# Four Points in the Plane

Find all configurations of four points in the plane, such that the pairwise distances between the points take at most two different values.

**SOLUTION**

All 5 configurations are shown below: a square, a rhombus with 60°-120°-60°-120°, an equilateral triangle with its center, an isosceles triangle with 75°-75°-30° and its center, and a quadrilateral with 75°-150°-75°-150°.

**email**.

Here is the solution:

The problem states there are four points such that the distances between any two points of these points can only be one of two different values. If we call the four points A, B, C, and D, then this problem can be stated as the line segments AB, BC, CD, DA, AC, and BD take on only two unique values.

The shape ABCD must have equal sides, or AB = BC = CD = DA. We know this because if you take any quadrilateral that does not have equal sides, then it is not possible to make both diagonal lengths equal to one of the side lengths. This narrows it down to a rhombus.

Further, since the distances between the points can only be one of two values, we know that the diagonals are equal to each other. This makes the shape a square.

Hi Jim. You are correct that a square is a shape that satisfies the conditions. However, there are some other configurations for the points that also work:)