{
  "index": "1962-A-3",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "3. In a triangle \\( A B C \\) in the Euclidean plane, let \\( A^{\\prime} \\) be a point on the segment from \\( B \\) to \\( C, B^{\\prime} \\) a point on the segment from \\( C \\) to \\( A \\), and \\( C^{\\prime} \\) a point on the segment from \\( A \\) to \\( B \\) such that\n\\[\n\\frac{A B^{\\prime}}{B^{\\prime} C}=\\frac{B C^{\\prime}}{C^{\\prime} A}=\\frac{C A^{\\prime}}{A^{\\prime} B}=k\n\\]\nwhere \\( k \\) is a positive constant. Let \\( \\Delta \\) be the triangle formed by parts of the segments obtained by joining \\( A \\) and \\( A^{\\prime}, B \\) and \\( B^{\\prime} \\), and \\( C \\) and \\( C^{\\prime} \\). Prove that the areas of the triangles \\( \\Delta \\) and \\( A B C \\) are in the ratio.\n\\[\n\\frac{(k-1)^{2}}{k^{2}+k+1}\n\\]",
  "solution": "First Solution. Use barycentric coordinates for the triangle \\( A B C \\).\n\\[\n\\begin{aligned}\nA & =\\langle 1,0,0\\rangle \\\\\nB & =\\langle 0,1,0\\rangle \\\\\nC & =\\langle 0,0,1\\rangle \\\\\nA^{\\prime} & =\\frac{1}{k+1}\\langle 0, k, 1\\rangle \\\\\nB^{\\prime} & =\\frac{1}{k+1}\\langle 1,0, k\\rangle \\\\\nC^{\\prime} & =\\frac{1}{k+1}\\langle k .1,0\\rangle .\n\\end{aligned}\n\\]\n\nThe equations of \\( B B^{\\prime} \\) and \\( C C^{\\prime} \\) are \\( ==k x \\) and \\( x=k y \\), respectively. If their intersection \\( P \\) is given by \\( \\langle\\lambda, \\mu, \\nu\\rangle \\), then \\( \\lambda=k \\mu, \\nu=k \\lambda \\), and \\( \\lambda+\\mu+\\nu=1 \\). So\n\\[\nP=\\frac{1}{k^{2}+k+1}\\left\\langle k, 1, k^{2}\\right\\rangle .\n\\]\n\nBy symmetry\n\\[\n\\begin{array}{l}\nQ=\\frac{1}{k^{2}+k+1}\\left\\langle k^{2}, k, 1\\right\\rangle \\\\\nR=\\frac{1}{k^{2}+k+1}\\left\\langle 1, k^{2}, k\\right\\rangle .\n\\end{array}\n\\]\n\nThen\n\\[\n\\begin{aligned}\n\\frac{\\operatorname{area} P Q R}{\\operatorname{area} A B C} & =\\frac{1}{\\left(k^{2}+k+1\\right)^{3}}\\left|\\begin{array}{lll}\nk & 1 & k^{2} \\\\\nk^{2} & k & 1 \\\\\n1 & k^{2} & k\n\\end{array}\\right| \\\\\n& =\\frac{k^{6}-2 k^{3}+1}{\\left(k^{2}+k+1\\right)^{3}}=\\frac{\\left(k^{3}-1\\right)^{2}}{\\left(k^{2}+k+1\\right)^{3}}=\\frac{(k-1)^{2}}{k^{2}+k+1} .\n\\end{aligned}\n\\]\n\nSecond Solution. A synthetic solution may be of interest.\n\\[\n\\begin{array}{l} \n\\frac{\\Delta A Q C}{\\triangle A Q C^{\\prime}} \\\\\n=\\frac{Q C}{Q C^{\\prime}}=\\frac{\\Delta B Q C}{\\Delta B Q C^{\\prime}} \\\\\n\\therefore \\quad \\frac{\\Delta B Q C}{\\Delta A Q C}=\\frac{\\Delta B Q C^{\\prime}}{\\Delta A Q C^{\\prime}}=\\frac{B C^{\\prime}}{A C^{\\prime}}=k\n\\end{array}\n\\]\n\nSimilarly\n\\[\n\\begin{aligned}\n& \\frac{\\Delta A Q C}{\\Delta A^{\\prime} Q C}=\\frac{A Q}{A^{\\prime} Q}=\\frac{\\Delta B A Q}{\\Delta B A^{\\prime} Q} \\\\\n\\therefore \\quad & \\frac{\\Delta A Q B}{\\Delta A Q C}=\\frac{\\Delta A^{\\prime} Q B}{\\Delta A^{\\prime} Q C}=\\frac{A^{\\prime} B}{A^{\\prime} C}=\\frac{1}{k} .\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\frac{\\Delta A Q B+\\Delta B Q C+\\Delta A Q C}{\\Delta A Q C}=\\frac{\\Delta A B C}{\\Delta A Q C}=1+k+\\frac{1}{k}=\\frac{1+k+k^{2}}{k} .\n\\]\n\nBy cyclic symmetry we have\n\\[\n\\frac{\\Delta A Q C}{\\Delta A B C}=\\frac{\\Delta B R A}{\\Delta A B C}=\\frac{\\Delta C P B}{\\Delta A B C}=\\frac{k}{1+k+k^{2}} .\n\\]\n\nNow \\( \\triangle P Q R=\\triangle A B C-\\triangle A Q C-\\triangle B R A-\\triangle C P B \\), so\n\\[\n\\frac{\\Delta P Q R}{\\Delta A B C}=1-3 \\frac{k}{1+k+k^{2}}=\\frac{(1-k)^{2}}{1+k+k^{2}}\n\\]",
  "vars": [
    "A",
    "B",
    "C",
    "P",
    "Q",
    "R",
    "x",
    "y",
    "\\\\lambda",
    "\\\\mu",
    "\\\\nu",
    "\\\\Delta"
  ],
  "params": [
    "k"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "vertexa",
        "B": "vertexb",
        "C": "vertexc",
        "P": "pointp",
        "Q": "pointq",
        "R": "pointr",
        "x": "coordx",
        "y": "coordy",
        "\\lambda": "lambdaa",
        "\\mu": "mucoeff",
        "\\nu": "nucalc",
        "\\Delta": "areaoperator",
        "k": "ratiofactor"
      },
      "question": "3. In a triangle \\( vertexa vertexb vertexc \\) in the Euclidean plane, let \\( vertexa^{\\prime} \\) be a point on the segment from \\( vertexb \\) to \\( vertexc, vertexb^{\\prime} \\) a point on the segment from \\( vertexc \\) to \\( vertexa \\), and \\( vertexc^{\\prime} \\) a point on the segment from \\( vertexa \\) to \\( vertexb \\) such that\n\\[\n\\frac{vertexa vertexb^{\\prime}}{vertexb^{\\prime} vertexc}=\\frac{vertexb vertexc^{\\prime}}{vertexc^{\\prime} vertexa}=\\frac{vertexc vertexa^{\\prime}}{vertexa^{\\prime} vertexb}=ratiofactor\n\\]\nwhere \\( ratiofactor \\) is a positive constant. Let \\( areaoperator \\) be the triangle formed by parts of the segments obtained by joining \\( vertexa \\) and \\( vertexa^{\\prime}, vertexb \\) and \\( vertexb^{\\prime} \\), and \\( vertexc \\) and \\( vertexc^{\\prime} \\). Prove that the areas of the triangles \\( areaoperator \\) and \\( vertexa vertexb vertexc \\) are in the ratio.\n\\[\n\\frac{(ratiofactor-1)^{2}}{ratiofactor^{2}+ratiofactor+1}\n\\]",
