## Incommensurability

### Regular Decagon

Let $$[ABCDEFGHIJ]$$ be a regular decagon with centre $$O$$ such that the lengths of the side and the longest diagonal are $$l$$ and $$d$$, respectively. Let $$[AF]$$, $$[CH]$$ and $$[EJ]$$ be three diagonals passing through the centre of the decagon and $$[AH]$$ and $$[BE]$$ two of the second shortest diagonals. Consider the points $$K$$ and $$L$$, produced by the intersection between the diagonals $$[AH]$$ and $$[BE]$$ with the diagonals $$[EJ]$$ and $$[CH]$$, respectively. We have $J\hat{A}L=J\hat{A}H=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ},$ $L\hat{J}A=E\hat{J}A=\frac{1}{2}.4.36\,^{\circ}=72\,^{\circ}$ and $J\hat{L}A=180\,^{\circ}-J\hat{A}L-L\hat{J}A=72\,^{\circ}=L\hat{J}A,$ so $$[AJL]$$ is an isosceles triangle, with $$\overline{AL}=\overline{AJ}=l$$. Analogously, $$[BCK]$$ is also isosceles, with $$\overline{BK}=\overline{BC}=l$$. Additionally, since these triangles are congruent, we have $$\overline{JL}=\overline{CK}$$. The triangle $$[LAO]$$ is also isosceles, since $L\hat{A}O=H\hat{A}F=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ}$ and $L\hat{O}A=J\hat{O}A=36\,^{\circ}=L\hat{A}O,$ so $$\overline{LO}=\overline{AL}=l$$ and $\overline{KO}=\overline{CO}-\overline{CK}=\overline{JO}-\overline{JL}=\overline{LO}=l.$ Therefore, $\overline{AB}=\overline{BK}=\overline{KO}=\overline{OL}=\overline{LA}=l.$ On the other hand, we have $L\hat{A}B=H\hat{A}B=\frac{1}{2}.6.36\,^{\circ}=108\,^{\circ},$ $A\hat{B}K=A\hat{B}E=\frac{1}{2}.6.36\,^{\circ}=108\,^{\circ},$ $B\hat{K}O=180\,^{\circ}-B\hat{K}C=108\,^{\circ},$ $K\hat{O}L=C\hat{O}J=3.36\,^{\circ}=108\,^{\circ}$ and $O\hat{L}A=180\,^{\circ}-J\hat{L}A=108\,^{\circ}.$ Therefore, the polygon $$[ABKLO]$$ has all its sides equal and all its interior angles equal, so it is a regular pentagon with side $$\overline{AB}=l$$ and whose diagonal is $$\overline{AO}=\frac{1}{2}\overline{AF}=\frac{1}{2}d$$. If the longest diagonal of $$[ABCDEFGHIJ]$$ and its side were commensurable, that is, if the ratio $$\frac{d}{l}$$ were a rational number, then the ratio $$\frac{\frac{1}{2}d}{l}=\frac{1}{2}.\frac{d}{l}$$ would also be rational and the diagonal of $$[ABKLO]$$ would be commensurable with its side, which we have already seen is not the case. Therefore, the longest diagonal and the side of the decagon $$[ABCDEFGHIJ]$$ are incommensurable magnitudes.

Remark: actually, it can be proved that the ratio between the longest diagonal and the side of a $$2n$$-sides regular polygon, with $$n$$ an odd number larger than $$3$$, is always double the ratio between the longest diagonal and the side of a $$n$$-sides regular polygon. The above case corresponds to taking $$n=5$$. To visualize the next case, consult this app.