2022年中考总结复习冲刺练：初中数学“最值问题”集锦 .pdf

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1、中考总结复习冲刺练：“最值问题”集锦平面几何中的最值问题 01 几何的定值与最值 07 最短路线问题 14 对称问题 18 巧作“对称点”妙解最值题 22 数学最值题的常用解法26 求最值问题29 有理数的一题多解34 4 道经典题37 平面几何中的最值问题在平面几何中，我们常常遇到各种求最大值和最小值的问题，有时它和不等式联系在一起，统称最值问题如果把最值问题和生活中的经济问题联系起来，可以达到最经济、最节约和最高效率下面介绍几个简例在平面几何问题中，当某几何元素在给定条件变动时，求某几何量（如线段的长度、图形的面积、角的度数）的最大值或最小值问题，称为最值问题。最值问题的解决方法通常有两种

2、：（1）应用几何性质：三角形的三边关系：两边之和大于第三边，两边之差小于第三边；两点间线段最短；连结直线外一点和直线上各点的所有线段中，垂线段最短；定圆中的所有弦中，直径最长。运用代数证法：运用配方法求二次三项式的最值；运用一元二次方程根的判别式。例 1、A、B两点在直线 l 的同侧，在直线 L 上取一点 P，使 PA+PB 最小。分析：在直线 L 上任取一点 P，连结 A P，BP，在ABP 中 AP+BP AB，如果 AP+BP AB,则 P必在线段 AB上，而线段 AB与直线 L 无交点，所以这种思路错误。取点 A关于直线 L 的对称点 A，则 AP AP，在ABP中 AP+BPAB,当

3、 P移到 AB 与直线 L 的交点处 P点时 AP+BPAB，所以这时PA+P B最小。1 已知 AB是半圆的直径，如果这个半圆是一块铁皮，ABDC 是内接半圆的梯形，试问怎样剪这个梯形，才能使梯形ABDC 的周长最大(图 391)？分析 本例是求半圆 AB的内接梯形的最大周长，可设半圆半径为R 由于 ABCD，必有 AC=BD 若设 CD=2y，AC=x，那么只须求梯形ABDC 的半周长 u=x+y+R的最大值即可解 作 DE AB于 E，则x2=BD2=AB BE 2R(R-y)2R2-2Ry，所以所以求 u 的最大值，只须求-x2+2Rx+2R2最大值即可-x2+2Rx+2R2=3R2-

4、(x-R)23R2，上式只有当 x=R时取等号，这时有文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2

5、F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G

6、2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R1

7、0J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6

8、F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W

9、2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1

10、R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2文档编码:CB6D1R10J4U7 HI4E4Z6F2F7 ZG2V2A3W2G2所以2y=R=x 所以把半圆三等分，便可得到梯形两个顶点C，D，这时，梯形的底角恰为60和 1202.如图 392是半圆与矩形结合而成的窗户，如果窗户的周长为8 米(m)，怎样才能得出最大面积，使得窗户透光最好

11、？分析与解设 x 表示半圆半径，y 表示矩形边长 AD，则必有2x+2y+x=8，若窗户的最大面积为S，则把代入有即当窗户周长一定时，窗户下部矩形宽恰为半径时，窗户面积最大3.已知 P点是半圆上一个动点，试问 P在什么位置时，PA+PB 最大(图 393)？文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编

12、码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G

13、1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10

14、A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4

15、U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6

16、X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC1

17、0J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H

18、3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1分析与解因为 P点是半圆上的动点，当P近于 A或 B时，显然 PA+PB 渐小，在极限状况(P 与 A重合时)等于 AB 因此，猜想 P在半圆弧中点时，PA+PB 取最大值设 P为半圆弧中点，连 PB，PA，延长 AP到 C，使 PC=PA，连 CB，则 CB是切线为了证 PA+PB 最大，我们在半圆弧上另取一点P，连 PA，PB，延长 AP 到 C，使 PC=BP，连 CB，CC，则 PC B=PBC=PCB=45，所以 A，B，C，C四点共圆，所以 CC A=CBA=90，所以在 ACC 中，AC A