      "solution": "First Solution. Use barycentric coordinates for the triangle \\( vertexa vertexb vertexc \\).\n\\[\n\\begin{aligned}\nvertexa & =\\langle 1,0,0\\rangle \\\\\nvertexb & =\\langle 0,1,0\\rangle \\\\\nvertexc & =\\langle 0,0,1\\rangle \\\\\nvertexa^{\\prime} & =\\frac{1}{ratiofactor+1}\\langle 0, ratiofactor, 1\\rangle \\\\\nvertexb^{\\prime} & =\\frac{1}{ratiofactor+1}\\langle 1,0, ratiofactor\\rangle \\\\\nvertexc^{\\prime} & =\\frac{1}{ratiofactor+1}\\langle ratiofactor .1,0\\rangle .\n\\end{aligned}\n\\]\n\nThe equations of \\( vertexb vertexb^{\\prime} \\) and \\( vertexc vertexc^{\\prime} \\) are \\( ==ratiofactor coordx \\) and \\( coordx=ratiofactor coordy \\), respectively. If their intersection \\( pointp \\) is given by \\( \\langle lambdaa, mucoeff, nucalc\\rangle \\), then \\( lambdaa=ratiofactor mucoeff, nucalc=ratiofactor lambdaa \\), and \\( lambdaa+mucoeff+nucalc=1 \\). So\n\\[\npointp=\\frac{1}{ratiofactor^{2}+ratiofactor+1}\\left\\langle ratiofactor, 1, ratiofactor^{2}\\right\\rangle .\n\\]\n\nBy symmetry\n\\[\n\\begin{array}{l}\npointq=\\frac{1}{ratiofactor^{2}+ratiofactor+1}\\left\\langle ratiofactor^{2}, ratiofactor, 1\\right\\rangle \\\\\npointr=\\frac{1}{ratiofactor^{2}+ratiofactor+1}\\left\\langle 1, ratiofactor^{2}, ratiofactor\\right\\rangle .\n\\end{array}\n\\]\n\nThen\n\\[\n\\begin{aligned}\n\\frac{\\operatorname{area} pointp pointq pointr}{\\operatorname{area} vertexa vertexb vertexc} & =\\frac{1}{\\left(ratiofactor^{2}+ratiofactor+1\\right)^{3}}\\left|\\begin{array}{lll}\nratiofactor & 1 & ratiofactor^{2} \\\\\nratiofactor^{2} & ratiofactor & 1 \\\\\n1 & ratiofactor^{2} & ratiofactor\n\\end{array}\\right| \\\\\n& =\\frac{ratiofactor^{6}-2 ratiofactor^{3}+1}{\\left(ratiofactor^{2}+ratiofactor+1\\right)^{3}}=\\frac{\\left(ratiofactor^{3}-1\\right)^{2}}{\\left(ratiofactor^{2}+ratiofactor+1\\right)^{3}}=\\frac{(ratiofactor-1)^{2}}{ratiofactor^{2}+ratiofactor+1} .\n\\end{aligned}\n\\]\n\nSecond Solution. A synthetic solution may be of interest.\n\\[\n\\begin{array}{l} \n\\frac{areaoperator vertexa pointq vertexc}{\\triangle vertexa pointq vertexc^{\\prime}} \\\\\n=\\frac{pointq vertexc}{pointq vertexc^{\\prime}}=\\frac{areaoperator vertexb pointq vertexc}{areaoperator vertexb pointq vertexc^{\\prime}} \\\\\n\\therefore \\quad \\frac{areaoperator vertexb pointq vertexc}{areaoperator vertexa pointq vertexc}=\\frac{areaoperator vertexb pointq vertexc^{\\prime}}{areaoperator vertexa pointq vertexc^{\\prime}}=\\frac{vertexb vertexc^{\\prime}}{vertexa vertexc^{\\prime}}=ratiofactor\n\\end{array}\n\\]\n\nSimilarly\n\\[\n\\begin{aligned}\n& \\frac{areaoperator vertexa pointq vertexc}{areaoperator vertexa^{\\prime} pointq vertexc}=\\frac{vertexa pointq}{vertexa^{\\prime} pointq}=\\frac{areaoperator vertexb vertexa pointq}{areaoperator vertexb vertexa^{\\prime} pointq} \\\\\n\\therefore \\quad & \\frac{areaoperator vertexa pointq vertexb}{areaoperator vertexa pointq vertexc}=\\frac{areaoperator vertexa^{\\prime} pointq vertexb}{areaoperator vertexa^{\\prime} pointq vertexc}=\\frac{vertexa^{\\prime} vertexb}{vertexa^{\\prime} vertexc}=\\frac{1}{ratiofactor} .\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\frac{areaoperator vertexa pointq vertexb+areaoperator vertexb pointq vertexc+areaoperator vertexa pointq vertexc}{areaoperator vertexa pointq vertexc}=\\frac{areaoperator vertexa vertexb vertexc}{areaoperator vertexa pointq vertexc}=1+ratiofactor+\\frac{1}{ratiofactor}=\\frac{1+ratiofactor+ratiofactor^{2}}{ratiofactor} .\n\\]\n\nBy cyclic symmetry we have\n\\[\n\\frac{areaoperator vertexa pointq vertexc}{areaoperator vertexa vertexb vertexc}=\\frac{areaoperator vertexb pointr vertexa}{areaoperator vertexa vertexb vertexc}=\\frac{areaoperator vertexc pointp vertexb}{areaoperator vertexa vertexb vertexc}=\\frac{ratiofactor}{1+ratiofactor+ratiofactor^{2}} .\n\\]\n\nNow \\( \\triangle pointp pointq pointr=\\triangle vertexa vertexb vertexc-\\triangle vertexa pointq vertexc-\\triangle vertexb pointr vertexa-\\triangle vertexc pointp vertexb \\), so\n\\[\n\\frac{areaoperator pointp pointq pointr}{areaoperator vertexa vertexb vertexc}=1-3 \\frac{ratiofactor}{1+ratiofactor+ratiofactor^{2}}=\\frac{(1-ratiofactor)^{2}}{1+ratiofactor+ratiofactor^{2}}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "marmalade",
        "B": "pendulum",
        "C": "topazstone",
        "P": "butterfly",
        "Q": "telescope",
        "R": "chandelier",
        "x": "windchime",
        "y": "scytheblade",
        "\\lambda": "sunflower",
        "\\mu": "raincloud",
        "\\nu": "brainstorm",
        "\\Delta": "overlook",
        "k": "radishseed"
      },
      "question": "3. In a triangle \\( marmalade pendulum topazstone \\) in the Euclidean plane, let \\( marmalade^{\\prime} \\) be a point on the segment from \\( pendulum \\) to \\( topazstone, pendulum^{\\prime} \\) a point on the segment from \\( topazstone \\) to \\( marmalade \\), and \\( topazstone^{\\prime} \\) a point on the segment from \\( marmalade \\) to \\( pendulum \\) such that\n\\[\n\\frac{marmalade pendulum^{\\prime}}{pendulum^{\\prime} topazstone}=\\frac{pendulum topazstone^{\\prime}}{topazstone^{\\prime} marmalade}=\\frac{topazstone marmalade^{\\prime}}{marmalade^{\\prime} pendulum}=radishseed\n\\]\nwhere \\( radishseed \\) is a positive constant. Let \\( overlook \\) be the triangle formed by parts of the segments obtained by joining \\( marmalade \\) and \\( marmalade^{\\prime}, pendulum \\) and \\( pendulum^{\\prime} \\), and \\( topazstone \\) and \\( topazstone^{\\prime} \\). Prove that the areas of the triangles \\( overlook \\) and \\( marmalade pendulum topazstone \\) are in the ratio.\n\\[\n\\frac{(radishseed-1)^{2}}{radishseed^{2}+radishseed+1}\n\\]",
