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.