Post by telemeter on Mar 19, 2020 16:53:39 GMT
I for some reason liked doing this question, so share the answer, though I cannot match the formatting masterclass shown in the published answers
i) Let a polygon have n sides and as per the hint be divided into triangles whose vertices meet at an internal point
By (9) of p279, Angle excess AE = k(Area) where k is curvature. Here k=1 as it is we are on the unit sphere
The angle sum of the n triangles AS(triangles) = n*pi + n*AE
The AS(polygon) = AS (triangles) - 2*pi. (we deduct the sum of the angles at the internal point as they are part of the AS(triangle) but not part of AS(polygon))
Thus, AS(polygon) = n*pi + n*AE(triangles) - 2*pi
But n*AE(triangles) = Area (polygon) by (9) p279, so
AS (polygon) = (n-2)*pi + A(Pn) or A(Pn) = AS(polygon) - (n-2)*pi as required. (1)
ii)
Assume there are F polygons of differing number of sides n(i) (i=1 to F)
Consider the edges. Each edge is shared by two polygons. Therefore the number of edges, E = 0.5 * sum (n(i)) (i=1 to F) (2)
Now, considering the vertices, each vertex will be the meeting point of internal angles of polygons that sum to 2*pi. Thus the total angle sum of all the polygons, if there are V vertices
AS (all polygons) = V * 2 * pi
Applying (1) to all the polygons and adding then up...
4*pi = V *2*pi - [sum (n(i) (i= 1 to F) - 2F] * pi
Substituting from (2)
4*pi = V*2*pi - 2*E *pi + 2*F*pi
Dividing by 2*pi and rearranging...
F-E+V = 2 as required.
telemeter