2008 AMC 12A Problem 25

Attempt Problem 25 of the 2008 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12A solutions, or check the answer key.

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25.

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}}

Answer: D

Concepts:complex number De Moivre’s Theorem

Difficulty rating: 2440

Small hint:

Write

z_n = a_n + b_n i;

the recurrence becomes

z_{n+1} = z_n(\sqrt{3} + i)

Big hint:

Since

\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ),

apply De Moivre’s theorem to

(\sqrt{3} + i)^{99}

Solution:

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.

Problem 25 in other years