How many ways can four points be arranged in a plane so that the six distances between pairs of points take on only two different values?

(In reply to by )

CeeAnne-

I made the same mistake when I first read Steve's post. First I thought he was constructing points D and D' in such a way that D' was coincident with A, and D resulted in your rhombus, but then I realized that I was mis-reading the condition AB = AD as AB = BD

Assuming the shorter edge (a) is 1 unit long, the longer edge (b) is:

Case 1 Equilateral triangle and centroid: (centroid to vertex) a = 1; (vertex to vertex) b = √3

Case 2 Rhombus: (outside edge and short diagonal) a = 1; (long diagonal) b= √3

Case 3 Square: (edge) a = 1; (diagonal) b = √2

Case 4 Trapezoid based on regular pentagon: (edge) a = 1; (diagonal) b = Φ = (1 + √5)/2

Case 5 "kite": (BD and CD) a = 1; (all others) b = 2

Case 6 isoscelese triangle and centroid (base and centroid to verteces) a = 1; (legs) b= √4.25



Every time I edit, I lose my special characters "Φ" and "√"
Edited on November 22, 2004, 1:21 am

Posted by TomM on 2004-11-22 01:07:22