Question Number 198828 by a.lgnaoui last updated on 24/Oct/23
$$\:\:\boldsymbol{\mathrm{Show}}\:\boldsymbol{\mathrm{th}{at}}: \\ $$$$\:\boldsymbol{{Area}}\left(\boldsymbol{{blue}}\:\boldsymbol{{circle}}\right)=\boldsymbol{{Area}}\left(\boldsymbol{{Green}}\:\:\boldsymbol{{circle}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by a.lgnaoui last updated on 24/Oct/23
Answered by mr W last updated on 25/Oct/23
Commented by mr W last updated on 25/Oct/23
Commented by mr W last updated on 25/Oct/23
$${AD}=\mathrm{2}{R}+{p} \\ $$$${BD}={R}−{p} \\ $$$${CD}={R}+{p} \\ $$$${AB}=\sqrt{\mathrm{2}}{R} \\ $$$${BC}={R} \\ $$$${AC}=\sqrt{{R}^{\mathrm{2}} +\left(\mathrm{2}{R}\right)^{\mathrm{2}} }=\sqrt{\mathrm{5}}{R} \\ $$$$\mathrm{cos}\:\alpha=\frac{\left(\sqrt{\mathrm{2}}{R}\right)^{\mathrm{2}} +\left({R}−{p}\right)^{\mathrm{2}} −\left(\mathrm{2}{R}+{p}\right)^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}{R}\left({R}−{p}\right)} \\ $$$$\:\:\:\:=−\frac{{R}+\mathrm{6}{p}}{\mathrm{2}\sqrt{\mathrm{2}}\left({R}−{p}\right)}=−\frac{\mathrm{1}+\mathrm{6}\lambda}{\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{1}−\lambda\right)} \\ $$$${with}\:\lambda=\frac{{p}}{{R}} \\ $$$$\mathrm{cos}\:\beta=\frac{{R}^{\mathrm{2}} +\left({R}−{p}\right)^{\mathrm{2}} −\left({R}+{p}\right)^{\mathrm{2}} }{\mathrm{2}{R}\left({R}−{p}\right)} \\ $$$$\:\:\:\:=\frac{{R}−\mathrm{4}{p}}{\mathrm{2}\left({R}−{p}\right)}=\frac{\mathrm{1}−\mathrm{4}\lambda}{\mathrm{2}\left(\mathrm{1}−\lambda\right)} \\ $$$$\alpha+\beta=\pi+\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=−\frac{\mathrm{1}+\mathrm{6}\lambda}{\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{1}−\lambda\right)}×\frac{\mathrm{1}−\mathrm{4}\lambda}{\mathrm{2}\left(\mathrm{1}−\lambda\right)}−\frac{\sqrt{\mathrm{8}\left(\mathrm{1}−\lambda\right)^{\mathrm{2}} −\left(\mathrm{1}+\mathrm{6}\lambda\right)^{\mathrm{2}} }}{\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{1}−\lambda\right)}×\frac{\sqrt{\mathrm{4}\left(\mathrm{1}−\lambda\right)^{\mathrm{2}} −\left(\mathrm{1}−\mathrm{4}\lambda\right)^{\mathrm{2}} }}{\mathrm{2}\left(\mathrm{1}−\lambda\right)} \\ $$$$−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{24}\lambda^{\mathrm{2}} −\mathrm{2}\lambda−\mathrm{1}}{\:\mathrm{4}\sqrt{\mathrm{2}}\left(\mathrm{1}−\lambda\right)^{\mathrm{2}} }−\frac{\sqrt{\mathrm{3}\left(\mathrm{7}−\mathrm{28}\lambda−\mathrm{28}\lambda^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} \right)}}{\:\mathrm{4}\sqrt{\mathrm{2}}\left(\mathrm{1}−\lambda\right)^{\mathrm{2}} } \\ $$$$\mathrm{28}\lambda^{\mathrm{2}} −\mathrm{10}\lambda+\mathrm{3}=\sqrt{\mathrm{3}\left(\mathrm{7}−\mathrm{28}\lambda−\mathrm{28}\lambda^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} \right)} \\ $$$$\mathrm{4}\left(\lambda−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{112}\lambda^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=\mathrm{1}\:\Rightarrow{rejected} \\ $$$$\Rightarrow\mathrm{112}\lambda^{\mathrm{2}} =\mathrm{3}\:\Rightarrow\lambda=\sqrt{\frac{\mathrm{3}}{\mathrm{112}}}=\frac{\sqrt{\mathrm{21}}}{\mathrm{28}} \\ $$$$\Rightarrow{p}=\frac{\sqrt{\mathrm{21}}{R}}{\mathrm{28}}\:\checkmark \\ $$$${in}\:{similar}\:{way}\:{we}\:{can}\:{get}\:{q}=\frac{\sqrt{\mathrm{21}}{R}}{\mathrm{28}}. \\ $$$${p}={q}=\frac{\sqrt{\mathrm{21}}{R}}{\mathrm{28}}\: \\ $$$$\Rightarrow{blue}\:{circle}={green}\:{circle} \\ $$