1、2016 AMC 12BProblem 1What is the value of when ?Solution By: DragonflyWe find that is the same as , since a number to the power of is just the reciprocal of that number. We then get the equation to beWe can then simplify the equation to get Problem 2The harmonic mean of two numbers can be calculated
2、 as twice their product divided by their sum. The harmonic mean of and is closest to which integer?SolutionBy: dragonflySince the harmonic mean is times their product divided by their sum, we get the equationwhich is thenwhich is finally closest to .Problem 3Let . What is the value of ?SolutionBy: d
3、ragonflyFirst of all, lets plug in all of the s into the equation.Then we simplify to getwhich simplifies intoand finally we get Problem 4The ratio of the measures of two acute angles is , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum
4、 of the degree measures of the two angles?SolutionBy: dragonflyWe set up equations to find each angle. The larger angle will be represented as and the larger angle will we represented as , in degrees. This implies thatandsince the larger the original angle, the smaller the complement.We then find th
5、at and , and their sum is Problem 5The War of started with a declaration of war on Thursday, June , . The peace treaty to end the war was signed days later, on December , . On what day of the week was the treaty signed?SolutionBy: dragonflyTo find what day of the week it is in days, we have to divid
6、e by to see the remainder, and then add the remainder to the current day. We get that has a remainder of 2, so we increase the current day by to get Problem 6All three vertices of lie on the parabola defined by , with at the origin and parallel to the -axis. The area of the triangle is . What is the
7、 length of ?SolutionBy: Albert471Plotting points and on the graph shows that they are at and , which is isosceles. By setting up the triangle area formula you get: Making x=4, and the length of is , so the answer is .Problem 7Josh writes the numbers . He marks out , skips the next number , marks out
8、 , and continues skipping and marking out the next number to the end of the list. Then he goes back to the start of his list, marks out the first remaining number , skips the next number , marks out , skips , marks out , and so on to the end. Josh continues in this manner until only one number remai
9、ns. What is that number?SolutionBy Albert471Following the pattern, you are crossing out.Time 1: Every non-multiple of Time 2: Every non-multiple of Time 3: Every non-multiple of Following this pattern, you are left with every multiple of which is only .Problem 8A thin piece of wood of uniform densit
10、y in the shape of an equilateral triangle with side length inches weighs ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of inches. Which of the following is closest to the weight, in ounces, of the second piece?
11、Solution 1By: dragonflyWe can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use.We can then solve the equation to get which is closest to Solution 2Another approach to this problem, very similar to the previous one but perhaps explaine
12、d more thoroughly, is to use proportions. First, since the thickness and density are the same, we can set up a proportion based on the principle that , thus .However, since density and thickness are the same and (recognizing that the area of an equilateral triangle is ), we can say that .Then, by in
13、creasing s by a factor of , is increased by a factor of , thus or .Problem 9Carl decided to fence in his rectangular garden. He bought fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly yards between neighboring posts. T
14、he longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carls garden?SolutionBy Albert471To start, use algebra to determine the number of posts on each side. You have (the long sides count for beca
15、use there are twice as many) (each corner is double counted so you must add ) Making the shorter end have , and the longer end have . . Therefore, the answer is Problem 10A quadrilateral has vertices , , , and , where and are integers with . The area of is . What is ?Solution 1By distance formula we
16、 have . SImplifying we get . Thus and have to be a factor of 8. The only way for them to be factors of and remain integers is if and . So the answer is Solution by I_Dont_Do_MathSolution 2Solution by e_power_pi_times_iBy the Shoelace Theorem, the area of the quadrilateral is , so . Since and are int
17、egers, and , so .Problem 11How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line , the line and the line SolutionSolution by e_power_pi_times_i(Note: diagram is needed)If we draw a picture showin
18、g the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is squares , and the limit for the -value is squares. First we count the squares. In the back row, there are squares with length because generates squares from to , and continuing on
19、we have , , and for -values for , , and in the equation . So there are squares with length in the figure. For squares, each square takes up un left and un up. Squares can also overlap. For squares, the back row stretches from to , so there are squares with length in a by box. Repeating the process,
20、the next row stretches from to , so there are squares. Continuing and adding up in the end, there are squares with length in the figure. Squares with length in the back row start at and end at , so there are such squares in the back row. As the front row starts at and ends at there are squares with
21、length . As squares with length would not fit in the triangle, the answer is which is .Problem 12All the numbers are written in a array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corner
22、s add up to . What is the number in the center?SolutionSolution by Mlux: Draw a matrix. Notice that no adjacent numbers could be in the corners since two consecutive numbers must share an edge. Now find 4 nonconsecutive numbers that add up to . Trying works. Place each odd number in the corner in a
23、clockwise order. Then fill in the spaces. There has to be a in between the and . There is a between and . The final grid should similar to this.is in the middle.Solution 2If we color the square like a chessboard, since the numbers altrenate between even and odd, and there are five odd numbers and fo
24、ur even numbers, the odd numbers must be in the corners/center, while the even numbers must be on the edges. Since the odd numbers add up to 25, and the numbers in the corners add up to 18, the number in the center must be 25-18=7Problem 13Alice and Bob live miles apart. One day Alice looks due nort
25、h from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is from Alices position and from Bobs position. Which of the following is closest to the airplanes altitude, in miles?SolutionLets set the alti
26、tude = z, distance from Alice to airplanes ground position (point right below airplane)=y and distance from Bob to airplanes ground position=xFrom Alices point of view, . . So, From Bobs point of view, . . So, We know that + = Solving the equation (by plugging in x and y), we get z= = about 5.5.So,
