Original title: Symmetry breaking in the polygonal Szegö-Weinberger inequality as $p\to1^+$: the longest shortest-fence quadrilateral
Beniamin Bogosel, Dorin Bucur, Ilaria Fragalà · September 2026
Imagine you have a piece of land with a set area. You want to split it in half using the shortest fenceThe minimal length of a closed curve that can enclose a given shape. For convex shapes, this is the perimeter. possible. Pólya's problemA classic problem in geometry: among all shapes with a given area, find the one that has the longest possible shortest fence (the minimal length of a fence that can enclose it). asks: which shape forces you to build the longest fence? In this paper, researchers study this problem for polygons with a fixed number of sides. For three-sided shapes, an equilateral triangle is the clear winner. You might expect a four-sided shape to favor a square because of its symmetry. However, the researchers discovered that symmetry breaks. The best quadrilateral is actually an isosceles trapeziumA four-sided shape with one pair of parallel sides and the other two sides equal in length., not a square. An elementary equation for its base angle uniquely defines this optimal shape, up to scaling and rigid movements. The researchers proved this using analytical reasoning alongside strict interval-arithmetic calculations.
Imagine you have a fixed area of land and you want to build the longest possible fence around it. You might think a square would give a long fence, but actually a specific trapezoid gives an even longer fence. In this paper, the researchers found that for quadrilaterals, the shape that requires the longest fence is not the square but a particular isosceles trapezium.
This result challenges the intuition that the most symmetric shape is always optimal in geometry. It provides a concrete example of symmetry breaking in a classic optimization problem, which could influence how mathematicians think about shape optimization for polygons with more sides.
The result is limited to quadrilaterals; polygons with more sides are not considered. The proof relies on interval-arithmetic computations, which, while rigorous, are not purely analytical. Additionally, the problem is restricted to convex polygons.
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Symmetry breaking in the polygonal Szegö-Weinberger inequality as $p\to1^+$: the longest shortest-fence quadrilateral — Beniamin Bogosel, Dorin Bucur, Ilaria Fragalà
Abstract licence: CC0
Explanation generated by Synthesis from the abstract — it is not the original paper. This is a preprint, so it hasn't necessarily been peer reviewed.