2006年AMC8试题及答案

合集下载

2006年AMC8试题答案

This Solutions Pamphlet gives at least one solution for each problem on this year’s exam and shows that all the problems can be solved using material normally as-sociated with the mathematics curriculum for students in eighth grade or below. These solutions are by no means the only ones possible, nor are they necessarily superior to others the reader may devise.We hope that teachers will share these solutions with their students. However, the publication, reproduction, or communication of the problems or solutions of the AMC 8 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination at any time via copier, telephone, e-mail, World Wide Web or media of any type is a violation of the competition rules.Correspondence about the problems and solutions should be addressed to:Ms. Bonnie Leitch , AMC 8 Chair / bleitch@548 Hill Avenue, New Braunfels, TX 78130Orders for prior year Exam questions and Solutions Pamphlets should be addressed to:Attn: Publications American Mathematics Competitions University of Nebraska-Lincoln P .O. Box 81606Lincoln, NE 68501-1606Copyright © 2006, The Mathematical Association of AmericaT he M aTheMaTical a ssociaTion of a MericaAmerican Mathematics Competitions 22nd AnnualAMC 8(American Mathematics Contest 8)Solutions PamphletTuesday, NOVEMBER 14, 20061.(D)Mindy’s total was approximately 2+5+10=$17.2.(C)On the AMC 8a student’s score is the number of problems answered cor-rectly.So Billy’s score is 13.Because there is no penalty for guessing,if he wants to increase his score,he probably should ﬁll in the last ﬁve answers.3.(A)When Elisa started,she completed a lap in 2510=2.5minutes.Now she cancomplete a lap in 2412=2minutes.She has improved her lap time by 2.5−2=0.5or 12minute.4.(B)Ignore the number of complete revolutions because they do not aﬀect direc-tion.One-fourth of the distance around the circle clockwise from west is north.Three-fourths of the distance counterclockwise around the circle from north is east.Chenille’s spinner points east.5.(D)Divide the larger square into 8congruent triangles,as shown,4of which make up the smaller square.ABD The area of the smaller square is 48or 12of the area of the larger square,so thearea of the smaller square is equal to 30.6.(C)11422442The perimeter is 4+2+1+4+2+4+1+2=20inches.OREach rectangle has perimeter =2l +2w =2(4)+2(2)=8+4=12inches.When the two rectangles are positioned to form the T,a two-inch segment of each rectangle is inside the T and is not on the perimeter of the T.So the perimeter of the T is 2(12)−2(2)=24−4=20inches.7.(B)Because circumference C =2πr and circle Y has circumference 8π,itsradius is 8π2π=4.Because area A =πr 2and circle Z has area 9π,its radius is √9=3.Ordering the radii gives 3<π<4,so the circles in ascending order ofradii length are Z ,X and Y .8.(E)Because 200−96=104of those surveyed were male,104−26=78of those surveyed are male listeners.Listen Don’t Listen Total Male7826104Female583896Total 13664200The percentage of males surveyed who listen to KAMC is 78104×100%=75%.9.(C)Note that in each fraction,the numerator is the same as the denominatorin the next fraction,so they divide.The product of ¡32×¡4¡3×¡5¡4×...×2006$$$2005=20062=1003.10.(A)When the area of a rectangle is 12square units and the sides are integers,the factors of 12are the possible lengths of the sides.In point form,the side lengths could be (1,12),(2,6),(3,4),(4,3),(6,2)and (12,1).Only graph A ﬁts these points.11.(C)The sum of the digits of a two-digit number is at most 9+9=18.Thismeans the only possible perfect square sums are 1,4,9and 16.Each square has the following two-digit possibilities:1:104:40,31,22,139:90,81,72,63,54,45,36,27,1816:97,88,79There are 17two-digit numbers in all.12.(D)Note that 70%of 10is 7,80%of 20is 16and 90%of 30is 27.Antonetteanswers 7+16+27=50problems correctly out of 60problems in all.Her overall score is 5060or 83.3%.13.(D)Between 8:30and 9:00AM Cassie travels 6miles.At 9:00Cassie and Brianare only 56miles apart.After 9:00,because they are both biking towards each other,the distance between them decreases at the rate of 12+16=28miles per hour.At that rate,it will take them 5628=2hours to meet.So they will meetat 11:00AM.14.(B)Bob takes 45−30=15more seconds per page than Chandra.So thediﬀerence in their total reading times is 760·15=11,400seconds.Bob will spend 11,400more seconds reading than Chandra.15.(C)The ratio of time it takes Bob to read a page to the time it takes Chandrato read a page is45:30or3:2,so Bob should read23of the number of pagesthat Chandra reads.Divide the book into5parts,each with7605=152pages.Chandra will read theﬁrst3·152=456pages,while Bob reads the last2·152= 304pages.ORIf Chandra reads x pages,she will read for30x seconds.Bob has to read760−x pages,and this takes him45(760−x)seconds.Because Chandra and Bob read the same amount of time,30x=45(760−x).Solving for x,30x=45·760−45x,75x=45·760,x=45·76075=456.So Chandra will read theﬁrst456pages.16.(E)The least common multiple of20,45and30is22·32·5=ing theLCM,in180seconds Alice reads18020=9pages,Chandra reads18030=6pagesand Bob reads18045=4pages.Together they read a total of19pages in180seconds.The total number of seconds each reads is76019·180=7200.17.(B)Because the sum of a number from spinner Q and a number from spinner Ris always odd,the sum of the numbers on the three spinners will be odd exactly when the number from spinner P is even.Because2is the only even number onspinner P,the probability of getting an odd sum is13.18.(D)Four black andﬁve white squares are visible on each of the six faces of thecube.So59of the surface will be white.19.(D)Because triangles ABD and ECD are congruent and triangle ABC isisosceles,EC=AB=BC=11.That means BD=112or5.5.20.(C)Each of the six players played5games,and each game involved two players.So there were6·52=15games.Helen,Ines,Janet,Kendra and Lara won a totalof4+3+2+2+2=13games,so Monica won15−13=2games.21.(A)Using the volume formula lwh=V,the volume of water in the aquariumis100×40×37=148,000cm3.When the rock is put in,the water and the rock will occupy a box-shaped region with volume148,000+1000=149,000cm3.The volume of the water and the rock is100×40×h,where h is the new height of the water.The new volume=4000h=149,000cm3,so the new height ish=1490004000=37.25cm.After adding the rock,the water rises37.25−37=0.25cm.ORBecause the shape of the rock is irrelevant,we may assume that the rock is shaped like a rectangular box with base measuring100cm×40cm and height hing the volume formula,100×40×h=1000,so h=1000100×40=0.25cm.When the rock is put into the aquarium,the water level will rise by0.25cm.22.(D)If the lower cells contain A,B and C,then the second row will containA+B and B+C,and the top cell will contain A+2B+C.To obtain the smallest sum,place1in the center cell and2and3in the outer ones.The top number will be7.For the largest sum,place9in the center cell and7and8in the outer ones.This top number will be33.The diﬀerence is33−7=26. 23.(A)The counting numbers that leave a remainder of4when divided by6are4,10,16,22,28,34,....The counting numbers that leave a remainder of3when divided by5are3,8,13,18,23,28,33,....So28is the smallest possible number of coins that meets both conditions.Because4×7=28,there are no coins left when they are divided among seven people.ORIf there were two more coins in the box,the number of coins would be divisible by both6and5.The smallest number that is divisible by6and5is30,so the smallest possible number of coins in the box is28.24.(A)We can decompose CDCD into CD×100+CD=CD(101).That meansthat A=1and B=0.The sum is1+0=1.25.(B)There are one odd and two even numbers showing.Because all primesother than2are odd and the sum of an even number and an odd number is odd,the common sum must be odd.That means2must be opposite59and the common sum is2+59=61.The other two hidden numbers are61−44=17and61−38=23.The average of2,17and23is2+17+233=423=14.TheAmerican Mathematics Competitionsare Sponsored byThe Mathematical Association of AmericaThe Akamai FoundationContributorsAmerican Mathematical Association of Two Year CollegesAmerican Mathematical SocietyAmerican Society of Pension ActuariesAmerican Statistical AssociationArt of Problem SolvingCanada/USA MathcampCanada/USA MathpathCasualty Actuarial SocietyClay Mathematics InstituteConsortium for Mathematics & its ApplicationsInstitute for Operations Research and the Management SciencesL. G. Balfour CompanyMu Alpha ThetaNational Council of Teachers of MathematicsNational Assessment & TestingPedagoguery Software Inc.Pi Mu EpsilonSociety of ActuariesU.S.A. Math Talent SearchW. H. Freeman and CompanyWolfram Research Inc.。

全国中学生生物学联赛试题及答案及美国数学竞赛8年级(AMC8)真题(附答案)(电脑版)

全国中学生生物学联赛试题注意事项：1．所有试题使用2B铅笔在机读卡上作答；2．试题按学科分类，单选和多选题混排，多选题答案完全正确才可得分3．纸质试卷72题，电子试卷48题，共计l20题；4．答题时间120分钟。

一、细胞生物学、生物化学、微生物学16题1．癌细胞与正常细胞的不同之处在于A．癌细胞不能合成DNA B．癌细胞被锁定在细胞周期中的S期C．癌细胞能持续分裂尽管彼此紧密相接D．癌细胞始终处于细胞周期中的分裂期2．人的肌肉组织分为快缩纤维和慢缩纤维两种，快缩纤维负责剧烈运动如举重，短跑，易产生酸痛感觉；慢缩纤维负责慢跑，游泳等有氧运动。

以下关于慢缩纤维和快缩纤维的描述，哪个是正确的A．快缩纤维含有的线粒体多，有氧呼吸能产生大量乳酸和ATP供能B．慢缩纤维含有的线粒体多，有氧呼吸不产生乳酸，产生的ATP也少C．快缩纤维含有的线粒体少，主要依靠糖酵解产生ATP供能，因此产生大量乳酸D．慢缩纤维含有的线粒体多，主要依靠糖酵解产生ATP供能3．在光合作用中参与电子传递的载体是A．叶绿素B．磷酸烯醇式丙酮酸C．NADH D．NADPH4．肽链生物合成时，信号肽A．是线粒体的定位信号B．将新生肽链导入内质网C．控制蛋白质分子的最终构象D．处于肽链的C末端5．原核细胞的特征包括A．无核糖体B．无转录后修饰C．无有氧呼吸D．无染色体6．以下糖类中属于寡糖的有(多选2分)A．甘露糖B．麦芽糖C．半乳糖D．蔗糖E．糖原7．以下关于蛋白质变性后所产生现象的描述不正确的是：A．溶解度降低B．疏水侧链基团被包裹在分子内部C．形成沉淀D．失去结晶能力8．真菌中的半知菌是指。

