See http://www.geometrie.tuwien.ac.at/havlicek/pub/hoehen.pdf Hans Havlicek, Gunter Weiß "Altitudes of a Tetrahedron and Traceless Quadratic Forms" Tech. Univ. Vienna (no date?) The altitudes of the tetrahedron lie on a quadric regulus, and by counting freedoms they must satisfy one further constraint: that the hyperboloid is "equilateral". This turns out to mean simply that it is _traceless_, that is reducible to the form a^2 x^2 + b^2 y^2 - c^2 z^2, where c^2 = a^2 + b^2. Geometrically, the asymptotic cone apparently comprises a regulus of orthogonal triplets of generator lines. Also the hyperboloid is swept out by a second "conjugate" regulus; the perpendiculars at the faces' orthocentres apparently belong to this other regulus, so lying on the same hyperboloid and meeting all the altitudes. Pity I had to slog through figuring most of this out for myself, before being in a position to do a sufficiently specific search to discover it was already well-known! Fred Lunnon On 11/1/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
Ho-ho! Suppose we are given (Pluecker coordinates of) altitudes K,L,M,N of a tetrahedron, and want to solve for faces (projective) E,F,G,H.
The 3-vector directions of paired altitudes and faces are proportional, so we can immediately set them equal (freedom 8). The 3-vector moments of the altitudes now give 12 linear equations for the "constant" components Eo,Fo,Go,Ho of the faces, with coefficients cubic in their direction components; as polynomials these can be computed to have rank 4, so can in general be solved uniquely [also consistently, since the tetrahedron exists].
Therefore the altitudes must also have freedom 12, the quadric freedom 8, and it cannot be constrained to be only a hyperboloid of rotation (freedom 7), let alone one with conical angle pi/4 (freedom 6).
The mystery deepens ... what single Euclidean invariant could possibly apply to the quadric? Perhaps Steiner was completely mistaken, and the constraint is after all concerned with (say) the cross-ratio of the spacing of the lines along the regulus?
The Pluecker vector for N in terms of faces' components is shown below: moment first, direction second, each with components ordered x,y,z.
N = [
+ Gy Fx Hy Eo – Gy Fo Hy Ex – Fy Eo Hy Gx + Ey Hy Fo Gx + Fy Ex Go Hy – Ey Hy Fx Go + Gz Fx Hz Eo – Gz Fo Hz Ex – Fz Eo Hz Gx + Ez Hz Fo Gx + Fz Ex Go Hz – Ez Hz Fx Go,
– Fx Eo Hx Gy + Ex Hx Fo Gy + Gx Fy Hx Eo – Gx Fo Hx Ey – Ex Hx Fy Go + Fx Ey Go Hx + Gz Fy Hz Eo – Gz Fo Hz Ey – Fz Eo Hz Gy + Ez Hz Fo Gy + Fz Ey Go Hz – Ez Hz Fy Go,
– Fx Eo Hx Gz + Ex Hx Fo Gz + Gx Fz Hx Eo – Gx Fo Hx Ez – Ex Hx Fz Go + Fx Ez Go Hx – Fy Eo Hy Gz + Ey Hy Fo Gz + Gy Fz Hy Eo – Gy Fo Hy Ez – Ey Hy Fz Go + Fy Ez Go Hy,
– Hx (Gz Ex Fy – Gz Ey Fx – Gy Ex Fz + Gy Ez Fx + Gx Ey Fz – Gx Ez Fy),
– Hy (Gz Ex Fy – Gz Ey Fx – Gy Ex Fz + Gy Ez Fx + Gx Ey Fz – Gx Ez Fy),
– Hz (Gz Ex Fy – Gz Ey Fx – Gy Ex Fz + Gy Ez Fx + Gx Ey Fz – Gx Ez Fy)
]
Fred Lunnon
On 11/1/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
Richard,
I had confused myself again: of course the constraint is not a projective invariant, since altitudes are invariant only under similarity! Then "hyperboloid of rotation" becomes the natural interpretation, and the additional fixed conical angle pi/4 also looks plausible.
However: freedom of hyperboloid axis line 4, centre point 1, radius 1, (cone angle 1); of 4 altitude lines along regulus 4. So total freedom of altitudes would then be 4+1+1+4 = 10, less by 2 than the tetrahedron's!
Anyway, I shall go away to think up a lazy brute-force way to decide the matter algebraically.
Fred
On 10/31/11, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
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.