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)
图
2
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