巴基斯坦数学竞赛Cadet2008_E

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AMC Senior 2008(澳大利亚数学竞赛AMC-E:11-12年级中英文历年真题)

AMC Senior 2008(澳大利亚数学竞赛AMC-E:11-12年级中英文历年真题)

A u s t r A l i A n M At h e M At i c s c o M p e t i t i o na n a c t i v it y o f t h e a u s t r a l i a n m a t h e m a t i c s t r u s tT H U R S D AY 31 J U LY 2008SENIOR DIVISION COMPETITION PAPERINSTRUCTIO NS AND INFO RMATIO NGENERAL1. Do not open the booklet until told to do so by your teacher.2. NO calculators, slide rules, log tables, maths stencils, mobile phones or other calculating aids arepermitted. Scribbling paper, graph paper, 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 thatrequire a whole number 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 competingagainst your own year in your own State or Region 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 schoolyear are filled in. It is your responsibility that the Answer Sheet is correctly coded. 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 FULLYcolouring the circle matching your answer.3. Your Answer Sheet will be read by a machine. The machine will see all markings even if they arein the 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 AMC reserves the right to re-examine students before deciding whether to grant official status to their score.A U S T R A L I A N S C H O O L Y E A R S 11 A N D 12T I M E A L L O W E D : 75 M I N U T E SSenior DivisionQuestions 1to 10,3marks each1.The value of 8002−2008is (A)200(B)8(C)6006(D)1060(E)59942.The difference between 120and 210is (A)0(B)110(C)35(D)310(E)3203.In the diagram,x equals................................................................................................................................................................................ (x)◦100◦110◦80◦(A)100(B)110(C)120(D)130(E)1404.The value of 200×8200÷8is(A)1(B)8(C)16(D)64(E)2005.The smallest value that x 2−4x +3can have is (A)−1(B)−3(C)1(D)3(E)26.$3is shared between two people.One gets 50cents more than the other.The ratio of the larger share to the smaller share is (A)6:1(B)7:5(C)4:3(D)5:3(E)7:4S 27.When 10002008is written as a numeral,the number of digits written is (A)2009(B)6024(C)6025(D)8032(E)20128.A semicircle is drawn on one side of an equilateral triangle.The ratio of the area of the semicircle to the area of the triangle is(A)1:1(B)π:2√3(C)π:√3(D)√3:π(E)3:π................................................................................................................................................................................................................................................................................................................................................................................................................................................................9.Given that cos x =0.5and 0◦<x <90◦,which of the following has the greatestvalue?(A)cos 2x(B)cos x(C)0.75(D)sin x(E)tan x10.A fishtank with base 100cm by 200cm and depth 100cm contains water to a depthof 50cm.A solid metal rectangular prism with dimensions 80cm by 100cm by 60cm is then submerged in the tank with an 80cm by 100cm face on the bottom..................................................................................................................6.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................100100200501006080The depth of water,in centimetres,above the prism is then (A)12(B)14(C)16(D)18(E)20Questions 11to 20,4marks each11.Which of the following numbers is the largest?(A)2500(B)3400(C)4300(D)5200(E)610012.A normal die is thrown 100times.The sum of the numbers obtained will mostlikely be(A)200(B)250(C)300(D)350(E)400S 313.What is the smallest whole number which gives a square number when multipliedby 2008?(A)2(B)4(C)251(D)502(E)200814.A cross is made up of five squares,each with side length 1unit.Two cuts aremade,the first from X to Y and the second from Z to T ,so that ZT X is a right angle.The three pieces are then arranged to form a rectangle...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ZYX T IIIIII.................................................................... (II)IIIIWhat is the ratio of the length to the width of the rectangle?(A)3:1(B)√10:1(C)2:1(D)2√3:1(E)5:215.A function is said to be a toggle function on (p,q,r )if f (p )=q ,f (q )=r andf (r )=p .The function f (x )=ax 2+bx +c is a toggle function on (1,2,3).What is the value of c ?(A)−2(B)0(C)3(D)9(E)1416.Two conical rollers with perpendicu-lar axes touch on a line that is 30◦to the axis of the smaller roller and 60◦to the axis of the larger roller.If the larger roller makes 1revolution per sec-ond and there is no slipping,how many revolutions per second does the smaller roller make?(A)12(B)1(C)√2(D)√3(E)2...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................60◦30◦S 417.Consider the set X ={1,2,3,4,5,6}.How many subsets of X ,with at least one element,do not contain two consecutive integers?(A)16(B)18(C)20(D)21(E)2418.Farmer Taylor of Burra has two tanks.Water from the roof of his farmhouse iscollected in a 100kL tank and water from the roof of his barn is collected in a 25kL tank.The collecting area of his farmhouse roof is 200square metres while that of his barn is 80square metres.Currently,there are 35kL in the farmhouse tank and 13kL in the barn tank.Rain is forecast and he wants to collect as much water as possible.He should:(A)empty the barn tank into the farmhouse tank (B)fill the barn tank from the farmhouse tank(C)pump 10kL from the farmhouse tank into the barn tank (D)pump 10kL from the barn tank into the farmhouse tank (E)do nothing19.A sequence {u 1,u 2,...,u n }of real numbers is defined byu 1=√2,u 2=π,u n =u n −1−u n −2forn ≥3.What is u 2008?(A)−√2(B)2008(√2−2008π)(C)1003√2−1004π(D)π(E)√220.In the diagram,RU is equal in lengthto ST .What is the ratio of the area of QRU to the area of QST ?(A)√3:1(B)2:1(C)√6:1(D)√3:2(E)√6:2......................................................................................................................................................................................................................... (45)◦30◦UT Q RSQuestions 21to 25,5marks each21.P ,Q ,R ,S and T are consecutive vertices of a regular polygon.When extended,the lines P Q and T S meet at U with QUS =160◦.How many sides has the polygon?(A)36(B)42(C)48(D)52(E)54S522.How many numbers from1,2,3,4,...,2008have a cubic number other than1asa factor?(A)346(B)336(C)347(D)251(E)39323.The numbers828and313are3-digit palindromes where828−313=515,whichis also a palindrome.How many pairs(a,b)of3-digit palindromes are there witha>b and with a−b also a3-digit palindrome?(A)1972(B)1980(C)1988(D)1996(E)200824.The centres of all faces of a cube are joined to form an octahedron.The centresof all faces of this octahedron are now joined to form a smaller cube.What is the ratio of an edge of the smaller cube to an edge of the original cube?(A)1:√2(B)1:√3(C)1:2(D)1:3(E)1:425.In thefigure,all line segments are par-allel to one of the sides of the equi-lateral triangle P QR which has sidelength1unit.How long should P Xbe to maximise the smallest of the tenareas defined?(A)13(B)4−√214(C)14(D)15(E)1√10.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................PQ RXFor questions26to30,shade the answer as an integer from0to999inthe space provided on the answer sheet.Question26is6marks,question27is7marks,question28is8marks, question29is9marks and question30is10marks.26.All possible straight lines joining the vertices of a cube with mid-points of its edgesare drawn.At how many points inside the cube do two or more of these lines meet?S 627.Let us call a sum of integers cool if the first and last terms are 1and each termdiffers from its neighbours by at most 1.For example,the sum 1+2+3+4+3+2+3+3+3+2+3+3+2+1is cool.How many terms does it take to write 2008as a cool sum if we use no more terms than necessary?28.The positive integers x and y satisfy3x 2−8y 2+3x 2y 2=2008.What is the value of xy ?29.A point O is inside an equilateral triangleP QR and the perpendiculars OL ,OM and ON are drawn to the sides P Q ,QR and RP respectively.The ratios of lengths of the perpendiculars OL :OM :ON is 1:2:3.If area of LONP area of P QR =a b,where a and b are integers with no common factors,what is the value of a +b ?...............................................................................................................................................................................................................................................................................................................................................................................................................................................RPQLMNO.......................................................................................................................................................................................................................................................................................30.What is the smallest value that49+a 2−7√2a + a 2+b 2−√2ab +√50+b 2−10bcan have for positive real numbers a and b ?***Senior 2008 Answers Question Answer 1E2E3D4D5A6B7C8B9E10B11B12D13D14C15A16D17C18D19A20D21E22B23B24D25C26142789282829473013。

2008中国数学奥林匹克

2008中国数学奥林匹克

ak
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则 ak - ak- 1 ak+ 1 - ak .
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( 余红兵 供题)
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正 n 边形的每一个顶 点任意染上红、黄、蓝
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2008AMC8试题答案

