2020年度美国数学竞赛AMC12 A卷(带答案)
2020年AMC12B竞赛真题及答案
2015 AMC 12B竞赛真题Problem 1What is the value of ?Problem 2Marie does three equally time-consuming tasks in a row without taking breaks. She begins the first task at 1:00 PM and finishes the second task at 2:40 PM. When does she finish the third task?Problem 3Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100, and one of the numbers is 28. What is the other number?Problem 4David, Hikmet, Jack, Marta, Rand, and Todd were in a 12-person race with 6 other people. Rand finished 6 places ahead of Hikmet. Marta finished 1 place behind Jack. David finished 2 places behind Hikmet. Jack finished 2 places behind Todd. Todd finished 1 place behind Rand. Marta finished in 6th place. Who finished in 8th place?Problem 5The Tigers beat the Sharks 2 out of the 3 times they played. They then played more times, and the Sharks ended up winning at least 95% of all the games played. What is the minimum possible value for ?Problem 6Back in 1930, Tillie had to memorize her multiplication facts from to . The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?A regular 15-gon has lines of symmetry, and the smallest positive angle for which it has rotational symmetry is degrees. What is ?Problem 8What is the value of ?Problem 9Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a player knocks the bottle off the ledgeis , independently of what has happened before. What is the probability that Larry wins the game?Problem 10How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?Problem 11The line forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?Problem 12Let , , and be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation ?Problem 13Quadrilateral is inscribed in a circle withand . What is ?A circle of radius 2 is centered at . An equilateral triangle with side 4 has a vertex at . What is the difference between the area of the region that lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle?Problem 15At Rachelle's school an A counts 4 points, a B 3 points, a C 2 points, and a D 1 point. Her GPA on the four classes she is taking is computed as the total sum of points divided by 4. She is certain that she will get As in both Mathematics and Science, and at least a C in each of English and History. She thinks she has achance of getting an A in English, and a chance of getting a B. In History, shehas a chance of getting an A, and a chance of getting a B, independently ofwhat she gets in English. What is the probability that Rachelle will get a GPA of at least 3.5?Problem 16A regular hexagon with sides of length 6 has an isosceles triangle attached to each side. Each of these triangles has two sides of length 8. The isosceles triangles are folded to make a pyramid with the hexagon as the base of the pyramid. What is the volume of the pyramid?Problem 17An unfair coin lands on heads with a probability of . When tossed times, theprobability of exactly two heads is the same as the probability of exactly three heads. What is the value of ?Problem 18For every composite positive integer , define to be the sum of the factors inthe prime factorization of . For example, because the primefactorization of is , and . What is the range of the function , ?Problem 19In , and . Squares and are constructed outside of the triangle. The points , , , and lie on a circle. What is the perimeter of the triangle?Problem 20For every positive integer , let be the remainder obtained when is divided by 5. Define a functionrecursively as follows: What is ?Problem 21Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of ?Problem 22Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same chair and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?Problem 23A rectangular box measures , where , , and are integers and. The volume and the surface area of the box are numerically equal. How many ordered triples are possible?Problem 24Four circles, no two of which are congruent, have centers at , , , and , and points and lie on all four circles. The radius of circle is times the radius of circle , and the radius of circle is times the radius of circle . Furthermore, and . Let be the midpoint of . What is ?Problem 25A bee starts flying from point . She flies inch due east to point . For , once the bee reaches point , she turns counterclockwise and then flies inches straight to point . When the bee reaches she is exactlyinches away from , where , , and are positive integers and and are not divisible by the square of any prime. What is ?2015 AMC 12B竞赛真题答案1.C2.b3.a4.b5.b6.a7.d8.d9.c 10.c 11.e 12.d 13.b 14.d 15.d 16.c 17.d 18.d 19.c 20.b 21.d 22.d 23.b 24.d 25.b。
2020年AMC12真题及答案
2016 AMC12 AProblem 1What is the value of ?SolutionProblem 2For what value of does ?SolutionProblem 3The remainder can be defined for all realnumbers and with by where denotes the greatest integer less than or equal to . What is the value of ?SolutionProblem 4The mean, median, and mode of the data values are all equal to . What is the value of ?SolutionProblem 5Goldbach's conjecture states that every even integer greater than 2 can be written as the sum of two prime numbers (for example, ). So far, no one has been able to prove that the conjecture is true, and no one has found a counterexample to show that the conjecture is false. What would a counterexample consist of?SolutionProblem 6A triangular array of coins has coin in the first row, coins in the second row, coins in the third row, and so on up to coins in the th row. What is the sum of the digits of ?SolutionProblem 7Which of these describes the graph of ?SolutionProblem 8What is the area of the shaded region of the given rectangle?SolutionProblem 9The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the otherfour small squares as shown. The common side length is , where and are positive integers. What is ?SolutionProblem 10Five friends sat in a movie theater in a row containing seats, numbered to from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?SolutionProblem 11Each of the students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. Thereare students who cannot sing, students who cannot dance, and students who cannot act. How many students have two of these talents?SolutionProblem 12In , , , and . Point lies on ,and bisects . Point lies on , and bisects . The bisectors intersect at . What is the ratio : ?SolutionProblem 13Let be a positive multiple of . One red ball and green balls are arranged in a line inrandom order. Let be the probability that at least of the green balls are on the sameside of the red ball. Observe that and that approaches as grows large. What is the sum of the digits of the least value of such that ?SolutionProblem 14Each vertex of a cube is to be labeled with an integer from through , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?SolutionProblem 15Circles with centers and , having radii and , respectively, lie on the same side of line and are tangent to at and , respectively, with between and . The circle with center is externally tangent to each of the other two circles. What is the area of triangle ?SolutionProblem 16The graphs of and are plotted on the same set of axes. How many points in the plane with positive -coordinates lie on two or more of the graphs?SolutionProblem 17Let be a square. Let and be the centers, respectively, of equilateral triangles with bases and each exterior to the square. What is the ratio of the area of square to the area of square ?SolutionProblem 18For some positive integer the number has positive integer divisors,including and the number How many positive integer divisors does thenumber have?SolutionProblem 19Jerry starts at on the real number line. He tosses a fair coin times. When he gets heads, he moves unit in the positive direction; when he gets tails, he moves unit in the negative direction. The probability that he reaches at some time during this processis where and are relatively prime positive integers. What is (For example, he succeeds if his sequence of tosses is )SolutionProblem 20A binary operation has the properties that andthat for all nonzero real numbers and (Here the dot represents the usual multiplication operation.) The solution to the equation can bewritten as where and are relatively prime positive integers. What isSolutionProblem 21A quadrilateral is inscribed in a circle of radius Three of the sides of this quadrilateral have length What is the length of its fourth side?SolutionProblem 22How many ordered triples of positive integerssatisfy and ?SolutionProblem 23Three numbers in the interval are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area? SolutionProblem 24There is a smallest positive real number such that there exists a positive realnumber such that all the roots of the polynomial are real. In fact, for this value of the value of is unique. What is the value ofSolutionProblem 25Let be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with digits. Every time Bernardo writes a number, Silvia erases the last digits of it. Bernardo then writes the next perfect square, Silvia erases the last digits of it, and this process continues until thelast two numbers that remain on the board differ by at least 2. Let be the smallestpositive integer not written on the board. For example, if , then the numbers that Bernardo writes are , and the numbers showing on the board after Silviaerases are and , and thus . What is the sum of the digitsof ?2016 AMC 12A Answer Key1 B2 C3 B4 D5 E6 D7 D8 D9 E10 B11 E12 C13 A14 C15 D16 D17 B18 D19 B20 A21 E22 A23 C24 B25 E。
amc12知识点
amc12知识点在美国每年举行的学术类竞赛中，American Mathematics Competition（AMC）12级考试是一场非常重要的测试，其考试内容涵盖了高中数学课程的多个知识点，考生在考试时需要掌握的相关知识点如下：一、代数：代数是AMC12级考试的主要考察内容之一，考生在此考试中需要掌握的知识点主要包括：1、方程的求解：包括一元二次方程，一元三次方程，多项式方程以及配方方程等。
2、分式和集合：包括有理数，有理数分式，有理数函数，有理数方程，组合数，集合性质和集合运算等。
3、代数定义与定理：包括一元、多元函数的定义，函数空间，逻辑与数学等定义，以及组合数学，对数，矩阵，向量，可解性，泰勒展开式，方程的几何意义等定理。
4、几何：几何考试中的知识点包括点、线、面的定义，平面几何元素的定义，平面几何的公式，几何定理，空间几何元素的定义，空间几何的公式和定理等。
5、抽象代数：涉及的知识内容包括：群的定义、子群的定义、群的最基本性质、群的性质、群的等价定义、群的（减去空集）最大子群、有序群、群上函数、群的直和和拆分、群的素性以及群的素元性等。
二、分析：分析在AMC12级考试中也是非常重要的考试内容，需要掌握的知识点包括：1、实数的定义：定义实数的上下界，定义实数的连续性，定义实数的可穿透性，定义实数的有理性，定义实数的完全性，定义无穷大和无穷小等。
