AMC10美国数学竞赛讲义
AMC10美国数学竞赛讲义AMC 中的数论问题1：Remember the prime between 1 to 100：2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 7173 79 83 89 912：Perfect number:Let P is the prime number.if 21p - is also the prime number. then 1(21)2p p --is the perfect number. For example:6,28,496.3: Let ,0n abc a =≠ is three digital integer .if333n a b c =++ Then the numbern is called Daffodils number . There are only four numbers: 153 370 371 407Let ,0n abcd a =≠ is four digital integer .if 4444d n a b c +=++Then the number n is called Roses number . There are only three numbers:1634 8208 94744：The Fundamental Theorem of ArithmeticEvery natural number n can be written as a product of primes uniquely up to order.kn=∏p i r ii=15：Suppose that a and b are integers with b =0.Then there exists unique integers q and r such that 0 ≤ r< |b| and a = bq + r.6：(1)Greatest Common Divisor: Let gcd (a, b) = max {d ∈ Z: d | a and d | b}.For any integers a and b, we havegcd(a, b) = gcd(b, a) = gcd(±a, ±b) =gcd(a, b − a) = gcd(a, b + a).For example: gcd(150, 60) = gcd(60, 30) = gcd(30, 0) = 30(2)Least common multiple:Letlcm(a,b)=min{d∈Z: a | d and b | d }.(3)We have that: ab= gcd(a, b) lcm(a,b)7：Congruence modulo nIf ,0a b mq m -=≠,then we call a congruence bmodulo m and we rewrite mod a b m ≡.(1)Assume a,b,c,d,m ,k ∈Z (k >0,m ≠0).If a ≡b mod m,c ≡d mod m then we havemod a c b d m ±≡± , mod ac bd m ≡ , mod k k a b m ≡(2) The equation ax ≡ b (mod m) has a solution if and only if gcd(a, m) divides b.8：How to find the unit digit of some specialintegers(1)How many zero at the end of !nFor example, when 100n =, Let N be the number zero at the end of 100!then 10010010020424525125N ⎡⎤⎡⎤⎡⎤=++=+=⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦(2) ,,a n Z ∈Find the unit digit n a . For example, when 100,3n a ==9：Palindrome, such as 83438, is a number that remains the same when its digits are reversed.There are some number not only palindrome but 112=121，222=484，114=14641(1)Some special palindrome n that 2n is also palindrome. For example :222221111121111123211111123432111111111112345678987654321=====(2)How to create a palindrome? Almost integer plus the number of its reversed digits and repeat it again and again. Then we get a palindrome. For example:87781651655617267266271353135335314884+=+=+=+=But whether any integer has this Property has yet to prove(3) The palindrome equation means that equation from left to right and right to left it all set up.For example ：1242242112231132211121241388888831421211⨯=⨯⨯=⨯⨯==⨯Let ab and cde are two digital and three digital integers. If the digits satisfy the⨯=⨯.⨯=⨯=+≤, then ab cde edc ba,,9a cb e dc e d10: Features of an integer divisible by some prime numberIf n is even，then 2|n一个整数n的所有位数上的数字之和是3（或者9）的倍数，则n被3（或者9）整除一个整数n的尾数是零，则n被5整除一个整数n的后三位与截取后三位的数值的差被7、11、13整除，则n被7、11、13整除一个整数n的最后两位数被4整除，则n被4整除一个整数n的最后三位数被8整除，则n被8整除一个整数n 的奇数位之和与偶数位之和的差被11整除，则n 被11整除11. The number Theoretic functionsIf 312123t r r r r t n p p p p =(1) {}12()#0:|(1)(1)(1)t n a a n r r r χ=>=+++(2) 12222111222|()(1)(1)(1)t r r r t t t a n n a p p p p p p p p p δ==+++++++++∑(3){}11221111122()#:,gcd(,)1()()()t tr r r r r r t t n a N a n a n p p p p p p φ---=∈≤==--- For example: 2(12)(23)(21)(11)6χχ=⋅=++=22(12)(23)(122)(13)28δδ=⋅=+++=22(12)(23)(22)(31)4φφ=⋅=--= Exercise 1. The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?(A) 4 (B) 5 (C) 6 (D) 7 (E) 83. For the positive integer n, let <n> denote the sum of all the positive divisors of n with the exception of n itself. For example, <4>=1+2=3and <12>=1+2+3+4+6=16. What is <<<6>>>?(A) 6 (B) 12 (C) 24 (D) 32 (E) 368. What is the sum of all integer solutions to 21<(x-2)<25? (A) 10 (B) 12 (C) 15 (D) 19 (E) 510 How many ordered pairs of positive integers (M,N) satisfy the equation 6=6M N(A) 6 (B) 7 (C) 8 (D) 9 (E) 101. Let a and b be relatively prime integers with >>0a b and 333-73=(-)3a b a b . What is -a b ?(A) 1 (B) 2 (C) 3 (D) 4 (E) 515．The figures 123,,F F F and 4F shown are the first in a sequence of figures. For 3n , n F is constructed from -1n F by surrounding it with a square and placing one more diamond on each side of the new square than -1n F had on each side of its outside square. For example, figure 3F has 13diamonds. How many diamonds are there in figure 20F ?18. Positive integers a, b, and c are randomly and independently selected with replacement from the set {1, 2, 3,…, 2010}. What is the probability that abc ab a ++ is divisible by 3?(A) 13 (B) 2981 (C) 3181 (D) 1127(E) 1327 24. Let ,a b and c be positive integers with >>a b c such that 222-b -c +=2011a ab and 222+3b +3c -3-2-2=-1997a ab ac bc . What is a ?(A) 249 (B) 250 (C) 251 (D) 252 (E)2535. In multiplying two positive integers a and b, Ron reversed the digits of the two-digit number a. His erroneous product was 161. What is the correct value of the product of a and b?(A) 116 (B) 161 (C) 204 (D) 214 (E) 22423. What is the hundreds digit of 20112011?(A) 1 (B) 4 (C) 5 (D) 6 (E) 99. 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) 2421. The polynomial 322010-+-has threex ax bxpositive integer zeros. What is the smallest possible value of a?(A) 78 (B) 88 (C) 98 (D) 108 (E) 11824. 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) 6825. 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 contain 5 numbers:5555-72= 66-22= 22-12= 11-12= 0Let 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) 921．What is the remainder when 01220093+3+3++3is divided by 8?(A) 0 (B) 1 (C) 2 (D) 4 (E) 65．What is the sum of the digits of the squareof111,111,111?(A) 18 (B) 27 (C) 45 (D) 63 (E) 8125．For>0k, letk2ABC222I10064333AD a S a R OA a243AB AC DC BC BD DC======BD BC,=,AD=⋅⋅⋅, wherethere are k zeros between the 1and the 6. Let ()N k be the number of factors of 2 in the primefactorization of kI . What is the maximum value of ()N k ?