2010 AMC 10B 13

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2019年 AMC 10B真题含答案

2019AMC10BProblem1Alicia had two containers.The first was full of water and the second was empty.She poured all the water from the first container into the second container,at which point the second containerwas full of water.What is the ratio of the volume of the first container to the volume of the second container?Alicia有两个容器，第一个里面的水是满的，第二个是空的；她将第一个容器中的所有水倒入第二个容器中，此时第二个容器中的水是满的，较小容器的容积与较大容器的容积之比是多少？Problem2Consider the statement,"If is not prime,then is prime."Which of the following valuesof is a counterexample to this statement?考虑论断：“如果n不是质数，那么n−2就是质数。

”以下的哪个n值是此论断的反例？Problem3In a high school with students,of the seniors play a musical instrument,while of thenon-seniors do not play a musical instrument.In all,of the students do not play a musicalinstrument.How many non-seniors play a musical instrument?在一所有500名学生的高中，40%的高三学生会演奏乐器，而30%的非高三学生不会演奏乐器。

2010AMC10B试题及答案解析

2010 AMC 10B1 . What is ?SolutionWe first expand the first term, simplify, and then compute to get an answer of .2 、Makarla attended two meetings during her -hour work day. The firstmeeting took minutes and the second meeting took twice as long.What percent of her work day was spent attending meetings?SolutionThe percentage of her time spent in meetings is the total amount of time spent in meetings divided by the length of her workday.The total time spent in meetings isTherefore, the percentage is3、A drawer contains red, green, blue, and white socks with at least 2 of each color. What is the minimum number of socks that must be pulled from the drawer to guarantee a matching pair?SolutionAfter you draw socks, you can have one of each color, so (according tothe pigeonhole principle), if you pull then you will be guaranteed a matching pair.4 、For a real number , define to be the average of and . Whatis ?SolutionThe average of two numbers, and , is defined as . Thus theaverage of and would be . With that said, we need to find thesum when we plug, , and into that equation. So:.5 、A month with days has the same number of Mondays andWednesdays.How many of the seven days of the week could be the first day of this month?Solution（B）. 36 、A circle is centered at , is a diameter and is a point on thecircle with . What is the degree measure of ?SolutionAssuming the reader is not readily capable to understand how willalways be right, the I will continue with an easily understandable solution.Since is the center, are all radii, they are congruent.Thus, and are isosceles triangles. Also, note thatand are supplementary, then . Sinceis isosceles, then . They also sum to , so eachangle is .7 、A triangle has side lengths , , and . A rectangle has width andarea equal to the area of the rectangle. What is the perimeter of this rectangle?SolutionThe triangle is isosceles. The height of the triangle is therefore given byNow, the area of the triangle isWe have that the area of the rectangle is the same as the area of the triangle, namely 48. We also have the width of the rectangle: 4.The length of the rectangle therefor is:The perimeter of the rectangle then becomes:The answer is:8 、A ticket to a school play cost dollars, where is a whole number. A group of 9th graders buys tickets costing a total of $, and a group of 10th graders buys tickets costing a total of $. How many values for are possible?SolutionWe see how many common integer factors 48 and 64 share. Of the factors of 48 - 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48; only 1, 2, 4, 8, and 16are factors of 64. So there are possibilities for the ticket price.9 、Lucky Larry's teacher asked him to substitute numbers for , , ,, and in the expression and evaluate the result.Larry ignored the parenthese but added and subtracted correctly and obtained the correct result by coincidence. The number Larry sustitued for , , , and were , , , and , respectively. What number didLarry substitude for ?SolutionSimplify the expression . I recommend to start with the innermost parenthesis and work your way out.So you get:Henry substituted with respectively.We have to find the value of , such that(the same expression withoutparenthesis).Substituting and simplifying we get:So Henry must have used the value for .Our answer is:10、Shelby drives her scooter at a speed of miles per hour if it is notraining, and miles per hour if it is raining. Today she drove in the sunin the morning and in the rain in the evening, for a total of miles inminutes. How many minutes did she drive in the rain?SolutionWe know thatSince we know that she drove both when it was raining and when it was not and that her total distance traveled is miles.We also know that she drove a total of minutes which is of an hour.We get the following system of equations, where is the time traveled when it was not raining and is the time traveled when it was raining:Solving the above equations by multiplying the second equation by 30 and subtracting the second equation from the first we get:We know now that the time traveled in rain was of an hour, which is minutesSo, our answer is:11 、A shopper plans to purchase an item that has a listed price greater than $and can use any one of the three coupns. Coupon A gives off the listed price, Coupon B gives $off the listed price, andCoupon C gives off the amount by which the listed price exceeds $.Let and be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is −SolutionLet the listed price be , whereCoupon A saves us:Coupon B saves us:Coupon C saves us:Now, the condition is that A has to be greater than or equal to either B or C which give us the following inequalities:We see here that the greatest possible value for p is and the smallest isThe difference between and isOur answer is:12 、At the beginning of the school year, of all students in Mr. Wells'math class answered "Yes" to the question "Do you love math", and answered "No." At the end of the school year, answered "Yes"and answerws "No." Altogether, of the students gave a differentanswer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values ofSolutionThe minimum possible value occurs when of the students whooriginally answered "No." answer "Yes." In this case,The maximum possible value occurs when of the students whooriginally answered "Yes." answer "No." and the of the students whooriginally answered "No." answer "Yes." In this case,Subtract to obtain an answer of13 、What is the sum of all the solutions of ?SolutionCase 1:Case 1a:Case 1b:Case 2:Case 2a:Case 2b:Since an absolute value cannot be negative, we exclude . Theanswer is14 、The average of the numbers and is . What isSolutionWe must find the average of the numbers from to and in terms of. The sum of all these terms is . We must divide this by the total number of terms, which is . We get: . Thisis equal to , as stated in the problem. We have: .We can now cross multiply. This gives:This gives us our answer.15 、On a -question multiple choice math contest, students receivepoints for a correct answer, points for an answer left blank, andpoint for an incorrect answer. Jesse’s total score on the contest was .What is the maximum number of questions that Jesse could have answered correctly?SolutionLet be the amount of questions Jesse answered correctly, be theamount of questions Jesse left blank, and be the amount of questions Jesse answered incorrectly. Since there were questions on the contest,. Since his total score was , . Also,. We can substitute this inequality into the previous equation to obtain another inequality:. Since is an integer, themaximum value for is .16 、A square of side length and a circle of radius share the samecenter. What is the area inside the circle, but outside the square?Solution(B)17 、Every high school in the city of Euclid sent a team of students to amath contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed th and th, respectively. How many schools are in the city?SolutionLet the be the number of schools, be the number of contestants, andbe Andrea's score. Since the number of participants divided by three isthe number of schools, . Andrea received a higher scorethan her teammates, so . Since is the maximum possiblemedian, then is the maximum possible number ofparticipants. Therefore, . This yields the compound inequality: . Since a set with an even number of elements has a median that is the average of the two middle terms, anoccurrence that cannot happen in this situation, cannot be even.is the only other option.18 、Positive integers , , and are randomly and independentlyselected with replacement from the set . What is the probability that is divisible by ?Solution（E）13/2719 、A circle with center has area . Triangle is equilateral,is a chord on the circle, , and point is outside .What is the side length of ?Solution(B)620 、Two circles lie outside regular hexagon . The first istangent to , and the second is tangent to . Both are tangent tolines and . What is the ratio of the area of the second circle to thatof the first circle?SolutionA good diagram here is very helpful.The first circle is in red, the second in blue. With this diagram, we can see that the first circle is inscribed in equilateral triangle while thesecond circle is inscribed in . From this, it's evident that the ratio ofthe red area to the blue area is equal to the ratio of the areas of triangles toSince the ratio of areas is equal to the square of the ratio of lengths, we know our final answer is From the diagram, we can see that thisisThe letter answer is D21 、A palindrome between and is chosen at random. What isthe probability that it is divisible by ?SolutionView the palindrome as some number with form (decimalrepresentation): . But because the number is a palindrome, . Recombining this yields .1001 is divisible by 7, which means that as long as , thepalindrome will be divisible by 7. This yields 9 palindromes out of 90() possibilities for palindromes. However, if , then this givesanother case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to22 、Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?SolutionWe can count the total number of ways to distribute the candies (ignoring the restrictions), and then subtract the overcount to get the answer.Each candy has three choices; it can go in any of the three bags.Since there are seven candies, that makes the total distributionsTo find the overcount, we calculate the number of invalid distributions: the red or blue bag is empty.The number of distributions such that the red bag is empty is equal to ,since it's equivalent to distributing the 7 candies into 2 bags.We know that the number of distributions with the blue bag is empty will be the same number because of the symmetry, so it's also .The case where both the red and the blue bags are empty (all 7 candies are in the white bag) are included in both of the above calculations, and this case has only distribution.The total overcount isThe final answer will beThat makes the letter choice C23 、The entries in a array include all the digits from through ,arranged so that the entries in every row and column are in increasing order. How many such arrays are there?Solution（D）6024 、A high school basketball game between the Raiders and Wildcatswas tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than points. What was the total number of points scoredby the two teams in the first half?SolutionRepresent the teams' scores as: andWe have Manipulating this, we can get, orSince both are increasing sequences, . We can check cases up tobecause when , we get . When▪▪▪Checking each of these cases individually back into the equation, we see that only when a=5 and n=2, we get an integer value for m, which is 9. The original question asks for the first half scores summed, so we must find25 、Let , and let be a polynomial with integer coefficients suchthat, and.What is the smallest possible value of ?SolutionThere must be some polynomial such thatThen, plugging in values of we getThus, the least value ofmust be the . Solving, we receive , so our answer is.。

2000-2012美国AMC10中文版试题及答案

2000到2012年AMC10美国数学竞赛0 0P 0 A 0 B 0 C 0D 0 全美中学数学分级能力测验(AMC 10)2000年第01届美国AMC10 (2000年2月日时间75分钟)1. 国际数学奥林匹亚将于2001年在美国举办，假设I 、M 、O 分别表示不同的正整数，且满足I ⨯M ⨯O =2001，则试问I +M +O 之最大值为。

(A) 23 (B) 55 (C) 99 (D) 111 (E) 6712. 2000(20002000)为。

(A) 40 (B) 50 (C) 55 (D) 60 (E) 754. Candra 每月要付给网络公司固定的月租费及上网的拨接费，已知她12月的账单为12.48元，而她1月的账单为17.54元，若她1月的上网时间是12月的两倍，试问月租费是元。

(A) 2.53 (B) 5.06 (C) 6.24 (D) 7.42 (E) 8.775. 如图M ，N 分别为PA 与PB 之中点，试问当P 在一条平行AB 的直在线移动时，下列各数值有项会变动。

(a) MN 长 (b) △P AB 之周长 (c) △P AB 之面积 (d) ABNM 之面积(A) 0项 (B) 1项 (C) 2项 (D) 3项 (E) 4项 6. 费氏数列是以两个1开始，接下来各项均为前两项之和，试问在费氏数列各项的个位数字中，最后出现的阿拉伯数字为。

(A) 0 (B) 4 (C) 6 (D) 7 (E) 97. 如图，矩形ABCD 中，AD =1，P 在AB 上，且DP 与DB 三等分∠ADC ，试问△BDP 之周长为。

