{
  "index": "2013-A-6",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Define a function $w: \\mathbb{Z} \\times \\mathbb{Z} \\to \\mathbb{Z}$\nas follows. For $\\left| a \\right|, \\left| b \\right| \\leq 2$,\nlet $w(a,b)$ be as in the table shown; otherwise, let $w(a,b) = 0$.\n\\begin{center}\n\\begin{tabular}{|cc|r|r|r|r|r|}\n\\hline\n\\multicolumn{2}{|c|}{\\multirow{2}{*}{$w(a,b)$}} & \\multicolumn{5}{|c|}{$b$} \\\\\n&  & -2 & -1 & 0 & 1 & 2 \\\\\n\\hline\n& -2 & -1 & -2 & 2 & -2 & -1 \\\\\n& -1 & -2 & 4 & -4 & 4 & -2 \\\\\n$a$ & 0 & 2 & -4 & 12 & -4 & 2 \\\\\n& 1 & -2 & 4 & -4 & 4 & -2 \\\\ \n& 2 &  -1 & -2 & 2 & -2 & -1 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\nFor every finite subset $S$ of $\\mathbb{Z} \\times \\mathbb{Z}$,\ndefine\n\\[\nA(S) = \\sum_{(\\mathbf{s}, \\mathbf{s}') \\in S \\times S} w(\\mathbf{s} - \\mathbf{s}').\n\\]\nProve that if $S$ is any finite nonempty subset of $\\mathbb{Z} \\times \\mathbb{Z}$, then $A(S) > 0$.\n(For example, if $S = \\{(0,1), (0,2), (2,0), (3,1)\\}$, then the terms in $A(S)$ are $12, 12, 12, 12, 4, 4, 0, 0, 0,0,-1,-1,-2,-2,-4,-4$.)",
  "solution": "(by Harm Derksen)\nConsider the generating functions\n\\begin{align*}\nf(x,y) &= \\sum_{(a,b) \\in S} x^a y^b, \\\\\ng(x,y) &= \\sum_{(a,b) \\in \\mathbb{Z}^2} w(a,b) x^a y^b.\n\\end{align*}\nThen $A(S)$ is the constant coefficient of the Laurent polynomial\n$h(x,y) = f(x,y) f(x^{-1}, y^{-1}) g(x,y)$. We may compute this coefficient by averaging over unit circles:\n\\begin{align*}\n(2 \\pi)^2 A(S) &= \\int_0^{2\\pi} \\int_0^{2\\pi} h(e^{is}, e^{it})\\,dt\\,ds \\\\\n&= \\int_0^{2\\pi} \\int_0^{2\\pi} \\left| f(e^{is}, e^{it}) \\right|^2 g(e^{is}, e^{it}) \\,dt\\,ds.\n\\end{align*}\nConsequently, it is enough to check that $g(e^{is}, e^{it})$ is a nonnegative real number for all $s,t \\in \\mathbb{R}$. But\n$g(e^{is}, e^{it}) = 16 G(\\cos s,\\cos t)$ for\n\\[\nG(z,w) = zw + z^2 + w^2 - z^2 w - zw^2 - z^2w^2.\n\\]\nIf $z,w \\in [-1,1]$ and $zw \\geq 0$, then\n\\[\nG(z,w) = zw(1-zw) + z^2(1-w) + w^2(1-z) \\geq 0.\n\\]\nIf $z,w \\in [-1,1]$ and $zw \\leq 0$, then\n\\[\nG(z,w) = (z+w)^2 - zw(1+z)(1+w) \\geq 0.\n\\]\nHence $g(e^{is},e^{it}) \\geq 0$ as desired.",
  "vars": [
    "w",
    "a",
    "b",
    "A",
    "S",
    "x",
    "y",
    "f",
    "g",
    "h",
    "s",
    "t",
    "z",
    "G"
  ],
  "params": [],
  "sci_consts": [
    "e",
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "w": "weightfn",
        "a": "firstidx",
        "b": "secondidx",
        "A": "totalsum",
        "S": "subsetset",
        "x": "axisvarx",
        "y": "axisvary",
        "f": "genfunone",
        "g": "genfuntwo",
        "h": "prodpoly",
        "s": "anglelong",
        "t": "anglewide",
        "z": "cosinevar",
        "G": "helperfun"
      },
