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  "body": "\n## Statement\n\nLet $w \u003e 0$, and let $K$ be a plane convex body of constant width $w$.\nSo $K$ is bounded by a [[curve of constant width|B4YD1ZXRVH]] $w$.\nThen the area of $K$ is at least $\\frac{\\pi - \\sqrt{3}}{2} w^2$.\nEquality holds if and only if $K$ is a [[Reuleaux triangle|KMNV11B91V]] of width $w$.\n\n## Remarks\n\nOne classical route approximates $K$ by Reuleaux polygons of width $w$, which are bodies of constant width bounded by finitely many circular arcs of radius $w$.\nFor Reuleaux polygons one shows that the area can be lowered by deforming the polygon towards a Reuleaux triangle, and a limit argument finishes the proof.\nExample: a disc of diameter $w$ has area $\\frac{\\pi}{4} w^2 \\approx 0.785 \\, w^2$, while the Reuleaux triangle has area $\\frac{\\pi - \\sqrt{3}}{2} w^2 \\approx 0.705 \\, w^2$.\nThe constant width hypothesis is needed: a segment of length $w$ has diameter $w$ and area $0$.\n\n## Explanations\n\n### L1 human\n\nAmong all shapes that have the same width $w$ in every direction, the Reuleaux triangle uses the least area.\nIt is the \"thinnest\" such shape, while the round disc is the \"fattest\".\n\n### L2 human\n\nThe maximum among bodies of constant width $w$ is the disc, which follows from [[Barbier's theorem|T5D9RSE0X2]] and the isoperimetric inequality, since all such bodies have perimeter $\\pi w$.\nThe minimum is much harder, and the Blaschke–Lebesgue theorem identifies it.\nIn three dimensions the question of the body of constant width with least volume is open.\n",
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