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Question Number 47697 by ajfour last updated on 13/Nov/18
Commented by ajfour last updated on 13/Nov/18
The blue sphere touches the xy, yz, zx, and the triangular plate, while the smaller pink sphere touches the yz, zx, larger sphere, and the triangular plate. Find their radii R, and r in terms of a,b,and c.
$${The}\:{blue}\:{sphere}\:{touches}\:{the} \\ $$$${xy},\:{yz},\:{zx},\:{and}\:{the}\:{triangular} \\ $$$${plate},\:{while}\:{the}\:{smaller}\:{pink} \\ $$$${sphere}\:{touches}\:{the}\:{yz},\:{zx},\:{larger} \\ $$$${sphere},\:{and}\:{the}\:{triangular}\:{plate}. \\ $$$${Find}\:{their}\:{radii}\:\boldsymbol{{R}},\:{and}\:\boldsymbol{{r}}\:{in}\:{terms} \\ $$$${of}\:\boldsymbol{{a}},\boldsymbol{{b}},{and}\:\boldsymbol{{c}}. \\ $$
Answered by ajfour last updated on 13/Nov/18
let center of blue sphere B(R,R,R) eq. of △plate : (x/a)+(y/b)+(z/c)=1 Normal to △plate is n^� =(i^� /a)+(j^� /b)+(k^� /c) let blue sphere touch the △plate at T ⇒ r_T ^� =R(i^� +j^� +k^� )+ (R/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))((i^� /a)+(j^� /b)+(k^� /c)) let (√((1/a^2 )+(1/b^2 )+(1/c^2 ))) = ∣n^� ∣=l As r_T ^� lies on △plate ⇒ R[(1/a)(1+(1/(al)))+(1/b)(1+(1/(bl)))+(1/c)(1+(1/(cl)))]=1 ⇒ R =(1/(((1/a)+(1/b)+(1/c))+(√((1/a^2 )+(1/b^2 )+(1/c^2 ))))) Let center of pink sphere be P (r,r,z). As BP should be R+r ⇒ 2(R−r)^2 +(z−R)^2 = (R+r)^2 ⇒ 2((R/r)−1)^2 +((z/r)−(R/r))^2 = ((R/r)+1)^2 ⇒ (z/r)=(R/r)+(√(((R/r)+1)^2 −2((R/r)−1)^2 )) ......(i) And ⊥ distance of P from plate is r. Let smaller sphere touch the plate at S. r_S ^� = (ri^� +rj^� +zk^� )+ (r/l)((i^� /a)+(j^� /b)+(k^� /c)) and S lies on the plate, therefore r[(1/a)(1+(1/(al)))+(1/b)(1+(1/(bl)))+(1/c)((z/r)+(1/(cl)))]= 1 ⇒ r = (1/((1/a)+(1/b)+(1/c)((z/r))+l)) ⇒ l+(1/a)+(1/b)+(1/c)((z/r))=(1/r) Now using (i) l+(1/a)+(1/b)+(1/c)[(R/r)+(√(((R/r)+1)^2 −2((R/r)−1)^2 )) ]=(1/r) let (1/p) = l+(1/a)+(1/b) = (1/R)−(1/c) ; then (1/p)+(1/c)[(R/r)+(√(((R/r)+1)^2 −2((R/r)−1)^2 )) ]=(1/r) ⇒((R/r)+1)^2 −2((R/r)−1)^2 = [c((1/r)−(1/p))−(R/r)]^2 ⇒ ((6R)/r)−1−2((R/r))^2 =c^2 ((1/r)−(1/p))^2 −((2cR)/r)((1/r)−(1/p)) let (1/r) = t t^2 (c^2 −2cR+2R^2 )−t(((2c^2 )/p)+((2cR)/p)−6R) +((c^2 /p^2 )+1) = 0 __________________________ ⇒ (1/r) = ((((c^2 /p)+((cR)/p)−3R)±(√(((c^2 /p)+((cR)/p)−3R)^2 −(c^2 −2cR+2R^2 )((c^2 /p^2 )+1))))/(c^2 −2cR+2R^2 )) with (1/p) =(1/a)+(1/b)+(√((1/a^2 )+(1/b^2 )+(1/c^2 ))) and R =(1/(((1/a)+(1/b)+(1/c))+(√((1/a^2 )+(1/b^2 )+(1/c^2 ))))) . __________________________
