2008 AMC 12B Problems and Solution
2005 AMC 12A Problems and Solution
2005 AMC 12A Problems and Solution Problem 1Two is of and of . What is ?Solution. Problem 2The equations and have the sam e solution. What is the value of ?SolutionProblem 3A rectangle with diagonal length is twice as long as it is wide. What is the area of the rectangle?SolutionLet be the width, so the length is . By the Pythagorean Theorem,. The area of the rectangle is.Problem 4A store norm ally sells windows at $each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How much will they save if they purchase the windows together rather than separately?SolutionFor windows, the store offers a discount of (floor function). Davereceives a discount of and Doug receives a disc ount of. These am ount to dollars in discounts. Together, they receivea discount of , so they save .Problem 5The average (m ean) of 20 numbers is 30, and the average of 30 other num bers is 20. What is the average of all 50 numbers?SolutionThe sum of the first 20 numbers is and the sum of the other 30 numbers is. Hence the overall average is .Problem 6Josh and Mike live 13 miles apart. Yesterday, Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. Whenthey m et, Josh had ridden for twice the length of tim e as Mike and at four-fifths of Mike's rate. How m any miles had Mike ridden when they m et?SolutionLet be the distances traveled by Josh and Mike, respectively, and let bethe tim e and rate of Mike. Using , we have that and. Then.Problem 7Square is inside the square so that each side of can beextended to pass through a vertex of . Square has side lengthand . What is the area of the inner square ?SolutionArguable the hardest part of this question is to visualize the diagram. Since each side of can be extended to pass through a vertex of , we realizethat must be tilted in such a fashion. Let a side of be .Notice the right triangle (in blue) with legs and hypotenuse. By the Pythagorean Theorem, we have. Thus,Problem 8Let , and be digits withWhat is ?SolutionClearly the two quantities are both integers, so we check the prime factorization of. It is easy to see now that works, so theanswer is .Problem 9There are two values of for which the equation has only one solution for . What is the sum of these values of ?SolutionSolution 1A quadratic equation always has two roots, unless it has a double root. That m eans we can write the quadratic as a square, and the coefficients 4 and 9 suggest this.Completing the square, , so. The sum of these is .Solution 2Another method would be to use the quadratic form ula, since our coefficient isgiven as 4, the coefficient is and the constant term is . Hence,Because we want only a single solution for, the determinant m ust equal 0. Therefore, we can write whichfactors to ; using Vieta's formulas we see that the sum of thesolutions for is the opposite of the coeffi cient of , or .Using the discriminant, the result m ust equal .Therefore, or , giving a sum of .Problem 10A wooden cube units on a side is painted red on all six faces and then cut into unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is ?SolutionThere are sides total on the unit cubes, and are painted red.Problem 11How many three-digit numbers satisfy the property that the m iddle digit is the average of the first and the last digits?SolutionSolution 1Let the digits be so that . In order for this to be an integer,and have to have the sam e parity. There are possibilities for , and for .depends on the value of both and and is unique for each . Thus ouranswer is .Thus, the three digits form an arithmetic sequence.▪If the num bers are all the sam e, then there are possible three-digit numbers.▪If the numbers are different, then we count the number of strictly increasingarithmetic sequences between and and multiply by 2 for the decreasing ones:This gives us . However, the question asks for three-digit numbers, so we have to ignore the four sequences starting with . Thus our answeris .Problem 12A line passes through and . How m any other points withinteger coordinates are on the line and strictly between and ?SolutionFor convenience’s sake, we can transform to the origin and to (thisdoes not change the problem). The line has the equation.The coordinates are integers if , so the values of are , with a totalof coordinates.Problem 13The regular 5-point star is drawn and in each vertex, there is a number.Each and are chosen such that all 5 of them cam e from set. Each letter is a different number (so one possible way is). Let be the sum of the num bers onand , and so forth. If and form an arithmetic sequence(not necessarily in increasing order), find the value of .Solution. The sumwill always be , so the arithm etic sequence has a sum of . Since is the middle term, it must be theaverage of the five num bers, of .Problem 14On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?SolutionThere are dots total. Casework:▪The dot is rem oved from an even face. There is a chance ofthis happening. Then there are 4 odd faces, giving us a probability of .▪The dot is rem oved from an odd face. There is a chance ofthis happening. Then there are 2 odd faces, giving us a probability of .Thus the answer is .Problem 15Let be a diam eter of a circle and be a point on with . Letand be points on the circle such that and is a second diameter.What is the ratio of the area of to the area of ?SolutionSolution 1Notice that the bases of both triangles are diameters of the circle. Hence the ratio ofthe areas is just the ratio of the heights of the triangles, or (is the foot of the perpendicular from to ).Call the radius . Then , . Using the PythagoreanTheorem in , we get .Now we have to find . Notice , so we can write theproportion:By the Pythagorean Theorem in , we have.Our answer is .Solution 2Let the centre of the circle be .Note that .is midpoint of .is midpoint of Area of Area of Area ofArea of .Problem 16Three circles of radius are drawn in the first quadrant of the -plane. The first circle is tangent to both axes, the second is tangent to the first circle and the -axis, and the third is tangent to the first circle and the -axis. A circle of radius istangent to both axes and to the second and third circles. What is ?SolutionWithout loss of generality, let . Draw the segm ent between the center of thethird circle and the large circle; this has length . We then draw the radius of the large circle that is perpendicular to the x-axis, and draw the perpendicular from this radius to the center of the third circle. This gives us a right triangle with legsand hypotenuse. The Pythagorean Theorem yields:Quite obviously , so and .Problem 17A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure 1. The cube is then cut in the sam e m anner along the dashed lines shown in Figure 2. This creates nine pieces. What is the volum e of the piece that contains vertex ?SolutionIt is a pyramid, so .Problem 18Call a number "prime-looking" if it is composite but not divisible by 2, 3, or 5. The three sm allest prime-looking numbers are 49, 77, and 91. There are 168 prime numbers less than 1000. How many prime-looking numbers are there less than 1000?SolutionThe given states that there are prime numbers less than , which is a fact we m ust som ehow utilize. Since there seem s to be no easy way to directly calculate the number of "prime-looking" numbers, we can apply com plementary counting.We can split the num bers from to into several groups:. Hence, the num ber of prime-looking numbers is(note that are primes).We can calculate using the Principle of Inclusion-Exclusion: (thevalues of and their intersections can be found quite easily)Substituting, we find that our answer is .Problem 19A faulty car odom eter proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005, how many miles has the car actually traveled?SolutionWe find the number of numbers with a and subtract from. Quick counting tells us that there are numbers with a 4 in the hundreds place, numberswith a 4 in the tens place, and numbers with a 4 in the units place (counting). Now we apply the Principle of Inclusion-Exclusion. There are numberswith a 4 in the hundreds and in the tens, and for both the other two intersections.The intersection of all three sets is just . So we get:Alternatively, consider that counting without the number is almost equivalent tocounting in base ; only, in base , the number is not counted. Since is skipped,the sym bol represents miles of travel, and we have traveled miles. Bybasic conversion, .Problem 20For each in , defineLet , and for each integer . Forhow m any values of in is ?SolutionFor the two functions and ,wecan see that as long as is between and , will be in the right domain.Therefore, we don't need to worry about the dom ain of . Also, every tim e wechange , the final equation will be in a different form and thus we will get adifferent value of x. Every tim e we have two choices for ) and altogether wehave to choose tim es. Thus, .Problem 21How many ordered triples of integers, with , , and ,satisfy both and ?SolutionCasework upon :▪: Then . Thus we get .▪: Then . Thus we get .▪: Then the exponent of becom es huge, and since there is noway we can satisfy the second condition. Hence we have two ordered triples . Problem 22A rectangular box is inscribed in a sphere of radius . The surface area of is 384,and the sum of the lengths of its 12 edges is 112. What is ?SolutionThe box P has dimensions , , and . Therefore,▪▪Now we m ake a formula for . Since the diam eter of the sphere is the spacediagonal of the box,▪We square :▪We get thatProblem 23Two distinct num bers and are chosen randomly from the set .What is the probability that is an integer?SolutionLet , so . Define , ; then , so .Here we can just make a table and count the num ber of values of per value of .The largest possible value of is 12, and we get.The total number of ways to pick two distinct numbers is ,so we get a probability of .Problem 24Let . For how m any polynomials does thereexist a polynomial of degree 3 such that ?SolutionSince has degree three, then has degree six. Thus, hasdegree six, so must have degree two, since has degree three.Hence, we conclude , , and must each be , , or . Since aquadratic is uniquely determined by three points, there can bedifferent quadratics after each of the values of , , and are chosen.However, we have included which are not quadratics. Nam ely,Clearly, we could not have included any other constant functions. For any linearfunction, we have . Again, it is pretty obvious that we have not included any other linear functions. Therefore, the desired answer is.Problem 25Let be the set of all points with coordinates , where and are eachchosen from the set . How many equilateral triangles have all their vertices in ?SolutionSolution 1 (non-rigorous)For this solution, we will just find as many solutions as possible, until it becom es intuitive that there are no m ore triangles left.Take an unit cube. We try to m ake three of its vertices form an equilateral triangle. This we find is possible by taking any vertex, and connecting the three adjacentvertices into a triangle. This triangle will have a side length of ; a quick further examination of this cube will show us that this is the only possible side length. Each of these triangles is determined by one vertex of the cube, so in one cube we have 8 equilateral triangles. We have 8 unit cubes, and then the entire cube, giving us 9cubes and equilateral triangles.It m ay be tem pting to connect the centers of the faces and to call that a cube, but a quick look at this tells us that that figure is actually an octahedron.Now, we look for any additional equilateral triangles. Since the space diagonals of the cube cannot m ake an equilateral triangle, we will assume symmetry in the cube.A bit m ore searching shows us that connecting the midpoints of three non-adjacent, non-parallel edges gives us m ore equilateral triangles.Notice that picking these three edges leaves two vertices alone, and that picking any two opposite vertices determine two equilateral triangles. Hence there areadditional equilateral triangles, for a total of .Solution 2 (rigorous)The three dim ensional distance form ula shows that the lengths of the equilateraltriangle must be , which yields the possible edge lengths ofSom e casework shows that are the only lengths that work, from which we can use the sam e counting argument as above.See Math Jam solution.。
2008 AMC 12A Problems(答案)
2008 AMC 12A ProblemsProblem 1A bakery owner turns on his doughnut machine at . At themachine has completed one third of the day's job. At what time will the doughnut machine complete the job?SolutionThe machine completes one-third of the job in hours. Thus,the entire job is completed in hours.Since the machine was started at , the job will be finished hours later, at. The answer is .Problem 2What is the reciprocal of ?Solution.Problem 3Suppose that of bananas are worth as much as oranges. How many orangesare worth as much as of bananas?SolutionIf , then.Problem 4Which of the following is equal to the productSolutionSolution 1.Solution 2Notice that everything cancels out except for in the numerator and in thedenominator.Thus, the product is , and the answer is .Problem 5Suppose that is an integer. Which of the following statements must be true about ?SolutionFor to be an integer, mustbe even, but not necessarily divisible by . Thus, the answer is .Problem 6Heather compares the price of a new computer at two different stores. Storeoffers off the sticker price followed by a rebate, and store offers offthe same sticker price with no rebate. Heather saves by buying the computer at store instead of store . What is the sticker price of the computer, in dollars?SolutionSolution 1Let the sticker price be .The price of the computer is at store , and at store .Heather saves at store , so .Solving, we find , and the thus answer is .Solution 2The in store is better than the additional off at store .Thus the off is equal to , and therefore the sticker price is .Problem 7While Steve and LeRoy are fishing 1 mile from shore, their boat springs a leak, and water comes in at a constant rate of 10 gallons per minute. The boat will sink if it takes in more than 30 gallons of water. Steve starts rowing toward the shore at a constant rate of 4 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?SolutionIt will take of an hour orminutes to get to shore.Since only gallons of water can enter the boat, only net gallons can enterthe boat per minute. Sincegallons of water enter the boat each minute, LeRoy must bailgallons per minute.Problem 8What is the volume of a cube whose surface area is twice that of a cube with volume 1?SolutionA cube with volume has a side of lengthand thus a surface area of.A cube whose surface area is has a side of length and avolume of .Problem 9Older television screens have an aspect ratio of . That is, the ratio of the widthto the height is . The aspect ratio of many movies is not , so they aresometimes shown on a television screen by "letterboxing" - darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of and is shown on an older television screen with a -inchdiagonal. What is the height, in inches, of each darkened strip?SolutionLet the width and height of the screen be and respectively, and let the widthand height of the movie be and respectively.By the Pythagorean Theorem, the diagonal is . So.Since the movie and the screen have the same width, .Thus, the height of each strip is .Problem 10Doug can paint a room in hours. Dave can paint the same room in hours. Dougand 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 ?SolutionDoug can paint of a room per hour, Dave can paint of a room in an hour, and the time they spend working together is .Since rate times time gives output,Problem 11Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the visible numbers have the greatest possiblesum. What is that sum?SolutionTo maximize the sum of the faces that are showing, we can minimize the sum ofthe numbers of the faces that are not showing.The bottom cubes each have a pair of opposite faces that are covered up. Whenthe cube is folded, ; ; and are opposite pairs. Clearlyhas the smallest sum.The top cube has 1 number that is not showing. The smallest number on a face is .So, the minimum sum of the unexposed faces is . Since the sum ofthe numbers on all the cubes is , the maximumpossible sum of visible numbers is .Problem 12A function has domain and range. (The notation denotes.) What are the domain and range, respectively, of the functiondefined by ?Solutionis defined if is defined. Thus the domain is all.Since , . Thusis the range of .Thus the answer is .Problem 13Points and lie on a circle centered at , and . A second circle isinternally 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?SolutionLet be the center of the small circle with radius , and let be the point wherethe small circle is tangent to . Also, let be the point where the small circle istangent to the big circle with radius .Then is a right triangle, and a triangle at that. Therefore,.Since , we have , or, or .Then the ratio of areas will be squared, or .Problem 14What is the area of the region defined by the inequality?Area is invariant under translation, so after translating left and up units, wehave the inequalitywhich forms a diamond centered at the origin and vertices at .Thus the diagonals are of length and . Using the formula , theanswer is .Problem 15Let . What is the units digit of ?Solution.So, . Since is a multiple of four and the unitsdigit of powers of two repeat in cycles of four, .Therefore, . So the units digit is .Problem 16The numbers , , and are the first three terms of anarithmetic sequence, and the term of the sequence is . What is ?Solution 1Let and .The first three terms of the arithmetic sequence are , , and, and the term is .Thus, .Since the first three terms in the sequence are , , and , the th termis .Thus the term is .Solution 2If , , and are in arithmetic progression, then ,, and are in geometric progression. Therefore,Therefore, , , therefore the 12th term in the sequence isProblem 17Let be a sequence determined by the rule if is evenand if is odd. For how many positive integers is it true that is less than each of , , and ?SolutionAll positive integers can be expressed as , , , or , whereis a nonnegative integer.▪ If , then .▪ If , then ,, and.▪If, then.▪ If , then,, and.Since , every positive integerwill satisfy.Since one fourth of the positive integers can be expressed as,where is a nonnegative integer, the answer is.Problem 18Triangle, 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?SolutionWithout loss of generality, let be on the axis, be on the axis, and be onthe axis, and let have respective lengths of 5, 6, and 7. Letdenote the lengths of segments respectively. Then by thePythagorean Theorem, so ; similarly,and . Since , , and are mutually perpendicular, thetetrahedron's volume is which is answer choice C.Problem 19In the expansion of what isthe coefficient of ?SolutionLet and . We areexpanding .Since there are terms in , there are ways to choose one term fromeach . The product of the selected terms is for some integer between andinclusive. For each , there is one and only one in . Since there isonly one way to choose one term from each to get a product of , there areways to choose one term from each and one term from to get aproduct of . Thus the coefficient of the term is .Problem 20Triangle has , , and . Point is on , andbisects the right angle. The inscribed circles of and have radiiand , respectively. What is ?SolutionBy the Angle Bisector Theorem, By Law of Sines on, Since the areaof a triangle satisfies , where the inradius and the semiperimeter,we have Since and share the altitude (to ),their areas are the ratio of their bases, or The semiperimetersare and . Thus,Problem 21A permutation of is heavy-tailed if. What is the number of heavy-tailed permutations?SolutionThere are total permutations.For every permutation such that , there isexactly one permutation such that . Thus it suffices to count thepermutations such that ., , and are the only combinations of numbers that can satisfy .There are combinations of numbers, possibilities of which side of the equation isand which side is , and possibilities for rearrangingand . Thus, there are permutations such that.Thus, the number of heavy-tailed permutations is .Problem 22A round table has radius . Six rectangular place mats are placed on the table. Eachplace mat has width and length as shown. They are positioned so that each mathas 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 ?SolutionSolution 1 (trigonometry)Let one of the mats be , and the center be as shown:Since there are mats, is equilateral. So, . Also,.By the Law of Cosines:.Since must be positive, .Solution 2 (without trigonometry)Draw and as in the diagram. Draw the altitude from to and call theintersectionAs proved in the first solution, . That makes atriangle, so andSince is a right triangle,Solving for givesProblem 23The solutions of the equation are the vertices of a convex polygon in the complex plane. What is the area of the polygon?SolutionLooking at the coefficients, we are immediately reminded of the binomial expansionof .Modifying this slightly, we can write the given equation as:We can apply a translation of and a rotation of(both operations preserve area) to simplify the problem:Because the roots of this equation are created by rotating radians successively about the origin, the quadrilateral is a square.We know that half the diagonal length of the square isTherefore, the area of the square isProblem 24Triangle has and . Point is the midpoint of . What isthe largest possible value of ?SolutionLet . Then , and sinceand , we haveWith calculus, taking the derivative and setting equal to zero will give the maximum value of . Otherwise, we can apply AM-GM:Thus, the maximum is at .Problem 25A sequence , , , of points in the coordinate plane satisfiesfor .Suppose that . What is ?SolutionThis sequence can also be expressed using matrix multiplication as follows:.Thus, is formed by rotating counter-clockwise about the originbyand dilating the point's position with respect to the origin by a factor of .So, starting with and performing the above operationstimes inreverse yields.Rotating clockwise by yields . A dilation by a factor ofyields the point .Therefore, .。
2004 AMC 12A Problems and Solution
2004 AMC 12A Problems and Solution Problem 1Alicia earns dollars per hour, of which is deducted to pay local taxes. Howmany cents per hour of Alicia's wages are used to pay local taxes?Solution20 dollars is the sam e as 2000 cents, and of 2000 iscents .Problem 2On the AMC 12, each correct answer is worth points, each incorrect answer isworth points, and each problem left unanswered is worth points. If Charlynleaves of the problems unanswered, how m any of the rem aining problemsmust she answer correctly in order to score at least ?SolutionShe gets points for the problems she didn't answer. She must getproblems right to score at least 100.Problem 3For how m any ordered pairs of positive integers is ?SolutionEvery integer value of leads to an integer solution for Since must be positive,Also, Since must be positive,This leaves values for y, which m ean there are solutions to theequationProblem 4Bertha has daughters and no sons. Som e of her daughters have daughters, andthe rest have none. Bertha has a total of daughters and granddaughters, and nogreat-granddaughters. How many of Bertha's daughters and grand-daughters have no children?SolutionSince Bertha has 6 daughters, Bertha has granddaughters, of whichnone have daughters. Of Bertha's daughters, have daughters, sodo not have daughters.Therefore, of Bertha's daughters and granddaughters, do not havedaughters .ORDraw a tree diagram and see that the answer can be found in the sum of 6 + 6 granddaughters, 5 + 5 daughters, and 4 more daughters.Problem 5The graph of the line is shown. Which of the following is true?SolutionIt looks like it has a slope of and is shifted up.Problem 6Let , , , ,and . Which of the following is the largest?SolutionAfter comparison, is the largest.Problem 7A gam e is played with tokens according to the following rules. In each round, the player with the most tokens gives one token to each of the other players and also places one token into a discard pile. The gam e ends when some player r uns out oftokens. Players , and start with , and tokens, respectively. Howmany rounds will there be in the game?SolutionLook at a set of 3 rounds, where the players have , , and tokens. Each of the players will gain two tokens from the others and give away 3 tokens, so overall, each player will lose 1 token.Therefore, after 12 sets of 3 rounds, or 36 rounds, the players will have 3, 2, and 1 tokens, repectively. After 1 m ore round, player will give away his last 3 tokensand the gam e will stop .Problem 8In the overlapping triangles and sharing common side ,and are right angles, , , , and andintersect at . What is the difference between the areas of and?SolutionSolution 1Since and , . By alternate interior angles and AA~,we find that , with side length ratio . Their heights also havethe sam e ratio, and since the two heights add up to , we have thatand . Subtracting the areas,.Solution 2Let represent the area of figure . Note thatand ..Problem 9A com pany sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars would increase sales. If the diam eter of the jars is increased bywithout altering the volume, by what percent m ust the height be decreased?SolutionWhen the diam eter is increased by , it is increased by , so the area of the baseis increased by .To keep the volum e the sam e, the height m ust be of the original height,which is a reduction .Problem 10The sum of consecutive integers is . What is their m edian?SolutionThe m edian of a sequence is the m iddle number of the sequence when t he sequence is arranged in order. Since the integers are consecutive, the m edian is also themean, so the m edian is .Problem 11The average value of all the pennies, nickels, dimes, and quarters in Paula's purse iscents. If she had one m ore quarter, the average value would be cents. Howmany dimes does she have in her purse?SolutionSolution 1Let the total value (in cents) of the coins Paula has originally be , and the numberof coins she has be . Then and . Substitutingyields . It is easy to see now that Paulahas 3 quarters, 1 nickel, so she has dimes.Solution 2If the new coin was worth 20 cents, adding it would not change the m ean at all. The additional 5 cents raise the m ean by 1, thus the new number of coins m ust be 5.Therefore initially there were 4 coins worth a total of cents. As in the previous solution, we conclude that the only way to get 80 cents using 4 coins is 25+25+25+5.Problem 12Let and . Points and are on the line , andand intersect at . What is the length of ?SolutionThe equation of can be found using points to be. Similarily, has the equation. These two equations intersectthe line at and . Using the distance formula or righttriangles, the answer is .Problem 13Let be the set of points in the coordinate plane, where each of and maybe , , or . How many distinct lines pass through at least two m embers of ?SolutionSolution 1Let's count them by cases:▪Case 1: The line is horizontal or vertical, clearly .▪Case 2: The line has slope , with through and additional ones one unit above or below those. These total .