Question Number 175717 by ajfour last updated on 05/Sep/22
Answered by mr W last updated on 05/Sep/22
$$\left({R}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +\left(\frac{{R}}{\mathrm{2}}+{r}\right)^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} +\mathrm{3}{Rr}−\frac{\mathrm{3}{R}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{r}}{{R}}=\sqrt{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{232} \\ $$
Commented by mr W last updated on 05/Sep/22
Commented by ajfour last updated on 05/Sep/22
$${Excellent}\:{notice}\:{sir}! \\ $$
Commented by ajfour last updated on 05/Sep/22
$$\frac{{R}}{{r}}=\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}}=\frac{\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}\right)}{\mathrm{3}}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}+\mathrm{2} \\ $$$$=\frac{\mathrm{4}×\mathrm{1}.\mathrm{732}}{\mathrm{3}}+\mathrm{2}=\frac{\mathrm{6}.\mathrm{928}}{\mathrm{3}}+\mathrm{2}\approx\mathrm{4}.\mathrm{31} \\ $$
Commented by Tawa11 last updated on 05/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$