2008AMC1012A-TM-problems
2008 AMC 12A Problems(答案)
2008 AMC 12A ProblemsProblem 1A bakery owner turns on his doughnut machine at . At themachine has completed one third of the day's job. At what time will the doughnut machine complete the job?SolutionThe machine completes one-third of the job in hours. Thus,the entire job is completed in hours.Since the machine was started at , the job will be finished hours later, at. The answer is .Problem 2What is the reciprocal of ?Solution.Problem 3Suppose that of bananas are worth as much as oranges. How many orangesare worth as much as of bananas?SolutionIf , then.Problem 4Which of the following is equal to the productSolutionSolution 1.Solution 2Notice that everything cancels out except for in the numerator and in thedenominator.Thus, the product is , and the answer is .Problem 5Suppose that is an integer. Which of the following statements must be true about ?SolutionFor to be an integer, mustbe even, but not necessarily divisible by . Thus, the answer is .Problem 6Heather compares the price of a new computer at two different stores. Storeoffers off the sticker price followed by a rebate, and store offers offthe same sticker price with no rebate. Heather saves by buying the computer at store instead of store . What is the sticker price of the computer, in dollars?SolutionSolution 1Let the sticker price be .The price of the computer is at store , and at store .Heather saves at store , so .Solving, we find , and the thus answer is .Solution 2The in store is better than the additional off at store .Thus the off is equal to , and therefore the sticker price is .Problem 7While Steve and LeRoy are fishing 1 mile from shore, their boat springs a leak, and water comes in at a constant rate of 10 gallons per minute. The boat will sink if it takes in more than 30 gallons of water. Steve starts rowing toward the shore at a constant rate of 4 miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking?SolutionIt will take of an hour orminutes to get to shore.Since only gallons of water can enter the boat, only net gallons can enterthe boat per minute. Sincegallons of water enter the boat each minute, LeRoy must bailgallons per minute.Problem 8What is the volume of a cube whose surface area is twice that of a cube with volume 1?SolutionA cube with volume has a side of lengthand thus a surface area of.A cube whose surface area is has a side of length and avolume of .Problem 9Older television screens have an aspect ratio of . That is, the ratio of the widthto the height is . The aspect ratio of many movies is not , so they aresometimes shown on a television screen by "letterboxing" - darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of and is shown on an older television screen with a -inchdiagonal. What is the height, in inches, of each darkened strip?SolutionLet the width and height of the screen be and respectively, and let the widthand height of the movie be and respectively.By the Pythagorean Theorem, the diagonal is . So.Since the movie and the screen have the same width, .Thus, the height of each strip is .Problem 10Doug can paint a room in hours. Dave can paint the same room in hours. Dougand Dave paint the room together and take a one-hour break for lunch. Let be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by ?SolutionDoug can paint of a room per hour, Dave can paint of a room in an hour, and the time they spend working together is .Since rate times time gives output,Problem 11Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the visible numbers have the greatest possiblesum. What is that sum?SolutionTo maximize the sum of the faces that are showing, we can minimize the sum ofthe numbers of the faces that are not showing.The bottom cubes each have a pair of opposite faces that are covered up. Whenthe cube is folded, ; ; and are opposite pairs. Clearlyhas the smallest sum.The top cube has 1 number that is not showing. The smallest number on a face is .So, the minimum sum of the unexposed faces is . Since the sum ofthe numbers on all the cubes is , the maximumpossible sum of visible numbers is .Problem 12A function has domain and range. (The notation denotes.) What are the domain and range, respectively, of the functiondefined by ?Solutionis defined if is defined. Thus the domain is all.Since , . Thusis the range of .Thus the answer is .Problem 13Points and lie on a circle centered at , and . A second circle isinternally tangent to the first and tangent to both and . What is the ratio of the area of the smaller circle to that of the larger circle?SolutionLet be the center of the small circle with radius , and let be the point wherethe small circle is tangent to . Also, let be the point where the small circle istangent to the big circle with radius .Then is a right triangle, and a triangle at that. Therefore,.Since , we have , or, or .Then the ratio of areas will be squared, or .Problem 14What is the area of the region defined by the inequality?Area is invariant under translation, so after translating left and up units, wehave the inequalitywhich forms a diamond centered at the origin and vertices at .Thus the diagonals are of length and . Using the formula , theanswer is .Problem 15Let . What is the units digit of ?Solution.So, . Since is a multiple of four and the unitsdigit of powers of two repeat in cycles of four, .Therefore, . So the units digit is .Problem 16The numbers , , and are the first three terms of anarithmetic sequence, and the term of the sequence is . What is ?Solution 1Let and .The first three terms of the arithmetic sequence are , , and, and the term is .Thus, .Since the first three terms in the sequence are , , and , the th termis .Thus the term is .Solution 2If , , and are in arithmetic progression, then ,, and are in geometric progression. Therefore,Therefore, , , therefore the 12th term in the sequence isProblem 17Let be a sequence determined by the rule if is evenand if is odd. For how many positive integers is it true that is less than each of , , and ?SolutionAll positive integers can be expressed as , , , or , whereis a nonnegative integer.▪ If , then .▪ If , then ,, and.▪If, then.