{
  "index": "1940-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG",
    "GEO"
  ],
  "difficulty": "",
  "question": "11. From any point \\( (a, b) \\) in the Cartesian plane, show that (i) three normals, real or imaginary, can be drawn to the parabola \\( y^{2}=4 p x \\); (ii) these are real and distinct if \\( 4(2 p-a)^{3}+27 p b^{2}<0 \\); (iii) two of them coincide if \\( (a, b) \\) lies on the curve \\( 27 p y^{2}=4(x-2 p)^{3} \\); (iv) all three coincide only if \\( (a, b) \\) is the point \\( (2 p, 0) \\).",
  "solution": "Solution. The slope of the parabola at the point \\( (x, y) \\) is \\( 2 p / y \\). Hence the line joining \\( (a, b) \\) to \\( (x, y) \\) is normal to the parabola at \\( (x, y) \\) if and only if\n\\[\n(y-b)=-\\frac{y}{2 p}(x-a)=-\\frac{y}{2 p}\\left(\\frac{y^{2}}{4 p}-a\\right)\n\\]\n(The case \\( y=0 \\) is also covered by this equation.) This reduces to\n\\[\ny^{3}+4 p(2 p-a) y-8 p^{2} b=0\n\\]\n\nThis cubic will have three roots if counted with multiplicity, the roots may be real or complex. Since (real or complex) values of \\( y \\) correspond one to one with (real or imaginary) points on the parabola, part (i) of the problem has been established.\n\nA real cubic equation of the form \\( y^{3}+A y+B=0 \\) will have all real roots if and only if \\( \\Delta=27 B^{2}+4 A^{3} \\leq 0 \\). Two roots will coincide if and only if \\( \\Delta=0 \\), and all three roots will coincide if and only if \\( A=B=0 \\).\n\nIn the present case, \\( A=4 p(2 p-a), B=-8 p^{2} b \\). So \\( \\Delta=64 p^{3}[4(2 p \\) \\( \\left.-a)^{3}+27 p b^{2}\\right] \\). The problem apparently intends that we take \\( p>0 \\) (as is usual with the standard form of the parabola). With this assumption, \\( \\Delta \\) has the same sign as\n\\[\n\\Delta^{\\prime}=4(2 p-a)^{3}+27 p b^{2}\n\\]\n\nThe roots of (1), and hence the points of the parabola, will be real and distinct if and only if \\( \\Delta^{\\prime}<0 \\). Two normals will coincide if and only if \\( \\Delta^{\\prime}=0 \\); that is, \\( (a, b) \\) lies on the curve \\( 27 p y^{2}=4(x-2 p)^{3} \\). Finally, all three normals will coincide if and only if \\( 4 p(2 p-a)=0 \\) and \\( 8 p^{2} b=0 \\), i.e., if and only if \\( (a, b)=(2 p, 0) \\).\n\nThe algebraic facts stated above for the roots of \\( y^{3}+A y+B=0 \\) can be derived by the methods of calculus as follows.\n\nThe graph of \\( y^{3}+A y \\) will be increasing with strictly positive slope if \\( A>0 \\), so any equation \\( y^{3}+A y=-B \\) will have just one real root and therefore two complex roots. Thus the roots are distinct if \\( A>0 \\).\n\nIf \\( A=0 \\), the equation will have one real and two complex roots for \\( B \\neq 0 \\), and a triple root at 0 if \\( B=0 \\) also.\n\nFor \\( A<0 \\), the graph has turning points at\n\\[\n\\left(-\\sqrt{-\\frac{A}{3}},-\\frac{2}{3} A \\sqrt{-\\frac{A}{3}}\\right) \\quad \\text { and } \\quad\\left(+\\sqrt{-\\frac{A}{3}}, \\frac{2}{3} A \\sqrt{-\\frac{A}{3}}\\right) .\n\\]\n\nIt is then clear that the equation will have three distinct real roots if\n\\[\n|B|<\\left|\\frac{2}{3} A \\sqrt{-\\frac{A}{3}}\\right|,\n\\]\none double and one single root if\n\\[\n|B|=\\left|\\frac{2}{3} A \\sqrt{-\\frac{A}{3}}\\right|,\n\\]\nand one real and two complex roots if\n\\[\n|B|>\\left|\\frac{2}{3} A \\sqrt{-\\frac{A}{3}}\\right| .\n\\]\n\nSquaring these relations we can write them in terms of \\( \\Delta=27 B^{2}+4 A^{3} \\), and we have\n\\[\n\\begin{aligned}\n\\Delta<0 \\Rightarrow \\text { three distinct real roots, } \\\\\n\\Delta=0 \\Rightarrow \\text { (at least) two equal real roots, } \\\\\n\\Delta>0 \\Rightarrow \\text { one real and two complex roots, } \\\\\nA=B=0 \\Leftrightarrow \\text { a triple root. }\n\\end{aligned}\n\\]\n\nThese facts are often derived algebraically by showing that \\( -\\Delta \\), which is called the discriminant of the cubic, is the product of the squares of the differences of the roots. See any text on the theory of equations.",
  "vars": [
    "x",
    "y",
    "A",
    "B",
    "\\\\Delta"
  ],
  "params": [
    "a",
    "b",
    "p"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xcoord",
        "y": "ycoord",
        "A": "cubiccoef",
        "B": "cubiconst",
        "\\\\Delta": "discrim",
        "a": "pointax",
        "b": "pointay",
        "p": "paraparam"
      },
      "question": "11. From any point \\( (pointax, pointay) \\) in the Cartesian plane, show that (i) three normals, real or imaginary, can be drawn to the parabola \\( ycoord^{2}=4 paraparam xcoord \\); (ii) these are real and distinct if \\( 4(2 paraparam-pointax)^{3}+27 paraparam pointay^{2}<0 \\); (iii) two of them coincide if \\( (pointax, pointay) \\) lies on the curve \\( 27 paraparam ycoord^{2}=4(xcoord-2 paraparam)^{3} \\); (iv) all three coincide only if \\( (pointax, pointay) \\) is the point \\( (2 paraparam, 0) \\).",
