2020美赛E题

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美赛竞赛试题

美赛竞赛试题

1.2013 MCM A: The Ultimate Brownie Pan2013 MCM B: Water, Water, Everywhere2013 ICM: Network Modeling of Earth's Health2. 2012 MCM A: The Leaves of a Tree2012 MCM B: Camping along the Big Long River2012 ICM: Modeling for Crime Busting3. 2011 MCM A: Snowboard Course2011 MCM B: Repeater Coordination2011 ICM: How environmentally and economically sound are electric vehicles? Is their widespread use feasible and practical4. 2010 MCM A: The Sweet Spot2010 MCM A: The Sweet Spot2010 ICM: The Great Pacific Ocean Garbage Patch5. 2009 MCM A: Designing a Traffic Circle2009 MCM B: Energy and the Cell Phone2009 ICM: Creating Food Systems: Re-Balancing Human-Influenced Ecosystems6. 2008 MCM A: Take a Bath2008 MCM B: Creating Sudoku Puzzles2008 ICM: Finding the Good in Health Care Systems7. 2007 MCM A: Gerrymandering2007 MCM B: The Airplane Seating Problem2007 ICM: Organ Transplant: The Kidney Exchange Problem8. 2006 MCM A: Positioning and Moving Sprinkler Systems for Irrigation2006 MCM B: Wheel Chair Access at Airports2006 ICM: Trade-offs in the fight against HIV/AIDS9. 2005 MCM A: Flood Planning2005 MCM B: Tollbooths2005 ICM: Nonrenewable Resources10. 2004 MCM A: Are Fingerprints Unique?2004 MCM B: A Faster QuickPass System2004 ICM: To Be Secure or Not to Be?11. 2003 MCM A: The Stunt Person2003 MCM B: Gamma Knife Treatment Planning2003 ICM: Aviation Baggage Screening Strategies: To Screen or Not to Screen, that is the Question12. 2002 MCM A: Wind and Waterspray2002 MCM B: Airline Overbooking2002 ICM: Scrub Lizards13. 2001 MCM A: Choosing a Bicycle Wheel2001 MCM B: Escaping a Hurricane's Wrath (An III Wind...)14. 2000 MCM A: Air Traffic Control2000 MCM B: Radio Channel Assignments2000 ICM: Elephants: When is Enough, Enough?15. 1999 MCM A: Deep Impact1999 MCM B: Unlawful Assembly1999 ICM: Ground Pollution16. 1998 MCM A: MRI Scanners1998 MCM B: Grade Inflation17. 1997 MCM A: The Velociraptor Problem1997 MCM B: Mix Well for Fruitful Discussions 18. 1996 MCM A: Submarine Tracking1996 MCM B: Paper Judging19. 1995 MCM A: Helix Construction1995 MCM B: Faculty Compensation20. 1994 MCM A: Concrete Slab Floors1994 MCM B: Network Design21. 1993 MCM A: Optimal Composting1993 MCM B: Coal-Tipple Operations22. 1992 MCM A: Air-Traffic-Control Radar Power1992 MCM B: Emergency Power Restoration23. 1991 MCM A: Water Tank Flow1991 MCM B: The Steiner Tree Problem24. 1990 MCM A: The Brain-Drug Problem1990 MCM B: Snowplow Routing25. 1989 MCM A: The Midge Classification Problem1989 MCM B: Aircraft Queueing26. 1988 MCM A: The Drug Runner Problem1988 MCM B: Packing Railroad Flatcars27. 1987 MCM A: The Salt Storage Problem1987 MCM B: Parking Lot Design28. 1986 MCM A: Hydrographic Data1986 MCM B: Emergency-Facilities Location29. 1985 MCM A: Animal Populations1985 MCM B: Strategic Reserve Management。

