2015 AMC 12A 美国数学竞赛 2015年 12A

2008年美国数学竞赛-AMC-12AUSA AMC12A 20081.A bakery owner turns on his doughnut machine at 8:30 AM. At 11:10 AM themachine has completed one third of the day's job. At what time will the doughnut machine complete the job?1:50 PM 3:00 PM 3:30 PM 4:30 PM 5:50 PM2.What is the reciprocal of ?3.Suppose that of bananas are worth as much as oranges. How manyoranges are worth as much is of bananas?4.Which of the following is equal to the product5.Suppose thatis an integer. Which of the following statements must be true about ?6.Heather compares the price of a new computer at two different stores. StoreA offers off the sticker price followed by a rebate, and storeB offersoff the same sticker price with no rebate. Heather saves by buying thecomputer at store A instead of store B. What is the sticker price of the computer, in dollars?7.While Steve and LeRoy are fishing mile from shore, their boat springs a leak,and water comes in at a constant rate of gallons per minute. The boat will sink if it takes in more than gallons of water. Steve starts rowing toward the shore at a constant rate of miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking?8.What is the volume of a cube whose surface area is twice that of a cube withvolume ?9.Older television screens have an aspect ratio of . That is, the ratio of thewidth to the height is . The aspect ratio of many movies is not , so they are sometimes shown on a television screen by'letterboxing' - darkening strips of equal height at the top and bottom of the screen, as shown. Supposea movie has an aspect ratio of and is shown on an older television screenwith a -inch diagonal. What is the height, in inches, of each darkened strip?10.Doug can paint a room in hours. Dave can paint the same room in hours.Doug and Dave paint the room together and take a one-hour break for lunch.Let 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 ?11.Three cubes are each formed from the pattern shown. They are then stackedon a table one on top of another so that the visible numbers have the greatest possible sum. What is that sum?12.A function has domain and range . (The notation denotes.) What are the domain and range, respectively, of the function defined by ?13.Points and lie on a circle centered at , and . A second circleis internally tangent to the first and tangent to both and . What is the ratio of the area of the smaller circle to that of the larger circle?14.What is the area of the region defined by the inequality15.Let . What is the units digit of ?16.The numbers , , and are the first three terms of anarithmetic sequence, and the term of the sequence is . What is ?17.Let be a sequence of integers determined by the rule ifis even and if is odd. For how many positive integersis it true that is less than each of , , and ?18.Triangle , with sides of length , , and , has one vertex on the positive-axis, one on the positive -axis, and one on the positive -axis. Let be the origin. What is the volume of tetrahedron ? 19.In the expansion of,what is the coefficient of ?20.Triangle has , , and . Point is on , andbisects the right angle. The inscribed circles of and have radii and , respectively. What is ?21.A permutation of is heavy-tailed if. What is the number of heavy-tailed permutations?22.A round table has radius . Six rectangular place mats are placed on the table.Each place mat has width and length as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length . Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is ?23.The solutions of the equation are the vertices of aconvex polygon in the complex plane. What is the area of the polygon?24.Triangle has and . Point is the midpoint of . What is the largest possible value of ?25.A sequence , , , of points in the coordinate plane satisfiesfor .Suppose that . What is ?。

2019AMC12AProblem1The area of a pizza with radius inches is percent larger than the area of a pizza withradius inches.What is the integer closest to?半径为4英寸的比萨饼面积比半径为3英寸的比萨饼面积大的百分数是N。

问最接近N的整数是什么？P roblem2Suppose is of.What percent of is?假设a是b的150%。

那么a的百分之多少是3b?Problem3A box contains red balls,green balls,yellow balls,blue balls,white balls,and black balls.What is the minimum number of balls that must be drawn from the box withoutreplacement to guarantee that at least balls of a single color will be drawn?一个盒子中有28个红球，20个绿球，19个黄球，13个蓝球，11个白球和9个黑球。

为了保证至少取出15个单一颜色的球,在不允许放回重取的情况下，必须从盒子里取出的球的数量最少是多少个？Problem4What is the greatest number of consecutive integers whose sum is?最多可以有多少个连续整数，它们的总和是45?Two lines with slopes and intersect at.What is the area of the triangle enclosed by thesetwo lines and the line?两条斜率分别为12和2的直线相交于(2,2)。

