does there exist a compact, contractible subset of $\mathbb R^2$ that cannot be dissected into $4$ parts of equal area by perpendicular lines?

Question

There was a question asked here, to show that a convex subset of $\mathbb R^2$ can be cut into $4$ pieces of equal area. In an attempt to prove it, I tried to more or less apply a "ham sandwich approach" twice and deduce the answer.

The problem was that although I could find two perpendicular lines that bisected the area of any compact set, I couldn't guarantee that these lines simultaneously divided the set into $4$ equal parts.

A friend of mine suggested considering a continuous map $f: S^1 \times K\to \mathbb R/(x \sim-x) ???$ (we couldn't decide on a co-domain) that took a line given by some angle with respect to the $x$-axis and a point in our convex subset $K$, and outputted the area swept out by a counterclockwise $90^{\circ}$ rotation about this line. Ultimately, we wished to look at the homology of $K$ since it is trivial and derive a contradiction. Yet this doesn't really use enough about convexity, only that $K$ is contractible.

my question: I'm not looking for an answer to the original question, but instead I'm asking if someone can exhibit a compact, contractible subset of $\mathbb R^2$ that cannot be dissected into $4$ equal parts by two perpendicular lines.

Edit 1: Since I deleted my answer, here is my "first impulse," which ended up being incorrect:

Step 1: I claim that a single line can bisect the area of a compact set into two pieces of equal area.

Since we know our set is compact, it is bounded and closed so we take the minimal and maximal $y$ coordinate of our set: $y_{min}$ and $y_{max}$. Then, all we want to do is consider a constant line (parallel to the $x$ axis) that moves up slowly, starting from the least y value and ending at the maximal one.

In other words, some function $\mathbb R^2 \times [0,1] \to \mathbb > R^2$ given by $(x,y,t) \mapsto (x,y_{min}+(y_{max}-y_{min})\cdot t)$. Notice that $(x,y,0)=(x,y_{min})$ and $(x,y,1)= (x,y_{max})$

Now, we consider a new function $g$ whose value at some time $t$ is exactly the area of $X$ under the line (Here, you can take the integral to be the area).

$g$ is continuous, just because the area in any $\epsilon$ amount of time is bounded by the extremal $x$-values.

Now, since the values of $g$ vary between zero and the area of the whole set, we apply the intermediate value theorem, to deduce that one half of the area will be attained at some time $t \in [0,1]$.

step 2: we do the same thing, except we take a vertical line and vary it. These two lines together will dissect our compact set into $4$ parts, and they will be perpendicular.

Answer 1

POSSIBLE SPOILER

These are my thoughts so far on a possible solution to the more general problem. Don't read if you want to avoid polluting your own thoughts. I won't say here whether the remainder of this contains a complete solution -- you'll have to read to find out.

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Consider this sketch of the situation:

The region is blue; a bounding box for the region is drawn in black and parameterized by $uv$ coordinates; the crossing lines are red, and divide the plane into four quadrants named $A, B, C, D$, and the crossing lines are specified by three numbers: the $uv$ coordinates of the crossing, and $s$, which is proportional the angle from the ray that divides the $A$ and $D$ quadrants and the positive $x$-axis. We take this angle divided by $\pi/2$ to get a number $s$ between $0$ and $1$.

Now the set of all possible "crosses" corresponds to $(u,v,s)$ triples with all coordinates ranging from $0$ to $1$. The quadrants themselves are also functions of these coordinates, and I hope it's clear that $$ D(u, v, 1) = A(u, v, 0) $$ because the $D$ quadrant, when the cross is rotated by $\pi/2$ ends up sitting where the $A$ quadrant was before rotation.

