A3. Let $A, B, C$ and $D$ be the vertices of the square, and let $E$ and $F$ be the vertices of the white triangle (see the figure).

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-25.jpg?height=291&width=303&top_left_y=1936&top_left_x=888)

Let $x = |A E|$. Since the right triangles $A E D$ and $F C D$ have one side equal to the side of the square and have the same area, $|C F| = |A E| = x$. If we consider that the triangles $A E D$ and $E B F$ also have the same area, we get $\frac{2 \cdot x}{2} = \frac{(2 - x) \cdot (2 - x)}{2}$. We rearrange the equation to $x^{2} - 6x + 4 = 0$ and solve it as a quadratic equation to get $x = 3 \pm \sqrt{5}$. Since $x < 2$, the correct solution is $x = 3 - \sqrt{5}$. The area of the white triangle is therefore

$$
p = 2^{2} - 3 \cdot \frac{2 \cdot (3 - \sqrt{5})}{2} = 3 \sqrt{5} - 5
$$
A3. A square with a side length of 2 is divided into 4 triangles (see the figure). All 3 shaded triangles have the same area. What is the area of the white triangle?
(A) $\frac{1+\sqrt{5}}{2}$
(B) $\frac{8}{5}$
(C) 2
(D) $3 \sqrt{5}-5$
(E) $6-2 \sqrt{5}$

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-10.jpg?height=209&width=251&top_left_y=1706&top_left_x=1656)