Difference between revisions of "2019 AMC 10B Problems/Problem 18"
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Henry decides one morning to do a workout, and he walks <math>\tfrac{3}{4}</math> of the way from his home to his gym. The gym is <math>2</math> kilometers away from Henry's home. At that point, he changes his mind and walks <math>\tfrac{3}{4}</math> of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks <math>\tfrac{3}{4}</math> of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked <math>\tfrac{3}{4}</math> 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 <math>A</math> kilometers from home and a point <math>B</math> kilometers from home. What is <math>|A-B|</math>? | Henry decides one morning to do a workout, and he walks <math>\tfrac{3}{4}</math> of the way from his home to his gym. The gym is <math>2</math> kilometers away from Henry's home. At that point, he changes his mind and walks <math>\tfrac{3}{4}</math> of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks <math>\tfrac{3}{4}</math> of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked <math>\tfrac{3}{4}</math> 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 <math>A</math> kilometers from home and a point <math>B</math> kilometers from home. What is <math>|A-B|</math>? | ||
− | <math>\textbf{(A) } \frac{2}{3} \qquad \textbf{(B) } 1 \qquad \textbf{(C) } | + | <math>\textbf{(A) } \frac{2}{3} \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{6}{5} \qquad \textbf{(D) } \frac{5}{4} \qquad \textbf{(E) } \frac{3}{2}</math> |
==Solution 1== | ==Solution 1== |
Revision as of 17:41, 6 September 2020
Contents
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 1
Let the two points that Henry walks in between be and
, with
being closer to home. As given in the problem statement, the distances of the points
and
from his home are
and
respectively. By symmetry, the distance of point
from the gym is the same as the distance from home to point
. Thus,
. In addition, when he walks from point
to home, he walks
of the distance, ending at point
. Therefore, we know that
. By substituting, we get
. Adding these equations now gives
. Multiplying by
, we get
, so
.
Solution 2 (not rigorous)
We assume that Henry is walking back and forth exactly between points and
, with
closer to Henry's home than
. Denote Henry's home as a point
and the gym as a point
. Then
and
, so
. Therefore,
.
Solution 3 (not rigorous; similar to 2)
Since Harry is very close to walking back and forth between two points, let us denote closer to his house, and
closer to the gym. Then, let us denote the distance from
to
as
. If Harry was at
and walked
of the way, he would end up at
, vice versa. Thus we can say that the distance from
to the gym is
the distance from
to his house. That means it is
. This also applies to the other side. Furthermore, we can say
+
+
=
. We solve for
and get
. Therefore, the answer is
.
~aryam
Video Solution
For those who want a video solution: https://youtu.be/45kdBy3htOg
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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