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![](https://rs.olm.vn/images/avt/0.png?1311)
\(T=\overrightarrow{GA}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)+\overrightarrow{GB}.\overrightarrow{CA}+\overrightarrow{GC}.\overrightarrow{AB}\)
\(=\overrightarrow{AB}\left(\overrightarrow{GC}-\overrightarrow{GA}\right)+\overrightarrow{AC}\left(\overrightarrow{GA}-\overrightarrow{GB}\right)\)
\(=\overrightarrow{AB}\left(\overrightarrow{GC}+\overrightarrow{AG}\right)+\overrightarrow{AC}\left(\overrightarrow{GA}+\overrightarrow{BG}\right)\)
\(=\overrightarrow{AB}.\overrightarrow{AC}+\overrightarrow{AC}.\overrightarrow{BA}\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\overrightarrow{GA}.\overrightarrow{GB}+\overrightarrow{GB}.\overrightarrow{GC}+\overrightarrow{GC}.\overrightarrow{GA}\)
\(0=\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)^2=GA^2+GB^2+GC^2+2S\Rightarrow S=-\dfrac{GA^2+GB^2+GC^2}{2}\)
\(GA^2+GB^2+GC^2=\dfrac{4}{9}\left(m_a^2+m_b^2+m_c^2\right)\\ =\dfrac{4}{9}\left(\dfrac{2AB^2+2AC^2-BC^2+2BC^2+2AC^2-AB^2+2AB^2+2BC^2-AC^2}{4}\right)\\ =\dfrac{AB^2+AC^2+BC^2}{3}=\dfrac{29}{3}\)
\(\Rightarrow S=-\dfrac{29}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta đã biết nếu G' là trọng tâm tam giác ABC thì:
\(\overrightarrow{G'A}+\overrightarrow{G'B}+\overrightarrow{G'C}=\overrightarrow{0}\).
Gỉa sử có điểm G thỏa mãn: \(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\).
Ta sẽ chứng minh \(G\equiv G'\).
Thật vậy:
\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)
\(\Leftrightarrow3\overrightarrow{GG'}+\overrightarrow{G'A}+\overrightarrow{G'B}+\overrightarrow{G'C}=\overrightarrow{0}\)
\(\Leftrightarrow3\overrightarrow{GG'}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{GG'}=\overrightarrow{0}\).
Vậy \(G\equiv G'\).
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo tính chất trọng tâm tam giác ta luôn có:
\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\Rightarrow\overrightarrow{GA}=-\overrightarrow{GB}-\overrightarrow{GC}\)
Thế vào đẳng thức giả thiết ta được:
\(BC.\left(-\overrightarrow{GB}-\overrightarrow{GC}\right)+AC.\overrightarrow{GB}+AB.\overrightarrow{GC}=\overrightarrow{0}\)
\(\Rightarrow\left(AC-BC\right)\overrightarrow{GB}=\left(BC-AB\right)\overrightarrow{GC}\) (1)
Mà \(\overrightarrow{GB};\overrightarrow{GC}\) không phải 2 vecto cùng phương
\(\Rightarrow\left(1\right)\) xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}AC-BC=0\\BC-AB=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}AC=BC\\AB=BC\end{matrix}\right.\)
\(\Rightarrow AB=AC=BC\) \(\Rightarrow\Delta ABC\) là tam giác đều
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi M là trung điểm BC \(\Rightarrow\overrightarrow{MB}+\overrightarrow{MC}=\overrightarrow{0}\)
Ta có:
\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{GM}+\overrightarrow{MA}+\overrightarrow{GM}+\overrightarrow{MB}+\overrightarrow{GM}+\overrightarrow{MC}=\overrightarrow{0}\)
\(\Leftrightarrow3\overrightarrow{GM}+\overrightarrow{MA}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{GM}=\dfrac{1}{3}\overrightarrow{AM}\)
\(\Leftrightarrow\overrightarrow{GA}+\overrightarrow{AM}=\dfrac{1}{3}\overrightarrow{AM}\)
\(\Leftrightarrow\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}\)
\(\Rightarrow G\) là trọng tâm tam giác ABC
![](https://rs.olm.vn/images/avt/0.png?1311)
Kéo dài AG lấy E sao cho AG=GE
\(2\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GB}=\overrightarrow{GE}+\overrightarrow{GB}=\overrightarrow{AG}+\overrightarrow{GB}=\overrightarrow{AB}\)
\(\overrightarrow{GI}=\overrightarrow{IA}\Rightarrow6\overrightarrow{GI}=3\overrightarrow{GA}\)
\(\overrightarrow{AB}+\overrightarrow{AC}+3\overrightarrow{GA}=\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GA}=\overrightarrow{GE}+\overrightarrow{GA}=\overrightarrow{AG}+\overrightarrow{GA}=\overrightarrow{0}\)
\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\Rightarrow\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)^2=0\)
\(\Rightarrow-2\left(\overrightarrow{GA}.\overrightarrow{GB}+\overrightarrow{GB}.\overrightarrow{GC}+\overrightarrow{GC}.\overrightarrow{GA}\right)=GA^2+GB^2+GC^2\)
\(\Rightarrow\overrightarrow{GA}.\overrightarrow{GB}+\overrightarrow{GB}.\overrightarrow{GC}+\overrightarrow{GC}.\overrightarrow{GA}=-\frac{1}{2}\left(\frac{2}{3}m_a^2+\frac{2}{3}m_b^2+\frac{2}{3}m_c^2\right)\)
\(=-\frac{1}{6}\left(AB^2+BC^2+CA^2\right)\)
Hình như đề bài sai dấu?