2013年美赛真题题目

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2013美国数学竞赛AMC8题目及答案

2.is the value of ?friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $ to cover her portion of the total bill. What was the total bill?is in the grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds.Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, . What is the missing number in the top row?and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?fair coin is tossed 3 times. What is the probability of at least two consecutive heads?Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer?is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?11.Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walkedless time on the treadmill. How many minutes less?12.At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pairof sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at halfthe regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save?13.WhenClara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score.By which of the following might her incorrect sum have differed from the correct one?14.Let the two digits be and .The correct score was . Clara misinterpreted it as . The difference between the two is which factors into .Therefore, since the difference is a multiple of 9, the only answer choice that is a multiple of 9 is .15.If , , and , what is the product of , , and ?16.A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of -graders to -graders is , and the the ratio of -graders to -graders is . What is the smallest number of students that could be participating in the project?17.The sum of six consecutive positive integers is 2013. What is the largest of these six integers?18.Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?--Arpanliku 16:22, 27 November 2013 (EST) Courtesy of19.Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget andCassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowestscore in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest?20.A rectangle is inscribed in a semicircle with longer side on the diameter. What is the area of the semicircle?21.Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?22.Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?23.Angle of is a right angle. The sides of are the diameters of semicircles as shown. The area of the semicircle on equals , and the arc of the semicircle on has length . What is the radius of the semicircle on ?24.Squares , , and are equal in area. Points and are the midpoints of sides and , respectively. What is the ratio of the area of the shaded pentagon to the sum of the areas of the three squares?25.A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are inches, inches, and inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?1.50% off price of half a pound of fish is $3, so the 100%, or the regular price, of a half pound of fish is$6. Consequently, if half a pound of fish costs $6, then a whole pound of fish is dollars.that we can pair up every two numbers to make a sum of 1:Therefore, the answer is .of her seven friends paid to cover Judi's portion. Therefore, Judi's portion must be . Since Judi was supposed to pay of the total bill, the total bill must be .median here is obviously less than the mean, so option (A) and (B) are out.Lining up the numbers (5, 5, 6, 8, 106), we see that the median weight is 6 pounds.The average weight of the five kids is .Therefore, the average weight is bigger, by pounds, making the answer .1: Working BackwardsLet the value in the empty box in the middle row be , and the value in the empty box in the top row be . isthe answer we're looking for.It follows that , so .Solution 2: Jumping Back to the StartAnother way to do this problem is to realize what makes up the bottommost number. This method doesn't work quite as well for this problem, but in a larger tree, it might be faster. (In this case, Solution 1 would be faster since there's only two missing numbers.)We see that , making .Again, let the value in the empty box in the middle row be , and the value in the empty box in the top row be . is the answer we're looking for. We can write some equations:Now we can substitute into the first equation using the two others:Trey saw , then he saw .2 minutes and 45 seconds can also be expressed as seconds.Trey's rate of seeing cars, , can be multiplied by on the top and bottom (and preserve the same rate):.It follows that the most likely number of cars is .Solution 2minutes and seconds is equal to .Since Trey probably counts around cars every seconds, there are groups of cars that Trey most likely counts.Since , the closest answer choice is .,there are ways to flip the coins, in order.The ways to get two consecutive heads are HHT and THH.The way to get three consecutive heads is HHH.Therefore, the probability of flipping at least two consecutive heads is .is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters,However, because the first term is and not , the solution to the problem is10.To find either the LCM or the GCF of two numbers, always prime factorize first.The prime factorization of .The prime factorization of .Then, find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is ). Multiply all of these to get 5940.For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. = 18.Thus the answer ==.we note that .We start off with a similar approach as the original solution. From the prime factorizations, the GCF is .It is a well known fact that . So we have,.Dividing by yields .Therefore,.11.We use that fact that . Let d= distance, r= rate or speed, and t=time. In this case, let represent the time.On Monday, he was at a rate of . So,.For Wednesday, he walked at a rate of . Therefore,.On Friday, he walked at a rate of . So,.Adding up the hours yields + + =.We now find the amount of time Grandfather would have taken if he walked at per day. Set up the equation,.To find the amount of time saved, subtract the two amounts: - = . To convert this to minutes, we multiply by .Thus, the solution to this problem is12.First, find the amount of money one will pay for three sandals without the discount. We have .Then, find the amount of money using the discount:.Finding the percentage yields .To find the percent saved, we have13.Let the two digits be and .The correct score was . Clara misinterpreted it as . The difference between the two is which factors into .Therefore, since the difference is a multiple of 9, the only answer choice that is a multiple of 9 is .14.The probability that both show a green bean is . The probability that both show a red bean is . Therefore the probability is15.Therefore,.Therefore,.To most people, it would not be immediately evident that , so we can multiply 6's until we get the desired number: ,so .Therefore the answer is .16.Solution 1: AlgebraWemultiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:Therefore, the ratio of 8th graders to 7th graders to 6th graders is . Since the ratio is in lowest terms, the smallest number of students participating in the project is .Solution 2: FakesolvingThe number of 8th graders has to be a multiple of 8 and 5, so assume it is 40 (the smallest possibility).Then there are 6th graders and 7 th graders. The numbers of students is17.Solution 1The mean of these numbers is . Therefore the numbers are , so the answer isSolution 2Let the number be . Then our desired number is .Our integers are , so we have that .Solution 3Let the first term be . Our integers are . We have,18.Solution 1There are cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are cubes. Hence, the answer is .Solution 2We can just calculate the volume of the prism that was cut out of the original box. Each interior side ofthe fort will be feet shorter than each side of the outside. Since the floor is foot, the height will be feet.So the volume of the interior box is .The volume of the original box is . Therefore, the number of blocks contained in the fort is .19.If Hannah did better than Cassie, there would be no way she could know for sure that she didn't get the lowest score in the class. Therefore, Hannah did worse than Cassie. Similarly, if Hannah did worse than Bridget, there is no way Bridget could have known that she didn't get the highest in the class. Therefore, Hannah did better than Bridget, so our order is .20.A semicircle has symmetry, so the center is exactly at the midpoint of the 2 side on the rectangle, making the radius, by the Pythagorean Theorem, . The area is .21. The number of ways to get from Samantha's house to City Park is , and the number of ways to get from CityPark to school is . Since there's one way to go through City Park (just walking straight through), the number of different ways to go from Samantha's house to City Park to school .22.There are vertical columns with a length of toothpicks, and there are horizontal rows with a length of toothpicks. An effective way to verify this is to try a small case, . a grid of toothpicks. Thus, our answer is .23.Solution 1If the semicircle on AB were a full circle, the area would be 16pi. Therefore the diameter of the first circle is 8. The arc of the largest semicircle would normally have a complete diameter of 17. The Pythagorean theorem says that the other side has length 15, so the radius is .Solution 2We go as in Solution 1, finding the diameter of the circle on AC and AB. Then, an extended version of the theorem says that the sum of the semicircleson the left is equal to the biggest one, so the area of the largest is , and the middle one is , so the radius is .24.First let (where is the side length of the squares) for simplicity. We can extend until it hits the extensionof . Call this point . The area of triangle then is The area of rectangle is . Thus, our desired area is .Now, the ratio of the shaded area to the combined area of the three squares is .Solution 2Let the side length of each square be .Let the intersection of and be .Since , . Since and are vertical angles, they are congruent. We also have by definition.So we have by congruence. Therefore,.Since and are midpoints of sides, . This combined with yields .The area of trapezoid is .The area of triangle is .So the area of the pentagon is .The area of the squares is .Therefore,.Solution 3Let the intersection of and be .Now we have and .Because both triangles has a side on congruent squares therefore .Because and are vertical angles .Also both and are right angles so .Therefore by AAS(Angle, Angle, Side).Then translating/rotating the shaded into the position ofSo the shaded area now completely covers the squareSet the area of a square asTherefore,.25.Solution 1The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses inches, and it gains inches on B. So, the departure from the length of the track means that the answer is .Solution 2The total length of all of the arcs is . Since we want the path from the center, the actual distance will be shorter. Therefore, the only answer choice less than is . This solution may be invalid because the actual distance can be longer if the path the center travels is on the outside of the curve, as it is in the middle bump.。

