2004美国数学竞赛题更正

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。

amc12数学题篇一：AMC12数学题是指美国数学竞赛（AMC）12级的数学题目。

AMC12是一项面向高中生的数学竞赛，旨在鼓励学生对数学的深入学习和探索。

这些数学题目涵盖了广泛的数学概念和技巧，并要求学生通过逻辑推理和解题技巧来解决问题。

AMC12数学题的难度较高，常常需要学生具备扎实的数学基础和解题能力。

这些题目涉及的数学领域包括代数、几何、数论和概率等。

许多题目都要求学生运用多个数学概念进行综合分析和解答。

因此，参加AMC12数学竞赛是考验学生数学能力和解题能力的良好机会。

解答AMC12数学题需要学生具备良好的数学思维和解题技巧。

学生需要能够正确理解问题并找出解决问题的方法。

在解答过程中，学生需要通过分析和推理来得出结论，同时要注意细节和逻辑的严谨性。

有时候，解答AMC12数学题还需要一些创造性思维和巧妙的数学变换。

参加AMC12数学竞赛对学生有很多好处。

首先，它可以提高学生的数学能力和解题能力，培养学生的数学思维和分析能力。

其次，参加竞赛可以增强学生对数学的兴趣和热爱，激发学生对数学的探索欲望。

此外，竞赛还可以培养学生的竞争意识和团队合作精神。

最重要的是，参加AMC12数学竞赛可以为学生在大学申请中增加一份优势，显示出学生在数学方面的才能和成就。

总之，AMC12数学题是一系列挑战性的数学问题，要求学生具备扎实的数学基础和解题能力。

通过参加AMC12数学竞赛，学生能够提高数学能力，培养数学思维，并为将来的学术和职业发展奠定坚实的基础。

篇二：AMC12（美国数学竞赛12级）是由美国数学协会举办的一项数学竞赛，面向高中学生。

这项竞赛旨在鼓励学生培养解决复杂数学问题的能力，并提供一个展示自己数学才能的平台。

AMC12数学题目通常涵盖广泛的数学领域，包括代数、几何、概率与统计、数论等。

这些题目要求学生在有限的时间内思考、分析和解决问题，考察他们的数学思维能力、推理能力和解决问题的能力。

举一个例子，假设有一道AMC12的几何问题：如图所示，一个直径为12的圆与一个半径为8的圆相切。

2.3.What is the value of ?4.Eight 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 $2.50 to cover her portion of the total bill. What was the total bill?5.Hammie 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?6.The 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?7.Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 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?8.A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?9.The 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?10.What 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 walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less 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 pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the 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. When Clara 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 answerchoice 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 of Lord.of.AMC19. Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score 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 sidesand , 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 theball travels over the course from A to B?1.2.The 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.3.Notice that we can pair up every two numbers to make a sum of 1:Therefore, the answer is .4.Each 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 .5.The 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.6.Solution 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 . is the answer we're looking for.We see that , making .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.)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:7.If 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 .8.First, 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 .9.This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that .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 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 answerchoice 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: AlgebraWe multiply 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 7th 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 arecubes. 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 of the 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 inthe 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 toget from City Park 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, i.e. 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 Pythagoreantheorem 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 semicircles on the left is equal to the biggest one, so the area of thelargest 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 theextension of . 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 thecombined area of the three squares is .Let 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, .Let 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 . Thissolution 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.古希腊哲学大师亚里士多德说：人有两种，一种即“吃饭是为了活着”,一种是“活着是为了吃饭”.一个人之所以伟大，首先是因为他有超于常人的心。

美国数学建模竞赛题目1985年：A题：动物群体的管理B题：战略物资储备的管理问题1986年：A题：海底地型测量问题B题：应急设施的优化选址问题1987年：A题：堆盐问题（盐堆稳定性问题）B题：停车场安排问题1988年：A题：确定毒品走私船位置B题：平板列车车厢的优化装载1989年：A题：蠓虫识别问题；最佳分类与隔离B题：飞机排队模型1990年：A题：脑中多巴胺的分布B题：铲雪车的路径与效率问题1991年：A题：估计水塔的水流量B题：通信网络费用问题1992年：A题：雷达系统的功率与设计式样B题：紧急修复系统的研制1993年：A题：堆肥问题B题：煤炭装卸场的最优操作1994年：A题：保温房屋设计问题B题：计算机网络的最小接通时间1996年：A题：大型水下物体的探测B题：快速遴选优胜者问题1997年：A题：恐龙捕食问题B题：会议混合安排问题1998年：A题：MRI图象处理问题B题：分数贬值问题1999年：A题：小星体撞击地球问题B题：公用设施的合法容量问题C题：确定环境污染的物质、位置、数量和时间的问题2000年：A题：空间交通管制B题：无线电信道分配C题：大象群落的兴衰2001年：A题：选择自行车车轮B题：逃避飓风怒吼C题：我们的水系-不确定的前景2002年：A题：风和喷水池B题：航空公司超员订票C题：如果我们过分扫荡自己的土地，将会失去各种各样的蜥蜴。

2003年：A题：特技演员B题：Gamma刀治疗方案C题：航空行李的扫描对策2004年：A题：指纹是独一无二的吗？B题：更快的快通系统C题：安全与否？2005年：A题：flood planningB题：tollboothsC题: Nonrenewable Resources2006年：A题：Positioning and Moving SprinklerSystems for IrrigationB题：Wheel Chair Access at AirportsC题：Trade-offs in the fight againstHIV/AIDS2007年：A题：GerrymanderingB题：The Airplane Seating ProblemC题：Organ Transplant: The Kidney Exchange Problem2008年：A题：Take a BathB题：Creating Sudoku PuzzlesC题：Finding the Good in Health Care Systems2009年：A题：Designing a Traffic CircleB题：Energy and the Cell PhoneC题：Creating Food Systems: Re-Balancing Human-Influenced Ecosystems。

