{
  "index": "1972-A-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "\\text { A-4. Of all ellipses inscribed in a square, show that the circle has the maximum perimeter. }",
  "solution": "A-4 Let the square of sidelength \\( 2 R \\) have the vertices \\( ( \\pm R \\sqrt{2}, 0) \\) and \\( (0, \\pm R, \\overline{2}) \\). The ellipse\n\\[\n\\frac{x^{\\llcorner }}{a^{2}}+\\frac{y^{\\llcorner }}{b^{2}}=1\n\\]\nwith \\( 0 \\leqq b \\leqq a \\leqq R \\sqrt{2} \\) has the line \\( x+y=R \\sqrt{2} \\) as a tangent if and only if the quadratic equation \\( x^{2} / a^{2}+(R \\sqrt{2}-x)^{2} / b^{2}=1 \\) has a double root. It can be verified that its discriminant vanishes if and only if \\( a^{2}+b^{2}=2 R^{2} \\). As \\( a \\) varies from \\( R \\) to \\( R \\sqrt{2} \\) and \\( b \\) varies from \\( R \\) to 0 , the curve (1) varies from the circle of radius \\( R \\) through all the non-circular ellipses inscribed in the square to the degenerate \"flat\" ellipse lying on the \\( x \\)-axis.\n\nLet \\( 4 L \\) denote the length of the ellipse \\( x=a \\cos t, y=b \\sin t, 0 \\leqq t \\leqq 2 \\pi \\). Then\n\\[\n\\begin{array}{c}\nL=\\int_{0}^{\\pi / 2}\\left[a^{2} \\sin ^{2} t+b^{2} \\cos ^{2} t\\right]^{\\frac{1}{2}} d t=\\int_{0}^{\\pi / 2}\\left[\\frac{1}{2} a^{2}(1-\\cos 2 t)+\\frac{1}{2} b^{2}(1+\\cos 2 t)\\right]^{\\frac{1}{2}} d t \\\\\n=\\int_{0}^{\\pi / 2}\\left[R^{2}-\\frac{1}{2} c^{2} \\cos 2 t\\right]^{\\frac{1}{2}} d t\n\\end{array}\n\\]\nwhere \\( c^{2}=a^{2}-b^{2} \\). The last integral we split into one from 0 to \\( \\pi / 4 \\) and one from \\( \\pi / 4 \\) to \\( \\pi / 2 \\), and in the latter we substitute \\( t=\\pi / 2-t^{\\prime} \\), obtaining\n\\[\nL=\\int_{0}^{\\pi / 4}\\left\\{\\left[R^{2}-\\frac{1}{2} c^{2} \\cos 2 t\\right]^{\\frac{1}{2}}+\\left[R^{2}+\\frac{1}{2} c^{2} \\cos 2 t\\right]^{\\frac{1}{2}}\\right\\} d t\n\\]\n\nNote that \\( \\cos 2 t>0 \\) for \\( 0 \\leqq t<\\pi / 4 \\).\nNow the function \\( f(u)=(p-u)^{\\frac{1}{2}}+(p+u)^{\\frac{1}{2}} \\) decreases in the interval \\( 0 \\leqq u \\leqq p \\), because \\( 2 f^{\\prime}(u)=-(p-u)^{-\\frac{1}{2}}+(p+u)^{-\\frac{1}{2}}<0 \\) for \\( 0<u<p \\). Thus the integral in (2) as a function of \\( c \\) has its largest value when \\( c=0 \\), that is, for the inscribed circle.\n\nTo show that an ellipse inscribed in the square must have its axes along the diagonals of the square, we choose the square as having sides \\( u= \\pm R \\) and \\( v= \\pm R \\) and the ellipse as having the equation\n\\[\nA u^{2}+B u v+C v^{2}+D u+E v+F=0\n\\]\nwhere\n\\[\n4 A C-B^{2}>0\n\\]\n\nTaking the \"highest,\" \"lowest,\" \"rightest,\" and 'leftest'\" points on the ellipse, we see that all four sides of the square must be tangents to the ellipse.\n\nThe line \\( u=R \\) is a tangent if and only if the equation \\( C v^{2}+(B R+E) v \\) \\( +\\left(A r^{2}+D r+F\\right)=0 \\) has a double root or\n\\[\n(B R+E)^{2}-4 C\\left(A R^{2}+D R+F\\right)=0\n\\]\n\nThe corresponding conditions for \\( u=-R, v=R \\) and \\( v=-R \\) are\n\\[\n(-B R+E)^{2}-4 C\\left(A R^{2}-D R+F\\right)=0\n\\]\n\\[\n(B R+D)^{2}-4 A\\left(C R^{2}+E R+F\\right)=0\n\\]\nrespectively. Subtract (2) from (3) and divide by \\( 4 R \\); this gives\n\\[\n2 C D-B E=0\n\\]\n\nSimilarly, from (4) and (5),\n\\[\n-B D+2 A E=0\n\\]\n\nBy (6), (7) and (1), \\( D=E=0 \\). Therefore (2) and (4) become\n\\[\nB^{2} R^{2}-4 A C R^{2}-4 C F=0, B^{2} R^{2}-4 A C R^{2}-4 A F=0\n\\]\nrespectively. Since \\( F \\neq 0 \\), we have \\( A=C \\); this means that the ellipse has its axes along the lines \\( u \\pm v=0 \\).",
  "vars": [
    "x",
    "y",
    "t",
    "u",
    "v"
  ],
  "params": [
    "A",
    "B",
    "C",
    "D",
    "E",
    "F",
    "L",
    "R",
    "a",
    "b",
    "c",
    "r",
    "p",
    "f"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horizcoor",
        "y": "verticoor",
        "t": "paramangl",
        "u": "tiltedhori",
        "v": "tiltedvert",
        "A": "coeffalpha",
        "B": "coeffbravo",
        "C": "coeffcharlie",
        "D": "coeffdelta",
        "E": "coeffecho",
        "F": "coefffoxtrot",
        "L": "arclength",
        "R": "halfsidelg",
        "a": "semimajor",
        "b": "semiminor",
        "c": "focalparam",
        "r": "radiusvar",
        "p": "midvalue",
        "f": "funcexpr"
      },
      "question": "\\text { A-4. Of all ellipses inscribed in a square, show that the circle has the maximum perimeter. }",
