## Incommensurability

### Regular Decagon

Consider a regular decagon 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 circumscribing the polygon, corresponding to an arc measuring $$\frac{1}{10}.360\,^{\circ}=36\,^{\circ}$$. Choose the point $$E$$ on the diagonal $$[AD]$$ such that $$\overline{AE}=\overline{AB}=l$$. Then, as $$[AEB]$$ is isosceles and $E\hat{A}B=D\hat{A}B=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ},$ we get $A\hat{E}B=A\hat{B}E=\frac{1}{2}.(180\,^{\circ}-36\,^{\circ})=72\,^{\circ}.$ On the other hand, $C\hat{B}E=C\hat{B}A-A\hat{B}E=\frac{1}{2}.8.36\,^{\circ}-72\,^{\circ}=144\,^{\circ}-72\,^{\circ}=72\,^{\circ}=A\hat{E}B,$ so the diagonal $$[AD]$$ is parallel to the side $$[BC]$$. Now choose the point $$F$$ on the diagonal $$[AD]$$ such that $$\overline{EF}=\overline{BC}=l$$. Then, since $$[EF]$$ is parallel to $$[BC]$$, $$[EBCF]$$ is a parallelogram, so $D\hat{F}C=D\hat{E}B=180\,^{\circ}-A\hat{E}B=108\,^{\circ}.$ Since $A\hat{C}D=\frac{1}{2}.7.36\,^{\circ}=126\,^{\circ}>108\,^{\circ}=D\hat{F}C,$ we choose the point $$G$$ in the half-line $$DC$$ such that $$D\hat{F}G=126\,^{\circ}$$. As $$A\hat{D}C=G\hat{D}F$$ and $$A\hat{C}D=D\hat{F}G$$, we get that the triangles $$[ACD]$$ and $$[DFG]$$ are similar, so $$[DG]$$ is the second shortest diagonal of a regular decagon with side $$[DF]$$. On the other hand, we have $C\hat{G}F=D\hat{G}F=D\hat{A}C=\frac{1}{2}.36\,^{\circ}=18\,^{\circ}$ and $C\hat{F}G=D\hat{F}G-D\hat{F}C=126\,^{\circ}-108\,^{\circ}=18\,^{\circ},$ so $$[FCG]$$ is isosceles, with $$\overline{CG}=\overline{CF}$$. We also have $F\hat{D}C=A\hat{D}C=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ}$ and $F\hat{C}D=C\hat{G}F+C\hat{F}G=36\,^{\circ},$ so $$[DFC]$$ is also isosceles, with $$\overline{CF}=\overline{DF}$$. Therefore, we can construct a regular decagon passing through the points $$D$$, $$F$$ and $$G$$, and whose side and second shortest diagonal have lengths given by: $l_{1}=\overline{DF}=\overline{AD}-\overline{AE}-\overline{EF}=d-2l$ $d_{1}=\overline{DG}=\overline{DC}+\overline{CG}=\overline{DC}+\overline{DF}=l+(d-2l)=d-l$

As $$d=mx$$ and $$l=nx$$, we have: $l_{1}=mx-2nx=(m-2n)x=n_{1}x$ $d_{1}=mx-nx=(m-n)x=m_{1}x,$ where $$n_{1}=m-2n$$ and $$m_{1}=m-n$$ are two positive integer numbers, with $$m_{1}<m$$. Proceeding in the same way, we obtain a new regular decagon 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 $$m_{2}=m_{1}-n_{1}$$  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 second shortest diagonal and the side of the regular decagon are incommensurable magnitudes.