2009 AMC 12B Problems(答案)

2009 AMC 12B Problems( 答案 八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案 ) Problem 1 Each morning of her five-day workweek, Jane bought either a 50-cent muffin or a 75-cent bagel. Her total cost for the week was a whole number of dollars, How many bagels did she buy? Solution The only combination of five items with total cost a whole number of dollars is 3 muffins and bagels. The answer is . Problem 2 Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms. How many cans of paint did she use for the 25 rooms? Solution Losing three cans of paint corresponds to being able to paint five fewer rooms. So . The answer is . Problem 3 Twenty percent off 60 is one-third more than what number? Solution Twenty percent less than 60 is . One-third more than a number n is . Therefore and the number is . The answer is . Problem 4 A 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. The parallel sides of the trapezoid have lengths and meters. What fraction of the yard is occupied by the flower beds? Solution Each 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 the yard occupied by the flower beds is . The answer is . Problem 5 Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages? Solution The age of each person is a factor of . So the twins could be years of age and, consequently Kiana could be 128, 32, 8 or 2 years old, respectively. Because Kiana is younger than her brothers, she must be 2 years old. So the sum of their ages is . The answer is . Problem 6 By inserting parentheses, it is possible to give the expression several values. How many different values can be obtained? Solution The three operations can be performed on any of orders. However, if the addition 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 check that the values of the four expressions are in fact all distinct. So the answer is , which is choice . Problem 7 In a certain year the price of gasoline rose by during January, fell by during 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 Solution Let be the price at the beginning of January. The price at the end of March was Because the price at the of April was , the price decreased by during April, and the percent decrease was So to the nearest integer is . The answer is . Problem 8 When a bucket is two-thirds full of water, the bucket and water weigh kilograms. When the bucket is one-half full of water the total weight is kilograms. In terms of and , what is the total weight in kilograms when the bucket is full of water? Solution Solution 1 Let 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 2 Imagine 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 3 We 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 9 Triangle has vertices , , and , where is on the line . What is the area of ? Solution Solution 1 Because the line is parallel to , the area of is independent of the location of on that line. Therefore it may be assumed that is . In that case the triangle has base and altitude , so its area is . Solution 2 The base of the triangle is . Its altitude is the distance between the point and the parallel line , which is . Therefore its area is . The answer is . Problem 10 A 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? Solution Solution 1 The clock will display the incorrect time for the entire hours of and . So the 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 not display correctly are and and . This amounts to fifteen of the sixty possible minutes for any given hour. Hence the fraction of the day that the clock shows the correct time is . The answer is . Solution 2 The 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 a number 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 are correct times. Therefore the required fraction is . Problem 11 On 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 without removing 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? Solution On Monday, day 1, the birds find quart of millet in the feeder. On Tuesday they find quarts of millet. On Wednesday, day 3, they find quarts of millet. The number of quarts of millet they find on day is The birds always find quart of other seeds, so more than half the seeds are millet if , that is, when . Because and , this will first occur on day which is . The answer is . Problem 12 The fifth and eighth terms of a geometric sequence of real numbers are and respectively. What is the first term? Solution Let the th term of the series be . Because it follows that and the first term is . The answer is . Problem 13 Triangle has and , and the altitude to has length . What is the sum of the two possible values of ? Solution Let be the foot of the altitude to . Then and . Thus or . The sum of the two possible values is . The answer is . Problem 14 Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from to , divides the entire region into two regions of equal area. What is ? Solution Solution 1 For the shaded area is at most , which is too little. Hence , and therefore 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 with vertices , , and . Its area is obviously , therefore the area of 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 2 The total area is 5, so the area of the shaded area is . If we add a unit square in the lower right corner, the area is . Therefore , or . Therefore . Problem 15 Assume . Below are five equations for . Which equation has the largest solution ? 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 for the option for which is smallest. And as is an increasing function, this is the option for which is smallest. We now get the following easier problem: Given that , find the smallest value 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, we can take the square root and get . Thus the answer is . Problem 16 Trapezoid has , , , and . The ratio is . What is ? Solution Solution 1 Extend and to meet at . Then Thus is isosceles with . Because , it follows that the triangles and are similar. Therefore so Solution 2 Let be the intersection of and the line through parallel to By constuction and ; it follows that is the bisector of the angle . So by the Angle Bisector Theorem we get The answer is . Problem 17 Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the 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? Solution Solution 1 There are two possible stripe orientations for each of the six faces of the cube, so there 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 in a 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 