      "solution": "Solution. The slope of the parabola at the point \\( (xcoord, ycoord) \\) is \\( 2 paraparam / ycoord \\). Hence the line joining \\( (pointax, pointay) \\) to \\( (xcoord, ycoord) \\) is normal to the parabola at \\( (xcoord, ycoord) \\) if and only if\n\\[\n(ycoord-pointay)=-\\frac{ycoord}{2 paraparam}(xcoord-pointax)=-\\frac{ycoord}{2 paraparam}\\left(\\frac{ycoord^{2}}{4 paraparam}-pointax\\right)\n\\]\n(The case \\( ycoord=0 \\) is also covered by this equation.) This reduces to\n\\[\nycoord^{3}+4 paraparam(2 paraparam-pointax) ycoord-8 paraparam^{2} pointay=0\n\\]\n\nThis cubic will have three roots if counted with multiplicity, the roots may be real or complex. Since (real or complex) values of \\( ycoord \\) correspond one to one with (real or imaginary) points on the parabola, part (i) of the problem has been established.\n\nA real cubic equation of the form \\( ycoord^{3}+cubiccoef ycoord+cubiconst=0 \\) will have all real roots if and only if \\( discrim=27 cubiconst^{2}+4 cubiccoef^{3} \\leq 0 \\). Two roots will coincide if and only if \\( discrim=0 \\), and all three roots will coincide if and only if \\( cubiccoef=cubiconst=0 \\).\n\nIn the present case, \\( cubiccoef=4 paraparam(2 paraparam-pointax),\\; cubiconst=-8 paraparam^{2} pointay \\). So \\( discrim=64 paraparam^{3}[4(2 paraparam - pointax)^{3}+27 paraparam pointay^{2}] \\). The problem apparently intends that we take \\( paraparam>0 \\) (as is usual with the standard form of the parabola). With this assumption, \\( discrim \\) has the same sign as\n\\[\ndiscrim^{\\prime}=4(2 paraparam-pointax)^{3}+27 paraparam pointay^{2}\n\\]\n\nThe roots of (1), and hence the points of the parabola, will be real and distinct if and only if \\( discrim^{\\prime}<0 \\). Two normals will coincide if and only if \\( discrim^{\\prime}=0 \\); that is, \\( (pointax, pointay) \\) lies on the curve \\( 27 paraparam ycoord^{2}=4(xcoord-2 paraparam)^{3} \\). Finally, all three normals will coincide if and only if \\( 4 paraparam(2 paraparam-pointax)=0 \\) and \\( 8 paraparam^{2} pointay=0 \\), i.e., if and only if \\( (pointax, pointay)=(2 paraparam, 0) \\).\n\nThe algebraic facts stated above for the roots of \\( ycoord^{3}+cubiccoef ycoord+cubiconst=0 \\) can be derived by the methods of calculus as follows.\n\nThe graph of \\( ycoord^{3}+cubiccoef ycoord \\) will be increasing with strictly positive slope if \\( cubiccoef>0 \\), so any equation \\( ycoord^{3}+cubiccoef ycoord=-cubiconst \\) will have just one real root and therefore two complex roots. Thus the roots are distinct if \\( cubiccoef>0 \\).\n\nIf \\( cubiccoef=0 \\), the equation will have one real and two complex roots for \\( cubiconst \\neq 0 \\), and a triple root at 0 if \\( cubiconst=0 \\) also.\n\nFor \\( cubiccoef<0 \\), the graph has turning points at\n\\[\n\\left(-\\sqrt{-\\frac{cubiccoef}{3}},-\\frac{2}{3} cubiccoef \\sqrt{-\\frac{cubiccoef}{3}}\\right) \\quad \\text { and } \\quad\\left(+\\sqrt{-\\frac{cubiccoef}{3}}, \\frac{2}{3} cubiccoef \\sqrt{-\\frac{cubiccoef}{3}}\\right) .\n\\]\n\nIt is then clear that the equation will have three distinct real roots if\n\\[\n|cubiconst|<\\left|\\frac{2}{3} cubiccoef \\sqrt{-\\frac{cubiccoef}{3}}\\right|,\n\\]\none double and one single root if\n\\[\n|cubiconst|=\\left|\\frac{2}{3} cubiccoef \\sqrt{-\\frac{cubiccoef}{3}}\\right|,\n\\]\nand one real and two complex roots if\n\\[\n|cubiconst|>\\left|\\frac{2}{3} cubiccoef \\sqrt{-\\frac{cubiccoef}{3}}\\right| .\n\\]\n\nSquaring these relations we can write them in terms of \\( discrim=27 cubiconst^{2}+4 cubiccoef^{3} \\), and we have\n\\[\n\\begin{aligned}\ndiscrim<0 \\Rightarrow \\text { three distinct real roots, } \\\\\ndiscrim=0 \\Rightarrow \\text { (at least) two equal real roots, } \\\\\ndiscrim>0 \\Rightarrow \\text { one real and two complex roots, } \\\\\ncubiccoef=cubiconst=0 \\Leftrightarrow \\text { a triple root. }\n\\end{aligned}\n\\]\n\nThese facts are often derived algebraically by showing that \\( -discrim \\), which is called the discriminant of the cubic, is the product of the squares of the differences of the roots. See any text on the theory of equations."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lanterns",
        "y": "dolphins",
        "A": "orchards",
        "B": "meanders",
        "\\Delta": "galaxies",
        "a": "sunflower",
        "b": "pinecones",
        "p": "rainstorm"
      },
      "question": "11. From any point \\( (sunflower, pinecones) \\) in the Cartesian plane, show that (i) three normals, real or imaginary, can be drawn to the parabola \\( dolphins^{2}=4 rainstorm lanterns \\); (ii) these are real and distinct if \\( 4(2 rainstorm-sunflower)^{3}+27 rainstorm pinecones^{2}<0 \\); (iii) two of them coincide if \\( (sunflower, pinecones) \\) lies on the curve \\( 27 rainstorm dolphins^{2}=4(lanterns-2 rainstorm)^{3} \\); (iv) all three coincide only if \\( (sunflower, pinecones) \\) is the point \\( (2 rainstorm, 0) \\).",
      "solution": "Solution. The slope of the parabola at the point \\( (lanterns, dolphins) \\) is \\( 2 rainstorm / dolphins \\). Hence the line joining \\( (sunflower, pinecones) \\) to \\( (lanterns, dolphins) \\) is normal to the parabola at \\( (lanterns, dolphins) \\) if and only if\n\\[\n(dolphins-pinecones)=-\\frac{dolphins}{2 rainstorm}(lanterns-sunflower)=-\\frac{dolphins}{2 rainstorm}\\left(\\frac{dolphins^{2}}{4 rainstorm}-sunflower\\right)\n\\]\n(The case \\( dolphins=0 \\) is also covered by this equation.) This reduces to\n\\[\ndolphins^{3}+4 rainstorm(2 rainstorm-sunflower) dolphins-8 rainstorm^{2} pinecones=0\n\\]\n\nThis cubic will have three roots if counted with multiplicity, the roots may be real or complex. Since (real or complex) values of \\( dolphins \\) correspond one to one with (real or imaginary) points on the parabola, part (i) of the problem has been established.