
Re: A Theorem concerning the Trisectors of a Triangle
Posted:
Sep 17, 1998 1:59 PM


John Conway wrote,
> But as to a version for tetrahedra, I can't think of any reasonable >type of solidangle quadrisection that could even take part in a >meaningful statement, let alone a true one! [Let me say that although >lots of trianglegeometry does admit extensions to tetrahedra, there's >lots that doesn't even among the very simple stuff  for instance the >general tetrahedron doesn't have an orthocenter.]
Hmmm. Well, I was shooting from the hip, really all I'm capable of, being largely ignorant of all these triangle centers, which seem to form such a rich subject.
For a meaningful solidangle quadrisection, all I can imagine at this point, is, fixing one's attention upon a single vertex of a tetrahedron, let the three faces surrounding be bisected at that vertex. Then the three lines of bisection, with the three edges of the tetrahedron, define four solid angles meeting at the original vertex: three on "the exterior," and a single one interior.
Completing the construction, one obtains a dissection of the original tetrahedron into 11 tetrahedra, one on each of the six original edges, four hidden but touching each original vertex, and one central tetrahedron, touching the centroid of each face. Four solid angles meet at each vertex. What might one have to do, to make the four solid angles equal? Intercept each tetrahedral region by a sphere on the common vertex, and give up on "bisecting" the angles of the original tetrahedron, allowing the divisions to fluctuate until the volumes of the tetrahedral regions meeting at a vertex are equal, or ...?
Russell Towle
Russell Towle Giant Gap Press: books on California history, digital topographic maps P.O. Box 141 Dutch Flat, California 95714  Voice: (916) 3892872 email: rustybel@foothill.net 