2 Without 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 are painted 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 the probabilities sum . Solution 3 There 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 total probability is . The answer is . Solution 4 Consider 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 encircling the 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 stripes out 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 connected to make a continuous stripe. There is a probability of that they are aligned, so there is a probability of that there is a continuous stripe. Problem 18 Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the same line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing 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? Solution After 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 the picture is . The answer is . Problem 19 For each positive integer , let . What is the sum of all values of that are prime numbers? Solution Solution 1 To find the answer it was enough to play around with . One can easily find that is a prime, then becomes negative for between and , and then is again a prime number. And as is already the largest option, the answer must be . Solution 2 We 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 as squares 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. For the second factor is positive and the first one is negative, hence is not a prime. The remaining cases are and . In both cases, is indeed a prime, and their sum is . Solution 3 Instead 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 the exception of the term. In fact, we have . But then we 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. Problem 20 A convex polyhedron has vertices , and edges. The polyhedron is cut by planes in such a way that plane cuts only those edges that meet at vertex . In addition, no two planes intersect inside or on . The cuts produce pyramids and a new polyhedron . How many edges does have? Solution Solution 1 Each edge of is cut by two planes, so has vertices. Three edges of meet at each vertex, so has edges. Solution 2 At 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 total number of endpoints of the original edges, which is . A middle portion of each original edge is also present in , so has edges. Solution 3 Euler's Polyhedron Formula applied to gives , where F is the number 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 Polyhedron Formula applied to gives , where is the number of edges of . Subtracting the first equation from the second gives , whence . The answer is . Problem 21 Ten women sit in seats in a line. All of the get up and then reseat themselves using all seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated? Solution Let 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 the rightmost 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 the other women. If woman sits on seat , we only have one option for woman : she must take seat , all the other seats are too far for her. We are left with women to sitting on seats to , and there are clearly valid arrangements of these. We get the recurrence . (Hence is precisely the -th Fibonacci number.) Using this recurrence we can easily compute that . Problem 22 Parallelogram has area . Vertex is at and all other vertices are in the first quadrant. Vertices and are lattice points on the lines and for some integer , respectively. How many such parallelograms are there? Solution Solution 1 The area of any parallelogram can be computed as the size of the vector product of and . In our setting where , , and this is simply . In other words, we need to count the triples of integers where , and . These can be counted as follows: We have identical red balls (representing powers of ), blue balls (representing powers of ), and three labeled urns (representing the factors , , and ). The red balls can be distributed in ways, and for each of these ways, the blue balls can then also be distributed in ways. (See Distinguishability for a more detailed explanation.) Thus there are exactly ways how to break into three positive integer factors, and for each of them we get a single parallelogram. Hence the number of valid parallelograms is . Solution 2 Without the vector product the area of can be computed for example as follows: If and , then clearly . Let , and be the orthogonal projections of , , and onto the axis. Let denote the area of the polygon . We can then compute: Problem 23 A region in the complex plane is defined by A complex number is chosen uniformly at random from . What is the probability that is also in ? Solution We 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 of the 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 do this graphically: Coordinate axes are dashed, is shown in red, in green and their intersection is yellow. The intersections of the boundary of and are obviously at and 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 all four 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 the first quadrant and there the probability is . This saves us the work of first multiplying and then dividing by .) Problem 24 For how many values of in is ? Note: The functions and denote inverse trigonometric functions. Solution First of all, we have to agree on the range of and . This should have been a part of the problem statement -- but as it is missing, we will assume the most common 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 ) one can 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 has exactly one root on . For we have . Hence is negative on the entire interval . Now note that . Hence for we have , and we 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 such that . 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 one is in , and the last two are in . Problem 25 The set is defined by the points with integer coordinates, , . How many squares of side at least have their four vertices in ? Solution We 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 the intermediate 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 total 6 (3,3) 1 1 1 7 (3,3) 1 4 4 8 (3,3) 1 4 4 8 (3,4) 1 4 4 8 (4,4) 3 1 3 9 (3,3) 1 4 4 9 (3,4) 1 8 8 9 (4,4) 3 4 12 10 (3,3) 1 4 4 10 (3,4) 1 8 8 10 (3,5) 1 4 4 10 (4,4) 3 4 12 10 (4,5) 3 4 12 10 (5,5) 5 1 5 11 (4,4) 3 4 12 11 (4,5) 3 8 24 11 (5,5) 5 4 20 12 (5,5) 5 4 20 12 (5,6) 5 4 20 12 (6,6) 7 1 7 13 (6,6) 7 4 28 14 (7,7) 9 1 9 Summing the last column, we get that the answer is .