      "question": "Define a function $\\weightfn: \\mathbb{Z} \\times \\mathbb{Z} \\to \\mathbb{Z}$\nas follows. For $\\left| \\firstidx \\right|, \\left| \\secondidx \\right| \\leq 2$,\nlet $\\weightfn(\\firstidx,\\secondidx)$ be as in the table shown; otherwise, let $\\weightfn(\\firstidx,\\secondidx) = 0$.\n\\begin{center}\n\\begin{tabular}{|cc|r|r|r|r|r|}\n\\hline\n\\multicolumn{2}{|c|}{\\multirow{2}{*}{$\\weightfn(\\firstidx,\\secondidx)$}} & \\multicolumn{5}{|c|}{$\\secondidx$} \\\\\n&  & -2 & -1 & 0 & 1 & 2 \\\\\n\\hline\n& -2 & -1 & -2 & 2 & -2 & -1 \\\\\n& -1 & -2 & 4 & -4 & 4 & -2 \\\\\n$\\firstidx$ & 0 & 2 & -4 & 12 & -4 & 2 \\\\\n& 1 & -2 & 4 & -4 & 4 & -2 \\\\ \n& 2 &  -1 & -2 & 2 & -2 & -1 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\nFor every finite subset $\\subsetset$ of $\\mathbb{Z} \\times \\mathbb{Z}$,\ndefine\n\\[\n\\totalsum(\\subsetset) = \\sum_{(\\mathbf{anglelong}, \\mathbf{anglelong}') \\in \\subsetset \\times \\subsetset} \\weightfn(\\mathbf{anglelong} - \\mathbf{anglelong}').\n\\]\nProve that if $\\subsetset$ is any finite nonempty subset of $\\mathbb{Z} \\times \\mathbb{Z}$, then $\\totalsum(\\subsetset) > 0$.\n(For example, if $\\subsetset = \\{(0,1), (0,2), (2,0), (3,1)\\}$, then the terms in $\\totalsum(\\subsetset)$ are $12, 12, 12, 12, 4, 4, 0, 0, 0,0,-1,-1,-2,-2,-4,-4$.)",
      "solution": "(by Harm Derksen)\nConsider the generating functions\n\\begin{align*}\n\\genfunone(\\axisvarx,\\axisvary) &= \\sum_{(\\firstidx,\\secondidx) \\in \\subsetset} \\axisvarx^{\\firstidx} \\axisvary^{\\secondidx}, \\\\\n\\genfuntwo(\\axisvarx,\\axisvary) &= \\sum_{(\\firstidx,\\secondidx) \\in \\mathbb{Z}^2} \\weightfn(\\firstidx,\\secondidx) \\axisvarx^{\\firstidx} \\axisvary^{\\secondidx}.\n\\end{align*}\nThen $\\totalsum(\\subsetset)$ is the constant coefficient of the Laurent polynomial\n$\\prodpoly(\\axisvarx,\\axisvary) = \\genfunone(\\axisvarx,\\axisvary) \\genfunone(\\axisvarx^{-1}, \\axisvary^{-1}) \\genfuntwo(\\axisvarx,\\axisvary)$. We may compute this coefficient by averaging over unit circles:\n\\begin{align*}\n(2 \\pi)^2 \\totalsum(\\subsetset) &= \\int_0^{2\\pi} \\int_0^{2\\pi} \\prodpoly(e^{i\\anglelong}, e^{i\\anglewide})\\,d\\anglewide\\,d\\anglelong \\\\\n&= \\int_0^{2\\pi} \\int_0^{2\\pi} \\left| \\genfunone(e^{i\\anglelong}, e^{i\\anglewide}) \\right|^2 \\genfuntwo(e^{i\\anglelong}, e^{i\\anglewide}) \\,d\\anglewide\\,d\\anglelong.\n\\end{align*}\nConsequently, it is enough to check that $\\genfuntwo(e^{i\\anglelong}, e^{i\\anglewide})$ is a nonnegative real number for all $\\anglelong,\\anglewide \\in \\mathbb{R}$. But\n$\\genfuntwo(e^{i\\anglelong}, e^{i\\anglewide}) = 16 \\helperfun(\\cos \\anglelong,\\cos \\anglewide)$ for\n\\[\n\\helperfun(\\cosinevar,\\weightfn) = \\cosinevar\\weightfn + \\cosinevar^2 + \\weightfn^2 - \\cosinevar^2 \\weightfn - \\cosinevar \\weightfn^2 - \\cosinevar^2\\weightfn^2.\n\\]\nIf $\\cosinevar,\\weightfn \\in [-1,1]$ and $\\cosinevar\\weightfn \\geq 0$, then\n\\[\n\\helperfun(\\cosinevar,\\weightfn) = \\cosinevar\\weightfn(1-\\cosinevar\\weightfn) + \\cosinevar^2(1-\\weightfn) + \\weightfn^2(1-\\cosinevar) \\geq 0.\n\\]\nIf $\\cosinevar,\\weightfn \\in [-1,1]$ and $\\cosinevar\\weightfn \\leq 0$, then\n\\[\n\\helperfun(\\cosinevar,\\weightfn) = (\\cosinevar+\\weightfn)^2 - \\cosinevar\\weightfn(1+\\cosinevar)(1+\\weightfn) \\geq 0.\n\\]\nHence $\\genfuntwo(e^{i\\anglelong},e^{i\\anglewide}) \\geq 0$ as desired."
