AMC12AProblemandSolution.doc

2017 AMC 12A2017 AMC 12A2017 AMC 12A problems and solutions. The test was held on February 7, 2017.Problem 1Pablo buys popsicles for his friends. The store sells single popsicles for  each, 3-popsicle boxes for , and 5-popsicle boxes for . What is the greatest number of popsicles that Pablo can buy with ?SolutionBy the greedy algorithm, we can take two 5-popsicle boxes and one 3-popsicle box with . To prove that this is optimal, consider an upper bound as follows: at the rate of  per 5 popsicles, we can get  popsicles, which is less than 14. .Problem 2The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?SolutionLet  be our two numbers. Then . Thus,..Problem3Ms. Carroll promised that anyone who got all the multiple choice questions right on the upcoming exam would receive an A on the exam. Which one of these statements necessarily follows logically?SolutionTaking the contrapositive of the statement "if he got all of them right, he got an A" yields "if he didn't get an A, he didn't get all of them right", yielding the answer .Problem 4Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?SolutionLet  represent how far Jerry walked, and  represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides,  Since Silvia walked the diagonal, she walked the hypotenuse of a 45, 45, 90 triangle with leg length 1. Thus,  We can then take  .Problem 5At a gathering of  people, there are  people who all know each other and  people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?SolutionLet the group of people who all know each other be , and let the group of people who know no one be . Handshakes occur between each pair  such that  and , and between each pair of members in . Thus, the answer isSolution - Complementary CountingThe number of handshakes will be equivalent to the difference between the number of total interactions and the number of hugs, which are  and , respectively. Thus, the total amount of handshakes is Problem6Joy has  thin rods, one each of every integer length from  through . She places the rods with lengths , , and on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?SolutionThe quadrilateral cannot be a straight line. Thus, the fourth side must be longer than  and shorter than  = 25. This means Joy can use the 19 possible integer rod lengths that fall into . However, she has already used the rods of length cm and  cm so the answer is   Problem7Define a function on the positive integers recursively by ,  if  is even, and  if  is odd and greater than . What is ?SolutionThis is a recursive function, which means the function is used to evaluate itself. To solve this, we must identify the base case, . We also know that when  is odd, . Thus we know that . Thus we know that n will always be odd in the recursion of , and we add  each recursive cycle, which there are  of. Thus the answer is , which is answer .Problem8The region consisting of all points in three-dimensional space within  units of line segment  has volume . What is the length ?SolutionLet the length  be . Then, we see that the region is just the union of the cylinder with central axis  and radius  and the two hemispheres connected to each face of the cylinder (also with radius ). Thus the volume isProblem9Let  be the set of points  in the coordinate plane such that two of the three quantities , , and  are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of ?SolutionIf the two equal values are  and , then . Also,  because 3 is the common value. Solving for , we get . Therefore the portion of the line  where  is part of . This is a ray with an endpoint of .Similar to the process above, we assume that the two equal values are  and . Solving the equation  then . Also,  because 3 is the common value. Solving for , we get . Therefore the portion of the line  where  is also part of . This is another ray with the same endpoint as the above ray: .If  and  are the two equal values, then . Solving the equation for , we get . Also  because  is one way to express the common value (using  as the common value wor。