{
  "index": "1940-A-8",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "8. A triangle is bounded by the lines\n\\[\n\\begin{array}{c}\nA_{1} x+B_{1} y+C_{1}=0, \\quad A_{2} x+B_{2} y+C_{2}=0 \\\\\nA_{3} x+B_{3} y+C_{3}=0\n\\end{array}\n\\]\n\nShow that the area, disregarding sign, is",
  "solution": "Solution. Let \\( L_{i} \\) be the line with equation \\( A_{i} x+B_{i} y+C_{i}=0 \\), and let \\( \\left(x_{i}, y_{i}\\right) \\) be the vertex of the triangle opposite \\( L_{i} \\). Let\n\\[\nM=\\left(\\begin{array}{lll}\nA_{1} & B_{1} & C_{1} \\\\\nA_{2} & B_{2} & C_{2} \\\\\nA_{3} & B_{3} & C_{3}\n\\end{array}\\right)\n\\]\nand\n\\[\nX=\\left(\\begin{array}{ccc}\nx_{1} & x_{2} & x_{3} \\\\\ny_{1} & y_{2} & y_{3} \\\\\n1 & 1 & 1\n\\end{array}\\right]\n\\]\n\nThen the area of the triangle is \\( \\frac{1}{2}|\\operatorname{det} X| \\).\nSince the point \\( \\left(x_{i}, y_{i}\\right) \\) is on the side \\( L_{j} \\) for \\( i \\neq j \\), we find\n\\[\nM X=\\left(\\begin{array}{lll}\nd_{1} & 0 & 0 \\\\\n0 & d_{2} & 0 \\\\\n0 & 0 & d_{3}\n\\end{array}\\right]\n\\]\nwhere the diagonal entries remain to be determined.\nIf we solve the system of equations\n\\[\nM\\left(\\begin{array}{l}\nx \\\\\ny \\\\\nz\n\\end{array}\\right]=\\left(\\begin{array}{c}\nd_{1} \\\\\n0 \\\\\n0\n\\end{array}\\right]\n\\]\nby Cramer's rule, we find\n\\[\nz \\operatorname{det} M=d_{1}\\left|\\begin{array}{ll}\nA_{2} & B_{2} \\\\\nA_{3} & B_{3}\n\\end{array}\\right|\n\\]\n\nBut from (1) we know that the solution is\n\\[\n\\left(\\begin{array}{c}\nx_{1} \\\\\ny_{1} \\\\\n1\n\\end{array}\\right] .\n\\]\n\nHence,\n\\[\nd_{1}=\\frac{\\operatorname{det} M}{\\left|\\begin{array}{ll}\nA_{2} & B_{2} \\\\\nA_{3} & B_{3}\n\\end{array}\\right| .}\n\\]\n\nSimilarly,\n\\[\n\\begin{array}{l}\nd_{2}=\\frac{\\operatorname{det} M}{\\left|\\begin{array}{ll}\nA_{3} & B_{3} \\\\\nA_{1} & B_{1}\n\\end{array}\\right|} \\\\\nd_{3}=\\frac{\\operatorname{det} M}{\\left|\\begin{array}{ll}\nA_{1} & B_{1} \\\\\nA_{2} & B_{2}\n\\end{array}\\right| .}\n\\end{array}\n\\]\n\nTherefore, from (1),\n\\[\n(\\operatorname{det} M)(\\operatorname{det} X)=\\frac{(\\operatorname{det} M)^{3}}{\\left|\\begin{array}{ll}\nA_{2} & B_{2} \\\\\nA_{3} & B_{3}\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nA_{3} & B_{3} \\\\\nA_{1} & B_{1}\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nA_{1} & B_{1} \\\\\nA_{2} & B_{2}\n\\end{array}\\right|}\n\\]\n\nNow \\( \\operatorname{det} M \\neq 0 \\) is precisely the condition that the given lines form a triangle, so\n\\[\n\\begin{array}{c}\n\\text { Area of triangle }=\\frac{1}{2}|\\operatorname{det} X| \\\\\n= \\pm \\frac{(\\operatorname{det} M)^{2}}{2\\left|\\begin{array}{ll}\nA_{2} & B_{2} \\\\\nA_{3} & B_{3}\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nA_{3} & B_{3} \\\\\nA_{1} & B_{1}\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nA_{1} & B_{1} \\\\\nA_{2} & B_{2}\n\\end{array}\\right|}\n\\end{array}\n\\]",
  "vars": [
    "x",
    "y",
    "z",
    "x_1",
    "y_1",
    "x_2",
    "y_2",
    "x_3",
    "y_3",
    "x_i",
    "y_i",
    "d_1",
    "d_2",
    "d_3",
    "i",
    "j",
    "M",
    "X",
    "L_i",
    "L_j"
  ],
  "params": [
    "A_1",
    "A_2",
    "A_3",
    "A_i",
    "B_1",
    "B_2",
    "B_3",
    "B_i",
    "C_1",
    "C_2",
    "C_3",
    "C_i"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "y": "variabley",
        "z": "variablez",
        "x_1": "vertexxone",
        "y_1": "vertexyone",
        "x_2": "vertexxtwo",
        "y_2": "vertexytwo",
        "x_3": "vertexxthree",
        "y_3": "vertexythree",
        "x_i": "vertexxgen",
        "y_i": "vertexygen",
        "d_1": "diagonalone",
        "d_2": "diagonaltwo",
        "d_3": "diagonalthree",
        "i": "indexi",
        "M": "matrixm",
        "X": "matrixx",
        "L_i": "boundaryi",
        "L_j": "boundaryj",
