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Question Number 166036 by mr W last updated on 12/Feb/22

Commented by mr W last updated on 12/Feb/22

$i f t h e c h o r d s o n a q u a t e r c i r c l e h a v e$ $t h e l e n g t h a, b, c r e s p e c t i v e l y, f i n d$ $t h e r a d i u s R o f t h e q u a t e r c i r c l e i n$ $t e r m s o f a, b, c .$
Commented by ajfour last updated on 13/Feb/22

$g r e a t!$
	Answered by Rasheed.Sindhi last updated on 12/Feb/22
	$A Try a^2 =R^2 +R^2 −2.R.R.cos α { ((a^2 =2R^2 (1−cos α))),((b^2 =2R^2 (1−cos β))),((c^2 =2R^2 (1−cos γ))) :} ; α+β+γ=90 R^2 =(a^2 /(2(1−cos α)))=(b^2 /(2(1−cos β)))=(c^2 /(2(1−cos γ))) 3R^2 =((a^2 +b^2 +c^2 )/(2(3−(cos α+cos β+cos γ)))) 3R^2 =((a^2 +b^2 +c^2 )/(2(3−(cos α+cos β+cos(90−(α+β))))) 3R^2 =((a^2 +b^2 +c^2 )/(2(3−(cos α+cos β+sin(α+β))))) R^2 =((a^2 +b^2 +c^2 )/(6(3−(cos α+cos β+sin(α+β))))) I think R is also dependant on α,β$
	$A T r y$ $a^{2} = R^{2} + R^{2} - 2 . R . R . \cos α$ ${\begin{cases} a^{2} = {2 R}^{2} (1 - \cos α) \\ b^{2} = {2 R}^{2} (1 - \cos β) \\ c^{2} = {2 R}^{2} (1 - \cos γ) \end{cases}; α + β + γ = 90$ $R^{2} = \frac{a^{2}}{2 (1 - \cos α)} = \frac{b^{2}}{2 (1 - \cos β)} = \frac{c^{2}}{2 (1 - \cos γ)}$ ${3 R}^{2} = \frac{a^{2} + b^{2} + c^{2}}{2 (3 - (\cos α + \cos β + \cos γ))}$ ${3 R}^{2} = \frac{a^{2} + b^{2} + c^{2}}{2 (3 - (\cos α + \cos β + \cos (90 - (α + β)))}$ ${3 R}^{2} = \frac{a^{2} + b^{2} + c^{2}}{2 (3 - (\cos α + \cos β + \sin (α + β)))}$ $R^{2} = \frac{a^{2} + b^{2} + c^{2}}{6 (3 - (\cos α + \cos β + \sin (α + β)))}$ $I t h i n k R i s a l s o d e p e n d a n t o n α, β$
	Commented by mr W last updated on 12/Feb/22

	$y o u h a v e$ $\sin \frac{α}{2} = \frac{a}{2 R} \Rightarrow \frac{α}{2} = \sin^{- 1} \frac{a}{2 R}$ $\sin \frac{β}{2} = \frac{b}{2 R} \Rightarrow \frac{β}{2} = \sin^{- 1} \frac{b}{2 R}$ $\sin \frac{γ}{2} = \frac{c}{2 R} \Rightarrow \frac{γ}{2} = \sin^{- 1} \frac{c}{2 R}$ $\frac{α}{2} + \frac{β}{2} + \frac{γ}{2} = \frac{π}{4}$ $\Rightarrow \sin^{- 1} \frac{a}{2 R} + \sin^{- 1} \frac{b}{2 R} + \sin^{- 1} \frac{c}{2 R} = \frac{π}{4}$ $t h e r e i s a n u n i q u e s o l u t i o n f o r R,$ $b u t w e c a n n o t s o l v e i t i n e x a c t f o r m$ $i n t e r m s o f a, b, c .$
	Commented by mahdipoor last updated on 12/Feb/22
	$⇒0=cos90=cos(α+β+λ)= cosα.cosβ.cosλ−sinβ.sinα.cosλ− sinλ.sinα.cosβ−sinλ.sinβ.cosα , u^2 =2R^2 (1−cosv) ⇒ { ((cosv=1−(u/( (√2)R)))),((sinv=(√((√2)(u/R)−(u^2 /(2R^2 )))))) :}$
	$\Rightarrow 0 = c o s 90 = c o s (α + β + λ) =$ $c o s α . c o s β . c o s λ - s i n β . s i n α . c o s λ -$ $s i n λ . s i n α . c o s β - s i n λ . s i n β . c o s α$ $,$ $u^{2} = 2 R^{2} (1 - c o s v)$ $\Rightarrow {\begin{cases} c o s v = 1 - \frac{u}{\sqrt{2} R} \\ s i n v = \sqrt{\sqrt{2} \frac{u}{R} - \frac{u^{2}}{2 R^{2}}} \end{cases}$
	Commented by Rasheed.Sindhi last updated on 12/Feb/22