2、极限：涵盖了极限的定义，极限的性质，极限的解释，一阶导数，二阶导数，多元函数的一二阶导数，函数的极值，梯形公式，距离内切圆等知识点。
3、函数：函数是AMC12级考试中的重要考点，涉及的知识有定义域，值域，单调性，非单调性，函数的增减性，函数的最大值与最小值，函数的偶函数，奇函数，奇偶函数，偶函数的性质，非减函数，反函数，函数的奇元性，函数的唯一性，函数的可导性，函数的可积性，函数的可微性，函数的积分，积分函数的性质，函数的连续性，函数的紧致性，函数的序号，函数的反向性，函数的一致性，函数的可算性，函数的多项式，函数的可分解性，函数的完全性，函数可数性，函数的定义域，函数的轴对称，函数的完整性，函数的稳定性，函数的对称性，函数的半轴对称，函数的单调性等内容。
赏析几道美国AMC12数学竞赛题
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2021年美国数学竞赛AMC12A 真题加答案解析
2021 AMC 12A Peeyush Pandaya et al.February 20211 Answer1.B2. D3. D4.D5.E Key6.C7.D8.C9.C10.E11.C12.A13.B14.E15. D16.C17.D18.E19.C20.B21.A22.D23.D24.D25.E2 Problems and SolutionsProblem 1. What is the value of21+2+3- (2¹+2²+23)?(A)0 (B)50 (C)52 (D)54 (E)57Solution.2⁶-(2¹+2²+23)=64-(2+4+8)=64-14=(B)50Problem 2.Under what conditionsisva²+62=a+btrue,whereaand bare real numbers?(A)It is never true(B) It is true if and only ifab=0(C) It is true if and only ifa+b≥0(D)It is true ifandonlyifab=0anda+b≥0(E)It is always trueSolution. It is clear that both sides of the equation must be nonnegative.Squaring,a²+b²=a²+2ab+b2→ab=0The answer is (D)Problem 3.The sum of two natural numbers is 17,402.One of the two numbers is divisible by 10.If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?(A)10,272 (B)11,700 (C)13,362 (D)14,238 (E)15,426Solution. Let the first number mentioned be 10n;the second is n. Then10n+n=17,402,from which it follows thatProblem 4. Tom has a collection of 13 snakes,4 of which are purple and 5 of which are happy. He observes that●all of his happy snakes can add,·none of his purple snakes can subtract●all of his snakes that can't subtract also can't add.Which of these conclusions can be drawn about Tom's snakes?Solution. Together, the second and third conditions imply that none of Tom's purple snakes can add.Thus,(D) is correct: happy snakes are not purple.Problem 5.When a student multiplied the number 66 by the repeating decimal,where a and b are digits, he did not notice the notation and just multiplied 66 times 1.ab. Later he found that his answer is 0.5 less than the correct answer. What is the 2-digit integer gb? (A)15 (B)30 (C)45 (D)60 (E)75Solution. The student computed 66 ; the correct answer is 66 . Thus,Problem 6.A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is . When 4 black cards are added to the deck, the probability of choosing red becomes . How many cards were in the deck originally?(A)6 (B)9 (C) 12 (D)15 (E)18Solution. If the deck begins with x red cards and 3x cards in total, thenProblem 7.What is the least possible value of(ry- 1)²+(x+y)2 for real numbersa and y?Solution. We have(ry- 1)2+(x+y)2=(ry)2-2ry+1+x²+2ry+v²=x2v²+z²+y2+1= (r²+1)(y²+1),which achieves a minimum of (D)1 atx=y=0.D Do D ₁ D ₂D ₃ D ₁ D ₅D ₆ D ₇ D ₈ D, D1o Problem 8.A sequence of numbers isdefinedbyDo=0,D ₁=0,Dz=1,andDn=Dn- 1+Dn-3 forn≥3.What are the parities(evenness oroddness)of the triple of numbers(D2021,D2022,D2023), whereE denotes even and O denotes odd?Solution.0/10 01 D2+Do=1+0=1 D ₃+Di=1+0=1D ₄+Dz=1+1=0 Ds+D ₃=0+1=1D ₆+D4=1+1=0D-+Ds=0+0=0D ₈+D ₆=0+1=1Dg+D ₇=1+0=1We can see that the pattern repeats in cycles of length7.and as 2021=5 mod7,we have D2021= Ds,D2022=D6,D2023=D7→(C)(E,O,E) Problem 9.Which of the following is equivalent to(2+3)(2²+3²)(2⁴+34)(2⁸+3⁸)(216+316)(2³²+3³2)(264+364)?(A)3127+2127 (B)3127+2127+2.363+3.263 (C)3128-2128 (D)3128+2128 (E)5127Solution.(3-2)(2+3)(2²+3²)(2⁴+3⁴)(2⁸+3⁸)(216+316)(2³²+3³2)(264+364)=(3²-2²)(2²+3²)(2⁴+3⁴)(2⁸+3⁸)(216+316)(232+3³2)(264+364)=(3⁴-2⁴)(2⁸+3⁸)(216+316)(232+332)(264+364)=(C)3128-2129=(364-264)(264+364)Problem 10.Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are 3cm and 6cm. Into each cone is dropped a spherical marble of radius lcm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level the narrow cone to the rise of the liquid level in the wide cone?(A)1:1 (B)47:43 (C) 2:1 (D)40:13 (E)4:1Solution. The two cones have equal volume, so the height of the first is times that of the second. Since the volumes increase by equal proportions, the heights increase by equal proportions. Thus, the ratio of the rise in liquid levels is (E)4:1Problem 11.A laser is placed at the point(3,5). The laser beam travels in a straight rry wants the beam to hit and bounce off the y-axis, then hit and bounce off the x-axis,then hit the point (7,5). What is the total distance the beam will travel along this path?(A)2√10 (B)5√2(C)10√2(D)15√2 (E)10√5Solution. Reflect about the y-axis then the z-axis. It is well-known that the image under the two reflections must be a straight line.The answer isv(3-(-7))²+(5-(-5)= (C) 10√2Problem 12.All the roots of polynomialz6- 10z ⁵+Az ⁴+Bz³+Cz²+Dz+16are positiveintegers, possibly repeated. What is the value of B?(A)-88 (B)-80 (C)-64 (D)-41 (E)-40Solution. By Vieta's,the sum of the 6 roots is 10 and the product is 16,hence they are all powers of 2. It is not hard to find that the only working unordered sextuple is(2,2,2,2,1,1). As(z-2)4=24-8z3+24z2-32z+16 and(z- 1)2=z2-2z+1.the z3 coefficient is -8.1+24 · (-2)+(-32) · 1= (B)-88Problem 13.Of the following complex numbers z, which has the property that z5 has the greatest real part?(A)-2 (B)-√3+i (C)-√2+√2i Solution. The magnitude of each complex number is the same, so it suffices to look at the argu- ment. The angles are π, ,and ,which after raising to the 5th power give π, and . We seek the angle that reaches farthest to the right(smallest argument),which is . Thus, our answer is (B)-√3+iProblem 14.What is the value of(A)21 (B)100logs3 (C)200log35 (D)2,200 (E)21,000Solution.And,Therefore, their product is210logs3.100log35=(E)21,000(D)- 1+√3i (E)2iProblem 15.A choir director must select a group of singers from among his 6 tenors and 8 basses. The only requirements are that the difference between the numbers of tenors and basses must be a multiple of 4,and the group must have at least one singer.Let be the number of groups that could be selected. What is the remainder when N is divided by 100?(A)47 (B)48 (C)83 (D)95 (E)96Solution. Suppose we mark down(1) the tenors that are in the group,and(2)the basses that aren't in the group. Then we necessarily mark down a number of people that is a multiple of 4. This is also sufficient; we mark down some people numbering a multiple of 4, then select the marked tenors and unmarked basses to form our choir. Clearly, we just mark down at least one person. The answer is thus(D)95Problem 16.In the following list of numbers, the integer n appears times in the list for l≤n≤200.1,2,2,3,3,3,4,4,4,4,.…,200,200,...200.What is the median of the numbers in this list?(A)100.5 (B)134 (C)142 (D)150.5 (E)167Solution.For general n, we have numbers. We want to approximate a such that is close to . Since the formula is a quadratic in n and we are halving this value,we can find that a is approximately .Plugging in n = 200,this is about 100√2,or 141.Of the answer choices, (C)142 is the closest, and indeed it is our answer.To verify, we can see that and ), so clearly 142 works.Problem 17.Trapezoid ABCD has ABICD,BC=CD=43,and AD1BD.Let O be the intersection of the diagonals AC and BD,and let P be the midpoint of BD.Given that OP=11, the length AD can be written in the form myn,where m and n are positive integers and n is not divisible by the square of any prime. What is m+n?(A)65 (B)132 (C)157 (D)194 (E)215Solution. Let M be the intersection of CPand AB.Since DCBMisakite,andCMIBD,we have MP1PB,and by considering the homothethy taking △MBD to △ABD with scale factor 2,we can see that M is the midpoint of AB.In particular,we haveSince AD 1BD,we have AD1DOandthusZADO=90°,andasCD=CB,wehaveCP1BD and ZCPD=2CPO=90°.Also,ZAOD=ZCOP,so △AOD~ △COP.Therefore,so DO=22.Thus,AD=√AB²-BD²=√86²-66²=4√ 190→m+n= (D)194The desired answer isProblem 18.Let f be afunction defined on the set of positive rational numbers with the property that f(a ·b)=f(a)+f(b)for all positive rational numbersa and b.Suppose that falso has the property that f(p)= pfor every prime numberp.For which of the following numbers zis f(x)<0?(A)整(B) (C) (D) (E) 51Solution. Note that f(a ·1)=f(a)+f(1)= f(1)=0,andIn particular, it follows by induction that f(p*)= kp for each k ∈Z.Thus,(A) f(2-5. 17)=-5·2+17=7(B) f(2-4. 11)=-4·2+11=3(C) f(3-2.7)=-2.3+7=1(D)f(2- 1.3- 1.7)=- 1.2+(- 1) ·3+7=2(E)f(52.11- 1)=2.5- 11=- 125The answer is (E)11Problem 19.How many solutions does the equation sinclosed interval [0,π]?(A)0 (B)1 (C) 2 (D) 3 (E)4Solution. Note on the interval , the left-hand side is negative while the right-hand side is positive.We thus restrict our attention to ]. The arguments cosx and sz are both between 0 and .ForsinA=cosBinA,B ∈[0,],we must have . This implies sinz+cosz=1,hence . There are (C) 2 solutions.Problem 20.Suppose that on a parabola with vertexV and focus F there exists a point Asuch that AF=20 and AV=21.What is the sum of all possible values of the length FV?(A)13 (B) 40 (C) (D)14 (E)Solution. Let the directrix be the x-axis,F=(0,2d),V=(0,d),A=(x,y),andB=(0,y)for some d>0.By the definition of a parabola,y=20.We compute x in two ways:x²=AF²-BF²=20²-|20-2d|2=AV²-BV²=21²-|20-d²Subtracting,O=20²-21²+(20-d)²-(20-2d)2=3d²-40d+41.The sum of all possible values ofProblem 21.The five solutions to the equation(z- 1)(z²+2z+4)(z²+4z+6)=0may be written in theformzk+ykiforl≤k≤5,where xk and yk are real.Let E be the unique ellipse that passes through the points(Ti,yi),(x2,32),(r3,Y3),(x4,y4)and(xs,ys).The eccentricity ofE can be written in the form ,where m and n are relatively prime positive integers. What is m+n?(Recall that the eccentrictiy of an ellipse E is the ratio,where 2a is the length of the major axis ofE and 2c is the distance between its two foci.)(A)7 (B)9 (C)11 (D)13 (E)15Solution. The roots of the polynomial arez=1,z=- 1±i√3,andz=-2±i√2,hence the five points onE are(1,0),( - 1,土√3),( - 2,±√2) .By symmetry through the x-axis, the ellipse is of the formE: a(x-r)²+by²=1.We then have the relationsa(1-r)²=1a(1+r)²+3b=1a(2+r)²+2b=1.Eliminating b from the latter two,1=3[a(2+r)²+2]-2[a(1+r)²+36]=a(r²+8r+10),henceit follows that and so the eccentricity is 1/√6. The requested sum is1+6= (A)7Problem 22.Suppose that the roots of the polynomialP(x)=x³+ax²+bx+care cos 2π,COS 47 and cos ,where angles are in radians. What is abc?