4. Two adjacent faces of a tetrahedron, representing equilateral triangles with side length 3, form a dihedral angle of 30 degrees. The tetrahedron is rotated around the common edge of these faces. Find the maximum area of the projection of the rotating tetraheron onto the plane containing the given edge. (12 points)
Solution. Let the area of each of the given faces be $S$. If the face is located 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 $0<\varphi<60^{\circ}$, the area of the projection is $\Pi=S \cos \varphi<S$.

![](https://cdn.mathpix.com/cropped/2024_05_06_66ff5449eae3e24affe1g-09.jpg?height=251&width=408&top_left_y=797&top_left_x=1475)

When rotated by an angle $60^{\circ}<\varphi<90^{\circ}$, the area of the projection is

$\Pi=S \cos \varphi+S \cos \psi=S \cos \varphi+S \cos (\pi-\pi / 6-\varphi)=S \cos \varphi+S \cos \left(\frac{5 \pi}{6}-\varphi\right)$.

$\Pi^{\prime}=S\left(-\sin \varphi+\sin \left(\frac{5 \pi}{6}-\varphi\right)\right), \Pi^{\prime}=0$ when $\varphi=5 \pi / 12$. The maximum of the function in the considered

interval is reached at $\varphi=\pi / 3$ with the area $\Pi=2 S \cos \left(\frac{5 \pi}{12}\right)<S$, $75 \pi / 12 \quad \triangle$ since $\cos \left(\frac{5 \pi}{12}\right)<\cos \left(\frac{4 \pi}{12}\right)=\cos \left(\frac{\pi}{3}\right)=\frac{1}{2}$.

When rotated by an angle $90^{\circ}<\varphi<150^{\circ}$, the area of the projection is $\Pi=S \cos (\pi / 2-\varphi)=S \sin \varphi<S$. At $\varphi=5 \pi / 6$ the area $\Pi=S$.

Answer. $\Pi=S=\frac{3 \sqrt{3}}{4}$.