On Nov 28, 2006, at 10:07 AM, Michael Kleber wrote:
I wonder whether this might be a universal construction for the case where there are three planar arrangements. That is, given (almost) any triangle ABC in the plane, Bill's method allows you to identify three points Z1, Z2, Z3 such that the three Z's distances to A,B,C are permutations of one another.
The construction I described actually gives 3 triples of Z's, since it depends on choosing two sides of the triangle, to do 180 degree rotations around. Michael's rephrasing brings it back closer to Dan's construction: given a triangle, draw the 3 medians, and draw their perpendicular lines at their feet. These three lines form another triangle, similar to the first. Any vertex of this larger triangle is the Z1 of a triple, where Z2 and Z3 are obtained by reflecting it across two of the medians. You can't do any more reflections unless Z2 say is another apex of the second triangle, but that only happens when the original triangle is isosceles.
(Bill says that sometimes he can produce a Z4 also, but that's beside the point here.) Is there only one such set of Z-distances for each ABC, or might there be more? How about for generic ABC?
No, the 4th configuration I was talking about didn't come from a Z4: in this 4th configuration, there would be no triangle congruent to ABC, these lengths would be rearranged linearly.
Anyway, now we're back to the problems Ed sent out two days ago:
Q1. Are there 6 integer road lengths that lead to distinct town
configurations?
Q2. Are there 10 road lengths that lead to distinct [five-]town
configurations?
Here's how you can get arbitrarily many towns with the same set of distances, generalizing Dan's construction: Let rho: E^2 -> E^2 be reflection through the origin, and sigma be reflection in the x-axis. Let X be any collection of points such that sigma(X) = X Y be any collection such that rho(Y) = Y, such that X U Y has no symmetry. Claim: for any subset Z of the y-axis, the distances among X U Y U Z and X U Y U rho(Z) are equal. That's because rho(Z) = sigma(Z). There are clearly some more variations of this, e.g. using other symmetries, as well as other geomtries and other dimensions --- it would be intriguing to get the full picture --- but I should stop for now! Bill