Question Number 225756 by ajfour last updated on 09/Nov/25
Commented by mr W last updated on 11/Nov/25
$${applying}\:{descartes}\:{theorem}\:{twice}: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{r}}}=\frac{\mathrm{1}}{\:\sqrt{{R}}}+\frac{\mathrm{1}}{\:\sqrt{{R}}}=\frac{\mathrm{2}}{\:\sqrt{{R}}}\:\Rightarrow{r}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{a}}}=\frac{\mathrm{1}}{\:\sqrt{{r}}}+\frac{\mathrm{1}}{\:\sqrt{{R}}}=\frac{\mathrm{3}}{\:\sqrt{{R}}} \\ $$$$\Rightarrow{a}=\frac{{R}}{\mathrm{9}} \\ $$
Commented by mr W last updated on 11/Nov/25
Commented by ajfour last updated on 11/Nov/25
$${Thank}\:{you}\:{sir} \\ $$