The circle centered at point $A$ with radius $19$ and the circle centered at point $B$ with radius $32$ are both internally tangent to a circle centered at point $C$ with radius $100$ such that point $C$ lies on segment $\overline{AB}$. Point $M$ is on the circle centered at $A$ and point $N$ is on the circle centered at $B$ such that line $MN$ is a common internal tangent of those two circles. Find the distance $MN$.
[img]https://cdn.artofproblemsolving.com/attachments/3/d/1933ce259c229d49e21b9a2dcadddea2a6b404.png[/img]
1. **Identify the given information and set up the problem:**
   - Circle centered at \( A \) with radius \( 19 \).
   - Circle centered at \( B \) with radius \( 32 \).
   - Both circles are internally tangent to a larger circle centered at \( C \) with radius \( 100 \).
   - Point \( C \) lies on segment \( \overline{AB} \).
   - Line \( MN \) is a common internal tangent to the circles centered at \( A \) and \( B \).

2. **Determine the distances between the centers:**
   - Since the circles are internally tangent to the larger circle, the distance from \( A \) to \( C \) is \( 100 - 19 = 81 \).
   - Similarly, the distance from \( B \) to \( C \) is \( 100 - 32 = 68 \).

3. **Set up the coordinate system:**
   - Place \( C \) at the origin \((0, 0)\).
   - Let \( A \) be at \((81, 0)\) and \( B \) be at \((-68, 0)\).

4. **Find the distance \( AB \):**
   \[
   AB = 81 + 68 = 149
   \]

5. **Determine the intersection point \( D \) of \( MN \) and \( AB \):**
   - Let \( AD = x \).
   - By similarity of triangles, \( BD = \frac{32}{19}x \).

6. **Set up the equation for the total length of \( AB \):**
   \[
   x + \frac{32}{19}x = 149
   \]
   \[
   x \left(1 + \frac{32}{19}\right) = 149
   \]
   \[
   x \left(\frac{51}{19}\right) = 149
   \]
   \[
   x = \frac{149 \cdot 19}{51} = \frac{2831}{51}
   \]

7. **Calculate \( BD \):**
   \[
   BD = \frac{32}{19} \cdot \frac{2831}{51} = \frac{32 \cdot 2831}{19 \cdot 51}
   \]

8. **Use the Pythagorean theorem to find \( MN \):**
   - The length of \( MN \) is the distance between the points of tangency on the circles centered at \( A \) and \( B \).
   - Using similar triangles, the length of \( MN \) can be found by:
   \[
   MN = \frac{51}{19} \sqrt{\left(\frac{2831}{51}\right)^2 - 19^2}
   \]
   \[
   MN = \frac{51}{19} \sqrt{\left(\frac{2831}{51}\right)^2 - 19^2}
   \]
   \[
   MN = \frac{51}{19} \sqrt{\left(\frac{2831}{51}\right)^2 - 19^2}
   \]
   \[
   MN = \sqrt{149^2 - 51^2}
   \]
   \[
   MN = \sqrt{22201 - 2601}
   \]
   \[
   MN = \sqrt{19600}
   \]
   \[
   MN = 140
   \]

The final answer is \( \boxed{140} \).