Two equilateral triangles are contained in a square whose side length is $2\sqrt{3}$. The bases of these triangles are opposite sides of the square, and their intersection is a rhombus. The area of the rhombus can be expressed in the from $a+b\sqrt{c}$ for integers $a,b,$ and $c$ such that $c$ square-free. Determine $-a+b+c$.

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?

---

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.

Two equilateral triangles are contained in a square whose side length is $2\sqrt{3}$. The bases of these triangles are opposite sides of the square, and their intersection is a rhombus. The area of the rhombus can be expressed in the form $a + b\sqrt{c}$ for integers $a, b,$ and $c$ such that $c$ is square-free. Determine $-a + b + c$.

Solve a geometry problem involving two equilateral triangles inscribed in a square of side length 2√3. Learn how to calculate the area of the rhombus formed by the triangles' overlap using diagonal properties and equilateral triangle height. Suitable for high school students in Grades 10-12.

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