Prove there is a line that cuts the area in half

Question

Suppose you are given two compact, convex sets A and B in the plane, prove there exists a line such that the area of A and B is simultaneously divided in half.

Can you help me with this proof? What I think I have to do is give a fuction that measures the area and then use intermediate value theorem, but I don't know how to give this function explicitly, I'm a third semester undergrad, so I cannot use very advanced tools.

Thanks for any suggestions.

Answer 1

This is only a rough sketch of an idea, but I think it should work.

Put some kind of Cartesian coordinates down on the plane. For any $\theta \in [0, \pi]$,consider also a coordinate system obtained by rotation through this angle about the origin. Let $f_A$ and $f_B$ be functions defined by the property: the line defined by $x=f_A(\theta)$ in the corresponding coordinate system divides A into two parts of equal areas. Analogously for $B$.

Now set a function $g = f_A - f_B$ at every $\theta$. Since $g(\pi) = -g(0)$, the intermediate value theorem will show it is possible.

Answer 2

Okay, this is convoluted so bear with me.

For any point x in the plane you can construct a line, L, from the origin of the plane through x. For each real number r (positive or negative) you can construct a line perpendicular to L that is r distance from the origin. Precisely one of these lines at a specific distance r from the origin will cut A is half. (Basically, by the intermediate value theorem by considering a function that measures what portion of A is cut by the perpendicular line r distance from the origin. The is contionous and goes from 0 to 100% and is monotonic so there is exactly one, r, where the result is 50%.)

Thus we can define f: Plane -> Real Numbers where f(x) is the real number r where the perpendicular line cuts A in half. This is a continuous function. Define g: Plane -> Real Numbers where g(x) is the real number r where the perpendicular line cuts B in half. Define h(x) = f(x) - g(x).

By the fixed point theorem there must be a point y in the plane where h(y) is 0. Then f(y) = g(x) = some real number r. Find the line that is that is r distance from the origin and perpendicular to the line from y to the origin.
This line will cut both A and B in half.

Answer 3

Here is a proof via the Borsuk Ulam theorem for two dimensions. In fact it uses nothing about convexity, but lets keep compactness for now (this proof works for a broader class of subsets.)

Lemma: ( 2d Borsuk Ulam) Every continuous map $f:S^2 \to \mathbb R^2$, where $f(-\mathbf{x})=-f(\mathbf{x})$ has a zero.

$ax+by=c$ is a line granted that $a,b$ are not simultaneously zero.

The equation can be scaled so that the triple $(a,b,c) \in S^2$, by dividing by the appropriate factor.

Consider a "half space'' determined by a line. Given a point $(a,b,c) \in S^2$, we define $$H^{+}(\mathbf{u}):=\{(x,y) \in \mathbb R^2 \mid ax+by \leq c \}$$

In the most natural way, we extend this definition to include the ``lines at infinity:'' $$H^{+}((0,0,1))= \mathbb R^2 \, \, \,\,\,\,, H^{+}((0,0,-1))=\emptyset$$

Hence, Our "test space'' of all lines becomes the whole sphere $S^2$.

Consider the function $f:S^2 \to \mathbb R^2$ given by $f(\mathbf{x})=(A_1(H^{+}(\mathbf{x}),A_2H^{+}(\mathbf{x}))$**, where $A_1,A_2$ are area functions for the two blobs.

The first thing to notice here, is that $$H^{+}(a,b,c)=\{(x,y) \in \mathbb R^2 \mid ax+by \leq c \},$$ while $$H^{+}(-a,-b,-c)=\{(x,y) \in \mathbb R^2 \mid ax+by \geq c \}.$$

Then, a solution to our problem is a root of $$g(\mathbf{x})=f(\mathbf{x})-f(-\mathbf{x})=0.$$

The Borsuk-Ulam theorem says this exists!

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proof of Lemma: If $f:S^2 \to \mathbb R^2$ is an odd function such that is nonvanishing, we can define $g(x):=f(x)/\|f(x)\|$, where $g:S^2 \to S^1$. However, since this is an odd map, it descends to $ \mathbb RP^2 \to \mathbb RP^1 \cong S^1$, so $g_*:\pi_1(\mathbb R P^2) \to \pi_1(S^1)$ must be trivial, and hence there is a lift to $\mathbb R$, making the map $g_*$ nullhomotopic, contradicting the fact that $g(-x)=-g(x)$.

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** We need to show that this assignment is continuous. Let's make these area functions formal by claiming them to be lebesgue measures on compact sets $\mu_1$ and $\mu_2$. It will suffice to show that for any $u_n \to u \in S^2$, we also have that $f(u_i) \to u$. Note that for some $x \in \mathbb R^2$ that is not on the boundary $\partial H^{+}(u)$ (which has measure zero since it is a line), we will have for sufficiently large $n$, $x \in H^{+}(u_n) \iff x \in H^{+}(u)$. This in turn means that the characteristic function $\chi_u$ on $H^{+}(U)$, we have that $\chi_{u_n} \to \chi(u)$ almost everywhere.

Hence by the compactness assumption, we can apply the dominated convergence theorem:

$$\mu_i(H^{+}u_n)=\lim_{n \to \infty}\int \chi_{ u_n} d \mu_i\to \int \chi_u d \mu_i=\mu_i(H^+(u)),$$

as desired.