AMC12美国数学竞赛 2012-2014

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2014AMC12-B

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 12 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. SCORING: You will receive 6 points for each correct answer, 1.5 points for each problem left unanswered, and 0 points for each incorrect answer.5. No aids are permitted other than scratch paper, graph paper, rulers, compass, protractors, and erasers. No calculators are allowed. 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 75 minutes to complete the test.9. When you finish the exam, sign your name in the space provided on the Answer Form.© 2014 Mathematical Association of AmericaThe Committee on the American Mathematics Competitions (CAMC) reserves the right to re-examine students before deciding whether to grant official status to their scores. The CAMC also reserves the right to disqualify all scores from a school if it is determined that the required security procedures were not followed.Students who score 100 or above or finish in the top 5% on this AMC 12 will be invited to take the 32nd annual American Invitational Mathematics Examination (AIME) on Thursday, March 13, 2014 or Wednesday, March 26, 2014. More details about the AIME and other information are on the back page of this test booklet.The publication, reproduction or communication of the problems or solutions of the AMC 12 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 problems in paper or electronic form including posting on web-pages for educational use is granted without fee provided that copies are notmade or distributed for profit or commercial advantage and that copies bear the copyright notice.**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 FEBRUARY 19, 2014. Nothing is needed from inside this package until February 19.2. Your PRINCIPAL or VICE-PRINCIPAL must verify on the AMC 12 CERTIFICATION FORM (found in the Teachers’ Manual) that you followed all rules associated with the conduct of the exam.3. The Answer Forms must be mailed by trackable mail to the AMC office no later than 24 hours following the exam.4. 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, email, internet or media of any type is a violation of the competition rules.2014AMC 12 B DO NOT OPEN UNTIL wEDNEsDAy, fEbrUAry 19, 2014The American Mathematics Competitionsare Sponsored byThe Mathematical Association of America – MAA The Akamai Foundation ContributorsAcademy of Applied sciences – AAs American Institute of Mathematics - AIM American Mathematical Association of Two-year Colleges – AMATyC ........................................................... American Mathematical society – AMs ............................................................................................................. American statistical Association – AsA ......................................................................................................... Art of Problem solving – Association of symbolic Logic - AsL Awesome Math Casualty Actuarial society – CAs ................................................................................................................... Conference board of the Mathematical sciences - CbMs The D.E. shaw Group ................................................................................................................................... IDEA Jane street Capital Math for America Math Training Center Mu Alpha Theta – MAT......................................................................................................................... Pi Mu Epsilon – PME ................................................................................................................................. society for Industrial and Applied Math – sIAM ................................................................................................ w.H. freeman 1.Leah has13coins,all of which are pennies and nickels.If she had one morenickel than she has now,then she would have the same number of pennies and nickels.In cents,how much are Leah’s coins worth?(A)33(B)35(C)37(D)39(E)412.Orvin went to the store with just enough money to buy30balloons.Whenhe arrived he discovered that the store had a special sale on balloons:buy1balloon at the regular price and get a second at13oﬀthe regular price.Whatis the greatest number of balloons Orvin could buy?(A)33(B)34(C)36(D)38(E)393.Randy drove theﬁrst third of his trip on a gravel road,the next20miles onpavement,and the remaining one-ﬁfth on a dirt road.In miles,how long was Randy’s trip?(A)30(B)40011(C)752(D)40(E)30074.Susie pays for4muﬃns and3bananas.Calvin spends twice as much payingfor2muﬃns and16bananas.A muﬃn is how many times as expensive as a banana?(A)32(B)53(C)74(D)2(E)1345.Doug constructs a square window using8equal-size panes of glass,as shown.The ratio of the height to width for each pane is5:2,and the borders around and between the panes are2inches wide.In inches,what is the side length of the square window?(A)26(B)28(C)30(D)32(E)346.Ed and Ann both have lemonade with their lunch.Ed orders the regular size.Ann gets the large lemonade,which is50%more than the regular.After bothconsume34of their drinks,Ann gives Ed a third of what she has left,and2additional ounces.When theyﬁnish their lemonades they realize that they both drank the same amount.How many ounces of lemonade did they drink together?(A)30(B)32(C)36(D)40(E)507.For how many positive integers n is n30−nalso a positive integer?(A)4(B)5(C)6(D)7(E)88.In the addition shown below A,B,C,and D are distinct digits.How manydiﬀerent values are possible for D?ABBCB+BCADADBDDD(A)2(B)4(C)7(D)8(E)99.Convex quadrilateral ABCD has AB=3,BC=4,CD=13,AD=12,and ∠ABC=90◦,as shown.What is the area of the quadrilateral?(A)30(B)36(C)40(D)48(E)58.510.Danica drove her new car on a trip for a whole number of hours,averaging55miles per hour.At the beginning of the trip,abc miles was displayed on the odometer,where abc is a3-digit number with a≥1and a+b+c≤7.At the end of the trip,the odometer showed cba miles.What is a2+b2+c2?(A)26(B)27(C)36(D)37(E)4111.A list of11positive integers has a mean of10,a median of9,and a unique modeof8.What is the largest possible value of an integer in the list?(A)24(B)30(C)31(D)33(E)3512.A set S consists of triangles whose sides have integer lengths less than5,andno two elements of S are congruent or similar.What is the largest number of elements that S can have?(A)8(B)9(C)10(D)11(E)1213.Real numbers a and b are chosen with1<a<b such that no triangle withpositive area has side lengths1,a,and b or1b ,1a,and1.What is the smallestpossible value of b?(A)3+√32(B)52(C)3+√52(D)3+√62(E)314.A rectangular box has a total surface area of94square inches.The sum of thelengths of all its edges is48inches.What is the sum of the lengths in inches of all of its interior diagonals?(A)8√3(B)10√2(C)16√3(D)20√2(E)40√215.When p= 6k=1k ln k,the number e p is an integer.What is the largest powerof2that is a factor of e p?(A)212(B)214(C)216(D)218(E)22016.Let P be a cubic polynomial with P(0)=k,P(1)=2k,and P(−1)=3k.Whatis P(2)+P(−2)?(A)0(B)k(C)6k(D)7k(E)14k17.Let P be the parabola with equation y =x 2and let Q =(20,14).There are realnumbers r and s such that the line through Q with slope m does not intersect P if and only if r <m <s .What is r +s ?(A)1(B)26(C)40(D)52(E)8018.The numbers 1,2,3,4,5are to be arranged in a circle.An arrangement isbad if it is not true that for every n from 1to 15one can ﬁnd a subset of the numbers that appear consecutively on the circle that sum to n .Arrangements that diﬀer only by a rotation or a reﬂection are considered the same.How many diﬀerent bad arrangements are there?(A)1(B)2(C)3(D)4(E)519.A sphere is inscribed in a truncated right circular cone as shown.The volumeof the truncated cone is twice that of the sphere.What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?(A)32(B)1+√52(C)√3(D)2(E)3+√5220.For how many positive integers x is log 10(x −40)+log 10(60−x )<2?(A)10(B)18(C)19(D)20(E)inﬁnitely many21.In theﬁgure,ABCD is a square of side length1.The rectangles JKHG andEBCF are congruent.What is BE?(A)12(√6−2)(B)14(C)2−√3(D)√36(E)1−√2222.In a small pond there are eleven lily pads in a row labeled0through10.A frogis sitting on pad1.When the frog is on pad N,0<N<10,it will jump topad N−1with probability N10and to pad N+1with probability1−N10.Eachjump is independent of the previous jumps.If the frog reaches pad0it will be eaten by a patiently waiting snake.If the frog reaches pad10it will exit the pond,never to return.What is the probability that the frog will escape being eaten by the snake?(A)3279(B)161384(C)63146(D)716(E)1223.The number2017is prime.Let S= 62k=02014k.What is the remainder whenS is divided by2017?(A)32(B)684(C)1024(D)1576(E)201624.Let ABCDE be a pentagon inscribed in a circle such that AB=CD=3,BC=DE=10,and AE=14.The sum of the lengths of all diagonals ofABCDE is equal to mn ,where m and n are relatively prime positive integers.What is m+n?(A)129(B)247(C)353(D)391(E)42125.What is the sum of all positive real solutions x to the equation2cos(2x)cos(2x)−cos2014π2x=cos(4x)−1?(A)π(B)810π(C)1008π(D)1080π(E)1800πWRITE TO US!Correspondence about the problems and solutions for this AMC 12and orders for publications should be addressed to:American Mathematics Competitions1740 Vine StreetLincoln, NE 68508-1228Phone 402-472-2257 | Fax 402-472-6087 | amcinfo@The problems and solutions for this AMC 12 were prepared by the MAA’s Committee on theAMC 10 and AMC 12 under the direction of AMC 12 Subcommittee Chair:Prof. Bernardo M. Abrego2014 AIMEThe 32nd annual AIME will be held on Thursday, March 13, with the alternate on Wednesday, March 26. It is a 15-question, 3-hour, integer-answer exam. You will be invited to participate only if you score 120 or above or finish in the top 2.5% of the AMC 10, or if you score 100 or above or finish in the top 5% of the AMC 12. T op-scoring students on the AMC 10/12/AIME will be selected to take the 43rd Annual USA Mathematical Olympiad (USAMO) on April 29-30, 2014. The best way to prepare for the AIME and USAMO is to study previous exams. Copies may be ordered as indicated below.PUBLICATIONSA complete listing of current publications, with ordering instructions, is at our web site:/math-competitions American Mathematics Competitions。