(C) (D) (E)Solution. Recall1+e2m/7+ …+e12mi/7=0→e2mi/7+e4xi/7+e6mi/7=- 1/2,and in particular caNote are solutions to the equation cosb+cos28+cos30=- 1/2,so lettingx=co sθimplies thathas roots ,COS 4π7Thus, the polynomial in question is and the requested answer isRemark. Perhaps it is easier to motivate the solution as follows.Lett=e2mi/7andx=t+t- 1= 2cos2π/7.Thenx²-2=t²+t-2andx³-3x=t³+t-3.Moreover,t⁶+t⁵+ …+1=0impliest³+t²+t+t- 1+t-2+t-3=- 1,i.e.x+(x²-2)+(r³-3r) has root 2cos2π/7.It certainly seems logical that the Galois conjugates of 2cos would be 2cos and 2cos (especially given the phrasing of the problem),so simply replacex → to get the desired polynomial form.Let w = e2ix/7.Note thatLet these be r,s,t respectively. By Vieta's formulas, note that the desired quantity is(-rst)(rs+st+tr)(-r-s-t)=(r+s+t)(rs+st+tr)(rst).Note that 1+w+ …+w⁶=0.We haveThen,FinallyMultiplying yields the answer of (D)1 32Solution Manual 2021AMC12AProblem 23.Frieda the frog begins a sequence of hops on a 3×3 grid of squares,moving one square on each hop and choosing at random the direction of each hop up,down, left,or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid,she ”wraps around”and jumps to the opposite edge.For example if Frieda begins in the center square and makes two hops ”up”,the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge,landing in the bottom row middle square.Suppose Frieda starts from the center square, makes at most four hops at random,and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?(A)G (B) (D)强(E)8Solution. We complementary count,and determine the probability we never reach a corner square. Denote by A the center,and B a square adjacent to the center. Then the first hop lands on B.● If the second hop lands on A(with probability ,then the third hop lands on B always,andthere is a chance the fourth hop lands on a non-corner square. The probability in this case is● If the second hop lands on B, then there is a chance the third hop lands on A,and achance the third hop lands on B.In the former subcase, the fourth hop always lands on a non-corner square, and in the latter subcase, there is a chance the fourth hop lands on a non-corner square. The probability in this case isThe requested probability isProblem 24.SemicircleT has diameter AB of length14.Circle Ωlies tangent to AB at a point P and intersectsI at pointsQandR.IfQR=3√3and ZQPR=60°,then the area of △PQR is ,where a and c are relatively prime positive integers, and b is a positive integer not divisible by the square of any prime. What isa+b+c?(A)110 (B)114 (C)118 (D)122 (E)126Solution. First, by Extended Law of Sines,we have that the radius of Q is .Let M be the midpoint of QR,O be the center ofT,and X be the center of 2.Since △OQR is isosceles, OM is perpendicular to QR. Thus we have that X lies on OM since it is the circumcenter of △puting lengths, we hav thagorean theorem on △ORM andfrom isosceles triangle XQR. Thus and OP=4 from Pythagorean theorem on △OXP.To find [PQR],we will find the height from P to QR.Let the foot of the perpendicular fromX to the P-altitude be D. Since PDⅡOM,we know that △XDP~△OPX.This means that . Now note that the bottom portion of the P-altitude after subtractingPD is equal to XM,so the height of the triangle is . The area is simply2021 AMC 12A11 Solution ManualProblem 25.Let d(n)denote the number of positive integers that divide n,including l and n. For example,d(1)=1,d(2)=2,and d(12)=6.(This function is known as the divisor function.) LetThere is a unique positive integer N such that f(N)> f(n)for all positive integers n≠N.What is the sum of the digits of N?(A)5 (B)6 (C)7 (D)8 (E) 9Solution. Letn=II;p',wherepi=2,Pz=3,etc.are the primes in increasing order and e; are nonnegative integers. ThenIt is equivalent to maximize Thus, it remains to find the optimal e; for each i. We go term-by-term, noting that it is only necessary to check until the expression first decreases, as exponentials increase more quickly than polynomials.ei 0 1 2 3 4 0 1 2 3 0 1 2 0 1 2 ((ez+1)3)/p⁸1³/20=1 23/2¹=433/2²=6.25 4³/2³=85³/2⁴<8 13/3⁰=1 23/31≈2.67 3³/32=3 43/3³<3 13/5⁰=1 2³/5¹=1.6 3³/5²=1.08 13/70=1 23/71≈1.14 3³/7²<1Note that we do not need to check p≥11,ase;=1yields which is suboptimal.Thus,the answer isN=23.32.5.7=2520,which has a Remark:It is sufficient to stop at p=3,for 3²|Nleaves N among the answer choices.digit sum of (E)9only one possibility for the digit sum ofi 1 1 1 1 1 2 2 2 2 3 3 3 4 4 4。
AMC美国数学竞赛AMCB试题及答案解析
AMC美国数学竞赛AMCB试题及答案解析A M C美国数学竞赛A M C B试题及答案解析 The latest revision on November 22, 20202003 AMC 10B1、Which of the following is the same as2、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 cost3、The sum of 5 consecutive even integers is less than the sum of the rst consecutive odd counting numbers. What is the smallest of the even integers4、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 gure. She plants one flower per square foot in each region. Asters cost 1 each, begonias each, cannas 2 each, dahlias each, and Easter lilies 3 each. What is the least possible cost, in dollars, for hergarden5、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 of feet 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.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 ofthe following7、The symbolism denotes the largest integer not exceeding . For example. , and . Compute.8、The second and fourth terms of a geometric sequence are and . Which of the following is a possible first term9、Find the value of that satisfies the equation10、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 increased11、A line with slope intersects a line with slope at the point. What is the distance between the -intercepts of these two lines12、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 origin al portion.13、Let denote the sum of the digits of the positive integer . For example, and . For how many two-digit values of is14、Given that , where both and are positive integers, find the smallest possible value for .15、There are players in a singles tennis tournament. The tournament is single elimination, meaning that a player who losesa 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 is16、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.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 ll 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 radius18、What is the largest integer that is a divisor offor all positive even integers19、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 semicircles20、In rectangle , and . Points and are onso that and . Lines and intersect at . Find the area of .21、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 regardlessof the color you pulled out. What is the probability that all beadsin the bag are red after three such replacements22、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 occur23、A regular octagon has an area of one square unit.What is the area of the rectangle24、The rst four terms in an arithmetic sequence are , , , and , in that order. What is the fth term25、How many distinct four-digit numbers are divisible by and haveas their last two digits。
美国高中数学测验 AMC12之机率问题(下)
美国高中数学测验 AMC12之机率问题(下)洪伟诚 . 李俊贤 . 蔡诚祐 . 何家兴 . 张福春关键词：高中数学几何机率前述两节的问题皆建立在样本点个数为可数时的情况,接下来将介绍一不可数的无穷样本空间S且利用此空间的一些几何测量m(S),例如长度、面积、或者体积,来求A事件的机率。
而A事件的机率可用A事件之几何测量与样本空间S之几何测量的比例来计算,其形式有下列三种:P(A)=A的长度S的长度或P(A)=A的面积S的面积或P(A)=A的体积S的体积注:我们必须假设一不可数无穷样本空间S满足均匀性质,这样才能做以上的几何机率。
几何测量—长度所谓长度的几何测量,表示其无穷样本空间可用一线段、数线或是时间轴⋯等表示,则考虑某事件的机率时,只需探讨此事件所占的线段(或数线)与样本空间相对的长度比值即可,下列为利用长度测量来求的机率问题。
例1.(1972 AMC12 #17) 随机将一条线切为两段,试问较长的一段至少是较短的一段的x倍(其中x¸1)的机率为多少?•(B)2x•(C)1x+1•(D)1x•(E)2x+1解:(E)假设线段AB被切为两段,若较长一段(标记为l1,长度为sx)为较短一段(标记为l2,长度为s)的x倍,则线段AB总长度为(x+1)s,因此切点会落于l2中的机率为\dfrac{1}{x+1}。
但因线段的切法可能为(l1,l2)或是(l2,l21)两种情况,如下图故所求机率为2x+1。
例2.(2007 AMC12B #13)有一交通号志以下列的循环重复的运作:绿灯30秒,然后黄灯3秒,之后再转红灯30秒。
利亚随机挑选三秒钟的区间去注视号志灯,试问号志灯在转换颜色时,利亚正在注视的机率为多少?•(A)163•(B)121•(C)110•(D)17解:(D)由题意知,交通号志运作一循环需时63秒,而若利亚所注视的时间在当绿灯转变为黄灯、黄灯转变为红灯或红灯转变为绿灯的前三秒钟内,则利亚会看到号志颜色正在转变,如下图所示故所求机率为3+3+363=963=17例3.(2009 AMC12B #18) 瑞吉儿与罗伯特在一圆形的跑道上跑步,其中瑞吉儿以逆时钟方向跑且跑完一圈需时90秒,而罗伯特以顺时钟方向跑且跑完一圈需时80秒。
AMC12真题及答案精编版
2016 AMC12 AProblem 1What is the value of ?SolutionProblem 2For what value of does ?SolutionProblem 3The remainder can be defined for all real numbers and with bywhere denotes the greatest integer less than or equal to . What is the value of ?SolutionProblem 4The mean, median, and mode of the data values are all equal to . What is the value of ?SolutionProblem 5Goldbach's conjecture states that every even integer greater than 2 can be written as the sum of two prime numbers (for example, ). So far, no one has been able to prove that the conjecture is true, and no one has found a counterexample to show that the conjecture is false. What would a counterexample consist of?SolutionProblem 6A triangular array of coins has coin in the first row, coins in the second row, coins in the third row, and so on up to coins in the th row. What is the sum of the digits of ?SolutionProblem 7Which of these describes the graph of ?SolutionProblem 8What is the area of the shaded region of the given rectangle?SolutionProblem 9The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the otherfour small squares as shown. The common side length is , where and are positive integers. What is ?SolutionProblem 10Five friends sat in a movie theater in a row containing seats, numbered to from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?SolutionProblem 11Each of the students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. Thereare students who cannot sing, students who cannot dance, and students who cannot act. How many students have two of these talents?SolutionProblem 12In , , , and . Point lies on ,and bisects . Point lies on , and bisects . The bisectors intersect at . What is the ratio : ?SolutionProblem 13Let be a positive multiple of . One red ball and green balls are arranged in a line in random order. Let be the probability that at least of the green balls are on the same side of the red ball. Observe that and that approaches as grows large. What is the sum of the digits of the least value of such that ?SolutionProblem 14Each vertex of a cube is to be labeled