\(\left\{{}\begin{matrix}x+2y-2=0\\2x+y+1=0\end{matrix}\right.\) \(\Rightarrow A\left(-\frac{4}{3};\frac{5}{3}\right)\)
Gọi \(\left\{{}\begin{matrix}B\left(2-2b;b\right)\\C\left(c;-2c-1\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MB}=\left(1-2b;b-2\right)\\\overrightarrow{MC}=\left(c-1;-2c-3\right)\end{matrix}\right.\)
Do \(M\in BC\Rightarrow\frac{1-2b}{c-1}=\frac{b-2}{-2c-3}\) \(\Leftrightarrow3bc+7b-5=0\) \(\Rightarrow c=\frac{-7b+5}{3b}\) (1)
\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(\frac{10}{3}-2b;b-\frac{5}{3}\right)\\\overrightarrow{AC}=\left(c+\frac{4}{3};-2c-\frac{8}{3}\right)\end{matrix}\right.\) mà AB=AC
\(\Rightarrow\left(\frac{10}{3}-2b\right)^2+\left(b-\frac{5}{3}\right)^2=\left(c+\frac{4}{3}\right)^2+\left(2c+\frac{8}{3}\right)^2\)
\(\Leftrightarrow3b^2-10b+3=3c^2+8c\) (2)
Thế (1) vào (2) ta được:
\(9b^4-30b^3+16b^2+30b-25=0\)
\(\Leftrightarrow\left(b-1\right)\left(b+1\right)\left(9b^2-30b+25\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}b=1\\b=-1\\b=\frac{5}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}c=-\frac{2}{3}\\c=-4\\c=-\frac{4}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}C\left(-\frac{2}{3};\frac{1}{3}\right)\\C\left(-4;7\right)\\C\left(-\frac{4}{3};\frac{5}{3}\right)\equiv A\left(l\right)\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}A\left(-\frac{4}{3};\frac{5}{3}\right)\\C\left(-\frac{2}{3};\frac{1}{3}\right)\end{matrix}\right.\) gọi \(D\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AD}=\left(x+\frac{4}{3};y-\frac{5}{3}\right)\\\overrightarrow{CD}=\left(x+\frac{2}{3};y-\frac{1}{3}\right)\end{matrix}\right.\)
\(\Rightarrow P=\overrightarrow{DA}.\overrightarrow{DC}=\overrightarrow{AD}.\overrightarrow{CD}=\left(x+\frac{4}{3}\right)\left(x+\frac{2}{3}\right)+\left(y-\frac{5}{3}\right)\left(y-\frac{1}{3}\right)\)
\(P=x^2+2x+\frac{8}{9}+y^2-2y+\frac{5}{9}\)
\(P=\left(x+1\right)^2+\left(y-1\right)^2-\frac{5}{9}\ge-\frac{5}{9}\)
\(\Rightarrow P_{min}=-\frac{5}{9}\) khi \(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\) hay \(D\left(-1;1\right)\)
TH2: bạn tự giải, thật ra D luôn là trung điểm AC