2013年美赛A题翻译题目

2013 MCM A 终极布朗尼锅
当在一个矩形锅里烘烤，热量被集中在四个角上，在角上的产品会煮过头（边缘部位程度会小一些）。

在一个圆形的锅里，热量会均匀分布在整个锅的外缘部位并且产品也不会被煮过头。

然而，因为大多数的烤箱的形状都是矩形的，用圆锅对于利用锅的内部空间是没有效率的。

建立一个模型，以显示热量穿过锅的外边缘上的分布，对于不同形状-矩形和圆形以及他们之间形状的锅。

假设：
1、矩形烤箱的宽长比是W/L.
2、每个锅的面积是A
3、烤箱中初始的两个隔板是均匀分布的
建立一个模型，可以用来选择最优的锅的形状，满足以下的条件：
1.烤箱中可以容纳锅的最大数量N。

2.使锅的热量H最大化均匀分布
3.优化组合（1）和（2）中的条件，他们的权重分配是p和（1-p），来说明在不同的W/L和p的值下，结果是怎么变化的。

除了你的MCM格式的解决方案，准备一到两页的广告片新布朗尼美食杂志突出自己的设计和结果。

2013美赛题目中文翻译

A当在一个方形平底锅烘烤时，由于热量集中在四个角上，在角上的（和在伸展度小的边缘）食物就会过度烹饪，圆形平底锅热量平均分配在整个外缘，在边上的食物不会过度烹饪。

然而，由于大部分烤箱都是方形，考虑到烤箱空间的使用，圆形锅不会有效的利用烤箱的空间。

Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes - rectangular to circular and other shapes in between.目的：建立一个模型对于不同形状锅的边缘的热量分布，方形到圆形或他们中任一形状。

Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.2. Each pan must have an area of A.3. Initially two racks in the oven, evenly spaced. 假设方形烤箱的宽长比是W/L每个锅都有一个A型区域首先，烤箱有两个烤架，并且均匀分布。

Develop a model that can be used to select the best type of pan (shape) under the following conditions:1. Maximize number of pans that can fit in the oven (N)2. Maximize even distribution of heat (H) for the pan 3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different values of W/L and p.在如下条件下建立一个模型，用来选择在形状上最好类型的锅。

13年数学美国赛建模题目中英文

PROBLEM A: The Ultimate Brownie PanWhen baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient with respect to using the space in an oven.Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes - rectangular to circular and other shapes in between.Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.2. Each pan must have an area of A.3. Initially two racks in the oven, evenly spaced.Develop a model that can be used to select the best type of pan (shape) under the following conditions:1. Maximize number of pans that can fit in the oven (N)2. Maximize even distribution of heat (H) for the pan3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different values of W/L and p.In addition to your MCM formatted solution, prepare a one to two page advertising sheet for the new Brownie Gourmet Magazine highlighting your design and results.烘烤时，在一个长方形的锅热集中在4角和产品得到了在角落（以及在较小程度上的边缘处）。

2013年美国数学建模大赛MCM试题(附中文翻译)

2013 MCM ProblemsPROBLEM A:The Ultimate Brownie PanWhen baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient with respect to using the space in an oven. Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes - rectangular to circular and other shapes in between.Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.2. Each pan must have an area of A.3. Initially two racks in the oven, evenly spaced.Develop a model that can be used to select the best type of pan (shape) under the following conditions:1. Maximize number of pans that can fit in the oven (N)2. Maximize even distribution of heat (H) for the pan3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different values of W/L and p.In addition to your MCM formatted solution, prepare a one to two page advertising sheet for the new Brownie Gourmet Magazine highlighting your design and results.当在矩形盘子中烘烤食物时，热量会集中于四个角。

