A square plate has side length $L$ and negligible thickness. It is laid down horizontally on a table and is then rotating about the axis $\overline{MN}$ where $M$ and $N$ are the midpoints of two adjacent sides of the square. The moment of inertia of the plate about this axis is $kmL^2$, where $m$ is the mass of the plate and $k$ is a real constant. Find $k$. 

[color=red]Diagram will be added to this post very soon. If you want to look at it temporarily, see the PDF.[/color]

[i]Problem proposed by Ahaan Rungta[/i]
1. **Moment of Inertia about the Diagonal:**
   - First, we need to find the moment of inertia of the square plate about its diagonal. By the Perpendicular Axis Theorem, the moment of inertia of a planar object about an axis perpendicular to the plane (let's call it the z-axis) is the sum of the moments of inertia about two perpendicular axes (x and y) lying in the plane and intersecting at a point.
   - For a square plate of side length \( L \) and mass \( m \), the moment of inertia about an axis through its center and parallel to one of its sides is \( \frac{1}{6}mL^2 \).
   - Since the square is symmetric, the moment of inertia about the other axis through the center and parallel to the other side is also \( \frac{1}{6}mL^2 \).
   - Therefore, the moment of inertia about the z-axis (perpendicular to the plane and passing through the center) is:
     \[
     I_z = I_x + I_y = \frac{1}{6}mL^2 + \frac{1}{6}mL^2 = \frac{1}{3}mL^2
     \]
   - The moment of inertia about the diagonal (which is in the plane) is half of \( I_z \) due to symmetry:
     \[
     I_{\text{diagonal}} = \frac{1}{2}I_z = \frac{1}{2} \times \frac{1}{3}mL^2 = \frac{1}{6}mL^2
     \]

2. **Using the Parallel Axis Theorem:**
   - The axis \(\overline{MN}\) is parallel to the diagonal and passes through the midpoints of two adjacent sides of the square. The distance from the center of mass (which is at the intersection of the diagonals) to the axis \(\overline{MN}\) is \( \frac{L}{2} \).
   - By the Parallel Axis Theorem, the moment of inertia about an axis parallel to one passing through the center of mass is given by:
     \[
     I = I_{\text{cm}} + md^2
     \]
     where \( I_{\text{cm}} \) is the moment of inertia about the center of mass and \( d \) is the distance from the center of mass to the new axis.
   - Here, \( I_{\text{cm}} = I_{\text{diagonal}} = \frac{1}{6}mL^2 \) and \( d = \frac{L}{2} \):
     \[
     I_{\overline{MN}} = \frac{1}{6}mL^2 + m \left( \frac{L}{2} \right)^2 = \frac{1}{6}mL^2 + m \frac{L^2}{4} = \frac{1}{6}mL^2 + \frac{1}{4}mL^2
     \]
   - Simplifying the expression:
     \[
     I_{\overline{MN}} = \frac{1}{6}mL^2 + \frac{1}{4}mL^2 = \frac{2}{12}mL^2 + \frac{3}{12}mL^2 = \frac{5}{12}mL^2
     \]

3. **Conclusion:**
   - The moment of inertia of the plate about the axis \(\overline{MN}\) is \( \frac{5}{12}mL^2 \).
   - Therefore, the constant \( k \) is \( \frac{5}{12} \).

The final answer is \( \boxed{ k = \frac{5}{12} } \)