Difference between revisions of "2019 AMC 10B Problems/Problem 18"
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==Solution== | ==Solution== | ||
Let the two points that Henry walks in-between be <math>A</math> and <math>B</math>, with <math>A</math> being closer to home. In addition, let the distance that the points <math>A</math> and <math>B</math> are from his home be <math>a</math> and <math>b</math>, respectively. By symmetry, the distance from point <math>B</math> is from the gym is the same as the distance from home to point <math>A</math>. Thus, <math>a = 2 - b</math>. In addition, the distance that he walks as he repeatedly heads to home and then to the gym is <math>b - a</math>. Therefore, we are looking for the value of <math>b - a</math>. When he walks from point <math>B</math> to home, he walks <math>\frac{3}{4}</math> of the distance, ending at point <math>A</math>. Therefore, we know that <math>b - a = \frac{3}{4} \cdot b</math>. Similarily, we know <math>b - a = \frac{3}{4} \cdot (2 - a)</math>. Adding these equations, we get <math>2(b - a) = \frac{3}{4} \cdot (2 + b - a)</math>. Multiplying by <math>4</math>, we get <math>8(b - a) = 6 + 3(b - a)</math>, so <math>b - a = \frac{6}{5} = 1 \frac{1}{5}</math>. | Let the two points that Henry walks in-between be <math>A</math> and <math>B</math>, with <math>A</math> being closer to home. In addition, let the distance that the points <math>A</math> and <math>B</math> are from his home be <math>a</math> and <math>b</math>, respectively. By symmetry, the distance from point <math>B</math> is from the gym is the same as the distance from home to point <math>A</math>. Thus, <math>a = 2 - b</math>. In addition, the distance that he walks as he repeatedly heads to home and then to the gym is <math>b - a</math>. Therefore, we are looking for the value of <math>b - a</math>. When he walks from point <math>B</math> to home, he walks <math>\frac{3}{4}</math> of the distance, ending at point <math>A</math>. Therefore, we know that <math>b - a = \frac{3}{4} \cdot b</math>. Similarily, we know <math>b - a = \frac{3}{4} \cdot (2 - a)</math>. Adding these equations, we get <math>2(b - a) = \frac{3}{4} \cdot (2 + b - a)</math>. Multiplying by <math>4</math>, we get <math>8(b - a) = 6 + 3(b - a)</math>, so <math>b - a = \frac{6}{5} = 1 \frac{1}{5}</math>. | ||
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==See Also== | ==See Also== |
Revision as of 12:22, 15 February 2019
Problem
Henry decides one morning to do a workout, and he walks of the way from his home to his gym. The gym is kilometers away from Henry's home. At that point, he changes his mind and walks of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point kilometers from home and a point kilometers from home. What is ?
Solution
Let the two points that Henry walks in-between be and , with being closer to home. In addition, let the distance that the points and are from his home be and , respectively. By symmetry, the distance from point is from the gym is the same as the distance from home to point . Thus, . In addition, the distance that he walks as he repeatedly heads to home and then to the gym is . Therefore, we are looking for the value of . When he walks from point to home, he walks of the distance, ending at point . Therefore, we know that . Similarily, we know . Adding these equations, we get . Multiplying by , we get , so .
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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