▪Case 3: The only remaining lines pass through two points, a vertex and a non-vertex point on the opposite side. Thus we have each vertex pairing up with twopoints on the two opposites sides, giving lines.These add up to .Solution 2There are ways to pick two points, but we've clearly overcounted all of the lines which pass through three points. In fact, each line which passes throughthree points will have been counted tim es, so we have to subtract foreach of these lines. Quick counting yields horizontal, vertical, and diagonallines, so the answer is distinct lines.Problem 14A sequence of three real numbers forms an arithmetic progression with a first term of . If is added to the second term and is added to the third term, the threeresulting numbers form a geometric progression. What is the sm allest possible value for the third term in the geom etric progression?SolutionLet be the common difference. Thenare the term s of the geom etric progression. Since the m iddle term is the geom etric m ean of the other two term s,. The sm allestpossible value occurs when , and the third term is .Problem 15Brenda and Sally run in opposite directions on a circular track, starting atdiametrically opposite points. They first m eet after Brenda has run meters. Theynext m eet after Sally has run meters past their first m eeting point. Each girl runs at a constant speed. What is the length of the track in m eters?SolutionSolution 1Call the length of the race track . When they m eet at the first m eeting point,Brenda has run m eters, while Sally has run meters. By the secondmeeting point, Sally has run meters, while Brenda has run meters.Since they run at a constant speed, we can set up a proportion:. Cross-m ultiplying, we get that .Solution 2The total distance the girls run between the start and the first m eeting is one half of the track length.The total distance they run between the two m eetings is the track lengt h.As the girls run at constant speeds, the interval between the m eetings is twice as long as the interval between the start and the first m eeting.Thus between the m eetings Brenda will run meters. Therefore thelength of the track is metersProblem 16The set of all real numbers for whichis defined is . What is the value of ?SolutionWe know that the dom ain of , where is a constant, is . So. By the definition of logarithms, we then have. Then and.Problem 17Let be a function with the following properties:, and, for any positive integer .What is the value of ?Solution. Problem 18Square has side length . A sem icircle with diameter is constructedinside the square, and the tangent to the semicircle from intersects side at .What is the length of ?SolutionSolution 1Let the point of tangency be . By the TwoTangent Theorem and . Thus . The Pythagorean Theorem on yieldsHence .Solution 2Clearly, . Thus, the sides of right triangle are in arithmeticprogression. Thus it is similar to the triangle and since ,.Problem 19Circles and are externally tangent to each other, and internally tangent tocircle . Circles and are congruent. Circle has radius and passes throughthe center of . What is the radius of circle ?SolutionSolution 1Note that since is the center of the larger circle of radius . Usingthe Pythagorean Theorem on ,Now using the Pythagorean Theorem on ,Substituting ,Solution 2We can apply Descartes' Circle Formula.The four circles have curvatures , and .We haveSimplifying, we getProblem 20Select numbers and between and independently and at random, and let betheir sum. Let and be the results when and , respectively, are roundedto the nearest integer. What is the probability that ?SolutionSolution 1Casework:1.. The probability that and is . Notice thatthe sum ranges from to with a symmetric distribution across, and we want . Thus the chance is .2.. The probability that and is , but now, which m akes autom atically. Hence the chanc e is.3.. This is the sam e as the previous case.4.. We recognize that this is equivalent to the first case.Our answer is .Solution 2Use areas to deal with this continuous probability problem. Set up a unit square with values of on x-axis and on y-axis.If then this will work because . Similarly ifthen this will work because in order for this to happen, and are eachgreater than making , and . Each of these triangles in theunit square has area of 1/8.The only case left is when . Then each of and must be 1 and 0, in anyorder. These cut off squares of area 1/2 from the upper left and lower right corners of the unit square.Then the area producing the desired result is 3/4. Since the area of the unit squareis 1, the probability is .Problem 21If , what is the value of ?SolutionThis is an infinite geom etric series, which sum s to. Using the formula.Problem 22Three m utually tangent spheres of radius rest on a horizontal plane. A sphere of radius rests on them. What is the distance from the plane to the top of the largersphere?SolutionThe height from the center of the bottom sphere to the plane is , and from the center of the top sphere to the tip is . We now need the vertical height of thecenters. If we connect the centers, we get a triangular pyramid with an equilateral triangle base. The distance from the vertex of the equilateral triangle to its centroidcan be found by s to be .By the Pythagorean Theorem, we have . Addingthe heights up, we get .Problem 23A polynomialhas real coefficients with and distinct com plex zeroes, with and real, , andWhich of the following quantities can be a nonzero number?SolutionWe have to evaluate the answer choices and use process of elimination:▪: We are given that , so . If one of the roots is zero,then .▪: By Vieta's formulas, we know that is the sum of all of the rootsof . Since that is real, , and , so .▪: All of the coefficients are real. For sake of contradiction suppose none ofare zero. Then for each complex root , its com plex conjugateis also a root. So the roots should pair up, but we have an odd num ber of im aginary roots! This gives us the contradiction, and therefore the product is equal to zero.▪: We are given that . Since the coefficients are real, it follows that if a root is complex, its conjugate is also a root; and the sum of the imaginary parts of com plex conjugates is zero. Hence the RHS is zero.There is, however, no reason to believe that should be zero (in fact, thatquantity is , and there is no evidence that is a root of ).Problem 24A plane contains points and with . Let be the union of all disks ofradius in the plane that cover . What is the area of ?SolutionAs the red circles m ove about segment , they cover the area we are looking for.On the left side, the circle must m ove around pivoted on . On the right side, thecircle m ust m ove pivoted on However, at the top and bottom, the circle must lieon both A and B, giving us our upper and lower bounds.This egg-like shape is .The area of the region can be found by dividing it into several sectors, nam elyProblem 25For each integer , let denote the base-number . The productcan be expressed as , where and are positive integers and is as sm all as possible. What is the value of ?SolutionThis is an infinite geom etric series with common ratio and initial term, so.Alternatively, we could have used the algebraic m anipulation for repeatingdecim als,Telescoping,Som e factors cancel, (after all, isn't one of the answer choices)Since the only factor in the num erator that goes into is , is minimized.Therefore the answer is .。
2009 AMC 12B Problems and Solution
2009 AMC 12B Problems and Solution Problem 1Each morning of her five-day workweek, Jane bought either a -cent muffin or a-cent bagel. Her total cost for the week was a whole number of dollars. How manybagels did she buy?SolutionThe only combination of five items with total cost a whole number of dollars is 3muffins and bagels. The answer is .Problem 2Paula the painter had just enough paint for identically sized rooms. Unfortunately,on the way to work, three cans of paint fell off her truck, so she had only enough paint for rooms. How many cans of paint did she use for the rooms?SolutionLosing three cans of paint corresponds to being able to paint five fewer rooms. So. The answer is .Problem 3Twenty percent off is one-third more than what number?Twenty percent less than 60 is . One-third more than a number n is .Therefore and the number is . The answer is .Problem 4A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. Theparallel sides of the trapezoid have lengths and meters. What fraction of theyard is occupied by the flower beds?SolutionEach triangle has leg length meters and area square meters. Thus the flower beds have a total area of 25 square meters. The entire yard has length 25 m and width 5 m, so its area is 125 square meters. The fraction of theyard occupied by the flower beds is . The answer is .Problem 5Kiana has two older twin brothers. The product of their ages is . What is the sumof their three ages?The age of each person is a factor of . So the twins could beyears of age and, consequently Kiana could be 128, 32, 8 or 2 years old, respectively. Because Kiana is younger than her brothers, shemust be 2 years old. So the sum of their ages is . The answer is.Problem 6By inserting parentheses, it is possible to give the expression several values. How many different values can be obtained?SolutionThe three operations can be performed on any of orders. However, if theaddition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to checkthat the values of the four expressions are in fact alldistinct. So the answer is , which is choice .Problem 7In a certain year the price of gasoline rose by during January, fell byduring February, rose by during March, and fell by during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is ?Let be the price at the beginning of January. The price at the end of March wasBecause the price at the of April was , the pricedecreased by during April, and the percent decrease wasSo to the nearest integer is . The answer is.Problem 8When 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 ofand , what is the total weight in kilograms when the bucket is full of water?SolutionSolution 1Let be the weight of the bucket and let be the weight of the water in a full bucket.Then we are given that and . Hence , so. Thus . Finally .The answer is .Solution 2Imagine that we take three buckets of the first type, to get rid of the fraction. We will have three buckets and two buckets' worth of water.On the other hand, if we take two buckets of the second type, we will have two buckets and enoung water to fill one bucket.The difference between these is exactly one bucket full of water, hence the answer is .Solution 3We are looking for an expression of the form .We must have , as the desired result contains exactly one bucket. Also,we must have , as the desired result contains exactly one bucket of water.At this moment, it is easiest to check that only the options (A), (B), and (E) satisfy, and out of these only (E) satisfies the second equation.Alternately, we can directly solve the system, getting and .Problem 9Triangle has vertices , , and , where is on the line. What is the area of ?SolutionSolution 1Because the line is parallel to , the area of is independent ofthe location of on that line. Therefore it may be assumed that is . In thatcase the triangle has base and altitude , so its area is .Solution 2The base of the triangle is . Its altitude is the distancebetween the point and the parallel line , which is .Therefore its area is . The answer is .Problem 10A particular -hour digital clock displays the hour and minute of a day.Unfortunately, whenever it is supposed to display a , it mistakenly displays a .For example, when it is 1:16 PM the clock incorrectly shows 9:96 PM. What fraction of the day will the clock show the correct time?SolutionSolution 1The clock will display the incorrect time for the entire hours of and . Sothe correct hour is displayed of the time. The minutes will not display correctly whenever either the tens digit or the ones digit is a , so the minutes that will notdisplay correctly are and and . This amounts to fifteen of the sixty possible minutes for any given hour. Hence the fraction of the daythat the clock shows the correct time is . The answeris .Solution 2The required fraction is the number of correct times divided by the total times. There are 60 minutes in an hour and 12 hours on a clock, so there are 720 total times.We count the correct times directly; let a correct time be , where is anumber from 1 to 12 and and are digits, where . There are 8 values of that will display the correct time: 2, 3, 4, 5, 6, 7, 8, and 9. There are five values of that will display the correct time: 0, 2, 3, 4, and 5. There are nine values of that will display the correct time: 0, 2, 3, 4, 5, 6, 7, 8, and 9. Therefore there arecorrect times.Therefore the required fraction is .Problem 11On Monday, Millie puts a quart of seeds, of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds withoutremoving any seeds that are left. Each day the birds eat only of the millet in the feeder, but they eat all of the other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?SolutionOn Monday, day 1, the birds find quart of millet in the feeder. On Tuesday theyfind quarts of millet. On Wednesday, day 3, they findquarts of millet. The number of quarts of millet they find on day isThe birds always find quart of other seeds, so more than half the seeds are milletif , that is, when . Because and, this will first occur on day which is . The answer is.Problem 12The fifth and eighth terms of a geometric sequence of real numbers are andrespectively. What is the first term?SolutionLet the th term of the series be . Because it followsthat and the first term is . The answer is .Problem 13Triangle has and , and the altitude to has length .What is the sum of the two possible values of ?SolutionLet be the foot of the altitude to . Then and. Thus or. The sum of the two possible values is. The answer is .Problem 14Five unit squares are arranged in the coordinate plane as shown, with the lower leftcorner at the origin. The slanted line, extending from to , divides the entire region into two regions of equal area. What is ?SolutionSolution 1For the shaded area is at most , which is too little. Hence , andtherefore the point is indeed inside the shaded part, as shown in the picture. Then the area of the shaded part is one less than the area of the triangle withvertices , , and . Its area is obviously , therefore the areaof the shaded part is .The entire figure has area , hence we want the shaded part to have area .Solving for , we get . The answer is .Solution 2The total area is 5, so the area of the shaded area is . If we add a unit square inthe lower right corner, the area is . Therefore , or .Therefore .Problem 15Assume . Below are five equations for . Which equation has the largestsolution ?Solution(B) Intuitively, will be largest for that option for which the value in the parentheses is smallest.Formally, first note that each of the values in parentheses is larger than . Now,each of the options is of the form. This can be rewritten as. As, we have . Thus is the largest forthe option for which is smallest. And as is an increasing function,this is the option for whichis smallest.We now get the following easier problem: Given that , find the smallestvalue in the set.Clearly is smaller than the first and the third option.We have , dividing both sides by we get .And finally, , therefore , and as both sides are positive, wecan take the square root and get.Thus the answer is .Problem 16Trapezoid has ,,, and.The ratiois. What is?Solution Solution 1Extendandto meet at. ThenThus is isosceles with . Because , it follows that thetriangles and are similar. ThereforesoSolution 2Let be the intersection of and the line through parallel to Byconstuction and ; it follows that is the bisector of theangle . So by the Angle Bisector Theorem we getThe answer is . Problem 17Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of its opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?SolutionSolution 1There are two possible stripe orientations for each of the six faces of the cube, sothere are possible stripe combinations. There are three pairs of parallel faces so, if there is an encircling stripe, then the pair of faces that do not contribute uniquely determine the stripe orientation for the remaining faces. In addition, the stripe on each face that does not contribute may be oriented in either of two different ways. So a total of stripe combinations on the cube result ina continuous stripe around the cube. The required probability is .Here's another way similar to this:So there are choices for the stripes as mentioned above. Now, let's just consider the "view point" of one of the faces. We can choose any of the 2 orientation for the stripe (it can go from up to down, or from right to left). Once that orientation is chosen, each of the other faces that contribute to that loop only have 1 choice, which is to go in the direction of the loop. That gives us a total count of 2 possibilities for any one of the faces. Since there are six faces, and this argument is valid for all of them, we conclude that there are 2(6) = 12 total ways to have the stripe. Therefore, the probability is 12/64 = 3/16.Solution 2Without loss of generality, orient the cube so that the stripe on the top face goes from front to back. There are two mutually exclusive ways for there to be an encircling stripe: either the front, bottom and back faces are painted to complete an encircling stripe with the top face's stripe or the front, right, back and left faces arepainted to form an encircling stripe. The probability of the first case is ,and the probability of the second case is . The cases are disjoint, so theprobabilities sum .Solution 3There are three possible orientations of an encircling stripe. For any one of these to appear, the stripes on the four faces through which the continuous stripe is to pass must be properly aligned. The probability of each such stripe alignment is. Since there are three such possibilities and they are disjoint, the totalprobability is . The answer is .Solution 4Consider a vertex on the cube and the three faces that are adjacent to that vertex. If no two stripes on those three faces are aligned, then there is no stripe encirclingthe cube. The probability that the stripes aren't aligned is , since for each alignment of one stripe, there is one and only one way to align the other two stripesout of four total possibilities. therefore there is a probability of that two stripes are aligned.Now consider the opposing vertex and the three sides adjacent to it. Given the two connected stripes next to our first vertex, we have two more that must be connectedto make a continuous stripe. There is a probability of that they arealigned, so there is a probability of that there is a continuous stripe.Problem 18Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every seconds, and Robert runs clockwise and completes a lapevery seconds. Both start from the start line at the same time. At some randomtime between minutes and minutes after they begin to run, a photographerstanding inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?SolutionAfter 10 minutes (600 seconds), Rachel will have completed 6 laps and be 30 seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in 22.5 seconds, she will be in the picture between 18.75 seconds and 41.25 seconds of the tenth minute. After 10 minutes Robert will have completed 7 laps and will be 40 seconds past the starting line. Because Robert runs one-fourth of a lap in 20 seconds, he will be in the picture between 30 and 50 seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between 30 and 41.25 seconds of the tenth minute. So the probability that both runners are in thepicture is . The answer is .Problem 19For each positive integer , let . What is the sum of allvalues of that are prime numbers?SolutionSolution 1To find the answer it was enough to play around with . One can easily find thatis a prime, then becomes negative for between and , and thenis again a prime number. And as is already the largestoption, the answer must be .Solution 2We will now show a complete solution, with a proof that no other values are prime.Consider the function , then obviously .The roots of are:We can then write , and thus.We would now like to factor the right hand side further, using the formula. To do this, we need to express both constants assquares of some other constants. Luckily, we have a pretty good idea how they look like.We are looking for rational and such that .Expanding the left hand side and comparing coefficients, we get and. We can easily guess (or compute) the solution , .Hence , and we can easily verify that also.We now know the complete factorization of :As the final step, we can now combine the factors in a different way, in order to get rid of the square roots.We have, and.Hence we obtain the factorization .For both terms are positive and larger than one, hence is not prime. Forthe second factor is positive and the first one is negative, hence isnot a prime. The remaining cases are and . In both cases, isindeed a prime, and their sum is . Solution 3Instead of doing the hard work, we can try to guess the factorization. One good approach:We can make the observation that looks similar to with theexception of the term. In fact, we have . But thenwe notice that it differs from the desired expression by a square:.Now we can use the formula to obtain the same factorization as in the previous solution, without all the work.Solution 4After arriving at the factorization , a more mathematical approach would be to realize that the second factor is alwayspositive when is a positive integer. Therefore, in order for to be prime, the first factor has to be .We can set it equal to 1 and solve for :Substituting these values into the second factor and adding would give the answer. Problem 20A convex polyhedron has vertices , and edges. The polyhedronis cut by planes in such a way that plane cuts only those edgesthat meet at vertex . In addition, no two planes intersect inside or on . The cutsproduce pyramids and a new polyhedron . How many edges does have?SolutionSolution 1Each edge of is cut by two planes, so has vertices. Three edges of meetat each vertex, so has edges.Solution 2At each vertex, as many new edges are created by this process as there are original edges meeting at that vertex. Thus the total number of new edges is the totalnumber of endpoints of the original edges, which is . A middle portion of eachoriginal edge is also present in , so has edges.Solution 3Euler's Polyhedron Formula applied to gives , where F is thenumber of faces of . Each edge of is cut by two planes, so has vertices.Each cut by a plane creates an additional face on , so Euler's PolyhedronFormula applied to gives , where is the number ofedges of . Subtracting the first equation from the second gives ,whence . The answer is .Problem 21Ten women sit in seats in a line. All of the get up and then reseat themselvesusing all seats, each sitting in the seat she was in before or a seat next to the oneshe occupied before. In how many ways can the women be reseated?SolutionLet be the answer for women, we want to find .Clearly . Now let . Let the row of seats go from left to right.Label both the seats and the women to , going from left to right. Consider therightmost seat. Which women can sit there after the swap? It can either be woman or woman , as for any other woman the seat is too far.If woman stays in her seat, there are exactly valid arrangements of theother women. If woman sits on seat , we only have one option forwoman : she must take seat , all the other seats are too far for her. We areleft with women to sitting on seats to , and there are clearlyvalid arrangements of these.We get the recurrence . (Hence is precisely the -thFibonacci number.) Using this recurrence we can easily compute that .Problem 22Parallelogramhas area . Vertex is atand all othervertices are in the first quadrant. Vertices andare lattice points on the linesandfor some integer, respectively. How many suchparallelograms are there?Solution Solution 1The area of any parallelogram can be computed as the size of the vectorproduct ofand.In our setting where, , and this is simply.In other words, we need to count the triples of integerswhere ,and.These can be counted as follows: We have identical red balls (representing powersof ), blue balls (representing powers of ), and three labeled urns (representingthe factors , , and ). The red balls can be distributed in ways, andfor each of these ways, the blue balls can then also be distributed in ways. (SeeDistinguishability for a more detailed explanation.)Thus there are exactlyways how to breakinto three positiveinteger factors, and for each of them we get a single parallelogram. Hence thenumber of valid parallelograms is.Solution 2Without the vector product the area of can be computed for example asfollows: If and , then clearly . Let, and be the orthogonal projections of , ,and onto the axis. Let denote the area of the polygon . We can thencompute:Problem 23A region in the complex plane is defined byA complex number is chosenuniformly at random from . What is the probability that is also in ?SolutionWe can directly compute.This number is in if and only if and at the same time. This simplifies to and .Let , and let denote the area ofthe region . Then obviously the probability we seek is . All we need to do is to compute the area of the intersection of and . It is easiest to dothis graphically:Coordinate axes are dashed, is shown in red, in green and their intersection isyellow. The intersections of the boundary of and are obviously atand at .Hence each of the four red triangles is an isosceles right triangle with legs long ,and hence the area of a single red triangle is . Then the area of allfour is , and therefore the area of is . Then the probability we seek is.(Alternately, when we got to the point that we know that a single red triangle is , we can directly note that the picture is symmetric, hence we can just consider thefirst quadrant and there the probability is . This saves us the work of first multiplying and then dividing by .)Problem 24For how many values of in is ? Note: Thefunctions and denote inverse trigonometricfunctions.SolutionFirst of all, we have to agree on the range of and . This should have beena part of the problem statement -- but as it is missing, we will assume the mostcommon definition: and .Hence we get that , thus our equation simplifies to.Consider the function . We are looking for roots of on.By analyzing properties of and (or by computing the derivative of ) onecan discover the following properties of :▪.▪is increasing and then decreasing on .▪is decreasing and then increasing on .▪is increasing and then decreasing on .For we have . Hence hasexactly one root on .For we have . Hence isnegative on the entire interval .Now note that . Hence for we have , andwe can easily check that as well.Thus the only unknown part of is the interval . On this interval, is negative in both endpoints, and we know that it is first increasing and then decreasing. Hence there can be zero, one, or two roots on this interval.To prove that there are two roots, it is enough to find any from this interval suchthat .A good guess is its midpoint, , where the function has its local maximum. We can evaluate:.Summary: The function has roots on : the first one is , the second oneis in , and the last two are in .Problem 25The set is defined by the points with integer coordinates, ,. How many squares of side at least have their four vertices in ?SolutionWe need to find a reasonably easy way to count the squares.First, obviously the maximum distance between two points in the same quadrant is, hence each square has exactly one vertex in each quadrant.Given any square, we can circumscribe another axes-parallel square around it. In the picture below, the original square is red and the circumscribed one is blue.Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it?Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations.The size of the blue square can range from to , and for theintermediate sizes there is more than one valid placement. We will now examine the cases one after another. Also, we can use symmetry to reduce the number of cases.size upper_right solutions symmetries total6 (3,3) 1 1 17 (3,3) 1 4 48 (3,3) 1 4 48 (3,4) 1 4 48 (4,4) 3 1 39 (3,3) 1 4 49 (3,4) 1 8 89 (4,4) 3 4 1210 (3,3) 1 4 410 (3,4) 1 8 810 (3,5) 1 4 410 (4,4) 3 4 1210 (4,5) 3 4 1210 (5,5) 5 1 511 (4,4) 3 4 1211 (4,5) 3 8 2411 (5,5) 5 4 2012 (5,5) 5 4 2012 (5,6) 5 4 2012 (6,6) 7 1 713 (6,6) 7 4 2814 (7,7) 9 1 9Summing the last column, we get that the answer is .。
2012 AMC 12B Problems and Solution
The original expressionnow becomes, whereis a positive constant, hence the answer is(A).