▪ If , then,, and.Since , every positive integerwill satisfy.Since one fourth of the positive integers can be expressed as,where is a nonnegative integer, the answer is.Problem 18Triangle, with sides of length , , and , has one vertex on the positive-axis, one on the positive -axis, and one on the positive -axis. Let be the origin .What is the volume of tetrahedron?SolutionWithout loss of generality, let be on the axis, be on the axis, and be onthe axis, and let have respective lengths of 5, 6, and 7. Letdenote the lengths of segments respectively. Then by thePythagorean Theorem, so ; similarly,and . Since , , and are mutually perpendicular, thetetrahedron's volume is which is answer choice C.Problem 19In the expansion of what isthe coefficient of ?SolutionLet and . We areexpanding .Since there are terms in , there are ways to choose one term fromeach . The product of the selected terms is for some integer between andinclusive. For each , there is one and only one in . Since there isonly one way to choose one term from each to get a product of , there areways to choose one term from each and one term from to get aproduct of . Thus the coefficient of the term is .Problem 20Triangle has , , and . Point is on , andbisects the right angle. The inscribed circles of and have radiiand , respectively. What is ?SolutionBy the Angle Bisector Theorem, By Law of Sines on, Since the areaof a triangle satisfies , where the inradius and the semiperimeter,we have Since and share the altitude (to ),their areas are the ratio of their bases, or The semiperimetersare and . Thus,Problem 21A permutation of is heavy-tailed if. What is the number of heavy-tailed permutations?SolutionThere are total permutations.For every permutation such that , there isexactly one permutation such that . Thus it suffices to count thepermutations such that ., , and are the only combinations of numbers that can satisfy .There are combinations of numbers, possibilities of which side of the equation isand which side is , and possibilities for rearrangingand . Thus, there are permutations such that.Thus, the number of heavy-tailed permutations is .Problem 22A round table has radius . Six rectangular place mats are placed on the table. Eachplace mat has width and length as shown. They are positioned so that each mathas two corners on the edge of the table, these two corners being end points of the same side of length . Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is ?SolutionSolution 1 (trigonometry)Let one of the mats be , and the center be as shown:Since there are mats, is equilateral. So, . Also,.By the Law of Cosines:.Since must be positive, .Solution 2 (without trigonometry)Draw and as in the diagram. Draw the altitude from to and call theintersectionAs proved in the first solution, . That makes atriangle, so andSince is a right triangle,Solving for givesProblem 23The solutions of the equation are the vertices of a convex polygon in the complex plane. What is the area of the polygon?SolutionLooking at the coefficients, we are immediately reminded of the binomial expansionof .Modifying this slightly, we can write the given equation as:We can apply a translation of and a rotation of(both operations preserve area) to simplify the problem:Because the roots of this equation are created by rotating radians successively about the origin, the quadrilateral is a square.We know that half the diagonal length of the square isTherefore, the area of the square isProblem 24Triangle has and . Point is the midpoint of . What isthe largest possible value of ?SolutionLet . Then , and sinceand , we haveWith calculus, taking the derivative and setting equal to zero will give the maximum value of . Otherwise, we can apply AM-GM:Thus, the maximum is at .Problem 25A sequence , , , of points in the coordinate plane satisfiesfor .Suppose that . What is ?SolutionThis sequence can also be expressed using matrix multiplication as follows:.Thus, is formed by rotating counter-clockwise about the originbyand dilating the point's position with respect to the origin by a factor of .So, starting with and performing the above operationstimes inreverse yields.Rotating clockwise by yields . A dilation by a factor ofyields the point .Therefore, .。
2010 AMC 10A 试题及答案解析
2010 AMC 10A problems and solutions.The test was held on February 8, 2010. The first link contains the full set of test problems. The rest contain each individual problem and its solution.Problem 1Mary’s top book shelf holds five books with the follow ing widths, incentimeters: , , , , and . What is the average book width, in centimeters?SolutionTo find the average, we add up the widths , , , , and , to get a total sum of . Since there are books, the average book width isThe answer is .Problem 2Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?SolutionLet the length of the small square be , intuitively, the length of the big square is . It can be seen that the width of the rectangle is .Thus, the length of the rectangle is times large as the width. The answer is .Problem 3Tyrone had marbles and Eric had marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?SolutionLet be the number of marbles Tyrone gave to Eric. Then,. Solving for yields and . The answer is .Problem 4A book that is to be recorded onto compact discs takes minutes to read aloud. Each disc can hold up to minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?SolutionAssuming that there were fractions of compact discs, it would take CDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have minutes on each of the 8 discs. The answer is .Problem 5The area of a circle whose circumference is is . What is the value of ?SolutionIf the circumference of a circle is , the radius would be . Since the area of a circle is , the area is . The answer is . Problem 6For positive numbers and the operation is defined asWhat is ?Solution. Then, is The answer isProblem 7Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?SolutionCrystal first runs North for one mile. Changing directions, she runs Northeast for another mile. The angle difference between North and Northeast is 45 degrees. She then switches directions to Southeast, meaning a 90 degree angle change. The distance now from travelling North for one mile, and her current destination is miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to , which is equal to . The answer isTony works hours a day and is paid $per hour for each full year of his age. During a six month period Tony worked days and earned $. How old was Tony at the end of the six month period?SolutionTony worked hours a day and is paid dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is years old, he gets dollars a day. We also know that he worked days and earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. Because he earned dollars, we know that he was for some period of time, but not the whole time, because then the money earned would be greater than or equal to . This is why he was when he began, but turned sometime in the middle and earned dollars in total. So the answer is .The answer is . We could find out for how long he was and . . Then isand we know that he was for days, and for days. Thus, the answer is .Problem 9A palindrome, such as , is a number that remains the same when its digits are reversed. The numbers and are three-digit and four-digit palindromes, respectively. What is the sum of the digits of ?Solutionis at most , so is at most . The minimum value ofis . However, the only palindrome between and is , which means that must be .It follows that is , so the sum of the digits is .Marvin had a birthday on Tuesday, May 27 in the leap year . In what year will his birthday next fall on a Saturday?Solution(E) 2017There are 365 days in a non-leap year. There are 7 days in a week. Since 365 = 52 * 7 + 1 (or 365 is congruent to 1 mod 7), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.For example: 5/27/08 Tue 5/27/09 WedHowever, a leap year has 366 days, and 366 = 52 * 7 + 2. So the same date (after February) moves "forward" two days in the subsequent year, if that year is a leap year.For example: 5/27/11 Fri 5/27/12 SunYou