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求几何最值问题是初中数学的重要内容.因与实际应用联系密切,同时又应用三角形、四边形、多边形、圆和三角函数以及代数中的函数知识,所以知识的综合性与解题的灵活性较强.下面举例探究. 一、利用对称变换求最值例1 如图,正方形ABCD的边长为3,点E在BC上,且BE=2,点P在BD上,则PE+PC的最小值为__. 思路分析:因A、C关于BD对称,连结AE交BD于P,则PE+PC=PE+P以=AE最短. 解:连结AE,AE2=AB2+BE2=32+22=13∵AE=13~(1/2)
The problem of finding the geometrical minimum value is an important content of junior high school mathematics. Because it is closely related to practical applications, while applying triangles, quadrilaterals, polygons, circles, and trigonometric functions and function knowledge in algebra, the comprehensiveness of knowledge and the flexibility of problem solving Stronger. The following example to explore. First, the use of symmetrical transformation to find the best example 1 As shown in the figure, the side length of the square ABCD is 3, the point E is on the BC, and BE=2, the point P is on the BD, then the PE+PC The minimum value is __. Idea analysis: Because A and C are symmetric about BD, link AE to BD to P, then PE+PC=PE+P to AE is the shortest. Solution: AE, AE2=AB2+BE2=32+ 22=13∵AE=13~(1/2)