2014年中考数学压轴题精编--广东篇(试题及答案)

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2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

1 2014年中考数学压轴题精编—广东篇

1.广东省(中山市、汕头市、东莞市等)如图,已知P 是线段AB 上的任意一点(不含端点A ,B ),分别以AP 、BP 为斜边在AB 的同侧作等腰直角△APD 和△BPE ,连接AE 交PD 于点M ,连接BD 交PE 于点N .

(1)求证:①MN ∥AB ;②MN 1=AP

1+BP 1; (2)若AB =4,当点P 在AB 上运动时,求MN

的取值范围.

1.解:(1)①证明:∵△APD 和△BPE 都是等腰直角三角形,∴∠DAP =∠EPB =45°

∴AD ∥PE ,∴∠DAM =∠PEM ,∠ADM =∠EPM

∴△DAM ∽△PEM ,∴AD : PE =AM : ME 同理可得PD : BE =PN :

NE , ∵AD =PD ,BE =PE ,∴AM : ME =PN :

NE ∴MN ∥AP ,即MN ∥AB ·············································································· 3分

②证明:∵MN ∥AB ,∴∠PMN =∠ACP =45°,∠PNM =∠BPE =45°

∴∠PMN =∠PNM =45°,∴△PMN 是等腰直角三角形

∴PM =2

2MN ∵∠APM =∠ABE =45°,∠P AM =∠BAE (公共角)

∴△APM ∽△ABE ,∴PM : BE =AP : AB =AP :(

AP +BP ) ∴22MN : 2

2BP =AP :(

AP +BP )[来源:Z_xx_d5f7dfa4240c844768eaee0f] 整理得:MN 1=AP 1+BP

1 ·········································································· 6分 (2)解:∵MN 1=AP 1+BP

1 ∴MN =BP AP BP AP + · =AB BP AP · =AB AP AB AP )(- =AB 1[-(AP -21AB )2+41AB 2] ∴MN ≤41AB (当AP =21AB 时,MN 取得最大值为4

1AB ) ∵AB =4,∴MN ≤1,又∵P 不与A ,B 重合,∴AP <AB ,∴MN >0

∴0<MN

≤1 ································································································· 9分

2.广东省(中山市、汕头市、东莞市等)如图(1),(2)所示,矩形ABCD 的边长AB =6,BC =4,点F 在DC 上,DF =2.动点M 、N 分别从点D 、B 同时出发,沿射线DA 、线段BA 向点A 的方向运动(点M 可运动到DA 的延长线上),当动点N 运动到点A 时,M 、N 两点同时停止运动.连接FM 、MN 、FN ,当F 、N 、M 不在同一直线时,可得△FMN ,过△FMN 三边的中点作△PQW .设动点M 、N 的速度都是1个单位/秒,M 、N 运动的时间为x 秒.试解答下列问题:

A P D

B N M E A P D B N M

E

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

2

(1)说明△FMN ∽△QWP ;

(2)设0≤x ≤4(即M 从D 到A 运动的时间段).试问x 为何值时,△PQW 为直角三角形?当x 在何范围时,△PQW 不为直角三角形?

(3)问当x 为何值时,线段MN 最短?求此时MN 的值.

2.解:(1)由题意知:PW ∥MN ,PQ ∥FN ,∴∠MNF =∠PWF ,∠PWF =∠WPQ

∴MNF =∠WPQ

同理可得:∠NFM =∠PQW 或∠NMF =∠PWQ

∴△FMN ∽△QWP ························································································· 2分 (2)∵△FMN ∽△QWP ,∴当△FMN 为直角三角形时,△PQW 也为直角三角形

当0≤x ≤4时,若∠NMF =90°,则有MF 2+MN 2=FN 2

即(

x

2+2 2

)+(

4-x

)2+(

6-x

)2=(

4-x

)2+4

2

,此方程无实数解

若∠MNF =90°,则有MN 2+FN 2=MF 2

即(

4-x

)2+(

6-x

)2+(

4-x

)2+4

2=

x

2+2 2

解得x =4或x =10(舍去) ···················································· 3分 若∠NFM =90°,则有MF 2+FN 2=MN 2

即(

x

2+2 2

)+(

4-x

)2+4

2=(

4-x

)2+(

6-x

)2

解得x =3

4

·············································································· 4分

所以,当x =34

或x =4时,△PQW 为直角三角形 ··············· 5分

当0≤x <34,3

4

<x <4时,△PQW 不为直角三角形 ········ 6分

(3)在0≤x ≤4时间段,当x =4时,点M 与点A 重合,线段MN 最短,此时MN =2

·········································································································· 7分

当4<x ≤6时,如图(2)

∵MN 2=AM 2+AN 2=(

x -4

)2+(

6-x

)

2

=2x

2

-20x +52

=2(

x -5

)2

+2 ··················································································· 8分

∴当x =5时,MN 2

有最小值2,即线段MN 最短,此时MN =2

图(1)

图(2)

图(1)

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

3 综上所述,当x =5时,线段MN 最短,此时MN =2 ···························· 9分

3.(广东省广州市)如图,⊙O 的半径为1,点P 是⊙O 上一点,弦AB 垂直平分线段OP ,点D 是弧 ⌒APB

上的任一点(与端点A 、B 不重合),DE ⊥AB 于点E ,以D 为圆心、DE 长为半径作⊙D ,分别过点A 、B 作⊙D 的切线,两条切线相交于点C .