with an integer from through , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?SolutionProblem 15Circles with centers and , having radii and , respectively, lie on the same side of line and are tangent to at and , respectively, with between and . Thecircle with center is externally tangent to each of the other two circles. What is the area of triangle ?SolutionProblem 16The graphs of and are plotted on the same set of axes. How many points in the plane with positive -coordinates lie on two or more of the graphs?SolutionProblem 17Let be a square. Let and be the centers, respectively, of equilateral triangles with bases and each exterior to the square. What is the ratio of the area of square to the area of square ?SolutionProblem 18For some positive integer the number has positive integer divisors,including and the number How many positive integer divisors does thenumber have?SolutionProblem 19Jerry starts at on the real number line. He tosses a fair coin times. When he gets heads, he moves unit in the positive direction; when he gets tails, he moves unit in the negative direction. The probability that he reaches at some time during this processis where and are relatively prime positive integers. What is (For example, he succeeds if his sequence of tosses is )SolutionProblem 20A binary operation has the properties that andthat for all nonzero real numbers and (Here the dot represents the usual multiplication operation.) The solution to the equation can bewritten as where and are relatively prime positive integers. What isSolutionProblem 21A quadrilateral is inscribed in a circle of radius Three of the sides of this quadrilateral have length What is the length of its fourth side?SolutionProblem 22How many ordered triples of positive integerssatisfy and ?SolutionProblem 23Three numbers in the interval are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?SolutionProblem 24There is a smallest positive real number such that there exists a positive realnumber such that all the roots of the polynomial are real. In fact, for this value of the value of is unique. What is the value ofSolutionProblem 25Let be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with digits. Every time Bernardo writes a number, Silvia erases the last digits of it. Bernardo then writes the next perfect square, Silvia erases the last digits of it, and this process continues until thelast two numbers that remain on the board differ by at least 2. Let be the smallestpositive integer not written on the board. For example, if , then the numbers that Bernardo writes are , and the numbers showing on the board after Silviaerases are and , and thus . What is the sum of the digitsof ?2016 AMC 12A Answer Key1 B2 C3 B4 D5 E6 D7 D8 D9 E10 B11 E12 C13 A14 C15 D16 D17 B18 D19 B20 A21 E22 A23 C24 B25 E。
美国数学竞赛2020年真题AMC10 A(真题加详解)
2020 AMC 10A Solution Problem1What value of satisfiesSolutionAdding to bothsides, . Problem2The numbers and have an average (arithmetic mean) of . What is the average of and ?SolutionThe arithmetic mean of the numbers and is equalto . Solving for , we get . Dividing by to find the average of the twonumbers and gives .Problem3Assuming , , and , what is the value in simplest form of the following expression?SolutionNote that is times .Likewise, is times and is times . Therefore, the product of the given fractionequals .Problem4A driver travels for hours at miles per hour, during which her cargets miles per gallon of gasoline. She is paid per mile, and her onlyexpense is gasoline at per gallon. What is her net rate of pay, in dollars per hour, after this expense?SolutionSince the driver travels 60 miles per hour and each hour she uses 2 gallons of gasoline, she spends $4 per hour on gas. If she gets $0.50 per mile, then she gets $30 per hour of driving. Subtracting the gas cost, her net rate of pay perhour is .Problem5What is the sum of all real numbers for whichSolution 1Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.Case 1:The equation yields , which is equalto . Therefore, the two values for the positive caseis and .Case 2:Similarly, taking the nonpositive case for the value inside the absolute value notation yields . Factoring and simplifyinggives , so the only value for this case is .Summing all the values results in . Solution 2We have theequations and .Notice that the second is a perfect square with a double root at , and the first has real roots. By Vieta's, the sum of the roots of the first equationis .Problem6How many -digit positive integers (that is, integers between and , inclusive) having only even digits are divisible bySolutionThe ones digit, for all numbers divisible by 5, must be either or . However, from the restriction in the problem, it must be even, giving us exactly one choice () for this digit. For the middle two digits, we may choose any even integerfrom , meaning that we have total options. For the first digit, we follow similar intuition but realize that it cannot be , hence giving us 4 possibilities. Therefore, using the multiplication rule, weget . ~ciceroniiProblem7The integers from to inclusive, can be arranged to form a -by-square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?SolutionWithout loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by is the total value per row. The sum of the integersis , and the common sum is .Solution 2Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get as our answer. ~BaolanProblem8What is the value ofSolution 1Split the even numbers and the odd numbers apart. If we group every 2 evennumbers together and add them, we get a total of . Summing the odd numbers is equivalent to summing the first 100 odd numbers, which is equal to . Adding these two, we obtain the answerof .Solution 2 (bashy)We can break this entire sum down into integer bits, in which the sum is , where is the first integer in this bit. We can find that the first sum of every sequence is , which we plug in for the bits in the entire sequenceis , so then we can plug it into the first term of every sequence equation we gotabove , and so the sum of every bit is , and we only found the value of , the sum of the sequenceis .-middletonkidsSolution 3Another solution involves adding everything and subtracting out what is not needed. The first step involvessolving. To do this, we can simply multiply and and divide by to getus . The next step involves subtracting out the numbers with minus signs. We actually have to do this twice, because we need to take out the numbers we weren’t supposed to add and then subtract them from the problem. Then, we can see that from to , incrementing by , there are numbers that we have to subtract. To do this we can do times divided by , and then we can multiply by , because we are counting by fours, not ones. Our answer will be , but remember, we have to do this twice. Once we do that,we will get . Finally, we just have to do , and our answer is .—Solution 4In this solution, we group every 4 terms. Our groups shouldbe: , ,, .... We add them together to get this expression: . This can be rewrittenas . We add this toget . ~BaolanSolution 5We can split up this long sum into groups of four integers. Finding the first few sums, we have that , , and . Notice that this is an increasing arithmetic sequence, with a common difference of . We can find the sum of the arithmetic sequence by finding the average of the first and last terms, and then multiplying by the number of terms in the sequence. The first term is , or , the last term is , or , and thereare or terms. So, we have that the sum of the sequenceis , or . ~Arctic_BunnySolution 3Taking the average of the first and last terms, and , we have that the mean of the set is . There are 5 values in each row, column or diagonal, so thevalue of the common sum is , or . ~Arctic_Bunny, edited by KINGLOGICProblem9A single bench section at a school event can hold either adults or children. When bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value ofSolutionThe least common multiple of and is . Therefore, there mustbe adults and children. The total number of benchesis .Solution 2This is similar to Solution 1, with the same basic idea, but we don't need to calculate the LCM. Since both and are prime, their LCM must be theirproduct. So the answer would be . ~Baolan Problem10Seven cubes, whose volumes are , , , , , , and cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?Solution 1The volume of each cube follows the pattern of ascending, for isbetween and .We see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the cube (which is just ). The sides areas can be measured as thesum , giving us . Structurally, if we examine the tower from the top, we see that it really just forms a square of area . Therefore, wecan say that the total surface area is . Alternatively, for the area of the tops, we could have found thesum , giving us as well.~ciceroniiSolution 2It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7, inclusive.First, we will calculate the total surface area of the cubes, ignoring overlap. This valueis. Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of theoverlapped parts of the cubes is thus equal to . Subtracting the overlapped surface area from the total surface area, weget . ~emerald_blockSolution 3 (a bit more tedious than others)It can be seen that the side lengths of the cubes using cube roots are all integers from to , inclusive.Only the cubes with side length and have faces in the surface area and the rest have . Also, since thecubes are stacked, we have to find the difference betweeneach and side length as ranges from to.We then come up withthis:.We then add all of this and get .Problem 11What is the median of the following list of numbersSolution 1We can see that is less than 2020. Therefore, there are ofthe numbers after . Also, there are numbers that are under and equal to . Since is equal to , it, with the other squares, willshift our median's placement up . We can find that the median of the whole set is , and gives us . Our answeris .