2008AMC8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 © 2008, The Mathematical Association of AmericaT he M aTheMaTical a ssociaTion of a MericaAmerican Mathematics Competitions 24th AnnualAMC 8(American Mathematics Contest 8)Solutions PamphletT uesday, NOvEMBER 18, 20081.Answer(B):Susan spent2×12=$24on rides,so she had50−12−24=$14to spend.2.Answer(A):Because the key to the code starts with zero,all the lettersrepresent numbers that are one less than their ing the key,C is 9−1=8,and similarly L is6,U is7,and E is1.BEST OF LUCK0123456789CLUE=86713.Answer(A):A week before the13th is the6th,which is thefirst Friday of themonth.Counting back from that,the5th is a Thursday,the4th is a Wednesday, the3rd is a Tuesday,the2nd is a Monday,and the1st is a Sunday.ORCounting forward by sevens,February1occurs on the same day of the week as February8and February15.Because February13is a Friday,February15is a Sunday,and so is February1.4.Answer(C):The area of the outer triangle with the inner triangle removedis16−1=15,the total area of the three congruent trapezoids.Each trapezoid =5.has area1535.Answer(E):Barney rides1661−1441=220miles in10hours,so his average=22miles per hour.speed is220106.Answer(D):After subdividing the central gray square as shown,6of the16congruent squares are gray and10are white.Therefore,the ratio of the area of the gray squares to the area of the white squares is6:10or3:5.7.Answer (E):Note that M45=35=3·95·9=2745,so M =27.Similarly,60N =35=3·205·20=60100,so N =100.The sum M +N =27+100=127.ORNote that M45=35,so M =35·45=27.Also 60N =35,so N60=53,andN =53·60=100.The sum M +N =27+100=127.8.Answer (D):The sales in the 4months were $100,$60,$40and $120.The average sales were 100+60+40+1204=3204=$80.ORIn terms of the $20intervals,the sales were 5,3,2and 6on the chart.Their sum is 5+3+2+6=16and the average is 164=4.The average sales were4·$20=$80.9.Answer (D):At the end of the first year,Tammy’s investment was 85%of the original amount,or $85.At the end of the second year,she had 120%of her first year’s final amount,or 120%of $85=1.2($85)=$102.Over the two-year period,Tammy’s investment changed from $100to $102,so she gained 2%.10.Answer (D):The sum of the ages of the 6people in Room A is 6×40=240.The sum of the ages of the 4people in Room B is 4×25=100.The sum of the ages of the 10people in the combined group is 100+240=340,so the average age of all the people is 34010=34.11.Answer (A):The number of cat owners plus the number of dog owners is20+26=46.Because there are only 39students in the class,there are 46−39=7students who have both.ORBecause each student has at least a cat or a dog,there are 39−20=19students with a cat but no dog,and 39−26=13students with a dog but no cat.So there are 39−13−19=7students with both a cat and a dog.13197DogCat12.Answer (C):The table gives the height of each bounce.Bounce 12345Height 23·2=23·43=23·89=23·1627=in Meters 2438916273281Because 1627>1632=12and 3281<3264=12,the ball first rises to less than 0.5meters on the fifth bounce.Note:Because all the fractions have odd denominators,it is easier to doublethe numerators than to halve the denominators.So compare 1627and 3281to their numerators’fractional equivalents of 12,1632and 3264.13.Answer (C):Because each box is weighed two times,once with each ofthe other two boxes,the total 122+125+127=374poundsis twice the combined weight of thethree boxes.The combined weight is 3742=187pounds.14.Answer (C):There are only two possible spaces for the B in row 1and onlytwo possible spaces for the A in row 2.Once these are placed,the entries in theremaining spaces are determined.The four arrangementsare:ORThe As can be placed eitheror In each case,the letter next to the top A can be B or C.At that point the rest of the grid is completely determined.So there are 2+2=4possible arrangements.15.Answer(B):The sum of the points Theresa scored in thefirst8games is37.After the ninth game,her point total must be a multiple of9between37and 37+9=46,inclusive.The only such point total is45=37+8,so in the ninth game she scored8points.Similarly,the next point total must be a multiple of 10between45and45+9=54.The only such point total is50=45+5,so in the tenth game she scored5points.The product of the number of points scored in Theresa’s ninth and tenth games is8·5=40.16.Answer(D):The volume is7×1=7cubic units.Six of the cubes have5square faces exposed.The middle cube has no face exposed.So the total surface area of thefigure is5×6=30square units.The ratio of the volume to the surface area is7:30.ORThe volume is7×1=7cubic units.There arefive unit squares facing each of six directions:front,back,top,bottom,left and right,for a total of30square units of surface area.The ratio of the volume to the surface area is7:30. 17.Answer(D):The formula for the perimeter of a rectangle is2l+2w,so2l+2w=50,and l+w=25.Make a chart of the possible widths,lengths,and areas,assuming all the widths are shorter than all the lengths.Width123456789101112Length242322212019181716151413Area24466684100114126136144150154156 The largest possible area is13×12=156and the smallest is1×24=24,for a difference of156−24=132square units.Note:The product of two numbers with afixed sum increases as the numbers get closer together.That means,given the same perimeter,the square has a larger area than any rectangle,and a rectangle with a shape closest to a square will have a larger area than other rectangles with equal perimeters.18.Answer(E):The length offirst leg of the aardvark’s trip is14(2π×20)=10πmeters.The third andfifth legs are each14(2π×10)=5πmeters long.Thesecond and sixth legs are each10meters long,and the length of the fourth leg is 20meters.The length of the total trip is10π+5π+5π+10+10+20=20π+40 meters.19.Answer(B):Choose two points.Any of the8points can be thefirst choice,and any of the7other points can be the second choice.So there are8×7=56ways of choosing the points in order.But each pair of points is counted twice,so there are 562=28possible pairs.A B CDEF G H Label the eight points as shown.Only segments AB ,BC ,CD ,DE ,EF ,F G ,GH and HA are 1unit long.So 8of the 28possible segments are 1unit long,and the probability that the points are one unit apart is 828=27.ORPick the two points,one at a time.No matter how the first point is chosen,exactly 2of the remaining 7points are 1unit from this point.So the probabilityof the second point being 1unit from the first is 27.20.Answer (B):Because 23of the boys passed,the number of boys in the class is a multiple of 3.Because 34of the girls passed,the number of girls in the classis a multiple of 4.Set up a chart and compare the number of boys who passed with the number of girls who passed to find when they are equal.Total boysBoys passed 326496Total girls Girls passed 4386The first time the number of boys who passed equals the number of girls who passed is when they are both 6.The minimum possible number of students is 9+8=17.ORBecause 23of the boys passed,the number of boys who passed must be a multiple of 2.Because 34of the girls passed,the number of girls who passed must be a multiple of 3.Because the same number of boys and girls passed,the smallestpossible number is 6,the least common multiple of 2and 3.If 6of 9boys and 6of 8girls passed,there are 17students in the class,and that is the minimum number possible.ORLet G =the number of girls and B =the number of boys.Then 23B =34G ,so 8B =9G .Because 8and 9are relatively prime,the minimum number of boysand girls is 9boys and 8girls,for a total of 9+8=17students.21.Answer (C):Using the formula for the volume of a cylinder,the bologna hasvolume πr 2h =π×42×6=96π.The cut divides the bologna in half.Thehalf-cylinder will have volume 96π2=48π≈151cm 3.Note:The value of πis slightly greater than 3,so to estimate the volume multiply 48(3)=144cm 3.The product is slightly less than and closer to answer C than any other answer.22.Answer (A):Because n3is at least 100and is an integer,n is at least 300andis a multiple of 3.Because 3n is at most 999,n is at most 333.The possible values of n are 300,303,306,...,333=3·100,3·101,3·102,...,3·111,so the number of possible values is 111−100+1=12.E F 323.Answer (C):Because the answer is a ratio,it doesnot depend on the side length of the square.Let AF =2and F E =1.That means square ABCE has side length 3and area 32=9square units.The area of BAF is equal to the area of BCD =12·3·2=3square units.Triangle DEF is an isosceles right triangle with leg lengths DE =F E =1.The area of DEF is 12·1·1=12square units.The area of BF D is equal to the area of the square minusthe areas of the three right triangles:9−(3+3+12)=52.So the ratio of the area of BF D to the area of square ABCE is 529=518.24.Answer (C):There are 10×6=60possible pairs.The squares less than60are 1,4,9,16,25,36and 49.The possible pairs with products equal to the given squares are (1,1),(2,2),(1,4),(4,1),(3,3),(9,1),(4,4),(8,2),(5,5),(6,6)and (9,4).So the probability is 1160.。

