One of the problems I am grading this week is as follows: Given a simply connected bounded domain $\Omega$ on $\mathbb{R}^{2}$, prove that there exist a line that separates it into two parts of equal area and perimeter.

This problem can be resolved using intermediate value theorem. However the mathematical writing is a bit complicated. I found students gave erroneous arguments all over the place. Here is my question: Is there a clear, transparent way to prove this at a level the first time real analysis learner would understand? My own proof used some topology concepts, and I do not think I can explain it to the student really well. So I want to ask others for ideas.

A correct proof I have seen is to prove such a line dividing $\Omega$ into equal area exists for any angle $\theta$ by shifting the line. Similarly a line dividing $\Omega$ into equal perimeter exists. So if one take the equation for the difference of the two lines $h(\theta)=f(\theta)-g(\theta)$, there must exist some $\theta$ such that the two lines are the same since rotating changes the sign. But I feel this proof is not really clear. The difference of lines is ambiguous unless one interpret it as difference of equations, and the shifting process for the equation is not so clear. Even though I graded 5/5 for this student, I am wondering if there are clearer better proofs I can explain during office hours.

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    $\begingroup$ I would say that the problem itself is ill-posed. The perimeter of a simply connected bounded subset of $\mathbb R^2$ is not usually well-defined. Even if you specify that its boundary be rectifiable, you still have problems if the line coincides with a straight section of the boundary of non-zero length $-$ which part is credited with the length? Even if you sort that out, you can't use the intermediate value theorem, because the length of the boundary on one side of a line doesn't vary continuously as the line moves. $\endgroup$ – TonyK Nov 16 '13 at 12:17
  • $\begingroup$ @TonyK: Yes, that is a problem. Say the shape is a snowfalke or something. But at here we are assuming $\Omega$ is given and the perimeter is well defined. $\endgroup$ – Bombyx mori Nov 16 '13 at 12:28
  • $\begingroup$ Well you should say that then! But the proof is still invalid. $\endgroup$ – TonyK Nov 16 '13 at 12:29
  • $\begingroup$ @TonyK: You mean the student's proof? It is the closest to "correct" among the ones I have graded. By my standards it is not a totally correct proof, but I think it is okay for the professor. $\endgroup$ – Bombyx mori Nov 16 '13 at 12:36
  • $\begingroup$ As I said, you can't use the intermediate value theorem to show that a line dividing $\Omega$ into equal perimiters exists, because that function is not continuous (when the line is parallel to a straight-line segment of the perimeter). The whole concept is flawed. $\endgroup$ – TonyK Nov 16 '13 at 12:38

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