2004 AMC 10BProblem 1Each row of the Misty Moon Amphitheater has 33 seats. Rows 12 through 22 are reserved for a youth club. How many seats are reserved for this club?SolutionThere are rows of seats, giving seats.Problem 2How many two-digit positive integers have at least one 7 as a digit?SolutionTen numbers () have as the tens digit. Nine numbers () have it as the ones digit. Number is in both sets. Thus the result is .Problem 3At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made 48 free throws. How many free throws did she make at the first practice?SolutionAt the fourth practice she made throws, at the third one it was , then we get throws for the second practice, and finally throws at the first one.Problem 4A standard six-sided die is rolled, and P is the product of the five numbers that are visible. What is the largest number that is certain to divide P?Solution 1The product of all six numbers is . The products of numbers that can be visible are , , ..., . The answer to this problem is their greatest common divisor -- which is , where is the least common multiple of . Clearly and the answer is .Solution 2Clearly, can not have a prime factor other than , and .We can not guarantee that the product will be divisible by , as the number can end on the bottom.We can guarantee that the product will be divisible by (one of and will always be visible), but not by .Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by . This is the most we can guarantee, as when the is on the bottom side, the two visible even numbers are and , and their product is not divisible by .Hence .SolutionProblem 5In the expression , the values of , , , and are , , , and , although not necessarily in that order. What is the maximum possible value of the result?SolutionIf or , the expression evaluates to .If , the expression evaluates to .Case remains.In that case, we want to maximize where .Trying out the six possibilities we get that the best one is, where .Problem 6Which of the following numbers is a perfect square?SolutionUsing the fact that , we can write:▪▪▪▪▪Clearly is a square, and as , , and are primes, none of the other four are squares.Problem 7On a trip from the United States to Canada, Isabella took U.S.dollars. At the border she exchanged them all, receiving Canadian dollars for every U.S. dollars. After spending Canadian dollars, she had Canadian dollars left. What is the sum of the digits of ?SolutionSolution 1Isabella had Canadian dollars. Setting up an equation we get , which solves to , and the sum of digits of isSolution 2Each time Isabelle exchanges U.S. dollars, she gets Canadian dollars and Canadian dollars extra. Isabelle received a total of Canadian dollars extra, therefore she exchanged U.S. dollars times. Thus .Problem 8Minneapolis-St. Paul International Airport is 8 miles southwest of downtown St. Paul and 10 miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis?SolutionThe directions "southwest" and "southeast" are orthogonal. Thus the described situation is a right triangle with legs 8 miles and 10 miles long. The hypotenuse length is , and thus the answeris .Without a calculator one can note that . Problem 9A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle?SolutionThe area of the circle is , the area of the square is .Exactly of the circle lies inside the square. Thus the total area is.Problem 10A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains cans, how many rows does it contain?SolutionThe sum of the first odd numbers is . As in our case , we have .Problem 11Two eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?SolutionSolution 1We have , hence if at least one of the numbers is , the sum is larger. There such possibilities.We have .For we already have , hence all other cases are good.Out of the possible cases, we found that in the sum is greater than or equal to the product, hence in it is smaller. Therefore the answer is .Solution 2Let the two rolls be , and .From the restriction:Since and are non-negative integers between and , either , , orif and only if or .There are ordered pairs with , ordered pairs with , and ordered pair with and . So, there areordered pairs such that .if and only if and or equivalently and . This gives ordered pair .So, there are a total of ordered pairs with .Since there are a total of ordered pairs , there are ordered pairs with .Thus, the desired probability is .Problem 12An annulus is the region between two concentric circles. The concentric circles in the figure have radii and , with . Let be a radius of the larger circle, let be tangent to the smaller circle at , and let be the radius of the larger circle that contains . Let , , and . What is the area of the annulus?SolutionThe area of the large circle is , the area of the small one is , hence the shaded area is .From the Pythagorean Theorem for the right triangle we have , hence and thus the shaded area is . Problem 13In the United States, coins have the following thicknesses: penny, mm; nickel, mm; dime, mm; quarter, mm. If a stack of these coins is exactly mm high, how many coins are in the stack?SolutionAll numbers in this solution will be in hundreds of a millimeter.The thinnest coin is the dime, with thickness . A stack of dimes has height .The other three coin types have thicknesses , , and . By replacing some of the dimes in our stack by other, thicker coins, we can clearly create exactly all heights in the set.If we take an odd , then all the possible heights will be odd, and thus none of them will be . Hence is even.If the stack will be too low and if it will be too high. Thus we are left with cases and .If the possible stack heights are , with the remaining ones exceeding .Therefore there are coins in the stack.Using the above observation we can easily construct such a stack. A stack of dimes would have height , thus we need to add . This can be done for example by replacing five dimes by nickels (for ), and one dime by a penny (for ).Problem 14A bag initially contains red marbles and blue marbles only, with moreblue than red. Red marbles are added to the bag until only of the marbles in the bag are blue. Then yellow marbles are added to the baguntil only of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue?SolutionWe can ignore most of the problem statement. The only important information is that immediately before the last step blue marbles formed of the marbles in the bag. This means that there were blue and other marbles, for some . When we double the number ofblue marbles, there will be blue and other marbles, hence blue marbles now form of all marbles in the bag.Problem 15Patty has coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have cents more. How much are her coins worth?SolutionSolution 1She has nickels and dimes. Their total cost iscents. If the dimes were nickels and vice versa, she would havecents. This value should be cents more than the previous one. We get , which solves to . Her coins are worth .Solution 2Changing a nickel into a dime increases the sum by cents, and changing a dime into a nickel decreases it by the same amount. As the sum increased by cents, there are more nickels than dimes. As the total count is , this means that there are nickels and dimes.Problem 16Three circles of radius are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?SolutionThe situation in shown in the picture below. The radius we seek is . Clearly . The point is clearly the center of the equilateral triangle , thus is of the altitude of this triangle. We get that . Therefore the radius we seek is.WARNING. Note that the answer does not correspond to any of the five options. Most probably there is a typo in option D.Problem 17The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?SolutionSolution 1If Jack's current age is , then Bill's current age is .In five years, Jack's age will be and Bill's age will be .We are given that . Thus .For we get . For and the value is not an integer, and for it is more than . Thus the only solution is , and the difference in ages is .Solution 2Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.The age difference is , hence it is a multiple of . Thus Bill's current age modulo must be .Thus Bill's age is in the set .As Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options .Checking each of them, we see that only works, and gives the solution .Problem 18In the right triangle , we have , , and . Points , , and are located on , , and , respectively, so that , , and . What is the ratio of the area of to that of ?SolutionFirst of all, note that , and therefore.Draw the height from onto as in the picture below:Now consider the area of . Clearly the triangles and are similar, as they have all angles equal. Their ratio is ,hence . Now the area of can be computed as= . Similarly we can find that as well.Hence , and the answer is .Problem 19In the sequence , , , , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is. What is the term in this sequence?SolutionSolution 1We already know that , , , and . Let's compute the next few terms to get the idea how the sequence behaves. We get ,, , and so on.We can now discover the following pattern: and . This is easily proved by induction. It follows that.Solution 2Note that the recurrence can be rewritten as.Hence we get that and also From the values given in the problem statement we see that .From we get that .From we get that .Following this pattern, we get.Problem 20In points and lie on and , respectively. If and intersect at so that and , what isSolutionSolution (Triangle Areas)We use the square bracket notation to denote area.Without loss of generality, we can assume . Then , and . We have , so we need to find the area of quadrilateral .Draw the line segment to form the two triangles and . Let , and . By considering trianglesand , we obtain , and by considering triangles and , we obtain . Solving, we get , , so the area of quadrilateral is .ThereforeSolution (Mass points)The presence of only ratios in the problem essentially cries out for mass points.As per the problem, we assign a mass of to point , and a mass of to . Then, to balance and on , has a mass of .Now, were we to assign a mass of to and a mass of to , we'd have . Scaling this down by (to get , which puts and in terms of the masses of and ), we assign a mass of to and a mass of to .Now, to balance and on , we must give a mass of . Finally, the ratio of to is given by the ratio of the mass of tothe mass of , which is .Solution (Coordinates)Affine transformations preserve ratios of distances, and for any pair of triangles there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any , and we just need to compute it for any single triangle.We can choose the points , , and . This way we will have , and . The situation is shown in the picture below:The point is the intersection of the lines and . The points on the first line have the form , the points on the second line have the form . Solving for we get , hence.The ratio can now be computed simply by observing the coordinates of , , and :Problem 21Let ; ; and ; ; be two arithmetic progressions. The set is the union of the first terms of each sequence. How many distinct numbers are in ?SolutionThe two sets of terms are and.Now . We can compute. We will now find .Consider the numbers in . We want to find out how many of them lie in . In other words, we need to find out the number of valid values of for which .The fact "" can be rewritten as ", and ".The first condition gives , the second one gives .Thus the good values of are , and their count is .Therefore , and thus .Problem 22A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?SolutionThis is obviously a right triangle. Pick a coordinate system so that the right angle is at and the other two vertices are at and .As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at .The radius of the inscribed circle can be computed using the well-known identity , where is the area of the triangle and its perimeter. In our case, and , thus . As the inscribed circle touches both legs, its center must be at .The distance of these two points is then.Problem 23Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?SolutionLabel the six sides of the cube by numbers to as on a classic dice. Then the "four vertical faces" can be: , , or .Let be the set of colorings where are all of the same color, similarly let and be the sets of good colorings for the other two sets of faces.There are possible colorings, and there are goodcolorings. Thus the result is . We need to compute .Using the Principle of Inclusion-Exclusion we can writeClearly , as we have two possibilities for the common color of the four vertical faces, and two possibilities for each of the horizontal faces.What is ? The faces must have the same color, and at the same time faces must have the same color. It turns out thatthe set containing just the two cubes where all six faces have the same color.Therefore , and the result is .Problem 24In we have , , and . Point is on the circumscribed circle of the triangle so that bisects . What is the value of ?SolutionProblem 25A circle of radius is internally tangent to two circles of radius at points and , where is a diameter of the smaller circle. What is the area of the region, shaded in the picture, that is outside the smaller circle and inside each of the two larger circles?SolutionThe area of the small circle is . We can add it to the shaded region, compute the area of the new region, and then subtract the area of the small circle from the result.Let and be the intersections of the two large circles. Connect them to and to get the picture below:Now obviously the triangles and are equilateral with side .Take a look at the bottom circle. The angle is , hence the sector is of the circle. The same is true for the sector of the bottom circle, and sectors and of the top circle.If we now sum the areas of these four sectors, we will almost get the area of the new shaded region - except that each of the two equilateral triangles will be counted twice.Hence the area of the new shaded region is, and the area of the original shared region is .- 21 -。

美国大联盟数学竞赛答题填写要求美国大联盟数学竞赛分成不同的年级组，答题卡的要求略有不同。

3-9年纪年级学生使用答题说明如下：必须使用2B铅笔在答题卡上填涂。

答题卡填涂要求：1.机读卡基本信息填写用笔不限，准考证号码和作答区域必须使用2B铅笔进行填涂，如需修改，一定用橡皮擦拭干净，不要留下原填涂痕迹，以免造成误判。

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3.涂卡说明：（1）姓名：填写考生姓名，不能写昵称；学校：填写考生的公立学校；年级：填写考生参赛的年级。

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数学竞赛典型题目（一）1.(2004美国数学竞赛)设n a a a ,,,21 是整数列,并且他们的最大公因子是1.令S 是一个整数集,具有性质:（1）),,2,1(n i S a i =∈（2） }),,2,1{,(n j i S a a j i ∈∈-,其中j i ,可以相同（3）对于S y x ∈,，若S y x ∈+，则S y x ∈-证明：S 为全体整数的集合。

2．(2004美国数学竞赛)c b a ,,是正实数，证明：3252525)()3)(3)(3(c b a c c b b a a ++≥+-+-+-3．(2004加拿大数学竞赛)T 为1002004的所有正约数的集合，求集合T 的子集S 中的最大可能的元素个数。

其中S 中没有两个元素，一个是另一个的倍数。

4．(2004英国数学竞赛)证明：存在一个整数n 满足下列条件：（1）n 的二进制表达式中恰好有2004个1和2004个0；（2）2004能整除n .5．(2004英国数学竞赛)在0和1之间，用十进制表示为 21.0a a 的实数x 满足：在表达式中至多有2004个不同的区块形式，)20041(20031≤≤++k a a a k k k ，证明：x 是有理数。

6．(2004亚太地区数学竞赛)求所有由正整数组成的有限非空数集S ，满足：如果S n m ∈,，则S n m n m ∈+),( 7．(2004亚太地区数学竞赛)平面上有2004个点，并且无三点共线，S 为通过任何两点的直线的集合。

证明：点可以被染成两种颜色使得两点同色当且仅当S 中有奇数条直线分离这两点。

8．(2004亚太地区数学竞赛)证明：)()!1(*2N n n n n ∈⎥⎦⎤⎢⎣⎡+-是 偶数。

9．(2004亚太地区数学竞赛)z y x ,,是正实数，证明：)(9)2)(2)(2(222zx yz xy z y x ++≥+++10．（2003越南数学竞赛）函数f 满足)0(2sin 2cos )(cot π<<+=x x x x f ，令 )11)(1()()(≤≤--=x x f x f x g ，求)(x g 在区间]1,1[-的上最值。

2000到20XX年AMC10美国数学竞赛0 0P 0 A 0 B 0 C 0D 0 全美中学数学分级能力测验(AMC 10)2000年 第01届 美国AMC10 (2000年2月 日 时间75分钟)1. 国际数学奥林匹亚将于 在美国举办，假设I 、M 、O 分别表示不同的正整数，且满足I ⨯M ⨯O =2001，则试问I +M +O 之最大值为 。

(A) 23 (B) 55 (C) 99 (D) 111 (E) 6712. 2000(20002000)为 。

(A) 20002001 (B) 40002000 (C) 20004000 (D) 40000002000 (E) 200040000003. Jenny 每天早上都会吃掉她所剩下的聪明豆的20%，今知在第二天结束时，有32颗剩下，试问一开始聪明豆有 颗。

(A) 40 (B) 50 (C) 55 (D) 60 (E) 754. Candra 每月要付给网络公司固定的月租费及上网的拨接费，已知她12月的账单为12.48元，而她1月的账单为17.54元，若她1月的上网时间是12月的两倍，试问月租费是 元。

(A) 2.53 (B) 5.06 (C) 6.24 (D) 7.42 (E) 8.775. 如图M ，N 分别为PA 与PB 之中点，试问当P 在一条平行AB 的直 在线移动时，下列各数值有 项会变动。

(a) MN 长 (b) △P AB 之周长 (c) △P AB 之面积 (d) ABNM 之面积(A) 0项 (B) 1项 (C) 2项 (D) 3项 (E) 4项 6. 费氏数列是以两个1开始，接下来各项均为前两项之和，试问在费氏数列各项的个位数字中， 最后出现的阿拉伯数字为 。

(A) 0 (B) 4 (C) 6 (D) 7 (E) 97. 如图，矩形ABCD 中，AD =1，P 在AB 上，且DP 与DB 三等分∠ADC ，试问△BDP 之周长为 。

