Take a plane curve $\gamma$ and a disk of fixed radius whose center moves along $\gamma$. Suppose that $\gamma$ always cuts the disk in two simply connected regions of equal area. Is it true that $\gamma$ must be a straight line?



Does this property characterize straight lines in the plane?
It seems $\gamma:\mathbb{R}\to\mathbb{R}^2$ (I assume $\gamma$ is injective and continuous) is indeed a line. My argument is very similar to the [one](https://mathoverflow.net/a/377074/172802) by Ilkka Törmä (I thought I could write a shorter one, but it grew longer than expected). As in their answer, I will take the radius from the question to be $1$ and $D(p)$ will denote the disk of radius $1$ centered at $p$.


First suppose that $\gamma$ is bounded. Consider the point $p\_M=(x\_M,y\_M)\in\overline{\gamma(\mathbb{R})}$, with $x\_M$ being the maximal $x$ coordinate of all points of $\overline{\gamma(\mathbb{R})}$, and $y\_M$ the maximal $y\in\mathbb{R}$ such that $(x\_M,y)\in\overline{\gamma(\mathbb{R})}$. Then $D(p\_M)\setminus\overline{\gamma(\mathbb{R})}$ has a connected component of area $>\frac{\pi}{2}$, and the same is true of $D(p)\setminus\overline{\gamma(\mathbb{R})}$ for points $p$ in $\gamma$ close to $p\_M$, so $\gamma$ cannot be as in the question.


So we can assume that $\gamma$ is not bounded. This implies that for any $p\in\mathbb{R}^2$, $\gamma^{-1}(D(p))$ is a union of bounded intervals, and if $p\in\gamma(\mathbb{R})$, as explained in the other [answer](https://mathoverflow.net/a/377074/172802), $\gamma^{-1}(D(p))$ is just one bounded open interval, because if not $\gamma$ would divide $D(p)$ into more than $2$ connected components. Using this, it is not hard to prove by contradiction that $\gamma^{-1}(\overline{D(p)})$ is bounded for all $p\in\gamma(\mathbb{R})$.


**Lemma 1:** For any real numbers $a<b$, $\gamma([a,b])$ is contained in the convex closure of $X:=\gamma([a,b])\cap\left(\overline{D(\gamma(a))}\cup\overline{D(\gamma(b))}\right)$.


*Proof:* Note first that, for any point $p\in\gamma([a,b])\setminus X$, the interval $\gamma^{-1}(D(p))$ is strictly contained in $[a,b]$, so $D(p)$ only intersects $\gamma(\mathbb{R})$ in $\gamma([a,b])$. Also note that, as $X$ is compact, the convex closure of $X$ is the intersection of all closed half planes containing it.


So if the lemma was not true, there would be some closed half plane $H$ containing $X$ but not containing $\gamma([a,b])$. Composing with an isometry if necessary, we can suppose $H=\{(x,y)\in\mathbb{R}^2;x\leq0\}$. Now consider the point $p\_M=(x\_M,y\_M)\in\gamma([a,b])$ given by $x\_M=\max\{x\in\mathbb{R};(x,y)\in\gamma([a,b])\}$ and $y\_M=\max\{y\in\mathbb{R};(x\_M,y)\in\gamma(a,b)\}$. Then $D(p\_M)\setminus\gamma([a,b])$ has a connected component of area $>\frac{\pi}{2}$, but $p\_M\not\in H$, so $p\_M\not\in X$, so $D(p\_M)$ only intersects $\gamma(\mathbb{R})$ in some interval strictly contained in $\gamma([a,b])$. So $D(p\_M)\setminus\gamma(\mathbb{R})$ has a connected component of area $>\frac{\pi}{2}$, a contradiction. $\square$


**Lemma 2:** $\gamma(\mathbb{R})$ is contained between two parallel lines $l,l'$.


*Proof:* For each $v\in\mathbb{S}^1$ let $R\_v$ be the region between the two lines that have direction $v$ and are at distance $2$ of $\gamma(0)$; more precisely, $p\in R\_v$ iff $d(\gamma(0),p)^2-\left(\langle v,p-\gamma(0)\rangle\right)^2\leq4$. Consider the directions $v\_n=\frac{\gamma(n)-\gamma(-n)}{|\gamma(n)-\gamma(-n)|}$. Note that for all naturals $n$, by the previous lemma $\gamma(0)$ is in the convex hull of $\overline{D(\gamma(-n))}\cup\overline{D(\gamma(n))}$, so $\overline{D(\gamma(-n))}\cup\overline{D(\gamma(n))}$ is contained in $R\_{v\_n}$, so by the previous lemma again, $\gamma([-n,n])$ is contained in $R\_{v\_n}$. Then taking a limit point $v$ of the $v\_n$, $\gamma(\mathbb{R})$ has to be contained in $R\_v$: this is because for any $x\in\mathbb{R}$, $\gamma(x)$ is in $\bigcap\_{n\geq|x|}R\_{v\_n}$, that is, $d(\gamma(0),\gamma(x))^2-\left(\langle v\_n,\gamma(x)-\gamma(0)\rangle\right)^2\leq4$ for all $n\geq|x|$, so $d(\gamma(0),\gamma(x))^2-\left(\langle v,\gamma(x)-\gamma(0)\rangle\right)^2\leq4$, that is, $\gamma(x)$ is in $R\_v$. $\square$


