2015 AMC 10A 考题及答案
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2015INTERMEDIATE DIVISIONAUSTRALIAN S CHOOL YEARS 9 and 10TIME ALLOWED: 75 MINUTESINSTRUCTIONS AND INFORMATIONGENERAL1. Do not open the booklet until told to do so by your teacher.2. NO calculators, maths stencils, mobile phones or other calculating aids are permitted. Scribbling paper, graphpaper, ruler and compasses are permitted, but are not essential. 3. Diagrams are NOT drawn to scale. They are intended only as aids.4. There are 25 multiple-choice questions, each with 5 possible answers given and 5 questions that require awhole number answer between 0 and 999. The questions generally get harder as you work through the paper. There is no penalty for an incorrect response.5. This is a competition not a test; do not expect to answer all questions. You are only competing against yourown year in your own country/Australian state so different years doing the same paper are not compared. 6. Read the instructions on the answer sheet carefully. Ensure your name, school name and school year areentered. It is your responsibility to correctly code your answer sheet. 7. When your teacher gives the signal, begin working on the problems.THE ANSWER SHEET 1. Use only lead pencil.2. Record your answers on the reverse of the answer sheet (not on the question paper) by FULLY colouring thecircle matching your answer.3. Your answer sheet will be scanned. The optical scanner will attempt to read all markings even if they are inthe wrong places, so please be careful not to doodle or write anything extra on the answer sheet. If you want to change an answer or remove any marks, use a plastic eraser and be sure to remove all marks and smudges.INTEGRITY OF THE COMPETITIONThe AMT reserves the right to re-examine students before deciding whether to grant official status to their score.©AMT P ublishing 2015 AMTT liMiTed Acn 083 950 341A ustrAliAn M AtheMAtics c oMpetitionsponsored by the c oMMonweAlth b AnkAn AcTiviTy of The AusTrAliAn MATheMATics TrusTNAMEYEA TEACHE RA u s T r A l i A n M A T h e M A T i c s T r u s TIntermediate Division Questions1to10,3marks each1.What is the area of this triangle in squarecentimetres?(A)10(B)12(C)14(D)7(E)62cm2.A movie lasts for213hours.The movie is shown in two equal sessions.For how many minutes does each session last?(A)85(B)70(C)80(D)65(E)753.If p=11and q=−4,then p2−q2equals(A)105(B)137(C)117(D)115(E)944.2015−20.15equals(A)1984.85(B)1995.15(C)1994.85(D)1995.85(E)2035.155.What is the value of2015twenty-cent coins?(A)$2015(B)$107.50(C)$17.50(D)$403(E)$436.Ana,Ben,Con,Dan and Eve are sitting around a ta-ble in that order.Ana calls out the number1,thenBen calls out the number2,then Con calls out thenumber3,and so on.After a person calls out a num-ber,the next person around the table calls out thenext number.Anyone who calls out a multiple of7must immediatelyleave the table.Who is the last person remaining at the table?(A)Ana(B)Ben(C)Con(D)Dan(E)Eve7.On a farm the ratio of horses to cows is3:2and ratio of cows to goats is4:3.The ratio of goats to horses is(A)5:7(B)3:8(C)3:5(D)5:18(E)1:28.Warren the window washer starts on the38thfloor of a building that has12windowsperfloor.He washes all of the windows on eachfloor before moving down to thefloor below.Whichfloor is Warren on after he has washed141windows?