
Re: A Theorem concerning the Trisectors of a Triangle
Posted:
Sep 20, 1998 12:44 PM


On Sat, 19 Sep 1998, Russell Towle wrote:
> Douglas Zare, on September 17th 1998, wrote: > > >If one trisects the dihedral angles as you suggest, do the opposite > >endpoints of the rays of vertices of Morley triangles ever/always form the > >vertices of an ideal regular icosahedron?
I'm afraid I don't know exactly what this conjecture was, but if "regular icosahedron" means anything like what it usually means, it has no chance of being true, since the field generated by the coordinates of all the points involved would then contain root5, which for instance is not contained in the field generated by the trigonometric functions of onethird of the dihedral angle of the regular tetrahedron.
> That is such a fascinating conjecture! I myself did not think of trisecting > the dihedral angles of a tetrahedron; this may indeed may be the best 3D > analogy to trisecting the interior angles of a triangle.
It would have been fascinating were it true! Yes, I agree that trisecting the dihedral angles might be the best way to go. > No, I did not think of this; instead, upon the analogy to a plane angle > being trisected into three "regions of equal area" (supposing the regions > were bounded by an arc of a circle centered upon the vertex), I strove to > find a way to divide the solid angle at one vertex of a tetrahedron into > "four regions of equal volume." > > If one bisects each interior angle of the four triangles bounding a > tetrahedron, then the bisectors meet in an interior point of each triangle > (is this the "centroid"?).
No it isn't. The four bisecting planes meet in the incenter (the center of the inscribed sphere) of the tetrahedron, so the four points you describe are the Cevian feet of the incenter. Now the centroids of the faces are the Cevian feet not of the incenter, but of the tetrahedron's own centroid.
In the 2dimensional case, the Cevian feet of the centroid are the midpoints of the sides, whereas those of the incen  they are not any kind of "center" of the sides, since their position depends also on the omitted vertex.
things that aren't "> centroids"
> alone define the vertices of a smaller tetrahedron. > If we name the vertices of the initial tetrahedron V1, V2, V3, V4, and the > four centroids of its faces C1,C2, C3, C4, then we can dissect it into > eleven smaller tetrahedra as follows: > > > 1. {C1, C2, C3, C4}. > > 2. {C1, C2, C3, V4}. > 3. {C1, C2, C4, V3}. > 4. {C1, C3, C4, V2}. > 5. {C2, C3, C4, V1}. > > 6. {C1, C2, V3, V4}. > 7. {C2, C3, V1, V4}. > 8. {C3, C4, V1, V2}. > 9. {C1, C3, V2, V4}. > 10. {C1, C4, V2, V3}. > 11. {C2, C4, V1, V3}. > > The question is, whether in general it is even possible for the four > tetrahedra which share one of the vertices of {V1, V2, V3, V4}, to be of > equal volume. For instance, numbers 5, 7, 8, and 11 of the list above share > vertex V1. Perhaps it is true that the *regions they represent* do have > equal volumes; for if one takes a section of the large tetrahedron, {V1, > V2, V3, V4},by a plane containing the triangle {C2, C3, C4}, and allows it > to cut off the four tetrahedra meeting at V1, then, I think, the four > triangles in the section are similar, and have equal area, and the height > to V1 is the same for all of them. Maybe. > > > > 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 >  > > >

