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Question Number 204197 by ajfour last updated on 08/Feb/24

	Answered by mr W last updated on 08/Feb/24

	$\sin \frac{θ}{2} = \frac{a}{R - a}$ $O C = O B + B C$ $\frac{R}{\cos θ} = \sqrt{{(R + 1)}^{2} - 1^{2}} + \frac{1}{\tan (\frac{\frac{π}{2} - θ}{2})}$ $\frac{R}{1 - 2 \sin^{2} \frac{θ}{2}} = \sqrt{R (R + 2)} + \frac{1 + \tan \frac{θ}{2}}{1 - \tan \frac{θ}{2}}$ $\frac{R}{1 - 2 \sin^{2} \frac{θ}{2}} = \sqrt{R (R + 2)} + \frac{\cos \frac{θ}{2} + \sin \frac{θ}{2}}{\cos \frac{θ}{2} - \sin \frac{θ}{2}}$ $\frac{R}{1 - 2 {(\frac{a}{R - a})}^{2}} = \sqrt{R (R + 2)} + \frac{\sqrt{1 - {(\frac{a}{R - a})}^{2}} + \frac{a}{R - a}}{\sqrt{1 - {(\frac{a}{R - a})}^{2}} - \frac{a}{R - a}}$ $\Rightarrow \frac{R {(R - a)}^{2}}{R^{2} - 2 R a - a^{2}} = \sqrt{R (R + 2)} + \frac{\sqrt{R (R - 2 a)} + a}{\sqrt{R (R - 2 a)} - a}$ $e x a m p l e s :$ $a = 2 \Rightarrow R \approx 6.7890, θ \approx 49.3693 °$ $a = 3 \Rightarrow R \approx 12.9166, θ \approx 35.2184 °$
	Commented by mr W last updated on 08/Feb/24

	Commented by ajfour last updated on 08/Feb/24

	$T h a n k y o u s i r, y o u g o t m o r e c o m p a c t .$
	Commented by mr W last updated on 08/Feb/24

	$t h a n k s s i r!$ $p l e a s e h a v e a l o o k a t Q 204078 .$
	Answered by ajfour last updated on 08/Feb/24
	$centre of largest (here) circle be origin. y=mx line eq. m=(1/(tan θ)) at y=R, x=Rtan θ sin (θ/2)=(a/(R−a)) tan θ=(√((1/({1−2((a/(R−a)))^2 }^2 ))−1)) P[(R+1)cos φ, (R+1)sin φ] R−(R+1)sin φ=1 (R+1)cos φ=(√(4R)) P lies on ine through rightmost corner and bisector of angle there. (R+1)sin φ−R ={(R+1)cos φ−Rtan θ}tan ((π/4)−(θ/2)) ⇒ −1=(((2(√R)−Rtan θ)(1−tan θ))/((1+tan θ))) ⇒ 1+tan θ=(tan θ−1))(2(√R)−Rtan θ) ⇒ Rtan^2 θ−(R+2(√R)−1)tan θ +(1+2(√R))=0 tan θ=(((R+2(√R)−1))/(2R))±(√(((R+2(√R)−1)^2 −4R(1+2(√R)))/(4R^2 ))) also tan θ=(√((1/({1−2((a/(R−a)))^2 }^2 ))−1))$
	$c e n t r e o f l a r g e s t (h e r e) c i r c l e$ $b e o r i g i n .$ $y = m x l i n e e q . m = \frac{1}{\tan θ}$ $a t y = R, x = R \tan θ$ $\sin \frac{θ}{2} = \frac{a}{R - a}$ $\tan θ = \sqrt{\frac{1}{{1 - 2 {(\frac{a}{R - a})}^{2}}^{2}} - 1}$ $P [(R + 1) \cos ϕ, (R + 1) \sin ϕ]$ $R - (R + 1) \sin ϕ = 1$ $(R + 1) \cos ϕ = \sqrt{4 R}$ $P l i e s o n i n e t h r o u g h r i g h t m o s t$ $c o r n e r a n d b i s e c t o r o f a n g l e t h e r e .$ $(R + 1) \sin ϕ - R$ $= {(R + 1) \cos ϕ - R \tan θ} \tan (\frac{π}{4} - \frac{θ}{2})$ $\Rightarrow$ $- 1 = \frac{(2 \sqrt{R} - R \tan θ) (1 - \tan θ)}{(1 + \tan θ)}$ $\Rightarrow$ $1 + \tan θ = (\tan θ - 1)) (2 \sqrt{R} - R \tan θ)$ $\Rightarrow$ $R \tan^{2} θ - (R + 2 \sqrt{R} - 1) \tan θ$ $+ (1 + 2 \sqrt{R}) = 0$ $\tan θ = \frac{(R + 2 \sqrt{R} - 1)}{2 R} \pm \sqrt{\frac{{(R + 2 \sqrt{R} - 1)}^{2} - 4 R (1 + 2 \sqrt{R})}{4 R^{2}}}$ $a l s o \tan θ = \sqrt{\frac{1}{{1 - 2 {(\frac{a}{R - a})}^{2}}^{2}} - 1}$
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