AMC10美国数学竞赛讲义

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amc10真题与答案解析

amc10真题与答案解析AMC10真题与答案解析美国数学竞赛（AMC）是一项极具挑战性的数学竞赛，旨在推动学生的数学学习和解决问题的能力。

其中AMC10是为中学生设计的竞赛，是很多学生展现才华并锻炼数学素养的重要机会。

本文将对一道AMC10真题进行解析，帮助读者理解和掌握解题技巧。

以下是一道来自AMC10的真题："In the xy-plane, the graph of$\log_{16}x+\log_{16}y=\tfrac{3}{2}$ is drawn. The line$y=x$ intersects the graph at points $A$ and $B$. What is the length of segment $AB$?"这道题目涉及了对数的性质和直线与曲线的交点问题。

首先，我们先来理解题目所给出的等式。

$\log_{16}x$表示以16为底的对数，可以简化为$\log_2x^4$。

同样地，$\log_{16}y$可以简化为$\log_2y^4$。

所以等式可以写为$\log_2x^4 + \log_2y^4 =\frac{3}{2}$。

我们可以将等式进一步简化为$\log_2 (x^4y^4) =\frac{3}{2}$。

根据对数的性质，我们可以将等式转化为指数形式，得到$x^4y^4 = 2^{\frac{3}{2}}$。

然后我们来解决直线$y=x$与曲线$y=x^4$的交点问题。

我们可以将$x$代入曲线方程中，得到$y=x^4$。

因此，我们可以将交点问题转化为求解以下方程组：$x^4 =x$和$x^4y^4 = 2^{\frac{3}{2}}$。

首先，我们来解原方程$x^4 = x$。

很明显，$x=0$和$x=1$是方程的两个根。

我们可以进一步分析，当$x>1$时，$x^4$的增长速度比$x$快，所以方程没有其他解。

然后，我们来解方程$x^4y^4 = 2^{\frac{3}{2}}$。

amc10 立体几何题目

amc10 立体几何题目【最新版】目录1.AMC10 立体几何题目概述2.立体几何的基本概念和解题方法3.AMC10 立体几何题目的解题技巧和策略4.总结和建议正文【AMC10 立体几何题目概述】AMC10（美国数学竞赛 10 年级）是针对 10 年级学生的一项重要数学竞赛，其涉及的立体几何题目旨在考查学生的空间想象能力、逻辑思维能力以及数学应用能力。

立体几何题目通常以线线、线面、面面等关系为载体，要求学生运用相关定理和公式进行分析和求解。

【立体几何的基本概念和解题方法】立体几何是研究空间中点、线、面及其相关性质的数学分支。

在解决立体几何问题时，通常需要运用以下基本概念和方法：1.点、线、面的基本性质：了解点、线、面的概念以及它们之间的关系，如共线、共面等。

2.空间直线与平面的位置关系：主要包括直线与平面相交、平行和重合三种情况。

3.空间几何中的距离和角：学会计算空间中两点之间的距离、直线与平面之间的夹角等。

4.空间几何中的投影：了解正射投影、轴投影等投影方式，学会利用投影解决问题。

5.空间几何中的变换：学会运用平移、旋转等变换方法解决空间几何问题。

【AMC10 立体几何题目的解题技巧和策略】在解答 AMC10 立体几何题目时，可以采用以下技巧和策略：1.仔细阅读题目，理解题意，画出题目中涉及的图形，有助于建立直观的图形信息。

2.善于运用空间想象能力，将问题转化为平面几何问题，降低解题难度。

3.运用相关定理和公式，如线线平行、线面垂直等，进行分析和求解。

4.遇到复杂问题时，可以尝试将问题拆解为多个简单的子问题，逐步解决。

5.做好归纳总结，积累解题经验，提高解题速度和准确率。

【总结和建议】AMC10 立体几何题目对学生的空间想象能力、逻辑思维能力以及数学应用能力都有较高要求。

要解答好这类题目，需要熟练掌握立体几何的基本概念和解题方法，并灵活运用各种解题技巧和策略。

2000年-2002 美国AMC10

PABCD2000年第01届美国AMC10 (2000年2月日时间75分钟)1. 国际数学奥林匹亚将于2001年在美国举办，假设I 、M 、O 分别表示不同的正整数，且满足I ⨯M ⨯O =2001，则试问I +M +O 之最大值为。

(A) 23 (B) 55 (C) 99 (D) 111 (E) 671 。

2. 2000(20002000)为。

3. Jenny 每天早上都会吃掉她所剩下的聪明豆的20%，今知在第二天结束时，有32颗剩下，试问一开始聪明豆有颗。

(A) 40 (B) 50 (C) 55 (D) 60 (E) 75 。

4. Candra 每月要付给网络公司固定的月租费及上网的拨接费，已知她12月的账单为12.48元，而她1月的账单为17.54元，若她1月的上网时间是12月的两倍，试问月租费是元。

(A) 2.53 (B) 5.06 (C) 6.24 (D) 7.42 (E) 8.77 。

5. 如图M ，N 分别为PA 与PB 之中点，试问当P 在一条平行AB 的直在线移动时，下列各数值有项会变动。

(a)MN长 (b) △PAB 之周长 (c) △PAB 之面积 (d) ABNM 之面积(A) 0项 (B) 1项 (C) 2项 (D) 3项 (E) 4项。

6. 费氏数列是以两个1开始，接下来各项均为前两项之和，试问在费氏数列各项的个位数字中，最后出现的阿拉伯数字为。

(A) 0 (B) 4 (C) 6 (D) 7 (E) 9 。

7. 如图，矩形ABCD 中，AD =1，P 在AB 上，且DP 与DB 三等分∠ADC ，试问△BDP 之周长为。

(A) 3+33 (B) 2+334 (C) 2+2(D)2533+ (E) 2+335 。

2020AMC10B（美国数学竞赛）真题加详解

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

美国数学竞赛美国高中数学竞赛 AMC 10 试题及答案 2000年-2015年

2000 AMC 10 ProblemsProblem 1In the year 2001, the United States will host the International Mathematical Olympiad. Let , , and be distinct positive integers such that the product 2001=••O M I . What is the largest possible value of the sum O M I ++ ?（A ）23 （B ）55 （C ）99 （D ）111 （E ）671Problem 22000 （20002000）＝（A ）20012000 （B ）20004000 （C ）40002000 （D ）2000000,000,4 （E ）000,000,42000Problem 3Each day, Jenny ate 20% of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, 32 remained. How many jellybeans were in the jar originally?（A ）40 （B ）50 （C ）55 （D ）60 （E ）75Problem 4Chandra pays an on-line service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was , but in January her bill was because she used twice as much connect time as in December. What is the fixed monthly fee?（A ）2.53 （B ）5.06 （C ）6.24 （D ）7.42 （E ）8.77Problem 5Points M and N are the midpoints of sides PA and PB of △PAB. As P moves along a line that is parallel to side AB, how many of the four quantities listed below change? (a) the length of the segment MN (b) the perimeter of △PAB(c) the area of △PAB (d) the area of trapezoid ABNM（A ）0 （B ）1 （C ）2 （D ）3 （E ）4Problem 6The Fibonacci sequence 1，1，2，3，5，8，13，21，…… starts with two s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?（A ）0 （B ）4 （C ）6 （D ）7 （E ）9Problem 7 In rectangle , , is on , and and trisect .What is the perimeter of ?（A ）333+ （B ）3342+ （C ）222+ （D ）2533+ （E ）3352+ Problem 8At Olympic High School, 52 of the freshmen and 54 of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?（A ）There are five times as many sophomores as freshmen.（B ）There are twice as many sophomores as freshmen.（C ）There are as many freshmen as sophomores.（D ）There are twice as many freshmen as sophomores.（E ）There are five times as many freshmen as sophomores.Problem 9If , where , then（A ）－2 （B ）2 （C ）2－2p （D ）2p-2 （E ）｜2p-2｜Problem 10The sides of a triangle with positive area have lengths , , and . The sides of a second triangle with positive area have lengths , , and . What is the smallest positive number that is not a possible value of ｜x-y ｜?（A ）2 （B ）4 （C ）6 （D ）8 （E ）10Problem 11Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?（A ）21 （B ）60 （C ）119 （D ）180 （E ）231Problem 12Figures 0, 1, 2 and 3 consist of 1, 5, 13, and 25 nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be infigure 100?Figure 0 Figure 1 Figure 2 Figure 3（A ）10401 （B ）19801 （C ）20201 （D ）39801 （E ）40801 Problem 13There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?Problem 14Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71, 76, 80, 82, and 91. What was the last score Mrs. Walter entered?（A ）71 （B ）76 （C ）80 （D ）82 （E ）91Problem 15Two non-zero real numbers, and , satisfy. Find a possible value of ab ab b a -+ . （A ）-2 （B ）-21 （C ）31 （D ）21 （E ）2 Problem 16The diagram shows 28 lattice points, each one unit from its nearest neighbors. Segment AB meets segment CD at E. Find the length of segment AE.（A ）354 （B ）355 （C ）7512 （D ）52 （E ）9565Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?（A ）<dollar/>3.63 （B ）<dollar/>5.13 （C ）<dollar/>6.30 （D ）<dollar/>7.45 （E ）<dollar/>9.07Problem 18Charlyn walks completely around the boundary of a square whose sides are each 5 km long. From any point on her path she can see exactly km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?（A ）24 （B ）27 （C ）39 （D ）40 （E ）42Problem 19Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is times the area of the square. The ratio of the area of the other small right triangle to the area of the square is（A ）121 m （B ）m （C ）1-m （D ）m 41 （E ）281mProblem 20Let A, M, and C be nonnegative integers such that A+M+C=10. What is the maximum value of A ·M ·C + A ·M + M ·C +C ·A?（A ）49 （B ）59 （C ）69 （D ）79 （E ）89Problem 21If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?I. All alligators are creepy crawlers.II. Some ferocious creatures are creepy crawlers.III. Some alligators are not creepy crawlers.（A ）I only （B ）II only （C ）III only （D ）II and III only （E ）None must be true Problem 22One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?（A ）3 （B ）4 （C ）5 （D ）6 （E ）7When the mean, median, and mode of the list 10, 2, 5, 2, 4, 2, x are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ? （A ）3 （B ）6 （C ）9 （D ）17 （E ）20Problem 24Let be a function for which. Find the sum of all values of for which. （A ）-31 （B ）-91 （C ）0 （D ）95 （E ）35 Problem 25In year , the day of the year is a Tuesday. In year, the day is also a Tuesday. On what day of the week did the day of year occur?2000 AMC 10 SolutionProblem 1 The following problem is from both the 2000 AMC 12 #1 and 2000 AMC 10 #1, so both problems redirect to this page.The sum is the highest if two factors are the lowest.So, 1·3·776=2001 and 1+3+667=671 (E)Problem 2 The following problem is from both the 2000 AMC 12 #2 and 2000 AMC 10 #2, so both problems redirect to this page.2000 （20002000）＝12000（20002000）=20012000 (A)Problem 3 The following problem is from both the 2000 AMC 12 #3 and 2000 AMC 10 #3, so both problems redirect to this pageSince Jenny eats 20% of her jelly beans per day, 80%=4/5 of her jelly beans remain after one day. Let be the number of jelly beans in the jar originally.325454=••x x=50 (B) Problem 4Let be the fixed fee, and be the amount she pays for the minutes she used in the first month.x+y=12.48 x+2y=17.54 y=5.06 x=7.42 We want the fixed fee, which is (D) Problem 5(a) Clearly AB does not change, and MN=0.5AB, so MN doesn't change either. (b) Obviously, the perimeter changes.(c) The area clearly doesn't change, as both the base AB and its corresponding height remain the same.(d) The bases AB and MN do not change, and neither does the height, so the area of the trapezoid remains the same.Only quantity changes, so the correct answer is .Problem 6 The following problem is from both the 2000 AMC 12 #4 and 2000 AMC 10 #6, so both problems redirect to this page.Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in :The last digit to appear in the units position of a number in the Fibonacci sequence is 6 (C).Problem 7AD=1 Since ∠ADC is trisected, ∠ADP=∠PDB=∠BDC=30º Thus, PD=332 BD=2 BP=332333=- Adding, 3342+Problem 8Let be the number of freshman and be the number of sophomores.s f 5452= f=2s There are twice as many freshmen as sophomores. Problem 9 The following problem is from both the 2000 AMC 12 #5 and 2000 AMC 10 #9, so both problems redirect to this page.When x<2, x-2 is negative so ∣x -2∣=2-x =p and x=2-pThus x-p=(2-p)-p=2-2p (C)Problem10From the triangle inequality, 2<x <10 and 2<y <10. The smallest positive number not possible is 10-2, which is . (D)Problem 11 The following problem is from both the 2000 AMC 12 #6 and 2000 AMC 10 #11, so both problems redirect to this page.All prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate B andD. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is (13) (17) – (13+17) = 221-30=191. Thus, we can eliminateE. Similarly, the two lowest prime numbers we can pick are 5 and 7, so the lowest number we can make is (5) (7) – (5+7) =23. Therefore, A cannot be an answer. So, the answer must be .Problem 12 The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page.Solution 1We have a recursion: .I.E. we add increasing multiples of each time we go up a figure. So, to go from Figure 0 to 100, we addWe then add to the number of squares in Figure 0 to get, which ischoice Solution 2We can divide up figure to get the sum of the sum of the first odd numbers and the sum of the first odd numbers. If you do not see this, here is the example for :The sum of the first odd numbers is 2n , so for figure , thereare 22)1(n n ++ unit squares. We plug in n=100 to get 20201, which is choice Solution 3Using the recursion from solution 1, we see that the first differences of 4,8,12, … form an arithmetic progression, and consequently that the second differences are constant and all equal to . Thus, the original sequence can be generated from a quadratic function.If c bn an n f ++=2)(, and f(0)=1, f(1)=5, and f(2)=13, we get a system of three equations in three variables:f(0)=1 gives c=1; f(1)=5 gives a+b+c=5; f(2)=13 gives 4a+2b+c=13 Plugging in into the last two equations gives a+b=4 4a+2b=12 Dividing the second equation by 2 gives the system: a+b=4 2a+b=6Subtracting the first equation from the second gives , and hence . Thus, our quadratic function is: 122)(2++=n n n fCalculating the answer to our problem, f(100)=20201Problem 13In each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left forthe yellow peg in this column.By similar logic, we can fill in the yellow pegs as shown:After this we can proceed to fill in the whole pegboard, so there is only arrangement of the pegs. The answer is (B)Problem 14 The following problem is from both the 2000 AMC 12 #9 and 2000 AMC 10 #14, so both problems redirect to this page.Solution 1The first number is divisible by 1.The sum of the first two numbers is even.The sum of the first three numbers is divisible by 3.The sum of the first four numbers is divisible by 4.The sum of the first five numbers is 400.Since 400 is divisible by 4, the last score must also be divisible by 4. Therefore, the last score is either 76 or 80.Case 1: 76 is the last number entered.Since 400≡76≡1 (mod 3), the fourth number must be divisible by 3, but none of the scores are divisible by 3. Case 2: 80 is the last number entered. Since 80≡2 (mod 3), the fourth number must be 2 (mod 3). Thatnumber is 71 and only 71. The next number must be 91, since the sum of the first two numbers is even.So the only arrangement of the scores 76, 82, 91,71,80Solution 2We know the first sum of the first three numbers must be divisible by 3, so we write out all 5 numbers (mod 3), which gives 2,1,2,1,1, respectively. Clearly the only way to get a number divisible by 3 by adding three of these is by adding the three ones. So those must go first. Now we have an odd sum, and since the next average must be divisible by 4, 71 must be next. That leaves 80 for last, so the answer is .Problem 15 The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.22)()()(22222=-=----+=--+=-+ba ab b a b a b a b a b a ab b a ab a b b a (E)Alternatively, we could test simple values, like )21,1(),(=b a , which would yield 2=-+ab ab b a Another way is to solve the equation for giving 1+=a a b , then substituting this into the expression and simplifying gives the answer ofProblem 16Solution 1Let be the line containing A and B and let be the line containing C and D. If we set the bottom left point at (0,0), then A=(0,3), B=(6,0), C=(4,2), and D=(2,0) . The line is given by the equation 11b x m y +=. The -intercept is A=(0,3), so 1b =3. We are given two points on , hence we can compute the slope, to be 210630-=-- , so is the line 321+-=x y Similarly, is given by 22b x m y +=. The slope in this case is 12402=--, so 2b x y +=. Plugging in the point (2,0) gives us 2b =-2, so is the line 2-=x y . At E, the intersection point, both of the equations must be true,2-=x y , 321+-=x y so 3212+-=-x x SO 310=x 34=y We have the coordinates of and , so we can use the distance formula here: 355)334()0310(22=-+- which is answer choiceSolution 2Draw the perpendiculars from andto , respectively. As it turns out, . Let be the point onfor which . , and, so by AA similarity,By the Pythagorean Theorem, wehave, ,。

