\section*{Problem 13}

What is the smallest number of tetrahedrons into which a cube can be partitioned?


\section*{Solution}

Answer: 5.

Tetrahedral faces are triangular, so each cube face requires at least two tetrahedral faces. So at least 12 tetrahedral faces are needed in all. At most three faces of a tetrahedron can be mutually orthogonal (and no two faces can be parallel), so at most 3 faces from each tetrahedron can contribute towards these 12 . So we require at least 4 tetrahedra to provide the cube faces. But these tetrahedra each have volume at most \(1 / 6(1 / 3 \times\) face area \(x 1\), and face area is at most \(1 / 2\) ). So if we have only 4 tetrahedra in total then their total volume is less than the cube's volume. Contradiction. Hence we need at least 5 tetrahedra.

It can be done with 5: lop off 4 non-adjacent corners to leave a tetrahedron. More precisely, take the cube as \(\mathrm{ABCDA} \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C} \mathrm{D}^{\prime}\) ' with \(\mathrm{ABCD}\) horizontal, \(\mathrm{A}\) ' directly under \(\mathrm{A}, \mathrm{B}^{\prime}\) directly under \(\mathrm{B}\) and so on. Then the five tetrahedra are \(\mathrm{AA}^{\prime} \mathrm{BD}, \mathrm{CC}^{\prime} \mathrm{BC}, \mathrm{DD}^{\prime} \mathrm{A}^{\prime} \mathrm{C}^{\prime}, \mathrm{BB}^{\prime} \mathrm{A}^{\prime} \mathrm{C}^{\prime}, \mathrm{BDA}^{\prime} \mathrm{C}^{\prime}\).

