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amc考试题及答案1. 题目：一个圆的半径是5厘米，求这个圆的面积。
答案：圆的面积公式为 \( A = \pi r^2 \)，其中 \( r \) 是圆的半径。
将半径 \( r = 5 \) 厘米代入公式，得到 \( A = \pi \times5^2 = 25\pi \) 平方厘米。
2. 题目：如果一个数列的前三项是2，4，8，求这个数列的第四项。
答案：这个数列是一个等比数列，公比为2（因为 \( 4/2 = 2 \) 和\( 8/4 = 2 \)）。
所以第四项是 \( 8 \times 2 = 16 \)。
3. 题目：一个长方体的长、宽、高分别是10厘米、8厘米和6厘米，求这个长方体的体积。
答案：长方体的体积公式为 \( V = l \times w \times h \)，其中\( l \) 是长，\( w \) 是宽，\( h \) 是高。
将长 \( l = 10 \)厘米、宽 \( w = 8 \) 厘米、高 \( h = 6 \) 厘米代入公式，得到\( V = 10 \times 8 \times 6 = 480 \) 立方厘米。
4. 题目：一个等差数列的前两项分别是3和7，求这个数列的第五项。
答案：等差数列的公差 \( d \) 可以通过第二项减去第一项得到，即\( d = 7 - 3 = 4 \)。
等差数列的第 \( n \) 项公式为 \( a_n =a_1 + (n-1)d \)，其中 \( a_1 \) 是第一项。
将 \( n = 5 \)，\( a_1 = 3 \)，\( d = 4 \) 代入公式，得到 \( a_5 = 3 + (5-1)\times 4 = 3 + 16 = 19 \)。
5. 题目：如果一个函数 \( f(x) = 2x^2 - 3x + 5 \)，求 \( f(2) \) 的值。
答案：将 \( x = 2 \) 代入函数 \( f(x) = 2x^2 - 3x + 5 \)，得到 \( f(2) = 2 \times 2^2 - 3 \times 2 + 5 = 8 - 6 + 5 = 7 \)。
amc10英语真题及答案新课标AMC10是美国数学竞赛（American Mathematics Competitions）的10年级级别，面向10年级及以下的学生。
以下是一套模拟的AMC10英语真题及答案，请注意，这只是一个示例，并非真实的AMC10题目。
AMC10 英语真题及答案Part 1: Multiple Choice1. The word "innovate" is most closely related to which of the following?A. InnovatorB. InventionC. InventionistD. InnovatoryAnswer: D. Innovatory2. Which of the following is the correct spelling of the word meaning "to make something new"?A. InnovateB. InnovateC. InnovateD. InnovateAnswer: A. Innovate3. The phrase "a leap of faith" is used to describe:A. A large jumpB. A risky decisionC. A new religionD. A sudden increaseAnswer: B. A risky decision4. In the sentence "The company is looking to streamline its operations," the word "streamline" means:A. To make more expensiveB. To make more efficientC. To make more complicatedD. To make more visibleAnswer: B. To make more efficient5. The word "altruistic" is an antonym for:A. SelfishB. AltruismC. AltruisticallyD. AltruistAnswer: A. SelfishPart 2: Fill in the Blanks6. The scientist was awarded the Nobel Prize for his _______ contributions to the field of physics.Answer: innovative7. The _______ of the old building was a significantachievement for the preservation society.Answer: renovation8. The _______ of the new policy was met with mixed reactions from the public.Answer: implementation9. The _______ of the company's profits was due to a series of successful marketing campaigns.Answer: increase10. The _______ of the ancient ruins provided valuable insights into the history of the civilization.Answer: excavationPart 3: Reading ComprehensionRead the following passage and answer the questions that follow.Passage:In recent years, there has been a significant increase in the number of people who are interested in sustainable living. This trend has led to the development of various eco-friendly products and practices. One such practice is the use of solar panels to generate electricity. Solar panels are becoming more popular due to their ability to harness the power of the sun and convert it into usable energy.Questions:11. What is the main topic of the passage?Answer: Sustainable living and the use of solar panels.12. Why are solar panels becoming more popular?Answer: Because they can harness the power of the sun and convert it into usable energy.13. What is the trend mentioned in the passage?Answer: An increase in the number of people interested in sustainable living.Part 4: Vocabulary in Context14. The _______ of the old factory was a major concern for the environmentalists.Answer: pollution15. The _______ of the new technology was celebrated by the scientific community.Answer: advancement16. The _______ of the endangered species was a top priority for the conservation organization.Answer: preservation17. The _______ of the ancient artifact provided evidence ofa previously unknown civilization.Answer: discovery18. The _______ of the new policy was met with skepticism by some members of the community.Answer: enforcement请注意，AMC10是一个数学竞赛，通常不包含英语题目。
