Consider four points in the plane, located at the corners of a square:
(The points are shown as slightly large circles just so you can see them better. Imagine those are actual points, with no width or height). The first question we’re going to ask is: How many distinct distances between pairs of points are there, in this arrangement?
Let’s see what this means, exactly. We have four points. Picking any two of them, we get a cetain distance separating the two. We look at those distances and count how many distinct or different ones there are. So how many are there?
In fact, there are two distinct distances here: one spans an edge of the square, and the other distance is a diagonal of the square. Now this is a fairly special property of arranging four points in this way: there are only 2 distinct distances. So our challenge question for the day is:
Apart from the corners of a square, can you find other ways of arranging four points in the plane, such that there are only two distinct distances between them?
This post is based on a talk I gave at the San Jose Math Circle on 10/1/2008 and at the Bay Area Circle for Teachers. Thanks to Tatiana Shubin, Josh Zucker and Tom Davis for inviting me to participate at the circles.
The images were created using GeoGebra. Try it, it’s great.
Tags: Math Circles, SJMC
My 8-year old daughter came up with an additional solution. She made the points colinear. Your challenge does not seem to preclude that.
@Tumblemark: you can’t have four collinear points having just two distinct distances. If the points are ABCD in order, then AB < AC < AD.