{
  "index": "2017-B-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "A line in the plane of a triangle $T$ is called an \\emph{equalizer} if it divides $T$ into two regions having equal area and equal perimeter. Find positive integers $a>b>c$, with $a$ as small as possible, such that there exists a triangle with side lengths $a, b, c$ that has exactly two distinct equalizers.",
  "solution": "The desired integers are $(a,b,c) = (9,8,7)$.\n\nSuppose we have a triangle $T = \\triangle ABC$ with $BC=a$, $CA=b$, $AB=c$ and $a>b>c$.\nSay that a line is an \\textit{area equalizer} if it divides $T$ into two regions of equal area. A line intersecting $T$ must intersect two of the three sides of $T$. First consider a line intersecting the segments $AB$ at $X$ and $BC$ at $Y$, and let $BX=x$, $BY=y$. This line is an area equalizer if and only if $xy\\sin B = 2\\operatorname{area}(\\triangle XBY) = \\operatorname{area}(\\triangle ABC) = \\frac{1}{2}ac\\sin B$, that is, $2xy=ac$. Since $x \\leq c$ and $y \\leq a$, the area equalizers correspond to values of $x,y$ with $xy=ac/2$ and $x \\in [c/2,c]$. Such an area equalizer is also an equalizer if and only if $p/2=x+y$, where $p=a+b+c$ is the perimeter of $T$. If we write $f(x) = x+ac/(2x)$, then we want to solve $f(x) = p/2$ for $x \\in [c/2,c]$. Now note that $f$ is convex, $f(c/2) = a+c/2 > p/2$, and $f(c) = a/2+c < p/2$; it follows that there is exactly one solution to $f(x)=p/2$ in $[c/2,c]$.\nSimilarly, for equalizers intersecting $T$ on the sides $AB$ and $AC$, we want to solve $g(x) = p/2$ where $g(x) = x+bc/(2x)$ and $x \\in [c/2,c]$; since $g$ is convex and $g(c/2)<p/2$, $g(c) < p/2$, there are no such solutions.\n\n\nIt follows that if $T$ has exactly two equalizers, then it must have exactly one equalizer intersecting $T$ on the sides $AC$ and $BC$. Here we want to solve $h(x) = p/2$ where $h(x) = x+ab/(2x)$ and $x \\in [a/2,a]$. Now $h$ is convex and $h(a/2) > p/2$, $h(a) > p/2$; thus $h(x) = p/2$ has exactly one solution $x \\in [a/2,a]$ if and only if there is $x_0 \\in [a/2,a]$ with $h'(x_0) = 0$ and $h(x_0) = p/2$. The first condition implies $x_0 = \\sqrt{ab/2}$, and then the second condition gives $8ab = p^2$. Note that $\\sqrt{ab/2}$ is in $[a/2,a]$ since $a>b$ and $a<b+c<2b$.\n\n\nWe conclude that $T$ has two equalizers if and only if $8ab=(a+b+c)^2$. Note that $(a,b,c) = (9,8,7)$ works. We claim that this is the only possibility when $a>b>c$ are integers and $a \\leq 9$. Indeed, the only integers $(a,b)$ such that $2 \\leq b < a \\leq 9$ and $8ab$ is a perfect square are $(a,b) = (4,2)$, $(6,3)$, $(8,4)$, $(9,2)$, and $(9,8)$, and the first four possibilities do not produce triangles since they do not satisfy $a<2b$. This gives the claimed result.",
  "vars": [
    "a",
    "b",
    "c",
    "x",
    "y",
    "p",
    "T",
    "A",
    "B",
    "C",
    "X",
    "Y"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "sidemajor",
        "b": "sidemedium",
        "c": "sideminor",
        "x": "segmentone",
        "y": "segmenttwo",
        "p": "perimeter",
        "T": "trianglebody",
        "A": "vertexalpha",
        "B": "vertexbeta",
        "C": "vertexgamma",
        "X": "pointxi",
        "Y": "pointupsilon"
      },
      "question": "A line in the plane of a triangle $trianglebody$ is called an \\emph{equalizer} if it divides $trianglebody$ into two regions having equal area and equal perimeter. Find positive integers $sidemajor>sidemedium>sideminor$, with $sidemajor$ as small as possible, such that there exists a triangle with side lengths $sidemajor, sidemedium, sideminor$ that has exactly two distinct equalizers.",
