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Question Number 215400 by ajfour last updated on 05/Jan/25

Commented by ajfour last updated on 05/Jan/25

Correction I think- Find b, given a.

CorrectionIthink−Findb,givena.

Answered by mr W last updated on 06/Jan/25

Commented by mr W last updated on 06/Jan/25

r=1 α=(a/r), β=(b/r), γ=(c/γ) d=b−r+c−r=b+c−2r d^2 =b^2 +c^2 =b^2 +c^2 +4r^2 +2bc−4r(b+c) 2r^2 +bc−2r(b+c)=0 2+βγ−2(β+γ)=0 ⇒γ=((2(β−1))/(β−2)) x=a−(√(c^2 −b^2 )) y=b−(√(a^2 −d^2 ))=b−(√(a^2 −b^2 −c^2 )) [a−(√(c^2 −b^2 ))]^2 +[b−(√(a^2 −b^2 −c^2 ))]^2 =b^2 a^2 −b^2 =a(√(c^2 −b^2 ))+b(√(a^2 −b^2 −c^2 )) α(√(γ^2 −β^2 ))+β(√(α^2 −β^2 −γ^2 ))=α^2 −β^2 ⇒α(√(4(((β−1)/(β−2)))^2 −β^2 ))+β(√(α^2 −β^2 −4(((β−1)/(β−2)))^2 ))=α^2 −β^2

r=1α=ar,β=br,γ=cγd=b−r+c−r=b+c−2rd2=b2+c2=b2+c2+4r2+2bc−4r(b+c)2r2+bc−2r(b+c)=02+βγ−2(β+γ)=0⇒γ=2(β−1)β−2x=a−c2−b2y=b−a2−d2=b−a2−b2−c2[a−c2−b2]2+[b−a2−b2−c2]2=b2a2−b2=ac2−b2+ba2−b2−c2αγ2−β2+βα2−β2−γ2=α2−β2⇒α4(β−1β−2)2−β2+βα2−β2−4(β−1β−2)2=α2−β2

Commented by ajfour last updated on 06/Jan/25

Yeah this is finer, i could follow smoothly, Sir.

Yeahthisisfiner,icouldfollowsmoothly,Sir.

Answered by ajfour last updated on 05/Jan/25

I get underneath relation: [(√((((b−1)/((b/2)−1)))^2 −b^2 ))−a]^2 +[(√((((b−1)/((b/2)−1))+b−2)^2 −a^2 ))−b]^2 =a^2 +b^2

Igetunderneathrelation:[(b−1b2−1)2−b2−a]2+[(b−1b2−1+b−2)2−a2−b]2=a2+b2

Answered by mr W last updated on 05/Jan/25

Commented by mr W last updated on 05/Jan/25

r=1 c=(b/(cos θ)) a=b(tan θ+cos θ) d=(√(b^2 +c^2 ))=b(√(1+(1/(cos^2 θ)))) bc=(b+c+d)r ⇒b=(1+cos θ+(√(1+cos^2 θ)))r ⇒a=(tan θ+cos θ)(1+cos θ+(√(1+cos^2 θ)))r with 0<θ<(π/2)

r=1c=bcosθa=b(tanθ+cosθ)d=b2+c2=b1+1cos2θbc=(b+c+d)r⇒b=(1+cosθ+1+cos2θ)r⇒a=(tanθ+cosθ)(1+cosθ+1+cos2θ)rwith0<θ<π2

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