AMC8（美国数学竞赛）历年真题、答案及中英文解析艾蕾特教育的AMC8 美国数学竞赛考试历年真题、答案及中英文解析：AMC8-2020年：真题 --- 答案---解析（英文解析+中文解析）AMC8 - 2019年：真题----答案----解析（英文解析+中文解析）AMC8 - 2018年：真题----答案----解析（英文解析+中文解析）AMC8 - 2017年：真题----答案----解析（英文解析+中文解析）AMC8 - 2016年：真题----答案----解析（英文解析+中文解析）AMC8 - 2015年：真题----答案----解析（英文解析+中文解析）AMC8 - 2014年：真题----答案----解析（英文解析+中文解析）AMC8 - 2013年：真题----答案----解析（英文解析+中文解析）AMC8 - 2012年：真题----答案----解析（英文解析+中文解析）析）AMC8 - 2010年：真题----答案----解析（英文解析+中文解析）AMC8 - 2009年：真题----答案----解析（英文解析+中文解析）AMC8 - 2008年：真题----答案----解析（英文解析+中文解析）AMC8 - 2007年：真题----答案----解析（英文解析+中文解析）AMC8 - 2006年：真题----答案----解析（英文解析+中文解析）AMC8 - 2005年：真题----答案----解析（英文解析+中文解析）AMC8 - 2004年：真题----答案----解析（英文解析+中文解析）AMC8 - 2003年：真题----答案----解析（英文解析+中文解析）AMC8 - 2002年：真题----答案----解析（英文解析+中文解析）AMC8 - 2001年：真题----答案----解析（英文解析+中文解析）AMC8 - 2000年：真题----答案----解析（英文解析+中文解析）析）AMC8 - 1998年：真题----答案----解析（英文解析+中文解析）AMC8 - 1997年：真题----答案----解析（英文解析+中文解析）AMC8 - 1996年：真题----答案----解析（英文解析+中文解析）AMC8 - 1995年：真题----答案----解析（英文解析+中文解析）AMC8 - 1994年：真题----答案----解析（英文解析+中文解析）AMC8 - 1993年：真题----答案----解析（英文解析+中文解析）AMC8 - 1992年：真题----答案----解析（英文解析+中文解析）AMC8 - 1991年：真题----答案----解析（英文解析+中文解析）AMC8 - 1990年：真题----答案----解析（英文解析+中文解析）AMC8 - 1989年：真题----答案----解析（英文解析+中文解析）AMC8 - 1988年：真题----答案----解析（英文解析+中文解析）析）AMC8 - 1986年：真题----答案----解析（英文解析+中文解析）AMC8 - 1985年：真题----答案----解析（英文解析+中文解析）◆AMC介绍◆AMC（American Mathematics Competitions) 由美国数学协会（MAA）组织的数学竞赛，分为 AMC8 、 AMC10、 AMC12 。

