2017AMC10A真题与答案解析
【完整版】河北省衡水中学2017届高三第十次调研英语A卷试题(课标版卷)(含答案)
本试卷由四局部组成。
其中，第一、二局部和第三局部的第一节为选择题。
第三局部的第二节和第四局部为非选择题。
全卷总分值150分，考试用时120分钟。
考前须知：1.答选择题前，考生务必将自己的姓名、准考证号填写在答题卡上。
并将准考证号条码黏贴在答题卡上的指定位置。
用2B铅笔将答题卡上正确试卷类型后的方框涂黑。
2. 选择题的作答：每题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑。
写在试卷、草稿纸和答题卡上的非答题区域均无效。
3.非选择题的作答：用签字笔直接在答题卡上对应的答题区域内。
写在试卷、草稿纸和答题卡的非答题区域均无效。
4．考试完毕后，将本试卷和答题卡一并上交。
第一局部：听力〔共两节，总分值30分〕第一节〔共5小题；每题1.5分，总分值7.5分〕听下面5段对话。
每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最正确选项，并标在试卷的相应位置。
听完每段对话后，你都有10秒钟的时间来答复有关小题和阅读下一小题。
每段对话仅读一遍。
1.What is the man doing?A.Talking on the phone.B.Tidying up his bedroom.C.Reading a book.2.What does the woman think of the program?A.It can make a lot of money.B.It will limit the number of babies.C.It is helpful to many new parents.3.What does the man mean?A.The woman should give up the interview.B.The woman may get another chance.C.The woman can apply for the job again.4.What does the woman ask the man to do?A.Clean the room.B.Wash the clothes.C.Clean the dishes.5.How many people are there in the woman’s family？A.Four.B.Five.C.Six.第二节〔共15小题；每题1.5分，总分值22.5分〕听下面5段对话或独白。
AMC考题和答案解析
A M C考题和答案解析Document serial number【NL89WT-NY98YT-NC8CB-NNUUT-NUT108】2017 AMC 8 考题及答案Problem 1Which of the following values is largestProblem 2Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received 36 votes, then how many votes were cast all togetherProblem 3What is the value of the expressionProblem 4When is multiplied by 7,928,564 the product is closest to which of the followingProblem 5What is the value of the expressionProblem 6If the degree measures of the angles of a triangle are in the ratio , what is the degree measure of the largest angle of the triangleProblem 7Let be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor ofProblem 8Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tellshim, "My house number has two digits, and exactly three of the following four statements about it are true."(1) It is prime.(2) It is even.(3) It is divisible by 7.(4) One of its digits is 9.This information allows Malcolm to determine Isabella's house number. What is its units digitProblem 9All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could haveProblem 10A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selectedProblem 11A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floorProblem 12The smallest positive integer greater than 1 that leaves a remainderof 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbersProblem 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 winProblem 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 answersProblem 15In the arrangement of letters and numerals below, by how manydifferent 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.Problem 16In the figure below, choose point on so that and have equal perimeters. What is the area ofProblem 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 treasurechests 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 haveProblem 18In the non-convex quadrilateral shown below, is a right angle, , , , and .What is the area of quadrilateralProblem 19For any positive integer , the notation denotes the product of the integers through . What is the largest integer for whichis a factor of the sumProblem 20An integer between and , inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinctProblem 21Suppose , , and are nonzero real numbers, and . What are the possible value(s) forProblem 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 semicircleProblem 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 fourtripsProblem 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 phonecall from any of her grandchildrenProblem 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 shown2017 AMC 8 Answer Key1.A2.E3.C4.D5.B6.D7.A8.D9.D10.C11.C12.D13.B14.C15.D16.D17.C18.B19.D20.B21.A22.D23.C24.D25.B。
2017年全国高中数学联赛A卷试题和答案
2017年全国高中数学联赛A 卷一试一、填空题1.设)(x f 是定义在R 上的函数，对任意实数x 有1)4()3(-=-⋅+x f x f .又当70<≤x 时，)9(log )(2x x f -=，则)100(-f 的值为__________.2.若实数y x ,满足1cos 22=+y x ，则y x cos -的取值范围是__________.3.在平面直角坐标系xOy 中，椭圆C 的方程为1109:22=+y x ，F 为C 的上焦点，A 为C 的右顶点，P 是C 上位于第一象限内的动点，则四边形OAPF 的面积的最大值为__________. 4.若一个三位数中任意两个相邻数码的差不超过1，则称其为“平稳数”.平稳数的个数是 5.正三棱锥ABC P -中，1=AB ，2=AP ，过AB 的平面α将其体积平分，则棱PC 与平面α所成角的余弦值为__________.6.在平面直角坐标系xOy 中，点集}{1,0,1,),(-==y x y x K .在K 中随机取出三个点，则这三点中存在两点之间距离为5的概率为__________.7.在ABC ∆中，M 是边BC 的中点，N 是线段BM 的中点.若3π=∠A ，ABC ∆的面积为3，则AN AM ⋅的最小值为__________.8.设两个严格递增的正整数数列{}{}n n b a ,满足：20171010<=b a ，对任意正整数n ，有n n n a a a +=++12，n n b b 21=+，则11b a +的所有可能值为__________.二、解答题9.设m k ,为实数，不等式12≤--m kx x 对所有[]b a x ,∈成立.证明：22≤-a b .10.设321,,x x x 是非负实数，满足1321=++x x x ，求)53)(53(321321x x x x x x ++++的最小值和最大值.11.设复数21,z z 满足0)Re(1>z ，0)Re(2>z ，且2)Re()Re(2221==z z （其中)Re(z 表示复数z 的实部）.（1）求)Re(21z z 的最小值；（2）求212122z z z z --+++的最小值.2017年全国高中数学联赛A 卷二试一.如图，在ABC ∆中，AC AB =，I 为ABC ∆的内心，以A 为圆心，AB 为半径作圆1Γ，以I 为圆心，IB 为半径作圆2Γ，过点I B ,的圆3Γ与1Γ,2Γ分别交于点Q P ,（不同于点B ）.设IP 与BQ 交于点R .证明：CR BR ⊥二.设数列{}n a 定义为11=a ，Λ,2,1,,,,1=⎩⎨⎧>-≤+=+n n a n a n a n a a n n n n n .求满足20173≤<r a r 的正整数r 的个数.三.将3333⨯方格纸中每个小方格染三种颜色之一，使得每种颜色的小方格的个数相等.若相邻连个小方格的颜色不同，则称它们的公共边为“分隔边”.试求分隔边条数的最小值.四.设n m ,均是大于1的整数，n m ≥，n a a a ,,,21Λ是n 个不超过m 的互不相同的正整数，且n a a a ,,,21Λ互素.证明：对任意实数x ，均存在一个)1(n i i ≤≤，使得x m m x a i )1(2+≥，这里y 表示实数y 到与它最近的整数的距离.2017年全国高中数学联赛A 卷一试答案1.2.3.4.5.6.7.8.9.10.11.2017年全国高中数学联赛A卷二试答案一.二.三.四.。
2017年全国初中数学联合竞赛试题(pdf版,含答案)
D．4．
（）
若 (a,b, c) 为好数组，则 abc 2(a b c) 6c ，所以 ab 6 .显然， a 只能为 1 或 2.
