Fred, One source said it's got by rotating a rectangular hyperbola about its asymptotes, but they meant axes? Definitely not a hyperbolic paraboloid. Rotate a rect hyp about its `minor' (imaginary?) axis, giving a hyperboloid of one sheet. Asymptotic cone would have a vertical semi-angle pi/4. Does that use up all your degrees of freedom? I believe that the centre of the quadric is the Monge point of the tetrahedron. R. On Mon, 31 Oct 2011, Fred lunnon wrote:
Apologies for earlier multiple postings, resulting from Gmail server hiccoughs!
I've just noticed the following recent observation by Richard Guy --- "In 1827 Jacob Steiner noted that the altitudes ofa tetrahedron are the generators of an equilateral hyperboloid."
I haven't yet discovered exactly what "equilateral" means in this context --- possibly a hyperbolic paraboloid? --- but it looks as though Steiner knew the answer to the problem below a while back. WFL
On 10/31/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
A tetrahedron has freedom 12, since each its 4 vertex points has freedom 3; but 4 lines on a quadric regulus have freedom 4 + 9 = 13.
Therefore there must be a single further constraint on the 4 altitudes, besides their lying on a quadric. What could this be?
Regarding the altitudes as 4 points on the Grassman quadric (4 quadratic constraints) in projective 5-space, the quadric constraint is equivalent to their also lying on a plane (3 further linear constraints).
The missing constraint must surely be (skew-)symmetric in the altitudes, and rational involving quadratic or quartic polynomial; it must also be projectively (semi-)invariant, at any rate with respect to 3-space.
One (slightly asymmetric) possibility might have been the "cross-ratio" of lines K,L,M,N defined by (K^L)(M^N) / (K^M)(L^N), where "^" denotes the Grassman wedge inner product: however this turns out to be the squared ratio of two polynomials, each quartic in the faces' components.
Anyone got any more suggestions?
Fred Lunnon
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