{
  "index": "1970-A-2",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "A-2. Consider the locus given by the real polynomial equation\n\\[\nA x^{2}+B x y+C y^{2}+D x^{3}+E x^{2} y+F x y^{2}+G y^{3}=0\n\\]\nwhere \\( B^{\\mathbf{s}}-4 A C<0 \\). Prove that there is a positive number \\( \\delta \\) such that there are no points of the locus in the punctured disk\n\\[\n0<x^{2}+y^{2}<\\delta^{2} .\n\\]",
  "solution": "A-2 Let \\( (x, y)=(r \\cos \\theta, r \\sin \\theta), r>0 \\), be a point of the locus. Then\n\\[\nr=\\frac{\\left|A \\cos ^{2} \\theta+B \\sin \\theta \\cos \\theta+C \\sin ^{2} \\theta\\right|}{\\left|D \\cos ^{3} \\theta+E \\cos ^{2} \\theta \\sin \\theta+F \\cos \\theta \\sin ^{2} \\theta+G \\sin ^{3} \\theta\\right|}\n\\]\n\nThe denominator of (1) is less than or equal to \\( |D|+|E|+|F|+|G| \\), whereas the numerator has a positive minimum\n\\[\nN=\\frac{|A+C|-\\sqrt{(A-C)^{2}+B^{2}}}{2}\n\\]\nsince \\( B^{2}<4 A C \\). Therefore\n\\[\nr \\geqq \\frac{N}{|D|+|E|+|F|+|G|}=\\delta\n\\]\nand there are no points of the locus within \\( 0<r<\\delta \\).\nAlternate Solution: Set \\( H(x, y) \\) equal to the polynomial on the left hand side of the given equation. The standard theory for maxima or minima of functions of two variables can be used together with the condition \\( B^{2}<4 A C \\) to show that \\( H(x, y) \\) has a local maximum or a local minimum at \\( (0,0) \\).",
  "vars": [
    "x",
    "y",
    "r",
    "\\\\theta"
  ],
  "params": [
    "A",
    "B",
    "C",
    "D",
    "E",
    "F",
    "G",
    "\\\\delta",
    "N",
    "H"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xcoord",
        "y": "ycoord",
        "r": "radius",
        "\\theta": "anglevar",
        "A": "firstcof",
        "B": "secondco",
        "C": "thirdcof",
        "D": "fourthco",
        "E": "fifthcoe",
        "F": "sixthcoe",
        "G": "seventhc",
        "\\delta": "deltaval",
        "N": "numerlow",
        "H": "polyfunc"
      },
      "question": "A-2. Consider the locus given by the real polynomial equation\n\\[\nfirstcof xcoord^{2}+secondco xcoord ycoord+thirdcof ycoord^{2}+fourthco xcoord^{3}+fifthcoe xcoord^{2} ycoord+sixthcoe xcoord ycoord^{2}+seventhc ycoord^{3}=0\n\\]\nwhere \\( secondco^{\\mathbf{s}}-4 firstcof thirdcof<0 \\). Prove that there is a positive number \\( deltaval \\) such that there are no points of the locus in the punctured disk\n\\[\n0<xcoord^{2}+ycoord^{2}<deltaval^{2} .\n\\]",
      "solution": "A-2 Let \\( (xcoord, ycoord)=(radius \\cos anglevar, radius \\sin anglevar), radius>0 \\), be a point of the locus. Then\n\\[\nradius=\\frac{\\left|firstcof \\cos ^{2} anglevar+secondco \\sin anglevar \\cos anglevar+thirdcof \\sin ^{2} anglevar\\right|}{\\left|fourthco \\cos ^{3} anglevar+fifthcoe \\cos ^{2} anglevar \\sin anglevar+sixthcoe \\cos anglevar \\sin ^{2} anglevar+seventhc \\sin ^{3} anglevar\\right|}\n\\]\n\nThe denominator of (1) is less than or equal to \\( |fourthco|+|fifthcoe|+|sixthcoe|+|seventhc| \\), whereas the numerator has a positive minimum\n\\[\nnumerlow=\\frac{|firstcof+thirdcof|-\\sqrt{(firstcof-thirdcof)^{2}+secondco^{2}}}{2}\n\\]\nsince \\( secondco^{2}<4 firstcof thirdcof \\). Therefore\n\\[\nradius \\geqq \\frac{numerlow}{|fourthco|+|fifthcoe|+|sixthcoe|+|seventhc|}=deltaval\n\\]\nand there are no points of the locus within \\( 0<radius<deltaval \\).