成都中考数学压轴题专项讲解
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1.解:(1)延长AC 至点E ,使CE =CA ,连接BE
∵C 为OB 中点,∴△BCE ≌△OCA ∴BE =OA ,∠E =∠OAC ∴BE ∥OA ,∴△APD ∽△EPB
∴EP AP =EB AD 又∵D 为OA 中点,OA =OB ,∴EP AP =AO
AD =21
∴
EP AP =AP
PC AP +2=21,∴PC AP
=2 ·
·················· 3分 (2)延长AC 至点H ,使CH =CA ,连结BH
∵C 为OB 中点,∴△BCH ≌△OCA
∴∠CBH =∠O =90︒,BH =OA
由AO AD =41,设AD =t ,OD =3t ,则BH =OA =OB =4t
在Rt △BOD 中,BD =2243)()(t t +=5t
∵OA ∥BH ,∴△HBP ∽△ADP ∴DP BP
=AD BH =t
t 4=4 ∴BP =4PD =5
4
BD =4t ,∴BH =BP ···················· 6分 ∴tan ∠BPC =tan ∠H =BH BC =t
t 42=21
············································ 7分
(3)tan ∠BPC =
n
n
········································································ 10分 2.解:(1)∵抛物线y 1=ax
2-2ax +b 经过A (-1,0),C (0,2
3
)两点
∴⎩⎪⎨⎪⎧a +2a +b =0b =2
3
∴⎩
⎪⎨⎪⎧a =-21b =2
3 ······················· 2分 ∴抛物线的解析式为y 1=-2
1x
2+x +23 ·············· 3分
(2)作MN ⊥AB ,垂足为N
由y 1=-2
1x
2+x +23
易得M (1,2),N (1,0),A (-1,0),B (3,0)
∴AB =4,MN =BN =2,MB =22,∠MBN =45︒ 根据勾股定理有BM 2-BN 2=PM 2-PN
2
∴(22)2-2
2=PM 2-(1-x )2 ① ··········· 5分 又∠MPQ =45︒=∠MBP ,∴△MPQ ∽△MBP
∴PM 2
=MQ ·MB =22
y 2·22 ② ··········· 6分
由①②得y 2=21x
2-x +2
5
∵0≤x <3,∴y 2与x 的函数关系式为y 2=21x
2-x +2
5
(0≤x <3) ······· 7分
(3)四边形EFHG 可以为平行四边形,m 、n 之间的数量关系是m +n =2(0≤m ≤2,且m ≠1)
A B
C
D P O
E
D
C O
P
H
A
B
∵点E 、G 是抛物线y 1=-
2
1x
2+x +23
分别与直线x =m ,x =n 的交点
∴点E 、G 坐标为E (m ,-21m
2+m +23),G (m ,-2
1n
2+n +23
)
同理,点F 、H 坐标为(m ,21m
2-m +25),H (n ,21n
2-n +2
5
)
∴EF =21m
2-m +25-(-21m
2+m +23
)=m
2-2m +1 ······················ 9分
GH =
21n
2-n +25-(-2
1n
2+n +23
)=n
2-2n +1 ···························· 10分
∵四边形EFHG 是平行四边形,∴EF =GH
∴m
2-2m +1=n
2-2n +1,∴(m +n -2)(m -n )=0 ························ 11分
由题意知m ≠n ,∴m +n =2(0≤m ≤2,且m ≠1)
因此,四边形EFHG 可以为平行四边形,m 、n 之间的数量关系是m +n =2(0≤m ≤2,且m ≠1)
3.解:(1)∵四边形ABCD 为正方形,∴AB =BC
∵BG ⊥AP ,AG =GE ,∴AB =BE
∴BE =BC ·················································································· 3分 (2)过点D 作DH ⊥AE 于H
∵BN 平分∠CBE ,∴∠EBN =∠CBN ∵AB =BE ,∴∠BEN =∠BAP
∵BG ⊥AP ,∠ABP =90°,∴∠BAP =∠PBG ∴∠BEN =∠PBG
∵∠BNG =∠BEN +∠EBN ,∴∠BNG =∠GBN ∴BG =NG
∴BN =2NG ·············································································· 5分 ∵DH ⊥AE ,∠DAB =90°,∴∠BAG =∠ADH 又AB =DA ,∴△BAG ≌△ADH ∴DH =AG ,BG =AH =GN ,∴DH =HN ∴DN =2DH =2AG
∴BN +DN =2AN ········································································ 8分 (3)CE =
10
2 ················································································· 10分 P A
B G
N
E
H
图1