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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:
Line 34: Line 34:
  
 
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 02:19, 16 January 2018

Problem

A bee starts flying from point $P_0$. She flies $1$ inch due east to point $P_1$. For $j \ge 1$, once the bee reaches point $P_j$, she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$. When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$, where $a$, $b$, $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$ ?

$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$

Solution 1

Let $x = e^{i \pi / 6}$, a $30^\circ$ counterclockwise rotation centered at the origin. Notice that $P_k$ on the complex plane is:

\[1 + 2x + 3x^2 + \cdots + (k+1)x^k\]

We need to find the magnitude of $P_{2015}$ on the complex plane. This is an arithmetic/geometric series.

\begin{align*} S &= 1 + 2x + 3x^2 + \cdots + 2015x^{2014} \\ xS &= x + 2x^2 + 3x^3 + \cdots + 2015x^{2015} \\ (1-x)S &= 1 + x + x^2 + \cdots + x^{2014} - 2015x^{2015} \\ S &= \frac{1 - x^{2015} }{(1-x)^2} - \frac{2015x^{2015}}{1-x} \end{align*}

We want to find $|S|$. First, note that $x^{2015} = x^{11} = x^{-1}$ because $x^{12} = 1$. Therefore

\[S = \frac{1 - \frac{1}{x}}{(1-x)^2} - \frac{2015}{x(1-x)} = -\frac{1}{x(1-x)} - \frac{2015}{x(1-x)} = -\frac{2016}{x(1-x)}.\]

Hence, since $|x|=1$, we have $|S| = \frac{2016}{|1-x|}.$

Now we just have to find $|1-x|$. This can just be computed directly:

\[1 - x = 1 - \frac{\sqrt{3}}{2} - \frac{1}{2}i\]

\[|1-x|^2 = \left(1 - \sqrt{3} + \frac{3}{4} \right) + \frac{1}{4} = 2 - \sqrt{3} = {\left( \frac{\sqrt{6}-\sqrt{2}}{2} \right)}^2\]

\[|1-x| = \frac{\sqrt{6} - \sqrt{2}}{2}.\]

Therefore $|S| = 2016 \cdot \frac{2}{\sqrt{6} -\sqrt{2}} = 2016 \left( \frac{\sqrt{6} + \sqrt{2}}{2} \right) = 1008 \sqrt{2} + 1008 \sqrt{6}.$

Thus the answer is $1008 + 1008 + 2 + 6 = \boxed{\textbf{(B)}\; 2024}.$

Solution 2

Here is an alternate solution that does not use complex numbers:

We will calculate the distance from $P_{2015}$ to $P_0$ using the Pythagorean theorem. Assume $P_0$ lies at the origin, so we will calculate the distance to $P_{2015}$ 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:

$x=1\cos{0}+2\cos {30}+3\cos {60}+4\cos {90}+5\cos {120}+\cdot \cdot \cdot+2011\cos{180}+2012\cos {210}+2013\cos{240}+2014\cos{270}+2015\cos{300}$

A movement of $p$ units at $q$ degrees is the same thing as a movement of $-p$ units at $q-180$ degrees, so we can adjust all the cosines with arguments greater than 180 as follows:

$x=1\cos{0}+2\cos {30}+3\cos {60}+4\cos {90}+5\cos {120}+6\cos{150}-7\cos{0}-8\cos{30}-\cdot \cdot \cdot -2015 \cos{120}$

Now we group terms with like-cosines and factor out the cosines:

$x=(1-7+13-\cdot \cdot \cdot +2005-2011)\cos{0}+(2-8+14- \cdot \cdot \cdot +2006-2012)\cos{30}+\cdot \cdot \cdot +(5-11+17- \cdot \cdot \cdot +2008-2014)\cos{120}+(6-12+18- \cdot \cdot \cdot -2004+2010)\cos{150}$

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 $\frac {336}{2}$ pairs of $-6$. So each sum evaluates to $168\cdot -6=-1008$, except the very last sum, which has 167 pairs of $-6$ and an extra 2010, so it evaluates to $167\cdot -6+2010=1008$. Plugging in these values:

$x=-1008\cos{0}-1008\cos{30}-1008\cos{60}-1008\cos{90}-1008\cos{120}+1008\cos{150}$ $x=1008(-1-\frac{\sqrt{3}}{2}-\frac{1}{2}-0+\frac{1}{2}-\frac{\sqrt{3}}{2})=-1008(1+\sqrt{3})$

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:

$y=-1008\sin{0}-1008\sin{30}-1008\sin{60}-1008\sin{90}-1008\sin{120}+1008\sin{150}$

$y=1008(0-\frac{1}{2}-\frac{\sqrt{3}}{2}-1-\frac{\sqrt{3}}{2}+\frac{1}{2})=-1008(1+\sqrt{3})$

The last step is to use the Pythagorean to find the distance from $P_0$. This distance is given by:

$\sqrt{x^2+y^2}=\sqrt{(-1008(1+\sqrt{3}))^2+(-1008(1+\sqrt{3}))^2}=\sqrt{2\cdot 1008^2 \cdot (1+\sqrt{3})^2}=1008(1+\sqrt{3})\sqrt{2}$

Multiplying out, we have $1008\sqrt{2}+1008\sqrt{6}$, so the answer is $1008+2+1008+6= \boxed {\bold {(B)}\; 2024}$.

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

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