Question Number 48653 by ajfour last updated on 26/Nov/18
Commented by ajfour last updated on 26/Nov/18
$${Find}\:\:\frac{{r}}{{R}}.\:{Smaller}\:{sphere} \\ $$$$\:{inscribed}\:{inside}\:{tetrahedron} \\ $$$${and}\:{larger}\:{one}\:{circumscribing}\:{the} \\ $$$${tetrahedron}\:\left({regular}\right). \\ $$
Answered by mr W last updated on 26/Nov/18
$${let}\:{a}={edge}\:{length} \\ $$$$\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{3}} \\ $$$$\sqrt{{a}^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{3}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{3}} \\ $$$$\frac{\mathrm{2}{R}}{{a}}=\frac{{a}}{\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{3}}} \\ $$$$\Rightarrow{R}=\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{4}} \\ $$$$\frac{\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{3}}−{r}}{{r}}=\frac{\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{6}}}=\mathrm{3} \\ $$$$\Rightarrow{r}=\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{12}} \\ $$$$\Rightarrow\frac{{r}}{{R}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by ajfour last updated on 26/Nov/18
$${Thank}\:{you}\:{Sir}. \\ $$$${And}\:{if}\:{tetrahedron}\:{is}\:{replaced} \\ $$$${by}\:{a}\:{cube},\:\:{then}\:\:{r}/{R}\:=\:\mathrm{1}/\sqrt{\mathrm{3}}\:. \\ $$