Is 'small enough' ellipse projected on a surface of a sphere convex? By ellipse I mean a set of points 'C' with a constant sum |AC| + |BC|, A and B are the centers. By 'small enough' I mean that the radii fits into 90 degrees (I think it is not convex once you make it large enough, though the limit is probably more like 180 degrees).


It seems to me that it is indeed convex, but is there some simple proof? The mathematics I tried to do usually ends up as f(x)=arccos(a(x)) + arccos(b(x)) and it isn't quite easy to prove that that the function has a reasonable shape when a(x) is decreasing and b(x) is increasing. Is there some easy proof I have overlooked?


By convex I mean that any shortest line connecting the points on the ellipse is 'inside' the ellipse (i.e. the distance |AX| + |XB| is smaller or equal then the distance defining the ellipse for any point X of the line).


Update: I think I eventually found a solution; triangle inequality works for these 'small enough' triangles. Geometrically the problem can be somewhat shuffled, in the end I have to prove that a triangle that is 'inside' another triangle is indeed smaller; triangle inequality combined with the way of computing a distance on a sphere will do the trick.



Is ellipse on a sphere convex? (proof)
Yes it is. After central projection on the plane (Klein model for sphere) you obtain usual ellipse.


Also you can show it using triangle inequality. All proofs from euclidean plane works. For example this one:
Suppose $F\_1$ and $F\_2$ foci of the ellipse. Take any two points $A$ and $B$ inside and reflect $F\_2$ with respect to the line $AB$. New point denote by $F\_2'$.
Take any point $X$ on the segment $AB$. Suppose ray $F\_1X$ intersect the segment $F\_2'A$ (the case $F\_2'B$ is the same) in the point $Y$. We have, 
$$F\_1X+F\_2X=F\_1X+F\_2'X< F\_1X+XY+YF\_2'=F\_1Y+YF\_2'< F\_1A+AY+YF\_2'=F\_1A+AF\_2'$$