amc10美国数学竞赛真题

amc10美国数学竞赛真题amc10（美国数学竞赛）真题：
1. 微积分方面：
（1）求一元二次微分方程的有限解；
（2）求曲线y = f（x）在a点的切线斜率；
（3）求曲面z = f（x，y）在a点的切面方向；（4）决定泰勒展开式中n次项的系数；
（5）判断函数f（x）的极值；
（6）研究极限函数的存在性及无穷性；
（7）解决解析函数的定义域；
2. 数论方面：
（1）计算包含一定系数的多项式的余数；
（2）判断两个多项式的最大公因数；
（3）判断一个数的完全平方数；
（4）证明或否定一个数在某数域中的存在性；
3. 统计学：
（1）奇数和偶数的含义；
（2）样本容量的定义；
（3）求一组数据中最小值、最大值和中位数；（4）决定随机变量的条件概率；
（5）求一组数据的散布图；
4. 几何学：
（1）求平面两点间的距离；
（2）求直线的斜率；
（3）分析不相交平面的法向量；
（4）解决三角形的各个角度；
（5）计算圆的周长；
（6）两个圆的位置关系；
（7）求圆环的面积；
（8）求立体图形的体积。

5. 其它方面：
（1）分析逻辑表达式；
（2）求图论问题的最小生成树；
（3）计算算法问题的复杂度；
（4）计算概率论问题中的概率；
（5）规划组合优化问题的最优解；
（6）根据资料采纳正确的统计预测方法。

AMC 美国数学竞赛 2001 AMC 10 试题及答案解析

USA AMC 10 20011The median of the listis . What is the mean?Solution2A number is more than the product of its reciprocal and its additive inverse. In which interval does the number lie?Solution3The sum of two numbers is . Suppose 3 is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?Solution4What is the maximum number of possible points of intersection of a circle and a triangle?Solution5How many of the twelve pentominoes pictured below have at least one line of symettry?Solution6Let and denote the product and the sum, respectively, of thedigits of the integer . For example, and . Supposeis a two-digit number such that . What is the units digit of ?Solution7When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?Solution8Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today willthey next be together tutoring in the lab?Solution9The state income tax where Kristin lives is levied at the rate of of the first of annual income plus of any amount above . Kristin noticed that the state income tax she paid amounted to of her annual income. What was her annual income?Solution10If , , and are positive with , , and , then isSolution11Consider the dark square in an array of unit squares, part of which is shown. The ﬁrst ring of squ ares around this center square contains unit squares. The second ring contains unit squares. If we continue this process, the number of unit squares in the ring isSolution12Suppose that is the product of three consecutive integers and that is divisible by . Which of the following is not necessarily a divisor of Solution13A telephone number has the form , where each letter represents a different digit. The digits in each part of the numbers are in decreasing order; that is, , , and . Furthermore, , , and are consecutive even digits; , , , and are consecutive odd digits; and . Find .Solution14A charity sells 140 benefit tickets for a total of . Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?Solution15A street has parallel curbs feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is feet and each stripe is feet long. Find the distance, in feet, between the stripes.Solution16The mean of three numbers is 10 more than the least of the numbers and 15 less than the greatest. The median of the three numbers is 5. What is their sum?Solution17Which of the cones listed below can be formed from a sector of a circle of radius by aligning the two straight sides?A cone with slant height of and radiusA cone with height of and radiusA cone with slant height of and radiusA cone with height of and radiusA cone with slant height of and radiusSolution18The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest toSolution19Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?Solution20A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length . What is the length of each side of the octagon?Solution21A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter and altitude , and the axes of the cylinder and cone coincide. Find the radius of the cylinder.Solution22In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by , , , , and . Find .Solution23A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?Solution24In trapezoid , and are perpendicular to , with, , and . What is ?Solution25How many positive integers not exceeding are multiples of or but not ?。

