Theorem. Let P be a convex polyhedron with V vertices, E edges, and F faces. then
We will start the proof by choosing a point C inside P. Since P is convex, the line segment joining C to any point inside the polyhedron P, or on P itself, lies entirely within P. Next we choose a radius R so large that the sphere with center C and radius R contains the polyhedron P. We will map the polyhedron onto the sphere using central projection from C. This means that for each point on the polyhedron, we take the line from C through the point, and map the point to the intersection of this line with the sphere.
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 V vertices, E edges and F faces, just like P does. Furthermore, since the center of the sphere was chosen inside P, the spherical polyhedron covers the entire sphere. Hence the spherical polyhedron is a division of the sphere into F disjoint spherical polygons, which we will call Q1, ... , QF.
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 Qi. Actually we will use the extension of Girard's Theorem to spherical polygons. Let ei denote the number of sides of Qi. Then
sum of angles of Qi = (ei - 2) + area(Qi)/R2
Summing this over the faces we get
sum of angles of Qi = (ei-2) +area (Qi) /R2
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.
(ei-2) =ei -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
ei = 2 E.
The second sum is simply
2 = 2 F
Finally, since the polygons are disjoint and cover the entire sphere we have
area(Qi) /R2 = area(S)/R2 = 4.
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.|
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