Romanian Mathematical Olympiad

a) Applying Menelaus' theorem in the triangle $ABE$ with transversal $Q - R - C$, we get that $\frac{AQ}{QB} \cdot \frac{BC}{CE} \cdot \frac{ER}{RA} = 1$, therefore $\frac{AQ}{QB} = \frac{CE}{BC} \cdot \frac{RA}{ER}$. Similarly, applying Menelaus' theorem in the triangle $AEC$ with transversal $P - R - D$, we get that $\frac{AP}{PC} = \frac{DE}{CD} \cdot \frac{RA}{ER}$. We have:
$$PQ \parallel BC \Leftrightarrow \frac{AQ}{QB} = \frac{AP}{PC} \Leftrightarrow \frac{CE}{BC} = \frac{DE}{CD}.$$
This relation does not depend on the position of point $R$ on segment ($AE$), but only on the positions of points $D$ and $E$ on segment ($BC$). It follows that, if there is a remarkable point on the segment ($AE$), then any point of the segment ($AE$) is remarkable.

b) Let's denote by $x, y$ and $z$ the lengths of the segments $BD, DE$ and $EC$, respectively. According to the above, there is a remarkable point on the segment ($AE$) if and only if
$$\frac{x+y+z}{z} = \frac{y+z}{y} \Leftrightarrow \frac{x+y}{z} = \frac{z}{y} \Leftrightarrow z^2 = y^2 + xy.$$
In the same way, there is a remarkable point on the segment ($AD$) if and only if $x^2 = y^2 + yz$.
Subtracting these equalities, it turns out that $z^2 - x^2 = y(x - z)$. In order not to have contrary signs in the two members, the condition yields $x = z$. We deduce that $x^2 - xy - y^2 = 0$, or $t^2 - t - 1 = 0$, where $t = \frac{x}{y} > 0$.
The only positive solution of this equation is $t = \varphi$, hence the requirement of the problem.