to solve the problem, we start by noting that four points (a, b, c, d) have 6 distinct pairwise distances: 1, 2, 3, 4, 5, 6. given (ab=6) (the longest distance), we need to find (cd).
key observations:
- (a) and (b) are the farthest apart ((ab=6)). for any other point (p), (ap pb \geq 6) (triangle inequality).
- for point (c), possible pairs ((ac, bc)) from 1-5 must sum to at least 6. a valid pair is ((ac=5, bc=1)) (sum=6, collinear with (a) and (b)).
- remaining distances: 2, 3, 4. for point (d), ((ad=2, bd=4)) (sum=6, collinear with (a) and (b)).
coordinate setup:
assign (a(0,0)) and (b(6,0)):
- (c) is at (5) (since (ac=5), (bc=1)).
- (d) is at (2) (since (ad=2), (bd=4)).
(cd = |5-2| = 3).
answer: (\boxed{3})
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