2007AMC10B答案和部分问题参考解答

AMC10B 2007 勘误一. 上次练习中第21题结论: “of the square ”没印出来.21. Right △ABC has AB =3,BC =4,and AC =5.Square XYZW is inscribed in △ABC with X and Y on AC ,W on AB , and Z on BC ( )(A) 23 (B) 3760 (C) 712 (D) 1323(E) 2二. 第24题印漏了.24. A player chooses one of the numbers 1 through 4. After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered 1 through 4. If the number chosen appears on the bottom of exactly one die after it is rolled, then the player wins $1. If the number chosen appears on the bottom of both of the dice, then the player wins $2. If the number chosen does not appear on the bottom of either of the dice, the player loses $1. What is the expected return to the player, in dollars, for one roll of the dice? ( )(A) -81 (B) -161(C) 0(D) 161 (E) 81三. 第25题关键词: “gcd ”错, 应该为: “ged ”.以上问题请同学们重新练习,再参考答案和解答!AMC10B 2007 答案1(E) 2(E) 3(B) 4(D) 5(D) 6(D) 7(E) 8(D) 9(D) 10(A) 11(C) 12(D) 13(D) 14(C) 15(D) 16(C) 17(D) 18(B) 19(C) 20(C) 21(B) 22(B) 23(C) 24(E) 25(A)AMC10B 2007部分题参考解答18T.[译题]: 一个半径为1的圆被4个半径均为r 的圆环绕, ) (A)2 (B)1＋2 (C)6 (D) 3 (E) 2＋2 (分析): 设半径均为r 四个圆的圆心分别为A 、B 、C 、D .半径为1的圆的圆心为O.则正方形ABCD 中: O 为中心, AC =2r+2, AD =2r∴2r+2=2(2r) ∴ r=2+1 选(B )19T.[译题]: 一个如图所示的轮子旋转两次, 并且指针随机地指向的数字被记录下来. 被4除的余数和第二个数字被5除的余数分别作为列数和行数, 对应标示出棋盘的一格.求: 这对数标示黑格的机率. ( )(A)31 (B)94 (C)21 (D)95 (E)32(分析): 1与9、2与6、3与7被4整除余数依次为的列数1、2 、3的机率相等, 均为31. 而每列白格和黑格的数目一样, 所以从列数上合格与白格被标示机率一样.1与6、2与7、3、9被4整除余数依次为1、2 、3、4, 相应机率依次为:31、31、61、61. 而第一行和第三行均为两个黑格一个白格, 第二行和第四行均为两个白格一个黑格, 黑格相应机率为32、31、32、31.综上, 最终黑格被标示机率为: (61+61)⨯32+(61+61)⨯31+61⨯32+61⨯31=21选(C )20T[译题]: 25个小正方块排成5⨯5的大正方形, 从中选出3个小方块, 且每两块均不同行不同列. 求不同顺序的选法数有多少种? ( ) (A) 100 (B) 125 (C) 600 (D) 2300 (E) 3600(分析) 给每个小方块按行列编号: 第i 行, 第j 列(i , j =1,2,3,4,5).则小方块的行号不同, 方块不同; 小方块的列号不同, 方块也不同.选出3个不同行的方法数如下:(1) 选第一块的行有5种; (2)上一块的行不能再选, 第二块的行有4种选法; (3) 上两块的行不能再选, 第三块的行有3种选法; ∴不同顺序的3个不同行的方法数为: 5⨯4⨯3=60种.而选出3个不同列的方法数与选3个不同行的方法数基本一样: 5⨯4⨯3=60种但3个小方块不同顺序仅由行或列之一确定, 在选行时已定顺序, 则列无需再有顺序. 不同列的方法数5⨯4⨯3=60中, 每六种实际为为同一种. 例如: 532, 623, 325, 352, 253, 235 实际为为同一种 .∴不同行的方法数为: 60/6=10种 . 综上, 不同顺序的选法数有60⨯10=100种. 选(C )21T[译题]: 直角△ABC 的边AB =3, BC =4, 且AC =5. 正方形XYAW 内接于△ABC , 且点X 、Y 在线段 AC 上, 点W 在线段BC 上. (分析) 设正方形XYZW 边长为a .易知RT ⊿BWZ ∽RT ⊿XA W ∽RT ⊿BAC ∴W Z BW =AC BA =53, W X W A =BC AC =45∴ BW =53a , AW =45a又BW +AW =AB =3 ∴ 3=53a +45a ∴a =376022T[译题]: 一个底面为正方形的锥体被一个与底面平行且距离为2的平面分成两个部分, 其中顶部被分割出的小锥的表面积为原来的大锥体的表面积的一半. 求原锥体的高为多少?( )(A) 2 (B) 2＋2 (C) 12 (分析) 锥PO 1与锥PO 相似.∴PO PO 1=SS 1=21 ∴PO O O 1=PO PO PO 1-=212- ∴ PO =1221-OO =1222-=22()12+=4+22 选(E )23T[译题]: 定义n 为满足如下条件的最小正整数: 能被4和9整除, 且十进制形式之下至少含一个数字4和9. 求n 的后四位上数字. ( )(A) 4444 (B) 4494 (C) 4944 (D) 9444 (E) 9944(分析) ∵9|n ∴n 的各位数字之和能被9整除, 又其至少含一个4, 则共至少有9个4. (含4的个数为9的倍数). ∵4|n ∴n 的后两位数字对应两位数能被4整除. ∴只能为44. 以上条件之下n 的最小值为4444444944 ∴4944为所求. 选(D ).24T[译题]: 一个玩家从1至4中选定一个数字, 选定后, 摇下两个四面上分别标有1至4个点的正四面体骰子(如下图)两个骰子. 摇完后, 若其中恰有一个骰子底面的数字为选定数,则玩家赢得1元, 若两个骰子的底面数字都为选定数字, 则玩家赢得2元. 若两个骰子的底面数字都不是选定数字, 则玩家输1元. 撒完一次骰子,玩家的赢钱期望值为多少?(A) -81 (B) -161(C) 0(D)161 (E) 81(分析) 玩家赢1元的概率P(ξ=1)=2⨯41⨯43=83玩家赢1元的概率P(ξ=2)=41⨯41=161, 玩家输1元的概率P(ξ=-1)=43⨯43=169. ∴玩家赢钱的期望为: E ξ=1⨯83+2⨯161+(-1)⨯ 169= -161选 (B )25T. [译题]: 正整数对(a ,b )满足如下条件: 它们最大公因数为1, 且b a +ab914为整数. 这样的数对有多少对?( )(A) 4 (B) 6 (C) 9 (D) 12 (E) 无穷多对(分析) 设b a +ab 914= k (k ∈N *), 则易整理得9a 2-9kab +14b 2=0 . 视其为关于a 的二次方程,有正整数解. ∴∆=9b 2(9k 2-56)为完全平方数. 设9k 2-56=m 2 (m ∈N *) , 变形有 (3k -m)(3k +m)=56 ∴(3k -m)(3k +m)=1⨯56=2⨯28=4⨯14=8⨯7 . 由6| (3k -m)+(3k +m)知:3k -m=2且3k +m=28; 或3k -m=4且3k +m=14 .∴(a ,b )=(1,3)或(2,3)、(7,3)、(14,3) 验证这四组解均合题意. 选 ( A )。

2009 AMC 10B 试题及答案解析

2009 AMC 10B1 、Each morning of her five-day workweek, Jane bought either a 50-cent muffin or a 75-cent bagel. Her total cost for the week was a whole number of dollars, How many bagels did she buy?SolutionThe only combination of five items with total cost a whole number ofdollars is 3 muffins and bagels. The answer is .2 、Which of the following is equal to ?SolutionMultiplying the numerator and the denumerator by the same value does not change the value of the fraction. We can multiply both by ,getting .Alternately, we can directly compute that the numerator is , thedenumerator is , and hence their ratio is .3 、Paula the painter had just enough paint for identically sizedrooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for rooms. How many cansof paint did she use for the rooms?SolutionLosing three cans of paint corresponds to being able to paint fivefewer rooms. So . The answer is .4 、A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths and meters. What fraction of the yard is occupied by theflower beds?SolutionEach triangle has leg length meters and areasquare meters. Thus the flower beds have a total area of 25 square meters. The entire yard has length 25 m and width 5 m, so its area is 125 square meters. The fraction of the yard occupied by theflower beds is . The answer is .5 、Twenty percent less than 60 is one-third more than what number?SolutionTwenty percent less than 60 is . One-third more than anumber n is . Therefore and the number is . The answeris .6 、Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?SolutionThe age of each person is a factor of . So the twins could beyears of age and, consequently Kiana could be 128, 32, 8 or 2 years old, respectively. Because Kiana is younger than her brothers, she must be 2 years old. So the sum of their agesis . The answer is .7 、By inserting parentheses, it is possible to give the expressionseveral values. How many different values can beobtained?SolutionThe three operations can be performed on any of orders.However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values ofthe four expressions are in fact all distinct. Sothe answer is , which is choice .8 、In a certain year the price of gasoline rose by during January,fell by during February, rose by during March, and fell byduring April. The price of gasoline at the end of April was the sameas it had been at the beginning of January. To the nearest integer, what isSolutionLet be the price at the beginning of January. The price at the end ofMarch was Because the price at the of April was , the price decreased by during April, and the percent decreasewas So to the nearest integer is . Theanswer is .9 、Segment and intersect at , as shown,, and . What is the degree measureof ?Solutionis isosceles, hence .The sum of internal angles of can now be expressed as, hence , and each of the other twoangles is .Now we know that .Finally, is isosceles, hence each of the two remaining angles(and ) is equal to .10 、A flagpole is originally meters tall. A hurricane snaps theflagpole at a point meters above the ground so that the upper part, still attached to the stump, touches the ground meter away from thebase. What is ?SolutionThe broken flagpole forms a right triangle with legs and , andhypotenuse . The Pythagorean theorem now states that, hence , and .(Note that the resulting triangle is the well-known righttriangle, scaled by .)11 、How many -digit palindromes (numbers that read the samebackward as forward) can be formed using the digits , , , , , ,?SolutionA seven-digit palindrome is a number of the form . Clearly,must be , as we have an odd number of fives. We are then left with. Each of the permutations of the set will give us one palindrome.12、Distinct points , , , and lie on a line, with. Points and lie on a second line, parallel to thefirst, with . A triangle with positive area has three of the sixpoints as its vertices. How many possible values are there for the area of the triangle?SolutionConsider the classical formula for triangle area: . Each of the triangles that we can make has exactly one side lying on one of the two parallel lines. If we pick this side to be the base, the height will always be the same - it will be the distance between the two lines.Hence each area is uniquely determined by the length of the base. And it can easily be seen, that the only possible base lengths are , ,and . Therefore there are only possible values for the area.(To be more precise in the last step, the possible base lengths are, , and .)13 、As shown below, convex pentagon has sides ,, , , and . The pentagon is originallypositioned in the plane with vertex at the origin and vertex on thepositive -axis. The pentagon is then rolled clockwise to the rightalong the -axis. Which side will touch the point on the-axis?SolutionThe perimeter of the polygon is . Hence as we rollthe polygon to the right, every units the side will be the bottomside.We have . Thus at some point in time we will get thesituation when and is the bottom side. Obviously, atthis moment .After that, the polygon rotates around until point hits the axis at.And finally, the polygon rotates around until point hits the axisat . At this point the side touches the point .14 、On Monday, Millie puts a quart of seeds, of which are millet,into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only of the millet in the feeder, but they eat all ofthe other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?SolutionOn Monday, day 1, the birds find quart of millet in the feeder. OnTuesday they find quarts of millet. On Wednesday, day 3,they find quarts of millet. The number of quarts of millet they find on day isThe birds always find quart of other seeds, so more than half theseeds are millet if , that is, when . Becauseand , this will first occur on daywhich is . The answer is .15 、When a bucket is two-thirds full of water, the bucket and water weigh kilograms. When the bucket is one-half full of water the total weight is kilograms. In terms of and , what is the total weight inkilograms when the bucket is full of water?SolutionSolution 1Let be the weight of the bucket and let be the weight of the waterin a full bucket. Then we are given that and .Hence , so . Thus .Finally . The answer is .Solution 2Imagine that we take three buckets of the first type, to get rid of the fraction. We will have three buckets and two buckets' worth of water. On the other hand, if we take two buckets of the second type, we will have two buckets and enoung water to fill one bucket.The difference between these is exactly one bucket full of water, hence the answer is .Solution 3We are looking for an expression of the form .We must have , as the desired result contains exactly onebucket. Also, we must have , as the desired result contains exactly one bucket of water.At this moment, it is easiest to check that only the options (A), (B), and (E) satisfy , and out of these only (E) satisfies the secondequation.Alternately, we can directly solve the system, getting and.16 、Points and lie on a circle centered at , each of andare tangent to the circle, and is equilateral. The circleintersects at . What is ?SolutionSolution 1As is equilateral, we have , hence. Then , and from symmetry we have. Finally this gives us .We know that , as lies on the circle. From we alsohave , Hence , therefore, and .Solution 2As in the previous solution, we find out that .Hence and are both equilateral.We then have , hence is the incenter of, and as is equilateral, is also its centroid. Hence, and as , we have ,therefore , and as before we conclude that .17、Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extendingfrom to , divides the entire region into two regions of equalarea. What is ?SolutionSolution 1For the shaded area is at most , which is too little. Hence, and therefore the point is indeed inside the shaded part,as shown in the picture.Then the area of the shaded part is one less than the area of thetriangle with vertices , , and . Its area is obviously, therefore the area of the shaded part is .The entire figure has area , hence we want the shaded part to havearea . Solving for , we get . The answer is .Solution 2The total area is 5, so the area of the shaded area is . If we add aunit square in the lower right corner, the area is . Therefore, or . Therefore .18 、Rectangle has and . Point is themidpoint of diagonal , and is on with . What is thearea of ?SolutionBy the Pythagorean theorem we have , hence .The triangles and have the same angle at and a rightangle, thus all their angles are equal, and therefore these two triangles are similar.The ratio of their sides is , hence the ratio of their areas is.And as the area of triangle is , the area of triangleis .19 、A particular -hour digital clock displays the hour and minute ofa day. Unfortunately, whenever it is supposed to display a , itmistakenly displays a . For example, when it is 1:16 PM the clockincorrectly shows 9:96 PM. What fraction of the day will the clock show the correct time?SolutionSolution 1The clock will display the incorrect time for the entire hours ofand . So the correct hour is displayed of the time. The minutes willnot display correctly whenever either the tens digit or the ones digit is a , so the minutes that will not display correctly areand and . This amounts to fifteen of the sixty possibleminutes for any given hour. Hence the fraction of the day that theclock shows the correct time is . The answeris .Solution 2The required fraction is the number of correct times divided by the total times. There are 60 minutes in an hour and 12 hours on a clock, so there are 720 total times.We count the correct times directly; let a correct time be , whereis a number from 1 to 12 and and are digits, where . Thereare 8 values of that will display the correct time: 2, 3, 4, 5, 6, 7, 8, and 9. There are five values of that will display the correct time: 0, 2, 3, 4, and 5. There are nine values of that will display the correct time: 0, 2, 3, 4, 5, 6, 7, 8, and 9. Therefore there arecorrect times.Therefore the required fraction is .20 、Triangle has a right angle at , , and . Thebisector of meets at . What is ?Solution、By the Pythagorean Theorem, . The Angle Bisector Theorem now yields that21 、What is the remainder when is divided by 8?SolutionSolution 1The sum of any four consecutive powers of 3 is divisible byand hence is divisible by 8. Thereforeis divisible by 8. So the required remainder is . Theanswer is .Solution 2We have . Hence for any we have, and then . Therefore our sum gives the same remainder modulo as. There are terms in the sum, hencethere are pairs , and thus the sum is.22 、A cubical cake with edge length inches is iced on the sides andthe top. It is cut vertically into three pieces as shown in this top view, where is the midpoint of a top edge. The piece whose top is trianglecontains cubic inches of cake and square inches of icing. What is?SolutionLet's label the points as in the picture above. Let be the area of. Then the volume of the corresponding piece is . This cake piece has icing on the top and on the vertical side thatcontains the edge . Hence the total area with icing is. Thus the answer to our problem is, and all we have to do now is to determine . Solution 1Introduce a coordinate system where , and.In this coordinate system we have , and the line has theequation .As the line is orthogonal to , it must have the equationfor some suitable constant . As this line contains thepoint , we have .Substituting into , we get , and then .We can note that in is the height from onto , hence itsarea is , and therefore the answer is.Solution 2Extend to intersect at :It is now obvious that is the midpoint of . (Imagine rotating thesquare by clockwise around its center. This rotation willmap the segment to a segment that is orthogonal to ,contains and contains the midpoint of .)From we can compute that .Observe that and have the same angles and thereforethey are similar. The ratio of their sides is .Hence we have , and .Knowing this, we can compute the area of as.Finally, we compute , andconclude that the answer is .You could also notice that the two triangles in the original figure are similar.Solution 3Use trigonometry.The length of and is and respectively. So ,and .From the right-angled triangle , the hypotenuse, So, andKnowing this, . So we proceed as follows:So the answer is .Note that we didn't use a calculator, but we used trigonometric identities23 、Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the same line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?SolutionAfter 10 minutes (600 seconds), Rachel will have completed 6 laps and be 30 seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in 22.5 seconds, she will be in the picture between 18.75 seconds and 41.25 seconds of the tenth minute. After 10 minutes Robert will have completed 7 laps and will be 40 seconds past the starting line. Because Robert runs one-fourth of a lap in 20 seconds, he will be in the picture between 30 and 50 seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between 30 and 41.25 seconds of the tenth minute. So theprobability that both runners are in the picture is .The answer is .24 、The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with trapezoids, let bethe angle measure in degrees of the larger interior angle of the trapezoid. What is ?SolutionExtend all the legs of the trapezoids. They will all intersect in the middle of the bottom side of the picture, forming the situation shown below.Each of the angles at has . From , the size of thesmaller internal angle of the trapezoid is , hence thesize of the larger one is .Proof that all the extended trapezoid legs intersect in the same point: It is sufficient to prove this for any pair of neighboring trapezoids. For two neighboring trapezoids, the situation is symmetric according to their common leg, therefore the extensions of both outside legs intersect the extension of the common leg in the same point, q.e.d. Knowing this, we can now easily see that the intersection point must be on the bottom side of our picture, as it lies on the bottom leg of the rightmost trapezoid. And by symmetry the point must be in the center of this side.25 、Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?SolutionSolution 1There are two possible stripe orientations for each of the six faces ofthe cube, so there are possible stripe combinations. There are three pairs of parallel faces so, if there is an encircling stripe, then the pair of faces that do not contribute uniquely determine the stripeorientation for the remaining faces. In addition, the stripe on each face that does not contribute may be oriented in either of two different ways. So a total of stripe combinations on the cube resultin a continuous stripe around the cube. The required probability is.Here's another way similar to this:So there are choices for the stripes as mentioned above. Now, let'sjust consider the "view point" of one of the faces. We can choose any of the 2 orientation for the stripe (it can go from up to down, or from right to left). Once that orientation is chosen, each of the other faces that contribute to that loop only have 1 choice, which is to go in the direction of the loop. That gives us a total count of 2 possibilities for any one of the faces. Since there are six faces, and this argument is valid for all of them, we conclude that there are 2(6) = 12 total ways to have the stripe. Therefore, the probability is 12/64 = 3/16. Solution 2Without loss of generality, orient the cube so that the stripe on the top face goes from front to back. There are two mutually exclusive ways for there to be an encircling stripe: either the front, bottom and back faces are painted to complete an encircling stripe with the top face's stripe or the front, right, back and left faces are painted to form anencircling stripe. The probability of the first case is , and theprobability of the second case is . The cases are disjoint, sothe probabilities sum .Solution 3There are three possible orientations of an encircling stripe. For any one of these to appear, the stripes on the four faces through which the continuous stripe is to pass must be properly aligned. The probabilityof each such stripe alignment is . Since there are three such- 21 -possibilities and they are disjoint, the total probability is .The answer is. Solution 4 Consider a vertex on the cube and the three faces that are adjacent to that vertex. If no two stripes on those three faces are aligned, then there is no stripe encircling the cube. The probability that the stripes aren't aligned is , since for each alignment of one stripe, there is one and only one way to align the other two stripes out of four total possibilities. therefore there is a probability of that two stripes are aligned.Now consider the opposing vertex and the three sides adjacent to it. Given the two connected stripes next to our first vertex, we have two more that must be connected to make a continuous stripe. There is a probability ofthat they are aligned, so there is a probability ofthat there is a continuous stripe.。