美赛题目总结

美赛题目总结

年份题号短名名称题型学科2014A直行与超车评价靠右行驶的政策是否正确评价类交通B大学教练评价问题设计类研究数据网络C社交网络的影响力评价研究者想在浩如烟海的信息中提取有2013A烤盘问题B水资源问题C地球健康网络建模2012A树叶问题叶子形状与树形、叶子总质量的问题设计类植物学= = B野营规划河流的trip设计、最大载客量计算规划类集合中元素的排设计类网络C犯罪克星罪犯网络,通过已知信息网络来分析2011A滑雪滑道设计如何使垂直距离最大,在空气中的扭设计类物理学设计类无线电B通讯频道问题为了防止两个频谱之间的干扰,可以C电动汽车前景问题2010A棒球棒最佳击球点问题B犯罪学(连环谋杀案搜索)问题C大太平洋塑料垃圾带2009A环岛交通管理B能源和手机C创建食物系统——重建受到人类影响的生态系统2008A球温度的上升而导致的北极冰盖的融化对陆地的影响B研制构成不同难度的数独智力游戏的算法C医保系统评价2007A不公正的选区划分B飞机就座问题C器官移植:肾交换问题2006A喷灌系统管理与移动问题B机场轮椅使用问题C抗击艾滋病的协调问题主要方法元胞自动机模拟交通流,计算评价变量网络:基于几个数据特征(中心化,连接,传递),构建相关向量,PCA阅读文献归纳总结、对阴影覆盖面积建模、自相似性、光合作用碳同化量模拟出速度、旅程中各种参数;基于队伍数量、营地数量、各种参数进行建模机器学习:基于每个节点的性质来判断是否是同谋者或leader受力分析最重要,模拟出滑道的样子并计算各种参数构建两层的网络、图论、Shannon’s information theory、naïve solution、利用Voronoi diagram构建算法用Voronoi diagram构建算法。

24美赛e题解题思路

24美赛e题解题思路

24美赛e题解题思路全文共四篇示例,供读者参考第一篇示例:美国大学生数学建模竞赛(MCM/ICM)是全球最具影响力的学术比赛之一,每年吸引着来自世界各地的顶尖学生参赛。

24美赛e题作为MCM/ICM比赛中的经典难题之一,一直备受关注和研究。

本文将从不同角度出发,简要介绍24美赛e题的解题思路,希望能对大家参与比赛、提高建模能力有所帮助。

我们需要了解24美赛e题的具体内容。

这道题目要求建立一个数学模型,分析对于一定间隔时间内结成的高速公路内部关键路段的平均速度。

参赛者需要考虑不同车辆的速度分布、路段长度、车辆之间的安全距离等多个因素,并提出合理的解决方案。

在解题过程中,首先需要明确题目所需研究的实际问题,确定建模的范围和目标。

然后,可以尝试构建数学模型,针对题目中提到的各种因素进行量化分析。

可以考虑利用微积分、概率统计等数学知识,结合计算机模拟和数据分析技术,分析车辆的速度分布、车流密度、路段拥堵情况等情况。

接着,可以根据建立的数学模型进行数值模拟和实验验证,调整参数、优化模型。

通过不断的尝试和实践,逐步完善模型,提高解决问题的准确性和有效性。

可以参考国内外相关领域的研究成果,借鉴其他学者的思路和方法,不断拓展思路,提高模型的创新性和实用性。

在参与24美赛e题解题时,还需要注重团队合作和沟通交流。

比赛中,每个队员可以负责不同部分的工作,共同协作完成建模和分析工作。

通过团队合作,有效分工,相互协调,密切配合,相信一定可以取得更好的成绩。

解决24美赛e题需要充分利用数学建模、数据分析和计算机技术,兼顾理论研究和实际应用,注重团队合作和创新思维。

希望大家在参加比赛时能够充分发挥自己的智慧和创造力,不断提高建模能力,获得更多的知识和经验。

祝愿大家在24美赛e题的解题过程中取得优异的成绩,为巩固和提高中国的建模实力做出贡献!第二篇示例:第24届美国数学建模竞赛(MCM)的E题是一个涉及到环境科学和气候变化的问题,要求参赛者根据提供的数据和情景,对于气候变化和海平面上升的影响进行定量分析和模拟。

