From: edgar@dizzy.mps.ohio-state.edu (G. A. Edgar) Newsgroups: sci.math Subject: Re: Anybody prove Morley's Theorem (geometry)? Date: Wed, 30 Aug 1995 08:19:43 -0400 In article , "Dr P.R. Brown" wrote: >Take any triangle and trisect >its angles. These trisectors . >meet at points A, B and C. .... >Then triangle ABC is equilateral. . . . . > . . . . >This conjecture is quoted without . . . . >proof in Steinhaus's . . . . >"Mathematical Snapshots". He . . . . >does give an ancient reference . A C . >(J.M. Child, Math Gazette 11 . . . . >(1923) p171) which I have been . . B . . >unable to locate. . . . . . . > . . . . . . >Can anyone give a proof of what .. . . . >is referred to as Morley's ........................................ >Theorem? It looks intriguingly >easy. Thanks. Phil There is a discussion (and proof) in Coxeter's book, INTRODUCTION TO GEOMETRY. He does remark that straightforward approaches to the problem don't seem to work. -- Gerald A. Edgar edgar@math.ohio-state.edu Department of Mathematics The Ohio State University Columbus, OH 43210 ============================================================================== Newsgroups: sci.math From: thf@luna3.lc.att.com (Thomas Foregger) Subject: Re: Anybody prove Morley's Theorem (geometry)? Date: Mon, 4 Sep 1995 15:41:24 GMT An elementary proof of Morley's theorem is given in Mathemetical Gems, by Ross Honsberger, c. 1973. He attributes it to M.T. Naraniengar (pub. in 1909) and says: "the simplicity of which, to this day, has never been surpassed at the elementary level of exposition". The proof is a little over 2 pages. tom foregger -- Thomas H. Foregger thf@garage.att.com ============================================================================== Newsgroups: sci.math From: jfran@world.std.com (Joseph A Francoeur) Subject: Conway's Proof of Morley's Theorem Date: Wed, 6 Sep 1995 01:04:52 GMT Someone asked about proofs of Morley's Theorem recently. I remembered downloading a proof given by John Conway. I found it and am enclosing it below. I started going through it, but haven't finished reading it to see if I'm convinced. Any comments on this proof? Joe ----------------------------------------------------------------------- Xref: world geometry.puzzles:340 Path: world!forum.swarthmore.edu!gateway From: conway@math.Princeton.EDU (John Conway) Newsgroups: geometry.puzzles Subject: Re: Dictionary Date: 15 Feb 1995 23:10:24 -0500 Organization: Forum news/mail gateway Lines: 74 Sender: daemon@forum.swarthmore.edu Distribution: inet Message-ID: <9502160331.AA02375@broccoli.princeton.edu> NNTP-Posting-Host: forum.swarthmore.edu I have the undisputedly simplest proof of Morley's Trisector Theorem. Here it is: Let your triangle have angles 3a,3b,3c and let x* mean x + pi/3, so that a + b + c = 0*. Then triangles with angles 0*,0*,0* a,b*,c* a*,b,c* a*,b*,c a**,b,c a,b**,c a,b,c** exist abstractly, since in every case the angle-sum is pi. Build them on a scale defined as follows: 0*,0*,0* - this is equilateral - make it have edge 1 a,b*,c* - make the edge joining the angles b* and c* have length 1 - similarly for a*,b,c* and a*,b*,c a,b**,c (and the other two like it) - let me draw this one: X A Y Z C Let the angles at A,X,C be a,b**,c, and draw lines from X cutting AC at angle b* in the two senses, so forming an equilateral triangle XYZ. Choose the scale so that XY and XZ are both 1. Now just fit all these 7 triangles together! They'll form a figure like:- B P Q X A Y Z C (in which the points X,Y should really be omitted). The points Y,Z are what I meant. To make it a bit more clear, let me say that the angles of APX are a (at A), b* (at P), c* (at X). Why do they all fit together? Well, at each internal vertex, the angles add up to 2pi, as you'll easily check. And two coicident edges have either both been declared to have length 1, or are like the common edge AXZ of sorry - AX of triangles APX and AXC. But APX is congruent to the subtriangle AZX of AXC, since PX = ZX = 1, PAX = ZAX = a, and APX = AZX = b*. So the figure formed by these 7 triangles is similar to the one you get by trisecting the angles of your given triangle, and therefore in that triangle the middle subtriangle must also be equilateral. John Conway ============================================================================== From: cet1@cus.cam.ac.uk (Chris Thompson) Newsgroups: sci.math Subject: Re: Simson's Line Date: 3 Sep 1995 15:25:33 GMT Keywords: Simson line Morley triangle In article , ram@tiac.net (robert a. moeser) writes: [...] |> but! the really strange thing is that if the point X is dragged |> around the circle and Simson's line is continuously recomputed |> to form a locus by "envelope", a curious figure emerges. |> |> it looks like a three-cornered hat, kind of a triangle with curved |> sides. i am sure this has a name. the three curvy sides are each |> tangent to a side of the triangle. now, no matter how the original |> triangle is drawn, if the radius of the circle is held constant, |> the size of the "three-cornered hat" is also constant, although |> its orientation changes. it is also "equilateral". The envelope is a "deltoid", a 3-cusped hypocycloid. It is indeed intriguing that that a figure with three-fold rotational symmetry emerges from an arbitrary triangle. Those following another sci.math thread will, perhaps, not be suprised to learn that that there is a relationship to another such figure, the (equilateral) Morley triangle formed by the intersections of the trisectors of the angles of the triangle. The tangents to the deltoid (i.e. particular Simson lines) at its cusps are parallel to the medians of the Morley triangle, those at the points of the deltoid nearest the circumcentre are parallel to its sides. Chris Thompson Email: cet1@cus.cam.ac.uk