Response from Prof. B.

You have set up, and partially answered, an interesting problem, and you are right that I did not make it easy to see how to solve it. That would not have been of interest to "the readership", that mythical body of Scientific American readers I was not allowed to challenge in any serious way.

As it happens, a solution can be found using yet another algebraic manipulation a little bit more complicated than simply subtracting off rational numbers from each of the coordinates to put one vertex at the origin. The idea is that we can rotate the figure about the origin by some amount so that the vertex (a,b) is sent to a point (r,0) on the x-axis. Here r^2 must equal a^2 + b^2 since a rotation preserves length. The rule for that rotation can be written down explicitly: x' = ax/r + by/r, y' = -bx/r + ay/r. Multiplying through by r gives rx' = ax + by, ry' = -bx + ay. As a check, note that if x = a and y = b, then rx' = a^2 + b^2 = r^2 so x' = r as desired, and ry' = -ba + ab = 0 so y' = 0 as desired. Now look at what happens to (c,d). The point (c,d) is rotated to (x',y') with rx' = ac + bd and ry' = -bc + ad. If a,b,c, and d are all rational numbers, then rx' and ry' would also be rational numbers. But once we have (0,0) and (r,0) as two vertices of an equilateral triangle, the third vertex is either (r/2, sqrt(3)r/2) or (r/2,-sqrt(3)r/2). The first coordinates are all right, since we would have r/2 = rx' so x' = 1/2. However we would also have sqrt(3)r/2 = ry' (or -ry'), and that would mean that -bc + ad = sqrt(3)/2 and that is a contradiction since we know from prior experience that sqrt(3) is not a rational number.

In summary, the hypothesis that (0,0), (a,b), and (c,d) are vertices of an equilateral triangle with all coordinates rational leads to the conclusion that there is a rational number whose square is 3. Since this is impossible, there is no such equilateral triangle in the coordinate plane with vertices given by pairs of numbers with rational coordinates.

Isn't is nice, therefore, that we can get vertices with rational coordinates if we let ourselves go up one higher dimension, to get (1,0,0),(0,1,0), (0,0,1) for example? We could go up to four-space to get a similar set of vertices for the 3-simplex, but we don't have to since (1,1,1), (1, 0,0), (0,1,0),(0,0,1) will work. Is that just an accident? How high a dimension do we have to enter to find rational coordinates for the five vertices of a 4-simplex? Good luck on that one.

Good work.