	${Y o u}^{'} r e r i g h t m r W s i r, I c o n c l u d e d$ $s o b e c s u s e I s a w t h a t \cos α + \cos β + \sin (α + β)$ $i s n o t c o n s t a n t! B u t i n s p i t e t h i s$ $i t m a y b e d e p e n d a n t u p o n a, b & c .$
	Commented by mr W last updated on 13/Feb/22

	$w e c a n t r a n s f o r m t h e e q u a t i o n t o :$ $4 R^{6} - 4 (a^{2} + b^{2} + c^{2}) R^{4} - 6 \sqrt{2} {a b c R}^{3} + (a^{4} + b^{4} + c^{4}) R^{2} + \sqrt{2} a b c (a^{2} + b^{2} + c^{2}) R + a^{2} b^{2} c^{2} = 0$ $b u t s t i l l n o t s o l v a b l e .$
	Commented by Rasheed.Sindhi last updated on 12/Feb/22

	$N i c e r e l a t i o n s i r!$
	Commented by ajfour last updated on 13/Feb/22
	$θ+ψ=(π/4)−φ cos (θ+ψ)=(1/( (√2))){cos φ+sin φ} (√(1−p^2 ))(√(1−r^2 ))−pr=(1/( (√2))){(√(1−q^2 ))+q} (1−p^2 )(1−r^2 )+((1−q^2 )/2) −(√2)(√(1−q^2 ))(√(1−p^2 ))(√(1−r^2 )) =p^2 r^2 +(q^2 /2)+(√2)pqr ⇒ { (3/2)−(p^2 +r^2 +q^2 )−(√2)pqr}^2 =2(1−p^2 )(1−q^2 )(1−r^2 ) ⇒ (9/4)+(p^2 +q^2 +r^2 )^2 +2p^2 q^2 r^2 −3(p^2 +r^2 +q^2 )−3(√2)pqr +2(√2)pqr(p^2 +q^2 +r^2 ) =2−p^2 q^2 r^2 +2(p^2 q^2 +q^2 r^2 +r^2 p^2 ) ⇒ p^4 +q^4 +r^4 +(1/4)+3p^2 q^2 r^2 +2(√2)pqr(p^2 +q^2 +r^2 ) −3(p^2 +q^2 +r^2 )=0 say (1/(2R))=t ⇒ (3a^2 b^2 c^2 )t^6 +2(√2)abc(a^2 +b^2 +c^2 )t^5 +(a^4 +b^4 +c^4 )t^4 −3(a^2 +b^2 +c^2 )t^2 +(1/4)=0$
	$θ + ψ = \frac{π}{4} - ϕ$ $\cos (θ + ψ) = \frac{1}{\sqrt{2}} {\cos ϕ + \sin ϕ}$ $\sqrt{1 - p^{2}} \sqrt{1 - r^{2}} - p r = \frac{1}{\sqrt{2}} {\sqrt{1 - q^{2}} + q}$ $(1 - p^{2}) (1 - r^{2}) + \frac{1 - q^{2}}{2}$ $- \sqrt{2} \sqrt{1 - q^{2}} \sqrt{1 - p^{2}} \sqrt{1 - r^{2}}$ $= p^{2} r^{2} + \frac{q^{2}}{2} + \sqrt{2} p q r$ $\Rightarrow {\frac{3}{2} - (p^{2} + r^{2} + q^{2}) - \sqrt{2} p q r}^{2}$ $= 2 (1 - p^{2}) (1 - q^{2}) (1 - r^{2})$ $\Rightarrow \frac{9}{4} + {(p^{2} + q^{2} + r^{2})}^{2} + 2 p^{2} q^{2} r^{2}$ $- 3 (p^{2} + r^{2} + q^{2}) - 3 \sqrt{2} p q r$ $+ 2 \sqrt{2} p q r (p^{2} + q^{2} + r^{2})$ $= 2 - p^{2} q^{2} r^{2} + 2 (p^{2} q^{2} + q^{2} r^{2} + r^{2} p^{2})$ $\Rightarrow$ $p^{4} + q^{4} + r^{4} + \frac{1}{4} + 3 p^{2} q^{2} r^{2}$ $+ 2 \sqrt{2} p q r (p^{2} + q^{2} + r^{2})$ $- 3 (p^{2} + q^{2} + r^{2}) = 0$ $s a y \frac{1}{2 R} = t \Rightarrow$ $(3 a^{2} b^{2} c^{2}) t^{6} + 2 \sqrt{2} a b c (a^{2} + b^{2} + c^{2}) t^{5}$ $+ (a^{4} + b^{4} + c^{4}) t^{4}$ $- 3 (a^{2} + b^{2} + c^{2}) t^{2} + \frac{1}{4} = 0$
