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Question Number 213021 by Spillover last updated on 28/Oct/24

Commented by Ghisom last updated on 29/Oct/24

$r_{1} = 1 - \sqrt{2} + \sqrt{2 - \sqrt{2}} \approx . 351153302$ $r_{2} = \frac{(1 - \sqrt{2 - \sqrt{2}}) \sqrt{2}}{2} \approx . 165910681$ $\frac{r_{1}}{r_{2}} = 2 - \sqrt{2} + 2 \sqrt{2 - \sqrt{2}} \approx 2.11652017$
	Answered by a.lgnaoui last updated on 29/Oct/24

	$r_{1} = \frac{\sqrt{6 - 4 \sqrt{2}}}{1 + \sqrt{6 - 4 \sqrt{2}}}$ $r_{2} = \frac{\sqrt{55 - 36 \sqrt{2}} - 3 \sqrt{6 - 4 \sqrt{2}}}{1 + \sqrt{6 - 4 \sqrt{2}}}$ $then \frac{r_{1}}{r_{2}} = \frac{\sqrt{6 - 4 \sqrt{2}}}{\sqrt{55 - 36 \sqrt{2}} - 3 \sqrt{6 - 4 \sqrt{2}}}$
	Commented by a.lgnaoui last updated on 29/Oct/24

	Commented by a.lgnaoui last updated on 29/Oct/24

	Commented by Ghisom last updated on 29/Oct/24

	$the circles have no common point$ $the big circle touches the skewed line at$ $F \approx (\begin{matrix} . 324 \\ . 217 \end{matrix}) while the small circle touches$ $this line at G \approx (\begin{matrix} . 319 \\ . 229 \end{matrix})$ $∣ F G ∣\approx . 0137$ $\Rightarrow your result is wrong$
	Answered by mr W last updated on 29/Oct/24

	Commented by mr W last updated on 29/Oct/24

	$\frac{b}{a} = \sqrt{2} \Rightarrow b = \frac{\sqrt{2}}{\sqrt{2} + 1}$ $\frac{A B}{1} = \frac{b}{a} \Rightarrow A B = \sqrt{2}$ $E C = c - r_{1} + b - r_{1} = b + c - 2 r_{1}$ ${E C}^{2} = {(2 r_{1})}^{2} + {(c - b)}^{2}$ $4 r_{1}^{2} + {(c - b)}^{2} = {(b + c - 2 r_{1})}^{2}$ $\Rightarrow r_{1} = \frac{b c}{b + c}$ $\frac{c}{b} = \frac{\sqrt{2} + 2 r_{1}}{\sqrt{2}} = 1 + \sqrt{2} \times \frac{b c}{b + c} = 1 + \sqrt{2} \times \frac{\sqrt{2}}{\sqrt{2} + 1} \times \frac{(\frac{c}{b})}{1 + (\frac{c}{b})}$ ${(\frac{c}{b})}^{2} - 2 (\sqrt{2} - 1) (\frac{c}{b}) - 1 = 0$ $\Rightarrow \frac{c}{b} = \sqrt{2} - 1 + \sqrt{2 (2 - \sqrt{2})}$ $\frac{r_{1}}{r_{2}} = \frac{c}{a} = \frac{b}{a} \times \frac{c}{b} = \sqrt{2} [\sqrt{2} - 1 + \sqrt{2 (2 - \sqrt{2})}]$ $= 2 - \sqrt{2} + 2 \sqrt{2 - \sqrt{2}} \approx 2.116520$
	Commented by TonyCWX08 last updated on 31/Oct/24

	$W h a t A p p d o y o u u s e t o d r a w d i a g r a m s l i k e t h i s ?$ $A n d w h y I {c a n}^{'} t s e n d p i c t u r e s ?$
	Commented by mr W last updated on 31/Oct/24

	$i {d o n}^{'} t u s e a s p e c i a l a p p f o r d r a w i n g .$ $e v e r y s m a r t p h o n e h a s a n a p p f o r$ $e d i t i n g f o t o s . y o u c a n d r a w l i n e s,$ $m a k e f r e e - h a n d s k e t c h e t c .$ $t h e r e i s a g o o d a p p c a l l e d L e k h f o r$ $d r a w i n g d i a g r a m s, s e e p o s t s f r o m$ $a j f o u r s i r .$ $f o l l o w i n g p i c t u r e s h o w s h o w o n e$ $c a n u p l o a d a p i c t u r e a s a n s w e r o r a s$ $c o m m e n t . i f i t {d o e s n}^{'} t w o r k w i t h$ $y o u r s m a r t p h o n e, t h e n y o u s h o u l d$ $c o n t a c t t h e d e v e l o p e r T i n k u T a r a .$
	Commented by mr W last updated on 31/Oct/24

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