Is it possible to place a point on each edge of the cube so that the volume of the convex hull of the 12 points is exactly half the volume of the cube?
Solution. Let the edge of the cube be 1 unit long. On each edge, take a segment of length $a$ such that equal-length segments run into each vertex of the cube. This way, we can reach the convex polyhedron from the cube by cutting off the pyramids that form at the corners of the cube. Two types of pyramids are formed, four of each type. The volume of the polyhedron will be half the volume of the cube if the total volume of the pyramids is also exactly half the volume of the cube.
![](https://cdn.mathpix.com/cropped/2024_05_02_b99ea79ccb8a00369f6cg-1.jpg?height=1222&width=612&top_left_y=385&top_left_x=754)
It follows that:

The volume of one type of pyramid is $V_{A}=\frac{a^{3}}{6}$, and the other is $V_{B}=\frac{(1-a)^{3}}{6}$. Since the total volume of the pyramids must be $\frac{1}{2}$,

$$
4 \cdot V_{A}+4 \cdot V_{B}=\frac{1}{2}, \quad \text { from which } \quad 12 a^{2}-12 a+1=0
$$

$a_{1,2}=\frac{3 \pm \sqrt{6}}{6}$, the roots obtained are positive numbers less than 1 (their sum is naturally 1). Therefore, by choosing $a$ as $a_{1}$, we arrive at a suitable convex polyhedron: the 12 points can be placed on the edges of the cube in accordance with the requirements. ( $a=a_{2}$ could also be chosen, but this would result in a congruent solution.)