美赛历年题目_pdf

美赛历年题目_pdf

马剑整理历年美国大学生数学建模赛题目录MCM85问题-A 动物群体的管理 (3)MCM85问题-B 战购物资储备的管理 (3)MCM86问题-A 水道测量数据 (4)MCM86问题-B 应急设施的位置 (4)MCM87问题-A 盐的存贮 (5)MCM87问题-B 停车场 (5)MCM88问题-A 确定毒品走私船的位置 (5)MCM88问题-B 两辆铁路平板车的装货问题 (6)MCM89问题-A 蠓的分类 (6)MCM89问题-B 飞机排队 (6)MCM90-A 药物在脑内的分布 (6)MCM90问题-B 扫雪问题 (7)MCM91问题-B 通讯网络的极小生成树 (7)MCM 91问题-A 估计水塔的水流量 (7)MCM92问题-A 空中交通控制雷达的功率问题 (7)MCM 92问题-B 应急电力修复系统的修复计划 (7)MCM93问题-A 加速餐厅剩菜堆肥的生成 (8)MCM93问题-B 倒煤台的操作方案 (8)MCM94问题-A 住宅的保温 (9)MCM 94问题-B 计算机网络的最短传输时间 (9)MCM-95问题-A 单一螺旋线 (10)MCM95题-B A1uacha Balaclava学院 (10)MCM96问题-A 噪音场中潜艇的探测 (11)MCM96问题-B 竞赛评判问题 (11)MCM97问题-A Velociraptor(疾走龙属)问题 (11)MCM97问题-B为取得富有成果的讨论怎样搭配与会成员 (12)MCM98问题-A 磁共振成像扫描仪 (12)MCM98问题-B 成绩给分的通胀 (13)MCM99问题-A 大碰撞 (13)MCM99问题-B “非法”聚会 (14)MCM2000问题-A空间交通管制 (14)MCM2000问题-B: 无线电信道分配 (14)MCM2001问题- A: 选择自行车车轮 (15)MCM2001问题-B 逃避飓风怒吼(一场恶风...) .. (15)MCM2001问题-C我们的水系-不确定的前景 (16)MCM2002问题-A风和喷水池 (16)MCM2002问题-B航空公司超员订票 (16)MCM2002问题-C (16)MCM2003问题-A: 特技演员 (18)MCM2003问题-B: Gamma刀治疗方案 (18)MCM2003问题-C航空行李的扫描对策 (19)MCM2004问题-A:指纹是独一无二的吗? (19)MCM2004问题-B:更快的快通系统 (19)MCM2004问题-C安全与否? (19)MCM2005问题A.水灾计划 (19)MCM2005B.Tollbooths (19)MCM2005问题C:不可再生的资源 (20)MCM2006问题A: 用于灌溉的自动洒水器的安置和移动调度 (20)MCM2006问题B: 通过机场的轮椅 (20)MCM2006问题C : 抗击艾滋病的协调 (21)MCM2007问题B :飞机就座问题 (24)MCM2007问题C:器官移植:肾交换问题 (24)MCM2008问题A:给大陆洗个澡 (28)MCM2008问题B:建立数独拼图游戏 (28)MCM85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。

