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Question Number 43023 by ajfour last updated on 06/Sep/18

Commented by ajfour last updated on 06/Sep/18

$F i n d l o c a t i o n o f c e n t r e (0, h) a n d$ $r a d i u s R o f c i r c l e s u c h t h a t$ $a l l f o u r b o u n d e d a r e a s$ $O \overset{⌢}{A D}, \overset{⌢}{A P B A}, A \overset{⌢}{B C} \overset{⌢}{D A}, & \overset{⌢}{C Q D C}$ $a r e e q u a l .$ $D e t e r m i n e R, h i n t e r m s o f θ .$
	Answered by MrW3 last updated on 07/Sep/18

	Commented by MrW3 last updated on 07/Sep/18

	$R \sin γ = h \sin θ$ $\Rightarrow \sin γ = \frac{h}{R} \sin θ$ $w i t h λ = \frac{h}{R}$ $\Rightarrow γ = \sin^{- 1} (λ \sin θ)$ $α = γ - θ$ $β = γ + θ$ $Δ_{O M D} = \frac{h}{2} \times R \sin α$ $Δ_{O M C} = \frac{h}{2} \times R \sin β$ $A_{O A D} = 2 \times \frac{h}{2} \times R \sin α - α R^{2} = \frac{π R^{2}}{3}$ $\Rightarrow \frac{h}{R} \sin α - α = \frac{π}{3}$ $\Rightarrow λ \sin [\sin^{- 1} (λ \sin θ) - θ] - [\sin^{- 1} (λ \sin θ) - θ] = \frac{π}{3} . . . (i)$ $A_{O B C} = 2 \times \frac{h}{2} \times R \sin β + β R^{2} = \frac{2 π R^{2}}{3}$ $\Rightarrow \frac{h}{R} \sin β + β = \frac{2 π}{3}$ $\Rightarrow λ \sin [\sin^{- 1} (λ \sin θ) + θ] + [\sin^{- 1} (λ \sin θ) + θ] = \frac{2 π}{3} . . . (i i)$ $f r o m (i) a n d (i i) w e g e t λ a n d θ .$ $I f o u n d a s o l u t i o n :$ $λ = \frac{h}{R} \approx 5.763$ $θ \approx 2.64 °$
	Commented by ajfour last updated on 07/Sep/18

	$V e r y N i c e S i r, h a l f t h e t a s k i s$ $d o n e . R e n e D e s c a r t e s s a i d,^{'} G i v e$ $n a m e s t o a l l t h e {l i n e s}^{'} (i n G e o m e t r y) .$
	Commented by ajfour last updated on 07/Sep/18

	$W o n d e r f u l & B e a u t i f u l S i r .$ $(I h a d s u s p e c t e d w h a t i h a d a s k e d$ $t o d e t e r m i n e, i s e e i t c l e a r e r n o w)!$ $S o θ a c c e p t s j u s t o n e v a l u e, u n d e r$ $t h e i m p o s e d c o n d i t i o n . . i t m e a n s .$
	Commented by MrW3 last updated on 07/Sep/18

	$t h a n k y o u s i r f o r t h i s n i c e q u e s t i o n .$
	Commented by MrW3 last updated on 07/Sep/18

	$f o r a m o m e n t I w a s u n s u r e i f s u c h a$ $c o n f i g u r a t i o n i s p o s s i b l e a t a l l . n o w$ $I k n o w t h e r e i s o n e a n d o n l y o n e s o l u t i o n .$
	Commented by ajfour last updated on 07/Sep/18

	$A V e r y w i s e a n d p r a c t i c a l s o l u t i o n$ $t o o, S i r .$
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