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

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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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==Alternate Solution==
 
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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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We want <math>2 + 4 + 6 + 8 + ... + 2n = 2015</math>, but it does not work out cleanly. Instead, we get that <math>2 + 4 + 6 + ... + 2(44) = 1980</math>, which means that there are <math>35</math> extra steps past adding <math>-44</math> in the <math>x</math>-direction (and the final number we add in the <math>y</math>-direction is <math>-44</math>).
 
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So <math>p_{2015} = (0+1-2+3-4+5...-44+35, 0+1-2+3-4+5...-44)</math>.
 
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We can group <math>1-2+3-4+5...-44</math> as <math>(1-2)+(3-4)+(5-6)+...+(43-44) = -22</math>.
 
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Thus <math>p_{2015} = \boxed{\textbf{(D)}\; (13, -22)}</math>.
 
  
 
==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 11:06, 26 July 2017

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

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 $(-22+35,-22) = (13,-22) \implies \boxed{\textbf{(D)} (13, -22)}$

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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