若 a ＝2，由 ab 6 可得 b 2 或 3， b 2 时可得 c 4 ， b 3 时可得 c 5 （不是整数）； 2
若 a ＝1，则 bc 2(1 b c) ，于是可得 (b 2)(c 2) 6 ，可求得 (a,b, c) ＝（1，3，8）或（1，4，
若 b ＝6，则 (a 9)2 0 ，解得 a 9 ，此时 c 18 .
2017 年全国初中数学联合竞赛试题参考答案及评分标准 第 4 页（共 7 页）
因此， a 9 ， b ＝6， c 18 ，故 a2 b2 c2 ＝441.
5．设 O 是四边形 ABCD 的对角线 AC 、BD 的交点，若 BAD ACB 180，且 BC 3，AD 4 ，
（2）以 a 2,b 3,c 4 为边长可以构成三角形，但以 a2 4,b2 9,c2 16 为边长的三角形不存在；
（3）因为 a b c ，所以 | a b | 1 a b 1,| b c | 1 b c 1,| c a | 1 a c 1 ，故三条边中 | c a | 1 大于或等于其余两边，而（| a b | 1）（| b c | 1）（a b 1）（b c 1）=a c 11 a c 1 | c a | 1 ，故
2
4
2
设 m 是最接近 n 的整数，则| m n | 1 ， m 1. 2
易知：当 m 1时，| m n | 1 (m 1)2 n (m 1)2 m2 m 1 n m2 m 1 .
2017年全国高中数学联赛A卷第10题思路与详解
2017年全国高中数学联赛A 卷第10题思路江西师大附中汪师林题目：设123,,x x x 是非负实数，1231x x x ++=，求()3212313535x x x x x x ⎛⎫++++ ⎪⎝⎭的最大值和最小值（20分）分析与解答: 首先看到这个题目，123,,x x x 是非负实数，立马可以得知，在最大值或者最小值的时候，一定至少有一个变量是等于0的，否则最大值和最小值都在变量全部大于0取到的话，那么123,,x x x 是非负实数这个条件我们就可以改为123,,x x x 都为正实数，则题目相当于多了一个无用的条件，这个在竞赛中是不允许的。
（1） 最小值的求解是显然的，运用一个柯西不等式就可以一步到位()()232123112335135x x x x x x x x x ⎛⎫++++≥++= ⎪⎝⎭， 而且当()123,,(1,0,0)x x x =及其循环时()32123135135x x x x x x ⎛⎫++++= ⎪⎝⎭所以最小值为1.（2） 做最大值的时候，首先第一步也是猜，要么是两个相等，一个为0，要么是三个全部相等。
当12313x x x ===时，()32123123353515x x x x x x ⎛⎫++++= ⎪⎝⎭；当1231,02x x x ===时，()321231435353x x x x x x ⎛⎫++++= ⎪⎝⎭；当1321,02x x x ===时，()321231935355x x x x x x ⎛⎫++++= ⎪⎝⎭；当2311,02x x x ===时，()32123116353515x x x x x x ⎛⎫++++= ⎪⎝⎭；通过上述实验可知，当1321,02x x x ===时，()3212313535x x x x x x ⎛⎫++++ ⎪⎝⎭是可以取得最大值的，而且我们发现在()12335x x x ++中，13,x x 的系数比为1比5，在32135x x x ⎛⎫++ ⎪⎝⎭中，13,x x 的系数比为5比1，我们是要在13x x =的时候取得最大值，所以我们需要它们的系数比最后要相同，而20x =，0乘以任何数都等于0，所以2x的系数可以不需要太多限制，为了达到达到13,x x 的系数相同，这里我们需要引进一个系数参数k ，()()321231312123231212323123535353535(1)435(251)(1)(91)435x x x x x x x x xk x x x k k k x x x k x x x k k k k x k x k x k k k k ⎛⎫++++ ⎪⎝⎭⎛⎫=++++ ⎪⎝⎭⎛⎫≤+++++ ⎪⎝⎭+++⎛⎫=++ ⎪⎝⎭（1） 式子要想将积的形式转换成和的形式，常用的要么是柯西不等式，要么是均值不等式，这里选的是均值不等式。
-年amc 10a和b竞赛真题及答案(英文版)
2010-2015年AMC 10A和B竞赛真题及答案（英文版）
你好，目前，只分享了2010-2013年AMC 10A和B竞赛真题及答案（英文版），2014-2015年的暂时还没有上传，等采纳后再私信我吧！需要的话，我可以将Word文档中的原图文件一同发给你，求采纳，毕竟我花了差不多一个下午才整理完，谢谢！(@_@)
2010年AMC 10Aห้องสมุดไป่ตู้赛真题及答案（英文版）
2010年AMC 10B竞赛真题及答案（英文版）
2011年AMC 10A竞赛真题及答案（英文版）
2011年AMC 10B竞赛真题及答案（英文版）
Problem 4
2012年AMC 10A竞赛真题及答案（英文版）
go去wentgone
2012年AMC 10B竞赛真题及答案（英文版）
get得到gotgot
sink下沉sank / sunksunk / sunken
give给gavegiven
hit打hithit
不规则动词表
2013年AMC 10A竞赛真题及答案（英文版）
mistake误认mistookmistaken
grow成长grewgrown
think思考thoughtthought
2017 AMC 10A
2017 AMC 10A Problems and SolutionsProblem 1What is the value of?（A）70 （B）97 （C）127 （D）159 （E）729Problem 2Pablo buys popsicles for his friends. The store sells single popsicles for $1each, 3-popsicle boxes for $2each, and5-popsicle boxes for$3. What is the greatest number of popsicles that Pablo can buy with$8?（A）8 （B）11 （C）12 （D）13 （E）15Problem3Tamara has three rows of two-feet by-feet flower beds in her garden. The beds are separated and also surrounded by-foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?（A）72 （B）78 （C）90 （D）120 （E）150Problem 4Mia is “helping” her mom pick up 30toys that are strewn on the floor. Mia’s mom manages to put 3toys into the toy box every 30 seconds, but each time immediately after those 30 seconds have elapsed, Mia takes 2toys out of the box. How much time, in minutes, will it take Mia and her mom to put all 30 toys into the box for the first time?（A）13.5 （B）14 （C）14.5 （D）15 （E）15.5Problem 5The sum of two nonzero real numbers is 4 times their product. What is the sumof the reciprocals of the two numbers?（A）1 （B）2 （C）4 （D）8 （E）12Problem 6Ms. Carroll promised that anyone who got all the multiple choice questions right on the upcoming exam would receive an A on the exam. Which of these statements necessarily follows logically?Problem 7Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?（A）30% （B）40% （C）50% （D）60% （E）70%Problem 8At a gathering of 30people, there are 20 people who all know each other and 10 people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?（A）240 （B）245 （C）250 （D）480 （E）490Problem 9Minnie rides on a flat road at20kilometers per hour (kph), downhill at30kph, and uphill at5kph. Penny rides on a flat road at kph, downhill at40kph, and uphill at kph. Minnie goes from town to town, a distance of km all uphill, then from town to town, a distance of km all downhill, and then back to town, a distance of km on the flat. Penny goes the other way around using the same route. How many more minutes does it take Minnie to complete the45-km ride than it takes Penny?（A）45 （B）60 （C）65 （D）90 （E）95Problem 10Joy has30thin rods, one each of every integer length from1cm through30cm. She places the rods with lengths cm,cm, and cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can shechoose as the fourth rod?（A）16 （B）17 （C）18 （D）19 （E）20Problem 11The region consisting of all points in three-dimensional space within units of line segment has volume. What is the length?（A）6 （B）12 （C）18 （D）20 （E）24Problem 12Let be a set of points in the coordinate plane such that two of the three quantities and are equal and the third of the three quantities is no greater than this common value. Which of the following is a correct description forProblem 13Define a sequence recursively by and the remainder when is divided by3,for all Thus the sequence starts What is（A）6 （B）7 （C）8 （D）9 （E）10Problem 14Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was dollars. The cost of his movie ticket was 20% of the difference between and the cost of his soda, while the cost of his soda was 5% of the difference between and the cost of his movie ticket. To the nearest whole percent, what fraction of did Roger pay for his movie ticket and soda?