2013年全国初中数学竞赛预赛试题答案及详解

6745123第2题图1.解：因为a、b都是有理数，且(1)7()0aab，所以10a,且0ab，得1,1ab，所以1ab．故选A.2.解：若将图中标有1的面去掉，则标有2、3、4、5、6、7的六个面恰好是正方体的一种展开图，其中标有3和6的面是对面；只看题图最下面一行，标有3和1的面应是对面，所以重叠的两个面是标有1和6的面，应选B．3.解：（1）∵AD∥BC，∴∠DAE=111222DACACB，∴①正确；（2）∵AD∥BC，∴△AOD∽△COB，∴ADAOCBCO，∴②正确；（3）∠AEB=∠DAE+∠ADB=∠DAE+∠CBD=1()2，∴③正确；（4）∵∠BAC=180()，只有当AB∥DC时，∠ACD=180()才能成立．∴④不正确.综上，应选B.4.解：由图象可知当12yy时，3x，当01y时，1x，所以当012yy时，13x.故应选C．5.解：如右图，解：由33axax，得(1)(3)0ax，由不等式的解集为3x，知30x，所以10a，得1a．故应选B．6.解：从点A出发，每次向上或向右走一步，到达每一点的最短路径条数如图中所标数字，如：到达点P、Q的最短路径条数分别为2和3.以此类推，到达点B的最短路径条数为35条.选D.7.解：原式=22211.228.解：由图象可知1k为负数，2k、3k为正数，不妨取x=1，代入解析式，显然点2(1,)Ak在点3(1,)Bk的正下方，所以320kk，又1k为负数，所以123kkk．题号123456答案ABBCBD题号789101112答案22211.22123kkk0.24cm4120139.解：摸出的2个球都是黑球的概率是2116515，所以摸出的球颜色一样的概率是113155.10.解：在Rt△ADC中，∠A=30°，得ACDC21，同理BCEC21,所以4212121ABBCACECDC(cm)．11.解：由题意得2272mmm，解得120,8mm，当10m时，原方程无实数根，当28m时，原方程有两个不相等的实数根，所以8164m.12解：令1111456670a，则原式=1671a13a113671a×a=211132013671aaa2113671aaa=12013.13.【答案】设这个单位参加健身操比赛的职工有y人，6人、5人、4人一列分别可以整排a、b、c列，则62524yabc．（a、b、c是正整数）∴6252,624.abac①②·················································4分由②，得62312(1).422aaaac因为c为正整数，可令12,am所以21,am（m是正整数）③将③代入①，得6(21)252.mb∴122102(1).55mmmb··················································7分因为b为正整数，可令15,mn所以51,mn（n是正整数）④将④代入③，得2(51)1101.ann···································11分∴626(101)2608.yann（n是正整数）.当n=1时，y有最小值52.即参加比赛列队的至少有52人.········14分14.【答案】（1）∵E、B、C、H、F在同一圆上，且∠EBC=90°，∴∠EHC=90°，∠EFC=90°．·················································2分又∠FBC=∠HBC=45°，∴CF=CH．······································4分∵∠HBF+∠HCF=180°，∴∠HCF=90°．·······························6分∴四边形EFCH是正方形．···················································8分（2）∵∠GHB+∠GCB=180°，图③FEDBAC图④FEDBAC∴∠GHB=90°，由（1）知∠CHE=90°，∴∠CHG+∠CHB=∠EHB+∠CHB．∴∠CHG=∠EHB．∴CG=BE=x，∴DG=1DCCGx．·······························12分∴△CGH中，CG边上是高为11(1).22DGx∴211111(1).224216yxxx····································15分当x=12时，y有最大值116．·················································16分15.【答案】（1）当∠ACB为直角时，△ABC为直角三角形，b=0，AD=AC=BD=a．········································································2分（2）当∠ACB为锐角时，如图③，作∠DAE=45°，AE和BC的延长线相交于点E，过点C作CF⊥AE于点F．则△CEF和△ADE都是等腰直角三角形.设ADDEx，CFEFm.则2AEx.∴2AFxm.··4分∵∠FAC+∠CAD=45°，∠DAB+∠CAD=45°，∴∠FAC=∠DAB.又∵∠AFC=∠ADB=90°，∴△FAC∽△DAB.……………………6分∴.FAFCDADB即2.xmmxa解得2.axmxa∴2222axaxCEEFxaxa.·····················8分∵CECDDEAD,∴2axbxxa.···························10分整理得2()0xabxab.解得2216,2abababx22262abababx（舍去）.·······················································································12分（3）当∠ACB为钝角时，如图④，作∠DAE=45°，AE和BC的延长线相交于点E，过点C作CF⊥AE于点F．与（1）中的求法类似，可设ADDEx，CFEFm，则2AFxm.同（1）中的理由，得△FAC∽△DAB，2axCExa.∵ADDECECD,∴2axxbxa.·······························16分整理，得2()0xabxab，解得226.2abababx…17分综上，AD的长为a或2262ababab或2262ababab或2262ababab．······························································18分