Equilateral triangle $ABC$ has side length $2$. A semicircle is drawn with diameter $BC$ such that it lies outside the triangle, and minor arc $BC$ is drawn so that it is part of a circle centered at $A$. The area of the “lune” that is inside the semicircle but outside sector $ABC$ can be expressed in the form $\sqrt{p}-\frac{q\pi}{r}$, where $p, q$, and $ r$ are positive integers such that $q$ and $r$ are relatively prime. Compute $p + q + r$.
[img]https://cdn.artofproblemsolving.com/attachments/7/7/f349a807583a83f93ba413bebf07e013265551.png[/img]
1. **Calculate the area of the semicircle with diameter \( BC \):**
   - The side length of the equilateral triangle \( ABC \) is given as \( 2 \). Therefore, the diameter of the semicircle is also \( 2 \).
   - The radius \( r \) of the semicircle is \( \frac{2}{2} = 1 \).
   - The area of the semicircle is given by:
     \[
     \text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (1)^2 = \frac{\pi}{2}
     \]

2. **Calculate the area of sector \( BAC \):**
   - Since \( ABC \) is an equilateral triangle, the angle \( \angle BAC \) is \( 60^\circ \) or \( \frac{\pi}{3} \) radians.
   - The radius of the circle centered at \( A \) passing through \( B \) and \( C \) is the side length of the triangle, which is \( 2 \).
   - The area of sector \( BAC \) is given by:
     \[
     \text{Area of sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} (2)^2 \left( \frac{\pi}{3} \right) = 2 \cdot \frac{\pi}{3} = \frac{2\pi}{3}
     \]

3. **Calculate the area of triangle \( ABC \):**
   - The area of an equilateral triangle with side length \( s \) is given by:
     \[
     \text{Area of } \triangle ABC = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} (2)^2 = \frac{\sqrt{3}}{4} \cdot 4 = \sqrt{3}
     \]

4. **Calculate the area of the lune:**
   - The area of the lune is the area of the semicircle minus the area of the sector \( BAC \) plus the area of triangle \( ABC \):
     \[
     \text{Area of lune} = \text{Area of semicircle} - (\text{Area of sector} - \text{Area of } \triangle ABC)
     \]
     \[
     \text{Area of lune} = \frac{\pi}{2} - \left( \frac{2\pi}{3} - \sqrt{3} \right)
     \]
     \[
     \text{Area of lune} = \frac{\pi}{2} - \frac{2\pi}{3} + \sqrt{3}
     \]
     \[
     \text{Area of lune} = \sqrt{3} + \left( \frac{3\pi}{6} - \frac{4\pi}{6} \right)
     \]
     \[
     \text{Area of lune} = \sqrt{3} - \frac{\pi}{6}
     \]

5. **Express the area in the form \( \sqrt{p} - \frac{q\pi}{r} \):**
   - Comparing \( \sqrt{3} - \frac{\pi}{6} \) with \( \sqrt{p} - \frac{q\pi}{r} \), we identify \( p = 3 \), \( q = 1 \), and \( r = 6 \).

6. **Compute \( p + q + r \):**
   \[
   p + q + r = 3 + 1 + 6 = 10
   \]

The final answer is \( \boxed{10} \)