2017年全国初中数学联合竞赛(四川初三初赛)试题参考答案及评分标准

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、B2、C3、D4、A5、B6、C二、填空题（本题满分28分，每小题7分）7、38、409、1910、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，则2k50，25k0，············································（5分）两式相减，得(k5)(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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DFFC的值。解：设OEa，则OAOBOCOD2a，CE5a，因为△COE∽△CGD，所以COCGOEGDCECD，所以CGCOCDCE245aaa855a，所以EGCGCE355a。················································（10分）过G作GH⊥AB于点H，则△COE∽△GHE，所以COGHOEHECEGE53，所以GH65a，HE35a，于是AH185a。HFGEDCOAB由△AOF∽△AHG，得OFHGAOAH59，所以OF23a。·······························································（20分）从而DF43a，FC83a。所以DFFC12。······························（25分）法2：连结BG、AC。设OEBEa，则OAOBOCOD2a，CE5a，因为△ACE∽△GBE，所以ACGBAEGECEBE，又AC22a，所以BG85a。·····························································（10分）在Rt△ABG中，由勾股定理可得AG22ABBG228165aa725a。又因为△AOF∽△AGB，所以AOAGOFBG，所以OFAOBGAG825725aaa23a。····································（20分）从而DF43a，FC83a。所以DFFC12。······························（25分）法3：设OEa，则OAOBOCOD2a，CE5a，因为△COE∽△CGD，所以COCGOEGDCECD，所以DGOECDCE45aaa455a，CGCOCDCE245aaa855a，··········································（10分）过G作GH⊥CD于点H，则△DGH∽△DCG，所以DGDCDHDGGHCG，可得DH2DGDC21654aa45a，GHDGCGDC23254aa85a，由△AOF∽△GHF，得AOGHOFFH，所以OFFH54，又OFFH2a45a65a，所以OF23a。·······························································（20分）FGEDCOABHFGEDCOAB从而DF43a，FC83a。所以DFFC12。······························（25分）五、（本大题满分25分）13、已知a、b、c、d、x、y、z、w是互不相等的非零实数，且222222abaybx222222bcbzcy222222cdcwdzabcdxyzw．求22ax22by22cz22dw的值．解：将每个已知分式分子分母交换，得222222aybxab222222bzcybc222222cwdzcdxyzwabcd．（）··············（5分）即22xa22yb22yb22zc22zc22wd，易得xa±zc，yb±wd．·········（10分）设zcu，wdv，当xau，ybv或xau，ybv时，则由22zc22wdxyzwabcd得，u2v2u2v2．显然不可能成立，舍去。··················································（15分）当xau，ybv或xau，ybv时，则由22zc22wdxyzwabcd得，u2v2u2v2．···································（20分）而22ax22by22cz22dw2(21u21v)22222uvuv2．····················（25分）