        "A_1": "coeffaone",
        "A_2": "coeffatwo",
        "A_3": "coeffathree",
        "A_i": "coeffagen",
        "B_1": "coeffbone",
        "B_2": "coeffbtwo",
        "B_3": "coeffbthree",
        "B_i": "coeffbgen",
        "C_1": "coeffcone",
        "C_2": "coeffctwo",
        "C_3": "coeffcthree",
        "C_i": "coeffcgen"
      },
      "question": "8. A triangle is bounded by the lines\n\\[\n\\begin{array}{c}\ncoeffaone variablex+coeffbone variabley+coeffcone=0, \\quad coeffatwo variablex+coeffbtwo variabley+coeffctwo=0 \\\\\ncoeffathree variablex+coeffbthree variabley+coeffcthree=0\n\\end{array}\n\\]\n\nShow that the area, disregarding sign, is",
      "solution": "Solution. Let \\( boundaryi \\) be the line with equation \\( coeffagen variablex+coeffbgen variabley+coeffcgen=0 \\), and let \\( \\left(vertexxgen, vertexygen\\right) \\) be the vertex of the triangle opposite \\( boundaryi \\). Let\n\\[\nmatrixm=\\left(\\begin{array}{lll}\ncoeffaone & coeffbone & coeffcone \\\\\ncoeffatwo & coeffbtwo & coeffctwo \\\\\ncoeffathree & coeffbthree & coeffcthree\n\\end{array}\\right)\n\\]\nand\n\\[\nmatrixx=\\left(\\begin{array}{ccc}\nvertexxone & vertexxtwo & vertexxthree \\\\\nvertexyone & vertexytwo & vertexythree \\\\\n1 & 1 & 1\n\\end{array}\\right]\n\\]\n\nThen the area of the triangle is \\( \\frac{1}{2}|\\operatorname{det} matrixx| \\).\nSince the point \\( \\left(vertexxgen, vertexygen\\right) \\) is on the side \\( boundaryj \\) for \\( indexi \\neq j \\), we find\n\\[\nmatrixm matrixx=\\left(\\begin{array}{lll}\ndiagonalone & 0 & 0 \\\\\n0 & diagonaltwo & 0 \\\\\n0 & 0 & diagonalthree\n\\end{array}\\right]\n\\]\nwhere the diagonal entries remain to be determined.\nIf we solve the system of equations\n\\[\nmatrixm\\left(\\begin{array}{l}\nvariablex \\\\\nvariabley \\\\\nvariablez\n\\end{array}\\right]=\\left(\\begin{array}{c}\ndiagonalone \\\\\n0 \\\\\n0\n\\end{array}\\right]\n\\]\nby Cramer's rule, we find\n\\[\nvariablez \\operatorname{det} matrixm=diagonalone\\left|\\begin{array}{ll}\ncoeffatwo & coeffbtwo \\\\\ncoeffathree & coeffbthree\n\\end{array}\\right|\n\\]\n\nBut from (1) we know that the solution is\n\\[\n\\left(\\begin{array}{c}\nvertexxone \\\\\nvertexyone \\\\\n1\n\\end{array}\\right] .\n\\]\n\nHence,\n\\[\ndiagonalone=\\frac{\\operatorname{det} matrixm}{\\left|\\begin{array}{ll}\ncoeffatwo & coeffbtwo \\\\\ncoeffathree & coeffbthree\n\\end{array}\\right| .}\n\\]\n\nSimilarly,\n\\[\n\\begin{array}{l}\nDiagonaltwo=\\frac{\\operatorname{det} matrixm}{\\left|\\begin{array}{ll}\ncoeffathree & coeffbthree \\\\\ncoeffaone & coeffbone\n\\end{array}\\right|} \\\\\nDiagonalthree=\\frac{\\operatorname{det} matrixm}{\\left|\\begin{array}{ll}\ncoeffaone & coeffbone \\\\\ncoeffatwo & coeffbtwo\n\\end{array}\\right| .}\n\\end{array}\n\\]\n\nTherefore, from (1),\n\\[\n(\\operatorname{det} matrixm)(\\operatorname{det} matrixx)=\\frac{(\\operatorname{det} matrixm)^{3}}{\\left|\\begin{array}{ll}\ncoeffatwo & coeffbtwo \\\\\ncoeffathree & coeffbthree\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\ncoeffathree & coeffbthree \\\\\ncoeffaone & coeffbone\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\ncoeffaone & coeffbone \\\\\ncoeffatwo & coeffbtwo\n\\end{array}\\right|}\n\\]\n\nNow \\( \\operatorname{det} matrixm \\neq 0 \\) is precisely the condition that the given lines form a triangle, so\n\\[\n\\begin{array}{c}\n\\text { Area of triangle }=\\frac{1}{2}|\\operatorname{det} matrixx| \\\\\n= \\pm \\frac{(\\operatorname{det} matrixm)^{2}}{2\\left|\\begin{array}{ll}\ncoeffatwo & coeffbtwo \\\\\ncoeffathree & coeffbthree\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\ncoeffathree & coeffbthree \\\\\ncoeffaone & coeffbone\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\ncoeffaone & coeffbone \\\\\ncoeffatwo & coeffbtwo\n\\end{array}\\right|}\n\\end{array}\n\\]\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "thunderbolt",
