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2015 AMC 12B Problems/Problem 25

Revision as of 23:22, 3 March 2015 by Fclvbfm934 (talk | contribs) (→‎Solution)

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-degree angle). We're going to toss this onto the complex plane. Notice that $P_k$ on the complex plane is: \[1 + 2x + 3x^2 + \cdots + kx^k\] Therefore, 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 of all, note that $x^{2015} = x^{11} = x^{-1}$ because $x^{12} = 1$. So then: \[S = \frac{1 - \frac{1}{x}}{(1-x)^2} - \frac{2015}{x(1-x)} = -\frac{1}{x(1-x)} - \frac{2015}{x(1-x)}\] So therefore $S = -\frac{2016}{x(1-x)}$. Let us now find $|S|$: \[|S| = 2016 \left| \frac{1}{1-x} \right|\] We dropped the $x$ term because $|x| = 1$. Now we just have to find $|1-x|$. This can just be computed directly. $1 - x = 1 - \frac{\sqrt{3}}{2} - \frac{1}{2}i$ so $|1-x|^2 = (1 + \frac{3}{4} - \sqrt{3}) + \frac{1}{4} = 2 - \sqrt{3}$. Notice that $\sqrt{2 - \sqrt{3}} = \frac{\sqrt{6} - \sqrt{2}}{2}$, so we take the reciprocal of that to get $|S| = 2016 \cdot \frac{2}{\sqrt{6} -\sqrt{2}}$ which evaluates to be $|S| = 2016 \cdot \left( \frac{\sqrt{6} + \sqrt{2}}{2} \right)$ so we have $|S| = 1008 \sqrt{2} + 1008 \sqrt{6}$ so our answer is $1008 + 1008 + 2 + 6 = \boxed{(B) 2024}$

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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