Problem 7
Small lights are hung on a string 6 inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of 2 red lights followed by 3 green lights. How many feet separate the 3rd red light and the 21st red light?
Solution
Solution 1
Multiplying 18 and 2 by 4 we get 72 and 8 students and rabbits respectively. Subtracting 8 from 72 we get
Solution 2
In each class, there aremore students than rabbits. So for all classrooms, the difference between students and rabbits is
Note:1 foot is equal to 12 inches.
Solution
We know the repeating section is made of 2 red lights and 3 green lights. The 3rd red light would appear in the 2nd section of this pattern, and the 21st red light would appear in the 11th section. There would then be a total of 44 lights in between the 3rd and 21st red light, translating to 45 6-inch gaps. Since it wants the answer in feet, so the answer issince it wants the answer in feet.
2007 AMC 12B Problems and Solution
2007 AMC 12B Problems and Solution Problem 1Isabella's house has 3 bedroom s. Each bedroom is 12 feet long, 10 feet wide, and 8 feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy 60 square feet in each bedroom. How many square feet of walls m ust be painted?SolutionThere are four walls in each bedroom, since she can't paint floors or ceilings. So we calculate the num ber of square feet of wall there is in one bedroom:We havethree bedrooms, so she must paint square feet of wall.Problem 2A college student drove his com pact car 120 miles home for the weekend and averaged 30 miles per gallon. On the return trip the student drove his parents' SUV and averaged only 20 miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?SolutionThe trip was miles long and took gallons. Therefore,the average m ileage wasThe point is the center of the circle circum scribed about triangle , withand , as shown. What is the degree m easure of ?SolutionProblem 4At Frank's Fruit Market, 3 bananas cost as m uch as 2 apples, and 6 apples cost as much as 4 oranges. How many oranges cost as m uch as 18 bananas?Solution18 bananas cost the sam e as 12 apples, and 12 apples cost the sam e as 8 oranges,so 18 bananas cost the sam e as oranges.The 2007 AMC 12 contests will be scored by awarding 6 points for each correct response, 0 points for each incorrect response, and 1.5 points for each problem left unanswered. After looking over the 25 problems, Sarah has decided to attem pt the first 22 and leave the last 3 unanswered. How many of the first 22 problems must she solve correctly in order to score at least 100 points?SolutionShe must get at least points, and that can only be possible byanswering at least questions correctly.Problem 6Triangle has side lengths , , and . Two bugs startsimultaneously from and crawl along the sides of the triangle in oppositedirections at the sam e speed. They m eet at point . What is ?SolutionOne bug goes to . The path that he takes is units long. The lengthof isProblem 7All sides of the convex pentagon are of equal length, and. What is the degree m easure of ?SolutionSince and are right angles, and equals , is a square, and is5. Since and are also 5, triangle is equilateral. Angle is thereforeProblem 8Tom's age is years, which is also the sum of the ages of his three children. His ageyears ago was twice the sum of their ages then. What is ?SolutionTom's age years ago was . The ages of his three children years ago wassince there are three people. If his age years ago was twice the sum ofthe children's ages then,Problem 9A function has the property that for all real numbers .What is ?SolutionProblem 10Som e boys and girls are having a car wash to raise m oney for a class trip to China.Initially % of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then % of the group are girls. How many girls were initially in thegroup?SolutionIf we let be the num ber of people initially in the group, the is the number of girls. If two girls leave and two boys arrive, the number of people in the group is stillbut the num ber of girls is . Since only of the group are girls,The num ber of girls isProblem 11The angles of quadrilateral satisfy . What isthe degree m easure of , rounded to the nearest whole number?SolutionThe sum of the interior angles of any quadrilateral isProblem 12A teacher gave a test to a class in which of the students are juniors and areseniors. The average score on the test was . The juniors all received the sam e score, and the average score of the seniors was . What score did each of thejuniors receive on the test?SolutionWe can assum e there are people in the class. Then there will be junior andseniors. The sum of everyone's scores is Since the average score ofthe seniors was the sum of all the senior's scores is The only score that has not been added to that is the junior's score, which isProblem 13A traffic light runs repeatedly through the following cycle: green for seconds,then yellow for seconds, and then red for seconds. Leah picks a randomthree-second tim e interval to watch the light. What is the probability that the color changes while she is watching?SolutionThe traffic light runs through a second cycle.Letting reference the m oment it turns green, the light changes at threedifferent tim es: , , andThis m eans that the light will change if the beginning of Leah's interval lies in, orThis gives a total of seconds out ofProblem 14Point is inside equilateral . Points , , and are the feet of theperpendiculars from to , , and , respectively. Given that ,, and , what is ?SolutionDrawing , , and , is split into three sm aller triangles. Thealtitudes of these triangles are given in the problem as , , and .Summing the areas of each of these triangles and equating it to the area of the entire triangle, we get:where is the length of a sideProblem 15The geom etric series has a sum of , and the terms involving odd powers of have a sum of . What is ?SolutionSolution 1The sum of an infinite geom etric series is given by where is the first term and is the common ratio.In this series,The series with odd powers of is given asIt's sum can be given byDoing a little algebraSolution 2The given series can be decomposed as follows:Clearly . We obtain that, hence .Then from we get , and thus . Problem 16Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?SolutionA tetrahedron has 4 sides. The ratio of the number of faces with each color must be one of the following:, , , orThe first ratio yields appearances, one of each color.The second ratio yields appearances, three choices for the first color, andtwo choices for the second.The third ratio yields appearances since the two colors areinterchangeable.The fourth ratio yields appearances. There are three choices for the first color, andsince the second two colors are interchangeable, there is only one distinguishable pair that fits them.The total is appearancesSolution 2Every colouring can be represented in the form, where is the num ber of white faces, is the number of red faces, and is the number of blue faces. Everydistinguishable colouring pattern can be represented like this in exactly one way, and every ordered whole number triple with a total sum of 4 represents exactly one colouring pattern (if two tetrahedra have rearranged colours on their faces, it is always possible to rotate one so that it m atches the other).Therefore, the number of colourings is equal to the num ber of ways 3 distinguishable nonnegative integers can add to 4. If you have 6 cockroaches in a row, this number is equal to the num ber of ways to pick two of the cockroaches to eat for dinner (because the rem aining cockroaches in between are separated in to three sections with a non-negative number of cockroaches each), which isProblem 17If is a nonzero integer and is a positive number such that , what isthe m edian of the set ?SolutionNote that if is positive, then, the equation will have no solutions for . Thisbecom es more obvious by noting that at , . The LHS quadraticfunction will increase faster than the RHS logarithmic function, so they will never intersect.This puts as the sm allest in the set since it m ust be negative.Checking the new equation:Near , but at ,This implies that the solution occurs som ewhere in between:This also implies thatThis m akes our set (ordered)The m edian isProblem 18Let , , and be digits with . The three-digit integer lies one third of theway from the square of a positive integer to the square of the next larger integer. The integer lies two thirds of the way between the sam e two squares. What is?SolutionThe difference between and is given byThe difference between the two squares is three tim es this amount orThe difference between two consecutive squares is always an odd number, therefore is odd. We will show that must be 1. Otherwise we would belooking for two consecutive squares that are at least 81 apart. But already theequation solves to , and has m ore than threedigits.The consecutive squares with common difference are and .One third of the way between them is and two thirds of the way is .This gives , , .Problem 19Rhombus , with side length , is rolled to form a cylinder of volum e bytaping to . What is ?SolutionWhere andProblem 20The parallelogram bounded by the lines , , , andhas area . The parallelogram bounded by the lines ,, , and has area . Given that , , , andare positive integers, what is the sm allest possible value of ?SolutionThis solution is incomplete. You can help us out by completing it.Plotting the parallelogram on the coordinate plane, the 4 corners are at. Because , wehave that or that , whichgives (consider a hom othety, or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by , it follows that thestretch along the diagonal is ). The area of triangular half of the parallelogram onthe right side of the y-axis is given by , so substituting:Thus , and we verify that , will give us a minimum value for . Then.Solution 2This solution is incomplete. You can help us out by completing it.The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the linesand . Now, thearea of the parallelogram contained by is the former is equal to the area of arectangle with sides and , , and the area contained bythe latter is . Thus, and must be even if the form erquantity is to equal . so is a m ultiple of . Putting this alltogether, the minimal solution for , so the sum is . Problem 21The first positive integers are each written in base . How m any of thesebase-representations are palindromes? (A palindrome is a number that reads the sam e forward and backward.)SolutionAll numbers of six or less digits in base 3 have been written.The form of each palindrome is as follows1 digit -2 digits -3 digits -4 digits -5 digits -6 digits -Where are base 3 digitsSince , this gives a total ofpalindromes so far.7 digits - , but not all of the num bers are less thanCase:All of these numbers are less than giving more palindromesCase: ,All of these numbers are also small enough, giving more palindromesCase: ,It follows that , since any other would make the value too large. This leavesthe number as . Checking each value of d, all of the three are sm all enough, so that gives more palindromes.Summing our cases there areProblem 22Two particles m ove along the edges of equilateral in the directionstarting simultaneously and moving at the sam e speed. Onestarts at , and the other starts at the midpoint of . The midpoint of the linesegm ent joining the two particles traces out a path that encloses a region . Whatis the ratio of the area of to the area of ?SolutionFirst, notice that each of the midpoints of ,, and are on the locus.Suppose after som e time the particles have each been displaced by a short distance, to new positions and respectively. Consider and drop aperpendicular from to hit at . Then, and .From here, we can use properties of a triangle to determine thelengths and as m onomials in . Thus, the locus of the midpoint will be linear between each of the three special points m entioned above. It follows that the locus consists of the only triangle with those three points as vertices. Com paring inradii between this "midpoint" triangle and the original triangle, the area containedby must be of the total area.Problem 23How many non-congruent right triangles with positive integer leg lengths have areas that are num erically equal to tim es their perimeters?SolutionUsing Euclid's formula for generating primit ive triples: , ,where and are relatively prime positive integers, exactly one of which being even.Since we do not want to restrict ourselves to only primitives, we will add a factor ofk. , ,Now we do som e casework.Forwhich has solutions , , ,Removing the solutions that do not satisfy the conditions of Euclid's formula, theonly solutions are andForhas solutions , , both of which are valid.Forhas solutions , of which only is valid.Forhas solution , which is valid.This m eans that the solutions for aresolutionsProblem 24How many pairs of positive integers are there such that andis an integer?SolutionCombining the fraction, must be an integer.Since the denominator contains a factor of ,Rewriting as for some positive integer , we can rewrite the fraction asSince the denominator now contains a factor of , we get.But since , we must have , and thus .For the original fraction simplifies to .For that to be an integer, must divide , and therefore we must have. Each of these values does indeed yield an integer.Thus there are four solutions: , , , and the answer isProblem 25Points and are located in 3-dim ensional space withand .The plane of is parallel to . What is the area of ?SolutionLet , and . Since , we could let ,, and . Now to get back to we need another vertex. Now if we look at this configuration as if it was two dim ensions, we would see a square missing a side if we don't draw . Now we can bend thesethree sides into an equilateral triangle, and the coordinates change: ,, , , and . Checking for all the requirements, they are all satisfied. Now we find the area of triangle . Itis a triangle, which is an isosceles right triangle. Thus the area of it is.。
AMC2008
AMC2008年真题Problem 1Susan had dollars to spend at the carnival. She spent dollars on food and twice as much on rides. How many dollars did she have left to spend?Problem 2The ten-letter code represents the ten digits , in order. What4-digit number is represented by the code word ?Problem 3If February is a month that contains Friday the , what day of the week is February 1?Problem 4In the figure, the outer equilateral triangle has area , the inner equilateral triangle has area , and the three trapezoids are congruent. What is the area of one of the trapezoids?Problem 5Barney Schwinn notices that the odometer on his bicycle reads , a palindrome, becauseit reads the same forward and backward. After riding more hours that day and the next,he notices that the odometer shows another palindrome, . What was his average speed in miles per hour?In the figure, what is the ratio of the area of the gray squares to the area of the white squares?Problem 7If , what is ?Problem 8Candy sales from the Boosters Club from January through April are shown. What were theaverage sales per month in dollars?Problem 9In Tycoon Tammy invested dollars for two years. During the the first year her investment suffered a loss, but during the second year the remaining investmentshowed a gain. Over the two-year period, what was the change in Tammy's investment?The average age of the people in Room A is . The average age of the people in RoomB is . If the two groups are combined, what is the average age of all the people?Problem 11Each of the students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and students have a cat. How manystudents have both a dog and a cat?Problem 12A ball is dropped from a height of meters. On its first bounce it rises to a height of meters.It keeps falling and bouncing to of the height it reached in the previous bounce. On which bounce will it not rise to a height of meters?Problem 13Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than pounds or more than pounds. So the boxes are weighed in pairs in every possible way. The results are, and pounds. What is the combined weight in pounds of the three boxes?Three , three , and three are placed in the nine spaces so that each row andcolumn contain one of each letter. If is placed in the upper left corner, how manyarrangements are possible?Problem 15In Theresa's first basketball games, she scored and points. In her ninthgame, she scored fewer than points and her points-per-game average for the nine gameswas an integer. Similarly in her tenth game, she scored fewer than points and herpoints-per-game average for the games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?Problem 16A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units?Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?Problem 18Two circles that share the same center have radii meters and meters. An aardvark runs along the path shown, starting at and ending at . How many meters does theaardvark run?Problem 19Eight points are spaced around at intervals of one unit around a square, as shown.Two of the points are chosen at random. What is the probability that the two points are oneunit apart?The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?Problem 21Jerry cuts a wedge from a -cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?Problem 22For how many positive integer values of are both and three-digit whole numbers?In square , and . What is the ratio of the area ofto the area of square ?Problem 24Ten tiles numbered through are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?Problem 25Margie's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Approximately what percent of the design is black?。
2007 AMC 12A Problems and Solution
2007 AMC 12A Problems and Solution Problem 1One ticket to a show costs at full price. Susan buys 4 tickets using a coupon thatgives her a 25% discount. Pam buys 5 ti ckets using a coupon that gives her a 30% discount. How m any m ore dollars does Pam pay than Susan?Solution= the am ount Pam spent = the am ount Susan spent▪▪Pam pays 10 m ore dollars than SusanProblem 2An aquarium has a rectangular base that m easures 100 c m by 40 c m and has a height of 50 cm. It is filled with water to a height of 40 cm. A brick with a rectangular base that m easures 40 cm by 20 c m and a height of 10 cm is placed in the aquarium.By how many centim eters does the water rise?SolutionThe brick has volume . The base of the aquarium has area . Forevery inch the water rises, the volum e increases by ; therefore, when thevolume increases by , the water level risesProblem 3The larger of two consecutive odd integers is three tim es the sm aller. What is their sum?SolutionSolution 1 Let be the sm aller term. Then▪Thus, the answer is▪By trial and error, 1 and 3 work. 1+3=4.Problem 4Kate rode her bicycle for 30 minutes at a speed of 16 m ph, then walked for 90 minutes at a speed of 4 mph. What was her overall average speed in miles per hour?Solution▪▪Problem 5Last year Mr. Jon Q. Public received an inheritance. He paid in federal taxes onthe inheritance, and paid of what he had left in state taxes. He paid a total of $for both taxes. How m any dollars was his inheritance?SolutionAfter paying his taxes, he has of the inheritance left. Since isof the inheritance, the whole inheritance is .Problem 6Triangles and are isosceles with and . Pointis inside triangle , angle measures 40 degrees, and anglemeasures 140 degrees. What is the degree m easure of angle ?SolutionWe angle chase, and find out that:▪▪▪Problem 7Let , and be five consecutive term s in an arithmetic sequence, andsuppose that . Which of or can be found?SolutionLet be the common difference between the term s.▪▪▪▪▪, so . But we can't find any m ore variables,because we don't know what is. So the answer is .Problem 8A star-polygon is drawn on a clock face by drawing a chord from each number to thefifth number counted clockwise from that number. That is, chords are drawn from12 to 5, from 5 to 10, from 10 to 3, and so on, ending back at 12. What is the degreemeasure of the angle at each vertex in the star polygon?We look at the angle between 12, 5, and 10. It subtends of the circle, ordegrees (or you can see that the arc is of the right angle). Thus, the angle at each vertex is an inscribed angle subtending degrees, making the answerProblem 9Yan is som ewhere between his home and the stadium. To get to the stadium he c an walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 tim es as fast as he walks, and both choices require the sam e am ount of tim e. What is the ratio of Yan's distance from his home to his distance from the stadium?Let the distance from Yan's initial position to the stadium be and the distance fromYan's initial position to home be . We are trying to find , and we have thefollowing identity given by the problem:Thus and the answer isProblem 10A triangle with side lengths in the ratio is inscribed in a circle with radius 3. What is the area of the triangle?SolutionSince 3-4-5 is a Pythagorean triple, the triangle is a right triangle. Since the hypotenuse is a diameter of the circum circle, the hypotenuse is . Then theother legs are and . The area isProblem 11A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hund reds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the term s 247, 475, and 756 and end with the term 824. Let be the sum of all theterm s in the sequence. What is the largest prim e factor that always divides ?SolutionA given digit appears as the hundreds digit, the tens digit, and the units digit of a term the sam e number of tim es. Let be the sum of the units digits in all the term s.Then , so must be divisible by . To see that it neednot be divisible by any larger prime, the sequence gives.Problem 12Integers and , not necessarily distinct, are chosen independently and atrandom from 0 to 2007, inclusive. What is the probability that is even?SolutionThe only tim es when is even is when and are of the sam e parity. Thechance of being odd is , so it has a probability of being even.Therefore, the probability that will be even is .A piece of cheese is located at in a coordinate plane. A m ouse is atand is running up the line. At the point the m ouse starts getting farther from the cheese rather than closer to it. What is ?SolutionWe are trying to find the foot of a perpendicular from to .Then the slope of the line that passes through the cheese and is the negativereciprocal of the slope of the line, or . Therefore, the line is . Thepoint where and intersect is , and.Problem 14Let a, b, c, d, and e be distinct integers such thatWhat is ?SolutionIf 45 is expressed as a product of five distinct integer factors, t he absolute value ofthe product of any four it as least , so no factor can have an absolute value greater than 5. Thus the factors of the given expression are five ofthe integers . The product of all six of these is , so thefactors are -3, -1, 1, 3, and 5. The corresponding values of a, b, c, d, and e are 9, 7, 5, 3, and 1, and their sum is 25 (C).The set is augmented by a fifth elem ent , not equal to any of the other four. The m edian of the resulting set is equal to its m ean. What is the sum of all possible values of ?SolutionThe m edian must either be or . Casework:▪Median is : Then and .▪Median is : Then and .▪Median is : Then and .All three cases are valid, so our solution is .Problem 16How many three-digit numbers are com posed of three distinct digits such that one digit is the average of the other two?SolutionWe can find the num ber of increasing arithm etic sequences of length 3 possible from0 to 9, and then find all the possible permutations of these sequences.This gives us a total of sequences. There are to permutethese, for a total of .However, we note that the conditions of the problem require three-digit numbers,and hence our numbers cannot start with zero. There are numbers whichstart with zero, so our answer is .Problem 17Suppose that and . What is ?SolutionWe can m ake use the of the Pythagorean identities: square both equations and add them up:This is just the cosine difference identity, which sim plifies toProblem 18The polynomial has real coefficients, andWhat isA fourth degree polynomial has four roots. Since the coefficients are real, the remaining two roots m ust be the com plex conjugates of the two given roots, nam ely. Now we work backwards for the polynomial:Thus our answer is .Problem 19Triangles and have areas and respectively, withand What is the sum of all possible x-coordinates of ?Solution 1From, we have that the height of is. Thus lies on the lines .using 45-45-90 triangles, so in we have that. The slope of is , so the equation of the line is. The point lies onone of two parallel lines that are units away from. Now take anarbitrary point on the line and draw the perpendicular to one of the parallel lines; then draw a line straight down from the sam e arbitrary point. These form a45-45-90 , so the straight line down has a length of . Now wenote that the y-intercept of the parallel lines is either units above or below they-intercept of line ; hence the equation of the parallel lines is.We just need to find the intersections of these two lines and sum up the values ofthe x-coordinates. Substituting the into , we get.Solution 2We are finding the intersection of two pairs of parallel lines, which will form aparallelogram. The centroid of this parallelogram is just the intersection of am d, which can easily be calculated to be . Now the sum of thex-coordinates is just .Problem 20Corners are sliced off a unit cube so that the six faces each becom e regular octagons. What is the total volum e of the rem oved tetrahedra?SolutionSince the sides of a regular polygon are equal in length, we can call each side . Examine one edge of the unit cube: each contains two slanted diagonal edges of anoctagon and one straight edge. The diagonal edges form righttriangles, making the distance on the edge of the cube . Thus, ,and .Each of the cut off corners is a pyramid, whose volume can be calculated by. Use the base as one of the three congruent isosceles triangles, with the height being one of the edges of the pyramid that sits on the edges of the cube. Theheight is . The base is a with leg of length ,making its area . Plugging this in, we get that the areaof one of the tetrahedra is . Since thereare 8 removed corners, we get an answer ofProblem 21The sum of the zeros, the product of the zeros, and the sum of the coeffi cients of thefunction are equal. Their common value must also be which of the following?SolutionBy Vieta's formulas, the sum of the roots of a quadratic equation is , the productof the zeros is , and the sum of the coefficients is . Setting equal the firsttwo tells us that . Thus, , so thecommon value is also equal to the coefficient of .To disprove the others, note that:▪: then , which is not necessarily true.▪: the y-intercept is , so , not necessarily true.▪: an x-intercept of the graph is a root of the polynomial, but this excludes the other root.