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 18Each vertex of a cube is to be labeled with an integer through , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?Solution 1First of all, the adjacent faces have the same sum , because , , so now consider the (the two sides which are parallel but not on same face of the cube); they must have the same sum value too. Now think about the extreme condition 1 and 8, if they are not sharing the same side, which means they would become endpoints of , we should have , but no solution for , contradiction.Now we know and must share the same side, which sum is , the also must have sum of , same thing for the other two parallel sides.Now we have parallel sides . thinking about endpoints number need to have a sum of . It is easy to notice only and would work.So if we fix one direction or all other parallel sides must lay in one particular direction. or Now, the problem is same as the problem to arrange points in a two-dimensional square. which is =Solution 2Again, all faces sum to If are the vertices next to one, then the remaining vertices are Now it remains to test possibilities. Note that we must have Without loss of generality, let Does not work. Works. Works. Does not work. Does not work. Does not work. Works.So our answer is Solution 3We know the sum of each face is If we look at an edge of the cube whose numbers sum to , it must be possible to achieve the sum in two distinct ways, looking at the two faces which contain the edge. If and were on the same face, it is possible to achieve the desired sum only with the numbers and since the values must be distinct. Similarly, if and were on the same face, the only way to get the sum is with and . This means that and are not on the same edge as , or in other words they are diagonally across from it on the same face, or on the other end of the cube.Now we look at three cases, each yielding two solutions which are reflections of each other:1) and are diagonally opposite on the same face. 2) is diagonally across the cube from , while is diagonally across from on the same face. 3) is diagonally across the cube from , while is diagonally across from on the same face.This means the answer is Problem 19In rectangle and . Point between and , and point between and are such that . Segments and intersect at and , respectively. The ratio can be written as where the greatest common factor of and is What is ?Solution 1Use similar triangles. Our goal is to put the ratio in terms of . Since Similarly, . This means that . As and are similar, we see that . Thus . Therefore, so Solution 2Coordinate Bash: We can set coordinates for the points. and . The line 's equation is , line 's equation is , and line 's equation is . Adding the equations of lines and , we find that the coordinates of are . Furthermore we find that the coordinates of are . Using the Pythagorean Theorem, we get that the length of is , and the length of is The length of . Then The ratio Then and is and , respectively. The problem tells us to find , so An alternate solution is to perform the same operations, but only solve for the x-coordinates. By similar triangles, the ratios will be the same.Solution 3Extend to meet at point . Since and , by similar triangles and . It follows that . Now, using similar triangles and , . WLOG let . Solving for gives and . So our desired ratio is and .Solution 4Mass Points: Draw line segment , and call the intersection between and point . In , observe that and . Using mass points, find that . Again utilizing , observe that and . Use mass points to find that . Now, draw a line segment with points ,,, and ordered from left to right. Set the values ,, and . Setting both sides segment equal, we get . Plugging in and solving gives , ,. The question asks for , so we add to and multiply the ratio by to create integers. This creates . This sums up to Solution 5 (Cheap Solution)Use your ruler (you should probably recommended you bring ruler and protractor to AMC10 tests) and accurately draw the diagram as one in solution 1, then measure the length of the segments, you should get a ratio of being , multiplying each side by the result 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，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数学竞赛真题以及官方解答吧：Problem 16A triangle with vertices , , and is reflected about the -axis, then the image is rotated counterclockwise about the origin by to produce . Which of the following transformations will return to ? counterclockwise rotation about the origin by . clockwise rotation about the origin by . reflection about the -axis reflection about the line reflection about the -axis.SolutionConsider a point . Reflecting it about the -axis will map it to , and rotating it counterclockwise about the origin by will map it to . The operation that undoes this is a reflection about the , so the answer is .Problem 17Let be a positive multiple of . One red ball and green balls are arranged in a line in random order. Let be the probability that at least of the green balls are on the same side of the red ball. Observe that and that approaches as grows large. What is the sum of the digits of the least value of such that ?Solution 1Let . Then, consider blocks of green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the positions between the green balls to insert the red ball. Less than of the green balls will be on the same side of the red ball if the red ball is inserted in the middle block of balls, and there are positions where this happens. Thus, , soMultiplying both sides of the inequality by , we haveand by the distributive property,Subtracting on both sides of the inequality gives usTherefore, , so the least possible value of is . The sum of the digits of is .Solution 2 (Pattern Solution)Let , (Given)Let , Let , Notice that the fraction can be written as Now it's quite simple to write the inequality as We can subtract on both sides to obtain Dividing both sides by , we derive . (Switch the inequality sign when dividing by )We then cross multiply to get Finally we get To achieve So the sum of the digits of = Solution 3We are trying to find the number of places to put the red ball, such that of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with : Trying a few values, we see that the ball "works" in places to and spaces to . This is a total of spaces, over a total possible places to put the ball. So: And we know that the next value is what we are looking for, so , and the sum of it's digits 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，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数学竞赛真题以及官方解答吧：Problem 14How many ways are there to write as the sum of twos and threes, ignoring order? (For example, and are two