Question

已知A,B,C为平面上不共线的三点,点O在三角形ABC内,且向量OA+3向量OB+5向量OC=0.

Answer 1

设向量OB'=3OB,OC'=5OC,则

OA+OB'+OC'=0,

∴O是△AB'C'的重心,

∴S△AOB'=S△B'OC'=S△C'OA,设为1,

S△AOB/S△AOB'=OB/OB'=1/3,

S△BOC/S△B'OC'=OB/OB'*OC/OC'=1/15,

S△COA/S△C'OA=OC/OC'=1/5,

∴S△ABC=1/3+1/15+1/5=9/15=3/5,

∴S△AOB/S△ABC=(1/3)/(3/5)=5/9.