（A）9% （B）19% （C）22% （D）23% （E）25%Problem 15Chloéchooses a real number uniformly at random from the interval. Independently, Laurent chooses a real number uniformly at random from the interval . What is the probability that Laurent's number is greater than Chloé's number?Problem 16There are 10 horses, named Horse 1, Horse 2,, Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse runs one lap in exactly minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time, in minutes, at which all 10 horses will again simultaneously be at the starting point is. Let be the least time, in minutes, such that at least 5 of the horses are again at the starting point. What is the sum of the digits of?（A）2 （B）3 （C）4 （D）5 （E）6Problem 17Distinct points,,,lie on the circle and have integer coordinates. The distances and are irrational numbers. What is the greatest possible value of the ratio?Problem 18Amelia has a coin that lands heads with probability 1/3, and Blaine has a coin that lands on heads with probability 2/5. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is, where and are relatively prime positive integers. What is?（A）1 （B）2 （C）3 （D）4 （E）5Problem 19Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of 5 chairs under these conditions?（A）12 （B）16 （C）28 （D）32 （E）40Problem 20Let S(n)equal the sum of the digits of positive integer. For example,S(1507)=13. For a particular positive integer,S(n)=1274. Which ofthe following could be the value of?（A）1 （B）3 （C）12 （D）1239 （E）1265Problem 21A square with side length is inscribed in a right triangle with sides of length 3,4, and5so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length is inscribed in another right triangle with sides of length 3,4, and so that one side of the square lies on the hypotenuse of the triangle. What is?Problem 22Sides and of equilateral triangle are tangent to a circle at points and respectively. What fraction of the area of lies outside the circle?Problem 23How many triangles with positive area have all their vertices at points inthe coordinate plane, where and are integers between1and5, inclusive? （A）2128 （B）2148 （C）2160 （D）2200 （E）2300Problem 24For certain real numbers,, and, the polynomialhas three distinct roots, and each root of is also a root of the polynomialWhat is?（A）-9009 （B）-8008 （C）-7007 （D）-6006 （E）-5005Problem 25How many integers between 100and999, inclusive, have the property that some permutation of its digits is a multiple of11 between 100and 999For example, both121 and 211have this property.（A）226 （B）243 （C）270 （D）469 （E）4862017 AMC 10A SolutionsProblem 1Solution 1Notice this is the term in a recursive sequence, defined recursivelyas Thus:Solution 2Starting to compute the inner expressions, we see the results are 1,3 ,7,15,…. This is always1less than a power of2. The only admissible answer choice by this rule is thus C127.Solution 3Working our way from the innermost parenthesis outwards and directly computing, we have C.Problem 2$3boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying2, we have $2left. We cannot buy a third $3box, so we opt for the $2box instead (since it has a higher popsicles/dollar ratio than the $1pack). We're now out of money. We bought 5+5+3=13popsicles, so the answer is D13.Problem3Finding the area of the shaded walkway can be achieved by computing the total area of Tamara's garden and then subtracting the combined area of her six flower beds.Since the width of Tamara's garden contains three margins, the total width is feet.Similarly, the height of Tamara's garden is feet. Therefore, the total area of the garden is square feet.Finally, since the six flower beds each have an area of square feet, the area we seek is, and our answer isSolution1Every 30 seconds 3-2=1 toys are put in the box, so after seconds there will be 27 toys in the box. Mia's mom will then put 3 toys into to the box and we have our total amount of time to seconds, which equals 14 minutes. B14Solution 2Though Mia's mom places 3 toys every 30 seconds, Mia takes out 2 toys right after. Therefore, after 30 seconds, the two have collectively placed 1 toy into the box. Therefore by 13.5 minutes, the two would have placed 27 toys into the box. Therefore, at 14 minutes, the two would have placed 30 toys into the box. Though Mia may take 2 toys out right after, the number of toys in the box first reaches 30 by 14 minutes. B14Problem 5Solution1Let the two real numbers be . We are given that anddividing both sides by , ，Note: we can easily verify that this is the correct answer; for example, 1/2 and 1/2 work, and the sum of their reciprocals is 4.Solution 2Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction. See for yourself. And by looking into fractions, we immediately see that and 1would fit the rule. C14Solution 3Rewriting the given statement: "if someone got all the multiple choice questions right on the upcoming exam then he or she would receive