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By Igor Pak

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Denote by X ′ the clockwise rotation of X around z by an angle π/3. Clearly, area(X) = area(X ′ ). Thus for every z in the interior of an edge in X, polygons X and X ′ intersect at z and at least one other point v. Denote by u the counterclockwise rotation of v around z. Then (z, u, v) is the desired inscribed triangle. The same argument works when z is a vertex as in the proposition. Indeed, if z = xi and ∠ xi−1 xi xi+1 ≥ π/3, then the interiors of the corresponding regions A and A′ intersect.

Jordan curve C and an inscribed square. implies the claim for self-intersecting closed polygons as well, since taking any simple cycle in it suffices. 1 generalizes to higher dimensions. The answer is yes, but the proof is more delicate. In three dimensions this is called the Kakutani theorem; we prove it in the next section. In fact, much of the next section is based on various modifications and generalizations of the Kakutani theorem. 2. Inscribing triangles is easy. , X = ∂A. We say that an equilateral triangle is inscribed into X if there exist three distinct points y1 , y2, y3 ∈ X such that |y1y2 | = |y1 y3 | = |y2y3 |.

2 (Dyson). Let P ⊂ R3 be a convex polytope containing the origin: O ∈ P − ∂P . Then there exists a centrally symmetric square inscribed into P . Proof. Let us prove that the surface S = ∂P of the polytope P contains a centrally symmetric polygon Q. 1. Let P ′ be a reflection of P in O, and let S ′ = ∂P ′ . Since O ∈ P , then P ∩ P ′ = ∅, and thus either P ′ = P or S ∩ S ′ = ∅. Indeed, if S ∩ S ′ = ∅, then one of the polytopes would be contained in another and thus has a greater volume. Assume for now that the intersection W = P ∩ P ′ is one-dimensional, and moreover a union of non-intersecting polygons Q1 , .

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