Question

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 $|A-B|?$

$\frac{2}{3}$

$1$

$1 \frac{1}{5}$

$1 \frac{1}{4}$

$1 \frac{1}{2}$

Solution:

Suppose he starts $x$ miles from home and then goes in the direction of the gym before coming back. The current distance between him and the gym is $2-x,$ so he would have a distance of $\frac{2-x}4$ miles from the gym after walking. This would put him $2-\frac {2-x}4 = 1.5 + \frac x4$ miles from home.

Then, when he walks home, he becomes $\frac{1.5 + \frac x4}4 = \frac 38 + \frac x {16}$ miles from home.

If Henry is at one of the points, then his position would remain the same, so $x= \frac 38 +\frac x{16}.$ As such, we know that $16=6x+x$ $x=0.4.$ Thus, one of the steady state points is $0.4$ miles from the house.

Using similar logic, but starting $x$ miles from the gym and then going home, and then to the gym, we get that another point is $1.6$ miles from the house.

Therefore, the points are $0.4$ and $1.6$ miles, so their difference is $1.2$ miles.

Thus, the correct answer is C .

2019 AMC 10B Problem 18

Solution:

Problem 18 in Other Years