A．没有光合作用B．菌丝没有横隔C．没有发现有性生殖阶段D．不能运动。

9．关于维生素A的生理功能，下面的描述中哪个是错误的?A．抑制胆碱酯酶活性B．构成视觉细胞感光物质的成分C．参与上皮组织细胞膜糖蛋白合成D．严重缺乏时导致夜盲症10．磷酸戊糖途径的发生部位在A．细胞质B．线粒体C．叶绿体D．细胞膜11．在C4植物循环中，CO2进入叶肉细胞被固定的最初产物是A．甘油酸-3-磷酸B．苹果酸C．草酰乙酸D．丙酮酸12．在等电聚焦电泳过程中，随着蛋白质样品的迁移，电流的变化为A．越变越大，当样品到达其等电点位置时，电流达到最大值B．越变越小，当样品到达其等电点位置时，电流达到最小值，接近于零C．基本不变，为一恒定值D．不确定13．利用酵母菌进行乙醇发酵时若通入氧气，会导致菌株对葡萄糖利用速度降低，甚至停止生成乙醇，这种现象称为A．呼吸抑制效应B．巴斯德效应C．分子氧效应D．葡萄糖效应14．巴斯德设计的曲颈瓶实验，可以(单选1分)A．证明酒是酵母菌发酵获得B．否定自然发生学说的论点C．说明培养基灭菌后的作用D．验证某细菌是不能培养的15．营养缺陷型菌株是指(单选1分)A．不需要添加生长因子可在基础培养基上生长的菌株B．不需要添加生长因子可在丰富培养基上生长的菌株C．因突变需提供某种营养才能在基础培养基上生长的菌株D．因自发或诱发突变而导致的可抵抗环境不良因素的菌株16．以下哪类化合物属于微生物的次生代谢产物(多选2分)A．脂肪B．色素C．抗生素D．蛋白质E．毒素二、植物和动物的解剖、生理、组织和器官18题17．草履虫、水螅、乌贼、蟾蜍受到刺激后，均可从体内发出一些物质以攻击或防御敌害，在他们身体上，发出这些物质的结构是A．刺丝泡、刺细胞、墨囊、耳后腺B．刺丝泡、刺丝囊、外套腔、唾液腺C．表膜泡、刺丝囊、墨囊、唾液腺D．表膜泡、刺细胞、外套腔、耳后腺18．在动物卵裂时期，由于不同动物受精卵内卵黄多少及其在卵内分布的不同，卵裂方式也有很大差异，海胆、沙蚕、昆虫、乌贼的卵裂方式依次分别为(单选1分) A．完全均等卵裂(等裂)、表面卵裂、螺旋形卵裂、盘裂B．螺旋形卵裂、完全均等卵裂(等裂)、盘裂、表面卵裂C．螺旋形卵裂、完全均等卵裂(等裂)、表面卵裂、盘裂D．完全均等卵裂(等裂)、螺旋形卵裂、表面卵裂、盘裂19．不同动物类群具有独特的特征，现存棘皮动物、海绵动物、哺乳动物、鸟类所特有的特征依次为A．水管系、水沟系、下颌为单一齿骨、羽毛B．后口、水沟系、胎生、飞翔C．后口、骨针、胎生、羽毛D．水管系、骨针、下颌为单一齿骨、飞翔20．节肢动物类群很多，不同类群的排泄器官亦有差异，节肢动物门甲壳纲动物的排泄器官有(多选l分)A．基节腺B．触角腺C．颚腺D．马氏管21．家鸽的一侧体动脉弓退化，雌家鸽的一侧卵巢和输卵管也退化了，退化的这些器官是(单选1分)A．左体动脉弓和右侧的卵巢、输卵管B．左体动脉弓和左侧的卵巢、输卵管C．右体动脉弓和左侧的卵巢、输卵管D．右体动脉弓和右侧的卵巢、输卵管22．在海滨潮间带经常可以见到石鳖和沙蚕，以下不属于它们共同特征的是A．以裂体腔法形成真体腔B．后肾型排泄系统C．具有担轮幼虫期D．开管式循环系统23．以下哪项不是文昌鱼的特征A．具有脊索，背神经管，鳃裂B．有分节的肌肉，有哈氏窝C．有头，有心脏D．有特化的口器24．一家饭店涉嫌出售野生鸟类，检查人员在检查时发现了一种鸟类的足，三趾向前一趾向后，后趾与前面三趾在同一平面上，趾长，基部有蹼相连，这种鸟类是A．鹈形目B．鹳形目C．雁形目D．鹤形目25．以下哪组元素在植物体内参与氧化还原反应(单选2分)A．钼镍铜铁B．铁铜镁钼C．钙镁镍铜D．锰镍镁铜26．盐胁迫条件下，较耐盐的禾本科植物大麦可以通过将盐分局域于以下部位来缓解盐分对植物生长造成的危害(多选l分)A．根系B．幼叶C．叶鞘D．老叶27．关于植物的种子，下列描述正确的是(多选2分)A．种子由胚珠发育而来B．种子表面均有种孔、种脐和种脊的结构C．种子中的胚乳多来源于受精后的中央细胞，也有来自于雌配子体的细胞D．胚是休眠的幼小孢子体E．无胚乳种子在发育过程中没有胚乳形成28．有关被子植物花的叙述，下列哪一个是错误的(单选2分)A．花是适应于繁殖功能的变态短枝B．花托、花萼和花冠被认为是叶的变态C．雄蕊和雌蕊也被认为是叶的变态D．花托、花被、雄蕊和雌蕊均有茎的顶端分生组织产生29．玉米干旱缺水时叶片的内卷主要是失水造成的A．叶肉细胞B．叶表皮的毛状体C．位于上表皮的泡状(运动)细胞D．位于下表皮的泡状(运动)细胞30．有关C4植物，以下说法中正确的是(多选2分)A．叶解剖结构中可观察到“花环结构”B．光合作用CO2的初步固定和同化在不同细胞中进行C．光合作用CO2的初步固定和同化在同一细胞中进行D．在一定范围的强光、高温条件下光合效率高31．心肌细胞有效不应期的长短主要取决于A．静息电位水平B．0期去极化的速度C．阈电位水平D．平台期的长短32．血液中CO2分压升高使呼吸运动加强的最主要途径是(单选2分)A．直接刺激脑桥的呼吸相关神经元B．直接刺激延髓呼吸中枢的神经元C．刺激中枢化学感受器D．刺激颈动脉体和主动脉体感受器33．当去甲肾上腺素与β受体结合时，下列哪一种肌肉收缩或收缩加强(单选1分) A．心室肌B．子宫平滑肌C．小肠平滑肌D．血管平滑肌E．支气管平滑肌34．下列哪种因素可引起人尿量的明显增加的(多选2分)A．献血200ml后B．饮用清水1000ml后C．静脉注射神经垂体激素D．饮用生理盐水100ml后三、动物行为学、生态学15题35．如果一项研究，专注于了解不同生态因子对生物的影响，及生物对它们的耐受，那么这个研究属于哪一层次上的研究A．个体生态学B．种群生态学C．群落生态学D．生态系统生态学36．关于高等动物种群中性别比例，下面论述中错误的是A．大多数种群倾向于使出生性比趋近于l：1 B．老年组往往雌性多于雄性C．出生的时候，往往雄性多于雌性D．种群性比与世代长度直接相关37．社会性寄生是指A．寄生在动物社会中是普遍现象B．寄生只发生在特定社会等级的动物中C．社会性昆虫中发生的寄生行为D．强迫寄主动物为其提供食物或其他利益38．关于外来物种，以下论述错误的是A．所有的外来物种都是入侵种，都是有害的B．外来物种可以依靠风、鸟、昆虫等自然因素入侵C．有些外来物种是人类有意引入的D．入侵物种可能对生态系统造成长久的破坏39．适合度是指A．动物单一行为的适应性B．动物调整自己的行为以适合于生活在当时的环境C．动物的总体繁殖成功性D．最适合动物生活习性、满足营养需求的食物40．以下哪种情况不属于动物的行为节律A．候鸟随季节的迁徙B．哺乳动物晨昏活动习性C．细菌生长速度随营养物浓度起落而快慢变化D．招潮蟹的活动随潮汐变化而变化41．动物的生长和发育是需要一定温度的，下列哪个说法是正确的(单选2分) A．外界温度的高低直接决定了动物机体的体温，进而影响其生长发育B．当外界温度低于某一温度时，昆虫就停止生长发育，这一温度阈值称为发育起点温度C．动物的发育速度总是随环境温度的增高而加快的D．昆虫发育的有效积温是发育历期乘以发育期的平均温度，然后求和42．下列有关水生群落演替的说法中哪个是错误的A．水生群落的演替一般会依次经历裸底期、浮水植物期、沉水植物期、挺水植物期、湿生草本植物期等阶段B．在这一演替过程中池底逐渐变浅，最终向陆地变化C．挺水植物根系往往较发达，可以使水底迅速增高D．浮水植物的叶子漂浮在水面，影响到水下光照，不利于沉水植物生长43．关于固定行为型，下述论述正确的是(多选2分)A．固定行为型被特定的外部刺激所释放B．每一个物种都有物种特异的固定行为型C．固定行为型一旦释放就会持续到底D．固定行为型是一种先天行为44．在动物行为学研究中，严格定义行为类型是研究工作的基础。

