## Spherical Construction Problem

On the surface of a sphere, and given a length x (on the sphere), construct (using compasses and a ruler which is curved to draw great circles on the sphere) a regular quadrilateral on the surface of the sphere. The diagram shows what we are attempting. We start with one segment x. As you can see, the vertex angles are not 90 degrees, and they will vary depending upon the length of x (or the size of the sphere). Also, a line segment (part of a great circle) connecting the midpoints of two opposite sides of our quadrilateral is somewhat greater than x. Try to invent a construction, before reading further. See Geometric Constructions.

Solution:

Draw the perpendicular bisector of AB. Let's call the midpoint of AB M. Through some point N on this perpendicular bisector (MN) draw a line perpendicular to MN. This line meets our original line (AB) at two points (the poles of line MN), one of which we will label O. Draw AP=AO so that P is on line MN. Draw lines AP and BP. On line PA, we can draw segment AD=AB, and on PB draw BC=AB. Draw segment CD, and we are done.

Is there some way to construct this quadrilateral without going all the way out to a couple of the poles? We can do it by trial and error, getting closer and closer to our quadrilateral. But that is not how constructing is done. Let's say that we try it on a sphere of unknown size. The size of the vertex angles could be any value from 90 degrees (if x=0) to 180 degrees (if x is 1/4 the circumference of the sphere). We cannot even construct this angle, without knowing the size of our sphere. So we do have to figure out where the poles of a line are, in some way.

See Non-Euclidean Geometries. The above construction tells us how to draw a regular quadrilateral, given the length of one side, in Elliptical (Riemann) Geometry. We have similar poles there. There are no such poles in Hyperbolic (Lobachevsky) Geometry. So maybe our construction is impossible there? Of course, in Euclidean Geometry, our vertex angles are 90 degrees. And we can easily construct the appropriate square.