In a cube, the edges starting from one vertex define a corner tetrahedron each. What volume of the cube is occupied by the points that are located in the common part of at least two corner tetrahedra?
Each vertex of the cube is associated with (exactly) one corner tetrahedron. Let's call the vertex of the corner tetrahedron from which three mutually perpendicular edges start the main vertex. The main vertices of two corner tetrahedra can be two opposite vertices of the cube, or two opposite vertices of a face, or the two endpoints of a cube edge. The first case is illustrated in diagram 1./a, the second in diagram $1 /$ b. It is obvious that in these cases the volume of the common part of the corner tetrahedra is zero. Therefore, if we consider three or more corner tetrahedra, the volume of their common part is always zero, since from three (or more) main vertices, two can always be selected that are not adjacent vertices of the cube.

It is then sufficient to determine the volume of the common part of two corner tetrahedra whose main vertices are adjacent cube vertices. In diagram $1 / c$, we can see such tetrahedra. The two main vertices are $A$ and $D$. The intersecting edges of these two tetrahedra are the face diagonals of the cube, which intersect at the midpoints of the corresponding faces. Therefore, the common part is a tetrahedron whose two vertices are the two endpoints of a cube edge, $A$ and $D$ on our diagram, and the other two vertices are the midpoints of the faces intersecting the cube edge, $I$ and $J$. If the edge of the cube is $a$, then the area of the triangle $A I D$ is $\frac{a^{2}}{4}$, and the height of the tetrahedron corresponding to this face is $\frac{a}{2}$, so the volume of the common part of the two corner tetrahedra is $\frac{1}{3} \cdot \frac{a^{2}}{4} \cdot \frac{a}{2}=\frac{a^{3}}{24}$. A common part with a volume of $\frac{a^{3}}{24}$ is associated with each edge, and the set of common points of two common parts is at most one edge. Therefore, the sought volume is $12 \cdot \frac{a^{3}}{24}=\frac{a^{3}}{2}$, half the volume of the cube.

Péter Frenkel (Budapest, Fazekas M. Gyak. Gimn., II. o. t.)

Note: The problem can be solved similarly for any parallelepiped.
![](https://cdn.mathpix.com/cropped/2024_05_02_42ec7e03d6a6f908a2f8g-1.jpg?height=316&width=1158&top_left_y=990&top_left_x=470)