2006全国初中数学联赛试题及答案全

2006年全国初中数学联合竞赛试题第一试一、选择题：（本题满分42分，每小题7分）1．已知四边形ABCD 为任意凸四边形，E 、F 、G 、H 分别是边AB 、BC 、CD 、DA 的中点用S 、p 分别表示四边形ABCD 的面积和周长；S1、p1，分别表示四边形EFGH 的面积和周长．设111,p p k S S k ==．则下面关于1k k 、的说法中，正确的是（）A ．1k k 、均为常值．B ．k 为常值，1k 不为常值． C.k 不为常值，1k 为常值． D.1k k 、均不为常值．2．已知m 为实数，且ααcos sin 、是关于x 的方程0132=+-mx x 的两根．则4sin α+α4cos 的值为（） A.92． B.31 ． C.97 ． D.1．3．关于x 的方程a x x =-|1|2仅有两个不同的实根．则实数a 的取值范围是（）A.a ＞0．B.a≥4．C.2＜a ＜4．D.0＜a ＜4． 4．设.,02,0222a bc c ab a b >=+->则实数c b a 、、的大小关系是 ( )A.a c b >> ．B.b a c >> ．C.c b a >> ．D.c a b >> ．5．b a 、为有理数，且满足等式324163++⨯=+b a ,则b a +的值为 ( )A.2．B.4．C.6．D.8．6．将满足条件“至少出现一个数字0且是4的倍数的正整数”从小到大排成一列数：20，40，60，80，100，104，…．则这列数中的第158个数为 ( )A ．2000．B ．2004．C ．2008．D ．2012．二、填空题：（本题满分28分，每小题7分）7．函数2008||20062+-=x x y 的图像与x 轴交点的横坐标之和等于． 8．在等腰ABC Rt ∆中，AC ＝BC ＝1，M 是BC 的中点，CE ⊥AM 于点E ，交AB 于点F ，则S △MBF ＝．9．使16)8(422+-++x x 取最小值的实数x 的值为．10．在平面直角坐标系中，正方形OABC 的顶点坐标分别为O(0，0)、A(100，0)、B(100，100)、C(0，100)．若正方形0ABC 内部(边界及顶点除外)一格点P 满足PO C PAB PBC PO A S S S S ∆∆∆∆⋅=⋅.就称格点P 为“好点”．则正方形OABC 内部好点的个数为．注：所谓格点，是指在平面直角坐标系中横、纵坐标均为整数的点．第二试（A ）一、（本题满分20分）已知关于x 的一元二次方程0)994()32(222=++++++b a x b a x 无相异两实根．则满足条件的有序正整数组)(b a ，有多少组?二、（本题满分25分）如图，D 为等腰△ABC 底边BC 的中点，E 、F 分别为AC 及其延长线上的点．已知∠EDF ＝90°．ED ＝DF ＝1，AD ＝5．求线段BC 的长．三、（本题满分25分）如图，在平行四边形ABCD 中，∠A 的平分线分别与BC 、DC 的延长线交于点E 、F ，点O 、O1分别为△CEF 、△ABE 的外心．求证： (1)O 、E 、O1三点共线；(2).21ABC OBD ∠=∠ ．第二试（B ）一、（本题满分20分）题目与（A ）卷第一题相同．二、（本题满分25分）题目与（A ）卷第二题相同．三、（本题满分25分）如图，在平行四边形ABCD 中，∠A 的平分线分别与BC 、DC 的延长线交于点E 、F ，点O 、O1分别为△CEF 、△ABE 的外心．(1)求证：O 、E 、01三点共线；(2)若,70o ABC =∠求OBD ∠的度数．第二试（C ）一、（本题满分20分）题目与（A ）卷第二题相同．二、（本题满分25分）题目与（B ）卷第三题相同．三、（本题满分25分）设p 为正整数，且2≥p ．在平面直角坐标系中，点),0(p A 和点)0,(p B 的连线段通过1-p 个格点,),1,1(1 -p C )1,1(,).,(1---p C i p i C p i ．证明： (1)若p 为质数，则在原点O(0，0)与点),(i p i C i-的连线段)1,,2,1(.-=p i OC i 上除端点外无其他格点；(2)若在原点O(0，0)与点),(i p i C i -的连线段)1,,2,1(-=p i OC i 上除端点外无其他格点，则p 为质数． 2007年全国初中数学联合竞赛试题第一试一、选择题：（本题满分42分，每小题7分）1. 已知z y x ,,满足x z z y x +=-=532，则zy y x 25+-的值为（） A ．1. B ．31. C ．31-. D ．21. 2．当x 分别取值20071，20061，20051，…，21，1，2，…，2005，2006，2007时，计算代数式2211xx +-的值，将所得的结果相加，其和等于（） A ．－1. B ．1. C ．0. D ．2007.3. 设c b a ,,是△ABC 的三边长，二次函数2)2(2b a cx x ba y ----=在1=x 时取最小值b 58-，则△ABC 是（） A ．等腰三角形. B ．锐角三角形. C ．钝角三角形. D ）直角三角形.4. 已知锐角△ABC 的顶点A 到垂心H 的距离等于它的外接圆的半径，则∠A 的度数是（）A ．30°.B ．45°.C ．60°.D ．75°.5．设K 是△ABC 内任意一点，△KAB 、△KBC 、△KCA 的重心分别为D 、E 、F ，则ABC DEF S S △△:的值为（）A ．91.B ．92.C ．94.D ．32. 6．袋中装有5个红球、6个黑球、7个白球，从袋中摸出15个球，摸出的球中恰好有3个红球的概率是（）A ．101.B ．51.C ．103.D ．52. 二、填空题：（本题满分28分，每小题7分）1. 设121-=x ，a 是x 的小数部分，b 是x -的小数部分，则=++ab b a 333___ ． 2. 对于一切不小于2的自然数n ，关于x 的一元二次方程22(2)20x n x n -+-=的两个根记作n n b a ,（2≥n ），则)2)(2(122--b a )2)(2(133--+b a +)2)(2(120072007--+b a = ． 3. 已知直角梯形ABCD 的四条边长分别为6,10,2====AD CD BC AB ，过B 、D 两点作圆,与BA 的延长线交于点E ，与CB 的延长线交于点F ，则BF BE -的值为．4．若64100+a 和64201+a 均为四位数，且均为完全平方数，则整数a 的值是．第二试（A ）一、（本题满分20分）设n m ,为正整数，且2≠m ，如果对一切实数t ，二次函数mt x mt x y 3)3(2--+=的图象与x 轴的两个交点间的距离不小于2t n +，求n m ,的值.二、（本题满分25分）如图，四边形ABCD 是梯形，点E 是上底边AD 上一点，CE 的延长线与BA 的延长线交于点F ，过点E 作BA 的平行线交CD 的延长线于点M ，BM 与AD 交于点N .证明：∠AFN =∠DME .三、（本题满分25分）已知a 是正整数，如果关于x 的方程056)38()17(23=--+++x a x a x 的根都是整数，求a 的值及方程的整数根.第二试（B ）一、（本题满分20分）设n m ,为正整数，且2≠m ，二次函数mt x mt x y 3)3(2--+=的图象与x 轴的两个交点间的距离为1d ，二次函数nt x n t x y 2)2(2+-+-=的图象与x 轴的两个交点间的距离为2d .如果21d d ≥对一切实数t 恒成立，求n m ,的值.二、（本题满分25分）题目与（A ）卷第二题相同.三、（本题满分25分）设a 是正整数，二次函数a x a x y -+++=38)17(2，反比例函数xy 56=，如果两个函数的图象的交点都是整点（横坐标和纵坐标都是整数的点），求a 的值. A B C D EF M N P第二试（C ）一、（本题满分20分）题目与（B ）卷第一题相同.二、（本题满分25分）题目与（A ）卷第二题相同.三、（本题满分25分）设a 是正整数，如果二次函数a x a x y 710)232(22-+++=和反比例函数x a y 311-=的图象有公共整点（横坐标和纵坐标都是整数的点），求a 的值和对应的公共整点. 2008年全国初中数学联合竞赛试题第一试一、选择题：（本题满分42分，每小题7分）1．设213a a +=，213b b +=，且a b ≠，则代数式2211a b+的值为（） A. 5. B.7. C .9. D.11.2．如图，设AD ，BE ，CF 为三角形ABC 的三条高，若6AB =，5BC =，3EF =，则线段BE 的长为（）A.185.B.4.C.215.D.245. 3．从分别写有数字1，2，3，4，5的5张卡片中任意取出两张，把第一张卡片上的数字作为十位数字，第二张卡片上的数字作为个位数字，组成一个两位数，则所组成的数是3的倍数的概率是（）A.15.B.310.C.25.D.12. 4．在△ABC 中，12ABC ∠=︒，132ACB ∠=︒，BM 和CN 分别是这两个角的外角平分线，且点,M N 分别在直线AC 和直线AB 上，则（）A.BM CN >.B.BM CN =.C.BM CN <.D.BM 和CN 的大小关系不确定.5．现有价格相同的5种不同商品，从今天开始每天分别降价10％或20％，若干天后，这5种商品的价格互不相同，设最高价格和最低价格的比值为r ，则r 的最小值为（）A.39()8.B.49()8.C.59()8.D.98. 6．已知实数,x y 满足22(2008)(2008)2008x x y y ----=，则223233x y x y -+-2007-的值为（）A.2008-.B.2008.C.1-.D.1.二、填空题：（本题满分28分，每小题7分）1．设51a -=，则5432322a a a a a a a +---+=-.2．如图，正方形ABCD 的边长为1，,M N 为BD 所在直线上的两点，且5AM =，135MAN ∠=︒，则四边形AMCN 的面积为3．已知二次函数2y x ax b =++的图象与x 轴的两个交点的横坐标分别为m ，n ，且1m n +≤.设满足上述要求的b 的最大值和最小值分别为p ，q ，则p q +=4．依次将正整数1，2，3，…的平方数排成一串：149162536496481100121144…，排在第1个位置的数字是1，排在第5个位置的数字是6，排在第10个位置的数字是4，排在第2008个位置的数字是 .第二试（A ）一、（本题满分20分）已知221a b +=，对于满足条件01x ≤≤的一切实数x ，不等式(1)(1)()0a x x ax bx b x bx ------≥ 恒成立.当乘积ab取最小值时，求,a b 的值.二、（本题满分25分）如图，圆O 与圆D 相交于,A B 两点，BC 为圆D 的切线，点C 在圆O 上，且AB BC =.（1）证明：点O 在圆D 的圆周上.（2）设△ABC 的面积为S ，求圆D 的的半径r 的最小值.三、（本题满分25分）设a 为质数，b 为正整数，且29(2)509(4511)a b a b +=+，求a ，b 的值.第二试（B ）一、（本题满分20分）已知221a b +=，对于满足条件1,0x y xy +=≥的一切实数对(,)x y ，不等式220ay xy bx -+≥恒成立.当乘积ab 取最小值时，求,a b 的值.二、（本题满分25分）题目与（A ）卷第二题相同.三、（本题满分25分）题目与（A ）卷第三题相同. 第二试（C ）一、（本题满分20分）题目与（B ）卷第一题相同.二、（本题满分25分）题目与（A ）卷第二题相同.三、（本题满分25分）设a 为质数，,b c 为正整数，且满足29(22)509(41022511)2a b c a b c b c ⎧+-=+-⎨-=⎩ ,求()a b c +的值. 2009年全国初中数学联合竞赛试题第一试一、选择题：（本题满分42分，每小题7分）1.设1a =，则32312612a a a +--=（）A.24.B. 25.C. 10. D. 12.2．在△ABC 中，最大角∠A 是最小角∠C 的两倍，且AB ＝7，AC ＝8，则BC ＝（）A.103．用[]x 表示不大于x 的最大整数，则方程22[]30x x --=的解的个数为（）A.1.B. 2.C. 3.D. 4.4．设正方形ABCD 的中心为点O ，在以五个点A 、B 、C 、D 、O 为顶点所构成的所有三角形中任意取出两个，它们的面积相等的概率为（） A.314. B. 37. C. 12. D. 47. 5．如图，在矩形ABCD 中，AB ＝3，BC ＝2，以BC 为直径在矩形内作半圆，自点A 作半圆的切线AE ，则sin ∠CBE ＝（）23. C. 13. 6．设n 是大于1909的正整数，使得19092009n n--为完全平方数的n 的个数是（） A.3. B. 4. C. 5. D. 6.二、填空题（本题满分28分，每小题7分）1．已知t 是实数，若,a b 是关于x 的一元二次方程2210x x t -+-=的两个非负实根，则22(1)(1)a b --的最小值是____________.2．设D 是△ABC 的边AB 上的一点，作DE//BC 交AC 于点E ，作DF//AC 交BC 于点F ，已知△ADE 、△DBF 的面积分别为m 和n ，则四边形DECF 的面积为______.3．如果实数,a b 满足条件221a b +=，22|12|21a b a b a -+++=-，则a b +=_ ____. 4．已知,a b是正整数，且满足是整数，则这样的有序数对(,)a b 共有_____对. 第二试（A ）一、（本题满分20分）已知二次函数2(0)y x bx c c =++<的图象与x 轴的交点分别为A 、DCB ，与y 轴的交点为C.设△ABC 的外接圆的圆心为点P.（1）证明：⊙P 与y 轴的另一个交点为定点.（2）如果AB 恰好为⊙P 的直径且2ABC S △＝，求b 和c 的值.二、（本题满分25分）设CD 是直角三角形ABC 的斜边AD 上的高，1I 、2I 分别是△ADC 、△BDC 的内心，AC ＝3，BC ＝4，求1I 2I .三、（本题满分25分）已知,,a b c 为正数，满足如下两个条件：32a b c ++=14b c a c a b a b c bc ca ab +-+-+-++=.第二试（B ）一、（本题满分20分）题目与（A ）卷第一题相同.二、（本题满分25分）已知△ABC 中，∠ACB ＝90°，AB 边上的高线CH 与△ABC 的两条内角平分线 AM 、BN 分别交于P 、Q 两点.PM 、QN 的中点分别为E 、F.求证：EF ∥AB.三、（本题满分25分）题目与（A ）卷第三题相同. 第二试（C ）一、（本题满分20分）题目与（A ）卷第一题相同.二、（本题满分25分）题目与（B ）卷第二题相同.三、（本题满分25分）已知,,a b c 为正数，满足如下两个条件：32a b c ++=14b c a c a b a b c bc ca ab +-+-+-++=.2010年全国初中数学联合竞赛试题第一试一、选择题：（本题满分42分，每小题7分）1. 若,,a b c 均为整数且满足1010()()1a b a c -+-=，则||||||a b b c c a -+-+-=（）A ．1.B ．2.C ．3.D ．4.2．若实数,,a b c满足等式3||6b =，9||6b c =，则c 可能取的最大值为（）A ．0.B ．1.C ．2.D ．3.N A B3．若b a ,是两个正数，且,0111=+-+-ab b a 则（） A ．103a b <+≤. B ．113a b <+≤. C ．413a b <+≤. D ．423a b <+≤. 4．若方程2310x x --=的两根也是方程420x ax bxc +++=的根，则2a b c +-的值为（）A ．－13.B ．－9.C ．6.D ． 0.5．在△ABC 中，已知︒=∠60CAB ，D ，E 分别是边AB ，AC 上的点，且︒=∠60AED ，CE DB ED =+，CDE CDB ∠=∠2,则=∠DCB （）A ．15°.B ．20°.C ．25°.D ．30°.6．对于自然数n ，将其各位数字之和记为n a ，如2009200911a =+++=，201020103a =+++=，12320092010a a a a a +++++=（）A ．28062.B ．28065.C ．28067.D ．28068.二、填空题：（本题满分28分，每小题7分）1．已知实数,x y 满足方程组3319,1,x y x y ⎧+=⎨+=⎩则22x y += .2．二次函数c bx x y ++=2的图象与x 轴正方向交于A ，B 两点，与y 轴正方向交于点C ．已知AC AB 3=，︒=∠30CAO ，则c = ．3．在等腰直角△ABC 中，AB ＝BC ＝5，P 是△ABC 内一点，且PAPC ＝5，则PB ＝_____．4．将若干个红、黑两种颜色的球摆成一行，要求两种颜色的球都要出现，且任意中间夹有5个或10个球的两个球必为同一种颜色的球.按这种要求摆放，最多可以摆放______个球.第二试（A ）一、（本题满分20分）设整数,,a b c （a b c ≥≥）为三角形的三边长，满足22213a b c ab ac bc ++---=，求符合条件且周长不超过30的三角形的个数.二、（本题满分25分）已知等腰三角形△ABC 中，AB ＝AC ，∠C 的平分线与AB 边交于点P ，M 为△ABC 的内切圆⊙I 与BC 边的切点，作MD//AC ，交⊙I 于点D.证明：PD 是⊙I 的切线.三、（本题满分25分）已知二次函数2y x bx c =+-的图象经过两点P (1,)a ，Q (2,10)a . N（1）如果,,a b c 都是整数，且8c b a <<，求,,a b c 的值.（2）设二次函数2y x bx c =+-的图象与x 轴的交点为A 、B ，与y 轴的交点为C.如果关于x 的方程20x bx c +-=的两个根都是整数，求△ABC 的面积.第二试（B ）一、（本题满分20分）设整数,,a b c 为三角形的三边长，满足22213a b c ab ac bc ++---=，求符合条件且周长不超过30的三角形的个数（全等的三角形只计算1次）.二、（本题满分25分）题目与（A ）卷第二题相同.三、（本题满分25分）题目与（A ）卷第三题相同.第二试（C ）一、（本题满分20分）题目与（B ）卷第一题相同.二、（本题满分25分）题目与（A ）卷第二题相同.三、（本题满分25分）设p 是大于2的质数，k 为正整数．若函数4)1(2-+++=p k px x y 的图象与x 轴的两个交点的横坐标至少有一个为整数，求k 的值．2012年全国初中数学联合竞赛试题第一试一、选择题：（本题满分42分，每小题7分）1．已知，，，那么的大小关系是（）A. B. C. D.2．方程的整数解的组数为（）A．3. B．4. C．5. D．6.3．已知正方形ABCD的边长为1，E为BC边的延长线上一点，CE＝1，连接AE，与CD交于点F，连接BF并延长与线段DE交于点G，则BG的长为（）A． B． C． D．4．已知实数满足，则的最小值为（）A．. B．0. C．1. D．.5．若方程的两个不相等的实数根满足，则实数的所有可能的值之和为（）A．0. B．. C．. D．.6．由1，2，3，4这四个数字组成四位数（数字可重复使用），要求满足.这样的四位数共有（）A．36个. B．40个. C．44个. D．48个.二、填空题：（本题满分28分，每小题7分）1．已知互不相等的实数满足，则．2．使得是完全平方数的整数的个数为．3．在△ABC中，已知AB＝AC，∠A＝40°，P为AB上一点，∠ACP＝20°，则＝．4．已知实数满足，，，则＝．第二试（A）一、（本题满分20分）已知直角三角形的边长均为整数，周长为30，求它的外接圆的面积.二、（本题满分25分）如图，PA为⊙O的切线，PBC为⊙O的割线，AD⊥OP于点D.证明：.三、（本题满分25分）已知抛物线的顶点为P，与轴的正半轴交于A、B（）两点，与轴交于点C，PA是△ABC的外接圆的切线.设M，若AM//BC，求抛物线的解析式.第二试（B）一、（本题满分20分）已知直角三角形的边长均为整数，周长为60，求它的外接圆的面积.二、（本题满分25分）如图，PA为⊙O的切线，PBC为⊙O的割线，AD⊥OP于点D，△ADC的外接圆与BC的另一个交点为E.证明：∠BAE＝∠ACB.三、（本题满分25分）题目与（A）卷第三题相同.第二试（C）一、（本题满分20分）题目与（B）卷第一题相同.二、（本题满分25分）题目与（B）卷第二题相同.三、（本题满分25分）已知抛物线的顶点为P ，与轴的正半轴交于A 、B（）两点，与轴交于点C ，PA 是△ABC 的外接圆的切线.将抛物线向左平移个单位，得到的新抛物线与原抛物线交于点Q ，且∠QBO ＝∠OBC.求抛物线的解析式.2011年四川初中数学联赛初赛试题一、选择题:（本题满分42分，每小题7分） 1．如果a ，b ，c 是三个任意的数，那么2b a +，2c b +，2ac +这三个数一定（） A.都是整数. B.都不是整数. C.至多有两个整数. D.至少有一个整数. 2.关于x 的方程m x x =+-1||22恰好有3个不同的实数根，则实数m 的值等于（） A.1-. B.0. C.1. D.2.3.ABC ∆中，BAC ∠的平分线交BC 于D ，若BD AB BC +=，︒=∠30C ，则B ∠的度数等于（）A. 45.B. 60.C. 75.D. 90.4.在1，2，3，…，100这100个数之间添上（99个）“+”号或“－”号，使算式的代数和为4150，则“－”号至少可添的个数是（）A.4.B.5.C.6.D.7.5.点P 是矩形ABCD 内部的一点，满足6=PA ，8=PB ，10=PC ，则PD 等于（） A.25. B.35. C.26. D.28.6.设正数a 、b 、c 、x 、y 、z 满足c by ax =+，a cx bz =+，b az cy =+，则以a 、b 、c 为边长的三角形一定是（）A.锐角三角形.B.直角三角形.C.钝角三角形.D.形状不等确定. 二、填空题:（本题满分28分，每小题7分） 1.已知131+=a ，131-=b ，则baa b +的值为 . 