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

Latest revision as of 15:57, 9 July 2024

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

Solution 2

First, some intuition. The given recurrence relation: $a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}$ looks much like the tangent addition formula. So, we let $a_n=\tan{\theta_n}$ . Then, we get: $\tan{\theta_n}=\tan{(\theta_{n-1}+\theta_{n-2})}$ This gives us: $\theta_n \equiv \theta_{n-1}+\theta_{n-2} \pmod{180}$ Now, observe that $\theta_1=45$ $\theta_2=30$ We know that this sequence of values for $\theta_n$ will repeat eventually because it is mod $180$ . It is just a matter of when, so we start bashing:

$\theta_3=75$ $\theta_4=105$ $\theta_5=0$ $\theta_6=105$ $\theta_7=105$ $\theta_8=30$ $\theta_9=135$ $\theta_{10}=165$ $\theta_{11}=120$ $\theta_{12}=105$ $\theta_{13}=45$ $\theta_{14}=150$ $\theta_{15}=15$ $\theta_{16}=165$ $\theta_{17}=0$ $\theta_{18}=165$ $\theta_{19}=165$ $\theta_{20}=150$ $\theta_{21}=135$ $\theta_{22}=105$ $\theta_{23}=60$ $\theta_{24}=165$ $\theta_{25}=45$ $\theta_{26}=30$ And there is the repetition. So, this series has a period of 24. $2009 \equiv 17 \pmod{24}$ , so $|a_{2009}|=|\tan{\theta_{17}}|=|\tan{0}|=|0|= \boxed{\textbf{(A)}\ 0}$ ~Firebolt360

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 $b_{17} = 0 \Rightarrow \boxed{\text{A}}$ .

Video Solution

https://www.youtube.com/watch?v=tic0ehB6TU4

@@ Line 8: / Line 8: @@
 <math>\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 + \sqrt3</math>
-== Solution ==
+== Solution 1==
 Consider another sequence <math>\{\theta_1, \theta_2, \theta_3...\}</math> such that <math>a_n = \tan{\theta_n}</math>, and <math>0 \leq \theta_n < 180</math>.
@@ Line 30: / Line 30: @@
 Our answer is <math>|a_{2009}| = \boxed{\textbf{(A)}\ 0}</math>.
+==Solution 2==
+First, some intuition. The given recurrence relation: <math>a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}</math> looks much like the tangent addition formula. So, we let <math>a_n=\tan{\theta_n}</math>. Then, we get:
+<cmath>\tan{\theta_n}=\tan{(\theta_{n-1}+\theta_{n-2})}</cmath>
+This gives us:
+<cmath>\theta_n \equiv \theta_{n-1}+\theta_{n-2} \pmod{180}</cmath>
+Now, observe that
+<cmath>\theta_1=45</cmath>
+<cmath>\theta_2=30</cmath>
+We know that this sequence of values for <math>\theta_n</math> will repeat eventually because it is mod <math>180</math>.
+It is just a matter of when, so we start bashing:
+<cmath>\theta_3=75</cmath>
+<cmath>\theta_4=105</cmath>
+<cmath>\theta_5=0</cmath>
+<cmath>\theta_6=105</cmath>
+<cmath>\theta_7=105</cmath>
+<cmath>\theta_8=30</cmath>
+<cmath>\theta_9=135</cmath>
+<cmath>\theta_{10}=165</cmath>
+<cmath>\theta_{11}=120</cmath>
+<cmath>\theta_{12}=105</cmath>
+<cmath>\theta_{13}=45</cmath>
+<cmath>\theta_{14}=150</cmath>
+<cmath>\theta_{15}=15</cmath>
+<cmath>\theta_{16}=165</cmath>
+<cmath>\theta_{17}=0</cmath>
+<cmath>\theta_{18}=165</cmath>
+<cmath>\theta_{19}=165</cmath>
+<cmath>\theta_{20}=150</cmath>
+<cmath>\theta_{21}=135</cmath>
+<cmath>\theta_{22}=105</cmath>
+<cmath>\theta_{23}=60</cmath>
+<cmath>\theta_{24}=165</cmath>
+<cmath>\theta_{25}=45</cmath>
+<cmath>\theta_{26}=30</cmath>
+And there is the repetition. So, this series has a period of 24. <math>2009 \equiv 17 \pmod{24}</math>, so <math>|a_{2009}|=|\tan{\theta_{17}}|=|\tan{0}|=|0|= \boxed{\textbf{(A)}\ 0}</math> ~Firebolt360
+==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 [[Trigonometric_identities#Angle_Addition/Subtraction_Identities|Tangent Addition Formula]].
+Since <math>a_1 = 1</math>, let <math>a_1</math> be <math>\tan{45}</math>. Similarly, let <math>a_2</math> be <math>\tan{30}</math>. Then the formula for <math>a_3</math> reads
+<center><cmath>a_3 = \frac {\tan{45} + \tan{30} }{1 - \tan{45}*\tan{30}}.</cmath></center>
+But from the [[Trigonometric_identities#Angle_Addition/Subtraction_Identities|Tangent Addition Formula]] we know that this is just the formula for <math>\tan{(45+30)}</math> or <math>\tan{75}</math>, meaning <math>a_3 = \tan{75}</math>. So, the sequence becomes a sort of [[Fibonacci sequence]]  with angle measures. We continue to sum angle measures, like so:
+<math>a_4 = \tan{(75+30)} = \tan{105}</math>
+<math>a_5 = \tan{(105+75)} = \tan{180}</math>
+<math>a_6 = \tan{(180+105)} =</math> ... wait a minute! <math>\tan{180} = \tan{0}</math>!
+So now we have
+<math>a_5 = \tan{0}</math>
+<math>a_6 = \tan{(105+0)} = \tan{105}</math>
+<math>a_7 = \tan{(105+105)} = \tan{210}</math>
+<math>a_8 = \tan{(210+105)} = </math> .... 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 <math>15</math>. So, let's express the angle measures as multiples of <math>15</math>.
+Let <math>b_n = \frac{\arctan{(a_n)}}{15}</math>.
+(Basically, <math>b_n</math> is the angle measure of the corresponding <math>a_n,</math> divided by <math>15</math>)
+Now we have
+<math>b_1 = 3</math>
+<math>b_2 = 2</math>
+<math>b_3 = 5</math>
+<math>b_4 = 7</math>
+<math>b_5 = 12</math>
+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> (<math>\equiv</math> denotes modular equivalence in this context). 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 <math>24</math> terms. The problem asks us for the value of <math>a_{2009}</math>. Let's whip out the <math>\bmod</math> function again.
+<math>2009 \bmod 24 = 17</math>
+So we want the value of the <math>17</math>th number in the sequence, and looking back at it (hopefully you wrote it down somewhere) we find that <math>b_{17} = 0 \Rightarrow \boxed{\text{A}}</math>.
+== Video Solution ==
+https://www.youtube.com/watch?v=tic0ehB6TU4
 == See also ==
@@ Line 35: / Line 149: @@
 [[Category:Introductory Algebra Problems]]
+[[Category:Introductory Trigonometry Problems]]
+{{MAA Notice}}

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