{
  "index": "2007-A-2",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "Find the least possible area of a convex set in the plane that\nintersects both branches of the hyperbola $xy = 1$ and both branches of\nthe hyperbola $xy = -1$. (A set $S$ in the plane is called \\emph{convex}\nif for any two points in $S$ the line segment connecting them is\ncontained in $S$.)",
  "solution": "The minimum is 4, achieved by the square with vertices $(\\pm 1, \\pm 1)$.\n\n\\textbf{First solution:}\nTo prove that 4 is a lower bound, let $S$ be a convex set of the desired\nform. Choose $A,B,C,D \\in S$ lying on the branches of the two hyperbolas,\nwith $A$ in the upper right\nquadrant, $B$ in the upper left, $C$ in the lower left, $D$ in the lower right.\nThen the area of the quadrilateral $ABCD$ is a lower bound for\nthe area of $S$.\n\nWrite $A = (a,1/a)$,\n$B = (-b,1/b)$, $C = (-c,-1/c)$, $D = (d, -1/d)$ with $a,b,c,d > 0$.\nThen the area of the quadrilateral $ABCD$ is\n\\[\n\\frac{1}{2}(a/b + b/c + c/d + d/a + b/a + c/b + d/c + a/d),\n\\]\nwhich by the arithmetic-geometric mean inequality is at least\n$4$.\n\n\\textbf{Second solution:}\nChoose $A,B,C,D$ as in the first solution.\nNote that both the hyperbolas and the area of the convex hull of $ABCD$ are\ninvariant under the transformation $(x,y) \\mapsto (xm, y/m)$ for any\n$m>0$. For $m$ small, the counterclockwise angle from the line $AC$ to\nthe line $BD$ approaches 0; for $m$ large, this angle approaches $\\pi$.\nBy continuity, for some $m$ this angle becomes $\\pi/2$, that is,\n$AC$ and $BD$ become perpendicular. The area of $ABCD$\nis then $AC \\cdot BD$.\n\nIt thus suffices to note that $AC \\geq 2 \\sqrt{2}$ (and similarly for $BD$).\nThis holds because if we draw the tangent lines to the hyperbola $xy=1$\nat the points $(1,1)$ and $(-1,-1)$, then $A$ and $C$ lie outside the region\nbetween these lines. If we project the segment $AC$ orthogonally\nonto the line $x=y=1$, the resulting projection has length at\nleast $2 \\sqrt{2}$,\nso $AC$ must as well.\n\n\\textbf{Third solution:}\n(by Richard Stanley)\nChoose $A,B,C,D$ as in the first solution. Now fixing $A$ and $C$, move $B$ and\n$D$ to the points at which the tangents to the curve are parallel to the line\n$AC$. This does not increase the area of the quadrilateral $ABCD$ (even if\nthis quadrilateral is not convex).\n\nNote that $B$ and $D$ are now diametrically opposite; write $B = (-x, 1/x)$\nand $D = (x, -1/x)$. If we thus repeat the\nprocedure, fixing $B$ and $D$ and moving $A$ and $C$ to the points where the\ntangents are parallel to $BD$, then $A$ and $C$ must move to $(x, 1/x)$\nand $(-x,-1/x)$, respectively, forming a rectangle of area 4.\n\n\\textbf{Remark:}\nMany geometric solutions are possible. An example suggested by David Savitt\n(due to Chris Brewer):\nnote that $AD$ and $BC$ cross the\npositive and negative $x$-axes, respectively, so the convex hull of $ABCD$\ncontains $O$. Then check that the area of triangle $OAB$ is at least 1, et cetera.",
  "vars": [
    "x",
    "y",
    "a",
    "b",
    "c",
    "d",
    "m"
  ],
  "params": [
    "S",
    "A",
    "B",
    "C",
    "D",
    "O"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "a": "alphaone",
        "b": "betatwo",
        "c": "charlie",
        "d": "deltaone",
        "m": "scalefactor",
        "S": "setplane",
        "A": "pointone",
        "B": "pointtwo",
        "C": "pointthree",
        "D": "pointfour",
        "O": "originpt"
      },
      "question": "Find the least possible area of a convex set in the plane that\nintersects both branches of the hyperbola $abscissa ordinate = 1$ and both branches of\nthe hyperbola $abscissa ordinate = -1$. (A set $setplane$ in the plane is called \\emph{convex}\nif for any two points in $setplane$ the line segment connecting them is\ncontained in $setplane$.)",