(A)25th(B)24th(C)28th(D)27th(E)26th9.A packet of lollies contains5blue lollies,15yellow lollies and some red lollies.One-third of the lollies are red.What fraction of the lollies are yellow?(A)13(B)56(C)12(D)16(E)2310.The diagram shows two small squares in opposite corners ofa large square.The squares have sides of length1cm,2cmand7cm.What is the area of the shaded pentagon?(A)18cm2(B)16cm2(C)22cm2(D)24cm2(E)20cm2Questions 11to 20,4marks each11.Jenna measures three sides of a rectangle and gets a total of 80cm.Dylan measuresthree sides of the same rectangle and gets a total of 88cm.What is the perimeter of the rectangle?(A)112cm(B)132cm(C)96cm(D)168cm(E)156cm12.A bar-tailed godwit was recorded by satellite tag in 2007tohave flown 11500km in eight days.On average,approximately how many kilometres per hour is that?(A)120(B)6(C)1(D)24(E)6013.A cube has the letters A,C,M,T,H and S on its six faces.Here are two views ofthis cube.CMAA MTWhich one of the following could be a third view of the same cube?(A)MHT(B)ATC(C)T SC(D)HTA (E)SCM 14.Two ordinary dice are rolled.The two resulting numbers are multiplied together tocreate a score.The probability of rolling a score that is a multiple of six is(A)16(B)512(C)14(D)13(E)1218.A strip of paper1cm wide is folded4times to make a regular octagon as shown.If the ends of the strip meet exactly when folded,how many centimetres long is the strip?√2(B)8(C)4+4√2(D)16(E)16−4√2(A)819.The country of Numismatica has six coins of thefollowing denominations:1cent,2cents,4cents, 10cents,20cents and40cents.Using the coins in my pocket,I can pay exactly for any amount up to and including200cents.What is the smallest number of coins I could have?(A)12(B)10(C)11(D)9(E)820.What fraction of the large triangle is shaded?(A)16(B)13(C)49(D)12(E)25Questions21to25,5marks each21.A student noticed that in a list offive integers,the mean,median and mode wereconsecutive integers in ascending order.What is the largest range possible for these five integers?(A)5(B)9(C)8(D)7(E)622.The square P QRS has sides of length2units andJ is the midpoint of P S.The line QJ intersectsthe diagonal P R at L.The length of LP is(A)√23(B)√33(C)√22(D)2√33(E)2√23SQ211PRJ23.For each integer from0to999,Andr´e wrote down the sum of its digits.What is theaverage of the numbers that Andr´e wrote down?(A)13.5(B)15(C)12(D)12.5(E)10.524.Max’s journey around this grid starts on a grid pointon side AB.He visits a grid point on each of sides BC,CD and DA in order before returning to his starting point, forming a quadrilateral.Max does not visit corner points A,B,C or D.How many journeys are possible which are not rect-angles?(Note that a square is a rectangle.)(A)256(B)252(C)64(D)248(E)76C B25.It takes Nicolai one and a half hours to paint the walls of a room and two hours topaint the ceiling.Elena needs exactly one hour to paint the walls of the same room and one hour to paint the ceiling.If Nicolai and Elena work together,what is the shortest possible time in minutes in which they can paint the walls and the ceiling of the room?