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Question Number 166417 by ajfour last updated on 19/Feb/22

	Answered by mr W last updated on 20/Feb/22

	Commented by mr W last updated on 20/Feb/22

	$O G = R - r$ $O F = 3 r$ ${(R - r)}^{2} = {(3 r)}^{2} + r^{2}$ $9 r^{2} + 2 R r - R^{2} = 0$ $\Rightarrow r = \frac{(\sqrt{10} - 1) R}{9}$ $\tan α = \frac{1}{3} \Rightarrow α = \tan^{- 1} \frac{1}{3}$ $β = \frac{π}{4} - α$ $O A = (1 + \sqrt{2}) r$ ${A G}^{2} = {(1 + \sqrt{2})}^{2} r^{2} + {(\sqrt{10} r)}^{2} - 2 (1 + \sqrt{2}) \sqrt{10} r^{2} \cos β$ ${A G}^{2} = (13 + 2 \sqrt{2} - 2 (1 + \sqrt{2}) \sqrt{10} \frac{1}{\sqrt{2}} (\frac{3}{\sqrt{10}} + \frac{1}{\sqrt{10}})) r^{2}$ ${A G}^{2} = (5 - 2 \sqrt{2}) r^{2}$ $A G = \sqrt{5 - 2 \sqrt{2}} r ✓$ $\frac{\sin φ}{\sqrt{10} r} = \frac{\sin β}{\sqrt{5 - 2 \sqrt{2}} r}$ $\sin φ = \frac{\sqrt{10}}{\sqrt{5 - 2 \sqrt{2}}} \times \frac{1}{\sqrt{2}} (\frac{3}{\sqrt{10}} - \frac{1}{\sqrt{10}})$ $\sin φ = \frac{\sqrt{2}}{\sqrt{5 - 2 \sqrt{2}}} \Rightarrow φ = π - \sin^{- 1} \frac{\sqrt{2}}{\sqrt{5 - 2 \sqrt{2}}}$ $\sin γ = \frac{1}{\sqrt{5 - 2 \sqrt{2}}} \Rightarrow γ = \sin^{- 1} \frac{1}{\sqrt{5 - 2 \sqrt{2}}}$ $δ = π - φ - γ = \sin^{- 1} \frac{\sqrt{2}}{\sqrt{5 - 2 \sqrt{2}}} - \sin^{- 1} \frac{1}{\sqrt{5 - 2 \sqrt{2}}}$ $\cos δ = \frac{\sqrt{(5 - 2 \sqrt{2} - 2) (5 - 2 \sqrt{2} - 1)}}{5 - 2 \sqrt{2}} + \frac{\sqrt{2}}{5 - 2 \sqrt{2}}$ $\cos δ = \frac{\sqrt{2} (1 + \sqrt{10 - 7 \sqrt{2}})}{5 - 2 \sqrt{2}}$ $\sin δ = \frac{\sqrt{2 (5 - 2 \sqrt{2} - 1)} - \sqrt{5 - 2 \sqrt{2} - 2}}{5 - 2 \sqrt{2}}$ $\sin δ = \frac{2 \sqrt{2 - \sqrt{2}} + 1 - \sqrt{2}}{5 - 2 \sqrt{2}}$ $O B = (1 + \sqrt{2}) r + A D \cos δ$ $B D = A D \sin δ$ ${O B}^{2} + {B D}^{2} = {O D}^{2} = R^{2}$ ${[(1 + \sqrt{2}) r + A D \cos δ]}^{2} + {(A D \sin δ)}^{2} = R^{2}$ ${(A D)}^{2} + 2 (1 + \sqrt{2}) \cos δ A D + {(1 + \sqrt{2})}^{2} - R^{2} = 0$ ${(\frac{A D}{r})}^{2} + 2 (1 + \sqrt{2}) \cos δ (\frac{A D}{r}) + {(1 + \sqrt{2})}^{2} - {(\frac{9}{\sqrt{10} - 1})}^{2} = 0$ ${(\frac{A D}{r})}^{2} + 2 (1 + \sqrt{2}) \frac{\sqrt{2} (1 + \sqrt{10 - 7 \sqrt{2}})}{5 - 2 \sqrt{2}} (\frac{A D}{r}) + {(1 + \sqrt{2})}^{2} - {(\frac{9}{\sqrt{10} - 1})}^{2} = 0$ ${(\frac{A D}{r})}^{2} + \frac{2 (14 + 9 \sqrt{2}) (1 + \sqrt{10 - 7 \sqrt{2}})}{17} (\frac{A D}{r}) - 2 (4 + \sqrt{10} - \sqrt{2}) = 0$ $\frac{A D}{r} = - \frac{(14 + 9 \sqrt{2}) (1 + \sqrt{10 - 7 \sqrt{2}})}{17} + \sqrt{\frac{{(14 + 9 \sqrt{2})}^{2} {(1 + \sqrt{10 - 7 \sqrt{2}})}^{2}}{17^{2}} + 2 (4 + \sqrt{10} - \sqrt{2})}$ $\frac{A D}{r} = \frac{\sqrt{2722 + 578 \sqrt{10} - 312 \sqrt{2} + (716 + 504 \sqrt{2}) \sqrt{10 - 7 \sqrt{2}}} - (14 + 9 \sqrt{2}) (1 + \sqrt{10 - 7 \sqrt{2}})}{17}$ $A r e a o f t r i a n g l e$ $A = {(A D)}^{2} \sin δ \cos δ$ $\approx 1.5956802 r^{2}$ $\approx 0.0921051 R^{2}$ $\approx 5.8947276$
	Commented by Tawa11 last updated on 20/Feb/22