▪: the m ean of the x-intercepts will be the sum of the roots of the quadratic divided by 2.Problem 22For each positive integer , let denote the sum of the digits of For howmany values of isSolutionSolution 1For the sake of notation let . Obviously .Then the m aximum value of is when , and the sumbecom es . So the minimu m bound is . We do casework upon thetens digit:Case 1: . Easy to directly disprove.Case 2: . , and ifand otherwise.Subcase a:. Thisexceeds our bounds, so no solution here.Subcase b:.First solution.Case 3: . , and if andotherwise.Subcase a:.Second solution.Subcase b:. Third solution.Case 4: . But , and the these clearly sum to .Case 5: . So and (recall that ), and. Fourth solution.In total we have solutions, which are and .Solution 2Clearly, . We can break this up into three cases:Case 1:Inspection gives .Case 2: , ,If you set up an equation, it reduces towhich has as its only solution satisfying the constraints , .Case 3: , ,This reduces to. The only two solutions satisfying the constraints for thisequation are , and , .The solutions are thus and the answer is .Solution 3As in Solution 1, we note that and .Obviously, .As , this m eans that , or equivalently that.Thus . For each possible we get three possible .(E. g., if , then is a number such that and ,therefore .)For each of these nine possibilities we compute as andcheck whether .We'll find out that out of the 9 cases, in 4 the value has the correct sum of digits.This happens for .Problem 23Square has area and is parallel to the x-axis. Vertices , andare on the graphs of and respectively. WhatisSolutionLet be the x-coordinate of and , and be the x-coordinate of and be they-coordinate of and . Then and. Since the distance between and is, we have , yielding .However, we can discard the negative root (all three logarithmic equations areunderneath the line and above when is negative, hence we can't squeeze in a square of side 6). Thus .Substituting back, , so .Problem 24For each integer , let be the number of solutions to the equationon the interval . What is ?SolutionSolution 1By looking at various graphs, we obtain that, for m ost of the graphsHowever, when , the middle apex of the sine curve touches the sinecurve at the top only one tim e (instead of two), so we get here .Solution 2So if and only ifor .The first occurs whenever , or for som enonnegative integer . Since , . So there are solutions inthis case.The second occurs whenever , or for som e nonnegativeinteger . Here so that there are solutions here.However, we overcount intersections. These occur wheneverwhich is equivalent to dividing . Ifis even, then is odd, so this never happens. If ,then there won't be intersections either, since a multiple of 8 can't divide a number which is not even a multiple of 4.This leaves . In this case, the divisibility becom es dividing. Since and are relatively prime (subtracting twice thesecond num ber from the first gives 1), must divide . Since ,. Then there is only one intersection, nam ely when.Therefore we find is equal to , unless, in which case it is one less, or . The problem may then be finishedas in Solution 1.Problem 25Call a set of integers spacy if it contains no m ore than one out of any threeconsecutive integers. How many subsets of including the empty set, are spacy?SolutionSolution 1Let denote the num ber of spacy subsets of . We have.The spacy subsets of can be divided into two groups:▪those not containing . Clearly .▪those containing . We have , since rem ovingfrom any set in produces a spacy set with all elem ents at m ost equal to andeach such spacy set can be constructed from exactly one spacy set in .Hence,From this recursion, we find thatSolution 2Since each of the elem ents of the subsets m ust be spaced at least two apart, a divider counting argument can be used.From the set we choose at m ost four numbers. Let those numbers be represented by balls. Between each of the balls there are at leasttwo dividers. So for example, o | | o | | o | | o | | represents .For subsets of size there must be dividers between the balls, leavingdividers to be be placed in spots betweenthe balls. The number of way this can be done is.Therefore, the number of spacy subsets is.Solution 3A shifting argument is also possible, and is similiar in spirit to Solut ion 2. Clearly wecan have at m ost elem ents. Given any arrangment, we subract from theelem ent in our subset, when the elem ents are arranged in increasing order.This creates a bijection with the num ber of size subsets of the set of the firstpositive integers. For instance, the arrangment o | | o | | o | | | o | corresponds to the arrangment o o o | o |. Notice that there is no longer any restriction on consectutive num bers. Therefore, we can easily plug in the possibleintegers 0, 1, 2, 3, 4, 5 for :In general, the number of subsets of a set with elem ent and with no consecutivenumbers is .Solution 4As a last resort, we can brute force the result by repeated casework. Luckily, 12 is not a very large number, so solving it this way is still possible.。
2008 AMC 12B Problems(答案)
2008 AMC 12B ProblemsProblem 1A basketball player made baskets during a game. Each basket was worth either or points. How many different numbers could represent the total points scored by the player?SolutionIf the basketball player makes three-point shots and two-point shots, he scorespoints. Clearly every value of yields a different number of total points. Since he can make any number ofthree-point shots between and inclusive, the number of different point totals is .Problem 2A block of calendar dates is shown. The order of the numbers in the second row is to be reversed. Then the order of the numbers in the fourth row is to be reversed. Finally, the numbers on each diagonal are to be added. What will be the positive difference between the two diagonal sums?SolutionAfter reversing the numbers on the second and fourth rows, the block will look like this:页脚内容1The difference between the two diagonal sums is:.Problem 3A semipro baseball league has teams with players each. League rules state that a player must be paid at leastdollars, and that the total of all players' salaries for each team cannot exceed dollars. What is the maximum possiblle salary, in dollars, for a single player?SolutionWe want to find the maximum any player could make, so assume that everyone else makes the minimum possible and that the combined salaries total the maximum ofThe maximum any player could make is dollars (answer choice C)Problem 4On circle , points and are on the same side of diameter , , and . What is the ratio of the area of the smaller sector to the area of the circle?页脚内容2Solution.Since a circle has , the desired ratio is .Problem 5A class collects dollars to buy flowers for a classmate who is in the hospital. Roses cost dollars each, and carnations cost dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly dollars?SolutionThe class could send just carnations (25 of them). They could also send 22 carnations and 2 roses, 19 carnations and 4 roses, and so on, down to 1 carnation and 16 roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), which is answer choice C.Problem 6页脚内容3Postman Pete has a pedometer to count his steps. The pedometer records up tosteps, then flips over toon the next step. Pete plans to determine his mileage for a year. On January Pete sets the pedometer to . During the year, the pedometer flips from to forty-four times. On December the pedometer reads . Pete takes steps per mile. Which of the following is closest to the number of miles Pete walked during the year?SolutionEvery time the pedometer flips, Pete has walked steps. Therefore, Pete has walked a total ofsteps, which is miles, which is closest to answer choice A.Problem 7For real numbers and , define . What is ?SolutionProblem 8页脚内容4Pointsand lie on . The length of is times the length of , and the length of is times thelength of . The length of is what fraction of the length of ?SolutionSince and , .Since and , .Thus, .Problem 9Points and are on a circle of radius and . Point is the midpoint of the minor arc . What is the length of the line segment ?SolutionTrig Solution:Let be the angle that subtends the arc AB. By the law of cosines,页脚内容5The half-angle formula says that, which is answer choice A.Other SolutionDefine D as the midpoint of AB, and R the center of the circle. R, C, and D are collinear, and since D is the midpoint of AB, , and so . Since , , and soProblem 10Bricklayer Brenda would take hours to build a chimney alone, and bricklayer Brandon would take hours to build it alone. When they work together they talk a lot, and their combined output is decreased by bricks per hour. Working together, they build the chimney in hours. How many bricks are in the chimney?SolutionLet be the number of bricks in the house.Without talking, Brenda and Brandon lay and bricks per hour respectively, so together they layper hour together.Since they finish the chimney in hours, . Thus, .页脚内容6Problem 11A cone-shaped mountain has its base on the ocean floor and has a height of 8000 feet. The top of the volume of the mountain is above water. What is the depth of the ocean at the base of the mountain in feet?SolutionIn a cone, radius and height each vary inversely with increasing height (i.e. the radius of the cone formed by cutting off the mountain at feet is half that of the original mountain). Therefore, volume varies as the inverse cube of increasing height (expressed as a percentage of the total height of cone):Plugging in our given condition,, answer choice A.Problem 12For each positive integer , the mean of the first terms of a sequence is . What is the th term of the sequence?SolutionLetting be the nth partial sum of the sequence:页脚内容7The only possible sequence with this result is the sequence of odd integers.Problem 13Vertex of equilateral is in the interior of unit square . Let be the region consisting of all points inside and outside whose distance from is between and . What is the area of ?Problem 14A circle has a radius of and a circumference of . What is ?SolutionLet be the circumference of the circle, and let be the radius of the circle.Using log properties, and .Since , .Problem 15 (no solution)页脚内容8On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the12 triangles have no points in common. Let be the region formed by the union of the square and all the triangles, and be the smallest convex polygon that contains . What is the area of the region that is inside but outside ?Baidu查的答案答案是A，其实画个图就清楚了，边长为1的正方形（unit square) 连同周边12个正三角形组成一个新的边长为2的正方形，要使包在正方形外面的八边形面积最小，只有A是正确的，BCEDE的话面积都比1/4的时候大。
2011 AMC 12B Problems and Solution
2011 AMC 12B Problems and Solution Problem 1What isSolutionAdd up the numbers in each fraction to get , which equals . Doingthe subtraction yieldsProblem 2Josanna's test scores to date are , , , , and . Her goal is to raise hertest average at least points with her next test. What is the minimum test score shewould need to accomplish this goal?SolutionTake the average of her current test scores, which isThis m eans that she wants her test average after the sixth test to be Let be thescore that Josanna receives on her sixth test. Thus, our equation isProblem 3LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paiddollars and Bernardo had paid dollars, where . How m any dollars mustLeRoy give to Bernardo so that they share the costs equally?SolutionThe total am ount of m oney that was spent during the trip was So eachperson should pay if they were to share the costs equally. Because LeRoyhas already paid dollars of his part, he still has to payProblem 4In multiplying two positive integers and , Ron reversed the digits of the two-digitnumber . His erroneous product was 161. What is the correct value of the product of and ?SolutionTaking the prime factorization of reveals that it is equal to Therefore, theonly ways to represent as a product of two positive integers is andBecause neither nor is a two-digit number, we know that and areand Because is a two-digit number, we know that a, with its two digits reversed,gives Therefore, and Multiplying our two correct values of andyieldsProblem 5Let be the second sm allest positive integer that is divisible by every positiveinteger less than . What is the sum of the digits of ?Solutionmust be divisible by every positive integer less than , or and . Eachnumber that is divisible by each of these is is a multiple of their least commonmultiple. , so each num ber divisible by these is a multiple of . The sm allest m ultiple of is clearly , so the second sm allestmultiple of is . Therefore, the sum of the digits of isProblem 6Two tangents to a circle are drawn from a point . The points of contact anddivide the circle into arcs with lengths in the ratio . What is the degree m easureof ?SolutionIn order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d.Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:1/2 (216°-144°) = 1/2 (72°)Problem 7Let and be two-digit positive integers with m ean . What is the m aximumvalue of the ratio ?SolutionIf and have a m ean of , then and . To m aximize ,we need to m aximize and minimize . Since they are both two-digit positiveintegers, the m aximum of is which gives . cannot be decreasedbecause doing so would increase , so this gives the m aximum value of , which isProblem 8Keiko walks once around a track at exactly the sam e constant speed every day. The sides of the track are straight, and the ends are sem icircles. The track has widthmeters, and it takes her seconds longer to walk around the outside edge of thetrack than around the inside edge. What is Keiko's speed in meters per second?SolutionTo find Keiko's speed, all we need to find is the difference between the distance around the inside edge of the track and the distance around the outside edge of the track, and divide it by the differenc e in tim e it takes her for each distance. We aregiven the difference in tim e, so all we need to find is the difference between the distances.The track is divided into lengths and curves. The lengths of the track will exhibit no difference in distance between the inside and outside edges, so we only need to concern ourselves with the curves.The curves of the track are semicircles, but since there are two of them, we can consider both of the at the sam e time by treating them as a single circle. We need to find the difference in the circum ferences of the inside and outside edges of the circle.The form ula for the circum ference of a circle is where is the radiusof the circle.Let's define the circum ference of the inside circle as and the circum ference of theoutside circle as .If the radius of the inside circle () is , then given the thickness of the track is 6meters, the radius of the outside circle () is .Using this, the difference in the circum ferences is:is the difference between the inside and outside lengths of the track. Divided by the tim e differential, we get:Problem 9Two real numbers are selected independently and at random from the interval. What is the probability that the product of those num bers is greater than zero?SolutionFor the product to be greater than zero, we m ust have either both numbers negative or both positive.Both numbers are negative with a chance.Both numbers are positive with a chance.Therefore, the total probability is and we are done.Problem 10Rectangle has and . Point is chosen on side sothat . What is the degree m easure of ?SolutionSince , hence . Therefore. ThereforeProblem 11A frog located at , with both and integers, makes successive jumps oflength and always lands on points with integer coordinates. Suppose that the frogstarts at and ends at . What is the sm allest possible number of jumps the frog m akes?SolutionSince the frog always jumps in length and lands on a lattice point, the sum of itscoordinates m ust change either by (by jumping parallel to the x- or y-axis), or byor (based off the 3-4-5 right triangle).Because either , , or is always the change of the sum of the coordinates, the sum of the coordinates will always change from odd to even or vice versa. Thus, itis impossible for the frog to go from to in an even number of moves.Therefore, the frog cannot reach in two m oves.However, a path is possible in 3 m oves: from to to to .Thus, the answer is .Problem 12A dart board is a regular octagon divided into regions as shown below. Suppose t hat a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?SolutionLet's assum e that the side length of the octagon is . The area of the center squareis just . The triangles are all triangles, with a side length ratio of. The area of each of the identical triangles is , so thetotal area of all of the triangles is also . Now, we must find the area of all of the 4 identical rectangles. One of the side lengths is and the other side length is, so the area of all of the rectangles is . The ratio of the area ofthe square to the area of the octagon is . Cancelling from thefraction, the ratio becom es . Multiplying the numerator and thedenominator each by will cancel out the radical, so the fraction is nowProblem 13Brian writes down four integers whose sum is . The pairwisepositive differences of these num bers are and . What is the sum of thepossible values of ?SolutionAssum e that results in the greatestpairwise difference, and thus it is . This m eans . must be inthe set . The only way for 3 num bers in the set to add up to 9 is if they are. , and then must be the rem aining two numbers which are and. The ordering of must be either or .Case 1Case 2The sum of the two w's isProblem 14A segm ent through the focus of a parabola with vertex is perpendicular toand intersects the parabola in points and . What is ?SolutionName the directrix of the parabola . Define to be the distance between apoint and a line .Now we remember the geom etric definition of a parabola: given any line (calledthe directrix) and any point (called the focus), the parabola corresponding to thegiven directrix and focus is the locus of the points that are equidistant from and. Therefore . Let this distance be . Now note that , so. Therefore . We now use the PythagoreanTheorem on triangle ; . Similarly,. We now use the Law of Cosines:This shows that the answer is .Problem 15How many positive two-digit integers are factors of ?SolutionFrom repeated application of difference of squares:Aplying sum of cubes:A quick check shows is prime. Thus, the only factors to be concerned about are, since multiplying by will make any factor too large.Multiply by or will give a two digit factor; itself will also work. The nextsm allest factor, , gives a three digit number. Thus, there are factors which aremultiples of .Multiply by or will also give a two digit factor, as well as itself. Highernumbers will not work, giving an additional factors.Multiply by or for a two digit factor. There are no mare factors to check, asall factors which include are already counted. Thus, there are an additionalfactors.Multiply by or for a two digit factor. All higher factors have been countedalready, so there are more factors.Thus, the total num ber of factors isProblem 16Rhombus has side length and . Region consists of all pointsinside of the rhombus that are closer to vertex than any of the other threevertices. What is the area of ?SolutionSuppose that is a point in the rhombus and let be the perpendicularbisector of . Then if and only if is on the sam e side of as .The line divides the plane into two half-planes; let be the half-planecontaining . Let us define similarly and . Then is equal to. The region turns out to be an irregular pentagon. We can m ake it easier to find the area of this region by dividing it into four triangles:Since and are equilateral,contains , contains and , and contains . Thenwith and soand has area .Problem 17Let , andfor integers . What is the sum of the digits of ?SolutionProof by induction that :ForAssum e is true for n:Therefore, if it is true for n, then it is true for n+1; since it is also true for n = 1, it is true for all positive integers n., which is the 2011-digitnumber 8888 (8889)The sum of the digits is 8 tim es 2010 plus 9, orProblem 18A pyramid has a square base with side of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?SolutionWe can use the Pythagorean Theorem to split one of the triangular faces into two30-60-90 triangles with side lengths and .Next, take a cross-section of the pyram id, forming a triangle with the top of the triangle and the midpoints of two opposite sides of the square base.This triangle is isosceles with a base of 1 and two sides of length .The height of this triangle will equal the height of the pyram id. To find this height,split the triangle into two right triangles, with sides and .Sorry, the GeoGebra Applet could not be started. Please m ake sure that Java 1.4.2 (or later) is installed and activated. (click here to install Java now)The cube, touching all four triangular faces, will form a similar pyramid which sits ontop of the cube. If the cube has side length , the pyramid has side length .Thus, the height of the cube plus the height of the sm aller pyramid equals the height of the larger pyramid..side length of cube.Problem 19A lattice point in an -coordinate system is any point where both and areintegers. The graph of passes through no lattice point withfor all such that . What is the m aximum possible value of ?SolutionIt is very easy to see that the in the graph does not impact whether it passes through lattice.We need to m ake sure that cannot be in the form of for , otherwise the graph passes through lattice point at . We only need to worryabout very close to , , will be the only case we need to worryabout and we want the minimu m of those, clearly for , the sm allest is, so answer isProblem 20Triangle has , and . The points , andare the midpoints of , and respectively. Let be the intersectionof the circum circles of and . What is ?Solution 1Answer: (C)Let us also consider the circum circle of .Note that if we draw the perpendicular bisector of each side, we will have thecircum center of which is , Also, since .is cyclic, similarly, and are also cyclic. With this, we knowthat the circum circles of , and all intercept at , so is.The question now becom es calculate the sum of distance from each vertices to the circum center.We can do it will coordinate geometry, note that because ofbeing circum center.Let , , ,Then is on the line and also the line with slope and passes through.SoandSolution 2Consider an additional circum circle on . After drawing the diagram, it isnoticed that each triangle has side values: , , . Thus they are congruent, and their respective circum circles are. By inspection, we see that , , andare the circum diameters, and so they are congruent. Therefore, the solution can be found by calculating one of these circum diam eters and multiplying it by a factor of . We can find the circumradius quite easily with the form ula, s.t. and R is the circumradius.Since :After a few algebraic m anipulations:.Problem 21The arithm etic m ean of two distinct positive integers and is a two-digit integer. The geom etric m ean of and is obtained by reversing the digits of the arithm eticmean. What is ?SolutionAnswer: (D)for som e ,.Note that in order for x-y to be integer, has to be for some perfectsquare . Since is at m ost , orIf , , if , . In AMC, we are done. Otherwise,we need to show that is impossible.->, or or and , , respectively.And since , , , but there is no integer solution for , .Problem 22Let be a triangle with sides , and . For , ifand , and are the points of tangency of the incircle of to the sides, and , respectively, then is a triangle with side lengths ,and , if it exists. What is the perim eter of the last triangle in the sequence ?SolutionAnswer: (D)Let , , andThen , andThen , ,Hence:Note that and for , I claim that it is true for all , assum e for induction that it is true for som e , thenFurthermore, the average for the sides is decreased by a factor of 2 each tim e.So is a triangle with side length , ,and the perimeter of such isNow we need to find what fails the triangle inequality. So we need to find the lastsuch thatFor , perimeter isProblem 23A bug travels in the coordinate plane, moving only along the lines that are paralle lto the -axis or -axis. Let and . Consider all possible paths of the bug from to of length at m ost . How m any points with integercoordinates lie on at least one of these paths?SolutionAnswer: (C)If a point satisfy the property that ,then it is in the desire range because is the shortest path fromto , and is the shortest path from toIf , then satisfy the property. there arelattice points here.else let (and for it is symmetrical,,So for , there are lattice points,for , there are lattice points,etc.For , there are lattice points.Hence, there are a total of lattice points. Problem 24Let . What is the minimum perimeter am ong all the -sided polygons in the complex plane whose vertices are preciselythe zeros of ?SolutionAnswer: (B)First of all, we need to find all such thatSo ororNow we have a solution at if we look at them in polar coordinate, further m ore, the 8-gon is symm etric (it is an equilateral octagon) . So we only need to find the side length of one and multiply by .So answer distance from toSide lengthHence, answer is .Problem 25For every and integers with odd, denote by the integer closest to . Forevery odd integer , let be the probability thatfor an integer randomly chosen from the interval . What is theminimum possible value of over the odd integers in the interval?SolutionAnswer:First of all, you have to realize thatifthenSo, we can consider what happen in and it will repeat. Also since range ofis to , it is always a m ultiple of . So we can just consider for.LET be the fractional part functionThis is an AMC exam, so use the given choices wisely. With the given choices, andthe previous explanation, we only need to consider , , , .For , . 3 of the that should consider lands in here.For , , then we needelse for , , then we needFor ,So, for the condition to be true, . ( , no worry for therounding to be ), so this is always true.For , , so we want , orFor k = 67,For k = 69,etc.We can clearly see that for this case, has the m inimu m, which is .Also, .So for AMC purpose, answer is (D).Now, let's say we are not given any answer, we need to consider .I claim thatIf got round down, then all satisfy the condition along withbecause if and , so m ustand for , it is the sam e as ., which m akes.If got round up, then all satisfy the condition along withbecause if andCase 1)->Case 2)->and for , since is odd,->->, and is prime so or , which is not in this set, which m akes.Now the only case without rounding, . It must be true.。
2002 AMC 12B Problems and Solution
