2015 AMC 10B Problems/Problem 24

Revision as of 14:20, 17 January 2018 by Jskalarickal (talk | contribs) (Solution 2)

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

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 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 $p_{n^2-1}$ is the bottom right corner of the $n$x$n$ 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 1x1 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 45x45 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 (ProblemsAnswer KeyResources)
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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