89. In a regular tetrahedron, points $M$ and $N$ are the midpoints of opposite edges. The projection of the tetrahedron onto a plane parallel to $M N$ is a quadrilateral with area $S$, one of whose angles is $60^{\circ}$. Find the surface area of the tetrahedron.
89. Let $x$ be the edge of the tetrahedron, $|M N|=-\frac{x}{\sqrt{2}}$. If the edge, the midpoint of which is $M$, forms an angle $\alpha$ with the given plane, then the opposite edge forms an angle $\frac{\pi}{2}-\alpha$. The projection of the tetrahedron onto this plane is an isosceles trapezoid with bases $x \cos \alpha$ and $x \sin \alpha$ and the distance between the bases equal to $\frac{x}{\sqrt{2}}$.
![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-062.jpg?height=456&width=732&top_left_y=405&top_left_x=267)

Fig. 9.

Thus, $S=\frac{x^{2}}{2 \sqrt{2}}(\cos \alpha+\sin \alpha)$. Moreover, according to the condition, the angle at the larger base is $60^{\circ}$, from which $|\cos \alpha-\sin \alpha|=\sqrt{\frac{2}{3}}$.

Answer: $3 S \sqrt{2}$.