~aryamSolution 2As we are trying to find the median of a -term set, we must find the average of the th and st terms.Since is slightly greater than , we know thatthe perfect squares through are less than , and the rest aregreater. Thus, from the number to the number , thereare terms. Since is less than and less than , we will only need to consider theperfect square terms going down from the th term, , after going down terms. Since the th and st terms areonly and terms away from the th term, we can simplysubtract from and from to get the two terms, whichare and . Averaging the two, weget ~emerald_blockSolution 3We want to know the th term and the th term to get the median. We know thatSo numbers are in between to .So the sum of and will result in , which means that is the th number.Also, notice that , which is larger than .Then the th term will be , and similarlythe th term will be .Solving for the median of the two numbers, we getProblem12Triangle is isoceles with .Medians and are perpendicular to each other,and . What is the area ofSolution 1Since quadrilateral has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also notethat has the area of triangle by similarity,so Thus,Solution 2 (Trapezoid)We know that , and since the ratios of its sidesare , the ratio of of their areas is .If is the area of , then trapezoid is the area of .Let's call the intersection of and . Let .Then . Since , and are heights of triangles and , respectively. Both of these triangles have base .Area ofArea ofAdding these two gives us the area of trapezoid , whichis .This is of the triangle, so the area of the triangleis ~quacker88, diagram by programjames1 Solution 3 (Medians)Draw median .Since we know that all medians of a triangle intersect at the incenter, we know that passes through point . We also know that medians of a triangle divide each other into segments of ratio . Knowing this, we can seethat , and since the two segments sumto , and are and , respectively.Finally knowing that the medians divide the triangle into sections of equal area, finding the area of is enough. .The area of . Multiplying this by givesus~quacker88Solution 4 (Triangles)We knowthat , , so .As , we can seethat and with a side ratio of .So , .With that, we can see that , and the area oftrapezoid is 72.As said in solution 1, .-QuadraticFunctions, solution 1 by ???Solution 5 (Only Pythagorean Theorem)Let be the height. Since medians divide each other into a ratio, and the medians have length 12, wehave and . From righttriangle ,so . Since is a median, . From righttriangle ,which implies . Bysymmetry .Applying the Pythagorean Theorem to righttriangle gives,so . Then the areaof isSolution 6 (Drawing)(NOT recommended) Transfer the given diagram, which happens to be to scale, onto a piece of a graph paper. Counting the boxes should give a reliable result since the answer choices are relatively far apart. -LingjunSolution 7Given a triangle with perpendicular medians with lengths and , the area will be .Solution 8 (Fastest)Connect the line segment and it's easy to seequadrilateral has an area of the product of its diagonals dividedby which is . Now, solving for triangle could be an option, but the drawing shows the area of will be less than the quadrilateral meaning the the area of is less than but greater than , leaving onlyone possible answer choice, .Problem 13A frog sitting at the point begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length , and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square withvertices and . What is the probability that thesequence of jumps ends on a vertical side of the squareSolutionDrawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of thishappening is . If the frog goes to the right, it will be in the center ofthe square at , and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is . The probability of this happening is .If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it tohit a vertical wall is . Because there's a chance of the frog going up and down, the total probability for this case is and summing up all thecases,Solution 2Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have "succeeded", while B means that we have a half chance, wecompute .We get , or-yeskaySolution 3If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes . Since it starts on , there is a chance (up,down, or right) it will reach a diagonal on the first jump and chance (left) it will reach the vertical side. The probablity of landing on a verticalis - Lingjun.Solution 4 (Complete States)Let denote the probability of the frog's sequence of jumps ends with ithitting a vertical edge when it is at . Note that by reflective symmetry over the line .Similarly, , and .Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from thatpoint:We have a system of equationsin variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equationgivesPlugging in the third equation into thisgivesNext, plugging in the second and third equation into the first equationyieldsNow plugging in (*) into this, wegetProblem14Real numbers and satisfy and . What is the value ofSolutionContinuing tocombineFrom the givens, it can be concluded that .Also, This meansthat . Substituting this informationinto , wehave . ~PCChess Solution 2As above, we need to calculate . Note that are the roots of andso and .Thuswhere and as in the previous solution. Thus the answer is .Solution 3Note that Now, we only need to find the values of andRecall that andthat We are able to solve thesecond equation, and doing so gets us Plugging this into the first equation, we getIn order to find the value of we find a common denominator so that we can add them together. This getsus Recallingthat and solving this equation, weget Plugging this into the first equation, wegetSolving the original equation, weget ~emerald_blockSolution 4 (Bashing)This is basically bashing using Vieta's formulas to find and (which I highly do not recommend, I only wrote this solution for fun).We use Vieta's to find a quadratic relating and . We set and to be the roots of the quadratic (because , and ). We can solve the quadratic to get theroots and . and are "interchangeable", meaning that it doesn't matter which solution or is, because it'll return the same result when plugged in. So we plug in for and andget as our answer.~BaolanSolution 5 (Bashing Part 2)This usually wouldn't work for most problems like this, but we're lucky that we can quickly expand and factor this expression in this question.We first change the original expression to ,because . This is equalto. We can factor andreduce to. Now our expression is just . Wefactor to get . So the answer would be .Solution 6 (Complete Binomial Theorem)We first simplify the expression to Then, we can solve for and given the system of equations in the problem.Since we can substitute for . Thus, this becomes theequation Multiplying both sides by , weobtain or By the quadratic formula we obtain . We also easily find thatgiven , equals the conjugate of . Thus, plugging our valuesin for and , our expression equalsBy the binomial theorem, we observe that every second terms of theexpansions and will cancel out (since a positive plus a negative of the same absolute value equals zero). We also observe that the other terms notcanceling out are doubled when summing the expansions of . Thus, our expression equals whichequals which equals .Problem15A positive integer divisor of is chosen at random. The probability that thedivisor chosen is a perfect square can be expressed as , where and are relatively prime positive integers. What is ?SolutionThe prime factorization of is . This yields a totalof divisors of In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that and can not be in the prime factorization of a perfect square because there is only one of each in Thus, thereare perfect squares. (For , you can have , , , , , or 0 s, etc.) The probability that the divisor chosen is a perfect squareisProblem16A point is chosen at random within the square in the coordinate plane whose vertices are and . Theprobability that the point is within units of a lattice point is . (A point is a lattice point if and are both integers.) What is to the nearest tenthSolution 1DiagramDiagram by MathandSki Using AsymptoteNote: The diagram represents each unit square of thegiven square.SolutionWe consider an individual one-by-one block.If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius , the area covered by the circles should be . Because of this, and the fact that there are four circles, we writeSolving for , we obtain , where with , we get , and from here, we simplify and seethat ~CrypthesTo be more rigorous, note that since if thenclearly the probability is greater than . This would make sure the above solution works, as if there is overlap with thequartercircles.Solution 2As in the previous solution, we obtain the equation , which simplifies to . Since is slightly more than , is slightly less than . We notice that is slightly morethan , so is roughly ~emerald_blockSolution 3 (Estimating)As above, we find that we need to estimate .Note that we can approximate andso .And so our answer is .Problem 17Define How many integers are there such that ?Solution 1Notice that is a product of many integers. We either need one factor to be 0 or an odd number of negative factors.Case 1: There are 100 integers for whichCase 2: For there to be an odd number of negative factors, must be between an odd number squared and an even number squared. This means that thereare total possible values of . Simplifying, thereare possible numbers.Summing, there are total possible values of . ~PCChess Solution 2Notice that is nonpositive when isbetween and , and , and (inclusive), whichmeans that the amount of valuesequals.This reducesto~ZericSolution 3 (end behavior)We know that is a -degree function with a positive leading coefficient. Thatis, .Since the degree of is even, its end behaviors match. And since theleading coefficient is positive, we know that both ends approach as goes in either direction.So the first time is going to be negative is when it intersects the -axis atan -intercept and it's going to dip below. This happens at , which is the smallest intercept.However, when it hits the next intercept, it's going to go back up again into positive territory, we know this happens at . And when it hits , it's going to dip back into negative territory. Clearly, this is going to continue to snake around the intercepts until .To get the amount of integers below and/or on the -axis, we simply need to count the integers. For example, the amount of integers in betweenthe interval we got earlier, we subtract and addone. integers, so there are four integers in this interval that produce a negative result.Doing this with all of the other intervals, we have. Proceed with Solution 2. ~quacker88Problem18Let be an ordered quadruple of not necessarily distinct integers, each one of them in the set For how many such quadruples is it truethat is odd? (For example, is one such quadruple, because is odd.)SolutionSolution 1 (Parity)In order for to be odd, consider parity. We must have (even)-(odd) or (odd)-(even). There are ways to pick numbers to obtain an even product. There are ways to obtain an odd product. Therefore, the total amount of ways to make oddis .