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

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But wait... we're dealing with the <math>\tan</math> function, which has a period (recurrence rate) of <math>2\pi</math> or <math>180^{\circ}</math>. Since we divided the angle measures by <math>15</math>, the period is now <math>12</math> (which aligns with what we got earlier: <math>a_5 = 0</math>). This means that we can reduce the terms of the sequence based on the <math>\bmod</math> function, which returns the remainder after dividing by a certain amount. So now we can say <math>b_n \equiv b_n \bmod{12}</math>. Now we continue our sequence:
 
But wait... we're dealing with the <math>\tan</math> function, which has a period (recurrence rate) of <math>2\pi</math> or <math>180^{\circ}</math>. Since we divided the angle measures by <math>15</math>, the period is now <math>12</math> (which aligns with what we got earlier: <math>a_5 = 0</math>). This means that we can reduce the terms of the sequence based on the <math>\bmod</math> function, which returns the remainder after dividing by a certain amount. So now we can say <math>b_n \equiv b_n \bmod{12}</math>. Now we continue our sequence:
 +
 +
<math>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</math> (It's starting to repeat now)
 +
 +
So <math>a_1  = 3, a_25 = 3, a_49 = 3</math>, and so on. The sequence repeats every 24 terms
  
 
== See also ==
 
== See also ==

Revision as of 10:49, 30 July 2020

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

Solution 2

(This is for the brute force users; kudos to the very intuitive solution above)

First we interpret the formula:

"The next term in the sequence is equivalent to the sum of the previous two divided by 1 minus their product"

Next, we work out the first few terms of the sequence:


$a_3 = \frac{1 + \frac{ \sqrt{3} } {3} } {1-\frac{\sqrt{3}}{3}}$


$\Rightarrow \frac{3+\sqrt{3}}{3-\sqrt{3}}$


$\Rightarrow 2 + \sqrt{3}$


$a_4 = \frac{\frac{\sqrt{3}}{3}+2+\sqrt{3}}{1-\frac{\sqrt{3}(2+\sqrt{3})}{3}}$


$\Rightarrow \frac{6+4\sqrt{3}}{-2\sqrt{3}}$


$\Rightarrow -(2+\sqrt{3})$


So at this point, our sequence reads


$1,\: \frac{\sqrt{3}}{3}, \: 2 + \sqrt{3}, \: -(2+\sqrt{3})$


Now for $a_5$..... but wait! the numerator of the next term is equal to $2 + \sqrt{3} + -(2+\sqrt{3})$..... $\Rightarrow 0$. So as long as the denominator isn't $0$ (which we can quickly verify), $a_5 = 0$. Now our sequence is


$1,\: \frac{\sqrt{3}}{3}, \: 2 + \sqrt{3}, \: -(2+\sqrt{3}), \: 0$


Solving for $a_6$:


$\frac{-(2+\sqrt{3})+0}{1-0}$


$\Rightarrow -(2+\sqrt{3})$


Our sequence is now


$1,\: \frac{\sqrt{3}}{3}, \: 2 + \sqrt{3}, \: -(2+\sqrt{3}), \: 0, \: -(2+\sqrt{3})$


Solving in this way until $a_$ (Error compiling LaTeX. Unknown error_msg)


The problem asks us for the value of $a_{2009}$, and since $2009$ is an odd number, we know that $a_{2009} = 0 \Rightarrow \boxed{\text{E}}$.


Clarification of Solution 1

(While I did say the first solution is intuitive, it's kinda hard for people with just a basic knowledge of math, and no knowledge of proofing symbols to understand.)


Notice that the formula 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 reads


\[a_3 = \frac {\tan{45} + \tan{30} }{1 - \tan{45}*\tan{30}}.\]


But from the Tangent Addition Formula we know that this is the formula for $\tan{(45+30)}$ or $\tan{75}$, meaning $a_3 = \tan{75}$. So, the sequence is simply the sum of the two 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 $15$. 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 $b_n \equiv b_n \bmod{12}$. 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

See also

2009 AMC 12A (Problems • Answer Key • Resources)
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