AO＝√３

$${「AO=\sqrt{3}」}$$

$${半径\sqrt{3}の円を外接円とする5角形ABCDE}$$

$${中心O}$$

$${AOとBEの交点P}$$

$${\angle{AOB}=\angle{COD}=\angle{EOA}=72^{\circ}}$$

$${\triangle{AOB}\equiv\triangle{COD}\equiv\triangle{EOA}}$$

$${AB=CD=EA=2}$$

$${BE=\frac{4\sqrt{2}}{\sqrt{3}}}$$

$${AP=\frac{2}{\sqrt{3}}\quad\quad{}PO=\frac{1}{\sqrt{3}}\quad\quad{}AO=\sqrt{3}}$$

$${\triangle{BOC}\equiv\triangle{DOE}の2等辺3角形と仮定すると、}$$

$${BとCの円周上の中点FはEOの延長線との交点}$$

$${DとEの円周上の中点GはBOの延長線との交点}$$

$${FG=\frac{4\sqrt{2}}{\sqrt{3}}}$$

$${\because}$$

$${108^{\circ}:72^{\circ}=54^{\circ}:36^{\circ}=3:2}$$

$${\angle{GBC}=\angle{OCB}=54^{\circ}}$$

$${\angle{BFG}=90^{\circ}}$$

$${\angle{EGB}=\angle{FBG}=72^{\circ}}$$

$${\angle{BGF}=18^{\circ}}$$

$${\angle{BOF}=36^{\circ}}$$

$${\angle{FED}=\angle{ODE}=54^{\circ}}$$

$${\angle{EGF}=90^{\circ}}$$

$${\angle{BFE}=\angle{GEF}=72^{\circ}}$$

$${\angle{EFG}=18^{\circ}}$$

$${\angle{EOG}=36^{\circ}}$$

$${\therefore}$$

$${\angle{BOC}=\angle{DOE}=72^{\circ}}$$

$${\angle{ABC}=\angle{BCD}=\angle{CDE}=\angle{DEA}=\angle{EAB}=108^{\circ}}$$

$${AB=BC=CD=DE=EA=2}$$

$${A(0,\sqrt{3})}$$

$${B(-\frac{2\sqrt{2}}{\sqrt{3}},\frac{1}{\sqrt{3}})}$$

$${C(-1,-\sqrt{2})}$$

$${D(1,-\sqrt{2})}$$

$${E(\frac{2\sqrt{2}}{\sqrt{3}},\frac{1}{\sqrt{3}})}$$

$${1^2+(\sqrt{2})^2=(\sqrt{3})^2}$$

$${半径\sqrt{2}の円を内接円とする。}$$

$${正5角形の黄金率\phi=\frac{2\sqrt{2}}{\sqrt{3}}である。}$$

$${矛盾はない。}$$

$${cf.}$$

$${Pを求めると、}$$

$${正5角形ABCDEの作図ができる。}$$