2012美国数学竞赛AMC12B

2012 AMC 12B ProblemsProblem 1Each third-grade classroom at Pearl Creek Elementary has 18 students and 2 pet rabbits. How many more students than rabbits are there in all 4 of the third-grade classrooms?SolutionProblem 2A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?SolutionProblem 3For a science project, Sammy observed a chipmunk and squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?SolutionProblem 4Suppose that the euro is worth 1.30 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?SolutionProblem 5Two integers have a sum of 26. when two more integers are added to the first two, the sum is 41. Finally, when two more integers are added to the sum of the previous 4 integers, the sum is 57. What is the minimum number of even integers among the 6 integers?SolutionProblem 6In order to estimate the value of where and are real numbers with , Xiaoli rounded up by a small amount, rounded down by the same amount, and then subtracted her rounded values. Which of the following statements is necessarily correct?SolutionProblem 7Small lights are hung on a string 6 inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of 2 red lights followed by 3 green lights. How many feet separate the 3rd red light and the 21st red light?Note: 1 foot is equal to 12 inches.SolutionProblem 8A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?SolutionProblem 9It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How seconds would it take Clea to ride the escalator down when she is not walking?SolutionProblem 10What is the area of the polygon whose vertices are the points of intersection of the curves and ?SolutionProblem 11In the equation below, and are consecutive positive integers, and , , and represent number bases: What is ?SolutionProblem 12How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?SolutionProblem 13Two parabolas have equations and , where , , , and are integers, eachchosen independently by rolling a fair six-sided die. What is the probability that the parabolas will have a least one point in common?SolutionProblem 14Bernardo and Silvia play the following game. An integer between 0 and 999 inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she addes 50 to it and passes the result to Bernardo. The winner is the last person who produces a number less than 1000. Let N be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of N?SolutionProblem 15Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?SolutionProblem 16Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those girls but disliked by the third. In how many different ways is this possible?SolutionProblem 17Square lies in the first quadrant. Points and lie on lines and , respectively. What is the sum of the coordinates of the center of the square ?SolutionProblem 18Let be a list of the first 10 positive integers such that for each either oror both appear somewhere before in the list. How many such lists are there?SolutionProblem 19A unit cube has vertices and . Vertices , , and are adjacent to , and forvertices and are opposite to each other. A regular octahedron has one vertex in each of the segments , , , , , and . What is the octahedron's side length?SolutionProblem 20A trapezoid has side lengths 3, 5, 7, and 11. The sums of all the possible areas of the trapezoid can be written in the form of , where , , and are rational numbers and and are positive integers not divisible by the square of any prime. What is the greatest integer less than or equal to ?SolutionProblem 21Square is inscribed in equiangular hexagon with on , on , and on . Suppose that , and . What is the side-length of the square?SolutionProblem 22A bug travels from to along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?SolutionProblem 23Consider all polynomials of a complex variable, , where and are integers, , and the polynomial has a zero with What is the sum of all values over all the polynomials with these properties?SolutionProblem 24\item Define the function on the positive integers by setting and if is the prime factorization of , then For every , let. For how many in the range is the sequenceunbounded?Note: A sequence of positive numbers is unbounded if for every integer , there is a member of the sequence greater than .Problem 25\item Let . Let be the set of all right triangles whose vertices are in . For every right triangle with vertices , , and in counter-clockwise order and right angle at , let . What is。

历届美国数学建模竞赛赛题(汉语版)

历届美国数学建模竞赛赛题, 1985－2006AMCM1985问题-A 动物群体的管理AMCM1985问题-B 战购物资储备的管理AMCM1986问题-A 水道测量数据AMCM1986问题-B 应急设施的位置AMCM1987问题-A 盐的存贮AMCM1987问题-B 停车场AMCM1988问题-A 确定毒品走私船的位置AMCM1988问题-B 两辆铁路平板车的装货问题AMCM1989问题-A 蠓的分类AMCM1989问题-B 飞机排队AMCM1990问题-A 药物在脑内的分布AMCM1990问题-B 扫雪问题AMCM1991问题-A 估计水塔的水流量AMCM1992问题-A 空中交通控制雷达的功率问题AMCM1992问题-B 应急电力修复系统的修复计划AMCM1993问题-A 加速餐厅剩菜堆肥的生成AMCM1993问题-B 倒煤台的操作方案AMCM1994问题-A 住宅的保温AMCM1994问题-B 计算机网络的最短传输时间AMCM1995问题-A 单一螺旋线AMCM1995问题-B A1uacha Balaclava学院AMCM1996问题-A 噪音场中潜艇的探测AMCM1996问题-B 竞赛评判问题AMCM1997问题-A Velociraptor(疾走龙属)问题AMCM1997问题-B为取得富有成果的讨论怎样搭配与会成员AMCM1998问题-A 磁共振成像扫描仪AMCM1998问题-B 成绩给分的通胀AMCM1999问题-A 大碰撞AMCM1999问题-B “非法”聚会AMCM1999问题- C 大地污染AMCM2000问题-A空间交通管制AMCM2000问题-B: 无线电信道分配AMCM2000问题-C：大象群落的兴衰AMCM2001问题- A: 选择自行车车轮AMCM2001问题-B：逃避飓风怒吼（一场恶风…）AMCM2001问题-C我们的水系-不确定的前景AMCM2002问题-A风和喷水池AMCM2002问题-B航空公司超员订票AMCM2002问题-C蜥蜴问题AMCM2003问题-A: 特技演员AMCM2003问题-C航空行李的扫描对策AMCM2004问题-A：指纹是独一无二的吗？AMCM2004问题-B：更快的快通系统AMCM2004问题-C：安全与否？AMCM2005问题-A：.水灾计划AMCM2005问题-B：TollboothsAMCM2005问题-C：.Nonrenewable ResourcesAMCM2006问题-A：用于灌溉的自动洒水器的安置和移动调度AMCM2006问题-B：通过机场的轮椅AMCM2006问题-C：在与HIV/爱滋病的战斗中的交易AMCM85问题-A 动物群体的管理在一个资源有限，即有限的食物、空间、水等等的环境里发现天然存在的动物群体。

