Re: tetrahedron with given altitudes

I don't have a formal proof of sufficiency, but I do know how to come
arbitrarily close to the boundary condition -- consider a very flat
tetrahedron with one vetrex just out of the plane of the other three.

A formal proof would likely proceed aloing these lines: Let the largest
face (face 1) be an equilateral triangle with sides s, thus S_1 =
s^2*sqrt(3)/4. Assume h_m/h_1 (m = 2, 3, 4) are given parameters all
exceeding 1 and express S_m/S_1 accordingly. Since S_1 is fixed you can
then express the FACE altitudes in terms of the h_m/h_1 ratios, thus
fixing all degrees of freedom associatged with the tetrahedron. I would
guess that you can solve for a real tetrahedron whenever the necessary
condition identified above (the "3-dimensional triangle equality," if
you will) holds. If that can be shown, then we can access any ratio of
altitudes satisfying the 3-dimensikonal triangle ineqiality, proving
that inequality sufficient as well as necessary. But I haven't checcked
out the math to make sure things are kosher.

--OL

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