      "solution": "The minimum is 4, achieved by the square with vertices $(\\pm 1, \\pm 1)$.\n\n\\textbf{First solution:}\nTo prove that 4 is a lower bound, let $setplane$ be a convex set of the desired\nform. Choose $pointone, pointtwo, pointthree, pointfour \\in setplane$ lying on the branches of the two hyperbolas,\nwith $pointone$ in the upper right\nquadrant, $pointtwo$ in the upper left, $pointthree$ in the lower left, $pointfour$ in the lower right.\nThen the area of the quadrilateral $pointone pointtwo pointthree pointfour$ is a lower bound for\nthe area of $setplane$.\n\nWrite $pointone = (alphaone,1/alphaone)$,\n$pointtwo = (-betatwo,1/betatwo)$, $pointthree = (-charlie,-1/charlie)$, $pointfour = (deltaone, -1/deltaone)$ with $alphaone, betatwo, charlie, deltaone > 0$.\nThen the area of the quadrilateral $pointone pointtwo pointthree pointfour$ is\n\\[\n\\frac{1}{2}(alphaone/betatwo + betatwo/charlie + charlie/deltaone + deltaone/alphaone + betatwo/alphaone + charlie/betatwo + deltaone/charlie + alphaone/deltaone),\n\\]\nwhich by the arithmetic-geometric mean inequality is at least\n$4$.\n\n\\textbf{Second solution:}\nChoose $pointone, pointtwo, pointthree, pointfour$ as in the first solution.\nNote that both the hyperbolas and the area of the convex hull of $pointone pointtwo pointthree pointfour$ are\ninvariant under the transformation $(abscissa, ordinate) \\mapsto (abscissa scalefactor, ordinate/scalefactor)$ for any\n$scalefactor>0$. For $scalefactor$ small, the counterclockwise angle from the line $pointonepointthree$ to\nthe line $pointtwopointfour$ approaches 0; for $scalefactor$ large, this angle approaches $\\pi$.\nBy continuity, for some $scalefactor$ this angle becomes $\\pi/2$, that is,\n$pointonepointthree$ and $pointtwopointfour$ become perpendicular. The area of $pointone pointtwo pointthree pointfour$\nis then $pointonepointthree \\cdot pointtwopointfour$.\n\nIt thus suffices to note that $pointonepointthree \\geq 2 \\sqrt{2}$ (and similarly for $pointtwopointfour$).\nThis holds because if we draw the tangent lines to the hyperbola $abscissa ordinate=1$\nat the points $(1,1)$ and $(-1,-1)$, then $pointone$ and $pointthree$ lie outside the region\nbetween these lines. If we project the segment $pointonepointthree$ orthogonally\nonto the line $abscissa=ordinate=1$, the resulting projection has length at\nleast $2 \\sqrt{2}$,\nso $pointonepointthree$ must as well.\n\n\\textbf{Third solution:}\n(by Richard Stanley)\nChoose $pointone, pointtwo, pointthree, pointfour$ as in the first solution. Now fixing $pointone$ and $pointthree$, move $pointtwo$ and\n$pointfour$ to the points at which the tangents to the curve are parallel to the line\n$pointonepointthree$. This does not increase the area of the quadrilateral $pointone pointtwo pointthree pointfour$ (even if\nthis quadrilateral is not convex).\n\nNote that $pointtwo$ and $pointfour$ are now diametrically opposite; write $pointtwo = (-abscissa, 1/abscissa)$\nand $pointfour = (abscissa, -1/abscissa)$. If we thus repeat the\nprocedure, fixing $pointtwo$ and $pointfour$ and moving $pointone$ and $pointthree$ to the points where the\ntangents are parallel to $pointtwopointfour$, then $pointone$ and $pointthree$ must move to $(abscissa, 1/abscissa)$\nand $(-abscissa,-1/abscissa)$, respectively, forming a rectangle of area 4.\n\n\\textbf{Remark:}\nMany geometric solutions are possible. An example suggested by David Savitt\n(due to Chris Brewer):\nnote that $pointonepointfour$ and $pointtwo pointthree$ cross the\npositive and negative $abscissa$-axes, respectively, so the convex hull of $pointone pointtwo pointthree pointfour$\ncontains $originpt$. Then check that the area of triangle $originptpointonepointtwo$ is at least 1, et cetera."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marigold",
