Spherical Geometry Proof Help Wanted

Ultros88 · #1 $Old$ 08-01-2008, 03:03 PM

How would one go about proving that each point on a sphere not within an n-sided convex spherical polygon nor its antipode is contained in (n-2) lunes defined by the angles of said polygon?

I think it has something do with the fact that the two great circles that intersect in a vertex include two of the other points of the polygon and thus the lunes described by the angles at those other two points are disjoint with those of the vertex under consideration. I can't seem to seal the deal though and I could use some help.

Thanks,
Ultros

What I'm trying to do at the end of it all is to prove that the sum of the angles of the polygon = pi*(n-2) + Area(polygon)/R^2. I can get everything except the crucial step of why there is this factor of pi*(n-2).

TwistedOne151 · #2 $Old$ 08-06-2008, 10:54 PM

Well, $\pi(n-2)$ is the sum of the interior angles of an n-sided polygon in a plane (see limit of your formula as $R\to\infty$ ), so that might have something to do with it. I'm not sure if that helps, but does that give you any ideas?

--Kevin C.

Ultros88 · #3 $Old$ 08-11-2008, 11:15 AM

Let the angles of a polygon be $a_1, a_2, ... a_n$ . Then the polygon may be divided into $(n-2)$ triangles. Let the sum of the angles of each triangle be denoted by $T_1, T_2, ... T_{n-2}$ . Let the area of each triangle be denoted by $A_1, A_2, ... A_{n-2}$ . Then $\sum_{i=1}^{n-2} T_i =$ sum of angles of polygon, and $\sum_{i=1}^{n-2} A_i =$ area of polygon. Since $T_i = \pi + \frac{1}{R^2} A_i$ , summing over all the $(n-2)$ triangles gives $\sum_{i=1}^{n} a_i = (n-2)\pi + \frac{1}{R^2} Area(polygon)$ .