目录2000 年美国大学生数学建模竞赛MCM、ICM 试题 (3)2000 MCM A: Air Traffic Control (3)2000 MCM B: Radio Channel Assignments (3)2000 ICM: Elephants: When is Enough, Enough? (5)2001 年美国大学生数学建模竞赛MCM、ICM 试题 (7)2001 MCM A: Choosing a Bicycle Wheel (7)2001 MCM B: Escaping a Hurricane's Wrath (An Ill Wind...). (8)2001 ICM: Our Waterways - An Uncertain Future (10)2002 年美国大学生数学建模竞赛MCM、ICM 试题 (14)2002 MCM A: Wind and Waterspray (14)2002 MCM B: Airline Overbooking (14)2002 ICM: Scrub Lizards (15)2003 年美国大学生数学建模竞赛MCM、ICM 试题 (19)2003 MCM A: The Stunt Person (19)2003 MCM B: Gamma Knife Treatment Planning (19)2003 ICM: Aviation Baggage Screening Strategies: To Screen or Not to Screen, that is the Question (20)2004 年美国大学生数学建模竞赛MCM、ICM 试题 (24)2004 MCM A: Are Fingerprints Unique? (24)2004 MCM B: A Faster QuickPass System (24)2004 ICM: To Be Secure or Not to Be? (24)2005 年美国大学生数学建模竞赛MCM、ICM 试题 (25)2005 MCM A: Flood Planning (25)2005 MCM B: Tollbooths (25)2005 ICM: Nonrenewable Resources (25)2006 年美国大学生数学建模竞赛MCM、ICM 试题 (27)2006 MCM A: Positioning and Moving Sprinkler Systems for Irrigation (27)2006 MCM B: Wheel Chair Access at Airports (27)2006 ICM: Trade-offs in the fight against HIV/AIDS (28)2007 年美国大学生数学建模竞赛MCM、ICM 试题 (32)2007 MCM A: Gerrymandering (32)2007 MCM B: The Airplane Seating Problem (32)2007 ICM: Organ Transplant: The Kidney Exchange Problem (33)2008 年美国大学生数学建模竞赛MCM、ICM 试题 (38)2008 MCM A: Take a Bath (38)2008 MCM B: Creating Sudoku Puzzles (38)2008 ICM: Finding the Good in Health Care Systems (38)2009 年美国大学生数学建模竞赛MCM、ICM 试题 (40)2009 MCM A: Designing a Traffic Circle (40)2009 MCM B: Energy and the Cell Phone (40)2009 ICM: Creating Food Systems: Re-Balancing Human-Influenced Ecosystems41 2010年美国大学生数学建模竞赛 MCM、ICM 试题 (42)2010 MCM A: The Sweet Spot (42)2010 MCM B: Criminology (43)2010 ICM: The Great Pacific Ocean Garbage Patch (44)2011年美国大学生数学建模竞赛 MCM、ICM 试题 (45)2011 MCM A: Snowboard Course (45)2011 MCM B: Repeater Coordination (45)2011 ICM: Environmentally and Economically Sound (46)2012年美国大学生数学建模竞赛 MCM、ICM 试题 (48)2012 MCM A: The Leaves of a Tree (48)2012 MCM B: Camping along the Big Long River (50)2012 ICM: Modeling for Crime Busting (51)2013年美国大学生数学建模竞赛 MCM、ICM 试题 (59)2013 MCM A: The Ultimate Brownie Pan (59)2013 MCM B: Water, Water, Everywhere (61)2013 ICM: NetworkModeling of Earth's Health (62)2000 年美国大学生数学建模竞赛MCM、ICM 试题2000 MCM A: Air Traffic ControlTo improve safety and reduce air traffic controller workload, the Federal Aviation Agency (FAA) is considering adding software to the air traffic control system that would automatically detect potential aircraft flight path conflicts and alert the controller. To that end, an analysit at the FAA has posed the following problems.Requirement A: Given two airplanes flying in space, when should the air traffic controller consider the objects to be too close and to require intervention? Requirement B: And airspace sector is the section of three-dimensional airspace that one air traffic controller controls. Given any airspace sector, how do we measure how complex it is from an air traffic workload perspective? To what extent is complexity determined by the number of aircraft simultaneously passing through that sector1.at any one instant?2.during any given interval of time?3.during a particular time of day?How does the number of potential conflicts arising during those periods affect complexity? Does the presence of additional software tools to automatically predict conflicts and alert the controller reduce or add to this complexity? In addition to the guidelines for your report, write a summary (no more than two pages) that the FAA analyst can present to Jane Garvey, the FAA Administrator, to defend your conclusions.2000 MCM B: Radio Channel AssignmentsWe seek to model the assignment of radio channels to a symmetric network of transmitter locations over a large planar area, so as to avoid interference. One basic approach is to partition the region into regular hexagons in a grix (honeycomb-style), as shown in Figure 1, where a transmitter is located at the center of each hexagon.An interval of the frequency spectrum is to be alloted for transmitter frequencies. The interval will be divided into regularly spaced channels, which we represent by integers 1,2,3, … . Each transmitter wil be assigned one positive integer channel. The same channel can be used at many locations, provided that interference from nearby transmitters is avoided.Our goal is to minimize the width of the interval in the frequency spectrum that is needed to assugn channels subject to some constraints. This is achieved with the concept of a span. The span is the minimum, over all assignments satisfying the constraints, of the largest channel used at any location. It is not required that every channel smaller than the span be used in an assignment that attains the span.Let s be the length of a side of one of the hexagons. We concentrate on the case that there are two levels of interference.Requirement A: There are several contrainsts on the frequency assignments. First, no two transmitters within distance 4s of each other can be given the same channel. Second, due to spectral spreading, transmitters within distance 2s of each other must not be given the same or adjacent channels: Their channels must differ by at least 2. Under these contraints, what can we say about the span in Figure 1?Requirement B: Repeat Requirement A, assuming the grid in the example spreads arbitrarily far in all directions.Requirement C: Repeat Requirements A and B, except assume now more generally that channels for transmitters within distance 2s differ by at least some given integer k, while those at distance at most 4s must still differ by at least one. What cna we say about the span and about efficient strategies for designing assignments, as a function of k?Requirement D: Consider generalizations of the problem, such as several levels of interference or irregular transmitter placements. What other factors may be important to consider?Requirement E: Write an article (no more than 2 pages) for the local newspaper explaining your findings.2000 ICM: Elephants: When is Enough, Enough?“Ultimately, if a habitat is undesirably changed by elephants, then their removal should be considered -even by culling.”National Geographic (Earth Almanac) –December 1999 A large National Park in South Africa contains approximately 11,000 elephants. Management policy requires a healthy environment that can maintain a stable herf of 11,000 elephants. Each year park rangers count the elephant population. During the past 20 years whole herds have been removed to keep the population as close to 11,000 as possible. The process involved shooting (for the most part) and occasionally relocating approximately 600 to 800 elephants per year.Recently, there has been a public outcry against the shooting of these elephants. In addition, it is no longer feasible to relocate even a small population of elephants each year. A contraceptive dart, however, has been developed that can prevent a mature elephant cow from conceiving for a period of two years.Here is some information about eh elephants in the Park:∙There is very little emigration of immigration of elephants.∙The gender ratio is very close to 1:1 and control measures have endeavored to maintain parity.∙The gender ratio of newborn calves is also about 1:1. Twins are born about 1.35% of the time.∙Cows first conceive between the ages of 10 and 12 and produce, on average, a calf every 3.5 years until they reach an age of about 60.Gestation is approximately 22 months.∙The contraceptive dart causes an elephant cow to come into oestrus every month (but not conceiving). Elephants usually have courtship only once in 3.5 years, so the monthly cycle can cause additional stress.∙ A cow can be darted every year without additional detrimental effects. A mature elephant cow will not be able to conceive for 2 years after thelast darting.∙Between 70% and 80% of newborn calves survive to age 1 year.Thereafter, the survival rate is uniform across all ages and is very high(over 95%), until about age 60; it is a good assumption that elephantsdie before reading age 70.There is no hunting and negligible poaching in the Park.The park management has a rough data file of the approximate ages and gender of the elephants they have transported out of the region during the past 2 years. This data is available on website: icm2000data.xls. Unfortunately no data is available for the elephants that have been shot or remain in the Park.Your overall task is to develop and use models to investigate how the contraceptive dart might be used for population control. Specifically:Task 1: Develop and use a model to speculate about the likely survival rate for elephants aged 2 to 60. Also speculate about the current age structure of the elephant population.Task 2: Estimate how many cows would need to be darted each year to keep the population fixed at approximately 11,000 elephants. Show how the uncertainty in the data at your disposal affects your estimate. Comment on any changes in the age structure of the population and how this might affect tourists. (You may want to look ahead about 30-60 years.)Task 3: If it were feasible to relocate between 50 and 300 elephants per year, how would this reduce the number of elephants to be darted? Comment on the trade-off between darting and relocation.Task 4: Some opponents of darting argue that if there were a sudden loss of a large number of elephants (due to disease or uncontrolled poaching), even if darting stopped immediately, the ability of the population to grow again would be seriously impeded. Investigate and respond to this concer.Task 5: The management in the Park is skeptical about modeling. In particular, they argue that a lack of complete data makes a mockery of any attempt to use models to guide their decision. In addition to your technical report, include a carefully crafted report (3-page maximum) written explicitly for the park management that responds to their concerns and provides advice. Also suggest ways to increase the park managers confidence in your model and your conclusions.Task 6: If your model works, other elephant parks in Africa would be interested in using it. Prepare a darting plan for parks of various sizes (300-25,000 elephants), with slightly different survival rates and transportation possibilities.2001 年美国大学生数学建模竞赛MCM、ICM 试题2001 MCM A: Choosing a Bicycle WheelCyclists have different types of wheels they can use on their bicycles. The two basic types of wheels are those constructed using wire spokes and those constructed of a solid disk (see Figure 1) The spoked wheels are lighter, but the solid wheels are more aerodynamic. A solid wheel is never used on the front for a road race but can be used on the rear of the bike.Professional cyclists look at a racecourse and make an educated guess as to what kind of wheels should be used. The decision is based on the number and steepness of the hills, the weather, wind speed, the competition, and other considerations. The director sportif of your favorite team would like to have a better system in place and has asked your team for information to help determine what kind of wheel should be used for a given course.Figure 1: A solid wheel is shown on the left and a spoked wheel is shown on the right.The director sportif needs specific information to help make a decision and has asked your team to accomplish the tasks listed below. For each of the tasks assume that the same spoked wheel will always be used on the front but there is a choice of wheels for the rear.Task 1. Provide a table giving the wind speed at which the power required for a solid rear wheel is less than for a spoked rear wheel. The table should include the wind speeds for different road grades startingfrom zero percent to ten percent in one percent increments. (Roadgrade is defined to be the ratio of the total rise of a hill divided by thelength of the road. If the hill is viewed as a triangle, the grade is the sine of the angle at the bottom of the hill.) A rider starts at the bottom of the hill at a speed of 45 kph, and the deceleration of the rider is proportionalto the road grade. A rider will lose about 8 kph for a five percent grade over 100 meters.∙Task 2. Provide an example of how the table could be used for a specific time trial course.