27、answer is solution by sudeepnaralaSolution 2Non-trig solution by e_power_pi_times_iSet the distance from Alices and Bobs position to the point directly below the airplane to be and , respectively. From the Pythagorean Theorem, . As both are triangles, the altitude of the airplane can be expressed as
28、 or . Solving the equation , we get . Plugging this into the equation , we get , or ( cannot be negative), so the altitude is , which is closest to Problem 14The sum of an infinite geometric series is a positive number , and the second term in the series is . What is the smallest possible value of S
29、olutionThe second term in a geometric series is , where is the common ratio for the series and is the first term of the series. So we know that and we wish to find the minimum value of the infinite sum of the series. We know that: and substituting in , we get that . From here, you can either use cal
30、culus or AM-GM.Calculus: Let , then . Since and are undefined . This means that we only need to find where the derivative equals , meaning . So , meaning thatAM-GM For 2 positive real numbers and , . Let and . Then: . This implies that . or . Rearranging : . Thus, the smallest value is .Solution 2A
31、simple approach is to initially recognize that and . We know that , since the series must converge. We can start by observing the greatest answer choice, 4. Therefore, , because that would make , which would make the series exceed 4. In order to minimize both the initial term and the rest of the ser
32、ies, we can recognize that is the opitimal ratio, thus the answer is .Problem 15All the numbers are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to th
33、e three faces that include that vertex. What is the greatest possible value of the sum of these eight products?SolutionFirst assign each face the letters . The sum of the product of the faces is . We can factor this into which is the product of the sum of each pair of opposite faces. In order to max
34、imize we use the numbers or .Problem 16In how many ways can be written as the sum of an increasing sequence of two or more consecutive positive integers?SolutionWe proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number
35、 of consecutive numbers.For the first case, we can cleverly choose the convenient form of our sequence to bebecause then our sum will just be . We now have and will have a solution when is an integer, namely when is a divisor of 345. We check that work, and no more, because does not satisfy the requ
36、irements of two or more consecutive integers, and when equals the next biggest factor, , there must be negative integers in the sequence. Our solutions are .For the even cases, we choose our sequence to be of the form:so the sum is . In this case, we find our solutions to be .We have found all 7 sol
37、utions and our answer is .Solution 2The sum from to where and are integers and isLet and If we factor into all of its factor groups we will have several ordered pairs where The number of possible values for is half the number of factors of which is However, we have one extraneous case of because her
38、e, and we have the sum of one consecutive number which is not allowed by the question.Thus the answer is .Problem 17In shown in the figure, , , , and is an altitude. Points and lie on sides and , respectively, so that and are angle bisectors, intersecting at and , respectively. What is ?SolutionGet
39、the area of the triangle by herons formula:Use the area to find the height AH with known base BC:Apply angle bisector theorem on triangle and triangle , we get and , respectively. From now, you can simply use the answer choices because only choice has in it and we know that the segments on it all ha
40、ve integral lengths so that will remain there. However, by scaling up the length ratio: and . we get .Problem 18What is the area of the region enclosed by the graph of the equation SolutionConsider the case when , .Notice the circle intersect the axes at points and . Find the area of this circle in
41、the first quadrant. The area is made of a semi-circle with radius of and a triangle:Because of symmetry, the area is the same in all four quadrants. The answer is Problem 19Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips a fair coin repeatedly until he gets his
42、 first head, at which point he stops. What is the probability that all three flip their coins the same number of times?Solution 1By: dragonflyWe can solve this problem by listing it as an infinite geometric equation. We get that to have the same amount of tosses, they have a chance of getting all he
43、ads. Then the next probability is all of them getting tails and then on the second try, they all get heads. The probability of that happening is .We then get the geometric equationAnd then we find that equals to because of the formula of the sum for an infinite series, .Solution 2Call it a “win“ if
44、the boys all flip their coins the same number of times, and the probability that they win is . The probability that they win on their first flip is . If they dont win on their first flip, that means they all flipped tails (which also happens with probability ) and that their chances of winning have
45、returned to what they were at the beginning. This covers all possible sequences of winning flips. So we haveSolving for gives .Problem 20A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won games and lost games; there were no ties. How
46、 many sets of three teams were there in which beat , beat , and beat SolutionWe use complementary counting. Firstly, because each team played other teams, there are teams total. All sets that do not have beat , beat , and beat have one team that beats both the other teams. Thus we must count the num
47、ber of sets of three teams such that one team beats the two other teams and subtract that number from the total number of ways to choose three teams.There are ways to choose the team that beat the two other teams, and to choose two teams that the first team both beat. This is sets. There are sets of
48、 three teams total. Subtracting, we obtain as our answer.Problem 21Let be a unit square. Let be the midpoint of . For let be the intersection of and , and let be the foot of the perpendicular from to . What isSolution(By Qwertazertl)We are tasked with finding the sum of the areas of every where is a
49、 positive integer. We can start by finding the area of the first triangle, . This is equal to . Notice that since triangle is similar to triangle in a 1 : 2 ratio, must equal (since we are dealing with a unit square whose side lengths are 1). is of course equal to as it is the mid-point of CD. Thus, the area of the first triangle is .The second triangle has a base equal to that of (see that ) and using the same similar triangle logic as with the first triangle, we find the area to be . If we continue and test the next few triangles, we will find that the sum of all
