2013年AMC_12真题(A)
2013E试卷(澳大利亚数学竞赛AMC-E：11-12年级中英文历年真题)
THURSDAY 1 AUGUST 2013SENIOR DIVISIONAUSTRALIAN S CHOOL YEARS 11 and 12TIME ALLOWED: 75 MINUTES©AMT P ublishing 2013 AMTT liMiTed Acn 083 950 341A ustrAliAn M AtheMAtics c oMpetitionsponsored by the c oMMonweAlth b AnkAn AcTiviTy of The AusTrAliAn MATheMATics TrusTNAMEYEAR TEACHERA u s T r A l i A n M A T h e M A T i c s T r u s T姓 名： 年 级： 监考老师：意事项一般规定1.未获监考老师许可之前不可翻开此测验题本。
2.各种通讯器材一律不得携入考场，不准使用电子计算器、计算尺、对数表、数学公式等计算器具。
作答时可使用直尺与圆规，以及两面全空白的草稿纸。
3.题目所提供之图形只是示意图，不一定精准。
4.最前25题为选择题，每题有五个选项。
最后题要求填入的答案为000至999的正整数。
题目一般而言是依照越来越难的顺序安排，对于错误的答案不会倒扣分数。
5.本活动是数学竞赛而不同于学校测验，别期望每道题目都会作。
考生只与同地区同年级的其它考生评比，因此不同年级的考生作答相同的试卷将不作评比。
6.请依照监考老师指示，谨慎地在答案卡上填写您的基本数据。
若因填写错误或不详所造成之后果由学生自行负责。
7.进入试场后，须等待监考老师宣布开始作答后，才可以打开题本进行答题。
作答须知1.限用B 或2B 铅笔填写答案。
2.请用B 或2B 铅笔在答案卡上（不是在题本上）将您认为正确选项的圆圈涂满。
3.您的答案卡将由计算机阅卷，为避免计算机误判，请不要在答案卡上其它任何地方涂划任何记号。
填写答案卡时，若需要修改，可使用软性橡皮小心擦拭，并确定答案卡上无残留痕迹。
美国数学竞赛AMC题目及答案
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。
2013年美赛真题题目
现在需要他们的解决方案文件太solutions@为Word或PDF附件的电子邮件提交电子副本（汇总表和解决方案）队（由学生或者指导教师）。
COMAP的提交截止日期为2013年2月4日美国东部时间下午8:00，必须在收到您的电子邮件。
主题行COMAP是你的控制示例：COMAP 11111点击这里下载PDF格式的完整的竞赛说明。
点击这里下载Microsoft Word中的格式汇总表的副本。
*请务必变更控制之前选择打印出来的页面的数量和问题。
团队可以自由选择之间MCM问题MCM问题A，B或ICM问题C.COMAP镜像站点：更多：/undergraduate/contests/mcm/MCM：数学建模竞赛ICM：交叉学科建模竞赛2013年赛题MCM问题问题A：终极布朗尼潘当在一个矩形的锅热烘烤时的4个角落中浓缩，并在拐角处（以及在较小程度上在边缘处）：产品会过头。
在一个圆形盘的热量被均匀地分布在整个外缘和在边缘处的产品不过头。
然而，因为大多数烤炉使用圆形平底锅的形状是矩形的是效率不高的相对于使用在烘箱中的空间。
开发一个模型来显示横跨平底锅平底锅不同形状 - 矩形之间的圆形和其他形状的外边缘的热量分布。
假设1。
的宽度与长度之比的W / L的形状是矩形的烘箱。
2。
- 每个盘必须具有的区域A的3。
最初，两个机架在烤箱，间隔均匀。
建立一个模型，可用于选择最佳的泛类型（形状）在下列情况下：1。
适合在烤箱的锅，可以最大限度地提高数（N）2。
最大限度地均匀分布热量（H），泛3。
优化的组合的条件（1）和（2）式中的权重p和（为1 - p）被分配的结果来说明如何随不同的值的W / L和p。
在除了MCM格式解决方案中，准备一到两页的广告片的新布朗尼美食杂志突出自己的设计和结果。
问题B：水，水，无处不在新鲜的白开水是在世界大部分地区的发展限制约束。
建立一个数学模型，为确定有效的，可行的和具有成本效益的水资源战略于2013年，以满足预计的用水需求，从下面的列表]中选择一个国家，到2025年，确定最佳的水战略。
2013年美国大学生数学建模大赛A题 一等奖
最终的布朗尼蛋糕盘Team #23686 February 5, 2013摘要Summary/Abstract为了解决布朗尼蛋糕最佳烤盘形状的选择问题，本文首先建立了烤盘热量分布模型，解决了烤盘形态转变过程中所有烤盘形状热量分布的问题。
又建立了数量最优模型，解决了烤箱所能容纳最大烤盘数的问题。
然后建立了热量分布最优模型，解决了烤盘平均热量分布最大问题。
最后，我们建立了数量与热量最优模型，解决了选择最佳烤盘形状的问题。
模型一：为了解决烤盘形态转变过程中所有烤盘形状热量分布的问题，我们假设烤盘的任意一条边为半无限大平板，结合第三边界条件下非稳态导热公式，建立了不同形状烤盘的热量分布模型，模拟出不同形状烤盘热量分布图。
最后得到结论：在烤盘由多边形趋于圆的过程中，烤焦的程度会越来越小。
模型二：为了解决烤箱所能容纳最大烤盘数的问题，本文建立了随烤箱长宽比变化下的数量最优模型。
求解得到烤盘数目N 随着烤箱长宽比和烤盘边数n 变化的函数如下：AL W L W cont cont cont N 4n2nsin 1222⎪⎭⎫ ⎝⎛⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛+⋅--=π模型三：本文定义平均热量分布H 为未超过某一温度时的非烤焦区域占烤盘边缘总区域的百分比。
为了解决烤盘平均热量分布最大问题，本文建立了热量分布最优模型，求解得到平均热量分布随着烤箱长宽比和形状变化的函数如下：n sin n cos -n 2nsin 22ntan1H ππδπδπ⎪⎪⎪⎪⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛⋅-=A结论是：当烤箱长宽比为定值时，正方形烤盘在烤箱中被容纳的最多，圆形烤盘的平均热量分布最大。
当烤盘边数为定值时，在长宽比为1:1的烤箱中被容纳的烤盘数量最多，平均热量分布H 最大。
模型四：通过对函数⎪⎭⎫ ⎝⎛n ,L W N 和函数⎪⎭⎫⎝⎛n ,L W H 作无量纲化处理，结合各自的权重p 和()p -1，本文建立了数量和热量混合最优模型，得到烤盘边数n 随p值和LW的函数。
amc数学竞赛试题
amc数学竞赛试题
以下是一道AMC数学竞赛试题：
题目：一个圆锥的高为12厘米，半径为6厘米。
将该圆锥按
照高度为1:2:3的比例划分成三个部分。
求最小那部分的体积。
解析：首先我们需要确定三个部分的高度分别是多少。
根据比例，第一个部分的高度为2厘米，第二个部分的高度为4厘米，第三个部分的高度为6厘米。
接下来，我们需要确定三个部分的半径是多少。
由于圆锥的半径与高成比例，我们可以得到第一个部分的半径为2厘米，第二个部分的半径为4厘米，第三个部分的半径为6厘米。
最后，我们可以计算出三个部分的体积。
第一个部分的体积为
1/3 * π * (2^2) * 2 = 8/3π厘米^3，第二个部分的体积为1/3 * π
* (4^2) * 4 = 64/3π厘米^3，第三个部分的体积为1/3 * π * (6^2) * 6 = 72π厘米^3。
因此，最小的部分的体积为8/3π厘米^3。
2013届12月联考英语试题参考答案
Living a low-carbon style of life is playing a more and more important role in energy saving and environmental protection. To fulfill it, we should bear the following in mind. First, it's important to turn off taps and lights and recycle everything recyclable in our daily life. What's more, while sticking to the habit of setting the temperature of your air conditioner at above 26℃ in summer and eating fruits and vegetables in seasons, we should also climb stairs under the fifth floor instead of taking lifts. Let's all take action now and contribute to the prosperity of our mother－the earth.
Thank you!
This passage is about a little hero. Being an orphan, he lived a hard life in early years. He helped the captain and his soldiers but lost his er listening to “A Little Hero”, I was moved and encouraged .The little hero was homeless. He had neither parents nor brothers nor sisters. He lived a poor and hard life by working for another family in his early years. He didn’t study at school, but he loved his country. Compared with the little hero, I had a happy childhood. I have a happy family with many good books,
2013美国数学竞赛12年级试题A
INSTRUCTIONS1. DO NOT OPEN THIS BOOKLET UNTIL YOUR PROCTOR TELLS YOU.2. This is a twenty-five question multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.3. Mark your answer to each problem on the AMC 12 Answer Form with a #2 pencil. Check the blackened circles for accuracy and erase errors and stray marks completely. Only answers properly marked on the answer form will be graded.4. SCORING: You will receive 6 points for each correct answer, 1.5 points for each problem left unanswered, and 0 points for each incorrect answer.5. No aids are permitted other than scratch paper, graph paper, rulers, compass, protractors, and erasers. No calculators are allowed. No problems on the test will require the use of a calculator.6. Figures are not necessarily drawn to scale.7. Before beginning the test, your proctor will ask you to record certain information on the answer form.8. When your proctor gives the signal, begin working on the problems. You will have 75 minutes to complete the test.9. When you finish the exam, sign your name in the space provided on the Answer Form.