∙Task 3. Determine if the table is an adequate means for deciding on the wheel configuration and offer other suggestions as to how to make this decision.2001 MCM B: Escaping a Hurricane's Wrath (An Ill Wind...)Evacuating the coast of South Carolina ahead of the predicted landfall of Hurricane Floyd in 1999 led to a monumental traffic jam. Traffic slowed to a standstill on Interstate I-26, which is the principal route going inland from Charleston to the relatively safe haven of Columbia in the center of the state. What is normally an easy two-hour drive took up to 18 hours to complete. Many cars simply ran out of gas along the way. Fortunately, Floyd turned north and spared the state this time, but the public outcry is forcing state officials to find ways to avoid a repeat of this traffic nightmare.The principal proposal put forth to deal with this problem is the reversal of traffic on I-26, so that both sides, including the coastal-bound lanes, have traffic headed inland from Charleston to Columbia. Plans to carry this out have been prepared (and posted on the Web) by the South Carolina Emergency Preparedness Division. Traffic reversal on principal roads leading inland from Myrtle Beach and Hilton Head is also planned.A simplified map of South Carolina is shown. Charleston has approximately 500,000 people, Myrtle Beach has about 200,000 people, and another 250,000 people are spread out along the rest of the coastal strip. (More accurate data, if sought, are widely available.)The interstates have two lanes of traffic in each direction except in the metropolitan areas where they have three. Columbia, another metro area of around 500,000 people, does not have sufficient hotel space to accommodate the evacuees (including some coming from farther north by other routes), so some traffic continues outbound on I-26 towards Spartanburg; on I-77 north to Charlotte; and on I-20 east to Atlanta. In 1999, traffic leaving Columbia going northwest was moving only very slowly. Construct a model for the problem to investigate what strategies may reduce the congestion observed in 1999. Here are the questions that need to be addressed:1.Under what conditions does the plan for turning the two coastal-boundlanes of I-26 into two lanes of Columbia-bound traffic, essentiallyturning the entire I-26 into one-way traffic, significantly improveevacuation traffic flow?2.In 1999, the simultaneous evacuation of the state's entire coastal regionwas ordered. Would the evacuation traffic flow improve under analternative strategy that staggers the evacuation, perhapscounty-by-county over some time period consistent with the pattern of how hurricanes affect the coast?3.Several smaller highways besides I-26 extend inland from the coast.Under what conditions would it improve evacuation flow to turn around traffic on these?4.What effect would it have on evacuation flow to establish moretemporary shelters in Columbia, to reduce the traffic leaving Columbia?5.In 1999, many families leaving the coast brought along their boats,campers, and motor homes. Many drove all of their cars. Under whatconditions should there be restrictions on vehicle types or numbers ofvehicles brought in order to guarantee timely evacuation?6.It has been suggested that in 1999 some of the coastal residents ofGeorgia and Florida, who were fleeing the earlier predicted landfalls ofHurricane Floyd to the south, came up I-95 and compounded the traffic problems. How big an impact can they have on the evacuation trafficflow? Clearly identify what measures of performance are used tocompare strategies. Required: Prepare a short newspaper article, not to exceed two pages, explaining the results and conclusions of your study to the public.Clearly identify what measures of performance are used to compare strategies. Required: Prepare a short newspaper article, not to exceed two pages, explaining the results and conclusions of your study to the public.2001 ICM: Our Waterways - An Uncertain FutureZebra mussels, Dreissena polymorpha, are small, fingernail-sized, freshwater mollusks unintentionally introduced to North America via ballast water from a transoceanic vessel. Since their introduction in the mid 1980s, they have spread through all of the Great Lakes and to an increasing number of inland waterways in the United States and Canada. Zebra mussels colonize on various surfaces,such as docks, boat hulls, commercial fishing nets, water intake pipes and valves, native mollusks and other zebra mussels. Their only known predators, some diving ducks, freshwater drum, carp, and sturgeon, are not numerous enough to have a significant effect on them. Zebra mussels have significantly impacted the Great Lakes ecosystem and economy. Many communities are trying to control or eliminate these aquatic pests. SOURCE: Great Lakes Sea Grant Network /.Researchers are attempting to identify the environmental variables related to the zebra mussel infestation in North American waterways. The relevant factors that may limit or prevent the spread of the zebra mussel are uncertain. You will have access to some reference data to include listings of several chemicals and substances in the water system that may affect the spread of the zebra mussel throughout waterways. Additionally, you can assume individual zebra mussels grow at a rate of 15 millimeters per year with a life span between 4 - 6 years. The typical mussel can filter 1 liter of water each day.Requirement A: Discuss environmental factors that could influence the spread of zebra mussels.Requirement B: Utilizing the chemical data provided at:ap/undergraduate/contests/icm/imagesdata/LakeAChem1.xls, and the mussel population data provided at:ap/undergraduate/contests/icm/imagesdata/LakeAPopulation 1.xls model the population growth of zebra mussels in Lake A. Be sure to review the Information about the collection of the zebra mussel data. Requirement C: Utilizing additional data on Lake A from another scientist provided at :ap/undergraduate/contests/icm/imagesdata/LakeAChem2.xls and additional mussel population data provided at:ap/undergraduate/contests/icm/imagesdata/LakeAPopulation 2.xls corroborate the reasonableness of your model from Requirement B. As a result of this additional data, adjust your earlier model. Analyze the performance of your model. Discuss the sensitivity of your model. Requirement D: Utilizing the Chemical data from two lakes (Lake B and Lake C) in the United States provided atap/undergraduate/contests/icm/imagesdata/LakeB.xls and ap/undergraduate/contests/icm/imagesdata/LakeC.xls determine if these lakes are vulnerable to the spread of zebra mussels. Discuss your prediction.Requirement E: The community in the vicinity of Lake B (in requirement D) is considering specific policies for the de-icing of roadways near the lake duringthe winter season. Provide guidance to the local government officials regarding a policy on “de-icing agents.”In your guidance include predictions on the long-term impact of de-icing on the zebra mussel population. Requirement F: It has been recommended by a local community in the United States to introduce round goby fish. Zebra mussels are not often eaten by native fish species so they represent a dead end ecologically. However, round gobies greater than 100 mm feed almost exclusively on zebra mussels. Ironically, because of habitat destruction, the goby is endangered in its native habitat of the Black and Caspian Seas in Russia. In addition to your technical report, include a carefully crafted report (3-page maximum) written explicitly for the local community leaders that responds to their recommendation to introduce the round goby. Also suggest ways to help reduce the growth of the mussel within and among waterways.Information about the collection of the zebra mussel dataThe developmental state of the Zebra mussel is categorized by three stages: veligers (larvae), settling juveniles, and adults. Veligers (microscopic zebra mussel larvae) are free-swimming, suspended in the water for one to three weeks, after which they begin searching for a hard surface to attach to and begin their adult life. Looking for zebra mussel veligers is difficult because they are not easily visible by the naked eye. Settled juvenile zebra mussels can be felt on smooth surfaces like boats and motors. An advanced zebra mussel infestation can cover a surface, even forming thick mats sometimes reaching very high densities. The density of juveniles was determined along the lake using three 15×15 cm settling plates. The top plate remained in the water for the entire sampling season (S - seasonal) to estimate seasonal accumulation. The middle and bottom plates are collected after specific periods (A –alternating ) of time denoted by “Lake Days”in the data files.The settling plates are placed under the microscope and all juveniles on the undersides of the plate are counted and densities are reported as juveniles/m^2.2002 年美国大学生数学建模竞赛MCM、ICM 试题2002 MCM A: Wind and WatersprayAn ornamental fountain in a large open plaza surrounded by buildings squirts water high into the air. On gusty days, the wind blows spray from the fountain onto passersby. The water-flow from the fountain is controlled by a mechanism linked to an anemometer (which measures wind speed and direction) located on top of an adjacent building. The objective of this control is to provide passersby with an acceptable balance between an attractive spectacle and a soaking: The harder the wind blows, the lower the water volume and height to which the water is squirted, hence the less spray falls outside the pool area. Your task is to devise an algorithm which uses data provided by the anemometer to adjust the water-flow from the fountain as the wind conditions change.2002 MCM B: Airline OverbookingYou're all packed and ready to go on a trip to visit your best friend in New York City. After you check in at the ticket counter, the airline clerk announces that your flight has been overbooked. Passengers need to check in immediately to determine if they still have a seat.Historically, airlines know that only a certain percentage of passengers who have made reservations on a particular flight will actually take that flight. Consequently, most airlines overbook-that is, they take more reservations than the capacity of the aircraft. Occasionally, more passengers will want to take a flight than the capacity of the plane leading to one or more passengers being bumped and thus unable to take the flight for which they had reservations. Airlines deal with bumped passengers in various ways. Some are given nothing, some are booked on later flights on other airlines, and some are given some kind of cash or airline ticket incentive.Consider the overbooking issue in light of the current situation: Less flights by airlines from point A to point B Heightened security at and around airports Passengers' fear Loss of billions of dollars in revenue by airlines to dateBuild a mathematical model that examines the effects that different overbooking schemes have on the revenue received by an airline company in order to find an optimal overbooking strategy, i.e., the number of people by which an airline should overbook a particular flight so that the company's revenue is maximized. Insure that your model reflects the issues above, andconsider alternatives for handling “bumped”passengers. Additionally, write a short memorandum to the airline's CEO summarizing your findings and analysis.2002 ICM: Scrub LizardsThe Florida scrub lizard is a small, gray or gray-brown lizard that lives throughout upland sandy areas in the Central and Atlantic coast regions of Florida. The Florida Committee on Rare and Endangered Plants classified the scrub lizard as endangered.You will find a fact sheet on the Florida Scrub Lizard at/undergraduate/contests/mcm/contests/2002/problem s/icm2002data/scrublizard.pdfThe long-term survival of the Florida scrub lizard is dependent upon preservation of the proper spatial configuration and size of scrub habitat patches.Task 1: Discuss factors that may contribute to the loss of appropriate habitat for scrub lizards in Florida. What recommendations would you make to the state of Florida to preserve these habitats and discuss obstacles to the implementation of your recommendations?Task 2: Utilize the data provided in Table 1 to estimate the value for Fa (the average fecundity of adult lizards); Sj (the survivorship of juvenile lizards- between birth and the first reproductive season); and Sa (the average adult survivorship).Table 1Summary data for a cohort of scrub lizards captured and followed for 4 consecutive years. Hatchling lizards (age 0) do not produce eggs during the summer they are born. Average clutch size for all other females is proportional to body size according to the function y = 0.21*(SVL)-7.5, where y is the clutch size and SVL is the snout-to-vent length in mm.Year Age Total NumberLivingNumber of LivingFemalesAvg. Female Size(mm)1 0 972 495 30.32 1 180 92 45.83 2 20 11 55.84 3 2 2 56.0Task 3: It has been conjectured that the parameters Fa , Sj , and Sa , are related to the size and amount of open sandy area of a scrub patch. Utilize the data provided in Table 2 to develop functions that estimate Fa, Sj , and Sa for different patches. In addition, develop a function that estimates C, the carrying capacity of scrub lizards for a given patch.Table 2Summary data for 8 scrub patches including vital rate data for scrub lizards. Annual female fecundity (Fa), juvenile survivorship (Sj), and adult survivorship (Sa) are presented for each patch along with patch size and the amount of open sandy habitat.Patch Patch Size (ha) Sandy Habitat (ha) Fa Sj Sa Density (lizards/ha)a 11.31 4.80 5.6 0.12 0.06 58b 35.54 11.31 6.6 0.16 0.10 60c 141.76 51.55 9.5 0.17 0.13 75d 14.65 7.55 4.8 0.15 0.09 55e 63.24 20.12 9.7 0.17 0.11 80f 132.35 54.14 9.9 0.18 0.14 82g 8.46 1.67 5.5 0.11 0.05 40h 278.26 84.32 11.0 0.19 0.15 115Task 4: There are many animal studies that indicate that food, space, shelter, or even reproductive partners may be limited within a habitat patch causing individuals to migrate between patches. There is no conclusive evidence on why scrub lizards migrate. However, about 10 percent of juvenile lizards do migrate between patches and this immigration can influence the size of the population within a patch. Adult lizards apparently do not migrate. Utilizing the data provided in the histogram below estimate the probability of lizards surviving the migration between any two patches i and patch j.Table 3HistogramMigration data for juvenile lizards marked, released, and recaptured up to 6 months later. Surveys for recapture were conducted up to 750m from release sites.Task 5: Develop a model to estimate the overall population size of scrub lizards for the landscape given in Table 3. Also, determine which patches are suitable for occupation by scrub lizards and which patches would not support a viable population.Patch size and amount of open sandy habitat for a landscape of 29 patches located on the Avon Park Air Force Range. See:/undergraduate/contests/icm/2002problem/map.jpg for a map of the landscape.Patch Identification Patch Size (ha) Sandy Habitat (ha)1 13.66 5.382 32.74 11.913 1.39 0.234 2.28 0.765 7.03 3.626 14.47 4.387 2.52 1.998 5.87 2.499 22.27 8.44。