        "y": "marshland",
        "z": "peppermint",
        "x_1": "sandcastle",
        "y_1": "sunflower",
        "x_2": "lighthouse",
        "y_2": "raincloud",
        "x_3": "breadcrumb",
        "y_3": "paintbrush",
        "x_i": "meadowlark",
        "y_i": "dragonfly",
        "d_1": "sailorman",
        "d_2": "parchment",
        "d_3": "snowflake",
        "i": "i",
        "j": "woodpecker",
        "M": "zephyrwind",
        "X": "stardusts",
        "L_i": "moonlight",
        "L_j": "firebreath",
        "A_1": "ivorytower",
        "A_2": "coppermine",
        "A_3": "silverspear",
        "A_i": "brasshorn",
        "B_1": "stonewheel",
        "B_2": "amberfield",
        "B_3": "crystalweb",
        "B_i": "pewterleaf",
        "C_1": "rubyforest",
        "C_2": "onyxgarden",
        "C_3": "goldcascade",
        "C_i": "quartzwing"
      },
      "question": "8. A triangle is bounded by the lines\n\\[\n\\begin{array}{c}\nivorytower\\, thunderbolt+stonewheel\\, marshland+rubyforest=0, \\quad coppermine\\, thunderbolt+amberfield\\, marshland+onyxgarden=0 \\\\\nsilverspear\\, thunderbolt+crystalweb\\, marshland+goldcascade=0\n\\end{array}\n\\]\n\nShow that the area, disregarding sign, is",
      "solution": "Solution. Let \\( moonlight \\) be the line with equation \\( brasshorn\\, thunderbolt+pewterleaf\\, marshland+quartzwing=0 \\), and let \\( \\left(meadowlark, dragonfly\\right) \\) be the vertex of the triangle opposite \\( moonlight \\). Let\n\\[\nzephyrwind=\\left(\\begin{array}{lll}\nivorytower & stonewheel & rubyforest \\\\\ncoppermine & amberfield & onyxgarden \\\\\nsilverspear & crystalweb & goldcascade\n\\end{array}\\right)\n\\]\nand\n\\[\nstardusts=\\left(\\begin{array}{ccc}\nsandcastle & lighthouse & breadcrumb \\\\\nsunflower & raincloud & paintbrush \\\\\n1 & 1 & 1\n\\end{array}\\right]\n\\]\n\nThen the area of the triangle is \\( \\frac{1}{2}|\\operatorname{det} stardusts| \\).\nSince the point \\( \\left(meadowlark, dragonfly\\right) \\) is on the side \\( firebreath \\) for \\( i \\neq woodpecker \\), we find\n\\[\nzephyrwind\\, stardusts=\\left(\\begin{array}{lll}\nsailorman & 0 & 0 \\\\\n0 & parchment & 0 \\\\\n0 & 0 & snowflake\n\\end{array}\\right]\n\\]\nwhere the diagonal entries remain to be determined.\nIf we solve the system of equations\n\\[\nzephyrwind\\left(\\begin{array}{l}\nthunderbolt \\\\\nmarshland \\\\\npeppermint\n\\end{array}\\right]=\\left(\\begin{array}{c}\nsailorman \\\\\n0 \\\\\n0\n\\end{array}\\right]\n\\]\nby Cramer's rule, we find\n\\[\npeppermint \\operatorname{det} zephyrwind = sailorman \\left|\\begin{array}{ll}\ncoppermine & amberfield \\\\\nsilverspear & crystalweb\n\\end{array}\\right|\n\\]\n\nBut from (1) we know that the solution is\n\\[\n\\left(\\begin{array}{c}\nsandcastle \\\\\nsunflower \\\\\n1\n\\end{array}\\right] .\n\\]\n\nHence,\n\\[\nsailorman = \\frac{\\operatorname{det} zephyrwind}{\\left|\\begin{array}{ll}\ncoppermine & amberfield \\\\\nsilverspear & crystalweb\n\\end{array}\\right| .}\n\\]\n\nSimilarly,\n\\[\n\\begin{array}{l}\nparchment = \\frac{\\operatorname{det} zephyrwind}{\\left|\\begin{array}{ll}\nsilverspear & crystalweb \\\\\nivorytower & stonewheel\n\\end{array}\\right|} \\\\\nsnowflake = \\frac{\\operatorname{det} zephyrwind}{\\left|\\begin{array}{ll}\nivorytower & stonewheel \\\\\ncoppermine & amberfield\n\\end{array}\\right| .