2023AMC10美国数学竞赛A卷1. A cell phone plan costs $20 each month, plus 5¢ per text message sent, plus 10¢ for each minute used over 30 hours. In January Michelle sent 100 text messages and talked for 30.5 hours. How much did she have to pay?(A) $24.00 (B) $24.50 (C) $25.50 (D) $28.00 (E) $30.002. A small bottle of shampoo can hold 35 milliliters of shampoo, Whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?(A) 11 (B) 12 (C) 13 (D) 14 (E) 153. Suppose [a b] denotes the average of a and b, and {a b c} denotes the average of a, b, and c. What is {{1 1 0} [0 1] 0}?(A) (B)(C)(D) (E)4. Let X and Y be the following sums of arithmetic sequences:X= 10 + 12 + 14 + …+ 100.Y= 12 + 14 + 16 + …+ 102.What is the value of(A) 92 (B) 98 (C) 100 (D) 102 (E) 1125. At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of 12, 15, and 10 minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?(A) 12 (B) (C) (D) 13 (E) 146. Set A has 20 elements, and set B has 15 elements. What is the smallest possible number of elements in A∪B, the union of A and B?(A) 5 (B) 15 (C) 20 (D) 35 (E) 3007. Which of the following equations does NOT have a solution?(A) (B) (C)(D) (E)8. Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?(A) 20 (B) 30 (C) 40 (D) 50 (E) 609. A rectangular region is bounded by the graphs of the equations y=a, y=-b, x=-c, and x=d, where a, b, c, and d are all positive numbers. Which of the following represents the area of this region?(A) ac + ad + bc + bd (B) ac – ad + bc – bd (C) ac + ad – bc – bd(D) –ac –ad + bc + bd (E) ac – ad – bc + bd10. A majority of the 20 students in Ms. Deameanor’s class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $17.71. What was the cost of a pencil in cents?(A) 7 (B) 11 (C) 17 (D) 23 (E) 7711. Square EFGH has one vertex on each side of square ABCD. Point E is on AB with AE=7·EB. What is the ratio of the area of EFGH to the area of ABCD?(A) (B)(C)(D) (E)12. The players on a basketball team made some three-point shots, some two-point shots, some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team’s total score was 61 points. How many free throws did they make?(A) 13 (B) 14 (C) 15 (D) 16 (E) 1713. How many even integers are there between 200 and 700 whose digits are alldifferent and come from the set {1, 2, 5, 7, 8, 9}?(A) 12 (B)20 (C)72 (D) 120 (E) 20014. A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle’s circumference?(A) (B)(C)(D) (E)15. Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged55 miles per gallon. How long was the trip in miles?(A) 140 (B) 240 (C) 440 (D) 640 (E) 84016. Which of the following in equal to(A) (B) (C) (D) (E)17. In the eight-term sequence A, B, C, D, E, F, G, H, the value of C is 5 and the sum of any three consecutive terms is 30. What is A + H?(A) 17 (B) 18 (C) 25 (D) 26 (E) 4318. Circles A, B, and C each have radius 1. Circles A and B share one point of tangency. Circle C has a point of tangency with the midpoint of AB. What is the area inside Circle C but outside Circle A and Circle B?(B) (C) (D) (E)(A)19. In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2023, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town’s popu lation during this twenty-year period?(A) 42 (B) 47 (C) 52 (D) 57 (E) 6220. Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?(A) (B) (C) (D) (E)21. Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?(A) (B) (C) (D) (E)22. Each vertex of convex pentagon ABCDE is to be assigned a color. There are 6 colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?(A) 2500 (B) 2880 (C) 3120 (D) 3250 (E) 375023. Seven students count from 1 to 1000 as follows:·Alice says all the numbers, except she skips the middle number in each consecutive group of thre e numbers. That is Alice says 1, 3, 4, 6, 7, 9, …, 997, 999, 1000.·Barbara says all of the numbers that Alice doesn’t say, except she also skips the middle number in each consecutive grope of three numbers.·Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers. ·Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.·Finally, George says the only number that no one else says.What number does George say?(A) 37 (B) 242 (C) 365 (D) 728 (E) 99824. Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?(A) (B) (C) (D) (E)an integer. A point X in the interior of R is25. Let R be a square region andcalled n-ray partitional if there are n rays emanating from X that divide R into N triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?(A) 1500 (B) 1560 (C) 2320 (D) 2480 (E) 25002023AMC10美国数学竞赛A卷1. 某通讯公司手机每月基本费为20美元, 每传送一则简讯收 5美分（一美元=100 美分）。