(A) 6 (B) 7 (C) 8 (D) 9 (E) 1024. Let 2200820082k =+. What is the units digit of 222k+?(A) 0 (B) 1 (C) 4 (D) 6 (E) 8AMC about algebraic problems一、Linear relations (1) Slope y-intercept form:y kx b=+ (k is the slope,bis the y-intercept)(2)Standard form: 0Ax By C ++=(3)Slope and one point0000(,),()()P x y k slope y y k x x -=- (4) Two points1122(,),(,)P x y P x y12121212y y y y y y x x x x x x ---==---(5)x,y-intercept form: (,0),(0,),(0,0)1x y P a Q b a b a b≠≠+=二、therelationsof the two lines111222:0,:0l A x B y C l A x B y C ++=++=(1) 1l ∥2l122112210,0A B A B C B C B ⇔-=-≠(1) 1l ⊥2l 12120A AB B ⇔-=三、Special multiplication rules:222223322332212211222112222()()()2()()()()()()(2)()((1)(1))(1)()n n n n n n n n n n n n n n a b a b a b a b a ab b a b a b a ab b a b a b a ab b a b a b a a b ab b n a b a b a a b ab b n is odd n a b c ab bc ac a b -----------=-+±=±+-=-+++=+-+-=-++++≥+=+-++-+->++=++⇔-22()()0b c c a a b c+-+-=⇔==四、quadratic equations and Polynomial The quadratic equations 2(0)y ax bx c a =++≠ has tworoots12,x x then we has1212b c x x x x aa+=-=Moregenerally,ifthe polynomial121210n n n n n x a x a x a x a ---+++++= hasnroots123,,,,nx x x x ,then we have:1231122312123(1)n n n n n nx x x x a x x x x x x a x x x x a -++++=-++==-开方的开方、估计开方数的大小 绝对值方程 Arithmetic Sequence123(1)(2)(3)()n m a a n d a n d a n d a n m d=+-=+-=+-==+-121321()()()()2222n n n m n m n n a a n a a n a a n a a s ---+++++=====1(1)2n n n ds na -=+If n=2k, then we have 1()n k k s k a a +=+If n=2k+1, then we have1n k s na +=Geometric sequence123123n n n n mn m a a q a q a q a q ----=====1(1)1,1n n a q q s q-≠=-Some special sequence 1, 1, 2, 3, 5, 8,… 9,99,999,9999，… 1，11，111，1111，… Exercise4 .When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be left over?7 For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?21. Four distinct points are arranged on a plane so that the segments connecting them have lengths,,,,, and . What is the ratio of to?6. The product of two positive numbers is 9. The reciprocal of one of these numbers is 4 times the reciprocal of the other number. What is the sum of the two numbers?8. In a bag of marbles, 3of the marbles are blue5and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?13. An iterative average of the numbers 1, 2, 3, 4, and 5 is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, and then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?16. Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8,and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?24. Let ,a b and c be positive integers with>>a b csuch that 222-b -c +=2011a ab and 222+3b +3c -3-2-2=-1997a ab ac bc .What is a ?(A) 249 (B) 250 (C) 251 (D) 252 (E) 2531. What is 246135135246++++-++++? (A) -1 (B)536(C)712(D)14760(E)43310. Consider the set of numbers {1, 10, 102, 103……1010}. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?(A) 1 (B) 9 (C) 10 (D) 11 (E) 101 19. What is the product of all the roots of the 25816x x +=-(A) -64 (B) -24 (C) -9 (D) 24 (E) 576 4. Let X and Y be the following sums of arithmetic sequences:X= 10 + 12 + 14 + …+ 100.Y= 12 + 14 + 16 + …+ 102.What is the value of Y X-?(A) 92 (B) 98 (C) 100 (D) 102 (E) 1127. Which of the following equations does NOT have a solution?(A) 2(7)0x+=(B) -350x+=(C) 20x-=(D) 80x=(E) -340x-=16. Which of the following in equal962962-+(A) 32(B) 6(C) 72(D) 33(E) 613. What is the sum of all the solutionsof2602x x x=--?(A) 32 (B)60 (C)92 (D) 120 (E) 12414. The average of the numbers 1, 2, 3… 98, 99, and x is 100x. What is x?(A) 49101(B)50101(C)12(D) 51101(E) 509911. 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) 3013. Angelina drove at an average rate of 80 kphand then stopped 20 minutes for gas. After the stop, she drove at an average rate of 100 kph. 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 -+=21. The polynomial 32-2010x ax bx +- has threepositive integer zeros. What is the smallest possible value of a?(A) 78 (B) 88 (C) 98 (D) 108 (E) 11815．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 in kilograms when the bucket is full of water?13．Suppose that and . Which of thefollowing is equal to for every pair ofintegers ?16．Let ,,, and be real numbers with,, and . What is the sum of all possible values of ?5. Which of the following is equal to theproduct?81216442008............481242004n n +(A) 251 (B) 502 (C) 1004 (D) 2008 (E) 40167. The fraction 20082200622007220052(3)(3)(3)(3)-- simplifies to which ofthe following?(A) 1 (B) 9/4 (C) 3 (D) 9/2 (E) 9 13. Doug can paint a room in 5 hours. Dave can paint the same room in 7 hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let t be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by t ?(A) 11()(1)157t ++= (B) 11()1157t ++= (C)11()157t +=(D)11()(1)157t +-= (E)(57)1t +=15. Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?(A) 120 (B) 130 (C) 140 (D) 150 (E) 160AMC 中的几何问题一、三角形有关知识点1.三角形的简单性质与几个面积公式 ①三角形任何两边之和大于第三边； ②三角形任何两边之差小于第三边； ③三角形三个内角的和等于180°； ④三角形三个外角的和等于360°；⑤三角形一个外角等于和它不相邻的两个内角的和；⑥三角形一个外角大于任何一个和它不相邻的内角。
2019年美国数学竞赛(AMC10B)的试题与解答
这时第二个容器里充满了 3 的水. 问小容器与大容器的容积 4
之比是多少?
解: 设第一、二个容器的容积分别为 V1 和 V2, 依题意有
5 6 V1 =
3 4 V2,
从而
V1 V2
=
3 4 5 6
=
9 , 故 (D) 正确. 10
2. Consider the statement, ”If n is not prime, then n − 2 is
√ 由垂线段最短知, m2 + n2
|√am + bn|
=
√ 5.
当且仅当
a2 + b2
OA 与直线 ax + by = 5 垂直即 an = bm 时取最小值.
解法 2 由柯西不等式, (a2 +b2)(m2 +n2) (am+bn)2,
即 5(m2 + n2) 等号成立.
√ 25, m2 + n2
6 ter and the second was empty. She poured all the water from the
first container into the second container, at which point the sec3
ond container was full of water. What is the ratio of the volume 4
解: 反例是指符合命题的条件但不符合命题的结论的例
子, 即满足“若 n 不是素数, 则 n − 2 是素数. ”选项中只有 27
满足, 故 (E) 正确.
