But hold on - so far our unspoken assumption has been that we are working in Euclidean geometry. In non-Euclidean geometry, the angles of a triangle add up to less than 180° (in hyperbolic, negative-curvature geometry), or more than 180° (elliptic, or positive-curvature geometry) - and the more so, in both cases, the larger the area of the triangle. The same will go for for the internal angles of a pentagon - so what we need to get a pentagon with angles of 109.471° is elliptic geometry. Altogether, the angles will exceed 540° by 7.355° or 0.128 of a radian, implying that the area of the pentagon must be 0.128 times the square of the radius of curvature - or just 0.128 if we take the radius of curvature as one unit.
We can obtain the length of the side of the pentagon, by doing a bit of spherical trigonometry (which applies in elliptic geometry). The second fundamental formula of spherical trigonometry goes as follows:
cos(A) = - cos(B) cos(C) + sin(B) sin(C) cos(a).
Here A, B, C are the three angles of a triangle, and a is the length of the side opposite A (again we assume the radius of curvature is one unit). The formula is usually thought of as determining the angle A when a, B and C are known. However, it can also be used to determine a when A, B and C are known. In the present case, A is 72°, B and C are each α/2. We obtain
cos(a) = ( cos(72) + cos²(α/2))/sin²(α/2) = 0.9635
hence a = 0.2709.
In flat space, the volume of a dodecahedron whose side is a, is (2 + 3.5 φ) times a³ (here φ stands for the golden number), giving us in the present case 0.1524. This is of course only an approximation (an underestimate, in fact), for the dodecahedron about which we are now talking exists in positively curved space.
It is interesting to compare this with the volume comprised in a 3-sphere of unit radius, which is just 2π²: enough to fit in approximately 2π²/0.1524 ≈ 129 of our dodecahedra.
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