2.3.What is the value of ?4.Eight 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 $2.50 to cover her portion of the total bill. What was the total bill?5.Hammie 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?6.The 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?7.Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 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?8.A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?9.The 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?10.What 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 walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less 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 pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the 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. When Clara 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 answerchoice 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 of Lord.of.AMC19. Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score 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 sidesand , 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 theball travels over the course from A to B?1.2.The 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.3.Notice that we can pair up every two numbers to make a sum of 1:Therefore, the answer is .4.Each 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 .5.The 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.6.Solution 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 . is the answer we're looking for.We see that , making .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.)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:7.If 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 .8.First, 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 .9.This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that .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 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 answerchoice 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: AlgebraWe multiply 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 7th 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 arecubes. 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 of the 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 inthe 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 toget from City Park 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, i.e. 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 Pythagoreantheorem 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 semicircles on the left is equal to the biggest one, so the area of thelargest 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 theextension of . 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 thecombined area of the three squares is .Let 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, .Let 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 . Thissolution 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.古希腊哲学大师亚里士多德说：人有两种，一种即“吃饭是为了活着”,一种是“活着是为了吃饭”.一个人之所以伟大，首先是因为他有超于常人的心。

2004年全国硕士研究生入学统一考试数学三试题一、填空题（本题共6小题，每小题4分，满分24分. 把答案填在题中横线上） (1) 若，则a =______，b =______. (2) 设函数f (u , v )由关系式f [xg (y ) , y ] = x + g (y )确定，其中函数g (y )可微，且g (y ) ≠ 0，则.(3) 设，则.(4) 二次型的秩为 . (5) 设随机变量服从参数为的指数分布, 则_______.(6) 设总体服从正态分布, 总体服从正态分布,和分别是来自总体和的简单随机样本, 则.二、选择题（本题共6小题，每小题4分，满分24分. 每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内） (7) 函数在下列哪个区间内有界. (A) (-1 , 0).(B) (0 , 1).(C) (1 , 2).(D) (2 , 3). [ ](8) 设f (x )在(-∞ , +∞)内有定义，且， ，则(A) x = 0必是g (x )的第一类间断点. (B) x = 0必是g (x )的第二类间断点.(C) x = 0必是g (x )的连续点.(D) g (x )在点x = 0处的连续性与a 的取值有关. [ ] (9) 设f (x ) = |x (1 - x )|，则(A) x = 0是f (x )的极值点，但(0 , 0)不是曲线y = f (x )的拐点. (B) x = 0不是f (x )的极值点，但(0 , 0)是曲线y = f (x )的拐点.5)(cos sin lim0=--→b x ae xx x 2fu v∂=∂∂⎪⎩⎪⎨⎧≥-<≤-=21,12121,)(2x x xe x f x 212(1)f x dx -=⎰213232221321)()()(),,(x x x x x x x x x f ++-++=X λ=>}{DX X P X ),(21σμN Y ),(22σμN 1,,21n X X X 2,,21n Y Y Y X Y 12221112()()2n n i j i j X X Y Y E n n ==⎡⎤-+-⎢⎥⎢⎥=⎢⎥+-⎢⎥⎢⎥⎣⎦∑∑2)2)(1()2sin(||)(---=x x x x x x f a x f x =∞→)(lim ⎪⎩⎪⎨⎧=≠=0,00,)1()(x x x f x g(C) x = 0是f (x )的极值点，且(0 , 0)是曲线y = f (x )的拐点.(D) x = 0不是f (x )的极值点，(0 , 0)也不是曲线y = f (x )的拐点. [ ](10) 设有下列命题：(1) 若收敛，则收敛.(2) 若收敛，则收敛.(3) 若，则发散. (4) 若收敛，则，都收敛.则以上命题中正确的是 (A) (1) (2). (B) (2) (3). (C) (3) (4). (D) (1) (4). [ ] (11) 设在[a , b]上连续，且，则下列结论中错误的是 (A) 至少存在一点，使得> f (a ). (B) 至少存在一点，使得> f (b ). (C) 至少存在一点，使得.(D) 至少存在一点，使得= 0.[ D ](12) 设阶矩阵与等价, 则必有(A) 当时, . (B) 当时, .(C) 当时, . (D) 当时, . [ ](13) 设阶矩阵的伴随矩阵 若是非齐次线性方程组 的互不相等的解，则对应的齐次线性方程组的基础解系 (A) 不存在. (B) 仅含一个非零解向量.(C) 含有两个线性无关的解向量. (D) 含有三个线性无关的解向量.[ ](14) 设随机变量服从正态分布, 对给定的, 数满足,若, 则等于 (A) . (B) . (C) . (D) . [ ]三、解答题(本题共9小题，满分94分. 解答应写出文字说明、证明过程或演算步骤.) (15) (本题满分8分)求. ∑∞=-+1212)(n n n u u ∑∞=1n n u ∑∞=1n n u ∑∞=+11000n n u 1lim1>+∞→n n n u u ∑∞=1n n u ∑∞=+1)(n n n v u ∑∞=1n n u ∑∞=1n n v )(x f '0)(,0)(<'>'b f a f ),(0b a x ∈)(0x f ),(0b a x ∈)(0x f ),(0b a x ∈0)(0='x f ),(0b a x ∈)(0x f n A B )0(||≠=a a A a B =||)0(||≠=a a A a B -=||0||≠A 0||=B 0||=A 0||=B n A ,0*≠A 4321,,,ξξξξb Ax =0=Ax X )1,0(N )1,0(∈ααu αu X P α=>}{αx X P =<}|{|x 2αu 21αu-21αu -αu -1)cos sin 1(lim 2220xxx x -→(16) (本题满分8分)求，其中D所围成的 平面区域(如图).(17) (本题满分8分) 设f (x ) , g (x )在[a , b ]上连续，且满足，x ∈ [a , b )，.证明：.(18) (本题满分9分) 设某商品的需求函数为Q = 100 - 5P ，其中价格P ∈ (0 , 20)，Q 为需求量. (I) 求需求量对价格的弹性(> 0)；(II) 推导(其中R 为收益)，并用弹性说明价格在何范围内变化时， 降低价格反而使收益增加. (19) (本题满分9分) 设级数的和函数为S (x ). 求：(I) S (x )所满足的一阶微分方程； (II) S (x )的表达式. (20)(本题满分13分)设, , , ,试讨论当为何值时,(Ⅰ) 不能由线性表示;(Ⅱ) 可由唯一地线性表示, 并求出表示式;(Ⅲ) 可由线性表示, 但表示式不唯一, 并求出表示式. (21) (本题满分13分) 设阶矩阵⎰⎰++Dd y y x σ)(2222122=⎰⎰≥x axadt t g dt t f )()(⎰⎰=bab adt t g dt t f )()(⎰⎰≤bab adx x xg dx x xf )()(d E d E )1(d E Q dPdR-=d E )(864264242864+∞<<-∞+⋅⋅⋅+⋅⋅+⋅x x x x T α)0,2,1(1=T ααα)3,2,1(2-+=T b αb α)2,2,1(3+---=Tβ)3,3,1(-=b a ,β321,,αααβ321,,αααβ321,,αααn. (Ⅰ) 求的特征值和特征向量;(Ⅱ) 求可逆矩阵, 使得为对角矩阵. (22) (本题满分13分)设，为两个随机事件,且, , , 令求(Ⅰ) 二维随机变量的概率分布; (Ⅱ) 与的相关系数 ; (Ⅲ) 的概率分布.(23) (本题满分13分)设随机变量的分布函数为其中参数. 设为来自总体的简单随机样本,(Ⅰ) 当时, 求未知参数的矩估计量;(Ⅱ) 当时, 求未知参数的最大似然估计量; (Ⅲ) 当时, 求未知参数的最大似然估计量.2004年考研数学（三）真题解析一、填空题（本题共6小题，每小题4分，满分24分. 把答案填在题中横线上） (1) 若，则a =，b =.【分析】本题属于已知极限求参数的反问题. 【详解】因为，且，所以，得a = 1. 极限化为，得b = -4. 因此，a = 1，b = -4.⎪⎪⎪⎪⎪⎭⎫⎝⎛=111 b b b bb b A A P AP P 1-A B 41)(=A P 31)|(=AB P 21)|(=B A P ⎩⎨⎧=不发生，，发生，A A X 0,1⎩⎨⎧=.0,1不发生，发生，B B Y ),(Y X X Y XY ρ22Y X Z +=X ⎪⎩⎪⎨⎧≤>⎪⎭⎫ ⎝⎛-=，，，αx αx x αβαx F β0,1),,(1,0>>βαn X X X ,,,21 X 1=αβ1=αβ2=βα5)(cos sin lim0=--→b x ae xx x 14-5)(cos sin lim0=--→b x ae xx x 0)(cos sin lim 0=-⋅→b x x x 0)(lim 0=-→a e x x 51)(cos lim )(cos sin lim00=-=-=--→→b b x xxb x a e x x x x【评注】一般地，已知＝ A ， (1) 若g (x ) → 0，则f (x ) → 0；(2) 若f (x ) → 0，且A ≠ 0，则g (x ) → 0.(2) 设函数f (u , v )由关系式f [xg (y ) , y ] = x + g (y )确定，其中函数g (y )可微，且g (y ) ≠ 0，则.【分析】令u = xg (y )，v = y ，可得到f (u , v )的表达式，再求偏导数即可. 【详解】令u = xg (y )，v = y ，则f (u , v ) =，所以，，.(3) 设，则.【分析】本题属于求分段函数的定积分，先换元：x - 1 = t ，再利用对称区间上奇偶函数的积分性质即可.【详解】令x - 1 = t ，＝.【评注】一般地，对于分段函数的定积分，按分界点划分积分区间进行求解. (4) 二次型的秩为 2 .【分析】二次型的秩即对应的矩阵的秩, 亦即标准型中平方项的项数, 于是利用初等变换或配方法均可得到答案. 【详解一】因为于是二次型的矩阵为 ,)()(limx g x f )()(22v g v g vu f '-=∂∂∂)()(v g v g u+)(1v g u f =∂∂)()(22v g v g v u f '-=∂∂∂⎪⎩⎪⎨⎧≥-<≤-=21,12121,)(2x x xe x f x 21)1(221-=-⎰dx x f ⎰⎰⎰--==-121121221)()()1(dt x f dt t f dx x f 21)21(0)1(12121212-=-+=-+⎰⎰-dx dx xe x 213232221321)()()(),,(x x x x x x x x x f ++-++=213232221321)()()(),,(x x x x x x x x x f ++-++=323121232221222222x x x x x x x x x -++++=⎪⎪⎪⎭⎫ ⎝⎛--=211121112A由初等变换得 ,从而 , 即二次型的秩为2.【详解二】因为,其中 .所以二次型的秩为2.(5) 设随机变量服从参数为的指数分布, 则. 【分析】 根据指数分布的分布函数和方差立即得正确答案. 【详解】 由于, 的分布函数为 故. 【评注】本题是对重要分布, 即指数分布的考查, 属基本题型.(6) 设总体服从正态分布, 总体服从正态分布,和 分别是来自总体和的简单随机样本, 则.【分析】利用正态总体下常用统计量的数字特征即可得答案.【详解】因为 , , 故应填 .⎪⎪⎪⎭⎫ ⎝⎛--→⎪⎪⎪⎭⎫ ⎝⎛---→000330211330330211A 2)(=A r 213232221321)()()(),,(x x x x x x x x x f ++-++=323121232221222222x x x x x x x x x -++++=2322321)(23)2121(2x x x x x -+++=2221232y y +=,21213211x x x y ++=322x x y -=X λ=>}{DX X P e121λDX =X ⎩⎨⎧≤>-=-.0,0,0,1)(x x e x F x λ=>}{DX X P =≤-}{1DX X P =≤-}1{1λX P )1(1λF -e1=X ),(21σμN Y ),(22σμN 1,,21n X X X 2,,21n Y Y Y X Y 22121212)()(21σn n Y Y X X E n j j n i i =⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡-+-+-∑∑==2121])(11[1σX X n E n i i =--∑=2122])(11[2σY Y n E n j j =--∑=2σ【评注】本题是对常用统计量的数字特征的考查.二、选择题（本题共6小题，每小题4分，满分24分. 每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内） (7) 函数在下列哪个区间内有界. (A) (-1 , 0).(B) (0 , 1).(C) (1 , 2).(D) (2 , 3). [ A ]【分析】如f (x )在(a , b )内连续，且极限与存在，则函数f (x )在(a , b )内有界.【详解】当x ≠ 0 , 1 , 2时，f (x )连续，而，，，，，所以，函数f (x )在(-1 , 0)内有界，故选(A).【评注】一般地，如函数f (x )在闭区间[a , b ]上连续，则f (x )在闭区间[a , b ]上有界；如函数f (x )在开区间(a , b )内连续，且极限与存在，则函数f (x )在开区间(a , b )内有界.(8) 设f (x )在(-∞ , +∞)内有定义，且，，则 (A) x = 0必是g (x )的第一类间断点. (B) x = 0必是g (x )的第二类间断点.(C) x = 0必是g (x )的连续点.(D) g (x )在点x = 0处的连续性与a 的取值有关. [ D ] 【分析】考查极限是否存在，如存在，是否等于g (0)即可，通过换元， 可将极限转化为.【详解】因为= a (令)，又g (0) = 0，所以，当a = 0时，，即g (x )在点x = 0处连续，当a ≠ 0时，，即x = 0是g (x )的第一类间断点，因此，g (x )在点x = 0处的连续性与a 的取值有关，故选(D).【评注】本题属于基本题型，主要考查分段函数在分界点处的连续性. (9) 设f (x ) = |x (1 - x )|，则(A) x = 0是f (x )的极值点，但(0 , 0)不是曲线y = f (x )的拐点. (B) x = 0不是f (x )的极值点，但(0 , 0)是曲线y = f (x )的拐点. (C) x = 0是f (x )的极值点，且(0 , 0)是曲线y = f (x )的拐点.(D) x = 0不是f (x )的极值点，(0 , 0)也不是曲线y = f (x )的拐点.[ C ]2)2)(1()2sin(||)(---=x x x x x x f )(lim x f a x +→)(lim x f b x -→183sin )(lim 1-=+-→x f x 42sin )(lim 0-=-→x f x 42sin )(lim 0=+→x f x ∞=→)(lim 1x f x ∞=→)(lim 2x f x )(lim x f a x +→)(lim x f b x -→a x f x =∞→)(lim ⎪⎩⎪⎨⎧=≠=0,00,)1()(x x xf xg )(lim 0x g x →xu 1=)(lim 0x g x →)(lim x f x ∞→)(lim )1(lim )(lim 0u f x f x g u x x ∞→→→==xu 1=)0()(lim 0g x g x =→)0()(lim 0g x g x ≠→【分析】由于f (x )在x = 0处的一、二阶导数不存在，可利用定义判断极值情况，考查f (x )在x = 0的左、右两侧的二阶导数的符号，判断拐点情况.