If a rectangle is inscribed in an equilateral triangle of side length $$2\sqrt{2}$$ as shown in the figure, then the square of the largest area of such a rectangle is _________.

Given:

Side of equilateral triangle,

$$s=2\sqrt2$$

Height of the triangle:

$$H=\frac{\sqrt3}{2}s$$

$$=\frac{\sqrt3}{2}(2\sqrt2)$$

$$=\sqrt6$$

Base of triangle:

$$B=2\sqrt2$$

Let the inscribed rectangle have:

- width $$w$$

- height $$h$$

Using similar triangles,

$$\frac{w}{B}=\frac{H-h}{H}$$

Hence,

$$w=B\left(1-\frac{h}{H}\right)$$

Area of rectangle:

$$A=wh$$

$$=Bh\left(1-\frac{h}{H}\right)$$

$$=B\left(h-\frac{h^2}{H}\right)$$

Differentiate with respect to $$h$$ :

$$\frac{dA}{dh}=B\left(1-\frac{2h}{H}\right)$$

For maximum area,

$$\frac{dA}{dh}=0$$

$$1-\frac{2h}{H}=0$$

$$h=\frac{H}{2}$$

Now,

$$w=B\left(1-\frac12\right)=\frac{B}{2}$$

Therefore maximum area is

$$A_{\max}=\frac{B}{2}\cdot\frac{H}{2}$$

$$=\frac{BH}{4}$$

But,

$$\frac12 BH$$

is the area of the triangle.

Hence,

$$A_{\max}=\frac12(\text{Area of triangle})$$

Area of equilateral triangle:

$$=\frac{\sqrt3}{4}s^2$$

$$=\frac{\sqrt3}{4}(2\sqrt2)^2$$

$$=\frac{\sqrt3}{4}\cdot8$$

$$=2\sqrt3$$

Thus,

$$A_{\max}=\frac{2\sqrt3}{2}$$

$$=\sqrt3$$

Therefore,

$$(A_{\max})^2=(\sqrt3)^2$$

$$=3$$

Hence, the required answer is

$$\boxed{3}$$