      "solution": "A-4 Let the square of sidelength \\( 2 halfsidelg \\) have the vertices \\( ( \\pm halfsidelg \\sqrt{2}, 0) \\) and \\( (0, \\pm halfsidelg, \\overline{2}) \\). The ellipse\n\\[\n\\frac{horizcoor^{\\llcorner }}{semimajor^{2}}+\\frac{verticoor^{\\llcorner }}{semiminor^{2}}=1\n\\]\nwith \\( 0 \\leqq semiminor \\leqq semimajor \\leqq halfsidelg \\sqrt{2} \\) has the line \\( horizcoor+verticoor=halfsidelg \\sqrt{2} \\) as a tangent if and only if the quadratic equation \\( horizcoor^{2} / semimajor^{2}+(halfsidelg \\sqrt{2}-horizcoor)^{2} / semiminor^{2}=1 \\) has a double root. It can be verified that its discriminant vanishes if and only if \\( semimajor^{2}+semiminor^{2}=2 halfsidelg^{2} \\). As \\( semimajor \\) varies from \\( halfsidelg \\) to \\( halfsidelg \\sqrt{2} \\) and \\( semiminor \\) varies from \\( halfsidelg \\) to 0 , the curve (1) varies from the circle of radius \\( halfsidelg \\) through all the non-circular ellipses inscribed in the square to the degenerate \"flat\" ellipse lying on the \\( horizcoor \\)-axis.\n\nLet \\( 4 arclength \\) denote the length of the ellipse \\( horizcoor=semimajor \\cos paramangl, verticoor=semiminor \\sin paramangl, 0 \\leqq paramangl \\leqq 2 \\pi \\). Then\n\\[\n\\begin{array}{c}\narclength=\\int_{0}^{\\pi / 2}\\left[semimajor^{2} \\sin ^{2} paramangl+semiminor^{2} \\cos ^{2} paramangl\\right]^{\\frac{1}{2}} d paramangl=\\int_{0}^{\\pi / 2}\\left[\\frac{1}{2} semimajor^{2}(1-\\cos 2 paramangl)+\\frac{1}{2} semiminor^{2}(1+\\cos 2 paramangl)\\right]^{\\frac{1}{2}} d paramangl \\\\\n=\\int_{0}^{\\pi / 2}\\left[halfsidelg^{2}-\\frac{1}{2} focalparam^{2} \\cos 2 paramangl\\right]^{\\frac{1}{2}} d paramangl\n\\end{array}\n\\]\nwhere \\( focalparam^{2}=semimajor^{2}-semiminor^{2} \\). The last integral we split into one from 0 to \\( \\pi / 4 \\) and one from \\( \\pi / 4 \\) to \\( \\pi / 2 \\), and in the latter we substitute \\( paramangl=\\pi / 2-paramangl^{\\prime} \\), obtaining\n\\[\narclength=\\int_{0}^{\\pi / 4}\\left\\{\\left[halfsidelg^{2}-\\frac{1}{2} focalparam^{2} \\cos 2 paramangl\\right]^{\\frac{1}{2}}+\\left[halfsidelg^{2}+\\frac{1}{2} focalparam^{2} \\cos 2 paramangl\\right]^{\\frac{1}{2}}\\right\\} d paramangl\n\\]\n\nNote that \\( \\cos 2 paramangl>0 \\) for \\( 0 \\leqq paramangl<\\pi / 4 \\).\nNow the function \\( funcexpr(tiltedhori)=(midvalue-tiltedhori)^{\\frac{1}{2}}+(midvalue+tiltedhori)^{\\frac{1}{2}} \\) decreases in the interval \\( 0 \\leqq tiltedhori \\leqq midvalue \\), because \\( 2 funcexpr^{\\prime}(tiltedhori)=-(midvalue-tiltedhori)^{-\\frac{1}{2}}+(midvalue+tiltedhori)^{-\\frac{1}{2}}<0 \\) for \\( 0<tiltedhori<midvalue \\). Thus the integral in (2) as a function of \\( focalparam \\) has its largest value when \\( focalparam=0 \\), that is, for the inscribed circle.\n\nTo show that an ellipse inscribed in the square must have its axes along the diagonals of the square, we choose the square as having sides \\( tiltedhori= \\pm halfsidelg \\) and \\( tiltedvert= \\pm halfsidelg \\) and the ellipse as having the equation\n\\[\ncoeffalpha\\, tiltedhori^{2}+coeffbravo\\, tiltedhori\\, tiltedvert+coeffcharlie\\, tiltedvert^{2}+coeffdelta\\, tiltedhori+coeffecho\\, tiltedvert+coefffoxtrot=0\n\\]\nwhere\n\\[\n4\\, coeffalpha\\, coeffcharlie-coeffbravo^{2}>0\n\\]\n\nTaking the \"highest,\" \"lowest,\" \"rightest,\" and \"leftest\" points on the ellipse, we see that all four sides of the square must be tangents to the ellipse.\n\nThe line \\( tiltedhori=halfsidelg \\) is a tangent if and only if the equation \\( coeffcharlie\\, tiltedvert^{2}+(coeffbravo\\, halfsidelg+coeffecho)\\, tiltedvert +\\left(coeffalpha\\, radiusvar^{2}+coeffdelta\\, radiusvar+coefffoxtrot\\right)=0 \\) has a double root or\n\\[\n(coeffbravo\\, halfsidelg+coeffecho)^{2}-4\\, coeffcharlie\\left(coeffalpha\\, halfsidelg^{2}+coeffdelta\\, halfsidelg+coefffoxtrot\\right)=0\n\\]\n\nThe corresponding conditions for \\( tiltedhori=-halfsidelg, tiltedvert=halfsidelg \\) and \\( tiltedvert=-halfsidelg \\) are\n\\[\n(-coeffbravo\\, halfsidelg+coeffecho)^{2}-4\\, coeffcharlie\\left(coeffalpha\\, halfsidelg^{2}-coeffdelta\\, halfsidelg+coefffoxtrot\\right)=0\n\\]\n\\[\n(coeffbravo\\, halfsidelg+coeffdelta)^{2}-4\\, coeffalpha\\left(coeffcharlie\\, halfsidelg^{2}+coeffecho\\, halfsidelg+coefffoxtrot\\right)=0\n\\]\nrespectively. Subtract (2) from (3) and divide by \\( 4 halfsidelg \\); this gives\n\\[\n2\\, coeffcharlie\\, coeffdelta-coeffbravo\\, coeffecho=0\n\\]\n\nSimilarly, from (4) and (5),\n\\[\n-coeffbravo\\, coeffdelta+2\\, coeffalpha\\, coeffecho=0\n\\]\n\nBy (6), (7) and (1), \\( coeffdelta=coeffecho=0 \\). Therefore (2) and (4) become\n\\[\ncoeffbravo^{2}\\, halfsidelg^{2}-4\\, coeffalpha\\, coeffcharlie\\, halfsidelg^{2}-4\\, coeffcharlie\\, coefffoxtrot=0,\\quad coeffbravo^{2}\\, halfsidelg^{2}-4\\, coeffalpha\\, coeffcharlie\\, halfsidelg^{2}-4\\, coeffalpha\\, coefffoxtrot=0\n\\]\nrespectively. Since \\( coefffoxtrot \\neq 0 \\), we have \\( coeffalpha=coeffcharlie \\); this means that the ellipse has its axes along the lines \\( tiltedhori \\pm tiltedvert=0 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sandstone",
        "y": "driftwood",
        "t": "trelliswork",
        "u": "buttercup",
        "v": "dragonfly",
        "A": "blueberry",
        "B": "rainstorm",
        "C": "lighthouse",
