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Question-19915

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

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