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

Revision as of 10:42, 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 above 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:

@@ Line 86: / Line 86: @@
-At this point you should be getting suspicious, because some terms are repeating. Notice that the formula asks for a sum and then later a product of two terms, neither of which depend on order (commutative and associative properties). So that means if <math>a_n = x</math>, and <math>a_{n+1} = y</math>, then we would get the same thing if <math>a_n = y</math> and <math>a_{n+1} = x</math> (Basically, as long the combination of the previous two elements is the same, we should get the same result). So <math>a_7 = 0</math>, <math>a_8 = -(2+\sqrt{3})</math>, <math>a_9 = 0</math>, and so on.
+Solving in this way until <math>a_</math>
-We can generalize this to say
-"For all even <math>n</math> such that <math>n \ge 4</math>, <math>a_n = -(2+\sqrt{3})</math>"
+The problem asks us for the value of <math>a_{2009}</math>, and since <math>2009</math> is an odd number, we know that <math>a_{2009} = 0 \Rightarrow \boxed{\text{E}}</math>.
+==Clarification of Solution 1==
+(While I did say the above 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 [[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 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  the formula for <math>\tan{(45+30)}</math> or <math>\tan{75}</math>, meaning <math>a_3 = \tan{75}</math>. So, the sequence is simply the sum of the two 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>
-"For all odd <math>n</math> such that <math>n \ge 5</math>, <math>a_n = 0</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>. 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>. Now we continue our sequence:
-The problem asks us for the value of <math>a_{2009}</math>, and since <math>2009</math> is an odd number, we know that <math>a_{2009} = 0 \Rightarrow \boxed{\text{E}}</math>.
 == See also ==

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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