2002 AMC 12B Problems and Solution Problem 1The arithm etic m ean of the nine numbers in the setis a -digit number , all of whose digits aredistinct. The num ber does not contain the digitSolutionWe wish to find , or. This does not have the digit 0,so the answer isProblem 2What is the value ofwhen ?SolutionBy the distributive property,Problem 3For how m any positive integers is a prime number?SolutionFactoring, we get . Exactly of andmust be and the other a prim e number. If , then , and, which is not prime. On the other hand, if , then ,and , which is a prime number. The answer is .Problem 4Let be a positive integer such that is an integer. Which of the following statem ents is not true:SolutionSince ,From which it follows that and . Thus the answer is.Problem 5Let and be the degree m easures of the five angles of a pentagon.Suppose that and and form an arithmeticsequence. Find the value of .SolutionThe sum of the degree measures of the angles of a pentagon (as a pentagon can besplit into triangles) is . If we let, it follows thatNote that since is the middle term of an arithmetic sequence wit h an odd num ber of term s, it is simply the average of the sequence.Problem 6Suppose that and are nonzero real numbers, and that the equationhas solutions and . Then the pair isSolutionSolution 1Since , it follows bycom paring coefficients that and that . Since is nonzero, ,and . Thus .Solution 2Another method is to use Vieta's formulas. The sum of the solutions to this polynomial is equal to the opposite of the coefficient, since the leading coefficientis 1; in other words, and the product of the solutions is equal to theconstant term (i.e, ). Since is nonzero, it follows that andtherefore (from the first equation), . Hence,The product of three consecutive positive integers is tim es their sum. What is thesum of their squares?SolutionLet the three consecutive positive integers be , , and . So,. Rearranging and factoring,, so . Hence, the sum of the squares is.Problem 8Suppose July of year has five Mondays. Which of the following must occur fivetim es in August of year ? (Note: Both months have 31 days.)SolutionIf there are five Mondays, there are only three possibilities for their dates:, , and .In the first case August starts on a Thursday, and there are five Thursdays, Fridays, and Saturdays in August.In the second case August starts on a Wednesday, and there are five Wednesdays, Thursdays, and Fridays in August.In the third case August starts on a Tuesday, and there are five Tuesdays, Wednesdays, and Thursdays in August.The only day of the week that is guaranteed to appear five tim es is therefore.If are positive real numbers such that form an increasingarithmetic sequence and form a geometric sequence, then isSolutionSolution 1We can let a=1, b=2, c=3, and d=4.Solution 2As is a geom etric sequence, let and for som e .Now, is an arithmetic sequence. Its difference is . Thus.Comparing the two expressions for we get . The positive solution is, and .Solution 3Letting be the common difference of the arithm etic progression, we have, , . We are given that = , orCross-m ultiplying, we getSo .How m any different integers can be expressed as the sum of three distinct m embersof the set ?SolutionEach number in the set is congruent to 1 m odulo 3. Therefore, the sum of any three numbers is a multiple of 3. We can m ake all multiples of three between 1+4+7=12 (the minimu m sum) and 13+16+19=48 (the m aximum sum), inclusive. There areintegers we can form.Problem 11The positive integers and are all prime numbers. The sum ofthese four prim es isSolutionSince and must have the sam e parity, and since there is only oneeven prime number, it follows that and are both odd. Thus one ofis odd and the other even. Since , it follows that(as a prime greater than ) is odd. Thus , and areconsecutive odd primes. At least one of is divisible by , fromwhich it follows that and . The sum of these num bers is thus , aprime, so the answer is .Problem 12For how m any integers is the square of an integer?SolutionSolution 1Let , with (note that the solutions do not give anyadditional solutions for ). Then rewriting, . Since, it follows that divides . Listing the factors of , wefind that are the only solutions (respectively yielding).Solution 2For and the fraction is negative, for it is not defined, and forit is between 0 and 1.Thus we only need to examine and .For and we obviously get the squares and respectively.For prime the fraction will not be an integer, as the denominator will not contain the prime in the numerator.This leaves , and a quick substitution shows that out ofthese only and yield a square.Problem 13The sum of consecutive positive integers is a perfect square. The sm allest possible value of this sum isSolutionLet be the consecutive positive integers. Their sum,, is a perfect square. Since is a perfect square, itfollows that is a perfect square. The sm allest possible such perfect square iswhen , and the sum is .Problem 14Four distinct circles are drawn in a plane. What is the m aximum number of points where at least two of the circles intersect?SolutionFor any given pair of circles, they can intersect at m ost tim es. Since there arepairs of circles, the m aximum number of possible intersections is. We can construct such a situation as below, so the answer is .Problem 15How many four-digit numbers have the property that the three-digit numberobtained by removing the leftm ost digit is one ninth of ?SolutionLet , such that . Then. Since , fromwe have three-digit solutions, and the answer is . Problem 16Juan rolls a fair regular octahedral die marked with the numbers through . ThenAm al rolls a fair six-sided die. What is the probability that the product of the two rolls is a multiple of 3?SolutionSolution 1On both dice, only the faces with the numbers are divisible by . Letbe the probability that Juan rolls a or a , and thatAm al does. By the Principle of Inclusion-Exclusion,Alternatively, the probability that Juan rolls a m ultiple of is , and the probabilitythat Juan does not roll a multiple of but Amal does is . Thus thetotal probability is .Solution 2The probability that neither Juan nor Amal rolls a m ultiple of is ; usingcom plementary counting, the probability that at least one does is.Problem 17Andy’s lawn has twice as m uch area as Beth’s lawn and three tim es as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to m ow their lawns at the sam e tim e, who will finish first?SolutionWe say Andy's lawn has an area of . Beth's lawn thus has an area of , andCarlos's lawn has an area of .We say Andy's lawn mower cuts at a speed of . Carlos's cuts at a speed of , andBeth's cuts at a speed .Each person's lawn is cut at a speed of , so Andy's is cut in tim e, as is Carlos's.Beth's is cut in , so the first one to finish is .Problem 18A point is randomly selected from the rectangular region with vertices. What is the probability that is closer to the origin thanit is to the point ?SolutionThe region containing the points closer to than to is bounded by theperpendicular bisector of the segm ent with endpoints . Theperpendicular bisector passes through midpoint of , which is ,the center of the unit square with coordinates . Thus, itcuts the unit square into two equal halves of area . The total area of therectangle is , so the area closer to the origin than to and in the rectangle is. The probability is .Problem 19If and are positive real numbers such thatand , then isSolutionAdding up the three equations gives. Subtracting each of the above equations from this yields, respectively,. Taking their product,. Problem 20Let be a right-angled triangle with . Let and be themidpoints of legs and , respectively. Given that and ,find .SolutionLet , . By the Pythagorean Theorem onrespectively,Summing these gives .By the Pythagorean Theorem again, we haveProblem 21For all positive integers less than , letCalculate .SolutionSince , it follows thatThus .Problem 22For all integers greater than , define . Letand . Then equalsSolutionBy the change of base form ula, . ThusIn , we have and . Side and the m edian from tohave the sam e length. What is ?SolutionLet be the foot of the m edian from to , and we let . Thenby the Law of Cosines on , we haveSince , we can add these two equations and getHence and .Alternate SolutionFrom Stewart's Theorem, we haveSimplifying, we getA convex quadrilateral with area contains a point in its interior suchthat . Find the perimeter of .SolutionWe have (This is true for any convex quadrilateral: split the quadrilateral along and then using the triangle areaformula to evaluate and ), with equality only if . By the triangle inequality,with equality if lies on and respectively. ThusSince we have the equality case, at point , as shown below.By the Pythagorean Theorem,The perimeter of is .Problem 25Let , and let denote the set of points in the coordinateplane such that The area of isclosest toSolutionThe first condition gives us thatwhich is a circle centered at with radius. The second condition gives us thatThus eitherorEach of those lines passes through and has slope , as shown above.Therefore, the area of is half of the area of the circle, which is.。
美国数学竞赛AMC8 -- 2008年真题解析(英文解析+中文解析)
美国数学竞赛AMC8 – 2008年真题解析(英文解析+中文解析)Problem 1Answer: BSolution:50-12-24=14中文解析：总共花的钱是：12+12*2=36元。
剩余50-36=14元。
答案是BProblem 2Answer: ASolution:We can derive that c=8,L=6, U=7,and E=1. Therefore, the answer is 8671.中文解析：这10个字母的对应关系是： B -0；E-1; S-2; ......K -9. 按照这个对应关系：C-8，L-6，U-7，E-1. 即8671. 答案是A。
Problem 3Answer: ASolution:We can go backwards by days, but we can also backwards by weeks. If we go backwards by weeks, we see that February 6 is a Friday. If we now go backwards by days, February 1 is a Sunday.中文解析：13日是周五，则13-7=6，即6日也是周五，则倒推2月1日是周日。
答案是A。
Problem 4Answer: CSolution:The area outside the small triangle but inside the large triangle is 16-1=15. This is equally distributed between the three trapezoids. Each trapezoid has an area of 15/3=5.中文解析：大三角形的面积等于小的等边三角形的面积加上3个梯形的面积。
据此，三个梯形的面积是16-1=15. 每个梯形的面积是15/3=5. 答案是C。
2006 AMC 12B Problems and Solution
2006 AMC 12B Problems and Solution Problem 1What is ?Solutionif n is even and if n is odd. So we haveProblem 2For real numbers and , define . What is ?SolutionProblem 3A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?SolutionProblem 4Mary is about to pay for five items at the grocery store. The prices of the items are, , , , and . Mary will pay with a twenty-dollar bill.Which of the following is closest to the percentage of the that she will receive in change?SolutionThe total price of the items isProblem 5John is walking east at a speed of 3 miles per hour, while Bob is also walking east, but at a speed of 5 miles per hour. If Bob is now 1 mile west of John, how many minutes will it take for Bob to catch up to John?SolutionThe speed that Bob is catching up to John is miles per hour. Since Bob isone mile behind John, it will take of an hour to catch up to John. Problem 6Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade.There are calories in grams of lemonade. There arecalories in grams of lemonade.Problem 7Mr. and Mrs. Lopez have two children. When they get into their family car, two people sit in the front, and the other two sit in the back. Either Mr. Lopez or Mrs. Lopez must sit in the driver's seat. How many seating arrangements are possible?SolutionFirst, we seat the children.The first child can be seated in spaces.The second child can be seated in spaces.Now there are ways to seat the adults.Alternative solution:If there was no restriction, there would be 4!=24 ways to sit. However, only 2/4 ofthe people can sit in the driver's seat, so our answer isProblem 8The lines and intersect at the point . What is ?Problem 9How many even three-digit integers have the property that their digits, read left to right, are in strictly increasing order?SolutionSolution 1Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 8, the ones digit cannot be even).If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit. If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit. If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,and so on.So, the answer is .The last digit is 4, 6, or 8.If the last digit is , the possibilities for the first two digits correspond to 2-elementsubsets of .Thus the answer is .Problem 10In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?SolutionIf the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have:Now, since we want the greatest perimeter, we want the greatest integer x, and if then . Then, the first side haslength , the second side has length , the third side has length , andso the perimeter is .Problem 11Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?Joe has 2 ounces of cream, as stated in the problem.JoAnn had 14 ounces of liquid, and drank of it. Therefore, she drank of hercream, leaving her .Problem 12The parabola has vertex and -intercept , where . What is ?SolutionSubstituting , we find that , so ourparabola is .The x-coordinate of the vertex of a parabola is given by .Additionally, substituting , we find that. Since it is given that , then .Problem 13Rhombus is similar to rhombus . The area of rhombus is24, and . What is the area of rhombus ?SolutionThe ratio of any length on ABCD to a corresponding length on BFDE squared is equal to the ratio of their areas. Since , and are equilateral.DB, which is equal to AB, is the diagonal of rhombus ABCD. Therefore,. DB and AC are the longer diagonal of rhombuses BEDF andABCD, respectively. So the ratio of their areas is or . One-third the area ofABCD is equal to 8. So the answer is CProblem 14Elmo makes sandwiches for a fundraiser. For each sandwich he uses globs ofpeanut butter at cents per glob and blobs of jam at cents per glob. The cost ofthe peanut butter and jam to make all the sandwiches is . Assume that ,and are all positive integers with . What is the cost of the jam Elmo uses tomake the sandwiches?SolutionFrom the given, we know that(The numbers are in cents)since , and since is an integer, then or . It iseasily deduced that is impossible to make with and integers, so and. Then, it can be guessed and checked quite simply that if and, then . The problem asks for the total cost ofjam, or cents, orProblem 15Circles with centers and have radii 2 and 4, respectively, and are externallytangent. Points and are on the circle centered at , and points and are onthe circle centered at , such that and are common external tangents tothe circles. What is the area of hexagon ?SolutionDraw the altitude from onto and call the point . Because andare right angles due to being tangent to the circles, and the altitude creates as a right angle. is a rectangle with bisecting . The lengthis and has a length of , so by pythagorean's, is ., which is half the area of the hexagon, sothe area of the entire hexagon isSolution 2and are congruent right trapezoids with legs and and withequal to . Draw an altitude from to either or , creating a rectangle withwidth and base , and a right triangle with one leg , the hypotenuse , and theother . Using the Pythagorean theorem, is equal to , and is also equal tothe height of the trapezoid. The area of the trapezoid is thus ,and the total area is two trapezoids, or .Problem 16Regular hexagon has vertices and at and , respectively.What is its area?SolutionTo find the area of the regular hexagon, we only need to calculate the side length. Drawing in points , , and , and connecting and with an auxiliary line, wesee two 30-60-90 triangles are formed.Points and are a distance of apart. Half of thisdistance is the length of the longer leg of the right triangles. Therefore, the sidelength of the hexagon is .The apothem is thus , yielding an area of.Problem 17For a particular peculiar pair of dice, the probabilities of rolling , , , , and oneach die are in the ratio . What is the probability of rolling a total of on the two dice?SolutionThe probability of getting an on one of these dice is .The probability of getting on the first and on the second die is . Similarlywe can express the probabilities for the other five ways how we can get a total .(Note that we only need the first three, the other three are symmetric.) Summing these, the probability of getting a total is:Problem 18An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?SolutionLet the starting point be . After steps we can only be in locationswhere . Additionally, each step changes the parity of exactly onecoordinate. Hence after steps we can only be in locations where iseven. It can easily be shown that each location that satisfies these two conditions is indeed reachable.Once we pick , we have valid choices for , giving atotal of possible positions.Problem 19Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?SolutionFirst, The number of the plate is divisible by 9 and in the form of aabb, abba or abab We can conclude straight away that a+b is 9 using the 9 divisibility rule After trying various numbers, and reaching five I can observe that b=0 or 5. If b=5, the number is not divisible by 2 If b=0, the number must be 9900, and is not divisible by 8 Answer B 5Problem 20Let be chosen at random from the interval . What is the probability that? Here denotes the greatest integer that is less than or equal to .SolutionLet be an arbitrary integer. For which do we have ?The equation can be rewritten as . The second onegives us . Combining these, we get that both hold at the sametime if and only if .Hence for each integer we get an interval of values for which. These intervals are obviously pairwise disjoint.For any the corresponding interval is disjoint with , so it does not contribute to our answer. On the other hand, for any the entire interval isinside . Hence our answer is the sum of the lengths of the intervals for .For a fixed the length of the interval is .This means that our result is . Problem 21Rectangle has area . An ellipse with area passes through andand has foci at and . What is the perimeter of the rectangle? (The area of anellipse is where and are the lengths of the axes.)SolutionThis solution needs a picture. Please help add it.Let the rectangle have side lengths and . Let the axis of the ellipse on which thefoci lie have length , and let the other axis have length . We haveFrom the definition of an ellipse, . Also,the diagonal of the rectangle has length . Comparing the lengths of the axes and the distance from the foci to the center, we haveSince , we now know and because, or one-fourth of the rectangle's perimeter, we multiply by four to get ananswer of .Problem 22Suppose , and are positive integers with , and, where and are integers and is not divisible by . What isthe smallest possible value of ?SolutionClearly, the power of that divides is larger or equal than the power of whichdivides it. Hence we are trying to minimize the power of that will divide .Consider . Each fifth term is divisible by , each -th one by ,and so on. Hence the total power of that divides is . (For anyonly finitely many terms in the sum are non-zero.)In our case we have , so the largest power of that will be less than is atmost . Therefore the power of that divides is equal to. The same is true for and .Intuition may now try to lure us to split into as evenly as possible,giving and . However, this solution is not optimal.To see how we can do better, let's rearrange the terms as follows:The idea is that the rows of the above equation are roughly equal to , , etc.More precisely, we can now notice that for any positive integers we canwrite in the form , , , where allare integers and .It follows that andHence we get that for any positive integers we haveTherefore for any the result is at least.If we now show how to pick so that we'll get the result , we will be done.Consider the row with in the denominator. We need to achieve sum in this row,hence we need to make two of the numbers smaller than . Choosingdoes this, and it will give us the largest possible remainders for andin the other three rows, so this is a pretty good candidate. We can computeand verify that this triple gives the desired result .Problem 23Isosceles has a right angle at . Point is inside , such that, , and . Legs and have length ,where and are positive integers. What is ?SolutionUsing the Law of Cosines on , we have:Using the Law of Cosines on , we have:Now we use .Note that we know that we want the solution with since we know that. Thus, .Problem 24Let be the set of all points in the coordinate plane such that and. What is the area of the subset of for which?SolutionWe start out by solving the equality first.We end up withthree lines that matter: , , and .We plot these lines below. Note that by testing thepoint , we can see that we want the area of the pentagon. We cancalculate that by calculating the area of the sqaure and then subtracting the area of the 3 triangles. (Note we could also do this by adding the areas of the isosceles triangle in the bottom left corner and the rectangle with the previous triangle'shypotenuse as the longer side.)Problem 25A sequence of non-negative integers is defined by the rulefor . If , and , how many different values of are possible?SolutionWe say the sequence completes at if is the minimal positive integer such that. Otherwise, we say does not complete.Note that if , then for all , and does not divide ,so if , then does not complete. (Also, cannot be 1 in thiscase since does not divide , so we do not care about these at all.)From now on, suppose .We will now show that completes at for some . We will do this with 3lemmas.Lemma: If , and neither value is , then.Proof: There are 2 cases to consider.If , then , and . Soand .If , then , and . So and .In both cases, , as desired.If , then . Moreover, if instead we have for some , then .By the way is constructed in the problem statement, having two equal consecutive terms implies that divides every term in the sequence. So and , so , so . For the proof of the secondresult, note that if , then , so by the first result we just proved, .completes at for some .Suppose completed at some or not at all. Then by the second lemma and the fact that neither nor are , none of the pairscan have a or be equal to . So the first lemma implies so, a contradiction. Hence completes at for some .Now we're ready to find exactly which values of we want to count.Let's keep in mind that and that is odd. We have twoIf is odd, then is even, so is odd, so is odd, so is even, and this pattern must repeat every three terms because of the recursive definition of , so the terms of reduced modulo 2 are so is odd and hence (since if completes at , then must be or for all ).If is even, then is odd, so is odd, so is even, so is odd, and this pattern must repeat every three terms, so the terms of reduced modulo 2 are so is even, and hence .We have found that is true precisely when and is odd.There are numbers less than and relatively prime to it (is a 1-1 correspondence between the odd and even numbers less than and relatively prime to . So our final answer is , or .。
2001 AMC 12 Problems and Solution
2001 AMC 12 Problems and Solution Problem 1The sum of two numbers is . Suppose is added to each number and then each ofthe resulting numbers is doubled. What is the sum of the final two numbers?SolutionSuppose the two numbers are and , with . Then the desired sum is, which is answer . Problem 2Let and denote the product and the sum, respectively, of the digits of theinteger . For example, and . Suppose is a two-digitnumber such that . What is the units digit of ?SolutionDenote and as the tens and units digit of , respectively. Then . Itfollows that , which implies that . Since ,. So the answer is .Problem 3The 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 thatthe state income tax she paid amounted to of her annual income. What was her annual income?$$$$$SolutionLet the income amount be denoted by .We know that .We can now try to solve for :So the answer isProblem 4The mean of three numbers is more than the least of the numbers and lessthan the greatest. The median of the three numbers is . What is their sum?SolutionLet be the mean of the three numbers. Then the least of the numbers isand the greatest is . The middle of the three numbers is the median, 5. So, which implies that . Hence, the sum ofthe three numbers is , and the answer is .Problem 5What is the product of all positive odd integers less than 10000?SolutionProblem 6A telephone number has the form , where each letter representsa different digit. The digits in each part of the number are in decreasing order; that is, , , and . Furthermore, , , andare consecutive even digits; , , , and are consecutive odd digits; and. Find .SolutionThe last four digits are either or , and the other odd digit (or )must be , , or . Since , that digit must be . Thus the sum ofthe two even digits in is . must be , , or , whichrespectively leave the pairs and , and , or and , as the two even digits in. Only and has sum , so is , and the required first digit is 8, sothe answer is .Problem 7A charity sells benefit tickets for a total of $. Some tickets sell for full price(a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?