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.