2003A M C10 B 1、Which of the following is the same asSolution2、Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs more than a pink pill, and Al’s pills cost a total of for the two weeks. How much does one green pill costSolution3、The sum of 5 consecutive even integers is less than the sum of the rst consecutive odd counting numbers. What is the smallest of the even integersSolution4、Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the gure. She plants one flower per square foot in each region. Asters cost 1 each, begonias 1.50 each, cannas 2 each, dahlias 2.50 each, and Easter lilies 3 each. What is the least possible cost, in dollars, for her gardenSolution5、Moe uses a mower to cut his rectangular -foot by -foot lawn. The swath he cuts is inches wide, but he overlaps each cut by inches tomake sure that no grass is missed. He walks at the rate of feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawnSolution.6、Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is . The horizontal length of a “-inch” television screen is closest, in inches, to which of the followingSolution7、The symbolism denotes the largest integer not exceeding . For example. , and . ComputeSolution.8、The second and fourth terms of a geometric sequence are and . Which of the following is a possible first termSolution9、Find the value of that satisfies the equationSolution10、Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times is the number of possible license plates increasedSolution11、A line with slope intersects a line with slope at the point . What is the distance between the -intercepts of these two linesSolution12、Al, Betty, and Clare split among them to be invested in different ways. Each begins with a different amount. At the end of one year they have a total of . Betty and Clare have both doubled their money, whereas Al has managed to lose . What was Al’s original portionSolution.13、Let denote the sum of the digits of the positive integer . For example, and . For how many two-digit values of isSolution14、Given that , where both and are positive integers, find the smallest possible value for .Solution15、There are players in a singles tennis tournament. The tournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest players are given a bye, and the remaining players are paired off to play. After each round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played isSolution16、A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that the restaurant should offer so that a customer could have a different dinner each night in the yearSolution.17、An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly ll the cone. Assume that the melted ice cream occupies of the volume of the frozen ice cream. What is the ratio of the cone’s height to its radiusSolution18、What is the largest integer that is a divisor offor all positive even integersSolution19、Three semicircles of radius are constructed on diameter of a semicircle of radius . The centers of the small semicircles divide into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicirclesSolution20、In rectangle , and . Points and are on so that and . Lines and intersect at . Find the area of .Solution21、A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacementsSolution22、A clock chimes once at minutes past each hour and chimes on the hour according to the hour. For example, at 1 PM there is one chime and at noon and midnight there are twelve chimes. Starting at 11:15 AM on February , , on what date will the chime occurSolution23、A regular octagon has an area of one square unit. What is the area of the rectangleSolution24、The rst four terms in an arithmetic sequence are , , , and, in that order. What is the fth termSolution25、How many distinct four-digit numbers are divisible by and have as their last two digitsSolution。
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 。
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?There are rows of seats, giving seats.Problem 2How many two-digit positive integers have at least one 7 as a digit?Ten 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 made48 free throws. How many free throws did she make at the first practice?At 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 .Problem 5In the expression , the values of , , , and are , , , and , although not necessarily in that order. What is the maximum possible value of the result?If 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?Using 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 ?Solution 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?The directions "southwest" and "southeast" are orthogonal. Thus the described situation is a right triangle with legs 8 miles and 10 mileslong. The hypotenuse length is , and thus the answer is .