4. A paper equilateral triangle of side length 2 on a table has vertices labeled $A, B, C$. Let $M$ be the point on the sheet of paper halfway between $A$ and $C$. Over time, point $M$ is lifted upwards, folding the triangle along segment $B M$, while $A, B$, and $C$ remain on the table. This continues until $A$ and $C$ touch. Find the maximum volume of tetrahedron $A B C M$ at any time during this process.
Answer: $\frac{\sqrt{3}}{6}$
View triangle $A B M$ as a base of this tetrahedron. Then relative to triangle $A B M$, triangle $C B M$ rotates around segment $B M$ on a hinge. Therefore the volume is maximized when $C$ is farthest from triangle $A B M$, which is when triangles $A B M$ and $C B M$ are perpendicular. The volume in this case can be calculated using the formula for the volume of a tetrahedron as $\frac{1}{6} \cdot 1 \cdot 1 \cdot \sqrt{3}=\frac{\sqrt{3}}{6}$.