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

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

@@ Line 75: / Line 75: @@
 The answer is <math>x+y=\frac{1}{2^{97}}-\frac{1}{2^{98}}=\boxed{\textbf{(D) }\frac{1}{2^{98}}}</math>..
+==Solution 3==
+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
+<cmath> z_{99} = \frac{1}{2}e^{-i\pi/6}z_{100}</cmath>
+<cmath> z_{98} = \frac{1}{2^2}e^{2(-i\pi/6)}z_{100}</cmath>
+<cmath> \vdots</cmath>
+<cmath> z_{1} = \frac{1}{2^{99}}e^{99(-i\pi/6)}z_{100}</cmath>
+<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>
+Hence, <math>a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \boxed{\mathbf{(B)}\frac{1}{2^{98}}}</math> ~ brainfertilzer.
 ==See Also==
 {{AMC12 box|year=2008|ab=A|num-b=24|after=Last question}}
 {{MAA Notice}}

2008 AMC 12A (Problems • Answer Key • Resources)
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