2010 AMC 12AProblem 11. What is ?Solution.Problem 2A ferry boat shuttles tourists to an island every hour starting at 10 AM until its last trip, which starts at 3 PM. One day the boat captain notes that on the 10 AM trip there were 100 tourists on the ferry boat, and that on each successive trip, the number of tourists was 1 fewer than on the previous trip. How many tourists did the ferry take to the island that day?SolutionIt is easy to see that the ferry boat takes trips total. The total number of people taken to the island isProblem 3Rectangle , p of . Square shares of . W?SolutionSolution 1Let , le, a.Solution 2The answer does not change if we shift to coincide with , and add new horizontal lines to divide intThis helps us to see that an, w. H.Problem 4If , tSolutionis negative, so we can just place a negative value into each expression and find the one that is positive. Suppose we use .Obviously only isProblem 5Halfway through a 100-shot archery tournament, Chelsea leads by 50 points. For each shot a bullseye scores 10 points, with other possible scores being 8, 4, 2, and 0 points. Chelsea always scores at least 4 points on each shot. If Chelsea's next shots are bullseyes she will be guaranteed victory. What is the minimum value for ?SolutionLet be the number of points Chelsea currently has. In order to guarantee victory, we must consider the possibility that the opponent scores the maximum amount of points by getting only bullseyes.The lowest integer value that satisfies the inequality is .Problem 6A , sare reversed. The numbers and arrespectively. What is the sum of the digits of ?Solutionis at most , s is . T is . However, the only palindrome between an is , which means thatmu.It follows that is , s.Problem 7Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?SolutionThe water tower holds timTherefore, Logan should make his tower times shorter than the actual tower. This is me.Problem 8Triangle ha. L and be on an, rthat . Let be the intersection of segments an, a suppose that is ?SolutionLet .Since , t is triangle, soProblem 9A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?SolutionSolution 1Imagine making the cuts one at a time. The first cut removes a box . T second cut removes two boxes, each of dimensions , adoes the same as the second cut, on the last two faces. Hence the total volume of all cuts is .Therefore the volume of the rest of the cube is .Solution 2We can use Principle of Inclusion-Exclusion to find the final volume of the cube.There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has cuas the central cucorrect result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.Hence the total volume of the cuts is . Therefore the volume of the rest of the cube is . Solution 3We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.Each edge can be seen as a box, and each corner can be seen as a bo.Solution 4First, you can find the volume, which is 27. Now, imagine there are three prisms of dimensions 2 x 2 x 3. Now subtract the prism volumes from 27. We have -9. From here we add two times 2^3, because we over-removed (LOL). This is 16 - 9 = 7 (A).Problem 10The first four terms of an arithmetic sequence are , , , a. Wthe terSolutionan ar.The common difference is . The first term is and the terProblem 11The solution of the equation ca. What is ?SolutionThis problem is quickly solved with knowledge of the laws of exponents and logarithms.Since we are looking for the base of the logarithm, our answer is .Problem 12In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Fouramphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.Brian: "Mike and I are different species."Chris: "LeRoy is a frog."LeRoy: "Chris is a frog."Mike: "Of the four of us, at least two are toads."How many of these amphibians are frogs?SolutionSolution 1We can begin by first looking at Chris and LeRoy.Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog.Clearly, Chris and LeRoy are different species, and so we have at least frog out of the two of them.Now suppose Mike is a toad. Then what he says is true because we already have toads. However, if Brian is a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction.Therefore, Mike must be a frog. His statement must be false, which means that there is at most toad. Since either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false.Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have frogs total.Solution 2Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.As Mike is a frog, his statement is false, hence there is at most one toad.As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.Hence we must have one toad and three frogs.Problem 13For how many integer values of do the graphs of an no intersect?SolutionThe image below shows the two curves for . T, which is clearly a circle with radius , and the red curve is a part of the curve .In the special case th, a, tpoint is on the red curve as well, hence they intersect.The case is : the blue curve remains the same and the redcurve is flipped according to the axis. Hence we just need to focus on .Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as approaches 0, approaches . Hence the red curve intersects the blue one if and only if it contains a point whose distance from the origin is at most .At this point we can guess that on the red curve the point where isclosest to the origin, and skip the rest of this solution.For an exact solution, fix and consider any point onfrom the origin is . To minimize this distance, it is enough to minimize . Bthat this value is at least , a, i.e., .Now recall that the red curve intersects the blue one if and only if its closest point is at most from the origin. We just computed that the distance between the origin and the closest point on the red curve is . Tintegers such that .Clearly the only such integer is , han. T values.Problem 14Nondegenerate ha is , and . What is the smallest possible value of the perimeter?SolutionBy the Angle Bisector Theorem, we know that . Ipossible integer values for AB and BC (the measures of AD and DC, respectively), then , contradicting the Triangle Inequality. If weuse the next lowest values ( a),satisfied. Therefore, our answer is , or choice .Problem 15A coin is altered so that the probability that it lands on heads is less than and when the coin is flipped four times, the probability of an equal number of heads and tailsis . WSolutionLet be the probability of flipping heads. It follows that the probability of flipping tails is .The probability of flipping heads and tails is equal to the number of ways to flip it times the product of the probability of flipping each coin.As for the desired probability both and are nonnegative, we only need to consider the positive root, henceApplying the quadratic formula we get that the roots of this equation are . Athe probability of heads is less than , we get that the answer is .Problem 16Bernardo randomly picks 3 distinct numbers from the set an arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set 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?SolutionWe can solve this by breaking the problem down into cases and adding up the probabilities.Case : Bernardo picks . If Bernardo picks a then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a is .Case : Bernardo does not pick . Since the chance of Bernardo picking is , the probability of not picking is .If Bernardo does not pick 9, then he can pick any number from to . Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger.Ignoring the for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers.We get this probability to beProbability of Bernardo's number being greater isFactoring the fact that Bernardo could've picked a but didn't:Adding up the two cases we getProblem 17Equiangular hexagon ha an. The area of is ofis the sum of all possible values of ?SolutionIt is clear that isthat . Therefore, the area of is .If we extend , an so an me, an meet at ,and an me, we find that hexagon isequilateral triangle of side length antriangles, , an, o. The area of is therefore.Based on the initial conditions,Simplifying this gives us . Bsum of the possible value of is .Problem 18A 16-step path is to go from to wi-coordinate or the -coordinate by 1. How many such paths stay outside or on the boundary of the square , at each step?SolutionBrute Force SolutionThe number of ways to reach any point onways to reach plu. Urecursion, we can draw the diagram and label each point with the number of ways to reach it and go up until we reach the end. Luckily, the figure is not so big that this is too time-consuming or difficult to do.For example:etc.We soon reachCombinatorial Solution 1By symmetry we only need to count the paths that go through the second quadrant (, ).For each of these paths, let be. Cand the previous point on such path has to be .Fix the value of . There are wa to, a wa to . Hence for we paths, for wepaths, and for wepaths. This gives us pahence the total number of paths is .Combinatorial Solution 2Each path that goes through the second quadrant must pass through exactly one of the points , , a.There is exactly path of the first kind, paths of the second kind, andpaProblem 19Each of 2010 boxes in a line contains a single red marble, and for , the box in the po white marbles. Isabella begins at the first boxand successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let bestops after drawing exactly marbles. What is the smallest value of for which ?SolutionThe probability of drawing a white marble from box is . Tdrawing a red marble from box is .The probability of drawing a red marble at box is thereforeIt is then easy to see that the lowest integer value of that satisfies the inequality is .Problem 20Arithmetic sequences an haand for. What is the largest possible value of ?SolutionSolution 1Since an ha, weach sequence aswhere and () aSinceit is easy to see that.Hence, we have to find the largest such that and arThe prime factorization of is . Whave a product ofand soon find that the largest va for the pair , alargest value is .Solution 2As above, let an for some . Now we get , hence. Therefore div. A as the second term is greater than the first one, we only have to consider the options .For we th another itFor we such that . Nthat must be divisible by . We can start looking for the solution by trying the possible values for , a we, w.Hence is . (There is no need to check anProblem 21The graph of lies above the lineexcept at three values of , where the graph and the line intersect. What is the largest of these values?SolutionSolutionThe values in which intersect atare the same as the zeros of .Since there are zeros and the function is never negative, all zeros must be double roots because the function's degree is .Suppose we let , , and be the roots of this function, and letbe the cubic polynomial with roots , , and .In order to find we must first expand out the terms of .[Quick note: Since we don't know , , and , we really don't even need the last 3 terms of the expansion.]All that's left is to find the largest root of .Problem 22What is the minimum value of ?SolutionSolution 1If we graph each term separately, we will notice that all of the zeros occur at , where is any integer from to , inThe minimum value occurs where the sum of the slopes is at a minimum, since it is easy to see that the value will be increasing on either side. That means theminimum must happen at some .The sum of the slope at isNow we want to minimize . T an, wmeans the slope is where .We can now verify that both an yie.Solution 2Rewrite the given expression as follows:Imagine the real line. For eachim boys standing at the coordinate . We now need to place a girl on the real line in such a way that the sum of her distances from all the boys is minimal, and we need to compute this sum.Note that there are boys in total. Let's label them from 1 (the only boy placed at ) to (the last boy placed at .Clearly, the minimum sum is achieved if the girl's coordinate is the median of the boys' coordinates. To prove this, place the girl at the median coordinate. If you nowmove her in any direction by any amount , there will be bomoves away from this boy. For each of the remaining boys, she moves at most closer, hence the total sum of distances does not decrease.Hence the optimal solution is to place the girl at the median coordinate. Or, more precisely, as is even, we can place her anywhere on the segment formed by boyand boy : bsegment the sum of distances is the same.By trial and error, or by solving the quadratic equation wethat boy number is and the next boy is the one placed at . H. Common part of both solutionsTo find the minimum, pick . Note that the terms to ar negative, and the terms to arandHence the total sum of distances is .Problem 23The number obtained from the last two nonzero digits of is . What is ?SolutionWe will use the fact that for any integer ,First, we find that the number of factors of in is. Let . T we want is therefore the last two digits of , o. Since there is clearly an excess of factors of 2, weknow that , s.If we divide by by in , w as where where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form is , and every number in the form is .The number caUsing the identity at the beginning of the solution, we can reduce toUsing the fact that (oif you have your powers of 2 memorized), we can deduce that. T. Finally, combining with the fact that yie.Problem 24Let . The intersection of the domain of wi is disjoint open intervals. What is ?SolutionThe question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them.We note that since all of the facundefined where the inside of the logarithm is equal to or less than .First, let us find the number of zeros of the inside of the logarithm.After counting up the number of zeros for each factor and eliminating the excess cases we get ze intIn order to find which intervals are negative, we must first realize that at every zero of each factor, the sign changes. We also have to be careful, as some zeros are doubled, or even tripled, quadrupled, etc.The first interval is is negative. Continuing the pattern and accounting for doubled roots (which do not flipsign), we realize that there are negative intervals from to . Since the function is symmetric, we know that there are also negative intervals from to .And so, the total number of disjoint open intervals isProblem 25Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?SolutionIt should first be noted that given any quadrilateral of fixed side lengths, the angles can be manipulated so that the quadrilateral becomes cyclic.Denote , , , and as the integer side lengths of the quadrilateral. Without loss of generality, let .Since , the Triangle Inequality implies that .We will now split into cases.Case : (side lengths are equal)Clearly there is only way to select the side lengths , and no matter how the sides are rearranged only unique quadrilateral can be formed.Case : or(side lengths are equal)If side lengths are equal, then each of those side lengths can only be integers from to ex(because that is counted in the first case). Obviously there is still only unique quadrilateral that can be formed from one set of side lengths, resulting in a total of quadrilaterals.Case : (pairs of side lengths are equal)and can be any integer from to , a and can be any integer from to . However, a single set of side lengths can form different cyclic quadrilaterals (a rectangle and a kite), so the total number of quadrilaterals for this case is.Case : or or(side lengths are equal)If the equal side lengths are each , then the other sides must each be , wwe have already counted in an earlier case. If the equal side lengths are each , there is possible set of side lengths. Likewise, for side lengths of there are sets. Continuing this pattern, we find a total ofsets of side lengths. (Be VERY careful when adding up the total for this case!) For each set of side lengths, there are possible quadrilaterals that can be formed, sot the total number of quadrilaterals for this case is .Case : (no side lengths are equal) Using the same counting principles starting from an, wnumber of possible side lengths is . T wa side lengths, but there is only unique quadrilateral for rotations, so the number ofquadrilaterals for each set of side lengths is . Tquadrilaterals is .And so, the total number of quadrilaterals that can be made is.。

