Girard's Theorem:
On a Sphere, Area(Triangle) = Radius^{2} x AngleExcess
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Proof
Consider the white triangle \(\sf T \) on the sphere shown above. Girard's Theorem gives a formula for the area of \(\sf T \). The key to understanding the derivation is the configuration of the three great circles on the sphere, as shown on this figure. There is no difficulty understanding what you see there. What might cause problems is what the configuration looks like on the other side of the sphere. However this figure is a WebGL applet and you can rotate it by clicking and dragging the mouse starting anywhere on the figure. You can also click on the Make Translucent checkbox to make the sphere translucent.
We label the vertices of \(\sf T \) by \(\sf R \), \(\sf G \), and \(\sf B \), and the corresponding angles of \(\sf T \) by \(\sf r,\) \(\sf g \), and \(\sf b \). The letters stand for red, green, and blue, and, for example, the vertex \(\sf R \) is the vertex of \(\sf T \) where \(\sf T \) is opposite a red triangle. The angles at \(\sf R \) in the white triangle \(\sf T \) and in the red triangle are opposite angles and therefore are equal. Their value will be denoted by \(\sf r \). In fact \(\sf R \) is the vertex of two congruent lunes, one of which consists of the red triangle and a gray triangle, and the other of which contains the white triangle and another red triangle. We refer to these two lunes as the red lunes. We denote by \(\sf L_r' \) the red lune which does not contain \(\sf T \), and by \(\sf L_r \) the red lune which does contain \(\sf T \). In exactly the same way we see that \(\sf G \) is the vertex of two congruent, green lunes --- \(\sf L_g \) which contains \(\sf T \), and \(\sf L_g' \) which does not contain \(\sf T \), and the vertex \(\sf B \) is the vertex of two congruent, blue lunes --- \(\sf L_b \) which contains \(\sf T \), and \(\sf L_b' \) which does not contain \(\sf T \).
If you rotate the sphere you will also see a gray triangle that looks pretty much the same as \(\sf T \). This is the antipodal triangle \(\sf T' \). Its vertices are \(\sf R' \), \(\sf G' \), and \(\sf B', \) which are the points antipodal to \(\sf R \), \(\sf G \), and \(\sf B \) respectively. Since \(\sf T \) and \(\sf T' \) are images of each other under the antipodal map, which is an isometry, they have the same area.
It is important to understand the situation of each pair of like colored lunes. Concentrate on the two blue lunes, \(\sf L_b \) and \(\sf L_b' \). Notice the white triangle \(\sf T \) is part of the lune \(\sf L_b \) and the gray triangle \(\sf T' \), which is antipodal to \(\sf T \), is part of \(\sf L_b' \). Examination of the other pairs of lunes reveals that the lunes \(\sf L_g \) and \(\sf L_r \) also contain \(\sf T \), while \(\sf L_g' \) and \(\sf L_r' \) contain \(\sf T'. \)
To sum up, the six lunes \(\sf L_r \), \(\sf L_r' \), \(\sf L_g \), \(\sf L_g' \), \(\sf L_b \), and \(\sf L_b' \), have the following properties:
- The triangle \(\sf T \) is contained in each of the three lunes \(\sf L_r \), \(\sf L_g \), \(\sf L_b \), and in no others.
- The antipodal triangle \(\sf T' \) is contained in each of the three lunes \(\sf L_r' \), \(\sf L_g' \), \(\sf L_b' \), and in no others.
- Every point of the sphere which is not in \(\sf T \) or \(\sf T' \) is contained in precisely one of the lunes.
We can sum up the bulleted points by saying that the six lunes cover the entire sphere with the points in \(\sf T \) and \(\sf T' \) covered two additional times. Therefore when we add up the areas of the lunes we have
\(\sf \qquad \text{area}(L_r) + \text{area}(L_g) + \text{area}(L_b) + \text{area}(L_r') + \text{area}(L_g') + \text{area}(L_b') = \text{area(sphere)} + 2 \; \text{area}(T) + 2 \; \text{area}(T') . \)
Into this equation we substitute the formulas for the area of a lune, and the surface area of a sphere of radius \(\sf R \). Finally, using the fact that \(\sf T \) and \(\sf T' \) have the same area, we get
\(\sf \qquad 2 \; R^2 r + 2 \; R^2 g + 2 \; R^2 b + 2 \; R^2 r + 2 \; R^2 g + 2 \; R^2 b = 4 \pi R^2 + 4 \; \text{area}(T) . \)
Next, solving for the area of \(\sf T \), and collecting terms this becomes
\(\sf \qquad \text{area}(T) = R^2 ( r + g + b - \pi ) . \)
The sum of the angles minus \(\sf \pi \) radians is called the angle excess.
This last formula is called Girard's formula, and the result of the formula is called Girard's Theorem.
The proof of this classical theorem follows closely the proof given by John Polking.
Updated 2020 August 29