can keep count forward to find that the first time this date falls on a Saturday is in 2017:5/27/13 Mon 5/27/14 Tue 5/27/15 Wed 5/27/16 Fri 5/27/17 Sat Problem 11The length of the interval of solutions of the inequality is . What is ?SolutionSince we are given the range of the solutions, we must re-write the inequalities so that we have in terms of and .Subtract from all of the quantities:Divide all of the quantities by .Since we have the range of the solutions, we can make them equal to .Multiply both sides by 2.Re-write without using parentheses.Simplify.We need to find for the problem, so the answer isProblem 12Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?SolutionThe water tower holds times more water than Logan's miniature. Therefore, Logan should make his towertimes shorter than the actual tower. This ismeters high, or choice .Problem 13Angelina drove at an average rate of kph and then stopped minutes for gas. After the stop, she drove at an average rate of kph. Altogether she drove km in a total trip time of hours including the stop. Which equation could be used to solve for the time in hours that she drove before her stop?SolutionThe answer is （）because she drove at kmh for hours (the amount of time before the stop), and 100 kmh for because she wasn't driving for minutes, or hours. Multiplying by gives the total distance, which is kms. Therefore, the answer isProblem 14Triangle has . Let and be on and , respectively, such that . Let be the intersection of segments and , and suppose that is equilateral. What is ?SolutionLet .Since ,Problem 15In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.Brian: "Mike and I are different species."Chris: "LeRoy is a frog."LeRoy: "Chris is a frog."Mike: "Of the four of us, at least two are toads."How many of these amphibians are frogs?SolutionSolution 1We can begin by first looking at Chris and LeRoy.Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog.Clearly, Chris and LeRoy are different species, and so we have at least frog out of the two of them.Now suppose Mike is a toad. Then what he says is true because we already have toads. However, if Brian is a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction.Therefore, Mike must be a frog. His statement must be false, which means that there is at most toad. Since either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false.Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have frogs total.Solution 2Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.As Mike is a frog, his statement is false, hence there is at most one toad.As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad. Hence we must have one toad and three frogs.Problem 16Nondegenerate has integer side lengths, is an angle bisector, , and . What is the smallest possible value of the perimeter?SolutionBy the Angle Bisector Theorem, we know that . If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then , contradicting the Triangle Inequality. If we use the next lowest values (and ), the Triangle Inequality is satisfied. Therefore, our answer is , or choice .Problem 17A solid cube has side length inches. A -inch by -inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?SolutionSolution 1Imagine making the cuts one at a time. The first cut removes a box . The second cut removes two boxes, each of dimensions, and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is .Therefore the volume of the rest of the cube is.Solution 2We can use Principle of Inclusion-Exclusion to find the final volume of the cube.There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has cubic inches. However, we can not just sum their volumes, as the central cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.Hence the total volume of the cuts is.Therefore the volume of the rest of the cube is.Solution 3We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.Each edge can be seen as a box, and each corner can be seen as a box..Problem 18Bernardo randomly picks 3 distinct numbers from the setand arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?SolutionWe can solve this by breaking the problem down into cases and adding up the probabilities.Case : Bernardo picks . If Bernardo picks a then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a is .Case : Bernardo does not pick . Since the chance of Bernardo picking is , the probability of not picking is .If Bernardo does not pick 9, then he can pick any number from to . Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger.Ignoring the for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers.We get this probability to beProbability of Bernardo's number being greater isFactoring the fact that Bernardo could've picked a but didn't:Adding up the two cases we getProblem 19Equiangular hexagon has side lengthsand . The area of is of the area of the hexagon. What is the sum of all possible values of ?SolutionSolution 1It is clear that is an equilateral triangle. From the Law of Cosines, we get that . Therefore, the area of is .If we extend , and so that and meet at , and meet at , and and meet at , we find that hexagon is formed by taking equilateral triangle of side length and removing three equilateral triangles, , and , of side length . The area of is therefore.Based on the initial conditions,Simplifying this gives us . By Vieta's Formulas we know that the sum of the possible value of is .Solution 2As above, we find that the area of is .We also find by the sine triangle area formula that, and thusThis simplifies to.Problem 20A fly trapped inside a cubical box with side length meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?SolutionThe distance of an interior diagonal in this cube is and the distance of a diagonal on one of the square faces is . It would not make sense if the fly traveled an interior diagonal twice in a row, as it would return to the point it just came from, so at most the final sum can only have 4 as the coefficient of . The other 4 paths taken can be across a diagonal on one of the faces, so the maximum distance traveled is.Problem 21The polynomial has three positive integer zeros. What is the smallest possible value of ?SolutionBy Vieta's Formulas, we know that is the sum of the three roots of the polynomial . Also, 2010 factors into. But, since there are only three roots to the polynomial, two of the four prime factors must be multiplied so that we are left with three roots. To minimize , and should be multiplied, which means will be and the answer is .Problem 22Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point insidethe circle. How many triangles with all three vertices in the interior of the circle are created?SolutionTo choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. Therefore, the answer is which is equivalent to 28,Problem 23Each of 2010 boxes in a line contains a single red marble, and for , the box in the position also contains white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let be the probability that Isabella stops afterdrawing exactly marbles. What is the smallest value of for which ?SolutionThe probability of drawing a white marble from box is . Theprobability of drawing a red marble from box is .The probability of drawing a red marble at box is thereforeIt is then easy to see that the lowest integer value of that satisfies the inequality is .Problem 24The number obtained from the last two nonzero digits of is equal to . What is ?SolutionWe will use the fact that for any integer ,First, we find that the number of factors of in is equal to. Let . The we want is therefore the last two digits of , or . Since there is clearly an excess of factors of 2, we know that , so it remains to find .If we divide by by taking out all the factors of in , we canwrite as where where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form is replaced by , and every number in the form is replaced by .The number can be grouped as follows:Using the identity at the beginning of the solution, we can reducetoUsing the fact that (or simply the fact that if you have your powers of 2 memorized), we can deduce that . Therefore.Finally, combining with the fact that yields.Problem 25Jim starts with a positive integer and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with , then his sequence contains numbers:Let be the smallest number for which Jim’s sequence has numbers. What is the units digit of ?SolutionWe can find the answer by working backwards. We begin with on the bottom row, then the goes to the right of the equal's sign in the row above. We find the smallest value for whichand , which is .We repeat the same procedure except with for the next row and for the row after that. However, at the fourth row, wesee that solving yields , in which case it would be incorrect since is not the greatest perfect square less than or equal to . So we make it a and solve . We continue on using this same method where we increase the perfect square until can be made bigger than it. When we repeat this until we have rows, we get:Hence the solution is the last digit of , which is .。
2008-AMC10-Bsolutions
T he MATheMATICAL ASSOCIATION Of AMeRICAAmerican Mathematics Competitions9th Annual American Mathematics Contest 10AMC 10Contest BSolutions PamphletWednesday, FEBRUARY 27, 2008This Pamphlet gives at least one solution for each problem on this year’s contest and shows that all problems can be solved without the use of a calculator. When more than one solution is provided, this is done to illustrate a significant contrast in methods, e.g., algebraic vs geometric, computational vs conceptual, elementary vs advanced. These solu-tions are by no means the only ones possible, nor are they superior to others the reader may devise.We hope that teachers will inform their students about these solutions, both as illustrations of the kinds of ingenuity needed to solve nonroutine problems and as examples of good mathematical exposition. However,the publication, reproduction or communication of the problems or solutions of the AMC 10 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via copier, telephone, e-mail, WorldWide Web or media of any type during this period is a violation of the competition rules.After the contest period, permission to make copies of problems in paper or electronic form including posting on web-pages for educational use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear the copyright notice.Correspondence about the problems/solutions for this AMC 10 and orders for any publications should be addressed to:American Mathematics CompetitionsUniversity of Nebraska, P.O. Box 81606, Lincoln, NE 68501-1606Phone: 402-472-2257; Fax: 402-472-6087; email: amcinfo@ The problems and solutions for this AMC 10 were prepared by the MAA’s Committee on the AMC 10 and AMC 12 under the direction of AMC 10 Subcommittee Chair:Chair: LeRoy Wenstrom, Columbus, MSCopyright © 2008, The Mathematical Association of America1.Answer (E):The number of points could be any integer between 5·2=10and 5·3=15,inclusive.The number of possibilities is 15−10+1=6.123489101115161718222324252.Answer (B):The two sums are 1+10+17+22=50and4+9+16+25=54,so the positive difference between the sumsis 54−50=4.Query:If a different 4×4block of dates had been chosen,theanswer would be unchanged.Why?3.Answer (D):The properties of exponents imply that3x √x = x ·x 12 13= x 32 13=x 12.4.Answer (C):A single player can receive the largest possible salary only when the other 20players on the team are each receiving the minimum salary of $15,000.Thus the maximum salary for any player is $700,000−20·$15,000=$400,000.5.Answer (A):Note that (y −x )2=(x −y )2,so(x −y )2$(y −x )2=(x −y )2$(x −y )2= (x −y )2−(x −y )2 2=02=0.6.Answer (C):Because AB +BD =AD and AB =4BD ,it follows thatBD =15·AD .By similar reasoning,CD =110·AD .ThusBC =BD −CD =15·AD −110·AD =110·AD.7.Answer (C):The side length of the large triangle is 10times the side length of each small triangle,so the area of the large triangle is 102=100times the area of each small triangle.8.Answer (C):The total cost of the carnations must be an even number of dollars.The total number of dollars spent is the even number 50,so the number of roses purchased must also be even.In addition,the number of roses purchased cannot exceed 503.Therefore the number of roses purchased must be one of theeven integers between 0and 16,inclusive.This gives 9possibilities for the number of roses purchased,and consequently 9possibilities for the number of bouquets.9.Answer (A):The quadratic formula implies that the two solutions arex 1=2a +√4a 2−4ab 2a and x 2=2a −√4a 2−4ab 2a,so the average is 12(x 1+x 2)=122a 2a +2a 2a =1.OR The sum of the solutions of a quadratic equation is the negative of the coefficient of the linear term divided by the coefficient of the quadratic term.In this case the sum of the solution is −(−2a )a=2.Hence the average of the solutions is1.10.Answer (A):Let O be the center of the circle,and let D be the intersection of OC and AB .Be-cause OC bisects minor arc AB ,OD is a perpen-dicular bisector of chord AB .Hence AD =3,andapplying the Pythagorean Theorem to ADO yieldsOD =√52−33=4.Therefore DC =1,and apply-ing the Pythagorean Theorem to ADC yields AC =√32+12=√10.11.Answer (B):Note that u 5=2u 4+9and 128=u 6=2u 5+u 4=5u 4+18.Thus u 4=22,and it follows that u 5=2·22+9=53.12.Answer (A):During the year Pete takes44×105+5×104=44.5×105steps.At 1800steps per mile,the number of miles Pete walks is44.5×10518×10=44.518×103≈2.5×103=2500.13.Answer (B):Because the mean of the first n terms is n ,their sum is n 2.Therefore the n th term is n 2−(n −1)2=2n −1,and the 2008th term is 2·2008−1=4015.14.Answer (B):Because OAB is a 30–60–90◦triangle,we have BA =5√33.Let A and B be the images of A and B ,respectively,under the rotation.ThenB =(0,5),B A is horizontal,and B A =BA=5√33.Hence A is in thesecond quadrant andA =−53√3,5.15.Answer(A):By the Pythagorean Theorem we have a2+b2=(b+1)2,soa2=(b+1)2−b2=2b+1.Because b is an integer with b<100,a2is an odd perfect square between1and 201,and there are six of these,namely,9,25,49,81,121,and169.Hence a must be3,5,7,9,11,or13,and there are6triangles that satisfy the given conditions.16.Answer(A):If one die is rolled,3of the6possible numbers are odd.If twodice are rolled,18of the36possible outcomes