{
  "index": "1948-B-1",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "1. Let \\( f(x) \\) be a cubic polynomial with roots \\( x_{1}, x_{2} \\), and \\( x_{3} \\). Assume that \\( f(2 x) \\) is divisible by \\( f^{\\prime}(x) \\) and compute the ratios \\( x_{1}: x_{2}: x_{3} \\).",
  "solution": "Solution. Let \\( f(x)=x^{3}+a x^{2}+b x+c \\). Since \\( f(2 x) \\) is divisible by \\( f^{\\prime}(x) \\), we have\n\\[\n8 x^{3}+4 a x^{2}+2 b x+c=\\left(3 x^{2}+2 a x+b\\right)(p x+q)\n\\]\nfor some \\( p \\) and \\( q \\). Comparing coefficients we find\n\\[\n\\begin{array}{l}\n3 p=8, \\quad 2 a p+3 q=4 a \\\\\nb p+2 a q=2 b, \\quad q b=c\n\\end{array}\n\\]\nfrom which it follows that\n\\[\np=8 / 3, \\quad q=-4 a / 9, \\quad b=4 a^{2} / 3, \\quad \\text { and } \\quad c=-16 a^{3} / 27\n\\]\n\nHence\n\\[\nf(x)=x^{3}+a x^{2}+\\frac{4}{3} a^{2} x-\\frac{16}{27} a^{3}\n\\]\n\nNow if \\( a=0 \\), all the roots of \\( f(x)=0 \\) are zero and their ratios are undefined; so we assume from now on that \\( a \\neq 0 \\). Set \\( w=3 x / a \\) and consider\n\\[\n\\begin{aligned}\nF(w) & =27 f(x)=27 f\\left(\\frac{a w}{3}\\right)=a^{3}\\left(w^{3}+3 w^{2}+12 w-16\\right) \\\\\n& =a^{3}(w-1)\\left(w^{2}+4 w+16\\right) .\n\\end{aligned}\n\\]\n\nThe roots of \\( F(w)=0 \\) are \\( 1,-2 \\pm 2 \\sqrt{3} i \\). The roots of \\( f(x)=0 \\) have the same ratios as the roots of \\( F(w)=0 \\), so with suitable numbering we have\n\\[\nx_{1}: x_{2}: x_{3}=1:(-2+2 \\sqrt{3} i):(-2-2 \\sqrt{3} i)\n\\]",
  "vars": [
    "x",
    "w",
    "x_1",
    "x_2",
    "x_3",
    "f",
    "F"
  ],
  "params": [
    "a",
    "b",
    "c",
    "p",
    "q"
  ],
  "sci_consts": [
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "w": "auxiliaryw",
        "x_1": "rootone",
        "x_2": "roottwo",
        "x_3": "rootthree",
        "f": "cubicfun",
        "F": "scaledfun",
        "a": "coeffa",
        "b": "coeffb",
        "c": "coeffc",
        "p": "factorp",
        "q": "factorq"
      },
      "question": "1. Let \\( cubicfun(variablex) \\) be a cubic polynomial with roots \\( rootone, roottwo \\), and \\( rootthree \\). Assume that \\( cubicfun(2 variablex) \\) is divisible by \\( cubicfun^{\\prime}(variablex) \\) and compute the ratios \\( rootone: roottwo: rootthree \\).",
      "solution": "Solution. Let \\( cubicfun(variablex)=variablex^{3}+coeffa\\,variablex^{2}+coeffb\\,variablex+coeffc \\). Since \\( cubicfun(2 variablex) \\) is divisible by \\( cubicfun^{\\prime}(variablex) \\), we have\n\\[\n8\\,variablex^{3}+4\\,coeffa\\,variablex^{2}+2\\,coeffb\\,variablex+coeffc=\\left(3\\,variablex^{2}+2\\,coeffa\\,variablex+coeffb\\right)(factorp\\,variablex+factorq)\n\\]\nfor some \\( factorp \\) and \\( factorq \\). Comparing coefficients we find\n\\[\n\\begin{array}{l}\n3\\,factorp=8, \\quad 2\\,coeffa\\,factorp+3\\,factorq=4\\,coeffa \\\\\ncoeffb\\,factorp+2\\,coeffa\\,factorq=2\\,coeffb, \\quad factorq\\,coeffb=coeffc\n\\end{array}\n\\]\nfrom which it follows that\n\\[\nfactorp=\\frac{8}{3}, \\quad factorq=-\\frac{4\\,coeffa}{9}, \\quad coeffb=\\frac{4\\,coeffa^{2}}{3}, \\quad \\text { and } \\quad coeffc=-\\frac{16\\,coeffa^{3}}{27}\n\\]\n\nHence\n\\[\ncubicfun(variablex)=variablex^{3}+coeffa\\,variablex^{2}+\\frac{4}{3}\\,coeffa^{2}\\,variablex-\\frac{16}{27}\\,coeffa^{3}\n\\]\n\nNow if \\( coeffa=0 \\), all the roots of \\( cubicfun(variablex)=0 \\) are zero and their ratios are undefined; so we assume from now on that \\( coeffa \\neq 0 \\). Set \\( auxiliaryw=3\\,variablex / coeffa \\) and consider\n\\[\n\\begin{aligned}\nscaledfun(auxiliaryw) & =27\\,cubicfun(variablex)=27\\,cubicfun\\left(\\frac{coeffa\\,auxiliaryw}{3}\\right)=coeffa^{3}\\left(auxiliaryw^{3}+3\\,auxiliaryw^{2}+12\\,auxiliaryw-16\\right) \\\\\n& =coeffa^{3}(auxiliaryw-1)\\left(auxiliaryw^{2}+4\\,auxiliaryw+16\\right) .\n\\end{aligned}\n\\]\n\nThe roots of \\( scaledfun(auxiliaryw)=0 \\) are \\( 1,-2 \\pm 2 \\sqrt{3} i \\). The roots of \\( cubicfun(variablex)=0 \\) have the same ratios as the roots of \\( scaledfun(auxiliaryw)=0 \\), so with suitable numbering we have\n\\[\nrootone: roottwo: rootthree = 1 : (-2+2 \\sqrt{3} i) : (-2-2 \\sqrt{3} i)\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sandstone",
        "w": "caravansary",
        "x_1": "marigold",
        "x_2": "peppermint",
        "x_3": "chandelier",
        "f": "lighthouse",
        "F": "applecart",
        "a": "porpoise",
        "b": "sunflower",
        "c": "raincloud",
        "p": "harmonica",
        "q": "blackbird"
      },
      "question": "Problem:\n<<<\n1. Let \\( lighthouse(sandstone) \\) be a cubic polynomial with roots \\( marigold, peppermint \\), and \\( chandelier \\). Assume that \\( lighthouse(2 sandstone) \\) is divisible by \\( lighthouse^{\\prime}(sandstone) \\) and compute the ratios \\( marigold: peppermint: chandelier \\).\n>>>",
      "solution": "Solution:\n<<<\nSolution. Let \\( lighthouse(sandstone)=sandstone^{3}+porpoise sandstone^{2}+sunflower sandstone+raincloud \\). Since \\( lighthouse(2 sandstone) \\) is divisible by \\( lighthouse^{\\prime}(sandstone) \\), we have\n\\[\n8 sandstone^{3}+4 porpoise sandstone^{2}+2 sunflower sandstone+raincloud=\\left(3 sandstone^{2}+2 porpoise sandstone+sunflower\\right)(harmonica sandstone+blackbird)\n\\]\nfor some \\( harmonica \\) and \\( blackbird \\). Comparing coefficients we find\n\\[\n\\begin{array}{l}\n3 harmonica=8, \\quad 2 porpoise harmonica+3 blackbird=4 porpoise \\\\\nsunflower harmonica+2 porpoise blackbird=2 sunflower, \\quad blackbird sunflower=raincloud\n\\end{array}\n\\]\nfrom which it follows that\n\\[\nharmonica=8 / 3, \\quad blackbird=-4 porpoise / 9, \\quad sunflower=4 porpoise^{2} / 3, \\quad \\text { and } \\quad raincloud=-16 porpoise^{3} / 27\n\\]\n\nHence\n\\[\nlighthouse(sandstone)=sandstone^{3}+porpoise sandstone^{2}+\\frac{4}{3} porpoise^{2} sandstone-\\frac{16}{27} porpoise^{3}\n\\]\n\nNow if \\( porpoise=0 \\), all the roots of \\( lighthouse(sandstone)=0 \\) are zero and their ratios are undefined; so we assume from now on that \\( porpoise \\neq 0 \\). Set \\( caravansary=3 sandstone / porpoise \\) and consider\n\\[\n\\begin{aligned}\napplecart(caravansary) & =27 lighthouse(sandstone)=27 lighthouse\\left(\\frac{porpoise caravansary}{3}\\right)=porpoise^{3}\\left(caravansary^{3}+3 caravansary^{2}+12 caravansary-16\\right) \\\\\n& =porpoise^{3}(caravansary-1)\\left(caravansary^{2}+4 caravansary+16\\right) .