【详解】设0 < δ < 1，当x ∈ (-δ , 0) ⋃ (0 , δ)时，f (x ) > 0，而f (0) = 0，所以x = 0是f (x )的极小值点. 显然，x = 0是f (x )的不可导点. 当x ∈ (-δ , 0)时，f (x ) = -x (1 - x )，， 当x ∈ (0 , δ)时，f (x ) = x (1 - x )，，所以(0 , 0)是曲线y = f (x )的拐点. 故选(C).【评注】对于极值情况，也可考查f (x )在x = 0的某空心邻域内的一阶导数的符号来判断. (10) 设有下列命题：(1) 若收敛，则收敛.(2) 若收敛，则收敛.(3) 若，则发散. (4) 若收敛，则，都收敛.则以上命题中正确的是 (A) (1) (2). (B) (2) (3). (C) (3) (4). (D) (1) (4). [ B ]【分析】可以通过举反例及级数的性质来说明4个命题的正确性. 【详解】(1)是错误的，如令，显然，分散，而收敛.(2)是正确的，因为改变、增加或减少级数的有限项，不改变级数的收敛性.(3)是正确的，因为由可得到不趋向于零(n → ∞)，所以发散. (4)是错误的，如令，显然，，都发散，而收敛. 故选(B).【评注】本题主要考查级数的性质与收敛性的判别法，属于基本题型.(11) 设在[a , b]上连续，且，则下列结论中错误的是 (A) 至少存在一点，使得> f (a ). (B) 至少存在一点，使得> f (b ).(C) 至少存在一点，使得.02)(>=''x f 02)(<-=''x f ∑∞=-+1212)(n n n u u ∑∞=1n n u ∑∞=1n n u ∑∞=+11000n n u 1lim1>+∞→n n n u u ∑∞=1n n u ∑∞=+1)(n n n v u ∑∞=1n n u ∑∞=1n n v nn u )1(-=∑∞=1n n u ∑∞=-+1212)(n n n u u 1lim1>+∞→n n n u u n u ∑∞=1n n u n v n u n n 1,1-==∑∞=1n n u ∑∞=1n n v ∑∞=+1)(n n n v u )(x f '0)(,0)(<'>'b f a f ),(0b a x ∈)(0x f ),(0b a x ∈)(0x f ),(0b a x ∈0)(0='x f(D) 至少存在一点，使得= 0.[ D ]【分析】利用介值定理与极限的保号性可得到三个正确的选项，由排除法可选出错误选项. 【详解】首先，由已知在[a , b]上连续，且，则由介值定理，至少存在一点，使得；另外，，由极限的保号性，至少存在一点使得，即. 同理，至少存在一点使得. 所以，(A) (B) (C)都正确，故选(D).【评注】 本题综合考查了介值定理与极限的保号性，有一定的难度. (12) 设阶矩阵与等价, 则必有(A) 当时, . (B) 当时, .(C) 当时, . (D) 当时, . [ D ] 【分析】 利用矩阵与等价的充要条件: 立即可得.【详解】因为当时, , 又 与等价, 故, 即, 故选(D). 【评注】本题是对矩阵等价、行列式的考查, 属基本题型.(13) 设阶矩阵的伴随矩阵 若是非齐次线性方程组 的互不相等的解，则对应的齐次线性方程组的基础解系 (A) 不存在. (B) 仅含一个非零解向量.(C) 含有两个线性无关的解向量. (D) 含有三个线性无关的解向量. [ B ] 【分析】 要确定基础解系含向量的个数, 实际上只要确定未知数的个数和系数矩阵的秩. 【详解】 因为基础解系含向量的个数=, 而且根据已知条件 于是等于或. 又有互不相等的解,即解不惟一, 故. 从而基础解系仅含一个解向量, 即选(B).【评注】本题是对矩阵与其伴随矩阵的秩之间的关系、线性方程组解的结构等多个知识点的综合考查.(14) 设随机变量服从正态分布, 对给定的, 数满足,若, 则等于 (A) . (B) . (C) . (D) . [ C ]【分析】 利用标准正态分布密度曲线的对称性和几何意义即得.【详解】 由, 以及标准正态分布密度曲线的对称性可得),(0b a x ∈)(0x f )(x f '0)(,0)(<'>'b f a f ),(0b a x ∈0)(0='x f 0)()(lim )(>--='+→ax a f x f a f a x ),(0b a x ∈0)()(00>--ax a f x f )()(0a f x f >),(0b a x ∈)()(0b f x f >n A B )0(||≠=a a A a B =||)0(||≠=a a A a B -=||0||≠A 0||=B 0||=A 0||=B A B )()(B r A r =0||=A n A r <)(A B n B r <)(0||=B n A ,0*≠A 4321,,,ξξξξb Ax =0=Ax )(A r n -⎪⎩⎪⎨⎧-<-===.1)(,0,1)(,1,)(,)(*n A r n A r n A r n A r ,0*≠A )(A r n 1-n b Ax =1)(-=n A r A *A X )1,0(N )1,0(∈ααu αu X P α=>}{αx X P =<}|{|x 2αu 21αu-21αu -αu -1αx X P =<}|{|. 故正确答案为(C). 【评注】本题是对标准正态分布的性质, 严格地说它的上分位数概念的考查.三、解答题(本题共9小题，满分94分. 解答应写出文字说明、证明过程或演算步骤.) (15) (本题满分8分)求. 【分析】先通分化为“”型极限，再利用等价无穷小与罗必达法则求解即可.【详解】 =. 【评注】本题属于求未定式极限的基本题型，对于“”型极限，应充分利用等价无穷小替换来简化计算.(16) (本题满分8分) 求，其中D 是由圆和所围成的平面区域(如图).【分析】首先，将积分区域D 分为大圆减去小圆，再利用对称性与极坐标计算即可.【详解】令，由对称性，..所以，. 【评注】本题属于在极坐标系下计算二重积分的基本题型，对于二重积分，经常利用对称性21}{αx X P -=>)cos sin 1(lim 2220xxx x -→0xx xx x x x x x x 2222202220sin cos sin lim )cos sin 1(lim -=-→→346)4(21lim 64cos 1lim 44sin 212lim 2sin 41lim 22020304220==-=-=-→→→→x x x x x x x x x x x x x x 0⎰⎰++Dd y y x σ)(22422=+y x 1)1(22=++y x }4|),{(221≤+=y x y x D }1)1(|),{(222≤++=y x y x D }1)1(|),{(},4|),{(222221≤++=≤+=y x y x D y x y x D 0=⎰⎰Dyd σ⎰⎰⎰⎰⎰⎰+-+=+21222222D D Dd y x d y x d y x σσσ⎰⎰⎰⎰--=θπππθθcos 20223220220dr r d dr r d )23(916932316-=-=ππ)23(916)(22-=++⎰⎰πσDd y y x及将一个复杂区域划分为两个或三个简单区域来简化计算. (17) (本题满分8分) 设f (x ) , g (x )在[a , b ]上连续，且满足，x ∈ [a , b )，.证明：.【分析】令F (x ) = f (x ) - g (x )，，将积分不等式转化为函数不等式即可.【详解】令F (x ) = f (x ) - g (x )，，由题设G (x ) ≥ 0，x ∈ [a , b ]， G (a ) = G (b ) = 0，. 从而，由于 G (x ) ≥ 0，x ∈ [a , b ]，故有 ，即.因此.【评注】引入变限积分转化为函数等式或不等式是证明积分等式或不等式的常用的方法. (18) (本题满分9分) 设某商品的需求函数为Q = 100 - 5P ，其中价格P ∈ (0 , 20)，Q 为需求量. (I) 求需求量对价格的弹性(> 0)；(II) 推导(其中R 为收益)，并用弹性说明价格在何范围内变化时，降低价格反而使收益增加. 【分析】由于> 0，所以；由Q = PQ 及可推导 . 【详解】(I) . (II) 由R = PQ ，得.又由，得P = 10.⎰⎰≥x axadt t g dt t f )()(⎰⎰=bab adt t g dt t f )()(⎰⎰≤bab adx x xg dx x xf )()(⎰=xadt t F x G )()(⎰=xadt t F x G )()()()(x F x G ='⎰⎰⎰⎰-=-==bab aba babadx x G dx x G x xG x xdG dx x xF )()()()()(0)(≤-⎰badx x G 0)(≤⎰ba dx x xF ⎰⎰≤babadx x xg dx x xf )()(d E d E )1(d E Q dPdR-=d E d E dP dQ Q P E d =dPdQQ P E d =)1(d E Q dPdR-=PPdP dQ Q P E d -==20)1()1(d E Q dP dQ Q P Q dP dQ P Q dP dR -=+=+=120=-=PPE d当10 < P < 20时，> 1，于是，故当10 < P < 20时，降低价格反而使收益增加.【评注】当> 0时，需求量对价格的弹性公式为. 利用需求弹性分析收益的变化情况有以下四个常用的公式：，，， (收益对价格的弹性). (19) (本题满分9分) 设级数的和函数为S (x ). 求：(I) S (x )所满足的一阶微分方程； (II) S (x )的表达式.【分析】对S (x )进行求导，可得到S (x )所满足的一阶微分方程，解方程可得S (x )的表达式.【详解】(I) ， 易见 S (0) = 0，.因此S (x )是初值问题的解.(II) 方程的通解为d E 0<dPdRd E dPdQQ P dP dQ Q P E d -==Qdp E dR d )1(-=Q E dp dR d )1(-=p E dQ dR d)11(-=d E EpER-=1)(864264242864+∞<<-∞+⋅⋅⋅+⋅⋅+⋅x x x x +⋅⋅⋅+⋅⋅+⋅=864264242)(864x x x x S +⋅⋅+⋅+='642422)(753x x x x S )642422(642 +⋅⋅+⋅+=x x x x )](2[2x S x x +=0)0(,23=+='y x xy y 23x xy y +=']2[3C dx e x e y xdx xdx +⎰⎰=⎰-，由初始条件y(0) = 0，得C = 1.故，因此和函数.【评注】本题综合了级数求和问题与微分方程问题，2002年考过类似的题. (20)(本题满分13分)设, , , ,试讨论当为何值时,(Ⅰ) 不能由线性表示;(Ⅱ) 可由唯一地线性表示, 并求出表示式;(Ⅲ) 可由线性表示, 但表示式不唯一, 并求出表示式.【分析】将可否由线性表示的问题转化为线性方程组是否有解的问题即易求解. 【详解】 设有数使得. (*) 记. 对矩阵施以初等行变换, 有. (Ⅰ) 当时, 有.可知. 故方程组(*)无解, 不能由线性表示. (Ⅱ) 当, 且时, 有22212x Ce x +--=12222-+-=x e x y 12)(222-+-=x e x x S T α)0,2,1(1=T ααα)3,2,1(2-+=T b αb α)2,2,1(3+---=Tβ)3,3,1(-=b a ,β321,,αααβ321,,αααβ321,,αααβ321,,αααβαk αk αk =++332211,,,321k k k βαk αk αk =++332211),,(321αααA =),(βA ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-+---+-=323032221111),(b a a b a βA ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---→000101111b a b a 0=a ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---→10001001111),(b βA ),()(βA r A r ≠β321,,ααα0≠a b a ≠, 方程组(*)有唯一解：, , ．此时可由唯一地线性表示, 其表示式为． (Ⅲ) 当时, 对矩阵施以初等行变换, 有， , 方程组(*)有无穷多解， 其全部解为, , ， 其中为任意常数．可由线性表示, 但表示式不唯一, 其表示式为． 【评注】本题属于常规题型, 曾考过两次(1991, 2000).(21) (本题满分13分) 设阶矩阵.(Ⅰ) 求的特征值和特征向量;(Ⅱ) 求可逆矩阵, 使得为对角矩阵.【分析】这是具体矩阵的特征值和特征向量的计算问题, 通常可由求解特征方程和齐次线性方程组来解决. 【详解】 (Ⅰ) 当时，⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---→000101111),(b a b a βA ⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡-→0100101011001a a 3),()(==βA r A r ak 111-=a k 12=03=k β321,,ααα211)11(αaαa β+-=0≠=b a ),(βA ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---→000101111),(b a b a βA ⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡--→0000111011001a a 2),()(==βA r A r a k 111-=c a k +=12c k =3c β321,,ααα321)1()11(αc αc aαa β+++-=n ⎪⎪⎪⎪⎪⎭⎫⎝⎛=111 b b b bb b A A P AP P 1-0||=-A E λ0)(=-x A E λ 10≠b＝ ,得的特征值为，． 对，解得，所以的属于的全部特征向量为 (为任意不为零的常数)．对，得基础解系为，，．故的属于的全部特征向量为111||---------=-λb b bλb b b λA E λ 1)]1(][)1(1[------n b λb n λA b n λ)1(11-+=b λλn -===12 b n λ)1(11-+=⎪⎪⎪⎪⎪⎭⎫ ⎝⎛---------=-b n b b b b n b b b b n A E λ)1()1()1(1 →⎪⎪⎪⎪⎪⎭⎫ ⎝⎛---------)1(111)1(111)1(n n n →⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛------------0000111111111111 n n n →⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛---------0000111111111111 n n n →⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛---000000001111n n n n n →⎪⎪⎪⎪⎪⎪⎭⎫⎝⎛---0000110010101001Tξ)1,,1,1,1(1 =A 1λTk ξk )1,,1,1,1(1 =k b λ-=12⎪⎪⎪⎪⎪⎭⎫ ⎝⎛---------=-b b b b b b b b b A E λ 2→⎪⎪⎪⎪⎪⎭⎫ ⎝⎛000000111 T ξ)0,,0,1,1(2 -=T ξ)0,,1,0,1(3 -=T n ξ)1,,0,0,1(,-= A 2λ(是不全为零的常数)． 当时，,特征值为，任意非零列向量均为特征向量．(Ⅱ) 当时，有个线性无关的特征向量，令，则当时，，对任意可逆矩阵, 均有．【评注】本题通过考查矩阵的特征值和特征向量而间接考查了行列式的计算, 齐次线性方程组的求解和矩阵的对角化等问题, 属于有一点综合性的试题. 另外,本题的解题思路是容易的, 只要注意矩阵中含有一个未知参数, 从而一般要讨论其不同取值情况. (22) (本题满分13分)设，为两个随机事件,且, , , 令求(Ⅰ) 二维随机变量的概率分布; (Ⅱ) 与的相关系数 ;(Ⅲ) 的概率分布.【分析】本题的关键是求出的概率分布，于是只要将二维随机变量的各取值对转化为随机事件和表示即可．【详解】 (Ⅰ) 因为 ， 于是 ， 则有 ， ， ， n n ξk ξk ξk +++ 3322n k k k ,,,32 20=b n λλλλA E λ)1(100010001||-=---=- 11===n λλ 10≠b A n ),,,(21n ξξξP =⎪⎪⎪⎪⎪⎭⎫ ⎝⎛---+=-b b b n AP P 11)1(1120=b E A =P E AP P =-1A B 41)(=A P 31)|(=AB P 21)|(=B A P ⎩⎨⎧=不发生，，发生，A A X 0,1⎩⎨⎧=.0,1不发生，发生，B B Y ),(Y X X Y XY ρ22Y X Z +=),(Y X ),(Y X A B 121)|()()(==A B P A P AB P 61)|()()(==B A P AB P B P 121)(}1,1{====AB P Y X P 61)()()(}0,1{=-====AB P A P B A P Y X P 121)()()(}1,0{=-====AB P B P B A P Y X P，( 或)，即的概率分布为：(Ⅱ)方法一：因为，，，，，，，，所以与的相关系数．方法二：X, Y的概率分布分别为X 0 1 Y 0 1P P则，，DY=, E(XY)=,故，从而(Ⅲ) 的可能取值为：0，1，2 ．，，，即的概率分布为：0 1 232)]()()([1)(1)(}0,0{=-+-=⋃-=⋅===ABPBPAPBAPBAPYXP32121611211}0,0{=---===YXP),(YX41)(==APEX61)(==BPEY121)(=XYE41)(2==APEX61)(2==BPEY163)(22=-=EXEXDX165)(22=-=EYEYDY241)(),(=-=EXEYXYEYXCovX Y1515151),(==⋅=DYDXYXCovρXY4341656161,41==EYEX163=DX365121241)(),(=⋅-=EYEXXYEYXCov.1515),(=⋅=DYDXYXCovXYρZ32}0,0{}0{=====YXPZP41}1,0{}0,1{}1{===+====YXPYXPZP121}1,1{}2{=====YXPZPZ【评注】本题考查了二维离散随机变量联合概率分布，数字特征和二维离散随机变量函数的分布等计算问题，属于综合性题型 (23) (本题满分13分)设随机变量的分布函数为其中参数. 设为来自总体的简单随机样本,(Ⅰ) 当时, 求未知参数的矩估计量;(Ⅱ) 当时, 求未知参数的最大似然估计量; (Ⅲ) 当时, 求未知参数的最大似然估计量.【分析】本题是一个常规题型, 只要注意求连续型总体未知参数的矩估计和最大似然估计都须已知密度函数, 从而先由分布函数求导得密度函数. 【详解】 当时, 的概率密度为(Ⅰ) 由于令, 解得 , 所以, 参数的矩估计量为 . (Ⅱ) 对于总体的样本值, 似然函数为当时, , 取对数得 ,对求导数，得X ⎪⎩⎪⎨⎧≤>⎪⎭⎫ ⎝⎛-=，，，αx αx x αβαx F β0,1),,(1,0>>βαn X X X ,,,21 X 1=αβ1=αβ2=βα1=αX ⎪⎩⎪⎨⎧≤>=+，，，101,),(1x x x ββx f β⎰⎰+∞++∞∞--=⋅==11,1);(ββdx x βx dx βx xf EX βX ββ=-11-=X Xββ1-=X XβX n x x x ,,,21 ∏=+⎪⎩⎪⎨⎧=>==ni i βn ni n i x x x x βαx f βL 1121.,0),,,2,1(1,)();()(其他 ),,2,1(1n i x i =>0)(>βL ∑=+-=ni ixββn βL 1ln )1(ln )(ln β， 令， 解得 ，于是的最大似然估计量为．( Ⅲ) 当时, 的概率密度为对于总体的样本值, 似然函数为当时, 越大，越大, 即的最大似然估计值为, 于是的最大似然估计量为 ．∑=-=ni i x βnβd βL d 1ln )]([ln 0ln )]([ln 1=-=∑=ni i x βnβd βL d ∑==ni ixnβ1ln β∑==ni ixnβ1ln ˆ2=βX ⎪⎩⎪⎨⎧≤>=，，，αx αx x αβx f 0,2),(32X n x x x ,,,21 ∏=⎪⎩⎪⎨⎧=>==ni i n nn i n i αx x x x ααx f βL 13212.,0),,,2,1(,)(2);()(其他 ),,2,1(n i αx i =>α)(αL α},,,min{ˆ21n x x x α=α},,,min{ˆ21n X X X α=。

