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Difference between revisions of "2015 AMC 10B Problems/Problem 24"

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<math> \textbf{(A) } (-22,-13)\qquad\textbf{(B) } (-13,-22)\qquad\textbf{(C) } (-13,22)\qquad\textbf{(D) } (13,-22)\qquad\textbf{(E) } (22,-13) </math>
 
<math> \textbf{(A) } (-22,-13)\qquad\textbf{(B) } (-13,-22)\qquad\textbf{(C) } (-13,22)\qquad\textbf{(D) } (13,-22)\qquad\textbf{(E) } (22,-13) </math>
  
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==Solution==
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==Solution 1==
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===Solution 1===
 
 
The first thing we would do is track Aaron's footsteps:
 
The first thing we would do is track Aaron's footsteps:
  
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If we substitute <math>k = 22</math> into the equation:  <math>44(45) = 1980 < 2015</math>.  So he has <math>35</math> moves to go.  This makes him end up at <math>(-22+35,-22) = (13,-22) \implies \boxed{\textbf{(D)} (13, -22)}</math>.
 
If we substitute <math>k = 22</math> into the equation:  <math>44(45) = 1980 < 2015</math>.  So he has <math>35</math> moves to go.  This makes him end up at <math>(-22+35,-22) = (13,-22) \implies \boxed{\textbf{(D)} (13, -22)}</math>.
  
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===Solution 2===
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==Solution 2==
 
We are given that Aaron starts at <math>(0, 0)</math>, and we note that his net steps follow the pattern of <math>+1</math> in the <math>x</math>-direction, <math>+1</math> in the <math>y</math>-direction, <math>-2</math> in the <math>x</math>-direction, <math>-2</math> in the <math>y</math>-direction, <math>+3</math> in the <math>x</math>-direction, <math>+3</math> in the <math>y</math>-direction, and so on, where we add odd and subtract even.
 
We are given that Aaron starts at <math>(0, 0)</math>, and we note that his net steps follow the pattern of <math>+1</math> in the <math>x</math>-direction, <math>+1</math> in the <math>y</math>-direction, <math>-2</math> in the <math>x</math>-direction, <math>-2</math> in the <math>y</math>-direction, <math>+3</math> in the <math>x</math>-direction, <math>+3</math> in the <math>y</math>-direction, and so on, where we add odd and subtract even.
  
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Thus <math>p_{2015} = \boxed{\textbf{(D)}\; (13, -22)}</math>.
 
Thus <math>p_{2015} = \boxed{\textbf{(D)}\; (13, -22)}</math>.
  
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===Solution 3===
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==Solution 3==
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Looking at his steps, we see that he walks in a spiral shape. At the 8th step, he is on the bottom right corner of the 3x3 square centered on the origin. On the 24th step, he is on the bottom right corner of the 5x5 square centered at the origin. It seems that the <math>p_{n^2-1}</math> is the bottom right corner of the <math>n</math>x<math>n</math> square. This makes sense since, after <math>n^2-1</math>, he has been on n^2 dots, including the point <math>p_{0}</math>. Also, this is only for odd <math>n</math>, because starting with the 1x1 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 45x45 square. This point is <math>\frac{45-1}{2}=22</math> over to the right, and therefore 22 down, so <math>p_{2024}=(22, -22). Since </math>p_{2024}<math> is 9 ahead of </math>p_{2015}<math>, we go back 9 spaces to </math>\boxed{\textbf{(D)}\; (13, -22)}$.
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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>.
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 +
==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>
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~Technodoggo
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==Solution 5 (Similar to Solution 1)==
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The first thing is to see the amount of footsteps at points <math>(k,k)</math>
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We see that at <math>(1,1)</math> he has taken 2 footsteps.
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To get to <math>(2,2)</math>, Aaron takes <math>2</math> steps West, <math>2</math> steps South, <math>3</math> steps East, and <math>3</math> steps North.
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 +
We can think about this in the following way. For Aaron to get from <math>(k,k)</math> to <math>(k+1,k+1)</math>, he takes <math>2k+2k+(2k+1)+(2k+1)</math> steps = <math>8k+2</math> steps
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We can find the total steps taken to get to a point <math>(k+1,k+1)</math>
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<math>2+10+18... +8k+2</math> = <math>(4k+2)(k+1</math>) works as a formula
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We find that to get to <math>(22,22)</math> this is 2+10+18...+ (4(21)+2)(21)= 1892. We then simply go <math>88</math> steps to get to <math>(-22,-22)</math>
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and <math>35</math> more to get to the 2015th step at <math>(13,-22)</math> or answer D
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==Solution 6==
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If we graph <math>p_n</math> for a few values of <math>n</math>, we begin to see a particular pattern for <math>p_n</math> when <math>n</math> is a perfect square. For example, <math>p_0=(0,0)</math>, <math>p_4=(-1,1)</math>, <math>p_{16}=(2,-2)</math>, and so on. Similarly, <math>p_1=(1,0)</math>, <math>p_9=(2,-1)</math>, <math>p_{25}=(3,-2)</math>, and so on. Using this, we can generate a pattern for even and odd squares: <cmath>p_{(2n)^2}=(-n, n)</cmath> <cmath>p_{(2n+1)^2}=(n+1,-n)</cmath>
 +
 