2012美国大学生数学建模竞赛_MCM_A题

SummaryMany scholars conclude that leaf shape is highly related with the veins. Based on this theory, we assume the leaf growth in each direction satisfies a function. For the leaves in the same tree, the parameters are different; for those of separate trees, the function mode is different. Thus the shape of leaf differs from that of another. In the end of section 3, we simulate one growing period and depict the leaf shape.Through thousands of years of evolution, the leaves find various wa ys to make a full use of natural resources, including minimizing overlapping individual shadows. In order to find the main factors promoting the evolution of leaves, we analyze the distribution of adjacent leaves and the equilibrium point of photosynthesis and respiration. Besides, we also make a coronary hierarchical model and transmission model of the solar radiation to analyze the influence of the branches.As to the tree structure and the leaf shape, first we consider one species. Different tree shapes have different space which is built up by the branch quality and angle, effect light distribution, ventilation and humidity and concentration of CO2 in the tree crown. These are the factors which affect the leaf shape according to the model in section 1. Here we analyze three typical tree shapes: Small canopy shape, Open center shape and Freedom spindle shape, which can be described by BP network and fractional dimension model. We find that the factors mainly affect the function of Sthat affects the additional leaf area. Factors are assembled in different ways to create different leaf shapes. So that the relationship between leaf shape and tree profile/branching structure is proved.Finally we develop a model to calculate the leaf mass from the basic formula of . By adjusting the crown of a tree to a half ellipsoid, we first define thefunction of related factors,such as the leaf density and the effective ratio of leaf area. Then we develop the model using calculus. With this model, weapproximately evaluate the leaf mass of a middle-sized tree is 141kg.Dear editor,How much the leaves on a tree weigh is the focus of discussion all the time. Our team study on the theme following the current trend and we find something interesting in the process.The tree itself is component by many major elements. In our findings, we analyze the leaf mass with complicated ones, like leaf shape, tree structure and branch characteristics, which interlace with each other.With the theory that leaf shape is highly related with the veins, we assume theleaf growth in each direction satisfies a function . For the leaves in thesame tree, the parameters are different; for those of separate trees, the function mode is di fferent. That’s why no leaf shares the same shape. Also, we simulate one growing period and depict the leaf shape.In order to find the main factors promoting the evolution of leaves, we analyze the distribution of adjacent leaves and the equilibrium point of photosynthesis and respiration. Besides, we also make a coronary hierarchical model and transmission model of the solar radiation to analyze the influence of the branches.As to the tree structure and the leaf shape, different tree shapes have differe nt space which is built up by the branch quality and angle, effect light distribution, ventilation and humidity and concentration of CO2 in the tree crown that affect leaf shapes. Here we analyze three typical tree shapes which can be described by BP network and fractional dimension model. We find that the factors mainly affect the function of Sthat affects the additional leaf area. Factors are assembled in different ways to create different leaf shapes. So that the relationship between leaf shape and tree profile or branching structure is proved.Finally we develop the significant model to calculate the leaf mass from thebasic formula of . By adjusting the crown of a tree to a half ellipsoid,we first define the function of related factors and then we develop the model using calculus. With this model, we approximately evaluate the leaf mass of a middle-sized tree is 141kg.We are greatly appreciated that if you can take our findings into consideration. Thank you very much for your precious time for reading our letter.Yours sincerely,Team #14749Contents1. Introduction (4)2. Parameters (4)3. Leaves have their own shapes (5)3.1 Photosynthesis is important to plants (5)3.2 How leaves grow? (6)3.3 Build our model (7)3.4 A simulation of the model (10)4. Do the shapes maximize exposure? (14)4.1 The optimum solution of reducing overlapping shadows (14)4.1.1 The distribution of adjacent leaves (14)4.1.2 Equilibrium point of photosynthesis and respiration (15)4.2 The influence of the “volume” of a tree and its branches (17)4.2.1 The coronary hierarchical model (17)4.2.2 Spatial distribution model of canopy leaf area (18)4.2.3 Transmission model of the solar radiation (19)5. Is leaf shape related to tree structure? (20)5.1 The experiment for one species (21)5.2 Different tree shapes affect the leaf shapes (23)5.2.1 The light distribution in different shapes (23)5.2.2 Wind speed and humidity in the canopy (23)5.2.3 The concentration of carbon dioxide (24)5.3 Conclusion and promotion (25)6. Calculus model for leaf mass (26)6.1 How to estimate the leaf mass? (26)6.2 A simulation of the model (28)7. Strengths and Weakness (29)7.1 Strengths (29)7.2 Weaknesses (30)8. Reference (30)1. IntroductionHow much do the leaves on a tree weigh? Why do leaves have the various shapes that they have? How might one estimate the actual weight of the leaves? How might one classify leaves?We human-beings have never stopped our steps on exploring the natural world. But, as a matter of fact, the answer to those questions is still unresolved. Many scientists continue to study on this area. Recently , Dr. Benjamin Blonder (2010) achieved a new breakthrough on the venation networks and the origin of the leaf econo mics spectrum. They defined a standardized set of traits – density , distance and loopiness and developed a novel quantitative model that uses these venation traits to model leaf-level physiology .Now, it is commonly thought that there are four key leaf functional traits related to leaf economics: net carbon assimilation rate, life span, leaf mass per area ratio and nitrogen content.2. Parametersthe area a leaf grows decided by photosynthesisthe additional leaf area in one growing periodthe leaf growing obliquity Pthe total photosynthetic rate 0d R the dark respiration rate of leavesn P the net photosynthetic rateh the height of the canopyd the distance between two branchesdi the illumination intensity of scattered light from a given directionthe solar zenith angleh the truck highh the crown high13. Leaves have their own shapes3.1 Photosynthesis is important to plantsIt is widely accepted that two leaves are different, no matter where they are chosen from; even they are from the very tree. To understand how leaves grow is helpful to answer why leaves have the various shapes that they have.The canopy photosynthesis and respiration are the central parts of most biophysical crop and pasture simulation models. In most models, the acclamatory responses of protein and the environmental conditions, such as light, temperature and CO2 concentration, are concerned[1].In 1980, Farquhar et al developed a model named FvCB model to describe photosynthesis[2]:The FvCB model predicts the net assimilation rate by choosing the minimum between the Rubisco-limited net photosynthetic rate and the electron transport-limited net photosynthetic rate.Assume A n, A c, A j are the symbols for net assimilation rate, the Rubisco-limited net photosynthetic rate and the electron transport-limited net photosynthetic rate respectively, and the function can be described as:(1)(2)where and are the intercellular partial pressures of CO2 and O2,respectively, and are the Michaelis–Menten coefficients ofRubisco for CO2and O2, respectively, is the CO2compensation point inthe absence of (day respiration in andis the photosystem II electron transport rate that is used for CO2fixation and photorespiration[3][4].We apply the results of this model to build the relationship between the photosynthesis and the area a leaf grows during a period of time. It can be released as:(3)and are the area of the target leaf and the period of time it grows.is a function which can transfer the amount ofCO2into the area the leaf grows and the are parameters which affect S p. S p can be used as a constraint condition in our model.3.2 How leaves grow?As the collocation of computer hardware and software develops, people can refer to bridging biology, morphogenesis, applied mathematics and computer graphics to simulate living organisms[5], thus how to model leaves is of great challenge. In 2001, Dengler and Kang[6]brought up the thought that leaf shape is highly related to venation patterns. Recently, Runions[7] brought up a method to portray the leaf shape by analyzing venation patterns. Together with the Lindenmayer system (L-system), an advanced venation model can adjust the growth better that it solved the problem occurred in the previous model that the secondary veins are retarded.We knowleaves have various shapes.For example, leaves can be classified in to simple leaves which have an undivided blade and compound leaves whose blade is divided into two or more distinctleaflets such as the Fabaceae. As to the shape of a leaf, it may have marginal dentations of the leaf blades or not, and like a palm with various fingers or an elliptical cake. Judd et al defined a set of terms which describe the shape of leaves as follows [8]:We chose entire leaves to produce this model as a simplification. What’s more, they confirmed again that the growth of venations relates with that of the leaf.To disclose this relationship, Relative Elementary Rate of Growth (RERG) can be introduced to depict leaves growth [9]. RERG is defined as the growth rate per distance, in the definitive direction l at a point p of the growing object, yielding(4)Considered RERG , the growth patterns of leaves are also different. Roth-Nebelsick et al brought up four styles in their paper [10]:3.3 Build our modelWe chose marginal growth to build our model. Amid all above-mentioned studies, weFigure 3.1 Terms pertinent to the description of leaf shapes.Figure 3.2 A sample leaf (a) and the results of its: (b) marginal growth, (c) uniform isotropic (isogonic) growth, (d) uniform anisotropic growth, and (e) non-uniform anisotropic growth.Figure 3.3 The half of a leaf is settled in x-y plane like this with primary vein overlapping x-axis. The leaf grows in the direction of .assume that the leaf produce materials it needs to grow by photosynthesis to expand its leaf area from its border and this process is only affected by what we have discussed in the previous section about photosynthesis. The border can be infinitesimally divided into points. Set as the angle between the x-axis and thestraight line connecting the grid origin and one point on the curve,andas the growth distance in the direction of . To simplify the model, we assume that the leaf grows symmetrical. We put half of the leaf into the x-y plane and make the primary vein overlap x-axis.This is how we assume the leaf grows.In one circle of leaf growth, anything that photosynthesis provided transfers into theadditional leaf area, which can be described as:(5)while in the figure.In this case, we can simulate leaf growth thus define the leaf shape by using iterative operations the times N a leaf grow in its entire circle ①.① For instance, if the vegetative circle of a leaf is 20 weeks on average, the times of iterative operations N can beset as 20 when we calculate on a weekly basis.Figure 3.4 The curves of the adjacent growing period and their relationship.First, we pre-establish the border shape of a leaf in the x-y plane, yielding . where , the relevant satisfies:(6)(7) In the first growing period, assume , the growth distance in the direction of ,satisfies:(8)(9) It releases the relationship of the coordinates in the adjacent growing period. In this case, we can use eq.(9)to predict the new border of the leaf after one period of growth②:(10)And the average simple recursions are③:②That means, in the end of period 1.③As we both change the x coordinate and the y coordinate, in the new period, these two figures relate through those in the last period in the functions.(11)After simulate the leaf borders of the interactive periods, use definite integral ④ to settle parametersin the eq.(8) then can be calculated in each direction of , thus the exact shape of a leaf in the next period is visible.When the number of times N the leaf grows in its life circle applies above-mentioned recursions to iterate N times and the final leaf shape can be settled.By this model, we can draw conclusions about why leaves have different shapes. For the leaves on the same tree, they share the same method of expansion which can be described as the same type of function as Eq.(8). The reason why they are different, not only in a sense of big or small, is that in each growing period they acquire different amount of materials used to expand its own area. In a word, the parameters in the fixedly formed eq.(8) are different for any individual leaf on the same tree. For the leaves of different tree species, the corresponding forms of eq.(8) are dissimilar. Some are linear, some are logarithmic, some are exponential or mixtures of that, which settle the totally different expansion way of leaf, are related with the veins. On that condition, the characters can be divided by a more general concept such as entire or toothed.3.4 A simulation of the modelWe set the related parameters by ourselves to simulate the shape of a leaf and to express the model better.First we initialize the leaf shape by simulating the function of a leaf border at the④The relationship must meet eq.(5).Figure 3.5 and the recurrence relations.beginning of growing period 1 in the x-y plane. By observation, we assume that themovement of the initial leaf border satisfied:(12)Suppose the curve goes across the origin of coordinates, then the constraint conditionscan be:(13)Thus the solution to eq.(13) is:(14)By using Mathematica we calculatewhereandthe area of the half leaf is: (15)Wesettleaccording to the research by S. V . Archontoulisin et al [11] in eq.(3). On a weekly basis, theparameter. In eq.(15), we have .Figure 3.6When assume ,thus constraint condition eq.(5) becomes: (16)When eq.(8) is linear and after referring to Runions’s paper, we assume that Y -valuedecreases when X-value increases, which means the leaf grows faster at the end of theprimary vein. If the grow rate at the end of the primary vein is 0,as ,eq.(8) can be described as: (17)where b is decided by eq.(16).To settle the value of b , we calculate multiple sets of data by Excel then use a planecurve to trace them and get the approximation of b . In this method, we use grid toapproximate .Apparently,is monotone.When b is 2: The square of each square is 0.0139cm 2, and the total number of the squares in theadditional area is about 240. SoFigure 3.7When b is 2.5:Thesquare of each squareis 0.0240cm 2, and the total number of the squares in theadditional area is about 201. SoWhen b is 3:The square of each square is 0.0320cm 2, and the total number of the squares in the additional area is about 193. SoAfter comparing, we can draw a conclusion that fit eq.(16)best; accordingly, in period 1: (18)In each growing period, may be different for the amount of material produced isFigure 3.9Figure 3.8related with various factors, such as the change of relative location and CO 2 or O 2concentration, and other reasons. By using the same method, the leaf shape in period2 or other period can be generated on the basis of the previous growing period untilthe end of its life circle.What’s more, when eq.(8) is remodeled, the corresponding leaf shape can be changed.With , the following figure shows the transformationof a leaf shape in the firstgrowing period with the samesettled above and we can see that the shape will be dissected in the end.4. Do the shapes maximize exposure?4.1 The optimum solution of reducing overlapping shadows4.1.1 The distribution of adjacent leavesV ein is the foundation of the leaves. With the growing of veins, the leaves graduallyexpand around. The distribution of main and lateral veins plays an important decisiverole in the shapes of leaves. The scientists created a mathematical model which usesthree decisive factors - the relationship between the rate of photosynthesis, leaf life,carbon consumption or nitrogen consumption, to simulate the leaves’ shape. Becauseof carbon consumption is a constant for one tree, and we take the neighboring leavesin the same growth cycle to observe. So, we can only focus on one factor - the rate ofphotosynthesis.Figure 3.10The shape differs from figure 3.7-3.9, as the kind offunction of is different. In this case, it is a cubic model while a linear model in figure 3.7-3.9.Through the observation of dicotyledon, leaves on a branch will grow in a staggered way that can reduce the overlapping individual shadows of adjacent leaves and make them get more sunlight (Figure 4.1).Figure 4.1 The rotation distribution of leavesBase on the similar environment, we assume that adjacent leaves nearly have the same shape. From the perspective of looking down, the leaves grow from a point on the branch. So, we can simplify the vertical view of leaves as a circle of which the radius is the length of a vein which is represented with r. The width of the leaf is represented with w. The angle between two leaves is represented with β(Figure4.2).Figure 4.2 The vertical view of leavesThe leaves should use the space as much as possible, and for the leaf with one main vein, oval is the best choice. In general, βis between 15°and 90°. In this way, effectively reduce the direct overlapping area. According to the analysis of the first question, r and w are determined by the rate of photosynthesis and respiration. Besides, the width of leaf is becoming narrower when the main vein turns to be thinner.4.1.2 Equilibrium point of photosynthesis and respirationThe organism produced by photosynthesis firstly satisfies needs of leaf itself. Then the remaining organism delivered to the root to meet the growth needs of the tree. As we know, respiration needs to consume organism. If the light is not sufficient, organism produced by photosynthesis may no longer be able to afford the materials required for the growth of leaves. There should be an equilibrium point so as toprevent the leaf is behindhand in its circumstances.The formula [12] that describes the photosynthetic rate in response to light intensity with gradual exponential growth index can be expressed as:(19) Where P is total photosynthetic rate, m ax P is maximum photosynthetic rate ofleaves, a is initial solar energy utilization and I is photosynthetic photon quanta flux density .The respiration rate is affected by temperature ，using the formal of index to describe as follows:025102T d d R R -=⋅ (20)Where T is temperature and 0d R is the dark respiration rate of leaves, when the temperature is 25 degrees. As a model parameter, 0d R can be determined bynonlinear fitting. Therefore, the net photosynthetic rate can be expressed in the indexform as follows:(21)Where n P is the net photosynthetic rate, which does not include the concentration ofcarbon dioxide and other factors. The unit of n P is 21m ol m s μ--⋅⋅.We assume that the space for leaf growth is limited, the initial area of the leaf is 0S . At this point, the entire leaf happens to be capable of receiving sunlight. If the leaf continue to grow, some part of the leaf will be in the shadows and the area in the shadows is represented with x . Ignore the fluctuation cycle of photosynthesis and respiration, on average, the duration of photosynthesis is six hours per day . In the meanwhile, the respiration is ongoing all the time. When the area increased to S , we established an equation as follows:(22)where μ is the remaining organism created by the leaf per day . And from where 0μ=, which means the organism produced by photosynthesis has all been broken down completely in the respiration, we get the equilibrium point.4n n d S P x P R ⋅=+ (23) The proportion of the shaded area in the total area:4nn d P xS P R =+ (24)Cite an example of oak trees, we found the following data (Figure 4.3), which shows the relationship between net photosynthesis and dark respiration during 160 days. Thehabitat of the trees is affected by the semi-humid monsoon climate.Figure 4.3 The diurnal rate of net photosynthesis and dark respiration [13]In general, 217n P m ol m s μ--=⋅⋅，214d R m ol m s μ--=⋅⋅. Taking the given numbersinto the equation, we can get the result.0025xS =From the result, we can see that in the natural growth of leaves, with the weakening of photosynthesis, the leaves will naturally stop growing once they come across the blade between blocked. Therefore, the leaves can always keep overlapping individual shadows about 25%, so as to maximize exposure.4.2 The influence of the “volume” of a tree and its branches4.2.1 The coronary hierarchical modelMr. Bōken and Dr. J. Fischer (1987) found that in order to adequate lighting, leaves have different densities and the branches is distributed according to certain rules. They observed tropical plants in Miami, found that the ratio of main branch and two side branches is 1:0.94:0.87, and the angles between them are 24.4°and 36.9°. According to the computer simulation, the two angles can maximize the exposure of leaves.For simply, we use hemisphere to simulate the shape of the canopy. According to the light transmittance rate, we can divide the canopy into outer and inner two layers(Figure 4.3). The volume of the hemisphere depends on the size and distribution of branches.Figure 4.3 The coronary hierarchical modelThe angle between the branches will affect the depth of penetration of sun radiation and the distribution of leaves. Besides, thickness of the branches will affect the transfer of nutrients to the leaves.4.2.2 Spatial distribution model of canopy leaf areaAssume that the distribution of leaves is uniform in the section xz but not uniform in the y direction, as the figure 4.3. Then, we can get the formula of leaf area index (LAI) as follows [14]:/20/21(,)h d d dz x z dx LAI d -∂=⎰⎰(25)Where h is the height of the canopy , d is the distance between two branches, and (,)x z ∂ is the leaf area density function of the micro-body at the point (,)x z . x a is the function of leaf area which means the distribution of cross-section of the X direction, and Its value is a dimensionless, defined as follows: /2/21()d x d a x dx LAI d -=⎰ (26)()()x s a x d LAI C x =⋅⋅ (27)For the canopy which is not uniform in the horizontal direction, the distribution of its leaves is not entirely clear. We use s C to represent the distribution function of thedensity of leaves. s C can be expressed as a quadratic function or a Gaussiandistribution function.4.2.3 Transmission model of the solar radiationAssume that the attenuation of the solar radiation accords with the law of Beer-Lambert. The attenuation value at a point of canopy where the light arrives with a certain angle of incidence and azimuth is proportional to the length of the path, and the length can be calculated by Goudriaan function. Approximate function of the G function is shown as follows [15]：00(12)cos G G k G θ=+- (28)where k is a parameter decided by different plants.Take a micro unit in the canopy . Direct sunlight intercepted in this micro unit can be expressed in the following form:()(,)b b L dI I z G z dLAI =-Θ (29)Where b I means the direct sunlight, Θ is the solar zenith angle and L dL A I is the LAI on the path of light.Figure 4.4 Micro unit of the canopyThrough the integral and the chain rule, we can get the transmittance at the point (',')x z as follows:(,)[()]tan sin ()(',')exp()()h b b z G z a x z a z I x z I h ΘΘ∆Φ+=-⎰ (30)Average direct transmission rate:/2/21(')(',')'d b b d t z t x z dx d --=⎰(31)Different with the direct light, the scattered light in all directions is intercepted by theleaf surface from the upper hemisphere. The irradiance d dI of scattering at the point(',')x z can be expressed as follows:'cos d d b dI i t d ωθ=⋅⋅ (32) where d i is the illumination intensity of scattered light from a given direction.The transmission rate of the sun scattered light is as follows:2/2'00(',')1sin cos ()d b d I x z t d d I h ππθθθϕπ=⎰⎰ (33) Average scattering transmission rate:/2/21(')(',')'d d d d t z t x z dx d --=⎰(34)Global solar radiation reaching the canopy with a given depth z :(')()[(1)(')(')]d b d d I z I h k t z k t z ---=-+ (35)According to eq.(30) and eq.(33), the maximum depth the solar radiation can reach has a major link with h and θ. The shape of leaves have a relationship with photosynthesis. So, the h and θ of branches do have an influence on the leaves. Results showed that the light level and light utilization of high stem and open ce nter shape as well as small and sparse canopy shape were better than others ．Double canopy shape ，spindle shape and center shape took second place ，while big canopy shape had the lowest light distribution [16]．5. Is leaf shape related to tree structure?Due to internal and external factors, there are many kinds of tree shapes in the nature. For example, the apple's tree shape is semi-ellipsoidal, the willow's is hemispherical, the peach's likes a cup, the pine's likes cone and so on. The shape of their leaves varies. The leaf shape of apple is oval, pine's is needle, and the Indus's is palm. Is leaf shape related to tree shape? Even for the same species, there are many kinds of tree shapes. For instance, Small canopy shape, Open center shape, Freedom spindle shape, high stem and open center shape, Double canopy shape and so on. The sizes of leaves are different. Does the tree profile/branching structure effects the leaf shape?5.1 The experiment for one speciesTo solve this problem, we first consider the relationship of one tree species such as apple, which is semi-ellipsoidal. According to the second question, the hemispherical model is further extended to the semi-ellipsoidal model. Every tree individual shows irregularly because of natural and man-made factors, which reflects on the diversity of the crown in the canopy height direction of the changing relationship between the crown and canopy height. Figure 5.1 shows the relationship.Figure5.1 Schematic diagram of tree crown contourWhere 0h means the truck high, 1h means the crown high, 8710,,,,d d d d ⋅⋅⋅ meanscrown diameter. Tree shape with the scale change can be characterized in fractal dimension. Across the same scales, a fixed fractal dimension indicates the boundary shape's self-similarity; on different scales, the change of fractal dimension means that different processes or limiting factor has superiority. (Wiens, 1989)[17] According to the application of BP network and fractional dimension, we describe the tree shapes.[18]According to Li Guodong and Zhang Junke's work which detected and evaluated in different tree shapes of ‘Fuji’ apple. We choose three kinds of tree shapes to study; they are Small canopy shape, Open center shape and Freedom spindle shape as figure5.2 shows.。