2.如图，矩形ABCD 中，8=AB ，6=AD ，将BDC ∆沿BD 对折为BDE ∆，再将点B 对折与点A 重合，则折痕MN 的长度为 .3.若方程0132=+-x x 的两根也是方程024=+-q px x 的根，则()11q p +的个位数字是．4.在正方形ABCD 中，P 、Q 分别是BC 、CD 上的点，满足︒=∠20BAP ，︒=∠45PAQ ，则AQP ∠的度数为 .三、（本题满分20分）已知抛物线()02a c bx ax y ++=与直线()412k x k y --=．无论k 取任何实数，此抛物线与直线都只有一个公共点，求抛物线的解析式.DM CBNE A四、（本题满分25分）如图，ABC ∆与ADE ∆都是等腰直角三角形，其中 90=∠=∠DAE BAC ，点M 是线段BE 的中点，求证：DC AM ⊥．五、（本题满分25分）已知a 为实数，若关于x 的方程0143||214442=-+-+a x x x x 有实数解，求实数a 的取值范围.2011年四川初中数学联赛决赛试题一、选择题：（本题满分42分，每小题7分）1．一个凸多边形的每一个内角都等于150°，则这个凸多边形所有对角线的条数总共有（） A ．42条. B ．54条. C ．66条. D ．78条.2．如图，矩形ABCD 的对角线相交于O ，AE 平分∠BAD 交BC 于E ．若∠CAE＝15°，则∠BOE ＝（） A ．30°. B ．45°. C ．60°. D ．75°. 3．设方程()()0x a x b x ---=的两根是c ，d ，则方程()()0x c x d x --+=的根分别是（）A ．a ，b.B ．－a ，－b.C ．c ，d.D ．－c ，－d. 4．若不等式2133x x a -+-≤有解，则实数a 的最小值是（）A ．1.B ．2.C ．4.D ．6.5．若一个三角形的任意两条边都不相等，则称它为“不规则三角形”．用一个正方体上的任意三个顶点构成的所有三角形中，“不规则三角形”的个数是（） A ．18 B ．24 C ．30 D ．36.6．不定方程2225x y -=的正整数解（x ，y ）的组数是（） A ．0组. B ．2组. C ．4组. D ．无穷多组．二、填空题:（本大题满分28分，每小题7分）1．二次函数22y x ax =-+的图象关于直线x=1对称，则y 的最小值是__________． 2．已知1a ，则20122011201022a a a +-的值为_____________．3．已知△ABC 中，AB，BC ＝6，CAM 是BC 的中点，过点B 作AM 延长线的垂线，垂足为D ，则线段BD 的长度是_______________．4．一次棋赛，有n 个女选手和9n 个男选手参赛，每位选手都与其余10n －1个选手各对局一次．计分方式为：胜者得2分，负者得0分，平局各得1分．比赛结束后统计发现，所有男选手的得分总和是所有女选手得分总和的4倍．则n 的所有可能值是__________．三、（本题满分20分）已知x 1，x 2是关于x 的一元二次方程22(31)210x a x a +-+-=的两个实数根，使得1212(3)(3)80x x x x --=-成立．求实数a 的所有可能值．DM CBEAO EDCBA四、（本题满分25分）抛物线2y ax bx c =++的图象与x 轴有两个交点M （x 1，0），N （x 2，0），且经过点A （0，1），其中0<x 1<x 2．过点A 的直线l 与x 轴交于点C ，与抛物线交于点B （异于点A ），满足△CAN 是等腰直角三角形，且S △BMN =52S △AMN ．求该抛物线的解析式．五、（本题满分25分）如图，AD 、AH 分别是△ABC （其中AB>AC）的角平分线、高线，M 是AD 的中点．△MDH 的外接圆交CM 于E ．求证：∠AEB ＝90°．2012年四川初中数学联赛初赛试题一、选择题:（本题满分42分，每小题7分） 1．已知关于x 的方程3x+a=0的根比关于x 的方程5x －a=0的根大2，那么a 的值为（）A ．415-. B.415. C.41-. D.45. 2．设a a 312=+，b b 312=+且a ≠b ，则代数式2211ba +的值为（）A.5.B.7.C.9.D.11.3．如图，直线AB ，CD 相交于点O ，∠AOD=30°，半径为1cm 的⊙P 的圆心在射线OA 上，且与点O 得距离为8cm.如果⊙P 以1cm/秒的速度沿由A 到B 的方向移动，那么⊙P 与直线CD 相切所需的时间为（）秒A.6.B.8.C.10.D.6或10. 4．已知7=a,70=b,则9.4等于（）A.10b a +. B.10a b -. C.a b . D.10ab.5．已知0221≠+=+b ab a ，则b a 为（）A.－1.B.1.C.2.D.4.6．如图所示，在梯形ABCD 中，AB ∥CD,AC 交BD 于O,MON ∥AB,且MON 分别交AD 、BC 于M 、N ，则CDMNAB MN +等于（ A.1. B. 2. C.3. D.4.二、填空题：（本题满分28分，每小题7分） 1．有一列数，按1,2,3,4,3,2,1,2,3,4,3,2,1,2，…的规律排列，那么从左往右数，第2012个位置上的数是 .EHMDCBA2．若函数y=kx与函数y=2x的图象交于A、C两点，AB垂直x轴于B，则△ABC的面积为 .3．如图，在平面上将△ABC绕点B旋转到△A′BC′的位置时，AA′∥BC，∠ABC=70°，则∠CBC′= .4．如图，大圆O的直径AB=12cm，分别以OA，OB为直径作圆1O和圆2O，并在圆O与圆1O和圆2O的空隙间作两个等圆圆3O和圆4O，这些圆相互内切或外切，则四边形1423OO O O的面积为 cm2.三、（本题满分20分）如图，一次函数y=－2x+8的图象与两坐标轴分别交于P、Q两点，在线段PQ上有一点A，过A点分别作两坐标轴的垂线，垂足分别为B、C.(1)若矩形ABOC的面积为4，求A点坐标；(2)若点A在线段PQ上移动，求矩形ABOC面积的最大值.四、（本题满分25分）如图，在△ABC中，D为AC边上一点，且AD=DC+CB,过D作AC的垂线交△ABC的外接圆于M，过M作AB的垂线MN，交圆于N，求证：MN为△ABC外接圆的直径.五、（本题满分25分）已知方程组⎪⎩⎪⎨⎧=+=++=++azxyazxyzxyzyx2的所有各组解（x,y,z）都是由正实数组成的，其中a是参数.试求a的取值范围.一、2012年四川初中数学联赛决赛试题一、选择题：(每小题7分，共42分)1．若－3＜x＜－1，则化简2|1|x-+得( )A．1－x. B．－3＋x. C．3－x. D．3＋x.2．若抛物线y＝x2－4x＋m的顶点在x轴上，则m的值是( )A．0. B．1. C．2. D．4.3．菱形ABCD的边长为1，面积为79，则AC＋BD的值是( )A．43. B．169. C．83. D．329.4．在凸四边形ABCD中，AB＝2AD，BC＝1，∠ABC＝∠BCD＝60°，∠ADC＝90°，则AB的长度是( )A．. B．.C．2. D．3.5．一个活动小组，如果有5个13岁的成员退出，或者有5个17岁的人员加入(两种情况不同时发生)，其成员的平均年龄都增加1岁，则这个活动小组原有成员的人数是( ) A ．10. B ．12. C ．14. D ．16.6．一个正整数，如果它顺着数和倒着数都是一样的，则称这个数为“回文数”．比如：1、11、121都是回文数，而110则不是回文数，将所有“回文数”从小到大排成一列：1、2、…、9、11、22、…，则第2012个“回文数”是( )A ．1011101.B ．1013101.C ．1021201.D ．1030301. 二、填空题：(每小题7分，共28分)1．设1x 、2x 是方程x2－2x －m ＝0的两根，且122x x +＝0，则m 的值是＿＿＿＿＿． 2．在△ABC 中，∠ACB ＝45°，D 是AB 边上异于A 、B 两点的任意一点，△ABC 、△ADC 和△BDC 的外接圆圆心分别为O 、1O 、2O ，则∠12O OO 的度数等于＿＿＿＿．3．已知a ，b 为正实数，m 为正整数，且满足14,48,a b ab m +≤⎧⎨≥+⎩则m 的值是＿＿＿＿＿．4．在一次球类比赛中有8个队参赛，每两队要进行一场比赛，胜一场得2分，平一场得1分，负一场得0分．一个队要确保进入前四名(即积分至少要超过其他四个队)，则他的积分最少是＿＿＿＿＿＿．三、（本题满分20分）已知抛物线2y x =与直线(2)(21)y k x k =+--．(1)求证：无论k 为什么实数，该抛物线与直线恒有两个不同的交点；(2)设该抛物线与直线的两个不同的交点分别为A(1x ，1y )，B(2x ，2y )，若1x ，2x 均为整数，求实数k 的值．四、（本题满分25分）如图，已知⊙A 与⊙B 相交于C 、D 两点，延长AC 交⊙B 于E ，延长BC 交⊙A 于F ．求证：C 是△DEF 的内心．五、（本题满分25分）将10，11，12，…，98，99这90个正整数写在黑板上，擦去其中的n 个数，可使黑板上剩下的所有数的乘积的个位数是1，求n 的最小值．二、 2013年四川初中数学联赛初赛试题一、选择题：（本题满分42分，每小题7分）1．已知10x ，则2x ，x ，1x的大小关系是（）A ．21x xx B ．21x x x C ．21x x x D ．21x x x2．如图，正方形ABCD ，点P 是对角线AC 上一点，连接BP ，过P 作PQ ⊥BP ，PQ 交CD 于Q ，若AP ＝CQ ＝2，则正方形ABCD 的面积为A ．642B ．16C ．1282D ．323．若实数a ，b 满足2220ba b ，则a 的取值范围是（） A ． a ≤－1 B ：a ≥－1 C ：a ≤1 D ：a ≥14．如图，在四边形ABCD 中，∠B=135°，∠C=120°，，BC=33，CD=6，则AD 边的长为（）A ．B ．C ．D ．5．方程1137x y 的正整数解(,)x y 的组数是（） A ．0B ．1C ．3D ．5 6．已知实数,,x y z 满足1x y z y z z x x y ，则222x y z y z z x x y 的值是（） A ．1 B ．0 C ．1D ．2二、填空题：（本题满分28分,每小题7分）1．x 是正整数，○x 表示x 的正约数个数，则③×④÷⑥等于． 2．草原上的一片青草，到处长得一样密一样快，70头牛在24天内可以吃完这片青草，30头牛在60天内可以吃完这片青草，则20头牛吃完这片青草需要的天数是． 3．如图，在平行四边形ABCD 中，M 、N 分别是BC 、DC 的中点，AM=4，AN=3，且角MAN=60°，则AB 的长是．4、小明将1，2，3，…，n 这n 个数输入电脑求其平均值，当他认为输完时，电脑上只显示输入(1)n 个数，且平均值为30.75，假设这(1)n 个数输入无误，则漏输入的一个数是．三、（本题满分20分）解方程2|21|20x x ．四、（本题满分25分）如图，圆内接四边形ABCD 中，CB CD ，求证：CA 2－CB 2＝AB ×AD ；五、（本题满分25分）已知二次函数2yaxbx c 和一次函数ybx ，其中a 、b 、c 满足a b c ，0a b c ．(a 、b 、c ∈R )．（1）求证：两函数的图象有两个不同的交点A 、B ；（2）过（1）中的两点A 、B 分别作x 轴的垂线，垂足为A 1、B 1．求线段A1B 1的长的取值范围．2006年全国初中数学联合竞赛试题答案第一试一、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）7．0 8．1129．8310．197第二试(A)一、（本题满分20分）解：由题可得二、（本题满分25分）三、（本题满分25分）解：第二试（B）一、（本题满分20分）题目与（A）卷第一题相同二、（本题满分20分）题目与（A）卷第二题相同三、（本题满分25分）解：第二试（C）一、（本题满分20分）题目与（A）卷第二题相同二、（本题满分20分）题目与（B）卷第三题相同三、（本题满分25分）解：2007年全国初中数学联合竞赛试题答案第一试一、选择题：（本题满分42分，每小题7分）1.B2.C3.D4.C5.A6.B（解析：1.由x z z y x +=-=532得x z x y 23,3==，所以31333525=+-=+-x x x x z y y x ，故选B. 注：本题也可用特殊值法来判断.2．因为=+-++-222211)1(1)1(1n n n n 011112222=+-++-n n n n ，即当x 分别取值n 1，n n (为正整数）时，计算所得的代数式的值之和为0；而当1=x 时，0111122=+-.因此，当x 分别取值20071，20061，20051，…，21，1，2，…，2005，2006，2007时，计算所得各代数式的值之和为0.故选C.3. 由题意可得⎪⎪⎩⎪⎪⎨⎧-=----=---,5822,1)2(2b b a c b a b a c 即⎪⎩⎪⎨⎧==+,53,2b c a c b 所以b c 53=，b a 54=，因此222b c a =+，所以△ABC 是直角三角形. 故选D.4. 锐角△ABC 的垂心在三角形内部，如图，设△ABC 的外心为O ，D 为BC 的中点，BO 的延长线交⊙O 于点E ，连CE 、AE ，则CE //AH ，AE //CH ，则OD CE AH OB 2===，所以∠OBD ＝30°，∠BOD ＝60°，所以∠A ＝∠BOD ＝60°.故选C.5． A.分别延长KD 、KE 、KF ，与△ABC 的三边AB 、BC 、CA 交于点M 、N 、P ，由于D 、E 、F 分别为△KAB 、△KBC 、△KCA 的重心，易知M 、N 、P 分别为AB 、BC 、CA 的中点，所以ABC MNP S S △△41=.易证△DEF ∽△MNP ，且相似比为3:2，所以MNP DEF S S △△2)32(=ABC S △4194⋅=ABC S △91=.所以:DEF S △19ABC S =△.故选A. 6．设摸出的15个球中有x 个红球、y 个黑球、z 个白球，则z y x ,,都是正整数，且7,6,5≤≤≤z y x ，15=++z y x .因为13≤+z y ，所以x 可取值2，3，4，5.当2=x 时，只有一种可能，即7,6==z y ；当3=x 时，12=+z y ，有2种可能，7,5==z y 或6,6==z y ；当4=x 时，11=+z y ，有3种可能，7,4==z y 或6,5==z y 或5,6==z y ；当5=x 时，10=+z y ，有4种可能，7,3==z y 或6,4==z y 或5,5==z y 或4,6==z y .因此，共有1＋2＋3＋4＝10种可能的摸球结果，其中摸出的球中恰好有3个红球的结果有2种，所以所求的概率为51102=.故选B.）二、填空题：（本题满分28分，每小题7分） 1.1 2. 10034016- 3.4 4.7 （解析：1.∵12121+=-=x ，而3122<+<，∴122-=-=x a . 又∵12--=-x ，而2123-<--<-，∴22)3(-=---=x b .∴1=+b a ，∴=++ab b a 333=++-+ab b ab a b a 3))((221)(3222=+=++-b a ab b ab a . 2.由根与系数的关系得2+=+n b a n n ，22n n a b n ⋅=-，所以 =--)2)(2(n n b a (2-n n b a 4)++n n b a 222(2)42(1)n n n n =--++=-+，则11111()(2)(2)2(1)21n n a b n n n n =-=----++， )2)(2(122--b a )2)(2(133--+b a +)2)(2(120072007--+b a ＝11111111111003()()()()22334200720082220084016⎡⎤--+-++-=--=-⎢⎥⎣⎦. 3.延长CD 交⊙O 于点G ，设DG BE ,的中点分别为点N M ,，则易知DN AM =.因为10==CD BC ，由割线定理，易证DG BF =，所以42)(2)(2==-=-=-=-AB AM BM DN BM DG BE BF BE .4．设264100m a =+，264201n a =+，则100,32<≤n m ，两式相减得))((10122m n m n m n a -+=-=，因为101是质数，且101101<-<-m n ，所以101=+m n ，故1012-=-=n m n a .代入264201n a =+，整理得020*******=+-n n ，解得59=n ，或343=n （舍去）.所以171012=-=n a .）第二试（A ）一、（本题满分20分）解：因为一元二次方程03)3(2=--+mt x mt x 的两根分别为mt 和3-，所以二次函数mt x mt x y 3)3(2--+=的图象与x 轴的两个交点间的距离为3mt + （5分）由题意，32mt t n +≥+，即22(3)(2)mt t n +≥+，即222(4)(64)90m t m n t n -+-+-≥（10分）由题意知，042≠-m ，且上式对一切实数t 恒成立，所以⎪⎩⎪⎨⎧≤----=∆>-,0)9)(4(4)46(,042222n m n m m （15分） 22,4(6)0,m mn >⎧⇒⇒⎨-≤⎩⎩⎨⎧=>,6,2m n m 所以⎩⎨⎧==,2,3n m 或⎩⎨⎧==.1,6n m （20分）二、（本题满分25分）证明:设MN 与EF 交于点P ，∵NE //BC ， ∴△PNE ∽△PBC ，∴PCPE PB PN =, ∴PC PN PE PB ⋅=⋅.(5分)又∵ME //BF ，∴△PME ∽△PBF ，∴PF PE PB PM =, ∴PF PM PE PB ⋅=⋅.(10分)∴PF PM PC PN ⋅=⋅,故PFPC PN PM =(15分) 又∠FPN =∠MPE ，∴△PNF ∽△PMC ，∴∠PNF ＝∠PMC ，∴NF//MC(20分)∴∠ANF ＝∠EDM.又∵ME//BF ，∴∠FAN ＝∠MED. A B C D E FM N P∴∠ANF ＋∠FAN ＝∠EDM ＋∠MED ，∴∠AFN=∠DME.(25分)三、（本题满分25分）解：观察易知，方程有一个整数根11=x ，将方程的左边分解因式，得[]056)18()1(2=+++-x a x x （5分）因为a 是正整数，所以关于x 的方程056)18(2=+++x a x （1）的判别式0224)18(2>-+=∆a ，它一定有两个不同的实数根.而原方程的根都是整数，所以方程（1）的根都是整数，因此它的判别式224)18(2-+=∆a 应该是一个完全平方数.（10分）设22224)18(k a =-+（其中k 为非负整数），则224)18(22=-+k a ，即 224)18)(18(=-+++k a k a .（15分）显然k a ++18与k a -+18的奇偶性相同，且1818≥++k a ，而8284562112224⨯=⨯=⨯=，所以⎩⎨⎧=-+=++,218,11218k a k a 或⎩⎨⎧=-+=++,418,5618k a k a 或⎩⎨⎧=-+=++,818,2818k a k a 解得⎩⎨⎧==,55,39k a 或⎩⎨⎧==,26,12k a 或⎩⎨⎧==,10,0k a 而a 是正整数，所以只可能⎩⎨⎧==,55,39k a ⎩⎨⎧==.26,12k a （20分）当39=a 时，方程（1）即056572=++x x ，它的两根分别为1-和56-.此时原方程的三个根为1，1-和56-.当12=a 时，方程（1）即056302=++x x ，它的两根分别为2-和28-.此时原方程的三个根为1，2-和28-.（25分）第二试（B ）一、（本题满分20分）解：因为一元二次方程03)3(2=--+mt x mt x 的两根分别为mt 和3-，所以31+=mt d ；一元二次方程02)2(2=+-+-nt x n t x 的两根分别为t 2和n -，所以n t d +=22.（5分）所以，21d d ≥22)2()3(23n t mt n t mt +≥+⇔+≥+⇔09)46()4(222≥-+-+-⇔n t n m t m （1）（10分）由题意知，042≠-m ，且（1）式对一切实数t 恒成立，所以⎪⎩⎪⎨⎧≤----=∆>-,0)9)(4(4)46(,042222n m n m m （15分） 22,4(6)0,m mn >⎧⇒⇒⎨-≤⎩⎩⎨⎧=>,6,2m n m 所以⎩⎨⎧==,2,3n m ⎩⎨⎧==.1,6n m （20分）二、（本题满分25分）题目与（A ）卷第二题相同.三、（本题满分25分）解：联立方程组⎪⎩⎪⎨⎧=-+++=,56,38)17(2x y a x a x y 消去y 得a x a x -+++38)17(2x56=，即056)38()17(23=--+++x a x a x ，分解因式得[]056)18()1(2=+++-x a x x （1）（5分）显然11=x 是方程（1）的一个根，（1，56）是两个函数的图象的一个交点.因为a 是正整数，所以关于x 的方程056)18(2=+++x a x （2）的判别式0224)18(2>-+=∆a ，它一定有两个不同的实数根.（10分）而两个函数的图象的交点都是整点，所以方程（2）的根都是整数，因此它的判别式224)18(2-+=∆a 应该是一个完全平方数.设22224)18(k a =-+（其中k 为非负整数），则224)18(22=-+k a ，即 224)18)(18(=-+++k a k a .（15分）显然k a ++18与k a -+18的奇偶性相同，且1818≥++k a ，而8284562112224⨯=⨯=⨯=，所以⎩⎨⎧=-+=++,218,11218k a k a 或⎩⎨⎧=-+=++,418,5618k a k a 或⎩⎨⎧=-+=++,818,2818k a k a 解得⎩⎨⎧==,55,39k a 或⎩⎨⎧==,26,12k a 或⎩⎨⎧==,10,0k a 而a 是正整数，所以只可能⎩⎨⎧==,55,39k a 或⎩⎨⎧==.26,12k a （20分）当39=a 时，方程（2）即056572=++x x ，它的两根分别为1-和56-，此时两个函数。

