Problem 2. A triangular pyramid, all edges of which are 6 cm long, stands on a flat table. The pyramid is rolled over its edges 6 times in such a way that one of its vertices remains stationary, and it is not rolled over the same edge twice in a row. Find the length of the trajectory along which the mobile vertex of the pyramid moves during these rolls.
Answer: $\left(\pi-\arccos \frac{1}{3}\right) \cdot 12 \sqrt{3}$ cm.

Solution. Let the pyramid be $A B C D$. Suppose $A$ is the fixed vertex, and vertex $D$ does not touch the table surface initially. Let the first rolling be over the edge $A B$. During this rolling, vertex $D$ moves along an arc of a circle with the center at the midpoint $H$ of edge $A B$ and a radius equal to the slant height of the pyramid, which is $3 \sqrt{3}$ cm. Using the cosine rule in triangle $C D H$, we find that $\cos \angle C H D = \frac{1}{3}$, so the central angle subtended by the arc along which point $D$ moves is $\pi - \arccos \frac{1}{3}$. Therefore, the length of this arc is $\left(\pi - \arccos \frac{1}{3}\right) \cdot 3 \sqrt{3}$ cm.

During the second rolling (over the edge $A D$), point $D$ remains stationary, during the third rolling (over the edge $A B$), it travels along an arc of the same length as during the first rolling, and the next three rollings occur similarly to the first three, after which the pyramid returns to its initial position. In this process, point $D$ travels a trajectory of length $4 \cdot \left(\pi - \arccos \frac{1}{3}\right) \cdot 3 \sqrt{3}$ cm.