Question Number 97174 by mathocean1 last updated on 06/Jun/20
![Exercise_(−) ABC is a triangle. AB=AC=2 and BC=2(√2). I is midle of [BC]. J is a point such as AJ^(→) =(2/3)AI^(→) . J is the center of gravity of the triangle. 1)a) we define the set(T) of ∀ point M of plane: AM^2 +BM^2 +CM^2 =8. Show that BM^2 +CM^2 =2IM^2 +4 and AM^2 +2IM^2 =3JM^2 +(4/3) b)deduct that AM^2 +BM^2 +CM^2 =3JM^2 +((16)/3) c)Deduct the nature of (T)](https://www.tinkutara.com/question/Q97174.png)
$$\underset{−} {{Exercise}} \\ $$$${ABC}\:{is}\:{a}\:{triangle}.\:{AB}={AC}=\mathrm{2}\:{and} \\ $$$${BC}=\mathrm{2}\sqrt{\mathrm{2}}.\:{I}\:{is}\:{midle}\:{of}\:\left[{BC}\right]. \\ $$$${J}\:{is}\:{a}\:{point}\:{such}\:{as}\:\overset{\rightarrow} {{AJ}}=\frac{\mathrm{2}}{\mathrm{3}}\overset{\rightarrow} {{AI}}.\:{J}\:{is} \\ $$$${the}\:{center}\:{of}\:{gravity}\:{of}\:{the}\:{triangle}. \\ $$$$\left.\mathrm{1}\left.\right){a}\right)\:{we}\:{define}\:{the}\:{set}\left({T}\right)\:{of}\:\forall\:{point}\:{M} \\ $$$${of}\:{plane}: \\ $$$${AM}^{\mathrm{2}} +{BM}^{\mathrm{2}} +{CM}^{\mathrm{2}} =\mathrm{8}. \\ $$$${Show}\:{that}\:{BM}^{\mathrm{2}} +{CM}^{\mathrm{2}} =\mathrm{2}{IM}^{\mathrm{2}} +\mathrm{4} \\ $$$${and}\:{AM}^{\mathrm{2}} +\mathrm{2}{IM}^{\mathrm{2}} =\mathrm{3}{JM}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\left.{b}\right){deduct}\:{that}\: \\ $$$${AM}^{\mathrm{2}} +{BM}^{\mathrm{2}} +{CM}^{\mathrm{2}} =\mathrm{3}{JM}^{\mathrm{2}} +\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\left.{c}\right){Deduct}\:{the}\:{nature}\:{of}\:\left({T}\right) \\ $$