1\以AB,AC为基底AM=1/2mAB+1/2nACAN=1/2AB+1/2ACA,B,C三点共线,就是AN与AM平行AM=kAN所以,带入 k=m,k=n所以m=n2\MN=AN-AM=1/2(1-m)AB+1/2(1-m)AC|MN|^2=1/2[(1-m)^2+(1-n)^2+(1-m)(1-n)] =1/2[1-mn](mm有最大值)MN最小值根号3/8