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While featherbrainedly doodling the other day I noticed that it's probably impossible to draw a convex polygon with a greater perimeter then that of the square around it.

Can someone find a counterexample or maybe even find a proof?

  • To clarify: Are you asking if a convex polygon, inscribed within a square, always has the smaller perimeter? – Semiclassical Jul 22 '14 at 21:20
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    This question was beaten to death in http://mathoverflow.net/questions/71502/circumference-of-convex-shapes. – Moishe Kohan Jul 23 '14 at 05:07
  • @studiosus Well, at some MO-like level, and in greater generality... By the way, there's at least one solution that is specific to having a square (or at least rectangle)... –  Jul 23 '14 at 05:17

2 Answers2

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If a convex set is contained in another set, the convex set has smaller perimeter.

This can be understood by considering the projection map on the convex set (i.e. the map which sends a point to the closer point on the convex set). This is a map which does not increase distances and sends the perimeter of the enclosing set onto the perimeter of the convex set.

A general formal proof of this fact could be obtained by means of the following facts:

  1. The projection onto a convex set is a $1$-lipschitz map
  2. The measure of the image of a set does not increase through a $1$-lipschitz map
  3. The projection of the boundary of a set containing a convex set contains the whole boundary of the convex set.

For your special case (i.e. polygons) you can manage to get an elementary proof by noticing that the projection of a segment on any line is not longer than the segment itself. By subdividing the edges of the outer polygon so that every edge gets projected on a single edge of the inner convex set, you should manage to prove the result.

Emanuele Paolini
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  • Yes, OK $-$ but a justification would be nice. Or a reference, at least. – TonyK Jul 22 '14 at 21:01
  • A Proof of this conjecture? – Nathaniel Bubis Jul 22 '14 at 21:01
  • What's "the projection map", please? – TonyK Jul 22 '14 at 21:03
  • The projection map is the map which sends every point of the plane into the closest point on the convex set. – Emanuele Paolini Jul 22 '14 at 21:05
  • That seems to work! (But shouldn't your 'into' be 'onto'?) – TonyK Jul 22 '14 at 21:11
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    Well thank you! But I guess for me it will be easier to find a proof for this special case than to usably understand what "1-lipschitz-map" is supposed to mean :) –  Jul 22 '14 at 21:29
  • @Levix: A map $f: X\to Y$ between two metric spaces is called $L$-Lipschitz if $d(f(x),f(x'))\le L d(x,x')$ for all $x, x'\in X$. This is a very standard terminology. 1-Lipschitz maps are also called "short maps" or "distance non-increasing maps". – Moishe Kohan Jul 23 '14 at 04:52
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I'll rephrase the argument by Emanuele Paolini without "$1$-Lipschitz" and projection. Place a bomb inside of the convex polygon. When it goes off, the polygon breaks apart at the vertices, with every side of it flying out in the direction in which it was facing: I drew arrows to show the directions in which two of the sides go.

enter image description here

Flying out, each side takes out a chunk of the boundary of the square. Use some geometry to prove that

  1. The chunk taken out by each side is at least as long as the side itself.
  2. The chunks taken out by different sides do not overlap.

Conclusion follows from 1 and 2.

  • By the way, nothing here requires the outer polygon to be a square. –  Jul 23 '14 at 04:40