AMC12 2014AProblem 1What isSolutionAt the theater children get in for half price. The price for adult tickets and child tickets is . How much would adult tickets and child tickets cost?SolutionWalking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?SolutionSuppose that cows give gallons of milk in days. At this rate, how many gallons of milk will cows give in days?SolutionOn an algebra quiz, of the studentsscored points, scored points, scored points, and the rest scored points. What is the difference between the mean and median score of the students' scores on this quiz?SolutionThe difference between a two-digit number and the number obtained by reversing its digits is times the sum of the digits of either number. What is the sum of the two digit number and its reverse?SolutionThe first three terms of a geometric progression are , , and . What is the fourth term?SolutionA customer who intends to purchase an appliance has three coupons, only one of which may be used:Coupon 1: off the listed price if the listed price is at leastCoupon 2: dollars off the listed price if the listed price is at leastCoupon 3: off the amount by which the listed price exceedsFor which of the following listed prices will coupon offer a greater price reduction than either coupon or coupon ?Five positive consecutive integers starting with have average . What is the average of consecutive integers that start with ?SolutionThree congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length . The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles?SolutionDavid drives from his home to the airport to catch a flight. He drives miles in the first hour, but realizes that he will be hour late if he continues at this speed. He increases his speed by miles per hour for the rest of the way to the airport and arrives minutes early. How many miles is the airport from his home?SolutionTwo circles intersect at points and . The minor arcs measure on one circle and on the other circle. What is the ratio of the area of the larger circle to the area of the smaller circle?A fancy bed and breakfast inn has rooms, each with a distinctive color-coded decor. One day friends arrive to spend the night. There are no other guests that night. The friends can room in any combination they wish, but with no morethan friends per room. In how many ways can the innkeeper assign the guests to the rooms?SolutionLet be three integers such that is an arithmetic progressionand is a geometric progression. What is the smallest possible value of ?SolutionA five-digit palindrome is a positive integer with respective digits , where is non-zero. Let be the sum of all five-digit palindromes. What is the sum of the digits of .SolutionThe product , where the second factor has digits, is an integer whose digits have a sum of . What is ?SolutionA rectangular box contains a sphere of radius and eight smaller spheres of radius . The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is ?SolutionThe domain of the function is an interval of length , where and are relatively prime positive integers. What is ?SolutionThere are exactly distinct rational numbers such that andhas at least one integer solution for . What is ?SolutionIn , , , and . Points and lieon and respectively. What is the minimum possible valueof ?SolutionFor every real number , let denote the greatest integer not exceeding , and let The set of all numbers suchthat and is a union of disjoint intervals. What is the sum of the lengths of those intervals?SolutionThe number is between and . How many pairs ofintegers are there such that andSolutionThe fraction where is the length of the period of the repeating decimal expansion. What is the sum ?SolutionLet , and for , let . For how many values of is ?The parabola has focus and goes through the points and . For how many points with integer coefficients is it truethat ?AMC 12 2013AProblem 1Square has side length . Point is on , and the areaof is . What is ?SolutionA softball team played ten games, scoring , and runs. They lost by one run in exactly five games. In each of the other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?SolutionA flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?SolutionWhat is the value ofSolutionTom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $, Dorothy paid $, and Sammy paid $. In order to share the costs equally, Tom gave Sammy dollars, and Dorothy gave Sammy dollars. What is ?SolutionIn a recent basketball game, Shenille attempted only three-point shots andtwo-point shots. She was successful on of her three-point shots and of her two-point shots. Shenille attempted shots. How many points did she score?SolutionThe sequence has the property that every term beginning with the third is the sum of the previous two. That is,Suppose that and . What is ?SolutionGiven that and are distinct nonzero real numbers such that , what is ?SolutionIn , and . Points and are onsides , , and , respectively, such that and are parallelto and , respectively. What is the perimeter of parallelogram ?SolutionLet be the set of positive integers for which has the repeating decimal representation with and different digits. What is the sum of the elements of ?SolutionTriangle is equilateral with . Points and are on and points and are on such that both and are parallel to . Furthermore, triangle and trapezoids and all have the same perimeter. What is ?SolutionThe angles in a particular triangle are in arithmetic progression, and the side lengths are . The sum of the possible values of x equals where , and are positive integers. What is ?SolutionLet points and .Quadrilateral is cut into equal area pieces by a line passing through . This line intersects at point , where these fractions are in lowest terms. What is ?SolutionThe sequence, , , ,is an arithmetic progression. What is ?SolutionRabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?Solution, , are three piles of rocks. The mean weight of the rocks in is pounds, the mean weight of the rocks in is pounds, the mean weight of the rocks in the combined piles and is pounds, and the mean weight of the rocks in the combined piles and is pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles and ?SolutionA group of pirates agree to divide a treasure chest of gold coins among themselves as follows. The pirate to take a share takes of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the pirate receive?SolutionSix spheres of radius are positioned so that their centers are at the vertices of a regular hexagon of side length . The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?SolutionIn , , and . A circle with center andradius intersects at points and . Moreover and have integer lengths. What is ?SolutionLet be the set . For , define to mean thateither or . How many ordered triples of elements of have the property that , , and ?SolutionConsider. Which of the following intervals contains ?SolutionA palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10 with no leading zeros. A 6-digit palindrome is chosen uniformly at random. What is the probability that is also a palindrome?Solutionis a square of side length . Point is on such that . The square region bounded by is rotated counterclockwise with center , sweeping out a region whose area is , where , , and are positive integers and . What is ?SolutionThree distinct segments are chosen at random among the segments whoseend-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?SolutionLet be defined by . How many complexnumbers are there such that and both the real and the imaginary parts of are integers with absolute value at most ?AMC12 2012AProblem 1A bug crawls along a number line, starting at . It crawls to , then turns around and crawls to . How many units does the bug crawl altogether?SolutionCagney can frost a cupcake every seconds and Lacey can frost a cupcake every seconds. Working together, how many cupcakes can they frostin minutes?SolutionA box centimeters high, centimeters wide, and centimeters long canhold grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold grams of clay. What is ?SolutionIn a bag of marbles, of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?SolutionA fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad?SolutionThe sums of three whole numbers taken in pairs are , , and . What is the middle number?SolutionMary divides a circle into sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?SolutionAn iterative average of the numbers , , , , and is computed in the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?SolutionA year is a leap year if and only if the year number is divisible by (such as ) or is divisible by but not by (such as ). The anniversary of the birth of novelist Charles Dickens was celebrated on February , , a Tuesday. On what day of the week was Dickens born?SolutionA triangle has area , one side of length , and the median to that side of length . Let be the acute angle formed by that side and the median. Whatis ?SolutionAlex, Mel, and Chelsea play a game that has rounds. In each round there is a single winner, and the outcomes of the rounds are independent. For each round the probability that Alex wins is , and Mel is twice as likely to win as Chelsea. What is the probability that Alex wins three rounds, Mel wins two rounds, and Chelsea wins one round?SolutionA square region is externally tangent to the circle withequation at the point on the side . Vertices and are on the circle with equation . What is the side length of this square?SolutionPaula the painter and her two helpers each paint at constant, but different, rates. They always start at , and all three always take the same amount of time to eat lunch. On Monday the three of them painted of a house, quittingat . On Tuesday, when Paula wasn't there, the two helpers paintedonly of the house and quit at . On Wednesday Paula worked by herself and finished the house by working until . How long, in minutes, was each day's lunch break?SolutionThe closed curve in the figure is made up of congruent circular arcs each of length , where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side . What is the area enclosed by the curve?SolutionA square is partitioned into unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is the rotated clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability that the grid is now entirely black?SolutionCircle has its center lying on circle . The two circles meet at and . Point in the exterior of lies on circle and , ,and . What is the radius of circle ?SolutionLet be a subset of with the property that no pair of distinct elements in has a sum divisible by . What is the largest possible size of ?SolutionTriangle has , , and . Let denote the intersection of the internal angle bisectors of . What is ?SolutionAdam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?SolutionConsider the polynomialThe coefficient of is equal to . What is ?SolutionLet , , and be positive integers with such thatWhat is ?SolutionDistinct planes intersect the interior of a cube . Let be the unionof the faces of and let . The intersection of and consists of the union of all segments joining the midpoints of every pair of edges belonging to the same face of . What is the difference between the maximum and minimum possible values of ?SolutionLet be the square one of whose diagonals hasendpoints and . A point is chosen uniformly at random over all pairs of real numbers and suchthat and . Let be a translated copyof centered at . What is the probability that the square region determinedby contains exactly two points with integer coefficients in its interior?SolutionLet be the sequence of real numbers definedby , and in general,Rearranging the numbers in the sequence in decreasing order produces anew sequence . What is the sum of all integers , , such thatSolutionLet where denotes the fractional part of . Thenumber is the smallest positive integer such that the equationhas at least real solutions. What is ? Note: the fractional part of is a real number such that and is an integer.2014A1.C2.B3.B4.A5.C6.D7.A8.C9.B10.B11.C12.D13.B14.C15.B16.D17.A18.C19.E20.D21.A22.B23.B24.C25.B 2013A1. E2. C3. E4. C5. B6. B7. C8. D9. C10. D11. C12. A13. B14. B15. D16. E17. D18. B19. D20. B21. A22. E23. C24. E25. A2012A1. E2. D3. D4. C5. D6. D7. C8. C9. A10. D11. B12. D13. D14. E15. A16. E17. B18. A19. B20. B21. E22. C23. C24. C25. C。

