**Theorem. **Let ** P** be a
convex polyhedron with

We will start the proof by choosing a point ** C**
inside

This is illustrated in the accompanying figure. The original polyhedron is indicated in red, and the blue lines show how the vertices are mapped to the sphere. The black lines are reference great circles on the sphere, and the purple is the image of the polyhedron.

A good way to visualize the central projection is to consider what
happens if we put a light at the center of the sphere. Then the
result of central projection is the
shadow of the polyhedron on the sphere.

It is important to understand what
happens to an edge of the polyhedron under central projection. An
edge is a segment of a line. That line and the center determine a
unique plane. The line segment from ** C** to
any point of the edge lies completely in this plane, and the plane
intersects the sphere in a great circle. Hence the image of an edge
is a segment of a great circle on the sphere. This means that each
face of the polyhedron is mapped into a spherical polygon, and the
polyhedron is mapped onto a spherical polyhedron which is simply a
curved image of the original. (In the figure this is the purple
configuration.) The spherical polyhedron has

The second figure shows the marking of the sphere determined by the polyhedron in the first figure.

Let's apply
Girard's Theorem to the polygon
** Q_{i}**. Actually we will use the
extension of Girard's Theorem to
spherical polygons. Let

sum of
angles of ** Q_{i} = (e_{i} -
2)** +
area

Summing this over the faces we get

sum of angles of
** Q_{i} = (e_{i}-2)**
+area

We will examine each of these sums. In each case we will be able to find the sum by geometric means.

As complicated as it looks,
the first sum is just the sum of all of the angles in the spherical
polyhedron. Let's reorder
the sum. Instead of grouping the angles by the face they belong to,
let's group them by their vertex.
Since the spherical polyhedron covers the sphere, at any vertex the
angles with that vertex fill out the entire 2 radians. Thus the angles at each vertex
contribute 2 to the sum, and multiplying by the
number of vertices we see that the first sum is
equal to 2 ** V**.

We split the second sum into two sums.

** (e_{i}-2)**
=

The first sum is times the total number of all of the edges of all of the faces. Notice that each edge of the spherical poyhedron separates two faces. Since we are summing over the faces, each edge of the polyhedron is counted twice in this sum. Therefore

** e_{i}
** =
2

The second sum is simply

`2 = 2 F`

Finally, since the polygons are disjoint and cover the entire sphere we have

area** (Q_{i})**
/

Putting this all together we get

`2
V = 2 E
- 2 F + 4.
`

Dividing by ** 2**,
and rearranging we get
Euler's formula

`V - E + F = 2.`

The previous section discusses some consequences of Girard's Theorem. | |

Table of Contents. |

url: http://math.rice.edu/~pcmi/sphere/gos6.html John C. Polking <polking@rice.edu> Last modified: Thu Apr 15 09:22:04 Central Daylight Time 1999