      "solution": "The desired integers are $(sidemajor,sidemedium,sideminor) = (9,8,7)$.\\n\\nSuppose we have a triangle $trianglebody = \\triangle vertexalpha vertexbeta vertexgamma$ with $vertexbeta vertexgamma=sidemajor$, $vertexgamma vertexalpha=sidemedium$, $vertexalpha vertexbeta=sideminor$ and $sidemajor>sidemedium>sideminor$.\\nSay that a line is an \\textit{area equalizer} if it divides $trianglebody$ into two regions of equal area. A line intersecting $trianglebody$ must intersect two of the three sides of $trianglebody$. First consider a line intersecting the segments $vertexalpha vertexbeta$ at $pointxi$ and $vertexbeta vertexgamma$ at $pointupsilon$, and let $vertexbeta pointxi=segmentone$, $vertexbeta pointupsilon=segmenttwo$. This line is an area equalizer if and only if $segmentone segmenttwo\\sin vertexbeta = 2\\operatorname{area}(\\triangle pointxi vertexbeta pointupsilon) = \\operatorname{area}(\\triangle vertexalpha vertexbeta vertexgamma) = \\frac{1}{2}sidemajor sideminor\\sin vertexbeta$, that is, $2segmentone segmenttwo=sidemajor sideminor$. Since $segmentone \\leq sideminor$ and $segmenttwo \\leq sidemajor$, the area equalizers correspond to values of $segmentone,segmenttwo$ with $segmentone segmenttwo=sidemajor sideminor/2$ and $segmentone \\in [sideminor/2,sideminor]$. Such an area equalizer is also an equalizer if and only if $perimeter/2=segmentone+segmenttwo$, where $perimeter=sidemajor+sidemedium+sideminor$ is the perimeter of $trianglebody$. If we write $f(segmentone) = segmentone+sidemajor sideminor/(2segmentone)$, then we want to solve $f(segmentone) = perimeter/2$ for $segmentone \\in [sideminor/2,sideminor]$. Now note that $f$ is convex, $f(sideminor/2) = sidemajor+sideminor/2 > perimeter/2$, and $f(sideminor) = sidemajor/2+sideminor < perimeter/2$; it follows that there is exactly one solution to $f(segmentone)=perimeter/2$ in $[sideminor/2,sideminor]$.\\nSimilarly, for equalizers intersecting $trianglebody$ on the sides $vertexalpha vertexbeta$ and $vertexalpha vertexgamma$, we want to solve $g(segmentone) = perimeter/2$ where $g(segmentone) = segmentone+sidemedium sideminor/(2segmentone)$ and $segmentone \\in [sideminor/2,sideminor]$; since $g$ is convex and $g(sideminor/2)<perimeter/2$, $g(sideminor) < perimeter/2$, there are no such solutions.\\n\\n\\nIt follows that if $trianglebody$ has exactly two equalizers, then it must have exactly one equalizer intersecting $trianglebody$ on the sides $vertexalpha vertexgamma$ and $vertexbeta vertexgamma$. Here we want to solve $h(segmentone) = perimeter/2$ where $h(segmentone) = segmentone+sidemajor sidemedium/(2segmentone)$ and $segmentone \\in [sidemajor/2,sidemajor]$. Now $h$ is convex and $h(sidemajor/2) > perimeter/2$, $h(sidemajor) > perimeter/2$; thus $h(segmentone) = perimeter/2$ has exactly one solution $segmentone \\in [sidemajor/2,sidemajor]$ if and only if there is $segmentone_{0} \\in [sidemajor/2,sidemajor]$ with $h'(segmentone_{0}) = 0$ and $h(segmentone_{0}) = perimeter/2$. The first condition implies $segmentone_{0} = \\sqrt{sidemajor sidemedium/2}$, and then the second condition gives $8sidemajor sidemedium = perimeter^2$. Note that $\\sqrt{sidemajor sidemedium/2}$ is in $[sidemajor/2,sidemajor]$ since $sidemajor>sidemedium$ and $sidemajor<sidemedium+sideminor<2sidemedium$.\\n\\n\\nWe conclude that $trianglebody$ has two equalizers if and only if $8sidemajor sidemedium=(sidemajor+sidemedium+sideminor)^2$. Note that $(sidemajor,sidemedium,sideminor) = (9,8,7)$ works. We claim that this is the only possibility when $sidemajor>sidemedium>sideminor$ are integers and $sidemajor \\leq 9$. Indeed, the only integers $(sidemajor,sidemedium)$ such that $2 \\leq sidemedium < sidemajor \\leq 9$ and $8sidemajor sidemedium$ is a perfect square are $(sidemajor,sidemedium) = (4,2)$, $(6,3)$, $(8,4)$, $(9,2)$, and $(9,8)$, and the first four possibilities do not produce triangles since they do not satisfy $sidemajor<2sidemedium$. This gives the claimed result."
