Difference between revisions of "2002 AIME I Problems/Problem 12"
Mathgeek2006 (talk | contribs) m (→Solution) |
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Iterating <math>F</math> we get: | Iterating <math>F</math> we get: | ||
− | < | + | <cmath> |
\begin{align*} | \begin{align*} | ||
F(z) &= \frac{z+i}{z-i}\\ | F(z) &= \frac{z+i}{z-i}\\ | ||
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F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \frac{2z}{2} = z. | F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \frac{2z}{2} = z. | ||
\end{align*} | \end{align*} | ||
− | </ | + | </cmath> |
From this, it follows that <math>z_{k+3} = z_k</math>, for all <math>k</math>. Thus | From this, it follows that <math>z_{k+3} = z_k</math>, for all <math>k</math>. Thus | ||
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Thus <math>a+b = 1+274 = \boxed{275}</math>. | Thus <math>a+b = 1+274 = \boxed{275}</math>. | ||
+ | |||
== See also == | == See also == | ||
{{AIME box|year=2002|n=I|num-b=11|num-a=13}} | {{AIME box|year=2002|n=I|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:35, 13 March 2015
Problem
Let for all complex numbers , and let for all positive integers . Given that and , where and are real numbers, find .
Solution
Iterating we get:
From this, it follows that , for all . Thus
Thus .
See also
2002 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.