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On the isodiametric and isominwidth inequalities for planar bisections. (English) Zbl 1469.52003

Let \(C\) be a compact set in \(\mathbb{R}^2\), and let us denote its area, diameter, and width by \(a(C)\), \(d(C)\), and \(w(C)\), respectively. Given a planar convex body \(K\), let us consider the set \(B(K)\) of all pairs of compact sets \((K_1,K_2)\), such that each pair is a bisection of \(K\) determined by a curve. The authors prove that \[ \tfrac{4}{\sqrt{3}} W(K)^2 \leq a(K) \leq 2 \operatorname{arctan}(\tfrac{3}{4}) D(K)^2, \] where \[ D(K)=\inf_{(K_1,K_2)\in B(K)} \max\{d(K_1),d(K_2)\}, \] and \[ W(K)=\inf_{(K_1,K_2)\in B(K)} \max\{w(K_1),w(K_2)\}. \]
The equality on the left is achieved if and only if \(K\) is a equilateral triangle. The equality on the right is achieved if and only if \(K\) is the centrally symmetric convex body divided into two regions of equal area presented as Example 2.3 in [C. Miori et al., J. Math. Anal. Appl. 300, No. 2, 464–476 (2004; Zbl 1080.52002)]. The proofs in this paper are not based on those of Miori et al.
Let us consider a position of \(K\) such that the supremum of \(\frac{a(\phi(K))}{D(\phi(K))^2}\) is achieved, where \(\phi\) runs the endomorphisms of \(\mathbb{R}^2\). Then \(\frac{\sqrt{3}}{4} D(K)^2\) is an lower bound for \(a(K)\), and it can be improved to \(\frac{\sqrt{3}}{2} D(K)^2\) if \(K\) is centrally symmetric.
In a similar sense, if \(K\) is in an optimal supremum position for \(\frac{a(\phi(K))}{W(\phi(K))^2}\), then \(a(K)\leq 4\; W(K)^2\), and the equality is achieved if and only if \(K\) is a square.

MSC:

52A10 Convex sets in \(2\) dimensions (including convex curves)
52A40 Inequalities and extremum problems involving convexity in convex geometry

Citations:

Zbl 1080.52002
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References:

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