© 2013 Mathematical Association of AmericaThe Committee on the American Mathematics Competitions (CAMC) reserves the right to re-examine students before deciding whether to grant official status to their scores. The CAMC also reserves the right to disqualify all scores from a school if it is determined that the required security procedures were not followed.Students who score 120 or above or finish in the top 2.5% on this AMC 10 will be invited to take the 31st annual American Invitational Mathematics Examination (AIME) on Thursday, March 14, 2013 or Wednesday, April 3, 2013. More details about the AIME and other information are on the back page of this test booklet.The publication, reproduction or communication of the problems or solutions of the AMC 12 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via copier, telephone, e-mail, World Wide Web or media of any type during this period is a violation of the competition rules. After the contest period, permission to make copies of problems in paper or electronic form including posting on web-pages for educational use is granted without fee provided that copies are notmade or distributed for profit or commercial advantage and that copies bear the copyright notice.**Administration On An Earlier Date Will Disqualify Your School’s Results**1. All information (Rules and Instructions) needed to administer this exam is contained in the TEACHERS’ MANUAL, which is outside of this package. PLEASE READ THE MANUAL BEFORE FEBRUARY 5, 2013. Nothing is needed from inside this package until February 5.2. Your PRINCIPAL or VICE-PRINCIPAL must verify on the AMC 12 CERTIFICATION FORM (found in the Teachers’ Manual) that you followed all rules associated with the conduct of the exam.3. The Answer Forms must be mailed by trackable mail to the AMC office no later than 24 hours following the exam.4. The publication, reproduction or communication of the problems or solutions of this test during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination at any time via copier, telephone, email, internet or media of any type is a violation of the competition rules.2013AMC 12 A DO NOT OPEN UNTIL TUEsDAy, fEbrUAry 5, 2013The American Mathematics Competitionsare Sponsored byThe Mathematical Association of America – MAA The Akamai Foundation ContributorsAcademy of Applied Sciences – AAs American Mathematical Association of Two-Year Colleges – AMATyC ................................................ American Mathematical Society – AMs ..................................................................................................... American Statistical Association – AsA ............................................................................................... Art of Problem Solving – Awesome Math Casualty Actuarial Society – CAs .......................................................................................................... D.E. Shaw & Co. ........................................................................................................................... Delta Airlines................................................................................................................................... Jane Street Capital Math For America Mu Alpha Theta – MAT ................................................................................................................ National Council of Teachers of Mathematics – NCTM ............................................................................ Pi Mu Epsilon – PME ......................................................................................................................... Society for Industrial and Applied Math (SIAM) 1.Square ABCD has side length 10.Point E is on BC ,and the area of ABE is 40.What is BE ?(A)4(B)5(C)6(D)7(E)8E 2.A softball team played ten games,scoring 1,2,3,4,5,6,7,8,9,and 10runs.They lost by one run in exactly five games.In each of their other games,they scored twice as many runs as their opponent.How many total runs did their opponents score?(A)35(B)40(C)45(D)50(E)553.A flower bouquet contains pink roses,red roses,pink carnations,and red carna-tions.One third of the pink flowers are roses,three fourths of the red flowers are carnations,and six tenths of the flowers are pink.What percent of the flowers are carnations?(A)15(B)30(C)40(D)60(E)704.What is the value of 22014+2201222014−22012?(A)−1(B)1(C)53(D)2013(E)240245.Tom,Dorothy,and Sammy went on a vacation and agreed to split the costs evenly.During their trip Tom paid $105,Dorothy paid $125,and Sammy paid $175.In order to share the costs equally,Tom gave Sammy t dollars,and Dorothy gave Sammy d dollars.What is t −d ?(A)15(B)20(C)25(D)30(E)356.In a recent basketball game,Shenille attempted only three-point shots and two-point shots.She was successful on20%of her three-point shots and30%of her two-point shots.Shenille attempted30shots.How many points did she score?(A)12(B)18(C)24(D)30(E)367.The sequence S1,S2,S3,...,S10has the property that every term beginningwith the third is the sum of the previous two.That is,S n=S n−2+S n−1for n≥3.Suppose that S9=110and S7=42.What is S4?(A)4(B)6(C)10(D)12(E)168.Given that x and y are distinct nonzero real numbers such that x+2x =y+2y,what is xy?(A)14(B)12(C)1(D)2(E)49.In ABC,AB=AC=28and BC=20.Points D,E,and F are on sides AB,BC,and AC,respectively,such that DE and EF are parallel to AC and AB, respectively.What is the perimeter of parallelogram ADEF?(A)48(B)52(C)56(D)60(E)7210.Let S be the set of positive integers n for which1n has the repeating decimalrepresentation0.ab=0.ababab...,with a and b different digits.What is the sum of the elements of S?