2006年全国高中数学联赛试题及详细解析

一、选择题（本题满分36分，每小题6分）1. 已知△ABC ，若对任意R t ∈≥-，则△ABC 一定为A ．锐角三角形 B. 钝角三角形 C. 直角三角形 D. 答案不确定【答案】（）2. 设2log (21)log 2 1x x x x +->-，则x 的取值范围为A ．112x << B ．1, 12x x >≠且 C ． 1x > D ． 01x <<【答案】（） 5. 设()322()log 1f x x x x =+++，则对任意实数,a b ，0a b +≥是()()0f a f b +≥的A. 充分必要条件B. 充分而不必要条件C. 必要而不充分条件D. 既不充分也不必要条件【答案】（） 6. 数码1232006,,,,a a a a 中有奇数个9的2007位十进制数12320062a a a a 的个数为A ．200620061(108)2+ B ．200620061(108)2- C ．20062006108+ D ．20062006108-【答案】（）二、填空题（本题满分54分，每小题9分）7. 设x x x x x f 44cos cos sin sin )(+-=，则)(x f 的值域是。

8. 若对一切θ∈R ，复数(cos )(2sin )i z a a θθ=++-的模不超过2，则实数a 的取值范围为.9. 已知椭圆221164x y +=的左右焦点分别为1F 与2F ，点P 在直线l ：80x -++=上. 当12F PF ∠取最大值时，比12PF PF 的值为.10. 底面半径为1cm 的圆柱形容器里放有四个半径为21cm 的实心铁球，四个球两两相切，其中底层两球与容器底面相切. 现往容器里注水，使水面恰好浸没所有铁球，则需要注水cm 3. 11. 方程20062420042005(1)(1)2006xx x x x +++++=的实数解的个数为.12. 袋内有8个白球和2个红球，每次从中随机取出一个球，然后放回1个白球，则第4次恰好取完所有红球的概率为.三、解答题（本题满分60分，每小题20分） 15. 设2()f x x a =+. 记1()()f x f x =，1()(())n n f x f f x -=2,3,n =，，{}R (0)2n M a n f =∈≤对所有正整数，. 证明：⎥⎦⎤⎢⎣⎡-=41 ,2M .2006年全国高中数学联合竞赛加试试卷（考试时间：上午10：00—12：00）一、以B 0和B 1为焦点的椭圆与△AB 0B 1的边AB i 交于C i （i =0，1）。

2006年AMC8试题

(A) 12 (B) 16 (C) 20 (D) 22 (E) 24
7 Circle X has a radius of π. Circle Y has a circumference of 8π. Circle Z has an area of 9π. List the circles in order from smallest to largest radius.
(A) 425 (B) 444 (C) 456 (D) 484 (E) 506
16 Problems 14, 15 and 16 involve Mrs. Reed’s English assignment.
A Novel Assignment
The students in Mrs. Reed’s English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
USA
AMC 8 2006
1 Mindy made three purchases for $1.98, $5.04 and $9.89. What was her total, to the nearest dollar?
(A) $10 (B) $15 (C) $16 (D) $17 (E) $18
3 Elisa swims laps in the pool. When she ﬁrst started, she completed 10 laps in 25 minutes.