\n\nA real cubic equation of the form \\( dolphins^{3}+orchards \\, dolphins+meanders=0 \\) will have all real roots if and only if \\( galaxies=27 \\, meanders^{2}+4 \\, orchards^{3} \\leq 0 \\). Two roots will coincide if and only if \\( galaxies=0 \\), and all three roots will coincide if and only if \\( orchards=meanders=0 \\).\n\nIn the present case, \\( orchards=4 rainstorm(2 rainstorm-sunflower), \\; meanders=-8 rainstorm^{2} pinecones \\). So \\( galaxies=64 rainstorm^{3}[4(2 rainstorm-sunflower)^{3}+27 rainstorm \\, pinecones^{2}] \\). The problem apparently intends that we take \\( rainstorm>0 \\) (as is usual with the standard form of the parabola). With this assumption, \\( galaxies \\) has the same sign as\n\\[\n\\galaxies^{\\prime}=4(2 rainstorm-sunflower)^{3}+27 rainstorm \\, pinecones^{2}\n\\]\n\nThe roots of (1), and hence the points of the parabola, will be real and distinct if and only if \\( galaxies^{\\prime}<0 \\). Two normals will coincide if and only if \\( galaxies^{\\prime}=0 \\); that is, \\( (sunflower, pinecones) \\) lies on the curve \\( 27 rainstorm dolphins^{2}=4(lanterns-2 rainstorm)^{3} \\). Finally, all three normals will coincide if and only if \\( 4 rainstorm(2 rainstorm-sunflower)=0 \\) and \\( 8 rainstorm^{2} pinecones=0 \\), i.e., if and only if \\( (sunflower, pinecones)=(2 rainstorm, 0) \\).\n\nThe algebraic facts stated above for the roots of \\( dolphins^{3}+orchards \\, dolphins+meanders=0 \\) can be derived by the methods of calculus as follows.\n\nThe graph of \\( dolphins^{3}+orchards \\, dolphins \\) will be increasing with strictly positive slope if \\( orchards>0 \\), so any equation \\( dolphins^{3}+orchards \\, dolphins=-meanders \\) will have just one real root and therefore two complex roots. Thus the roots are distinct if \\( orchards>0 \\).\n\nIf \\( orchards=0 \\), the equation will have one real and two complex roots for \\( meanders \\neq 0 \\), and a triple root at 0 if \\( meanders=0 \\) also.\n\nFor \\( orchards<0 \\), the graph has turning points at\n\\[\n\\left(-\\sqrt{-\\frac{orchards}{3}},-\\frac{2}{3} \\, orchards \\sqrt{-\\frac{orchards}{3}}\\right) \\quad \\text { and } \\quad\\left(+\\sqrt{-\\frac{orchards}{3}}, \\frac{2}{3} \\, orchards \\sqrt{-\\frac{orchards}{3}}\\right) .\n\\]\n\nIt is then clear that the equation will have three distinct real roots if\n\\[\n|meanders|<\\left|\\frac{2}{3} \\, orchards \\sqrt{-\\frac{orchards}{3}}\\right|,\n\\]\none double and one single root if\n\\[\n|meanders|=\\left|\\frac{2}{3} \\, orchards \\sqrt{-\\frac{orchards}{3}}\\right|,\n\\]\nand one real and two complex roots if\n\\[\n|meanders|>\\left|\\frac{2}{3} \\, orchards \\sqrt{-\\frac{orchards}{3}}\\right| .\n\\]\n\nSquaring these relations we can write them in terms of \\( galaxies=27 \\, meanders^{2}+4 \\, orchards^{3} \\), and we have\n\\[\n\\begin{aligned}\n\\galaxies<0 &\\Rightarrow \\text { three distinct real roots, } \\\\\n\\galaxies=0 &\\Rightarrow \\text { (at least) two equal real roots, } \\\\\n\\galaxies>0 &\\Rightarrow \\text { one real and two complex roots, } \\\\\norchards=meanders=0 &\\Leftrightarrow \\text { a triple root. }\n\\end{aligned}\n\\]\n\nThese facts are often derived algebraically by showing that \\( -\\galaxies \\), which is called the discriminant of the cubic, is the product of the squares of the differences of the roots. See any text on the theory of equations."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "A": "negativefactor",
        "B": "variablefactor",
        "\\\\Delta": "equilibrium",
        "a": "terminalvalue",
        "b": "centralvalue",
        "p": "randomvariable"
      },
      "question": "From any point \\( (terminalvalue, centralvalue) \\) in the Cartesian plane, show that (i) three normals, real or imaginary, can be drawn to the parabola \\( horizontalaxis^{2}=4 randomvariable verticalaxis \\); (ii) these are real and distinct if \\( 4(2 randomvariable-terminalvalue)^{3}+27 randomvariable centralvalue^{2}<0 \\); (iii) two of them coincide if \\( (terminalvalue, centralvalue) \\) lies on the curve \\( 27 randomvariable horizontalaxis^{2}=4(verticalaxis-2 randomvariable)^{3} \\); (iv) all three coincide only if \\( (terminalvalue, centralvalue) \\) is the point \\( (2 randomvariable, 0) \\).",
      "solution": "Solution. The slope of the parabola at the point \\( (verticalaxis, horizontalaxis) \\) is \\( 2 randomvariable / horizontalaxis \\). Hence the line joining \\( (terminalvalue, centralvalue) \\) to \\( (verticalaxis, horizontalaxis) \\) is normal to the parabola at \\( (verticalaxis, horizontalaxis) \\) if and only if\n\\[\n(horizontalaxis-centralvalue)=-\\frac{horizontalaxis}{2 randomvariable}(verticalaxis-terminalvalue)=-\\frac{horizontalaxis}{2 randomvariable}\\left(\\frac{horizontalaxis^{2}}{4 randomvariable}-terminalvalue\\right)\n\\]\n(The case \\( horizontalaxis=0 \\) is also covered by this equation.) This reduces to\n\\[\nhorizontalaxis^{3}+4 randomvariable(2 randomvariable-terminalvalue) horizontalaxis-8 randomvariable^{2} centralvalue=0\n\\]\n\nThis cubic will have three roots if counted with multiplicity, the roots may be real or complex. Since (real or complex) values of \\( horizontalaxis \\) correspond one to one with (real or imaginary) points on the parabola, part (i) of the problem has been established.\n\nA real cubic equation of the form \\( horizontalaxis^{3}+negativefactor horizontalaxis+variablefactor=0 \\) will have all real roots if and only if \\( equilibrium=27 variablefactor^{2}+4 negativefactor^{3} \\leq 0 \\). Two roots will coincide if and only if \\( equilibrium=0 \\), and all three roots will coincide if and only if \\( negativefactor=variablefactor=0 \\).