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Question Number 50561 by ajfour last updated on 17/Dec/18

Commented by ajfour last updated on 17/Dec/18

$F i n d r a d i u s r o f d a s h e d c i r c l e s$ $(a l l e q u a l), g i v e n n a n d R .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 18/Dec/18

	$j o i n i n g t h e c e n t r e s o f n c i r c l e s m a k e a [c i r c l e$ $o f r a d i u s = R + r$ $j o i n e a c h c e n t r e o f c i r c l e o f r a d i u s r w i t h$ $c e n t r e o f c i r c l e o f r a d i u s R$ $(n) θ = 2 π$ $θ = \frac{2 π}{n}$ $c o s θ = \frac{{(R + r)}^{2} + {(R + r)}^{2} - {(2 r)}^{2}}{2 (R + r) (R + r)}$ ${(2 r)}^{2} = 2 {(R + r)}^{2} - 2 {(R + r)}^{2} c o s θ$ ${(2 r)}^{2} = 2 {(R + r)}^{2} [1 - c o s θ]$ ${(2 r)}^{2} = 4 {(R + r)}^{2} {s i n}^{2} (\frac{θ}{2})$ $2 r = 2 (R + r) s i n (\frac{θ}{2})$ $r = (R + r) s i n (\frac{θ}{2})$ $r - r s i n (\frac{θ}{2}) = R s i n (\frac{θ}{2})$ $r = \frac{R s i n (\frac{θ}{2})}{1 - s i n (\frac{θ}{2})} = \frac{R s i n (\frac{π}{n})}{1 - s i n (\frac{π}{n})}$ $p l s c h e c k . .$
	Commented by ajfour last updated on 18/Dec/18

	$T h a n k s T a n m a y S i r .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 18/Dec/18

	$t h a n k y o u s i r . . . l d t m e r e c t i f y . .$
	Answered by mr W last updated on 17/Dec/18

	$l e t λ = \frac{r}{R}$ $\sin θ = \frac{r}{r + R} = \frac{λ}{1 + λ}$ $\Rightarrow λ = \frac{\sin θ}{1 - \sin θ} = \frac{1}{1 - \sin θ} - 1$ $2 n θ = 2 π$ $\Rightarrow θ = \frac{π}{n}$ $\Rightarrow λ = \frac{1}{1 - \sin \frac{π}{n}} - 1$ $============$ $λ = \frac{1}{1 - \sin \frac{π}{n}} - 1 ⩽ 1$ $\sin \frac{π}{n} ⩽ \frac{1}{2}$ $\frac{π}{n} ⩽ \frac{π}{6}$ $\Rightarrow n ⩾ 6$
	Commented by mr W last updated on 17/Dec/18

	$I h a v e n o t i c e d i t, b u t I c a n n o t s o l v e i t .$
	Commented by ajfour last updated on 17/Dec/18

	$G o o d S i r, T h a n k s .$ $h o w a b o u t r a d i u s o f n e x t l a y e r$ $c i r c l e s ?!$
	Commented by mr W last updated on 17/Dec/18

	$n = 5 : r_{5} = R (\frac{1}{1 - \sin \frac{π}{5}} - 1) = (\frac{\sqrt{10 + 10 \sqrt{5}}}{5} - 1) R$ $n = 4 : r_{4} = R (\frac{1}{1 - \sin \frac{π}{4}} - 1) = (1 + 2 \sqrt{2}) R$ $n = 3 : r_{3} = R (\frac{1}{1 - \sin \frac{π}{3}} - 1) = (3 + 2 \sqrt{3}) R$
	Commented by ajfour last updated on 17/Dec/18

	$n e x t L A Y E R C i r c l e s, S i r ?$
	Commented by mr W last updated on 18/Dec/18
	$let s=radius of circles from next layer (√((s+r)^2 −s^2 ))+R+r=(s/(tan θ))=(s/(tan (π/n))) (√(r(2s+r)))=(s/(tan (π/n)))−R−r (s^2 /(tan^2 (π/n)))−2[(((R+r))/(tan (π/n)))+r]s+R(R+2r)=0 s=(((((R+r))/(tan (π/n)))+r+(√([(((R+r))/(tan (π/n)))+r]^2 −((R(R+2r))/(tan^2 (π/n))))))/(1/(tan^2 (π/n)))) ⇒s=tan (π/n) {R+(1+tan (π/n))r+(√([R+(1+tan (π/n))r]^2 −R(R+2r)))}$
	$l e t s = r a d i u s o f c i r c l e s f r o m n e x t l a y e r$ $\sqrt{{(s + r)}^{2} - s^{2}} + R + r = \frac{s}{\tan θ} = \frac{s}{\tan \frac{π}{n}}$ $\sqrt{r (2 s + r)} = \frac{s}{\tan \frac{π}{n}} - R - r$ $\frac{s^{2}}{\tan^{2} \frac{π}{n}} - 2 [\frac{(R + r)}{\tan \frac{π}{n}} + r] s + R (R + 2 r) = 0$ $s = \frac{\frac{(R + r)}{\tan \frac{π}{n}} + r + \sqrt{{[\frac{(R + r)}{\tan \frac{π}{n}} + r]}^{2} - \frac{R (R + 2 r)}{\tan^{2} \frac{π}{n}}}}{\frac{1}{\tan^{2} \frac{π}{n}}}$ $\Rightarrow s = \tan \frac{π}{n} {R + (1 + \tan \frac{π}{n}) r + \sqrt{{[R + (1 + \tan \frac{π}{n}) r]}^{2} - R (R + 2 r)}}$
	Commented by ajfour last updated on 17/Dec/18

	$H e i g h t s! s o q u i c k s i r, s o m e t y p o s,$ $i g u e s s . .$
	Commented by ajfour last updated on 17/Dec/18

	Commented by mr W last updated on 17/Dec/18

	$n o! n o t s u c h a n i n t e g e r n e x i s t s .$ $n \approx 8.84$
	Commented by ajfour last updated on 17/Dec/18

	$t h a n k s s i r .$
	Answered by Smail last updated on 18/Dec/18

	$θ = \frac{2 π}{n}$ $s i n (\frac{θ}{2}) = \frac{r}{R + r} \Leftrightarrow r (1 - s i n (\frac{θ}{2}) = R s i n (\frac{θ}{2})$ $r = \frac{R s i n (\frac{θ}{2})}{1 - s i n (\frac{θ}{2})} = \frac{R}{c s c (\frac{θ}{2}) - 1}$ $r = \frac{R}{c s c (\frac{π}{n}) - 1}$ $W h e n n = 6, r s h o u l d e q u a l R$ $r_{6} = \frac{R}{c s c (\frac{π}{6}) - 1} = \frac{R}{2 - 1} = R$
	Commented by Smail last updated on 18/Dec/18

	Commented by Smail last updated on 18/Dec/18

	Commented by Smail last updated on 18/Dec/18

	$n = 6 s o r = R$
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