美国高中学生数学竞赛题
amc12数学题
amc12数学题篇一：AMC12数学题是指美国数学竞赛（AMC）12级的数学题目。
AMC12是一项面向高中生的数学竞赛，旨在鼓励学生对数学的深入学习和探索。
这些数学题目涵盖了广泛的数学概念和技巧，并要求学生通过逻辑推理和解题技巧来解决问题。
AMC12数学题的难度较高，常常需要学生具备扎实的数学基础和解题能力。
这些题目涉及的数学领域包括代数、几何、数论和概率等。
许多题目都要求学生运用多个数学概念进行综合分析和解答。
因此，参加AMC12数学竞赛是考验学生数学能力和解题能力的良好机会。
解答AMC12数学题需要学生具备良好的数学思维和解题技巧。
学生需要能够正确理解问题并找出解决问题的方法。
在解答过程中，学生需要通过分析和推理来得出结论，同时要注意细节和逻辑的严谨性。
有时候，解答AMC12数学题还需要一些创造性思维和巧妙的数学变换。
参加AMC12数学竞赛对学生有很多好处。
首先，它可以提高学生的数学能力和解题能力，培养学生的数学思维和分析能力。
其次，参加竞赛可以增强学生对数学的兴趣和热爱，激发学生对数学的探索欲望。
此外，竞赛还可以培养学生的竞争意识和团队合作精神。
最重要的是，参加AMC12数学竞赛可以为学生在大学申请中增加一份优势，显示出学生在数学方面的才能和成就。
总之，AMC12数学题是一系列挑战性的数学问题，要求学生具备扎实的数学基础和解题能力。
通过参加AMC12数学竞赛，学生能够提高数学能力，培养数学思维，并为将来的学术和职业发展奠定坚实的基础。
篇二：AMC12（美国数学竞赛12级）是由美国数学协会举办的一项数学竞赛，面向高中学生。
这项竞赛旨在鼓励学生培养解决复杂数学问题的能力，并提供一个展示自己数学才能的平台。
AMC12数学题目通常涵盖广泛的数学领域，包括代数、几何、概率与统计、数论等。
这些题目要求学生在有限的时间内思考、分析和解决问题，考察他们的数学思维能力、推理能力和解决问题的能力。
举一个例子，假设有一道AMC12的几何问题：如图所示，一个直径为12的圆与一个半径为8的圆相切。
美国数学竞赛试题-部分
2000AMC12ProblemsProblem1In the year,the United States will host the International Mathematical Olympiad.Let and be distinct positive integers such that the product.What is the largest possible value of the sum?Problem2Problem3Each day,Jenny ate of the jellybeans that were in her jar at the beginning of that day. At the end of the second day,remained.How many jellybeans were in the jar originally?Problem4The Fibonacci sequence starts with two1s,and each term afterwards is the sum of its two predecessors.Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?Problem5If where thenProblem6Two different prime numbers between and are chosen.When their sum is subtracted from their product,which of the following numbers could be obtained?Problem7How many positive integers have the property that is a positive integer?Problem8Figures,,,and consist of,,,and non-overlapping squares.If the pattern continued,how many non-overlapping squares would there be in figure?Problem9Mrs.Walter gave an exam in a mathematics class of five students.She entered the scores in random order into a spreadsheet,which recalculated the class average after each score was entered.Mrs. Walter noticed that after each score was entered,the average was always an integer.The scores (listed in ascending order)were71,76,80,82,and91.What was the last score Mrs.Walters entered?Problem10The point is reflected in the-plane,then its image is rotatedby about the-axis to produce,and finally,is translated by5units in the positive-direction to produce.What are the coordinates of?Problem11Two non-zero real numbers,and satisfy.Which of the following is a possible value of?Problem12Let A,M,and C be nonnegative integers such that.What is the maximum value of+++?Problem13One morning each member of Angela’s family drank an8-ounce mixture of coffee with milk.The amounts of coffee and milk varied from cup to cup,but were never zero.Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee.How many people are in the family?Problem14When the mean,median,and modeof the listare arranged in increasing order,they form a non-constant arithmetic progression.What is the sum of all possible real values of?Problem15Let be a function for which.Find the sum of all values of for which.Problem16A checkerboard of rows and columns has a number written in each square,beginning in the upper left corner,so that the first row is numbered,the second row, and so on down the board.If the board is renumbered so that the left column,top to bottom, is,the second column and so on across the board,some squares have the same numbers in both numbering systems.Find the sum of the numbers in these squares (under either system).Problem17A centered at has radius and contains the point.The segment is tangent to the circle at and.If point lies on and bisects,thenProblem18In year,the day of the year is a Tuesday.In year,the day is also a Tuesday.On what day of the week did th day of year occur?Problem19triangle,,,.Let denote the midpointof and let denote the intersection of with the bisector of angle.Which of the following is closest to the area of the triangle?Problem20If and are