\n\nIn the present case, \\( negativefactor=4 randomvariable(2 randomvariable-terminalvalue),\\; variablefactor=-8 randomvariable^{2} centralvalue \\). So \\( equilibrium=64 randomvariable^{3}[4(2 randomvariable \\) \\( \\left.-terminalvalue)^{3}+27 randomvariable centralvalue^{2}\\right] \\). The problem apparently intends that we take \\( randomvariable>0 \\) (as is usual with the standard form of the parabola). With this assumption, \\( equilibrium \\) has the same sign as\n\\[\nequilibrium^{\\prime}=4(2 randomvariable-terminalvalue)^{3}+27 randomvariable centralvalue^{2}\n\\]\n\nThe roots of (1), and hence the points of the parabola, will be real and distinct if and only if \\( equilibrium^{\\prime}<0 \\). Two normals will coincide if and only if \\( equilibrium^{\\prime}=0 \\); that is, \\( (terminalvalue, centralvalue) \\) lies on the curve \\( 27 randomvariable horizontalaxis^{2}=4(verticalaxis-2 randomvariable)^{3} \\). Finally, all three normals will coincide if and only if \\( 4 randomvariable(2 randomvariable-terminalvalue)=0 \\) and \\( 8 randomvariable^{2} centralvalue=0 \\), i.e., if and only if \\( (terminalvalue, centralvalue)=(2 randomvariable, 0) \\).\n\nThe algebraic facts stated above for the roots of \\( horizontalaxis^{3}+negativefactor horizontalaxis+variablefactor=0 \\) can be derived by the methods of calculus as follows.\n\nThe graph of \\( horizontalaxis^{3}+negativefactor horizontalaxis \\) will be increasing with strictly positive slope if \\( negativefactor>0 \\), so any equation \\( horizontalaxis^{3}+negativefactor horizontalaxis=-variablefactor \\) will have just one real root and therefore two complex roots. Thus the roots are distinct if \\( negativefactor>0 \\).\n\nIf \\( negativefactor=0 \\), the equation will have one real and two complex roots for \\( variablefactor \\neq 0 \\), and a triple root at 0 if \\( variablefactor=0 \\) also.\n\nFor \\( negativefactor<0 \\), the graph has turning points at\n\\[\n\\left(-\\sqrt{-\\frac{negativefactor}{3}},-\\frac{2}{3} negativefactor \\sqrt{-\\frac{negativefactor}{3}}\\right) \\quad \\text { and } \\quad\\left(+\\sqrt{-\\frac{negativefactor}{3}}, \\frac{2}{3} negativefactor \\sqrt{-\\frac{negativefactor}{3}}\\right) .\n\\]\n\nIt is then clear that the equation will have three distinct real roots if\n\\[\n|variablefactor|<\\left|\\frac{2}{3} negativefactor \\sqrt{-\\frac{negativefactor}{3}}\\right|,\n\\]\none double and one single root if\n\\[\n|variablefactor|=\\left|\\frac{2}{3} negativefactor \\sqrt{-\\frac{negativefactor}{3}}\\right|,\n\\]\nand one real and two complex roots if\n\\[\n|variablefactor|>\\left|\\frac{2}{3} negativefactor \\sqrt{-\\frac{negativefactor}{3}}\\right| .\n\\]\n\nSquaring these relations we can write them in terms of \\( equilibrium=27 variablefactor^{2}+4 negativefactor^{3} \\), and we have\n\\[\n\\begin{aligned}\nequilibrium<0 \\Rightarrow \\text { three distinct real roots, } \\\\\nequilibrium=0 \\Rightarrow \\text { (at least) two equal real roots, } \\\\\nequilibrium>0 \\Rightarrow \\text { one real and two complex roots, } \\\\\nnegativefactor=variablefactor=0 \\Leftrightarrow \\text { a triple root. }\n\\end{aligned}\n\\]\n\nThese facts are often derived algebraically by showing that \\( -equilibrium \\), which is called the discriminant of the cubic, is the product of the squares of the differences of the roots. See any text on the theory of equations."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "A": "mpldqezi",
        "B": "vstokhya",
        "\\Delta": "rcqusbne",
        "a": "fnobrikc",
        "b": "telawmzi",
        "p": "goshtvra"
      },
      "question": "11. From any point \\( (fnobrikc, telawmzi) \\) in the Cartesian plane, show that (i) three normals, real or imaginary, can be drawn to the parabola \\( hjgrksla^{2}=4 goshtvra qzxwvtnp \\); (ii) these are real and distinct if \\( 4(2 goshtvra-fnobrikc)^{3}+27 goshtvra telawmzi^{2}<0 \\); (iii) two of them coincide if \\( (fnobrikc, telawmzi) \\) lies on the curve \\( 27 goshtvra hjgrksla^{2}=4(qzxwvtnp-2 goshtvra)^{3} \\); (iv) all three coincide only if \\( (fnobrikc, telawmzi) \\) is the point \\( (2 goshtvra, 0) \\).",
      "solution": "Solution. The slope of the parabola at the point \\( (qzxwvtnp, hjgrksla) \\) is \\( 2 goshtvra / hjgrksla \\). Hence the line joining \\( (fnobrikc, telawmzi) \\) to \\( (qzxwvtnp, hjgrksla) \\) is normal to the parabola at \\( (qzxwvtnp, hjgrksla) \\) if and only if\n\\[\n(hjgrksla-telawmzi)=-\\frac{hjgrksla}{2 goshtvra}(qzxwvtnp-fnobrikc)=-\\frac{hjgrksla}{2 goshtvra}\\left(\\frac{hjgrksla^{2}}{4 goshtvra}-fnobrikc\\right)\n\\]\n(The case \\( hjgrksla=0 \\) is also covered by this equation.) This reduces to\n\\[\nhjgrksla^{3}+4 goshtvra(2 goshtvra-fnobrikc) hjgrksla-8 goshtvra^{2} telawmzi=0\n\\]\n\nThis cubic will have three roots if counted with multiplicity, the roots may be real or complex. Since (real or complex) values of \\( hjgrksla \\) correspond one to one with (real or imaginary) points on the parabola, part (i) of the problem has been established.\n\nA real cubic equation of the form \\( hjgrksla^{3}+mpldqezi hjgrksla+vstokhya=0 \\) will have all real roots if and only if \\( rcqusbne=27 vstokhya^{2}+4 mpldqezi^{3} \\leq 0 \\). Two roots will coincide if and only if \\( rcqusbne=0 \\), and all three roots will coincide if and only if \\( mpldqezi=vstokhya=0 \\).