        "y": "pinecone",
        "a": "sailboat",
        "b": "drumstick",
        "c": "lampshade",
        "d": "riverbank",
        "m": "goldfinch",
        "S": "mapleleaf",
        "A": "chrysalis",
        "B": "thunderbolt",
        "C": "watermelon",
        "D": "honeysuckle",
        "O": "cloudberry"
      },
      "question": "Find the least possible area of a convex set in the plane that\nintersects both branches of the hyperbola $marigold pinecone = 1$ and both branches of\nthe hyperbola $marigold pinecone = -1$. (A set $mapleleaf$ in the plane is called \\emph{convex}\nif for any two points in $mapleleaf$ the line segment connecting them is\ncontained in $mapleleaf$.)",
      "solution": "The minimum is 4, achieved by the square with vertices $(\\pm 1, \\pm 1)$.\n\n\\textbf{First solution:}\nTo prove that 4 is a lower bound, let $mapleleaf$ be a convex set of the desired\nform. Choose $chrysalis, thunderbolt, watermelon, honeysuckle \\in mapleleaf$ lying on the branches of the two hyperbolas,\nwith $chrysalis$ in the upper right\nquadrant, $thunderbolt$ in the upper left, $watermelon$ in the lower left, $honeysuckle$ in the lower right.\nThen the area of the quadrilateral $chrysalis thunderbolt watermelon honeysuckle$ is a lower bound for\nthe area of $mapleleaf$.\n\nWrite $chrysalis = (sailboat,1/sailboat)$,\n$thunderbolt = (-drumstick,1/drumstick)$, $watermelon = (-lampshade,-1/lampshade)$, $honeysuckle = (riverbank, -1/riverbank)$ with $sailboat,drumstick,lampshade,riverbank > 0$.\nThen the area of the quadrilateral $chrysalis thunderbolt watermelon honeysuckle$ is\n\\[\n\\frac{1}{2}(sailboat/drumstick + drumstick/lampshade + lampshade/riverbank + riverbank/sailboat + drumstick/sailboat + lampshade/drumstick + riverbank/lampshade + sailboat/riverbank),\n\\]\nwhich by the arithmetic-geometric mean inequality is at least\n$4$.\n\n\\textbf{Second solution:}\nChoose $chrysalis, thunderbolt, watermelon, honeysuckle$ as in the first solution.\nNote that both the hyperbolas and the area of the convex hull of $chrysalis thunderbolt watermelon honeysuckle$ are\ninvariant under the transformation $(marigold,pinecone) \\mapsto (marigold goldfinch, pinecone/goldfinch)$ for any\n$goldfinch>0$. For $goldfinch$ small, the counterclockwise angle from the line $chrysalis watermelon$ to\nthe line $thunderbolt honeysuckle$ approaches 0; for $goldfinch$ large, this angle approaches $\\pi$.\nBy continuity, for some $goldfinch$ this angle becomes $\\pi/2$, that is,\n$chrysalis watermelon$ and $thunderbolt honeysuckle$ become perpendicular. The area of $chrysalis thunderbolt watermelon honeysuckle$\nis then $chrysalis watermelon \\cdot thunderbolt honeysuckle$.\n\nIt thus suffices to note that $chrysalis watermelon \\geq 2 \\sqrt{2}$ (and similarly for $thunderbolt honeysuckle$).\nThis holds because if we draw the tangent lines to the hyperbola $marigold pinecone=1$\nat the points $(1,1)$ and $(-1,-1)$, then $chrysalis$ and $watermelon$ lie outside the region\nbetween these lines. If we project the segment $chrysalis watermelon$ orthogonally\nonto the line $marigold=pinecone=1$, the resulting projection has length at\nleast $2 \\sqrt{2}$,\nso $chrysalis watermelon$ must as well.\n\n\\textbf{Third solution:}\n(by Richard Stanley)\nChoose $chrysalis, thunderbolt, watermelon, honeysuckle$ as in the first solution. Now fixing $chrysalis$ and $watermelon$, move $thunderbolt$ and\n$honeysuckle$ to the points at which the tangents to the curve are parallel to the line\n$chrysalis watermelon$. This does not increase the area of the quadrilateral $chrysalis thunderbolt watermelon honeysuckle$ (even if\nthis quadrilateral is not convex).