      "solution": "First Solution. Use barycentric coordinates for the triangle \\( marmalade pendulum topazstone \\).\n\\[\n\\begin{aligned}\nmarmalade & =\\langle 1,0,0\\rangle \\\\\npendulum & =\\langle 0,1,0\\rangle \\\\\ntopazstone & =\\langle 0,0,1\\rangle \\\\\nmarmalade^{\\prime} & =\\frac{1}{radishseed+1}\\langle 0, radishseed, 1\\rangle \\\\\npendulum^{\\prime} & =\\frac{1}{radishseed+1}\\langle 1,0, radishseed\\rangle \\\\\ntopazstone^{\\prime} & =\\frac{1}{radishseed+1}\\langle radishseed .1,0\\rangle .\n\\end{aligned}\n\\]\n\nThe equations of \\( pendulum pendulum^{\\prime} \\) and \\( topazstone topazstone^{\\prime} \\) are \\( ==radishseed windchime \\) and \\( windchime=radishseed scytheblade \\), respectively. If their intersection \\( butterfly \\) is given by \\( \\langle sunflower, raincloud, brainstorm\\rangle \\), then \\( sunflower=radishseed raincloud, brainstorm=radishseed sunflower \\), and \\( sunflower+raincloud+brainstorm=1 \\). So\n\\[\nbutterfly=\\frac{1}{radishseed^{2}+radishseed+1}\\left\\langle radishseed, 1, radishseed^{2}\\right\\rangle .\n\\]\n\nBy symmetry\n\\[\n\\begin{array}{l}\ntelescope=\\frac{1}{radishseed^{2}+radishseed+1}\\left\\langle radishseed^{2}, radishseed, 1\\right\\rangle \\\\\nchandelier=\\frac{1}{radishseed^{2}+radishseed+1}\\left\\langle 1, radishseed^{2}, radishseed\\right\\rangle .\n\\end{array}\n\\]\n\nThen\n\\[\n\\begin{aligned}\n\\frac{\\operatorname{area} butterfly\\ telescope\\ chandelier}{\\operatorname{area} marmalade\\ pendulum\\ topazstone} & =\\frac{1}{\\left(radishseed^{2}+radishseed+1\\right)^{3}}\\left|\\begin{array}{lll}\nradishseed & 1 & radishseed^{2} \\\\\nradishseed^{2} & radishseed & 1 \\\\\n1 & radishseed^{2} & radishseed\n\\end{array}\\right| \\\\\n& =\\frac{radishseed^{6}-2 radishseed^{3}+1}{\\left(radishseed^{2}+radishseed+1\\right)^{3}}=\\frac{\\left(radishseed^{3}-1\\right)^{2}}{\\left(radishseed^{2}+radishseed+1\\right)^{3}}=\\frac{(radishseed-1)^{2}}{radishseed^{2}+radishseed+1} .\n\\end{aligned}\n\\]\n\nSecond Solution. A synthetic solution may be of interest.\n\\[\n\\begin{array}{l} \n\\frac{overlook marmalade telescope topazstone}{\\triangle marmalade telescope topazstone^{\\prime}} \\\\\n=\\frac{telescope topazstone}{telescope topazstone^{\\prime}}=\\frac{overlook pendulum telescope topazstone}{overlook pendulum telescope topazstone^{\\prime}} \\\\\n\\therefore \\quad \\frac{overlook pendulum telescope topazstone}{overlook marmalade telescope topazstone}=\\frac{overlook pendulum telescope topazstone^{\\prime}}{overlook marmalade telescope topazstone^{\\prime}}=\\frac{pendulum topazstone^{\\prime}}{marmalade topazstone^{\\prime}}=radishseed\n\\end{array}\n\\]\n\nSimilarly\n\\[\n\\begin{aligned}\n& \\frac{overlook marmalade telescope topazstone}{overlook marmalade^{\\prime} telescope topazstone}=\\frac{marmalade telescope}{marmalade^{\\prime} telescope}=\\frac{overlook pendulum marmalade telescope}{overlook pendulum marmalade^{\\prime} telescope} \\\\\n\\therefore \\quad & \\frac{overlook marmalade telescope pendulum}{overlook marmalade telescope topazstone}=\\frac{overlook marmalade^{\\prime} telescope pendulum}{overlook marmalade^{\\prime} telescope topazstone}=\\frac{marmalade^{\\prime} pendulum}{marmalade^{\\prime} topazstone}=\\frac{1}{radishseed} .\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\frac{overlook marmalade telescope pendulum+overlook pendulum telescope topazstone+overlook marmalade telescope topazstone}{overlook marmalade telescope topazstone}=\\frac{overlook marmalade pendulum topazstone}{overlook marmalade telescope topazstone}=1+radishseed+\\frac{1}{radishseed}=\\frac{1+radishseed+radishseed^{2}}{radishseed} .\n\\]\n\nBy cyclic symmetry we have\n\\[\n\\frac{overlook marmalade telescope topazstone}{overlook marmalade pendulum topazstone}=\\frac{overlook pendulum chandelier marmalade}{overlook marmalade pendulum topazstone}=\\frac{overlook topazstone butterfly pendulum}{overlook marmalade pendulum topazstone}=\\frac{radishseed}{1+radishseed+radishseed^{2}} .\n\\]\n\nNow \\( \\triangle butterfly\\ telescope\\ chandelier=\\triangle marmalade\\ pendulum\\ topazstone-\\triangle marmalade\\ telescope\\ topazstone-\\triangle pendulum\\ chandelier\\ marmalade-\\triangle topazstone\\ butterfly\\ pendulum \\), so\n\\[\n\\frac{overlook butterfly telescope chandelier}{overlook marmalade pendulum topazstone}=1-3 \\frac{radishseed}{1+radishseed+radishseed^{2}}=\\frac{(1-radishseed)^{2}}{1+radishseed+radishseed^{2}}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "voidpoint",
        "B": "nullshape",
        "C": "blankzone",
        "P": "farregion",
        "Q": "awayplace",
        "R": "noncorner",
        "x": "wideaxis",
        "y": "broadline",
        "\\lambda": "staticval",
        "\\mu": "fixednum",
        "\\nu": "rigidpar",
        "\\Delta": "roundfigure",
        "k": "zeroconst"
      },
