Given any triangle, through each vertex draw the external angle bisector at that vertex. Show that the new triangle that has as its vertices the three pairwise intersection points of these is always an acute triangle (all three angles strictly less than 90 degrees).

Extra Credit: Extended to meet the new triangle, the internal angle bisectors of the given triangle are what with respect to the new triangle?

Say a, b, c are the interior angles of the original triangle, you can easily calculate that the angles of the new triangle are (a+b)/2, (a+c)/2 and (b+c)/2. These cannot be more than 90 degrees if this holds for a,b,c.

Side not: When iterating the process of creating "external angles triangles" the shape of the triangles gets closer and closer to a regular triangle as can be seen by the above formulas.

Extra credit: Hmmm, one can immediately see the internal bisectors are perpendicular to the new triangle. Is that the answer Richard is looking for or is there more to it?

BTW, I wonder how one could construct the interior triangle, when only the exterior one is given... Any ideas?

Posted by JLo on 2006-07-21 09:49:43