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Question Number 74527 by Maclaurin Stickker last updated on 25/Nov/19

Commented by Maclaurin Stickker last updated on 25/Nov/19

$I n t h e f i g u r e d e t e r m i n e t h e r a d i u s$ $o f t h e s m a l l e s t c i r c u m f e r e n c e a s a$ $f u n c t i o n o f t h e r a d i u s R o f t h e q u a d r a n t .$
	Answered by ajfour last updated on 25/Nov/19

	Commented by ajfour last updated on 25/Nov/19
	$let R=2 OQ=2−a (2−a)^2 −a^2 =(1+a)^2 −(1−a)^2 ⇒ 4−4a= 4a ⇒ a=(1/2) now sin α=(1/3) {(2−b)cos γ−a}^2 + {(2−a)cos α−(2−b)sin γ}^2 =(a+b)^2 .........(I) and (2−b)^2 sin^2 γ+{(2−b)cos γ−1}^2 = (1+b)^2 .....(II) ⇒ (2−b)^2 −(1+b)^2 +1= 2(2−b)cos γ ⇒ cos γ= ((2−3b)/(2−b)) now from (I) (2−a)^2 cos^2 α+a^2 +(2−b)^2 −(a+b)^2 = 2a(2−b)cos γ +2(2−a)(2−b)cos αsin γ ⇒ (9/4)×(8/9)+(1/4)+(2−b)^2 −(1/4)−b−b^2 = (2−3b)+2(√2)×(√((2−b)^2 −(2−3b)^2 )) (4−2b)^2 = 8(8b−8b^2 ) ⇒ 68b^2 −80b+16= 0 ⇒ 17b^2 −20b+4=0 ⇒ b= ((20−8(√2))/(34)) = ((10−4(√2))/(17)) (b/R) = ((5−2(√2))/(17)) = (1/(5+2(√2))) . as it was assumed R=2 .$
	$l e t R = 2$ $O Q = 2 - a$ ${(2 - a)}^{2} - a^{2} = {(1 + a)}^{2} - {(1 - a)}^{2}$ $\Rightarrow 4 - 4 a = 4 a$ $\Rightarrow a = \frac{1}{2} n o w \sin α = \frac{1}{3}$ ${(2 - b) \cos γ - a}^{2} +$ ${(2 - a) \cos α - (2 - b) \sin γ}^{2} = {(a + b)}^{2}$ $. . . . . . . . . (I) a n d$ ${(2 - b)}^{2} \sin^{2} γ + {(2 - b) \cos γ - 1}^{2}$ $= {(1 + b)}^{2} . . . . . (I I)$ $\Rightarrow$ ${(2 - b)}^{2} - {(1 + b)}^{2} + 1 = 2 (2 - b) \cos γ$ $\Rightarrow \cos γ = \frac{2 - 3 b}{2 - b}$ $n o w f r o m (I)$ ${(2 - a)}^{2} \cos^{2} α + a^{2} + {(2 - b)}^{2} - {(a + b)}^{2}$ $= 2 a (2 - b) \cos γ$ $+ 2 (2 - a) (2 - b) \cos α \sin γ$ $\Rightarrow$ $\frac{9}{4} \times \frac{8}{9} + \frac{1}{4} + {(2 - b)}^{2} - \frac{1}{4} - b - b^{2}$ $= (2 - 3 b) + 2 \sqrt{2} \times \sqrt{{(2 - b)}^{2} - {(2 - 3 b)}^{2}}$ ${(4 - 2 b)}^{2} = 8 (8 b - 8 b^{2})$ $\Rightarrow 68 b^{2} - 80 b + 16 = 0$ $\Rightarrow 17 b^{2} - 20 b + 4 = 0$ $\Rightarrow b = \frac{20 - 8 \sqrt{2}}{34} = \frac{10 - 4 \sqrt{2}}{17}$ $\frac{b}{R} = \frac{5 - 2 \sqrt{2}}{17} = \frac{1}{5 + 2 \sqrt{2}} .$ $a s i t w a s a s s u m e d R = 2 .$
	Commented by mr W last updated on 25/Nov/19

	$p l e a s e c h e c k s i r :$ ${(2 - b)}^{2} - {(1 + b)}^{2} + 1 = 2 (2 - b) \cos γ$ $\Rightarrow \cos γ = \frac{2 - 3 b}{2 - b}$
	Commented by ajfour last updated on 25/Nov/19

	$m a n y m a n y t h a n k s s i r!$
	Answered by mr W last updated on 25/Nov/19

	Commented by mr W last updated on 25/Nov/19

	$O B = R - a$ $O C = R - b$ $B C = a + b$ $D B = \frac{R}{2} + a$ $D C = \frac{R}{2} + b$ $\sin α = \frac{a}{R - a}$ $\cos β = \frac{{(R - b)}^{2} + \frac{R^{2}}{4} - {(\frac{R}{2} + b)}^{2}}{2 (R - b) \frac{R}{2}} = \frac{R - 3 b}{R - b}$ $\cos γ = \frac{{(R - a)}^{2} + {(R - b)}^{2} - {(a + b)}^{2}}{2 (R - a) (R - b)}$ $= \frac{R^{2} - R (a + b) - a b}{(R - a) (R - b)}$ $\cos (β + γ) = \frac{{(R - a)}^{2} + \frac{R^{2}}{4} - {(\frac{R}{2} + a)}^{2}}{2 (R - a) \frac{R}{2}} = \frac{R - 3 a}{R - a}$ $\cos (β + γ) = \sin α$ $\frac{R - 3 a}{R - a} = \frac{a}{R - a}$ $R^{2} - 5 R a + 4 a^{2} = 0$ $(R - 4 a) (R - a) = 0$ $\Rightarrow R = 4 a \Rightarrow a = \frac{R}{4}$ $\cos (β + γ) = \frac{1}{3}$ $\Rightarrow γ = \cos^{- 1} \frac{1}{3} - β$ $\cos γ = \frac{1}{3} \cos β + \frac{2 \sqrt{2}}{3} \sin β$ $\cos γ = \frac{3 R - 5 b}{3 (R - b)}$ $\cos β = \frac{R - 3 b}{R - b} \Rightarrow \sin β = \frac{2 \sqrt{(R - 2 b) b}}{R - b}$ $\frac{3 R - 5 b}{2 (R - b)} = \frac{R - 3 b}{3 (R - b)} + \frac{4 \sqrt{2 (R - 2 b) b}}{3 (R - b)}$ $R - b = 2 \sqrt{2 b (R - 2 b)}$ $R^{2} - 2 R b + b^{2} = 8 b (R - 2 b)$ $R^{2} - 10 R b + 17 b^{2} = 0$ $\Rightarrow b = \frac{R}{5 + 2 \sqrt{2}} \approx 0.12774 R$
	Commented by Maclaurin Stickker last updated on 25/Nov/19

	$A n o t h e r p e r f e c t a n s w e r . T h a n k y o u .$
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