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Difference between revisions of "2008 AMC 12A Problems/Problem 25"

(New page: ==Problem== 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 <math>(a_{n + 1}, b_{n + 1})...)
 
(Standardized answer choices)
Line 6: Line 6:
 
Suppose that <math>(a_{100},b_{100}) = (2,4)</math>.  What is <math>a_1 + b_1</math>?
 
Suppose that <math>(a_{100},b_{100}) = (2,4)</math>.  What is <math>a_1 + b_1</math>?
  
−
<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>
+
<math>\mathrm{(A)}\ -\frac{1}{2^{97}}\qquad\mathrm{(B)}\ -\frac{1}{2^{99}}\qquad\mathrm{(C)}\ 0\qquad\mathrm{(D)}\ \frac{1}{2^{98}}\qquad\mathrm{(E)}\ \frac{1}{2^{96}}</math>
  
 
==Solution==  
 
==Solution==  

Revision as of 01:01, 26 April 2008

Problem

A sequence $(a_1,b_1)$, $(a_2,b_2)$, $(a_3,b_3)$, $\ldots$ of points in the coordinate plane satisfies

$(a_{n + 1}, b_{n + 1}) = (\sqrt {3}a_n - b_n, \sqrt {3}b_n + a_n)$ for $n = 1,2,3,\ldots$.

Suppose that $(a_{100},b_{100}) = (2,4)$. What is $a_1 + b_1$?

$\mathrm{(A)}\ -\frac{1}{2^{97}}\qquad\mathrm{(B)}\ -\frac{1}{2^{99}}\qquad\mathrm{(C)}\ 0\qquad\mathrm{(D)}\ \frac{1}{2^{98}}\qquad\mathrm{(E)}\ \frac{1}{2^{96}}$

Solution

This sequence can also be expressed using matrix multiplication as follows:

$\left[ \begin{array}{c} a_{n+1} \\ b_{n+1} \end{array} \right] = \left[ \begin{array}{cc} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{array} \right] \left[ \begin{array}{c} a_{n} \\ b_{n} \end{array} \right] = 2 \left[ \begin{array}{cc} \cos 30^\circ & -\sin 30^\circ \\ \sin 30^\circ & \ \cos 30^\circ \end{array} \right] \left[ \begin{array}{c} a_{n} \\ b_{n} \end{array} \right]$.

Thus, $(a_{n+1} , b_{n+1})$ is formed by rotating $(a_n , b_n)$ counter-clockwise about the origin by $30^\circ$ and dilating the point's position with respect to the origin by a factor of $2$.

So, starting with $(a_{100},b_{100})$ and performing the above operations $99$ times in reverse yields $(a_1,b_1)$.

Rotating $(2,4)$ clockwise by $99 \cdot 30^\circ \equiv 90^\circ$ yields $(4,-2)$. A dilation by a factor of $\frac{1}{2^{99}}$ yields the point $(a_1,b_1) = \left(\frac{4}{2^{99}}, -\frac{2}{2^{99}} \right) = \left(\frac{1}{2^{97}}, -\frac{1}{2^{98}} \right)$.

Therefore, $a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \frac{1}{2^{98}} \Rightarrow D$.

See Also

2008 AMC 12A (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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All AMC 12 Problems and Solutions
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