have odd sums.In each of thesecases,the probability of an odd sum is12.If no die is rolled,the sum is0,whichis not odd.The probability that no die is rolled is equal to the probability thatboth coin tosses are tails,which is(12)2=14.Thus the requested probability is1−14·12=38.17.Answer(B):The responses on these three occasions,in order,must beYNN,NYN,or NNY,where Y indicates approval and N indicates disapproval.The probability of each of these is(0.7)(0.3)(0.3)=0.063,so the requested probability is3(0.063)=0.189.18.Answer(B):Let n be the number of bricks in the chimney.Then thenumber of bricks per hour Brenda and Brandon can lay working alone is n9andn 10,respectively.Working together they can lay(n9+n10−10)bricks in an hour,or5 n9+n10−10bricks in5hours to complete the chimney.Thus5 n9+n10−10=n,and the number of bricks in the chimney is n=900.ORSuppose that Brenda can lay x bricks in an hour and Brandon can lay y bricks in an hour.Then the number of bricks in the chimney can be expressed as9x,10y ,or 5(x +y −10).The equality of these expressions leads to the system of equations4x −5y =−50−5x +5y =−50.It follows that x =100,so the number of bricks in the chimney is 9x =900.19.Answer (E):The portion of each end of the tank thatis under water is a circular sector with two right trianglesremoved as shown.The hypotenuse of each triangle is 4,and the vertical leg is 2,so each is a 30–60–90◦triangle.Therefore the sector has a central angle of 120◦,and thearea of the sector is 120360·π(4)2=163π.The area of each triangle is 12(2) 2√3 ,so the portion of each end that is underwater has area 163π−4√3.The length of the cylinder is 9,so the volume of the water is 9 163π−4√3 =48π−36√3.20.Answer (B):Of the 36possible outcomes,the four pairs (1,4),(2,3),(2,3),and (4,1)yield a sum of 5.The six pairs (1,6),(2,5),(2,5),(3,4),(3,4),and (4,3)yield a sum of 7.The four pairs (1,8),(3,6),(3,6),and (4,5)yield a sum of 9.Thus the probability of getting a sum of 5,7,or 9is (4+6+4)/36=7/18.Note:The dice described here are known as Sicherman dice.The probability of obtaining each sum between 2and 12is the same as that on a pair of standard dice.21.Answer (C):Let the women be seated first.The first woman may sit inany of the 10chairs.Because men and women must alternate,the number of choices for the remaining women is 4,3,2,and 1.Thus the number of possible seating arrangements for the women is 10·4!=240.Without loss of generality,suppose that a woman sits in chair 1.Then this woman’s spouse must sit in chair 4or chair 8.If he sits in chair 4then the women sitting in chairs 7,3,9,and 5must have their spouses sitting in chairs 10,6,2,and 8,respectively.If he sits in chair 8then the women sitting in chairs 5,9,3,and 7must have their spouses sitting in chairs 2,6,10,and 4,respectively.So for each possible seating arrangement for the women there are two arrangements for the men.Hence,there are 2·240=480possible seating arrangements.22.Answer (C):There are 6!/(3!2!1!)=60distinguishable orders of the beadson the line.To meet the required condition,the red beads must be placed inone of four configurations:positions 1,3,and 5,positions 2,4,and 6,positions 1,3,and 6,or positions 1,4,and 6.In the first two cases,the blue bead can be placed in any of the three remaining positions.In the last two cases,the blue bead can be placed in either of the two adjacent remaining positions.In each case,the placement of the white beads is then determined.Hence there are 2·3+2·2=10orders that meet the required condition,and the requestedprobability is 1060=16.23.Answer (B):Because the area of the border is half the area of the floor,thesame is true of the painted rectangle.The painted rectangle measures a −2by b −2feet.Hence ab =2(a −2)(b −2),from which 0=ab −4a −4b +8.Add 8to each side of the equation to produce8=ab −4a −4b +16=(a −4)(b −4).Because the only integer factorizations of 8are8=1·8=2·4=(−4)·(−2)=(−8)·(−1),and because b >a >0,the only possible ordered pairs satisfying this equation for (a −4,b −4)are (1,8)and (2,4).Hence (a,b )must be one of the two ordered pairs (5,12),or (6,8).A B CD M AB C D O 24.Answer (C):Let M be on the same side of line BC asA ,such that BMC is equilateral.Then ABM andMCD are isosceles with ∠ABM =10◦and ∠MCD =110◦.Hence ∠AMB =85◦and ∠CMD =35◦.There-fore∠AMD =360◦−∠AMB −∠BMC −∠CMD=360◦−85◦−60◦−35◦=180◦.It follows that M lies on AD and ∠BAD =∠BAM =85◦.ORLet ABO be equilateral as shown.Then∠OBC =∠ABC −∠ABO =70◦−60◦=10◦.Because ∠BCD =170◦and OB =BC =CD ,the quadrilateral BCDO is a parallelogram.ThusOD=BC=AO and AOD is isosceles.Letα=∠ODA=∠OAD.The sum of the interior angles of ABCD is360◦,so we have360=(α+60)+70+170+(α+10)andα=25.Thus∠DAB=60+α=85◦.25.Answer(B):Number the pails consecutively so that Michael is presently atpail0and the garbage truck is at pail1.Michael takes200/5=40seconds to walk between pails,so for n≥0he passes pail n after40n seconds.The truck takes20seconds to travel between pails and stops for30seconds at each pail.Thus for n≥1it leaves pail n after50(n−1)seconds,and for n≥2it arrives at pail n after50(n−1)−30seconds.Michael will meet the truck at pail n if and only if50(n−1)−30≤40n≤50(n−1)or,equivalently,5≤n≤8.50time (sec)distance (ft)200truckMichaelHence Michaelfirst meets the truck at pail5after200seconds,just as the truck leaves the pail.He passes the truck at pail6after240seconds and at pail7 after280seconds.Finally,Michael meets the truck just as it arrives at pail8 after320seconds.These conditions imply that the truck is ahead of Michael between pails5and6and that Michael is ahead of the truck between pails7 and8.However,the truck must pass Michael at some point between pails6and 7,so they meet a total offive times.TheAmerican Mathematics Competitionsare Sponsored byThe Mathematical Association of AmericaThe Akamai FoundationContributorsAmerican Mathematical Association of Two Year CollegesAmerican Mathematical SocietyAmerican Society of Pension ActuariesAmerican Statistical AssociationArt of Problem SolvingAwesome MathCanada/USA MathcampCanada/USA MathpathCasualty Actuarial SocietyClay Mathematics InstituteIDEA MathInstitute for Operations Research and the Management SciencesL. G. Balfour CompanyMu Alpha ThetaNational Assessment & TestingNational Council of Teachers of MathematicsPedagoguery Software Inc.Pi Mu EpsilonSociety of ActuariesU.S.A. Math Talent SearchW. H. Freeman and CompanyWolfram Research Inc.。
AMC10的真题答案及中文翻译
AMC10的真题答案及中文翻译AMC10的真题及中文翻译1、One ticket to a show costs $20 at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?(A) 2 (B) 5 (C) 10 (D) 15 (E) 20中文：一张展览票全价为20美元。
Susan用优惠券买4张票打七五折。
Pam用优惠券买5张票打七折。
Pam比Susan多花了多少美元？2、An aquarium has a rectangular base that measures 100cm by 40cm and has a height of 50cm. It is filled with water to a height of 40cm. A brick with a rectangular base that measures 40cm by 20cm and a height of 10cm is placed in the aquarium. By how many centimeters does that water rise?(A) 0.5 (B) 1 (C) 1.5 (D) 2 (E)2.5中文：一个养鱼缸有100cm×40cm的底，高为50cm。
它装满水到40cm的高度。
把一个底为40cm×20cm，高为10cm的砖块放在这个养鱼缸里。
鱼缸里的水上升了多少厘米？3、The larger of two consecutive odd integers is three times the smaller. What is their sum?(A) 4 (B) 8 (C) 12 (D) 16 (E) 20中文：2个连续的奇整数中较大的数是较小的数的3倍。