        "D": "quesadilla",
        "E": "marshmallow",
        "F": "blacksmith",
        "L": "salamander",
        "R": "carpentry",
        "a": "ploughshare",
        "b": "cantilever",
        "c": "parchment",
        "r": "rainwater",
        "p": "goldsmith",
        "f": "candlewick"
      },
      "question": "\\text { A-4. Of all ellipses inscribed in a square, show that the circle has the maximum perimeter. }",
      "solution": "A-4 Let the square of sidelength \\( 2 carpentry \\) have the vertices \\( ( \\pm carpentry \\sqrt{2}, 0) \\) and \\( (0, \\pm carpentry, \\overline{2}) \\). The ellipse\n\\[\n\\frac{sandstone^{\\llcorner }}{ploughshare^{2}}+\\frac{driftwood^{\\llcorner }}{cantilever^{2}}=1\n\\]\nwith \\( 0 \\leqq cantilever \\leqq ploughshare \\leqq carpentry \\sqrt{2} \\) has the line \\( sandstone+driftwood=carpentry \\sqrt{2} \\) as a tangent if and only if the quadratic equation \\( sandstone^{2} / ploughshare^{2}+(carpentry \\sqrt{2}-sandstone)^{2} / cantilever^{2}=1 \\) has a double root. It can be verified that its discriminant vanishes if and only if \\( ploughshare^{2}+cantilever^{2}=2 carpentry^{2} \\). As \\( ploughshare \\) varies from \\( carpentry \\) to \\( carpentry \\sqrt{2} \\) and \\( cantilever \\) varies from \\( carpentry \\) to 0 , the curve (1) varies from the circle of radius \\( carpentry \\) through all the non-circular ellipses inscribed in the square to the degenerate \"flat\" ellipse lying on the \\( sandstone \\)-axis.\n\nLet \\( 4 salamander \\) denote the length of the ellipse \\( sandstone=ploughshare \\cos trelliswork, driftwood=cantilever \\sin trelliswork, 0 \\leqq trelliswork \\leqq 2 \\pi \\). Then\n\\[\n\\begin{array}{c}\nsalamander=\\int_{0}^{\\pi / 2}\\left[ploughshare^{2} \\sin ^{2} trelliswork+cantilever^{2} \\cos ^{2} trelliswork\\right]^{\\frac{1}{2}} d\\,trelliswork=\\int_{0}^{\\pi / 2}\\left[\\frac{1}{2} ploughshare^{2}(1-\\cos 2 trelliswork)+\\frac{1}{2} cantilever^{2}(1+\\cos 2 trelliswork)\\right]^{\\frac{1}{2}} d\\,trelliswork \\\\\n=\\int_{0}^{\\pi / 2}\\left[carpentry^{2}-\\frac{1}{2} parchment^{2} \\cos 2 trelliswork\\right]^{\\frac{1}{2}} d\\,trelliswork\n\\end{array}\n\\]\nwhere \\( parchment^{2}=ploughshare^{2}-cantilever^{2} \\). The last integral we split into one from 0 to \\( \\pi / 4 \\) and one from \\( \\pi / 4 \\) to \\( \\pi / 2 \\), and in the latter we substitute \\( trelliswork=\\pi / 2-trelliswork^{\\prime} \\), obtaining\n\\[\nsalamander=\\int_{0}^{\\pi / 4}\\left\\{\\left[carpentry^{2}-\\frac{1}{2} parchment^{2} \\cos 2 trelliswork\\right]^{\\frac{1}{2}}+\\left[carpentry^{2}+\\frac{1}{2} parchment^{2} \\cos 2 trelliswork\\right]^{\\frac{1}{2}}\\right\\} d\\,trelliswork\n\\]\n\nNote that \\( \\cos 2 trelliswork>0 \\) for \\( 0 \\leqq trelliswork<\\pi / 4 \\).\nNow the function \\( candlewick(buttercup)=(goldsmith-buttercup)^{\\frac{1}{2}}+(goldsmith+buttercup)^{\\frac{1}{2}} \\) decreases in the interval \\( 0 \\leqq buttercup \\leqq goldsmith \\), because \\( 2 candlewick^{\\prime}(buttercup)=-(goldsmith-buttercup)^{-\\frac{1}{2}}+(goldsmith+buttercup)^{-\\frac{1}{2}}<0 \\) for \\( 0<buttercup<goldsmith \\). Thus the integral in (2) as a function of \\( parchment \\) has its largest value when \\( parchment=0 \\), that is, for the inscribed circle.\n\nTo show that an ellipse inscribed in the square must have its axes along the diagonals of the square, we choose the square as having sides \\( buttercup= \\pm carpentry \\) and \\( dragonfly= \\pm carpentry \\) and the ellipse as having the equation\n\\[\nblueberry buttercup^{2}+rainstorm buttercup dragonfly+lighthouse dragonfly^{2}+quesadilla buttercup+marshmallow dragonfly+blacksmith=0\n\\]\nwhere\n\\[\n4 blueberry lighthouse-rainstorm^{2}>0\n\\]\n\nTaking the \"highest,\" \"lowest,\" \"rightest,\" and 'leftest'\" points on the ellipse, we see that all four sides of the square must be tangents to the ellipse.\n\nThe line \\( buttercup=carpentry \\) is a tangent if and only if the equation \\( lighthouse dragonfly^{2}+(rainstorm carpentry+marshmallow) dragonfly +\\left(blueberry rainwater^{2}+quesadilla rainwater+blacksmith\\right)=0 \\) has a double root or\n\\[\n(rainstorm carpentry+marshmallow)^{2}-4 lighthouse\\left(blueberry carpentry^{2}+quesadilla carpentry+blacksmith\\right)=0\n\\]\n\nThe corresponding conditions for \\( buttercup=-carpentry, dragonfly=carpentry \\) and \\( dragonfly=-carpentry \\) are\n\\[\n(-rainstorm carpentry+marshmallow)^{2}-4 lighthouse\\left(blueberry carpentry^{2}-quesadilla carpentry+blacksmith\\right)=0\n\\]\n\\[\n(rainstorm carpentry+quesadilla)^{2}-4 blueberry\\left(lighthouse carpentry^{2}+marshmallow carpentry+blacksmith\\right)=0\n\\]\nrespectively. Subtract (2) from (3) and divide by \\( 4 carpentry \\); this gives\n\\[\n2 lighthouse quesadilla-rainstorm marshmallow=0\n\\]\n\nSimilarly, from (4) and (5),\n\\[\n-rainstorm quesadilla+2 blueberry marshmallow=0\n\\]\n\nBy (6), (7) and (1), \\( quesadilla=marshmallow=0 \\). Therefore (2) and (4) become\n\\[\nrainstorm^{2} carpentry^{2}-4 blueberry lighthouse carpentry^{2}-4 lighthouse blacksmith=0, \\; rainstorm^{2} carpentry^{2}-4 blueberry lighthouse carpentry^{2}-4 blueberry blacksmith=0\n\\]\nrespectively. Since \\( blacksmith \\neq 0 \\), we have \\( blueberry=lighthouse \\); this means that the ellipse has its axes along the lines \\( buttercup \\pm dragonfly=0 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalpo",
