Difference between revisions of "2015 AMC 12B Problems/Problem 25"
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==Solution== | ==Solution== | ||
− | Let <math>x = e^{i \pi / 6}</math> (a 30 | + | Let <math>x = e^{i \pi / 6}</math> (a <math>30^\circ</math> angle). We're going to toss this onto the complex plane. Notice that <math>P_k</math> on the complex plane is: |
+ | |||
<cmath>1 + 2x + 3x^2 + \cdots + kx^k</cmath> | <cmath>1 + 2x + 3x^2 + \cdots + kx^k</cmath> | ||
+ | |||
Therefore, we just want to find the magnitude of <math>P_{2015}</math> on the complex plane. This is an arithmetic/geometric series. | Therefore, we just want to find the magnitude of <math>P_{2015}</math> on the complex plane. This is an arithmetic/geometric series. | ||
+ | |||
<cmath>\begin{align*} S &= 1 + 2x + 3x^2 + \cdots + 2015x^{2014} \\ | <cmath>\begin{align*} S &= 1 + 2x + 3x^2 + \cdots + 2015x^{2014} \\ | ||
xS &= x + 2x^2 + 3x^3 + \cdots + 2015x^{2015} \\ | xS &= x + 2x^2 + 3x^3 + \cdots + 2015x^{2015} \\ | ||
(1-x)S &= 1 + x + x^2 + \cdots + x^{2014} - 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*} </cmath> | S &= \frac{1 - x^{2015} }{(1-x)^2} - \frac{2015x^{2015}}{1-x} \end{align*} </cmath> | ||
+ | |||
We want to find <math>|S|</math>. First of all, note that <math>x^{2015} = x^{11} = x^{-1}</math> because <math>x^{12} = 1</math>. So then: | We want to find <math>|S|</math>. First of all, note that <math>x^{2015} = x^{11} = x^{-1}</math> because <math>x^{12} = 1</math>. So then: | ||
+ | |||
<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)}</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)}</cmath> | ||
+ | |||
So therefore <math>S = -\frac{2016}{x(1-x)}</math>. Let us now find <math>|S|</math>: | So therefore <math>S = -\frac{2016}{x(1-x)}</math>. Let us now find <math>|S|</math>: | ||
+ | |||
<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 | + | |
+ | 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: | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | Thus the answer is <math>1008 + 1008 + 2 + 6 = \boxed{\textbf{(B)}\; 2024}</math> | ||
--fclvbfm934 | --fclvbfm934 |
Revision as of 00:44, 6 March 2015
Problem
A bee starts flying from point . She flies inch due east to point . For , once the bee reaches point , she turns counterclockwise and then flies inches straight to point . When the bee reaches she is exactly inches away from , where , , and are positive integers and and are not divisible by the square of any prime. What is ?
Solution
Let (a angle). We're going to toss this onto the complex plane. Notice that on the complex plane is:
Therefore, we just want to find the magnitude of on the complex plane. This is an arithmetic/geometric series.
We want to find . First of all, note that because . So then:
So therefore . Let us now find :
We dropped the term because . Now we just have to find . This can just be computed directly:
Therefore
Thus the answer is
--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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.