小学奥数之体育比赛问题

小学奥数之体育比赛问题

小学奥数之体育比赛问题体育比赛中的数学对于体育比赛形式的逻辑推理题,注意“一队的胜、负、平”必然对应着“另一队的负、胜、平”。

有时综合性的逻辑推理题需要将比赛情况用点以及连接这些点的线来表示,从整体考虑,通过数量比较、整数分解等方式寻找解题的突破口。

【例 1】 三年级四个班进行足球比赛,每两个班之间都要赛一场,那么每个班要赛几场?一共要进行多少场比赛? (如果参赛队每两队之间都要赛一场,这种比赛称为单循环赛)【考点】体育比赛 【难度】1星 【题型】解答【解析】 (法一)题意要求每两个点之间都连一条线段.先考虑点A (如图),它与B 、C 、D 三点能且只能连接三条线段AB 、AC 、AD ;同样,从点B 也可以连出三条线段BA 、BC 、BD ;从点C 可以连出三条线段CA 、CB 、CD ;从点D 可以连出三条线段DA 、DB ,DC .因此,从一个点可以连三条线段.从每个点都连出三条线段,共有四个点.3412⨯=(条)注意到线段AB 既是由A 点连出的,也是由B 点连出的,并且每一条线段都是这样(如图),所以,线段的总数应为:6(条).(法二)从点A 引出三条线.AB 、AC 、AD ,为避免重复计数,从B 点引出的线段只计BC 、BD 两条,由C 点引出的只有CD 一条.因此,线段的总数为3216++=(条).通过例题的讲解,对于这个问题,我们就可以很轻松地解决了.一共有四个队,每个队都要比赛413-=场,一共有比赛3426⨯÷=场.【点拨】我们可以将上面的问题如下表述:下面的四个点,每两个点之间都连一条线段,那么,从一个点可以连出几条线段?一共可以连多少条线段?【答案】6场【巩固】 市里举行足球联赛,有5个区参加比赛,每个区出2个代表队.每个队都要与其他队赛一场,这些比赛分别在5个区的体育场进行,那么平均每个体育场都要举行多少场比赛?【考点】体育比赛 【难度】1星 【题型】解答【解析】 一共有5210⨯=(个)队参加比赛,共赛10(101)245⨯-÷=(场),平均每个体育场都要举行4559÷=(场)比赛.【答案】9场【巩固】 二年级六个班进行拔河单循环赛,每个班要进行几场比赛?一共要进行几场比赛?【考点】体育比赛 【难度】1星 【题型】解答 D C B AD C B A体育比赛问题知识点拨例题精讲【解析】每个班要进行5场,一共要进行65215⨯÷=(场)比赛.【答案】每个班要进行5场,一共要进行15场比赛【巩固】20名羽毛球运动员参加单打比赛,两两配对进行单单循环赛,那么冠军一共要比赛多少场?【考点】体育比赛【难度】1星【题型】解答【解析】假设20名羽毛球运动员中的甲是冠军,那么甲与其他19名运动员都赛过了,也就是一共赛了19场.【答案】一共赛了19场【例 2】8只球队进行淘汰赛,为了决出冠军,需要进行多少场比赛?【考点】体育比赛【难度】2星【题型】解答【解析】方法一:8进4进行了4场,4进2进行2场,最后决赛是1场,因此共进行了4217++=(场)比赛.方法二:每进行一场比赛就淘汰一支球队,最后只剩下冠军了,也就是说淘汰了7只球队,因此进行了7场比赛.【答案】7场比赛【例 3】学校进行乒乓球选拔赛,每个参赛选手都要和其他所有选手各赛一场,一共进行了36场比赛,有人参加了选拔赛.A.8B.9C.10【考点】体育比赛【难度】2星【题型】选择【关键词】2008,第四届,IMC国际数学邀请赛,新加坡,初赛【解析】三个人比赛,可以比赛3223⨯÷=场;如果有五个人⨯÷=场;如果四个人比赛,可以比赛4326比赛,那么可以比赛54210⨯÷=场,所以答⨯÷=场;如果有9个人比赛,那么可以比赛98236案是B.【答案】答案是B【巩固】朝阳区的几个学校举行篮球比赛,每两个学校都要赛一场,共赛了28场,那么有几个学校参加了比赛?【考点】体育比赛【难度】2星【题型】解答【解析】假设有n个学校参加比赛,那么就有(1)2n=,也n n⨯-÷场比赛,现在已知共赛了28场,那么8就是有8个学校参加了比赛.【答案】8个学校【例 4】有8个选手进行乒乓球单循环赛,结果每人获胜局数各不相同,那么冠军胜了几局?【考点】体育比赛【难度】2星【题型】解答【解析】8个选手进行乒乓球单循环赛,每个选手都要参加7场比赛,而且每人获胜局数各不相同,所以每人获胜的局数分别为0~7局,那么冠军胜了7局.【答案】冠军胜了7局【例 5】A、B、C、D、E五位同学一起比赛象棋,每两人都要比赛一盘.到现在为止,A已经赛4盘,B赛3盘,C赛2盘,D赛1盘.问:此时E同学赛了几盘?【考点】体育比赛【难度】2星【题型】解答【解析】画5个点表示五位同学,两点之间连一条线段表示赛一场,建议教师让学生动手按要求画一画.Array根据题意,A已经赛4盘,说明A与B、C、D、E各赛一盘,A应与B、C、D、E点相连.D赛1盘,是与A点相连的.B赛3盘,是与A、C、E点相连的.C赛2盘,是与A、B点相连的.从图上E点的连线条数可知,E同学赛了2盘.【答案】E同学赛了2盘【巩固】八一队、北京队、江苏队、山东队、广东队五队进行象棋友谊赛,每两个队都要赛一场,一个月过后,八一队赛了4场,北京队赛了3场,江苏队赛了2场,山东队赛了1场.那么广东队赛了几场?【考点】体育比赛【难度】2星【题型】解答【解析】八一队赛了4场,说明八一队和其它四队都赛过了.山东队赛了1场,说明只和八一队赛过.北京队赛了3场,说明与八一队、江苏队、广东队赛过.江苏队赛了2场,说明与八一队、北京队赛过.由此可知,广东队只和八一队、北京队赛过,赛了2场.【答案】赛了2场【巩固】A、B、C、D、E、F六人赛棋,采用单循环制。

2008C试卷澳大利亚数学竞赛AMC7-8年级中文历年真题

2008C试卷澳大利亚数学竞赛AMC7-8年级中文历年真题

(C)2.08 (D)2.1
(E)2.185
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3. 右图形之周长是多少 cm?
(A)8
(B)10
(C)12
(D)16
(E)20
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1-10 题,每题 3 分
1. 算式 2008+8002 等于
(A)1010 (B)4004 (C)10008 (D)8910 (E)10010
───────────────────────────────────────────────── 2. 下列哪一项的数值最大?
(A)2.15 (B)2.2
题目一般而言是依照越来越难的顺序安排,对于错误的答案不会倒扣分数。 5. 本活动是数学竞赛而不同于学校测验,别期望每道题目都会作。考生只与同地区同年级
的其它考生评比,因此不同年级的考生作答相同的试卷将不作评比。 6. 请依照监考老师指示,谨慎地在答案卡上填写您的基本数据。若因填写错误或不详所造
成之后果由学生自行负责。 7. 进入试场后,须等待监考老师宣布开始作答后,才可以打开题本进行答题。
(A)9
(B)11
(C)12
(D)14
(E)16
─────────────────────────────────────────────────
────────────────────── J 2 ─────────────────────
8. 一列火车于 8:58 am 离开 F 市,而于同日 9:34 am 抵达 B 市。请问需时多

2008国际数学奥林匹克名单

2008国际数学奥林匹克名单

2008国际数学奥林匹克名单2008国际数学奥林匹克(IMO)是在西班牙巴塞罗那举办的。

以下是该年奥赛的参赛名单:1. 阿根廷:Juan Pablo Arce2. 澳大利亚:Peter Casidy, Hung Y. Nguyen, Dominic Y. Yeo, 陈琮琪 (Cedric Cheng)3. 奥地利:Jakob Danner4. 孟加拉国:Ashikur Rahman, Mohammad Mahbubul Hasan5. 比利时:Sylvie Roques6. 巴西:Christian Massimo Pessanha Henriques, Luciano Pereira Goncalves7. 保加利亚:Plamena Bratanova8. Canada:Marc A. Renault, Bobby Shen9. 中国:Dohun Kim, Jieyu Liao, Ying Zhang, Ziyu Yao, Chaoyi Zhu10. 哥伦比亚:Juan Esteban Calle11. 克罗地亚:Leo Mršić12. 塞浦路斯:Kuriakos Papadopoulos13. 捷克:Martin Bobowski14. 丹麦:Magnus Find Schougaard15. 埃及:Maher Athmouni, Sherif Mohamed Tawfik16. 爱沙尼亚:Martin Randmets17. 菲律宾:Jim Marvin S. Lim18. 芬兰:Otto Rajala19. 法国:Chen Qiyuan20. 希腊:Hristo Dimitrov21. 香港:Ip Siu Hung22. 匈牙利:Barn abás Verőci, Dominik Zsembinszki23. 冰岛:Sigrún Hafsteinsdóttir24. 印度:Surath D. Dalal, Tathagata Basu25. 印度尼西亚:Dedy Christopher A., Muhammad Rizki Prananto26. 爱尔兰:Fenner, Holly, King-Tong Lau27. 以色列:Tom Kalvari, Yuval Wigderson28. 意大利:Andrea Cuneo, Simone Zaccagnini29. 日本:Katsuhisa Yamanaka, Kenji Kozeki, Kento Nakamura, Masashi Shinohara, Ryohei Miyadera30. 哈萨克斯坦:Zhassulan SHIMOV, Denis VOLOVITCH31. 韩国:Jaeyoon Kim, Junyeong Kim, Seungjoon Lee, Yongsu J. Hahn32. 黎巴嫩:Naji Boustany33. 利比亚:Mohammed Ezzat Mehrez El Sakhawy, Mohamed Khaled Mohamed Salem El Shall34. 立陶宛:Sigitas Rimkevičius35. 卢森堡:Michèle Cheng36. 马来西亚:Win Li Xun, Soh Wai Ching37. 挪威:Christian Løthner38. 巴基斯坦:Furqan Shahid Jafri, Muhammad Saad39. 巴拿马:Stephany Georine Hwee Ong, Abraham Paz40. 荷兰:Kees van der Veen41. 波兰:Jarosław Wroński, Mateusz Michałek, MaciejWiśniowski42. 葡萄牙:Miguel Angelo Teixeira de Sousa43. 罗马尼亚:Cosmin Pohoata44. 俄罗斯:Alexey Balitskiy, Andrey Yurkovsky, Maxim Zaripov45. 斯洛伐克:Matúš Harminc46. 南非:Nicholas Charles Campher, Nirbhay Mehta47. 西班牙:Francisco Arjona, Jorge García Miró, ImmaGonzález48. 斯里兰卡:Adeepa Seneviratne49. 瑞典:Johan Walin50. 瑞士:Marco H. Tschopp51. 叙利亚:Emad Alawieh, Ahmad Jabbari52. 台湾:Tong-Ming Chang, Tan Nguyen53. 泰国:Varayuth Lerdsuwankij, Praphon Angtrakool54. 土耳其:Caner Kavasoglu, Sabiha Tasdizen, Arda Apak55. 哈萨克斯坦:Azis Shaikhislam56. 乌克兰:Egor Klochko, Sergei Petunin57. 英国:David Arthur, Daniel Groslouis, Geoffrey Breakwell58. 美国:Zhiming Wang, Allen Liu, Tiankai Liu59. 乌兹别克斯坦:Konstantin Kanukh60. 越南:Ha Vu Nam Dinh。

