A rectangular cuboid has edges of 3, 4, and 6 units. Can the cuboid be placed in such a way that the contour of its orthogonal projection onto a given plane is a square?
Let the projection plane be denoted by $S$. Place the cuboid on the plane $S$ such that the 6-unit edge is perpendicular to $S$. Then rotate the cuboid around one of its 4-unit edges by an angle $\alpha$.

We drew a projection of the described position on a plane perpendicular to the axis of rotation. In the diagram, the projection of the axis of rotation is $B$, and the projection of the plane $S$ is $S'$. Use the additional notations in the diagram.

The contour of the perpendicular projection of the cuboid onto $S$ will be a square if $x+y=4$. The equality of the angles denoted by $\alpha$ implies that the triangles $AEB$ and $BFC$ are similar, and the ratio of similarity is $3:6$. Therefore, $CF=2x$, and by the Pythagorean theorem, $4x^2 + y^2 = 36$. Since $x+y=4$, we obtain the quadratic equation for $x$: $5x^2 - 8x - 20 = 0$, from which $x = \frac{8 \pm \sqrt{464}}{10}$. Considering that $x > 0$, we have $x = \frac{4 + \sqrt{116}}{5}$. Then, $\cos \alpha = \frac{4 + \sqrt{116}}{15}$, so $\alpha \approx 10^\circ$.

There is no other solution to the problem, because when rotating around the 3-unit edge, with similar notations,

\[
x + y = 4 \cdot \cos \alpha + 6 \cdot \sin \alpha = 4 \cdot (\cos \alpha + \sin \alpha) + 2 \cdot \sin \alpha > 3
\]

for every acute angle $\alpha$, since $\cos \alpha + \sin \alpha > 1 \quad (\alpha > 0)$; when rotating around the 6-unit edge, the maximum value of $x + y$ is only $\sqrt{9 + 16} = 5$.

Gergely Szabó (Debrecen, Fazekas M. Gymnasium, III. grade)

Note. Ádám Kiss and László Szobonya proved that every cuboid has (at most two) positions such that the contour of its perpendicular projection onto a given plane is a square.

![](https://cdn.mathpix.com/cropped/2024_05_02_d89c918791735026f9efg-1.jpg?height=648&width=590&top_left_y=934&top_left_x=754)