2006年全国高中数学联赛试题及答案

2006年全国高中数学联合竞赛试题1. 已知△ABC ，若对任意R t ∈，BA tBC AC -≥，则△ABC 一定为( )A ．锐角三角形 B. 钝角三角形 C. 直角三角形 D. 答案不确定2. 设2log (21)log 2 1x x x x +->-，则x 的取值范围为( )A ．112x <<B ．1, 12x x >≠且 C ． 1x > D ． 01x <<3. 已知集合{}50A x x a =-≤，{}06>-=b x x B ,N b a ∈,，且{}2,3,4A B N ⋂⋂=，则整数对()b a ,的个数为( ) A. 20 B. 25 C. 30 D. 424. 在直三棱柱111A B C ABC -中，2BAC π∠=，11AB AC AA ===. 已知Ｇ与Ｅ分别为11A B 和1CC 的中点，Ｄ与Ｆ分别为线段AC 和AB 上的动点（不包括端点）. 若GD EF ⊥，则线段DF 的长度的取值范围为( )A. 1⎫⎪⎭B.1, 25⎡⎫⎪⎢⎣⎭C. 1,⎡⎣D. 5.设(32()log f x x x =+，则对任意实数,a b ，0a b +≥是()()0f a f b +≥的( )A. 充分必要条件B. 充分而不必要条件C. 必要而不充分条件D. 既不充分也不必要条件6. 数码1232006,,,,a a a a 中有奇数个9的2007位十进制数12320062a a a a 的个数为( )A ．200620061(108)2+B ．200620061(108)2- C ．20062006108+ D ．20062006108-二.填空题（本题满分54分，每小题9分）7. 设x x x x x f 44cos cos sin sin )(+-=，则)(x f 的值域是。

8. 若对一切θ∈R ，复数(cos )(2sin )i z a a θθ=++-的模不超过2，则实数a 的取值范围为______.9. 已知椭圆221164x y +=的左右焦点分别为1F 与2F ，点P 在直线l：80x ++=上. 当12F PF ∠取最大值时，比12PF PF 的值为_______. 10. 底面半径为1cm 的圆柱形容器里放有四个半径为21cm 的实心铁球，四个球两两相切，其中底层两球与容器底面相切. 现往容器里注水，使水面恰好浸没所有铁球，则需要注水_______.11.方程20062420042005(1)(1)2006x x x x x +++++= 的实数解的个数为12.袋内有8个白球和2个红球，每次从中随机取出一个球，然后放回1个白球，则第4次恰好取完所有红球的概率为 .三. 解答题（本题满分60分，每小题20分） 13. 给定整数2n ≥，设 ),(000y x M 是抛物线12-=nx y 与直线x y =的一个交点.试证明对于任意正整数m ，必存在整数2k ≥，使),(00m m y x 为抛物线12-=kx y 与直线x y =的一个交点.14. 将2006表示成5个正整数12345,,,,x x x x x 之和. 记15i j i j S x x ≤<≤=∑. 问：当12345,,,,x x x x x 取何值时，S 取到最大值；进一步地，对任意1,5i j ≤≤有2i j x x -≤，当12345,,,,x x x x x 取何值时，S 取到最小值. 说明理由.15. 设 2()f x x a =+. 记1()()f x f x =，1()(())n n f x f f x -=2,3,n = ，，{}R (0)2n M a n f =∈≤对所有正整数，. 证明：⎥⎦⎤⎢⎣⎡-=41 ,2M .二○○五年全国高中数学联合竞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。

2001-2006-AMC8-中文试题和答案

2001 年美国AMC8 (2001年月日时间40分钟)1. 卡西的商店正在制作一个高尔夫球奖品。

他必须给一颗高尔夫球面上的300个小凹洞着色，如果他每着色一个小凹洞需要2秒钟，试问共需多分钟才能完成他的工作。

(A) 4 (B) 6 (C) 8 (D) 10 (E) 12 。

2. 我正在思考两个正整数，它们的乘积是24且它们的和是11，试问这两个数中较大的数是什么。

(A) 3 (B) 4 (C) 6 (D) 8 (E) 12 。

3. 史密斯有63元，艾伯特比安加多2元，而安加所有的钱是史密斯的三分之一，试问艾伯特有元。

(A) 17 (B) 18 (C) 19 (D) 21 (E) 23 。

4. 在每个数字只能使用一次的情形下，将1，2，3，4及9作成最小的五位数，且此五位数为偶数，则其十位数字为。

(A) 1 (B) 2 (C) 3 (D) 4 (E) 9 。

5. 在一个暴风雨的黑夜里，史努比突然看见一道闪光。

10秒钟后，他听到打雷声音。

声音的速率是每秒1088呎，但1哩是5280呎。

若以哩为单位的条件下，估计史努比离闪电处的距离最接近下列何者。

(A) 1 (B) 1 (C) 2 (D) 2 (E) 3 。

6. 在一笔直道路的一旁有等间隔的6棵树。

第1棵树与第4棵树之间的距离是60呎。

试问第1棵树到最后一棵树之间的距离是呎。

(A) 90 (B) 100 (C) 105 (D) 120 (E) 140 。

问题7、8、9请参考下列叙述：主题：竞赛场所上的风筝展览7. 葛妮芙为提升她的学校年度风筝奥林匹亚竞赛的质量，制作了一个小风筝与一个大风筝，并陈列在公告栏展览，这两个风筝都如同图中的形状，葛妮芙将小风筝张贴在单位长为一吋(即每两点距离一吋)的格子板上，并将大风筝张贴在单位长三吋(即每两点距离三吋)的格子板上。

试问小风筝的面积是平方吋。

(A) 21 (B) 22 (C) 23 (D) 24 (E) 25。

2006年AMC8试题

INSTRUCTIONS1. DO NOT OPEN THIS BOOKLET UNTIL YOUR PROCTOR TELLS YOU.2. This is a twenty-five question multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.3. Mark your answer to each problem on the AMC 8 Answer Form with a #2 pencil. Check the blackened circles for accuracy and erase errors and stray marks completely. Only answers properly marked on the answer form will be graded.4. There is no penalty for guessing. Your score on this test is the number of correct answers.5. No aids are permitted other than scratch paper, graph paper, rulers, erasers, and cal-culators that are accepted for use on the SAT. No problems on the test will require the use of a calculator.6.Figures are not necessarily drawn to scale.7. Before beginning the test, your proctor will ask you to record certain information on the answer form.8. When your proctor gives the signal, begin working on the problems. You will have 40 minutes to complete the test.9. When you finish the exam, sign your name in the space provided on the Answer Form.The Committee on the American Mathematics Competitions reserves the right to re-examine students before deciding whether to grant official status to their scores. The Committee also reserves the right to disqualify all scores from a school if it determines that the required security procedures were not followed.The publication, reproduction or communication of the problems or solutions of the AMC 8 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via copier, telephone, e-mail, World Wide Web or media of any type during this period is a violation of the competition rules. After the contest period, permission to make copies of individual problems in paper or electronic form including posting on web-pages for educational use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear the copyright notice.Cop yright © 2006, The Mathematical Association of AmericaT he M aTheMaTical a ssociaTion of a MericaAmerican Mathematics Competitions 22nd AnnualAMC 8(American Mathematics Contest 8)T uesday, NOVEMBER 14, 20061.Mindy made three purchases for $1.98,$5.04and $9.89.What was her total,to the nearest dollar?(A)$10(B)$15(C)$16(D)$17(E)$182.On the AMC 8contest Billy answers 13questions correctly,answers 7questions incorrectly and doesn’t answer the last 5.What is his score?(A)1(B)6(C)13(D)19(E)263.Elisa swims laps in the pool.When she ﬁrst started,shecompleted 10laps in 25minutes.Now she can ﬁnish 12lapsin 24minutes.By how many minutes has she improved herlap time?(A)12(B)34(C)1(D)2(E)34.Initially,a spinner points west.Chenille moves it clockwise214revolutions and then counterclockwise 334revolutions.In what direction does the spinner point after the twomoves?(A)north (B)east (C)south (D)west (E)northwest 5.Points A ,B ,C and D are midpoints of the sides of the larger square.If the larger square has area 60,what is the area of the smaller square?ABCD (A)15(B)20(C)24(D)30(E)406.The letter T is formed by placing two2inch×4inch rectangles next to eachother,as shown.What is the perimeter of the T,in inches?(A)12(B)16(C)20(D)22(E)247.Circle X has a radius ofπ.Circle Y has a circumference of8π.Circle Z has anarea of9π.List the circles in order from smallest to largest radius.(A)X,Y,Z(B)Z,X,Y(C)Y,X,Z(D)Z,Y,X(E)X,Z,Y8.The table shows some of the results of a survey by radiostation KAMC.What percentage of the males surveyedlisten to the station?Listen Don’t Listen TotalMale?26?Female58?96Total13664200(A)39(B)48(C)52(D)55(E)759.What is the product of32×43×54...×20062005?(A)1(B)1002(C)1003(D)2005(E)200610.Jorge’s teacher asks him to plot all the ordered pairs(w,l)of positive integersfor which w is the width and l is the length of a rectangle with area12.What should his graph look like?(A)(B)(C)(D)(E)11.How many two-digit numbers have digits whose sum is a perfect square?(A)13(B)16(C)17(D)18(E)1912.Antonette gets70%on a10-problem test,80%on a20-problem test and90%on a30-problem test.If the three tests are combined into one60-problem test, which percent is closest to her overall score?(A)40(B)77(C)80(D)83(E)8713.Cassie leaves Escanaba at8:30AM heading for Marquette on her bike.Shebikes at a uniform rate of12miles per hour.Brian leaves Marquette at9:00 AM heading for Escanaba on his bike.He bikes at a uniform rate of16miles per hour.They both bike on the same62-mile route between Escanaba and Marquette.At what time in the morning do they meet?(A)10:00(B)10:15(C)10:30(D)11:00(E)11:30Problems14,15and16involve Mrs.Reed’s English assignment.A Novel AssignmentThe students in Mrs.Reed’s English class are reading the same760-page novel.Three friends,Alice,Bob and Chandra,are in the class.Alice reads a page in 20seconds,Bob reads a page in45seconds and Chandra reads a page in30 seconds.14.If Bob and Chandra both read the whole book,Bob willspend how many more seconds reading than Chandra?(A)7,600(B)11,400(C)12,500(D)15,200(E)22,80015.Chandra and Bob,who each have a copy of the book,decide that they can savetime by“team reading”the novel.In this scheme,Chandra will read from page 1to a certain page and Bob will read from the next page through page760,ﬁnishing the book.When they are through they will tell each other about the part they read.What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?(A)425(B)444(C)456(D)484(E)50616.Before Chandra and Bob start reading,Alice says she would like to team readwith them.If they divide the book into three sections so that each reads for the same length of time,how many seconds will each have to read?(A)6400(B)6600(C)6800(D)7000(E)720017.Jeﬀrotates spinners P,Q and R and adds the resulting numbers.What is theprobability that his sum is an odd number?P Q R(A)14(B)13(C)12(D)23(E)3418.A cube with3-inch edges is made using27cubes with1-inch edges.Nineteenof the smaller cubes are white and eight are black.If the eight black cubes are placed at the corners of the larger cube,what fraction of the surface area of the larger cube is white?(A)19(B)14(C)49(D)59(E)192719.Triangle ABC is an isosceles triangle with AB=BC.Point D is the midpointof both BC and AE,and CE is11units long.Triangle ABD is congruent to triangle ECD.What is the length of BD?A BDCE(A)4(B)4.5(C)5(D)5.5(E)620.A singles tournament had six players.Each player played every other playeronly once,with no ties.If Helen won4games,Ines won3games,Janet won 2games,Kendra won2games and Lara won2games,how many games did Monica win?(A)0(B)1(C)2(D)3(E)421.An aquarium has a rectangular base that measures100cm by40cm and has aheight of50cm.The aquarium isﬁlled with water to a depth of37cm.A rock with volume1000cm3is then placed in the aquarium and completely submerged.By how many centimeters does the water level rise?(A)0.25(B)0.5(C)1(D)1.25(E)2.522.Three diﬀerent one-digit positive integers are placed in the bottom row of cells.Numbers in adjacent cells are added and the sum is placed in the cell above them.In the second row,continue the same process to obtain a number in the top cell.What is the diﬀerence between the largest and smallest numbers possible in the top cell?(A)16(B)24(C)25(D)26(E)3523.A box contains gold coins.If the coins are equally divided amongsix people,four coins are left over.If the coins are equally dividedamongﬁve people,three coins are left over.If the box holds thesmallest number of coins that meets these two conditions,howmany coins are left when equally divided among seven people?(A)0(B)1(C)2(D)3(E)524.In the multiplication problem below,A,B,C and D are diﬀerent digits.Whatis A+B?ABA×CDCDCD(A)1(B)2(C)3(D)4(E)925.Barry wrote6diﬀerent numbers,one on each side of3cards,and laid the cardson a table,as shown.The sums of the two numbers on each of the three cards are equal.The three numbers on the hidden sides are prime numbers.What is the average of the hidden prime numbers?445938(A)13(B)14(C)15(D)16(E)17SOLUTIONSYour School Manager has been sent at least one copy of the 2006 AMC 8 Solutions Pamphlet. It is meant to be loaned to students (but not duplicated).WRITE TO USComments about the problems and solutions for this AMC 8 should be addressed to:Ms. Bonnie Leitch, AMC 8 Chair / bleitch@548 Hill Avenue, New Braunfels, TX 78130Comments about administrative arrangements should be addressed to:MAA American Mathematics Competitions / amcinfo@American Mathematics Competitions, University of Nebraska-LincolnP.O. Box 880658, Lincoln, NE 68588-0658AMC 10 & AMC 12The AMC 10 and AMC 12 are 25-question, 75-minute contests with 5 choices of answers for each problem (A through E). Schools with high scoring students on the AMC 8 will receive an Invitation Brochure for the 2007 AMC 10. The best way to prepare for these upper level contests is to study exams from previous years. Orders for all publications listed below should be addressed to:American Mathematics CompetitionsATTN: PublicationsP.O. Box 81606Lincoln, NE 68501-1606PUBLICATIONSA complete listing of current publications, with ordering instructions, is at our web site:/amc.**Administration On An Earlier Date Will Disqualify Your School’s Results**1. All information (Rules and Instructions) needed to administer this exam is contained in the TEACHERS’ MANUAL, which is outside of this package. PLEASE READ THE MANUAL BEFORE NOVEMBER 14, 2006. Noth-ing is needed from inside this package until November 14.2. Your PRINCIPAL or VICE-PRINCIPAL must verify on the AMC 8 CER-TIFICATION FORM that you followed all rules associated with the conduct of the exam.3. The Answer Forms must be mailed First Class to the AMC office no later than 24 hours following the exam.4. THE AMC 8 IS TO BE ADMINISTERED DURING A CONVENIENT 40 MINUTE PERIOD. THE EXAM MAY BE GIVEN DURING A REGULAR MATH CLASS.5. The publication, reproduction or communication of the problems or solutions of this test during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination at any time via copier, telephone, e-mail, World Wide Web or media of any type is a violation of the competition rules.The American Mathematics Competitionsare Sponsored by The Mathematical Association of America –– MAA /The Akamai Foundation – /Contributors American Mathematical Association of Two Year Colleges ...................................................... /American Mathematical Society ........................................................................................... /American Society of Pension Actuaries ............................................................................... /American Statistical Association ....................................................................................... /Art of Problem Solving /Canada/USA Mathpath /Canada/USA Mathcamp /Casualty Actuarial Society ................................................................................................ /Clay Mathematics Institute ........................................................................................... /Consortium for Mathematics & its Applications /Institute for Operations Research and the Management Sciences .................................................. /L. G. Balfour Company /Mu Alpha Theta .................................................................................................... /National Assessment & Testing ................................................................../National Council of Teachers of Mathematics ...................................................................... /Pedagoguery Software Inc. /Pi Mu Epsilon ............................................................................................................. /Society of Actuaries .............................................................................................................. /U. S. A. Math Talent Search ............................................................................................ /W. H. Freeman and Company .................................................................................www. /Wolfram Research Inc. /2006AMC 8DO NOT OPEN UNTIL TUESDAY, NOVEMBER 14, 2006。

