O为△ABC内一点,且OA+2OB+3OC=0,则S△AOC:S△ABC=_.

问题描述:

O为△ABC内一点,且

O
A+2
O
B+3
O
C=
0
,则S△AOC:S△ABC=______.

延长OB至B',使OB'=2OB;延长OC至C',使OC'=3OC,则OA+OB′+OC′=0 ∴O是△AB′C′的重心∴S△AOC′=S△B′OC′=S△AOB′=13S△AB′C′,∵S△AOC=13S△AOC′,S△BOC=16S△B′′OC′,S△AOB=12S△AOB′,∴S△AOC...