Okay, I had another drip of time... WPT:
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.
Oof, of course; very nice. What's more, starting with a 3-4-5 and rotating about the midpoints of the length 3 and 5 sides gets you one extra rational (integer, even!) value; you end up with {3, 4, 5, 13, 5*sqrt(145), sqrt(7081)}. Jim Buddenhagen likewise mentioned four integer sides. Jim, is yours any smaller than this one? Going back further, I just reread Fred Lunnon's mail from two days ago:
Let the road lengths be arranged [AB, BC, CA, BD, AD, CD], where A,B,C,D denote cities. Then the follwoing two charts are distinct and planar, and both permute the same set of integers:
[5, 8, 13, 11, 9, 17] [5, 17, 13, 11, 9, 8]
I'm not sure why you decided it was somehow inferior, Fred! Yes, the first hexad has three cities collinear, but when handed a problem about distances along roads, nothing seems more likely than that some city is along the route between two others. Did anyone verify this, per Fred's request? It certainly seems like the prettiest configuration seen so far. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.