(A)11(B)44(C)110(D)143(E)15511.Triangle ABC is equilateral with AB =1.Points E and G are on AC and points D and F are on AB such that both DE and F G are parallel to BC .Furthermore,triangle ADE and trapezoids DF GE and F BCG all have the same perimeter.What is DE +F G ?(A)1(B)32(C)2113(D)138(E)5312.The angles in a particular triangle are in arithmetic progression,and the side lengths are 4,5,and x .The sum of the possible values of x equals a +√b +√c ,where a ,b ,and c are positive integers.What is a +b +c ?(A)36(B)38(C)40(D)42(E)4413.Let points A =(0,0),B =(1,2),C =(3,3),and D =(4,0).QuadrilateralABCD is cut into equal area pieces by a line passing through A .This line intersects CD at point (pq ,r s ),where these fractions are in lowest terms.What is p +q +r +s ?(A)54(B)58(C)62(D)70(E)7514.The sequencelog 12162,log 12x,log 12y,log 12z,log 121250is an arithmetic progression.What is x ?(A)125√3(B)270(C)162√5(D)434(E)225√615.Rabbits Peter and Pauline have three offspring—Flopsie,Mopsie,and Cotton-tail.These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child.It is not required that every store gets a rabbit.In how many different ways can this be done?(A)96(B)108(C)156(D)204(E)37216.A ,B ,and C are three piles of rocks.The mean weight of the rocks in A is 40pounds,the mean weight of the rocks in B is 50pounds,the mean weight of the rocks in the combined piles A and B is 43pounds,and the mean weight of the rocks in the combined piles A and C is 44pounds.What is the greatest possible integer value for the mean in pounds of the rocks in the combined pilesB andC ?(A)55(B)56(C)57(D)58(E)5917.A group of 12pirates agree to divide a treasure chest of gold coins among themselves as follows.The k th pirate to take a share takes k 12of the coins that remain in the chest.The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins.How many coins does the 12th pirate receive?(A)720(B)1296(C)1728(D)1925(E)385018.Six spheres of radius 1are positioned so that their centers are at the vertices of a regular hexagon of side length 2.The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon.An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere.What is the radius of this eighth sphere?(A)√2(B)32(C)53(D)√3(E)219.In ABC ,AB =86,and AC =97.A circle with center A and radius AB intersects BC at points B and X .Moreover BX and CX have integer lengths.What is BC ?(A)11(B)28(C)33(D)61(E)7220.Let S be the set {1,2,3,...,19}.For a,b ∈S ,define a b to mean that either 0<a −b ≤9or b −a >9.How many ordered triples (x,y,z )of elements of S have the property that x y ,y z ,and z x ?(A)810(B)855(C)900(D)950(E)98821.ConsiderA =log (2013+log (2012+log (2011+log(···+log (3+log 2)···)))).Which of the following intervals contains A ?(A)(log 2016,log 2017)(B)(log 2017,log 2018)(C)(log 2018,log 2019)(D)(log 2019,log 2020)(E)(log 2020,log 2021)22.A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10with no leading zeros.A 6-digit palindrome n is chosen uniformly at random.What is the probability that n 11is also a palindrome?(A)825(B)33100(C)720(D)925(E)113023.ABCD is a square of side length √3+1.Point P is on AC such that AP =√2.The square region bounded by ABCD is rotated 90◦counterclockwise with center P ,sweeping out a region whose area is 1c (aπ+b ),where a ,b ,and c are positive integers and gcd(a,b,c )=1.What is a +b +c ?(A)15(B)17(C)19(D)21(E)2324.Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon.What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?(A)553715(B)443572(C)111143(D)81104(E)22328625.Let f :C →C be defined by f (z )=z 2+iz +1.How many complex numbersz are there such that Im(z )>0and both the real and the imaginary parts of f (z )are integers with absolute value at most 10?(A)399(B)401(C)413(D)431(E)441WRITE TO US!Correspondence about the problems and solutions for this AMC 12and orders for publications should be addressed to:American Mathematics CompetitionsUniversity of Nebraska, P .O. Box 81606Lincoln, NE 68501-1606Phone 402-472-2257 | Fax 402-472-6087 | amcinfo@The problems and solutions for this AMC 12 were prepared by the MAA’s Committee on theAMC 10 and AMC 12 under the direction of AMC 12 Subcommittee Chair:Prof. Bernardo M. Abrego2013 AIMEThe 31st annual AIME will be held on Thursday, March 14, with the alternate on Wednesday, April 3. It is a 15-question, 3-hour, integer-answer exam. You will be invited to participate only if you score 120 or above or finish in the top 2.5% of the AMC 10, or if you score 100 or above or finish in the top 5% of the AMC 12. T op-scoring students on the AMC 10/12/AIME will be selected to take the 42nd Annual USA Mathematical Olympiad (USAMO) on April 30-May 1, 2013. The best way to prepare for the AIME and USAMO is to study previous exams. Copies may be ordered as indicated below.PUBLICATIONSA complete listing of current publications, with ordering instructions, is at our web site: American Mathematics Competitions。
——12013 AMC 10A Problems
