AMC美国高中数学竞赛难题精选
美国AMC8数学竞赛试题(含答案)
2001 年 美国AMC8 (2001年 月 日 时间40分钟)1. 卡西的商店正在制作一个高尔夫球奖品。
他必须给一颗高尔夫球面上的300个小凹洞着色， 如果他每着色一个小凹洞需要2秒钟，试问共需多 分钟才能完成他的工作。
(A) 4 (B) 6 (C) 8 (D) 10 (E) 12 。
2. 我正在思考两个正整数，它们的乘积是24且它们的和是11，试问这两个数中较大的数是什 么 。
(A) 3 (B) 4 (C) 6 (D) 8 (E) 12 。
3. 史密斯有63元，艾伯特比安加多2元，而安加所有的钱是史密斯的三分之一，试问艾伯特 有 元。
(A) 17 (B) 18 (C) 19 (D) 21 (E) 23 。
4. 在每个数字只能使用一次的情形下，将1，2，3，4及9作成最小的五位数，且此五位数为 偶数，则其十位数字为 。
(A) 1 (B) 2 (C) 3 (D) 4 (E) 9 。
5. 在一个暴风雨的黑夜里，史努比突然看见一道闪光。
10秒钟后，他听到打雷声音。
声音的速 率是每秒1088呎，但1哩是5280呎。
若以哩为单位的条件下，估计史努比离闪电处的距离 最接近下列何者 。
(A) 1 (B) 121 (C) 2 (D) 221 (E) 3 。
6. 在一笔直道路的一旁有等间隔的6棵树。
第1棵树与第4棵树之间的距离是60呎。
试问第1 棵树到最后一棵树之间的距离是 呎。
(A) 90 (B) 100 (C) 105 (D) 120 (E) 140 。
问题7、8、9请参考下列叙述：主题：竞赛场所上的风筝展览7. 葛妮芙为提升她的学校年度风筝奥林匹亚竞赛的质量，制作了一个小风筝与一个大风筝，并陈列在公告栏展览，这两个风筝都如同图中的形状，葛妮芙将小风筝张贴在单位长为一吋(即每两点距离一吋)的格子板上，并将大风筝张贴在单位长三吋(即每两点距离三吋)的格子板上。
试问小风筝的面积是 平方吋。
(A) 21 (B) 22 (C) 23 (D) 24 (E) 25 。
2020AMC10B（美国数学竞赛）真题加详解
2020AMC10B（美国数学竞赛）真题加详解2020 AMC 10B Solution Problem1What is the value ofSolutionWe know that when we subtract negative numbers, .The equation becomesProblem2Carl has cubes each having side length , and Kate has cubes each having side length . What is the total volume of these cubes?SolutionA cube with side length has volume , so of these will have a total volume of .A cube with side length has volume , so of these will have a total volume of .~quacker88Problem 3The ratio of to is , the ratio of to is , and the ratioof to is . What is the ratio of toSolution 1WLOG, let and .Since the ratio of to is , we can substitute in the value of toget .The ratio of to is , so .The ratio of to is then so our answeris ~quacker88Solution 2We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two., and since , we can link themtogether to get .Finally, since , we can link this again to get: ,so ~quacker88Problem4The acute angles of a right triangle are and , where andboth and are prime numbers. What is the least possible value of ?SolutionSince the three angles of a triangle add up to and one of the anglesis because it's a right triangle, .The greatest prime number less than is . If ,then , which is not prime.The next greatest prime number less than is . If ,then , which IS prime, so we have our answer ~quacker88 Solution 2Looking at the answer choices, only and are coprime to . Testing , the smaller angle, makes the other angle which is prime, therefore our answerisProblem5How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)SolutionLet's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.There are ways to order objects. However, since there's ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and ways to order the green tiles, we have to divide out these possibilities.~quacker88SolutionWe can repeat chooses extensively to find the answer. Thereare choose ways to arrange the brown tiles which is . Then from the remaining tiles there are choose ways to arrange the red tiles. And now from the remaining two tiles and two slots we can see there are two ways to arrange the purple and brown tiles, giving us an answerofProblem6Driving along a highway, Megan noticed that her odometershowed (miles). This number is a palindrome-it reads the same forward and backward. Then hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this -hour period?SolutionIn order to get the smallest palindrome greater than , we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.So we raise to the next largest value, , but obviously, that's not how place value works, so we're in the s now. To keep this a palindrome, our number is now .So Megan drove miles. Since this happened over hours, she drove at mph. ~quacker88 Problem7How many positive even multiples of less than are perfect squares?SolutionAny even multiple of is a multiple of , so we need to find multiples of that are perfect squares and less than . Any solution that we want will be in theform , where is a positive integer. The smallest possible value isat , and the largest is at (where the expression equals ). Therefore, there are a total of possible numbers.-PCChess Problem8 Points and lie in a plane with . How many locations forpoint in this plane are there such that the triangle with vertices , ,and is a right triangle with area square units?Solution 1There are options here:1. is the right angle.It's clear that there are points that fit this, one that's directly to the rightof and one that's directly to the left. We don't need to find the length, we just need to know that it is possible, which it is.2. is the right angle.Using the exact same reasoning, there are also solutions for this one.3. The new point is the right angle.