(1)求弦AB 的长; (2)判断∠ACB 是否为定值,若是,求出∠ACB 的大小;否则,请说明理由;

(3)记△ABC 的面积为S ,若2DE S

=34,求△ABC 的周长.

3.解:(1)如图,连结OA 、OB ,设OP 与AB 的交点为F ,则有OA =1

∵弦AB 垂直平分线段OP ,∴OF =21

OP =21

,AF =BF

在Rt △AOF 中,∵AF =2

2OF OA -=22211)( -=23

∴AB =2AF =3 ·························································································· 4分

(2)∠ACB 是定值 ······························································································· 5分

理由:由(1)易知∠AOB =120°

连结AD 、BD ,∵点D 为△ABC 的内心,∴∠CAB =2∠DAB ,∠CBA =2∠DBA

∵∠DAB +∠DBA =21

∠AOB =60°,∴∠CAB +∠CBA =120°

∴∠ACB =60°(定值) ················································································· 8分

(3)记△ABC 的周长为l ,设AC 、BC 与⊙D 的切点分别为G ,H ,连结DG ,DC ,DH

则有DG =DH =DE ,DG ⊥AC ,DH ⊥BC ∴S =S △ABD +S △ACD +S △BCD

=21

AB ·DE +21

AC ·DG +21

BC ·DH =21

(AB +BC +AC )·DE =21

l ·DE ∵2DE S

=34,∴2·21DE DE l =34,∴l =38DE

∵CG ,CH 是⊙D 的切线,∴∠GCD =21

∠ACB =30°

C

P D

O

B A E C

P D O

B

A E F G

H

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

4

∴在Rt △CGD 中,CG =

30tan DG =3

3

DE =3DE ,∴CH =CG =3DE 又由切线长定理可知AG =AE ,BH =BE

∴l =AB +BC +AC =32+32DE =

,解得DE =31

················ 13分

∴△ABC 的周长为

3

3

8 ············································································· 14分 4.(广东省广州市)如图,四边形OABC 是矩形,点A 、C 的坐标分别为(3,0),(0,1),点D 是线段BC 上的动点(与端点B 、C 不重合),过点D 作直线y =-

2

1

x +b 交折线OAB 于点E . (1)记△ODE 的面积为S ,求S 与b 的函数关系式;

(2)当点E 在线段OA 上时,若矩形OABC 关于直线DE 的对称图形为四边形O 1A 1B 1C 1,试探究四边形O 1A 1B 1C 1与矩形OABC 重叠部分的面积是否发生变化,若不变,求出该重叠部分的面积;若改变,请说明理由.

4.解:(1)由题意得点B 的坐标为(3,1)

若直线经过点A (3,0)时,则b =23

若直线经过点B (3,1)时,则b =

2

5 若直线经过点C (0,1)时,则b =1

①若直线与折线OAB 的交点在OA 上,即1<b ≤2

3

时,如图1 此时E (2b ,0) ∴S

21OE ·CO =2

1

×2b ×1=b ································································· 4分 ②若直线与折线OAB 的交点在BA 上,即23<b <2

5

时,如图2 此时E (3,b -

2

3

),D (2b -2,1) ∴S

=S 矩形OABC

-(S △COD +S △AOE +S △BDE )

=3-[21(2b -2)×1+21×3(b -23)+21

(5-2b )(25-b )]

2

5

b -b

2

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

5

∴S

=?????b (1<b ≤2

3

)25b -b

2

(1<b ≤23)

····································································· 8分

(2)如图3,设O 1A 1与CB 相交于点M ,OA 与C 1B 1相交于点N ,则矩形O 1A 1B 1C 1与矩形OABC

的重叠部分的面积即为四边形DNEM 的面积

由题意知,DM ∥NE ,DN ∥ME ,∴四边形DNEM 为平行四边形 根据轴对称知,∠MED =∠NED 又∠MDE =∠NED ,∴∠MED =∠MDE

∴MD =ME ,∴平行四边形DNEM 为菱形 ······················ 10分 过点D 作DH ⊥OA ,垂足为H 由题易知,tan ∠DEN =

2

1

,DH =1,∴HE =2 设菱形DNEM 的边长为a ,则在Rt △DHN 中,由勾股定理得: a

2=(2-a

)2+1

2

,∴a =

4

5 ········································································ 12分 ∴S 四边形DNEM

=NE ·DH =

45×1=4

5 ·························································· 13分 故矩形O 1A 1B 1C 1与矩形OABC 重叠部分的面积不发生变化,面积始终为

4

5

····································································································· 14分

5.(广东省深圳市)如图是一圆形纸片,AB 是直径,BC 是弦,将纸片沿弦BC 折叠后,劣弧BC 与AB 交

于点D ,得到BDC ︵

. (1)若BD ︵=CD ︵,求证:BDC ︵

必经过圆心O ;

(2)若AB =8,BD ︵=2CD ︵

,求BC 的长.