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We notice that <math>p_{2015}</math> is quite close to <math>p_{2025}=p_{45^2}</math>. By the same pattern for odd squares, <math>p_{2025}=(23,-22)</math>. <math>p_{2015}</math> is just <math>10</math> units to the left of <math>p_{2025}</math>. Therefore, <math>p_{2015}=\boxed{\textbf{(D) }(13,-22)}</math>.
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 +
~[https://artofproblemsolving.com/wiki/index.php/User:Sid2012 sid2012]
  
 
==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}}

Latest revision as of 22:36, 4 November 2024

Problem

Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin $p_0=(0,0)$ facing to the east and walks one unit, arriving at $p_1=(1,0)$. For $n=1,2,3,\dots$, right after arriving at the point $p_n$, if Aaron can turn $90^\circ$ left and walk one unit to an unvisited point $p_{n+1}$, he does that. Otherwise, he walks one unit straight ahead to reach $p_{n+1}$. Thus the sequence of points continues $p_2=(1,1), p_3=(0,1), p_4=(-1,1), p_5=(-1,0)$, and so on in a counterclockwise spiral pattern. What is $p_{2015}$?

$\textbf{(A) } (-22,-13)\qquad\textbf{(B) } (-13,-22)\qquad\textbf{(C) } (-13,22)\qquad\textbf{(D) } (13,-22)\qquad\textbf{(E) } (22,-13)$

Solution 1

The first thing we would do is track Aaron's footsteps:

He starts by taking $1$ step East and $1$ step North, ending at $(1,1)$ after $2$ steps and about to head West.

Then he takes $2$ steps West and $2$ steps South, ending at $(-1,-1$) after $2+4$ steps, and about to head East.

Then he takes $3$ steps East and $3$ steps North, ending at $(2,2)$ after $2+4+6$ steps, and about to head West.

Then he takes $4$ steps West and $4$ steps South, ending at $(-2,-2)$ after $2+4+6+8$ steps, and about to head East.

From this pattern, we can notice that for any integer $k \ge 1$ he's at $(-k, -k)$ after $2 + 4 + 6 + ... + 4k$ steps, and about to head East. There are $2k$ terms in the sum, with an average value of $(2 + 4k)/2 = 2k + 1$, so:

\[2 + 4 + 6 + ... + 4k = 2k(2k + 1)\]

If we substitute $k = 22$ into the equation: $44(45) = 1980 < 2015$. So he has $35$ moves to go. This makes him end up at $(-22+35,-22) = (13,-22) \implies \boxed{\textbf{(D)} (13, -22)}$.

Solution 2

We are given that Aaron starts at $(0, 0)$, and we note that his net steps follow the pattern of $+1$ in the $x$-direction, $+1$ in the $y$-direction, $-2$ in the $x$-direction, $-2$ in the $y$-direction, $+3$ in the $x$-direction, $+3$ in the $y$-direction, and so on, where we add odd and subtract even.