(Diagram temporarily removed due to asymptote error)The diagram looks something like this. We know that the altitude tobase must be since the area is . From here, we must see if there are valid triangles that satisfy the necessary requirements. First of all, because of the area.Next, from the Pythagorean Theorem.From here, we must look to see if there are valid solutions. There are multiple ways to do this:We know that the minimum value of iswhen . In this case, the equationbecomes , which is LESSthan . . The equationbecomes , which is obviously greater than . We canconclude that there are values for and in between that satisfy the Pythagorean Theorem.And since , the triangle is not isoceles, meaning we could reflectit over and/or the line perpendicular to for a total of triangles this case.Solution 2Note that line segment can either be the shorter leg, longer leg or thehypotenuse. If it is the shorter leg, there are two possible points for that cansatisfy the requirements - that being above or below . As such, thereare ways for this case. Similarly, one can find that there are also ways for point to lie if is the longer leg. If it is a hypotenuse, then thereare possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is .Problem9How many ordered pairs of integers satisfy theequationSolutionRearranging the terms and and completing the square for yields theresult . Then, notice that can onlybe , and because any value of that is greater than 1 will causethe term to be less than , which is impossible as must be real. Therefore, plugging in the above values for gives the ordered pairs , , , and gives a totalof ordered pairs.Solution 2Bringing all of the terms to the LHS, we see a quadraticequation in terms of . Applying the quadratic formula, weget In order for to be real, which it must be given the stipulation that we are seekingintegral answers, we know that the discriminant, must benonnegative. Therefore, Here, we see that we must split the inequality into a compound, resultingin .The only integers that satisfy this are . Plugging thesevalues back into the quadratic equation, we see that both produce a discriminant of , meaning that there is only 1 solution for . If , then the discriminant is nonzero, therefore resulting in two solutions for .Thus, the answer is .~TiblisSolution 3, x firstSet it up as a quadratic in terms of y:Then the discriminant is This will clearly only yield real solutionswhen , because it is always positive. Then . Checking each one: and are the same when raised to the 2020th power:This has only has solutions , so are solutions. Next, if :Which has 2 solutions, so andThese are the only 4 solutions, soSolution 4, y firstMove the term to the other side toget . Because for all , then . If or , the right side is and therefore . When , the right side become , therefore . Our solutions are , , , . There are solutions, so the answer is - wwt7535Problem 10A three-quarter sector of a circle of radius inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubicinches?SolutionNotice that when the cone is created, the radius of the circle will become the slant height of the cone and the intact circumference of the circle will become the circumference of the base of the cone.We can calculate that the intact circumference of the circle is . Since that is also equal to the circumference of the cone, the radius of the cone is . We also have that the slant height of the cone is . Therefore, we use the Pythagorean Theorem to calculate that the height of the coneis . The volume of the coneis -PCChessSolution 2 (Last Resort/Cheap)Using a ruler, measure a circle of radius 4 and cut out the circle and then the quarter missing. Then, fold it into a cone and measure the diameter to be 6cm . You can form a right triangle with sides 3, 4, and then through the Pythagorean theorem the height is found tobe . The volume of a cone is . Plugging in we findProblem11Ms. Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?SolutionWe don't care about which books Harold selects. We just care that Bettypicks books from Harold's list and that aren't on Harold's list.The total amount of combinations of books that Betty can selectis .There are ways for Betty to choose of the books that are on Harold's list.From the remaining books that aren't on Harold's list, thereare ways to choose of them.~quacker88Problem12The decimal representation of consists of a string of zeros after the decimal point, followed by a and then several more digits. How many zeros are in that initial string of zeros after the decimal point?Solution 1Now we do some estimation. Notice that , which meansthat is a little more than . Multiplying itwith , we get that the denominator is about . Notice that whenwe divide by an digit number, there are zeros before the first nonzero digit. This means that when we divide by the digit integer , there are zeros in the initial string after the decimal point. -PCChessSolution 2First rewrite as . Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to findthe number of digits in .and memming (alternatively use the factthat ),digits.Our answer is .Solution 3 (Brute Force)Just as in Solution we rewrite as We thencalculate entirely by hand, first doing then multiplying that product by itself, resulting in Because this is digits,after dividing this number by fourteen times, the decimal point is beforethe Dividing the number again by twenty-six more times allows a stringof zeroes to be formed. -OreoChocolateSolution 4 (Smarter Brute Force)Just as in Solutions and we rewrite as We can then look at the number of digits in powersof . , , , , ,, and so on. We notice after a few iterations that every power of five with an exponent of , the number of digits doesn't increase. This means should have digits since thereare numbers which are from to , or digits total. This means our expression can be written as , where is in therange . Canceling gives , or zeroes before the since the number should start on where the one would be in . ~aop2014 Solution 5 (Logarithms)Problem13Andy the Ant lives on a coordinate plane and is currently at facingeast (that is, in the positive -direction). Andy moves unit and thenturns degrees left. From there, Andy moves units (north) and thenturns degrees left. He then moves units (west) and againturns degrees left. Andy continues his progress, increasing his distance each time by unit and always turning left. What is the location of the point at which Andy makes the th leftturn?Solution 1You can find that every four moves both coordinates decrease by 2. Therefore, both coordinates need to decrease by two 505 times. You subtract, giving you theanswer of ~happykeeperProblem14As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?Solution 1Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, , since B is the center of the semicircle with radius 1 that C lies on, , since B is the center of the semicircle with radius 1 that A lies on,and , as a regular hexagon has angles of 120,and is half of any angle in this hexagon. Now, using the sinelaw, , so . Since the angles in a triangle sum to 180, is also 60. Therefore, is an equilateral triangle with side lengths of 1.Since the area of a regular hexagon can be found with the formula , where is the side length of the hexagon, the area of this hexagonis . Since the area of an equilateral triangle can be foundwith the formula , where is the side length of the equilateral triangle,the area of an equilateral triangle with side lengths of 1 is . Since the area of a circle can be found with the formula , the area of a sixthof a circle with radius 1 is . In each sixth of the hexagon, thereare two equilateral triangles colored white, each with an area of , and onesixth of a circle with radius 1 colored white, with an area of . The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixthof the hexagon is , which equals , and the total areacolored white is , which equals . Since the area colored gray equals the total area of the hexagon minus the area colored white,the area colored gray is , whichequals .Solution 2First, subdivide the hexagon into 24 equilateral triangles with side length1:Now note that the entire shadedregion is just 6 times this part:The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of: The arc that is not included has an area of:Hence, the area ofthe shaded region in that section is For a final areaof:Problem15Steve wrote the digits , , , , and in order repeatedly from left to right, forming a list of digits, beginning He thenerased every third digit from his list (that is, the rd, th, th, digits from the left), then erased every fourth digit from the resulting list (that is, the th, th, th, digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions ?Solution 1After erasing every third digit, the list becomes repeated. After erasing every fourth digit from this list, the listbecomes repeated. Finally, after erasing every fifth digit from this list, the list becomes repeated. Since this list repeats every digits andsince are respectively in we have that the th, th, and st digits are the rd, th, and thdigits respectively. It follows that the answer is~dolphin7Problem16Bela and Jenn play the following game on the closed interval of the real number line, where is a fixed integer greater than . They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in theinterval . Thereafter, the player whose turn it is chooses a real numberthat is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?SolutionNotice that to use the optimal strategy to win the game, Bela must select themiddle number in the range and then mirror whatever number Jennselects. Therefore, if Jenn can select a number within the range, so can Bela. Jenn will always be the first person to run out of a number to choose, so theanswer is .Solution 2 (Guessing)First of all, realize that the value of should have no effect on the strategy at all. This is because they can choose real numbers, not integers, so even if is odd, for