美国数学竞赛AMC12词汇Aabbreviation 简写符号；简写absolute error 绝对误差absolute value 绝对值accuracy 准确度acute angle 锐⾓acute-angled triangle 锐⾓三⾓形add 加addition 加法addition formula 加法公式addition law 加法定律addition law(of probability)（概率）加法定律additive property 可加性adjacent angle 邻⾓adjacent side 邻边algebra 代数algebraic 代数的algebraic equation 代数⽅程algebraic expression 代数式algebraic fraction 代数分式；代数分数式algebraic inequality 代数不等式algebraic operation 代数运算alternate angle （交）错⾓alternate segment 交错⼸形altitude ⾼；⾼度；顶垂线；⾼线ambiguous case 两义情况；⼆义情况amount 本利和；总数analysis 分析；解析analytic geometry 解析⼏何angle ⾓angle at the centre 圆⼼⾓angle at the circumference 圆周⾓angle between a line and a plane 直与平⾯的交⾓angle between two planes 两平⾯的交⾓angle bisection ⾓平分angle bisector ⾓平分线；分⾓线angle in the alternate segment 交错⼸形的圆周⾓angle in the same segment 同⼸形内的圆周⾓angle of depression 俯⾓angle of elevation 仰⾓angle of greatest slope 最⼤斜率的⾓angle of inclination 倾斜⾓angle of intersection 相交⾓；交⾓angle of rotation 旋转⾓angle of the sector 扇形⾓angle sum of a triangle 三⾓形内⾓和angles at a point 同顶⾓annum(X% per annum) 年（年利率X%）anti-clockwise direction 逆时针⽅向；返时针⽅向anti-logarithm 逆对数；反对数anti-symmetric 反对称apex 顶点approach 接近；趋近approximate value 近似值approximation 近似；略计；逼近Arabic system 阿刺伯数字系统arbitrary 任意arbitrary constant 任意常数arc 弧arc length 弧长arc-cosine function 反余弦函数arc-sin function 反正弦函数arc-tangent function 反正切函数area ⾯积arithmetic 算术arithmetic mean 算术平均；等差中顶；算术中顶arithmetic progression 算术级数；等差级数arithmetic sequence 等差序列arithmetic series 等差级数arm 边arrow 前号ascending order 递升序ascending powers of X X 的升幂associative law 结合律assumed mean 假定平均数assumption 假定；假设average 平均；平均数；平均值average speed 平均速率axiom 公理axis 轴axis of parabola 拋物线的轴axis of symmetry 对称轴Bback substitution 回代bar chart 棒形图；条线图；条形图；线条图base （1）底；（2）基；基数base angle 底⾓base area 底⾯base line底线base number 底数；基数base of logarithm 对数的底bearing ⽅位（⾓）；⾓⽅向（⾓）bell-shaped curve 钟形图bias 偏差；偏倚billion ⼗亿binary number ⼆进数binary operation ⼆元运算binary scale ⼆进法binary system ⼆进制binomial ⼆项式binomial expression ⼆项式bisect 平分；等分bisection method 分半法；分半⽅法bisector 等分线；平分线boundary condition 边界条件boundary line 界（线）；边界bounded 有界的bounded above 有上界的；上有界的bounded below 有下界的；下有界的bounded function 有界函数brace ⼤括号bracket 括号breadth 阔度broken line graph 折线图Ccalculation 计算calculator 计算器；计算器cancel 消法；相消canellation law 消去律capacity 容量Cartesian coordinates 笛卡⼉坐标Cartesian plane 笛卡⼉平⾯category 类型；范畴central line 中线central tendency 集中趋centre 中⼼；⼼centre of a circle 圆⼼centroid 形⼼；距⼼certain event 必然事件chance 机会change of base 基的变换change of subject 主项变换change of variable 换元；变量的换chart 图；图表checking 验算chord 弦chord of contact 切点弦circle 圆circular 圆形；圆的circular function 圆函数；三⾓函数circular measure 弧度法circumcentre 外⼼；外接圆⼼circumcircle 外接圆circumference 圆周circumradius 外接圆半径circumscribed circle 外接圆class 区；组；类class boundary 组界class interval 组区间；组距class limit 组限；区限class mark 组中点；区中点classification 分类clnometer 测斜仪clockwise dirction 顺时针⽅向closed convex region 闭凸区域closed interval 闭区间coefficient 系数coincide 迭合；重合collection of terms 并项collinear 共线collinear planes 共线⾯column (1)列；纵⾏；(2) 柱combination 组合common chord 公弦common denominator 同分母；公分母common difference 公差common divisor 公约数；公约common factor 公因⼦；公因⼦common logarithm 常⽤对数common multiple 公位数；公倍common ratio 公⽐common tangetn 公切commutative law 交换律comparable 可⽐较的compass 罗盘compass bearing 罗盘⽅位⾓compasses 圆规compasses construction 圆规作图complement 余；补余complementary angle 余⾓complementary event 互补事件complementary probability 互补概率completing the square 配⽅complex number 复数complex root 复数根composite number 复合数；合成数compound bar chart 综合棒形图compound discount 复折扣compound interest 复利；复利息computation 计算computer 计算机；电⼦计算器concave 凹concave downward 凹向下的D data 数据decagon ⼗边形decay 衰变decay factor 衰变因⼦decimal ⼩数decimal place ⼩数位decimal point ⼩数点decimal system ⼗进制decrease 递减decreasing function 递减函数；下降函数decreasing sequence 递减序列；下降序列decreasing series 递减级数；下降级数decrement 减量deduce 演绎deduction 推论deductive reasoning 演绎推理definite 确定的；定的distance 距离distance formula 距离公式distinct roots 相异根distincr solution 相异解distribution 公布distrivutive law 分配律divide 除dividend (1)被除数；(2)股息divisible 可整除division 除法division algorithm 除法算式divisor 除数；除式；因⼦divisor of zero 零因⼦dodecagon ⼗⼆边形dot 点double root ⼆重根due east/ south/ west /north 向东/ 南/ 西/ 北definiton 定义degree (1)度；(2)次degree of a polynomial 多项式的次数degree of accuracy 准确度degree of precision 精确度delete 删除；删去denary number ⼗进数denary scale ⼗进法denary system ⼗进制denominator 分母dependence (1)相关；(2)应变dependent event(s) 相关事件；相依事件；从属事件dependent variable 应变量；应变数depreciation 折旧descending order 递降序descending powers of X X的降序detached coefficients 分离系数(法)deviation 偏差；变差deviation from the mean 离均差diagonal 对⾓diagram 图；图表diameter 直径difference 差digit 数字dimension 量；量⽹；维(数)direct proportion 正⽐例direct tax, direct taxation 直接税direct variation 正变（分）directed angle 有向⾓directed number 有向数direction ⽅向；⽅位discontinuous 间断(的)；⾮连续(的)；不连续(的) discount 折扣discount per cent 折扣百分率discrete 分⽴；离散discrete data 离散数据；间断数据discriminant 判别式dispersion 离差displacement 位移disprove 反证Eedge 棱；边elimination 消法elimination method 消去法；消元法elongation 伸张；展empirical data 实验数据empirical formula 实验公式empirical probability 实验概率；经验概率enclosure 界限end point 端点entire surd 整⽅根equal 相等equal ratios theorem 等⽐定理equal roots 等根equality 等(式)equality sign 等号equation ⽅程equation in one unknown ⼀元⽅程equation in two unknowns (variables) ⼆元⽅程equation of a straight line 直线⽅程equation of locus 轨迹⽅程equiangular 等⾓(的)extreme value 极值equidistant 等距(的)equilaeral 等边(的)equilateral polygon 等边多边形equilateral triangle 等边三⾓形equivalent 等价(的)error 误差escribed circle 旁切圆estimate 估计；估计量Euler's formula 尤拉公式；欧拉公式evaluate 计值even function 偶函数even number 偶数evenly distributed 均匀分布的event 事件exact 真确exact solution 准确解；精确解；真确解exact value 法确解；精确解；真确解example 例excentre 外⼼exception 例外excess 起exclusive 不包含exclusive events 互斥事件exercise 练习expand 展开expand form 展开式expansion 展式expectation 期望expectation value, expected value 期望值；预期值experiment 实验；试验experimental 试验的experimental probability 实验概率exponent 指数express…in terms of….. 