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Question Number 163736 by HongKing last updated on 09/Jan/22

Answered by mr W last updated on 10/Jan/22

Commented by mr W last updated on 10/Jan/22

R=(a/(2 sin A)) Δ=((bc sin A)/2) r=((2Δ)/(a+b+c))=((bc sin A)/(a+b+c)) OI//AB ⇔ R cos A=r (a/(2 sin A)) cos A=((bc sin A)/(a+b+c)) cos A=((2bc sin^2 A)/(a(a+b+c))) cos A=((2bc(1−cos A)(1+cos A))/(a(a+b+c))) ((b^2 +c^2 −a^2 )/(2bc))=((2bc(1−((b^2 +c^2 −a^2 )/(2bc)))(1+((b^2 +c^2 −a^2 )/(2bc))))/(a(a+b+c))) b^2 +c^2 −a^2 =(([a^2 −(b−c)^2 ][(b+c)^2 −a^2 ])/(a(a+b+c))) b^2 +c^2 −a^2 =(([a^2 −(b−c)^2 ](b+c−a))/a) (b+c)a^2 −2bca−(b^2 −c^2 )(b−c)=0 a=((bc+(√(b^2 c^2 +(b+c)(b^2 −c^2 )(b−c))))/(b+c)) a=((bc+(√(b^2 c^2 +(b^2 −c^2 )^2 )))/(b+c)) ⇒a=((bc+(√(b^4 +c^4 −b^2 c^2 )))/(b+c)) ✓

R=a2sinAΔ=bcsinA2r=2Δa+b+c=bcsinAa+b+cOI//AB⇔RcosA=ra2sinAcosA=bcsinAa+b+ccosA=2bcsin2Aa(a+b+c)cosA=2bc(1−cosA)(1+cosA)a(a+b+c)b2+c2−a22bc=2bc(1−b2+c2−a22bc)(1+b2+c2−a22bc)a(a+b+c)b2+c2−a2=[a2−(b−c)2][(b+c)2−a2]a(a+b+c)b2+c2−a2=[a2−(b−c)2](b+c−a)a(b+c)a2−2bca−(b2−c2)(b−c)=0a=bc+b2c2+(b+c)(b2−c2)(b−c)b+ca=bc+b2c2+(b2−c2)2b+c⇒a=bc+b4+c4−b2c2b+c✓

Commented by Tawa11 last updated on 10/Jan/22

Great sir

Greatsir

Commented by HongKing last updated on 10/Jan/22

very nice my dear Sir thank you so much

verynicemydearSirthankyousomuch

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