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}
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$$\mathbb{M}^2$$
\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}
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$$\mathbb{M}^d$$
\end{document} be a
d
-dimensional Minkowski space, and
K
⊂ \documentclass{aastex}
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$$\mathbb{M}^d$$
\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:
Sd
−1
→ ℝ, for a convex body
K
with Euclidean surface area measure
dSK
(
u
), with ϕ(
u
) being
dSK
(
u
)-almost everywhere non-0, by the formula
B
↦ ∫
B
ϕ(
u
)
dSK
(
u
) (supposing that ϕ is integrable with respect to
dSK
(
u
)), for
B
⊂
Sd
−1
a Borel set, rather than the Euclidean surface area measure
B
↦ ∫
BdSK
(
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
dSK
(
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].
One-sided Lebesgue Bernoulli maps of the sphere of degreen2and2n2