	$Great sir .$
	Commented by ajfour last updated on 20/Feb/22
	$say (1+(1/( (√2))))r=βr=a R=(1+(√(10)))r=γr A(a,a) eq. of right circle (x−3r)^2 +(y−r)^2 =r^2 eq. of AD: y=m(x−a)+a ⊥ distance of AD from G is r. ((−r+m(3r−a)+a)/( (√(1+m^2 ))))=r ⇒ {−1+m(3−β)+β}^2 =1+m^2 from here we get m. m≈ 0.250425 δ=45°−tan^(−1) m tan δ=((1−m)/(1+m)) Now say AB=λr (a(√2)+λr)^2 +λ^2 r^2 tan^2 δ=R^2 =r^2 (1+(√(10)))^2 we get h from here. ⇒ (β(√2)+λ)^2 +(((1−m)/(1+m)))^2 λ^2 =γ^2 △=(((1−m)/(1+m)))λ^2 r^2 λ≈ 1.631526783 △≈ 1.59568 r^2 (...is what i get, Sir)!$
	$s a y (1 + \frac{1}{\sqrt{2}}) r = β r = a$ $R = (1 + \sqrt{10}) r = γ r$ $A (a, a)$ $e q . o f r i g h t c i r c l e$ ${(x - 3 r)}^{2} + {(y - r)}^{2} = r^{2}$ $e q . o f A D : y = m (x - a) + a$ $⊥ d i s t a n c e o f A D f r o m G i s r .$ $\frac{- r + m (3 r - a) + a}{\sqrt{1 + m^{2}}} = r$ $\Rightarrow {- 1 + m (3 - β) + β}^{2} = 1 + m^{2}$ $f r o m h e r e w e g e t m .$ $m \approx 0.250425$ $δ = 45 ° - \tan^{- 1} m$ $\tan δ = \frac{1 - m}{1 + m}$ $N o w s a y A B = λ r$ ${(a \sqrt{2} + λ r)}^{2} + λ^{2} r^{2} \tan^{2} δ = R^{2} = r^{2} {(1 + \sqrt{10})}^{2}$ $w e g e t h f r o m h e r e .$ $\Rightarrow {(β \sqrt{2} + λ)}^{2} + {(\frac{1 - m}{1 + m})}^{2} λ^{2} = γ^{2}$ $△ = (\frac{1 - m}{1 + m}) λ^{2} r^{2}$ $λ \approx 1.631526783$ $△ \approx 1.59568 r^{2}$ $(. . . i s w h a t i g e t, S i r)!$
	Commented by mr W last updated on 20/Feb/22

	$f i n e s i r!$ $c a n A \approx 1.5956802 r^{2} b e c o m f i r m e d ?$
	Commented by ajfour last updated on 20/Feb/22

	$y e a h, i t s a b s o l u t e l y c o n f i r m e d!$
	Commented by ajfour last updated on 20/Feb/22
	Awesome solution sir.
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