A square $ABCD$ has side length $ 1$. A circle passes through the vertices of the square. Let $P, Q, R, S$ be the midpoints of the arcs which are symmetrical to the arcs $AB$, $BC$, $CD$, $DA$ when reflected on sides $AB$, $B$C, $CD$, $DA$, respectively. The area of square $PQRS$ is  $a+b\sqrt2$, where $a$ and $ b$ are integers. Find the value of $a+b$.
[img]https://cdn.artofproblemsolving.com/attachments/4/3/fc9e1bd71b26cfd9ff076db7aa0a396ae64e72.png[/img]
1. **Identify the problem and given data:**
   - We have a square \(ABCD\) with side length \(1\).
   - A circle passes through the vertices of the square.
   - \(P, Q, R, S\) are the midpoints of the arcs symmetrical to the arcs \(AB\), \(BC\), \(CD\), \(DA\) when reflected on sides \(AB\), \(BC\), \(CD\), \(DA\), respectively.
   - We need to find the area of the square \(PQRS\) in the form \(a + b\sqrt{2}\) and determine \(a + b\).

2. **Determine the radius of the circle:**
   - Since the circle passes through the vertices of the square, the diameter of the circle is the diagonal of the square.
   - The diagonal of the square \(ABCD\) with side length \(1\) is \(\sqrt{2}\).
   - Therefore, the radius \(R\) of the circle is \(\frac{\sqrt{2}}{2}\).

3. **Find the coordinates of the midpoints of the arcs:**
   - The center of the circle is the center of the square, which is \(\left(\frac{1}{2}, \frac{1}{2}\right)\).
   - The midpoints of the arcs \(AB\), \(BC\), \(CD\), and \(DA\) are the points on the circle that are equidistant from the endpoints of these arcs.

4. **Reflect the midpoints on the sides of the square:**
   - Let \(M\) and \(N\) be the midpoints of the arcs \(\overarc{AB}\) and \(\overarc{CD}\), respectively.
   - The line \(\overline{MN}\) intersects \(\overline{AB}\) at \(K\).
   - Since \(\overline{MN}\) and \(\overline{AC}\) are diameters of the circle, their length is \(\sqrt{2}\).

5. **Calculate the length \(MK\):**
   - Let \(MK = x\).
   - Using the property of the circle, we have:
     \[
     x(\sqrt{2} - x) = \left(\frac{1}{2}\right)^2
     \]
   - Solving for \(x\):
     \[
     x(\sqrt{2} - x) = \frac{1}{4}
     \]
     \[
     x\sqrt{2} - x^2 = \frac{1}{4}
     \]
     \[
     x^2 - x\sqrt{2} + \frac{1}{4} = 0
     \]
     - Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
     \[
     x = \frac{\sqrt{2} \pm \sqrt{2 - 1}}{2} = \frac{\sqrt{2} \pm 1}{2}
     \]
     - Since \(x\) must be positive:
     \[
     x = \frac{\sqrt{2} - 1}{2}
     \]

6. **Calculate the length \(PR\):**
   - The length \(PR\) is given by:
     \[
     PR = \sqrt{2} - 4x = \sqrt{2} - 4 \left(\frac{\sqrt{2} - 1}{2}\right) = \sqrt{2} - 2(\sqrt{2} - 1) = 2 - \sqrt{2}
     \]

7. **Calculate the area of the square \(PQRS\):**
   - The area of the square \(PQRS\) is:
     \[
     \text{Area} = \left(\frac{PR}{\sqrt{2}}\right)^2 = \left(\frac{2 - \sqrt{2}}{\sqrt{2}}\right)^2 = \left(\frac{2\sqrt{2} - 2}{2}\right)^2 = (2 - \sqrt{2})^2 = 4 - 4\sqrt{2} + 2 = 6 - 4\sqrt{2}
     \]
   - Therefore, the area of the square \(PQRS\) is \(3 - 2\sqrt{2}\).

8. **Identify \(a\) and \(b\):**
   - Comparing \(3 - 2\sqrt{2}\) with \(a + b\sqrt{2}\), we have \(a = 3\) and \(b = -2\).

The final answer is \( \boxed{ a + b = 3 - 2 = 1 } \).