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

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<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>
 
<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>
  
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==Solution==  
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==Solution 1==  
 
This sequence can also be expressed using matrix multiplication as follows:  
 
This sequence can also be expressed using matrix multiplication as follows:  
  
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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>.
 +
 
 +
==Solution 2 (algebra)==
 +
Let <math>(x,y)=(a_1,b_1)</math>. Then, we can begin to list out terms as follows:
 +
 
 +
 
 +
 
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<math>(a_2,b_2)=(x\sqrt{3}-y,y\sqrt{3}+x)</math>
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 +
<math>(a_3,b_3)=(2x-2y\sqrt{3},2y+2x\sqrt{3})</math>
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 +
<math>(a_4,b_4)=(-8y,8x)</math>
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 +
 
 +
 
 +
We notice that the sequence follows the rule <math>(a_{n+3},b_{n+3})=(-2^3b_n,2^3a_n)</math>
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 +
We can now start listing out every third point, getting:
 +
 
 +
 
 +
 
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<math>(a_1,b_1)=(x,y)</math>
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<math>(a_4,b_4)=(-2^3y,2^3x)</math>
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<math>(a_7,b_7)=(-2^6x,-2^6y)</math>
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<math>(a_{10},b_{10})=(2^9y,-2^9x)</math>
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<math>(a_{13},b_{13})=(2^{12}x,2^{12}y)</math>
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 +
 
 +
 
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We can make two observations from this:
 +
 
 +
(1) In <math>a_n</math>, the coefficient of <math>x</math> and <math>y</math> is <math>2^{n-1}</math>
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(2) The positioning of <math>x</math> and <math>y</math>, and their signs, cycle with every <math>12</math> terms.
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 +
 
 +
 
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We know then that from (1), the coefficients of <math>x</math> and <math>y</math> in <math>(a_{100},b_{100})</math> are both <math>2^{99}</math>
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 +
We can apply (2), finding <math>100 \text{(mod }12)=4</math>, so the positions and signs of <math>x</math> and <math>y</math> are the same in <math>(a_{100},b_{100})</math> as they are in <math>(a_{4},b_{4})</math>.
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From this, we can get <math>(a_{100},b_{100})=(-2^{99}y,2^{99}x)</math>. We know that <math>(a_{100},b_{100})=(2,4)</math>, so we get the following:
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 +
 
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<math>-2^{99}y=2 \Rightarrow y=-\frac{1}{2^{98}}</math>
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<math>2^{99}x=4 \Rightarrow x=\frac{1}{2^{97}}</math>
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 +
 
 +
 
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The answer is <math>x+y=\frac{1}{2^{97}}-\frac{1}{2^{98}}=\boxed{\textbf{(D) }\frac{1}{2^{98}}}</math>..
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==Solution 3==
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 +
The ordered pairs and <math>\sqrt{3}</math>'s makes us think to use complex numbers. We have <math>(a_{n+1},b_{n+1}) = 2\left(\frac{\sqrt{3}}{2}a_n - \frac{1}{2}b_n, \frac{\sqrt{3}}{2}b_n + \frac{1}{2}a_n\right)</math>, so <math>a_{n+1} + b_{n+1}i = 2\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)(a_n + b_ni) = \frac{1}{2}e^{i\pi/6}(a_n + b_ni)</math>. Letting <math>z_n = a_n + b_ni</math> (so <math>z_{n+1} = a_{n+1} + b_{n+1}i</math>), we have <math>z_{n+1} = 2e^{i\pi/6}z_n</math>. Letting <math>n\rightarrow n-1</math>, we have <math>z_n = 2e^{i\pi/6}z_{n-1}</math>, so <math>z_{n-1} = \frac{1}{2}e^{-i\pi/6}z_n</math>. This is the reverse transformation. We have
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<cmath> z_{99} = \frac{1}{2}e^{-i\pi/6}z_{100}</cmath>
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<cmath> z_{98} = \frac{1}{2^2}e^{2(-i\pi/6)}z_{100}</cmath>
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<cmath> \vdots</cmath>
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<cmath> z_{1} = \frac{1}{2^{99}}e^{99(-i\pi/6)}z_{100}</cmath>
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<cmath> = \frac{1}{2^{99}}e^{-i\pi/2}z_{100} = -\frac{1}{2^{99}}i(2 + 4i) = \frac{1}{2^{97}} - \frac{1}{2^{98}}i.</cmath>
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Hence, <math>a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \boxed{\mathbf{(D)}\frac{1}{2^{98}}}</math> ~ brainfertilzer.
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==Video Solution==
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https://www.youtube.com/watch?v=_4UJzyBslFA
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==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}}

