75th Romanian Mathematical Olympiad

A point $M\in (AB)$ is uniquely determined by a real number $k\in (0,\infty )$ such that $\overline{AM}=k$, from which we get $$\overrightarrow{OM}=\frac{1}{k+1}\overrightarrow{OA}+\frac{k}{k+1}\overrightarrow{OB}.$$ To find the points $N$ and $P$ uniquely, we need to find $x,y\in (0,\infty )$ such that $\overrightarrow{ON}=x,\overrightarrow{OP}=y$, and $O$ is the centroid of triangle $MNP$. From this, we have: $$\overrightarrow{ON}=\frac{x}{x+1}\overrightarrow{OC}=-\frac{x}{x+1}\overrightarrow{OA}\text{and}\overrightarrow{OP}=\frac{y}{y+1}\overrightarrow{OD}=-\frac{y}{y+1}\overrightarrow{OB}.$$ Since $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are not collinear, $O$ is the centroid of triangle $MNP$ if and only if $\frac{x}{x+1}=\frac{1}{k+1}\iff x=\frac{1}{k}$ and $\frac{y}{y+1}=\frac{k}{k+1}\iff y=k$, meaning that point $N$ is uniquely determined by the ratio $x=\frac{1}{k}=\frac{ON}{NC}$, and point $P$ is uniquely determined by the ratio $y=k=\frac{OP}{PD}$, which completes the problem.