      "question": "3. In a triangle \\( voidpoint nullshape blankzone \\) in the Euclidean plane, let \\( voidpoint^{\\prime} \\) be a point on the segment from \\( nullshape \\) to \\( blankzone, nullshape^{\\prime} \\) a point on the segment from \\( blankzone \\) to \\( voidpoint \\), and \\( blankzone^{\\prime} \\) a point on the segment from \\( voidpoint \\) to \\( nullshape \\) such that\n\\[\n\\frac{voidpoint nullshape^{\\prime}}{nullshape^{\\prime} blankzone}=\\frac{nullshape blankzone^{\\prime}}{blankzone^{\\prime} voidpoint}=\\frac{blankzone voidpoint^{\\prime}}{voidpoint^{\\prime} nullshape}=zeroconst\n\\]\nwhere \\( zeroconst \\) is a positive constant. Let \\( roundfigure \\) be the triangle formed by parts of the segments obtained by joining \\( voidpoint \\) and \\( voidpoint^{\\prime}, nullshape \\) and \\( nullshape^{\\prime} \\), and \\( blankzone \\) and \\( blankzone^{\\prime} \\). Prove that the areas of the triangles \\( roundfigure \\) and \\( voidpoint nullshape blankzone \\) are in the ratio.\n\\[\n\\frac{(zeroconst-1)^{2}}{zeroconst^{2}+zeroconst+1}\n\\]",
      "solution": "First Solution. Use barycentric coordinates for the triangle \\( voidpoint nullshape blankzone \\).\n\\[\n\\begin{aligned}\nvoidpoint & =\\langle 1,0,0\\rangle \\\\\nnullshape & =\\langle 0,1,0\\rangle \\\\\nblankzone & =\\langle 0,0,1\\rangle \\\\\nvoidpoint^{\\prime} & =\\frac{1}{zeroconst+1}\\langle 0, zeroconst, 1\\rangle \\\\\nnullshape^{\\prime} & =\\frac{1}{zeroconst+1}\\langle 1,0, zeroconst\\rangle \\\\\nblankzone^{\\prime} & =\\frac{1}{zeroconst+1}\\langle zeroconst .1,0\\rangle .\n\\end{aligned}\n\\]\n\nThe equations of \\( nullshape nullshape^{\\prime} \\) and \\( blankzone blankzone^{\\prime} \\) are \\( broadline=zeroconst wideaxis \\) and \\( wideaxis=zeroconst broadline \\), respectively. If their intersection \\( farregion \\) is given by \\( \\langle staticval, fixednum, rigidpar\\rangle \\), then \\( staticval=zeroconst fixednum, rigidpar=zeroconst staticval \\), and \\( staticval+fixednum+rigidpar=1 \\). So\n\\[\nfarregion=\\frac{1}{zeroconst^{2}+zeroconst+1}\\left\\langle zeroconst, 1, zeroconst^{2}\\right\\rangle .\n\\]\n\nBy symmetry\n\\[\n\\begin{array}{l}\nawayplace=\\frac{1}{zeroconst^{2}+zeroconst+1}\\left\\langle zeroconst^{2}, zeroconst, 1\\right\\rangle \\\\\nnoncorner=\\frac{1}{zeroconst^{2}+zeroconst+1}\\left\\langle 1, zeroconst^{2}, zeroconst\\right\\rangle .\n\\end{array}\n\\]\n\nThen\n\\[\n\\begin{aligned}\n\\frac{\\operatorname{area} farregion awayplace noncorner}{\\operatorname{area} voidpoint nullshape blankzone} & =\\frac{1}{\\left(zeroconst^{2}+zeroconst+1\\right)^{3}}\\left|\\begin{array}{lll}\nzeroconst & 1 & zeroconst^{2} \\\\\nzeroconst^{2} & zeroconst & 1 \\\\\n1 & zeroconst^{2} & zeroconst\n\\end{array}\\right| \\\\\n& =\\frac{zeroconst^{6}-2 zeroconst^{3}+1}{\\left(zeroconst^{2}+zeroconst+1\\right)^{3}}=\\frac{\\left(zeroconst^{3}-1\\right)^{2}}{\\left(zeroconst^{2}+zeroconst+1\\right)^{3}}=\\frac{(zeroconst-1)^{2}}{zeroconst^{2}+zeroconst+1} .\n\\end{aligned}\n\\]\n\nSecond Solution. A synthetic solution may be of interest.\n\\[\n\\begin{array}{l} \n\\frac{roundfigure voidpoint awayplace blankzone}{\\triangle voidpoint awayplace blankzone^{\\prime}} \\\\\n=\\frac{awayplace blankzone}{awayplace blankzone^{\\prime}}=\\frac{roundfigure nullshape awayplace blankzone}{roundfigure nullshape awayplace blankzone^{\\prime}} \\\\\n\\therefore \\quad \\frac{roundfigure nullshape awayplace blankzone}{roundfigure voidpoint awayplace blankzone}=\\frac{roundfigure nullshape awayplace blankzone^{\\prime}}{roundfigure voidpoint awayplace blankzone^{\\prime}}=\\frac{nullshape blankzone^{\\prime}}{voidpoint blankzone^{\\prime}}=zeroconst\n\\end{array}\n\\]\n\nSimilarly\n\\[\n\\begin{aligned}\n& \\frac{roundfigure voidpoint awayplace blankzone}{roundfigure voidpoint^{\\prime} awayplace blankzone}=\\frac{voidpoint awayplace}{voidpoint^{\\prime} awayplace}=\\frac{roundfigure nullshape voidpoint awayplace}{roundfigure nullshape voidpoint^{\\prime} awayplace} \\\\\n\\therefore \\quad & \\frac{roundfigure voidpoint awayplace nullshape}{roundfigure voidpoint awayplace blankzone}=\\frac{roundfigure voidpoint^{\\prime} awayplace nullshape}{roundfigure voidpoint^{\\prime} awayplace blankzone}=\\frac{voidpoint^{\\prime} nullshape}{voidpoint^{\\prime} blankzone}=\\frac{1}{zeroconst} .\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\frac{roundfigure voidpoint awayplace nullshape+roundfigure nullshape awayplace blankzone+roundfigure voidpoint awayplace blankzone}{roundfigure voidpoint awayplace blankzone}=\\frac{roundfigure voidpoint nullshape blankzone}{roundfigure voidpoint awayplace blankzone}=1+zeroconst+\\frac{1}{zeroconst}=\\frac{1+zeroconst+zeroconst^{2}}{zeroconst} .\n\\]\n\nBy cyclic symmetry we have\n\\[\n\\frac{roundfigure voidpoint awayplace blankzone}{roundfigure voidpoint nullshape blankzone}=\\frac{roundfigure nullshape noncorner voidpoint}{roundfigure voidpoint nullshape blankzone}=\\frac{roundfigure blankzone farregion nullshape}{roundfigure voidpoint nullshape blankzone}=\\frac{zeroconst}{1+zeroconst+zeroconst^{2}} .\n\\]\n\nNow \\( \\triangle farregion awayplace noncorner=\\triangle voidpoint nullshape blankzone-\\triangle voidpoint awayplace blankzone-\\triangle nullshape noncorner voidpoint-\\triangle blankzone farregion nullshape \\), so\n\\[\n\\frac{roundfigure farregion awayplace noncorner}{roundfigure voidpoint nullshape blankzone}=1-3 \\frac{zeroconst}{1+zeroconst+zeroconst^{2}}=\\frac{(1-zeroconst)^{2}}{1+zeroconst+zeroconst^{2}}\n\\]"
