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

Revision as of 16:42, 6 September 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

Video Solution

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

@@ Line 6: / Line 6: @@
 ==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</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 - A = \frac{3}{4}(2 - A)</math>. Adding these equations now gives <math>2(B - A) = \frac{3}{4}(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} = \boxed{\textbf{(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>. 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 - A = \frac{3}{4}(2 - A)</math>. Adding these equations now gives <math>2(B - A) = \frac{3}{4}(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} = \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) } 1\frac{1}{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)==
-Since Harry 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 Harry 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>.
+Since Harry 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 Harry 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

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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