设A(x1，y1)，B(x2，y2)，C(x3，y3)抛物线焦点坐标F(1，0)，准线方程：x=-1∵FA+FB+FC=O∴点F是△ABC重心则x1+x2+x3=3 y1+y2+y3=0而|FA|=x1-(-1)=x1+1 |FB|=x2-(-1)=x2+1 |FC|=x3-(-1)=x3+1∴|FA|+|FB|+|FC|=x1+1+x2+1+x3+1=(x1+x2+x3)+3=3+3=6