AMC10美国数学竞赛讲义全
AMC10美国数学竞赛讲义全AMC 中的数论问题1：Remember the prime between 1 to 100：2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 91 2：Perfect number:Let is the prime number.if21p - is also the prime number. then1(21)2p p --is the perfect number. For example:6,28,496.3: Let ,0n abc a =≠ is three digital integer .if 333n a b c =++Then the number n is called Daffodils number . There are only four numbers: 153 370 371 407 Let ,0n abcd a =≠ is four digital integer .if 4444d n a b c +=++Then the number n is called Roses number . There are only three numbers: 1634 8208 94744：The Fundamental Theorem of ArithmeticEvery natural number can be written as a product of primes uniquely up to order.5：Suppose that a and b are integers with b =0. Then there exists unique integers q a d such that 0 ≤ < |b| a d a = bq + .6：(1)G eatest Commo D v so : Let gcd (a, b) = max {d ∈ Z: d | a a d d | b}. For any integers a and b, we havegcd(a, b) = gcd(b, a) = gcd(±a, ±b) = gcd(a, b ? a) = gcd(a, b + a). For example: gcd(150, 60) = gcd(60, 30) = gcd(30, 0) = 30 (2)Least commo mult le:Let lcm(a,b)=m {d∈Z: a | d a d b | d }. (3)We have that: ab= gcd(a, b) lcm(a,b) 7：Congruence modulo nIf ,0a b mq m -=≠,then we call a congruence b modulo m and we rewritemod a b m ≡.(1)Assume a b c d m Z 0 m 0 .If a b mod m c d mod m then we havemod a c b d m ±≡± , mod ac bd m ≡ , mod k k a b m ≡(2) The equat o ax ≡ b (mod m) has a solut o f a d o ly f gcd(a, m) d v des b.8：How to find the unit digit of some special integers (1)How many zero at the end of !nFor example, when 100n =, Let N be the number zero at the end of 100!then10010010020424525125N=++=+=(2) ,,a n Z ∈Find the unit digit n a . For example, when 100,3n a ==9：Palindrome, such as 83438, is a number that remains the same when its digits are reversed.There are some number not only palindrome but 2 2 ，222 ， (1)Some special palindrome n that 2n is also palindrome. For example :222221111121111123211111123432111111111112345678987654321=====(2)How to create a palindrome? Almost integer plus the number of its reverseddigits and repeat it again and again. Then we get a palindrome. For example: 87781651655617267266271353135335314884+=+=+=+=But whether any integer has this Property has yet to prove(3) The palindrome equation means that equation from left to right and right to left it all set up. For example ：1242242112231132211121241388888831421211====Let ab and cde are two digital and three digital integers. If the digits satisfy the,,9a c b e d c e d ?=?=+≤, then ab cde edc ba ?=?.10: Features of an integer divisible by some prime number If n is even ，then 2|n⼀个整数n 的所有位数上的数字之和是3（或者9）的倍数，则n 被3（或者9）整除⼀个整数n 的尾数是零，则n 被5整除⼀个整数n 的后三位与截取后三位的数值的差被7、11、13整除，则n 被7、11、13整除⼀个整数n 的最后两位数被4整除，则n 被4整除⼀个整数n 的最后三位数被8整除，则n 被8整除⼀个整数n 的奇数位之和与偶数位之和的差被11整除，则n 被11整除 11. The number Theoretic functions If 312123t rr rrt n p p p p =(1) {}12()#0:|(1)(1)(1)t n a a n r r r χ=>=+++(2) 12222111222|()(1)(1)(1)t r r r t t t a nn a p pp p p p p p p δ==+++++++++∑(3) {}11221111122()#:,gcd(,)1()()()t t r r rr rr t t n a N a n a n p p p p p p φ---=∈≤==---For example: 2(12)(23)(21)(11)6χχ=?=++= 22(12)(23)(122)(13)28δδ=?=+++= 22(12)(23)(22)(31)4φφ=?=--= Exercise1. The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?(A) 4 (B) 5 (C) 6 (D) 7(E) 83. For the positive integer n, let denote the sum of all the positive divisors of n with the exception of n itself. For example,<4>=1+2=3 and <12>=1+2+3+4+6=16. What is <<<6>>>?(A) 6 (B) 12 (C) 24 (D) 32 (E) 36 8. What is the sum of all integer solutions to 21<(x-2)<25? (A) 10 (B) 12(C) 15(D) 19(E) 5(A) 6 (B) 7 (C) 8 (D) 9 (E) 10(A) 1(B) 2(C) 3(D) 4(E) 515．The figures 123,,F F F and 4F shown are the first in a sequence of figures. For3n ≥, n F is constructed from -1n F by surrounding it with a square and placing one more diamond on each side of the new square than -1n F had on each side of its outside square. For example, figure 3F has 13 diamonds. How many diamonds are there in figure 20F ?18. Positive integers a, b, and c are randomly and independently selected with replacement from the set {1, 2, 3,…, 2010}. What is the probability that abc ab a ++ is divisible by 3? (A)13(B)2981(C)3181(D)1127(E)132724. Let ,a b and c be positive integers with >>a b c such that 222-b -c +=2011a ab and222+3b +3c -3-2-2=-1997a ab ac bc . What is a ?(A) 249 (B) 250 (C) 251 (D) 252 (E)2535. In multiplying two positive integers a and b, Ron reversed the digits of the two-digit number a. His erroneous product was 161. What is the correct value of the product of a and b?(A) 116 (B) 161 (C) 204 (D) 214 (E) 224 23. What is the hundreds digit of 20112011?(A) 1 (B) 4 (C) 5 (D) 6 (E) 99. A palindrome, such as 83438, is a number that remains the same when its digits are reversed. The numbers x and x+32 are three-digit and four-digit palindromes, respectively. What is the sum of the digits of x?(A) 20 (B) 21 (C) 22 (D) 23 (E) 2421. The polynomial 322010x ax bx -+- has three positive integer zeros. What is the smallest possible value of a?(A) 78 (B) 88 (C) 98 (D) 108 (E) 11824. The number obtained from the last two nonzero digits of 90! Is equal to n. What is n?(A) 12 (B) 32 (C) 48 (D) 52 (E) 6825. Jim starts with a positive integer n and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with n=55, then his sequence contain 5 numbers:55 55-72= 6 6-22= 2 2-12= 1 1-12= 0Let N be the smallest umbe fo wh ch J m’s seque ce has umbe s. What is the units digit of N?(A) 1 (B) 3 (C) 5 (D) 7 (E) 9 21．What is the remainder when 01220093+3+3++3is divided by 8?(A) 0 (B) 1 (C) 2 (D) 4 (E) 65．What is the sum of the digits of the square of 111,111,111? (A) 18 (B) 27(C) 45(D) 63(E) 81ABC 100643a S 2(A) 6 (B) 7(C) 8(D) 9(E) 1024. Let 2200820082k =+. What is the units digit of 222k +? (A) 0 (B) 1(C) 4(D) 6(E) 8AMC about algebraic problems⼀、Linear relations(1) Slope y-intercept form: y kx b =+ (k is the slope, b is the y-intercept) (2)Standard form: 0Ax By C ++= (3)Slope and one point 0000(,),()()P x y k slope y y k x x -=- (4) Two points 1122(,),(,)P x y P x y12121212y y y y y y x x x x x x ---==--- (5)x,y-intercept form: (,0),(0,),(0,0)1x yP a Q b a b a b≠≠+= ⼆、the relations of the two lines 111222:0,:0l A x B y C l A x B y C ++=++=(1) 1l ∥2l 122112210,0A B A B C B C B ?-=-≠ (1) 1l ⊥2l 12120A A B B ?-=三、Special multiplication rules:222223322332212211222112222()()()2()()()()()()(2)()((1)(1))(1)()n n n n n n n n n n n n n n a b a b a b a b a ab b a b a b a ab b a b a b a ab b a b a b a a b ab b n a b a b a a b ab b n is odd n a b c ab bc ac a b -----------=-+±=±+-=-+++=+-+-=-++++≥+=+-++-+->++=++?-22()()0b c c a a b c+-+-=?