(A)72(B)60(C)83(D)75(E)76For questions26to30,shade the answer as an integer from0to999in the space provided on the answer sheet.Question26is6marks,question27is7marks,question28is8marks, question29is9marks and question30is10marks.26.Mike has2015matches and uses them to builda triangular pattern like the one shown,but asbig as possible.How many matches does he haveleft over?27.How many positive integers n less than 2015have the property that 13+1ncan besimplified to a fraction with denominator less than n ?28.A rectangle has all sides of integer length.When 3units are added to the heightand 2units to the width,the area of the rectangle is tripled.What is the sum of the original areas of all such rectangles?29.At Berracan station,northbound trains arrive every three minutes starting at noonand finishing at midnight,while southbound trains arrive every five minutes starting at noon and finishing at midnight.Each day,I walk to Berracan station at a random time in the afternoon and wait for the first train in either direction.On average,how many seconds should I expect to wait?30.In a 14×18rectangle ABCD ,points P,Q,R and Sare chosen,one on each side of ABCD as pictured.The lengths AP ,P B ,BQ ,QC ,CR ,RD ,DS and SA are all positive integers and P QRS is a rectangle.What is the largest possible area that P QRS could have?DCB AQSIntermediate 2015 Answers Question Answer 1B2B3A4C5D6D7E8D9C10C11A12E13A14B15E16C17E18C19B20B21D22E23A24D25A2637272242844297430150。
2015AMC10AProblem1What is the value of的值是多少?Problem2A box contains a collection of triangular and square tiles.There are tiles in the box,containing edges total.How many square tiles are there in the box?一个盒子装有一堆三角形形状和正方形形状的瓷砖,盒子里总共有25块瓷砖,含有84条边。
那么盒子里有多少块正方形瓷砖?Problem3Ann made a3-step staircase using18toothpicks as shown in the figure.How many toothpicks does she need to add to complete a5-step staircase?Ann用18根牙签做了三阶楼梯,如下图所示,她还需要添加多少根牙签就可以做成五阶楼梯?Problem4Pablo,Sofia,and Mia got some candy eggs at a party.Pablo had three times as many eggs as Sofia, and Sofia had twice as many eggs as Mia.Pablo decides to give some of his eggs to Sofia and Mia so that all three will have the same number of eggs.What fraction of his eggs should Pablo give to Sofia?Pablo,Sofia和Mia在一个聚会上得到了一些糖果蛋。
Pablo的糖果蛋的数量是Sofia的3倍,Sofia的蛋的数量是Mia的2倍。
AMC10的真题答案及中文翻译AMC10的真题及中文翻译1、One ticket to a show costs $20 at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?(A) 2 (B) 5 (C) 10 (D) 15 (E) 20中文:一张展览票全价为20美元。
Susan用优惠券买4张票打七五折。
Pam用优惠券买5张票打七折。
Pam比Susan多花了多少美元?2、An aquarium has a rectangular base that measures 100cm by 40cm and has a height of 50cm. It is filled with water to a height of 40cm. A brick with a rectangular base that measures 40cm by 20cm and a height of 10cm is placed in the aquarium. By how many centimeters does that water rise?(A) 0.5 (B) 1 (C) 1.5 (D) 2 (E)2.5中文:一个养鱼缸有100cm×40cm的底,高为50cm。
它装满水到40cm的高度。
把一个底为40cm×20cm,高为10cm的砖块放在这个养鱼缸里。
鱼缸里的水上升了多少厘米?3、The larger of two consecutive odd integers is three times the smaller. What is their sum?(A) 4 (B) 8 (C) 12 (D) 16 (E) 20中文:2个连续的奇整数中较大的数是较小的数的3倍。
2015 年全国高中数学联合竞赛(A 卷)参考答案及评分标准一试说明:1.评阅试卷时,请依据本评分标冶填空题只设。
