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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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    $\begingroup$ You may find this answer on mathoverflow useful. $\endgroup$
    – Sasha
    Feb 23, 2013 at 13:42
  • $\begingroup$ 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.) $\endgroup$ Feb 23, 2013 at 14:04

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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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  • $\begingroup$ How do you define 'corners'? Perhaps $\gamma$ is tangent to $\Gamma$ at each intersection point... $\endgroup$
    – Jessica B
    Oct 12, 2016 at 16:52
  • $\begingroup$ @JessicaB If so then γ lies on a great circle. But you are right, the proof requires a bit of calculus --- it use that functions with vanishing derivative are constant. $\endgroup$ Oct 12, 2016 at 17:26
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It is sufficient to prove that the shortest curve connecting $N=(0,0,1)\in S^2$ with some other point $P\in S^2$ different from $(0,0,-1)$ is the meridian arc $\mu$ connecting $N$ with $P$.

Let $N$ be at latitude $\theta=0$ and $P$ at latitude $\theta=\theta_P\in\>]0,\pi[\>$, and assume that $$\gamma: t\mapsto x(t)\in S^2\quad(0\leq t\leq 1)$$ is a curve on $S^2$ with $x(0)=N$, $x(1)=P$. On the other hand, let $$\mu:\quad \theta\to m(\theta)\qquad(0\leq\theta\leq\theta_P)$$ be the meridian arc connecting $N$ with $P$. Consider an arbitrary subdivision $$0=\theta_0<\theta_1<\ldots< \theta_n=\theta_P$$ of the parameter interval, and put $m(\theta_k)=:m_k$ $(0\leq k\leq n)$. The $m_k$ determine a polygonal path $\mu'\subset{\mathbb R}^3$ approximating the arc $\mu$. This path has Euclidean length $$L(\mu')=\sum_{k=1}^n|m_k-m_{k-1}|\ .$$ For each $\theta_k$ there is a "last" point $x_k=x(t_k)$ on $\gamma$ having latitude $\theta_k$. The $x_k$ form a polygonal path $\gamma'$ inscribed in $\gamma$. Elementary geometry shows that $|m_k-m_{k-1}|\leq |x_k-x_{k-1}|$, so that $$L(\mu')\leq L(\gamma')\leq L(\gamma)\ ,$$ by definition of $L(\gamma)$. Since $\sigma'$ is an arbitrary polygonal approximation to $\mu$ we then also have $L(\mu)\leq L(\gamma)$.

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I've now written up a proof that appears in the March 2018 edition of The Mathematical Gazette. The main steps are:

  • Show that a strict version of the triangle inequality holds for straight lines (locally for spheres). This can be done using intuition or connecting to the Euclidean case.
  • Take a straight line and a geodesic connecting two (nearby) points.
  • Assume a point on the geodesic lies off the line. Form a straight-line-triangle, and split the geodesic in two.
  • Use rotation to move the two new sides of the triangle to match the old one, carrying the two pieces of geodesic along with them.
  • Use reflection to make two new geodesics on the same side of the original line.
  • Use the intermediate value theorem to find an intersection of these two geodesics, and use the triangle inequality to reach a contradiction.
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