AMC8(美国数学竞赛)历年真题、答案及中英文解析

AMC8（美国数学竞赛）历年真题、答案及中英文解析艾蕾特教育的AMC8 美国数学竞赛考试历年真题、答案及中英文解析：AMC8-2020年：真题 --- 答案---解析（英文解析+中文解析）AMC8 - 2019年：真题----答案----解析（英文解析+中文解析）AMC8 - 2018年：真题----答案----解析（英文解析+中文解析）AMC8 - 2017年：真题----答案----解析（英文解析+中文解析）AMC8 - 2016年：真题----答案----解析（英文解析+中文解析）AMC8 - 2015年：真题----答案----解析（英文解析+中文解析）AMC8 - 2014年：真题----答案----解析（英文解析+中文解析）AMC8 - 2013年：真题----答案----解析（英文解析+中文解析）AMC8 - 2012年：真题----答案----解析（英文解析+中文解析）析）AMC8 - 2010年：真题----答案----解析（英文解析+中文解析）AMC8 - 2009年：真题----答案----解析（英文解析+中文解析）AMC8 - 2008年：真题----答案----解析（英文解析+中文解析）AMC8 - 2007年：真题----答案----解析（英文解析+中文解析）AMC8 - 2006年：真题----答案----解析（英文解析+中文解析）AMC8 - 2005年：真题----答案----解析（英文解析+中文解析）AMC8 - 2004年：真题----答案----解析（英文解析+中文解析）AMC8 - 2003年：真题----答案----解析（英文解析+中文解析）AMC8 - 2002年：真题----答案----解析（英文解析+中文解析）AMC8 - 2001年：真题----答案----解析（英文解析+中文解析）AMC8 - 2000年：真题----答案----解析（英文解析+中文解析）析）AMC8 - 1998年：真题----答案----解析（英文解析+中文解析）AMC8 - 1997年：真题----答案----解析（英文解析+中文解析）AMC8 - 1996年：真题----答案----解析（英文解析+中文解析）AMC8 - 1995年：真题----答案----解析（英文解析+中文解析）AMC8 - 1994年：真题----答案----解析（英文解析+中文解析）AMC8 - 1993年：真题----答案----解析（英文解析+中文解析）AMC8 - 1992年：真题----答案----解析（英文解析+中文解析）AMC8 - 1991年：真题----答案----解析（英文解析+中文解析）AMC8 - 1990年：真题----答案----解析（英文解析+中文解析）AMC8 - 1989年：真题----答案----解析（英文解析+中文解析）AMC8 - 1988年：真题----答案----解析（英文解析+中文解析）析）AMC8 - 1986年：真题----答案----解析（英文解析+中文解析）AMC8 - 1985年：真题----答案----解析（英文解析+中文解析）◆AMC介绍◆AMC（American Mathematics Competitions) 由美国数学协会（MAA）组织的数学竞赛，分为 AMC8 、 AMC10、 AMC12 。

AMC_2006

有 30 道题目的测验卷答对了 90%.若将这三份试卷合成一份有 60 道题目的测验卷，他的答对率最接近百
分之（）.
om A.49 B.77 C.80 D.83 E.87 h.c m 13.小明早上 8:30 从甲地骑脚踏車向乙地出发，他的速率是每小时 12 公里.小华早上 9:00 从乙地骑 at .co 脚踏車向甲地出发，他的速率是每小时 16 公里.他们两人骑車在連接两地长 62 公里的同一条路径上.他们 00m th 在早上（）相遇. .1 0ma A.10:00 B.10:15 C.10:30 D.11:00 E.11:30 www .10 下列第 14 题，第 15 题及第 16 题是关于锐德老师所规定的英文作业.
圣才学习网
7.圆 X 的半径为 π，圆 Y 的周长为 8π，圆 Z 的面积为 9π，则三个圆依半径由小到大排列的顺序为（）. ww 100 A.X，Y，Z B.Z，X，Y C.Y，X，Z D.Z，Y，X E.X，Z，Y 网w w. 8.下表为 KAMC 广播电台收听率调查的一些结果.在受调查的男性中有收听的占百分之（）.
25.贝克将六个相异的数分别写在三张卡片的两面上，并将卡片放在桌上，如图所示.如果每张卡片两 .10 ma 面上的数之和都相等，且被盖住的三个数都是质数，则被盖住的三个质数的平均值是（）.
中华数学竞赛网www 华数学竞赛网www.100 A.13 B.14 C.15 D.16 E.17 中
中华数学竞赛网
圣才学习网
中华数学竞赛网
圣才学习网
A.0 B.1 C.2 D.3 E.5 24.在下面的直式乘法运算中 A，B，C，D 为相异的数字.之值是（）. A.1 B.2 C.3 D.4 E.9

2006年第二学期八年级数学竞赛试题(含答案)-

二○○六学年第二学期八年级数学竞赛试题（本卷满分120分，考试用时120分钟）一、选择题（本题有6小题，每题5分，共30分）1、在关于x 1、x2、x 3的方程组中，已知P 1＞P 2＞P 3那么x 1、x 2、x 3的大小顺序是（）A. x 1＞x 2＞x 3B. x 2＞x 3＞x 1C. x 3＞x 1＞x 2D. x 2＞x 1＞x 3 2、六个全等的直角三角形拼接成如图的花环状图案，ABCDEF 和A 1B 1C 1D 1E 1F 1都是正六边形，已知∠ABB 1＝90°AB ＝3，则正六边形A 1B 1C 1D 1E 1F 1的边长为（） A.2B. 1C.D. 3、一个均匀的立方体六个面上分别标有1、2、3、4、5、6，右图是这个立方体表面的展开图，抛掷这个立方体，则朝上一面上的数恰好等于朝下一面上的数的的概率是（）A.B. C. D.4、甲、乙两个工程队完成某项工程，首先是甲单独做了10天，然后，乙队加入合做，完成剩下的全部工程，设工程总量为单位1，工程进度满足如图所示的函数关系，那么实际完成这项工程所用的时间比由甲单独完成这项工程所需的时间少（）A. 2天B. 3天C. 4天D. 5天5、如图A 、B 、C 是4×4网格中的格点，网格中的每个小正方形的边长为1，则点C 到线段AB 的距离为（） A.2B.1052C.1054D.10516、正实数a 、b 、c 满足a ＋b ＋c ＝1，设m ＝13+a ＋13+b ＋13+c ，则（） A. m ＞4B. m ＝4C. m ＜4D.m 与4的大小关系不确定x 1＋x 2＝P 1x 2＋x 3＝P 2 x 3＋x 1＝P 3212221ABDEFA B C D E F 111111(第2题图)123546(第3题图)61312132(第4题图)A(第5题图)二、填空题（本题有6小题，每题5分，共30分）7、如图，锐角△ABC 中，AD 和CE 分别是BC 和AB 边上的高，若AD 与CE夹的锐角为54º，则∠BAC ＋∠BCA 的大小是。

AMC8数学竞赛试题及详解

Problem 3What is the value of the expression ?化简的标准和顺序Problem 6If the degree measures of the angles of a triangle are in the ratio , what is the degree measure of the largest angle of the triangle?角度制与弧度制Problem 7Let be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of ?整除特征Problem 10A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?离散概率Problem 11A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?tiles英[taɪlz]美[taɪlz]n. 瓦片，瓷砖( tile的名词复数); 扁平的小棋子;Problem 13Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?Problem 14Chloe and Zoe are both students in Ms. Demeanor's math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only of the problems she solved alone, but overall of her answers were correct. Zoe had correct answersto of the problems she solved alone. What was Zoe's overall percentage of correct answers? C设数Problem 15In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path allows only moves from oneletter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.乘法原理加法原理Problem 16In the figure below, choose point on so that and have equal perimeters. What is the area of ?D如果在AC上呢Problem 17Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?盈亏问题图示法Problem 18In the non-convex quadrilateral shown below, is a rightangle, , , , and .What is the area of quadrilateral ?勾股定理与逆定理“凸”定义Problem 19For any positive integer , the notation denotes the product of the integers through . What is the largest integer for which is a factor of the sum ?Problem 20An integer between and , inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?5*8*8*7Problem 21Suppose , , and are nonzero real numbers, and . What are the possible value(s) for ?可能性列全或者变个形Problem 22In the right triangle , , , and angle is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?相切，连接切点和圆心。

美国数学竞赛AMC8 -- 2006年真题解析(英文解析+中文解析)

美国数学竞赛AMC8 – 2006年真题解析(英文解析+中文解析) \Problem 1Answer: DSolution:The three prices round to $2, $5, and $10, which has a sum of 17.中文解析：三件商品价格先近似取整，然后求和：2+5+10=17. 答案是D。

Problem 2Answer: CSolution:As the AMC 8 only rewards 1 point for each correct answer, everything is irrelevant except the number Billy answered correctly，13.中文解析：正确的题目每题1分，错误或没做的题目都是0分，做对13题的得分应该是13. 答案是C。

Problem 3Answer: ASolution:When Elisa started, she finished a lap in 25/10=2.5 minutes. Now, she finishes a lap is 24/12=2 minutes. The difference is 2.5-2=0.5中文解析：开始25分钟游10圈，平均2.5分钟游1圈。

后来24分钟游12圈，平均2分钟游1圈。

速度从2.5分钟提高到2分钟，提高了0.5分钟，即1/2 分钟。

答案是A。

Problem 4Answer: BSolution:If the spinner goes clockwise 2+1/4 revolutions and then counterclockwise 3+3/4 revolutions, it ultimately goes counterclockwise 1+1/2 which brings the spinner pointing east.中文解析：最初方向指向西，转整数圈不改变指针方向。

AMC简介及AMC8试题

参加美国数学测试（竞赛）活动，用美国原题（全是英文）、由美国老师评分、排序、评奖，只要参加都有全英文证书（成绩不同，内容不一样）。

世界统一每学年一次考试，共25道选择题，40分钟完成。

通过活动感受西方数学文化，判断自己数学的国际水平，同时测试自己的英语水平，世界近100个国家（地区）承认这个成绩，正是因为有这样的意义，所以自报名以来，得到了知道这项信息的有关家长、同学的高度重视和积极参与。

American Mathematics Competition又称为American High School Mathematics Examination美国中学数学科考试（AHSME），是由美国数学协会（Mathematics Association of America）于1950年成立，目前总部设于美国内布拉斯加大学林肯校区（University Of Nebraska-Lincoln），是美国数学协会（Mathematics Association of America）的直属机构。

AMC美国数学竞赛在1985年时，增加了初中数学科的检定考试American Junior High School Mathematics Examination（AJHSME）、每年仅在北美地区，正式登记应试的学生就超过600,000人次，也因此AMC是世界上目前信度和效度最高的数学科试题。

