Difference between revisions of "2015 AMC 10B Problems/Problem 24"
Jskalarickal (talk | contribs) (→Solution 3) |
(???) |
||
(15 intermediate revisions by 15 users not shown) | |||
Line 4: | Line 4: | ||
<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> | ||
− | + | ==Solution 1== | |
− | |||
− | |||
The first thing we would do is track Aaron's footsteps: | The first thing we would do is track Aaron's footsteps: | ||
Line 23: | Line 21: | ||
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>. | ||
− | + | ==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. | ||
Line 34: | Line 32: | ||
Thus <math>p_{2015} = \boxed{\textbf{(D)}\; (13, -22)}</math>. | Thus <math>p_{2015} = \boxed{\textbf{(D)}\; (13, -22)}</math>. | ||
− | + | ==Solution 3== | |
− | Looking at his steps, we see that he walks in a spiral shape. At the | + | 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}} |
Latest revision as of 17:42, 13 August 2023
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
.
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.