Good question. I don't know where you're getting the "5" below. There are 6! permutations of distances, 4! permutations of cities, leaving 30 possibilities to try that could potentially give different geometric arrangements. No doubt there are ways to further limit it, but it's not obvious to me what's obvious to whom about it. Dan's ambiguity involves a configuration where A, 1/2(B + C), D forms a right angle. The 6 distances can be thought of as edges of a tetrahedron; Dan's ambiguity interchanges two edges that have a vertex in common. I tried this out with Geometer's Sketchpad: draw 4 points, the 6 edges connecting them, and the 6 midpoints, and the 6 circles with the lines as diameters. Color opposite edges and their circles with like colors, e.g. R, G, B. You want to arrange so red circles pass through red midpoints, etc. You can readily arrange both red circles to pass through red midpoints, or one red circle to pass through a red midpoint and one green circle to pass through a green midpoint. This gives at least 3 configurations with the same set of distances. For one approximate solution, GSP measures with no duplicate distances. I don't see why you couldn't have more, but it's hard to verify this with this simple GSP experiment, since one would expect it to happen on a codimension 3 set, but you can only control 2 parameters at a time by moving a vertex with the mouse. Bill On Nov 26, 2006, at 10:01 PM, Fred lunnon wrote:
So why stop at ambiguity? In principle, it might be possible to choose the 6 (real) edge lengths to provide up to 5 essentially different distance charts. However ...
Conjecture: if there are 3 or more assignments of 6 given lengths to edges, all consistent with planarity, then at least one pair of edges has equal lengths; the resulting symmetries render all charts indistinguishable.
Now why this might be true? WFL
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun