4. What is the maximum area of the orthogonal projection of a unit cube onto a plane

Answer. $S_{\max }=2 \sqrt{3}$.
The center of a cube is its center of symmetry. The edges of the cube can be divided into three classes, with each class containing edges that are parallel to each other. When orthogonally projecting the cube onto a plane in any direction, all edges of the same class are depicted as parallel segments. The result of the projection is a convex centrally symmetric polygon, the sides of which are parallel to no more than three directions. Such a polygon can be either a parallelogram or a centrally symmetric hexagon. A projection in the form of a parallelogram (which will always be a rectangle) can be considered a degenerate case of a projection in the form of a hexagon, where two pairs of sides complement each other to form two segments.

![](https://cdn.mathpix.com/cropped/2024_05_06_d8efe70e6e1aeb491532g-3.jpg?height=688&width=657&top_left_y=250&top_left_x=445)

Let \( ABCA'B'C' \) be such a hexagon. Denote \( P \) as the point symmetric to point \( A' \) relative to the midpoint of segment \([BC]\). The quadrilateral \( BA'C'P \) is a parallelogram because its diagonals bisect each other at their point of intersection. Therefore, \([PC] \parallel [BA'] \parallel [AB']\) and \(|PC| = |BA'| = |AB']\), making the quadrilateral \( PCB'A \) a parallelogram, and points \( P \) and \( B' \) are symmetric relative to the midpoint of segment \([AC]\). Similarly, the quadrilateral \( AC'VB \) is a parallelogram, and points \( P \) and \( C' \) are symmetric relative to the midpoint of segment \([AB]\). Consequently, the area of the hexagon is twice the area of triangle \( ABC \). When orthogonally projecting an arbitrary polygon, the area of the projection does not exceed the area of the polygon itself. Equality is possible only when the direction of projection is perpendicular to the plane of the polygon. Therefore, the maximum area of the projection of the cube is in the direction of its diagonal. It is given by \( S_{\max} = 2 \cdot \sqrt{2} \cdot \sqrt{2} \cdot \frac{\sqrt{3}}{2} = 2 \sqrt{3} \).