Winter Mathematical Competition

Let $P$ and $Q$ be the centers of the faces $BC{C}_1{B}_1$ and $DC{C}_1{D}_1$ and let $\alpha =(APQ)$ (Fig. 1). Since $PQ\parallel BD$ the plane $\alpha$ meets the plane $(ABCD)$ at the line through $A$ which is parallel to $BD$. We denote by $T$ and $S$ the intersection points of this line with the lines $CB$ and $CD$, respectively.

The lines $TP$ and $SQ$ meet the edge $C{C}_1$ at the intersection point $R$ of $\alpha$ and $C{C}_1$. Let $M=TR\cap B{B}_1$ and $N=SR\cap D{D}_1$. Then the intersection of $\alpha$ and the surface of the cube is the quadrilateral $AMRN$. (It is easy to see that it is a rhombus.) It is clear that $BT=BA=1$. Hence $B$ and $M$ are the midpoints of $TC$ and $TR$, respectively. Then $\frac{BM}{RC}=\frac{1}{2}$ and since $BM=R{C}_1=1-RC$, we find $BM=\frac{1}{3}$. Analogously $DN=\frac{1}{3}$.

Denote by $V$ the volume of the polytope cut from the cube by the planes $(ABCD)$ and $(AMRN)$ (Fig. 2). Let $A_2$ and $C_2$ be the intersection points of $A{A}_1$ and $C{C}_1$ with the plane through $MN$ and parallel to $(ABCD)$. Then $R{C}_2=RC-C{C}_2=MB=A{A}_2=\frac{1}{3}$ and hence the tetrahedra $NM{C}_{2}R$ and $NM{A}_{2}A$ have equal volumes. This shows that $V=V_{ABCD{A}_{2}M{C}_{2}N}=\frac{1}{3}$. Hence the required ratio is $1:2$.

Fig. 1

Fig. 2