ATRACTOR

Consider a regular decagon for which the points \(A\), \(B\), \(C\), \(D\), \(E\) and \(F\) are consecutive vertices, \(d\) is the length of its longest 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 to the longest diagonal and the side of the decagon. Note that each side is a chord of the circle circumscribing the polygon, corresponding to an arc measuring \(\frac{1}{10}.360\,^{\circ}=36\,^{\circ}\). Choose the points \(G\) and \(H\) on the diagonal \([CF]\) such that \(\overline{CG}=\overline{GH}=l\). Then, as \([CDG]\) is isosceles and \[G\hat{C}D=F\hat{C}D=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ},\] we get \[C\hat{D}G=C\hat{G}D=\frac{1}{2}.(180\,^{\circ}-36\,^{\circ})=72\,^{\circ}.\] On the other hand, \[E\hat{D}G=E\hat{D}C-C\hat{D}G=\frac{1}{2}.8.36\,^{\circ}-72\,^{\circ}=144\,^{\circ}-72\,^{\circ}=72\,^{\circ}=C\hat{G}D,\] so the diagonal \([CF]\) is parallel to the side \([DE]\). Then, as \([GH]\) is parallel to \([DE]\) and \(\overline{GH}=\overline{DE}\), we have that \([GDEH]\) is a parallelogram, where \(\overline{GD}=\overline{HE}\) and \(H\hat{E}D=180\,^{\circ}-E\hat{D}G\). Since \(\overline{OC}=\overline{OF}\), \([OCF]\) is an isosceles triangle, with \[O\hat{C}F=O\hat{F}C=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ}.\] We then have that \[C\hat{O}G=C\hat{O}D=36\,^{\circ}=O\hat{C}G,\] so \([OCG]\) is an isosceles triangle, with \(\overline{OG}=\overline{CG}=l\). Therefore, \[\overline{GD}=\overline{OD}-\overline{OG}= \frac{1}{2}d-l.\]

Let \(l\) be the point of intersection of the line \(EH\) with the diagonal \([AF]\). Since \(\overline{OD}=\overline{OE}\), \([ODE]\) is an isosceles triangle, with \[O\hat{D}E=D\hat{E}O=\frac{1}{2}(180\,^{\circ}-D\hat{O}E)=\frac{1}{2}(180\,^{\circ}-36\,^{\circ})=72\,^{\circ}.\] Therefore, \[I\hat{E}D=H\hat{E}D=180\,^{\circ}-E\hat{D}G=180\,^{\circ}-O\hat{D}E=108\,^{\circ},\] \[I\hat{E}F=F\hat{E}D-I\hat{E}D=\frac{1}{2}.8.36\,^{\circ}-108\,^{\circ}=36\,^{\circ},\] \[I\hat{F}E=A\hat{F}E=\frac{1}{2}.4.36\,^{\circ}=72\,^{\circ}\] and \[F\hat{I}E=180\,^{\circ}-I\hat{E}F-I\hat{F}E=72\,^{\circ}=I\hat{F}E,\] so \([IEF]\) is an isosceles triangle, with \(\overline{IE}=\overline{FE}=l\). Therefore we get \[\overline{IH}=\overline{IE}-\overline{HE}=\overline{IE}-\overline{GD}=l-\left(\frac{1}{2}d-l\right)=2l-\frac{1}{2}d.\] We also have that \(\overline{OE}=\overline{OF}\), so \([OEF]\) is an isosceles triangle, with \[O\hat{E}F=E\hat{F}O=A\hat{F}E=72\,^{\circ},\] whence \[O\hat{E}I=O\hat{E}F-I\hat{E}F=72\,^{\circ}-36\,^{\circ}=36\,^{\circ}=I\hat{O}E\] and \([IOE]\) is also an isosceles triangle, with \(\overline{IO}=\overline{IE}=l\).

