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Question Number 215094 by ajfour last updated on 28/Dec/24

Commented by ajfour last updated on 28/Dec/24
A question discussed Statics+Geometry https://youtu.be/vktsHDGgBlw?si=z6XC7xw6G8m6XjrS
Commented by mr W last updated on 28/Dec/24

Commented by mr W last updated on 28/Dec/24

$\frac{\sin (\frac{π}{2} - α - θ)}{\sin α} = \frac{\sin (α + θ)}{\sin (\frac{π}{2} - α)}$ $\Rightarrow \tan (α + θ) = \frac{1}{\tan α} = \tan (\frac{π}{2} - α)$ $\Rightarrow α + θ = \frac{π}{2} - α \Rightarrow θ = \frac{π}{2} - 2 α$
Commented by Frix last updated on 29/Dec/24

$\frac{R^{2}}{r^{2}} = 12 + 4 \sqrt{3}$ $(Simply had forgotten to \sqrt{. . .})$
Commented by ajfour last updated on 28/Dec/24
Explain me the first line itself sir.
Commented by mr W last updated on 28/Dec/24

$\frac{\sin (\frac{π}{2} - α - θ)}{\sin α} = \frac{C D}{A C} = \frac{C D}{B C} = \frac{\sin (α + θ)}{\sin (\frac{π}{2} - α)}$
Commented by ajfour last updated on 28/Dec/24

$Y e a h$ $θ_{a j f o u r} + θ_{m r w} = 0$ $w e h a v e j u s t m a r k e d d i f f e r e n t l y .$ $w e h a v e s a m e s i t u a t i o n . f o r e x a m p l e$ $c h e c k f o r α = 60 ° .$
Commented by ajfour last updated on 28/Dec/24

$T h a n k s s i r F r i x, y o u c a m e b a c k$ $a f t e r r e a l l y s o m e t i m e .$
Commented by mr W last updated on 29/Dec/24

$i n t h i s c a s e w e c a n e v e n ‘ ‘ {s e e}^{″} t h e$ $r e s u l t o b v i o u s l y :$
Commented by mr W last updated on 29/Dec/24

Commented by ajfour last updated on 29/Dec/24
https://youtu.be/m_z86aXCFIo?si=vpI3_hoVBncvRfE_
	Answered by mr W last updated on 29/Dec/24

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

	$\sin θ = \frac{r}{R} = \frac{1}{k}$ $A B = 2 \sqrt{R^{2} - r^{2}}$ $O B = \sqrt{R^{2} - r^{2}} + \sqrt{{(\sqrt{2} R)}^{2} - r^{2}} = \sqrt{R^{2} - r^{2}} + \sqrt{2 R^{2} - r^{2}}$ $\cos 2 θ = \frac{4 (R^{2} - r^{2}) + R^{2} - r^{2} + 2 R^{2} - r^{2} + 2 \sqrt{(R^{2} - r^{2}) (2 R^{2} - r^{2})} - R^{2}}{4 \sqrt{R^{2} - r^{2}} (\sqrt{R^{2} - r^{2}} + \sqrt{2 R^{2} - r^{2}})}$ $= \frac{3 k^{2} - 3 + \sqrt{(k^{2} - 1) (2 k^{2} - 1)}}{2 \sqrt{k^{2} - 1} (\sqrt{k^{2} - 1} + \sqrt{2 k^{2} - 1})}$ $\frac{3 k^{2} - 3 + \sqrt{(k^{2} - 1) (2 k^{2} - 1)}}{2 \sqrt{k^{2} - 1} (\sqrt{k^{2} - 1} + \sqrt{2 k^{2} - 1})} = 1 - \frac{2}{k^{2}}$ $\Rightarrow \frac{k^{2} + 4}{k^{2} - 4} = \sqrt{\frac{2 k^{2} - 1}{k^{2} - 1}}$ $\Rightarrow k^{4} - 24 k^{2} + 32 = 0$ $\Rightarrow k^{2} = 4 (3 + \sqrt{7})$ $\frac{R}{r} = k = 2 \sqrt{3 + \sqrt{7}} \approx 4.752158$
	Commented by mr W last updated on 29/Dec/24

	Commented by Frix last updated on 29/Dec/24
	$Let R=1 ⇒ C_1 : (x−1)^2 +(y−1)^2 =1 C_2 : (x−1)^2 +(y−1)^2 =r^2 L_1 : y=((1+(√(2p−p^2 )))/p)x L_2 : y=((√(2p−p^2 ))/p)x+1 C_2 ∩L_2 ={T_2 } ⇒ p=2−2r^2 C_2 ∩L_1 ={T_1 } ⇒ r=2(√(3+(√7)))=4.75215...$
	$Let R = 1$ $\Rightarrow$ $C_{1} : {(x - 1)}^{2} + {(y - 1)}^{2} = 1$ $C_{2} : {(x - 1)}^{2} + {(y - 1)}^{2} = r^{2}$ $L_{1} : y = \frac{1 + \sqrt{2 p - p^{2}}}{p} x$ $L_{2} : y = \frac{\sqrt{2 p - p^{2}}}{p} x + 1$ $C_{2} \cap L_{2} = {T_{2}} \Rightarrow p = 2 - 2 r^{2}$ $C_{2} \cap L_{1} = {T_{1}} \Rightarrow r = 2 \sqrt{3 + \sqrt{7}} = 4.75215 . . .$
	Commented by mr W last updated on 29/Dec/24
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