Last time I was asking about cyclic polygons with rational vertices and rational edge lengths, like the one below.
It is easy to find all points on the unit circle with rational coordinates as (cs(t), sn(t)) with rational t, using
Moreover, any pair of such points can be rotated into each other using a rational rotation R(t). Here t is again a rational parameter and not the angle of rotation. So we are interested in finding superrational rotations R(t) that are rational and displace points on the unit circle by a rational distance.
Surprisingly, the solution is quite simple: A rotation is superrational if and only if it is the square of a rational rotation.
Let’s first assume that u is rational so that R(u) is rational. The square R(u)² is then also rational. We can determine how it displaces points on the unit circle as follows: Suppose R(u) maps (1,0) into the point (c,s), which is rational. As R(Then R(u)² maps (c,-s) into (c,s), and the distance between these two points is obviously 2s, R(u)² is indeed superrational.
Now assume that R(v) is superrational, and write it as R(u)². We need to show that u is rational. Again let (c,s) be the image of (1,0) under R(u). As R(v) is superrational, we see that at least 2s and hence s must be rational, as above. Now we compute, using the matrix form of R(u), the image of (1,0) under R(u)² as (c²-s², 2cs). As R(v) is rational, 2cs must be rational. As we already know that s is rational, it follows that also c is rational. Thus R(u) is a rational rotation, and we are done.
This shows that superrational rotations form a group, namely the group of squares of all rational rotations. And this implies that a superrational cyclic polygon has necessarily all its diagonals rational…
What’s next? Probably some dreary landscapes from wintry Indiana or heart warming pictures of tea leaves. But of course, there are more questions: Can we also have superrational spherical polyhedra? I don’t know yet…