        "y": "horizontal",
        "t": "stillness",
        "u": "constants",
        "v": "steadfast",
        "A": "independent",
        "B": "unrelated",
        "C": "unaffected",
        "D": "staticity",
        "E": "instabile",
        "F": "variablex",
        "L": "shortness",
        "R": "thinness",
        "a": "concaveax",
        "b": "convexax",
        "c": "converged",
        "r": "edgeless",
        "p": "variance",
        "f": "rigidity"
      },
      "question": "\\text { A-4. Of all ellipses inscribed in a square, show that the circle has the maximum perimeter. }",
      "solution": "A-4 Let the square of sidelength \\( 2 thinness \\) have the vertices \\( ( \\pm thinness \\sqrt{2}, 0) \\) and \\((0, \\pm thinness, \\overline{2})\\). The ellipse\n\\[\n\\frac{verticalpo^{\\llcorner }}{concaveax^{2}}+\\frac{horizontal^{\\llcorner }}{convexax^{2}}=1\n\\]\nwith \\( 0 \\leqq convexax \\leqq concaveax \\leqq thinness \\sqrt{2} \\) has the line \\( verticalpo+horizontal=thinness \\sqrt{2} \\) as a tangent if and only if the quadratic equation \\( verticalpo^{2} / concaveax^{2}+(thinness \\sqrt{2}-verticalpo)^{2} / convexax^{2}=1 \\) has a double root. It can be verified that its discriminant vanishes if and only if \\( concaveax^{2}+convexax^{2}=2 thinness^{2} \\). As \\( concaveax \\) varies from \\( thinness \\) to \\( thinness \\sqrt{2} \\) and \\( convexax \\) varies from \\( thinness \\) to 0, the curve (1) varies from the circle of radius \\( thinness \\) through all the non-circular ellipses inscribed in the square to the degenerate \"flat\" ellipse lying on the \\( verticalpo \\)-axis.\n\nLet \\( 4 shortness \\) denote the length of the ellipse \\( verticalpo=concaveax \\cos stillness, \\; horizontal=convexax \\sin stillness, \\; 0 \\leqq stillness \\leqq 2 \\pi \\). Then\n\\[\n\\begin{array}{c}\nshortness=\\int_{0}^{\\pi / 2}\\left[concaveax^{2} \\sin ^{2} stillness+convexax^{2} \\cos ^{2} stillness\\right]^{\\frac{1}{2}} d stillness=\\int_{0}^{\\pi / 2}\\left[\\frac{1}{2} concaveax^{2}(1-\\cos 2 stillness)+\\frac{1}{2} convexax^{2}(1+\\cos 2 stillness)\\right]^{\\frac{1}{2}} d stillness \\\\\n=\\int_{0}^{\\pi / 2}\\left[thinness^{2}-\\frac{1}{2} converged^{2} \\cos 2 stillness\\right]^{\\frac{1}{2}} d stillness\n\\end{array}\n\\]\nwhere \\( converged^{2}=concaveax^{2}-convexax^{2} \\). The last integral we split into one from 0 to \\( \\pi / 4 \\) and one from \\( \\pi / 4 \\) to \\( \\pi / 2 \\), and in the latter we substitute \\( stillness=\\pi / 2-stillness^{\\prime} \\), obtaining\n\\[\nshortness=\\int_{0}^{\\pi / 4}\\left\\{\\left[thinness^{2}-\\frac{1}{2} converged^{2} \\cos 2 stillness\\right]^{\\frac{1}{2}}+\\left[thinness^{2}+\\frac{1}{2} converged^{2} \\cos 2 stillness\\right]^{\\frac{1}{2}}\\right\\} d stillness\n\\]\n\nNote that \\( \\cos 2 stillness>0 \\) for \\( 0 \\leqq stillness<\\pi / 4 \\).\nNow the function \\( rigidity(constants)=(variance-constants)^{\\frac{1}{2}}+(variance+constants)^{\\frac{1}{2}} \\) decreases in the interval \\( 0 \\leqq constants \\leqq variance \\), because \\( 2 rigidity^{\\prime}(constants)=-(variance-constants)^{-\\frac{1}{2}}+(variance+constants)^{-\\frac{1}{2}}<0 \\) for \\( 0<constants<variance \\). Thus the integral in (2) as a function of \\( converged \\) has its largest value when \\( converged=0 \\), that is, for the inscribed circle.\n\nTo show that an ellipse inscribed in the square must have its axes along the diagonals of the square, we choose the square as having sides \\( constants= \\pm thinness \\) and \\( steadfast= \\pm thinness \\) and the ellipse as having the equation\n\\[\nindependent\\,constants^{2}+unrelated\\,constants\\,steadfast+unaffected\\,steadfast^{2}+staticity\\,constants+instabile\\,steadfast+variablex=0\n\\]\nwhere\n\\[\n4\\,independent\\,unaffected-unrelated^{2}>0\n\\]\n\nTaking the \"highest,\" \"lowest,\" \"rightest,\" and 'leftest'\" points on the ellipse, we see that all four sides of the square must be tangents to the ellipse.