Question Number 201526 by 281981 last updated on 08/Dec/23
Answered by AST 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)](https://www.tinkutara.com/question/Q201531.png)
$${WLOG},{let}\:{O}\:{be}\:{the}\:{origin};\:{g}=\frac{{a}+{b}+{c}}{\mathrm{3}} \\ $$$$\mid{o}−{a}\mid=\mid{o}−{b}\mid=\mid{o}−{c}\mid={R} \\ $$$$\Rightarrow\left({o}−{a}\right)\left(\overset{−} {{o}}−\overset{−} {{a}}\right)={R}^{\mathrm{2}} \Rightarrow{a}\overset{−} {{a}}={R}^{\mathrm{2}} \Rightarrow\overset{−} {{a}}=\frac{{R}^{\mathrm{2}} }{{a}} \\ $$$${Then}\:\mid{OG}\mid^{\mathrm{2}} =\left({g}−{o}\right)\left(\overset{−} {{g}}−\overset{−} {{o}}\right)={g}\overset{−} {{g}} \\ $$$$=\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)\left(\frac{{R}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)}{\mathrm{3}}\right)=\frac{{R}^{\mathrm{2}} \left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)}{\mathrm{9}{abc}} \\ $$$$\mid{b}−{a}\mid^{\mathrm{2}} +\mid{b}−{c}\mid^{\mathrm{2}} +\mid{c}−{a}\mid^{\mathrm{2}} \\ $$$$={R}^{\mathrm{2}} \left({b}−{a}\right)\left(\frac{{a}−{b}}{{ab}}\right)+{R}^{\mathrm{2}} \left({b}−{c}\right)\left(\frac{{c}−{b}}{{bc}}\right)+{R}^{\mathrm{2}} \left({c}−{a}\right)\left(\frac{{a}−{c}}{{ac}}\right) \\ $$$$=\frac{{R}^{\mathrm{2}} }{{abc}}\left[{c}\left({b}−{a}\right)\left({a}−{b}\right)+{a}\left({b}−{c}\right)\left({c}−{b}\right)+{b}\left({c}−{a}\right)\left({a}−{c}\right)\right] \\ $$$$=\frac{{r}^{\mathrm{2}} }{{abc}}\left[\left\{−\mathrm{3}{abc}+\mathrm{3}{abc}−{a}^{\mathrm{2}} {c}−{b}^{\mathrm{2}} {c}−{ab}^{\mathrm{2}} −{c}^{\mathrm{2}} {a}+\left({abc}−{a}^{\mathrm{2}} {b}\right)−{c}^{\mathrm{2}} {b}+\mathrm{6}{abc}\right]\right. \\ $$$$=\frac{−{R}^{\mathrm{2}} }{{abc}}\left[{a}^{\mathrm{2}} {b}+{abc}+{a}^{\mathrm{2}} {c}+{ab}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}+{abc}+{abc}+{bc}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}−\mathrm{9}{abc}\right] \\ $$$$=\frac{−{r}^{\mathrm{2}} \left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)}{{abc}}+\mathrm{9}{r}^{\mathrm{2}} \\ $$$$\Rightarrow\mid{BA}\mid^{\mathrm{2}} +\mid{AC}\mid^{\mathrm{2}} +\mid{CB}\mid^{\mathrm{2}} =−\mathrm{9}\mid{OG}\mid^{\mathrm{2}} +\mathrm{9}{R}^{\mathrm{2}} \\ $$$$\left.\Rightarrow\mathrm{3}\right) \\ $$