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "mfdbeuto",
        "P": "vclrqwzi",
        "Q": "snoptyaf",
        "R": "dmhclruo",
        "x": "tigbqsev",
        "y": "pkjrnoda",
        "\\lambda": "wexhupiz",
        "\\mu": "bozfarmt",
        "\\nu": "ysrplcjd",
        "\\Delta": "urezhvak",
        "k": "lofansie"
      },
      "question": "3. In a triangle \\( qzxwvtnp hjgrksla mfdbeuto \\) in the Euclidean plane, let \\( qzxwvtnp^{\\prime} \\) be a point on the segment from \\( hjgrksla \\) to \\( mfdbeuto, hjgrksla^{\\prime} \\) a point on the segment from \\( mfdbeuto \\) to \\( qzxwvtnp \\), and \\( mfdbeuto^{\\prime} \\) a point on the segment from \\( qzxwvtnp \\) to \\( hjgrksla \\) such that\n\\[\n\\frac{qzxwvtnp hjgrksla^{\\prime}}{hjgrksla^{\\prime} mfdbeuto}=\\frac{hjgrksla mfdbeuto^{\\prime}}{mfdbeuto^{\\prime} qzxwvtnp}=\\frac{mfdbeuto qzxwvtnp^{\\prime}}{qzxwvtnp^{\\prime} hjgrksla}=lofansie\n\\]\nwhere \\( lofansie \\) is a positive constant. Let \\( urezhvak \\) be the triangle formed by parts of the segments obtained by joining \\( qzxwvtnp \\) and \\( qzxwvtnp^{\\prime}, hjgrksla \\) and \\( hjgrksla^{\\prime} \\), and \\( mfdbeuto \\) and \\( mfdbeuto^{\\prime} \\). Prove that the areas of the triangles \\( urezhvak \\) and \\( qzxwvtnp hjgrksla mfdbeuto \\) are in the ratio.\n\\[\n\\frac{(lofansie-1)^{2}}{lofansie^{2}+lofansie+1}\n\\]",
      "solution": "First Solution. Use barycentric coordinates for the triangle \\( qzxwvtnp hjgrksla mfdbeuto \\).\n\\[\n\\begin{aligned}\nqzxwvtnp & =\\langle 1,0,0\\rangle \\\\\nhjgrksla & =\\langle 0,1,0\\rangle \\\\\nmfdbeuto & =\\langle 0,0,1\\rangle \\\\\nqzxwvtnp^{\\prime} & =\\frac{1}{lofansie+1}\\langle 0, lofansie, 1\\rangle \\\\\nhjgrksla^{\\prime} & =\\frac{1}{lofansie+1}\\langle 1,0, lofansie\\rangle \\\\\nmfdbeuto^{\\prime} & =\\frac{1}{lofansie+1}\\langle lofansie .1,0\\rangle .\n\\end{aligned}\n\\]\n\nThe equations of \\( hjgrksla hjgrksla^{\\prime} \\) and \\( mfdbeuto mfdbeuto^{\\prime} \\) are \\( ==lofansie tigbqsev \\) and \\( tigbqsev=lofansie pkjrnoda \\), respectively. If their intersection \\( vclrqwzi \\) is given by \\( \\langle wexhupiz, bozfarmt, ysrplcjd\\rangle \\), then \\( wexhupiz=lofansie bozfarmt, ysrplcjd=lofansie wexhupiz \\), and \\( wexhupiz+bozfarmt+ysrplcjd=1 \\). So\n\\[\nvclrqwzi=\\frac{1}{lofansie^{2}+lofansie+1}\\left\\langle lofansie, 1, lofansie^{2}\\right\\rangle .\n\\]\n\nBy symmetry\n\\[\n\\begin{array}{l}\nsnoptyaf=\\frac{1}{lofansie^{2}+lofansie+1}\\left\\langle lofansie^{2}, lofansie, 1\\right\\rangle \\\\\ndmhclruo=\\frac{1}{lofansie^{2}+lofansie+1}\\left\\langle 1, lofansie^{2}, lofansie\\right\\rangle .\n\\end{array}\n\\]\n\nThen\n\\[\n\\begin{aligned}\n\\frac{\\operatorname{area} vclrqwzi snoptyaf dmhclruo}{\\operatorname{area} qzxwvtnp hjgrksla mfdbeuto} & =\\frac{1}{\\left(lofansie^{2}+lofansie+1\\right)^{3}}\\left|\\begin{array}{lll}\nlofansie & 1 & lofansie^{2} \\\\\nlofansie^{2} & lofansie & 1 \\\\\n1 & lofansie^{2} & lofansie\n\\end{array}\\right| \\\\\n& =\\frac{lofansie^{6}-2 lofansie^{3}+1}{\\left(lofansie^{2}+lofansie+1\\right)^{3}}=\\frac{\\left(lofansie^{3}-1\\right)^{2}}{\\left(lofansie^{2}+lofansie+1\\right)^{3}}=\\frac{(lofansie-1)^{2}}{lofansie^{2}+lofansie+1} .\n\\end{aligned}\n\\]\n\nSecond Solution. A synthetic solution may be of interest.\n\\[\n\\begin{array}{l} \n\\frac{urezhvak qzxwvtnp snoptyaf mfdbeuto}{\\triangle qzxwvtnp snoptyaf mfdbeuto^{\\prime}} \\\\\n=\\frac{snoptyaf mfdbeuto}{snoptyaf mfdbeuto^{\\prime}}=\\frac{urezhvak hjgrksla snoptyaf mfdbeuto}{urezhvak hjgrksla snoptyaf mfdbeuto^{\\prime}} \\\\\n\\therefore \\quad \\frac{urezhvak hjgrksla snoptyaf mfdbeuto}{urezhvak qzxwvtnp snoptyaf mfdbeuto}=\\frac{urezhvak hjgrksla snoptyaf mfdbeuto^{\\prime}}{urezhvak qzxwvtnp snoptyaf mfdbeuto^{\\prime}}=\\frac{hjgrksla mfdbeuto^{\\prime}}{qzxwvtnp mfdbeuto^{\\prime}}=lofansie\n\\end{array}\n\\]\n\nSimilarly\n\\[\n\\begin{aligned}\n& \\frac{urezhvak qzxwvtnp snoptyaf mfdbeuto}{urezhvak qzxwvtnp^{\\prime} snoptyaf mfdbeuto}=\\frac{qzxwvtnp snoptyaf}{qzxwvtnp^{\\prime} snoptyaf}=\\frac{urezhvak hjgrksla qzxwvtnp snoptyaf}{urezhvak hjgrksla qzxwvtnp^{\\prime} snoptyaf} \\\\\n\\therefore \\quad & \\frac{urezhvak qzxwvtnp snoptyaf hjgrksla}{urezhvak qzxwvtnp snoptyaf mfdbeuto}=\\frac{urezhvak qzxwvtnp^{\\prime} snoptyaf hjgrksla}{urezhvak qzxwvtnp^{\\prime} snoptyaf mfdbeuto}=\\frac{qzxwvtnp^{\\prime} hjgrksla}{qzxwvtnp^{\\prime} mfdbeuto}=\\frac{1}{lofansie} .\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\frac{urezhvak qzxwvtnp snoptyaf hjgrksla+urezhvak hjgrksla snoptyaf mfdbeuto+urezhvak qzxwvtnp snoptyaf mfdbeuto}{urezhvak qzxwvtnp snoptyaf mfdbeuto}=1+lofansie+\\frac{1}{lofansie}=\\frac{1+lofansie+lofansie^{2}}{lofansie} .\n\\]\n\nBy cyclic symmetry we have\n\\[\n\\frac{urezhvak qzxwvtnp snoptyaf mfdbeuto}{urezhvak qzxwvtnp hjgrksla mfdbeuto}=\\frac{urezhvak hjgrksla dmhclruo qzxwvtnp}{urezhvak qzxwvtnp hjgrksla mfdbeuto}=\\frac{urezhvak mfdbeuto vclrqwzi hjgrksla}{urezhvak qzxwvtnp hjgrksla mfdbeuto}=\\frac{lofansie}{1+lofansie+lofansie^{2}} .\n\\]\n\nNow \\( \\triangle vclrqwzi snoptyaf dmhclruo=\\triangle qzxwvtnp hjgrksla mfdbeuto-\\triangle qzxwvtnp snoptyaf mfdbeuto-\\triangle hjgrksla dmhclruo qzxwvtnp-\\triangle mfdbeuto vclrqwzi hjgrksla \\), so\n\\[\n\\frac{urezhvak vclrqwzi snoptyaf dmhclruo}{urezhvak qzxwvtnp hjgrksla mfdbeuto}=1-3 \\frac{lofansie}{1+lofansie+lofansie^{2}}=\\frac{(1-lofansie)^{2}}{1+lofansie+lofansie^{2}}\n\\]"
