67th Romanian Mathematical Olympiad

If $\frac{D^{\prime }M}{D^{\prime }C}=\frac{DN}{D{A}^{\prime }}=\frac{1}{3}$, then $M$ and $N$ are the centroids of the triangles $D{D}^{\prime }C$ and, respectively, $AD{D}^{\prime }$, hence the points $C^{\prime }$, $M$, $P$ and $A$, $N$, $P$ are collinear,

where $P$ is the midpoint of the side $D{D}^{\prime }$. From the triangle $AP{C}^{\prime }$, $\frac{PM}{P{C}^{\prime }}=\frac{PN}{PA}$, hence $MN\parallel A{C}^{\prime }$. From $C{D}^{\prime }\perp (AD{C}^{\prime })$ and $D{A}^{\prime }\perp (A{D}^{\prime }{C}^{\prime })$ follows $A{C}^{\prime }\perp D^{\prime }C$ and $A{C}^{\prime }\perp A^{\prime }D$. This shows that $MN$ is the common perpendicular of the lines $C{D}^{\prime }$ and $D{A}^{\prime }$.



For the converse, if $MN$ is the common perpendicular of the two lines, due to the uniqueness of this line, it must coincide with the line from a), hence $\frac{D^{\prime }M}{D^{\prime }C}=\frac{DN}{D{A}^{\prime }}=\frac{1}{3}$.