2013数学建模美赛题目(中文版)[1]

2013 ICM问题problem A:当用矩形平底锅高温加热物品时，热量一般集中于4个角落，因而在角落的物品会被焙烧过度（较小程度在角落的物品一部分会被焙烧过度）。

当用一个圆锅加热物品时，热量是均匀分布在整个外缘，因而物品不会在边缘被焙烧过度。

然而，大多数烤箱是长方形的，而圆型的锅被认为效率低的。

建立一个模型以显示不同形状如矩形圆形或者其他介于两者之间的形状的锅在整个外缘的热量分布。

假设1长方形烤箱的宽/长=W/L ；2 每个锅的面积是确定的常熟A；3 最初，烤箱里的烤架两两之间间隔均匀。

建立一个模型，该模型可用于在下列条件之下选择最佳形状的锅：1烤箱中，锅数量（N）最大；2均匀分布的热量（H）最大的锅；3 优化组合条件1和条件2，以比重p和（1-p）的不同分配来说明结果与W/L 和p的不同值的关系。

problem B :对世界来说，新鲜的水资源是限制发展的制约因素。

对2013年建立一个确实有效的，可行的和具有成本效益的水资源战略数学模型，以满足2025年[从下面的列表选择一个国家]预计的用水需求,并确定最佳水资源战略。

尤其是，你的数学模型必须解决水的存储，运动，盐碱化和保护等问题。

如果可能的话，用你的模型，探讨经济，物理和环境对于你的战略的影响。

提供一个非技术性的文件，向政府领导介绍你的方法，介绍其可行性和成本，以及为什么它是“最好的的水战略选择。

”国家有：美国，中国，俄罗斯，埃及，沙特阿拉伯3.网络建模的地球的健康背景:社会是感兴趣的发展和使用模型来预测生物和环境卫生条件我们的星球。

许多科学研究认为越来越多的压力在地球的环境和生物吗系统,但是有很少的全球模型来测试这些索赔。

由联合国支持的年生态系统评估综合报告》显示,近三分之二的地球的维持生命的生态系统——包括干净的水,纯净的空气,和稳定的气候-正在退化,被不可持续的使用。

人类是归咎于很多这次的损坏。

不断飙升的要求食品、新鲜水、燃料和木材有贡献到戏剧性的环境变化,从森林砍伐,空气,土地和水的污染。

2013美国大学生数学建模竞赛题目和翻译和A题图解

2013年美赛A题

2013年美赛A题最终的布朗尼锅摘要关键字：目录引言题目背景近年来，电烤箱普遍采用远红外加热技术，使电烤箱的技术含量增加，耗能降低，深受广大用户的欢迎。

利用红外线加热物体，就是利用辐射波长与物体接收波长一致时，物体吸收大量的红外能，从而加剧物体内部的分子运动，使之加热升温。

加热时间短，能耗低，使用方便。

但是，当我们使用矩形烤盘烘烤食物时，热传导方程，加上一些边界条件，导致方形烤盘热量集中在的四个角上，因此四个角上的物体会因过度受热（以及在较小程度的边缘处）而变焦。

如果用圆形烤盘，热量会平均分布在整个外围边缘，在外围的物体就不会过度受热。

然而，由于大多数的烤箱都是矩形的，所以用圆形的烤盘就不能较好的利用烤箱的空间。

给烘烤食物的朋友带来了很大的不便。

为什么角部的食物肉容易烤焦，以及选择哪种形状的烤盘，，这是令人很费解的问题。

电烤箱工作原理电烤箱利用电热元件所发出的辐射热来烘烤食品，利用它我们可以制作烤鸡、烤鸭、烘烤面包、糕点等。

根据烘烤食品的不同需要，电烤箱的温度一般可在50-250℃范围内调节。

电烤箱主要由箱体、电热元件、调温器、定时器和功率调节开关等构成。

其箱体主要由外壳、中隔层、内胆组成三层结构，在内胆的前后边上形成卷边，以隔断腔体空气；在外层腔体中充填绝缘的膨胀珍珠岩制品，使外壳温度大大减低；同时在门的下面安装弹簧结构，使门始终压紧在门框上，使之有较好的密封性。

电烤箱的加热方式可分为面火（上加热器加热）、底火（下加热器加热）和上下同时加热三种。

电烤箱技术参数温度范围室温-200℃（300℃）温度稳定度±0.5℃温度分布均匀度±2℃（特佳）排气烟道叶片式设计可调出风量符号和定义l：多边形边长L: 多边形周长k: 周长与面积的比G：单位圆的周长C: 单位椭圆周长a: 椭圆的长半轴b: 椭圆的短半轴假设1、烤箱内温度同一层表分布均匀且稳定2、烤箱内风扇使空气及时流通3、假设烤盘之间相互不影响4、假设各层之间相互不影响5、假设烤盘的深度影响忽略不计6、假设烤箱内垂直分布的热辐射场为递增针对问题1问题1的说明针对问题1，考虑同一层烤架上温度稳定且分布均匀，我们提出了两个模型。

美赛历年题目2005—2013

MCM 2013 A题：最佳巧克力蛋糕烤盘当你使用一个矩形的烤盘烘烤食物时，热量会集中在烤盘的四个角落，于是角落处的食物就会被烤糊（烤盘边缘处也有类似情形，但程度轻一些）。