example, they can still go halfway. Similarly, there is no reason the strategy would change when .So we are left with (A) and (B). From here it is best to try out random numbers and try to find the strategy that will let Bela win, but if you can't find it, realize thatit is more likely the answer is since Bela has the first move and thus has more control.Problem17There are people standing equally spaced around a circle. Each person knows exactly of the other people: the people standing next to her or him, as well as the person directly across the circle. How many ways are there forthe people to split up into pairs so that the members of each pair know each other?SolutionLet us use casework on the number of diagonals.Case 1: diagonals There are ways: either pairs with , pairs with , and so on or pairs with , pairs with , etc.Case 2: diagonal There are possible diagonals to draw (everyone else pairs with the person next to them.Note that there cannot be 2 diagonals.Case 3: diagonalsNote that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction.Case 4: diagonals There is way to do this.Thus, in total there are possible ways. Problem18An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?SolutionLet denote that George selects a red ball and that he selects a blue one. Now, in order to get balls of each color, he needs more of both and .There are 6cases:(wecan confirm that there are only since ). However we canclump , ,and together since they are equivalent by symmetry.andLet's find the probability that he picks the balls in the order of .。
美国AMC10中文版试题及答案
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 。
AMC美国数学竞赛AMC.B 试题及答案解析
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。
amc10美国数学竞赛真题
amc10美国数学竞赛真题amc10（美国数学竞赛）真题：
1. 微积分方面：
（1）求一元二次微分方程的有限解；
（2）求曲线y = f（x）在a点的切线斜率；
（3）求曲面z = f（x，y）在a点的切面方向；（4）决定泰勒展开式中n次项的系数；
（5）判断函数f（x）的极值；
（6）研究极限函数的存在性及无穷性；
（7）解决解析函数的定义域；
2. 数论方面：
（1）计算包含一定系数的多项式的余数；
（2）判断两个多项式的最大公因数；
（3）判断一个数的完全平方数；
（4）证明或否定一个数在某数域中的存在性；
3. 统计学：
（1）奇数和偶数的含义；
（2）样本容量的定义；
（3）求一组数据中最小值、最大值和中位数；（4）决定随机变量的条件概率；
（5）求一组数据的散布图；
4. 几何学：
（1）求平面两点间的距离；
（2）求直线的斜率；
（3）分析不相交平面的法向量；
（4）解决三角形的各个角度；
（5）计算圆的周长；
（6）两个圆的位置关系；
（7）求圆环的面积；
（8）求立体图形的体积。
5. 其它方面：
（1）分析逻辑表达式；
（2）求图论问题的最小生成树；
（3）计算算法问题的复杂度；
（4）计算概率论问题中的概率；
（5）规划组合优化问题的最优解；
（6）根据资料采纳正确的统计预测方法。
美国数学测评AMC12A-2019(中英双语)
2019AMC 12A Problems1. The area of a pizza with radius 4 is N percent larger than the area of a pizza with radius 3 inches. What is the integer closest to N ?半径为4英寸的比萨饼面积比半径为3英寸的比萨饼面积大百分之N 。
最接近N 的整数是什么？A B C D E ()25()33()44()66()782.Suppose a is 150% of b . What percent of a is b 3?设a 是b 的150%。
那么a 的百分之多少是3b ?A B C D E +3()50()66()150()200()45023. A box contains 28 red balls, 20 green balls, 19 yellow balls, 13 blue balls, 11 white balls, and 9 black balls. What is the minimum number of balls that must be drawnfrom the box without replacement to guarantee that at least 15 balls of a single color will be drawn?一个盒子中有28个红球，20个绿球，19个黄球，13个蓝球，11个白球和9个黑球。
为了保证至少取出15个单一颜色的球，在不允许放回重取的情况下，最少须从盒子中取出多少个球？A B C D E ()75()76()79()84()914. What is the greatest number of consecutive integers whose sum is 45? 最多可以有多少个连续整数，它们的总和是45?A B C D E ()9()25()45()90()1205. Two lines with slopes21and 2 intersect at (2,2). What is the area of the triangle enclosed by these two lines and the line +=x y 10?两条斜率分别为21和2的直线相交于(2, 2)。
amc数学竞赛试题
amc数学竞赛试题 AMC数学竞赛试题是一种具有挑战性的数学竞赛，旨在激励学生对数学的兴趣和才华，并培养他们解决实际问题的能力。
AMC数学竞赛涵盖广泛的数学领域，是学生们展示数学知识和解题能力的重要场所。
AMC数学竞赛试题一般包括代数、几何、数据分析和概率等多个知识领域。
这些试题通常设计得非常巧妙，要求学生具备深入理解和分析问题的能力。
接下来，我们将通过一些典型的AMC数学竞赛试题，来了解一下其中的难点和解题思路。
首先，我们先来看一道代数题： 1. 给定实数x，如果满足方程|2x+1| + 3x = 5x，则x的值是？ 解析：首先我们将绝对值分成两种情况：当2x+1≥0时，|2x+1|=2x+1；当2x+1<0时，|2x+1|=-(2x+1)。
然后，我们依次将这两种情况带入原方程，进行化简，最后得到x=1/7。
接下来，我们来看一道几何题： 2. 如图，一个等边三角形内切在一个圆内，然后又在等边三角形内切一个圆，重复这样的操作无穷次。
求最后等边三角形的边长与最初等边三角形的边长的比值。
解析：设最初等边三角形的边长为a，我们可以发现不断嵌套下来，趋近于一个极限值。
假设极限值为L，那么根据相似三角形的性质，我们可以得到L=2L+a，解方程得到L=a/3。
因此，最后等边三角形的边长与最初等边三角形的边长的比值为1/3。
再来看一道数据分析题： 3. 有一组数据集{(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)}，其中横坐标表示时间，纵坐标表示速度。
根据这组数据集，我们能否得到一个准确的速度-时间图像？如果能，请画出图像；如果不能，请说明原因。
解析：我们可以计算出每个点的斜率，也就是速度，发现每个点的速度都是2。
因此，这组数据集得到的速度-时间图像是一个速度恒定的水平直线。
以上只是AMC数学竞赛试题的一小部分示例，它们展示了竞赛中可能遇到的不同类型的问题。
参加AMC数学竞赛不仅可以提高学生的数学水平，还可以锻炼他们的思维能力和解决问题的能力。
amc10典型题目
amc10典型题目AMC 10（美国数学竞赛10）是一项面向中学生的数学竞赛，旨在提高学生的数学解题能力和思维能力。
以下是一些典型的AMC 10题目，供你参考：1. 题目，一个正方形的边长为10。
从正方形的一个顶点出发，每次只能向右或向上移动一个单位。
求到达对角线上另一个顶点的最短路径长度。
解答，通过观察可以发现，每次移动都会使横坐标或纵坐标增加1，因此最短路径长度为10。
2. 题目，若a、b、c是满足a+b+c=10的正整数，且abc的最大值为多少？解答，根据AM-GM不等式，当a=b=c时，abc的值最大。
因此，a=b=c=10/3，所以abc的最大值为(10/3)^3。
3. 题目，已知等差数列的前四项之和为10，且公差为2，求该等差数列的第一项。
解答，设等差数列的第一项为a，则根据等差数列的性质，前四项之和为4a+6。
将其与题目中给出的和10相等，得到4a+6=10。
解方程可得a=1，所以该等差数列的第一项为1。
4. 题目，在平面直角坐标系中，点A(3, 4)和点B(-1, 2)的中点坐标为多少？解答，中点的横坐标等于两点横坐标之和的一半，中点的纵坐标等于两点纵坐标之和的一半。
因此，中点的坐标为((3 + (-1))/2, (4 + 2)/2)，即(1, 3)。
5. 题目，一个数与它的平方之和等于90，求这个数。
解答，设这个数为x，根据题目条件可得x + x^2 = 90。
解这个二次方程可得x = 9或x = -10。
