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Topic: A Theorem concerning the Trisectors of a Triangle
Replies: 24   Last Post: Nov 10, 1998 12:19 AM

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 Den Roussel Posts: 8 Registered: 12/6/04
Re: A Theorem concerning the Trisectors of a Triangle
Posted: Nov 10, 1998 12:19 AM

Thank you all for your contributions. Special thanks to Ray Roussel and
John
Conway.

The theorem has to be amended to read;

"The angular trisectors of a triangle cut the circumcircle at 18 points
which form 3 sets of 9 parallel lines centered on the circumcenter and
parallel to the Morley system."

This array is rich in equilateral triangles. Much order and symmetry can
be seen in the trisector indices. The intersection points of the Morley
system correspond to secant lines and chords in this system. When we
draw lines connecting various cuts, interesting things happen. I
especially like the triangle formed by connecting the auxillary
trisector cuts, a3a6-b3b6-c3c6. This forms a triangle similar to the
original. Likewise, for a1a4-b1b4-c1c4. The system is centered on the
circumcenter. The central triangle is b1c2-b3c4-b5c6. Curious that there
is no third vertex involved for what should be a critical triangle.

The 54 equilateral triangles among the 108 intersection points, as
charted by the Morley team at Abracadabri
good stuff here), are all accounted for in this configuration. All
equilateral triangles in the Morley system appear as equilateral
triangles in this system, yet there are some equilateral triangles in
this system that are not equilateral triangles in the Morley system.

There are 3 pairs of equilateral triangles on the circumcircle. These 3
pairs are special because when the circle is brought into play we only
need 1 node (i.e., the origin of a 60 degree axis, e.g., (a1a3a5)) to
produce an equilateral triangle on the circle. The chord connecting this
vertex to the corresponding vertex of its companion triangle (b2b4b6),
another 1 node creation, forms a line parallel to a Morley edge. These 9
chords in the 3 pairs of triangles form a miniature Morley grid of 3
sets of 3 parallel lines intersecting in sixes at 27 points. These
triangles might be considered as pure lighthouse triangles or as "les
etoiles pure", from theorems by Richard Guy and Andre Viricel.

For Divine reasons, trisections produce equilateral triangles.
Sometimes the converse is also true. Notice that:

"In a circle, if an equilateral triangle is erected on the middle third
of a trisected arc, then the other arc is also trisected."

This can be seen by extending the sides of the triangle to the circle to
form a pair of equilateral triangles.

A quick 3 stroke trisection results from this statement.

Stroke 1: Draw a circle of radius R.

Stroke 2: With center on the circle and radius <sqr(3)R, draw an arc
cutting the circle at A and B.

Stroke 3: With B as center and the same radius, draw an arc cutting the
circle at C and D and cutting the arc AB at E.

The angle ACD is trisected by the angle ECD.
The supplement of the trisected arc shows itself as the diagonal angle
in the pure isoceles trapezoid (3 equal sides). This is what we have
trisected. Of course you can NOT trisect a given angle.

This demonstrates that we can form pairs of equilateral triangles with
any two points on a circle as base, as long as the base is < sqr(3)R.
Our 18 cutting points provide us with many pairs of equilateral
triangles. The pairs formed on the chords (a1a2-a4a5), (b1b2-b4b5),
(c1c2-c4c5) have these chords parallel to the sides of the original
to the sides a,b,c of the original triangle and therefore, similar.

We may now have enough triangles to properly welcome in the centennary
year of Frank Morley's observation. It is probably best to stop here,
remarking that the original question is still open.

How is all this related to Morley's theorem?

Thanks again.

Den Roussel

Date Subject Author
9/12/98 Den Roussel
9/12/98 John Conway
9/13/98 Larry Cusick
10/27/98 John Conway
9/13/98 steve sigur
9/14/98 John Conway
9/15/98 John Conway
9/15/98 Richard Guy
9/15/98 John Conway
9/15/98 Richard Guy
9/15/98 Richard Guy
9/16/98 Floor van Lamoen
9/16/98 John Conway
9/17/98 Floor van Lamoen
9/17/98 John Conway
9/17/98 Floor van Lamoen
9/17/98 Russell Towle
9/17/98 John Conway
9/17/98 Russell Towle
9/17/98 Douglas J. Zare
9/19/98 Russell Towle
9/20/98 John Conway
9/20/98 John Conway
9/18/98 Antreas P. Hatzipolakis
11/10/98 Den Roussel