# Dividing open domains in $\mathbb R^2$ in parts of equal area

(i) Let $A$ be an open, bounded and connected subset of $\mathbb R^2$. Prove that for every given direction $d$, there exists a line parallel to $d$ that divides $A$ in two parts with the same area.

(ii) Show that if $A,B$ are open, bounded and connected subsets of $\mathbb R^2$ there exists a line which divides both $A$ and $B$ in two parts of equal area.

I'm stuck and I do not know how to begin. What should I do? The first statement if obvious if $A$ is a ball, but I do not know how to treat the general case.

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You can parametrise in a natural way the set of all line parallel to $d$ with a real parameter $x$, and associate a right and a left half-plane to each $x$ in a consistant manner. Then show that the map which associates at each $x$ the area of the intersection between $A$ and the right half-plane associate to $x$ is continuous. Then apply intermediate value theorem. –  Ahriman Aug 21 '12 at 12:53
en.wikipedia.org/wiki/Centroid –  Sean Eberhard Aug 21 '12 at 12:56
See also: en.wikipedia.org/wiki/Ham_sandwich_theorem for a variant. –  t.b. Aug 21 '12 at 14:39
@t.b. Why did you call the ham sandiwch theorem a variant? Isn't it a more general case of the above theorem? Or is there some difference I am missing? –  Jayesh Badwaik Sep 11 '12 at 11:47
@Jayesh: No subtle difference, except that it only generalizes (ii) and is much harder to prove in the generality given on Wikipedia. –  t.b. Sep 11 '12 at 11:51

Both of these rely on the intermediate value theorem.

(i) Define one direction normal to $d$ as up. Draw a line $l_0$ "below" $A$, and one line $l_1$ "above" $A$ (possible because $A$ is bounded) (both parallel to $d$),and call the distance between them $r$. You can define a function $f$ from the interval $[0, 1]$ into $\mathbb R$ taking a number $x$ to the area below the line parallel to $d$ and distance $rx$ above $l_1$. We now have $f(0) = 0$, $f(1) = |A|$, and it is continuous, so somewhere it has to equal $|A|/2$. That's where you put your line.

To solve (ii) we need to know that the line is unique. We therefore need to show that $f$ is monotone, and that at the point where $f(x)= |A|/2$, it is strictly increasing.

The first part is simple, since we only move the line in one direction, points of $A$ can only go from being over the line to being under the line as $x$ increases, so the calculated area cannot decrease.

The second part is a bit more tricky. I'll show the "top part", the bottom part is completely analogus. Given the point $x$ where $f(x) = |A|/2$, for any $\epsilon > 0$ (and small enough that $x+\epsilon < 1$) we must show that $f(x+\epsilon) > f(x)$. Since $A$ is open and connected there will be a point $p$ on the line corresponding to $x$ with $p\in A$ which is the centre of a ball contained in $A$ with radius less than $r\epsilon$. Half this ball will be above the $x$-line, but below the $x+\epsilon$-line, thus $f$ has increased between these two points.

We now know that the line is unique, and can carry on with (ii).

(ii) For each possible direction, find the line that divides $A$ into equal parts by (i). Remark that opposite directions yield the same line, since opposite directions are still parallell. Now pick a direction $d$. Define a direction normal to $d$ as up, and now define a function $g$ from the unit interval to $\mathbb R$ as follows: For a number $x$, turn the direction $d$ (and the orientation "up") a total of $\pi x$ radians counterclockwise and let the value of the function be how much area of $B$ is above the line with that direction, dividing $A$ in half. The key remark here is that $f(0) + f(1) = |B|$, since they are both calculated from the same direction $d$, but with the orientation "up" in opposite direction. You therefore have a function from the unit interval to $\mathbb R$ with $f(0) = |B|/2 + a$ and $f(1) = |B|/2 - a$ for some real number $a$. Thus it has to be $|B|/2$ somewhere in between, and you then have your line.

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The biggest hole in my answer is, I think, that I didn't show that the lines in (ii) vary continuously wrt the direction. It would take some effort and space to prove, and my answer is more than enough a block of text as is. –  Arthur Sep 5 '12 at 19:32
Thanks a lot for your answer. –  Romeo Sep 11 '12 at 11:53

Under the given conditions, the sets have a centre of gravity. One usually learns in multivariate calculus which integrals are used to compute them and the basic property that any line through them splits the areas as you require.

Then for (i) take a line through the centre, parallel to the given direction.

For (ii) take a line through both centres.

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Centre of gravity does not divide sets into equal area. Consider a dumbbell-shaped set with one small disk, one large disk, and a narrow strip connecting them. –  Erick Wong Sep 5 '12 at 19:25
Agreed, the center of gravity does not yield equal areas in general. –  Ralph Tandetzky Sep 6 '12 at 12:09