such ways.)Solution 1The amount of twos in our sum ranges from to , with differences of because .The possible amount of twos is .Solution 2You can also see that you can rewrite the word problem into an equation + = . Therefore the question is just how many multiples of subtracted from 2016 will be an even number. We can see that , all the way to , and works, with being incremented by 's.Therefore, between and , the number of multiples of is .Solution 3We can utilize the stars-and-bars distribution technique to solve this problem. We have 2 "buckets" in which we will distribute parts of our total sum, 2016. By doing this, we know we will have "total" answers. We want every third x and second y, so we divide our previous total by 6, which will result in . We have to round down to the nearest integer, and we have to add 2 because we did not consider the 2 solutions involving x or y being 0. So, .Problem 15Seven cookies of radius inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?SolutionThe big cookie has radius , since the center of the center cookie is the same as that of the large cookie. The difference in areas of the big cookie and the seven small ones is . The scrap cookie has this area, so its radius must 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，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数学竞赛真题以及官方解答吧：Problem 13Five friends sat in a movie theater in a row containing seats, numbered to from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?Solution 1Bash: we see that the following configuration works.Bea - Ada - Ceci - Dee - EdieAfter moving, it becomesAda - Ceci - Bea - Edie - Dee.Thus, Ada was in seat .Solution 2Process of elimination of possible configurations.Let's say that Ada=, Bea=, Ceci=, Dee=, and Edie=. Since moved more to the right than did left, this implies that was in a LEFT end seat originally:This is affirmed because , which there is no new seats uncovered. So are restricted to the same seats. Thus, it must be , and more specifically:So , Ada, was originally in seat .Solution 3The seats are numbered 1 through 5, so let each letter () correspond to a number. Let a move to the left be subtraction and a move to the right be addition.We know that . After everyone moves around, however, our equation looks like because and switched seats, moved two to the right, and moved 1 to the left.For this equation to be true, has to be -1, meaning moves 1 left from her original seat. Since is now sitting in a corner seat, the only possible option for the original placement of is in seat number .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，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数学竞赛真题以及官方解答吧：Problem 11Find the area of the shaded region.Solution 1The bases of these triangles are all , and their heights are , , , and . Thus, their areas are , , , and , which add to the area of the shaded region, which is .Solution 2Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. We can do this by splitting up the unshaded areas into various triangles and rectangles as shown.Notice that the two added lines bisect each of the sides of the large rectangle.Subtracting the unshaded area from the total area gives us , so the correct answer is .Solution 3Notice that we can graph this on the coordinate plane.The top-left shaded figure has coordinates of .Notice that we can apply the shoelace method to find the area of this polygon.We find that the area of the polygon is .However, notice that the two shaded regions are two congruent polygons.Hence, the total area is or .Problem 12Three distinct integers are selected at random between and , inclusive. Which of the following is a correct statement about the probability that the product of the three integers is odd?Solution 1For the product to be odd, all three factors have to be odd. The probability of this is ., but and are slightly less than . Thus, the whole product is slightly less than , so .Solution 2For the product to be odd, all three factors have to be odd. There are a total of ways to choose 3 numbers at random, and there are to choose 3 odd numbers. Therefore, the probability of choosing 3 odd numbers is . Simplifying this, we obtain , which is slightly less than , so 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，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数学竞赛真题以及官方解答吧：Problem 9A triangular array of coins has coin in the first row, coins in the second row, coins in the third row, and so on up to coins in the th row. What is the sum of the digits of ?SolutionWe are trying to find the value of such thatNoticing that we have so our answer is Notice that we were attempting to solve . Approximating , we were looking for a square that is close to, but less than, . Since , we see that is a likely candidate. Multiplying confirms that our assumption is correct.Problem 10A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is foot wide on all four sides. What is the length in feet of the inner rectangle?SolutionLet the length of the inner rectangle be .Then the area of that rectangle is .The second largest rectangle has dimensions of and , making its area . The area of the second shaded area, therefore, is .The largest rectangle has dimensions of and , making its area . The area of the largest shaded region is the largest rectangle minus the second largest rectangle, which is .The problem states that is an arithmetic progression, meaning that the terms in the sequence increase by the same amount each term.Therefore,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，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数学竞赛真题以及官方解答吧：Problem 7The mean, median, and mode of the data values are all equal to . What is the value of ?SolutionSince is the mean,Therefore, , so CheckOrder the list: . must be either or because it is both the median and the mode of the set. Thus is correct.Problem 8Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?SolutionIf you started backwards you would 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，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数学竞赛真题以及官方解答吧：Problem 5A rectangular box has integer side lengths in the ratio . Which of the following could be the volume of the box?SolutionLet the smallest side length be . Then the volume is .If , then Problem 6Ximena lists the whole numbers through once. Emilio copies Ximena's numbers, replacing each occurrence of the digit by the digit . Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?SolutionFor every tens digit 2, we subtract 10, and for every units digit 2, we subtract 1. Because 2 appears 10 times as a tens digit and 2 appears 3 times as a units digit, 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，更多学习资料请持续关注AMC数学竞赛网！
