Cho tứ diện $ABCD.$ Gọi $G$ là trọng tâm tam giác $BCD$ và $O$ là trung điểm $AG.$
a) Chứng minh: $3\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\overrightarrow{0}$
b) $M$ là điểm tùy ý. Chứng minh rằng: $3MA^2 + MB^2 + MC^2 + MD^2 = 6MO^2+3OA^2 + OB^2+ OC^2 + OD^2$
c) Tìm quỹ tích $M$ sao cho: $3MA^2 + MB^2 + MC^2 + MD^2 =k^2 (k$ là hằng số)
a/
$3\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}$ 
=$3\overrightarrow{OA}+\overrightarrow{OG}+\overrightarrow{GB}+\overrightarrow{OG}+\overrightarrow{GC}+\overrightarrow{OG}+\overrightarrow{GD}$ 
=$3(\overrightarrow{OA}+\overrightarrow{OG})+(\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD})=0$(đpcm)
b/
$3MA^2+MB^2+MC^2+MD^2$
$=3(\overrightarrow{MO}+\overrightarrow{OA})^2+(\overrightarrow{MO}+\overrightarrow{OB})^2+(\overrightarrow{MO}+\overrightarrow{OC})^2+(\overrightarrow{MO}+\overrightarrow{OD})^2$ 
$=6MO^2+3OA^2+OB^2+OC^2+OD^2+2\overrightarrow{MO}(3\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD})$ 
$=6MO^2+3OA^2+OB^2+OC^2+OD^2$ (đpcm)
c/
 $k^2=3MA^2+MB^2+MC^2+MD^2$
$\Leftrightarrow k^2=6MO^2+3OA^2+OB^2+OC^2+OD^2$
$\Leftrightarrow k^2=6MO^2+3OA^2+(\overrightarrow{OG}+\overrightarrow{GB})^2+(\overrightarrow{OG}+\overrightarrow{GC})^2+(\overrightarrow{OG}+\overrightarrow{GD})^2$ 
$\Leftrightarrow k^2=6MO^2+3OA^2+3OG^2+2\overrightarrow{OG}(\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD})$
$\Leftrightarrow k^2=6MO^2+3OA^2+3OG^2$ 
$\begin{cases}\overrightarrow{OA}=\frac{1}{2}\overrightarrow{GA} \\ \overrightarrow{OG}=\frac{1}{2}\overrightarrow{AG} \end{cases}$
$\Rightarrow k^2=6MO^2+\frac{3}{2}AG^2$
$\Leftrightarrow MO^2=\frac{1}{6}(k^2-\frac{3}{2}AG^2)$ 
$+k^2<3/2AG^2 $ Quỹ tích M thuộc rỗng 
$+k^2=3/2AG^2$  Quỹ tích M là O
$+k^2>3/2AG^2$ Quỹ tích M là đường  tròn tâm 0, bán kính R=$\sqrt{\frac{1}{6}(k^2-\frac{3}{2}AG^2)}$
Thế quỹ tích điểm M đâu ạ? –  Xusint 10-03-13 05:52 PM

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