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数学竞赛真题以及官方解答吧：Problem 25Let , where denotes the greatest integer less than or equal to . How many distinct values does assume for ?SolutionSince , we haveThe function can then be simplified intowhich becomesWe can see that for each value of k, can equal integers from 0 to k-1.Clearly, the value of changes only when x is equal to any of the fractions .So we want to count how many distinct fractions have the form where . We can find this easily by computingwhere is the Euler Totient Function. Basically counts the number of fractions with as its denominator (after simplification). This comes out to be .Because the value of is at least 0 and can increase 31 times, there are a total of different possible values of . 以上就是小编对AMC10数学竞赛真题以及解析的介绍，希望对你有所帮助，如果想了解更多关于AMC数学竞赛报考点、南京AMC数学竞赛培训、美国数学竞赛AMC有用吗以及AMC学习资料等信息请持续关注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数学竞赛真题以及官方解答吧：Problem 24How many four-digit integers , with , have the property that the three two-digit integers form an increasing arithmetic sequence? One such number is , where , , , and .SolutionThe numbers are and . Note that only can be zero, and that .To form the sequence, we need . This can be rearranged as . Notice that since the left-hand side is a multiple of , the right-hand side can only be or . (A value of would contradict .) Therefore we have two cases: and .Case 1If , then , so . This gives . If , then , so . This gives . If , then , so , giving . There is no solution for . Added together, this gives us answers for Case 1.Case 2This means that the digits themselves are in arithmetic sequence. This gives us answers, .Adding the two cases together, we find the answer to be . 以上就是小编对AMC10数学竞赛真题以及解析的介绍，希望对你有所帮助，如果想了解更多关于AMC数学竞赛报考点、南京AMC数学竞赛培训、美国数学竞赛AMC有用吗以及AMC学习资料等信息请持续关注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数学竞赛真题以及官方解答吧：Problem 22A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won games and lost games; there were no ties. How many sets of three teams were there in which beat , beat , and beat SolutionThere are teams. Any of the sets of three teams must either be a fork (in which one team beat both the others) or a cycle:But we know that every team beat exactly other teams, so for each possible at the head of a fork, there are always exactly choices for and . Therefore there are forks, and all the rest must be cycles.Thus the answer is which is . Problem 23In regular hexagon , points , , , and are chosen on sides , , , and respectively, so lines , , , and are parallel and equally spaced. What is the ratio of the area of hexagon to the area of hexagon ?Solution 1We draw a diagram to make our work easier:Assume that is of length . Therefore, the area of is . To find the area of , we draw , and find the area of the trapezoids and .From this, we know that . We also know that the combined heights of the trapezoids is , since and are equally spaced, and the height of each of the trapezoids is . From this, we know and are each of the way from to and , respectively. We know that these are both equal to .We find the area of each of the trapezoids, which both happen to be , and the combined area is .We find that is equal to . At this point, you can answer and move on with your test.Solution 2 (a lot faster than Solution 1)First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are small triangles in hexagon , and small triangles in the whole hexagon.Thus, the answer is . 以上就是小编对AMC10数学竞赛真题以及解析的介绍，希望对你有所帮助，如果想了解更多关于AMC数学竞赛报考点、南京AMC数学竞赛培训、美国数学竞赛AMC有用吗以及AMC学习资料等信息请持续关注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数学竞赛真题以及官方解答吧：Problem 21What is the area of the region enclosed by the graph of the equation Solution 1WLOG note that if a point in the first quadrant satisfies the equation, so do its corresponding points in the other three quadrants. Therefore, we can assume that , which implies that and , and multiply by at the end.We can rearrange the equation to get , which describes a circle with center and radius It's clear we now want to find the union of four circles with overlap.There are several ways to find the area, but note that if you connect to its other three respective points in the other three quadrants, you get a square of area , along with four half-circles of diameter , for a total area of which is .Solution 2Another way to solve this problem is using cases. Though this may seem tedious, we only have to do one case. The equation for this figure is To make this as easy as possible, we can make both and positive. Simplifying the equation for and being positive, we get the equation Using the complete the square method, we get Therefore, the origin of this section of the shape is at Using the equation we can also see that the radius has a length of .With this shape we see that this shape can be cut into a right triangle and a semicircle. The length of the hypotenuse of the triangle is so using special right triangles, we see that the area of the triangle is . The semicircle has the area of .But this is only case. There are cases in total so we have to multiply by .After multiplying, our answer is: 以上就是小编对AMC10数学竞赛真题以及解析的介绍，希望对你有所帮助，如果想了解更多关于AMC数学竞赛报考点、南京AMC数学竞赛培训、美国数学竞赛AMC有用吗以及AMC学习资料等信息请持续关注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数学竞赛真题以及官方解答吧：Problem 20A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius centered at to the circle of radius centered at . What distance does the origin , move under this transformation?Solution 1: AlgebraicThe center of dilation must lie on the line , which can be expressed . Also, the ratio of dilation must be equal to , which is the ratio of the radii of the circles. Thus, we are looking for a point such that (for the -coordinates), and . Solving these, we get and . This means that any point on the plane will dilate to the point , which means that the point dilates to . Thus, the origin moves units.Solution 2: GeometricUsing analytic geometry, we find that the center of dilation is at and the coefficient/factor is . Then, we see that the origin is from the center, and will be from it afterwards.Thus, it will move .Solution 3: Logic and GeometryUsing the ratios of radii of the circles, , we find that the scale factor is . If the origin had not moved, this indicates that the center of the circle would be , simply because of . Since the center has moved from to , we apply the distance formula and get: . 