Difference between revisions of "2019 AMC 10B Problems/Problem 18"
m |
Geometry285 (talk | contribs) (→Problem) |
||
(6 intermediate revisions by 4 users not shown) | |||
Line 3: | Line 3: | ||
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) } \frac{ | + | <math>\textbf{(A) } \frac{2}{3} \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 1 \frac{1}{5} \qquad \textbf{(D) } 1 \frac{1}{4} \qquad \textbf{(E) } 1 \frac{1}{2}</math> |
==Solution 1== | ==Solution 1== | ||
− | Let the two points that Henry walks in between be <math>P</math> and <math>Q</math>, with <math>P</math> being closer to home. As given in the problem statement, the distances of the points <math>P</math> and <math>Q</math> from his home are <math>A</math> and <math>B</math> respectively. By symmetry, the distance of point <math>Q</math> from the gym is the same as the distance from home to point <math>P | + | Let the two points that Henry walks in between be <math>P</math> and <math>Q</math>, with <math>P</math> being closer to home. As given in the problem statement, the distances of the points <math>P</math> and <math>Q</math> from his home are <math>A</math> and <math>B</math> respectively. By symmetry, the distance of point <math>Q</math> from the gym is the same as the distance from home to point <math>P</math>. |
− | ==Solution 2 ( | + | Thus, <math>A = 2 - B</math>. |
+ | |||
+ | In addition, when he walks from point <math>Q</math> to home, he walks <math>\frac{3}{4}</math> of the distance, ending at point <math>P</math>. Therefore, we know that <math>B - A = \frac{3}{4}B</math>. | ||
+ | |||
+ | By substituting, we get <math>B - (2-B) = \frac{3}{4}\cdot B</math> and we solve to get <math>B=\dfrac{8}{5}</math>, so <math>A=2-\dfrac{8}{5}=\dfrac{2}{5}</math>. | ||
+ | |||
+ | <math>|A-B|=\left|\dfrac{2}{5}-\dfrac{8}{5} \right|=\boxed{\textbf{(C) } \frac{6}{5}}</math>. | ||
+ | |||
+ | ==Solution 2 (Not Rigorous)== | ||
We assume that Henry is walking back and forth exactly between points <math>P</math> and <math>Q</math>, with <math>P</math> closer to Henry's home than <math>Q</math>. Denote Henry's home as a point <math>H</math> and the gym as a point <math>G</math>. Then <math>HP:PQ = 1:3</math> and <math>PQ:QG = 3:1</math>, so <math>HP:PQ:QG = 1:3:1</math>. Therefore, <math>|A-B| = PQ = \frac{3}{1+3+1} \cdot 2 = \frac{6}{5} = \boxed{\textbf{(C) } \frac{6}{5}}</math>. | We assume that Henry is walking back and forth exactly between points <math>P</math> and <math>Q</math>, with <math>P</math> closer to Henry's home than <math>Q</math>. Denote Henry's home as a point <math>H</math> and the gym as a point <math>G</math>. Then <math>HP:PQ = 1:3</math> and <math>PQ:QG = 3:1</math>, so <math>HP:PQ:QG = 1:3:1</math>. Therefore, <math>|A-B| = PQ = \frac{3}{1+3+1} \cdot 2 = \frac{6}{5} = \boxed{\textbf{(C) } \frac{6}{5}}</math>. | ||
==Solution 3 (not rigorous; similar to 2)== | ==Solution 3 (not rigorous; similar to 2)== | ||
− | Since | + | Since Henry is very close to walking back and forth between two points, let us denote <math>A</math> closer to his house, and <math>B</math> closer to the gym. Then, let us denote the distance from <math>A</math> to <math>B</math> as <math>x</math>. If Henry was at <math>B</math> and walked <math>\frac{3}{4}</math> of the way, he would end up at <math>A</math>, vice versa. Thus we can say that the distance from <math>A</math> to the gym is <math>\frac{1}{4}</math> the distance from <math>B</math> to his house. That means it is <math>\frac{1}{3}x</math>. This also applies to the other side. Furthermore, we can say <math>\frac{1}{3}x</math> + <math>x</math> + <math>\frac{1}{3}x</math> = <math>2</math>. We solve for <math>x</math> and get <math>x=\frac{6}{5}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) } \frac{6}{5}}</math>. |
~aryam | ~aryam | ||
+ | |||
+ | ==Solution 4 == | ||
+ | |||
+ | Let <math>A</math> have a distance of <math>x</math> from the home. Then, the distance to the gym is <math>2-x</math>. This means point <math>B</math> and point <math>A</math> are <math>\frac{3}{4} \cdot (2-x)</math> away from one another. It also means that Point <math>B</math> is located at <math>\frac{3}{4} (2-x) + x.</math> So, the distance between the home and point <math>B</math> is also <math>\frac{3}{4} (2-x) + x.</math> | ||
+ | |||
+ | It follows that point <math>A</math> must be at a distance of <math>\frac{3}{4} \left( \frac{3}{4} (2-x) + x \right)</math> from point <math>B</math>. However, we also said that this distance has length <math>\frac{3}{4} (2-x)</math>. So, we can set those equal, which gives the equation: <cmath>\frac{3}{4} \left( \frac{3}{4} (2-x) + x \right) = \frac{3}{4} (2-x).</cmath> | ||
+ | |||
+ | Solving, we get <math>x = \frac{2}{5}</math>. This means <math>A</math> is at point <math>\frac{2}{5}</math> and <math>B</math> is at point <math>\frac{3}{4} \cdot \frac{8}{5} + \frac{2}{5} = \frac{8}{5}.</math> | ||
+ | |||
+ | So, <math>|A - B| = \boxed{\frac{6}{5}}.</math> | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 13:08, 8 August 2021
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 and we solve to 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 Henry 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 Henry 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
Solution 4
Let have a distance of from the home. Then, the distance to the gym is . This means point and point are away from one another. It also means that Point is located at So, the distance between the home and point is also
It follows that point must be at a distance of from point . However, we also said that this distance has length . So, we can set those equal, which gives the equation:
Solving, we get . This means is at point and is at point
So,
Video Solution
For those who want a video solution: https://youtu.be/45kdBy3htOg
Video Solution 2
~IceMatrix
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.