初中竞赛几何问题四

锐角三角形ABC的顶点A与重心G的连线垂直于内心I与外心O连线的充要条件是:2/BC=1/AB+1/AC.证 设BC=a,CA=b,AB=c,2s=a+b+c.r为内切圆半径.延长BG交AC于N,内切圆切AC于P,切AB于Q.BO^2-ON^2=AO^2-ON^2=AN^2=b^2/4.BI^2=BQ^2+r^2=(s-b)^2+r^2;IN^2=PN^2+r^2=[b/2-(s-b)]^2+r^2=[(c-a)/2]^2+r^2.于是BG⊥IO<==>BN⊥IO<==>BI^2-IN^2=BO^2-ON^2<==>(s-b)^2+r^2-[(c-a)/2]^2-r^2=b^2/4<==>b(c+a)=ac<==>1/a+1/c=2/b. 收起