In a square $ABCD$ of diagonals $AC$ and $BD$, we call $O$ at the center of the square. A square $PQRS$ is constructed with sides parallel to those of $ABCD$ with $P$ in segment $AO, Q$ in segment $BO, R$ in segment $CO, S$ in segment $DO$.  If area of $ABCD$ equals two times the  area of $PQRS$,  and $M$ is the midpoint of the $AB$ side, calculate the measure of the angle $\angle AMP$. 
1. **Define the coordinates of the square \(ABCD\):**
   - Let \(A(0,0)\), \(B(1,0)\), \(C(1,1)\), and \(D(0,1)\).
   - The center \(O\) of the square \(ABCD\) is at \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

2. **Define the coordinates of the square \(PQRS\):**
   - Since \(PQRS\) is a smaller square with sides parallel to \(ABCD\) and centered at \(O\), and given that \(P\) is on segment \(AO\), \(Q\) on segment \(BO\), \(R\) on segment \(CO\), and \(S\) on segment \(DO\), we can denote the coordinates of \(P, Q, R, S\) as follows:
     - \(P(\lambda, \lambda)\)
     - \(Q(1-\lambda, \lambda)\)
     - \(R(1-\lambda, 1-\lambda)\)
     - \(S(\lambda, 1-\lambda)\)

3. **Calculate the side length of \(PQRS\):**
   - The side length of \(PQRS\) is \(PQ = Q(1-\lambda, \lambda) - P(\lambda, \lambda) = (1-2\lambda)\).

4. **Relate the areas of \(ABCD\) and \(PQRS\):**
   - The area of \(ABCD\) is \(1\) (since it is a unit square).
   - The area of \(PQRS\) is \((1-2\lambda)^2\).
   - Given that the area of \(ABCD\) is twice the area of \(PQRS\):
     \[
     1 = 2(1-2\lambda)^2
     \]
   - Solving for \(\lambda\):
     \[
     1 = 2(1-2\lambda)^2
     \]
     \[
     \frac{1}{2} = (1-2\lambda)^2
     \]
     \[
     1-2\lambda = \pm \frac{1}{\sqrt{2}}
     \]
     \[
     1-2\lambda = \pm \frac{\sqrt{2}}{2}
     \]
     \[
     2\lambda = 1 \mp \frac{\sqrt{2}}{2}
     \]
     \[
     \lambda = \frac{1}{2} \mp \frac{\sqrt{2}}{4}
     \]
   - Since \(\lambda\) must be between 0 and \(\frac{1}{2}\), we choose:
     \[
     \lambda = \frac{1}{2} - \frac{\sqrt{2}}{4}
     \]

5. **Calculate the angle \(\angle AMP\):**
   - The coordinates of \(M\) (the midpoint of \(AB\)) are \(\left(\frac{1}{2}, 0\right)\).
   - The coordinates of \(P\) are \(\left(\frac{1}{2} - \frac{\sqrt{2}}{4}, \frac{1}{2} - \frac{\sqrt{2}}{4}\right)\).
   - The slope of \(AM\) is:
     \[
     \text{slope of } AM = \frac{0 - 0}{\frac{1}{2} - 0} = 0
     \]
   - The slope of \(MP\) is:
     \[
     \text{slope of } MP = \frac{\left(\frac{1}{2} - \frac{\sqrt{2}}{4}\right) - 0}{\left(\frac{1}{2} - \frac{\sqrt{2}}{4}\right) - \frac{1}{2}} = \frac{\frac{1}{2} - \frac{\sqrt{2}}{4}}{\frac{1}{2} - \frac{\sqrt{2}}{4} - \frac{1}{2}} = \frac{\frac{1}{2} - \frac{\sqrt{2}}{4}}{-\frac{\sqrt{2}}{4}} = -\frac{2 - \sqrt{2}}{\sqrt{2}}
     \]
   - The angle \(\theta\) between \(AM\) and \(MP\) is given by:
     \[
     \tan(\theta) = \left| \frac{\text{slope of } MP - \text{slope of } AM}{1 + (\text{slope of } MP)(\text{slope of } AM)} \right| = \left| \frac{-\frac{2 - \sqrt{2}}{\sqrt{2}} - 0}{1 + 0} \right| = \left| -\frac{2 - \sqrt{2}}{\sqrt{2}} \right|
     \]
     \[
     \theta = \tan^{-1}\left(\frac{2 - \sqrt{2}}{\sqrt{2}}\right)
     \]
   - Simplifying further, we find that:
     \[
     \theta = \frac{45^\circ}{2}
     \]

The final answer is \(\boxed{\frac{45^\circ}{2}}\).