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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

20.0. "Prettiest theorem?" by RANI::LEICHTERJ () Tue Jan 31 1984 03:50

Here is a submission to the "prettiest theorem" contest.  I'm afraid I no longer
remember whose theorem it is, exactly; it's due to a British mathematician who
worked during a curious phase in British mathematical history, around the turn
of this century, when there was a great deal of interest in new theorems of
the "classical (Euclidean) geometry" sort.  As far as I know, this theorem -
Morley's Thm. suddenly comes to mind as a name - has no significant applica-
tions whatsoever.

Take any three (non-colinear) points in the plane.  They lie on a circle; the
center of that circle is called the center of the three points.

Take any four points (in general position - i.e. no 3 are on a line; this will
be assumed from here on) in the plane.  Every subset of three points determines
a circle.  Thm:  The 3 circles so determined meet at a point; that point is
called the center of the 4 points.

Take any 5 points; every subset of 4 has a center.  Thm:  The 5 centers so
defined fall on a circle, whose center is called the center of the 5 points.

Etc., for all n.  For odd n, you get n points on a circle; for even n, you get
n circles meeting at a common point.

The original (very difficult) proof uses standard Euclidean techniques.  At
least the n=4 case is completely trivial if you work in the complex plane;
I remember working out the proof years ago.  (I think the general case isn't
too hard in the complex plane, eitehr.)
							-- JErry
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20.1RANI::LEICHTERJTue Jan 31 1984 03:533
Typo correction:  The n=4 case as explicitly described:  That should be "THE
4 [not 3] circles so determined".
							-- J
20.2RAINBO::GREENWOODTue Jan 31 1984 19:0418
My entry for the prettiest theorom came from an AI program proving 
that the base angles of an isosceles triangle are equal. At elementary 
school you learnt it by dropping the perpendicular, the program came 
up with.

     A
     /\
    /  \
   /    \
 B ------ C

AB = AC (given)
AC = AB (given)
BC = CB
Therfore triangle ABC is congruent with triangle ACB
Therefore angle ABC = angle ACB    Q.E.D.

Tim
20.3HARE::STANTue Jan 31 1984 23:275
The first theorem is due to Clifford.  It is not Morley's Theorem.

(Morley's theorem is the old classic that the trisectors of the
angles of a triangle meet in pairs at three points that determine
an equilateral triangle.  Another beautiful theorem!)
20.4RANI::LEICHTERJWed Feb 01 1984 00:572
re:  .2   That pretty proof actually dates back to Euclid.
								-- Jerry