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?The 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?The 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?Solution 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?The area of the large circle is , the area of the small one is , hence the shaded area is .From the 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?All 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?We 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 of blue marbles, there will be blue and other marbles, hence bluemarbles 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?Solution 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?The 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?Solution 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 ?First 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?Solution 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 alsoFrom 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 is ?Solution (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 ?The 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?This 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?Label 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 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 ?Problem 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?The 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 .。
2003 AMC 10B1、Which of the following is the same asSolution2、Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs more than a pinkpill, and Al’s pills cost a total of for the two weeks. How much doesone green pill cost?Solution3、The sum of 5 consecutive even integers is less than the sum of thefirst consecutive odd counting numbers. What is the smallest of theeven integers?Solution4、Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost 1 each,begonias 1.50 each, cannas 2 each, dahlias 2.50 each, and Easterlilies 3 each. What is the least possible cost, in dollars, for hergarden?Solution5、Moe uses a mower to cut his rectangular -foot by -foot lawn.The swath he cuts is inches wide, but he overlaps each cut byinches to make sure that no grass is missed. He walks at the rate of feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn?Solution.6、Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is . The horizontal length ofa “-inch” television screen is closest, in inches, to which of thefollowing?Solution7、The symbolism denotes the largest integer not exceeding . Forexample. , and . ComputeSolution.8、The second and fourth terms of a geometric sequence are and .Which of the following is a possible first term?Solution9、Find the value of that satisfies the equationSolution10、Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times is the number of possible license plates increased?Solution11、A line with slope intersects a line with slope at the point .What is the distance between the -intercepts of these two lines?Solution12、Al, Betty, and Clare split among them to be invested indifferent ways. Each begins with a different amount. At the end of one year they have a total of . Betty and Clare have both doubledtheir money, whereas Al has managed to lose . What was Al’soriginal portion?Solution.13、Let denote the sum of the digits of the positive integer . Forexample, and . For how many two-digitvalues of is ?Solution14、Given that , where both and are positive integers,find the smallest possible value for .Solution15、There are players in a singles tennis tournament. Thetournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest players aregiven a bye, and the remaining players are paired off to play. Aftereach round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played isSolution16、A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that the restaurant should offer so that a customer could have a different dinner each night in the year ?Solution.17、An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies of the volume of the frozen ice cream. What is theratio of the cone’s height to its radius?Solution18、What is the largest integer that is a divisor offor all positive even integers ?Solution19、Three semicircles of radius are constructed on diameter of asemicircle of radius . The centers of the small semicircles divideinto four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside thesmaller semicircles?Solution20、In rectangle , and . Points and are onso that and . Lines and intersect at . Findthe area of .Solution21、A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?22、A clock chimes once at minutes past each hour and chimes onthe hour according to the hour. For example, at 1 PM there is one chime and at noon and midnight there are twelve chimes. Starting at11:15 AM on February , , on what date will the chimeoccur?Solution23、A regular octagon has an area of one square unit.What is the area of the rectangle ?Solution24、The first four terms in an arithmetic sequence are , , ,and , in that order. What is the fifth term?Solution25、How many distinct four-digit numbers are divisible by and haveas their last two digits?Solution。
