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  • 1 University of Technology Mathematical Faculty D-09107 Chemnitz Germany
  • 2 Hungarian Academy of Sciences A. Rényi Institute of Mathematics Pf. 127 H-1364 Budapest Hungary
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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

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, 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 Euclidean d -space ℝ d , R. Schneider [38] proved that if K ⊂ 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
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be a d -dimensional Minkowski space, and K
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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 ϕ: S d −1 → ℝ, for a convex body K with Euclidean surface area measure dS K ( u ), with ϕ( u ) being dS K ( u )-almost everywhere non-0, by the formula B ↦ ∫ B ϕ( u ) dS K ( u ) (supposing that ϕ is integrable with respect to dS K ( u )), for BS d −1 a Borel set, rather than the Euclidean surface area measure B ↦ ∫ 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 are a measure dS K ( u ) by the k -th area measure of K ( k with 1 ≦ kd − 2 an integer), the analogous result holds. We follow rather closely the proof for ℝ d , which is due to Schneider [38].

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