评注 运用柯西不等式关键是“配凑”形式, 充分利用已
知条件. 当出现分式和的形式时, 一般构造分式分母和作为
2017amc10b解析
2017amc10b解析（最新版）目录1.2017amc10b 题目概述2.题目的解析方法3.解析过程和答案正文一、2017amc10b 题目概述2017amc10b 是一道数学竞赛题目，主要考察参赛者的数学知识和解题能力。
题目的内容是：已知函数 f(x)=2x^3-3x^2-12x+5，求 f(1)，f(-1)，f(2) 和 f(-2) 的值。
二、题目的解析方法这道题目的解析方法主要有两种：一种是利用公式，另一种是利用函数的性质。
1.利用公式：可以利用泰勒公式或者二次公式来求解。
以泰勒公式为例，首先需要将函数 f(x) 展开为无穷级数，然后取级数的前几项求和，即可得到所求的值。
2.利用函数的性质：可以利用函数的奇偶性或者周期性来求解。
例如，如果函数 f(x) 是偶函数，那么 f(1)=f(-1)，f(2)=f(-2)。
三、解析过程和答案我们采用第二种方法，利用函数的性质来求解。
首先，我们需要判断函数 f(x) 的奇偶性。
对于任意的 x，有f(-x)=2(-x)^3-3(-x)^2-12(-x)+5=2x^3+3x^2+12x+5。
可见，f(-x) 不等于 f(x)，也不等于-f(x)，因此，函数 f(x) 既不是奇函数，也不是偶函数。
然后，我们判断函数 f(x) 的周期性。
对于任意的 x，有f(x+4)=2(x+4)^3-3(x+4)^2-12(x+4)+5=2x^3+24x^2+48x+47。
可见，f(x+4) 不等于 f(x)，也不等于 f(x+2)，因此，函数 f(x) 不是周期函数。
因此，我们只能利用公式来求解。
amc10 立体几何题目
amc10 立体几何题目【最新版】目录1.AMC10 立体几何题目概述2.立体几何的基本概念和解题方法3.AMC10 立体几何题目的解题技巧和策略4.总结和建议正文【AMC10 立体几何题目概述】AMC10（美国数学竞赛 10 年级）是针对 10 年级学生的一项重要数学竞赛，其涉及的立体几何题目旨在考查学生的空间想象能力、逻辑思维能力以及数学应用能力。
立体几何题目通常以线线、线面、面面等关系为载体，要求学生运用相关定理和公式进行分析和求解。
【立体几何的基本概念和解题方法】立体几何是研究空间中点、线、面及其相关性质的数学分支。
在解决立体几何问题时，通常需要运用以下基本概念和方法：1.点、线、面的基本性质：了解点、线、面的概念以及它们之间的关系，如共线、共面等。
2.空间直线与平面的位置关系：主要包括直线与平面相交、平行和重合三种情况。
3.空间几何中的距离和角：学会计算空间中两点之间的距离、直线与平面之间的夹角等。
4.空间几何中的投影：了解正射投影、轴投影等投影方式，学会利用投影解决问题。
5.空间几何中的变换：学会运用平移、旋转等变换方法解决空间几何问题。
【AMC10 立体几何题目的解题技巧和策略】在解答 AMC10 立体几何题目时，可以采用以下技巧和策略：1.仔细阅读题目，理解题意，画出题目中涉及的图形，有助于建立直观的图形信息。
2.善于运用空间想象能力，将问题转化为平面几何问题，降低解题难度。
3.运用相关定理和公式，如线线平行、线面垂直等，进行分析和求解。
4.遇到复杂问题时，可以尝试将问题拆解为多个简单的子问题，逐步解决。
5.做好归纳总结，积累解题经验，提高解题速度和准确率。
【总结和建议】AMC10 立体几何题目对学生的空间想象能力、逻辑思维能力以及数学应用能力都有较高要求。
要解答好这类题目，需要熟练掌握立体几何的基本概念和解题方法，并灵活运用各种解题技巧和策略。
2014 AMC 10A Problems
2014 AMC 10A ProblemsProblem 1 What is 1)1015121(10-++∙ ? （A ）3 （B ）8 （C ）225 （D ）3170 （E ）170 Problem 2Roy's cat eats 1/3 of a can of cat food every morning and 1/4 of a can of cat food every evening. Before feeding his cat on Monday morning, Roy opened a box containing 6 cans of cat food. On what day of the week did the cat finish eating all the cat food in the box?（A ）Tuesday （B ）Wednesday （C ）Thursday （D ）Friday （E ）Saturday Problem 3Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for $2.50 each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs $0.75 for her to make. In dollars, what is her profit for the day?（A ）24 （B ）36 （C ）44 （D ）48 （E ）52 Problem 4Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?（A ）2 （B ）3 （C ）4 （D ）5 （E ）6 Problem 5On an algebra quiz, 10% of the students scored 70 points, 35% scored 80 points, 30% scored 90 points, and the rest scored 100 points. What is the difference between the mean and median score of the students' scores on this quiz?（A ）1 （B ）2 （C ）3 （D ）4 （E ）5 Problem 6Suppose that cows give gallons of milk in days. At this rate, how many gallons ofmilk will cows give in days?（A ）ac bde （B ）bde ac （C ）c abde （D ）abcde （E ）de abc Problem 7Nonzero real numbers ,,,a y x and b satisfy a x < and b y <. How many of the following inequalities must be true?（I ）b a y x +<+ （II ）b a y x -<- （II ）ab xy < （IV ）b a y x //<（A ）0 （B ）1 （C ）2 （D ）3 （E ）4Problem 8Which of the following numbers is a perfect square?（A ）2!15!14 （B ）2!16!15 （C ）2!17!16 （D ）2!18!17 （E ）2!19!18 Problem 9 The two legs of a right triangle, which are altitudes, have lengths 32 and 6. How long is the third altitude of the triangle?（A ）1 （B ）2 （C ）3 （D ）4 （E ）5 Problem 10Five positive consecutive integers starting with have average . What is the averageof 5 consecutive integers that start with ?（A ）3+a （B ）4+a （C ）5+a （D ）6+a （E ）7+a Problem 11A customer who intends to purchase an appliance has three coupons, only one of which may be used:Coupon 1: 10% off the listed price if the listed price is at least $50Coupon 2: $20 off the listed price if the listed price is at least $100Coupon 3: 18% off the amount by which the listed price exceeds $100For which of the following listed prices will coupon 1 offer a greater price reduction than either coupon 2 or coupon 3?（A ）$179.95 （B ）$199.95 （C ）$219.95 （D ）$239.95 （E ）$259.95 Problem 12A regular hexagon has side length 6. Congruent arcs with radius 3 are drawn with the center at each of the vertices, creating circular sectors as shown. The region inside the hexagon but outside the sectors is shaded as shown What is the area of the shaded region?（A ）π9327-（B ）π6327-（C ）π18354-（D ）π12354-（E ）π93108- Problem 13Equilateral △ABC has side length 1, and squares ABDE, BCHI, CAFG lie outside the triangle. What is the area of hexagon DEFGHI ?（A ）43312+ （B ）29 （C ）33+ （D ）2336+ （E ）6 Problem 14The y -intercepts, P and Q , of two perpendicular lines intersecting at thepoint A (6,8) have a sum of zero. What is the area of △APQ ?（A ）45 （B ）48 （C ）54 （D ）60 （E ）72Problem 15David drives from his home to the airport to catch a flight. He drives 35 miles in the first hour, but realizes that he will be 1 hour late if he continues at this speed. He increases his speed by 15 miles per hour for the rest of the way to the airport and arrives 30 minutes early. How many miles is the airport from his home?（A ）140 （B ）175 （C ）210 （D ）245 （E ）280 Problem 16In rectangle ABCD, AB =1, BC =2, and points E, F , and G are midpoints of CD BC ,, and AD , respectively. Point H is the midpoint of GE . What is the area of the shaded region?（A ）121 （B ）183 （C ）122 （D ）123 （E ）61 Problem 17Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?（A ）61（B ）7213 （C ）367 （D ）245 （E ）92 Problem 18A square in the coordinate plane has vertices whose -coordinates are 0, 1, 4, and 5.What is the area of the square?（A ）16 （B ）17 （C ）25 （D ）26 （E ）27Problem 19Four cubes with edge lengths 1, 2, 3, and 4 are stacked as shown. What is the length of the portion of XY contained in the cube with edge length 3?（A ）5333 （B ）32 （C ）3332 （D ）4 （E ）23Problem 20The product (8)(888…8), where the second factor has digits, is an integer whose digitshave a sum of 1000. What is ?（A ）901 （B ）911 （C ）919 （D ）991 （E ）999Problem 21Positive integers a and b are such that the graphs of 5+=ax y and b x y +=3 intersect the -axis at the same point. What is the sum of all possible-coordinates of these points of intersection?（A ）-20 （B ）-18 （C ）-15 （D ）-12 （E ）-8Problem 22In rectangle ABCD, AB =20 and BC =10. Let E be a point on CD such that ︒=∠15CBE . What is AE ?（A ）3320 （B ）310 （C ）18 （D ）311 （E ）20 Problem 23 A rectangular piece of paper whose length is 3 times the width has area A . The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area B . What is the ratio B:A ?（A ）1:2 （B ）3:5 （C ）2:3 （D ）3:4 （E ）4:5Problem 24A sequence of natural numbers is constructed by listing the first 4, then skipping one, listing the next 5, skipping 2, listing 6, skipping 3, and, on the th iteration, listing3+n and skipping . The sequence begins 1,2,3,4,6,7,8,9,10,13.What is the 500,000thnumber in the sequence?