1一个猜想文[1]中提出了一个猜想,叙述如下:设单位圆的方程为(x-1)2+y2=1,圆心为A,与x轴交于O,B点.给定正数h(h<2),在OB上依次取点h,2h,3h,…,nh,显然n=[2/h](表示取整).过点ih(i=1,2,3,…,n)作x轴的垂线,记垂线在圆内线段为ai(i=1,2,3,…,n)(图1),则由射影定理或直接由圆方程线段ai的长度为ih(2-ih)~(1/2),故n条线段ai(i=1,2,3,…,n)的长度总和为 1 A conjecture in the conjecture [1] proposes a conjecture, which is described as follows: Let the equation of the unit circle be (x-1)2+y2=1, the center of the circle is A, and intersect with the x-axis at point O, B. The number h (h<2), in turn take points h, 2h, 3h,...,nh on OB, obviously n=[2/h] (representing rounding). Over the point ih (i=1, 2, 3, ..., n) for the vertical axis of the x-axis, note that the vertical line in the circle is ai(i=1,2,3,...,n) (Fig. 1), then by the projective theorem or directly by the circular equation ai The length is ih(2-ih)~(1/2), so the total length of the n line segments ai(i=1, 2, 3,...,n) is