# Construction of Paragon

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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.)

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Copyright © 1996-2018 Alexander Bogomolny### Construction of Paragon

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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 *parahexagon*. In these two particular cases (Varignon Parallelogram and Parahexagon), we have established the following:

### Theorem

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

### Proof

Let Q_{i} = (P_{i} + P_{i+1} + ... + P_{i + N/2 - 1})/(N/2), the barycenter of P_{i}P_{i+1} ... P_{i + N/2 - 1}, i = 1, ..., N. Obviously,

Q_{i}Q_{i+1} = (P_{i+1} + P_{i+2} + ... + P_{i+N/2})/(N/2) - (P_{i} + P_{i+1} + ... + P_{i+N/2-1})/(N/2) = (P_{i+N/2} - P_{i})/(N/2).

Similarly,

Q_{i+N/2}Q_{i+N/2+1} = (P_{i+N/2+1} + P_{i+N/2+2} + ... + P_{i+N})/(N/2) - (P_{i+N/2} + P_{i+N/2+1} + ... + P_{i+N-1})/(N/2) = (P_{i+N} - P_{i+N/2})/(N/2).

But, since P_{i+N} = P_{i}, Q_{i}Q_{i+1} and Q_{i+N/2}Q_{i+N/2+1} are equal except for the direction - they are parallel and equal in length.

### References

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

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Copyright © 1996-2018 Alexander Bogomolny