Geometric proof of the geodesics of a sphere?

Question

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.

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.

Answer 2

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 $L(\mu)\leq L(\gamma)$.

Answer 3

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.