So, maybe after a rotation, we can suppose that $\gamma(\mathbb{R})$ is contained between two horizontal lines. Now, let $[a,b]$ be the minimal closed interval such that $\gamma$ is contained in the horizontal strip $\{a\leq y\leq b\}$


**Lemma 3:** There is some $p\in \gamma(\mathbb{R})$ with $y$ coordinate $a$.


If this lemma is true and $\gamma$ satisfies the question, then $\gamma(\mathbb{R})$ has to contain all the line $\{y=a\}$: if not, there would be some $p=(x,a)\in\gamma(\mathbb{R})$ such that the interval $[x-1,x+1]\times\{a\}$ is not contained in $\gamma(\mathbb{R})$, so $D(p)\setminus\gamma(\mathbb{R})$ would have a connected component of area $>\frac{\pi}{2}$, a contradiction.


*Proof:* We will prove it by contradiction, suppose the lemma is not true. By minimality of the interval $[a,b]$, there is a sequence of points $\gamma(t\_n)=(x\_n,y\_n)$ of $\gamma(\mathbb{R})$ such that $y\_n\to a$. The falsehood of lemma 3 implies that $|t\_n|\to\infty$, and then, the facts that $|t\_n|\to\infty$ and that $\gamma^{-1}(D(\gamma(x)))$ is bounded for all $x$ implies that $|x\_n|\to\infty$ (if not, consider an accumulation point of $x\_n$). Also note that $\gamma^{-1}\left(\overline{D(\gamma(0))}\right)$ is bounded, so $\gamma(\mathbb{R})\cap\overline{D(\gamma(0))}$ is contained in some rectangle $R=[c,d]\times[a+\varepsilon,b]$ for some $\varepsilon>0$. Now we will prove that $\gamma([0,\infty))\subseteq\{y\geq a+\delta\}$ for some $\delta>0$ by dividing in two cases; the same proof works for $\gamma((-\infty,0])$, contradicting the minimality of $[a,b]$.


* Case $1$: There are no sequences $x\_n,y\_n,t\_n$ as in the paragraph above such that $t\_n\to\infty$ (that is, we don't always have $t\_n\to-\infty$). Then it is obvious that $\exists\delta>0$ such that $\gamma([0,\infty))\subseteq\{y\geq a+\delta\}$.
* Case $2$: There are sequences $t\_n,x\_n,y\_n$ as above with $t\_n\to\infty$. Then by lemma $1$, $\gamma([0,t\_n])$ is contained for all $n$ in the convex closure of $\gamma([0,t\_n])\cap\left(\overline{D(\gamma(0))}\cup\overline{D(\gamma(t\_n))}\right)\subseteq R\cup\overline{D(\gamma(t\_n))}$. Note that $R$ is a fixed rectangle with $y$-coordinates in $[a+\varepsilon,b]$, and $\overline{D(\gamma(t\_n))}$ is a sequence of disks of radius $1$
centered at points $(x\_n,y\_n)$, where $y\_n\in[a,b]$ $\forall n$ and $|x\_n|\to\infty$. So any fixed point $p=(x,y)$ with $y<a+\varepsilon$ cannot be in the convex hull of $R\cup\overline{D(\gamma(t\_n))}$ for big enough $n$: if that was the case (see $(\*)$ below), then there would be a segment $s$ from some point $r\in R$ to some point $q\in \overline{D(\gamma(t\_n))}$ passing through $p$, so the slope of the segments $qp$ and $pr$ would be the same. However, the slopes of segments from $p$ to points of $R$ are bounded below, while the slopes from $p$ to points of $\overline{D(\gamma(t\_n))}$ converge to $0$ when $n\to\infty$, so for big enough $n$, the segment $s$ cannot exist.
So any point $p=(x,y)\in\gamma([0,\infty))$ has to satisfy $y\geq a+\varepsilon$, as we wanted. $\square$


$(\*)$ I have used that if $X,Y$ are convex subsets of $\mathbb{R}^2$, then any point $p$ in the convex closure of $X\cup Y$ is in a segment between a point of $X$ and a point of $Y$: indeed, as $p$ is in the convex hull of $X,Y$, there are points $x\_i\in X,y\_j\in Y$ and coefficients $a\_i,b\_j\geq0$ with $\sum a\_i+\sum b\_j=1$ such that $p=\sum\_i a\_ix\_i+\sum\_j b\_jy\_j$. Suppose $\sum a\_i,\sum b\_j>0$, if not $p$ is in $X$ or in $Y$. So $p$ is in fact an affine combination of the point $\sum\_i \frac{a\_i}{\sum\_ka\_k}x\_i$ of $X$ and the point $\sum\_j\frac{b\_j}{\sum\_kb\_k}y\_j$ of $Y$: $p=\left(\sum\_ia\_i\right)\left(\sum\_i \frac{a\_i}{\sum\_ka\_k}x\_i\right)+\left(\sum\_jb\_j\right)\left(\sum\_j\frac{b\_j}{\sum\_kb\_k}y\_j\right)$.


