Abstract
Brunn in 1887 formulated a theorem on three parallel sections of a convex body with extreme sections of the same area, but not obtained from each other by a parallel shift, asserting that the area of the middle section is strictly larger, and correctly proved, as Minkowski noted, that only not less. The elimination of equality, which was still considered the most difficult in the theorem, has been proved up to the present time by many authors, using serious mathematics. The article proposes a fundamentally different geometric approach to the proof of the theorem, due to which, for the correct completion of Brunn’s original proof, one can restrict oneself to the elementary means available to schoolchildren, bypassing the difficulties with equality. The proposed reasoning extends to all dimensions, like the theorem itself, as pointed out by Brunn. Let, in the general case, 𝑉𝑛(𝑄) be the 𝑛-dimensional volume of the body 𝑄 ⊂ R𝑛, 𝐿0, 𝐿1 be parallel hyperplanes in R𝑛+1, containing respectively convex bodies 𝑃0, 𝑃1, and 𝐿 is a parallel hyperplane, located strictly between them, and 𝑃 is the intersection of 𝐿 with the convex hull 𝑃0 ∪ 𝑃1. Brunn’s theorem states that if 𝑃1 is not obtained from 𝑃0 by parallel translation and 𝑉𝑛(𝑃1) = 𝑉𝑛(𝑃0) = 𝑣 > 0, then 𝑉𝑛(𝑃) > 𝑣. In 1887, Brunn rigorously proved that 𝑉𝑛(𝑃) > 𝑣 using the effective trick of the division of the volumes 𝑃0, 𝑃1 by a hyperplane in R𝑛+1. In this article, this is called Brunn cuts. For the strictly inequality 𝑉𝑛(𝑃) > 𝑣, it remained a small perturbation go from the body 𝑃1 to another convex body 𝑃1, 𝑉𝑛( 𝑃1) = 𝑣 , so that 𝑉𝑛(𝑃) > 𝑉𝑛( 𝑃), where 𝑃 is a new section in the hyperplane 𝐿 arising after replacing 𝑃1 with 𝑃1. Since 𝑉𝑛( 𝑃) > 𝑣, then 𝑉𝑛(𝑃) > 𝑣. The easiest way is to replace 𝑃1 with 𝑃1 in the case of convex polytopes 𝑃0, which can approximate convex bodies arbitrarily close. The required replacement of 𝑃1 by 𝑃1 is quite simple, when 𝑛-dimensional simplices act as 𝑃0, into which the convex polytope can be split by Brunn cuts. Until now, the sufficiently naive natural geometric method outlined above has not been proposed for proving the strict inequality 𝑉𝑛(𝑃) > 𝑣, as it were head-on, due to the fact that initially the theorem was formulated not for convex polytopes 𝑃0, 𝑃1, but for arbitrary convex bodies. The main reason, according to the author, lies in the algebraic representation 𝑃 = (1 − 𝑡)𝑃0 + 𝑡𝑃1, where 𝑡 is the ratio of the distance from 𝐿0 to 𝐿 to the distance from 𝐿0 to 𝐿1, 0 𝑣 are significantly reduced. This article shows that in the proof of the theorem in an equivalent formulation, on the contrary, the space R𝑛 should be included in R𝑛+1 and use the original formulation of the theorem, when the main tool of the proof the elementary means are Brunn cuts. For the sake of fairness, it should be noted that numerous applications of this theorem, obtained by Minkowski and other authors, are connected precisely with its equivalent formulation, with mixed volumes, with algebraic representations 𝑃 = (1 − 𝑡)𝑃0 + 𝑡𝑃1, called Minkowski sums.
Highlights
Согласно п. 4.1.3 точку Bk′ j ∈ [Bi1, Bkj ] можно выбрать на столько близкой к вершине Bkj, что точка x останется внутри L︀−n′i1, из-за чего многогранник Q− будет относиться к классу j−1, что достаточно проверить в случае L︀n′i1 = Ln′i1
М. 1997, "Оптимизационная задача для неравенства Брунна-Минковского" , Труды МИАН, 218, Наука, М., стр
Summary
Брунн в 1887 году сформулировал теорему о трёх параллельных сечениях выпуклого тела с одинаковыми по площади крайними сечениями, но не получающимися друг из друга параллельным сдвигом, утверждающую, что площадь среднего сечения строго больше, а корректно доказал, как заметил Минковский, что только не меньше. В статье предлагается принципиально иной геометрический подход к доказательству теоремы, благодаря чему для корректного завершения исходного доказательства Брунна можно ограничиться элементарными средствами, доступными школьникам, минуя трудности с равенством, причём предлагаемые рассуждения распространяются на все размерности, как и сама теорема, на что указывал Брунн. Для строго неравенства Vn(P ) > v оставалось небольшим "шевелением" перейти от тела P1 к другому выпуклому телу P︀1, Vn(P︀1) = v, так, что. До настоящего времени не предлагался очерченный выше достаточно наивный естественный геометрический способ доказательства строгого неравенства Vn(P ) > v как бы в лоб может из-за того, что изначально теорема формулировалась не для выпуклых многогранников P0, P1, а для произвольных выпуклых тел. Что при доказательстве теоремы в эквивалентной формулировке следует, напротив, пространство Rn включать в Rn+1 и использовать изначальную формулировку теоремы, когда основным инструментом доказательства элементарными средствами становится рассечение.
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