2003年全国高中数学联赛有这样一个问题:一张纸上画有半径为R的圆O和圆内一定点A,且OA=a.折叠纸片,使圆周上某一点A′刚好与A点重合,这样的每一种折法,都留下一条直线折痕,当A′取遍圆周上所有点时,求所有折痕所在直线上点的集合.图1其解答如下:如图1,以O为原点,OA所在直线为x In the 2003 National High School Mathematics League, there was a problem that a piece of paper was drawn with a circle O with a radius R and a certain point A in the circle, and OA = a. folded paper so that a point A’ on the circumference coincided with point A Coincidence, each of these folds leaves a straight line crease. When A’ fetches all the points on the circumference, find the set of points on the straight line where all creases lie. The answer to Figure 1 is as follows: Figure 1 O is the origin and the OA is on the x