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Let $S$ be a simple closed curve in ${\Bbb R}^2$ enclosing a convex region $I$.

Must there exist a straight line which cuts $S$ into two pieces of equal length and also cuts $I$ into two regions of equal area? If so, how can such a line be found?

[If the answer is, "no, because the sandwich might have a pathological boundary", then please also consider the case of non-pathological sandwiches.]

MJD
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  • Do you require $S$ to be rectifiable, or does cutting $S$ into two parts of infinite length count as cutting it into parts of equal lengths? – Daniel Fischer Oct 02 '13 at 16:23
  • If you think that's the most interesting part of the question, then I will not spoil your fun by ruling it out. – MJD Oct 02 '13 at 16:24
  • I don't know what's the most interesting part. If the curve is rectifiable, it might be possible to argue by continuity. But maybe not. – Daniel Fischer Oct 02 '13 at 16:27
  • If the region is convex, as you state in the question, it must have a rectifiable boundary. Consider any other rectifiable curve $C$, say a circle, that contains the region $I$. The projection from $C$ to its closest point on $I$ is a non-expansive map because $I$ is convex, so the length of its image (i.e. the boundary of $I$) is less than the length of $C$, which was finite. Thus @Daniel Fischer's fun is spoiled unless you want to drop the convexity requirement. –  Oct 02 '13 at 17:05
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    @RahulNarain The convexity requirement was added later. That spoils the fun indeed. – Daniel Fischer Oct 02 '13 at 17:08
  • If you drop convexity, I think it is straightforward to construct an example where the length cannot be split by a line (because the 'splitting' line lies along an edge, creating some ambiguity about how the length is split)? – copper.hat Oct 02 '13 at 17:11
  • @RahulNarain: Is it 'well known' that a convex set has a rectifiable boundary? – copper.hat Oct 02 '13 at 17:13
  • @copper.hat: I don't know in general, but I've seen a couple of answers on this site demonstrating the proposition in different ways. Oops, the set being bounded as well as convex is important too, of course, which somehow I left out in my comment. –  Oct 02 '13 at 17:20
  • @RahulNarain: I have seen it in a few places, quoted in passing, but have never seen an accessible (to me) proof. – copper.hat Oct 02 '13 at 17:40

3 Answers3

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Let $$s\mapsto z(s)\qquad(0\leq s\leq L)$$ be the counterclockwise representation of the crust with respect to arc length. Hold a knive over the sandwich connecting the points $z(0)$ and $z(L/2)$, and assume that the area to the right of the knive is more than half of the sandwich. Now turn the knive slowly counterclockwise so that at all times it points from $z(s)$ to $z(s+L/2)$. When we arrive at $s=L/2$ we will have less than half of the sandwich to the right of the knive. By the intermediate value theorem there has to be a position $\sigma\in\ ]0,L/2[\ $ of the knive for which the area of the sandwich is exactly halved.

Christian Blatter
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I am not a mathematician, but here is a handwaving argument that the answer is "yes", which maybe someone else can flesh out. Any line going through the center of mass of the region will cut it in two. Take a north-south line going through this center and call its north intersection with S A and its south intersection B. Call this line L and call the length of the curve from A to B its "right length" and the length of the curve from B to A its "left length". If they're equal we're done. Otherwise, by symmetry say that the left length is smaller than the right length.

Now rotate the line 180 degrees to form L'. What was the left length of L is now the right length of L', and vice versa, so the left length of L' is larger than its right length. By the intermediate value theorem, when we rotated L and turned it into L', at some point the left length and right lengths must have been equal. That line is the desired line.

dfan
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  • I came up with that argument also, but then I wasn't sure that the required central point would necessarily exist. – MJD Oct 02 '13 at 16:53
  • If the set is convex, it is measurable. Since it is bounded, the average $\frac{1}{mC}\int_C x dx$ exists (assuming it is a non-trivial sandwich :-)). – copper.hat Oct 02 '13 at 16:56
  • I think the issue boils down to a definition of length and whether or not it is continuous in some sense. – copper.hat Oct 02 '13 at 16:58
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    "Any line going through the center of mass of the region will cut it in two." Yes, but the two regions may not have equal area. Consider a T shape where the upper bar is extremely long; a horizontal line through the center of mass will find almost all the area above it. What should work instead is to consider the one-parameter family of lines that divide the length into half (by connecting two points halfway around the perimeter) and consider how the area varies. –  Oct 02 '13 at 19:05
  • My understanding that any line going through the center of mass will divide the region into two equal areas; otherwise it would not be the center of mass (if you tried to balance the region on this line, it would fall over). In your case, I would think that the center of mass would be located just barely below the center of that long upper bar, and thus half of the total area would be above it. Am I missing something? – dfan Oct 02 '13 at 19:11
  • I am missing something! The center of mass is weighted by distance from the center. I like your modified solution. – dfan Oct 02 '13 at 19:12
  • Ah, this is just what I was worried about! I wasn't sure that the family of lines that would divide the area in half would be concurrent, and Rahul has pointed out that they might not be. – MJD Oct 02 '13 at 19:16
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  1. Consider S, a closed convex region in R^2. Let us first target bisecting the area. A line can be found in any direction n, which will bisect the region I. This is because: Consider a line which is placed outside S, then this line cuts the region I in a fraction 0:1. Now traverse this line continuosly, and parallel to itself, to other side of S. Now the line cuts the region I in ratio 1:0. Since this is a continuous traversion, there exists a position where the line cuts region in ratio 1:2.
    Now, this argument is true for any given direction n, and for any convex region. Thus there exists a line in every direction which will bisect a convex region I.

Now, for the length of S. Consider line l, which cuts the region I into half has a direction n as it's normal. Let a and b be the lengths of the curves which are towards n and -n respectively. And let d = a-b (difference). Now, we can turn the line continuously and get various n, by maintaining the condition that the line bisects the region I(from above). Thus when normal direction of the line is -n, then d = -(a-b).
Here, either (a-b) or -(a-b) is negative and other is positive. Thus, d has continuosly varied from positive to negative, and thus should have acquired 0(zero) at some position of n. Which states that a = b, and thus the lengths are equal.

Thus, we can find a line which will bisect area and perimeter of a convex region simultaneously.

Aditya
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