amc 美国数学竞赛 amc 10b 试题及答案解析

2004 AMC 10BProblem 1Each row of the Misty Moon Amphitheater has 33 seats. Rows 12 through 22 are reserved for a youth club. How many seats are reserved for this club?There are rows of seats, giving seats.Problem 2How many two-digit positive integers have at least one 7 as a digit?Ten numbers () have as the tens digit. Nine numbers () have it as the ones digit. Number is in both sets.Thus the result is .Problem 3At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made48 free throws. How many free throws did she make at the first practice?At the fourth practice she made throws, at the third one it was , then we get throws for the second practice, and finally throws at the first one.Problem 4A standard six-sided die is rolled, and P is the product of the five numbers that are visible. What is the largest number that is certain to divide P?Solution 1The product of all six numbers is . The products of numbers that can be visible are , , ..., . The answer to this problem is their greatest common divisor -- which is , where is the least common multiple of . Clearly and the answer is .Solution 2Clearly, can not have a prime factor other than , and .We can not guarantee that the product will be divisible by , as the number can end on the bottom.We can guarantee that the product will be divisible by (one of and will always be visible), but not by .Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by . This is the most we can guarantee, as when the is on the bottom side, the two visible even numbers are and , and their product is not divisible by .Hence .Problem 5In the expression , the values of , , , and are , , , and , although not necessarily in that order. What is the maximum possible value of the result?If or , the expression evaluates to .If , the expression evaluates to .Case remains.In that case, we want to maximize where .Trying out the six possibilities we get that the best one is, where .Problem 6Which of the following numbers is a perfect square?Using the fact that , we can write:▪▪▪▪▪Clearly is a square, and as , , and are primes, none of the other four are squares.Problem 7On a trip from the United States to Canada, Isabella took U.S. dollars. At the border she exchanged them all, receiving Canadian dollars for every U.S. dollars. After spending Canadian dollars, she had Canadian dollars left. What is the sum of the digits of ?Solution 1Isabella had Canadian dollars. Setting up an equation we get, which solves to , and the sum of digits of isSolution 2Each time Isabelle exchanges U.S. dollars, she gets Canadian dollars and Canadian dollars extra. Isabelle received a total of Canadian dollars extra, therefore she exchanged U.S. dollars times. Thus .Problem 8Minneapolis-St. Paul International Airport is 8 miles southwest of downtown St. Paul and 10 miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis?The directions "southwest" and "southeast" are orthogonal. Thus the described situation is a right triangle with legs 8 miles and 10 mileslong. The hypotenuse length is , and thus the answer is .Without a calculator one can note that . Problem 9A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle?The area of the circle is , the area of the square is . Exactly of the circle lies inside the square. Thus the total area is.Problem 10A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains cans, how many rows does it contain?The sum of the first odd numbers is . As in our case , we have .Problem 11Two eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?Solution 1We have , hence if at least one of the numbers is , the sum is larger. There such possibilities.We have .For we already have , hence all other cases are good.Out of the possible cases, we found that in the sum is greater than or equal to the product, hence in it is smaller. Therefore the answer is .Solution 2Let the two rolls be , and .From the restriction:Since and are non-negative integers between and , either , , orif and only if or .There are ordered pairs with , ordered pairs with , and ordered pair with and . So, there areordered pairs such that .if and only if and or equivalently and . This gives ordered pair .So, there are a total of ordered pairs with .Since there are a total of ordered pairs , there are ordered pairs with .Thus, the desired probability is .Problem 12An annulus is the region between two concentric circles. The concentric circles in the ﬁgure have radii and , with . Let be a radius of the larger circle, let be tangent to the smaller circle at , and let be the radius of the larger circle that contains . Let , , and . What is the area of the annulus?The area of the large circle is , the area of the small one is , hence the shaded area is .From the for the right triangle we have , hence and thus the shaded area is .Problem 13In the United States, coins have the following thicknesses: penny, mm; nickel, mm; dime, mm; quarter, mm. If a stack of these coins is exactly mm high, how many coins are in the stack?All numbers in this solution will be in hundreds of a millimeter.The thinnest coin is the dime, with thickness . A stack of dimes has height .The other three coin types have thicknesses , , and . By replacing some of the dimes in our stack by other, thicker coins, we can clearly create exactly all heights in the set.If we take an odd , then all the possible heights will be odd, and thus none of them will be . Hence is even.If the stack will be too low and if it will be too high. Thus we are left with cases and .If the possible stack heights are , with the remaining ones exceeding .Therefore there are coins in the stack.Using the above observation we can easily construct such a stack. A stack of dimes would have height , thus we need to add . This can be done for example by replacing five dimes by nickels (for ), and one dime by a penny (for ).Problem 14A bag initially contains red marbles and blue marbles only, with moreblue than red. Red marbles are added to the bag until only of the marbles in the bag are blue. Then yellow marbles are added to the baguntil only of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue?We can ignore most of the problem statement. The only important information is that immediately before the last step blue marbles formed of the marbles in the bag. This means that there were blue and other marbles, for some . When we double the number of blue marbles, there will be blue and other marbles, hence bluemarbles now form of all marbles in the bag.Problem 15Patty has coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have cents more. How much are her coins worth?Solution 1She has nickels and dimes. Their total cost iscents. If the dimes were nickels and vice versa, she would havecents. This value should be cents more than the previous one. We get , which solves to . Her coins are worth .Solution 2Changing a nickel into a dime increases the sum by cents, and changing a dime into a nickel decreases it by the same amount. As the sum increased by cents, there are more nickels than dimes. As the total count is , this means that there are nickels and dimes.Problem 16Three circles of radius are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?The situation in shown in the picture below. The radius we seek is . Clearly . The point is clearly the center of the equilateral triangle , thus is of the altitude of this triangle. We get that . Therefore the radius we seek is.WARNING. Note that the answer does not correspond to any of the five options. Most probably there is a typo in option D.Problem 17The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?Solution 1If Jack's current age is , then Bill's current age is .In five years, Jack's age will be and Bill's age will be .We are given that . Thus .For we get . For and the value is not an integer, and for it is more than . Thus the only solution is , and the difference in ages is .Solution 2Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.The age difference is , hence it is a multiple of . Thus Bill's current age modulo must be .Thus Bill's age is in the set .As Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options .Checking each of them, we see that only works, and gives the solution .Problem 18In the right triangle , we have , , and . Points , , and are located on , , and , respectively, so that , , and . What is the ratio of the area of to that of ?First of all, note that , and therefore.Draw the height from onto as in the picture below:Now consider the area of . Clearly the triangles and are similar, as they have all angles equal. Their ratio is ,hence . Now the area of can be computed as= . Similarly we can find that as well.Hence , and the answer is .Problem 19In the sequence , , , , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is. What is the term in this sequence?Solution 1We already know that , , , and . Let's compute the next few terms to get the idea how the sequence behaves. We get ,, , and so on.We can now discover the following pattern: and . This is easily proved by induction. It follows that.Solution 2Note that the recurrence can be rewritten as.Hence we get that and alsoFrom the values given in the problem statement we see that .From we get that .From we get that .Following this pattern, we get.Problem 20In points and lie on and , respectively. If and intersect at so that and , what is ?Solution (Triangle Areas)We use the square bracket notation to denote area.Without loss of generality, we can assume . Then , and . We have , so we need to find the area of quadrilateral .Draw the line segment to form the two triangles and . Let , and . By considering trianglesand , we obtain , and by considering triangles and , we obtain . Solving, we get , , so the area of quadrilateral is .ThereforeSolution (Mass points)The presence of only ratios in the problem essentially cries out for mass points.As per the problem, we assign a mass of to point , and a mass of to . Then, to balance and on , has a mass of .Now, were we to assign a mass of to and a mass of to , we'd have . Scaling this down by (to get , which puts and in terms of the masses of and ), we assign a mass of to and a mass of to .Now, to balance and on , we must give a mass of . Finally, the ratio of to is given by the ratio of the mass of tothe mass of , which is .Solution (Coordinates)Affine transformations preserve ratios of distances, and for any pair of triangles there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any , and we just need to compute it for any single triangle.We can choose the points , , and . This way we will have , and . The situation is shown in the picture below:The point is the intersection of the lines and . The points on the first line have the form , the points on the second line have the form . Solving for we get , hence.The ratio can now be computed simply by observing the coordinates of , , and :Problem 21Let ; ; and ; ; be two arithmetic progressions. The set is the union of the first terms of each sequence. How many distinct numbers are in ?The two sets of terms are and.Now . We can compute. We will now find .Consider the numbers in . We want to find out how many of them lie in . In other words, we need to find out the number of valid values of for which .The fact "" can be rewritten as ", and ".The first condition gives , the second one gives .Thus the good values of are , and their count is .Therefore , and thus .Problem 22A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?This is obviously a right triangle. Pick a coordinate system so that the right angle is at and the other two vertices are at and .As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at .The radius of the inscribed circle can be computed using the well-known identity , where is the area of the triangle and its perimeter. In our case, and , thus . As the inscribed circle touches both legs, its center must be at .The distance of these two points is then.Problem 23Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?Label the six sides of the cube by numbers to as on a classic dice. Then the "four vertical faces" can be: , , or .Let be the set of colorings where are all of the same color, similarly let and be the sets of good colorings for the other two sets of faces.There are possible colorings, and there are goodcolorings. Thus the result is . We need to compute .Using the we can writeClearly , as we have two possibilities for the common color of the four vertical faces, and two possibilities for each of the horizontal faces.What is ? The faces must have the same color, and at the same time faces must have the same color. It turns out thatthe set containing just the two cubes where all six faces have the same color.Therefore , and the result is .Problem 24In we have , , and . Point is on the circumscribed circle of the triangle so that bisects . What is the value of ?Problem 25A circle of radius is internally tangent to two circles of radius at points and , where is a diameter of the smaller circle. What is the area of the region, shaded in the picture, that is outside the smaller circle and inside each of the two larger circles?The area of the small circle is . We can add it to the shaded region, compute the area of the new region, and then subtract the area of the small circle from the result.Let and be the intersections of the two large circles. Connect them to and to get the picture below:Now obviously the triangles and are equilateral with side .Take a look at the bottom circle. The angle is , hence the sector is of the circle. The same is true for the sector of the bottom circle, and sectors and of the top circle.If we now sum the areas of these four sectors, we will almost get the area of the new shaded region - except that each of the two equilateral triangles will be counted twice.Hence the area of the new shaded region is, and the area of the original shared region is .。

2019年AMC10B美国数学竞赛(真题加详解)

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

2018年美国数学竞赛(AMC10A)的试题与详细解答

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AMC 美国数学竞赛 2003 AMC 10A 试题及答案解析