\n\nNote that $thunderbolt$ and $honeysuckle$ are now diametrically opposite; write $thunderbolt = (-marigold, 1/marigold)$\nand $honeysuckle = (marigold, -1/marigold)$. If we thus repeat the\nprocedure, fixing $thunderbolt$ and $honeysuckle$ and moving $chrysalis$ and $watermelon$ to the points where the\ntangents are parallel to $thunderbolt honeysuckle$, then $chrysalis$ and $watermelon$ must move to $(marigold, 1/marigold)$\nand $(-marigold,-1/marigold)$, respectively, forming a rectangle of area 4.\n\n\\textbf{Remark:}\nMany geometric solutions are possible. An example suggested by David Savitt\n(due to Chris Brewer):\nnote that $chrysalis honeysuckle$ and $thunderbolt watermelon$ cross the\npositive and negative $marigold$-axes, respectively, so the convex hull of $chrysalis thunderbolt watermelon honeysuckle$\ncontains $cloudberry$. Then check that the area of triangle $cloudberry chrysalis thunderbolt$ is at least 1, et cetera."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalcoordinate",
        "y": "horizontalcoordinate",
        "a": "sluggishrate",
        "b": "dormantfactor",
        "c": "inertmeasure",
        "d": "staticmagnitude",
        "m": "steadyscale",
        "S": "voidcollection",
        "A": "abysspoint",
        "B": "barrenpoint",
        "C": "chaospoint",
        "D": "desertpoint",
        "O": "infinitepoint"
      },
      "question": "Find the least possible area of a convex set in the plane that\nintersects both branches of the hyperbola $verticalcoordinatehorizontalcoordinate = 1$ and both branches of\nthe hyperbola $verticalcoordinatehorizontalcoordinate = -1$. (A set $voidcollection$ in the plane is called \\emph{convex}\nif for any two points in $voidcollection$ the line segment connecting them is\ncontained in $voidcollection$.)",
      "solution": "The minimum is 4, achieved by the square with vertices $(\\pm 1, \\pm 1)$.\n\n\\textbf{First solution:}\nTo prove that 4 is a lower bound, let $voidcollection$ be a convex set of the desired\nform. Choose $abysspoint,barrenpoint,chaospoint,desertpoint \\in voidcollection$ lying on the branches of the two hyperbolas,\nwith $abysspoint$ in the upper right\nquadrant, $barrenpoint$ in the upper left, $chaospoint$ in the lower left, $desertpoint$ in the lower right.\nThen the area of the quadrilateral $abysspoint barrenpoint chaospoint desertpoint$ is a lower bound for\nthe area of $voidcollection$.\n\nWrite $abysspoint = (sluggishrate,1/sluggishrate)$,\n$barrenpoint = (-dormantfactor,1/dormantfactor)$, $chaospoint = (-inertmeasure,-1/inertmeasure)$, $desertpoint = (staticmagnitude, -1/staticmagnitude)$ with $sluggishrate,dormantfactor,inertmeasure,staticmagnitude > 0$.\nThen the area of the quadrilateral $abysspoint barrenpoint chaospoint desertpoint$ is\n\\[\n\\frac{1}{2}(sluggishrate/dormantfactor + dormantfactor/inertmeasure + inertmeasure/staticmagnitude + staticmagnitude/sluggishrate + dormantfactor/sluggishrate + inertmeasure/dormantfactor + staticmagnitude/inertmeasure + sluggishrate/staticmagnitude),\n\\]\nwhich by the arithmetic-geometric mean inequality is at least\n$4$.\n\n\\textbf{Second solution:}\nChoose $abysspoint,barrenpoint,chaospoint,desertpoint$ as in the first solution.\nNote that both the hyperbolas and the area of the convex hull of $abysspoint barrenpoint chaospoint desertpoint$ are\ninvariant under the transformation $(verticalcoordinate,horizontalcoordinate) \\mapsto (verticalcoordinate steadyscale, horizontalcoordinate/steadyscale)$ for any\n$steadyscale>0$. For $steadyscale$ small, the counterclockwise angle from the line $abysspoint chaospoint$ to\nthe line $barrenpoint desertpoint$ approaches 0; for $steadyscale$ large, this angle approaches $\\pi$.