而全球进行同步测验的国家还有加拿大、新加坡、香港、日本、匈牙利、希腊、土耳其、法国、等二十余国。

此项测验已获美国中学校长推介为每年的主要活动之一。

AMC测验由试题研发、命制到统一阅卷等作业，完全委托素由数理闻名的内布拉斯加大学林肯校区University Of Nebraska-Lincoln数学系教授带领专家学者成立委员会全权负责。

该委员会成员皆来自全美一流学府，如麻省理工学院MIT、哈佛大学 Harvard、普林斯顿大学Princeton等名校，共同研究规划。

2006年全国高中数学联赛试题及解答

1≤i＜j≤5
Σ
xixj＝x1x2＋(x1＋x2)(x3＋x4＋x5)＋x3x4＋x3x5＋x4x5
同时有 S＝x1x2＋(x1＋x2)((x3＋x4＋x5)＋x3x4＋x3x5＋x4x5．于是有 S－S＝x1x2－x1x2＞0．这与 S 在 x1，
x2，x3，x4，x5 时取到最大值矛盾．所以必有|xi－xj|≤1，(1≤i，j≤5)．
x0
14．将 2006 表示成 5 个正整数 x1，x2，x3，x4，x5 之和．记 S＝ ⑴ 当 x1，x2，x3，x4，x5 取何值时，S 取到最大值；
1≤i＜j≤5
Σ
xixj．问：
⑵ 进一步地，对任意 1≤i，j≤5 有|xi－xj|≤2，当 x1，x2，x3，x4，x5 取何值时，S 取到最小值. 说明理由．解：(1) 首先这样的 S 的值是有界集，故必存在最大值与最小值。若 x1＋x2＋x3＋x4＋x5＝2006，且使 S＝
|→ DF |＝
3
2 t2 5t2 1 ＋t 2 ＝ 2 －4t2＋1＝
2 2 1 1 → 5ห้องสมุดไป่ตู้t2－ ) ＋，从而有 ≤ DF ＜1． 5 5 5
| |
5．设 f(x)＝x ＋log2(x＋ x ＋1)，则对任意实数 a，b，a＋b≥0 是 f(a)＋f(b)≥0 的
2
A. 充分必要条件 C. 必要而不充分条件
2
1 B．x＞且 x≠1 2
C． x＞1
D． 0＜x＜1
1 所以 x 的取值范围为 x＞且 x≠1． 2 3．已知集合 A＝{x|5x－a≤0}，B＝{x|6x－b＞0}，a，b∈N，且 A∩B∩N＝{2，3，4}，则整数对(a，b) 的个数为 A．20 答 C．解：5x－a≤0x≤ ；6x－b＞0x＞．要使 A∩B∩N＝{2，3，4}，则 5 6 B．25 C．30 D．42

2006年中国国家集训队选拔考试题解答(全)

显然 d 1(否则 a 1 a (mod n) 推出 n 1)。同时注意到(n) n ，所以 d n 。因此
(a, d ) 是原问题的另一个解并且1 d n 。重复上述过程我们就得到了一个无穷递降正整数序列，而这是不可能的，因此上面的假设是错误的，即没有 n 1的解。
3．设 a1, a2 ,, an 是给定的 n 个实数 (n 1) . 求证：存在实数 b1,b2 ,,bn 满足下列条件： (a)对任何的1 i n ， ai bi 是正整数;
a0
an
b0b1 (a0
a1
n1
)
k 0
bk
1 bk
1
，
n 1
a0
.
k 0 bk bk 1 b0b1 (a0 a1 )
()
令 xn bnbn1 , 则
x1
x2Βιβλιοθήκη xnb1b2 2
b2
b3 2
bn
bn1 2
3
3
b1 b2 bn1 (n 1) 2 4n 2 .
对任意正整数 k ，
共圆。
3．设 ai , bi (i 1,2,, n) 是有理数使得对任意的实数 x 都有
n
x 2 x 4 (ai x bi )2 . i 1
求 n 的最小可能值.
解：因
x2
x
4
x
1
2
3
2
12
1
2
1
2
,
2 2
2 2
故 n 5是可以的. 下证 n 4 不可以.
4
反证法设 x 2 x 4 (ai x bi )2 , ai , bi Q ， i 1
ci bi 1。由 B 的选择可知 (B) (C) ，即

1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。

Copyright © 2016 Art of Problem Solving Mindy made three purchases for dollars, dollars, and dollars. What was her total, tothe nearest dollar?The three prices round to , , and , which has a sum of2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byFirstQuestion Followed by Problem 21 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_1&oldid=55964"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingOn the AMC 8 contest Billy answers 13 questions correctly, answers 7 questions incorrectly and doesn'tanswer the last 5. What is his score?As the AMC 8 only rewards 1 point for each correct answer, everything is irrelevant except the number Billy answered correctly, .2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 1Followed by Problem 31 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_2&oldid=55965"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingElisa swims laps in the pool. When she first started, she completed 10 laps in 25 minutes. Now, she canfinish 12 laps in 24 minutes. By how many minutes has she improved her lap time?When Elisa started, she finished a lap in minutes. Now, she finishes a lap is minutes.The difference is .2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 2Followed by Problem 41 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_3&oldid=55966"SolutionSee AlsoCopyright © 2016 Art of Problem Solving Initially, a spinner points west. Chenille moves it clockwise revolutions and then counterclockwiserevolutions. In what direction does the spinner point after the two moves?If the spinner goes clockwise revolutions and then counterclockwise revolutions, it ultimately goes counterclockwise which brings the spinner pointing.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 3Followed by Problem 51 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_4&oldid=55967"SolutionSee AlsoPoints and are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?Drawing segments and , the number of triangles outside square is the same as the number of triangles inside the square. Thus areas must be equal so the area of is half the area of thelarger square which is .If the side length of the larger square is , the side length of the smaller square is . Therefore the area of the smaller square is , half of the larger square's area, .Thus, the area of the smaller square in the picture is .Copyright © 2016 Art of Problem Solving The letter T is formed by placing twoinch rectangles next to each other, as shown. What is theperimeter of the T, in inches?If the two rectangles were seperate, the perimeter would be . It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is .2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 5Followed by Problem 71 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_6&oldid=55969"SolutionSee AlsoCopyright © 2016 Art of Problem Solving Circle has a radius of . Circle has a circumference of . Circle has an area of . List thecircles in order from smallest to largest ing the formulas of circles, and , we find that circle has a radius of and circle has a radius of . Thus, the order from smallest to largest radius is.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 6Followed by Problem 81 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21• 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_7&oldid=55970"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingThe table shows some of the results of a survey by radiostation KAMC. What percentage of the males surveyedlisten to the station?Filling out the chart, it becomes Thus, the percentage of males surveyed that listen to the station is .2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 7Followed by Problem 91 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_8&oldid=55971"SolutionSee AlsoCopyright © 2016 Art of Problem Solving What is the product of?By telescoping, it's easy to see the sum becomes .2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 8Followed by Problem 101 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_9&oldid=55972"SolutionSee AlsoJorge's teacher asks him to plot all the ordered pairs of positive integers for which is the width and is the length of a rectangle with area 12. What should his graph look like?Copyright © 2016 Art of Problem SolvingThe length of the rectangle will relate invertly to the width, specifically using the theorem . Theonly graph that could represent a inverted relationship is . (The rest are linear graphs thatrepresent direct relationships)2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 9Followed by Problem 111 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_10&oldid=55973"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingHow many two-digit numbers have digits whose sum is a perfect square?There is integer whose digits sum to : .There are integers whose digits sum to : .There are integers whose digits sum to : .There are integers whose digits sum to :.Two digits cannot sum toor any greater square since the greatest sum of digits of a two-digit number is.Thus, the answer is.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 10Followed by Problem 121 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_11&oldid=80944"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingAntonette gets on a 10-problem test, on a 20-problem test and on a 30-problem test. Ifthe three tests are combined into one 60-problem test, which percent is closest to her overall score?Adding them up gets. The overall percentage correct would be.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 11Followed by Problem 131 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_12&oldid=55975"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingCassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian leaves Marquette at 9:00 AM heading for Escanaba on his bike. He bikes at a uniform rate of 16 miles per hour. They both bike on the same 62-mile route between Escanaba and Marquette. At what time inthe morning do they meet?If Cassie leavesan hour earlier then Brian, when Brian starts, the distance between them will be. Every hour, they will getmiles closer., so 2 hours from 9:00AM is when they meet, which is.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 12Followed by Problem 141 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_13&oldid=55976"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingProblems 14, 15 and 16 involve Mrs. Reed's English assignment.A Novel AssignmentThe students in Mrs. Reed's English class are reading the same -page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in seconds and Chandra reads a page inseconds.If Bob and Chandra both read the whole book, Bob will spend how many more seconds reading than Chandra?The information is the same for Problems 14,15, and 16. Therefore, we shall only use the information we need. All we need for this problem is that there's 760 pages, Bob reads a page in 45 seconds and Chandrareads a page in 30 seconds. A lot of people will find how long it takes Bob to read the book, how long it takes Chandra to read the book, and then find the seconds. However, if we just set up the expression, we can find an easier way.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 13Followed by Problem 151 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_14&oldid=55977"SolutionSee AlsoLet be the number of pages that Chandra reads.Distribute theAdd to both sidesDivide both sides by to make it easier to solveDivide both sides bySolution 2Bob and Chandra read at a rate of seconds per page, respectively. Simplifying that gets us Bob reads pages for every pages that Chandra reads. Therefore Chandra should read of the book. =$\boxed{\textbf{(C)} 456}See AlsoCopyright © 2016 Art of Problem SolvingProblems 14, 15 and 16 involve Mrs. Reed's English assignment.A Novel AssignmentThe students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have toread?The amount of pages Bob, Chandra, and Alice would read is in the ratio 4:6:9. Therefore, Bob, Chandra, and Alice read 160, 240, and 360 pages respectively. They would also be reading in the same amount of time because the ratio of pages read was based on the time it takes each of them to read a page. Therefore, the amount of seconds each person reads is.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 15Followed by Problem 171 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_16&oldid=55979"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingJeff rotates spinners ,andand adds the resulting numbers. What is the probability that his sum isan odd number?In order for Jeff to have an odd number sum, the numbers must either be Odd + Odd + Odd or Even + Even +Odd. We easily notice that we cannot obtain Odd + Odd + Odd because spinner contains only even numbers.Therefore we must work with Even + Even + Odd and spinner will give us one of our even numbers. We also see that spinner only contains odd, so spinner must give us our odd nmber. We still need one evennumber from spinner . There is only 1 even number: . Since spinning the required numbers are automatic onthe other spinners, we only have to find the probability of spinning a in spinner , which clearly is2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 16Followed by Problem 181 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24• 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_17&oldid=55980"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingA cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white?The surface area of the cube is . Each of the eight black cubes has 3 faces on the outside,making black faces. Therefore there are white faces. To find the probability,we evaluate.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 17Followed by Problem 191 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 •25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_18&oldid=55981"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingTriangleis an isosceles triangle with . Pointis the midpoint of bothand, and is 11 units long. Triangleis congruent to triangle . What is the length of?Since triangleis congruent to triangleand ,. Since,. Because pointis the midpoint of,.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 18Followed by Problem 201 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19• 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_19&oldid=55982"SolutionSee AlsoA singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how manygames did Monica win?Since there are 6 players, a total ofgames are played. So far,games finished (one person won from each game), so Monica needs to win.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 19Followed byProblem 211 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 •20• 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_20&oldid=81054"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingCopyright © 2016 Art of Problem Solving An aquarium has a rectangular base that measures cm by cm and has a height ofcm. The aquarium is filled with water to a depth of cm. A rock with volume is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?The water level will rise cm for every . Since is of , the waterwill rise 2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 20Followed by Problem 221 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22• 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_21&oldid=55984"SolutionSee AlsoThree different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process toobtain a number in the top cell. What is the difference between the largest and smallest numbers possible inthe top cell?If the lower cells contain and , then the second row will contain and , and the top cell will contain . To obtain the smallest sum, place in the center cell and and in the outer ones. The top number will be . For the largest sum, place in the center cell and and in the outer ones. This top number will be . The difference is .2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 21Followed by Problem 231 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 •22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_22&oldid=55985"SolutionSee AlsoThe counting numbers that leave a remainder of 4 when divided by 6 are The counting numbers that leave a remainder of 3 when divided by 5 are So 28 isthe smallest possible number of coins that meets both conditions. Because , there areIf there were two more coins in the box, the number of coins would be divisible by both 6 and 5. The smallest number that is divisible by 6 and 5 is , so the smallest possible number of coins in the box isand the remainder when divided by 7 is .The problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Copyright © 2016 Art of Problem Solving In the multiplication problem below , , , and are different digits. What is?, so . Therefore, and , so.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 23Followed by Problem 251 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 •24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_24&oldid=68712"SolutionSee AlsoCopyright © 2016 Art of Problem Solving Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides areprime numbers. What is the average of the hidden prime numbers?Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number would be to add another even number to 44, and a different one to 38. Since there is only one even prime (2), the middle card's hidden number cannot be anodd prime, and so must be even. Therefore, the middle card's hidden number must be 2, so the constant sum is . Thus, the first card's hidden number is , and the last card's hidden number is .Since the sum of the hidden primes is, the average of the primes is.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 24Followed by Last Problem1 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_25&oldid=55988"SolutionSee Also。