\n\\end{aligned}\n\\]\n\nThe roots of \\( applecart(caravansary)=0 \\) are \\( 1,-2 \\pm 2 \\sqrt{3} i \\). The roots of \\( lighthouse(sandstone)=0 \\) have the same ratios as the roots of \\( applecart(caravansary)=0 \\), so with suitable numbering we have\n\\[\nmarigold: peppermint: chandelier=1:(-2+2 \\sqrt{3} i):(-2-2 \\sqrt{3} i)\n\\]\n>>>"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "w": "stillvalue",
        "x_1": "lastpoint",
        "x_2": "middlepoint",
        "x_3": "firstpoint",
        "f": "constantfunc",
        "F": "steadyfunc",
        "a": "variable",
        "b": "shiftingnum",
        "c": "fixednum",
        "p": "unknownvalue",
        "q": "knownvalue"
      },
      "question": "1. Let \\( constantfunc(fixedvalue) \\) be a cubic polynomial with roots \\( lastpoint, middlepoint \\), and \\( firstpoint \\). Assume that \\( constantfunc(2 fixedvalue) \\) is divisible by \\( constantfunc^{\\prime}(fixedvalue) \\) and compute the ratios \\( lastpoint: middlepoint: firstpoint \\).",
      "solution": "Solution. Let \\( constantfunc(fixedvalue)=fixedvalue^{3}+variable fixedvalue^{2}+shiftingnum fixedvalue+fixednum \\). Since \\( constantfunc(2 fixedvalue) \\) is divisible by \\( constantfunc^{\\prime}(fixedvalue) \\), we have\n\\[\n8 fixedvalue^{3}+4 variable fixedvalue^{2}+2 shiftingnum fixedvalue+fixednum=\\left(3 fixedvalue^{2}+2 variable fixedvalue+shiftingnum\\right)(unknownvalue fixedvalue+knownvalue)\n\\]\nfor some \\( unknownvalue \\) and \\( knownvalue \\). Comparing coefficients we find\n\\[\n\\begin{array}{l}\n3\\,unknownvalue=8, \\quad 2\\,variable\\,unknownvalue+3\\,knownvalue=4\\,variable \\\\\nshiftingnum\\,unknownvalue+2\\,variable\\,knownvalue=2\\,shiftingnum, \\quad knownvalue\\,shiftingnum=fixednum\n\\end{array}\n\\]\nfrom which it follows that\n\\[\nunknownvalue=8/3, \\quad knownvalue=-4\\,variable/9, \\quad shiftingnum=4\\,variable^{2}/3, \\quad \\text { and } \\quad fixednum=-16\\,variable^{3}/27\n\\]\n\nHence\n\\[\nconstantfunc(fixedvalue)=fixedvalue^{3}+variable fixedvalue^{2}+\\frac{4}{3} variable^{2} fixedvalue-\\frac{16}{27} variable^{3}\n\\]\n\nNow if \\( variable=0 \\), all the roots of \\( constantfunc(fixedvalue)=0 \\) are zero and their ratios are undefined; so we assume from now on that \\( variable \\neq 0 \\). Set \\( stillvalue=3 fixedvalue/variable \\) and consider\n\\[\n\\begin{aligned}\nsteadyfunc(stillvalue) & =27\\,constantfunc(fixedvalue)=27\\,constantfunc\\left(\\frac{variable\\,stillvalue}{3}\\right)=variable^{3}\\left(stillvalue^{3}+3\\,stillvalue^{2}+12\\,stillvalue-16\\right) \\\\\n& =variable^{3}(stillvalue-1)\\left(stillvalue^{2}+4\\,stillvalue+16\\right).\n\\end{aligned}\n\\]\n\nThe roots of \\( steadyfunc(stillvalue)=0 \\) are \\( 1,-2 \\pm 2 \\sqrt{3} i \\). The roots of \\( constantfunc(fixedvalue)=0 \\) have the same ratios as the roots of \\( steadyfunc(stillvalue)=0 \\), so with suitable numbering we have\n\\[\nlastpoint: middlepoint: firstpoint=1:(-2+2 \\sqrt{3} i):(-2-2 \\sqrt{3} i)\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "w": "hjgrksla",
