Brazilian Math Olympiad

Let $ABCD$ and $EFGH$ be two opposite faces, $AE$, $BF$, $CG$ and $DH$ being edges of the cube, and let $X'$ be the orthogonal projection of point $X$ onto the plane. Notice that $\{A, G\}$, $\{B, H\}$, $\{C, E\}$ and $\{D, F\}$ are pairs of opposite vertices. Suppose, without loss of generality, that $A'$ lies on the boundary of the projection of the cube. Then, considering the symmetry of the cube around the center of the cube, its symmetric point $G'$ lies on the boundary as well. Two of the three neighboring vertices of $A$ are going to be neighbors of $A'$ in the projection (unless, say, face AEHD projects onto a line; but in this case we consider a degenerate vertex inside this line). Suppose without loss that these neighbors are $B'$ and $D'$. So $E'$ is inside the projection. Again by symmetry $H'$ and $F'$ lie on the boundary of the projection and $C'$ lies inside the projection. Finally, since $\overrightarrow{AE} = \overrightarrow{BF} = \overrightarrow{CG} = \overrightarrow{DH}$, the projection of the cube is $A'D'H'G'F'B'$.

The faces $ABCD$, $BCGF$ and $CDHG$ project onto the parallelograms (or line segments) $A'B'C'D$, $B'C'G'F$ and $C'D'H'G'$. Draw diagonals $B'D'$, $B'G'$ and $D'G'$. The area of the projection is then twice the area of the triangle $B'D'G'$, which is at most the area of triangle $BDG$. This triangle is equilateral with side $\sqrt{2}$, so the desired maximum is $2\frac{(\sqrt{2})^2\sqrt{3}}{4} = \sqrt{3}$.