Let \(J\) be the point on the segment \([IF]\) such that \[\overline{IJ}=\overline{IH}=2l-\frac{1}{2}d\] (note that \[\overline{IF}=\overline{OF}-\overline{OI}=\frac{1}{2}d-l>2l-\frac{1}{2}d,\] because \[\begin{array}{ccl} \frac{1}{2}d-l\leq2l-\frac{1}{2}d & \Leftrightarrow & d\leq3l\\ & \Leftrightarrow & \frac{d}{l}\leq3\\ & \Leftrightarrow & \frac{1}{\sin\frac{\pi}{10}}\leq3\\ & \Leftrightarrow & \sin\frac{\pi}{10}\geq\frac{1}{3} \end{array},\] and this is absurd, as \(\sin\frac{\pi}{10}=\frac{\sqrt{5}-1}{4}\approx0,31\)). Therefore, \([IHJ]\) is an isosceles triangle, with \[I\hat{H}J=I\hat{J}H=\frac{1}{2}(180\,^{\circ}-J\hat{I}H)=\frac{1}{2}(180\,^{\circ}-F\hat{I}E)=54\,^{\circ}.\] But then \[F\hat{J}H=180\,^{\circ}-I\hat{J}H=126\,^{\circ}\] and \[J\hat{F}H=O\hat{F}C=36\,^{\circ},\] so \([JFH]\) is similar to the triangle \([DCF]\), since \[D\hat{C}F=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ}=J\hat{F}H\] and \[C\hat{D}F=\frac{1}{2}.7.36\,^{\circ}=126\,^{\circ}=F\hat{J}H.\]

Therefore, we can construct a regular decagon passing through the points \(F\), \(J\) and \(H\), where \([FJ]\) is one of its sides and \([JH]\) is one of its second shortest diagonals. Note that one of the vertices of this new decagon lies on the diagonal \([DF]\) (more precisely, it is the point of intersection of \([DF]\) with \([IE]\), since \(J\hat{F}D=\frac{1}{2}.3.36\,^{\circ}\) and \[F\hat{H}E=F\hat{G}D=180\,^{\circ}-C\hat{G}D=108\,^{\circ}=\frac{1}{2}.6.36\,^{\circ})\] and another vertex lies on the side \([EF]\) (since \(J\hat{F}E=\frac{1}{2}.4.36\,^{\circ}\)). Let \(K\) and \(L\) be these two points, respectively. We have that \(K\hat{E}L=I\hat{E}F=36\,^{\circ}\) and \[K\hat{H}L=K\hat{L}H=\frac{1}{2}.36\,^{\circ}=18\,^{\circ},\] so \[E\hat{K}L=K\hat{H}L+K\hat{L}H=36\,^{\circ}=K\hat{L}E\] and \([KEL]\) is an isosceles triangle, with \(\overline{EL}=\overline{KL}=\overline{JF}\). Therefore, the lengths of the side and the longest diagonal are given by: \[l_{1}=\overline{JF}=\overline{OF}-\overline{OI}-\overline{IJ}=\frac{1}{2}d-l-\left(2l-\frac{1}{2}d\right)=d-3l\] \[d_{1}=\overline{LF}=\overline{EF}-\overline{EL}=\overline{EF}-\overline{JF}=l-(d-3l)=4l-d\]

As \(d=mx\) and \(l=nx\), we have: \[l_{1}=mx-3nx=(m-3n)x=n_{1}x\] \[d_{1}=4nx-mx=(4n-m)x=m_{1}x,\] where \(n_{1}=m-3n\) e \(m_{1}=4n-m\) are two positive integer numbers, with \(m_{1}<m\). Indeed, \[\begin{array}{ccl} m_{1}\geq m & \Rightarrow & 4n\geq2m\\ & \Rightarrow & 2n\geq m\\ & \Rightarrow & 2l\geq d\\ & \Rightarrow & \frac{d}{l}\leq2, \end{array}\] which is absurd, since \[\frac{d}{l}=\frac{1}{\sin\frac{\pi}{10}}=\frac{4}{\sqrt{5}-1}=\sqrt{5}+1>2.\]

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

Since we can go on indefinitely constructing new regular decagons, 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 longest diagonal and the side of the regular decagon are incommensurable magnitudes.