2009 AMC 12A Problems/Problem 25

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Problem

The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$ . For $n\ge1$ ,

$a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.$

What is $|a_{2009}|$ ?

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}...$

Solution

Consider another sequence $\{\theta_1, \theta_2, \theta_3...\}$ such that $a_n = \tan{\theta_n}$ , and $0 \leq \theta_n < 180$ .

The given recurrence becomes

$\begin{align*} a_{n + 2} & = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\\tan{\theta_{n + 2}} & = \frac {\tan{\theta_...$

It follows that $\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}$ . Since $\theta_1 = 45, \theta_2 = 30$ , all terms in the sequence $\{\theta_1, \theta_2, \theta_3...\}$ will be a multiple of $15$ .

Now consider another sequence $\{b_1, b_2, b_3...\}$ such that $b_n = \theta_n/15$ , and $0 \leq b_n < 12$ . The sequence $b_n$ satisfies $b_1 = 3, b_2 = 2, b_{n + 2} \equiv b_{n + 1} + b_n \pmod{12}$ .

As the number of possible consecutive two terms is finite, we know that the sequence $b_n$ is periodic. Write out the first few terms of the sequence until it starts to repeat.

$\{b_n\} = \{3,2,5,7,0,7,7,2,9,11,8,7,3,10,1,11,0,11,11,10,9,7,4,11,3,2,5,7,...\}$

Note that $b_{25} = b_1 = 3$ and $b_{26} = b_2 = 2$ . Thus $\{b_n\}$ has a period of $24$ : $b_{n + 24} = b_n$ .

It follows that $b_{2009} = b_{17} = 0$ and $\theta_{2009} = 15 b_{2009} = 0$ . Thus $a_{2009} = \tan{\theta_{2009}} = \tan{0} = 0.$

Our answer is $|a_{2009}| = \boxed{\textbf{(A)}\ 0}$ .

2009 AMC 12A (Problems • Resources)
Preceded by Problem 24	Followed by Last question
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2009 AMC 12A Problems/Problem 25

From AoPSWiki

Problem

Solution

See also