AMC 美国数学竞赛 2003 AMC 10B 试题及答案解析

2003 AMC 10B1、Which of the following is the same asSolution2、Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs more than a pink pill, and Al’s pills cost a total of for the two weeks. How much does one green pill cost?Solution3、The sum of 5 consecutive even integers is less than the sum of the ﬁrst consecutive odd counting numbers. What is the smallest of the even integers?Solution4、Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the ﬁgure. She plants one flower per square foot in each region. Asters cost 1 each, begonias 1.50 each, cannas 2 each, dahlias 2.50 each, and Easter lilies 3 each. What is the least possible cost, in dollars, for her garden? Solution5、Moe uses a mower to cut his rectangular -foot by -foot lawn. The swath he cuts is inches wide, but he overlaps each cut by inches to make sure that no grass is missed. He walks at the rate offeet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn?Solution.6、Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is . The horizontal length of a “-inch” television screen is closest, in inches, to which of the following?Solution7、The symbolism denotes the largest integer not exceeding . For example. , and . ComputeSolution.8、The second and fourth terms of a geometric sequence are and . Which of the following is a possible first term?Solution9、Find the value of that satisfies the equationSolution10、Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times is the number of possible license plates increased? Solution11、A line with slope intersects a line with slope at the point . What is the distance between the -intercepts of these two lines?Solution12、Al, Betty, and Clare split among them to be invested in different ways. Each begins with a different amount. At the end of one year they have a total of . Betty and Clare have both doubled their money, whereas Al has managed to lose . What was Al’s original portion?Solution.13、Let denote the sum of the digits of the positive integer . For example, and . For how many two-digit values of is ?Solution14、Given that , where both and are positive integers, find the smallest possible value for .Solution15、There are players in a singles tennis tournament. The tournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest players are given a bye, and the remaining players are paired off to play. After each round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played isSolution16、A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that the restaurant should offer so that a customer could have a different dinner each night in the year ?Solution.17、An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly ﬁll the cone. Assume that the melted ice cream occupies of the volume of the frozen ice cream. What is the ratio of the cone’s height to its radius?Solution18、What is the largest integer that is a divisor offor all positive even integers ?Solution19、Three semicircles of radius are constructed on diameter of a semicircle of radius . The centers of the small semicircles divide into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?Solution20、In rectangle , and . Points and are onso that and . Lines and intersect at . Find the area of .Solution21、A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?Solution22、A clock chimes once at minutes past each hour and chimes on the hour according to the hour. For example, at 1 PM there is one chime and at noon and midnight there are twelve chimes. Starting at 11:15 AM on February , , on what date will the chime occur?Solution23、A regular octagon has an area of one square unit. What is the area of the rectangle ?Solution24、The ﬁrst four terms in an arithmetic sequence are , , , and , in that order. What is the ﬁfth term?Solution25、How many distinct four-digit numbers are divisible by and have as their last two digits?Solution。

2011 AMC 10B Problems 详解16-25

A dart board is a regular octagon divided into regions as shown. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is probability that the dart lands within the center square?If the side lengths of the dart board and the side lengths of the center square are all then the side length of the legs of the triangles are .Use Geometric probability by putting the area of the desired region over the area of the entire region.In the given circle, the diameter is parallel to , and is parallel to . The angles and are in the ratio . What is the degree measure of angle ?We can let be and be because they are in the ratio . When an inscribed angle contains the diameter, the inscribed angle is a right angle. Therefore by triangle sumtheorem, and .because they are alternate interior angles and . Opposite angles in a cyclic quadrilateral are supplementary, so. Use substitution to getRectangle has and . Point is chosen on side so that . What is the degree measure of ?It is given that . Since and are alternate interior angles and , . Use the Base Angle Theorem to show . We know that is a rectangle, so it follows that . We notice that is atriangle, and. If we let be the measure of thenWhat is the product of all the roots of the equationFirst, square both sides, and isolate the absolute value.Solve for the absolute value and factor.However, this is not the final answer. Plug it back into the original equation to ensure it still works. Whether the number is positive or negative does not matter since the absolute value or square will cancel it out anyways.The roots of this equation are and and product isRhombus has side length and °. Region consists of all points inside the rhombus that are closer to vertex than any of the other three vertices. What is the area of ?Suppose that is a point in the rhombus and let bethe perpendicular bisector of . Then if and only if is on the same side of as . The line divides the plane into two half-planes; let be the half-plane containing . Let us define similarly and . Then is equal to . The region turns out to be anirregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:Since and are equilateral, contains , contains and , and contains .Then with and soand has area .Brian writes down four integers whose sum is . The pairwise positive differences of these numbers are and . What is the sum of the possible values for ?The largest difference, must be between andThe smallest difference, must be directly between two integers. This also means the differences directly between the other two should add up to The only remaining differences that would make this possible are and However, those two differences can't be right next to each other because they would make a difference of This means must be the difference between and We can express the possible configurations as the lines.If we look at the first number line, you can express as as and as Since the sum of all these integers equal ,You can do something similar to this with the second number line to find the other possible value ofThe sum of the possible values of isWhat is the hundreds digit of ?Since we knowTo compute this, write it as and use the binomial theorem.From then on the powerof is greater than and cancel out withTherefore, the hundreds digit isA lattice point in an -coordinate system is any point whereboth and are integers. The graph of passes through no lattice point with for all such that . What is the maximum possible value of ?We see that for the graph of to not pass through any lattice points its denominator must be greater the than 100. We see that the nearest fraction bigger than that does not have its denominator over 100 is 50/99.Let be a triangle with sides and . For ,if and and are the points of tangency of the incircleof to the sides and respectively, then is a triangle with side lengths and if it exists. What is the perimeter of the last triangle in the sequence ?By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites.Hence and and .Let and gives three equations:(where for the first triangle.)Solving gives:Subbing in gives that has sides of .can easily be derivied from this as the sides still differ by 1 hence the above solutions still work (now with ). All additional triangles will differ by one as the solutions above differ by one so this process can be repeated indefinately until the side lengths no longer form a triangle.Subbing in gives with sides .has sides .has sides .has sides .has sides .has sides .has sides .has sides .would have sides but these length do not make a triangle as .Hence the perimeter is。