分和香分两档;其他各题的评阅,请严格按照本评分标准的评分档次给分,不要增加其他中间档次.2.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,在评卷时可参考本评分标准适当划分档次评分,解答题中第9小题4分为一个档次,第10、11小题该分为一个档次,不要增加其他中间档次.一、填空题:本大题共8小题,每小题份分,满分64分.1.设b a ,为不相等的实数,若二次函数b ax x x f ++=2)(满足)()(b f a f =,则=)2(f 答案:4.解:由己知条件及二次函数图像的轴对称性,可得22a b a+=-,即20a b +=,所以(2)424f a b =++=.2.若实数α满足ααtan cos =,则αα4cos sin 1+的值为 . 答案:2. 解:由条件知,ααsin cos 2=,反复利用此结论,并注意到1sin cos 22=+αα,得)cos 1)(sin 1(sin sin sin cos cos sin 122224αααααααα-+=++=+ 2cos sin 22=-+=αα.3.已知复数数列{}n z 满足),2,1(1,111⋅⋅⋅=++==+n ni z z z n n ,则=2015z .答案:2015 + 1007i .解:由己知得,对一切正整数n ,有211(1)11(1)2n n n n z z n i z ni n i z i ++=+++=+++++=++, 于是201511007(2)20151007z z i i =+⨯+=+.4.在矩形ABCD 中,1,2==AD AB ,线段DC 上的动点P 与CB 延长线上的动点Q 满=,则PQ PA ⋅的最小值为 .答案34.解:不妨设 A ( 0 , 0 ) , B ( 2 , 0 ) , D ( 0 , l ) .设 P 的坐标为(t , l) (其中02t ≤≤),则由||||DP BQ =得Q 的坐标为(2,-t ),故(,1),(2,1)PA t PQ t t =--=---,因此,22133()(2)(1)(1)1()244PA PQ t t t t t t ⋅=-⋅-+-⋅--=-+=-+≥.当12t =时,min 3()4PA PQ ⋅=.5.在正方体中随机取三条棱,它们两两异面的概率为 . 答案:255.解:设正方体为ABCD-EFGH ,它共有12条棱,从中任意取出3条棱的方法共有312C =220种.下面考虑使3条棱两两异面的取法数.由于正方体的棱共确定3个互不平行的方向(即 AB 、AD 、AE 的方向),具有相同方向的4条棱两两共面,因此取出的3条棱必属于3个不同的方向.可先取定AB 方向的棱,这有4种取法.不妨设取的棱就是AB ,则AD 方向只能取棱EH 或棱FG ,共2种可能.当AD 方向取棱是EH 或FG 时,AE 方向取棱分别只能是CG 或DH .由上可知,3条棱两两异面的取法数为4×2=8,故所求概率为8222055=.6.在平面直角坐标系中,点集{}0)63)(63(),(≤-+-+y x y x y x 所对应的平面区域的面积为 . 答案:24.解:设1{(,)||||3|60}K x y x y =+-≤. 先考虑1K 在第一象限中的部分,此时有36x y +≤,故这些点对应于图中的△OCD 及其内部.由对称性知,1K 对应的区域是图中以原点O为中心的菱形ABCD 及其内部.同理,设2{(,)||3|||60}K x y x y =+-≤,则2K 对应的区域是图中以O 为中心的菱形EFGH 及其内部.由点集K 的定义知,K 所对应的平面区域是被1K 、2K 中恰好一个所覆盖的部分,因此本题所要求的即为图中阴影区域的面积S .由于直线CD 的方程为36x y +=,直线GH 的方程为36x y +=,故它们的交点P 的坐标为33(,)22.由对称性知,138842422CPG S S ∆==⨯⨯⨯=.7.设ω为正实数,若存在实数)2(,ππ≤<≤b a b a ,使得2sin sin =+b a ωω,则ω的取值范围为 . 答案:9513[,)[,)424w ∈+∞.解:2sin sin =+b a ωω知,1sin sin ==b a ωω,而]2,[,ππωωw w b a si ∈,故题目条件等价于:存在整数,()k l k l <,使得 ππππππw l k w 22222≤+≤+≤. ①当4w ≥时,区间]2,[ππw w 的长度不小于π4,故必存在,k l 满足①式. 当04w <<时,注意到)8,0(]2,[πππ⊆w w ,故仅需考虑如下几种情况:(i) ππππw w 2252≤<≤,此时21≤w 且45>w 无解;(ii) ππππw w 22925≤<≤,此时2549≤≤w ;(iii) ππππw w 221329≤<≤,此时29413≤≤w ,得4413<≤w .综合(i)、(ii)、(iii),并注意到4≥w 亦满足条件,可知9513[,)[,)424w ∈+∞.8.对四位数abcd ,若,,,d c c b b a ><>则称abcd 为P 类数,若d c c b b a <><,,,则称abcd 为Q 类数,则P 类数总量与Q 类数总量之差等于 .答案:285.解:分别记P 类数、Q 类数的全体为A 、B ,再将个位数为零的P 类数全体记为0A ,个位数不等于零的尸类数全体记为1A .对任一四位数1A abcd ∈,将其对应到四位数dcba ,注意到1,,≥><>d c c b b a ,故B dcba ∈.反之,每个B dcba ∈唯一对应于从中的元素abcd .这建立了1A 与B 之间的一一对应,因此有011()()||||||||||||N P N Q A B A A B A -=-=+-=.下面计算0||A 对任一四位数00A abc ∈, b 可取0, 1,…,9,对其中每个b ,由9≤<a b 及9≤<c b 知,a 和c 分别有b -9种取法,从而992200191019||(9)2856b k A b k ==⨯⨯=-===∑∑. 因此,()()285N P N Q -=. 三、解答题9.(本题满分16分)若实数c b a ,,满足cb ac b a 424,242=+=+,求c 的最小值. 解:将2,2,2abc分别记为,,x y z ,则,,0x y z >.由条件知,222,x y z x y z +=+=,故2222224()2z y x z y z y z y -==-=-+.8分因此,结合平均值不等式可得,4221111(2)244y y z y y y y +==++≥⋅=12分 当212y y =,即y =时,zx求).由于2log c z =,故c的最小值225log log 33=-.16分 10.(本题满分20分)设4321,,,a a a a 为四个有理数,使得:{}⎭⎬⎫⎩⎨⎧----=≤<≤3,1,81,23,2,2441j i aa ji,求4321a a a a +++的值. 解:由条件可知,(14)i j a a i j ≤<≤是6个互不相同的数,且其中没有两个为相反数,由此知,4321,,,a a a a 的绝对值互不相等,不妨设||||||||4321a a a a <<<,则||||(14)i j a a i j ≤<≤中最小的与次小的两个数分别是12||||a a 及13||||a a ,最大与次大的两个数分别是34||||a a 及24||||a a ,从而必须有121324341,81,3,24,a a a a a a a a ⎧=-⎪⎪⎪=⎨⎪=⎪=-⎪⎩ 10 分 于是2341112113,,248a a a a a a a =-===-. 