	Answered by ajfour last updated on 15/Feb/22
	$((abc)/(2R))=casin β=absin γ=bcsin α=p let ((abc)/(2r))=q q=casin 2β=absin 2γ=bcsin 2α α+β+γ=(π/4) sin 2β=cos (2α+2γ) (q/(ac))=(√(1−((q/(bc)))^2 ))(√(1−((q/(ab)))^2 )) −(q^2 /(b^2 ac)) ⇒ (q^2 /(a^2 c^2 ))+(q^4 /(b^4 a^2 c^2 ))+((2q^3 )/(a^2 b^2 c^2 )) =1−(q^2 /(a^2 b^2 c^2 ))(a^2 +c^2 )+(q^4 /(b^4 a^2 c^2 )) ⇒ 2q^3 +(a^2 +b^2 +c^2 )q^2 −a^2 b^2 c^2 =0 say we obtained q. Now (q/p)=2cos β ⇒ q^2 =4p^2 (1−(b^2 /(4R^2 )))=4p^2 (1−(p^2 /(a^2 c^2 ))) ⇒ p^4 −a^2 c^2 p^2 +((a^2 c^2 q^2 )/4)=0 p^2 =(((abc)/(2R)))^2 =((a^2 c^2 )/2)±(√(((a^2 c^2 )/4)(a^2 c^2 −q^2 ))) ⇒ R^2 =(b^2 /2)((q/(ac)))^2 {1∓(√(1−((q/(ac)))^2 ))}$
	$\frac{a b c}{2 R} = c a \sin β = a b \sin γ = b c \sin α = p$ $l e t \frac{a b c}{2 r} = q$ $q = c a \sin 2 β = a b \sin 2 γ = b c \sin 2 α$ $α + β + γ = \frac{π}{4}$ $\sin 2 β = \cos (2 α + 2 γ)$ $\frac{q}{a c} = \sqrt{1 - {(\frac{q}{b c})}^{2}} \sqrt{1 - {(\frac{q}{a b})}^{2}}$ $- \frac{q^{2}}{b^{2} a c}$ $\Rightarrow \frac{q^{2}}{a^{2} c^{2}} + \frac{q^{4}}{b^{4} a^{2} c^{2}} + \frac{2 q^{3}}{a^{2} b^{2} c^{2}}$ $= 1 - \frac{q^{2}}{a^{2} b^{2} c^{2}} (a^{2} + c^{2}) + \frac{q^{4}}{b^{4} a^{2} c^{2}}$ $\Rightarrow 2 q^{3} + (a^{2} + b^{2} + c^{2}) q^{2} - a^{2} b^{2} c^{2} = 0$ $s a y w e o b t a i n e d q .$ $N o w \frac{q}{p} = 2 \cos β$ $\Rightarrow q^{2} = 4 p^{2} (1 - \frac{b^{2}}{4 R^{2}}) = 4 p^{2} (1 - \frac{p^{2}}{a^{2} c^{2}})$ $\Rightarrow p^{4} - a^{2} c^{2} p^{2} + \frac{a^{2} c^{2} q^{2}}{4} = 0$ $p^{2} = {(\frac{a b c}{2 R})}^{2} = \frac{a^{2} c^{2}}{2} \pm \sqrt{\frac{a^{2} c^{2}}{4} (a^{2} c^{2} - q^{2})}$ $\Rightarrow R^{2} = \frac{b^{2}}{2} {(\frac{q}{a c})}^{2} {1 \mp \sqrt{1 - {(\frac{q}{a c})}^{2}}}$
	Commented by ajfour last updated on 15/Feb/22