\n\nThe line \\( constants=thinness \\) is a tangent if and only if the equation \\( unaffected\\,steadfast^{2}+(unrelated\\,thinness+instabile)\\,steadfast +(independent\\,edgeless^{2}+staticity\\,edgeless+variablex)=0 \\) has a double root or\n\\[\n(unrelated\\,thinness+instabile)^{2}-4\\,unaffected\\left(independent\\,thinness^{2}+staticity\\,thinness+variablex\\right)=0\n\\]\n\nThe corresponding conditions for \\( constants=-thinness, \\; steadfast=thinness \\) and \\( steadfast=-thinness \\) are\n\\[\n(-unrelated\\,thinness+instabile)^{2}-4\\,unaffected\\left(independent\\,thinness^{2}-staticity\\,thinness+variablex\\right)=0\n\\]\n\\[\n(unrelated\\,thinness+staticity)^{2}-4\\,independent\\left(unaffected\\,thinness^{2}+instabile\\,thinness+variablex\\right)=0\n\\]\nrespectively. Subtract (2) from (3) and divide by \\( 4 thinness \\); this gives\n\\[\n2\\,unaffected\\,staticity-unrelated\\,instabile=0\n\\]\n\nSimilarly, from (4) and (5),\n\\[\n-unrelated\\,staticity+2\\,independent\\,instabile=0\n\\]\n\nBy (6), (7) and (1), \\( staticity=instabile=0 \\). Therefore (2) and (4) become\n\\[\nunrelated^{2} thinness^{2}-4\\,independent\\,unaffected\\,thinness^{2}-4\\,unaffected\\,variablex=0, \\quad unrelated^{2} thinness^{2}-4\\,independent\\,unaffected\\,thinness^{2}-4\\,independent\\,variablex=0\n\\]\nrespectively. Since \\( variablex \\neq 0 \\), we have \\( independent=unaffected \\); this means that the ellipse has its axes along the lines \\( constants \\pm steadfast=0 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "t": "mndpvcia",
        "u": "ksbxqjre",
        "v": "lthwymod",
        "A": "zrfvopkia",
        "B": "glsemydtu",
        "C": "bnqazxwer",
        "D": "lmvkpstio",
        "E": "xtqwhsarc",
        "F": "pdohlwgye",
        "L": "rmcfujbka",
        "R": "nskgjqlwe",
        "a": "jratvchbo",
        "b": "gxdmruqsz",
        "c": "wplkzenya",
        "r": "yfgpdsmci",
        "p": "htzkncavo",
        "f": "osiuxlqen"
      },
      "question": "\\text { A-4. Of all ellipses inscribed in a square, show that the circle has the maximum perimeter. }",
      "solution": "A-4 Let the square of sidelength \\( 2 nskgjqlwe \\) have the vertices \\( ( \\pm nskgjqlwe \\sqrt{2}, 0) \\) and \\( (0, \\pm nskgjqlwe, \\overline{2}) \\). The ellipse\n\\[\n\\frac{qzxwvtnp^{\\llcorner }}{jratvchbo^{2}}+\\frac{hjgrksla^{\\llcorner }}{gxdmruqsz^{2}}=1\n\\]\nwith \\( 0 \\leqq gxdmruqsz \\leqq jratvchbo \\leqq nskgjqlwe \\sqrt{2} \\) has the line \\( qzxwvtnp+hjgrksla=nskgjqlwe \\sqrt{2} \\) as a tangent if and only if the quadratic equation \\( qzxwvtnp^{2} / jratvchbo^{2}+(nskgjqlwe \\sqrt{2}-qzxwvtnp)^{2} / gxdmruqsz^{2}=1 \\) has a double root. It can be verified that its discriminant vanishes if and only if \\( jratvchbo^{2}+gxdmruqsz^{2}=2 nskgjqlwe^{2} \\). As \\( jratvchbo \\) varies from \\( nskgjqlwe \\) to \\( nskgjqlwe \\sqrt{2} \\) and \\( gxdmruqsz \\) varies from \\( nskgjqlwe \\) to 0 , the curve (1) varies from the circle of radius \\( nskgjqlwe \\) through all the non-circular ellipses inscribed in the square to the degenerate \"flat\" ellipse lying on the \\( qzxwvtnp \\)-axis.\n\nLet \\( 4 rmcfujbka \\) denote the length of the ellipse \\( qzxwvtnp=jratvchbo \\cos mndpvcia, hjgrksla=gxdmruqsz \\sin mndpvcia, 0 \\leqq mndpvcia \\leqq 2 \\pi \\). Then\n\\[\n\\begin{array}{c}\nrmcfujbka=\\int_{0}^{\\pi / 2}\\left[jratvchbo^{2} \\sin ^{2} mndpvcia+gxdmruqsz^{2} \\cos ^{2} mndpvcia\\right]^{\\frac{1}{2}} d mndpvcia=\\int_{0}^{\\pi / 2}\\left[\\frac{1}{2} jratvchbo^{2}(1-\\cos 2 mndpvcia)+\\frac{1}{2} gxdmruqsz^{2}(1+\\cos 2 mndpvcia)\\right]^{\\frac{1}{2}} d mndpvcia \\\\\n=\\int_{0}^{\\pi / 2}\\left[nskgjqlwe^{2}-\\frac{1}{2} wplkzenya^{2} \\cos 2 mndpvcia\\right]^{\\frac{1}{2}} d mndpvcia\n\\end{array}\n\\]\nwhere \\( wplkzenya^{2}=jratvchbo^{2}-gxdmruqsz^{2} \\). The last integral we split into one from 0 to \\( \\pi / 4 \\) and one from \\( \\pi / 4 \\) to \\( \\pi / 2 \\), and in the latter we substitute \\( mndpvcia=\\pi / 2-mndpvcia^{\\prime} \\), obtaining\n\\[\nrmcfujbka=\\int_{0}^{\\pi / 4}\\left\\{\\left[nskgjqlwe^{2}-\\frac{1}{2} wplkzenya^{2} \\cos 2 mndpvcia\\right]^{\\frac{1}{2}}+\\left[nskgjqlwe^{2}+\\frac{1}{2} wplkzenya^{2} \\cos 2 mndpvcia\\right]^{\\frac{1}{2}}\\right\\} d mndpvcia\n\\]\n\nNote that \\( \\cos 2 mndpvcia>0 \\) for \\( 0 \\leqq mndpvcia<\\pi / 4 \\).\nNow the function \\( osiuxlqen(ksbxqjre)=(htzkncavo-ksbxqjre)^{\\frac{1}{2}}+(htzkncavo+ksbxqjre)^{\\frac{1}{2}} \\) decreases in the interval \\( 0 \\leqq ksbxqjre \\leqq htzkncavo \\), because \\( 2 osiuxlqen^{\\prime}(ksbxqjre)=-(htzkncavo-ksbxqjre)^{-\\frac{1}{2}}+(htzkncavo+ksbxqjre)^{-\\frac{1}{2}}<0 \\) for \\( 0<ksbxqjre<htzkncavo \\). Thus the integral in (2) as a function of \\( wplkzenya \\) has its largest value when \\( wplkzenya=0 \\), that is, for the inscribed circle.\n\nTo show that an ellipse inscribed in the square must have its axes along the diagonals of the square, we choose the square as having sides \\( ksbxqjre= \\pm nskgjqlwe \\) and \\( lthwymod= \\pm nskgjqlwe \\) and the ellipse as having the equation\n\\[\nzrfvopkia ksbxqjre^{2}+glsemydtu ksbxqjre lthwymod+bnqazxwer lthwymod^{2}+lmvkpstio ksbxqjre+xtqwhsarc lthwymod+pdohlwgye=0\n\\]\nwhere\n\\[\n4 zrfvopkia bnqazxwer-glsemydtu^{2}>0\n\\]\n\nTaking the \"highest,\" \"lowest,\" \"rightest,\" and 'leftest'\" points on the ellipse, we see that all four sides of the square must be tangents to the ellipse.