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Topic: Original and Modified Polygon Theorem
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steve gray

Posts: 17
Registered: 12/6/04
Original and Modified Polygon Theorem
Posted: Jan 27, 1995 6:29 PM
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Original Prime N-gon theorem:
Take a plane N-gon, G1, where N is prime. Let the vertices
of G1 be G11, ... G1N. Draw a second N-gon whose vertices are:
G2j = (G1j + G1(j+1))/2 (that is, vertices of G2 are the midpoints
of the sides of G1.) Draw a third N gon, G3, whose vertices are
given by G3j=(G2j + G2(j+2))/2. (This time instead of taking
midpoints of sides connecting successive vertices, we take midpoints
of lines connecting vertices separated by 1). Continue, each time
skipping one more vertex, until the last N gon, GN, is given by
GNj = (G(N-1)j + G(N-1)(j+N-1))/2.
Show that GN is similar to, oriented like, and
2^(1-N) the linear size of G1.
For N=3 this amounts to the original triangle G1, a first
construction on its midpoints giving G2, and a second construction
on the midpoints giving G3, which is 1/4 the size of G1.
For N=17 the size ratio is 1/65536, so if the final N-gon is
barely visible having sides .01", the starting N-gon is much larger
than the biggest piece of paper in the world.
Also the mere possibility of a proof using matrices
(as a means of manipulating vectors) shows that the N-gons do not
have to be planar, and in fact can be in any number of dimensions.
This theorem, which I found in 1961, is an interesting
combination of plane geometry and number properties, two fields
not easily connected even in this not very deep sense. It can also be
viewed as a theorem about products of certain matrices, but this
is maybe less interesting.

Modified Prime N-Gon Theorem (1963)
Instead of as above constructing polygon n+1 by averaging
two vertices of polygon N, use the average of 'b' vertices of polygon n,
having vertex indices in an arithmetic progression with difference 'd'.
For example if b=3 and d=2, vertex 'i' of the second n-gon would be
V_(n+1,i) =(1/3) (V_(n,i) + V_(n,i+2) + V_(n,i+4)). (Vertex indices
are taken mod(n).)
Theorem: If you do this n-1 times with d=1,2,...,n-1, the final
polygon is just the old, scaled by b^(1-n). For example if n=7 and b=3,
the final polygon is 3^6 = 729 times smaller than the original.
Example: consider a compound application of the above
construction, where as stated for each value of b, n-1 constructions
are done. Now apply this complete construction n-1 times, letting
b=2,3,...n-1. The size ratio from is (n-1)!^(n-1). For n=7 the ratio is
about 1 light year to 1 angstrom!
--
Steve Gray
Graphics and Geometry Consultant
Santa Monica CA





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