2002amc12a美国数学竞赛

2002 AMC 12A ProblemsProblem 1Compute the sum of all the roots ofSolutionProblem 2Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?SolutionProblem 3According to the standard convention for exponentiation,If the order in which the exponentiations are performed is changed, how many other values are possible?SolutionProblem 4Find the degree measure of an angle whose complement is 25% of its supplement.SolutionProblem 5Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.SolutionProblem 6For how many positive integersdoes there exist at least one positive integer n such that?infinitely manySolutionProblem 7Aarc of circle A is equal in length to aarc of circle B. What is the ratio of circle As area and circle Bs area?SolutionProblem 8Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Letbe the total area of the blue triangles,the total area of the white squares, andthe area of the red square. Which of the following is correct?SolutionProblem 9Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?SolutionProblem 10Sarah 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?SolutionProblem 11Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?SolutionProblem 12Both roots of the quadratic equationare prime numbers. The number of possible values ofisSolutionProblem 13Two different positive numbersandeach differ from their reciprocals by. What is?SolutionProblem 14For all positive integers, let. Let. Which of the following relations is true?SolutionProblem 15The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection isSolutionProblem 16Tina randomly selects two distinct numbers from the set, and Sergio randomly selects a number from the set. What is the probability that Sergios number is larger than the sum of the two numbers chosen by Tina?SolutionProblem 17Several sets of prime numbers, such asuse each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?SolutionProblem 18Letandbe circles defined byandrespectively. What is the length of the shortest line segmentthat is tangent toatand toat?SolutionProblem 19The graph of the functionis shown below. How many solutions does the equationhave?SolutionProblem 20Suppose thatandare digits, not both nine and not both zero, and the repeating decimalis expressed as a fraction in lowest terms. How many different denominators are possible?SolutionProblem 21Consider the sequence of numbers:For, the-th term of the sequence is the units digit of the sum of the two previous terms. Letdenote the sum of the firstterms of this sequence. The smallest value offor whichis:SolutionProblem 22Triangleis a right triangle withas its right angle, and. Letbe randomly chosen inside, and extendto meetat. What is the probability that?SolutionProblem 23In triangle, sideand the perpendicular bisector ofmeet in point, andbisects. Ifand, what is the area of triangle?SolutionProblem 24Find the number of ordered pairs of real numberssuch that.SolutionProblem 25The nonzero coefficients of a polynomialwith real coefficients are all replaced by their mean to form a polynomial. Which of the following could be a graph ofandover the interval?Solution答案：Problem 1SolutionSolution 1We expand to getwhich isafter combining like terms. Using the quadratic part ofVietas Formulas, we find the sum of the roots is.Solution 2Combine terms to get, hence the roots areand, thus our answer is.Problem 2SolutionWe work backwards; the number that Cindy started with is. Now, the correct result is. Our answer isProblem 3SolutionThe best way to solve this problem is by simple brute force.It is convenient to drop the usual way how exponentiation is denoted, and to write the formula as, wheredenotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:1.2.3.4.5.We can note that. Therefore options 1 and 2 are equal, and options 3 and 4 are equal. Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.Thus the only other result is, and our answer is.Problem 4SolutionProblem 5SolutionThe outer circle has radius, and thus area. The little circles have areaeach; since there are 7, their total area is. Thus, our answer isProblem 6SolutionSolution 1For anywe can pick, we get, therefore the answer is.Solution 2Another solution, slightly similar to this first one would be usingSimons Favorite Factoring Trick.Let, thenThis means that there are infinitely many numbersthat can satisfy the inequality. So the answer isProblem 7SolutionLetandbe the radii of circles A and B, respectively.It is well known that in a circle with radius r, a subtended arc opposite an angle ofdegrees has length.Using that here, the arc of circle A has length. The arc of circle B has length. We know that they are equal, so, so we multiply through and simplify to get. As all circles