Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Another simple-seeming theorem with no proof - yet?
Replies: 0

 Steve Gray Posts: 103 Registered: 7/6/07
Another simple-seeming theorem with no proof - yet?
Posted: Dec 15, 2007 11:37 AM

Five points in the plane in general convex position are labeled A-E.
No three lie in a line and no four lie in a circle. "Convex position"
means that a pentagon with vertices A-E 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 A-E 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 non-null; it may seem true but there are
counterexamples.