A solid cube of side length is removed from each corner of a solid cube of side length . How many edges does the remaining solid have?Euler's polyhedron f ormula that says that . We know that there originally faces on cube, and creates more. . I creates new vertices original ,yielding vertices. Thus, soTwo sides of a triangle have lengths and . T he length of the altitude to the third side is the average of the lengths of the altitudes to the tw o given sides. How long is the third side?The shortest side length has the longest altitude perpendicular to it. The average of the tw o altitudes given w ill be betw een the lengths of the two altitudes, therefore the length of the side perpendicular to that altitude w ill be betw een and . T he only ans w er c hoice that meets this requirement is .Let the height to the side of length 15 be , the height to the side of length 10 be , the area be , and the height to the unknown side be .Because the area of a triangle is , we get that and , s o, setting them equal, . Now, we know that . Substituting, w e get that.be .A triangle with vertices, , and is ref lected about the line to c reate a second triangle. What is the area of the union of the two triangles?Let be at , B be at , and be at . Reflecting over the line , we seethat , (as the x-coordinate of B is8), and .We see that if w e connect to , we get a line of length (betw een and ). T he area of is equalto .Now, let the point of intersection between and be . I f w e can j ust find the area of and subtract it from 16, w e are done.We realize that because the diagram is symmetric over , the intersection of lines and should intersect at an x-coordinate of . We know that the slope of is. Thus, we can represent the line goingthrough and as. Plugging in , we find that the y-coordinate of F is . T hus, the heightof is. Using the f ormula for the area of a triangle, the area of is.from .Daphne is visited periodically by her three best f riends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every f ourth day, and Claire visits every f ifth day. All three friends visited Daphne yesterday. How many days of the next 365-day period will exactly two f riends visit her?The 365-day time period can be split up into 60-day time periods, because, af ter days, all three of them visit again. (Least commonmultiple of , , and ) Y ou can find how many times each pair of visitors can meet by f inding the LCM of their visiting days and dividing that number by 60 Remember to subtract 1, because you do not wish to count the 60th day, when all three visit.A andB visit times. B andC visit times. C and A visit times.of visits per day is.Let points, , , and . Quadrilateral is cut into equal area pieces by a line passing through . This line intersects at point , where these fractions are in lowest terms. Whatis?First, various area formulas (shoelace, splitting, etc) allow us to find that . Therefore, each equal piece that the lineseparates into must have an area of .Call the point where the line through intersects . We know that . Furthermore, we knowthat , as. T hus, solving for , we find that , so . T his gives that the y coordinate of E is .Line CD can be expressed as, s o the c oordinate of E satis f ies. Solving f or , we findthat .that .In base , the number ends in the digit . I n base , on the other hand, the same number is written as and ends in the digit . For how many digits does the base--representation of end in the digit ?the integers suc that is of . Since , ithas factors. Since cannot equal or , asanswer is后边还有！！！2013 AMC 10A Problems/Problem 20ProblemA unit square is rotated about its center. What is the area of the region swept out by the interior of the square?First, w e need to see what this looks like. Below is a diagram.For this square with side length 1, the distance f rom center to vertex is , hence the area is composed of a semicircle of radius ,plus times a parallelogram w ith height and base . That is to say, the total areais.A group of pirates agree to divide a treasure c hest of gold coins among themselves as f ollow s. The pirate to take a sharetakes of the coins that remain in the chest. The number of coins initially in the chest is the smallest number f or which this arran gement will allow each pirate to receive a positive whole number of coins. How many coins does the pirate receive?Let be the number of coins. Af ter the pirate takes his share, of the original amount is left. Thus, w e know thatmust be an integer. Simplif ying, w e get. Now, the minimal x is the denominator of this fraction multiplied out, obviously. We mentioned bef ore that this product must be an integer. Specif ically, it is an integer and it is the amount thatthe pirate receives, as he receives all of what is remaining.know the denominator is cancelled out, so the number, .Six spheres of radius are positioned so that their centers are at the vertices of a regular hexagon of side length . The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?Set up an isosceles triangle betw een the center of the 8th sphere and two opposite ends of the hexagon. Then set up another t rianglebetw een the point of tangency of the 7th and 8th spheres, and the points of tangency betw een the 7th sphere and 2of the original spheres on opposite sides of the hexagon. Express each side length of the triangles in terms of r (the radius of sphere 8) and h (the height of the f irst triangle). You can then use Pythagorean Theorem to set up tw o equations f or the two triangles, and f ind the values of h and r.In , , and . A circle w ith center and radius intersects at points and . Moreover and have integer lengths. What is ?Let , , and meet the circle at and , with on . Then . Using the Pow er of a Point, we get that. We know that , and that by the triangleinequality on . Thus, we get thatLet represent , and let represent . Since the circle goes through and , . Then by Stew art's Theorem,(Since cannot be equal to , dividing both sides of the equation by is allow ed.)prime f actors of are , , and . Obviously, . In addition, by the TriangleInequality, , so . ore, must equal , and must equalCentral High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play tw o games against each of the other school's players. The