Question Number 164018 by cortano1 last updated on 13/Jan/22
Answered by mr W last updated on 13/Jan/22
Commented by cortano1 last updated on 13/Jan/22
$${superb}\:{solution} \\ $$
Commented by mr W last updated on 13/Jan/22
$${b}=\mathrm{2}{R} \\ $$$${a}=\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }+\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left(\mathrm{2}{R}−{r}\right)^{\mathrm{2}} } \\ $$$${a}=\mathrm{2}\sqrt{{Rr}}+\sqrt{\mathrm{3}{R}\left(\mathrm{2}{r}−{R}\right)} \\ $$$${BC}=\mathrm{2}\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }=\mathrm{2}\sqrt{{R}\left({R}+\mathrm{2}{r}\right)} \\ $$$${a}=\sqrt{{BC}^{\mathrm{2}} −{R}^{\mathrm{2}} }=\sqrt{\mathrm{4}{R}\left({R}+\mathrm{2}{r}\right)−{R}^{\mathrm{2}} }=\sqrt{{R}\left(\mathrm{3}{R}+\mathrm{8}{r}\right)} \\ $$$$\mathrm{2}\sqrt{{Rr}}+\sqrt{\mathrm{3}{R}\left(\mathrm{2}{r}−{R}\right)}=\sqrt{{R}\left(\mathrm{3}{R}+\mathrm{8}{r}\right)} \\ $$$${let}\:\lambda=\frac{{R}}{{r}} \\ $$$$\mathrm{2}+\sqrt{\mathrm{3}\left(\mathrm{2}−\lambda\right)}=\sqrt{\mathrm{3}\lambda+\mathrm{8}} \\ $$$$\mathrm{3}\left(\mathrm{2}−\lambda\right)=\mathrm{3}\lambda+\mathrm{8}+\mathrm{4}−\mathrm{4}\sqrt{\mathrm{3}\lambda+\mathrm{8}} \\ $$$$\mathrm{3}\left(\lambda+\mathrm{1}\right)=\mathrm{2}\sqrt{\mathrm{3}\lambda+\mathrm{8}} \\ $$$$\mathrm{9}\left(\lambda^{\mathrm{2}} +\mathrm{2}\lambda+\mathrm{1}\right)=\mathrm{4}\left(\mathrm{3}\lambda+\mathrm{8}\right) \\ $$$$\mathrm{9}\lambda^{\mathrm{2}} +\mathrm{6}\lambda−\mathrm{23}=\mathrm{0} \\ $$$$\lambda=\frac{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{1}}{\mathrm{3}}\:\checkmark \\ $$
Commented by Tawa11 last updated on 13/Jan/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$