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "sandstone",
        "b": "driftwood",
        "c": "hummingby",
        "x": "marigold",
        "y": "tortoise",
        "p": "beachcomb",
        "T": "raincloud",
        "A": "pinecone",
        "B": "woodpeck",
        "C": "starfruit",
        "X": "moonstone",
        "Y": "sunflower"
      },
      "question": "<<<\nA line in the plane of a triangle $raincloud$ is called an \\emph{equalizer} if it divides $raincloud$ into two regions having equal area and equal perimeter. Find positive integers $sandstone>driftwood>hummingby$, with $sandstone$ as small as possible, such that there exists a triangle with side lengths $sandstone, driftwood, hummingby$ that has exactly two distinct equalizers.\n>>>",
      "solution": "<<<\nThe desired integers are $(sandstone,driftwood,hummingby) = (9,8,7)$.\n\nSuppose we have a triangle $raincloud = \\triangle pinecone woodpeck starfruit$ with $woodpeck starfruit = sandstone$, $starfruit pinecone = driftwood$, $pinecone woodpeck = hummingby$ and $sandstone>driftwood>hummingby$. Say that a line is an \\textit{area equalizer} if it divides $raincloud$ into two regions of equal area. A line intersecting $raincloud$ must intersect two of the three sides of $raincloud$. First consider a line intersecting the segments $pinecone woodpeck$ at $moonstone$ and $woodpeck starfruit$ at $sunflower$, and let $woodpeck moonstone = marigold$, $woodpeck sunflower = tortoise$. This line is an area equalizer if and only if $marigold tortoise\\sin woodpeck = 2\\operatorname{area}(\\triangle moonstone woodpeck sunflower) = \\operatorname{area}(\\triangle pinecone woodpeck starfruit) = \\frac{1}{2}sandstone hummingby\\sin woodpeck$, that is, $2marigold tortoise = sandstone hummingby$. Since $marigold \\leq hummingby$ and $tortoise \\leq sandstone$, the area equalizers correspond to values of $marigold,tortoise$ with $marigold tortoise = sandstone hummingby/2$ and $marigold \\in [hummingby/2,hummingby]$. Such an area equalizer is also an equalizer if and only if $beachcomb/2 = marigold + tortoise$, where $beachcomb = sandstone+driftwood+hummingby$ is the perimeter of $raincloud$. If we write $f(marigold) = marigold + sandstone hummingby / (2 marigold)$, then we want to solve $f(marigold) = beachcomb/2$ for $marigold \\in [hummingby/2,hummingby]$. Now note that $f$ is convex, $f(hummingby/2) = sandstone + hummingby/2 > beachcomb/2$, and $f(hummingby) = sandstone/2 + hummingby < beachcomb/2$; it follows that there is exactly one solution to $f(marigold)=beachcomb/2$ in $[hummingby/2,hummingby]$.\nSimilarly, for equalizers intersecting $raincloud$ on the sides $pinecone woodpeck$ and $pinecone starfruit$, we want to solve $g(marigold) = beachcomb/2$ where $g(marigold) = marigold + driftwood hummingby /(2 marigold)$ and $marigold \\in [hummingby/2,hummingby]$; since $g$ is convex and $g(hummingby/2)<beachcomb/2$, $g(hummingby) < beachcomb/2$, there are no such solutions.