Let's analyze the problem systematically.

1. Square Dimensions and Triangle Heights:

   • We have a square with side length $2\sqrt{3}$.
   • Since the triangles are equilateral, the altitude (height) of each triangle can be calculated from the side length of the square, which serves as the side length of each triangle.
   • The altitude $h$ of an equilateral triangle with side length $s$ is given by: $h = \frac{\sqrt{3}}{2} s$
   • Here, $s = 2\sqrt{3}$, so: $h = \frac{\sqrt{3}}{2} \cdot 2\sqrt{3} = 3$
   • Therefore, the height of each equilateral triangle is $3$, and since the square has a side length of $2\sqrt{3}$, the two triangles overlap partially, forming a rhombus.

2. Position of the Rhombus:

   • The bases of the triangles lie on opposite sides of the square.
   • Since each triangle has a height of $3$, their top vertices meet halfway along the height of the square (which is $2\sqrt{3}$), intersecting at $1.5$ units from each base of the triangles.

3. Side Length of the Rhombus:

   • The intersection of the triangles forms a rhombus whose vertices are located at the midpoints of each triangle’s height.
   • The side length of the rhombus is therefore equal to the distance from the center of the square to a vertex of one of the triangles.

4. Calculating the Area of the Rhombus:

   • The area of the rhombus can be computed by finding the length of its diagonals. Here, one diagonal is equal to the side length of the square, $2\sqrt{3}$, and the other diagonal is twice the height from the base to the intersection point of the triangles, which is also $2\sqrt{3}$.
   • The area $A$ of a rhombus with diagonals $d_1$ and $d_2$ is: $A = \frac{1}{2} \times d_1 \times d_2$
   • Substituting $d_1 = 2\sqrt{3}$ and $d_2 = 2\sqrt{3}$: $A = \frac{1}{2} \times 2\sqrt{3} \times 2\sqrt{3} = \frac{1}{2} \times 12 = 6$

5. Expressing the Area in the Form $a + b\sqrt{c}$:

   • The area of the rhombus is $6$, which can be written as $6 + 0\sqrt{3}$.
   • Thus, $a = 6$, $b = 0$, and $c = 3$.

6. Calculating $-a + b + c$:

   • We need to find $-a + b + c$: $-a + b + c = -6 + 0 + 3 = -3$

Final Answer:

$\boxed{-3}$

Would you like further details on any part of the solution, or have questions?

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1. How is the altitude of an equilateral triangle derived?
2. Why does the intersection form a rhombus?
3. What is the significance of using diagonals for calculating a rhombus's area?
4. How does the placement of the triangles affect the geometry of the intersection?
5. Can the problem be extended to other polygons inscribed in a square?

Tip: Remember that the area of a rhombus is always half the product of its diagonals, which can simplify calculations in problems like these.