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 。
2021 AMC 12A Peeyush Pandaya et al.February 20211 Answer1.B2. D3. D4.D5.E Key6.C7.D8.C9.C10.E11.C12.A13.B14.E15. D16.C17.D18.E19.C20.B21.A22.D23.D24.D25.E2 Problems and SolutionsProblem 1. What is the value of21+2+3- (2¹+2²+23)?(A)0 (B)50 (C)52 (D)54 (E)57Solution.2⁶-(2¹+2²+23)=64-(2+4+8)=64-14=(B)50Problem 2.Under what conditionsisva²+62=a+btrue,whereaand bare real numbers?(A)It is never true(B) It is true if and only ifab=0(C) It is true if and only ifa+b≥0(D)It is true ifandonlyifab=0anda+b≥0(E)It is always trueSolution. It is clear that both sides of the equation must be nonnegative.Squaring,a²+b²=a²+2ab+b2→ab=0The answer is (D)Problem 3.The sum of two natural numbers is 17,402.One of the two numbers is divisible by 10.If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?(A)10,272 (B)11,700 (C)13,362 (D)14,238 (E)15,426Solution. Let the first number mentioned be 10n;the second is n. Then10n+n=17,402,from which it follows thatProblem 4. Tom has a collection of 13 snakes,4 of which are purple and 5 of which are happy. He observes that●all of his happy snakes can add,·none of his purple snakes can subtract●all of his snakes that can't subtract also can't add.Which of these conclusions can be drawn about Tom's snakes?Solution. Together, the second and third conditions imply that none of Tom's purple snakes can add.Thus,(D) is correct: happy snakes are not purple.Problem 5.When a student multiplied the number 66 by the repeating decimal,where a and b are digits, he did not notice the notation and just multiplied 66 times 1.ab. Later he found that his answer is 0.5 less than the correct answer. What is the 2-digit integer gb? (A)15 (B)30 (C)45 (D)60 (E)75Solution. The student computed 66 ; the correct answer is 66 . Thus,Problem 6.A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is . When 4 black cards are added to the deck, the probability of choosing red becomes . How many cards were in the deck originally?(A)6 (B)9 (C) 12 (D)15 (E)18Solution. If the deck begins with x red cards and 3x cards in total, thenProblem 7.What is the least possible value of(ry- 1)²+(x+y)2 for real numbersa and y?Solution. We have(ry- 1)2+(x+y)2=(ry)2-2ry+1+x²+2ry+v²=x2v²+z²+y2+1= (r²+1)(y²+1),which achieves a minimum of (D)1 atx=y=0.D Do D ₁ D ₂D ₃ D ₁ D ₅D ₆ D ₇ D ₈ D, D1o Problem 8.A sequence of numbers isdefinedbyDo=0,D ₁=0,Dz=1,andDn=Dn- 1+Dn-3 forn≥3.What are the parities(evenness oroddness)of the triple of numbers(D2021,D2022,D2023), whereE denotes even and O denotes odd?Solution.0/10 01 D2+Do=1+0=1 D ₃+Di=1+0=1D ₄+Dz=1+1=0 Ds+D ₃=0+1=1D ₆+D4=1+1=0D-+Ds=0+0=0D ₈+D ₆=0+1=1Dg+D ₇=1+0=1We can see that the pattern repeats in cycles of length7.and as 2021=5 mod7,we have D2021= Ds,D2022=D6,D2023=D7→(C)(E,O,E) Problem 9.Which of the following is equivalent to(2+3)(2²+3²)(2⁴+34)(2⁸+3⁸)(216+316)(2³²+3³2)(264+364)?(A)3127+2127 (B)3127+2127+2.363+3.263 (C)3128-2128 (D)3128+2128 (E)5127Solution.(3-2)(2+3)(2²+3²)(2⁴+3⁴)(2⁸+3⁸)(216+316)(2³²+3³2)(264+364)=(3²-2²)(2²+3²)(2⁴+3⁴)(2⁸+3⁸)(216+316)(232+3³2)(264+364)=(3⁴-2⁴)(2⁸+3⁸)(216+316)(232+332)(264+364)=(C)3128-2129=(364-264)(264+364)Problem 10.Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are 3cm and 6cm. Into each cone is dropped a spherical marble of radius lcm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level the narrow cone to the rise of the liquid level in the wide cone?(A)1:1 (B)47:43 (C) 2:1 (D)40:13 (E)4:1Solution. The two cones have equal volume, so the height of the first is times that of the second. Since the volumes increase by equal proportions, the heights increase by equal proportions. Thus, the ratio of the rise in liquid levels is (E)4:1Problem 11.A laser is placed at the point(3,5). The laser beam travels in a straight rry wants the beam to hit and bounce off the y-axis, then hit and bounce off the x-axis,then hit the point (7,5). What is the total distance the beam will travel along this path?