3D variant of Morley's theorem?
Jim Ferry
>
> Is there an analogue of Morley's theorem to dimension 3 (or higher)?
I don't know, but if there is, a way to find the Morley
tetrahedron might be as follows. According to
http://cedar.evansville.edu/~ck6/tcenters/recent/morley.html
the second Morley center of a triangle has trilinear coordinates
sec(A/3) : sec(B/3) : sec(C/3).
Thus, take a point with analogous tetralinear coordinates
(sec(A/n):sec(B/n):sec(C/n):sec(D/n) for n = 3 (or 4?)) and consider
the lines determined by this point and each of the vertices of a
tetrahedron. There are four degrees of freedom to pick a tetrahedron
with vertices on each of these lines, but being regular entails five
constraints. Hence, if a regular tetrahedron exists, it's special in
some way.
Note: I'm assuming that sec(A/3) etc. make sense--that there's a way
that geometers assign angles in 3 or more dimensions such that the
formulae incenter = 1:1:1, centroid = csc(A):csc(B):csc(C),
circumcenter = cos(A):cos(B):cos(C), orthocenter = sec(A):sec(B):sec(C),
etc. generalize.
| Jim Ferry | Center for Simulation |
+------------------------------------+ of Advanced Rockets |
| http://www.uiuc.edu/ph/www/jferry/ +------------------------+
| jferry@[delete_this]uiuc.edu | University of Illinois |
