On 11/29/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
Sadly nobody has answered the question that was actually asked below, so I'll take up the challenge. As remarked in my last posting, I've found a handful, with the added constraint that all 20 possible triangles formed from the distances are proper. The examples I gave there were [AB, BC, CA, BD, AD, CD] = [15, 19, 20, 18, 9, 13]; [15, 19, 20, 13, 26, 18].
Fred Lunnon
On 11/30/06, franktaw@netscape.net <franktaw@netscape.net> wrote:
Before people go too far looking for such charts, I have a question: is it possible to have 4 points in a plane, no 3 colinear, so that all 6 distances are integers? And with the additional constraint that no 2 distances are equal?
Franklin T. Adams-Watters
Your arrangements are planar (so solve Franklin T. Adams-Watters question). I didn't check carefully, but it appears that the (6 take 3) triangles formed by the distances are non-degenerate, but so far as I can tell neither arrangement can be rearranged in a planar way to give an integer solution to the ambiguous towns problem. I didn't check carefully, so maybe I am missing the point of these arrangements or something. JB