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

[s]Shortcut: no answer has $3$ in the numerator. 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, $2/2^{99}=\boxed{\textbf{(D) }\frac{1}{2^{98}}}$.[/s] Does not work as there are negative answers.

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

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
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