### Regular Octagon

Consider a regular octagon for which the points \(A\), \(B\), \(C\) and \(D\) are consecutive vertices, \(d\) is the length of its second shortest diagonal and \(l\) the length of its side. If we assume 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 second shortest diagonal and the side of the octagon. Note that each side is a chord of the circle circumscribing the polygon, corresponding to an arc measuring \(\frac{1}{8}.360\,^{\circ}=45\,^{\circ}\). Choose the point \(E\) on the diagonal \(\left[ AD\right] \) such that \(\overline{AE}=\overline{AB}=l\). Then, as \(\left[ AEB\right] \) is isosceles and \[E\hat{A}B=D\hat{A}B=\frac{1}{2}.2.45\,^{\circ}=45\,^{\circ},\] we get \[A\hat{E}B=A\hat{B}E=\frac{1}{2}.(180\,^{\circ}-45\,^{\circ})=67,5\,^{\circ}.\] On the other hand, \[C\hat{B}E=C\hat{B}A-A\hat{B}E=\frac{1}{2}.6.45\,^{\circ}-67,5\,^{\circ}=135\,^{\circ}-67,5\,^{\circ}=67,5\,^{\circ}=A\hat{E}B,\] so the diagonal \(\left[ AD\right] \) is parallel to the side \(\left[ BC\right] \). Now choose the point \(F\) on the diagonal \(\left[ AD\right] \) such that \(\overline{EF}=\overline{BC}=l\). Then, since \(\left[ EF\right] \) is parallel to \(\left[ BC\right] \), \([EBCF]\) is a parallelogram, so \[D\hat{F}C=D\hat{E}B=180\,^{\circ}-A\hat{E}B=112,5\,^{\circ}.\] Since \[A\hat{C}D=\frac{1}{2}.5.45\,^{\circ}=112,5\,^{\circ}=D\hat{F}C\] and \(A\hat{D}C=C\hat{D}F\), the triangles \(\left[ ADC\right] \) and \(\left[ CDF\right] \) are similar, so \(\left[ DC\right] \) is the second shortest diagonal of a regular octagon with side \(\left[ DF\right]\). This being so, we can construct a regular octagon passing through the points \(D\), \(F\) and \(C\), and whose side and second shortest diagonal have lengths given by: \[l_{1}=\overline{DF}=\overline{AD}-\overline{AE}-\overline{EF}=d-2l\] \[d_{1}=\overline{DC}=l\]

As \(d=mx\) and \(l=nx\), we have: \[l_{1}=mx-2nx=(m-2n)x=n_{1}x\] \[d_{1}=nx=m_{1}x,\] where \(n_{1}=m-2n\) and \(m_{1}=n\) are two positive integer numbers, with \(m_{1}<m\). Indeed, \[\begin{array}{ccl} m_{1}\geq m & \Rightarrow & n\geq m\\ & \Rightarrow & l\geq d, \end{array}\] which is absurd, since any diagonal in a regular polygon is always larger than its side.

Proceeding in the same way, we obtain a new regular octagon whose second shortest diagonal is \(d_{2}=m_{2}x\) and whose side is \(l_{2}=n_{2}x\), where \(n_{2}=m_{1}-2n_{1}\) and \(m_{2}=n_{1}\) are two positive integer numbers, with \(m_{2}<m_{1}<m\).

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