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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>?
  
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<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) } 1 \frac{1}{5} \qquad \textbf{(D) } 1 \frac{1}{4} \qquad \textbf{(E) } 1 \frac{1}{2}</math>
  
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==Solution==
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==Solution 1==
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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} = (C)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>.  
  
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Solution by greersc
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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|=\frac{6}{5}=\boxed{\textbf{(C) } 1 \frac{1}{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) } 1 \frac{1}{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) } 1\frac{1}{5}}</math>.
 +
 
 +
~aryam
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 +
==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| = \frac{6}{5}=\boxed{\textbf{(C) } 1\frac{1}{5}}.</math>
 +
 
 +
== Solution 5 (Rigorous) ==
 +
Let A be the point closer to Henry’s home, and B be the point closer to the gym. Define <math>(a_n)</math> to be the position of Henry after <math>2n</math> walks. Similarly, define <math>(b_n)</math> to be the position of Henry after <math>2n - 1</math> walks. Thus, <math>a_1 = \frac{1}{4} \cdot (\frac{3}{4} \cdot 2) = \frac{3}{8}</math> and <math>b_1 = \frac{3}{4} \cdot 2 = \frac{3}{2}</math>. We can also deduce that <cmath>a_n = \frac{1}{4} ( \frac{3}{4} (2 - a_{n-1}) + a_{n-1} ) = \frac{1}{16} a_{n-1} + \frac{3}{8}</cmath> (<math>2 - a_{n-1}</math> is Henry's distance to the gym, so we take <math>\frac{3}{4}</math> of that and add it to our original position. Then, we take <math>\frac{1}{4}</math> of that to obtain Henry's distance from home). Similarly, we can deduce that <cmath>b_n = \frac{3}{4} (2 - \frac{1}{4} b_{n-1}) + \frac{1}{4} b_{n-1} = \frac{1}{16} b_{n-1} + \frac{3}{2}</cmath> Now, we follow the standard procedure to convert this arithmetico geometric recursion into a closed form. Let <math>a_n - k = \frac{1}{16} (a_{n-1} -k)</math> for some constant <math>k</math>. Then, <math>a_n = \frac{1}{16} a_{n-1} + \frac{15}{16} k</math>. So, <math>\frac{1}{16} a_{n-1} + \frac{15}{16} k = \frac{1}{16} a_{n-1} + \frac{3}{8} \Rightarrow k = \frac{3}{8} \cdot \frac{16}{15} = \frac{2}{5}</math>. This means that <cmath>a_n - \frac{2}{5} = \frac{1}{16} (a_{n-1} - \frac{2}{5}) \Rightarrow a_n - \frac{2}{5} = (\frac{1}{16})^{n-1} (a_1 - \frac{2}{5}) = (\frac{1}{16})^{n-1} (\frac{3}{8} - \frac{2}{5}) = (\frac{1}{16})^{n-1} \cdot -\frac{1}{40} \Rightarrow a_n = \frac{2}{5} - \frac{1}{16^{n-1} \cdot 40}</cmath> Now, calculating <cmath>\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{2}{5} - \frac{1}{16^{n-1} \cdot 40} = \frac{2}{5} - \lim_{n \to \infty} \frac{1}{16^{n-1} \cdot 40} = \frac{2}{5} - 0 = \frac{2}{5} </cmath> Thus, <math>A = \frac{2}{5}</math>. Taking a similar process for <math>B</math>, we derive that <math>b_n = \frac{8}{5} - \frac{1}{16^{n-1} \cdot 10}</math>, so <math>B = \lim_{n \to \infty} \frac{8}{5} - \frac{1}{16^{n-1} \cdot 10} = \frac{8}{5}</math>. Finally, <math>|A-B| = |\frac{2}{5} - \frac{8}{5}| = \boxed{\frac{6}{5}}</math>.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:CrazyVideoGamez CrazyVideoGamez]
 +
 
 +
== Note ==
 +
This problem can be estimated by drawing out a diagram.
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/4WttvHavnkM?t=55
 +
 
 +
~ pi_is_3.14
 +
 
 +
== Video Solution by On The Spot STEM==
 +
https://youtu.be/45kdBy3htOg
 +
 
 +
== Video Solution by TheBeautyofMath==
 +
https://youtu.be/U5PjjZ-5MIQ
 +
 
 +
~IceMatrix
  
 
==See Also==
 
==See Also==

Latest revision as of 09:28, 15 July 2024

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 $|A-B|$?

