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?

Given at most two values for 6 distances, one value must occur at least 3 times.

If a value occurs 4 or more times, a equilateral triangle or rhombus is present.

Suppose we have an equilateral triangle. The fourth point can be the second distance from all three points, in the center of the equilateral triangle. If the fourth point is the first distance from a point, then there are three positions it can take, two on the perpendicular bisector of the other two points (the isosceles triangle with center and the kite), or it can form the two-equilateral triangle/rhombus. Suppose we have a rhombus; we have the previously mentioned equilateral rhombus or the square. Thus, the presence of an equilateral triangle or rhombus generates five arrangements.

What is left is the case where each distance occurs three times and no equilateral triangle occurs. Think of the four points chained by three equal length line segments. There are the “C” and  “Z” cases. For the “C”, the four points sit on a regular pentagon, yielding a lovely trapezoid. A little thought forces the “Z” case back to the equilateral rhombus.

Given four distinct points, this gives exactly 6 cases.

Posted by owl on 2004-11-22 03:09:59