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Question Number 199046 by mr W last updated on 27/Oct/23
Commented by mr W last updated on 27/Oct/23
as Q198763, but the balls are not on a table, but inside a large spherical bowl with radius R.
$${as}\:{Q}\mathrm{198763},\:{but}\:{the}\:{balls}\:{are}\:{not}\:{on} \\ $$$${a}\:{table},\:{but}\:{inside}\:{a}\:{large}\:{spherical} \\ $$$${bowl}\:{with}\:{radius}\:{R}. \\ $$
Answered by mr W last updated on 27/Oct/23
Commented by mr W last updated on 27/Oct/23
k=radius of the ball in the middle AM=R−r OM=R−k OA=k+r O′A=a=(r/(sin 22.5°))=(√(2(2+(√2))))r (√((k+r)^2 −a^2 ))=(R−k)−(√((R−r)^2 −a^2 )) R(R−r)−(R+r)k=(R−k)(√((R−r)^2 −a^2 )) (4Rr+a^2 )k^2 −2R(2Rr−2r^2 +a^2 )k+R^2 a^2 =0 k=((R(2Rr−2r^2 +a^2 )±(√(R^2 (2Rr−2r^2 +a^2 )^2 −R^2 a^2 (4Rr+a^2 ))))/(4Rr+a^2 )) ⇒(k/R)=((R+(1+(√2))r−(√(R^2 −2Rr−(1+(√2))^2 r^2 )))/(2R+(2+(√2))r)) or with λ=(r/R) ⇒(k/r)=((1+(1+(√2))λ−(√(1−2λ−(1+(√2))^2 λ^2 )))/(2λ+(2+(√2))λ^2 )) lim_(λ→0) ((k/r))=((1+(√2)+1)/2)=1+((√2)/2) this is the case when the balls are on a table.
$${k}={radius}\:{of}\:{the}\:{ball}\:{in}\:{the}\:{middle} \\ $$$${AM}={R}−{r} \\ $$$${OM}={R}−{k} \\ $$$${OA}={k}+{r} \\ $$$${O}'{A}={a}=\frac{{r}}{\mathrm{sin}\:\mathrm{22}.\mathrm{5}°}=\sqrt{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{r} \\ $$$$\sqrt{\left({k}+{r}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }=\left({R}−{k}\right)−\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${R}\left({R}−{r}\right)−\left({R}+{r}\right){k}=\left({R}−{k}\right)\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}{Rr}+{a}^{\mathrm{2}} \right){k}^{\mathrm{2}} −\mathrm{2}{R}\left(\mathrm{2}{Rr}−\mathrm{2}{r}^{\mathrm{2}} +{a}^{\mathrm{2}} \right){k}+{R}^{\mathrm{2}} {a}^{\mathrm{2}} =\mathrm{0} \\ $$$${k}=\frac{{R}\left(\mathrm{2}{Rr}−\mathrm{2}{r}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\pm\sqrt{{R}^{\mathrm{2}} \left(\mathrm{2}{Rr}−\mathrm{2}{r}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} −{R}^{\mathrm{2}} {a}^{\mathrm{2}} \left(\mathrm{4}{Rr}+{a}^{\mathrm{2}} \right)}}{\mathrm{4}{Rr}+{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{k}}{{R}}=\frac{{R}+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){r}−\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} {r}^{\mathrm{2}} }}{\mathrm{2}{R}+\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){r}} \\ $$$${or} \\ $$$${with}\:\lambda=\frac{{r}}{{R}} \\ $$$$\Rightarrow\frac{{k}}{{r}}=\frac{\mathrm{1}+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\lambda−\sqrt{\mathrm{1}−\mathrm{2}\lambda−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \lambda^{\mathrm{2}} }}{\mathrm{2}\lambda+\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\lambda^{\mathrm{2}} } \\ $$$$\underset{\lambda\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{k}}{{r}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{2}}=\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${this}\:{is}\:{the}\:{case}\:{when}\:{the}\:{balls}\:{are} \\ $$$${on}\:{a}\:{table}. \\ $$