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

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 $A=2-\dfrac{8}{5}=\dfrac{2}{5}$.

$|A-B|=\left|\dfrac{2}{5}-\dfrac{8}{5} \right|=\frac{6}{5}=\boxed{\textbf{(C) } 1 \frac{1}{5}}$.

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, $|A-B| = PQ = \frac{3}{1+3+1} \cdot 2 = \frac{6}{5} = \boxed{\textbf{(C) } 1 \frac{1}{5}}$.

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 $\boxed{\textbf{(C) } 1\frac{1}{5}}$.

~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, $|A - B| = \frac{6}{5}=\boxed{\textbf{(C) } 1\frac{1}{5}}.$

Solution 5 (Rigorous)

Let A be the point closer to Henry’s home, and B be the point closer to the gym. Define $(a_n)$ to be the position of Henry after $2n$ walks. Similarly, define $(b_n)$ to be the position of Henry after $2n - 1$ walks. Thus, $a_1 = \frac{1}{4} \cdot (\frac{3}{4} \cdot 2) = \frac{3}{8}$ and $b_1 = \frac{3}{4} \cdot 2 = \frac{3}{2}$. We can also deduce that \[a_n = \frac{1}{4} ( \frac{3}{4} (2 - a_{n-1}) + a_{n-1} ) = \frac{1}{16} a_{n-1} + \frac{3}{8}\] ($2 - a_{n-1}$ is Henry's distance to the gym, so we take $\frac{3}{4}$ of that and add it to our original position. Then, we take $\frac{1}{4}$ of that to obtain Henry's distance from home). Similarly, we can deduce that \[b_n = \frac{3}{4} (2 - \frac{1}{4} b_{n-1}) + \frac{1}{4} b_{n-1} = \frac{1}{16} b_{n-1} + \frac{3}{2}\] Now, we follow the standard procedure to convert this arithmetico geometric recursion into a closed form. Let $a_n - k = \frac{1}{16} (a_{n-1} -k)$ for some constant $k$. Then, $a_n = \frac{1}{16} a_{n-1} + \frac{15}{16} k$. So, $\frac{1}{16} a_{n-1} + \frac{15}{16} k = \frac{1}{16} a_{n-1} + \frac{3}{8} \Rightarrow k = \frac{3}{8} \cdot \frac{16}{15} = \frac{2}{5}$. This means that \[a_n - \frac{2}{5} = \frac{1}{16} (a_{n-1} - \frac{2}{5}) \Rightarrow a_n - \frac{2}{5} = (\frac{1}{16})^{n-1} (a_1 - \frac{2}{5}) = (\frac{1}{16})^{n-1} (\frac{3}{8} - \frac{2}{5}) = (\frac{1}{16})^{n-1} \cdot -\frac{1}{40} \Rightarrow a_n = \frac{2}{5} - \frac{1}{16^{n-1} \cdot 40}\] Now, calculating \[\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{2}{5} - \frac{1}{16^{n-1} \cdot 40} = \frac{2}{5} - \lim_{n \to \infty} \frac{1}{16^{n-1} \cdot 40} = \frac{2}{5} - 0 = \frac{2}{5}\] Thus, $A = \frac{2}{5}$. Taking a similar process for $B$, we derive that $b_n = \frac{8}{5} - \frac{1}{16^{n-1} \cdot 10}$, so $B = \lim_{n \to \infty} \frac{8}{5} - \frac{1}{16^{n-1} \cdot 10} = \frac{8}{5}$. Finally, $|A-B| = |\frac{2}{5} - \frac{8}{5}| = \boxed{\frac{6}{5}}$.

~CrazyVideoGamez

Note

This problem can be estimated by drawing out a diagram.

Video Solution by OmegaLearn

https://youtu.be/4WttvHavnkM?t=55

~ pi_is_3.14

Video Solution by On The Spot STEM

https://youtu.be/45kdBy3htOg

Video Solution by TheBeautyofMath

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

~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

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