Letting $Q$ denote one quarter of the area of the blue region, we can define $$F: [0, 1]\times [0, 1] \times [0, 1] \to \Bbb R^3: (u, v, s) \mapsto (a, b, c) $$ where $a$ is the area of the intersection of $A(u,v,s)$ with the blue region, minus Q. Thus if $A$ cuts out one quarter of the blue region, then $a$ is zero. Similar definitions hold for $b$ and $c$. And if $a, b, c$ are all zero, then $D$ intersects a quarter of the blue region too. I'm going to say that for "nice enough" blue regions of the plane, $a, b, c$ are all continuous functions of $u,v,s$. ("Nice enough" might rule out things like the complement of a cantor-like set, but almost certainly includes everything with the property that the boundary is a finite union of nonintersecting Jordan curves.)

What we're looking for is then a $(u,v,s)$ triple with $F(u,v,s) = (0,0,0)$.

The domain is a cube in 3-space. If I can show that its boundary maps to an essential cube in $ \Bbb R^3 - \{(0,0,0\}$, then I'll know that there's a "cross" that splits the region's area as required.

Let's look at $a(u, v, s)$ on the $s = 0$ square of the unit cube: That square corresponds to moving the center of the cross around the black box, oriented at angle 0. Starting at the $(0,0)$ and going counterclockwise, $a$ goes from $3Q$ to $-Q$ to $-Q$ to $-Q$ to $3Q$. $b$ goes from $-Q$ to $3Q$ to $-Q$ to $-Q$ to $3Q$. $C$ goes from $-Q$ to $-Q$ to $3Q$ to $-Q$ to $-Q$. You can plot this as a quad in $abc$ coordinates, and do the same for each of the other faces of the unit cube. With luck, it'll turn out to be obvious why this ends up defining an essential map of the cube to $\Bbb R^3 - \{(0,0,0)\}$, but I don't have time to draw it all out right now, so I'm abandoning this for now.

Answer 2

I believe the answer is no. Draw a vertical line $A$ that bisects the set and a horizontal line $B$ that bisects the set. If the quarters are equal we are done. Otherwise the pairs of opposite pieces have the same area. Rotate $A$ clockwise, sliding it to keep it dividing the area equally and similarly with $B$. The lines will move continuously as a function of the angle of rotation. If we started with the upper left quadrant too small, after we have rotated $\frac \pi 2$ the corresponding piece is the lower left quadrant, which is too large, so we must have passed through a point where the quadrants were equal.

Answer 3

Actually no real topological constraint is needed. Let $\mu$ be a finite borel measure on $\mathbb R^2$ such that any hyperplane has measure zero (absolutely continuous with respect to lebesgue measure (to get rid of continuity problems.) We will use the intermediate value theorem twice.

The Configuration Space: Consider the lines $ax+by=c$. To each point $(a,b) \in S^1$, we take $$H^+(a,b)=\{ (x,y)\in\mathbb{R}^2\mid ax+by \geq c\},$$

and choose $c$ so that $\mu(H^+(u))=\frac{1}{2}\mu(\mathbb R^2)$ (which is possible by the IVT.) We define $H^{-}$ by flipping the inequality, and note that $H^{+}(-u)=H^{-}(u)$ for all $u \in S^1$.

However, viewing $S^1 \subset \mathbb C$, we can consider multiplication by $i$ (just for ease of notation.) Then $H(u)$ and $H(it)$ are orthogonal and determine quadrants indexed by $(\mathbb Z_2)^2=\{\pm 1, \pm i\}$: which are the intersections $S_{1}=H^+(u)\cap H^+(is)$ etc. Note that this construction is equivariant with respect to rotation:$S_{1}(it)=S_i(u)$.

Because $H(u)$ and $H(iu )$ both bisect, $\mu(S_1(u))=\mu(S_{-1}(u))$ and $\mu(S_i(u))=\mu(S_{-i}(u))$. Hence, it will suffice to find some $u$ so that the measures of $S_1(u)$ and the measure of $S_i(u)$ are equal.

Consider the continuous map $f:S^1 \to \mathbb R$ given by $f(u):=S_1(u) -S_i(u)=S_1(u)-S_1(iu)$. However, $f(iu)=-f(u)$, so the intermediate value theorem ensures a solution.