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\(T=\overrightarrow{GA}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)+\overrightarrow{GB}.\overrightarrow{CA}+\overrightarrow{GC}.\overrightarrow{AB}\)
\(=\overrightarrow{AB}\left(\overrightarrow{GC}-\overrightarrow{GA}\right)+\overrightarrow{AC}\left(\overrightarrow{GA}-\overrightarrow{GB}\right)\)
\(=\overrightarrow{AB}\left(\overrightarrow{GC}+\overrightarrow{AG}\right)+\overrightarrow{AC}\left(\overrightarrow{GA}+\overrightarrow{BG}\right)\)
\(=\overrightarrow{AB}.\overrightarrow{AC}+\overrightarrow{AC}.\overrightarrow{BA}\)
\(=0\)
\(\overrightarrow{AB}=\overrightarrow{AG}+\overrightarrow{GB}=\overrightarrow{b}-\overrightarrow{a}\)
\(\overrightarrow{GC}=0-\overrightarrow{GA}-\overrightarrow{GB}=-\overrightarrow{a}-\overrightarrow{b}\)
\(\overrightarrow{BC}=\overrightarrow{BG}+\overrightarrow{GC}=-\overrightarrow{b}-\overrightarrow{a}-\overrightarrow{b}=-\overrightarrow{a}-2\overrightarrow{b}\)
\(\overrightarrow{CA}=\overrightarrow{CG}+\overrightarrow{GA}=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{a}=2\overrightarrow{a}+\overrightarrow{b}\)
\(\overrightarrow{AG}=\frac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\Rightarrow\overrightarrow{GA}=-\frac{1}{3}\overrightarrow{AB}-\frac{1}{3}\overrightarrow{AC}=-\frac{1}{3}\overrightarrow{a}-\frac{1}{3}\overrightarrow{b}\)
\(\Rightarrow m=n=-\frac{1}{3}\Rightarrow mn=\frac{1}{9}\)
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 \(G\equiv G'\).
\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\Rightarrow\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)^2=0\)
\(\Rightarrow-2\left(\overrightarrow{GA}.\overrightarrow{GB}+\overrightarrow{GB}.\overrightarrow{GC}+\overrightarrow{GC}.\overrightarrow{GA}\right)=GA^2+GB^2+GC^2\)
\(\Rightarrow\overrightarrow{GA}.\overrightarrow{GB}+\overrightarrow{GB}.\overrightarrow{GC}+\overrightarrow{GC}.\overrightarrow{GA}=-\frac{1}{2}\left(\frac{2}{3}m_a^2+\frac{2}{3}m_b^2+\frac{2}{3}m_c^2\right)\)
\(=-\frac{1}{6}\left(AB^2+BC^2+CA^2\right)\)
Hình như đề bài sai dấu?
Lời giải:
$G$ là trọng tâm tam giác $ABC$ thì ta có 1 bổ đề quen thuộc là:
$\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}$
$\Leftrightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{GC}=\overrightarrow{0}$
$\Rightarrow \overrightarrow{GC}=-(\overrightarrow{a}+\overrightarrow{b})$
Ta có:
\(\frac{1}{2}\overrightarrow{AB}-\overrightarrow{BC}=\frac{1}{2}(\overrightarrow{AG}+\overrightarrow{GB})-(\overrightarrow{BG}+\overrightarrow{GC})\)
\(=\frac{1}{2}(-\overrightarrow{a}+\overrightarrow{b})-[-\overrightarrow{b}-(\overrightarrow{a}+\overrightarrow{b})]\)
\(=\frac{\overrightarrow{a}}{2}+\frac{5\overrightarrow{b}}{2}\)
Theo tính chất trọng tâm ta luôn có:
\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{GC}=-\overrightarrow{GA}-\overrightarrow{GB}=-\overrightarrow{a}-\overrightarrow{b}\)
\(\Rightarrow m=n=-1\Rightarrow m+n=-2\)