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F là trung điểm AB \(\Rightarrow\overrightarrow{AF}=\dfrac{1}{2}\overrightarrow{AB}\) ; E là trung điểm AC \(\Rightarrow\overrightarrow{AE}=\dfrac{1}{2}\overrightarrow{AC}\)
Ta có EF song song BC (đường trung bình)
Mà D là trung điểm BC \(\Rightarrow\) I là trung điểm EF \(\Rightarrow AI\) là trung tuyến tam giác AEF
\(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AE}+\dfrac{1}{2}\overrightarrow{AF}\)
Theo tính chất trọng tâm:
\(\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AD}=\dfrac{2}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{2}{3}\left(\overrightarrow{AE}+\overrightarrow{AF}\right)=\dfrac{2}{3}\overrightarrow{AE}+\dfrac{2}{3}\overrightarrow{AF}\)
DE là đường trung bình tam giác ABC
\(\Rightarrow\overrightarrow{DE}=\dfrac{1}{2}\overrightarrow{BA}=-\dfrac{1}{2}\overrightarrow{AB}=-\overrightarrow{AE}\) hay \(\overrightarrow{DE}=-\overrightarrow{AE}+0.\overrightarrow{AF}\)
D là trung điểm BC \(\Rightarrow\overrightarrow{DC}=\dfrac{1}{2}\overrightarrow{BC}\)
\(\Rightarrow\overrightarrow{DC}=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{AC}=-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}=-\overrightarrow{AE}+\overrightarrow{AF}\)
1.
Gọi M là trung điểm BC thì theo tính chất trọng tâm: \(\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}=\dfrac{2}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)\)
\(\Rightarrow\overrightarrow{AG}=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\Rightarrow x+y=\dfrac{2}{3}\)
2.
\(CH=\dfrac{1}{2}BC=\dfrac{a}{2}\)
\(T=\left|\text{ }\overrightarrow{CA}-\overrightarrow{HC}\right|=\left|\overrightarrow{CA}+\overrightarrow{CH}\right|\)
\(\Rightarrow T^2=CA^2+CH^2+2\overrightarrow{CA}.\overrightarrow{CH}=a^2+\left(\dfrac{a}{2}\right)^2+2.a.\dfrac{a}{2}.cos60^0=\dfrac{7a^2}{4}\)
\(\Rightarrow T=\dfrac{a\sqrt{7}}{2}\)
3.
\(10< x< 100\Rightarrow10< 3k< 100\)
\(\Rightarrow\dfrac{10}{3}< k< \dfrac{100}{3}\Rightarrow4\le k\le33\)
\(\Rightarrow\sum x=3\left(4+5+...+33\right)=1665\)
Gọi M là trung điểm BC, theo tính chất trọng tâm:
\(\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}\)
Mà I là trung điểm AG \(\Rightarrow\overrightarrow{IG}=\dfrac{1}{2}\overrightarrow{AG}=\dfrac{1}{3}\overrightarrow{AM}\Rightarrow\overrightarrow{GI}=-\dfrac{1}{3}\overrightarrow{AM}\)
Lại có: M là trung điểm BC \(\Rightarrow\overrightarrow{MB}+\overrightarrow{MC}=\overrightarrow{0}\)
Nên ta có:
\(\overrightarrow{AB}+\overrightarrow{AC}+6\overrightarrow{GI}=\overrightarrow{AM}+\overrightarrow{MB}+\overrightarrow{AM}+\overrightarrow{MC}+6.\left(-\dfrac{1}{3}\right)\overrightarrow{AM}\)
\(=2\overrightarrow{AM}-2\overrightarrow{AM}=\overrightarrow{0}\) (đpcm)
Do G là trọng tâm tam giác
\(\Rightarrow\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AD}=\dfrac{2}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=\dfrac{1}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{CB}+\dfrac{1}{3}\overrightarrow{AC}\)
\(=\dfrac{2}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{CB}=-\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\)
Do I là trung điểm AG
\(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AG}=\dfrac{1}{2}\left(-\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\right)=-\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}\)
\(\overrightarrow{AK}=\dfrac{1}{5}\overrightarrow{AB}=\dfrac{1}{5}\left(\overrightarrow{AC}+\overrightarrow{CB}\right)=-\dfrac{1}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}\)
\(\overrightarrow{CI}=\overrightarrow{CA}+\overrightarrow{AI}=\overrightarrow{CA}-\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}\)
\(\overrightarrow{CK}=\overrightarrow{CA}+\overrightarrow{AK}=\overrightarrow{CA}-\dfrac{1}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}=\dfrac{4}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}\)
a: \(\overrightarrow{CA}+\overrightarrow{AB}+\overrightarrow{BC}\)
\(=\overrightarrow{CB}+\overrightarrow{BC}\)
\(=\overrightarrow{0}\)
b: \(\overrightarrow{AM}+\overrightarrow{AP}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\dfrac{1}{2}\cdot2\cdot\overrightarrow{AN}=\overrightarrow{AN}\)