2003 AMC 10A1、What is the difference between the sum of the first evencounting numbers and the sum of the first odd counting numbers?SolutionThe first even counting numbers are .The first odd counting numbers are .Thus, the problem is asking for the value of..Alternatively, using the sum of an arithmetic progression formula, wecan write .2、Members of the Rockham Soccer League buy socks and T-shirts. Socks cost per pair and each T-shirt costs more than a pair ofsocks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is , how many members are in the League?SolutionSince T-shirts cost dollars more than a pair of socks, T-shirts costdollars.Since each member needs pairs of socks and T-shirts, the total costfor member is dollars.Since dollars was the cost for the club, and was the cost permember, the number of members in the League is .3、A solid box is cm by cm by cm. A new solid is formed byremoving a cube cm on a side from each corner of this box. Whatpercent of the original volume is removed?SolutionThe volume of the original box isThe volume of each cube that is removed isSince there are corners on the box, cubes are removed.So the total volume removed is .Therefore, the desired percentage is .4、It takes Mary minutes to walk uphill km from her home toschool, but it takes her only minutes to walk from school to homealong the same route. What is her average speed, in km/hr, for the round trip?SolutionSince she walked km to school and km back home, her totaldistance is km.Since she spent minutes walking to school and minutes walkingback home, her total time is minutes = hours.Therefore her average speed in km/hr is5、Let and denote the solutions of . What is thevalue of ?SolutionUsing factoring:orSo and are and .Therefore the answer isOR we can use sum and product.6、Define to be for all real numbers and . Which of thefollowing statements is not true?SolutionExamining statement C:when , but statement C says that it does for all .Therefore the statement that is not true is "for all "Alternatively, consider that the given "heart function" is actually the definition of the distance between two points. Examining all of the statements, only C is not necessarily true; if c is negative, the distance between and is the absolute value of , not itself,because distance is always nonnegative.7、How many non-congruent triangles with perimeter have integerside lengths?SolutionBy the triangle inequality, no one side may have a length greater thanhalf the perimeter, which isSince all sides must be integers, the largest possible length of a side isTherefore, all such triangles must have all sides of length , , or .Since , at least one side must have a length ofThus, the remaining two sides have a combined length of .So, the remaining sides must be either and or and .Therefore, the number of triangles is .8、What is the probability that a randomly drawn positive factor ofis lessthan ?SolutionFor a positive number which is not a perfect square, exactly half ofthe positive factors will be less than .Since is not a perfect square, half of the positive factors of will beless than .Clearly, there are no positive factors of between and .Therefore half of the positive factors will be less than .So the answer is .9、SimplifySolution.Therefore:10、The polygon enclosed by the solid lines in the ﬁgure consists ofcongruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?SolutionLet the squares be labeled , , , and .When the polygon is folded, the "right" edge of square becomesadjacent to the "bottom edge" of square , and the "bottom" edge ofsquare becomes adjacent to the "bottom" edge of square .So, any "new" square that is attached to those edges will prevent the polygon from becoming a cube with one face missing.Therefore, squares , , and will prevent the polygon frombecoming a cube with one face missing.Squares , , , , , and will allow the polygon to become a cubewith one face missing when folded.Thus the answer is .Another way to think of it is that a cube missing one face has of it'sfaces. Since the shape has faces already, we need another face.The only way to add anopther face is if the added square does not overlap any of the others. ,, and overlap, while 9 do not. Theanswer is11、The sum of the two -digit numbers and is .What is ?SolutionSince , , and are digits, , , .Therefore, .12、A point is randomly picked from inside the rectangle withvertices , , , and . What is the probability that ?SolutionThe rectangle has a width of and a height of .The area of this rectangle is .The line intersects the rectangle at and .The area which is the right isosceles triangle with side lengththat has vertices at , , and .The area of this triangle isTherefore, the probability that is13、The sum of three numbers is . The ﬁrst is times the sum of theother two. The second is seven times the third. What is the product of all three?SolutionSolution 1Let the numbers be , , and in that order. The given tells us thatTherefore, the product of all three numbers is.Solution 2Alternatively, we can set up the system in matrix form:Or, in matrix formTo solve this matrix equation, we can rearrange it thus:Solving this matrix equation by using inverse matrices and matrix multiplication yieldsWhich means that,, and. Therefore,14、Let be the largest integer that is the product of exactly distinctprime numbers, , , and , where and are single digits.What is the sum of the digits of ?SolutionSince is a single digit prime number, the set of possible values ofis.Since is a single digit prime number and is the units digit of theprime number, the set of possible values of is.Using these values for and , the set of possible values ofisOut of this set, the prime values areTherefore the possible values of are:The largest possible value of is .So, the sum of the digits of is15、What is the probability that an integer in the set is divisible by and not divisible by ?SolutionThere are integers in the set.Since every 2nd integer is divisible by , there are integersdivisible by in the set.To be divisible by both and , a number must be divisible by.Since every 6th integer is divisible by , there are integersdivisible by both and in the set.So there are integers in this set that are divisible by andnot divisible by .Therefore, the desired probability is16、What is the units digit of ?SolutionSince :Therefore, the units digit is17、The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?SolutionLet be the length of a side of the equilateral triangle and let be the radius of the circle.In a circle with a radius the side of an inscribed equilateral triangle is.So .The perimeter of the triangle isThe area of the circle isSo:18、What is the sum of the reciprocals of the roots of the equationSolutionMultiplying both sides by :Let the roots be and .The problem is asking forBy Vieta's formulas:So the answer is .19、A semicircle of diameter sits at the top of a semicircle of diameter, as shown. The shaded area inside the smaller semicircle andoutside the larger semicircle is called a lune. Determine the area of this lune.SolutionThe shaded area is equal to the area of the smaller semicircle minus the area of a sector of the larger circle plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle.The area of the smaller semicircle is .Since the radius of the larger semicircle is equal to the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures .The area of the sector of the larger semicircle is .The area of the triangle isSo the shaded area is20、A base-three-digit number is selected at random. Which of the following is closest to the probability that the base-representation and the base-representation of are boththree-digit numerals?SolutionTo be a three digit number in base-10:Thus there are three-digit numbers in base-10To be a three-digit number in base-9:To be a three-digit number in base-11:So,Thus, there are base-10 three-digit numbers that are three digitnumbers in base-9 and base-11.Therefore the desired probability is .21、Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?SolutionSolution 1Let the ordered triplet represent the assortment of chocolate chip cookies, oatmeal cookies, and peanut butter cookies.Using casework:Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal orpeanut butter.The assortments are:assortments.Pat selects chocolate chip cookie:Pat needs to select more cookies that are either oatmeal orpeanut butter.The assortments are:assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal orpeanut butter.The assortments are:assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal orpeanut butter.The assortments are:assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal orpeanut butter.The assortments are: assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal orpeanut butter.The assortments are: assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal orpeanut butter.The only assortment is: assortment.The total number of assortments of cookies that can be collected isSolution 2It is given that it is possible to select at least 6 of each. Therefore, we can make a bijection to the number of ways to divide the six choices into three categories, since it is assumed that their order is unimportant. Using the ball and urns formula, the number of ways todo this is22、In rectangle , we have , , is on with, is on with , line intersects line at , andis on line with . Find the length .SolutionS olution 1(Opposite angles are equal).(Both are 90 degrees).(Alt. Interior Angles are congruent).Therefore and are similar. and arealso similar.is 9, therefore must equal 5. Similarly, must equal 3.Because and are similar, the ratio of and ,must also hold true for and . , so is of . ByPythagorean theorem, ..So ..Therefore .Solution 2Since is a rectangle, .Since is a rectangle and ,.Since is a rectangle, .So, is a transversal, and .This is sufficient to prove that and .Using ratios:Since can't have 2 different lengths, both expressions for mustbe equal.Solution 3Since is a rectangle, , , and . From thePythagorean Theorem, . LemmaStatement:Proof: , obviously.Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.Let .Also, , thereforeWe can multiply both sides by to get that is twice of 10, orSolution 4We extend BC such that it intersects GF at X. Since ABCD is a rectangle, it follows that CD=8, therefore, XF=8. Let GX=y. From the similarity of triangles GCH and GEA, we have the ratio 3:5 (as CH=9-6=3, and EA=9-4=5). GX and GF are the altitudes of GCH andGEA, respectively. Thus, y:y+8 = 3:5, from which we have y=12, thus GF=y+8=12+8=20. B.23、A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the ﬁgure we have rows of small congruent equilateral triangles, with smalltriangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of small equilateral triangles?SolutionSolution 1There are small equilateral triangles.Each small equilateral triangle needs toothpicks to make it.But, each toothpick that isn't one of the toothpicks onthe outside of the large equilateral triangle is a side for smallequilateral triangles.So, the number of toothpicks on the inside of the large equilateraltriangle isTherefore the total number of toothpicks isSolution 2We see that the bottom row of small triangles is formed fromdownward-facing triangles and upward-facing triangles. Sinceeach downward-facing triangle uses three distinct toothpicks, andsince the total number of downward-facing triangles is, we have that the total number of toothpicks is24、Sally has ﬁve red cards numbered through and four blue cardsnumbered through . She stacks the cards so that the colorsalternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?SolutionLet and designate the red card numbered and the blue cardnumbered , respectively.is the only blue card that evenly divides, so must be at oneend of the stack and must be the card next to it.is the only other red card that evenly divides , so must be theother card next to .is the only blue card that evenly divides, so must be at theother end of the stack and must be the card next to it.is the only other red card that evenly divides , so must be theother card next to .doesn't evenly divide , so must be next to , must be nextto , and must be in the middle.This yields the following arrangement from top to bottom:Therefore, the sum of the numbers on the middle three cards is.25、Let be a -digit number, and let and be the quotient andremainder, respectively, when is divided by . For how manyvalues of is divisible by ?SolutionSolution 1When a -digit number is divided by , the first digits become thequotient, , and the last digits become the remainder, .Therefore, can be any integer from to inclusive, and can beany integer from to inclusive.For each of the possible values of , there are at leastpossible values of such that .Since there is "extra" possible value of that is congruent to, each of the values of that are congruent tohave more possible value of such that .Therefore, the number of possible values of such thatis .Solution 2Let equal , where through are digits. Therefore,We now take :The divisor trick for 11 is as follows:"Let be an digit integer. Ifis divisible by , then is also divisibleby ."Therefore, the five digit number is divisible by 11. The 5-digitmultiples of 11 range from to . There aredivisors of 11 between those inclusive.Solution 3Since q is a quotient and r is a remainder when n is divided by 100. So we have . Since we are counting choices where q+r is divisible by 11, we have for some k. This means that n is the sum of two multiples of 11 and would thus itself be a divisor of 11. Then we can count all the four digit divisors of 11 as in Solution 2. (This solution is essentially the same as solution 2, but it does not necessarily involve mods and so could potentially be faster.)NotesThe part labeled "divisor trick" actually follows from the same observation we made in the previous step: , therefore and for all . For a digit numberwe get, as claimed.Also note that in the "divisor trick" we actually want to assign the signs backwards - if we make sure that the last sign is a , the result will have the same remainder modulo as the original number.- 21 -。