2020AMC10B(美国数学竞赛)真题加详解

2020AMC10B(美国数学竞赛)真题加详解

2020AMC10B(美国数学竞赛)真题加详解2020 AMC 10B Solution Problem1What is the value ofSolutionWe know that when we subtract negative numbers, .The equation becomesProblem2Carl has cubes each having side length , and Kate has cubes each having side length . What is the total volume of these cubes?SolutionA cube with side length has volume , so of these will have a total volume of .A cube with side length has volume , so of these will have a total volume of .~quacker88Problem 3The ratio of to is , the ratio of to is , and the ratioof to is . What is the ratio of toSolution 1WLOG, let and .Since the ratio of to is , we can substitute in the value of toget .The ratio of to is , so .The ratio of to is then so our answeris ~quacker88Solution 2We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two., and since , we can link themtogether to get .Finally, since , we can link this again to get: ,so ~quacker88Problem4The acute angles of a right triangle are and , where andboth and are prime numbers. What is the least possible value of ?SolutionSince the three angles of a triangle add up to and one of the anglesis because it's a right triangle, .The greatest prime number less than is . If ,then , which is not prime.The next greatest prime number less than is . If ,then , which IS prime, so we have our answer ~quacker88 Solution 2Looking at the answer choices, only and are coprime to . Testing , the smaller angle, makes the other angle which is prime, therefore our answerisProblem5How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)SolutionLet's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.There are ways to order objects. However, since there's ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and ways to order the green tiles, we have to divide out these possibilities.~quacker88SolutionWe can repeat chooses extensively to find the answer. Thereare choose ways to arrange the brown tiles which is . Then from the remaining tiles there are choose ways to arrange the red tiles. And now from the remaining two tiles and two slots we can see there are two ways to arrange the purple and brown tiles, giving us an answerofProblem6Driving along a highway, Megan noticed that her odometershowed (miles). This number is a palindrome-it reads the same forward and backward. Then hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this -hour period?SolutionIn order to get the smallest palindrome greater than , we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.So we raise to the next largest value, , but obviously, that's not how place value works, so we're in the s now. To keep this a palindrome, our number is now .So Megan drove miles. Since this happened over hours, she drove at mph. ~quacker88 Problem7How many positive even multiples of less than are perfect squares?SolutionAny even multiple of is a multiple of , so we need to find multiples of that are perfect squares and less than . Any solution that we want will be in theform , where is a positive integer. The smallest possible value isat , and the largest is at (where the expression equals ). Therefore, there are a total of possible numbers.-PCChess Problem8 Points and lie in a plane with . How many locations forpoint in this plane are there such that the triangle with vertices , ,and is a right triangle with area square units?Solution 1There are options here:1. is the right angle.It's clear that there are points that fit this, one that's directly to the rightof and one that's directly to the left. We don't need to find the length, we just need to know that it is possible, which it is.2. is the right angle.Using the exact same reasoning, there are also solutions for this one.3. The new point is the right angle.(Diagram temporarily removed due to asymptote error)The diagram looks something like this. We know that the altitude tobase must be since the area is . From here, we must see if there are valid triangles that satisfy the necessary requirements. First of all, because of the area.Next, from the Pythagorean Theorem.From here, we must look to see if there are valid solutions. There are multiple ways to do this:We know that the minimum value of iswhen . In this case, the equationbecomes , which is LESSthan . . The equationbecomes , which is obviously greater than . We canconclude that there are values for and in between that satisfy the Pythagorean Theorem.And since , the triangle is not isoceles, meaning we could reflectit over and/or the line perpendicular to for a total of triangles this case.Solution 2Note that line segment can either be the shorter leg, longer leg or thehypotenuse. If it is the shorter leg, there are two possible points for that cansatisfy the requirements - that being above or below . As such, thereare ways for this case. Similarly, one can find that there are also ways for point to lie if is the longer leg. If it is a hypotenuse, then thereare possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is .Problem9How many ordered pairs of integers satisfy theequationSolutionRearranging the terms and and completing