==四、quadratic equations and PolynomialThe quadratic equations 2(0)y ax bx c a =++≠ has two roots 12,x x then we has1212b c x x x x a a+=-=More generally, if the polynomial 121210nn n n n x a x a x a x a ---+++++= has nroots 123,,,,n x x x x ,then we have:1231122312123(1)n n n n n nx x x x a x x x x x x a x x x x a -++++=-++==-开⽅的开⽅、估计开⽅数的⼤⼩绝对值⽅程Arithmetic Sequence123(1)(2)(3)()n m a a n d a n d a n d a n m d =+-=+-=+-==+-121321()()()()2222n n n m n m n n a a n a a n a a n a a s ---+++++=====1(1)2n n n ds na -=+If n=2k, then we have 1()n k k s k a a +=+ If n=2k+1, then we have 1n k s na += Geometric sequence123123n n n n m n m a a q a q a q a q ----=====1(1)1,1n n a q q s q-≠=-Some special sequence , , 2, 3, 5, ,… 9,99,999,9999，… 1，11，111，1111，… Exercise4 .When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be left over?7 For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?21. Four distinct points are arranged on a plane so that the segments connecting them have lengths ,,,,, and . What is the ratio of to ?13. An iterative average of the numbers 1, 2, 3, 4, and 5 is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, and then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?16. Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?24. Let ,a b and c be positive integers with >>a b c such that 222-b -c +=2011a ab and222+3b +3c -3-2-2=-1997a ab ac bc . What is a ?(A) 249 (B) 250(C) 251(D) 252(E) 2531. What is246135135246++++-++++? (A) -1 (B) 536(C) 712(D)14760(E)43310. Consider the set of numbers {1, 10, 102, 103(010)}. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?(A) 1 (B) 9 (C) 10 (D) 11 (E) 101=(A) -64 (B) -24 (C) -9 (D) 24 (E) 5764. Let X and Y be the following sums of arithmetic sequences: X= 10 + 12 + + …+ 00. Y= 2 + + + …+ 02. What is the value of Y X -?(A) 92 (B) 98 (C) 100 (D) 102 (E) 1127. Which of the following equations does NOT have a solution?(A) 2(7)0x +=(B) -350x += 20=80= (E) -340x -=(A)(B)(C)2(D) (E) 613. What is the sum of all the solutions of 2602x x x =--?(A) 32 (B) 60 (C) 92 (D) 120 (E) 12414. The average of the numbers 1, 2, 3… 9 , 99, and x is 100x. What is x?(A)49101(B)50101(C)12 (D)51101(E)509911. The length of the interval of solutions of the inequality 23a x b ≤+≤ is 10.What is b-a?(A) 6 (B) 10 (C) 15 (D) 20 (E) 3013. Angelina drove at an average rate of 80 kph and then stopped 20 minutes for gas. After the stop, she drove at an average rate of 100 kph. Altogether she drove 250 km in a total trip time of 3 hours including the stop. Which equation could be used to solve for the time t in hours that she drove before her stop? (A) 880100()2503t t +-=(B) 80250t = (C) 100250t =(D) 90250t =(E) 880()1002503t t -+=21. The polynomial 32-2010x ax bx +- has three positive integer zeros. What is the smallest possible value of a?(A) 78 (B) 88 (C) 98 (D) 108 (E) 118 15．When a bucket is two-thirds full of water, the bucket and water weigh kilograms. When the bucket is one-half full of water the total weight is kilograms. In terms of and , what is the total weight in kilograms when the bucket is full of water?13．Suppose thatand. Which of the following is equal tofor every pair of integers16．Let ,,, and be real numbers with,, and. What is the sum of all possible values of5. Which of the following is equal to the product?81216442008............481242004n n +? (A) 251(B) 502(C) 1004(D) 2008 (E) 40167. The fraction 20082200622007220052(3)(3)(3)(3)-- simplifies to which of the following? (A) 1 (B) 9/4 (C) 3 (D) 9/2 (E) 913. Doug can paint a room in 5 hours. Dave can paint the same room in 7 hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let t be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by t ?(A) 11()(1)157t ++=(B) 11()1157t ++= (C) 11()157t +=(D) 11()(1)157t +-=(E) (57)1t +=15. Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?(A) 120 (B) 130 (C) 140 (D) 150 (E) 160AMC 中的⼏何问题⼀、三⾓形有关知识点1.三⾓形的简单性质与⼏个⾯积公式①三⾓形任何两边之和⼤于第三边；②三⾓形任何两边之差⼩于第三边；③三⾓形三个内⾓的和等于 0°；④三⾓形三个外⾓的和等于3 0°；⑤三⾓形⼀个外⾓等于和它不相邻的两个内⾓的和；⑥三⾓形⼀个外⾓⼤于任何⼀个和它不相邻的内⾓。
【国家自然科学基金】_过载保护_基金支持热词逐年推荐_【万方软件创新助手】_20140803
2013年 序号 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
科研热词 风力发电 过载保护 轻型高压直流输电 负荷减载 负刚度 线路过负荷 紧急控制 电磁环网 电池储能系统 电压源换流器 生机界面 潮流跟踪 暂态电压安全 接受腔 循环功率 并网逆变器 平滑时间常数 多端网络 发电机减出力 压力可变 功率限幅 功率转移比 低压减载 优化控制
推荐指数 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2010年 序号 1 2 3 4
科研热词 糖尿病 生物地质环境 地球化学元素 发病率
ቤተ መጻሕፍቲ ባይዱ推荐指数 1 1 1 1
2011年 序号 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
2012年 序号 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
2011年 科研热词 频率稳定 非线性接触 非同期并列 隔离 限流可调 重叠分解 部分元等效电路 过载控制 过载保护 过载 输电塔 负荷/频率控制(lfc) 耦合器 线性二次最优控制 直流稳压电源 疲劳 电路建模 电磁式 牺牲层技术 极限承载力 无接触 探针 性能测试 寄生参数 安全销 外旁通柜 备自投 增量加载 压力传感器 单片机 协调控制 区域备自投 准入控制 交直流系统 串供 三维微力 session粒度 mems传感器 flash crowd ansoft 推荐指数 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
实用文档之AMC10美国数学竞赛A卷附中文翻译和答案
实用文档之"2011AMC10美国数学竞赛A卷"1. A cell phone plan costs $20 each month, plus 5¢per text message sent, plus 10¢ for each minute used over 30 hours. In January Michelle sent 100 text messages and talked for 30.5 hours. How much did she have to pay?(A) $24.00 (B) $24.50 (C) $25.50 (D) $28.00 (E) $30.002. A small bottle of shampoo can hold 35 milliliters of shampoo, Whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?(A) 11 (B) 12 (C) 13 (D) 14 (E) 153. Suppose [a b] denotes the average of a and b, and {a b c} denotes the average of a, b, and c. What is {{1 1 0} [0 1] 0}?(A) 29(B)518(C)13(D) 718(E) 234. Let X and Y be the following sums of arithmetic sequences:X= 10 + 12 + 14 + …+ 100.Y= 12 + 14 + 16 + …+ 102.What is the value of Y X -?(A) 92(B) 98 (C) 100 (D) 102 (E) 1125. At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of 12, 15, and 10 minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?(A) 12 (B) 373 (C) 887 (D) 13 (E) 146. Set A has 20 elements, and set B has 15 elements. What is the smallest possible number of elements in A ∪B, the union ofA and B?(A) 5(B) 15 (C) 20 (D) 35 (E) 3007. Which of the following equations does NOT have a solution?(A) 2(7)0x += (B) -350x += (C) 20=(D) 80x-==(E) -3408. Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?(A) 20 (B) 30 (C) 40 (D) 50 (E) 609. A rectangular region is bounded by the graphs of the equations y=a, y=-b, x=-c, and x=d, where a, b, c, and d are all positive numbers. Which of the following represents the area of this region?(A) ac + ad + bc + bd (B) ac – ad + bc – bd (C) ac + ad – bc – bd(D) –ac –ad + bc + bd (E) ac – ad – bc + bd10. A majority of the 20 students in Ms. Deameanor’s class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than thenumber of pencils each student bought, and the total cost of all the pencils was $17.71. What was the cost of a pencil in cents?(A) 7(B) 11 (C) 17 (D) 23 (E) 7711. Square EFGH has one vertex on each side of square ABCD. Point E is on AB with AE=7·EB. What is the ratio of the area of EFGH to the area of ABCD?(A)4964 (B) 2532 (C) 78 (D) 8 (E)412. The players on a basketball team made some three-point shots, some two-point shots, some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team’s total score was 61 points. How many free throws did they make?(A) 13(B) 14 (C) 15 (D) 16 (E) 1713. How many even integers are there between 200 and 700 whose digits are all different and come from the set {1, 2, 5, 7, 8, 9}?(A) 12 (B)20 (C)72 (D) 120 (E) 20014. A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle’s circumference?(A) 136(B)112(C)16(D) 14(E) 51815. Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?(A) 140 (B) 240 (C) 440 (D) 640 (E) 84016. Which of the following in equal to(A)(B) (C) 2 (D) (E)617. In the eight-term sequence A, B, C, D, E, F, G , H, the value of C is 5 and the sum of any three consecutive terms is 30. What is A + H?