Let $\Omega_1$ and $\Omega_2$ be two circles in the plane.  Suppose the common external tangent to $\Omega_1$ and $\Omega_2$ has length $2017$ while their common internal tangent has length $2009$.  Find the product of the radii of $\Omega_1$ and $\Omega_2$.

[i]Proposed by David Altizio[/i]
1. Let the radii of the circles \(\Omega_1\) and \(\Omega_2\) be \(r_1\) and \(r_2\) respectively. Let the distance between the centers of the circles be \(d\).

2. The length of the common external tangent between two circles is given by:
   \[
   \sqrt{d^2 - (r_1 + r_2)^2} = 2017
   \]
   Squaring both sides, we get:
   \[
   d^2 - (r_1 + r_2)^2 = 2017^2
   \]

3. The length of the common internal tangent between two circles is given by:
   \[
   \sqrt{d^2 - (r_1 - r_2)^2} = 2009
   \]
   Squaring both sides, we get:
   \[
   d^2 - (r_1 - r_2)^2 = 2009^2
   \]

4. We now have two equations:
   \[
   d^2 - (r_1 + r_2)^2 = 2017^2 \quad \text{(1)}
   \]
   \[
   d^2 - (r_1 - r_2)^2 = 2009^2 \quad \text{(2)}
   \]

5. Subtract equation (2) from equation (1):
   \[
   (d^2 - (r_1 + r_2)^2) - (d^2 - (r_1 - r_2)^2) = 2017^2 - 2009^2
   \]
   Simplifying, we get:
   \[
   -(r_1 + r_2)^2 + (r_1 - r_2)^2 = 2017^2 - 2009^2
   \]
   \[
   -(r_1^2 + 2r_1r_2 + r_2^2) + (r_1^2 - 2r_1r_2 + r_2^2) = 2017^2 - 2009^2
   \]
   \[
   -2r_1r_2 - 2r_1r_2 = 2017^2 - 2009^2
   \]
   \[
   -4r_1r_2 = 2017^2 - 2009^2
   \]

6. Calculate \(2017^2 - 2009^2\):
   \[
   2017^2 - 2009^2 = (2017 - 2009)(2017 + 2009) = 8 \times 4026 = 32208
   \]

7. Substitute back into the equation:
   \[
   -4r_1r_2 = 32208
   \]
   \[
   r_1r_2 = \frac{32208}{-4} = 8052
   \]

The final answer is \(\boxed{8052}\).