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Question Number 81193 by ajfour last updated on 10/Feb/20

Commented by ajfour last updated on 10/Feb/20

$I f b o t h c o l o u r e d r e g i o n s h a v e e q u a l$ ${(a r e a s)}_{m a x i m u m}, d e t e r m i n e α .$
Commented by jagoll last updated on 10/Feb/20

$w a w$
Commented by ajfour last updated on 14/Feb/20

	Answered by mr W last updated on 10/Feb/20

	Commented by mr W last updated on 10/Feb/20

	$R = r a d i u s o f s e m i - c i r c l e$ $R^{2} = a^{2} + {(\frac{a}{\tan α} + b - R)}^{2}$ $b^{2} + 2 (\frac{a}{\tan α} - R) b + \frac{a^{2}}{\sin^{2} α} - \frac{2 a R}{\tan α} = 0$ $b = R - \frac{a}{\tan α} + \sqrt{{(R - \frac{a}{\tan α})}^{2} - \frac{a^{2}}{\sin^{2} α} + \frac{2 a R}{\tan α}}$ $\Rightarrow b = R + \sqrt{R^{2} - a^{2}} - \frac{a}{\tan α}$ $d = 2 (\frac{a}{\sin α} - R \cos α)$ ${(c + R \sin α)}^{2} + {(\frac{d}{2})}^{2} = R^{2}$ ${(c + R \sin α)}^{2} + {(\frac{a}{\sin α} - R \cos α)}^{2} = R^{2}$ $c^{2} + 2 R \sin α c + \frac{a (a - R \sin 2 α)}{\sin^{2} α} = 0$ $\Rightarrow c = \frac{\sqrt{R^{2} \sin^{4} α - a (a - R \sin 2 α)}}{\sin α} - R \sin α$ $A r e a = a b = c d$ $l e t λ = \frac{a}{R}$ $P = \frac{A r e a}{R^{2}} = λ (1 + \sqrt{1 - λ^{2}} - \frac{λ}{\tan α}) = 2 (\frac{λ}{\sin α} - \cos α) [\frac{\sqrt{\sin^{4} α - λ (λ - \sin 2 α)}}{\sin α} - \sin α]$ $n o w i t i s t o f i n d t h e m a x i m u m o f$ $P = λ (1 + \sqrt{1 - λ^{2}} - \frac{λ}{\tan α})$ $u n d e r t h e c o n d i t i o n$ $λ (1 + \sqrt{1 - λ^{2}} - \frac{λ}{\tan α}) = \frac{(2 λ - \sin 2 α) [\sqrt{\sin^{4} α - λ (λ - \sin 2 α)} - \sin^{2} α]}{\sin^{2} α}$ $P_{m a x} \approx 0.3795 a t α \approx 27.01 °, λ \approx 0.5943$
	Commented by mr W last updated on 10/Feb/20

	Commented by mr W last updated on 10/Feb/20

	Commented by mr W last updated on 10/Feb/20

	$E x p l a n a t i o n o f m e t h o d i u s e d :$ $r e d c u r v e s h o w s t h e c o n d i t i o n$ $λ (1 + \sqrt{1 - λ^{2}} - \frac{λ}{\tan α}) = \frac{(2 λ - \sin 2 α) [\sqrt{\sin^{4} α - λ (λ - \sin 2 α)} - \sin^{2} α]}{\sin^{2} α}$ $t h e g r e e n c u r v e s r e p r e s e n t t h e e q u a t i o n$ $F (λ, α) = λ (1 + \sqrt{1 - λ^{2}} - \frac{λ}{\tan α}) - P = 0$ $w i t h d i f f e r e n t v a l u e s o f P :$ $c u r v e 1 : w i t h P w h i c h f u l f i l l s t h e c o n d i t i o n$ $c u r v e 3 : w i t h P w h i c h {d o e s n}^{'} t f u l f i l l t h e c o n d i t i o n$ $c u r v e 2 : w i t h P_{m a x} w h i c h f u l f i l l s t h e c o n d i t i o n$
	Commented by ajfour last updated on 10/Feb/20

	$T h a n k s S i r, f o r e v e r y t h i n g,$ $b u t i s t i l l s h a l l t r y a t a n$ $a n a l y t i c a l w a y t o f e t c h t h e$ $a n s w e r, w i s h m e g o o d l u c k . .$
	Commented by mr W last updated on 10/Feb/20

	$y e s, i d o s i r! i w i s h e d a l s o a b e t t e r a n d$ $a n a l y t i c a l w a y, b u t h a d n o i d e a .$
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