Difference between revisions of "2009 AMC 12A Problems/Problem 25"
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<center><cmath>\begin{align*} a_{n + 2} & = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \ | <center><cmath>\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}} & = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \ | ||
− | \tan{\theta_{n + 2}} & = \tan | + | \tan{\theta_{n + 2}} & = \tan(\theta_{n + 1} + \theta_n) \end{align*}</cmath></center> |
It follows that <math>\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}</math>. Since <math>\theta_1 = 45, \theta_2 = 30</math>, all terms in the sequence <math>\{\theta_1, \theta_2, \theta_3...\}</math> will be a multiple of <math>15</math>. | It follows that <math>\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}</math>. Since <math>\theta_1 = 45, \theta_2 = 30</math>, all terms in the sequence <math>\{\theta_1, \theta_2, \theta_3...\}</math> will be a multiple of <math>15</math>. |
Revision as of 10:13, 11 April 2009
Problem
The first two terms of a sequence are and . For ,
What is ?
Solution
Consider another sequence such that , and .
The given recurrence becomes
It follows that . Since , all terms in the sequence will be a multiple of .
Now consider another sequence such that , and . The sequence satisfies .
As the number of possible consecutive two terms is finite, we know that the sequence is periodic. Write out the first few terms of the sequence until it starts to repeat.
Note that and . Thus has a period of : .
It follows that and . Thus
Our answer is .
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 |