}\n\\end{array}\n\\]\n\nTherefore, from (1),\n\\[\n(\\operatorname{det} zephyrwind)(\\operatorname{det} stardusts)=\\frac{(\\operatorname{det} zephyrwind)^{3}}{\\left|\\begin{array}{ll}\ncoppermine & amberfield \\\\\nsilverspear & crystalweb\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nsilverspear & crystalweb \\\\\nivorytower & stonewheel\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nivorytower & stonewheel \\\\\ncoppermine & amberfield\n\\end{array}\\right|}\n\\]\n\nNow \\( \\operatorname{det} zephyrwind \\neq 0 \\) is precisely the condition that the given lines form a triangle, so\n\\[\n\\begin{array}{c}\n\\text { Area of triangle }=\\frac{1}{2}|\\operatorname{det} stardusts| \\\\\n= \\pm \\frac{(\\operatorname{det} zephyrwind)^{2}}{2\\left|\\begin{array}{ll}\ncoppermine & amberfield \\\\\nsilverspear & crystalweb\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nsilverspear & crystalweb \\\\\nivorytower & stonewheel\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nivorytower & stonewheel \\\\\ncoppermine & amberfield\n\\end{array}\\right|}\n\\end{array}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalco",
        "y": "horizontalco",
        "z": "planefixer",
        "x_1": "verticalone",
        "y_1": "horizontalone",
        "x_2": "verticaltwo",
        "y_2": "horizontaltwo",
        "x_3": "verticalthree",
        "y_3": "horizonthree",
        "x_i": "verticalindex",
        "y_i": "horizontalindex",
        "d_1": "maximalone",
        "d_2": "maximaltwo",
        "d_3": "maximalthree",
        "i": "finalindex",
        "j": "startindex",
        "M": "minuteness",
        "X": "knownstuff",
        "L_i": "pointindex",
        "L_j": "pointstart",
        "A_1": "constantone",
        "A_2": "constanttwo",
        "A_3": "constantthree",
        "A_i": "constantindex",
        "B_1": "variableone",
        "B_2": "variabletwo",
        "B_3": "variablethree",
        "B_i": "variableindex",
        "C_1": "changingone",
        "C_2": "changingtwo",
        "C_3": "changingthree",
        "C_i": "changingindex"
      },
      "question": "8. A triangle is bounded by the lines\n\\[\n\\begin{array}{c}\nconstantone\\ verticalco+variableone\\ horizontalco+changingone=0, \\quad constanttwo\\ verticalco+variabletwo\\ horizontalco+changingtwo=0 \\\\\nconstantthree\\ verticalco+variablethree\\ horizontalco+changingthree=0\n\\end{array}\n\\]\n\nShow that the area, disregarding sign, is",
      "solution": "Solution. Let \\( pointindex \\) be the line with equation \\( constantindex verticalco+variableindex horizontalco+changingindex=0 \\), and let \\( \\left(verticalindex, horizontalindex\\right) \\) be the vertex of the triangle opposite \\( pointindex \\). Let\n\\[\nminuteness=\\left(\\begin{array}{lll}\nconstantone & variableone & changingone \\\\\nconstanttwo & variabletwo & changingtwo \\\\\nconstantthree & variablethree & changingthree\n\\end{array}\\right)\n\\]\nand\n\\[\nknownstuff=\\left(\\begin{array}{ccc}\nverticalone & verticaltwo & verticalthree \\\\\nhorizontalone & horizontaltwo & horizonthree \\\\\n1 & 1 & 1\n\\end{array}\\right]\n\\]\n\nThen the area of the triangle is \\( \\frac{1}{2}|\\operatorname{det} knownstuff| \\).\nSince the point \\( \\left(verticalindex, horizontalindex\\right) \\) is on the side \\( pointstart \\) for \\( finalindex \\neq startindex \\), we find\n\\[\nminuteness\\,knownstuff=\\left(\\begin{array}{lll}\nmaximalone & 0 & 0 \\\\\n0 & maximaltwo & 0 \\\\\n0 & 0 & maximalthree\n\\end{array}\\right]\n\\]\nwhere the diagonal entries remain to be determined.\nIf we solve the system of equations\n\\[\nminuteness\\left(\\begin{array}{l}\nverticalco \\\\\nhorizontalco \\\\\nplanefixer\n\\end{array}\\right]=\\left(\\begin{array}{c}\nmaximalone \\\\\n0 \\\\\n0\n\\end{array}\\right]\n\\]\nby Cramer's rule, we find\n\\[\nplanefixer \\,\\operatorname{det} minuteness=maximalone\\left|\\begin{array}{ll}\nconstanttwo & variabletwo \\\\\nconstantthree & variablethree\n\\end{array}\\right|\n\\]\n\nBut from (1) we know that the solution is\n\\[\n\\left(\\begin{array}{c}\nverticalone \\\\\nhorizontalone \\\\\n1\n\\end{array}\\right] .\n\\]\n\nHence,\n\\[\nmaximalone=\\frac{\\operatorname{det} minuteness}{\\left|\\begin{array}{ll}\nconstanttwo & variabletwo \\\\\nconstantthree & variablethree\n\\end{array}\\right| .