Let $ABC$ be a triangle such that $AB = 7$, $BC = 8$, and $CA = 9$. There exists a unique point $X$ such that $XB = XC$ and $XA$ is tangent to the circumcircle of $ABC$. If $XA = \tfrac ab$, where $a$ and $b$ are coprime positive integers, find $a + b$.

[i]Proposed by Alexander Wang[/i]
1. **Identify the point \( X \)**:
   - \( X \) is the point such that \( XB = XC \) and \( XA \) is tangent to the circumcircle of \( \triangle ABC \). This implies that \( X \) is the excenter opposite \( A \).

2. **Use the Law of Cosines to find \( \cos \angle BAC \)**:
   \[
   \cos \angle BAC = \frac{b^2 + c^2 - a^2}{2bc} = \frac{7^2 + 9^2 - 8^2}{2 \cdot 7 \cdot 9} = \frac{49 + 81 - 64}{126} = \frac{66}{126} = \frac{11}{21}
   \]

3. **Find \( \cos \angle BMC \)**:
   - Since \( \angle BMC = 180^\circ - \angle BAC \), we have:
   \[
   \cos \angle BMC = -\cos \angle BAC = -\frac{11}{21}
   \]

4. **Use the Law of Cosines in \( \triangle BMC \)**:
   \[
   BC^2 = BM^2 + MC^2 - 2 \cdot BM \cdot MC \cdot \cos \angle BMC
   \]
   Given \( BC = 8 \), we have:
   \[
   8^2 = BM^2 + MC^2 + 2 \cdot BM \cdot MC \cdot \frac{11}{21}
   \]
   Since \( BM = MC \) (because \( X \) is the excenter and \( XB = XC \)):
   \[
   64 = 2BM^2 \left(1 + \frac{11}{21}\right) = 2BM^2 \cdot \frac{32}{21}
   \]
   Solving for \( BM \):
   \[
   64 = 2BM^2 \cdot \frac{32}{21} \implies BM^2 = \frac{64 \cdot 21}{64} = 21 \implies BM = \sqrt{21}
   \]

5. **Find the radius \( R \) of the circumcircle**:
   - Using the formula for the circumradius \( R \):
   \[
   R = \frac{abc}{4K}
   \]
   where \( K \) is the area of \( \triangle ABC \). Using Heron's formula:
   \[
   s = \frac{a + b + c}{2} = \frac{7 + 8 + 9}{2} = 12
   \]
   \[
   K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12(12-7)(12-8)(12-9)} = \sqrt{12 \cdot 5 \cdot 4 \cdot 3} = \sqrt{720} = 12\sqrt{5}
   \]
   Thus,
   \[
   R = \frac{7 \cdot 8 \cdot 9}{4 \cdot 12\sqrt{5}} = \frac{504}{48\sqrt{5}} = \frac{21\sqrt{5}}{10}
   \]

6. **Find \( \tan \angle AOX \)**:
   - Since \( \angle AOX = 2 \angle AMX \) and using the double angle formula:
   \[
   \cos \angle AOX = 2 \cos^2 \angle AMX - 1
   \]
   Given \( \cos \angle AMX = \frac{2\sqrt{5}}{\sqrt{21}} \):
   \[
   \cos \angle AOX = 2 \left(\frac{2\sqrt{5}}{\sqrt{21}}\right)^2 - 1 = 2 \cdot \frac{20}{21} - 1 = \frac{40}{21} - 1 = \frac{19}{21}
   \]
   \[
   \sin \angle AOX = \sqrt{1 - \cos^2 \angle AOX} = \sqrt{1 - \left(\frac{19}{21}\right)^2} = \sqrt{\frac{441 - 361}{441}} = \sqrt{\frac{80}{441}} = \frac{4\sqrt{5}}{21}
   \]
   \[
   \tan \angle AOX = \frac{\sin \angle AOX}{\cos \angle AOX} = \frac{\frac{4\sqrt{5}}{21}}{\frac{19}{21}} = \frac{4\sqrt{5}}{19}
   \]

7. **Calculate \( XA \)**:
   \[
   XA = R \cdot \tan \angle AOX = \frac{21\sqrt{5}}{10} \cdot \frac{4\sqrt{5}}{19} = \frac{21 \cdot 4 \cdot 5}{10 \cdot 19} = \frac{420}{190} = \frac{42}{19}
   \]

8. **Find \( a + b \)**:
   - Here, \( a = 42 \) and \( b = 19 \), so:
   \[
   a + b = 42 + 19 = 61
   \]

The final answer is \( \boxed{61} \)