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| − | ==Problem==
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| − | A sequence <math>(a_1,b_1)</math>, <math>(a_2,b_2)</math>, <math>(a_3,b_3)</math>, <math>\ldots</math> of points in the coordinate plane satisfies
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| − | <math>(a_{n + 1}, b_{n + 1}) = (\sqrt {3}a_n - b_n, \sqrt {3}b_n + a_n)</math> for <math>n = 1,2,3,\ldots</math>.
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| − | Suppose that <math>(a_{100},b_{100}) = (2,4)</math>. What is <math>a_1 + b_1</math>?
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| − | <math>\textbf{(A)}\ - \frac {1}{2^{97}} \qquad \textbf{(B)}\ - \frac {1}{2^{99}} \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \frac {1}{2^{98}} \qquad \textbf{(E)}\ \frac {1}{2^{96}}</math>
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| − | ==Solution==
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| − | This sequence can also be expressed using matrix multiplication as follows:
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| − | <math>\left[ \begin{array}{c} a_{n+1} \\ b_{n+1} \end{array} \right] = \left[ \begin{array}{cc} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{array} \right] \left[ \begin{array}{c} a_{n} \\ b_{n} \end{array} \right] = 2 \left[ \begin{array}{cc} \cos 30^\circ & -\sin 30^\circ \\ \sin 30^\circ & \ \cos 30^\circ \end{array} \right] \left[ \begin{array}{c} a_{n} \\ b_{n} \end{array} \right]</math>.
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| − | Thus, <math>(a_{n+1} , b_{n+1})</math> is formed by rotating <math>(a_n , b_n)</math> counter-clockwise about the origin by <math>30^\circ</math> and dilating the point's position with respect to the origin by a factor of <math>2</math>.
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| − | So, starting with <math>(a_{100},b_{100})</math> and performing the above operations <math>99</math> times in reverse yields <math>(a_1,b_1)</math>.
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| − | Rotating <math>(2,4)</math> clockwise by <math>99 \cdot 30^\circ \equiv 90^\circ</math> yields <math>(4,-2)</math>. A dilation by a factor of <math>\frac{1}{2^{99}}</math> yields the point <math>(a_1,b_1) = \left(\frac{4}{2^{99}}, -\frac{2}{2^{99}} \right) = \left(\frac{1}{2^{97}}, -\frac{1}{2^{98}} \right)</math>.
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| − | Therefore, <math>a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \frac{1}{2^{98}} \Rightarrow D</math>.
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| − | ==See Also==
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| − | {{AMC12 box|year=2008|ab=A|num-b=23|num-a=25}}
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