\n\nIn the present case, \\( mpldqezi=4 goshtvra(2 goshtvra-fnobrikc), \\; vstokhya=-8 goshtvra^{2} telawmzi \\). So \\( rcqusbne=64 goshtvra^{3}[4(2 goshtvra-fnobrikc)^{3}+27 goshtvra telawmzi^{2}] \\). The problem apparently intends that we take \\( goshtvra>0 \\) (as is usual with the standard form of the parabola). With this assumption, \\( rcqusbne \\) has the same sign as\n\\[\nrcqusbne^{\\prime}=4(2 goshtvra-fnobrikc)^{3}+27 goshtvra telawmzi^{2}\n\\]\n\nThe roots of (1), and hence the points of the parabola, will be real and distinct if and only if \\( rcqusbne^{\\prime}<0 \\). Two normals will coincide if and only if \\( rcqusbne^{\\prime}=0 \\); that is, \\( (fnobrikc, telawmzi) \\) lies on the curve \\( 27 goshtvra hjgrksla^{2}=4(qzxwvtnp-2 goshtvra)^{3} \\). Finally, all three normals will coincide if and only if \\( 4 goshtvra(2 goshtvra-fnobrikc)=0 \\) and \\( 8 goshtvra^{2} telawmzi=0 \\), i.e., if and only if \\( (fnobrikc, telawmzi)=(2 goshtvra, 0) \\).\n\nThe algebraic facts stated above for the roots of \\( hjgrksla^{3}+mpldqezi hjgrksla+vstokhya=0 \\) can be derived by the methods of calculus as follows.\n\nThe graph of \\( hjgrksla^{3}+mpldqezi hjgrksla \\) will be increasing with strictly positive slope if \\( mpldqezi>0 \\), so any equation \\( hjgrksla^{3}+mpldqezi hjgrksla=-vstokhya \\) will have just one real root and therefore two complex roots. Thus the roots are distinct if \\( mpldqezi>0 \\).\n\nIf \\( mpldqezi=0 \\), the equation will have one real and two complex roots for \\( vstokhya \\neq 0 \\), and a triple root at 0 if \\( vstokhya=0 \\) also.\n\nFor \\( mpldqezi<0 \\), the graph has turning points at\n\\[\n\\left(-\\sqrt{-\\frac{mpldqezi}{3}},-\\frac{2}{3} mpldqezi \\sqrt{-\\frac{mpldqezi}{3}}\\right) \\quad \\text { and } \\quad\\left(+\\sqrt{-\\frac{mpldqezi}{3}}, \\frac{2}{3} mpldqezi \\sqrt{-\\frac{mpldqezi}{3}}\\right) .\n\\]\n\nIt is then clear that the equation will have three distinct real roots if\n\\[\n|vstokhya|<\\left|\\frac{2}{3} mpldqezi \\sqrt{-\\frac{mpldqezi}{3}}\\right|,\n\\]\none double and one single root if\n\\[\n|vstokhya|=\\left|\\frac{2}{3} mpldqezi \\sqrt{-\\frac{mpldqezi}{3}}\\right|,\n\\]\nand one real and two complex roots if\n\\[\n|vstokhya|>\\left|\\frac{2}{3} mpldqezi \\sqrt{-\\frac{mpldqezi}{3}}\\right| .\n\\]\n\nSquaring these relations we can write them in terms of \\( rcqusbne=27 vstokhya^{2}+4 mpldqezi^{3} \\), and we have\n\\[\n\\begin{aligned}\nrcqusbne<0 &\\Rightarrow \\text { three distinct real roots, } \\\\\nrcqusbne=0 &\\Rightarrow \\text { (at least) two equal real roots, } \\\\\nrcqusbne>0 &\\Rightarrow \\text { one real and two complex roots, } \\\\\nmpldqezi=vstokhya=0 &\\Leftrightarrow \\text { a triple root. }\n\\end{aligned}\n\\]\n\nThese facts are often derived algebraically by showing that \\( -rcqusbne \\), which is called the discriminant of the cubic, is the product of the squares of the differences of the roots. See any text on the theory of equations."
    },
    "kernel_variant": {
      "question": "Fix a negative real constant c and consider the left–opening parabola  \n\n  Γ ⊂ ℝ² :  y² = 4 c x   (c < 0).  \n\nFor an arbitrary point P =(u , v) ∈ ℝ² investigate the (real or complex) straight lines that are normal to Γ and pass through P.\nPut  \n\n  Δ(P) = 4(2c − u)³ + 27 c v²                (the Δ–function)  \n\nand keep the notation throughout.\n\nFive assertions lead from the elementary “three–normals’’ fact to a rich\npicture involving projective duality, evolutes, singularity theory and small\nanalytic perturbations.\n\n(i) (Existence in the complex projective line)  \nShow that, counted with multiplicity, exactly three complex normal lines pass through P; equivalently, prove that the feet ζ of those normals are precisely the three (possibly coincident, possibly non–real) roots of  \n\n  ζ³ + 4 c(2c − u) ζ − 8 c² v = 0.         (★)\n\n(ii) (Evolute of the parabola)  \nProve that the cubic curve  \n\n  H : 27 c y² = 4(x − 2c)³             (♣)\n\nis the evolute of Γ, i.e. the locus of the centres of curvature of Γ.  \nIn particular, show that for every Q ∈ H there exists a unique circle having third–order contact with Γ and whose centre is Q.  \nCheck explicitly that the cusp of H is the point (2c,0).\n\n(iii) (Real analytic stratification and an algebraic area invariant)  \n(a) Let D(P) be the discriminant of (★).  Show that  \n\n  D(P) = −64 c³ Δ(P)\n\nand hence, because c < 0, D(P) and Δ(P) have the same sign.  Conclude the\nreality pattern of the normals:\n\n  Δ(P) > 0 ⇔ three distinct real normals,  \n  Δ(P) = 0 ⇔ at least two coincident real normals,  \n  Δ(P) < 0 ⇔ exactly one real normal.\n\n(b) Let Y₁,Y₂,Y₃ be the (complex) feet of the three normals from P (counting multiplicities) and set  \n\n  A(P) = ½ det\n  ⎛⎝1 x₁ y₁\n   1 x₂ y₂\n   1 x₃ y₃⎞⎠            (†)\n\n(the oriented area of the triangle determined by the feet).  \nProve the identity  \n\n  A(P)² = – c Δ(P).            (‡)\n\nIn particular, Δ(P)=0 ⇔ A(P)=0 ⇔ the three feet are collinear ⇔ P ∈ H.\n\n(iv) (Projective dual curve of Γ)  \nEmbed the discussion in the complex projective plane ℙ²(ℂ).  \nShow that the set Γ* of all (complex) normal lines to Γ is the rational cuspidal cubic  \n\n  c s³ + 2 c s t² + u t² = 0    in dual (Plücker) coordinates [s:t:u]≠0,  (Δ)\n\nand that the naturally induced normal map  \n\n  ν : Γ → Γ*, X ↦ normal line at X,\n\nhas degree three.  