Commented by mr W last updated on 27/Oct/23
Commented by mr W last updated on 27/Oct/23
Answered by ajfour last updated on 27/Oct/23
Commented by ajfour last updated on 28/Oct/23
p=(a/R) , q=(b/R) , a=kb ⇒ p=kq (R−a)sin θ(sin (π/8))=a say sin (π/8)=λ and since a=kb (1−kq)λsin θ=kq .....(i) {(R−b)−(R−a)cos θ}^2 +(R−a)^2 sin^2 θ=(a+b)^2 ⇒ (R−a)^2 +(R−b)^2 −2(R−b)(R−a)cos θ =(a+b)^2 2R^2 −2R(a+b)−2ab =2(R−b)(R−a)cos θ or ⇒ 1−q(k+1)−kq^2 =(1−q)(1−kq)cos θ .....(ii) Now (1−kq)^2 cos^2 θ+(1−kq)^2 sin^2 θ =(1−kq)^2 ⇒ (({1−q(k+1)−kq^2 }^2 )/((1−q)^2 ))+((kq^2 )/λ^2 )=(1−kq)^2 ⇒ {(1−q)−kq(1+q)}^2 +((kq^2 (1−q)^2 )/λ^2 ) =(1+k^2 q^2 −2kq)(1−q)^2 ⇒ say (1/λ)=μ 4q^2 k^2 −2(1−q)(1+q)k +μ^2 q(1−q)^2 +2(1−q)^2 k=0 ⇒ 4qk^2 −4(1−q)k+μ^2 (1−q)^2 =0 k=(((1−q)/(2q))){1±(√(1−(q/λ^2 )))} (a/b)=(((R−b)/(2b))){1−(√(1−(b/(Rsin^2 (π/8)))))} If R→∞ (a/b)=lim_(R→∞) (((R−b)/(2b)))((b/(2Rsin^2 (π/8)))) =(1/(2(1−cos (π/4))))=(1/(2(1−(1/( (√2)))))) =(1/( (√2)((√2)−1)))=(((√2)+1)/( (√2)))=1+((√2)/2)≈ 1.707 but a<b ??????
$${p}=\frac{{a}}{{R}}\:\:\:,\:\:{q}=\frac{{b}}{{R}}\:\:\:,\:\:\:\:{a}={kb} \\ $$$$\Rightarrow\:\:{p}={kq} \\ $$$$\left({R}−{a}\right)\mathrm{sin}\:\theta\left(\mathrm{sin}\:\frac{\pi}{\mathrm{8}}\right)={a} \\ $$$${say}\:\mathrm{sin}\:\frac{\pi}{\mathrm{8}}=\lambda\:{and}\:\:\:{since}\:\:{a}={kb} \\ $$$$\left(\mathrm{1}−{kq}\right)\lambda\mathrm{sin}\:\theta={kq}\:\:\:\:…..\left({i}\right) \\ $$$$\left\{\left({R}−{b}\right)−\left({R}−{a}\right)\mathrm{cos}\:\theta\right\}^{\mathrm{2}} \\ $$$$+\left({R}−{a}\right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta=\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\:\left({R}−{a}\right)^{\mathrm{2}} +\left({R}−{b}\right)^{\mathrm{2}} −\mathrm{2}\left({R}−{b}\right)\left({R}−{a}\right)\mathrm{cos}\:\theta \\ $$$$\:\:\:=\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}{R}\left({a}+{b}\right)−\mathrm{2}{ab} \\ $$$$\:\:\:\:\:=\mathrm{2}\left({R}−{b}\right)\left({R}−{a}\right)\mathrm{cos}\:\theta \\ $$$${or}\:\: \\ $$$$\Rightarrow\:\:\mathrm{1}−{q}\left({k}+\mathrm{1}\right)−{kq}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\left(\mathrm{1}−{q}\right)\left(\mathrm{1}−{kq}\right)\mathrm{cos}\:\theta\:\:\:\:\:…..