Question Number 203370 by mr W last updated on 17/Jan/24
Commented by mr W last updated on 17/Jan/24
$${Q}\mathrm{203319} \\ $$
Answered by mr W last updated on 17/Jan/24
$$\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${small}\:{tetrahedron}\:{PQRS}: \\ $$$${edge}\:{length}\:{a}=\mathrm{2}{r} \\ $$$${height}\:{h}=\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{3}}=\frac{\mathrm{2}\sqrt{\mathrm{6}}{r}}{\mathrm{3}} \\ $$$$ \\ $$$${big}\:{tetrahedron}\:{ABCD}: \\ $$$${height}\:{H}={r}+{h}+{h}_{\mathrm{1}} \\ $$$${with}\:{h}_{\mathrm{1}} =\frac{{r}}{\mathrm{cos}\:\theta}=\mathrm{3}{r} \\ $$$${edge}\:{length}\:{b} \\ $$$${H}=\frac{\sqrt{\mathrm{6}}{b}}{\mathrm{3}}={r}+\frac{\mathrm{2}\sqrt{\mathrm{6}}{r}}{\mathrm{3}}+\mathrm{3}{r} \\ $$$$\Rightarrow{b}=\mathrm{2}\left(\sqrt{\mathrm{6}}+\mathrm{1}\right){r} \\ $$
Commented by mr W last updated on 18/Jan/24
$${or}\:{method}\:\mathrm{2}: \\ $$$$\frac{{H}}{\mathrm{4}}−\frac{{h}}{\mathrm{4}}={r} \\ $$$$\Rightarrow{H}={h}+\mathrm{4}{r} \\ $$$$\frac{{b}}{{a}}=\frac{{H}}{{h}}=\mathrm{1}+\frac{\mathrm{4}{r}}{{h}}=\mathrm{1}+\sqrt{\mathrm{6}} \\ $$$$\Rightarrow{b}=\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{6}}\right){r} \\ $$