当使用一个圆形烤盘时，热量会均匀地分布在整个边缘上，就不会再有边缘上烤糊的现象发生。

然而，由于大多数烤箱内部是矩形的，如果使用圆形烤盘，就不能充分利用烤箱的内部空间了。

建立一个模型，来描述热量在不同形状的烤盘表面的分布。

这些形状包括矩形、圆形以及两者之间的过渡形状。

假设，1、矩形烤箱的宽长比为 W/L。

2、每个烤盘的面积为A。

3、先考虑烤箱内有两个搁架且间隔均匀的情形。

建立一个模型用以选择满足下列条件的最佳烤盘的形状：(1)、使得烤箱中可以容纳的烤盘数量(N)最大。

(2)、使得烤盘上的热量分布(H)最均匀。

3、综合(1)、(2)两个条件，并且为(1)、(2)分别设置权值p和(1-p)，寻求最优。

然后描述结果随着 W/L 和 p 的值的变化是如何变化的。

除了撰写 MCM 论文之外，你还要为新的一期巧克力蛋糕美食杂志准备一个一至两页的广告，阐述你的设计和结果的亮点所在。

MCM 2013 B题：水，水，无处不在淡水资源是世界上许多地方持续发展的限制因素。

建立数学模型来确定一个有效的，可行的，低成本的2013年用水计划，来满足某国(从下方的列表中选择一个国家)未来(2025年)的用水需求，并确定最优的淡水分配计划。

