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Question Number 125508 by ajfour last updated on 11/Dec/20

Commented by ajfour last updated on 11/Dec/20

$F i n d R .$
	Answered by ajfour last updated on 11/Dec/20

	Commented by ajfour last updated on 11/Dec/20
	$let ∠C=2θ AE=AF+EF Rtan ((π/4)+θ)=tan ((π/4)+θ)+2(√R) .....(I) CE=ED+DC Rcot 2θ=2(√R)+cot θ ....(II) let tan θ=m from (I): R=1+2(√R)(((1−m)/(1+m))) ⇒ ((R−1)/(2(√R)))=((1−m)/(1+m)) m=((2(√R)+1−R)/(2(√R)+R−1)) and from (II): R=2(√R)(((2m)/(1−m^2 )))+(2/(1−m^2 )) ⇒ R{1−(((2(√R)+1−R)/(2(√R)+R−1)))^2 }=4(√R)(((2(√R)+1−R)/(2(√R)+R−1)))+2 ⇒ 4R(√R)(R−1)=2(√R){4R−(R−1)^2 } +4R+(R−1)^2 +4(√R)(R−1) say R=x 4x^2 (√x)−4x(√x)=16x(√x)−6(√x)+4x +(x−1)^2 −2x^2 (√x) ⇒ 6(√x)(x^2 +1)=20x(√x)+(x+1)^2 ⇒ x≈ 3.56918 (Sir, i think just this answer is valid)$
	$l e t ∠ C = 2 θ$ $A E = A F + E F$ $R \tan (\frac{π}{4} + θ) = \tan (\frac{π}{4} + θ) + 2 \sqrt{R}$ $. . . . . (I)$ $C E = E D + D C$ $R \cot 2 θ = 2 \sqrt{R} + \cot θ . . . . (I I)$ $l e t \tan θ = m$ $f r o m (I) :$ $R = 1 + 2 \sqrt{R} (\frac{1 - m}{1 + m})$ $\Rightarrow \frac{R - 1}{2 \sqrt{R}} = \frac{1 - m}{1 + m}$ $m = \frac{2 \sqrt{R} + 1 - R}{2 \sqrt{R} + R - 1}$ $a n d f r o m (I I) :$ $R = 2 \sqrt{R} (\frac{2 m}{1 - m^{2}}) + \frac{2}{1 - m^{2}}$ $\Rightarrow$ $R {1 - {(\frac{2 \sqrt{R} + 1 - R}{2 \sqrt{R} + R - 1})}^{2}} = 4 \sqrt{R} (\frac{2 \sqrt{R} + 1 - R}{2 \sqrt{R} + R - 1}) + 2$ $\Rightarrow$ $4 R \sqrt{R} (R - 1) = 2 \sqrt{R} {4 R - {(R - 1)}^{2}}$ $+ 4 R + {(R - 1)}^{2} + 4 \sqrt{R} (R - 1)$ $s a y R = x$ $4 x^{2} \sqrt{x} - 4 x \sqrt{x} = 16 x \sqrt{x} - 6 \sqrt{x} + 4 x$ $+ {(x - 1)}^{2} - 2 x^{2} \sqrt{x}$ $\Rightarrow 6 \sqrt{x} (x^{2} + 1) = 20 x \sqrt{x} + {(x + 1)}^{2}$ $\Rightarrow x \approx 3.56918$ $(S i r, i t h i n k j u s t t h i s a n s w e r$ $i s v a l i d)$
	Commented by mr W last updated on 11/Dec/20

	$y e s, y o u a r e r i g h t s i r!$
	Answered by mr W last updated on 11/Dec/20

	Commented by mr W last updated on 11/Dec/20

	$\sin α = \frac{R - 1}{R + 1}$ $\sin β = \frac{1}{R + 1}$ $γ = α - β = \frac{π}{2} - 2 α$ $\Rightarrow 3 α = \frac{π}{2} + β$ $\Rightarrow \sin (3 α) = \sin (\frac{π}{2} + β) = \cos β$ $\Rightarrow \sin α (3 - 4 \sin^{2} α) = \cos β$ $\Rightarrow (\frac{R - 1}{R + 1}) [3 - 4 \times {(\frac{R - 1}{R + 1})}^{2}] = \frac{\sqrt{{(R + 1)}^{2} - 1}}{R + 1}$ $\Rightarrow (R - 1) [3 - 4 \times {(\frac{R - 1}{R + 1})}^{2}] = \sqrt{R (R + 2)}$ $\Rightarrow R \approx 2.3981 o r 3.5692$
	Commented by ajfour last updated on 11/Dec/20

	$T h a n k s S i r, n i c e w a y!$
	Answered by islom last updated on 11/Feb/21

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