Find all configurations of 4 points in a plane with only 2 distinct values for distances between pairs of points.

From Mathematical Mind-Benders (160 pages, 2007) by Peter Winkler. The book does not contain a proof that only six configurations exist.

Diagram thanks to AffineMess.

1. AAAAAA: No possible configuration.
2. AAAAAB: The only possible configuration is two equilateral triangles sharing an edge.
3. AAAABB: Two cases (resulting in 3 configurations):
   1. All distances from some point to the other 3 points are A. Then some 3 points form an equilateral triangle. Let XYZ be the equilateral triangle. Draw a circle with center X and radius XY. The fourth point W must lie at the intersection of the perpendicular bisector of YZ with the circle. There are two such points. Both are valid configurations.
   2. No point has all 3 distances to other points as A. Then the four points must form a rhombus. Since the other two distances are BB, the configuration must be a square.
4. AAABBB: Two cases (resulting in 2 configurations):
   1. AAA corresponds to an equilateral triangle. The BBB corresponds to distances of the fourth point to the three points in the triangle. So the fourth point must lie at the intersection of the perpendicular bisectors of the edges of the equilateral triangle, which also happens to be centroid.
   2. AAA corresponds to edges { WX, XY, YZ } while BBB corresponds to edges { XZ, ZW, WY }. This configuration has a unique solution: 4 out of 5 points of a regular pentagon.

Notes: Detailed explanation of the solution: here (at AffineMess).