本帖最后由 realnumber 于 2013-11-10 20:02 编辑 第一问,设正方体棱长为1,则有$A(0,0,0),E(1,0.5,0),F(0,0.5,1),P(x,y,z)$
因为$t\vv{EF}=\vv{EP} $,得到$P(1-t,0.5,t)$,因为PM与$CC_1$垂直,所以M(1,1,t)
如此得$\abs{AP}+\abs{PM}=\sqrt{(1-t)^2+0.25+t^2}+\sqrt{t^2+0.25}$,大约是$f(t)_{min}=f(t_0)≈f(0.27679)=1.49326$...最小.
求导后可得$t_0$是方程$8t^3+3t-1=0$的根. |