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I have seen the standard variational proof that great circles are the geodesics on the $2$-sphere. Do you know a purely geometric proof of this fact, not involving calculus of variations or differential geometry?

It seems like it could be possible to provide a more elementary proof. If you disagree, please explain why.

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You may find this answer on mathoverflow useful. –  Sasha Feb 23 '13 at 13:42
    
And in a comment to the accepted answer to the question @Sasha is referring to, mention is made of the triangle inequality for spherical triangles. It should be possible to make a proof using it in the same way you would for the Euclidean plane. (Of course, the triangle inequality for spherical triangles may fail if one side of the triangle spans more than $180^\circ$ of a great circle, so you have to take that into account.) –  Harald Hanche-Olsen Feb 23 '13 at 14:04

1 Answer 1

Let us assume that existence of geodesic it trivial; yet assume that is it obvious that a geodesic can not have corners.

If you agree, then draw a great circle $\Gamma$ through two points of geodesic $\gamma$. If $\gamma\not\subset\Gamma$, reflect the part of $\gamma$ which lies on one side from $\Gamma$. You get a new geodesic, say $\gamma'$, and it has corners, a contradiction.

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