## Task B-4.7.

In an equilateral triangle $A B C$, point $D$ is on side $\overline{A B}$, point $E$ is on side $\overline{B C}$, and point $F$ is on side $\overline{A C}$. Given that $|A D|=3$, $|C E|=2$, and $\varangle D F E=60^{\circ}$. If the area of triangle $D E F$ is $\frac{3}{2} \sqrt{3}$, determine the area of triangle $A B C$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_c317f9bfd22736a3ecebg-36.jpg?height=480&width=536&top_left_y=568&top_left_x=680)

Let $x$ be the side of triangle $ABC$ and the other sides as shown in the diagram. By considering the areas of the triangles in the diagram, we get the following sequence of equalities:

$P(ABC) = P(ADF) + P(DBE) + P(CFE) + P(DEF)$

$\frac{x^2 \sqrt{3}}{4} = \frac{1}{2} |AD| \cdot |AF| \sin 60^\circ + \frac{1}{2} |DB| \cdot |BE| \sin 60^\circ + \frac{1}{2} |CE| \cdot |CF| \sin 60^\circ + \frac{3}{2} \sqrt{3}$

$\frac{x^2 \sqrt{3}}{4} = \frac{3 \sqrt{3}}{4} y + (x-3)(x-2) \frac{\sqrt{3}}{4} + (x-y) \frac{\sqrt{3}}{2} + \frac{3}{2} \sqrt{3}$

$x^2 = 3y + (x-3)(x-2) + 2(x-y) + 6$

$x^2 = 3y + x^2 - 5x + 6 + 2x - 2y + 6$

$y = 3x - 12 (*)$

The angle $\angle DFC$ is an exterior angle of triangle $ADF$, so we have:

$\angle DFC = \angle DAF + \angle FDA = 60^\circ + \angle FDA$,

and since $\angle DFC = 60^\circ + \angle EFC$, it follows that $\angle EFC = \angle FDA$.

Thus, triangles $ADF$ and $CFE$ have two congruent angles and are similar by the AA similarity criterion. 1 point

Therefore, $\frac{3}{x-y} = \frac{y}{2}$, which implies $xy - y^2 = 6$.

Substituting $y$ from equation $(*)$ into this expression, we get:

$x(3x - y) - (3x - 12)^2 = 6$

$3x^2 - 12x - 9x^2 + 72x - 144 = 6$

$6x^2 - 60x + 150 = 0$

$x^2 - 10x + 25 = 0$.

The solution to this equation is $x = 5$.

Finally, $P(ABC) = \frac{25 \sqrt{3}}{4}$.

## Note:

If a student has written the cosine rule for each of the four triangles but has not derived a relationship between $x$ and $y$, they can receive only 1 point for all the written cosine rules. If they derived a correct relationship between $x$ and $y$ but did not recognize the similarity and did not calculate $x$ and $y$, they receive 3 points.