## Incommensurability

### Regular Decagon

Let $$[ABCDEFGHIJ]$$ be a regular decagon with centre $$O$$ such that the lengths of the side, the longest diagonal and the second shortest diagonal are $$l$$, $$d$$ and $$d'$$, respectively. Let $$[AF]$$ and $$[CH]$$ be two diagonals passing through the centre of the decagon, $$[AD]$$ the second shortest diagonal joining the vertices $$A$$ and $$D$$, and $$K$$ the point of intersection of the diagonal $$[AD]$$ with the diagonal $$[CH]$$. Then, $C\hat{D}K=C\hat{D}A=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ}$ and $K\hat{C}D=H\hat{C}D=\frac{1}{2}.4.36\,^{\circ}=72\,^{\circ},$ so that $C\hat{K}D=180\,^{\circ}-C\hat{D}K-K\hat{C}D=72\,^{\circ}$ and $$[KCD]$$ is an isosceles triangle, with $$\overline{KD}=\overline{CD}=l$$. In addition, $O\hat{A}K=F\hat{A}D=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ},$ $A\hat{K}O=C\hat{K}D=72\,^{\circ}$ and $A\hat{O}K=180\,^{\circ}-O\hat{A}K-A\hat{K}O=72\,^{\circ},$ so that $$[AOK]$$ is an isosceles triangle, with $$\overline{AK}=\overline{AO}$$. So, the length of the longest diagonal is given by: $d=\overline{AF}=2\overline{AO}=2\overline{AK}=2\left(\overline{AD}-\overline{KD}\right)=2(d'-l)$

Hence we have: $\frac{d}{l}=2\left(\frac{d'}{l}-1\right)$

Equivalently, we have: $\frac{d'}{l}=\frac{1}{2}.\frac{d}{l}+1$

Therefore, $$\frac{d}{l}$$ is a rational number if and only if $$\frac{d'}{l}$$ is also a rational number, which we already saw is not the case. So, the longest diagonal and the side of the decagon $$[ABCDEFGHIJ]$$ are incommensurable magnitudes.