特别的，你的数学模型必须包括储存、运输、淡化和节水等环节。

如果可能的话，用你的模型来讨论你的计划对经济，自然和环境的影响。

提供一个非技术性的意见书给政府领导概述你的方法，以及方法的可行性和成本，以及它为什么是“最好的用水计划的选择”。

国家：美国、中国、俄罗斯、埃及或者沙特阿拉伯。

ICM 2013 C题：地球健康的网络建模背景：全社会都在关注如何研究与应用模型来预测我们地球的生物和环境的健康状况。

许多科学研究表明地球的环境和生物系统所面对的压力正在增加,但是能够验证这一观点的全局性模型却很少。

2013美国数学竞赛12年级B卷试题

INSTRUCTIONS1. DO NOT OPEN THIS BOOKLET UNTIL YOUR PROCTOR TELLS YOU.2. This is a twenty-five question multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.3. Mark your answer to each problem on the AMC 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.© 2013 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 31st annual American Invitational Mathematics Examination (AIME) on Thursday, March 14, 2013 or Wednesday, April 3, 2013. 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 not made ordistributed 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 20, 2013. Nothing is needed from inside this package until February 20.2. Your PRINCIPAL or VICE-PRINCIPAL must verify on the AMC 12 CERTIFICATION FORM (found in the T eachers’ 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, e-mail, internet or media of any type is a violation of the competition rules.2013AMC 12 B DO NOT OPEN UNTIL WEDNEsDAy, fEbrUAry 20, 2013The American Mathematics Competitionsare Sponsored byThe Mathematical Association of America – MAA The Akamai Foundation ContributorsAcademy of Applied Sciences – AAs American Mathematical Association of Two-Year Colleges – AMATyC ...................................................... American Mathematical Society – AMs ........................................................................................................... American Statistical Association – AsA ...................................................................................................... Art of Problem Solving – Awesome Math Casualty Actuarial Society – CAs ................................................................................................................ D.E. Shaw & Co. ................................................................................................................................. Delta Airlines Jane Street Math For America Mu Alpha Theta – MAT ....................................................................................................................... National Council of Teachers of Mathematics – NCTM ................................................................................... Pi Mu Epsilon – PME ............................................................................................................................... Society for Industrial and Applied Math - SIAM ............................................................................................ 1.On a particular January day,the high temperature in Lincoln,Nebraska,was16degrees higher than the low temperature,and the average of the high and low temperatures was3◦.In degrees,what was the low temperature in Lincoln that day?(A)−13(B)−8(C)−5(D)−3(E)112.Mr.Green measures his rectangular garden by walking two of the sides andﬁnds that it is15steps by20steps.Each of Mr.Green’s steps is2feet long.Mr.Green expects a half a pound of potatoes per square foot from his garden.How many pounds of potatoes does Mr.Green expect from his garden?(A)600(B)800(C)1000(D)1200(E)14003.When counting from3to201,53is the51st number counted.When countingbackwards from201to3,53is the n th number counted.What is n?(A)146(B)147(C)148(D)149(E)1504.Ray’s car averages40miles per gallon of gasoline,and Tom’s car averages10miles per gallon of gasoline.Ray and Tom each drive the same number of miles.What is the cars’combined rate of miles per gallon of gasoline?(A)10(B)16(C)25(D)30(E)405.The average age of33ﬁfth-graders is11.The average age of55of their parentsis33.What is the average age of all of these parents andﬁfth-graders?(A)22(B)23.25(C)24.75(D)26.25(E)286.Real numbers x and y satisfy the equation x2+y2=10x−6y−34.What isx+y?(A)1(B)2(C)3(D)6(E)87.Jo and Blair take turns counting from1to one more than the last number saidby the other person.Jo starts by saying“1”,so Blair follows by saying“1,2”.Jo then says“1,2,3”,and so on.What is the53rd number said?(A)2(B)3(C)5(D)6(E)88.Line 1has equation3x−2y=1and goes through A=(−1,−2).Line 2hasequation y=1and meets line 1at point B.Line 3has positive slope,goes through point A,and meets 2at point C.The area of ABC is3.What is the slope of 3?(A)23(B)34(C)1(D)43(E)329.What is the sum of the exponents of the prime factors of the square root of thelargest perfect square that divides12!?(A)5(B)7(C)8(D)10(E)1210.Alex has75red tokens and75blue tokens.There is a booth where Alex cangive two red tokens and receive in return a silver token and a blue token,and another booth where Alex can give three blue tokens and receive in return a silver token and a red token.Alex continues to exchange tokens until no more exchanges are possible.How many silver tokens will Alex have at the end?(A)62(B)82(C)83(D)102(E)10311.Two bees start at the same spot andﬂy at the same rate in the followingdirections.Bee A travels1foot north,then1foot east,then1foot upwards, and then continues to repeat this pattern.Bee B travels1foot south,then1 foot west,and then continues to repeat this pattern.In what directions are the bees traveling when they are exactly10feet away from each other?(A)A east,B west(B)A north,B south(C)A north,B west(D)A up,B south(E)A up,B west12.Cities A,B,C,D,and E are connected by roadsAB,AD, AE,BC,BD,CD, andDE.How many diﬀerent routes are there from A to B that use each road exactly once?(Such a route will necessarily visit some cities more than once.)(A)7(B)9(C)12(D)16(E)1813.The internal angles of quadrilateral ABCD form an arithmetic progression.Tri-angles ABD and DCB are similar with∠DBA=∠DCB and∠ADB=∠CBD.Moreover,the angles in each of these two triangles also form an arithmetic pro-gression.In degrees,what is the largest possible sum of the two largest angles of ABCD?(A)210(B)220(C)230(D)240(E)25014.Two non-decreasing sequences of nonnegative integers have diﬀerentﬁrst terms.Each sequence has the property that each term beginning with the third is the sum of the previous two terms,and the seventh term of each sequence is N.What is the smallest possible value of N?(A)55(B)89(C)104(D)144(E)27315.The number2013is expressed in the form2013=a1!a2!···a m!b1!b2!···b n!,where a1≥a2≥···≥a m and b1≥b2≥···≥b n are positive integers and a1+b1is as small as possible.What is|a1−b1|?(A)1(B)2(C)3(D)4(E)516.Let ABCDE be an equiangular convex pentagon of perimeter 1.The pairwise intersections of the lines that extend the sides of the pentagon determine a ﬁve-pointed star polygon.Let s be the perimeter of this star.What is the diﬀerence between the maximum and the minimum possible values of s ?(A)0(B)12(C)√5−12(D)√5+12(E)√517.Let a ,b ,and c be real numbers such thata +b +c =2,anda 2+b 2+c 2=12.What is the diﬀerence between the maximum and minimum possible values of c ?(A)2(B)103(C)4(D)163(E)20318.Barbara and Jenna play the following game,in which they take turns.A number of coins lie on a table.When it is Barbara’s turn,she must remove 2or 4coins,unless only one coin remains,in which case she loses her turn.When it is Jenna’s turn,she must remove 1or 3coins.A coin ﬂip determines who goes ﬁrst.Whoever removes the last coin wins the game.Assume both players use their best strategy.Who will win when the game starts with 2013coins and when the game starts with 2014coins?(A)Barbara will win with 2013coins,and Jenna will win with 2014coins.(B)Jenna will win with 2013coins,and whoever goes ﬁrst will win with 2014coins.(C)Barbara will win with 2013coins,and whoever goes second will win with 2014coins.(D)Jenna will win with 2013coins,and Barbara will win with 2014coins.(E)Whoever goes ﬁrst will win with 2013coins,and whoever goes second will win with 2014coins.19.In triangle ABC ,AB =13,BC =14,and CA =15.Distinct points D ,E ,and F lie on segments BC ,CA ,and DE ,respectively,such that AD ⊥BC ,DE ⊥AC ,and AF ⊥BF .The length of segment DF can be written as m n ,where m and n are relatively prime positive integers.What is m +n ?(A)18(B)21(C)24(D)27(E)3020.For135◦<x<180◦,points P=(cos x,cos2x),Q=(cot x,cot2x),R=(sin x,sin2x),and S=(tan x,tan2x)are the vertices of a trapezoid.What is sin(2x)?(A)2−2√2(B)3√3−6(C)3√2−5(D)−34(E)1−√321.Consider the set of30parabolas deﬁned as follows:all parabolas have as focusthe point(0,0)and the directrix lines have the form y=ax+b with a and b integers such that a∈{−2,−1,0,1,2}and b∈{−3,−2,−1,1,2,3}.No three of these parabolas have a common point.How many points in the plane are on two of these parabolas?(A)720(B)760(C)810(D)840(E)87022.Let m>1and n>1be integers.Suppose that the product of the solutions forx of the equation8(lognx)(log m x)−7log n x−6log m x−2013=0 is the smallest possible integer.What is m+n?(A)12(B)20(C)24(D)48(E)27223.Bernardo chooses a three-digit positive integer N and writes both its base-5and base-6representations on a ter LeRoy sees the two numbers Bernardo has written.Treating the two numbers as base-10integers,he adds them to obtain an integer S.For example,if N=749,Bernardo writes the numbers10,444and3,245,and LeRoy obtains the sum S=13,689.For how many choices of N are the two rightmost digits of S,in order,the same as those of2N?(A)5(B)10(C)15(D)20(E)2524.Let ABC be a triangle where M is the midpoint of AC,and CN is the anglebisector of∠ACB with N on AB.Let X be the intersection of the median BM and the bisector CN.In addition BXN is equilateral and AC=2.What is BN2?(A)10−6√27(B)29(C)5√2−3√38(D)√26(E)3√3−4525.Let G be the set of polynomials of the formP(z)=z n+c n−1z n−1+···+c2z2+c1z+50, where c1,c2,...,c n−1are integers and P(z)has n distinct roots of the form a+ib with a and b integers.How many polynomials are in G?(A)288(B)528(C)576(D)992(E)1056WRITE TO US!Correspondence about the problems and solutions for this AMC 12and orders for publications should be addressed to:American Mathematics CompetitionsUniversity of Nebraska, P .O. Box 81606Lincoln, NE 68501-1606Phone 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. Abrego2013 AIMEThe 31st annual AIME will be held on Thursday, March 14, with the alternate on Wednesday, April 3. 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 42nd Annual USA Mathematical Olympiad (USAMO) on April 30 - May 1, 2013. 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: American Mathematics Competitions。

2013年AMC 10B竞赛真题及答案(英文版)

