Using Polyhedrons to Define Maximum Volumes

Abstract

D. L. Carleton is a former staff member of Santa Fe Junior College in Gainesville, Florida, where he had taught mathematics from 1969 to 1971. He is continuing his studies now at the University of Texas at Austin, where he is doing research in the relationships of mathematics and art. In the plane, we know that the isosceles triangle of maximum area which can be inscribed in a circle of given radius is actually an equilateral triangle. Similarly, the rectangle of maximum area which can be inscribed in a circle of given radius is a square. In these two cases, the maximum areas are defined by regular polygons. When we consider this observation together with the fact that these two polygons generate a right circular cone and a right circular cylinder, an interesting question arises. Can the right circular cone of maximum volume and the right circular cylinder of maximum volume which can be inscribed in a sphere of given radius be defined in terms of regular polygons? The purpose of this article is to answer this question in the affirmative. In particular, this article will: (1) use the regular tetrahedron to define the right circular cone of maximum volume which can be inscribed in a sphere of given radius; and (2) use the regular octahedron to define the right circular cylinder of maximum volume which can be inscribed in a sphere of given radius. We will verify statement (1) by showing that the vertex angle of the cone of maximum volume is the dihedral angle of the regular tetrahedron. One way to find the value of this angle is to express the volume of the cone as a function of its vertex angle and the radius of the sphere, where the radius is treated as a constant. Figure 1 is a cross-section of a right circular cone inscribed in a sphere, where the vertex angle of the cone is 0, the center of the sphere is 0, and the length of the radius of the sphere is R. The radius of the base of the cone and the height of the cone are AC and BC, respectively. Since the triangle AOB is an isosceles triangle, the angle BA O is equal to 0 /2 which means that the angle A OC is equal to