}\n\\]\n\nSimilarly,\n\\[\n\\begin{array}{l}\nmaximaltwo=\\frac{\\operatorname{det} minuteness}{\\left|\\begin{array}{ll}\nconstantthree & variablethree \\\\\nconstantone & variableone\n\\end{array}\\right|} \\\\\nmaximalthree=\\frac{\\operatorname{det} minuteness}{\\left|\\begin{array}{ll}\nconstantone & variableone \\\\\nconstanttwo & variabletwo\n\\end{array}\\right| .}\n\\end{array}\n\\]\n\nTherefore, from (1),\n\\[\n(\\operatorname{det} minuteness)(\\operatorname{det} knownstuff)=\\frac{(\\operatorname{det} minuteness)^{3}}{\\left|\\begin{array}{ll}\nconstanttwo & variabletwo \\\\\nconstantthree & variablethree\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nconstantthree & variablethree \\\\\nconstantone & variableone\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nconstantone & variableone \\\\\nconstanttwo & variabletwo\n\\end{array}\\right|}\n\\]\n\nNow \\( \\operatorname{det} minuteness \\neq 0 \\) is precisely the condition that the given lines form a triangle, so\n\\[\n\\begin{array}{c}\n\\text { Area of triangle }=\\frac{1}{2}|\\operatorname{det} knownstuff| \\\\\n= \\pm \\frac{(\\operatorname{det} minuteness)^{2}}{2\\left|\\begin{array}{ll}\nconstanttwo & variabletwo \\\\\nconstantthree & variablethree\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nconstantthree & variablethree \\\\\nconstantone & variableone\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nconstantone & variableone \\\\\nconstanttwo & variabletwo\n\\end{array}\\right|}\n\\end{array}\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "bvdqkplo",
        "x_1": "nmpqlsrt",
        "y_1": "wzcxtfgh",
        "x_2": "ldkpaeor",
        "y_2": "smbtrnqa",
        "x_3": "gcrfploi",
        "y_3": "hvdnqwer",
        "x_i": "frtgmzcl",
        "y_i": "pmxldkoi",
        "d_1": "kqnvzmsa",
        "d_2": "tbhplcso",
        "d_3": "rvwankde",
        "i": "i",
        "j": "zmxprlga",
        "M": "plkstnra",
        "X": "wotynqce",
        "L_i": "kdhfjsla",
        "L_j": "uqtzmgwa",
        "A_1": "ovntpsrq",
        "A_2": "camprnld",
        "A_3": "wxslopqr",
        "A_i": "nflbdxra",
        "B_1": "tgadnqwe",
        "B_2": "rqbmklsa",
        "B_3": "ycnwplrt",
        "B_i": "sdfkplma",
        "C_1": "pqrsnvmc",
        "C_2": "axjtrqlw",
        "C_3": "mfptgvoe",
        "C_i": "kzbrplao"
      },
      "question": "8. A triangle is bounded by the lines\n\\[\n\\begin{array}{c}\novntpsrq qzxwvtnp+tgadnqwe hjgrksla+pqrsnvmc=0, \\quad camprnld qzxwvtnp+rqbmklsa hjgrksla+axjtrqlw=0 \\\\\nwxslopqr qzxwvtnp+ycnwplrt hjgrksla+mfptgvoe=0\n\\end{array}\n\\]\n\nShow that the area, disregarding sign, is",
      "solution": "Solution. Let \\( kdhfjsla \\) be the line with equation \\( nflbdxra qzxwvtnp+sdfkplma hjgrksla+kzbrplao=0 \\), and let \\( \\left(frtgmzcl, pmxldkoi\\right) \\) be the vertex of the triangle opposite \\( kdhfjsla \\). Let\n\\[\nplkstnra=\\left(\\begin{array}{lll}\novntpsrq & tgadnqwe & pqrsnvmc \\\\\ncamprnld & rqbmklsa & axjtrqlw \\\\\nwxslopqr & ycnwplrt & mfptgvoe\n\\end{array}\\right)\n\\]\nand\n\\[\nwotynqce=\\left(\\begin{array}{ccc}\nnmpqlsrt & ldkpaeor & gcrfploi \\\\\nwzcxtfgh & smbtrnqa & hvdnqwer \\\\\n1 & 1 & 1\n\\end{array}\\right]\n\\]\n\nThen the area of the triangle is \\( \\frac{1}{2}|\\operatorname{det} wotynqce| \\).\nSince the point \\( \\left(frtgmzcl, pmxldkoi\\right) \\) is on the side \\( uqtzmgwa \\) for \\( i \\neq zmxprlga \\), we find\n\\[\nplkstnra wotynqce=\\left(\\begin{array}{lll}\nkqnvzmsa & 0 & 0 \\\\\n0 & tbhplcso & 0 \\\\\n0 & 0 & rvwankde\n\\end{array}\\right]\n\\]\nwhere the diagonal entries remain to be determined.