(A)2√10 (B)5√2(C)10√2(D)15√2 (E)10√5Solution. Reflect about the y-axis then the z-axis. It is well-known that the image under the two reflections must be a straight line.The answer isv(3-(-7))²+(5-(-5)= (C) 10√2Problem 12.All the roots of polynomialz6- 10z ⁵+Az ⁴+Bz³+Cz²+Dz+16are positiveintegers, possibly repeated. What is the value of B?(A)-88 (B)-80 (C)-64 (D)-41 (E)-40Solution. By Vieta's,the sum of the 6 roots is 10 and the product is 16,hence they are all powers of 2. It is not hard to find that the only working unordered sextuple is(2,2,2,2,1,1). As(z-2)4=24-8z3+24z2-32z+16 and(z- 1)2=z2-2z+1.the z3 coefficient is -8.1+24 · (-2)+(-32) · 1= (B)-88Problem 13.Of the following complex numbers z, which has the property that z5 has the greatest real part?(A)-2 (B)-√3+i (C)-√2+√2i Solution. The magnitude of each complex number is the same, so it suffices to look at the argu- ment. The angles are π, ,and ,which after raising to the 5th power give π, and . We seek the angle that reaches farthest to the right(smallest argument),which is . Thus, our answer is (B)-√3+iProblem 14.What is the value of(A)21 (B)100logs3 (C)200log35 (D)2,200 (E)21,000Solution.And,Therefore, their product is210logs3.100log35=(E)21,000(D)- 1+√3i (E)2iProblem 15.A choir director must select a group of singers from among his 6 tenors and 8 basses. The only requirements are that the difference between the numbers of tenors and basses must be a multiple of 4,and the group must have at least one singer.Let be the number of groups that could be selected. What is the remainder when N is divided by 100?(A)47 (B)48 (C)83 (D)95 (E)96Solution. Suppose we mark down(1) the tenors that are in the group,and(2)the basses that aren't in the group. Then we necessarily mark down a number of people that is a multiple of 4. This is also sufficient; we mark down some people numbering a multiple of 4, then select the marked tenors and unmarked basses to form our choir. Clearly, we just mark down at least one person. The answer is thus(D)95Problem 16.In the following list of numbers, the integer n appears times in the list for l≤n≤200.1,2,2,3,3,3,4,4,4,4,.…,200,200,...200.What is the median of the numbers in this list?(A)100.5 (B)134 (C)142 (D)150.5 (E)167Solution.For general n, we have numbers. We want to approximate a such that is close to . Since the formula is a quadratic in n and we are halving this value,we can find that a is approximately .Plugging in n = 200,this is about 100√2,or 141.Of the answer choices, (C)142 is the closest, and indeed it is our answer.To verify, we can see that and ), so clearly 142 works.Problem 17.Trapezoid ABCD has ABICD,BC=CD=43,and AD1BD.Let O be the intersection of the diagonals AC and BD,and let P be the midpoint of BD.Given that OP=11, the length AD can be written in the form myn,where m and n are positive integers and n is not divisible by the square of any prime. What is m+n?(A)65 (B)132 (C)157 (D)194 (E)215Solution. Let M be the intersection of CPand AB.Since DCBMisakite,andCMIBD,we have MP1PB,and by considering the homothethy taking △MBD to △ABD with scale factor 2,we can see that M is the midpoint of AB.In particular,we haveSince AD 1BD,we have AD1DOandthusZADO=90°,andasCD=CB,wehaveCP1BD and ZCPD=2CPO=90°.Also,ZAOD=ZCOP,so △AOD~ △COP.Therefore,so DO=22.Thus,AD=√AB²-BD²=√86²-66²=4√ 190→m+n= (D)194The desired answer isProblem 18.Let f be afunction defined on the set of positive rational numbers with the property that f(a ·b)=f(a)+f(b)for all positive rational numbersa and b.Suppose that falso has the property that f(p)= pfor every prime numberp.For which of the following numbers zis f(x)<0?(A)整(B) (C) (D) (E) 51Solution. Note that f(a ·1)=f(a)+f(1)= f(1)=0,andIn particular, it follows by induction that f(p*)= kp for each k ∈Z.Thus,(A) f(2-5. 17)=-5·2+17=7(B) f(2-4. 11)=-4·2+11=3(C) f(3-2.7)=-2.3+7=1(D)f(2- 1.3- 1.7)=- 1.2+(- 1) ·3+7=2(E)f(52.11- 1)=2.5- 11=- 125The answer is (E)11Problem 19.How many solutions does the equation sinclosed interval [0,π]?(A)0 (B)1 (C) 2 (D) 3 (E)4Solution. Note on the interval , the left-hand side is negative while the right-hand side is positive.We thus restrict our attention to ]. The arguments cosx and sz are both between 0 and .ForsinA=cosBinA,B ∈[0,],we must have . This implies sinz+cosz=1,hence . There are (C) 2 solutions.Problem 20.Suppose that on a parabola with vertexV and focus F there exists a point Asuch that AF=20 and AV=21.What is the sum of all possible values of the length FV?