Okay, I think I have one. The six integer distances are {1449, 2706, 2775, 4097, 5402, 6253}. More constructively, consider the isosceles triangle ABC, A=(0,0), B=(2640, 594), and C=(4680, 4147), with edge lengths {2706, 4097, 6253}. Then the points D1=(0, 1449) and D2=(2640, -855) are both at distances {1449, 2775, 5402} from ABC's vertices. Yes, the fact that all points invovled have integer coordinates is indeed a result of how I found it. I'll describe my search below, but it's kind of boring, so let me first point out that (a) there's no reason to believe that this is the minimal possible solution, and (b) I still want to know whether there's a solution that doesn't preserve a triangle (which my search technique could never find). I tried a search based on a geometric construction that I at first thought was the one Dan suggested last weeked, but rereading I realized it's not. Oh, but it is one half of Bill Thurston's method that gives triply-planar distance sets. Consider a triangle PQR. Let M be the midpoint of QR, and let L be the line though M perpendicular to the median PM. Then any point S on L is the fourth point (along with P,Q,R) of an ambiguous set, because P can be replaced by its reflection through M. So of course you start with an integer-sided triangle, but the distance d(Q,S) is unsurprisingly ugly-- it involves nested square roots, where the inner one is the square root of the altitude from R to PQ. Aha, says I: what if we restrict our search to cases where that altitude is also an integer? Which is to say, suppose our original triangle PQR was made out of two Pythagorean triangles glued together along a common leg. We can then keep pushing our luck, and demand that PQS likewise be made of two Pythagoreans, glued along the altitude from S to PQ. This is a space of configurations amenable to dumb computer search. It's easy to list all Pythagorean triples containing a given leg (of which there are only finitely many, of course). Gluing distinct pairs, it's easy to list all integer-sided triangles with given integer altitude (um, "height", I guess I should say). Now search over all pairs of heights (r,s) for two such glued triangles with those heights and the same base PQ. Put them on top of each other in all possible ways, and test for two things: a) The median PM is perpendicular to SM (distance-ambiguous), b) The one unconstrained distance, d(R,S), is an integer. Of course d(R,S) is already the square root of an integer. Configurations satisfying (a) -- which give rise to ambiguous sets of six distances, at least 5 integral -- aren't too hard to come by. The first one, with max(r,s)=176 and distances {100, 220, 20 sqrt(58), 75, 35, 185}, shows up immediately, then come three with max(r,s)=192, etc. I left this to run overnight, and the 544th set, with max(r,s)=4680, had an integer sixth distance. --Michael -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.