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Prove/Disprove:

$\forall S' \subset S\subset\mathbb R^2$ such that $S',S$ are convex and have finite area, the perimeter of $S'$ is smaller than the perimeter of $S$.

e.g. $S$ could be the unit square and the claim suggests that any convex shape inside the square has perimeter $\leq 4$.

user26857
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R B
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1 Answers1

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Yes, and it works in $d$-dimensions with $d-1$-dimensional area of the boundary instead.

Successively chop off portions from $S$ thereby reducing its size of the boundary by using supporting hyperplanes to $S'$. We get smaller and smaller figures $S_i$ still containing $S'$.

Imagine cutting a young coconut, the hard core being $S'$.

Note: we only need $S'$ convex.

orangeskid
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  • Does it work if $S'$ is not a polygon? If $S'$ is a circle when you'll need infinite number of chopping steps and I'm not sure it works. – R B Sep 29 '14 at 11:37
  • @R B: It would work in the limit, but then you need continuity for the perimeter or boundary size as function of the set. Another method bypassing these choppings is to consider projections onto a variable hyperplane and take the area, then integrate and get the boundary area. – orangeskid Sep 29 '14 at 11:44
  • @orangeskid, your approach with projections sounds interesting! Could you elaborate to make it clear how it works? – ByteEater Oct 22 '21 at 00:11
  • @ByteEater: Think you are slicing a piece of an apple. The new will have a smaller area ( you are substituting a curved piece with a straight piece). Do this several times till you reach the convex core inside ( in the limit at least, if the convex core is not a polytope). That is the intuition. – orangeskid Oct 22 '21 at 00:18
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    @ByteEater: Oh, about the projection method: the area of a convex set is proportional ( positive constant is not relevant here) to the integral of the areas of all the projections onto planes of direction $\perp s$, where $s \in S^2$. In the plane ( with the perimeter), it's the Crofton formula, works in higher dimensions too https://en.wikipedia.org/wiki/Crofton_formula – orangeskid Oct 22 '21 at 00:35
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    @ByteEater: Crofton formula works for regions that are not nec convex. But then we need to count the foldings in projections. But it's based on the formula for a flat surfaces, and approximating a surface (convex or not) with a polytope ( maybe not connected). (some care is required when approximating-- see Schwarz area example). – orangeskid Oct 22 '21 at 00:43