故2231412113{,}{,24}{2,}82a a a a a a =--=--,15分结合1a Q ∈,只可能114a =±.由此易知,123411,,4,642a a a a ==-==-或者123411,,4,642a a a a =-==-=.检验知这两组解均满足问题的条件. 故123494a a a a +++=±. 20 分 11.(本题满分20分)设21,F F 分别为椭圆1222=+y x 的左右焦点,设不经过焦点1F 的直线l 与椭圆交于两个不同的点B A ,,焦点2F 到直线l 的距离为d ,如果11,,BF l AF 的斜率依次成等差数列,求d 的取值范围.解:由条件知,点1F 、2F 的坐标分别为(-1, 0)和(l, 0) .设直线l 的方程为y kx m =+,点A 、B 的坐标分别为11(,)x y 和22(,)x y ,则12,x x 满足方程22()12x kx m ++=,即 222(21)4(22)0k x kmx m +++-=.由于点A 、B 不重合,且直线l 的斜率存在,故12,x x 是方程①的两个不同实根,因此有①的判别式22222(4)4(21)(22)8(21)0km k m k m ∆=-⋅+⋅-=+->,即2221k m +>.②由直线11,,BF l AF 的斜率1212,,11y y k x x ++依次成等差数列知,1212211y yk x x +=++,又1122,y kx m y kx m =+=+,所以122112()(1)()(1)2(1)(1)kx m x kx m x k x x +++++=++,化简并整理得,12()(2)0m k x x -++=.假如m k =,则直线l 的方程为y kx k =+,即 z 经过点1F (-1, 0),不符合条件. 因此必有1220x x ++=,故由方程①及韦达定理知,1224()221kmx x k =-+=+,即12m k k=+.③ 由②、③知,222121()2k m k k +>=+,化简得2214k k>,这等价于||2k >. 反之,当,m k满足③及||2k >l 必不经过点1F (否则将导致m k =,与③矛盾), 而此时,m k 满足②,故l 与椭圆有两个不同的交点A 、B ,同时也保证了1AF 、1BF 的斜率存在(否则12,x x 中的某一个为- l ,结合1220x x ++=知121x x ==-,与方程①有两个不同的实根矛盾).10分点2F (l , 0)到直线l: y kx m =+的距离为211|2|(2)22d k kk ==+=+.注意到||2k >t =t ∈,上式可改写为 21313()()222t d t t t=⋅+=⋅+.考虑到函数13()()2f t t t=⋅+在上上单调递减,故由④得,(1)f d f <<,即2)d ∈.20 分加试1.(本题满分40分)设)2(,,,21≥⋅⋅⋅n a a a n 是实数,证明:可以选取{}1,1,,,21-∈⋅⋅⋅n εεε,使得))(1()()(122121∑∑∑===+≤+ni i i n i i ni i a n a a ε.证法一:我们证明:2[]222111[]2()(1)()n n n n i i j i n i i i j a a a n a ====⎛⎫ ⎪+-≤+ ⎪ ⎪⎝⎭∑∑∑∑,① 即对1,2,,[]2n i =,取1i ε=,对[]1,,2ni n =+,取1i ε=-符合要求.(这里,[]x 表示实数x 的整数部分.) 10分事实上,①的左边为2222[][][]222111[]1[]1[]122222n n n n n n i j i j i j n n n i i i j j j a a a a a a ====+=+=+⎛⎫⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪ ⎪++-=+ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭∑∑∑∑∑∑ []2221[]122222n n i j n i j n n a n a ==+⎛⎫⎛⎫⎛⎫⎡⎤⎡⎤ ⎪ ⎪≤+- ⎪⎢⎥⎢⎥ ⎪ ⎪⎣⎦⎣⎦⎝⎭ ⎪ ⎪⎝⎭⎝⎭∑∑(柯西不等式)30分 []2221[]1212222n n i j n i j n n a a ==+⎛⎫⎛⎫⎛+⎫⎡⎤⎡⎤ ⎪ ⎪=+ ⎪⎢⎥⎢⎥ ⎪ ⎪⎣⎦⎣⎦⎝⎭ ⎪⎪⎝⎭⎝⎭∑∑(利用122n n n +⎡⎤⎡⎤-=⎢⎥⎢⎥⎣⎦⎣⎦) []2221[]12(1)n n i j n i j n a n a ==+⎛⎫⎛⎫ ⎪ ⎪≤++ ⎪ ⎪⎪ ⎪⎝⎭⎝⎭∑∑(利用[]x x ≤) 21(1)()ni i n a =≤+∑.所以 ① 得证,从而本题得证.证法二:首先,由于问题中12,,,n a a a 的对称性,可设12n a a a ≥≥≥.此外,若将12,,,n a a a 中的负数均改变符号,则问题中的不等式左边的21)(∑=n i i a 不减,而右边的21ni i a=∑不变,并且这一手续不影响1i ε=±的选取,因此我们可进一步设120n a a a ≥≥≥≥. 10分引理:设120n a a a ≥≥≥≥,则1110(1)ni i i a a -=≤-≤∑.事实上,由于1(1,2,,1)i i a a i n +≥=-,故当n 是偶数时,1123411(1)()()()0ni i n n i a a a a a a a --=-=-+-++-≥∑,11232111(1)()()ni i n n n i a a a a a a a a ---=-=------≤∑.当n 是奇数时,11234211(1)()()()0ni i n n n i a a a a a a a a ---=-=-+-++-+≥∑,1123111(1)()()ni i n n i a a a a a a a --=-=-----≤∑.引理得证. 30 分回到原题,由柯西不等式及上面引理可知22122211111(1)(1)n n n ni i i i i i i i i a a n a a n a -====⎛⎫⎛⎫⎛⎫+-≤+≤+ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭∑∑∑∑,这就证明了结论. 40分证法三:加强命题:设12,,,n a a a ⋅⋅⋅(2n ≥)是实数,证明:可以选取12,,,{1,1}n εεε⋅⋅⋅∈-,使得 2221111()()()()n nn i i i i i i i a a n a n ε===+≤+∑∑∑.证明 不妨设22212n a a a ≥≥⋅⋅⋅≥,以下分n 为奇数和n 为偶数两种情况证明.当n 为奇数时,取12121n εεε-==⋅⋅⋅==,13221n n n εεε++==⋅⋅⋅==-,于是有12221112()[()()]n nni i jn i i j a a a -+===+-∑∑∑12221122[()+()]n ni jn i j a a -+===∑∑1222112112()+2()()22n n i j n i j n n a n a -+==--≤⋅⋅-∑∑(应用柯西不等式).1222112(1)()+(1)()n ni jn i j n a n a -+===-+∑∑ ①另外,由于22212n a a a≥≥⋅⋅⋅≥,易证有122211211(1)(1)n n i j n i j a a n n -+==+≥-∑∑,因此,由式①即得到1222112(1)()+(1)()n nijn i j n a n a -+==-+∑∑211()()n i i n a n =≤+∑,故n 为奇数时,原命题成立,而且由证明过程可知,当且仅当12121n εεε-==⋅⋅⋅==,13221n n n εεε++==⋅⋅⋅==-,且12n a a a ==⋅⋅⋅=时取等号.