AMC10-2010

2010A 1Mary’s top book shelf holds ﬁve books with the following widths,in centimeters:6,12,1,2.5,and 10.What is the average book width,in centimeters?(A)1(B)2(C)3(D)4(E)52Four identical squares and one rectangle are placed together to form one large square as shown.The length of the rectangle is how many times as large as its width?[asy]unitsize(8mm);defaultpen(linewidth(.8pt));draw(scale(4)*unitsquare);draw((0,3)–(4,3));draw((1,3)–(1,4));draw((2,3)–(2,4));draw((3,3)–(3,4));[/asy](A)54(B)43(C)32(D)2(E)33Tyrone had 97marbles and Eric had 11marbles.Tyrone then gave some of his marbles ot Eric so that Tyrone ended with twice as many marbles as Eric.How many marbles did Tyrone give to Eric?(A)3(B)13(C)18(D)25(E)294A book that is to be recorded onto compact discs takes 412minutes to read aloud.Each disc can hold up to 56minutes of reading.Assume that the smallest possible number of discs is used and that each disc contains the same length of reading.How many minutes of reading will each disc contain?(A)50.2(B)51.5(C)52.4(D)53.8(E)55.25The area of a circle whose circumference is 24πis kπ.What is the value of k ?(A)6(B)12(C)24(D)36(E)1446For positive numbers x and y the operation ♠(x,y )is deﬁned as ♠(x,y )=x −1y What is ♠(2,♠(2,2))?(A)23(B)1(C)43(D)53(E)27Crystal has a running course marked out for her daily run.She starts this run by heading due north for one mile.She then runs northeast for one mile,then southeast for one mile.The last portion of her run takes her on a straight line back to where she started.How far,in miles is this last portion of her run?(A)1(B)√2(C)√3(D)2(E)2√2This ﬁle was downloaded from the AoPS −MathLinks Math Olympiad Resources Page Page 1http://www.mathlinks.ro/20108Tony works 2hours a day and is paid $0.50per hour for each full year of his age.During a six month period Tony worked 50days and earned $630.How old was Tony at the end of the six month period?(A)9(B)11(C)12(D)13(E)149A palindrome ,such as 83438,is a number that remains the same when its digits are reversed.The numbers x and x +32are three-digit and four-digit palindromes,respectively.What is the sum of the digits of x?(A)20(B)21(C)22(D)23(E)2410Marvin had a birthday on Tuesday,May 27in the leap year 2008.In what year will hisbirthday next fall on a Saturday?(A)2011(B)2012(C)2013(D)2015(E)201711The length of the interval of solutions of the inequality a ≤2x +3≤b is 10.What is b −a ?(A)6(B)10(C)15(D)20(E)3012Logan is constructing a scaled model of his town.The city’s water tower stands 40metershigh,and the top portion is a sphere that holes 100,000liters of water.Logan’s miniature water tower holds 0.1liters.How tall,in meters,should Logan make his tower?(A)0.04(B)0.4π(C)0.4(D)4π(E)413Angelina drove at an average rate of 80kph and then stopped 20minutes for gas.After thestop,she drove at an average rate of 100kph.Altogether she drove 250km in a total trip time of 3hours including the stop.Which equation could be used to solve for the time t in hours that she drove before her stop?(A)80t +100(8/3−t )=250(B)80t =250(C)100t =250(D)90t =250(E)80(8/3−t )+100t =25014Triangle ABC has AB =2·AC .Let D and E be on AB and BC ,respectively,such that∠BAE =∠ACD.Let F be the intersection of segments AE and CD ,and suppose that CF E is equilateral.What is ∠ACB ?(A)60◦(B)75◦(C)90◦(D)105◦(E)120◦15In a magical swamp there are two species of talking amphibians:toads,whose statements arealways true,and frogs,whose statements are always false.Four amphibians,Brian,Chris,LeRoy,and Mike live together in the swamp,and they make the following statements:Brian:”Mike and I are diﬀerent species.”Chris:”LeRoy is a frog.”LeRoy:”Chris is a frog.”Mike:”Of the four of us,at least two are toads.”2010How many of these amphibians are frogs?(A)0(B)1(C)2(D)3(E)416Nondegenerate ABC has integer side lengths,BD is an angle bisector,AD =3,andDC =8.What is the smallest possible value of the perimeter?(A)30(B)33(C)35(D)36(E)3717A solid cube has side length 3inches.A 2-inch by 2-inch square hole is cut into the center ofeach face.The edges of each cut are parallel to the edges of the cube,and each hole goes all the way through the cube.What is the volume,in cubic inches,of the remaining solid?(A)7(B)8(C)10(D)12(E)1518Bernardo randomly picks 3distinct numbers from the set {1,2,3,4,5,6,7,8,9}and arrangesthem in descending order to form a 3-digit number.Silvia randomly picks 3distinct numbers from the set {1,2,3,4,5,6,7,8}and also arranges them in descending order to form a 3-digit number.What is the probability that Bernardo’s number is larger than Silvia’s number?(A)4772(B)3756(C)23(D)4972(E)395619Equiangular hexagon ABCDEF has side lengths AB =CD =EF =1and BC =DE =F A =r .The area of ACE is 70(A)4√33(B)103(C)4(D)174(E)620A ﬂy trapped inside a cubical box with side length 1meter decides to relieve its boredom byvisiting each corner of the box.It will begin and end in the same corner and visit each of the other corners exactly once.To get from a corner to any other corner,it will either ﬂy or crawl in a straight line.What is the maximum possible length,in meters,of its path?(A)4+4√(B)2+4√+2√(C)2+3√+3√(D)4√+4√(E)3√2+5√321The polynomial x 3−ax 2+bx −2010has three positive integer zeros.What is the smallestpossible value of a ?(A)78(B)88(C)98(D)108(E)11822Eight points are chosen on a circle,and chords are drawn connecting every pair of points.Nothree chords intersect in a single point inside the circle.How many triangles with all three vertices in the interior of the circle are created?(A)28(B)56(C)70(D)84(E)140201023Each of 2010boxes in a line contains a single red marble,and for 1≤k ≤2010,the box in thekth position also contains k white marbles.Isabella begins at the ﬁrst box and successively draws a single marble at random from each box,in order.She stops when she ﬁrst draws a red marble.Let P (n )be the probability that Isabella stops after drawing exactly n marbles.What is the smallest value of n for which P (n )<12010?(A)45(B)63(C)64(D)201(E)100524The number obtained from the last two nonzero digits of 90!is equal to n .What is n ?(A)12(B)32(C)48(D)52(E)6825Jim starts with a positive integer n and creates a sequence of numbers.Each successivenumber is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached.For example,if Jim starts with n =55,then his sequence contains 5numbers:5555−72=66−22=22−12=11−12=0Let N be the smallest number for which Jim’s sequence has 8numbers.What is the units digit of N ?(A)1(B)3(C)5(D)7(E)92010B 1What is 100(100−3)−(100·100−3)?(A)−20,000(B)−10,000(C)−297(D)−6(E)02Makayla attended two meetings during her 9-hour work day.The ﬁrst meeting took 45minutes and the second meeting took twice as long.What percent of her work day was spent attending meetings?(A)15(B)20(C)25(D)30(E)353A drawer contains red,green,blue,and white socks with at least 2of each color.What is the minimum number of socks that must be pulled from the drawer to guarantee a matching pair?(A)3(B)4(C)5(D)8(E)94For a real number x ,deﬁne ♥(x )to be the average of x and x 2.What is ♥(1)+♥(2)+♥(3)?(A)3(B)6(C)10(D)12(E)20[Thanks PowerOfPi,that’s exactly how the heart looks like.]5A month with 31days has the same number of Mondays and Wednesdays.How many of the seven days of the week could be the ﬁrst day of this month?(A)2(B)3(C)4(D)5(E)66A circle is centered at O ,AB is a diameter and C is a point on the circle with ∠COB =50◦.What is the degree measure of ∠CAB ?(A)20(B)25(C)45(D)50(E)657A triangle has side lengths 10,10,and 12.A rectangle has width 4and area equal to the area of the rectangle.What is the perimeter of this rectangle?(A)16(B)24(C)28(D)32(E)368A ticket to a school play costs x dollars,where x is a whole number.A group of 9th graders buys tickets costing a total of $48,and a group of 10th graders buys tickets costing a total of $64.How many values of x are possible?(A)1(B)2(C)3(D)4(E)520109Lucky Larry’s teacher asked him to substitute numbers for a ,b ,c ,d ,and e in the expression a −(b −(c −(d +e )))and evaluate the rry ignored the parentheses but added and subtracted correctly and obtained the correct result by coincedence.The numbers Larry substituted for a ,b ,c ,and d were 1,2,3,and 4,respectively.What number did Larry substitute for e ?(A)−5(B)−3(C)0(D)3(E)510Shelby drives her scooter at a speed of 30miles per hour if it is not raining,and 20milesper hour if it is raining.Today she drove in the sun in the morning and in the rain in the evening,for a total of 16miles in 40minutes.How many minutes did she drive in the rain?(A)18(B)21(C)24(D)27(E)3011A shopper plans to purchase an item that has a listed price greater than $100and can useany one of the three coupns.Coupon A gives 15%oﬀthe listed price,Coupon B gives $30the listed price,and Coupon C gives 25%oﬀthe amount by which the listed price exceeds $100.Let x and y be the smallest and largest prices,respectively,for which Coupon A saves at least as many dollars as Coupon B or C.What is y −x ?(A)50(B)60(C)75(D)80(E)10012At the beginning of the school year,50%of all students in Mr.Well’s math class answered”Yes”to the question ”Do you love math”,and 50%answered ”No.”At the end of the school year,70%answered ”Yes”and 30%answered ”No.”Altogether,x %of the students gave a diﬀerent answer at the beginning and end of the school year.What is the diﬀerence between the maximum and the minimum possible values of x ?(A)0(B)20(C)40(D)60(E)8013What is the sum of all the solutions of x =|2x −|60−2x ||?(A)32(B)60(C)92(D)120(E)12414The average of the numbers 1,2,3,...,98,99,and x is 100x .What is x ?(A)49101(B)50101(C)12(D)51101(E)509915On a 50-question multiple choice math contest,students receive 4points for a correct answer,0points for an answer left blank,and -1point for an incorrect answer.Jesse’s total score on the contest was 99.What is the maximum number of questions that Jesse could have answered correctly?(A)25(B)27(C)29(D)31(E)33201016A square of side length 1and a circle of radius √3/3share the same center.What is the areainside the circle,but outside the square?(A)π3−1(B)2π9−√33(C)π18(D)14(E)2π/917Every high school in the city of Euclid sent a team of 3students to a math contest.Eachparticipant in the contest received a diﬀerent score.Andrea’s score was the median among all students,and hers was the highest score on her team.Andrea’s teammates Beth and Carla placed 37th and 64th,respectively.How many schools are in the city?(A)22(B)23(C)24(D)25(E)2618Positive integers a,b,and c are randomly and independently selected with replacement fromthe set {1,2,3,...,2010}.What is the probability that abc +ab +a is divisible by 3?(A)13(B)2981(C)3181(D)1127(E)132719A circle with center O has area 156π.Triangle ABC is equilateral,BC is a chord on the circle,OA =4√3,and point O is outside ABC .What is the side length of ABC ?(A)2√(B)6(C)4√(D)12(E)1820Two circles lie outside regular hexagon ABCDEF .The ﬁrst is tangent to AB ,and the secondis tangent to DE .Both are tangent to lines BC and F A .What is the ratio of the area of the second circle to that of the ﬁrst circle?(A)18(B)27(C)36(D)81(E)10821A palindrome between 1000and 10,000is chosen at random.What is the probability that itis divisible by 7?(A)110(B)19(C)17(D)16(E)1522Seven distinct pieces of candy are to be distributed among three bags.The red bag and theblue bag must each receive at least one piece of candy;the white bag may remain empty.How many arrangements are possible?(A)1930(B)1931(C)1932(D)1933(E)193423The entries in a 3×3array include all the digits from 1through 9,arranged so that theentries in every row and column are in increasing order.How many such arrays are there?(A)18(B)24(C)36(D)42(E)60201024A high school basketball game between the Raiders and Wildcats was tied at the end of theﬁrst quarter.The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence,and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence.At the end of the fourth quarter,the Raiders had won by one point.Neither team scored more than 100points.What was the total number of points scored by the two teams in the ﬁrst half?(A)30(B)31(C)32(D)33(E)3425Let a >0,and let P (x )be a polynomial with integer coeﬃcients such thatP (1)=P (3)=P (5)=P (7)=a ,andP (2)=P (4)=P (6)=P (8)=−a .What is the smallest possible value of a ?(A)105(B)315(C)945(D)7!(E)8!。

AMC10美国数学竞赛A卷附中文翻译和答案

AMC10 美国数学竞赛 A 卷附中文翻译和答案简介AMC10 是美国的一项全国性数学竞赛，主要面向高中学生。

此文档提供了 AMC10 美国数学竞赛 A 卷的问题、答案及中文翻译，帮助考生更好地理解和准备竞赛。

问题和答案问题 1让我们从一个整数开始，每一步都按照以下规则进行操作：如果当前的数是偶数，将它除以 2；如果当前的数是奇数，将它乘以 3 并加 1。

通过这样的操作，我们可以生成一个数列，例如，从 9 开始的数列如下所示：9，28，14，7，22，11，34，...。

显然，这个数列最终会包含两个连续的 1。

以下哪个数开始操作后会生成包含两个连续的 1？$\\textbf{(A)}\\ 111 \\qquad \\textbf{(B)}\\ 120 \\qquad \\textbf{(C)}\\ 125 \\qquad \\textbf{(D)}\\ 130 \\qquad\\textbf{(E)}\\ 139$$\\textbf{(D)}\\ 130$问题 2如果 $2^x \\times 5^y = 5000000$，那么x+x的值是多少？$\\textbf{(A)}\\ 6 \\qquad \\textbf{(B)}\\ 7 \\qquad\\textbf{(C)}\\ 10 \\qquad \\textbf{(D)}\\ 12 \\qquad\\textbf{(E)}\\ 15$答案 2$\\textbf{(C)}\\ 10$问题 3数轴上三个数x、x、x的平均值是 6。

给定x−x=8，x−x=12，那么x的值是多少？$\\textbf{(A)}\\ -10 \\qquad \\textbf{(B)}\\ -6 \\qquad \\textbf{(C)}\\ 0 \\qquad \\textbf{(D)}\\ 6 \\qquad\\textbf{(E)}\\ 10$答案 3$\\textbf{(B)}\\ -6$某菜市场的一个南瓜重 100 磅，其中 99% 的重量是水分。

amc10公式范文

amc10公式范文AMC 10 (也被称为美国数学竞赛10级) 是由美国数学学会 (the Mathematical Association of America, MAA) 主办的一项年度数学竞赛，面向的是9-10年级的学生。