\n\nIt follows that if $raincloud$ has exactly two equalizers, then it must have exactly one equalizer intersecting $raincloud$ on the sides $starfruit pinecone$ and $woodpeck starfruit$. Here we want to solve $h(marigold) = beachcomb/2$ where $h(marigold) = marigold + sandstone driftwood /(2 marigold)$ and $marigold \\in [sandstone/2,sandstone]$. Now $h$ is convex and $h(sandstone/2) > beachcomb/2$, $h(sandstone) > beachcomb/2$; thus $h(marigold) = beachcomb/2$ has exactly one solution $marigold \\in [sandstone/2,sandstone]$ if and only if there is $marigold_0 \\in [sandstone/2,sandstone]$ with $h'(marigold_0) = 0$ and $h(marigold_0) = beachcomb/2$. The first condition implies $marigold_0 = \\sqrt{sandstone driftwood/2}$, and then the second condition gives $8 sandstone driftwood = beachcomb^2$. Note that $\\sqrt{sandstone driftwood/2}$ is in $[sandstone/2,sandstone]$ since $sandstone>driftwood$ and $sandstone<driftwood+hummingby<2 driftwood$.\n\nWe conclude that $raincloud$ has two equalizers if and only if $8 sandstone driftwood=(sandstone+driftwood+hummingby)^2$. Note that $(sandstone,driftwood,hummingby) = (9,8,7)$ works. We claim that this is the only possibility when $sandstone>driftwood>hummingby$ are integers and $sandstone \\leq 9$. Indeed, the only integers $(sandstone,driftwood)$ such that $2 \\leq driftwood < sandstone \\leq 9$ and $8 sandstone driftwood$ is a perfect square are $(sandstone,driftwood) = (4,2)$, $(6,3)$, $(8,4)$, $(9,2)$, and $(9,8)$, and the first four possibilities do not produce triangles since they do not satisfy $sandstone<2driftwood$. This gives the claimed result.\n>>>"
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "miniside",
        "b": "maxiside",
        "c": "hugeside",
        "x": "fullspan",
        "y": "entiregap",
        "p": "interior",
        "T": "squarefig",
        "A": "voidpoint",
        "B": "blanknode",
        "C": "nullvertex",
        "X": "outerpoint",
        "Y": "farpoint"
      },
      "question": "A line in the plane of a triangle $squarefig$ is called an \\emph{equalizer} if it divides $squarefig$ into two regions having equal area and equal perimeter. Find positive integers $miniside>maxiside>hugeside$, with $miniside$ as small as possible, such that there exists a triangle with side lengths $miniside, maxiside, hugeside$ that has exactly two distinct equalizers.",
      "solution": "The desired integers are $(miniside,maxiside,hugeside) = (9,8,7)$.\\n\\nSuppose we have a triangle $squarefig = \\triangle voidpoint blanknode nullvertex$ with $blanknodenullvertex=miniside$, $nullvertexvoidpoint=maxiside$, $voidpointblanknode=hugeside$ and $miniside>maxiside>hugeside$.\\n\\nSay that a line is an \\textit{area equalizer} if it divides $squarefig$ into two regions of equal area. A line intersecting $squarefig$ must intersect two of the three sides of $squarefig$. First consider a line intersecting the segments $voidpointblanknode$ at $outerpoint$ and $blanknodenullvertex$ at $farpoint$, and let $blanknodeouterpoint=fullspan$, $blanknodefarpoint=entiregap$. This line is an area equalizer if and only if $fullspan\\,entiregap\\sin blanknode = 2\\operatorname{area}(\\triangle outerpoint farpoint blanknode) = \\operatorname{area}(\\triangle voidpoint blanknode nullvertex) = \\frac{1}{2}miniside hugeside\\sin blanknode$, that is, $2\\,fullspan\\,entiregap = miniside hugeside$. Since $fullspan \\le hugeside$ and $entiregap \\le miniside$, the area equalizers correspond to values of $fullspan,entiregap$ with $fullspan\\,entiregap=miniside hugeside/2$ and $fullspan \\in [hugeside/2,hugeside]$. Such an area equalizer is also an equalizer if and only if $interior/2 = fullspan+entiregap$, where $interior = miniside+maxiside+hugeside$ is the perimeter of $squarefig$.