AMC12是针对高中学生的数学测验，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。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：Problem 9Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a player knocks the bottle off the ledge is , independently of what has happened before. What is the probability that Larry wins the game?SolutionSolution 1If Larry wins, he either wins on the first move, or the third move, or the fifth move, etc. Let represent "player wins", and represent "player loses". Then the events corresponding to Larry winning are Thus the probability of Larry winning isThis is a geometric series with ratio , hence the answer is .Solution 2Break the problem up into two separate cases: (a) Larry wins on the first throw or (b) Larry wins after the first throw.a: The probability that Larry wins on the first throw is .b: The probability that Larry wins after the first throw is half the probability that Julius wins because it only occurs half the time. This probability is , where is the probability that Larry wins.Therefore, . This equation can be solved for to find that the probability that Larry wins is .Problem 10How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?SolutionSince we want non-congruent triangles that are neither isosceles nor equilateral, we can just list side lengths with . Furthermore, "positive area" tells us that and the perimeter constraints means .There are no triangles when because then must be less than , implying that , contrary to .When , similar to above, must be less than , so this leaves the only possibility . This gives 3 triangles within our perimeter constraint.When , can be or , which gives triangles . Note that is a right triangle, so we get rid of it and we get only 2 triangles.All in all, this gives us triangles.以上就是小编对AMC12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！
AMC12是针对高中学生的数学测验，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。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：Problem 7A regular 15-gon has lines of symmetry, and the smallest positive angle for which it has rotational symmetry is degrees. What is ?SolutionFrom consideration of a smaller regular polygon with an odd number of sides (e.g. a pentagon), we see that the lines of symmetry go through a vertex of the polygon and bisect the opposite side. Hence , the number of sides / vertices. The smallest angle for a rotational symmetry transforms one side into an adjacent side, hence , the number of degrees between adjacent sides. Therefore the answer is .Problem 8What is the value of ?Solution 1Solution 2We can rewrite as as . Thus, 以上就是小编对AMC12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！
AMC12是针对高中学生的数学测验，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。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：Problem 5The Tigers beat the Sharks 2 out of the 3 times they played. They then played more times, and the Sharks ended up winning at least 95% of all the games played. What is the minimum possible value for ?SolutionThe ratio of the Shark's victories to games played is . For to be at its smallest, the Sharks must win all the subsequent games because . Then we can write the equationCross-multiplying yields , and we find that .Problem 6Back in 1930, Tillie had to memorize her multiplication facts from to . The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?SolutionThere are a total of products, and a product is odd if and only if both its factors are odd. There are odd numbers between and , namely hence the number of odd products is . Therefore the answer is .以上就是小编对AMC12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！
AMC12是针对高中学生的数学测验，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。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：Problem 3Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100, and one of the numbers is 28. What is the other number?SolutionLet be the number written two times, and the number written three times. Then . Plugging in doesn't yield an integer for , so it must be that , and we get . Solving for , we obtain .Problem 4David, Hikmet, Jack, Marta, Rand, and Todd were in a 12-person race with 6 other people. Rand finished 6 places ahead of Hikmet. Marta finished 1 place behind Jack. David finished 2 places behind Hikmet. Jack finished 2 places behind Todd. Todd finished 1 place behind Rand. Marta finished in 6th place. Who finished in 8th place?Solution 1Let denote any of the 6 racers not named. Then the correct order following all the logic looks like:Clearly the 8th place runner is .Solution 2We can list these out vertically to ensure clarity, starting with Marta and working from there.Thus our answer is .以上就是小编对AMC12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！
