The shortest curve that divides the unit sphere into two regions of equal area How to prove that if a simple closed curve $\gamma$ divides the unit sphere into two regions of same
area, then its length is not smaller than $2\pi$. The hint is to consider a reflection of the sphere,
then to show that must intersection with its reflection. But I don't know what to do.

If $\gamma$ is the curve, then assume that $(1)$ there is $x,\ -x\in \gamma$. Clearly the length of $  \gamma $ is greater than $2\pi$.
If not, then assume that $\gamma$ encloses a domain $D$, whose area is $2\pi$. When $A$ is an (isometric) antipodal map on the unit sphere, then $A(D)$ still has the area $2\pi$. When $D$ intersect $A(D)$, then $A(\gamma),\ \gamma$ have nonempty intersection. Hence $D,\ A(D)$ has empty intersection.
Hence unit sphere has area $> 2{\rm area}\ D=4\pi$, which is a
contradiction. Hence we can assume $(1)$.
