Đáp án C

\(a,\) \(\overrightarrow{IA}=2\overrightarrow{IB}-4\overrightarrow{IC}\)

\(\overrightarrow{IA}=2\overrightarrow{IB}-2\overrightarrow{IC}-2\overrightarrow{IC}=2\overrightarrow{CB}-2\overrightarrow{IC}\)

\(=2\left(\overrightarrow{AB}-\overrightarrow{AC}\right)-2\left(\overrightarrow{AC}-\overrightarrow{AI}\right)\)

\(\overrightarrow{IA}=2\overrightarrow{AB}-2\overrightarrow{AC}-2\overrightarrow{AC}+2\overrightarrow{AI}\)

\(\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}\)

\(b,\overrightarrow{IJ}=\overrightarrow{AJ}-\overrightarrow{AI}=\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}=\dfrac{4}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(1\right)\)

\(\overrightarrow{JG}=\overrightarrow{AG}-\overrightarrow{AJ}=\dfrac{2}{3}\overrightarrow{AM}-\dfrac{2}{3}\overrightarrow{AB}\)\((\) \(\) \(M\)  \(trung\) \(điểm\) \(BC)\)

\(\overrightarrow{JG}=\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{3}-\dfrac{2}{3}\overrightarrow{AB}=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=-\dfrac{1}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\overrightarrow{IJ}=-4\overrightarrow{JG}\Rightarrow I,J,G\) \(thẳng\) \(hàng\)

1.

\(\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{AI}=\overrightarrow{0}\Leftrightarrow\overrightarrow{IB}+\overrightarrow{AC}=0\)

\(\Leftrightarrow\overrightarrow{IB}=\overrightarrow{CA}\)

\(\Rightarrow\) I là 1 đỉnh của hình bình hành ABIC

2.

Gọi N là trung điểm AB \(\Rightarrow\overrightarrow{AN}=\dfrac{1}{2}\overrightarrow{AB}\)

\(\overrightarrow{MA}+\overrightarrow{BM}+2\overrightarrow{MC}=\overrightarrow{0}\Leftrightarrow\overrightarrow{BA}+2\overrightarrow{MC}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{MC}=\dfrac{1}{2}\overrightarrow{AB}\Leftrightarrow\overrightarrow{MC}=\overrightarrow{AN}\)

\(\Rightarrow\) M là 1 đỉnh của hình bình hành ANCM

\(\overrightarrow{MA}+\overrightarrow{MB}=\overrightarrow{\left.0\right.}\)

=>\(\overrightarrow{MA}=-\overrightarrow{MB}\)

=>M là trung điểm của AB

\(2\cdot\overrightarrow{NA}+\overrightarrow{NC}=\overrightarrow{0}\)

=>\(\overrightarrow{NC}=-2\cdot\overrightarrow{NA}=2\cdot\overrightarrow{AN}\)

=>N nằm giữa A và C sao cho NC=2NA

\(\overrightarrow{AI}=\overrightarrow{AM}+\overrightarrow{MI}=\frac12\cdot\overrightarrow{AB}+\frac12\cdot\overrightarrow{MN}\)

\(=\frac12\cdot\overrightarrow{AB}+\frac12\left(\overrightarrow{MA}+\overrightarrow{AN}\right)=\frac12\cdot\overrightarrow{AB}+\frac12\cdot\left(-\frac12\cdot\overrightarrow{AB}+\frac13\cdot\overrightarrow{AC}\right)\)

\(=\frac14\cdot\overrightarrow{AB}+\frac16\cdot\overrightarrow{AC}\)

\(\overrightarrow{DB}=k\cdot\overrightarrow{DC}\)

=>\(\overrightarrow{BD}=-k\cdot\overrightarrow{DC}\)

=>\(\overrightarrow{BD}+\overrightarrow{DC}=-k\cdot\overrightarrow{DC}+\overrightarrow{DC}=\overrightarrow{DC}\left(-k+1\right)\)

=>\(\overrightarrow{BC}=\left(-k+1\right)\cdot\overrightarrow{DC}\)

=>\(\frac{\overrightarrow{BD}}{\overrightarrow{BC}}=\frac{-k\cdot\overrightarrow{DC}}{\left(-k+1\right)\cdot\overrightarrow{DC}}=\frac{-k}{-k+1}=\frac{k}{k-1}\)

=>\(\overrightarrow{BD}=\frac{k}{k-1}\cdot\overrightarrow{BC}\)

\(\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AB}+\frac{k}{k-1}\cdot\overrightarrow{BC}\)

\(=\overrightarrow{AB}+\frac{k}{k-1}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\left(1-\frac{k}{k-1}\right)\cdot\overrightarrow{AB}+\frac{k}{k-1}\cdot\overrightarrow{AC}\)

\(=\frac{-1}{k-1}\cdot\overrightarrow{AB}+\frac{k}{k-1}\cdot\overrightarrow{AC}\)

Để A,I,D thẳng hàng thì \(\frac{-1}{k-1}:\frac14=\frac{k}{k-1}:\frac16\)

=>\(\frac{-4}{k-1}=\frac{6k}{k-1}\)

=>6k=-4

=>\(k=-\frac23\)

a) \(2\overrightarrow{IA}-\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{0}\Rightarrow2\overrightarrow{IA}-\overrightarrow{IA}-\overrightarrow{AB}+\overrightarrow{IA}+\overrightarrow{AC}=\overrightarrow{0}\)

\(\Rightarrow2\overrightarrow{AI}=\overrightarrow{AC}-\overrightarrow{AB}\Rightarrow\overrightarrow{AB}+2\overrightarrow{AI}=\overrightarrow{AC}\). Từ đó suy ra cách dựng điểm I:

A B C I

b) Với cách lấy điểm I như trên, ta có điểm I cố định. Khi đó MN đi qua I, thật vậy:

\(\overrightarrow{MN}=2\overrightarrow{MA}-\overrightarrow{MB}+\overrightarrow{MC}=2\overrightarrow{MI}+2\overrightarrow{IA}-\overrightarrow{MI}-\overrightarrow{IB}+\overrightarrow{MI}+\overrightarrow{IC}\)

\(=2\overrightarrow{MI}+\left(2\overrightarrow{IA}-\overrightarrow{IB}+\overrightarrow{IC}\right)=2\overrightarrow{MI}\)

Suy ra I là trung điểm MN hay MN đi qua điểm I cố định (đpcm).

c) \(\overrightarrow{MP}=\frac{1}{2}\overrightarrow{MB}+\frac{1}{2}\overrightarrow{MN}=\overrightarrow{MA}+\frac{1}{2}\overrightarrow{MC}\)

Đặt K là điểm sao cho \(\overrightarrow{KA}+\frac{1}{2}\overrightarrow{KC}=\overrightarrow{0}\Rightarrow\hept{\begin{cases}K\in\left[AC\right]\\KA=\frac{1}{2}KC\end{cases}}\)tức K xác định

Khi đó \(\overrightarrow{MP}=\overrightarrow{MK}+\overrightarrow{KA}+\frac{1}{2}\overrightarrow{MK}+\frac{1}{2}\overrightarrow{KC}=\frac{3}{2}\overrightarrow{MK}\), suy ra MP đi qua K cố định (đpcm).