Construction of Paragon
What is it about?
A Mathematical Droodle


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(The applet allows one to toy with N-gons, N even. To change N, click a little off its vertical center line.)

Explanation

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Copyright © 1996-2018 Alexander Bogomolny

Construction of Paragon


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

By analogy with the parahexagon, let's call an N-gon (N even) with opposite sides parallel and equal Paragon. I do not believe there is an established terminology for such polygons. In all likelihood, the term has been coined by Kasner and Newman in their Mathematics and the Imagination. Eves designates such polygons parpolygons. For N = 4, we have a parallelogram and, for N = 6, a parahexagon. In these two particular cases (Varignon Parallelogram and Parahexagon), we have established the following:

Theorem

For an arbitrary N-gon, N even, let Qi, 1, ..., N, denote the barycenter of N/2 successive vertices Pi, Pi+1, ..., Pi + N/2 - 1, where the indices are cyclic, i.e. computed modulo N. Then the N-gon Q1...QN is a Paragon.

Proof

Let Qi = (Pi + Pi+1 + ... + Pi + N/2 - 1)/(N/2), the barycenter of PiPi+1 ... Pi + N/2 - 1, i = 1, ..., N. Obviously,

QiQi+1 = (Pi+1 + Pi+2 + ... + Pi+N/2)/(N/2) - (Pi + Pi+1 + ... + Pi+N/2-1)/(N/2) = (Pi+N/2 - Pi)/(N/2).

Similarly,

Qi+N/2Qi+N/2+1 = (Pi+N/2+1 + Pi+N/2+2 + ... + Pi+N)/(N/2) - (Pi+N/2 + Pi+N/2+1 + ... + Pi+N-1)/(N/2) = (Pi+N - Pi+N/2)/(N/2).

But, since Pi+N = Pi, QiQi+1 and Qi+N/2Qi+N/2+1 are equal except for the direction - they are parallel and equal in length.

References

  1. H. Eves, A Survey of Geometry, Allyn and Bacon, 1972
  2. E. Kasner and J. Newman in their Mathematics and the Imagination, Dover Publications (March 28, 2001)

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