以………表达expression 式；数式extension 外延；延长；扩张；扩充exterior angle 外⾓external angle bisector 外分⾓external point of division 外分点extreme point 极值点Fface ⾯factor 因⼦；因式；商factor method 因式分解法factor theorem 因⼦定理；因式定理factorial 阶乘factorization 因⼦分解；因式分解factorization of polynomial 多项式因式分解FALSE 假(的)feasible solution 可⾏解；容许解Fermat’s last theorem 费尔马最后定理Fibonacci number 斐波那契数；黄⾦分割数Fibonacci sequence 斐波那契序列fictitious mean 假定平均数figure (1)图(形)；(2)数字finite 有限finite population 有限总体finite sequence 有限序列finite series 有限级数first quartile 第⼀四分位数first term ⾸项fixed deposit 定期存款fixed point 定点flow chart 流程图foot of perpendicular 垂⾜for all X 对所有Xfor each /every X 对每⼀Xform 形式；型formal proof 形式化的证明format 格式；规格formula(formulae) 公式four rules 四则four-figure table 四位数表fourth root 四次⽅根fraction 分数；分式fraction in lowest term 最简分数fractional equation 分式⽅程fractional index 分数指数fractional inequality 分式不等式free fall ⾃由下坠frequency 频数；频率frequency distribution 频数分布；频率分布frequency distribution table 频数分布表frequency polygon 频数多边形；频率多边形frustum 平截头体function 函数function of function 复合函数；迭函数functional notation 函数记号Ggain 增益；赚；盈利gain per cent 赚率；增益率；盈利百分率game （1）对策；（2）博奕general form ⼀般式；通式general solution 通解；⼀般解general term 通项geoborad ⼏何板geometric mean ⼏何平均数；等⽐中项geometric progression ⼏何级数；等⽐级数geometric sequence 等⽐序列geometric series 等⽐级数geometry ⼏何；⼏何学given 给定；已知golden section 黄⾦分割grade 等级gradient (1)斜率；倾斜率；(2)梯度grand total 总计graph 图像；图形；图表graph paper 图表纸graphical method 图解法graphical representation 图⽰；以图样表达graphical solution 图解greatest term 最⼤项greatest value 最⼤值grid lines ⽹⽹格线group 组；grouped data 分组数据；分类数据grouping terms 并项；集项growth 增长growth factor 增长因⼦Hhalf closed interval 半闭区间half open interval 半开区间head 正⾯（钱币）height ⾼（度）hemisphere 半球体；半球heptagon 七边形Heron's formula 希罗公式hexagon 六边形higher order derivative ⾼阶导数highest common factor(H.C.F) 最⼤公因⼦；最⾼公因式；最⾼公因⼦Hindu-Arabic numeral 阿刺伯数字histogram 组织图；直⽅图；矩形图horizontal ⽔平的；⽔平horizontal line 横线；⽔平线hyperbola 双曲线hypotenuse 斜边Iidentical 全等；恒等identity 等（式）identity relation 恒等关系式if and only if/iff 当且仅当；若且仅若if…., then 若….则；如果…..则illustration 例证；说明image 像点；像imaginary circle 虚圆imaginary number 虚数imaginary root 虚根implication 蕴涵式；蕴含式imply 蕴涵；蕴含impossible event 不可能事件improper fraction 假分数inclination 倾⾓；斜⾓inclined plane 斜⾯included angle 夹⾓included side 夹边inclusive 包含的；可兼的inconsistent 不相的(的)；不⼀致(的) increase 递增；增加increasing function 递增函数interior angles on the same side of the transversal 同旁内⾓interior opposite angle 内对⾓internal bisector 内分⾓internal division 内分割internal point of division 内分点inter-quartile range 四分位数间距intersect 相交intersection (1)交集；(2)相交；(3)交点interval 区间intuition 直观invariance 不变性invariant (1)不变的；(2)不变量；不变式inverse 反的；逆的inverse circular function 反三⾓函数inverse cosine function 反余弦函数inverse function 反函数；逆函数inverse problem 逆算问题inverse proportion 反⽐例；逆⽐例inverse sine function 反正弦函数inverse tangent function 反正切函数inverse variation 反变(分)；逆变(分)irrational equation ⽆理⽅程irrational number ⽆理数irreducibility 不可约性irregular 不规则isosceles triangle 等腰三⾓形increasing sequence 递增序列increasing series 递增级数increment 增量independence 独⽴；⾃变independent event 独⽴事件independent variable ⾃变量；独⽴变量indeterminate (1)不定的；(2)不定元；未定元indeterminate coefficient 不定系数；未定系数indeterminate form 待定型；不定型index,indices 指数；指index notation 指数记数法inequality 不等式；不等inequality sign 不等号infinite ⽆限；⽆穷infinite population ⽆限总体infinite sequence ⽆限序列；⽆穷序列infinite series ⽆限级数；⽆穷级数infinitely many ⽆穷多infinitesimal ⽆限⼩；⽆穷⼩infinity ⽆限(⼤)；⽆穷(⼤)initial point 始点；起点initial side 始边initial value 初值；始值input 输⼊input box 输⼊inscribed circle 内切圆insertion 插⼊insertion of brackets 加括号instantaneous 瞬时的integer 整数integral index 整数指数integral solution 整数解integral value 整数值intercept 截距；截段intercept form 截距式intercept theorem 截线定理interchange 互换interest 利息interest rate 利率interest tax 利息税interior angle 内⾓Jjoint variation 联变(分)；连变(分)Kknown ⼰知LL.H.S. 末项law 律；定律law of indices 指数律；指数定律law of trichotomy 三分律leading coefficient ⾸项系数least common multiple, lowest common multiple (L.C.M) 最⼩公倍数；最低公倍式least value 最⼩值lemma 引理length 长(度) letter ⽂字；字母like surd 同类根式like terms 同类项limit 极限line 线；⾏line of best-fit 最佳拟合line of greatest slope 最⼤斜率的直；最⼤斜率line of intersection 交线line segment 线段linear 线性；⼀次linear equation 线性⽅程；⼀次⽅程linear equation in two unknowns ⼆元⼀次⽅程；⼆元线性⽅程linear inequality ⼀次不等式；线性不等式linear programming 线性规划literal coefficient ⽂字系数literal equation ⽂字⽅程load 负荷loaded coin 不公正钱币loaded die 不公正骰⼦locus, loci 轨迹logarithm 对数logarithmic equation 对数⽅程logarithmic function 对数函数logic 逻辑logical deduction 逻辑推论；逻辑推理logical step 逻辑步骤long division method 长除法loss 赔本；亏蚀loss per cent 赔率；亏蚀百分率lower bound 下界lower limit 下限lower quartile 下四分位数lowest common multiple(L.C.M) 最⼩公倍数Mmagnitude 量；数量；长度；⼤⼩major arc 优弧；⼤弧major axis 长轴major sector 优扇形；⼤扇形major segment 优⼸形；⼤⼸形mantissa 尾数mantissa of logarithm 对数的尾数；对数的定值部many-sided figure 多边形marked price 标价mathematical induction 数学归纳法mathematical sentence 数句mathematics 数学maximize 极⼤maximum absolute error 最⼤绝对误差maximum point 极⼤点maximum value 极⼤值mean 平均(值)；平均数；中数mean deviation 中均差；平均偏差measure of dispersion 离差的量度measurement 量度median (1)中位数；(2)中线meet 相交；相遇mensuration 计量；求积法method ⽅法method of completing square 配⽅法method of substitution 代换法；换元法metric unit ⼗进制单位mid-point 中点mid-point formula 中点公式mid-point theorem 中点定理million 百万minimize 极⼩minimum point 极⼩点minimum value 极⼩值minor (1)⼦⾏列式；(2)劣；较⼩的minor arc 劣弧；⼩弧minor axis 短轴minor sector 劣扇形；⼩扇形minor segment 劣⼸形；⼩⼸形minus 减minute 分mixed number(fraction) 带分数modal class 众数组mode 众数model 模型monomial 单项式multinomial 多项式multiple 倍数multiple root 多重根multiplicand 被乘数multiplication 乘法multiplication law (of probability) (概率)乘法定律multiplicative property 可乘性multiplier 乘数；乘式multiply 乘mutually exclusive events 互斥事件mutually independent 独⽴; 互相独⽴mutually perpendicular lines 互相垂直Nn factorial n阶乘n th root n次根；n次⽅根natural number ⾃然数negative 负negative angle 负⾓negative index 负指数negative integer 负整数negative number 负数neighborhood 邻域net 净(值)n-gon n边形nonagon 九边形non-collinear 不共线non-linear ⾮线性non-linear equation ⾮线性⽅程non-negative ⾮负的non-trivial ⾮平凡的non-zero ⾮零normal (1)垂直的；正交的；法线的(2)正态的(3)正常的；正规的normal curve 正态分记伲怀１分记伲徽媲伲徽忧?normal distribution 正态分布，常态分布normal form 法线式notation 记法；记号number 数number line数线number pair 数偶number pattern 数型number plane 数平⾯number system 数系numeral 数字；数码numeral system 记数系统numerator 分⼦numerical 数值的；数字的numerical expression 数字式numerical method 计算⽅法；数值法Ooblique 斜的oblique cone 斜圆锥oblique triangle 斜三⾓形obtuse angle 钝⾓obtuse-angled