Consider a rectangle $ABCD$ with $BC = 2 \cdot AB$. Let $\omega$ be the circle that touches the sides $AB$, $BC$, and $AD$. A tangent drawn from point $C$ to the circle $\omega$ intersects the segment $AD$ at point $K$. Determine the ratio $\frac{AK}{KD}$.

[i]Proposed by Giorgi Arabidze, Georgia[/i]
1. **Define Variables and Relationships:**
   Let \( AB = x \) and \( BC = 2x \). Since \( BC = 2 \cdot AB \), the rectangle \(ABCD\) has dimensions \( x \) and \( 2x \).

2. **Circle \(\omega\) and Tangent Properties:**
   The circle \(\omega\) is inscribed in the right triangle \(AKCB\). Therefore, the tangents from a point to a circle are equal in length. Let \( AK = t \).

3. **Using Tangent Properties:**
   Since \(\omega\) touches \(AB\), \(BC\), and \(AD\), we have:
   \[
   AK + BC = AB + KC
   \]
   Substituting the known lengths:
   \[
   t + 2x = x + KC \implies KC = t + x
   \]

4. **Applying the Pythagorean Theorem:**
   In \(\triangle DKC\), we apply the Pythagorean theorem:
   \[
   KC^2 = DC^2 + DK^2
   \]
   Here, \( DC = AB = x \) and \( DK = AD - AK = 2x - t \). Substituting \( KC = t + x \):
   \[
   (t + x)^2 = x^2 + (2x - t)^2
   \]

5. **Expanding and Simplifying:**
   Expanding both sides:
   \[
   t^2 + 2tx + x^2 = x^2 + 4x^2 - 4tx + t^2
   \]
   Simplifying:
   \[
   t^2 + 2tx + x^2 = 5x^2 - 4tx + t^2
   \]
   Canceling \( t^2 \) from both sides:
   \[
   2tx + x^2 = 5x^2 - 4tx
   \]
   Combining like terms:
   \[
   6tx = 4x^2
   \]
   Solving for \( t \):
   \[
   t = \frac{2x}{3}
   \]

6. **Finding the Ratio \(\frac{AK}{KD}\):**
   Now, \( AK = t = \frac{2x}{3} \) and \( KD = AD - AK = 2x - \frac{2x}{3} = \frac{6x - 2x}{3} = \frac{4x}{3} \).

   Therefore, the ratio is:
   \[
   \frac{AK}{KD} = \frac{\frac{2x}{3}}{\frac{4x}{3}} = \frac{2}{4} = \frac{1}{2}
   \]

The final answer is \( \boxed{ \frac{1}{2} } \)