We have selected 5 vertices of a cube. What fraction of the cube's volume can the volume determined by these vertices be?
Let the two opposite faces of the cube be $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, and let the edge length of the cube be chosen as the unit. Out of the five vertices of the solid, at least three are located on the same face of the cube. Without loss of generality, we can assume that these three vertices are $A, B$, and $C$. Depending on the placement of the other two vertices, we need to distinguish several cases.

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\text { Figure 1 }
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If $D$ is also a vertex of the solid, then choosing any of $A^{\prime}, B^{\prime}, C^{\prime}$, or $D^{\prime}$ as the fifth vertex results in a pyramid with a square base (Figure 1), whose height is equal to the edge of the cube, so its volume is $\frac{1 \cdot 1 \cdot 1}{3}$, which is one-third of the volume of the cube.

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\text { Figure 2 }
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If $D$ is not among the selected vertices, then we have chosen two out of $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$. If $D^{\prime}$ is not among them, then either there is a face whose all four vertices have been selected, leading us to the previously obtained square-based pyramid, or the vertices $A^{\prime}, C^{\prime}$ have been chosen (Figure 2). This solid is congruent to the other two, where $A^{\prime}$ and $D^{\prime}$, or $C^{\prime}$ and $D^{\prime}$ are the vertices on the base. Let's examine, for example, the solid $A B C A^{\prime} C^{\prime}$ (Figure 2). The points $A, C, C^{\prime}, A^{\prime}$ lie in the same plane, forming the four vertices of a rectangle, and the solid itself is a pyramid with a rectangular base. The distance of the fifth vertex from the base is equal to half the face diagonal of the cube, so the volume of the pyramid is: $\frac{1}{3}\left(\sqrt{2} \cdot 1 \cdot \frac{\sqrt{2}}{2}\right)=\frac{1}{3}$, which is also one-third of the volume of the cube.

# 1988-01-022-1.eps 

Figure 3

There is one more case to examine, when the selected vertices are $A, B, C, D^{\prime}$, and $B^{\prime}$ (Figure 3). In this case, the solid consists of two tetrahedra joined along the $A C B^{\prime}$ triangular face. The line $B D^{\prime}$ is perpendicular to the plane of the $A C B^{\prime}$ triangle, so the sum of the heights of the two tetrahedra to the $A C B^{\prime}$ face is exactly $B D^{\prime}=\sqrt{3}$, the space diagonal of the cube. Each side of the $A C B^{\prime}$ triangle is a face diagonal, so its area is $(\sqrt{2})^{2} \cdot \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$. The combined volume of the two tetrahedra is thus $\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{3}=\frac{1}{2}$, which is half the volume of the cube.

Therefore, the volume of the solid determined by the five vertices is either half or one-third of the volume of the cube.