-MidnightSolution 2 (Basically Solution 1 but more in depth)Consider parity. We need exactly one term to be odd, one term to be even. Because of symmetry, we can set to be odd and to be even, then multiply by If is odd, both and must be odd, therefore thereare possibilities for Consider Let us say that is even. Then there are possibilities for However, can be odd, in which case we have more possibilities for Thus there are ways forus to choose and ways for us to choose Therefore, also consideringsymmetry, we have total values ofSolution 3 (Complementary Counting)There are 4 ways to choose any number independently and 2 ways to choose any odd number independently. To get an even products, wecount: , which is . The number of ways to get an odd product can be counted like so: , which is , or . So, for oneproduct to be odd the other to be even: (order matters). ~ Anonymous and Arctic_BunnySolution 4 (Solution 3 but more in depth)We use complementary counting: If the difference is even, then we can subtract those cases. There are a total of cases.For an even difference, we have (even)-(even) or (odd-odd).From Solution 3:"There are 4 ways to choose any number independently and 2 ways to choose any odd number independently. even products:(number)*(number)-(odd)*(odd): . odd products: (odd)*(odd): ."With this, we easily calculate . Problem19As shown in the figure below, a regular dodecahedron (the polyhedron consisting of congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?DiagramSolution 1Since we start at the top face and end at the bottom face without moving from the lower ring to the upper ring or revisiting a face, our journey must consist of the top face, a series of faces in the upper ring, a series of faces in the lower ring, and the bottom face, in that order.We have choices for which face we visit first on the top ring. From there, we have choices for how far around the top ring we go before movingdown: or faces around clockwise, or faces aroundcounterclockwise, or immediately going down to the lower ring without visiting any other faces in the upper ring.We then have choices for which lower ring face to visit first (since every upper-ring face is adjacent to exactly lower-ring faces) and then once again choices for how to travel around the lower ring. We then proceed to the bottom face, completing the trip.Multiplying together all the numbers of choices we have, weget .Solution 2Swap the faces as vertices and the vertices as faces. Then, this problem is the same as 2016 AIME I #3which had an answerof .Problem20Quadrilateral satisfiesand Diagonals and intersect at point and What is the area of quadrilateralSolution 1 (Just Drop An Altitude)It's crucial to draw a good diagram for this one.Since and , we get . Now weneed to find to get the area of the whole quadrilateral. Drop an altitudefrom to and call the point of intersection . Let .Since , then . By dropping this altitude, we can also see two similar triangles, and .Since is , and , we get that . Now, if we redraw another diagram just of , we getthat . Now expanding, simplifying, and dividing by the GCF, we get . This factorsto . Since lengths cannot be negative, .Since , .So(I'm very sorry if you're a visual learner but now you have a diagram by ciceronii) ~ Solution by Ultraman~ Diagram by ciceroniiSolution 2 (Pro Guessing Strats)We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of thelegs is a multiple of 20. Guess that the legs are equal to and , and because the hypotenuse is 20 we get . Testing small numbers, we get that when and , is indeed a square. The area of the triangle is thus 60, so the answer is .~tigershark22 ~(edited by HappyHuman)Solution 3 (coordinates)Let the pointsbe , , , ,and , respectively. Since lies on line , we know that . Furthermore, since , lies on the circle with diameter , so . Solving for and with these equations, we get the solutions and . We immediately discardthe solution as should be negative. Thus, we concludethat.Solution 4 (Trigonometry)Let and Using Law of Sineson we get and LoSon yieldsDivide the two toget Now,and solve the quadratic, taking the positive solution (C is acute) toget Soif then and By Pythagorean Theorem, and the answeris(This solution is incomplete, can someone complete it please-Lingjun) ok Latex edited by kc5170We could use the famous m-n rule in trigonometry in triangle ABC with Point E [Unable to write it here.Could anybody write the expression] We will find that BD is angle bisector of triangle ABC(because we will get tan (x)=1) Therefore by converse of angle bisector theorem AB:BC = 1:3. By using phythagorean theorem we have values of AB and AC. AB.AC = 120. Adding area of ABC and ACD Answer••360Problem21There exists a unique strictly increasing sequence of nonnegativeintegers suchthat What isSolution 1First, substitute with . Then, the given equationbecomes . Now consider only . Thisequals . Notethat equals , since the sum of a geometricsequence is . Thus, we can see that forms the sum of 17 different powers of 2. Applying the same method to eachof , , ... , , we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us . But we must count also the term. Thus, Our answeris .~seanyoon777Solution 2(This is similar to solution 1) Let . Then, . The LHS canbe rewrittenas。
美国数学竞赛AMC题目及答案
2.3.What is the value of ?4.Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?5.Hammie is in the grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?6.The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, . What is the missing number in the top row?7.Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?8.A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?9.The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer?10.What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?11. Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?12. At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save?13. When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?14. Let the two digits be and .The correct score was . Clara misinterpreted it as . The difference between the two is which factors into . Therefore, since the difference is a multiple of 9, the only answerchoice that is a multiple of 9 is .15. If , , and , what is the product of , , and ?16. A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of -graders to -graders is , and the the ratio of -graders to -graders is . What is the smallest number of students that could be participating in the project?17. The sum of six consecutive positive integers is 2013. What is the largest of these six integers?18. Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?--Arpanliku 16:22, 27 November 2013 (EST) Courtesy of Lord.of.AMC19. Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest?20. A rectangle is inscribed in a semicircle with longer side on the diameter. What is the area of the semicircle?21. Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?22. Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?23. Angle of is a right angle. The sides of are the diameters of semicircles as shown. The area of the semicircle on equals , and the arc of the semicircle on has length .What is the radius of the semicircle on ?24. Squares , , and are equal in area. Points and are the midpoints of sidesand , respectively. What is the ratio of the area of the shaded pentagon to the sum of the areas of the three squares?25. A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are inches, inches, and inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of theball travels over the course from A to B?1.2.The 50% off price of half a pound of fish is $3, so the 100%, or the regular price, of a half pound of fish is $6. Consequently, if half a pound of fish costs $6, then a whole pound of fish is dollars.3.Notice that we can pair up every two numbers to make a sum of 1:Therefore, the answer is .4.Each of her seven friends paid to cover Judi's portion. Therefore, Judi's portion must be . Since Judi was supposed to pay of the total bill, the total bill must be .5.The median here is obviously less than the mean, so option (A) and (B) are out.Lining up the numbers (5, 5, 6, 8, 106), we see that the median weight is 6 pounds.The average weight of the five kids is .Therefore, the average weight is bigger, by pounds, making the answer.6.Solution 1: Working BackwardsLet the value in the empty box in the middle row be , and the value in the empty box in the top row be . is the answer we're looking for.We see that , making .It follows that , so .Solution 2: Jumping Back to the StartAnother way to do this problem is to realize what makes up the bottommost number. This method doesn't work quite as well for this problem, but in a larger tree, it might be faster. (In this case, Solution 1 would be faster since there's only two missing numbers.)Again, let the value in the empty box in the middle row be , and the value in the empty box in the top row be . is the answer we're looking for.We can write some equations:Now we can substitute into the first equation using the two others:7.If Trey saw , then he saw .2 minutes and 45 seconds can also be expressed as seconds.Trey's rate of