Question-140373

Question Number 140373 by mr W last updated on 06/May/21

Commented by mr W last updated on 06/May/21

$${find}\:{the}\:{radius}\:{of}\:{the}\:{smallest}\: \\ $$$${hollow}\:{sphere}\:{which}\:{can}\:{hold} \\ $$$${four}\:{balls}\:{with}\:{radii}\:{a},{b},{c}\:{and}\:{d}\: \\ $$$${respectively}\:{inside}. \\ $$

Commented by ajfour last updated on 08/May/21

$${Very}\:{beautiful}\:{Sir},\:{i}\:{intend}\:{to} \\ $$$${attempt}\:{in}\:{a}\:{day}\:{or}\:{two}.. \\ $$

Commented by mr W last updated on 08/May/21

((1/a)+(1/b)+(1/c)+(1/d)−(1/R))^2 =3((1/a^2 )+(1/b^2 )+(1/c^2 )+(1/d^2 )+(1/R^2 )) (1/R^2 )+((1/a)+(1/b)+(1/c)+(1/d))(1/R)+((1/a^2 )+(1/b^2 )+(1/c^2 )+(1/d^2 )−(1/(ab))−(1/(ac))−(1/(ad))−(1/(bc))−(1/(bd))−(1/(cd))) (1/R)=(1/2)[(√(6((1/(ab))+(1/(ac))+(1/(ad))+(1/(bc))+(1/(bd))+(1/(cd)))−3((1/a^2 )+(1/b^2 )+(1/c^2 )+(1/d^2 ))))−((1/a)+(1/b)+(1/c)+(1/d)) ⇒R=(2/( (√(6((1/(ab))+(1/(ac))+(1/(ad))+(1/(bc))+(1/(bd))+(1/(cd)))−3((1/a^2 )+(1/b^2 )+(1/c^2 )+(1/d^2 ))))−((1/a)+(1/b)+(1/c)+(1/d))))

$$\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{\mathrm{1}}{{d}}−\frac{\mathrm{1}}{{R}}\right)^{\mathrm{2}} =\mathrm{3}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }+\frac{\mathrm{1}}{{d}^{\mathrm{2}} }+\frac{\mathrm{1}}{{R}^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{1}}{{R}^{\mathrm{2}} }+\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{\mathrm{1}}{{d}}\right)\frac{\mathrm{1}}{{R}}+\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }+\frac{\mathrm{1}}{{d}^{\mathrm{2}} }−\frac{\mathrm{1}}{{ab}}−\frac{\mathrm{1}}{{ac}}−\frac{\mathrm{1}}{{ad}}−\frac{\mathrm{1}}{{bc}}−\frac{\mathrm{1}}{{bd}}−\frac{\mathrm{1}}{{cd}}\right) \\ $$$$\frac{\mathrm{1}}{{R}}=\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt{\mathrm{6}\left(\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{ac}}+\frac{\mathrm{1}}{{ad}}+\frac{\mathrm{1}}{{bc}}+\frac{\mathrm{1}}{{bd}}+\frac{\mathrm{1}}{{cd}}\right)−\mathrm{3}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }+\frac{\mathrm{1}}{{d}^{\mathrm{2}} }\right)}−\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{\mathrm{1}}{{d}}\right)\right. \\ $$$$\Rightarrow{R}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}\left(\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{ac}}+\frac{\mathrm{1}}{{ad}}+\frac{\mathrm{1}}{{bc}}+\frac{\mathrm{1}}{{bd}}+\frac{\mathrm{1}}{{cd}}\right)−\mathrm{3}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }+\frac{\mathrm{1}}{{d}^{\mathrm{2}} }\right)}−\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{\mathrm{1}}{{d}}\right)} \\ $$

Answered by ajfour last updated on 07/May/21

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