941

已知正方形 ${\displaystyle ABCD}$ 的边长为 ${\displaystyle 1}$, 当每个 ${\displaystyle {\lambda }_i(i=1,2,3,4,5,6)}$ 取遍 ${\displaystyle \pm 1}$ 时，求 ${\displaystyle |{\lambda }_1\overrightarrow{AB}+{\lambda }_2\overrightarrow{BC}+{\lambda }_3\overrightarrow{CD}+{\lambda }_4\overrightarrow{DA}+{\lambda }_5\overrightarrow{AC}+{\lambda }_6\overrightarrow{BD}|}$ 的最值.

一方面原式 ${\displaystyle \ge 0}$, 可以取 ${\displaystyle ({\lambda }_1,{\lambda }_2,{\lambda }_3,{\lambda }_4,{\lambda }_5,{\lambda }_6)=(-1,1,1,1,1,1)}$.
另一方面$$\begin{array}{rl}\text{原式}=& |({\lambda }_1-{\lambda }_3+{\lambda }_5-{\lambda }_6)\overrightarrow{AB}+({\lambda }_2-{\lambda }_4+{\lambda }_5+{\lambda }_6)\overrightarrow{AD}|\\ =& \sqrt{({\lambda }_1-{\lambda }_3+{\lambda }_5-{\lambda }_6{)}^2+({\lambda }_2-{\lambda }_4+{\lambda }_5+{\lambda }_6{)}^2},\end{array}$$
不妨设${\displaystyle {\lambda }_5={\lambda }_6=1}$(因为这个式子中${\displaystyle {\lambda }_5-{\lambda }_6,{\lambda }_5+{\lambda }_6}$地位等价，可以让${\displaystyle {\lambda }_5,{\lambda }_6}$同号), 则$$\begin{array}{r}\text{原式}=\sqrt{({\lambda }_1-{\lambda }_3{)}^2+({\lambda }_2-{\lambda }_4+2{)}^2}\le \sqrt{2^2+4^2}=2\sqrt{5},\end{array}$$
可取${\displaystyle ({\lambda }_1,{\lambda }_2,{\lambda }_3,{\lambda }_4,{\lambda }_5,{\lambda }_6)=(1,1,-1,-1,1,1)}$.
综上，原式最小值是${\displaystyle 0}$, 最大值是${\displaystyle 2\sqrt{5}}$.