2013 AMC 10B Problems

2013 AMC 10B ProblemsProblem 1 What is642531531642++++-++++? （A ）-1 （B ）365 （C ）127 （D ）2049 （E ）343 Problem 2Mr. Green measures his rectangular garden by walking two of the sides and finding that it is 15 steps by 20 steps. Each of Mr. Green's steps is 2 feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?（A ）600 （B ）800 （C ）1000 （D ）1200 （E ）1400 Problem 3On a particular January day, the high temperature in Lincoln, Nebraska, was 16 degrees higher than the low temperature, and the average of the high and the low temperatures was 3º. In degrees, what was the low temperature in Lincoln that day?（A ）-13 （B ）-8 （C ）-5 （D ）-3 （E ）11Problem 4When counting from 3 to 201, 53 is the st 51 number counted. When counting backwards from 201 to 3, 53 is the th n number counted. What is ?（A ）146 （B ）147 （C ）148 （D ）149 （E ）150 Problem 5Positive integers and are each less than 6. What is the smallest possible value forab a -2 ?（A ）-20 （B ）-15 （C ）-10 （D ）0 （E ）2Problem 6The average age of 33 fifth-graders is 11. The average age of 55 of their parents is 33. What is the average age of all of these parents and fifth-graders?（A ）22 （B ）23.25 （C ）24.75 （D ）26.25 （E ）28 Problem 7Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?（A ）33 （B ）23 （C ）1 （D ）2 （E ）2Ray's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?（A ）10 （B ）16 （C ）25 （D ）30 （E ）40Problem 9Three positive integers are each greater than 1, have a product of 27000, and are pairwise relatively prime. What is their sum?（A ）100 （B ）137 （C ）156 （D ）160 （E ）165Problem 10A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?（A ）10 （B ）15 （C ）20 （D ）25 （E ）30Problem 11Real numbers x and y satisfy the equation 3461022--=+y x y x . What is y x + ?（A ）1 （B ）2 （C ）3 （D ）6 （E ）8Problem 12Let S be the set of sides and diagonals of a regular pentagon. A pair of elements of S are selected at random without replacement. What is the probability that the two chosen segments have the same length?（A ）52 （B ）94 （C ）21 （D ）95 （E ）54 Problem 13Jo and Blair take turns counting from 1 to one more than the last number said by the other person. Jo starts by saying "1", so Blair follows by saying "1,2" . Jo then says "1,2,3" , and so on. What is the 53rd number said?（A ）2 （B ）3 （C ）5 （D ）6 （E ）8Problem 14Define. Which of the following describes the set of points forwhich ? （A ）a finite set of points （B ）one line （C ）two parallel lines（D ）two intersecting lines （E ）three linesA wire is cut into two pieces, one of length and the other of length . The piece oflength is bent to form an equilateral triangle, and the piece of length is bent to form aregular hexagon. The triangle and the hexagon have equal area. What is a/b?（A ）1 （B ）26 （C ）3 （D ）2 （E ）223 Problem 16In triangle ABC , medians AD and C E intersect at P , PE =1.5, PD =2, and DE =2.5. What is the area of AEDC ?（A ）13 （B ）13.5 （C ）14 （D ）14.5 （E ）15Problem 17Alex has 75 red tokens and 75 blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?（A ）62 （B ）82 （C ）83 （D ）102 （E ）103Problem 18The number 2013 has the property that its units digit is the sum of its other digits, that is 2+0+1=3. How many integers less than 2013 but greater than 1000 share this property?（A ）33 （B ）34 （C ）45 （D ）46 （E ）58Problem 19The real numbers a b c ,,form an arithmetic sequence with 0≥≥≥c b a . The quadratic c bx ax ++2 has exactly one root. What is this root?（A ）347-- （B ）32-- （C ）-1 （D ）32+- （E ）347+- Problem 20The number 2013 is expressed in the form !!...!!!...!20132121n m b b b a a a =, where m a a a ≥≥≥...21 And n b b b ≥≥≥...21 are positive integers and 11b a + is as small as possible. What is 11b a - ?（A ）1 （B ）2 （C ）3 （D ）4 （E ）5Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is . What is the smallestpossible value of N?（A ）55 （B ）89 （C ）104 （D ）144 （E ）273Problem 22The regular octagon ABCDEFGH has its center at J . Each of the vertices and the center are to be associated with one of the digits 1 through 9, with each digit used once, in such a way that the sums of the numbers on the lines AJE, BJF , CJG , and DJH are all equal. In how many ways can this be done?（A ）384 （B ）576 （C ）1152 （D ）1680 （E ）3456Problem 23In triangle ABC , AB =13, BC =14, and CA =15. Distinct points D, E , and F lie on segments CA BC ,, and DE , respectively, such that ,,AC DE BC AD ⊥⊥ and BF AF ⊥. The length of segment DF can be written as n m /, where and are relatively primepositive integers. What is n m + ?（A ）18 （B ）21 （C ）24 （D ）27 （E ）30Problem 24A positive integer is nice if there is a positive integerwith exactly four positive divisors(including 1 and ) such that the sum of the four divisors is equal to . How manynumbers in the set {2010, 2011, 2012, …, 2019} are nice?（A ）1 （B ）2 （C ）3 （D ）4 （E ）5Problem 25Bernardo chooses a three-digit positive integer N and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer S . For example, if N =749, Bernardo writes the numbers 10,444 and 3,245, and LeRoy obtains the sum S =13,689. For how many choices of N are the two rightmost digits of S , in order, the same as those of 2N ?（A ）5 （B ）10 （C ）15 （D ）20 （E ）252013 AMC 10B SolutionsProblem 1This expression is equivalent to 127129912=- Problem 2The following problem is from both the 2013 AMC 12B #2 and 2013 AMC 10B #2Since each step is 2 feet, his garden is 30 by 40 feet. Thus, the area of 30(40)=1200 square feet. Since he is expecting 1/2 of a pound per square foot, the totalamount of potatoes expected is 1200*1/2=600. Problem 3The following problem is from both the 2013 AMC 12B #1 and 2013 AMC 10B #3Let L be the low temperature. The high temperature is L +16. The average is32)16(=++L L . Solving for L , we get L =-5. Problem 4The following problem is from both the 2013 AMC 12B #3 and 2013 AMC 10B #4Note that is equal to the number of integers between 53 and 201, inclusive. Thus,149153201=+-=n . Problem 5Factoring the equation gives )2(b a - . From this we can see that to obtain the least possible value, b -2should be negative, and should be as small as possible. To doso, should be maximized. Because b -2 is negative, we should maximize the positivevalue of as well. The maximum values of both and are 5, so the answer is5(2-5)=-15. Problem 6The following problem is from both the 2013 AMC 12B #5 and 2013 AMC 10B #6The sum of the ages of the fifth graders is 33*11, while the sum of the ages of the parents is 55*33. Therefore, the total sum of all their ages must be 2178, and given33+55=88 people in total, their average age is 2178/88=99/4=24.75. .Problem 7If there are no two points on the circle that are adjacent, then the triangle would be equilateral. If the three points are all adjacent, it would be isosceles. Thus, the onlypossibility is two adjacent points and one point two away. Because one of the sides of this triangle is the diameter, the opposite angle is a right angle. Also, because the two adjacent angles are one sixth of the circle apart, the angle opposite them is thirty degrees. This is a 30—60—90 triangle. If the original six points are connected, a regular hexagon is created. This hexagon consists of six equilateral triangles, so the radius is equal to one of its side lengths. The radius is 1, so the side opposite the thirty degree angle in the triangle isalso 1. From rules with 30—60—90 triangles, the area is 2/32/31=∙ Problem 8The following problem is from both the 2013 AMC 12B #4 and 2013 AMC 10B #8Let both Ray and Tom drive 40 miles. Ray's car would require 40/40=1 gallon of gas and Tom's car would require 40/10=4gallons of gas. They would have driven a total of40+40=80 miles, on 1+4=5 gallons of gas, for a combined rate ofProblem 9The prime factorization of 27000 is 333532∙∙. These three factors are pairwise relativelyprime, so the sum is 160532333=++ Problem 10Call the number of two point shots attempted and the number of three point shotsattempted. Because each two point shot is worth two points and the team made 50% and each three point shot is worth 3 points and the team made 40%, 54)3(4.0)2(5.0=+y x or 542.1=+y x . Because the team attempted 50% more two point shots then threes, y x 5.1=. Substituting y 5.1 for in the first equation gives542.15.1=+y y , so 20=y Problem 11The following problem is from both the 2013 AMC 12B #6 and 2013 AMC 10B #11If we complete the square after bringing the and terms to the other side, we get0)3()5(22=++-y x . Squares of real numbers are nonnegative, so we need both 2)5(-x and 2)3(+y to be 0 which only happens when 5=x and 3-=y . Therefore,2=+y x Problem 12In a regular pentagon, there are 5 sides with the same length, and 5 diagonals with the same length. Picking an element at random will leave 4 elements with the same length as the element picked, with 9 total elements remaining. Therefore, the probability is 4/9.Problem 13The following problem is from both the 2013 AMC 12B #7 and 2013 AMC 10B #13We notice that the number of numbers said is incremented by one each time; that is, Josays one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns," 1+2+3+4+5+6+7+8+9=45 numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, as 53-45=8. Since we're starting from 1each time, the 53rd number said will be 8. Problem 14and . Therefore, we have the equation2222yx x y xy y x -=-. Factoring out a -1 gives )(2222xy y x xy y x --=-. Factoring both sides further, )()(y x xy y x xy --=-. It follows that if 0,0==y x , or 0)(=-y x , both sides of the equation equal 0. By this, there are 3 lines (0,0==y x , ory x = ) so the answer is three lines. Problem 15Solution 1Using the formulas for area of a regular triangle )43(2s and regular hexagon )233(2s and plugging 3a and 6b into each equation, you find that 24336322b a =.Simplifying this, you get 26=b a . Solution 2The regular hexagon can be broken into 6 small equilateral triangles, each of which is similar to the big equilateral triangle. The big triangle's area is 6 times the area of one of the little triangles. Therefore each side of the big triangle is 6 times the side of thesmall triangle. The desired ratio is 26663=. Problem 16Solution 1Let us use mass points: Assign B mass 1. Thus, because E is the midpoint of AB , A also has a mass of 1. Similarly, C has a mass of 1. D and E each have a mass of 2 because they are between B and C and A and B respectively. Note that the mass of D is twice the mass of A , so AP must be twice as long as PD . PD has length 2, so AP has length 4 and AD has length 6. Similarly, CP is twice PE and PE =1.5, so CP =3 and CE =4.5. Now note that triangle PED is a 3-4-5 right triangle with the right angle DPE . This means that the quadrilateral AEDC is a kite. The area of a kite is half the product of the diagonals, AD and CE . Recall that they are 6 and 4.5 respectively, so the area of AEDC is6*4.5/2=13.5. Solution 2Note that triangle DPE is a right triangle, and that the four angles that have point P are allright angles. Using the fact that the centroid (P ) divides each median in a 2:1 ratio, AP =4 and CP =3. Quadrilateral AEDC is now just four right triangles. The area is5.1325.1223345.14=∙+∙+∙+∙ Solution 3From the solution above, we can find the length of the diagonals to be 6 and 4.5. Now, since AEDC is a trapezoid, we use the area formula to find that the total area is5.1325.4*6= Problem 17The following problem is from both the 2013 AMC 12B #10 and 2013 AMC 10B #17 Solution 1We can approach this problem by assuming he goes to the red booth first. You start with 75R and 75B and at the end of the first booth, you will have 1R and 112B and 37S. We now move to the blue booth, and working through each booth until we have none left, wewill end up with: 1R, 2B and 103S. So, the answer is 103. Solution 2Let denote the number of visits to the first booth and denote the number of visits to thesecond booth. Then we can describe the quantities of his red and blue coins as follows: 752),(++-=y x y x R , 753),(+-=y x y x B . There are no legal exchanges when he has fewer than 2 red coins and fewer than 3 blue coins, namely when he has 1 red coin and 2 blue coins. We can then create a system of equations: 7521++-=y x , 7532+-=y x . Solving yields 59=x and 44=y . Since he gains one silver coin pervisit to each booth, he has 1035944=+=+y x silver coins in total. Problem 18First, note that the only integer 20132000<≤x is 2002. Now let's look at all numbers where 20001000<<x . Let the hundreds digit be 0. Then, the tens andunits digit can be 01, 12, 23, …, 89, which is 9 possibilities. We notice as the hundreds digit goes up by one, the number of possibilities goes down by one. Thus, the number ofProblem 19Solution 1It is given that 02=++c bx ax has 1 real root, so the discriminant is zero, or ac b 42=. Because a, b, c are in arithmetic progression, ,b c a b -=-or 2c a b +=. We need to find the unique root, or a b 2- (discriminant is 0). From ac b 42=, we have b c a b 22-=-.Ignoring the negatives, we have .1114224144124+===+==+++ca c c a c a c a c c a c cbc Fortunately, finding c a /is not very hard. Plug in 2c a b += to ac b 42=, we have ,16222ac c ac a =++ or 01422=+-c ac a , and dividing by 2c gives,01)(14)(2=+-c a c a so 347219214±=±=c a . But 1347<-, violating the assumptionthat c a ≥. Therefore,347+=ca . Plugging this in, we have 323211414-=+=+c a . Butwe need the negative of this, so the answer is . Solution 2Note that we can divide the polynomial by to make the leading coefficient 1 sincedividing does not change the roots or the fact that the coefficients are in an arithmetic sequence. Also, we know that there is exactly one root so this equation is must be of theform 2222)(r rx x r x +-=- where 0212≥≥-≥r r . We now use the fact that the coefficients are in an arithmetic sequence. Note that in any arithmetic sequence, the average is equal to the median. Thus, r r 412-=+ and 32±-=r . Since 21r >, we easily see that r has to be between 1 and 0. Thus, we can eliminate 32-- and areleft with 32+- as the answer. Solution 3Given that 02=++c bx ax has only 1 real root, we know that the discriminant must equal 0, or that ac b 42=. Because the discriminant equals 0, we have that the root of the quadratic is a b r 2/-=. We are also given that the coefficients of the quadratic are in arithmetic progression, where 0≥≥≥c b a . Letting the arbitrary difference equal variable , we have that d b a += and that d b c -=. Plugging those two equationsinto ac b 42=, we have )(4222d b b -= which yields 2243d b =. Isolating , we have 2/3b d =. Substituting that in for in d b a +=, we get )231(23+=+=b b b a . Once again, substituting that in for in a b r 2/-=, we have32321)1(223+-=+-=+-=b b r . The answer is Solution 4Let the double root be . Then by the arithmetic progression and Vieta's,ac a b a b c b b a -=-⇒-=-1 3201422122±-=⇒=++⇒--=+r r r r r rWe see 100≤≤⇒≤≤ab a b , and so we want 120≤-≤r . Note that since 1324)32(20≥+=---≤ and 1324)32(20≤-=+--≤, we can concludethat 32+-=r and the answer is .Problem 20The following problem is from both the 2013 AMC 12B #15 and 2013 AMC 10B #20The prime factorization of 2013 is 61·11·3. To have a factor of 61 in the numerator and to minimize11,a a must equal 61. Now we notice that there can be no prime which is not a factor of 2013 such that 611<<p b , because this prime will not be represented in the denominator, but will be represented in the numerator. The highest lessthan 61 is 59, so there must be a factor of 59 in the denominator. It follows that591=b (to minimize 1b as well), so the answer is 25961=- to express 2013 with )59,61(),(11=b a is !10!20!59!11!19!612013∙∙∙∙= Problem 21The following problem is from both the 2013 AMC 12B #14 and 2013 AMC 10B #21 Let the first two terms of the first sequence be 1x and 2x and the first two of the second sequence be 1y and 2y . Computing the seventh term, we see that 21218585y y x x +=+. Note that this means that 1x and 1y must have the same value modulo 8. To minimize, let one of them be 0; WLOG assume that 01=x . Thus, the smallest possible value of 1y is 8; since the sequences are non-decreasing 82≥y . To minimize, let 82=y .Thus, 10464408521=+=+y y . Problem 22First of all, note that J must be 1, 5, or 9 to preserve symmetry. We also notice that A+E=B+F=C+G=D+H.WLOG assume that J =1. Thus the pairs of vertices must be 9 and 2, 8 and 3, 7 and 4, and 6 and 5. There are 4!=24 ways to assign these to the vertices. Furthermore, there are 1624= ways to switch them (i.e. do 2 9 instead of 9 2).Thus, there are 16(24)=384 ways for each possible J value. There are 3 possible J valuesthat still preserve symmetry: 384(3)=1152. Solution 2As in solution 1, J must be 1, 5, or 9 giving us 3 choices. Additionally A+E=B+F=C+G=D+H . This means once we choose J there are 8 remaining choices. Going clockwise from A we count, 8 possibilities for A . Choosing A also determines E which leaves 6 choices for B , once B is chosen it also determines F leaving 4 choices for C . Once C is chosen it determines G leaving 2 choicesfor D . Choosing D determines H , exhausting the numbers. To get the answer we multiply2*4*6*8*3=1152. Problem 23The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, Solution 1Since 90=∠=∠ADB AFB , quadrilateral ABDF is cyclic. It follows that ABF ADE ∠=∠. In addition, since 90=∠=∠AED AFB , triangles ABF and ADE are similar. It follows that )5/3)(13(),5/4)(13(=BF AF . By Ptolemy , we have )5/3)(13)(12()5/4)(13)(5(13=+DF .Cancelling 13, the rest is easy. We obtain 215165/16=+⇒=DF . Solution 2Using the similar triangles in triangle ADC gives AE =48/5 and DE =36/5. Quadrilateral ABDF is cyclic, implying that ︒=∠+∠180DFA B . Therefore, EFA B ∠=∠, and triangles AEF and ADB are similar. Solving the resulting proportion gives EF =4. Therefore,DF=ED-EF =16/5. Problem 24A positive integer with only four positive divisors has its prime factorization in the form of b a *, where and are both prime positive integers or 3c where is a prime. One caneasily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of b a *. The four factors of this number would be 1, ,,b a and ab . The sum of these would be 1+++b a ab , which can be factored into the form )1)(1(++b a . Easily we can see that now we can take cases again.Case 1: Either or is 2.If this is true then we have to have that one of )1(+a or )1(+b is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either 132016- or 132010- is a prime. We see that in this case none of them work. Case 2: Both and are odd primes.This implies that both )1(+a and )1(+b are even which implies that in this case the number must be divisible by 4. This leaves only 2012 and 2016. 2012=4*503 so we have that a factor of 2 must go to both )1(+a and )1(+b . So we have that )1(+a and )1(+b equal the numbers (502+1)(3+1), but 502 is not an odd prime, so 2012 does not work. 2016=4*504, so we have (503+1)(3+1). 503 and 3 are both oddprimes, so 2016 is a solution. Thus the answer is 1. Problem 25The following problem is from both the 2013 AMC 12B #23 and 2013 AMC 10B #25,First, we can examine the units digits of the number base 5 and base 6 and eliminatesome possibilities.Say that )6(mod a N ≡ also that )5(mod b N ≡Substituting these equations into the question and setting the units digits of 2N and Sequal to each other, it can be seen thatb a =, and 5<b , so),6(mod a N ≡40),30(mod )5(mod ≤≤=⇒≡a a N a N 。

