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2009 AMC 12A Problems/Problem 25

Revision as of 15:28, 10 May 2019 by Asdf334 (talk | contribs)

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 $|a_{2009}|$?

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

Solution 1

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_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\ \tan{\theta_{n + 2}} & = \tan(\theta_{n + 1} + \theta_n) \end{align*}

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

Note

It is not actually difficult to list out the terms until it repeats. You will find that the period is 24 starting from term 2.

See also

2009 AMC 12A (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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All AMC 12 Problems and Solutions

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