    },
    "kernel_variant": {
      "question": "Let $n\\ge 2$ and let  \n\\[\n\\mathcal S=A_{0}A_{1}\\dots A_{n}\n\\]\nbe an $n$–simplex in the Euclidean space $\\mathbb R^{\\,n}$  \n(all indices are understood modulo $n+1$).\n\nFix a positive constant $\\lambda\\neq 1$.  \nOn every edge $A_{i}A_{i+1}$ choose the interior point  \n\\[\nA_{i}^{\\prime}\\in A_{i}A_{i+1}\\qquad (0)\n\\]\nso that  \n\\[\n\\lvert A_{i}A_{i}^{\\prime}\\rvert : \\lvert A_{i}^{\\prime}A_{i+1}\\rvert=\\lambda:1.\\qquad (1)\n\\]\n\nWith respect to the barycentric coordinates $(x_{0},\\dots,x_{n})$ of $\\mathcal S$ introduce the hyper–planes  \n\\[\nH_{i}:=\\bigl\\{x\\mid \\lambda x_{i}=x_{i+1}\\bigr\\},\\qquad i=0,1,\\dots ,n. \\qquad (2)\n\\]\n\nObserve that $H_{i}$ contains the vertex $A_{i-1}$ as well as the point $A_{i}^{\\prime}$ introduced in $(1)$.\n\n1. Prove that the $n+1$ hyper–planes $H_{0},\\dots ,H_{n}$ are in general position and therefore determine a unique inner $n$–simplex; denote it by $\\mathcal T$.\n\n2. Show that for every positive $\\lambda\\neq 1$\n\\[\n\\frac{\\operatorname{Vol}_{n}(\\mathcal T)}{\\operatorname{Vol}_{n}(\\mathcal S)}\n   =\\frac{\\lvert \\lambda-1\\rvert^{\\,n}}{\\lambda^{\\,n}+\\lambda^{\\,n-1}+\\dots +\\lambda+1}. \\qquad (3)\n\\]\n\n3. Verify the limiting cases  \n\\[\n\\lim_{\\lambda\\to 1^{+}}\\operatorname{Vol}_{n}(\\mathcal T)=0,\n\\qquad\n\\lim_{\\lambda\\to\\infty}\\operatorname{Vol}_{n}(\\mathcal T)=\\operatorname{Vol}_{n}(\\mathcal S). \\qquad (4)\n\\]\n\n(The description $(2)$ is equivalent to saying that $H_{i}$ is the unique hyper–plane through $A_{i-1}$ together with the $n-1$ points $A_{i}^{\\prime},A_{i+1}^{\\prime},\\dots ,A_{i+n-2}^{\\prime}$.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Step 1 – Barycentric description of the data  \nAssign the standard barycentric coordinates to the vertices of $\\mathcal S$,\n\\[\nA_{0}\\equiv e_{0},\\; A_{1}\\equiv e_{1},\\dots ,A_{n}\\equiv e_{n},\n\\qquad\\sum_{j=0}^{n}x_{j}=1.\\tag{5}\n\\]\nOn the edge $A_{i}A_{i+1}$ the point that fulfils $(1)$ is\n\\[\nA_{i}^{\\prime}\n      =\\bigl(0,\\dots ,0,\\ \\tfrac{1}{1+\\lambda},\\ \\tfrac{\\lambda}{1+\\lambda},0,\\dots ,0\\bigr),\\tag{6}\n\\]\nthe two non–zero coordinates occupying the $i$–th and $(i+1)$–st positions.  \nConsequently $A_{i}^{\\prime}$ and $A_{i-1}$ satisfy $\\lambda x_{i}=x_{i+1}$, hence $(2)$ does describe the desired hyper–planes.\n\nStep 2 – The vertices of the inner simplex  \nFor $j=0,1,\\dots ,n$ let $V_{j}$ be the common intersection of all hyper–planes except $H_{j}$, i.e.\n\\[\n\\lambda x_{i}=x_{i+1}\\quad\\text{for every } i\\neq j,\\tag{7}\n\\]\ntogether with $(5)$.  \nBecause the situation is cyclic we treat $j=0$ and obtain the others by rotation.\n\nWriting $(7)$ for $i=1,\\dots ,n$ gives\n\\[\nx_{2}=\\lambda x_{1},\\;x_{3}=\\lambda x_{2}=\\lambda^{2}x_{1},\\dots ,\nx_{n}=\\lambda^{\\,n-1}x_{1},\\;x_{0}=\\lambda x_{n}=\\lambda^{\\,n}x_{1}.\\tag{8}\n\\]\nInsert $(8)$ into $(5)$,\n\\[\n1=x_{1}\\bigl(1+\\lambda+\\dots +\\lambda^{\\,n}\\bigr)=x_{1}\\Lambda,\n\\qquad\\Lambda:=\\lambda^{\\,n}+\\dots +\\lambda+1.\\tag{9}\n\\]\nHence\n\\[\nx_{1}=\\frac{1}{\\Lambda},\\quad\nx_{k}=\\frac{\\lambda^{\\,k-1}}{\\Lambda}\\;(2\\le k\\le n),\\quad\nx_{0}=\\frac{\\lambda^{\\,n}}{\\Lambda}.\\tag{10}\n\\]\nTherefore\n\\[\nV_{0}=\\frac{\\bigl(\\lambda^{\\,n},1,\\lambda,\\lambda^{2},\\dots ,\\lambda^{\\,n-1}\\bigr)}{\\Lambda}.\\tag{11}\n\\]\nA cyclic permutation of the coordinates yields the remaining vertices,\n\\[\nV_{j}\n  =\\frac{\n   \\bigl(\\lambda^{\\,n-j+1},\\dots ,\\lambda^{\\,n},1,\\lambda,\\dots ,\\lambda^{\\,n-j}\\bigr)}\n   {\\Lambda},\n\\qquad j=0,\\dots ,n.\\tag{12}\n\\]\nThe un–normalised vectors in $(12)$ are consecutive cyclic shifts of one another and form the rows of a circulant matrix; they are linearly independent, proving that the hyper–planes are in general position (Part 1).\n\nStep 3 – The determinant of the affine map $e_{i}\\mapsto V_{i}$  \nLet $C$ be the $(n+1)\\times (n+1)$ circulant matrix whose first row is\n\\[\n(1,\\lambda,\\lambda^{2},\\dots ,\\lambda^{\\,n}).\n\\]\nDenote by $W$ the matrix whose rows are the un–scaled vectors in $(12)$.  $W$ differs from $C$ only by a cyclic permutation of the columns, i.e. an $(n+1)$–cycle; hence\n\\[\n\\det W = (-1)^{\\,n}\\det C.\\tag{13}\n\\]\n\nThe eigen–values of $C$ are\n\\[\n\\mu_{k}\n   =\\sum_{j=0}^{n}\\lambda^{\\,j}\\omega_{k}^{\\,j}\n   =\\frac{(\\lambda\\omega_{k})^{\\,n+1}-1}{\\lambda\\omega_{k}-1}\n   =\\frac{\\lambda^{\\,n+1}-1}{\\lambda\\omega_{k}-1},\n\\qquad\n\\omega_{k}=e^{2\\pi i k/(n+1)},\\ k=0,\\dots ,n.\\tag{14}\n\\]\nTherefore\n\\[\n\\det C\n   =\\prod_{k=0}^{n}\\mu_{k}\n   =\\frac{(\\lambda^{\\,n+1}-1)^{\\,n+1}}\n          {\\prod_{k=0}^{n}(\\lambda\\omega_{k}-1)}.