14. What is the smallest number of triangular pyramids (tetrahedrons) into which a cube can be divided?
14. It is easy to see that the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ can be divided into five tetrahedra: if we cut off the tetrahedra $B A C B_{1}$ and $D A C D_{1}$, as well as the tetrahedra $A_{1} B_{1} D_{1} A$ and $C_{1} B_{1} D_{1} C$, we will have one more (fifth) tetrahedron $A C B_{1} D_{1}$ left (Fig. 51; the tetrahedron $A C B_{1} D_{1}$ is even a regular one). It is more difficult to establish that the cube cannot be divided into fewer than five tetrahedra.

Indeed, suppose that the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ is divided into some number of tetrahedra. In this case, the face $A B C D$ of the cube is divided into parts that are faces of at least two tetrahedra (the square $A B C D$ can be divided into two or more triangles), and the sum of the areas

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-082.jpg?height=335&width=371&top_left_y=492&top_left_x=762)

Fig. 51. of the bases of these tetrahedra is equal to $a^{2}$, and the height of each of them does not exceed $a$; therefore, the sum of the volumes of the tetrahedra adjacent to the face $A B C D$ of the division does not exceed $\frac{1}{3} a^{2} \cdot a = \frac{a^{3}}{3}$. Similarly, the face $A_{1} B_{1} C_{1} D_{1}$ of the cube is adjacent to at least two tetrahedra, and the total volume of these tetrahedra also does not exceed $\frac{a^{3}}{3}$. Since no tetrahedron can simultaneously have faces that are part of the square $A B C D$ and part of the square $A_{1} B_{1} C_{1} D_{1}$ (since no tetrahedron has parallel faces!), we already have at least four tetrahedra, and the total number of tetrahedra in the division is not yet exhausted, since the total volume of the specified tetrahedra does not exceed $\frac{a^{3}}{3} + \frac{a^{3}}{3} = \frac{2}{3} a^{3}$, which is less than the volume $a^{3}$ of the cube. Hence, the number of tetrahedra in the division cannot be less than five.