an A on the exam." If that someone is Lewis the statement becomes: "if Lewis got all the multiple choice questions right, then he got an A on the exam." The contrapositive: "If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong (did not get all of them right)" must also be true leaving B as the correct answer. B is also equivalent to the contrapositive of the original statement, which implies that it must be true, so the answer is B Problem 7Let represent how far Jerry walked, and represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides,Since Silvia walked the diagonal, she walked the hypotenuse of a 45, 45, 90 triangle with leg length 1. Thus,We can thentakeProblem 8Solution 1Each one of the ten people has to shake hands with all the other people they don’t know. So. From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands, or. Thus the answer is.Solution 2We can also use complementary counting. First of all,handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from 435 to find the handshakes. Hugs only happen between the 20 peoplewho know each other, so there are hugs..Solution 3We can focus on how many handshakes the 10 people get.The 1st person gets 29 handshakes.2nd gets 28......And the 10th receives 20 handshakes.We can write this as the sum of an arithmetic sequence.Therefore, the answer isProblem 9The distance from town to town is km uphill, and since Minnie rides uphill at a speed of kph, it will take her hours. Next, she will ride from town to town, a distance of km all downhill. Since Minnie rides downhill at a speed of kph, it will take her half an hour. Finally, she rides from town back to town, a flat distance of km. Minnie rides on a flat road at kph, so this will take her hour. Her entire trip takes her hours. Secondly, Penny will go from town to town, a flat distance of km. Since Penny rides on a flat road at kph, it will take her2/3of an hour. Next Penny will go from town to town, which is uphill for Penny. Since Penny rides at a speed of kph uphill, and town and are km apart, it will take her hours. Finally, Penny goes from Town back to town, a distance of km downhill. Since Penny rides downhill at kph, it will only take her 1/4of an hour. In total, it takes her 29/12hours, which simplifies to hours and minutes. Finally, Penny's Hour Minute trip was minutesless than Minnie's Hour Minute TripProblem 10The triangle inequality generalizes to all polygons, so and to get. Now, we know that there are numbers between and exclusive, but we must subtract to account for the 2 lengths already used that are between those numbers, which gives 19-2=17 BProblem 11In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within units of a point would be a sphere with radius. However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end.We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal):, where is equal to the length of our line segment. Solving, we find that.Problem 12Problem 13A pattern starts to emerge as the function is continued. The repeating pattern is The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which isProblem 14Let = cost of movie ticket, Let = cost of sodaWe can create two equations:Problem 15 Solution 1Solution 2Solution3Scale down by 2017to get that Chloe picks from and Laurent picks from. There are an infinite number of cases for the number that Chloe picks, but they are all centered around the average of. Therefore, Laurent has a range of to to pick from, on average, which is a length of out of a total length of. Therefore, the probability is 1.5/2=3/4 CProblem 16Solution 1If we have horses,, then any number that is a multiple of the all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that. Finally,.Solution 2We are trying to find the smallest number that has one-digit divisors. Therefore we try to find the LCM for smaller digits, such as,,, or. We quickly consider since it is the smallest number that is the LCM of,,and. Since has single-digit divisors, namely,,,, and, our answer isProblem 17Because,,, and are integers there are only a few coordinates that actually satisfy the equation. The coordinates are and We want to maximize and minimize They also have to be the square root of something, because they are both irrational. The greatest value of happens when it and are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be (-4,3)and (3,-4)because the two points are almost across from each other. The least value of is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example,is (3,4)and is (4,3)They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point (3,4)than (4,3)Using the distanceformula, we get that is and that isProblem 18Solution 1Solution 2Problem 19Solution 1For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E. We can split this problem up into two cases:A sits on an edge seat.Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of.A does not sit in an edge seat.In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are seatings in this case.Adding up all the cases, we have.Solution 2Label the seats through. The number of ways to seat Derek and Eric in the five seats with no restrictions is. The number of ways to seat Derek and Eric such that they sit next to each other is(which can be figure out quickly), so the number of ways such that Derek and Eric don't sit next to each other is. Note that once Derek and Eric are seated, there are three cases.The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us ways.Another possible case is if Derek and Eric seat in seats and in some order. There are 2 possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are ways to do this. So the second case gives us total ways for the second case.The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats is available. There are ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot not sit in one of the two consecutive available seats without sitting next to Bob and Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us ways.So in total there are 12+16=28. So our answer is.Problem 20Solution 1Solution 2Solution 3Problem 21Problem 22Problem 23Problem 24 Solution 1Solution 2Solution 3Problem 25Solution 1Let the three-digit number be :If a number is divisible by , then the difference between the sums of alternating digits is a multiple of .There are two cases: andWe now proceed to break down the cases. Note: let so that we avoid counting the same permutations and having to subtract them later.Solution 2We note that we only have to consider multiples of 11 and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of 11 has:Solution 3We can overcount and then subtract. We know there are multiples of. We can multiply by for each permutation of these multiples. (Yet some multiples don't have 6)Now divide by, because if a number with digits,, and is a multiple of 11, then is also a multiple of 11 so we have counted the same permutations twice.Basically, each multiple of 11 has its own 3 permutations (say has and whereas has and). We know that each multiple of 11 has at least 3 permutations because it cannot have 3 repeating digits.Hence we have permutations without subtracting for overcounting. Now note that we overcounted cases in which we have 0's at the start of each number. So, in theory, we could just answer and move on.If we want to solve it, then we continue.We overcounted cases where the middle digit of the number is 0 and the last digit is 0.Note that we assigned each multiple of 11 3 permutations.The last digit is gives possibilities where we overcounted by permutation for each of.The middle digit is 0 gives 8 possibilities where we overcount by 1.andSubtracting gives.Now, we may ask if there is further overlap (I.e if two of and and were multiples of) Thankfully, using divisibility rules, this can never happen as taking the divisibility rule mod 11 and adding we get that,, or is congruent to. Since are digits, this can never happen as none of them can equal 11 and they can't equal 0 as they are the leading digit of a 3 digit number in each of the cases.2017 AMC 10A Answer Key1. C2. D3. B4. B5. C6. B7. A8. B9. C10. B11. D12. E13. D14. D15. C16. B17. D18. D19. C20. D21. D22. E23. B24. C25. A。
AMC 10A 试题及答案解析
2006 AMC 10A1、Sandwiches at Joe's Fast Food cost each and sodas cost each. How many dollars will it cost to purchase 5 sandwiches and 8 sodas? SolutionThe sandwiches cost dollars. The sodas cost dollars. In total, the purchase costs dollars. The answer is .2、Define . What is ?SolutionBy the definition of , we have . Then. The answer is .3、The ratio of Mary's age to Alice's age is . Alice is years old. How old is Mary?SolutionLet be Mary's age. Then . Solving for , we obtain . The answer is .4 、A digital watch displays hours and minutes with and . What is the largest possible sum of the digits in the display? SolutionFrom the greedy algorithm, we have in the hours section and in the minutes section.5 、Doug and Dave shared a pizza with equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was , and there was an additional cost of for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each paid for what he had eaten. How many more dollars did Dave paythan Doug?SolutionDave and Doug paid dollars in total. Doug paid for threeslices of plain pizza, which cost . Dave paid dollars. Dave paid more dollars than Doug. The answer is .6、What non-zero real value for satisfies ?Solution7 、The rectangle is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is ?SolutionTaking the seventh root of both sides, we get . Simplifying the LHS gives , which then simplifies to . Thus, , and the answer is .8、A parabola with equation passes through the points and . What is ?SolutionSubstitute the points (2,3) and (4,3) into the given equation for (x,y).Then we get a system of two equations:Subtracting the first equation from the second we have:Then using in the first equation:is the answer.Alternatively, notice that since the equation is that of a monicparabola, the vertex is likely . Thus, the form of the equation of the parabola is . Expanding this out, we find that .9 、How many sets of two or more consecutive positive integershave a sum of 15?SolutionNotice that if the consecutive positive integers have a sum of 15, then their average (which could be a fraction) must be a divisor of15. If the number of integers in the list is odd, then the averagemust be either 1, 3, or 5, and 1 is clearly not possible. The other two possibilities both work:▪▪If the number of integers in