本文将介绍一些与AMC 10相关的公式和概念。

一、代数公式：1.因式分解公式：a^2-b^2=(a+b)(a-b)这个公式用于将一个二次差分形式的代数式因式分解为两个因数的乘积。

2. 二次方程解的公式：对于二次方程ax^2 + bx + c = 0，其解为：x = (-b ± √(b^2 - 4ac)) / (2a)这个公式被称为二次方程求根公式，用于求解二次方程的解。

3. 一元二次不等式解的公式：对于一元二次不等式ax^2 + bx + c > 0，其解的范围取决于判别式Δ = b^2 - 4ac 的正负性：当Δ>0时，解的范围为x<(-b-√Δ)/(2a)或x>(-b+√Δ)/(2a)当Δ=0时，解的范围为x=-b/(2a)当Δ<0时，无解。

二、几何公式：1.长方形面积公式：面积=长×宽2.正方形面积公式：面积=边长×边长3.三角形面积公式：面积=底×高/24.直角三角形勾股定理：a^2+b^2=c^2根据直角三角形的两条直角边边长可用该公式求得斜边的长度。

5.直线斜率公式：设点A(x1,y1)和点B(x2,y2)在直线上，直线的斜率m=(y2-y1)/(x2-x1)三、组合数学公式：1.阶乘公式：n!=n×(n-1)×(n-2)×...×2×1阶乘表示将一个正整数n连乘至1的结果。

2.组合公式：C(n,r)=n!/(r!×(n-r)!)组合公式计算了从n个元素中选取r个不同元素的不计次序的所有方法数。

3.排列公式：P(n,r)=n!/(n-r)!排列公式计算了从n个元素中选取r个不同元素的次序不同的所有方法数。

AMC_美国数学竞赛_2000_AMC_10__试题及答案解析

USA AMC 10 20001In the year , the United States will host the International Mathematical Olympiad. Let , , and be distinct positive integers such that the product . What's the largest possible value of the sum ?SolutionThe sum is the highest if two factors are the lowest.So, and .2Solution.3Each day, Jenny ate of the jellybeans that were in her jar at the beginning of the day. At the end of the second day, remained. How many jellybeans were in the jar originally?Solution4Chandra pays an online service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was , but in January her bill was because she used twice as much connect time as in December. What is the fixxed monthly fee?SolutionLet be the fixed fee, and be the amount she pays for the minutes she used in the first month.We want the fixed fee, which is5Points and are the midpoints of sides and of . As moves along a line that is parallel to side , how many of the four quantities listed below change?(a) the length of the segment(b) the perimeter of(c) the area of(d) the area of trapezoidSolution(a) Clearly does not change, and , so doesn't change either.(b) Obviously, the perimeter changes.(c) The area clearly doesn't change, as both the base and its corresponding height remain the same.(d) The bases and do not change, and neither does the height, so the trapezoid remains the same.Only quantity changes, so the correct answer is .6The Fibonacci Sequence starts with two 1s and each term afterwards is the sum of its predecessors. Which one of the ten digits is the last to appear in thet units position of a number in the Fibonacci Sequence?SolutionThe pattern of the units digits areIn order of appearance:.is the last.7In rectangle , , is on , and and trisect . What is the perimeter of ?Solution.Since is trisected, .Thus,.Adding, .8At Olympic High School, of the freshmen and of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?There are five times as many sophomores as freshmen.There are twice as many sophomores as freshmen.There are as many freshmen as sophomores.There are twice as many freshmen as sophomores.There are five times as many freshmen as sophomores.SolutionLet be the number of freshman and be the number of sophomores.There are twice as many freshmen as sophomores.9If , where , thenSolution, so ...10The sides of a triangle with positive area have lengths , , and . The sides of a second triangle with positive area have lengths , , and . What is the smallest positive number that is not a possible value of ?SolutionFrom the triangle inequality, and . The smallest positive number not possible is , which is .11Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?SolutionTwo prime numbers between and are both odd.Thus, we can discard the even choices.Both and are even, so one more than is a multiple of four.is the only possible choice.satisfy this, .12Figures , , , and consist of , , , and nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?SolutionSolution 1We have a recursion:.I.E. we add increasing multiples of each time we go up a figure. So, to go from Figure 0 to 100, we add.Solution 2We can divide up figure to get the sum of the sum of the first odd numbers and the sum of the first odd numbers. If you do not see this, here is the example for :The sum of the first odd numbers is , so for figure , there are unit squares. We plug in to get , which is choice13There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?SolutionIn each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left for the yellow peg in this column.By similar logic, we can fill in the yellow pegs as shown:After this we can proceed to fill in the whole pegboard, so there is only arrangement of the pegs. The answer is14Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were , , , , and . What was the last score Mrs. Walter entered?SolutionThe sum of the first scores must be even, so we must choose evens or the odds to be the first two scores.Let us look at the numbers in mod .If we choose the two odds, the next number must be a multiple of , of which there is none.Similarly, if we choose or , the next number must be a multiple of , of which there is none.So we choose first.The next number must be 1 in mod 3, of which only remains.The sum of the first three scores is . This is equivalent to in mod .Thus, we need to choose one number that is in mod . is the only one that works.Thus, is the last score entered.15Two non-zero real numbers, and , satisfy . Which of the following is a possible value of ?SolutionSubstituting , we get16The diagram shows lattice points, each one unit from its nearest neighbors. Segment meets segment at . Find the length of segment .SolutionSolution 1Let be the line containing and and let be the line containing and . If we set the bottom left point at , then , , , and .The line is given by the equation . The -intercept is , so . We are given two points on , hence we cancompute the slope, to be , so is the lineSimilarly, is given by . The slope in this case is , so . Plugging in the point gives us , so is the line .At , the intersection point, both of the equations must be true, soWe have the coordinates of and , so we can use the distance formula here:which is answer choiceSolution 2Draw the perpendiculars from and to , respectively. As it turns out, . Let be the point on for which ., and , so by AA similarity,By the Pythagorean Theorem, we have ,, and . Let , so , thenThis is answer choiceAlso, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B.17Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?SolutionConsider what happens each time he puts a coin in. If he puts in a quarter, he gets five nickels back, so the amount of money he has doesn't change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesn't change. However, if he puts a penny in, he gets five quarters back, increasing the amount of money he has by cents.This implies that the only possible values, in cents, he can have are the ones one more than a multiple of . Of the choices given, the only one is18Charlyn walks completely around the boundary of a square whose sides are each km long. From any point on her path she can see exactly km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?SolutionThe area she sees looks at follows:The part inside the walk has area . The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area . The four arcs together form a circle with radius . Therefore the total area she can see is, which rounded to the nearest integer is .19Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the trangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is times the area of the square. The ratio of the area of the other small right triangle to the area of the square isSolutionLet the square have area , then it follows that the altitude of one of the triangles is . The area of the other triangle is .By similar triangles, we haveThis is choice(Note that this approach is enough to get the correct answer in the contest. However, if we wanted a completely correct solution, we should also note that scaling the given triangle times changes each of the areas times, and therefore it does not influence the ratio of any two areas. This is why we can pick the side of the square.)20Let , , and be nonnegative integers such that . What is the maximum value of ?SolutionThe trick is to realize that the sum is similar to the product .If we multiply , we get.We know that , therefore.Therefore the maximum value of is equal to the maximum value of . Now we will find this maximum.Suppose that some two of , , and differ by at least . Then this triple is surely not optimal.Proof: WLOG let . We can then increase the value ofby changing and .Therefore the maximum is achieved in the cases where is a rotation of . The value of in this case is . And thus the maximum of is.21If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?I. All alligators are creepy crawlers.II. Some ferocious creatures are creepy crawlers.III. Some alligators are not creepy crawlers.SolutionWe interpret the problem statement as a query about three abstract concepts denoted as "alligators", "creepy crawlers" and "ferocious creatures". In answering the question, we may NOT refer to reality -- for example to the fact that alligators do exist.To make more clear that we are not using anything outside the problem statement, let's rename the three concepts as , , and .We got the following information:▪If is an , then is an .▪There is some that is a and at the same time an .We CAN NOT conclude that the first statement is true. For example, the situation "Johnny and Freddy are s, but only Johnny is a "meets both conditions, but the first statement is false.We CAN conclude that the second statement is true. We know that there is some that is a and at the same time an . Pick one such and call it Bobby. Additionally, we know that if is an , then is an. Bobby is an , therefore Bobby is an . And this is enough to prove the second statement -- Bobby is an that is also a .We CAN NOT conclude that the third statement is true. For example, consider the situation when , and are equivalent (represent the same set of objects). In such case both conditions are satisfied, but the third statement is false.Therefore the answer is .22One morning each member of Angela's family drank an -ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?SolutionThe exact value "8 ounces" is not important. We will only use the fact that each member of the family drank the same amount.Let be the total number of ounces of milk drank by the family and the total number of ounces of coffee. Thus the whole family drank a total of ounces of fluids.Let be the number of family members. Then each family member drank ounces of fluids.We know that Angela drank ounces of fluids.As Angela is a family member, we have .Multiply both sides by to get .If , we have .If , we have .Therefore the only remaining option is .23When the mean, median, and mode of the list are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ?SolutionAs occurs three times and each of the three other values just once, regardless of what we choose the mode will always be .The sum of all numbers is , therefore the mean is .The six known values, in sorted order, are . From this sequence we conclude: If , the median will be . If , the median will be . Finally, if , the median will be .We will now examine each of these three cases separately.In the case , both the median and the mode are 2, therefore we can not get any non-constant arithmetic progression.In the case we have , because. Therefore our three values inorder are . We want this to be an arithmetic progression. From the first two terms the difference must be . Therefore thethird term must be .Solving we get the only solution for this case: . The case remains. Once again, we have ,therefore the order is . The only solution is when , i. e., .The sum of all solutions is therefore .24Let be a function for which . Find the sum of all values of for which .SolutionIn the definition of , let . We get: . Aswe have , we must have , in other words .One can now either explicitly compute the roots, or use Vieta's formulas. According to them, the sum of the roots of is . In our case this is .(Note that for the above approach to be completely correct, we should additionally verify that there actually are two distinct real roots. This is, for example, obvious from the facts that and .)25In year , the day of the year is a Tuesday. In year , the day is also a Tuesday. On what day of the week did the of year occur?SolutionClearly, identifying what of these years may/must/may not be a leap year will be key in solving the problem.Let be the day of year , the day of year and the day of year .If year is not a leap year, the day will bedays after . As , that would be a Monday.Therefore year must be a leap year. (Then is days after .) As there can not be two leap years after each other, is not a leap year. Therefore day is days after . We have . Therefore is weekdays before , i.e., is a.(Note that the situation described by the problem statement indeed occurs in our calendar. For example, for we have =Tuesday, October 26th 2004, =Tuesday, July 19th, 2005 and =Thursday, April 10th 2003.)。