seeing cars, , can be multiplied by on the top and bottom (and preserve the same rate):. It follows that the most likely number of cars is . Solution 2minutes and seconds is equal to .Since Trey probably counts around cars every seconds, there are groups of cars that Trey most likely counts. Since , the closest answer choice is .8.First, there are ways to flip the coins, in order.The ways to get two consecutive heads are HHT and THH.The way to get three consecutive heads is HHH.Therefore, the probability of flipping at least two consecutive heads is .9.This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that .However, because the first term is and not , the solution to the problem is10. To find either the LCM or the GCF of two numbers, always prime factorize first.The prime factorization of .The prime factorization of .Then, find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is ). Multiply all of these to get 5940.For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. = 18.Thus the answer = = .We start off with a similar approach as the original solution. From the prime factorizations, the GCF is .It is a well known fact that . So we have,.Dividing by yields .Therefore, .11. We use that fact that . Let d= distance, r= rate or speed, and t=time. In this case, let represent the time.On Monday, he was at a rate of . So, .For Wednesday, he walked at a rate of . Therefore, .On Friday, he walked at a rate of . So, .Adding up the hours yields + + = .We now find the amount of time Grandfather would have taken if he walked at per day. Set up the equation, .To find the amount of time saved, subtract the two amounts: - = . To convert this to minutes, we multiply by .Thus, the solution to this problem is12. First, find the amount of money one will pay for three sandals without the discount. We have.Then, find the amount of money using the discount: .Finding the percentage yields .To find the percent saved, we have13. Let the two digits be and .The correct score was . Clara misinterpreted it as . The difference between the two is which factors into . Therefore, since the difference is a multiple of 9, the only answerchoice that is a multiple of 9 is .14. The probability that both show a green bean is . The probability that both show a red bean is . Therefore the probability is15.Therefore, .Therefore, .To most people, it would not be immediately evident that , so we can multiply 6's until we get the desired number:, so .Therefore the answer is .16. Solution 1: AlgebraWe multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:Therefore, the ratio of 8th graders to 7th graders to 6th graders is . Since the ratio is in lowest terms, the smallest number of students participating in the project is .Solution 2: FakesolvingThe number of 8th graders has to be a multiple of 8 and 5, so assume it is 40 (the smallest possibility). Then there are 6th graders and 7th graders. The numbers of students is17. Solution 1The mean of these numbers is . Therefore the numbers are, so the answer isSolution 2Let the number be . Then our desired number is .Our integers are , so we have that.Solution 3Let the first term be . Our integers are . We have,18. Solution 1There are cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there arecubes. Hence, the answer is .Solution 2We can just calculate the volume of the prism that was cut out of the original box. Each interior side of the fort will be feet shorter than each side of the outside. Since the floor is foot, the height will be feet. So the volume of the interior box is .The volume of the original box is . Therefore, the number of blocks contained inthe fort is .19. If Hannah did better than Cassie, there would be no way she could know for sure that she didn't get the lowest score in the class. Therefore, Hannah did worse than Cassie. Similarly, if Hannah did worse than Bridget, there is no way Bridget could have known that she didn't get the highest in the class.Therefore, Hannah did better than Bridget, so our order is .20.A semicircle has symmetry, so the center is exactly at the midpoint of the 2 side on the rectangle, making the radius, by the Pythagorean Theorem, . The area is .21.The number of ways to get from Samantha's house to City Park is , and the number of ways toget from City Park to school is . Since there's one way to go through City Park (just walking straight through), the number of different ways to go from Samantha's house to City Park to school .22. There are vertical columns with a length of toothpicks, and there are horizontal rows with a length of toothpicks. An effective way to verify this is to try a small case, i.e. a grid of toothpicks.Thus, our answer is .23. Solution 1If the semicircle on AB were a full circle, the area would be 16pi. Therefore the diameter of the first circle is 8. The arc of the largest semicircle would normally have a complete diameter of 17. The Pythagoreantheorem says that the other side has length 15, so the radius is .Solution 2We go as in Solution 1, finding the diameter of the circle on AC and AB. Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of thelargest is , and the middle one is , so the radius is .24.First let (where is the side length of the squares) for simplicity. We can extend until it hits theextension of . Call this point . The area of triangle then is The area of rectangle is . Thus, our desired area is . Now, the ratio of the shaded area to thecombined area of the three squares is .Let the side length of each square be .Let the intersection of and be .Since , . Since and are vertical angles, they are congruent. We also have by definition.So we have by congruence. Therefore, .Since and are midpoints of sides, . This combined with yields.The area of trapezoid is .The area of triangle is .So the area of the pentagon is .The area of the squares is .Therefore, .Let the intersection of and be .Now we have and .Because both triangles has a side on congruent squares therefore .Because and are vertical angles .Also both and are right angles so .Therefore by AAS(Angle, Angle, Side) .Then translating/rotating the shaded into the position ofSo the shaded area now completely covers the squareSet the area of a square asTherefore, .25. Solution 1The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses inches, and it gains inches on B.So, the departure from the length of the track means that the answer is . Solution 2The total length of all of the arcs is . Since we want the path from the center, the actual distance will be shorter. Therefore, the only answer choice less than is . Thissolution may be invalid because the actual distance can be longer if the path the center travels is on the outside of the curve, as it is in the middle bump.古希腊哲学大师亚里士多德说：人有两种，一种即“吃饭是为了活着”,一种是“活着是为了吃饭”.一个人之所以伟大，首先是因为他有超于常人的心。
2020年AMC12真题及答案
2016 AMC12 AProblem 1What is the value of ?SolutionProblem 2For what value of does ?SolutionProblem 3The remainder can be defined for all realnumbers and with by where denotes the greatest integer less than or equal to . What is the value of ?SolutionProblem 4The mean, median, and mode of the data values are all equal to . What is the value of ?SolutionProblem 5Goldbach's conjecture states that every even integer greater than 2 can be written as the sum of two prime numbers (for example, ). So far, no one has been able to prove that the conjecture is true, and no one has found a counterexample to show that the conjecture is false. What would a counterexample consist of?SolutionProblem 6A triangular array of coins has coin in the first row, coins in the second row, coins in the third row, and so on up to coins in the th row. What is the sum of the digits of ?SolutionProblem 7Which of these describes the graph of ?SolutionProblem 8What is the area of the shaded region of the given rectangle?SolutionProblem 9The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the otherfour small squares as shown. The common side length is , where and are positive integers. What is ?SolutionProblem 10Five friends sat in a movie theater in a row containing seats, numbered to from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?SolutionProblem 11Each of the students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. Thereare students who cannot sing, students who cannot dance, and students who cannot act. How many students have two of these talents?SolutionProblem 12In , , , and . Point lies on ,and bisects . Point lies on , and bisects . The bisectors intersect at . What is the ratio : ?SolutionProblem 13Let be a positive multiple of . One red ball and green balls are arranged in a line inrandom order. Let be the probability that at least of the green balls are on the sameside of the red ball. Observe that and that approaches as grows large. What is the sum of the digits of the least value of such that ?SolutionProblem 14Each vertex of a cube is to be labeled with an integer from through , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?SolutionProblem 15Circles with centers and , having radii and , respectively, lie on the same side of line and are tangent to at and , respectively, with between and . The circle with center is externally tangent to each of the other two circles. What is the area of triangle ?SolutionProblem 16The graphs of and are plotted on the same set of axes. How many points in the plane with positive -coordinates lie on two or more of the graphs?SolutionProblem 17Let be a square. Let and be the centers, respectively, of equilateral triangles with bases and each exterior to the square. What is the ratio of the area of square to the area of square ?SolutionProblem 18For some positive integer the number has positive integer divisors,including and the number How many positive integer divisors does thenumber have?SolutionProblem 19Jerry starts at on the real number line. He tosses a fair coin times. When he gets heads, he moves unit in the positive direction; when he gets tails, he moves unit in the negative direction. The probability that he reaches at some time during this processis where and are relatively prime positive integers. What is (For example, he succeeds if his sequence