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Question Number 199922 by Mingma last updated on 11/Nov/23

Answered by deleteduser1 last updated on 11/Nov/23

WLOG,let C be the origin and let AD coincide with the real axis;then a=a^− ;b=−b^− ;d=d^− ;e=−e^− ((b−a)/(−b−a))=((a−e)/(−e−a))⇒−be+2a^2 =+eb⇒a^2 =be ((b−a)/(−b−a))=((b−d)/(d+b))⇒b^2 −ad=−b^2 +ad⇒b^2 =ad ((b−a)/(−b−a))=((h−a)/(h^− −a))⇒bh^− −ah^− =−bh+2ab−ah ⇒h^− (b−a)=−h(b+a)+2ab...(i) (h/h^− )=((a−b)/(−b−a))⇒−h(b+a)=h^− (a−b)...(ii) (i)+(ii)⇒0=−2h(b+a)+2ab⇒h=((ab)/(b+a)) ((i−o)/(i^− −o^− ))=^? ((e−d)/(d+e))⇔((((ab)/(2b+2a))−((a+b)/2))/(((−ab)/(−2b+2a))−((a−b)/2)))=(((a^2 /b)−(b^2 /a))/((a^3 +b^3 )/(ab))) ⇔((([a^2 +b^2 +ab])/((b+a)))/(([a^2 +b^2 −ab])/((a−b))))=(((a−b)(a^2 +b^2 +ab))/((a+b)(a^2 +b^2 −ab)))⇔0=0 ⇒((i−o)/(i^− −o^− ))=((e−d)/(d+e))⇒IO⊥ED

WLOG,letCbetheoriginandletADcoincidewiththerealaxis;thena=a−;b=−b−;d=d−;e=−e−b−a−b−a=a−e−e−a⇒−be+2a2=+eb⇒a2=beb−a−b−a=b−dd+b⇒b2−ad=−b2+ad⇒b2=adb−a−b−a=h−ah−−a⇒bh−−ah−=−bh+2ab−ah⇒h−(b−a)=−h(b+a)+2ab...(i)hh−=a−b−b−a⇒−h(b+a)=h−(a−b)...(ii)(i)+(ii)⇒0=−2h(b+a)+2ab⇒h=abb+ai−oi−−o−=?e−dd+e⇔ab2b+2a−a+b2−ab−2b+2a−a−b2=a2b−b2aa3+b3ab⇔[a2+b2+ab](b+a)[a2+b2−ab](a−b)=(a−b)(a2+b2+ab)(a+b)(a2+b2−ab)⇔0=0⇒i−oi−−o−=e−dd+e⇒IO⊥ED

Commented by Mingma last updated on 11/Nov/23

Very elegant, sir!

Answered by ajfour last updated on 11/Nov/23

let H be origin. I(0,b) C(0,2b) A(−r+a, 0) B(r+a, 0) d=(2r)(((2b)/(r−a))) e=(2r)(((2b)/(r+a))) slope of DE m_(CD) =((d−e)/(2r))=((4ab)/(r^2 −a^2 )) but (√((r−a)(r+a)))=2b ⇒ m_(CD) =(a/b) slope of OI =m_(OI) =−(b/a) =−(1/m_(CD) ) ⇒ OI ⊥ CD

letHbeorigin.I(0,b)C(0,2b)A(−r+a,0)B(r+a,0)d=(2r)(2br−a)e=(2r)(2br+a)slopeofDEmCD=d−e2r=4abr2−a2but(r−a)(r+a)=2b⇒mCD=abslopeofOI=mOI=−ba=−1mCD⇒OI⊥CD

Commented by Mingma last updated on 11/Nov/23

Nice one, sir!

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