## Incommensurability

### Regular Dodecagon

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.