\nBy continuity, for some $steadyscale$ this angle becomes $\\pi/2$, that is,\n$abysspoint chaospoint$ and $barrenpoint desertpoint$ become perpendicular. The area of $abysspoint barrenpoint chaospoint desertpoint$\nis then $abysspoint chaospoint \\cdot barrenpoint desertpoint$.\n\nIt thus suffices to note that $abysspoint chaospoint \\geq 2 \\sqrt{2}$ (and similarly for $barrenpoint desertpoint$).\nThis holds because if we draw the tangent lines to the hyperbola $verticalcoordinatehorizontalcoordinate=1$\nat the points $(1,1)$ and $(-1,-1)$, then $abysspoint$ and $chaospoint$ lie outside the region\nbetween these lines. If we project the segment $abysspoint chaospoint$ orthogonally\nonto the line $verticalcoordinate=horizontalcoordinate=1$, the resulting projection has length at\nleast $2 \\sqrt{2}$,\nso $abysspoint chaospoint$ must as well.\n\n\\textbf{Third solution:}\n(by Richard Stanley)\nChoose $abysspoint,barrenpoint,chaospoint,desertpoint$ as in the first solution. Now fixing $abysspoint$ and $chaospoint$, move $barrenpoint$ and\n$desertpoint$ to the points at which the tangents to the curve are parallel to the line\n$abysspoint chaospoint$. This does not increase the area of the quadrilateral $abysspoint barrenpoint chaospoint desertpoint$ (even if\nthis quadrilateral is not convex).\n\nNote that $barrenpoint$ and $desertpoint$ are now diametrically opposite; write $barrenpoint = (-verticalcoordinate, 1/verticalcoordinate)$\nand $desertpoint = (verticalcoordinate, -1/verticalcoordinate)$. If we thus repeat the\nprocedure, fixing $barrenpoint$ and $desertpoint$ and moving $abysspoint$ and $chaospoint$ to the points where the\ntangents are parallel to $barrenpoint desertpoint$, then $abysspoint$ and $chaospoint$ must move to $(verticalcoordinate, 1/verticalcoordinate)$\nand $(-verticalcoordinate,-1/verticalcoordinate)$, respectively, forming a rectangle of area 4.\n\n\\textbf{Remark:}\nMany geometric solutions are possible. An example suggested by David Savitt\n(due to Chris Brewer):\nnote that $abysspoint desertpoint$ and $barrenpoint chaospoint$ cross the\npositive and negative $x$-axes, respectively, so the convex hull of $abysspoint barrenpoint chaospoint desertpoint$\ncontains $infinitepoint$. Then check that the area of triangle $infinitepoint abysspoint barrenpoint$ is at least 1, et cetera."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "a": "mvdcnrke",
        "b": "pfkjlwzt",
        "c": "rgnsfqya",
        "d": "ltxpmhqo",
        "m": "swzqfeyn",
        "S": "xjbrundl",
        "A": "bphtsova",
        "B": "fzldmqer",
        "C": "tvsnkgha",
        "D": "wycrseup",
        "O": "nsfqmpad"
      },
      "question": "Find the least possible area of a convex set in the plane that\nintersects both branches of the hyperbola $qzxwvtnphjgrksla = 1$ and both branches of\nthe hyperbola $qzxwvtnphjgrksla = -1$. (A set $xjbrundl$ in the plane is called \\emph{convex}\nif for any two points in $xjbrundl$ the line segment connecting them is\ncontained in $xjbrundl$.)",
      "solution": "The minimum is 4, achieved by the square with vertices $(\\pm 1, \\pm 1)$.\n\n\\textbf{First solution:}\nTo prove that 4 is a lower bound, let $xjbrundl$ be a convex set of the desired\nform. Choose $bphtsova,fzldmqer,tvsnkgha,wycrseup \\in xjbrundl$ lying on the branches of the two hyperbolas,\nwith $bphtsova$ in the upper right\nquadrant, $fzldmqer$ in the upper left, $tvsnkgha$ in the lower left, $wycrseup$ in the lower right.\nThen the area of the quadrilateral $bphtsova fzldmqer tvsnkgha wycrseup$ is a lower bound for\nthe area of $xjbrundl$.\n\nWrite $bphtsova = (mvdcnrke,1/mvdcnrke)$,\n$fzldmqer = (-pfkjlwzt,1/pfkjlwzt)$, $tvsnkgha = (-rgnsfqya,-1/rgnsfqya)$, $wycrseup = (ltxpmhqo, -1/ltxpmhqo)$ with $mvdcnrke,pfkjlwzt,rgnsfqya,ltxpmhqo > 0$.