AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：Problem 4Four siblings ordered an extra large pizza. Alex ate , Beth , and Cyril of the pizza. Dan got the leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed? SolutionLet the pizza have slices, since the least common multiple of . Therefore, Alex ate slices, Beth ate slices, and Cyril ate slices. Dan must have eaten slices. In decreasing order, we see the answer is .Problem 5David, Hikmet, Jack, Marta, Rand, and Todd were in a -person race with other people. Rand finished places ahead of Hikmet. Marta finished place behind Jack. David finished places behind Hikmet. Jack finished places behind Todd. Todd finished place behind Rand. Marta finished in th place. Who finished in th place?SolutionBecause Marta was th, Jack was th, so Todd was rd. Thus, Rand was nd and the 8th place finisher was Problem 6Marley practices exactly one sport each day of the week. She runs three days a week but never on two consecutive days. On Monday she plays basketball and two days later golf. She swims and plays tennis, but she never plays tennis the day after running or swimming. Which day of the week does Marley swim?SolutionMarley does basketball on Monday and golf on Wednesday. Since there are four consecutive days between Wednesday and Monday exclusive, she must run on Tuesday. Thus, she must also run on Thursday and Saturday or Friday and Sunday. She can't run on Thursday and Saturday because she would have to do tennis after running, which is not allowed. Thus, she runs on Friday and Sunday, so she must do tennis on Thursday, so she swims on 以上就是小编对AMC10数学竞赛试题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：Problem1What is the value of ?SolutionProblem2Marie does three equally time-consuming tasks in a row without taking breaks. She begins the first task at 1:00 PM and finishes the second task at 2:40 PM. When does she finish the third task?SolutionMarie does her work twice in hour and minutes. Therefore, one task should take minutes to finish. minutes after PM is PM, so our answer is Problem3Isaac has written down one integer two times and another integer three times. The sum of the five numbers is , and one of the numbers is . What is the other number?SolutionLet the first number be and the second be . We have . We are given one of the numbers is . If were to be , would not be an integer, thus . , which gives .以上就是小编对AMC10数学竞赛试题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC 8数学竞赛专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC 8的官方真题以及官方解答吧：Problem 23Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?SolutionIt is well known that Distance=SpeedTime. In the question, we want distance. From the question, we have that the time is (One hour). By the equation derived from Distance=SpeedTime, we have Speed=Distance/Time, so the speed is mile/mins. Because we want the distance, we multiply the time and speed together yielding . The minutes cancel out, so now we have as our distance for the first day. The distance for the following days are:We then start our trial and error: The factors of are . We plug each of those numbers in for , and we get that is 5, soProblem 24Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?Solution 1In days, there aredays without calls. Note that in the last five days of the year, day and also do not have any calls, as they are not multiples of , , or . Thus our answer is .Alternatively, there are days without calls. Multiplying the fractions in this order prevents partial days, as is a multiple of , is a multiple of and is a multiple of .Solution 2We use Principle of Inclusion and Exclusion. There are days in the year, and we subtract the days that she gets at least phone call, which isTo this result we add the number of days where she gets at least phone calls in a day because we double subtracted these days. This number isWe now subtract the number of days where she gets three phone calls, which is . Therefore, our answer is.Problem 25In the figure shown, and are line segments each of length 2, and . Arcs and are each one-sixth of a circle with radius 2. What is the area of the region shown?SolutionLet the centers of the circles containing arcs and be and , respectively. Extend and to and , and connect point with point .We can clearly see that is an equilateral triangle, because two of its angles are , which is the degree measure of a circle. The area of the figure is equal to minus the combined area of the sectors of the circles. Using the area formula for an equilateral triangle, where is the side length of the equilateral triangle, is The combined area of the sectors is , which is Thus, our final answer is 以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC 8数学竞赛专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC 8的官方真题以及官方解答吧：Problem 21Suppose , , and are nonzero real numbers, and . What are the possible value(s) for ?Solution 1There are cases to consider:Case : of , , and are positive and the other is negative. WLOG assume that and are positive and is negative. In this case, we have thatCase : of , , and are negative and the other is positive. WLOG assume that and are negative and is positive. In this case, we have thatIn both cases, we get that the given expression equals .Solution 2Assuming numbers:WLOG and (Other numbers can apply for and as long as their sum is .) . Then plug and into the given equation. The result is always .Problem 22In the right triangle , , , and angle is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?SolutionWe can reflect triangle over line This forms the triangle and a circle out of the semicircle. Let us call the center of the circle We can see that Circle is the incircle of We can use the formula for finding the radius of the incircle to solve this problem. The area of is The semiperimeter is Simplifying Our answer is therefore Solution 2We immediately see that , and we label the center of the semicircle . Drawing radius with length such that is perpendicular to , we immediately see that because of congruence, so and . By similar triangles