THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Let $O$ be the midpoint of $[A^{\prime }C]$.

a) Triangle $M{A}^{\prime }C$ is isosceles, hence $MO\perp A^{\prime }C$. Similarly $P{A}^{\prime }C$ is isosceles, therefore $PO\perp A^{\prime }C$, yielding $A^{\prime }C\perp (PMO)$, hence $A^{\prime }C\perp MP$.



b) Let $S$ be the midpoint of $[MP]$. The triangles $M{A}^{\prime }C$ and $P{A}^{\prime }C$ are congruent, yielding $[PO]=[MO]$. Then $OS\perp MP$, hence $OS$ is the distance between the lines $MP$ and $A^{\prime }C$. We have $MC=\frac{a\sqrt{5}}{2}$, therefore $MO=\frac{a\sqrt{2}}{2}$. Next, $MP=\frac{a\sqrt{6}}{2}$, and from the right triangle $OSM$, we obtain $OS=\frac{a\sqrt{2}}{4}$.