
Re: A Theorem concerning the Trisectors of a Triangle
Posted:
Sep 15, 1998 1:03 PM


On Tue, 15 Sep 1998, Richard Guy wrote:
> Start from the lighthouse theorem.
Even I am not quite sure what this is, so I'm sure that many geometryforum subscribers won't. But I think it's that if the beams of three lighthouses all rotate at the same uniform rate, then the triangle they form remains similar to itself, and rotates at the same rate (while also expanding or contracting). Is this right, Richard?
> This gives 9 potential Morley points for each pair of lighthice, BC, > CA, AB. However, their (only 9 distinct) sides are (segments of) all 18 > Morley triangles.
... I've edited out quite a lot ...
> What I want to do is to show that the 3 sets of 9 lines from > the lighthouse triangles are parallel (easy) and coincide in threes > (not immediately obvious?) But I (have yet to) think (enough to > show that) there is a proof for all 18 Morley triangles in one bang, > and that Roussel's triangle is also in the picture. > > I meant to have thought this out more carefully, but found myself > in a stream of (un?)consciousness. Can JHC make any sense of this?
I can't yet, but am certainly going to try to. I wasn't even aware that the sides of the 18 Morley triangles coincide(d) in sets of 3 (if I understand you correctly). You have the advantage over me in having a picture that shows them all.
Let me say that there are in fact 18 Roussel triangles, not just one. I don't know to what extent their edges or vertices coincide (if at all).
After disproving my "general parallelism" conjecture, I made the new one that (again replacing the trisectors by arbitrary isogonal pairs of lines to the vertices) the Rousseltype triangle would always be similar to the Morleytype one; but this too has failed (if my calculations are correct). Nevertheless I'm sure that there is something new and interesting to be found here, and I'm still thinking!
John Conway