\nIf we solve the system of equations\n\\[\nplkstnra\\left(\\begin{array}{l}\nqzxwvtnp \\\\\nhjgrksla \\\\\nbvdqkplo\n\\end{array}\\right]=\\left(\\begin{array}{c}\nkqnvzmsa \\\\\n0 \\\\\n0\n\\end{array}\\right]\n\\]\nby Cramer's rule, we find\n\\[\nbvdqkplo \\operatorname{det} plkstnra=kqnvzmsa\\left|\\begin{array}{ll}\ncamprnld & rqbmklsa \\\\\nwxslopqr & ycnwplrt\n\\end{array}\\right|\n\\]\n\nBut from (1) we know that the solution is\n\\[\n\\left(\\begin{array}{c}\nnmpqlsrt \\\\\nwzcxtfgh \\\\\n1\n\\end{array}\\right] .\n\\]\n\nHence,\n\\[\nkqnvzmsa=\\frac{\\operatorname{det} plkstnra}{\\left|\\begin{array}{ll}\ncamprnld & rqbmklsa \\\\\nwxslopqr & ycnwplrt\n\\end{array}\\right| .}\n\\]\n\nSimilarly,\n\\[\n\\begin{array}{l}\ntbhplcso=\\frac{\\operatorname{det} plkstnra}{\\left|\\begin{array}{ll}\nwxslopqr & ycnwplrt \\\\\novntpsrq & tgadnqwe\n\\end{array}\\right|} \\\\\nrvwankde=\\frac{\\operatorname{det} plkstnra}{\\left|\\begin{array}{ll}\novntpsrq & tgadnqwe \\\\\ncamprnld & rqbmklsa\n\\end{array}\\right| .}\n\\end{array}\n\\]\n\nTherefore, from (1),\n\\[\n(\\operatorname{det} plkstnra)(\\operatorname{det} wotynqce)=\\frac{(\\operatorname{det} plkstnra)^{3}}{\\left|\\begin{array}{ll}\ncamprnld & rqbmklsa \\\\\nwxslopqr & ycnwplrt\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nwxslopqr & ycnwplrt \\\\\novntpsrq & tgadnqwe\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\novntpsrq & tgadnqwe \\\\\ncamprnld & rqbmklsa\n\\end{array}\\right|}\n\\]\n\nNow \\( \\operatorname{det} plkstnra \\neq 0 \\) is precisely the condition that the given lines form a triangle, so\n\\[\n\\begin{array}{c}\n\\text { Area of triangle }=\\frac{1}{2}|\\operatorname{det} wotynqce| \\\\\n= \\pm \\frac{(\\operatorname{det} plkstnra)^{2}}{2\\left|\\begin{array}{ll}\ncamprnld & rqbmklsa \\\\\nwxslopqr & ycnwplrt\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\nwxslopqr & ycnwplrt \\\\\novntpsrq & tgadnqwe\n\\end{array}\\right| \\cdot\\left|\\begin{array}{cc}\novntpsrq & tgadnqwe \\\\\ncamprnld & rqbmklsa\n\\end{array}\\right|}\n\\end{array}\n\\]"
    },
    "kernel_variant": {
      "question": "Let n (≥2) be a positive integer.  \nIn ℝⁿ with Cartesian coordinates x=(x₁,…,xₙ) consider the n+1 distinct affine hyperplanes  \n\n  Πᵢ : a_{i1}x₁ + a_{i2}x₂ + … + a_{in}xₙ + bᵢ = 0  (i = 0,1,…,n),\n\nwritten compactly as  \n\n  Πᵢ : Aᵢ·x + bᵢ = 0 with Aᵢ=(a_{i1},…,a_{in})∈ℝⁿ .\n\nAssume  \n\n1. (General position) every choice of n of the hyperplanes meets in a unique point;  \n2. the full collection {Π₀,…,Πₙ} encloses a non–empty, bounded region S – necessarily an n-simplex.\n\nIntroduce the (n+1)×(n+1) matrix  \n\n  T =\n  ⎡ A₀ b₀ ⎤  \n  ⎢ A₁ b₁ ⎥  \n  ⎢ ⋮  ⋮ ⎥ , row i being (a_{i1},…,a_{in},bᵢ),                       (1)\n  ⎣ Aₙ bₙ ⎦  \n\nand let A be the (n+1)×n matrix obtained from T by deleting its last column.  \nFor i=0,…,n denote by\n\n  Δᵢ := det A_{(i)}\n\nthe determinant of the n×n matrix A_{(i)} obtained from A by removing the i-th row.  \n(The general-position hypothesis implies Δᵢ≠0.)\n\nProve that the (unsigned) volume of the simplex S is\n\n       |det T|ⁿ  \n  Vol(S)= ─────────────────────────────────── .                      (2)  \n       n!·|Δ₀Δ₁⋯Δₙ|\n\nFor n=2 the formula yields the classical expression for the area of a triangle through the three bounding lines, and for n=3 it gives the volume of a tetrahedron bounded by four planes.",
      "solution": "Step 1.  Vertices in homogeneous form  \nFor each i let Pᵢ be the vertex of S opposite the face Πᵢ; equivalently,\n\n  Aⱼ·Pᵢ + bⱼ = 0  (j≠i).                                        (3)\n\nWrite Pᵢ as a column vector and collect the (n+1) homogeneous columns\n\n  X := [ P₀ P₁ … Pₙ ; 1 1 … 1 ] ∈ M_{n+1}(ℝ);                     (4)\n\nthe (n+1)×(n+1) matrix X has j-th column (Pⱼ,1)ᵗ.\n\nStep 2.  Diagonalisation  \nBecause Pᵢ lies on Πⱼ for j≠i but, in general, not on Πᵢ, we obtain\n\n  T·X = diag(d₀,d₁,…,dₙ) =: D,                                    (5)\n\nwhere dᵢ := Aᵢ·Pᵢ + bᵢ (non–zero by boundedness of S).\n\nStep 3.  