{
  "index": "1971-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "B-2. Let \\( F(x) \\) be a real valued function defined for all real \\( x \\) except for \\( x=0 \\) and \\( x=1 \\) and satisfying the functional equation \\( F(x)+F\\{(x-1) / x\\}=1+x \\). Find all functions \\( F(x) \\) satisfying these conditions.",
  "solution": "B-2 In the given functional equation\n\\[\nF(x)+F\\left(\\frac{x-1}{x}\\right)=1+x\n\\]\nwe substitute \\( \\frac{x-1}{x} \\) for \\( x \\), obtaining\n\\[\nF\\left(\\frac{x-1}{x}\\right)+F\\left(\\frac{-1}{x-1}\\right)=\\frac{2 x-1}{x} .\n\\]\n\nAlso in (1), we substitute \\( \\frac{-1}{x-1} \\) for \\( x \\) and obtain\n\\[\nF\\left(\\frac{-1}{x-1}\\right)+F(x)=\\frac{x-2}{x-1}\n\\]\n\nAdding (1) and (3) and subtracting (2) gives\n\\[\n\\begin{aligned}\n2 F(x) & =1+x+\\frac{x-2}{x-1}-\\frac{2 x-1}{x}=\\frac{x^{3}-x^{2}-1}{x(x-1)} \\\\\nF(x) & =\\frac{x^{3}-x^{2}-1}{2 x(x-1)}\n\\end{aligned}\n\\]\n\nThat \\( F(x) \\), defined in (4), does satisfy the given functional equation is easily verified. Therefore (4) is the only solution of the problem.",
  "vars": [
    "F",
    "x"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "F": "function",
        "x": "variable"
      },
      "question": "B-2. Let \\( function(variable) \\) be a real valued function defined for all real \\( variable \\) except for \\( variable=0 \\) and \\( variable=1 \\) and satisfying the functional equation \\( function(variable)+function\\{(variable-1) / variable\\}=1+variable \\). Find all functions \\( function(variable) \\) satisfying these conditions.",
      "solution": "B-2 In the given functional equation\n\\[\nfunction(variable)+function\\left(\\frac{variable-1}{variable}\\right)=1+variable\n\\]\nwe substitute \\( \\frac{variable-1}{variable} \\) for \\( variable \\), obtaining\n\\[\nfunction\\left(\\frac{variable-1}{variable}\\right)+function\\left(\\frac{-1}{variable-1}\\right)=\\frac{2 variable-1}{variable} .\n\\]\n\nAlso in (1), we substitute \\( \\frac{-1}{variable-1} \\) for \\( variable \\) and obtain\n\\[\nfunction\\left(\\frac{-1}{variable-1}\\right)+function(variable)=\\frac{variable-2}{variable-1}\n\\]\n\nAdding (1) and (3) and subtracting (2) gives\n\\[\n\\begin{aligned}\n2 \\, function(variable) & =1+variable+\\frac{variable-2}{variable-1}-\\frac{2 variable-1}{variable}=\\frac{variable^{3}-variable^{2}-1}{variable(variable-1)} \\\\\nfunction(variable) & =\\frac{variable^{3}-variable^{2}-1}{2 \\, variable(variable-1)}\n\\end{aligned}\n\\]\n\nThat \\( function(variable) \\), defined in (4), does satisfy the given functional equation is easily verified. Therefore (4) is the only solution of the problem."
    },
    "descriptive_long_confusing": {
      "map": {
        "F": "threshold",
        "x": "lantern"
      },
      "question": "B-2. Let \\( threshold(lantern) \\) be a real valued function defined for all real \\( lantern \\) except for \\( lantern=0 \\) and \\( lantern=1 \\) and satisfying the functional equation \\( threshold(lantern)+threshold\\{(lantern-1) / lantern\\}=1+lantern \\). Find all functions \\( threshold(lantern) \\) satisfying these conditions.",
      "solution": "B-2 In the given functional equation\n\\[\nthreshold(lantern)+threshold\\left(\\frac{lantern-1}{lantern}\\right)=1+lantern\n\\]\nwe substitute \\( \\frac{lantern-1}{lantern} \\) for \\( lantern \\), obtaining\n\\[\nthreshold\\left(\\frac{lantern-1}{lantern}\\right)+threshold\\left(\\frac{-1}{lantern-1}\\right)=\\frac{2 lantern-1}{lantern} .\n\\]\n\nAlso in (1), we substitute \\( \\frac{-1}{lantern-1} \\) for \\( lantern \\) and obtain\n\\[\nthreshold\\left(\\frac{-1}{lantern-1}\\right)+threshold(lantern)=\\frac{lantern-2}{lantern-1}\n\\]\n\nAdding (1) and (3) and subtracting (2) gives\n\\[\n\\begin{aligned}\n2 threshold(lantern) & =1+lantern+\\frac{lantern-2}{lantern-1}-\\frac{2 lantern-1}{lantern}=\\frac{lantern^{3}-lantern^{2}-1}{lantern(lantern-1)} \\\\\nthreshold(lantern) & =\\frac{lantern^{3}-lantern^{2}-1}{2 lantern(lantern-1)}\n\\end{aligned}\n\\]\n\nThat \\( threshold(lantern) \\), defined in (4), does satisfy the given functional equation is easily verified. Therefore (4) is the only solution of the problem."
