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

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==Solution 4 ==
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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>
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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) 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>
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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>
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So, </math>|A - B| = \boxed{\frac{6}{5}}.$
  
 
==Video Solution==
 
==Video Solution==

Revision as of 08:36, 23 December 2020

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 $B - A = \frac{3}{4}(2 - A)$. Adding these equations now gives $2(B - A) = \frac{3}{4}(2 + B - A)$. Multiplying by $4$, we get $8(B - A) = 6 + 3(B - A)$, so $B - A = \frac{6}{5} = \boxed{\textbf{(C) } \frac{6}{5}}$.

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 Harry 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 Harry 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: <cmath>\frac{3}{4} \left( \frac{3}{4} (2-x) + x \right) = \frac{3}{4} (2-x).</cmath>

Solving, we get$ (Error compiling LaTeX. Unknown error_msg)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 (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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