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Question Number 164468 by ajfour last updated on 17/Jan/22

	Answered by mr W last updated on 18/Jan/22

	Commented by mr W last updated on 18/Jan/22

	$\sin α = \frac{r}{R} = λ$ $\sin β = \frac{R - 2 r}{R} = 1 - 2 λ$ $\sin γ = \frac{3 r - R}{R - r} = \frac{3 λ - 1}{1 - λ}$ $γ = α + β$ $\sin γ = \sin (α + β)$ $\frac{3 λ - 1}{1 - λ} = λ \sqrt{1 - {(1 - 2 λ)}^{2}} + (1 - 2 λ) \sqrt{1 - λ^{2}}$ $\frac{3 λ - 1}{1 - λ} = 2 λ \sqrt{λ (1 - λ)} + (1 - 2 λ) \sqrt{1 - λ^{2}}$ $λ = \frac{r}{R} \approx 0.4362$
	Commented by ajfour last updated on 19/Jan/22

	$T h a n k y o u S i r . I n d e e d c o r r e c t .$
	Commented by mr W last updated on 19/Jan/22

	$i {d i d n}^{'} t t r y i f e x a c t s o l u t i o n c a n b e$ $o b t a i n e d .$
	Commented by mr W last updated on 18/Jan/22

	Answered by ajfour last updated on 20/Jan/22

	Commented by ajfour last updated on 21/Jan/22

	$T h a n k y o u v e r y v e r y m u c h, s i r .$
	Commented by ajfour last updated on 21/Jan/22

	$\tan (\frac{90 ° - θ}{2}) = \frac{r}{R - r + \sqrt{R^{2} - r^{2}}} = p$ $\sec θ = \frac{R - r}{3 r - R} = \frac{t - 1}{3 - t} (\forall t = \frac{R}{r} > 2)$ $\Rightarrow \frac{1 - \tan \frac{θ}{2}}{1 + \tan \frac{θ}{2}} = p$ $\tan \frac{θ}{2} = \frac{1 - p}{1 + p}$ $s a y \frac{R}{r} = t \Rightarrow \tan \frac{θ}{2} = \frac{1 - (\frac{1}{t - 1 + \sqrt{t^{2} - 1}})}{1 + (\frac{1}{t - 1 + \sqrt{t^{2} - 1}})}$ $= \frac{\sqrt{t^{2} - 1} + t - 2}{t + \sqrt{t^{2} - 1}} = \frac{\sec θ - 1}{\tan θ}$ $& \sec^{2} θ = 1 + \tan^{2} θ = {(\frac{t - 1}{3 - t})}^{2}$ $\frac{\sqrt{t^{2} - 1} + t - 2}{t + \sqrt{t^{2} - 1}} = \frac{(\frac{2 t - 4}{3 - t})}{\frac{\sqrt{4 t - 8}}{3 - t}} = \sqrt{t - 2}$ $\Rightarrow (t - 2 + \sqrt{t^{2} - 1}) (t - \sqrt{t^{2} - 1}) = \sqrt{t - 2}$ $l e t t - \sqrt{t^{2} - 1} = p$ $\Rightarrow t + \sqrt{t^{2} - 1} = \frac{1}{p}$ $& 2 t = p + \frac{1}{p}$ $\Rightarrow 2 {(\frac{1}{p} - 2)}^{2} p^{2} = p + \frac{1}{p} - 4$ $\Rightarrow 2 + 8 p^{2} - 8 p = p + \frac{1}{p} - 4$ $\Rightarrow 8 p^{3} - 9 p^{2} + 6 p - 1 = 0$ $p \approx 0.22961$ $\frac{1}{t} = \frac{2 p}{(p^{2} + 1)} \approx 0.4362 .$
	Commented by mr W last updated on 21/Jan/22

	$g r e a t! o n e c a n e v e n g e t e x a c t v a l u e o f$ $p!$
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