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

Revision as of 00:58, 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:

$1 + 2x + 3x^2 + \cdots + kx^k$

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:

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

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

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

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

--fclvbfm934

@@ Line 29: / Line 29: @@
 <cmath>1 - x = 1 - \frac{\sqrt{3}}{2} - \frac{1}{2}i</cmath>
-<cmath>|1-x|^2 = \left(1 + \frac{3}{4} - \sqrt{3} \right) + \frac{1}{4} = 2 - \sqrt{3} = {\left( \frac{\sqrt{6}-\sqrt{2}}{2} \right)}^2.</cmath>
+<cmath>|1-x|^2 = \left(1 + \frac{3}{4} - \sqrt{3} \right) + \frac{1}{4} = 2 - \sqrt{3} = {\left( \frac{\sqrt{6}-\sqrt{2}}{2} \right)}^2</cmath>
+<cmath>|1-x| = \frac{\sqrt{6} - \sqrt{2}}{2}.</cmath>
 Therefore
-<cmath>|S| = \frac{2016}{\sqrt{2 - \sqrt{3}}} = 2016 \cdot \frac{2}{\sqrt{6} -\sqrt{2}} = 2016 \left( \frac{\sqrt{6} + \sqrt{2}}{2} \right) = 1008 \sqrt{2} + 1008 \sqrt{6}.</cmath>
+<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>

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

Revision as of 00:58, 6 March 2015

Problem

Solution

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