\section*{Problem 2 - 021132}

Given a triangle \(A B C\). A parallel to side \(B C\) is drawn, intersecting sides \(A B\) and \(A C\) at points \(D\) and \(E\), respectively.

In what ratio does \(D\) divide side \(A B\), if the perimeters of triangles \(A D E\) and \(A B C\) are in the same ratio as the area of triangle \(A D E\) to the area of trapezoid \(D B C E\)?
}

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1692.jpg?height=323&width=351&top_left_y=1403&top_left_x=858)

Let \(k \equiv \frac{A D}{A B}\). Then, it also holds that \(k=\frac{A E}{A C}=\frac{D E}{B C}\), since \(A B C D E\) is a figure of the Thales' theorem due to \(D E \| B C\). With the usual abbreviations \(B C \equiv a, C A \equiv b\) and \(A B \equiv c\), it is given by the condition that

\[
\frac{D E+E A+A D}{a+b+c}=\frac{k a+k b+k c}{a+b+c}=k=\frac{[A D E]}{[D B C E]}
\]

where \([X Y Z]\) denotes the area of \(\triangle X Y Z\).

Since \([D B C E]=[A B C]-[A D E]\) and \(\triangle A D E\) is derived from \(\triangle A B C\) by a central dilation with factor \(k\), it follows that \([A D E]=k^{2}[A B C]\). This leads to the equation

\[
k=\frac{[A D E]}{[A B C]-[A D E]}=\frac{k^{2}}{1-k^{2}} \quad \Rightarrow \quad k^{2}+k-1=0
\]

The solution to this quadratic equation, given \(0<k<1\), is \(k=\frac{1}{2}(\sqrt{5}-1) \approx 0.618\), which is the ratio of the golden section. Point \(D\) thus divides the side \(A B\) in the ratio

\[
\frac{k}{1-k}=\frac{\frac{1}{2}(\sqrt{5}-1)}{1-\frac{1}{2}(\sqrt{5}-1)}=\frac{\sqrt{5}-1}{3-\sqrt{5}}=\frac{(\sqrt{5}-1)(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}=\frac{1}{2}(1+\sqrt{5})
\]

Adapted from \([2]\)