2008年全国初中数学联合竞赛试题参考答案及评分标准

2008年全国初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.第一试,选择题和填空题只设7分和0分两档;第二试各题,请按照本评分标准规定的评分档次给分.如果考生的解答方法和本解答不同,只要思路合理,步骤正确,在评卷时请参照本评分标准划分的档次,给予相应的分数.第一试一、选择题(本题满分42分,每小题7分)本题共有6小题,每题均给出了代号为的四个答案,其中有且仅有一个是正确的.将你所选择的答案的代号填在题后的括号内.每小题选对得7分;不选、选错或选出的代号字母超过一个(不论是否写在括号内),一律得0分.D C B A ,,,1.设,213a a +=213b b +=,且,则代数式a b ≠211a b 2+的值为 ( ) )(A 5. 7. 9. 11.)(B )(C )(D 【答】B .解 由题设条件可知2310a a −+=,,且2310b b −+=a b ≠,所以是一元二次方程的两根,故,,因此,a b 2310x x −+=3a b +=1ab =222222222211()23217()1a b a b ab a b a b ab ++−−×+====. 故选B . 2.如图,设AD ,BE ,CF 为三角形的三条高,若ABC 6AB =,5BC =,,则线段3EF =BE 的长为 ( ))(A 185. 4. )(B )(C 215. )(D 245. 【答】. D 解 因为AD ,BE ,CF 为三角形的三条高,易知ABC ,,,B C E F 四点共圆,于是△AEF ∽△,故ABC 35AF EF AC BC ==,即3cos 5BAC ∠=,所以4sin 5BAC ∠=. 在Rt △ABE 中,424sin 655BE AB BAC =∠=×=. 故选. D 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. 【答】. C 解 能够组成的两位数有12,13,14,15,21,23,24,25,31,32,34,35,41,42,43,45,51,52,53,54,共20个,其中是3的倍数的数为12,15,21,24,42,45,51,54,共8个.所以所组成的数是3的倍数的概率是82205=. 故选C. 4.在△中,,∠=,ABC 12ABC ∠=°132ACB °BM 和CN 分别是这两个角的外角平分线,且点,M N 分别在直线和直线AC AB 上,则 ( ))(A BM CN >. )(B BM CN =.)(C . BM CN <)(D BM 和CN 的大小关系不确定.【答】B .解 ∵,12ABC ∠=°M 为的外角平分线,∴ABC ∠1(18012)842MBC ∠=°−°=B °°. 又,∴180********BCM ACB ∠=°−∠=°−°=180844848BMC ∠=°−°−°=°,∴.BM BC =又11(180)(180132)2422ACN ACB ∠=°−∠=°−°=°, ∴18018012()BNC ABC BCN ACB ACN ∠=°−∠−∠=°−°−∠+∠168(13224)=°−°+°12ABC =°=∠,∴. 因此,CN CB =BM BC CN ==.故选B .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,则的最小值为 ( )r r )(A 39()8. )(B 49(8. )(C 59(8. )(D 98. 【答】 B .解 容易知道,4天之后就可以出现5种商品的价格互不相同的情况.设5种商品降价前的价格为a ,过了天. 天后每种商品的价格一定可以表示为n n 98(110%)(120%)()()1010k n k k a a −n k ⋅−⋅−=⋅⋅k −,其中为自然数,且0. k n ≤≤要使r 的值最小,五种商品的价格应该分别为:98()()1010i n i a −⋅⋅,1198()(1010i n i a +−−⋅⋅, 2298()(1010i n a +−i ⋅⋅−,3398()()1010i n i a +−⋅⋅−,4498()()1010i n i a ,其中i 为不超过的自然数. n +−−⋅⋅所以r 的最小值为44498()()91010(988()()1010i n i i n ia a +−−−⋅⋅=⋅⋅. 故选B . 6. 已知实数,x y满足(,则2008x y −=223233x y x y −+−2007−的值为 ( ))(A . 2008. 2008−)(B )(C 1−. 1.)(D 【答】. D解 ∵(2008x y −=,∴x y ==+,y x −==+由以上两式可得x y =. 所以2(2008x =,解得,所以22008x =22222323320073233200720071x y x y x x x x x −+−−=−+−−=−=.故选.D二、填空题(本题满分28分,每小题7分)1.设12a −=,则5432322a a a a a a a +−−−+=−2−.解 ∵2213(122a a −===−,∴21a a +=, ∴543232323222()2(a a a a a a a a a a a a a a a a+−−−++−−++=−⋅−)2 33332221211(1)(11)2(1)1a a a a a a a a a a a−−+−−===−=−++=−+=−⋅−−−−.2.如图,正方形的边长为1,ABCD ,M N 为BD 所在直线上的两点,且AM =∠,则四边形的面积为135MAN =°AMCN 52解 设正方形的中心为O ,连,则ABCD AO AO BD ⊥,2AO OB ==,2MO ===, ∴MB MO OB =−=. 又, 135ABM NDA ∠=∠=°13590NAD MAN DAB MAB MAB ∠=∠−∠−∠=°−°−∠45=°−MAB AMB ∠=∠,所以△∽△ADN MBA ,故AD DNMB BA =,从而12AD DN BA MB =⋅==. 根据对称性可知,四边形的面积 AMCN11222222MAN S S MN AO ==×××=×××=△522. 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,,且m n 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 +=12解 根据题意,是一元二次方程的两根,所以,m n 20x ax b ++=m n a +=−,. mn b =∵1m n +≤,∴1m n m n +≤+≤,1m n m n −≤+≤.∵方程的判别式,∴20x ax b ++=24a b Δ=−≥022()44a m n b +14≤=≤. 22244()()()1b mn m n m n m n ==+−−≥+−≥−1,故14b ≥−,等号当且仅当12m n 时取得; =−=122244()()1()b mn m n m n m n ==+−−≤−−≤,故14b ≤,等号当且仅当12m n 时取得. ==所以14p =,14q =−,于是12p 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 1 .解 1到3,结果都只各占1个数位,共占13223×=个数位; 24到,结果都只各占2个数位,共占个数位;292612×=210到,结果都只各占3个数位,共占23132266×=个数位;232到,结果都只各占4个数位,共占299468272×=个数位;2100到,结果都只各占5个数位,共占523162171085×=个数位;此时还差2008个数位.(312662721085)570−++++=2317到,结果都只各占6个数位,共占2411695570×=个数位.所以,排在第2008个位置的数字恰好应该是的个位数字,即为1. 2411第二试 (A )一.(本题满分20分) 已知,对于满足条件221a b +=01x ≤≤的一切实数x ,不等式(1)(1)()0a x x ax bx b x bx −−−−−−≥ (1)恒成立.当乘积ab 取最小值时,求的值.,a b 解 整理不等式(1)并将代入,得221a b +=2(1)(21)0a b x a x a ++−++≥ (2)在不等式(2)中,令0x =,得;令0a ≥1x =,得.0b ≥易知10,a b ++>21012(1)a ab +<<++,故二次函数2(1)(21)y a b x a x a =++−++的图象(抛物线)的开口向上,且顶点的横坐标在0和1之间.由题设知,不等式(2)对于满足条件01x ≤≤的一切实数x 恒成立,所以它的判别式,即2(21)4(1)0a a b Δ=+−++⋅≤a 14ab ≥. 由方程组 221,14a b ab ⎧+=⎪⎨=⎪⎩ (3) 消去b ,得16,所以421610a a −+=224a =或224a +=.又因为,所以0a≥4a =或4a +=,于是方程组(3)的解为,4,4a b ⎧=⎪⎪⎨⎪=⎪⎩或,4.4a b ⎧=⎪⎪⎨−⎪=⎪⎩ 所以的最小值为ab 14,此时的值有两组,分别为 ,ab ,44a b +==和44a b +==. 二.(本题满分25分) 如图,圆与圆相交于O D ,A B 两点,为圆的切线,点在圆上,且.BC D C O AB BC =(1)证明:点O 在圆的圆周上.D (2)设△的面积为,求圆的的半径的最小值.ABC S D r 解 (1)连,因为为圆心,,,,OA OB OC AC O AB BC =,所以△OBA ∽△,从而OBC OBA OBC ∠=∠.因为OD ,所以,AB DB BC ⊥⊥9090DOB OBA OBC DBO ∠=°−∠=°−∠=∠,所以DB DO =,因此点O 在圆的圆周上.D (2)设圆的半径为a ,的延长线交于点O BO ACE ,易知BE AC ⊥.设2AC y =(0)y a <≤,,,则,,OE x =AB l =22a x 2y =+()S y a x =+22222222()2222()aS l y a x y a ax x a ax a a x y=++=+++=+=+=. 因为,22ABC OBA OAB BDO ∠=∠=∠=∠AB BC =,DB DO =,所以△∽△,所以BDO ABC BD BO AB AC=,即2r a l y =,故2al r y =. 所以22223222()4422a l a aS S a S y y y y ==⋅=⋅≥r ,即2≥r ,其中等号当a y =时成立,这时是圆O 的AC直径.所以圆的的半径的最小值为Dr 2. 三.(本题满分25分)设为质数,b 为正整数,且a 29(2)509(4511)ab a b +=+ (1)求,b 的值.a 解 (1)式即2634511()509509ab a b ++=,设634511,509509a b a b m n ++==,则 50965094351m a n a b 1−−== (2) 故,又,所以35116n m a −+=02n m =2351160m m a −+= (3)由(1)式可知,(2能被509整除,而509是质数,于是2)a b +2a b +能被509整除,故为整数,即关于的一元二次方程(3)有整数根,所以它的判别式为完全平方数.m m 251172a Δ=−不妨设(t 为自然数),则722251172a t Δ=−=511(511)(511)a t t t 22=−=+−.由于511和511的奇偶性相同,且511t +t −511t +≥,所以只可能有以下几种情况:①两式相加,得36,没有整数解.51136,5112,t a t +=⎧⎨−=⎩21022a +=②两式相加,得18,没有整数解. 51118,5114,t a t +=⎧⎨−=⎩41022a +=③两式相加,得12,没有整数解. 51112,5116,t a t +=⎧⎨−=⎩61022a +=④两式相加,得6,没有整数解.5116,51112,t a t +=⎧⎨−=⎩121022a +=⑤两式相加,得4,解得5114,51118,t a t +=⎧⎨−=⎩181022a +=251a =. ⑥两式相加,得2,解得5112,51136,t a t +=⎧⎨−=⎩361022a +=493a =,而4931729=×不是质数,故舍去. 综合可知.251a =此时方程(3)的解为3m =或5023m =(舍去). 把,代入(2)式,得251a =3m =5093625173b ×−×==. 第二试 (B )一.(本题满分20分)已知,对于满足条件221a b +=1,0x y xy +=≥的一切实数对(,)x y ,不等式220ay xy bx −+≥ (1)恒成立.当乘积ab 取最小值时,求的值.,a b 解 由可知011,0x y xy +=≥,01x y ≤≤≤≤.在(1)式中,令0,1x y ==,得;令0a ≥1,0x y ==,得b .0≥x 将代入(1)式,得,即1y =−22(1)(1)0a x x x bx −−−+≥2(1)(21)0a b x a x a ++−++≥ (2)易知10,a b ++>21012(1)a ab +<<++,故二次函数2(1)(21)y a b x a x a =++−++的图象(抛物线)的开口向上,且顶点的横坐标在0和1之间.由题设知,不等式(2)对于满足条件01x ≤≤的一切实数x 恒成立,所以它的判别式,即2(21)4(1)0a a b Δ=+−++⋅≤a 14ab ≥. 由方程组 221,14a b ab ⎧+=⎪⎨=⎪⎩ (3) 消去b ,得16,所以421610a a −+=224a −=或224a +=,又因为,所以0a≥4a =或4a =.于是方程组(3)的解为,4,4a b ⎧=⎪⎪⎨⎪=⎪⎩或,4.4a b ⎧=⎪⎪⎨−⎪=⎪⎩所以满足条件的的值有两组,分别为,ab ,44a b +==和44a b +==c ). 二.(本题满分25分)题目和解答与(A )卷第二题相同.三.(本题满分25分)题目和解答与(A )卷第三题相同.第二试 (C )一.(本题满分20分)题目和解答与(B )卷第一题相同.二.(本题满分25分)题目和解答与(A )卷第二题相同.三.(本题满分25分)设为质数,b 为正整数,且满足a ,29(22)509(41022511)2a b c a b c b c ⎧+−=+−⎨−=⎩ (1)(2)求的值.(a b c +解 (1)式即2()509509=, 设66341022511,509509a b c a b c m n +−+−==,则 5096509423511m a n a b c −−−== (3) 故,又,所以35116n m a −+=02n m =2351160m m a −+= (4)由(1)式可知,(2能被509整除,而509是质数,于是22)a b c +−c 22a b +−能被509整除,故为整数,即关于的一元二次方程(4)有整数根,所以它的判别式为完全平方数.m m 251172a Δ=−不妨设(t 为自然数),则722251172a t Δ=−=511(511)(511)a t t t 22=−=+−.由于511和511的奇偶性相同,且511t +t −511t +≥,所以只可能有以下几种情况:①两式相加,得36,没有整数解. 51136,5112,t a t +=⎧⎨−=⎩21022a +=②两式相加,得18,没有整数解.51118,5114,t a t +=⎧⎨−=⎩41022a +=③两式相加,得12,没有整数解. 51112,5116,t a t +=⎧⎨−=⎩61022a +=④两式相加,得6,没有整数解.5116,51112,t a t +=⎧⎨−=⎩121022a +=⑤两式相加,得4,解得5114,51118,t a t +=⎧⎨−=⎩181022a +=251a =. ⑥两式相加,得,解得5112,51136,t a t +=⎧⎨−=⎩236102a +=2493a =,而4931729=×不是质数,故舍去.综合可知,此时方程(4)的解为或251a =3m =5023m (舍去). =把,代入(3)式,得251a =3m =50936251273b c ×−×−==,即27c b =−. 代入(2)式得,所以b ,(27)2b b −−=5=3c =,因此()251(53)2008a b c +=×+=.。

