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Consider the black triangle T on the sphere to the left. We will be deriving a formula for the area of 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 java applet and you can rotate it by clicking and dragging the mouse starting anywhere on the figure.
We will label the vertices of T by R, G, and B, and the corresponding angles of T by r, g, and b. The letters stand for red, green, and blue, and, for example, the vertex R is the vertex of T where T is opposite a red triangle. The angles at R in the black triangle T and in the red triangle are opposite angles and therefore are equal. Their value will be denoted by r. In fact 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 black triangle and another red triangle. We will refer to these two lunes as the red lunes. We will denote by Lr' the red lune which does not contain T, and by Lr the red lune which does contain T. In exactly the same way we see that G is the vertex of two congruent, green lunes --- Lg which contains T, and Lg' which does not contain T, and the vertex B is the vertex of two congruent, blue lunes --- Lb which contains T, and Lb' which does not contain T.
If you rotate the sphere you will also see a gray triangle that looks pretty much the same as T. This is the antipodal triangle T'. Its vertices are R', G', and B', which are the points antipodal to R, G, and B respectively. Since T and 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, Lb and Lb'. They are shown in isolation in the applet to the right. Notice the black triangle T is part of the lune Lb and the gray triangle T', which is antipodal to T, is part of Lb'. Examination of the other pairs of lunes reveals that the lunes Lg and Lr also contain T, while Lg' and Lr' contain T'.
To sum up, the six lunes Lr, Lr', Lg, Lg', Lb, and Lb', have the following properties:
Understanding the proof of Girard's Theorem comes down to understanding the configuration of the triangle and the six lunes, and verifying the three bulleted points. Hopefully the applets on this page are helpful. However, by far the best way to visualize the six lunes is by physical experimentation with an actual sphere. Get a beach ball about 8 to 12 inches in diameter. Draw a triangle T on it. Then carefully extend each side of the triangle to a complete great circle. It will be noticed that these great circles intersect on the other side of the sphere and form another triangle T' which is the antipodal image of T. Thus T' is congruent to T and consequently has the same area.
Suppose that the three angles of T are R, G, and B. At each of these vertices there are two lunes of the appropriate angle that meet. One of them contains T. Call this lune Lr, Lg, and Lb as the case may be. Denote the other lune, which does not contain T by Lr', Lg', and Lb'. Hatch the two lunes Lr and Lr' with a distinctive color or marking (such as little circles). Hatch the lunes Lg and Lg' with a different color or marking, and use yet a third for the lunes Lb and Lb'.
Now by examining the beach ball you will be able to verify the three bulleted points.
We can sum up the bulleted points by saying that the six lunes cover the entire sphere with the points in T and T' covered two additional times. Therefore when we add up the areas of the lunes we have
.Into this equation we substitute the formulas for the area of a lune, and the surface area of a sphere of radius R. Finally, using the fact that T and T' have the same area, we get
.Next, solving for the area of T, and collecting terms this becomes
.This last formula is called Girard's formula, and the result of the formula is called Girard's Theorem.
We get an interesting variant if we solve for the sum of the angles:
.Both formulas are interesting. The first emphasizes the area of the spherical triangle, and the second emphasizes the sum of the angles of the spherical triangle. For comparison with planar geometry, the second is especially interesting because it says precisely how much the sum of the angles of a spherical triangle exceeds two right angles, the sum of the angles for a planar triangle. That the difference involves the area of the sphere is a remarkable departure from what we would expect from our knowledge of plane geometry.
Exercise: Find the formula for the result of Girard's Theorem when the angles are measured in degrees instead of radians.
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The next section discusses some consequences of Girard's Theorem. |
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The previous section discusses area on the sphere. |
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Table of Contents. |