因为题目中没有指定数的范围，所以这两个解都是可能的。
以上是一些AMC 10的典型题目，希望对你有所帮助。
请注意，这只是一小部分题目，实际的AMC 10考试还包含更多类型的数学问题。
赏析几道美国AMC12数学竞赛题
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美国高中数学测验 AMC12之机率问题(下)
美国高中数学测验 AMC12之机率问题(下)洪伟诚 . 李俊贤 . 蔡诚祐 . 何家兴 . 张福春关键词：高中数学几何机率前述两节的问题皆建立在样本点个数为可数时的情况,接下来将介绍一不可数的无穷样本空间S且利用此空间的一些几何测量m(S),例如长度、面积、或者体积,来求A事件的机率。
而A事件的机率可用A事件之几何测量与样本空间S之几何测量的比例来计算,其形式有下列三种:P(A)=A的长度S的长度或P(A)=A的面积S的面积或P(A)=A的体积S的体积注:我们必须假设一不可数无穷样本空间S满足均匀性质,这样才能做以上的几何机率。
几何测量—长度所谓长度的几何测量,表示其无穷样本空间可用一线段、数线或是时间轴⋯等表示,则考虑某事件的机率时,只需探讨此事件所占的线段(或数线)与样本空间相对的长度比值即可,下列为利用长度测量来求的机率问题。
例1.(1972 AMC12 #17) 随机将一条线切为两段,试问较长的一段至少是较短的一段的x倍(其中x¸1)的机率为多少?•(B)2x•(C)1x+1•(D)1x•(E)2x+1解:(E)假设线段AB被切为两段,若较长一段(标记为l1,长度为sx)为较短一段(标记为l2,长度为s)的x倍,则线段AB总长度为(x+1)s,因此切点会落于l2中的机率为\dfrac{1}{x+1}。
但因线段的切法可能为(l1,l2)或是(l2,l21)两种情况,如下图故所求机率为2x+1。
例2.(2007 AMC12B #13)有一交通号志以下列的循环重复的运作:绿灯30秒,然后黄灯3秒,之后再转红灯30秒。
利亚随机挑选三秒钟的区间去注视号志灯,试问号志灯在转换颜色时,利亚正在注视的机率为多少?•(A)163•(B)121•(C)110•(D)17解:(D)由题意知,交通号志运作一循环需时63秒,而若利亚所注视的时间在当绿灯转变为黄灯、黄灯转变为红灯或红灯转变为绿灯的前三秒钟内,则利亚会看到号志颜色正在转变,如下图所示故所求机率为3+3+363=963=17例3.(2009 AMC12B #18) 瑞吉儿与罗伯特在一圆形的跑道上跑步,其中瑞吉儿以逆时钟方向跑且跑完一圈需时90秒,而罗伯特以顺时钟方向跑且跑完一圈需时80秒。
2021年美国数学竞赛(AMC10A)的试题与解答
2021年美国数学竞赛(AMC10A)的试题与解答华南师范大学数学科学学院(510631)李湖南1.算式(22−2)−(32−3)+(42−4)的值是多少?(A)1(B)2(C)5(D)8(E)12解直接计算,原式=2−6+12=8,故(D)正确.2.Portia高中的学生人数是Lara高中的3倍,这两所高中总共有2600名学生.问Portia高中有多少名学生?(A)600(B)650(C)1950(D)2000(E)2050解依题意可知,Portia高中占总人数的34,即有2600×34=1950名学生,故(C)正确.3.两个自然数之和是17402.这两个数中的一个可以被10整除,如果去掉该数的个位数字则得到另外一个数.问这两个数的差是多少?(A)10272(B)11700(C)13362(D)14238(E)15426解设第一个数为x,则第二个数为x10,已知x+x10=17402,解得x=15820,于是两数之差为x−x10=910x=14238,故(D)正确.4.一辆小车冲下山坡,它第一秒移动了5英寸,并且速度不断加快.在每个连续的1秒时间间隔内,它都比前1秒多移动7英寸.小车用了30秒到达山脚.问它一共行进了多少英寸?(A)215(B)360(C)2992(D)3195(E)3242解小车每秒移动的距离是个等差数列,首项为5,公差为7,第30秒移动了208英寸,从而距离和为(5+208)×30÷2=3195英寸,故(D)正确.5.一个共有k>12名学生的班级进行测验的平均分为8分,其中12名学生的测验平均分是14分.问其余学生的测验平均分如何用k来表示?(A)14−8k−12(B)8k−168k−12(C)1412−8k(D)14(k−12)k2(E)14(k−12)8k解所有学生的分数总和为8k,其中12名学生的分数和为168,剩下k−12名学生的分数和为8k−168,故平均分为8k−168k−12,(B)正确.6.Chantal和Jean从山路的起点开始向消防塔徒步旅行.Jean背着一个沉重的背包,走得较慢.Chantal开始以每小时4英里的速度行走,走到路程一半,山路变得非常陡峭,Chantal的速度减慢到每小时2英里.到达塔后,她立即掉头,以每小时3英里的速度沿着陡峭的山路向下走,她在路程的一半处遇见了Jean.从出发到他们相遇,Jean的平均速度是每小时多少英里?(A)1213(B)1(C)1312(D)2413(E)2解设半程为x英里,Jean所用的时间与Chantal一样,均为x4+x2+x3=13x12小时,于是Jean的平均速度为x(13x)/12=1213英里/小时,故(A)正确.7.汤姆有13条蛇,其中4条是紫色的,5条是快乐的.他观察发现:他的所有快乐的蛇都能做加法,他的紫色的蛇不会做减法,而且他所有不会做减法的蛇也不会做加法.关于汤姆的蛇,可以得出以下哪个结论?(A)紫色的蛇可以做加法(B)紫色的蛇是快乐的(C)能做加法的蛇是紫色的(D)快乐的蛇不是紫色的(E)快乐的蛇不能做减法解紫色的蛇不会做减法,从而也不会做加法,因此也不是快乐的.故快乐的蛇都不是紫色的,(D)正确.8.一名学生在用66乘以如下的循环小数时, 1.abab···=1.˙a˙b,其中a和b是数字.他没有注意到循环小数的标识,而只是做了66乘以1.ab.后来他发现他的答案比正确答案小0.5.问两位整数ab是多少?(A)15(B)30(C)45(D)60(E)75解由题意可得,0.5=66×(1.˙a˙b−1.ab)=66×0.00˙a˙b= 0.66×ab99,解得ab=75,故(E)正确.9.对于实数x和y,(xy−1)2+(x+y)2的最小可能值是多少?(A)0(B)14(C)12(D)1(E)2解(xy−1)2+(x+y)2=x2y2+x2+y2+1 1,此时x=y=0,故(D)正确.10.算式(2+3)(22+32)(24+34)(28+38)(216+ 316)(232+332)(264+364)与下面哪个表达式相等?(A)3127+2127(B)3127+2127+2×363+3×263(C)3128−2128(D)3128+2128(E)5127解连续使用平方差公式,可得原式=(3−2)(3+2)(32+22)(34+24)(38+28)(316+216)(332+232)(364+264)=3128−2128,故(C)正确.11.选择下面哪个整数b 为基数,可以使得b 进制数2021b −221b 不能被3整除?(A)3(B)4(C)6(D)7(E)8解由于2021b −221b =2×b 3−2×b 2=2b 2(b −1),当b =3,4,6,7时,该数均能被3整除,故(E)正确.12.如图所示,两个顶点朝下的正圆锥包含相同量的液体.液体顶部表面的半径分别为3厘米和6厘米.在每个圆锥体中放入一个半径为1厘米的球形弹子,它沉入底部,完全浸没,没有任何液体溢出.问窄圆锥内液面上升的高度与宽圆锥内液面上升的高度之比是多少?(A)1:1(B)47:43(C)2:1(D)40:13(E)4:1解设液体的体积为V ,左右两边液面高度分别为h 1,h 2,则V =13π32·h 1=13π62·h 2,因而h 1=4h 2.设放入弹子的体积为V 0=43π,左右两边液面上升高度分别为∆h 1,∆h 2,顶部液面的半径分别为r 1,r 2,则有V +V 0=13πr 21·(h 1+∆h 1)=13πr 22·(h 2+∆h 2),3r 1=h 1h 1+∆h 1,6r 2=h 2h 2+∆h 2,解得r 1r 2=12,∆h 1∆h 2=41,故(E)正确.13.四面体ABCD 中,各边长为AB =2,AC =3,AD =4,BC =√13,BD =2√5和CD =5,问它的体积是多少?(A)3(B)2√3(C)4(D)3√3(E)6解如图所示,由于AB 2+AC 2=BC 2,AB 2+AD 2=BD 2,AC 2+AD 2=CD 2,可得AB,AC,AD 互相垂直,即ABCD 是个直四面体.故V ABCD =13S ∆ABD ·AC =13·12×2×4·3=4,(C)正确.14.多项式z 6−10z 5+Az 4+Bz 3+Cz 2+Dz +16的根都是正整数,有可能重复.问B 的取值是多少?(A)−88(B)−80(C)−64(D)−41(E)−40解设多项式的根为x 1,x 2,x 3,x 4,x 5,x 6,其中均为正整数,则z 6−10z 5+Az 4+Bz 3+Cz 2+Dz +16=(z −x 1)(z −x 2)···(z −x 6),可得 x 1+x 2+x 3+x 4+x 5+x 6=10,x 1x 2x 3x 4x 5x 6=16,−∑i<j<kx i x j x k =B,解得x 1=x 2=1,x 3=x 4=x 5=x 6=2,于是B =−(C 14·12·2+C 12·C 24·1·22+C 34·23)=−88,故(A)正确.15.A,B,C 和D 的值从{1,2,3,4,5,6}中不重复地选取(即没有两个字母的取值相同),使得两条曲线y =Ax 2+B 和y =Cx 2+D 相交的不同取值方式有多少种?(不考虑曲线列出的顺序,例如,A =3,B =2,C =4,D =1与A =4,B =1,C =3,D =2被认为是相同的)(A)30(B)60(C)90(D)180(E)360解要使得y =Ax 2+B 和y =Cx 2+D 相交,则方程Ax 2+B =Cx 2+D 有解,即有(A −C )x 2=D −B ,此时A −C,D −B 的符号一样.任取A,C ∈{1,2,3,4,5,6},则D,B ∈{1,2,3,4,5,6}−{A,C },大小关系须一致,则有2C 26C 24=180种选择;另外不考虑曲线的顺序,故不同的取值方式共有180÷2=90种,(C)正确.16.在下面的数据列表中,对于1 n 200,整数n 出现了n 次.1,2,2,3,3,3,4,4,4,4,···,200,200,···,200.问这组数据列表中的中位数是多少?(A)100.5(B)134(C)142(D)150.5(E)167解这列数共有1+2+3+ (200)12×200×201=20100个,中位数是第10050与第10051个数的平均值.先求得12n (n +1) 10050的最大值为n max =141,且12n max (n max +1)=10011.这说明第10050和第10051个数均为142,故中位数为142,(C)正确.17.在梯形ABCD 中,AB //CD ,BC =CD =43,并且AD ⊥BD .设O 是对角线AC 和BD 的交点,P 是BD的中点.已知OP =11,AD 的长度可以表示成m √n ,其中m 和n 是正整数,并且n 不能被任何质数的平方所整除.问m +n 的值是多少?