\left({ii}\right) \\ $$$${Now} \\ $$$$\:\left(\mathrm{1}−{kq}\right)^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta+\left(\mathrm{1}−{kq}\right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}−{kq}\right)^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\frac{\left\{\mathrm{1}−{q}\left({k}+\mathrm{1}\right)−{kq}^{\mathrm{2}} \right\}^{\mathrm{2}} }{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} }+\frac{{kq}^{\mathrm{2}} }{\lambda^{\mathrm{2}} }=\left(\mathrm{1}−{kq}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\left\{\left(\mathrm{1}−{q}\right)−{kq}\left(\mathrm{1}+{q}\right)\right\}^{\mathrm{2}} +\frac{{kq}^{\mathrm{2}} \left(\mathrm{1}−{q}\right)^{\mathrm{2}} }{\lambda^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:=\left(\mathrm{1}+{k}^{\mathrm{2}} {q}^{\mathrm{2}} −\mathrm{2}{kq}\right)\left(\mathrm{1}−{q}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{say}\:\:\frac{\mathrm{1}}{\lambda}=\mu \\ $$$$\mathrm{4}{q}^{\mathrm{2}} {k}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−{q}\right)\left(\mathrm{1}+{q}\right){k} \\ $$$$\:\:\:+\mu^{\mathrm{2}} {q}\left(\mathrm{1}−{q}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}−{q}\right)^{\mathrm{2}} {k}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{4}{qk}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}−{q}\right){k}+\mu^{\mathrm{2}} \left(\mathrm{1}−{q}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${k}=\left(\frac{\mathrm{1}−{q}}{\mathrm{2}{q}}\right)\left\{\mathrm{1}\pm\sqrt{\mathrm{1}−\frac{{q}}{\lambda^{\mathrm{2}} }}\right\} \\ $$$$\frac{{a}}{{b}}=\left(\frac{{R}−{b}}{\mathrm{2}{b}}\right)\left\{\mathrm{1}−\sqrt{\mathrm{1}−\frac{{b}}{{R}\mathrm{sin}\:^{\mathrm{2}} \frac{\pi}{\mathrm{8}}}}\right\} \\ $$$${If}\:\:{R}\rightarrow\infty \\ $$$$\frac{{a}}{{b}}=\underset{{R}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{R}−{b}}{\mathrm{2}{b}}\right)\left(\frac{{b}}{\mathrm{2}{R}\mathrm{sin}\:^{\mathrm{2}} \frac{\pi}{\mathrm{8}}}\right) \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\right)}=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}=\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\approx\:\mathrm{1}.\mathrm{707} \\ $$$${but}\:\:{a}<{b}\:\:?????? \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 28/Oct/23
you want to find a from R, b. how is to find b from R, a?
$${you}\:{want}\:{to}\:{find}\:\boldsymbol{{a}}\:{from}\:\boldsymbol{{R}},\:\boldsymbol{{b}}. \\ $$$${how}\:{is}\:{to}\:{find}\:\boldsymbol{{b}}\:{from}\:\boldsymbol{{R}},\:\boldsymbol{{a}}? \\ $$
Commented by ajfour last updated on 28/Oct/23
I thought lets inquire with what radius balls to surround with, given the single blue ball inside a sphere.. (⌣^(• •) )
$${I}\:{thought}\:{lets}\:{inquire}\:{with}\:{what} \\ $$$${radius}\:{balls}\:{to}\:{surround}\:{with},\: \\ $$$${given}\:{the}\:{single}\:{blue}\:{ball}\:{inside}\:{a}\:{sphere}..\: \\ $$$$\left(\overset{\bullet\:\bullet} {\smile}\right) \\ $$
Commented by mr W last updated on 28/Oct/23
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