    },
    "descriptive_long_confusing": {
      "map": {
        "w": "sandpaper",
        "a": "turntable",
        "b": "doorknob",
        "A": "raincloud",
        "S": "colosseum",
        "x": "pinecrown",
        "y": "lighthawk",
        "f": "stingward",
        "g": "marblegum",
        "h": "snowglove",
        "s": "nightfern",
        "t": "chocowire",
        "z": "ambervine",
        "G": "driftwood"
      },
      "question": "Define a function $sandpaper: \\mathbb{Z} \\times \\mathbb{Z} \\to \\mathbb{Z}$\nas follows. For $\\left| turntable \\right|, \\left| doorknob \\right| \\leq 2$,\nlet $sandpaper(turntable,doorknob)$ be as in the table shown; otherwise, let $sandpaper(turntable,doorknob) = 0$.\n\\begin{center}\n\\begin{tabular}{|cc|r|r|r|r|r|}\n\\hline\n\\multicolumn{2}{|c|}{\\multirow{2}{*}{$sandpaper(turntable,doorknob)$}} & \\multicolumn{5}{|c|}{$doorknob$} \\\\\n&  & -2 & -1 & 0 & 1 & 2 \\\\\n\\hline\n& -2 & -1 & -2 & 2 & -2 & -1 \\\\\n& -1 & -2 & 4 & -4 & 4 & -2 \\\\\n$turntable$ & 0 & 2 & -4 & 12 & -4 & 2 \\\\\n& 1 & -2 & 4 & -4 & 4 & -2 \\\\ \n& 2 &  -1 & -2 & 2 & -2 & -1 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\nFor every finite subset $colosseum$ of $\\mathbb{Z} \\times \\mathbb{Z}$,\ndefine\n\\[\nraincloud(colosseum) = \\sum_{(\\mathbf{nightfern}, \\mathbf{nightfern}') \\in colosseum \\times colosseum} sandpaper(\\mathbf{nightfern} - \\mathbf{nightfern}').\n\\]\nProve that if $colosseum$ is any finite nonempty subset of $\\mathbb{Z} \\times \\mathbb{Z}$, then $raincloud(colosseum) > 0$.\n(For example, if $colosseum = \\{(0,1), (0,2), (2,0), (3,1)\\}$, then the terms in $raincloud(colosseum)$ are $12, 12, 12, 12, 4, 4, 0, 0, 0,0,-1,-1,-2,-2,-4,-4$.)",
      "solution": "(by Harm Derksen)\nConsider the generating functions\n\\begin{align*}\nstingward(pinecrown,lighthawk) &= \\sum_{(turntable,doorknob) \\in colosseum} pinecrown^{turntable} lighthawk^{doorknob}, \\\\\nmarblegum(pinecrown,lighthawk) &= \\sum_{(turntable,doorknob) \\in \\mathbb{Z}^2} sandpaper(turntable,doorknob) pinecrown^{turntable} lighthawk^{doorknob}.\n\\end{align*}\nThen $raincloud(colosseum)$ is the constant coefficient of the Laurent polynomial\n$snowglove(pinecrown,lighthawk) = stingward(pinecrown,lighthawk)\\, stingward(pinecrown^{-1}, lighthawk^{-1})\\, marblegum(pinecrown,lighthawk)$. We may compute this coefficient by averaging over unit circles:\n\\begin{align*}\n(2 \\pi)^2 raincloud(colosseum) &= \\int_0^{2\\pi} \\int_0^{2\\pi} snowglove(e^{i\\,nightfern}, e^{i\\,chocowire})\\,dchocowire\\,dnightfern \\\\\n&= \\int_0^{2\\pi} \\int_0^{2\\pi} \\left| stingward(e^{i\\,nightfern}, e^{i\\,chocowire}) \\right|^2\\, marblegum(e^{i\\,nightfern}, e^{i\\,chocowire}) \\,dchocowire\\,dnightfern.\n\\end{align*}\nConsequently, it is enough to check that $marblegum(e^{i\\,nightfern}, e^{i\\,chocowire})$ is a nonnegative real number for all $nightfern,chocowire \\in \\mathbb{R}$. But\n$marblegum(e^{i\\,nightfern}, e^{i\\,chocowire}) = 16\\, driftwood(\\cos nightfern,\\cos chocowire)$ for\n\\[\ndriftwood(ambervine,sandpaper) = ambervine\\,sandpaper + ambervine^2 + sandpaper^2 - ambervine^2 sandpaper - ambervine sandpaper^2 - ambervine^2 sandpaper^2.\n\\]\nIf $ambervine,sandpaper \\in [-1,1]$ and $ambervine\\,sandpaper \\geq 0$, then\n\\[\ndriftwood(ambervine,sandpaper) = ambervine\\,sandpaper(1-ambervine\\,sandpaper) + ambervine^2(1-sandpaper) + sandpaper^2(1-ambervine) \\geq 0.\n\\]\nIf $ambervine,sandpaper \\in [-1,1]$ and $ambervine\\,sandpaper \\leq 0$, then\n\\[\ndriftwood(ambervine,sandpaper) = (ambervine+sandpaper)^2 - ambervine\\,sandpaper(1+ambervine)(1+sandpaper) \\geq 0.\n\\]\nHence $marblegum(e^{i\\,nightfern},e^{i\\,chocowire}) \\geq 0$ as desired."