We want $2 + 4 + 6 + 8 + ... + 2n = 2015$, but it does not work out cleanly. Instead, we get that $2 + 4 + 6 + ... + 2(44) = 1980$, which means that there are $35$ extra steps past adding $-44$ in the $x$-direction (and the final number we add in the $y$-direction is $-44$).

So $p_{2015} = (0+1-2+3-4+5...-44+35, 0+1-2+3-4+5...-44)$.

We can group $1-2+3-4+5...-44$ as $(1-2)+(3-4)+(5-6)+...+(43-44) = -22$.

Thus $p_{2015} = \boxed{\textbf{(D)}\; (13, -22)}$.

Solution 3

Looking at his steps, we see that he walks in a spiral shape. At the $8$th step, he is on the bottom right corner of the $3\times 3$ square centered on the origin. On the $24$th step, he is on the bottom right corner of the $5\times 5$ square centered at the origin. It seems that the $p_{n^2-1}$ is the bottom right corner of the $n\times n$ square. This makes sense since, after $n^2-1$, he has been on $n^2$ dots, including the point $p_0$. Also, this is only for odd $n$, because starting with the $1\times 1$ square, we can only add one extra set of dots to each side, so we cannot get even $n$. Since $45^2=2025$, $p_{2024}$ is the bottom right corner of the $45\times 45$ square. This point is $\frac{45-1}{2}=22$ over to the right, and therefore $22$ down, so $p_{2024}=(22, -22)$. Since $p_{2024}$ is $9$ ahead of $p_{2015}$, we go back $9$ spaces to $\boxed{\textbf{(D)}\; (13, -22)}$.

Solution 4 (similar to 3)

We call $p_0$ the first point, $p_1$ 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 $3^2=9,$ he is at $(1,-1)$; at step $5^2=25,$ he is at $(2,-2)$; and so on. We see that at step $n^2$ where $n$ is an odd number, he is at $(x,-x)$ where $x=\dfrac{n-1}2.$ The closest $n^2$ to $2016$ (we want $p_{2015},$ which is the $2016$th step) is $45^2=2025,$ so we let $n=45.$ On the $45^2=2025$th step, $x=\dfrac{45-1}2=22,$ so he is at $(22,-22).$ We see that he approached $(22,-22)$ from the left, so backtrack $2025-2016=9$ steps to the left (i.e. subtracting $9$ from the x-coordinate). Thus, we have $(22-9,-22)=\boxed{\textbf{(D)}~(13,-22).}$ ~Technodoggo

Solution 5 (Similar to Solution 1)

The first thing is to see the amount of footsteps at points $(k,k)$

We see that at $(1,1)$ he has taken 2 footsteps.

To get to $(2,2)$, Aaron takes $2$ steps West, $2$ steps South, $3$ steps East, and $3$ steps North.

We can think about this in the following way. For Aaron to get from $(k,k)$ to $(k+1,k+1)$, he takes $2k+2k+(2k+1)+(2k+1)$ steps = $8k+2$ steps

We can find the total steps taken to get to a point $(k+1,k+1)$

$2+10+18... +8k+2$ = $(4k+2)(k+1$) works as a formula

We find that to get to $(22,22)$ this is 2+10+18...+ (4(21)+2)(21)= 1892. We then simply go $88$ steps to get to $(-22,-22)$

and $35$ more to get to the 2015th step at $(13,-22)$ or answer D

Solution 6

If we graph $p_n$ for a few values of $n$, we begin to see a particular pattern for $p_n$ when $n$ is a perfect square. For example, $p_0=(0,0)$, $p_4=(-1,1)$, $p_{16}=(2,-2)$, and so on. Similarly, $p_1=(1,0)$, $p_9=(2,-1)$, $p_{25}=(3,-2)$, and so on. Using this, we can generate a pattern for even and odd squares: \[p_{(2n)^2}=(-n, n)\] \[p_{(2n+1)^2}=(n+1,-n)\]

We notice that $p_{2015}$ is quite close to $p_{2025}=p_{45^2}$. By the same pattern for odd squares, $p_{2025}=(23,-22)$. $p_{2015}$ is just $10$ units to the left of $p_{2025}$. Therefore, $p_{2015}=\boxed{\textbf{(D) }(13,-22)}$.

~sid2012

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

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