\n\nThe line \\( ksbxqjre=nskgjqlwe \\) is a tangent if and only if the equation \\( bnqazxwer lthwymod^{2}+(glsemydtu nskgjqlwe+xtqwhsarc) lthwymod+\\left(zrfvopkia yfgpdsmci^{2}+lmvkpstio yfgpdsmci+pdohlwgye\\right)=0 \\) has a double root or\n\\[\n(glsemydtu nskgjqlwe+xtqwhsarc)^{2}-4 bnqazxwer\\left(zrfvopkia nskgjqlwe^{2}+lmvkpstio nskgjqlwe+pdohlwgye\\right)=0\n\\]\n\nThe corresponding conditions for \\( ksbxqjre=-nskgjqlwe, lthwymod=nskgjqlwe \\) and \\( lthwymod=-nskgjqlwe \\) are\n\\[\n(-glsemydtu nskgjqlwe+xtqwhsarc)^{2}-4 bnqazxwer\\left(zrfvopkia nskgjqlwe^{2}-lmvkpstio nskgjqlwe+pdohlwgye\\right)=0\n\\]\n\\[\n(glsemydtu nskgjqlwe+lmvkpstio)^{2}-4 zrfvopkia\\left(bnqazxwer nskgjqlwe^{2}+xtqwhsarc nskgjqlwe+pdohlwgye\\right)=0\n\\]\nrespectively. Subtract (2) from (3) and divide by \\( 4 nskgjqlwe \\); this gives\n\\[\n2 bnqazxwer lmvkpstio-glsemydtu xtqwhsarc=0\n\\]\n\nSimilarly, from (4) and (5),\n\\[\n-glsemydtu lmvkpstio+2 zrfvopkia xtqwhsarc=0\n\\]\n\nBy (6), (7) and (1), \\( lmvkpstio=xtqwhsarc=0 \\). Therefore (2) and (4) become\n\\[\nglsemydtu^{2} nskgjqlwe^{2}-4 zrfvopkia bnqazxwer nskgjqlwe^{2}-4 bnqazxwer pdohlwgye=0,\\quad glsemydtu^{2} nskgjqlwe^{2}-4 zrfvopkia bnqazxwer nskgjqlwe^{2}-4 zrfvopkia pdohlwgye=0\n\\]\nrespectively. Since \\( pdohlwgye \\neq 0 \\), we have \\( zrfvopkia=bnqazxwer \\); this means that the ellipse has its axes along the lines \\( ksbxqjre \\pm lthwymod=0 \\)."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 2$ be an integer and set  \n\\[\n\\mathcal C_{n}:=\\bigl\\{x=(x_{1},\\dots ,x_{n})\\in\\mathbb R^{n}\\;:\\;|x_{i}|\\le 1\\text{ for }1\\le i\\le n\\bigr\\},\n\\]\nthe closed unit $n$–cube centred at the origin.\n\nFor every invertible matrix $A\\in\\mathbb R^{\\,n\\times n}$ define the ellipsoid  \n\\[\n\\mathcal E(A):=\\bigl\\{Au\\;:\\;u\\in\\mathbb R^{n},\\ \\lVert u\\rVert_{2}\\le 1\\bigr\\} \\tag{1}\n\\]\nand denote by $S(A)$ the $(n-1)$–dimensional Hausdorff measure of $\\partial\\mathcal E(A)$.\n\nThe ellipsoid $\\mathcal E(A)$ is called \\emph{admissible} if  \n\n(i) $\\mathcal E(A)\\subset\\mathcal C_{n}$, and  \n\n(ii) $\\mathcal E(A)$ is tangent to each of the $2n$ facets $x_{i}=\\pm 1$ of $\\mathcal C_{n}$  \n(the inclusion in (i) follows from (ii), but we state both for emphasis).\n\na) Prove the isoperimetric–type inequality  \n\\[\nS(A)\\le S(I_{n})=\\lvert S^{\\,n-1}\\rvert , \\tag{2}\n\\]\nfor every admissible matrix $A$.\n\nb) Show that equality in \\textup{(2)} holds if and only if $A$ is orthogonal, i.e. $AA^{\\top}=I_{n}$.  \nHence, among all ellipsoids that are contained in the cube and tangent to every facet, the Euclidean unit ball is the unique (up to rotation) maximiser of the surface area.",
      "solution": "Throughout write  \n\\[\nG:=AA^{\\top}\\in\\mathbb R^{\\,n\\times n},\\qquad G\\text{ symmetric positive–definite}. \\tag{3}\n\\]\nThen  \n\\[\n\\mathcal E(A)=\\bigl\\{x\\in\\mathbb R^{n}\\;:\\;x^{\\top}G^{-1}x\\le 1\\bigr\\}. \\tag{4}\n\\]\n\nStep 0.  Containment from tangency via support functions.  \nFor a convex body $K$ the support function is $h_{K}(\\xi):=\\sup_{x\\in K}x\\cdot\\xi$.\nFor the cube one has  \n\\[\nh_{\\mathcal C_{n}}(\\xi)=\\sum_{i=1}^{n}|\\,\\xi_{i}\\,|. \\tag{5}\n\\]\nFor the ellipsoid (4) the support function equals  \n\\[\nh_{\\mathcal E(A)}(\\xi)=\\sqrt{\\xi^{\\top}G\\,\\xi}. \\tag{6}\n\\]\nBecause of tangency we have $h_{\\mathcal E(A)}(\\pm e_{i})=1$ for every $i$.  \nNow $\\pm e_{i}$ are the extreme points of the cross–polytope  \n\\(\n\\bigl\\{\\xi\\in\\mathbb R^{n}\\;:\\;h_{\\mathcal C_{n}}(\\xi)=1\\bigr\\},\n\\)\nand both $h_{\\mathcal E(A)}$ and $h_{\\mathcal C_{n}}$ are convex and positively\nhomogeneous of degree $1$.  Therefore  \n\\[\nh_{\\mathcal E(A)}(\\xi)\\le h_{\\mathcal C_{n}}(\\xi)\\quad\\forall\\,\\xi\\in\\mathbb R^{n},\n\\]\nwhich is equivalent to $\\mathcal E(A)\\subset\\mathcal C_{n}$.\n\nStep 1.  Tangency forces unit diagonal.  \nFix $k\\in\\{1,\\dots ,n\\}$ and maximise $e_{k}\\cdot x$ subject to $x^{\\top}G^{-1}x=1$.\nA Lagrange multiplier calculation yields the maximiser  \n\\[\nx^{\\ast}=\\frac{Ge_{k}}{\\sqrt{e_{k}^{\\top}Ge_{k}}},\\qquad\n\\max=\\sqrt{e_{k}^{\\top}Ge_{k}}.\n\\]\nBecause the facet $x_{k}=1$ is tangent, the maximum equals $1$, hence  \n\\[\nG_{kk}=1\\qquad(1\\le k\\le n). \\tag{7}\n\\]\n\nStep 2.  Surface area of $\\mathcal E(A)$.  \nLet $\\sigma$ denote the surface measure on $S^{\\,n-1}$.\nThe smooth map  \n\\[\nF:S^{\\,n-1}\\longrightarrow\\partial\\mathcal E(A),\\qquad F(u)=Au, \\tag{8}\n\\]\nhas $(n-1)$–Jacobian (see e.g. Lee, \\emph{Riemannian Manifolds}, Lemma 15.19)  \n\\[\n\\operatorname{Jac}_{n-1}F(u)=|\\det A|\\,\\|A^{-{\\!\\top}}u\\|_{2}. \\tag{9}\n\\]\nHence  \n\\[\n\\begin{aligned}\nS(A)&=\\int_{S^{\\,n-1}}|\\det A|\\,\\|A^{-{\\!