    },
    "descriptive_long_misleading": {
      "map": {
        "w": "emptiness",
        "a": "stagnant",
        "b": "frozenvar",
        "A": "negativesum",
        "S": "wholeset",
        "x": "outputval",
        "y": "horizval",
        "f": "staticform",
        "g": "stagnantmap",
        "h": "fragment",
        "s": "lengthval",
        "t": "widthval",
        "z": "imagival",
        "G": "trivialmap"
      },
      "question": "Define a function $emptiness: \\mathbb{Z} \\times \\mathbb{Z} \\to \\mathbb{Z}$\nas follows. For $\\left| stagnant \\right|, \\left| frozenvar \\right| \\leq 2$,\nlet $emptiness(stagnant,frozenvar)$ be as in the table shown; otherwise, let $emptiness(stagnant,frozenvar) = 0$.\n\\begin{center}\n\\begin{tabular}{|cc|r|r|r|r|r|}\n\\hline\n\\multicolumn{2}{|c|}{\\multirow{2}{*}{$emptiness(stagnant,frozenvar)$}} & \\multicolumn{5}{|c|}{$frozenvar$} \\\\\n&  & -2 & -1 & 0 & 1 & 2 \\\\\n\\hline\n& -2 & -1 & -2 & 2 & -2 & -1 \\\\\n& -1 & -2 & 4 & -4 & 4 & -2 \\\\\n$stagnant$ & 0 & 2 & -4 & 12 & -4 & 2 \\\\\n& 1 & -2 & 4 & -4 & 4 & -2 \\\\ \n& 2 &  -1 & -2 & 2 & -2 & -1 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\nFor every finite subset $wholeset$ of $\\mathbb{Z} \\times \\mathbb{Z}$,\ndefine\n\\[\nnegativesum(wholeset) = \\sum_{(\\mathbf{s}, \\mathbf{s}') \\in wholeset \\times wholeset} emptiness(\\mathbf{s} - \\mathbf{s}').\n\\]\nProve that if $wholeset$ is any finite nonempty subset of $\\mathbb{Z} \\times \\mathbb{Z}$, then $negativesum(wholeset) > 0$.\n(For example, if $wholeset = \\{(0,1), (0,2), (2,0), (3,1)\\}$, then the terms in $negativesum(wholeset)$ are $12, 12, 12, 12, 4, 4, 0, 0, 0,0,-1,-1,-2,-2,-4,-4$.)",
      "solution": "(by Harm Derksen)\nConsider the generating functions\n\\begin{align*}\nstaticform(outputval,horizval) &= \\sum_{(stagnant,frozenvar) \\in wholeset} outputval^{stagnant} horizval^{frozenvar}, \\\\\nstagnantmap(outputval,horizval) &= \\sum_{(stagnant,frozenvar) \\in \\mathbb{Z}^2} emptiness(stagnant,frozenvar) outputval^{stagnant} horizval^{frozenvar}.\n\\end{align*}\nThen $negativesum(wholeset)$ is the constant coefficient of the Laurent polynomial\n$fragment(outputval,horizval) = staticform(outputval,horizval)\\,staticform(outputval^{-1},horizval^{-1})\\,stagnantmap(outputval,horizval)$. We may compute this coefficient by averaging over unit circles:\n\\begin{align*}\n(2 \\pi)^2\\,negativesum(wholeset) &= \\int_0^{2\\pi} \\int_0^{2\\pi} fragment(e^{i\\,lengthval}, e^{i\\,widthval})\\,d\\!widthval\\,d\\!lengthval \\\\\n&= \\int_0^{2\\pi} \\int_0^{2\\pi} \\left| staticform(e^{i\\,lengthval}, e^{i\\,widthval}) \\right|^2\n      stagnantmap(e^{i\\,lengthval}, e^{i\\,widthval}) \\,d\\!widthval\\,d\\!lengthval.\n\\end{align*}\nConsequently, it is enough to check that $stagnantmap(e^{i\\,lengthval}, e^{i\\,widthval})$ is a nonnegative real number for all $lengthval,widthval \\in \\mathbb{R}$. But\n$stagnantmap(e^{i\\,lengthval}, e^{i\\,widthval}) = 16\\,trivialmap(\\cos lengthval,\\cos widthval)$ for\n\\[\ntrivialmap(imagival,emptiness) = imagival\\,emptiness + imagival^2 + emptiness^2 - imagival^2\\,emptiness - imagival\\,emptiness^2 - imagival^2\\,emptiness^2.\n\\]\nIf $imagival,emptiness \\in [-1,1]$ and $imagival\\,emptiness \\ge 0$, then\n\\[\ntrivialmap(imagival,emptiness) = imagival\\,emptiness(1-imagival\\,emptiness) + imagival^2(1-emptiness) + emptiness^2(1-imagival) \\ge 0.\n\\]\nIf $imagival,emptiness \\in [-1,1]$ and $imagival\\,emptiness \\le 0$, then\n\\[\ntrivialmap(imagival,emptiness) = (imagival+emptiness)^2 - imagival\\,emptiness(1+imagival)(1+emptiness) \\ge 0.\n\\]\nHence $stagnantmap(e^{i\\,lengthval},e^{i\\,widthval}) \\ge 0$ as desired."