the square for yields theresult . Then, notice that can onlybe , and because any value of that is greater than 1 will causethe term to be less than , which is impossible as must be real. Therefore, plugging in the above values for gives the ordered pairs , , , and gives a totalof ordered pairs.Solution 2Bringing all of the terms to the LHS, we see a quadraticequation in terms of . Applying the quadratic formula, weget In order for to be real, which it must be given the stipulation that we are seekingintegral answers, we know that the discriminant, must benonnegative. Therefore, Here, we see that we must split the inequality into a compound, resultingin .The only integers that satisfy this are . Plugging thesevalues back into the quadratic equation, we see that both produce a discriminant of , meaning that there is only 1 solution for . If , then the discriminant is nonzero, therefore resulting in two solutions for .Thus, the answer is .~TiblisSolution 3, x firstSet it up as a quadratic in terms of y:Then the discriminant is This will clearly only yield real solutionswhen , because it is always positive. Then . Checking each one: and are the same when raised to the 2020th power:This has only has solutions , so are solutions. Next, if :Which has 2 solutions, so andThese are the only 4 solutions, soSolution 4, y firstMove the term to the other side toget . Because for all , then . If or , the right side is and therefore . When , the right side become , therefore . Our solutions are , , , . There are solutions, so the answer is - wwt7535Problem 10A three-quarter sector of a circle of radius inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubicinches?SolutionNotice that when the cone is created, the radius of the circle will become the slant height of the cone and the intact circumference of the circle will become the circumference of the base of the cone.We can calculate that the intact circumference of the circle is . Since that is also equal to the circumference of the cone, the radius of the cone is . We also have that the slant height of the cone is . Therefore, we use the Pythagorean Theorem to calculate that the height of the coneis . The volume of the coneis -PCChessSolution 2 (Last Resort/Cheap)Using a ruler, measure a circle of radius 4 and cut out the circle and then the quarter missing. Then, fold it into a cone and measure the diameter to be 6cm . You can form a right triangle with sides 3, 4, and then through the Pythagorean theorem the height is found tobe . The volume of a cone is . Plugging in we findProblem11Ms. Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?SolutionWe don't care about which books Harold selects. We just care that Bettypicks books from Harold's list and that aren't on Harold's list.The total amount of combinations of books that Betty can selectis .There are ways for Betty to choose of the books that are on Harold's list.From the remaining books that aren't on Harold's list, thereare ways to choose of them.~quacker88Problem12The decimal representation of consists of a string of zeros after the decimal point, followed by a and then several more digits. How many zeros are in that initial string of zeros after the decimal point?Solution 1Now we do some estimation. Notice that , which meansthat is a little more than . Multiplying itwith , we get that the denominator is about . Notice that whenwe divide by an digit number, there are zeros before the first nonzero digit. This means that when we divide by the digit integer , there are zeros in the initial string after the decimal point. -PCChessSolution 2First rewrite as . Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to findthe number of digits in .and memming (alternatively use the factthat ),digits.Our answer is .Solution 3 (Brute Force)Just as in Solution we rewrite as We thencalculate entirely by hand, first doing then multiplying that product by itself, resulting in Because this is digits,after dividing this number by fourteen times, the decimal point is beforethe Dividing the number again by twenty-six more times allows a stringof zeroes to be formed. -OreoChocolateSolution 4 (Smarter Brute Force)Just as in Solutions and we rewrite as We can then look at the number of digits in powersof . , , , , ,, and so on. We notice after a few iterations that every power of five with an exponent of , the number of digits doesn't increase. This means should have digits since thereare numbers which are from to , or digits total. This means our expression can be written as , where is in therange . Canceling gives , or zeroes before the since the number should start on where the one would be in . ~aop2014 Solution 5 (Logarithms)Problem13Andy the Ant lives on a coordinate plane and is currently at facingeast (that is, in the positive -direction). Andy moves unit and thenturns degrees left. From there, Andy moves units (north) and thenturns degrees left. He then moves units (west) and againturns degrees left. Andy continues his progress, increasing his distance each time by unit and always turning left. What is the location of the point at which Andy