Latest revision as of 10:58, 9 December 2022

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 1

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

Solution 2 (algebra)

Let $(x,y)=(a_1,b_1)$. Then, we can begin to list out terms as follows:


$(a_2,b_2)=(x\sqrt{3}-y,y\sqrt{3}+x)$

$(a_3,b_3)=(2x-2y\sqrt{3},2y+2x\sqrt{3})$

$(a_4,b_4)=(-8y,8x)$


We notice that the sequence follows the rule $(a_{n+3},b_{n+3})=(-2^3b_n,2^3a_n)$

We can now start listing out every third point, getting:


$(a_1,b_1)=(x,y)$

$(a_4,b_4)=(-2^3y,2^3x)$

$(a_7,b_7)=(-2^6x,-2^6y)$

$(a_{10},b_{10})=(2^9y,-2^9x)$

$(a_{13},b_{13})=(2^{12}x,2^{12}y)$


We can make two observations from this:

(1) In $a_n$, the coefficient of $x$ and $y$ is $2^{n-1}$

(2) The positioning of $x$ and $y$, and their signs, cycle with every $12$ terms.


We know then that from (1), the coefficients of $x$ and $y$ in $(a_{100},b_{100})$ are both $2^{99}$

We can apply (2), finding $100 \text{(mod }12)=4$, so the positions and signs of $x$ and $y$ are the same in $(a_{100},b_{100})$ as they are in $(a_{4},b_{4})$.

From this, we can get $(a_{100},b_{100})=(-2^{99}y,2^{99}x)$. We know that $(a_{100},b_{100})=(2,4)$, so we get the following:


$-2^{99}y=2 \Rightarrow y=-\frac{1}{2^{98}}$

$2^{99}x=4 \Rightarrow x=\frac{1}{2^{97}}$


The answer is $x+y=\frac{1}{2^{97}}-\frac{1}{2^{98}}=\boxed{\textbf{(D) }\frac{1}{2^{98}}}$..

Solution 3

The ordered pairs and $\sqrt{3}$'s makes us think to use complex numbers. We have $(a_{n+1},b_{n+1}) = 2\left(\frac{\sqrt{3}}{2}a_n - \frac{1}{2}b_n, \frac{\sqrt{3}}{2}b_n + \frac{1}{2}a_n\right)$, so $a_{n+1} + b_{n+1}i = 2\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)(a_n + b_ni) = \frac{1}{2}e^{i\pi/6}(a_n + b_ni)$. Letting $z_n = a_n + b_ni$ (so $z_{n+1} = a_{n+1} + b_{n+1}i$), we have $z_{n+1} = 2e^{i\pi/6}z_n$. Letting $n\rightarrow n-1$, we have $z_n = 2e^{i\pi/6}z_{n-1}$, so $z_{n-1} = \frac{1}{2}e^{-i\pi/6}z_n$. This is the reverse transformation. We have \[z_{99} = \frac{1}{2}e^{-i\pi/6}z_{100}\] \[z_{98} = \frac{1}{2^2}e^{2(-i\pi/6)}z_{100}\] \[\vdots\] \[z_{1} = \frac{1}{2^{99}}e^{99(-i\pi/6)}z_{100}\] \[= \frac{1}{2^{99}}e^{-i\pi/2}z_{100} = -\frac{1}{2^{99}}i(2 + 4i) = \frac{1}{2^{97}} - \frac{1}{2^{98}}i.\]

Hence, $a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \boxed{\mathbf{(D)}\frac{1}{2^{98}}}$ ~ brainfertilzer.

Video Solution

https://www.youtube.com/watch?v=_4UJzyBslFA


See Also

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