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Difference between revisions of "2009 AMC 12A Problems/Problem 25"

(written by kitsune)
 
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<center><cmath>\begin{align*} a_{n + 2} & = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\
 
<center><cmath>\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}} & = \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*}</cmath></center>
+
\tan{\theta_{n + 2}} & = \tan(\theta_{n + 1} + \theta_n) \end{align*}</cmath></center>
  
 
It follows that <math>\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}</math>. Since <math>\theta_1 = 45, \theta_2 = 30</math>, all terms in the sequence <math>\{\theta_1, \theta_2, \theta_3...\}</math> will be a multiple of <math>15</math>.
 
It follows that <math>\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}</math>. Since <math>\theta_1 = 45, \theta_2 = 30</math>, all terms in the sequence <math>\{\theta_1, \theta_2, \theta_3...\}</math> will be a multiple of <math>15</math>.

Revision as of 10:13, 11 April 2009

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

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

See also

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