[ Parallel Design ]

Find the maximum value of the area of the orthogonal projection of a rectangular parallelepiped with dimensions $a, b$, and $c$ onto some plane.

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Let's first prove the following statement. If the projection of vertex A of the parallelepiped $A B C D A 1 B 1 C 1 D 1$ onto a certain plane lies inside the projection of triangle $A 1 B D$ onto this plane, then the area of the projection of the parallelepiped is twice the area of the projection of triangle $A 1 B D$. Let $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}, A 1^{\prime}, B 1^{\prime}, C 1^{\prime}$ and $D 1^{\prime}$ be the parallel projections of the vertices $A, B, C, D, A 1, B 1, C 1$ and $D 1$ of the parallelepiped $A B C D A 1 B 1 C 1 D 1$, (Fig.1) and point $A^{\prime}$ lies inside triangle $A 1^{\prime} B^{\prime} D^{\prime}$. Since parallel projection preserves the parallelism of lines, the quadrilaterals $A^{\prime} B^{\prime} B 1^{\prime} A 1^{\prime}, A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ and $A^{\prime} D^{\prime} D 1^{\prime} A 1^{\prime}$ are parallelograms. They are divided by their diagonals $B^{\prime} A 1^{\prime}, B^{\prime} D^{\prime}$ and $D^{\prime} A 1^{\prime}$ into equal triangles. Since point $A^{\prime}$ lies inside triangle $A 1^{\prime} B^{\prime} D^{\prime}$, each pair of these parallelograms has exactly one common side, and their union (hexagon $A 1^{\prime} B 1^{\prime} B^{\prime} C^{\prime} D^{\prime} D 1^{\prime}$) is the projection of the given parallelepiped onto plane $\alpha$. Therefore,

$$
S^{\prime}=S_{A 1^{\prime} B 1^{\prime} B^{\prime} C^{\prime} D^{\prime} D 1^{\prime}}=2 S_{\Delta A 1^{\prime} B^{\prime} D^{\prime}}
$$

Now let $A^{\prime} B^{\prime} C^{\prime} D^{\prime} A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be the orthogonal projection of the rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$ ($A B=a, A D=b, A A 1=c$) onto a certain plane. Denote by $\phi$ the angle between this plane and the plane of triangle $A 1 B D$. According to the previously proven result,

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-44.jpg?height=71&width=1154&top_left_y=1294&top_left_x=452)

where equality is achieved if $\phi=0^{o}$, i.e., when the projection plane is parallel to the plane of triangle $A 1 B D$. To find the area of triangle $A 1 B D$, drop the perpendicular $A 1 H$ from vertex $A 1$ to line $B D$ (Fig.2). By the theorem of three perpendiculars, $A H$ is the height of the right triangle $A B D$. Therefore,

$$
\begin{aligned}
& A H=\frac{A B \cdot A D}{B D}=\frac{\frac{a b}{\sqrt{a^{2}+b^{2}}}}{} \\
& A 1 H=\sqrt{A_{1} A^{2}+A H^{2}}=\sqrt{c^{2}+\frac{a^{2} b^{2}}{a^{2}+b^{2}}}=\frac{\sqrt{a^{2} b^{2}+a^{2} c^{2}+b^{2} c^{2}}}{\sqrt{a^{2}+b^{2}}} \\
& S_{\triangle A 1 B D}=\frac{1}{2} B D \cdot A 1 H=\frac{1}{2} \sqrt{a^{2}+b^{2}} \cdot \frac{\sqrt{a^{2} b^{2}+a^{2} c^{2}+b^{2} c^{2}}}{\sqrt{a^{2}+b^{2}}} = \\
& =\frac{1}{2} \sqrt{a^{2} b^{2}+a^{2} c^{2}+b^{2} c^{2}} \\
& \text{Therefore,} \\
& S_{\max }=2 S_{\triangle A 1 B D}=\sqrt{a^{2} b^{2}+a^{2} c^{2}+b^{2} c^{2}}
\end{aligned}
$$

It is clear that if the orthogonal projection of point A lies outside the orthogonal projection of triangle $A 1 B D$ onto a certain plane, then the area of the orthogonal projection of the parallelepiped onto this plane will be less than the one found.

## Answer

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