makes the th leftturn?Solution 1You can find that every four moves both coordinates decrease by 2. Therefore, both coordinates need to decrease by two 505 times. You subtract, giving you theanswer of ~happykeeperProblem14As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?Solution 1Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, , since B is the center of the semicircle with radius 1 that C lies on, , since B is the center of the semicircle with radius 1 that A lies on,and , as a regular hexagon has angles of 120,and is half of any angle in this hexagon. Now, using the sinelaw, , so . Since the angles in a triangle sum to 180, is also 60. Therefore, is an equilateral triangle with side lengths of 1.Since the area of a regular hexagon can be found with the formula , where is the side length of the hexagon, the area of this hexagonis . Since the area of an equilateral triangle can be foundwith the formula , where is the side length of the equilateral triangle,the area of an equilateral triangle with side lengths of 1 is . Since the area of a circle can be found with the formula , the area of a sixthof a circle with radius 1 is . In each sixth of the hexagon, thereare two equilateral triangles colored white, each with an area of , and onesixth of a circle with radius 1 colored white, with an area of . The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixthof the hexagon is , which equals , and the total areacolored white is , which equals . Since the area colored gray equals the total area of the hexagon minus the area colored white,the area colored gray is , whichequals .Solution 2First, subdivide the hexagon into 24 equilateral triangles with side length1:Now note that the entire shadedregion is just 6 times this part:The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of: The arc that is not included has an area of:Hence, the area ofthe shaded region in that section is For a final areaof:Problem15Steve wrote the digits , , , , and in order repeatedly from left to right, forming a list of digits, beginning He thenerased every third digit from his list (that is, the rd, th, th, digits from the left), then erased every fourth digit from the resulting list (that is, the th, th, th, digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions ?Solution 1After erasing every third digit, the list becomes repeated. After erasing every fourth digit from this list, the listbecomes repeated. Finally, after erasing every fifth digit from this list, the list becomes repeated. Since this list repeats every digits andsince are respectively in we have that the th, th, and st digits are the rd, th, and thdigits respectively. It follows that the answer is~dolphin7Problem16Bela and Jenn play the following game on the closed interval of the real number line, where is a fixed integer greater than . They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in theinterval . Thereafter, the player whose turn it is chooses a real numberthat is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?SolutionNotice that to use the optimal strategy to win the game, Bela must select themiddle number in the range and then mirror whatever number Jennselects. Therefore, if Jenn can select a number within the range, so can Bela. Jenn will always be the first person to run out of a number to choose, so theanswer is .Solution 2 (Guessing)First of all, realize that the value of should have no effect on the strategy at all. This is because they can choose real numbers, not integers, so even if is odd, for example, they can still go halfway. Similarly, there is no reason the strategy would change when .So we are left with (A) and (B). From here it is best to try out random numbers and try to find the strategy that will let Bela win, but if you can't find it, realize thatit is more likely the answer is since Bela has the first move and thus has more control.Problem17There are people standing equally spaced around a circle. Each person knows exactly of the other people: the people standing next to her or him, as well as the person directly across the circle. How many ways are there forthe people to split up into pairs so that the members of each pair know each other?SolutionLet us use casework on the number of diagonals.Case 1: diagonals There are ways: either pairs with , pairs with , and so on or pairs with , pairs with , etc.Case 2: diagonal There are possible diagonals to draw (everyone else pairs with the person next to them.Note that there cannot be 2 diagonals.Case 3: diagonalsNote that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction.Case 4: diagonals There is way to do this.Thus, in total there are possible ways. Problem18An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?SolutionLet denote that George selects a red ball and that he selects a blue one. Now, in order to get balls of each color, he needs more of both and .There are 6cases:(wecan confirm that there are only since ). However we canclump , ,and together since they are equivalent by symmetry.andLet's find the probability that he picks the balls in the order of .。

