Detailed Solutions of Problems 18 and 21 on the 2017 AMC 10 A (also known as Problems 15 and 19 on the 2017 AMC 12 A) Henry Wan, Ph.D. We have developed a Solutions Manual that contains detailed solutions for all 25 Problems on the 2017 AMC 10A. The Manual also includes many new problems we proposed based on the 2017 AMC 10A. This file only demonstrates detailed solutions of two typical problems -- Problems 18 and 21 on the 2017 AMC 10A (also known as Problems 15 and 19 on the 2017 AMC 12A). Part I The 2017 AMC 10A Problem 18 is the same as the following 3 problems: 2015 AMC 12B Problem 9 2016 AMC 12B Problem 19 1981 AHSME Problem 26 2017 AMC 10A Problem 18 Amelia has a coin that lands heads with probability, and Blaine has a coin that lands on heads with probability. Amelia and Blaine alternately toss their coins until someone gets a head; the Email: chiefmathtutor@gmail.com Page 1
first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is, where and are relatively prime positive integers. What is? Solution 1 Let be the probability that Amelia wins. Note that Amelia can win on her first turn by getting a head; if she and Blaine both miss, then it is as if started the game all over. More detailedly, we check some cases. If Amelia gets a head on her first toss, then she wins and the game ends; this happens with probability. Otherwise, Blaine gets a chance to play; this happens (when Amelia got a tail) with probability. If Blaine gets a chance to play, she wins with probability, or resets the game with probability. Thus, Solving gives and so Solution 2 Amelia Head (Win) Tail (Loss) Blaine Email: chiefmathtutor@gmail.com Page 2
Let be the probability that Amelia wins. For Amelia to win on the first round, she must toss a head. The probability of this occurring is For Amelia to win on the second round, she must toss a tail, then Blaine has tossed a tail, after that Amelia tosses a head, which occurs with probability This pattern of possible success continues. Amelia wins in the third round with probability in the fourth with probability and so on. The probability that Amelia wins is the sum of probabilities that she wins in each of the individual rounds, that is, This is the sum of an infinite geometric series with the first term, Using the formula for the sum of an infinite geometric series, we have Thus, Hence, Email: chiefmathtutor@gmail.com Page 3
OR Note that which is the same as the equation obtained in Solution 1. 2015 AMC 12B Problem 9 Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a player knocks the bottle off the ledge is, independently of what has happened before. What is the probability that Larry wins the game? 2016 AMC 12B Problem 19 Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips a fair coin repeatedly until he gets his first head, at which point he stops. What is the probability that all three flip their coins the same number of times? 1981 AHSME Problem 26 Email: chiefmathtutor@gmail.com Page 4
Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is, independent of the outcome of any other toss.) Generalization Based on the 2017 AMC 10A Problem 18, we raise the following new problems: Alex has a coin that lands heads with probability, Bob has a coin that lands on heads with probability, and Carl has a coin that lands on heads with probability. Alex, Bob, and Carl take turns flipping a coin with Alex flipping first, Bob second, and Carl third. The winner is the one who first obtains a head. All coin tosses are independent. What is the probability that Alex wins? Email: chiefmathtutor@gmail.com Page 5
Part II The first part of the 2017 AMC 10A Problem 21/2017 AMC 12A Problem 19 is exactly same as the 2007 AMC 10B Problem 21 2017 AMC 10A Problem 21/2017 AMC 12A Problem 19 A square with side length is inscribed in a right triangle with sides of length 3, 4, and 5 so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length is inscribed in another right triangle with sides of length 3, 4, and 5 so that one side of the square lies on the hypotenuse of the triangle. What is? Solution Draw the first right triangle and the square, where and. Let. Then. Note that So is a 3-4-5 right triangle with Thus, Email: chiefmathtutor@gmail.com Page 6
which gives Now draw the second right triangle and the square as shown below. Let. Since and are both 3-4-5 right triangles. We have and so Thus, Solving for gets Therefore, Email: chiefmathtutor@gmail.com Page 7
2007 AMC 10B # 21 Right has and. Square is inscribed in with and on, on and on. What is the side length of the square? Based on the 2017 AMC 10A Problem 21/2017 AMC 12A Problem 19, we propose the following new problems: 1. The first square with diagonal length is inscribed in a right triangle with sides of length 5, 12, and 13 so that one vertex of the square coincides with the right-angle vertex of the triangle. The second square with diagonal length is inscribed in the same right triangle so that one side of the square lies on the hypotenuse of the triangle. What is? 2. A square is inscribed in a right triangle with sides of length 3, 4, and 5 so that one vertex of the square coincides with the right-angle vertex of the triangle. A square is inscribed in the same right triangle so that one side of the square lies on the hypotenuse of the triangle. What is the distance between the centers of two squares? 3. Three semicircles with radii and are inscribed in a right triangle with sides of length 3, 4, and 5 so that their diameters are on the leg with length 3, the leg with length 4, and the hypotenuse, respectively. What is the ratio? Answers: The diameter of the semicircle on the leg with length 3: Email: chiefmathtutor@gmail.com Page 8
The diameter of the semicircle on the leg with length 4 The diameter of the semicircle on the hypotenuse The ratio of the radii is 4. Right has and. Three equilateral triangles are inscribed in such that the first triangle has one side on the leg with length 3, the second triangle has a side on the leg with length 4, and the third triangle has one side on the hypotenuse. What is the ratio of side lengths of the first, second, and third triangles? Answers: Having one side on the leg with length 3: Having one side on the leg with length 4: Having one side on the hypotenuse Email: chiefmathtutor@gmail.com Page 9
5. What is the area of the largest equilateral triangle inscribed in a right triangle with sides of length 3, 4, and 5 so that three vertices of the equilateral triangle are, respectively, on the three sides of the right triangle? Email: chiefmathtutor@gmail.com Page 10