### 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.