positive numbers satisfyingThen what is the value of latex?Problem21Through a point on the hypotenuse of right triangle,lines are drawn parallel to the legs of the triangle so that the triangle is divided into asquare and two smaller right triangles.The area of one of the two small right triangles times the area of the square.The ratio of the area of the other small right triangle to the area of the square isProblem22The graph below shows a portion of the curve defined by the quarticpolynomial.Which of the following is the smallest?Problem23Professor Gamble buys a lottery ticket,which requires that he pick six different integers from through,inclusive.He chooses his numbers so that the sum of the base-ten logarithms of his six numbers is an integer.It so happens that the integers on the winning ticket have the same property—the sum of the base-ten logarithms is an integer.What is the probability that Professor Gamble holds the winning ticket?Problem24If circular arcs and centers at and,respectively,then there exists a circletangent to both and,and to.If the length of is,then the circumference of the circle isProblem25Eight congruent Equilateral triangle each of a different color,are used to construct a regular octahedron.How many distinguishable ways are there to construct the octahedron?(Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)Problem6For how many positive integers does there exist at least one positive integer n such that?infinitely manyProblem7A arc of circle A is equal in length to a arc of circle B.What is the ratio of circle A's area and circle B's area?Problem8Betsy designed a flag using blue triangles,small white squares,and a red center square,as shown. Let be the total area of the blue triangles,the total area of the white squares,and the area of the red square.Which of the following is correct?Problem9Jamal wants to save30files onto disks,each with1.44MB space.3of the files take up0.8MB, 12of the files take up0.7MB,and the rest take up0.4MB.It is not possible to split a file onto2different disks.What is the smallest number of disks needed to store all30files?Problem10Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size.She then pours half the coffee from the first cup to the second and,after stirring thoroughly,pours half the liquid in the second cup back to the first.What fraction of the liquid in the first cup is now cream?Problem11Mr.Earl E.Bird gets up every day at8:00AM to go to work.If he drives at an average speed of40miles per hour,he will be late by3minutes.If he drives at an average speed of60milesper hour,he will be early by3minutes.How many miles per hour does Mr.Bird need to drive to get to work exactly on time?Problem12Both roots of the quadratic equation are prime numbers.The number of possible values of isProblem13Two different positive numbers and each differ from their reciprocals by.What is?Problem14For all positive integers,let.Let. Which of the following relations is true?Problem15The mean,median,unique mode,and range of a collection of eight integers are all equal to8. The largest integer that can be an element of this collection isProblem16Tina randomly selects two distinct numbers from the set{1,2,3,4,5},and Sergio randomly selects a number from the set{1,2,...,10}.What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?Problem17Several sets of prime numbers,such as use each of the nine nonzero digits exactly once.What is the smallest possible sum such a set of primes could have?Problem18Let and be circles definedby and respectively.What is the length of the shortest line segment that is tangent to at and to at?Problem19The graph of the function is shown below.How many solutions does theequation have?Problem20Suppose that and are digits,not both nine and not both zero,and the repeatingdecimal is expressed as a fraction in lowest terms.How many different denominators are possible?Problem21Consider the sequence of numbers:For,the-th term of the sequence is the units digit of the sum of the two previous terms.Let denote the sum of the first terms of this sequence.The smallest value of for which is:Problem22Triangle is a right triangle with as its right angle,, and.Let be randomly chosen inside,and extend to meet at. What is the probability that?Problem23In triangle,side and the perpendicular bisector of meet in point,and bisects.If and,what is the area of triangle?SAT II数学词汇表代数部分1.