Problem:
Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin p0=(0,0) facing to the east and walks one unit, arriving at p1=(1,0). For n=1,2,3,…, right after arriving at the point pn, if Aaron can turn 90∘ left and walk one unit to an unvisited point pn+1, he does that. Otherwise, he walks one unit straight ahead to reach pn+1. Thus the sequence of points continues p2=(1,1),p3=(0,1),p4=(−1,1),p5=(−1,0), and so on in a counterclockwise spiral pattern. What is p2015?
Answer Choices:
A. (−22,−13)
B. (−13,−22)
C. (−13,22)
D. (13,−22)
E. (22,−13)
Solution:
Note that for any natural number k, when Aaron reaches point (k,−k), he will have just completed visiting all of the grid points within the square with vertices at (k,−k),(k,k),(−k,k), and (−k,−k). Thus the point (k,−k) is equal to p(2k+1)2−1. It follows that p2024=p(2⋅22+1)2−1=(22,−22). Because 2024−2015=9, the point p2015=(22−9,−22)=(D)(13,−22).
The problems on this page are the property of the MAA's American Mathematics Competitions