2013美国数学竞赛10年级试题B

INSTRUCTIONS1. DO NOT OPEN THIS BOOKLET UNTIL YOUR PROCTOR TELLS YOU.2. This is a twenty-five question multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.3. Mark your answer to each problem on the AMC 10 Answer Form with a #2 pencil. Check the blackened circles for accuracy and erase errors and stray marks completely. Only answers properly marked on the answer form will be graded.4. SCORING: You will receive 6 points for each correct answer, 1.5 points for each problem left unanswered, and 0 points for each incorrect answer.5. No aids are permitted other than scratch paper, graph paper, rulers, compass, protractors, and erasers. No calculators are allowed. No problems on the test will require the use of a calculator.6. Figures are not necessarily drawn to scale.7. Before beginning the test, your proctor will ask you to record certain information on the answer form.8. When your proctor gives the signal, begin working on the problems. You will have 75 minutes to complete the test.9. When you finish the exam, sign your name in the space provided on the Answer Form.© 2013 Mathematical Association of AmericaThe Committee on the American Mathematics Competitions (CAMC) reserves the right to re-examine students before deciding whether to grant official status to their scores. The CAMC also reserves the right to disqualify all scores from a school if it is determined that the required security procedures were not followed.Students who score 120 or above or finish in the top 2.5% on this AMC 10 will be invited to take the 31st annual American Invitational Mathematics Examination (AIME) on Thursday, March 14, 2013 or Wednesday, April 3, 2013. More details about the AIME and other information are on the back page of this test booklet.The publication, reproduction or communication of the problems or solutions of the AMC 10 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via copier, telephone, e-mail, World Wide Web or media of any type during this period is a violation of the competition rules. After the contest period, permission to make copies of problems in paper or electronic form including posting on web-pages for educational use is granted without fee provided that copies are not made ordistributed for profit or commercial advantage and that copies bear the copyright notice.**Administration On An Earlier Date Will Disqualify Your School’s Results**1. All information (Rules and Instructions) needed to administer this exam is contained in the TEACHERS’ MANUAL, which is outside of this package. PLEASE READ THE MANUAL BEFORE FEBRUARY 20, 2013. Nothing is needed from inside this package until February 20.2. Your PRINCIPAL or VICE-PRINCIPAL must verify on the AMC 10 CERTIFICATION FORM (found in the T eachers’ Manual) that you fol-lowed all rules associated with the conduct of the exam.3. The Answer Forms must be mailed by trackable mail to the AMC office no later than 24 hours following the exam.4. The publication, reproduction or communication of the problems or solutions of this test during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination at any time via copier, telephone, e-mail, internet or media of any type is a violation of the competition rules.2013AMC 10 B DO NOT OPEN UNTIL WEDNEsDAy, fEbrUAry 20, 2013The American Mathematics Competitionsare Sponsored byThe Mathematical Association of America – MAA The Akamai Foundation ContributorsAcademy of Applied Sciences – AAs American Mathematical Association of Two-Year Colleges – AMATyC ...................................................... American Mathematical Society – AMs ........................................................................................................... American Statistical Association – AsA ...................................................................................................... Art of Problem Solving – Awesome Math Casualty Actuarial Society – CAs ................................................................................................................ D.E. Shaw & Co. ................................................................................................................................. Delta Airlines ....................................................................................................................................... Jane Street Math For America Mu Alpha Theta – MAT ....................................................................................................................... National Council of Teachers of Mathematics – NCTM ................................................................................... Pi Mu Epsilon – PME ............................................................................................................................... Society for Industrial and Applied Math - SIAM ............................................................................................ 1.What is 2+4+61+3+5−1+3+52+4+6?(A)−1(B)536(C)712(D)4920(E)4332.Mr.Green measures his rectangular garden by walking two of the sides and ﬁnds that it is 15steps by 20steps.Each of Mr.Green’s steps is 2feet long.Mr.Green expects a half a pound of potatoes per square foot from his garden.How many pounds of potatoes does Mr.Green expect from his garden?(A)600(B)800(C)1000(D)1200(E)14003.On a particular January day,the high temperature in Lincoln,Nebraska,was 16degrees higher than the low temperature,and the average of the high and low temperatures was 3◦.In degrees,what was the low temperature in Lincoln that day?(A)−13(B)−8(C)−5(D)−3(E)114.When counting from 3to 201,53is the 51st number counted.When counting backwards from 201to 3,53is the n th number counted.What is n ?(A)146(B)147(C)148(D)149(E)1505.Positive integers a and b are each less than6.What is the smallest possible value for 2·a −a ·b ?(A)−20(B)−15(C)−10(D)0(E)26.The average age of 33ﬁfth-graders is 11.The average age of 55of their parents is 33.What is the average age of all of these parents and ﬁfth-graders?(A)22(B)23.25(C)24.75(D)26.25(E)287.Six points are equally spaced around a circle of radius 1.Three of these points are the vertices of a triangle that is neither equilateral nor isosceles.What is the area of this triangle?(A)√33(B)√32(C)1(D)√2(E)28.Ray’s car averages40miles per gallon of gasoline,and Tom’s car averages10miles per gallon of gasoline.Ray and Tom each drive the same number of miles.What is the cars’combined rate of miles per gallon of gasoline?(A)10(B)16(C)25(D)30(E)409.Three positive integers are each greater than1,have a product of27,000,andare pairwise relatively prime.What is the sum of these integers?(A)100(B)137(C)156(D)160(E)16510.A basketball team’s players were successful on50%of their two-point shots and40%of their three-point shots,which resulted in54points.They attempted 50%more two-point shots than three-point shots.How many three-point shots did they attempt?(A)10(B)15(C)20(D)25(E)3011.Real numbers x and y satisfy the equation x2+y2=10x−6y−34.What isx+y?(A)1(B)2(C)3(D)6(E)812.Let S be the set of sides and diagonals of a regular pentagon.A pair of elementsof S are selected at random without replacement.What is the probability that the two chosen segments have the same length?(A)25(B)49(C)12(D)59(E)4513.Jo and Blair take turns counting from1to one more than the last number saidby the other person.Jo starts by saying“1”,so Blair follows by saying“1,2”.Jo then says“1,2,3”,and so on.What is the53rd number said?(A)2(B)3(C)5(D)6(E)814.Deﬁne a♣b=a2b−ab2.Which of the following describes the set of points(x,y)for which x♣y=y♣x?(A)aﬁnite set of points(B)one line(C)two parallel lines(D)two intersecting lines(E)three lines15.A wire is cut into two pieces,one of length a and the other of length b.Thepiece of length a is bent to form an equilateral triangle,and the piece of lengthb is bent to form a regular hexagon.The triangle and the hexagon have equalarea.What is ab?(A)1(B)√62(C)√3(D)2(E)3√2216.In ABC,medians AD and CE intersect at P,P E=1.5,P D=2,andDE=2.5.What is the area of AEDC?CB(A)13(B)13.5(C)14(D)14.5(E)1517.Alex has75red tokens and75blue tokens.There is a booth where Alex cangive two red tokens and receive in return a silver token and a blue token,and another booth where Alex can give three blue tokens and receive in return a silver token and a red token.Alex continues to exchange tokens until no more exchanges are possible.How many silver tokens will Alex have at the end?(A)62(B)82(C)83(D)102(E)10318.The number2013has the property that its units digit is the sum of its otherdigits,that is2+0+1=3.How many integers less than2013but greater than1000share this property?(A)33(B)34(C)45(D)46(E)5819.The real numbers c,b,a form an arithmetic sequence with a≥b≥c≥0.Thequadratic ax2+bx+c has exactly one root.What is this root?(A)−7−4√3(B)−2−√3(C)−1(D)−2+√3(E)−7+4√3 20.The number2013is expressed in the form2013=a1!a2!···a m!1!b2!···b n!,where a1≥a2≥···≥a m and b1≥b2≥···≥b n are positive integers and a1+b1is as small as possible.What is|a1−b1|?(A)1(B)2(C)3(D)4(E)521.Two non-decreasing sequences of nonnegative integers have diﬀerentﬁrst terms.Each sequence has the property that each term beginning with the third is the sum of the previous two terms,and the seventh term of each sequence is N.What is the smallest possible value of N?(A)55(B)89(C)104(D)144(E)27322.The regular octagon ABCDEF GH has its center at J.Each of the verticesand the center are to be associated with one of the digits1through9,with each digit used once,in such a way that the sums of the numbers on the lines AJE, BJF,CJG,and DJH are equal.In how many ways can this be done?(A)384(B)576(C)1152(D)1680(E)3456CD G23.In triangle ABC ,AB =13,BC =14,and CA =15.Distinct points D ,E ,and F lie on segments BC ,CA ,and DE ,respectively,such that AD ⊥BC ,DE ⊥AC ,and AF ⊥BF .The length of segment DF can be written as m n ,where m and n are relatively prime positive integers.What is m +n ?(A)18(B)21(C)24(D)27(E)3024.A positive integer n is nice if there is a positive integer m with exactly four positive divisors (including 1and m )such that the sum of the four divisors is equal to n .How many numbers in the set {2010,2011,2012,...,2019}are nice?(A)1(B)2(C)3(D)4(E)525.Bernardo chooses a three-digit positive integer N and writes both its base-5and base-6representations on a ter LeRoy sees the two numbers Bernardo has written.Treating the two numbers as base-10integers,he adds them to obtain an integer S .For example,if N =749,Bernardo writes the numbers 10,444and 3,245,and LeRoy obtains the sum S =13,689.For how many choices of N are the two rightmost digits of S ,in order,the same as those of 2N ?(A)5(B)10(C)15(D)20(E)25WRITE TO US!Correspondence about the problems and solutions for this AMC 10and orders for publications should be addressed to:American Mathematics CompetitionsUniversity of Nebraska, P .O. Box 81606Lincoln, NE 68501-1606Phone 402-472-2257 | Fax 402-472-6087 | amcinfo@The problems and solutions for this AMC 10 were prepared by the MAA’s Committee on theAMC 10 and AMC 12 under the direction of AMC 10 Subcommittee Chair:Dr. Leroy Wenstrom2013 AIMEThe 31st annual AIME will be held on Thursday, March 14, with the alternate on Wednesday, April 3. It is a 15-question, 3-hour, integer-answer exam. You will be invited to participate only if you score 120 or above or finish in the top 2.5% of the AMC 10, or if you score 100 or above or finish in the top 5% of the AMC 12. T op-scoring students on the AMC 10/12/AIME will be selected to take the 42nd Annual USA Mathematical Olympiad (USAMO) on April 30 - May 1, 2013. The best way to prepare for the AIME and USAMO is to study previous exams. Copies may be ordered as indicated below.PUBLICATIONSA complete listing of current publications, with ordering instructions, is at our web site: American Mathematics Competitions。

amc10b 2023 题目pdf

vb6程序设计手册1. VB6基础知识与语法1.1 VB6概述与历史：1.1.1 VB6简介：Visual Basic 6.0（VB6）是一种基于事件驱动的编程语言，由微软开发，用于Windows应用程序的开发。

