## Incommensurability

### Regular Decagon

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.