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

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<cmath>|S| = 2016 \left| \frac{1}{1-x} \right|</cmath>
 
<cmath>|S| = 2016 \left| \frac{1}{1-x} \right|</cmath>
 
We dropped the <math>x</math> term because <math>|x| = 1</math>. Now we just have to find <math>|1-x|</math>. This can just be computed directly. <math>1 - x = 1 - \frac{\sqrt{3}}{2} - \frac{1}{2}i</math> so <math>|1-x|^2 = (1 + \frac{3}{4} - \sqrt{3}) + \frac{1}{4} = 2 - \sqrt{3}</math>. Notice that <math>\sqrt{2 - \sqrt{3}} = \frac{\sqrt{6} - \sqrt{2}}{2}</math>, so we take the reciprocal of that to get <math>|S| = 2016 \cdot \frac{2}{\sqrt{6} -\sqrt{2}}</math> which evaluates to be <math>|S| = 2016 \cdot \left( \frac{\sqrt{6} + \sqrt{2}}{2} \right)</math> so we have <math>|S| = 1008 \sqrt{2} + 1008 \sqrt{6}</math> so our answer is <math>1008 + 1008 + 2 + 6 = \boxed{(B) 2024}</math>
 
We dropped the <math>x</math> term because <math>|x| = 1</math>. Now we just have to find <math>|1-x|</math>. This can just be computed directly. <math>1 - x = 1 - \frac{\sqrt{3}}{2} - \frac{1}{2}i</math> so <math>|1-x|^2 = (1 + \frac{3}{4} - \sqrt{3}) + \frac{1}{4} = 2 - \sqrt{3}</math>. Notice that <math>\sqrt{2 - \sqrt{3}} = \frac{\sqrt{6} - \sqrt{2}}{2}</math>, so we take the reciprocal of that to get <math>|S| = 2016 \cdot \frac{2}{\sqrt{6} -\sqrt{2}}</math> which evaluates to be <math>|S| = 2016 \cdot \left( \frac{\sqrt{6} + \sqrt{2}}{2} \right)</math> so we have <math>|S| = 1008 \sqrt{2} + 1008 \sqrt{6}</math> so our answer is <math>1008 + 1008 + 2 + 6 = \boxed{(B) 2024}</math>
 +
 +
--fclvbfm934
  
 
==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 23:23, 3 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-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 $1008 + 1008 + 2 + 6 = \boxed{(B) 2024}$

--fclvbfm934

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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