\nAlternate Solution: Set \\( polyfunc(xcoord, ycoord) \\) equal to the polynomial on the left hand side of the given equation. The standard theory for maxima or minima of functions of two variables can be used together with the condition \\( secondco^{2}<4 firstcof thirdcof \\) to show that \\( polyfunc(xcoord, ycoord) \\) has a local maximum or a local minimum at \\( (0,0) \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "shoreline",
        "y": "driftwood",
        "r": "sandgrain",
        "\\\\theta": "sunhorizon",
        "A": "compassrose",
        "B": "shipanchor",
        "C": "windcurrent",
        "D": "tidalcrest",
        "E": "coralridge",
        "F": "seagullcry",
        "G": "lighthouse",
        "\\\\delta": "mistyharbor",
        "N": "barnacle",
        "H": "horizonline"
      },
      "question": "A-2. Consider the locus given by the real polynomial equation\n\\[\ncompassrose shoreline^{2}+shipanchor shoreline driftwood+windcurrent driftwood^{2}+tidalcrest shoreline^{3}+coralridge shoreline^{2} driftwood+seagullcry shoreline driftwood^{2}+lighthouse driftwood^{3}=0\n\\]\nwhere \\( shipanchor^{\\mathbf{s}}-4 compassrose windcurrent<0 \\). Prove that there is a positive number \\( mistyharbor \\) such that there are no points of the locus in the punctured disk\n\\[\n0<shoreline^{2}+driftwood^{2}<mistyharbor^{2} .\n\\]",
      "solution": "A-2 Let \\( (shoreline, driftwood)=(sandgrain \\cos sunhorizon, sandgrain \\sin sunhorizon), sandgrain>0 \\), be a point of the locus. Then\n\\[\nsandgrain=\\frac{\\left|compassrose \\cos ^{2} sunhorizon+shipanchor \\sin sunhorizon \\cos sunhorizon+windcurrent \\sin ^{2} sunhorizon\\right|}{\\left|tidalcrest \\cos ^{3} sunhorizon+coralridge \\cos ^{2} sunhorizon \\sin sunhorizon+seagullcry \\cos sunhorizon \\sin ^{2} sunhorizon+lighthouse \\sin ^{3} sunhorizon\\right|}\n\\]\n\nThe denominator of (1) is less than or equal to \\( |tidalcrest|+|coralridge|+|seagullcry|+|lighthouse| \\), whereas the numerator has a positive minimum\n\\[\nbarnacle=\\frac{|compassrose+windcurrent|-\\sqrt{(compassrose-windcurrent)^{2}+shipanchor^{2}}}{2}\n\\]\nsince \\( shipanchor^{2}<4 compassrose windcurrent \\). Therefore\n\\[\nsandgrain \\geqq \\frac{barnacle}{|tidalcrest|+|coralridge|+|seagullcry|+|lighthouse|}=mistyharbor\n\\]\nand there are no points of the locus within \\( 0<sandgrain<mistyharbor \\).\nAlternate Solution: Set \\( horizonline(shoreline, driftwood) \\) equal to the polynomial on the left hand side of the given equation. The standard theory for maxima or minima of functions of two variables can be used together with the condition \\( shipanchor^{2}<4 compassrose windcurrent \\) to show that \\( horizonline(shoreline, driftwood) \\) has a local maximum or a local minimum at \\( (0,0) \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "r": "diameter",