triangle 钝⾓三⾓形octagon ⼋边形octahedron ⼋⾯体odd function 奇函数odd number 奇数one-one correspondence ⼀⼀对应open interval 开区间open sentence 开句operation 运算opposite angle 对⾓opposite interior angle 内对⾓opposite side 对边optimal solution 最优解order (1)序；次序；(2)阶；级ordered pair 序偶origin 原点outcome 结果output 输出overlap 交迭；相交Pparabola 拋物线parallel 平⾏(的)parallel lines 平⾏(直线) parallelogram 平⾏四边形parameter 参数；参变量partial fraction 部分分数；分项分式polar coordinate system 极坐标系统polar coordinates 极坐标pole 极polygon 多边形polyhedron 多⾯体polynomial 多项式polynomial equation 多项式⽅程positive 正positive index 正指数positive integer 正整数positive number 正数power (1)幂；乘⽅；(2)功率；(3)检定⼒precise 精密precision 精确度prime 素prime factor 质因⼦；质因素prime number 素数；质数primitive (1)本原的；原始的；(2)原函数principal (1)主要的；(2)本⾦prism 梭柱(体)；⾓柱(体)prismoid 平截防庾短?probability 概率problem 应⽤题produce 延长product 乘积；积product rule 积法则profit 盈利profit per cent 盈利百分率profits tax 利得税progression 级数proof 证(题)；证明proper fraction 真分数property 性质property tax 物业税proportion ⽐例proportional 成⽐例protractor 量⾓器pyramid 棱锥(体)；⾓锥(体) Pythagoras’Theorem 勾股定理Pythagorean triplet 毕⽒三元数组partial sum 部分和partial variation 部分变(分)particular solution 特解Pascal’s triangle 帕斯卡斯三⾓形pattern 模型；规律pegboard 有孔版pentadecagon ⼗五边形pentagon 五边形per cent 百分率percentage 百分法；百分数percentage decrease 百分减少percentage error 百分误差percentage increase 百分增加percentile 百分位数perfect number 完全数perfecr square 完全平⽅perimeter 周长；周界period 周期periodic function 周期函数permutation 排列perpendicular 垂线；垂直(于) perpendicular bisector 垂直平分线；中垂线perpendicular line 垂直线pictogram 象形图pie chart 饼图；圆瓣图pinboard 钉板place holder 补位数字place value 位值plan (1)平⾯图；(2)计划plane 平⾯plane figure 平⾯图形plot 绘图plus 加point 点point circle 点圆point of contact 切点point of division 分点point of intersection 交点point-slope form 点斜式polar axis 极轴polar coordinate plane 极坐标平⾯polar coordinate 极坐标系统Qquadrant 象限quadratic equation ⼆次⽅程(式) quadratic formula ⼆次公式quardratic function ⼆次函数quadratic inequality ⼆次不等式quadratic polynomial 四边形quantity 数量quartile 四分位数quotient 商；商式RR.H.S 右radian 弧度radian measure 弧度法radical 根式；根号；根数radius, radii 半径random 随机random experiment 随机试验random number 随机数range 值域；区域；范围；极差；分布域rate 率；利率ratio ⽐; ⽐率rational expression 有理式；有理数式rational function 有理函数rational index 有理数指数rational number 有理数rationalization 有理化raw data 原始数据raw score 原始分（数）real axis 实轴real number 实数real root 实根reason 理由reciprocal 倒数rectangle 长⽅形；矩形rectangular block 长⽅体rectangular coordinate plane 直⾓坐标平⾯rectangular coordinates 直⾓坐rectilinear figure 直线图形recurrent 循环的recurring decimal 循环⼩数reduce 简化reducible 可约的；可化简的reference angle 参考⾓reflex angle 优⾓；反⾓region 区域regular 正；规则regular polygon 正多边形reject 舍去；否定relation 关系；关系式relative error 相对误差remainder 余数；余式；剩余remainder term 余项remainder theorem 余式定理removal of brackets 撤括号；去括号repeated trials 重复试验resolve 分解revolution 旋转；周转rhombus 菱形right angle 直⾓right circular cone 直⽴圆锥（体）right circular cylinder 直⽴圆柱（体）right prism 直⽴棱柱；直⽴⾓柱(体)right pyramid 直⽴棱锥；直⽴⾓锥(体)right-angled triangle 直⾓⼆⾓形root 根rotation 旋转round angle 周⾓rounded number 舍数rounding(off) 舍⼊；四舍五⼊row ⾏；棋⾏rule 规则；法(则)ruler 直尺Ssalaries tax 俸税sample 抽样；样本sample space 样本空间satisfy 满⾜；适合scale ⽐例尺；标度；图尺scalene triangle 不等边三⾓形；不规则三⾓形scientific notation 科学记数法solution of triangle 三⾓形解法solve 解special angle 特殊⾓；特别⾓speed 速率sphere 球形；球⾯square (1)平⽅；(2)正⽅形square bracket ⽅括号square number 正⽅形数；平⽅数square root 平⽅根；⼆次根standard deviation 标准差；标准偏离secant 割second 秒second quartile 第⼆四分位数(1)截⾯；截线；(2)截点section (1)截⾯；截线；(2)截点section formula 截点公式sector 扇式segment 段；节segment of a circle ⼸形selling price 售价semi-circle 半圆semi-vertical angle 半顶⾓sentence 句；语句sequence 序列series 级数set square 三⾓尺；三⾓板shaded portion 有阴影部分shape 形状side 边；侧sign 符号；记号signed number 有符号数significant figure 有效数字similar 相似similar figures 相似图形similar triangles 相似三⾓形similarity 相似(性)simple equation 简易⽅程simple interest 单利；单利息simplify 简化simultaneous equations 联⽴⽅程simultaneous inequalities 联⽴不等式simultaneous linear equations in two unknowns 联合⼆次线性⽅程式sine 正弦sine formula 正弦公式slant edge 斜棱slant height 斜⾼slope 斜率；斜度；倾斜；坡度slope-intercept form 斜率截距式；斜截式solid ⽴体；固体soild with uniform corss-section 有均匀横切⾯的⽴体solution 解；解法solution of equation ⽅程解Uuniform ⼀致(的)；均匀(的)uniform cross-section 均匀横切⾯uniform speed 匀速率uniformly distributed 均匀分布unique solution 唯⼀解uniqueness 唯⼀性unit 单位unit area 单位⾯积unit circle 单位圆unit volume 单位体积unknown 未知数；未知量unlike 异类项upper bound 上界upper limit 上限upper quartile 上四分位数Vvalue 值variable 变项；变量；元；变元；变数variable speed 可变速率variance ⽅差variation 变数；变分verify 证明；验证vertex, vertices 顶(点)；极点vertical 铅垂；垂直vertical angle 顶⾓vertical line 纵线；铅垂vertically opposite angles 对顶⾓volume 体积Wweight (1)重量；(2)权weighted average, weighted mean 加权平均数whole number 整数；完整数width 阔度without loss of generality 不失⼀般性X x-axis x轴x-coordinate x坐标x-intercept x轴截距Yy-axis y轴y-coordinate y坐标y-intercept y轴截距Zzero 零zero factor 零因⼦zeros of a function 函数零值统计学population 母体sample 样本census 普查sampling 抽样quantitative 量的qualitative/categorical 质的discrete 离散的continuous 连续的population parameters 母体参数sample statistics 样本统计量descriptive statistics 叙述统计学inferential/inductive statistics 推论/归纳统计学levels of measurement 衡量尺度nominal scale 名⽬尺度ordinal scale 顺序尺度interval scale 区间尺度ratio scale ⽐例尺度frequency distribution 次数分配relative frequency 相对次数range 全距class midpoint 组中点class limits 组限class boundaries 组界class width 组距cumulative frequency (以下) 累加次数decumulative frequency 以上累加次数histogram 直⽅图pie chart 饼图ogive 肩形图frequency polygon 多边形图cumulative frequency polygon 累加次数多边形图box plot 盒须图stem and leaf plot 枝叶图measures of central tendency 中央趋势量数mean 平均数median 中位数mode 众数location measures 位置量数percentile 百分位数quartile 四分位数decile ⼗分位数dispersion measures 分散量数range 全距interquartile-range IQR 四分位距mean absolute deviation 平均绝对离差variance 变异数standard deviation 标准差coefficient of variation 变异系数left-skewed 左偏negative-skewed 负偏right-skewed 右偏positive-skewed 正偏contingency table 列联表sampling distribution (of a statistic)(某个统计量的) 抽样分布point estimate 点估计值point estimator 点估计式unbiased estimator 不偏点估计式efficient estimator 有效点估计式consistent estimator ⼀致点估计式confidence level 信赖⽔准confidence interval 信赖区间null hypothesis 虚⽆假设alternative hypothesis 对⽴假设left-tailed test 左尾检定right-tailed test 右尾检定two-tailed test 双尾检定test statistic 检定统计量critical value 临界值。