    },
    "garbled_string": {
      "map": {
        "w": "qzxwvtnp",
        "a": "hjgrksla",
        "b": "vrgploni",
        "A": "mzjxcklq",
        "S": "cpydrsom",
        "x": "sbgvkilp",
        "y": "dftnqwer",
        "f": "lhmkdvza",
        "g": "bqtwsneo",
        "h": "mkpvradl",
        "s": "radjovkn",
        "t": "ufqneizl",
        "z": "ksmbltay",
        "G": "vclqprme"
      },
      "question": "Define a function $qzxwvtnp: \\mathbb{Z} \\times \\mathbb{Z} \\to \\mathbb{Z}$\nas follows. For $\\left| hjgrksla \\right|, \\left| vrgploni \\right| \\leq 2$,\nlet $qzxwvtnp(hjgrksla,vrgploni)$ be as in the table shown; otherwise, let $qzxwvtnp(hjgrksla,vrgploni) = 0$.\n\\begin{center}\n\\begin{tabular}{|cc|r|r|r|r|r|}\n\\hline\n\\multicolumn{2}{|c|}{\\multirow{2}{*}{$qzxwvtnp(hjgrksla,vrgploni)$}} & \\multicolumn{5}{|c|}{$vrgploni$} \\\\\n&  & -2 & -1 & 0 & 1 & 2 \\\\\n\\hline\n& -2 & -1 & -2 & 2 & -2 & -1 \\\\\n& -1 & -2 & 4 & -4 & 4 & -2 \\\\\n$hjgrksla$ & 0 & 2 & -4 & 12 & -4 & 2 \\\\\n& 1 & -2 & 4 & -4 & 4 & -2 \\\\ \n& 2 &  -1 & -2 & 2 & -2 & -1 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\nFor every finite subset $cpydrsom$ of $\\mathbb{Z} \\times \\mathbb{Z}$,\ndefine\n\\[\nmzjxcklq(cpydrsom) = \\sum_{(\\mathbf{radjovkn}, \\mathbf{radjovkn}') \\in cpydrsom \\times cpydrsom} qzxwvtnp(\\mathbf{radjovkn} - \\mathbf{radjovkn}').\n\\]\nProve that if $cpydrsom$ is any finite nonempty subset of $\\mathbb{Z} \\times \\mathbb{Z}$, then $mzjxcklq(cpydrsom) > 0$.\n(For example, if $cpydrsom = \\{(0,1), (0,2), (2,0), (3,1)\\}$, then the terms in $mzjxcklq(cpydrsom)$ are $12, 12, 12, 12, 4, 4, 0, 0, 0,0,-1,-1,-2,-2,-4,-4$.)",
      "solution": "(by Harm Derksen)\nConsider the generating functions\n\\begin{align*}\nlhmkdvza(sbgvkilp,dftnqwer) &= \\sum_{(hjgrksla,vrgploni) \\in cpydrsom} sbgvkilp^{hjgrksla} dftnqwer^{vrgploni}, \\\\\nbqtwsneo(sbgvkilp,dftnqwer) &= \\sum_{(hjgrksla,vrgploni) \\in \\mathbb{Z}^2} qzxwvtnp(hjgrksla,vrgploni) sbgvkilp^{hjgrksla} dftnqwer^{vrgploni}.\n\\end{align*}\nThen $mzjxcklq(cpydrsom)$ is the constant coefficient of the Laurent polynomial\n$mkpvradl(sbgvkilp,dftnqwer) = lhmkdvza(sbgvkilp,dftnqwer) lhmkdvza(sbgvkilp^{-1}, dftnqwer^{-1}) bqtwsneo(sbgvkilp,dftnqwer)$. We may compute this coefficient by averaging over unit circles:\n\\begin{align*}\n(2 \\pi)^2 mzjxcklq(cpydrsom) &= \\int_0^{2\\pi} \\int_0^{2\\pi} mkpvradl(e^{i radjovkn}, e^{i ufqneizl})\\,d ufqneizl\\,d radjovkn \\\\\n&= \\int_0^{2\\pi} \\int_0^{2\\pi} \\left| lhmkdvza(e^{i radjovkn}, e^{i ufqneizl}) \\right|^2 bqtwsneo(e^{i radjovkn}, e^{i ufqneizl}) \\,d ufqneizl\\,d radjovkn.\n\\end{align*}\nConsequently, it is enough to check that $bqtwsneo(e^{i radjovkn}, e^{i ufqneizl})$ is a nonnegative real number for all $radjovkn,ufqneizl \\in \\mathbb{R}$. But\n$bqtwsneo(e^{i radjovkn}, e^{i ufqneizl}) = 16 vclqprme(\\cos radjovkn,\\cos ufqneizl)$ for\n\\[\nvclqprme(ksmbltay,qzxwvtnp) = ksmbltay qzxwvtnp + ksmbltay^{2} + qzxwvtnp^{2} - ksmbltay^{2} qzxwvtnp - ksmbltay qzxwvtnp^{2} - ksmbltay^{2} qzxwvtnp^{2}.