Triangle $ABC$ is equilateral with side length $12$. Point $D$ is the midpoint of side $\overline{BC}$. Circles $A$ and $D$ intersect at the midpoints of side $AB$ and $AC$. Point $E$ lies on segment $\overline{AD}$ and circle $E$ is tangent to circles $A$ and $D$. Compute the radius of circle $E$.

[i]2015 CCA Math Bonanza Individual Round #5[/i]
1. **Identify the key points and distances in the equilateral triangle:**
   - Triangle \(ABC\) is equilateral with side length \(12\).
   - Point \(D\) is the midpoint of side \(\overline{BC}\), so \(BD = DC = \frac{12}{2} = 6\).
   - The length of \(AD\) can be found using the Pythagorean theorem in triangle \(ABD\):
     \[
     AD = \sqrt{AB^2 - BD^2} = \sqrt{12^2 - 6^2} = \sqrt{144 - 36} = \sqrt{108} = 6\sqrt{3}
     \]

2. **Determine the positions of the midpoints of sides \(AB\) and \(AC\):**
   - Let \(M\) and \(N\) be the midpoints of \(AB\) and \(AC\) respectively.
   - Since \(M\) and \(N\) are midpoints, \(AM = MB = AN = NC = \frac{12}{2} = 6\).

3. **Analyze the intersection of circles centered at \(A\) and \(D\):**
   - Circle centered at \(A\) with radius \(6\) (since it passes through \(M\) and \(N\)).
   - Circle centered at \(D\) with radius \(6\) (since it passes through \(M\) and \(N\)).

4. **Locate point \(E\) on segment \(\overline{AD}\):**
   - Point \(E\) lies on segment \(\overline{AD}\) such that circle \(E\) is tangent to both circles centered at \(A\) and \(D\).
   - Let the radius of circle \(E\) be \(r\).

5. **Calculate the distance between the centers of the circles:**
   - The distance between the centers of circles \(A\) and \(D\) is \(AD = 6\sqrt{3}\).

6. **Determine the tangency condition:**
   - Since circle \(E\) is tangent to both circles, the sum of the radii of circle \(E\) and the other two circles must equal the distance between the centers:
     \[
     r + 6 + r + 6 = 6\sqrt{3}
     \]
     Simplifying, we get:
     \[
     2r + 12 = 6\sqrt{3}
     \]
     Solving for \(r\):
     \[
     2r = 6\sqrt{3} - 12
     \]
     \[
     r = 3\sqrt{3} - 6
     \]

The final answer is \(\boxed{3\sqrt{3} - 6}\).