当n 为偶数时,取1221n εεε==⋅⋅⋅==,24221n n n εεε++==⋅⋅⋅==-,于是有2222112()[()()]n nni i j n i i j a a a +===+-∑∑∑22222122[()+()]n ni j n i j a a +===∑∑2222122()+2()()22nn i j n i j n n a n a +==≤⋅⋅-∑∑(应用柯西不等式).222212[()+()]n nijn i j n a a +===∑∑22111()()()nn ii i i n a n a n ===≤+∑∑,故n 为偶数时,原命题也成立,而且由证明过程可知,当且仅当120n a a a ==⋅⋅⋅==时取等号,若12,,,n a a a ⋅⋅⋅不全为零,则取不到等号.综上,联赛加试题一的加强命题获证. 2.(本题满分40分)设{},,,,21n A A A S ⋅⋅⋅=其中n A A A ,,,21⋅⋅⋅是n 个互不相同的有限集合)2(≥n ,满足对任意的S A A j i ∈,,均有S A A j i ∈ ,若2min 1≥=≤≤i ni A k ,证明:存在i ni A x 1=∈ ,使得x 属于n A A A ,,,21⋅⋅⋅中的至少kn个集合.证明:不妨设1||A k =.设在12,,,n A A A 中与1A 不相交的集合有s 个,重新记为12,,,s B B B ,设包含1A 的集合有t 个,重新记为12,,,t C C C .由已知条件,1()i B A S ∈,即112(){,,,}i t B A C C C ∈,这样我们得到一个映射12121:{,,,}{,,,},()s t i i f B B B C C C f B B A →=. 显然f 是单映射,于是,s t ≤. 10 分设112{,,,}k A a a a =.在n A A A ,,,21⋅⋅⋅中除去12,,,s B B B ,12,,,t C C C 后,在剩下的n s t --个集合中,设包含i a 的集合有i x 个(1i k ≤≤),由于剩下的n s t --个集合中每个集合与从的交非空,即包含某个i a ,从而12k x x x n s t +++≥--. 20 分不妨设11max i i k x x ≤≤=,则由上式知i n s tx k --≥,即在剩下的n s t --个集合中,包含1a的集合至少有n s tk--个.又由于),,2,1(1t i C A i ⋅⋅⋅=⊆,故12,,,t C C C 都包含1a ,因此包含1a 的集合个数至少为(1)n s t n s k t n s tt k k k---+---+=≥(利用2k ≥) nk ≥(利用s t ≤). 40 分 3.(本题满分50分)如图,ABC ∆内接于圆O ,P 为BC 弧上一点,点K 在AP 上,使得BK 平分ABC ∠,过C P K ,,三点的圆Ω与边AC 交于D ,连接BD 交圆Ω于E ,连接PE ,延长交AB 于F ,证明:FCB ABC ∠=∠2.证法一:设CF 与圆Q 交于点L (异于C),连接PB 、PC 、 BL 、KL .注意此时C 、D 、L 、K 、E 、P 六点均在圆Ω上,结合A 、 B 、P 、C 四点共圆,可知∠FEB=∠DEP=180°-∠DCP=∠ABP=∠FBP ,因此△FB E ∽△FPB ,故FB 2=FE ·FP .10分又由圆幂定理知,FE ·FP= FL ·FC ,所以FB 2=FL ·FC . 从而△FBL ∽△FCB .因此, ∠FLB=∠FBC=∠APC=∠KPC=∠FLK, 即B 、K 、L 三点共线. 30 分再根据△FBL ∽△FCB 得,∠FCB=∠FBL=12∠ABC, 即∠ABC=2∠FCB .证法二:设CF 与圆Ω交于点L (异于C).对圆内接广义六边形DCLKPE 应用帕斯卡定理可知, DC 与KP 的交点A 、CL 与PE 的交点F 、LK 与ED 的交点了共线,因此B ’是AF 与ED 的交点,即B ’=B .所以B 、K 、L 共线.10分根据A 、B 、P 、C 四点共圆及L 、K 、P 、C 四点共圆,得 ∠ABC=∠APC=∠FLK=∠FCB+∠LBC,又由BK 平分∠ABC 知,∠FBL=12∠ABC ,从而 ∠ABC=2∠FCB .4.(本题满分50分)求具有下述性质的所有正整数k :对任意正整数n 都有1)1(2+-n k 不整除!)!(n kn . 解:对正整数m ,设2()v m 表示正整数m 的标准分解中素因子2的方幂,则熟知2(!)()v m m S m =-,①这里()S m 表示正整数m 在二进制表示下的数码之和.由于1)1(2+-n k 不整除()!!kn n ,等价于2()!()(1)!kn v k n n ≤-,即22(()!)(!)kn v kn n v n -≥-,进而由①知,本题等价于求所有正整数k ,使得()()S kn S n ≥对任意正整数n 成立. 10分我们证明,所有符合条件的k 为2(0,1,2,)aa =.一方面,由于(2)()aS n S n =对任意正整数n 成立,故2ak =符合条件. 20 分另一方面,若k 不是2的方幂,设2,0,ak q a q =⋅≥是大于1的奇数.下面构造一个正整数n ,使得()()S kn S n <.因为()(2)()aS kn S q S qn <⋅=, 因此问题等价于我们选取q 的一个倍数m ,使得()()m S m S q <. 由(2,q )=l ,熟知存在正整数u ,使得21(mod )uq ≡.(事实上,由欧拉定理知,u 可以取()q ϕ的.)设奇数q 的二进制表示为1212222,0,2t a a at a a a t +++=<<<≥.取1122222t t a a tu aa-+++++,则()S m t =,且2(21)0(mod )t a tu m q q =+-≡.我们有1(1)02121211212(122)12t t ttu uu t a a lu a u t ul m q q q q q -+-=---=++⋅=+⋅+++=+⋅∑由于2102u uq -<<,故正整数21u q -的二进制表示中的最高次幂小于u ,由此易知,对任意整数,(01)i j i j t ≤<≤-,数212t u iu a q +-⋅与212tu ju a q+-⋅的二进制表示中没有相同的项.又因为0i a >,故212(0,1,,1)tu lu a l t q +-⋅=-的二进制表示中均不包含1,故由②可知21()1()()u m S S t t S m q q-=+⋅>=, 因此上述选取的m 满足要求.综合上述的两个方面可知,所求的k 为2(0,1,2,)aa =.50分。
