Difference between revisions of "2015 AMC 10B Problems/Problem 24"
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==Solution 3== | ==Solution 3== | ||
Looking at his steps, we see that he walks in a spiral shape. At the <math>8</math>th step, he is on the bottom right corner of the <math>3\times 3</math> square centered on the origin. On the <math>24</math>th step, he is on the bottom right corner of the <math>5\times 5</math> square centered at the origin. It seems that the <math>p_{n^2-1}</math> is the bottom right corner of the <math>n\times n</math> square. This makes sense since, after <math>n^2-1</math>, he has been on <math>n^2</math> dots, including the point <math>p_0</math>. Also, this is only for odd <math>n</math>, because starting with the <math>1\times 1</math> square, we can only add one extra set of dots to each side, so we cannot get even <math>n</math>. Since <math>45^2=2025</math>, <math>p_{2024}</math> is the bottom right corner of the <math>45\times 45</math> square. This point is <math>\frac{45-1}{2}=22</math> over to the right, and therefore <math>22</math> down, so <math>p_{2024}=(22, -22)</math>. Since <math>p_{2024}</math> is <math>9</math> ahead of <math>p_{2015}</math>, we go back <math>9</math> spaces to <math>\boxed{\textbf{(D)}\; (13, -22)}</math>. | Looking at his steps, we see that he walks in a spiral shape. At the <math>8</math>th step, he is on the bottom right corner of the <math>3\times 3</math> square centered on the origin. On the <math>24</math>th step, he is on the bottom right corner of the <math>5\times 5</math> square centered at the origin. It seems that the <math>p_{n^2-1}</math> is the bottom right corner of the <math>n\times n</math> square. This makes sense since, after <math>n^2-1</math>, he has been on <math>n^2</math> dots, including the point <math>p_0</math>. Also, this is only for odd <math>n</math>, because starting with the <math>1\times 1</math> square, we can only add one extra set of dots to each side, so we cannot get even <math>n</math>. Since <math>45^2=2025</math>, <math>p_{2024}</math> is the bottom right corner of the <math>45\times 45</math> square. This point is <math>\frac{45-1}{2}=22</math> over to the right, and therefore <math>22</math> down, so <math>p_{2024}=(22, -22)</math>. Since <math>p_{2024}</math> is <math>9</math> ahead of <math>p_{2015}</math>, we go back <math>9</math> spaces to <math>\boxed{\textbf{(D)}\; (13, -22)}</math>. | ||
+ | |||
+ | ==Solution 4 (similar to 3)== | ||
+ | We call <math>p_0</math> the first point, <math>p_1</math> the second, and so on. As in Solution 3, we see that he walks in a counterclockwise spiral. We see that his path is traced out by a series of squares with odd-length sides that contain each other. At step <math>3^2=9,</math> he is at <math>(1,-1)</math>; at step <math>5^2=25,</math> he is at <math>(2,-2)</math>; and so on. We see that at step <math>n^2</math> where <math>n</math> is an odd number, he is at <math>(x,-x)</math> where <math>x=\dfrac{n-1}2.</math> The closest <math>n^2</math> to <math>2016</math> (we want <math>p_{2015},</math> which is the <math>2016</math>th step) is <math>45^2=2025,</math> so we let <math>n=45.</math> On the <math>45^2=2025</math>th step, <math>x=\dfrac{45-1}2=22,</math> so he is at <math>(22,-22).</math> We see that he approached <math>(22,-22)</math> from the left, so backtrack <math>2025-2016=9</math> steps to the left (i.e. subtracting <math>9</math> from the x-coordinate). Thus, we have <math>(22-9,-22)=\boxed{\textbf{(D)}~(13,-22).}</math> | ||
+ | ~Technodoggo | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|num-b=23|num-a=25}} | {{AMC10 box|year=2015|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:49, 5 August 2023
Contents
[hide]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 .
This is just beautiful!!! - sleepypuppy
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 .
Solution 4 (similar to 3)
We call the first point, the second, and so on. As in Solution 3, we see that he walks in a counterclockwise spiral. We see that his path is traced out by a series of squares with odd-length sides that contain each other. At step he is at ; at step he is at ; and so on. We see that at step where is an odd number, he is at where The closest to (we want which is the th step) is so we let On the th step, so he is at We see that he approached from the left, so backtrack steps to the left (i.e. subtracting from the x-coordinate). Thus, we have ~Technodoggo
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.