
Re: A Theorem concerning the Trisectors of a Triangle
Posted:
Sep 15, 1998 2:27 PM


Sorry that that went to lots of people other than DR and JHC, especially if they were not in need of education, but as there may be people out there in need of enlightenment, let me say that the Lighthouse theorem (which I haven't seen in print) states, in general, that n lines through A, making angles k pi / n with each other, intersect n similar lines through B in n^2 points which are the vertices of n regular ngons. If n is odd, they are homothetic. If n is even, they split into two homothetic sets of n/2 ngons, with members of different sets at an angle pi / n with one another. Hint for proof: the circumcircles of the polygons pass through A and B. The case n = 3 is especially relevant to the Morley triangle theorem. Three lighthouse triangles from A and B, three from B and C, and three from C and A, *** if they are chosen at moments when the lighthouse beams are angle trisectors of A, B and C *** will be found to have their 27 vertices, lying in sixes on 3 sets of 3 parallel lines (at angles pi / 3 with one another) You can choose one line from each set in 3 x 3 x 3 = 27 ways. 9 of the triangles are the lighthouse triangles you started with. The other 18 are genuine Morley triangles. The thing I find confusing is this: Suppose a trisector of angle B + 2j pi meets the appropriate trisector of C + 2k pi in the point Ajk, and that Bki, Cij are similarly defined, for 1 < i, j, k < 3, then we have the same 27 points, but Ajk Bki Cij is a Morley triangle just if i + j + k is not congruent to 1 mod 3. If i + j + k is congruent to 1, then you get some crazy, nonequilateral triangle. I know that this is so by drawing, and I vaguely understand the Galois theory behind it, in that, among the cube roots of unity, at a critical point, you only need the complex, quadratic, ones. Trisecting 2 angles solves 2 cubics, but to make the third one `fit' you have to make the right choice and the group is of order 3 x 3 x 2. R.
On Tue, 15 Sep 1998, John Conway wrote:
> 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

