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Question Number 201526 by 281981 last updated on 08/Dec/23

Answered by deleteduser1 last updated on 08/Dec/23

WLOG,let O be the origin; g=((a+b+c)/3) ∣o−a∣=∣o−b∣=∣o−c∣=R ⇒(o−a)(o^− −a^− )=R^2 ⇒aa^− =R^2 ⇒a^− =(R^2 /a) Then ∣OG∣^2 =(g−o)(g^− −o^− )=gg^− =(((a+b+c)/3))(((R^2 ((1/a)+(1/b)+(1/c)))/3))=((R^2 (a+b+c)(ab+bc+ca))/(9abc)) ∣b−a∣^2 +∣b−c∣^2 +∣c−a∣^2 =R^2 (b−a)(((a−b)/(ab)))+R^2 (b−c)(((c−b)/(bc)))+R^2 (c−a)(((a−c)/(ac))) =(R^2 /(abc))[c(b−a)(a−b)+a(b−c)(c−b)+b(c−a)(a−c)] =(r^2 /(abc))[{−3abc+3abc−a^2 c−b^2 c−ab^2 −c^2 a+(abc−a^2 b)−c^2 b+6abc] =((−R^2 )/(abc))[a^2 b+abc+a^2 c+ab^2 +b^2 c+abc+abc+bc^2 +c^2 a−9abc] =((−r^2 (a+b+c)(ab+bc+ca))/(abc))+9r^2 ⇒∣BA∣^2 +∣AC∣^2 +∣CB∣^2 =−9∣OG∣^2 +9R^2 ⇒3)

WLOG,letObetheorigin;g=a+b+c3∣o−a∣=∣o−b∣=∣o−c∣=R⇒(o−a)(o−−a−)=R2⇒aa−=R2⇒a−=R2aThen∣OG∣2=(g−o)(g−−o−)=gg−=(a+b+c3)(R2(1a+1b+1c)3)=R2(a+b+c)(ab+bc+ca)9abc∣b−a∣2+∣b−c∣2+∣c−a∣2=R2(b−a)(a−bab)+R2(b−c)(c−bbc)+R2(c−a)(a−cac)=R2abc[c(b−a)(a−b)+a(b−c)(c−b)+b(c−a)(a−c)]=r2abc[{−3abc+3abc−a2c−b2c−ab2−c2a+(abc−a2b)−c2b+6abc]=−R2abc[a2b+abc+a2c+ab2+b2c+abc+abc+bc2+c2a−9abc]=−r2(a+b+c)(ab+bc+ca)abc+9r2⇒∣BA∣2+∣AC∣2+∣CB∣2=−9∣OG∣2+9R2⇒3)

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