Question

如图,M是Rt△ABC斜边CB中点,点P在AB上,且AP：PB=1:2,联结PM,QM⊥PM于M,交AC于Q点,求AQ/

Answer 1

设|BC|=a,|AC|=b,AQ/QC=m,则向量MQ=mMC+(1-m)MA,

AP/AB=1/3,∴MP=(1/3)MB+(2/3)MA,

MP⊥MQ,CM=MB,MA^2=MB^2=a^2/4,

∠AMB=2∠C,cosAMB=2(cosC)^2-1=2b^2/a^2-1,

∴0=MP*MQ=2(1-m)/3*MA^2+MA*[(2m/3)MC+(1-m)/3*MB]+(m/3)MB*MC

=2(1-m)/3*MA^2+MA*(1-3m)/3*MB-(m/3)MB^2

=(2-3m)/3*MA^2+(1-3m)/3*MA^2*cosAMB

∴2-3m+(1-3m)[2b^2/a^2-1]=0,

2b^2/a^2+1=6mb^2/a^2,

m=(a^2+2b^2)/(6b^2),为所求.