From: xpolakis@REMOVE.hol.CAPS.gr (Antreas P. Hatzipolakis)
Newsgroups: sci.math
Subject: Don Newman's Proof of Morley's Theorem
Date: 12 Jan 1999 19:21:06 -0500
Keywords: Nice proofs of Morley's Theorem

I read that:

 In a Temple Seminar (Spring 1995), Don Newman gave a beautiful proof
of
 Morley's trisection theorem that is probably in God's book.
  
http://www.math.temple.edu/~zeilberg/mamarim/mamarimhtml/morely.html

Does anyone know this proof? (Is it published? In the "Math. Intell."
perhaps?)

Antreas
==============================================================================

From: jpr2718@aol.com (Jpr2718)
Newsgroups: sci.math
Subject: Re: Don Newman's Proof of Morley's Theorem
Date: 14 Jan 1999 22:10:56 GMT

.>Don Newman gave a beautiful proof
>of
> Morley's trisection theorem

See Alex Bogomolny's most recent column at the MAA web page,  www.maa.org,
Columns, for a link to this proof, and more discussion of Morley's Theorem. 
Actually his last three columns deal with Morley's Theorem, and apparantly
there's more to come.  According to him, D.J.Newman's proof appeared in the
Mathematical Entertainments section of The Mathematical Intelligencer, vol. 18,
No. 1 (1996), p 31-32.

John

==============================================================================

Newsgroups: sci.math
From: eclrh@sun.leeds.ac.uk (Robert Hill)
Subject: Re: Don Newman's Proof of Morley's Theorem
Date: Thu, 14 Jan 1999 17:52:23 +0000 (GMT)

In article <vnbfahxzkvvw@forum.swarthmore.edu>, xpolakis@REMOVE.hol.CAPS.gr (Antreas P. Hatzipolakis) writes:
> I read that:
> 
>  In a Temple Seminar (Spring 1995), Don Newman gave a beautiful proof
> of
>  Morley's trisection theorem that is probably in God's book.
>   
> http://www.math.temple.edu/~zeilberg/mamarim/mamarimhtml/morely.html
> 
> Does anyone know this proof? (Is it published? In the "Math. Intell."
> perhaps?)

I don't know the answers to your questions, but two or three years ago 
somebody posted to sci.math a simple proof of Morley's theorem due to
J.H. Conway, which works by taking an equilateral triangle of unit side,
and 6 other triangles of prescribed sides and/or angles, and showing that
they must fit together to form the Morley diagram.

Here is my rephrasing of Conway's proof.

In order to prove the theorem it is enough to show that every triangle
is similar to some triangle for which the theorem is true.  We'll do a
bit more.  For any triangle ABC and any equilateral triangle PQR, we'll
construct a triangle XYZ, similar to ABC, and having PQR as its Morley
triangle.

Let the angles of ABC be 3a, 3b, 3c, where a + b + c = 60 degrees.
Construct X on the side of QR away from P, so that angles

        QRX = 60 + b,   RQX = 60 + c.

Similarly construct Y, Z, away from Q, R respectively, so that

        PRY = 60 + a,  RPY = 60 + c,  PQZ = 60 + a,  QPZ = 60 + b.

Then I claim that XYZ is the required triangle.

For a start, we obviously have

        QXR = a,  PXR = b,  PZQ = c.

The main thrust of the proof is to show that RXY = a.  If we can show
that, then similar arguments will show that QXZ = a, RYX = b, etc,
proving both that the lines XR etc are trisectors and also that 
the triangles XYZ and ABC are equiangular and hence similar.

To show that RXY = a, we'll work in the triangle RXY.  We'll need its
sides RX, RY.  Without loss of generality the sides of PQR are 1.
Then by the sine rule in QRX,

       RX/sin(60+c) = 1/sin(a),

and similarly in PRY, so    

       RX = sin(60+c)/sin(a),  RY = sin(60+c)/sin(b).

We could now get XY by the cosine rule in triangle RXY (knowing angle XRY),
and then use the sine rule to get angle RXY, but the following is easier.
Let a line through X, near XY, making an angle a with XR, meet RY at Y'.  
Then XY'R = b, so we can compute RY' by the sine rule in RXY', and
we find that Y' coincides with Y.  Therefore RXY = a and the proof is complete.  

-- 
Robert Hill

University Computing Service, Leeds University, England

"Though all my wares be trash, the heart is true."
  - John Dowland, Fine Knacks for Ladies (1600)