Difference between revisions of "2015 AMC 12B Problems/Problem 25"
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<math>\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048</math> | <math>\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048</math> | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>x = e^{i \pi / 6}</math>, a <math>30^\circ</math> counterclockwise rotation centered at the origin. Notice that <math>P_k</math> on the complex plane is: | Let <math>x = e^{i \pi / 6}</math>, a <math>30^\circ</math> counterclockwise rotation centered at the origin. Notice that <math>P_k</math> on the complex plane is: | ||
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Thus the answer is <math>1008 + 1008 + 2 + 6 = \boxed{\textbf{(B)}\; 2024}.</math> | Thus the answer is <math>1008 + 1008 + 2 + 6 = \boxed{\textbf{(B)}\; 2024}.</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Here is an alternate solution that does not use complex numbers: | ||
+ | |||
+ | We will calculate the distance from <math>P_{2015}</math> to <math>P_0</math> using the Pythagorean theorem. Assume <math>P_0</math> lies at the origin, so we will calculate the distance to <math>P_{2015}</math> by calculating the distance traveled in the x-direction and the distance traveled in the y-direction. We can calculate this by summing each movement: | ||
+ | |||
+ | <math>x=1\cos{0}+2\cos {30}+3\cos {60}+4\cos {90}+5\cos {120}+\cdot \cdot \cdot+2011\cos{180}+2012\cos {210}+2013\cos{240}+2014\cos{270}+2015\cos{300}</math> | ||
+ | |||
+ | A movement of <math>p</math> units at <math>q</math> degrees is the same thing as a movement of <math>-p</math> units at <math>q-180</math> degrees, so we can adjust all the cosines with arguments greater than 180 as follows: | ||
+ | |||
+ | <math>x=1\cos{0}+2\cos {30}+3\cos {60}+4\cos {90}+5\cos {120}+6\cos{150}-7\cos{0}-8\cos{30}-\cdot \cdot \cdot -2015 \cos{120}</math> | ||
+ | |||
+ | Now we group terms with like-cosines and factor out the cosines: | ||
+ | |||
+ | <math>x=(1-7+13-\cdot \cdot \cdot +2005-2011)\cos{0}+(2-8+14- \cdot \cdot \cdot +2006-2012)\cos{30}+\cdot \cdot \cdot +(5-11+17- \cdot \cdot \cdot +2008-2014)\cos{120}+(6-12+18- \cdot \cdot \cdot -2004+2010)\cos{150}</math> | ||
+ | |||
+ | Each sum in the parentheses has 336 terms (except the very last one, which has 335), so by pairing each term, we can see that there are <math>\frac {336}{2}</math> pairs of <math>-6</math>. So each sum evaluates to <math>168\cdot -6=-1008</math>, except the very last sum, which has 167 pairs of <math>-6</math> and an extra 2010, so it evaluates to <math>167\cdot -6+2010=1008</math>. Plugging in these values: | ||
+ | |||
+ | <math>x=-1008\cos{0}-1008\cos{30}-1008\cos{60}-1008\cos{90}-1008\cos{120}+1008\cos{150}</math> | ||
+ | <math>x=1008(-1-\frac{\sqrt{3}}{2}-\frac{1}{2}-0+\frac{1}{2}-\frac{\sqrt{3}}{2})=-1008(1+\sqrt{3})</math> | ||
+ | |||
+ | Now that we have how far was traveled in the x-direction, we need to find how far was traveled in the y-direction. Using the same logic as above, we arrive at the sum: | ||
+ | |||
+ | <math>y=-1008\sin{0}-1008\sin{30}-1008\sin{60}-1008\sin{90}-1008\sin{120}+1008\sin{150}</math> | ||
+ | |||
+ | <math>y=1008(0-\frac{1}{2}-\frac{\sqrt{3}}{2}-1-\frac{\sqrt{3}}{2}+\frac{1}{2})=-1008(1+\sqrt{3})</math> | ||
+ | |||
+ | The last step is to use the Pythagorean to find the distance from <math>P_0</math>. This distance is given by: | ||
+ | |||
+ | <math>\sqrt{x^2+y^2}=\sqrt{(-1008(1+\sqrt{3}))^2+(-1008(1+\sqrt{3}))^2}=\sqrt{2\cdot 1008^2 \cdot (1+\sqrt{3})^2}=1008(1+\sqrt{3})\sqrt{2}</math> | ||
+ | |||
+ | Multiplying out, we have <math>1008\sqrt{2}+1008\sqrt{6}</math>, so the answer is <math>1008+2+1008+6= \boxed {\bold {(B)}\; 2024}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2015|ab=B|after=Last Problem|num-b=24}} | {{AMC12 box|year=2015|ab=B|after=Last Problem|num-b=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 03:19, 16 January 2018
Contents
Problem
A bee starts flying from point . She flies
inch due east to point
. For
, once the bee reaches point
, she turns
counterclockwise and then flies
inches straight to point
. When the bee reaches
she is exactly
inches away from
, where
,
,
and
are positive integers and
and
are not divisible by the square of any prime. What is
?
Solution 1
Let , a
counterclockwise rotation centered at the origin. Notice that
on the complex plane is:
We need to find the magnitude of on the complex plane. This is an arithmetic/geometric series.
We want to find . First, note that
because
. Therefore
Hence, since , we have
Now we just have to find . This can just be computed directly:
Therefore
Thus the answer is
Solution 2
Here is an alternate solution that does not use complex numbers:
We will calculate the distance from to
using the Pythagorean theorem. Assume
lies at the origin, so we will calculate the distance to
by calculating the distance traveled in the x-direction and the distance traveled in the y-direction. We can calculate this by summing each movement:
A movement of units at
degrees is the same thing as a movement of
units at
degrees, so we can adjust all the cosines with arguments greater than 180 as follows:
Now we group terms with like-cosines and factor out the cosines:
Each sum in the parentheses has 336 terms (except the very last one, which has 335), so by pairing each term, we can see that there are pairs of
. So each sum evaluates to
, except the very last sum, which has 167 pairs of
and an extra 2010, so it evaluates to
. Plugging in these values:
Now that we have how far was traveled in the x-direction, we need to find how far was traveled in the y-direction. Using the same logic as above, we arrive at the sum:
The last step is to use the Pythagorean to find the distance from . This distance is given by:
Multiplying out, we have , so the answer is
.
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.