AMC12是针对高中学生的数学测验，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。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：Problem 1What is the value of ?SolutionProblem 2Marie 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?SolutionThe first two tasks took minutes. Thus, each task takes minutes. So the third task finishes at minutes .以上就是小编对AMC12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！接下来小编就为大家讲一下美国NRC表示：数字、几何、空间及测量是早期学数学的重点的具体内容：美国CCSS（共同核心州标准）是一个非常值得参考的素材，它不仅给出了详细的标准，而且提供了完整的教育理念和教育方法支撑。上次我在公众号分享了其中各年级数学教学标准一览图（在公众号后台回复“CCSS”获取）。这里我为大家选取了美国CCSS中小学数学教学标准的主旨阐述《Toward Greater Focus and Coherence》部分，可以让我们更直观地了解到CCSS数学标准对我们学习数学、进行数学拓展的可参考性。儿童早期阶段的数学经验应该集中在（1）数字（包括整数、运算和关系）和（2）几何学，空间关系和测量中，（他们的）数学学习时间要更多用于数量而不是其他主题上。数学教学的过程和目标应该整合到这些内容领域。—Mathematics Learning in Early Childhood, National Research Council, 2009（香港，韩国和新加坡的）综合标准有许多特征可以为美国K-6数学标准的发展提供国际化基准。首先，综合标准将数学的早期学习集中在数字、测量和几何关系，而不重视数据分析，并且几乎不涉及代数。其次，香港的1~3年级标准里将一半的时间放在数字上，剩余的时间用于几何和测量的教学。— Ginsburg, Leinwand and Decker, 2009由于[美国]教科书中的数学概念往往较弱，因此教学演示变得比理想中更加机械（死板）。我们研究了美国使用的传统和非传统教科书，发现两者都存在这种概念上的缺陷。— Ginsburg et al., 2005有很多方法来组织课程。现在（组织课程）面临的挑战不再是避免扭曲数学和让学生失去兴趣。— Steen, 2007十多年来，高效国家的数学教育研究指出，美国的数学课程必须变得更聚焦和更连贯，以提高该国的数学成就。为履行共同标准（common standard）的承诺，标准必须解决课程只有广度没有深度（“a mile wide and an inch deep. ”）的问题。这些标准是对这一挑战的实质性回答。重要的是要认识到“较少的标准”（fewer standards）不能替代重点标准。通过使用广义的一般性陈述很容易实现“较少的标准” （fewer standards），但共同标准旨在清晰和具体。评估一套标准的连贯性（coherence）比评估其重点（focus）更难。 William Schmidt和Richard Houang（2002）曾经说过如何去评估内容标准和课程是连贯的：“（一套标准要）随着时间的推移逐渐表现出一系列符合逻辑并经过思考的主题，且在适当的时候能够反映出该科目内容来源的顺序和层次。也就是说，学生学习什么和如何去学不仅要反映该知识点属于某个学科的主题，还要反映出该知识点在该学科中组织和产生的关键思想。这意味着为了保持连贯性，一套内容标准必须从描述详细情况（例如，整数的含义和运算，包括简单的数学事实和与整数和分数相关的常规计算过程）演变为展现该学科内在的更深层结构。 这些更深层次的结构可以作为连接细节的手段（比如理解有理数及其性质）。 ”共同标准努力遵循这样的设计，不仅强调关键概念的概念性理解，还要根据不断重组原则，如位值原理或运算性质，来构建这些概念。此外，数学标准体系中概述的“主题和表现顺序”也必须尊重学生的学习方式。 正如Confrey（2007）指出的，发展“有序的障碍和挑战对于那些缺乏通过认真学习来获得理解的学生而言，将是不幸的，也是不明智的。”认识到这一点，共同标准的发展史从基于研究的学习过程开始，其中详细介绍了现今已知的关于学生 ”数学知识、技能和理解随着时间的推移而发展”的理论。以上就是小编了解到的关于美国NRC表示：数字、几何、空间及测量是早期学数学的重点的具体内容，更多AMC资讯和真题，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：Problem 23Let be a positive integer greater than 4 such that the decimal representation of ends in zeros and the decimal representation of ends in zeros. Let denote the sum of the four least possible values of . What is the sum of the digits of ?Solution 1A trailing zero requires a factor of two and a factor of five. Since factors of two occur more often than factors of five, we can focus on the factors of five. We make a chart of how many trailing zeros factorials have:We first look at the case when has zero and has zeros. If , has only zeros. But for , has zeros. Thus, and work.Secondly, we look at the case when has zeros and has zeros. If , has only zeros. But for , has zeros. Thus, the smallest four values of that work are , which sum to . The sum of the digits of is Solution 2By Legendre's Formula and the information given, we have that .Trivially, it is obvious that as there is no way that if , would have times as many zeroes as .First, let's plug in the number We get that , which is obviously not true. Hence, After several attempts, we realize that the RHS needs to more "extra" zeroes than the LHS. Hence, is greater than a multiple of .Very quickly, we find that the least are ..Problem 24Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin facing to the east and walks one unit, arriving at . For , right after arriving at the point , if Aaron can turn left and walk one unit to an unvisited point , he does that. Otherwise, he walks one unit straight ahead to reach . Thus the sequence of points continues , and so on in a counterclockwise spiral pattern. What is ?Solution 1The first thing we would do is track Aaron's footsteps:He starts by taking step East and step North, ending at after steps and about to head West.Then he takes steps West and steps South, ending