Question

正方体中,点M.N.P,Q分别在棱AA1,D1C1,BC,AB上,且AM=3MA1,D1N=2NC1,BP=PC,AQ=

Answer 1

设正方体棱长为L

AM=3MA1,D1N=2NC1,BP=PC,AQ=mQB

AM=3L/4 L MA1=L/4 D1N=2L/3 NC1=L/3 BP=PC=L/2 AQ=mL/(m+1) QB=L/(m+1)

延长NM交平面ABCD于点R 过点R作AD的垂线交于S

易得 RS=(AM/MA1)D1N=3D1N =2L SA=(AM/MA1)AD=3AD=3L

PQR 共线(两平面交线）

PB/AS=QB/(RS+AQ)

得L/2/(3L)=(L/(m+1))/(2L+mL/(m+1))

2+m/(m+1)=6/(m+1) 2m+2+m=6 m=4/3