5.解:(1)过C 作CE ⊥AB 于E ,连接CA 、CD

∵∠CDA =∠BCD +∠CBD =21BDC ︵=2

1BmC ︵

=∠CAD

∴AC =CD ∵BD ︵=CD ︵

,∴BD =CD ,∴BCD =∠CBD

∴∠CDA =2∠CBD ,∴∠CAD =2∠CBD ∵AB 是直径,∴∠ACB =90°

∴∠CAD +∠CBD =90°,∴2∠CBD +∠CBD =90°

图3

1

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

6

∴∠CBD =30°,∴∠CDA =60°

∴△CAD 是等边三角形,∴AD =CD

∴AD =BD ,∴BDC ︵

必经过圆心O ····························································· 4分

(2)∵AC =CD ,∴AC ︵=CD ︵

∵BD ︵=2CD ︵,∴AC ︵=21BD ︵=31BDC ︵=31BmC ︵=41

半圆

连接CO ,则∠OPC =4

1

×180°=45°

∴CE =OE =

22OC =2

2×4=22 ∴BE =OE +OB =22+4

∴BC =2222422 )()(

++=224+ ·

··············································· 7分

6.(广东省深圳市)如图,抛物线y =ax

2

+c (a >0)经过梯形ABCD 的四个顶点,梯形的底AD 在x 轴上,其中A (-2,0),B (-1,-3). (1)求抛物线的解析式;

(2)点M 为y 轴上任意一点,当点M 到A 、B 两点的距离之和为最小时,求此时点M 的坐标;

(3)在第(2)问的结论下,抛物线上的点P 使S △PAD =4S △ABM 成立,求点

P 的坐标.

6.解:(1)∵抛物线y =ax

2

+c 经过A (-2,0),B (-1,-3)

∴?????4a +c =0a +c =-3 解之得:?

????a =1

c =-4 ∴抛物线的解析式为y =x

2

-4 ····································· 3分

(2)如图1,连接BD ,交y 轴于点M ,则点M 就是所求作的点

设BD 的解析式为y =kx +b

则?????2k +b =0-k +b =-3 ∴?

????k =1b =-2 ∴BD 的解析式为y =x -2,令x =0,则y =-2

∴M (0,-2) ······························································································· 5分 (3)如图2,连接AM ,设BC 与y 轴的交点为N ,则N (0,-3)

由(2)知,OM =OA =OD =2,∠AMB =90°

图1

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

∴BN =MN =1,AM =22,BM =2 ∴S △ABM =

2

1

AM ·BM =22×2=2 设P (x ,x

2

-4),依题意有:2

1

AD ·|

x

2-4|=4×2 即

2

1

×4·|

x

2-4|=4×2,整理得:|

x

2-4|=4 解之得:x 1=22,x 2=22-,x 3=0

故符合条件的P 点有三个:P 1(22,4),P 2(22-,4),P 3(0,-4)

······································································································· 9分

7.(广东省深圳市)如图1,以点M (-1,0)为圆心的圆与y 轴、x 轴分别交于点A 、B 、C 、D ,直线y =-

33x -3

3

5与⊙M 相切于点H ,交x 轴于点E ,交y 轴于点F . (1)请直接写出OE 、⊙M 的半径r 、CH 的长;

(2)如图2,弦HQ 交x 轴于点P ,且DP : PH =3 :

2,求cos ∠QHC 的值;

(3)如图3,点K 为线段EC 上一动点(不与E 、C 重合),连接BK 交⊙M 于点T ,弦AT 交x 轴于点N .是否存在一个常数a ,始终满足MN ·MK =a ,如果存在,请求出a 的值;如果不存在,请说明理由.

7.解:(1)OE =5,r =2,CH =2 ························································· 3(2)如图2,连接QC 、QD ,则∠CQD =90°

∠PHC =∠PDQ ,∠CPH =∠QPD ∴△CPH ∽△QPD ,∴

CH QD =

PH

DP

即2

QD =23

,∴QD =3

∵CD =4,∴cos ∠QHC =cos ∠QDC =

4

3

······················· 6(3)如图3,连接AK 、AM ,延长AM ,与圆交于点G ,连接TG ,则∠GTA =90°

∴∠1+∠2=90°

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

8 ∵∠3=∠2,∴∠1+∠3=90°

∵∠4+∠3=90°,∠5=∠4,∴∠1=∠5

又∠NMA =∠AMK ,∴△NMA ∽△AMK ∴AM MN =MK

AM ,即MN ·MK =AM 2=4 故存在常数a ,始终满足MN ·MK =a ,常数a =4

············································································· 9分

8.(广东省珠海市)如图,△ABC 内接于⊙O ,AB =6,AC =4,D 是AB 边上一点,P 是优弧BAC 的中点,连结P A 、PB 、PC 、PD . (1)当BD 的长度为多少时,△P AD 是以AD 为底边的等腰三角形?并证明; (2)若cos ∠PCB =55,求P A 的长.