以上就是小编对AMC10数学竞赛真题以及解析的介绍，希望对你有所帮助，如果想了解更多关于AMC数学竞赛报考点、南京AMC数学竞赛培训、美国数学竞赛AMC有用吗以及AMC学习资料等信息请持续关注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数学竞赛真题以及官方解答吧：Problem 19Rectangle has and . Point lies on so that , point lies on so that , and point lies on so that . Segments and intersect at and , respectively. What is the value of ?Solution 1 (Answer Choices)Since the opposite sides of a rectangle are parallel and due to vertical angles, . Furthermore, the ratio between the side lengths of the two triangles is . Labeling and , we see that turns out to be equal to . Since the denominator of must now be a multiple of 7, the only possible solution in the answer choices is .Solution 2 (Coordinate Geometry)First, we will define point as the origin. Then, we will find the equations of the following three lines: , , and . The slopes of these lines are , , and , respectively. Next, we will find the equations of , , and . They are as follows:After drawing in altitudes to from , , and , we see that because of similar triangles, and so we only need to find the x-coordinates of and .Finding the intersections of and , and and gives the x-coordinates of and to be and . This means that . Now we can find Solution 3 (Similar Triangles)Extend to intersect at . Letting , we have that Then, notice that and . Thus, we see that and Thus, we see that 以上就是小编对AMC10数学竞赛真题以及解析的介绍，希望对你有所帮助，如果想了解更多关于AMC数学竞赛报考点、南京AMC数学竞赛培训、美国数学竞赛AMC有用吗以及AMC学习资料等信息请持续关注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数学竞赛真题以及官方解答吧：Problem 18In how many ways can be written as the sum of an increasing sequence of two or more consecutive positive integers?Solution 1Factor .Suppose we take an odd number of consecutive integers, with the median as . Then with . Looking at the factors of , the possible values of are with medians as respectively.Suppose instead we take an even number of consecutive integers, with median being the average of and . Then with . Looking again at the factors of , the possible values of are with medians respectively.Thus the answer is .Solution 2We need to find consecutive numbers (an arithmetic sequence that increases by ) that sums to . This calls for the sum of an arithmetic sequence given that the first term is , the last term is and with elements, which is: .So, since it is a sequence of consecutive numbers starting at and ending at . We can now substitute with . Now we substiute our new value of into to get that the sum is .This simplifies to . This gives a nice equation. We multiply out the 2 to get that . This leaves us with 2 integers that multiplies to which leads us to think of factors of . We know the factors of are: . So through inspection (checking), we see that only and work. This gives us the answer of ways. 以上就是小编对AMC10数学竞赛真题以及解析的介绍，希望对你有所帮助，如果想了解更多关于AMC数学竞赛报考点、南京AMC数学竞赛培训、美国数学竞赛AMC有用吗以及AMC学习资料等信息请持续关注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数学竞赛真题以及官方解答吧：Problem 15All the numbers are written in a array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to . What is the number in the center?Solution 1 - Trial and ErrorQuick testing shows thatis a valid solution. , and the numbers follow the given condition. The center number is found to be .Solution 2First let the numbers bewith the numbers around the outsides and in the middle. We see that the sum of the four corner numbers is . If we switch and , then the corner numbers will add up to and the consecutive numbers will still be touching each other. The answer is .Solution 3Consecutive numbers share an edge. That means that it is possible to walk from to by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity:But there are only four even numbers in the set, so the five darker squares must contain the odd numbers, which sum to Therefore if the sum of the numbers in the corners is , the number in the center must be , which is answer . Problem 16The sum of an infinite geometric series is a positive number , and the second term in the series is . What is the smallest possible value of Solution 1The sum of an infinite geometric series is of the form:where is the first term and is the ratio whose absolute value is less than 1.We know that the second term is the first term multiplied by the ratio. In other words:Thus, the sum is the following:Since we want the minimum value of this expression, we want the maximum value for the denominator, . The maximum x-value of a quadratic with negative is .Plugging into the quadratic yields:Therefore, the minimum sum of our infinite geometric sequence is . (Solution by akaashp11)Solution 2After observation we realize that in order to minimize our sum with being the reciprocal of r. The common ratio has to be in the form of with being an integer as anything more than divided by would give a larger sum than a ratio in the form of .The first term has to be , so then in order to minimize the sum, we have minimize .The smallest possible value for such that it is an integer that's greater than is . So our first term is and our common ratio is . Thus the sum is or . Solution 2 by No_OneSolution 3We can see that if is the first term, and is the common ratio between each of the terms, then we can getAlso, we know that the second term can be expressed as notice if we multiply by , we would getThis quadratic has real solutions if the discriminant is greater than or equal to zero, orThis yields that or . However, since we know that has to be positive, we can safely conclude that the minimum possible value of is . 以上就是小编对AMC10数学竞赛真题以及解析的介绍，希望对你有所帮助，如果想了解更多关于AMC数学竞赛报考点、南京AMC数学竞赛培训、美国数学竞赛AMC有用吗以及AMC学习资料等信息请持续关注AMC数学竞赛网。
