4. Two adjacent faces of a tetrahedron, which are isosceles right triangles with a hypotenuse of 2, form a dihedral angle of 60 degrees. The tetrahedron is rotated around the common edge of these faces. Find the maximum area of the projection of the rotating tetrahedron onto the plane containing

![](https://cdn.mathpix.com/cropped/2024_05_06_66ff5449eae3e24affe1g-31.jpg?height=243&width=396&top_left_y=724&top_left_x=1481)
the given edge. (12 points)

Solution. Let the area of each of the given faces be \( S \). If the face lies in the plane of projection, then the projection of the tetrahedron is equal to the area of this face \( \Pi = S \).

When rotated by an angle \( 0 < \varphi < 30^\circ \), the area of the projection is \( \Pi = S \cos \varphi < S \).

![](https://cdn.mathpix.com/cropped/2024_05_06_66ff5449eae3e24affe1g-31.jpg?height=263&width=394&top_left_y=1128&top_left_x=1482)

When rotated by an angle \( 30^\circ < \varphi < 90^\circ \), the area of the projection is

\[
\Pi = S \cos \varphi + S \cos \psi = S \cos \varphi + S \cos \left(\pi - \frac{\pi}{3} - \varphi\right) = S \cos \varphi + S \cos \left(\frac{2\pi}{3} - \varphi\right).
\]

\[
\Pi' = S \left(-\sin \varphi + \sin \left(\frac{2\pi}{3} - \varphi\right)\right), \quad \Pi' = 0 \text{ when } \varphi = \frac{\pi}{3}.
\]

The maximum of the function in the considered interval is achieved at

![](https://cdn.mathpix.com/cropped/2024_05_06_66ff5449eae3e24affe1g-31.jpg?height=165&width=371&top_left_y=1702&top_left_x=1508)

\[
\varphi = \frac{\pi}{3}, \quad \Pi = 2 S \cos \left(\frac{\pi}{3}\right) = 2 S \cdot \frac{1}{2} = S.
\]

When rotated by an angle \( 90^\circ < \varphi < 120^\circ \), the area of the projection is \( \Pi = S \cos \left(\frac{\pi}{2} - \varphi\right) = S \sin \varphi < S \).

When \( \varphi = \frac{2\pi}{3} \), the area \( \Pi = S \).
Answer. $\Pi=S=\frac{1}{2} \cdot \sqrt{2}^{2}=1$.