中考二轮专题复习：第6课时探索性问题

第二轮复习 探索性问题

Ⅰ、综合问题精讲：

探索性问题是指命题中缺少一定的条件或无明确的结论，需要经过推断，补充并加以证明的题型．探索性问题一般有三种类型：（1）条件探索型问题；（2）结论探索型问题；（3）探索存在型问题．条件探索型问题是指所给问题中结论明确，需要完备条件的题目；结论探索型问题是指题目中结论不确定，不唯一，或题目结论需要类比，引申推广，或题目给出特例，要通过归纳总结出一般结论；探索存在型问题是指在一定的前提下，需探索发现某种数学关系是否存在的题目．

探索型问题具有较强的综合性，因而解决此类问题用到了所学过的整个初中数学知识．经常用到的知识是：一元一次方程、平面直角坐标系、一次函数与二次函数解析式的求法（图象及其性质）、直角三角形的性质、四边形（特殊）的性质、相似三角形、解直

角三角形等．其中用几何图形的某些特殊性质：勾股定理、相似三角形对应线段成比例等来构造方程是解决问题的主要手段和途径．因此复习中既要重视基础知识的复习，又要加强变式训练和数学思想方法的研究，切实提高分析问题、解决问题的能力．

Ⅱ、典型例题剖析

【例1】如图2－6－1，已知抛物线的顶点为A(O ，1)，矩形CDEF 的顶点C 、F 在抛物线

上，D 、E 在x 轴上，CF 交y 轴于点B(0，2)，且其面积为8．

(1)求此抛物线的解析式；

(2)如图2－6－2，若P 点为抛物线上不同于A 的一点，连结PB 并延长交抛物线于

点Q ，过点P 、Q 分别作x 轴的垂线，垂足分别为S 、R ．

①求证：PB ＝PS ；

②判断△SBR 的形状；

③试探索在线段SR 上是否存在点M ，使得以点P 、S 、M 为顶点的三角形和以点Q 、R 、M 为顶点的三角形相似，若存在，请找出M 点的位置；若不存在，请说明理由．

⑴解：方法一：∵B 点坐标为(0，2)，∴OB ＝2，

∵矩形CDEF 面积为8，∴CF=4.

∴C 点坐标为(一2，2)．F 点坐标为(2，2)。

设抛物线的解析式为2

y ax bx c =++．

其过三点A(0，1)，C(-2．2)，F(2，2)。 得1242242x a b c a b c =??=-+??=++?

解得1,0,14a b c === ∴此抛物线的解析式为21

14y x =+ 方法二：∵B 点坐标为(0，2)，∴OB ＝2，

∵矩形CDEF 面积为8， ∴CF=4.

∴C 点坐标为(一2，2)。

根据题意可设抛物线解析式为2y ax c =+。

其过点A(0，1)和C(-2．2)

124c a c

=??=+? 解得1,14a c == 此抛物线解析式为21

14y x =+

(2)解：

①过点B 作BN BS ⊥，垂足为N ．

∵P 点在抛物线y=214x +l 上．可设P 点坐标为21(,1)4a a +．∴PS ＝2114a +，OB ＝NS ＝2，BN ＝a 。∴PN=PS —NS=2114

a - 在Rt PNB 中．

PB 2＝222222211

(1)(1)44PN BN a a a +=-+=+ ∴PB ＝PS ＝2114a +

②根据①同理可知BQ ＝QR 。

∴12∠=∠，

又∵ 13∠=∠，

∴23∠=∠，

同理∠SBP ＝∠B

∴2523180∠+∠=?

∴5390∠+∠=?∴90SBR ∠=?.