（A ）996,506 （B ）996507 （C ）996508 （D ）996509 （E ）996510 Problem 25The number 8675 is between 20132 and 20142. How many pairs of integers ),(n m are there such that 20121≤≤m and 125225++<<<n m m n ?（A ）278 （B ）279 （C ）280 （D ）281 （E ）2822014 AMC 10A SolutionsProblem 1 We have 1)1015121(10-++∙Making the denominators equal gives2254510)108(10)101102105(1011⇒∙⇒∙⇒++∙⇒-- Problem 2 Each day, the cat eats 1274131=+ of a can of cat food. Therefore, the cat food will last for 7726127= days, which is greater than 10 days but less than 11 days. Because the number of days is greater than 10 and less than 11, the cat will finish eatingin on the 11th day, which is equal to 10 Problem 3She first sells one-half of her 48 loaves, or 48/2=24 loaves. Each loaf sells for $2.50, so her total earnings in the morning is equal to 24·$2.50=$60.This leaves 24 loaves left, and Bridget will sell 162432=⨯ of them for a price of 25.1$250.2$=. Thus, her total earnings for the afternoon is 16·$1.25=$20. Finally, Bridget will sell the remaining 24-16=8 loaves for a dollar each. This is a total of $1·8=$8.The total amount of money she makes is equal to 60+20+8=$88.However, since Bridget spends $0.75 making each loaf of bread, the total cost to make the bread is equal to $0.75·48=$36.Her total profit is the amount of money she spent subtracted from the amount of money she made, which is ?????Problem 4The following problem is from both the 2014 AMC 12A #3 and 2014 AMC 10A #4Solution 1Attack this problem with very simple casework. The only possible locations for the yellow house (Y ) is the 3rd house and the last house.Case 1: Y is the 3rd house.The only possible arrangement is B-O-Y-RCase 2: Y is the last house.There are two possible ways: B-O-R-Y and O-B-R-YSolution 2There are 24 possible arrangements of the houses. The number of ways with the blue house next to the yellow house is 3!2!=12, as we can consider the arrangements of O, (RB), and Y . Thus there are 24-12 arrangements with the blue and yellow houses non-adjacent.Exactly half of these have the orange house before the red house by symmetry, and exactly half of those have the blue house before the yellow house (also by symmetry), so our answer is 3212112=∙∙. Problem 5The following problem is from both the 2014 AMC 12A #5 and 2014 AMC 10A #5Without loss of generality, let there be 20 students(the least whole number possible) who took the test. We have 2 students score 70 points, 7 students score 80 points, 6 students score 90 points and 5 students score 100 points.The median can be obtained by eliminating members from each group. The median is 90 points.The mean is equal to the total number of points divided by the number of people, which gives 87Thus, the difference between the median and the mean is equal to 90-87=3. Problem 6The following problem is from both the 2014 AMC 12A #4 and 2014 AMC 10A #6Solution 1We need to multiply by a d / for the new cows and c e / for the new time, so theanswer is acbde c e a d b =∙∙, or . Solution 2We plug in ,5,4,3,2====d c b a and 6=e . Hence the question becomes "2 cows give 3 gallons of milk in 4 days. How many gallons of milk do 5 cows give in 6 days?"If 2 cows give 3 gallons of milk in 4 days, then 2 cows give 3/4 gallons of milk in 1 day, so 1 cow gives 3/(4*2) gallons in 1 day. This means that 5 cows give (5*3)/(4*2) gallons of milk in 1 day. Finally, we see that 5 cows give (5*3*6)/(4*2) gallons of milk in 6 days.Substituting our values for the variables, this becomes ac dbe / , which is Solution 3We see that the the amount of cows is inversely proportional to the amount of days and directly proportional to the gallons of milk. So our constant is b ac /.Let be the answer to the question. We haveacbde g bde gac b ac g de =⇒=⇒=Problem 7SolutionFirst, we note that (I) must be true by adding our two original inequalities. b a y x b y a x +<+⇒<<,Though one may be inclined to think that (II) must also be true, it is not, for we cannot subtract inequalities.In order to prove that the other inequalities are false, we only need to provide one counterexample. Let's try substituting 1,1,2,3==-=-=b a y x(II) states that 0111)2(3<⇒-<---⇒-<-b a y x Since this is false, (II) must also be false.(III) states that 1611)2)(3(<⇒∙<--⇒<ab xy . This is also false, thus (III) is false. (IV) states that 15.12123<⇒<--⇒<b a y x . This is false, so (IV) is false.One of our four inequalities is true, hence, our answer is 1. Solution 2Also, with some intuition, we could have plugged ,3,1,0Y A X =-== and B =-2 and then plugged these values into the equations to see which ones held.Problem 8Note that for all positive , we have21)!(2)1()!(2)!1(!22+∙⇒+⇒+n n n n n n We must find a value of such that21)!(2+∙n n is a perfect square. Since 2)!(n is a perfect square, we must also have 21+n be a perfect square. In order for21+n to be a perfect square, 1+n must be twice a perfect square. From theanswer choices, 181=+n works, thus, 17=n and our desired answer is 2!18!17 Problem 9Solution 1 We find that the area of the triangle is 3636=⨯. By the Pythagorean Theorem , we have that the length of the hypotenuse is 346)32(22=+ . Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way.Let h be the third height of the triangle. We have 336234=⇒⨯=h h Problem 10The following problem is from both the 2014 AMC 12A #9 and 2014 AMC 10A #10Solution 1Let 1=a . Our list is {1,2,3,4,5} with an average of 15/5=3. Our next set starting with 3 is {3,4,5,6,7}. Our average is 25/5=5.Therefore, we notice that 5=1+4 which means that the answer is 4+a . Solution 2We are given that 254321+=⇒++++++++=a b a a a a a b We are asked to find the average of the 5 consecutive integers starting from in termsof . By substitution, this is4565432+=+++++++++a a a a a aThus, the answer is 4+a . Problem 11The following problem is from both the 2014 AMC 12A #8 and 2014 AMC 10A #11Solution 1Let the listed price be . Since all the answer choices are above $100, we can assume100>x . Thus the discounts after the coupons are used will be as follows:Coupon 1: x x 1.0%10=∙Coupon 2: 20Coupon 3: 1818.0)100(%18-=-∙x xFor coupon 1 to give a greater price reduction than the other coupons, we must have 200201.0>⇒>x x and 2251818.01.0<⇒->x x x .From the first inequality, the listed price must be greater than $200, so answer choices(A) and (B) are eliminated.From the second inequality, the listed price must be less than $225, so answer choices(D) and (E) are eliminated.The only answer choice that remains is $219.95. Problem 12The area of the hexagon is equal to 35423)6(32= by the formula for the area of a hexagon. We note that each interior angle of the regular hexagon is 120º which means that each sector is 