\nThen the area of the quadrilateral $bphtsova fzldmqer tvsnkgha wycrseup$ is\n\\[\n\\frac{1}{2}(mvdcnrke/pfkjlwzt + pfkjlwzt/rgnsfqya + rgnsfqya/ltxpmhqo + ltxpmhqo/mvdcnrke + pfkjlwzt/mvdcnrke + rgnsfqya/pfkjlwzt + ltxpmhqo/rgnsfqya + mvdcnrke/ltxpmhqo),\n\\]\nwhich by the arithmetic-geometric mean inequality is at least\n$4$.\n\n\\textbf{Second solution:}\nChoose $bphtsova,fzldmqer,tvsnkgha,wycrseup$ as in the first solution.\nNote that both the hyperbolas and the area of the convex hull of $bphtsova fzldmqer tvsnkgha wycrseup$ are\ninvariant under the transformation $(qzxwvtnp,hjgrksla) \\mapsto (qzxwvtnpswzqfeyn, hjgrksla/swzqfeyn)$ for any\n$swzqfeyn>0$. For $swzqfeyn$ small, the counterclockwise angle from the line $bphtsova tvsnkgha$ to\nthe line $fzldmqer wycrseup$ approaches 0; for $swzqfeyn$ large, this angle approaches $\\pi$.\nBy continuity, for some $swzqfeyn$ this angle becomes $\\pi/2$, that is,\n$bphtsova tvsnkgha$ and $fzldmqer wycrseup$ become perpendicular. The area of $bphtsova fzldmqer tvsnkgha wycrseup$\nis then $bphtsova tvsnkgha \\cdot fzldmqer wycrseup$.\n\nIt thus suffices to note that $bphtsova tvsnkgha \\geq 2 \\sqrt{2}$ (and similarly for $fzldmqer wycrseup$).\nThis holds because if we draw the tangent lines to the hyperbola $qzxwvtnphjgrksla=1$\nat the points $(1,1)$ and $(-1,-1)$, then $bphtsova$ and $tvsnkgha$ lie outside the region\nbetween these lines. If we project the segment $bphtsova tvsnkgha$ orthogonally\nonto the line $qzxwvtnp=hjgrksla=1$, the resulting projection has length at\nleast $2 \\sqrt{2}$,\nso $bphtsova tvsnkgha$ must as well.\n\n\\textbf{Third solution:}\n(by Richard Stanley)\nChoose $bphtsova,fzldmqer,tvsnkgha,wycrseup$ as in the first solution. Now fixing $bphtsova$ and $tvsnkgha$, move $fzldmqer$ and\n$wycrseup$ to the points at which the tangents to the curve are parallel to the line\n$bphtsova tvsnkgha$. This does not increase the area of the quadrilateral $bphtsova fzldmqer tvsnkgha wycrseup$ (even if\nthis quadrilateral is not convex).\n\nNote that $fzldmqer$ and $wycrseup$ are now diametrically opposite; write $fzldmqer = (-qzxwvtnp, 1/qzxwvtnp)$\nand $wycrseup = (qzxwvtnp, -1/qzxwvtnp)$. If we thus repeat the\nprocedure, fixing $fzldmqer$ and $wycrseup$ and moving $bphtsova$ and $tvsnkgha$ to the points where the\ntangents are parallel to $fzldmqer wycrseup$, then $bphtsova$ and $tvsnkgha$ must move to $(qzxwvtnp, 1/qzxwvtnp)$\nand $(-qzxwvtnp,-1/qzxwvtnp)$, respectively, forming a rectangle of area 4.\n\n\\textbf{Remark:}\nMany geometric solutions are possible. An example suggested by David Savitt\n(due to Chris Brewer):\nnote that $bphtsova wycrseup$ and $fzldmqer tvsnkgha$ cross the\npositive and negative $qzxwvtnp$-axes, respectively, so the convex hull of $bphtsova fzldmqer tvsnkgha wycrseup$\ncontains $nsfqmpad$. Then check that the area of triangle $nsfqmpad bphtsova fzldmqer$ is at least 1, et cetera."
    },
    "kernel_variant": {
      "question": "Let n ≥ 2 be an integer and write, for x=(x₁,…,xₙ)∈ℝⁿ,  \n\n  Σ₊ := {x : x₁x₂⋯xₙ =  1},  Σ₋ := {x : x₁x₂⋯xₙ = –1}.  \n\nBoth hypersurfaces consist of 2ⁿ⁻¹ connected components, one in every open orthant.  \nA convex body K⊂ℝⁿ is called unconditional if, together with any point  \n(x₁,…,xₙ)∈K, it contains all 2ⁿ points obtained by changing the signs of the\ncoordinates.  (Equivalently: K is symmetric with respect to each coordinate hyper-plane.)\n\nWe say that K dominates the two hypersurfaces if it meets every component of Σ₊∪Σ₋.\nFor an unconditional body this is equivalent to K∩Σ₊≠∅.\n\nAmong all unconditional convex bodies that dominate Σ₊∪Σ₋ determine  \n\na) the minimum possible n-dimensional volume  \n\n  Vₙ := min{ Volₙ(K) : K unconditional, K∩Σ₊≠∅ };  \n\nb) all bodies for which that minimum is attained.",
