Dividing a region in two regions of the same area

Question

Let $\mathscr C$ be a Jordan curve, i.e. a non-self-intersecting continuous loop in the plane. We call $\mathcal R$ the region inside $\mathscr C$ (the region $\mathcal R$ is well-defined because of the Jordan curve theorem).

The curve $\mathscr C$ and the region $\mathcal R$ should look like something like this:

Let $P$ be a point of the plane.

Does there exist a line $\mathscr D$ such that $P\in \mathscr D$ and $\mathscr D$ divides the region $\mathcal R$ in two smaller regions of the same area?

If such a line exists, when is it unique?

For the first question, I would tend to think the answer is yes, since the curve $\mathscr C$ is continuous and because of the intermediate value theorem.

For the second question, we know that the line $\mathscr D$ can not be unique if $P$ is inside $\mathcal R$.

For example, if $\mathscr C$ is a circle and $P$ is its center, every line $\mathscr D$ works:

I do think the line $\mathscr D$ is unique if $P$ is outside the convex hull of $\mathcal R$.

Answer 1

I will give you an answer to a very related question. This can be found in Frédéric Testard's book $\textit{Analyse mathématique - la maîtrise de l'implicite}$.

A subset $X$ of the plane will be said strongly connected if you cannot find two parallel lines such that above one of them, the area of $X$ is non-zero, below the other one, the area of $X$ is non-zero, but between the two lines, the area of $X$ is zero.

We will also say that a line $D$ is of direction $\theta$ if $(\cos(\theta),\sin(\theta))$ is a director vector of $D$.

We can now give the

$\textbf{Theorem.}$ Let $A$ be a non-empty, compact and strongly connected subset of the plane with strictly positive measure. Then for each $\theta$, there exists a unique line $D_{\theta}$ which divides $A$ into two regions of the same area.

$\textit{Proof.}$ Let's $a$ be the area of $A$.

Since the problem is rotation-invariant, we just need to show the result for $\theta = 0$. A line of direction $0$ has an equation of type $y=c$ for a certain $c$.

That's why, for every $c$, we will note $\mathcal{A}_+(c) = \{(x,y) \in A,\ y \geq c\}$ and $\Phi (c)$ the area of $\mathcal{A}_+(c)$. We now want to show that there exists a unique $c$ such that $\Phi(c) = \frac{a}{2}$.

First, we show the existence of such a $c$. For $A$ is compact, and so bounded, we can consider an $M$ such that for every $(x,y)$, if $(x,y) \in A$, then $|x| \leq M$ and $|y| \leq M$. For $\Phi(-M) = a$ and $\Phi(M) = 0$, we just have to see that $\Phi$ is continuous to conclude with the intermediate-values theorem.

In fact, it is easy to see that $\Phi$ is $2M$-lipschitzian : if $c'>c$, then the part of $A$ between the lines of respective equations $y=c'$ and $y=c$ is included in $[-M,M] \times [c,c']$ and this rectangle is of area $2M(c'-c)$. Then we have $$0 \leq \Phi(c) - \Phi(c') \leq 2M(c' -c).$$

Note that for the proof of this existence, we didn't need to use the strongly connexion hypothesis : that one is useful only for the unicity part, that we write now :

Suppose that you have $c_1 > c_2$ such that $\Phi(c_1) = \Phi(c_2) = \frac{a}{2}$. Then the area of $\mathcal{A}_+(c_1)$ is the same as the area of $\mathcal{A}_+(c_2)$ and equals $\frac{a}{2}$. Taking the complementary, you have the area of $\mathcal{A}_-(c_1)$ (trivial notation) which equals $\frac{a}{2}$ too. But since $c_1 \neq c_2$, you have $\mathcal{A}_+(c_2) \cap \mathcal{A}_-(c_1) = \emptyset$ and so, the area of $\mathcal{A}_+(c_2) \cup \mathcal{A}_-(c_1)$ is $a$. This contradicts the strongly-connexion hypothesis. $\square$

$\textbf{Remark}$ This theorem can be used to prove the following fact : if you fix $A$ and $B$ two non-empty compact and connected sets of the plane with strictly positive measure, then there exist a line which divides $A$ in two regions of the same area and $B$ in two regions of the same area.

Answer 2

Look at pages 76/77. This is the answer for your question.