To solve this problem, let's start by analyzing the information given:

Step 1: Understanding the problem

We have a quadrilateral $PQRS$ inscribed in a circle with center $O$. The length $PQ = \sqrt{4 + 2\sqrt{2}}$, and the angle $\angle POQ = 135^\circ$. We need to find the maximum possible area of $PQRS$.

The area of a cyclic quadrilateral (a quadrilateral inscribed in a circle) can be calculated using Brahmagupta's formula, but we need to express it in a form involving the given information.

Step 2: Analyzing the given information

Given $\angle POQ = 135^\circ$, we need to understand the relationship between the sides and the angles.

The length $PQ = \sqrt{4 + 2\sqrt{2}}$ suggests we can rewrite this using trigonometric identities.

Let $x = \sqrt{4 + 2\sqrt{2}}$. We can express this as: $x^2 = 4 + 2\sqrt{2}$ This expression is related to cosine values, possibly involving the double angle formulas, but let's move forward assuming this is accurate and continue.

Step 3: Maximum Area

For a cyclic quadrilateral, the maximum area occurs when the quadrilateral becomes a rectangle. In this case, $PQRS$ could become a rectangle when $\angle POQ = 135^\circ$, corresponding to the two diagonals of the rectangle intersecting at an angle of $45^\circ$ (since diagonals of a rectangle bisect each other and are equal, forming an isosceles triangle).

Given $PQ = \sqrt{4 + 2\sqrt{2}}$, let's focus on the possible configurations that maximize the area.

Step 4: Simplify the formula for area

For a rectangle inscribed in a circle, the area is maximized when the diagonals are perpendicular.

$\text{Area} = \frac{1}{2} \times \text{Diagonal 1} \times \text{Diagonal 2}$

But for a maximum area scenario involving a cyclic quadrilateral, the configuration where opposite angles sum to $180^\circ$ (supplementary) gives us the optimal setup.

Using the cyclic property and given the maximum area:

$\text{Area}_{\text{max}} = \text{length of diagonal 1} \times \text{length of diagonal 2} \times \frac{1}{2}$

Since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$,

$\text{Area}_{\text{max}} = \frac{x^2}{2} = \frac{4 + 2\sqrt{2}}{2} = 2 + \sqrt{2}$

Step 5: Final Calculation

This expression must be of the form $\frac{a + \sqrt{b}}{c}$ where $\gcd(a, b, c) = 1$.

Here, $a = 2$, $b = 2$, and $c = 1$.

Thus, $a + b + c = 2 + 2 + 1 = 5$.

Final Answer: $a + b + c = 5$

Do you have any questions or want more details?

Here are 5 related questions:

1. How can Brahmagupta's formula be used for any cyclic quadrilateral?
2. What are the properties of cyclic quadrilaterals in a circle?
3. Can you calculate the area of a quadrilateral with a different angle between the diagonals?
4. What are the special cases of quadrilaterals inscribed in a circle?
5. How does the relationship between the diagonals of a rectangle relate to the circle's radius?

Tip: The diagonals of a rectangle inscribed in a circle always intersect at a right angle, simplifying the maximum area calculations.