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
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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
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From: conway@math.Princeton.EDU (John Conway)
Newsgroups: geometry.puzzles
Subject: Re: Dictionary
Date: 15 Feb 1995 23:10:24 -0500
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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
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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