Question Number 173032 by mnjuly1970 last updated on 05/Jul/22
Answered by mr W last updated on 05/Jul/22
Commented by mr W last updated on 05/Jul/22
$${AC}=\sqrt{\left({r}+{R}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} }=\sqrt{{r}\left({r}+\mathrm{2}{R}\right)} \\ $$$${OC}={r}+\sqrt{{r}\left({r}+\mathrm{2}{R}\right)} \\ $$$${OB}^{\mathrm{2}} ={R}^{\mathrm{2}} +\left(\frac{{r}}{\:\sqrt{\mathrm{2}}}+\sqrt{\left({r}+{R}\right)^{\mathrm{2}} −\left({R}−\frac{{r}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} ={R}^{\mathrm{2}} +\left({r}+\sqrt{{r}\left({r}+\mathrm{2}{R}\right)}\right)^{\mathrm{2}} \\ $$$$\frac{{r}}{\:\sqrt{\mathrm{2}}}+\sqrt{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){rR}+\frac{{r}^{\mathrm{2}} }{\mathrm{2}}}={r}+\sqrt{{r}\left({r}+\mathrm{2}{R}\right)} \\ $$$$\sqrt{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){rR}+{r}^{\mathrm{2}} }=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){r}+\sqrt{\mathrm{2}{r}\left({r}+\mathrm{2}{R}\right)} \\ $$$$\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){rR}+{r}^{\mathrm{2}} =\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} {r}^{\mathrm{2}} +\mathrm{2}{r}\left({r}+\mathrm{2}{R}\right)+\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){r}\sqrt{\mathrm{2}{r}\left({r}+\mathrm{2}{R}\right)} \\ $$$$\sqrt{\mathrm{2}}{R}−\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){r}=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{2}{r}\left({r}+\mathrm{2}{R}\right)} \\ $$$${R}=\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){r} \\ $$$$\frac{{R}}{{r}}=\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\approx\mathrm{1}.\mathrm{172} \\ $$
Commented by Tawa11 last updated on 06/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$