1.1.2 发展历史：VB6于1998年发布，成为当时Windows 平台上最受欢迎的应用程序开发工具之一。

1.2 VB6语法要点：1.2.1 事件驱动编程：VB6采用事件驱动的编程模型，程序通过对用户或系统事件的响应来执行相应的代码。

1.2.2 对象导向特性：VB6支持对象导向编程，允许开发者创建和操作各种对象，提高代码的模块化和可维护性。

1.3 VB6程序结构：1.3.1 模块与过程：VB6程序由模块组成，模块包含过程（Sub 和Function）以执行特定任务。

1.3.2 窗体与控件：窗体是VB6应用程序的可视化界面，控件用于与用户交互，例如按钮、文本框等。

1.4 数据类型与变量：1.4.1 基本数据类型：VB6包括整数、浮点数、字符串等基本数据类型，每种类型都有其特定的用途。

1.4.2 变量声明与作用域：学习如何声明变量以及它们的作用域是VB6编程的基础。

1.5 异常处理与调试：1.5.1 错误处理机制：VB6提供了异常处理机制，通过On Error 语句来处理运行时错误。

1.5.2 调试工具：学习使用VB6集成的调试工具，如断点、监视窗口等，提高程序调试效率。

2. VB6高级特性与设计模式2.1 ADO数据库编程：2.1.1 数据库连接与操作：学习如何使用VB6中的ActiveX Data Objects（ADO）来连接和操作数据库。

2.1.2 数据绑定：探讨在VB6中如何实现数据绑定，将数据库数据与用户界面关联起来。

2.2 多线程与异步编程：2.2.1 多线程概念：理解多线程编程的基本概念，以提高程序的并发性。

2.2.2 异步编程模式：学习在VB6中实现异步编程，以避免阻塞用户界面的情况。

2020AMC10B（美国数学竞赛）真题加详解

2020AMC10B（美国数学竞赛）真题加详解2020 AMC 10B Solution Problem1What is the value ofSolutionWe know that when we subtract negative numbers, .The equation becomesProblem2Carl has cubes each having side length , and Kate has cubes each having side length . What is the total volume of these cubes?SolutionA cube with side length has volume , so of these will have a total volume of .A cube with side length has volume , so of these will have a total volume of .~quacker88Problem 3The ratio of to is , the ratio of to is , and the ratioof to is . What is the ratio of toSolution 1WLOG, let and .Since the ratio of to is , we can substitute in the value of toget .The ratio of to is , so .The ratio of to is then so our answeris ~quacker88Solution 2We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two., and since , we can link themtogether to get .Finally, since , we can link this again to get: ,so ~quacker88Problem4The acute angles of a right triangle are and , where andboth and are prime numbers. What is the least possible value of ?SolutionSince the three angles of a triangle add up to and one of the anglesis because it's a right triangle, .The greatest prime number less than is . If ,then , which is not prime.The next greatest prime number less than is . If ,then , which IS prime, so we have our answer ~quacker88 Solution 2Looking at the answer choices, only and are coprime to . Testing , the smaller angle, makes the other angle which is prime, therefore our answerisProblem5How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)SolutionLet's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.There are ways to order objects. However, since there's ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and ways to order the green tiles, we have to divide out these possibilities.~quacker88SolutionWe can repeat chooses extensively to find the answer. Thereare choose ways to arrange the brown tiles which is . Then from the remaining tiles there are choose ways to arrange the red tiles. And now from the remaining two tiles and two slots we can see there are two ways to arrange the purple and brown tiles, giving us an answerofProblem6Driving along a highway, Megan noticed that her odometershowed (miles). This number is a palindrome-it reads the same forward and backward. Then hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this -hour period?SolutionIn order to get the smallest palindrome greater than , we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.So we raise to the next largest value, , but obviously, that's not how place value works, so we're in the s now. To keep this a palindrome, our number is now .So Megan drove miles. Since this happened over hours, she drove at mph. ~quacker88 Problem7How many positive even multiples of less than are perfect squares?SolutionAny even multiple of is a multiple of , so we need to find multiples of that are perfect squares and less than . Any solution that we want will be in theform , where is a positive integer. The smallest possible value isat , and the largest is at (where the expression equals ). Therefore, there are a total of possible numbers.-PCChess Problem8 Points and lie in a plane with . How many locations forpoint in this plane are there such that the triangle with vertices , ,and is a right triangle with area square units?Solution 1There are options here:1. is the right angle.It's clear that there are points that fit this, one that's directly to the rightof and one that's directly to the left. We don't need to find the length, we just need to know that it is possible, which it is.2. is the right angle.Using the exact same reasoning, there are also solutions for this one.3. The new point is the right angle.(Diagram temporarily removed due to asymptote error)The diagram looks something like this. We know that the altitude tobase must be since the area is . From here, we must see if there are valid triangles that satisfy the necessary requirements. First of all, because of the area.Next, from the Pythagorean Theorem.From here, we must look to see if there are valid solutions. There are multiple ways to do this:We know that the minimum value of iswhen . In this case, the equationbecomes , which is LESSthan . . The equationbecomes , which is obviously greater than . We canconclude that there are values for and in between that satisfy the Pythagorean Theorem.And since , the triangle is not isoceles, meaning we could reflectit over and/or the line perpendicular to for a total of triangles this case.Solution 2Note that line segment can either be the shorter leg, longer leg or thehypotenuse. If it is the shorter leg, there are two possible points for that cansatisfy the requirements - that being above or below . As such, thereare ways for this case. Similarly, one can find that there are also ways for point to lie if is the longer leg. If it is a hypotenuse, then thereare possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is .Problem9How many ordered pairs of integers satisfy theequationSolutionRearranging the terms and and completing the square for yields theresult . Then, notice that can onlybe , and because any value of that is greater than 1 will causethe term to be less than , which is impossible as must be real. Therefore, plugging in the above values for gives the ordered pairs , , , and gives a totalof ordered pairs.Solution 2Bringing all of the terms to the LHS, we see a quadraticequation in terms of . Applying the quadratic formula, weget In order for to be real, which it must be given the stipulation that we are seekingintegral answers, we know that the discriminant, must benonnegative. Therefore, Here, we see that we must split the inequality into a compound, resultingin .The only integers that satisfy this are . Plugging thesevalues back into the quadratic equation, we see that both produce a discriminant of , meaning that there is only 1 solution for . If , then the discriminant is nonzero, therefore resulting in two solutions for .Thus, the answer is .~TiblisSolution 3, x firstSet it up as a quadratic in terms of y:Then the discriminant is This will clearly only yield real solutionswhen , because it is always positive. Then . Checking each one: and are the same when raised to the 2020th power:This has only has solutions , so are solutions. Next, if :Which has 2 solutions, so andThese are the only 4 solutions, soSolution 4, y firstMove the term to the other side toget . Because for all , then . If or , the right side is and therefore . When , the right side become , therefore . Our solutions are , , , . There are solutions, so the answer is - wwt7535Problem 10A three-quarter sector of a circle of radius inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubicinches?SolutionNotice that when the cone is created, the radius of the circle will become the slant height of the cone and the intact circumference of the circle will become the circumference of the base of the cone.We can calculate that the intact circumference of the circle is . Since that is also equal to the circumference of the cone, the radius of the cone is . We also have that the slant height of the cone is . Therefore, we use the Pythagorean Theorem to calculate that the height of the coneis . The volume of the coneis -PCChessSolution 2 (Last Resort/Cheap)Using a ruler, measure a circle of radius 4 and cut out the circle and then the quarter missing. Then, fold it into a cone and measure the diameter to be 6cm . You can form a right triangle with sides 3, 4, and then through the Pythagorean theorem the height is found tobe . The volume of a cone is . Plugging in we findProblem11Ms. Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?SolutionWe don't care about which books Harold selects. We just care that Bettypicks books from Harold's list and that aren't on Harold's list.The total amount of combinations of books that Betty can selectis .There are ways for Betty to choose of the books that are on Harold's list.From the remaining books that aren't on Harold's list, thereare ways to choose of them.~quacker88Problem12The decimal representation of consists of a string of zeros after the decimal point, followed by a and then several more digits. How many zeros are in that initial string of zeros after the decimal point?Solution 1Now we do some estimation. Notice that , which meansthat is a little more than . Multiplying itwith , we get that the denominator is about . Notice that whenwe divide by an digit number, there are zeros before the first nonzero digit. This means that when we divide by the digit integer , there are zeros in the initial string after the decimal point. -PCChessSolution 2First rewrite as . Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to findthe number of digits in .and memming (alternatively use the factthat ),digits.Our answer is .Solution 3 (Brute Force)Just as in Solution we rewrite as We thencalculate entirely by hand, first doing then multiplying that product by itself, resulting in Because this is digits,after dividing this number by fourteen times, the decimal point is beforethe Dividing the number again by twenty-six more times allows a stringof zeroes to be formed. -OreoChocolateSolution 4 (Smarter Brute Force)Just as in Solutions and we rewrite as We can then look at the number of digits in powersof . , , , , ,, and so on. We notice after a few iterations that every power of five with an exponent of , the number of digits doesn't increase. This means should have digits since thereare numbers which are from to , or digits total. This means our expression can be written as , where is in therange . Canceling gives , or zeroes before the since the number should start on where the one would be in . ~aop2014 Solution 5 (Logarithms)Problem13Andy the Ant lives on a coordinate plane and is currently at facingeast (that is, in the positive -direction). Andy moves unit and thenturns degrees left. From there, Andy moves units (north) and thenturns degrees left. He then moves units (west) and againturns degrees left. Andy continues his progress, increasing his distance each time by unit and always turning left. What is the location of the point at which Andy makes the th leftturn?Solution 1You can find that every four moves both coordinates decrease by 2. Therefore, both coordinates need to decrease by two 505 times. You subtract, giving you theanswer of ~happykeeperProblem14As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?Solution 1Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, , since B is the center of the semicircle with radius 1 that C lies on, , since B is the center of the semicircle with radius 1 that A lies on,and , as a regular hexagon has angles of 120,and is half of any angle in this hexagon. Now, using the sinelaw, , so . Since the angles in a triangle sum to 180, is also 60. Therefore, is an equilateral triangle with side lengths of 1.Since the area of a regular hexagon can be found with the formula , where is the side length of the hexagon, the area of this hexagonis . Since the area of an equilateral triangle can be foundwith the formula , where is the side length of the equilateral triangle,the area of an equilateral triangle with side lengths of 1 is . Since the area of a circle can be found with the formula , the area of a sixthof a circle with radius 1 is . In each sixth of the hexagon, thereare two equilateral triangles colored white, each with an area of , and onesixth of a circle with radius 1 colored white, with an area of . The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixthof the hexagon is , which equals , and the total areacolored white is , which equals . Since the area colored gray equals the total area of the hexagon minus the area colored white,the area colored gray is , whichequals .Solution 2First, subdivide the hexagon into 24 equilateral triangles with side length1:Now note that the entire shadedregion is just 6 times this part:The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of: The arc that is not included has an area of:Hence, the area ofthe shaded region in that section is For a final areaof:Problem15Steve wrote the digits , , , , and in order repeatedly from left to right, forming a list of digits, beginning He thenerased every third digit from his list (that is, the rd, th, th, digits from the left), then erased every fourth digit from the resulting list (that is, the th, th, th, digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions ?Solution 1After erasing every third digit, the list becomes repeated. After erasing every fourth digit from this list, the listbecomes repeated. Finally, after erasing every fifth digit from this list, the list becomes repeated. Since this list repeats every digits andsince are respectively in we have that the th, th, and st digits are the rd, th, and thdigits respectively. It follows that the answer is~dolphin7Problem16Bela and Jenn play the following game on the closed interval of the real number line, where is a fixed integer greater than . They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in theinterval . Thereafter, the player whose turn it is chooses a real numberthat is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?SolutionNotice that to use the optimal strategy to win the game, Bela must select themiddle number in the range and then mirror whatever number Jennselects. Therefore, if Jenn can select a number within the range, so can Bela. Jenn will always be the first person to run out of a number to choose, so theanswer is .Solution 2 (Guessing)First of all, realize that the value of should have no effect on the strategy at all. This is because they can choose real numbers, not integers, so even if is odd, for example, they can still go halfway. Similarly, there is no reason the strategy would change when .So we are left with (A) and (B). From here it is best to try out random numbers and try to find the strategy that will let Bela win, but if you can't find it, realize thatit is more likely the answer is since Bela has the first move and thus has more control.Problem17There are people standing equally spaced around a circle. Each person knows exactly of the other people: the people standing next to her or him, as well as the person directly across the circle. How many ways are there forthe people to split up into pairs so that the members of each pair know each other?SolutionLet us use casework on the number of diagonals.Case 1: diagonals There are ways: either pairs with , pairs with , and so on or pairs with , pairs with , etc.Case 2: diagonal There are possible diagonals to draw (everyone else pairs with the person next to them.Note that there cannot be 2 diagonals.Case 3: diagonalsNote that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction.Case 4: diagonals There is way to do this.Thus, in total there are possible ways. Problem18An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?SolutionLet denote that George selects a red ball and that he selects a blue one. Now, in order to get balls of each color, he needs more of both and .There are 6cases:(wecan confirm that there are only since ). However we canclump , ,and together since they are equivalent by symmetry.andLet's find the probability that he picks the balls in the order of .。

2019年AMC10B美国数学竞赛(真题加详解)