the list is even, then the average will have a . The only possibility is , from which we get:▪Thus, the correct answer is 3, answer choice .10、For how many real values of is an integer?SolutionFor to be an integer, must be a perfect square. Since can't be negative, .The perfect squares that are less than or equal to are, so there are values for . Since every value of gives one and only one possible value for , the number of values of is .11、Which of the following describes the graph of the equation?SolutionExpanding the left side, we haveThus there are two lines described in this graph, the horizontal line and the vertical line . Thus, our answer is .12 、Rolly wishes to secure his dog with an 8-foot rope to a square shed that is 16 feet on each side. His preliminary drawings are shown.Which of these arrangements gives the dog the greater area to roam, and by how many square feet?SolutionLet us first examine the area of both possible arrangements. The rope outlines a circular boundary that the dog may dwell in. Arrangement I allows the dog square feet of area. Arrangement II allows square feet plus a little more on the top part of the fence. So we already know that Arrangement II allows more freedom - only thing left is to find out how much. The extraarea can be represented by a quarter of a circle with radius 4. So the extra area is . Thus the answer is .13 、A player pays to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)SolutionThere are possible combinations of 2 dice rolls. The only possible winning combinations are , and . Since there are winning combinations and possible combinations of dice rolls, the probability of winning is .Let be the amount won in a fair game. By the definition of a fair game,.Therefore, .14 、A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?SolutionThe inside diameters of the rings are the positive integers from 1 to 18. The total distance needed is the sum of these values plus 2 for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series, the answer is.Alternatively, the sum of the consecutive integers from 3 to 20 is. However, the 17 intersections between the rings must be subtracted, and we also get .15、Odell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they pass each other?SolutionSolutionSince , we note that Odell runs one lap in minutes, while Kershaw also runs one lap in minutes. They take the same amount of time to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there arelaps run by both, or meeting points .16 、A circle of radius 1 is tangent to a circle of radius 2. The sides of are tangent to the circles as shown, and the sides and are congruent. What is the area of ?SolutionNote that . Using the first pair of similar triangles, we write the proportion:By the Pythagorean Theorem we have that . Now using ,The area of the triangle is.17、In rectangle , points and trisect , and points and trisect . In addition, . What is the area of quadrilateral shown in the figure?SolutionSolution 1It is not difficult to see by symmetry that is a square.Draw . Clearly . Then is isosceles, and is a . Hence , and . There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.Solution 2Draw the lines as shown above, and count the squares. There are 12, so we have .18、A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. These six characters may appear in any order, except that the two letters must appear next to each other. How many distinct license plates are possible?SolutionThere are ways to choose 4 digits.There are ways to choose the 2 letters.For the letters to be next to each other, they can be the 1st and 2nd, 2nd and 3rd, 3rd and 4th, 4th and 5th, or the 5th and 6th characters. So, there are choices for the position of the letters.Therefore, the number of distinct license plates is .19、How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression?SolutionThe sum of the angles of a triangle is degrees. For an arithmetic progression with an odd number of terms, the middle term is equal to the average of the sum of all of the terms, making itdegrees. The minimum possibly value for the smallest angle is and the highest possible is (since the numbers are distinct), so there are possibilities .20 、Six distinct positive integers are randomly chosen between 1 and 2006, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of 5?SolutionFor two numbers to have a difference that is a multiple of 5, the numbers must be congruent (their remainders after division by 5 must be the same).are the possible values of numbers in . Since there are only 5 possible values in and we are picking numbers, by the Pigeonhole Principle, two of the numbers must be congruent .Therefore the probability that some pair of the 6 integers has a difference that is a multiple of 5 is .21 、How many four-digit positive integers have at least one digit that is a 2 or a 3?SolutionSince we are asked for the number of positive 4-digit integers with at least 2 or 3 in it, we can find this by finding the total number of 4-digit integers and subtracting off those which do not have any 2s or 3s as digits.The total number of 4-digit integers is , since we have 10 choices for each digit except the first (which can't be 0). Similarly, the total number of 4-digit integers without any 2 or 3 is .Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their decimal representation is22 、Two farmers agree that pigs are worth dollars and that goats are worth dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?SolutionThe problem can be restated as an equation of the form , where is the number of pigs, is the number of goats, and is the positive debt. The problem asks us to find the lowest x possible. p and g must be integers, which makes the equation a Diophantine equation. The Euclidean algorithm tells us that there are integer solutions to the Diophantine equation , where is the greatest common divisor of and , and no solutions for any smaller . Therefore, the answer is the greatestcommon divisor of 300 and 210, which is 30,Alternatively, note that is divisible by 30 no matter what and are, so our answer must be divisible by 30. In addition, three goats minus two pigs give us exactly. Since our theoretical best can be achived, it must really be the best, and the answer is . debt that can be resolved.23 、Circles with centers and have radii 3 and 8, respectively. A common internal tangent intersects the circles at and , respectively. Lines and intersect at , and . What is ?Solutionand are vertical angles so they are congruent, as are angles and (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, .By the Pythagorean Theorem, line segment . The sides are proportional, so . This makes and.24 、Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?SolutionWe can break the octahedron into two square pyramids by cutting it along a plane perpendicular to one of its internal diagonals. The cube has edges of length one so the edge of the octahedron is of length . Then the square base of the pyramid has area. We also know that the height of the pyramid is half the height of the cube, so it is . The volume of a pyramid with base area and height is so each of the pyramids has volume . The whole octahedron is twice this volume, so.25 、A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?SolutionLet us count the good paths. The bug starts at an arbitrary vertex, moves to a neighboring vertex (3 ways), and then to a new neighbor (2 more ways). So, without loss of generality, let the cube have vertex such that and are two opposite faces with above and above . The bug starts at and moves first to , then to .From this point, there are two cases.Case 1: the bug moves to . From , there is only one good move available, to . From , there are two ways to finish the trip, either by going or . So there are 2 good paths in this case.Case 2: the bug moves to . Case 2a: the bug moves . In this case, there are 0 good paths because it will not be possible to visit both and without double-visiting some vertex. Case 2b: the bug moves . There is a unique good path in this case, .Thus, all told we have 3 good paths after the first two moves, for a total of good paths. There were possible paths the bug could have taken, so the probability a random path is good is the ratio of good paths to total paths, , which is answer choice .。
2016第17届AMC10A试题及详解(繁体中文)
這三種不同顏色的區域面積成等差數列。
最內層長方形的寬為 1 公尺， 其他兩種色彩較暗區域距離較淺顏色四邊的寬也都是 1 公尺。 試問最內層長方形的長是多少公尺？ (A) 1 (B) 2 (C) 4 (D) 6 (E) 8 【2016AMC10A】
11
1
1
1
答：(B)
解：白色區1 x 、淺灰區 3 x 2 1 x 2 x 6 、深灰區 5 x 4 3 x 2 2 x 14 三種不同顏色的區域面積成等差數列，故 x 2 x 14 2 2 x 6 x 2
【2016AMC10A】
答：(A)
解：令 N 5 t ，將全部位置編號為 1～ 5 t 1
當紅球位於 1～ 2 t 1 號，或 3 t 1 ～ 5 t 1 號時，至少有 3 綠球在紅球同一側
5
故P
N
2 t 12
5t 1
321 400
5t
依此繼續排放，直到第 N 列排放 N 個銅板恰好排完。試問 N 各個位數的數字和為多少？
(A) 6 (B) 7 (C) 8 (D) 9 (E) 10
【2016AMC10A】
答： (D)
解：
N
N 2
1
2016
N
63 ，則
N
各個位數的數字和為 6
3
9
10. 有一塊由三種不同顏色製成的地毯，如圖所示。
100
【2016AMC10A】
2. 滿足10 x 100 2 x 1000 5 的 x 之值為？ (A) 1 (B) 2 (C) 3 (D) 4 (E) 5
2017年第10届全国中学数理化竞赛7年级数学答案(初赛)
第十届全国中学生数理化学科能力展示活动七年级数学答案一、选择题（每题6分，共48分）1. C.解析 由图形可得各月的平均最低气温都在0 ℃以上，A 正确；7月的平均温差约为10 ℃，而1月的平均温差约为5℃，故B 正确；3月和11月的平均最高气温都在10 ℃左右，基本相同，D 正确；故C 错误．2. D.解析 四人所知只有自己看到、老师所说及最后小明说的话．小明不知道自己的考场，则小红、小刚中必有一个在校内、一个在校外（若两个都在校内，小明会知道自己的考场；两个都在校外亦然）.小红看了小刚的考场，知道自己的考场。
小丽看小明，小明、小丽中也为一个校内，一个校外，小丽能知道自己的考场．3. B.解析 因为每次都把油箱加满，第二次加了48升油，说明这段时间总耗油量为48升，而行驶的路程为35 600－35 000＝600(千米)，故每100千米平均耗油量为48÷6＝8(升)．4.B.解析 译：某个计算器，只有二个按键[+1]和[2⨯]，当你按下一个键后，计算器自动显示结果，例如，如果计算器最初显示9，当你按下[+1]，屏幕将显示为10；你再按下[2⨯]，显示为20.从最初显示1开始，如果你要得到200，那最少的按键次数为（ ）要得到最少次数，则20025222=÷÷÷，1222)111(25+⨯⨯⨯++=或1222)121(25+⨯⨯⨯+⨯=，均为6次，所以一共要用6+3=9次。
5.C ．6.D.解析 ①80＋60×2＝200<230，80＋80＝160<180，①符合；②64＋60×2＝184<230，100＋80＝180，②符合；③45×2＋60×2＝210<230，45×2＋80＝170<180，③符合；④餐桌长为60＋2×30＝120，120＋80＝200<230，60＋60×2＝180，④符合．所以四种规格的餐桌都符合要求．7.A.解析 由题意知，被截去部分纸环的个数为5n ＋3.8. A.解析 由前4行特点，归纳可得：若a nm ＝(a ，b )，则a ＝m ，b ＝n －m ＋1，所以a nm ＝(m ，n －m ＋1)．二、填空题（每题8分，共32分） 9..23解析 23100log 1000log 1000log 1010100==10. 110．11. 1.23⨯101612.)0,2(解析 由()()()0,5q p,☉2,1=得，所以⎩⎨⎧⎩⎨⎧-==⇒=+=-210252q p q p q p ()()()()()0,22-1⊕2,1q p,⊕2,1==，. 三解答题（共40分）13．（13分）解 设运输路程为S （km ），使用汽车、火车、飞机三种运输工具运输时各自的总费用分别为y 1(元)、y 2(元)、y 3(元). 则由题意,)2(,)(21S v m b y S v m a m v S aS y +=+=+=S vm c y 10(3+=,………3分 S vm b a y y ]2)[(21+-=-，由a >b ,各字母均为正值，所以y 1–y 2>0，即y 2<y 1. ………3分 由y 3–y 2=［(c –b )–v m 52］S .令y 3–y 2>0,由c >b 及每字母都是正值，得c >b +vm 52. ………3分 所以，当c >b +vm 52时y 2<y 3,由y 2<y 1即y 2最小，………2分 当b <a <c <b +v m 52时，y 3<y 2<y 1,y 3最小. ………2分14.（13分）解；，一个数是大于中至少有一个数是小于说明00,,a 0c b c b a =++;;;a ,0a c b b c a c b c b a -=+-=+-=+=++可知：由 ………3分 ;1c -a 00b 0a =-+--+-=cb b a xc 则＜，＜，＞若………3分 ;1c a 00b 0a -=-+--+-=cb b a xc 则＞，＜，＞若………3分 1±=x 综上所述：；时，当120172017119=+-=x x x ………2分 ………2分 ；时，当403320172017119=+--=x x x15.（14分）解 从A 城出发到达B 城的路线分成两类：1）经过O 城。
