12. Among all orthogonal projections of a regular tetrahedron onto different planes, find the one that has the largest area.
12. The projection of any polyhedron is a polygon, the vertices of which are the projections of the vertices of the polyhedron. Since a tetrahedron has four vertices, depending on the choice of the projection plane, its projection can be either a triangle or a quadrilateral. (It cannot be a segment or a point, as a regular tetrahedron is not a planar figure.) Let's consider both cases.

$1^{\circ}$. A regular tetrahedron is projected into a triangle that coincides with the projection of one of its faces, for example

![](https://cdn.mathpix.com/cropped/2024_05_21_d3653343768e91d6de9ag-061.jpg?height=455&width=562&top_left_y=598&top_left_x=346)

Fig. 45.

$A B C$ (Fig. 45). Let the vertices $A, B, C$ of the face be projected into the vertices $A^{\prime}, B^{\prime}, C^{\prime}$ of the projection. By the known formula for the area of the projection (Theorem of Problem 3), we have

$$
S_{\triangle A^{\prime} B^{\prime} C^{\prime}}=S_{\triangle A B C} \cdot \cos \varphi,
$$

where $\varphi$ is the angle between the plane of the face and the plane of the projection. The area of the projection will be the largest when $\cos \varphi=1, \varphi=0$ (i.e., when the plane of the face $A B C$ is parallel to the plane of the projection). In this case,

$$
S_{\triangle A^{\prime} B^{\prime} C^{\prime}}=S_{\triangle A B C}=\frac{a^{2} \sqrt{3}}{4}
$$

where $a$ is the edge of the tetrahedron.
$2^{\circ}$. A regular tetrahedron is projected into a quadrilateral (Fig. 46). The formula for the area of a quadrilateral is known:

![](https://cdn.mathpix.com/cropped/2024_05_21_d3653343768e91d6de9ag-062.jpg?height=497&width=591&top_left_y=300&top_left_x=313)

Fig. 46.

$S=\frac{1}{2} b c \sin \varphi$, where $b$ and $c$ are the diagonals of the quadrilateral, and $\Psi$ is the angle between them. But the diagonals of the quadrilateral are the projections of the edges of the tetrahedron $A B C D$. Therefore, $b=a \cos \theta_{1}$ and $c=a \cos \theta_{2}$, where $\theta_{1}$ and $\theta_{2}$ are the angles of the edges with the plane of the projection. The area $S=\frac{1}{2} a^{2} \cos \theta_{1} \cos \theta_{2} \sin \psi$ will be the largest when $\cos \theta_{1}=\cos \theta_{2}=\sin \varphi=1$, i.e., when $\theta_{1}=\theta_{2}=0$, $\varphi=\frac{\pi}{2}$.

To obtain the projection of the tetrahedron with the largest area, we construct a cube with side $\frac{a \sqrt{2}}{2}$ and inscribe a regular tetrahedron with edge $a$ in it (Fig. 47). Obviously, the projection of this tetrahedron onto any face of the cube will be a square. The angle between its diagonals is a right angle, and the diagonals themselves are equal to the edges of the tetrahedron. Thus, the largest area of the quadrilateral projection of the tetrahedron is $\frac{a^{2}}{2}$. This is greater than $\frac{a^{2} \sqrt{3}}{4}$, i.e., than the largest area of its triangular projection.

Therefore, among all orthogonal projections of a regular tetrahedron, the largest area is that of a square, which is obtained by projecting the tetrahedron onto a plane parallel to any two of its skew edges.

![](https://cdn.mathpix.com/cropped/2024_05_21_d3653343768e91d6de9ag-062.jpg?height=375&width=352&top_left_y=984&top_left_x=721)

Fig. 47.