2016AMC12AProblem1What is the value of?表达式的值是多少？Problem2For what value of does?的值为多少时，？Problem3The remainder can be defined for all real numbers and with bywhere denotes the greatest integer less than or equal to.What is the value of所有实数和的余数定义为:，其中，且表示取小于或等于的最大整数，的值是多少？Problem4The mean,median,and mode of the data values are all equal to.What is the value of?60，100，x，40，50，200，90这7个数字的平均值、中位数和众数都等子x，那么x的值是？Problem5Goldbach's conjecture states that every even integer greater than2can be written as the sum of two prime numbers(for example,).So far,no one has been able to prove that theconjecture is true,and no one has found a counterexample to show that the conjecture is false.What would a counterexample consist of?哥德巴赫猜想的内容是：任何大于2的偶数都可以写成2个质数之和。

（例如，2016=13+2003)。

到目前为止，还没人能够证明这个猜想是正确的，也没人能够找到一个反例证明这个猜想是错的。

This Pamphlet gives at least one solution for each problem on this year’s contest and shows that all problems can be solved without the use of a calculator. When more than one solution is provided, this is done to illustrate a significant contrast in methods, e.g., algebraic versus geometric, computational versus conceptual, elementary versus advanced. These solutions are by no means the only ones possible, nor are they superior to others the reader may devise.We hope that teachers will inform their students about these solutions, both as illustrations of the kinds of ingenuity needed to solve nonroutine problems and as examples of good mathematical exposition. However, the publication, reproduction or communication of the problems or solutions for this contest during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination at any time via copier, telephone, email, internet, or media of any type is a violation of the competition rules.Correspondence about the problems/solutions for this AMC 12 and orders for any publications should be addressed to:MAA American Mathematics CompetitionsAttn: Publications, PO Box 471, Annapolis Junction, MD 20701Phone 800.527.3690 | Fax 240.396.5647 | amcinfo@The problems and solutions for this AMC 12 were prepared byMAA’s Subcommittee on the AMC10/AMC12 Exams, under the direction of theco-chairs Jerrold W. Grossman and Silvia Fernandez.© 2016 Mathematical Association of AmericaTheMAA American Mathematics Competitionsare supported byThe Akamai FoundationAmerican Mathematical SocietyAmerican Statistical AssociationArt of Problem SolvingCasualty Actuarial SocietyCollaborator’s Circle Conference Board of the Mathematical SciencesThe D.E. Shaw GroupExpiiIDEA MATHJane Street CapitalMathWorksMu Alpha ThetaNational Council of Teachers of MathematicsSimons FoundationSociety for Industrial and Applied MathematicsStar LeagueTudor Investment CorporationTwo Sigma。

2012AMC12AProblem1A bug crawls along a number line,starting at.It crawls to,then turns around and crawls to. How many units does the bug crawl altogether?一个虫子从这个点开始沿着数轴爬行，它先爬到，然后掉头再爬到，那么这只虫子总共爬了多长？Problem2Cagney can frost a cupcake every seconds and Lacey can frost a cupcake every seconds.Working together,how many cupcakes can they frost in minutes?Cagney每20秒可以给一个蛋糕上霜，Lacey每30秒可以给一个蛋糕上霜，若他俩合作，那么5分钟内可以给多少个蛋糕上霜？Problem3A box centimeters high,centimeters wide,and centimeters long can hold grams of clay.Asecond box with twice the height,three times the width,and the same length as the first box can hold grams of clay.What is?一个2厘米高，3厘米宽，5厘米长的盒子能够装40克的泥土。

第二个盒子高度和宽度分别是前一个盒子的2倍和3倍，长度和前一个盒子一样长，可以装n克的泥土，问n是多少？In a bag of marbles,of the marbles are blue and the rest are red.If the number of red marbles is doubled and the number of blue marbles stays the same,what fraction of the marbles will be red?一包玻璃球有是蓝色的，剩下的都是红色的，如果红色玻璃球的数量翻倍而蓝色的个数保持不变，那么最后总数的几分之几将是红色？Problem5A fruit salad consists of blueberries,raspberries,grapes,and cherries.The fruit salad has a totalof pieces of fruit.There are twice as many raspberries as blueberries,three times as many grapesas cherries,and four times as many cherries as raspberries.How many cherries are there in the fruit salad?一种水果沙拉由蓝莓、覆盆子、葡萄和樱桃组成，这种水果沙拉总共有280个水果，其中覆盆子的个数是蓝莓的2倍，葡萄是樱桃的3倍，樱桃是覆盆子的4倍，问这种水果沙拉里有多少个樱桃?Problem6The sums of three whole numbers taken in pairs are,,and.What is the middle number?三个整数两两求和结果分别为12，17和19，问中间那个数是多少?Problem7Mary divides a circle into sectors.The central angles of these sectors,measured in degrees,are allintegers and they form an arithmetic sequence.What is the degree measure of the smallest possible sector angle?玛丽把一个圆分成了12个扇形，这些扇形的圆心角的度数都是整数且形成一个等差数列，那么最小的那个扇形角度最小可能是多少度？An iterative average of the numbers,,,,and is computed in the following way.Arrange thefive numbers in some order.Find the mean of the first two numbers,then find the mean of that with the third number,then the mean of that with the fourth number,and finally the mean of that with the fifth number.What is the difference between the largest and smallest possible values that can be obtained using this procedure?数字1，2，3，4，5的迭代平均值由以下方式计算，把这5个数字以某种顺序排列，先计算前2个数的平均值，再算出这个平均值和第三个数的平均值，然后再算出新得到的平均值和第四个数的平均值，最后算出这个平均值和第五个数的平均值。

