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

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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>.
 
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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==Video Solution==
 
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https://www.youtube.com/watch?v=XO-lwBWP2tE
 
 
==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 07:42, 30 September 2022

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 $\boxed{\textbf{(D)}\; (13, -22)}$.

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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