amc12考试内容

AMC竞赛是近年来热度比较高的竞赛，含金量也是非常高的，对于学生个人能力提升也有着很大的帮助!本文我们就详细介绍一下AMC12竞赛详情，一起来看看吧!美国高中数学竞赛由美国数学协会(MAA)组织，每年举办，针对12年级及以下学生(对应国内高一和高二学生)的数学竞赛，第一届AMC12最早追溯到于1950年举办的美国高中生数学考试AHSME，2000年AHSME正式改名为AMC12, 2002年开始AMC12分A赛和B赛，分别于每年的2月初和2月中举行，参赛者任选其中一项参加。

1.考试形式AMC12竞赛考试时间75分钟，一共25道选择题，问题的难度随着考试进行逐渐加大。

满分150分，2008年起，AMC12不再允许使用计算器。

AMC12的评分方式为：答对得6分，答错不得分，不答得1.5分，在2002-2006年间，不答的2.5分，2002年前，不答得2分。

AMC12参赛者的优胜者可受邀参加美国数学邀请赛(AIME)。

2.考试安排AMC12通常在每年11月举办AMC 10/12 A卷▶早期注册截止日期：2021年9月24日▶常规注册截止日期：2021年10月15日▶延迟注册截止日期：2021年10月22日▶比赛日期：2021年11月10日上午8点(美东时间) AMC 10/12 B卷▶早期注册截止日期：2021年10月1日▶常规注册截止日期：2021年10月22日▶延迟注册截止日期：2021 年10 月29 日▶比赛日期：2021年11月16日上午8点(美东时间) 3.考试内容AMC12 的考试内容包括(但不局限于)整数、分数、小数、百分数、比例、数论、日常的几何、面积、体积、概率及统计、逻辑推理等。

不需要任何微积分知识4.AMC12知识点分布在AMC10基础上新增：进阶代数：复杂不等式、调和不等式、轮换不等式、柯西不等式;复杂函数问题，反函数和符合函数，三角函数和差化积、积化和差，万能公式;复数，复平面，欧拉公式，蒂莫夫公式;数学归纳法、复杂数列和极限。

2014AMC12-Asolutions

1 1 1 1
3
Thus the third term is correctly given as r · 3 3 = 3− 6 · 3 3 = 3 6 , and the fourth 1 1 1 term is r · 3 6 = 3− 6 · 3 6 = 30 = 1. 8. Answer (C): Let P > 100 be the listed price. Then the price reductions in dollars are as follows:
P Coupon 1: 10 Coupon 2: 20 18 Coupon 3: 100 (P − 100)
P > 20, that Coupon 1 gives a greater price reduction than coupon 2 when 10 is, P > 200. Coupon 1 gives a greater price reduction than coupon 3 when P 18 10 > 100 (P − 100), that is, P < 225. The only choice that satisﬁes these inequalities is $219.95.
10. Answer (B): Let h represent the altitude of each of the isosceles triangles from the base on the equilateral triangle. Then the area of each of the congruent isosceles triangles 1 is 1 triangles is the 2 · 1 · h = 2 h. The sum of the areas of the three isosceles √ √ 1 same as the area of the equilateral triangle, so 3 · 1 h = 3 and h = 1 2 4 6 3. As a consequence, the Pythagorean Theorem implies that the side length of the isosceles triangles is 1 2