8.解:(1)当BD =AC =4时,△P AD 是以AD 为底边的等腰三角形 ∵P 是优弧BAC 的中点,∴PB ︵=PC ︵

∴PB =PC

∵BD =AC =4,∠PBD =∠PCA

∴△PBD ≌△PCA

∴P A =PD ,即△P AD 是以AD 为底边的等腰三角形

·········································································································· 4分

(2)由(1)可知,当BD =4时,PD =P A ,AD =AB -BD =6-4=2

过点P 作PE ⊥AD 于E ,则AE =

2

1AD =1 ∵∠PCB =∠P AD

∴cos ∠P AD =cos ∠PCB =PA AE =55 ∴P A =5 ······································································································ 9分

9.(广东省珠海市)如图,平面直角坐标系中有一矩形ABCO (O 为原点),点A 、C 分别在x 轴、y 轴上,且C 点坐标为(0,6).将BCD 沿BD 折叠(D 点在OC 边上),使C 点落在OA 边的E 点上,并将BAE 沿BE 折叠,恰好使点A 落在BD 的点F 上.

(1)直接写出∠ABE 、∠CBD 的度数,并求折痕BD 所在直线的函数解析式;

(2)过F 点作FG ⊥x 轴,垂足为G ,FG 的中点为H ,若抛物线y =ax

2+bx +c 经过B 、H 、D 三点,求抛

物线的函数解析式;

(3)若点P 是矩形内部的点,且点P 在(2)中的抛物线上运动(不含B 、D 点),过点P 作PN ⊥BC 分别交BC 和BD 于点N 、M ,设h =PM -MN ,试求出h 与P 点横坐标x 的函数解析式,并画出该函数的简图,

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

9 分别写出使PM <MN 、PM =MN 、PM >MN 成立的x 的取值范围.

9.解:(1)∠ABE =∠CBD =30° ······················································································· 1分

在△ABE 中,AB =6

BC =BE = 30cos AB =34

CD =BC ·tan30°=34×33

=4

∴OD =OC -CD =6-4=2 ∴B (34,6),D (0,2) 设BD 所在直线的函数解析式为y =kx +b 则???34k +b =6b =2 ∴?????k =3

3

b =2

∴BD 所在直线的函数解析式为y =33

x +2 ·············································· 3分

(2)∵EF =EA =AB ·tan30°=32,∠FEG =180°-∠FEB -∠AEB =60°

又∵FG ⊥OA

∴FG =EF ·sin60°=3,GE =EF ·cos60°=3,OG =OA -AE -GE =3

又H 为FG 中点

∴H (3,23

) ···························································································· 4分

∵抛物线y =ax 2

+bx +c 经过B (34,6)、H (3,23)、D (0,2)三点

∴?????48a +34+c =63a +3b +c =23c =2 ∴?????a =6

1

b =-

3

3c =2

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

10 ∴抛物线的解析式为y =6

1x 2-33x +2 ····················································· 5分 (3)∵PM =(33x +2)-(61x 2-33x +2)=-6

1x

2+332x MN =6-(33x +2)=4-33x h =PM -MN =(-61x 2+332)-(4-33x )=-6

1x

2+3x -4 ················ 6分 由-61x 2+3x -4得x 1=32,x 2=34 该函数的简图如图所示: ········································7分

当0<x <32时,h <0,即PM <MN

当x =32时,h =0,即PM =MN 当32<x <34时,h >0,即PM >MN ·············9分

10.(广东省佛山市)一般来说,依据数学研究对象本质属性的相同点和差异点,将数学对象分为不同种类的数学思想叫做“分类”的思想;将事物进行分类,然后对划分的每一类分别进行研究和求解的方法叫做“分类讨论”的方法.请依据分类的思想和分类讨论的方法解决下列问题:

如图,在△ABC 中,∠ACB >∠ABC .

(1)若∠BAC 是锐角,请探索在直线AB 上有多少个点D ,能保证△ACD ∽△ABC (不包括全等)?

(2)请对∠BAC 进行恰当的分类,直接写出每一类在直线AB 上能保证△ACD ∽△ABC (不包括全等)的

点D 的个数.

11.解:(1)(ⅰ)如图①,若点D 在线段AB 上

由于∠ACB >∠ABC ,可以作一个点D 满足∠ACD =∠ABC

使得△ACD ∽△ABC ····················································································· 2分

(ⅱ)如图②,若点D 在线段AB 的延长线上

则∠ACD >∠ACB >∠ABC ,与条件矛盾

因此,这样的点D 不存在 ············································································ 4分

(ⅲ)如图③,若点D 在线段AB 的反向延长线上

由于∠BAC 是锐角,则∠BAC <90°<∠CAD

不可能有△ACD ∽△ABC

因此,这样的点D 不存在 ············································································ 6分

A B C

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

11 综上所述,这样的点D 有一个 ···································································· 8分

注:(ⅲ)中用“∠CAD 是钝角,△ABC 中只可能∠ACB 是钝角,而∠CAD >∠ACB ”说

明不存在点D 亦可

(2)若∠BAC 为锐角,由(1)知,这样的点D 有一个 ····································· 9分

若∠BAC 为直角,这样的点D 有两个 ······················································ 10分

若∠BAC 为钝角,这样的点D 有一个 ······················································ 11分

注:(2)的第一个解答不写不扣分,第二个解答回答“这样的点D 有一个”给1分

12.(广东省茂名市)如图,在直角坐标系xO y 中,正方形OCBA 的顶点A 、C 分别在y 轴、x 轴上,点B

坐标为(6,6),抛物线y =ax 2+bx +c 经过A 、B 两点,且3a -b =-1.

