Difference between revisions of "2009 AMC 12A Problems/Problem 2"
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| − | + | \frac 53 | |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
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| + | This is choice <math>\boxed{\text{C}}</math>. | ||
| + | |||
| + | Interesting sidenote: The continued fraction <math>1 + \frac {1}{1 + \frac {1}{1 + 1....}}</math> is equal to the golden ratio, or <math>\frac{1+\sqrt{5}}{2}</math>. | ||
== See Also == | == See Also == | ||
| + | {{AMC10 box|year=2009|ab=A|num-b=2|num-a=4}} | ||
{{AMC12 box|year=2009|ab=A|num-b=1|num-a=3}} | {{AMC12 box|year=2009|ab=A|num-b=1|num-a=3}} | ||
| − | {{ | + | {{MAA Notice}} |
Latest revision as of 08:19, 8 June 2021
- The following problem is from both the 2009 AMC 12A #2 and 2009 AMC 10A #3, so both problems redirect to this page.
Problem
Which of the following is equal to 
?
Solution
We compute:
This is choice 
.
Interesting sidenote: The continued fraction 
is equal to the golden ratio, or 
.
See Also
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 10 Problems and Solutions | ||
| 2009 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