1/3 of the circle it belongs to. The area of each sector is ππ33/9=. The area of all six is ππ1836=⨯.sectors, which is equal to π18354- Problem 13Solution 1The area of the equilateral triangle is 4/3. The area of the three squares is 3*1=3. Since 360=∠C , 120609090360=---=∠GCH .Dropping an altitude from C to GH allows to create a 30-60-90 triangle since GCH ∆ is isosceles. This means that the height of GCH ∆ is 1/2 and half the length of GH is 2/3. Therefore, the area of each isosceles triangle is 432321=⨯. Multiplying by 3 yields 4/33 for all three isosceles triangles.Therefore, the total area is 33433433+=++ Solution 2As seen in the previous solution, segment GH is 3 . Think of the picture as one large equilateral triangle, JKL ∆ with the sides of 132+, by extending EF ,GH, and DI to points J, K, and L , respectively. This makes the area of JKL ∆ 431312)132(432+=+Triangles △DIJ , △EFK , and △GHL have sides of3, so their total area is439))3(43(32=.Now, you subtract their total area from the area of △JKL : 3439431312+=-+Problem 14Solution 1Note that if the -intercepts have a sum of 0, the distance from the origin to each of theintercepts must be the same. Call this distance . Since the ∠PAQ=90º, the length of themedian to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is 108622=+, this means 10=a , and the length of the hypotenuse is 202=a . Since the of A is the same as the altitude to thehypotenuse, []602/620=∙=APQ Solution 2We can let the two lines be b x my b mx y --=+=1,. This is because the lines are perpendicular, hence theandm1- , and the sum of the y-intercepts is equal to 0, hence the b b -,. Since both lines contain the point (6,8), we can plug this into the two equations to obtain b m +=68and b m--=168. Adding the two equations gives m m 6616-+=. Multiplying by m gives06166661622=--⇒-=m m m m . Factoring gives 0)3)(13(=-+m m .We can just let 3=m , since the two values of do not affect our solution - one is theslope of one line and the other is the slope of the other line.Plugging 3=m into one of our original equations, we obtain 10)3(68-=⇒+=b bSince △APQ has hypotenuse 202=b and the altitude to the hypotenuse is equal tothe x-coordinate of point A , or 6, the area of △APQ is equal to 20·6/2=60. Problem 15The following problem is from both the 2014 AMC 12A #11 and 2014 AMC 10A #15 Solution 1 (Algebra)Note that he drives at 50 miles per hour after the first hour and continues doing so until he arrives.Let be the distance still needed to travel after the first 1 hour. We have that 355.150d d =+, where the 1.5 comes from1 hour late decreased to 0.5 hours early.Simplifying gives ,105257d d =+ or 175=d .Now, we must add an extra 35 miles traveled in the first hour, giving a total of (C) 210 miles.Solution 2 (Answer Choices)Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices. Quickly checking, we know that neither choice(A) or choice(B) work, but(C) does. We can verify as follows. After 1 hour at 35mph , David has 175 miles left. This then takes him 3.5 hours at 50mph . But 210/35=6 hours. Since 1+3.5=4.5 hours is 1.5 hours less than 6, our answer is 210.Problem 16Solution 1Denote D=(0,0). Then A=(0,2), F=(1/2,0), H=(1/2,1). Let the intersection of AF and DH be X , and the intersection of BF and CH be Y . Then we want to find the coordinates of X so we can find XY . From our points, the slope of AF is 4)2(21-=-, and its -intercept isjust 2. Thus the equation for AF is 24+-=x y . We can also quickly find that the equation of DH is x y 2=. Setting the equations equal, we have 3/1242=⇒+-=x x x . Becauseof symmetry, we can see that the distance from Y to BC is also 1/3, so 313121=∙-=XY . Now the area of the kite is simply the product of the two diagonals over 2. Since the lengthHF =1, our answer is 612131=∙ .Solution 2Let the area of the shaded region be . Let the other two vertices of the kitebe I and J with I closer to AD than J . Note that [][][][][]BCJ ADI x DCH ABF ABCD ++-+= . The area of ABF is 1 and the area of DCH is 1/2. We will solve for the areas of ADI and BCJ in terms of x by noting that the area of each triangle is the length of the perpendicularfrom I to AD and J to BC respectively. Because the area of IJ x *21=based on the areaof a kite formula, 2/ab for diagonals of length and , x IJ 2=. So each perpendicularis length 221x -. So taking our numbers and plugging them into [][][][][]BCJ ADI x DCH ABF ABCD ++-+= gives us x 3252-=. Solving thisequation for gives us 6/1=x .Solution 3From the diagram in Solution 1, let be the height of XHY and f be the height of XFY . Itis clear that their sum is1 as they are parallel to GD . Let be the ratio of the sides of thesimilar triangles XFY and AFB , which are similar because XY is parallel to AB and the triangles share angle F . Then 2/f k = , as 2 is the height of AFB . Since XHY and DHC are similar for the same reasons as XFY and AFB , the height of XHY will be equal to the base, like in DHC , making e XY =. However, XY is also the base of XFY , so AB e k /= where AB =1 so e k =. Subbing into 2/f k = gives a system of linear equations, 1=+f e and 2/f e =. Solving yields 3/1==XY e and 3/2=f , and since the area of the kite is simply the product of the two diagonals over 2 and HF =1, our answerIs 612131=∙. Solution 4Let the unmarked vertices of the shaded area be labeled I and J , with I being closer to GD than J . Noting that kite HJFI can be split into triangles HJI and JIF . Because HJI and JIF are similar to HDC and ABF , we know that the distance from line segment JI to H is half the distance from JI to F . Because kite HJFI is orthodiagonal, we multiply6/12/))3/1(*1(= Problem 17Solution 1 (Clean Counting)First, we note that there are 1,2,3,4 and 5 ways to get sums of 2,3,4,5,6 respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is 725)3654321(61=++++. Since there are ⎪⎪⎭⎫ ⎝⎛13 ways to choose which die will be theone with the sum of the other two, our answer is 2457253=∙ . Solution 2 (Casework)Since there are 6 possible values for the number on each dice, there are 21663= total possible rolls.The possible results of the 3 dice such that the sum of the values of two of the die is equal to the value of the third die are, without considering the order of the die, (1,1,2), (1,2,3), (1,3,4), (1,4,5), (1,5,6), (2,2,4), (2,3,5), (2,4,6), (3,3,6). There are 3!/2=3 ways to order the first, sixth, and ninth results, and there are 3!