at ) after steps, and about to head East.Then he takes steps East and steps North, ending at after steps, and about to head West.Then he takes steps West and steps South, ending at after steps, and about to head East.From this pattern, we can notice that for any integer he's at after steps, and about to head East. There are terms in the sum, with an average value of , so:If we substitute into the equation: . So he has moves to go. This makes him end up at .Solution 2We are given that Aaron starts at , and we note that his net steps follow the pattern of in the -direction, in the -direction, in the -direction, in the -direction, in the -direction, in the -direction, and so on, where we add odd and subtract even.We want , but it does not work out cleanly. Instead, we get that , which means that there are extra steps past adding in the -direction (and the final number we add in the -direction is ).So .We can group as .Thus .Solution 3Looking at his steps, we see that he walks in a spiral shape. At the th step, he is on the bottom right corner of the square centered on the origin. On the th step, he is on the bottom right corner of the square centered at the origin. It seems that the is the bottom right corner of the square. This makes sense since, after , he has been on n^2 dots, including the point . Also, this is only for odd , because starting with the square, we can only add one extra set of dots to each side, so we cannot get even . Since , is the bottom right corner of the square. This point is over to the right, and therefore down, so . Since is ahead of , we go back spaces to .Problem 25A rectangular box measures , where , , and are integers and . The volume and the surface area of the box are numerically equal. How many ordered triples are possible?SolutionThe surface area is , the volume is , so .Divide both sides by , we have:First consider the bound of the variable . Since we have , or .Also note that , we have . Thus, , so .So we have or .Before the casework, let's consider the possible range for if .From , we have . From , we have . Thus When , , so . The solutions we find are , for a total of solutions.When , , so . The solutions we find are , for a total of solutions.When , , so . The only solution in this case is .When , is forced to be , and thus .Thus, our answer is Simplification of SolutionStart right after the Solution says is . We can say , where .Notice that ! This is our key step. Then we can say , . If we clear the fraction about b and c (do the math), our immediate result is that . Realize also that .Now go through cases for and you end up with the same result. However, now you don't have to guess solutions. For example, when , then and .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：Problem 22In the figure shown below, is a regular pentagon and . What is ?Solution 1Triangle is isosceles, so . since is also isosceles. Using the symmetry of pentagon , notice that . Therefore, .Since ,.So, since must be greater than 0.Notice that .Therefore, Solution 2 (Trigonometry)Note that since is a regular pentagon, all of its interior angles are . We can say that pentagon is also regular by symmetry. So, all of the interior angles of are . Now, we can angle chase and use trigonometry to get that , , and . Adding these together, we get that . Because calculators were not permitted in the 2015 AMC 10B, we can not use a calculator to find out which of the options is equal to , but we can find that this is closest to .Solution 3When you first see this problem you can't help but see similar triangles. But this shape is filled with triangles throwing us off. First, let us write our answer in terms of one side length. I chose to write it in terms of so we can apply similar triangles easily. To simplify the process lets write as .First what is in terms of , also remember : = Next, find in terms of , also remember : = So adding all the we get . Now we have to find out what x is. For this, we break out a bit of trig. Let's look at By the law of sines:Now by the double angle identities in trig. substituting inA good thing to memorize for AMC and AIME is the exact values for all the nice sines and cosines. You would then know that: =so now we know:Substituting back into we get 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！