and , we see that .Solution 3Let the center of the semicircle be . Let the point of tangency between line and the semicircle be . Angle is common to triangles and . By tangent properties, angle must be degrees. Since both triangles and are right and share an angle, is similar to . The hypotenuse of is , where is the radius of the circle. (See for yourself) The short leg of is . Because ~ , we have and solving gives Solution 4Let the tangency point on be . NoteBy Power of a Point,Solving for gives以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC 8数学竞赛专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC 8的官方真题以及官方解答吧：Problem 19For any positive integer , the notation denotes the product of the integers through . What is the largest integer for which is a factor of the sum ?Solution 1Factoring out , we have . Next, has factors of . Now has factors of , so there are a total of factors of .Solution 2The number of 's in the factorization of is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by , until you can't divide by anymore. Factorizing , you get . To find the number of trailing zeroes in 98!, we do . Now since has 4 zeroes, we add to get factors of .Problem 20An integer between and , inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?SolutionThere are options for the last digit, as the integer must be odd. The first digit now has options left (it can't be or the same as the last digit). The second digit also has options left (it can't be the same as the first or last digit). Finally, the third digit has options (it can't be the same as the three digits that are already chosen).Since there are total integers, our answer is以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC 8数学竞赛专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC 8的官方真题以及官方解答吧：Problem 17Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?SolutionWe can represent the amount of gold with and the amount of chests with . We can use the problem to make the following equations:Therefore, This implies that We therefore have So, our answer is .Problem 18In the non-convex quadrilateral shown below, is a right angle, , , , and .What is the area of quadrilateral ?SolutionWe first connect point with point .We can see that is a 3-4-5 right triangle. We can also see that is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of is , and the area of the smaller 3-4-5 triangle is . Thus, the area of quadrialteral is 以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC 8数学竞赛专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC 8的官方真题以及官方解答吧：Problem 15In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path allows only moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.SolutionNotice that the is adjacent to s, each is adjacent to s, and each is adjacent to s. Thus, the answer is Solution 2There are three different kinds of paths that are on this diagram. The first kind is when you directly count A, M, C in a straight line. The second is when you count A, turn left or right to get M, then go straight to count M and C. The third is the one where you start with A, move forward to count M, turn left or right to count C, then move straight again to get 8.There are 8 paths for each kind of path, making for paths.Problem 16In the figure below, choose point on so that and have equal perimeters. What is the area of ?Solution 1Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that is the length of the line from B to D. So, the perimeter of would be , while the perimeter of would be . Notice that we can find out from these two equations. We can find out that , so that means that the area of Solution 2We know that the perimeters of the two small triangles are and . Setting both equal and using , we have and . Now, we simply have to find the area of . Since , we must have . Combining this with the fact that , we get Solution 3Since point is on line , it will split it into and . Let and . Triangle has side lengths and triangle has side lengths . Since both perimeters are equal, we have the equation . Eliminating and solving the resulting linear equation gives . Draw a perpendicular from point to . Call the point of intersection . Because angle is common to both triangles and , and both are right triangles, both are similar. The hypotenuse of triangle is 2, so the altitude must be Because and share the same altitude, the height of therefore must be . The base of is 4, so Solution 4Using any preferred method, realize . Since we are given a 3-4-5 right triangle, we know the value of . Since we are given , apply the Sine Area Formula to get .以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC 8数学竞赛专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC 8的官方真题以及官方解答吧：Problem 13Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?SolutionGiven games, there must be a total of wins and losses. Hence, where is Kyler's wins. , so our final answer is Problem 14Chloe and Zoe are both students in Ms. Demeanor's math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only of the problems she solved alone, but overall of her answers were correct. Zoe had correct answers to of the problems she solved alone. What was Zoe's overall percentage of correct answers?Solution 1Let the number of questions that they solved alone be . Let the percentage of problems they correctly solve together be %. As given,Hence, .Zoe got problems right out of . Therefore, Zoe got percent of the problems correct.Solution 2Assume the total amount of problems is per half homework assignment, since we are dealing with percentages, and no values. Then, we know that Chloe got problems correct by herself, and got problems correct overall. We also know that Zoe had problems she did alone correct. We can see that the total amount of correct problems Chloe had when Zoe and she did the homework together is , which is the total amount of problems she got correct, subtracted by the number of correct problems she did alone. Therefore Zoe has problems out of problems correct. This is percent.以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！