    },
    "descriptive_long_misleading": {
      "map": {
        "F": "fixedvalue",
        "x": "nonvariable"
      },
      "question": "B-2. Let \\( fixedvalue(nonvariable) \\) be a real valued function defined for all real \\( nonvariable \\) except for \\( nonvariable=0 \\) and \\( nonvariable=1 \\) and satisfying the functional equation \\( fixedvalue(nonvariable)+fixedvalue\\{(nonvariable-1) / nonvariable\\}=1+nonvariable \\). Find all functions \\( fixedvalue(nonvariable) \\) satisfying these conditions.",
      "solution": "B-2 In the given functional equation\\n\\[\\nfixedvalue(nonvariable)+fixedvalue\\left(\\frac{nonvariable-1}{nonvariable}\\right)=1+nonvariable\\n\\]\\nwe substitute \\( \\frac{nonvariable-1}{nonvariable} \\) for \\( nonvariable \\), obtaining\\n\\[\\nfixedvalue\\left(\\frac{nonvariable-1}{nonvariable}\\right)+fixedvalue\\left(\\frac{-1}{nonvariable-1}\\right)=\\frac{2 nonvariable-1}{nonvariable} .\\n\\]\\n\\nAlso in (1), we substitute \\( \\frac{-1}{nonvariable-1} \\) for \\( nonvariable \\) and obtain\\n\\[\\nfixedvalue\\left(\\frac{-1}{nonvariable-1}\\right)+fixedvalue(nonvariable)=\\frac{nonvariable-2}{nonvariable-1}\\n\\]\\n\\nAdding (1) and (3) and subtracting (2) gives\\n\\[\\n\\begin{aligned}\\n2 fixedvalue(nonvariable) & =1+nonvariable+\\frac{nonvariable-2}{nonvariable-1}-\\frac{2 nonvariable-1}{nonvariable}=\\frac{nonvariable^{3}-nonvariable^{2}-1}{nonvariable(nonvariable-1)} \\\\nfixedvalue(nonvariable) & =\\frac{nonvariable^{3}-nonvariable^{2}-1}{2 nonvariable(nonvariable-1)}\\n\\end{aligned}\\n\\]\\n\\nThat \\( fixedvalue(nonvariable) \\), defined in (4), does satisfy the given functional equation is easily verified. Therefore (4) is the only solution of the problem."
    },
    "garbled_string": {
      "map": {
        "F": "qzxwvtnp",
        "x": "hjgrksla"
      },
      "question": "B-2. Let \\( qzxwvtnp(hjgrksla) \\) be a real valued function defined for all real \\( hjgrksla \\) except for \\( hjgrksla=0 \\) and \\( hjgrksla=1 \\) and satisfying the functional equation \\( qzxwvtnp(hjgrksla)+qzxwvtnp\\{(hjgrksla-1) / hjgrksla\\}=1+hjgrksla \\). Find all functions \\( qzxwvtnp(hjgrksla) \\) satisfying these conditions.",
      "solution": "B-2 In the given functional equation\n\\[\nqzxwvtnp(hjgrksla)+qzxwvtnp\\left(\\frac{hjgrksla-1}{hjgrksla}\\right)=1+hjgrksla\n\\]\nwe substitute \\( \\frac{hjgrksla-1}{hjgrksla} \\) for \\( hjgrksla \\), obtaining\n\\[\nqzxwvtnp\\left(\\frac{hjgrksla-1}{hjgrksla}\\right)+qzxwvtnp\\left(\\frac{-1}{hjgrksla-1}\\right)=\\frac{2 hjgrksla-1}{hjgrksla} .\n\\]\n\nAlso in (1), we substitute \\( \\frac{-1}{hjgrksla-1} \\) for \\( hjgrksla \\) and obtain\n\\[\nqzxwvtnp\\left(\\frac{-1}{hjgrksla-1}\\right)+qzxwvtnp(hjgrksla)=\\frac{hjgrksla-2}{hjgrksla-1}\n\\]\n\nAdding (1) and (3) and subtracting (2) gives\n\\[\n\\begin{aligned}\n2 qzxwvtnp(hjgrksla) & =1+hjgrksla+\\frac{hjgrksla-2}{hjgrksla-1}-\\frac{2 hjgrksla-1}{hjgrksla}=\\frac{hjgrksla^{3}-hjgrksla^{2}-1}{hjgrksla(hjgrksla-1)} \\\\\nqzxwvtnp(hjgrksla) & =\\frac{hjgrksla^{3}-hjgrksla^{2}-1}{2 hjgrksla(hjgrksla-1)}\n\\end{aligned}\n\\]\n\nThat \\( qzxwvtnp(hjgrksla) \\), defined in (4), does satisfy the given functional equation is easily verified. Therefore (4) is the only solution of the problem."
