2017年全国初中数学联合竞赛(四川初赛)
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2017年全国初中数学联合竞赛(四川初赛)
试题参考答案及评分标准
说明:评阅试卷时,请依据本评分标准.选择题和填空题只设7分和0分两档;解答题,请严格按照本评分标准规定的评分档次给分,不要再增加其他中间档次.如果考生的解答方法和本解答不同,只要思路合理,步骤正确,在评卷时请参照本评分标准划分的档次,给予相应的分数.
一、选择题(本题满分42分,每小题7分)
1、B
2、C
3、D
4、A
5、B
6、C
二、填空题(本题满分28分,每小题7分)
7、3 8、40o9、1
9
10、4
三、(本大题满分20分)
11、已知关于x的一元二次方程x2-kx+5=0与x2+5x-k=0只有一个公共的实根,求关于x的方程|x2+kx|=|k|所有的实根之和。
解:设x2-kx+5=0与x2+5x-k=0的公共实根为α,
则α 2-kα+5=0,α 2+5α-k=0, ············································(5分)
两式相减,得(k+5)α -(k+5) =0.·········································(10分)
因为当k=-5时两方程相同,有两个公共的实根,不合题意.所以k≠-5.
因此α =1.从而求得k=6.················································(15分)
所以方程|x2+kx|=|k|即为|x2+6x|=6,x2+6x=6或x2+6x=-6.
显然两方程都有实根,因此方程所有实根之和是-12. ············(20分)
四、(本大题满分25分)
12、如图,已知圆O的直径AB与CD互相垂直,E为OB中点,CE的延长线交
圆O于G,AG交CD于F,求DF
FC
的值。
解:设OE=a,则OA=OB=OC=OD=2a,CE
a,
因为△COE∽△CGD,
所以CO
CG
=
OE
GD
=
CE
CD
,
所以CG=CO CD
CE
⋅
a,
所以EG=CG-CE
a。 ················································(10分)
过G作GH⊥AB于点H,则△COE∽△GHE,
所以CO
GH
=
OE
HE
=
CE
GE
=
5
3
,
所以GH=6
5
a,HE=
3
5
a,于是AH=
18
5
a。
由△AOF∽△AHG,得OF
HG
=
AO
AH
=
5
9
,
所以OF=2
3 a。·······························································(20分)
从而DF=4
3
a,FC=
8
3
a。所以
DF
FC
=
1
2
。 ······························(25分)
法2:连结BG、AC。设OE=BE=a,则OA=OB=OC=OD=2a,CE
a,
因为△ACE∽△GBE,所以AC
GB
=
AE
GE
=
CE
BE
,又AC=
a,
所以BG
。·····························································(10分)
在Rt△ABG中,由勾股定理可得
AG
a。
又因为△AOF∽△AGB,所以AO
AG
=
OF
BG
,
所以OF=AO BG
AG
⋅
=
2
3
a。 ····································(20分)
从而DF=4
3
a,FC=
8
3
a。所以
DF
FC
=
1
2
。 ······························(25分)
法3:设OE=a,则OA=OB=OC=OD=2a,CE
a,
因为△COE∽△CGD,
所以CO
CG
=
OE
GD
=
CE
CD
,
所以DG=OE CD
CE
⋅
a,
CG=CO CD
CE
⋅
a,··········································(10分)
过G作GH⊥CD于点H,则△DGH∽△DCG,
所以DG
DC
=
DH
DG
=
GH
CG
,
可得DH=
2
DG
DC
=
2
16
5
4
a
a
=
4
5
a,GH=
DG CG
DC
⋅
=
2
32
5
4
a
a
=
8
5
a,
由△AOF∽△GHF,得AO
GH
=
OF
FH
,
所以OF
FH
=
5
4
,又OF+FH=2a-
4
5
a=
6
5
a,