match takes place in six rounds, w ith three games played simultaneously in each round. In how many diff erent w ays can the match be scheduled?Credit to the Math Jam f or this solutionLet us label the players of the first team , , and , and those of the second team, , , and .Let us f irst consider how to organize A's matches, , , , , , and . Because w e have three duplicates, there are ways to organize A's matches.Now, consider and . WLOG assume that A's matches were , as w e w ill multiply by the end anyw ays, and that, in the first round, played and played .There are two cases.1. plays again in the second round (and plays in the second round)In this case, the rest of the matches are f orced, as must play in both of rounds and (as it has already played twice) and same with rounds and with and . Thus, there is only one option.2. plays in the second round (and plays in the second round)In this case, can play in either round or and in either round or , so there are options.Thus, with playing in the f irst round, there are options. Multiplying this by for the case where plays in the first round, we get options.multiply waysAll 20 diagonals are draw n in a regular octagon. At how many distinct points in the interior of the octagon (not on the boundary) do tw o or more diagonals intersect?I f you draw a good diagram like the one below, it is easy to see that there are, points.Solution 2 (elimination)Let the number of intersections be. We know that, as every 4 points forms aquadrilateral with intersecting diagonals. However, four diagonals intersect in the center, so we need to subtract from this count.. One might be tempted to choose 65at this point, but one then sees that diagonals like AD, CG, and BE all intersect at the same point.There are of this type with three diagonals intersecting at the same point. We need tosubtract of the(one is kept as the actual intersection), so we get。
明志教育：AMC10、12考试的A和B卷是什么？
明志教育：AMC10/12考试的A/B卷是神马？
大家都知道，AMC10/12考试一直分有A、B卷，但是很多考生都不明白为什么会分两个卷呢？那我们中国的考生要考哪个呢？
AMC试卷命制、批阅地点
AMC总部现设在美国加州內布拉斯加大学林肯校区，AMC测试的试题研发、命制到统一阅卷等工作由AMC主试委员会全权负责。
该委员会由以数理闻名的内布拉斯加大学林肯分校数学系的专家和来自其它全美一流学府，如麻省理工学院（MIT）、哈佛大学（Harvard）、普林斯顿大学（Princeton）等名校的专家组成。
美国区考试
AMC10和AMC12每年会进行效力相同、难度相近的两次考试，这两次考试的试卷即A卷与B卷
两次考试的时间相差一周，美国的考生可以两次考试都参加（例如2017年AMC10/12 A试的考试时间为2月7日，B试的考试时间为2月15日）
考试的形式可以是
AMC10A+AMC10B（两次都考AMC10)
AMC12A+AMC12B（两次都考AMC12)
AMC10A+AMC12B （第一次考AMC10 第二次考AMC12）
AMC12A+AMC10B （第一次考AMC12 第二次考AMC10）
中国区考试
AMC10/12考试时间每年都在中国时间的2月初到2月中旬，与中国的春节时间重合，所以中国区的考试只选择与春节不冲突的一次考试，默认为A试，但是当A试时间与春节冲突（如2017年春节与A试冲突），则中国举办B试。
以上就是AMC10/12分A/B卷考试的具体情况。
想了解更多留学资讯请访问，明志教育——更高效的留学英语培训！。
2013A试卷AMC-A：澳大利亚数学3-4年级中文历年真题
THURSDAY 1 AUGUST 2013MIDDLE PRIMARY DIVISIONAUSTRALIAN S CHOOL YEARS 3 & 4TIME ALLOWED: 60 MINUTESINSTRUCTIONS AND INFORMATION GENERAL 1. Do not open the booklet until told to do so by your teacher. 2. You may use any teaching aids normally available in your classroom, such as MAB blocks, counters, currency, calculators, play money etc. You are allowed to work on scrap paper and teachers may explain the meaning of words in the paper. 3. Diagrams are NOT drawn to scale. They are intended only as aids. 4. There are 25 multiple-choice questions, each with 5 possible answers given and 5 questions that require a whole number answer between 0 and 999. The questions generally get harder as you work through the paper. There is no penalty for an incorrect response. 5. This is a competition not a test; do not expect to answer all questions. You are only competing against your own year in your own State or Region so different years doing the same paper are not compared. 6. Read the instructions on the answer sheet carefully. Ensure your name, school name and school year are entered. It is your responsibility to correctly code your answer sheet. 7. When your teacher gives the signal, begin working on the problems. THE ANSWER SHEET 1. Use only lead pencil.2. Record your answers on the reverse of the answer sheet (not on the question paper) by FULLY colouring the circle matching your answer.3. Your answer sheet will be scanned. The optical scanner will attempt to read all markings even if they are in the wrong places, so please be careful not to doodle or write anything extra on the answer sheet. If you want to change an answer or remove any marks, use a plastic eraser and be sure to remove all marks and smudges.INTEGRITY OF THE COMPETITIONThe AMT reserves the right to re-examine students before deciding whether to grant official status to their score.©AMT P ublishing 2013 AMTT liMiTed Acn 083 950 341A ustrAliAn M AtheMAtics c oMpetitionsponsored by the c oMMonweAlth b AnkAn AcTiviTy of The AusTrAliAn MATheMATics TrusT姓 名： 年 级：监考老师：意事项 一般规定1.未获监考老师许可之前不可翻开此测验题本。
2013年AMC_12真题(B)
8. Line 1 has equation 3x − 2y = 1 and goes through A = (−1, −2). Line 2 has equation y = 1 and meets line 1 at point B. Line 3 has positive slope, goes through point A, and meets 2 at point C. The area of △ABC is 3. What is the slope of 3 ? (A)
USA AMC 12/AHSME 2013 B
1. On a particular January day, the high temperature in Lincoln, Nebraska, was 16 degrees higher than the low temperature, and the average of the high and low temperatures was 3 . In degrees, what was the low temperature in Lincoln that day? (A) − 13 (B) − 8 (C) − 5 (D) 3 (E) 11
16. Let ABCDE be an equiangular convex pentagon of perimeter 1. The pairwise intersections of the lines that extend the side of the pentagon determine a five-pointed star polygon. Let s be the perimeter of the star. What is the difference between the maximum and minimum possible perimeter of s?
