= Solution
Because $k>0$ and the original smooth closed curve has periodic Frenet data, $\arctan(\tau/k)$ is a continuous real-valued periodic function. Therefore $\int k_g\,d\bar s=-\oint d\arctan(\tau/k)=0$. The simple closed normal curve divides the sphere into two disks by the Jordan theorem. Apply <Gauss-Bonnet theorem> to either disk, choosing its consistent boundary orientation. Its <Gaussian curvature> is one and its total boundary curvature remains zero. Thus each region has area $2\pi$, proving \b[the two regions have equal area].
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