Question Number 139957 by cherokeesay last updated on 02/May/21
Commented by mr W last updated on 02/May/21
$${R}={r}+\frac{\mathrm{2}}{\mathrm{3}}×\sqrt{\mathrm{3}}{r} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\mathrm{1}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$
Commented by Dwaipayan Shikari last updated on 02/May/21
$${I}\:{have}\:{done}\:{in}\:{the}\:{same}\:{way}\:.{But}\:{what}\:{theorem}\:{you}\:{used}\: \\ $$$${below}\:{sir}?\:{Kissing}\:{Circle}\:{Theorem}? \\ $$
Commented by mr W last updated on 02/May/21
$${Descartes}\:{theorem} \\ $$
Commented by cherokeesay last updated on 03/May/21
$${thank}\:{you}\:{sir}\:! \\ $$
Answered by mr W last updated on 02/May/21
$$\left(\frac{\mathrm{1}}{{r}_{\mathrm{1}} }+\frac{\mathrm{1}}{{r}_{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{\mathrm{3}} }−\frac{\mathrm{1}}{{R}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{r}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{\mathrm{2}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{\mathrm{3}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{{R}^{\mathrm{2}} }\right) \\ $$$$\left(\frac{\mathrm{3}}{{r}}−\frac{\mathrm{1}}{{R}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{3}}{{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{{R}^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{9}}{{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{{R}^{\mathrm{2}} }−\frac{\mathrm{6}}{{rR}}=\frac{\mathrm{6}}{{r}^{\mathrm{2}} }+\frac{\mathrm{2}}{{R}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{3}{R}^{\mathrm{2}} }{{r}^{\mathrm{2}} }−\frac{\mathrm{6}{R}}{{r}}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\mathrm{1}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\approx\mathrm{2}.\mathrm{1547} \\ $$