2013年美国数学竞赛AMC10B真题What is ?Problem 2Mr. Green measures his rectangular garden by walking two of the sides and finding that it is steps by steps. Each of Mr. Green's steps is feet long. Mr. Green expects a half a pound of potatoes persquare foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?Problem 3On a particular January day, the high temperature in Lincoln, Nebraska, was degrees higher than the low temperature, and the average of the high and the low temperatures was . In degrees, whatwas the low temperature in Lincoln that day?Problem 4When counting from to , is the number counted. When counting backwardsfrom to , is the number counted. What is ?Problem 5Positive integers and are each less than . What is the smallest possible value for ?The average age of 33 fifth-graders is 11. The average age of 55 of their parents is 33. What is the average age of all of these parents and fifth-graders?Problem 7Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of atriangle that is neither equilateral nor isosceles. What is the area of this triangle?Problem 8Ray's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of milesper gallon of gasoline?Problem 9Three positive integers are each greater than , have a product of , and are pairwise relativelyprime. What is their sum?Problem 10A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?Real numbers and satisfy the equation . What is ?Problem 12Let be the set of sides and diagonals of a regular pentagon. A pair of elements of are selected atrandom without replacement. What is the probability that the two chosen segments have the samelength?Problem 13Jo and Blair take turns counting from to one more than the last number said by the other person. Jostarts by saying "", so Blair follows by saying "" . Jo then says "" , and so on. What is the53rd number said?Problem 14Define . Which of the following describes the set of points forwhich ?Problem 15A wire is cut into two pieces, one of length and the other of length . The piece of length is bent to form an equilateral triangle, and the piece of length is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is ?In triangle , medians and intersect at , , , and . What isthe area of ?Problem 17Alex has red tokens and blue tokens. There is a booth where Alex can give two red tokens andreceive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until nomore exchanges are possible. How many silver tokens will Alex have at the end?Problem 18The number has the property that its units digit is the sum of its other digits, thatis . How many integers less than but greater than share this property?Problem 19The real numbers form an arithmetic sequence with . Thequadratic has exactly one root. What is this root?Problem 20The number is expressed in the formwhere and are positive integers and is as small aspossible. What is ?Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is . What is the smallest possible value of N?Problem 22The regular octagon has its center at . Each of the vertices and the center are to beassociated with one of the digits through , with each digit used once, in such a way that the sums ofthe numbers on the lines , , , and are all equal. In how many ways can this bedone?In triangle , , , and . Distinct points , , and lie onsegments , , and , respectively, such that , , and . Thelength of segment can be written as , where and are relatively prime positive integers.What is ?Problem 24A positive integer is nice if there is a positive integer with exactly four positive divisors (including and ) such that the sum of the four divisors is equal to . How many numbers in theset are nice?Bernardo chooses a three-digit positive integer and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer . For example, if ,Bernardo writes the numbers and , and LeRoy obtains the sum . For how many choices of are the two rightmost digits of , in order, the same as those of ?Answer Key1. C2. A3. C4. D5. B6. C7. B8. B9. D10.C11.B12.B13.E14.E15.B16.B17.E18.D19.D20.B21.C22.C23.B24.A25.E。

2013 MCM 数学建模美赛试题

2013 Contest ProblemsMCM PROBLEMSPROBLEM A: The Ultimate Brownie PanWhen baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient with respect to using the space in an oven.Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes - rectangular to circular and other shapes in between.Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.2. Each pan must have an area of A.3. Initially two racks in the oven, evenly spaced.Develop a model that can be used to select the best type of pan (shape) under the following conditions:1. Maximize number of pans that can fit in the oven (N)2. Maximize even distribution of heat (H) for the pan3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different valuesof W/L and p.In addition to your MCM formatted solution, prepare a one to two page advertising sheet for the new Brownie Gourmet Magazine highlighting your design and results.PROBLEM B: Water, Water, EverywhereFresh water is the limiting constraint for development in much of the world. Build a mathematical model for determining an effective, feasible, and cost-efficient water strategy for 2013 to meet the projected water needs of [pick one country from the list below] in 2025, and identify the best water strategy. In particular, your mathematical model must address storage andmovement; de-salinization; and conservation. If possible, use your model to discuss the economic, physical, and environmental implications of your strategy. Provide a non-technical position paper to governmental leadership outlining your approach, its feasibility and costs, and why it is the “best water strategy choice.”Countries: United States, China, Russia, Egypt, or Saudi Arabia。

2013年美赛c题生态系统评价英文版.

For office use onlyT1________________ T2________________ T3________________T4________________Team Control Number15879Problem ChosenAFor office use onlyF1 ________________F2 ________________F3 ________________F4 ________________ Small Leaf, Big WorldAbstract地球的生态状况越来越受到人们的关注,但评价地球这个大生态系统的健康状况一直是个难题。

本文我们主要通过主成分分析分析各国家(节点的健康状况,利用有向复杂网络构建了一个可以联系各节点的网络,并对地球健康状况进行GM(1,1预测来分析复杂的地球生态系统。

地球是一个庞大而复杂的生态系统,就像一个网络。

对此,我们选取国家作为我们分析地球这个复杂网络的节点,因为地球上的国家较多,我们通过考虑地理位置和国家发展状况选取典型的国家,最终确定21个国家作为我们分析的节点。

对于各个节点健康状况,我们根据生态学的相关知识确定了社会、经济、环境和人这四个因素范围内的11个地球健康指标,并在世界银行数据库查得21个国家的11个指标在1991年到2010年这20年的数据,以及世界总的相关数据。

我们分析这个复杂的地球生态系统主要有以下三个方面:(1对每个节点的健康指数求解。

我们采用主成分分析法对21个节点以及世界的11个指标进行综合分析,把11个指标抽象成5个主成分进行健康指数的求解,我们以2000年的数据用MATLAB进行编程求解,结果显示五个主成分的累积贡献率高达92%,结果可靠。

我们同时对各个节点的综合值Z(健康指数进行排名,排名靠前的是美国、澳大利亚、加拿大等发达国家,排名靠后的是Morocco、Egypt Arab Rep. China、Kenya等发展中国家且生态保护较差的国家。

