직사각형의 대각선의 길이

$\mathrm{\angle} ADB = \angle DBC = \angle BAE = \angle D EG$이므로 이 각을 $\theta$라 하고 $\displaystyle \mathrm{\overline{BD}}$$= x$라 하면 $\displaystyle \mathrm{\overline{AB}}$$= x \sin \theta$, $\displaystyle \mathrm{\overline{BE}}$$\displaystyle = \mathrm{\overline{AB}} \sin \theta$$= x \sin ^{2} \theta$, $\displaystyle \mathrm{\overline{EF}}$$\displaystyle = \mathrm{\overline{BE}} \sin \theta$$= x \sin ^{3} \theta$ $\displaystyle \mathrm{\overline{AD}}$$= x \cos \theta$, $\displaystyle \mathrm{\overline{DE}}$$\displaystyle = \mathrm{\overline{AD}} \cos \theta$$= x \cos ^{2} \theta$, $\displaystyle \mathrm{\overline{EG}}$$\displaystyle = \mathrm{\overline{DE}} \cos \theta$$= x \cos ^{3} \theta$ $\therefore a = x \sin ^{3} \theta$, $b = x \cos ^{3} \theta$ $\displaystyle a ^{\frac{2}{3}} + b ^{\frac{2}{3}}$$\displaystyle = x ^{\frac{2}{3}} \sin ^{2} \theta + x ^{\frac{2}{3}} \cos ^{2} \theta$$\displaystyle = x ^{\frac{2}{3}}$ (∵ $\sin ^{2} \theta + \cos ^{2} \theta = 1$) 이므로 $\displaystyle x = \left( a ^{\frac{2}{3}} + b ^{\frac{2}{3}} \right) ^{\frac{3}{2}}$