2019 AMC 10B Problems/Problem 1The following problem is from both the 2019 AMC 10B #1 and 2019 AMC 12B #1, so both problems redirect to this page.ProblemAlicia had two containers. The first was full of water and the second was empty. She poured all the water from the first container into the secondcontainer, at which point the second container was full of water. What is the ratio of the volume of the first container to the volume of the secondcontainer?Solution 1Let the first jar's volume be and the second's be . It is giventhat . We find thatWe already know that this is the ratio of the smaller to the larger volumebecause it is less thanSolution 2We can set up a ratio to solve this problem. If is the volume of the firstcontainer, and is the volume of the second container, then:Cross-multiplying allows us to get . Thus the ratio of the volume of the first container to the second containeris .~IronicNinjaSolution 3An alternate solution is to plug in some maximum volume for the firstcontainer - let's say , so there was a volume of in the first container, and then the second container also has a volume of , so youget . Thus the answer is .2019 AMC 10B Problems/Problem 2The following problem is from both the 2019 AMC 10B #2 and 2019 AMC 12B #2, so both problems redirect to this page.ProblemConsider the statement, "If is not prime, then is prime." Which of the following values of is a counterexample to this statement?SolutionSince a counterexample must be value of which is not prime, must be composite, so we eliminate and . Now we subtract from the remaining answer choices, and we see that the only time is not prime iswhen .2019 AMC 10B Problems/Problem 3 ProblemIn a high school with students, of the seniors play a musical instrument, while of the non-seniors do not play a musical instrument. Inall, of the students do not play a musical instrument. How many non-seniors play a musical instrument?Solution 1of seniors do not play a musical instrument. If we denote as the numberof seniors, thenThus there are non-seniors. Since 70% of the non-seniorsplay a musical instrument, .~IronicNinjaSolution 2Let be the number of seniors, and be the number of non-seniors.ThenMultiplying both sides by gives usAlso, because there are 500 students in total.Solving these system of equations give us , .Since of the non-seniors play a musical instrument, the answer issimply of , which gives us .Solution 3 (using the answer choices)We can clearly deduce that of the non-seniors do play an instrument, but,since the total percentage of instrument players is , the non-senior population is quite low. By intuition, we can therefore see that the answer isaround or . Testing both of these gives us the answer . 2019 AMC 10B Problems/Problem 4 ProblemAll lines with equation such that form an arithmeticprogression pass through a common point. What are the coordinates of that point?Solution 1If all lines satisfy the condition, then we can just plug in values for , ,and that form an arithmetic progression. Let's use , , ,and , , . Then the two lines we get are:Use elimination to deduce and plug this into one of the previous line equations. We get Thus the common point is .~IronicNinjaSolution 2We know that , , and form an arithmetic progression, so if the commondifference is , we can say Now wehave , and expandinggives Factoringgives . Since this must always be true (regardless of the values of and ), we musthave and , so and the common point is .2019 AMC 10B Problems/Problem 5 ProblemTriangle lies in the first quadrant. Points , , and are reflected across the line to points , , and , respectively. Assume that none of the vertices of the triangle lie on the line . Which of the following statements is not always true?Triangle lies in the first quadrant.Triangles and have the same area.The slope of line is .The slopes of lines and are the same.Lines and are perpendicular to each other.SolutionLet's analyze all of the options separately.: Clearly is true, because a point in the first quadrant will have non-negative - and -coordinates, and so its reflection, with the coordinates swapped, will also have non-negative - and -coordinates.: The triangles have the same area,since and are the same triangle (congruent). More formally, we can say that area is invariant under reflection.: If point has coordinates , then will have coordinates .The gradient is thus , so this is true. (We know since the question states that none of the points , , or lies on the line , so there is no risk of division by zero).: Repeating the argument for , we see that both lines have slope , so this is also true.: By process of elimination, this must now be the answer. Indeed, ifpoint has coordinates and point has coordinates ,then and will, respectively, have coordinates and . The product of the gradientsof and is , so in fact these lines are never perpendicular to each other (using the "negative reciprocal" condition for perpendicularity).Thus the answer is .CounterexamplesIf and , then the slopeof , , is , while the slope of , ,is . is the reciprocal of , but it is not the negative reciprocal of . To generalize, let denote thecoordinates of point , let denote the coordinates of point ,let denote the slope of segment , and let denote the slope of segment . Then, the coordinates of are , andof are . Then, ,and .If and , , and in these cases, the condition is false.2019 AMC 10B Problems/Problem 6The following problem is from both the 2019 AMC 10B #6 and 2019 AMC 12B #4, so both problems redirect to this page.ProblemThere is a real such that .What is the sum of the digits of ?Solution 1Solving by the quadraticformula,(since clearly ). The answer is therefore .~IronicNinjaSolution 2Dividing both sidesby givesSince is non-negative, . The answer is .Solution 3Dividing both sides by as beforegives . Now factorout , giving . By considering the prime factorization of , a bit of experimentation givesus and , so , so the answeris .2019 AMC 10B Problems/Problem 7The following problem is from both the 2019 AMC 10B #7 and 2019 AMC 12B #5, so both problems redirect to this page.ProblemEach piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either pieces of red candy, pieces of green candy, pieces of blue candy, or pieces of purple candy. A piece of purple candy costs cents. What is the smallest possible value of ?Solution 1If he has enough money to buy pieces of red candy, pieces of green candy, and pieces of blue candy, then the smallest amount of money hecould have is cents. Since a piece of purple candy costs cents, the smallest possible valueof is .~IronicNinjaSolution 2We simply need to find a value of that is divisible by , , and .Observe that is divisible by and , but not . is divisible by , , and , meaning that we have exact change (in this case, cents) to buy each type of candy, so the minimum valueof is .2019 AMC 10B Problems/Problem 8ProblemThe figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?Solution 1We notice that the square can be split into congruent smaller squares, with the altitude of the equilateral triangle being the side of this smaller square. Therefore, the area of each shaded part that resides within a square is the total area of the square subtracted from each triangle (which has already been split in half). When we split an equilateral triangle in half, we gettwo triangles. Therefore, the altitude, which is also the side length of one of the smaller squares, is . We can then compute the areaof the two triangles as .The area of the each small squares is the square of the side length,i.e. . Therefore, the area of the shaded region in each of the four squares is .Since there are of these squares, we multiply this by toget as our answer.Solution 2We can see that the side length of the square is by considering thealtitude of the equilateral triangle as in Solution 1. Using the Pythagorean Theorem, the diagonal of the square isthus . Because of this, the height of one ofthe four shaded kites is . Now, we just need to find the length of that kite. By the Pythagorean Theorem again, this lengthis . Nowusing , the area of one of the four kitesis . 2019 AMC 10B Problems/Problem 9 ProblemThe function is defined by for all real numbers , where denotes the greatest integer less than or equal to the real number . What is the range of ?Solution 1There are four cases we need to consider here.Case 1: is a positive integer. Without loss of generality, assume . Then .Case 2: is a positive fraction. Without loss of generality, assume .Then .Case 3: is a negative integer. Without loss of generality, assume . Then .Case 4: is a negative fraction. Without loss of generality, assume . Then .Thus the range of the function is .~IronicNinja, edited by someone elseSolution 2It is easily verified that when is an integer, is zero. We therefore need only to consider the case when is not an integer.When is positive, , soWhen is negative, let be composed of integer part and fractional part (both ):Thus, the range of f is .Note: One could solve the case of as a negative non-integer in thisway:2019 AMC 10B Problems/Problem 10The following problem is from both the 2019 AMC 10B #10 and 2019 AMC 12B #6, so both problems redirect to this page.ProblemIn a given plane, points and are units apart. How manypoints are there in the plane such that the perimeterof is units and the area of is square units?Solution 1Notice that whatever point we pick for , will be the base of thetriangle. Without loss of generality, letpoints and be and , since for any other combination of points, we can just rotate the plane to makethem and under a new coordinate system. When we pickpoint , we have to make sure that its -coordinate is , because that's the only way the area of the triangle can be .Now when the perimeter is minimized, by symmetry, we put in the middle, at . We can easily see that and will bothbe . The perimeter of this minimal triangleis , which is larger than . Since the minimum perimeter is greater than , there is no triangle that satisfies the condition, givingus .~IronicNinjaSolution 2Without loss of generality, let be a horizontal segment of length .Now realize that has to lie on one of the lines parallel to andvertically units away from it. But is already 50, andthis doesn't form a triangle. Otherwise, without loss ofgenerality, . Dropping altitude , we have a righttriangle with hypotenuse and leg , which is clearly impossible, again giving the answer as .2019 AMC 10B Problems/Problem 11 ProblemTwo jars each contain the same number of marbles, and every marble is either blue or green. In Jar the ratio of blue to green marbles is , and the ratio of blue to green marbles in Jar is . There are green marbles in all. How many more blue marbles are in Jar than in Jar ?SolutionCall the number of marbles in each jar (because the problem specifies that they each contain the same number). Thus, is the number of green marbles in Jar , and is the number of green marbles in Jar .Since , we have , so thereare marbles in each jar.Because is the number of blue marbles in Jar , and is the number of blue marbles in Jar , there are more marbles in Jar than Jar . This means the answer is .2019 AMC 10B Problems/Problem 12 ProblemWhat is the greatest possible sum of the digits in the base-seven representation of a positive integer less than ?Solution 1Observe that . To maximize the sum of the digits, we want as many s as possible (since is the highest value in base ), and this will occur with either of the numbers or . Thus, the answeris .~IronicNinja, edited by some peopleNote: the number can also be , which will also give the answer of . Solution 2Note that all base numbers with or more digits are in fact greaterthan . Since the first answer that is possible using a digit number is , we start with the smallest base number that whose digits sum to ,namely . But this is greater than , so we continue bytrying , which is less than 2019. So the answer is .2019 AMC 10B Problems/Problem 13The following problem is from both the 2019 AMC 10B #13 and 2019 AMC 12B #7, so both problems redirect to this page.ProblemWhat is the sum of all real numbers for which the median of thenumbers and is equal to the mean of those five numbers?SolutionThe mean is .There are three possibilities for the median: it is either , , or .Let's start with .has solution , and the sequenceis , which does have median , so this is a valid solution.Now let the median be .gives , so the sequence is , which has median , so this is not valid.Finally we let the median be ., and the sequence is , which has median . This case is therefore again not valid.Hence the only possible value of is2019 AMC 10B Problems/Problem 14 ProblemThe base-ten representationfor is , where , ,and denote digits that are not given. What is ?Solution 1We can figure out by noticing that will end with zeroes, as there are three s in its prime factorization. Next, we use the fact that is a multiple of both and . Their divisibility rules (see Solution 2) tell usthat and that . By inspection, we see that is a valid solution. Therefore the answer is .Solution 2 (similar to Solution 1)We know that and are both factors of . Furthermore, we knowthat , because ends in three zeroes (see Solution 1). We can simply use the divisibility rules for and for this problem to find and . For to be divisible by , the sum of digits must simply be divisible by .Summing the digits, we get that must be divisible by . Thisleaves either or as our answer choice. Now we test for divisibility by . For a number to be divisible by , the alternating sum must be divisibleby (for example, with the number , ,so is divisible by ). Applying the alternating sum test to this problem, we see that must be divisible by 11. By inspection, we can see that this holds if and . The sumis .2019 AMC 10B Problems/Problem 15ProblemRight triangles and , have areas of 1 and 2, respectively. A side of iscongruent to a side of , and a different side of is congruent to a differentside of . What is the square of the product of the lengths of the other (third)side of and ?Solution 1First of all, let the two sides which are congruent be and , where . The only way that the conditions of the problem can be satisfied is if is the shorter leg of and the longer leg of , and is the longer leg of and thehypotenuse of .Notice that this means the value we are looking for is the squareof , which is just . The area conditions give us twoequations: and .This means that and that .Taking the second equation, we get , sosince , .Since , we get .The value we are looking for is just so the answer is .Solution bySolution 2Like in Solution 1, we have and .Squaring both equations yields and .Let and . Then ,and , so .We are looking for the value of , so the answeris .Solution 3Firstly, let the right triangles be and ,with being the smaller triangle. As in Solution 1,let and . Additionally,let and .We are given that and , sousing , we have and . Dividing the twoequations, we get = , so .Thus is a right triangle, meaningthat . Now by the Pythagorean Theoremin ,.The problem requires the square of the product of the third side lengths of each triangle, which is . By substitution, we seethat = . We alsoknow.Since we want , multiplying both sides by getsus . Now squaringgives .2019 AMC 10B Problems/Problem 16ProblemIn with a right angle at , point lies in the interior of andpoint lies in the interior of so that and the ratio . What is the ratioSolution 1Without loss of generality, let and . Let and .As and areisosceles, and .Then , so isa triangle with .Then , and is a triangle.In isosceles triangles and , drop altitudesfrom and onto ; denote the feet of these altitudesby and respectively. Then by AAA similarity, so we get that ,and . Similarly we get ,and .Solution 2Let , and . (For thissolution, is above , and is to the right of ). Also let ,so , whichimplies . Similarly, , whichimplies . This further impliesthat .Now we seethat. Thus is a right triangle, with side lengths of , , and (by the Pythagorean Theorem, or simply the Pythagorean triple ).Therefore (by definition), ,and . Hence (by thedouble angle formula), giving .By the Law of Cosines in , if , wehaveNow . Thus theanswer is .~IronicNinjaSolution 3Draw a nice big diagram and measure. The answers to this problem are not very close, so it is quite easy to get to the correct answer by simply drawing a diagram. (Note: this strategy should only be used as a last resort!)2019 AMC 12B Problems/Problem 13(Redirected from 2019 AMC 10B Problems/Problem 17)The following problem is from both the 2019 AMC 10B #17 and 2019 AMC 12B #13, so both problems redirect to this page.ProblemA red ball and a green ball are randomly and independently tossed into binsnumbered with the positive integers so that for each ball, the probability that it is tossed into bin is for What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?Solution 1By symmetry, the probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher-numbered bin. Clearly, the probability of both landing in the same binis (by the geometric series sum formula). Therefore the other two probabilities have to bothbe .Solution 2Suppose the green ball goes in bin , for some . The probability of this occurring is . Given that this occurs, the probability that the red ball goesin a higher-numbered bin is (by thegeometric series sum formula). Thus the probability that the green ball goesin bin , and the red ball goes in a bin greater than , is . Summing from to infinity, we getwhere we again used the geometric series sum formula. (Alternatively, if this sum equals , then by writing out the terms andmultiplying both sides by , we see , which gives .) Solution 3The probability that the two balls will go into adjacent binsisby the geometric series sum formula. Similarly, the probability that the two balls will go into bins that have a distance of from each otheris(again recognizing a geometric series). We can see that each time we add a bin between the two balls, the probability halves. Thus, our answeris , which, by the geometric series sum formula,is .-fidgetboss_4000Solution 4 (quick, conceptual)Define a win as a ball appearing in higher numbered box.Start from the first box.There are possible results in the box: Red, Green, Red and Green, ornone, with an equal probability of for each. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution, so if is the probability that Red wins, we canwrite : there is a probability that "Red" wins immediately,a probability in the cases "Green" or "Red and Green", and in the "None"case (which occurs with probability), we then start again, giving the sameprobability . Hence, solving the equation, we get . Solution 5Write out the infinite geometric series as , . To find the probablilty that red goes in a higher-numbered bin than green, we can simply remove all odd-index terms (i.e term , term , etc.), and then sum theremaining terms - this is in fact precisely equivalent to the method of Solution2. Writing this out as another infinite geometric sequence, we are leftwith . Summing, we get2019 AMC 10B Problems/Problem 18 ProblemHenry decides one morning to do a workout, and he walks of the way from his home to his gym. The gym is kilometers away from Henry's home. At thatpoint, he changes his mind and walks of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks ofthe distance from there back toward the gym. If Henry keeps changing his mindwhen he has walked of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point kilometers from home and a point kilometers fromhome. What is ?Solution 1Let the two points that Henry walks in between be and , with being closer to home. As given in the problem statement, the distances of thepoints and from his home are and respectively. By symmetry, the distance of point from the gym is the same as the distance from home to point . Thus, . In addition, when he walks from point to home, he walks of the distance, ending at point . Therefore, we knowthat . By substituting, we get .Adding these equations now gives . Multiplying by , we get ,so .Solution 2 (not rigorous)We assume that Henry is walking back and forth exactly betweenpoints and , with closer to Henry's home than . Denote Henry's home as a point and the gym as a point .Then and ,so .Therefore,. 2019 AMC 10B Problems/Problem 19The following problem is from both the 2019 AMC 10B #19 and 2019 AMC 12B #14, so both problems redirect to this page.ProblemLet be the set of all positive integer divisors of How many numbers are the product of two distinct elements ofSolutionThe prime factorization of is . Thus, we choose twonumbers and where and, whose product is ,where and .Notice that this is analogous to choosing a divisorof , whichhas divisors. However, some of thedivisors of cannot be written as a product of two distinct divisorsof , namely: , , , and . The last twocannot be so written because the maximum factor of containingonly s or s (and not both) is only or . Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require or . This gives candidatenumbers. It is not too hard to show that every number of the form ,where , and are not both or , can be written asa product of two distinct elements in . Hence the answer is . 2019 AMC 10B Problems/Problem 20The following problem is from both the 2019 AMC 10B #20 and 2019 AMC 12B #15, so both problems redirect to this page.ProblemAs shown in the figure, line segment is trisected bypoints and so that Three semicirclesof radius and have their diameters on and are tangent to line at and respectively. A circle ofradius has its center on The area of the region inside the circle butoutside the three semicircles, shaded in the figure, can be expressed in the form where and are positive integersand and are relatively prime. What is ?SolutionDivide the circle into four parts: the top semicircle (); the bottom sector (), whose arc angle is because the large circle's radius is and the short length (the radius of the smaller semicircles) is , givinga triangle; the triangle formed by the radii of and the chord (), and the four parts which are the corners of a circle inscribedin a square (). Then the area is (in , wefind the area of the shaded region above the semicircles but below the diameter, and in we find the area of the bottom shaded region).The area of is .The area of is .For the area of , the radius of , and the distance of (the smaller semicircles' radius) to , creates two triangles,so 's area is .The area of is .Hence, finding , the desired areais , so the answeris2019 AMC 10B Problems/Problem 21 ProblemDebra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?SolutionWe firstly want to find out which sequences of coin flips satisfy the given condition. For Debra to see the second tail before the seecond head, her first flip can't be heads, as that would mean she would either end with double tails before seeing the second head, or would see two heads before she sees two tails. Therefore, her first flip must be tails. The shortest sequence of flips by which she can get two heads in a row and see the second tail before she sees the secondhead is , which has a probability of . Furthermore, she can prolong her coin flipping by adding an extra , which itself has aprobability of . Since she can do this indefinitely, this gives an infinite geometric series, which means the answer (by the geometric series sum formula)is .Solution 2 (Easier)Note that the sequence must start in THT, which happens with probability. Now, let be the probability that Debra will get two heads in a row after flippingTHT. Either Debra flips two heads in a row immediately (probability ), or flips a head and then a tail and reverts back to the "original position" (probability ). Therefore, , so , so our final answeris . -Stormersyle get rect2019 AMC 10B Problems/Problem 22The following problem is from both the 2019 AMC 10B #22 and 2019 AMC 12B #19, so both problems redirect to this page.ProblemRaashan, Sylvia, and Ted play the following game. Each starts with . A bell rings every seconds, at which time each of the players who currently have money simultaneously chooses one of the other two playersindependently and at random and gives to that player. What is theprobability that after the bell has rung times, each player willhave ? (For example, Raashan and Ted may each decide to give to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which pointRaashan will have , Sylvia will have , and Ted will have , and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their to, and the holdings will be the same at the end of the second round.)SolutionOn the first turn, each player starts off with . Each turn after that, there are only two possibilities: either everyone stays at , which we will writeas , or the distribution of moneybecomes in some order, which we writeas . We will consider these two states separately.In the state, each person has two choices for whom to give their dollar to, meaning there are possible ways that the money canbe rearranged. Note that there are only two ways that we canreach again:1. Raashan gives his money to Sylvia, who gives her money to Ted, who gives his money to Raashan.2. Raashan gives his money to Ted, who gives his money to Sylvia, who gives her money to Raashan.Thus, the probability of staying in the state is , while theprobability of going to the state is (we can check that the 6 other possibilities lead to )In the state, we will label the person with as person A, the person with as person B, and the person with as person C.Person A has two options for whom to give money to, and person B has 2 options for whom to give money to, meaning there are。

