Several messages collected. I also owe JHC a more coherent message about this, which I'll endeavor to compile in the next few hours. [turned out to be minutes -- hope I haven't gone off half-cock again.]
Start from the lighthouse theorem. This gives 9 potential Morley points for each pair of lighthice, BC, CA, AB. I say `potential' because the 9 lighthouse triangles are not Morley triangles. However, their (only 9 distinct) sides are (segments of) all 18 Morley triangles. Go back to the lighthouse theorem, just for B and C, say. The 9 points are seen, 3 in each of equally spaced photographs, to form 3 homothetic equilateral triangles whose circumcircles form a coaxal system thru B and C. Note that the circumcircle ABC belongs to this system, and one should keep in mind the points on it where the various lighthice are shining.
Calculate the angles that the sides of the lighthouse triangles make with BC, when the beams make angles beta, gamma, say, with BC. For the moment, these angles are measured in the same sense. In fact, in the Morley case, the lighthice are rotating in opposite senses, and you get a paradoxical situation in that you can't have opposite senses at B and C, at C and A, and at A and B. However, on the one hand it doesn't matter, since I only took 3 snapshots (this nonsense is easily turned into a formal proof, using nothing more than `angles in the same segment'), not a movie (though such a movie exists in the video that Andy made for me), and I can't tell from (equally spaced) snapshots, which way the lighthice were turning. On the other hand, suppose they were turning in opposite senses. The illuminated spots are describing rectangular hyperbolas which pass through B and C. As we threw in the circumcircle before we can throw in any rectangular hyperbola through A, B, C and H.
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?
On Tue, 15 Sep 1998, John Conway wrote:
> On 12 Sep 1998, Den Roussel wrote (and I've edited for greater clarity): > > > If the Angular trisectors of a triangle are produced to the Circumcircle, > > then the Chords determined by the pairs of trisectors adjacent to the > > edges form an Equilateral triangle. > > > > It has been suggested that this result might follow from Morleys > > theorem. So far, I have been unable to make this connection. Any > > thoughts ? > > As I've already remarked, this is indeed connected to Morley's theorem, > in that Den Roussel's triangle is parallel to Morley's. However, I can > produce no deduction of this result from Morley's theorem that's any > simpler than an outright proof of it not using that theorem. > > I conjectured that the parallelism might continue to hold if the > trisectors of the typical angle (say B) were replaced by arbitrary > pairs of isogonal lines (ie., B B1 and B B2 such that angles A B B1 > and C B B2 are equal). The main purpose of this note is to announce > that I've now disproved this conjecture (by replacing the Morley triangle > by the Brocard "minor triangle"). > > However, I'm still very hopeful that there will exist SOME > wonderful generalization, because this construction is near others > that have interesting properties. For instance, it is known that > if A(P), B(P), C(P) are the points where the Cevians of P hit > the circumcircle again, then the triangle they form is similar to > the pedal triangle of P (whose vertices A[P], B[P], C[P] are > where the normals from P hit the sides). > > John Conway