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

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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>.  
 
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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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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Shortcut: no answer has <math>3</math> in the denominator. So the point cannot have orientation <math>(2,4)</math> or <math>(-2,-4)</math>. Also there are no negative answers. Any other non-multiple of <math>90^\circ</math> rotation of <math>30n^\circ</math> would result in the need of radicals. So either it has orientation <math>(4,-2)</math> or <math>(-4,2)</math>. Both answers add up to <math>2</math>. Thus, <math>2/2^99=\boxed{\textbf{(D) }\frac{1}{2^98}</math>.
  
 
==See Also==  
 
==See Also==  
 
{{AMC12 box|year=2008|ab=A|num-b=24|after=Last question}}
 
{{AMC12 box|year=2008|ab=A|num-b=24|after=Last question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:39, 13 February 2016

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

Shortcut: no answer has $3$ in the denominator. So the point cannot have orientation $(2,4)$ or $(-2,-4)$. Also there are no negative answers. Any other non-multiple of $90^\circ$ rotation of $30n^\circ$ would result in the need of radicals. So either it has orientation $(4,-2)$ or $(-4,2)$. Both answers add up to $2$. Thus, $2/2^99=\boxed{\textbf{(D) }\frac{1}{2^98}$ (Error compiling LaTeX. Unknown error_msg).

See Also

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

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