        "\\theta": "directionless",
        "A": "variableone",
        "B": "variabletwo",
        "C": "variablethree",
        "D": "variablefour",
        "E": "variablefive",
        "F": "variablesix",
        "G": "variableseven",
        "\\delta": "infinite",
        "N": "maximumvalue",
        "H": "linearfunc"
      },
      "question": "A-2. Consider the locus given by the real polynomial equation\n\\[\nvariableone\\ verticalaxis^{2}+variabletwo\\ verticalaxis\\ horizontalaxis+variablethree\\ horizontalaxis^{2}+variablefour\\ verticalaxis^{3}+variablefive\\ verticalaxis^{2}\\ horizontalaxis+variablesix\\ verticalaxis\\ horizontalaxis^{2}+variableseven\\ horizontalaxis^{3}=0\n\\]\nwhere \\( variabletwo^{\\mathbf{s}}-4\\ variableone\\ variablethree<0 \\). Prove that there is a positive number \\( infinite \\) such that there are no points of the locus in the punctured disk\n\\[\n0<verticalaxis^{2}+horizontalaxis^{2}<infinite^{2} .\n\\]",
      "solution": "A-2 Let \\( (verticalaxis, horizontalaxis)=(diameter \\cos directionless, diameter \\sin directionless), diameter>0 \\), be a point of the locus. Then\n\\[\ndiameter=\\frac{\\left|variableone \\cos ^{2} directionless+variabletwo \\sin directionless \\cos directionless+variablethree \\sin ^{2} directionless\\right|}{\\left|variablefour \\cos ^{3} directionless+variablefive \\cos ^{2} directionless \\sin directionless+variablesix \\cos directionless \\sin ^{2} directionless+variableseven \\sin ^{3} directionless\\right|}\n\\]\nThe denominator of (1) is less than or equal to \\( |variablefour|+|variablefive|+|variablesix|+|variableseven| \\), whereas the numerator has a positive minimum\n\\[\nmaximumvalue=\\frac{|variableone+variablethree|-\\sqrt{(variableone-variablethree)^{2}+variabletwo^{2}}}{2}\n\\]\nsince \\( variabletwo^{2}<4 variableone variablethree \\). Therefore\n\\[\ndiameter \\geqq \\frac{maximumvalue}{|variablefour|+|variablefive|+|variablesix|+|variableseven|}=infinite\n\\]\nand there are no points of the locus within \\( 0<diameter<infinite \\).\n\nAlternate Solution: Set \\( linearfunc(verticalaxis, horizontalaxis) \\) equal to the polynomial on the left hand side of the given equation. The standard theory for maxima or minima of functions of two variables can be used together with the condition \\( variabletwo^{2}<4 variableone variablethree \\) to show that \\( linearfunc(verticalaxis, horizontalaxis) \\) has a local maximum or a local minimum at \\( (0,0) \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "r": "bdjfkepl",
        "\\theta": "mvncxero",
        "A": "qprlsxwt",
        "B": "fhdmzaoilu",
        "C": "lqkstmve",
        "D": "rzkfgnow",
        "E": "ghsalnuv",
        "F": "wbxqzpeo",
        "G": "tmdirfla",
        "\\delta": "alwopsnf",
        "N": "vczopral",
        "H": "ygrkdslm"
      },