\\n\\nIf we write $f(fullspan) = fullspan + miniside hugeside/(2\\,fullspan)$, then we want to solve $f(fullspan) = interior/2$ for $fullspan \\in [hugeside/2,hugeside]$. Now note that $f$ is convex, $f(hugeside/2) = miniside + hugeside/2 > interior/2$, and $f(hugeside) = miniside/2 + hugeside < interior/2$; it follows that there is exactly one solution to $f(fullspan)=interior/2$ in $[hugeside/2,hugeside]$.\\n\\nSimilarly, for equalizers intersecting $squarefig$ on the sides $voidpointblanknode$ and $voidpointnullvertex$, we want to solve $g(fullspan) = interior/2$ where $g(fullspan) = fullspan + maxiside hugeside/(2\\,fullspan)$ and $fullspan \\in [hugeside/2,hugeside]$; since $g$ is convex and $g(hugeside/2)<interior/2$, $g(hugeside) < interior/2$, there are no such solutions.\\n\\nIt follows that if $squarefig$ has exactly two equalizers, then it must have exactly one equalizer intersecting $squarefig$ on the sides $voidpointnullvertex$ and $blanknodenullvertex$. Here we want to solve $h(fullspan) = interior/2$ where $h(fullspan) = fullspan + miniside maxiside/(2\\,fullspan)$ and $fullspan \\in [miniside/2,miniside]$. Now $h$ is convex and $h(miniside/2) > interior/2$, $h(miniside) > interior/2$; thus $h(fullspan) = interior/2$ has exactly one solution $fullspan \\in [miniside/2,miniside]$ if and only if there is $fullspan_0 \\in [miniside/2,miniside]$ with $h'(fullspan_0) = 0$ and $h(fullspan_0) = interior/2$. The first condition implies $fullspan_0 = \\sqrt{miniside maxiside/2}$, and then the second condition gives $8\\,miniside maxiside = interior^2$. Note that $\\sqrt{miniside maxiside/2}$ is in $[miniside/2,miniside]$ since $miniside>maxiside$ and $miniside<maxiside+hugeside<2\\,maxiside$.\\n\\nWe conclude that $squarefig$ has two equalizers if and only if $8\\,miniside maxiside=(miniside+maxiside+hugeside)^2$. Note that $(miniside,maxiside,hugeside)=(9,8,7)$ works. We claim that this is the only possibility when $miniside>maxiside>hugeside$ are integers and $miniside \\le 9$. Indeed, the only integers $(miniside,maxiside)$ such that $2 \\le maxiside < miniside \\le 9$ and $8\\,miniside maxiside$ is a perfect square are $(miniside,maxiside) = (4,2)$, $(6,3)$, $(8,4)$, $(9,2)$, and $(9,8)$, and the first four possibilities do not produce triangles since they do not satisfy $miniside<2\\,maxiside$. This gives the claimed result."
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "nvdprkse",
        "x": "mpqslwzn",
        "y": "trbgcavf",
        "p": "skljvqno",
        "T": "fgbrqzna",
        "A": "ufypdnrm",
        "B": "kzmtwgha",
        "C": "rdnqlsvi",
        "X": "bytjagoe",
        "Y": "cvxspelu"
      },
      "question": "A line in the plane of a triangle $fgbrqzna$ is called an \\emph{equalizer} if it divides $fgbrqzna$ into two regions having equal area and equal perimeter. Find positive integers $qzxwvtnp>hjgrksla>nvdprkse$, with $qzxwvtnp$ as small as possible, such that there exists a triangle with side lengths $qzxwvtnp, hjgrksla, nvdprkse$ that has exactly two distinct equalizers.",