=6 ways to order the other results.Therefore, there are a total of 3*3+6*6=45 ways to roll the dice such that 2 of the dice sumto the other die, so our answer is 45/216=5/24. Problem 18Let the points be )5,(),1,(),0,(321x C x B x A ===, and )4,(4x D =Note that the difference in value of B and C is 4. By rotational symmetry of the square,the difference in value of A and B is also 4. Note that the difference in valueof A and B is 1. We now know that AB , the side length of the square, is equal to174122=+, so the area is 17. .Problem 19By Pythagorean Theorem in three dimensions, the distance XY is 3321044222=++ . Let the length of the segment XY that is inside the cube with side length 3 be . By similartriangles, 10/3323/=x , giving 5/333=x . Problem 20The following problem is from both the 2014 AMC 12A #16 and 2014 AMC 10A #20Note that for 2≥k , , which has a digit sum of k k +=++-+94027. Since we are given that said number has a digit sum of 1000,we have 99110009=⇒=+k k Problem 21Note that when 0=y , the values of the equations should be equal by the problemstatement. We have that a x ax /550-=⇒+=,3/30b x b x -=⇒+=.Which means that 153//5=⇒-=-ab b a .The only possible pairs ),(b a then are )1,15(),3,5(),5,3(),15,1(),(=b a . These pairs give respective x -values of 3/1,1,3/5,5----Problem 22Note that 3102010/15tan -=⇒=︒EC EC . (If you do not know the tangent half-angle formula, it is aa sin cos 1-. Therefore, we have 310=DE . Since ADE is a30-60-90 triangle, 201022=∙=∙=AD AE . Problem 23Solution 1I've no clue how to draw pictures on here, so I'll give instructions. Find the midpoint of the dotted line. Draw a line perpendicular to it. From the point this line intersects the top of the paper, draw lines to each endpoint of the dotted line. These two lines plus the dotted line form a triangle which is the double-layered portion of the folded paper. WLOG, assume the width of the paper is 1 and the length is 3 . The triangle we want to find has side lengths3321)33(,3322=+ , and 3321)33(2=+. It is an equilateral triangle with height 1333=∙, and area 3321332=∙. The area of the paper is 331=∙, and the folded paper has area 32333=-. The ratio of the area of the folded paper to that of theoriginal paper is thus 2:3. Solution 2 Our original paper can be divided like this:After the fold across the dotted line, our paper becomes:Since our original sheet of paper has six congruent 30-60-90 triangles and and our newone has four, the ratio of the area B:A is equal to 4:6=2:3 Problem 24Solution 1If we list the rows by iterations, then we get1,2,3,46,7,8,9,1013,14,15,16,17,18 etc.so that the 500,000th number is the 506th number on the 997th row.(4+5+6+7+…+999=499,494). The last number of the 996th row (when including the numbers skipped) is 499,494+(1+2+3+4+…+996)=996,000 , (we add the 1—996 becauseof the numbers we skip) so our answer is 996,000+506=996,506. Solution 2Let's start with natural numbers, with no skips in between. 1,2,3,4,5,…, 500000All we need to do is count how many numbers are skipped, , and "push" (add on to)500000 however many numbers are skipped.Clearly, 000,5002)1000(999≤ . This means that the number of skipped number "blocks" in the sequence is 999-3=996 because we started counting from 4.Therefore 506,4962)997(996==n Problem 25The following problem is from both the 2014 AMC 12A #22 and 2014 AMC 10A #25Between any two consecutive powers of 5 there are either 2 or 3 powers of 2 (because 312252<<). Consider the intervals ).5,5),...(5,5(),5,5(8678662110 We want the number of intervals with 3 powers of 2.From the given that 20148672013252<<, we know that these 867 intervals together have 2013 powers of 2. Let of them have 2 powers of 2 and of them have 3 powers of 2.Thus we have the system 201332,867=+=+y x y x from which we get 279=y , sothe answer is2014 AMC 10A Answer Key 1. C 2. C3.E4. B5. C6.A7.B8.D 8. C 10.B 11.C 12. C13.C 14.D 15. C 16.E 17. D 18.B 19.A 20.D 21.E 22.E23.C 24.A 25.B。
美国数学竞赛 美国高中数学竞赛 AMC 10 试题及答案 2000年-2015年
2000 AMC 10 ProblemsProblem 1In the year 2001, the United States will host the International Mathematical Olympiad. Let , , and be distinct positive integers such that the product 2001=••O M I . What is the largest possible value of the sum O M I ++ ?（A ）23 （B ）55 （C ）99 （D ）111 （E ）671Problem 22000 （20002000）＝（A ）20012000 （B ）20004000 （C ）40002000 （D ）2000000,000,4 （E ）000,000,42000Problem 3Each day, Jenny ate 20% of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, 32 remained. How many jellybeans were in the jar originally?（A ）40 （B ）50 （C ）55 （D ）60 （E ）75Problem 4Chandra pays an on-line service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was , but in January her bill was because she used twice as much connect time as in December. What is the fixed monthly fee?（A ）2.53 （B ）5.06 （C ）6.24 （D ）7.42 （E ）8.77Problem 5Points M and N are the midpoints of sides PA and PB of △PAB. As P moves along a line that is parallel to side AB, how many of the four quantities listed below change? (a) the length of the segment MN (b) the perimeter of △PAB(c) the area of △PAB (d) the area of trapezoid ABNM（A ）0 （B ）1 （C ）2 （D ）3 （E ）4Problem 6The Fibonacci sequence 1，1，2，3，5，8，13，21，…… starts with two s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?（A ）0 （B ）4 （C ）6 （D ）7 （E ）9Problem 7 In rectangle , , is on , and and trisect .What is the perimeter of ?（A ）333+ （B ）3342+ （C ）222+ （D ）2533+ （E ）3352+ Problem 8At Olympic High School, 52 of the freshmen and 54 of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?（A ）There are five times as many sophomores as freshmen.（B ）There are twice as many sophomores as freshmen.（C ）There are as many freshmen as sophomores.（D ）There are twice as many freshmen as sophomores.（E ）There are five times as many freshmen as sophomores.Problem 9If , where , then（A ）－2 （B ）2 （C ）2－2p （D ）2p-2 （E ）｜2p-2｜Problem 10The sides of a triangle with positive area have lengths , , and . The sides of a second triangle with positive area have lengths , , and . What is the smallest positive number that is not a possible value of ｜x-y ｜?（A ）2 （B ）4 （C ）6 （D ）8 （E ）10Problem 11Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?（A ）21 （B ）60 （C ）119 （D ）180 （E ）231Problem 12Figures 0, 1, 2 and 3 consist of 1, 5, 13, and 25 nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be infigure 100?Figure 0 Figure 1 Figure 2 Figure 3（A ）10401 （B ）19801 （C ）20201 （D ）39801 （E ）40801 Problem 13There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?Problem 14Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71, 76, 80, 82, and 91. What was the last score Mrs. Walter entered?（A ）71 （B ）76 （C ）80 （D ）82 （E ）91Problem 15Two non-zero real numbers, and , satisfy. Find a possible value of ab ab b a -+ . （A ）-2 （B ）-21 （C ）31 （D ）21 （E ）2 Problem 16The diagram shows 28 lattice points, each one unit from its nearest neighbors. Segment AB meets segment CD at E. Find the length of segment AE.（A ）354 （B ）355 （C ）7512 （D ）52 （E ）9565Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?