Slovenija 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 $DEF$: $$\frac{4^2\cdot \sqrt{3}}{4}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}-\frac{3^2\cdot \sqrt{3}}{4}=\frac{3\sqrt{3}}{4}$$

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Alternative solution.

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 $$|DG|\cdot |FG|=\frac{3\sqrt{3}}{4}.$$

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Alternative solution.

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, $$p_{DEF}=2p_{DFG}=2\cdot \frac{1{\sqrt{3}}^2\cdot \sqrt{3}}{4}=\frac{3\sqrt{3}}{4}.$$