Stereographic Projection
Page 1
Page 2
Page 3
Page 4
Page 5
Page 6
Page 7
Page 8

Stereographic Projection is a map from a sphere to a plane where points on the sphere are projected along a line from the North Pole of the sphere onto the plane tangent to the sphere's south pole.

Actually, it doesn't matter if the plane is tangent to the sphere, but it must be parallel to this tangent plane and lie below the North Pole.

The great thing about stereographic projection is that it preserves angles (i.e. is a conformal mapping), and maps circles on the sphere to circles on the plane.

I want to show that the image of a circle on the sphere under stereographic projection is circle on the plane.

To do this, we can use the following chain of facts, starting with the fact that for the angle marked above is a right angle, no matter where you put the point P on the circle. I will not prove this, but you can find Euclid's Proof here:

Euclid's Elements Proposition 31

Try moving around the point in the picture to see for yourself!

Now we can use this fact to show that for a circle, the ratio PQ/AQ = QB/PQ, i.e. triangle PAQ is similar to triangle PQB.

Another way to write this is PQ^2 = (AQ)*(QB).

We also want to show that only for points on a circle is this true. So if we have a set of points P s.t. for A and B in the plane, PQ^2 = (AQ)*(QB), then we know these points form a circle. (Q is the point at which a line from P perpendicularily intersects the line defined by A & B.)

This is not hard to do if we think of the pythagorian theorem on this picture.

So now let us turn back to our projection. We want to show that the image of the circle on the sphere projected onto the plane is a circle.

Next we create another plane perpendicular to the axial plane we sliced everything with, so that the following similar triangles are created in the axial plane:

Next we create another plane perpendicular to the axial plane we sliced everything with, so that the following similar triangles are created in the axial plane:

So the goal is to show the projection in the plane is a circle, and we're trying to do that by using the fact that for a circle, PQ^2 = (AQ)*(BQ).

We already know the red shape is a circle since it is in the cone and parallel to the one on the sphere. So we already know that PQ^2 = (A'Q)*(B'Q).

So we'll be done if we can show (AQ)*(BQ) = (A'Q)*(B'Q).

So the goal is to show the projection in the plane is a circle, and we're trying to do that by using the fact that for a circle, PQ^2 = (AQ)*(BQ).

We already know the red shape is a circle since it is in the cone and parallel to the one on the sphere. So we already know that PQ^2 = (A'Q)*(B'Q).

So we'll be done if we can show (AQ)*(BQ) = (A'Q)*(B'Q).

And we can use the similar triangles to show this!

This figure should show that (AQ)*(BQ) = (A'Q)*(B'Q). So PQ^2 = (AQ)*(BQ) and the image is indeed a circle!

Special thanks to all the organizers and fellow students, all of whom were willing to help me answer my many questions about all this stuff.