2012 AMC 12A 详解

2012 AMC 12AProblem 1A bug crawls along a number line, starting at -2. It crawls to -6, then turns around and crawls to 5. How many units does the bug crawl altogether?SolutionCrawling from -2 to -6 takes it a distance of 4 units. Crawling from -6 to 5 takes it a distance of 11 units. Add 4 and 11 to getProblem 2Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?SolutionCagney can frost one in 20 seconds, and Lacey can frost one in 30 seconds. Working together, they can frost one in seconds. In 300 seconds (5 minutes), they can frost .Problem 3A box centimeters high, centimeters wide, and centimeters long can hold grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold grams of clay. What is ?SolutionThe first box has volume , and the second has volume. The second has a volume that is times greater, so it holds grams.Problem 4In a bag of marbles, of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?SolutionAssume that there are 5 total marbles in the bag. The actual number does not matter, since all we care about is the ratios, and the only operation performed on the marbles in the bag is doubling.There are 3 blue marbles in the bag and 2 red marbles. If you double the amount of red marbles, there will still be 3 blue marbles but now there will be 4 red marbles.Thus, the answer is .Problem 5A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad?SolutionSo let the number of blueberries be the number of raspberries be the number of grapes be and finally the number of cherries beObserve that since there are pieces of fruit,Since there are twice as many raspberries as blueberries,The fact that there are three times as many grapes as cherries implies,Because there are four times as many cherries as raspberries, we deduce the following:Note that we are looking for So, we try to rewrite all of the other variables in terms of The third equation gives us the value of in terms of already. Wedivide the fourth equation by to get that Finally, substituting this value of into the first equation provides us with the equation and substituting yields:Multiply this equation by to get:Problem 6The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?SolutionLet the three numbers be equal to , , and . We can now write three equations:Adding these equations together, we get thatandSubstituting the original equations into this one, we findTherefore, our numbers are 12, 7, and 5. The middle number isProblem 7Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?SolutionIf we let be the smallest sector angle and be the difference between consecutive sector angles, then we have the angles . Use the formula for the sum of an arithmetic sequence and set it equal to 360, the number of degrees in a circle.All sector angles are integers so must be a multiple of 2. Plug in even integers for starting from 2 to minimize We find this value to be 4 and the minimum value ofto beProblem 8An iterative average of the numbers 1, 2, 3, 4, and 5 is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?SolutionThe minimum and maximum can be achieved with the orders and .The difference between the two isProblem 9A year is a leap year if and only if the year number is divisible by 400 (such as 2000) or is divisible by 4 but not 100 (such as 2012). The 200th anniversary of the birth of novelist Charles Dickens was celebrated on February 7, 2012, a Tuesday. On what day of the week was Dickens born?SolutionEach year we go back is one day back, because . Each leap year we go back is two days back, since . A leap year is GENERALLYevery four years, so 200 years would have = leap years, but the problem points out that 1900 does not count as a leap year.This would mean a total of 150 regular years and 49 leap years, so= days back. Since , four days back from Tuesday would beProblem 10A triangle has area , one side of length , and the median to that side of length . Let be the acute angle formed by that side and the median. What is ?SolutionSolution 1is the side of length , and is the median of length . The altitude ofto is because the 0.5(altitude)(base)=Area of the triangle. is . To find , just use opposite over hypotenuse with the right triangle . Thisis equal to .Solution 2It is a well known fact that a median divides the area of a triangle into two smaller triangles of equal area. Therefore, the area of in the above figure.Expressing the area in terms of , . Solving for gives . .Problem 11Alex, Mel, and Chelsea play a game that has rounds. In each round there is a single winner, and the outcomes of the rounds are independent. For each round theprobability that Alex wins is , and Mel is twice as likely to win as Chelsea. What is the probability that Alex wins three rounds, Mel wins two rounds, and Chelsea wins one round?SolutionIf is the probability Mel wins and is the probability Chelsea wins, and . From this we get and . For Alex to win three, Mel to win two, and Chelsea to win one, in that order, is . Multiply this by the number of permutations (orders they can win) which isProblem 12A square region is externally tangent to the circle with equationat the point on the side . Vertices and are on the circle with equation . What is the side length of this square?SolutionThe circles have radii of and . Draw the triangle shown in the figure above and write expressions in terms of (length of the side of the square) for the sides of the triangle. Because is the radius of the larger circle, which is equal to , we can write the Pythagorean Theorem.Use the quadratic formula.Problem 13Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?SolutionLet Paula work at a rate of , the two helpers work at a combined rate of , and the time it takes to eat lunch be , where and are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:Adding the second and third equations together gives us. Subtracting the first equation from this new onegives us , and solving for in terms of gives . We can plug this into the second equation:We can then subtract this from the third equation:Therefore . Plugging this into our third equation then gives usSince we let be in hours, the lunch break then takes minutes, which leads us to the correct answer of .Problem 14The following problem is from both the 2012 AMC 12A #14 and 2012 AMC 10A #18, so both problems redirect to this page.The closed curve in the figure is made up of 9 congruent circular arcs each of length, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?SolutionDraw the hexagon between the centers of the circles, and compute its area. Then add the areas of the three sectors outside thehexagon () and subtract the areas of the three sectors inside the hexagon () to get the area enclosed in the curved figure , which is .Problem 15The following problem is from both the 2012 AMC 12A #15 and 2012 AMC 10A #20, so both problems redirect to this page.A x square is partitioned into unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is then rotated clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability the grid is now entirely black?SolutionFirst, the middle square is always black because it is rotated onto itself. Then we can consider the corners and edges separately. Let's first just consider the number of ways we can color the corners. There is case with all black squares. There are four cases with one white square and all work. There are six cases with two white squares, but only the with the white squares opposite from each other work. There are no cases with three white squares or four white squares. Then the total number of ways to color the corners is . In essence, the edges work the same way, so there are also ways to color them. The number of ways to fit the conditions over the number of ways to color the squares isProblem 16Circle has its center lying on circle . The two circles meet at and . Point in the exterior of lies on circle and , , and . What is the radius of circle ?SolutionSolution 1Let denote the radius of circle . Note that quadrilateral is cyclic. By Ptolemy's Theorem, we have and . Let t be the measure of angle . Since , the law of cosines on triangle gives us . Again since is cyclic, the measure of angle . We apply the law of cosines to triangle so that. Sincewe obtain . Butso that . .Solution 2Let us call the the radius of circle , and the radius of . Considerand . Both of these triangles have the same circumcircle (). From the Extended Law of Sines, we see that . Therefore,. We will now apply the Law of Cosines to andand get the equations,,respectively. Because , this is a system of two equations and two variables. Solving for gives . .Problem 17Let be a subset of with the property that no pair of distinct elements in has a sum divisible by . What is the largest possible size of ?SolutionOf the integers from to , there are six each of . We can create several rules to follow for the elements in subset . No element can be iff there is an element that is . No element can beiff there is an element that is . Ignoring those that are , we can get a subset with elements. Considering , there can be one element that is so because it will only be divisible by if paired with anotherelement that is . The final answer is .Problem 18Triangle has , , and . Let denote the intersection of the internal angle bisectors of . What is ?SolutionInscribe circle of radius inside triangle so that it meets at ,at , and at . Note that angle bisectors of triangle are concurrent at the center of circle . Let , and . Note that , and . Hence , , and. Subtracting the last 2 equations we have and adding this to the first equation we have .By Herons formula for the area of a triangle we have that the area of triangleis . On the other hand the area is given by. Then so that .Since the radius of circle is perpendicular to at , we have by the pythagorean theorem so that .Problem 19The following problem is from both the 2012 AMC 12A #19 and 2012 AMC 10A #23, so both problems redirect to this page.ProblemAdam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?SolutionNote that if is the number of friends each person has, then can be any integer from to , inclusive.Also note that the cases of and are the same, since a map showing a solution for can correspond one-to-one with a map of a solution for by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with when compared to configurations of . Thus, we have two cases to examine, and , and we count each of these combinations twice.For , if everyone has exactly one friend, that means there must be pairs of friends, with no other interconnections. The first person has choices for a friend. There are people left. The next person has choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are configurations with .For , there are two possibilities. The group of can be split into two groups of , with each group creating a "friendship triangle." The first person hasways to pick two friends from the other five, while the other three are forced together. Thus, there are triangular configurations.However, the group can also from a "friendship hexagon", with each person sitting on a vertex, and each side representing the two friends that person has. The firstperson may be seated anywhere on the hexagon WLOG. This person haschoices for the two friends on the adjoining vertices. Each of the three remaining people can be seated "across" from one of the original three people, forming a different configuration. Thus, there are hexagonal configurations, and in total configurations for .As stated before, has configurations, and has configurations. This gives a total of configurations, which is optionProblem 20Consider the polynomialThe coefficient of is equal to . What is ?SolutionEvery term in the expansion of the product is formed by taking one term from each factor and multiplying them all together. Therefore, we pick a power of or a power of from each factor.Every number, including , has a unique representation by the sum of powers of two, and that representation can be found by converting a number to its binary form., meaning.Thus, the term was made by multiplying from thefactor, from the factor, and so on. The only numbers not used are , , and .Thus, from the factors, , , and were chosen as opposed to , and .Thus, the coefficient of the term is . So the answer is .Problem 21The following problem is from both the 2012 AMC 12A #21 and 2012 AMC 10A #24, so both problems redirect to this page.ProblemLet , , and be positive integers with such thatand . What is ?SolutionAdd the two equations..Now, this can be rearranged:and factored:, , and are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that ., since is the biggest difference. It is impossible to determine by inspection whether or , or whether or .We want to solve for , so take the two cases and solve them each for an expression in terms of . Our two cases are or. Plug these values into one of the original equations to see if we can get an integer for ., after some algebra, simplifies to . 2021 is not divisible by 7, so is not an integer.The other case gives , which simplifies to . Thus, and the answer isProblem 22Distinct planes intersect the interior of a cube . Let be the unionof the faces of and let . The intersection of and consists of the union of all segments joining the midpoints of every pair of edges belonging to the same face of . What is the difference between the maximum and minimum possible values of ?SolutionProblem 23Let be the square one of whose diagonals has endpoints and. A point is chosen uniformly at random over all pairs of real numbers and such that and . Let be a translated copy of centered at . What is the probability that the square regiondetermined by contains exactly two points with integer coefficients in its interior?SolutionSolution 1We first notice that for the translated square to contain two points with integer coordinates (lattice points) in its interior, these two points must be adjacent. This can be shown by considering the diagonal of . The diagonal is, which is the length of the diagonal of a unit square. Becausesquare is not parallel to the axis, square cannot two points that are not adjacent.Because we have showed that the two lattice points contained in must be adjacent, let us consider the unit square with vertices and . Let us first consider only two vertices, and . We want to find the area of the region within that the point will create the translation of , such that it covers both and . By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.For to contain the point , must be inside square . Similarly, forto contain the point , must be inside a translated square with center at , which we will call . Therefore, the area we seek is Area.To calculate the area, we notice that Area Area by symmetry. Let . Let be the midpoint of , and along the line . Let be the intersection of and within , and be the intersection ofand outside . Therefore, the area we seek is Area. Because all have coordinate , they are collinear. Noting that the side length of and is (as shown above), we also see that, so . If follows thatand. Therefore, the area is Area.Because there are three other regions in the unit square that we need to count, the total area of within such that contains two adjacent lattice points is .By periodicity, this probability is the same if , where and . Therefore, the answer is .Note: the ranges of and in the problem are arbitrary as long as the maximum and minimum of the range are integers.Problem 24Let be the sequence of real numbers defined by, and in general,Rearranging the numbers in the sequence in decreasing order produces a new sequence . What is the sum of all integers , , such thatSolutionProblem 25Let where denotes the fractional part of . The numberis the smallest positive integer such that the equation has at least real solutions. What is ? Note:the fractional part of is a real number such that and is an integer.Solution2012 AMC 12A Answer Key1. E2. D3. D4. C5. D6. D7. C8. C9. A10. D11. B12. D13. D14. E15. A16. E17. B18. A19. B20. B21. E22. C23. C24. C25. C。