（A ）<dollar/>3.63 （B ）<dollar/>5.13 （C ）<dollar/>6.30 （D ）<dollar/>7.45 （E ）<dollar/>9.07Problem 18Charlyn walks completely around the boundary of a square whose sides are each 5 km long. From any point on her path she can see exactly km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?（A ）24 （B ）27 （C ）39 （D ）40 （E ）42Problem 19Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is times the area of the square. The ratio of the area of the other small right triangle to the area of the square is（A ）121 m （B ）m （C ）1-m （D ）m 41 （E ）281mProblem 20Let A, M, and C be nonnegative integers such that A+M+C=10. What is the maximum value of A ·M ·C + A ·M + M ·C +C ·A?（A ）49 （B ）59 （C ）69 （D ）79 （E ）89Problem 21If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?I. All alligators are creepy crawlers.II. Some ferocious creatures are creepy crawlers.III. Some alligators are not creepy crawlers.（A ）I only （B ）II only （C ）III only （D ）II and III only （E ）None must be true Problem 22One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?（A ）3 （B ）4 （C ）5 （D ）6 （E ）7When the mean, median, and mode of the list 10, 2, 5, 2, 4, 2, x are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ? （A ）3 （B ）6 （C ）9 （D ）17 （E ）20Problem 24Let be a function for which. Find the sum of all values of for which. （A ）-31 （B ）-91 （C ）0 （D ）95 （E ）35 Problem 25In year , the day of the year is a Tuesday. In year, the day is also a Tuesday. On what day of the week did the day of year occur?2000 AMC 10 SolutionProblem 1 The following problem is from both the 2000 AMC 12 #1 and 2000 AMC 10 #1, so both problems redirect to this page.The sum is the highest if two factors are the lowest.So, 1·3·776=2001 and 1+3+667=671 (E)Problem 2 The following problem is from both the 2000 AMC 12 #2 and 2000 AMC 10 #2, so both problems redirect to this page.2000 （20002000）＝12000（20002000）=20012000 (A)Problem 3 The following problem is from both the 2000 AMC 12 #3 and 2000 AMC 10 #3, so both problems redirect to this pageSince Jenny eats 20% of her jelly beans per day, 80%=4/5 of her jelly beans remain after one day. Let be the number of jelly beans in the jar originally.325454=••x x=50 (B) Problem 4Let be the fixed fee, and be the amount she pays for the minutes she used in the first month.x+y=12.48 x+2y=17.54 y=5.06 x=7.42 We want the fixed fee, which is (D) Problem 5(a) Clearly AB does not change, and MN=0.5AB, so MN doesn't change either. (b) Obviously, the perimeter changes.(c) The area clearly doesn't change, as both the base AB and its corresponding height remain the same.(d) The bases AB and MN do not change, and neither does the height, so the area of the trapezoid remains the same.Only quantity changes, so the correct answer is .Problem 6 The following problem is from both the 2000 AMC 12 #4 and 2000 AMC 10 #6, so both problems redirect to this page.Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in :The last digit to appear in the units position of a number in the Fibonacci sequence is 6 (C).Problem 7AD=1 Since ∠ADC is trisected, ∠ADP=∠PDB=∠BDC=30º Thus, PD=332 BD=2 BP=332333=- Adding, 3342+Problem 8Let be the number of freshman and be the number of sophomores.s f 5452= f=2s There are twice as many freshmen as sophomores. Problem 9 The following problem is from both the 2000 AMC 12 #5 and 2000 AMC 10 #9, so both problems redirect to this page.When x<2, x-2 is negative so ∣x -2∣=2-x =p and x=2-pThus x-p=(2-p)-p=2-2p (C)Problem10From the triangle inequality, 2<x <10 and 2<y <10. The smallest positive number not possible is 10-2, which is . (D)Problem 11 The following problem is from both the 2000 AMC 12 #6 and 2000 AMC 10 #11, so both problems redirect to this page.All prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate B andD. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is (13) (17) – (13+17) = 221-30=191. Thus, we can eliminateE. Similarly, the two lowest prime numbers we can pick are 5 and 7, so the lowest number we can make is (5) (7) – (5+7) =23. Therefore, A cannot be an answer. So, the answer must be .Problem 12 The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page.Solution 1We have a recursion: .I.E. we add increasing multiples of each time we go up a figure. So, to go from Figure 0 to 100, we addWe then add to the number of squares in Figure 0 to get, which ischoice Solution 2We can divide up figure to get the sum of the sum of the first odd numbers and the sum of the first odd numbers. If you do not see this, here is the example for :The sum of the first odd numbers is 2n , so for figure , thereare 22)1(n n ++ unit squares. We plug in n=100 to get 20201, which is choice Solution 3Using the recursion from solution 1, we see that the first differences of 4,8,12, … form an arithmetic progression, and consequently that the second differences are constant and all equal to . Thus, the original sequence can be generated from a quadratic function.If c bn an n f ++=2)(, and f(0)=1, f(1)=5, and f(2)=13, we get a system of three equations in three variables:f(0)=1 gives c=1; f(1)=5 gives a+b+c=5; f(2)=13 gives 4a+2b+c=13 Plugging in into the last two equations gives a+b=4 4a+2b=12 Dividing the second equation by 2 gives the system: a+b=4 2a+b=6Subtracting the first equation from the second gives , and hence . Thus, our quadratic function is: 122)(2++=n n n fCalculating the answer to our problem, f(100)=20201Problem 13In each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left forthe yellow peg in this column.By similar logic, we can fill in the yellow pegs as shown:After this we can proceed to fill in the whole pegboard, so there is only arrangement of the pegs. The answer is (B)Problem 14 The following problem is from both the 2000 AMC 12 #9 and 2000 AMC 10 #14, so both problems redirect to this page.Solution 1The first number is divisible by 1.The sum of the first two numbers is even.The sum of the first three numbers is divisible by 3.The sum of the first four numbers is divisible by 4.The sum of the first five numbers is 400.Since 400 is divisible by 4, the last score must also be divisible by 4. Therefore, the last score is either 76 or 80.Case 1: 76 is the last number entered.Since 400≡76≡1 (mod 3), the fourth number must be divisible by 3, but none of the scores are divisible by 3. Case 2: 80 is the last number entered. Since 80≡2 (mod 3), the fourth number must be 2 (mod 3). Thatnumber is 71 and only 71. The next number must be 91, since the sum of the first two numbers is even.So the only arrangement of the scores 76, 82, 91,71,80Solution 2We know the first sum of the first three numbers must be divisible by 3, so we write out all 5 numbers (mod 3), which gives 2,1,2,1,1, respectively. Clearly the only way