Slovenian Mathematical Olympiad, 2008

We have $|DA| = |DB| = \frac{|AB|}{2} = 2$. The angles of the triangle $DBE$ measure $60^\circ$, $90^\circ$ and $30^\circ$, so $DBE$ is one half of an equilateral triangle. This implies $|BE| = \frac{|BD|}{2} = 1$. A similar argument for the triangle $ADF$ shows that this triangle is congruent to the triangle $BDE$. The lengths of the sides $DE$ and $DF$ are $\frac{\sqrt{3}}{2} \cdot |BD| = \sqrt{3}$, so the area of each of the two triangles is $\frac{\sqrt{3}}{2}$.

In the triangle $ECF$ the lengths of the sides are $|CF| = |CA| - |AF| = 3$ and $|CE| = |CB| - |EB| = 3$, so this is an isosceles triangle with the angle of $60^\circ$ at the apex. The triangle $ECF$ is therefore equilateral. Its area is $\frac{3^2 \cdot \sqrt{3}}{4}$.

By subtracting the areas of triangles $DBE$, $ADF$ and $ECF$ from the area of the triangle $ABC$ we can obtain the area of the triangle $DEF$:
$$\frac{4^2 \cdot \sqrt{3}}{4} - \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} - \frac{3^2 \cdot \sqrt{3}}{4} = \frac{3\sqrt{3}}{4}$$

We see that $|DA| = |DB| = \frac{|AB|}{2} = 2$. The angles of triangles $AFD$ and $BED$ are equal to $60^\circ$, $90^\circ$ and $30^\circ$ and these two triangles agree in the lengths of the hypotenuses, so they are congruent. This implies $|DE| = |DF|$, so $DEF$ is an isosceles triangle. Let $G$ be the intersection of segments $CD$ and $EF$. Since $\angle FDG = \angle GDE = 60^\circ$, $DG$ is the altitude in the triangle $DEF$. Triangles $DGE$ and $DGF$ are congruent and their angles are $60^\circ$, $90^\circ$ and $30^\circ$.

Triangles $ADC$, $AFD$ and $DGF$ are similar. From the first two we get $\frac{|AD|}{|AC|} = \frac{|AF|}{|AD|}$, so $|AF| = 1$. Similarly, $\frac{|AD|}{|DC|} = \frac{|AF|}{|FD|}$, implies $|FD| = \sqrt{3}$.

Since $ADC$ and $DGF$ are similar, we have $\frac{|AD|}{|AC|} = \frac{|DG|}{|DF|}$, so $|DG| = \frac{\sqrt{3}}{2}$. Likewise, $\frac{|AD|}{|DC|} = \frac{|DG|}{|GF|}$ implies $|GF| = \frac{3}{2}$.

The area of the triangle $DEF$ is equal to the sum of the areas of triangles $FGD$ and $DGE$, so
$$|DG| \cdot |FG| = \frac{3\sqrt{3}}{4}.$$

Obviously, $|DA| = |DB| = \frac{|AB|}{2} = 2$. Triangles $AFD$ and $BED$ are congruent since they both have angles equal to $60^\circ$, $90^\circ$ and $30^\circ$ and the lengths of the two hypotenuses are also equal. So, $|DE| = |DF| = \sqrt{3}$. Let $G$ be the midpoint of the segment $EF$. Since $EF$ is parallel to $AB$, right triangles $FGD$ and $EGD$ are congruent and $\angle DFG = \angle DEG = 30^\circ$. Thus,
$$p_{DEF} = 2p_{DFG} = 2 \cdot \frac{1\sqrt{3}^2 \cdot \sqrt{3}}{4} = \frac{3\sqrt{3}}{4}.$$