      "solution": "Throughout Volₙ denotes n-dimensional Lebesgue measure and ℝⁿ₊ the positive\northant.\n\nStep 1.  A canonical positive point.  \nIf K is unconditional and dominates Σ₊∪Σ₋, then there is a point  \n\n  p = (p₁,…,pₙ) ∈ K∩ℝⁿ₊   with p₁p₂⋯pₙ = 1.                (1.1)\n\nFix such a point for the remainder of the proof.\n\nStep 2.  The 2ⁿ sign variations and their convex hull.  \nUnconditionality implies that the 2ⁿ points  \n\n  p^{ε} := (ε₁p₁,…,εₙpₙ), ε_i ∈ {−1,+1},                      (2.1)\n\nall belong to K.  Let  \n\n  B(p) := conv{p^{ε} : ε∈{−1,+1}ⁿ}.                           (2.2)\n\nLemma 2.1   B(p) is the axis–parallel box  \n\n  B(p)=∏_{i=1}^{n}[−p_i,\\,p_i].\n\nProof.  The box in (2.3) is convex and contains every vertex p^{ε}, hence\ncontains B(p).  Conversely, given any x=(x₁,…,xₙ) with −p_i≤x_i≤p_i for each i,\nwrite  \n\n  x_i = λ_i p_i + (1−λ_i)(−p_i), λ_i:=½(1+x_i/p_i)∈[0,1].\n\nDefine weights w_{ε}:=∏_{i=1}^{n}λ_i^{(1+ε_i)/2}(1−λ_i)^{(1−ε_i)/2}.\nThey satisfy w_{ε}≥0 and Σ_{ε}w_{ε}=1, while Σ_{ε}w_{ε}p^{ε}=x.\nThus x∈B(p), proving the equality. ∎\n\nStep 3.  A universal lower bound for Volₙ(K).  \nBecause B(p) ⊂ K by convexity,\n\n  Volₙ(K) ≥ Volₙ(B(p)) = ∏_{i=1}^{n}2p_i.                    (3.1)\n\nUsing (1.1) we compute Volₙ(B(p))=2ⁿ(p₁p₂⋯pₙ)=2ⁿ, whence\n\n  Volₙ(K) ≥ 2ⁿ.                                               (3.2)\n\nConsequently Vₙ ≥ 2ⁿ.\n\nStep 4.  Sharpness.  \nFor any vector a=(a₁,…,aₙ) with a_i>0 and Π a_i = 1, the box  \n\n  C(a):=∏_{i=1}^{n}[−a_i,a_i]                                (4.1)\n\nis unconditional, contains the point a∈Σ₊, and therefore dominates\nΣ₊∪Σ₋.  Its volume equals Volₙ(C(a)) = 2ⁿ, so the lower bound (3.2)\nis attained.  Hence\n\n  Vₙ = 2ⁿ.                                                    (4.2)\n\nStep 5.  Characterisation of all minimisers.  \nAssume Volₙ(K)=2ⁿ.  With (3.1) we have Volₙ(K)=Volₙ(B(p)),\nbut B(p)⊂K.  Since both sets are convex and B(p) is full-dimensional,\nequality of volumes forces K=B(p).  Writing a_i:=p_i, we obtain\nΠ a_i=1 and K=C(a).\n\nConversely every box C(a) with Π a_i=1 attains the minimum.\nTherefore\n\n  𝓜ₙ = {∏_{i=1}^{n}[−a_i,a_i] : a_i>0, Π_{i=1}^{n}a_i=1}.    (5.1)\n\nStep 6.  Final answers.  \na) Vₙ = 2ⁿ.  \nb) The complete set of minimising bodies is 𝓜ₙ given in (5.1). ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.803957",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the problem is lifted from 2-dimensions (area) to arbitrary n-dimensions (volume).  \n• Many more branches: instead of 4 branches of two plane hyperbolas, we now have 2ⁿ branches of two n-dimensional hypersurfaces.  \n• Additional constraint: the centroid of the body is fixed at the origin, forcing the use of symmetry and support-function arguments.  \n• Deeper theory: the solution invokes Brunn–Minkowski, symmetrisation, support planes, zonotopes and a multidimensional AM–GM argument, rather than a single elementary inequality.  \n• Characterisation of all minimisers, not just the minimal value, requires an extremal-volume argument via Aleksandrov–Fenchel, adding another advanced layer.\n\nThus the enhanced variant is substantially more technical and conceptually richer than both the original olympiad problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let n ≥ 2 be an integer and write, for x=(x₁,…,xₙ)∈ℝⁿ,  \n\n  Σ₊ := {x : x₁x₂⋯xₙ =  1},  Σ₋ := {x : x₁x₂⋯xₙ = –1}.  \n\nBoth hypersurfaces consist of 2ⁿ⁻¹ connected components, one in every open orthant.  \nA convex body K⊂ℝⁿ is called unconditional if, together with any point  \n(x₁,…,xₙ)∈K, it contains all 2ⁿ points obtained by changing the signs of the\ncoordinates.  (Equivalently: K is symmetric with respect to each coordinate hyper-plane.)\n\nWe say that K dominates the two hypersurfaces if it meets every component of Σ₊∪Σ₋.\nFor an unconditional body this is equivalent to K∩Σ₊≠∅.\n\nAmong all unconditional convex bodies that dominate Σ₊∪Σ₋ determine  \n\na) the minimum possible n-dimensional volume  \n\n  Vₙ := min{ Volₙ(K) : K unconditional, K∩Σ₊≠∅ };  \n\nb) all bodies for which that minimum is attained.",
