What is the maximum area of the orthogonal projection of a rectangular prism with dimensions $3 \times 4 \times 12$?
Solution. In orthogonal projection, the image of a convex set is a convex set, and the images of parallel and equal-length segments are also parallel and of equal length (the images of non-parallel but equal-length segments are generally not of equal length). Therefore, if we project the rectangular parallelepiped $\mathcal{T}$, the projection will be a convex planar figure that is centrally symmetric and is bounded by the projections of certain edges of $\mathcal{T}$. The edges of $\mathcal{T}$ can be divided into 3 groups, each containing 4 parallel and equal-length edges. Since a convex planar figure can have at most two sides parallel to the same line, the orthogonal projection of $\mathcal{T}$ is a centrally symmetric hexagon, the bounding edges of which are the projections of two suitable edges from each of the three groups, and the opposite sides are the projections of the edges of $\mathcal{T}$ of different lengths (Figure 1, $P^{\prime}$ denotes the orthogonal projection of point $P$). If the projection is along a line that is not parallel to any face of $\mathcal{T}$, then the projection is a true hexagon; if the projection is along a line that is parallel to a face of $\mathcal{T}$, then the hexagon is degenerate (as shown in Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_229a715491165d9ca123g-1.jpg?height=377&width=593&top_left_y=641&top_left_x=755)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_229a715491165d9ca123g-1.jpg?height=330&width=550&top_left_y=1163&top_left_x=777)

We can choose the labeling of the vertices of $\mathcal{T}$ such that $A^{\prime}$ and $G^{\prime}$ are not vertices of the orthogonal projection. Then the (possibly degenerate) hexagonal projection $B^{\prime} C^{\prime} D^{\prime} H^{\prime} E^{\prime} F^{\prime}$ is divided into parallelograms by the segments $A^{\prime} B^{\prime}$, $A^{\prime} D^{\prime}$, and $A^{\prime} E^{\prime}$. The diagonals $B^{\prime} D^{\prime}$, $D^{\prime} E^{\prime}$, and $E^{\prime} B^{\prime}$ of these parallelograms bisect their areas, so the area of the triangle $B^{\prime} D^{\prime} E^{\prime}$ is exactly half the area of the hexagon that is the projection of $\mathcal{T}$.

The sides of the triangle $B D E$ are face diagonals of $\mathcal{T}$. The lengths of the face diagonals, based on the Pythagorean theorem, are

$$
e=\sqrt{3^{2}+4^{2}}=5, \quad f=\sqrt{3^{2}+12^{2}}=\sqrt{153} \quad \text { and } \quad g=\sqrt{4^{2}+12^{2}}=\sqrt{160}
$$

The area of the triangle $B^{\prime} D^{\prime} E^{\prime}$ cannot be greater than the area $t$ of the triangle $B D E$, which can be calculated using Heron's formula:

$$
\begin{aligned}
t & =\sqrt{\frac{e+f+g}{2} \cdot \frac{e+f-g}{2} \cdot \frac{e-f+g}{2} \cdot \frac{-e+f+g}{2}}= \\
& =\frac{1}{4} \sqrt{\left((e+f)^{2}-g^{2}\right)\left(g^{2}-(e-f)^{2}\right)}= \\
& =\frac{1}{4} \sqrt{(10 \sqrt{153}+18)(10 \sqrt{153}-18)}=\frac{1}{4} \sqrt{15300-324}=6 \sqrt{26}
\end{aligned}
$$

Therefore, the area of the orthogonal projection of $\mathcal{T}$ cannot be greater than $12 \sqrt{26}$, which can be achieved if the rectangular parallelepiped is projected onto a plane parallel to two face diagonals of $\mathcal{T}$ that share a common vertex.