If $\phi$ is the Golden Ratio, we know that $\frac1\phi = \phi - 1$.  Define a new positive real number, called $\phi_d$, where $\frac1{\phi_d} = \phi_d - d$ (so $\phi = \phi_1$).  Given that $\phi_{2009} = \frac{a + \sqrt{b}}{c}$, $a, b, c$ positive integers, and the greatest common divisor of $a$ and $c$ is 1, find $a + b + c$.
1. Let $\phi_d = x$. We are given the equation:
   \[
   \frac{1}{\phi_d} = \phi_d - d
   \]
   Substituting $x$ for $\phi_d$, we get:
   \[
   \frac{1}{x} = x - d
   \]

2. Multiply both sides by $x$ to clear the fraction:
   \[
   1 = x^2 - dx
   \]

3. Rearrange the equation to standard quadratic form:
   \[
   x^2 - dx - 1 = 0
   \]

4. To solve for $x$, use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -d$, and $c = -1$:
   \[
   x = \frac{-(-d) \pm \sqrt{(-d)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}
   \]
   Simplify the expression:
   \[
   x = \frac{d \pm \sqrt{d^2 + 4}}{2}
   \]

5. Since we are looking for the positive real number $\phi_d$, we take the positive root:
   \[
   \phi_d = \frac{d + \sqrt{d^2 + 4}}{2}
   \]

6. For $d = 2009$, we have:
   \[
   \phi_{2009} = \frac{2009 + \sqrt{2009^2 + 4}}{2}
   \]

7. Calculate $2009^2 + 4$:
   \[
   2009^2 = 2009 \times 2009 = 4036081
   \]
   \[
   2009^2 + 4 = 4036081 + 4 = 4036085
   \]

8. Thus, $\phi_{2009}$ is:
   \[
   \phi_{2009} = \frac{2009 + \sqrt{4036085}}{2}
   \]

9. Given that $a = 2009$, $b = 4036085$, and $c = 2$, and noting that the greatest common divisor of $a$ and $c$ is 1, we find:
   \[
   a + b + c = 2009 + 4036085 + 2 = 4038096
   \]

The final answer is $\boxed{4038096}$