2015 年全国高中数学联合竞赛参考答案及评分标准一试一、填空题:本大题共8小题,每小题8分,满分64分.1.设b a ,为不相等的实数,若二次函数b ax x x f ++=2)(满足)()(b f a f =,则=)2(f 答案:4.解:由己知条件及二次函数图像的轴对称性,可得22a b a+=-,即20a b +=,所以(2)424f a b =++=.2.若实数α满足ααtan cos =,则αα4cos sin 1+的值为 . 答案:2. 解:由条件知,ααsin cos 2=,反复利用此结论,并注意到1sin cos 22=+αα,得)cos 1)(sin 1(sin sin sin cos cos sin 122224αααααααα-+=++=+ 2cos sin 22=-+=αα.3.已知复数数列{}n z 满足),2,1(1,111⋅⋅⋅=++==+n ni z z z n n ,其中i 为虚数单位,n z 表示n z 的共轭复数,则=2015z .答案:2015 + 1007i .解:由己知得,对一切正整数n ,有211(1)11(1)2n n n n z z n i z ni n i z i ++=+++=+++++=++, 于是201511007(2)20151007z z i i =+⨯+=+.4.在矩形ABCD 中,1,2==AD AB ,边DC 上(包含点D 、C )的动点P 与CB 延长线上(包含点B )的动点Q =,则PQ PA ⋅的最小值为 . 答案34. 解:不妨设 A ( 0 , 0 ) , B ( 2 , 0 ) , D ( 0 , l ) .设 P 的坐标为(t , l) (其中02t ≤≤),则由||||DP BQ =得Q 的坐标为(2,-t ),故(,1),(2,1)PA t PQ t t =--=---,因此,22133()(2)(1)(1)1()244PA PQ t t t t t t ⋅=-⋅-+-⋅--=-+=-+≥.当12t =时,min 3()4PA PQ ⋅=.5.在正方体中随机取三条棱,它们两两异面的概率为 . 答案:255.解:设正方体为ABCD-EFGH ,它共有12条棱,从中任意取出3条棱的方法共有312C =220种.下面考虑使3条棱两两异面的取法数.由于正方体的棱共确定3个互不平行的方向(即 AB 、AD 、AE 的方向),具有相同方向的4条棱两两共面,因此取出的3条棱必属于3个不同的方向.可先取定AB 方向的棱,这有4种取法.不妨设取的棱就是AB ,则AD 方向只能取棱EH 或棱FG ,共2种可能.当AD 方向取棱是EH 或FG 时,AE 方向取棱分别只能是CG 或DH .由上可知,3条棱两两异面的取法数为4×2=8,故所求概率为8222055=.6.在平面直角坐标系xOy 中,点集{}0)63)(63(),(≤-+-+y x y x y x 所对应的平面区域的面积为 . 答案:24.解:设1{(,)||||3|60}K x y x y =+-≤. 先考虑1K 在第一象限中的部分,此时有36x y +≤,故这些点对应于图中的△OCD 及其内部.由对称性知,1K 对应的区域是图中以原点O为中心的菱形ABCD 及其内部.同理,设2{(,)||3|||60}K x y x y =+-≤,则2K 对应的区域是图中以O 为中心的菱形EFGH 及其内部.由点集K 的定义知,K 所对应的平面区域是被1K 、2K 中恰好一个所覆盖的部分,因此本题所要求的即为图中阴影区域的面积S .由于直线CD 的方程为36x y +=,直线GH 的方程为36x y +=,故它们的交点P 的坐标为33(,)22.由对称性知,138842422CPG S S ∆==⨯⨯⨯=.7.设ω为正实数,若存在实数)2(,ππ≤<≤b a b a ,使得2sin sin =+b a ωω,则ω的取值范围为 . 答案:9513[,)[,)424w ∈+∞.解:2sin sin =+b a ωω知,1sin sin ==b a ωω,而]2,[,ππωωw w b a si ∈,故题目条件等价于:存在整数,()k l k l <,使得 ππππππw l k w 22222≤+≤+≤. ① 当4w ≥时,区间]2,[ππw w 的长度不小于π4,故必存在,k l 满足①式. 当04w <<时,注意到)8,0(]2,[πππ⊆w w ,故仅需考虑如下几种情况:(i) ππππw w 2252≤<≤,此时21≤w 且45>w 无解;(ii) ππππw w 22925≤<≤,此时2549≤≤w ;(iii) ππππw w 221329≤<≤,此时29413≤≤w ,得4413<≤w .综合(i)、(ii)、(iii),并注意到4≥w 亦满足条件,可知9513[,)[,)424w ∈+∞.8.对四位数abcd (9d ,0,91≤≤≤≤c b a ,),若,,,d c c b b a ><>则称abcd 为P 类数;若d c c b b a <><,,,则称abcd 为Q 类数,用N(P)和N(Q)分别表示P 类数与Q 类数的个数,则N(P)-N(Q)的值为 .答案:285.解:分别记P 类数、Q 类数的全体为A 、B ,再将个位数为零的P 类数全体记为0A ,个位数不等于零的尸类数全体记为1A .对任一四位数1A abcd ∈,将其对应到四位数dcba ,注意到1,,≥><>d c c b b a ,故B dcba ∈.反之,每个B dcba ∈唯一对应于从中的元素abcd .这建立了1A 与B 之间的一一对应,因此有011()()||||||||||||N P N Q A B A A B A -=-=+-=.下面计算0||A 对任一四位数00A abc ∈, b 可取0, 1,…,9,对其中每个b ,由9≤<a b 及9≤<c b 知,a 和c 分别有b -9种取法,从而992200191019||(9)2856b k A b k ==⨯⨯=-===∑∑. 因此,()()285N P N Q -=.二、解答题:本大题共3小题,满分56分,解答应写出文字说明、证明过程或演算步骤。
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2015年GCT英语真题及答案解析
2015年GCT语文真题及答案解析
2015年GCT考试逻辑真题及答案。
2015 AMC 10A 考题及答案Problem 1What is the value ofProblem 2A box contains a collection of triangular and square tiles. There are tiles in the box, containing edges total. How many square tiles are there in the box?Problem 3Ann made a 3-step staircase using 18 toothpicks as shown in the figure. How many toothpicks does she need to add to complete a 5-step staircase?Problem 4Pablo, Sofia, and Mia got some candy eggs at a party. Pablo had three times as many eggs as Sofia, and Sofia had twice as many eggs as Mia. Pablo decides to give some of his eggs to Sofia and Mia so that all threewill have the same number of eggs. What fraction of his eggs should Pablo give to Sofia?Problem 5Mr. Patrick