Consequently, every line of Γ* not passing through the cusp [0:0:1] is normal to Γ at three (possibly coincident) points.\n\n(v) (Stability under analytic perturbations)  \nLet f be a real analytic function with f(0)=f′(0)=0 and consider the ε–perturbed parabola  \n\n  Γ_ε : y² = 4c x + ε f(x)  (ε ∈ ℝ, |ε| « 1).\n\nFix P∈ℝ² and keep Δ(P) as above.\n\n(a) If Δ(P) ≠ 0 (i.e. P ∉ H) prove that there exists ε₀>0 such that for |ε|<ε₀ the equation of the normals to Γ_ε through P still possesses exactly three distinct complex solutions; the reality pattern is unchanged.\n\n(b) If Δ(P)=0 but P ≠ (2c,0) (P a regular point of H) show that the triple of normals splits analytically: there is an explicit real–analytic function g(P) (computed in the solution) such that  \n\n  Δ_ε(P) = Δ(P) + ε g(P) + O(ε²) = ε g(P) + O(ε²).\n\nFor ε g(P)>0 the three perturbed normals are real and distinct, whereas for ε g(P)<0 only one real normal survives.\n\n(c) Analyse separately the cusp P₀=(2c,0): prove that Δ_ε(P₀)=O(ε²) and describe the local unfolding of the fivefold contact between Γ and its evolute at P₀.\n\n\n\n──────────────────────────────────────────────",
      "solution": "Throughout c<0 is fixed and\n\n  X(y) = (y²/(4c) , y)  (y ∈ ℂ)\n\nparametrises Γ.\n\n(i) Number of complex normals  \nThe tangent vector at X(y) is T(y) = (y/(2c) , 1).\nA line through P = (u,v) is normal at X(y) iff (P−X(y))·T(y)=0, i.e.  \n\n  (u − y²/4c)(y/2c) + (v − y) = 0  \n ⇔ y³ + 4 c(2c − u) y − 8 c² v = 0,  \n\nwhich is exactly (★).  As the cubic is monic it has three roots\nζ₁,ζ₂,ζ₃ in ℂ, counted with multiplicity, and each root produces one normal line through P. Hence exactly three complex normals exist.\n\n(ii) The evolute  \n1. Curvature and normal.  With parameter y one has\n\n x′ = y/(2c), y′ = 1, x″ = 1/(2c), y″ = 0,\n\nso the curvature is\n\n κ(y)= |x′y″−y′x″| / (x′²+y′²)^{3/2}\n    = (1/(2|c|)) · (8|c|³)/(y²+4c²)^{3/2}\n    = 4|c|² /(y²+4c²)^{3/2}.                     (1)\n\nThe unit normal (obtained by rotating the unit tangent through +π/2) is  \n\n n(y)= ( 2c , –y ) / √(y²+4c²).                   (2)\n\n2. Centre of curvature.  \nIts position vector is\n\n C(y)=X(y)+κ(y)^{-1} n(y)\n    =(y²/(4c)+ (y²+4c²)/(2c), y−y(y²+4c²)/(4c²))             (3)\n    =( 2c + 3y²/(4c) ,  −y³/(4c²) ).                          (4)\n\n3. Elimination of y.  \nSetting (x,y) = C(y) and eliminating y between (4) gives\n\n 27 c y² = 4(x − 2c)³,\n\nwhich is (♣).  Each y corresponds to one point of H and vice-versa, hence H is the evolute of Γ.  By construction the circle with centre C(y) and radius ρ(y)=1/κ(y) satisfies third–order contact with Γ at X(y) and is unique.\n\n(iii) Reality pattern and the algebraic area invariant  \n\n(a) Reality.  \nWrite (★) as Y³ + A Y + B = 0 with\n\n A = 4c(2c − u), B = −8c² v.\n\nThe discriminant of a depressed cubic is\n\n D = −4A³ − 27B² = −64 c³ Δ(P).                     (5)\n\nBecause c<0, the factor −64 c³ is positive, hence sign(D)=sign(Δ).  \nFor a monic cubic:\n\n D>0  ⇔ three distinct real roots,  \n D=0  ⇔ at least two coincident real roots,  \n D<0  ⇔ exactly one real root.\n\nThis gives the announced classification in terms of Δ(P).\n\n(b) Algebraic area.  \nLet y₁,y₂,y₃ be the roots of (★).\nFrom Vieta,\n\n y₁+y₂+y₃ = 0, \n y₁y₂+y₂y₃+y₃y₁ = A, \n y₁y₂y₃ = −B.                                      (6)\n\nPut x_i = y_i²/(4c).  As in the original computation\n\n det(1,x_i,y_i) = −(y₁−y₂)(y₂−y₃)(y₃−y₁)/(4c).    (7)\n\nHence\n\n A(P)² = 1/(64 c²) · (y₁−y₂)²(y₂−y₃)²(y₃−y₁)²\n        = D /(64 c²)  (using (5))                  (8)\n        = – c Δ(P).                                (‡)\n\nBecause –c > 0, Δ(P)=0 ⇔ A(P)=0, so the three feet are collinear\nexactly when P lies on the evolute H.\n\n(iv) The projective dual (normal) curve Γ*  \n\n1. Plücker coordinates of a normal.  \nThe affine normal at X(y) has equation\n\n (y − y) = −y/(2c) (x − y²/4c)\n ⇔ (y/(2c)) x − y + (−y − y³/(8c²)) = 0.\n\nUp to scale the Plücker coordinates are therefore  \n\n [s : t : u] = [ y : 2c : −2c y − y³/(4c) ].          (9)\n\n2. Eliminating y gives\n\n c s³ + 2c s t² + u t² = 0,                          (Δ)\n\nan irreducible rational cubic with a unique cusp at [0:0:1].  \nHence Γ* is rational and cuspidal; the map y ↦ (9) has degree three, so\na generic line of Γ* is normal to Γ at three points.\n\n(v) Analytic stability under small perturbations  \n\nLet F_ε(x,y)=y²−4c x−ε f(x).  A point (x,y) lies on Γ_ε⇔F_ε=0, and\n∇F_ε=(−4c−ε f′(x), 2y).\nRepeating the calculation in (i) with these quantities produces\n\n y³ + 4c(2c − u) y − 8c² v + ε G₁(y;P) + ε² G₂(y;P,ε) = 0,   (★_ε)\n\nwhere G₁ is a polynomial of degree ≤2 in y whose coefficients are\nreal–analytic in P and depend on f,f′,f″.  The discriminant of (★_ε)\nis an analytic function of ε:\n\n Disc_ε(P) = Δ(P) + ε g(P) + O(ε²).                (10)\n\nA direct differentiation of the classical discriminant formula yields\n\n g(P)= –32 c (2c−u) f′(u) – 27 c v² f″(u).           (11)\n\n(a) If Δ(P)≠0.  \nThen Disc_ε(P)=Δ(P)+O(ε) never vanishes for |ε|<ε₀ small, so the\nthree normals remain simple; their real/complex nature does not change.\n\n(b) If Δ(P)=0 and P≠(2c,0).  \nNow Disc_ε(P)=ε g(P)+O(ε²).  For g(P)≠0 the sign of ε g(P) decides:\n\n ε g(P) > 0 ⇔ Disc_ε > 0 ⇔ three distinct real normals,  \n ε g(P) < 0 ⇔ Disc_ε < 0 ⇔ only one real normal.\n\n(c) The cusp P₀=(2c,0).  \nHere y=0 is a triple root of the unperturbed cubic; a short computation\nshows Disc_ε(P₀)=O(ε²).  Under a generic perturbation Γ and its evolute\nsplit into a 1+2 configuration, modelling the A₂–singularity.\n\nThus each item of the corrected statement is proved.\n\n\n\n──────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.376137",
