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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