2013美国数学竞赛12年级B卷试题
INSTRUCTIONS1. DO NOT OPEN THIS BOOKLET UNTIL YOUR PROCTOR TELLS YOU.2. This is a twenty-five question multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.3. Mark your answer to each problem on the AMC 12 Answer Form with a #2 pencil. Check the blackened circles for accuracy and erase errors and stray marks completely. Only answers properly marked on the answer form will be graded.4. SCORING: You will receive 6 points for each correct answer, 1.5 points for each problem left unanswered, and 0 points for each incorrect answer.5. No aids are permitted other than scratch paper, graph paper, rulers, compass, protractors, and erasers. No calculators are allowed. No problems on the test will require the use of a calculator.6. Figures are not necessarily drawn to scale.7. Before beginning the test, your proctor will ask you to record certain information on the answer form.8. When your proctor gives the signal, begin working on the problems. You will have 75 minutes to complete the test.9. When you finish the exam, sign your name in the space provided on the Answer Form.© 2013 Mathematical Association of AmericaThe Committee on the American Mathematics Competitions (CAMC) reserves the right to re-examine students before deciding whether to grant official status to their scores. The CAMC also reserves the right to disqualify all scores from a school if it is determined that the required security procedures were not followed.Students who score 100 or above or finish in the top 5% on this AMC 12 will be invited to take the 31st annual American Invitational Mathematics Examination (AIME) on Thursday, March 14, 2013 or Wednesday, April 3, 2013. More details about the AIME and other information are on the back page of this test booklet.The publication, reproduction or communication of the problems or solutions of the AMC 12 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via copier, telephone, e-mail, World Wide Web or media of any type during this period is a violation of the competition rules. After the contest period, permission to make copies of problems in paper or electronic form including posting on web-pages for educational use is granted without fee provided that copies are not made ordistributed for profit or commercial advantage and that copies bear the copyright notice.**Administration On An Earlier Date Will Disqualify Your School’s Results**1. All information (Rules and Instructions) needed to administer this exam is contained in the TEACHERS’ MANUAL, which is outside of this package. PLEASE READ THE MANUAL BEFORE FEBRUARY 20, 2013. Nothing is needed from inside this package until February 20.2. Your PRINCIPAL or VICE-PRINCIPAL must verify on the AMC 12 CERTIFICATION FORM (found in the T eachers’ Manual) that you followed all rules associated with the conduct of the exam.3. The Answer Forms must be mailed by trackable mail to the AMC office no later than 24 hours following the exam.4. The publication, reproduction or communication of the problems or solutions of this test during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination at any time via copier, telephone, e-mail, internet or media of any type is a violation of the competition rules.2013AMC 12 B DO NOT OPEN UNTIL WEDNEsDAy, fEbrUAry 20, 2013The American Mathematics Competitionsare Sponsored byThe Mathematical Association of America – MAA The Akamai Foundation ContributorsAcademy of Applied Sciences – AAs American Mathematical Association of Two-Year Colleges – AMATyC ...................................................... American Mathematical Society – AMs ........................................................................................................... American Statistical Association – AsA ...................................................................................................... Art of Problem Solving – Awesome Math Casualty Actuarial Society – CAs ................................................................................................................ D.E. Shaw & Co. ................................................................................................................................. Delta Airlines Jane Street Math For America Mu Alpha Theta – MAT ....................................................................................................................... National Council of Teachers of Mathematics – NCTM ................................................................................... Pi Mu Epsilon – PME ............................................................................................................................... Society for Industrial and Applied Math - SIAM ............................................................................................ 1.On a particular January day,the high temperature in Lincoln,Nebraska,was16degrees higher than the low temperature,and the average of the high and low temperatures was3◦.In degrees,what was the low temperature in Lincoln that day?(A)−13(B)−8(C)−5(D)−3(E)112.Mr.Green measures his rectangular garden by walking two of the sides andfinds that it is15steps by20steps.Each of Mr.Green’s steps is2feet long.Mr.Green expects a half a pound of potatoes per square foot from his garden.How many pounds of potatoes does Mr.Green expect from his garden?(A)600(B)800(C)1000(D)1200(E)14003.When counting from3to201,53is the51st number counted.When countingbackwards from201to3,53is the n th number counted.What is n?(A)146(B)147(C)148(D)149(E)1504.Ray’s car averages40miles per gallon of gasoline,and Tom’s car averages10miles per gallon of gasoline.Ray and Tom each drive the same number of miles.What is the cars’combined rate of miles per gallon of gasoline?(A)10(B)16(C)25(D)30(E)405.The average age of33fifth-graders is11.The average age of55of their parentsis33.What is the average age of all of these parents andfifth-graders?(A)22(B)23.25(C)24.75(D)26.25(E)286.Real numbers x and y satisfy the equation x2+y2=10x−6y−34.What isx+y?