        "was_fixed": false,
        "difficulty_analysis": "• Part (i) is essentially the original statement but couched in projective-complex language to force the solver to manipulate homogeneous objects and multiplicities.  \n\n• Part (ii) introduces differential geometry (curvature, evolute, osculating circle) and demands non-trivial elimination to obtain equation (♣).  \n\n• Part (iii) links classical discriminant theory with a geometric invariant (oriented area), requiring skill in both algebraic symmetrics and planar geometry.  \n\n• Part (iv) moves the problem into the dual projective plane, requiring knowledge of Plücker coordinates, rational parametrisations, and singularity analysis of the cuspidal cubic.  \n\n• Part (v) adds an analytic-perturbation layer: the solver must produce a stability argument via the Implicit-Function Theorem and compute the first variation of the discriminant.\n\nAll these layers greatly exceed the scope of the original exercise, demand several advanced techniques (differential geometry, algebraic geometry, singularity theory, real analytic perturbation), and intertwine them in a single coherent problem, making the enhanced variant significantly harder than both the original and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Fix a negative real constant c and consider the left–opening parabola  \n\n  Γ ⊂ ℝ² :  y² = 4 c x   (c < 0).  \n\nFor an arbitrary point P =(u , v) ∈ ℝ² investigate the (real or complex) straight lines that are normal to Γ and pass through P.\nPut  \n\n  Δ(P) = 4(2c − u)³ + 27 c v²                (the Δ–function)  \n\nand keep the notation throughout.\n\nFive assertions lead from the elementary “three–normals’’ fact to a rich\npicture involving projective duality, evolutes, singularity theory and small\nanalytic perturbations.\n\n(i) (Existence in the complex projective line)  \nShow that, counted with multiplicity, exactly three complex normal lines pass through P; equivalently, prove that the feet ζ of those normals are precisely the three (possibly coincident, possibly non–real) roots of  \n\n  ζ³ + 4 c(2c − u) ζ − 8 c² v = 0.         (★)\n\n(ii) (Evolute of the parabola)  \nProve that the cubic curve  \n\n  H : 27 c y² = 4(x − 2c)³             (♣)\n\nis the evolute of Γ, i.e. the locus of the centres of curvature of Γ.  \nIn particular, show that for every Q ∈ H there exists a unique circle having third–order contact with Γ and whose centre is Q.  \nCheck explicitly that the cusp of H is the point (2c,0).\n\n(iii) (Real analytic stratification and an algebraic area invariant)  \n(a) Let D(P) be the discriminant of (★).  Show that  \n\n  D(P) = −64 c³ Δ(P)\n\nand hence, because c < 0, D(P) and Δ(P) have the same sign.  Conclude the\nreality pattern of the normals:\n\n  Δ(P) > 0 ⇔ three distinct real normals,  \n  Δ(P) = 0 ⇔ at least two coincident real normals,  \n  Δ(P) < 0 ⇔ exactly one real normal.\n\n(b) Let Y₁,Y₂,Y₃ be the (complex) feet of the three normals from P (counting multiplicities) and set  \n\n  A(P) = ½ det\n  ⎛⎝1 x₁ y₁\n   1 x₂ y₂\n   1 x₃ y₃⎞⎠            (†)\n\n(the oriented area of the triangle determined by the feet).  \nProve the identity  \n\n  A(P)² = – c Δ(P).            (‡)\n\nIn particular, Δ(P)=0 ⇔ A(P)=0 ⇔ the three feet are collinear ⇔ P ∈ H.\n\n(iv) (Projective dual curve of Γ)  \nEmbed the discussion in the complex projective plane ℙ²(ℂ).  \nShow that the set Γ* of all (complex) normal lines to Γ is the rational cuspidal cubic  \n\n  c s³ + 2 c s t² + u t² = 0    in dual (Plücker) coordinates [s:t:u]≠0,  (Δ)\n\nand that the naturally induced normal map  \n\n  ν : Γ → Γ*, X ↦ normal line at X,\n\nhas degree three.  Consequently, every line of Γ* not passing through the cusp [0:0:1] is normal to Γ at three (possibly coincident) points.\n\n(v) (Stability under analytic perturbations)  \nLet f be a real analytic function with f(0)=f′(0)=0 and consider the ε–perturbed parabola  \n\n  Γ_ε : y² = 4c x + ε f(x)  (ε ∈ ℝ, |ε| « 1).\n\nFix P∈ℝ² and keep Δ(P) as above.\n\n(a) If Δ(P) ≠ 0 (i.e. P ∉ H) prove that there exists ε₀>0 such that for |ε|<ε₀ the equation of the normals to Γ_ε through P still possesses exactly three distinct complex solutions; the reality pattern is unchanged.\n\n(b) If Δ(P)=0 but P ≠ (2c,0) (P a regular point of H) show that the triple of normals splits analytically: there is an explicit real–analytic function g(P) (computed in the solution) such that  \n\n  Δ_ε(P) = Δ(P) + ε g(P) + O(ε²) = ε g(P) + O(ε²).\n\nFor ε g(P)>0 the three perturbed normals are real and distinct, whereas for ε g(P)<0 only one real normal survives.\n\n(c) Analyse separately the cusp P₀=(2c,0): prove that Δ_ε(P₀)=O(ε²) and describe the local unfolding of the fivefold contact between Γ and its evolute at P₀.\n\n\n\n──────────────────────────────────────────────",