	Commented by mr W last updated on 15/Feb/22

	$t h a n k s s i r!$ $c a n y o u p l e a s e s h o w i n a d i a g r a m$ $w h a t a r e α, β, γ ?$
	Commented by mr W last updated on 15/Feb/22

	Commented by mr W last updated on 15/Feb/22

	$a r e y o u s u r e s i r ?$ $i h a v e d o u b t i f t h i s w a y i s c o r r e c t .$ $b c \sin α = c a \sin β = a b \sin γ = \frac{a b c}{2 R} ✓$ $w i t h 2 α + 2 β + 2 γ = \frac{π}{2} ✓$ $b c \sin θ = c a \sin φ = a b \sin ϕ = \frac{a b c}{2 r} ✓$ $w i t h 2 θ + 2 φ + 2 ϕ = π ✓$ $w e c a n s a y θ + φ + ϕ = 2 (α + β + γ) .$ $b u t i t h i n k w e c a n n o t a s s u m e (s e t)$ $θ = 2 α, φ = 2 β, ϕ = 2 γ .$ $a c t u a l l y \sin θ = \frac{a}{2 r}, \sin α = \frac{a}{2 R}$ $i f θ = 2 α, t h e n$ $\sin^{2} θ = \sin^{2} 2 α = 1 - \cos^{2} 2 α = 1 - {(1 - 2 \sin^{2} α)}^{2}$ ${(\frac{a}{2 r})}^{2} = 1 - {(1 - \frac{a^{2}}{2 R^{2}})}^{2} . . . (I)$ $s i m i l a r l y, f r o m φ = 2 β a n d ϕ = 2 γ$ ${w e}^{'} l l g e t$ ${(\frac{b}{2 r})}^{2} = 1 - {(1 - \frac{b^{2}}{2 R^{2}})}^{2} . . . (I I)$ ${(\frac{c}{2 r})}^{2} = 1 - {(1 - \frac{c^{2}}{2 R^{2}})}^{2} . . . (I I I)$ $(I), (I I), (I I I) a r e n o t c o n s i s t e n t .$ $t h a t m e a n s w e c a n n o t s e t$ $θ = 2 α, φ = 2 β, ϕ = 2 γ a t t h e s a m e t i m e .$ $b c \sin α = c a \sin β = a b \sin γ = \frac{a b c}{2 R} ✓$ $⇏ b c \sin 2 α \overset{?}{=} c a \sin 2 β \overset{?}{=} a b \sin 2 γ \overset{?}{=} \frac{a b c}{2 r}$
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