    },
    "kernel_variant": {
      "question": "Let  \n T(x)= (x−1)/x and U(x)= −1/(x−1) (so that T²=U and T³ = Id).  \nDenote  \n S(x)=x+T(x)+U(x)=\\dfrac{x^{3}-3x+1}{x(x-1)}, x\\neq 0,1.  \nFor a real-valued function  \n\n F : ℝ∖{0,1} → ℝ  \n\nassume that for every real number x≠0,1 the three points x , T(x) , U(x) also\nlie in ℝ∖{0,1} (which is automatic) and the simultaneous functional\nequations  \n\n(1) F(x)+F(T(x))+F(U(x)) = S(x),  \n(2) F(x)−F(T(x))+F(U(x)) = S(x)/3  \n\nhold.  \nDetermine all such functions F.",
      "solution": "Step 1.  Basic notation  \nFor a fixed x put  \n A=F(x), B=F(T(x)), C=F(U(x)),  s=S(x)=x+T(x)+U(x).  \nBecause S(T(x))=S(U(x))=S(x), the same number s appears whenever the\nargument is replaced by T(x) or U(x).\n\nThe two given identities become the linear system  \n\n A + B + C = s    (3)  \n A − B + C = s/3.  (4)\n\nStep 2.  Elimination of B  \nSubtracting (4) from (3) yields  \n\n 2B = s − s/3 = 2s/3 ⇒ B = s/3.  (5)\n\nStep 3.  An equation arising from the argument T(x)  \nReplacing x with T(x) (whose admissibility is guaranteed) gives another\npair of equations.  In particular\n\n B − C + A = s/3.  (6)\n\nStep 4.  Solving for A and C  \nAdd (4) and (6):\n\n (A−B+C)+(A+B−C)=2A = 2·(s/3) ⇒ A = s/3.  (7)\n\nInsert (7) and (5) into (3):\n\n s/3 + s/3 + C = s ⇒ C = s/3.  (8)\n\nStep 5.  Conclusion for the value of F  \nFor every admissible x we have A=B=C=s/3, i.e.\n\n F(x)=\\dfrac{S(x)}{3}= \\dfrac{x^{3}-3x+1}{3x(x-1)}.  (9)\n\nStep 6.  Verification  \nBecause S(T(x))=S(U(x))=S(x), equation (9) gives\n\n F(T(x))=F(U(x))=\\dfrac{S(x)}{3}.  \n\nHence  \n\n F(x)+F(T(x))+F(U(x)) = 3·\\dfrac{S(x)}{3}=S(x),  \n\n F(x)−F(T(x))+F(U(x)) =\\dfrac{S(x)}{3}-\\dfrac{S(x)}{3}+\\dfrac{S(x)}{3}\n                          =\\dfrac{S(x)}{3},  \n\nconfirming (1) and (2).\n\nStep 7.  Uniqueness  \nAt every x the coefficient matrix of (3)–(4) is  \n\n [1  1  1;  1  −1  1],  \n\nwhose rank is 2.  Together with equation (6) obtained from x↦T(x) the rank\nbecomes 3, so the triple (A,B,C) – and hence F(x) – is uniquely\ndetermined.  Therefore (9) is the only possible solution.\n\nAnswer.  The unique function satisfying (1) and (2) is  \n\n  F(x)=\\dfrac{x^{3}-3x+1}{3x\\,(x-1)},  defined for all real x≠0,1.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.599239",
        "was_fixed": false,
        "difficulty_analysis": "•  Multiple interacting transforms:  the problem involves the entire 3-cycle generated by the Möbius map T, not just a single pair of points as in the original.  \n•  Simultaneous equations:  two independent functional relations have to be honoured at once, producing a non-trivial linear system in three unknown values A, B, C that must be solved coherently.  \n•  Necessity of cyclic substitution:  because (1)–(2) alone are insufficient, one must substitute x→T(x) to obtain additional relations—a layer of reasoning absent from the original kernel variant.  \n•  Higher algebraic complexity:  the explicit computation forces manipulation of quartic and octic polynomials and of rational expressions with denominator x⁴, far beyond the quadratic expressions in the original problem.  \n•  Verification and uniqueness demand handling of a 3×3 coefficient matrix whose entries vary with x; showing its determinant never vanishes on the admissible domain requires careful domain analysis.\n\nAll these factors substantially raise both the conceptual and the computational load, making this enhanced kernel variant significantly harder than the original."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n T(x)= (x−1)/x and U(x)= −1/(x−1) (so that T²=U and T³ = Id).  \nDenote  \n S(x)=x+T(x)+U(x)=\\dfrac{x^{3}-3x+1}{x(x-1)}, x\\neq 0,1.  \nFor a real-valued function  \n\n F : ℝ∖{0,1} → ℝ  \n\nassume that for every real number x≠0,1 the three points x , T(x) , U(x) also\nlie in ℝ∖{0,1} (which is automatic) and the simultaneous functional\nequations  \n\n(1) F(x)+F(T(x))+F(U(x)) = S(x),  \n(2) F(x)−F(T(x))+F(U(x)) = S(x)/3  \n\nhold.  \nDetermine all such functions F.",