(1)求a ,b ,c 的值;

(2)如果动点E 、F 同时分别从点A 、点B 出发,分别沿A →B 、B →C 运动,速度都是每秒1个单位长度,

当点E 到达终点B 时,点E 、F 随之停止运动.设运动时间为t 秒,△EBF 的面积为S .

①试求出S 与t 之间的函数关系式,并求出S 的最大值;

②当S 取得最大值时,在抛物线上是否存在点R ,使得以点E 、B 、R 、F 为顶点的四边形是平行四边形?如果存在,求出点R 的坐标;如果不存在,请说明理由.

12.解:(1)由已知A (0,6)、B (6,6)在抛物线上

得方程组:?????c =636a +6b +c =63a -b =-1

········································································· 1分

A B C D 图① A B C D 图② A B C D 图②

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

解得:???a =-

9

1b =32c =6

·························································································· 3分

(2)①运动开始t 秒时,EB =6-t ,BF =t

S =21EB ·BF =21(6-t )t =-2

1t

2

+3t ··············· 4分

∵S =-21t

2+3t =-21(t -3)2+2

9

∴当t =3时,S 有最大值2

9

······························ 5分

②当S 取得最大值时,由①知t =3,所以BF =3,CF =3,EB =6-3=3 若存在某点R ,使得以E 、B 、R 、F 为顶点的四边形是平行四边形, 则FR 1=EB 且FR 1∥EB ,即可得R 1为(9,3)、(3,3) ·························· 6分

或者ER 2=BF 且ER 2∥BF ,可得R 2为(3,9) ·········································· 7分

再将所求得的三个点代入y =-91x

2+3

2

x +6,可知只有点(9,3)在抛物线上,因此抛物

线上存在点R 1(9,3),使得四边形EBRF 为平行四边形 ························· 8分

13.(广东省茂名市)已知⊙O 1的半径为R ,周长为C .

(1)在⊙O 1内任意作三条弦,其长分别是l 1、l 2、l 3.求证:l 1+l 2+l 3<C ; (2)如图,在直角坐标系xO y 中,设⊙O 1的圆心O 1(R ,R ). ①当直线l :y =x +b (b >0)与⊙O 1相切时,求b 的值; ②当反比例函数y =x

k

(k >0)的图象与⊙O 1有两个交点时,求k 的取值范围.

13.(1)证明:∵l 1≤2R ,l 1≤2R ,l 1≤2R

∴l 1+l 2+l 3≤3×2R <π×2R =C ······························································· 2分 ∴l 1+l 2+l 3<C ·························································································· 3分

(2)解:①如图,根据题意可知⊙O 1与与x 轴、y 轴分别相切

设直线l 与⊙O 1相切于点M ,则O 1M ⊥l ,过点O 1作直线NH ⊥x 轴,与l 交于点N ,与x 轴交于点H

(备用图)

(备用图)

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

13

∵直线l 与x 轴、y 轴分别交于点E (-b ,0)、F (0,b ) ∴OE =OF =b ,∴∠NEO =45°,∴∠ENO 1=45° 在Rt △O 1MN 中,O 1N =O 1M ÷sin45°=2R

∴点N 的坐标为N (R ,2R +R ) ···························· 4分 把点N 坐标代入y =kx +b 得:2R +R =R +b ,解得:b =2R

··························································································②如图,设经过点O 、O 1的直线交⊙O 1于点A 、D

则由已知,直线OO 1:y =x 圆与反比例函数图象的对称轴

当反比例函数y =x k

的图象与⊙O 1直径AD 相交时(点A 、D 则反比例函数y =x k

的图象与⊙O 1有两个交点

过点A 作AB ⊥x 轴交x 轴于点B ,过O 1作O 1C ⊥x 轴于点C 则OO 1=O 1C ÷sin45o

=2R ,OA =2R +R ∴OB =AB =OA ·sin45o =(2R +R )·22=R +2

2R ∴A (R +

22R ,R +22R ),将点A 的坐标代入y =x

k

解得:k =(2

3

+2)R 2 ············································································ 6分

同理可求得点D 的坐标为D (R -

22R ,R -2

2R ) 将点D 的坐标代入y =x k ,解得:k =(23

-2)R 2 ······························· 7分

∴当反比例函数y =x

k

(k >0)的图象与⊙O 1有两个交点时,k 的取值范围是:

(23-2)R 2< k

<(2

3

+2)R 2 ····························································· 8分

14.(广东省湛江市)如图,在△ABC 中,以AB 为直径的⊙O 交BC 于点P ,PD ⊥AC 于点D ,且PD 与⊙

O 相切.

(1)求证:AB =AC ;

(2)若BC =6,AB =4,求CD 的值.

14.(1)证明:连接OP

∵PD

与⊙O 相切,∴OP ⊥PD ∵AC ⊥PD ,∴OP ∥AC

B B

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

14 ∵OP =OA =OB =21AB ,∴OP =2

1AC ∴AB =AC ···························································6分

(2)解:连接AP

∵AB =AC ,P 为BC 的中点,∴AP ⊥BC

在Rt △CDP 与Rt △CP A 中,∠C =∠C

∴Rt △CDP ∽Rt △CP A ,∴PC CD =AC

PC ∵BC =6,AB =4,∴PC =3,AC =4,∴3CD =43 ∴CD =4

9 ································································································ 12分

15.(广东省湛江市)如图,在平面直角坐标系中,点B 的坐标为(-3,-4),线段OB 绕原点逆时针旋转

后与x 轴的正半轴重合,点B 的对应点为点A .