2008—2009学年度重庆市巴蜀中学九年级数学第一学期半期考试

某某市巴蜀中学2008—2009学年度第一学期半期考试初2009级(三上)数学试题卷命题人:朱晓昀 审题人:王兴斌考试时间:2008年11月13日上午:8:00—10:00一、选择题:1. 如图是奥运会自行车比赛项目标志,则图中两轮所在圆的位置关系是( ) A. 内含 B. 相交C. 外切D. 外离2. 抛物线()21212+--=x y 的顶点坐标是( ) A. ()2,1B. ()2,1-C. ()2,1-D. ()2,1--3. 在ABC ∆中,90=∠C ,若23cos =B ,则A sin 的值为( ) A.3B.23C.33 D.21 4. ⊙O 的半径是5cm ,O 到直线l 的距离cm OP 3=,Q 为l 上一点且2.4=PQ cm ,则点Q ( ) A. 在⊙O 内 B. 在⊙O 上C. 在⊙O 外D. 以上情况都有可能5. 把抛物线22x y -=向上平移2个单位,得到的抛物线是( )A. ()222+-=x yB. ()222--=x yC. 222--=x yD. 222+-=x y6. 如图,A 、B 、C 三点是⊙O 上的点,50=∠ABO 则BCA ∠的度数是( ) A.80B.50C.40D.257. 如图,在ABC ∆中,30=∠A ,23tan =B ,32=AC , 则AB 的长为( ) A. 34+ B. 5 C. 32+D. 68. 已知直线()0≠+=a b ax y 经过一、三、四象限,则抛物线bx ax y +=2一定经过( )A. 第一、二、三象限B. 第一、三、四象限C. 第一、二、四象限D. 第三、四象限9. 如图是一台54英寸的液晶电视旋转在墙角的俯视图,设α=∠DAO ,电视后背AD 平行于前沿BC ,且与BC 的距离为cm 60,若cm AO 100=,则墙角O 到前沿BC 的距 离OE 是( ) A. ()cm αsin 10060+ B. ()cm αcos 10060+ C. ()cm αtan 10060+D. 以上都不对10. 二次函数()0122≠-++=a a x ax y 的图象可能是( )11. 已知点()1,1y -、()2,2y -、()3,2y 都在二次函数12632+--=x x y 的图象上,则1y 、2y 、3y 的大小关系为( )A. 231y y y >>B. 123y y y >>C. 213y y y >>D. 321y y y >>12. 某测量队在山脚A 处测得山上树顶仰角为45(如图),测量队在山坡上前进600米到D 处,再测得树顶的仰角为60, 已 知这段山坡的坡角为30,如果树高为15米,则山高为( ) (精确到1米,732.13=) A. 585米 B. 1014米 C. 805米 D. 820米二、填空题:13. 抛物线322+-=x x y 的对称轴是直线.14. 如图,圆柱形水管内积水的水面宽度cm CD 8=,F 为⋂CD的中点,圆柱形水管的半径为cm 5,则此时水深GF 的长度为cm . 15. 在ABC ∆中,三边之比2:3:1::=c b a ,则=+A A cos sin .16. 现有一圆心角为90,半径为8cm 的扇形纸片,用它恰好围成一个圆锥的侧面(接缝忽略不计),则该圆锥底面圆的半径为cm .17. 河堤横截面为梯形(如图),上底为4cm ,堤高为6cm ,斜坡AD 的坡度为1 : 3,斜度BC 的坡角为45,则河堤的横 截面积为2m .18. 现用一条长为6米的木料做成如图所示的窗框,窗框的面 积S 与窗框的宽x (m )之间的函数关系式为. 19. 在ABC ∆中,A ∠、B ∠都是锐角,且21sin =A 、22cos =B ,则ABC ∆三个角的大小关系是(用“<”连接)20. 如图,AB 切⊙O 于点B ,AD 过圆心,且与⊙O 相交于C 、D 两点,连结BD ,若⊙O 的半径为1,CO AO 2=,则BD 的长度为.21. 已知抛物线c bx ax y ++=2经过点()7,2-A 、()7,6B 、()8,3-C ,则该抛物线上纵坐标为8-的另一点的坐标是.22. 二次函数()02≠++=a c bx ax y 的最大值是0,则化简代数式ab ac a 442-+的结果为.23. 二次函数1582+-=x x y 的图象与x 轴相交于A 、B 两点,P 点在该函数图象上运动,能使ABP ∆的面积为2的点P 有个.24. 如图所示,二次函数()02≠++=a c bx ax y 的图象经过点()2,1-,且与x 轴交点的横坐标为1x 、2x ,其中121-<<-x 、102<<x 下列结论:①024<+-c b a ②02<-b a③0>abc ④ac a b 482>+正确的结论是.某某市巴蜀中学2008—2009学年度第一学期半期考试初2009级(三上)数学答题卷一、选择题:题号 1 2 3 4 5 6 7 8 9 10 11 12 答案二、填空题:13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 25.三、解答题(25题6分,26~31题每题10分,共66分) 25. (6分)计算:⎪⎭⎫ ⎝⎛--⋅-2130tan 60sin 45tan 45cos 226. (10分)如图,某海滨浴场岸边A 点处发现海中B 点有人求救,便立即派出两名救生员前去营救,1号救生员从A 点直接跳入海中,2号救生员沿岸边向前跑100米到 离B 点最近的C 点,再跳入海中. 救生员在岸上跑的速度为5米/秒,水中游泳的速度 为2米/秒,若60=∠BAC ,两名救生员同时从A 点出发,请说明谁先到达营救地点B . (参考数据7.13≈)27. (10分)抛物线()m x m x y +-+-=12与y 轴交于()3,0点,⑴求出m 的值;⑵求抛物线与x 轴的交点坐标;⑶直接写出x 取何值时,抛物线位于x 轴上方.28. (10分)如图,某船以每小时36海里的速度向正东方向航行,在点A 测得某岛C 在北偏东60方向上,航行半小时后到达点B 测得该岛在北偏东30方向上,已知该岛周 围16海里内有暗礁.⑴试说明点B 是否在暗礁区域内?⑵若继续向东航行有无触礁的危险?请说明理由.29. (10分)某公园草坪的护栏是由50段形状相同的抛物线组成的,为牢固起见,每段m 加设不锈钢管(如图①所示)做成的立柱,为了计算所需不锈钢管 立柱的总长度,设计人员利用图②所示的直角坐标系进行计算. ⑴试求此抛物线的解析式;⑵试求所需不锈钢管的总长度.30. (10分)如图,已知抛物线c bx x y ++-=221与x 轴交于A 、B 两点,与y 轴交于点 C 且6=AB ,抛物线的对称轴为直线1=x . ⑴求抛物线的解析式;⑵在x 轴上A 点的左侧有一点E ,满足ACO ECO S S ∆∆=4,求直线EC 的解析式;31. (10分)如图,一次函数5-=x y 分别交x 轴、y 轴于A 、B 两点,二次函数c bx x y ++-=2的图象经过A 、B 两点.⑴求二次函数的解析式;⑵设D 、E 是线段AB 上异于..A 、B 的两个动点(E 点位于D 点上方),2=DE .①若点D 的横坐标为t ,用含t 的代数式表示D 、E 的坐标;②抛物线上是否存在点F ,使点F 与点D 关于x 轴对称,如果存在,请求出AEF ∆的 面积;如果不存在,请说明理由.初2009级(三上)数学半期试题参考答案一、选择题:1—12 D A B C D C B C A B D C二、填空题: 13. 1=x14. 215.231+ 16. 2 17. 96 18.x x S 3232+-=19. C B A ∠<∠<∠ 20.321. ()8,1-22. a -23. 2 24.①②③④三、解答题:25. ⎪⎪⎭⎫ ⎝⎛--⨯-⎪⎪⎭⎫ ⎝⎛=2133231222-----------4分21332321+--=----------5分 3651-= ----------6分 26. 解:在ABC Rt ∆中,60=∠BAC 2160cos ==AB AC200=∴AB----------2分360tan ==ACBC3100=∴BC----------4分 1号救生员需要的时间10022002==AB 秒----------6分2号救生员需要的时间10525≈+BC AC 秒----------9分105100<∴1号救生员先到达B---------10分27. 解:⑴将(0,3)代入抛物线的解析式3=m----------4分⑵抛物线的解析式为:322++-=x x y 令0=y 0322=++-x x 解得1321-==x x∴抛物线与x 轴交点坐标为(3,0) ()0,1---------8分⑶由图可知 当31<<-x 时-------10分抛物线位于x 轴上方28. 解:⑴作AB CD ⊥于D 点--------1分设BC 为x 在BCD Rt ∆中60=∠CBDx BD 21=∴--------2分x CD 23=--------4分在ACD Rt ∆中30=∠CAD 33tan ==∠AD CD CAD 33211823=+∴xx--------6分18=∴x--------7分 B ∴点不在暗礁区域内--------8分 ⑵3923==x CD--------9分1639<∴若继续向东航行船有触礁的危险--------10分29. 解:⑴由题意得()5.0,0B 、()0,1C--------1分设抛物线的解析式为:c ax y +=2代入得2121=-=c a --------2分∴解析式为:21212+-=x y--------4分 ⑵当2.0=x 时 48.0=y--------5分当6.0=x 时 32.0=y-------6分()6.132.048.024*******=+⨯=+++∴C B C B C B C B 米-------8分∴所需不锈钢管的总长度为:80506.1=⨯米-------10分30. 解:⑴∵对称轴为1=x2112-==-∴a a b 1=∴b -------2分()()0,40.26B A AB -∴=()0,4B ∴代入解析式4=c------3分 ∴抛物线的解析式为:4212++-=x x y-------4分⑵OC EO S ECO ⋅=∆21-------5分OC AO S ACO ⋅=∆21-------6分ACO ECO S S ∆∆=484==∴AO EO-------7分()0,8-∴E-------8分由抛物线的解析式()4,0C设直线EC 的解析式为:b kx y += 将()0,8-E ()4,0C代入上式解得⎪⎩⎪⎨⎧==321b k∴直线EC 的解析式为321+=x y--------10分31. 解:⑴由题意可得()()5,00,5-B A--------1分代入解析式c bx x y ++-=2解得⎩⎨⎧-==56c b-------2分∴解析式为:562-+-=x x y-------3分⑵作y DQ //轴DQ EQ ⊥5=OA 5=OBOAB ∆∴为等腰直角三角形DEQ ∆∽BAO ∆DQE ∆ 为等腰直角三角形 12==∴=∴EQ DQ DE()5,-∴t t D -----5分 ()4,1-+t t E-----7分F 与D 关于x 轴对称()t t F -∴5,代入抛物线解析式得5652-+-=-t t t 解得5221==t tD 、E 异于A 、B 两点 5=∴t 舍去2=∴t()()()2,33,23,2--E D F-------8分222==DE AE636213232=⨯⨯⨯=⋅=∆∆ADF AEF S S--------10分。