基础add，plus加subtract减difference差multiply times乘product积divide除divisible可被整除的divided evenly被整除dividend被除数divisor因子，除数quotient商remainder余数factorial阶乘power乘方radical sign,root sign根号round to四舍五入to the nearest四舍五入2.有关集合union并集proper subset真子集solution set解集3.有关代数式、方程和不等式algebraic term代数项like terms,similar terms同类项numerical coefficient数字系数literal coefficient字母系数inequality不等式triangle inequality三角不等式range值域original equation原方程equivalent equation同解方程等价方程linear equation线性方程(e.g.5x+6=22)4.有关分数和小数proper fraction真分数improper fraction假分数mixed number带分数vulgar fraction，common fraction普通分数simple fraction简分数complex fraction繁分数numerator分子denominator分母(least)common denominator(最小)公分母quarter四分之一decimal fraction纯小数infinite decimal无穷小数recurring decimal循环小数tenths unit十分位5.基本数学概念arithmetic mean算术平均值weighted average加权平均值geometric mean几何平均数exponent指数，幂base乘幂的底数,底边cube立方数，立方体square root平方根cube root立方根common logarithm常用对数digit数字constant常数variable变量inverse function反函数complementary function余函数linear一次的，线性的factorization因式分解absolute value绝对值，e.g.｜-32｜=32round off四舍五入6.有关数论natural number自然数positive number正数negative number负数odd integer奇整数,odd number奇数even integer,even number偶数integer,whole number整数positive whole number正整数negative whole number负整数consecutive number连续整数real number,rational number实数,有理数irrational(number)无理数inverse倒数composite number合数e.g.4,6,8,9,10,12,14,15……reciprocal 倒数common divisor公约数multiple倍数(least)common multiple(最小)公倍数(prime)factor(质)因子common factor公因子prime number质数e.g.2,3,5,7,11,13,15……ordinary scale,decimal scale十进制nonnegative非负的tens十位units个位mode众数median中数common ratio公比7.数列arithmetic progression(sequence)等差数列geometric progression(sequence)等比数列8.其它approximate近似(anti)clockwise(逆)顺时针方向cardinal基数ordinal序数direct proportion正比distinct不同的estimation估计，近似parentheses括号proportion比例permutation排列combination组合table表格trigonometric function三角函数unit单位,位几何部分1.所有的角alternate angle内错角corresponding angle同位角vertical angle对顶角central angle圆心角interior angle内角exterior angle外角supplementary angles补角complementary angle余角adjacent angle邻角acute angle锐角obtuse angle钝角right angle直角round angle周角straight angle平角included angle夹角2.所有的三角形equilateral triangle等边三角形scalene triangle不等边三角形isosceles triangle等腰三角形right triangle直角三角形oblique斜三角形inscribed triangle内接三角形3.有关收敛的平面图形，除三角形外semicircle半圆concentric circles同心圆quadrilateral四边形pentagon五边形hexagon六边形heptagon七边形octagon八边形nonagon九边形decagon十边形polygon多边形parallelogram平行四边形equilateral等边形plane平面square正方形，平方rectangle长方形regular polygon正多边形rhombus菱形trapezoid梯形4.其它平面图形arc弧line,straight line直线line segment线段parallel lines平行线segment of a circle弧形5.有关立体图形cube立方体，立方数rectangular solid长方体regular solid/regular polyhedron正多面体circular cylinder圆柱体cone圆锥sphere球体solid立体的6.有关图形上的附属物altitude高depth深度side边长circumference,perimeter周长radian弧度surface area表面积volume体积arm直角三角形的股cross section横截面center of acircle圆心chord弦radius半径angle bisector角平分线diagonal对角线diameter直径edge棱face of a solid立体的面hypotenuse斜边included side夹边leg三角形的直角边median of a triangle三角形的中线base底边，底数(e.g.2的5次方，2就是底数) opposite直角三角形中的对边midpoint中点endpoint端点vertex(复数形式vertices)顶点tangent切线的transversal截线intercept截距7.有关坐标coordinate system坐标系rectangular coordinate直角坐标系origin原点abscissa横坐标ordinate纵坐标Number line数轴quadrant象限slope斜率complex plane复平面8.其它plane geometry平面几何trigonometry三角学bisect平分circumscribe外切inscribe内切intersect相交perpendicular垂直Pythagorean theorem勾股定理congruent全等的multilateral多边的其它相关词汇cent美分penny一美分硬币nickel5美分硬币dime一角硬币dozen打(12个)score廿(20个)Centigrade摄氏Fahrenheit华氏quart夸脱gallon加仑(1gallon=4quart)yard码meter米micron微米inch英寸foot英尺minute分(角度的度量单位，60分=1度) square measure平方单位制cubic meter立方米pint品脱(干量或液量的单位)。
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。
2019年美国数学竞赛(AMC10B)的试题与解答
这时第二个容器里充满了 3 的水. 问小容器与大容器的容积 4
之比是多少?
