Definition
Given some length $w\in\mathbb R$, I'm interested in closed convex sets $S$ of points with the following properties:
- For all pairs of points from $S$, the distance between them will be no more than $w$: $$\forall x,y\in S:d(x,y)\le w$$
- For every point outside $S$, there is some point in $S$ which is at a distance of more than $w$: $$\forall x\in S^c\;\exists y\in S:d(x,y)>w$$
In the plane
In the plane, these requirements should describe a curve of constant width $w$. (If not, please point out my mistake, as my question is based on the assumption that my above requirements are equivalent to that.) According to MathWorld, the curve of constant width with minimal area is the Reuleaux triangle with an area of $\frac12(\pi-\sqrt3)\,d^2\approx0.705\,d^2$, and that of maximal area is the circle with $\frac14\pi\,d^2\approx0.785\,d^2$. I'm willing to trust that statement.
On the sphere
But how about other geometries, spherical geometry in particular? The usual definition of width in terms of parallel tangents won't work since there are no parallel geodesics on the sphere. But the above definition should still work on the unit sphere. For example, consider $w=\frac\pi2$. The equivalent of the Reuleaux triangle would be a simple spherical triangle with three right angles, since every point on a boundary is $\frac\pi2$ away from the opposite corner. Its area is $\frac12\pi\approx1.571$. The circle with diameter $\frac\pi2$ (and hence radius $\frac\pi4$) has area
$$\int_{\frac12\sqrt2}^2 2\pi\,\mathrm dx =(2-\sqrt2)\pi\approx1.840$$
Question
Again the circle has larger area than this “spherical Reuleaux triangle”. But is the circle still maximal, and the Reuleaux triangle still minimal, or are there other shapes to consider in this case?
What would be the shape with maximal area, for any given $w$ in general, and for $w=\frac\pi2$ in particular? The geometry should always be on the unit sphere.