If seven vertices of a hexahedron lie on a sphere, then so does the eighth vertex.

Question

I'm trying to prove https://imomath.com/index.cgi?page=inversion (Problem 11) by projective geometry:

If seven vertices of a (quadrilaterally-faced) hexahedron lie on a sphere, then so does the eighth vertex.

Is it suffice to prove: If seven vertices of a (quadrilaterally-faced) hexahedron lie on a quadric surface, then so does the eighth vertex?

The projective proposition (B) is easy to prove, much easier than the original (sphere) one (A).

I believe B implies A in this case, but I don't think it as easy as "A is a special case of B, so B implies A".

Think about the converse of Pascal's theorem: ..., then 6 vertices lie on a conic. We can't change it to "..., then 6 vertices lie on a circle". The correct one should be: "..., and 5 vertices lie on a circle, then so does the 6th vertex".

But this problem is a slight different than the converse of Pascal's theorem (the circle version): 5 vertices determine a conic, but 7 vertices is not enough to determine a quadratic surface.

Answer 1

I think that OP's strategy can be made to work.

The following is excerpted from Salmon, Analytical Geometry of Three Dimensions, pg 130, Article 131.

Given seven points common to a series of quadrics, then an eighth point common to the whole system is determined.

For let $U, V, W$ be three quadrics, each of which passes through the seven points, then $U +\lambda V +\mu W$ may represent any quadric which passes through them; for the constants $\lambda,\mu$ may be so determined that the surface shall pass through any two other points, and may in this way be made to coincide with any given quadric through the seven points. But $U +\lambda V +\mu W$ represents a surface passing through all points common to $U, V, W,$ and since these intersect in eight points, it follows that there is a point, in addition to the seven given, which is common to the whole system of surfaces.

Let's refer to the set of all quadrics $U +\lambda V +\mu W$ as the net $N$ generated by $U, V, W$. The net $N$, which contains all quadrics passing through the eight intersection points of $U,V,W$ will in particular contain the three degenerate quadrics consisting of two planes corresponding to opposite faces of the given hexahedron. The intersection of the three degenerate quadrics will be the vertices of the hexahedron, and thus all eight hexahedron vertices lie on all members of $N$.

If I've argued this correctly I think we can state that:

Theorem 1: For a quadric $q$ and a hexahedron $h$, if seven vertices of $h$ lie upon $q$ then so does the eighth vertex.

Since a sphere is a special case of a quadric, the theorem holds for a sphere and a hexahedron.