2019 AMC 10B Problems/Problem 18


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


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 unlike all the other solutions)

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, $|A-B| = |\frac{2}{5} - \frac{8}{5}| = \boxed{\frac{6}{5}}$.


Video Solution by OmegaLearn


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Video Solution

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

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See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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