Let $ABC$ be a triangle with $AB = 13$, $BC = 14$, $CA = 15$. Let $H$ be the orthocenter of $ABC$. Find the radius of the circle with nonzero radius tangent to the circumcircles of $AHB$, $BHC$, $CHA$.
1. **Identify the given elements and required circle:**
   - We are given a triangle \(ABC\) with sides \(AB = 13\), \(BC = 14\), and \(CA = 15\).
   - \(H\) is the orthocenter of \(\triangle ABC\).
   - We need to find the radius of the circle that is tangent to the circumcircles of \(\triangle AHB\), \(\triangle BHC\), and \(\triangle CHA\).

2. **Calculate the circumradius \(R\) of \(\triangle ABC\):**
   - Using the formula for the circumradius \(R\) of a triangle with sides \(a\), \(b\), and \(c\):
     \[
     R = \frac{abc}{4K}
     \]
     where \(K\) is the area of the triangle.
   - First, calculate the area \(K\) using Heron's formula:
     \[
     s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = 21
     \]
     \[
     K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84
     \]
   - Now, calculate \(R\):
     \[
     R = \frac{13 \cdot 14 \cdot 15}{4 \cdot 84} = \frac{2730}{336} = \frac{65}{8}
     \]

3. **Understand the relationship between the circles:**
   - The circumcircles of \(\triangle AHB\), \(\triangle BHC\), and \(\triangle CHA\) are congruent to the circumcircle of \(\triangle ABC\).
   - The desired circle \(\Omega\) is tangent to these three circumcircles.

4. **Determine the radius of the desired circle \(\Omega\):**
   - The common external tangents of the circumcircles of \(\triangle AHB\), \(\triangle BHC\), and \(\triangle CHA\) meet at points that form a triangle similar to \(\triangle ABC\) but scaled by a factor of 2.
   - Therefore, the radius of \(\Omega\) is half the radius of the circumcircle of \(\triangle ABC\).

5. **Calculate the radius of \(\Omega\):**
   - Since the radius of the circumcircle of \(\triangle ABC\) is \(\frac{65}{8}\), the radius of \(\Omega\) is:
     \[
     \frac{1}{2} \times \frac{65}{8} = \frac{65}{16}
     \]

The final answer is \(\boxed{\frac{65}{16}}\).