      "question": "A-2. Consider the locus given by the real polynomial equation\n\\[\nqprlsxwt qzxwvtnp^{2}+fhdmzaoilu qzxwvtnp hjgrksla+lqkstmve hjgrksla^{2}+rzkfgnow qzxwvtnp^{3}+ghsalnuv qzxwvtnp^{2} hjgrksla+wbxqzpeo qzxwvtnp hjgrksla^{2}+tmdirfla hjgrksla^{3}=0\n\\]\nwhere \\( fhdmzaoilu^{\\mathbf{s}}-4 qprlsxwt lqkstmve<0 \\). Prove that there is a positive number \\( alwopsnf \\) such that there are no points of the locus in the punctured disk\n\\[\n0<qzxwvtnp^{2}+hjgrksla^{2}<alwopsnf^{2} .\n\\]",
      "solution": "A-2 Let \\( (qzxwvtnp, hjgrksla)=(bdjfkepl \\cos mvncxero, bdjfkepl \\sin mvncxero), bdjfkepl>0 \\), be a point of the locus. Then\n\\[\nbdjfkepl=\\frac{\\left|qprlsxwt \\cos ^{2} mvncxero+fhdmzaoilu \\sin mvncxero \\cos mvncxero+lqkstmve \\sin ^{2} mvncxero\\right|}{\\left|rzkfgnow \\cos ^{3} mvncxero+ghsalnuv \\cos ^{2} mvncxero \\sin mvncxero+wbxqzpeo \\cos mvncxero \\sin ^{2} mvncxero+tmdirfla \\sin ^{3} mvncxero\\right|}\n\\]\n\nThe denominator of (1) is less than or equal to \\( |rzkfgnow|+|ghsalnuv|+|wbxqzpeo|+|tmdirfla| \\), whereas the numerator has a positive minimum\n\\[\nvczopral=\\frac{|qprlsxwt+lqkstmve|-\\sqrt{(qprlsxwt-lqkstmve)^{2}+fhdmzaoilu^{2}}}{2}\n\\]\nsince \\( fhdmzaoilu^{2}<4 qprlsxwt lqkstmve \\). Therefore\n\\[\nbdjfkepl \\geqq \\frac{vczopral}{|rzkfgnow|+|ghsalnuv|+|wbxqzpeo|+|tmdirfla|}=alwopsnf\n\\]\nand there are no points of the locus within \\( 0<bdjfkepl<alwopsnf \\).\nAlternate Solution: Set \\( ygrkdslm(qzxwvtnp, hjgrksla) \\) equal to the polynomial on the left hand side of the given equation. The standard theory for maxima or minima of functions of two variables can be used together with the condition \\( fhdmzaoilu^{2}<4 qprlsxwt lqkstmve \\) to show that \\( ygrkdslm(qzxwvtnp, hjgrksla) \\) has a local maximum or a local minimum at \\( (0,0) \\)."
    },
    "kernel_variant": {
      "question": "Fix an integer n ≥ 3.  \nLet A be an n × n real symmetric positive–definite matrix and denote its smallest\neigen-value by   \n\n  λ := λ_min(A) > 0.\n\nLet   \n\n  C(x₁,…,x_n)= ∑_{|α|=3} c_α x^α (homogeneous of degree 3),  \n  D(x₁,…,x_n)= ∑_{|β|=4} d_β x^β (homogeneous of degree 4),\n\nwhere α=(α₁,…,α_n), |α|=α₁+⋯+α_n, x^α=x₁^{α₁}⋯x_n^{α_n}, and analogously for β.  \nPut  \n\n  M₃ := max_{|α|=3}|c_α|  (if C≡0 set M₃:=0),  \n  M₄ := max_{|β|=4}|d_β|  (if D≡0 set M₄:=0).\n\nConsider the real algebraic hypersurface in ℝⁿ given by  \n\n  F(x)=xᵀAx + C(x) + D(x)=0.   (★)\n\n1.  Prove that there exists ε>0 — depending only on λ, M₃, M₄ and n — such that the set of real points of (★) does not intersect the punctured ball  \n\n  0 < ‖x‖₂ < ε.\n\n2.  Show that the explicit quantity  \n\n  ε := min { λ / (2 n^{3/2} M₃) (if M₃>0),  √[ λ / (2 n² M₄) ] (if M₄>0) }  (†)\n\nis admissible, with the convention that a missing entry in the minimum is ignored (equivalently, it is understood to be ∞).  \nIn particular, if C≡0 (M₃=0) the first entry in (†) is dropped, and if D≡0 (M₄=0) the second entry is dropped.  When both C≡0 and D≡0 the hypersurface (★) reduces to the empty set {0}, and any ε>0 works.",