Computing dᵢ with Cramer’s rule  \nFix i.  Equation (5) means\n\n  T (Pᵢ,1)ᵗ = dᵢ eᵢ,                                              (6)\n\neᵢ being the i-th standard basis vector of ℝ^{n+1}.  \nApply Cramer’s rule to the last (n+1) unknown of the linear system (6): replace\nthe last column of T (not the i-th!) by the right–hand side vector dᵢeᵢ.  \nExpanding this determinant along that last column gives  \n\n  dᵢ (–1)^{i+n+1} det A_{(i)} = det T.                           (7)\n\nHence  \n\n  dᵢ = (–1)^{i+n+1} det T / Δᵢ.                                  (8)\n\nTaking absolute values – which is all that will survive in the final formula – we have  \n\n  |dᵢ| = |det T| / |Δᵢ|.                                          (9)\n\nStep 4.  Determinant of X  \nFrom (5) we compute\n\n  |det T|·|det X| = |det D| = ∏_{i=0}^{n}|dᵢ|\n           = (|det T|)^{\\,n+1} / (|Δ₀Δ₁⋯Δₙ|).                (10)\n\nBecause det T ≠ 0 (again by general position), divide both sides of (10) by |det T|:\n\n  |det X| = (|det T|)^{\\,n} / |Δ₀Δ₁⋯Δₙ|.                          (11)\n\nStep 5.  Volume of an n-simplex  \nThe oriented n-volume of a simplex whose homogeneous vertex matrix is X equals det X / n!; discarding sign we obtain\n\n  Vol(S) = |det X| / n!\n       = |det T|ⁿ / [ n! · |Δ₀Δ₁⋯Δₙ| ],                        (12)\n\nwhich is exactly formula (2). □",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.373167",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher Dimension: The problem jumps from 2-dimensional triangles to n-dimensional simplices bounded by n+1 hyperplanes, forcing competitors to work with (n+1)×(n+1) determinants and n-fold Cramer arguments.\n\n2. Additional Structures:  The formulation uses homogeneous coordinates, diagonal matrix factorizations, and the general position hypothesis, all absent from the original.\n\n3. Deeper Theory:  Competitors must generalise the “area = ½|det X|” fact to “volume = |det X|/n!”, understand how removing one row of a rectangular matrix yields square “opposite minors”, and keep track of orientation signs through formula (9).\n\n4. Multi-step Proof:  The solution demands  \n (a) identification of the vertex set via intersections of hyperplanes;  \n (b) construction of the matrix relation T·X = D;  \n (c) a Cramer-rule computation of each diagonal entry;  \n (d) a determinant identity linking T, X, and the minors;  \n (e) the volume deduction.  \nEach step is itself non-trivial in n variables.\n\n5. Non-pattern-matching:  The contestant cannot rely on a memorised 2-D determinant trick; instead they must discover and manage a web of (n+1) matrices, n+1 minors, and an exponent n on det T, all while keeping the algebra consistent across arbitrary dimension.\n\nConsequently the enhanced variant is substantially more intricate and technically demanding than both the original triangle problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let n (≥2) be a positive integer.  \nIn ℝⁿ with Cartesian coordinates x=(x₁,…,xₙ) consider the n+1 distinct affine hyperplanes  \n\n  Πᵢ : a_{i1}x₁ + a_{i2}x₂ + … + a_{in}xₙ + bᵢ = 0  (i = 0,1,…,n),\n\nwritten compactly as  \n\n  Πᵢ : Aᵢ·x + bᵢ = 0 with Aᵢ=(a_{i1},…,a_{in})∈ℝⁿ .\n\nAssume  \n\n1. (General position) every choice of n of the hyperplanes meets in a unique point;  \n2. the full collection {Π₀,…,Πₙ} encloses a non–empty, bounded region S – necessarily an n-simplex.\n\nIntroduce the (n+1)×(n+1) matrix  \n\n  T =\n  ⎡ A₀ b₀ ⎤  \n  ⎢ A₁ b₁ ⎥  \n  ⎢ ⋮  ⋮ ⎥ , row i being (a_{i1},…,a_{in},bᵢ),                       (1)\n  ⎣ Aₙ bₙ ⎦  \n\nand let A be the (n+1)×n matrix obtained from T by deleting its last column.  \nFor i=0,…,n denote by\n\n  Δᵢ := det A_{(i)}\n\nthe determinant of the n×n matrix A_{(i)} obtained from A by removing the i-th row.  \n(The general-position hypothesis implies Δᵢ≠0.)\n\nProve that the (unsigned) volume of the simplex S is\n\n       |det T|ⁿ  \n  Vol(S)= ─────────────────────────────────── .                      (2)  \n       n!·|Δ₀Δ₁⋯Δₙ|\n\nFor n=2 the formula yields the classical expression for the area of a triangle through the three bounding lines, and for n=3 it gives the volume of a tetrahedron bounded by four planes.",
