Question Number 175476 by ajfour last updated on 31/Aug/22
Commented by ajfour last updated on 31/Aug/22
$${A}\:{cone}\:{has}\:{an}\:{inscribed}\:{cube}\:{resting} \\ $$$${at}\:{its}\:{base}.\:{Atop}\:{the}\:{cube}\:{inscribed} \\ $$$${in}\:{the}\:{cone}\:{is}\:{a}\:{sphere}\:{of}\:{radius}\:{r}. \\ $$$${Find}\:{r}\:{in}\:{terms}\:{of}\:{R},\:{H}\:\left({cone}'{s}\right. \\ $$$$\left.{radius}\:{and}\:{height}\right). \\ $$
Answered by mr W last updated on 01/Sep/22
Commented by mr W last updated on 01/Sep/22
$${a}={edge}\:{length}\:{of}\:{cube} \\ $$$$\mathrm{tan}\:\theta=\frac{{H}}{{R}}=\frac{{a}}{{R}−\frac{{a}}{\:\sqrt{\mathrm{2}}}} \\ $$$$\frac{{H}}{{R}}=\frac{{a}}{{R}−\frac{{a}}{\:\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow{a}=\frac{{R}}{\frac{{R}}{{H}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \\ $$$${r}=\frac{{a}}{\:\sqrt{\mathrm{2}}}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\frac{{a}}{\:\sqrt{\mathrm{2}}}×\:\frac{\frac{{H}}{\:\sqrt{{H}^{\mathrm{2}} +{R}^{\mathrm{2}} }}}{\mathrm{1}+\frac{{R}}{\:\sqrt{{H}^{\mathrm{2}} +{R}^{\mathrm{2}} }}} \\ $$$$\Rightarrow{r}=\frac{{RH}}{\left(\frac{\sqrt{\mathrm{2}}{R}}{{H}}+\mathrm{1}\right)\left(\sqrt{{H}^{\mathrm{2}} +{R}^{\mathrm{2}} }+{R}\right)} \\ $$
Commented by ajfour last updated on 01/Sep/22
$${wow},\:{fantastic},\:{thanks}\:{Sir}. \\ $$
Commented by Tawa11 last updated on 02/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$