67th Romanian Mathematical Olympiad

a) In the right triangle $ABD$, $AM\perp BD$, so $AM=\sqrt{2}$, that is the triangle $A^{\prime }MA$ is isosceles. The same goes for the triangle $C^{\prime }CN$. It follows $\angle A^{\prime }MA=\angle C^{\prime }NC=45^{\circ }$. Since $CN\parallel AM$ and $A{A}^{\prime }\parallel C{C}^{\prime }$, the parallel from $A^{\prime }$ to $N{C}^{\prime }$ is coplanar with $AM$, hence the angle of the lines $A^{\prime }M$ and $C^{\prime }N$ has $180^{\circ }-45^{\circ }-45^{\circ }=90^{\circ }$.

b) Let $CM\cap AD=\{M^{\prime }\}$, $AN\cap BC=\{N^{\prime }\}$, $P$ and $Q$ be the midpoints of the segments $[N^{\prime }{C}^{\prime }]$, respectively $[A^{\prime }{M}^{\prime }]$ and $\{T\}=A^{\prime }C\cap A{C}^{\prime }$. Then $BD=3$, $DM=MN=NB=1$, $D{M}^{\prime }=\frac{\sqrt{3}}{2}$. $M^{\prime }$ and $N^{\prime }$ are the midpoints of the edges $[AD]$ respectively $[BC]$, hence $QT\parallel CM\parallel AN\parallel PT$, that is $P,T,Q$ are collinear.

Therefore, the intersection of the planes $(A^{\prime }MC)$ and $(AN{C}^{\prime })$ is line $PQ$. Let $O$ be the midpoint of the segment $[BD]$ and $R\in AN$, $S\in CM$ be the intersections of the perpendicular from $O$ on $AN$ (and on $CM$). The angle of the planes $(A^{\prime }MC)$ and $(AN{C}^{\prime })$ is $\angle RTS$. Now $C{M}^{\prime }=\frac{3\sqrt{3}}{2}$, $RS=\frac{\sqrt{6}}{3}$ (compute in two ways the area of $A{N}^{\prime }C{M}^{\prime }$), $OT=\frac{\sqrt{2}}{2}$ and it follows that the triangle $RTS$ is equilateral, hence the required measure is $60^{\circ }$.