of tosses is )SolutionProblem 20A binary operation has the properties that andthat for all nonzero real numbers and (Here the dot represents the usual multiplication operation.) The solution to the equation can bewritten as where and are relatively prime positive integers. What isSolutionProblem 21A quadrilateral is inscribed in a circle of radius Three of the sides of this quadrilateral have length What is the length of its fourth side?SolutionProblem 22How many ordered triples of positive integerssatisfy and ?SolutionProblem 23Three numbers in the interval are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area? SolutionProblem 24There is a smallest positive real number such that there exists a positive realnumber such that all the roots of the polynomial are real. In fact, for this value of the value of is unique. What is the value ofSolutionProblem 25Let be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with digits. Every time Bernardo writes a number, Silvia erases the last digits of it. Bernardo then writes the next perfect square, Silvia erases the last digits of it, and this process continues until thelast two numbers that remain on the board differ by at least 2. Let be the smallestpositive integer not written on the board. For example, if , then the numbers that Bernardo writes are , and the numbers showing on the board after Silviaerases are and , and thus . What is the sum of the digitsof ?2016 AMC 12A Answer Key1 B2 C3 B4 D5 E6 D7 D8 D9 E10 B11 E12 C13 A14 C15 D16 D17 B18 D19 B20 A21 E22 A23 C24 B25 E。
2020年美国数学邀请赛(AIMEⅡ)试题与解答
上海中学数学• 2020年第7 — 8期912020年美国数学邀请赛（AIME I )试题与解答200001 上海市格致中学马陆陆殷琦涛摘要：原定于2020年3月26日举行的美国数学邀请赛（A I M E n )，因受疫情影响，推迟到2020 年6月6日进行，采取线上答题的方式.笔者翻译该邀请赛试题，并提供详细解答.关键词：美国数学邀请赛（A I M E n );试题；解答1. Find the number o f ordered p a i r s o f p o s i t i v e integers {m ^n ) such t h a t m 2 n = 2020.翻译：求满足m 、= 202°的有序正整数对(m ，r z ) 的个数.解：由 m 2 r z = 2020 = 24° X 52。
，设 m56，r z =2<^，其中(2,6,〇^/6]>1，从而262 +〔 = 40,26+3 = 20.在2a +c = 40中，可令a = 0，l，2,…，20,则该方 程有21组自然数解.在26+3 = 20中，可令6 = 0, 1，2,…，10,得该方程有11组自然数解，因此有序正 整数对的个数为21X 11 = 231.2. Let F be a point chosen uniformly a t random i n th e i n t e r i o r o f the u n i t square with v e r t i c e s at(0,0)，（l，0)，（l，l) and(0, 1 )• The p r o b a b i l i t y t h a t the slope o f the l i n e determined by P and the point (*，吾）i s greater than 去 can be written a s —,where m and n are r e l a t i v e l y prime p o s i t i v e integers. Find m + w.翻译:在以（0,0)，（1，0)，（1，1)和（0，1)为顶点 的单位正方形内随机选取一个点P.若由点P 和点(吾，I )所确定的直线的斜率大于+的概率为其中m 和r ?是互质的正整数.求m + 77.解：如图1，过斜率为去的直线= 士.r + &,交正方形区域边界于点A (0，h )，直线1616CD:.r = |•交正方形区域边界于C (|~，0)，D (音，1)，设 £：(1，1).图1四边形Q B _E D内，所以7=S 四边形〇4Q c + S 四边形oet’D =丄 r J_-U 3、 5,1 ,5,7. 3 _ 43 m t . (r^+j ). y +y . . y —顶，因此 m +171.3. The value o f x t h at s a t i s f i e s log2r 320 =l〇g 2.r +332020 C an be written aswhere m and n ar enr e l a t i v e l y prime p o s i t i v e integers. Find m -\~n .翻译：满足1叩2' 32°=1呢2'+33_的.r 的值可写成^■的形式，其中是互质的正整数.求w + n.解：依条件有翌 l 〇g23 = ^|l 〇g23(.r^0，一3)，.r x ~r 6所以一=—，解得.r =：p ^，故 w + n =103..r .r 十 3 1004. Triangles A A B C and A A ，B ，C ， l i e i n the coordinate plane with v e r t i c e s A (0,0) , B (0,12), C (16,0),A ，(24,18),B ，(36,18),C ，(24,2). A rotat i o n o f m degrees clockwise around the point (.r ,y ) where 0 < w < 180, w i l l transform A A B C t o/S A 'B 'C '. Find m -\-.x -\-y .翻译：A A B C 和A A 'B 'C '位于坐标平面上，顶 点坐标为 A (0,0)，B (0,12)，C (16,0)，A '(24，18), 以（36，18),(^(24,2).绕点Cr，3〇顺时针旋转w 度， 其中0<w <180,可将A A B C 变换为A A 'B 'C '•求解法1:根据条件，旋转中心P (.r ，：y)在A A '，B B '，C C '中垂线上.A A '的中垂线为4.r + 3y —75 = 0，B B '的中垂线为6.r+y —123 = 0，C C '的中垂线为 4.r+：y_81 = 0,这三条直线交于点（21，一 3)，所以^ =21，_y = —3•在 A A P v V 中，丨 A 4’ 丨= 30,丨 P A | =|P A '|=15/^,从而 w ° = Z A P A ’ = 90°d(I 所=90, 故 w+.r + j y =108.解法2:如图2,因为丄A 'B '，所以旋转角为 90°，即w = 90•旋转中心P (.r，WA对应于复数一.r—：y i ，P/V 对应于复数 24 —.r+(18 —：y)i ，所以[24 —.r 若P Q 斜率大于备，则P 落在四边形或+(18 — y )i ]i =—〇■ —yi，于是有匕一18=—’，解得L\ 24—.r=—y92上海中学数学• 2020年第7 — 8期，所以 w +.r+_y=108.•V = —3解：令= 5/…，则= 100，f l 2 = 1〇5，q …«-2 + 1+ 1…-2 "i "a nI +1a…-2 <2；i -3f l ”-2 +1-3 + 1 =^n -2<3n -3(77>5)，所以 u …}是以5为周期的周期数列.而5350?103 七=^，所以心。
2020年美国湾区数学竞赛试题word
2020美国湾区数学竞赛1.如图下图,一个梯形被分成7个宽度相等的条带,求阴影部分面积与梯形总面积的比值?2.安娜、鮑勃、西莉亚和大卫四个人交换了一些钱,对于任意两个人,恰好一个人给了另一个人钱.例如,西莉亚可以给大卫一些钱,但是大卫不会给西莉亚钱。
最终，每个人的收支平衡(意味着没有人赚钱或亏钱)(a)所给的钱的金额有没有可能分别为10、20、30、40、50、60?(b)所给的钱的金额有没有可能分别为20、30、40、50、60、70? 对于每一问,如果你的答案是“可能”,则请描述谁给谁多少钱来说明情况可能发生；如果你的答案是“不可能”,则请证明。
3.请尽可能多地求出满足下列方程的实数n。
20002120192202012120002201912020-++-+-=-++-+-x x x x x x K K4.有一张形状为等边三角形的纸,将它沿着虚线折叠,如下左图所示沿着虚线折叠每一个角,得到一个均匀四层厚的新作品,如下右图所示。
每个区域中的数字表示该处的厚度(以纸张的层数为单位).我们刚刚看到了一个例子,说明如何将一个平面图形折叠成厚度均匀的物品,本题要求你举出其它几个例子.在任何一种情况下,你都可以沿着任何 条线折叠，被折叠的不同部分的形状可以不同。
假设纸张可以折叠任意次数且不会被撕裂或者因为太厚而无法折叠.如果需要,你可以使用以下工具:一把神器的尺子,你可以用它画过任意两点的直线,并且你可以把一条线段分成任意多个相等的部分。
一种直角三角板，你可以过一点作一条直线的垂线。
5.已知整数202020是91的倍数对于任意正整数n,求如何将n 个2插入到数字202020中,使得得到的(n+6)位数也是91的倍数.例如,当n=5时,22020220222是一个可行的方法。
6.对于△ABC,设M为边BC上一点,O为△ABM外接圆的圆心，k为过点A、M的圆且其圆心在BC上。
直线MO再次交k于K.求证:对于任意满足条件的点M,直线BK为一条定直线。
2020美赛A题
2020美赛A题翻译：向北迁徙全球海洋温度影响某些海洋生物的栖息地质量。
当温度变化太大以至于无法继续生长时，这些物种便开始寻找其他更适合其现在和将来的生活和生殖成功的栖息地。
其中一个明显的例子就是美国缅因州的龙虾种群，该种群正缓慢地向北迁徙到加拿大，那里较低的海洋温度提供了更合适的栖息地。
这种地理上的种群迁移会严重破坏依赖海洋生物物种稳定性的公司的生计。
您的团队已被苏格兰北大西洋渔业管理协会聘为顾问。
如果全球海洋温度升高，该财团希望更好地了解与苏格兰鲱鱼和鲭鱼从其目前在苏格兰附近的栖息地迁徙有关的问题。
这两种鱼类是苏格兰渔业的重要经济来源。
鲱鱼和鲭鱼种群位置的变化可能使以苏格兰为基地的小型捕捞公司在经济上造成不确定风险，后者使用没有船上制冷的渔船来捕捞鲜鱼并将其运送到苏格兰渔港的市场。
要求1.建立一个数学模型，以识别未来50年内这两种鱼类最可能的位置，假设水温将发生足够的变化以导致种群移动。
2.根据海水温度变化的速度，使用您的模型预测最佳情况、最坏情况和最有可能经过的时间，直到这些种群距离小渔业公司太远以至于如果小渔业公司继续在其当前位置外作业将一无所获。
3.根据您的预测分析，这些小型捕捞公司是否应该改变其经营方式？•a.如果是，请使用您的模型为小型捕捞公司识别和评估实用且经济上有吸引力的策略。
您的策略应考虑但不限于现实的选择，包括：o将部分或全部捕捞公司的资产从苏格兰港口的当前位置迁移到两个鱼类种群都迁徙的附近；o使用一定比例的小型渔船，这些渔船可以在没有陆上支持的情况下运行一段时间，同时仍确保渔获物的新鲜度和高质量。
o您的团队可以识别和模拟的其他可能的选项。
•b.如果您的团队拒绝进行任何更改，请根据建模结果来说明拒绝的原因，因为建模结果与您的团队所做的假设有关。
4.使用您的模型来解决:如果有一部分渔业移至另一个国家的领海时您的建议受到的影响。
5.除了技术报告外，还要为 Hook Line and Sinker 杂志准备一份长达两页的文章，以帮助渔民了解问题的严重性以及您提出的解决方案将如何改善他们的未来的业务前景。
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AMC2020 AProblem 1Carlos took of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?Problem 2The acronym AMC is shown in the rectangular grid below with grid lines spaced unit apart. In units, what is the sum of the lengths of the line segments that form the acronym AMCProblem 3A driver travels for hours at miles per hour, during which her car gets miles per gallon of gasoline. She is paid per mile, and her only expense is gasoline at per gallon. What is her net rate of pay, in dollars per hour, after this expense?Problem 4How many -digit positive integers (that is, integersbetween and , inclusive) having only even digits are divisible byProblem 5The integers from to inclusive, can be arranged to form a -by- square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?Problem 6In the plane figure shown below, of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetryProblem 7Seven cubes, whose volumes are , , , , , ,and cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?Problem 8What is the median of the following list of numbersProblem 9How many solutions does the equation have on the intervalProblem 10There is a unique positive integer suchthat What is the sum of the digits ofProblem 11A frog sitting at the point begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length , and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square withvertices and . What is the probability that the sequence of jumps ends on a vertical side of the squareProblem 12Line in the coordinate plane has the equation . This line is rotated counterclockwise about the point to obtain line . What is the -coordinate of the -intercept of lineProblem 13There are integers , , and , each greater than 1, suchthat for all . What is ?Problem 14Regular octagon has area . Let be the area of quadrilateral . What isProblem 15In the complex plane, let be the set of solutions to and let be the set of solutions to . What is the greatest distance between a point of and a point ofA point is chosen at random within the square in the coordinate plane whose vertices are and . The probability that the point is within units of a lattice point is .(A point is a lattice point if and are both integers.) What is to the nearest tenthProblem 17The vertices of a quadrilateral lie on the graph of , and the -coordinates of these vertices are consecutive positive integers.The area of the quadrilateral is . What is the -coordinate of the leftmost vertex?Problem 18Quadrilateral satisfies , and . Diagonals and intersect at point ,and . What is the area of quadrilateral ?There exists a unique strictly increasing sequence of nonnegativeintegers such thatWhat isProblem 20Let be the triangle in the coordinate plane withvertices , , and . Consider the following five isometries (rigid transformations) of the plane: rotations of , , and counterclockwise around the origin, reflection across the -axis, and reflection across the -axis. How many ofthe sequences of three of these transformations (not necessarily distinct) will return to its original position? (For example, a rotation, followed by a reflection across the -axis, followed by a reflection across the -axis will return to its original position, but a rotation, followed by a reflection across the-axis, followed by another reflection across the -axis will not return to its original position.)How many positive integers are there such that is a multiple of , and the least common multiple of and equals times the greatest common divisor of andProblem 22Let and be the sequences of real numbers suchthat for all integers , where . WhatisProblem 23Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly . Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?Suppose that is an equilateral triangle of side length , with the property that there is a unique point inside the triangle such that , , and . What isProblem 25The number , where and are relatively prime positive integers, has the property that the sum of all realnumbers satisfying is , where denotes the greatest integer less than or equal to and denotes the fractional part of . What is2020 AMC 12A Answer Key 1. C2. C3. E4. B5. C6. D7. B8. C9. E10.E11.B12.B13.B14.B15.D16.B17.D18.D19.C20.A21.D22.B23.A24.B25.C。