Bigl(F^2(\theta+\pi)-f^2(\theta+\pi)\Bigr). $$


Thus, the domain (\ref{1}) satisfies $\boldsymbol{C}$ iff the function $ G(\theta)=F^2(\theta)-f^2(\theta)$ is $\pi$-periodic.  Note that


$$F(\theta)= \sqrt{f^2(\theta)+ G(\theta)}.$$

If we choose

$$f(\theta)=  e^{\sin \theta},\;\; G(\theta)=e^{\cos 2\theta},$$

 then we obtain the domain bounded by the two closed curves in the   figure below. This domain obviously violates the  symmetry condition $\boldsymbol{S}$.