(A)1(B)2(C)3(D)6(E)87.Jo and Blair take turns counting from1to one more than the last number saidby the other person.Jo starts by saying“1”,so Blair follows by saying“1,2”.Jo then says“1,2,3”,and so on.What is the53rd number said?(A)2(B)3(C)5(D)6(E)88.Line 1has equation3x−2y=1and goes through A=(−1,−2).Line 2hasequation y=1and meets line 1at point B.Line 3has positive slope,goes through point A,and meets 2at point C.The area of ABC is3.What is the slope of 3?(A)23(B)34(C)1(D)43(E)329.What is the sum of the exponents of the prime factors of the square root of thelargest perfect square that divides12!?(A)5(B)7(C)8(D)10(E)1210.Alex has75red tokens and75blue tokens.There is a booth where Alex cangive two red tokens and receive in return a silver token and a blue token,and another booth where Alex can give three blue tokens and receive in return a silver token and a red token.Alex continues to exchange tokens until no more exchanges are possible.How many silver tokens will Alex have at the end?(A)62(B)82(C)83(D)102(E)10311.Two bees start at the same spot andfly at the same rate in the followingdirections.Bee A travels1foot north,then1foot east,then1foot upwards, and then continues to repeat this pattern.Bee B travels1foot south,then1 foot west,and then continues to repeat this pattern.In what directions are the bees traveling when they are exactly10feet away from each other?(A)A east,B west(B)A north,B south(C)A north,B west(D)A up,B south(E)A up,B west12.Cities A,B,C,D,and E are connected by roadsAB,AD, AE,BC,BD,CD, andDE.How many different routes are there from A to B that use each road exactly once?(Such a route will necessarily visit some cities more than once.)(A)7(B)9(C)12(D)16(E)1813.The internal angles of quadrilateral ABCD form an arithmetic progression.Tri-angles ABD and DCB are similar with∠DBA=∠DCB and∠ADB=∠CBD.Moreover,the angles in each of these two triangles also form an arithmetic pro-gression.In degrees,what is the largest possible sum of the two largest angles of ABCD?(A)210(B)220(C)230(D)240(E)25014.Two non-decreasing sequences of nonnegative integers have differentfirst terms.Each sequence has the property that each term beginning with the third is the sum of the previous two terms,and the seventh term of each sequence is N.What is the smallest possible value of N?(A)55(B)89(C)104(D)144(E)27315.The number2013is expressed in the form2013=a1!a2!···a m!b1!b2!···b n!,where a1≥a2≥···≥a m and b1≥b2≥···≥b n are positive integers and a1+b1is as small as possible.What is|a1−b1|?(A)1(B)2(C)3(D)4(E)516.Let ABCDE be an equiangular convex pentagon of perimeter 1.The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon.Let s be the perimeter of this star.What is the difference between the maximum and the minimum possible values of s ?(A)0(B)12(C)√5−12(D)√5+12(E)√517.Let a ,b ,and c be real numbers such thata +b +c =2,anda 2+b 2+c 2=12.What is the difference between the maximum and minimum possible values of c ?(A)2(B)103(C)4(D)163(E)20318.Barbara and Jenna play the following game,in which they take turns.A number of coins lie on a table.When it is Barbara’s turn,she must remove 2or 4coins,unless only one coin remains,in which case she loses her turn.When it is Jenna’s turn,she must remove 1or 3coins.A coin flip determines who goes first.Whoever removes the last coin wins the game.Assume both players use their best strategy.Who will win when the game starts with 2013coins and when the game starts with 2014coins?(A)Barbara will win with 2013coins,and Jenna will win with 2014coins.(B)Jenna will win with 2013coins,and whoever goes first will win with 2014coins.(C)Barbara will win with 2013coins,and whoever goes second will win with 2014coins.(D)Jenna will win with 2013coins,and Barbara will win with 2014coins.(E)Whoever goes first will win with 2013coins,and whoever goes second will win with 2014coins.19.In triangle ABC ,AB =13,BC =14,and CA =15.Distinct points D ,E ,and F lie on segments BC ,CA ,and DE ,respectively,such that AD ⊥BC ,DE ⊥AC ,and AF ⊥BF .The length of segment DF can be written as m n ,where m and n are relatively prime positive integers.What is m +n ?(A)18(B)21(C)24(D)27(E)3020.For135◦<x<180◦,points P=(cos x,cos2x),Q=(cot x,cot2x),R=(sin x,sin2x),and S=(tan x,tan2x)are the vertices of a trapezoid.What is sin(2x)?(A)2−2√2(B)3√3−6(C)3√2−5(D)−34(E)1−√321.Consider the set of30parabolas defined as follows:all parabolas have as focusthe point(0,0)and the directrix lines have the form y=ax+b with a and b integers such that a∈{−2,−1,0,1,2}and b∈{−3,−2,−1,1,2,3}.No three of these parabolas have a common point.How many points in the plane are on two of these parabolas?(A)720(B)760(C)810(D)840(E)87022.Let m>1and n>1be integers.Suppose that the product of the solutions forx of the equation8(lognx)(log m x)−7log n x−6log m x−2013=0 is the smallest possible integer.What is m+n?(A)12(B)20(C)24(D)48(E)27223.Bernardo chooses a three-digit positive integer N and writes both its base-5and base-6representations on a ter LeRoy sees the two numbers Bernardo has written.Treating the two numbers as base-10integers,he adds them to obtain an integer S.For example,if N=749,Bernardo writes the numbers10,444and3,245,and LeRoy obtains the sum S=13,689.For how many choices of N are the two rightmost digits of S,in order,the same as those of2N?(A)5(B)10(C)15(D)20(E)2524.Let ABC be a triangle where M is the midpoint of AC,and CN is the anglebisector of∠ACB with N on AB.Let X be the intersection of the median BM and the bisector CN.In addition BXN is equilateral and AC=2.What is BN2?(A)10−6√27(B)29(C)5√2−3√38(D)√26(E)3√3−4525.Let G be the set of polynomials of the formP(z)=z n+c n−1z n−1+···+c2z2+c1z+50, where c1,c2,...,c n−1are integers and P(z)has n distinct roots of the form a+ib with a and b integers.How many polynomials are in G?(A)288(B)528(C)576(D)992(E)1056WRITE TO US!Correspondence about the problems and solutions for this AMC 12and orders for publications should be addressed to:American Mathematics CompetitionsUniversity of Nebraska, P .O. Box 81606Lincoln, NE 68501-1606Phone 402-472-2257 | Fax 402-472-6087 | amcinfo@The problems and solutions for this AMC 12 were prepared by the MAA’s Committee on theAMC 10 and AMC 12 under the direction of AMC 12 Subcommittee Chair:Prof. Bernardo M. Abrego2013 AIMEThe 31st annual AIME will be held on Thursday, March 14, with the alternate on Wednesday, April 3. It is a 15-question, 3-hour, integer-answer exam. You will be invited to participate only if you score 120 or above or finish in the top 2.5% of the AMC 10, or if you score 100 or above or finish in the top 5% of the AMC 12. T op-scoring students on the AMC 10/12/AIME will be selected to take the 42nd Annual USA Mathematical Olympiad (USAMO) on April 30 - May 1, 2013. The best way to prepare for the AIME and USAMO is to study previous exams. Copies may be ordered as indicated below.PUBLICATIONSA complete listing of current publications, with ordering instructions, is at our web site: American Mathematics Competitions。