are similar to one another, the ratio of the areas is just the square of the ratios of the radii, so our answer is.Problem 8SolutionThe blue thats touching the center red square makes up 8 triangles, or 4 squares. Each of the corners is 2 squares and each of the edges is 1, totaling 12 squares. There are 12 white squares, thus we have.Problem 9SolutionA 0.8 MB file can either be on its own disk, or share it with a 0.4 MB. Clearly it is not worse to pick the second possibility. Thus we will have 3 disks, each with one 0.8 MB file and one 0.4 MB file.We are left with 12 files of 0.7 MB each, and 12 files of 0.4 MB each. Their total size isMB. The total capacity of 9 disks isMB, hence we need at least 10 more disks. And we can easily verify that 10 disks are indeed enough: six of them will carry two 0.7 MB files each, and four will carry three 0.4 MB files each.Thus our answer is.Problem 10SolutionWe will simulate the process in steps.In the beginning, we have: ounces of coffee in cup ounces of cream in cupIn the first step we pourounces of coffee from cupto cup, getting: ounces of coffee in cup ounces of coffee andounces of cream in cupIn the second step we pourounce of coffee andounces of cream from cupto cup, getting: ounces of coffee andounces of cream in cup the rest in cupHence at the end we haveounces of liquid in cup, and out of theseounces is cream. Thus the answer isProblem 11SolutionSolution 1Let the time he needs to get there in be t and the distance he travels be d. From the given equations, we know thatand. Setting the two equal, we haveand we findof an hour. Substituting t back in, we find. From, we find that r, and our answer, is.Solution 2Since either time he arrives at is 3 minutes from the desired time, the answer is merely theharmonic meanof 40 and 60. The harmonic mean of a and b is. In this case, a and b are 40 and 60, so our answer is, so.Solution 3A more general form of the argument in Solution 2, with proof:Letbe the distance to work, and letbe the correct average speed. Then the time needed to get to work is.We know thatand. Summing these two equations, we get:.Substitutingand dividing both sides by, we get, hence.(Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighed sum in step two, and hence obtain a weighed harmonic mean in step three.)Problem 12SolutionConsider a general quadratic with the coefficient ofbeingand the roots beingand. It can be factored aswhich is just. Thus, the sum of the roots is the negative of the coefficient ofand the product is the constant term. (In general, this leads toVietas Formulas).We now have that the sum of the two roots iswhile the product is. Since both roots are primes, one must be, otherwise the sum would be even. That means the other root isand the product must be. Hence, our answer isProblem 13SolutionEach of the numbersandis a solution to.Hence it is either a solution to, or to. Then it must be a solution either to, or to.There are in total four such values of, namely.Out of these, two are positive:and. We can easily check that both of them indeed have the required property, and their sum is.Problem 14SolutionFirst, note that.Using the fact that for any base we have, we get thatProblem 15SolutionAs the unique mode is, there are at least twos.As the range isand one of the numbers is, the largest one can be at most.If the largest one is, then the smallest one is, and thus the mean is strictly larger than, which is a contradiction.If the largest one is, then the smallest one is. This means that we already know four of the values:,. Since the mean of all the numbers is, their sum must be. Thus the sum of the missing four numbers is. But ifis the smallest number, then the sum of the missing numbers must be at least, which is again a contradiction.If the largest number is, we can easily find the solution. Hence, our answer is.NoteThe solution foris, in fact, unique. As the median must be, this means that both theand thenumber, when ordered by size, must bes. This gives the partial solution. For the mean to beeach missing variable must be replaced by the smallest allowed value.Problem 16SolutionSolution 1This is not too bad using casework.Tina gets a sum of 3: This happens in only one wayand Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here.Tina gets a sum of 4: This once again happens in only one way. Sergio can choose a number from 5 to 10, so 6 ways here.Tina gets a sum of 5: This can happen in two waysand. Sergio can choose a number from 6 to 10, so 2ways here.Tina gets a sum of 6: Two ways hereand. Sergio can choose a number from 7 to 10, sohere.Tina gets a sum of 7: Two ways hereand. Sergio can choose from 8 to 10, soways here.Tina gets a sum of 8: Only one way possible). Sergio