Let equilateral triangle $\vartriangle ABC$ be inscribed in a circle $\omega_1$ with radius $4$. Consider another circle $\omega_2$ with radius $2$ internally tangent to $\omega_1$ at $A$. Let $\omega_2$ intersect sides $\overline{AB}$ and $\overline{AC}$ at $D$ and $E$,  respectively, as shown in the diagram. Compute the area of the shaded region.
[img]https://cdn.artofproblemsolving.com/attachments/a/5/8255cdb8b041719d735607da8139aa4016375d.png[/img]
1. **Calculate the area of the circular segment \( BC \):**

   The area of the circular segment \( BC \) can be found by subtracting the area of the equilateral triangle \( \triangle ABC \) from the area of the sector of the circle that subtends the arc \( BC \).

   - The area of the sector subtending \( \angle BAC = 120^\circ \) is:
     \[
     \text{Area of sector} = \frac{120^\circ}{360^\circ} \cdot \pi \cdot 4^2 = \frac{1}{3} \cdot 16\pi = \frac{16\pi}{3}
     \]

   - The area of the equilateral triangle \( \triangle ABC \) with side length \( s \) can be calculated using the formula:
     \[
     \text{Area of } \triangle ABC = \frac{\sqrt{3}}{4} s^2
     \]
     Since the radius of the circumscribed circle is \( 4 \), the side length \( s \) of the equilateral triangle is:
     \[
     s = 4\sqrt{3}
     \]
     Therefore, the area of \( \triangle ABC \) is:
     \[
     \text{Area of } \triangle ABC = \frac{\sqrt{3}}{4} (4\sqrt{3})^2 = \frac{\sqrt{3}}{4} \cdot 48 = 12\sqrt{3}
     \]

   - The area of the circular segment \( BC \) is:
     \[
     \text{Area of segment } BC = \frac{16\pi}{3} - 12\sqrt{3}
     \]

2. **Calculate the area of the shape formed by subtracting the circular segment \( DE \) from the trapezoid \( BDEC \):**

   - The area of the trapezoid \( BDEC \) can be found by subtracting the area of the equilateral triangle \( \triangle ADE \) from the area of the larger equilateral triangle \( \triangle ABC \).

   - The area of the equilateral triangle \( \triangle ADE \) with side length \( s' \) can be calculated using the same formula:
     \[
     \text{Area of } \triangle ADE = \frac{\sqrt{3}}{4} s'^2
     \]
     Since the radius of the inscribed circle \( \omega_2 \) is \( 2 \), the side length \( s' \) of the equilateral triangle \( \triangle ADE \) is:
     \[
     s' = 2\sqrt{3}
     \]
     Therefore, the area of \( \triangle ADE \) is:
     \[
     \text{Area of } \triangle ADE = \frac{\sqrt{3}}{4} (2\sqrt{3})^2 = \frac{\sqrt{3}}{4} \cdot 12 = 3\sqrt{3}
     \]

   - The area of the circular segment \( DE \) can be found similarly to the segment \( BC \):
     \[
     \text{Area of sector} = \frac{120^\circ}{360^\circ} \cdot \pi \cdot 2^2 = \frac{1}{3} \cdot 4\pi = \frac{4\pi}{3}
     \]
     \[
     \text{Area of } \triangle ADE = 3\sqrt{3}
     \]
     \[
     \text{Area of segment } DE = \frac{4\pi}{3} - 3\sqrt{3}
     \]

   - The area of the shape formed by subtracting the circular segment \( DE \) from the trapezoid \( BDEC \) is:
     \[
     \text{Area of trapezoid } BDEC = 12\sqrt{3} - 3\sqrt{3} = 9\sqrt{3}
     \]
     \[
     \text{Area of shape} = 9\sqrt{3} - \left(\frac{4\pi}{3} - 3\sqrt{3}\right) = 9\sqrt{3} - \frac{4\pi}{3} + 3\sqrt{3} = 12\sqrt{3} - \frac{4\pi}{3}
     \]

3. **Combine the areas to find the total area of the shaded region:**

   \[
   \text{Total area} = \left(\frac{16\pi}{3} - 12\sqrt{3}\right) + \left(12\sqrt{3} - \frac{4\pi}{3}\right) = \frac{16\pi}{3} - \frac{4\pi}{3} = 4\pi
   \]

The final answer is \(\boxed{4\pi}\)