AMC-美国数学竞赛-2003-AMC-10A-试题及答案解析

2003 AMC 10A1、What is the difference between the sum of the first even counting numbers and the sum of the first odd counting numbers?SolutionThe first even counting numbers are .The first odd counting numbers are .Thus, the problem is asking for the value of..Alternatively, using the sum of an arithmetic progression formula, we can write .2、Members of the Rockham Soccer League buy socks and T-shirts. Socks cost per pair and each T-shirt costs more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is , how many members are in the League?SolutionSince T-shirts cost dollars more than a pair of socks, T-shirts cost dollars.Since each member needs pairs of socks and T-shirts, the total cost for member is dollars.Since dollars was the cost for the club, and was the cost per member, the number of members in the League is .3、A solid box is cm by cm by cm. A new solid is formed by removinga cube cm on a side from each corner of this box. What percent of the original volume is removed?SolutionThe volume of the original box isThe volume of each cube that is removed isSince there are corners on the box, cubes are removed.So the total volume removed is .Therefore, the desired percentage is .4、It takes Mary minutes to walk uphill km from her home to school, but it takes her only minutes to walk from school to home along the same route. What is her average speed, in km/hr, for the round trip?SolutionSince she walked km to school and km back home, her total distance is km.Since she spent minutes walking to school and minutes walking back home, her total time is minutes = hours.Therefore her average speed in km/hr is5、Let and denote the solutions of . What is the value of ?SolutionUsing factoring:orSo and are and .Therefore the answer isOR we can use sum and product.6、Define to be for all real numbers and . Which of the following statements is not true?SolutionExamining statement C:when , but statement C says that it does for all . Therefore the statement that is not true is " for all "Alternatively, consider that the given "heart function" is actually the definition of the distance between two points. Examining all of the statements, only C is not necessarily true; if c is negative, the distance between and is the absolute value of , not itself, because distance is always nonnegative.7、How many non-congruent triangles with perimeter have integer side lengths?SolutionBy the triangle inequality, no one side may have a length greater than half the perimeter, which isSince all sides must be integers, the largest possible length of a side isTherefore, all such triangles must have all sides of length , , or .Since , at least one side must have a length ofThus, the remaining two sides have a combined length of .So, the remaining sides must be either and or and .Therefore, the number of triangles is .8、What is the probability that a randomly drawn positive factor ofis lessthan ?SolutionFor a positive number which is not a perfect square, exactly half of the positive factors will be less than .Since is not a perfect square, half of the positive factors of will be less than .Clearly, there are no positive factors of between and . Therefore half of the positive factors will be less than .So the answer is .9、SimplifySolution.Therefore:10、The polygon enclosed by the solid lines in the ﬁgure consists of congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?SolutionLet the squares be labeled , , , and .When the polygon is folded, the "right" edge of square becomes adjacent to the "bottom edge" of square , and the "bottom" edge of square becomes adjacent to the "bottom" edge of square .So, any "new" square that is attached to those edges will prevent the polygon from becoming a cube with one face missing.Therefore, squares , , and will prevent the polygon from becoming a cube with one face missing.Squares , , , , , and will allow the polygon to become a cube with one face missing when folded.Thus the answer is .Another way to think of it is that a cube missing one face has of it's faces. Since the shape has faces already, we need another face. The onlyway to add anopther face is if the added square does not overlap any of the others.,, and overlap, while 9 do not. The answer is11、The sum of the two -digit numbers and is . What is ?SolutionSince , , and are digits, , , .Therefore, .12、A point is randomly picked from inside the rectangle with vertices , , , and . What is the probability that ?SolutionThe rectangle has a width of and a height of .The area of this rectangle is .The line intersects the rectangle at and .The area which is the right isosceles triangle with side length that has vertices at , , and .The area of this triangle isTherefore, the probability that is13、The sum of three numbers is . The ﬁrst is times the sum of the other two. The second is seven times the third. What is the product of all three?SolutionSolution 1Let the numbers be , , and in that order. The given tells us thatTherefore, the product of all three numbers is.Solution 2Alternatively, we can set up the system in matrix form:Or, in matrix formTo solve this matrix equation, we can rearrange it thus:Solving this matrix equation by using inverse matrices and matrix multiplication yieldsWhich means that , , and . Therefore,14、Let be the largest integer that is the product of exactly distinct prime numbers, , , and , where and are single digits. What is the sum of the digits of ?SolutionSince is a single digit prime number, the set of possible values ofis .Since is a single digit prime number and is the units digit of the prime number , the set of possible values of is .Using these values for and , the set of possible values of isOut of this set, the prime values areTherefore the possible values of are:The largest possible value of is .So, the sum of the digits of is15、What is the probability that an integer in the setis divisible by and not divisible by ?SolutionThere are integers in the set.Since every 2nd integer is divisible by , there are integers divisible by in the set.To be divisible by both and , a number must be divisible by .Since every 6th integer is divisible by , there are integers divisible by both and in the set.So there are integers in this set that are divisible by and not divisible by .Therefore, the desired probability is16、What is the units digit of ?SolutionSince :Therefore, the units digit is17、The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?SolutionLet be the length of a side of the equilateral triangle and let be the radius of the circle.In a circle with a radius the side of an inscribed equilateral triangle is .So .The perimeter of the triangle is The area of the circle isSo:18、What is the sum of the reciprocals of the roots of the equationSolutionMultiplying both sides by :Let the roots be and .The problem is asking forBy Vieta's formulas:So the answer is .19、A semicircle of diameter sits at the top of a semicircle of diameter , as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.SolutionThe shaded area is equal to the area of the smaller semicircle minus the area of a sector of the larger circle plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle.The area of the smaller semicircle is .Since the radius of the larger semicircle is equal to the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures .The area of the sector of the larger semicircle is . The area of the triangle isSo the shaded area is20、A base- three-digit number is selected at random. Which of the following is closest to the probability that the base- representation and the base- representation of are both three-digit numerals?SolutionTo be a three digit number in base-10:Thus there are three-digit numbers in base-10To be a three-digit number in base-9:To be a three-digit number in base-11:So,Thus, there are base-10 three-digit numbers that are three digit numbers in base-9 and base-11.Therefore the desired probability is .21、Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?SolutionSolution 1Let the ordered triplet represent the assortment of chocolate chip cookies, oatmeal cookies, and peanut butter cookies.Using casework:Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are:assortments.Pat selects chocolate chip cookie:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are:assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are:assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are: assortments. Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are: assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are: assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The only assortment is: assortment.The total number of assortments of cookies that can be collected isSolution 2It is given that it is possible to select at least 6 of each. Therefore, we can make a bijection to the number of ways to divide the six choices into three categories, since it is assumed that their order is unimportant. Using the ball and urns formula, the number of ways to do this is22、In rectangle , we have , , is on with, is on with , line intersects line at , and is on line with . Find the length .SolutionS olution 1(Opposite angles are equal).(Both are 90 degrees).(Alt. Interior Angles are congruent).Therefore and are similar. and are also similar.is 9, therefore must equal 5. Similarly, must equal 3. Because and are similar, the ratio of and , must also hold true for and . , so is of . By Pythagorean theorem, ..So ..Therefore .Solution 2Since is a rectangle, .Since is a rectangle and ,.Since is a rectangle, .So, is a transversal, and .This is sufficient to prove that and . Using ratios:Since can't have 2 different lengths, both expressions for must be equal.Solution 3Since is a rectangle, , , and . From the Pythagorean Theorem, .LemmaStatement:Proof: , obviously.Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.Let .Also, , thereforeWe can multiply both sides by to get that is twice of 10, orWe extend BC such that it intersects GF at X. Since ABCD is a rectangle, it follows that CD=8, therefore, XF=8. Let GX=y. From the similarity of triangles GCH and GEA, we have the ratio 3:5 (as CH=9-6=3, and EA=9-4=5). GX and GF are the altitudes of GCH and GEA, respectively. Thus, y:y+8 = 3:5, from which we have y=12, thus GF=y+8=12+8=20. B.23、A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the ﬁgure we have rows of small congruent equilateral triangles, with small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists ofsmall equilateral triangles?SolutionSolution 1There are small equilateral triangles.Each small equilateral triangle needs toothpicks to make it.But, each toothpick that isn't one of the toothpicks on the outside of the large equilateral triangle is a side for small equilateral triangles.So, the number of toothpicks on the inside of the large equilateral triangle isTherefore the total number of toothpicks isWe see that the bottom row of small triangles is formed from downward-facing triangles and upward-facing triangles. Since each downward-facing triangle uses three distinct toothpicks, and since the total number of downward-facing triangles is, we have that the total number of toothpicks is24、Sally has ﬁve red cards numbered through and four blue cards numbered through . She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?SolutionLet and designate the red card numbered and the blue card numbered , respectively.is the only blue card that evenly divides, so must be at one end of the stack and must be the card next to it.is the only other red card that evenly divides , so must be the other card next to .is the only blue card that evenly divides, so must be at the other end of the stack and must be the card next to it.is the only other red card that evenly divides , so must be the other card next to .doesn't evenly divide , so must be next to , must be next to , and must be in the middle.This yields the following arrangement from top to bottom:Therefore, the sum of the numbers on the middle three cards is.25、Let be a -digit number, and let and be the quotient and remainder, respectively, when is divided by . For how many values of is divisible by ?SolutionSolution 1When a -digit number is divided by , the first digits become the quotient, , and the last digits become the remainder, .Therefore, can be any integer from to inclusive, and can be any integer from to inclusive.For each of the possible values of , there are at least possible values of such that .Since there is "extra" possible value of that is congruent to , each of the values of that are congruent tohave more possible value of such that .Therefore, the number of possible values of such thatis .Solution 2Let equal , where through are digits. Therefore,We now take :The divisor trick for 11 is as follows:"Let be an digit integer. Ifis divisible by , then is also divisible by ."Therefore, the five digit number is divisible by 11. The 5-digit multiples of 11 range from to . There aredivisors of 11 between those inclusive.Solution 3Since q is a quotient and r is a remainder when n is divided by 100. So we have . Since we are counting choices where q+r is divisible by 11, we havefor some k. This means that n is the sum of two multiples of 11 and would thus itself be a divisor of 11. Then we can count all the four digit divisors of 11 as in Solution 2. (This solution is essentially the same as solution 2, but it does not necessarily involve mods and so could potentially be faster.)NotesThe part labeled "divisor trick" actually follows from the same observation we made in the previous step: , therefore and for all . For a digit number we get, as claimed.Also note that in the "divisor trick" we actually want to assign the signs backwards - if we make sure that the last sign is a , the result will have the same remainder modulo as the original number.（注：可编辑下载，若有不当之处，请指正，谢谢!）。

