Question

Una sucesión $(a_1, b_1), (a_2, b_2), (a_3, b_3), \ldots$ de puntos en el plano coordenado satisface $\begin{aligned} &(a_{n+1}, b_{n+1}) \\ &= \left(\sqrt{3}\,a_n - b_n,\; \sqrt{3}\,b_n + a_n\right) \\ &\quad (n = 1, 2, 3, \ldots) \end{aligned}$ Supongamos que $(a_{100}, b_{100}) = (2, 4).$ ¿Cuánto vale $a_1 + b_1$ ?

A sequence $(a_1, b_1), (a_2, b_2), (a_3, b_3), \ldots$ of points in the coordinate plane satisfies $\begin{aligned} &(a_{n+1}, b_{n+1}) \\ &= \left(\sqrt{3}\,a_n - b_n,\; \sqrt{3}\,b_n + a_n\right) \\ &\quad (n = 1, 2, 3, \ldots) \end{aligned}$ Suppose that $(a_{100}, b_{100}) = (2, 4).$ What is $a_1 + b_1?$

$-\dfrac{1}{2^{97}}$

$-\dfrac{1}{2^{99}}$

$0$

$\dfrac{1}{2^{98}}$

$\dfrac{1}{2^{96}}$

Solución:

Sea $z_n = a_n + b_n i.$ Entonces $\begin{aligned} z_{n+1} &= (\sqrt{3}\,a_n - b_n) \\ &\quad {}+ (\sqrt{3}\,b_n + a_n)i \\ &= (a_n + b_n i)(\sqrt{3} + i), \end{aligned}$ así que $z_{n+1} = z_n(\sqrt{3} + i)$ y $z_{100} = z_1(\sqrt{3} + i)^{99}.$

Como $\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ),$ el teorema de De Moivre da $(\sqrt{3} + i)^{99}$ $= 2^{99}(\cos 2970^\circ + i\sin 2970^\circ).$ Como $2970^\circ$ es coterminal con $90^\circ,$ esto es igual a $2^{99} i.$

Así $2 + 4i = z_1 \cdot 2^{99} i,$ de modo que $z_1 = \dfrac{2 + 4i}{2^{99} i} = \dfrac{4 - 2i}{2^{99}}.$

Entonces $a_1 = \tfrac{4}{2^{99}}$ y $b_1 = -\tfrac{2}{2^{99}},$ así que $a_1 + b_1 = \dfrac{2}{2^{99}} = \dfrac{1}{2^{98}}.$

Por lo tanto, D es la respuesta correcta.

Let $z_n = a_n + b_n i.$ Then $\begin{aligned} z_{n+1} &= (\sqrt{3}\,a_n - b_n) \\ &\quad {}+ (\sqrt{3}\,b_n + a_n)i \\ &= (a_n + b_n i)(\sqrt{3} + i), \end{aligned}$ so $z_{n+1} = z_n(\sqrt{3} + i)$ and $z_{100} = z_1(\sqrt{3} + i)^{99}.$

Since $\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ),$ De Moivre's theorem gives $(\sqrt{3} + i)^{99}$ $= 2^{99}(\cos 2970^\circ + i\sin 2970^\circ).$ As $2970^\circ$ is coterminal with $90^\circ,$ this equals $2^{99} i.$

Thus $2 + 4i = z_1 \cdot 2^{99} i,$ so $z_1 = \dfrac{2 + 4i}{2^{99} i} = \dfrac{4 - 2i}{2^{99}}.$

Then $a_1 = \tfrac{4}{2^{99}}$ and $b_1 = -\tfrac{2}{2^{99}},$ so $a_1 + b_1 = \dfrac{2}{2^{99}} = \dfrac{1}{2^{98}}.$

Thus, D is the correct answer.

2008 AMC 12A Problema 25

Solución:

El Problema 25 en otros años