Element number of the Platonic solids (Q964203)

Given a set of polyhedra, \(\Sigma\), an element set for \(\Sigma\) is defined as a set \(\Omega\) of polyhedra such that each polyhedron \(P\in\Sigma\) may be represented as the union of finite number of polyhedra \(Q_i\in \Omega\) with mutually disjoint interiors. For instance, a unit cube may be dissected into one right tetrahedron and four congruent tetrahedrons, \textsl{corners}, with equilateral bases and with right angles at vertices; moreover, an octahedron may be represented as the union of 8 corners. So if \(\Sigma\) is the set of right tetrahedra, cubes and octahedra, then the right tetrahedra and corners form an element set for \(\Sigma\). The main question is to construct a smallest element set for a given set of polyhedra. The authors consider the set \(\Sigma_P\) of the Platonic solids and construct explicitly an element set \(\Omega_P\) for \(\Sigma_P\), which consists of four elements -- right tetrahedra, \textsl{equihepta}, \textsl{golden tetra} and \textsl{roofs}. The main result states that four is the minimal cardinality of the element sets for the Platonic solids. Besides, it is announced that if \(\Sigma\) is the set of parallelohedra, then there exists an element set for \(\Sigma\) which consists of a single element.