I owe the group(s) a screed about the Lighthouse Theorem. Here's about 1% of what is known. Lighthouse theorem for k = 3: Two lighthouses each have a doubly infinite beam, rotating once per minute, but not necessarily in phase. Photos, taken at 20 second (or 10 second!) intervals, when superimposed, show 3 equally inclined beams through each lighthouse. These illuminate (intersect in) 9 points which are the vertices of 3 equilateral triangles. It's tempting to say that these are homothetic, but it's more perspicuous to say that they are successively rotated through pi/3. Conway has called these three triangles the Guy Faux triangles. The Morley Miracle is that we now have the complete Morley triangle theorem without even having a triangle to start with! A reasonably direct, and quite elementary synthetic proof of the whole kaboodle follows from `angles in the same segment' (often known, I'm told, as the `butterfly theorem'). The nine edges of the Guy Faux triangles form 3 sets of 3 parallel lines, equally inclined, forming 27 equilateral triangles, 3 triangles sharing each of 27 points. 18 of the triangles are Morley triangles for a triangle which it's left to the reader to construct. It will be found that the 27 points lie on 18 lines other than the edges of the Morley triangles. These are trisectors of angles of a triangle ABC where B and C are the original lighthouses, with two more pairs of lighthouses at A and B and at C and A. The 9 non-Morley triangles form 3 sets of 3 Guy Faux triangles, one set for each pair of lighthouses. Lighthouse Theorem for general k. k equally inclined (multiples of pi/k) beams from each of two lighthouses intersect in k^2 points which are the vertices of k regular k-gons, inclined at multiples of pi/k with each other. So, if k is odd, they appear to be homothetic, but if k is even, there are two different orientations. E.g., for k = 4, you get two pairs of parallel squares, inclined at pi/4 with each other. It's worth looking at the lighthouse theorem for k = 2, yielding 4 orthocentric points, and giving a hint to the fact that the 72 Morley triangles of the triangles ABC, BCH, CAH, ABH (H being the orthocentre of ABC) are all `homothetic'. Note: these 4 triangles have the same 9-point circle. A remarkable theorem of Bremner, Goggins, Guy & Guy is that, apart from a couple of exceptions (equilateral triangle, and a family of Pythagorean triangles), if a triangle with rational edges has a Morley triangle with a rational edge-length, then all 72 triangles have rational edge-length. [{\it Acta Arith.}, {\bf93}(2000) 177--187; {\it MR} {\bf2001f}:51024] I haven't managed a complete description of the general situation, but an outline is as follows: The k k-gons each have (k choose 2) edges (join any pair of vertices). So there are k sets of (k choose 2) edges, inclined at a multiple of pi/k with each other. We can choose 2 sets of parallels in (k choose 2) ways, so the number of points of intersection is (k choose 2)^3. These lie on other lines, which (at present) are too numerous to mention [when k = 3, k = (k choose 3), and this obscures the generality] [perhaps an OEIS editor will compose a remark to go with the other comments to sequence A059827 ?]. But with k = 6 a good deal can be seen, [k = 2 is included, so beams are perpendicular to other beams, bringing in the orthocentre], since we now not only have 72 Morley triangles, but also (I believe) 72 more for each of the 12 triangles IBC, where I is any of the 4 incentres and B, C are any pair of vertices of some triangle. I'm not sure (yet) whether one needs to take a variety of positions for A --- perhaps Io, Ia, Ib, Ic (apologies that I don't know the official notation) or H may (also) qualify. One may also be able to say something about Malfatti. R. On Tue, 4 Nov 2003, jpehrmfr wrote:
Dear Hyacinthists with the 108 common points of the trisectors of ABC, we can form 54 equilateral triangles. 27 are homothetic with the standard Morley triangle 18 other ones have their sidelines parallel to three trisectors of the same vertex of ABC But what about the direction of the sidelines of the 9 other ones? Friendly. Jean-Pierre
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