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Another simpleseeming theorem with no proof  yet?
Posted:
Dec 15, 2007 11:37 AM


Five points in the plane in general convex position are labeled AE. No three lie in a line and no four lie in a circle. "Convex position" means that a pentagon with vertices AE is convex by the standard definition. Draw all 10 distinct circles (ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE) through these points. Call this configuration X(5,10). Each point AE in X(5,10) is assigned a "depth" d, according to how many circles strictly enclose it. (It is true but not obvious that d(A)+d(B)+d(C)+d(D)+d(E) = 10.) Relabel the points in clockwise or counterclockwise order such that d(A)=0 and d(B)=4. Show that this is always possible, and further that d(C)=1, d(D)=2, and d(E)=3. (The sequence d(A),
,d(E) is 0,4,1,2,3, NOT 0,4,3,2,1.) Finally, show that any configuration X(5,10) is isomorphic to any other. Here isomorphic means that any X(5,10) can be transformed to be geometrically similar to any other with no points crossing any circles, with reflection allowed. In other words, topologically there is exactly one X(5,10) configuration. I have established this result by means of many experiments but have not been able to prove it. Do not try to prove that the intersection of all 10 circle interiors is nonnull; it may seem true but there are counterexamples.



