Difference between revisions of "2019 AMC 10B Problems/Problem 18"

(Video Solution Added)
(23 intermediate revisions by 13 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) } 1\frac{1}{5} \qquad \textbf{(D) } 1\frac{1}{4} \qquad \textbf{(E) } 1\frac{1}{2}</math>
+
<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==
+
==Solution 1==
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>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>.  
 +
 
 +
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>.
 +
 
 +
==Solution 3 (not rigorous; similar to 2)==
 +
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
 +
 
 +
==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==
 
For those who want a video solution: https://youtu.be/45kdBy3htOg
 
For those who want a video solution: https://youtu.be/45kdBy3htOg
 +
 +
==Video Solution 2==
 +
https://youtu.be/U5PjjZ-5MIQ
 +
 +
~IceMatrix
 +
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2019|ab=B|num-b=17|num-a=19}}
 
{{AMC10 box|year=2019|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:31, 1 May 2021

Problem

Henry decides one morning to do a workout, and he walks $\tfrac{3}{4}$ of the way from his home to his gym. The gym is $2$ kilometers away from Henry's home. At that point, he changes his mind and walks $\tfrac{3}{4}$ of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks $\tfrac{3}{4}$ of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked $\tfrac{3}{4}$ 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 $A$ kilometers from home and a point $B$ kilometers from home. What is $|A-B|$?

$\textbf{(A) } \frac{2}{3} \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{6}{5} \qquad \textbf{(D) } \frac{5}{4} \qquad \textbf{(E) } \frac{3}{2}$

Solution 1

Let the two points that Henry walks in between be $P$ and $Q$, with $P$ being closer to home. As given in the problem statement, the distances of the points $P$ and $Q$ from his home are $A$ and $B$ respectively. By symmetry, the distance of point $Q$ from the gym is the same as the distance from home to point $P$.

Thus, $A = 2 - B$.

In addition, when he walks from point $Q$ to home, he walks $\frac{3}{4}$ of the distance, ending at point $P$. Therefore, we know that $B - A = \frac{3}{4}B$.

By substituting, we get $B - (2-B) = \frac{3}{4}\cdot B$ and we solve to get $B=\dfrac{8}{5}$, so $A=2-\dfrac{8}{5}=\dfrac{2}{5}$.

$|A-B|=\left|\dfrac{2}{5}-\dfrac{8}{5} \right|=\boxed{\textbf{(C) } \frac{6}{5}}$.

Solution 2 (Not Rigorous)

We assume that Henry is walking back and forth exactly between points $P$ and $Q$, with $P$ closer to Henry's home than $Q$. Denote Henry's home as a point $H$ and the gym as a point $G$. Then $HP:PQ = 1:3$ and $PQ:QG = 3:1$, so $HP:PQ:QG = 1:3:1$. Therefore, $|A-B| = PQ = \frac{3}{1+3+1} \cdot 2 = \frac{6}{5} = \boxed{\textbf{(C) } \frac{6}{5}}$.

Solution 3 (not rigorous; similar to 2)

Since Henry is very close to walking back and forth between two points, let us denote $A$ closer to his house, and $B$ closer to the gym. Then, let us denote the distance from $A$ to $B$ as $x$. If Henry was at $B$ and walked $\frac{3}{4}$ of the way, he would end up at $A$, vice versa. Thus we can say that the distance from $A$ to the gym is $\frac{1}{4}$ the distance from $B$ to his house. That means it is $\frac{1}{3}x$. This also applies to the other side. Furthermore, we can say $\frac{1}{3}x$ + $x$ + $\frac{1}{3}x$ = $2$. We solve for $x$ and get $x=\frac{6}{5}$. Therefore, the answer is $\boxed{\textbf{(C) } \frac{6}{5}}$.

~aryam

Solution 4

Let $A$ have a distance of $x$ from the home. Then, the distance to the gym is $2-x$. This means point $B$ and point $A$ are $\frac{3}{4} \cdot (2-x)$ away from one another. It also means that Point $B$ is located at $\frac{3}{4} (2-x) + x.$ So, the distance between the home and point $B$ is also $\frac{3}{4} (2-x) + x.$

It follows that point $A$ must be at a distance of $\frac{3}{4} \left( \frac{3}{4} (2-x) + x \right)$ from point $B$. However, we also said that this distance has length $\frac{3}{4} (2-x)$. So, we can set those equal, which gives the equation: \[\frac{3}{4} \left( \frac{3}{4} (2-x) + x \right) = \frac{3}{4} (2-x).\]

Solving, we get $x = \frac{2}{5}$. This means $A$ is at point $\frac{2}{5}$ and $B$ is at point $\frac{3}{4} \cdot \frac{8}{5} + \frac{2}{5} = \frac{8}{5}.$

So, $|A - B| = \boxed{\frac{6}{5}}.$

Video Solution

For those who want a video solution: https://youtu.be/45kdBy3htOg

Video Solution 2

https://youtu.be/U5PjjZ-5MIQ

~IceMatrix

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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. AMC logo.png