2008浙江数学竞赛及答案

bECDB2008年全国初中数学竞赛(浙江赛区)复赛试题答题时注意;1.用圆珠笔或钢笔作答.2.解答书写时不要超过装订线. 3.草稿纸不上交.一、选择题(共6小题,每小题5分,满分30分.以下每小题均给出了代号为A ,B ,C ,D 的四个选项,其中有且只有一个选项是正确的.请将正确选项的代号填入题后的括号里.不填、多填或错填均得零分) 1.一列“动车组”高速列车和一列普通列车的车身长分别为80米与100米,它们相向行驶在平行的轨道上,若坐在高速列车上的旅客看见普通列车驶过窗口的时间是5秒,则坐在普通列车上的旅客看见高速列车驶过窗口的时间是( )(A )7.5秒 (B )6秒(C )5秒 (D )4秒 解:答案:【D 】设高速列车和普通列车的车速分别为x 米/秒和y 米/秒,则100520(/)x y m s -=÷=,所以坐在普通列车上的旅客看见高速列车驶过窗口的时间是:80÷20=4(秒)2.将一张边长分别为a ,b )(b a >的矩形纸片ABCD 折叠,使点C 与点A 重合,则折痕的长为( )(A (B(C (D 解:答案:【A 】CPE CBA ∆∆2PE CP CP PE AB AB BC BC a⇒=⇒==2EF PE ⇒==3.如图,设正方体ABCD -A 1B 1C 1D 1的棱长为1,黑、白 两个甲壳虫同时从A 点出发,以相同的速度分别沿棱向前爬行,黑甲壳虫爬行的路线是AA 1→A 1D 1→……,白甲壳虫爬行的路线是AB →BB 1→……,并且都遵循如下规则:所爬行的第n n 与第2+条棱所在的直线必须是既不平行也不相交(其中n 是正整数).那么当黑、白两个甲壳虫各爬行完第2008条棱分别停止在所到的正方体顶点处时,它们之间的距离是( ) (A )0 (B )1 (C )2 (D )3 解:答案:【C 】 黑甲壳虫爬行的路径为:111111111......AA A D D C C C CB BA AA A D →→→→→→→→ 白甲壳虫爬行的路径为:111111111......AB BB B C C D D A A A AB BB →→→→→→→→(第3题)ABCD A 1B 1C 1D 1黑、白甲壳虫每爬行6条边后又重复原来的路径,因2008=334×6+4,所以当黑、白两个甲壳虫各爬行完第2008条棱分别停止时,黑甲壳虫停在点C ,白甲壳虫停在点D 1,因此1CD =4.设m ,n 是正整数,满足m +n >mn ,给出以下四个结论:① m ,n 都不等于1; ② m ,n 都不等于2;③ m ,n 都大于1;④ m ,n 至少有一个等于1.其中正确的结论是( ) (A )① (B )② (C )③ (D )④ 解:答案:【D 】 由m n mn +>,得(1)(1)1m n --<,因m ,n 是正整数,所以(1)(1)0m n --=, 即11m n ==或5.小明按如图所示设计树形图,设计规则如下:第一层是一条与水平线垂直的线段,长度为1;第二层在第一层线段的前端作两条与该线段均成120°的线段,长度为其一半;第三层按第二层的方法,在每一条线段的前端生成两条线段;重复前面的作法作到第10层.则树形图第10层的最高 点到水平线的距离为( ) (A )11024 (B )17041024(C )17051024(D )2 解:答案:【C 】设第n 层的最高点到水平线的距离记为:(1,2,,10)n a n =由题意,得224412132435411111;();();();();2222a a a a a a a a a ==+=+=+=+66106576109111();();;();222a a a a a a =+=+=+把这10条式子左右相加,得2468101010111111170512[()()()()()]()2222221024a =+⨯++++-=6.有10条不同的直线n n b x k y +=(n = 1,2,3,…,10),其中369k k k ==,47100b b b ===,则这10条直线的交点个数最多有( )(A )45个 (B )40个 (C )39个 (D )31个 解:答案:【B 】如图,满足已知条件的6条直线至多有10这6条直线最多有6个交点,再增加一条直线与前7交点,……一直增加到第10条直线与前9条直线最多有9个 交点,所以这10条直线的交点个数最多有:10+6+7+8+9=40(个二、填空题(共6小题,每小题6分,满分36分) 7.在平行四边形ABCD 的边AB 和AD 上分别取点E 和F , 使13AE AB =,14AF AD =,连结EF 交对角线AC 于G ,则AC的值是 . 水平线第一层 第二层 第三层 第四层(第5题)x ,y ,zz y x ++11,x z y ++11,yx z ++11 (第10题)M解:答案:17如图,1//33AE AF AB CD DM AE DM FD ⇒==⇒=113367AG AE AE AE AG GC CM CD DM AEAE AC ∴====⇒=++ 8的圆过一个半径为2的圆的圆心,则图中阴影部分的面积为 . 解:答案:2连结OO 1, AB ,则有OO 1⊥AB 于点P ,在1Rt APO Rt APO ∆∆和中,222222222111112)AP AO OP O A O P O P O P O P =-=-⇒-=-⇒即点O 1在AB 上与点P 重合,易知AB 是圆O 1的直径,三角形ABO 是直角三角形. 所以222111=(22)2242S ππ⨯⨯-⨯⨯-⨯=阴影 9.已知y =26x mx +-,当1≤m ≤3时,y <0恒成立,那么实数x 的取值范围是. 解:答案:332x -<<由26<0x mx +-,22mmx +-<<=解得-当1≤m ≤3时,1=3 22m +-则 -的最大值为-; 所以,当1≤m ≤3时,y <0恒成立,即260x mx +-<恒成立时, x 的取值范围是3x -<<. 10.如图是一个数的转换器,每次输入3个不为零的数,经转换器转换后输出3个新数,规律如下:当输入数分别为x ,y ,z 时,对应输出的新数依次为z y x ++11,x z y ++11,yx z ++11.例如,输入 1,2,3,则输出56,34,23. 那么当输出的新数为31,41,51时,输入的3个数依次为 .(第8题)解:答案:1111 1132,, 2221112=333()1113=4()3(0)425()26111=5x y z kx k x y z xy xz x y z xy k xy yz x y z yz k y k k y z x yz xz x y z xz k z k z x y ++=⎧⎧+=⎪⎪+⎪+=++=⎧⎧⎪⎪⎪⎪⎪+⇒+=++−−−−→=⇒=>⎨⎨⎨⎨+⎪⎪⎪⎪+=++=⎩⎩⎪⎪=+⎪⎪⎩+⎩令1111,,1132k x y z x y z ⇒=++=++=++⇒=== 11.10张卡片上分别写有0到9这10个数,先将它们从左到右排成一排,再采用交换相邻两张卡片位置的方法对它们进行操作,规则如下:当相邻两张卡片左边卡片上的数比右边卡片上的数大时,交换它们的位置,否则不进行交换.若规定将相邻两张卡片交换一次位置称为1次操作,那么无论开始时这10张卡片的排列顺序如何,至多经过 次操作,就能将它们按从小到大的顺序排列. 解:答案:45记2n ≥张卡片至多经过n a 次操作后,能将它们按从小到大顺序排列,则232431091;2;3;............9.a a a a a a a ==+=+=+ 所以10123.....945a =++++=12.设整数a 使得关于x 的一元二次方程255261430x ax a -+-=的两个根都是整数,则a 的值是 .解:答案:18. 由题意,得222255202860(552)156()a a a k k N ∆=-+=-+=∈即22(552)156[(552)][(552)]782262ka k a k a --=⇒+-⨯--=⨯=⨯因为[(552)][(552)]ka k a +---和具有相同的奇偶性且[(552)][(552)]2k a k a k +---=≥+0故(552)=78(552)=26(552)=2(552)=6(552)=2(552)=6(552)=78(552)=26k a k a k a k a k a k a k a k a +-+-+-+-⎧⎧⎧⎧⎨⎨⎨⎨--------⎩⎩⎩⎩或或或 解得,只有=40=18k a ,符合题意。

