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Question-47697




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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