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Difference between revisions of "2015 AMC 12B Problems/Problem 25"

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<cmath>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)}.</cmath>
 
<cmath>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)}.</cmath>
  
−
Hence, since <math>|x|=1</math>, we have
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Hence, since <math>|x|=1</math>, we have <math>|S| = \frac{2016}{|1-x|}.</math>
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<cmath>|S| = \frac{2016}{|1-x|}.</cmath>
 
  
 
Now we just have to find <math>|1-x|</math>. This can just be computed directly:
 
Now we just have to find <math>|1-x|</math>. This can just be computed directly:
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<cmath>|1-x| = \frac{\sqrt{6} - \sqrt{2}}{2}.</cmath>
 
<cmath>|1-x| = \frac{\sqrt{6} - \sqrt{2}}{2}.</cmath>
  
−
Therefore
+
Therefore <math>|S| = 2016 \cdot \frac{2}{\sqrt{6} -\sqrt{2}} = 2016 \left( \frac{\sqrt{6} + \sqrt{2}}{2} \right) = 1008 \sqrt{2} + 1008 \sqrt{6}.</math>
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<cmath>|S| = 2016 \cdot \frac{2}{\sqrt{6} -\sqrt{2}} = 2016 \left( \frac{\sqrt{6} + \sqrt{2}}{2} \right) = 1008 \sqrt{2} + 1008 \sqrt{6}.</cmath>
 
  
 
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>

Revision as of 08:53, 6 March 2015

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

Let $x = e^{i \pi / 6}$ (a $30^\circ$ angle). We're going to toss this onto the complex plane. Notice that $P_k$ on the complex plane is:

\[1 + 2x + 3x^2 + \cdots + kx^k\]

We just want 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 $1008 + 1008 + 2 + 6 = \boxed{\textbf{(B)}\; 2024}.$

See Also

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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All AMC 12 Problems and Solutions

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