(A)65(B)132(C)157(D)194(E)215解如图所示,延长CP 交AB 于点E ,由于∆CBD 是等腰三角形,P 是BD 中点,从而CP⊥BD ,进而CE //DA ,于是AECD 是个平行四边形,得EA =CD =43.又∠CBD =∠CDB =∠ABD ,可得∆CBP =∆EBP ,即得BE =BC =43,于是AB =86.再根据∆CDO ∆ABO ,有OD OB =CD AB =12,即OD OB =BP −11BP +11=12,解得BP =33,从而AD =CE =2CP =2√BC 2−BP 2=2√432−332=4√190,故m +n =194,(D)正确.18.令f 是一个定义在正有理数集合上的函数,它具有性质:对于所有的正有理数a 和b ,f (ab )=f (a )+f (b ).假设f 还具有性质:对于每一个质数p ,f (p )=p .问以下哪个数x 满足f (x )<0?(A)1732(B)1116(C)79(D)76(E)2511解f (1)=f (1·1)=2f (1)⇒f (1)=0,f (1)=f (b ·1b )=f (b )+f (1b )=0⇒f (1b)=−f (b ),对于任意正有理数b ;f (a b )=f (a ·1b )=f (a )+f (1b)=f (a )−f (b ),对于任意正有理数a 和b ;f (p n )=f (p ·p ·····p n 个)=nf (p ),对于任意质数p ,n ∈N .因此,f (2511)=f (25)−f (11)=2f (5)−f (11)=2·5−11=−1<0,故(E)正确.19.由x 2+y 2=3|x −y |+3|x +y |的图像所界定的图形的面积是m +nπ,其中m 和n 是整数.问m +n 是多少?(A)18(B)27(C)39(D)45(E)54解如图所示,对点(x,y )的位置进行讨论:(1)在第一象限:当y x 时,方程为x 2+y 2=3(x −y )+3(x +y ),化简可得(x −3)2+y 2=9,它是一个八分之一圆;当y >x 时,方程可化为x 2+(y −3)2=9,也是一个八分之一圆;(2)在第二象限:当y −x 时,方程可化为x 2+(y −3)2=9;当y <−x 时,方程可化为(x +3)2+y 2=9;(3)在第三象限:当y x 时,方程可化为(x +3)2+y 2=9;当y <x 时,方程可化为x 2+(y +3)2=9;(4)在第四象限:当y −x 时,方程可化为(x −3)2+y 2=9;当y <−x 时,方程可化为x 2+(y +3)2=9.综上可得,该图像由四个半径为3的半圆封闭而成,面积为62+2·π·32=36+18π,故m +n =36+18=54,(E)正确.20.数列1,2,3,4,5有多少种重新排列的方式,使得没有连续三项是递增的,也没有连续三项是递减的?(A)10(B)18(C)24(D)32(E)44解由题意可知,符合条件的数列中,连续两项的单调性只能是:增减增减,或减增减增.用排列表示即有:13254,14253,14352,15243,15342;21435,21534,23154,24153,24351,25143,25341;31425,31524,32415,32514,34152,34251,35142,35241.如果对集合{1,2,3,4,5}做一个置换1234554321 ,所得排列仍然符合条件,即以5,4开头的排列和以1,2开头的排列一样多.故所有符合条件的数列有(5+7)×2+8=32个,(D)正确.21.设ABCDEF 是等角六边形,由直线AB ,CD 和EF 所组成的三角形面积为192√3,由直线BC ,DE 和F A 所组成的三角形的面积是324√3.六边形ABCDEF 的周长可用m +n √p 表达,其中m ,n 和p 是正整数,并且p不能被任何质数的平方整除.问m +n +p 的值是多少?(A)47(B)52(C)55(D)58(E)63解如图所示,分别向两边延长各边,交于点G ,H ,I ,J ,K ,L ,由题意可得,S ∆GHI =192√3,S ∆JKL =324√3.由于六边形的每个内角都是120◦,因此图中所有三角形的内角都是60◦,即所有三角形都是等边三角形.于是,S ∆GHI =√34IG 2=√34(a +b +f )2=192√3,解得a +b +f =16√3;同理S ∆JKL =√34(c +d +e )2=324√3,得c +d +e =36.故六边形的周长为a +b +c +d +e +f =36+16√3,即m +n +p =36+16+3=55,(C)正确.22.Hiram 的代数笔记有50页,打印在25张纸上;第一张纸包括第1和第2页,第二张纸包括第3和第4页,以此类推.有一天,他去午餐前把笔记本放在桌子上,室友决定从笔记中间借几页.当Hiram回来时,他发现他的室友从笔记中拿走了连续的若干张纸,并且所有剩余纸张上页码的平均值正好是19.问有多少张纸被借走了?(A)10(B)13(C)15(D)17(E)20解设笔记被借走了x张纸,分别是从第a张纸到第a+x−1张纸,其中a+x−1 25,页码正好是从2a−1到2(a+x−1),页码和为(2a−1)+2a+···+2(a+x−1)=(2a−1)+2(a+x−1)2·2x=(4a+2x−3)x,而所有页码和为1+2+···+50=1275,从而剩下的页码和为19(50−2x)=1275−(4a+2x−3)x,整理得(4a+2x−41)x=325,解得x=13,a=10,故(B)正确.23.青蛙Frieda在一个3×3的方格表上开始一系列跳跃,每次跳跃都随机选择一个方向—–向上、向下、向左或向右,从一个方格移动到旁边的方格.她不能斜着跳,当跳跃的方向会使得Frieda离开方格表时,她会“绕个圈”,跳到相对的另一边.例如,如果Frieda从中心方格开始,向上跳跃两次,第一次跳跃后她将位于最上面一行的中间方格,第二次跳跃将使得Frieda跳到相对的边,落在最下面一行的中间方格.假设Frieda从中心方格出发,最多随机跳跃四次,并且当到达角落方格时就停止跳跃.问她在四次跳跃中到达角落方格的概率是多少?(A)916(B)58(C)37(D)2532(E)1316解记P(n)为第n次到达角落方格的概率,显然P(1)=0;第一次跳跃可朝四个方向,不妨设第一次跳跃向右,则当第二次跳跃向上或向下时,可到达角落方格,概率为24,即P(2)=12;若第二次跳跃向左,概率为14,即回到出发点S,此时第四次跳跃到达角落方格的概率为1 4×P(2)=18;若第二次跳跃向右,概率为14,此时到了S左边的方格,则第三次跳跃向上或向下可到达角落方格,即P(3)=14×12=18;若第三次跳跃向左,概率为14,则回到S右边的方格,第四次跳跃向上或向下可到达角落方格,概率为14×14×12=132,从而P(4)=18+132=532.而其它情况均四次到达不了角落方格.故所求概率为4∑i=1P(i)=0+12+18+532=2532,(D)正确.24.设a是正实数,考虑由(x+ay)2=4a2和(ax−y)2=a2组成的四边形的内部.对所有的a>0而言,这个区域的面积怎样用a来表示?(A)8a2(a+1)2(B)4aa+1(C)8aa+1(D)8a2a2+1(E)8aa2+1解如图所示,四边形由四条直线y=−1ax±2,y=ax±a围成,由于斜率之积为−1,从而四边形是个矩形.设直线y=−1ax+2与x轴负方向的夹角为θ,则tanθ=1a,矩形的长为4cosθ,宽为2cosθ,即所求面积为S=8cos2θ.于是S=4·2cos2θ=4(1+cos2θ)=4(1+1−tan2θ1+tan2θ)=4·21+tan2θ=8·11+1a2=8a21+a2.故(D)正确.25.将3枚不可区分的红色筹码,3枚不可区分的蓝色筹码和3枚不可区分的绿色筹码分别放入3×3方格表的各个小方格中,使得无论是垂直方向还是水平方向,都没有两个相同颜色的筹码相邻,问共有多少种放法?(A)12(B)18(C)24(D)30(E)36解首先考虑中心方格,不妨设放入红色筹码,分两种情况:(1)其余两个红色筹码在一边,四个方向均可:那剩下6个方格只能按以下方式放入,∆放一种颜色,空格放另一种颜色.此时有4×2=8种放法;(2)其余两个红色筹码各在一边,即在对角线上,两个方向均可:那剩下6个方格也只能按以下方式放入,∆放一种颜色,空格放另一种颜色.此时有2×2=4种放法.综上,中心方格放入红色筹码,共有12种放法.如果放入蓝色筹码或绿色筹码,也是一样,故所有的放法有3×12=36种,(E)正确.。
amc 8 逻辑推理题
amc 8 逻辑推理题
AMC 8是美国数学竞赛(American Mathematics Competition)的初级组别，主要面向8年级及以下的学生。
逻辑推理题是AMC 8中常见的一种题型，考察学生的逻辑推理能力。
逻辑推理题一般会给出一些信息，然后根据这些信息来推导出结论。
这些信息可能是关于几何图形、数列、逻辑关系等方面的，而结论则通常是根据这些信息进行推理得出的。
以下是一个AMC 8逻辑推理题的示例：
题目：有5个不同颜色的球，红、蓝、黄、绿、紫。
把它们放入3个不同的盒子里，每个盒子至少有一个球。
现在知道红球和蓝球一定在同一个盒子里，紫球不能和绿球放在同一个盒子里，黄球不能和蓝球放在同一个盒子里。
请问，如何将这5个球放入3个盒子中？
根据题目中的信息，我们可以进行以下推理：
1. 红球和蓝球一定在同一个盒子里，所以我们可以将它们放入第一个盒子。
2. 紫球不能和绿球放在同一个盒子里，所以紫球只能放入第二个盒子或第三个盒子。
3. 黄球不能和蓝球放在同一个盒子里，所以黄球只能放入第二个盒子或第三个盒子。
4. 每个盒子至少有一个球，所以如果紫球和黄球分别放入了第二个盒子和第三个盒子，那么第一个盒子中只能是红球和蓝球。
5. 剩下的绿球只能放入第二个盒子或第三个盒子。
因此，我们得到以下可能的方案：
* 第一个盒子：红球、蓝球；
* 第二个盒子：紫球、绿球；
* 第三个盒子：黄球。
或者
* 第一个盒子：红球、蓝球；
* 第二个盒子：黄球、绿球；
* 第三个盒子：紫球。
历届美国中学生数学竞赛试题及解答
历届美国中学生数学竞赛试题及解答
下面是近几届美国中学生数学竞赛试题及解答：
一、2019：
1、求解下列不等式（2 + 2·3² < 5·3）？
答案：2 + 2·3² < 5·3
2、如何快速计算三角形面积？
答案：三角形面积可以用公式：S=½·a·h，其中a为三角形的底边，h 为三角形高度。
二、2018：
1、求复数（1+i）³的模和辐角？
答案：复数（1+i）³的模为2，辐角为2π/3。
2、什么是余弦定理？
答案：余弦定理是一种三角形的定理，它规定了三角形的两条边的长度和它们之角之间的关系：C²=a² +b² -2ab·cosC。
三、2017：
1、如何快速求解圆面积？
答案：圆面积可以用公式：S=πr²，其中r为圆半径。
2、求解下列方程（x²-5x=3）？答案：x²-5x=3， x=2 或 3。
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。
2018年全美数学竞赛amc8第13题
2018年全美数学竞赛amc8第13题
2018年全美数学竞赛AMC8第13题
在2018年全美数学竞赛AMC8中，第13题是一个比较有意思的
数学问题，它考察了考生对于几何图形的理解和计算能力。