(A) 17(B) 18 (C) 25 (D) 26 (E) 4318. Circles A, B, and C each have radius 1. Circles A and B share one point of tangency. Circle C has a point of tangency with the midpoint of AB. What is the area inside Circle C but outside Circle A and Circle B? (A) 32π- (B) 2π (C) 2 (D) 34π (E) 12π+19. In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfectsquare. Which of the following is closest to the percent growth of the town’s population during this twenty-year period? (A) 42 (B) 47 (C) 52 (D) 57 (E) 6220. Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?(A) 16(B) 15(C) 14(D) 13(E) 1221. Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins.A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?(A) 711(B) 913(C) 1115(D) 1519(E) 151622. Each vertex of convex pentagon ABCDE is to be assigned a color. There are 6 colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?(A) 2500 (B) 2880 (C) 3120 (D) 3250 (E) 375023. Seven students count from 1 to 1000 as follows:·Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is Alice says 1, 3, 4, 6, 7, 9, …, 997, 999, 1000.·Barbara says all of the numbers that Alice doesn’t say, except she also skips the middle number in each consecutive grope of three numbers.·Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.·Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs inthe alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.·Finally, George says the only number that no one else says. What number does George say?(A) 37(B) 242 (C) 365 (D) 728 (E) 99824. Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?(A)112 (B) 12 (C) (D) 16 (E)25. Let R be a square region and 4n an integer. A point X in the interior of R is called n-ray partitional if there are n rays emanating from X that divide R into N triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?(A) 1500(B) 1560 (C) 2320 (D) 2480 (E)25002011AMC10美国数学竞赛A卷1. 某通讯公司手机每个月基本费为20美元, 每传送一则简讯收 5美分（一美元=100 美分）。
2008 AMC10A中文解析
AMC10A 2008 精品解析Q1. A bakery owner turns on his doughnut machine at 8:30 AM. At 11:10 AM the machine has completed one third of the day’s job. At what time will the doughnut machine complete the job?A) 1:50 PM B) 3:00 PM C)3:30 PM D)4:30 PM E) 5:50 PM 翻译：面包店老板在早上8:30打开他的甜甜圈机。
上午11:10，机器完成了一天工作的三分之一。
甜甜圈机什么时候能够完成工作?答案：D解析：由题意，从8:30到11:10完成了工作的三分之一，共花费了160分钟，若完成整项工作，需要花费480分钟，在4:30 PM 完成，故选D.Q2. A square is drawn inside a rectangle. The ratio of the width of the rectangle to a side of the square is 2 : 1. The ratio of the rectangle’s length to its width is 2 : 1. What percent of the rectangle’s area is inside the square?A) 12.5 B) 25 C)50 D)75 E)87.5 翻译：在矩形内画一个正方形。
矩形的宽度与正方形的边长之比是2:1。
这个矩形的长宽比是2:1。
问正方形面积是矩形面积的百分之多少?答案：A解析：设正方形的边长为a ，则由题意，矩形的宽为2a ，长为4a ，所以正方形的面积为2a ，矩形的面积为28a ，正方形面积是矩形面积的百分之12.5，故选A.Q3. For the positive integer n , let n denote the sum of all the positive divisors of n with the exception of n itself. For example,4123=+= and 121+2+3+4+6=16=， What is 6?A) 6 B) 12 C)24 D)32 E)36 翻译：对于正整数n ，n 表示除n 本身外所有正因数的和。
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
2007 AMC 10 B, Problem #25— 2007 AMC 12 B, Problem #24— 14b “Play with the restriction: a b + 9a is an integer.”
Solution Answer (A): Let u = a/b. Then the problem is equivalent to finding all positive rational numbers u 14 = k, for some integer k. This equation is equivalent to 9u2 − 9uk + 14 = 0, whose such that u + 9 u solutions are √ 9k ± 81k2 − 504 k 1 u= 9k2 − 56. = ± 18 2 6 √ Hence u is rational if and only if 9k2 − 56 is rational, which is true if and only if 9k2 − 56 is a perfect square. Suppose that 9k2 − 56 = s2 for some positive integer s. Then (3k − s)(3k + s) = 56. The only factors of 56 are 1, 2, 4, 7, 8, 14, 28, and 56, so (3k − s, 3k + s) is one of the ordered pairs (1, 56), (2, 28), (4, 14), or (7, 8). The cases (1, 56) and (7, 8) yield no integer solutions. The cases (2, 28) and (4, 14) yield k = 5 and k = 3, respectively. If k = 5, then u = 1/3 or u = 14/3. If k = 3, then u = 2/3 or u = 7/3. Therefore there are four pairs (a, b) that satisfy the given conditions, namely (1, 3), (2, 3), (7, 3), and (14, 3). OR a 14b = k, in two different forms. First, multiply both sides by b and subtract Rewrite the equation + b 9a a to obtain 14b2 = bk − a. 9a Because a, b, and k are integers, 14b2 must be a multiple of a, and because a and b have no common factors greater than 1, it follows that 14 is divisible by a. Next, multiply both sides of the original equation by 9a and subtract 14b to obtain 9a2 = 9ak − 14b. b This shows that 9a2 is a multiple of b, so 9 must be divisible by b. Thus if (a, b) is a solution, then b = 1, 3, or 9, and a = 1, 2, 7, or 14. This gives a total of twelve possible solutions (a, b), each of which can be checked quickly. The only such pairs for which a 14b + b 9a is an integer are when (a, b) is (1, 3), (2, 3), (7, 3), or (14, 3).
AMC 10 Student Practice Questions The larger two consecutive odd integers is three You will find these and additional problemsof for the AMC 10 and AMC 12 on AMC’s web site: /amc, available from the current and previous AMC 10/12 Teacher Manuals, (/amc/e-exams/e6-amc12/archive12.shtml) or times smaller. What is their sum? from our Problems page archives the (/amc/a-activities/a7-problems/problem81012archive.shtml).
she must solve at least 16 of the first 22 problems correctly. This would give her a score of 100.5.
Difficulty: Medium-easy NCTM Standard: Problem Solving Standard: solve problems that arise in mathematics and in other contexts. Classification: Calculus and Analysis > Inequalities Number Theory > Arithmetic > Addition and Subtraction
x + 2. So x + 2 = 3x, from which x = 1. Thus the two integers are 1 and 3, and their sum is 4. Solution Answer (A):
Difficulty: Easy + 2. So x Algebra +2 = 3 x, from which x and = 1analyze . Thusmathematical the two integers are 1 NCTMxStandard: Standard: represent situations and structures symbols. 3,using and algebraic their sum is 4. Classification: Number Theory > Integers > Odd number
15
AMC 10 Student Practice Questions continued
The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is 20, the father is 48 years old, and the average age of the mother and children is 16. How many children are in the family?
Solution Answer (D): Sarah will receive 4.5 points for the three questions she leaves unanswered, so she must earn at least 100 − 4.5 = 95.5 points on the first 22 problems. Because 15 < 95.5 < 16, 6
(A) 2 (B) 3 (C) 4 (D) 5 (E) 6
2007 AMC 10 A, Problem #10— “Set up equations with two variables, and use the conditions above to eliminate one. ”
Solution Answer (E): Let N represent the number of children in the family and T represent the sum of the ages of all the family members. The average age of the members of the family is 20, and the average age of the members when the 48-year-old father is not included is 16, so 20 = This implies that 20N + 40 = T so 20N + 40 = 16N + 64. Hence 4N = 24 and N = 6.
(A) 13 (B) 14 (C) 15 (D) 16 (E) 17
2007 AMC 10 B, Problem #6— 2007 AMC 12 B, Problem #5— “Sarah must earn at least 95.5 points on the first 22 problems.”
XIV. Classroom Accessories