19、C，即 PA+PB PA+P B4 如图 394，在直角 ABC中，AD是斜边上的高，M，N分别是 ABD，ACD的内心，直线 MN交 AB，AC于 K，L求证：SABC2SAKL证 连结 AM，BM，DM，AN，DN，CN 因为在 ABC中，A=90，AD BC于 D，所以 ABD=DAC，ADB=ADC=90 因为 M，N分别是 ABD 和ACD 的内心，所以1=2=45，3=4，所以ADN BDM，又因为 MDN=90=ADB，所以MDN BDA，所以BAD=MND 由于 BAD=LCD，所以MND=LCD，所以 D，C，L，N四点共圆，所以ALK=NDC=45 同理，AKL=1=45，

20、所以 AK=AL 因为AKM ADM，所以AK=AD=AL而而从而文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10

21、A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4

22、U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6

23、X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC1

24、0J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H

25、3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编

26、码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1所以 SABCSAKL5.如图 395已知在正三角形ABC内(包括边上)有两点 P，Q 求证：PQAB

27、 证 设过 P，Q的直线与 AB，AC分别交于 P1，Q1，连结 P1C，显然，PQ P1Q1因为 AQ1P1+P1Q1C=180，所以 AQ1P1和P1Q1C中至少有一个直角或钝角若AQ1P190，则 PQ P1Q1AP1AB；若P1Q1C 90，则 PQ P1Q1P1C 同理，AP1C和BP1C中也至少有一个直角或钝角，不妨设 BP1C90，则 P1C BC=AB 对于 P，Q两点的其他位置也可作类似的讨论，因此，PQ AB 6.设ABC 是边长为 6 的正三角形，过顶点A引直线 l，顶点 B，C到 l 的距离设为 d1，d2，求 d1+d2的最大值(1992 年上海初中赛题)解 如图 3

28、96，延长 BA到 B，使 AB=AB，连 BC，则过顶点 A的直线l 或者与 BC相交，或者与 BC相交以下分两种情况讨论(1)若 l 与 BC相交于 D，则所以文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8

29、G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G1

30、0A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ

31、4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q

32、6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC

33、10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6

34、H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1只有

35、当 l BC时，取等号(2)若 l 与 BC相交于 D，则所以上式只有 l BC时，等号成立7.如图 397已知直角 AOB 中，直角顶点 O在单位圆心上，斜边与单位圆相切，延长 AO，BO分别与单位圆交于C，D试求四边形 ABCD 面积的最小值解 设O与 AB相切于 E，有 OE=1，从而即AB 2当 AO=BO 时，AB有最小值 2从而文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编

36、码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G

37、1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10

38、A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4

39、U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6

40、X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC1

41、0J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H

42、3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1所以，当 AO=OB 时，四边形 ABCD 面积的最小值为几何的定值与最值几何中的定值问题，是指变动的图形中某些几何元素的几何量保持不变，或几何元素间的某些几何性质或位置关系不变的一类问题，解几何定值问题的基本方法是：分清问题的定量及变量，运用特殊位置、极端位置，直接计算等方法，先探求出定值，再给出证明几何中的最值问题是指在一定的条件下，求平面几何图形中某个确定的量(如线段长度、角度大小、图形面积)等的最大值或最小值，求几

43、何最值问题的基本方法有：1特殊位置与极端位置法；2几何定理(公理)法；3数形结合法等注：几何中的定值与最值近年广泛出现于中考竞赛中，由冷点变为热点 这是由于这类问题具有很强的探索性(目标不明确)，解题时需要运用动态思维、数形结合、特殊与一般相结合、逻辑推理与合情想象相结合等思想方法【例题就解】文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6

44、X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC1

45、0J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H

46、3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编

47、码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G

48、1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10

49、A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4

50、U9H5Q6X1 ZC10J1J6H3K1文档编码:CX8G1P1G10A8 HZ4U9H5Q6X1 ZC10J1J6H3K1【例 1】如图，已知 AB=10，P是线段 AB上任意一点，在 AB的同侧分别以AP和 PB为边作等边 APC和等边 BPD，则 CD长度的最小值为思路点拨如图，作 CC AB于 C，DD AB于 D，DQ CC，CD2=DQ2+CQ2，DQ=21AB一常数，当 CQ越小，CD越小，本例也可设 AP=x，则 PB=x10，从代数角度探求CD的最小值注：从特殊位置与极端位置的研究中易得到启示，常能找到解题突破口，特殊位置与极端位置是指：(1)中点处、垂直位置关系等；(2)