[ Regular Tetrahedron ]

In a regular tetrahedron, points $M$ and $N$ are the midpoints of opposite edges. The orthogonal projection of the tetrahedron onto a plane parallel to the line $M N$ is a quadrilateral with area $S$, one of whose angles is $60^{\circ}$. Find the surface area of the tetrahedron.
Let $ABCD$ be the given regular tetrahedron (Fig. 1), $M$ and $N$ be the midpoints of its opposite edges $AB$ and $CD$ respectively. We will assume that the plane onto which the tetrahedron is projected passes through the line $MN$. The orthogonal projections of the vertices $A, B, C$, and $D$ onto this plane are denoted as $A1, B1, C1$, and $D1$ respectively. Then $M$ is the midpoint of $A1B1$, and $N$ is the midpoint of $C1D1$. It is known that the segment connecting the midpoints of opposite edges of a regular tetrahedron is perpendicular to them. Therefore, by the theorem of three perpendiculars, $MN \perp A1B1$ and $MN \perp C1D1$. Thus, $A1B1 \| C1D1$, and the quadrilateral $A1B1C1D1$ is an isosceles trapezoid with bases $A1B1$ and $C1D1$ and an acute angle of $60^\circ$ (Fig. 2). Let $\angle C1 = \angle D1 = 60^\circ$. If $a$ is the edge of the regular tetrahedron, then $MN = \frac{a \sqrt{2}}{2}$. If $\alpha$ is the angle between the line $AB$ and the plane of projection, then since $AB \perp CD$, the angle between the line $CD$ and the plane of projection is $90^\circ - \alpha$. Therefore,

$$
A1B1 = a \cos \alpha, \quad C1D1 = a \cos (90^\circ - \alpha) = a \sin \alpha
$$

According to the problem statement,

$$
\frac{1}{2}(A1B1 + C1D1) \cdot MN = S, \text{ or } \frac{1}{4} a^2 (\sin \alpha + \cos \alpha) \cdot \sqrt{2} = S
$$

from which

$$
a^2 = \frac{2 S \sqrt{2}}{\sin (2 + \cos \theta(r)}
$$

It remains to find $\sin \alpha + \cos \alpha$. For this, draw a line through point $B1$ parallel to $MN$ until it intersects $C1D1$ at point $E$. Then

$$
\begin{gathered}
EC1 = \frac{1}{2}(C1D1 - A1B1) = \frac{2}{2}(\sin \alpha - \cos \alpha) \\
MN = B1E = EC1 \operatorname{tg} 60^\circ
\end{gathered}
$$

Thus,

$$
\frac{a \sqrt{2}}{2} = \frac{a \sqrt{3}}{2} (\sin \alpha - \cos \alpha), \text{ or } \sin \alpha - \cos \alpha = \sqrt{\frac{2}{3}}
$$

By squaring, we find that

$$
2 \sin \alpha \cos \alpha = \frac{1}{3}
$$

therefore,

$$
\begin{gathered}
(\sin \alpha + \cos \alpha)^2 = 1 + 2 \sin \alpha \cos \alpha = \frac{4}{3}, \\
\sin \alpha + \cos \alpha = \frac{2}{\sqrt{3}}.
\end{gathered}
$$

$$
a^2 = \frac{2 S \sqrt{2}}{\sin (2 + \cos \sqrt{2}} = S \sqrt{6}
$$

The total surface area of the tetrahedron is

$$
4 \cdot \frac{a^2 \sqrt{3}}{4} = a^2 \sqrt{3} = 3 S \sqrt{2}
$$

## Answer

$3 S \sqrt{2}$.