      "solution": "The desired integers are $(qzxwvtnp,hjgrksla,nvdprkse) = (9,8,7)$.\\n\\nSuppose we have a triangle $fgbrqzna = \\triangle ufypdnrm kzmtwgha rdnqlsvi$ with $kzmtwghardnqlsvi=qzxwvtnp$, $rdnqlsvi ufypdnrm=hjgrksla$, $ufypdnrm kzmtwgha=nvdprkse$ and $qzxwvtnp>hjgrksla>nvdprkse$.\\nSay that a line is an \\textit{area equalizer} if it divides $fgbrqzna$ into two regions of equal area. A line intersecting $fgbrqzna$ must intersect two of the three sides of $fgbrqzna$. First consider a line intersecting the segments $ufypdnrm kzmtwgha$ at $bytjagoe$ and $kzmtwghardnqlsvi$ at $cvxspelu$, and let $kzmtwgha bytjagoe=mpqslwzn$, $kzmtwgha cvxspelu=trbgcavf$. This line is an area equalizer if and only if $mpqslwzn trbgcavf\\sin kzmtwgha = 2\\operatorname{area}(\\triangle bytjagoe kzmtwgha cvxspelu) = \\operatorname{area}(\\triangle ufypdnrm kzmtwgha rdnqlsvi) = \\frac{1}{2}qzxwvtnp nvdprkse\\sin kzmtwgha$, that is, $2 mpqslwzn trbgcavf = qzxwvtnp nvdprkse$. Since $mpqslwzn \\leq nvdprkse$ and $trbgcavf \\leq qzxwvtnp$, the area equalizers correspond to values of $mpqslwzn, trbgcavf$ with $mpqslwzn trbgcavf = qzxwvtnp nvdprkse/2$ and $mpqslwzn \\in [nvdprkse/2,nvdprkse]$. Such an area equalizer is also an equalizer if and only if $skljvqno/2 = mpqslwzn + trbgcavf$, where $skljvqno = qzxwvtnp + hjgrksla + nvdprkse$ is the perimeter of $fgbrqzna$. If we write $f(mpqslwzn) = mpqslwzn + qzxwvtnp nvdprkse/(2 mpqslwzn)$, then we want to solve $f(mpqslwzn) = skljvqno/2$ for $mpqslwzn \\in [nvdprkse/2,nvdprkse]$. Now note that $f$ is convex, $f(nvdprkse/2) = qzxwvtnp + nvdprkse/2 > skljvqno/2$, and $f(nvdprkse) = qzxwvtnp/2 + nvdprkse < skljvqno/2$; it follows that there is exactly one solution to $f(mpqslwzn)=skljvqno/2$ in $[nvdprkse/2,nvdprkse]$.\\nSimilarly, for equalizers intersecting $fgbrqzna$ on the sides $ufypdnrm kzmtwgha$ and $ufypdnrm rdnqlsvi$, we want to solve $g(mpqslwzn) = skljvqno/2$ where $g(mpqslwzn) = mpqslwzn + hjgrksla nvdprkse/(2 mpqslwzn)$ and $mpqslwzn \\in [nvdprkse/2,nvdprkse]$; since $g$ is convex and $g(nvdprkse/2)<skljvqno/2$, $g(nvdprkse) < skljvqno/2$, there are no such solutions.\\n\\n\\nIt follows that if $fgbrqzna$ has exactly two equalizers, then it must have exactly one equalizer intersecting $fgbrqzna$ on the sides $ufypdnrm rdnqlsvi$ and $kzmtwghardnqlsvi$. Here we want to solve $h(mpqslwzn) = skljvqno/2$ where $h(mpqslwzn) = mpqslwzn + qzxwvtnp hjgrksla/(2 mpqslwzn)$ and $mpqslwzn \\in [qzxwvtnp/2,qzxwvtnp]$. Now $h$ is convex and $h(qzxwvtnp/2) > skljvqno/2$, $h(qzxwvtnp) > skljvqno/2$; thus $h(mpqslwzn) = skljvqno/2$ has exactly one solution $mpqslwzn \\in [qzxwvtnp/2,qzxwvtnp]$ if and only if there is $mpqslwzn_0 \\in [qzxwvtnp/2,qzxwvtnp]$ with $h'(mpqslwzn_0) = 0$ and $h(mpqslwzn_0) = skljvqno/2$. The first condition implies $mpqslwzn_0 = \\sqrt{qzxwvtnp hjgrksla/2}$, and then the second condition gives $8 qzxwvtnp hjgrksla = skljvqno^2$. Note that $\\sqrt{qzxwvtnp hjgrksla/2}$ is in $[qzxwvtnp/2,qzxwvtnp]$ since $qzxwvtnp>hjgrksla$ and $qzxwvtnp<hjgrksla + nvdprkse <2 hjgrksla$.\\n\\n\\nWe conclude that $fgbrqzna$ has two equalizers if and only if $8 qzxwvtnp hjgrksla = (qzxwvtnp + hjgrksla + nvdprkse)^2$. Note that $(qzxwvtnp,hjgrksla,nvdprkse) = (9,8,7)$ works. We claim that this is the only possibility when $qzxwvtnp>hjgrksla>nvdprkse$ are integers and $qzxwvtnp \\leq 9$. Indeed, the only integers $(qzxwvtnp,hjgrksla)$ such that $2 \\leq hjgrksla < qzxwvtnp \\leq 9$ and $8 qzxwvtnp hjgrksla$ is a perfect square are $(qzxwvtnp,hjgrksla) = (4,2)$, $(6,3)$, $(8,4)$, $(9,2)$, and $(9,8)$, and the first four possibilities do not produce triangles since they do not satisfy $qzxwvtnp<2 hjgrksla$. This gives the claimed result."