解: 设第一、二个容器的容积分别为 V1 和 V2, 依题意有
5 6 V1 =
3 4 V2,
从而
V1 V2
=
3 4 5 6
=
9 , 故 (D) 正确. 10
2. Consider the statement, ”If n is not prime, then n − 2 is
√ 由垂线段最短知, m2 + n2
|√am + bn|
=
√ 5.
当且仅当
a2 + b2
OA 与直线 ax + by = 5 垂直即 an = bm 时取最小值.
解法 2 由柯西不等式, (a2 +b2)(m2 +n2) (am+bn)2,
即 5(m2 + n2) 等号成立.
√ 25, m2 + n2
6 ter and the second was empty. She poured all the water from the
first container into the second container, at which point the sec3
ond container was full of water. What is the ratio of the volume 4
解: 反例是指符合命题的条件但不符合命题的结论的例
子, 即满足“若 n 不是素数, 则 n − 2 是素数. ”选项中只有 27
满足, 故 (E) 正确.
评注 运用柯西不等式关键是“配凑”形式, 充分利用已
知条件. 当出现分式和的形式时, 一般构造分式分母和作为
美国高中学生数学竞赛题
美国高中学生数学竞赛题1.(1995年文理)设(3x-1)6=a6x6+a5x5+a4x4+a3x3+a2x2+a1x+a0，求a6+a5+a4+a3+a2+a1+a0的值。
答案：64。
2.(1989年文)如果(1-2x)7=a0+a1x+a2x2+…+a7x7，那么a1+a2+…+a7的值等于()A.-2B.-1C.0D.2答案：(A)3.(1989年理)已知(1-2x)7=a0+a1x+a2x2+…+a7x7，那么a1+a2+…+a7=____。
答案：-2。
题源：(美28届10题)若(3x-1)7=a7x7+a6x6+…+a0，那么a7+a6+…+a0等于()A.0B.1C.64D.-64E.128答案：(E)改编点评：1题将指数7改为6，改为简答题；2题将底数(3x-1)改为(1-2x)，展开式改为x的升幂排列，所求结论中去掉了常数项a0，3题改编方法同2题，改为填空题。
4.(1990年文)已知f(x)=x5+ax3+bx-8，且f(-2)=10，那么f(2)等于()A.-26B.-18C.-10D.10答案：(A)题源：(美33届12题)设f(x)=ax7+bx3+cx-5，其中a.b和c是常数，如图f(-7)=7，那么f(7)等于()A.-17B.-7C.14D.21E.不能唯一确定答案：(A)改编点评：降低了次数，减少了一个字母系数，降低了难度。
5.(1990年文理)如果实数x、y满足等式(x-2)2+y2=3，那么的最大值是()A. B. C. D.答案：(D)题源：(美35届29题)在满足方程(x-3)2+(y-3)2=6的实数对(x，y)中，的最大值是()A.3+2B.2+C.3D.6E.6+2答案：(A)改编点评：圆方程中的圆心坐标、半径作了改变，题设的叙述方式也作了变化。
6.(1990年文理)函数y=+++的值域是()A.{-2，4}B.{-2，0，4}C.{-2，0，2，4}D.{-4，-2，0，4}答案：(B)题源：(美28届8题)非零实数的每一个三重组(a，b，c)构成一个数。
2019美国数学联赛高三组试题(2019AMC12A)
1、一个半径为4英尺的披萨的面积比一个半径为3英尺披萨大N ％，那个整数最接近于N ？ A 、25 B 、33 C 、44 D 、66 E 、782、如果a 是b 的150％，那么b 3是a 的百分之几？ A 、50B 、3266C 、150D 、200E 、4503、一个盒子里由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。
至少要从盒子里一次性取出多少个球，才能确保有其中有15个颜色相同的球。
A 、75 B 、76 C 、79 D 、84 E 、914、若干连续整数的和为45，这些连续整数中，最大的整数为n ，则n 的最大值是什么？ A 、9 B 、25 C 、45 D 、90 E 、1205、两条直线的斜率分别为2和21，相交于点()2,2，两条直线和直线10=+y x 所围成三角形的面积是多少？ A 、4B 、24C 、6D 、8E 、266、如下图所示，直线l 上有着有一个正方形和一条线段构成的无限多个一样的花纹。
下面四种平面变换中，有多少个是变换后所得的图形和原来图形相同的变换。
（1）绕着某个直线l 上的点的旋转变换； （2）某个沿着于直线l 平行方向的平移变换； （3）以直线l 为对称轴的反射变换；（4）以垂直于直线l 的直线为对称轴的反射变换 A 、0 B 、1 C 、2 D 、3E 、47、Melenie 计算2019年365天的日期的平均数μ，中位数M ，和众数。