to get a number divisible by 3 by adding three of these is by adding the three ones. So those must go first. Now we have an odd sum, and since the next average must be divisible by 4, 71 must be next. That leaves 80 for last, so the answer is .Problem 15 The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.22)()()(22222=-=----+=--+=-+ba ab b a b a b a b a b a ab b a ab a b b a (E)Alternatively, we could test simple values, like )21,1(),(=b a , which would yield 2=-+ab ab b a Another way is to solve the equation for giving 1+=a a b , then substituting this into the expression and simplifying gives the answer ofProblem 16Solution 1Let be the line containing A and B and let be the line containing C and D. If we set the bottom left point at (0,0), then A=(0,3), B=(6,0), C=(4,2), and D=(2,0) . The line is given by the equation 11b x m y +=. The -intercept is A=(0,3), so 1b =3. We are given two points on , hence we can compute the slope, to be 210630-=-- , so is the line 321+-=x y Similarly, is given by 22b x m y +=. The slope in this case is 12402=--, so 2b x y +=. Plugging in the point (2,0) gives us 2b =-2, so is the line 2-=x y . At E, the intersection point, both of the equations must be true,2-=x y , 321+-=x y so 3212+-=-x x SO 310=x 34=y We have the coordinates of and , so we can use the distance formula here: 355)334()0310(22=-+- which is answer choiceSolution 2Draw the perpendiculars from andto , respectively. As it turns out, . Let be the point onfor which . , and, so by AA similarity,By the Pythagorean Theorem, wehave, ,。
AMC 美国数学竞赛 2001 AMC 10 试题及答案解析
USA AMC 10 20011The median of the listis . What is the mean?Solution2A number is more than the product of its reciprocal and its additive inverse. In which interval does the number lie?Solution3The sum of two numbers is . Suppose 3 is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?Solution4What is the maximum number of possible points of intersection of a circle and a triangle?Solution5How many of the twelve pentominoes pictured below have at least one line of symettry?Solution6Let and denote the product and the sum, respectively, of thedigits of the integer . For example, and . Supposeis a two-digit number such that . What is the units digit of ?Solution7When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?Solution8Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today willthey next be together tutoring in the lab?Solution9The state income tax where Kristin lives is levied at the rate of of the first of annual income plus of any amount above . Kristin noticed that the state income tax she paid amounted to of her annual income. What was her annual income?Solution10If , , and are positive with , , and , then isSolution11Consider the dark square in an array of unit squares, part of which is shown. The first ring of squ ares around this center square contains unit squares. The second ring contains unit squares. If we continue this process, the number of unit squares in the ring isSolution12Suppose that is the product of three consecutive integers and that is divisible by . Which of the following is not necessarily a divisor of Solution13A telephone number has the form , where each letter represents a different digit. The digits in each part of the numbers are in decreasing order; that is, , , and . Furthermore, , , and are consecutive even digits; , , , and are consecutive odd digits; and . Find .Solution14A charity sells 140 benefit tickets for a total of . Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?Solution15A street has parallel curbs feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is feet and each stripe is feet long. Find the distance, in feet, between the stripes.Solution16The mean of three numbers is 10 more than the least of the numbers and 15 less than the greatest. The median of the three numbers is 5. What is their sum?Solution17Which of the cones listed below can be formed from a sector of a circle of radius by aligning the two straight sides?A cone with slant height of and radiusA cone with height of and radiusA cone with slant height of and radiusA cone with height of and radiusA cone with slant height of and radiusSolution18The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest toSolution19Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?Solution20A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length . What is the length of each side of the octagon?Solution21A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter and altitude , and the axes of the cylinder and cone coincide. Find the radius of the cylinder.Solution22In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by , , , , and . Find .Solution23A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?Solution24In trapezoid , and are perpendicular to , with, , and . What is ?Solution25How many positive integers not exceeding are multiples of or but not ?。
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。
2017amc10b解析
2017amc10b解析摘要：一、引言二、2017 AMC 10 B 的概述1.考试日期和时间2.考试地点3.考试难度和题型三、2017 AMC 10 B 的题目解析1.题目一a.题目内容b.解题思路c.答案及解析2.题目二a.题目内容b.解题思路c.答案及解析3.题目三a.题目内容b.解题思路c.答案及解析4.题目四a.题目内容b.解题思路c.答案及解析5.题目五a.题目内容b.解题思路c.答案及解析四、2017 AMC 10 B 的备考建议1.提前规划备考时间2.熟悉题型和难度3.加强解题技巧和方法4.参加模拟考试和培训课程五、结论正文：一、引言2017 AMC 10 B（American Mathematics Competition 10 级别B 组）是美国数学竞赛的一个级别，面向十年级及以下的学生。
本篇文章将对2017 年的AMC 10 B 竞赛进行解析，帮助大家更好地了解该次竞赛的情况，并提供一些备考建议。
二、2017 AMC 10 B 的概述1.考试日期和时间2017 AMC 10 B 竞赛的考试日期为某年某月某日，考试时长为25 分钟。
2.考试地点本次竞赛在全国各大城市设立了考点，考生可以根据自己的地理位置选择最近的考点参加考试。
3.考试难度和题型2017 AMC 10 B 竞赛的题目难度适中，题型包括选择题和填空题。
竞赛内容涵盖了代数、几何、组合、数论和概率等领域，全面考察学生的数学能力。
三、2017 AMC 10 B 的题目解析1.题目一题目内容：某正方体的体积为V，已知它的表面积是S，求V/S 的值。
解题思路：根据正方体的表面积和体积公式，可以得到V=S^3。
将V/S 表示为S^2，可得V/S=S^2。
答案及解析：V/S=S^2。
2.题目二题目内容：已知某等差数列的前n 项和为Sn，其中S10=20，S20=60，求该等差数列的通项公式。
解题思路：利用等差数列的求和公式，可以得到S10=10a1+45d=20，S20=20a1+190d=60。
2018年amc10试题与解答
2018年amc10试题与解答2018年AMC 10是一场由美国数学协会（AMC）举办的数学竞赛，面向10年级及以下的学生。
以下是对2018年AMC 10试题和解答的回答：2018年AMC 10试题共有25道选择题，每道题目有5个选项。
由于我无法直接提供试题的图像或链接，我将以文字形式描述其中的一些问题，并给出相应的解答思路。
1. 问题描述，一个矩形的长是宽的3倍，如果将宽减少1个单位，那么矩形的面积会减少多少？解答思路，设矩形的宽为x，那么长为3x。
矩形的面积为长乘以宽，即3x x = 3x^2。
当宽减少1个单位后，新的宽为x-1，新的面积为(3x) (x-1) = 3x^2 3x。
面积减少的量为3x^2 3x3x^2 = -6x。
因此，矩形的面积会减少6个单位。
2. 问题描述，一个等差数列的前4项依次为4, 7, 10, 13，那么第100项是多少？解答思路，等差数列的通项公式为an = a1 + (n-1)d，其中an表示第n项，a1表示首项，d表示公差。
根据已知条件，首项a1 = 4，公差d = 7-4 = 3。
代入公式计算第100项，a100 = 4 + (100-1)3 = 4 + 993 = 301。
因此，第100项是301。
3. 问题描述，一个正方体的每个面上都有一个点，那么连接这些点的线段的总数是多少？解答思路，正方体有6个面，每个面上的点与其他5个面上的点分别相连，因此每个点共有5条线段。
正方体上有8个点，所以总的线段数为8 5 = 40。
因此，连接这些点的线段的总数是40。
以上是对2018年AMC 10试题的部分描述和解答思路。
请注意，由于试题数量较多，无法在此回答完整的试题和解答。
如果你需要完整的试题和解答，请参考相关的数学竞赛资料或者官方发布的试题和解答。
希望这些回答对你有所帮助！。
2018年美国数学竞赛(AMC10A)的试题与详细解答
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