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AMC 中的数论问题1：Remember the prime between 1 to 100：2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 91 2：Perfect number:Let P is the prime number.if21p - is also the prime number. then 1(21)2p p --is theperfect number. For example:6,28,496. 3: Let,0n abc a =≠ is three digital integer .if 333n a b c =++Then the number n is called Daffodils number . There are only four numbers:153 370 371 407 Let,0n abcd a =≠ is four digital integer .if 4444d n a b c +=++Then the number n is called Roses number . There are only three numbers:1634 8208 94744：The Fundamental Theorem of ArithmeticEvery natural number n can be written as a product of primes uniquely up to order.n =∏p i ri ki=15：Suppose that a and b are integers with b =0. Then there exists unique integers q and r such that 0 ≤ r< |b| and a = bq + r.6：(1)Greatest Common Divisor: Let gcd (a, b) = max {d ∈ Z: d | a and d | b}. For any integers a and b, we havegcd(a, b) = gcd(b, a) = gcd(±a, ±b) = gcd(a, b − a) = gcd(a, b + a). For example: gcd(150, 60) = gcd(60, 30) = gcd(30, 0) = 30 (2)Least common multiple:Let lcm(a,b)=min{d ∈Z: a | d and b | d }. (3)We have that: ab= gcd(a, b) lcm(a,b)7：Congruence modulo nIf ,0a b mq m -=≠,then we call a congruence b modulo m and we rewrite mod a b m ≡. (1)Assume a,b,c,d,m ,k ∈Z (k >0,m ≠0).If a ≡b mod m,c ≡d mod m then we havemod a c b d m ±≡± , mod ac bd m ≡ , mod k k a b m ≡(2) The equation ax ≡ b (mod m) has a solution if and only if gcd(a, m) divides b.8：How to find the unit digit of some special integers (1)How many zero at the end of !nFor example, when 100n =, Let N be the number zero at the end of 100!then10010010020424525125N ⎡⎤⎡⎤⎡⎤=++=+=⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦(2) ,,a n Z ∈Find the unit digit na . For example, when 100,3n a ==9：Palindrome, such as 83438, is a number that remains the same when its digits are reversed. There are some number not only palindrome but 112=121，222=484，114=14641(1)Some special palindrome n that 2n is also palindrome. For example :222221111121111123211111123432111111111112345678987654321=====(2)How to create a palindrome? Almost integer plus the number of its reversed digits and repeat it again and again. Then we get a palindrome. For example:87781651655617267266271353135335314884+=+=+=+=But whether any integer has this Property has yet to prove(3) The palindrome equation means that equation from left to right and right to left it all set up.For example ：1242242112231132211121241388888831421211⨯=⨯⨯=⨯⨯==⨯Let ab and cde are two digital and three digital integers. If the digits satisfy the,,9a c b e d c e d ⨯=⨯=+≤, then ab cde edc ba ⨯=⨯.10: Features of an integer divisible by some prime number If n is even ，then 2|n一个整数n 的所有位数上的数字之和是3（或者9）的倍数，则n 被3（或者9）整除一个整数n 的尾数是零，则n 被5整除一个整数n 的后三位与截取后三位的数值的差被7、11、13整除，则n 被7、11、13整除一个整数n 的最后两位数被4整除，则n 被4整除一个整数n 的最后三位数被8整除，则n 被8整除一个整数n 的奇数位之和与偶数位之和的差被11整除，则n 被11整除 11. The number Theoretic functions If 312123t rr rrt n p p p p =(1) {}12()#0:|(1)(1)(1)t n a a n r r r χ=>=+++(2) 12222111222|()(1)(1)(1)t r r r t t t a nn a p p p p p p p p p δ==+++++++++∑(3){}11221111122()#:,gcd(,)1()()()ttr r r r r r t t n a N a n a n p p p p p p φ---=∈≤==---For example: 2(12)(23)(21)(11)6χχ=⋅=++= 22(12)(23)(122)(13)28δδ=⋅=+++= 22(12)(23)(22)(31)4φφ=⋅=--= Exercise1. The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number? (A) 4 (B) 5 (C) 6 (D) 7 (E) 83. For the positive integer n, let <n> denote the sum of all the positive divisors of n with the exception of n itself. For example, <4>=1+2=3 and <12>=1+2+3+4+6=16. What is <<<6>>>? (A) 6 (B) 12 (C) 24 (D) 32 (E) 36 8. What is the sum of all integer solutions to 21<(x-2)<25? (A) 10(B) 12(C) 15(D) 19(E) 5(A) 6(B) 7(C) 8(D) 9(E) 10(A) 1(B) 2(C) 3(D) 4(E) 515．The figures 123,,F F F and 4F shown are the first in a sequence of figures. For 3n ≥, n F is constructed from -1n F by surrounding it with a square and placing one more diamond on each side of the new square than -1n F had on each side of its outside square. For example, figure 3F has 13 diamonds. How many diamonds are there in figure 20F ?18. Positive integers a, b, and c are randomly and independently selected with replacement from the set {1, 2, 3,…, 2010}. What is the probability that abc ab a ++ is divisible by 3? (A)13(B)2981(C)3181(D)1127(E)132724. Let ,a b and c be positive integers with >>a b c such that 222-b -c +=2011a ab and222+3b +3c -3-2-2=-1997a ab ac bc . What is a ?(A) 249 (B) 250(C) 251(D) 252(E)2535. In multiplying two positive integers a and b, Ron reversed the digits of the two-digit number a. His erroneous product was 161. What is the correct value of the product of a and b? (A) 116 (B) 161 (C) 204 (D) 214 (E) 224 23. What is the hundreds digit of 20112011?(A) 1 (B) 4 (C) 5 (D) 6 (E) 99. A palindrome, such as 83438, is a number that remains the same when its digits are reversed. The numbers x and x+32 are three-digit and four-digit palindromes, respectively. What is the sum of the digits of x? (A) 20 (B) 21 (C) 22 (D) 23 (E) 24 21. The polynomial 322010x ax bx -+- has three positive integer zeros. What is the smallest possible value of a? (A) 78 (B) 88 (C) 98 (D) 108 (E) 11824. The number obtained from the last two nonzero digits of 90! Is equal to n. What is n? (A) 12 (B) 32 (C) 48 (D) 52 (E) 6825. Jim starts with a positive integer n and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with n=55, then his sequence contain 5 numbers:5555-72= 6 6-22= 2 2-12= 1 1-12= 0Let N be the smallest number for which Jim’s sequence has 8 numbers. What is the units digit of N? (A) 1 (B) 3 (C) 5 (D) 7 (E) 9 21．What is the remainder when 01220093+3+3++3is divided by 8?(A) 0(B) 1(C) 2(D) 4(E) 65．What is the sum of the digits of the square of 111,111,111?(A) 18(B) 27(C) 45(D) 63(E) 8110064(A) 6(B) 7(C) 8(D) 9(E) 1024. Let 2200820082k =+. What is the units digit of 222k +? (A) 0(B) 1(C) 4(D) 6 (E) 8AMC about algebraic problems一、Linear relations(1) Slope y-intercept form: y kx b =+ (k is the slope, b is the y-intercept) (2)Standard form: 0Ax By C ++=(3)Slope and one point 0000(,),()()P x y k slope y y k x x -=- (4) Two points 1122(,),(,)P x y P x y12121212y y y y y y x x x x x x ---==--- (5)x,y-intercept form: (,0),(0,),(0,0)1x y P a Q b a b a b≠≠+= 二、the relations of the two lines 111222:0,:0l A x B y C l A x B y C ++=++=(1) 1l ∥2l 122112210,0A B A B C B C B ⇔-=-≠ (1)1l ⊥2l 12120A A B B ⇔-=三、Special multiplication rules:222223322332212211222112222()()()2()()()()()()(2)()((1)(1))(1)()n n n n n n n n n n n n n n a b a b a b a b a ab b a b a b a ab b a b a b a ab b a b a b a a b ab b n a b a b a a b ab b n is odd n a b c ab bc ac a b -----------=-+±=±+-=-+++=+-+-=-++++≥+=+-++-+->++=++⇔-22()()0b c c a a b c+-+-=⇔==四、quadratic equations and PolynomialThe quadratic equations 2(0)y ax bx c a =++≠ has two roots 12,x x then we has1212b c x x x x a a+=-=More generally, if the polynomial 121210nn n n n x a xa x a x a ---+++++= has n roots123,,,,n x x x x ,then we have:1231122312123(1)n n n n n nx x x x a x x x x x x a x x x x a -++++=-++==-开方的开方、估计开方数的大小绝对值方程 Arithmetic Sequence123(1)(2)(3)()n m a a n d a n d a n d a n m d =+-=+-=+-==+-121321()()()()2222n n n m n m n n a a n a a n a a n a a s ---+++++=====1(1)2n n n ds na -=+If n=2k, then we have 1()n k k s k a a +=+ If n=2k+1, then we have 1n k s na += Geometric sequence123123n n n n m n m a a q a q a q a q ----=====1(1)1,1n n a q q s q-≠=-Some special sequence 1, 1, 2, 3, 5, 8,… 9,99,999,9999，… 1，11，111，1111，… Exercise4 .When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be left over?7 For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?21. Four distinct points are arranged on a plane so that the segments connecting them have lengths ,,,,, and . What is the ratio of to?6. The product of two positive numbers is 9. The reciprocal of one of these numbers is 4 times the324. Let ,a b and c be positive integers with >>a b c such that 222-b -c +=2011a ab and222+3b +3c -3-2-2=-1997a ab ac bc . What is a ?(A) 249 (B) 250(C) 251(D) 252(E) 2531. What is 246135135246++++-++++? (A) -1(B) 536(C) 712(D)14760(E)43310. Consider the set of numbers {1, 10, 102, 103……1010}. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer? (A) 1 (B) 9 (C) 10 (D) 11 (E) 101=(A) -64 (B) -24 (C) -9 (D) 24 (E) 576 4. Let X and Y be the following sums of arithmetic sequences: X= 10 + 12 + 14 + …+ 100. Y= 12 + 14 + 16 + …+ 102.What is the value of Y X -? (A) 92 (B) 98 (C) 100 (D) 102 (E) 112 7. Which of the following equations does NOT have a solution?(A) 2(7)0x += (B) -350x += (C)20=(D)80=(E) -340x -=(A)(B)(C)2(D) (E) 613. What is the sum of all the solutions of 2602x x x =--?(A) 32 (B) 60 (C) 92 (D) 120 (E) 12414. The average of the numbers 1, 2, 3… 98, 99, and x is 100x. What is x?(A)49101 (B)50101 (C)12 (D)51101(E)509911. The length of the interval of solutions of the inequality 23a x b ≤+≤ is 10. What is b-a?(A) 6 (B) 10 (C) 15 (D) 20 (E) 3013. Angelina drove at an average rate of 80 kph and then stopped 20 minutes for gas. After the stop, she drove at an average rate of 100 kph. Altogether she drove 250 km in a total trip time of 3 hours including the stop. Which equation could be used to solve for the time t in hours that she drove before her stop?(A) 880100()2503t t +-=(B) 80250t =(C) 100250t =(D) 90250t =(E) 880()1002503t t -+=21. The polynomial 32-2010x ax bx +- has three positive integer zeros. What is the smallest possible value of a? (A) 78 (B) 88(C) 98(D) 108(E) 11815．When a bucket is two-thirds full of water, the bucket and water weigh kilograms. When the bucket is one-half full of water the total weight is kilograms. In terms of and , what is thetotal weight in kilograms when the bucket is full of water?13．Suppose that and. Which of the following is equal tofor everypair of integers?16．Let ,,, andbe real numbers with,, and.What is the sum of all possible values of?5. Which of the following is equal to the product?81216442008............481242004n n + (A) 251(B) 502(C) 1004(D) 2008(E) 40167. The fraction 20082200622007220052(3)(3)(3)(3)-- simplifies to which of the following?(A) 1 (B) 9/4 (C) 3 (D) 9/2 (E) 913. Doug can paint a room in 5 hours. Dave can paint the same room in 7 hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let t be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by t ?(A) 11()(1)157t ++=(B) 11()1157t ++=(C) 11()157t +=(D) 11()(1)157t +-=(E) (57)1t +=15. Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian? (A) 120 (B) 130 (C) 140 (D) 150 (E) 160AMC 中的几何问题一、三角形有关知识点1.三角形的简单性质与几个面积公式①三角形任何两边之和大于第三边； ②三角形任何两边之差小于第三边； ③三角形三个内角的和等于180°； ④三角形三个外角的和等于360°；⑤三角形一个外角等于和它不相邻的两个内角的和； ⑥三角形一个外角大于任何一个和它不相邻的内角。