AMC 10B 试题及答案解析

2010 AMC 10B1 . What is ?SolutionWe first expand the first term, simplify, and then compute to get an answer of .2 、Makarla attended two meetings during her -hour work day. The first meeting took minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?SolutionThe percentage of her time spent in meetings is the total amount of time spent in meetings divided by the length of her workday.The total time spent in meetings isTherefore, the percentage is3、A drawer contains red, green, blue, and white socks with at least 2 of each color. What is the minimum number of socks that must be pulled from the drawer to guarantee a matching pair?SolutionAfter you draw socks, you can have one of each color, so (according to the pigeonhole principle), if you pull then you will be guaranteed a matching pair.4 、For a real number , define to be the average of and . What is ?SolutionThe average of two numbers, and , is defined as . Thus the average of and would be . With that said, we need to find the sum when we plug, , and into that equation. So:.5 、A month with days has the same number of Mondays and Wednesdays.How many of the seven days of the week could be the first day of this month?Solution（B）. 36 、A circle is centered at , is a diameter and is a point on the circle with . What is the degree measure of ?SolutionAssuming the reader is not readily capable to understand how will always be right, the I will continue with an easily understandable solution. Since is the center, are all radii, they are congruent. Thus, and are isosceles triangles. Also, note thatand are supplementary, then . Since is isosceles, then . They also sum to , so each angle is .7 、A triangle has side lengths , , and . A rectangle has width and area equal to the area of the rectangle. What is the perimeter of this rectangle?SolutionThe triangle is isosceles. The height of the triangle is therefore given byNow, the area of the triangle isWe have that the area of the rectangle is the same as the area of the triangle, namely 48. We also have the width of the rectangle: 4.The length of the rectangle therefor is:The perimeter of the rectangle then becomes:The answer is:8 、A ticket to a school play cost dollars, where is a whole number. A group of 9th graders buys tickets costing a total of $, and a group of 10th graders buys tickets costing a total of $. How many values for are possible?SolutionWe see how many common integer factors 48 and 64 share. Of the factors of 48 - 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48; only 1, 2, 4, 8, and 16 are factors of 64. So there are possibilities for the ticket price.9 、Lucky Larry's teacher asked him to substitute numbers for , , , , and in the expression and evaluate the result. Larry ignored the parenthese but added and subtracted correctly andobtained the correct result by coincidence. The number Larry sustitued for , , , and were , , , and , respectively. What number did Larry substitude for ?SolutionSimplify the expression . I recommend to start with the innermost parenthesis and work your way out.So you get:Henry substituted with respectively.We have to find the value of , such that(the same expression without parenthesis).Substituting and simplifying we get:So Henry must have used the value for .Our answer is:10、Shelby drives her scooter at a speed of miles per hour if it is not raining, and miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of miles in minutes. How many minutes did she drive in the rain?SolutionWe know thatSince we know that she drove both when it was raining and when it was not and that her total distance traveled is miles.We also know that she drove a total of minutes which is of an hour. We get the following system of equations, where is the time traveled when it was not raining and is the time traveled when it was raining:Solving the above equations by multiplying the second equation by 30 and subtracting the second equation from the first we get:We know now that the time traveled in rain was of an hour, which is minutesSo, our answer is:11 、A shopper plans to purchase an item that has a listed price greater than $and can use any one of the three coupns. Coupon A givesoff the listed price, Coupon B gives $off the listed price, and Coupon C gives off the amount by which the listed price exceeds $.Let and be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is −?SolutionLet the listed price be , whereCoupon A saves us:Coupon B saves us:Coupon C saves us:Now, the condition is that A has to be greater than or equal to either B or C which give us the following inequalities:We see here that the greatest possible value for p is and the smallest isThe difference between and isOur answer is:12 、At the beginning of the school year, of all students in Mr. Wells' math class answered "Yes" to the question "Do you love math", andanswered "No." At the end of the school year, answered "Yes" and answerws "No." Altogether, of the students gave a differentanswer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of ?SolutionThe minimum possible value occurs when of the students who originally answered "No." answer "Yes." In this case,The maximum possible value occurs when of the students who originally answered "Yes." answer "No." and the of the students who originally answered "No." answer "Yes." In this case,Subtract to obtain an answer of13 、What is the sum of all the solutions of ?SolutionCase 1:Case 1a:Case 1b:Case 2:Case 2a:Case 2b:Since an absolute value cannot be negative, we exclude . The answer is14 、The average of the numbers and is . What is ?SolutionWe must find the average of the numbers from to and in terms of . The sum of all these terms is . We must divide this by the total number of terms, which is . We get: . This is equal to , as stated in the problem. We have: . We can now cross multiply. This gives:This gives us our answer.15 、On a -question multiple choice math contest, students receive points for a correct answer, points for an answer left blank, and point for an incorrect answer. Jesse’s total score on the contest was . What is the maximum number of questions that Jesse could have answered correctly?SolutionLet be the amount of questions Jesse answered correctly, be the amount of questions Jesse left blank, and be the amount of questions Jesse answered incorrectly. Since there were questions on the contest, . Since his total score was , . Also,. We can substitute this inequality into the previous equation to obtain another inequality:. Since is an integer, the maximum value for is .16 、A square of side length and a circle of radius share the same center. What is the area inside the circle, but outside the square?Solution(B)17 、Every high school in the city of Euclid sent a team of students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed th and th, respectively. How many schools are in the city?SolutionLet the be the number of schools, be the number of contestants, and be Andrea's score. Since the number of participants divided by three isthe number of schools, . Andrea received a higher score than her teammates, so . Since is the maximum possible median, then is the maximum possible number ofparticipants. Therefore, . This yields thecompound inequality: . Since a set with an even number of elements has a median that is the average of the two middle terms, an occurrence that cannot happen in this situation, cannot be even.is the only other option.18 、Positive integers , , and are randomly and independently selected with replacement from the set . What is the probability that is divisible by ?Solution（E）13/2719 、A circle with center has area . Triangle is equilateral,is a chord on the circle, , and point is outside . What is the side length of ?Solution(B)620 、Two circles lie outside regular hexagon . The first is tangent to , and the second is tangent to . Both are tangent to lines and . What is the ratio of the area of the second circle to that of the first circle?SolutionA good diagram here is very helpful.The first circle is in red, the second in blue. With this diagram, we can see that the first circle is inscribed in equilateral triangle while the second circle is inscribed in . From this, it's evident that the ratio of the red area to the blue area is equal to the ratio of the areas of triangles toSince the ratio of areas is equal to the square of the ratio of lengths, weknow our final answer is From the diagram, we can see that this isThe letter answer is D21 、A palindrome between and is chosen at random. What is the probability that it is divisible by ?SolutionView the palindrome as some number with form (decimal representation): . But because the number is a palindrome, . Recombining this yields . 1001 is divisible by 7, which means that as long as , the palindrome will be divisible by 7. This yields 9 palindromes out of 90 () possibilities for palindromes. However, if , then this givesanother case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to22 、Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?SolutionWe can count the total number of ways to distribute the candies (ignoring the restrictions), and then subtract the overcount to get the answer.Each candy has three choices; it can go in any of the three bags.Since there are seven candies, that makes the total distributionsTo find the overcount, we calculate the number of invalid distributions: the red or blue bag is empty.The number of distributions such that the red bag is empty is equal to , since it's equivalent to distributing the 7 candies into 2 bags.We know that the number of distributions with the blue bag is empty will be the same number because of the symmetry, so it's also .The case where both the red and the blue bags are empty (all 7 candies are in the white bag) are included in both of the above calculations, and this case has only distribution.The total overcount isThe final answer will beThat makes the letter choice C23 、The entries in a array include all the digits from through ,arranged so that the entries in every row and column are in increasing order. How many such arrays are there?Solution（D）6024 、A high school basketball game between the Raiders and Wildcatswas tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than points. What was the total number of points scored by the two teams in the first half?SolutionRepresent the teams' scores as: andWe have Manipulating this, we can get, orSince both are increasing sequences, . We can check cases up to because when , we get . When▪▪▪Checking each of these cases individually back into the equation, we see that only when a=5 and n=2, we get an integer value for m, which is 9. The original question asks for the first half scores summed, so we must find25 、Let , and let be a polynomial with integer coefficients such that, and.What is the smallest possible value of ?SolutionThere must be some polynomial such thatThen, plugging in values of we getThus, the least value of must be the . Solving, we receive , so our answer is .。

2010 AMC10A试题(含答案)

2010 AMC10A ProblemsQ1. Mary’s top book shelf holds ﬁve books with the follo wing widths, in centimeters:6,12, 1, 2.5, and 10. What is the average book width, in centimeters?A) 1 B) 2 C) 3 D) 4 E)5Q2. Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?A)54B)43C)32D) 2 E)3Q3.Tyrone had 97 marbles and Eric had 11 marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?A)3 B)13 C)18 D) 25 E)29 Q4.A book that is to be recorded onto compact discs takes 412 minutes to read aloud. Each disc can hold up to 56 minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?A)50.2 B)51.5 C)52.4 D) 53.8 E)55.2Q5.The area of a circle whose circumference is 24π is kπ. What is the value of k?A)6 B)12 C)24 D) 36 E)144Q6. For positive numbers x and y the operation ♠(x, y) is deﬁned as1(,)x y xy♠=-What is ♠(2, ♠(2, 2))?A)23B)1 C)43D)53E)2Q7.Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?A)1 B) D)2 E)Q8.Tony works 2 hours a day and is paid $0.50 per hour for each full year of his age. During a six month period Tony worked 50 days and earned $630. How old was Tony at the end of the six month period?A)9 B)11 C)12 D) 13 E)14Q9.A palindrome, such as 83438, is a number that remains the same when its digits are reversed. The numbers x and x + 32 are three -digit and four -digit palindromes, respectively. What is the sum of the digits of x ?A)20 B)21 C)22 D) 23 E)24Q11.The length of the interval of solutions of the inequality 23a x b ≤+≤is 10.What is b a -？A)6 B)10 C)15 D) 20 E)30Q12.Logan is constructing a scaled model of his town. The city’s water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan’s miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?A)0.04 B)0.4π C)0.4 D) 4πE)4Q13.Angelina drove at an average rate of 80 kmh and then stopped 20 minutes for gas. After the stop, she drove at an average rate of 100 kmh. Altogether she drove 250 km in a total trip time of 3 hours including the stop. Which equation could be used to solve for the time t in hours that she drove before her stop? A)880100()2503t t +-= B)80250t = C)100250t = D)90250t = E)880()1002503t t -+=Q14.Triangle ABC has 2AB AC =⋅. Let D and E be on AB and BC , respectively, such that ∠BAE = ∠ACD . Let F be the intersection of segments AE and CD , and suppose that △CFE is equilateral. What is ∠ACB ?A)60° B)75° C)90° D) 105° E)120°Q15.In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.Brian: “Mike and I are diﬀerent species.”Chris: “LeRoy is a frog.”LeRoy: “Chris is a frog.”Mike: “Of the four of us, at least two are toads.”How many of these amphibians are frogs?A)0 B)1 C)2 D) 3 E)4Q16.Nondegenerate △ABC has integer side lengths, BD is an angle bisector, AD = 3, and DC = 8. What is the smallest possible value of the perimeter?A)30 B)33 C)35 D) 36 E)37 Q17.A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?A)7 B)8 C)10 D) 12 E)15Q18.Bernardo randomly picks 3 distinct numbers from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set {1, 2, 3, 4, 5, 6, 7, 8} and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo’s number is larger than Silvia’s number?A)4772B)3756C)23D)4972E)3956Q19.Equiangular hexagon ABCDEF has side lengths AB = CD = EF = 1 and BC = DE = F A = r. The area of △ACE is 70% of the area of the hexagon. What is the sum of all possible values of r?A)3B)103C)4 D)174E) 6Q20.A ﬂy trapped inside a cubical box with side length 1 meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either ﬂy or crawl in a straight line. What is the maximum possible length, in meters, of its path?A)4+B)2+C)2+D)E)Q21.The polynomial 322010-+-has three positive integer roots. What isx ax bxthe smallest possible value of a?A)78 B)88 C)98 D)108 E)118Q22.Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?A) 28 B) 56 C) 70 D) 84 E) 140Q23.Each of 2010 boxes in a line contains a single red marble, and for 1 ≤k≤2010, the box in the k th position also contains k white marbles. Isabella begins at the ﬁrst box and successively draws a single marble at random from each box, in order. She stops when she ﬁrst draws a red marble. Let P(n) be the probability that Isabella stops after drawing exactly n marbles. What is the smallest value of n for which1()P n<?2010A) 45 B) 63 C) 64 D) 201 E) 1005Q24.The number obtained from the last two nonzero digits of 90! is equal to n. What is n?A) 12 B) 32 C) 48 D) 52 E) 68Q25.Jim starts with a positive integer n and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with n = 55, then his sequence contains 5 numbers:Let N be the smallest number for which Jim’s sequence has 8 numbers. What is the units digit of N?A) 1 B) 3 C) 5 D) 7 E) 9AMC10A 2010中文解析Q1. Mary’s top book shelf holds ﬁve books with the follo wing widths, in centimeters:6,12, 1, 2.5, and 10. What is the average book width, in centimeters?A) 1 B) 2 C) 3 D) 4 E)5翻译：Mary最上面的书架上有五本书，宽度分别为: 6厘米、1厘米、2.5厘米和10厘米。

2021年Spring AMC10B真题含答案

2021Spring AMC10BProblem1How many integer values of satisfy?有多少个整数值满足？Problem2What is the value of?算式的值是多少？Problem3In an after-school program for juniors and seniors,there is a debate team with an equal number of students from each class on the team.Among the students in the program,of the juniorsand of the seniors are on the debate team.How many juniors are in the program?在面向高二和高三学生的某课外课程中，有一个辩论队，每个年级参加辩论队的人数相同。

已知在参加这个课程的28名同学中，有25％的高二学生和10％的高三学生参加了辩论队。

问这个课程中有多少名高二的学生？At a math contest,students are wearing blue shirts,and another students are wearing yellowshirts.The students are assigned into pairs.In exactly of these pairs,both students arewearing blue shirts.In how many pairs are both students wearing yellow shirts?在一次数学竞赛中，57名学生穿着蓝色衬衫，另75外名学生穿着黄色衬衫。

132名学生被分成了66对。

这其中恰好有23对，每对的两名学生都穿着蓝色衬衫。

问两名学生都穿着黄色衬衫的对有多少个？Problem5The ages of Jonie's four cousins are distinct single-digit positive integers.Two of the cousins'ages multiplied together give,while the other two multiply to.What is the sum of the ages of Jonie's four cousins?Jonie的四个表兄弟的年龄是不同的一位正整数。

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Problem
What is the sum of all the solutions of ?
方程所有根的和？
Solution
We evaluate this in cases:我们分几种情况来讨论：
Case 1
情况1: X<30
When we are going to have . When we are going to have and when we are going to
have . Therefore we
have .
当X<30时，则；当X>0，我们就可以去除绝对值符号：，而如果X<0，则，
因此这时可简化：
进而：
Subcase 1
When we are going to have . When this happens, we can express as . Therefore we
get . We check if is in the domain of the numbers that we put into this subcase, and it is, since . Therefore is one possible solution.
子情况：
当，则，这时，＝
因此：
检验：是否在子情况的值域内。

因为所以20是它的一个根。

Subcase 2
When we are going to have , therefore can be expressed in the form . We have the
equation . Since is less
than , is another possible solution.
子情况2：
当，则因此＝，则
因12〈15，12是另一个可能的根
Case 2 :
When , . When we can express this in the form . Therefore we have . This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we have
情况2：
当，则，因此：
We have now evaluated all the cases, and found the solution to
be which have a sum of 我们检验所有的根，其可能根的集合为故和为92。