usaco2024年12月铜题解全文共四篇示例，供读者参考第一篇示例：USACO（美国计算机奥林匹克竞赛）是美国著名的计算机竞赛活动，旨在激发学生对计算机科学的兴趣，并培养他们解决问题和编程的能力。

每年举办多次比赛，分为四个级别：铜牌、银牌、金牌和铂金。

其中铜牌级别是最基础的级别，适合初学者和有限的编程经验的学生参加。

2024年12月的USACO铜牌题目共有3个问题，分别为"Cow Radio"、"Sleepy Cow Sorting"和"Mooyomooyo"。

这些问题涉及到不同的编程知识和技巧，下面我们来解析这些问题的解题思路和方法。

1. Cow Radio这个问题描述了一头母牛在牛棚里玩耍，她想要选择一首歌曲来听，希望通过换频道来找到最喜欢的歌曲。

每首歌曲都有一个唯一的频道编号和长度，母牛可以通过加或减频道来切换歌曲。

给定母牛当前所在的频道和要切换的频道，需要计算出播放完所有歌曲需要的最少时间。

解题思路：首先需要计算出母牛当前所在频道和目标频道之间的距离，然后根据每首歌曲的长度来判断是否需要调整频道。

最后将所有歌曲的长度相加即可得到最少时间。

2. Sleepy Cow Sorting这个问题描述了一些牛在一行上排队睡觉，但它们总是在不断地调整位置，直到所有牛都按照顺序排好。

需要计算最少需要多少次调整位置才能使所有牛按照从小到大的顺序排列。

解题思路：可以通过编写一个排序算法来模拟牛的位置调整过程，每次调整位置时计数器加一，直到所有牛都按照顺序排列为止。

最后输出计数器的值即为最少次数。

3. Mooyomooyo这个问题描述了一个由"0"和"1"组成的矩阵，其中相邻的"1"可以被合并为一个整体。

给定一个矩阵和一个整数K，需要将所有相邻的"1"合并后，将矩阵中大于等于K个连续的"1"替换为"0"。

全美数学竞赛流程全美数学竞赛（AMC）是美国数学协会举办的一项重要数学比赛活动，旨在激发学生对数学的兴趣和热爱，提高数学解决问题的能力。

通过AMC的选拔，优秀的学生有机会代表自己所在的学校或地区参加更高级别的数学竞赛，包括AIME（美国初级数学邀请赛）和USAMO （美国数学奥林匹克竞赛）等。

本文将详细介绍全美数学竞赛的流程和重要环节。

一、比赛阶段全美数学竞赛主要分为AMC 10和AMC 12两个阶段，分别针对初中和高中的学生。

每个阶段的竞赛内容相对应，并在时间和难度上区分。

1. AMC 10AMC 10是针对初中学生的比赛，共有25道选择题，时间限制为75分钟。

题目涵盖了初中各个数学领域，包括代数、几何、数论、概率等。

考生需要通过选择正确的答案，每题得分为1分，错选或不选均不扣分。

AMC 10的难度适中，旨在考察学生的数学基础和解题能力。

2. AMC 12AMC 12是针对高中学生的比赛，同样是25道选择题，时间限制也为75分钟。

AMC 12相对于AMC 10难度更高，题目内容更加复杂和挑战性，要求考生具备更高的数学思维和解题能力。

得分方式和AMC 10相同，每题得1分，错选或不选不扣分。

二、报名及参赛资格1. 报名学生可以通过学校或教育机构的组织报名参加AMC。

一般来说，学校会提供报名表格供学生填写并缴纳报名费用。

报名费用相对较低，以鼓励更多学生积极参与数学竞赛。

2. 参赛资格AMC针对初中和高中的学生开放报名，参赛学生需符合相应年级要求。

AMC 10主要面向9年级及以下的学生，AMC 12则面向9年级及以上的学生。

在同一个学年中，学生只能参加其中一个阶段的比赛。

获得AMC的优胜者有资格晋级更高级别竞赛，例如AIME和USAMO。

三、考试及阅卷AMC比赛在规定的时间和地点进行，具体考试日期会提前告知参赛学生。

考试使用标准化的试卷，并由指定的监考人员负责监督考试过程。

考试结束后，学生将试卷交回，并由专业人员进行阅卷和评分。