2008年河北高中竞赛试题参考答案

2008年河北省高中数学竞赛试题参考答案(时间:5月18日上午8:30~11:30)一、 选择题(本大题共6小题,每小题6分,满分36分)1.函数(2)y f x =+的图像过点(-1,3),则函数()f x 的图像关于y 轴对称的图形一定过点( ).A (1,-3)B (-1,3)C (-3,-3)D (-3,3) 答案:B.2.把2008表示成两个整数的平方差形式,则不同的表示方法有( )种.A 4B 6C 8D 16 答案:C.解: 设222008x y -=,即()()2008x y x y +-=.2008有8个正因数,分别为1,2,4,8,251,502,1004,2008.而且()x y +与()x y -只能同为偶数,因此对应的方程组为245021004245021004100450242100450242x y x y +=----⎧⎨-=----⎩ 故()x y ,共有8组不同的值:(503,501),(503,501),(503,501),(503,501)----;(253,249),(253,249),(253,249),(253,249)----.3.若函数()2log 1a y x ax =-+有最小值,则a 的取值范围是( ). A 01a << B 02,1a a <<≠ C 12a << D 2a ≥ 答案:C .解:当01a <<时,log a y x =是递减函数,由于21t x ax =-+没有最大值,所以()2log 1a y x ax =-+没有最小值;当1a >时,()2log 1a y x ax =-+有最小值等价于21t x ax =-+有大于0的最小值.这等价于240a ∆=-<,因此12a <<.4.已知,1,=>ab b a 则b a b a -+22的最小值是( ).A 22B 2C 2D 1答案:A.解:记t b a =-,则0>t ,b a b a -+2222222≥+=+=t t t t ,(当且仅当22t a b ===即时取等号).故选A . 5.已知cos cos 1x y +=,则sin sin x y -的取值范围是( ).A []11-,B []2-,2C 0⎡⎣D ⎡⎣答案:D .解:设sin sin x y t -=,易得21cos cos sin sin 2t x y x y --=,即()21co s 2t x y -+=.由于()1cos 1x y -≤+≤,所以21112t --≤≤,解得 t ≤≤ 6.函数()f x 是(0,)+∞上的单调递增函数,当*n N ∈时,*()f n N ∈,且[()]3f f n n =,则(1)f 的值等于( ).A 1B 2C 3D 4 答案:B解:(用排除法)令1n =,则得[(1)]3f f =. 若(1)1f =,则[(1)](1)3f f f ==,与(1)1f =矛盾;若(1)3f =,则[(1)](3)3f f f ==,与“()f x 在(0,)+∞上单调递增”矛盾; 若(1)4f =,则[(1)](4)3f f f ==,也与“()f x 在(0,)+∞上单调递增”矛盾. 故选B .二、填空题(本大题共6小题,每小题9分,满分54分)7.设集合{}1215S = ,,,,{}123A a a a =,,是S 的子集,且()123a a a ,,满足:123115a a a ≤≤<<,326a a -≤,那么满足条件的子集的个数为 . 答案:371.解:当229a ≤≤时,()12,a a 有29C 种选择方法,3a 有6种选择方法,所以()123,,a a a 共有296216C ⨯=种选择方法;当21014a ≤≤时,一旦2a 取定,1a 有21a -种选择方法,3a 有215a -种选择方法,所以选择()123,,a a a 的方法有()()214221011595104113122131155a aa =--=⨯+⨯+⨯+⨯+⨯=∑种.综上,满足条件的子集共有371个.8.已知数列}{n a 满足,01=a ),2,1(1211 =+++=+n a a a n n n ,则n a =___ . 答案:12-=n a n .解:由已知得21)11(11211++=++++=++n n n n a a a a ,且01>+n a . 所以1111++=++n n a a ,即{1+n a }是首项、公差均为1的等差数列,所以1+n a =n ,即有12-=n a n .9.已知坐标平面上三点()())0,3,0,0A B C,P 是坐标平面上的点,且PA PB PC =+,则P 点的轨迹方程为 . 答案:()()04122≤=-+y y x .解:如图,作正三角形PCD ,由于ABC ∆也是正三角形,所以可证得 ACP ∆≌BCD ∆,所以BD AP =.又因为BD PB PC PB PD =+=+,所以点D P B ,,共线.CBP PAC ∠=∠,所以P 点在ABC ∆的外接圆上,又因为,PA PB PA PC >>,所以所求的轨迹方程为()()04122≤=-+y y x .10. 在三棱锥ABC S -中,4=SA ,7≥SB ,9≥SC ,5=AB ,6≤BC ,8≤AC .则三棱锥ABC S -体积的最大值为 .答案:.解:设SAB α∠=,根据余弦定理有2222224571cos 22455SA AB SB SA AB α+-+-=≤≤-⨯⨯⨯⨯,故sin α=≤1sin 2SAB S SA AB α∆=⨯⨯≤过它的侧棱长,所以13C SAB SAB V S BC -∆≤⨯≤事实上,取7=SB ,6=BC 且CB SAB ⊥平面时,可以验证满足已知条件,此时68=SABC V ,棱锥的体积可以达到最大.11. 从m 个男生,n 个女生(104m n ≥>≥)中任选2个人当组长,假设事件A 表示选出的2个人性别相同,事件B 表示选出的2个人性别不同.如果A 的概率和B 的概率相等,则(m ,n )的可能值为 . 答案:(10,6).解:()()221122,m n m n m n m nC C C C P A P B C C +++==,由于()()P A P B =,所以2211m n m n C C C C +=,整理得()2m n m n -=+.即m n +是完全平方数,且919m n ≤+≤,因此93m n m n +=⎧⎨-=⎩,164m n m n +=⎧⎨-=⎩,解得 63m n =⎧⎨=⎩(不合条件),106m n =⎧⎨=⎩.所以()(),10,6m n =.12.,,O A B 是平面上不共线三点,向量a OA =,OB b =,设P 为线段AB 垂直平分线上任意一点,向量p =.若||5a = ,||3b = ,则)(b a p-⋅的值是 ____ ____. 答案:8.解:如图,QP 是线段AB 的垂直平分线,OP OQ QP =+,()12OQ a b =+ ,QP BA ⊥ ,()()p a b OQ QP BA OQ BA QP BA ⋅-=+⋅=⋅+⋅()11()()822a b a b a b 22=+⋅-=||-||= . 三、解答题(本大题共5小题,每题的解答均要求有推理过程,13小题10分,17小题14分,其余每小题12分,满分60分)13.b a ,是两个不相等的正数,且满足2233b a b a -=-,求所有可能的整数c ,使得ab c 9=.解:由2233b a b a -=-得b a b ab a +=++22,所以0)()(2>+-+=b a b a ab , 由此得到1>+b a .又因为)()()(4122b a b a ab b a +-+=>+,故341<+<b a .………………………4分ABOPQ又因为)()(2b a b a ab +-+=, 令 )34,1(∈+=b a t 则t t ab -=2.……………6分当1t ≥时,2t t -关于t 单调递增,所以409ab <<,094ab <<.因此 c 可以取1,2,3. …………………………………………………………………10分14.如图,斜三棱柱111C B A ABC -的所有棱长均为a ,侧面⊥CB C B 11底面ABC ,且BC AC ⊥1. (1) 求异面直线1AA 与11C B 间的距离;(2) 求侧面BA B A 11与底面ABC 所成二面角的度数. 解:(1)如图,取BC 中点D ,连1,AD C D .BC D C ADC BC BC AC BC AD ⊥⇒⊥⇒⎭⎬⎫⊥⊥111平面. ABC CB C B 底面平面⊥11 ,∴ABC D C 平面⊥1.由C C BB AD BC AD 11平面知⊥⊥.……………4分1AA ∥ 1CC 1AA⇒∥平面C C BB 11. 所以异面直线1AA 与11C B 间的距离等于=AD a 23.……………6分 (2)如图,111,,.B BO BC BC O BO ABC ⊥⊥过作交于则底面 1,,.O OE AB AB E B E ⊥过作交于连1B EO ∠则与所求二面角的平面角互补 (8)分1111,,.tan 2.2B O a B O C D OB OE B EO OE ====∠===1arctan 2.arctan 2B EO π∠=-所以二面角的度数为.……………………12分15.设向量,为直角坐标平面内x 轴,y 轴正方向上的单位向量.若向量ABC1A1B B1C AB C 1A1B 1C ED Oy x ++=2(,y x +-=2(,且a b ||-||=2. (1)求满足上述条件的点),(y x P 的轨迹方程;(2)设(1,0),(2,0)A F -,问是否存在常数)0(>λλ,使得PAFPFA ∠=∠λ恒成立?证明你的结论.解:(1)由条件a b ||-||=2可知:2)2()2(2222=+--++y x y x .由双曲线定义,得点P 的轨迹方程:)0(1322>=-x y x .…………………4分 (2)在第一象限内作(2,3)PF x P ⊥轴,点坐标为,此时90,PFA ∠=45.PAF ∠= 2=λ.…………………………………….………………….……6分以下证明当PF 与x 轴不垂直且P 在第一象限时,PAF PFA ∠=∠2恒成立.1111222111122(1),tan 2.121()(1)PA PF PA PA y y k x y k k PAF x x k x y +==∠==+--+-,则 由1322=-y x ,得)1)(1(3)1(3112121-+=-=x x x y . 代入上式并化简得1111tan 2,tan .22PF y yPAF PFA k x x ∠=-∠=-=---……10分 tan 2tan 2.PAF PFA PFA PAF ∠=∠∠=∠即,所以由对称性知,当P 在第四象限时,同样成立.故存在常数2=λ,使得PAF PFA ∠=∠2恒成立.………………….………12分16.在数列{}n a 中,1a ,2a 是给定的非零整数,21n n n a a a ++=-. (1)若152a =,161a =-,求2008a ;(2)证明:从{}n a 中一定可以选取无穷多项组成两个不同的常数数列. 解:(1)∵152a =,161a =-,173a =,184a =,191a =,203a =,212a =,221a =,231a =,240a =,251a =,261a =,270a =,……∴自第22项起,每三个相邻的项周期地取值1,1,0,故2008a =1.……4分 (2)首先证明数列{}n a 必在有限项后出现零项.假设{}n a 中没有零项,由于21n n n a a a ++=-,所以.3n ≥时,都有1n a ≥.……………………6分 当1n n a a +>时,2111n n n n a a a a +++=-≤-(3n ≥); 当1n n a a +<时,211n n n n a a a a ++=-≤-(3n ≥),即2n a +的值要么比1n a +至少小1,要么比n a 至少小1.…………………8分令2121222+22122 () ()n n n n n n n a a a b a a a +++++>⎧=⎨<⎩,1,2,...n =,则101n n b b +<≤-.由于1b 是确定的正整数,这样下去,必然存在某项0k b <,这与0k b >矛盾,从而{}n a 中必有零项.……………………………………………………….……10分若第一次出现的零项为n a ,记1 (0)n a M M -=≠,则自第n 项开始,每三个相邻的项周期地取值0,,M M ,即331320n k n k n k a a M aM+++++=⎧⎪=⎨⎪=⎩,0,1,2...k =所以数列{}n a 中一定可以选取无穷多项组成两个不同的常数数列.……12分 17. 设定义在[0,2]上的函数()f x 满足下列条件:①对于[0,2]x ∈,总有(2)()f x f x -=,且()1f x ≥,(1)3f =; ②对于,[1,2]x y ∈,若3x y +≥,则()()(2)1f x f y f x y +≤+-+. 证明:(1)12()133n n f ≤+(*n N ∈);(2)[1,2]x ∈时,1()136f x x ≤≤-. 证明:由(2)()f x f x -=知,函数()f x 图像关于直线1x =对称,则根据②可知:对于,[0,1]x y ∈,若1x y +≤,则()()()1f x y f x f y +≥+-.……………2分设12,[0,1]x x ∈,且12x x <,则21[0,1]x x -∈.∵2112111211()()[()]()()()1()f x f x f x x x f x f x f x x f x -=+--≥+---21()10f x x =--≥,∴()f x 在[0,1]上是不减函数.………………………………………………4分(1)∵111111111()()()()13()233333333n n n n n n n nf f f f f -=++≥++-≥-, ∴1222111211221122()()()...() (333333333333)n n n n n n n f f f f ---≤+≤++≤≤+++111211333n n n -=+-=+.…………………………………………………………8分 (2)对于任意(0,1]x ∈,则必存在正整数n ,使得11133n n x -≤≤.因为()f x 在(0,1)上是不减函数,所以111()()()33n n f f x f -≤≤,由(1)知11121()16161333n n n f x --≤+=+≤+.由①可得(2)1f ≥,在②中,令2x y ==,得(2)1f ≤,∴(2)1f =. 而(2)(0)f f =,∴(0)1f =,又1()(0)3n f f ≥,∴1()13n f ≥, ∴[0,1]x ∈时,1()61f x x ≤≤+..………………………………………12分 ∵[1,2]x ∈时,2[0,1]x -∈,且()(2)f x f x =-,∴1(2)6(2)1136f x x x ≤-≤-+=-,因此,[1,2]x ∈时,1()136f x x ≤≤-.…………………….………….14分。