        "x_1": "pqnrcvmd",
        "x_2": "lzkhfedu",
        "x_3": "smtgakvf",
        "f": "dgnrplse",
        "F": "trbqhxui",
        "a": "vkwsiejd",
        "b": "fjlqprza",
        "c": "umrycvao",
        "p": "gcznfeas",
        "q": "oxtewklm"
      },
      "question": "1. Let \\( dgnrplse(qzxwvtnp) \\) be a cubic polynomial with roots \\( pqnrcvmd, lzkhfedu \\), and \\( smtgakvf \\). Assume that \\( dgnrplse(2 qzxwvtnp) \\) is divisible by \\( dgnrplse^{\\prime}(qzxwvtnp) \\) and compute the ratios \\( pqnrcvmd: lzkhfedu: smtgakvf \\).",
      "solution": "Solution. Let \\( dgnrplse(qzxwvtnp)=qzxwvtnp^{3}+vkwsiejd qzxwvtnp^{2}+fjlqprza qzxwvtnp+umrycvao \\). Since \\( dgnrplse(2 qzxwvtnp) \\) is divisible by \\( dgnrplse^{\\prime}(qzxwvtnp) \\), we have\n\\[\n8 qzxwvtnp^{3}+4 vkwsiejd qzxwvtnp^{2}+2 fjlqprza qzxwvtnp+umrycvao=\\left(3 qzxwvtnp^{2}+2 vkwsiejd qzxwvtnp+fjlqprza\\right)(gcznfeas qzxwvtnp+oxtewklm)\n\\]\nfor some \\( gcznfeas \\) and \\( oxtewklm \\). Comparing coefficients we find\n\\[\n\\begin{array}{l}\n3 gcznfeas=8, \\quad 2 vkwsiejd gcznfeas+3 oxtewklm=4 vkwsiejd \\\\\nfjlqprza gcznfeas+2 vkwsiejd oxtewklm=2 fjlqprza, \\quad oxtewklm fjlqprza=umrycvao\n\\end{array}\n\\]\nfrom which it follows that\n\\[\ngcznfeas=8 / 3, \\quad oxtewklm=-4 vkwsiejd / 9, \\quad fjlqprza=4 vkwsiejd^{2} / 3, \\quad \\text { and } \\quad umrycvao=-16 vkwsiejd^{3} / 27\n\\]\nHence\n\\[\ndgnrplse(qzxwvtnp)=qzxwvtnp^{3}+vkwsiejd qzxwvtnp^{2}+\\frac{4}{3} vkwsiejd^{2} qzxwvtnp-\\frac{16}{27} vkwsiejd^{3}\n\\]\nNow if \\( vkwsiejd=0 \\), all the roots of \\( dgnrplse(qzxwvtnp)=0 \\) are zero and their ratios are undefined; so we assume from now on that \\( vkwsiejd \\neq 0 \\). Set \\( hjgrksla=3 qzxwvtnp / vkwsiejd \\) and consider\n\\[\n\\begin{aligned}\ntrbqhxui(hjgrksla) & =27 dgnrplse(qzxwvtnp)=27 dgnrplse\\left(\\frac{vkwsiejd hjgrksla}{3}\\right)=vkwsiejd^{3}\\left(hjgrksla^{3}+3 hjgrksla^{2}+12 hjgrksla-16\\right) \\\\\n& =vkwsiejd^{3}(hjgrksla-1)\\left(hjgrksla^{2}+4 hjgrksla+16\\right) .\n\\end{aligned}\n\\]\nThe roots of \\( trbqhxui(hjgrksla)=0 \\) are \\( 1,-2 \\pm 2 \\sqrt{3} i \\). The roots of \\( dgnrplse(qzxwvtnp)=0 \\) have the same ratios as the roots of \\( trbqhxui(hjgrksla)=0 \\), so with suitable numbering we have\n\\[\npqnrcvmd: lzkhfedu: smtgakvf=1:(-2+2 \\sqrt{3} i):(-2-2 \\sqrt{3} i)\n\\]"
    },
    "kernel_variant": {
      "question": "Let f(x) be a cubic polynomial with leading coefficient 4 whose three roots x₁ , x₂ , x₃ are not all equal to 0.  Suppose that f(3x) is divisible by the derivative f '(x).  Determine the ratio of the roots x₁ : x₂ : x₃ (up to a common non–zero factor).",
