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  "body": "\n## Statement\n\nLet $C \\subset \\R^2$ be a Jordan curve, that is, a [[closed curve|045YFKRW0P]] in the plane that does not meet itself.\nThen $C$ contains four points that are the vertices of a square.\n\n## Remarks\n\nThe conjecture is open in general.\nIt is known for many classes of curves, among them convex curves, polygons, and curves that are smooth enough.\nFor smooth Jordan curves, Joshua Greene and Andrew Lobb proved more: such a curve contains the vertices of a rectangle of every prescribed aspect ratio.\nThe difficulty for general continuous curves is that limits of inscribed squares of approximating smooth curves may shrink to a single point.\n\n## Explanations\n\n### L1 human\n\nDraw any closed loop on paper that never crosses itself, however wiggly.\nThe question is whether one can always find four points on the loop that form the corners of a square.\nFor a circle this is easy, and for many nice curves it has been proved, but for arbitrary continuous loops nobody knows.\n",
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      "first_stated": "1911",
      "posed_by": "Otto Toeplitz"
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    "name": "Inscribed square problem",
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  "msc": [
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  "name": "Inscribed square problem",
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  "source": "https://github.com/tamnd/math2/blob/main/nodes/J72/J72DEZHNAW.md",
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