Problem:
Henry decides one morning to do a workout, and he walks 43​ 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 43​ of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks 43​ of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked 43​ 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∣?
Answer Choices:
A. 32​
B. 1
C. 151​
D. 141​
E. 121​
Solution:
By symmetry Henry's walks will converge toward walking between two points, one at a distance x from the gym and the other at the same distance x from his home. Because Henry would be 2−x kilometers from home when he is closest to the gym and also because his trip toward home would take him to 41​ this distance from home, x=41​(2−x). Solving this yields x=52​. Therefore, Henry's walks will approach 2−2⋅52​=151​ kilometers in length.
OR
If Henry is ak​ kilometers from home after his k th walk toward the gym and bk​ kilometers from home after his k th walk toward home, then a0​=b0​=0, and for k≥1,
ak​=bk−1​+43​(2−bk−1​)=23​+41​bk−1​
and
bk​=41​ak​=83​+161​bk−1​
Iterating shows that the sequence (bk​) converges to
B=83​+83​⋅161​+83​⋅(161​)2+⋯=83​⋅1−161​1​=52​
from which it then follows that (ak​) converges to A=23​+41​⋅52​=58​. The requested absolute difference is ∣∣∣∣∣​58​−52​∣∣∣∣∣​=(C)151​​.
OR
Let xk​ denote Henry's distance from home after his k th walk. The following formulas give the value of xk​ :
xk​=52​−5⋅4k2​ when k is even
and
xk​=58​−5⋅4k2​ when k is odd
To prove this by mathematical induction, first note that indeed x0​= 52​−5⋅402​=0 and x1​=58​−5⋅412​=23​=43​⋅2. Then for even values of k≥2, Henry was heading home, so
xk​=41​xk−1​=41​(58​−5⋅4k−12​)=52​−5⋅4k2​
and for odd values of k≥3, Henry was heading toward the gym, so
xk​​=xk−1​+43​(2−xk−1​)=23​+41​xk−1​=23​+41​(52​−5⋅4k−12​)=58​−5⋅4k2​.​
As k approaches infinity, these values rapidly converge to 52​ and 58​, respectively, so Henry is essentially walking back and forth between two points that are 58​−52​=(C)151​​ kilometers apart.
The problems on this page are the property of the MAA's American Mathematics Competitions