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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

$a s Q 198763, b u t t h e b a l l s a r e n o t o n$ $a t a b l e, b u t i n s i d e a l a r g e s p h e r i c a l$ $b o w l w i t h r a d i u s R .$
	Answered by mr W last updated on 27/Oct/23

	Commented by mr W last updated on 27/Oct/23

	$k = r a d i u s o f t h e b a l l i n t h e m i d d l e$ $A M = R - r$ $O M = R - k$ $O A = k + r$ $O^{'} A = a = \frac{r}{\sin 22.5 °} = \sqrt{2 (2 + \sqrt{2})} r$ $\sqrt{{(k + r)}^{2} - a^{2}} = (R - k) - \sqrt{{(R - r)}^{2} - a^{2}}$ $R (R - r) - (R + r) k = (R - k) \sqrt{{(R - r)}^{2} - a^{2}}$ $(4 R r + a^{2}) k^{2} - 2 R (2 R r - 2 r^{2} + a^{2}) k + R^{2} a^{2} = 0$ $k = \frac{R (2 R r - 2 r^{2} + a^{2}) \pm \sqrt{R^{2} {(2 R r - 2 r^{2} + a^{2})}^{2} - R^{2} a^{2} (4 R r + a^{2})}}{4 R r + a^{2}}$ $\Rightarrow \frac{k}{R} = \frac{R + (1 + \sqrt{2}) r - \sqrt{R^{2} - 2 R r - {(1 + \sqrt{2})}^{2} r^{2}}}{2 R + (2 + \sqrt{2}) r}$ $o r$ $w i t h λ = \frac{r}{R}$ $\Rightarrow \frac{k}{r} = \frac{1 + (1 + \sqrt{2}) λ - \sqrt{1 - 2 λ - {(1 + \sqrt{2})}^{2} λ^{2}}}{2 λ + (2 + \sqrt{2}) λ^{2}}$ $\lim_{λ \to 0} (\frac{k}{r}) = \frac{1 + \sqrt{2} + 1}{2} = 1 + \frac{\sqrt{2}}{2}$ $t h i s i s t h e c a s e w h e n t h e b a l l s a r e$ $o n a t a b l e .$
	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
	$p=(a/R) , q=(b/R) , a=kb ⇒ p=kq (R−a)sin θ(sin (π/8))=a say sin (π/8)=λ and since a=kb (1−kq)λsin θ=kq .....(i) {(R−b)−(R−a)cos θ}^2 +(R−a)^2 sin^2 θ=(a+b)^2 ⇒ (R−a)^2 +(R−b)^2 −2(R−b)(R−a)cos θ =(a+b)^2 2R^2 −2R(a+b)−2ab =2(R−b)(R−a)cos θ or ⇒ 1−q(k+1)−kq^2 =(1−q)(1−kq)cos θ .....(ii) Now (1−kq)^2 cos^2 θ+(1−kq)^2 sin^2 θ =(1−kq)^2 ⇒ (({1−q(k+1)−kq^2 }^2 )/((1−q)^2 ))+((kq^2 )/λ^2 )=(1−kq)^2 ⇒ {(1−q)−kq(1+q)}^2 +((kq^2 (1−q)^2 )/λ^2 ) =(1+k^2 q^2 −2kq)(1−q)^2 ⇒ say (1/λ)=μ 4q^2 k^2 −2(1−q)(1+q)k +μ^2 q(1−q)^2 +2(1−q)^2 k=0 ⇒ 4qk^2 −4(1−q)k+μ^2 (1−q)^2 =0 k=(((1−q)/(2q))){1±(√(1−(q/λ^2 )))} (a/b)=(((R−b)/(2b))){1−(√(1−(b/(Rsin^2 (π/8)))))} If R→∞ (a/b)=lim_(R→∞) (((R−b)/(2b)))((b/(2Rsin^2 (π/8)))) =(1/(2(1−cos (π/4))))=(1/(2(1−(1/( (√2)))))) =(1/( (√2)((√2)−1)))=(((√2)+1)/( (√2)))=1+((√2)/2)≈ 1.707 but a<b ??????$
	$p = \frac{a}{R}, q = \frac{b}{R}, a = k b$ $\Rightarrow p = k q$ $(R - a) \sin θ (\sin \frac{π}{8}) = a$ $s a y \sin \frac{π}{8} = λ a n d s i n c e a = k b$ $(1 - k q) λ \sin θ = k q . . . . . (i)$ ${(R - b) - (R - a) \cos θ}^{2}$ $+ {(R - a)}^{2} \sin^{2} θ = {(a + b)}^{2}$ $\Rightarrow$ ${(R - a)}^{2} + {(R - b)}^{2} - 2 (R - b) (R - a) \cos θ$ $= {(a + b)}^{2}$ $2 R^{2} - 2 R (a + b) - 2 a b$ $= 2 (R - b) (R - a) \cos θ$ $o r$ $\Rightarrow 1 - q (k + 1) - {k q}^{2}$ $= (1 - q) (1 - k q) \cos θ . . . . . (i i)$ $N o w$ ${(1 - k q)}^{2} \cos^{2} θ + {(1 - k q)}^{2} \sin^{2} θ$ $= {(1 - k q)}^{2}$ $\Rightarrow$ $\frac{{1 - q (k + 1) - {k q}^{2}}^{2}}{{(1 - q)}^{2}} + \frac{{k q}^{2}}{λ^{2}} = {(1 - k q)}^{2}$ $\Rightarrow {(1 - q) - k q (1 + q)}^{2} + \frac{{k q}^{2} {(1 - q)}^{2}}{λ^{2}}$ $= (1 + k^{2} q^{2} - 2 k q) {(1 - q)}^{2}$ $\Rightarrow s a y \frac{1}{λ} = μ$ $4 q^{2} k^{2} - 2 (1 - q) (1 + q) k$ $+ μ^{2} q {(1 - q)}^{2} + 2 {(1 - q)}^{2} k = 0$ $\Rightarrow$ $4 {q k}^{2} - 4 (1 - q) k + μ^{2} {(1 - q)}^{2} = 0$ $k = (\frac{1 - q}{2 q}) {1 \pm \sqrt{1 - \frac{q}{λ^{2}}}}$ $\frac{a}{b} = (\frac{R - b}{2 b}) {1 - \sqrt{1 - \frac{b}{R \sin^{2} \frac{π}{8}}}}$ $I f R \to \infty$ $\frac{a}{b} = \lim_{R \to \infty} (\frac{R - b}{2 b}) (\frac{b}{2 R \sin^{2} \frac{π}{8}})$ $= \frac{1}{2 (1 - \cos \frac{π}{4})} = \frac{1}{2 (1 - \frac{1}{\sqrt{2}})}$ $= \frac{1}{\sqrt{2} (\sqrt{2} - 1)} = \frac{\sqrt{2} + 1}{\sqrt{2}} = 1 + \frac{\sqrt{2}}{2} \approx 1.707$ $b u t a < b ? ? ? ? ? ?$
	Commented by mr W last updated on 28/Oct/23

	$y o u w a n t t o f i n d a f r o m R, b .$ $h o w i s t o f i n d b f r o m R, a ?$
	Commented by ajfour last updated on 28/Oct/23

	$I t h o u g h t l e t s i n q u i r e w i t h w h a t$ $r a d i u s b a l l s t o s u r r o u n d w i t h,$ $g i v e n t h e s i n g l e b l u e b a l l i n s i d e a s p h e r e . .$ $(\overset{∙ ∙}{⌣})$
	Commented by mr W last updated on 28/Oct/23
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