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Original and Modified Polygon Theorem
Posted:
Jan 27, 1995 6:29 PM


Original Prime Ngon theorem: Take a plane Ngon, G1, where N is prime. Let the vertices of G1 be G11, ... G1N. Draw a second Ngon 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(N1)j + G(N1)(j+N1))/2. Show that GN is similar to, oriented like, and 2^(1N) 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 Ngon is barely visible having sides .01", the starting Ngon 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 Ngons 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 NGon 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 ngon 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 n1 times with d=1,2,...,n1, the final polygon is just the old, scaled by b^(1n). 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, n1 constructions are done. Now apply this complete construction n1 times, letting b=2,3,...n1. The size ratio from is (n1)!^(n1). For n=7 the ratio is about 1 light year to 1 angstrom!  Steve Gray Graphics and Geometry Consultant Santa Monica CA