(1)直接写出点A 的坐标,并求出经过A 、O 、B 三点的抛物线的解析式;

(2)在抛物线的对称轴上是否存在点C ,使BC +OC 的值最小?若存在,求出点C 的坐标;若不存在,

请说明理由;

(3)如果点P 是抛物线上的一个动点,且在x 轴的上方,当点P 运动到什么位置时,△P AB 的面积最

大?求出此时点P 的坐标和△P AB 的最大面积.

15.解:(1)A (5,0) ······································································································· 1分

由抛物线经过点O ,可设抛物线的解析式为y =ax 2

+bx ··························· 2分 把A (5,0)、B (-3,-4)代入y =ax 2+bx ,得:

?????25a +5b =09a -3b =-4 解得?????a =-61b =6

5 ·································································· 4分 ∴抛物线的解析式为y =-61x 2+6

5x ·························································· 5分 (2)如图①,∵y =-61x

2+65x =-61( x -25)2+24

25 ∴抛物线的对称轴是直线x =25,点O 、A 关于直线x =25对称

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

15 连接AB 交直线x =2

5于点C ,则点C 使BC +OC 的值最小 ···················· 6分 设直线AB 的解析式为y =kx +b

则?????5k +b =0-3k +b =-4 解得?????k =21b =-2

5 ∴直线AB 的解析式为y =21x -2

5 ························8分 把x =25代入y =21x -25,得y =-4

5 ∴点C 的坐标为(25,-45) ································9分 (3)如图②,过点P 作y 轴的平行线交AB 于点D ,设点P 的横坐标为x ,

则P (x ,-61x 2+65x ),D (x ,21x -2

5) ··············································· 10分 ∴S △P AB =S △P AD +S △PBD =2

1PD ·(x A -x B ) =2

1(y P -y D )(x A -x B ) =21[(-61x 2+65x )-(21x -25)]×[5-(-3)] =-32x 2+34x +10=-32(x -1 )2+332 ∴当x =1时,S △P AB 的最大值为332 ···················· 12分 把x =1代入y =-61x 2+65x ,得y =3

2 ∴此时点P 的坐标为(1,32) ·································································· 13分

16.(广东省肇庆市)已知二次函数y =x

2+bx +c +1的图象过点P (2,1).

(1)求证:c =-2b -4; (2)求bc 的最大值;

(3)若二次函数的图象与x 轴交于点A (x 1,0)、B (x 2,0),△ABP 的面积是

43,求b 的值.

16.(1)证明:将点P (2,1)代入y =x 2+bx +c +1得:1=2 2

+2b +c +1 ·················· 1分 整理得:c =-2b -4 ·················································································· 2分

(2)解:∵c =-2b -4,∴bc =b (-2b -4)=-2(b +1)2+2 ································· 4分

∵-2<0,∴当b =-1时,bc 有最大值2 ·············································· 5分

(3)解:由题意得:21AB ×1=4

3

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

16 ∴AB =| x 2-x 1|=23,即|

x 2-x 1|2=4

9 ··················································· 6分 亦即(x 1+ x 2)-4x 1x 2=49 ·········································································· 7分 由根与系数的关系得:x 1+

x 2=-b ,x 1x 2=c +1=-2b -4+1=-2b -3 ··································································································· 8分

代入(x 1+ x 2)-4x 1x 2=49得:(-b )2-4(-2b -3)=4

9 整理得:b 2+8b +4

39=0 ········································································· 9分 解得:b 1=-23,b

2=-213,经检验均合题意 ···································· 10分

17.(广东省清远市)在⊙O 中,点P 在直径AB 上运动,但与A 、B 两点不重合,过点P 作弦CE ⊥AB ,在AB ︵上任取一点D ,直线CD 与直线AB 交于点F ,弦DE 交直线AB 于点M ,连接CM .

(1)如图1,当点P 运动到与O 点重合时,求∠FDM 的度数;

(2)如图2、图3,当点P 运动到与O 点不重合时,求证:FM ·OB =DF ·MC .

17.(1)解:当点P 与点O 重合时(如图1)

∵CE 是直径,∴∠CDE =90° ·················································· 1分

∵∠CDE +∠FDM =180°,∴∠FDM =90° ···························· 2分

(2)证明:当点P 在OA 上运动时(如图2)

∵OP ⊥CE ,∴AC ︵=AE ︵=2

1CE ︵,CP =EP ∴CM =EM ,∴∠CMP =∠EMP

∵∠DMO =∠EMP ,∴∠CMP =∠DMO

∴∠CMP +∠DMC =∠DMO +∠DMC

∴∠DMF =∠CMO ································································ 3分

∵∠D 所对的弧是CE ︵,∠COM 所对的弧是AC ︵

图1 B O (P ) F D C E M 图2 A B O F D C E M P 图3 B O F D C E M P 图1 A B O (P ) F D C E M

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

17

∴∠D =∠COM ····································································· 4分

∴△DFM ∽△OCM ,∴OC DF =

MC FM

∴FM ·OC =DF ·MC

∵OB =OC ,∴FM ·OB =DF ·MC ······································· 5分 当点P 在OB 上运动时(如图3)

