笔者曾于1989年8月用较为简捷的方法(本人称之为线段投影法)解决了以下几何“难题”(读者可参阅[1],[2]),即问题假设P、Q、R分别是△ABC的三边BC、CA、AB上三点,且满足AQ+AR=BR+BP=CP+CQ=1/3,则PQ+QR+RP≥1/2.证明(线段投影法)如图,记BC=a,CA=b,AB=c,由已知有a+b+c=1.从R、Q向直线BC In August 1989, the author solved the following geometry “problem ” (readers can refer to [1], [2]) in a relatively simple method (I call it the line projection method) R are the three points BC, CA, AB on △ ABC respectively, and if AQ + AR = BR + BP = CP + CQ = 1/3, then PQ + QR + RP≥1 / Method) as shown in Figure, remember BC = a, CA = b, AB = c, known by a + b + c = 1 from R, Q to a straight line BC