\\top}}u\\|_{2}\\,d\\sigma(u)\\\\\n&=(\\det G)^{1/2}\\int_{S^{\\,n-1}}\\bigl(u^{\\top}G^{-1}u\\bigr)^{1/2}\\,d\\sigma(u). \\tag{10}\n\\end{aligned}\n\\]\nPut  \n\\[\nI(G):=\\int_{S^{\\,n-1}}\\bigl(u^{\\top}G^{-1}u\\bigr)^{1/2}\\,d\\sigma(u),\n\\qquad\\text{so}\\qquad S(A)=(\\det G)^{1/2}I(G). \\tag{11}\n\\]\n\nStep 3.  Two estimates.\n\n(Jensen). Because $t\\mapsto\\sqrt t$ is concave,  \n\\[\nI(G)\\le |S^{\\,n-1}|\\,\n\\sqrt{\\frac{1}{|S^{\\,n-1}|}\\int_{S^{\\,n-1}}u^{\\top}G^{-1}u\\,d\\sigma(u)}\n=|S^{\\,n-1}|\\sqrt{\\frac{\\operatorname{tr}G^{-1}}{n}}. \\tag{12}\n\\]\n\n(Algebraic inequality). We prove  \n\\[\n(\\det G)\\,\\operatorname{tr}G^{-1}\\le n. \\tag{13}\n\\]\nLet $\\lambda_{1},\\dots ,\\lambda_{n}>0$ be the eigenvalues of $G$.  \nFrom (7) we have the trace constraint $\\sum_{i=1}^{n}\\lambda_{i}=n$.\nDefine  \n\\[\n\\Phi(\\lambda):=\\log(\\det G)+\\log(\\operatorname{tr}G^{-1})\n          =\\sum_{i=1}^{n}\\log\\lambda_{i}+\\log\\Bigl(\\sum_{i=1}^{n}\\lambda_{i}^{-1}\\Bigr).\n\\]\nMaximising $\\Phi$ under $\\sum_{i}\\lambda_{i}=n$ via Lagrange multipliers gives\n\\(\n\\lambda_{1}=\\dots=\\lambda_{n}=1,\n\\)\nso\n\\(\n\\max(\\det G)\\,\\operatorname{tr}G^{-1}=n\n\\)\nand (13) follows.  Equality occurs exactly when all $\\lambda_{i}=1$ (for $n\\ge 3$; in the\ntwo–dimensional case the identity also holds but the uniqueness statement\nbelow will settle equality).\n\nCombining (11), (12) and (13) yields  \n\\[\nS(A)\\le |S^{\\,n-1}|, \\tag{14}\n\\]\nwhich proves (a).\n\nStep 4.  The equality case.  \nSuppose $S(A)=|S^{\\,n-1}|$.  Then equality holds in both (12) and (13).\n\n•  Equality in (12) (Jensen) requires the integrand to be $\\sigma$–a.e. constant, i.e.  \n\\[\nu^{\\top}G^{-1}u=\\lambda\\quad\\forall\\,u\\in S^{\\,n-1}. \\tag{15}\n\\]\nHence $G^{-1}=\\lambda I_{n}$ and $G=\\lambda^{-1}I_{n}$.\n\n•  Equality in (13) forces $\\lambda_{1}=\\dots=\\lambda_{n}$ (already granted) and, using $\\det G=\\lambda^{-n}$ together with $\\operatorname{tr}G^{-1}=n\\lambda$, gives\n\\(\n\\lambda^{-n}\\cdot n\\lambda=n\\Rightarrow\\lambda=1.\n\\)\nThus\n\\[\nG=I_{n}. \\tag{16}\n\\]\n\nTherefore $AA^{\\top}=I_{n}$, so $A$ is orthogonal.  \nConversely, any orthogonal $A$ is admissible and attains equality in (2).  \nThis completes part (b). \\qed",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.602881",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension:  The original two–dimensional question is replaced by an n–dimensional one, introducing eigen-values, determinants, traces and integration over S^{n−1}.  \n\n• Additional constraints:  Instead of four supporting lines we cope with 2n supporting hyperplanes, encoded through the matrix condition diag G = 1.  \n\n• Sophisticated structures:  The solution requires differential–geometric computation of the surface-area element of a linear image of the sphere (Step 2), Hadamard’s determinant inequality, Jensen’s inequality on the sphere, and eigen-value/minor techniques (Step 5).  \n\n• Deeper theory:  Matrix analysis and integration on manifolds replace the elementary trigonometric manipulation sufficient in 2-D.  \n\n• Multiple interacting concepts:  Optimisation combines (i) analytic expression of S(A), (ii) convexity/concavity tools, and (iii) linear-algebraic inequalities, all glued together to obtain the sharp bound and the uniqueness clause.\n\nAll of these layers render the enhanced variant substantially harder than both the original problem and the earlier kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let n ≥ 2 be an integer and put  \n  𝒞ₙ := { x = (x₁,…,xₙ) ∈ ℝⁿ : |xᵢ| ≤ 1 for 1 ≤ i ≤ n }  \n(the closed unit n–cube centred at the origin).\n\nFor every invertible real n × n matrix A define the ellipsoid  \n  𝔈(A) := { Au : u ∈ ℝⁿ and ‖u‖₂ ≤ 1 }.                               (1)\n\nWe call 𝔈(A) admissible provided that  \n(i)  𝔈(A) ⊂ 𝒞ₙ (the ellipsoid is contained in the cube), and  \n(ii)  𝔈(A) is tangent to each of the 2n facets xᵢ = ±1 of 𝒞ₙ.\n\nWrite S(A) for the (n – 1)–dimensional Hausdorff measure of ∂𝔈(A) (the surface area).\n\na) Show the isoperimetric–type bound  \n  S(A) ≤ S(Iₙ) = |S^{n−1}|                                             (2)  \nfor every admissible matrix A.\n\nb) Prove that equality in (2) holds precisely when AAᵀ = Iₙ, i.e. when A is orthogonal.  Equivalently, among all ellipsoids that are contained in the cube and tangent to every facet, the Euclidean unit ball is the unique (up to rotation) maximiser of the surface area.",
