2019 AMC 10B Problem 18

Problem 18 of 25IntermediateAlgebra

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|?

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Solution

Suppose a limiting homeward turnaround point is

x

kilometers from home. After Henry walks toward the gym, his position is

2-\frac{2-x}{4}=\frac32+\frac{x}{4}.

After he turns back toward home, one quarter of that distance remains, so his next homeward turnaround point is

\frac14\left(\frac32+\frac{x}{4}\right)=\frac38+\frac{x}{16}.

At the limiting point this two-step map leaves the position unchanged, so

x=\frac38+\frac{x}{16},

which gives

x=\frac25

. The other limiting point is the position reached after walking toward the gym:

2-\frac{2-\frac25}{4}=\frac85.

Therefore

|A-B|=\frac85-\frac25=\frac65=1\frac15.

Thus, the correct answer is C .

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