$${let}\:{center}\:{of}\:{blue}\:{sphere} \\ $$$${B}\left({R},{R},{R}\right) \\ $$$${eq}.\:{of}\:\bigtriangleup{plate}\::\:\:\:\frac{{x}}{{a}}+\frac{{y}}{{b}}+\frac{{z}}{{c}}=\mathrm{1} \\ $$$${Normal}\:{to}\:\bigtriangleup{plate}\:{is} \\ $$$$\:\:\:\:\bar {{n}}=\frac{\hat {{i}}}{{a}}+\frac{\hat {{j}}}{{b}}+\frac{\hat {{k}}}{{c}} \\ $$$${let}\:{blue}\:{sphere}\:{touch}\:{the}\:\bigtriangleup{plate} \\ $$$${at}\:{T}\: \\ $$$$\Rightarrow\:\bar {{r}}_{{T}} ={R}\left(\hat {{i}}+\hat {{j}}+\hat {{k}}\right)+ \\ $$$$\:\:\:\:\:\:\frac{{R}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}\left(\frac{\hat {{i}}}{{a}}+\frac{\hat {{j}}}{{b}}+\frac{\hat {{k}}}{{c}}\right) \\ $$$${let}\:\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}\:=\:\mid\bar {{n}}\mid={l} \\ $$$${As}\:\bar {{r}}_{{T}} \:{lies}\:{on}\:\bigtriangleup{plate} \\ $$$$\Rightarrow\:{R}\left[\frac{\mathrm{1}}{{a}}\left(\mathrm{1}+\frac{\mathrm{1}}{{al}}\right)+\frac{\mathrm{1}}{{b}}\left(\mathrm{1}+\frac{\mathrm{1}}{{bl}}\right)+\frac{\mathrm{1}}{{c}}\left(\mathrm{1}+\frac{\mathrm{1}}{{cl}}\right)\right]=\mathrm{1} \\ $$$$\Rightarrow\:{R}\:=\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}} \\ $$$${Let}\:{center}\:{of}\:{pink}\:{sphere}\:{be} \\ $$$$\:\:{P}\:\left({r},{r},{z}\right). \\ $$$${As}\:{BP}\:{should}\:{be}\:{R}+{r}\:\Rightarrow \\ $$$$\:\:\mathrm{2}\left({R}−{r}\right)^{\mathrm{2}} +\left({z}−{R}\right)^{\mathrm{2}} =\:\left({R}+{r}\right)^{\mathrm{2}} \:\: \\ $$$$\Rightarrow\:\mathrm{2}\left(\frac{{R}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} +\left(\frac{{z}}{{r}}−\frac{{R}}{{r}}\right)^{\mathrm{2}} =\:\left(\frac{{R}}{{r}}+\mathrm{1}\right)^{\mathrm{2}} \: \\ $$$$\Rightarrow\:\:\frac{{z}}{{r}}=\frac{{R}}{{r}}+\sqrt{\left(\frac{{R}}{{r}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{{R}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……\left({i}\right) \\ $$$${And} \\ $$$$\:\:\:\bot\:{distance}\:{of}\:{P}\:{from}\:{plate}\:{is}\:{r}. \\ $$$${Let}\:{smaller}\:{sphere}\:{touch}\:{the} \\ $$$${plate}\:{at}\:{S}. \\ $$$$\:\:\bar {{r}}_{{S}} \:=\:\left({r}\hat {{i}}+{r}\hat {{j}}+{z}\hat {{k}}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{r}}{{l}}\left(\frac{\hat {{i}}}{{a}}+\frac{\hat {{j}}}{{b}}+\frac{\hat {{k}}}{{c}}\right) \\ $$$${and}\:{S}\:{lies}\:{on}\:{the}\:{plate},\:{therefore} \\ $$$${r}\left[\frac{\mathrm{1}}{{a}}\left(\mathrm{1}+\frac{\mathrm{1}}{{al}}\right)+\frac{\mathrm{1}}{{b}}\left(\mathrm{1}+\frac{\mathrm{1}}{{bl}}\right)+\frac{\mathrm{1}}{{c}}\left(\frac{{z}}{{r}}+\frac{\mathrm{1}}{{cl}}\right)\right]=\:\mathrm{1} \\ $$$$\Rightarrow\:\:{r}\:=\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\left(\frac{{z}}{{r}}\right)+{l}} \\ $$$$\Rightarrow\:\:{l}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\left(\frac{{z}}{{r}}\right)=\frac{\mathrm{1}}{{r}} \\ $$$${Now}\:{using}\:\left({i}\right) \\ $$$$\:\:\:{l}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\left[\frac{{R}}{{r}}+\sqrt{\left(\frac{{R}}{{r}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{{R}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} }\:\right]=\frac{\mathrm{1}}{{r}} \\ $$$${let}\:\:\:\frac{\mathrm{1}}{{p}}\:=\:{l}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\:=\:\frac{\mathrm{1}}{{R}}−\frac{\mathrm{1}}{{c}}\:;\:{then} \\ $$$$\:\:\:\:\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{c}}\left[\frac{{R}}{{r}}+\sqrt{\left(\frac{{R}}{{r}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{{R}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} }\:\right]=\frac{\mathrm{1}}{{r}} \\ $$$$\Rightarrow\left(\frac{{R}}{{r}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{{R}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} \:=\:\left[{c}\left(\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{p}}\right)−\frac{{R}}{{r}}\right]^{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{6}{R}}{{r}}−\mathrm{1}−\mathrm{2}\left(\frac{{R}}{{r}}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}} −\frac{\mathrm{2}{cR}}{{r}}\left(\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{p}}\right) \\ $$$${let}\:\:\frac{\mathrm{1}}{{r}}\:=\:{t} \\ $$$$\:{t}^{\mathrm{2}} \left({c}^{\mathrm{2}} −\mathrm{2}{cR}+\mathrm{2}{R}^{\mathrm{2}} \right)−{t}\left(\frac{\mathrm{2}{c}^{\mathrm{2}} }{{p}}+\frac{\mathrm{2}{cR}}{{p}}−\mathrm{6}{R}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{c}^{\mathrm{2}} }{{p}^{\mathrm{2}} }+\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{r}}\:=\:\frac{\left(\frac{{c}^{\mathrm{2}} }{{p}}+\frac{{cR}}{{p}}−\mathrm{3}{R}\right)\pm\sqrt{\left(\frac{{c}^{\mathrm{2}} }{{p}}+\frac{{cR}}{{p}}−\mathrm{3}{R}\right)^{\mathrm{2}} −\left({c}^{\mathrm{2}} −\mathrm{2}{cR}+\mathrm{2}{R}^{\mathrm{2}} \right)\left(\frac{{c}^{\mathrm{2}} }{{p}^{\mathrm{2}} }+\mathrm{1}\right)}}{{c}^{\mathrm{2}} −\mathrm{2}{cR}+\mathrm{2}{R}^{\mathrm{2}} } \\ $$$$\:{with}\:\:\:\frac{\mathrm{1}}{{p}}\:=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }} \\ $$$${and}\:\:\:{R}\:=\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}\:\:. \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$
Commented by mr W last updated on 13/Nov/18
+ is for a sphere which contains the big sphere.
