A property of the bidimensional sphere

TL;DR

This paper investigates a geometric property of the bidimensional sphere, establishing that any open set with area greater than three-quarters of the sphere always contains the vertices of a regular tetrahedron, and proves this constant is optimal.

Contribution

The paper proves that the constant 3/4 is sufficient and necessary for open sets on a sphere to contain vertices of a regular tetrahedron.

Findings

Any open set with area > 3/4 of the sphere contains a regular tetrahedron's vertices.

The constant 3/4 is the best possible for this property.

Generalization of the property to broader geometric contexts.

Abstract

It is natural to ask for a reasonable constant k having the property that any open set of area greater than k on a bidimensional sphere of area 1 always contains the vertices of a regular tetrahedron. We shall prove that it is sufficient to take k=3/4. In fact we shall prove a more general result. The interested reader will not have any problem in establishing that 3/4 is the best constant with this property.