∴ △SBR 为直角三角形．

③方法一：设,PS b QR c ==，

∵由①知PS ＝PB ＝b ．QR QB c ==，PQ b c =+。∴222()()SR b c b c =+--

∴SR =。假设存在点M ．且MS ＝x ，别MR

＝x 。若使△PSM ∽△MRQ ，

则有b x =

20x bc -+=

∴12x x = ∴SR ＝

∴M 为SR 的中点. 若使△PSM ∽△QRM ，

则有b x =

。∴x =。

∴1MR c QB RO MS b BP OS ==-===。 ∴M 点即为原点O 。

综上所述，当点M 为SR 的中点时．?PSM ∽ΔMRQ ；当点M 为原点时，?PSM ∽?MRQ ．

方法二：若以P 、S 、M 为顶点的三角形与以Q 、M 、R 为顶点三角形相似，

∵90PSM MRQ ∠=∠=?，

∴有?PSM ∽?MRQ 和?PSM ∽△QRM 两种情况。

当?PSM ∽?MRQ 时．∠SPM ＝∠RMQ ，∠SMP ＝∠RQM ．

由直角三角形两锐角互余性质．知∠PMS+∠QMR ＝90°。∴90PMQ ∠=?。

取PQ 中点为N ．连结MN ．则MN ＝12PQ=1()2

QR PS +．

∴MN 为直角梯形SRQP 的中位线,

∴点M 为SR 的中点 当△PSM ∽△QRM 时，

RM QR QB MS PS BP ==。又RM RO MS OS

=，即M 点与O 点重合。∴点M 为原点O 。 综上所述，当点M 为SR 的中点时，?PSM ∽△MRQ ；当点M 为原点时，?PSM ∽△QRM 。

点拨：通过对图形的观察可以看出C 、F 是一对关于y 轴的对称点，所以（1）的关键是求出其中一个点的坐标

就可以应用三点式或 y=ax 2+c 型即可．而对于点 P 既然在抛物线上，所以就可以得到它的坐标为（a ，14

a 2+1）．这样再过点B 作BN ⊥PS ．得出的几何图形求出PB 、PS 的大小．最后一问的关键是要找出△PSM 与△MRQ 相似的条件．

【例2】探究规律：如图2－6－4所示，已知：直线m ∥n ，A 、B 为直线n 上两点，C 、P 为直线m 上两点．

（1）请写出图2－6－4中，面积相等的各对三角形；

（2）如果A 、B 、C 为三个定点，点P 在m 上移动，那么，无论P 点移动到任何位置，总有________与△ABC 的面积相等．理由是：_________________.

解决问题：如图 2－6－5所示，五边形 ABCDE 是张大爷十年前承包的一块土地的示意图，经过多年开垦荒地，现已变成如图2－6－6所示的形状，但承包土地与开垦荒地的分界小路（2－6－6中折线CDE ）还保留着；张大爷想过E 点修一条直路，直路修好后，要保持直路左边的土地面积与承包时的一样多，右边的土地面积与开垦的荒地面积一样多．请你用有关的几何知识，按张大爷的要求设计出修路方案（不计分界小路与直路的占地面积）．

（1）写出设计方案．并画出相应的图形；

（2）说明方案设计理由．

解：探究规律：（l ）△ABC 和△ABP ，△AOC 和△ BOP 、△CPA 和△CPB ．

（2）△ABP ；因为平行线间的距离相等，所以无论点P 在m 上移动到任何位置，总有△ABP 与△ABC 同底等高，因此，它们的面积总相等．

解决问题：⑴画法如图2－6－7所示．

连接EC ，过点D 作DF ∥EC ，交CM 于点F ，连接EF ，EF 即为所求直路位置．

⑵设EF 交CD 于点H ，由上面得到的结论可知：

S ΔECF =S ΔECD ，S ΔHCF =S ΔEDH ，所以S 五边形ABCDE =S 五边形ABCFE ，S 五边形EDCMN =S 四边形EFMN ．

点拨：本题是探索规律题，因此在做题时要从前边问题中总结出规律，后边的问题要用前边的结论去一做，所以要连接EC ，过D 作DF ∥EC ，再运用同底等高的三角形的面积相等．

【例3】如图2－6－8所示，已知抛物线的顶点为M （2，－4），且过点A(－1,5)，连结AM 交x 轴于点B ．

⑴求这条抛物线的解析式；

⑵求点 B 的坐标；

⑶设点P （x ，y ）是抛物线在x 轴下方、顶点 M 左方一段上的动点，连结 PO ，以P

为顶点、PQ 为腰的等腰三角形的另一顶点Q 在x 轴上，过Q 作x 轴的垂线交直线AM 于

点R ，连结PR ．设面 PQR 的面积为S ．求S 与x 之间的函数解析式；

⑷在上述动点P （x ，y ）中，是否存在使S ΔPQR =2的点？若存在，求点P 的坐标；若不

存在，说明理由．

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