By Edwin H. Spanier
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In particular therefore, the reﬂection group G is ﬁnite. We see from the Gauss–Bonnet theorem, proved in the next section, that the area of the triangle is π(1/p + 1/q + 1/r − 1), and hence that 1/p + 1/q + 1/r > 1. The only solutions here are: • • ( p, q, r) = (2, 2, n) with n ≥ 2. The area of ( p, q, r) = (2, 3, 3). The area of is π/6. is π/n. 34 SPHERIC AL GEOMETRY • • ( p, q, r) = (2, 3, 4). The area of ( p, q, r) = (2, 3, 5). The area of is π/12. is π/30. The fact that S 2 (of area 4π ) is tessellated by the images of under G then implies that G has order 4n, 24, 48 and 120 in these cases.
Show that the composite of the corresponding reﬂections Rl Rl is a rotation about P through an angle 2α. If l, l are parallel lines, show that the composite is a translation. Give an example of an isometry of R 2 which cannot be expressed as the composite of less than three reﬂections. Let R(P, θ ) denote the clockwise rotation of R 2 through an angle θ about a point P. If A, B, C are the vertices, labelled clockwise, of a triangle in R 2 , prove that the composite R(A, θ )R(B, φ)R(C, ψ) is the identity if and only if θ = 2α, φ = 2β and ψ = 2γ , where α, β, γ denote the angles at, respectively, the vertices A, B, C of the triangle ABC.
If two spherical line segments on S 2 meet at a point P (other than the north pole) at an angle θ , show that, under the stereographic projection map π , the corresponding segments of circles or lines in C meet at π(P), with the same angle and the same orientation. ] For every spherical triangle = ABC, show that a < b + c, b < c + a, c < a + b and a + b + c < 2π. Conversely, show that, for any three positive numbers a, b, c less than π satisfying the above conditions, we have cos(b + c) < cos a < cos(b − c), and that there is a spherical triangle (unique up to isometries of S 2 ) with those sides.