13年数学美国赛建模题目中英文
PROBLEM A: The Ultimate Brownie PanWhen baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient with respect to using the space in an oven.Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes - rectangular to circular and other shapes in between.Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.2. Each pan must have an area of A.3. Initially two racks in the oven, evenly spaced.Develop a model that can be used to select the best type of pan (shape) under the following conditions:1. Maximize number of pans that can fit in the oven (N)2. Maximize even distribution of heat (H) for the pan3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different values of W/L and p.In addition to your MCM formatted solution, prepare a one to two page advertising sheet for the new Brownie Gourmet Magazine highlighting your design and results.烘烤时，在一个长方形的锅热集中在4角和产品得到了在角落（以及在较小程度上的边缘处）。
amc12数学竞赛试题2023
amc12数学竞赛试题2023AMC12数学竞赛试题2023 AMC12数学竞赛是美国数学协会（AMC）举办的一项年度数学竞赛，旨在选拔优秀的高中生，通过解决一系列挑战性的数学问题来测试他们的数学能力。
下面将介绍2023年的AMC12数学竞赛试题。
第一题： 设正整数$a$、$b$满足$a+b=2023$，则$a \cdot b$的最大值是多少？第二题： 定义$S = 1 + 2 + 3 + \cdots + n$，其中$n$是一个正整数。
如果$S$的两位数形式为$xy$（$x$和$y$分别表示十位和个位数字），求$x-y$的值。
第三题： 若$f(x) = \frac{(x+1)(x+2)(x+3)(x+4)}{4}$，求$f(2023)-f(2022)$的值。
第四题： 已知整数$a$满足$2^{2023} \equiv a \pmod {100}$，求$a$的最小正整数值。
第五题： 在等边三角形$ABC$内部任取一点$P$，连接$PA, PB, PC$与边界交点分别为$D, E, F$，证明$AD+BE+CF$的最小值同时也是最大值，并找出该值。
第六题： 已知$0 < x < y < 1$，计算$(0.001+x)^{-1}-(0.001+y)^{-1}$的值。
第七题： 已知平面上的正方形$EFGH$的边长为$5$，点$O$为正方形中心。
若点$ABCD$为正方形边上的四个点，且满足$\angle EAB = \angle ECD = \frac{\pi}{6}$，求$\angle DOA$的度数。
第八题： 已知$1 \leq a,b,c,d \leq 9$，求满足$\frac{a}{b}+\frac{c}{d}=1$的所有不同的整数解$(a,b,c,d)$的个数。
第九题： 设函数$f(x) = \sum_{n=0}^{\infty} \frac{2^n}{n!}(x-1)^n$，求$f(2)$的值。
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1. Square ABCD has side length 10. Point E is on BC , and the area of △ ABE is 40. What is BE? (A)4 (B)5 (C)6 (D)7 (E)8
2. A softball team played ten games, scoring 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 runs. They lost by one run in exactly five games. In each of the other games, they scored twice as many runs as their opponent. How many total runs did their opponents score? (A) 35 (B) 40 (C) 45 (D) 50 (E) 55
2 2 y , what is xy? x y
8. Given that x and y are distinct nonzero real numbers such that x (A)
1 4
(B)
1 2
(C) 1
(D) 2
(E) 4
9. In △ABC, AB = AC = 28 and BC = 20. Points D, E, and F are on sides AB, BC , and
15. Rabbits Peter and Pauline have three offspringFlopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done? (A) 96 (B) 108 (C) 156 (D) 204 (E) 372
2
(B)
3 2
(C)
5 3
(D)
3
(E) 2
19. In △ABC, AB = 86, and AC = 97. A circle with center A and radius AB intersects BC at points B and X. Moreover BX and CX have integer lengths. What is BC? (A) 11 (B) 28 (C) 33 (D) 61 (E) 72
7. The sequence S1 , S 2 , S3 ,, S10 has the property that every term beginning with the third is the sum of the previous two. That is, for n ≥ 3. S n S n 2 S n 1 Suppose that S9 110 and S7 42 . What is S 4 ? (A) 4 (B) 6 (C) 10 (D) 12 (E) 16
4. What is the value of (A) − 1 (B) 1
22014 22012 ? 22014 22012
(C) 53
(D) 2013
(E) 24024
5. Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175. In order to share the costs equally, Tom gave Sammy t dollars, and Dorothy gave Sammy d dollars. What is t − d? (A) 15 (B) 20 (C) 25 (D) 30 (E) 35
6. In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on 20% of her three-point shots and 30% of her two-point shots. Shenille attempted 30 shots. How many points did she score? (A) 12 (B) 18 (C) 24 (D) 30 (E) 36
13. Let points A = (0, 0), B = (1, 2), C = (3, 3), and D = (4, 0). Quadrilateral ABCD is cut
p r into equal area pieces by a line passing through A. This line intersects CD at point q,s ,
piles A and B is 43 pounds, and the mean weight of the rocks in the combined piles A and C is 44 pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles B and C? (A) 55 (B) 56 (C) 57 (D) 58 (E) 59
18. Six spheres of radius 1 are positioned so that their centers are at the vertices of a regular hexagon of side length 2. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere? (A)
16. A, B, C are three piles of rocks. The mean weight of the rocks in A is 40 pounds, the mean weight of the rocks in B is 50 pounds, the mean weight of the rocks in the combined
17. A group of 12 pirates agree to divide a treasure chest of gold coins among themselves as follows. The k th pirate to take a share takes
k of the coins that remain in the chest. The 12
3. A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations? (A) 15 (B) 30 (C) 40 (D) 60 (E) 70
where these fractions are in lowest terms. What is p + q + r + s? (A) 54 (B) 58 (C) 62 (D) 70 (E) 75
14. The sequence log12 162, log12 x, log12 y, log12 z , log12 1250 is an arithmetic progression. What is x? (A) 125 3 (B) 270 (C) 162 5 (D) 434 (E) 225 6
AC , respectively, such that DE and EF are parallel to AC and AB, respectively.
What is the perimeter of parallelogram ADEF?
(A) 48
(B) 52
(C) 56
(D) 60
(E) 72
(A) 1
(B)
3 2
(C)
21 13
(D)
13 8
(E)
5 3
12. The angles in a particular triangle are in arithmetic progression, and the side lengths are 4, 5,x. The sum of the possible values of x equals a b c where a,b, and c are positive integers. What is a + b + c? (A) 36 (B) 38 (C) 40 (D) 42 (E) 44