      "solution": "Because the leading coefficient is 4 we may write\n  f(x)=4x^{3}+ax^{2}+bx+c,          a,b,c \\in \\mathbb C.\nConsequently\n  f'(x)=12x^{2}+2ax+b.\n\n1.  Impose the divisibility condition.\n    If f(3x) is divisible by f'(x) there exist constants p,q such that\n        f(3x)=(px+q)\\,f'(x).\n    Compute both sides:\n        f(3x)=4(3x)^{3}+a(3x)^{2}+b(3x)+c\n              =108x^{3}+9ax^{2}+3bx+c,\n        (px+q)f'(x)=(px+q)(12x^{2}+2ax+b)\n              =12p x^{3}+(12q+2ap)x^{2}+(2aq+bp)x+bq.\n    Match the coefficients of equal powers of x:\n        12p   =108,                                   (1)\n        12q+2ap= 9a,                                   (2)\n        2aq+bp = 3b,                                   (3)\n        bq     = c.                                    (4)\n\n2.  Solve for p,q,b,c in terms of a.\n        From (1)  p=108/12=9.\n        From (2)  12q+18a=9a  \\Rightarrow q=-\\tfrac34 a.\n        Insert p,q into (3):\n            2a(-\\tfrac34 a)+9b=3b\n            -\\tfrac32 a^{2}+9b=3b \\Rightarrow 6b=\\tfrac32 a^{2}\n            \\Rightarrow b=\\tfrac14 a^{2}.\n        From (4)  c=bq=\\tfrac14 a^{2}\\bigl(-\\tfrac34 a\\bigr)=-\\tfrac{3}{16}a^{3}.\n\n    Hence every cubic that fulfils the requirement (apart from the yet–to–be–considered case a=0) is\n        f(x)=4x^{3}+ax^{2}+\\tfrac14 a^{2}x-\\tfrac{3}{16}a^{3}.          (5)\n\n3.  Find the roots of (5).\n    Put  w=\\dfrac{4x}{a}\\;(a\\neq0) \\;\\Rightarrow\\; x=\\dfrac{a}{4}w.  Substituting in (5),\n        f\\!\\left(\\dfrac{a}{4}w\\right)=\\dfrac{a^{3}}{16}\\bigl(w^{3}+w^{2}+w-3\\bigr).\n    Thus the roots of f are in the same ratio as the roots of\n        g(w)=w^{3}+w^{2}+w-3.\n    Factor g:\n        g(w)=(w-1)(w^{2}+2w+3).\n    Therefore\n        w_{1}= 1,\n        w_{2}=-1+\\mathrm i\\sqrt2,\n        w_{3}=-1-\\mathrm i\\sqrt2.\n\n    Because x=\\dfrac{a}{4}w, the three roots of f are proportional to w_{1},w_{2},w_{3}.  Hence\n        x_{1}:x_{2}:x_{3}=1:(-1+\\mathrm i\\sqrt2):(-1-\\mathrm i\\sqrt2).\n\n4.  The excluded degenerate case.\n    If a=0, then equations (2)–(4) force b=c=0, so f(x)=4x^{3}.  All three roots are 0 and their ratio is indeterminate.  This situation is ruled out by the assumption that not all roots are 0, so the ratio obtained in step 3 is the unique answer.\n\nAnswer.\n  x₁ : x₂ : x₃ = 1 : (−1 + i√2) : (−1 − i√2).",
      "_meta": {
        "core_steps": [
          "Write general monic cubic f(x)=x^3+ax^2+bx+c and express divisibility f(2x)=f'(x)(px+q).",
          "Compare coefficients to solve for p, q, b, c in terms of a (up to an overall scale).",
          "Use a linear change of variable to clear fractions and simplify the cubic.",
          "Factor the simplified cubic and read off its three roots.",
          "State the proportionality (ratios) of the original roots from the factored form."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Constant by which the argument of f is magnified in the divisibility condition.",
            "original": "2 in  f(2x)"
          },
          "slot2": {
            "description": "Normalization choice that the leading coefficient of f(x) is 1 (monic cubic).",
            "original": "implicit factor 1 on x^3 term"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}