这道题
目可以帮助考生提高对于几何图形的认识，同时也可以锻炼他们的
逻辑思维和解题能力。
题目描述如下：在一个正方形的对角线上，取一点P，连接PA，PB两条线段，使得PA:PB=1:2。
如果正方形的边长为6，那么线段
PA的长度是多少？
解题思路如下： 1. 首先，我们可以利用正方形的对角线的性质，即对角线的长度等于边长的平方根乘以2，来求得对角线的长度。
即对角线的长度为6*√2。
2. 然后，我们可以利用PA:PB=1:2
的比例关系，来求得线段PA的长度。
假设PA的长度为x，则PB的
长度为2x。
根据比例关系，我们可以得到x+2x=6*√2，即
3x=6*√2，解得x=2*√2。
因此，线段PA的长度为2*√2。
通过解答这道题目，考生可以加深对于几何图形的认识，同时也可以提高解题的能力。
在平时的学习中，考生可以多做类似的几何题目，加强对于几何知识的掌握，提高解题的能力。
总之，这道题目不仅可以帮助考生提高数学解题能力，也可以帮助他们加深对于几何知识的理解。
希望考生在备战数学竞赛的过程中，能够认真对待每一道题目，不断提升自己的数学水平。
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F组题1.Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters ofa solution that is 48% acid. Jar C contains one liter of a solution that is acid.From jar C, liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that and are relatively prime positive integers, find.Answer:Solution:Omited.Resource: 2011 AIME I Problems12.Let be the line with slope that contains the point , and letbe the line perpendicular to line that contains the point . The originalcoordinate axes are erased, and line is made the -axis and line the -axis. Inthe new coordinate system, point is on the positive -axis, and point is on thepositive -axis. The point with coordinates in the original system hascoordinates in the new coordinate system. Find .Answer:Solution:Omited.Resource: 2011 AIME I Problems33. Suppose that a parabola has vertex and equation , where and is an integer. The minimum possible value of can bewritten in the form , where and are relatively prime positive integers. Find.Answer:Solution:Omited.Resource: 2011 AIME I Problems64. Suppose is in the interval and . Find .Answer:Solution:Omited.Resource: 2011 AIME I Problems95.For some integer , the polynomial has the three integer roots, , and . Find .Answer:Solution:Omited.Resource: 2011 AIME I Problems156. The sum of the first 2011 terms of a geometric sequence is 200. The sum of the first 4022 terms is 380. Find the sum of the first 6033 terms.Answer:the first terms is .Solution:Omited.Resource: 2011 AIME II Problems/Problem 57.Gary purchased a large beverage, but only drank of it, where and are relatively prime positive integers. If he had purchased half as much and drunk twiceas much, he would have wasted only as much beverage. Find .Answer:Solution:Omited.Resource: 2011 AIME II Problems/Problem 18. On square, point lies on side and point lies on side , so that. Find the area of the square .Answer:810Solution:Omited.Resource: 2011 AIME II Problems/Problem 29.The degree measures of the angles in a convex18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.Answer: The first term is thenSolution: The average angle in an 18-gon is . In an arithmetic sequence the average is the same as the median, so the middle two terms of the sequenceaverage to . Thus for some positive (the sequence is increasing and thusnon-constant) integer , the middle two terms are and .Since the step is the last term of the sequence is , which must beless than , since the polygon is convex. This gives , so the only suitable positive integer is 1.Resource: 2011 AIME II Problems/Problem 310.In triangle , . The angle bisector of intersects atpoint , and point is the midpoint of . Let be the point of the intersectionof and . The ratio of to can be expressed in the form , whereand are relatively prime positive integers. Find .Answer:.Solution:Omited.Resource: 2011 AIME II Problems/Problem 411. The sum of the first 2011 terms of a geometric sequence is 200. The sum of the first 4022 terms is 380. Find the sum of the first 6033 termsAnswer:542.Solution:Omited.Resource: 2011 AIME II Problems/Problem 512. Let . A real number is chosen at random from the interval. The probability that is equal to, where , , , , and are positive integers. Find.Answer:850Solution:Omited.Resource: 2011 AIME II Problems/Problem1513. Maya lists all the positive divisors of . She then randomly selects two distinct divisors from this list. Let be the probability that exactly one of the selected divisors is a perfect square. The probability can be expressed in theform , where and are relatively prime positive integers. Find .Answer:107Solution:Omited.Resource: 2010 AIME I Problems/Problem114. Suppose that and . The quantity can be expressed as arational number , where and are relatively prime positive integers. Find . Answer:529Solution:Omited.Resource: 2010 AIME I Problems/Problem315. Positive integers, , , and satisfy , ,and . Find the number of possible values of . Answer:501Solution:Omited.Resource: 2010 AIME I Problems/Problem516.Rectangle and a semicircle with diameter are coplanar and havenonoverlapping interiors. Let denote the region enclosed by the semicircle andthe rectangle. Line meets the semicircle, segment , and segment atdistinct points , , and, respectively. Line divides region into two regionswith areas in the ratio . Suppose that , , and .Then can be represented as , where and are positive integers andis not divisible by the square of any prime. Find .Answer:069Solution:Omited.Resource: 2010 AIME I Problems/Problem1317. In with , , and , let be a point onsuch that the incircles of and have equal radii. Let and bepositive relatively prime integers such that . Find .Answer:045Solution:Omited.Resource: 2010 AIME I Problems/Problem1518. The terms of an arithmetic sequence add to . The first term of the sequence is increased by , the second term is increased by , the third term is increased by ,and in general, the th term is increased by the th odd positive integer. The termsof the new sequence add to . Find the sum of the first, last, and middle terms ofthe original sequence.Answer:195Solution:Omited.Resource: 2012 AIME I Problems/Problem219. Let be the set of all perfect squares whose rightmost three digits in baseare . Let be the set of all numbers of the form , where is in . Inother words, is the set of numbers that result when the last three digits of eachnumber in are truncated. Find the remainder when the tenth smallest element ofis divided by .Answer:170Solution:Omited.Resource: 2012 AIME I Problems/Problem1020. Let be the set of all binary integers that can be written using exactly zerosand ones where leading zeros are allowed. If all possible subtractions are performed in which one element of is subtracted from another, find the number oftimes the answer is obtained.Answer:330Solution:Omited.Resource: 2012 AIME I Problems/Problem521. At each of the sixteen circles in the network below stands a student. A totalof coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally.Answer:280Solution:Omited.Resource: 2012 AIME I Problems/Problem722. Three concentric circles have radii and An equilateral triangle with one vertex on each circle has side length The largest possible area of the triangle canbe written as where and are positive integers, and arerelatively prime, and is not divisible by the square of any prime.FindAnswer:041Solution:Omited.Resource: 2012 AIME I Problems/Problem1323. Ana, Bob, and Cao bike at constant rates of meters per second, meters per second, and meters per second, respectively. They all begin biking at thesame time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point on the south edge of the field. Cao arrives at point at the same time that Ana and Bob arrive at for the first time. Theratio of the field's length to the field's width to the distance from point to the southeast corner of the field can be represented as , where , , and arepositive integers with and relatively prime. Find .Answer:061Solution:Omited.Resource: 2012 AIME II Problems/Problem424. In the accompanying figure, the outer square has side length . A secondsquare of side length is constructed inside with the same center as andwith sides parallel to those of . From each midpoint of a side of , segments aredrawn to the two closest vertices of . The result is a four-pointed starlike figure inscribed in . The star figure is cut out and then folded to form a pyramid withbase . Find the volume of this pyramid.Answer:750Solution:Omited.Resource: 2012 AIME II Problems/Problem425. Let be the complex number with and such that thedistance between and is maximized, and let . Find .Answer:750Solution:Omited.Resource: 2012 AIME II Problems/Problem426. Let be the increasing sequence of positive integers whose binaryrepresentation has exactly ones. Let be the 1000th number in . Find theremainder when is divided by .Answer:32Solution:Omited.Resource: 2012 AIME II Problems/Problem727. Let , and for , define . The value of that satisfies can be expressed in the form , where and are relatively prime positive integers. Find .Answer:008Solution:Omited.Resource: 2012 AIME II Problems/Problem1128. n a group of nine people each person shakes hands with exactly two of the other people from the group. Let be the number of ways this handshaking can occur.Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when is divided by .Answer: 160Solution:Omited.Resource: 2012 AIME II Problems/Problem1429. Find the number of five-digit positive integers, , that satisfy the following conditions:(a) the number is divisible by 5(b) the first and last digits of are equal, and(c) the sum of the digits of is divisible by 5Answer: 200Solution:Omited.Resource: 2013 AIME I Problems/Problem230. Ms. Math's kindergarten class has 16 registered students. The classroom has a very large number, N, of play blocks which satisfies the conditions:(a) If 16, 15, or 14 students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and(b) There are three integers such that when , ,or students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.Find the sum of the distinct prime divisors of the least possible value of N satisfying the above conditions.Answer: 148Solution:Omited.Resource: 2013 AIME I Problems/Problem11。