G是三角形ABO的重心,M是AB的中点,若PQ过三角形的重心G,且向量OP=mOA,OQ=nOB,求证（1/m）+（1/

G是三角形ABO
=> GA+GB+GO =0
M是AB的中
=> AM =MB
G is on PQ
OP = mOA
OQ = nOB
let |PG| :|GQ|= k
OG = (OP + kOQ)/(1+k)
= (mOA + knOB)/(1+k)
GA+GB+GO =0
(OA-OG) + (OB-OG) +GO =0
OA+OB=3OG
= 3(mOA + knOB)/(1+k)
=> 3m/(1+k) = 1 (1) and
3kn/(1+k)=1 (2)
(2)/(1)
kn/m =1
k = m/n (3)
sub (3) into (1)
3m/(1+m/n) =1
3mn/(m+n) =1
3mn =m+n
1/m + 1/n = 3

1、求向量GA+向量GB=2*向量GM
又因为G是△ABO的重心
所以向量GO=-2*向量GM
所以求向量GA+向量GB+向量GO=0
2、方一：特殊值
设p在A点，则Q在中点
所以m=1，n=1/2
所以
1/m + 1/n =3
方二：
显然OM = 1/2(a+b).因为G是△ABO的重心， 所以Ƚ...