In rectangle $ABCD$, point $M$ is the midpoint of $AB$ and $P$ is a point on side $BC$. The perpendicular bisector of $MP$ intersects side $DA$ at point $X$. Given that $AB = 33$ and $BC = 56$, find the least possible value of $MX$.

[i]Proposed by Michael Tang[/i]
1. **Identify the coordinates of the points:**
   - Let \( A = (0, 0) \), \( B = (33, 0) \), \( C = (33, 56) \), and \( D = (0, 56) \).
   - Since \( M \) is the midpoint of \( AB \), its coordinates are \( M = \left(\frac{33}{2}, 0\right) = \left(16.5, 0\right) \).
   - Let \( P \) be a point on \( BC \) with coordinates \( P = (33, y) \) where \( 0 \leq y \leq 56 \).

2. **Equation of the perpendicular bisector of \( MP \):**
   - The midpoint of \( MP \) is \( \left(\frac{16.5 + 33}{2}, \frac{0 + y}{2}\right) = \left(\frac{49.5}{2}, \frac{y}{2}\right) = \left(24.75, \frac{y}{2}\right) \).
   - The slope of \( MP \) is \( \frac{y - 0}{33 - 16.5} = \frac{y}{16.5} \).
   - The slope of the perpendicular bisector is \( -\frac{16.5}{y} \).

3. **Equation of the perpendicular bisector:**
   - Using the point-slope form of the line equation, the perpendicular bisector is:
     \[
     y - \frac{y}{2} = -\frac{16.5}{y} (x - 24.75)
     \]
     Simplifying, we get:
     \[
     y - \frac{y}{2} = -\frac{16.5}{y} x + \frac{16.5 \cdot 24.75}{y}
     \]
     \[
     \frac{y}{2} = -\frac{16.5}{y} x + \frac{16.5 \cdot 24.75}{y}
     \]
     \[
     y = -\frac{33}{y} x + \frac{33 \cdot 24.75}{y}
     \]

4. **Intersection with \( DA \):**
   - \( DA \) is the line \( x = 0 \).
   - Substituting \( x = 0 \) into the perpendicular bisector equation:
     \[
     y = \frac{33 \cdot 24.75}{y}
     \]
     \[
     y^2 = 33 \cdot 24.75
     \]
     \[
     y^2 = 816.75
     \]
     \[
     y = \sqrt{816.75}
     \]

5. **Calculate \( MX \):**
   - Since \( X \) is on the perpendicular bisector, \( MX = PX \).
   - The minimum value of \( PX \) occurs when \( P \) is at the midpoint of \( BC \), i.e., \( y = 28 \).
   - Thus, \( MX = PX = 33 \).

The final answer is \( \boxed{33} \).