2020美赛E题参考翻译

2020美赛E题参考翻译

2020ICM 周末1问题E :塑料溺水自20世纪50年代以来,塑料的制造因其多种用途而成倍增长,如食品包装、消费品、医疗器械和建筑。

虽然有很大的好处,但增加塑料生产的负面影响是令人关切的。

塑料产品不容易分解,很难处理,只有大约9%的塑料被回收[1],每年进入海洋的大约4-1200万吨塑料废物可以看到效果[1,2]。

塑料废物具有严重的环境后果,据预测,如果我们目前的趋势继续下去,到2050年,海洋中的塑料含量将超过鱼类[2]。

对海洋生物的影响已被研究[3],但对人类健康的影响尚未完全理解[4]单一用途和一次性塑料产品的兴起导致了致力于制造塑料废物的整个行业。

它还表明,产品有用的时间远远短于适当减少塑料浪费所需的时间。

因此,为了解决塑料废物问题,我们需要减缓塑料生产的速度,改进我们管理塑料废物的方法。

..您的团队已被国际塑料废物管理理事会(ICM )雇用,以解决这一不断升级的环境危机。

你必须制定一个计划,以显著减少,如果不消除,一次性使用和一次性塑料产品的浪费..∙开发一个模型,以估计单一用途或一次性塑料产品废物的最大水平,这些废物可以安全地减轻,而不会造成进一步的环境损害。

在许多因素中,您可能需要考虑这种废物的来源、当前废物问题的程度以及处理废物的资源的可用性。

∙讨论在多大程度上可以减少塑料废物,以达到环境安全水平。

这可能涉及考虑影响塑料废物水平的因素,包括但不限于单一用途或一次性塑料的来源和用途、塑料替代品的可用性、对公民生活的影响,或城市、区域、国家和大陆减少单一用途或一次性塑料的政策,以及此类政策的有效性。

这些因素可能因地区而异,因此考虑到特定区域的制约因素可能会使某些政策比其他政策更有效。

∙利用你的模型和讨论,为单一用途或一次性塑料产品的全球废物的最低可达到水平设定一个目标,并讨论实现这一水平的影响。

你可以考虑改变人类生活的方式,环境影响,或对数万亿美元塑料工业的影响。

∙虽然这是一个全球性问题,但其原因和影响在各国或各区域之间并不平等分布。

美赛金融题目

美赛金融题目

美赛金融题目美国大学生数学建模竞赛分为两种类型:MCM(Mathematical Contest In Modeling)和ICM(Interdisciplinary Contest In Modeling)两种类型竞赛采用统一标准进行,竞赛题目出来之后,参数队伍通过美赛官网进行选题,一共分为6种题型。

MCM是包括A、B、C三道题目:-- A题是指连续型(continuous),具体可以理解为是连续函数建立一类模型。

常用方法是微分方程,并多为“数值分析”领域的内容,需要熟练掌握偏微分方程以及精通将连续性方程离散化求解的编程能力。

-- B题是离散型(discrete)具体需要在编程上比较熟悉计算机的“算法与数据结构”。

-- C题是数据分析型(data insights),最好是有统计学、数理金融、量化分析相关背景的知识。

C题除了MATLAB、Python还可以是用无需编程的SPSS,也可能会用到R、STATS、SAS等统计软件。

ICM包括D、E、F三道题目:-- D题一般为运筹学或网络科学(operations research/network science),近几年网络科学是一个热门研究领域,算法、软件包括可视化的软件都很多。

-- E题是环境科学题(sustainability),大体上会集中在环境污染、资源短缺、可持续发展、生态保护等几个方面。

官方给出的六个问题的描述,对于选题有一定的提示作用,但是六道题目之间没有严格的模型划分,解答范围仍然很广,每道题所用模型、方法没有明显的界限。

因此小伙伴们在寒假先要摸清楚不同题目常用的模型算法、写作套路以及题目风格,比赛时根据自己队伍擅长的东西慎重选题。

02 近五年美赛A题特点小竞分析,在六道题目中,相对而言,A题是对同学们的数学模型素养以及建模能力要求较高的。

同时A题的专业性较强,非专业相关的同学对题目的理解较难。

例如2018年的A题:多跳HF无线电传播,这道赛题包含大量的专业名词,如电离层、最大可用频率、地面源、连续跳跃、局部介电常数等,大大增加了同学们对赛题的理解难度。