$$+\:{is}\:{for}\:{a}\:{sphere}\:{which}\:{contains}\:{the} \\ $$$${big}\:{sphere}. \\ $$
Commented by ajfour last updated on 13/Nov/18
But i have used R+r as the distance between their centers.
$${But}\:{i}\:{have}\:{used}\:{R}+{r}\:{as}\:{the} \\ $$$${distance}\:{between}\:{their}\:{centers}. \\ $$
Commented by mr W last updated on 13/Nov/18
I was wrong sir. It is for the sphere outside. This sphere touches the big one and the inclined plane at the same point.
$${I}\:{was}\:{wrong}\:{sir}.\:{It}\:{is}\:{for}\:{the}\:{sphere} \\ $$$${outside}.\:{This}\:{sphere}\:{touches}\:{the}\:{big} \\ $$$${one}\:{and}\:{the}\:{inclined}\:{plane}\:{at}\:{the}\:{same} \\ $$$${point}. \\ $$
Commented by MJS last updated on 14/Nov/18
just think what you set up before. you′ve been looking for spheres touching the blue sphere, the triangular plate and the two walls. the formulas don′t know the triangular plate ends at the floor and walls ⇒ it′s a plane ⇒ one sphere is the pink one in your picture, the other one′s under the blue one, it′s bigger than the blue one and still touching the two walls and − under the floor − the plane of the triangular plate you gave the formula for (1/r), r will be greater with the − and smaller with the +
$$\mathrm{just}\:\mathrm{think}\:\mathrm{what}\:\mathrm{you}\:\mathrm{set}\:\mathrm{up}\:\mathrm{before}.\:\mathrm{you}'\mathrm{ve}\:\mathrm{been} \\ $$$$\mathrm{looking}\:\mathrm{for}\:\mathrm{spheres}\:\mathrm{touching}\:\mathrm{the}\:\mathrm{blue}\:\mathrm{sphere}, \\ $$$$\mathrm{the}\:\mathrm{triangular}\:\mathrm{plate}\:\mathrm{and}\:\mathrm{the}\:\mathrm{two}\:\mathrm{walls}.\:\mathrm{the} \\ $$$$\mathrm{formulas}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{triangular}\:\mathrm{plate} \\ $$$$\mathrm{ends}\:\mathrm{at}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{and}\:\mathrm{walls}\:\Rightarrow\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{plane} \\ $$$$\Rightarrow\:\mathrm{one}\:\mathrm{sphere}\:\mathrm{is}\:\mathrm{the}\:\mathrm{pink}\:\mathrm{one}\:\mathrm{in}\:\mathrm{your}\:\mathrm{picture}, \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{one}'\mathrm{s}\:\mathrm{under}\:\mathrm{the}\:\mathrm{blue}\:\mathrm{one},\:\mathrm{it}'\mathrm{s}\:\mathrm{bigger} \\ $$$$\mathrm{than}\:\mathrm{the}\:\mathrm{blue}\:\mathrm{one}\:\mathrm{and}\:\mathrm{still}\:\mathrm{touching}\:\mathrm{the}\:\mathrm{two} \\ $$$$\mathrm{walls}\:\mathrm{and}\:−\:\mathrm{under}\:\mathrm{the}\:\mathrm{floor}\:−\:\mathrm{the}\:\mathrm{plane}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{triangular}\:\mathrm{plate} \\ $$$$\mathrm{you}\:\mathrm{gave}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{for}\:\frac{\mathrm{1}}{{r}},\:{r}\:\mathrm{will}\:\mathrm{be}\:\mathrm{greater} \\ $$$$\mathrm{with}\:\mathrm{the}\:−\:\mathrm{and}\:\mathrm{smaller}\:\mathrm{with}\:\mathrm{the}\:+ \\ $$
Commented by ajfour last updated on 14/Nov/18
yes sir, truly it is as you said.
$${yes}\:{sir},\:{truly}\:{it}\:{is}\:\:{as}\:{you}\:{said}. \\ $$

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