# 2009 AMC 12A Problems/Problem 25

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

## Clarification

(While the above solution is quite intuitive, it's kinda hard for people who don't know many proofing symbols to understand.)

Notice that the formula given in the problem looks incredibly similar to the Tangent Addition Formula. Since $a_1 = 1$, let $a_1$ be $\tan{45}$. Similarly, let $a_2$ be $\tan{30}$. Then the formula for $a_3$ reads $$a_3 = \frac {\tan{45} + \tan{30} }{1 - \tan{45}*\tan{30}}.$$

But from the Tangent Addition Formula we know that this is just the formula for $\tan{(45+30)}$ or $\tan{75}$, meaning $a_3 = \tan{75}$. So, the sequence becomes a sort of Fibonacci sequence with angle measures. We continue to sum angle measures, like so: $a_4 = \tan{(75+30)} = \tan{105}$ $a_5 = \tan{(105+75)} = \tan{180}$ $a_6 = \tan{(180+105)} =$ ... wait a minute! $\tan{180} = \tan{0}$!

So now we have $a_5 = \tan{0}$ $a_6 = \tan{(105+0)} = \tan{105}$ $a_7 = \tan{(105+105)} = \tan{210}$ $a_8 = \tan{(210+105)} =$ .... Ok this is getting tiring. Let's stop and think about whether we can simplify this. Notice that all of the angle measures are a multiple of $15$. So, let's express the angle measures as multiples of $15$.

Let $b_n = \frac{\arctan{(a_n)}}{15}$.

(Basically, $b_n$ is the angle measure of the corresponding $a_n,$ divided by $15$)

Now we have $b_1 = 3$ $b_2 = 2$ $b_3 = 5$ $b_4 = 7$ $b_5 = 12$

But wait... we're dealing with the $\tan$ function, which has a period (recurrence rate) of $2\pi$ or $180^{\circ}$. Since we divided the angle measures by $15$, the period is now $12$ (which aligns with what we got earlier: $a_5 = 0$). This means that we can reduce the terms of the sequence based on the $\bmod$ function, which returns the remainder after dividing by a certain amount. So now we can say $b_n \equiv b_n \bmod{12}$ ( $\equiv$ denotes modular equivalence in this context). Now we continue our sequence: $3, 2, 5, 7, 0, 7, 7, 2, 9, 11, 8, 7, 3, 10, 1, 11, 0, 11, 11, 10, 9, 7, 4, 11, 3, 2, 5$ (It's starting to repeat now)

So $a_1 = 3, a_{25} = 3, a_{49} = 3$, and so on. The sequence repeats every $24$ terms. The problem asks us for the value of $a_{2009}$. Let's whip out the $\bmod$ function again. $2009 \bmod 24 = 17$

So we want the value of the $17$th number in the sequence, and looking back at it (hopefully you wrote it down somewhere) we find that $b_{17} = 0 \Rightarrow \boxed{\text{E}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 