记12月1日为121，12月2日为122，12月28日为1228，7月31日为731。
设所有众数的平寻书为d ，下面那个选项是对的？ A 、M d <<μB 、μ<<d MC 、μ==M dD 、μ<<M dE 、M d <<μ8、在平面有四条不同的直线，这些直线恰好有N 个交点，那么N 所有可能取值之和是多？ A 、14 B 、16 C 、18 D 、19 E 、21 9、已知一个数列{}n a 满足11=a ，732=a ，并且12122-----⋅=n n n n n a a aa a 对于任意3≥n 恒成立，而且2019a 可以写成()*,N q p qp∈。
精选美国amc中文版试题及答案
2000到2012年A M C10美国数学竞赛P 0 A 0B 0 C0 D 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2001年在美国举办，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) △PAB 之周长 (c) △PAB 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 。
高中数学竞赛典型题目
数学竞赛典型题目（一）1.(美国数学竞赛)设n a a a ,,,21是整数列,并且他们的最大公因子是1.令S 是一个整数集,具有性质:（1）),,2,1(n i S a i （2）}),,2,1{,(n ji S a a ji,其中j i,可以相同（3）对于S y x,，若S yx，则Syx证明：S 为全体整数的集合。
2．(美国数学竞赛)c b a ,,是正实数，证明：3252525)()3)(3)(3(c b a ccbbaa3．(加拿大数学竞赛)T 为1002004的所有正约数的集合，求集合T 的子集S 中的最大可能的元素个数。
其中S 中没有两个元素，一个是另一个的倍数。
4．(英国数学竞赛)证明：存在一个整数n 满足下列条件：（1）n 的二进制表达式中恰好有2004个1和2004个0；（2）2004能整除n .5．(英国数学竞赛)在0和1之间，用十进制表示为21.0a a 的实数x 满足：在表达式中至多有2004个不同的区块形式，)20041(20031ka a a kkk ，证明：x 是有理数。
6．(亚太地区数学竞赛)求所有由正整数组成的有限非空数集S ，满足：如果S nm,，则Sn m n m),(7．(亚太地区数学竞赛)平面上有2004个点，并且无三点共线，S 为通过任何两点的直线的集合。
证明：点可以被染成两种颜色使得两点同色当且仅当S 中有奇数条直线分离这两点。
8．(亚太地区数学竞赛)证明：)()!1(*2N n nnn 是偶数。
9．(亚太地区数学竞赛)z y x ,,是正实数，证明：)(9)2)(2)(2(222zx yz xy zyx10．（越南数学竞赛）函数f 满足)0(2sin 2cos )(cot xx xx f ，令)11)(1()()(xx f x f x g ，求)(x g 在区间]1,1[的上最值。
11．（越南数学竞赛）定义17612)(,91524)(2323x xxx q x xxx p ，证明：（1）每个多项式都有三个不同的实根；（2）令A 为)(x p 的最大实根，B 为)(x q 的最大实根，证明：4322B A 12．（越南数学竞赛）令F 为所有满足R R f :且x x f f x f )]2([)3(对任意R x成立的函数f 的集合。
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美国高中学生数学竞赛题
1.(1995年文理)设(3x-1)6=a6x6+a5x5+a4x4+a3x3+a2x2+a1x+a0，求a6+a5+a4+a3+a2+a1+a0的值。
答案：64。
2.(1989年文)如果(1-2x)7=a0+a1x+a2x2+…+a7x7，那么a1+a2+…+a7的值等于()
A.-2
B.-1
C.0
D.2
答案：(A)
3.(1989年理)已知(1-2x)7=a0+a1x+a2x2+…+a7x7，那么a1+a2+…+a7=____。
答案：-2。
题源：(美28届10题)若(3x-1)7=a7x7+a6x6+…+a0，那么a7+a6+…+a0等于()
A.0
B.1
C.64
D.-64
E.128
答案：(E)
改编点评：1题将指数7改为6，改为简答题；2题将底数(3x-1)改为(1-2x)，展开式改为x的升幂排列，所求结论中去掉了常数项a0，3题改编方法同2题，改为填空题。
4.(1990年文)已知f(x)=x5+ax3+bx-8，且f(-2)=10，那么f(2)等于()
A.-26
B.-18
C.-10
D.10
答案：(A)
题源：(美33届12题)设f(x)=ax7+bx3+cx-5，其中a.b和c是常数，如图f(-7)=7，那么f(7)等于()
A.-17
B.-7
C.14
D.21
E.不能唯一确定
答案：(A)
改编点评：降低了次数，减少了一个字母系数，降低了难度。
5.(1990年文理)如果实数x、y满足等式(x-2)2+y2=3，那么的最大值是()
