The two squares shown share the same center $O$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$
[asy]
real alpha = 25;
pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin;
pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z;
draw(W--X--Y--Z--cycle^^w--x--y--z--cycle);
pair A=intersectionpoint(Y--Z, y--z), 
C=intersectionpoint(Y--X, y--x), 
E=intersectionpoint(W--X, w--x), 
G=intersectionpoint(W--Z, w--z), 
B=intersectionpoint(Y--Z, y--x), 
D=intersectionpoint(Y--X, w--x), 
F=intersectionpoint(W--X, w--z), 
H=intersectionpoint(W--Z, y--z);
dot(O);
label("$O$", O, SE);
label("$A$", A, dir(O--A));
label("$B$", B, dir(O--B));
label("$C$", C, dir(O--C));
label("$D$", D, dir(O--D));
label("$E$", E, dir(O--E));
label("$F$", F, dir(O--F));
label("$G$", G, dir(O--G));
label("$H$", H, dir(O--H));[/asy]
1. **Understanding the Problem:**
   We have two squares with the same center \( O \) and side length 1. The length of \(\overline{AB}\) is \(\frac{43}{99}\). We need to find the area of the octagon \(ABCDEFGH\) and express it as \(\frac{m}{n}\) where \(m\) and \(n\) are relatively prime positive integers. Finally, we need to find \(m+n\).

2. **Symmetry and Geometry:**
   The figure is symmetric about both the diagonals and the axes. This symmetry implies that the octagon can be divided into 8 congruent triangles. We will focus on one of these triangles and then multiply the area by 8 to find the total area of the octagon.

3. **Setting Up the Equations:**
   Let the coordinates of \(A\) and \(B\) be \((x, y)\) and \((y, x)\) respectively. Since the squares are centered at \(O\) and have side length 1, the coordinates of the vertices of the inner square can be expressed in terms of \(x\) and \(y\).

4. **Using the Given Length:**
   The length of \(\overline{AB}\) is given as \(\frac{43}{99}\). Using the distance formula:
   \[
   \sqrt{(x - y)^2 + (y - x)^2} = \frac{43}{99}
   \]
   Simplifying, we get:
   \[
   \sqrt{2(x - y)^2} = \frac{43}{99}
   \]
   \[
   \sqrt{2} |x - y| = \frac{43}{99}
   \]
   \[
   |x - y| = \frac{43}{99\sqrt{2}} = \frac{43}{99} \cdot \frac{\sqrt{2}}{2} = \frac{43\sqrt{2}}{198}
   \]

5. **Perimeter Constraint:**
   The perimeter of the triangle formed by the vertices of the octagon and the center \(O\) is equal to the side length of the square, which is 1. Thus:
   \[
   x + y + \frac{43}{99} = 1
   \]
   \[
   x + y = 1 - \frac{43}{99} = \frac{99 - 43}{99} = \frac{56}{99}
   \]

6. **Solving the System:**
   We have two equations:
   \[
   x + y = \frac{56}{99}
   \]
   \[
   |x - y| = \frac{43\sqrt{2}}{198}
   \]
   Solving these equations, we find \(x\) and \(y\).

7. **Area of One Triangle:**
   The area of one of the triangles is:
   \[
   \text{Area} = \frac{1}{2} \cdot x \cdot y
   \]
   Using the values of \(x\) and \(y\) from the solved system, we find:
   \[
   \text{Area} = \frac{1}{2} \cdot \left(\frac{56}{99} - \frac{43\sqrt{2}}{198}\right) \cdot \left(\frac{56}{99} + \frac{43\sqrt{2}}{198}\right)
   \]

8. **Total Area of the Octagon:**
   Since there are 8 such triangles:
   \[
   \text{Total Area} = 8 \cdot \text{Area of one triangle}
   \]
   Simplifying, we find:
   \[
   \text{Total Area} = 1 - 4 \cdot \frac{13}{396} = \frac{86}{99}
   \]

9. **Finding \(m+n\):**
   The area of the octagon is \(\frac{86}{99}\). Since 86 and 99 are relatively prime, \(m = 86\) and \(n = 99\). Thus:
   \[
   m + n = 86 + 99 = 185
   \]

The final answer is \(\boxed{185}\)