To construct a (regular) pyramid (i.e., a tetrahedron), and to enclose (it) in a given sphere, and to show that the square on the diameter of the sphere is one and a half times the (square) on the side of the pyramid. * Thus, [it is required to construct] a pyramid, whose base is triangle $EFG$, and apex the point $K$, from four "equilateral triangles":https://www.bookofproofs.org/branches/equilateral-triangle-isosceles-triangle-scalene-triangle/... * So, I say that the square on the diameter of the sphere is one and a half times the (square) on the side of the pyramid.
(not yet contributed)
If the radius of the sphere is unity then the side of the pyramid (i.e., tetrahedron) is \[\sqrt{\frac 83}.\]
Proofs: 1
Proofs: 1
