Shortest path on a sphere I'm quite a newbie in differential geometry. Calculus is not my cup of tea ; but I find geometrical proofs really beautiful. So I'm looking for a simple - by simple I mean with almost no calculus - proof that the shortest path between two points on a sphere is the arc of the great circle on which they lie. Any hint ? 

Edit: Or at least a reference ?
Here's a geometric observation that can hardly be called a "proof", but may be appealing nonetheless. 

If $p$ and $q$ are distinct points of the sphere $S^{n}$, if $C:[0, 1] \to S^{n}$ is a "shortest path" joining $p$ to $q$, and if $F:S^{n} \to S^{n}$ is a distance-preserving map fixing $p$ and $q$, then $F \circ C$ is also a shortest path (because the length of $F \circ C$ is equal to the length of $C$).

Assume $q \neq -p$. If you believe there exists a unique shortest path from $p$ to $q$, it's not difficult to see that the "short" great circle arc is the only candidate: Every point not on the great circle through $p$ and $q$ is moved by some isometry of the sphere that fixes $p$ and $q$.

If you're thinking specifically of $S^{2}$, reflection $F$ in the plane containing $p$, $q$, and the center of the sphere is an isometry, and $f(x) = x$ if and only if $x$ lies on the great circle through $p$ and $q$.

(A similar argument "justifies" that the shortest path between distinct points of the Euclidean plane is the line segment joining them.)
An image worth a thousand words....

a movie a thousand images, and an equation a thousand movies..... :)

Stare at the figure and try to draw a path shorter than that between
$A$ and $B$, or $B$ and $C$, or any of the fourth vertices with any of the 
other three.  As you said, you want a geometrical argument. It should not take time to convince you that the paths drawn on this figure (and the paths you can draw in your mind for the diagonals $A-C$ and $B-D$) are all connected to an arc that is centered at the origin. Those are parts of big circles.