The first two terms of a sequence are a_1 = 1 and a_2 = \dfrac1√3. For n ≥ 1, a_n+2 = \dfraca_n + a_n+11 - a_n a_n+1. What is |a_2009|?

2009 AMC 12A Problem 25 — Solution

2009 AMC 12A Problem 25

Below is the professionally curated solution for Problem 25 of the 2009 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12A solutions, or check the answer key.

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Concepts:trigonometric identity recursion modular arithmetic

Difficulty rating: 2520

25.

The first two terms of a sequence are $a_1 = 1$ and $a_2 = \dfrac{1}{\sqrt{3}}.$ For $n \ge 1,$ $a_{n+2} = \dfrac{a_n + a_{n+1}}{1 - a_n a_{n+1}}.$ What is $|a_{2009}|?$

$0$

$2 - \sqrt{3}$

$\dfrac{1}{\sqrt{3}}$

$1$

$2 + \sqrt{3}$

Solution:

The recursion is exactly the tangent addition formula, and $a_1 = \tan\dfrac{\pi}{4},$ $a_2 = \tan\dfrac{\pi}{6}.$

Writing $a_n = \tan\dfrac{\pi c_n}{12}$ with $c_1 = 3,$ $c_2 = 2,$ and $c_{n+2} \equiv c_n + c_{n+1} \pmod{12},$ the sequence $c_n$ is $3, 2, 5, 7, 0, 7, 7, 2, 9, 11, 8, 7, 3, 10, 1, 11, 0, 11, 11, 10, 9, 7, 4, 11, \ldots$ which is periodic with period $24.$

Since $2009 = 24\cdot 83 + 17,$ $c_{2009} = c_{17} = 0,$ so $a_{2009} = \tan 0 = 0$ and $|a_{2009}| = 0.$

Thus, the correct answer is A.

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