\n\\]\nIf $ksmbltay,qzxwvtnp \\in [-1,1]$ and $ksmbltay qzxwvtnp \\geq 0$, then\n\\[\nvclqprme(ksmbltay,qzxwvtnp) = ksmbltay qzxwvtnp (1-ksmbltay qzxwvtnp) + ksmbltay^{2}(1-qzxwvtnp) + qzxwvtnp^{2}(1-ksmbltay) \\geq 0.\n\\]\nIf $ksmbltay,qzxwvtnp \\in [-1,1]$ and $ksmbltay qzxwvtnp \\leq 0$, then\n\\[\nvclqprme(ksmbltay,qzxwvtnp) = (ksmbltay+qzxwvtnp)^{2} - ksmbltay qzxwvtnp (1+ksmbltay)(1+qzxwvtnp) \\geq 0.\n\\]\nHence $bqtwsneo(e^{i radjovkn},e^{i ufqneizl}) \\geq 0$ as desired."
    },
    "kernel_variant": {
      "question": "Let w : ℤ³ → ℤ be the convolution kernel whose non–zero values are  \n\n• w(0,0,0)=48;  \n\n• w(±1,0,0)=w(0,±1,0)=w(0,0,±1)=−13;  \n\n• w(±2,0,0)=w(0,±2,0)=w(0,0,±2)= 1;  \n\n• w(±1,±1,0)=w(±1,0,±1)=w(0,±1,±1)= 2  \n\n(the sign choices are independent in every occurrence); for all other lattice points set w=0.  \nFor a finite set S⊂ℤ³ put  \n\nA(S)=∑_{(s,s′)∈S×S} w(s−s′).\n\nProve that A(S)>0 for every non–empty finite subset S of ℤ³.",
      "solution": "Step 1.  A Fourier–analytic formulation.  \nFor a finite S write  \n\nf(x,y,z)=∑_{(a,b,c)∈S} x^{a}y^{b}z^{c}  (Laurent polynomial),  \n\nand let  \n\ng(x,y,z)=∑_{(a,b,c)∈ℤ³} w(a,b,c)x^{a}y^{b}z^{c}.  \n\nBecause convolution turns into multiplication of generating functions,  \n\nA(S)=([x^{0}y^{0}z^{0}] f(x,y,z)f(x^{-1},y^{-1},z^{-1})g(x,y,z)).          (1)\n\nAs in the original problem, the constant coefficient can be extracted by integrating over the 3–torus:\n\n(2π)³A(S)=∫_{0}^{2π}∫_{0}^{2π}∫_{0}^{2π}|f(e^{iα},e^{iβ},e^{iγ})|²·g(e^{iα},e^{iβ},e^{iγ}) dα dβ dγ.        (2)\n\nHence A(S)>0 once we know that \n\n(i) g(e^{iα},e^{iβ},e^{iγ}) is a non–negative real number for every (α,β,γ)∈ℝ³;  \n(ii) it is positive on a set of positive measure in the 3–torus.\n\nStep 2.  A closed form for g on the torus.  \nIntroduce\n\nΛ(α,β,γ)=6−(e^{iα}+e^{-iα}+e^{iβ}+e^{-iβ}+e^{iγ}+e^{-iγ})\n          =6−2( cos α+cos β+cos γ ).\n\nObserve that Λ≥0 for all (α,β,γ) because each cosine is at most 1.  A direct computation using the table for w (or, faster, using that our w was engineered as a polynomial in the discrete Laplacian) shows that\n\ng(x,y,z)=Λ(x,y,z)²+Λ(x,y,z).                                             (3)\n\nIndeed,\n\ng(x,y,z)= (6−(x+x^{-1}+y+y^{-1}+z+z^{-1}))²\n        +(6−(x+x^{-1}+y+y^{-1}+z+z^{-1})),\n\nand expanding (3) yields exactly the prescribed integer coefficients 48,−13,1,2.\n\nSubstituting x=e^{iα}, y=e^{iβ}, z=e^{iγ} gives\n\ng(e^{iα},e^{iβ},e^{iγ})=Λ(α,β,γ)·(Λ(α,β,γ)+1) ≥0,                (4)\n\nwith equality only when Λ=0, i.e. when cos α=cos β=cos γ=1, which forces (α,β,γ)≡(0,0,0) mod 2π.\n\nStep 3.  Strict positivity of the integral.  \nBecause S is finite and non–empty, f is a non–zero trigonometric (Laurent) polynomial.  Hence |f|² is continuous, non–negative and not identically zero, so its set of positivity has positive 3–dimensional measure.\n\nEquation (4) tells us that g is non–negative everywhere and strictly positive except at the single point (α,β,γ)=(0,0,0).  Therefore the integrand in (2) is non–negative everywhere and strictly positive on a set of positive measure (the intersection of “|f|²>0’’ with “Λ>0’’), whence the triple integral is strictly positive.  