Compute the sum of all the roots ofSolutionCindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?SolutionAccording to the standard convention forexponentiation,If the order in which the exponentiations are performed is changed, how many other values are possible?SolutionFind the degree measure of an angle whose complement is 25% of its supplement.SolutionEach of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.SolutionFor how many positive integers does there exist at least one positive integer n such that ?infinitely manySolutionA arc of circle A is equal in length to a arc of circle B. What is the ratio of circle A's area and circle B's area?SolutionBetsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let be the total area of the blue triangles, the total area ofthe white squares, and the area of the red square. Which of the following iscorrect?SolutionJamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?SolutionSarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?SolutionMr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?SolutionBoth roots of the quadratic equation are prime numbers. The number of possible values of isSolutionTwo different positive numbers and each differ from their reciprocals by . What is ?SolutionFor all positive integers , let . Let . Which of the following relations is true?SolutionThe mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection isSolutionTina randomly selects two distinct numbers from the set , and Sergiorandomly selects a number from the set . What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?SolutionSeveral sets of prime numbers, such as use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?SolutionLet and be circles definedby and respectively. What is the lengthof the shortest line segment that is tangent to at and to at ?SolutionThe graph of the function is shown below. How many solutions does the equation have?SolutionSuppose that and are digits, not both nine and not both zero, and the repeatingdecimal is expressed as a fraction in lowest terms. How many different denominators are possible?SolutionConsider the sequence of numbers: For , the -th term of the sequence is the units digit of the sum of the two previous terms.Let denote the sum of the first terms of this sequence. The smallest valueof for which is:SolutionTriangle is a right triangle with as its right angle, , and . Let be randomly chosen inside , and extend tomeet at . What is the probability that ?SolutionIn triangle , side and the perpendicular bisector of meet in point ,and bisects . If and , what is the area oftriangle ?SolutionFind the number of ordered pairs of real numbers suchthat .SolutionThe nonzero coefficients of a polynomial with real coefficients are all replaced by their mean to form a polynomial . Which of the following could be a graphof and over the interval ?Solution答案：Solution 1We expand to get whichis after combining like terms. Using the quadratic partof Vieta's Formulas, we find the sum of the roots is .Solution 2Combine terms to get ,hence the roots are and , thus our answer is . Problem 2We work backwards; the number that Cindy started with is . Now, the correct result is . Our answer isProblem 3The best way to solve this problem is by simple brute force.It is convenient to drop the usual way how exponentiation is denoted, and to writethe formula as , where denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:1.2.3.4.5.We can note that . Therefore options 1 and 2 are equal, and options 3 and 4 are equal. Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.Thus the only other result is , and our answer is .Problem 4Problem 5The outer circle has radius , and thus area . The little circles havearea each; since there are 7, their total area is . Thus, our answerisProblem 6Solution 1For any we can pick , we get , therefore the answeris .Solution 2Another solution, slightly similar to this first one would be using Simon's Favorite Factoring Trick.Let , thenThis means that there are infinitely many numbers that can satisfy the inequality.So the answer isProblem 7Let and be the radii of circles A and B, respectively.It is well known that in a circle with radius r, a subtended arc opposite an angleof degrees has length .Using that here, the arc of circle A has length . The arc of circle B has length . We know that they are equal, so , so wemultiply through and simplify to get . As all circles are similar to one another, the ratio of the areas is just the square of the ratios of the radii, so our answeris .Problem 8The blue that's touching the center red square makes up 8 triangles, or 4 squares. Each of the corners is 2 squares and each of the edges is 1, totaling 12 squares.There are 12 white squares, thus we have .Problem 9A 0.8 MB file can either be on its own disk, or share it with a 0.4 MB. Clearly it is not worse to pick the second possibility. Thus we will have 3 disks, each with one 0.8 MB file and one 0.4 MB file.We are left with 12 files of 0.7 MB each, and 12 files of 0.4 MB each. Their total size is MB. The total capacity of 9 disks is MB,hence we need at least 10 more disks. And we can easily verify that 10 disks are indeed enough: six of them will carry two 0.7 MB files each, and four will carry three 0.4 MB files each.Thus our answer is .Problem 10We will simulate the process in steps.In the beginning, we have:▪ ounces of coffee in cup▪ ounces of cream in cupIn the first step we pour ounces of coffee from cup to cup , getting:▪ ounces of coffee in cup▪ ounces of coffee and ounces of cream in cupIn the second step we pour ounce of coffee and ounces of cream from cup to cup , getting:▪ ounces of coffee and ounces of cream in cup▪the rest in cupHence at the end we have ounces of liquid in cup , and out ofthese ounces is cream. Thus the answer isProblem 11Solution 1Let the time he needs to get there in be t and the distance he travels be d. From thegiven equations, we know that and . Settingthe two equal, we have and we find of an hour. Substituting t back in, we find . From , we find that r, and our answer,is .Solution 2Since either time he arrives at is 3 minutes from the desired time, the answer is merely the harmonic mean of 40 and 60. The harmonic mean of a and bis . In this case, a and b are 40 and 60, so our answer is ,so .Solution 3A more general form of the argument in Solution 2, with proof:Let be the distance to work, and let be the correct average speed. Then the time needed to get to work is .We know that and . Summing these two equations, weget: .Substituting and dividing both sides by , we get ,hence .(Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighed sum in step two, and hence obtain a weighed harmonic mean in step three.)Problem 12Consider a general quadratic with the coefficient of being and the rootsbeing and . It can be factored as which isjust . Thus, the sum of the roots is the negative of the coefficient of and the product is the constant term. (In general, this leads to Vieta's Formulas).We now have that the sum of the two roots is while the product is . Since bothroots are primes, one must be , otherwise the sum would be even. That means theother root is and the product must be . Hence, our answer isProblem 13Each of the numbers and is a solution to .Hence it is either a solution to , or to . Then it must be asolution either to , or to .There are in total four such values of , namely .Out of these, two are positive: and . We can easily check that both of them indeed have the required property, and their sumis .Problem 14First, note that .Using the fact that for any base we have , we getthatProblem 15As the unique mode is , there are at least two s.As the range is and one of the numbers is , the largest one can be at most .If the largest one is , then the smallest one is , and thus the mean is strictlylarger than , which is a contradiction.If the largest one is , then the smallest one is . This means that we already knowfour of the values: , , , . Since the mean of all the numbers is , their summust be . Thus the sum of the missing four numbersis . But if is the smallest number, then the sum of the missing numbers must be at least , which is again a contradiction.If the largest number is , we can easily find the solution . Hence, our answer is .NoteThe solution for is, in fact, unique. As the median must be , this means thatboth the and the number, when ordered by size, must be s. This gives thepartial solution . For the mean to be each missing variablemust be replaced by the smallest allowed value.Problem 16Solution 1This is not too bad using casework.Tina gets a sum of 3: This happens in only one way and Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here.Tina gets a sum of 4: This once again happens in only one way . Sergio can choose a number from 5 to 10, so 6 ways here.Tina gets a sum of 5: This can happen in two ways and . Sergio canchoose a number from 6 to 10, so 2 ways here.Tina gets a sum of 6: Two ways here and . Sergio can choose a number from 7 to 10, so here.Tina gets a sum of 7: Two ways here and . Sergio can choose from 8 to 10, so ways here.Tina gets a sum of 8: Only one way possible ). Sergio chooses 9 or 10, so 2 ways here.Tina gets a sum of 9: Only one way . Sergio must choose 10, so 1 way.In all, there are ways. Tina chooses two distinctnumbers in ways while Sergio chooses a number in ways, so there are ways in all. Since , our answer is .Solution 2We want to find the average of the smallest possible chance of Sergio winning and the largest possible chance of Sergio winning. This is because the probability decreases linearly. The largest possibility of Sergio winning if Tina chooses a 1 anda 2. The chances of Sergio winning is then . The smallest possibility of Sergio winning is if Tina chooses a 4 and a 5. The chances of Sergio winning then is . Theaverage of and isProblem 17Neither of the digits , , and can be a units digit of a prime. Therefore the sum ofthe set is at least .We can indeed create a set of primes with this sum, for example the following setworks: .Thus the answer is .Problem 18(C) First examine the formula , for the circle . Itscenter, , is located at (10,0) and it has a radius of = 6. The next circle, usingthe same pattern, has its center, , at (-15,0) and has a radius of = 9. So wecan construct thisdiagram:Line PQ is tangent to both circles, so it forms a right angle with the radii (6 and 9). This, as well as the two vertical angles near O, prove triangles S QO and S PO similar by AA, with a scale factor of 6:9, or 2:3. Next, we must subdivide the line DD in a 2:3 ratio to get the length of the segments D O and D O. The total length is10 - (-15), or 25, so applying the ratio, D O = 15 and D O = 10. These are the hypotenuses of the triangles. We already know the length of D Q and DP, 9 and 6 (they're radii). So in order to find PQ, we must find the length of the longer legs of the two triangles and add them.Finally, the length of PQ is , or C.Problem 19First of all, note that the equation has two solutions: and .Given an , let . Obviously, to have , we need tohave , and we already know when that happens. In other words, thesolutions to are precisely the solutions to ( or ). Without actually computing the exact values, it is obvious from the graph that theequation has two and has four different solutions, giving us atotal of solutions.Problem 20The repeating decimal is equaltoWhen expressed in lowest terms, the denominator of this fraction will always be a divisor of the number . This gives us thepossibilities . As and are not both nine and not both zero, the denumerator can not be achieved, leaving us with possible denominators.(The other ones are achieved e.g. for equal to , , , , and , respectively.)Problem 21The sequence is infinite. As there are only pairs of digits, sooner or later a pairof consecutive digits will occur for the second time. As each next digit only depends on the previous two, from this point on the sequence will be periodic. (Additionally, as every two consecutive digits uniquely determine the previous one as well, the first pair of digits that will occur twice must be the first pair .) Hence it is a good idea to find the period. Writing down more terms of the sequence, we get:and we found the period. The length of the period is , and its sumis . Hence for each we have .We have and ,therefore . The rest can now be computed by hand, weget ,and , thus the answer isProblem 22Clearly and . Choose a and get a corresponding suchthat and . For we need , creating anisoclese right triangle with hyptonuse . Thus the point may only lie in the triangle . The probability of it doing so is the ratio of areas of to ,or equivalently, the ratio of to because the triangles have identicalaltitudes when taking and as bases. This ratio is equalto . Thus the answer isProblem 23Looking at the triangle , we see that its perpendicular bisector reaches the vertex, therefore hinting it is isoceles.Let , so that from given and the previous deducted.Then because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means and are similar,so .Then by using Heron's Formula on (with sides ), wehaveProblem 24Let be the magnitude of . Then the magnitudeof is , while the magnitude of is . We get that ,hence either or .For we get a single solution .Let's now assume that . Multiply both sides by . The left hand sidebecomes , the right hand sidebecomes . Hence the solutions for this case are precisely all the rd complex roots of unity, and there are of those.The total number of solutions is therefore .As in the other solution, split the problem into when and when .When and ,so we must have and hence . Since is restrictedto , can range from to inclusive, whichis values. Thus the total is . Problem 25(B) The sum of the coefficients of and of will be equal, so . Theonly answer choice with an intersection at is at (B). (The polynomials in thegraph are and .)。