AMC12美国数学竞赛2012-2014

AMC12 2014AProblem 1What isSolutionAt the theater children get in for half price. The price for adult tickets and child tickets is . How much would adult tickets and child tickets cost?SolutionWalking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?SolutionSuppose that cows give gallons of milk in days. At this rate, how many gallons of milk will cows give in days?SolutionOn an algebra quiz, of the studentsscored points, scored points, scored points, and the rest scored points. What is the difference between the mean and median score of the students' scores on this quiz?SolutionThe difference between a two-digit number and the number obtained by reversing its digits is times the sum of the digits of either number. What is the sum of the two digit number and its reverse?SolutionThe first three terms of a geometric progression are , , and . What is the fourth term?SolutionA customer who intends to purchase an appliance has three coupons, only one of which may be used:Coupon 1: off the listed price if the listed price is at leastCoupon 2: dollars off the listed price if the listed price is at leastCoupon 3: off the amount by which the listed price exceedsFor which of the following listed prices will coupon offer a greater price reduction than either coupon or coupon ?Five positive consecutive integers starting with have average . What is the average of consecutive integers that start with ?SolutionThree congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length . The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles?SolutionDavid drives from his home to the airport to catch a flight. He drives miles in the first hour, but realizes that he will be hour late if he continues at this speed. He increases his speed by miles per hour for the rest of the way to the airport and arrives minutes early. How many miles is the airport from his home?SolutionTwo circles intersect at points and . The minor arcs measure on one circle and on the other circle. What is the ratio of the area of the larger circle to the area of the smaller circle?A fancy bed and breakfast inn has rooms, each with a distinctive color-coded decor. One day friends arrive to spend the night. There are no other guests that night. The friends can room in any combination they wish, but with no morethan friends per room. In how many ways can the innkeeper assign the guests to the rooms?SolutionLet be three integers such that is an arithmetic progressionand is a geometric progression. What is the smallest possible value of ?SolutionA five-digit palindrome is a positive integer with respective digits , where is non-zero. Let be the sum of all five-digit palindromes. What is the sum of the digits of .SolutionThe product , where the second factor has digits, is an integer whose digits have a sum of . What is ?SolutionA rectangular box contains a sphere of radius and eight smaller spheres of radius . The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is ?SolutionThe domain of the function is an interval of length , where and are relatively prime positive integers. What is ?SolutionThere are exactly distinct rational numbers such that andhas at least one integer solution for . What is ?SolutionIn , , , and . Points and lieon and respectively. What is the minimum possible valueof ?SolutionFor every real number , let denote the greatest integer not exceeding , and let The set of all numbers suchthat and is a union of disjoint intervals. What is the sum of the lengths of those intervals?SolutionThe number is between and . How many pairs ofintegers are there such that andSolutionThe fraction where is the length of the period of the repeating decimal expansion. What is the sum ?SolutionLet , and for , let . For how many values of is ?The parabola has focus and goes through the points and . For how many points with integer coefficients is it truethat ?AMC 12 2013AProblem 1Square has side length . Point is on , and the areaof is . What is ?SolutionA softball team played ten games, scoring , and runs. They lost by one run in exactly five games. In each of the other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?SolutionA flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?SolutionWhat is the value ofSolutionTom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $, Dorothy paid $, and Sammy paid $. In order to share the costs equally, Tom gave Sammy dollars, and Dorothy gave Sammy dollars. What is ?SolutionIn a recent basketball game, Shenille attempted only three-point shots andtwo-point shots. She was successful on of her three-point shots and of her two-point shots. Shenille attempted shots. How many points did she score?SolutionThe sequence has the property that every term beginning with the third is the sum of the previous two. That is,Suppose that and . What is ?SolutionGiven that and are distinct nonzero real numbers such that , what is ?SolutionIn , and . Points and are onsides , , and , respectively, such that and are parallelto and , respectively. What is the perimeter of parallelogram ?SolutionLet be the set of positive integers for which has the repeating decimal representation with and different digits. What is the sum of the elements of ?SolutionTriangle is equilateral with . Points and are on and points and are on such that both and are parallel to . Furthermore, triangle and trapezoids and all have the same perimeter. What is ?SolutionThe angles in a particular triangle are in arithmetic progression, and the side lengths are . The sum of the possible values of x equals where , and are positive integers. What is ?SolutionLet points and .Quadrilateral is cut into equal area pieces by a line passing through . This line intersects at point , where these fractions are in lowest terms. What is ?SolutionThe sequence, , , ,is an arithmetic progression. What is ?SolutionRabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?Solution, , are three piles of rocks. The mean weight of the rocks in is pounds, the mean weight of the rocks in is pounds, the mean weight of the rocks in the combined piles and is pounds, and the mean weight of the rocks in the combined piles and is pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles and ?SolutionA group of pirates agree to divide a treasure chest of gold coins among themselves as follows. The pirate to take a share takes of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the pirate receive?SolutionSix spheres of radius are positioned so that their centers are at the vertices of a regular hexagon of side length . The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?SolutionIn , , and . A circle with center andradius intersects at points and . Moreover and have integer lengths. What is ?SolutionLet be the set . For , define to mean thateither or . How many ordered triples of elements of have the property that , , and ?SolutionConsider. Which of the following intervals contains ?SolutionA palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10 with no leading zeros. A 6-digit palindrome is chosen uniformly at random. What is the probability that is also a palindrome?Solutionis a square of side length . Point is on such that . The square region bounded by is rotated counterclockwise with center , sweeping out a region whose area is , where , , and are positive integers and . What is ?SolutionThree distinct segments are chosen at random among the segments whoseend-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?SolutionLet be defined by . How many complexnumbers are there such that and both the real and the imaginary parts of are integers with absolute value at most ?AMC12 2012AProblem 1A bug crawls along a number line, starting at . It crawls to , then turns around and crawls to . How many units does the bug crawl altogether?SolutionCagney can frost a cupcake every seconds and Lacey can frost a cupcake every seconds. Working together, how many cupcakes can they frostin minutes?SolutionA box centimeters high, centimeters wide, and centimeters long canhold grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold grams of clay. What is ?SolutionIn a bag of marbles, of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?SolutionA fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad?SolutionThe sums of three whole numbers taken in pairs are , , and . What is the middle number?SolutionMary divides a circle into sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?SolutionAn iterative average of the numbers , , , , and is computed in the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?SolutionA year is a leap year if and only if the year number is divisible by (such as ) or is divisible by but not by (such as ). The anniversary of the birth of novelist Charles Dickens was celebrated on February , , a Tuesday. On what day of the week was Dickens born?SolutionA triangle has area , one side of length , and the median to that side of length . Let be the acute angle formed by that side and the median. Whatis ?SolutionAlex, Mel, and Chelsea play a game that has rounds. In each round there is a single winner, and the outcomes of the rounds are independent. For each round the probability that Alex wins is , and Mel is twice as likely to win as Chelsea. What is the probability that Alex wins three rounds, Mel wins two rounds, and Chelsea wins one round?SolutionA square region is externally tangent to the circle withequation at the point on the side . Vertices and are on the circle with equation . What is the side length of this square?SolutionPaula the painter and her two helpers each paint at constant, but different, rates. They always start at , and all three always take the same amount of time to eat lunch. On Monday the three of them painted of a house, quittingat . On Tuesday, when Paula wasn't there, the two helpers paintedonly of the house and quit at . On Wednesday Paula worked by herself and finished the house by working until . How long, in minutes, was each day's lunch break?SolutionThe closed curve in the figure is made up of congruent circular arcs each of length , where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side . What is the area enclosed by the curve?SolutionA square is partitioned into unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is the rotated clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability that the grid is now entirely black?SolutionCircle has its center lying on circle . The two circles meet at and . Point in the exterior of lies on circle and , ,and . What is the radius of circle ?SolutionLet be a subset of with the property that no pair of distinct elements in has a sum divisible by . What is the largest possible size of ?SolutionTriangle has , , and . Let denote the intersection of the internal angle bisectors of . What is ?SolutionAdam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?SolutionConsider the polynomialThe coefficient of is equal to . What is ?SolutionLet , , and be positive integers with such thatWhat is ?SolutionDistinct planes intersect the interior of a cube . Let be the unionof the faces of and let . The intersection of and consists of the union of all segments joining the midpoints of every pair of edges belonging to the same face of . What is the difference between the maximum and minimum possible values of ?SolutionLet be the square one of whose diagonals hasendpoints and . A point is chosen uniformly at random over all pairs of real numbers and suchthat and . Let be a translated copyof centered at . What is the probability that the square region determinedby contains exactly two points with integer coefficients in its interior?SolutionLet be the sequence of real numbers definedby , and in general,Rearranging the numbers in the sequence in decreasing order produces anew sequence . What is the sum of all integers , , such thatSolutionLet where denotes the fractional part of . Thenumber is the smallest positive integer such that the equationhas at least real solutions. What is ? Note: the fractional part of is a real number such that and is an integer.2014A1.C2.B3.B4.A5.C6.D7.A8.C9.B10.B11.C12.D13.B14.C15.B16.D17.A18.C19.E20.D21.A22.B23.B24.C25.B 2013A1. E2. C3. E4. C5. B6. B7. C8. D9. C10. D11. C12. A13. B14. B15. D16. E17. D18. B19. D20. B21. A22. E23. C24. E25. A2012A1. E2. D3. D4. C5. D6. D7. C8. C9. A10. D11. B12. D13. D14. E15. A16. E17. B18. A19. B20. B21. E22. C23. C24. C25. C。