      "solution": "Write every non-zero x∈ℝⁿ uniquely as x=r ω with r=‖x‖₂>0 and ω∈S^{n−1}:={ω∈ℝⁿ:‖ω‖₂=1}.  Because C and D are homogeneous of degrees 3 and 4,\n\n  F(r ω)=r² Q(ω)+r³ C(ω)+r⁴ D(ω),                         (1)\n\nwhere Q(ω)=ωᵀAω, C(ω)=C(ω₁,…,ω_n) and D(ω)=D(ω₁,…,ω_n).\n\nStep 1. Angular estimates.  \nPositive-definiteness of A gives  \n\n  Q(ω) = ωᵀAω ≥ λ.                                          (2)\n\nPut S(ω):=∑_{i=1}^n |ω_i|.  By Cauchy–Schwarz, S(ω)≤√n, whence  \n\n  |C(ω)| ≤ M₃ S(ω)³ ≤ M₃ n^{3/2},                        (3)  \n  |D(ω)| ≤ M₄ S(ω)⁴ ≤ M₄ n².                             (4)\n\nStep 2. Radial inequality.  \nCombine (1)–(4):\n\n  F(r ω) ≥ r²λ − r³ n^{3/2}M₃ − r⁴ n²M₄.                (5)\n\nStep 3. Choosing ε.  \nFor a given r>0, define  \n\n  g(r):=r²λ − r³ n^{3/2}M₃ − r⁴ n²M₄.\n\nThe goal is to find ε such that g(r)>0 whenever 0<r<ε.  Note that g(r) is continuous, g(0)=0 and g′(0)=λ>0, so a suitable ε certainly exists; what remains is to verify that the explicit ε in (†) works.\n\n(a) If M₃>0 impose             r n^{3/2}M₃ < λ/2.  \n(b) If M₄>0 impose           r² n² M₄ < λ/2.\n\nWhenever a parameter is zero the corresponding inequality is vacuous.  \nBecause the left sides are increasing in r, the inequalities hold provided  \n\n  r < λ/(2 n^{3/2}M₃)  (if M₃>0),  \n  r < √( λ/(2 n² M₄) )  (if M₄>0).                        (6)\n\nDefine ε by (†); then every 0<r<ε satisfies all the relevant conditions (6).  \nSubstituting in (5) gives\n\n  F(r ω) > r²( λ − λ/2 − λ/2 ) = 0                     (7)\n\nfor all ω∈S^{n−1} and 0<r<ε.\n\nStep 4. Exclusion of real zeros.  \nIf 0<‖x‖₂=r<ε, writing x=r ω yields F(x)=F(r ω)>0 by (7), so no such x lies on the hypersurface (★).  Hence (★) has no real points in the punctured ball 0<‖x‖₂<ε.\n\nStep 5. Degenerate cases.  \n• If M₃=0 while M₄>0, the first entry in (†) is omitted; ε=√(λ/(2 n² M₄)) works by the same argument (inequality (a) is void).  \n• If M₄=0 while M₃>0, the second entry is omitted and ε=λ/(2 n^{3/2} M₃) suffices.  \n• If M₃=M₄=0, then F(x)=xᵀAx>0 for every x≠0, so the real zero-set of (★) is {0} and any ε>0 is admissible.  Consistent with the convention, (†) would read ε=min{∞,∞}=∞, leaving the choice of a finite ε arbitrary.\n\nTherefore the explicit ε prescribed in (†) — interpreted with the stated convention — is always admissible, completing the proof.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.588474",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: The problem is set in ℝⁿ with n≥3, instead of the original 2-dimensional setting.  \n• Additional polynomial degrees: Besides quadratic and cubic parts, a quartic homogeneous part is added, forcing simultaneous control of terms of three different orders.  \n• Spectral data: The argument relies on eigen-values of a symmetric matrix; competitors must recall and apply linear-algebraic facts (Rayleigh–Ritz bounds) that were irrelevant in the original problem.  \n• Uniform quantitative estimates: One must bound the cubic and quartic pieces uniformly over the entire unit sphere; this demands combinatorial counting of monomials (or a clever use of ℓ¹–ℓ^∞ bounds) rather than the single-angle analysis in two dimensions.  \n• Explicit ε: The task is not merely existential; an explicit lower bound in terms of λ, M₃, M₄ and n is requested, adding an optimization step.  \n• Interaction of multiple concepts: Competitors must blend homogeneous-polynomial scaling, eigenvalue estimates, and elementary but subtle inequalities to keep track of how three competing orders of r affect the sign of F.  \n\nThese layers of complexity collectively raise the problem well above the original’s level: the geometry is higher-dimensional, more algebraic data must be marshalled, and several advanced techniques—spectral estimates, uniform angular bounds, and quantitative inequality solving—are required to reach the conclusion."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer n ≥ 3.  \nLet A be an n × n real symmetric positive–definite matrix and denote its smallest\neigen-value by   \n\n  λ := λ_min(A) > 0.\n\nLet   \n\n  C(x₁,…,x_n)= ∑_{|α|=3} c_α x^α (homogeneous of degree 3),  \n  D(x₁,…,x_n)= ∑_{|β|=4} d_β x^β (homogeneous of degree 4),\n\nwhere α=(α₁,…,α_n), |α|=α₁+⋯+α_n, x^α=x₁^{α₁}⋯x_n^{α_n}, and analogously for β.  \nPut  \n\n  M₃ := max_{|α|=3}|c_α|  (if C≡0 set M₃:=0),  \n  M₄ := max_{|β|=4}|d_β|  (if D≡0 set M₄:=0).\n\nConsider the real algebraic hypersurface in ℝⁿ given by  \n\n  F(x)=xᵀAx + C(x) + D(x)=0.   (★)\n\n1.  Prove that there exists ε>0 — depending only on λ, M₃, M₄ and n — such that the set of real points of (★) does not intersect the punctured ball  \n\n  0 < ‖x‖₂ < ε.\n\n2.  Show that the explicit quantity  \n\n  ε := min { λ / (2 n^{3/2} M₃) (if M₃>0),  √[ λ / (2 n² M₄) ] (if M₄>0) }  (†)\n\nis admissible, with the convention that a missing entry in the minimum is ignored (equivalently, it is understood to be ∞).  \nIn particular, if C≡0 (M₃=0) the first entry in (†) is dropped, and if D≡0 (M₄=0) the second entry is dropped.  When both C≡0 and D≡0 the hypersurface (★) reduces to the empty set {0}, and any ε>0 works.",
      "solution": "Write every non-zero x∈ℝⁿ uniquely as x=r ω with r=‖x‖₂>0 and ω∈S^{n−1}:={ω∈ℝⁿ:‖ω‖₂=1}.  Because C and D are homogeneous of degrees 3 and 4,\n\n  F(r ω)=r² Q(ω)+r³ C(ω)+r⁴ D(ω),                         (1)\n\nwhere Q(ω)=ωᵀAω, C(ω)=C(ω₁,…,ω_n) and D(ω)=D(ω₁,…,ω_n).\n\nStep 1. Angular estimates.  \nPositive-definiteness of A gives  \n\n  Q(ω) = ωᵀAω ≥ λ.                                          (2)\n\nPut S(ω):=∑_{i=1}^n |ω_i|.  By Cauchy–Schwarz, S(ω)≤√n, whence  \n\n  |C(ω)| ≤ M₃ S(ω)³ ≤ M₃ n^{3/2},                        (3)  \n  |D(ω)| ≤ M₄ S(ω)⁴ ≤ M₄ n².                             (4)\n\nStep 2. Radial inequality.  \nCombine (1)–(4):\n\n  F(r ω) ≥ r²λ − r³ n^{3/2}M₃ − r⁴ n²M₄.                (5)\n\nStep 3. Choosing ε.  \nFor a given r>0, define  \n\n  g(r):=r²λ − r³ n^{3/2}M₃ − r⁴ n²M₄.\n\nThe goal is to find ε such that g(r)>0 whenever 0<r<ε.  