      "solution": "Throughout write  \n\n  G := A Aᵀ  ∈ ℝ^{n×n},                                                (3)\n\nso that G is symmetric and positive–definite.  The ellipsoid (1) can then be expressed as  \n\n  𝔈(A) = { x ∈ ℝⁿ : xᵀG^{-1}x ≤ 1 }.                                  (4)\n\nStep 1.  Tangency forces unit diagonal.  \nFix k ∈ {1,…,n}.  The extreme value of the k-th coordinate on the ellipsoid is obtained from the optimisation problem  \n\n  maximise e_kᵀx   subject to xᵀG^{-1}x = 1.                       (5)\n\nThe Lagrange multiplier computation gives the maximiser  \n  x* = G e_k / √(e_kᵀG e_k)  \nand the maximal value √(e_kᵀG e_k).  \nBecause of tangency (ii) this value equals 1, hence  \n\n  e_kᵀG e_k = G_{kk} = 1 for every k.                              (6)\n\nThus  \n\n  diag G = (1,…,1).                                                   (7)\n\n(Remark.  The inclusion condition (i) is **not** used here; (ii) alone is enough for (7).  Condition (i) will only play a rôle when we speak about admissible matrices.)\n\nStep 2.  A surface–area formula for 𝔈(A).  \nLet σ denote the standard surface measure on the unit sphere S^{n−1}.  \nThe map  \n\n  F : S^{n−1} → ∂𝔈(A), F(u)=Au,                                    (8)\n\nis a smooth parametrisation.  For each u ∈ S^{n−1} one has (see e.g. Lee, *Riemannian Manifolds*, Lemma 15.19)  \n\n  Jac_{n−1}F(u)=det A·‖A^{-ᵀ}u‖₂.                                (9)\n\nHence  \n\n  S(A)=∫_{S^{n−1}}det A·‖A^{-ᵀ}u‖₂ dσ(u)  \n         =(det G)^{1/2} ∫_{S^{n−1}}(uᵀG^{-1}u)^{1/2}dσ(u).            (10)\n\nIntroduce  \n\n  I(G):=∫_{S^{n−1}}(uᵀG^{-1}u)^{1/2} dσ(u), so S(A)=(det G)^{1/2}I(G). (11)\n\nStep 3.  First inequalities: Hadamard and Jensen.  \nBecause of (7) Hadamard’s determinant bound gives  \n\n  det G ≤ 1,                                                       (12)\n\nwith equality iff G = Iₙ.  \n\nFor the integral I(G) note that t ↦ √t is concave.  Applying Jensen to the probability measure σ/|S^{n−1}| we obtain  \n\n  I(G) ≤ |S^{n−1}|·√(⟨uᵀG^{-1}u⟩),                               (13)\n\nwhere ⟨·⟩ denotes averaging over S^{n−1}.  The standard identity ⟨uᵀMu⟩ = (tr M)/n yields  \n\n  I(G) ≤ |S^{n−1}|·√((tr G^{-1})/n).                              (14)\n\nCombining (11), (12) and (14) we obtain  \n\n  S(A) ≤ |S^{n−1}|·√{(det G)(tr G^{-1})/n}.                      (15)\n\nThus it suffices to prove the purely algebraic estimate  \n\n  (det G)(tr G^{-1}) ≤ n.                                          (16)\n\nStep 4.  Rewriting (det G)(tr G^{-1}).  \nFor a symmetric positive-definite matrix G let G^{ii} denote the determinant of the principal (n−1) × (n−1) minor obtained by deleting the i-th row and column.  Cramer’s rule gives  \n\n  (G^{-1})_{ii} = G^{ii}/det G.                                    (17)\n\nHence  \n\n  (det G)(tr G^{-1}) = ∑_{i=1}^{n}G^{ii}.                         (18)\n\nStep 5.  Bounding every principal minor.  \nBecause each reduced matrix still has diagonal entries equal to 1, Hadamard’s inequality implies  \n\n  G^{ii} ≤ 1 for all i.                                            (19)\n\nStep 6.  Completion of the estimate and of part (a).  \nSumming (19) and using (18) yields (16).  Substituting (16) back into (15) proves  \n\n  S(A) ≤ |S^{n−1}| = S(Iₙ),                                      (20)\n\nestablishing (2).\n\nStep 7.  The equality case.  \nSuppose S(A)=|S^{n−1}|.  Equality must then occur in every intermediate inequality:\n\n• From (12) we need det G = 1, whence by Hadamard G = Iₙ.  \n• Conversely, if G = Iₙ then (11) immediately gives S(A)=|S^{n−1}|.\n\nThus equality in (2) is equivalent to G = Iₙ.  But G = Iₙ means  \n\n  AAᵀ = Iₙ ⇔ A ∈ O(n).                                          (21)\n\nAny orthogonal matrix clearly satisfies (i) and (ii): 𝔈(A) is the Euclidean unit ball, which is contained in the cube and touches every facet at the points ±e₁,…,±eₙ.  Hence the Euclidean unit ball is the unique admissible ellipsoid of maximal surface area, and equality holds precisely for orthogonal A. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.482298",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension:  The original two–dimensional question is replaced by an n–dimensional one, introducing eigen-values, determinants, traces and integration over S^{n−1}.  \n\n• Additional constraints:  Instead of four supporting lines we cope with 2n supporting hyperplanes, encoded through the matrix condition diag G = 1.  \n\n• Sophisticated structures:  The solution requires differential–geometric computation of the surface-area element of a linear image of the sphere (Step 2), Hadamard’s determinant inequality, Jensen’s inequality on the sphere, and eigen-value/minor techniques (Step 5).  \n\n• Deeper theory:  Matrix analysis and integration on manifolds replace the elementary trigonometric manipulation sufficient in 2-D.  \n\n• Multiple interacting concepts:  Optimisation combines (i) analytic expression of S(A), (ii) convexity/concavity tools, and (iii) linear-algebraic inequalities, all glued together to obtain the sharp bound and the uniqueness clause.\n\nAll of these layers render the enhanced variant substantially harder than both the original problem and the earlier kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}