证法一:连结AC ,AE

∵OP ⊥CE ,∴BC ︵=BE ︵=21CE ︵

,CP =EP

∴CM =EM ,∴∠CMO =∠EMO

∵∠DMF =∠EMO ,∴∠DMF =∠CMO ································· 6分

∵∠CDE 所对的弧是CAE ︵,∠CAE 所对的弧是CE ︵

∴∠CDE +∠CAE =180°

∵∠CDE +∠FDM =180°,∴∠FDM =∠CAE

∵∠CAE 所对的弧是CE ︵,∠COM 所对的弧是BC ︵

∴∠CAE =∠COM

∴∠FDM =∠COM ·································································· 7分

∴△DFM ∽△OCM ,∴OC DF =

MC FM

∴FM ·OC =DF ·MC

∵OB =OC ,∴FM ·OB =DF ·MC ······································· 8分

证法二:∵OP ⊥CE ,∴BC ︵=BE ︵=21CE ︵,AC ︵=AE ︵=2

1CAE ︵

,CP =EP

∴CM =EM ,∴∠CMO =∠EMO

∵∠DMF =∠EMO ,∴∠DMF =∠CMO ···················································· 6分 ∵∠CDE 所对的弧是CAE ︵

∴∠CDE =CAE ︵

度数的一半=AC ︵

的度数=180°-BC ︵

的度数

∴∠FDM =180°-∠CDE =180°-(180°-BC ︵)=BC ︵

的度数

∵∠COM =BC ︵

的度数

∴∠FDM =∠COM ····················································································· 7分

∴△DFM ∽△OCM ,∴OC DF =

MC FM

∴FM ·OC =DF ·MC

∵OB =OC ,∴FM ·OB =DF ·MC ·························································· 8分

18.(广东省河源市、梅州市)如图,△ABC 中,点P 是边AC 上的一个动点,过P 作直线MN ∥BC ,设

图2

A

B

O

F

D

C

E

M P 图3

B O

F D C

E

M P

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

18 MN 交∠BCA 的平分线于点E ,交∠BCA 的外角平分线于点F .

(1)求证:PE =PF ;

(2)当点P 在边AC 上运动时,四边形BCFE 可能是菱形吗?说明理由;

(3)若在AC 边上存在点P ,使四边形AECF 是正方形,且

BC AP =23,求此时∠A 的大小.

18.(1)证明:∵CE 平分∠BCA ,∴∠BCE =∠PCE 又MN ∥BC ,∴∠BCE =∠PEC

∴∠PCE =∠PEC ,∴PE =PC ····················· 2分

同理可证PF =PC

∴PE =PF ····················································· 3分

(2)解:不可能 ·························································· 4分

理由如下: 方法1:∵由(1)可知,PE =PF =PC ,又PC +PF >CF

∴PE +PF >CF

即EF >CF ································································································· 5分

又菱形的四条边都相等,所以四边形BCFE 不可能是菱形 ··················· 6分

方法2:若四边形BCFE 为菱形,则BF ⊥CE

由(1)可知CF ⊥CE ··············································································· 5分

因为在平面内过同一点F 不可能有两条直线同垂直于一条直线,所以BF ⊥CE 不能成立,

所以四边形BCFE 不可能是菱形 ····························································· 6分

(3)解:若四边形AECF 是正方形,则AP =CP ,∠ACE =2

1∠ECF =45° ∵∠BCE =∠PCE ,∴∠BCA =90° ···························································· 7分 又∵BC AP =23,∴BC

AC =3 即tan ∠B =3 ·························································································· 8分

∴∠B =60°,∴∠A =30° ·········································································· 9分

19.(广东省河源市、梅州市)如图,直角梯形OABC 中,OC ∥AB ,C (0,3),B (4,1),以BC 为直径的圆交x 轴于E ,D 两点(D 点在E 点右方).

A B N F D

C E M P A B N

F D C E M P

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

19 (1)求点E 、D 的坐标;

(2)求过B 、C 、D 三点的抛物线的函数关系式;

(3)过B 、C 、D 三点的抛物线上是否存在点Q ,使△BDQ 是以BD 为直角边的直角三角形?若不存在,说明理由;若存在,求出点Q 的坐标.

19.解:(1)方法1:∵B (4,1),则A (4,0),设OD =x ,则DA =4-x

∵D 是以BC 为直径的圆与x 轴的交点,∴∠CDB =90°

∴∠ODC +∠BDA =90°

∵∠OCD +∠ODC =90°,∴∠OCD =∠BDA

∴Rt △OCD ∽Rt △ADB ∴OD OC =AB AD ,即x 3=14x

- ······································································ 1分

解得x 1=1,x 2=3

∴D (3,0),E (1,0) ··············································································· 2分

方法2:设BC 的中点为G ,过G 作GH ⊥x 轴于H ,连接GD 、GE ∵24

0+=2,21

3+=2 ∴G (2,2),∴H (2,0) ········································ 1分 ∵BC =22134)( - +=52,GH =2-0=2 又DG =EG =BG =21

BC =5

∴HD =EH =2225 -)(=1

∴D (3,0),E (1,0) ··············································································· 2分

