Prove that the sum of the angles of a triangle on a sphere is more than 180 degrees.

A related and easy problem is to prove that on a sphere the ratio of circumference of a circle to its radius is less than 2π. However, the one about triangles is trickier.

I'm sure that if you know a bit about non-euclidean geometry you can simply prove the theorem in all spaces with positive curvature. Then the sphere falls out as a special case. I don't know how to do that.

Here is another approach, which is not very rigorous, but pretty much good enough to convince me.

Take the longest side of the triangle and draw the perpendicular through it that crosses the opposite vertex. I should prove such a perpendicular exists, but won't. Now you have two right triangles. If you add up all the interior angles of the two right triangles, you get the interior angles of the big triangle, plus the two extra right angles. (draw your own picture. I don't feel like it.)

Now if the two right triangles both have the sum of their angles more than 180 degrees, added together they make something more than 360. Subtract the two right angles. You find that the big triangle also has its interior angles more than 180 degrees.

How do I know if one of those two right triangles has its interior angles more than 180 degrees? Drop a perpendicular from the hypotenuse to the right angle vertex, breaking the right triangle into two smaller right triangles. Now the same reasoning holds as before, and I only have to show that the smaller right triangles have interior angles that sum to more than 180. I can keep breaking the triangles down smaller and smaller, so all I have to show is that a very small right triangle has the sum of its interior angles greater than 180, and I'm done.

Now introduce coordinates on the sphere so that the small right triangle's right angle lines up with the intersection of the prime meridian and the equator. Then the length of the side on the equator is dφ, while the length of the side on the prime meridian is dθ. Finally, the metric on a sphere is ds

^{2}= dθ

^{2}+ sin

^{2}θ *dφ

^{2}, so that is also the length of the hypotenuse.

If I let θ = π/2, I have a normal right triangle in euclidean space. But θ = π/2 only right along the equator. Along the length of the hypotenuse, sin

^{2}θ < 1, meaning the hypotenuse is actually shorter than that of a right triangle.

The law of sines shows that the shorter the side of a triangle, the bigger the angles on at its endpoints. So the angles at the end of the hypotenuse are large than those of a triangle in euclidean space. The right triangle has a sum of its interior angles more than 180 degrees, and then so do all other triangles on the sphere.

I admit the last step is a bit of a stretch. The law of sines is really useless because it's true only in euclidean space, but for a very small triangle the space is a better and better approximation of euclidean.

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