2013年美赛MCM问题A翻译

2013年赛题
MCM问题
问题A：终极布朗尼平底锅
当在一个矩形的锅烘烤时，热度会集中在4个角，而且烘烤的产品会在4个角焙烤过度（以及较小程度上在边缘也有）。

在一个圆形盘的热量被均匀地分布在整个外缘，导致在边缘处的产品焙烤不足。

然而，由于大多数微波炉都是方形的，用圆锅放在里面的话空间利用率就不高。

开发一个模型来展示不同形状平底锅（方形到圆形以及介于之间的），沿平底锅边缘的热量分布。

假设
1．宽度与长度比率的W / L的形状是矩形的烘箱。

2．每个盘必须具有面积A。

3．最初，两个机架在烤箱，间隔均匀。

建立一个模型，可用于选择最佳的类型（形状）的平底锅，在下列情况下：1．适合在烤箱的锅，可以最大限度地提高数（N）
2．最大限度地均匀分布热量（H）
3．优化的组合的条件（1）和（2）式中的权重p和（为1 - p）被分配的结果来说明如何随不同的值的W / L和p。

在除了MCM格式解决方案中，准备一到两页的广告片的新布朗尼美食杂志突出自己的设计和结果。

2013年美赛a题资料

图5
体面相交于 M 、 N , 设 BP = x , M N = y , 则函数 y = f ( x ) 的图像大致是 ( ).
A 1 D1 交于点 P. 当 M 取不同的位置可确定不同的平面 ∀, 从而与定性分析显然只有当 P 移动到中心 O CD 有不同的交点 N , 直线 MN 与这 3 条异面直线都相交, 因此满足条件的直线有无数条. 定量分析建立空间直角坐标系如图 6, 设正方体的棱长为 1, 一直线与 A 1 D 1 、 EF 、 CD 分别交于点 P 、 H、 R, 设 P ( a, 0, 1 ) , H ( b, 1 - b,
( B) . 例4 ( 2008 年辽宁省高考试题 ) 在正方体 A B CD - A 1 B 1 C 1 D 1 中 , E 、 F 分别为棱 A A 1 , CC 1 的中点, 则在空间中与三条直线 A 1 D 1 、 EF 、 CD 都相交的直线 ( ). ( A ) 不存在 ( C ) 有且只有三条定性分析 ( 如图 5 ) 在 EF 上任意 ( B ) 有且只有两条 ( D ) 有无数条
++,
时, MN 有唯一的最大值, 淘汰选项 ( A) 、 ( C) , P 点移动时 , M 1 N 1 与 BP 1 的关系是线性的, BP1 与 BP 的关系也是线性的, 所以 MN = M 1 N 1 与 BP 的关系是线性的, 淘汰 ( D) , 选 ( B) . 定量分析设正方体的棱长为 1, M、 N 在 A B 和 BC 上的投影分别为 M 1, N 1, M1 N1 交 BD 于 P 1 , 2 2 2
张忠旺 6 时, S 取得最大值 5. 2
= 5. 且仅当 d 1 = d 2 =

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主题行
COMAP是你的控制
示例：COMAP 11111
点击这里下载PDF格式的完整的竞赛说明。

点击这里下载Microsoft Word中的格式汇总表的副本。

*请务必变更控制之前选择打印出来的页面的数量和问题。

团队可以自由选择之间MCM问题MCM问题A，B或ICM问题C.
COMAP镜像站点：更多：
/undergraduate/contests/mcm/
MCM：数学建模竞赛
ICM：交叉学科建模竞赛
2013年赛题
MCM问题
问题A：终极布朗尼潘
当在一个矩形的锅热烘烤时的4个角落中浓缩，并在拐角处（以及在较小程度上在边缘处）：产品会过头。

在一个圆形盘的热量被均匀地分布在整个外缘和在边缘处的产品不过头。

然而，因为大多数烤炉使用圆形平底锅的形状是矩形的是效率不高的相对于使用在烘箱中的空间。

开发一个模型来显示横跨平底锅平底锅不同形状 - 矩形之间的圆形和其他形状的外边缘的热量分布。

假设
1。

的宽度与长度之比的W / L的形状是矩形的烘箱。

2。

- 每个盘必须具有的区域A的
3。

最初，两个机架在烤箱，间隔均匀。

建立一个模型，可用于选择最佳的泛类型（形状）在下列情况下：
1。

适合在烤箱的锅，可以最大限度地提高数（N）
2。

最大限度地均匀分布热量（H），泛
3。

优化的组合的条件（1）和（2）式中的权重p和（为1 - p）被分配的结果来说明如何随不同的值的W / L和p。

在除了MCM格式解决方案中，准备一到两页的广告片的新布朗尼美食杂志突出自己的设计和结果。

问题B：水，水，无处不在
新鲜的白开水是在世界大部分地区的发展限制约束。

建立一个数学模型，为确定有效的，可行的和具有成本效益的水资源战略于2013年，以满足预计的用水需求，从下面的列表]中选择一个国家，到2025年，确定最佳的水战略。

