Question

cho ΔABC. tìm điểm M sao cho \(\overrightarrow{MA}.\overrightarrow{MB}+\overrightarrow{MB}.\overrightarrow{MC}+\overrightarrow{MC}.\overrightarrow{MA}\) đạt GTNN

Answer 1

Gọi G là trọng tâm tam giác ABC

\(\vec{MA}.\vec{MB}+\vec{MB}.\vec{MC}+\vec{MC}.\vec{MA}\)

\(=\dfrac{1}{2}\left(\vec{MA}+\vec{MB}+\vec{MC}\right)^2-\dfrac{1}{2}\left(MA^2+MB^2+MC^2\right)\)

\(\ge-\dfrac{1}{2}\left(MA^2+MB^2+MC^2\right)\)

\(=-\dfrac{1}{2}\left[\left(\vec{MG}+\vec{GA}\right)^2+\left(\vec{MG}+\vec{GB}\right)^2+\left(\vec{MG}+\vec{GC}\right)^2\right]\)

\(=-\dfrac{1}{2}\left[3MG^2+2\vec{MG}\left(\vec{GA}+\vec{GB}+\vec{GC}\right)+GA^2+GB^2+GC^2\right]\)

\(\ge-\dfrac{1}{2}\left(GA^2+GB^2+GC^2\right)\)

\(min=-\dfrac{1}{2}\left(GA^2+GB^2+GC^2\right)\Leftrightarrow M\equiv G\)