chooses 9 or 10, so 2 ways here.Tina gets a sum of 9: Only one way. Sergio must choose 10, so 1 way.In all, there areways. Tina chooses two distinct numbers inways while Sergio chooses a number inways, so there areways in all. Since, our answer is.Solution 2We want to find the average of the smallest possible chance of Sergio winning and the largest possible chance of Sergio winning. This is because the probability decreases linearly. The largest possibility of Sergio winning if Tina chooses a 1 and a 2. The chances of Sergio winning is then. The smallest possibility of Sergio winning is if Tina chooses a 4 and a 5. The chances of Sergio winning then is. The average ofandisProblem 17SolutionNeither of the digits, andcan be a units digit of a prime. Therefore the sum of the set is at least.We can indeed create a set of primes with this sum, for example the following set works:.Thus the answer is.Problem 18Solution(C)First examine the formula, for the circle. Its center, is located at (10,0) and it has a radius of= 6. The next circle, using the same pattern, has its center, at (-15,0) and has a radius of= 9. So we can construct this diagram:Line PQ is tangent to both circles, so it forms a right angle with the radii (6 and 9). This, as well as the two vertical angles near O, prove triangles SQO and SPO similar by AA, with a scale factor of 6:9, or 2:3. Next, we must subdivide the line DDin a 2:3 ratio to get the length of the segments DO and DO. The total length is 10 - (-15), or 25, so applying the ratio, DO =15and DO =10. These are the hypotenuses of the triangles. We already know the length of DQ and DP,9and6(theyre radii). So in order to find PQ, we must find the length of the longer legs of the two triangles and add them.Finally, the length of PQ is, orC.Problem 19SolutionFirst of all, note that the equationhas two solutions:and.Given an, let. Obviously, to have, we need to have, and we already know when that happens. In other words, the solutions toare precisely the solutions to (or).Without actually computing the exact values, it is obvious from the graph that the equationhas two andhas four different solutions, giving us a total ofsolutions.Problem 20SolutionThe repeating decimalis equal toWhen expressed in lowest terms, the denominator of this fraction will always be a divisor of the number. This gives us the possibilities. Asandare not both nine and not both zero, the denumeratorcan not be achieved, leaving us withpossible denominators.(The other ones are achieved e.g. forequal to, and, respectively.)Problem 21SolutionThe sequence is infinite. As there are onlypairs of digits, sooner or later a pair of consecutive digits will occur for the second time. As each next digit only depends on the previous two, from this point on the sequence will be periodic.(Additionally, as every two consecutive digits uniquely determine thepreviousone as well, the first pair of digits that will occur twice must be the first pair.)Hence it is a good idea to find the period. Writing down more terms of the sequence, we get:and we found the period. The length of the period is, and its sum is. Hence for eachwe have.We haveand, therefore. The rest can now be computed by hand, we get, and, thus the answer isProblem 22SolutionClearlyand. Choose aand get a correspondingsuch thatand. Forwe need, creating an isoclese right triangle with hyptonuse. Thus the pointmay only lie in the triangle. The probability of it doing so is the ratio of areas ofto, or equivalently, the ratio oftobecause the triangles have identical altitudes when takingandas bases. This ratio is equal to. Thus the answer isProblem 23SolutionLooking at the triangle, we see that its perpendicular bisector reaches the vertex, therefore hinting it is isoceles. Let, so thatfrom given and the previous deducted. Thenbecause any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That meansandaresimilar, so.Then by using Herons Formula on(with sides), we haveProblem 24SolutionLetbe the magnitude of. Then the magnitude ofis, while the magnitude ofis. We get that, hence eitheror.Forwe get a single solution.Lets now assume that. Multiply both sides by. The left hand side becomes, the right hand side becomes. Hence the solutions for this case are precisely all therd complex roots of unity, and there areof those.The total number of solutions is therefore.Solution 2As in the other solution, split the problem into whenand when. Whenand,so we must haveand hence. Sinceis restricted to,can range fromtoinclusive, which isvalues. Thus the total is.Problem 25Solution(B)The sum of the coefficients ofand ofwill be equal, so. The only answer choice with an intersection atis at(B). (The polynomials in the graph areand.)