\\tag{15}\n\\]\nUsing $\\prod_{k=0}^{n}(x-\\omega_{k})=x^{\\,n+1}-1$ with $x=\\lambda$ we obtain\n\\[\n\\prod_{k=0}^{n}(\\lambda\\omega_{k}-1)=(-1)^{\\,n}(\\lambda^{\\,n+1}-1).\\tag{16}\n\\]\nInsert $(16)$ into $(15)$ and then into $(13)$:\n\\[\n\\det W = (-1)^{\\,n}(-1)^{\\,n}(\\lambda^{\\,n+1}-1)^{\\,n}\n       =(\\lambda^{\\,n+1}-1)^{\\,n}.\\tag{17}\n\\]\n\nSince $\\lambda^{\\,n+1}-1=(\\lambda-1)\\Lambda$, it follows that\n\\[\n\\lvert\\det W\\rvert=\\lvert\\lambda-1\\rvert^{\\,n}\\Lambda^{\\,n}.\\tag{18}\n\\]\n\nTo pass from the rows of $W$ to the true vertices $V_{i}$ each coordinate is divided by $\\Lambda$; i.e.\\ every column is scaled by $\\Lambda^{-1}$.  Hence the Jacobian of the affine map $e_{i}\\mapsto V_{i}$ equals\n\\[\nJ=\\frac{\\det W}{\\Lambda^{\\,n+1}}.\\tag{19}\n\\]\nTaking absolute values and using $(18)$ gives\n\\[\n\\lvert J\\rvert=\\frac{\\lvert\\lambda-1\\rvert^{\\,n}}{\\Lambda}.\\tag{20}\n\\]\n\nStep 4 – The volume ratio (Part 2)  \nAn affine transformation in $\\mathbb R^{\\,n}$ multiplies $n$–dimensional volumes by $\\lvert J\\rvert$. Therefore\n\\[\n\\frac{\\operatorname{Vol}_{n}(\\mathcal T)}{\\operatorname{Vol}_{n}(\\mathcal S)}\n    =\\lvert J\\rvert\n    =\\frac{\\lvert\\lambda-1\\rvert^{\\,n}}\n           {\\lambda^{\\,n}+\\lambda^{\\,n-1}+\\dots +\\lambda+1},\\tag{21}\n\\]\nwhich is exactly the expression asserted in $(3)$.\n\nStep 5 – Limiting values (Part 3)  \n$\\bullet$ As $\\lambda\\to 1^{+}$ we have $\\lvert\\lambda-1\\rvert^{\\,n}\\to 0$ while $\\Lambda\\to n+1$, hence $\\operatorname{Vol}_{n}(\\mathcal T)\\to 0$.  \n\n$\\bullet$ When $\\lambda\\to\\infty$ we have $\\Lambda\\sim\\lambda^{\\,n}$, so the right–hand side of $(21)$ tends to $1$ and $\\mathcal T$ fills $\\mathcal S$ completely.\n\n\\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.535186",
        "was_fixed": false,
        "difficulty_analysis": "1.  Dimensional elevation – the problem is set in arbitrary dimension\n    \\(n\\ge 3\\), not merely in the plane.  Higher-dimensional barycentric\n    coordinates, circulant determinants and \\(n\\)-volumes are required.\n2.  Additional structures – the object of study is an inner\n    \\(n\\)-simplex cut out by \\(n+1\\) **hyperplanes**, not by three lines.\n3.  The proof demands mastery of\n    • multilinear algebra (determinants and Jacobians in dimension \\(n\\));  \n    • properties of circulant matrices and roots of unity;  \n    • generalised barycentric (affine) coordinates.  \n4.  The student must handle a non-trivial linear recurrence to obtain\n    the vertices, and then connect an algebraic factorisation\n    \\(\\lambda^{n+1}-1=(\\lambda-1)(\\lambda^{n}+\\dots +1)\\) with a\n    geometric quantity.  \n5.  The original problem (and its kernel variant) are recovered only as\n    the special case \\(n=2\\); all higher dimensions are genuinely new\n    and far more technically involved."
      }
    },
    "original_kernel_variant": {
      "question": "Let $n\\ge 2$ and let  \n\\[\n\\mathcal S=A_{0}A_{1}\\dots A_{n}\n\\]\nbe an $n$–simplex in the Euclidean space $\\mathbb R^{\\,n}$  \n(all indices are understood modulo $n+1$).\n\nFix a positive constant $\\lambda\\neq 1$.  \nOn every edge $A_{i}A_{i+1}$ choose the interior point  \n\\[\nA_{i}^{\\prime}\\in A_{i}A_{i+1}\\qquad (0)\n\\]\nso that  \n\\[\n\\lvert A_{i}A_{i}^{\\prime}\\rvert : \\lvert A_{i}^{\\prime}A_{i+1}\\rvert=\\lambda:1.\\qquad (1)\n\\]\n\nWith respect to the barycentric coordinates $(x_{0},\\dots,x_{n})$ of $\\mathcal S$ introduce the hyper–planes  \n\\[\nH_{i}:=\\bigl\\{x\\mid \\lambda x_{i}=x_{i+1}\\bigr\\},\\qquad i=0,1,\\dots ,n. \\qquad (2)\n\\]\n\nObserve that $H_{i}$ contains the vertex $A_{i-1}$ as well as the point $A_{i}^{\\prime}$ introduced in $(1)$.\n\n1. Prove that the $n+1$ hyper–planes $H_{0},\\dots ,H_{n}$ are in general position and therefore determine a unique inner $n$–simplex; denote it by $\\mathcal T$.\n\n2. Show that for every positive $\\lambda\\neq 1$\n\\[\n\\frac{\\operatorname{Vol}_{n}(\\mathcal T)}{\\operatorname{Vol}_{n}(\\mathcal S)}\n   =\\frac{\\lvert \\lambda-1\\rvert^{\\,n}}{\\lambda^{\\,n}+\\lambda^{\\,n-1}+\\dots +\\lambda+1}. \\qquad (3)\n\\]\n\n3. Verify the limiting cases  \n\\[\n\\lim_{\\lambda\\to 1^{+}}\\operatorname{Vol}_{n}(\\mathcal T)=0,\n\\qquad\n\\lim_{\\lambda\\to\\infty}\\operatorname{Vol}_{n}(\\mathcal T)=\\operatorname{Vol}_{n}(\\mathcal S). \\qquad (4)\n\\]\n\n(The description $(2)$ is equivalent to saying that $H_{i}$ is the unique hyper–plane through $A_{i-1}$ together with the $n-1$ points $A_{i}^{\\prime},A_{i+1}^{\\prime},\\dots ,A_{i+n-2}^{\\prime}$.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Step 1 – Barycentric description of the data  \nAssign the standard barycentric coordinates to the vertices of $\\mathcal S$,\n\\[\nA_{0}\\equiv e_{0},\\; A_{1}\\equiv e_{1},\\dots ,A_{n}\\equiv e_{n},\n\\qquad\\sum_{j=0}^{n}x_{j}=1.\\tag{5}\n\\]\nOn the edge $A_{i}A_{i+1}$ the point that fulfils $(1)$ is\n\\[\nA_{i}^{\\prime}\n      =\\bigl(0,\\dots ,0,\\ \\tfrac{1}{1+\\lambda},\\ \\tfrac{\\lambda}{1+\\lambda},0,\\dots ,0\\bigr),\\tag{6}\n\\]\nthe two non–zero coordinates occupying the $i$–th and $(i+1)$–st positions.  \nConsequently $A_{i}^{\\prime}$ and $A_{i-1}$ satisfy $\\lambda x_{i}=x_{i+1}$, hence $(2)$ does describe the desired hyper–planes.