2012美国数学建模比赛官方指南Instructions

Contest Rules, Registration and Instructions(All rules and instructions apply to both ICM and MCM contests, except where otherwise noted.)To participate in a contest, each team must be sponsored by a faculty advisor from its institution.Team Advisors: Please read these instructions carefully. It is your responsibility to make sure that teams are correctly registered and that all of the following steps required for participation in the contest are completed:Please print a copy of these contest instructions for reference before, during, and after the contest. Click here for the printer friendly version.COMAP is pleased to announce a new supplement to the MCM/ICM contest. Click here to read more details.I. BEFORE THE CONTEST BEGINS:A. RegistrationB. Choose your team membersII. AFTER THE CONTEST BEGINS:A. View the contest problems via the contest web siteB. Choose a problemC. Teams prepare solutionsD. Print Summary Sheet and Control SheetIII. BEFORE THE CONTEST ENDS:A. Send electronic copy of Solution Paper by emailIV. WHEN THE CONTEST ENDS:A. Prepare Solution PacketB. Mail Solution PacketV. AFTER THE CONTEST IS OVER:A. Confirm that your team’s solution was receivedB. Check contest resultsC. CertificatesD. PrizesIMPORTANT NOTES:z COMAP is the final arbiter of all rules and policies, and may disqualify or refuse to register any team that, in its sole discretion, does not follow these contest regulations and procedures.z If a team is caught violating the rules, the faculty advisor will not be permitted to advise another team for one year, and the advisor’s institution will be put on probation for one year.z If a team from the same institution is caught violating the rules a second time, then that school will not beallowed to compete for a period of at least one year.z All times given in these instructions are in terms of Eastern Standard Time (EST). (COMAP is located in the U.S. Eastern Time zone.)I. BEFORE THE CONTEST BEGINS:A. RegistrationAll teams must be registered before 2PM EST on Thursday, February 9, 2012. We recommend that all teams complete the registration process well in advance, since the registration system will not accept any newteam registrations after the deadline. COMAP will not accept late registrations for MCM/ICM 2012 under any circumstances. NO EXCEPTIONS WILL BE MADE.1.Register your team online via the contest web site: Go to/undergraduate/contests/mcm.a. If you are registering your first team for this year’s contest, click on Register for 2012 Contest on theleft-hand side of the screen.Enter all the required information, including your email address and contact information.IMPORTANT: Be sure to use a valid and current email address so that we can use it to contact you atany point before, during, and after the contest, if necessary.b. If you have already registered a team for this year’s contest and want to register a second team, click onAdvisor Login, then log in with the same email address and password that you used when youregistered your first team. Once you’re logged in, click on Register Another Team near the upper rightcorner of the page, then follow the instructions there.Note: An advisor may register no more than two (2) teams. If you already registered two teams, theRegister Another Team link will not appear. The system will not allow you to register more than twoteams.c. Although each advisor can register only two teams, there is no restriction on the number of advisors orteams that can register from any particular institution or department.2.Registration FeeA $100 registration fee per team is required.For an additional $100 fee per team, you can receive a Judges Commentary written specifically about your team’s paper.We accept payment with Mastercard or Visa only via our secure web site. We cannot accept other forms of payment. Our secure site will process your credit card payment, so your credit card number is protected. Our system will not store your credit card number after it processes your payment.3.After we receive approval from your financial institution (this takes only a few seconds), the system willissue a control number for your team. Your team is not officially registered until you have received a team control number. Print the page that displays your team control number: It is your only confirmation that your team has been registered. This page also lists the email address and password that you entered when registering; you will need this information to complete the contest procedures.You will NOT receive an email confirmation of your registration.4.If you need to change any of the information (name, address, contact information, etc.) that you specifiedwhen you registered, you can do so at any point before or during the contest by logging in to the contest web site with the same email address and password that you used when registering (click on the Advisor Login link on the left side of the screen). Once logged in, click on the Edit Advisor or Institution Data link near the upper right corner of the page.5.Check the contest web site regularly for any updated instructions or announcements about the contest.Except in extreme circumstances, COMAP will not send any confirmation, reminders, or announcements by email. All communication regarding the contest will be via the contest web site.6.You will return to the contest web site during the contest to enter and confirm information about your team,and to print out your team’s Control Sheet and Summary Sheets, which you will use when preparing your team’s solution packet. Details on these steps follow in the instructions below.B. Choose your team members:1.You must choose your team members before the contest begins at 8PM EST on Thursday February 9,2012. Once the contest begins you may not add or change any team members (you may, however, remove a team member, if he or she decides not to participate).2.Each team may consist of a maximum of three students.3.Each student may participate on only one team.4.Team members must be enrolled in school at the time of the contest, but they need not be full-time students.Team members must be enrolled at the same school as the advisor and other team members.II. AFTER THE CONTEST BEGINS:A. View the contest problems via the contest web site:Teams can view the contest problems via the contest web site when the contest begins at 8PM EST onThursday February 9, 2012:1.The contest problems will become available precisely at 8PM EST on Thursday February 9, 2012; teammembers can view them by visiting /undergraduate/contests/mcm. No passwordwill be needed to view the problems; simply go to the contest web site at or after 8PM EST on Thursday,February 9, 2012 and you will see a link to view the problems.2.If you cannot access our main web site at that time, go to our mirror site at ap-/mcm or click here. If you cannot access either site, there may be a problem with your localInternet connection. Contact your local Internet service provider to resolve the issue.B. Choose a problem:Each team must choose one problem according to the following rules:• MCM teams must choose either Problem A or Problem B; an MCM team may submit a solution to only one of the problems. (MCM teams should NOT choose Problem C.)• ICM teams must choose Problem C. There is no choice for ICM teams. (ICM teams should NOT choose Problem A or Problem B.)C. Teams prepare solutions:1.Teams may use any inanimate source of data or materials: computers, software, references, web sites,books, etc. ALL SOURCES USED MUST BE CREDITED. Failure to credit a source will result in a teambeing disqualified from the competition.2.Team members may not seek help from or discuss the problem with their advisor or anyone else, exceptother members of the same team. Input in any form from anyone other than student team members is strictly forbidden. This includes email, telephone contact, and personal conversation, communication via web chator other question-answer systems, or any other form of communication.3.Partial solutions are acceptable. There is no passing or failing cut-off score, and numerical scores will notbe assigned. The MCM/ICM contest judges are primarily interested in the team’s approach and methods.4.Summary Sheet: The summary is an essential part of your MCM/ICM paper. The judges placeconsiderable weight on the summary, and winning papers are often distinguished from other papers basedon the quality of the summary.To write a good summary, imagine that a reader will choose whether to read the body of the paper based on your summary: Your concise presentation in the summary should inspire a reader to learn about the detailsof your work. Thus, a summary should clearly describe your approach to the problem and, mostprominently, your most important conclusions. Summaries that are mere restatements of the contestproblem, or are a cut-and-paste boilerplate from the Introduction are generally considered to be weak.Each Summary Sheet should include:z Restatement and clarification of the problem: State in your own words what you are going to do.z Explain assumptions and rationale/justification: Emphasize the assumptions that bear on the problem.Clearly list all variables used in your model.z Include your model design and justification for type model used or developed.z Describe model testing and sensitivity analysis, including error analysis, etc.z Discuss the strengths and weaknesses of your model or approach.5.The judges will evaluate the quality of your writing in the Solution Paper:•Conciseness and organization are extremely important.•Key statements should present major ideas and results.•Present a clarification or restatement of the problem, as appropriate.•Present a clear exposition of all variables, assumptions, and hypotheses.•Present an analysis of the problem, including the motivation or justification for the model that is used.•Include a design of the model.•Discuss how the model could be tested, including error analysis and stability (conditioning, sensitivity, etc.).•Discuss any apparent strengths or weaknesses in your model or approach.6.Papers must be typed and in English.7.The solution must consist entirely of written text, and possibly figures, charts, or other written material, onpaper only. No non-paper support materials such as computer files or disks will be accepted.8.The Solution Paper must display the team control number and the page number at the top of everypage; for example, use the following page header on each page:Team # 321 Page 6 of 139.The names of the students, advisor, or institution should NOT appear on any page of the print solution orelectronic solution. The solution should not contain any identifying information other than the team controlnumber.10.Failure to adhere to any preparation rule is grounds for team disqualification.D. Print Summary Sheet and Control Sheet:After the contest begins at 8PM EST on Thursday February 9, 2012, and while the teams are preparing their solutions, the advisor should:1.Login to the contest web site (go to /undergraduate/contests/mcm. Click onAdvisor Login, then enter your email address and password).2.Enter the team member names and confirm that each name is spelled correctly. This determines how thenames will appear on the contest certificates. COMAP will not make any changes or reprintcertificates for any reason.3.Specify the problem that your team has chosen to solve.4.Print one copy of the Control Sheet.5.Print one copy of the team Summary Sheet.III. BEFORE THE CONTEST ENDS:A. Send electronic copy of Solution Paper by email:1.Each team is required to submit an electronic copy of its solution paper by email to solutions@.Any team member or the advisor may submit this email.a. Your email MUST be received at COMAP by the submission deadline of 8PM EST on February13, 2012.b. Failure by a team to submit a solution via email by 8PM EST on February 13, 2012 constitutes aviolation of the contest rules and will result in that team’s disqualification.c. No further modifications, enhancements, additions, or improvements may be made to the team’s solutionpaper after this email submission. Each team’s electronic submission will be cross-checked forconsistency with their paper submission. Any changes to the paper version will constitute a violation ofthe contest rules and may result in disqualification.2.In the subject line of your email write: COMAP and your team’s control number. For example:Subject: COMAP 2222e your team’s control number as the name of your file attachments.AP will accept only an Adobe PDF or Microsoft Word file of your solution. DO NOT includeprograms or software with your email as they will not be used in the judging process. Limit one solution per email. The names of the students, advisor, or institution should NOT appear on any page of theelectronic solution.5.See the rules for submitting two (2) paper copies and CD via mail below.IV. WHEN THE CONTEST ENDS:A. Prepare Solution Packet:When the contest ends at 8PM EST on February 13, 2012:1.Each team member must sign the Control Sheet to pledge that he or she abided by the contest rules andinstructions.2.Make two copies of your team’s Solution Paper.3.Staple the Summary Sheet on top of each of the Solution Papers.4.Staple the Control Sheet on top of one of the Solution Papers.5.Enclose an electronic copy (PDF or Word file) of your team’s Solution Paper on a CD-ROM. Programs andsoftware will not be used in the judging process. DO NOT include them on the CD.If you advise more than one team, please include both teams’ files on a single CD-ROM and label it withcontest, year, and both teams’ control numbers.Example:Year Contest Control Numbers2012 MCM/ICM 10004, 10005B. Mail Solution Packet:1.Mail your team’s complete Solution Packet to:MCM/ICM CoordinatorCOMAP, Inc.175 Middlesex Turnpike., Suite 3BBedford, MA 01730USAAP must receive your Solution Packet via mail on or before Wednesday February 22, 2012. It isyour responsibility to make sure that your team’s Solution Packet arrives at COMAP by this deadline. Useregistered or express mail, if necessary, to insure that it arrives at COMAP by Wednesday February 22,2012.AP will not accept late solutions under any circumstances.4.If you require confirmation that your paper was received by COMAP, send the packet via a carrier thatprovides package tracking. Due to the number of papers received, COMAP can not answer receipt inquiries or emails.V. AFTER THE CONTEST IS OVER:A. Confirm that your team’s solution was received:1.You may login to the contest web site using the Advisor Login link to verify that your team’s SolutionPacket was received at COMAP. After mailing your Solution Packet, please allow several days for us toprocess your packet before expecting to see this confirmation.B. Check contest results:1.Judging: Judging will be completed in March and the results will be posted on April 29, 2012. TheSolution Papers will be recognized as Unsuccessful, Successful Participant, Honorable Mention,Meritorious, Finalist, or Outstanding Winner.2.We will post the contest results on the web site as soon as they are available, so visit the contest web siteregularly to check for updates. It will take several weeks for the judges to evaluate the solutions and forCOMAP to process the results. Please do NOT call or email COMAP regarding contest results.C. Certificates:After the results are issued, each successfully participating team will receive a certificate of participation. The certificate will be mailed or emailed to the advisor at the address used during the registration process. All international teams will receive ONLY an electronic (PDF) certificate. Please allow several weeks after the results are posted to the contest web site to receive your certificate.D. Prizes:•The Institute for Operations Research and the Management Sciences (INFORMS) will designate an Outstanding team from each of the three problems as an INFORMS winner.•The Society for Industrial and Applied Mathematics (SIAM) will designate one Outstanding team from each problem as a SIAM winner.•The Mathematical Association of America (MAA) will designate one Outstanding team from each problem for the MCM as a MAA winner.•The Ben Fusaro Award, Typically, among the final MCM papers from which the Outstanding ones are selected is a paper that is especially creative but contains a flaw that prevents it from attaining theOutstanding designation. In accord with Ben’s wishes, the award will recognize such teams.。