By (2) this implies A(S)>0, completing the proof.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.827926",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension.  \nThe problem moves from ℤ² to ℤ³, so both the Fourier domain and the combinatorics of the kernel enlarge from a 2–torus to a 3–torus, adding an extra variable and significantly complicating any direct inspection or elementary enumeration approach.\n\n2. Larger support and many more weights.  \nInstead of a 3×3 table (9 numbers), the kernel now has support on 27 lattice points, with four different weight levels (48,−13,2,1).  Simple pattern spotting is impossible; one must understand how the coefficients were produced.\n\n3. Need for structural understanding.  \nThe solution hinges on recognising that w is a quadratic polynomial in the three–dimensional discrete Laplacian.  Without that observation the expansion of g and the verification of non–negativity are prohibitively long.  Participants must therefore combine discrete differential operators with Fourier analysis—material usually outside routine olympiad fare.\n\n4. Measure–theoretic argument.  \nTo transform non–negativity of g into strict positivity of A(S) one must note that g vanishes only on a set of measure 0 and combine this with the continuity and non–triviality of |f|².  This step forces competitors to mix harmonic analysis with elementary measure theory, none of which was required in the original task.\n\n5. Algebraic expansion in three variables.  \nEven after guessing (3), checking that it produces exactly the stated integer weights entails manipulating Laurent polynomials in three variables; the bookkeeping is far more intricate than in the 2–variable case.\n\nAll of these layers make the enhanced kernel variant substantially harder than both the original problem and the previously supplied kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let w : ℤ³ → ℤ be the convolution kernel whose non–zero values are  \n\n• w(0,0,0)=48;  \n\n• w(±1,0,0)=w(0,±1,0)=w(0,0,±1)=−13;  \n\n• w(±2,0,0)=w(0,±2,0)=w(0,0,±2)= 1;  \n\n• w(±1,±1,0)=w(±1,0,±1)=w(0,±1,±1)= 2  \n\n(the sign choices are independent in every occurrence); for all other lattice points set w=0.  \nFor a finite set S⊂ℤ³ put  \n\nA(S)=∑_{(s,s′)∈S×S} w(s−s′).\n\nProve that A(S)>0 for every non–empty finite subset S of ℤ³.",
      "solution": "Step 1.  A Fourier–analytic formulation.  \nFor a finite S write  \n\nf(x,y,z)=∑_{(a,b,c)∈S} x^{a}y^{b}z^{c}  (Laurent polynomial),  \n\nand let  \n\ng(x,y,z)=∑_{(a,b,c)∈ℤ³} w(a,b,c)x^{a}y^{b}z^{c}.  \n\nBecause convolution turns into multiplication of generating functions,  \n\nA(S)=([x^{0}y^{0}z^{0}] f(x,y,z)f(x^{-1},y^{-1},z^{-1})g(x,y,z)).          (1)\n\nAs in the original problem, the constant coefficient can be extracted by integrating over the 3–torus:\n\n(2π)³A(S)=∫_{0}^{2π}∫_{0}^{2π}∫_{0}^{2π}|f(e^{iα},e^{iβ},e^{iγ})|²·g(e^{iα},e^{iβ},e^{iγ}) dα dβ dγ.        (2)\n\nHence A(S)>0 once we know that \n\n(i) g(e^{iα},e^{iβ},e^{iγ}) is a non–negative real number for every (α,β,γ)∈ℝ³;  \n(ii) it is positive on a set of positive measure in the 3–torus.\n\nStep 2.  A closed form for g on the torus.  \nIntroduce\n\nΛ(α,β,γ)=6−(e^{iα}+e^{-iα}+e^{iβ}+e^{-iβ}+e^{iγ}+e^{-iγ})\n          =6−2( cos α+cos β+cos γ ).\n\nObserve that Λ≥0 for all (α,β,γ) because each cosine is at most 1.  A direct computation using the table for w (or, faster, using that our w was engineered as a polynomial in the discrete Laplacian) shows that\n\ng(x,y,z)=Λ(x,y,z)²+Λ(x,y,z).                                             (3)\n\nIndeed,\n\ng(x,y,z)= (6−(x+x^{-1}+y+y^{-1}+z+z^{-1}))²\n        +(6−(x+x^{-1}+y+y^{-1}+z+z^{-1})),\n\nand expanding (3) yields exactly the prescribed integer coefficients 48,−13,1,2.\n\nSubstituting x=e^{iα}, y=e^{iβ}, z=e^{iγ} gives\n\ng(e^{iα},e^{iβ},e^{iγ})=Λ(α,β,γ)·(Λ(α,β,γ)+1) ≥0,                (4)\n\nwith equality only when Λ=0, i.e. when cos α=cos β=cos γ=1, which forces (α,β,γ)≡(0,0,0) mod 2π.\n\nStep 3.  Strict positivity of the integral.  \nBecause S is finite and non–empty, f is a non–zero trigonometric (Laurent) polynomial.  Hence |f|² is continuous, non–negative and not identically zero, so its set of positivity has positive 3–dimensional measure.\n\nEquation (4) tells us that g is non–negative everywhere and strictly positive except at the single point (α,β,γ)=(0,0,0).  Therefore the integrand in (2) is non–negative everywhere and strictly positive on a set of positive measure (the intersection of “|f|²>0’’ with “Λ>0’’), whence the triple integral is strictly positive.  By (2) this implies A(S)>0, completing the proof.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.633004",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension.  \nThe problem moves from ℤ² to ℤ³, so both the Fourier domain and the combinatorics of the kernel enlarge from a 2–torus to a 3–torus, adding an extra variable and significantly complicating any direct inspection or elementary enumeration approach.\n\n2. Larger support and many more weights.  \nInstead of a 3×3 table (9 numbers), the kernel now has support on 27 lattice points, with four different weight levels (48,−13,2,1).  Simple pattern spotting is impossible; one must understand how the coefficients were produced.\n\n3. Need for structural understanding.  \nThe solution hinges on recognising that w is a quadratic polynomial in the three–dimensional discrete Laplacian.  Without that observation the expansion of g and the verification of non–negativity are prohibitively long.  Participants must therefore combine discrete differential operators with Fourier analysis—material usually outside routine olympiad fare.\n\n4. Measure–theoretic argument.  \nTo transform non–negativity of g into strict positivity of A(S) one must note that g vanishes only on a set of measure 0 and combine this with the continuity and non–triviality of |f|².  This step forces competitors to mix harmonic analysis with elementary measure theory, none of which was required in the original task.\n\n5. Algebraic expansion in three variables.  \nEven after guessing (3), checking that it produces exactly the stated integer weights entails manipulating Laurent polynomials in three variables; the bookkeeping is far more intricate than in the 2–variable case.\n\nAll of these layers make the enhanced kernel variant substantially harder than both the original problem and the previously supplied kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}