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Question Number 198763 by mr W last updated on 24/Oct/23

Commented by mr W last updated on 24/Oct/23

$8 s m a l l b a l l s a n d a b i g b a l l o n a$ $t a b l e t o u c h e a c h o t h e r .$ $i f t h e r a d i u s o f t h e s m a l l b a l l s i s r,$ $f i n d t h e r a d i u s o f t h e b i g b a l l .$
Commented by ajfour last updated on 24/Oct/23

$c o u l d i t b e \frac{R}{r} = 1 + \frac{1}{\sqrt{2}} \approx 1.707$
Commented by mr W last updated on 24/Oct/23

$y e s . i g o t t h e s a m e .$
	Answered by MM42 last updated on 24/Oct/23

	$α = 45^{0} & {(2 r)}^{2} = {(R + r)}^{2} + {(R + r)}^{2} - 2 {(R + r)}^{2} \times c o s 45$ ${(2 r)}^{2} = {(R + r)}^{2} (2 - \sqrt{2})$ $\Rightarrow R + r = \frac{2 r}{\sqrt{2 - \sqrt{2}}} \Rightarrow R = (\frac{2}{\sqrt{2 - \sqrt{2}}} - 1) r$ $R = (\sqrt{4 + 2 \sqrt{2}} - 1) r ✓$
	Commented by MM42 last updated on 24/Oct/23

	Answered by nikif99 last updated on 24/Oct/23

	$R + r = \frac{a}{2 \sin (\frac{180}{n})} = \frac{2 r}{2 \sin 22.5} \Rightarrow R = r (\frac{1}{\frac{\sqrt{2 - \sqrt{2}}}{2}} - 1) =$ $r (\frac{2}{\sqrt{2 - \sqrt{2}}} - 1) = r (\frac{2 \sqrt{2 + \sqrt{2}}}{\sqrt{2}} - 1) \Rightarrow$ $R = r (\sqrt{2} \cdot \sqrt{2 + \sqrt{2}} - 1) \approx 1.613 r$
	Commented by nikif99 last updated on 24/Oct/23

	Commented by mr W last updated on 24/Oct/23

	Commented by mr W last updated on 24/Oct/23

	$b e c a r e f u l l!$ $i n g r o u n d v i e w t h e b i g c i r c l e w i t h$ $r a d i u s R d o e s n t t o u c h t h e s m a l l$ $c i r c l e s w i t h r a d i u s r . t h e y o v e r l a p!$
	Commented by nikif99 last updated on 24/Oct/23

	$r i g h t . A d e t a i l n o t n o t i c e d .$
	Answered by mr W last updated on 24/Oct/23

	Commented by mr W last updated on 24/Oct/23

	$θ = \frac{360 °}{8 \times 2} = 22.5 °$ $1 - 2 \sin^{2} 22.5 ° = \cos 45 ° = \frac{\sqrt{2}}{2}$ $\Rightarrow \sin 22.5 ° = \frac{\sqrt{2 - \sqrt{2}}}{2}$ $a = \frac{r}{\sin θ} = \frac{2 r}{\sqrt{2 - \sqrt{2}}} = \sqrt{2 (2 + \sqrt{2})} r$ ${(R + r)}^{2} - {(R - r)}^{2} = a^{2}$ $4 R r = 2 (2 + \sqrt{2}) r^{2}$ $\Rightarrow \frac{R}{r} = 1 + \frac{\sqrt{2}}{2} \approx 1.707$ $◼$
	Commented by MM42 last updated on 24/Oct/23

	$S i r W$ $i n t h e t w o i m a g e g i v e n b y y o u,$ $t h e v a l u e o f ‘ ‘ a^{″} d o s e n o t m a t c h$
	Commented by mr W last updated on 24/Oct/23

	$b u t t h e y d o m a t c h! ‘ ‘ a^{″} i s t h e p r o j e c t i o n$ $o f O A i n \underset{―}{h o r i z o n t a l} d i r e c t i o n .$ $t h e f i r s t i m a g e i s a g r o u n d v i e w, i . e .$ $a v i e w f r o m t o p .$ $a \neq O A = O B$ $a = O^{'} A^{'} = O^{'} B^{'}$
	Commented by mr W last updated on 24/Oct/23

	Commented by BaliramKumar last updated on 24/Oct/23

	$you are right Sir$ We understood that these are all circles
	Commented by MM42 last updated on 24/Oct/23

	$\underset{―}{\underset{⏟}{⋚}}$
	Commented by MM42 last updated on 24/Oct/23

	$S i r W$ $w i t h y o u r e x p l a n t i o n, i u n d e r s t a n d .$ $t h a n k f u l .$ $\underset{―}{\underset{⏟}{⋚}}$
	Commented by mr W last updated on 24/Oct/23
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