AMC12历年试卷_2000-2008

USA
AMC 12 2000
1 In the year 2001, the United States will host the International Mathematical Olympiad. Let I, M, and O be distinct positive integers such that the product I · M · O = 2001. What’s the largest possible value of the sum I + M + O? (A)23 (B)55 (C)99 (D)111 (E)671
4 The Fibonacci Sequence 1, 1, 2, 3, 5, 8, 13, 21, . . . starts with two 1s and each term afterwards is the sum of its predecessors. Which one of the ten digits is the last to appear in thet units position of a number in the Fibonacci Sequence? (A)0 (B)4 (C)6 (D)7 (E)9
16 A checkerboard of 13 rows and 17 columns has a number written in each square, beginning in the upper left corner, so that the rst row is numbered 1, 2, . . . , 17, the second row 18, 19, . . . , 34, and so on down the board. If the board is renumbered so that the left column, top to bottom, is 1, 2, . . . , 13, the second column 14, 15, . . . , 26 and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system). (A) 222 (B) 333 (C) 444 (D) 555 (E) 666

2012年美国数学建模MCM题目(中英对照版)

2012 Contest ProblemsMCM PROBLEMSPROBLEM A: The Leaves of a Tree"How much do the leaves on a tree weigh?" How might one estimate the actual weight of the leaves (or for that matter any other parts of the tree)? How might one classify leaves? Build a mathematical model to describe and classify leaves. Consider and answer the following:• Why do leaves have the various shapes that they have?• Do the shapes “minimize” overlapping individual shadows that are cast, so as to maximize exposure? Does the distribution of leaves within the “volume” of the tree and its branches effect the shape?• Speaking of profiles, is leaf shape (general characteristics) related to tree profile/branching structure?• How would you estimate the leaf mass of a tree? Is there a correlation between the leaf mass and the size characteristics of the tree (height, mass, volume defined by the profile)?In addition to your one page summary sheet prepare a one page letter to an editor of a scientific journal outlining your key findings.“一棵树的叶子有多重？”怎么能估计树的叶子（或者树的任何其它部分）的实际重量？怎样对叶子进行分类？建立一个数学模型来对叶子进行描述和分类。

赏析几道美国AMC12数学竞赛题

－ ABy0 A2 + B2
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r1 ，r2 ，and r3 are rationalnumbers and n1 and n2 are positive integers not divisible by the square of a prime． What is the greatest integer less than or equal to r1 + r2 + r3 + n1 + n2 ？
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2012年美国数学奥林匹克(USAMO)试题及其解答

431 28 i=1 j=1
绿点相对位置关系是相同的。我们设这 28 个点依次为ak 1 、ak 2 、 … … 、ak 28 。
记Yi = �f�ak 1 + i�、f�ak 2 + i�、f�ak 3 + i�、 … … 、f�ak 28 + i�� ∩ �c1 、c2 、c3 、 … … 、c108 �
431 8 i=1 j=1
记Zi = �f�at 1 + i�、f�at 2 + i�、f�at 3 + i�、 … … 、f�at 8 + i�� ∩ �d1 、d2 、d3 、 … … 、d108 �
� card(Zi ) = � card ��f �at j + 1� 、f �at j + 2� 、f �at j + 3� 、 … … 、f �at j + 431�� ∩ �d1 、d2 、d3 、 … … 、d108 �� = 108 × 8
108×28 431
根据抽屉原理知，存在一个card� Yq � ≥ �
8 个蓝点的相对位置关系都是相同的。我们设这 8 个点依次为at 1 、at 2 、 … … 、at 8 。
� + 1 = 8。也就是说存在 8 个红点和 8 个绿点及

杏坛孔门
（i=1、2、3、……、431）。由于card ��f �at j + 1� 、f �at j + 2� 、f �at j + 3� 、 … … 、f �at j + 431∩d1、d2、d3、……、d108=108（j=1、2、3、……8），所以
+ 3pi −1 �ap i −1 − 1� + pi �ap i − 1�。由于（3pi −1 ，pi ）=1，。令 + 3pi −1 �ap i −1 − 1� +

2014年美国数学建模比赛B题(最佳教练)

F1________________ F2________________ F3________________ F4________________
B
Summary
2014 Mathematical Contest in Modeling (MCM) Summary Sheet
This paper mainly sets a model of selecting five “best all time college coach” in basketball, football and hockey respectively. The model is separated to six sub-problems and its solution in our paper:1)Selecting the best coach in one NCAA basketball season; 2)Screening the college coach at top-20 to top-30 in the past 100 years; 3)Further analysis to these selected coaches, and ranking the top-5; 4)Applying this model to football and hockey; 5)Gender impact analysis and time impact analysis; 6)Analyzing the reasons that some famous coach selected by some magazines and media are not on our ranking list. For sub-problem1:Identify four indicators, and confirm the weight of each indicator by Analytic Hierarchy Process, then transformer the indicator data into scores and add these scores multiplied by the weight of itself to get the final scores. The best coach in one season is the one with highest score. For sub-problem2: Use the method above to score for all of the coaches in the past 100 years , then select the top-20.For purpose of reducing the influence of subjective factors, we use Principal Component Analysis to get another ranking list, also, we select the top-20.And union these two top-20. For sub-problem3:Surveying these coaches in the union to learn how many NBA players they've respectively brought up. And regarding this as an indicator, then dividing all indicators into Experience,Leading ability and Ability of player cultivation three aspects, and using Multilevel Hierarchical Analysis to weight them. Next, score for each person by the Fuzzy Criterion of Composite and rank according to the scores. With the purpose of reducing the influence of subjective factors, using Grey Correlation Analysis to calculate the grey correlation degree between “perfect coach” and these coaches. After that, using t-test to judge whether the two results with significant differences. The conclusion obtained: There are no significant difference. Namely, this method is generality. For sub-problem4:Fine-tuning part of indicators and weights, then apply this model in football and hockey. For sub-problem5:As One-way Analysis of Variance used, and the conclusion obtained: gender doesn’t affect the overall score or the ranking of coaches, but female coach has unique advantages in women's basketball league. And the time has no significant effect on the comprehensive score ,but the most excellent coach were in the 70s. Combined with the history of NCAA development, we think that it is necessary to improve the weight of the indicator---champions that gotten recently, making this model better. .For sub-problem6: Comparing the data of their career with our standards to find the reasons.

2014AMC12A试题及简答(繁体中文)

度的和為何？
27 2
(C) 8 3
(D) 14
(E) 3 3 9
(A) 1
(B)
log 2015 log 2014
(C)
log 2014 log 2013
(D)
2014 2013
1
(E) 2014 2014
22. 已知 5867 是介於 2 2013 與 2 2014 之間。試問有多少組整數數對 m, n 滿足：
3
2.
3.
4.
5.
6.(D) 9ຫໍສະໝຸດ 3,6(E) 110
7.
若某等比數列的前三項為 3 ,
3 ，則第四項為何？
8.
9.
(B) 7 3 (C) 8 3 (D) 9 3 (E) 10 3 (A) 1 有位顧客想買某個器具，他有三張不同的折價券，但只可以用其中一張：折價券 1：如果訂價至少 50 美元，可以少付訂價的 10%；折價券 2：如果訂價至少 100 美元，可以少付 20 美元；折價券 3：如果訂價超過 100 美元，可以少付超過 100 美元部分的 18%。下列哪一個訂價，使用折價券 1 比使用折價券 2、折價券 3 能省更多的錢？ (A) 179.95 美元 (B) 199.95 美元 (C) 219.95 美元 (D) 239.95 美元 (E) 259.95 美元由 a 開始連續五個正整數，其平均數為 b。試問由 b 開始連續五個正整數的平均數為何？ (A) a 3 (B) a 4 (C) a 5 (D) a 6 (E) a 7
(A) 874
24. 設 f 0 x x x 100 x 100 ，並對所有的 n 1 ，令 f n x f n 1 x 1 。試問有多少個 x

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