$$$$$SolutionLet's multiply ticket costs by , then the half price becomes an integer, and thecharity sold tickets worth a total of dollars.Let be the number of half price tickets, we then have full price tickets. Thecost of full price tickets is equal to the cost of half price tickets.Hence we know that half price tickets cost dollars.Then a single half price ticket costs dollars, and this must be an integer.Thus must be a divisor of . Keeping in mind that , we arelooking for a divisor between and , inclusive.The prime factorization of is . We can easily find out thatthe only divisor of within the given range is .This gives us , hence there were half price tickets andfull price tickets.In our modified setting (with prices multiplied by ) the price of a half price ticket is. In the original setting this is the price of a full price ticket. Hencedollars are raised by the full price tickets.Problem 8Which of the cones listed below can be formed from a sector of a circle of radiusby aligning the two straight sides?SolutionThe blue lines will be joined together to form a single blue line on the surface of thecone, hence will be the of the cone.The red line will form the circumference of the base. We can compute its length and use it to determine the radius.The length of the red line is . This is the circumference of acircle with radius .Therefore the correct answer is .Problem 9Let be a function satisfying for all positive real numbers and . If, what is the value of ?Solution, so the answer is .Problem 10The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest toSolutionConsider any single tile:If the side of the small square is , then the area of the tile is , with coveredby squares and by pentagons. Hence exactly of any tile are covered bypentagons, and therefore pentagons cover of the plane. When expressed as apercentage, this is , and the closest integer to this value is .Problem 11A 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?SolutionImagine that we draw all the chips in random order, i.e., we do not stop when the last chip of a color is drawn. To draw out all the white chips first, the last chip left must be red, and all previous chips can be drawn in any order. Since there are 3 red chips, the probability that the last chip of the five is red (and so also the probabilitythat the last chip drawn in white) is .Problem 12How many positive integers not exceeding are multiple of or but not ?SolutionOut of the numbers to four are divisible by and three by , counting twice.Hence out of these numbers are multiples of or .The same is obviously true for the numbers to for any positive integer .Hence out of the numbers to there are numbers that aredivisible by or . Out of these , the numbers , , , , and aredivisible by . Therefore in the set there are preciselynumbers that satisfy all criteria from the problem statement.Again, the same is obviously true for the set for anypositive integer .We have , hence there are good numbers among thenumbers to . At this point we already know that the only answer that is stillpossible is , as we only have numbers left.By examining the remaining by hand we can easily find out that exactly of themmatch all the criteria, giving us good numbers.The parabola with equation and vertex is reflected aboutthe line . This results in the parabola with equation . Whichof the following equals ?SolutionWe write as (this is possible for any parabola). Then thereflection of is . Then we find .Since and , we have, so the answer is .Problem 14Given the nine-sided regular polygon , how manydistinct equilateral triangles in the plane of the polygon have at least two vertices inthe set ?SolutionEach of the pairs of vertices determines two equilateral triangles, one on each side of the segment. This would give us triangles. However, note that thereare three equilateral triangles that have all three vertices among the vertices of the polygon. These are the triangles , , and . We countedeach of these three times (once for each side). Hence we overcounted by , and thecorrect number of equilateral triangles is .An insect lives on the surface of a regular tetrahedron with edges of length 1. It wishes to travel on the surface of the tetrahedron from the midpoint of one edge to the midpoint of the opposite edge. What is the length of the shortest such trip? (Note: Two edges of a tetrahedron are opposite if they have no common endpoint.)SolutionGiven any path on the surface, we can unfold the surface into a plane to get a path of the same length in the plane. Consider the net of a tetrahedron in the picture below. A pair of opposite points is marked by dots. It is obvious that in the plane the shortest path is just a segment that connects these two points. Its length is thesame as the length of the tetrahedron's edge, i.e. .Problem 16A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?SolutionSolution 1Let the spider try to put on all things in a random order. Each of thepermutations is equally probable. For any fixed leg, the probability that he will firstput on the sock and only then the shoe is clearly . Then the probability that he willcorrectly put things on all legs is . Therefore the number of correct permutationsmust be .Solution 2Each dressing sequence can be uniquely described by a sequence containing twos, two s, ..., and two s -- the first occurrence of number means that the spiderputs the sock onto leg , the second occurrence of means he puts the shoe onto leg . If the numbers were all unique, the answer would be . However, since 8terms appear twice, the answer is .Problem 17A point is selected at random from the interior of the pentagon with vertices, , , , and . What is the probability that is obtuse?SolutionThe angle is obtuse if and only if lies inside the circle with diameter .(This follows for example from the fact that the inscribed angle is half of the central angle for the same arc.)The area of is , and the area of is.From the Pythagorean theorem the length of is , thus theradius of the circle is , and the area of the half-circle that is inside is.Therefore the probability that is obtuse is .Problem 18A circle centered at with a radius of 1 and a circle centered at with a radius of 4are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle isSolutionSolution 1In the triangle we have and , thus by thePythagorean theorem we have .We can now pick a coordinate system where the common tangent is the axis andlies on the axis. In this coordinate system we have and .Let be the radius of the small circle, and let be the -coordinate of its center .We then know that , as the circle is tangent to the axis. Moreover, the small circle is tangent to both other circles, hence we have and.We have and . Hence we getthe following two equations:Simplifying both, we getAs in our case both and are positive, we can divide the second one by the firstone to get .Now there are two possibilities: either , or . In the first case clearly , hence this is not the correct case. (Note: This case corresponds to theother circle that is tangent to both given circles and the axis - a large circle whosecenter is somewhere to the left of .) The second case solves to . We thenhave , hence .Solution 2The horizontal line is the equivalent of a circle of curvature , thus we can applyDescartes' Circle Formula.The four circles have curvatures , and .We haveSimplifying, we getObviously cannot equal , therefore .Problem 19The polynomial has the property that the mean of its zeros, the product of its zeros, and the sum of its coefficients are all equal. If the-intercept of the graph of is 2, what is ?SolutionWe are given . So the product of the roots is by Vieta's formulas.These also tell us that is the average of the zeros, so . We are also given that the sum of the coefficients is , so. So the answer is .Problem 20Points , , , and lie in the first quadrant and are the vertices of quadrilateral . The quadrilateral formed by joiningthe midpoints of , , , and is a square. What is the sum of the coordinates of point ?SolutionWe already know two vertices of the square: and.There are only two possibilities for the other vertices of the square: either they areand , or they are and . The second case would give usoutside the first quadrant, hence the first case is the correct one. As is themidpoint of , we can compute , and .Problem 21Four positive integers , , , and have a product of and satisfy:What is ?SolutionWe can rewrite the three equations as follows:Let . We get:Clearly divides . On the other hand, can not divide , as it then woulddivide . Similarly, can not divide . Hence divides both and . This leavesus with only two cases: and .The first case solves to , which gives us, but then . (We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by .)The second case solves to , which gives us a validquadruple , and we have . Problem 22In rectangle , points and lie on so that and isthe midpoint of . Also, intersects at and at . The area of therectangle is . Find the area of triangle .SolutionSolution 1Note that the triangles and are similar, as they have the same angles.Hence .Also, triangles and are similar, hence .We can now compute as . We have:▪.▪is of , as these two triangles have the same base , andis of , therefore also the height from onto is of the heightfrom . Hence .▪is of , as the base is of the base , and theheight from is of the height from . Hence .▪is of for similar reasons, hence .Therefore. Solution 2As in the previous solution, we note the similar triangles and prove that is inand in of .We can then compute that .As is the midpoint of , the height from onto is of the height fromonto . Therefore we have .Problem 23A polynomial of degree four with leading coefficient 1 and integer coefficients hastwo zeros, both of which are integers. Which of the following can also be a zero of the polynomial?SolutionLet the polynomial be and let the two integer zeros be and . We can thenwrite for some integers and .If a complex number with is a root of , it must be the root of, and the other root of must be .We can then write.We can now examine each of the five given complex numbers, and find the one forwhich the values and are integers. This is , for which wehave and.(As an example, the polynomial has zeroes , , and.)Problem 24In , . Point is on so that and. Find .SolutionWe start with the observation that , and.We can draw the height from onto . In the triangle , we have. Hence .By the definition of , we also have , therefore . Thismeans that the triangle is isosceles, and as , we must have.Then we compute , thus and thetriangle is isosceles as well. Hence .Now we can note that , hence also the triangle is isosceles and we have .Combining the previous two observations we get that , and as, this means that .Finally, we get .Problem 25Consider sequences of positive real numbers of the form in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of does the term 2001 appear somewhere in the sequence?SolutionIt never hurts to compute a few terms of the sequence in order to get a feel how itlooks like. In our case, the definition is that . Thiscan be rewritten as . We have and , and we compute:At this point we see that the sequence will become periodic: we have , , and each subsequent term is uniquely determined by the previous two.Hence if appears, it has to be one of to . As , we only have fourpossibilities left. Clearly for , and for . Theequation solves to , and the equation to .No two values of we just computed are equal, and therefore there are different values of for which the sequence contains the value .。
2005 AMC 12B Problems and Solution
2005 AMC 12B Problems and Solution Problem 1A scout troop buys candy bars at a price of five for dollars. They sell all thecandy bars at the price of two for dollar. What was their profit, in dollars?SolutionProblem 2A positive number has the property that of is . What is ?SolutionProblem 3Brianna is using part of the m oney she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her m oney to buy one third of the CDs. What fraction of her m oney will she have left after she buys all the CDs?SolutionLet Brianna's money. We have . Thus, themoney left over is , so the answer is .Problem 4At the beginning of the school year, Lisa's goal was to earn an A on at least ofher quizzes for the year. She earned an A on of the first quizzes. If she is toachieve her goal, on at m ost how many of the rem aining quizzes can she earn a grade lower than an A?SolutionLisa's goal was to get an A on quizzes. She already has A's onquizzes, so she needs to get A's on more. There arequizzes left, so she can afford to get less than an A on of them.Problem 5An -foot by -foot floor is tiles with square tiles of size foot by foot. Each tilehas a pattern consisting of four white quarter circles of radius foot centered at each corner of the tile. The rem aining portion of the tile is shaded. How m any square feet of the floor are shaded?SolutionThere are 80 tiles. Each tile has shaded. Thus:Problem 6In , we have and . Suppose that is a point online such that lies between and and . What is ?SolutionDraw height . We have that . From the Pythagorean Theorem,. Since , , and ,so .Problem 7What is the area enclosed by the graph of ?SolutionIf we get rid of the absolute values, we are left with the following 4 equations (usingthe logic that if , then is either or ):We can then put these equations in slope-intercept form in order to graph them.Now you can graph the lines to find the shape of the graph:We can easily see that it is a rhom bus with diagonals of and . The area is, orProblem 8For how m any values of is it true that the line passes through thevertex of the parabola ?We see that the vertex of the quadratic function is . They-intercept of the line is . We want to find the values (if any) suchthat . Solving for , the only values that satisfy this are and , so theanswer isProblem 9On a certain m ath exam, of the students got points, got points,got points, got points, and the rest got points. What is the differencebetween the m ean and the m edian score on this exam?SolutionTo begin, we see that the rem aining of the students got points. The smallestcase is that there are students, because , and. We see that students got points, students got points,students got points, students got points, and students got points. Themedian is , since the and term s in an ordered sequence are both .The m ean is . Thedifference between the m ean and m edian, therefore, is .Problem 10The first term of a sequence is . Each succeeding term is the sum of the cubesof the digits of the previous term s. What is the term of the sequence?Performing this operation several times yields the results of for the second term,for the third term, and for the fourth term. The sum of the cubes of the digitsof equal , a com plete cycle. The cycle is... excluding the first term, the ,, and term s will equal , , and , following the fourth term. Any termnumber that is equivalent to will produce a result of . It just sohappens that , which leads us to the answer of . Problem 11An envelope contains eight bills: ones, fives, tens, and twenties. Two bills aredrawn at random without replacem ent. What is the probability that their sum is $or more?SolutionThe only way to get a total of $or m ore is if you pick a twenty and another bill,or if you pick both tens. There are a total of ways to choosebills out of . There are ways to choose a twenty and som e other non-twenty bill.There is way to choose both twenties, and also way to choose both tens. Addingthese up, we find that there are a total of ways to attain a sum of or greater,so there is a total probability of .Problem 12The quadratic equation has roots twice those of , andnone of and is zero. What is the value of ?Let have roots and . Thenso and . Also, has roots and , soand and . Thus .Indeed, consider the quadratics . Problem 13Suppose that , , , ... , . What is?SolutionWe see that we can re-write , , , ... , asby using substitution. By using the properties ofexponents, we know that .Therefore, the answer isProblem 14A circle having center , with ,is tangent to the lines , and. What is the radius of this circle?SolutionLet be the radius of the circle. Draw the two radii that m eet the points of tangencyto the lines . We can see that a square is form ed by the origin, two tangency points, and the center of the circle. The side lengths of this square are and thediagonal is . The diagonal of a square is tim es the side length.Therefore, .Problem 15The sum of four two-digit numbers is . None of the eight digits is and no two ofthem are the sam e. Which of the following is not included among the eight digits?Solutioncan be written as the sum of eight two-digit numbers, let's say , , , and. Then . The last digit of is ,and won't affect the units digits, so must endwith . The sm allest value can have is , andthe greatest value is . Therefore, must equalor .Case 1:The only distinct positive integers that can add up to is . So,,,, and must include four of the five numbers . We have, or . We can add all of, and try subtracting one number to get to , but to noavail. Therefore, cannot add up to .Case 2:Checking all the values for ,,,and each individually may be tim e-consuming,instead of only having solution like Case 1. We can try a different approach bylooking at first. If ,, or . That m eans. We know, so the m issing digit isProblem 16Eight spheres of radius 1, one per octant, are each tangent to the coordinate planes. What is the radius of the sm allest sphere, centered at the origin, that contains these eight spheres?SolutionThe eight spheres are formed by shifting spheres of radius and centerin the directions. Hence, the centers of the spheres are . For asphere centered at the origin to contain all eight spheres, its radius must be greater than or equal to the longest distance from the origin to one of these spheres. Thislength is the sum of the distance from to the origin and the radius ofthe spheres, or . To verify this is the longest length, we can see from thetriangle inequality that the length from the origin to any other point on the spheresis strictly sm aller. Thus, the answer is .Problem 17How many distinct four-tuples of rational numbers are there with?SolutionUsing the laws of logarithms, the given equation becom esAs must all be rational, and there are no powers of or in ,. Then .Only the four-tuple satisfies the equation, so the answer is.Problem 18Let and be points in the plane. Define as the region in the firstquadrant consisting of those points such that is an acute triangle. What isthe closest integer to the area of the region ?SolutionFor angle and to be acute, must be between the two lines that areperpendicular to and contain points and . For angle to be acute, firstdraw a triangle with as the hypotenuse. Note cannot be insidethis triangle's circum scribed circle or else . Hence, the area of is thearea of the large triangle minus the area of the sm all triangle minus the area of thecircle, which is , which is approximately .The answer is .Problem 19Let and be two-digit integers such that is obtained by reversing the digits of .The integers and satisfy for som e positive integer . What is?Solutionlet , then where and are nonzero digits.By difference of squares,For this product to be a square, the factor of must be repeated in either or, and given the constraints it has to be . The factor of isalready a square and can be ignored. Now must be another square, andsince cannot be or greater then must equal or . If then, , , which is not a digit. Hence theonly possible value for is . Now we have ,, , then , , , , andProblem 20Let and be distinct elem ents in the setWhat is the minimum possible value ofSolutionThe sum of the set is , so if we could havethe sum in each set of parenthesis be then the minimum value would be. Considering the set of four term s containing , this sum could only be even if it had two or four odd term s. If it had all four odd terms then it would be, and with two odd term s then its minimum value is, so we cannot achieve two sum s of . The closest we couldhave to and is and , which can be achieved through and. So the m inimum possible value is .Problem 21A positive integer has divisors and has divisors. What is the greatestinteger such that divides ?SolutionIf has factors, then is a product of powers of (not necessarilydistinct) primes. When multiplied by , the am ount of factors of increased by, so there are possible powers of in the factorization of , andpossible powers of in the factorization of , which would be , , and .Therefore the highest power of that could divide is .Problem 22A sequence of com plex numbers is defined by the rulewhere is the complex conjugate of and . Suppose that and. How m any possible values are there for ?SolutionSince , let , where is an argument of . I will prove byinduction that , where .Base Case: trivialInductive Step: Suppose the formula is correct for , thenSincethe form ula is proven, where is an integer. Therefore,The value of only m atters m odulo. Since, k only needs to take values from 0 to ,so the answer isProblem 23Let be the set of ordered triples of real numbers for whichThere are real numbers andsuch that for all ordered triples in we haveWhat is the value ofSolutionCall and . Then, we note that which implies that. Therefore, . Let us note that. Since , we find that. Thus, . is the answer. Problem 24All three vertices of an equilateral triangle are on the parabola , and one of its sides has a slope of . The -coordinates of the three vertices have a sum of, where and are relatively prime positive integers. What is the value of?Let the points be , and . Using elementary calculus concepts andthe fact that they lie on ,= the slope of ,= the slope of ,= the slope of .So the value that we need to find is sim ply the sum of the slopes of the three sides of the triangle divided by . WLOG, let be the side that has the sm allest anglewith the positive -axis. Let be an arbitrary point with the coordinates . Letus translate the triangle so is at the origin. Then . Using the factthat the slope of a line is equal to the tangent of the angle form ed by the line and the positive x- axis, the value we now need to find is simply.Using , and basic trigonometric identities, this simplifies to , sothe answer isProblem 25Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Simultaneously and independently, each ant m oves from its vertex to one of the four adjacent verti ces, each with equal probability. What is the probability that no two ants arrive at the sam e vertex?Solution 1We approach this problem by counting the number of ways ants can do their desired migration, and then multiple this number by the probability that each case occurs.Let the octahedron be , with points coplanar. Then the antfrom and the ant from must m ove to plane . Suppose, without loss ofgenerality, that the ant from moved to point . Then, we m ust consider threecases.▪Case 1: Ant from point moved to pointOn the plane, points and are taken. The ant that moves to D can com e fromeither or . The ant that m oves to can com e from either or . Once thesetwo ants are fixed, the other two ants must migrate to the "poles" of the octahedron, points and . Thus, there are two degrees of freedom in deciding which antmoves to , two degrees of freedom in deciding which ant m oves to , and twodegrees of freedom in deciding which ant m oves to . Hence, there areways the ants can m ove to different points.▪Case 2: Ant from point moved to pointOn the plane, points and are taken. The ant that m oves to C m ust be from or, but the ant that m oves to must also be from or . The other two ants,originating from points and , m ust move to the poles. Therefore, there are twodegrees of freedom in deciding which ant m oves to and two degrees of freedom inchoosing which ant m oves to . Hence, there are ways the ants canmove to different points.▪Case 3: Ant from point moved to pointBy symm etry to Case 1, there are ways the ants can move to different points.Given a point , there is a total of ways the ants can m ove todifferent points. We oriented the square so that point was defined as the point towhich the ant from point moved. Since the ant from point can actually move tofour different points, there is a total of ways the ants can m ove to different points.Each ant acts independently, having four different points to choose from. Hence,each ant has probability of m oving to the desired location. Since there are sixants, the probability of each case occuring is . Thus, the desired answeris .Solution 2Let be the num ber of cycles of length the can be walked am ong the verticesof an octahedron. For example, would represent the num ber of ways in which an ant could navigate vertices and then return back to the original spot. Since anant cannot stay still, . We also easily see that .Now consider any four vertices of the octahedron. All four vertices will be connected by edges except for one pair. Let’s think of this as a square with one diagonal (from top left to bottom right).Suppose an ant m oved across this diagonal; then the ant at the other end can only move across the diagonal (which creates 2-cycle, bad) or it can m ove to another vertex, but then the ant at that vertex must m ove to the spot of the original ant (which creates 3-cycle, bad). Thus none of the ants can navigate the diagonal andcan either shift clockwise or counterclockwise, and so .For , consider an ant at the top of the octahedron. It has four choices.Afterwards, it can either travel directly to the bottom, and then it has ways backup, or it can travel along the sides and then go to the bottom, of which simplecounting gives us ways back up. Hence, this totals .Now, the number of possible ways is given by the sum of all possible cycles,where the coefficients represent the number of ways we can configure these cycles.To find , fix any face, there are adjacent faces to select from to com plete the cycle. From the four remaining faces there are only ways to create cycles, hence.To find , each cycle of faces is distinguished by their common edge, and there areedges, so .To find , each three-cycle is distinguished by the vertex, and there are edges.However, since the two three-cycles are indistinguishable, .Clearly . Finally,Each bug has possibilities to choose from, so the probability is .。
2010 AMC 12B Problems and Solution
