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Math Help - Spherical Geometry Proof Help Wanted

  1. #1
    Junior Member
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    Mar 2008
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    Cool Spherical Geometry Proof Help Wanted

    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).
    Last edited by Ultros88; August 1st 2008 at 04:50 PM. Reason: Clarification
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  2. #2
    Senior Member
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    Dec 2007
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    Anchorage, AK
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    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.
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  3. #3
    Junior Member
    Joined
    Mar 2008
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    Smile I got it.

    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) .
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