Let $ABCD$ be a square with side length $100$. Denote by $M$ the midpoint of $AB$. Point $P$ is selected inside the square so that $MP = 50$ and $PC = 100$. Compute $AP^2$.

[i]Based on a proposal by Amogh Gaitonde[/i]
1. **Identify the given information and setup the problem:**
   - Square \(ABCD\) with side length \(100\).
   - \(M\) is the midpoint of \(AB\), so \(AM = MB = 50\).
   - Point \(P\) inside the square such that \(MP = 50\) and \(PC = 100\).

2. **Analyze the geometric properties:**
   - Since \(M\) is the midpoint of \(AB\), \(M\) is at \((50, 0)\) in a coordinate system where \(A\) is at \((0, 0)\) and \(B\) is at \((100, 0)\).
   - \(P\) is such that \(MP = 50\), meaning \(P\) lies on a circle centered at \(M\) with radius \(50\).
   - \(P\) is also such that \(PC = 100\), meaning \(P\) lies on a circle centered at \(C\) with radius \(100\).

3. **Use the cyclic quadrilateral property:**
   - Note that \(CB = CP\) because \(CB\) is a tangent to the circle with center \(M\) and radius \(MA\), so is \(CP\).
   - Hence, \(MBCP\) is cyclic, and \(\angle AMP = \angle BCP\).

4. **Establish similarity of triangles:**
   - Since \(\triangle AMP \sim \triangle BCP\), the sides are proportional.
   - Let \(AP = x\). Then, by similarity, \(BP = 2x\).

5. **Apply the Pythagorean theorem:**
   - In \(\triangle ABP\), we have:
     \[
     AP^2 + BP^2 = AB^2
     \]
   - Substituting \(BP = 2x\) and \(AB = 100\):
     \[
     x^2 + (2x)^2 = 100^2
     \]
     \[
     x^2 + 4x^2 = 10000
     \]
     \[
     5x^2 = 10000
     \]
     \[
     x^2 = \frac{10000}{5}
     \]
     \[
     x^2 = 2000
     \]

The final answer is \(\boxed{2000}\).