论文部分内容阅读
图1模型如图1,在直线l的同侧有两点A、C,在直线l上找一点B使AB+BC的值最小.如图1,显然我们先找到点A关于直线l的对称点A′,连结A′C交直线l于点B,则此时AB+BC=A′C最小.证明简单,这里从略.生长点一一个动点图2例1(第16届希望杯赛题)如图2,正△ABC的边长为a,D是BC的中点,P是AC上的动点,连结PB和PD得到△PBD.求:(1)当点P运动到AC的中点时,△PBD的周长;(2)△PBD的周长的最小值.简析(1)略;(2)△PBD中,因为点B和点D是定点,所以BD的长度唯一确定,又正△ABC的边长为a,即BD=12a,所以若求△PBD的周长的最小值,只需求出PB+PD的最小值即可,此时已经
Figure 1 model shown in Figure 1, the same side of the straight line l has two points A, C, find a point on the straight line l B AB + BC minimum value shown in Figure 1, obviously we find the first point A symmetry about the straight line l Point A ’, link A’C cross-line l at point B, then AB + BC = A’C minimum. Proof is simple. Growing point one by one moving point Figure 2 Example 1 (16th hope 2), the side length of positive △ ABC is a, D is the midpoint of BC, P is the moving point on AC, and the PB and PD are connected to get △ PBD. (1) When point P moves to AC (2) △ PBD, because the point B and point D is a fixed point, so the length of the BD The only definite, and positive △ ABC side length a, that is, BD = 12a, so if you ask △ PBD the minimum length of the circumference, only need to find out the minimum PB + PD, this time has been