A3. A square with a side length of 2 is divided into 4 triangles, of which 2 are isosceles (see the figure). The area of one of the isosceles triangles is twice the area of the other isosceles triangle. What is the area of the shaded triangle?
(A) $\frac{2}{3}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{8}{13}$
(D) $2-\sqrt{2}$
(E) $\frac{3-\sqrt{2}}{2}$
A3. We draw the diagonal of the square and mark some points on the square with $A, B, C, D, E$ and $F$ (see the figure).

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-29.jpg?height=289&width=297&top_left_y=1529&top_left_x=891)

Let $x=|B F|$. Since the area of triangle $A C E$ is twice the area of triangle $A B C$, we have $|E F|=2|B F|=2 x$. Triangle $B F C$ is a right-angled and isosceles triangle, so $|B F|=$ $|F C|=|F A|=x$. Therefore, the area of triangle $A B C$ is $\frac{2 x \cdot x}{2}=x^{2}$, and the area of triangle $A C E$ is $\frac{2 x \cdot 2 x}{2}=2 x^{2}$. Since $3 x=|B E|=2 \sqrt{2}$, we have $x=\frac{2 \sqrt{2}}{3}$, and the area of the shaded triangle is

$$
p=\frac{1}{2}\left(2^{2}-x^{2}-2 x^{2}\right)=\frac{1}{2}\left(4-\frac{8}{9}-\frac{16}{9}\right)=\frac{2}{3}
$$