9.5: Ptolemy's Theorem
Ptolemy's Theorem
In our first application of the Gauss-Bonnet Formula, we consider triangular regions on the unit sphere. Specifically, we want to find the area of a triangle with interior angles α, β, and γ, and having geodesic sides. Ptolemy's Theorem states that the area of any triangle with geodesic sides can be expressed as the intersection of three double lunes.
Demonstration
The demo draws a geodesic triangle on the unit sphere whose three vertices are controlled by moveable hotspots. In the control panel, the three double lunes of angles α, β, and γ can be toggled on and off. Notice that when all three double lunes are toggled on, the sphere is completely covered once, except for two triangular regions, which are covered three times.
With this observation we can find an expression for the area of the geodesic triangle in terms of the areas of the double lunes.
∑(area of double lunes) = (area of sphere) + 4(area of triangle)
On the unit sphere the area of a double lune is equal to four times its angle. In this case, the angles of the three double lunes correspond to the three interior angles α, β, and γ of the geodesic triangle. Therefore,
4α + 4β + 4γ = 4π + 4 · Area(T), or
Area(T) = (α + β + γ) - π
We also can calculate the area of a geodesic triangle on the unit sphere using the Gauss-Bonnet Formula.
 TΚdA + ∂Tκgds + ∑ext.angles = 2π
Since the sides of the triangle are geodesics, κg = 0 over the entire boundary. So, the line integral term vanishes. We also know that the unit sphere has constant positive Gaussian curvature Κ = 1, which means that the total Gaussian curvature of the triangle is equal to its area. This gives us,
Area(T) + ∑ext.angles = 2π
If the interior angles of the triangle are α, β, and γ, then the sum of the exterior angles is
3π - (α + β + γ).
After subtituting this into the Gauss-Bonnet equation, we arrive at the same result for the area of the triangle:
Area(T) = (α + β + γ) - π
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