Let $\triangle ABC$ be a triangle with $AB=85$, $BC=125$, $CA=140$, and incircle $\omega$.  Let $D$, $E$, $F$ be the points of tangency of $\omega$ with $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ respectively, and furthermore denote by $X$, $Y$,  and $Z$ the incenters of $\triangle AEF$, $\triangle BFD$, and $\triangle CDE$, also respectively.  Find the circumradius of $\triangle XYZ$.

[i] Proposed by David Altizio [/i]
1. **Identify the sides of the triangle and the inradius:**
   Given the sides of $\triangle ABC$ are $AB = 85$, $BC = 125$, and $CA = 140$. We need to find the inradius $r$ of $\triangle ABC$.

2. **Calculate the semi-perimeter $s$ of $\triangle ABC$:**
   \[
   s = \frac{AB + BC + CA}{2} = \frac{85 + 125 + 140}{2} = 175
   \]

3. **Use Heron's formula to find the area $K$ of $\triangle ABC$:**
   \[
   K = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   where $a = 125$, $b = 140$, and $c = 85$.
   \[
   K = \sqrt{175(175-125)(175-140)(175-85)} = \sqrt{175 \cdot 50 \cdot 35 \cdot 90}
   \]
   Simplify the expression inside the square root:
   \[
   K = \sqrt{175 \cdot 50 \cdot 35 \cdot 90} = \sqrt{175 \cdot 50 \cdot 35 \cdot 90}
   \]
   \[
   K = \sqrt{(5^2 \cdot 7) \cdot (2 \cdot 5^2) \cdot (5 \cdot 7) \cdot (2 \cdot 3^2 \cdot 5)} = \sqrt{5^5 \cdot 7^2 \cdot 2^2 \cdot 3^2}
   \]
   \[
   K = 5^2 \cdot 7 \cdot 2 \cdot 3 = 25 \cdot 7 \cdot 2 \cdot 3 = 25 \cdot 42 = 1050
   \]

4. **Calculate the inradius $r$ using the formula $r = \frac{K}{s}$:**
   \[
   r = \frac{K}{s} = \frac{1050}{175} = 6
   \]

5. **Use Fact 5 to determine the circumradius of $\triangle XYZ$:**
   According to Fact 5, the circumradius of $\triangle XYZ$ is equal to the inradius of $\triangle ABC$.

6. **Conclusion:**
   The circumradius of $\triangle XYZ$ is $30$.

The final answer is $\boxed{30}$.