Good evening everyone.

Challenge problem:

*Arrange four points on a plane such that the distance between any 2 points is one of 2 possible lengths.*

There are supposedly at least 5 arrangements that satisfy this, but alas, I can only find the following 4:

1. One point on each corner of a square.

2. One point on each vertice of an equilateral triangle, and one point in the center of the triangle.

3. One point on each corner of a rhombus which has one diagonal congruent to it's sides. (can also be thought of as 2 equilateral triangles sharing a common side).

4. An equilateral triangle, with a fourth point centered on an arc drawn between two point of the triangle, the arc center (radius point) being the remaining point of the triangle.

Any other solutions completely elude me. I'll wait (with baited breath) to see what other solutions the brilliant minds that frequent this forum can come up with.

Thanks, and best regards, Hal