ATRACTOR

Consider a regular dodecagon for which the points \(A\), \(B\), \(C\) and \(D\) are consecutive vertices, \(d\) is the length of its second shortest diagonal and \(l\) is the length of the 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 for the second shortest diagonal and the side of the decagon. Note that each side is a chord of the circle with centre \(O\) circumscribing the polygon, corresponding to an arc measuring  \(\frac{1}{12}.360\,^{\circ}=30\,^{\circ}\). 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 \[E\hat{A}B=D\hat{A}B=\frac{1}{2}.2.30\,^{\circ}=30\,^{\circ},\] we get \[A\hat{E}B=A\hat{B}E=\frac{1}{2}.(180\,^{\circ}-30\,^{\circ})=75\,^{\circ}.\] On the other hand, \[C\hat{B}E=C\hat{B}A-A\hat{B}E=\frac{1}{2}.10.30\,^{\circ}-75\,^{\circ}=150\,^{\circ}-75\,^{\circ}=75\,^{\circ}=A\hat{E}B,\] 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 \(\left[ BC\right] \), \(\left[ EBCF\right] \) is a parallelogram, so \[D\hat{F}C=D\hat{E}B=180\,^{\circ}-A\hat{E}B=105\,^{\circ}.\] Since \[A\hat{C}D=\frac{1}{2}.9.30\,^{\circ}=135\,^{\circ}>105\,^{\circ}=D\hat{F}C,\] we want to have a point \(G\) on the half-line \(DC\) such that \(D\hat{F}G=135\,^{\circ}\). Let \(G\) be the point on the half-line \(DC\) such that \(D\hat{E}G=90\,^{\circ}\). Since \(\left[ AE\right] \) is parallel to \(\left[ BC\right] \) and \(\overline{AE}=\overline{BC}\), \(\left[ ABCE\right] \) is a parallelogram, so \(D\hat{E}C=D\hat{A}B=30\,^{\circ}\). Therefore we get \[E\hat{D}C=A\hat{D}C=\frac{1}{2}.2.30\,^{\circ}=30\,^{\circ}=D\hat{E}C\] and \(\left[ ECD\right] \) is an isosceles triangle, with \[\overline{DC}=\overline{EC}=\overline{AB}=l.\] Since \[D\hat{G}E=180\,^{\circ} - D\hat{E}G - E\hat{D}G=180\,^{\circ}-90\,^{\circ}-30\,^{\circ}=60\,^{\circ}\] and \[E\hat{C}G=D\hat{E}C+E\hat{D}C=60\,^{\circ},\] the triangle \(\left[ EGC\right] \) is equilateral, with \[\overline{EG}=\overline{CG}=\overline{EC}=l.\] Additionally, as \(\overline{EG}=l=\overline{EF}\), the right triangle \(\left[ FEG\right] \) is isosceles, so \[E\hat{G}F = E\hat{F}G=45\,^{\circ}.\] Therefore, \[D\hat{F}G = 180\,^{\circ} - E\hat{F}G=135\,^{\circ},\] as we wished.

Since \(A\hat{D}C = G\hat{D}F\) and \(A\hat{C}D = D\hat{F}G\), we have that the triangles \(\left[ ACD\right] \) and \(\left[ DFG\right] \) are similar, so \(\left[ DG\right] \) is the second shortest diagonal of a regular dodecagon with side \(\left[ DF\right] \). This being so, we can construct a regular dodecagon passing through the points \(D\), \(F\) and \(G\),  and whose side and second shortest diagonal have lengths given by: \[l_{1}=\overline{DF}=\overline{AD}-\overline{AE}-\overline{EF}=d-2l\] \[d_{1}=\overline{DG}=\overline{DC}+\overline{CG}=l+l=2l\]

As \(d=mx\) and \(l=nx\), we have: \[l_{1}=mx-2nx=(m-2n)x=n_{1}x\] \[d_{1}=2nx=m_{1}x\] where \(n_{1}=m-2n\) and \(m_{1}=2n\)  are two positive integer numbers, with \(m_{1}<m\). Indeed, \[\begin{array}{ccl} m_{1}\geq m & \Rightarrow & 2n\geq m\\ & \Rightarrow & 2l\geq d\\ & \Rightarrow & \frac{d}{l}\leq 2, \end{array},\] which is absurd, since \[\begin{array}{ccl} \frac{d}{l} & = & 1+2 \cos\frac{\pi}{12}\\ & > & 1+2 \cos\frac{\pi}{3}\\ & = & 1+2\frac{1}{2}\\ & = & 2. \end{array}\]  Proceeding in the same way, we are going to obtain a new regular dodecagon 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 \(m_{2}=2n_{1}\) are two positive integer numbers, with \(m_{2}<m_{1}<m \).

Since we can go on indefinitely constructing new regular dodecagons, 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 dodecagon are incommensurable magnitudes.