Perimeter of Convex Region in $S^2$

Question

Suppose I have a convex polygonal region (enclosed by a curve) entirely contained in a hemisphere of $S^2$. I have a hunch that the perimeter (length of enclosing curve) is at most $2\pi$. Is this obvious?

Answer 1

If $u_i$ is vertex in spherical $k$-gon $c$ in unit sphere of center $O$, then geodesic $u_iu_{i+1}$ and origin $O$ determine a plane Here from $k$ planes we have $2$-dimensional polyhedral surface $P$ whose only vertex is $O$ and whose edges are ray $Ou_i$

That is perimeter $l$ of curve $c$ is sum of face angles in $P$ at $O$ Intuitively we cut $P$ along $Ou_1$ so that we have a sector of central angle strictly less than $2\pi$ Here central angle, $l$, and sum of all face angles are equal

Answer 2

By Crofton's formula, length (a spherical curve $\gamma$) = $\pi$ times (average number of intersections $\gamma$ with equators). Since $\gamma$ is convex the average $\le 2$. Whence the result.

Answer 3

I will elaborate Petrunin's answer : First note the Crofton formula : $$\frac{1}{4}\int_{\mathbb{S}^2} \sharp (\gamma \cap \xi^\perp)d\xi = {\rm Length}\ \gamma$$ where $\xi^\perp =\{ v\in\mathbb{S}^2|v\perp \xi\}$ is a great circle (We can prove this considering area of lune in $\mathbb{S}^2$).

Here $\gamma$ is a convex polygon so that the number of intersections $\sharp(\gamma\cap\xi^\perp)$ is less than or equal to $2$, which completes the proof.