2010 AMC 12B Problems and Solution Problem 1Makarla attended two meetings during her -hour work day. The first meeting tookminutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?SolutionThe total number of minutes in here -hour work day is . The totalamount of time spend in meetings in minutes is . The answer isthenProblem 2A big is formed as shown. What is its area?SolutionWe find the area of the big rectangle to be , and the area of the smallerrectangle to be . The answer is then .Problem 3A ticket to a school play cost dollars, where is a whole number. A group of 9th graders buys tickets costing a total of $, and a group of 10graders buys ticketscosting a total of $. How many values for are possible?SolutionWe find the greatest common factor of and to be . The number of factors ofis which is the answer .Problem 4A month with days has the same number of Mondays and Wednesdays.Howmany of the seven days of the week could be the first day of this month?Solutionso the week cannot start with Saturday, Sunday, Tuesday or Wednesday as that would result in an unequal number of Mondays and Wednesdays.Therefore, Monday, Thursday, and Friday are valid so the answer is . Problem 5Lucky Larry's teacher asked him to substitute numbers for , , , , and in theexpression and evaluate the result. Larry ignored the parenthese but added and subtracted correctly and obtained the correct result by coincidence. The number Larry sustitued for , , , and were , , , and ,respectively. What number did Larry substitude for ?SolutionWe simply plug in the numbersProblem 6At the beginning of the school year, of all students in Mr. Wells' math classanswered "Yes" to the question "Do you love math", and answered "No." At theend of the school year, answered "Yes" and answerws "No." Altogether,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 ?SolutionClearly, the minimum possible value would be . The maximumpossible value would be . The difference is .Problem 7Shelby drives her scooter at a speed of miles per hour if it is not raining, andmiles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of miles in minutes. How many minutes did shedrive in the rain?SolutionLet be the time it is not raining, and be the time it is raining, in hours.We have the system: andSolving gives andWe want in minutes,Problem 8Every high school in the city of Euclid sent a team of students to a math contest.Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed th and th, respectively. How many schoolsare in the city?SolutionThere are schools. This means that there are people. Because no one's scorewas the same as another person's score, that means that there could only have been median score. This implies that is an odd number. cannot be less than ,because there wouldn't be a th place if there were. cannot be greater thaneither, because that would tie Andrea and Beth or Andrea's place would be worsethan Beth's. Thus, the only possible answer is .Problem 9Let be the smallest positive integer such that is divisible by , is a perfectcube, and is a perfect square. What is the number of digits of ?SolutionWe know that and . Cubing and squaring the equalitiesrespectively gives . Let . Now we know must be a perfect-th power because , which means that must be a perfect -thpower. The smallest number whose sixth power is a multiple of is , becausethe only prime factors of are and , and . Therefore our is equal tonumber , with digits .Problem 10The average of the numbers and is . What is ?SolutionWe first sum the first numbers: . Then, we knowthat the sum of the series is . Since the average is , and there areterms, we also find the sum to equal . Setting equal -. Thus, theanswer is .Problem 11A palindrome between and is chosen at random. What is the probabilitythat it is divisible by ?SolutionView the palindrome as some number with form (decimal representation):. But because the number is a palindrome,. Recombining this yields . 1001 is divisible by 7,which means that as long as , the palindrome will be divisible by 7. This yields 9 palindromes out of 90 () possibilities for palindromes. However, if, then this gives another case in which the palindrome is divisible by 7. Thisadds another 9 palindromes to the list, bringing our total toProblem 12For what value of doesSolutionProblem 13In , and . What is ?SolutionWe note that and . Therefore, there is no other way tosatisfy this equation other than making both and, since any other way would cause one of these values to become greater than 1, which contradicts our previous statement. From this we can easilyconclude that and and solving this system gives usand . It is clear that is a triangle with.Problem 14Let , , , , and be postive integers with and letbe the largest of the sum , , and . What is the smallestpossible value of ?SolutionWe want to try make , , , and as close as possible so that ,the maximum of these, if smallest.Notice that . In order to express as a sum ofnumbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as close as possible):or . We seethat in both cases, the value of is , so the answer is .Problem 15For how many ordered triples of nonnegative integers less than are thereexactly two distinct elements in the set , where ?SolutionWe have either , , or .For , this only occurs at . has only one solution, namely,. has five solutions between zero and nineteen,, and . has nineteen integer solutionsbetween zero and nineteen. So for , we haveordered triples.For , again this only occurs at . has nineteensolutions,has five solutions, and has one solution, so again we haveordered triples.For, this occurs at and . and both haveone solution while has fifteen solutions. andboth haveone solution, namely, and, while has twenty solutions. So wehaveordered triples.In total we haveordered triplesProblem 16Positive integers , , and are randomly and independently selected withreplacement from the set. What is the probability thatis divisible by ?SolutionThe value ofis arbitrary other than it is divisible by , so the setcan be grouped into threes.Obviously, if is divisible by (which has probability ) then the sum is divisible by. In the event that is not divisible by (which has probability , then the sum isdivisible by if, which is the same as.This only occurs when one of the factors or is equivalent toandthe other is equivalent to. All four events,, , and have a probability ofbecause the set is grouped in threes.In total the probability isProblem 17The entries in a array include all the digits from through , arranged so thatthe entries in every row and column are in increasing order. How many such arraysare there?SolutionThe first 4 numbers will form one of 3 tetris "shapes".First, let's look at the numbers that form a 2x2 block, sometimes called tetris:Second, let's look at the numbers that form a vertical "L", sometimes called tetris :Third, let's look at the numbers that form a horizontal "L", sometimes called tetris :Now, the numbers 6-9 will form similar shapes (rotated by 180 degrees, and anchored in the lower-right corner of the 3x3 grid).If you match up one tetris shape from the numbers 1-4 and one tetris shape from the numbers 6-9, there is only one place left for the number 5 to be placed.So what shapes will physically fit in the 3x3 grid, together?The answer is .Problem 18A frog makes jumps, each exactly meter long. The directions of the jumps arechosen independenly at random. What is the probability that the frog's final position is no more than meter from its starting position?SolutionSolution 1We will let the moves be complex numbers , , and , each of magnitude one. Thefrog starts on the origin. It is relatively easy to show that exactly one element in theset has magnitude less than orequal to . Hence, the probability is .Solution 2The first frog hop doesn't matter because no matter where the frog hops is lands on the border of the circle you want it to end in. The remaining places that the frog can jump to form a disk of radius 2 centered at the spot on which the frog first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps. The entirety of the circle you want the frog to land in is enclosed in this larger disk,so find the ratio of the two areas, which is .Problem 19A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than points. What was the total number of pointsscored by the two teams in the first half?SolutionLet be the quarterly scores for the Raiders. We know that The Raiders and Wildcats both scored the same number of points in the first quarter so letbe quarterly scores for the Raiders. The sum of the Raidersscores is and the sum of the Wildcats scores is . Now wecan narrow our search for the values of , and . Because points are always measured in positive integers, we can conclude that and are positive integers.We can also conclude that is a positive integer by writing down the equation:Because both the left hand and right hand sides must be integers, we have thatis a positive integer and due to our restrictions, it follows that must be a positive integer as well. Now we can start trying out some values of . We start atwhich gives that which is a contradiction. Next we try , whichgivesWe need the smallest multiple of (to satisfy the <100 condition) that is. We see that this is , and therefore and .So the Raiders first two scores were and and the Wildcats first two scores wereand .Problem 20A geometric sequence has , , and for somereal number . For what value of does ?SolutionBy defintion of a geometric sequence, we have . Since, we can rewrite this as .The common ratio of the sequence is , so we can writeSince , we have, whichis making our answer .Problem 21Let , and let be a polynomial with integer coefficients such that, and.What is the smallest possible value of ?SolutionThere must be some polynomial such thatThen, plugging in values of we getThus, the least value of mustbe the . Solving, we receive , so our answer is .Problem 22Let be a cyclic quadralateral. The side lengths of are distinctintegers less than such that . What is the largest possiblevalue of ?SolutionLet , , , and . We see that by the Law of Cosineson , we have . Also,. Now, we know that . Also, becauseis a cyclic quadrilateral, we must have that , so. Therefore, .Now, adding, we have .We now look at the equation . Suppose that . Then, we must haveeither or equal . Suppose that . We let and .Now, , so it is or . Problem 23Monic quadratic polynomial and have the property that haszeros at and , and has zeros atand . What is the sum of the minimum values of and?Solution. Notice that has roots ,so that the roots of are the roots of . For each individual equation, the sum of the roots will be (symmetry or Vieta's). Thus, wehave , or . Doing something similar forgives us . We now have. Since is monic, the roots ofare "farther" from the axis of symmetry than the roots of. Thus, we have , or. Adding these gives us ,or . Plugging this into , we get . The minimumvalue of is , and the minimum value of is . Thus, our answer is, or answer .Problem 24The set of real numbers for whichis the union of intervals of the form . What is the sum of the lengths of these intervals?SolutionBecause the right side of the inequality is a horizontal line, the left side can be translated horizontally by any value and the intervals will remain the same. Forsimplicity of calculation, we will find the intervals whereWe shall say that . has three vertical asymptotesat . As the sum of decreasing hyperbolas, the function is decreasing at all intervals. Values immediately to the left of each asymptote approach negative infinity, and values immediately to the right of each asymptote approach positiveinfinity. In addition, the function has a horizontal asymptote at . The function intersects at some point from to , and at some point from to, and at some point to the right of . The intervals where the function isgreater than are between the points where the function equals and the verticalasymptotes.If , , and are values of x where , then the sum of the lengths of theintervals is .And now our job is simply to find the sum of the roots of .soand the sum is the negative of the coefficient on the term, which is.For the AMC, one may note that the transformed inequality should not yield solutions that involve big numbers like 67 or 134, and immediately choose .Problem 25For every integer , let be the largest power of the largest prime thatdivides . For example . What is the largest integersuch that divides?SolutionBecause 67 is the largest prime factor of 2010, it means that in the primefactorization of , there'll be where is the desired value we are looking for. Thus, to find this answer, we need to look for the number of times 67 would be incorporated into the giant product. Any number of the formwould fit this form. However, this number tops at because 71 is ahigher prime than 67. itself must be counted twice because it's counted twice as a squared number. Any non-prime number that's less than 79 (and greater than 71) can be counted, and this totals 5. We have numbers (as 71 isn't counted - 1through 70), an additional (), and values just greater than but less than(72, 74, 75, 76, 77, and 78). Thus,Similar SolutionAfter finding the prime factorization of , divide by and adddivided by in order to find the total number of multiples of betweenand . Since ,, and are prime numbers greaterthan and less than or equal to , subtract from to get the answer.。
2010 AMC 12A Problems and Solution
2010 AMC 12A Problems and Solution Problem 1What 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 peopletaken to the island isProblem 3Rectangle , pictured below, shares of its area with square .Square shares of its area with rectangle . What is ?SolutionSolution 1Let , let , and let .Solution 2The answer does not change if we shift to coincide with , and add newhorizontal lines to divide into five equal parts:This helps us to see that and , where . Hence.Problem 4If , then which of the following must be positive?Solutionis negative, so we can just place a negative value into each expression and find the one that is positive. Suppose we use .Obviously only is positive.Problem 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 , such as 83438, is a number that remains the same when its digits are reversed. The numbers and are three-digit and four-digit palindromes,respectively. What is the sum of the digits of ?Solutionis at most , so is at most . The minimum value of is .However, the only palindrome between and is , which means thatmust be .It follows that is , so the sum of the digits is .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 times more water than Logan'sminiature. Therefore, Logan should make his tower times shorterthan the actual tower. This is meters high, or choice .Also, the fact that doesn't matter since only the ratios are important.Problem 8Triangle has . Let and be on and , respectively,such that . Let be the intersection of segments and ,and suppose that is equilateral. What is ?SolutionLet .Since , triangle is a triangle, soProblem 9A solid cube has side length inches. A -inch by -inch square hole is cut into thecenter 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 . Thesecond cut removes two boxes, each of dimensions , and the third cutdoes 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 cubic inches. However, we can not just sum theirvolumes, as the central cube is included in each of these three cuts. Toget the correct 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 abox..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 , , , and . Whatis the term of this sequence?Solutionand are consecutive terms, so the common difference is.The common difference is . The first term is and the term isProblem 11The solution of the equation can be expressed in the form . 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. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.Brian: "Mike and I are different species."Chris: "LeRoy is a frog."LeRoy: "Chris is a frog."Mike: "Of the four of us, at least two are toads."How many of these amphibians are frogs?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 exactly frog out of the two of them.Now suppose Mike is a toad. Then what he says is true because we already havetoads. 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 bea 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 havefrogs 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 and notintersect?SolutionThe image below shows the two curves for . The blue curve is ,which is clearly a circle with radius , and the red curve is a part of the curve.In the special case the blue curve is just the point , and as ,this point is on the red curve as well, hence they intersect.The case is symmetric to : 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 is always closest to the origin, and skip the rest of this solution.For an exact solution, fix and consider any point on the red curve. Itsdistance from the origin is . To minimize this distance, it is enough tominimize . By the Arithmetic Mean-Geometric Mean Inequality we getthat this value is at least , and that equality holds whenever , 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 originand the closest point on the red curve is . Therefore, we want to find all positiveintegers such that .Clearly the only such integer is , hence the two curves are only disjoint forand . This is a total of values.Solution 2:From the graph shown above, we see that there is a specific point closest to the center of the circle. Using some logic, we realize that as long as said furthest point is not inside or on the graph of the circle. This should be enough to conclude that the hyperbola does not intersect the circle.Therefore, for each value of k, we only need to check said value to determineintersection. Let said point, closest to the circle have coordinates derivedfrom the equation. Then, all coordinates that satisfyintersect the circle. Squaring, we findAfter multiplying through by and rearranging, wefind . We see this is a quadratic in and consider taking the determinant, which tells us that solutions are real when, after factoring:We plot this inequality on the number line to find it issatisfied for all values except: . We then eliminate 0 because it isextraneous as both and are points which coincide. Therefore,there are a total of values.Problem 14Nondegenerate has integer side lengths, is an angle bisector, ,and . What is the smallest possible value of the perimeter?SolutionBy the Angle Bisector Theorem, we know that . If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then , contradicting the Triangle Inequality. If weuse the next lowest values (and ), the Triangle Inequality issatisfied. 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 probaiblity of an equal number of heads and tailsis . What is the probability that the coin lands on heads?SolutionLet 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 ittimes the product of the probability of flipping each coin.As for the desired probability both and are nonnegative, we only need toconsider the positive root, henceApplying the quadratic formula we get that the roots of this equation are .As the probability of heads is less than , we get that the answer is . Problem 16Bernardo randomly picks 3 distinct numbers from the setand arranges them in descending order to form a 3-digit number. Silvia randomlypicks 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 theprobabilities.Case : Bernardo picks . If Bernardo picks a then it is guaranteed that hisnumber 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 , theprobability of not picking is .If Bernardo does not pick 9, then he can pick any number from to . SinceBernardo 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 thenumber 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 has side lengths and. The area of is of the area of the hexagon.What is the sum of all possible values of ?SolutionIt is clear that is an equilateral triangle. From the Law of Cosines, we getthat . Therefore, the area of is.If we extend , and so that and meet at , and meetat , and and meet at , we find that hexagon is formed bytaking equilateral triangle of side length and removing three equilateraltriangles, , and , of side length . The area of istherefore.Based on the initial conditions,Simplifying this gives us . By Vieta's Formulas we know that thesum of the possible value of is .Problem 18A 16-step path is to go from to with each step increasing either the -coordinate or the -coordinate by 1. How many such paths stay outside or on theboundary of the square , at each step?SolutionBrute Force SolutionThe number of ways to reach any point on the grid is equal to the number ofways to reach plus the number of ways to reach . Using this recursion, 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 the first point when it reaches . Clearlyand the previous point on such path has to be .Fix the value of . There are ways how the path can go from to, and then there are ways how the path can go from to .Hence for we get paths, for we getpaths, and for we getpaths. This gives us paths through the second quadrant, hence the total number of paths is .Combinatorial Solution 2Each path that goes through the second quadrant must pass through exactly one ofthe points , , and .There is exactly path of the first kind, paths of the second kind, andpaths of the third type. The conclusion remains the same. Problem 19Each of 2010 boxes in a line contains a single red marble, and for , the box in the position also contains white marbles. Isabella begins at the first boxand successively draws a single marble at random from each box, in order. Shestops when she first draws a red marble. Let be the probability that Isabella stops after drawing exactly marbles. What is the smallest value of for which?SolutionThe probability of drawing a white marble from box is . The probability ofdrawing 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 inequalityis .Problem 20Arithmetic sequences and have integer terms withand for some . What is the largest possible value of ?SolutionSolution 1Since and have integer terms with , we can write the terms of each sequence aswhere and () are the common differences of each, respectively.Sinceit is easy to see that.Hence, we have to find the largest such that and are both integers. The prime factorization of is . We list out all the possible pairs thathave a product ofand soon find that the largest value is for the pair , and so thelargest value is .Solution 2As above, let and for some .Now we get , hence. Therefore divides . Andas the second term is greater than the first one, we only have to consider theoptions .For we easily see that for the right side is less than and for anyother it is way too large.For we are looking for such that . Notethat must be divisible by . We can start looking for the solution by trying thepossible values for , and we easily discover that for we get, which has a suitable solution .Hence is the largest possible . (There is no need to check anymore.)Problem 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?SolutionThe values in which intersect atare the same as the zeros of .Since there are zeros and the function is never negative, all zeros must bedouble 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 expandout the terms of .[Quick note: Since we don't know , , and , we really don't even need the last 3terms 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 , inclusive.The 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 . The zeros occur at and , whichmeans the slope is where .We can now verify that both and yield .Solution 2Rewrite the given expression as follows:Imagine the real line. For eachimagine that there are 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'slabel 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 boys such that shemoves away from this boy. For each of the remaining boys, she moves at mostcloser, 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 : by extending the previous argument, anywhere on thissegment the sum of distances is the same.By trial and error, or by solving the quadratic equation we getthat boy number is the last boy placed at and the next boy is the one placedat . Hence the given expression is minimized for any . Common part of both solutionsTo find the minimum, we want to balance the expression so that it is neither top norbottom heavy. .Now that we know that the sum of the first 84 's is equivalent to the sum of 's 85to 119, we can plug either or to find the minimum.Note that the terms to are negative, and the terms toare positive. Hence we get: andHence the total sum of distances is.Solution 3Given that the minimum exists, we know that we want all s to cancel out. Thus, we want to find some such that. It thereforefollows that . This gives , from which it follows that .There are therefore positive ones and , or negative ones, so.Problem 23The number obtained from the last two nonzero digits of is equal to . What is?SolutionWe will use the fact that for any integer ,First, we find that the number of factors of in is equal to. Let . The we want is therefore the lasttwo digits of , or . If instead we find , we know that, what we are looking for, could be , ,, or . Only one of these numbers will be a multiple of four, and whichever one that is will be the answer, becausehas to be a multiple of 4.If we divide by by taking out all the factors of in , we can write aswhere where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form is replaced by , and every number in the form is replaced by .The number can be grouped as follows:Where the first line is composed of the numbers in that aren't multiples of five,the second line is the multiples of five and not 25 after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25.Using the identity at the beginning of the solution, we can reduce toUsing the fact that (or simply the fact thatif you have your powers of 2 memorized), we can deduce that. Therefore .Finally, combining with the fact that yields . Problem 24Let . The intersection ofthe domain of with the interval is a union of 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 factors are inside a logarithm, the function isundefined 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 zeros and intervals.In 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 obviously positive. This means the next interval 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 issymmetric, 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?Solution 1It 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 lossof 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 howthe 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 fromto except for (because that is counted in the first case). Obviously there is stillonly unique quadrilateral that can be formed from one set of side lengths, resultingin a total of quadrilaterals.Case : (pairs of side lengths are equal)and can be any integer from to , and likewise and can be any integer fromto . 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 areequal)If the equal side lengths are each , then the other sides must each be , whichwe 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, so the total number ofquadrilaterals for this case is .Case : (no side lengths are equal) Using the same countingprinciples starting from and eventually reaching , we find that thetotal number of possible side lengths is . There are ways to arrange the sidelengths, but there is only unique quadrilateral for rotations, so the number ofquadrilaterals for each set of side lengths is . The total number of quadrilaterals is .And so, the total number of quadrilaterals that can be made is.Solution 2As with solution we would like to note that given any quadrilateral we can changeits angles to make a cyclic one.Let be the sides of the quadrilateral.There are ways to partition . However, some of these will not bequadrilaterals since they would have one side bigger than the sum of the other three.This occurs when . For , . There are ways topartition . Since could be any of the four sides, we have counteddegenerate quadrilaterals. Similarly, there are , for other values of . Thus, there arenon-degenerate partitions of by the hockey stick theorem. However, foror , each quadrilateral is counted times, for eachrotation. Also, for , each quadrilateral is counted twice. Since thereis quadrilateral for which , and for which , thereare quads for which or .Thus there are total quadrilaterals.。
AMC的一些材料和题目08amc12