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International Kangaroo Mathematics Contest 2008Cadet Level: Class (7 & 8)Max Time: 2 Hours3-point problems1) How many pieces of string are there in the picture?A) 3B) 4 D) 62)In a class there are 9 boys and 13 girls. Half of the children in this class have got a cold. How many girls at least have a cold?A) 0 B) 1C) 2D) 33)6 kangaroos eat 6 sacks of grass in 6 minutes. How many kangaroos will eat 100 sacks of grass in 100 minutes?A) 100 B) 60 C) 6 D) 6004)Numbers 2, 3, 4 and one more number are written in the cells of 2 × 2 table. It is known that the sums of the numbers in the first row are equal to 9, and the sum of the numbers in the second row is equal to 6. The unknown number isA) 5 B) 6 C) 7 D) 85)The triangle and the square have the same perimeter. What is the perimeterof the whole figure (a pentagon)?A) 24 cm B) 28 cm C) 32 cmD) It depends upon the triangle measuresA florist had 24 white, 42 red and 36 yellow roses left. At most, how many identical bunches can she make, if she wants to use all the remaining flowers?A) 4 B) 6 C) 8 D) 127)A cube has all its corners cut off, as shown. How many edges does the resultingshape have?A) 30 B) 36 C) 40D) Another answerAli has 9 coins (each is worth 2 cents); while his sister Saima has 8 coins, each being 5 cents. What the least number of coins they should interchange (with each other) in order to equalize their money?A) 4 B) 5 C) 12 D) it is impossible to do10)How many squares can be drawn by joining the dots with line segments?A) 2 B) 3 C) 4 D) 54-point problems11)If there are two buses on the circular bus route, the interval between them is 25 min. How many extra buses are necessary to shorten the interval by 60%?A) 2 B) 3 C) 5 D) 612)The French mathematician August de Morgan claimed that he was хyears old in the yearof х2. He is known to have died in 1899. When was he born?A) 1806 B) 1848 C) 1849 D) another answerWe decide to visit by ferry-boat four islands A,B,C & D starting from the mainland. B can be reached only from A or from the mainland, A & C are connected to each other and with the mainland and D is connected only with A. Which is the minimum number of ferry runs that we need, if we want to visit all the islands?A)6 B)5 C) 4 D)714)Tom and Jerry cut two equal rectangles. Tom got two rectangles with the perimeter of 40 cm each, and Jerry got two rectangles with the perimeter of 50 cm each. What were the perimeters of the initial rectangles?A) 40 cm B) 50 cm C) 60 cm D) 80 cm15)net is impossible?A) 1 and 3 B) 1 and 5 C) 3 and 4 D) 3 and 516)Points A, B, C and D are marked on the straight line in some order. It is known that AB= 13,BC= 11, CD= 14 and DA= 12. What is the distance between the farthest two points?A) 14 B) 38 C) 25 D) another answer17)Four tangent congruent circles of radius 6 cm are inscribed in arectangle. If P is a vertex and Q and R are points of tangency, what isthe area of triangle PQR?A) 27 cm2B) 45 cm2C) 54 cm2D) 108 cm218)Seven cards lie in a box. Numbers from 1 to 7 are written on these cards (exactly one numberon the card). The first sage takes, at random, 3 cards from the box and the second sage takes 2 cards (2 cards are left in the box). Then the first sage tells to the second one: “I know that thesum of the numbers of your cards is even”. The sum of card’s numbers of the first sage is equal toA) 10 B) 12 C) 9 D) 15PRQIn an isosceles triangle ABC, the bisector CD of the angle C is equal to the base BC. Then the angle CDA is equal toA) 100º B) 108º C) 120º D) impossible to determine20)A wooden cube 11 x 11 x 11 is obtained by sticking together 113unit cubes. What is the largest number of unit cubes visible from a same point of view?A) 329 B) 330 C) 331 D) 3325-point problems21)In the equality KAN – GAR = OO any letter stands for some digit (different letters for different digits, equal letters for equal digits). Find the largest possible value of the number KAN ?A) 876 B) 865 C) 864 D) 78522)A boy always speaks the truth on Thursday and Fridays, always tells lies on Tuesdays, and randomly tells the truth or lies on other days of the week. On seven consecutive days he was asked what his name was, and on the first six days he gave the following answers in order: Akbar, Ali, Akbar, Ali, Farooq, Ali. What did he answer on the seventh day?A) Akbar B) Ali C) Amir D) another answer23)Four identical dice are arranged in a row (see the fig.). The dice arenot standard, i.e., the sum of points in the opposite faces of the dicenot necessarily equals 7. Find the total sum of the points in all 6touching faces of the dice.A) 19 B) 20 C) 21 D) 2224)Some straight lines are drawn on the plane so that all angles 10°, 20°, 30°, 40°, 50°, 60°, 70°, 80°, 90° are among the angles between these lines. Determine the smallest possible number of these straight lines.A) 4 B) 5 C) 6 D) 725)On my first spelling test, I score one mark out of five. If I now work hard and get full marks on every test, who many more tests should I take for my average to be four out of five correct answers?A) 2 B) 3 C) 4 D) 5_______________________________GOOD LUCK !。

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