      "solution": "Step 1.  Vertices in homogeneous form  \nFor each i let Pᵢ be the vertex of S opposite the face Πᵢ; equivalently,\n\n  Aⱼ·Pᵢ + bⱼ = 0  (j≠i).                                        (3)\n\nWrite Pᵢ as a column vector and collect the (n+1) homogeneous columns\n\n  X := [ P₀ P₁ … Pₙ ; 1 1 … 1 ] ∈ M_{n+1}(ℝ);                     (4)\n\nthe (n+1)×(n+1) matrix X has j-th column (Pⱼ,1)ᵗ.\n\nStep 2.  Diagonalisation  \nBecause Pᵢ lies on Πⱼ for j≠i but, in general, not on Πᵢ, we obtain\n\n  T·X = diag(d₀,d₁,…,dₙ) =: D,                                    (5)\n\nwhere dᵢ := Aᵢ·Pᵢ + bᵢ (non–zero by boundedness of S).\n\nStep 3.  Computing dᵢ with Cramer’s rule  \nFix i.  Equation (5) means\n\n  T (Pᵢ,1)ᵗ = dᵢ eᵢ,                                              (6)\n\neᵢ being the i-th standard basis vector of ℝ^{n+1}.  \nApply Cramer’s rule to the last (n+1) unknown of the linear system (6): replace\nthe last column of T (not the i-th!) by the right–hand side vector dᵢeᵢ.  \nExpanding this determinant along that last column gives  \n\n  dᵢ (–1)^{i+n+1} det A_{(i)} = det T.                           (7)\n\nHence  \n\n  dᵢ = (–1)^{i+n+1} det T / Δᵢ.                                  (8)\n\nTaking absolute values – which is all that will survive in the final formula – we have  \n\n  |dᵢ| = |det T| / |Δᵢ|.                                          (9)\n\nStep 4.  Determinant of X  \nFrom (5) we compute\n\n  |det T|·|det X| = |det D| = ∏_{i=0}^{n}|dᵢ|\n           = (|det T|)^{\\,n+1} / (|Δ₀Δ₁⋯Δₙ|).                (10)\n\nBecause det T ≠ 0 (again by general position), divide both sides of (10) by |det T|:\n\n  |det X| = (|det T|)^{\\,n} / |Δ₀Δ₁⋯Δₙ|.                          (11)\n\nStep 5.  Volume of an n-simplex  \nThe oriented n-volume of a simplex whose homogeneous vertex matrix is X equals det X / n!; discarding sign we obtain\n\n  Vol(S) = |det X| / n!\n       = |det T|ⁿ / [ n! · |Δ₀Δ₁⋯Δₙ| ],                        (12)\n\nwhich is exactly formula (2). □",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.322256",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher Dimension: The problem jumps from 2-dimensional triangles to n-dimensional simplices bounded by n+1 hyperplanes, forcing competitors to work with (n+1)×(n+1) determinants and n-fold Cramer arguments.\n\n2. Additional Structures:  The formulation uses homogeneous coordinates, diagonal matrix factorizations, and the general position hypothesis, all absent from the original.\n\n3. Deeper Theory:  Competitors must generalise the “area = ½|det X|” fact to “volume = |det X|/n!”, understand how removing one row of a rectangular matrix yields square “opposite minors”, and keep track of orientation signs through formula (9).\n\n4. Multi-step Proof:  The solution demands  \n (a) identification of the vertex set via intersections of hyperplanes;  \n (b) construction of the matrix relation T·X = D;  \n (c) a Cramer-rule computation of each diagonal entry;  \n (d) a determinant identity linking T, X, and the minors;  \n (e) the volume deduction.  \nEach step is itself non-trivial in n variables.\n\n5. Non-pattern-matching:  The contestant cannot rely on a memorised 2-D determinant trick; instead they must discover and manage a web of (n+1) matrices, n+1 minors, and an exponent n on det T, all while keeping the algebra consistent across arbitrary dimension.\n\nConsequently the enhanced variant is substantially more intricate and technically demanding than both the original triangle problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}