\section*{Problem 1 - 081031}

In a triangle \(\triangle A B C\), let \(A B=18 \mathrm{~cm} . \mathrm{To}\) this side, a parallel line is drawn inside the triangle, forming a trapezoid \(A B D E\) whose area \(F_{2}\) is one third of the area \(F_{1}\) of the triangle \(\triangle A B C\).

Calculate the length of the side \(D E\) of the trapezoid!
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The area \(F_{3}\) of the triangle \(\triangle C D E\) is obtained as the difference between the areas of the triangle \(\triangle A B C\) and the trapezoid \(A B D E\), that is, \(F_{3}=F_{1}-F_{2}=\frac{2}{3} \cdot F_{1}\).

Since the segments \(D E\) and \(A B\) are parallel, and \(E\) and \(A\) as well as \(D\) and \(B\) lie on a ray emanating from \(C\), the triangle \(\triangle C D E\) is derived from the triangle \(\triangle A B C\) through a central dilation with center \(C\). Let the dilation factor be \(k\). Then \(F_{3}=k^{2} \cdot F_{1}\), so \(k=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}\), and thus \(|D E|=k \cdot|A B|=\sqrt{6} \cdot 3 \text{ cm}\).