Let $ABC$ be a triangle with $AB=20$, $AC=34$, and $BC=42$.  Let $\omega_1$ and $\omega_2$ be the semicircles with diameters $\overline{AB}$ and $\overline{AC}$ erected outwards of $\triangle ABC$ and denote by $\ell$ the common external tangent to $\omega_1$ and $\omega_2$.  The line through $A$ perpendicular to $\overline{BC}$ intersects $\ell$ at $X$ and $BC$ at $Y$.  The length of $\overline{XY}$ can be written in the form $m+\sqrt n$ where $m$ and $n$ are positive integers.  Find $100m+n$.

[i]Proposed by David Altizio[/i]
1. **Set up the coordinate system:**
   Let $Y = (0, 0)$, $B = (-12, 0)$, $C = (30, 0)$, and $A = (0, 16)$. This setup places $A$ at $(0, 16)$, $B$ at $(-12, 0)$, and $C$ at $(30, 0)$.

2. **Determine the radii of the semicircles:**
   - The semicircle $\omega_1$ with diameter $\overline{AB}$ has radius $r_1 = \frac{AB}{2} = \frac{20}{2} = 10$.
   - The semicircle $\omega_2$ with diameter $\overline{AC}$ has radius $r_2 = \frac{AC}{2} = \frac{34}{2} = 17$.

3. **Find the centers of the semicircles:**
   - The center of $\omega_1$ is at the midpoint of $\overline{AB}$, which is $D = \left(\frac{-12 + 0}{2}, \frac{0 + 16}{2}\right) = (-6, 8)$.
   - The center of $\omega_2$ is at the midpoint of $\overline{AC}$, which is $E = \left(\frac{0 + 30}{2}, \frac{16 + 0}{2}\right) = (15, 8)$.

4. **Calculate the distance between the centers $D$ and $E$:**
   \[
   DE = \sqrt{(15 - (-6))^2 + (8 - 8)^2} = \sqrt{21^2} = 21
   \]

5. **Find the length of the common external tangent $\ell$:**
   The length of the common external tangent between two circles with radii $r_1$ and $r_2$ and distance $d$ between their centers is given by:
   \[
   \text{Length of tangent} = \sqrt{d^2 - (r_1 + r_2)^2}
   \]
   Here, $d = 21$, $r_1 = 10$, and $r_2 = 17$:
   \[
   \text{Length of tangent} = \sqrt{21^2 - (10 + 17)^2} = \sqrt{441 - 729} = \sqrt{-288}
   \]
   This calculation is incorrect because the distance $d$ should be greater than the sum of the radii for the external tangent to exist. Instead, we should use the correct formula for the external tangent:
   \[
   \text{Length of tangent} = \sqrt{d^2 - (r_1 - r_2)^2}
   \]
   \[
   \text{Length of tangent} = \sqrt{21^2 - (17 - 10)^2} = \sqrt{441 - 49} = \sqrt{392} = 14\sqrt{2}
   \]

6. **Determine the coordinates of $X$:**
   The line through $A$ perpendicular to $\overline{BC}$ intersects $\ell$ at $X$. Since $A$ is at $(0, 16)$ and the line is vertical, $X$ lies directly above or below $A$ on the y-axis. The length of the tangent from $A$ to $\ell$ is $14\sqrt{2}$, so $X$ is at $(0, 16 + 14\sqrt{2})$.

7. **Calculate the length of $\overline{XY}$:**
   \[
   XY = |16 + 14\sqrt{2} - 0| = 16 + 14\sqrt{2}
   \]

8. **Express $\overline{XY}$ in the form $m + \sqrt{n}$:**
   Here, $m = 16$ and $n = (14\sqrt{2})^2 = 392$.

9. **Calculate $100m + n$:**
   \[
   100m + n = 100 \cdot 16 + 392 = 1600 + 392 = 1992
   \]

The final answer is $\boxed{1992}$.