2020中美洲及加勒比数学奥林匹克试解答

2020中美洲及加勒比数学奥林匹克试解答

题目:探讨2020中美洲及加勒比数学奥林匹克试解答---1. 引言2020年的中美洲及加勒比数学奥林匹克试解答,是一个备受关注的事件。

在数学领域,这些试题代表着挑战和机遇。

本文将深入探讨这些试题,给您带来一场关于数学奥赛的知识盛宴。

2. 试题概述2020年中美洲及加勒比地区的数学奥林匹克试题,涵盖了代数、几何、概率等多个领域。

每道题目都设计严谨,考察了考生对数学知识的深度和灵活运用能力。

试题内容不仅注重基础知识的考察,更将数学与现实生活相结合,增加了题目的趣味性和挑战性。

3. 题目解析(1)代数题在代数题部分,考生需要熟练运用代数运算、方程式求解等技巧,解决问题。

一道题目要求计算多项式的值,另一道题目则考察了对齐次方程组的解法。

这些题目旨在考察考生对代数知识的掌握情况,并培养其逻辑推理能力。

(2)几何题几何题部分则考察了考生的空间想象能力和几何推理能力。

题目设计涉及了线段、角度、多边形等几何形状,要求考生在解题过程中画出清晰准确的图形,并通过严密的证明、推理解决问题。

(3)概率题概率题部分考察了考生对概率的理解和应用能力。

题目设计涉及了随机事件、概率计算、独立事件等内容,要求考生通过分析情况、计算概率,解决实际问题。

4. 试题意义2020年中美洲及加勒比数学奥林匹克试题的设计,旨在提高学生的数学综合能力,培养他们的逻辑思维和解决问题的能力。

通过解答这些试题,考生不仅能够巩固已学的数学知识,还能够开拓思维,拓展数学视野,提高解决实际问题的能力。

5. 个人观点作为一名数学爱好者,我认为2020年中美洲及加勒比数学奥林匹克试题的设计非常有价值。

这些试题既考察了考生的数学基础知识,又注重了数学在现实生活中的应用。

通过解答这些试题,我深切感受到数学的魅力和实用性,也在实践中不断提高自己的数学素养。

6. 总结2020年中美洲及加勒比数学奥林匹克试题的内容丰富多样,题目设计严谨合理,体现了数学的广度和深度。

解答这些试题不仅有助于考生巩固数学知识,还能够培养他们的逻辑思维和问题解决能力。

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对E题的理解:
目标一:建模分析一次性塑料垃圾的最大产量
Develop a model to estimate the maximum levels of single-use or
disposable plastic product waste (定语从句修饰waste)that can safely be mitigated without further environmental damage. 在可以安全分解的前提下,分析一次性塑料垃圾的最大产量。

You may need to consider,
among many factors, (要考虑下面几个因子)the source of this waste, the extent of the current waste problem, and the availability of resources to process the waste.
目标二:讨论减少多少塑料废物可以达到环境安全的目标
This may involve considering factors impacting the levels of plastic
waste to include, but not limited to(包括但不限于下面这些因素),
sources and uses of single-use or disposable plastics, the availability of
alternatives to plastics, the impact on the lives of citizens, or policies of
cities, regions, countries, and continents to decrease single-use or
disposable plastic and the effectiveness of such policies.
目标三:用1、2的结论确定一个一次性塑料最低水平的目标,分析影响。

(我认为这部分以分析吹水为主)
目标四:塑料问题的起因和结果在世界不同地区存在差异,探究公平性的问题,给出建议
目标五:给出目标和时间线,写两页纸大白话,考虑any circumstances that may accelerate or hinder the achievement of your target and timeline.
(突发事件这里可能用到建模)
The detail can be described by equation \eqref{eq:heat}:
\begin{equation}\label{eq:heat}
\frac{\partial u}{\partial t} - a^2 \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} +
\frac{\partial^2 u}{\partial z^2} \right) = f(x, y, z, t)
\end{equation}。

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