# From Arbitrary Hexagon to Regular One

### What Might This Be About?

24 April 2015, Created with GeoGebra

### Problem

Let $A_1A_2A_3A_4A_5A_6$ be a hexagon. For $i=1,\ldots,6,$ form linear combinations (where indices are taken modulo $6)$:

$\displaystyle B_{i}=\frac{A_{i}+2A_{i+1}+2A_{i+2}+A_{i+3}}{6}.$

Then $B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}$ is a parahexagon.

### Solution

$\begin{align}\displaystyle B_{i}B_{i+1}&=\frac{A_{i+1}+2A_{i+2}+2A_{i+3}+A_{i+4}}{6}-\frac{A_{i}+2A_{i+1}+2A_{i+2}+A_{i+3}}{6}\\ &=\frac{A_{i+3}+A_{i+4}-A_{i}-A_{i+1}}{6}. \end{align}$

Similarly,

$\begin{align}\displaystyle B_{i+3}B_{i+4}&=\frac{A_{i+3}+2A_{i+4}+2A_{i}+A_{i+1}}{6}-\frac{A_{i+3}+2A_{i+4}+2A_{i+5}+A_{i}}{6}\\ &=\frac{A_{i}+A_{i+1}-A_{i+3}-A_{i+4}}{6}. \end{align}$

Which shows that, as regular segments, $B_{i}B_{i+1}$ and $B_{i+3}B_{i+4}$ are equal and parallel. Since this is true for any $i,$ $B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}$ is indeed a parahexagon.

### Corollary

The centers of regular hexagons erected either internally or externally on the sides of $B_{1},B_{2},B_{3},B_{4},B_{5},B_{6}$ form a regular hexagon.

This a consequence of the above and the Napoleon-Barlotti theorem.

### Acknowledgment

The above construction of the regular hexagon is by Dao Thanh Oai (Vietnam) who posted it at the CutTheKnotMath facebook page.

### Extension

It was shown elsewhere, six points

$\displaystyle C_{i}=\frac{A_{i}+A_{i+1}+2A_{i+2}}{3},$ $i=1,\ldots,6$

also form a parahexagon.

Now, as is easily seen, it is possible to define addition of two parahexagons $X_{1}X_{2}X_{3}X_{4}X_{5}X_{6}$ and $Y_{1}Y_{2}Y_{3}Y_{4}Y_{5}Y_{6}$ by defining six vertices

$\displaystyle Z_{i}=\frac{X_{i}+Y_{i}}{2},$ $i=1,\ldots,6.$

Dao's statement is then obtained as the sum of $C_{1}C_{2}C_{3}C_{4}C_{5}C_{6}$ and $C_{2}C_{3}C_{4}C_{5}C_{6}C_{1}.$ In the same spirit, we may add $C_{1}C_{2}C_{3}C_{4}C_{5}C_{6}$ to $C_{3}C_{4}C_{5}C_{6}C_{1}C_{2}$ which results in a five-term expression:

$\displaystyle D_{i}=\frac{A_{i}+A_{i+1}+2A_{i+2}+A_{i+3}+A_{i+4}}{6},$ $i=1,\ldots,6.$

By adding already constructed parahexagon, it is easy to obtain new ones. We may expand the notion of the sum of two parahexagons to include taking an arbitrary linear combination of their vertices. E.g.,

$\displaystyle E_{i}=\frac{2A_{i}+5A_{i+1}+5A_{i+2}+3A_{i+4}}{15},$ $i=1,\ldots,6$

represents an additional parahexagon and, hence, a way to construct another (or two) regular hexagons from the same set of six points.

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