Question Number 221188 by ajfour last updated on 26/May/25
Commented by Frix last updated on 27/May/25
$$\left(\mathrm{1}\right)\:\mathrm{We}'\mathrm{re}\:\mathrm{looking}\:\mathrm{at}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{diagonally} \\ $$$$\mathrm{because}\:\mathrm{the}\:\mathrm{vertexes}\:\mathrm{touch}\:\mathrm{both}\:\mathrm{the}\:\mathrm{sphere} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{cone} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{I}\:\mathrm{think}\:\mathrm{we}\:\mathrm{need}\:\mathrm{1}\:\mathrm{more}\:\mathrm{information} \\ $$
Commented by ajfour last updated on 27/May/25
$${No},\:{take}\:{it}\:{front}\:{view}.\:{side}\:{length}\:{of} \\ $$$${cube}=\mathrm{1}.\:{Lets}\:{choose}\:{circular}\:{rim}\:{of} \\ $$$${cone}\:{to}\:{be}\:{diameter}\:{of}\:{larger}\:{sphere}. \\ $$
Answered by mr W last updated on 27/May/25
Commented by mr W last updated on 27/May/25
![side length of cube =2a base of cone at half heigth of cube tan β=(r/( (√2)a)) (√2)a tan 2β=(√2)a×(((2r)/( (√2)a))/(1−((r/( (√2)a)))^2 ))=((4a((r/a)))/(2−((r/( a)))^2 )) R=((4a((r/a)))/(2−((r/( a)))^2 ))+2a−(√(R^2 −((√2)a)^2 )) ⇒((4((r/a)))/(2−((r/( a)))^2 ))=(R/a)−2+(√(((R/a))^2 −2)) (√(R^2 −((√(R^2 −((√2)a)^2 ))−a)^2 ))=((((4a((r/a)))/(2−((r/( a)))^2 ))+a)/((4a((r/a)))/(2−((r/( a)))^2 )))×(√2)a ⇒1+2(√(((R/a))^2 −2))=2[1+((2−((r/a))^2 )/(4((r/a))))]^2 ⇒(R/a)=(√(2+(1/4){2[1+((2−((r/a))^2 )/(4((r/a))))]^2 −1}^2 )) ⇒(r/a)≈0.73205081, (R/a)=2.25 ⇒(R/r)≈3.07355716](https://www.tinkutara.com/question/Q221232.png)
$${side}\:{length}\:{of}\:{cube}\:=\mathrm{2}{a} \\ $$$${base}\:{of}\:{cone}\:{at}\:{half}\:{heigth}\:{of}\:{cube} \\ $$$$\mathrm{tan}\:\beta=\frac{{r}}{\:\sqrt{\mathrm{2}}{a}} \\ $$$$\sqrt{\mathrm{2}}{a}\:\mathrm{tan}\:\mathrm{2}\beta=\sqrt{\mathrm{2}}{a}×\frac{\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{2}}{a}}}{\mathrm{1}−\left(\frac{{r}}{\:\sqrt{\mathrm{2}}{a}}\right)^{\mathrm{2}} }=\frac{\mathrm{4}{a}\left(\frac{{r}}{{a}}\right)}{\mathrm{2}−\left(\frac{{r}}{\:{a}}\right)^{\mathrm{2}} } \\ $$$${R}=\frac{\mathrm{4}{a}\left(\frac{{r}}{{a}}\right)}{\mathrm{2}−\left(\frac{{r}}{\:{a}}\right)^{\mathrm{2}} }+\mathrm{2}{a}−\sqrt{{R}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}{a}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{4}\left(\frac{{r}}{{a}}\right)}{\mathrm{2}−\left(\frac{{r}}{\:{a}}\right)^{\mathrm{2}} }=\frac{{R}}{{a}}−\mathrm{2}+\sqrt{\left(\frac{{R}}{{a}}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$\sqrt{{R}^{\mathrm{2}} −\left(\sqrt{{R}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}{a}\right)^{\mathrm{2}} }−{a}\right)^{\mathrm{2}} }=\frac{\frac{\mathrm{4}{a}\left(\frac{{r}}{{a}}\right)}{\mathrm{2}−\left(\frac{{r}}{\:{a}}\right)^{\mathrm{2}} }+{a}}{\frac{\mathrm{4}{a}\left(\frac{{r}}{{a}}\right)}{\mathrm{2}−\left(\frac{{r}}{\:{a}}\right)^{\mathrm{2}} }}×\sqrt{\mathrm{2}}{a} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{2}\sqrt{\left(\frac{{R}}{{a}}\right)^{\mathrm{2}} −\mathrm{2}}=\mathrm{2}\left[\mathrm{1}+\frac{\mathrm{2}−\left(\frac{{r}}{{a}}\right)^{\mathrm{2}} }{\mathrm{4}\left(\frac{{r}}{{a}}\right)}\right]^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{R}}{{a}}=\sqrt{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{2}\left[\mathrm{1}+\frac{\mathrm{2}−\left(\frac{{r}}{{a}}\right)^{\mathrm{2}} }{\mathrm{4}\left(\frac{{r}}{{a}}\right)}\right]^{\mathrm{2}} −\mathrm{1}\right\}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{r}}{{a}}\approx\mathrm{0}.\mathrm{73205081},\:\frac{{R}}{{a}}=\mathrm{2}.\mathrm{25} \\ $$$$\Rightarrow\frac{{R}}{{r}}\approx\mathrm{3}.\mathrm{07355716} \\ $$
Commented by ajfour last updated on 29/May/25
$${Thanks}\:{sir}.\:{I}\:{will}\:{attempt}\:{by}\:{tomorrow}. \\ $$
Commented by Tawa11 last updated on 02/Jun/25
$$\mathrm{Sir},\:\mathrm{why}\:\mathrm{did}\:\mathrm{you}\:\mathrm{choose}\:\mathrm{2a}\:\mathrm{as}\:\mathrm{side}\:\mathrm{length}. \\ $$$$\mathrm{Can}\:\mathrm{the}\:\mathrm{side}\:\mathrm{length}\:\mathrm{also}\:\mathrm{be}\:\:\mathrm{1}? \\ $$
Commented by mr W last updated on 02/Jun/25
$${such}\:{that}\:{i}\:{can}\:{write}\:{a}\:{everywhere} \\ $$$${instead}\:{of}\:\frac{\mathrm{1}}{\mathrm{2}}.\:{that}'{s}\:{more}\:{convenient} \\ $$$${to}\:{me}.\: \\ $$
Commented by Tawa11 last updated on 02/Jun/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Commented by ajfour last updated on 02/Jun/25
$${In}\:{mrW}\:{sir}'{s}\:{answer}\:{you}\:{can}\:{put}\:\mathrm{2}{a}=\mathrm{1} \\ $$$${meaning}\:\:{a}=\mathrm{1}/\mathrm{2}=\mathrm{0}.\mathrm{5}\:\:\:{if}\:{you}\:{hsd}\:{taken} \\ $$$${a}\:{cube}\:{of}\:{side}\:{length}=\mathrm{1}\:\left({unity}\right). \\ $$