特别是，您的数学模型必须解决存储和运动，去盐碱化和保护。

如果可能的话，用你的模型，探讨经济，物理和环境的影响，你的战略。

提供一个非技术性的立场文件，政府领导介绍你的方法，其可行性和成本，以及为什么它是“最好的的水战略选择。

”
国家有：美国，中国，俄罗斯，埃及，沙特阿拉伯
ICM问题
问题C：地球的健康的网络建模
点击下面的标题以PDF格式下载2013年的ICM问题。

您的ICM提交应包括一个第1页汇总表和您的解决方案不能超过20页，最多21页。

2013年ICM问题
地球的健康的网络建模
背景：社会是有兴趣在开发和使用模型来预测我们这个星球的生物和环境的健康状况。

许多科学研究得出的结论是，有越来越大的压力对地球环境和生物系统，但也有极少数的全球模型，以测试这些要求。

联合国支持千年生态系统评估综合报告发现，近三分之二的地球的生命支持生态系统，包括干净的水，纯净的空气，稳定。

气候是降低不可持续的利用。

人类被指责为许多这种损伤。

飙升的食品，淡水，燃料和木材的需求作出了贡献剧烈的环境变化，滥伐森林，空气，土地和水的污染。

尽管当地的栖息地和区域进行了大量的研究因素的影响，目前的模式不充分告知决策者如何省政策可能会影响到整个地球健康的。

许多模型忽略了复杂的全球因素，是无法确定的远射潜在的政策的影响。

而科学家们认识到在无数的复杂关系和交叉效应环境和生物系统产生影响地球的生物圈，目前的模型往往忽略这些关系，或限制系统的连接。

该系统的复杂性体现在多个交互，反馈回路，紧急行为，和即将到来的状态的变化或临界点。

最
近22国际自然写的文章知名科学家题为“走近地球的生物圈”的状态转变列出了很多相关的问题需要科学模型的重要性。

预测潜在的行星卫生系统状态的变化。

本文提供
两个具体的量化建模的挑战，他们要求更好的预测模型：
1）为了提高生物通过全球性的拥抱复杂的模型，预测地球上的相互关联的系统，包括当地条件的影响全球性的系统，反之亦然。

2）确定因素，可能会产生不健康的全局状态变化，并显示如何使用有效的生态系统管理，以防止或限制这些即将发生的状态的变化。

研究的问题是，我们是否可以建立的全球车型使用本地或地球的健康状况，预测潜在状态的变化，并帮助区域组成部分决策者设计有效的政策，根据他们的潜在影响地球上的健康。

虽然许多警告标志出现，没有人知道地球是真正接近全球的临界点，如果这种极端的状态是不可避免的。

自然的文章和其他许多人指出，有几个重要的元素在地球的生态系统（例如，本地因素的影响，全球性的影响，多维的工作因素和关系，不同时间和空间尺度）。

也有许多其他因素，可以被包括在一个预测模型 - 人类人口，资源和压力栖息地，栖息地改造，能源消耗，气候变化，土地利用模式，污染，生物多样性，海洋化学，大气化学，政治;模式，如社会动荡和经济的不稳定。

古生物学家研究与模拟生态系统的行为和反应在以前的灾难性的状态转移基于历史的定性和定量信息，从而可以提供背景为未来的预测模型。

然而，应该指出的是，人类的影响有在我们目前的生物圈状况显着增加。

要求：
你是会员的国际联盟建模（ICM）将很快
举办了一个题为“网络和地球的健康”和您的研究
领导者要求你提前车间进行建模和分析。

他要求你的团队做到以下几点：
要求1：建立一个动态的全球网络模型的某些方面的地球确定此条件下的局部元素的健康（你发展的措施）（网络节点）和适当连接（网络链接），以追踪关系属性效果。

由于这些效果的动态性质是重要的，这网络模型必须包含一个动态的元素，允许该模型预测该卫生措施的未来状态。

例如，节点可以是国家，大陆，海洋，栖息地，或任何组合的这些或其他元素共同构成了全球性的模型。

您的链接可以代表节点或环境的影响，或流量或繁殖的物理元件（如污染），随着时间的推移。

您的健康措施可能是地球上的任何元素的条件，包括人口，生物，环境，社会，政治，物理和/或化学条件。

是确保你的模型中定义的所有元素，并解释你的科学依据网络的措施，节点实体和链路属性，造型决定。

确定一种方法来设置任何参数，并解释如何可以测试您的模型如果有足够的数据。

什么样的数据可以用来验证或验证你的模型的有效性？（注：如果您没有必要的数据，
确定参数进行验证，不扔出去的模型。

您的主管意识到，在这个阶段，良好的创意思想和理论的重要验证基于数据模型）。

请确保您包括人的因素在您的建模和解释人类行为和政府政策可能影响你的模型的结果。

要求2：运行你的模型看看它是如何预测未来的地球健康。

您可以需要估计的参数，你通常会确定的数据。

（请记住，这仅仅是测试和了解你的模型的元素，而不是使用预测或决策）。

各种因素模型产生？预测状态的变化或临界点，在地球的条件吗？能否提供改变当地条件的全球性后果的警告吗？难道通知重要政策的决策者？你考虑到人的因素在你的措施和网络属性？
要求3：使用网络建模的强大的元素之一是能够分析的网络结构。

网络性能可以帮助确定关键节点或在模型中的关系？如果是的话，执行这样的分析。

你的模型是如何敏感缺少的环节或不断变化的关系？你的模型使用反馈回路，需要考虑的不确定性？数据收集问题是什么？请问您的模型反应，政府的各项政策，并可能从而有助于规划？
要求4：写了20页的报告（汇总表不计算在20个
页），说明你的模型，它的潜力。

一定要详细的长处和模型的弱点。

你的上司会使用您的报告中的一大主题即将到来的车间，如果是适当的和有见地的行星健康建模，会问你在即将举行的研讨会。

好运气在你的网络建模工作！
潜在的有用的参考文献包括：
安东尼·D. Barnosky，霍尔迪Bascompte，伊丽莎白 A. Hadly，埃里克·L.伯洛，詹姆斯·H.布朗，
，韦恩MIKAEL Fortelius M.盖茨，约翰·哈特，艾伦·黑斯廷斯，巴勃罗·马尔凯，新D.
马丁内斯，阿恩Mooers，彼得Roopnarine，Geerat Vermeij，迷迭香，约翰·威廉姆斯
吉莱斯皮，贾斯汀Kitzes，查尔斯·马歇尔，尼古拉斯Matzke，大卫·P.明德尔，埃洛伊雷维利亚，
亚当·B·史密斯。

的“走近状态转变在地球的生物圈”。

自然，2012，486（7401）：52
DOI：10.1038/nature11018
Donella草甸，约根兰德，和丹尼斯·梅多斯。

“增长的极限”：30年
更新，2004年。

罗伯特·沃森（Robert Watson）和A.Hamid Zakri。

联合国千年生态系统评估综合报告，
联合国的报告，2005年。

美国加州大学 - 伯克利分校。

“即将引爆点的地球的证据。

”
科学日报，2012年6月6日。

网络。

2012年10月22日。