Câu 3
Gọi M là trung điểm BC và I là điểm nằm trên đoạn thẳng AM sao cho sao cho IM=2IA
\(\Rightarrow\overrightarrow{IM}+2\overrightarrow{IA}=\overrightarrow{0}\Rightarrow\dfrac{1}{2}\overrightarrow{IB}+\dfrac{1}{2}\overrightarrow{IC}+2\overrightarrow{IA}=\overrightarrow{0}\)
\(\Rightarrow4\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{0}\)
Ta có: \(AM=\dfrac{AB\sqrt{3}}{2}=\dfrac{a\sqrt{3}}{2}\Rightarrow IM=\dfrac{2}{3}AM=\dfrac{a\sqrt{3}}{3}\)
\(\Rightarrow IA=\dfrac{1}{3}AM=\dfrac{a\sqrt{3}}{6}\Rightarrow IA^2=\dfrac{a^2}{12}\)
\(IB=IC=\left(\dfrac{BC}{2}\right)^2+IM^2=\dfrac{7a^2}{12}\)
Ta có:
\(4MA^2+MB^2+MC^2=\dfrac{5a^2}{2}\)
\(\Leftrightarrow4\left(\overrightarrow{MI}+\overrightarrow{IA}\right)^2+\left(\overrightarrow{MI}+\overrightarrow{IB}\right)^2+\left(\overrightarrow{MI}+\overrightarrow{IC}\right)^2=\dfrac{5a^2}{2}\)
\(\Leftrightarrow6MI^2+4IA^2+IB^2+IC^2+2\overrightarrow{MI}\left(4\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}\right)=\dfrac{5a^2}{2}\)
\(\Leftrightarrow6MI^2+4IA^2+IB^2+IC^2=\dfrac{5a^2}{2}\)
\(\Leftrightarrow6MI^2=a^2\Rightarrow MI=\dfrac{a}{\sqrt{6}}\)
Quỹ tích M là đường tròn tâm I bán kính \(R=\dfrac{a}{\sqrt{6}}\)
