AMC12 2009 A

AMC12 2009 A · Q25

AMC12 2009 A · Q25. It mainly tests Sequences & recursion (algebra), Trigonometry (basic).

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 $|a_{2009}|$?

一个数列的前两项为 $a_1 = 1$ 和 $a_2 = \frac {1}{\sqrt3}$。对 $n\ge1$， \[a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.\] 求 $|a_{2009}|$。

(A) 0 0

(B) $2 - \sqrt{3}$ $2 - \sqrt{3}$

(D) 1 1

(E) $2 + \sqrt{3}$ $2 + \sqrt{3}$

Answer

Correct choice: (A)

正确答案：（A）

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

考虑另一个数列 $\{\theta_1, \theta_2, \theta_3...\}$，使得 $a_n = \tan{\theta_n}$，且 $0 \leq \theta_n < 180$。给定递推式变为 \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*} 因此 $\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}$。由于 $\theta_1 = 45, \theta_2 = 30$，数列 $\{\theta_1, \theta_2, \theta_3...\}$ 的所有项都是 $15$ 的倍数。再考虑数列 $\{b_1, b_2, b_3...\}$，使得 $b_n = \theta_n/15$，且 $0 \leq b_n < 12$。则 $b_n$ 满足 $b_1 = 3, b_2 = 2, b_{n + 2} \equiv b_{n + 1} + b_n \pmod{12}$。由于相邻两项的可能取值有限，数列 $b_n$ 必为周期数列。写出前若干项直到开始重复： $\{b_n\} = \{3,2,5,7,0,7,7,2,9,11,8,7,3,10,1,11,0,11,11,10,9,7,4,11,3,2,5,7,...\}$ 注意到 $b_{25} = b_1 = 3$ 且 $b_{26} = b_2 = 2$，因此 $\{b_n\}$ 的周期为 $24$：$b_{n + 24} = b_n$。于是 $b_{2009} = b_{17} = 0$，且 $\theta_{2009} = 15 b_{2009} = 0$。因此 $a_{2009} = \tan{\theta_{2009}} = \tan{0} = 0.$ 答案为 $|a_{2009}| = \boxed{\textbf{(A)}\ 0}$。

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