Difference between revisions of "2009 AMC 12A Problems/Problem 2"

Revision as of 08:18, 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 $1 + \frac {1}{1 + \frac {1}{1 + 1}}$ ?

$\textbf{(A)}\ \frac {5}{4} \qquad \textbf{(B)}\ \frac {3}{2} \qquad \textbf{(C)}\ \frac {5}{3} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$

Solution

We compute:

$\begin{align*} 1 + \frac {1}{1 + \frac {1}{1 + 1}} &= 1 + \frac {1}{1 + \frac {1}{1 + 1}} \\ &= 1 + \frac {1}{1 + \frac 12} \\ &= 1 + \frac {1}{\frac 32} \\ &= 1 + \frac 23 \\ &= \frac 53 \end{align*}$

This is choice $\boxed{\text{C}}$ .

Interesting sidenote: The continued fraction $1 + \frac {1}{1 + \frac {1}{1 + 1...}}$ is equal to the golden ratio, or $\frac{1+sqrt5}{2}$ .

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.

@@ Line 31: / Line 31: @@
 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+sqrt5}{2}</math>.
 == See Also ==

Page

Toolbox

Search

Difference between revisions of "2009 AMC 12A Problems/Problem 2"

Revision as of 08:18, 8 June 2021

Problem

Solution

See Also