2015 AMC 10B Problems/Problem 24
Problem
Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin facing to the east and walks one unit, arriving at . For , right after arriving at the point , if Aaron can turn left and walk one unit to an unvisited point , he does that. Otherwise, he walks one unit straight ahead to reach . Thus the sequence of points continues , and so on in a counterclockwise spiral pattern. What is ?
Solution 1
The first thing we would do is track Aaron's footsteps:
He starts by taking step East and step North, ending at after steps and about to head West.
Then he takes steps West and steps South, ending at ) after steps, and about to head East.
Then he takes steps East and steps North, ending at after steps, and about to head West.
Then he takes steps West and steps South, ending at after steps, and about to head East.
From this pattern, we can notice that for any integer he's at after steps, and about to head East. There are terms in the sum, with an average value of , so:
If we substitute into the equation: . So he has moves to go. This makes him end up at .
(Like: 3)
Solution 2
We are given that Aaron starts at , and we note that his net steps follow the pattern of in the -direction, in the -direction, in the -direction, in the -direction, in the -direction, in the -direction, and so on, where we add odd and subtract even.
We want , but it does not work out cleanly. Instead, we get that , which means that there are extra steps past adding in the -direction (and the final number we add in the -direction is ).
So .
We can group as .
Thus .
Solution 3
Looking at his steps, we see that he walks in a spiral shape. At the th step, he is on the bottom right corner of the square centered on the origin. On the th step, he is on the bottom right corner of the square centered at the origin. It seems that the is the bottom right corner of the square. This makes sense since, after , he has been on dots, including the point . Also, this is only for odd , because starting with the square, we can only add one extra set of dots to each side, so we cannot get even . Since , is the bottom right corner of the square. This point is over to the right, and therefore down, so . Since is ahead of , we go back spaces to .
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.