    },
    "kernel_variant": {
      "question": "A line in the plane of a triangle T is called an equalizer if it divides T into two regions that have both equal area and equal perimeter.  \n\nLet the side–lengths of the triangle be positive integers x < y < z (so z is the longest side).  Determine – with z as small as possible – all triples (x, y, z) for which there exists a triangle whose side-lengths are x, y, z and that possesses exactly two distinct equalizers.",
      "solution": "Throughout write △ABC with\n BC = a , CA = b , AB = c , with a > b > c > 0  (1)\nand total perimeter p = a + b + c.  \n(In the end z = a, y = b, x = c.)\n\nA straight line that meets the interior of △ABC intersects exactly two of its three sides, so possible equalizers fall into three geometric cases.\nFor each case we\n • choose signed lengths along the two sides that are cut,  \n • impose the equal–area condition, obtaining a relation of the form uv = constant,  \n • impose the equal–perimeter condition, which always turns out to be u + v = p ⁄ 2,  \n • obtain a single equation of the form\n   F(u) = u + K⁄u = p ⁄ 2   (2)\nwhose solutions inside a suitable closed interval correspond exactly to equalizers of that type.  Because F is strictly convex on (0, ∞), (2) has at most two solutions in any interval and the number of solutions is determined by the signs of F at the end–points.\n\n--------------------------------------------------\nCase 1 The line meets AB(=c) and AC(=b)\n--------------------------------------------------\nPlace points X ∈ AB and Y ∈ AC so that AX = x and AY = y.\n\nEqual area: xy = bc ⁄ 2.\nSince 0 ≤ x ≤ c and 0 ≤ y ≤ b, we must have x ∈ [ c ⁄ 2 , c ] and y = bc ⁄ (2x).\n\nEqual perimeter: x + y = p ⁄ 2.\nPutting these together we arrive at\n f(x) := x + bc ⁄ (2x) = p ⁄ 2 , x ∈ [c ⁄ 2 , c].\nBecause\n f(c ⁄ 2) = c ⁄ 2 + b    <    p ⁄ 2  [since a > b],\n f(c)      = c     + b ⁄ 2 <    p ⁄ 2  [since a > c],\nconvexity shows that the graph of f lies completely below the horizontal line y = p ⁄ 2, so\n Case 1 supplies NO equalizers.                         (3)\n\n--------------------------------------------------\nCase 2 The line meets AB(=c) and BC(=a)\n--------------------------------------------------\nIt is convenient to measure lengths from the common vertex B.\nPut X ∈ AB, Y ∈ BC with\n BX = u (0 ≤ u ≤ c) and BY = v (0 ≤ v ≤ a).  (4)\n\nEqual area.\nThe region adjacent to B is △BXY, whose sides BX = u and BY = v form the angle B.  Requiring\n area(△BXY) = ½·area(△ABC) = ¼·ac·sin B\nproduces\n uv = ac ⁄ 2.                                            (5)\nBecause v ≤ a, (5) forces u ≥ c ⁄ 2, so u ranges over the interval [c ⁄ 2 , c] and\n v = ac ⁄ (2u).\n\nEqual perimeter.\nPerimeter(△BXY) = u + v + XY ;\nperimeter of the complementary region = (c – u) + (a – v) + b + XY.  Equality of perimeters therefore gives\n u + v = p ⁄ 2.                                           (6)\nCombining (5) and (6) we obtain\n g(u) := u + ac ⁄ (2u) = p ⁄ 2 , u ∈ [c ⁄ 2 , c].\nThe end-point values are\n g(c ⁄ 2) = c ⁄ 2 + a  > p ⁄ 2  [since a > b],\n g(c)      = c     + a ⁄ 2 < p ⁄ 2  [since c < b].\nBy strict convexity there is exactly one root of g(u)=p/2 in (c ⁄ 2, c).  Hence\n Case 2 supplies EXACTLY ONE equalizer.                  (7)\n\n--------------------------------------------------\nCase 3 The line meets BC(=a) and CA(=b)\n--------------------------------------------------\nChoose X ∈ BC and Y ∈ CA so that CX = t and CY = s.\n\nEqual area: ts = ab ⁄ 2, whence t ∈ [a ⁄ 2 , a] and s = ab ⁄ (2t).\nEqual perimeter: t + s = p ⁄ 2,\nleading to\n h(t) := t + ab ⁄ (2t) = p ⁄ 2 , t ∈ [a ⁄ 2 , a].      (8)\nConvexity and the end-point values\n h(a ⁄ 2) = a ⁄ 2 + b  > p ⁄ 2,\n h(a)      = a     + b ⁄ 2 > p ⁄ 2                       (9)\nshow that (8) has either 0, 1, or 2 solutions.  There is exactly one solution iff (8) is satisfied at the sole critical point, which is at\n t₀ = √(ab ⁄ 2).  Evaluating we get\n h(t₀) = 2√(ab ⁄ 2).\nThus (8) has one root precisely when\n 2√(ab ⁄ 2) = p ⁄ 2    ⇔    8ab = p².                    (10)\nIf (10) holds, Case 3 contributes exactly one equalizer; otherwise it contributes none.\n\n--------------------------------------------------\nCounting equalizers\n--------------------------------------------------\nFrom (3), (7) and (10) the triangle has exactly two equalizers iff\n (i) 8ab = (a + b + c)², and\n (ii) a > b > c obey the triangle inequalities.         (11)\n\n--------------------------------------------------\nIntegral solutions with a minimal\n--------------------------------------------------\nWrite p = a + b + c = 2m.  Condition (11 i) becomes\n ab = m² ⁄ 2,\nso m is even; write m = 2n and obtain\n ab = 2n².                                               (12)\nTo minimise a (= longest side) we factor 2n² into unequal integers a > b such that a < b + c.  A very short search gives\n n = 1,2,3,4,5 → ab = 2, 8, 18, 32, 50   (no admissible pair),\n n = 6      → ab = 72.\nThe factorisation 72 = 9·8 yields (a, b) = (9, 8).  Then\n p = 2m = 4n = 24 and c = p – a – b = 24 – 9 – 8 = 7.\nAll triangle inequalities hold (8 > 7, 9 < 8 + 7), so (9, 8, 7) is feasible.\n\nFor a ≤ 9 this is the only possibility:  for a ≤ 8 condition (12) forces b ≤ 4, giving a ≥ 2b and thus violating a < b + c.  Therefore a = 9 is minimal and produces exactly one triple.\n\n--------------------------------------------------\nAnswer\n--------------------------------------------------\nWith the side-lengths listed in increasing order the unique triple is\n   (x, y, z) = (7, 8, 9).\nThe triangle with sides 7, 8, 9 possesses exactly two distinct equalizers, and no triangle whose longest side is smaller than 9 has this property.",
      "_meta": {
        "core_steps": [
          "Parameterize a candidate equalizer by its intersection distances (x,y) on two sides; translate the equal–area and equal–perimeter requirements into the system  xy = (product of the two intersected sides)/2  and  x + y = p/2.",
          "For each unordered pair of sides, reduce the system to a single equation of the form  F(x)=p/2  where  F(x)=x+k/x  (with k fixed by the chosen pair).",
          "Use convexity of F(x) and comparison of its endpoint values to count how many solutions each pair can contribute; deduce that only one specific pair can supply equalizers, and at most one line comes from that pair.",
          "Impose the additional condition that the unique solution occurs at the stationary point of F, giving  8ab = (a+b+c)^2.",
          "Solve the resulting Diophantine + triangle‐inequality constraints to find the minimal integral triple, yielding (9,8,7)."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Labeling convention: the longest, middle, and shortest sides are called a, b, c respectively (ordering a>b>c).  Renaming does not affect the argument.",
            "original": "a>b>c"
          },
          "slot2": {
            "description": "The particular side pair examined first (here AB–BC). Any of the three unordered pairs could be treated first without changing the logic.",
            "original": "(AB , BC)"
          },
          "slot3": {
            "description": "The ad-hoc upper bound a≤9 used in the final brute-force search for integral solutions; any larger bound or a different search strategy would leave the chain of reasoning intact.",
            "original": "a ≤ 9"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}