## Stereographic Projection is Conformal

In Hana Kitasei and Tim Reckart’s animated video, Summer Vacation, Angela says that stereographic projection preserves angles. Here’s a proof.

First, let’s recall what stereographic projection is. Angela uses stereographic projection to make a map of the earth. A light source at the North Pole projects the continents onto a flat screen tangent to the earth’s South pole. In other words, Angela is making a drawing of the earth’s surface as seen from the North Pole (you have to imagine that you can see through the earth) on a huge canvas.

In this scene from Summer Vacation, Angela projects the continents onto a flat screen tangent to the South pole via a light source at the North pole.

The process distorts distances and areas, particularly as you near the North Pole. However, Angela claimed that angles are preserved and we’re going to prove this now.

First, let’s observe that if we move the screen back and forth along the North-South axis (keeping it oriented perpendicular to that axis), the different maps obtained will all be similar to each other. Bringing the screen closer to the North Pole makes the whole map smaller and moving the screen further away enlarges everything. Since scaling preserves angles, if we can show that stereographic projection to any screen perpendicular to the North-South axis preserves angles, then we will have shown that Angela’s stereographic projection preserves angles. (Aside: Stereographic projection is usually done onto a screen that contains the equator.)

Next, let’s exploit the symmetry of the sphere. The whole setup is symmetric with respect to rotations around the North-South axis. So we only need to concern ourselves with the projection near one specific line of longitude. Pick a longitudinal line and select a point $P$ on it. Imagine two great circles passing through $P.$ They form some angle $\alpha.$ We have to show that the angle between the projections of these two great circles is also equal to $\alpha.$

Notice that if we’re able to show the result when one of the great circles is the line of longitude containing $P,$ then we’d be done because all other angles can be obtained as the difference between two angles which have this longitudinal line as a side. So let’s assume that one of the sides of our angle is the longitudinal line containing $P.$

The figure below shows the essentials from a vantage point hovering roughly over the North pole and looking down the North-South axis. We’ve indicated the angle $\alpha$ by lines tangent to (and in the same plane of) the great circles that define it. For reasons that will soon become clear, these two lines are used to form a (blue) right triangle with a third line that passes directly below the South Pole and is perpendicular to the plane of our longitudinal line. Let $P,$ $Q,$ and $R$ be the vertices of this right triangle where $R$ is the vertex at the right angle.

Let’s stereographically project angle $\alpha$ onto a screen perpendicular to the North-South axis at the point $R$:

(By the way, sometimes it’s more confusing to try to follow someone else’s geometric proof than it is to construct the proof for yourself. You might want to ignore the next 3 paragraphs and see if you can finish the proof on your own.)

We have to show that $\alpha = \beta$. Right triangles $PQR$ and $P'QR$ share the common side $QR,$ so if we can show that $PR = P'R,$ then we’d be done. We’re going to show that by showing that triangle $RP'P$ is isosceles.

Observe that angle $RPC$ is right because $RP$ is tangent to the longitudinal line. Thus, angles $RPP'$ and $CPN$ are complementary. Angle $P'RN$ is right because the screen was set up to be perpendicular to the North-South axis. Therefore, angles $RP'P$ and $RNP'$ are complementary too.

Because $CP$ and $CN$ are both radii of the earth, triangle $PCN$ is isosceles and angles $RNP'$ and $CPN$ are congruent. Since angles $RPP'$ and $RP'P$ are complementary to congruent angles, they must be congruent to each other. Therefore, triangle $RPP'$ is isosceles, $PR = P'R,$ and we’re done!