K. Zindler [47] and P.C. Hammer and T.J. Smith [19] showed the following: Let K be a convex body in the Euclidean plane such that any two boundary points p and q of K, that divide the circumference of K into two arcs of equal length, are antipodal. Then K is centrally symmetric. [19] announced the analogous result for any Minkowski plane M(2), with arc length measured in the respective Minkowski metric. This was recently proved by Y.D. Chai-Y.I. Kim [7] and G.Averkov [4]. On the other hand, for Euclide and-space R(d), R. Schneider [38] proved that if K subset of R(d) is a convex body, such that each shadow boundary of K with respect to parallel illumination halves the Euclidean surface area of K (for the definition of "halving" see in the paper), then K is centrally symmetric. (This implies the result from [19] for R(2).) We give a common generalization of the results of Schneider [38] and Averkov [4]. Namely, let M(d) be a d-dimensional Minkowski space, and K subset of M(d) be a convex body. If some Minkowskian surface area (e.g., Busemann's or Holmes-Thompson's) of K is halved by each shadow boundary of K with respect to parallel illumination, then K is centrally symmetric. Actually, we use little from the definition of Minkowskian surface area (s). We may measure "surface area" via any even Borel function phi: S(d-1) -> R, for a convex body K with Euclidean surface area measure dS(K)(u), with phi(u) being dS(K)(u)-almost everywhere non-0, by the formula B bar right arrow integral(B) phi(u)dS(K)(u) (supposing that phi is integrable with respect to dS(K)(u)), for B subset of S(d-1) a Borel set, rather than the Euclidean surface area measure B bar right arrow integral(B) dS(K)(u). The conclusion remains the same, even if we suppose surface area halving only for parallel illumination from almost all directions. Moreover, replacing the surface area measure dS(K)(u) by the k-th area measure of K (k with 1 <= k <= d - 2 an integer), the analogous result holds. We follow rather closely the proof for R(d), which is due to Schneider [38].