\n\nStep 2 – The vertices of the inner simplex  \nFor $j=0,1,\\dots ,n$ let $V_{j}$ be the common intersection of all hyper–planes except $H_{j}$, i.e.\n\\[\n\\lambda x_{i}=x_{i+1}\\quad\\text{for every } i\\neq j,\\tag{7}\n\\]\ntogether with $(5)$.  \nBecause the situation is cyclic we treat $j=0$ and obtain the others by rotation.\n\nWriting $(7)$ for $i=1,\\dots ,n$ gives\n\\[\nx_{2}=\\lambda x_{1},\\;x_{3}=\\lambda x_{2}=\\lambda^{2}x_{1},\\dots ,\nx_{n}=\\lambda^{\\,n-1}x_{1},\\;x_{0}=\\lambda x_{n}=\\lambda^{\\,n}x_{1}.\\tag{8}\n\\]\nInsert $(8)$ into $(5)$,\n\\[\n1=x_{1}\\bigl(1+\\lambda+\\dots +\\lambda^{\\,n}\\bigr)=x_{1}\\Lambda,\n\\qquad\\Lambda:=\\lambda^{\\,n}+\\dots +\\lambda+1.\\tag{9}\n\\]\nHence\n\\[\nx_{1}=\\frac{1}{\\Lambda},\\quad\nx_{k}=\\frac{\\lambda^{\\,k-1}}{\\Lambda}\\;(2\\le k\\le n),\\quad\nx_{0}=\\frac{\\lambda^{\\,n}}{\\Lambda}.\\tag{10}\n\\]\nTherefore\n\\[\nV_{0}=\\frac{\\bigl(\\lambda^{\\,n},1,\\lambda,\\lambda^{2},\\dots ,\\lambda^{\\,n-1}\\bigr)}{\\Lambda}.\\tag{11}\n\\]\nA cyclic permutation of the coordinates yields the remaining vertices,\n\\[\nV_{j}\n  =\\frac{\n   \\bigl(\\lambda^{\\,n-j+1},\\dots ,\\lambda^{\\,n},1,\\lambda,\\dots ,\\lambda^{\\,n-j}\\bigr)}\n   {\\Lambda},\n\\qquad j=0,\\dots ,n.\\tag{12}\n\\]\nThe un–normalised vectors in $(12)$ are consecutive cyclic shifts of one another and form the rows of a circulant matrix; they are linearly independent, proving that the hyper–planes are in general position (Part 1).\n\nStep 3 – The determinant of the affine map $e_{i}\\mapsto V_{i}$  \nLet $C$ be the $(n+1)\\times (n+1)$ circulant matrix whose first row is\n\\[\n(1,\\lambda,\\lambda^{2},\\dots ,\\lambda^{\\,n}).\n\\]\nDenote by $W$ the matrix whose rows are the un–scaled vectors in $(12)$.  $W$ differs from $C$ only by a cyclic permutation of the columns, i.e. an $(n+1)$–cycle; hence\n\\[\n\\det W = (-1)^{\\,n}\\det C.\\tag{13}\n\\]\n\nThe eigen–values of $C$ are\n\\[\n\\mu_{k}\n   =\\sum_{j=0}^{n}\\lambda^{\\,j}\\omega_{k}^{\\,j}\n   =\\frac{(\\lambda\\omega_{k})^{\\,n+1}-1}{\\lambda\\omega_{k}-1}\n   =\\frac{\\lambda^{\\,n+1}-1}{\\lambda\\omega_{k}-1},\n\\qquad\n\\omega_{k}=e^{2\\pi i k/(n+1)},\\ k=0,\\dots ,n.\\tag{14}\n\\]\nTherefore\n\\[\n\\det C\n   =\\prod_{k=0}^{n}\\mu_{k}\n   =\\frac{(\\lambda^{\\,n+1}-1)^{\\,n+1}}\n          {\\prod_{k=0}^{n}(\\lambda\\omega_{k}-1)}.\\tag{15}\n\\]\nUsing $\\prod_{k=0}^{n}(x-\\omega_{k})=x^{\\,n+1}-1$ with $x=\\lambda$ we obtain\n\\[\n\\prod_{k=0}^{n}(\\lambda\\omega_{k}-1)=(-1)^{\\,n}(\\lambda^{\\,n+1}-1).\\tag{16}\n\\]\nInsert $(16)$ into $(15)$ and then into $(13)$:\n\\[\n\\det W = (-1)^{\\,n}(-1)^{\\,n}(\\lambda^{\\,n+1}-1)^{\\,n}\n       =(\\lambda^{\\,n+1}-1)^{\\,n}.\\tag{17}\n\\]\n\nSince $\\lambda^{\\,n+1}-1=(\\lambda-1)\\Lambda$, it follows that\n\\[\n\\lvert\\det W\\rvert=\\lvert\\lambda-1\\rvert^{\\,n}\\Lambda^{\\,n}.\\tag{18}\n\\]\n\nTo pass from the rows of $W$ to the true vertices $V_{i}$ each coordinate is divided by $\\Lambda$; i.e.\\ every column is scaled by $\\Lambda^{-1}$.  Hence the Jacobian of the affine map $e_{i}\\mapsto V_{i}$ equals\n\\[\nJ=\\frac{\\det W}{\\Lambda^{\\,n+1}}.\\tag{19}\n\\]\nTaking absolute values and using $(18)$ gives\n\\[\n\\lvert J\\rvert=\\frac{\\lvert\\lambda-1\\rvert^{\\,n}}{\\Lambda}.\\tag{20}\n\\]\n\nStep 4 – The volume ratio (Part 2)  \nAn affine transformation in $\\mathbb R^{\\,n}$ multiplies $n$–dimensional volumes by $\\lvert J\\rvert$. Therefore\n\\[\n\\frac{\\operatorname{Vol}_{n}(\\mathcal T)}{\\operatorname{Vol}_{n}(\\mathcal S)}\n    =\\lvert J\\rvert\n    =\\frac{\\lvert\\lambda-1\\rvert^{\\,n}}\n           {\\lambda^{\\,n}+\\lambda^{\\,n-1}+\\dots +\\lambda+1},\\tag{21}\n\\]\nwhich is exactly the expression asserted in $(3)$.\n\nStep 5 – Limiting values (Part 3)  \n$\\bullet$ As $\\lambda\\to 1^{+}$ we have $\\lvert\\lambda-1\\rvert^{\\,n}\\to 0$ while $\\Lambda\\to n+1$, hence $\\operatorname{Vol}_{n}(\\mathcal T)\\to 0$.  \n\n$\\bullet$ When $\\lambda\\to\\infty$ we have $\\Lambda\\sim\\lambda^{\\,n}$, so the right–hand side of $(21)$ tends to $1$ and $\\mathcal T$ fills $\\mathcal S$ completely.\n\n\\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.445772",
        "was_fixed": false,
        "difficulty_analysis": "1.  Dimensional elevation – the problem is set in arbitrary dimension\n    \\(n\\ge 3\\), not merely in the plane.  Higher-dimensional barycentric\n    coordinates, circulant determinants and \\(n\\)-volumes are required.\n2.  Additional structures – the object of study is an inner\n    \\(n\\)-simplex cut out by \\(n+1\\) **hyperplanes**, not by three lines.\n3.  The proof demands mastery of\n    • multilinear algebra (determinants and Jacobians in dimension \\(n\\));  \n    • properties of circulant matrices and roots of unity;  \n    • generalised barycentric (affine) coordinates.  \n4.  The student must handle a non-trivial linear recurrence to obtain\n    the vertices, and then connect an algebraic factorisation\n    \\(\\lambda^{n+1}-1=(\\lambda-1)(\\lambda^{n}+\\dots +1)\\) with a\n    geometric quantity.  \n5.  The original problem (and its kernel variant) are recovered only as\n    the special case \\(n=2\\); all higher dimensions are genuinely new\n    and far more technically involved."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}