Given a polygon, connect the midpoints of the sides, in order, to create a new polygon. We'll call this the mediogon of the original polygon. There's a theorem that every mediogon of a quadrilateral is a parallelogram. Proof: opposing sides of the mediogon are parallel to a common diagonal of the original quadrilateral.

The generalization beyond quadrilaterals is: which n-gons are mediogons of some other n-gon? Using vector algebra, the following are very easy to prove:

  1. If n is even, then not all n-gons are mediogons (the class of mediogons can be described in a simple way), but each n-gon that is a mediogon is the mediogon of infinitely many n-gons.

  2. If n is odd, then every n-gon is a mediogon of a unique other n-gon.

The problem is, every time I said "n-gon" above, I'm allowing the possibility that the edges might cross, so it's not a simple n-gon.

Which n-gons are mediogons of some simple n-gon?

My conjecture is that for even n, since every mediogon is a mediogon of infinitely many polygons, one of those polygons is always simple. For odd n, some subset of n-gons must be mediogons of simple n-gons.

The problem is that I don't know any workable characterization of simple polygons. I came up with a rule for when two line segments cross, it's a pair of big cumbersome inequalities involving inner products. I don't think I can apply it to this problem.

  • $\begingroup$ That's a very nice and keen observation. +1. I too have some of my own other conjectures about mediogons. $\endgroup$ – 8hantanu Jan 20 '17 at 14:45
  • $\begingroup$ FWIW a very similar question was recently asked here (requires 10k rep to view). It is currently deleted, pending review of connection to some "MathPath entrance exam (whatever that is)". $\endgroup$ – dxiv Jan 20 '17 at 18:05

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