74th 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: $$PQ\parallel BC\iff \frac{AQ}{QB}=\frac{AP}{PC}\iff \frac{CE}{BC}=\frac{DE}{CD}.$$ 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 $$\frac{x+y+z}{z}=\frac{y+z}{y}\iff \frac{x+y}{z}=\frac{z}{y}\iff z^2=y^2+xy.$$ 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 $t=\phi$, hence the requirement of the problem.