(2)设过B 、C 、D 三点的抛物线的函数关系式为y =ax 2

+bx +c ,则

?????16a +4b +c =1

c =3

9a +3b +c =0

···························································································· 3分

解得???a =2

1

b =-2

5c =3

····························································································· 4分

∴所求抛物线的函数关系式为y =21

x 2-25x +3 ········································ 5分

(3)方法1:假设存在这样的点Q ,分两种情况讨论:

①当∠BDQ =90°时,由于∠BDC =90°,且点C 在抛物线上,故点Q 与点C 重合

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

20 ∴Q 1(0,3) ································································································· 7分

②当∠DBQ =90°时,过点B 作平行于DC 的直线BQ ,假设直线BQ 交抛物线于另一点Q 由D (3,0),C (0,3)可得直线DC 的解析式为y =-x +3

∵BQ ∥DC ,故可设直线BQ 为y =-x +m

将B (4,1)代入,得m =5(或将直线DC 向上平移2个单位与直线BQ 重合) ∴直线BQ 为y =-x +5

由?????y =-x +5

y =21x 2-25x +3 得

?????x =-1y =6 或 ?????x =4y =1 又点B (4,1),∴Q 2(-1,6)

故该抛物线上存在两点(0,3),(-1,6)满足条件 ··········· 9分

方法2:假设存在这样的点Q (x ,21x

2-2

5x +3) 过Q 作QN ⊥x 轴于N ,分两种情况讨论: ①当∠BDQ =90°时,则∠NDQ +∠BDA =90°

∵∠DNQ =∠BAD =90°,∴∠NDQ +∠NQD =90°

∴∠NQD =∠BDA ,∴△NDQ ∽△ABD ∴AD NQ =AB

ND ,即1325212+-x x =13x - ···················································· 6分 解得x 1=0,x 2=3

当x 1=0时,y 1=3;当x 2=3时,y 2=0

∴Q 1(0,3),Q 2(3,0)(与点D 重合,舍去) ······································· 7分

②当∠DBQ =90°时,则有DQ 2=BD 2+BQ 2

∵B (4,1),D (3,0),Q (x ,21x 2-2

5x +3) ∴DQ 2=(x -3)2+(21x 2-25x +3)2 BD 2=(4-3)2+(1-0)2

=2

BQ 2=(x -4)2+(21x 2-2

5x +3-1)2 ∴(x -3)2+(21x 2-25x +3)2=2+(x -4)2+(21x 2-25x +3-1)2 整理得:x 2

-3x -4=0,解得x 3=-1,x 4=4 ············································· 8分 当x 3=-1时,y 3=6;当x 4=4时,y 4=1

∴Q 3(-1,6),Q 4(4,1)(与点B 重合,舍去)

综上所述,抛物线上存在点Q 1(0,3)和Q 2(-1,6),使△BDQ 是以BD 为直角边的直角三角形 ··········································································································· 9分

2014年中考数学压轴题精编—广东篇

2014年中考数学压轴题精编—广东篇

21

20.(广东省高州市学科竞赛暨重点中学提前招生考试)已知抛物线y =ax

2

+bx +c (a ≠0)与x 轴交于A 、B 两点,顶点为C .

(1)当△ABC 为直角三角形时,求b 2

-4ac 的值;

(2)当△ABC 为等边三角形时,求b

2

-4ac 的值.

224.解:(1)设A (x 1,0),B (x 2,0)

当△ABC 为直角三角形时,显然∠ACB =90° 由抛物线的对称性可知△ABC 为等腰直角三角形 如图1,过C 作CD ⊥AB 于D ,则AB =2CD ∵抛物线与x 轴有两个交点,∴△=b

2

-4ac >0

AB =|x 1-x 2|=212214x x x x -+)( =a c a b 42--)(=2

24a ac

b -=︱︱

42a ac b -

CD =

442︱︱a ac b -

∵a ≠0,∴ac b 42

-=

242ac

b -

∵b

2-4ac ≠0,∴ac b 42-=2

∴b

2

-4ac =4 ······························································ 5分

(2)当△ABC 为等边三角形时,如图2,过C 作CD ⊥AB 于D

则CD =

2

3

AB 即

442ac b -=

ac b 42

3

2-,∴ac b 42-=32 ∴b

2

-4ac =12 ····························································································· 10分

20.(广东省高州市学科竞赛暨重点中学提前招生考试)已知一次函数y 1=2x ,二次函数y 2=mx

2

-3(m -1)x +2m -1的图象关于y 轴对称. (1)求二次函数y 2的解析式;

(2)是否存在二次函数y 3=ax

2

+bx +c ,其图象经过点(-5,2),且对于任意一个实数x ,这三个函数所对应的函数值y 1、y 2、y 3都有y 1≤y 3≤y 2成立?若存在,求出函数y 3的解析式;若不存在,请说明理由.

225.解:(1)∵二次函数y 2=mx

2

-3(m -1)x +2m -1的图象关于y 轴对称

∴3(m -1)=0,∴m =1 ∴y 2=x

2

+1

(2)存在满足条件的二次函数y 3

∵y 1-y 2=2x -(x

2+1)=-x

2+2x -1=-(x -1)2

≤0

∴对于任意一个实数x ,y 1≤y 2均成立

又二次函数y 3=ax

2

+bx +c 的图象经过点(-5,2)

2

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