Note that g(r) is continuous, g(0)=0 and g′(0)=λ>0, so a suitable ε certainly exists; what remains is to verify that the explicit ε in (†) works.\n\n(a) If M₃>0 impose             r n^{3/2}M₃ < λ/2.  \n(b) If M₄>0 impose           r² n² M₄ < λ/2.\n\nWhenever a parameter is zero the corresponding inequality is vacuous.  \nBecause the left sides are increasing in r, the inequalities hold provided  \n\n  r < λ/(2 n^{3/2}M₃)  (if M₃>0),  \n  r < √( λ/(2 n² M₄) )  (if M₄>0).                        (6)\n\nDefine ε by (†); then every 0<r<ε satisfies all the relevant conditions (6).  \nSubstituting in (5) gives\n\n  F(r ω) > r²( λ − λ/2 − λ/2 ) = 0                     (7)\n\nfor all ω∈S^{n−1} and 0<r<ε.\n\nStep 4. Exclusion of real zeros.  \nIf 0<‖x‖₂=r<ε, writing x=r ω yields F(x)=F(r ω)>0 by (7), so no such x lies on the hypersurface (★).  Hence (★) has no real points in the punctured ball 0<‖x‖₂<ε.\n\nStep 5. Degenerate cases.  \n• If M₃=0 while M₄>0, the first entry in (†) is omitted; ε=√(λ/(2 n² M₄)) works by the same argument (inequality (a) is void).  \n• If M₄=0 while M₃>0, the second entry is omitted and ε=λ/(2 n^{3/2} M₃) suffices.  \n• If M₃=M₄=0, then F(x)=xᵀAx>0 for every x≠0, so the real zero-set of (★) is {0} and any ε>0 is admissible.  Consistent with the convention, (†) would read ε=min{∞,∞}=∞, leaving the choice of a finite ε arbitrary.\n\nTherefore the explicit ε prescribed in (†) — interpreted with the stated convention — is always admissible, completing the proof.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.472543",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: The problem is set in ℝⁿ with n≥3, instead of the original 2-dimensional setting.  \n• Additional polynomial degrees: Besides quadratic and cubic parts, a quartic homogeneous part is added, forcing simultaneous control of terms of three different orders.  \n• Spectral data: The argument relies on eigen-values of a symmetric matrix; competitors must recall and apply linear-algebraic facts (Rayleigh–Ritz bounds) that were irrelevant in the original problem.  \n• Uniform quantitative estimates: One must bound the cubic and quartic pieces uniformly over the entire unit sphere; this demands combinatorial counting of monomials (or a clever use of ℓ¹–ℓ^∞ bounds) rather than the single-angle analysis in two dimensions.  \n• Explicit ε: The task is not merely existential; an explicit lower bound in terms of λ, M₃, M₄ and n is requested, adding an optimization step.  \n• Interaction of multiple concepts: Competitors must blend homogeneous-polynomial scaling, eigenvalue estimates, and elementary but subtle inequalities to keep track of how three competing orders of r affect the sign of F.  \n\nThese layers of complexity collectively raise the problem well above the original’s level: the geometry is higher-dimensional, more algebraic data must be marshalled, and several advanced techniques—spectral estimates, uniform angular bounds, and quantitative inequality solving—are required to reach the conclusion."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}