Completion of the proof of Brunn's theorem by elementary means

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, $ V_n (Q) $ be the $ n $-dimensional volume of the body $ Q \subset \mathbb{R} ^ n $, $ L_0, L_1 $ be parallel hyperplanes in $ \mathbb{R} ^ {n + 1} $, containing respectively convex bodies $ P_0, P_1 $, and $ L $ is a parallel hyperplane, located strictly between them, and $ P $ is the intersection of $ L $ with the convex hull $ P_0 \cup P_1 $. Brunn's theorem states that if $ P_1 $ is not obtained from $ P_0 $ by parallel translation and $ V_n (P_1) = V_n (P_0) = v> 0 $, then $ V_n (P)> v $. In 1887, Brunn rigorously proved that $ V_n (P) \geqslant v $ using the effective trick of the division of the volumes $ P_0, P_1 $ by a hyperplane in $ \mathbb {R} ^ {n + 1} $. In this article, this is called Brunn cuts. For the strictly inequality $ V_n (P)> v $, it remained a small "perturbation" go from the body $ P_1 $ to another convex body $ \widetilde {P} _1 $, $ V_n (\widetilde {P} _1) = v $ , so that $ V_n (P)> V_n (\widetilde {P}) $, where $ \widetilde {P} $ is a new section in the hyperplane $ L $ arising after replacing $ P_1 $ with $ \widetilde {P} _1 $. Since $ V_n (\widetilde {P}) \geqslant v $, then $ V_n (P)> v $. The easiest way is to replace $ P_1 $ with $ \widetilde {P} _1 $ in the case of convex polytopes $ P_0 $, which can approximate convex bodies arbitrarily close. The required replacement of $ P_1 $ by $ \widetilde {P} _1 $ is quite simple, when $ n $-dimensional simplices act as $ P_0 $, 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 $ V_n (P)> v $, as it were head-on, due to the fact that initially the theorem was formulated not for convex polytopes $ P_0, P_1 $, but for arbitrary convex bodies. The main reason, according to the author, lies in the algebraic representation $ P = (1-t) P_0 + tP_1 $, where $ t $ is the ratio of the distance from $ L_0 $ to $ L $ to the distance from $ L_0 $ to $ L_1 $, $ 0 <t <1 $. This leads to the temptation to go over in the proofs of the theorem from $ \mathbb {R} ^ {n + 1} $ to $ \mathbb {R} ^ n $ and use the equivalent statement of the theorem, assuming $ L_0 = L_1 = \mathbb {R} ^ n $. As a result, from the general situation, when $ L_0 \neq L_1 $, passed into the singularity $ L_0 = L_1 $, in the conditions of which the possibilities for attracting geometric intuition are significantly reduced and, as a consequence, the possibilities for simpler visual geometric justifications of the inequality $ V_n (P)> v $ are significantly reduced. This article shows that in the proof of the theorem in an equivalent formulation, on the contrary, the space $ \mathbb {R} ^ n $ should be included in $ \mathbb {R} ^ {n + 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) P_0 + tP_1 $, called Minkowski sums.