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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.

brainjam
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auntyellow
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    What does hexahedron mean in this case? Not all hexahedra have eight vertices. I must be missing something: Given some polyhedron with vertices on a sphere, nothing stops me from moving just one vertex a little bit "outwards" (in radial direction away from the center of the sphere) to get another polyhedron with the same number of faces, edges and vertices, but with one corner outside the ball. – Jeppe Stig Nielsen Mar 14 '24 at 09:18
  • @JeppeStigNielsen thank you for remind. Here hexahedron means quadrilaterally-faced hexahedron, and I edited the problem. – auntyellow Mar 14 '24 at 09:33
  • @JeppeStigNielsen the 8th vertex is fully restricted: just like parallelepiped, if 3 axis (4 vertices) are given, then all 8 vertices are fixed. In this case, if 3 axis (4 vertices) are given, another 3 vertices are semi-restricted (should be coplanar to 3 axis and on the sphere), and the last vertex is fully-restricted (should be coplanar to 3 faces). – auntyellow Mar 14 '24 at 09:41
  • (Ah, so when a vertex has three or more non-triangle faces (so quadrilaterals, pentagons, etc.) adjacent, then it is "fully-restricted"; you cannot move it even an infinitesimal bit in any direction without changing the topology of the polyhedron. On the other hand, I see that a hexahedron of this type can be a counter-example to the statement when worded without the quadrilaterally-faced part. First put all eight vertices on a sphere, then pick any vertex adjacent to a triangular face and move it on the line it is restricted to.) – Jeppe Stig Nielsen Mar 14 '24 at 15:01
  • I took the liberty of changing your last sentence to "not enough" – brainjam Mar 14 '24 at 17:00
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    In general a quadratic surface is defined by 9 points, right? Possibly barring some corner cases, I'd expect that your can always fit a quadratic surface to a given set of 8 points with 1 degree of freedom remaining in how you do that. So I don't see how lying on a quadratic surface is of much use to your scenario. Conversely 4 points define a spare, so 3 more points lying on the same sphere is a pretty strong condition. – MvG Mar 14 '24 at 19:43
  • @brainjam thank you for correcting my typo – auntyellow Mar 15 '24 at 14:45
  • @MvG "4 points define a sphere", you are right. Let's consider proposition A: If O is a vertex of the given quadrilateral hexahedron, and OA, OB, OC are 3 edgs, then these 4 vertices can determine a sphere. Another 3 points (e.g. D, E and F) are also on the sphere (this is a strong condition). A sphere is a special case of a quadratic surface, so according to proposition B, the 8th vertex G is also on that quadratic surface, i.e. the sphere. Does this enough to imply A? – auntyellow Mar 15 '24 at 15:00
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    No, not enough. When you know that the 8th point lies on a common quadratic surface with the other 7, you don't know that that surface is the sphere because there will be multiple surfaces satisfying this. – MvG Mar 15 '24 at 21:52

1 Answers1

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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.

brainjam
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