## Incommensurability

### Regular Hexagon

Let $$[ABCDEF]$$ be a regular hexagon, $$d$$ the length of its shortest diagonal and $$l$$ the length of its side. Assuming that $$d$$ and $$l$$ are commensurable, we have $$d=mx$$ and $$l=nx$$, where $$m$$ and $$n$$ are positive integers and $$x$$ is a common measure to the shortest diagonal and the side of the hexagon.

Let $$G$$ be the point on the diagonal $$\left[ AC\right]$$ such that $$\overline{CB}=\overline{CG}$$, $$I$$ the point of intersection of the line $$AB$$ with the line $$CD$$ and $$H$$ the point on the line segment $$\left[ AI\right]$$ such that $$H\hat{C}I=H\hat{C}A$$. Note, first, that the measure of the interior angle of a regular hexagon is $$120\,^{\circ}$$. Then, since $$A\hat{B}C=B\hat{C}D=120\,^{\circ}$$, we get $C\hat{B}I=B\hat{C}I=180\,^{\circ}-120\,^{\circ}=60\,^{\circ},$ so $$\left[ BCI\right]$$ is an equilateral triangle. As to the triangles $$\left[ CHI\right]$$ and $$\left[ CHG\right]$$, we have that $H\hat{C}I=H\hat{C}A=H\hat{C}G,$ $\overline{CI}=\overline{CB}=\overline{CG}$ and $$\left[ CH\right]$$ is a common side to both, so they are congruent, with $$\overline{HI}=\overline{HG}$$ and $C\hat{G}H=C\hat{I}H=C\hat{I}B=60\,^{\circ}.$ Therefore, $A\hat{G}H=180\,^{\circ}-60\,^{\circ}=120\,^{\circ},$ $H\hat{A}G=B\hat{A}C=\frac{1}{2}(180\,^{\circ} - A\hat{B}C)=30\,^{\circ}$ (note that $$\left[ ABC\right]$$ is isosceles, since $$\overline{AB}=\overline{BC}$$) and $A\hat{H}G=180\,^{\circ}-A\hat{G}H-H\hat{A}G=180\,^{\circ}-120\,^{\circ}-30\,^{\circ}=30\,^{\circ},$ so $$\left[ AGH\right]$$ is isosceles, with $$\overline{AG}=\overline{HG}$$. Therefore we can construct a new hexagon $$[AGHJKL]$$, passing through the points $$A$$, $$G$$ and $$H$$. Denoting by $$d_{1}$$ the length of its diagonal and $$l_{1}$$ the length of its side, we have: $l_{1}=\overline{AG}=\overline{AC}-\overline{GC}=\overline{AC}-\overline{BC}=d-l$ $d_{1}=\overline{AH}=\overline{AI}-\overline{HI}=\overline{AB}+\overline{BI}-\overline{HG}=\overline{AB}+\overline{BC}-\overline{AG}=2l-(d-l)=3l-d$

Since $$d=mx$$ and $$l=nx$$, we get: $l_{1}=mx-nx=(m-n)x=n_{1}x$ $d_{1}=3nx-mx=(3n-m)x=m_{1}x$ where $$n_{1}=m-n$$ and $$m_{1}=3n-m$$ are two positive integer numbers, with $$n_{1}<n$$. Indeed, $\begin{array}{ccl} n_{1}\geq n & \Rightarrow & m-n\geq n\\ & \Rightarrow & m\geq 2n\\ & \Rightarrow & mx \geq 2nx\\ & \Rightarrow & d \geq 2l, \end{array}$ which is absurd, since, in any triangle, each side has length less than the sum of the lengths of the other two sides; considering, for example, the triangle $$\left[ ABC\right]$$, we have $$d=\overline{AC}<\overline{AB}+\overline{BC}=2l$$.

Proceeding in the same way with the hexagon $$[AGHJKL]$$, we obtain a new regular hexagon whose diagonal is $$d_{2}=m_{2}x$$ and whose side is $$l_{2}=n_{2}x$$, where $$n_{2}=m_{1}-n_{1}$$ and $$m_{2}=3n_{1}-m_{1}$$ are two positive integer numbers, with $$n_{2}<n_{1}<n$$.

Since we can go on indefinitely building new regular hexagons, we obtain a strictly decreasing sequence of positive integer numbers: $n_{1}>n_{2}>n_{3}>n_{4}>...$ such that $$n_{i}<n,\:\forall i\in\mathbb{N},$$ which is absurd, as there cannot be an infinite number of distinct positive integers less than $$n$$ (there exist exactly $$n-1$$ such numbers: $$1$$, $$2$$, $$3$$,..., $$n-2$$ and $$n-1$$). Therefore, the shortest diagonal and the side of $$[ABCDEF]$$ are incommensurable magnitudes.