2008A 1A bakery owner turns on his doughnut machine at 8:30AM.At 11:10AM the machine hascompleted one third of the day’s job.At what time will the doughnut machine complete thejob?(A)1:50PM (B)3:00PM(C)3:30PM (D)4:30PM (E)5:50PM 2What is the reciprocal of 1223?(A)67(B)76(C)53(D)3(E)723Suppose that 23of 10bananas are worth as much as 8oranges.How many oranges are worth as much is 12of 5bananas?(A)2(B)52(C)3(D)72(E)44Which of the following is equal to the product 84.128.1612...4n 44n (20082004)?(A)251(B)502(C)1004(D)2008(E)40165Suppose that2x 3x6is an integer.Which of the following statements must be true about x ?(A)It is negative.(B)It is even,but not necessarily a multiple of 3.(C)It is a multiple of 3,but not necessarily even.(D)It is a multiple of 6,but not necessarily a multiple of 12.(E)It is a multiple of 12.6Heather compares the price of a new computer at two different stores.Store A offers 15%offthe sticker price followed by a $90rebate,and store B offers 25%offthe same sticker pricewith no rebate.Heather saves $15by buying the computer at store A instead of store B.What is the sticker price of the computer,in dollars?(A)750(B)900(C)1000(D)1050(E)1500This file was downloaded from the AoPS −MathLinks Math Olympiad Resources PagePage 1http://www.mathlinks.ro/20087While Steve and LeRoy are fishing 1mile from shore,their boat springs a leak,and watercomes in at a constant rate of 10gallons per minute.The boat will sink if it takes in morethan 30gallons of water.Steve starts rowing toward the shore at a constant rate of 4milesper hour while LeRoy bails water out of the boat.What is the slowest rate,in gallons perminute,at which LeRoy can bail if they are to reach the shore without sinking?(A)2(B)4(C)6(D)8(E)108What is the volume of a cube whose surface area is twice that of a cube with volume 1?(A)√2(B)2(C)2√2(D)4(E)89Older television screens have an aspect ratio of 4:3.That is,the ratio of the width to theheight is 4:3.The aspect ratio of many movies is not 4:3,so they are sometimes shown ona television screen by ’letterboxing’-darkening strips of equal height at the top and bottomof the screen,as shown.Suppose a movie has an aspect ratio of 2:1and is shown on an oldertelevision screen with a 27-inch diagonal.What is the height,in inches,of each darkenedstrip?[asy]unitsize(1mm);filldraw((0,0)–(21.6,0)–(21.6,2.7)–(0,2.7)–cycle,grey,black);filldraw((0,13.5)–(21.6,13.5)–(21.6,16.2)–(0,16.2)–cycle,grey,black);draw((0,0)–(21.6,0)–(21.6,16.2)–(0,16.2)–cycle);[/asy](A)2(B)2.25(C)2.5(D)2.7(E)310Doug can paint a room in 5hours.Dave can paint the same room in 7hours.Doug andDave 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 ofthe following equations is satisfied by t ?(A) 1517 (t 1)1(B) 1517 t 11(C) 1517 t 1(D) 1517(t 1)1(E)(57)t 111Three cubes are each formed from the pattern shown.They are then stacked on a table oneon top of another so that the 13visible numbers have the greatest possible sum.What isthat sum?[asy]unitsize(.8cm);pen p =linewidth(1);draw(shift(-2,0)*unitsquare,p);label(quot;1quot;,(-1.5,0.5));draw(shift(-1,0)*unitsquare,p);label(quot;2quot;,(-0.5,0.5));draw(unitsquare,p);label(quot;32quot;,(0.5,0.5));draw(shift(1,0)*unitsquare,p);label(quot;16quot;,(1.5,0.5));draw(shift(0,1)*unitsquare,p);la-bel(quot;4quot;,(0.5,1.5));draw(shift(0,-1)*unitsquare,p);label(quot;8quot;,(0.5,-0.5));[/asy](A)154(B)159(C)164(D)167(E)189200812A function f has domain [0,2]and range [0,1].(The notation [a,b ]denotes {x :a ≤x ≤b }.)What are the domain and range,respectively,of the function g defined by g (x )1f (x 1)?(A)[1,1],[1,0](B)[1,1],[0,1](C)[0,2],[1,0](D)[1,3],[1,0](E)[1,3],[0,1]13Points A and B lie ona circle centered at O ,and ∠AOB 60◦.A second circle is internallytangent to the first and tangent to both OA and OB .What is the ratio of the area of thesmaller circle to that of the larger circle?(A)116(B)19(C)18(D)16(E)1414What is the area of the region defined by the inequality |3x 18||2y 7|≤3?(A)3(B)72(C)4(D)92(E)515Let k 2008222008.What is the units digit of k 22k ?(A)0(B)2(C)4(D)6(E)816The numbers log(a 3b 7),log(a 5b 12),and log(a 8b 15)are the first three terms of an arithmeticsequence,and the 12th term of the sequence is log b n .What is n ?(A)40(B)56(C)76(D)112(E)14317Let a 1,a 2,...be a sequence of integers determined by the rule a n a n 1/2if a n 1is even anda n 3a n 11if a n 1is odd.For how many positive integers a 1≤2008is it true that a 1is less thaneach of a 2,a 3,and a 4?(A)250(B)251(C)501(D)502(E)100418Triangle ABC ,with sides of length 5,6,and 7,has one vertex on the positive x -axis,one onthe positive y -axis,and one on the positive z -axis.Let O be the origin.What is the volumeof tetrahedron OABC ?(A)√85(B)√90(C)√95(D)10(E)√10519In the expansion of1xx 2···x 27 1xx 2···x 14 2,what is the coefficient of x 28?(A)195(B)196(C)224(D)378(E)40520Triangle ABC has AC 3,BC 4,and AB 5.Point D is on AB ,and CD bisects the right angle.The inscribed circles of ADC and BCD have radii r a and r b ,respectively.What is r a /r b ?2008(A)128 10√2 (B)356 10√2 (C)114 10√2 (D)55610√2 (E)328 10√2 21A permutation (a 1,a 2,a 3,a 4,a 5)of (1,2,3,4,5)is heavy-tailed if a 1a 2<a 4a 5.What is thenumber of heavy-tailed permutations?(A)36(B)40(C)44(D)48(E)5222A round table has radius 4.Six rectangular place mats are placed on the table.Each placemat has width 1and length x as shown.They are positioned so that each mat has two cornerson the edge of the table,these two corners being end points of the same side of length x .Further,the mats are positioned so that the inner corners each touch an inner corner of anadjacent mat.What is x ?[asy]unitsize(4mm);defaultpen(linewidth(.8)+fontsize(8));draw(Circle((0,0),4));path mat=(-2.687,-1.5513)–(-2.687,1.5513)–(-3.687,1.5513)–(-3.687,-1.5513)–cycle;draw(mat);draw(rotate(60)*mat);draw(rotate(120)*mat);draw(rotate(180)*mat);draw(rotate(240)*mat);draw(rotate(300)*mat);label(quot;36;x36;quot;,(-2.687,0),E);label(quot;36;136;quot;,(-3.187,1.5513),S);[/asy](A)2√5√3(B)3(C)3√7√32(D)2√3(E)52√3223The solutions of the equation z 44z 3i 6z 24zii 0are the vertices of a convex polygon in thecomplex plane.What is the area of the polygon?(A)25/8(B)23/4(C)2(D)25/4(E)23/224Triangle ABC has ∠C 60◦and BC 4.Point D is the midpoint of BC .What is the largestpossible value of tan ∠BAD ?(A)√36(B)√33(C)√32√2(D)√34√23(E)125A sequence (a 1,b 1),(a 2,b 2),(a 3,b 3),...of points in the coordinate plane satisfies (a n 1,b n 1)(√3a n b n ,√3b n a n )for n 1,2,3,....Suppose that (a 100,b 100)(2,4).What is a 1b 1?(A)minus 1297(B)minus 1299(C)0(D)1298(E)12962008B 1A basketball player made 5baskets during a game.Each basket was worth either 2or 3points.How many different numbers could represent the total points scored by the player?(A)2(B)3(C)4(D)5(E)62A 4×4block of calendar dates is shown.The order of the numbers in the second row is tobe reversed.Then the order of the numbers in the fourth row is to be reversed.Finally,thenumbers on each diagonal are to be added.What will be the positive difference between thetwo diagonal sums?12348910111516171822232425(A)2(B)4(C)6(D)8(E)103A semipro baseball league has teams with 21players each.League rules state that a playermust be paid at least $15,000,and that the total of all players’salaries for each team cannotexceed $700,000.What is the maximum possiblle salary,in dollars,for a single player?(A)270,000(B)385,000(C)400,000(D)430,000(E)700,0004On circle O ,points C and D are on the same side of diameter AB ,∠AOC 30◦,and ∠DOB 45◦.What is the ratio of the area of the smaller sector COD to the area of the circle?[asy]unitsize(6mm);defaultpen(linewidth(0.7)+fontsize(8pt));pair C =3*dir (30);pair D =3*dir (135);pair A =3*dir (0);pair B =3*dir(180);pair O =(0,0);draw (Circle ((0,0),3));label (quot;36;C36;quot;,C,NE);label (quot;36;D36;quot;,D,NW);label (quot;36;B36;quot;,B,W);label (quot;36;A36;quot;,A,E);label (quot;36;O36;quot;,O,S);label (quot;36;45◦36;quot ;,(−0.3,0.1),W NW );label (quot ;36;30◦36;quot ;,(0.5,0.1),ENE );draw (A −−B );draw (O −−D );draw (O −−C );[/asy ](A)29(B)14(C)518(D)724(E)3105A class collects $50to buy flowers for a classmate who is in the hospital.Roses cost $3each,and carnations cost $2each.No other flowers are to be used.How many different bouquets could be purchased for exactly $50?(A)1(B)7(C)9(D)16(E)1720086Postman Pete has a pedometer to count his steps.The pedometer records up to 99999steps,then flips over to 00000on the next step.Pete plans to determine his mileage for a year.On January 1Pete sets the pedometer to 00000.During the year,the pedometer flips from 99999to 00000forty-four times.On December 31the pedometer reads 50000.Pete takes 1800steps per mile.Which of the following is closest to the number of miles Pete walked during the year?(A)2500(B)3000(C)3500(D)4000(E)45007For real numbers a and b ,define a $b (ab )2.What is (xy )2$(yx )2?(A)0(B)x 2y 2(C)2x 2(D)2y 2(E)4xy 8Points B and C lie on AD .The length of AB is 4times the length of BD ,and the length of AC is 9times the length of CD .The length of BC is what fraction of the length of AD ?(A)136(B)113(C)110(D)536(E)159Points A and B are on a circle of radius 5and AB 6.Point C is the midpoint of the minor arc AB .What is the length of the line segment AC ?(A)√10(B)72(C)√14(D)√15(E)410Bricklayer Brenda would take 9hours to build a chimney alone,and bricklayer Brandon would take10hours to build it alone.When they work together they talk a lot,and their combined output is decreased by 10bricks per hour.Working together,they build the chimney in 5hours.How many bricks are in the chimney?(A)500(B)900(C)950(D)1000(E)190011A cone-shaped mountain has its base on the ocean floor and has a height of 8000feet.The top 18of the volume of the mountain is above water.What is the depth of the ocean at the base of themountain,in feet?(A)4000(B)2000(4√2)(C)6000(D)6400(E)700012For each positive integer n ,the mean of the first n terms of a sequence is n .What is the 2008thterm of the sequence?(A)2008(B)4015(C)4016(D)4,030,056(E)4,032,06413Vertex E of equilateral ABE is in the interior of unit square ABCD .Let R be the region consisting of all points inside ABCD and outside ABE whose distance from AD is between 13and 23.What is the area of R ?(A)125√372(B)125√336(C)√318(D)3√39(E)√312200814A circle has a radius of log 10(a 2)and a circumference of log 10(b 4).What is log a b ?(A)14π(B)1π(C)π(D)2π(E)102π15On each side of a unit square,an equilateral triangle of side length 1is constructed.On each newside of each equilateral triangle,another equilateral triangle of side length 1is constructed.The interiors of the square and the 12triangles have no points in common.Let R be the region formed by the union of the square and all the triangles,and S be the smallest convex polygon that contains R .What is the area of the region that is inside S but outside R ?(A)14(B)√24(C)1(D)√3(E)2√316A rectangular floor measures a by b feet,where a and b are positive integers with b >a .An artistpaints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor.The unpainted part of the floor forms a border of width 1foot around the painted rectangle and occupies half of the area of the entire floor.How many possibilities are there for the ordered pair (a,b )?(A)1(B)2(C)3(D)4(E)517Let A ,B ,and C be three distinct points on the graph of yx 2such that line AB is parallel tothe x -axis and ABC is a right triangle with area 2008.What is the sum of the digits of the y -coordinate of C ?(A)16(B)17(C)18(D)19(E)2018A pyramid has a square base ABCD and vertex E .The area of square ABCD is 196,and theareas of ABE and CDE are 105and 91,respectively.What is the volume of the pyramid?(A)392(B)196√6(C)392√2(D)392√3(E)78419A function f is defined by f (z )(4i )z 2αzγfor all complex numbers z ,where αand γare complexnumbers and i 21.Suppose that f (1)and f (i )are both real.What is the smallest possible value of |α||γ|(A)1(B)√(C)2(D)2√(E)420Michael walks at the rate of 5feet per second on a long straight path.Trash pails are located every200feet along the path.A garbage truck travels at 10feet per second in the same direction as Michael and stops for 30seconds at each pail.As Michael passes a pail,he notices the truck ahead of him just leaving the next pail.How many times will Michael and the truck meet?(A)4(B)5(C)6(D)7(E)8200821Two circles of radius 1are to be constructed as follows.The center of circle A is chosen uniformlyand at random from the line segment joining (0,0)and (2,0).The center of circle B is chosen uniformly and at random,and independently of the first choice,from the line segment joining (0,1)to (2,1).What is the probability that circles A and B intersect?(A)2√24(B)3√328(C)2√212(D)2√34(E)4√33422A parking lot has 16spaces in a row.Twelve cars arrive,each of which requires one parking space,and their drivers chose spaces at random from among the available spaces.Auntie Em then arrives in her SUV,which requires 2adjacent spaces.What is the probability that she is able to park?(A)1120(B)47(C)81140(D)35(E)172823The sum of the base-10logarithms of the divisors of 10n is 792.What is n ?(A)11(B)12(C)13(D)14(E)1524Let A 0(0,0).Distinct points A 1,A 2,...lie on the x -axis,and distinct points B 1,B 2,...lie on thegraph of y √x .For every positive integer n ,A n 1B n A n is an equilateral triangle.What is the least n for which the length A 0A n ≥100?(A)13(B)15(C)17(D)19(E)2125Let ABCD be a trapezoid with AB CD ,AB 11,BC 5,CD 19,and DA 7.Bisectors of ∠A and∠D meet at P ,and bisectors of ∠B and ∠C meet at Q .What is the area of hexagon ABQCDP ?(A)28√3(B)30√3(C)32√3(D)35√3(E)36√3。
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2008 AMC 12B Problems and Solution Problem 1A basketball player made baskets during a game. Each basket was worth eitheror points. How many different numbers could represent the total points scored bythe player?SolutionIf the basketball player m akes three-point shots and two-point shots, hescores points. Clearly every value of yields a different number of total points. Since he can m ake any number of three-point shots betweenand inclusive, the number of different point totals is .Problem 2A block of calendar dates is shown. The order of the numbers in the second row is to be reversed. Then the order of the numbers in the fourth row is to be reversed. Finally, the numbers on each diagonal are to be added. What will be the positive difference between the two diagonal sums?SolutionAfter reversing the num bers on the second and fourth rows, the block will look like this:The difference between the two diagonal sums is:. Problem 3A sem ipro baseball league has team s with players each. League rules state that aplayer must be paid at least dollars, and that the total of all players' salariesfor each team cannot exceed dollars. What is the m aximum possible salary, in dollars, for a single player?SolutionWe want to find the m aximum any player could m ake, so assum e that everyone else makes the minimum possible and that the com bined salaries total the m aximum ofThe m aximum any player could m ake is dollarsProblem 4On circle , points and are on the sam e side of diameter , ,and . What is the ratio of the area of the sm aller sector to thearea of the circle?Solution.Since a circle has , the desired ratio is .Problem 5A class collects dollars to buy flowers for a classm ate who is in the hospital. Roses cost dollars each, and carnations cost dollars each. No other flowers are to beused. How many different bouquets could be purchased for exactly dollars?SolutionThe class could send carnations and no roses, carnations and roses,carnations and roses, and so on, down to carnation and roses. There are 9total possibilities (from 0 to 16 roses, incrementing by 2 at each step),Problem 6Postm an Pete has a pedom eter to count his steps. The pedom eter records up to steps, then flips over to on the next step. Pete plans to determine hismileage for a year. On January Pete sets the pedom eter to . During the year,the pedom eter flips from to forty-four tim es. On Decem ber thepedometer reads . Pete takes steps per m ile. Which of the following is closest to the num ber of m iles Pete walked during the year?SolutionEvery tim e the pedometer flips, Pete has walked steps. Therefore, Pete haswalked a total of steps, which ismiles, which is closest to answer choice .Problem 7For real numbers and , define . What is ?SolutionProblem 8Points and lie on . The length of is tim es the length of , and thelength of is tim es the length of . The length of is what fraction of thelength of ?SolutionSince and , .Since and , .Thus, .Problem 9Points and are on a circle of radius and . Point is the midpoint ofthe minor arc . What is the length of the line segment ?SolutionsSolution 1Let be the angle that subtends the arc . By the law of cosines,implies .The half-angle formula says that. The law of cosines tells us, which is answer choice.Solution 2Define as the midpoint of line segm ent , and the center of the circle. Then, , and are collinear, and since is the m idpoint of ,and so . Since , , and so.Problem 10Bricklayer Brenda would take hours to build a chimney alone, and bricklayerBrandon would take hours to build it alone. When they work together they talk alot, and their com bined output is decreased by bricks per hour. Working together,they build the chim ney in hours. How m any bricks are in the chim ney?SolutionLet be the number of bricks in the house.Without talking, Brenda and Brandon lay and bricks per hour respectively, sotogether they lay per hour together.Since they finish the chim ney in hours, . Thus,.Problem 11A cone-shaped m ountain has its base on the ocean floor and has a height of 8000feet. The top of the volum e of the mountain is above water. What is the depth of the ocean at the base of the mountain in feet?SolutionIn a cone, radius and height each vary inversely with increasing height (i.e. theradius of the cone form ed by cutting off the m ountain at feet is half that of the original m ountain). Therefore, volume varies as the inverse cube of increasing height (expressed as a percentage of the total height of cone):Plugging in our given condition,Problem 12For each positive integer , the m ean of the first term s of a sequence is . What is the th term of the sequence?SolutionLetting be the nth partial sum of the sequence:The only possible sequence with this result is the sequence of odd integers.Problem 13Vertex of equilateral triangle is in the interior of unit square . Letbe the region consisting of all points inside and outside whosedistance from is between and . What is the area of ?SolutionThe region is the shaded area:We can find the area of the shaded region by subtracting the pentagon from the middle third of the square. The area of the m iddle third of the square is. The pentagon can be split into a rectangle and an equilateraltriangle.The base of the equilateral triangle is and the height is .Thus, the area is .The base of the rectangle is and the height is the height of the equilateral triangleminus the height of the sm aller equilateral triangle. This is:Therefore, the area of the shaded region isProblem 14A circle has a radius of and a circum ference of . What is ?SolutionLet be the circum ference of the circle, and let be the radius of the circle.Using log properties, and.Since ,Problem 15On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let be the region formed by the union of the square and all thetriangles, and be the sm allest convex polygon that contains . What is the areaof the region that is inside but outside ?SolutionThe equilateral triangles form trapezoids withside lengths (half a unit hexagon) on each face of the unit square. The four triangles "in between" these trapezoids are isosceles triangles with two side lengthsand an angle of in between them, so the total area of these triangles (which isthe area of ) is, by the Law of Sines, which makes theanswer .Problem 16A rectangular floor measures by feet, where and are positive integers with. An artist paints a rectangle on the floor with the sides of the rectangle parallelto the sides of the floor. The unpainted part of the floor form s a border of width foot around the painted rectangle and occupies half of the area of the entire floor. Howmany possibilities are there for the ordered pair ?SolutionBy Simon's Favorite Factoring Trick:Since and are the only positive factorings of .or yielding solutions. Notice that because , the reversed pairs are invalid.Problem 17Let , and be three distinct points on the graph of such that line isparallel to the -axis and is a right triangle with area . What is thesum of the digits of the -coordinate of ?SolutionSupposing , is perpendicular to and, it follows, to the -axis,making a segm ent of the line x=m. But that would m ean that the coordinates ofare , contradicting the given that points and are distinct. So isnot . By a similar logic, neither is .This m eans that and is perpendicular to . So the slope of isthe negative reciprocal of the slope of , yielding.Because is the length of the altitude of triangle from, and isthe length of , the area of . Since, . Substituting,, whose digits sum to .Problem 18A pyramid has a square base and vertex . The area of square is, and the areas of and are and , respectively. What is thevolume of the pyramid?SolutionLet be the height of the pyramid and be the distance from to . The sidelength of the base is 14. The side lengths of and areand , respectively. We have a system s of equations through the Pythagorean Theorem:Setting them equal to each other and simplifying gives.Therefore, , and the volume of the pyramid is.Problem 19A function is defined by for all com plex numbers ,where and are complex numbers and . Suppose that and areboth real. What is the sm allest possible value of ?SolutionWe need only concern ourselves with the im aginary portions of and (both of which m ust be 0). These are:Since appears in both equations, we let it equal 0 to simplify the equations. This yields two single-variable equations. Equation 1 tells us that the im aginary partof must be , and equation 2 tells us that the real part of must be .Therefore, . There are no restrictions on , so to minimize 'sabsolute value, we let ..Problem 20Michael walks at the rate of feet per second on a long straight path. Trash pails arelocated every feet along the path. A garbage truck travels at feet per secondin the sam e direction as Michael and stops for seconds at each pail. As Michaelpasses a pail, he notices the truck ahead of him just leaving the next pail. How many tim es will Michael and the truck m eet?SolutionPick a coordinate system where Michael's starting pail is and the one where thetruck starts is . Let and be the coordinates of Michael and the truckafter seconds. Let be their (signed) distance after seconds.Meetings occur whenever . We have .The truck always m oves for seconds, then stands still for . During the firstseconds of the cycle the truck moves by meters and Michael by , henceduring the first seconds of the cycle increases by . During the remainingseconds decreases by .From this observation it is obvious that after four full cycles, i.e. at , we willhave for the first tim e.During the fifth cycle, will first grow from to , then fall from to .Hence Michael overtakes the truck while it is standing at the pail.During the sixth cycle, will first grow from to , then fall from to .Hence the truck starts moving, overtakes Michael on their way to the next pail, and then Michael overtakes the truck while it is standing at the pail.During the seventh cycle, will first grow from to , then fall from to. Hence the truck m eets Michael at the m oment when it arrives to the next pail.Obviously, from this point on will always be negative, m eaning that Michael isalready too far ahead. Hence we found all meetings.The m ovement of Michael and the truck is plotted below: Michael in blue, the truck in red.Problem 21Two circles of radius 1 are to be constructed as follows. The center of circle iscenter of circle is chosen uniformly and at random, and independently of the firstchoice, from the line segment joining to . What is the probability thatcircles and intersect?Two circles intersect if the distance between their centers is less than the sum oftheir radii. In this problem, and intersect iffIn other words, the two chosen X-coordinates must differ by no m ore than . To find this probability, we divide the problem into cases:1) is on the interval . The probability that falls within the desiredrange for a given is (on the left) (on the right) all over 2 (the range ofpossible values). The total probability for this range is the sum of all theseprobabilities of (over the range of ) divided by the total range of .Thus, the total possibility for this interval is.2) is on the interval . In this case, any value of will do, so theprobability for the interval is simply .3) is on the interval . This is identical, by symmetry, to case 1.The total probability is thereforeSolution 2Circles centered at and will overlap if and are closer to each other than ifthe circles were tangent. The circles are tangent when the distance between their centers is equal to the sum of their radii. Thus, the distance from to will be .Since and are separated by vertically, they m ust be separated byhorizontally. Thus, if , the circles intersect.Now, plot the two random variables and on the coordinate plane. Eachvariable ranges from to . The circles intersect if the variables are within ofeach other. Thus, the area in which the circles don't intersect is equal to the totalarea of two sm all triangles on opposite corners, each of area . We conclude the probability the circles intersectis:Problem 22A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?SolutionAuntie Em won't be able to park only when none of the four available spots touch. We can form a bijection between all such cases and the num ber of ways to pick four spots out of 13: since none of the spots touch, rem ove a spot from between each of the cars. From the other direction, given four spots out of 13, simply add a spot between each. So the probability she can park isProblem 23The sum of the base-logarithm s of the divisors of is . What is ?SolutionSolution 1Every factor of will be of the form. Using the logarithmicproperty , it suffi ces to count the total num ber of 2'sand 5's running through all possible . For every factor , there will beanother , so it suffices to count the total number of 2's occurring in all factors (because of this symmetry, the number of 5's will be equal). And since, the final sum will be the total num ber of 2'soccurring in all factors of .There are choices for the exponent of 5 in each factor, and for each of thosechoices, there are factors (each corresponding to a different expo nent of 2),yielding total 2's. The total number of 2's istherefore . Plugging in our answer choices into thisformula yields 11 (answer choice ) as the correct answer.Solution 2For every divisor of , , we have .There are divisors of that are . After casework on the parity of , we find that the answer is given by.Problem 24Let . Distinct points lie on the -axis, and distinct pointslie on the graph of . For every positive integer , isan equilateral triangle. What is the least for which the length ?SolutionLet . We need to rewrite the recursion into something m anageable.The two strange conditions, 's lie on the graph of and is anequilateral triangle, can be com pacted as follows:which uses , where is theheight of the equilateral triangle and therefore tim es its base.The relation above holds for and for , soOr,This implies that each segm ent of a successive triangle is morethan the last triangle. To find , we m erely have to plug in into theaforem entioned recursion and we have . Knowing that is , we candeduce that .Thus, , so. We want to find sothat . is our answer.Problem 25Let be a trapezoid with , , , , and. Bisectors of and meet at , and bisectors of and meet at. What is the area of hexagon ?SolutionDrop perpendiculars to from and , and call the intersectionsrespectively. Now, and. Thus, . We conclude and. To sim plify things even more, notice that, so .Also, So the areaof is:Over to the other side: is , and is therefore congruent to. So .The area of the hexagon is clearlyAlternate SolutionLet and meet at and , respectively.Since , , and they share , triangles andare congruent.By the same reasoning, we also have that triangles and are congruent. Hence, we have.If we let the height of the trapezoid be , we have.Thusly, if we find the height of the trapezoid and multiply it by 12, we will be done.Let the projections of and to be and , respectively.We have , , and.Therefore, . Solving this, we easily get that .Multiplying this by 12, we find that the area of hexagon is , whichcorresponds to answer choice .。