2019 AMC 10B Problem 18

P16-20~5 minPrint

Sequences & Series Linear Equations & Word Problems Set up the equation/formula and compute; no special trick needed Define states/recurrence and iterate

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 $|A-B|$ ?

Show solution without answering

Key insight

In the long run Henry bounces between two fixed turning points. Walking toward the gym from A must end at B, and walking home from B must end at A.

Solution

Measure every position in kilometers from home, so home is $0$ and the gym is $2$ .

Step 1: what one leg of the walk does.

Walking $\frac34$ of the way home from position $q$ : he covers $\frac34$ of the distance $q$ , so he ends at $q - \frac34 q = \frac{q}{4}$ .
Walking $\frac34$ of the way toward the gym from position $p$ : the distance left is $2 - p$ , so he ends at $p + \frac34(2 - p) = \frac{p}{4} + \frac32$ .

Step 2: watch the first few turns.

Start at $0$ . Toward the gym: $\frac32 = 1.5$ . Home: $\frac{1.5}{4} = 0.375$ . Gym: $\frac{0.375}{4} + 1.5 \approx 1.594$ . Home: $\approx 0.398$ . Gym: $\approx 1.600$ .
The turning points settle near $1.6$ and $0.4$ . Let us find them exactly.

Step 3: the settled points reproduce each other.

5. Call the point near home $A$ and the point near the gym $B$ . Walking toward the gym from $A$ lands on $B$ , and walking home from $B$ lands on $A$ :
$B = \frac{A}{4} + \frac32, \qquad A = \frac{B}{4}.$
6. Substitute $A = \frac{B}{4}$ into the first equation: $B = \frac{B}{16} + \frac32$ . So $\frac{15}{16}B = \frac32$ and $B = \frac32 \cdot \frac{16}{15} = \frac85$ .
7. Then $A = \frac{8/5}{4} = \frac25$ .
8. $|A - B| = \frac85 - \frac25 = \frac65 = 1\frac15$ .

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

Why this works

When a back-and-forth process settles down, the points it settles on must reproduce themselves: one step from $A$ gives $B$ , and one step from $B$ gives $A$ again. Writing that down gives two simple equations, which is much faster than following the walk forever. A few steps of the actual walk are still useful as a check on the numbers.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Measure positions in kilometers from home, with the gym at $2$ .

Walking $\tfrac34$ of the way toward the gym from position $p$ ends at $p + \tfrac34(2 - p) = \tfrac{p}{4} + \tfrac32$ .
Walking $\tfrac34$ of the way toward home from position $q$ ends at $\tfrac{q}{4}$ .

In the limit, Henry oscillates between a point $A$ (where he turns toward the gym) and a point $B$ (where he turns toward home). Each must produce the other:
$B = \frac{A}{4} + \frac32, \qquad A = \frac{B}{4}.$
Substituting the second into the first: $B = \frac{B}{16} + \frac32$ , so $\frac{15}{16}B = \frac32$ and $B = \frac{8}{5}$ . Then $A = \frac{2}{5}$ .

$|A - B| = \frac85 - \frac25 = \frac65 = 1\tfrac15.$

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

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Intermediate Counting & Probability, Chapter 10: Recursion
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)