A. B. C. D.
答案：(D)
题源：(美35届29题)在满足方程(x-3)2+(y-3)2=6的实数对(x，y)中，的最大值是()
A.3+2
B.2+
C.3
D.6
E.6+2
答案：(A)
改编点评：圆方程中的圆心坐标、半径作了改变，题设的叙述方式也作了变化。
6.(1990年文理)函数y=+++的值域是()
A.{-2，4}
B.{-2，0，4}
C.{-2，0，2，4}
D.{-4，-2，0，4}
答案：(B)
题源：(美28届8题)非零实数的每一个三重组(a，b，c)构成一个数。
如此构成的所有数的集是()
A.{0}
B.{-4，0，4}
C.{-4，-2，0，2，4}
D.{-4，-2，2，4}
E.这些都不对
答案：(B)
改编点评：将a、b、c改为三角函数，考查的知识面更广。
7.(1992年文理)长方体的全面积为11，十二条棱长度之和为24，则这个长方体的一条对角线长为()
A.2
B.
C.5
D.6
答案：(C)
题源：(美35届25题)一个长方体的表面积为22cm2，并且它的所有棱的总长度为24cm，那么它的对角线的长度(按cm计)是()
A. B. C. D. E.不能被唯一确定
答案：(D)
改编点评：作了两个方面的变化：将全面积22cm2改为11，去掉单位。
8.(1992年文理)如果函数f(x)=x2+bx+c对任意实数t都有f(2+t)=f(2-t)，那么()
A.f(2)＜f(1)＜f(4)
B.f(1)＜f(2)＜f(4)
C.f(2)＜f(4)＜f(1)
D.f(4)＜f(2)＜f(1)
题源：(美35届16题)函数f(x)对于一切实数x都满足f(2+x)=f(2-x)，如果方程f(x)=0恰好有四个不同的实根，那么这些根的和是()
A.0
B.2
C.4
D.6
E.8
答案：(E)
改编点评：题源是一个抽象函数问题，有一些难度，作为高考题是不合适的，8题主要采用题源中对称轴的表达方式，通过设计一个二次函数，结合对称性考查函数的单调性。
9.(1994年文理)已知sinθ+cosθ=，θ∈(0，π)，则ctgθ的值是____。
答案：-。
题源：(美29届15题)若sinx+cosx=，且0≤x≤π，那么tgx是()
A.-
B.-
C.
D.
答案：(A)
改编点评：x改为θ，将求正切值改为求余切值，改成填空题。
10.(1995年文理)等差数列{an}，{bn}的前n项和分别为Sn和Tn，若=，则等于()
A.1
B.
C.
D.
答案：(C)
题源：(美20届32题)设Sn和Tn分别为两个等差数列的前n项和，如果对所有的n，有=，则第一个数列与第二个数列的第十一项的比是()
A. B. C. D. E.不能确定
答案：(A)
改编点评：改变的比值，设问方式作了较大变化，将求某一项的比值改为求比值的极限。
11.(2002年文理)不等式(1+x)(1-|x|)＞0的解集是()
A.{x|0≤x＜1}
B.{x|x＜0且x≠-1}
C.{x|-1＜n＜1}
D.{x|x＜1且x≠-1}
答案：(D)
题源：(美27届7题)若x是实数，那么(1-|x|)(1+x)是正数的充分必要条件是() A.|x|＜1 B.x＜1 C.|x|＞1 D.x＜-1 E.x＜-1或-1＜x＜1
答案：(E)
改编点评：改变了设问方式，将求充要条件改为求不等式解集。