Question

如图所示,在△ABC中,点M是BC的中点,点N在边AC上,且AN=2NC,AM与BN相交于点P,求AP∶PM的值.

如图所示,在△ABC中,点M是BC的中点,点N在边AC上,且AN=2NC,AM与BN相交于点P,求AP∶PM的值.

Answer 1

4

设 =e 1 , =e 2 , 则 = + =-3e 2 -e 1 , =2e 1 +e 2 . ∵A,P,M和B,P,N分别共线,∴存在λ,μ∈R, 使 =λ =-λe 1 -3λe 2 , =μ =2μe 1 +μe 2 . 故 = - =(λ+2μ)e 1 +(3λ+μ)e 2 , 而 = + =2e 1 +3e 2 , ∴ ∴ ∴ = ,∴ = ,即AP∶PM=4.