      "solution": "Throughout c<0 is fixed and\n\n  X(y) = (y²/(4c) , y)  (y ∈ ℂ)\n\nparametrises Γ.\n\n(i) Number of complex normals  \nThe tangent vector at X(y) is T(y) = (y/(2c) , 1).\nA line through P = (u,v) is normal at X(y) iff (P−X(y))·T(y)=0, i.e.  \n\n  (u − y²/4c)(y/2c) + (v − y) = 0  \n ⇔ y³ + 4 c(2c − u) y − 8 c² v = 0,  \n\nwhich is exactly (★).  As the cubic is monic it has three roots\nζ₁,ζ₂,ζ₃ in ℂ, counted with multiplicity, and each root produces one normal line through P. Hence exactly three complex normals exist.\n\n(ii) The evolute  \n1. Curvature and normal.  With parameter y one has\n\n x′ = y/(2c), y′ = 1, x″ = 1/(2c), y″ = 0,\n\nso the curvature is\n\n κ(y)= |x′y″−y′x″| / (x′²+y′²)^{3/2}\n    = (1/(2|c|)) · (8|c|³)/(y²+4c²)^{3/2}\n    = 4|c|² /(y²+4c²)^{3/2}.                     (1)\n\nThe unit normal (obtained by rotating the unit tangent through +π/2) is  \n\n n(y)= ( 2c , –y ) / √(y²+4c²).                   (2)\n\n2. Centre of curvature.  \nIts position vector is\n\n C(y)=X(y)+κ(y)^{-1} n(y)\n    =(y²/(4c)+ (y²+4c²)/(2c), y−y(y²+4c²)/(4c²))             (3)\n    =( 2c + 3y²/(4c) ,  −y³/(4c²) ).                          (4)\n\n3. Elimination of y.  \nSetting (x,y) = C(y) and eliminating y between (4) gives\n\n 27 c y² = 4(x − 2c)³,\n\nwhich is (♣).  Each y corresponds to one point of H and vice-versa, hence H is the evolute of Γ.  By construction the circle with centre C(y) and radius ρ(y)=1/κ(y) satisfies third–order contact with Γ at X(y) and is unique.\n\n(iii) Reality pattern and the algebraic area invariant  \n\n(a) Reality.  \nWrite (★) as Y³ + A Y + B = 0 with\n\n A = 4c(2c − u), B = −8c² v.\n\nThe discriminant of a depressed cubic is\n\n D = −4A³ − 27B² = −64 c³ Δ(P).                     (5)\n\nBecause c<0, the factor −64 c³ is positive, hence sign(D)=sign(Δ).  \nFor a monic cubic:\n\n D>0  ⇔ three distinct real roots,  \n D=0  ⇔ at least two coincident real roots,  \n D<0  ⇔ exactly one real root.\n\nThis gives the announced classification in terms of Δ(P).\n\n(b) Algebraic area.  \nLet y₁,y₂,y₃ be the roots of (★).\nFrom Vieta,\n\n y₁+y₂+y₃ = 0, \n y₁y₂+y₂y₃+y₃y₁ = A, \n y₁y₂y₃ = −B.                                      (6)\n\nPut x_i = y_i²/(4c).  As in the original computation\n\n det(1,x_i,y_i) = −(y₁−y₂)(y₂−y₃)(y₃−y₁)/(4c).    (7)\n\nHence\n\n A(P)² = 1/(64 c²) · (y₁−y₂)²(y₂−y₃)²(y₃−y₁)²\n        = D /(64 c²)  (using (5))                  (8)\n        = – c Δ(P).                                (‡)\n\nBecause –c > 0, Δ(P)=0 ⇔ A(P)=0, so the three feet are collinear\nexactly when P lies on the evolute H.\n\n(iv) The projective dual (normal) curve Γ*  \n\n1. Plücker coordinates of a normal.  \nThe affine normal at X(y) has equation\n\n (y − y) = −y/(2c) (x − y²/4c)\n ⇔ (y/(2c)) x − y + (−y − y³/(8c²)) = 0.\n\nUp to scale the Plücker coordinates are therefore  \n\n [s : t : u] = [ y : 2c : −2c y − y³/(4c) ].          (9)\n\n2. Eliminating y gives\n\n c s³ + 2c s t² + u t² = 0,                          (Δ)\n\nan irreducible rational cubic with a unique cusp at [0:0:1].  \nHence Γ* is rational and cuspidal; the map y ↦ (9) has degree three, so\na generic line of Γ* is normal to Γ at three points.\n\n(v) Analytic stability under small perturbations  \n\nLet F_ε(x,y)=y²−4c x−ε f(x).  A point (x,y) lies on Γ_ε⇔F_ε=0, and\n∇F_ε=(−4c−ε f′(x), 2y).\nRepeating the calculation in (i) with these quantities produces\n\n y³ + 4c(2c − u) y − 8c² v + ε G₁(y;P) + ε² G₂(y;P,ε) = 0,   (★_ε)\n\nwhere G₁ is a polynomial of degree ≤2 in y whose coefficients are\nreal–analytic in P and depend on f,f′,f″.  The discriminant of (★_ε)\nis an analytic function of ε:\n\n Disc_ε(P) = Δ(P) + ε g(P) + O(ε²).                (10)\n\nA direct differentiation of the classical discriminant formula yields\n\n g(P)= –32 c (2c−u) f′(u) – 27 c v² f″(u).           (11)\n\n(a) If Δ(P)≠0.  \nThen Disc_ε(P)=Δ(P)+O(ε) never vanishes for |ε|<ε₀ small, so the\nthree normals remain simple; their real/complex nature does not change.\n\n(b) If Δ(P)=0 and P≠(2c,0).  \nNow Disc_ε(P)=ε g(P)+O(ε²).  For g(P)≠0 the sign of ε g(P) decides:\n\n ε g(P) > 0 ⇔ Disc_ε > 0 ⇔ three distinct real normals,  \n ε g(P) < 0 ⇔ Disc_ε < 0 ⇔ only one real normal.\n\n(c) The cusp P₀=(2c,0).  \nHere y=0 is a triple root of the unperturbed cubic; a short computation\nshows Disc_ε(P₀)=O(ε²).  Under a generic perturbation Γ and its evolute\nsplit into a 1+2 configuration, modelling the A₂–singularity.\n\nThus each item of the corrected statement is proved.\n\n\n\n──────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.324100",
        "was_fixed": false,
        "difficulty_analysis": "• Part (i) is essentially the original statement but couched in projective-complex language to force the solver to manipulate homogeneous objects and multiplicities.  \n\n• Part (ii) introduces differential geometry (curvature, evolute, osculating circle) and demands non-trivial elimination to obtain equation (♣).  \n\n• Part (iii) links classical discriminant theory with a geometric invariant (oriented area), requiring skill in both algebraic symmetrics and planar geometry.  \n\n• Part (iv) moves the problem into the dual projective plane, requiring knowledge of Plücker coordinates, rational parametrisations, and singularity analysis of the cuspidal cubic.  \n\n• Part (v) adds an analytic-perturbation layer: the solver must produce a stability argument via the Implicit-Function Theorem and compute the first variation of the discriminant.\n\nAll these layers greatly exceed the scope of the original exercise, demand several advanced techniques (differential geometry, algebraic geometry, singularity theory, real analytic perturbation), and intertwine them in a single coherent problem, making the enhanced variant significantly harder than both the original and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}