Search Results

You are looking at 1 - 2 of 2 items for

  • Author or Editor: Gennadiy Averkov x
Clear All Modify Search

Abstract  

A convex body K in ℝd is said to be reduced if the minimum width of each convex body properly contained in K is strictly smaller than the minimum width of K. We study the question of Lassak on the existence of reduced polytopes of dimension larger than two. We show that a pyramid of dimension larger than two with equal numbers of facets and vertices is not reduced. This generalizes the main result from [8].

Restricted access

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

\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \usepackage{bbm} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathbb{M}^2$$ \end{document}
, 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
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \usepackage{bbm} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathbb{M}^d$$ \end{document}
be a d -dimensional Minkowski space, and K
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \usepackage{bbm} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathbb{M}^d$$ \end{document}
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].

Restricted access