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Question Number 28516 by beh.i83417@gmail.com last updated on 26/Jan/18

Commented by beh.i83417@gmail.com last updated on 26/Jan/18

$\frac{r}{s} = ?$
	Answered by mrW2 last updated on 26/Jan/18

	Commented by mrW2 last updated on 26/Jan/18

	$\sin α = \frac{\sqrt{2} s}{r + s}$ $\Rightarrow \cos α = \sqrt{1 - {(\frac{\sqrt{2} s}{r + s})}^{2}} = \frac{\sqrt{r^{2} + 2 r s - s^{2}}}{r + s}$ $(r + s) \cos (\frac{π}{4} - α) = r - s$ $(r + s) \frac{1}{\sqrt{2}} (\cos α + \sin α) = r - s$ $(r + s) \frac{1}{\sqrt{2}} (\frac{\sqrt{r^{2} + 2 r s - s^{2}} + \sqrt{2} s}{r + s}) = r - s$ $\sqrt{r^{2} + 2 r s - s^{2}} = \sqrt{2} (r - 2 s)$ $r^{2} + 2 r s - s^{2} = 2 (r^{2} - 4 r s + 4 s^{2})$ $9 s^{2} - 10 r s + r^{2} = 0$ $(9 s - r) (s - r) = 0$ $\Rightarrow s = \frac{r}{9} o r s = r (n o t s u i t a b l e)$ $\Rightarrow \frac{r}{s} = 9$
	Answered by ajfour last updated on 26/Jan/18

	$l i n e t h r o u g h c e n t r e o f$ $t w o s m a l l c i r c l e s (t h a t t o u c h t h e$ $b i g g e r c i r c l e) b e S_{2} S_{3}$ $l e t m i d p o i n t o f S_{2} S_{3} b e M$ $d i s t a n c e o f c e n t r e o f b i g c i r c l e C$ $f r o m t h i s l i n e i s$ $C M = r \sqrt{2} - 2 \sqrt{2} s, {M S}_{3} = s \sqrt{2}$ ${M S}_{3}^{2} + {C M}^{2} = {C S}_{2}^{2} = {C S}_{3}^{2}$ $\Rightarrow 2 s^{2} + 2 {(r - 2 s)}^{2} = {(r + s)}^{2}$ $\Rightarrow 2 s^{2} + 2 r^{2} - 8 r s + 8 s^{2} = r^{2} + s^{2} + 2 r s$ $r^{2} - 10 r s + 9 s^{2} = 0$ $(r - 9 s) (r - s) = 0$ $\Rightarrow \frac{r}{s} = 9 .$
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