(A)13 (B) 40 (C) (D)14 (E)Solution. Let the directrix be the x-axis,F=(0,2d),V=(0,d),A=(x,y),andB=(0,y)for some d>0.By the definition of a parabola,y=20.We compute x in two ways:x²=AF²-BF²=20²-|20-2d|2=AV²-BV²=21²-|20-d²Subtracting,O=20²-21²+(20-d)²-(20-2d)2=3d²-40d+41.The sum of all possible values ofProblem 21.The five solutions to the equation(z- 1)(z²+2z+4)(z²+4z+6)=0may be written in theformzk+ykiforl≤k≤5,where xk and yk are real.Let E be the unique ellipse that passes through the points(Ti,yi),(x2,32),(r3,Y3),(x4,y4)and(xs,ys).The eccentricity ofE can be written in the form ,where m and n are relatively prime positive integers. What is m+n?(Recall that the eccentrictiy of an ellipse E is the ratio,where 2a is the length of the major axis ofE and 2c is the distance between its two foci.)(A)7 (B)9 (C)11 (D)13 (E)15Solution. The roots of the polynomial arez=1,z=- 1±i√3,andz=-2±i√2,hence the five points onE are(1,0),( - 1,土√3),( - 2,±√2) .By symmetry through the x-axis, the ellipse is of the formE: a(x-r)²+by²=1.We then have the relationsa(1-r)²=1a(1+r)²+3b=1a(2+r)²+2b=1.Eliminating b from the latter two,1=3[a(2+r)²+2]-2[a(1+r)²+36]=a(r²+8r+10),henceit follows that and so the eccentricity is 1/√6. The requested sum is1+6= (A)7Problem 22.Suppose that the roots of the polynomialP(x)=x³+ax²+bx+care cos 2π,COS 47 and cos ,where angles are in radians. What is abc?(C) (D) (E)Solution. Recall1+e2m/7+ …+e12mi/7=0→e2mi/7+e4xi/7+e6mi/7=- 1/2,and in particular caNote are solutions to the equation cosb+cos28+cos30=- 1/2,so lettingx=co sθimplies thathas roots ,COS 4π7Thus, the polynomial in question is and the requested answer isRemark. Perhaps it is easier to motivate the solution as follows.Lett=e2mi/7andx=t+t- 1= 2cos2π/7.Thenx²-2=t²+t-2andx³-3x=t³+t-3.Moreover,t⁶+t⁵+ …+1=0impliest³+t²+t+t- 1+t-2+t-3=- 1,i.e.x+(x²-2)+(r³-3r) has root 2cos2π/7.It certainly seems logical that the Galois conjugates of 2cos would be 2cos and 2cos (especially given the phrasing of the problem),so simply replacex → to get the desired polynomial form.Let w = e2ix/7.Note thatLet these be r,s,t respectively. By Vieta's formulas, note that the desired quantity is(-rst)(rs+st+tr)(-r-s-t)=(r+s+t)(rs+st+tr)(rst).Note that 1+w+ …+w⁶=0.We haveThen,FinallyMultiplying yields the answer of (D)1 32Solution Manual 2021AMC12AProblem 23.Frieda the frog begins a sequence of hops on a 3×3 grid of squares,moving one square on each hop and choosing at random the direction of each hop up,down, left,or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid,she ”wraps around”and jumps to the opposite edge.For example if Frieda begins in the center square and makes two hops ”up”,the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge,landing in the bottom row middle square.Suppose Frieda starts from the center square, makes at most four hops at random,and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?(A)G (B) (D)强(E)8Solution. We complementary count,and determine the probability we never reach a corner square. Denote by A the center,and B a square adjacent to the center. Then the first hop lands on B.● If the second hop lands on A(with probability ,then the third hop lands on B always,andthere is a chance the fourth hop lands on a non-corner square. The probability in this case is● If the second hop lands on B, then there is a chance the third hop lands on A,and achance the third hop lands on B.In the former subcase, the fourth hop always lands on a non-corner square, and in the latter subcase, there is a chance the fourth hop lands on a non-corner square. The probability in this case isThe requested probability isProblem 24.SemicircleT has diameter AB of length14.Circle Ωlies tangent to AB at a point P and intersectsI at pointsQandR.IfQR=3√3and ZQPR=60°,then the area of △PQR is ,where a and c are relatively prime positive integers, and b is a positive integer not divisible by the square of any prime. What isa+b+c?