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

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 $\frac{1+\sqrt{5}}{2}$.