      "solution": "Step 1.  Basic notation  \nFor a fixed x put  \n A=F(x), B=F(T(x)), C=F(U(x)),  s=S(x)=x+T(x)+U(x).  \nBecause S(T(x))=S(U(x))=S(x), the same number s appears whenever the\nargument is replaced by T(x) or U(x).\n\nThe two given identities become the linear system  \n\n A + B + C = s    (3)  \n A − B + C = s/3.  (4)\n\nStep 2.  Elimination of B  \nSubtracting (4) from (3) yields  \n\n 2B = s − s/3 = 2s/3 ⇒ B = s/3.  (5)\n\nStep 3.  An equation arising from the argument T(x)  \nReplacing x with T(x) (whose admissibility is guaranteed) gives another\npair of equations.  In particular\n\n B − C + A = s/3.  (6)\n\nStep 4.  Solving for A and C  \nAdd (4) and (6):\n\n (A−B+C)+(A+B−C)=2A = 2·(s/3) ⇒ A = s/3.  (7)\n\nInsert (7) and (5) into (3):\n\n s/3 + s/3 + C = s ⇒ C = s/3.  (8)\n\nStep 5.  Conclusion for the value of F  \nFor every admissible x we have A=B=C=s/3, i.e.\n\n F(x)=\\dfrac{S(x)}{3}= \\dfrac{x^{3}-3x+1}{3x(x-1)}.  (9)\n\nStep 6.  Verification  \nBecause S(T(x))=S(U(x))=S(x), equation (9) gives\n\n F(T(x))=F(U(x))=\\dfrac{S(x)}{3}.  \n\nHence  \n\n F(x)+F(T(x))+F(U(x)) = 3·\\dfrac{S(x)}{3}=S(x),  \n\n F(x)−F(T(x))+F(U(x)) =\\dfrac{S(x)}{3}-\\dfrac{S(x)}{3}+\\dfrac{S(x)}{3}\n                          =\\dfrac{S(x)}{3},  \n\nconfirming (1) and (2).\n\nStep 7.  Uniqueness  \nAt every x the coefficient matrix of (3)–(4) is  \n\n [1  1  1;  1  −1  1],  \n\nwhose rank is 2.  Together with equation (6) obtained from x↦T(x) the rank\nbecomes 3, so the triple (A,B,C) – and hence F(x) – is uniquely\ndetermined.  Therefore (9) is the only possible solution.\n\nAnswer.  The unique function satisfying (1) and (2) is  \n\n  F(x)=\\dfrac{x^{3}-3x+1}{3x\\,(x-1)},  defined for all real x≠0,1.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.479840",
        "was_fixed": false,
        "difficulty_analysis": "•  Multiple interacting transforms:  the problem involves the entire 3-cycle generated by the Möbius map T, not just a single pair of points as in the original.  \n•  Simultaneous equations:  two independent functional relations have to be honoured at once, producing a non-trivial linear system in three unknown values A, B, C that must be solved coherently.  \n•  Necessity of cyclic substitution:  because (1)–(2) alone are insufficient, one must substitute x→T(x) to obtain additional relations—a layer of reasoning absent from the original kernel variant.  \n•  Higher algebraic complexity:  the explicit computation forces manipulation of quartic and octic polynomials and of rational expressions with denominator x⁴, far beyond the quadratic expressions in the original problem.  \n•  Verification and uniqueness demand handling of a 3×3 coefficient matrix whose entries vary with x; showing its determinant never vanishes on the admissible domain requires careful domain analysis.\n\nAll these factors substantially raise both the conceptual and the computational load, making this enhanced kernel variant significantly harder than the original."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}