# 2008 AMC 12A Problems/Problem 25

## 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 