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2019 AMC 10B Problem 18

2019 AMC 10B Problem 18AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 3 of 5 · about 5 min · position P16-20

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 $|A-B|$ ?

A)
$\frac{2}{3}$
B)
$1$
C)
$1\frac{1}{5}$
D)
$1\frac{1}{4}$
E)
$1\frac{1}{2}$

Answer

Key insight

At the limit the two turning points reproduce each other: A = B/4 and B = A + (3/4)(2 - A); solving gives A = 2/5, B = 8/5.

Solution

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

Why this works

Each leg of the walk is a linear map on the position, and a "back and forth forever" pattern converges to the fixed points of the two-step map. Rather than computing many iterations, demand that the limiting points map to each other and solve the resulting system. The convergence is guaranteed because each map shrinks distances by a factor of $4$ .

Alternative approach

Iterate a few times: turning points are $\frac32, \frac38, \frac{51}{32}, \frac{51}{128}, \ldots$ , i.e. $1.5, 0.375, 1.594, 0.398, \ldots$ , visibly approaching $1.6$ and $0.4$ , whose difference is $1.2$ .

The trap

Using the first two turning points 3/2 and 3/8 (or the next pair) as A and B, instead of the limiting points.

Common mistakes

Using the first two turning points 3/2 and 3/8 (or the next pair) as A and B, instead of the limiting points.
Writing the gym-ward step as $\frac34 p$ (three quarters of the position) rather than $p + \frac34(2 - p)$ .

Techniques

Set up the equation/formula and compute; no special trick needed · Define states/recurrence and iterate