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

(added solution 2)
m (Fixed references to points A and B (whereas in the problem statement, these are distances))
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==Solution 1==
 
==Solution 1==
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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>A</math> and <math>B</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 substuting, 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 substuting, 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>.
  
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==Solution 2 (Not rigorous)==
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==Solution 2 (not rigorous)==
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We assume that Henry is walking back and forth exactly between points <math>A</math> and <math>B</math>, with <math>A</math> closer to Henry's home than <math>B</math>. Denote Henry's home as a point <math>H</math> and the gym as a point <math>G</math>. Then <math>HA:AB = 1:3</math> and <math>AB:BG = 3:1</math>, so <math>HA:AB:BG = 1:3:1</math>. Therefore, <math>|A-B| = AB = \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) } 1\frac{1}{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
 +
 
==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 18:44, 18 February 2019

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

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

Video Solution

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

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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