2017年全国高中数学联赛二试试题及答案解析.pdf
2010年全国高中数学联合竞赛加试 试题参考答案及评分标准（A 卷）说明：1. 评阅试卷时，请严格按照本评分标准的评分档次给分.2. 如果考生的解答方法和本解答不同，只要思路合理、步骤正确，在评卷时可参考本评分标准适当划分档次评分，10分为一个档次，不要增加其他中间档次。
一、（本题满分40分）如图，锐角三角形ABC 的外心为O ，K 是边BC 上一点（不是边BC 的中点），D 是线段AK 延长线上一点，直线BD 与AC 交于点N ，直线CD 与AB 交于点M ．求证：若OK ⊥MN ，则A ，B ，D ，C 四点共圆．证明：用反证法．若A ，B ，D ，C 不四点共圆，设三角形ABC 的外接圆与AD 交于点E ，连接BE 并延长交直线AN 于点Q ，连接CE 并延长交直线AM 于点P ，连接PQ ．因为2PK =P 的幂（关于⊙O ）+K 的幂（关于⊙O ） ()()2222PO rKOr =−+−，同理 ()()22222QK QO rKOr =−+−，所以 2222PO PK QO QK −=−，故 OK ⊥PQ ． （10分）由题设，OK ⊥MN ，所以PQ ∥MN ，于是AQ APQN PM=． ① 由梅内劳斯（Menelaus ）定理，得1NB DE AQBD EA QN⋅⋅=， ② 1MC DE APCD EA PM⋅⋅=． ③ 由①，②，③可得NB MCBD CD=， （30分） 所以ND MDBD DC=，故△DMN ∽ △DCB ，于是DMN DCB ∠=∠，所以BC ∥MN ，故OK ⊥BC ，即K 为BC 的中点，矛盾！从而,,,A B D C 四点共圆. （40分）注1：“2PK =P 的幂（关于⊙O ）+K 的幂（关于⊙O ）”的证明：延长PK 至点F ，使得PK KF AK KE ⋅=⋅， ④则P ，E ，F ，A 四点共圆，故PFE PAE BCE ∠=∠=∠，从而E ，C ，F ，K 四点共圆，于是PK PF PE PC ⋅=⋅， ⑤⑤-④，得 2PK PE PC AK KE =⋅−⋅=P 的幂（关于⊙O ）+K 的幂（关于⊙O ）． 注2：若点E 在线段AD 的延长线上，完全类似．二、（本题满分40分）设k 是给定的正整数，12r k =+．记(1)()()f r f r r r ==⎡⎤⎢⎥，()()l f r = (1)(()),2l f f r l −≥．证明：存在正整数m ，使得()()m f r 为一个整数．这里，x ⎡⎤⎢⎥表示不小于实数x 的最小整数，例如：112⎡⎤=⎢⎥⎢⎥，11=⎡⎤⎢⎥．证明：记2()v n 表示正整数n 所含的2的幂次．则当2()1m v k =+时，()()m f r 为整数．下面我们对2()v k v =用数学归纳法．当0v =时，k 为奇数，1k +为偶数，此时()111()1222f r k k k k ⎛⎞⎡⎤⎛⎞=++=++⎜⎟⎜⎟⎢⎥⎝⎠⎢⎥⎝⎠为整数． （10分)假设命题对1(1)v v −≥成立．对于1v ≥，设k 的二进制表示具有形式1212222v v v v v k αα++++=+⋅+⋅+"，FE Q PO NM KDC B A这里，0i α=或者1，1,2,i v v =++"． （20分)于是 ()111()1222f r k k k k ⎛⎞⎡⎤⎛⎞=++=++⎜⎟⎜⎟⎢⎥⎝⎠⎢⎥⎝⎠2122kk k =+++ 11211212(1)2()222v v v vv v v ααα−++++=+++⋅++⋅+++""12k ′=+， ①这里1121122(1)2()22v v v v v v v k ααα−++++′=++⋅++⋅+++""．显然k ′中所含的2的幂次为1v −．故由归纳假设知，12r k ′′=+经过f 的v 次迭代得到整数，由①知，(1)()v f r +是一个整数，这就完成了归纳证明． （40分） 三、（本题满分50分）给定整数2n >，设正实数12,,,n a a a "满足1,1,2,,k a k n ≤="，记12,1,2,,kk a a a A k n k+++==""．求证：1112nnk k k k n a A ==−−<∑∑． 证明：由01k a <≤知，对11k n ≤≤−，有110,0kni ii i k a k an k ==+<≤<≤−∑∑． （10分）注意到当,0x y >时，有{}max ,x y x y −<，于是对11k n ≤≤−，有11111kn n k i i i i k A A a a n k n ==+⎛⎞−=−+⎜⎟⎝⎠∑∑11111n ki i i k i a a n k n =+=⎛⎞=−−⎜⎟⎝⎠∑∑ 11111max ,n k i i i k i a a n k n =+=⎧⎫⎛⎞<−⎨⎬⎜⎟⎝⎠⎩⎭∑∑111max (),n k k nk n ⎧⎫⎛⎞≤−−⎨⎬⎜⎟⎝⎠⎩⎭1k n=−， （30分） 故111nnnk kn k k k k a AnA A ===−=−∑∑∑()1111n n nk n k k k AA A A −−===−≤−∑∑111n k k n −=⎛⎞<−⎜⎟⎝⎠∑12n −=． （50分） 四、（本题满分50分）一种密码锁的密码设置是在正n 边形12n A A A "的每个顶点处赋值0和1两个数中的一个，同时在每个顶点处涂染红、蓝两种颜色之一，使得任意相邻的两个顶点的数字或颜色中至少有一个相同．问：该种密码锁共有多少种不同的密码设置？解：对于该种密码锁的一种密码设置，如果相邻两个顶点上所赋值的数字不同，在它们所在的边上标上a ，如果颜色不同，则标上b ，如果数字和颜色都相同，则标上c ．于是对于给定的点1A 上的设置（共有4种），按照边上的字母可以依次确定点23,,,n A A A "上的设置．为了使得最终回到1A 时的设置与初始时相同，标有a 和b 的边都是偶数条．所以这种密码锁的所有不同的密码设置方法数等于在边上标记a ，b ，c ，使得标有a 和b 的边都是偶数条的方法数的4倍． （20分）设标有a 的边有2i 条，02n i ⎡⎤≤≤⎢⎥⎣⎦，标有b 的边有2j 条，202n i j −⎡⎤≤≤⎢⎥⎣⎦．选取2i 条边标记a 的有2in C 种方法，在余下的边中取出2j 条边标记b 的有22jn i C −种方法，其余的边标记c ．由乘法原理，此时共有2in C 22jn i C −种标记方法．对i ，j 求和，密码锁的所有不同的密码设置方法数为222222004n n i i j n n i i j C C −⎡⎤⎡⎤⎢⎥⎢⎥⎣⎦⎣⎦−==⎛⎞⎜⎟⎜⎟⎜⎟⎝⎠∑∑． ①这里我们约定001C =． （30分）当n 为奇数时，20n i −>，此时22221202n i j n i n i j C −⎡⎤⎢⎥⎣⎦−−−==∑． ② 代入①式中，得()()2222222221222000044222n n i n n i j i n i i n i n n i n n i j i i C C C C −⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎢⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦−−−−====⎛⎞⎜⎟==⎜⎟⎜⎟⎝⎠∑∑∑∑ 0022(1)(21)(21)nnkn kk n kk n n nn k k C C −−===+−=++−∑∑ 31n =+． （40分）当n 为偶数时，若2n i <，则②式仍然成立；若2ni =，则正n 边形的所有边都标记a ，此时只有一种标记方法．于是，当n 为偶数时，所有不同的密码设置的方法数为222222004n n i i j n n i i j C C −⎡⎤⎡⎤⎢⎥⎢⎣⎦⎣⎦−==⎛⎞⎜⎟=⎜⎟⎜⎟⎝⎠∑∑()122210412n i n i n i C ⎡⎤−⎢⎣⎦−−=⎛⎞⎜⎟×+⎜⎟⎜⎟⎝⎠∑ ()222124233n i n i n n i C ⎡⎤⎢⎣⎦−−==+=+∑．综上所述，这种密码锁的所有不同的密码设置方法数是：当n 为奇数时有31n+种；当n 为偶数时有33n+种． （50分）。
AMC10的真题答案及中文翻译
AMC10的真题及中文翻译1、One ticket to a show costs $20 at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?(A) 2 (B) 5 (C) 10 (D) 15 (E) 20中文：一张展览票全价为20美元。
Susan用优惠券买4张票打七五折。
Pam用优惠券买5张票打七折。
Pam比Susan多花了多少美元？2、An aquarium has a rectangular base that measures 100cm by 40cm and has a height of 50cm. It is filled with water to a height of 40cm. A brick with a rectangular base that measures 40cm by 20cm and a height of 10cm is placed in the aquarium. By how many centimeters does that water rise?(A) 0.5 (B) 1 (C) 1.5 (D) 2 (E)2.5中文：一个养鱼缸有100cm×40cm的底，高为50cm。
它装满水到40cm的高度。
把一个底为40cm×20cm，高为10cm的砖块放在这个养鱼缸里。
鱼缸里的水上升了多少厘米？3、The larger of two consecutive odd integers is three times the smaller. What is their sum?(A) 4 (B) 8 (C) 12 (D) 16 (E) 20中文：2个连续的奇整数中较大的数是较小的数的3倍。
美国数学邀请赛2017AMC10试题及中文翻译.doc
8th AMC 10 A 20071. One ticket to a show costs $20 at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?(A) 2 (B) 5 (C) 10 (D) 15 (E) 20中文：一张展览票全价为20美元。
Susan 用优惠券买4张票打七五折。
Pam 用优惠券买5张票打七折。
Pam 比Susan 多花了多少美元？2. Define a@b=ab - b 2 and a#b=a + b - ab 2. What is 6@26#2? (A)－12 (B)－14 (C) 18 (D) 14 (E) 12中文：定义a @ b=ab-b 2 ，a # b=a + b - ab 2。
求6@26#2的值。
3. An aquarium has a rectangular base that measures 100cm by 40cm and has a height of 50cm. It is filled with water to a height of 40cm. A brick with a rectangular base that measures 40cm by 20cm and a height of 10cm is placed in the aquarium. By how many centimeters does that water rise?(A) 0.5 (B) 1 (C) 1.5 (D) 2 (E)2.5中文：一个养鱼缸有100cm ×40cm 的底，高为50cm 。