(A)110 (B)114 (C)118 (D)122 (E)126Solution. First, by Extended Law of Sines,we have that the radius of Q is .Let M be the midpoint of QR,O be the center ofT,and X be the center of 2.Since △OQR is isosceles, OM is perpendicular to QR. Thus we have that X lies on OM since it is the circumcenter of △puting lengths, we hav thagorean theorem on △ORM andfrom isosceles triangle XQR. Thus and OP=4 from Pythagorean theorem on △OXP.To find [PQR],we will find the height from P to QR.Let the foot of the perpendicular fromX to the P-altitude be D. Since PDⅡOM,we know that △XDP~△OPX.This means that . Now note that the bottom portion of the P-altitude after subtractingPD is equal to XM,so the height of the triangle is . The area is simply2021 AMC 12A11 Solution ManualProblem 25.Let d(n)denote the number of positive integers that divide n,including l and n. For example,d(1)=1,d(2)=2,and d(12)=6.(This function is known as the divisor function.) LetThere is a unique positive integer N such that f(N)> f(n)for all positive integers n≠N.What is the sum of the digits of N?(A)5 (B)6 (C)7 (D)8 (E) 9Solution. Letn=II;p',wherepi=2,Pz=3,etc.are the primes in increasing order and e; are nonnegative integers. ThenIt is equivalent to maximize Thus, it remains to find the optimal e; for each i. We go term-by-term, noting that it is only necessary to check until the expression first decreases, as exponentials increase more quickly than polynomials.ei 0 1 2 3 4 0 1 2 3 0 1 2 0 1 2 ((ez+1)3)/p⁸1³/20=1 23/2¹=433/2²=6.25 4³/2³=85³/2⁴<8 13/3⁰=1 23/31≈2.67 3³/32=3 43/3³<3 13/5⁰=1 2³/5¹=1.6 3³/5²=1.08 13/70=1 23/71≈1.14 3³/7²<1Note that we do not need to check p≥11,ase;=1yields which is suboptimal.Thus,the answer isN=23.32.5.7=2520,which has a Remark:It is sufficient to stop at p=3,for 3²|Nleaves N among the answer choices.digit sum of (E)9only one possibility for the digit sum ofi 1 1 1 1 1 2 2 2 2 3 3 3 4 4 4。
2004 AMC12 试题1. 爱丽每小时的工资为美金20元，其中的1.45%要缴地方税，试问爱丽每小时的工资中要付地方税美金多少分？(美金1元二美金100分)(A) 0.0029 (B) 0.029 (C) 0.29 (D) 2.9 (E) 292. 于AMC2的测验中，每答对一题可得6分，答错可得0分，未作答可得2.5分.假设凯琳在25题中有8题未作答，她至少要答对几题，总分才会不低于100 分？(A) 11(B) 13(C) 14(D) 16(E) 173.滿足x 2y100的正整数序对(x, y)有多少組？(A) 33(B) 49(C) 50(D) 99(E) 1004.白婆婆有6个女儿、没有儿子，有些女儿也恰有6个女儿，其他的女儿没有孩子. 白婆婆有女儿及外孙女共30位，没有外曾孙女. 试问白婆婆的女儿及外孙女中有多少位没有女儿？(A) 22 (B) 23 (C) 24 (D) 25 (E) 265.直线y mx(A) mb 1 (B) 1 mb 0 (C) mb 0 (D) 0 mb 1 (E) mb 16.设u 2 20042005, V 20042005, W 2003 20042004, X 2 20042004Y 20042004, Z 20042003.試问以下何者值最大？(A) U V (B) V W (C) W X (D) X Y (E) Y Z7. 一种代币的游戏，其规那么如下：每回持有最多代币者须分给其他每一位参与游戏者一枚代币，并放一枚代币於回收桶中；当有一位游戏参与者没有代币时，那么游戏结束.假设A、B C三人玩此游戏，在游戏开始时分别持有15、14及13枚代币.试问游戏从开始到结束，共进行了多少回？(A) 36 (B) 37 (C) 38 (D) 39 (E) 409. 有一个将花生酱装在圆桶状瓶子内出售的公司 .市场研究建议瓶子较粗时可增加销售量.假设瓶子的直径增加25%,而体积仍维持不变，那么 瓶子的高度应减少百分之多少？(A) 10(B) 25(C) 36(D) 50(E) 6010. 有49个连续整数，它的和为75,則它们排在最中间的数为何？(A) 7(B) 72(C) 73(D) 74(E) 7511. 某国的硬币中有1分、5分、10分及25分四种，在保拉的皮包 內硬币的平均值为20分.假设再增加一枚25分的硬币，平均值則增为 21分.試问她的皮包內有多少枚10分的硬币？(A) 0(B) 1(C) 2(D) 3(E) 412. 设A (0, 9) , B (0,12).点A 、B 在直线y x 上, 且AA 与BB 交于点 C (2,8).試问AB的长度是多少？(A) 2(B) 2 2(C) 3(D) 22(E) 3“13. 以S 表示坐标平面上所有的点(a,b)所形成的集合，其中a,b 等于1, 0,或1.試问有多少条相异的直线至少通过集合 S 中的两个点？(A) 8(B) 20(C) 24(D) 27(E) 3614. 三个实数的数列形成一个等差数列，首项是9.假设将第二项加2、第 三项加20可使得这三个数变为等比数列，那么这个等比数列中第三项 最小可能是多少？(A) 1(B) 4(C) 36(D) 49(E) 8115.小美与小雯在一个圆形的跑道上向相反的方向跑 ，开始两人分别从8.如下图EAB 及 ABC 为直角，厢 4 , BC 6 , AE 8, AC 与BE 交于D 点.試问 差为多少？(A) 2(B) 4(C) 5 (D) 8 (E) 9ADE 与BDC 面积之圆形跑道直径的两端起跑.小美跑了 100公尺时她们第一次相遇；在 第一次相遇后小雯跑了 150公尺时她们第二次相遇.假设她们跑的速 度都分别维持固定不变，试问此圆形跑道的长度是多少公尺？(A) 250(B) 300(C) 350(D) 400(E) 50016.使函数 log 2004{log 2003{log 2002{log 2001 X}}}有定义的集合为 之值是多少？(A) 0(B) 20012002(C) 20022003(D) 20032004{xx c}.試问 c(E) 20012°°丹317.函数f 满足以下性质： (i) f(1) 1,且(ii)对任意的正整数n ,f(2n) n f(n).试问f(2100)之值为何？(A) 1(B) 299(C) 2100(D) 24950 18.如下图，ABCD 是边长2的正方形.在正方形 的内部作一个以AB 为直径的半圆，且自C 点引 此半圆的切线交AD 边于E 点•试问CE 的长度是 多少？(E) 5 ■.5(E) 29999(B) 5 (C) 6 (D)?19.如下图，A B C 三圆彼此外切且均内切于圆等，圆A 的半径1且通过圆D 的圆心.试问圆B 的半径是多少?D.B 、C 两圆相(A)3(B)于(C) 土 (D) |20.从0与1之间的数，随机独立取出两实数a 与b,并将a, b 之和记 作c.分别以A, B, C 表示最接近a, b, c 的整数.试问 的机率为何？(A) 1(B) !(C) 1(D) 2(E) 443 23421.假设n 02nCOS5,那么Cos2之值是多少？(A) 15 (B) I (C) ±55(D) ?(E)上5522.三个半径为1的球彼此外切且放置在一水平面上，一个半径为2 的大球放在它们的上面 .试问大球的最高点至平面的距离是多 少？C 2004 0,且 P (x)=0 有 2004个复数根 z a k b k i , 20042004b k 为实数，a ib i 0,且a kb k .k 1k 1试问以下哪一项可能不是 0 ?2004(A) C 0(B)C2003(C) b z b s L b s 004 (D)a kk 124. 设A 、B 为平面上的两点，其中AB 1.令S 为平面上所有半径是1且 能盖住线段AB 之圆的并集，則S 的面积是多少？(A) 2.3(B) -(C) 33(D)卫 3(E) 4 2 33 2325. 对每一个整数n 4,令a n 表示n 进位的循环小数0.133n .把乘积a 4a 5L a 99写成—的形式，其中m P 为正整数，且p 尽可能小.试问mp!之值是多少？(A) 98(B) 101(C) 132(D) 798(E) 962Q屈o J123 52(A) 3(B) 3 -(C) 3 -(D)2349(E) 3 2、223.多项式P(x) GX C 0 ,的系数都是实数, 1 k 2004,其中 a k ,2004(E)C kk 12004C 2004X 2003C 2003XL答案:1 ( E )2 ( C )3 ( B )4 ( E )5 ( B )6 ( A )7 ( B )8 ( B )9 ( C )10 ( C ) 11 ( A )12 ( B )13 ( B )14 ( A )15 ( C ) 16 ( B )17 ( D )18 ( D )19 ( D )20( E ) 21 ( D )22 ( B )23 ( E )24 ( C )25 ( E )。
