Question Number 19915 by ajfour last updated on 18/Aug/17
Commented by ajfour last updated on 18/Aug/17
$${Four}\:{equal}\:{spheres}\:{of}\:{radius}\:{r} \\ $$$${are}\:{externally}\:{tangent}\:{to}\:{the}\:{three} \\ $$$${others}.\:{Find}\:{the}\:{radius}\:{of}\:{a}\:{sphere} \\ $$$${tangent}\:{to}\:{all}\:{four}\:{spheres}\:{and} \\ $$$${containing}\:{them}\:{within}\:{it}. \\ $$
Commented by ajfour last updated on 18/Aug/17
Commented by ajfour last updated on 18/Aug/17
$$\:{The}\:{centres}\:{of}\:{the}\:{four}\:{spheres} \\ $$$$\:{be}\:{A},\:{B},\:{C},\:{D}.\:{They}\:{form}\:{a} \\ $$$${regular}\:{tetrahedron}. \\ $$$${Each}\:{edge}\:{of}\:{tetrahedron}\:{is}\:{then} \\ $$$${equal}\:{to}\:\mathrm{2}\boldsymbol{{r}}. \\ $$$$\:{O}\:\:{being}\:{the}\:{centre}\:{of}\:{the}\:{tetra}- \\ $$$${hedron},\:\:{OA}={OB}\:={x}\:\:\left({let}\right) \\ $$$$\:\:\:\:{and}\:\:\:{BC}=\mathrm{2}{r} \\ $$$$\:{BE}=\sqrt{{BC}^{\:\mathrm{2}} −{CE}^{\mathrm{2}} }=\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{r}^{\mathrm{2}} }={r}\sqrt{\mathrm{3}} \\ $$$${The}\:{foot}\:{of}\:{perpendicular}\:{from} \\ $$$${A}\:{to}\:{face}\:{BCD}\:{is}\:{at}\:{the}\:{centroid} \\ $$$${of}\:\bigtriangleup{BCD}. \\ $$$${Hence}\:\:\:{BF}=\frac{\mathrm{2}}{\mathrm{3}}\left({BE}\right)=\:\frac{\mathrm{2}}{\mathrm{3}}\left({r}\sqrt{\mathrm{3}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{2}{r}}{\sqrt{\mathrm{3}}}\:. \\ $$$${AF}^{\:\:\mathrm{2}} ={AB}^{\mathrm{2}} −{BF}^{\:\mathrm{2}} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} −\frac{\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{3}}\:=\frac{\mathrm{8}{r}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow\:\:\:\:\:{x}+{y}\:=\frac{\mathrm{2}\sqrt{\mathrm{2}}{r}}{\sqrt{\mathrm{3}}}\:\:\Rightarrow\:\:{y}=\frac{\mathrm{2}\sqrt{\mathrm{6}}{r}}{\mathrm{3}}−{x} \\ $$$${Also}\:\:\bigtriangleup{OBF}\:{is}\:{right}\:{angled},\:{so} \\ $$$$\:\:\:\:\:\:\:{x}^{\mathrm{2}} ={y}^{\mathrm{2}} +{BF}^{\:\:\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{x}^{\mathrm{2}} ={y}^{\mathrm{2}} +\frac{\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:{x}^{\mathrm{2}} =\left(\frac{\mathrm{2}\sqrt{\mathrm{6}}{r}}{\mathrm{3}}−{x}\right)^{\mathrm{2}} +\frac{\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow\:\:{x}^{\mathrm{2}} =\frac{\mathrm{8}{r}^{\mathrm{2}} }{\mathrm{3}}+{x}^{\mathrm{2}} −\frac{\mathrm{4}\sqrt{\mathrm{6}}{rx}}{\mathrm{3}}+\frac{\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow\:\:\:\:\:\frac{\mathrm{4}\sqrt{\mathrm{6}}{rx}}{\mathrm{3}}=\mathrm{4}{r}^{\mathrm{2}} \:\:{or}\:\:{x}=\frac{\mathrm{3}{r}}{\sqrt{\mathrm{6}}}\:=\frac{{r}\sqrt{\mathrm{3}}}{\sqrt{\mathrm{2}}}\:. \\ $$$${Radius}\:{of}\:{large}\:{sphere}\:{be}\:{R},\:{then} \\ $$$$\:\:\:\:\:\:{R}={x}+{r} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{{r}\sqrt{\mathrm{3}}}{\sqrt{\mathrm{2}}}+{r}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{R}=\:\frac{{r}}{\mathrm{2}}\left(\mathrm{2}+\sqrt{\mathrm{6}}\right)\:. \\ $$
Answered by 1kanika# last updated on 18/Aug/17
$$\mathrm{since}\:\mathrm{the}\:\mathrm{big}\:\mathrm{sphere}\:\mathrm{contains}\:\mathrm{all}\:\mathrm{four} \\ $$$$\mathrm{sphere}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{r}\:\mathrm{and}\:\mathrm{aldo}\:\mathrm{tangent}\: \\ $$$$\mathrm{to}\:\mathrm{all}\:\mathrm{four}\:\mathrm{spheres}\:,\:\mathrm{so}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{big}\:\mathrm{sphere} \\ $$
Answered by 1kanika# last updated on 18/Aug/17
$$\mathrm{is}\:\mathrm{2r}\:. \\ $$