teaches math to students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was . After he graded Payton's test, the test average became . What was Payton's score on the test?Problem 6The sum of two positive numbers is times their difference. What is the ratio of the larger number to the smaller number?Problem 7How many terms are there in the arithmetic sequence , , , . . .,, ?Problem 8Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be : ?Problem 9Two right circular cylinders have the same volume. The radius of the second cylinder is more than the radius of the first. What is the relationship between the heights of the two cylinders?Problem 10How many rearrangements of are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either or .Problem 11The ratio of the length to the width of a rectangle is : . If the rectangle has diagonal of length , then the area may be expressed as for some constant . What is ?Problem 12Points and are distinct points on the graph of. What is ?Problem 13Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?Problem 14The diagram below shows the circular face of a clock with radius cm and a circular disk with radius cm externally tangent to the clock face at o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?Problem 15Consider the set of all fractions where and are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by , the value of the fraction is increased by ?Problem 16If , and , what is the value of ?Problem 17A line that passes through the origin intersects both the line andthe line . The three lines create an equilateral triangle. What is the perimeter of the triangle?Problem 18Hexadecimal (base-16) numbers are written using numeric digits through as well as the letters through to represent through . Among the first positive integers, there are whose hexadecimal representation contains only numeric digits. What is the sum of the digits of ?Problem 19The isosceles right triangle has right angle at and area . The rays trisecting intersect at and . What is the area of ?Problem 20A rectangle with positive integer side lengths in has areaand perimeter . Which of the following numbers cannot equal ?NOTE: As it originally appeared in the AMC 10, this problem was stated incorrectly and had no answer; it has been modified here to be solvable. Problem 21Tetrahedron has , , , , , and . What is the volume of the tetrahedron?Problem 22Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?Problem 23The zeroes of the function are integers. What is the sum of the possible values of ?Problem 24For some positive integers , there is a quadrilateral with positive integer side lengths, perimeter , right angles at and , , and . How many different values of are possible?Problem 25Let be a square of side length . Two points are chosen at random on the sides of . The probability that the straight-line distance between the points is at least is , where , , and are positive integers with. What is ?2015 AMC 10A Answer Key1. C2. D3. D4. B5. E6. B7. B8. B9. D10.C11.C12.C13.C14.C15.B16.B17.D18.E19.D20.B (Note: This problem was originally stated incorrectly, and allcontestants received full credit regardless of their answer.)21.C22.A23.C24.B25.A。