Difference between revisions of "2015 AMC 10B Problems/Problem 24"
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Then he takes <math>2</math> steps West and <math>2</math> steps South, ending at <math>(-1,-1</math>) after <math>2+4</math> steps, and about to head East. | Then he takes <math>2</math> steps West and <math>2</math> steps South, ending at <math>(-1,-1</math>) after <math>2+4</math> steps, and about to head East. | ||
− | Then he takes <math>3</math> steps East and <math>3</math> steps North, ending at <math>(2,2)</math> after <math>2+4+6</math> steps, and about to head | + | Then he takes <math>3</math> steps East and <math>3</math> steps North, ending at <math>(2,2)</math> after <math>2+4+6</math> steps, and about to head . |
Then he takes <math>4</math> steps West and <math>4</math> steps South, ending at <math>(-2,-2)</math> after <math>2+4+6+8</math> steps, and about to head East. | Then he takes <math>4</math> steps West and <math>4</math> steps South, ending at <math>(-2,-2)</math> after <math>2+4+6+8</math> steps, and about to head East. |
Revision as of 11:46, 19 August 2015
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
(Used from 2015 AMC 10/12 B Math Jam)
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 .
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
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
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