Difference between revisions of "2019 AMC 10B Problems/Problem 18"
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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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+ | Solution by greersc | ||
==See Also== | ==See Also== |
Revision as of 16:42, 14 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
.
Solution by greersc
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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