Triangle $ABC$ is inscribed in a unit circle $\omega$. Let $H$ be its orthocenter and $D$ be the foot of the perpendicular from $A$ to $BC$. Let $\triangle XY Z$ be the triangle formed by drawing the tangents to $\omega$ at $A, B, C$. If $\overline{AH} = \overline{HD}$ and the side lengths of $\triangle XY Z$ form an arithmetic sequence, the area of $\triangle ABC$ can be expressed in the form $\tfrac{p}{q}$ for relatively prime positive integers $p, q$. What is $p + q$?

1. **Understanding the Problem:**
   - We have a triangle \( \triangle ABC \) inscribed in a unit circle \( \omega \).
   - \( H \) is the orthocenter of \( \triangle ABC \).
   - \( D \) is the foot of the perpendicular from \( A \) to \( BC \).
   - \( \triangle XYZ \) is formed by drawing tangents to \( \omega \) at \( A, B, C \).
   - Given \( \overline{AH} = \overline{HD} \) and the side lengths of \( \triangle XYZ \) form an arithmetic sequence.
   - We need to find the area of \( \triangle ABC \) in the form \( \frac{p}{q} \) and determine \( p + q \).

2. **Orthocenter Condition:**
   - Given \( \overline{AH} = \overline{HD} \), we know that \( H \) lies on the nine-point circle of \( \triangle ABC \), which is also the circumcircle of the medial triangle.
   - Reflecting \( H \) over \( D \) to \( H' \), we know \( H' \) lies on \( \omega \).

3. **Using Trigonometry and Power of a Point:**
   - Let \( h \) be the height from \( A \) to \( BC \).
   - Using the power of a point:
     \[
     BD \cdot DC = AD \cdot DH' \implies h \cot B \cdot h \cot C = h \cdot \frac{h}{2}
     \]
     \[
     \implies \tan B \cdot \tan C = 2
     \]

4. **Tangents and Inradius:**
   - Let \( I \) be the incenter of \( \triangle XYZ \).
   - Noting that \( \angle ZIA = \angle BCA \) and \( \angle YIA = \angle ABC \):
     \[
     \tan B \cdot \tan C = \frac{AZ}{1} \cdot \frac{AY}{1} = 2
     \]

5. **Side Lengths of \( \triangle XYZ \):**
   - Let \( XB = XC = x \), \( YA = YC = y \), and \( ZA = ZB = z \).
   - Given \( 2yz = 2 \), we have \( yz = 1 \).
   - The inradius of \( \triangle XYZ \) is \( \sqrt{\frac{xyz}{x+y+z}} \), which equals 1 (since the circumradius of \( \triangle ABC \) is 1):
     \[
     x + y + z = xyz \implies x + y + z = 2x \implies x = y + z
     \]

6. **Arithmetic Sequence Condition:**
   - Since the side lengths form an arithmetic sequence, we can assume \( x = y + z \).
   - Let \( 2z = x + y \):
     \[
     2z = x + y \implies 2z = 2y + z \implies z = 2y \implies z = 2, y = 1, x = 3
     \]

7. **Area Calculation:**
   - The side lengths \( x = 3 \), \( y = 1 \), \( z = 2 \) form a \( 3-4-5 \) triangle.
   - The area of \( \triangle ABC \) can be calculated by subtracting the outer three triangles from the total area:
     \[
     \text{Area of } \triangle ABC = \frac{6}{5}
     \]

The final answer is \( \boxed{11} \)