      "solution": "Throughout Volₙ denotes n-dimensional Lebesgue measure and ℝⁿ₊ the positive\northant.\n\nStep 1.  A canonical positive point.  \nIf K is unconditional and dominates Σ₊∪Σ₋, then there is a point  \n\n  p = (p₁,…,pₙ) ∈ K∩ℝⁿ₊   with p₁p₂⋯pₙ = 1.                (1.1)\n\nFix such a point for the remainder of the proof.\n\nStep 2.  The 2ⁿ sign variations and their convex hull.  \nUnconditionality implies that the 2ⁿ points  \n\n  p^{ε} := (ε₁p₁,…,εₙpₙ), ε_i ∈ {−1,+1},                      (2.1)\n\nall belong to K.  Let  \n\n  B(p) := conv{p^{ε} : ε∈{−1,+1}ⁿ}.                           (2.2)\n\nLemma 2.1   B(p) is the axis–parallel box  \n\n  B(p)=∏_{i=1}^{n}[−p_i,\\,p_i].\n\nProof.  The box in (2.3) is convex and contains every vertex p^{ε}, hence\ncontains B(p).  Conversely, given any x=(x₁,…,xₙ) with −p_i≤x_i≤p_i for each i,\nwrite  \n\n  x_i = λ_i p_i + (1−λ_i)(−p_i), λ_i:=½(1+x_i/p_i)∈[0,1].\n\nDefine weights w_{ε}:=∏_{i=1}^{n}λ_i^{(1+ε_i)/2}(1−λ_i)^{(1−ε_i)/2}.\nThey satisfy w_{ε}≥0 and Σ_{ε}w_{ε}=1, while Σ_{ε}w_{ε}p^{ε}=x.\nThus x∈B(p), proving the equality. ∎\n\nStep 3.  A universal lower bound for Volₙ(K).  \nBecause B(p) ⊂ K by convexity,\n\n  Volₙ(K) ≥ Volₙ(B(p)) = ∏_{i=1}^{n}2p_i.                    (3.1)\n\nUsing (1.1) we compute Volₙ(B(p))=2ⁿ(p₁p₂⋯pₙ)=2ⁿ, whence\n\n  Volₙ(K) ≥ 2ⁿ.                                               (3.2)\n\nConsequently Vₙ ≥ 2ⁿ.\n\nStep 4.  Sharpness.  \nFor any vector a=(a₁,…,aₙ) with a_i>0 and Π a_i = 1, the box  \n\n  C(a):=∏_{i=1}^{n}[−a_i,a_i]                                (4.1)\n\nis unconditional, contains the point a∈Σ₊, and therefore dominates\nΣ₊∪Σ₋.  Its volume equals Volₙ(C(a)) = 2ⁿ, so the lower bound (3.2)\nis attained.  Hence\n\n  Vₙ = 2ⁿ.                                                    (4.2)\n\nStep 5.  Characterisation of all minimisers.  \nAssume Volₙ(K)=2ⁿ.  With (3.1) we have Volₙ(K)=Volₙ(B(p)),\nbut B(p)⊂K.  Since both sets are convex and B(p) is full-dimensional,\nequality of volumes forces K=B(p).  Writing a_i:=p_i, we obtain\nΠ a_i=1 and K=C(a).\n\nConversely every box C(a) with Π a_i=1 attains the minimum.\nTherefore\n\n  𝓜ₙ = {∏_{i=1}^{n}[−a_i,a_i] : a_i>0, Π_{i=1}^{n}a_i=1}.    (5.1)\n\nStep 6.  Final answers.  \na) Vₙ = 2ⁿ.  \nb) The complete set of minimising bodies is 𝓜ₙ given in (5.1). ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.614945",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the problem is lifted from 2-dimensions (area) to arbitrary n-dimensions (volume).  \n• Many more branches: instead of 4 branches of two plane hyperbolas, we now have 2ⁿ branches of two n-dimensional hypersurfaces.  \n• Additional constraint: the centroid of the body is fixed at the origin, forcing the use of symmetry and support-function arguments.  \n• Deeper theory: the solution invokes Brunn–Minkowski, symmetrisation, support planes, zonotopes and a multidimensional AM–GM argument, rather than a single elementary inequality.  \n• Characterisation of all minimisers, not just the minimal value, requires an extremal-volume argument via Aleksandrov–Fenchel, adding another advanced layer.\n\nThus the enhanced variant is substantially more technical and conceptually richer than both the original olympiad problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}